Instructor's Manual to ACcolnpany
FOURTH EDITION
Fundamentals BRUCE R. MUNSON DONALD F. YOUNG Department of Aerospace Engineering and Engineering Mechanics
THEODORE H. OKIISHI Department of Mechanical Engineering Iowa State University Ames, Iowa, USA
John Wiley & Sons, Inc. New York
Chichester
Brisbane
Toronto
Singapore
TABLE OF CONTENTS
INTRODUCTION ................................................................................................................... 1 COMPUTER PROBLEMS .................................................................................................... 2 Standard Programs-File Names and Use .................................................................... 2 SOLUTIONS Chapter 1
Introduction................... .......... ............. ..................... ....................... 1-1
Chapter 2
Fluid Statics......... ..... ...... ........ ................ .......................................... 2-1
Chapter 3
Elementary Fluid Dynamics-Bernoulli Equation .......................... 3-1
Chapter 4
Fluid Kinematics... ...... .......... ......... ..... ................... .......................... 4-1
Chapter 5
Finite Control Volume Analysis ....................................................... 5-1
Chapter 6
Differential Analysis of Fluid Flow ................................................. 6-1
Chapter 7
Similitude, Dimensional Analysis, and Modeling ............ ............... 7-1
Chapter 8
Viscous Pipe Flow............................................ ................................ 8-1
Chapter 9
Flow Over Immersed Bodies ........................................................... 9-1
Chapter 10
Open-Channel Flow...... ...... ......... ....... ..................................... ...... 10-1
Chapter 11
Compressible Flow ......................................................................... 11-1
Chapter 12
Turbomachines ............. .................. ................................................ 12-1
Appendix A
Listing of Standard Programs .......................................................... A-I
INTRODUCTION
This manual contains solutions to the problems presented at the end of the chapters in the Fourth Edition of FUNDAMENTALS OF FLUID MECHANICS. It is our intention that the material in this manual be used as an aid in the teaching of the course. We feel quite strongly that problem solving is an essential ingredient in the process of understanding the variety of interesting concepts involved in fluid mechanics. This solutions manual is structured to enhance the learning process. Approximately 1220 problems are solved in a complete, detailed fashion with (in most cases) one problem per page. The problem statements and figures are included with the problem solutions to provide an easier and clearer understanding of the solution procedure. Except where a greater accuracy is warranted, all intermediate calculations and answers are given to three significant figures. Unless otherwise indicated in the problem statement, values of fluid properties used in the solutions are those given in the tables on the inside of the front cover of the text. Other fluid properties and necessary conversion factors are found in the tables of Chapter I or in the appendices. Some of the problems [those designed with an (*)] are intended to be solved with the aid of a programmable calculator or a computer. The solutions for each of these problems are presented in essentially the same format as for the non-computer problems. Where appropriate a graph of the results is also included. Further details concerning the computer and their solutions can be found in the following section entitled Computer Problems. In most chapters there are several problems [those designated with a (t)] that are "openended" problems and require critical thinking in that to work them one must make various assumptions and provide necessary data. There is not a unique answer to these problems. Since there are various ways that one may approach many of these problems and since specific values of data need to be assumed, looked up, or approximated, we have not included solutions to these problems in the manual. Providing solutions, we feel, would be counter to the rational for having these problems-we want students to realize that in the real world problems are not necessarily uniquely formulated to a have a specific answer.
One of the new features of the Fourth Edition of FUNDAMENTALS OF FLUID MECHANICS is the inclusion of new problems which refer to the fluid video segments contained in the E-book CD. These problems are clearly identified in the problem statement. Although it is not necessary to use the CD to solve these "videorelated" problems, it is hoped that the use of the CD will help students relate the analysis and solution of the problem to actual fluid mechanics phenomena.
Another new feature of the Fourth Edition is the inclusion of laboratory-related problems. In most chapters the last few problems are based on actual data from simple laboratory experiments. These problems are clearly identified by the "click here" words in the problem statement. This allows the user of the E-book CD to link to the complete problem statement and the EXCEL data for the problem. Copies of the problem statement, the original data, the EXCEL spread sheet calculations, and the resulting graphs are given in this solution manual. Considerable effort has been put forth to develop appropriate problems and to present their solutions in a manner that we feel is helpful to both instructors and students. Any comments or suggestions as to how we can improve this material are most welcome.
COMPUTER PROBLEMS As noted, problems designated with an (*) in the text are intended to be solved with the aid of a programmable calculator or computer. These problems typically involve solutions requiring repetitive calculations, iterative procedures, curve fitting, numerical integration, etc. Knowledge of advanced numerical techniques is not required. Solutions to all computer problems are included in the solutions manual. Although programs for many of these problems are written in the BASIC programming language, there are obviously several other math-solver or spreadsheet programs that can be used. A number of the solutions require the use of the same program, such as a program 'for curve fitting, or a numerical integration program, and these "standard" programs are included. For those requiring use of one of the standard programs, there is a statement in the problem solution which simply indicates the standard program used to solve the problem. A list of these standard programs, with their file names, follow. The actual programs are given in the appendix. Most of the standard programs are, of course, readily available in other math-solver or spreadsheet programs, and the student can simply use the programs with which they are most familiar.
Standard Programs-File Names and Use Curve Fitting EXPFIT.BAS
Determines the least squares fit for a function of the form y=ae bx
LINREG l.BAS
Determines the least squares y=bx Determines the least squares y=a+bx Determines the least squares y =do + d JX + d 2x2 + d 3x3 + ... Determines the least squares y=ax b
LINREG2.BAS POLREG.BAS POWERl.BAS
fit for a function of the form fit for a function of the form fit for a function of the form fit for a function of the form
Numerical Integration SIMPSON.BAS
Calculates the value of a definite integral over an odd number of equally spaced points using Simpson's rule
TRAPEZOLBAS
Calculates the value of a definite integral using the Trapezoidal Rule
Miscellaneous COLEBROO.BAS
Determines the friction factor for laminar or turbulent pipe flow with the Reynolds number and relative roughness specified (for turbulent flow the Colebrook formula, Eq. 8.35, is used)
CUBIC.BAS
Determines the real roots of a cubic equation
FAN_RAY.BAS
Calculates Fanno or Ray leigh flow parameters for an ideal gas with constant specific heat ratio (k> 1) for entered Mach number
ISENTROP.BAS
Calculates one-dimensional isentropic flow parameters for an ideal gas with constant specific heat ration (k> 1) for entered Mach number
SHOCK.BAS
Calculates normal-shock flow parameters for an ideal gas with constant specific heat ratio (k> 1) for entered upstream Mach number (Ma)
3
t. t
I
1..1 Detennine the dimensions. in both the FLT system and the MLT system, for (a) the product of mass times velocity, (b) the product of force times volume. and (c:) kinetic energy divided by area,
mASS
;( ve/oc;'& .:.
(;VI ) (L 7-
1 )
-
F .:. M L T-.2
Sinee.
Fr
=
( b)
./oree
J(
Y&/I/ml!
-
_
(~
)
F L3 (ML T-2.)(L3) _ /'1L if T-Z.
J::,;'e/:'G e ne r.!~ t:l
reL /'1T
/- I
-2.
/'2
1.2
Verify the dim~nsions, in both the FLT and MLT~ystems .. ofthe folioWing quantities which appear in Table 1.1: (a) angular velocity, (b) energy, (c) moment of inertia (area), (d) power, and (e) pressure.
= a 1'19 tI //1 r c/'spkce/?'J()~';' -time
( 0.)
(.b)
..!.
e he 1'"1:J ~ C.a.;aci +!J 01 b~cJ!1 1-0 do w()rk Since.
Wt?/'"K
= I()rce;(
d/sl-tll1tt:..)
~nerJ!J tJr
~if;,
;
FL
F _' /11 L T- 2
e. n erj tj ~ (M I- T -2) (L) == M L 2 T - 2 cc) /7l{pmfl1t 0/ inerlltt.~V'ea.) =
sec~l?d /nl'Jme/}f
. (1.:2-)(L~)
=
+-()rce
-
.£
,
~
LZ.
. L= F
=. L If
2
J..---------- - - - - - - - - - - - - - - - - - - - - - - -
/-2.
D/
t:lff?l
1.3 \. ~ Verify the dimensions, in both the FLT system and the MLT system, of the following quantities which appear in Table 1.1: (a) acceleration, (b) stress, (c) moment of a force, (d) volume, and (e) work.
a cc-e/e ro.:tt'tJl1 :::: ~ t-r-<
eS5
(C)
=
./C)Yce
/?1t:J/)')t"l1i ,,{
0. rea..
(£
Ve.JDC.I+~ .:=
+/me F. == L;" -
(-kyce
=
.force.K dlsftln('~
=f/1LT-VL ...: (a)
volume
(e)
Work -
Oen~f-h) 3.-:.
2
I1L T-
L3
--
!=L
/- '3
.-: 1= L Z
/''1
I I I
I
1.4
If P is a force and x a length, what are the dimensions (in the FLT system) of (a) dPI dx, (b) tf'Pldx\ and (c) JP dx?
dP
ra..)
-
(b)
d 3.f
dJC
dx:. (C)
-. --Lp- -. .
:::r
3
jPdx
!= L- 2
F -. -L3
-.
-"'
1= L-3
PL
I i
I I
I
/.5
I
1.5 If p is a pressure, V a velocity, and p a fluid density, what are the dimensions (in the MLT system) of (a) pip, (b) pVp, and (c) p/pV 2?
(a. )
1> _ --f
--.
f.1L-'T-Z.
(ML -3) (LT- I )
.
Z
--
'--_ _ _ _ _ _ _ _ _ ._........... _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _......J I-~
/. ID
I
1.6
If V is a velocity, fa length, and \I a fluid property having dimensions of UT-I, which of the following combinations are dimensionless: (a)
vr", (b) VC/',
v.R -V
(1:, )
(d)
V ).11
- L~ T-1 ,
. LOr"
(Lr')(L)
-
V 2 -z)
(C! )
I
V'" (d) VIM
V J. -zJ .:.. (L T -'j(L)f1. z r)
(a.)
j· 7
(e)
( dimension /ess)
(L'2. T I)
-
(L T-) "(L • r - I) ~
-
.
(LT - 1 )
{L )(L' r
-
')
mol dlm.nsienle,s)
L~r3
-l.
(oof dimfnsl'oIl!ess)
(not
L
dlfnen sion!e>s )
Dimensionless combinations of quan1.7 tities (commonly called di mensionless parameters) play an important role in flu id mechanics. Make up five possible dimensionless parameters
by using combinations of some of the quantities listed in Table 1.1.
Some possible
e" Q mpl e~ :
u C( e Ie r,,-/-'M " f 1m e ve /OCI f '1 frefllenc'j ;(
hme
(ve!oci+!j)
2.
/ t'179 f !? x.
-
•
-
(L r2)(T) (L
(rl){r) ..:. (LT - I)'"
•
=
rlJ
(11 LT -~
-
-
TO , ",
(L)( L r'-)
(F)(r)
-. L"T"
L"T
(j=){T) :. (1'7 zr:J(Lrj
F"i"TO
I deMif-') " velocil-j " len-P'4 --' (Mr 3)(LT - }(d =• M'L"T ' Mr ' 7-1 d'f nllr>1i< visUJ~if:J
-
1- 5
I
/.~
118 The force, P, that is exerted on a spherical particle moving slowly through a liquid is given by the equation P = 37CJlDV
where Jl is a fluid property (viscosity) having dimensions of FL -2T, D is the particle diameter, and V is the particle velocity. What are the dimensions of the constant, 37C? Would you classify this equation as a general homogeneous equation?
.p =- 37T;
==
[:?7TJ [pJ
37T 15
/.5
tt
d/men.510I1Je~s,
1 ene Y'(J/
C(nd
-the ~2t1a!-/{)I1·
hl/rn()~eneOU5 efJtAa..f-/on.
/- ~
yes.
/. 'I
I According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula h
=
(0.04 to 0.09){D / d)4V2 /2g
where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid
in any system of units? Explain.
~=
(O.OLf
gn= [D.O~
1-0 ('). {) 9)
(.!J )If 2.J ~~
O.O~ [tJ[i] [~:J[ t]
1-.
[L J== [O.OLf -1-0 0,07] [LJ Since eac.h hrf}z
li-t
the e$tt.a..f./~h must:
n4t1e
the
-the Cf!)I1~"'lfi I-erm (~. ~'f ~ ~. ~tj) rnusf /;~ climfns/f.,hless. Thus the e$ti/{,t/~H /.5 a. !J~n(lY~ I 1 h~mo1enet!Jvs €lttA..-6I4;;' .fh{(.i: IS 1I11//c/ IH CiI1.!! ~!:f5Iem
:Slll71e
d;'mel1$/tJh5
I
~f Un ,·f..:5.
t. /0
Yes.
I
1.10
The pressure difference, Ap, across a partial blockage in an artery (called a stenosis) is approximated by the equation
.1p = K!
pV (All D + K" A I
-
)2 p V-
1
1
where V is the blood velocity, Jl the blood vis-
cosity (FL -~T), p the blood density (ML -3), D the artery diameter, Ao the area of the unobstructed artery. and A I the area of the stenosis. Determine the dimensions of the constants K,. and K". Would this equation be valid in any system of units?
Since eac.h -terM mv.st h~lJe. the same dimensions; k'v Cll'ld Ku are dirnen5ionJe-:'5. Thu~.1 fhe efuafltJJI/ IS (;( ttener~1 h()f71~jel1eO"s e~ ua.l-;tJv, -tnCI'/- w{)uld be va/ic/ t'n Cfn!! C()tJ5isffnt sfjsl-em of U)1jf5. yes.
/-7
I. / /
J
I . II Assume that the speed of sound, c, in a fluid depends on an elastic modulus, Eu, with dimensions FL ~2, and the fluid density, p, in the form c = (Eu)"(p)h. If this is to be a dimensionally homogeneous equation, what are the values for a and h? Is your result consistent with the standard formula for the speed of sound? (See Eq. 1.19.)
0)
FPr ~ d)J11eY1~/Of1tt/I'1 h(!)mt1ef1eDIJ5 -€$ad
ea.ch +erm 'dlmeY15JO#.s, Thtl5, {!)'J1
the etua.t,bJ-f fntlS1- haf/(. -fJu 5f1/)/e -/ne Y"'9Jtf hand ~/de ()f. P~l OJ mus+ h~ve the dlmenslPA,s of- L 7-'. There /dYe) In
a-tb==o 2.},=-1
(i:1>
sa -/-1 's.f." C6"t/, ',,,()~
"n r)
.ta -f If b = - I (.£. :!iJ 1-, ~ I-y ~Y1 dJ/o'" "" L) a. =LZ. tlnt! /:; = - ).2. So
c = ~i0:
-tn..-f.
Thb re.5u
1+
/s
1
~nsisl-f"r /AI;-!/1
:5peed ()j2- 5DUJlJd.
YeS.
1- 'j
the, sblltlt/J'p
~rIl1U/A
-kr 17te
I,
/2.
I 1.12 A formula for estimating the volume rate of flow, Q. over the spillway of a dam is
Q
= C v28 B (H + V2/2g)3/2
where C is a constant. g the acceleration of gravity. B the spillway width. H the depth of water passing over the spillway. and V the velocity of water just upstream of the dam. Would this equation be valid in any system of units? Explain.
5/~ce ea.c;"
I:errn ,i1 ~e .e.Su.Lf/~H rnus-t- ha.ve +he SQ/7Ie dimellsi{)l/s -the ~11.sb1l/i C VI must:- he cilmeI15/!)/J )e~s. Thtls; -tnt!.. .et(f~tltJH is a ~-ene r-a I htPl1IP ,e/ledJ t(J eg Ua.,tIOJl -1'n¢,f WOf,{ /~ 411'1 e4)A~/sl:ent Set: of (,Iilif.s. Ye~.
be. v t).. //d
I»
/. / if
I
(c>-)
1.14- Make use of Table 1.3 to express the following quantities in SI units: (a) 10.2 in.lmin, (b) 4.81 slugs, (c) 3.02lb, (d) 73.1 ft/s2, (e) 0.0234 lb·s/ft2 •
1t),2
:;;'1 - (;0. 2 - i-.
;,;J (Z,S*;t/O-",:'.) ( ~;;n)
'a2. .;c It)
[ h) If. 9/ S/fA l' =
('I:?/
( ~ ) 3. tJ:L /b:::
(3. ~ Z / b ) (
Cd) 73. J :Efi
-3 /W1
s
=
sill!> ) (;. 'f$f' ;< I () If. If'If
f1 ).=:
tf. 32.
T
sju~) =
70, 2 ). ff
/3. If AI
:
ce) CJ, tJ23'1 Ib·s
~
ff~
(0. ~Z3'f ITt.)
('/,7.?1;tIO
N· -': ",.,1-
lb. s -ft'l-
-
I, /2
N·s M'J'l.
1-/0
)
/./.5'
I
(b)
1.15 Make use of Table 1.4 to express the following quantities in BG units: (a) 14.2 km, (b) 8.14 N/m 3 , (c) 1.61 kg/m\ (d) 0.0320 N·m/s, (e) 5.67 mm/hr.
o o.llf.
!!..3
,11'I'f
" (g. 'If ~
)
(~3U;(/O·3
':3 )
= 5'. IF)( 10'2
,,",,3
(
l I.
Cf Iff) )(. /0
-3
SJUjS)
W
=
~
~~
(d) 0.0320
N-1'H1
-S
--
(~, 0 j 20 N ~ I1f1
)
(7, 371P;( /V-I -il-·Ib ) oS
N·/'M
-
2.3b)(JD
- s: 17
)1.10
-2
-to
/-11
.{.f·/b
-
-1-1 ...5
oS
-
oS
Pt.
/. /(0
I
1.lG
Make use of Appendix A to express the following quantities in SI units: (a) 160 acre, (b) 742 Btu, (c) 240 miles, (d) 79.1 hp, (e) 60.3 OF.
IfpO a. ere
(6)
7tf2 137U
=
6'1-2 sru) (.°£,;
C~)
.2LjO int.'
=
(;'''10 tni ) (;'''Oq;(./(;.3 1"YY1,)::: 38iDX/oS"t?11
Cd)
71. / hp
(e)
I'n1L
Tc
=
0:
(7'i'./ hp )
l' ~1).3
k = /5",7
f) (
-
-r
32)
(7.'f5"7 X /02. (;{;) '" '=
273
/5.7 "C:: ::::),
1-/2
gr
1<
7.g3X/~5J
/./7
I 1.17 Clouds can weigh thousands of pounds due to their liquid water content. Often this content is measured in grams per cubic meter (glm3). Assume that a cumulus cloud occupies a volume of one cubic kilometer, and its liquid water content is 0.2 glm 3. (a) What is the volume of this cloud in cubic miles? (b) How much does the water in the cloud weigh in pounds?
U (;0'/111.1) (g, Z8'1 ~ ) (£ 2!b >fIb) 3
1M1= 3.281
J
t:)
0,2 £j 0 nn,,3
(h)
0 X -Vol"rn~ l d' =: jJ d = {0.2 ;'3 ){!D-
%J ==
"lJ =- (I. '( (,,2 = (I. "t,z
;( JD -3 ;;', )
X /D (. N )
;g. )(r.8/ ;) =f. UU/iJ-;;J
(10 1;m3) = /. '( ~2
X I DI,
N
(:1., 2tf8 x/D- -J& ) :::: ~, If! X JO f h 1
1- 13
S
1.18
1.18 For Table 1.3 verify the conversion relationships for: (a) area, (b) density, (c) velocity, and (d) specific weight. Use the basic conversion relationships: 1 ft = 0.3048 m; lib = 4.4482 N; and 1 slug = 14.594 kg.
(a)
I it 1..:
.ft'")f(a 301f.>') L
2/1?1 ,,-]
I-i ~
-ft 2 bJ
rnu//-'/0
Thus)
fo
(/
9.
= 0, () q 29{)
'2'i{)
£ - 2. +0
/H1
~
t!trJnvfrf
/ffI :2..
II;) /
slugs/.ft.3 b!:J 57 IS-If
Thus) mu/fipJ'j
E of 2. ;'0
CtJl'Jtlfrl
-to Ie? / /I'n ~ (I!)
If-
/
fj ) (~. 30'/; jJ)~
= (/
Thus.) muillpl!) Ills -I: 0 (d)
bIJ
3.0'le f - / -1-0
cOl1vert
/s.
/t11
I JIz - (I !l ') (If. 't'l12 !!..) [ I Ii 3 3 l If 3 - l' -It 3 ) l ~. /j, ( 0, "3 () Iff) /W1 3 J
-= TfJlAS)
fo
m
IA
IV /57, / ;;;;
If/pI:;
#/;m3
/ b/R ~
4
/-/if
b!:J /. 5'7/
}; -t 2
-10
t'e>ntlfY't
/,/9
.J
-
--
1..1 q
For Table 1.4 verify the conversion relationships for: (a) acceleration, (b) density. (c) pressure. and (d) volume f1owrate. Use the basic conversion relationships: 1 m = 3.2808 ft; 1 N = 0.22481 lb; and 1 kg = 0.068521 slug.
(a)
Thus) m""/+ipllj tt/ .J.t / .5 J.. (b)
I ~ ~ = (I ~3 ~
1111
""
-
I
040
. 1
')
x /0- 3
N
'2. () g r i. I D
-.2.
"='
Thu5) m/,.{lfip/~
N/rrn l
(3. ZFO~)3
-f1:: 3
J
/. qLfo E-3
~J1t/fri.
to
l (3.IlfOg) ft l J (M1.
2.
1.
Ik
f.t1b~
;;'.Ogq
E-l fo ~~n()fYt
/ h / f.t :L,
1::-0
7
3
==
T h US)
+(/
h,!j
tn1 2.
/'I't1 ?
1m,3
S l u ~~ f-t3
(I !:!. ) (O,2.2lfgl ~)f
I Ji ::
(d) /
T;
\ (
Th ~S.i m ul.f.i pJ'1 ~J/tt113 -1:0 S /u~/.ft 3. (C)
slugs) [
(0. oft> f/5:L/
(I ~) [cg, 1.KOS/~:l= rn f.,( I t
ifl':J
3
1»1 /5
b~
ft 3/s.
------------~~-
--------
/-/5
35". 3/
fr'
3. 531 E+ I -1:.0
rlOl1Vfyt
/.2..0
J
1.20 Water flows from a large drainage pipe at a rate of 3 1200 gal/min. What is this volume rate of flow in (a) m /s. (b) 3 liters/min. and (c) ft /s?
( ()...)
f./owrat e =
757 ;<. 10
/ Ii fer
(b) Since
/lowrfLte=
(C )
I I() W r
(I.
2
/i'Y7.3
.5
~
= / [) -3t1"/1
(7.57 ;'/6-
-:2
~.3)(/o3///.er.5)({Po.s) S
+. e. =
(7 S 7 )( J()- ~ if 3 )
:: 2. ~ 7
-I't J
s
-
I-/~
/H1 3
(3 S3 I
/'1?1/11
X J0
1.2 ,
1,,;2 /
A tank of oil has a mass of 3 0 slugs.
(a) Determine its weight in pounds and in newtons at the earth's surface. (b) What would be its
( t(.)
mass (in slugs) and its weight (in pounds) if located on the moon's surface where the gravitational attraction is approximately one-sixth that at the earth's surface? w.e i9 h i- .: ~. as.5 )(.
3
=
(3 0
- (30 ( b)
/h')
4
s.s
5
/uqs ) (
32.2
shillS) ('t. Sf
= 3 ()
5 J/A 9 S
;:)==
_o/~r;, 16
14 )("I.E! -f,,)-=
( /n1 ASS
dtJts
t}IJt-
,/Z'foN
dep~;1d
t!)1'1
JY'~ vihfitJl1ll / a ffrtu..J-if!)11 )
w.eijhi =
(30
s/uqS )
(32.~:Ef.. )
/ fa/
/b
;,:2 2 1.22 A certain object weighs 300 N at the earth's surface. Detennine the mass of the object (in kilograms) and its weight (in newtons) when located on a planet with an acceleration of gravity equal to 4.0 ft/S2.
9, 8/
'I: () ft Is :J. )
- (3tJ.(P
Jj. ) ( if. 0 ~) ((), 30'fg ; ; )
= 37.3 N
1-1:1
1.23 An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as Vlv'g'ii, where V is a velocity, g the acceleration of gravity, and r a length. Determine the value of the Froude number for V = 10 ft/s, g = 32.2 ft/s 2 , and r = 2 ft. Recalculate
In B 6
the Froude number using SI units for V, g, and e. Explain the significance of the results of these calculations.
tI/lits /
/.25"
;0
In
JI uni-t-s: V:: (to ft )(~. '3IJJfr S
~):: 3.06 ft
T
~;: 1',:g I ~
~
::: (~+t:) (0. "3 04-g ~ ):: -Fe
v =
--
y!~
Th e. Va /lle D I in cle;enciel7i
O. b I 0
t'l?1
1.25
d im-et1sjt'J n less parl!met ev un i t ~1 sl-em. of -the a.
1-/8
IS
1.2 4- The specific weight of a certain liquid is 85.3 lb/ft 3• Determine its density and specific gravity.
g5.3
d" ;0 -= -
1
56=
Ii? -.ft-3
2.&'5
'32,2 .pc
s I u 9.5 f-t3
5.2.
fJ
I.If;l.O @
f~c
-
2.~5
/. fi-
5/,,?.5 k-i
1.37
-
S/W9S ..ft.~
/, '25 1.25 A hydrometer is used to measure the specific gravity of liquids. (See Video V2.6.) For a certain liquid a hydrometer reading indicates a specific gravity of 1.15. What is the liquid's density and specific weight? Express your answer in SI units.
5G -= //5
-
(J
~D@'" °C
f /o/)o .k;'
1m 3
f== (I. /5) (I ()r;O :'3)
1-
/q
1150
~
h)?3
/.2 10 1 l. 2~
An open, rigid-walled, cylindrical tank contains 4 ft 3 of water at 40 of. Over a 24-hour period of time the water temperature varies from 40 of to 90 of. Make use of the data in Appendix B to determine how much the volume of water will change. For a tank diameter of 2 ft. would the corresponding change in water depth be very noticeable? Explain.
/)1QSS
of w~l:er = -V
X
t
Wheve ¥ /s the {/oh{rne and! 1he. deI15rfr:1. J/J1Ce.. -the. rnA$~ mU$1- Yefl1111M ~l1sfa)1i (/5 the -iempera.-tuye ehf/flqeJ
-tf x iJ 'fcc / 'io Ff~'11
-tI-)( ~
:=
'1~. (f~
p
(I )
(>
ra6)e B. J
/Hz o ~ r~"F = I. 13/ s~ TherekYt) (nil'! E $- (/)
+
.:!':!i~ )
/'
= ( if /t.3 )( I, 9'1"
1p ()
I. y"j J ~:;:3
Thus; the, I~crell~ In
Vt)
lumt:
,:s
'I: 1/Ji L - If. "00 -== The
If DIFb -Pi.]
.f-c3
chtlltfe Jil Wl..fey
cle;1Jt" 41) O. OJ;~
il-V-
Ai.:= a rea
:=. 71
0. 0/ i I:, .i
.pt
"o/tI~j fo
..ft3
(7f-1:)
.3
-3
== 5, '12 xlD
+t =~. ()7/~ in.
2-
Lf-
7h'5 ~/lJ4I/ e-hl(Hge tJ()l-lcet:l/;/~.
In def1h
would n~.J. he iJel"!1
AI0,
,4 S/;1hf/lj d.:PkY(~i.
b~ "h.fa;HfA If ~I'~c"f,i ("J(I;hf lJ!-wphr Jr Iur,r fflilJey 1ltQIf4t11s/-J-!1' 11J1~ 'J du e -10 t'h.e /rtc.t tho! 1Jtele is SIP/II e IIHcem,id]l 117- -!itt! fi,Jlr1h ~/;1;ln(~111 /'9l1Y'e of 1Jte..re. +tv" 1It//l(es,l lit'! ff;.(J ~()//,('h~Jt
'S
Vfi!,,(!
for' .l1)
f-I/; II
SPfls/fl';~..fl':J 7}"j unc..ryitlin-J.t;.
/-20
/,27?
I 1.2 ~
A liquid when poured into a graduated cylinder is found to weigh '8 N when occupying a volume of 500 ml (milliliters). Determine its specific weight, density, and specific gravity.
(=-
f= S6
gN
w~i~ht
.=
10. a
1/0/ tllYJ e
?!
c;.
-
/~
1.81
:3
JL
- J. ~3
/1113
hH
JfDC
-
x /0
3
-k ~ 1»1 3
57..
1
f ~o@
/ra;<.
/. b3 x / D
/0
/- 2/
3
~'f.
..fEg. ;m3
;m .?l
/. to 3
/,2Cj
1
I. '2. q The information on a can of pop indicates that the can contains 355 mL. The mass of a full can of pop is 0.369 kg while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at 20°C. Express your results in SI units.
y=
-h~/ we/fltf
v~/JlJf
()+- I-/UIC;
{/p/um~
t:J/ .fltlt'c/
= maSS x
(/ )
~. 3tf J# )(r.JI ~)::: d: 62 II
9- =
~f GIn:' C/. /53 IV r. / Vp/l(l'n~ ~ / "'-/1114::- c:i'S5 oX I~
wf,jhf
Th u~
I-r~1t7
-3L) (/0 / -3T/YYI3) =-
~:,-S- x/tJ
-(./YYI 3
E%. (/) ~. 153 II
3. "Z /II -
-
Cf77o!::! ;rna
tf 7 7013 r.8J~ oS;&.
~
rtf ~
1'm.3
j~o~
99/'
(J.
-
y /J?'13
Ulaler v
~t
20·C
(see.
- '17 g'13.'3 ·
oJ+z. iJ -
/J11)
/.J
~.J/e
(Jt.z. ~
::
B. 2
J~
ApjJfHd,X
f'/t. 2 ~ 1')n3
.
56
J])
= 0. qqg 2.
)
jj etJll11l"n;;" ~f 1AlS~ Jltl/IIR.I £" lOll/IV with 1ht)s~ ~y 11te P(J); sh()w.s 1Jt~.j iJ;~ ~~C;-hC 1A.)(',jhf~ c/tnS,fYI ifl1d ~eClhc' :Jr/tv,f, cf- iJre i",P are. all Sl'jhflfj Jp l
/-22.
/.30*1
1.30*
The variation in the density of water, p, with temperature, T, in the range 20°C :$ T :$ 60°C, is given in the following table. Density (kg/m')
1998.21997.11995.71994.11992.21990.21988.1
Temperature (0C)
I
20
I
25
I
I
30
35
1 40
1 45
1 50
Use these data to determine an empirical equation of the form p = c, + C2T + C3T1 which can be used to predict the density over the range indicated. Compare the predicted values with the data given. What is the density of water at .42.1°C?
To
S()/ve
1h:S
pr()~Jem use
POLRF6.
*************************************************** ** This program determines the least squares fit. ** ** for any order polynomial of the form: ** y = dO + dl*x + d2*x 2 + d3*x~3 + ... ** ** *************************************************** A
Enter number of terms in the polynomial: 3 Enter number of data points: 7 Enter data points (X , Yl ? 20,998.2 ? 25,997.1 ? 30,995.7 ? 35,994- . 1 ? tiO,992.2 ? 1±5.990.2
The coefficients of the polynomial are: d2 = -4,.0953E-03 d1 = -5.3332E-02 dO = +1.0009E+03 X
Y
+2.0000E+01 +2.5000E+01 +3.0000E+01 +3.5000E+01 +1±.0000E+01 +1±.5000E+01 +5.0000E+01
+9.9820E+02 +9.9710E+02 +9.9570E+02 +9.91±10E+02 +9.9220E+02 +9.9020E+02 +9.8810E+02
Tn US)
Y(predicted) +9.9825E+02 +9.9706E+02 +9.9566E+02 +9.91±07E+02 +9.9226E+02 +9.9026E+02 +9.8805E+02
f=== /00/ - O. OG"333 T - 0.00'1095 T:J. !Vote tl14t f (pJ'ecl'~fed) ~ l'n 9()OO Q9reemfl1t w;'1h f A t r = '1-2. / "C) ! = /00/- O.~S333 (Jf.Z. / DC) - (J. {)O tj.O?S (1f.2./ cc) ~
/-23
(gJ~~h).
/,32
I The density of oxygen contained in a tank is 2.0 kg/m 3 when the temperature is 25°C. Determine the gage pressure of the gas if the atmospheric pressure is 97 kPa.
1.32
p= f)/U
= (.2.0 #!. )(.)51.8 Ie~k.) [r.we r ,m)Kj -
-p
/.33
/5'5
i Pa.
(JCd/e): -1;,fibS I
-
(4 bS )
1:.4.rm I
::
/g5 J.~
- rt71e ~ = 5? k ~
I J.33 Some experiments are being conducted in a laboratory in which the air temperature is 27°C. and the atmospheric pressure is 14.3 psia. Determine the density of the air. Express your answers in slugs/ft3 and in kglm 3.
P=fJRT Tempera. fllYe,
-
0.00222
/ :: /0. Of) 222 SlIl9.s) ( c. R a
(s. /S¥ X !,,g! ) :: 2
.5LuS.l
1.14 1r
--:r£~
"
If)
In"
l.3 If A closed tank having a volume of 2 fe is filled with 0.30 lb of a gas. A pressure gage attached to the tank reads 12 psi when the gas temperat.ure is 80 of. There is some Question as to whether the gas in the tank is oxygen or helium. Which do you think it is? Explain how you arrived at your answer.
W~/ji, t = tJ. go IJ, ~)( (lo/ume (';2.2. ~) (z. ft3)
1=
1~~x/o
~ 1= -
Sin ce.
( J2 T
pre.sStlre
(tf/J (>;: + If.b~)
T =-
;0:::
Ii
llR
/
St.(9S
.;:-t;3
I't: 7 )
fS/~
~
/ if-: 7 f.J'/a )
.ft, //,,(jJ j
'"that:
b(
~~JUMFd
-3
(2/,,7
7
12 R.
(I)
3
lor- O)(jgel1 Ta ble I. 7 R :: /. 5"'S"1f X J~ ;€ = I, 2 if.Z X It) If Ii· J.j, ~y he/lto'n.
rr4'1'J?
ttinc1
0511/9 ' (),R.
Thus;
./r()1'}1
Ff.(J J I;:'
I
-!he 9tif
7./Z
::- 155'f X/~ 3
Is
s/tl1..:5
/t3 =
f.9XY'l/n
*5' i.x M
-3
!/u~ H3
he /11,lm ,tJ -
7. /2..
r- - /, 21f2 X/I)
A-
~mJJIIYJ51J1}
of 9t1S
-!7te
'1 q5
/?1 uS
i-
If
6/
1he.5e va/lies w/ fl,
/ rl
the -ban i.
be
1- 2 5
-tJ"e
/nell C.1I~...s
tlC/:t(o/ df!1>/~ 7h /I t 1it e
1.3G
A tire having a volume of 3 fe contains air at a gage pressure of 26 psi and a temperature of 70 oF. Determine the density of the air and the weight of the air contained in the tire.
t==
R.T
.:
wei!JH ::- !
~ 1P!l20 .,. / It, 7.f! ) (Ilflf In.2.) Jh.
(/7/~
I J1
.ft: "L
2.
h,/)' ) li(7~d;:+'ffPO)"R 5hlj'~ II ~
3- ;( ",,,Iume = (t..1fIf x/0
1-
2f.s.
3
~, "If)( /D
-3
I
s~
s!,,:) ('32.2 ::) ( ~.f.t~
/.37
I 1..'07 A rigid tank contains air at a pressure of 90 psia and a temperature of 60 oF. By how much will the pressure increase as the temperature is increased to 110°F?
-P::tRT J=oy a Y'lr/4 c.losed Jan./( V~/vme
4Y~
'DI1"iR~Z.
/!rIPt7I ct. /.
f
-the "';, rnpS5
,;fO
1=
4nd
~n5i:4nt-.
Thus",
(W/~ R etPI'I5rq"t)
-P, _ FL T, wht'f'e
"nil
-7;..
(I )
-A ~ fit) psia-) r;:: bO r -J- Jflt,D Ii
7i::
//()oF-+Jf6o = s-'lO~.
-b
l-l7
FY'~pr
ct. (j)
l<) (flJf~t.A.) = '18'. 7 OSLo..
~ = (/S7tJ r2. = 7i.. 7; 0 5z()-;e eJ
- S2.~ c ~.1
I .
I. 3 i
-'II:
I
"J .3X
Develop a computer program for calculating the density of an ideal gas when the gas pressure in pascals (abs). the temperature in degrees Celsius. and the gas constant in J/kg· K are specified.
;::';;1"
,dtlt/ ~115
lin
1::/RT
1=
~
"kr
t.)htY~ t iJ tlb.$l)/u/-(l. 1'1"!'SS"~) R 1h~ 9"S (}PIIS J-f,(1'1 i: I ClI'1~ is tlbsl)l",,~ -lempRrtlwre. Thus" ,-I 1'ht!. t~mJ~f-I4.t:UY~ {s
/ J1
4)C
7
-Inti'!
T =
~
{ +
273. JS
,4 spreadshp8 t (exCEL]
PY'()jY4h1
~,... C/4leulai-lIlj fJ
This program calculates the density of an ideal gas when the absolute pressure in Pascals, the temperature in degrees C, and the gas constant in J/kg-K are specified. To use, replace current values with desired values of temperature, pressure, and gas constant.
j.o/jfJLt)s.
I I
i
I A 8 CD! --+--------+--------~~~--~----~----~ Pressure, Temperature, Gas constant,. Density, : Pa 1.01 E+05
°C 15
J/kg· K 286.9
~----~--------~-------;
I
kg/m
3
I
1.23
Row 10
Formula: =A1 0/«81 0+273.15)*C1 0)
1
~xt¥rnp/e" taJcuLa..f-e I ~r P= 2.~o~ P()..) trhljJfrl.i:ure. ~O·CJ t1'1~ R:: 2..97 J/~. I~ I
A
8
C
Pressure, Temperature, Gas constant, Pa °C J/kg-K 20 287 2.00E+05 i
1-2.~
D
I
Density, 2.38
Row 10
I. 3f-- I
':'1.3 l ) Repeat Problem 1.38 for the case in which the pressure is given in psi (gage). the temperature in degrees Fahrenheit. and the gas constant in ft·lb/slug,oR.
/c/ea/
ql?
F(J)Y
9qS
1=fRT
/.)::: 1<; .::t
,
ThIl..5
/
Q"~~/lJ.fe +emptrai:tlre. f ~empYa-bo"e IH ";:: tfn" PYt'5~uve ,ft os£. -the" . -a
p
whtY.f!. I
T
/.s a
bs()/utt
= tI;= -r
prf.5SUYl'j t:i1'1I(
T
/
if5~. ~ 7
IS
r
f':: [
411r.1
I
p(1"£) -r t.ihtt (psia.)
.4 spreAdsheet. (t:XCEU pr()!f4m ~y C4/fL{J~t-Jir." This program calculates the density of an ideal gas when the gage pressure in psi, the atmospheric pressure in psia, the temperature in degrees F, and the gas constant in ft.lb/slug.deg R are specified. To use, replace current values with desired values of gage pressure, atmospheric pressure, temperature, and gas constant.
f
Jx 14LfJ;t~ In.
lallows.
~:~-~-~--------------~------~----+------4------~----~
i
E ABC D I Pressure, Temperature, Gas constant, Atm. Pressure, ' Density, 3 slugs/ft I of fi Ib/slug.oF psia psi
o
!
59
1716
14.7
i
0.00238
Row 12
~---l-------+_ _ _ _--+------il Formula: i=((A 12+D12)*144)/((C12)*(B12+459.67)) I
J? J(" rn)J Je'
ell /,uL~ -te fJ fi.J:""
~y
= 1'1-.7 P5L'(ij
P= LfOPJi.) tind
R= /7JI:, .fJ.t.lb/.sJU~'''~ ,
A I B I C i D j E . Pressure, Temperature, Gas constant, I Atm. Pressure, ! Density, psi I of ft Ib/slug of psia slugs/ft3
f-:::----
40
100
1716
14.7
J-
Zer
ifmprrIJ ture =/~~ ()t;
0.00820
I
i
Row 12~
___ _
/. '10
I
I.LfO Make use of the data in Appendix B to determine the dynamic viscosity of mercury at 75 of. Express your answer in BG units.
/-30
/.
~I
1. 4 J One type of capillary-tube viscometer is shown in Video V1.3 and in Fig. PI ~( . For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line. The kinematic viscosity, v, in m2/s is then obtained from the equation v = KR 4 t where K is a constant, R is the radius of the capillary tube in mm, and t is the drain time in seconds. When glycerin at 20 0 is used as a calibration fluid in a particular viscometer the drain time is 1,430 s. When a liquid having a density of 970 kg/m 3 is tested in the same viscometer the drain time is 900 s. What is the dynamic viscosity of this liquid?
Glass strengthening bridge
e
Capillary ---lr-+-.-li"\ tube
• FIGURE P1.41
~y ••
7J= !JCfxIP-~1s
~/tI~er/.n @ 20D[
!. / r X /1)-) hn"l-Is
::
Uc R Ij.) 0, ~30
s)
k R4-= 8. ~ 2 X If) -7 /}?12-1s 2.. IJ zu/d
win,
t::.
rODs
v= (3. $ ~ i/O- 7 /n1 "2./s 2.) (90
t) 5 )
=
-r-z/ =
(97 0 --k#~3) (7. If 'I x /0 -If fn1% )
D. 727 Im-S ~ = u,727
1-3/
I. ¥2
I J 042 The viscosity of a soft drink was determined by using a capillary tube viscometer similar to that shown in Fig. P 1.41 and Vidl'O V 1.3. For this device the kinematic viscosity, v, is directly proportional to the time, I, that it takes for a given amount of liquid to flow through a small capillary tube. That is, II = KI. The following data were obtained from regular pop and diet pop. The corresponding measured specific gravities are also given. Based on these data, by what percent is the absolute viscosity, J-l, of regular pop greater than that of diet pop?
Regular pop
Diet pop
I(S)
377.8
300.3
sa
1.044
1.003
- t
}-32.
J
1< /OD
1.13
I equation for the pouring time in seconds was t = I + 9 X 102" + 8 X I 0 3,,2 with" in m2/s. (a) Is this a general homogeneous equation? Explain. (b) Compare the time it would take to pour 100 ml of SAE 30 oil from a 150 ml beaker at O°C to the corresponding time at a temperature of 60°C. Make use of Fig. B.2 in Appendix B for viscosity data.
1. 43
The time, t, it takes to pour a liquid from a container depends on several factors, including the kinematic viscosity. ", of the liquid. (See Video V1.l.) In some laboratory tests various oils having the same density but different viscosities were poured at a fixed tipping rate from small 150 ml beakers. The time required to pour 100 ml of the oil was measured. and it was found that an approximate
-I:.
(a..)
=:
/
-t
fT] == [i ]
'I /o"l.-u J(
T
[tf;
+
9 X/OS -v
J
-t
(I)
2-
[3 x/oJ] [-¥.]
5/~c~ each +rn11 ;'n +he egutL.f:lbJ1 !1?"fs-t hftlle -t-he stlme d /1rJl'''''tM5 -tJte ~IJ 51-o",1-.s a..?petl r/n~ /rl 1J1e efllLa.:I:,clI
m u 51-
have
[)] :;; [TJ
dllnen ~/pif..s.I l [11.>< JD 1-J
~
'
e.. /
[.1::]
D~ X I D3J.:.
[ -3]
-b-
7htl..5) w; 1h a. c.hol1"~ I;' Ut1/fs /he J/,,/tI~ "I 7h~ C(J)115-fz1l1i5 wPt// tI e..l1l1l1fe q ntl -t7J/~ I S 11~t:- tt j-enent / horno jen e(J)tI.J. ..(2gaa-i'£;;J . ;\1.0. (j;)
Pr~m Ta /;Ie 8.2 /n A ppel1d1 x B (~r SAE:3'tJ ();/ @) O°c) -z/ = 2. 3 )( jtJ -,3 /Yn 2./s
(-for 5A-E~!) ()//
@
~DC)
-V = 'I: ~ x
/m2-1s
/o-s
l7f£. (; )
i::-
I
T-
C/Xj//
:3.1/ s @
(00° C
1+ I, 0'1-
5
)-33
(2.3X/D-~)-t
I. Lf4 I 1.44
The viscosity of a certain fluid is 5 x 1O-~ poise. Determine its viscosity in both SI and BG units.
/=
(5.>L/o-I(.,P~i~e)(IO-' ~~)= p~/$e
(/n
el
~
.
/. 'is
Frpl7?
~ 6/e
/.
If
~
-l
::: (5 X- /D - .!:!.:.!.. ) ( :/., o~q )(./ 0 /1'11 2
J 1.4S" The kinematic viscosity of oxygen at 20°C and a pressure of 150 kPa (abs) is 0.104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure.
vm'2.
- .s
-
*1.46 Auids for which the shearing stress, T, is not linearly related to the rate of shearing strain, 1', are designated as nonNewtonian fluids. Such fluids are commonplace and can exhibit unusual behavior as shown in Video V1.4. Some experimental data obtained for a particular non-Newtonian fluid at 80 of are shown below.
T(lb/ft2)-.J~ 2.11
I
l' (S-I)
01
1 7.82
50
100
I. 18.5 L31.7 I 150
I
200
Plot these data and fit a second-order polynomial to the data using a suitable graphing program. What is the apparent viscosity of this fluid when the rate of shearing strain is 70 s -I? Is this apparent viscosity larger or smaller than that for water at the same temperature? Shearing Rate of shearing stress, strain, 1/s Ib/sq ft
o
o
50 100 150 200
2.11 7.82 18.5 31.7
=
40
~...'
20
'r -
0
OO~8 i ±-q.Oill5-¥~.
! 30+--~!--~i--~i--~·~,--~i
+__+--_--+1_--:.1/fC-----1!----i
1../
U;
i
g' 10 +----l---j~--i~--t----rl------l
I
.~
:. U)
~~---+1---t---1i---------i1
0 ....- .....
o
50
100
150
200
250
Rate of shearing strain, 1/s
I \
~----------------------
-G'"
Table. S.I I~
1/0 •.5
A-PfHAti,x B) fl+z.O@$ODF = 1,7Q/XID +1;'-) q,,~ 5/l1ct.. WA.-tev is a. Newt:r;nl41f riu/d 171J~ l/a/ut. 1.5
/'lPf)f
/11~/eprl1dt"t ,,{
i .
TheiS", 17J~ t(l1KI1f)WfI nO!1-Newl:.t>Jltt1i-t /-/('ui/ hilS a mu(;.h /t'{Y''1fY VI:( J~ e. .
J-3S
/.47
I
1A7 Water flows near a flat surface and some measurements of the water velocity. u. parallel to the surface. at different heights, y, above the surface are obtained. At the surface y = O. After an analysis of the data. the lab technician reports that the velocity distribution in the range 0 < Y < 0.1 ft is given by the equation
u
= 0.81 + 9.2y + 4.1 x lOV
with u in ftls when y is in ft. (a) Do you think that this equation would be valid in any system of units? Explain. (b) Do you think this equation is correct? Explain. You may want to look at Vicko 1.2 to help you arrive at your answer. :J .3
U=
(a)
~.a/
+ q.2 :J -r tf..1 >(/0 !J
[Lr-]= ~.8il1-0.2][L] +- ft.JX}D~[L3J E'ach ferm ,A 1he ezua.;fJ~1I rn liS f h4 V! the ~(Jrne dl fYlfUI5I!)IIS. Thus,) the ~/J~t"J1t &. f / m1l51: hAlle dln1tl1~'IJII~ "f. /. . T-; 2. dl':"'(AS/Pj,~ c f T-~ t:I /ld 11-. / )( 1t;.3 dl",eA.5/PlIs D f L-2 T~I Slh,t!. 7?te. t~I1.si::4I1b /11 '1h~ ~!U4.tr~JI hf(~e c/lmpIIS/~;'.s '1J1elr
r.
Yll/ues
JtI/I/
Ch(Jl1f~ If)/tn a. Chll119E (rl /,411;f.5.
(b) E1Ji4tJP/IJ ~lIl1npt
No.
he t!lJrrRd ;5/nCti! a.t fj=o /A.:: ~.8J tils ) a. ntJlI-,eYb J/A/we w}",h wDulel V{'D.laJ~ .fne. '~o-sJlp'l e(JJl1d;i:/~)I.
NIJ-t t.Prrec-t.!
1-314
1.'11
I 1.-+1-1 Calculate the Reynolds numbers for the flow of water and for air through a 4-mm-diameter tube, if the mean velocity is 3 m/s and the temperature is 30°C in both cases (see Example 1.4). Assume the air is at standard atmospheric pressure.
B)~
9 '15'. 7 ;;;; 3 _
~
V D _ (rqs: 7
/
/;r a t ~ ~-I:
f :: Re
,h A-ppel1dJi
~#
I::: Re
3 o lf>C (-Ir()rn Ta~/e 8,2
a...t
Ftf)t" wILier
1!~J
(3
Lf) ( t'. 00 Lf ,'YY1)
7. 975' ;( IO-1f H,S
rrn
30' C ( 4~m Ta, ble
t. I (,fi
:!.
8. If /n
jA-::: I. 11.
=10°
00
:a.
4pperJdi)( ;<
ID - S
B) :
:s. . 752
=
1-37
/,.tIc:!
I
1. qq For air at standard atmospheric pressure the values of the constants that appear in the Sutherland equation (Eq. 1.10) are C = 1.458 x 10-" kg/(m· s· KI:) and 5 = 110.4 K. Use these values to predict the viscosity of air at 10 °C and 90 °C and compare with values given in Table 8.4 in Appendix B.
3
3
( TT.
T-tS
T= /O ·e.
T lo 'e ,
=
fl. "fF8
T'-
=
;(10- )
-5' N ., = 1.71.5" 10 "".,'"
.,. /I O.1f
B.It)/-'
T = '10'C _
/J o. If I<
S.Jt,- = %
( :1.83./51<)
). 1'3. 15' k
From Table
-t" .),7
-t"
=
'10'C
-t"
;'7~. W :
(f.If!JgXIO-') ( 3(,~./!ik.)
3 &, :1. I~- k.
r
1I O. If
Frc;m
/ -3~
31. -g. IS k )
3/, ;Z
==
-5" .2.13)(.10 NoS
-:;;;; ,
/.~o-tl'
1.5r)*
Use the values of viscosity of air given in Table B.4 at temperatures of 0, 20, 40, 60, 80, and 100°C to determine the constants C and S which appear in the Sutherland equation (Eq. 1.10). Compare your results with the values given in Problem l.lf'f. (Hint: Rewrite the equation in the form
=
(!)
T + S C C and plot 'P'2/ Ii versus T. From the slope and intercept of this curve C and S can be obtained.) T 312 Ii
(J)
T o
,173. IS"
0;.0
J./i3./6
-(i
I.f?;('JD /, Cj 7 ;( If) -tJ-
313.1!i
60
3~3.
80
3S3./~
J.o 7 ;( If)
I()~
373.lb
2../7.xJO
8
3. Si. JD
:2. 6'f~ ~ /0 fj z. 7sf X If) 8 ~. 963)L /0 8
I. 7/ ;C If) - b/. 'i Z X 10 -6'
'fi>
/;-
_3~ A- plot 0/ 1jP-
T~ [J<~(ljJ.,.s)]
I- (/V'S//tI1l.)
(k)
V.s.
T
I~
3.037 X 10 g l. ;; ~" X 1() 8 3. 322;( It) 8
-:,-
-5
Sh()Wn
b<-/ow.'
':i ::-:_~~==~ -==-~
~~~ Fi~ ~: ~ . ~~~;:~: j~~~=J~=::'~~-~J~~ :::::::r :.~.:~~: :-~:~:::-:.: ~-=~h:- .'.:...: =:.- :::. -=:=:: d:' ::~ =:-ir
.
.::!::
:::==;~ :~::=::~=: ;::.:!~:: ~:~~:~.~::1::= ::~= 1~>~: =:-~~~:~~-=:::~~~:: =~::=:~::::- ~:==-
3.0 X/{~f: ~~;-~cc;. ~
=:::..• 'j_' .. :_='::-: : f <
~~~,--: ~==-r~
·:1:·: ::'::=:=::~ :::: -: ~/~l~::: 1"'-::;" : .: :~:.:- ---... ~: :!::::~ 7~· :.:t:::. J : .(- -:::!:: ::j- (£-.-~; ....
::::l .•.•.. :~.-l··
.... ;:.: -::-:-
.
/. 50 jIi
I
(C~11 'i
)
5/~le. the dt-tA. P/Dt a..s flff approXJlt1l.te .rfrtli9Jtt ES' (/) C1i11 be. refrt!S'f111 kd "''I flh e8"'4..ti()~ of form !f::: b x. -t a..
!J
wher-I!
To
/V
obffllH
T3~ a 411d
) X"V T) j,
use
.b ""'
lie I
LJNRFG J.
tll1 d
a.N
.sIc.
po
K************************************************** ** This program determines the least squares f it. ** ** for a function of the form y = a + b * x ** *************************************************** Number of points: 6 Input X, Y .) '273.15,2. 640E8 293.15!2.758E8 '? 313.15,2.963E8 ,:"J 333 . 15,3.087E8 ? 353.15,3.206E8 ? 373.15,3.322E8 ?
a = +7.~~1E+07 b = +6.969E+05
Y(predicted)
X
Y
+2.7315E+02 +2.9315E+02 +3.1315E+02 +3.3315E+02 +3.5315E+02 +3.7315E+02
+2.6~00E+08
+2.6~76E+08
+2.7580E+08 +2.9630E+08 +3.0870E+08 +3.2060E+08 +3.3220E+08
+2.7869E+08 +2.9263E+08 +3.0657E+08 +3.2051E+08 +3.3~~~E+08
2C = a. = 7. JiJf/ X ID Qi/(i
Th,Se.
tv/ii?
1her-(~fe
[.It/lues
s=
ID7 /(
.kl' C f/11t1
il4/tltS
7
5
4te
In
91)t)d
rivtl1 il1 Problem /. '11 .
tl1f'femfllt
/.5'/
I 1.51 The viscosity of a fluid plays a very important role in determining how a fluid flows. (See Vieko V1.1.) The value of the viscosity depends not only on the specific fluid but also on the fluid temperature. Some experiments show that when a liquid, under the action of a constant driving pressure, is forced with a low velocity, V, through a small horizontal tube, the velocity is given by the equation V = K/,.,.. In this equation K is a constant for a given tube and pressure, and JJ is the dynamic viscosity. For a particular liquid of interest, the viscosity is given by Andrade's equation (Eq. 1.11) with D = 5 X 1O- 7 lb • s/ft2 and B = 4000 oR. By what percentage will the velocity increase as the liquid temperature is increased from 40 of to 100°F? Assume all other factors remain constant.
ell
I< -)AIfoo
(2.)
I<
': [b' _~~IOD IJ )J-'DOo
-'1
5~lD
e
(3)
/.52#
1.52.*
Use the value of the viscosity of water given in Table B.2 at temperatures of 0, 20, 40, 60, 80, and 100 DC to determine the constants D and B which appear in Andrade's equation (Eq. 1.11). Calculate the value of the viscosity at 50 DC and compare with the value given in Table B.2. (Hint: Rewrite the equation in the form In Jl
=
(B)
1
T + In D
and plot In Jl versus 11 T. From the slope and intercept of this curve Band D can be obtained. If a nonlinear curve fitting program is available the constants can be obtained directly from Eq. 1.11 without rewriting the equation.)
£"8"4 it;;" 1.11
DIn
be t.Jr;flf'i1
Ih;" ::- (/3) / w/th 1he cI~ia
tina
T ("() 0
~J'Om
T(k)
the .{;,rm
Ih
In..D
of
746ft!. 8.2 "
it (J./.sk 1.)
I/T(K)
3. b"l ;tID
:173. IS
-.3
I. 7K 7 .x' If) - J
.10
;'~3.1;
3. 'III xIf) -.3
/.Ib
'3 I 3. IG'
.1.1f3 x1D-
~6
333./6
yo I ()o
35'3. I£"
3. ooz xlO 3 t2.152, .x' 10-
373.
~. 'R~ .rlf)
),!,-
A- plot of In!-
VS.
(I)
liT
/. (!)Ol
x 10
-3
It} ~ - t.. "3Z 7 - I.. t/ob
3
6:.. 5"29 ;( I~ -1(0
- 7. -33 Y.
-3
~ ~'G'" X lo-if
-7.
-.I
.
IS
( Col1t) /-'1'2
3.
S"1f. 7
x/0 - If
2.81;-,;(10
-~
s hf){'()n be/f)w:
~70
- 7. c,'f'f -8.J7lf
/. 52 ~
I
(C£J" It)
51,,'ce
A
the dfti~
£1, (J)
plat as
tlflr()x,mll te sl,.Ai,1J, i b~ ".!~d .fo refyr.sfrli 1hese ddt/.,
(/117
B
To t)btlJ/H
f)
/In'f'
'11/
k-X?FI T,
lise
*************************************************** ** This program determines the least squares fit ** ** for a function of t.he form y = a * e ' b*x ** *************************************************** Number of points: 6 Input X, Y ? 3.661E-3,1.787E-3 ? 3.411E-3,1.002E-3 ? 3.193E-3,6.529E-4 ? 3.002E-3,4.665E-4 ? 2.832E-3,3.5~7E-~ ? 2.680E-3,2.818E-4
a = +1.767E-06 b +1.870E+03 X +3.6610E-03 +:3.4110E-03 +3.1930E-03 +3.0020E-03 +2.8320E-03 +2.6800E-03
So
13 -= b = I, f'J~ i< /0 i}ttrt.
~::
Ai
+1.6629E-03 +1.04-18E-03 +6.9298E-0,* +4.84-82E-04 +3.5277E-0l,t +2.6548E-04
I. 7' 7 X /D -, N·S I /1')1 2-
D =~:: ~d
Y(predictedl
Y +1.7870E-03 +1.0020E-03 +6.5290E-04+4.6650E-04 +3.5470E-04 +2.8180E-04
SOO{
;<=
/.7~7
x/a
-6
3
/(
-T
I! '10
e
(323,)5"1<»)
I. 7'7
;(.
-, e /()
1370 iJ23, )b-
1-'13
-
S.7~x)o
-it-
N.S//P1~
I. 53
I
1.5 ~ Crude oil having a viscosity of 9.52 X 10- 4 Ib·s/fe is contained between parallel plates. The bottom plate is fixed and upper plate moves when a force P is applied (see Fig. 1.3). If the distance between the two plates is 0.1 in., what value of P is required to translate the plate with a velocity of 3 ftl s? The effective area of the upper plate is 200 in. 2
I-'ll.{
/. 54 1.54 As shown in Video V1.2, the "no slip" condition means that a fluid "sticks" to a solid surface. This is true for both fixed and moving surfaces. Let two layers of t1uid be dragged along by the motion of an upper plate as shown in Fig. Pl.54. The bottom plate is stationary. The top fluid puts a shear stress on the upper plate, and the lower fluid puts a shear stress on the botton plate. Determine the ratio of these two shear stresses. Fluid 1
I-- 3 rnIs --i iJl
= 0.4 N • 51m 2
Fluid 2
f-02m/s..j
• FIGURE P1.54
n,r .f j{,lid '1j" ~r
I
h (~tJ;;Op
f_"
JU r 1TA Ct.
(6.L1 ~)(
N
20-
1m1.-
+I "'lei. 1;. ~ A (~) = (0.2 ~)( :.o:~) = bo-(h,,,, sur(.,,, mt
1.
T -b,p ;"";>-
(
~141""+'Ct
"O~fI'\ ~14 r...(.,(.,
1.55 There are many fluids that exhibit non-Newtonian behavior (see for example Video VI.4). For a given fluid the distinction between Newtonian and non-Newtonian behavior is usually based on measurements of shear stress and rate of shearing strain. Assume that the viscosity of blood is to be determined by measurements of shear stress, T, and rate of shearing strain, du/dy, obtained from a small blood sample tested in a suitable viscometer. Based on the data given below determine if the blood is a Newtonian or non-Newtonian fluid. Explain how you arrived at your answer. T(N/m 2)
0.04
0.12
2.10
du/dy ~-I) 2.25 4.50 11.25 22.5
Foy
C(!hISi::.tll1t:.
;=;:'1' -th~
Th~ ra /:./0 IS
du/dfj
15
~
tiAta., 9/Vt' 11
(lV,s/h1~) O. ~/78 P.OI3~ o. ~//)71 ~.()()Io
?dull,;
-t ilJ
.P/u/c/ in( ra.C/o of
Net.AJJ:()mQI1
a-
450
(/.0067
~.()Q5F
O.CIJ,5()
(),()()'f-7
~f1si""i "" ~ de're(Js~r q s the Ya.te t!;f shear/".,g stYII/n /n,yel/~l'!. Thll~ thiS F/w/d (;/tKJd) l,j ~ /7()I1- lIeu//;ol1l';' -fl t(/d. A- p/oi of 7hf! cltt(:a. .£S .sh/)(,Qb bt!/ow. ;:PI" A. f/ewI:rUII~H 71 u,C/ 1J1~ C-/.tl"'II-e WIJ)I/I,( b~ .a si:y~",h t /J~e UJI17t I( IllJpt!! (If / f() /.
noi.
tJ..
: -;< ,I ~- ~ ~;i:',; .~.... -
'"
.,
·:::T::::F~::-:-::~:::;V:~:;:'~#,>:>
1 • - ,>.' i:-:;; ~~:~!:~: <:: ::: ::::..: :: :;'t' •.• ~;::"
L-
_-'-!,_ '.
.-'
1
-
':
!
•• " ..
. ,~~:j :':':~ ~ ~::~; ::::t?~ ~ : : :~:~~ ~.:: ::::\:::::::: ;::: ;::IC~ = : :
_ ... ;.~:: ,,:,:,:,,~:::::, :;,;.;:;,,::::: ::, ~j~,,,:, :1': •.
I
' - -....
-
!
'I' ,
..• ,1' I I .':
i.-j
"
l"~::i::lll ! I II
i
i
:
r :
/0.0
I
ii' I
,1 -If i, ._+-'
ICO.(]
I
'
:
I
.1: I -:" '1'--1-1-f--, I,.
H; ,
HI'
I
1.56 A 40-lb, 0.8-ft-diameter, I-ft-tall cylindrical tank slides slowly down a ramp with a constant speed of 0.1 ftls as shown in Fig. P1.56. The uniform-thickness oil layer on the ramp has a viscosity of 0.2 lb . S/ft2. Determine the angle, 8, of the ramp.
• FIGURE P1.56
(I)
LJ
)
heve
a VI ~
t= (0.2 ~)(
V l~ the. Ve.IDC.d"'1 of- -b(A..,k.. b" ~ Tn j(:..k: n t5S f)f i I I a. 'jt..,.. t)
:.1 r ) #
.DOZ
'=..
F'V't?m l? ~ . CJ)
WO Ib) ~"YJ
f7 -
SI n f)
&
(t{) ~2..)(:q:)(O.8..ft)2
= 6. J2.5'1
= ,. Z 2.
D
/-1.f7
I
/.57
I. '57 A piston having a diameter of 5.48 in. and a length of 9.50 in. slides downward with a velocity V through a vertical pipe. The downward motion is resisted by an oil film between the piston and the pipe wall. The film thickness is 0.002 in., and the cylinder weighs 0.5 lb. Estimate V if the oil viscosity is 0.016 Ib·s/ft~. Assume the velocity distribution in the gap is linear.
2:fVerr" ... 1 nUS.)
wkev! aVId
=D
OW:. A
=
,I)-
L= .56
tA
rrDi
(v e 1t'>C.:~)
_
( +ilm1hlc.l::lle.5s)-
-th~t
1»= (I'-
I'W
1
~A
t-
~
't~
t )(1TDj)
fr
~
'¥
~
~
\\
i
f-
D
~
T P-
i
/.5'8
I
1. f) & A Newtonian fluid having a specific gravity of 0.92 and a kinematic viscosity of 4 X 10- 4 m2/s flows past a fixed surface. Due to the no-slip condition, the velocity at the fixed surface is zero (as shown in Video V1.2), and the velocity profile near the surface is shown in Fig. PI ~g. Determine the magnitude and direction of the shearing stress developed on the plate. Express your answer in terms of U and 0, with U and 0 expressed in units of meters· per second and meters, respectively.
)'
I I
~ _ ~L _
l( l)3
U- 2 0
2 0
(~:o)
dt.{ d!J
@ J=-O)
\
r~_---i \
u
\~--i
\'----i
~
• FIGURE P1.f>'B
?- 5(,/J'"loc~
U
I o
151
When a viscous fluid flows past a thin sharp-edged plate, a thin layer adjacent to the plate surface develops in which the velocity, u, changes rapidly from zero to the approach velocity, U, in a small distance, 8. This layer is called a boundary fayer. The thickness of this layer increases with the distance x along the plate as shown in Fig. PI.59. Assume that u = U y/8 and 8 = 3.5 V vx/ U where v is the kinematic viscosity of the fluid. Determine an expression for the force (drag) that would be developed on one side of the plate of length f and width b. Express your answer in terms of f, b, v, and p, where p is the fluid density.
U
r---. f-
f-f-'---~
\'
·
~
Boundary layer
-.1I==U
/
_---r--- ~- ---~.---I-
_---__e I _-
8 ~11 == U~ ~ t (5 ""----=.::=---L--IL---_ _ _ _ _ x
Plate width == b
tJ her~
.. ;
dA--(I )
Clntl
-Jl/ ~3. (f)
t{nd
IJ -
0,571
bf V-zJ1.U 3
I-50
1.601'
(a) Assume the velocity distribution is of the form
u = CIY
14.43 The coordinate Y is measured normal to the surface and u is the velocity parallel to the surface.
n()nh;'e4r
10 ()btllil'J
yejrl'.ss/~1'J progf'YIf'fl) such a,s
u;eIHc/fl1i::s (,
let/si s1"nrt's
, = 153
lind
JiVfl1)
s-'
C (~)
. SIJ1ce)
/f
~/J/)U/s
SAS- NLJN,)
C.1,.. Th;'.j pr(!)9fY1m pr{)duce.s est/males ~I /he. ,PIIY'lIl71ei:trs of II /?~111J11~4r
/=by iJle dttia..
rtl&r/el.
C 2Y)
and use a standard curve-fitting technique to determine the constants C I and C1 • (b) Make use of the results of part (a) to determine the magnitude of the shearing stress at the wall (y = 0) and at Y = 0.05 ft.
0.08
(~) Use
+
du
1:=~ d;
1114t
r=;-
Thus) at- the wal!
(~
t 3 C;z. :; l
(~=())
Ai
/-51
)
1.6 I The viscosity of liquids can be measured through the use of a rotating cylinder viscometer of the type illustrated in Fig. Pl.61. In this device the outer cylinder is fixed and the inner cylinder is rotated with an angular velocit)-" w. The torque :, required to develop w is measured and the vis. cosity is calculated from these two measurements. Develop an equation relating fl, w, 5", C) Ro and Ri • Neglect end effects and assume the velocity distribution in the gap is linear.
~~~ FIGURE P1.61
Tor't ue; t::J11
d
/nneJ-
d whi're.
r,
I~
C!j/Jnc/fr
'7: rr::.
~he"t;l1j sms.s
+()
due.
e!tltd..fr,
T dA
c/It = ~. de) 1.. Thus)
d'T=
~.
2-
J Ttit;
n d /-vrff lie. reg tI /re d to rtJ fa I: e ,nne", c'1/lntler i,S 2JT
top View
{J
J= 1
(J. "'"
ride
C'j Ilndrr
()
POI'
.2 TT R.t.''-
a Iln'ell!'
J. r
ve/oc./+:; distyibtl'l'/on
T=/-
R'W L
~ 7i R,~}. t
tV
Ro-RI.'
/-5'.2...
In
fhe gap
leMi fi.J )
/.bZ
I 1.62 The space between two 6-in. long concentric cylinders is filled with glycerin (viscosity = 8.5 X 10- 3 Ib·s/ft 2 ). The inner cylinder has a radius of 3 in. and the gap width between cylinders is 0.1 in. Determine the torque and the power required to rotate the inner cylinder at 180 rev Imin. The outer cylinder is fixed. Assume the velocity distribution in the gap to be linear.
Prl)/'/em /. " (, )
T =
02'ff
R,.3 ))A- W
:eo - /Ct..' W=
(; 80 !!.!. )(eillT ~ milt
rev
)(1 mlh) = blT
'0 s
.:l.7T (i£ft)3(-A ft:)(s,s)(/a- 3 !Jt.)(67T
vad
s
0/) = 0, qIf 'f It·l)'
( ~ -ft) 120
S/f]ce
pouJey
=
f()wer': (~tJif'fft'/h)(67T r;d)
/- '53
==
178
~.Ib
I.
(P3
1.63 One type of rotating cylinder viscometer, called a Stormer viscometer, uses a falling weight, 'lV, to cause the cylinder to rotate with an angular velocity, w, as illustrated in Fig. PI.6.3. For this device the viscosity, J.L, of the liquid is related to 'lV and w through [he equation 'lV = KJ.Lw, where K is a constant that depends only on the geometry (including the liquid depth) of the viscometer. The value of K is usually determined by using a calibration liquid (a liquid of known viscosity). (a) Some data for a particular Stormer viscometer, obtained using glycerin at 20°C as a calibration liquid, are given below. Plot values of the weight as ordinates and values of the angular velocity as abscissae. Draw the best curve through the plotted points and determine K for the viscometer. 'lV (lb)
2.20
w (rev/s)
5.49
Fixed outer cylinder
•
FIGURE P1.63
(b) A liquid of unknown viscosity is placed in the same viscometer used in part (a), and the data given below are obtained. Determine the viscosity of this liquid.
I
'lV (lb)
0.04 0.72
w (rev/s)
( Cl)
0.11 1.89
~:
5;;'re
0.33 5.44
I
0.44 7.42
-th~ ~/()fe
K)4u.J
IJ
s/t:Jpe = SO
I
0.22 3.73
~
tis. tu
UlriJe
tv
C~V)
s ) (16. YeV
.5 0 jJe
k=
-the czJ
We/b)
If)' =
fha..i
~I
(I)
/<-(~)
fi,y -rJ,e :J/~~en~ dai:a. (.see p/Dt: ~11 f')(?x/; page) ('/?Q.5rd ,t)11 a /ul.ri S,!ftllres ,f,:t Df the d~~) I.J S/CJj)e (J/,/ceni1):= O,.?9R
5In'~ IF. U tjl'lcerln)
=:
1<= ( h)
3.13X//}_zJb·s
it. ..
J~~ -tnel1
(),g9S Ib ...5 /'?v
;:P". the. un klJ~tVf1 filii (J d~-k.. sJ()P~
(b"~e#
tf}Jf
Gee. IJ/t)i GJI1 11t't.-t f4~~) the ~ legst, $8"II"S ~it ~I tH~ d/l,tA,.) IS
S/Db.# (tll1KdtJuJl1
r-
rill/g) =
/-51.f
O,CJ6o/
11:1'.$
~v
I, (P3
(~l1lt ) Thu~
I
/rpm
I (lIlJillPlIJlI
E"r.l/) S/tJff!!
fluid) =
1\
tJ.
tJ (p ~/
/2.,
Ib·5
-rev
7 H,2. rev
,.0
/-5,
1.6Y*
The following torque-angular velocity data were obtained with a rotating cylinder viscometer of the type described in Problem 1.61. Torque (ft-lb)
13.1 26.0 39.5 52.7 64.9 78.6
Angular velocity (rad/s)
1.0
2.0
3.0
4.0
5.0
6.0
For this viscometer Ro = 2.50 in., Ri = 2.45 in .. and = 5.00 in. Make use of these data and a standard curve-fitting program to determine the viscosity ofthe liquid contained in the viscometer.
r
The .fz,r~lIe J ~ -the .e$Ua. /;/~11 )
Jl'e/a.bed .J-o -the tlnJU/dY (/(!/t:Jcil-!1.; UJ.)
J'J
~--
(see so/tJl-,{)~ (1I'1d
a.
:;, 7T R,:
l!o -
3
)J
t.U
(/)
ftL'
Problem I. hI, ). Thus) /"y Ii flx'ed ,et:Jmef,.J /l1~CC).s;-f!:J J E~ ,/1) /05 ot 1h~ ~f'rn
..J.t>
'lIt/en
( !1 rv'J
y=hx IS
~,,~tr,1'J i
If.
j,
.fl/Utl /
Qni1
x'
r.."
W )
.fc
= :z Tr ~.:1). !=
().)
)2' (. WI
ih
LIAlfrF6
**************************************************~
**
This program determines the least squares fit ** ** for a function of the form y = b * x ** *************************************************** Number of points: 6 Input X, Y ? 1.0.13.1 '? 2.0,26 . 0 ~)
3.0,39.5
'? 4.0.52.7 ? 5.0,640.9 '? 6 . 0,78.6
b = +1.308E+Ol X
+1.0000E+OO +2.0000E+OO +3.0000E+OO +4,.OOOOE+OO +5.0000E+OO +6.0000E+OO
-Ft,Jb'5
Y +1.3100E+Ol +2.6000E+Ol +3.9500E+Ol +5.2700E+Ol +6.4,900E+01 +7.8600E+Ol
Y(predicted) +1.3082E+Ol +2.6165E+01 +3.921±7E+01 +5.2330E+Ol +6.51±12E+01 +7.81±95E+Ol
(C()I?'t ) /-5~
I.
(emit)
find
/
w/th
! =
(b) (
-/he
daf&"
r; g
='
~. -
f?.: ') .27T ~.3,R
Of ft.f/;'$ )
g/~tl1)
(.? 5"0 -
:I.. LJ.S-
ft-)
1'2.
~lT (~.'1~ 12-
_
ft) J (~It) 12.
/-57
rJ. 1f-5 IJ,·s -Fel.
I. 'S-
Rotating plate
1.65
A 12-in.-diameter circular plate is placed over a fixed bottom plate with a O.l-in. gap between the two plates filled with glycerin as shown in Fig. PI.6S. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.
0.1 In. gap
ili
dCiJ J c1u~ h s hellYJ~1j pI4+~ l.s~ e 1Uq / -1-"
kYO'lUi) pn
do;= ,. ~ ntrt
d It.: cJ
2.11" yo
dr,
C7J: "" T
Thel'5 J
Z.".. r d '"
ufl'!r. r
tlr
o
T,: f<-
sfrfSSl'.J
tdA
"(=
5;nce
FIGURE P1.65
~ ) 1/ fl.t/
"r
fA.
lIe/"'Jf.t J'5tY; h",-I-U/H (SlefijHye)
ellA _ V
'rW d;-I~T
-:: D. 0772
/1:. ·/t
I-58
I. ~ 7
J 1.(,1 A rigid-walled cubical container is completely filled with water at 40 of and sealed. The water is then heated to 100 oF. Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi.
5/~ce 1h~ t..Jd..frr
tnf4S)
YfMAll1S
~.1I::1 ~~
+
/j
/tJO
1If)/Ume
ttPl1s~fJ
(..pl71.l¥-) D
t1"~ 4-1' iJ
C.hlll1'1t2
111
ve)/um e 1"1 tva/-!'/'"
7h II~)
/, l''fb I. '/27
'/ir, I, / z.
& = V
If /11
tJ
sJuf,.S .ft.J
5'::'
-/
dp -~ d¥
~I/IJ UJS U.l/1J.. d1l- ::; 4 -v' 411 t:I. PI» -.:::= fJp 1hAf- 1hf CJtIlH'I€ />Y'e> 5 ",y~ r-e'gwred. ..f-r> ~nJ;,.es.s the Iva..-tey bltJt. Ie i-h
/1'/1/11.)
110 I~ ft1e...
,'.1
t1f= - r;tJ~/~~()f~L'){-O.OO~75) 2.o3iJD
3
pst...
1.(o~ In a test to determine the bulk modulus of a liquid it was found that as the absolute pressure was changed from 15 to 3000 psi the volume decreased from 10.240 to 10.138 in. 3 Determine the bulk modulus for this liquid.
d~
z A ¥ -=
I~, e1JfO - I~, 13~ :
t:J, jOJ,
I
Ii'l,
3
Ib
rJ9i'S ;;,. (
/,iD't
1t)2 In,7) /", ;l'l0 in.' C',
I 1.tJ1
Calculate the speed of sound in mt,s for (a) gasoline, (b) mercury, and (c) seawater.
( Eg.
I,/Q)
(a) f:by 7(is~l/ne,'
I, Lf5
~I'm s
T
1.70 Air is enclosed by a rigid cylinder containing a piston. A pressure gage attached to the cylinder indicates an initial reading of 25 psi. Determine the reading on the gage when the piston has compressed the air to one-third its original volume. Assume the compression process to be isothermal and the local atmospheric pressure to be 14.7 psi.
i~othermlJ/ ~&)mpY'ess/t:Jl1)
PC>t'
-Pl.'
..f',-.
= -fE
f!t Pf =~. 0/nc.e
f=
Where
~.f
f """
(.'~
f/nAI
f;. ~
m4SS J/t)ltll11
)
e.
1'h~reloye
1;.
= (3)[(:15 f-
t (p-;e)
/
'I.
7) p:s L' ftJ6s)) :::
=(i /9- 1'1. ~f'i
:-
/ni.J-,i.d state.. ~illie..
.
UI1A
It 7 /
J
1.1 I Often the assumption is made that the flow of a certain fluid can be considered as incompressible flow if the density of the fluid changes by less than 2%. If air is flowing through a tube such that the air pressure at one section is 9.0 psi and at a downstream section it is 8.6 psi at the same temperature. do you think that this flow could be considered an imcompressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressure.
h,y
1"0 therMo / CHfiA'It
-/;), f,
So
1},4.-t
I, ---f,
:-
--
111
den~;I!:J
-~P2-?z..~ )( I D 0
TAus
I
1.72.
J 1.72. Oxygen at 30°C and 300 kPa absolute pressure expands isotherrnalIy to an absolute pressure of lZ0 kPa. Determine the final density of the gas.
For
/sofh~rma/ ~. L
-
~.
...
~~
-
!j
ex.ftlI15i()YJ ) ,
u)),ere
I-
::t =t!LJl'Jsft/tJt:
~
t"'" ;'rll "V
t, 4/ s fa i-e
SIJ
1h4t
Pi1#
.fIn II / st:A..ie.
- 3.8/ ~3 /J?"I
/,73
J 1.73 Natural gas at 70 OF and standard atmospheric pressure of 14.7 psi is compressed isentropically to a new absolute pressure of 70 psi. Determine the final density and temperature of the gas.
For
;'sen irop'c~.
-p.-A "
::t
c~m?re~S/()11
J}
tVhere
1,*
~I
--P
=
;O~
,,; 'V
~t!/ns tQl1t
51) -thA-f
;'n;';';'/ 6~te cou/
.f 'V .f.J'ntl/ sta.te .
-f-
L
,
=
-,3
'I. 2S )( Jf)
S h~f5
-ft:'a C( J1
c/
7f
.....
-for
ft.R
--
7(P5
-
IIJ (7 c; 7ii: a.
('I. 25
I ) (
, ~
I/l.
If tf -:t:;"A-
;I. / ()- ~ S htf..s
h3
)
{3. O'If;( j 0
(J)R
~r
71:
7fD 5 oR - /f-IPt)
-
)
305 of
3
~b IJ:
sh",.liI~
-
)
/.7 if
1.7ll--
Compare the isentropic bulk modulus of air at 101 kPa (abs) with that of water at the same pressure.
( E' 'I. /,) 7 ))
J:C; r a l r
£ y ~ I<. f /=(; r
tva:te ".
E,,::
(
1t'3~ )
= (/, 'i-o ) (10/ x
Tr,; bJeo /, /, )
:1.)6;< It> "
,q
Thu.5 )
Ev (WIJ.,-ter) _
~, 15" ;< II)
£v (cur)
I, Ifl
9
Pa
X. 1~5'"/}
=
I. 75"
1.7.5' * Devel~p a -computer program for calculating the final gage pressure of gas when the initial gage pressure, initial and final volumes, atmospheric pressure, and the type of process (isothermal or isentropic) are specified. Use BG units. Check your program against the results obtained for Problem 1.70.
'4' J
r-o Y
C/!)11?
pye5SIol1
e1 Y
ex..pQI1JIOII)
?... = e04stoni.
!
wheye h=/ -ky isotho-mal process) and lOr 1.st'l1frt'Jllc proc.ess. Thus J
~.
;:.-*
where
/.'/1;
It:::
.Jj'e,;{tc. helL-/: va.!:'"
= !i:.
/;.-P.
In/ha'/ ~k.te I .f''V IiH~/ .str;le) So 1J1Ii't
if : (-J,:) "- f:: t.:
ml1ss
Vt!)/~lI1e
~ = Vt·
1hel1
~.
w he¥"e
v;,.) l'..f.g
T
~
~) tire
Thus) Ir~m S~
/1)
-Aa:eM :: { 'It. Vf
fh (! SWhSCh pi be. w y-; He!? as
t.Jhe Y'e CaM
(/ )
3
)-k
( ~! 1:.t"" ) T
reI-Frs 1::6 JaJe /J Y"e sst/re
if, = (~) -I. (~'j tt"J - ~t"" -t
(c~n It
)
(2 )
175
itt
100 110 120 130 1""0 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330 3l.!-O
I cls print "*********************************************************" print "** This program calculates the final gage pressure of **" print "** an ideal gas when the initial gage pressure in psi, **" print "** the initial volume, the final volume, the **" print "** atmospheric pressure in psi, and the type of **" print "** process (isothermal or isentropic) are specified **" print "*********************************************************" print input "Enter initial gage pressure in psi, Pi = ",p input "Enter initial volume, Vi = lI,vi input "Enter final volume, Vf = ",vf input "Enter atmospheric pressure in psi, Patm = ",patm pabsi=p+patm print:print "Enter type of process" print "0 Isothermal" print "1 : Isentropic" input pt print k=l if pt=l then input "Enter specific heat ratio, k = ",k pabsf=pabsi*(vi/vf)~k
pf=pabsf-patm print print using "The final gage pressure of the gas is Pf =
+#.####~~~~
psi";pf
********************************************************* ** This program calculates the final gage pressure of ** ** an ideal gas when the initial gage pressure in psi. ** ** the initial volume, the final volume, the ** ** atmospheric pressure in psi, and the type of ** ** process (isothermal or isentropic) are specified ** ********************************************************* Enter Enter Enter Ent.er
initial gage pressure in psi, Pi =ZS initial volume, Vi = 1 final volume, Vf = 0.3333 atmospheric pressure in psi, Patm = 1"".7
Enter type of process o Isothermal 1 Isentropic ? 0
The final gage pressure of the gas
lS
Pf =
+1.0~/E+02
psi
I. 7 G:.
I 1.7<; An important dimensionless parameter concerned with very high speed flow is the Mach number, defined as Vic, where V is the speed of the object such as an airplane or projectile, and c is the speed of sound in the fluid surrounding the object. For a projectile traveling at 800 mph through air at 50 of and standard atmospheric pressure, what is the value of the Mach number?
y Co
~.h)e.
B.-3
~al'r ~
In
50-F
Thu.s !v1I.G'r1 numblY -.
LO{P
1,77l I. 11 Jet airliners typically fly at altitudes between approximately 0 to 40,000 ft. Make use of the data in Appendix C to show on a graph how the speed of sound varies over this range.
V
c = Ie R.r t:;r 4< ::: lifO C= Fr~m
7
(EZ' I· 2o) If=- 17/t, 1i·1/,
Clf1d
s/,,!! .t)~
'19. tJ YT(~)
1a6/t: C. /
T= S'I. ~ C)
of-
1
4pp end':x
In
5J9°~
!fl;o :
C
at
.50
-thl.;/;
tI/1
(J /
-fi .f.u df!
til o +-t
::: I / /!. .f.i:s
5;'17/ ;/tfl '" Cd /CU/4,tltPIfS CiJn De mtule -hr (JiJ,n~ dlf/ltlt!t'~ tin' -the ,es~/f/;'! 1raJh is :J"/1()W)f b~/ow. Altitude, ft
Temp .•o F
Temp.," R
0 5000 10000 15000 20000 25000 30000 35000 40000
59 41.17 23.36 5.55 -12.26 -30.05 -47.83 -65.61 -69.7
519 501.17 483.36 465.55 447.74 429.95 412.17 394.39 390.3
c, fUs 1116 1097 1077 1057 1037 1016 995 973 968
1120 r--,.-----,---;---;----;----:---:-~__,
-~...... ~1080 I~
I
1100
;;g1060 (f) 1040
I.
I
I' L'" "-
1
""-~--:-----1f---J!---L-...---l
t---t---t---t--. .-
~1020 .t--r--+--+I_--+-~-~--+I---!--l , ----+~~,__LI-_+_I____i ~1 .r----+---+--_+_I 1 -1-
000
1.'
980
t----;---;---t---t---t---+---.......;.......:""~~---1
960
+-.---+----'----..:.---..:.---;--~_~---1
(f)
I
I'
o
5000
1
10000'
15000
I
20000
25000
Altitude, ft
30000
I.
r----
35000
40000
I.
I
73
1. 7 R When a fluid flows through a sharp bend. low pressures may develop in localized regions of the bend. Estimate the minimum absolute pressure (in psi) that can develop without ' causing cavitation if the fluid is water at 160 OF.
cC/J/i.faflon mtJ'1 (pee",,.
(/t2for ,res'Sure.
ir
=
whtn fhe I~cq/
!=Or waiel" aiif. 7Lf
pSI,'
pY{"$Stlfe
/(p() df'
(lj.~,."
e~t",ls the ?;bJl8,j
(IIbS)
Thus/
1.79 Estimate the minimum absolute pressure (in pascals) that can be developed at the inlet of a pump to avoid cavitation if the fluid is carbon tetrachloride at 20 °c.
Ca vi i-A.I/!)11 rnp'1 (pee tI r when fhe stlciion P;'(J$stlJle at- -tnt!.. pum,P inlet etttlA/.s the 1/a.,Pcr' fJY'es$ure.
t;r
ClJrhtm tei'('ac.J"Joy;d~
t2
t
2.0
IJ
C
...n
IV
rn ; n /m J,I J?1 !f".Rssure
/-70
= /3
~ R. (g,!;s)
/3 .Ie
Pa. (4.6S)
,~Apptw:l/J(B)
I, So
J I . ~D When water at 90°C flows through a converging section of pipe, the pressure is reduced in the direction of flow. Estimate the minimum absolute pressure that can develop without causing cavitation. Express your answer in both BG and S1 units.
('C/vif4tl{)" nUl'j cc.CClr /n 111e Ct'''J/er9'/~~ sec..-i-Idn ~ pile whel1 -rhe. pr~55«;,e tEf;aolS -th~ va.~J' fYe.J5'tlre. ;:-r/)/7'I 74lie B. 2 I;' I+!'ff"c/J( fj'
~r
wA,ter a t
.
9~ °C.I
1;::
70. / -h Po.... (ql,,,). Thu~
minimum pre~suv~ ::' 7()./ --k.?c.. (q/'5)
86 Hnifs
::: /0, 2 ps I.a
I 1.8/
A partially filled closed tank contains ethyl alcohol at 68 OF. If the air above the alcohol is evacuated what is the minimum absolute pressure that develops in the evacuated space?
f.iz
I
1.8Z Estimate the excess pressure inside a rain drop having a diameter of 3 mm.
().oo/5
sr
tln,fs.
= f;(). J x J~ 3 ::.. )(/. lj5"1; )( / J- ~ fl,.{ )
I11lnlmuM .P'fSJare
/.31
1/1
/1n
/-7f
I. 113
I. r~ A 12-mm diameter jet of water discharges vertically into the atmosphere. Due to surface tension the pressure inside the jet will be slightly higher than the surrounding atmospheric pressure. Determine this difference in pressure.
"Ft;/' erp,i J/bri/l"" fspe IIjure ).; SO
1(z~Ii/: cr(zJI.) -rnA i -t== 12 >' it; -.3 ~ 2:
=
12. 2
1'1V ex,t'SS f rfSSU re
Ii Sur-Hlle
1-72..
-ftHSIDIl
~'(,e:- cr 2. £~
/,8'-1 1. 'a Y. As shown in Vidl'O V1.5, surface tension forces can be strong enough to allow a double-edge steel razor blade to "float" on water, but a single-edge blade will sink. Assume that the surface tension forces act at an angle relative to the water surface as shown in Fig. PI ~~. (a) The mass of the double-edge blade is 0.64 x 10- 3 kg, and the total length of its sides is 206 mm. Determine the value of e required to maintain equilibrium between the blade weight and the resultant surface tension force. (b) The mass of the single-edge blade is 2.61 x 10- 3 kg, and the total length of its sides is 154 mm. Explain why this blade sinks. Support your answer with the necessary calculations.
Surface tension force
e
L FV€r+t '4 I
(a. )
• FIGURE
T
~
::.0
o.
ttJ ::
X
(Y'(l
blade
~
T:::- a- ><. Jenfn, of.
Ql;1d
( (), 10'1- ;( 10- 3-ka ) (U I I'tr./~.) =
~
tow
VW=Ts/n8 Luheye
p1.<64
fr. 3~ ;(
}O-2
.Jt. ) (IJ, ZO~
slqes
/In ) 5'111
e
:sin e- =- o. Lf-15 9 = :J... 4-.5 0
(b) For
blade
slnrle-edtje
'2J =
/yrl MAde
X
d- " (~2.I.1 ::
uYld
T
5111
x: 10().DZ~1s, N
3
-ka.J ('1. ~J I'M/~')
e :: (vx. J enJ1n of. /,lode ) ~I;' ::- (7.3LJ.x/o- Z
=
f7
Ntm) (O.15LtM1) '51Y1
O. 0 I J 3 '5/n
G
e
t>Y'aer +O~ h jq de +0 "-J./Da.i ~ -< T "SIn e. "StYlet.. rma)(Jf1'lUfn1 Value JoY" ~Ine IS \ I J'+- follows that '1.<.J > T St'n e and 'Sin9/
rn
It
/-73
I. 8'5
I 1.1'\5 To measure the water depth in a large open tank with opaque walls, an open vertical glass tube is attached to the side of the tank. The height of the water column in the tube is then used as a measure of the depth of water in the tank. (a) For a true water depth in the tank of 3 ft, make use of Eg. 1.22 (with () = 0°) to determine the percent error due to capillarity as the diameter of the glass tube is changed. Assume a water temperature of 80 oF. Show your results on a graph of percent error versus tube diameter, D, in the range 0.1 in. < D < 1.0 in. (b) If you want the error to be less than 1%, what is the smallest tube diameter allowed?
(a.)
The e;(ce~s he'jh t I h) CtlfA~ed bt 1h~ .sur~(t. ien>/~~ ,1
h= f:ipy-
tr:: 0 tv; f;,
zo-~~
'oR
b =.z. R.
D
h=
'fO-
0)
rD
/Dy-
PY'If)I?1 7i.J,/~ B./ln A-ppendl)( B 0-=
Jf.9Ix//J-3/bj.Pt Clnd
Th~~ .fr.9m
h ~f)::
1:1. (I) tf (Jf, tj I x: IO-J {(e>Z.lZ.
~l~ error
~t 'Z.2z. Jb/k~
r=
WtJ.-tev
-! )
fl. 1 q )(
(,n.) ~,) D I 2. I ~, /';-1::
h ~J
=.
froW! eq.l 1-) D /0 e yr" y =
f)
(E'Z. J.ll.)
.5'.ft; 1ha.~
-3
J0
.D ( I'n.) 100
)f..
0
3.7&f x J
3 D(J'n,)
-3
x I D0 ( 3 )
A- plot. t>.f ~ eY'r~r V-(i"S(,/S t"'be ShtJWI1 "n 111t /1f~t I"a'je,
( C!L;,/t. ) /-7Lf
C/'t:ll11et:er IS
/. 8S-
I
( Ccr/t.)
Diameter of tube, in. 0.1 0.15 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
% Error 1.26 0.84 0.63 0.42 0.32 0.25 0.21 0.18 0.16 0.14 0.13
"
1.50 : i
I
I,
... 0... ... W
1.00
~ 0
0.50
\
I
!
I
I
'{ I
I
I
I
I
i ~ ,
0.00
i
0.2
0
i I
0.4
For /ofo
i
I
, ! ,
0.6
I
eyrpy
J= D-=
;;'PII1 £Z. (3) t). /2b
/)(,'rJ.)
.
~./2.1D In.
/-7S
, i
:
I
..... 0.8
Tube diameter, in.
Values obtained from Eq. (3)
(1)
-
i
!
I
1
1
i
,,
i
I
1
,i
1.2
--
1.H6 Under the right conditions, it is possible, due to surface tension. to have metal objects float on water. (See Video V \5.) Consider placinf, a short length of a small diameter steel (sp. wt. = 490 lb/ft) rod on a surface of water. What is the maximum diameter that the rod can have before it will sink? Assume that the surface tension forces act vertically upward. Note: A standard paper clip has a diameter of 0.036 in. Partially unfold a paper clip and see if you can get it to float on water. Do the results of this experiment support your analysis?
cri..
rrL
I
grr
.
5 II
o. 0
S/nc-e
~
-3 r.L
X' I 0
;-"'l.
~ J '+ (n.
;st.andArd ~I:ee/
paptr c)lf
d/~met;('r ~f ". ~3" il1') wh/ch O.O'/'/- /n.) It sJ1()~J~ f/~4.t. iP f / I Vof v,' f ~ 111 I.S • Ye s .
J-7f6
IS
A-
hAS
~
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J.37
I 1. ~7 An open. clean glass tube. having a diameter of 3 mm. is inserted vertically into a dish of mercury at 20°C. How far will the column of mercury in the tube be depressed?
2 (}C&S
e
( ~g.
?rR
j. 22 )
-3
3.00 X
3. 0 0
I. g$
I
1)')1
ID
1m
t'YY1
1. gB
An open 2-mm-diameter tube is inserted into a pan of ethyl alcohol and a similar 4-mmdiameter tube is inserted into a pan of watef. In which tube will the height of the rise of the fluid column due to capillary actton be the greatest? Assume the angle of contact is the same for both tubes.
(Eg. (C/ / t~hp/ )
U (~dtph()/) '0 (WA if,)
(If
~ (tva tel")
a-( WA. tf'I") !"" ( ,,/tCh"/)
~
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1m""
)
IWIIW1
= (;.2.81-/0-'). ~)('/.r{)Xlo3~3)(#MAI'IIA) ( 7. 3lf)( JD-~ f;, (J,
7 g7
/-77
) (7. tlf X }f)3 ~3) (;). M1~
)
1. ~~ *
The capillary rise in a tube depends on the cleanliness of both the fluid and the tube. Typically, values of h are less than those predicted by Eq. 1.22 using values of (J and for clean fluids and tubes. Some measurements of the height, h, a water column rises in a vertical open tube of diameter, d, are given below. The water was tap water at a temperature of 60 of and no particular effort was made to clean the glass tube. Fit a curve
to these data and estimate the value of the product (J cos e. If it is assumed that (J has the value given in Table 1.5 what is the value of e? If it is assumed that is equal to 0° what is the value of
e
l7o/n
t. ~.
e
(J?
0.25 I 0.20 I 0.15 1 0 . 10 I 0.05 h (in.) 0.133 0.165 0.198 0.273 0.421 0.796
d (in.) 1 0 . 3
I
I. ')."L.
-P. =
2 O-d-U:;S
e (-k);
'f(j'
C: e ( -f ) S
d=l12. Thus; £j.(J} 1..5 ()t the.
i,:
/cYm
b d'
(Z)
d':::
b=
The ~I/ S /-III1't.; b) C41J b.c ()j,~/~et/ b'1 szuare''s fL't 6f 1J1.e, 911/"11 d ...-ba...
-P.
//I1t'l.Y
4
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!Po 80
O. () /65'0
(). t)z27S' (). b ~5"oK
120
(), 0"" 33
J.'fO
( CD!) t)
/-7'6
least
(J.. Ql1d lid).
(ft)
If. 0
J...
d
(C4;I1't)
To
"btfll~
b
LJNREG 1
11.5 e.
r
*************************************************** ** This program determines the least squares fit ** ** for a function of the form y = b * x: ** *************************************************** Number of points: 6 Input X, Y ? 4-0,0.01108 '7 4-8,0.01375 ? 60,0.01650 ? 80,0.02275 ? 120,0.03508 ? 24-0,0.06633
b = +2. 799E-04X +4-.0000E+01 +4-.8000E+01 +6.0000E+01 +8.0000E+01 +1.2000E+02 +2.4-000E+02
Thus,
C.L, 2.-
rt,
Y(predicted)
Y +1.1080E-02 +1. 3750E-02 +1.6500E-02 +2.2750E-02 +3.5080E-02 +6.6330E-02
rr e~Se =
h
+1.1195E-02 +1.34-34-E-02 +1.6792E-02 +2.2390E-02 +3.3584-E-02 +6.7169E-02
a
If
_ (,<..799 x )0· If It
"L.)~z. If ~J) 'i-
II
.3
0--= So {)3 Jt /0
Cc>J
e ::
/J,/fi
.3 lob
= 1-. 37 X jo
1hen
I
if: "17X/lJ
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s. tJ3X /o-J .1.l!-
=- o.g~r
.fi:
alld
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~
B=o
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rr=
= J 1.70 Cos
a =
/.0
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3
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d
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if, 37 XJO
3 --.. /.6 ,ft
.ft
1.90
Fluid Characterization by Use of a Stormer Viscometer
Objective:
As discussed in Section 1.6, some fluids can be classified as Newtonian fluids; others are non-Newtonian. The purpose of this experiment is to determine the shearing stress versus rate of strain characteristics of various liquids and, thus, to classify them as Newtonian or non-Newtonian fluids.
Equipment:
Stormer viscometer containing a stationary outer cylinder and a rotating, concentric inner cylinder (see Fig. P1.90); stop watch; drive weights for the viscometer; three different liquids (silicone oil, Latex paint, and corn syrup).
Experimental Procedure:
Fill the gap between the inner and outer cylinders with one of the three fluids to be tested. Select an appropriate drive weight (of mass m) and attach it to the end of the cord that wraps around the drum to which the inner cylinder is fastened. Release the brake mechanism to allow the inner cylinder to start to rotate. (The outer cylinder remains stationary.) After the cylinder has reached its steady-state angular velocity, measure the amount of time, t, that it takes the inner cylinder to rotate N revolutions. Repeat the measurements using various drive weights. Repeat the entire procedure for the other fluids to be tested.
Calculations: For each of the three fluids tested, convert the mass, m, of the drive weight to its weight, W = mg, where g is the acceleration of gravity. Also determine the angular velocity of the inner cylinder, w = Nit. Graph: For each fluid tested, plot the drive weight, W, as ordinates and angular velocity, w, as abscissas. Draw a best fit curve through the data. Results:
Note that for the flow geometry of this experiment, the weight, W, is proportional to the shearing stress, T, on the inner cylinder. This is true because with constant angular velocity, the torque produced by the viscous shear stress on the cylinder is equal to the torque produced by the weight (weight times the appropriate moment arm). Also, the angular velocity, w, is proportional to the rate of strain, dul dy. This is true because the velocity gradient in the fluid is proportional to the inner cylinder surface speed (which is proportional to its angular velocity) divided by the width of the gap between the cylinders. Based on your graphs, classify each of the three fluids as to whether they are Newtonian, shear thickening, or shear thinning (see Fig. 1.5).
Data:
To proceed, print this page for reference when you work the problem and click hl're to bring up an EXCEL page with the data for this problem.
Rotating inner cylinder Outer cylinder
Fluid
Ii FIGURE P1.90
(c On 't ) /- go
I
/.9'0
Solution for Problem 1.90: Fluid Characterization by Use of a Stormer Viscometer m, kg
N, revs
t, s
co, revls
W,N
59.3 66.0 64.2 35.0 31.7 31.0 17.4 18.8 26.0
0.07 0.18 0.37 0.57 0.76 0.97 1.15 1.33 1.54
0.20 0.49 0.98 1.47 1.96 2.45 2.94 3.43 3.92
28.2 27.5 27.2 25.7
0.04 0.07 0.15 0.31
0.49 0.98 1.96 3.92
32.7 20.2 32.2 47.3 37.2 29.8 24.6 20.1 34.0
0.06 0.10 0.16 0.21 0.27 0.34 0.41 0.50 0.59
0.20 0.29 0.39 0.49 0.59 0.69 0.78 0.88 0.98
Silicone Oil Data
0.02 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
4 12 24 20 24 30 20 25 40
Corn Syrup Data
0.05 0.10 0.20 0.40
1 2 4 8
Latex Paint Data
0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
2 2 5 10 10 10 10 10 20
/- 9 I
From the graphs: Silicone oil is Newtonian Corn Syrup is Newtonian Latex paint is shear thinning
co = Nit W=mg
( c~I") t )
!. 'to
4.50 4.00
·.
Problem 1.90 Weight, W, vs Angular Velocity, for Silicone Oil
Problem 1.90 Weight, W, vs Angular Velocity, for Corn Syrup
0)
~·-=~I
3.00 2.50
-----------1
~ 2.00
--_ .. _ - - - - - - - - - - - - - - - - j
z
1.50
4.50 ..,--.----
--~~~~---·---------~l
3.50
W=2.5~
0)
I
0)
4.00 3.50 3.00
z
2.50
~ 2.00
-- ------- --------..-----------1I
·W = 12.80)
------------_._---+---_. ---_.. - -
1.50
j
1.00
1.00
0.50
0.50
0.00 4 - - - - , . . - - - - - - ; - - - - , - - - - - 1 1.50 2.00 1.00 0.50 0.00
0.00 +-----r----r----,--------; 0.20 0.30 0.40 0.00 0.10
OJ,
+-~'-------~------~---------.-- ~
rev/s
OJ,
Problem 1.90 Weight, W, vs Angular Velocity, for Latex Paint
0)
-'1
1.20
z ~
rev/s
1.00
---..-----1
0.80
+-------~----~-~----1
0.60
---,1tfI""'------- ~-- ----------- ----- - --.-j
DAD
------
W
0.20 0.00 0.00
0.20
DAD 00 rev/s
J- 8'2.
= 1046600° 0.60
I
. . . _-\
707
0.80
..
I. 'f /
1.91
Capillary Thbe Viscometer
Objective;
The flowrate of a viscous fluid through a small diameter (capillary) tube is a function of the viscosity of the fluid. For the flow geometry shown in Fig. P1.91, the kinematic viscosity, v, is inversely proportional to the flowrate, Q. That is, v = KIQ, where K is the calibration constant for the particular device. The purpose of this experiment is to determine the value of K and to use it to determine the kinematic viscosity of water as a function of temperature.
Equipment:
Constant temperature water tank, capillary tube, thermometer, stop watch, graduated cylinder.
Experimental Procedure:
Adjust the water temperature to 15.6D C and determine the flowrate through the capillary tube by measuring the time, t, it takes to collect a volume, V, of water in a small graduated cylinder. Repeat the measurements for various water temperatures, T. Be sure that the water depth, h, in the tank is the same for each trial. Since the flowrate is a function of the depth (as well as viscosity), the value of K obtained will be valid for only that value of h.
Calculations; For each temperature tested, determine the flowrate, Q = Vlt. Use the data for the 15.6D C water to determine the calibration constant, K, for this device. That is, K = vQ, where the kinematic viscosity for 15.6°C water is given in Table 1.5 and Q is the measured flowrate at this temperature. Use this value of K and your other data to determine the viscosity of water as a function of temperature. Graph: Plot the experimentally determined kinematic viscosity, v, as ordinates and temperature, T, as abscissas. Results:
On the same graph, plot the standard viscosity-temperature data obtained from
Table B.2.
Data:
To proceed, print this page for reference when you work the problem and click hu!' to bring up an EXCEL page with the data for this problem.
T h
1 IIil FIGURE P1.91
/-93
I. f I
Solution for Problem 1.91: Capillary Tube Viscometer
V,ml
t, s
T, deg C
Q, mils
v, m"2/s
9.2 9.7 9.2 9.1 9.2
19.8 15.8 16.8 21.3 13.1 10.1 8.9
15.6 26.3 21.3 12.3 34.3
OA65
1.12E-06 8A9E-07 9.51 E-07 1.22E-06 7A2E-07 5.60E-07 5.10E-07
9A 9.1
0.614 0.548
OA27 0.702 0.931 1.022
50A 58.1
v
=KlQ
K
=v Q =1.12E-6 m"2/s * OA65 mils =5.21 E-7 m"2 ml/s"2
From Table B.2 T, deg C v, m"2/s 10 20 30 40 50 60
1.31 E-06 1.00E-06 8.01 E-07 6.58E-07 5.53E-07 4.75E-07
v (at 15.6 deg C), m"2/s 1.12E-06
K, m"2 ml/s"2 5.21 E-07
Problem 1.91 Viscosity, v, vs Temperature, T 1.5E-06
,......----------------~
1.0E-06
---------1
I !
-I------'k-----~-
! i
~ <
!
E
•
Experimental
I
[ - - From Table B.2: ' 5.0E-07
-I------~-----c--------'~~-------l
O.OE+OO
-I------,..------,------r-----i
o
20
40 T, deg C
/- 8"1
60
80
.2. I
I 2.1 The water level in an open standpipe is 80 ft above the ground. What is the static pressure at a fire hydrant that is connected to the standpipe and located at ground level? Express your answer in psi.
1>=
~~ -T-Po
Since.. -the ~.ftt"dPI'pe
l' = ~;;. J.;.
2.2.
~'!J
)
J~
o,Pen
I~ =0 I
(8() .ft)(/I ItL
""~/),."
'T"l.
)
I 2.2
Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure). As shown in Video \'2.1, such pressures are commonly measured with a mercury manometer. A typical value for this ratio for a human would be 120170, where the pressures are in mm Hg. (a) What would these pressures be in pascals? (b) If your car tire was inflated to 120 mm Hg. would it be sufficient for normal driving?
t= !'h (a.)
For
/.20 ~
HJ :
p=
(I-3J X
/O~ ) (0, /:J,Otm) = /~. O~ Pa.
l' = (; 33 X f03;!. XC;, 070,"",) '"
1'=
(/t..O;( I03!.)(f./fSIJXf b-'I ::~~ )
::- ~,~2.
5/~ce
is
a.
n"t
fjL
-t'jp/ca / -1-, r-e pve~s u ye ~u f+; d J el1 t
r. 31 ~ J?.
-fOr
~-I
105
/Jpr!114/
30-35 -p5(,,~ I Z()ItH",.. tlJ
dr I VI!1j .
2.:J
I 2.3 What pressure, expressed in pascals, will a skin diver be subjected to at a depth of YOm in seawater?
t : ~~ 4f-
-to
~
1Ju t'ff., e
t1, e
A = (.)
StJ
thllt
3
~::' (1 0 • Ix. /o.-it.. )(lfom-a) = 40,/-;(. 10 r~,
]
N ~
-"L
-
'1()'1"t/i
2.4
The two open tanks shown in Fig. P2.4 have the same bottom area, A, but different shapes. When the depth, h, of a liquid in the two tanks is the same, the pressure on the bottom of the two tanks will be the same in accordance with Eq. 2.7. However, the weight of the liquid in each of the tanks is different. How do you account for this apparent paradox?
-.-
..
.. c
_ .., _
lv'~'"' ' I; .....,.,
14.I,I
J?'
•
/
/
Area = A
Area = A
;=1) y fl!~ 1t/IfK ~ /flt tJ7e. I ;1Cj, ned wa //05,; 1h( pressure ()n the b" ~m i.s tlwt! fD fh~ we i,/I'! ".f -fl,e /J iJl1Q Ii The t-olumn t/J;'~(.f& ~iJ~tle -Ih~ bf)/Iotl1 as rlt~""11 iJ't 1h( cI(/SH~d Ild~ 111 71tt h11("~, 7h(s iJ the 'Am! w~'i)Jf AS thAI- ;'1' The /:(lI1K. un"1it 1'ht ~fyAi9l1f 5idts. 711 uS 1 -fh~
+tt.llt~ I j the 54f'ne. In -the bl1it. WJ7"h th~ Ihc..JIH~R wAils Il1clJl~1ed w{.(lIs) t!l5 11IusfrA.hef lit 1ht' f,8tO·-e.
t;.f-
i71~
PY'(S$I(/,'~ 01'1 7h( b~1I-01Yl 711~ e:tdcl.f'PIt#/ iv-t11hi 1:S j(,(pporl:ect blj th~
..f-wo
Air
2.5 Bourdon gages (see Video V2.2 and Fig. 2.13) are commonly used to measure pressure. When such a gage is attached to the closed water tank of Fig. P2. ~ the gage reads 5 psi. What is the absolute air pressure in the tank? Assume standard atmospheric pressure of 14.7 psi. 12 in. Bourdon gage
f=: r-h + R, ..{)
-
II-ale
../h
liz k)6;.
-))
4t) - lair
(Ii.
6 in .
_
/0/ . _
'(111'
•
FIGURE P2.5
In. -.f!tl2-
1;1/; -
.2.
~
1'1- 3 1~t.'tZ.
Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure at a depth of 5 km, assuming that seawater has a constant specific weight of 10.1 kN/m3? Express your answer in pascals and psi.
l.ti
P=-~J. +~
A-f -the :5urhtce
10
=0
1> =Ul)·J i. JD ;!; )(5" S
)(./0
~o !hAt 3
/WI)
= SO. S
= 50.S"
MP~
A/seJ )
f
=
(so. 5
7 320
p~iJ
~. 7
j
2.7 For the great depths that may be encountered in the ocean the compressibility of seawater may become an important consideration. (a) Assume that the bulk modulus for seawater is constant and derive a relationship between pressure and depth which takes into account the change in fluid density with depth. (b) Make use
(aJ
~::: -~ =-('3 Thus
~ :: - 1-
)
of part (a) to determine the pressure at a depth of 6 km assuming seawater has a bulk modulus of 2.3 x 109 Pa, and a density of 1030 kg/m3 at the surface. Compare this result with that obtained by assuming a constant density of 1030 kg/m3.
( £g.
2. If)
(I)
dr
be hre.. (Eg. 1.1'3 J
So
th.. t
a.t 1'=0
/e-:' 6
~
.so -inti/:
1>-
df =
~. 7
I
(C~n'i)
(b)
f~rt ~) )
Frt91??
-p = - Ev So
at A::
-thAt
t.
Jh, (/ -
::1. )
,~~
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)in
r~= - (:J..3 x IO".!t .-m2.
; . 3
;<. I tJ '1
...!:!.. ""' :r.
= &'1. Lf M P'I(c)
~I1S
h,.,.
tan t
dens N-!J
p= (/~ = 13- ~ =(I. ~3 ~/D3~)(tj,f/ ~)("~I/~) -
I
fo6. t, ;'1~
2.8 Blood pressure is commonly measured with a cuff placed around the arm, with the cuff pressure (which is a measure of the arterial blood pressure) indicated with a mercury manometer (see Video 2.1). A typical value for the maximum value of blood pressure (systolic pressure) is 120 mm Hg. Why wouldn't it be simpler, and cheaper; to use water in the manometer rather than mercury? Explain and support your answer with the necessary calculations.
-p =
~It
J; V"
/2, 0
/YK 1M
t -= >fh
I-J, :
:::: (/33 X Jb)
,!~ )(CJ. JZ 0
Mt )
- / /,. .() 4e P... ?;
(> /,
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I', () Q
3
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(J
A.
tV ;;;":a. IV
wa..-I:ey -
Col""., 11
J, It. 3 tWl
" ,,()Xlb ;;;;;
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CO/U"'''
1m 'Y~C,-ht41.
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In
-h1< 'Ttlan()l1?f..k..t- -th ~
htI9h/:.s w()u/d be
N'o. ,,2.-5'
t()tJ
hl~h anti.
2, Ii Two hemispherical shells are bolted together as shown in Fig. P2.9 . The resulting spherical container, which weighs 400 lb, is filled with mercury and supported by a cable as shown. The container is vented at the top. If eight bolts are symmetrically located around the circumference, what is the vertical force that each bolt must carry?
Cable
Sphere diameter = 3 ft
•
'1 ".. p.-y A,-.; \AI I+!
FIGURE
p2.Cf
hwce In ~ne h"JI:. frf-,surt. at mld-p}t/he
ai rnl d- p/~ne.. we.lgh i l i In-ert.ur!1 0';:' She II
CI r-fA AI
~ "'" w-eijJJt af j,tlt/Dnt
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hlJJf
()f
sheJ/
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L ~e"'hctl /
=
0
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g IX
f Ii
-r W#-j
1-
- O"u/f )(f D~
Ws +
'04*) (f lJ~
T
~
('too /J,)
=(8lf'l !,X ¥")(f) (3ft) ~ ~lf7 ~,)(~)(f)(Htf+ J,oo Ib ferlO
1.6
I
)./0
2.10
Develop an expression for the pressure variation in a liquid in which the specific weight increases with depth, h, as y = Kh + Yo, where K is a constant and Yo is the specific weight at the free surface.
:!1-_v dr -
( Eg . .2.4-)
Q
Le t So
-(heel:
tind
Qh&i
-R. :: 1:0 -
Z
d~ :-cli:
v
d.;1
~
I I
(Lont)
2.11. * In a certain liquid at rest, measurements of the specific weight at various depths show the following variation: h (ft)
o 10 20 30 40 50
60 70 80 90 100
Y (lb/ft3)
70 76 84 91 97 102
107 110 112 114 115
The depth, h = 0, corresponds to a free surface at atmospheric pressure. Determine, through numerical integration of Eq. 2.4, the corresponding variation in pressure, and show the results on a plot of pressure (in psf) versus depth (in feet).
D
/11/////1'1'
(I)
where CQI) ht. ;".teyrIl1e4 Ilvmprlcltl/y
depth -l,'. t/.J/ns
171ft
£!l.Illtl£J" (1)
~//~/I1J 1)f"oJrAIJ1.
(!Vof~: The. n"m,y/cil I /;'.ft"Y4;'1()~ elfl1 "Iso h< (1c(t!)rnp/lsju~4 1hroufh f'e;fq.J.e~ ~.se. of '/1Ie pY~9J1'''m T.R,lfPb"ro/).
~.J/ jIk
100 110 120 130 14,0 150 160 170 180 190 200 210 220 230 24,0 250 260 270 280 290 300 310 320 33C 34,0 350 360 370 380
cls print "*************************************************" print "** This program integrates Eq. 2.4, numerically **" print "** using the trapezoidal rule to obtain the **" print B** pressure at different depths **" print If*************************************************" print dim p(11),gamma(11) n=11 dh=10 p(1)=0 for i=l to n read gamma(il next i data 70,76,84,,91,97,102,107,110,112,114,,115 for i=2 to n s=(gamma(1)+gamma(i) )/2 im1=i-l for j=2 to im1 s=s+gamma(j) ne)':t j p(i)=dh*s next i ' 'Print the results print print If h (ft) Pressure (psf)" for i=l to n print using "###.# #####.#"; (i-1l*dh,p(i) 390 next i
Th ~
1,1( j~ fed f'esu/.J-$ ~rye.!p()ncl/".7 P/()t of fa
be/"w I 1/011, liS.
the..
tJlI1h
dfpTh.
~***********************************************~ Lt
This program integrates Eq. 2.4, numerically **
** using the trapezoidal rule to obtain the ** ** pressure at different depths ** * * ~c ** * * * * *' * * * ** * * * * * * *' * * * * * * ** * '* * '* * >I: * >I: * * * * * * '* '* '* '* * *10· 1.2
1.-1
(ft)
0.0 10.0 20.0 :30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0
Pressure (psf) 0.0 730.0 1530.0 2405.0 334,5.0 4,34,0.0 5385.0 64,70.0 7580.0 8710.0 9855.0
1. o·
~ O.B .!; Co
• O. B
m
Co. :J
•a
~ 0.4
Q.
0.2
0.0 0
20
40 Oapth,
60 h
(ft)
60
100
~.)2 "2. .12. The basic elements of a hydraulic press are shown in Fig. P2.12. The plunger has an area of 1 in. 2 , and a force, F I , can be applied to the plunger through a lever mechanism having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in?, what load, F2 , can be raised by a force of 30 lb applied to the lever'? Neglect the hydrostatic pressure variation.
Plunger
/ ""-F]
Hydraulic fluid
If. j..(')~(E. t:A.
o/.
P/un7fr"
'SIn ce.
~jI'ce,) F;)
F; :.. tAl
,Pt"'e SS{,{ r'e
pJu h 1f r
1//
COtel
4 ncJ
-th i"'ou 9h tJu t
('01
rI.42.,
-Ir:,
the.
Jevn-
V'esl-{/f.s
Fj = (?)(:6.o) = 2lfo II:; P-;...= plr 2 whe~e ? co·~
1i1e
i'e~fec.tJt/eI!1. SInce...
t{V'e4S
?
I
S
P1
. Is
"The..
-the.. elMS ~n t
()f
chq m be -yo )
7h e
1+/
1h,,-/;
~f
QI1t/
PJ5 foJ?)
{:,
:50
o.ff>Jj~cI
,3.o)b
::
-A1=2-
F::2- = ,401 6- t=;
2..
- (/~~ I~.~) (2'10 Jb)
3Gj
000
1.1:,
/ /11.
2.13 A 0.3-m-diameter pipe is connected to a 0.02-mdiameter pipe and both are rigidly held in place. Both pipes are horizontal with pistons at each end. If the space between the pistons is filled with water, what force will have to be applied to the larger piston to balance a force of 80 N applied to the smaller piston? Neglect friction.
~ ~
fA-I
!=s. = f A2 ThuS)
I I
--
i ..... ~-IO
I
.~
2. IS-
I 2.15 What would be the barometric pressure reading, in mm Hg, at an elevation of 4 km in the U.S. standard atmosphere? (Refer to Table C.2 in Appendix C.)
It t
C(
n
Ta b / e
eleva. -tJt)1'/ C.:;'
l'n
01
If J.m
P= ~. / ~ (p X /0
J
:!2-
( Ir~m
-7.7
PSt:
/1. Ppen dI xC) . S I H (. e.
1:> ==- ~~ ..:P... =-
d'
2. J(,
S--
~. J" ~ :I. 'D
If
AI
-::;-2-
J3~;(,/~3..!:L
AH'3
I 2.1 ~ An absolute pressure of 7 psia corresponds to what gage pressure for standard atmospheric pressure of 14.7 psia?
-
71'.)(.«-
)Lf.7/:lf.:
2-11
:::
2./7* *2.17 A Bourdon gage (see Fig. 2.13 and Video V2.2) is often used to measure pressure. One way to calibrate this type of gage is to use the arangement shown in Fig. P2.17 a. The container is filled with a liquid and a weight, W, placed on one side with the gage on the other side. The weight acting on the liquid through a O.4-in.-diameter opening creates a pressure that is transmitted to the gage. This arrangement, with a series of weights, can be used to determine what a change in the dial movement, (J, in Fig. P2.17b, corresponds to in terms of a change in pressure. For a particular gage, some data are given below. Based on a plot of these data, determine the relationship between (J and the pressure, p, where p is measured in psi?
W (lb) (J
(deg.)
0 0
2.00 40
1.04 20
3.23 60
4.05 80
5.24 100
Bourdon Gage
'IV
(a)
(b)
•
FIGURE P2.17
6.31 120
7 rt, ~(J") ( IN hfY~
P
(I)
p-, £)
Jj I;"
=o.()SZ2lii
fi,-."m R"'I'
So
1>
(j)
(~,;)
7.Cfb
-p (f'~t.) Theta, deg.
0 20 40 60 80 100 120
W,lb 0.00 1.04 2.00 3.23 4.05 5.24 6.31
.:-
o. 'II, e
W =0.0522
8.00 ,Q
e
I
]
100
150
6.00
~ 4.00
'Q)
~ 2.00
0.00 50
0
Theta, degrees
:L -1'2.
2.18
For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in barometers containing one of the following liquids: (a) mercury, (b) water. and (c) ethyl alcohol. Calculate the heights including the effect of vapor pressure, and compare the results with those obtained neglecting vapor pressure. Do these results support the widespread use of mercury for barometers? Why?
( 1h C./luhn1 va.f~Y r(A.brn) : yJ.. + w
?
he Jl'e.
f
"V
/,!A
( WI thou f
fYt'St;I/Ye.)
f
fJ'I/"
VttfD Y'
f Y~5.sU Y"e
)
(A. 6;J/'1) .. d' h
ftJY" 't'~.ss uY'e.
= /OIx.JtJ
(It) F~". mtY'cur'j:
3 /'I
-I N
101 )(.ID
- -I.(PXJD-l.. "",,:z. t111
133)( It)
'3 tv
;;~
J"3 3)(. 10"! .J!..
3..J:L /I't?~
(M.3
= 0,7511W1 t.b) (;r
/.VA.-ter:
It =
/o/x./b
= I (). J (()
Por
erf" J
a /Coho/ :
~
ttJ/;<.ID ;:
3 N -l. ;?H
-
'i. 30
/. )0 3
J
1.77 xIt; .Ji. AM
J:!..
1.
3
~ =
1. fo
,.,.,3
=
/)'Y1
3#
.{
/01
Co/urn
n
/S
renstPnt:lb/e.
;;. -ra
1'171"1..
/3,OI1?1
btlY'tlmeiers -fhe effect. 01 IItLlt/Y is n f'r//qi bfe / tlnd t:h~ v€$u/red he/9 hi 0/- The
For mere u r!)
Jb.J.!:!...
Ilt1
-
1??'1
)I..
7. 71f X I D3 31..3
11'H3
I ;J.. "3
/'tIA3
~
7. 7 If- .x. I 0 3 .!:L
--
X/0 3 J:L
/0,3 rm
311
-"""i -5.'1J<'/~ ".".., - ..
I/'H
}J
/01 'J. I D -;;-~
PWSf lI1"e
n1er(.11'!/
:I. I q
I
2. II:!
Aneroid barometers can be used to measure changes in altitude. If a barometer reads 30.1 in. Hg at one elevation, what has been the change in altitude in meters when the barometer reading is 28.3 in. Hg? Assume a standard atmosphere, and that Eq. 2.12 is applicable over the range of altitudes of interest.
( f'g.
:2. J:J. )
(Il
(2 )
/re;m £~. (J) -k ~bta (
..5u 10 Irac. t £1. (2.)
n)
l~-r( = ~ UfJ1- (rt )r] FDr
Tq,=2.89K)
I::
=
(5
tt, '!/ ~ sa..
"
-p,::
011-,#
00 "So
~
-P~
)
= 2?7 .::!Ct I'J Ii Ie;. If( (:'8'7 *fl< ) (~. 00#;57) ~
R.
&. = d U//th
f!).
==
[] )
/(:;) Je Pc.
J
)
) ::
f!J. I
q0
9.tl ~
52.
J,., =
f:;. = !~ ~ :: -fhen
E'fJ3)
.?".pm
~
~
- 1: J ::
).f~ /(
~ O~~5tJ !fn
-- 54-3
rm
[( lol I () I
~p~ )o.Ii~
.t PL
( qo.i. .v~ 101
-It.. Pa.
y,/qOJ
2.20
2.20 Pikes Peak near Denver, Colorado has an elevation of 14,110 ft. (a) Determine the pressure at this elevation, based on Eq. 2.12. (b) If the air is assumed to have a constant specific weight of 0.07647 Ib/ft3, what would the pressure be at this altitude? (c) If the_~ir is assumed to have a constant temperature of 59 OF what would the pressure be at this elevation? For all three cases assume standard atmospheric conditions at sea level (see Table 2.1).
(a)
Ihen
-L/; )
(abs)
-p -=
~ - tlJ,.
J.I/~. z ~l. [(. )
IO/.fo
J:!.
ft1-
-
(t). 67blf7 f;3)(;~
(o.bsJ
/lo.f.t)
2.21
2.21 Equation 2.12 provides the relationship . between pressure and elevation in the atmosphere for those regions in which the temperature varies linearly with elevation. Derive this equation and verify the value of the pressure given in
fOf
r~bf.lec~ in Appendix C r an ;~evation of 5 d
fJ
_
~
dl:
T - - R
T
~
Le t
-l-J
.ft,,,.
1;~t
Thus)
,p
i:
1:,
=- -
/?
~
t:PY"
I
d
(;(11
T= To..
-rs 7: .
di-
"7 i
?;.,
1::
L £. ~ ::'
-h. tiJ11 /()5a rithrn
~ [- ~ 1m (iA-~~i .1 h (J _ Rr4 0/
-P == ~
s. 'f 0
_L
It
j"". 7:2 ~ £
c.>
~
and
{Nf
-},.::.O )
h~tn
(I -
X I () If
$1
=
~ [ e. (To. -(3<) -h TJ
&-r) TA
atd D-t e3
L(
a.:tIO·H
':JleLdu
~
~)~
(~t.2.,2.)
!!.1
mr
-~~.-.-.--.=-.=.--.~---_--.J 1.-I<&.
2.22 2.2'2
As shown in Fig. 2.6 for the U.S. standard atmosphere, the troposphere extends to an altitude of 11 km where the pressure is 22.6 kPa (abs). In the next layer, called the stratosphere, the temperature remains constant at - 56.5 0c. Determine the pressure and density in this layer at an altitude of 15 km. Assume g = 9.77 m/s2 in your calculations. Compare your results with those given in Table C.2 in Appendix C.
h 1i
= 22. ~ ..lPa,
)
-r;:: - 5 '.5 ·C -r ;L 73. 15" = _
~ 77 70,..)( IS-x If/1M - 11;(. , /
/WI
[
-
/
~ I I
:; I (p. " 5' k .
(,1 S''l
tfk )(:l /1.. /, 5" /< )
)J
-k Pa. 1")./.x.JtJ
Lb/e. C.2 J~
A-ppeJ!1&//x.
L-/7
3 IV
::;;'l..
C)
h rJ;.
: J;1, II ~
pC(.
and
:<.23
+-1
2.23* Under normal conditions the temperature of the atmosphere decreases with increasing elevation. In some situations, however, a temperature inversion may exist so that the air temperature increases with elevation. A series of temperature probes on a mountain give the elevation-temperature data shown in the table below. If the barometric pressure at the base of the mountain is 12.1 psia, determine by means of numerical integration the pressure at the top of the mountain.
W; tb -the. .frmft ral", Yt. d~b...
Temperature CF) 50.1 (base) 55.2 60.3 62.6 67.0 68.4 70.0 69.5 68.0 67.1 (top)
Elevation (ft) 5000 5500 6000 6400 7100 7400 8200 8600 9200 9900
'9 W(11 fhe. ;tJ.J.e1rl/
be eV/J/u4Iell J111h1fnc"l1~ tls/llj
'H
B ~ . 2. '1 Ca i1
rteAPE"iol.
***************************************************** This program performs numerical integratior~ ** ** over a set of points using the Trapezoidal Rule ** *****************************************************
**
Enter number of data points: 10 Enter data points (X , Y) /Jeff : '? 5000,1.962E-3 7 5500,l.942E-3 '? 6000,l.923E-3 ? 6y,OO,l.915E-3 '? 7100.1.899E-3 ? 7{,J.OO.l.894E-3 ? 8200.1. 888E-3 ? 8600,l.890E-3 '? 9200,l.895E-3 7 9900,1.898E-3
The approximate value of the integral is: +9.34S2E+00
SODoft So
fhllt
( w/th
111 11ft
~
-
j ::
(~Z. 2
32.2.
4/52.
tlh4
tt. )(r. ~ t:) 3
17/ to ft:·/b I sIll 1--;< ~-Ig
R = 17/1, It., Jj,/sJu~. 42)<. )
-
- O. 175"i.f. ( COl?
t. )
(I)
2.231-1
( COI71; )
:t.f
Ie //()IUS
-h--PIJ1
1;'1,(/)
/() J1H
~ = (1'2. I ,s/a)
e
'A = /2.J PoJ L' a.
7).,lIf
- tJ,17S'1f.
==
10, Z
PSI.'a.
-
2.2 Y A U-tube manometer is connected to a closed tank containing air and water as shown in Fig. P2.2~. At the closed end of the manometer the air pressure is 16 psia. Determine the reading on the pressure gage for a differential reading of 4 ft on the manometer. Express your answer in psi (gage). Assume standard atmospheric pressure, and neglect the weight of the air columns in the manometer.
Closed valve
I-- Air pressure = 16 psia
T 2 ft
~
,1
Water
t
Gage fluid (y 90 Ibfft 3 )
Pressure gage
=
•
~
I~ (p ~ I~.1..
/i
FIG U REP 2 • 2\f-
. 2..)
-
Ib .!.!!: /'t. 7 -:1.. '0/fLf r J-l'y,. J f -n..
+ ( (p Z. ¥ -!ft:3 ) ( ~ f-t) -
(
I (J)
16)( / It
72.. -Ft."-
J u..u •
2.
TT'I?,
2.
)
-
'/-.
b7 rDSl. .
~.
Z.2.5" J I
i
Hemispherical dome
A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system as shown in Fig. P2.2S. The liquid in the top part of the piping system has a specific gravity of 0.8, and the remaining parts of the system are filled with water. If the pressure gage reading at A is 60 kPa, determine: (a) the pressure in pipe B, and (b) the pressure head, in millimeters of mercury, at the top ofthe dome (point C).
I
/
c
2.2.5'
/SG=O.8
4m
+ 3m
-1-
Water
2m
*
(a.)
1;;
-I-
[S G,)( ~:l.()) [3 ~)
+
~o (L/W1) :: 1>B
1>8 = ("o.J.p.. + (O.a)(r.II(//!.)(3",,)+
= (b )
~
::
(r.c~~1D3;;'.)(z1'M)
/03 ~Pa,
~2,0 (3hY1)
-pit -
- k0 .k Pa. - (9. ~ D X ID .;~ ) (:! '"" ) 3
I !
-It =
5f).
~ )( JD
..pc: !'jJ
3 N 1I?1'10
3 N Bo.t. xH -;;'l-
:::
/'33)( J0 3 .!:L
j
= t>. 230
-
tI,230m!
11H'3
ml
(la 3fW1,!:,
)
ht'\
2-2./
RJ
Water
=
2 30
1)1"1
/WI
2.2.(" Manometer fluid
2.26 For the stationary fluid shown in Fig. P2.26, the pressure at point B is 20 kPa greater than at point A. Determine the specific weight of the manometer fluid.
Density = 1500 kglm 3
•
Le-/;
r-",.,::
FIGURE P2.26
w-e/;ht t>11Y1~I'I()metel'
spe,l!,,,
dA = (SG)( ~
1.Jr. D f!'
'I- c.
) (,.) = (1.2) (IODD
fl"t"
'*~)('l.81!f.. )
..".
:: I I} i()()..!!
hr' 3
(fa ::
~ 3-
= (iSt>o
#!. )(Cj
.81 ;,.) == J Lfj
7()()
!
3
tA- = - ~ (2 hYI);- ~ ( 2m1) -t rB (~IW\) t. (n
13 -
I'm
'O~
= 7/
IV
J DO
;;3
2.
-2<.
Z. 2. 7 I
o
2.7..7 A U-tube mercury manometer is connected to a closed pressurized tank as illustrated in Fig. P2.27. If the air pressure is 2 psi, determine the differential reading, h. The specific weight of the air is negligible.
1'3"= 2 psi
Air
-'-, I
2ft
t
2ft
Water
-t-I
2ft
T
~ h
~ Mercury (SG
t,~ ~ =
f
~
'1
1 -
~
#,.0
'0JI.% 0 (If !-/J ~.J
-
~:LO
( it +
If (t)
::
fa/r
= ( '2. ¥f4J ) (If ./'.;) (/5. I,)({.2.lfAJ) - tz'If/c.
2.-2. 3
= 13.6) -
- tJ. 517 Ii
'------'
2.28 2.28 A suction cup is used to support a plate of weight W as shown in Fig. P2.28. For the conditions shown, determine
W.
.. FIGURE P2.28
f.o'rces ~I\ plo.;h-e.
FOr etu; I; hrlUM 0+
~ ~ ~,A. A, 1$ ar~A of. c.up
WheY'fo
FY()(n
~
manomet-er -
£1 }
a Vltl
P, (s
oil-tO
"
btlt
ne9a...4:\ve pr-tssur"e.
.QZ ULJ·/O·v\ :
~z. D (1. ~ H) t (s G) ('0~t. 0
t, ~
&(
[I. ~ ft
) ( (j.
~ .f-i:) .:: 0
- (1I)(0.4.ft) ]
~J uft -
(E){Mft)]
-= - Cfcr ~ ~~ Tht.ts/ /-Yom t=~. c. J}
OW " (q u :") (Tt ) ( O. 5'.f.r) 2.
2-2..,-/
=:
1 g. ~ IJ,
2.29 A piston having a cross-sectional area of 3 ft2 and negligible weight is located in a cylinder containing oil (SG = 0.9) as shown in Fig. P2.29. The cylinder is connected to a pressurized tank containing water and oil. A force, P, holds the piston in place. (a) Determine the required value of the force, P. (b) Determine the pressure head, expressed in feet of water, acting on the tank bottom.
T
+ t 2 ft
2ft
3 ft
L Tank bottom
( a...)
For I
\...C)nev~
-p,t+=p I \ .
"P1~.5
e$ l.l aJ-
L tP J1
&J i
FIGURE P2.29 (J )
presslAV'~ ac..-tlnq f>~
I_
1\
p/~ror1. l-t'mat1t!1M-eber-
1/-e.J.
1)1 T o~i I .so
•
.eblA~IL},rlilf'"
(Sft) - ~l' J
(2
H.):: ~I'r
iYzl4..t
P, = 11,r - ~ ~ (s+1:) ~; J (2 ft-) " (s ~.~)(I¥* ~:) - (0. q) (~1. ~~J) (5-H) t
J
T
(0. q)(~2.~~J (1-tt)
5~~ ~\.Th us fyf)m E'C£. () ) )
(b ")
\b ( 2.) - I l, to 0 \ 1 k~) 3 fi
P= (5"5"2
~bo&~ :- "RL'r
-1-
O~2.0 (3 +t)
-1-
~6; J ('t-fi)
" (5' ~:- )(i ~'t~:) ·d":2.4- ~3 (3 H) + ~.'1)~1.~~~~ft) )
-
J 13D
Ib
J;:a-
Ib
\ I 3D "h:a. :::
2.-25'
),31
2.31
The mercury manometer of Fig. P2.3 indicates a differential reading of 0.30 m when the pressure in pipe A is 50 mm Hg vacuum. Determine the pressure in pipe B.
Water
O.
m
Oil
~r 0.30 L . m
Mercury
FIGURE P2.3 \
-fa r Wh €Y'l.
(6.1> ....
'O.il
-R:: ,4 1'{:, =
T
6'JI.
~
O.
3D .. ) -
(/;3
t
!u.o ( 0.1."",) : ~
( IJ, fJ3() hrI )
- r~ (~, P30 ...
=- -
tJf3 (0.3",,) -
) -
~i I
((), Vfi ""')
+!U; (~, 3",,) ... (fu,/D./!;"",;
)(~,~30"") - (g9b~)(t1~~~) +(/~3~)~~MI) +
(r. 80 ~ y6.}~I&1)
2,32. Water
2.•U For the inclined-tube manometer of Fig. P2.32 the pressure in pipe A is 0.6 psi. The fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific gravity of 2.6. What is the pressure in pipe B corresponding to the differential reading shown? -.3 in.
+
FIGURE P2.32
-t !#,20
'0;1
(Ii H:)
P,e :: Pj1. " (0.6
~f ~ ft:) Sih3tJ°
~fec/f,c
the
/oJ
-
/Nfl"},
cr~t> (J~ ft) :: 18
-
~ the. ;tqJe .flu/d)
t
4'#1- ({ ft) .silt
:30 ~
/t,..)(N't!ji ) - (j.b)({,2/f ~,)a ft)(O.5-) :32.3
:: 32.3 Jbl.yt3-/I'I1I- til.· la· .= 0.22'-f
?s,:
IOpen
2. 3~
Compartments A and B of the tank shown in Fig. P2.3~ are closed and filled with air and a liquid with a specific gravIty e~ual to 0.6. Determine the manometer reading, h, if the ba:ometnc pressure is 14.7 psia and the pressure gage reads 0.5 pSI. The effect of the weight of the air is negligible.
Air
----.L 0.1 ft
Water
-t Liquid
A
AI> ~
'rllJh)
=
~/r
f. u (h)
-+
-
1;.; (~. J.ft)
Mercury (SG = 13.6)
:::0
~l /
= ( o. 5 ~.). ) (Ill-If
!ft.: )
(,2. lf .!Jt3
::
(SG = 0.6)
~t (tJ./ .pi)
f-
~l 0
-(-
B
-
-t {;
~.,)(t,1.1f- ~3) (~. J ft)
(().'){~Z·'f!lt3)
~. 2 i' it
2.-2.7
2.342.31}
Small differences in gas pressures are commonly measured with a micromanometer of the type illustrated in Fig. P2.31f. This device consists of two large reservoirs each having a crosssectional area, A" which are filled with a liquid having a specific weight, YI, and connected by a V-tube of cross-sectional area, A" containing a liquid of specific weight, Y2' When a differential gas pressure, PI - P2' is applied a differential reading, h, develops. It is desired to have this reading sufficiently large (so that it can be easily read) for small pressure differentials. Determine the relationship between h and PI - P2 when the area ratio A,I Ar is small, and show that the differential reading, h, can be magnified by making the difference in specific weights, Y2 - i'1, small. Assume that initially (withpI = P2) the fluid levels in the two reservoirs are equal.
Th __________~!_
12
FIGURE P2.3Lt-
~l..
1~
_1\_1
_ _II
- -- - r. At
I;'ih~/
l-e lit! I
J~;'h4/ I,ve/ /Dr grlit' 1-/,11 'd
When
ei
dl'fffnn/;Ja / pYt'$5Ure)
___
Vf5ervpj,. drt),P' b~ A c/J~kllrce.J Alll tU,(/ 111~ /7) til') " In et: fI- e $"Il t. i~)n be ~4'm e.s
~ ~ ~
(1, + -It - i Ja.) ~ J.
11 - f2 :: S/ nee
*
-the /,
f) J:dJ
LJl Ar ::-
/"
- d'/ ..£.
~
-t-
o
(2 A -I..
$/"nIl)J
T
-f:: - +2 :: (~ - " ) l /tlr1e
vt}ltI'RS
d/f-.feY'fnt/a 1s
if
pi
h 0;1,. - 6',
(-R.,
.
IJ
tIt"t; lell~1
T
~), ) =
111
+2
)
OJ
/ncomfY't'ssi /')e;>
_ At - A)-
:l~h«~
/hen
be ne,/Jec:'ted, Thus)
Cind
~,
-
:z J h
r-
we ~$$tI',H.
ritjJri l-eve/ r-is,s b~ Ah. Thus,
1he rntln6)meter /:Ire
-A 411: :2.
4Ai-· 15
~
d'2
i
,\ i.
I~ a.pp)/eH
It - Ii)
J '1\.
/ tift
term in EZ. (J)
2.3S
Open
2.J5 The cyclindrical tank with hemispherical ends shown in Fig. P2.35 contains a volatile liquid and its vapor. The liquid density is 800 kg/m3 , and its vapor density is negligible. The pressure in the vapor is 120 kPa Cabs), and the atmospheric pressure is 101 kPa Cabs). Determine: (a) the gage pressure reading on the pressure gage; and (b) the height, h. of the mercury manometer.
t
T
1m
h
t-
1
1m
l Mercury
•
(a)
dL = ~p. who
Le-{;
6
f \LiUJ'~
FIGURE P2.35
=(goo ~3 )(~.8 J;~): 1850;; 3
QVI~
i?va-poY'
(,119e.)::: I ZO .kPo- C~bs) -
J0 /
,k Pet (ttb$): I q ~ Po.
Thus)
~A5~ ::
-FVAPdY'
+ ~ (ll'M)
= ~'l )(\b3;;'~ 2 ~, ~
-k
1"
p~
{=O.ZOL/tM
(ll35"D~~ )(\fIM~
2.3'
J
2.3'Determine the elevation difference, Ah, between the water levels in the two open tanks shown in Fig. P2.36.
"
.2.37
/
I Oil density = 1.20 slugs/ft3_ _
i
~t
I-- 2-in. diameter
2:31 Water, oil, and salt water fill a tube as shown in Fig. P2.37. Determine the pressure at point I (inside the closed tube).
'--
-
-lOin. diameter
3 'ft
1.1'--- Salt water, ------'--- ~ (l) SG = 1.20
-~f
_Lei} 2ft
[~:~'~_C -p(S G)$4(+ ~U2.0 (3 Ii) I
-t
~"I (3 Ii)
i'
~2. 0 (2..rt) =0
W4+Vr
~ = (/.20 W. 2. 'f ~. )(3,f1:) - (;. 20 !.f/:)(az. 2 ';."{JIt)- ('~lfkJ(z.ft)
= _ I ~. I
iJ:,
.f·e·
2-30
Ocean surface
2.38
An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 m as, shown in Fig. P2.31. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates a differential reading of 735 mm Hg as illustrated. Based on these data what is the atmospheric pressure at the ocean surface?
I
735 mm
FIGURE P2.38
Pa.. "'" &lb~c/ute ~/j,.
Let:
/;'fMl -v ~ w-
1\1
Ins/de. shell =
pr"esS(Jre
Sit( rlae. e
at:1Y/1 f)sphent. fY'{,sSU re
Sf t't:.i Ii 'c.
uN, ;'11 t
0
f
1fuJ.
(~, 7(P5'1'W'l )
Selttc.J4.:/;e yo
f~~ .Jo
1h4t
~ihn =
t -
- (N3
(f
( / (), 'b ~) -t ~
d
Sw-
:f!. )ftJ. 7'b-_)
- (;I!J.I
-;;'-31
( (), 7" 5" ~ )
~) (;t>. 3b"") T ~~3 ~ )(".13f...)
2.3'7* Both ends of the U-tube mercury manometer of Fig. P2.3Cf are initially open to the atmosphere and under standard atmospheric pressure. When the valve at the top of the right leg is open the level of mercury below the valve is hi' After the valve is closed, air pressure is applied to the left leg. Determine the relationship between the differential reading on the manometer and the applied gage pressure, pg. Show on a plot how the differential reading varies with Pg for hi = 25, 50, 75, and 100 mm over the range o ::s Pg ::s 300 kPa. Assume that the temperature of the trapped air remains constant.
C /~seeJ.
114/(/(
FIGURE P2.3~
a.
e'lttr4
0. .tJ I..:'
-A _
'1
pyeSSlJre I
fj
I
.h
rtt.
#1
/j~: an,!
~
(I)
-t.
4f'e
tra.pped
:JItJe
fyeSSUYfs.
/;1'
.
IS() therm4 J
aIr
.:::E..:: ~ t1.5 .fqII i f .
~11.5ft1l1t
4' r
mas~
tJheY'e -V- Is (f/r II~/"me, / l' ,:S 4bS~/l(k /vt'SSI//I'e l (J;ul " Y'elfr 10 in/hal Clnd ft~iAI sfttff.5/ r-f~ft'c.hllelfj. TIHtS)
t
a:/:1I'1
ift'
=-
(~ -r
IrltlPul /11 r;g~i £tj,(Zj CQI1 btl 1t.I,.;ffen 4.5
for
41J!'
f,.t», ) ~
t : titm [ B$ ./3)
Es.
./) h =:
-t [p;
-r ( 2)
~.t. (A r~4. D/ I-~ b< ) leJ J if.:" /.
,£ ~. ~. _ An
/nflJ
lIJ1d
L.
::z..
(J)
t- -/;./-"" l-32
- I]
( 3)
E ~ Ii' tid) n
(it)
Cflll
eX.fv~ s~ed
b-e.
J
~
1]" (
(fj-R.)).- (t1.J.~+ -?; ;- ~f"", ) IJh
f"JI'YM
+
2.1 t.e a-u;
<1Hg.
Qt)d
-!he r()o.f.s cf
(-R.t..
i-
1hJ.S ~
-r ~i"") + ,.1
J; eV~)1(1J k
g"adr~II'c. e~IJ({i:";1( ave.
~a
IJ~ -the. ne.14-J.1 v<
It",
r,---(~-L-'-.,.-f;.-#-of--~-a.-f~-)-2.--2-18-i-,; (s ) ~ ~3
V SJ1H
IS
t1J
b'Ml~
~J 5/~'e. .A h =() h.,.
1
,4- pr"Jl'tI!Y1 /tJr ~mlu iJh (}s ~ htn't/~;' of f-y J. i !"/Iows (WItH ;1i,.."" /O/.,QP" qn' ~= 133:.B.Nk3 100 cis 110 print "***************************************************" 120 print "** This program calculates the lower root of a **" 130 print "** quadratic equation to give Dh (in m) for a **" 140 print "** range of gage pressure, Pg (in kPa) , and [or **" 145 print "** a set of different initial heights, hi (in m) **" 150 print "***************************************************11 160 print 162 dim dh(S) 164 patm=101 166 ghg=133 Dh(hi=O.OOO) Dh(hi=0.025) Dh(hi=0.050) 170 print " Pg Dh(hi=0.100) 180 for pg=O to 300 step 30 190 for i=O to S 195 hi=(i-1)*0.025 200 a=hi+(pg+patm)/(2*ghg) 210 dh(i):::a-(a~2-2*pg*hi/ghg)A.5 220 next i 230 print using 11####.# ###.##### ###.##### # # # , # # # It # It ; pg , dh ( 1 ) ,dh ( 2 ) ,dh ( 3 ) ,dh ( 4 ) , dh ( 5 ) 240 next pg
f1 =o.
114 'l'/oU..!
)'
Dh(hi=0.07S)
(C.()//t )
.1
I
~**************************************************
**
This program calculates the lower root of a ** ** quadratic equation to give Dh (in m) for a ** ** range of gage pressure, Pg (in kPa). and for ** ** a set of different initial heights, hi (in m) ** *************************************************** Pg 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 2'*0.0 270.0 300.0
Dhlhi=O.OOO) 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
Dhlhi=0.025) 0.00000 0.01101 0.01816 0.02313 0.02678 0.02956 0.03175 0.03353 0.03'*99 0.03621 0.03725
Dhlhi=0.050) -0.00000 0.02120 0.03538 0.0,*539 0.05280 0.058,*7 0.06295 0.06657 0.06956 0.07205 0.07'*18
Dhlhi=0.075) 0.00000 0.0306'* 0.05170 0.06681 0.07807 0.08673 0.09359 0.09913 0.10370 0.10753 0.11078
Dh(hi=0.100) 0.00000 0.03938 0.06716 0.08739 0.10258 0.11'*33 0.12365 0.13119 0.137'*1 0.1,*262 0.1,*70'*
0.20p-----------------~-----------------------------------, -
0 --
hi-O.OOO m
- 0 - - hl-0.025 m - b . - - hl-0.050 m - 0 - - hl-0.075 m - + - - hl-0.l00 m
0.15~----------------~
0.10
~
~
~
0.05
III
J:
O.OO~~~~---e----~----~--~~--~~---&----~----e----t
-0.05L-______ ________ __ ____________ ________________ ~
o
~
~
~
100 Gage Pressure,
200 Pg
(kPa)
~
300
SG = 1.10
2.40 A O.02-m-diameter manometer tube is connected to a 6-m-diameter full tank as shown in Fig. P2.40. Detennine the density of the unknown liquid in the tank.
4m
~
" __!IiIII-¥ Specific weight
= 25.0 kN/m 3
II FIGURE P2.40
LeI::
0':: ~R 14.71:. cl-
0-;. ::
UI7k.nrJwY/
.fluid tUt~
rI,/O)(f. 80 x /D3) ':' If). 8)( Jo 3 N /~3,
Thus,;
1~
-t
d' (1",,) -
r= /J -
\ 8. ~
cr
(~",,)
3
(zS.>W ;;. )
_
-
0;
(3",,) ::
;.. 1D3 .!!. 3 /WI
J ~.
q
3
N
.x 10 -;;-3
t - d-
2..-.3S-
=
/'f30
~ /WI 3
t;
2.4\ A 6-in.-diameter piston is located within a cylinder which is connected to a ~-in.-diameter inclined-tube manometer as shown in Fig. P2.4\. The fluid in the cylinde.· and the manometer is oil (specific weight = 59 Ib/ft 3 ). When a weight "W is placed on the top of the cylinder the fluid level in the manometer tube rises from point (1) to (2). How heavy is the weight? Assume that the change in position of the piston is negligible.
•
~/t)l1e let:- py",ss,,"G m4111?me iel'" -Rg"'Lt/~'1 b~c~""e.s
tv/in pi$t::I?"'1
- a::d/l, 1..,
~p
etdded
~ an p(
I
'::
g
m~1/14'me.j;el'"..e k~t"~J1
PI" 51,;( b I-t&tci
4il (-'-, +
E Z . (J) -fr~m
hce ~f f,·stol1 =
~
(I)
in c yot'IZ>8d
W A-p
+
SO
-\;../ A,
'::.
thlll-
W 'I: (~ If ,2
It) "-
~;J
~ It) ~JH
'3ofJ =0
(;t It)
S/H
0
3D to
- &9 ~3)(-{ ft )(()/~)
\AI =
;1.,Qo
Ap
I
wh e~e
t fJ~+OH )
hecl!>m es
Eg ,0) -k o.biol~
tine!
-to tI
1:0r '_ f.f - J:'" 'J (.J/.. ft.) ~/;' ~() ~ :. (!:II"
P2.41
sil? 30° .::: 0
'py.,SStI r~
~
()11
FIGURE
/1,
(2 )
2.4 Z The manometer fluid in the manometer of Fig. P2.4Z has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer.
T M=l-t-2c("
-.l
Gage fluid
(sG = 3.46)
•
~I"
FIG U REP 2.42.
+;,~ in;·fJill clPnli1l1NltJf:m :
r
~
(;l) r
0
~f
(2.) -
d;h~
(, ) ==
~
0)
where ~/I 1t'l1jtn,j I'(JA~ I;' R. tVhen ~ dec.re~st'.s fa ~ I Q H~ 1:, in~re/(se.5 +0 ~ I 11Je heljhfs Df t'he .flllid Caltlmns eh4n'je ts 'hown &Ji1 .fi:;w~e. h>)I- -tne ./;ht1/ ~()11./;9j,(rtll:l()·H "
p; + ~() (cl.-a.)
J~ bI:.~ct
~
1;g .c"2)
-1;/ t{.
Since I
=
1-
~~/ It}
"" (
fA. -~ :: (-
(J-r;;'~) -~o
(/+a.) =
t/
~f
-
-
~zo (q ) "
Pa - Pta'
(fIt) J
I. B f5L. /
q ;;;,'l. )b)(tIlflf h~ /n.~
~-fB :- - 0. q fSi.) tinA
)
rJ. ('2. If };1)
1hfre"r~ ~h
( 2)
(~ -p~)
-
(f~ ()
(.,,,) r
-
(/, 3
-/t,.~ )(IV-t; *~
q,.=
CI/Jc/
pl!J'
fum EZ' (l) -10 bb.bn'J1
( ?13 -18 I )
,
Jjf
-J-
= :2..ft + ;. 4:
-
(I-
)
3. if~ )
~.ft +:J. (1.03
~f = ~ lfb ~:J.O
H)
-
/. rJ3 It
-
-'I. Ob It
2.43
Determine the ratio of areas, All A 2 , of the two manometer legs of Fig. P2.43 if a change in pressure in pipe B of 0.5 psi gives a corresponding change of 1 in. in the level of the mercury in the right leg. The pressure in pipe A does not change.
Area = Al
.hI: ~ercury
FIGURE P2.43
For -the.. /n;t-141 C(!)nll'rwrlJf/~"
~
+
rT
18
when
111e lef-t 7)
FoY'
H1lfre) :
lJ:" ' L D (t..' +.411,... ) -!:/" . ,(iJh,') .. inc ~~ps~s the
Y'/9 I1t
t~/umn
Ct
must:
/ I;; /lId
(Sf'e
nje.s
YemA/h
rt): t/3
-!,.
("Iu",,, /rills
Su bfr", t
or
d;,
"LO
t:iZ.
- 0t:J.o
4HI-i
J
cI'sl-t:/I'ue) h, Sin(e. 1ne. I/o/lime tjf. 11te ~tf)I4S/:;4I1t ) rt, .A h = Az (4.. e>r A; At ::: h Q...
( ~, -r- lJ hI.' -,,) - ~
tJ)
d;si~""e ) a.
A
-the. .finAl (Onh;"r4t/~H) w/"ht /,y,ssure
fl4
(/)
(), J '
In
f3 eJlltll -b,
( J h.-A. - b ) - d:" ( t. ' +~)
~'()/
t.
P/3
J :
=..J.. I rl3
&171 EZ, (2) -fo ohio;",
(1:,) 7' ~;.
1:>,8 -
"3
?f,,;j
(tAo)
I
("
+h) -
b = (PR/ -~13)
-
d;/,
(If'{) :::
I~ (I/{)
t
!~ - ~~o
P/l/ -f.e = o. S" psi tl~tI a. = / ,'/1.),-t ~/IDWS thai b : (tJ.5 ft,..)(I'fIf!li: ) -r'l7 )t, )(,i It) + (tJ.gj('H ~,)M ,cr) 'Ktt7
It, - b'Z. If ~
= O. O~71/ I-t ::
/t. 7
{z.j
2.44
The inclined differential manometer of Fig. P2.44 contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain a brine (SG = 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, 8. FIGURE P2.44-
tdhtl1
1:;,. - P/3
,
in c.reAsed I::c ~ -
's
I
Pa me
le/i
Cr:;/qmn
IR//s
t<.
dlj'.f.an'e tt ClI1 P the r/9}rt Co/tlIY'n II'/Jes ~ dishtl"lce. b t:lj~l1f ) J 111~ Inclined ~be- 4$ .sh~UIi? /n .f,9£.{l"e. J::;v -tn,'.J +-/Yl4J ~nfi:juf'tJ.t/(;)1:
1ft + I
~,
1,/ - Psi
(I,,. +fA.) -I-
~c1l~ (It t-.b ./" e) -
(<1j, .. - 0;.'/11- ) (a
Th e eI,:P1e rent/oj read/nJ) 1J h)
-I-
h fJlh
lrbJ( -bs,." e) = +~
t9) ::: 0
(I)
111 e -tube is
tI )!)f19
~h=
Thus,; +rt)rn --h
'#
I
El.{))
-'Pia
'
+
( ~ ~ - 1',,1If
)(
A h 'S /n
~ )
=0
- (~'- Pra')
- (,), £fbI-
Thus )
2.45"
Determine the new differential reading along the inclined leg of the mercury manometer of Fig. P2.45", if the pressure in pipe A is decreased 10 kPa and the pressure in pipe B remains unchanged. The fluid in A has a specific gravity of 0.9 and the fluid in B is water.
FIGURE P2.4S
~". fh~ /n;/-/q/
~
A wh(;'"(
h,~~a tl(;rl
+ i' (0. t) + t'Jt
rnlP V-PS
t:t
C(J)n
A
fill up
3 /enftns eire A.
c/ 1'.5-k01Ce... J~)
14 I..
~
(().05
30") -
5111
tJ),e YI ~
m,
l'n
~/n 30 oj
~~O
!'lew !Y'~5S"'Y'e
the
EZ,C2.)
=
I
fA - fA::
/0
(~.08
t If4!.
In
1:.0
Ta.)
l2.)
A.
ob-N/~
~A
S/H
3 DO - ;Yft~ (5//1 30'TI) ;- O~O
-
10
ltPa..
0, 051./0
~N
--: M1
-
(J 3 3 ~ )(0, ~-T I) of-
1??1
New d/ffel"en-t14/ V'eudln1) 4h) me45tAY'eJ a/{)n1
to
«) -
8
Eg.{J)
( 0, q) ( tt. SJ ~) (0, 5" )
o
30' -/'
"
= 1,
from
a.. -
e'l I{a. I
/ e/-t CI::J/Utnl1 moves dlPWM
6"IJ., ( " ",;, JO' + O. oS sin
+
!:
a.
clec.l"'eases
(I)
ch$-hnce; a.) QJ'1e1 I"l!J hi ~''''ml1 as .s h~Wh /" fIJIIJI"~, Ft:>r the ·f.ti'1ll/ t~4tj"NI.. 6/~ ~
(tJ, J - a.
IS
~20 (0. ()8) ~ ~
~ 1- =
n
Cl
Sin '30 0 o,OSJ.jo (!:l,S'
+ 0, 05
4-
/I'YI -r t).05M1 i"
/Mc..J1I1eJ
+t.tl"e
1',5
a. O.
D5¥O tn1
= O,Z12
IW1
2.4~
Determine the change in the elevation of the mercury in the left leg of the manometer of Fig. P2.4f> as a result of an increase in pressure of 5 psi in pipe A while the pressure in pipe B remains constant. 1. m.
2
diameter
FIGURE
-fA
'tJ.l~O (-f!.) - 6'#3- (.1;.
-r
SIn
30°) -
P2.4~
~il (~) = ~
(I )
t..Uhet'~ Id/ /el1~1h.s 4re /n f.t. IJhen ~ /nc.retlses +0 1Jt' -the Ie.ft column ~/Is blj -the. dt's.J..q;tce; (1,) and 1ne rlJhi ~/"mJ1 m~l/e.s (
-rlnal
*:
Up
d/s~n~e. b)
-me
, t!4'I1.h!}HI""i;-JiJn.' I
+
n
Subtract
a;:
"~o
~b'
7£
( (I)
.p~ - ~
~
-t 4.. ) -
tr()rn "1-
J
3
FZ.ll)
t:lS
(a +
sh~ttll'J
~ :5 /n 30 D-r b
.f.o ()~iJ1
/n.) ~
a- = (; j n. ) '))
.b ::. If 4-
- ~/-j;A )
=
o#:lO - !Ut (!),
3 CJ 0 )
T
30 y.
.pi:
(3) -+ 4,;/ (2. )
(down)
:.
-
6
'30 )
~ 1:.!3
~i/ (bS/1') 3c~ =0
0/ IJZ'u'"d rnusf ~.e ~s.ftzl',.t A,
(i
=
oS /11
Irj",I"~. /;y 71l<.
~()/'I (!3.1:1. - b $JH
~~o (a.) - ~3 (li-rb5/~306)
Since -the I/()/ume
a..
;11 11Ie
a.
= A,z b)
( 2)
( 3)
rD-jd
2.4-1* Water initially fills the funnel and its connecting tube as shown in Fig. P2.47. Oil (SG = 0.85) is poured into the funnel until it reaches a level h > H 12 as indicated. Determine and plot the value of the rise in the water level in the tube, e, as a function of h for HI2 ~ h ~ H, with H = D = 2 ft and d = 0.1 ft.
T
______
H
I v'--.l
H -----,
h=2
~
4
Water
('2.)
Final
Initial
FIGURE P2.4-1
{= ~(lJ:r)-AD)T-1o
OJ
Oed
Tht
V()/lime
cf w~trY' mu.st
.E d"< /J f .,(.
A-l ~())
-ff/:z.
!i.1· t2·) CAn 3d'f.
toY'
3
12:: Pt..' = Do H
find
(DI.:)L it "'2.2.
E
-
-
t!Dl7ser/le~I
b€
H= 2
E~. (I)
-Pt
I
be.
D =2
w"iffen D2H
f t.
I
7r
-
1iJeyekt'e
(D~)';
3:z.
t/HU
~
- -8
_
(1111{
c
J) . .:: t.
(2)
12 2-
ct..s
(-P )~Jo
d = (). I
It )
3 ( 3)
411"
hec~d1es
;. :: 0. ~51t 1- O. IS ~D - /
(If )
be~m's
3
2.
(t;, 1ft)
P=
(2
ft)2
(1. f6)
8
1:
(1_ ().~3J.
) 3
(S)
IuI1(,t/~n ~{ 100 110 120 130 14,0 150 160 165 166 167 170 180 190 200 210 220 230 24,0
-f.
/.o/I()w~.
cl=: print 1(****************************************************" print "** This program solves iteratively a system of **11 print "** equations to calculate the elevation 1 (in ft) **" print. "** range of heights h (in ft) **" print "****************************************************" print print It h (ft) 1 (ft)" I for h=l 1=0 print using" ###. ### ###. ###"; 1. ,0. for h=1.10 to 2.01 step 0.10 1=0.0 las=l hO=(1-0.03*las)~(1/3)
1=0.85*h+0.15*hO-1 if abs(l-las/l»O.OOl then goto 190 print using" ###.### ###.###";h,l next h
To. bu/Ct ted dtl:l:a..
find
4(,
plot
of
tAe d4.ta
Clre .sh"IlJII
heJou).
**************************************************** ** This program solves iteratively a system of ** ** equations to calculate the elevation 1 (in ftl ** ** range of heights h (in ft) ** **************************************************** h (ft)
1 (ft)
1. 000
0.000
1.100 1.200 1.300 1.4,00 1.500 1. 600 1.700 1. 800 1.900 2.000
0.085 0.170 0.255 0.339 0.4,24, 0.509 0.59lJ. 0.679 0.76lJ. 0.84,9
1.0~------------------------------------------------,
0.8
-..... -....
o.e
0.41
0.2
1.2
i.e
1.4 h
).-¥3
(ft)
1.8
2.0
i-3ft-1
2.1.f 8 Concrete is poured into the forms as shown in Fig. P2.4e to produce a set of steps. Determine the weight of the sandbag needed to keep the bottomless forms from lifting off the. ground. The weight of the forms is 85 lb, and the specific weight of the concrete is 150 Ib/ft 3•
--==L 10 W FIGURE P2.1f8
FrfJm the. Iree- bfJJ'1 - dt'a. :JYI/m
"
([3:LP1=O ~+-~+'Ui-tA::;o
In.
tread
J~
~1J
(I)
i~A
tJhere .' ~ = we'flJt of 5t!Jldbtrg '1d" ~ lVu;ld of- C())lcreb
PJ1 : tve&Itt (/{ *l'Ins Ph : ?r~jII~
uft;"1
6"l!Dm ~lIrl4(i! due if) ~lfcyeie
A ::: I:Irelt ~I btP/folr) sur/ga
?rpm 1h~ d~
flU/A.'
1vc -- (/~P !t3)( ~f ~HCf'~) . ' . =-;; Jl!)/ L'L) [(I/);i1.f,z'f',h.) + (/0/".)(16",.) 7- fi()",.)(~ tn.J) (IS-O :n:. Us
Thus)
fr~m
. ~
l.37T;
/ f!-ff .!11·
.f-C.~
F!. OJ
~ = (jOt;~,.) (7. 5" H. 2) ~ 1&,5 1.J,
-
ISOt)
It - '65/.J
2A9 A square 3 m X 3 m gate is ·}ocated in the 45° sloping side of a dam. Some measurements indicate that the resultant force of the water on the gate is 500 kN. (a) Determine the pressure at the bottom of the gate. (b) Show on a sketch where this force acts.
(a)
~':.
ir'hc.A ::- (~. ~D ~) (-/tc.)(5'1nl JC3,,")
SDO-k. N
*
f
~ c.. =: S. 10 7 /YYt
~
= ~ (~(. + I. SI'tY\ )( 5111 Il-S·)
bofh,ft'\
_ (ct. Kl> :~ ) ( ~. (..1 ~ = G,".O~ 1.;1(.("
lj : R.
~c. A
':1c.,
=
~It = ==
~'.D~g
=
tl1'\""'
(j, )
+ I.Sht\ sin Lf~)
+ Yc..
~c:.
1:
s~'~ ~ S~
5'.fgJIM sin 4-5°
..L )3 12. (3",,)(3M\
(8.021ft" ) (31M <3.11
-
g.02
NY\
8,D2,*, :' ().l> q3Siv.. + 8.0 Z,w.
f
'i.5/r&4 )
IWo
/ / /
~
1.5" I'tI\ + 6. 0935 ~
= \. 5"'i
('In
~m
top of. ~ cs.1:!
2.6D
I r-0.lm~
SO
An inverted O.l-m-diameter circular cylinder is partially filled with water and held in place as shown in Fig. P2.50. A force of20 N is needed to pull the flat plate from the cylinder. Determine the air pressure within the cylinder. The plate is not fastened to the cylinder and has negligible mass.
2
Air
T
Water
0.2 m
Plate
1 F= 20 N
...pA .t.4l~lt!*l
h> y e ZU'I //bY/u m
2.
-Ph
F:-erfJCa I .:: 0
-+
ZtJ N =~
( N.u -that.
-pA-)~
be
a
pr'es~tlre
fYJtJJt.
'~u"I-/~/ pre. ~fllye.J
(J .I
(t). ltttf ) = ~ ~. -+ aIr 1J.z,~
Th{Js)
1>
ZtJ IV
-faIr = =
r
~.j,,") ~
Lf 5' /0
.!!.. /1112.
2-'1~
(iJ ~ j. /~ of.!!.. ) (t;. 2M') /1')1.3
-
-
If. 5" /
.J. ~
d.?!
I
T :'::", 1i:9_J "'-f. ,.l~;:;:P_IUg_-.[ f;:~:,::,~~~-::::~:::,;:
f
2.5/ A large, open tank contains water and is connected to a 6-ft diameter conduit as shown in Fig. P2.5) . A circular plug is used to seal the conduit. Determine the magnitude, direction, and location of the force of the water on the plug.
-,t_
..J..
,--
'
L::;"';'!
: ,"
•
r ...1..
I
FIGURE P2.S\
wheYt! :: G-ft) If + (/2 It) 11 flU)" {he
';;rce
01
Ii
.).~ ZfJo/J:, acfs
wp,i:ey Stlrhrce surface ~5
/"L
t:lh d
IS
--
12./ r
R
/21'i.fi: b!/t:J1d the pet/end, Ci;(/p r k fh~ pJllf
5h&(//)1.
;. - '17
2.52
Cable
2.52 A homogeneous. 4-ft-wide, 8-ft-Iong rectangular gate weighing 800 lb is held in place by a horizontal flexible cable as shown in Fig. P2.S2. Water acts against the gate which is hinged at point A. Friction in the hinge is negligible. Determine the tension in the cable.
Water
Th f,{!JJ
'i< =U2,lf !J)("4Jr;,;'h·){tftdft)
= 3390 If, To
/ tJ~4,te
Fte ) I. 1l C.
~Ie :: IJ, A So
ihd-t:
117. ::
-r '1,
I~ (ifl-t)(I.,ft)
where.
'1e. =31i:
:3
-I-
3
f-I::
(g h) (,ftJlIffl.)
- if. 0 II:
ef" i // 6;illlYl ) 210)1 tin
d
:::'0
T (3Ii)rs/~ 6~' = 'iJ ['1.r;t)(~()Sbb()) + ~ (2 +t:)
T=
(sbb/b)('f.ft)~S"D~)+ ea'lb Ij,) (2ft.) (g .ft ) (~t'11 boo)
/3S{) /J.,
2-'18
2.54-
An area in the form of an isosceles triangle with a base width of 6 ft and an altitude of 8 ft lies in the plane forming one wall of a tank which contains a liquid having a specific weight of 79.8 lb/ft). The side slopes upward making an angle of 60° with the horizontal. The base of the: triangle is horizontal and the vertex is above the base. Determine the resultant force the fluid exerts on the area when the fluid depth is 20 ft above ' the base of the triangular area. Show, with the aid of a sketch. where the center of pressure is located.
~c :(/f~D') ft JO. t.r3 ft
.::
.-f~ -
~ ~ ~ J., -
-&) tt
U
-rC
A
S/~ jp~ =
0
(]If. 8 ~3
f
PY'~5SI1 Jl'e.
.,l M 3
1
t-t) ~ t. (~){ "h d SI
0
H)
33;QooJb
a{../-s
1hr~ufh
c/l, fql1~e
t:J{
~
t:)
$liJ (Po~
~I-
i'Ae.
.fr/~if,le.
;;. -Lf'f
as
Ii - ;'p, t. Ii :: ;.If'1 /-1: 5h~U,)n in ~Jetc.h.
2.55
Solve Problem 2.Stt if the isosceles triangle is replaced with a right triangle having the same base width and altitude.
-- -
1
FI< = 3~) 9()o/b ~
1=
J, If q Ii
(see se>/ut-i,;;, ~ Prt;j,len? 2. '5"4 )
L';jc.
(Eg.
'Ie.- It _
( to It)
:J. (
2,2.0)
t fi) 2. -=
2, ~ '7
The 4rce.
J
~
t,4)(!')rdi"",l-e.r
-Pt
e:.tcl-s "/h;'''''ff h '1'hc cen.f-er t:J/ 1'1"eSStlrc w/tu ~~ = :l, (J 7 f t: Q#f~ .!J J = .:J, 'I- 9 f-t (5 -t!e. .ske I-c.h ) ,
J
z.s~
Open
2.56 A tanker truck carries water, and the cross section of the truck's tank is shown in Fig. P2.S6. Determine the magnitude of the force of the water against the vertical front end of the tank.
•
PR,::
'(f
FIGURE P2.56
he. A
Brea./( area. It'lto 3 P&\ . . i:-s a s shewn. Few' a~a.. ([) :
i=k I ='0 h~I At. I :: 5/'(I(t
~2.lf k3)( 1)(~.ct) ({ )(2+1: ,( lfi:) /1, F,~ I = F~ 3 1'J"el1 333
1="~3 :::
3-33 11,
~r tfret< @: ~ "2.. ~
?> hi! z.. Pt.,.
=(~ 2. Lt tt3') (1H) ('t-+t t::A. ~
tr t-t )
L F" .,. ~ "I
:. 2.
-
f.
'Z..
(333Ib)
2~ fDb
-t
02. 4- ~~) (z.f.t ) (~H
Ib
2.-5"/
,c.
't-H.)
2.57 2.'57 Two square gates close two openings in a conduit connected to an open tank of water as shown in Fig. P2.57. When the water depth, h, reaches 5 m it is desired that both gates open at the same time. Determine the weight of the homogeneous horizontal gate and the horizontal force, R, acting on the vertical gate that is required to keep the gates closed until this depth is reached. The weight of the vertical gate is negligible, and both gates are hinged at one end as shown. Friction in the hinges is negligible.
T 1
Horizontal gate, 4m x 4m
h
R
_
~
Water L -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Vertical gate, 4m x4m Hinge
),,, 1"17(:)11111 /
L MJ.J
1ha.i OW
=0
~Jure
.pA
=
P
/j tnt' tvp.-br fr~Sr/lye ~11 fh~
~~m SI/f"face .
r = ~/.It. (J (2.h!'J
..-6
11141:-
1v
N
=(9!OO;;3
tlfy-t/ c~d
!;r
fk =I'A~A Se>
)(z~) (tf/M
J<
Ji.M)
314
-I
rttie)
Luhey~
ht :,
7 hn
r'ff"!!--R.
7'ha. t.
1;.:: (?;p~ 1-3 )(7 )('l-tm X Jf~) trn
=- IIt)o.JGN To /1/ ett k FICL~c.
U
J~ -=. Y"A
-+~,
=
~ (ifltH ) (If/11( )
~-----~ ~ ( 7 """ (If,.,., , I1H )
J
If
~ u/'II'br/um
2: MIi
R=
:::'0
(; / IJ~
;
So
I< IV ) (rAn
1Yz41:. - 7.1 'IIIIK )
J..-S2..
11'W1 = 7.1'l1M4
2.$8Open to atmosphere
2.58 The rigid gate, OAB, of Fig. P2S8 is hinged at 0 and rests against a rigid support at B. What minimum horizontal force, P, is required to hold the gate closed if its width is 3 m? Neglect the weight of the gate and friction in the hinge The back of the gate is exposed to the atmosphere.
1
3m
il-'
0/.';,', .. ·/t
-Hi~-.
Water-
-
4m
';':.:'-':-. -.:- ::;.-: ..: B
A
. - .. -. .'. p:z-.-::;-.:r;::,.~_--1..._p
F., = Yh ',I A ~
=(q fpo!..
/J'Y13
tJheyf:.
) (5;w, )('ftm
".~_:~.~::.\/-:;-~-::: <::'r- 2 m--1
he. I = 5""1 •
X 3 1m)
FIG U REP 2.58
o
- s'i'8xl{)SN
hCl. = ~o "'fhAt
;;3 )(1/h1)(2~
;;.:: (1fOO To
The
'I. 12
0)
1~I
Ixc
=
!:fcI A,
Ii at:--Is L H0 =-0
-krce
F; (6: 2.6, 7~ SD
ie,
-
4i 17re
-
~ ~)
III---~"'"
() 1
, X
~~)
/ (
-t
o;
7M-1
p
;( /O~ N
/oc",-if:.
~
/ 2.
~"""
(S""m?)
Cenhr
-r
Ii
1714.i
p=
2.-53
) ( if;ffl )
('1M
3
.(-i;Jn)
0/ 1J,e
(J".,)
=:
-+
01.5'1 '2.'S'f The massless, 4-ft-wide gate shown in Fig. P2.59 pivots about the frictionless hinge O. It is held in place by the 2000 lb counterweight, W. Determine the water depth, h.
Gate - + - I I I
Pivot 0 Width
hC :: .b.2.
For'
ezu"~'br;"m)
::EM f;. d
()
So
=()
= %) (3A:)
-;-;'4 i (
2~&J() Ii; ) (3 H)
(0/.1z, ~)(!J... ~) It It) 2-
-It. 3
,=
(3){Z. 0(1)<>
It ) (3 f+)
(~Z.¥~,Yf )('rR)
i
==
5'. lift/:
= 4 ft
2.60 *
A 200-lb homogeneous gate of 10-ft. width and 5-ft length is hinged at point A and held in place by a 12-ft-Iong brace as shown in Fig. P2.60. As the bottom of the brace is moved to the right, the water level remains at the top of the gate. The line of action of the force that the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted on the gate by the brace as a function of the angle of the gate, 0, for 0 :5 0 :5 90°. (b) Repeat the calculations for the case in which the weight of! the gate is negligible. Comment on the results as 0--+0.
(a.)
tn,
HY
FIGURE P2.60
free- j,()dy • dlR~rllm 0/ tlte
(St*~ rl,lfl'tl))
f}4 -fe
Z t=;. =0 00
thAt '}. (*) +
Also"
q,v (f et> s e) ~ (1=;
J. ~/h e ..511, ~
f:.R.:: w hert!
W-
'6 (
=
J. L
w-
~
sa ~e T
~/;'
=
(
V' A 2. 0 ,(G.>
Few
r;
wulln. Thus) I:!.
_) <\AI Wj S I" B + '2 ~.s
F,.,= 8
'O-::'fDZ.'f 161ft))
CBJ {J ""
Cl;s
e
-=
(II
Clln
.
b.e
Fa 1 ( (ss ~ J/~ () l' J/' " ( (: zw-)
I' AII
t;~
e
(,{~
Wf, /fflt
('.{)s
e)
~ ~ -
( Z )
9
1=5ft/
(6,2. If !:))(Sft)l(10jt) &;
(Fa jl ;HJ) (j f()J a) (/ )
r (J ~/~'~ )(,Q w)
OW!
~s ¢; 51"!? T 51 H tP
.,.
( a s.slllfl/~ ~ h/l11e. t:{ ntl end /) f brllle II + .slllIIe elfIN. 'h~11 )
that
.50
~
{'J(sl. 9)
~) (1 , I ~ e)
-=- L .5/n tf;
o!'c A
the
I.J
UJ.5
hi" (; r
= ( CDn't )
~ ~() () (r)s
-hi" G
9 frill /}
f' I () 0
+ ~/H
4
( J)
~.bO 1-1
(~,,'t
)
SInce
51;'
t/;
SII1r/ c( h
d.
Clift
~ /....
::
-S-
::
12..
511"}
e
)
S/~e
?
~j/ ~ ?liI! 11 l) I t!iJh ~~ b( II.5f'd .fD tlel-ermHIe... Fe
proJ ftlm
foY'
C/J IUtllllulJ
L= /2 It
f:: s-ft
~
4.s
It
r/e.J.en11lI'lt4. 711 tiS, £3' (3) ~Y'
t{
JJi/!J.1
&.
I-uI1 C.N()·11 of e-
A
~"'Pu.,/;fY
~/j()l().s •
100 cis 110 print n**********************************************************" 120 print "** Variation of the resultant Fb as a function of theta **n 130 print n**********************************************************n 1/,r0 print 150 print n Theta (deg) Fb (lbs) (w=100 lbs) Fb (lbs) (w=O lbs)" 160 pi=4.0*atn(1.0) 170 for theta=pi/2 to pi/36 step -pi/36 180 sph=5/12*sin(theta) 190 phi=atn(sph/(1-sph~2)A(O.5)) 200 fbl={2600*tan{theta)+100)/(cos(phi)*tan(theta)+sin(phi)) 210 fb2=2600*tan(theta)/(cos(phi)*tan(theta)+sin(phi)) 220 print using It ###.# ####.# ####.#";theta*180/p i,fbl,fb2 230 next theta
Tabu/flied data..
rhe (b)
h//6'/A.)lh~
CIne!
a.
,lot of the data.
af'e
.!IJlfrJ
~I"}
pa..je.
h;r OW=O lEg. (3) reduces t~
-
.J
the
(/11(/
(w/·tn
£' l!J
((.s
w t:I
G!;s ¢
frJn& -r
5/)1
¢'
jC/me ,ProJY'lIm t:lS WIIS u.sed In f4rita) set: fS"4/ -t-o "'!freJ) (!.All be used ~ t!)bt4/h ftlnt;t/~h ~f e. T4J:,t(14~~ data.. qllt/ ~
f/r;t t!)f the d~i£ ~Ye J/;fYl ~11 the
.f,p//!')/P1i1!J p~~e.
********************************************************** ** Variation of the resultant Fb as a function of theta ** ********************************************************** Theta (deg)
Fb (lbs)
(w=100 lbs) 2860.1 2757.4: 2659.4: 2567.0 24:80.9 24:01. 6 2329.4: 2264:.8 2208.0 2159.6 2120.0 2090.0 2071. 3 2066.4: 2081.1 2128.8 224:9.8 264:6.3
90.0 85.0 80.0 75.0 70.0 65.0 60.0 55.0 50.0 4:5.0 4:0.0 35.0 30.0 2:,.0 20.0 15.0 10.0 5.0
*
10 4.5
Fb (Ibs) (w=O lbs) 2860.1 274:8.1 :264:1.5 :254:0.9 24:4:6.7 2359.2 2278.8 2205.4: 2139.0 2079.6 2027.1 1981.2 194:1.9 1909.0 1882.2 1861.6 184:7.0 1838.2
3
Legend
G
o
h
1:.
w-100 w-O
4.0
\
3.5
\
OJ
\\
.n
.....
3.0
ri
LL
\
\
2.5
\
2.0
o
20
40
60
Theta.
deg
( t.(}1)'t) .2-57
BO
lbs lbs
A-s
r; ~ 0
me
F-,:
e
If
"/I~II)J thAt. U.s
Th US
I
tis
CIIJ?"~
/la/lit eJ{ 17.3 2 (Pot)
tps ¢
Clll1
det~Y1mn e,,( !rtJlI1
E!. (~))
e
1T!1I f) rS/",
f -- V1- sli/'~
"::
pi / -- hS:) ~/"6
B ~0 ::
I f 11-0 J.b
/ + ..£' /2.
IJ/I~ Y'lIStili mellllS '/h,t: ~r' is Indeferft,,'n~ft! I but ft,y tll1~ "//n*, WI I / QfPt()lIc.h / ? J/.O /b.
Pht{'S/{II//'1
~ =. () I the sn",f/
I,
Vlllue.
lJtl/"e () / ~ ~I
eI
~
2.til An open tank has a vertical partition and on one side contains gasoline with a density p = 700 kg/m3 at a depth of 4 m, as shown in Fig. P2.61. A rectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h. will the gate start to open?
•
F=R j
~,. ..t" ~
-::.
A. ~ +0
'} refers
W neV't
:J4S6i1l1
e.
~; = (700 ~.)(q. 8/ ~) (ZNO) (tf,.. ~ 11 0
== t-
~l.cr
LU he V'e.
F; ;: RwWher~
1..0.
N
X 10 J
z.,..)
II 0 k N
==
1+ l4If e .(e rs toW cc....-h}l' .
0.Lv'" )"
-:.
FIGURE P2.61
w-
('t.(?O)c'ID3N)(.h)(2~~h) /)tt :i
n
f~
2.
dtpTh ~t wo...~V'.
~kr" : (~. g0 ~ Ir/ )~ 4 h:>.... e~u',ll bfl ~m )
2 M \.t So
:'0
-/11«,t
~ "" 1.",,::'
Tn 11.5) ~"'cA
(
1=/26 if!
q. 8DX I03 ) (h '1 ) [ ~ ) h:: 3. ~5 fYYl
wh,c.~ I~ th<.
{ l w d" .Q... = '3 a" 01 1, = 3 N't
= (I I
0
~ I ()
\,~~h~~ vAlwf -h,v
3
h.
N) ( -\ N'\)
2.'-2
A gate having the shape shown in Fig. P2.62 is located in the vertical side of an open tank containing water. The gate is mounted on a horizontal shaft. (a) When the water level is at the top of the gate, determine the magnitude of the fluid force on the rectangular portion of the gate above the shaft and the magnitude of the fluid force on the semicircular portion of the gate below the shaft. (b) For this same fluid depth determine the moment of the force acting on the semicircular portion of the gate with respect to an axis which coincides with the shaft.
1
Water
6m
~l
~ Side view of gate
(aJ
r:; r
rec.-t.al1'1l1/a,. f()rfu,'II)
(f)::r J- he A S"
~he f"
At
~ ""'
=
"fna.t:
(Fi<)y::(ri~b';.3)(3"""')(~~;('"",,) - /~6,t)~N I
he
~I'M
::
+
'fie 31i
~f'M -t If (~m,)
:=
= 1. Z1 M1
31T
/ 6J0
J? N
(h)
7.3~1"n1
ThUfJ,
tn ~mt /I i
~; fh reSfec.1
11 =(~)5, )( :: (;O/D
-4
s h~.f.f:.J M)
(7. 3' I'M
-
".
ao ~ )
x JD3N ) (i. 3'1'M)
= I. 37 X / ~ I:. Iv· /WI
2.63 A 6 ft X 6 ft square gate is free to pivot about the frictionless hinge shown in Fig. P2.63. In general, a force, P, is needed to keep the gate from rotating. Detennine the depth, h, for the situation when P = O.
•
eO ,;,I,'en um 21'1).J. =0
1=;,.,. ThllJJ
~r
~
p= ()
/NO"
fd
to,Pass tnrPJl9h 1ne /J 1n9.e) ,', e,) fJl?:::?' s.f.t + h hlft/(
t:J(e
Qnl{
.
y", A!:f~ -=
WIt;,
3.sR+h 6.5 it:.
_ -A
h T 3 f.I::
('~tJ ((,. ft))
(I, t 3ft) (tft,t. b-H)
-
h = 3. ()t> -tt
2.-(,.'
-+-
FIGURE P2.63
2.64 A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point 0, as shown in Fig. P2.64. The horizontal portion of the gate covers a I-ftdiameter drain pipe which contains air at atmospheric pressure. Determine the minimum water depth, h, at which the gate will pivot to allow water to flow into the pipe.
Width = 4 ft Right-angle gate
Hinge
pipe
.. FIGURE P2.64
POt' egu;/, 'hrJum
:2 1'10 I=::
~J
>< .
I=R,
=0
J. I ::
I=R
z.
1"
)(12.
(I )
).,
t:='J(L
~
OJ!'
17< J ='r heI A-,
~)ll.~
=(~z.lf ~,)(f) (If-Pc ~ h) == 12.5'
1,,~
h'l-
I;y -the t:,rt~ ~n -tnt! h~rJ~pl1ta J ~YI::J~YI 0 ~ ~ sa, l::e (whu. lt JJ hajq~ced b~ P~tS5"H/·t. ~~ bDTh 5Jaej e"cepi lo~
tnf. t.lre~ f)f 'lrre.
p/pe)
t=;~: tf htr;.)(lh)2::: aZ.Lf ::: Lf'l.O h Thus, Pr8m ~~. 0)
:3 )(h)(![)(!1i)).
~ U1.5h;2)(f) = r;.1.t>J,)(3+t) /AJ/'f;.
h::: /. rF fi
)..,
g
dnPl
1.=3t1: z
2.~s
2.65 The specific weight, 'Y, of the static liquid layer shown in Fig. P2.65 increases linearly with depth. At the free surface 3 'Y = 70 lb/ft3 , and at the bottom of the layer 'Y = 95 lb/ft • Make use of Eq. 2.4 to determine the pressure at the bottom of the layer.
=951b1ft3
z=o •
dp _ di:
-
FIGURE P2.65
( ~~. 2..lf)
-~
0= 'f5- 12.£' 2-/nat () c=-z.
fl'
=
1b. 1trJ17I
1t,lhm
=-
--
thot"fl,m
=-
irs
-12.H) d:t
6
2. [r,~- It'~-r9. z ()
['1Sb') J.b /~5" .ft,..
t... '1.5' { 2
)'j
2.0
2.'~*
An open rectangular settling tank contains a liquid suspension that at a given time has a specific weight that.varies approximately with depth according to the following data:
II (m)
o 0.4 1.2 1.6
., ? ".3.6
The depth II = 0 corresponds to the free surface. Determine. by means of numerical integration, the magnitude and location of the resultant force that the liquid suspension exerts on a vertical wall of the tank that is 6 m wide. The depth of fluid in the tank is 3.6 m.
10.0 10.1 10.2 10.6 11.3
0.8
12.3 12.7 12.9 13.0 13.1
1.4 2.8
'17
The
mAA",IslI/t. .,
-lIt/Itt .~~ce / ,fi..... ,
01 71te
"ff;III1' ", SlimIY/ I11 1 1i1~ d,I=HY'fIlf.,it/ ~"'ces a,chhlj fJlI Pte. h~I'I'o"'';'/ ~l-rlp ShtPWII
(!/In k
,,,; 1I1e. frrt.ll'e. /I
Whel'"e
To
?
~~
-
-
,-:0
1-:,..... 1'''-' 1'-""'"-:,/-:,/,,I-:-j""'-'-:-_L-:-'L-:-h-,.!'L f (). -t
Thus) J..I
Fk=jd{-bj1>Jh ()
r-_ -
OJ
T lot
L
0
the press lire at (j.eft;, ~ . wc uSe. Gr 2.'1
'oJ
iii,''' P
cjj; =-K d't
((HI
IN,:;'"
d'i= -dh
?(A.)~ E1 IJ 4.t,0I1 PrOfj'yl" m 100 110 120 130 1lJ,0
150 160 170 180 190 200 210 220 230 2lJ,0
250 260 270
(2)
ti 111{
{
1 ~ d-P.
(Z )
CIJ It be If} JeJ rll ff ~ 1/ " m eY'I ~ q / f;, "': /t" 111 (! kIf()uJI it" 11.51;'9 11ze. Vlly,'u·holl I; (f MJITh h :;llIel1.
cls print "*************************************************" print "** This program integrates Eq. 2.4 numerically **" print "** using the t:r:apezoidal rule to obtain the **" print "** pressure at different depths **" print "************:t:************************************" print dim p(10),gamma(10) n=10 dh=0.4 p(l)=O for i=l to n read gamma(i) next 1 data 10.0,10.1,10.2,10.6,11.3,12.3,12.7,12.9,13.0,13.1 for i=2 to n s=(gamma(1)+gamma(i))/2 iml=i-1 ( Ct>l'Jt )
280 290 300 310 320 330 3~0
350 360 370 380 390
for j=2 to im1 s=s+ganuna(j) next j p(i)=dh*s next i ' 'Print the results print. print" h (m) Pressure (kPa)" for i=1 to n print using "###.# ###.##";(i-1)*dh,p(i) next i
K************************************************ ** This program integrates Eq. 2.~ numerically ** ** using the trapezoidal rule to obtain the ** ** pressure at different depths ** ************************************************* h (m) 0.0
Pressure (kPal 0.00
O.l±
l±.02
0.8 1.2
12.2~
1.6
16.62
2.0 2.4 2.8
21.34-
OJ .:....
36.6~
3.6
41.86
'J
...J •
B.08
26.3~
31. 46
~gfl4. t/(;11
(I J CilI1 ,,()/AJ
b4 Ihk!hlkJ lJ("(mer;c,,li~ US/'n~ TIGItPEfO/
***************************************************** This program performs numerical integrat.ion ** ** over a set of points using the Trapezoidal Rule ** *****************************************************
**
Enter number of data points: 10 Enter data points (X , y) ? 0.0,0.00
?
O.4,~.02
? 0.8,8.08 ? 1.2,12.2~ ? 1.6,16.62 ?
2.0,21.34
? 2.4,26.34
7 2.8,31.46 ? 3.2.36.64 ? 3.6,41.86
The approximate value of the integral is: +7.1068E+Ol ( Ci>!1 'i
) 2-(ps
(Cd)tJl )
w;-n
II
j,?d~ =7/.07~ ()
A'H
~ = (~hrI ) ( 7). 07
To
t.j
z {, --k Iv'
.
sum h1~l'I'1fl1.f:s (I/'(Jlli a.x.I'.s !;;rmed b!l l"trj'S~C.t:'()H of
lie.
/"c4k
#).:
IIfl'h{~/ Wf/I II~~ I(/;"H
~ f,e
Sf,( Y'
rn" s,
/Au;,.
It I>
= b
d -f.
( .3)
o
il?.fejY4f)I ~f CAli be. dekY""~t4 41111 Eo . (3) /~l-ef,.,,/~4 "l/mer"clI//~ us/nj T,eA-PErIJ/. TIIJ,"/"#I r~.!tI/1s ql'e ,iflflt k/"w.
The
~****************************************************
** This program performs numerical integration ** ** over a set of points using the Trapezoidal Rule ** ***************************************************** Ent.er number of data points: 10 Enter data points (X , Y) Note: ? 0.0 to. 000 ') O.y,.1.608
YtV ~(}f
:1 O.8,6,L16lJ. '? 1.2,1"',688 ? 1.6.26.592 ? 2.0,l:t-2.680 ? 2.4,63.216 ? 2.8,88.088 7 3.2.117.24:8
? 2.6.150.696 The approximate value of the integralls: +1.7y,37E+02
TIlliS)
IN/tA
J
H
-I-fd-t. "
/ 7'1-. If -J, /(
0
if /DI/()WJ
,£ = /?
1r~1J1 Ii,!.
h
(3)
1'h1. t
l~I"h -
{ fD 1m )(/7'1: If ,AI)
?,,,, .A)/
fi
The Y'e.suJitlni ~f'te.
(j
c.ls 2. '1-1,
MI
be low fluId
:::
2. 'fie 1m
sur-fa c.e ,
Water
2.~"1
The inclined face AD of the tank of Fig. P2 .•1 is a plane surface containing a gate ABC, which is hinged along line Be. The shape of the gate is sho\\l1 in the plan view. If the tank contains water, determine the magnitude of the force that the water exerts on the gate.
y'
y'
=
4(x,)2
Plan of
gate x'
FIGURE P2. (p 7 /' ./
{.() he re.
A=
=
i'
[).;<.'
d!j ,
(".
j;{{) I/? d;/
=
-H1""')
ff({f}T =~'A,' o
/'1
'deI ,4 = f~' t:f d A:: o
I, d :J
.t. !:J,I(.
""' I
=
c)
I I
'1, .::
lind
J'n ce.
{c. =
'fG
$,1'1
30
0 )
~ = ~2.1f ~~ )(3"ti:)(J/~30tJ)( ~" .prl.) _ Sfet
/b
~2m-j 2.68 Dams can vary from very large structures with curved faces holding back water to great depths, as shown in Video V2.3, to relatively small structures with plane faces as shown in Fig. P2.68. Assume that the concrete dam shown in Fig. P2.68 weighs 23.6 kN/m3 and rests on a solid foundation. Determine the minimum coefficient of friction between the dam and the foundation required to keep the dam from sliding at the water depth shown. You do not need to consider possible uplift along the base. Base your analysis on a unit length of the dam.
1
. 5m 4m
If-~---6 m----"! FIGURE P2.1o
~
K /,(! A
:=
.4 - (
where
If ""'" ) SI.3°
(I
)
:S/].t
/=Or eZ 1I;//6r/ufYI ) F;<. =0
z:
or
fie
/f-/jt!); .:50
2..
SII1
s/.3
II
= F;
=
)/11
FfJ :. 0
1h()./;
N
=C>v<../
-t-
't
!=Je )(
C6S
51.3
V()/r.lm~
0
where.
DI- UDrlCrei:e)
Thus) N
= (~:3. ~ ~~ )(2 0
tm 3)
=
'1
(I ()() Ie. N)
C~s
S" I. 3 0
(/~O-RN)
Sin
Sf. 3
5'""'3 'f ..It.. N
= s 3 If -k N 0
-
I
2.b'f ""
Water backs up behind a concrete dam as shown in Fig. P2.~~. Leakage under the foundation gives a pressure distribution under the dam as indicated. If the water depth, h, is too great, the dam will topple over about its toe (point A). For the dimensions given, determine the maximum water depth for the following widths of the dam: r = 20, 30:- 40, 50, and 60 ft. Base your analysis on a unit length of the dam. The specific weight of the concrete is 150 Ib/ft3.
1
80 ft
PB = 'Yh
l· FIGURE P2.lDQ
/-\ hfe- b~d~- dla~r({m ot the dct~ 15 ~howl"l in -the +t~CAV'e. a.t 1J1e I'i~lti)w;,e",.e.: ~I = (lev l.(nl'+ lel?~11t)
'6;"-
F.3 =
(CS~'" '({~T :z.
F.'2. = 'c5 ~,::
)
Q
(~T) (~r ') = 7 Sihe
-t
,.
~-a "2-
':13
G=
5i"e
Z
I
'12,=-
3
+elt')
3
~~/{Jer
(~,.
)
SI'w8
the.
pfeSSwve
·
dlS.fV',J,,,,,~,o~
sW/
On
The. 10 ~~ :
1
~ = (~-ti)J Summ In~ mt)meY\~ a~ot.(.i AJ
~~ ~ 3
= FI (~) + (
f!c (~~) It. '
~----------------~--------------~~~ '--------------------------------~ '2..- f.:. ?
( CD" 't 5
0 f;,4 t
) .J. ) Fr (.. ) .,. ~ (tft Fj 5,,6>1-, '.fl( j.1f)~ of exl/I'fSSIDlf5
1f3 =
j. (
*
-r
7~ )
-i-r-1t
Fe> I( egui iij,.,./"W1 of fhe dqm I L: fi1A :: 0 I So
'J
':1, - '\;v' (;ft)
-
(=2.
1h~t
'12. + ~ t:13 =0
OJ
/ft 3 ) crt. = 150Iklft-', (/1111 ;'T':./tJ oft) then: F, = 3/.2 i ~= {gf!)ooJ. F-:. = 312. ':I 10/3 :z. Slit G :z si", 9 F..:: 3/. 2. (.f. ~ ,0) .i. '13 ~ P. (if.,. 1J():r (2.~ -t-I(Jj J).. 3
and
w'Jth
'If::: fe,2.lfl b 2.
0
::I
-R.1"-R..r
Su h:st, /£l.tJDJ( (3/. do. 1
01
)t~ ) r
-
1h~se
ex..preSS/O/LS
l:Z. (J)
I;,k
~ I eJds )
((,,""O~) (; Y-) - (3~~:){-~~~~) ['3/.2
aT/O);] [~~t :::~Q
.P. I
C411 be
Thus;, .for
tL
c,4>rld,f/{;H
-hl11~=?o/,f1 Qnd
A
3(...(t-t I O)
t:j"UH
CiPmtufty pl"0J)"qm
for
B
Ef,(2)
detfnn/ll l1iy
(
C~t1t :2-70
)
J =0
de,t.ermuucJ
/YOHf
the
~()JlluJ
kV h. h ~v t:t 7 , l/el1
).
~J/()WS
J. 6Cf /1&
I
100 110 120 130 110,0 150 160 170 180 190 200 210 220 230 210,0 250
c12 print "*******************************************************" print "1I:* This program solves a cubic equation to determine **" print "** the maximum water, h, depth for a series of dam **" print. "** widths, 1 **" print "*******************************************************" print. print" Dam width, 1 (ft) Maximum depth, h (ft)" for 1=20 to 60 step 10 theta=atn(80/1) h=O hp=h h=(3/31.2*(3896*1-2+10y,OO/(sin(theta)-2)-20.8*1-2*hp) )-(1/3) if abs(l-hp/h)O.OOl goto 210 print using" ##.# ##.#";l,h next. 1
~r
the dtlm w/dths s!..f'c.I'f,Pd The
ClY'e
glYfn
/AI/tf1"hs
fI14X Imum
WAifY
depth.s
i'}Jat .fr:,j, -the +1<10 /aY'~e~t dtll?1 -the tvaier wt)"'/,f ~vPJrf/~tV fh~ d~m h~fa't' IC lV~l"q'
be/"u). ,t10t(
+Otf/~ .
¥****************************************************** )\,.. This program solves a cubic equation t.o determine ct.1I :1<* the maximum water, h, depth for a series of dam ** ** 'i"idths. 1 ** ******************************************************* D~n
width, 1 (ft) 20.0 30.0 40.0 50.0 60.0
Maximum depth. h (ft) 48.2 61.1 71. 8 81. 1
89.2
2-71
:;.70
2.70 A 4-m-long curved gate is located in the side of a reservoir containing water as shown in Fig. P2.70. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. Will this force pass through point A? Explain.
II FIGURE P2.70
J~ II' E'~ u; I, 'bY'/un'I
rJ(. ::. r:;.:: ~ L
I~
I
--~
0
~ l(!,c '- A:& = '( {'11ft +J.!1rC )(1,-" .: ~ .. )
f:u = (9. HD ~ )(7shfI. ) (12 /WI ,.)
= ~8' 2. -Ie tJ
Slmll~.,.It-:J)
L
F~
=-0
whe/l"f
r, = [3' (6~)](311M
'
xLfht1) = (Cf.FO/lt/J )(6~)(/2tm")
~ = ~ 11- = (9. eo ~)(q i1 M4 3 ) Thus}
(/V/).;.~: The.
f:,," (r.80
~)[7l
/IH'3
",,3 .,.'l7f ",,3] " 1113
!;Ne ~I W4/;fr tOn ,j4~ IN til be. 111111- ~h~wn tDn h9ul"e.
d';'ecbtJ"n
c/ 4//
tt>ppes/!c
C/;lleYf"'';'/~/ Ior~es
I~ dll'~cf-~i,., etch;'f
CtlJI'vetl SUY'~c.e l.s ~r,Pl'''''41&U'4/1' ~ St.lr/ttce./ ctl'1d in! yes III 14111 /1'1 ",.sl- ?IH5 i11rlP"''1h the ';'+~Y.rt'~I-I"';, ~yce$ which I~ ~i; tt!)Jnf A-.
Ye s .
2..-72
).IJ
+0)
~'1 th~
"t11f/l'(ft,"'f!!)
01
all 7/lelf!.
2·71 Z.il The air pressure in the top of the two liter pop bottle shown in Vidt'o V2,4 and Fig. P2.7 I is 40 psi, and the pop depth is lOin. The bottom of the bottle has an irregular shape with a diameter of 4.3 in. (a) If the bottle cap has a diameter of I in. what is magnitude of the axial force required to hold the cap in place? (b) Determine the force needed to secure the bottom 2 inches of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c) By how much does the weight of the pop increase the pressure 2 inches above the bottom? Assume the pop has the same specific weight as that of water.
1 in. diameter Pair
4.3 in. diameter
• FIG U REP 2 .1
I
3I. Lf /j,
(a.J
L F.VfY+1 c..J. F.'51 d
~
::.
=0
FI = (pr'e~surt @ X.
2,;".
58\
h,HoWl)
(A yeo... )
- (Lfo ~: )(.~) ('1.3
=
QbolllZ
~------------------
iny
11,
(c)
lnCYea.5e.
U1
pY'e~suY'.e
\ e$S 1\1cH1
\
010
J-73
crue .J.o 0
=40 psi
f
tU
r
w~'\9ht :: pr'eSSu ye).
).- 72 1.n
Hoover Dam (see Vidl'O 1.3) is the highest archgravity type of dam in the United States. A cross section of the dam is shown in Fig. P2.72(a). The walls of the canyon in which the dam is located are sloped, and just upstream of the dam the vertical plane shown in Figure P2.72(b) approximately represents the cross section of the water acting on the dam. Use this vertical cross section to estimate the resultant horizontal force of the water on the dam, and show here this force acts.
1 . . - - - - 880
." .
:~,.;t£.'i··~~iAiW':h'i?t.;.·
ft - - - 1
\- 290 It ..\
y
~6601t-......,
(b)
(a)
Ii FIGURE P2.72
!3re~t tlvea 17:; y
S pa~.fs
/'rJfD
(I,S
~ho{J)n.
aYea.. / .'
liz! = /)hA = ~2.1(. 'b y~ )(715".f-I:'ji) {t.qS- Ii)[7/fift) q
::: l.:r7 .x/O /.6 }:?,y
arell
Fo -:: FI?
3 :
".3
For-
I
= 157 X If) if
J.b
Clrea. Z:
/=,e~~ (0),,- A~ = u'1.iI !t3){l;.) (7Isft) (290 1i)(7Jsfi) - if. l3 X / D'f / b ~
J
-t
~z. -t Fo1'.3 : /'57;( JD ~ I.J
-I-
'I: ~3 / J~ f /;;
-t
/.!;-7x JD 9 Job
7, 77 ;( JD' /j, Since -the.. /7J()mefl / 01 111(. rt's/J/1zurt ~yce a~lIf 'the. b4.se ~f the. dam t7l us t be ega-a J -fz, iht! rn(!)m~Y1 f..s due. tt:> F;.,) 17
.2, 72
I
(~tJl'}
'i )
Pi xci" ~
a:::
(1 )615 ft;)
T
F?/-D f7;5"H) t
r; )(7/5"H)
(/..f7 ;( Il/ h)(~
)[715"It) +(;f. LJilb qli'fJ. )(?J~f-f.) .ff'/. 57t /D liE)(71OIt) q
7.77 X J'/J'
-
~
LJ Ob
i);
·f-/;
(hils) 1h~ r~.>ul/rll1.f:- hfJ;'''I()t1fal -hvc~ ~11 -tJ,e dam
7 77 )( I () 9)1:,
a cJ-/nj
-'fot
t>f 1he dqm a/Pili 17te tl~is
It 0f
/J)
P Irm? fh( bl/5t! ~'f/11fl1e 11''1 t>/ the "rea. (,I.
2.73
50 kPa ,
2.73
-"
.
A plug in the bottom of a pressurized tank is conical in shape as shown in Fig. P2. 73. The air pressure is 50 kPa and the liquid in the tank has a specific weight of 27 kN I m 3 • Determine the magnitude, direction, and line of action of the force exerted on the curved surface of the cone within the tank due to the 50-kPa pressure and the liquid.
Air
I
Liquid
3m
kL..--~
FIGURE P2.73
e$ u. ;; I '/:n" /u m
Fo r
L 50
)
f:ver bI c.a / =- 0 I
in"t
F;. = -?a, irA Wher{
F"
10$
'\.v' The. force. The
CdI"( e~frl:s
ot. the .flu/ci, A ls~) ~ . A = (50!1i )(-r;)(d:1) flO'
= (50 CHI
ilN r-t'l---,
1-
d.
.fon 30·:=
2.
-.le.P4. ) (~ ) (I. J~£rrt1) =' S ~ ,LI·-k.N
~ I
d
PI
= :L -I:~" 30 • .:
/.
/55' I\IM
Thus) 5'1.4- ..k..N + and
-In e.
and
I~
it' "'c. d,,.~c.I-~,{
I£) ..,
111 e.
/Z8--kN
75'. 'f,k, N c...on e.
vey.flclI/l!:1
h 1(5
a.
d~W}1W4Y'~1
2-7'-
rn
t:t, n
Q/~I'ff
I
.f.1A e/-e.
the
~.f.. /2. 8' --k IV C-!Jne. ~)(.;~.
2.7J.f. 2.7 If A 12-in.-diameter pipe contains a gas under a pressure of 140 psi. If the pipe wall thickness is i-in., what is the average circumferential stress developed in the pipe wall?
i=;" e 6u/ }/}/"), 1"1 'um ()f
(ieI'
4
un;! /"'fllt
pip c.) )
the.
Or
rr
=
·At/~
1?. ( I 'fa
-
=
~
-
Jh
I'I? .~
) (G. I~.
I '"'Ii/ /11, . 33~o
pst.
)
)
.
2.-77
J.7E The concrete (specific weight = 150 Ib/ft3) seawall of Fig. P2.75 has a curved surface and restrains seawater at a depth of 24 ft. The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A).
2.75
\---15
·1
ft
FIGURE P2.75
The (!)I')
(HI
01 tne
Com pDnen +s
The (pllli fire If t:ll1d W -!-he fljuY't. where .
F;= ?f~(.A
(Ie ·/-JiI,
fllJ/d .ft,rce (J
S
S
h()WI1
('If.OIt3)(;J.~t)(2'tliJ{l+t)
==
= J~Lf()oJ;
Clnd
T
.4-15 0 )
W -=
~-if-
:l'f-Ij
1
To d~.feY'rHJne -¥ (s e~ .It 9 lire
/J. =-
-hne/ IIrea B CD. Thus) .fo ,,;,l1f) ~
!;~ '/-!1 ) dJi. = ((,2; - ".;;;x"') dx o Jo '
:: [~ Jt x t(11
d
WI
-p.L Thus)
th
XI::
= .4;<
0,; x'~lQ
rl20
I f-l:
I.
A = 175'" H:2.
'7~-
h
) (
1 '15
/j,
l
f:: o
;t.(.
=
dA
=
It 3)
= I~
1}ud:
_
J7S
Zt!)O
lob J&"
i1~'f-'J) x dx
/~ (~) t
So
3
To / ~cllie Cft1ff'()/~/ 0/ A :
'-<. A =-
=~
( /'I()';'~: All l-tn(ths I~
)to
)
--
Jt./)
().
.jj.1
2 (Vj"Z;,) ¥t
¥x - a, 2 / ) d><
4.11
tt
=
II: )
2.7 b
A cylindrical tank with its axis horizontal has a diameter of 2.0 m and a length of 4.0 m. The ends of the tank are vertical planes. A vertical, O.l-m-diameter pipe is connected to the top of the tank. The tank and the pipe are filled with ethyl alcohol to a level of 1.5 m above the top of the tank. Determine the resultant force of the alcohol on one end of the tank and show where it acts.
1
r~r------'
'------.
2..0tfll
whe~e ie = /. SO
!;J-1tY) 1-
~
I. /)/ffIt = 2. 5'"."
-tnA t
fR : - (7,7 if~
)
(2,5"
~ )(;:)(-2. OAM) '- = ~~. j ~ JI
AIs())
( ;}. S-NY!
)(r;) (Z~ )
Thus) the restl//q"t "wee
a11 v{ be jpw
~C.f.s
a ellS fttl1ce
of
~en feY'
0
I
fRill<
?
hilS
"f
e h c/
q
rn4'Jl1lfude
-
~ ':f~ L/.)O/ ; .
2.-7'1
"I
= :<. /, O/M
6~. 8-k)/
- :l. S"D hI1
=
O.
/O[)/'IH
~.
77 2.17 If the tank ends in Problem 2.7b are hemispherical. what is the magnitude of the resultant horizontal force of the alcohol on one of
the curved ends?
~or e 3";)/j,,.u;'m
J
;;;, = J:;." / /
(U.,
;;,w,,)
~
1 -I 1
I
s(>/u 11,,0 (." hOn)Mfe, I !ere", SC/me 4S .f,:,,.. PNblem 2.1'0.
SInce.
:< ./8 2.7 &
Imagine the tank of Problem 2.7(. split
by a horizontal plane. Determine the magnitude of the resultant force of the alcohol on the bottom half of the tank.
(onsJde, t>. fne- b(K/!rdl;'f~ Df Dojl.,m 11.1/ 01- "",,,); [Sp, H~~Y'<")
where:
1> ~
fr~'suY< D/ ~/u;tI CJH h";~M/.1 1./ ~ l1 ...e. .1 I1.N,MI" I pl'H"-
OW ~ wel9i1f of. volu",. ./ l/"i4
IH
h./f' of !
I=f ~
f'£'Sulk,,1
~".ce
Ii". .eSu;;'bNvm
fi:
e"",y,t.tI by *,,,1:
(rf'Hr
0"
1'1.,4
-10 "nohlt", 2.80 /;,
#"i< d"ne_SicnS J,
jA 'l'W = d' (I.~""./"")(:l"" d",,) + ?r [U;)(2~)(~ ...
(-7.71t ~ )( ~O",,3T ~7T,,")) Tht-lS, fo~(e t>{ alcohol verhcqll'j do UJn 1()4 rd.
DM
= 203J,.IV
+P.1k - 203 -k IV
dlYe~+ed
)J
l.llf A closed tank is filled with . wat~r and has a 4-ftdiameter hemispherical dome as shown m Fig. P2.79. A U- I~ be manometer is connected 10 the tank . Determine the vemcal force of the water on the dome if the differential manometer reading is 7 ft and the air pressure al the upper end of the manometer is 12.6 psi.
~/"
epidi/; r,uJ'YI,
Fver-nCa 1 I
2-
~o
t-pA
tit .. t
;So
Fp: -plr - 1v Fe,
l' Fr"lf)m
-
-the de>me ex.ri-.
I~
pl"e H"". .. af
I':'
-n..
1h. 1-0/4 h,.u of. 171( dt>mf.
Oil
1'he m {( nPIli e te Y')
~ So
.
(I )
-I
~f (7.ft.) - d"J./." (If ft) = P
-thAi;
f
~.~ )(J'l'f ~: ) + (~.)('2¥~.)(71t)-(iZ. ¥~.)Htt)
= (;2./.
= Z ~ 8~
Ib
k'"
fr4>m F'f, .r;) w) n,
1/0IufYI •• 1 Sphfr. =
Fe, = f2 Ho ~~ V;;Ylfft l 35: )
/00
:r (d/~m (~/Y' ).3
- t [:- (iHf)J (~2. 4 ~.)
Jb
The -Crt<. 1hot- 1he v.rl-I<..I ~r(. -tn.+.::~",)(",e!.rl~~_~"!!.h_7h=,,=-....!!d.~om~e,--I~ 3S; /00 Ib t 2-V 1
2..80 2.S0 If the bottom of a pop bottle similar to that shown in Fig. P2.71 and in Vid(>() V2,4 were changed so that it was hemispherical, as in Fig. P2.80, what would be the magnitude, line of action, and direction of the resultant force acting on the hemispherical bottom? The air pressure in the top of the bottle is 40 psi. and the pop has approximately the same specific gravity as that of water. Assume that the volume of pop remains at 2 liters. 4.3-in. diameter
11 FIGURE P2.80
t;Y'(<(
~ W
We I ;h t V()/UhI('
0
o/- pcp
i;o
dU(
~ pdp
= z.
b~
SlAfp"r.J:rd
bo}h,r>1 t
k'f"Ct!
Qlj,- ft"tssur~
= ?r
I, f.ey.s
)( v0 I~ mof c I pof Pdp 1\ .f.t" ) == (2..x. J0- ?'m'I 3) x (.3 S 3 ( x I {) ;;;) :
(II
O.
07 oj, ..f.t3
Thus1 ~11'l E"1. (I) W.el!hi- #/-
)'IP/, :
FDYc~ c1/1e to
til".
(6z.,. R,)(()·~7~b Pt 9) = Lf. '+J /b pr~JS/lre
:::-
.J.... -/;11" x pr()c)'~c..1Pt;I, aveA •a.{... YH'It1"pheYt(a/
= (LtD
~.~ ) (r){~.31·n.)
::: 58 J
Re~~1 t.At1t force.
=
J
Lf.'l-I J;, + 5~J
L l.J. OOJl1!)1')'1
2-
b
Jj, -
b-2b
II,
Tn e rt'su) tQ If t: ~Y(' t. /5 dI tee. ted () t vtica J1lf down lOtI rd) CI n d due .J.o slfmme tY'1 1 /t aC':f3 ()Il 1Jt(. hem/5pheriC41 bt!'
mm
a/~n~ -/)Ie. verft'c~d t2J(.jj
()fhle bot+fe.
2.31 2.81 Three gates of negligible weight are used to hold back water in a channel of width h as shown in Fig. P2.81 . The force of the gate against the block for gate (b) is R. Detennine (in tenns of R) the force against the blocks for the other two gates.
Fi>r C4$e (.J:, )
411'
(c)
(b)
(a)
• FIG U REP 2 . 81
~ = (rhe A:: d'l4 'J/(:: -f: J..
~b
)(hxj,j =
Thus;
:FMJ.I=o
h R = (~ l ) Ii
"/h,d
:>0
i R:: ( ~ It. ) ( }::J, ) _ "Yhlb R..-3 For
case
(el.)
On
0)
free- bt)c? - c/utfr41?J
F;. =tY~.b ( f-rrJh1 1fI~"Vt')
IJI?::-
11"
t:?
s),owl1
4114
~.::
::
th"t
.....
6)( -Val J' [1T~)~~)J
Fa
7TJh 2 j,
Z /vIH
=0
'&J (~ - ~;)
1-
T F/<.,
lie, ~D
--
.......
--t 1 ':
Ihu$;
ft.!1
!k~~) - ~h
2.-3.3
2.% /
I
((.~nr.)
.tf I> II~IU.J
1k, i
Fa :: :r),:Lb (t. 3f~)
FrtJl1'I ~f' OJ
J-h2.£ =-5R./ th".s
F/3 = 1.17 R. 4~
j;r C!ttse
(C).J
-the
r:...e,
/f:,Y(f!
-kl" -the !ree- bal~ - dt".J'rl//J1 .sh~Jl4l1t) 1h~
tJl'7
hI Yl1e ~n If(
fh e
f;, <. m "me" i
~ Y()III1I1
!f(; 1/ 28
t!fc
-h /I. ()n b~ fltul1 />4 rl ()/ JIlI:e
-
(~)(-f xb)
-t
h
~(it~j~~h Fr6Jm
f8~(4r426)~:) =fir6'///, j;:'~. rJ)
J'h"}, h ;;; 5 e I -fh u.>
Fa = -;,e
=
!),
irS/<.
-,
Thus ) 2}1U=0
iJr
I7eJ
~
-ah Lf
1
"
~l1fr":J/.(k
-d:(h )f-zh).3
)./.'"
""
t (~Jfh)(f x b) :: -ph2.b
:::
he
,
I--~
eu",.ve~ sec,.,T::It;11 p~s.Jes 1h;"u,Jr
1ht'I"e-hYe d()es /1ld.
lif..: ~ he It
3~
I
Fa
~z
!fR.1.
1
2,12
I 2. g2. A 3 ft X 3 ft X 3 ft wooden cube (specific weight = 37 Ib/ft 3 ) floats in a tank of water. How much of the cube extends above the water surface? If the tank were pressurized so that the air pressure at the water surface was increased to 1.0 psi, how much of the cube would extend above the water surface? Explain how you arrived at your answer.
h
e $ U~ /, 'br /U n1 )
y
z: Frrr+IC'1fJ : So
-n;
--_ --+-1"W~ -~ i~
(J
-thAt ~
- FI3
3.?l:-h
--=k
ThUS)
(57
i!'t
(3 .ft)3: (2,1. If 1t3 )(3It-h) (3ft i3{t)
3 )
{= j, 22 -!-/;
1:1 -the
(IIi-
tnt r(
/:;'1 / P~/ ~f the frt'$j",r~ tin
WI III R
bt
11() C hllnfe 1/1-
IIIt.Yf4!fs
/h( he,!Ai
ttlb~ ((1111/" tht. IdAter ..5'tlrhct. The Il1trellst'tI /z:,r("~ ()/1 tAt -t-~f 0/ 'i7te e"b~ ,; bp/4lfCeA b'l
"', tiP I
Jurhc~
pressure til- the W~';"r sur/lice
-force
fft5SIIYt
0}1
the btFmm
b/
1ht ttt6e
051 flee
'$ rfih,j/Jf;jff;d -11lr~"!h~lJ,i 1lJe .f/,,;'ri .
2 -8"S-
Iht'
2.83 The homogeneous timber AB of Fig. P2.83 is 0.15 m by 0.35 m in cross section. Determine the specific weight of the timber and the tension in the rope.
• FIG U REP 2 . 83
W= ~ Jf
~
is tHe. ~fe,jfj,- w(l~ht af The -h'mh.eY' al-lp( -tf' /~ ;fs lIa/"me Thus; WheV'(
(0. J5~ = ()"Z, t
\/J:; }(
)(
~.3'
/It4
~
/01lM)
~or e~u'd /brJum )
IM A- =0 So
/ha.t
Vi Cit'"
(10 3,1WI
)
(t?,5"Z, ~)( S"1'fYI)
c'o s 0(
= Fe
) Cos J..
= ( P·IfU) d'Uz.o )C yo ~
~ = (,?,Jf2t»( UD ~3 (tl. 7"1.;-)
:50
( 8:,
) (
)
~ 1 ""
( 5" h" )
trt,d:
T
=
~
- \1/= 6,lfzorrn J )('1.80
~~) - (o.{j]"tn13)(~.27~,) = 1ZLflV
Z.iJ.f.
I 2.R4 When the Tucurui dam was constructed in northern Brazil, the lake that was created covered a large forest of valuable hardwood trees. It was found that even after 15 years underwater the trees were perfectly preserved and underwater logging was started. During the logging process a tree is selected, trimmed, and anchored with ropes to prevent it from shooting to the surface like a missile when cut. Assume that a typical large tree can be approximated as a truncated cone with a base diameter of 8 ft, a top diameter of 2 ft, and a height of 100ft. Determine the resultant vertical force that the ropes must resist when the completely submerged tree is cut. The specific gravity of the wood is approximately 0.6.
h;Y
es /!J ], fA
rJ
Z. p:ver·hctC
U(YI
J
:::.0 J
so -tn I., -I::
T== Fa-'W
(/)
/7;y a. fruflcabd ~()J1e)
0/"me:'
j.b (IJ Z T Yj Yi. r;
Whe~ :
~ bl(S( Y'tJ..d I
Y'l. ::
h~ Thus)
-t Jl'z.3. )
us
i:cp rll d I u.s heIgh i.
¥.
= (-rr) (IDbh)[Ctt.fl:/+ ('fit J( 1ft) +ve~ 3 _ 2200
0:'L~
7Y~e
J(
'V
Fa;\.l 1-
we,~ It
t.
bt.t,,~an I: ft;rCt!
Uul]
-Prj
4'-+~~ =- (CJ.,){t,z.1f n~~)(l.Z-ooft.~\) = 81-I If,,ol~
E'l. {J)
T ==
2v
/ g~ I)()O / b - 9 A) Lj()a Ii,
2.86 An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle as shown in Video V2.S and Fig. P2.86. The amount of air in the tube has been adjusted so that it just floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon.
-~~Air
Test tube
Water
•
Plastic bottle
FIGURE P2.g("
"
.
: ' • .L--_--..lI..,•
.......
,
.'
2.87 The hydrometer shown in Vidl'O \'2.6 and Fig. P2.S7 has a mass of 0.045 kg and the cross-sectional area of its stem is 290 mm 2: Determine the distance between graduations (on the stem) for specific gravities of 1.00 and 0.90.
•
When
h!Jc/~metfY
ine
fiJl'lI/!he
2: F:v(yii h,r
i.s 1/"t1.tll1'i ;';'s we/rAt; w)
.f"Y'~e.
butP!1lt"i
Q~e e$u(J/ ~/J1C~
::;0 (4 /
5G -=
IJw"t/ wi1h
~I 01"
(SG, )
I
o. q )
=W
(~zo)-t7 =w
t ; t.JJ,. 0 @ If·C)
( J.Jhfre
~Jml J4rJIj ) ft,.,. Ih~~d w/th
Sq1,. -=
I+3.D
/.0)
(SG~ ) (~~O )~ = W ClI'1t1 ~wbfr~c.t/r1~ '1}1, e$uah6?l1s !:J~eJd.s
-if t:J
S/~C~
FIGURE P2.~7
:l.
V..!
-
(SGJ('O'tI-:z.o)
-t1 -it = LJ~
.45
tGJ
_ (tJ."lfsJ1 )(r·E'JTa- J
I()-,"" ~ ('I. 81 ;< 10:J!; ) -2
I. 72 ;(. I 0
I't'Y\
=
¥
N
(ll"ftt.
jubmp~ed vt>}t..Ime..
1
As ¥~hCl ')
Jf.
QI14
'W (SC71.)(Y~~O)
-(-~_/_-[ S~, (290
Let As "" s+em
[
cr-
-0-.
/ 7, Z "" m1
2.SS An L-shaped rigid gate is hinged at one end and is located between partitions in an open tank containing water as shown in Fig. P2.88. A block of concrete (y = 150 Ib/fi') is to be hung from the horizontal portion of the gate. Determine the required volume of the block so that the re-action of the gate on the partition at A is zero when the water depth is 2 ft above the hinge. The gate is 2 ft wide with a negligible weight, anrl t.he hinge is smooth.
•
r-;y
FIGURE P2.8a
esu;/d",U/11)
I. MU
=-0
.56 ~1:
~ J- 1
-t
~ J..2- ~
T J.. 1.
(I)
T
where,'
~I
'1= ~h,-f, :: ('2,~ ~,) (.3t1:)(zA4=~ z.hJ : 7 Lf-9 1.1,
1=1
J:;.. :: ?r hi! Z ,4z =~ 2. ~ ~3 ) ('t4t) (Zft "- If:tJ L:f RI
- fCf 8 1.1, = r,<.c.. ~c.. A
-t
J.. ( )
~
:: c:.
.Q I = ~ ~ I - z.
+'.f
.=
7.+t
17..
(2.fi) -t.3 (.3.ft) (2.~t I. z. ft) j. JJ
T
'j
H. =
fi - z. t-t. :: \. \ \ -tt
J.l. = / H Thlls) fr,nn
e:a~. (J J
(t?f8 J1,) (j ft) = T (I fiJ T= 18.30 11
(7'f'f IJ,) (1./1
fC)
-t-
f;r ~ncre~ hJocJt..) L t=""V(lo't. c. 1.:.0 0 'r ~ c. ~ T l' Fe c.. s" th,td: -Y: -= I B 30 I~ l' 6'"u, Jfc.. "
'-
41:= c..
J-.{)
I gj 0 Jb
2.-10
3.1/
+t.
2. 81 I
2.89 When a hydrometer (see Fig. P2.87 and Video V2.6) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface? The hydrometer weighs 0.042 lb.
Whln f'/,( h~drometfY IS I-J,,4.'l:Jr,!J /1-5 weI1ht.) 'h.J) J,i hAi4J1teei. ht;1 -fhe bu /)~I(" t- k;yc. e) 13 , J=o r
r-
um
e~ /Ai J Ibn
I
L~F
=0
Vfy/-t~al
IhJl$1
~y fA)A,l-ev ~=~
(~t-o ) -v;-
(I )
= 1;)
whevt!. -yt J~iu'd
Is tne.
C~mb,nJn~
cIs. (IJ
JuJ,merged volume.. w(1l1 1l1e new
(s 6) ( 5Uz,o ) 1{ :,ew
(rfJ.2.() ) -r:-
11I1I((Z)
wJin
~
::: (S Q ) ('ritz,() ) 4f
2-
~ = -¥, ~
U)
56
1-1/
J. '/0 2.90 The thin-walled. I-m-diameter tank of Fig. P2.90 is closed at one end and has a mass of 90 kg. The open end of the tank is lowered into the water and held in the position shown by a steel block having a density of 7840 kg/m3. Assume that the air that is trapped in the tank is compressed at a constant temperature. Determine: (a) the reading on the pressure gage at the top of the tank. and (b) the volume of the steel block.
Air
Water 3.0 m
~ FIGURE P2.90
Let 1f-:: At
~
(Jhd
1;:: ~ ( -£. - lJ. to ) T}lIAS J
~ = At- ~ Smce.
fL' =-
::-
~~th»
-+..-+/'.. l.
1.
:l -t-
"
-b'-. l..
lfilf~
./..11 9. go -:;; ~
or
/()/-iPIL..
9.8/J ~ hf
Cf.7 I ~ 1.71 -r
=
A--t:(3 )
1:1-l1l'i ::
-
0.'
/Itt
ItJl
) G 1t-
.3 0.1
::
+tt»rJ ~6'
~ PI)..
I
3 ('0 I .4..p~ ')
V(r.7} )1+ ~ (~{),9)
(/ )
if. (1tA.9<)
=0
0
:L
Th us.l
U)
«II
(p.-I-; _0.,,) { -
;-
n~ 2. + (
{wlter~
f:}.
fat_
-{ 1
't"
::' .2. -12
::'
:2. S-3hYl
e '6l1; / ,'hY" ~,.". c/ La 9 Y'II m ) J
(b) F; Y'
0
I i'/11 it. (s@e Ir(~- b~df:J -
T:: ~ At -
~
5h~ /
6/()c,k
N
·hr(l1~
we /'111 t /
t'I J;/ #(
.k ,. .
y
Thus) -¥,-= s
-
T
0, 208
3 I'm
),,-93
;;, 'II
~J
* An inverted hollow cone is pushed into the water as is shown in Fig. P2.91. Determine the distance, C, that the water rises in the cone as a function of the depth, d, of the lower edge of the cone. Plot the results for 0 :;; d :;; H, when H is equal to 1 m. Assume the temperature of the air within the cone remains constant.
2.91
! T·'2 /!cr\l~ .~
H
w",,·
Le d
Open end
-----+----
~
f-R--I
FIGURE P2.9\
I-emfJe~4.l:uye
~.
¥
whev-e
15 the.
.fln/ll :s--ta ~
1:..::
J
Vi
Cf;tr1pt'es>/o~
= ~
(1/; V,,/"'lne.
,
ClIfU
~
lM
.from E
t. (J )
t (7r ;<211)
=
41-""" .3
wh/c.h
5ImfJ,·he.s
1=d-
tJ, e.
.
tell 11, 1J1
"Ir
fire.
O!Jne,.J
(I) "
an" I n! Ie", +r, ,;;,.fuj/ 111. f (.s('t! f'JI-(~):
= ~ (d -,fl) +
and
fa.~
if =- 'J(H/)~>(H-.l) : :r:Y(H-Jl
11 = 7f R 'H Th (,is,
I
-v;
respecltve/!J . .:t".f ~D flow.}
;{i
0
[6' (d-.L) + tim<] :r P: t (1+-.1)
3
-1-0
1'1-- [(!-i/-I]
(.3 )
WheY'e 1. all" d 4Ve J ~ I"n1 e. .f.e Y' S . I-t ~(!JmflilfY ji:Jt"tJJNlm k,//OIlJS ~r Cti/c.J(IaII~ t:>
f- c/.
I
( C(f)11
t )
~.91
"I
C~1i 't
(
:00 110 120 130 135 1'"'0 150 160 170 180 185 190 200 210 220 230
)
print "**********************************************" print "** This program solves iteratively a fourt.h **" print "** order equation to give the water rise, **" print "** 1, as a function of t.he depth. ,j **" print "**** ** **** *********** ********** ** *** **** **t:*** II print. print" Depth, d (m) Water rlse, 1 (m)" for d=O.O to 1.01 step 0.1 1=0.0 if d=O then goto 220 lp=l 1=1-((d-lp)/10.3+1)-(-1/3)
if abs(l-lp/l)O.OOl goto 190 print using" #.### next d
#.####":d.l
*********************************************~
** This program solves iteratively a fourth ** ** order equation to give the water rise. ** ** 1, as a function of the depth, d ** ********************************************** Depth, d (m) 0.000 0.100 0.200 0.300 O. '.,!-OO
0.500 0.600 0.700 0.800 0.900 1.000
Water rise, 1 (m) 0.0000 0.0031 0.0062 0.0092 0.0122 0.0152 0.0182 0.0211 0.0239 0.0268 0.0296
"10- 2
//
3~----------------------------------------------~
:§:
2
.....
.; In ·M
L L
'"
1
~
o ./ 0.0
0.2
0 . .:1 Depth,
0.6 d
(m)
0.8
1.0
~.92.
I 2.92
An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at55 mi/hr. As the truck slows uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration?
:: d:t ::: cJ~
.//1'111 / we/~c"~ - I~' ~/.f J ve/~,;..J::, -61 ,;, ~
I;' h
,./Il,
I
o -
9. rl ~ +
O,50Z C>
2.93 A 5-gal. cylindrical open container with a bottom area of 120 in. 2 is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft/s2. (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: ] gal = 23] in. 3 )
T xl illill {
..po
I'
I, '11
-a. A = -t (120;~
VDll-tl'Yle.. ,1. )
:
(b ~tA I) ( ~ ~'al;I1~)
{, = C/. ~ 3
iJ, -
1'11.
f ( ~ -t Qi ) -{
- (::1.4-
11
T
sluq.s)( ft~
, £E $4
32 2
1""
3~t )( 1.'3 ft) 52.
12.
= (6)
Fr~m
frte- bcd!! -dla9NlfYJ
~:
--
or
CtJnta I ~e r )
1J, A
LEJ 1
(~i. r ft'l. Jb.) ( · "1)(' 't4-"~:+-I.' ) /2~ Ih,
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An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, what is the, maximum horizontal acceleration (along the long axis of the tank) that can develop before the gasoline would begin to spill?
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2 •.95 If the tank of Problem 2. 'ftt slides down a frictionless plane that is inclined at 30° with the horizontal, determine the angle the free surface· makes with the horizontal.
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2. Cf~ A closed cylindrical tank that is 8 ft in diameter and 24 ft long is completely filled with gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal surface. Determine the pressure difference between the ends (along the long axis of the tank) when the truck undergoes an acceleration of 5 ft/ S2.
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The open U-tube of Fig. P2.0!1 is partially filled with a liquid. When this device is accelerated with a horizontal acceleration, a, a differential reading, h, develops between the manometer legs which are spaced a distance r. apart. Determine the relationship between a, r, and h.
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2. q'ii' An open I-m-diameter tank contains water at a depth of 0.1 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank.
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The V-tube of Fig. P2.Q'1 is partially filled with water and rotates around the axis aa. Determine the angular velocity that will cause the water to start to vaporize at the bottom of the tube (point A).
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2.100 The U-tube of Fig. P2.100 contains mercury and rotates about the off-center axis a-a. At rest, the depth of mercury in each leg is 150 mm as illustrated. Determine the angular velocity for which the difference in heights between the two legs is 75 mm.
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A closed, OA-m-diameter cylindrical tank is completely filled with oil (SG = 0.9) and rotates about its vertical longitudinal axis with an angular velocity of 40 rad/s. Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis.
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2.102
Force Needed to Open a Submerged Gate
Objective:
A gate, hinged at the top, covers a hole in the side of a water filled tank as shown in Fig. P2.102 and is held against the tank by the water pressure. The purpose of this experiment is to compare the theoretical force needed to open the gate to the experimentally measured force.
Equipment:
Rectangular tank with a rectangular hole in its side; gate that covers the hole and is hinged at the top; force transducer to measure the force needed to open the gate; ruler to measure the water depth.
Experimental Procedure:
Measure the height, H, and width, b, of the hole in the tank and the distance, L, from the hinge to the point of application of the force, F, that opens the gate. Fill the tank with water to a depth h above the bottom of the gate. Use the force transducer to detennine the force, F, needed to slowly open the gate. Repeat the force measurements for various water depths.
Calculations:
For arbitrary water depths, h, detennine the theoretical force, F, needed to open the gate by equating the moment about the hinge from the water force on the gate to the moment produced by the applied force, F.
Graph: Plot the experimentally detennined force, F, needed to open the gate as ordinates and the water depth, h, as abscissas. Results:
On the same graph, plot the theoretical force as a function of water depth.
Data:
To proceed, print this page for reference when you work the problem and click heft' to bring up an EXCEL page with the data for this problem.
If!i FiGURE P2.. 102
(Con t)
~. / ()Z
(ef!JrJ't ) Solution for Problem 2.102: Force Needed to Open a Submerged Gate
L, in. 5.5
H, in. 6.0
h, in. 21.1 18.5 16.2 14.5 12.8 11.1 10.1 7.4
F,lb 10.1 8.9 7.6 6.7 5.8 4.7 4.3 2.9
Since h > H, A F
y, Ib/W'3
b, in. 4.0
Ixc , ft"4 0.003472
62.4
F 1 ,lb 15.69 13.43 11.44 9.97 8.49 7.02 6.15 3.81
Yr - Yc, ft 0.0138 0.0161 0.0189 0.0217 0.0255 0.0309 0.0352 0.0568
d,ft 0.264 0.266 0.269 0.272 0.276 0.281 0.285 0.307
F,lb 9.03 7.80 6.71 5.91 5.11 4.30 3.83 2.55
=H*b =constant and Ixc =b*H"3/12 =constant.
=F1*d/L, where F1 =y*(h - H/2)* A, d =H/2 + (Yr - Yc), and Yr - Yc =Ixc/(h - H/2)* A Problem 2.102 Force, F, vs Water Depth, h
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2./03
2.103
Hydrostatic Force on a Submerged Rectangle
Objective:
A quarter-circle block with a vertical rectangular end is attached to a balance beam as shown in Fig. P2.103. Water in the tank puts a hydrostatic pressure force on the block which causes a clockwise moment about the pivot point. This moment is balanced by the counterclockwise moment produced by the weight placed at the end of the balance beam. The purpose of this experiment is to determine the weight, W, needed to balance the beam as a function of the water depth, h.
Equipment:
Balance beam with an attached quarter-circle, rectangular cross-section block; pivot point directly above the vertical end of the beam to support the beam; tank; weights; ruler.
Experimental Procedure: Measure the inner radius, R1, outer radius, R2, and width, h, of the block. Measure the length, L, of the moment arm between the pivot point and the weight. Adjust the counter weight on the beam so that the beam is level when there is no weight on the beam and no water in the tank. Hang a known mass, m, on the beam and adjust the water level, h, in the tank so that the beam again becomes level. Repeat with different masses and water depths. Calculations: For a given water depth, h, determine the hydrostatic pressure force, FR = yhcA, on the vertical end of the block. Also determine the point of action of this force, a distance YR - Yc below the centroid of the area. Note that the equations for FR and YR - Yc are different when the water level is below the end of the block (h < R2 - R1) than when it is above the end of the block (h > R2 - Rl). For a given water depth, determine the theoretical weight needed to balance the beam by summing moments about the pivot point. Note that both FR and W produce a moment. However, because the curved sides of the block are circular arcs centered about the pivot point, the pressure forces on the curved sides of the block (which act normal to the sides) do not produce any moment about the pivot point. Thus the forces on the curved sides do not enter into the moment equation.
Graph:
Plot the experimentally determined weight, W, as ordinates and the water depth,
h, as abscissas.
Result:
On the same graph plot the theoretical weight as a function of water depth.
Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem. Pivot point
Counter weight
Quarter-circle block
)..-/03
III FIGURE P2.103
2. / P3 J
Solution for Problem 2.103: Hydrostatic Force on a Submerged Rectangle
W
R1, in. 5.0
R2, in. 9.0
m, kg 0.00 0.02 0.04 0.06 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.35 0.40 0.45 0.50 0.55
h, in. 0.00 1.11 1.58 1.92 2.51 2.76 2.99 3.20 3.41 3.60 3.80 3.99 4.17 4.33 4.50 4.95 5.39 5.83 6.27 6.70
L, in. 12.0
b, in. 3.0 Experimental W,lb 0.00 0.04 0.09 0.13 0.22 0.26 0.31 0.35 0.40 0.44 0.48 0.53 0.57 0.62 0.66 0.77 0.88 0.99 1.10 1.21
= 32.2 ftlsA2 * (m kg * 6.825E-2 slug/kg)
FR,lb 0.00 0.07 0.14 0.20 0.34 0.41 0.48 0.55 0.63 0.70 0.78 0.86 0.94 1.01 1.08 1.28 1.47 1.66 1.85 2.04
=R2 - (h/3)
Forh>R 2 -R 1: FR = y*(h - (R2 - R1)/2)*(R2 - R1)*b d = R2 - (R2 - R1 )/2 + (Yr - Yc) Yr -Yc =Ixc/hc *A Ixc = b*(R2 - R1 )A3/12 = 0.000771 ft A4 hc = h - (R 2 - R1)/2 A
y, Ib/ftA3
Yr -Yc, ft
d,ft 0.750 0.719 0.706 0.697 0.680 0.673 0.667 0.661 0.655 0.650 0.644 0.639 0.634 0.631 0.628 0.621 0.616 0.612 0.609 0.607
0.0512 0.0476 0.0444 0.0376 0.0328 0.0290 0.0260 0.0236
62.4 Theoretical W,lb 0.000 0.048 0.095 0.139 0.232 0.278 0.323 0.367 0.413 0.456 0.504 0.551 0.597 0.637 0.680 0.794 0.905 1.016 1.127 1.236
Sum moments about pivot to give W*L
For h < R2 - R1: FR =y*(h/2)*h*b d
g, ftlsA2 32.2
=b*(R2 -R1)
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2. /03
Problem 2.103 Weight, W, vs Water Depth, h
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Theoretical •
Experimental
2. /o/f
2.104
Vertical Uplift Force on an Open-Bottom Box with Slanted Sides
Objective:
When a box or form as shown in Fig. P2.104 is filled with a liquid, the vertical force of the liquid on the box tends to lift it off the surface upon which it sits, thus allowing the liquid to drain from the box. The purpose of this experiment is to determine the minimum weight, W, needed to keep the box from lifting off the surface.
Equipment:
An open-bottom box that has vertical side walls and slanted end walls; weights; ruler; scale.
Experimental Procedure:
Determine the weight, Wbox , of the empty box and measure its length, L, width, b, wall thickness, t, and the angle of the ends, e. Set the box on a smooth surface and place a known mass, m, on it. Slowly fill the box with water and note the depth, h, at which the net upward water force is equal to the total weight, W + Wbox , where W = mg. This condition will be obvious because the friction force between the box and the surface on which it sits will be zero and the box will "float" effortlessly along the surface. Repeat for various masses and water levels.
Calculations: For an arbitrary water depth, h, determine the theoretical weight, W, needed to maintain equilibrium with no contact force between the box and the surface below it. This can be done by equating the total weight, W + Wbox , to the net vertical hydrostatic pressure force on the box. Calculate this vertical pressure force for two different situations. (1) Assume the vertical pressure force is the vertical component of the pressure forces acting on the slanted ends of the box. (2) Assume the vertical upward force is that from part (1) plus the pressure force acting under the sides and ends of the box because of the finite thickness, t, of the box walls. This additional pressure force is assumed to be due to an average pressure of Pavg = -yh/2 acting on the "foot print" area of the box walls. Plot the experimentally determined total weight, W + Wbox , as ordinates and the water depth, h, as abscissas.
Graph:
Results:
On the same graph plot two theoretical total weight verses water depth curvesone involving only the slanted-end pressure force, and the other including the slanted end and the finite-thickness wall pressure forces.
Data:
To proceed, print this page for reference when you work the problem and click hac to bring up an EXCEL page with the data for this problem .
. , / Footprint of box
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III FIGURE P2.104
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Solution for Problem 2.104: Vertical Uplift Force on an Open-Bottom Box with Slanted Sides
8, deg
W
45
L, in. 10.3
b, in. 4.0
m, kg 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
h, in. 2.06 2.23 2.42 2.53 2.67 2.81 2.94 3.06 3.16
Experimental W + W box , Ib 0.942' 1.052 1.162 1.272 1.382 1.491 1.601 1.711 1.821
t, in. 0.25
Wbox,lb 0.942
h, in. 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25
Theory 1 W + W box , Ib 0.000 0.009 0.036 0.081 0.144 0.226 0.325 0.442 0.578 0.731 0.903 1.092 1.300 1.526
y, Ib/ftA3 62.4
Ib/ftA2 0.00 0.65 1.30 1.95 2.60 3.25 3.90 4.55 5.20 5.85 6.50 7.15 7.80 8.45
Pavg,
Theory 2 W + W box , Ib 0.000 0.047 0.111 0.194 0.295 0.414 0.551 0.706 0.879 1.070 1.279 1.506 1.752 2.015
=g*m =32.2 ftlsA2 * (m kg * 6.825E-2 slug/kg)
Theory 1. Including only the slanted-end pressure force: W + W box = y*Vol Vol = b*h*h Theory 2. Including the slanted-end pressure force and the finite-thickness wall pressure force: W + W box y*Vol + Pavg *A Pavg
A
= =0.5*y*h
=(b + 2*t)*(L + 2*tlsin8) - b*L =8.33 in. A2 = 0.0579 ftA2
2- /11.
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Problem 2.104 Total Weight, W + Wbox , vs Water Depth, h
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2.105
Air Pad Lift Force
Objective: As shown in Fig. P2.105, it is possible to lift objects by use of an air pad consisting of an inverted box that is pressurized by an air supply. If the pressure within the box is large enough, the box will lift slightly off the surface, air will flow under its edges, and there will be very little frictional force between the box and the surface. The purpose of this experiment is to detennine the lifting force, W, as a function of pressure, p, within the box. Equipment:
Inverted rectangular box; air supply; weights; manometer.
Experimental Procedure: Connect the air source and the manometer to the inverted square box. Detennine the weight, Wbo,,' of the square box and measure its length and width, L, and the wall thickness, t. Set the inverted box on a smooth surface and place a known mass, m, on it. Increase the air flowrate until the box lifts off the surface slightly and "floats" with negligible frictional force. Record the manometer reading, h, under these conditions. Repeat the measurements with various masses. Calculations: Determine the theoretical weight that can be lifted by the air pad by equating the total weight, W + Wbox , to the net vertical pressure force on the box. Here W = mg. Calculate this pressure force for two different situations. (1) Assume the pressure force is equal to the area of the box, A = L2, times the pressure, p = 'Ymh, within the box, where I'm is the specific weight of the manometer fluid. (2) Assume that the net pressure force is that from part (1) plus the pressure force acting under the edges of the box because of the finite thickness, t, of the box walls. This additional pressure force is assumed to be due to an average pressure of Pavg = 'Ymh/2 acting on the "foot print" area of the box walls, 4t(L + t). Graph: Plot the experimentally determined total weight, W + Wbox , as ordinates and the pressure within the box, P, as abscissas. Results: On the same graph, plot two theoretical total weight verses pressure curvesone involving only the pressure times box area pressure force, and the other including the pressure times box area and the finite-thickness wall pressure forces. Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem.
Weight
T h
1 IN-----------L-----------m
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2. I/)5" I
Solution for Problem 2.105: Air Pad Lift Force
W
L, in. 7.5
t, in. 0.25
m, kg 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
h, in. 0.54 0.64 0.74 0.82 0.94 1.04 1.12 1.23 1.32 1.42 1.52 1.63 1.72 1.83 1.96 2.06 2.12 2.23 2.32
YH20, Ib/ft"3
Wbox,lb 1.25
62.4 Experiment W + W box , Ib 1.25 1.47 1.69 1.91 2.13 2.35 2.57 2.79 3.01 3.23 3.45 3.67 3.89 4.11 4.33 4.55 4.77 4.99 5.21
p,lb/ft"2 2.81 3.33 3.85 4.26 4.89 5.41 5.82 6.40 6.86 7.38 7.90 8.48 8.94 9.52 10.19 10.71 11.02 11.60 12.06
Theory 1 W + W box , Ib 1.10 1.30 1.50 1.67 1.91 2.11 2.28 2.50 2.68 2.88 3.09 3.31 3.49 3.72 3.98 4.18 4.31 4.53 4.71
Theory 2 W + W box , Ib 1.17 1.39 1.61 1.78 2.04 2.26 2.43 2.67 2.87 3.08 3.30 3.54 3.73 3.97 4.26 4.47 4.60 4.84 5.04
= g*m = 32.2 ftls"2 * (m kg * 6.825E-2 slug/kg)
Theory 1. Involving only the pressure times the box area: W + W box = p*L"2
P = YH2o*h Theory 2. Involving the pressure times the box area plus the average pressure times the edge area: W + W box = p*L "2 + (p/2)*((L + 2t)"2 - L"2)
2. lOS-
Problem 2.105 Total Weight, W + Wbox, vs Pressure, p
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3./ 3.1
Water flows steadily through the variable area horizontal pipe shown in Fig. P3.1. The velocity is given by V = 10(1 + x)i ft/s, where x is in feet. Viscous effects are neglected. (a) Determine the pressure gradient, iJpl ax, (as a function of x) needed to produce this flow. (b) If the pressure at section (1) is 50 psi, determine the ,pressure at (2) by: (i) integration of the pressure gradient obtained in (a); (ii) application of the Bernoulli equation.
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'1
or ~ == SO pSJ' - /'1'1 (.3 +.;} )@ ( -)-'f;~i") = SO -/0. I :::3Q,9 E/ ///) fJl + 2"I r01/2. ~!7 _~ +.1- 1/2. ,jh ~ - 7 l." VI -+0 ~I -1'2. :2.. ('Y2. +(/ ~2. " or Wlr ~/-~:J. }1:7
A. .::: fJ; +-f f (V, Thvs;
'2. _
2_
~L)
wh ere V;; 10 (1 of 0) == loll 'vi ::: /0 (/+.3) ::: '/-0# "
..J.. I/. OLL .s~) II ,. LL 2.)JL ( / ·ft~ ) EO pSI, +:z. (••7T 773 "/ t 0 -,fl s::Z 71f'/-;":/. = 3 P, P PJ/ j
3-/
3.2
I Repeat Problem 3.1 if the pipe is vertical with the flow down,
3.2
, :.
.~
W
\~
(a)
-
asine - M- = f v*f
if = -p V¥S + If Thvs J
{j
r
*;% -f Vif +0 -: - f (10
-*::: -L9~ ~:"(IO#)2(f+X) + - -/9~ (IT x) .... 62.Jf
(b) (i)
if
N
=:
-/9'1 (t +x) + If)..'! .so
62.¥
~
~
'"
f)
T2.
.
= .5 0 PSII
t60-£
l.
;:=
J;;;/fX)
+62.1'
Jdx
XJ:::O
I If 2-
)
/'1-'1 i/J,'J.)
4-/.2fJS/
f1 +J..p~:J.. +'2i =;2. of d p~ 2. +!" 22. PI' w/I/J alld 10/::: /f'J(1 +0) :;:; /0# J L{::: /O(ff3) f{
+0'
wNh X in feet
(..3 :z.) Ih ( I ft lb ( +~ fi'- i~ifii." + 6:L.'f (3) ff2.
- 19'1 .3
= .50 -/0,/+/, g (il)
J
(J +x)) (J 0)
#p f, :;SOptl/
or
-
e:::: - qOO and v=io(J tX)!I
with
'&/::: 0,)
Z2. :::-311
=-~IJ.:fi J
:::1/ + i. f (14"'-~2.) - 02 = SO !s/ +i (I.?/f :jp-)(J01_ lftJ '") - t2.Y. #3 (-.3 ft) 2-
:: ,1./ /. 'lfJS /
3-Z
..3 . .3 3.3 An incompressible fluid with density p flows steadily past the object shown in Video V3.3 and Fig. P3.3. The fluid velocity along the horizontal dividing streamline (-00 ::s; x ::s; -a) is found to be V = Vo (1 + a/x). where a is the radius of curvature of the front of the object and Vo is the upstream velocity. (a) Determine the pressure gradient along this streamline. (b) If the upstream pressure is Po. integrate the pressure gradient to obtain the pressure p(x) for -00 ::s; x ::s; -a. (c) Show from the result of part (b) that the pressure at the stagnation point (x = -a) is Po + pV~/2. as expected from the Bernoulli equation.
.x
•
rQ)
t- ~ V.g. =-
where
41 g::-
Thvs J
==
FIGURE P3.3
v;: Va (J. + f) VA
fL·
x2. .
or
¥s=: ~=-e~(J~+)(-4l):: f
(b)
x
X
{dp =f #dx = faltj {f,. + ?)J~ po
-00
or
I
X
P-fo =: eaVo 2.[- f
a - 2:I XI
-00
(c) From part (h)) when X=-a
1'/ : ; 1'. - fit 11. i qf& ] 2
[-
xc-a
Frt)m the Bern()vll/ eqvatiDn
where
I
V, =V
=::
Yo (J +f-la») :: 0
)(~-a
-rhll~ '"
=:
PrJ +i f Vo.2. 4S expected.
3-3
3.4 What pressure gradient along the streamline, dpf ds, is required to accelerate water in a horizontal pipe at a rate of30 mfs2?
where e:::o and
VPs- ::: a.s ::: ThvsJ
M- : : - ('as:;::
- 999.!;!:s (30~) :;:: -30Poo(~)/m
or ~ dS
3.5
= - 30.0 kPa/m .
I 3.5 At a given location the air speed is 20 m/s and the pressure gradient along the streamline is 100 N/m3 • Estimate the air speed at a point 0.5 m further along the streamline.
If or
neqlecf qrQv/fy ~v
)
1$ :: IS
D
I
vlY C)S
or
N
-(jS = -Ioo~ = .... '1-.07 -:sI m 3 /(1 • 23 l!!L m3 )(,-o!!!.) s
Thus)
JV~ ¥S 6S - (-if,07-}) (O,511?) = -2.V 3 .qz) so/hat V+crV:::20~ -2.03: or V~/i.o'; 3.6
I
3.6 What pressure gradient along the streamline, dp/ds, is required to accelerate water upward in a vertical pipe at a rate of 30 ft/ S2? What is the answer if the flow is downward?
where e == qo ~ for flP Ilow }
e -= _qOO
and vjf: ~ ::: 3.0 fi
Ti1f1S, for /Jp flow
~ - - 62,'1-(1) W Ih c}.s o.n" for Jown f/()I;f/ d.(l = -6).,'1 (-I) lP Ib 7.S
for down flow)
-
1.9'1
- -12.o.6(-!fp.)/ff ::: -0. 83S:e;f ~~(30 it) s.2-
.sJup (.3°7ft)
/,9# ff.6
3-if
3.7 , to obtain the "Bernoulli equation" for this compressible flow as [n/(n - l)]plp + V 2 12 + gz = constant.
3.7
Consider a compressible fluid for which the pressure and density are related by pip" = Co, where n and Co are constants. Integrate the equation of motion along the streamline, Eq. 3.6,
~
f +.f- +~ r
= constant 010n9 a stream line
so that
or
ThlJs; S ~r
-nn
-I
n
n -I
3-.5
3.8
The Bernoulli equation is valid for steady, inviscid, incompressible flows with constant acceleration of gravity. Consider flow on a planet where the acceleration of gravity varies with height so that g = go - cz, where go and z are constants. Integrate "F = rna" along a streamline to obtain the equivalent of the Bernoulli equation for this flow.
From
Z b&: : 6m as one
df + f pd (V2)
of-
(2.)
S
Jm (I)
obfains
0 til where 't::: f'J
(see EI.(.3.$)
Thus
rlp
J
+ rJ.
(f pV2) + P(go - c 1) d1 ~O
(2.)
J
or by infe9r4f in9 from (I) to(:;.):
~
(1)
S"f +Sd(tfV'-) ~ f> f (~() -C2)J;z=:O
W
til
or f{ -p, + f
(/)
p(v:- ~~) .,. P'}o(~,.-l,) -
fpc (2').2. -z/-J
=0
Thvs, fJ + t pV2. +fi, z - f f C 2~ -; cons/alii a/oil' a .sfreamline.
3-6
3.9
J 3.9
Consider a compressible liquid that has a constant bulk modulus. Integrate "F = rna" along a streamline to obtain the equivalent of the Bernoulli equation for this flow. Assume steady, inviscid flow.
From Ef. .3.6
dp +fpdrv1.)
+odz~O
where
and
(:: e9
dp :: £~
1f
INhere
Ell ::: hulK 1Y}()t/"lvs ::: cOMI4nt
Th-us} (Alon,
Q
£.,.1f +:i-pd(V''J +fgrie::o £'V
~
ef,.
£v {
+
(see E~. 1.13)
sfream} ine;
d (f Vtl.) +~ dl :: 0 v,.
12.
or
which can. he infe9rate d he~ween (I) ana (2.) T() '1 ille
belween pomfs
~ T fdaV~)+ f9d,Z ~O
f'
V,
1/
or
"Ev[ fr2 - i] t[V,.2..- '/,1.] +~[Zl-Za ~O t
Hence:
?z - ~
+
r-
:=
cunsl.nf 410"'1
Q
slreamline
I
3,10
3.10
Water flows around the vertical two-dimensional bend with circular streamlines and constant velocity as shown in Fig. P3.1O. If the pressure is 40 kPa at point (1), determine the pressures at points (2) and (3). -Assume that the velocity profile is uniform as indicated.
4m
FIGURE P3.10
-a' d~ _y, _ e'C Tn on - "1( ,hilS; w/fh '1< -::;
(1)
wNh ~::/
OlJrP
V == /om/.s
6 - /}
()r
Ilz. =<2..,0
kPIJJ
alJrJ,
wlfh f; :: '1-01<14 4nd
fl,3
==
~m
S
/J ::: 'If) kP~ - 9.I()XIO:J~ (;;'/1/) -999
or
-t!s (10-1)2. /1 (-9-)
1.3 : : - 20. / k PiN
3-8
I
.3.12
'14
3.12. Water in a container and air in a tornado flow in horizontal circular streamlines of radius r and speed Vas shown in Video V3.2 and Fig. P3.12..Determine the radial pressure gradient. ap/ar. needed for the following situations: (a) The fluid is water with r = 3 in. and V = 0.8 ftls. (b) The fluid is air with r = 300 ft and V = 200 mph.
~--+-x
-
• FIGURE P3.J2
Fo"r ourved sfreamJ/ne,s, -
~
'"
~ + if ~
and
fn =- fr-
J
or wilh ~ =0 (horilDnf4! sfrea/lllines)J 1?::: r J
this
becDmes
!Ie :: iJ-" r (;/.r
With r:::
CA)
r!£
J,'r ::
(6)
/.9'1-
J
ff and V= 0,8 #- und wafer
*¥ (Ii
(O,81})2.
fI)
s//J9s
:: ~. 97 fI··.s~
(p = /,9'f ~ )J
= ~,97 Jb3
f/
Jt
W'i-Ih r = 300fT and V=- zoo tnph( 88 oS h ) = 2 '13 fts t,omp
and air ( f:: 0.00238 rkr
d:r
.:!~~s )
J
~ (11:)2
.
~ O,OO:2.3a~ 2'13..$. == 0.681 slv9,S2. 068/ 300 f1 ft"·,s ::, I
3-'1
/b
W
.3 ./3 3.B As shown in Fig. P3.13 and Video \'3.2. the swirling motion of a liquid can cause a depression in the free surface. Assume that an inviscid liquid in a tank with an R = 1.0 ft radius is rotated sufficiently to produce a free surface that is h = 2.0 ft below the liquid at the edge of the tank at a position r = 0.5 ft from the center of the tank. Also assume that the liquid velocity is given by V = K/r. where K is a constant. (a) Show that h = j(2 [(1/,-2) - (1/R2)]/(2g). (b) Determine the value of K for this problem.
(a) -
v:L
t"
Thvs
f"R
clp
#,:,:
(R
(fo
, j
Of'
==
eK'- ~
• FIGURE P3.13
J
~f
r Bvi fo::: 0hand f:::O at r ThvsJ ,. F
oh:: - ef [p -f,.J
0/1
the free
stJrface.
or sInce
(I)
(b) With h:: 2 f( R:: I if and r I
_ 2
ff -
K2.
o.s 1/ £qn. (jJ 9/ve.s
[I
2 (32..2 fils>') lJo.s tW
or K::: 6.55
==
-
#2.
3-/0
(f
~)L
]
3./1f 3.1#
Water flows from the faucet on the first floor of the building shown in Fig. P3.11f with a maximum velocity of 20 ft/s. For steady inviscid flow, determine the maximum water velocity from the basement faucet and from the faucet on the second floor (assume each floor is 12 ft tall).
(3)
(I)
\' = 20 ft/s
f~"
~=======r=12 ft
•
and or
P3.11f
wIth /J ::fll =() a/ld
oJ;l 2. (2 oS)
VI= 20 #
(froe J
Jet)
2/ =-If.
If
+ /b H
2'.3 ::: /6 It
~ ="';').02. _ 2(.32..2)(lJ..) == ,; _ 373'
ImjJoss/b/el #0 //tJlJ/
2. (32 . Z Of'
FIGURE
¥:J
+ if It :::
\0 2
~
2- (32.2
!h.)
from .secf)/Jd
{/rJf)fI felVest.
3.15
3.15 Water flows from a pop bottle that has holes in it as shown in Video V3.5 and Fig. P3.1S. Two streams coming from holes located distances hi and hl below the free surface intersect at a distance L from the side of the bottle. If viscous effects are ne~ligible and the flow is quasi-steady, show that L = 2(hl~)1 . Compare this result with experimental data measured from the paused video for which the holes are 2 inches apart.
~-O
• FIGURE P3.1.5
For sfeady /nviscid I/o~ Ihe velocities of the horiz on fa I jets of water at pOinfs 0) and (;z.) are obtained from ihe Bernovlli erv41ion 4S: fo + -J: f
Vr/".f 0 Zo::: A + -1 f ~z.+o2, ~ 12. +if 1Iz:L+!'Z:J. where
fJo~f,==f2.::0; Vo==Oj zo==h2.j 2,
:::.hz.-h,; and 22.:: 0
X=VI: Z
Thus ~ = r2~h,'" and ~:: y2!h:
(I)
J
Once olltside the fqnKJ qrtJvify is the on~ force on fhe fluid parTicles. Hence the hOf'iiontQ J compone"f of ve/ocify remlJillS cons/ant J bvi fhe parlicJes
~2= h_t~t2
o .........----x
accelerate downward wah fhe acceJert.lfion ct.f 9rav/1y. Thus I for a {II/itA par/ide Ihat exiled fhe lank t seconds Q90 at z;:h /-1 fol/()w.s theff
h - ±~ l ~ By e/il1Jind1in9 " ihi.s 9/ves the pafl ide pafh ((,~ e. the X :::
Vt and
2:::
T
%
h:z..:: Z
shape a/the VltAler jet Q.s Z =:. h - g X2 iV'
Thvs" the shapes()/ fhe flllo Water jets are as shown
Z=
0
in the fi9J1re.
3-/2
-=~~----~--------x
.3 ,J5
I (c on 'I ) X-L tJl1d ~::Z3..
The sfreams intersect when
71Jl),sJ
fromEf"s.(l.)tJnd(.3~
- _1-L2. (h 2. -hI )_LL'l. ~ ~2. 2 ~2.
L=12(~-h'~ W-1;L From Eqn. I
-v,2. -
0) J
I
/
~2.
::
2Jh , -
J
_
21h~ -
2, [-1- _.J..] .J.
hi
h2.
-rhvs by comh/nilJi
["'f/ls. (If) and (s)
L: Jz(kh,)j
(h.. -h,)
J
Y
9' 7
21 hi h2.
=
we ohtain
2yh h. 1··
3-/3
=
fhz.. -hi) 2j n,h:;.
(S)
.3 ./6
3.16 A 100 ftJs jet of air flows past a ball as shown in Video V3.1 and Fig. P3.16. When the ball is not centered in the jet, the air velocity is greater on the side of the ball near the jet center [point (1)] than it is on the other side of the balI [point (2)]. Determine the pressure difference, P2 - PI' across the ball if VI = 140 ftJs and V2 = 110 ftJs. Neglect gravity and viscous effects.
v1 = 110
V= 100 IUs
(Lf)
The Bernoulli e9uafion from point (3) fo (2.) and (1/-) to (I) with 9rQvily neq/ecfed 'lives I
{J.3 "':2 f ~ ::; 2
p:;. + ..L2 f> V:z.
2.
But {1.3 -== 1'1:::0 and
tllld
~.::- ~
• FIGURE P3.16
f'f of J..2. P1/2 VII- =:: P,
1.
-#- ,.
Pv," I
Thvs} e 'len fholJ.~h poinls (I) and (~) are nul 0" the same streamline} I
fl
-L
,,2
+ :i f V,
:-
~ of 2.
\/2
f V2.
3-/~
3,/8
I 3. 18 A fire hose nozzle has a diameter of H in. According to some fire codes, the nozzle must be capable of delivering at least 2.50 gallmin. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this fiowrate?
Thus; fJl:;::
4-[V;.~- v:
2-]
where alld
.so
/J
iAqt w/IA =
f ~f
i (/.9~ ~;P)[ 80.7). -1I.3/f 2.] ~
::: 6/90 J1L f12. == Jf.3.0 psi
3"'/5
3.19 Water flowing from the O.75-in.-diameter outlet shown in Video V8.6 and Fig. P3.19 rises 2.8 inches above the outlet. Determine the flowrate.
• FIGURE P3.19
The flowr~te is Q::: A, V; 1 where trom the Bernoulli efvalion I?J.M: +EI ::: I!!= ~ .,. E2. o + ~1 t +2.1 ThtJ~
with' fl
=: ~
~ ~I ~ ~ ~ 0
V; = {21 2;;. = {2(.32.2fi/.s:l.) So that 2. Q::: AI ~ = I(O~~SH) (3,88 ¥)
we Obtain ::: 3.88ffl.s
(2,8/12)ft
:: 0,0//9 f
.3
3.20 3.20 Pop (with the same properties as water) flows from a 4--in. diameter pop container that contains three holes as shown in Fig. P3.20 (see Video .t5). The diameter of each fluid stream is 0.15 in., and the distance between holes is 2 in. If viscous effects are negligible and quasi-steady conditions are assumed, detennine the time at which the pop stops draining from the top hole. Assume the pop surface is 2 in. above the top hole when t = O. Compare your results with the time you measure from the video.
• FIGURE P3.20 Q:: Q/ + Q,. .,. Q.a ::: -lJr -df where \,(1 /J.::: ~·A· and III :: fJ:r. :: A.3 :: f (o/I Ii) '}. ,~ :: Y2a (/' h· A· I
J,
= /,227 X /0-'11/2AT:: ~((fn)'-;: 0.087.3 fl2.
(i /If/J 2.".1)
where i is the time /1 lake fur fhe fros .s()rI~e fo re40h fhe IIf/Df' hole
or
t
(h~o),
L
AT [ dh ::: A, Y~1 ((h +Vh+L +fh+2J.) o
L
dh
O. OB 7 3 1 1 ' " (
_
- (1.22.7 ~ 10" fJ'J. )K2.)(.3:J., 2.
-rhll~
ff/s'-)]~t (Yh +Yh fL +1(h +2.L ) o
L
j :: 88.7 ( d h J «(h .fYh+L- +YhfZL)
where L:: -k i-l == 0.1667 ff
o
Nofe: Will; L If) leef) flJl'.r crvtJJf iOf) 9illes
Since fherB is
flO
closed form
3-/7
t in seCl)l'Idr.
3.1- 0
I (con 'i ) The nvtrJerica/ valve of lhe i"fe9ra I is obfailu}d by vsin9 the trapezoidal rule since the closeJ form analytical so/uil'on is nof 9/ven jn illfe9f'al fa hies. The EXcel- spread sheelv.red for this is 9irJefJ be/ow. L
, :;; ,88.7
Sf(h) Jh
J
where [(h)::: (VE -I- WL ./-1(hiii.)
o
:=88.7[ih(f,.t{+,)(hi t/-h,)] h, in. 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
Thus, t
f(h) , 11ft1/2 h,ft 0.0000 1.015 0.0083 0.914 0.0167 0.870 0.0250 0.837 0.0333 0.810 0.0417 0.786 0.0500 0.764 0.0583 0.745 0.0667 0.728 0.0750 0.712 0.0833 0.697 0.0917 0.684 0.1000 0.671 0.1083 0.659 0.1167 0.647 0.1250 0.637 0.1333 0.627 0.1417 0.617 0.1500 0.608 0.1583 0.599 0.1667 0.591 Sum of column integral
=
«(88.7
frd~.,tom1=
(1/2)*(fi + fi+1)*(hi+1 - hi), ft112
0.00804 0.00743 0.00711 0.00686 0.00665 0.00646 0.00629 0.00614 0.00600 0.00587 0.00575 0.00564 0.00554 0.00544 0.00535 0.00526 0.00518 0.00510 0.00503 0.00496
=
=88.7*0.12011 =10.7 s
3-/8
0.12011
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
/0.75
--L
J.21 Water flowing from a pipe or a tank is acted upon by gravity and follows a curved trajectory as shown in Fig. P3.21 and Videos V3.5 and V4.3. A simple flow meter can be constructed as shown in Fig. P3.21. A point gage mounted a distance L from the end of the horizontal pipe is adjusted to indicate that the top of the water stream is a distance x below the outlet of the pipe. Show that the flowrate from this pipe of diameter D is given by Q = 'TT'D2Lgl/2j(2S/2 XI/2).
~I'
• FIGURE P3.21
The only force aclin9 of allY w4fer parfiele ill -Ihe free Jet is Ihal d()e 10 9ravily - the parfic/e:S wel9hf. Thv.s J for the X - YQxes shown
i) :: a tf
",fl..
ana ~~:::o whicA for a parhcle .sfarfi»" dP'
I
x r:: y~ 0 ttl I::: 0 9iv e X;::: t1-t2. and yo:: Vi £//mt'nale l to 9ive fhe wafer Irajec7ury as af
x~{~(~t
or
x==r~t:
Thvs wifh y::::1.. : J
V c V-/x·'" Land Q::: AV:= fj)"v :: 7T D~L Vj/(2 51:J,
3-/9
Vi)
- - - . I...
.3 ."1.
I 3.22
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 220 mph? 2
2.
.I!.!-+XL ~ V2. r 2j +Z I = h. ~ 2.1
..J. Z
~
2.
=Z2
,/;::6StrJph (88fj ) =«5.3!117=0)
\4 =-0 6om,"
or Ii =- ~ (2.38X/O- ~~ig)(95.3#/·= /0.8 ~ 3
If V;::
2.20mph ( B8
12. : : f
3.23
~)
60 mph
(2.38 X/O-.3
::: 3
~3 !f
fhen J
~:fi) (3:L3 fj) 2. =
/2.'1
~
T3.23
A differential pressure gage attached to a Pitot-static tube (see Video \'3.4) is calibrated to give speed rather than the difference between the stagnation and static pressures. The calibration is done so that the speed indicated on the gage is the actual fluid speed if the fluid flowing past the Pitot-static tube is air at standard sea level conditions. Assume the same device is used in water and the gage indicates a speed of 200 knots. Detennine the water speed.
Af :: tpV2. In air Afail'::: J
f
(0. 00'-38
~~~.s ){2ookn ofs)f/.
In w4fer.i AIl ::: ..1 (/ 9/f S/()fJJ.) ( V)2 rW41er 2 - ' ff3
Ali- :: A fwater t(O,OOJ-31 )(ZtJO/' ::: i (I.9~) V.2
so thai w/lh
or V=
7.0/
knofs
A
J
,3,2Jf
I
3.,..
A 40-mph wind blowing past your house speeds up as it flows up and over the roof. If elevation effects are negligible, determine (a) the pressure at the point on the roof where the speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psia. Would this effect tend to push the roof down against the house, or would it tend to lift the roof? (b) Determine the pressure on a window facing the wind if the window is assumed to be a stagnation point.
~V2. (2.)
-o
(3)
to (:1-) : I, +i p~:z. = /2. + 1: P~2.
a) Thvs, from (JJ
b) From (I) +(13):
(J~ == PI
Since ~'='O"
Z
+1 pU or (J.J - P, :: -t pv,:L =- f (0,00').38 :~"/f) (S8.7il/s;2 ::: ~/O ~
3-2.1
v, ~
(I)
.3.2.S .~.25 Water flows steadily downward through the pipe shown in Fig. P3.2S. Viscous effects are negligible, and the pressure gage indicates the pressure is zero at point (I). Detennine the flowrate and the pressure at point (2).
•
A/so)
f=.
+Z
Jr
-I2.
]'z.2. ~I
=- .fL l:
~z
. . ~£
'~,.
where f,:;: 0 4nrJ sInce II, ~,42 TlJv~ 2:J. - 2,
~
-
if
0r
FIGURE P3.2S
Ii
l; ~ -2 ff
(Jr
12 :: -2. ft (62,tf Jjp) :: - /25 ~
3-22
{QI/olIIs
fha-r
Vz. V, #
3,26 3.2~ Small-diameter, high-pressure liquid jets can be used to cut various materials as shown in Fig. P3.26. If viscous effects are negligible, estimate the pressure needed to produce a O.lO-mm-diameter water jet with a speed of 700 m/s. Determine the flowrate.
:/:L
r
\.{ 2.
+ ~I + Z,
Thll.s
(J~
=V
~ == t
$~
2-
+
f~
2.
=
.,.;$J.
where
~ -zO J 2, ~ Z2.J
t ev,.2. = t ('9'1-i)(70()~)2
Also) Q = ~ A.z = 700 ~ [¥ (/O-i'1h)2] = S.$() x /o-~ .!fi
3-23
and fJ2- =0 = 2.9-5')(/0
5
.lsJ.
3.2-7 3.27 Air is drawn into a wind tunnel used for testing automobiles as shown in Fig. P3.27. (a) Determine the manometer reading. h, when the velocity in the test section is 60 mph. Note that there is a I-in. column of oil on the water in the manometer. (b) Determine the difference between the stagnation pressure on the front of the automobile and the pressure in the test section.
~
(I)
~
•
~
•
Wind tunnel 60 mph ~
~
('-)
FIGURE P3.27
(a)
(b)
f2.
~ Jt ::: 4' k z,.,.
1- Z.
7 where
Z:L ::: Z.J 41114
Thvs
-,
~::' ()
ty ~1; = If .J
.f Z.?l .f-!2.
2-
or
A - (J2. ~ f f 142.::: -f (0.00]..38 -§.~~.s )( eN fj)2 == 9.:z.z 1-.
-'"
-....
"'"
.3.1-8
I
3.28
A loon is a diving bird equally at home "flying" in the air or water. What swimming velocity under water will produce a dynamic pressure equal to that when it flies in the air at 40 mph?
or
3-25
3.29 A large open tank contains a layer of oil floating on water as shown in Fig. P3.29. The flow is steady and inviscid. (a) Determine the height. h. to which the water will rise. (b) Determine the water velocity in the pipe. (c) Determine the pressure in the horizontal pipe.
(c)
3-26
r
p
4m
.3.30
3.3.0
I
Water flows through the pipe contrac-
tion shown in Fig. P3.30. For the given O.2-m
I
0 .2 m
difference in manometer level , determine the flowrate as a function of the diameter of the small pipe, D.
FIGURE P3.30
./!L + ~
\1,2. + Z,
~?
~ =1;>'1 but f!, '"
=.if + ziV.' + Z,1
or w/lh z / =ih ani 11 =0
(fi/') ,
~k and /':z. ~ rh2
so
Thlls,
~ =rrZ-?-o.-;""/ ="';2.1(0.').)'
3-27
that 17-12. '" ((fir-h.) = 0.').0-
3.31
•t
0.2 m
3.31 Water flows through the pipe contraction shown in Fig. P3.31. For the given O.2-m difference in the manometer level, determine the flowrate as a function of the diameter of the small pipe, D.
FIGURE P3.31
1
1/
VI
or
or
Q_
0.0156/)
2-
- Y(0.1)'1 - l/'{
3-28
~ ==
0.2. (2-$)
[ (!ff)'" -I]
= (0.1) D V.I
3.3.2 3.32 Water flows through the pipe contraction shown in Fig. P3.3-2. For the given O.2-m difference in the manometer level, determine the flowra1;e as a function-of the diameter of the small pipe, D.
+ 0.2 m
• FIGURE P3.32
12- +X1 t it :LI
Z :: {J:J,. I
..,
-1-]1: +£. ~
2-
where Z,:: Z2. t4nrJ ~ -:::0. ThtJSJ
kr +ff = b z.,f r Bi/t IJ!-:::x and X
+¥ :: Il'"
t'f::: D,2m +x
O.2m
V, == f-------2-1 (0.')..",) Thus
Sf)
-Ihat
+X or =
(2 ('1.81 /§i) (0. :1-/11))k'2.:::
1.98-fL
J
Q ::; II, Vt ::: f (o.linl- (I.?!.p.) ~
0.01,56!f-3
3-1-9
for olJy D
3.33
J (3)
•
3.33 The.speed of an airplane through the air is obtained by use of a Pitot-static tube that measures the difference between the stagnation and static pressures. (See Video V3.4.) Rather than indicating this pressure difference (psi or N/m2) directly, the indicator is calibrated in speed (mph or knots). This calibration is done using the density of standard sea level air. Thus, the air speed displayed (termed the indicated air speed) is the actual air speed only at standard sea level conditions. If the aircraft is flying at an altitude of 20,000 ft and the indicated air speed is 220 knots, what is the actual air speed?
...... (4) r--
v.==V I
PI
:=
f
V2 ::: 0
.
(1)
V p
~~ooo or
3-30
•
~
•
(2)
3.3Lf
1 1
3.3'1
Streams of water from two tanks impinge upon each other as shown in Fig. ·P3.3'1. If viscous effects are negligible and point A is a stagnation point, determine the height h.
h
Free jets
I
t8ft
20 ft
1
~ FIGURE P3.3Jf
.()- ~~ z. - ~/I .xl ~ -II( + ').1 + 2. ,,- T + ~I' +.c/J
where Ii. =-0.) ~:::- 0 J
2',.:::- h+~()rl
~ =0.1 a/Jd ~ = 2.() If
Thus
J
or
h +2. 0 II ::: .If; f- 2 () If
h == i!f
(I)
II/so; fJ, 'h2. 4" + '-I + Z,
~+
.:: i'
+ 1'1
VtJ..2. Z
').j
TAv.sJ
.If::: 'f +z, h
=-
If +z, -
-219
Z.19
~
3-31
3.35
I :US A O.15-m-diameter pipe discharges into a a.IO-m-diameter pipe. Detennine the velocity head in each pipe if they are carrying 0.12 ro 3/s of kerosene.
m' O./Z-S m \I, "..Q. - 1£( • = 6.79$ I /II 'I'
O,/sm)
and
",3
0.12.
;s-
f (o.lom)'
" 15.271}
Thvs, \/"
v,
(';.7?f!1/ _ -:"::;':':"'-£"f" 2. 35m.
"1 -
2. (9.9I-f;,)
alld
v. '
2, "
(ls.'Z.7!}l
2(9;8/~) =
II." 111
3-n
3.36
J
ICf'
3.36
Water flows upward through a variable area pipe with a constant flowrate, Q, as shown in Fig. P3.36. If viscous effects are negligible, determine the diameter, D(z), in terms of D\ if the pressure is to remain constant throughout the pipe. That is, p(z) = Pl.
I .1
Z~_D(Z)!
l-Dl_L FIGURE P3.36
or will; I' = fJ; and 2.
2-
V, -V =2<;i!
and V=-f Thus)
'f~)2. (JfQ)2.. (TiD/~ 2pE TtI)2-
or ..l...
DIII
I
-11
3-33
.....
z, =0
I
tQ
J
(1)
..3.37
I
3.37 Water flows steadily with negligible viscous effects through the pipe shown in Fig. P3.37. Determine the diameter, D, of the pipe at the outlet (a free jet) if the velocity there is 20 ft/s.
rOpen V = 20 ftls
*
1
15 ft :::; h
-111.5 I,. do""'" fen • FIGURE P3.37
or
V, == 8.83 !1fhtff =F /). 2.(2.0 if) r:.z.
BtJf AI Vj = A:l ~ .1t D'- (8• i3 H) 'f-I S
or
So
oS
~
4. =( 8~~3) ,.(ff fI) == 0.0831 ff::
O. '197 in.
3,38
J 3.38 The circular stream of water from a faucet is observed to taper from a diameter of 20 rnm to 10 IJ1m in a distance of 50 cm. Determine the flowrate.
I!!- + ~
~ 2. +Z
')..~
I
T
'-1
where A ;::f2, ': 0 J Z,-- =0 and II Q v.: - () 7f;
VI -
J
r
= -1'2._ + v,,2. + E
2. -
0.5f)
2 J
Z, ==
:!
m
,
(%t + 21i!, = ("*J' or Q [(.;~*~)r = , =
tI
II:J. fii2, -(/J:J./A)'J.
or Since
112 - (- A. )2. we obfRin A,-o,
Q-
('-) D2. ;:: O. 0/0", Q
liz
~~
A
z
f
=2,5Lf
I
_ 11.( J -(D.zID,)7J - 'I- 0,010n, 1I2jz/
)2.[
2(9.eI'f;.)(O,SOfIJ1 ,_( 0.0/0
~If
f), fJ2.0 )
_¥
m3
x/a S
3-35
,
D = o.O:Lom
L.
o. son
":;) (/J
2
3.3 'I
3.3Q Water is siphoned from the tank shown in Fig. P3.39. The water barometer indicates a reading of 30.2 ft. Determine the maximum value of h allowed without cavitation occurring. Note that the pressure of the vapor in the closed end of the barometer equals the vapor pressure.
Closed end
r
(0) 3 in. diameter
30.2 ft
5 in. diameter
FIGURE P3..3Q 2.-
ft t Vi r 2.1
+Z
where jJ/ -=0 J
I
bllf
::;-0 J fJz
Z, =°,22 ::0
Thv,s,
o-
V;
fVlltPfJl'
0'
21
f/lapor ~
lIe/)c8.;
o = -30.:2. if l{::
3 q. .!'
~
Since or
+ ~.t of 6 II :1-1
!I11.3 = Vz A;z ,
'vi = 1'I-,21J
HiJwever) All
VJZ
zi +z, :::
~ f-
o
LJ.a ./L::!.
~
+-
V: +28
or ~ = Y2-?h
~
ThtJ.s J I if. 2 f1 s =
ft
+ '(,,2. + 6 f f
10 +.30, '). II r ::11 or .since ft:::: {Ji/(/.por J
Tf;vs J
=Ivaplr
';2. (32.Zli ) h ;/ s~
or
3-36
h::::3. 13 ff
I
=
-30,')..
It
.3. 'f0 3.40 An inviscid fluid flows steadily along the stagnation streamline shown in Fig. P3.40 and Vidt'H \l3_l, starting with
speed Vo far upstream of the object. Upon leaving the stagnation point, point ( I), the fluid speed along the surface of the object is assumed to be given by V"" 2 Vo sin 8, where (I is the angle indicated. At what angular position, 61, should a hole be drilled to glve a pressure difference of PI - P2 = pV~/2? Gravity is negligible.
Vo ----;;(0)
•
~ 11.'" ~ IlL I.+:z.f· "'/t+2.PV/
where V,
L
=0
Thvs,
f, - P..
::f.. +~:z.f~v.
FIGURE P3.40
'i p (v,.'" - W'J ~ d(' ~L
d
sO fh"f if
f, - f1,. ='
d r v/ then v.. " V.
Th"lis: ~ '" 2 II
ntlJc8;
V. sih ~ () __
L
'"
Vo
or
. "
..L
Stn~==2.
30' =
.3.'fI .l.41 A eelain vacuum cleaner can create a vacuum of2 kPa just inside the hose. What is the velocity of me air inside the
hose?
OJ
V,:
fl +f p~ 2. "f:z. + t p where II ::0, ~ '" 0 .so fhal
/ / / / / 1/ / ///111/7
~ !I.'" f",=-:z.f2.
flence J
3.JL .J. k ~ -2.XIO 1112. ~ -2.(1.~3!J3) ~
•
3-37
3.4f2
J
3.42 Water from a faucet fills a 16-oz glass (volume = 28.9 in. 3 ) in 10 s. If the diameter of the jet leaving the faucet is 0.60 in., what is the diameter of the jet when it strikes the water surface in the glass which is positioned 14 in. below the faucet?
= 0.852 it ...s HenceJ __------------------~--, 1:1. 2 tf )(111fl + (0. BS2.s) Vz. = 2 ( 3.2..:2.-;2. n:. T 2 (3.2..2 {tao) )
But, 11, ~ 112 Vz. %
SI)
8.7/
Ii .s
fhal
~
or
=
k 7 )2-(0 BS2 D - (Ji) V,2. f)'/ -- (0.8.1111 F+
2.-
LO ,·n)
.0
•
:: 0.188 il'l.
.$
3,113
I
3.43 A smooth plastic, lO-m-long garden hose with an inside diameter of 20 mm is used to drain a wading pool as is shown in Fig. P3.43. If viscous effects are neglected, what is the flowrate from the pool?
0.2 m : (I) ;;~ :.;;~::;:: C"" 0.2*3 ~ '/h ~/" /////////////~ 1-1
FIGURE P3.¥3
./!!- + '0:L + i!
zg Thus} ~
where
I
~=f2~(Z,-Z2)i -
11:r /3.. == 0 J
2 2. :: - O.23m.I
(2. (9.8/ ~)(O,2 m -
(- O.'].3hJ)~
~Q.!1l -- 2 .7.s
or
~I
Q = fJ2. ~ =-f(o.02-0ml·(2.90!P-)
= '1.1/ X/O'f./ff-3
k 2.
:: O.
and
')../1'1
~I =0
3. iflf
I
3.11-'1Carbon dioxide flows at a rate of 1.5 ft 3 / s from a 3in. pipe in which the pressure and temperature are 20 psi (gage) and 120 of into a 1.5-in. pipe. If viscous effects are neglected and incompressible conditions are assumed, determine the pressure in the smaller pipe.
o·(I) D - ':I,,,' . I-~
Tillis, f:;. :::; fJ, + i
.:: 2.
f ( ~ 2. -
Ib ero 7["
~'")
I"
- .53. I f[z. ==
If, 2.J 82-71f2.
(Jr ~
::; 19. 63 psi 9 Q g e
3-39
::_---.....:.-=c:.:..~):....;,o D2. ;: I,S/I').
3. 45
I r--
3.1f5
1
Oil of specific gravity 0.83 flows in the . pipe shown in Fig. P3.IfS If viscous effects are neglected, what is the flowrate?
4 in:=
Water~ ~ (tAl ~ -. f1in
TVt
7" LJ,
112.
+ ..!.L + :z ~i
~I
V/,
1)
~
= T -r~ L
....
-1--::2
~2
i ll--;a=o.83
tl or-
FIGURE P3. Jf.5
where
z, ~ Z:;.
h
~ (.liT
)1
()1
and V, ~ 0
Thus,
V;- _ 1'1 -I':l.
21 -
(I)
)"
but,
fl., :: ~ +rJ :::; If + oj
and
fJ:z.:: 't (J, +h) -
i}n h -f fll-
Thv!'J
ft - fz
:: ( ~ -l') h
Comhine £,05.
(I) aIJrk (2.)
(2)
10 obtain
,h
.1 ' ./ (.62,.'ffl3 )'.!i:r/l v.. = V2~V\·-") = v2.!(1- 1) h =1 2(32·1.t/-.)~O.83(6';fff.r J 6.."1 1
or
Vz = 2.lo!f
ThtJ-s, Q
=1l2. ~ =-f(!r.n)2.(z.lo!j)
=0.183 ~
"'?-ifo
3.'f6
.3.16
Water flows steadily from a large open tank and discharges into the atmosphere through a 3-in.-diameter pipe as shown in Fig. P3.46. Determine the diameter, d, in the narrowed section of the pipe at A if the pressure gages at A and B indicate the same pressure.
p~ + ~ e~2 + ~Zlf == f:J. + if '6.2. +i'z~
where
Z2.::
0
and 1:l.~O
14 =1'"
Thus.! since
13 + t e~2 + i'r.1I-
HOllleve~
J
== f: pVs2.
fl + ie~2..,.
so that
0)
r2, := fL +ipli2. +~E2..1 where It -::'11- ~ ~~Z2. #0
t. pv,.2. ;: ~i!,
or Vz. ""
Y2fz,' =y;.,;; =(2.(32.. 2 ~~}(16 fI)]~ 32.di/.s
Bvl f.3 + 1: P~2. t rZ.3 ~ f~ + i p\{2. f r 22. "hers ~ ~ ~ since 112. ~1l.3 ThllsJ
1.3::; -rZ3:: -(16+Q)fI(6,..Jl-l bltl3 ) From Efs. OJ ana (,.): -1,s&0
jtJ. f i
(I.9'1-
~ -/Sto l/,lfl:J.
.
~') '42.
:=
1::(1,9'1- ::~9f) (:l21/i})2-
or ~ ;:: ~t./
11/0$
Since A~ ~ =,42. V,. if follows 1h~f
¥ a~~ or
=1/):1.2. v,.
d '" L>.. V~'
'" (3io.)
3,..1111.r -~1t--I.~ ,/
~.I T /0$
= 2.5 0 in.
3-if/
(2)
3,'1-7
3.47
I Determine the flowrate through the pipe in Fig. P3.47.
FIGURE P3,,1f7
where z, =- Z2. and l.'2.::: 0
=
m
2.20s
Tho.s J Q== II, ~ = -: (0,08 m)2. ('1. .20 !f) :::
0.0/11
if
.3. '1-8 3.48 Water flows steadily with negligible viscous effects through the pipe shown in Fig. P3.48. It is known that the 4In. diameter section of thin-walled tubing will collapse if the pressure within it becomes less than 10 psi below atmospheric pressure. Determine the maximum value that h can have without causing collapse of the tubing.
•
or
h=
/,3/
fl
FIGURE P3.4B
3 . Ifq
I 3. ljq For the pipe enlargement shown in Fig. P3.lf'1. the pressures at sections (1) and (2) are 56.3 and 58.2 psi, respectively. Detennine the weight flow rate (Ibis) of the gasoline in the pipe.
<21
FIGURE P3./fQ
or
~( 1-(4;f) or
Vt~2.I.'f!}
Thvs, O'Q =
tUld
=
~=!l,~=:e/';f'f(21.'f~/) '"
If2.S"#-, (0.'1-90.[/) = ZO.8!E.
==.$=
0.'190
if
3.50
I 3.50
Water is pumped from a lake through an 8-in. pipe at a rate of 10 (1 3/ 5. If viscous effects are negligible, what is the pressure in the suction pipe (the pipe between the lake and the pump) at an elevalion 6 ft above the lake?
ill e
'I
0J
ilL " 6.0
(/o ij!)
1T(l£fIY
3 -'15
=
II
28.6!f
.3.5/ b
=width =0.06 m
Air flows through a Venturi channel of rectangular cross section as shown in Video V3.6 and Fig. P3.51 The .(1) constant width of the channel is 0.06 m and the height at the exit is 0.04 m. Compressibility and viscous effects are negligible. (a) Determine the flowrate when water is drawn up 004 13'::: 0.10 m 0.10 m in a small tube attached to the static pressure tap at . m 1 the throat where the channel height is 0.02 m. (b) Determine the channel height, hz, at section (2) where, for the same flowrate as in part (a), the water is drawn up 0.05 m. (c) Determine the pressure needed at section (I) to produce this Water flow. • FIG U REP 3 . 51
Free jet
3.SJ
Air
~
·(If)
0.04 m
(n
(2.)
(b)
~2.
t; +;.; 1- +i.f Ih 2
Whel'€
:=
From parr (a)
.f-,.
'1,.2
--.,...-- +
2(q.81!;,)
1'1..0/3
8 uf
Vi AA ::: V'I,44'
(c) AIsoJ
-r
\h :z.
LJ I
-I-
Bvf since {JI ::: f/f
e
2 (Q,8'fi)
2. :=
U()X'J!.3 (o.asml
I or IV2:::
3/ D•
r .,j
sm
ihai r::
(u.,!-) (O.06m) (0. oIf. m) or
fJ!t.
Jj :: ?
+
~ :z.
-ii
h;. ::O.02.53m
where fJ/f -:::() and AI ~ :: t9~ ~
/)1:::: (0.0'1-111)( 0.0&,,) ::: 0
I:;. =-~.oJ
= - lI90 m:z.
(~3.1 ~)2.
:::
.so
(36 ..s.!JL) (o.o6m)h,. (3)
OJ
V.f::: 23.,:s
ThllsJ £1n. (~) becomes - Jf90
='
m
II I
A
.3Ji
lip then V; ~
~ alia EfP. (3) r/ve..s
.3.52 3.52 An inviscid, incompressible liquid flows steadily from the large pressurized tank shown in Fig. P.3.S2. The velocity at the exit is 40 ftls. Determine the specific gravity of the liquid in the tank.
•
FIGURE P3.52
3.53 3.53 Air (assumed frictionless and incompressible) flows steadily through the device shown in Fig. P3.53. The exit velocity is 100 ft/s. and the differential pressure across the nozzle is 6Ib/ft2• (a) Determine the reading. H, for the water-filled manometer attached to the Pitot tube. (b) Determine the diameter, d. of the nozzle.
• FIGURE P3.S3
fJl + rz, +fpV;2-:: A +Z:). ft:pv,.:z.
(a)
where ~ ::: :z2.
J "
:: () J
ona l{::: ()
Thus;
/;1. : : dH.1c1 1/ so ij, aT (62,1/- ~) H i ((). 00:1-38 ~if) (100 #iJ. or fI == p,. -: : 1: p~ t.
J
/;
vi
=:
O. /9/ ff
Hence 61f-,. +t(O,OO').38sX'f)(I/-If."'Ij.)~;: 1(0.00:;"30:'$)~.2 J
or
£i
~::= 83.7 s
so
th4f with III/ V¥ -:: Q4' :' ~
J
-fd~ (83.7#) ::: 0.79s or
d:::; o. /09 ff
3-¥1
,3.5 if 3.54
The center pivot irrigation system shown in Fig. P3.54 is to provide uniform watering of the entire circular field. Water flows through the common supply pipe and out through 10 evenly spaced nozzles. Water from each nozzle is to cover a strip 30 feet wide as indicated. If viscous effects are negligible, determine the diameter of each nozzle, d i , i = 1 to 10, in terms of the diameter, d!(), of the nozzle at the outer end of the arm. Supply pipe
• It o
I
,
r,
I
r2.
•
I
!
I
I
T3
I
r;...,
!
I
r,'
tV~ • fa I
r
r, I
I
Nozzle
I
r,o
(Z)
= (rl.·2.-t; ...:)
(3)
(3fJ02. -'-70"")
f rJ,o -
ai
_
-
,"]4
r,·2.· rI-I t /7/00 J
These reslJ/fs are r;il/en
in
the f.hle.
3-'f9
I.
r:4 H
I
30
0.229
2
60
0.3'17
.3
90
0.51.3
11-
,,-0
().607
5
/50
0.688
6 7
/80
0.7&/
210
0.82.7
a
').11-0
o. seq
q
270
O,Q'l-6
10
300
/,00
x=L
3.5.5 Air flows steadily through a converging-diverging re. tctangular channel of constant width as shown in Fig. P3.55 and Video V3.6. The height of the channel at the exit and the exit velocity are Ho and Vo, respectively. The channel is to be shaped so that the distance, d, that water is drawn up into tubes attached to static pressure taps along the channel wall is linear with distance along the channel. That is, d = (dmax/L) x, where L is the channel length and dmax is the maximum water depth (at the minimum channel height; x = L). Determine the height, H(x), as a function of x and the other important parameters.
x-----"x = 0
;;Q~~l~~~~~;;;::lZ~r~=:;;:,;;..;n~
;: ; ,:
.£.....
III FIGURE P3.55
1-+ z'( -1-1. Pvz. ~ If) 1- Zo 0 4- t (lVo z. where -Z cEo
J
{Jo::: 0 J fJ::: - ~LO rJ.
7hvs
J
},If
- 0H,.o
J "''ltIX
X+
Bvf
AV :::: A0 Vo I}
- ~.~ or
J
()
...L :2.
wners p::: air dells/I,
= - ~D
PV2. ::: 2:I f Yo .z. \I
- Ao V,0 -- 7T flo r V-,CJ
II
Yo
f .s 0 J.1.. In"
~rflX +tp(!J; v.t« i ev,"
H/Ho vs xlL
. - -.
Q
::J: ::J:
1
0.8
0.6
0.4 xlL
3-50
0.2
o
0
*.3,56
I
*3.56 Air flows through a horizontal pipe of variable diameter, D = D(x), at a rate of 1.5 ft3/ S. The static pressure distribution obtained from a set of 12 static pressure taps along the pipe wall is as shown below. Plot the pipe shape, D(x), if the diameter at x = 0 is 1, 2, or 3 in. Neglect viscous and compressibility effects.
x (in.)
p (in. H 2O)
0 I 2 3 4 5 6
1.00 0.72 0.16 -0.96 -0.31 0.27 0.39
x (in.)
p (in. H 2O)
7 8 9 10 11 12
fI~ f D. (f-P. D-V
0.44 0.51 0.65 0.78 0.90 1.00
L .-
ail'
-
p 1) J
_J
J..-x ........
t
-.....
_:Ih water rN.z.O
If+ ¥(+zo =- f +lj +Z
where
J
Zo"::Z
n~ £ 2 V= 1'i + 2('9-/)' 'llilh v.::: Q = 1.5 ~ V0 P D ~ :It D2. If 0 and _ 62. If -Ita . fo-f= ~.ao (ho-h) - 12. 1 (1 m. -h) LJ nence
J
WI'/'L II
V= [(.J,iJ.)2.+
D:
-3.
p::: 2.38 X/O
1}.
j [3.&5 Do'"
Also AV:::Q or IllV:::(f
so
J
- [¥(J'#fq~ -
if~]~ ,... D-- [ 7l'V D-=
7T.$
t382
[3Dl65 t'l370(J-h)J~l ~
D2 0
c:
.£l whereD.",fI S
0
J
s.20(J-h)
~Jw,:jh h"'in.
11 we 11'oOTain
pf3
10.'+ (/-h) k2 =
2.38 x /,,3
- 1.91 -
~u:2.
-l- '1-370 (j-h)
fhtrl
1.38; VV
-
(J)
J
or when comhined wilh
Et{,{/J
.
H where Do'V ff h,., /1). J
J
+J
Plot D= /)(X) with Do:= if J and f fl) vsiIJ9 fhe va/lies of /,::/'(X) fro/TJ the lahle. Nofe : h i.s the SQIIJ(; a I/(J On. liz. 0)'1 in Ihe tab/e. AII EXCEL pro9r4m W~J' used If) ob/fA/1i fhe fol/oUl/II, re.rvl/u.
(con'-fJ 3-5/
(3.)
~3,.s6
r (con 'f) 0, ft (Do = 1/4 ft) 0.24996 0.20277 0.16776 0.13999 0.15299 0.17245 0.17841 0.18123 0.18558 0.19616 0.20944 0.22710 0.24996
x, in. 0 1 2 3 4 5 6 7 8 9 10 11 12
0, ft (Do = 1/6 ft) 0.16664 0.15733 0.14435 0.12870 0.13667 0.14649 0.14902 0.15015 0.15179 0.15537 0.15911 0.16300 0.16664
0, ft (Do = 1/12 ft) p, in. H2O 0.08332 1.00 0.08299 0.72 0.08234 0.16 0.08112 -0.96 0.08182 -0.31 0.08247 0.27 0.39 0.08260 0.44 0.08266 0.08274 0.51 0.08291 0.65 0.08306 0.78 0.08320 0.90 0.08332 1.00
,
Dvsx 0.30 -
j I I
i
0.25
~---
",
0.20
= c ~
0.15
.
·------1 ,
I
~I
:I
9
I
i
I
. ..
I
-~
. ,....' .......... "
..... .,.
...
i .. ••_ .. -1- .:
.
-------,~-.;;.~~ .",-
- -
'
... '" ... t ,._._JI H
__•
~
-
I I
I
--1
I
I I
I I
0
2
,
~!
I
0.00
-------.1
I
i i i I
, ,
-
!
0.10 - --.-~.--~+-----,
I
-1 I
_-!'-
'-":
0.05
,I
I
I
4
6 x, in.
3-052-
8
10
12
- - - 0 - -0 -0
= 1/4 ft '1 =1/6 ft I =1/12 ft
3.57
I 357
The vent on the lank shown in Fig. P3.57 is closed
and the tank pressurized to incr~se the f1owrate. What pressure, PI' is needed to produce twice the flowrate of that when the vent is open'?
FIGURE P3.57
Wi/;' ihe veil! open:
b
+~
•
r 2.!
+2
TIIVS, 11,' z , = -Zj L
'
A.
1-:' -I-&Z +2..
= J,;. r Z?
.
~~'------'
or
To hove dovUe fhe f/gwrafe /'liM Ihe veil! clfMed (j'l 10):
• ./.Ju, ==..Y:. , ~
where fot'th;.s uue
ThVJ} Il
(SO.8!t/
t2.'1''fP
2.(.32.2 Si)
(< ~',J,) t/ofl ==
or
It ==
fl
Ih . /87671'- = 1:l.OpS,
3-53
1Iz.: 2. (:z.s./ffi) ~ 6
.>0.8
II
- 3.581 3.58
Water flows steadily through the large tanks shown in Fig. P3 ..sS. Determine the water depth, h A •
A
For sfeady (low)
QI/- ~!9~
~
w/fA
Also;
-1.1-+ ~:L +z r ')..1 I
.so fhaf
V.2.
=,f2fh//
Thus}
1i2.~ ==cr~
or
f(O.03hJi";2(9.81f;.)~j
3-5'1-
= O.O/J-31}3
/
0.5 in. Hg vacuum
0.6-in. dja~eter
3.5'1
Air at 80 of and 14.7 psia flows into the tank shown in Fig. P3.59. Determine the flow rate in ft 3/s.lb/s. and slugs/so Assume incompressible flow.
(I) Q
•
~
+
.(2.)
t
~ump FIGURE P3.Sq
3-55
3.60
I
'J
(0)
.-.-•.-:::-'.' •.'.'.'-:-' :j:-
3. 60 Water flows from a large tank as shown in Fig. P3. 6o.. Atmospheric pressure is 14.5 psia and the vapor pressure is 1.60 psia. If viscous effects are neglected. at what height, h. will ~avi tation begin? To avoid cavitation should the' value of DI be mcrea~ed or dec reased'! To avoid cavitation should the val ue of D~ be increased or decreased? Explain.
FIGURE P3. 60
where f10 ; 1'1. 5psio. ,/'t ~ 1.60 pSia, E.; h, Z, ~o, and V. ~O (I)
How.ver, A, V, "11~ V,
Or
11
=(
where
z: r~ wah
( D.)'"h.
= - D,
(2)
Ih( 1'f1fWin?) ( IIf.5 -./.60 ) In?
'h [
6:z..'f N'
e::J -'J .. •
,1
From £y. (J) il is seen fhaf h increases in increasing D, and. c1ecreo.in9 D:z.. Thvs, fo avoid cavilaf;.n (,; ~. to have h slnall enot/9h) D, should be increq.sec1 and 4. decreased.
3-56
(3)
3.61
Water flows into the sink shown in Fig. P3.61 at a rate of 2 gal! min. If the drain is closed, the water will eventually flow through the overflow drain holes rather than over the edge of the sink. How many OA-in.-diameter drain holes are needed to ensure that the water does not overflow the sink? Neglect viscous effects.
=fy +if +%2. V1 .. V2,~, Also) Q ::: n A2. 14 :: n ~
, where
It "0, V, =0, Inti. .21"ti fJ.. =0
·=[2(32..2. Jj) ( 1+/~2. ff)] ~ =
2 .5i'
f d; V:z.
J
ff.
where n:: IJlllflher of PfJles reqlJired,l Q2. := o. 'I in; and. C;::: c,nfraction coel. = 0.61 (see ri'1.3.19)
ThlJs~ with J Q 2 w,( I ""'" ) (~31 in. )
=
111111
60.s
19GJ
(_
I fl~ ):::
17'-8 m.3
3-57
~, ~tx/O".J fi.! .&
)
3.62
What pressure, PI' is needed to produce a flowrate of 0.09 ft 3/s from the tank shown in Fig. P3.62?
3.(i2
FIGURE P3.62
If + :i +
Z2. "
1- +{f +j!~
where -/'2.:: f1 + ~ h J
"Z2 :: 3.6 fl J Z3 =0
ThusJ
and
~::tO
11 + Ooh
o
where
or
31.8-¥
o( 2~
\lg2.
or fJl ==
- Z2. ) -
t"h
== (1./
(6J..1f
Ib
7i;»
[ (3
I • 8 £:!. ~ ) 2-
2.(32.. 2.
- '12.5 ~ (2,0{-0
7'1-6
~
- 5./8 ps/
3-S8
fJ3 -: 0
~~)
3.63 , 3.63
Laboratories containing dangerous materials are often kept at a pressure slightly less than ambient pressure so that contaminants can be filtered through an exhaust system rather than leaked through cracks around doors, etc. If the pressure in such a room is 0.1 in. of water below that of the surrounding rooms, with what velocity will air enter the room through an opening? Assume viscous effects are negligible.
~~
v,~o
V2. (1)rL ........-. ..----.-..--.
(I) -/71111111/77 -
I { vise oV.s e (feefs
tJ,,~
v.2. I,. + ..L,.. f v,. I, +'2I P,::
neqligihle
1/2.
J
'"
I '
here V, ~ 0 I
4fJd
It -/2 :: i'azo h
or
p, - f:J.::: (~fI)('2.1f {t3) ;;0.520 ~
Thus
J
~ ==[ 2 (prp,.J ] ~ == [ 2.
e]
2 (0.520 Ib/I/1.) O.OO:J.39SIIJ1s/113
3-SQ
]~ -:=
20.9
Ji .s
3. 61f
I
3.6/f Water is siphoned from the tank shown in Fig. P3.6'f. Determine the ftowrate from the tank and the pressures at points (1), (2), and (3) if viscous effects are negligible. (1)
•
FIGURE P3.6.1f
fJo
of
~
'v" 2. + =z
~~.L.-o
::
-(1,- +
r
For t';:: 5 and 1'0::: 0 J Va i!.0 = + Es or
:;2.
"
(I)
r
Of' 6 ::
~'I;; ~
fJs =::0 flus become.s 2 ~ ( Eo - Z.s)' =1,--2-(3-.2.-2_~:~:--)-(3-ff--"')
=-0)
Vs"";
I
= /3.'1
-,-L I
~.2.. + 2. f '
~
n/Js)
l
Q:: /}s 'Is ::: .;-( -Ii: fl)'L(I.3,q fj) = 0.303 ~ Fro/IJ £r.(I)
with
l-::/ a/Jd
t{:::o J
:Ij
~
= o(Zo-Z,)
~ =(62,Lf ffa)( 8 (I) :7l9?fA
i,
Fro/1J E'[ II) wifh /=2 J LO = + E:z. where 1J2. ~ ::: flo!' ~ ~ ~'SJ/Jce f}2::' 115 if f()llows that L{:::~ or ;; = =20-Z,5 .j.
2;
Thus}
1- =
~2.
Z!o -22. -
:2.1 =Eo-%:z. -
~!1.
,.;
=Zo -22. -(Zo-Es ) =i!
or
:5
-22-
11 :: t(zs -~2.) =(62,1- -/fo)(s if) = 3/2-ffo. From
Er{ (J)
wifli t''''3J
20::
where fla ~ :: lis ~ Since 1).3:::115 if fo/lows
Thvs J
:fi; = Z!~ -2'3 -
thai
2.
~
Jf + ;:f+Z.3
~2.
~ t:: Vs or ~;
~ 2.
=-
2..1 =&0 -Zs
2-
:
Zo-E3 - ; ; :: %. -2.3 -(Zo-Es) :; Zs -
z,3
o~ =t(zs -%3) =(62.'1' ~~)(-.3fI) -/B7~ :0
3- 60
3.65
J
3.65 Redo Problem 3. 61f if a l-in.-diameter nozzle is placed at the end of the tube.
~)
fJo
i' +
\4, 2.
_
2.7
+Zo -
For l ~ 5 :Eo =-
2.
y~
Thus J Q =A.s
..t2.
\/,.7.
!0::: 0
and -/b =0 J or 11'
._/
Ter t -
fJ.s
J
•
S
(I)
,/°°)
=0
fhls becomes
~ =- 12.9{~.-:Z5)1= {2.(3Zo2fj-.J (3ff/ :::
~ == T
r
•
~ +?-9 +%1
(1)
4-
/.3.'/
(1)2. ii) _ :£:L3 -;z:ff (/3.9 oS - O.07S8 oS
From E'f(!) w/fh i=/ and
it;::o
J
A:::
a(zo-Z,)
=(6Z./f-#r3)(8f-l) == '1991-2-
Fro/h £i/o (J) wilh i where Since \J _
V.z -
Thus) fJ:;. ~
~2
~ + -¥!i + ~2.
i!o::
J
112 V:z. :: A5 ~ A:z. == (~y2.A5 = (t)~.s =LJ.As iT follows that .L ,/ J!:l _ J- (-1- 1/)2. _ J ~2 _ -L ( z ) ¥
V.s
or 2~
-
29
If
~
-
7D~ - /6 Zo -
oS
2-
= Zo -Z2 - ~
-:=
tI
or
2'0 - Z2. - -!r(:ZO-Z5)= afl-!t(3ft) == 7. 81 ff
Ih ) -Ii =(62 JI- ff.i) (7. 81 PI-
From fro (I) w/lh i = 3 where f}.3 ~ e.As or s/nce 113.=!J2.
J
::. 'f$ 8 20 ==
Ib 7P-
:t; ~; +Z.3 oj.
Vs
then ~::: ~. and
~2
2;
~2.
=
/
2.~ = 16 (zo-2~J
Thu,s ) ~ ~2. - ; ~ L O - Z 3 - 2$ :::Zo-~3-
.-l.. J /6 (Zo-Z5) = -76 (3 {-i) = - /~ PI
or fj
77fi.
Jl;..
3
== (62.Jf fP.) (-Tl"H)= - II.
_~-61
Ih
I 3.66
Determine the manometer reading, h, for the flow shown in Fig. P3.66.
T
0.37 m
h
1--I--H--~
1 . . . -_
... ----,,M'·-~ ---
a08m ... -~ dlameter(J}
====
'c:::=:=l
I
0.05 m diameter
Free jet
..
FIGURE P3.66
where
Z, =:Ez , ~ -:: 01 and
Vz = 0
Thus)
f, : : f:;. !lowe Ver, so Ihot
f1:::: O'h and 122.::: 0' (0.37m)
h = 0.37 m
3 -6z.
3.67
I
t
.-.,t::::;'" (f)
£1
3.67 The specific gravity of the manometer fluitl shown in Fig. P3.67 is 1.07. Determine the volume flowrate, Q, if the flow is inviscid and incompressible and the flowing fluid is (a) water, (b) gasoline, or (c) air at standard conditions.
o.~
(2.)
diameter
~
m
-~-
10 mm
4,..--' ....
0.09-m
t
20 mm=h
J ...,.j
~
FIGURE P3. 67
fluid fa)
Wc.i-ler (b) ~a.so//ne (e)
.
air
kN O'J m3
m3
Q)s
-.3
9.80
/.06 X /0
6,61
.3 . 02. X /0-.3
/2 X/o- 3
O.UB
3-63
3.68 3.68
IP-4 fuel (SC = 0.77) flows through the Venturi meter shown in Fig. P3.68 with a velocity of 15 ftls in the 6-in. pipe. If viscous effects are negligible determine the elevation, h, of the fuel in the open tube connected to the throat of the Venturi meter.
v=
15 fUs
FIGURE 1'3.68
where and
Z/::O
If:::
J
Z2. ::::
.f} f-l,
15 {-I/.s
or
1!f=-7. S3 ff Buf if=-h so 3.69
that h == Z 53 ff
3.69
Repeat Problem 3.68 if the flowing fluid is water rather than IP-4 fuel.
Note from the solution fo Prob lem 3.68
fhai fhe
value of 0 is /Jot needed, Thus; h = 7. S3 ff eft her waf er 0 r J P- if rIJeI.
3-6Lf
for
(I)
3.70 3.70 Air at standard conditions flows through the cylindrical drying stack shown in Fig. P3.70. If viscous effects are negligible and the inclined water-filled manometer reading is 20 mm as indicated, detennine the flowrate.
FIGURE P3.10,
However) or PI
By
fJz +~J2. +
Ymh == {1- '0(1- h-~)
where
;12. ""( ~ -I)h +i
h =(-zomm)sil)l. (2)
comhinilJg £rt's~(!) and (2)
1;/2. = (!p- -/)h or II, VI
=
2
~
~ ( /5 ~ - J) h _ -
Thus} Q::: AI 'It ==.:If D/-I/;
2 (9. 8/ ~)
9. 8tJ X/0 3
,*3
/2 .o=!b
--------~--/~5~~--~-------
2.35
~
==f (2",)2.(2,35.r;)
3-65
====
.3.71 3.71 Water, considered an inviscid. incompressible fluid, flows steadily as shown in Fig. P3.71. Determine h.
• FIGURE P3.71
+l-pv,2. ::: f:z. f rzz.+l:pV/1. IIi where ~ '" 0 %;L::.3 (I, Va '" 0, and ~ '" -1;:= ; ;f/J'"
f,
-I-
l~
I
= S. 09
f1
oS'
IhvsJ
1'1 + t (1,9'1 !'ff;) (£.()9 fj l' := 12. or " - f"
=: /
t:J.
+ 62.
oil-if, (3 f.I)
-Jf,.
(I)
B,,1 from fhe h14f1omelerJ (JJ -¥rJ-l3flJ or f,-d2,1f
.J.
a(h+/.) ::: Iz
-fp (3ft) ~62.Jf ~ h = /,3-
lienee,
f, : :
f% + le7 -
62.'fh
wlll~1J
whslJ C()/JIbilJDd wilh £t. II) ,ive.r
It.. + 117 -J2.'fh - f1~ -:: / bJ. Qr
h=
OJI-ooff
3-66
3. 72.
I
\1
(/)
~ ::----:::::--_-:..:-:---::=:= :..::=::-- -
3.72 Determine the flowrate through the submerged orifice shown in Fig. P3.72 if the contraction coefficient is Cc = 0.63.
--:-::=:=::----:..:-:-:-::=: 1('2.)
6ft
(.3)
=f='f
4ft
3-in.
diameter ...
2ft
·f
/
FIGURE P3. 7~
where
3-67
fJl:::O J ~:::O J zl::.'1f~ Z2.=O J qnd =2 ff
'l
3.73
3.7;3 Determine the flowrate through the Venturi meter shown in Fig. P3.73 if ideal conditions exist. 'Y = 9.1 kN/m 3
FIGURE P3.73
.£J..+ ~l +Z =.:b:. + v;.2 + Z '0 2.i I K' 2-9 2. ThlJs) tl+
r
(~rtv; '2-~
(73S
-s50)kPa
2(9.8/~) (9.1~) m
"
/9 him )If I - t-I 3/mm
3-68
== ZI,5!}-
I
3.7'1-
\ 3.74
For what flowrate through the Venturi meter of Prob.
3.73 will cavitation begin if PI ., 275 kPa gage, atmospheric pressure is 101 kPa (abs), and the vapor pressure is 3.6 kPa (ab')?
\
t '---.)~,k'----1\ I
Q
- - :)31rm.cn
.(') 19mm /
\
')' _ 9.1 kN/m3 /
(I)
.£!.. + WI. +.i? =.:IJ,.
r
~'J
+
It..\ z
r 2.7
'
2
Thvs, wl/h A, l'J' ;: 112 Vt.
where Z," 1!:2. and
)
~ =-3.6 KI'~
11 =(pH/OI)kfQ(Qbsl = 376 kPa labs)
or Vz " 30.6 -!}
Thvs, Q=/12 v.. =v.".Ll"
2.
v,"" ' )" m) -3 m 3 :z. = -:y:- (0.01'1 m (30.65 =8.68 X/O "S
3-69
3.75 3.75 What diameter orifice hole, d, is needed if under ideal conditions the ftowrate through the orifice meter of Fig. P3.7S is to be 30 gal/min of seawater with PI - P2 = 2.37 Ib/in.2? The contraction coefficient is assumed to be 0.63.
FIGURE P3.75
2. ( Ib) (.3.06 .fi) + 2. (32 .2 .£i) 2.37 X11I-'11p· .s oS 2. 6'1: 0 J.li . ft3
or
v,. == IB~8 fj
Thvs J since
f d2. ~ if {QI/ows fh47 ~Ji=[ 17'¥ x(O,63)(l8,Bf) 0.0668 .~·e J~ o.08'f7ff J
Q = 112. Vz = ~
d. =[
'IQ 7l'Cc
V,.
3-70
:=
1.016 //1,
.3.76 3.76 An ancient device for measuring time i~ shown in Fig. P3.76. The axisymmetric vessel shaped so that the water level falls at a constant rate. Determine the shape of the vessel, R :::J R(z), if the water level is to decrease at a rate of 0.10 m/hr and the drain hole is 5.0 mm in diameter. The device is to operate for 12 hr without needing refilling. Make a scale drawing of the shape of the vessel.
is I
-!l~~J.o-mm diameter
,
~, l~
FIGURE P3.76
the flow l.s assumed fo he tj /I tJ. oS i - .s Ie Q rly, jf
z, = z./ Thus) \42. = Z?
'vj2. t
2?
and Z2- ~o
which} if ~« ~ (i, e. R» s, OhJhJ~ become.s
z
'4 =y2IJZ' Since AI Vt "Az ~
and l{ = I~I = 0./ T,.(-3,/6~~.s) == 2.. 7B X /0- ~ 5
we ob1ain 1T R2 (2,78 X10
5
.p-) = -f-(o.oosm)2.
where Rand z are
J2 ('1.8 J1£;.) Z
J
~m
Thu.s,
B = o. qqB zu;'
or z. m
R.m
0 0.02 0.05 0.12 0.22 0.32 0.42 0.52 0.62 0.72 0.82 0.92 1.02 1.12 1.22
0 0.375 0.472 0.587 0.683 0.751 0.803 0.847 0.886 0.919 0.950 0.977 1.003 1.027 1.049
Vessel Radius, R, vs Water Level, Z
1.0 0.8 E N
0.6 0.4 0.2 0.0 0.0
0.2
0.4
0.6 R,m
3-71
0.8
1.0
3.7B* A spherical tank of diameter D has a drain hole of diameter d at its bottom . A vent at the top of the tank maintains atmospheric pressure within the tank. The flow is quasisteady and inviscid and the tank is full of water initially . Determine the water depth as a function of time , /! = h(t) , and plot graphs of h(t) for tank diameters of 1, 5, 10, and 20 ft if d = 1 in.
~t~\ZI =
1+ ~i+i!z
where fJ; ='0 J fJz" 0 J z/ "'h, Zz ='0 and Thvs) . Vz jh' which when cumbined IIIdh
=12.
At ~ =' Az ~
-II/I/t =11:.y7.1l or - rr r:1.4f = ¥dzY:l.jh' where R:1.:= rZ +(h -R)' ~ wdh
R,=4- =radivs of lank
Thvs)
r=v R'--(h-R;'-'
so
h-R
flJflt
;t '"
-[R'--(h-RJ'J 4; 1/2.1h or (h 3/z, _ 2Bh li) dh = d2.fi Jt t
h
Yh%.-2Rh'1.)dh
=
dZ.pfdt
(I)
w:;
£s.{/)
beco~i9.s
whid elln 6e infe9rafed from the il'll/tfil limo aAd deplh (t=o, h '" 2.R) fg dII or6drdry f;m~ aMdepll! (UI) as
2R 0 or . d2.1/2i 1;(H~i_(2.Rt2)_::R(h3/z._(2Rf2) = t
:7
Use d= ,i ff (wi j=32.2 ~ and pM !J=h(I) fu/' valves of R = o..s J 2.S J..s:; Q/ld III II !Vole: .It is easier to solve £r. (2.) as t s t (h) rafhe r
fhl/n h =hit)
3-72.
?IPes
(2)
J.71A. I (con'/.)
Results of an EXCEL Program to calculate h(t) from Eqn. (2): D=1ft t, s h,ft 0.00 1.000 0.09 0.950 0.35 0.900 0.77 0.850 1.34 0.800 2.05 0.750 2.89 0.700 3.84 0.650 4.91 0.600 6.06 0.550 7.30 0.500 8.60 0.450 9.94 0.400 11.31 0.350 12.69 0.300 14.06 0.250 15.37 0.200 16.61 0.150 0.100 17.72 18.62 0.050 19.14 0.000
D=5ft t, S h,ft 0 5.000 5 4.750 19 4.500 43 4.250 75 4.000 114 3.750 161 3.500 215 3.250 274 3.000 339 2.750 408 2.500 481 2.250 556 2.000 632 1.750 710 1.500 786 1.250 859 1.000 929 0.750 990 0.500 1041 0.250 1070 0.000
3-7.3
D = 10 ft t, s h,ft 0 10.00 28 9.50 110 9.00 242 8.50 422 8.00 647 7.50 913 7.00 1216 6.50 1552 6.00 1917 5.50 2308 5.00 2718 4.50 3143 4.00 3577 3.50 4014 3.00 4445 2.50 4862 2.00 5253 1.50 5603 1.00 5889 0.50 6053 0.00
D = 20 ft t, S h,ft 0 20 158 19 620 18 1370 17 2390 16 3661 15 5163 14 6876 13 8778 12 10846 11 13055 10 15376 9 17782 8 20237 7 22706 6 25144 5 27502 4 29714 3 31695 2 33311 1 34239 0
Water Depth vs Time D=1ft
Water Depth vs Time D=5ft
1.0
5,----:--~-----,--..,..........-...,.----.
0.9 0.8
I
!
0.7 0.6 _ 0.5 -= .c 0.4
:-i-=:.---"<=~ -=~--=:~---'----i--~---l
----t--
3
--~-
+---t---T---+----r---r-----+----1
-= 3+---~--~--~---+-------~
.c
2
+-_~--4-~~-~_--+--~
2 +----+----'-
0.3 0.2 0.1
O+---+--+--+---+---;-..L--!
0.0
o
10
5
15
20
o
200
400
600
t,s
t,s
Water Depth vs Time D 20 ft
Water Depth vs Time D 10 ft
=
=
10
,----~-----,-----r--,
20,------..,..........----~---------~
18
9
8
800 1000 1200
-t\---~-.
----:-----1I !
~:~--~--~~----~:--~~
-1---\_ _-
7 6
12
5
-= 10 .c
4
3
2
+------~--~-~--+-_4
,
,
--II
:~--~--~--~--~ 2
o +-------~----~r_-----4--~ 6000 4000 2000 o
I
--I
O+----~----_r--~~~~
o
10000
20000 t,s
t,s
3-79
I
-+----1
+--~--'<+--
30000
40000
3.7q* An inexpensive timer is to be made from a funnel as indicated in Fig. P3.7C(. The funnel is filled to the top with water and the plug is removed at time t = 0 to allow the water to run . out. Marks are to be placed on the wall of the funnel indicating the time in 15-s intervals, from o to 3 min (at which time the funnel becomes empty). If the funnel outlet has a diameter of d = 0.1 in., draw to scale the funnel with the timing marks for funnels with angles of 8 = 30,45, and 60°. Repeat the problem if the diameter is changed to 0.05 in.
r ~9V: +2
.£L + VI 2. + Z ~ 2'1 I
Plug
FIGURE P3.7q
= -J2.,. +
where "'::: 0
~
(I) ;J.
:if =- oJ and ~:::!*- « ~
J
22 ::: OJ
=:
0
J
11
-t
ThIlS, if R » \{ ~ ';2tjh' which when combined wl1h
-IJ,-# :::A2./zgh'
-7TR
or
where R =h ftlne Th!1 s J £'1' (I) become.s
or
Ct.1II
of t~o as
t
Yii Sdl: J(h3/3.'d' . n = _ l/-d1.tlJn"8 L
ho
--.J (I)
- h2. tan1-e ~ == /-V2.gh'
9
h
-l ~~ A~ :::A~ li 9ille.s
h
1!f =fd:;i~9h'
2
h / 2 dh = ~"f{8 di which III)
I
r - R--..t
2-
or 1f
De infeprqied From h =ho
[5~
54] -_-
h - h.
d 2{ij
'f
flJll~() t
0
ThvsJ
.
h= [h
S
5/2 0
-
d2. V2i"'t] 8 Ttln2.e
.
.2/5
Since h ==0 wnen if (ol/ows thaTJ
},
t = 3m//)
(2)
== /80.s
!J) (180 s ) . ho ~. = .8 +an 28 ~ w/fh which when combined £'1' (2.) gives :::[ S d 212 (32'~#:l./iIBOS)J (I _ --L)Z/S h 5A
sd2.1(2(32. 2
8 tan2.8
or
h = /5.2 (
rJ. fan
¥/.s
e)
J
.
t
(1-/eO) (con'i) 3-75
180
2-,1;
where
h- ft
(3)
J
d~nJ and t~s
The results of an EXCEL Program using Eqn. (3) to calculate h as a function of t are shown below. The time interval markings for the sixfunnels are shown in the figures on the following page. d=0.1 in.,8=30deg t, s h,ft 0 0.512 15 0.495 30 0.476 45 0.456 60 0.435 75 0.413 90 0.388 105 0.361 120 0.330 135 0.294 150 0.250 165 0.190 180 0.000
d = 0.05 in., t, s 0 15 30 45 60 75 90 105 120 135 150 165 180
e == 30 deg h,ft 0.294 0.284 0.273 0.262 0.250 0.237 0.223 0.207 0.190 0.169 0.144 0.109 0.000
d = 0.1 in., e == 45 deg t, s h,ft 0 0.330 15 0.319 30 0.307 45 0.294 60 0.281 75 0.266 90 0.250 105 0.232 120 0.213 135 0.190 150 0.161 165 0.122 180 0.000
d = 0.1 in., t, s 0 15 30 45 60 75 90 105 120 135 150 165 180
d = 0.05 in., e == 45 deg t, s h,ft 0 0.190 15 0.183 30 0.176 45 0.169 60 0.161 75 0.153 90 0.144 105 0.134 120 0.122 135 0.109 150 0.093 165 0.070 180 0.000
d = 0.05 in., t, s 0 15 30 45 60 75 90 105 120 135 150 165 180
(colli) 3-76
e == 45 deg h,ft 0.213 0.205 0.198 0.190 0.181 0.171 0.161 0.150 0.137 0.122 0.104 0.079 0.000
e == 45 deg h,ft 0.122 0.118 0.114 0.109 0.104 0.098 0.093 0.086 0.079 0.070 0.060 0.045 0.000
six funnels (d ~ O.oS in. or
: i [/ I ' , ! Z i i _ I : Y ; , ;. -h.. " i!: ' -c-+ 1 -i:; 1-+-+-i--1C-+-i--+-#:-++-r+7!V~!+-t-+-'r-+-t .'" ':?-+-t-J-----r-t-';, ~. --l-+- H---+-.L-+ I: ,I I i / ' , .v, I /, -H-+-~t-:.'. ,: ~+-
0•
'.' I l.i 1 .----rrl
I'
/,
I ii'
~..i-H ~;-
, v,v,, 'I/~.
if
y-: ,I
IV
I
I~
.Y
1/111 ;
~ /'
vr· I!./'
I!
:: :
I '
+
~-t-i. , .. , I 'I
I,
II
I
3-77
"
:+r-~'+'~
.'
.
+++ .
j-L
~ ~.' ...l.. f-C--,.'1-~- ...:......:iT,. . •'
3.80 3.80 The surface area, A, of the pond shown in Fig. P3. SO varies with the water depth, h, as shown in the table. At time t = 0 a valve is opened and the pond is allowed to drain through a pipe of diameter D. If viscous effects are negligible and quasi steady conditions are assumed, plot the water depth as a function of time from when the valve is opened (t = 0) until the pond is drained for pipe diameters of D = 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0 ft. Assume h = 18 ft at t = O. h (ft)
D
FIGURE P3.80
A [acres (1 acre = 43,560 ft2)]
o
o
2
0.3 0.5 0.8 0.9 1.1 1.5 1.8 2.4 2.8
4 6 8 10 12 14 16 18
12.::: 0 J z; = h J 22 =-311 and ~ ;: - j <.< ~ which when combined with A~ =112. ~ where -Ii =0 J
OJ
Note:
t
~
-2-
D
( con/f) 3-78
( cO/J'I)
An EXCEL Program using a trapezoidal integration approzimation was used to calculate the results shown below. D
=0.5 ft
D
=1.0 ft
D
=1.5 ft
D
=2.0 ft
D
=2.5 ft
D
=3.0 ft
h,ft
A, acres
A, ft2
t,s
t, s
t,s
18 16 14 12 10 8 6 4 2 0
t,s
t,s
2.8 2.4 1.8 1.5 1.1 0.9 0.8 0.5 0.3 0
t, s
121968 104544 78408 65340 47916 39204 34848 21780 13068 0
0 32181 59530 82354 101536 117506 132412 145035 153988 157704
0 8045 14882 20589 25384 29377 33103 36259 38497 39426
0 3576 6614 9150 11282 13056 14712 16115 17110 17523
0 2011 3721 5147 6346 7344 8276 9065 9624 9857
0 1287 2381 3294 4061 4700 5296 5801 6160 6308
0 894 1654 2288 2820 3264 3678 4029 4277 4381
The graph for D
=1 ft is shown below.
The shape of the curve is the same for any D.
Water Depth vs Time for D 1 ft
=
20 18 16
~
~
14 12 ~ 10
~
~ ....
I~
~.....
.c
8 6
4
~ ~
'\..
2
1...
o o
10000
20000
t, s
30000
40000
3,8 I
I
J.8'
Water flows through the branching pipe shown in Fig. P3.81. If viscous effects are negligible, determine the pressure at section (2) and the pressure at section (3).
A3
= 0.035 m2
~_------, ;:3 =
10 m
(3)
V2 A2
= 14 m/s = 0.03 m2
(2)
10
(7-):
where
or
(1,. ::: 2,52 XIO
s
Z,:: ~2. -= 0
Qllri
.f-,. : : 2S2 kP.
Alon9 fhe sfreamline from (0 10 (3):
t +if + ~1
Z,
= 7' t 4 + 2,3 where ~'Il
~&
Q3 : : A.J v., =~ .. Q;l =CV, - At. V2.
Sf)
Since
rt, =Q,. +Q3 then
thai
~::: Q,:A1.V2. _ 1111% -O.03","(/¥1II1,s) =16,t,.!R-
A.3
-
TIllIS) £'1' flJ becomes 3 .300 X 10
or
(wtlh
ZJ ~O J Z,J <# 10 11J)
#/.,2 + flo 1II/.s)'J.
9.801.,03 /'11m3
I.J::: 1,lifX/OS f;.
~
0,035 / ) ' J 2 . .
, _
{J3
2 (tl. 81 rnl.s2.} - 9.f}ox/ul Nlm 3
: : JIJfkPa
3-80
+ (JI•. tWJIs)'J.. +-Iom 2-('1,f)1 mIs')
(I)
3.6:2.. Water flows through the horizontal branching pipe shown in Fig. P3.82.at a rate of 10 ft 3 /s. If viscous effects are negligible, detennine the water speed at section (2), the pressure at section (3), and the ftowrate at section (4).
From (I) 10
-
A3
=0.2 ft2
V3
= 20 ft/s
"------..-
(2.):
From (!) +0 (3) ; (Jo~l
_
2(32..2 f{)
or
Ih
~ ::: 1/50 ff'- ::: 7.98PSI
Also} Q~ :::: Q, - Q,. - Q3 or
() U{1f
:::
IP lOs-
:::
Q, - II:;. \{ - A3 ~
tf ) o.o7fI'J. (2.9.0-:s
3
f fi) - 3, 9 7 Lst - 0.2 f'.L2. l. l20 oS
.
3.83
T
3.83
Water flows from a large tank through a large pipe that splits into two smaller pipes as shown in Fig. P3.83. If viscous effects are negligible, determine the flowrate from the tank and the pressure at point (1).
7m
L
0.02 m
FIGURE P3.,B3
t:; + If
f Zo '"
n~
:If 4- lJ':.,. 4.
where A" 0 I 1'2. ,,0
Vo '" 0
I
J
zb" 7/1)
~d~=~
J
Ik ::';21 (Zo -Z20)' = f 2. (9.8Ilfo.) (7- ~)ml
::::: 7, 67.1f
!Similarly
~ '~ Ir+-7-7-(:2-0 --Z-~---,)I = ,;2 ('1.8/ fA) (7 m)' = / /. 7.1f
Thvs J
Q:: Qz.'" Q3
::
flt\;2.
-I-
fD;~:J.
or Q:: f[(O.03h1)"(7.67~) .J.(o.02.m)~(1I.7.p.)J ::: 9./ fJx /o-.3.!J!
Also) -h. t r
Vg=z'f'
~I
70
-
.z;
-
or
.fl. Vt'- +'ZI r + ~'i
uJhere
"':"
or
f1 ::: 5 7. 9 kPo.
3-81-
0
and 3
\/,:: Q == 9.I()xlo-
I
_0~[z0 - 2QV, 2] =7.fo)l./O n J:L. f1 m3 . T 3
2/::::
I
7m -
s (0. 0.5 11J)2
-5 (Jf.63!f- )2.] (0 0(.')
-;z;; 2
.l!J!
h8Js~
=- ~63.1!L •
AI =s: 7?X/O.¢4 m
.s
3,8Lf 20 ft
3•. 8/f Water flows through the horizontal Y-fitting shown in Fig. P3.'6/f. If the flowrate and pressure in pipe (1) are QI = 2.3 ft3/ S and PI = 50 Ib/in. 2• determine the pressures, P2 and P3' in pipes (2) and (3) under the assumption that the flowrate divides evenly between pipes (2) and (3).
Ql-"
FIGURE P3.9/f
~(2) Q2
3-83
3.85 r 3.85 Water flows from the pipe shown in Fig. P3.85 as a free jet and strikes a circular flat plate. The flow geometry shown is axisymmetrical. Determine the flowrate and the manometer reading, H.
tH 1-0.1 m I Diameter
-1-
.I
1 -II (2)
--~~r«~=~ \\ I
0.4 mm
0.2 m
i. 'J III r(/) i,
111
O.Ol-m
diameter
PIPe/blJ
tQ FIGURE P3.85
where 7l4.h
11. If
Hence Eq. l
0)
D2.
(I) \I V2
I
qive.s
(J.6ov"f ·:: ~2. ~ 2,(fl.81-!k) (0. 2./IJ) or v" = I.Sq.lf .so fhtrf r; .: ~1 v,. = 71 (0,/ PJ)(t;.X!O-IIIh ) (I.sql}) :: 2.00X/O-'" J
3- 8Jf
.p-3
3.86 3. 86 Air, assumed incompressible and inviscid, flows into the outdoor cooking grill through nine holes of O.40-in. diameter as shown in Fig. P3.96 . If a flowrate of 40 in. 3/s into the grill is required to maintain the correct cooking conditions, determine the pressure within the grill near the holes.
FIGURE P3.86 9 holes, each DAD-in. diameter
Q = qA.z v.z
where
Thvs~
~ = ,..
-¥Q
Q
-
9~
971D,2. :2
A/so J
£!. + Z'VilJ2.+ ~ I =..ti:. + )f' K' -=z
V2.'2. -1-:12. where rLJI "-9
Thlls,
f:;. = - f PV: = - i (2.38 x/o 3
3-85
,
=0
J
z:: Z and ~::o I ") I
.s};r-) (2..Qlf!j.) 2. = - /.03 x/o-2. #1.
3.87 3.87 A conical plug is used to regulate the air flow from the pipe shown in Fig. P3.87 . The air leaves the edge of the cone with a uniform thickness of 0.02 m. If viscous effects are negligible and the flowrate is 0.50 m3 /s, determine the pressure within the pipe.
v
Q '" 0.50 m3 /s --I~~
-~+-
FIGURE P3.87
3.8a
r3. 98
An air cushion vehicle is supported by forcing air into the chamber created by a skirt around the periphery of the vehicle as shown in Fig. P3. Be. The air escapes through the 3-in. clearance between the lower end of the skirt and the ground (or water). Assume the vehicle weighs 10,000 lb and is essentially rectangular in shape, 30 bySO ft. The volume of the chamber is large enough so that the kinetic energy of the air within the chamber is negligible. Determine the f1owrate, Q, needed to support the vehicle. If the
ground clearance were reduced to 2 in., what f10wrate would be needed? If the vehicle weight were reduced to 5000 Ib and the ground clearance maintained at 3 in., what f10wrate would be needed? Fan
Skirt
,J
\.rr
~
P
11 ..",
L~ 3 in.
Ao= (30·f/) (50!+) ::: 1500 fll.
\,42. + %
~1
2
where
4.::: 0
F~
J
~():: and Z()-;:22. OJ
\/_ ~ 2.W
,qfJf h::: fjrovnrl clearance if follows fha-l
V2. -
Wah Q = fJ
.'M
'///////////////////////////////////////////////t//////7,1/,0
and
\/_';26'
.¥ -
2
or
V2. -
~ == 2h (L+ b) ~
where L =50 tI and b:: 30il
Thus) I Q= 2h(soff +30f.l)i(JsOOfI1)(:'~8X/O-3 .
or
¥
~) fI.i
'
h Yw' where h.-v fJ and W-/6 Thus) If h = /~ ff and W=/g 001) /6 the/} q:::: 3000 !f if h ;: ~ ff and W=/~ooo I~ fhe/) Q::: 2.000 .if ====== and if h = -Ii ff and 'II = 5 OOO /i:J the/) Q = 2/20 .fj-~ Q::::
IIq.8
Vehicle
where W= vehlf;!e weigh!
Also} so fha+
~~~
'If
/
FIGURE P3.89
To svpporf fhe load
.I!E... + \to:L +.z. :::: -A- + t 2g 0 r
Q
(
J
J
J
J
3.8Q 3. e'l A small card is placed on top of a spool as shown in Fig. P3.eQ . It is not possible to blow the card off the spool by blowing air through the hole in the center of the spool. The harder one blows, the harder the card "sticks" to the spool. In fact, by blowing hard enough it is possible to keep the card against the spool with the spool turned upside down. (Note: It may be necessary to use a thumb tack to prevent the card from sliding from the spool.) Explain this phenomenon.
•
FIG U REP 3 • eq
As fhe air flows r4dially ovfward in fhe 90P he/ween the cord and fhe spool if .slow.s down since fhe Flow area ilJcreases will? r; the radial dis/once from fhe center. (J =
2-71' r
hV
J
or
V=
That IS
').!h r
J
( see the fi'jvre).
h
If visco[).s effecf.r are nfJ/ imporfanfJ :2.. #.: +X : : cOl/slanl ;::: {JexH -I- 'lexif r '-I(f 29'
{J '"
2.
/ fQ Z
2.
if ('iex/I - V ) ) where
Bvf rex/I> r so fhat
f
~ 0,
(I)
I exit VI
II
if
or since lex/I:::O (a free jel)
~
r1/ / /! /
then
follows fhal
rexif I r..
,_1
from Er- (/)
~Xlt
_
2._
Q
2.[
V -t2rrh)
J
~t
_
J..
ri]
wlthi/) fhe 9a/. The card is svcked 1J9oi/ls! fhe spool, Tl;e harrier one blows fhrIJ{)flA fhe sfJ()o/( /atyBr())) fhe /rJrger fhe V(JCIIIIII'I IJnd fhe harder fh~ CQrd is !here /s a J
held rJ9aiIJsi the spool.
3-88
Vaclltlm
3.1:(0
I £ech
sz
*
3.90 Water flows over a weir plate (see Video V 10.7) which t'~(/f ~-Q~-: --'-"'" f has a parabolic opening as shown in Fig. P3.90. That is, the 'Z 2)::- H opening in th~ weir plate ~as a width CHIP, where C is a cont,' ,,~ stant. Detennme the funcbonal dependence of the flowrate on f. the head, Q = Q(H). ZeD
D\
f
Q == Ll dll where Thai is, frl1m
IJ.
is
1/
t.t
r
{()nc1;OA () h.
Ijf + #2, '"
IF #.%j
+'E:z.
wilh
!t=CHlI2=f I \-,-'
~'-:J
] _f / '"
"tAR
/If =H-Z'J t{
If::o
=ll
(,'free jei,
alld, ZJ. ::: II-h
or 2. (H-~/)+lj+z,::O Tnvs)
U1-
+0/ +(II-h)
__
II =
Vl,h +Jlt2. ~ f'J-,:h il ~ is "s/lla//
IJ
Also
dA):# C fftJ,r (i. e. dll:: 0 dl fflr Z${). till ;&cffii fIr ;~II ) bH
Q= f'~I'1/h cW rJl where p~o
ThlJs) f¥ =:
H
So
fhai
J
h::: H-z.
H
c1};j f1-Z/l--2--""1; dl
J
where pH
0
S1{iiHI71i:: f[(z--!) YHZ -Zo." + (~t.sin-I[(Z--f)/(H/J.)fl o
zaO
whjGh reduces If)!
~:::
¥ Vii H2.
Thai is Q- H:L
AIfernative/y J Q::: Vt9 where the QveraJe ve/Qc/IJ is fJrPjXll'liPIlII! fo 1if (t·.e. V'" Vi-,,#) IJlJti fhe 1,14'/ flow area., is pl'rJfJOn/olJa/ fo H~ (t·. e. A - HN (C HI'..) ::: c H.3h. ). TlJv~ Q ¥2-~H- (CH*) :: C V,.i' H2'V
That Is) ~'V HZ 4S ohfained above.
3.91 A weir (see Video VI 0.7) of trapezoidal cross section is used to measure the flowrate in a channel as shown in Fig. P3.91. If the flowrate is Qo when H = e/2, what flowrate is expected when H = e?
FIGURE P3.Q(
II V where if t'.s eXp'ecferl Mai V is (J. function of the head. fl. Thai is I V-,J2,Hi Also, from fhegeomelry A=fH(t+J,.) where~::=.t+2.Hff/173o" Thus A::: /I (i + H11/1'}.300) so fha1 T Q == C; 1'"1'" (J.f H tan30 H~/:1. where q ;s a con.stanf
Q ==
J
J
G
)
Lei Qo = f/oWfa1e when. jJ::: and Ql::: {Iowrale when H=1 Thus, Qo _ C;p:j (1-+-
q; - c, Vii (P.
4-
f ftln
l3,t
0)(i);" _
(J + ifan300) I, fan 31l) (£)3/2. - (I + tQn 30°) (2~/2.) :; O,2.Sq 30
3.92
Water flows down the sloping ramp shown in Fig. P3. q2. with negligible viscous effects. The flow is uniform at sections (1) and (2). For the conditions given show that three solutions for the downstream depth, h2 , are obtained by use of the Bernoulli and continuity equations. However, show that only two of these solutions are realistic. Determine these values.
FIGURE P3.Q2.
(I)
:£L + Vl +:r
= -Ii + V2.2. + Z2
21 I 1 29 fJ/so) A, V; :: A:t ~ or H V -..hLv, - (/N)(IO.s):; /0 2 - h,. I h2. h2. Thus) Ect. (I) becomes Jj. 2. ( /0 )2. ( 10 ;s) + 3 ff _ hi. ~
2 (32.2
or 6 'f. if
where f' =0 , and
Z2
=n2
~ -::0 J Z, :: 3 ill
- 2(32.2 ~J
¥:z.)
h: - h: ~ 2 93
100 -::: 0
By v.sin, a root findin~ pro9ram the fhree roofs erllalion ~re fovnd fo b8;
to this
o()bic
h:;. = 0.630 f1 h2 = if. IJf8 fl
Or
h2 = a ne9tt1il/e roof
C /early if is not po.ssihle (physicql/y) If) hlJve h2. <0 -rhus, h,.:::: 0.630 H or h2. = Jf.'f8 ft
3.Q3 J 3. q 3
The flow rate in a water channel is sometimes determined by use of a device called a Venturi flume. As shown in Fig. P3.Q,3 , this device consists simply of a hump on the bottom of the channel. If the water surface dips a distance of 0.07 rn for the conditions shown, what is the flowrate per width of the channel? Assume the velocity is uniform and viscous effects are negligible.
T
0.2 m
FIGURE P3. q3 (I)
w/fh"11 =0 J fJ2. =0 J E, == 1.2mJ
ond 22 ::: /.. 2m -
0.07117
= /. /.9 m
= I, 29 ~
or [(1.2.9)2. -I] w'" = 2. (q.8/~) (1,2. - /.13) m or
\1 =I,'I38~
Hence, Cf =h, ~ =(1. 'I38.f) (1,2m);:: I. 73
¥
3.9'f
Water flows in a rectangular channel that is 2.0 m wide as shown in Fig. P3. qlf . The upstream depth is 70 mm. The water surface rises 40 mm as it passes over a portion where the channel bottom rises 10 mm. If viscous effects are negligible, what is the ftowrate?
FIGURE P3. q'f
where
f'l:::0, 12.:::0 , Z,:::: O.07R'1, and Z2.;:; (o,()/ +o,/o)rn ~ O,lIm
Also) A,W:::A:l~ or \I
hi \I,
O.07m \/
h:t I = 0,/0 m VJ Thus, &t. (I) become.s V2:::
II
== 0·7-'(1
[J-O.72.]\/,'-=2(Q.81*)(O,II-o.o7)m
or
Hence} Q=A/~;:: (O,07m)(2.0m) (J.2lf!f):: O.17q.
3-93
y,== /.2'1~
(J)
3.QS
I (1)
3.95
1
Water flows under the inclined sluice gate shown in Fig. P3.95. Determine the flowrate if the gate is 8 ft wide.
~'~~'.. -:~''''
-
... --'--~
6ft
j
wher~
11
#
(J I
fo :: I)
#
~+6H = Jt.t'f-f "-I ,..,. B(IT A, V, :: II,. V2-) Qr
(/)
\/_ A,
_ 611" II - 7iT v, :: 6 v, Hence., £". tI) become.s 1f1 VI
Jt" + 6ff ;;: f1Y~Y/·t I If
'-,
"1
~1_/]~2 ::: 2 (32.Z.!f,.)(6-I)ff Hence
Z, ~ IfT,
alit/. Zz. I ff
Thlls J
Y2. -
J
J
Q:::./1, V, :: ~H (et~)(.3.0.3!j.)
~ ;;:;.3.03#
Q
.1
=I~S.!f
3.96
Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m'/s and a pressure of 200 kPa at an elevation of 25 m. Determine the velocity head and pressure head at elevations of 20 and 55 m.
= il. ==
D=0.15", -
",g
0. -::s 2. V nLI:a If (O,15m) 2
IJI p()int (0): ~ == (1J,31fl an d
'"I
m = Y'o :::- ~ = //, 3 -..5 .... 1/
== 6,5/ m
2(Q,8I/fo)
\,h2 A)/ ~2. .u:.!L+-L+z ::~+....l..+z 't 29 0 ~ ~D I
m3
= ~5tn
70 =20m
~_
orA _ 200 ~ r - 9.80!y:£
~
•
(I)
•
(0)
(1= 200kPa
(
,\
+ 25 - 20) m == 25.'1 In
3.t17
Draw the energy line and hydraulic grade line for the flow shown in Problem 3.6/f.
~)
(1)
•
FIGURE P3.6'1-
furbines} Me e/Jer9Y line (EL) is hori:zo/Jla/ af tJlJ elevation 01 the free s(Jr~7ce. T/;e /;ytirduhc grade line (H6L) is ()ne ve/oc/I; head lowe!',) ~welJ w/IJ file fJipe ou tie f . Since Ihe Iluid ve/ocily is cOlis/tint Ihr1)v9h,vr lIJe pipe wilh = :3 r~ the fd//ow/IJJ is Qbfained: For /lJviscid {low w/IIJ no pfllnjJs or I
-fi
3-96
3.98
I
'V (0) ..:----:::;:-----:----~::-- --
3.98 Draw the energy line and the hydraulic grade line for the flow of Problem 3.60
FIGURE P3.60
For il)lI/scid flow wilh no ptJmps or turbines, the energy line is horj2()nfo~ a disflJnce /; above
fhe ovf/et . From Problem
3.60
we obf"in h :: /. 7'1 fl..
The hydrolJlic qrade line is ~~ below fhe energy line, sl"rftng af Ihe free surface where Vo::: 0 and ending at fhe pipe exil where fJ2. =0 and =h. IN poinl (I) the pressure he(ui i.s fJ'/l = (2..98 - 1'1-.5) jff.:z. ( /~:;A?-) /62Jf = - 2.6.S I~ and £/ =0. 2-
;f
In the ~;n. pipe :z. If ~2. V -.¥.-J =( D:l.,) ~
Ia
Vj = /12. ~/Il.1 'I-
=
=(t) ~
so
fIJaf
If
(.A) h =(..£-) (/.7'1 fO = O. //2 (I
29 D3 ~9 D3 If The corresponding EL and HGL are drawl) 10 scale below, ~
V
(0)
~
_____
li:: 0./11 II
-----Jr~~' --;:=-_=-=_-==-~Ener91 LliJe 4(3) , .l/). 'Cf) Z! 0 , t 1 ,,,2) pipe eenler line: : 2:!. = h = 1.7'1 It 1::
'
I
I ,
I I
I
I
I
'
, , I
,
I I
,
,
:1- = -26.afj
HydralJlic Grorie --: I Line (HGL) :, ,I ,
I
, ,
I
,
I
,
,
'
,
r
I ,
, I
r
,
I
,
!..J--1
3-'17
ZQ ,
(EL)
3. qq
J
l 2-in.-diameter
hOS~
3. q q
Draw the energy line and hydraulic grade line for the flow shown in Problem 3.65.
J,lln. ~~
.
7. ~ · --L
2 ft (0) '\/ (3)t-
t=-=::-J:§:j::e:-= B ft
(S), (If)
(1)
•
lumps or furbine.s) fhe e/)~r}' li/)e (£1-) elevafio/J ()llhe free sur/dee. TlJe~ hy/rdv/Ie
For il'Jlliscid flow wilh is. horizo/J/tll, af
(J/)
IJIJ
9rade line (/l6L) ;0$ one veloc'fJ heod lower. SInce 2Ys ;: 3 It if follows fhal fhe !leL pa.rses Il)/'f)u,h tlJe f.i;; u( fhlIlOZ2/8. I//so, since
~ '" ~:s =(*r~ if Mlol/IS Ihal
-;.r~.7- = (l);D)'" zg~2 = (t) (3fl) :: II-
Ihe velocil, head is c()ns/olll
3 III
Sf)
ff . T/;r()u,houf
the pipe
fh4f the fQ/lolllilJg /.S obTtJ/IJeri:
3.10'0 * Water flows up the ramp shown in Fig. P3.100 with negligible viscous losses. The upstream depth and velocity are maintained at hi = 0.3 m and VI = 6 m/s. Plot a graph of the downstream depth, h2' as a function of the ramp height, H, for 0 s:,.H s:, 2 m. Note that for each value of H there are three solutions, not all of which are realistic.
III
o
+ ~ 2. + Z I =.:t!2: + 2.q
At 11 :::,42. U.
I//.so J
fJ,
\J -
~
V'%'2.
2q
lui
(J
.so fhat
-
h2.
=
(1.8 )'J. ~ + (f'~h3.)
or
( 61}i' +2. (9. 81 ':2.) ( 0.3 - f/- h,Jm which CdlJ be writ/en as ~ (2_13.5
1'2. ~O
z,2.:::
J
2/ ::: o.3m,
fI +I; 2.
1~II..JeJ"'e h2,"''''
",rl", .
become~
J
h: -
fJI ~O J
where
+ i!.:;..
n2.
Thvs £r- (I) : ; + 0.3 m
(I)
(O,31TJ)(6~) - ~
II -
~ Yj -
V2. -
FIGURE P3.100
-R)h: +0.//'5/
wdli L{ =' b~ J
=(- ~~ )'J-q;
=0
(2)
For 0 ~ f1~ 2m solve Ert. (2) {or h2.
Ra-lher than SO/Vifl1 a Cf/b/c e9{)ahon for h,. (?llIe II) con directly solve for 1/ f9/llen 171-)' From £r. (1-) : 1/=2./3.5 _l
"2.
J
_ 0,/651
fJ 9raph of £'1,
(3)
h"J.2.. (2)
one
or (3) /s 9/l/en on the {pi/ow/f)! pa98.
( con'-I)
3-qq
The results of an EXCEL Program to calculate H for given values of h2 are shown below.
h2' m 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1
H,m 0.001 0.703 0.975 1.076 1.098 1.077 1.031 0.970 0.899 0.820 0.737 0.651 0.562 0.471 0.378 0.284 0.189 0.094 -0.002
Water Depth vs Elevation Change
2.2 2.0
I
.~--------1
+--~~---~----------
-~
1.8
I
1.6 1.4 E 1.2 N .c 1.0
----I
0.8 0.6
I I
0.4 t====~:::::::::~=---...-=::::::--~--~---t-
-------- ------ --·----rI -----1
1
0.2
J
i I
I
0.0 0.0
0.2
0.4
0.6
0.8
J.oqg
1.0,'
1.2
H,m
For H ~ I,OQ8 m fhere are no real J posilil/e roofs of Ef(. (2.). Thai is J for fhe (jive/J upstream condilions (V; == 6 f- and hI::: O.3/n) We must have 11<: /,098 m. II w()IJld nol be possible fo hove the flow 90 vp a ramp of 9rB41er hei9)rI fhalJ fhis w/-/huv/ /ncreas/f)1} e/lher I1lJnd/or hi . The two P(J,s.s;;/g waler de,fJ;.r for a 9ive/J HtJfQ plo/leri be/ow.
3-100
3, /01 I 3.101
Pressure Distribution between 1\vo Circular Plates
Objective:
According to the Bernoulli equation, a change in velocity can cause a change in pressure. Also, for an incompressible flow, a change in flow area causes a change in velocity. The purpose of this experiment is to determine the pressure distribution caused by air flowing radially outward in the gap between two closely spaced flat plates as shown in Fig. P3.101.
Equipment:
Air supply with a flow meter; two circular flat plates with static pressure taps at various radial locations from the center of the plates; spacers to maintain a gap of height b between the plates; manometer; barometer; thermometer.
Experimental Procedure:
Measure the radius, R, of the plates and the gap width, b, between them. Adjust the air supply to provide the desired, constant flowrate, Q, through the inlet pipe and the gap between the flat plates. Attach the manometer to the static pressure tap located a radial distance r from the center of the plates and record the manometer reading, h. Repeat the pressure measurements (for the same Q) at different radial locations. Record the barometer reading, H atm , in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law.
Calculations: Use the manometer readings to obtain the experimentally determined pressure distribution, p = p(r), within the gap. That is, p = -"Imh, where "1m is the specific weight of the manometer fluid. Also use the Bernoulli equation (Ph + V 2/2g = constant) and the continuity equation (AV = constant, where A = 27Trb) to determine the theoretical pressure distribution within the gap between the plates. Note that the flow at the edge of the plates (r = R) is a free jet (p = 0). Also note that an increase in r causes an increase in A, a decrease in V, and an increase in p. Graph:
Plot the experimentally measured pressure head, and radial location, r, as abscissas.
ph,
in feet of air as ordinates
Results:
On the same graph, plot the tJleoretical pressure head distribution as a function of radial location.
Data:
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r- _
1 h
rT 1 ________- - -___ -1 ~ir.-,.__ vBlli·_-'" I I
circular plates
r -----
II1II
II-Water
/
•
~-----R-----4~1 1
tQ
•
(con'/: ) 3-/01
FIGURE P3.101
3,/0 I
(conJl) Solution for Problem 3.101: Pressure Distribuition between Two Circular Plates
P
Q, ft"3/s 0.879
R, in. 5.0
r, in. 0.7 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
h, in. -9.05 -6.02 -2.02 -0.96 -0.48 -0.24 -0.13 -0.03 -0.01 0.00
b, in. 0.125
H alm , in. Hg 29.09
T, deg F 83
Experiment ply, ft -663.75 -441.52 -148.15 -70.41 -35.20 -17.60 -9.53 -2.20 -0.73 0.00
YH2D, Ib/ft"3 62.4 Theory V,ftls ply, ft 220.8 -740.7 161.2 -387.2 107.4 -163.1 80.6 -84.7 64.5 -48.4 53.7 -28.7 46.0 -16.8 40.3 -9.1 35.8 -3.8 32.2 0.0
=Palm/RT where
=
=
Palm YHg*H alm 847Iblft"3*(29.09/12ft) R = 1716 ft Ib/slug deg R T 83 + 460 543 deg R
=
Thus, P
ply
=20531b/ft"2
=
=0.00220 slug/ft"3 and Y = p*g =0.00220*32.2 =0.0709 Ib/ft"3
=YH2D*h/y
v = Q/(27trb) = 0.879 ft"s/(2*3.1415*(O.125/12)ft*r) r I
Problem 3.101 Pressure Head, ply, vs Radial Position, r
-200
/!
4::
-
-;.. -400
• -
:
Q.
Experimental Theoretical
, ,
-600
I
41 f I
I
-800 0.0
2.0
4.0
I
6.0
r, in. IL-______________________________________________________I
3-/02
3. 102,
3.102
Calibration of a Nozzle Flow Meter
Objective: As shown in Section 3.6.3 of the text, the volumetric f]owrate, Q, of a given fluid through a nozzle flow meter is proportional to the square root of the pressure drop across the meter. Thus, Q = Khl/2, where K is the meter calibration constant and h is the manometer reading that measures the pressure drop across the meter (see Fig. P3.102). The purpose of this experiment is to determine the value of K for a given nozzle flow meter. Equipment:
Pipe with a nozzle flow meter; variable speed fan; exit nozzle to produce a uniform jet of air; Pi tot static tube; manometers; barometer; thermometer.
Adjust the fan speed control to give the desired flowrate, Q. Record the flow meter manometer reading, h, and the Pitot tube manometer reading, H. Repeat the measurements for various fan settings (i.e., flowrates). Record the nozzle exit diameter, d. Record the barometer reading, Hatm, in inches of mercury and the air temperature, T, so that the air density can be calculated from the perfect gal law.
Experimental Procedure:
Calculations: For each fan setting determine the f]owrate, Q = VA, where V and A are the air velocity at the exit and the nozzle exit area, respectively. The velocity, V, can be determined by using the Bernoulli equation and the Pitot tube manometer data, H (see Equation 3.16). Plot flowrate, Q, as ordinates and flow meter manometer reading, h, as abscissas on a log-log graph. Draw the best-fit straight line with a slope of Yz through the data.
Graph:
Results: Use your data to determine the calibration constant, K, in the flow meter equation Q = Kh 1(2. Data:
To proceed, print this page for reference when you work the problem and click here
to bring up an EXCEL page with the data for this problem.
Pitot tube
Flow meter maometer Water
Pitot static tube
Air
meter
Exit noule
3-/03
•
FIGURE P3.102
Solution for Problem 3.102: Calibration of a Nozzle Flow Meter
d, in. 1.169
Hatm , in. Hg T, deg F 75 29.01
H, in. 5.6 5.4 5.2 4.9 4.7 4.3 3.9 3.6 3.1 2.7 2.3 2.0 1.5 1.1 0.6
h, in. 11.6 11.1 10.7 10.1 9.6 8.8 7.9 7.2 6.1 5.4 4.5 3.8 2.9 2.1 1.0
~P,
Ib/ftA2
29.1 28.1 27.0 25.5 24.4 22.4 20.3 18.7 16.1 14.0 12.0 10.4 7.8 5.7 3.1
V, fUs 162 159 156 151 148 142 135 130 120 112 104 97 84 72 53
Q, ftA3/s
1.20 1.18 1.16 1.13 1.10 1.06 1.00 0.97 0.90 0.84 0.77 0.72 0.62 0.53 0.39
P = Patm /RT where Patm
R T
= YHg *H atm = 847 Ib/ftA3*(29.01/12 ft) = 2048 Ib/ft"2
= 1716 ft Ib/slug deg R
=75 + 460 =535 deg R
Thus, P = 0.00223 slug/ft"3
V = (2*~p/p)1/2 Q
=AV where A = nd 2 /4 = n*(1.169/12 ft)"2/4 = 7.45E-3 ftA2
3
From the graph, Q = K h 1/2 = 0.358 h 1/2 where Q is in ft /s and h is in in. Thus, K
=0.358 ft3 /(s*in. 1/2 )
3-/01/-
Problem 3.102 Flow Rate, Q, vs Manometer Reading, h
"i
i
i
iii
I
I
I
.v
I
, i
i
!,' , I :
I
:!!
i
i
0.1
', I'
I '
I
!
I
I,
Iii I ! I'
II iii
i
'1
I'
I
;1
111
ill:
+-------~~~~~----~~--~~~~
1
10
100
h, in.
3-/05
•
Experimental I
3,103
3.103
Pressure Distribution in a Two-Dimensional Channel
Objective: According to the Bernoulli equation, a change in velocity can cause a change in pressure. Also, for an incompressible flow, a change in flow area causes a change in velocity. The purpose of this experiment is to determine the pressure distribution caused by air flowing within a two-dimensional, variable area channel as shown in Fig. P3.103. Equipment:
Air supply with a flow meter; two-dimensional channel with one curved side and one flat side; static pressure taps at various locations along both walls of the channel; ruler; manometer; barometer; thermometer.
Measure the constant width, b, of the channel and the channel height, y, as a function of distance, x, along the channel. Adjust the air supply to provide the desired, constant flowrate, Q, through the channel. Attach the manometer to the static pressure tap located a distance, x, from the origin and record the manometer reading, h. Repeat the pressure measurements (for the same Q) at various locations on both the flat and the curved sides of the channel. Record the barometer reading, Halm , in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law.
Experimental Procedure:
Calculations: Use the manometer readings, h, to calculate the pressure within the channel, P = "Imh, where "1m is the specific weight of the manometer fluid. Convert this pressure into the pressure head, ph, where "I = gp is the specific weight of air. Also use the Bernoulli equation (Ph + V2/2g = constant) and the continuity equation (A V = Q, where A = yb) to determine the theoretical pressure distribution within the channel. Note that the air leaves the end of the channel (x = L) as a free jet (p = 0). Graph: Plot the experimentally determined pressure head, ph, as ordinates and the distance along the channel, x, as abscissas. There will be two curves-one for the curved side of the channel and another for the flat side. Results:
On the same graph, plot the theoretical pressure distribution within the channel.
Data:
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Static pressure taps
~--~+-----L----------~'I
•
3-106
FIGURE P3.103
(c.on'/) o ution for Problem 3.103: Pressure Distribution in a Two-Dimensional Channel
b, in. 2.0
Q, ft"3/s 1.32
Halm , in. Hg 28.96
T, deg F 71
x, in.
y, in.
0.75 2.50 4.00 4.63 5.38 8.14 10.75 13.25 15.78 21.75
2.00 2.00 1.28 1.05 1.05 1.29 1.54 1.77 2.00 2.00
h, in. flat side 0.28 0.21 -0.42 -0.77 -1.01 -0.63 -0.32 -0.15 -0.05 0.00
h, in. curved side 0.31 0.37 0.03 -1.63 -1.05 -0.62 -0.31 -0.15 0.00 0.00
L, in. 21.75 Experimental
ply, ft
Theory ply, ft
ply, ft
flat side curved side 20.2 22.3 15.1 26.6 -30.2 2.3 -55.5 -117.4 -72.7 -75.6 -45.4 -44.7 -23.0 -22.3 -10.8 -10.8 -3.6 0.0 0.0 0.0
0.0 0.0 -50.5 -92.2 -92.2 -49.2 -24.1 -9.7 0.0 0.0
P = Palm /RT where
= YHg*H alm = 847 Ib/ft"3*(28.96/12 ft) = 2044 Ib/ft"2 R = 1716 ft Ib/slug deg R T = 71 + 460 531 deg R Palm
=
Thus, P
=0.00224 slug/ft"3 and y = p*g =0.00224 slug/ft"3*(32.2 ftlsll2) = 0.0722 Ib/ft"3
ply =YH20 *h/y Theoretical:
ply =V ex /12g - V 2/2g where V = Q/A = Q/(b*y) and Vex~ = Q/Aex~ = (1.32 ftIl3/s/)*(2 *2/144 ft1l2)
=47.5 ftls
Problem 3.103 Pressure Head, ply, vs Distance, x
40 20
-~--J
0
.~~--:-~~~~~~
-20
= .:i:Q.
-40 -60
~~--------j-----------------r=-=J .--~~~--t-------T-------r-~~--
-80 -100 -120
~•......- Experimental,
I
J
!
flat side
- 1 - - - - - - - - + - - - - - - - + - - - - - - .. - Experimental, curved side --Theoretical
-140 0
5
15
10 x, in.
3-/07
20
25
.3, /oLf
3.104
Sluice Gate Flowrate
Objective: The flowrate of water under a sluice gate as shown in Fig. P3.104 is a function of the water depths upstream and downstream of the gate. The purpose of this experiment is to compare the theoretical flowrate with the experimentally determined flowrate. Equipment:
Flow channel with pump and control valve to provide the desired flowrate in the channel; sluice gate; point gage to measure water depth; float; stop watch.
Experimental Procedure:
Adjust the vertical position of the sluice gate so that the bottom of the gate is the desired distance, a, above the channel bottom. Measure the width, b, of the channel (which is equal to the width of the gate). Turn on the pump and adjust the control valve to produce the desired water depth upstream of the sluice gate. Insert a float into the water upstream of the gate and measure the water velocity, VI> by recording the time, t, it takes the float to travel a distance L. That is, VI = Lit. Use a point gage to measure the water depth, z\> upstream of the gate. Adjust the control valve to produce various water depths upstream of the gate and repeat the measurements. For each water depth used, determine the flowrate, Q, under the sluice gate by using the continuity equation Q = AI VI = b ZI VI' Use the Bernoulli and continuity equations to determine the theoretical flowrate under the sluice gate (see Equation 3.21). For these calculations assume that the water depth downstream of the gate, Z2, remains at 61 % of the distance between the channel bottom and the bottom of the gate. That is, Z2 = O.61a.
Calculations:
Graph: ZI>
Plot the experimentally determined flowrate, Q, as ordinates and the water depth, upstream of the gate as abscissas.
Results:
On the same graph, plot the theoretical flowrate as a function of water depth upstream of the gate.
Data:
To proceed, print this page for reference when you work the problem and click III're to bring up an EXCEL page with the data for this problem .
•
3-1 0 8
FIGURE P3.104
3./0/f
I (conJ/) Solution for problem 3.104: Sluice Gate Flowrate
a, in. 1.2
b, in. 6.0
ft 0.183 0.267 0.343 0.453 0.569 0.725 0.877
L, ft 4.0 Experimental V 1 , ftls Q, ft"3/s 0.087 0.952 0.800 0.107 0.769 0.132 0.645 0.146 0.625 0.178 0.571 0.207 0.465 0.204
t, S 4.2 5.0 5.2 6.2 6.4 7.0 8.6
Zl,
Theoretical Q, ft"3/s 0.091 0.114 0.132 0.155 0.175 0.200 0.222
Experimental: V 1 = Ut
= V 1 bz 1
Q
Theoretical: Q = b*Z23/2*(2*g)1I2*[((Zl/z2) -1)/(1 - (z2/z1)2)]1/2 where Z2 = 0.61*a
Problem 3.104 Flow Rate, Q, vs Depth, Z1
1.00
-r=======:r====r==:r==r==+=+==l=+=I i
1
', I
I !
-
I
CO')
<
= d
0.10
.-.
1
I
!
,i
i
:
!
--------+--1 -----'-----1
tn
I
•
i
,
;
,i.-~
i~l!
i
~~l 1
!
--t,~-
•
i
,
•
-
,
;
I
I
I
I
!
i
i i
,
r r
-I I ! I 1
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;
1
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I
Experimental Theoretical
!
I
0.01
I
I T
i
!
I
1
i
i
1
I
i I 1
I
,
I
i
-r
I
I
I
,I
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i
I I !
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1
1
0.1 Z1'
ft
I 3 -/0'1
/1./ 1}te velocity. field. of a flow is given by V = (x - 8)j + 5zk ft/s, where x, y, and z are in feet. Determine the fluid speed at the origin (x = y:::;: Z = 0) and on the y axis (x = z = 0). 4.1
+ 2)i +
(3y
IJ..::=
3Yf2.
J
v=x-8, W=Sz
=1«2.
Thl/s) af X= y:::r ~O V + V2. + W,,' and Of) the line x;: Z =OJ V =/(3 y +2)2. +(_8)").1 :::: J~9-y-2.-t-/2-y-+-68-"
J2~ +(-Sl ' = e.2s1j
=
#-
where y--ft
if. '14.2 A flow can be visualized by plotting the velocity field as velocity vectors at representative locations in the flow as shown in Video V4.1 and Fig. E4.1. Consider the velocity field given in polar coordinates by vr = -lOlr and Vo = 101r. This flow approximates a fluid swirling into a sink as shown in Fig. P4.2. Plot the velocity field at locations given by r = I, 2, and 3 with e = 0, 30, 60, and 90 deg.
Vr
W;1h tV";;:: -/ olr and N8 ::: lo/r then V-=Jrv;.'-+IV92. ::; I(-Io/r)% + (Jo/r)1 = I'I-;!.'IThe QI191e ex hetllleen the radial J/re{;/ion and the velocify vee/or is r;iven hy tan eX
=
filii -AIr
lolr
;:: _(-Iolr)
• FIGURE P4.2
=I
ThvsJ ()( = ~s for anv rJ B / ~·.e. the velocify vector is alway oriente') 0
3
8=60
Nofe: VIS
independ'snt
of 9.
V=II/.III 1ft r::: I
If-I
o /fS" re/4live
1orodia/ lines)
r
#.3 4.3
The velocity field of a flow is given by V = 20yl(.\"2 + y2)l'2i - '20xl(x 2 + )'2)1121 fils, where x and yare in feet. Determine the fluid speed at points along the x axis; along the y axis.
Also, fan
y
-2.0X
e =-tr-=
(xl.
What is the angle between the velocity vector and the x axis at points (x, y) = (5,0), (5, 5), and (0, 5)?
+y2.)J..2.
2 off .;s
(0,5)
20Y
(x'J. t y2.) Ji.
or
fane ==
-
f (5,0)
Thus J for (xJy) =(sJ 0) fan e ::: - 00 or
For
~~----~--=-....;"".--
e = - qoo
(~y) =(5,5)
tQn8=-/ or B=-'fsl!)
for (XJY) =(0,.5) fan () = 0 or B==
X
2.0 ff ,s
~v u
ot)
4.4 The x and y components of a velocity field are given by u = x - y and v = x 2y - 8. Detennine the location of any stagnation points in the flow field. That is, at what point(s) is the velocity zero?
v== 0
provided fhQf bolh u =0 and v::: o. ThIJs f).:: X - Y::: 0 or X := y anti V = x~ - B ::: 0 or x2y == 8 By combinin9 ob/ain X J :- 8 or X::: 2. Since X:::. y N follow..! fh41 y=2. also. Thl/.s (x) y) == (2.,2.) 1
Jf-2
~,5
45 The x and y components of velocity for a two-dimensional flow are u = 3 ft/s and v = 9x 2 ftls. where x is in feet. Determine the equation for the streamlines and graph representative streamlines in the upper half plane.
Ii = 3
and If:::: 9x2. so fhai Sfre4f)1//nes are 9/ven hy
4¥ ~ -f ~ 2/ :: 3x2 Or
Jdy =: J.3x2.dx
Thvs y:: X.3 +C J where c is ac()llsfanT. J
Repre.renfalirJe .sfream/ines correspondil'l9 care shoVln he/ow.
70 differenT va/VB.! of
=x 3 + C
Streamlines y
c=/OO
C=.5o
c= 0 I
I
-J i Ic= -so
100
50
I
-+-I-'--I------+--
o ~~~--r_~r--,--~--~~~--~~~~ o 1 2 3 4 5 -5 -4 -3 -2 -1 x
~-.3
~. 6
I
4. 6 Show that the streamlines for a flow whose velocity components are u = c( x 2 - y2) and v = - 2cxy, where c is a constant, are given by the equation x2y - y3/3 = constant. At which point (points) is the flow parallel to the y axis? At which point .(points) is the fluid stationary?
U=C(X2..-y2.)
J
*"f
V=-2cxy
Sireamlines given ny y"{(x) Qre -such fhal Consider +he {une/ion x'l.y - -f = t:.ofJsf, !VoTe; If i.s nof easy fo wr/le fhis eXf/iviily as y ~ f (x) However J we can differentials Ef. (JJ ff) 9ille 2xyr/x +x'-dy - y'1.rly :;;: 0 I or (x2.._y2.) dy +2xy dx =0 Thvs.l fhe //ne.s /n fhe x- y plane given
dL (IX ~,e.
==
- ;XY) or for allY cons Ian I (X _y'1.
fhe fun6fion x2y - ~
by £1, (I)
3
=consf. represenfs
The flow is par4/Je/ fo ihe
rlus
OCC{)rs
x- axis
whelJ eifher X=0
or
y:::O
.
The flow i.s para/Ie J fo The y-axi.s when This occurs who tJ X = ± Y The livid has z.ero ve/oc/I)' at x:: y
J
J
t:e-.;
*
=: 00 J
=0
=L. U
the .sfreafllines
*::0
wASil
a slope
lIx:: C-(2.X~ _y2. xy )
CJ
of +he 9iv81J flow.
hdl/O
(I)
or v=:O.
fhe X-ax/s
the
y-qX~
or
IJ.. ::::0.
Of
~. 7
I
4. 7 The velocity field of a flow is given by U =- - V~y/(x~ + y2)112 and v = V ox/(x 2 + y2)II2, where Vo is a constant. Where in the flow field is the speed equal to Vo? Determine equation of the streamlines and discuss the various characteristics of this flow.
11 = -
v.o (Xl. +y y"J.'fo-
V ;: _IU V
2.
-+-
V
= V, 0
X (X2-+yl.)V2.
,. [\/2(y2. + X '-) 1~ = .
V 2. :=
Yo
+ y2.)
(Xl.
oS 0
fh41
\t 0
Thus, V = ~ fhrou?houl the entire {low fielt1 Sfream lines IJro 9iveIJ by
i1... dx = LIJ. = 2-y fo give
or
- ydy = xdx which can be infe9rafed
X2. t y2. =cons!.
Thus J fhe flvid {low wifh circular .sfream liIJes and the speed is constanT fhrollghout.
,+-5
.If.B 4.8 Water flows from a rotating lawn sprinkler as shown in Video V4.6 and Figure P4.8. The end of the sprinkler arm moves with a speed of wR, where w = 10 radls is the angular velocity of the sprinkler arm and R = 0.5 ft is its radius. The water exits the nozzle with a speed of V = 10 ftls
• FIGURE P4.S
leave.s the nozzle wilh Q Ve10 ClfY 0 f V= If) fIls af all QI191e of ,gOO rela live 10 the radial direvlion - for an obserJler ri,)/nfj on the sprinKler 4rb1. Th/s is fhe rel41ille velocity. II.! shown ill the .skefch, fhe sprillk/PI 4rll1 has a c;"c/lmf~enci41 veloc/Iy 0 f RfP -::: O. b fI (10 rad/s) =: s His. TlJe IJbsolv/e ve10cily J Va; 4.!' observed hy a person sfanJilJr. on -Ihe lawn is the vee/or SlJm of re/t'J1/ve velocily and the nO"j"Jle ve/oeily. From fhe 9(;omell'Y of tne fi9vre :
(a) Wafer
i a. n 0( =
,
.
10 Sin 30
-.5 ::::. 0
10 Ct'J~300
Tho! i.s ex, == 0 (,',e,) the ahsolvfe wafer velov/ly is in the r4rJial
)
o
R=o,s
w;:;/o
rJ.irecfion. Since there is no force ac+i"1 fhe wafer after iT leavesJ "he W41er p4f'h'c/e-s
on
coni/nile f~ move in The raJial direc,lion. Th~ the palh1ine..r are .s1rai9hf raJiallines. (b)
TAe sh4pe of the water stream «1 a 1irJen installf (t:e"a ~fJ4P .shot" of fhe W4fer) can he ob-lained 4S follolVs, COf).rider the wafer .sfream emaIJaliA9 from +he end af-lhe nO"/.i/e al r::R (md (}::oat -lime /;=:()
(con'i) '/-6
-¥. 8
(con II)
that lefl from -Ihe no~~/e t seconds 490 d'd so when the nozzle wasaf f) ~ wi. SinGe /'0 /.t::0 the parth/e.s in sfrai1h1J ratii4/ pAths w/lh Va~."" A particle in -Ihi.s
sfream
speed ~ (.see pa,.{ (al)) fhis parliole is af a dis/once of r = R +Va { from .Jhe / orl9tn.
~"t ser;o~Js 070
;-~e=wl ...
Thlls the s1reanJ sho.pe js
e t t::
~
i=o
) fJ)
J
r = R +Va t and
~
fA)
J
or by eli,.,,;ncdiJ19 i
r::Rf(-i5-)e For the qiven data wi1h Va -;: V&0.s30' =0 0 1j-) (.os30 CI/Jd I.U == If) rad/s this hecomes r = o.s +0. 866 f:) where r,v ff alld ()~ rad. 'his .sfream shape i.s plofled beJoU/.
0 :::
8. Db ~ (seeparl(4J)
J
---2
------------------~--~--~---xJ~ OJ
#-7
~.9 '''4.9 Consider a ball thrown with initial speed Vo at an angle of 0 as shown in Fig. P4.9a. As discussed in beginning physics. if friction is negligible the path that the ball takes is given by 2
y = (tan O)x - [g/(2
V0 COS
2
O)Jx
2
That is. y = CIX + c&x 2, where Cl and C2 are constants. The path is a parabola. The pathline for a stream of water leaving a small nozzle is shown in Fig. P4.9b and Video \'4.3. The coordinates for this water stream are given in the following table. (a) Use the given data to determine appropriate values for C 1 and C2 in the above equation and, thus, show that these water particles also follow a parabolic pathline. (b) Use your values of c, and C2 to determine the speed of the water, Vo. leaving the nozzle.
)'
x, in.
y, in.
o
0 0.13 0.16 0.13 0.00 -0.20 -0.53 -0.90 -1.43
0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.00
-
........... ,
,
,, ,
x \ \
(a) )'
"'-,
",\
., ,
'
'
\\
(b)
•
FIGURE P4.9
An EX eEL Pttt79f!tJl!J W4S v.red to p/of the x-y d4itl d/Jd to fif a .seco"J order curve to fhe data. Tile results are .showfJ he/oUl. y vs x for Water Stream
0.04 - , - - - - - . , - - - - - . , - - - - - - - - - l 0.02 -~~~~~ ~J
o
-0.02 :::~ -0.04 >-0. 06 -0.08
.. -~-.~-. ~-~--~--PIIIo~-.---~-~-...
I i
--~--~-.~---~--~~~~
+----~--!------,-------!---~~ 2
~-----"C -8.4987x + 0.7n 15x
-0.1 -0.12 +----.,..-----,---------,.--+-----1 o 0.05 0.1
x, ft
C,:::: O. 7//5 ::::
14nB
or e::: 3S,~ I
and
c,. ::: - 8. 'f?11 =: 0"" 2.
'10
.32.. ~ = 2. (8.'f187) cos'"(.3S,rJ
Thll~ '10 -:: 1.69 fj
JI.-e
'I-~ 10
I 4. t 0 The x and y components of a velocity field are given by u = :l-y and v = -xy2. Determine the equation for the streamlines of this flow and compare with those in Example 4.2. Is the flow in this problem the same as that in Example 4.2? Explain.
Stream Jines are 9t"ven by ~J:: == J!. ::; _ X y2. __ .XJ X U x2.y - X d :: - ~ or .-t which cal} he iIJ1e9raied as: x
y
s* =-J~
Thus, In y = -Iny + c) where C /s ~ CIJI7.ff4nl.
Thfls) X y ::: C
No fe: These streofllline.s are the So.lhS shape (same Ilf/ow paifenn as in £xample ~ 2 - but fhe veJoc/fy fields are dt'lferel'r/. However lhe ralios -£ are the same: .;f.:; _ X y2. :: _ t J
x"Y
w
and
~
u
:=
(Volt) (-y) (voll) (x)
x
:= _
Xx
y ".-
4.11
In addition to the customary horizontal velocity components of the air in the atmosphere (the "wind"), there often are vertical air currents (thermals) caused by buoyant effects due to uneven heating of the air as indicated in Fig. P4.12. Assume that the velocity field in a certain region is approximated by u = Un' v = Va (1 - y/h) for o < y < h, and u = un' v = 0 for y > h. Plot the shape of the streamline that passes through the origin for values of lio/vo = 0.5, 1, and 2.
I
~
/
/'"
I
J. I
I o
/
X
FIGURE P4.12
.so
fhai streamlines x
= ~fdx o
Nofe: The lower limds () f infe1raJiol) (x=o) y:::o) insure fhal fhis eqvaf lon is for fhe streamline fhrov'Ih fhe ori9in.
Thi.s
sfream I//JB
X ==-h
i)
(~) In (J -
i~ ploffed befow.
y/h vs xlh 1 ~
O.8
-
0.6
.c
..
.... .. _p'.
,#
/'
-~
.- •• _M -:..
:
. . . . . t'"'"' I ..,.,
...r ---
+-------+--#~-,
,
,
,/
.......r..,
-.-'-""-
, ,
'I
I
----j--r-
>.
0.4
:
,/
0.2
-~~~
-uO/vO=2
"~~--+--------'---~--I I
-1
1
0~--~----1---~-----r----r---~
o
0.5
1
1.5
2
xlh
Jf-/O
2.5
3
-
-uOlvO = 1
- - - uO/vO = 0.5
y .."....
/7f'
4.13* Repeat Problem 4.12 using the same information except that u = UoY / II for 0 ::; y ::; II rather than u = U(). Use values of uo/vo = 0,0.1, 0.2, 0.4, 0.6, 0.8, and 1.0.
I
Uo ~
/
I
A I
I
,
/
//
0
tor o
, V= Va (I-i) for y '" h are given by
ll= UhY
1>:(J
{
x
=..::L IJ.
(h~ y)
Y
= vo(J-{) = Vo U
D
fiy =
1. h
IX~ dx
wtth
or
(h-y)
Y
Ua
X
.sfreafh//ne.s
X::::O
when y=o
This ifJ1e'lf'ales 117 'live
-~ -h In( h-y):h In (h) = -to; X
or
f=r&.,)[ln~)- iJ
"*
Thi.s sTreamline is plolled he/ow for 0 ~ ~ I J wilh ~ = 0 J 0.1 0.2. O. 'I- O. /) I o. B Qnri I. 0 The VIJ Itie s were calc tJ/oted ana plotted ()S ifJ? 41) £XC£L PrtJ'1J'i1J11. J
J
J
J
y/h vs xlh 1 _
0.9
0.8 0.7 0.6
-
O.C
D,/
~~-"~=---
------- ---------- I
,, - -- ------------- -- ---I I
~-I--
--~--1
to. 5 -l--'--l---1.f-,~ 0.4 0.3
!
-j
-P.I-'-J(-q--~--~- -~~~~~---~-WJIj'---~~--~__i____--~~~~_l______-
-~--~-----------------:
, i
0.2
-
0.1
------~--------
--~~----~-.----
----;i i
O+-----------T-----------~----------r_--------~
a
0.5
1
xlh
'1-//
1.5
2
'f.II/-
I A velocity field is given by u = cx 2 and v = cy2, where c is a constant. Determine the x and y components of the acceleration. At what point (points) in the flow field is the acceleration zero?
4.14
dlJ. =rr
Ox
dU
dlJ,
2.
+l1.Tx +VTy - (CX'")(')..CX) ::;2c X
and ay
=1f +u.¥X+v:; ::
(cy2.)(2-c.y)
= 2C2. y 3
Thus) a=axl+ayJ =0 af (X1Y) -::(0)0)
"".15
I .1.1.5 V
=
A three-dimensional velocity field is given by u = x 2 , -2xy, and w = x + y. Determine the acceleration vector.
JU Lt1 JU = If + U DX +~ 19
QX
=X2.(2X) Cly :::
+jJI
nO!J,
of)lf
JP:. 1i
~ 2X 3
J/IT dAl. + IV ry aN:. rr + It d)(
- X%.(-2y) +(-2.XY){-).X) -= 2X'y
= 1f +IJ¥f +1'1' 1'1 -fiJI W = x2.(J) +(-2-xy)0) xz. -:J.xy
Q1.
'=-
Ihvs
J
~
3/1
.1
A
2
I'
11. - 2 X t + 2. X Y I +(x ... 2XY) k
.3
If. ./7
-4.11
The velocity of air in the diverging pipe shown in Fig. P4.17 is given by VI = 4t ftf sand Vo = 21 ftf s, where t is in seconds. (a) Determine the local acceleration at points (I) and (2). (b) Is the average convective acceleration between these two points negative, zero, or positive? Explain.
#-
b) convecfive acceleration a/on9 the pipe : : U where IJ. >0. IN aIJr lime' ~.... < Vt . r/;ps he/ween OJ a"d(:J.) J J JI£ ~ V2.- ~ /0 J
JX"'"
1
~
¥X
/lel1cc", 11. c::: 0 or is neq4live.
the tlver41e conl/eet/ve acceleration
If. {8
f
a
4. I Water flows through a constant d!arneter pipe with a unifonn velocity given by V = (8/t + 5)j mis, where t is in seconds. Detennine the acceleration at time t = 1, 2, and lOs. ~
a: : Tf + .-V.vV..... ~V
thi.s
becomes
cV
~V)"
Wi Ih _ dV
U == 0
a = -rr + v Jy J - n J ThlJs, a=-81./fi af t=/.s a = -2.0 J.f.. af t =2 oS and ......110
(
A
__
s
V -(T
J
~
1\
m
t2. J S'-
a =-O.OBJ~ at t::::/O..s
4. Iq When a valve is opened, the velocity of water in a certain pipe is given by u = 10(1 - e-'), v = 0, and w = 0, where u is in ft/ sand t is in seconds. Detennine the maximum velocity and maximum acceleration of the water.
/
+.5) ':
J
w-=o
~,zo ---_._-
~-
'-_ ...
-
4.20*
Water flows through a pipe with V = u(t)i where the approximate measured values of u(t) are shown in the table. Plot the acceleration as a function of time for 0 ~ t ~ 20 s. Plot the acceleration as a function of time if all of the values of u(t) are increased by a factor of 2; by a factor of 5.
a= elt~v +V'VV
t (s)
u (ft/s)
t (s)
u (ft/s)
0 1.8 3.1 4.0 5.5 6.9 8.1 10.0
0 1.7 3.2 3.8 4.6 5.8 6.3 7.1
11.2 12.3 13.9 15.0 16.4 17.5 18.4 20.0
8.1 8.4 8.3 8.1 7.9 7.0 6.6 5.7
~
thi.s becomes ~ (~U. ~IJ.)I\ a:: Tf+urx l
u=u(t) ~ tL =rr
1./1
or
v=:O
J
w=-o
)
1Jl ~t
/J "'x -
(I)
The IJ. ~U(f) ,rttph Qnd fhe hesf [// clIbic erlJatiofJ was ploNeJ ilsifl9 In £XCEJ.. Pr091'4RJ.
Thvs w/lh J
fix
:=
IJ.::: -0.0012.-/ 3
:: -
1~
0.0031
-
if fo//ows 11141
O,OI/2tJ. +0,97.5"
flo - O,01.Z.lft + 0.97.51
jt
.shown helolJl
J
where tl'VS
This acce/fJI'4/ion is a/so p/ofleri he/QUI,
Note Ih4-f if U increMes by fA Ihe ()'cae lera lion J tI)(.
.:=
If
J
fAvlol' does also.
01 K ti, e, K::: 2. ,,, K:::S)J
Water Speed vs Time u = -0.00121
9 B
3
-
Acceleration vs Time
0.0112r + 0.97561
5 4 3 2
-
7 N
6
1 0 ;c 1"11 -1 -2 -3
~
~ 5 :; 4 3 2 1
-4
0
-5 0
5
10 t, s
15
20
0
5
10
t, S
15
20
Jf.2.1
I
4.2
The fluid velocity along the x axis shown in Fig. P4.21 changes from 6 mls at point A to IS mls at point B. It is also known that the velocity is a linear function of distance along the streamline. Determine the acceleration at points A, B, and C. Assume steady flow.
VA
=6
VB
m/s
~
• A
• B
• C
too5m~ //// ,///////"" ,.
0.1
= 19 m/s
"'I
~
x
m----l
FIGURE P4.2.1
a.-
~
oV
-
-.
= 1T + V-VV
U==U(X)
v=O
)
J
and
w=O
fhi.s becomes
a =(!t +u ~J)t = u ~~ t
(I)
Since U is a /ineqr ftlnclion ()f X J cons/an/.s C, J C1. are given as: and m
0r
Thus J (). =( /20 X + 6 ) s with x~m From Efj. (I)
a=
U
~~ L -= (/20X + 6
or for XII =:0
J
~= 7~O
for X8-::: 0,0.5m)
and. l' rf!Jr Xc. == 0, / m
)
J
aB= a~c =-
(j,=
IJftiOtp.
Am 21 6 0 l':S'J-
Lf-/6
where fhe
I.JA " 6 :: Cz. ltD
=
19 =0,/ C, .r C2.,
C, ;:;: J20
P. (/20 :5) t r 1i
c,X +c2,
J
C2.::
6.
r
R ~~~==:-,.....----x
4.2'1
=
0 x = fl4
When a fluid flows into a round pipe as shown in Fig.
P4.22, viscous effects may cause the velocity profile to change from a-uniform profile (V = Voi) at the entrance of the pipe to a parabolic profile {V = 2Vo [1 - (r/R)2]i} atx = Velocity
,
0.75 R
,,~x =fl2
iI~ I~~'\\
e.
profiles for various values of x are as indicated in the figure. Use this graph to show that a fluid particle moving along the centerline (r = 0) experiences an acceleration, but a particle close to the edge of the pipe (r ""'" R) experiences a deceleration. Does a particle traveling along the line r = 0.5 R experience an acceleration or deceleration, or both? Explain.
0.5 R
I I
0.25 R
I
CL
I I
I
x = 3Cl4
I I I
,
x = t
O~--~----~~--~--~~--~u
(b)
FIGURE P4.2 Z ..,).
DV
..,). a =F +V'VV
~ J
v=o
J
and
w=O
we obtain ~
a:=
aU
~
dU
where ax == Jf + uTi
Ox t
:=
dU
UR
a)
Along the centerline (r =0) we finri li >0 and ~ >0 Thus J Qx >0 on r::: O.
b)
/Veo.r the pipe wall (r ~ R) we find Ii >0 btA <0 (i.e.; fhe ve/oc/ly changes from (j.::: Va fo u.c: Vo (J.s X increase!;.) Th v.s J ax < 0 for r ~ R ,
c)
R we r/nd u >0 and >0 'leaf fhe pipe en/ranee) buf 0 elsewhere. This is indi cofed /n fhe t;9IJre beJow.
-if
For r::
f
fx
#..c.
Th{)~JJ
U
I
I
J
ax >0 neQr the en/ranee and
-I -
I
for r = 1~ ~
Ox <0
elsewhere
r== iR
o~~x ¥x->o ~ <0 x = 0
x = I' (a)
tf -/7
y f-o------el2-----l·1
4.23
As a valve is opened, water flows through the diffuser shown in Fig. P4.23 at an increasing flowrate so that the velocity along the centerline is given by V = ui = V o(1 - reI) (1 - xII' )i, where uo, c, and are constants. Determine the acceleration as a function of x and t. If Vo = 10 ftls and = 5 ft, what value of c (other than c = 0) is needed to make the acceleration zero for any x at t = 1 s? Explain how the acceleration can be zero if the flow rate is increasing with time.
e
u
----i~~
u
=
u
=
1
-VoG - e-et)
L _ _~~~~~_~2_~~~ x
VOG - e- c0
e
~
FIGURE P4.23 ~
Q =iY.. +V-VV ~t this becomes ~ (~U JM)/\ a::: rt t u 1)( t == Cl~ t
J
v::O
J
and
w::O
1\
af t:: 1 s we mils! htlve with Vo::: /0 Q/Jd 1.::: 5 [c i(;t _ .:;. (/_ e- ct )2J :: 0 I fax
ce
-c
s:
0 for any
10 (
-5/-e
X
_C)2
=0
For fhe above condl'fionJ fhe /()co,/I/ccelerafio/J (ft- >0) is precisely Do/alJced hy fhe Colll/eelitlc decelero/i on (it :~ <.0), The {Iowrote if)crea.re.s wilh lime J /;vi fhe I/Vid flows 10 an area of lower ve/oc/Iy.
4-/8
4.24 A,fluid flows along the x axis with a velocity given by V = (x/t)i, where x is in feet and t in seconds. (a) Plot the speed for 0 :s x:s 10 ft and t = 3 s. (b) Plot the speed for x = 7 ft and 2 :s t :s 4 s. (c) Detennine the local and convective acceleration. (d) Show that the acceleration of any fluid particle in the flow is zero. (e) Explain physically how the velocity of a particle in this unsteady flow remains constant throughout itf. motion.
tf
tJ,J fp.s (0)
U. -;;; :
(b)
For
1j
X == 7
so at t:: 3.s
If J
J
Ii:;:
f !f
=3.s
2. 0
U:;.2 --.sfl
t
3
t
10
0
3
(e)
JU x It = --p.
an
JU X(')_ Urx =T T -
d
U, fp5
X
/;'"
2-
X= 7 f+
Fi9,2 ~
a=1¥
(d) For Qny fluid particle +V,VV whi c.n with v:;: 0 J W -::: 0 becomes ...... (dlJ. (j(J,)J'\ a :- "IT +uTx l
_ (
-
X + X ) -1 -7272- {
:=
°0
0
(e) The {Jarfie/c.s flow Inlo area-s of
hl9het ve/oc/ly (see r;.~./),
hoi af 0"1 9iven loca/Ion fhe veJooifj i.s de6rea.sifJ9 i/J Ilme (see Fig- 2J. For The 9iven ve lo&/fy field the local qnd COhflBr;fil/C acceleratio"s ore e'lv41 and Oll()s,ie) 1iflifJ9 '2 e/,O
acee / era/ion fhl'otJ9n ~vl.
4.25 A hydraulic jump is a rather sudden change in depth of a liquid layer as it flows in an open channel as shown in Fig. P4.25 and Vitleo V 10.6. In a relatively short distance (thickness = e) the liquid depth changes from ZI to Z2' with a corresponding change in velocity from VI to V2• If VI = 1.20 ftls, V2 = 0.30 ft/s, and e = 0.02 ft, estimate the average deceleration of the liquid as it flows across the hydraulic jump. How many g's deceleration does this represent?
.
Hydraulic jump
1'1 FIGURE P4.25 .-S
JV
Ii -:: T{ of V·vV .>
-
SO
A .'" w/fA -V:: IJ.(XJi J a =Ill. X
Will1 ouf knDWin9 fhe acftJa/ ve/oc/Iy Ji.flr;/'u/i()11 the (Jcceleraliol) can he approximated tis tlx
~ fA
.w ~ i (V,+~) (~i"') :; ; =
ThllsJ
- /.05
1(1.2.0
+0,3 0)
-33.,{!z.
J
if
JIJ,
1\
::t(,TXt
t( == IJ(X~
)!
( O. 30 -1.;).0 O.Oloft
'I-:k6 I -t.U.
A fluid panicle flowing
along a stagnation stream·
line. as shown in Vick,) Y4.:: and Fig. P4.26. slows down as it Stagnation pornt. t::: 0 approaches the stagnation point. Measurements of the dye flow ~ ~~J(f~~¥ FI Id It I in the video indicate that the location of a panicle starting on i,; ~:tt~" ";.i;oro. ~ v / U PC! Ie e the stagnation streamline a distance s "" 0.6 ft upstream of the 'tfi~ s stagnation point at ( "" 0 is given approximately by s =O.6e- o.s" , where t is in seconds and s is in ft. (a) Determine the speed of • FI G U REP 4 . 2 6 a fluid particle as a function o(time, Vpanicl.Jt), as it flows along the sleamline. (b) Determine the speed of the fluid as a func· tion of position along the streamline. V"" V(s). (e) Determine the flu id acceleration along the streamli ne as a function of position. a, :::: a,(s).
i
(,,) Vlilh s" 0.6 e-o.s if follows Ihat II ds ( ) _O.Sf ~ ~"J
P',.TfC e "-rr ,,~
(b)
~
0.6 -o.s e
_o.st
"-0.3 e
His
From pari (a),
V"
(-0.5) [0_6
_0.51]
e
_o.sl
where .s" 0.6 e
Thvs I
v~ (-0 ..»[5], or
V= -0.s5
[lis whero s-ff
dV
(e) For sfeadl {lOlli, I?r" V 4i
Thvs, wilh V= - 0.5s pnd ~ « Qs
" (-o.s 5) (-0.5) "
0.255
-
O,S,
fils' where
$-
H
No Ie : For S >0, Os iJ pari/lve - the parficlt:r acee/eralion is /0 /he ri'llrl. S i"ce fhe paMic/e is mavifltj fa the left. a plJS//ive as fof' fj,is ca.re implies -fhal Ihe p",.fide is dece/era/in,! (qS ,I m~f be for Ihis
sfa? nalion point lJow 1.
'f-:J./
¥.271
4~1. 7
A nozzle is designed to accelerate the fluid from VI to V2 in a linear fashion . That is, V = ax + b. where a and b are constants. If the flow is constant with VI = 10 m/ s at XI = 0 and V 2 - 25 m/ s al X2 = I m. delennine the local acceleralion.
the conveclive acceleration, and the acceleration of the fluid at poinls (I) and (2).
Wiih u =ax +b ) V=O J and w=o fhe accelerafion can be wriHen as IV, (I) h a = Ox t were Ox = /J.1X • Since /J. = ~ = IO.p af x =0 and Ii- =v.. =25.p. al x =1 we oMain 10 = 0 + b 25 = a +b So thai a = 15 and b = 10 Thai i.s, /J. = (!5X+10).If ; where x- m , so th.i from EZ'O)
ax
= (/5X+ IO )f (15'f) = (t1.5XTISO)
'/-22
f,.
4.21.) Repeat Problem 4.27 with the assumption that the flow is not steady, but at the time when VI = 10 m/s and V2 = 25 mis, it is known that aVl/ar = 20 m/s 2 and aV2/ar = 60 m/s 2•
.-
~
JV
~
~
Wifh U=fJ.(x)l:) J v::: o J and w =:0 the acce Iera f ,'on a =Tf +V·vV can be wriffen a.s a :: Ox f where Ox:::: +Ii J w/lh /.J.:::; a(f)x +btfJ. (I) III fhe given lime (fc:io) u:: ~ = IO.qz. ~.d X ==0 (JnJ IJ == ~ ==25 11 aT X:=/m Thus) 10 = 0 + bOo) 25 :: aCto) t ha,,) so fhai aCt,,)::: /s tJlld !JII,) == 10 II/sf) af {:::to J == ~~~L :: 20 ~ at X=:(}
#- ¥X
#-
and
~/J. = JV2. _ 0f 0 !l!... Tt .s:J- a f X-I - m oMeI : "'If,. I 'lese Qre
-rr -
(Jcce /er"f ions
The convective accelerofion al X::O (Ef - 0) i.s u¥x= (oxtb) (a) =(15(O)+IO)~/5 vJ'h//e a f '1= / if i.s
~):::ISO?
u~ :: (/5 (I) +JO)!J-(ls f) == 375ii
The fllJid accelerofion ,d t:::to Is
a = (#- +I.l ~)t and
== (2.0+ ISO)'( ~
= 170t
a =( 60+ 37s )t ';2. ::: 'f.9st-jt af X'::lm
~ af
I (J f time t= to
x::()
/
OC"
9.3 0 v
-1.30 An incompressible fluid flows past a turbine blade as shown in Fig. P4.30a and Vidt'o \'4.5. Far upstream and downstream of the blade the velocity is Vo. Measurements show that the velocity of the fluid along streamline A-F near the blade is as indicated in Fig. P4.30b. Sketch the streamwise component of acceleration, as, as a function of distance, s, along the streamline. Discuss the important characteristics of your result.
1.5 Vo
Vo
0.5 Vo 0 1\
Vis
D
C
E
.I'
(b)
Vo
II FIGURE
(a)
tA.s ::
B
.~
---1
s
A
A
where froRJ the fit/fire
P4.30
of V =VIs) The (vl/cfion
1! has the foJ/olNin1 Sh4I'S. F
.s
F
The ffwd decelerates from 1/ 10 C) 4cce/er4les {rum c If} IJ) and fhe decelerates a94JI; from [) 10 F. the nel acceler4li(}17 (rpm Atf) F is zero (i. 8/ ~::: V, =- VF ),
*
4.31 Air flows steadily through a variable area pipe with a velocity of V = u(x)i ft/s, where the approximate measured values of u(x) are given in the table. Plot the acceleration as a function of x for 0 :S X :S 12 in. Plot the acceleration if the flowrate is increased by a factor of N (i.e., the values of u are increased by a factor of N), for N = 2, 4, 10.
Since /). ~U(x)
x (in.)
x (in.)
u (fils)
o
10.0
7
10.2
8 9
1 2
13.0 20.1
3 4 5 6
20.1 17.4 13.5 11.9 10.3 10.0
10
28.3
11 12 13
28.4 25.8
u (ft/s)
10.0
-
v==o) and w=o /1 follows fha! (j= :: +v·vV simplil/e.s fo 11 = ax t where (Ix -::: U The valves U are given in fhe faMe ; the correspondin9 values of ~~ C4n be tJbt"ined by an approximale numerical diFreren-fialion o.s qiven in Pro9rom P'f#31 shoJ¥fJ he/ow, J
¥X
#-
No Ie fhat since (/x ~ f). if fo//ows thol and increose from u. 10 N U Increases Ihe occelerafiol) from Ox If) 2-00 110 120 125 130 135 1~0
150 160 170 180 200 210 220 230 2~0
300 310 320 330 3~0
350 360
in ve/ocIl, ;V2dx
cls open "prnl! for output as #1 dim u(l~), n(~) u(l)=lO.O : u(2)=10.2 : u(3)=13.0 : u(~)=20.1 : u(5)=28.3 u(6)=28.~ : u(7)=25.S : u(S)=20.1 : u(9)=17.~ : u(10)=13.5 u(11)=11.9 : u(12)=10.3 : u(13)=10.0 : u(1~)=10.0 n(l)=l : n(2)=2 : n(3)=~ : n(~)=10 print#l, "************************************************" print#l, "** This program calculates the acceleration **" print#l, "** as a function of position. **" print.# 1, "********************************************* ***" print#l, " I! for i = 1 to ~ print#l, II " print.#l, using "For N = ##";n(i) print#l, 11 x, in. u, ft/s a, ft/s2" for j = 1 to 13 a = n(i)-2*((u(j+l) + u(j))/2)*{(u(j+l) - u(j))/(1/12)) uavg = (u(j+ll + u(j))/2 x = j - 0.5 print#l, using "###.# ###.# +#.###----";x,uavg,a next j next i
Nole fhQf aHhou9h Ihe velocIty· da/a J U :: fJ. (X)/ o/fJerJrs fo he qulfe "smoolh '; the occelerat;on resvll J ox:::: I.i ~ is somewhaT irrerju/ar (especia/~ for x> 7 in.). I
(I)
~***********************************************
**
**
This program calculates the acceleration as a function of position.
**
**
************************************************ For N = x, in. 0.5 1.5 2. :> 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11. 5 12.5
For N = 10 x. In. u, ft/s, 0.5 10.1 1.5 11.6 16.5 2.5 24,.2 3.5 4.5 28.3 5.5 27.1 6.5 23.0 7.5 18.8 15.4, 8.5 9.5 12.7 10.5 11.1 11. 5 10.1 12.5 10.0
1 a, ft/s2 +2.424E+01 +3.898E+02 +1.410E+03 +2.381E+03 +3.402E+01 -8.455E+02 -1.570E+03 -6.075E+02 -7.231E+02 -2.438E+02 -2.131E+02 -3.654,E+01 +O.OOOE+OO
u, ft/s, 10.1 11. 6
16.5 24,.2 28.3 27.1 23.0 18.8 15.4, 12.7 11.1
10.1 10.0
a, ft/s2 +2.4,2l"E+03 +3.898E+04, +1.l,,10E+05 +2.381E+05 +3.l,,02E+03 -8.l"55E+0l,, -1.570E+05 -6.075E+04, -7.231E+0l" -2.438E+0l" -2.131E+0'" -3.654E+03 +O.OOOE+OO
30
u"
ff/.s 2.0
10
5
10
Xl in.
15
3000
aXJ £l1.s2-
2.000
1000
-
----
--
f
-/-
N=F
.------ ---,---------------- -:-------~-----,------~--
-.
---:-----
---
--------.- ---------
--- - - - . ---
-------~' --.------~.----.-
/.
0 -1000 -2~
~~---+-'------~,....------'J5--Xi-in;
___ ~\~~/O-.---- ________,______
'1.32. J
4.32. Assume the temperature of the exhaust in an exhaust pipe can be approximated by T = To(1 + ae-bx)[l + c cos(wt)], where To = 100 °e, a = 3, b = 0.03 m- 1 , c = 0.05, and w = 100 rad/s. If the exhaust speed is a constant~ mis, determine the time rate of change of temperature of the fluid particles at x = 0 and x = 4 m when t = o.
Since u-== 2 DT Dt
~
'OT
T
'V
=rr+ V' V
T
J
v;O
J
and w=O if fol/ow.s +h~+ ~T
oT
IJT
dT
oT
'dT
=rt+lJ.rxf-V 1y tWn-=-rr+UdX
Thus I = To (I +a ibx)(_c fA) sin(4Ji)+u.T,; (t+c c.o.s(wt»)(-ab e-b:x)
gr
When t;:: 0
:
%f = - ab 7; (I + c) e- bx or w/llJ fhe given data., %f == -(3) (0.03 ;h)( 2.~ )(100 (/+0.05) i o.o3X =-/B.q e:C where x,vm IJ
J
DC)
O 03X •
Thus, and
Iff-;-
J
/B.q:j- at
X;::O
J
t=:o
DT °c Df = -/6.8 s af x;: Lf »J J t =0
/f-27
4.33* As is indicated in Fig. P4.33, the speed of exhaust in a car's exhaust pipe varies in time and distance because of the periodic nature of the. engine's operation and the damping effect with; distance from the engine. Assume that the speed. is given by V = V o[1 + ae- bx sin(wt)], where· Vo = 8 fps. a = 0.05. b = 0.2 ft-I, and w = 50 rad/s. Calculate and plot the fluid acceleration at x = 0, 1, 2. 3, 4, and 5 ft for 0 :5 t:5 n/25 s.
Since .-
fj,=U(X,i)
_.-v =
~ aV a = Tf + V •V
J\
+ ae- bx sin(wtl)
FIGURE P4.33
v=o J and w=o if fo/low.5 thaf
J
Qx I.
v = vorl
,
h
DUo
au
were Ox:: Tf + U IX
(I)
bX
ThIJS, w/lh U:: Vo [I + a e- Sin (wt)] Ef. (/) 9ives Ox = Va aw e- bx co.s(wt) + Vo[I+ a bx.sin (fAJi)] Va a. (-b) e-bxsin(wt)
e-
Vo a e bX [ tQ COoS ((pt) - Vo b Sin (4JtJ (J ta e- bXsin (wi »)] W/lh Va:: 8!j J a;:; 0.05 , b =:: 0.2. ft and I» =50 !.1 d =
J
fh is becomes ax == O. If e- o.2.X [so cos (sot) - 1.6 sin (Eot) (J +0.0,5
where t~.s and x- ff PI07 tlx from £'1- (:J.) (or o:s t ~ if
s
w/lh
X =0) J) 2.,3, if,
AIJ £xce/ Prorflam Wa.r vs-eci 10 (;4/clI/4Ie ax frolll ore shown en Ih9 /Jext p4,e,
(con 'I)
eo. 2xsin (sot))J f}a. PJ
£r (~).
tlnrl51!,
The re.rvlfs
'I: 33-
{con 'I) t, s 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060 0.065 0.070 0.075 0.080 0.085 0.090 0.095 0.100 0.105 0.110 0.115 0.120 0.125
x=Oft 20.00 19.22 17.24 14.18 10.24 5.67 0.74 -4.23 -8.93 -13.08 -16.42 -18.73 -19.89 -19.81 -18.51 -16.06 -12.61 -8.37 -3.62 1.36 6.26 10.77 14.61 17.54 19.38 20.01
Acceleration at various x locations, fUs"2 x=1ft x=2ft x= 3 ft x=4ft 16.37 13.41 10.98 8.99 15.73 12.88 10.55 8.64 14.11 11.56 9.46 7.75 11.61 9.51 7.79 6.38 8.39 6.87 5.63 4.61 4.65 3.81 3.12 2.55 0.61 0.51 0.42 0.34 -3.46 -2.83 -2.31 -1.89 -7.31 -5.98 -4.90 -4.01 -10.71 -8.76 -7.17 -5.87 -13.44 -11.00 -9.01 -7.37 -15.34 -12.56 -10.28 -8.42 -16.29 -13.33 -10.92 -8.94 -16.22 -13.28 -10.87 -8.90 -15.15 -12.41 -10.16 -8.32 -13.14 -10.76 -8.81 -7.21 -10.32 -8.45 -6.91 -5.66 -6.85 -5.61 -4.59 -3.76 -2.96 -2.42 -1.98 -1.62 1.12 0.92 0.75 0.62 5.13 4.20 3.44 2.82 8.82 7.22 5.92 4.84 11.96 9.80 8.02 6.57 14.36 11.76 9.63 7.88 15.87 12.99 10.64 8.71 16.38 13.41 10.98 8.99
x=5ft 7.36 7.07 6.34 5.22 3.77 2.09 0.28 -1.55 -3.28 -4.81 -6.04 -6.89 -7.32 -7.29 -6.81 -5.90 -4.63 -3.07 -1.32 0.51 2.31 3.97 5.38 6.45 7.13 7.36
Acceleration, ax, vs Time, t
20 15 10 5 N
tn
¢:! ~
co
0 -5
-.~----'---
-10
~-----.--------------
-15 -20+-----~----~----~--~--r_----~----~
0.00
0.02
0.04
0.06
t,
S
1/--29
0.08
0.10
0.12
4.34
A gas flows along the x-axis with a speed of V = 5x m/s and a pressure of p = IOx 2 N/m 2 , where x is in meters. (a) Determine the time rate of change of pressure at the fixed location x = I. (b) Determine the time rate of change of pres-
sure for a fluid particle flowing past x = I. (c) Explain without using any equations why the answers to parts (a) and (b) are different.
*
2
a) Since fJ:::: /0 X /i follows IhQf
= 0 for a 1/
x.
b) 'Wifh fj,:::5x J /tr:::O )P=:OJ and rll==jox2.d follows
~1
::
*
~
+u~ +/Ir ~
-rhvs ~fl ==
/00
+jJr
~ :::
fha.f
11.* :: (SXlf)(2oX1.) == IOOX2. ~/.s
;;(6
Xc:/m
c)
For this s lea rJy flow fhe press/Jre at a poinT is cO/Jsiallf (p4rl (",)~ blli 1he pressure fOr a 9i1l811 parfie Ie chafJ?eJ; with lime (,~flf (bJ) heca/)se fhe pafl1 (~/e flow.: into a hir;hel' pross fire r e9io/J.
4.35 4.35 The temperature distribution in a fluid is given by T = lOx + 5y, where x and y are the horizontal and vertical coordinates in meters and T is in degrees centigrade. Determine the time ra.te of chan~e of temperature of a fluid particle traveling (a) honzontally WIth u = 20 mis, v = 0 or (b) vertically with u = 0, v = 20 m/ s.
DT
Dt ==
ThIlS,
and
~T JT JT rr + IJ IX + V dY
J
h
Jf
were 1T :::: 0
and v~OJ then ~ :::u{f:::(2.o.t;-)(IO !.#r)
if
IJ,::;2o'f
If
'" T iJC PC u =0 anJ v=: 20 7) fhel1 ft-;; vfr7 : : (:LO!)(5'-;;;) -::: jOIJ ~
. U6 At the top of its trajectory, the stream of water shown in Fig. P4.36 and Yid,'o V4.3 flows with a horizontal velocity of 1.80 ft/s. The radius of curvature of its streamline at that point is approximately 0.10 ft. Determine the normal component of acceleration at that location.
a:n ==
2.ij2.
VI 1?
= (1.8 ) o,/off
= 3 2.~
s2.·
1./-30
=:200!f
ffi = 0.10 It
m FIGURE P4.36
~37
4.37 As shown in Video V4.2 and Fig. P431, a flying airplane produces swirling flow near the end of its wings. In certain circumstances this flow can be approximated by the velocity field u = -Ky/(;r? + y2) and v = Kx/(;r? + l), where K is a constant depending on various parameters associated with the airplane (Le., its weight, speed, etc.) and x and yare measured from the center of the swirl. (a) Show that for this flow the velocity is inversely proportional to the distance from the origin. That is, V = KI(;r? + yl)ln. (b) Show that the streamlines are circles.
• FIG U REP 4 .37
or
V-= Ji r
J
Kx (x'".f y2.)
(b) Sfreamline.s are (piJen by
-Ky (x 1 +y'")
t
y1. :::
or X2. -l-
which when inferraled qives - f X1. +C'.J where t; is I-i cOlJrfalJ/.
y"J, :: Con.sfanf
1f-3/
_
x
-y
'1-.38 4.3){ Assume that the streamlines for the wingtip vortices from an airplane (see Fig. P4.37 and Vidt'(j V4.1) can be approximated by circles of radius r and that the speed is V = K/r. where K is a constant. Determine the streamline acceleration. as. and the normal acceleration. an. for this flow .
• FIGURE P4.13
a - V!lias ~
where
SlfJce
Thvs
J
f4.s -::
IJ/so
£
'i 2- _ (Klr):1 ::: K -3 r "-Y[- -r
Q
J
_
-
V-J: - r
J
if ~(J
4.3 q A fluid flows past a sphere with an upstream velocity of Va = 40 mls as shown in Fig. P4.3Q. From a more advanced theory it is found that the speed of the fluid along the front part of the sphere is V = ~Va sin Determine the streamwise and normal components of acceleration at point A if the radius of the sphere is a = 0.20 m.
e.
FIGURE P4.3Q
v= ~
Va
sin e == -j: (~O ';) sine == (6 . 0)2. m1
an == X1(2. == and ~v
0 Sin 1./-0 O.2m
. ) ~v
CIs ::: V15 == (60 smf) dS
From £~,(J) J
j ~ :: io s::: a8 = 0.1- e I1J
60
SJ::::: 7/f'fO
J
cos fJ
where
sine
f
(I)
~ s
JV
~v ~e
J.5 ::: Ie TI dB -
/lIsa J where e- raJ, J so f hQ +r.s - ~ Thus, for B= 'fOD as:: (60 sinIfOO,T)(6Dco.sIfOO~)(o.~m) ::: 886o~ J
4.40* For flow past a sphere as discussed iI} Problem 4.39, plot a graph of the streamwise acceleration, as> the normal acceleration, a,,, and the magnitude of the acceleration as a function of 0 for 0 ::; 0 ::; 90° with Vo = 50 fils and a = 0.1, 1.0, and 10 ft. Repeat for Vo = 5 fils. At what point is the acceleration a maximum; a minimum?
_r _(tVaasinet-_ Jfa 9 '10 a n
-1'(-
dV
and G.s = V Ts
~V de =V18 E
J
2
•
Sin
2.8
(I)
oV
3
where T8:: '".2 Vo cosfJ and or gjl =..L
Thlls as = (f Va sine )(t Yo co.s8) f = J
~5
ff
.s::: a ()
a sinO cos e
(2)
lienee The ma9ndIJde of the acee/era/ion i.s I i 9Vc 2. I . 'I 9 ~ 2. J I I = y an2. +a.: = Jl.ao Ysin e +sin2.8 (,os2.8 = If aD sine Vsir/e +co.s~e
a
i
or
2-
urI = ¥.i
(3)
Thvs,
sinB
Ia/min = 0
tlf 8=0 J la/mIX = ~
af fj =90
lin bxcel Pr09fafn"W4s wed to caJGt6'i4iB tl.s a/J 4/ld C{ frllm £flns. (fJ/)'~ afJa (1'J. The l'eJ()/ls are shown ie/(JJ4I~ 7lJe resilils tor ofher V()'/V8s Od'e simIlar ,llhe f4,,/pr Vo '"/a is 4CG()vnieri /I/", The folroWif9 derfa e,
is f~f Vo:::5 fils 1
Q::: /
ff
deg an, ftls2 as, ftls2 a, ftls2
o 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
0.0 0.4 1.7 3.8 6.6 10.0 14.1 18.5 23.2 28.1 33.0 37.7 42.2 46.2 49.7 52,5 54.6 55.8 56.3
0.0 4.9 9.6 14.1 18.1 21.5 24.4 26.4 27,7 28.1 27.7 26.4 24.4 21.5 18.1 14.1 9.6 4.9 0.0
0.0 4.9 9.8 14.6 19.2 23.8 28.1 32.3 36.2 39.8 43.1 46.1 48.7 51.0 52.9 54.3 55.4 56.0 56.3
Acceleration vs Angular position
Va ~ 5
fils
a :: IN 50
T---~~-4--~~~---~
N
I/)
1 -normal accel, ftsQ2 i
¢:! 40
1-
o
10 20 30 40 50 60 70 80 90
e,deg
-
-streamwise accel, ' fUs A2 - accel, ftlsA2
0
i
r
If. Iff
4.4 J
A fluid flows past a circular cylinder of radius a with an upstream speed of V o as shown in Fig. P4.41. A more advanced theory indicates that if viscous effects are negligible, the velocity of the fluid along the surface of the cylinder is given by V = 2 Vo sin e. Determine the streamline and normal components of acceleration on the surface of the cylinder as a function of Va' a, and
v
FIGURE P4.41
e.
an--
V2
(2.
_
f( -
and
0.s=
6V
V().f=
-
Vo.sin8/- _ 'tVQ.2. ",·n2. n Q a ~ r:7
V dV
~e
~B Ts
J
======== where ~~:::: 2 Yo cos e and lJB or rs
.s =: a8
I
Thus
=-L-
J
"
as::: (2'Vo sine)(ZVo cos(i)+ = ~ sinO cose
~.tj2"
I
4.4,2*
Use the results of Problem 4.41 to plot graphs of as and an for 0 :5 :5 90° with V o = 10 m/s and a = 0.01, 0.10, 1.0, and 10.0 m.
e
Fro/?? Problem if. 'II
Th ese
J
C1n:= If.cJIo:l S ifl e C/nd t1.s ::
results with Vc,-.::/o§! and a
"al4,/I. sine cos().
== 0.0/ J 0./ 0 ) /.0.1
and /0. Om
are plotted below. e,
deg
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
a =0.01 m a =0.10 m a =1.0 m a =10 m 2 2 2 as, fUs as, fUs as, fUs as, fUs 2
0 3473 6840 10000 12856 15321 17321 18794 19696 20000 19696 18794 17321 15321 12856 10000 6840 3473 0
0 347 684 1000 1286 1532 1732 1879 1970 2000 1970 1879 1732 1532 1286 1000 684 347 0
0 35 68 100 129 153 173 188 197 200 197 188 173 153 129 100 68 35 0
0.00 3.47 6.84 10.00 12.86 15.32 17.32 18.79 19.70 20.00 19.70 18.79 17.32 15.32 12.86 10.00 6.84 3.47 0.00
(con't, 11--35
a =0.10 m a =0.10 m a =1.0 m a =10 m an, fUs 2 an. fUs 2 an, fUs 2 an, fUs 2
0 304 1206 2679 4679 7144 10000 13160 16527 20000 23473 26840 30000 32856 35321 37321 38794 39696 40000
0 30 121 268 468 714 1000 1316 1653 2000 2347 2684 3000 3286 3532 3732 3879 3970 4000
0 3 12 27 47 71 100 132 165 200 235 268 300 329 353 373 388 397 400
0.00 0.30 1.21 2.68 4.68 7.14 10.00 13.16 16.53 20.00 23.47 26.84 30.00 32.86 35.32 37.32 38.79 39.70 40.00
(CO!) 'I)
100000
ES==:::E==t~
100000
B=:====:E===l
1000
t~_=~~~'=-=-=~~~=-,.-]
10000
1000 T--- -:-. --:---±~ I
N
T·-----·--~-
.Ie
E
,\
I '
- - - a =0.01 m
•
- - - a = 0.01 m
N
.\ .
-
--~;---'------~-
~.....---r-+:...,I------t---t---i
"
-
- a = 0.10 m
- - - - - a = 1.0 m - - - -a =10. m
.Ie
I
i- •
:, i
E
- - - - - a = 1.0 m
----a=10m
100
100
10
10
f-.r'--C/~~~~-~~~~-~.-~
:1
.
f--+<---_-;-~_+______i_----+----i--
II
r o
50
e,deg
100
---a=0.1m
o
50
e, deg
100
j
4.4.3
Determine the x and y components of acceleration for the flow given in Problem 4.6. If c > 0, is the particle at point x = xo > 0 and y = 0 accelerating or decelerating? Explain. Repeat if Xo < O.
Since 11.== c(x"J._ y 2.) and v= -2 c xy if fo//ow.s fhaf ~ A A a = 0xt +(JyJ
Clx ==1t +u-¥x +
or
Ox
h were
J
v-W == c(x2--y')(~cx) +(-.2cxy)(-2CY)
=2c 2 X (x2.+y2)
and
Oy == ~ + u1f + v¥y0"
=c(x2.-y2.) (-2CY) +(-2cxy )(-2cx)
Qy == 2 c 2. y ( x" +y 2. )
For
X=X o
lJ.:::cx2. o
an d
and
y::::O We obfain:
vr::O
J
Ox= 2C2. x:
J
Oy =0
Thlls J wilh c>o and Xo>O ;i foJJows
With
C >0
thai
u >0 J Ox >OJ i.e.~
the
flviJ l.s acceleratinq. and xo~o if foJlo~.s fhat u > 0 J ax
4-37
-fM
Water flows through the curved hose shown in Fig. P4.44 with an increasing speed of V = lOt ft/s, where t is in seconds. For t = 2 s determine (a) the component of acceleration along the streamline, (b) the component of acceleration normal to the streamline, and (c) the net acceleration (magnitude and direction).
o:.JI.
a)
J
hlJ f ~ dS =0
1
fdr a/I t. b) On:::
f
2-
or
an} ::- .5"(")'- == 2.0if,. t =2.;
c)
I}{ t ::= 2 s
or J
a~J
==
Ii = Cis f
+
all
n =-
/0; .,. 2()
n~
[2. Q.s + all"]~ ::: [1. 10 + 20 2.]~ == 2 2. -i' ?ff and
e :: arcfo,n(M-) ; 63. ¥fJ
tj-38
= 20 It
4.45 Water flows steadily through the funnel shown in Fig. P4.45. Throughout most of the funnel the flow is approximately radial (along rays from 0) with a velocity of V = clr2, where r is the radial coordinate and c is a constant. If the velocity is 0.4 mls when r = 0.1 m, determine the acceleration at points A and B.
o FIGURE P4.4S
'1(-:::00 (i.e) fhe sfreaml ine.s arB sfraigh t )
Thvs J
2.
_ (C )( 2..C) 2C Qs-- ]"'2. -? :::: rs o,I"m
At point _A2.!(Lfx/o-.3 sm )2. :::: 3 2. 0 l!!3
a.s -
(O.lm).b
's:L
0,11n
O,lm
2.
rOB = (0,1) +(0.06) :: O. /167m
o
2-
4.4·6 Water flows through the slit at the bottom of a twodimensional water trough as shown in Fig. P4.46. Throughout most of the trough the flow is approximately radial (along rays from 0) with a velocity of V = clr, where r is the radial coordinate and c is a constant. If the velocity is 0.04 mls when r = 0.1 m, determine the acceleration at points A and B. FIGURE )'4.46
a~ :: an n +lls .s A.
J\ J
h V2. were CAn = 1? =: 0 since
1?:= OD (,,~e~ the sfream Jine.s
are h V C n/so) a.s::: VT.S = - V1F Were ::: -;: Since V== o.oq.!J- when r:::. O.lm if follow.s 1hat -.3 c::: Vr == ( o.oIfIp)(O.lm)::: 'fx/o-3:z. or V:::: 'f-x/o r ~v
lJ
~v
I
-r,
Thvs,
as = -( ~ )0;1) =
IN poin! 1/: (J ~
~
(lfxlO
m2.
~)
f2-
(O.Bm't
At point B : ~::;
-.3
-.3
= 3.13 x/o-5 4oS
m2. ~
(ifx/o :$") (0. '}..rn) 3
sfraiqhf)
-3
= 2. oox /0
m
:sa.
!!l. r oS J whert~ c;;
rJ
Tn
thlf7
I
4.,47
Air flows from a pipe into the region between two parallel circular disks as shown in Fig. P4.1f7. The fluid velocity in the gap between the disks is closely approximated by V = VoRlr, where R is the radius of the disk, r is the radial coordinate, and Vo is the fluid velocity at the edge of the disk. Determine the acceleration for r = 1, 2, or 3 ft if Vo = 5 ft/s and R = 3 ft.
DX
7
J\
S/nce
Va::
A
5*
= -8,33~
4-JfI
~vo
V
1(:: 00 (t', eJ +he sireamJine.s
are .straighT. )
and R:=3f.J J V== Irs ~ J where rrvff
I
Q.s
~
J
Thvs _(YoR)(_ VoR) __ VrfR 2 _ _ a.s - r /,,2 r3 - , fJl r:: / ff J (l.s == - 225 !is.
Af r=3 ff J
~,r
r
FIGURE P4. 47
I
¥r
'I
R
~ t ~PiPe
h Cln :::: 'ItV2 :: 0 since a = an n + Qs S were Also, as = V ¥S : : V where V:::: V~R -
5
(
f+)"( )23£1 .. :=_
5':$
;3 fl3
'),.2,s 3
r
!
4. '1-8 Air flows from a pipe into the region between a circular disk and a cone as shown in Fig. P4.49. The fluid velocity in the gap between the disk and the cone is closely approximated by V = VaR21r2, where R is the radius of the disk, r is the radial coordinate, and Va is the fluid velocity at the edge of the disk. Determine the acceleration for,. = 0.5 and 2 ft if Va = 5 ftl sand R = 2 ft.
r----R-----+<.! FIGURE P4.4S 2 • h V a :::: an n + CI.s.s , were Cln == "1( -:: 0 smce aV av ,were h V----;:r \b R2. AJ so) O.s -- Vr.s - V1"F -
A
A
Thvs ~(~) (_ 2. Vo3 R2.) = _ as r2. r
2. '" ~ RI/-
r- 5
lit r ::o.sH J O.s = -2.5)600 ~
4-42.
:::
-
'f<::co (e:e'.l fhe .s+ream/ine.s are .sfra iq hi) .fl
2.
2-(5 s) (20ft) .s .cl"s
r
TT
.
If
__ -
800
r
5
where
.fi s rNf-j
4.4Q Water flows through a duct of square cross section as shown in Fig. P4.4Qwith a constant, uniform velocity of V = 20 m/s. Consider fluid particles that lie along line A-B at time t = O. Determine the position of these particles, denoted by line A' -B', when t = 0.20 s. Use the volume of fluid in the region between lines A-B and A'-B' to determine the flowrate in the duct. Repeat the problem for fluid particles originally along line C -D; along line £-F. Compare your three answers.
B
A
D
B'
A'
F
C
E
FIGURE P4.4Q
Since V is cons/ani Ii) time and sptlce J all parfic/e.s on line IJB move a disfal/ce f -:: V At :d(:2o-!;) (o.2.s) =Lf/J'J from t ==0 10 f~ O.2.s 3 Thv-s, the volume of ABA'O' l.s ~8A~' = (o.Stn)~(LfIn) :::/.00 m so thQt 3 = ~8t~1 = 1.00 m ::: S.O ~3 Q 0.2s
$
Simdar/y from t==o fot==o.2..s fhe {Ivirl a/ol/9 JilJes CD and EF move fo C'/)' and E'r;respecfiveV. II/so, ~DC~/:::: ~F$~I ::: ~BJ981 so fhat weobfat'n Q::: ft : : 5.fJ!f re9ardlcss which line we consiclel'. '1-.50 20 m/s
B
4.50
Repeat Problem 4.4 qif the velocity profile is linear fromlO to 20 mls across the duct as shown in Fig. P4.50.
A'
10 m/s
FI GURE P4.S 0
From
t ==0 fo
I:~ 0.2$
the parficle imtially ai B fravels a distance 1B : : Ve At ::: (20.!p- ) (o.2.s) ='f-m as Sholll/} wht'/e one af II/ravels a Jisiance If/ ~ ~ At r:::~o-¥) (o.'1.s) == ~m. Since the '1e/ocdy prutlle is linear i/ne 198 remaiIJSsfraI9h.. blll ..h11.s..;..10Iinell.ll~ TAvs !he VOIllfll8 of flvid cros.silJ9 the in//;4/ IIIIe III ;'s J
t:SB1A' :::
so that IJ:: If
i
(iA of J8 )
~BB~I = - .6t
II ::; i: (2 m +'1-111) (0. Sin /. ;: 0.7£ ",3
t'
__F"-----ft J
O.7S m O.1-s
:::
3.75
d oS
F.Or any cvrved line £F (which /noves fo £'FI)
E
£'
~FF~' :: ~BB'R' sO thai the same I/oillme f/oUlrtJie) Q, is obfained for allY l/nB cunsidered. lJ-'f3
?': 5 I
I
r~---------~Control
1-
-~.!"r_~~: /Sluice gate
surface
I
In the region just downstream ~f a slui~e ?at~, the w~- '~~i: Vb = 3 ftlS~1 t~r may develop, a r~ver~e flow reg~on as IS. mdlcated m,' ..•.. ..' ~-------:-:::i=--7-~ ,.J ~~'"i=~.---': ~ FIg. P4.51 and Vldl'O \141.:-. The velocIty profile IS assumed to I ., '\' ".~. ~.. A~ .. ' 1.8 ft consist of two uniform regions, one with velocity V" = 10 fps , ' ~ ~': ' . ' ! and the other with V/J = 3 fps. Determine the net flowrate of \................ ' . '. --..... ·ct,..;:::~ ......· 1.2 ft water across. the porti?n of the control surface at section (2) if .... r7~/. --",. ~ ;,-:t-?,; the channel IS 20 ft wIde. (1) (2)
"'.51
VV. .
Va = 10 ftls
FIGURE P4.5'
= (10
Q-
if) (J.2.ft)(:J.olf)
_ /32
-
L{.. s 2.
/4.52
fI.3
oS
At time t = 0 the valve on an initially empty (perfect vacuum, p = 0) tank is opened and air rushes in. If the tank has a volume of -Vo and the density of air within the tank increases bt FOr J ] so
t ~o
cJM
p::. Po [I - e -bt
Thvs} df == po ~ b e
-(3!f )(J,Bff)(;'()ffJ
as p = p.,(1 - e- bl ), where b is a constant, determine the time rate of change of mass within the tank.
fhai M::: mtJss of air /n tanK =
eVa :::f'oVo [I-e- bt ]
4.5'1- : 4.5+
Air enters an elbow with a uniform speed of 10 mls as shown in Fig. P4.S4. At the exit of the elbow the velocity profile is not uniform. In fact, there is a region of separation or reverse flow. The fixed control volume ABeD coincides with the system at time t = O. Make a sketch to indicate (a) the system at time t = 0.01 sand (b) the fluid that has entered and exited the control volume in that time period.
Control volume
c 15 m/s
From i =0 to t:::: O.O/s parlic/es IIJ BJ ~ D and. E /!J()ve the followin9 distances: ~::: ~ dt -:: (10.!Jl- ) (o.o/s) :: 0./ m :: OD J
JB -:: Va Ji:: (5!j) (o,o/.s) =0.05 m 6c~
'!cit:: (J5~) (o.OJS)
=0./517)
and
J
a =0 E
Thv.s fluid. a/on9 lines /JIJ fill;' ana B'E'c' shown be/()w, J
1J1Jd.
BEe ori9in(J/~ moves 10 lines
O./m
-1
~
D ,\:_ I.. '\I J)I ...... -..;;
- - - - .sysfem at 1::::0
~
- - - - system
at
i :::0.0/£
//// {Iu/J -IhaT exHed con1rol volume
\ \ \ \ flvid tho" entered. cunlrol volvme
4-Jf5
'1-.55
J
.f.55 A layer of oil flows down a vertical plate as shown in Fig. P4.55 with a velocity of V = (Vo/h 2) (2Ju - .xl) j where Vo and h are constants. (a) Show that the fluid sticks to the plate and that the shear stress at the edge of the layer (x = h) is zero. (b) Determine the flowrate across surfaceAB. Assume the width of the plate is b. (Note: The velocity profile for laminar flow in a pipe has a similar shape. See Vidt.'o V6.(t.)
Plate ~
A----~---~~~-~-----B
a)
fir
==~(2h)(-X2.)
Thus) /1T
I = i£- (0 - 0)
Oil
and
== 0
1+----,.- h - -
X==O
1i/=JJ{r[2h -2xj '"' ()
' / =,# x=:h x=h
x=h
f/ence) fhe flvid sficKs 10 the p/tJfe and there is no shear slress al .Jhe free S{)rf4c.e.
x=n
b) IilIiB
'"
h
JIlT dlJ:= J/If" bJx =f.}a (2hx _X2) b a'x x=o
0
or
Vo b [h X2. -:3J x3 / Q118 -- V
h
()
hb -- 1::. 3 \I Vo
If. 56 V1
4.56 Water flows in the branching pipe shown in Fig. P4.S6 with uniform velocity at each inlet and outlet. The fixed control volume indicated coincides with the system at time t = 20 s. Make a sketch to indicate (a) the boundary of the system at time t = 20.2 s, (b) the fluid that left the control volume during that 0.2-s interval, and (c) the fluid that entered the control volume during that time interval.
V3
=
2.5 m/s
- - - Control volume
FIGURE P4.56
V/.s consf4nf J the flvrd trovels a disitmce i == VJt fime At, Thus -'t:::: Vt of :: (2.J}!-) ('20. -2.0)05 = O./ftn 12.. = V:l. Jt = (1 -P) (zo, - ')..o)s :::: O,2m and 1,3;;: ~ U = ( 2.5 !f) (').0. - ')..o)..s = 0.50 m The sy.sfem of f == 2-O.2.~ and fhe {Iuid fhal has eIJlercd or exiled. the conlrol volume are indicafed in fhe {''lure below. Since
J
-
-
-
confroJ volume
------- -- sysfem at i=20,2s
117
=
2 m/s
*S7 Two liquids with different densities and viscosities fill the gap between parallel plates as shown in Fig. P4.57. The bottom plate is fixed; the top plate moves with a speed of 2 ft/s. The velocity profile consists of two linear segements as indicated. The fixed control volume ABeD coincides with the system at time t = O. Make a sketch to indicate (a) the system at time t = 0.1 s and (b) the fluid that has entered and exited the . control volume in that time period. ..LSi
II FIGURE P4.57
The flv/J af y ~ - 0.'1-11 (Ihe hoflom pia Ie ) relJlaifls slafiP/JdflY. AI y::: 0 fhe flvid speed is I,.s fflr .$'f) -fhaf af fjlbe t ~ 0,10$ ;/ A4s fYJDved 10 -Ihe rifJhl 4 d;slance X:: Vt:: /.s # (o,/s) ::: o.lsll. In Ihe sahJe I;",e periOd Ihe fop p/IJ'/e anrllhe fluid sff)ci If) il has /hoI/sri a di.rt4fJCe X ~ 2. !f (OJls) -;:: 0.2 If. Since fhe velOCity prof/Ie is piec,ewire //neat) Ihe enJ.r () f the sysfeM 1I1t// /hove.,jt7 IhtJr Jines- At; 4"rJ, Be ~emain straiqh-l, This is ilJdicated in the.skeloh below. y
~2H
41#
rAl ID ID I AI!" r--~~-= --- ~-~--~ f/viJfhq,t
-:1 rlIJidfh4+ entered con1(IfJ I
.---r_~/: V/:
'
VO/fllYJe
ex/fed vo/Vti/e
co"'~1
,X r //
l-r.
.-J\I
(f
I-
~; 0, IS \' 0, IS ft B ~; --= -:....=- -=-=-- =-= --=-- -=-=- --=-- d Co ~ C I
-
-
-
cOlliro/ vO/llf11e ----- - - - - - sysfem af I ~o,J.s
~,se
4.5 Q Water is squirted from a syringe with a speed of V = 5 m/s by pushing in the plunger with a speed of Vp = 0.03 m/s as shown in Fig. P4.S8. The surface of the deforming control volume consists of the sides and end of the cylinder and the end of the plunger. The system consists of the water in the syringe at t = 0 when the plunger is at section (1) as shown. Make a sketch to indicate the control surface and the system when t = 0.5 s.
.. ".' ~ . ' V
(1)
r--
O.08
= 5 m/s
I m--J
FIGURE P4.E9
DIJriIJ9 the i == 0.5s lime irderval fhe p/lJnger moves ~:::: Vp Jt :::O.OJ5/r). and fhe waferiIJ/lio/1y af Ihe exd moves 12.::: V6i ::::2..!5m. The corresponrJifJ9 confrol surfaces and systems af t=:O t1nd t= O.ES S hOWl) in fhe fl9vre be/rJU/.
"
.....
,:
....
,
,-:. : .. .,' '...... 'r:::·.' :.....:.::.,:/J "
..
110..
..
~~.:.~.;t!!.. =:.-~L.~;"T"':'"",'T.-•• ":'":-:-••- ..........-.- ••- .- "••-- .-.,'-',-"j.,..-:-:••""".-'• . . . ,.,..... ~ -.~••~~-:-.~ ..... " ~
..•..... :, ..................... : .... ,
l- "~~:....:. fI) I..
2 . .5
m
O.06.5m
O.OBm
control Vo/()me af t= 0.5.$
system 41
f=0.55
~
/1-, s9 4.59 Water enters a 5-ft-wide. loft-deep channel as shown in Fig. P4.59. Across the inlet the water velocity is 6 ft/s in the center portion of the channel and I ft/s in the remainder of it. Farther downstream the water flows at a uniform 2 ft/s velocity across the entire channel. The fixed control volume ABeD coincides with the system at time t = O. Make a sketch to indicate (a) the system at time t = 0.5 s and (b) the fluid that has entered and exited the control volume in that time period.
6fUsl----
", .".......................... Control surface
II FIGURE P4.59
DtJrin9 the l :::0,5 s I/m8 1,,/el'lItJ/ the Ilv/rJ ff;a! was a/()119 //n8 Be Af lime t=o h4J' m()vea I() /he rl?111 III dirfaJlce J. ;: Vi::: 2 f1 ((j,ss) ::: / If, Si/IJilri"IyJ porfions oflhe fluid a/on9 Iills AD have fHPved J.::: 1# (o,s-.sJ ::= (),sll ()nd 1.::: b!/- (0,£ s) ~ ~ fl. TIJ/s 4.5'.[(//118$ fhe and,
/!!
6 ij flvid streallJs d() !'Jof nl/x Dr IAlcrmillf/R t:/vril'J91lJe (),£ s time lidel'va/. See ftf;tll'e be/ow.
~J
A
B
I:::: - --=-=- =-=-- -=-= - ~ --:-:. - =-=-- - -r..:. -; ~: ••
•
I
-
"1
t
_
_
_
_
_
_
_
~
-
-
"
l'
.. , J
-
.'-"~-
r···':
:"'1 • • 1::.= _ _ _ _ _ _ _ _ _ 1> \-D
----------
-
-
_
-
• I
-
f/vid fH41
exiler./,
cOfJ/~/
VOIV/H8
'I
.',1 ~ I
_":'_1
-c
Iff
..
;.'
f>.:------.-..:· ....... o.,sH J
"',:---
~r' I \'. '.' ~
_
""'1
Vo/u,"e
,
',t
flfJid tha14f;: en fe red. ~ :;fr:·:..:... ~ coniroJ 1:-:",. ':- ,~ ,.-..I.
..
8'
c
I
fixed con/I'o/ volt/me
___________ sysfem af 1::0.5s
4--50
'1-.60 1 4,60
Water flows through the 2-m-wide rectangular channel shown in Fig. P4.60 with a uniform velocity of 3 m/s. (a) Directly integrate Eq. 4.16 with b = I to determine the mass flowrate (kg/s) across section CD of the control volume. (b) Repeat part (a) with b = 1/ p, where p is the density. Explain the physical interpretation of the answer to part (b).
D
A
~
V= 3 m/s
r+
B
C
Control surface
a)
Bow == S rb v·~ rAil
/) D
i I
cSouf
-"
A
Wilh b =/ and V·n
==
VC()s{) Ihis becomes / ; ~n c
Bf = SoV cos f) dA =-~V cose SJ,/} O/)
v_. .
c/)'
CD
=pV c~B
ACJ)
IIcD ::: 1, (2"') _(0,5 Ih)(').fYJ ) - -
where
J
CfJ,r8
b) With b::: Bout
J/p
S
£f'O)
~ v·~ dfl CD
becomes
=J Vcosf) dlJ:::
Vcose lie/)
Gl>
=(3~)cosf) (~();8)n/·
m3
:: 3.00-:S
r
Wilh b'" Vp = ~) = !:~ if {,fj,w.s fhal '8 ~ V,//lJfle' vol • (i.e:; b = -!a.r,s) .so fha! f V'n till ::: B,uI refJI'8seIJIs fhe f/O/{)1fI9 f/ollr""fe C",3/s) frolll fhe cOld!'(;! 1f)/PIfIO.
ii-51
(I)
'f . (, I
I 15 ft/s
4.61 The wind blows across a field with an appro;(i matc velocity profile as shown in Fig. P4.6 1. Use Eq. 4. 16 wi th the parameter b equal to the velocity to determine the momentum fl owrale across the vertica l surface A- B. which is of un it depth
inlo the paper.
• FIGURE P4.61
y"20fi
- f ~Vt)[ MH'] (lWely y.O o
-#
Buf, V= ;~y for o~y~loff(i.•. , V;o,,+y,Oj V='S'~alrlo) and V; Is1f For y",o(l
- - fr[ J(1fyfdy SostJy]
Thus,
,0
2D
'
+
o
10
=
fF[2.1.s
s
fl + 2~ YI ] 0
10
=0.00238 S;7s [750 :( + 2250
~'],
If-52
,.
0
5·/ S.l Water flows into a sink as shown in Vidt'H V5.1 and Fig. P5.1 at a rate of 2 gallons per minute. Determine the average velocity through each of the three 0.4 in. diameter overflow holes if the drain is closed and the water level in the sink remains constant. Three O.4-in. diameter overflow Q = 2 gal/min
•
for
the
FIGURE PS.1
confrD/ vo/vme
indt'caTeJ, 0 .. 00'1-'1-£
lj-.1 /,70 1i s
5-1
5.2 Various types of attachments can be used with the shop vac shown in Video V5.:. Two such attachments are shown in Fig. PS .2-a nozzle and a brush. The flowrate is I ftl/s. (a) Determine the average velocity through the nozzle entrance, V•. (b) Assume the air enters the brush attachment in a radial di· rection all around the brush with a velocity profile that varies linearly from 0 to Vb along the length of the bristles as shown in the figure . .D~temtine the value of Vb'
• FIGURE PS.2
rlJ ='41z
(a)
Thvs, II, V, =' Qz
Of'
so
v"
= "f S. 8
~=
.fj
lind Q3': ~ 113 where Q" =' / .fl' s ayerage velocify at (3) = Vb
±
113 =' 7T lJ, h3 Thvs,
-} v;, [17 (ft Fil (# fi l] Yo
='
2M
=/
#-
5 - '2-..
i/,
Of'
5.3 Water flows into a rain gutter on a house as shown in Fig. PS.3 and in Vidl'O \'1.0.3 at a rate of 0.0040 ft 3/s per foot of length of the gutter. At the beginning of the gutter (x = 0) the water depth is zero. (a) If the water flows with a velocity of 1.0 ftls throughout the entire gutter, determine an equation for the water depth, h, as a function of location, x. (b) At what location will the gutter overflow?
(a) Ft7V the,
Q in
==
(Cl1M
h::
.shown
IV)
-the sketch abdVe
Qo~f
(O.ooLJ-O ~/1) (x ~o
VO/lAme.
if)
0.012 X
h :: 2. ff J~
So
.L f+
_
~, 012
X
I)"
x -...
10.g
f+
5'_ 3
5. if
I
; /
~b
504 . Air ~ows steadily between two cross sections in a long. straight section of O.l-m inside-diameter pipe. The static temperature and pressure at each section are indicated in Fig. P5.4. If the average air velocity at section (1) is 205 mis, determine the average air velocity at section (2).
Section (1)
PI = 77 kPa (abs)
P2 = 45 kPa (abs) T2 = 240 K
TI = 268 K VI
=205 rnIs
II FIGURE PS.4
This
C/;1aly.fis
For
steady
Is sl-"'i/(J~ ~ flte. dJ1e ()f' Cxtll11pJe 5:2.
/kw i>elweeY7 $ecl-/tJns
(I) and (2)
WI,.. - I'h I or .oA V :: / .).
,2;'"
~ 1/
A. V
I
I
Thu5 V=
--
(I)
2..
ul1der +J,e UJi1dihilns of f~i.s
AS$um/"!J t-haf behllj/e~
a~
we
~~~ !i~. ~
~
C owrb/n/nJ we. gef
Vz. = p,
('deal 9a5
Ihe (2.)
P T .,.
U$e
ClIT
date (E~. /.8) 10 get-
etjutrHoYl of
,.,
ie/eal gt/5'
Cin
,Pl7Jblel11/
f
Etjs.
~
'3/1- t.f 7j !!!. 5
I
V,
anti 2.
411&1
observ/~
:. (11k fa.(abJ )] ('2'1-0 J< ) [ ,/5 --APt;. (u105 )](268 k)
Mal
(-Z(}5
~)
5.5
The wind blows through a 7 ft x 10 ft garage door opening with a speed of 5 ft/ s as shown in Fig. P5.5. Determine the average speed, V, of the air through the two 3 ft X 4 ft openings in the windows.
!').:')
OS.~ I!",: Tits.
16 ft
10 ft
.
1~~1:30' II-·- - - - - 2 2 f t - - - - - - !
•
roy-
sfeady
or
Q.9a~e ad"Y' A V -'Jt1tr1fe 1'Jf?Y'Mt%/ -It, a r ,1U'a.9e. 4"OY' c)(J
v:
hie ()..~je ~~eJ) Aa/t.aAe. VMt"",,~
$0
V:'~
t1 df +r,
w~"drJw
FIGURE PS.S
V + A,
V
iJlmeifMI
fhe ~/r
III1'"4.'It
boY'
tI
=
5.6 5.6 A hydroelectric turbine passes If million gal/min through its blades. If the average velocity of the flow in the circular cross-section conduit leadin2: to the turbine is not to exceed 30 ft/s. determine the minim~m allowable diameter of the conduit.
I=or
A
IncompreSSible
CtJndJlif
flow
through fhe conduif a#d fur6ine.
v
C (HId 1.1if
ana
-~oYldui-f
d CtJMu/f d
Cf)ndui f
::
I~rs ff
5-5
4 )( /0 6~ min
( Lj. ) (/,
)
ft/'
5.7
I
5.7 The cross-sectional area of the test section of a large water tunnel is 100 ft2. For a test velocity of 50 ftls, what volume flow rate capacity in gal/min is needed?
~~AV
Q:= OO(){fJ.)(50
£! )(7.~P: ~ ff-]
S
)1/ / ) (t"O ~ m/n
5-7
·' ,
5.8 A hydraulic jump (see Video \,10.5) is in place downstream from a spill-way as indicated in Fig. P5.8. Upstream of the jump, the depth of the stream is 0.6 ft and ~he average stream velocity is 18 ft/ s. Just downstream of the Jump, the average stream velocity is 3.4 ft/s. Calculate the depth of the stream, h, just downstream of the jump.
(2.)
, I
'h
•
Ft,y
5feady
,;" coYJlfpfesj;6le
FIGURE
flow belween
sech'()n~ (/ )QJ/1d(Z)
Q, = 0.2. Or
V,-A ,=
Thus
~ hi and
h= 2.
- h. _
~
l{
( / 8 !j.) (0. 6 1+) = (3,tf ~-f)
5-$
~ 3.4ft/s
3, /8
-ff
conf"ol V"/U1't1.e,
5. q
A water jet pump (see Fig. P5. q) involves a jet cross section area of 0.01 m 2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross section area associated with the jet and entrained streams is 0.075 m2 • These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 .m/s through a cross section area of 0.075 m2 • Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/ s.
V3 .:: 6 V, :: r----'--;f---;P--~. 30 m/s jet I su+io,,~) I
FIGURE P5.
q
For sfeady in compre$sib/e flow -I-hl'ough +he.. COfJ~/ volume
0-,
+
Q.2. = QJ
or
-
V, AI +
Q:7.
Thus
Q2
-
'V;A 3
-
V,A,
/50 trIer)
s
m/s
5./0
5.10
Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min (see Fig. PS.lO). If the level of the water in the tank remains constant, calculate the average velocity of the flow leaving the tank through an 8-in.inside diameter pipe.
--------
Section (2)
Q2 = 250
gal/min~
Section (3)
FIGURE P5.10
For :'feady and /11 compress/hIe flow
0. 3
= QJ + Q2,.
or-/
1;d2 -"3
L.f
250 ~I'J'I1 ')(231
/
/
Vi.) gal A60 ~AIz !:!:..3 V.i.
;n.
J?'11;,
If
f+ ~ fuel in
5./1 5.11
At cruise conditions, air flows into a jet engine at a steady rate of 65 lbm/s. Fuel enters the engine at a steady rate of(!J.60 Ibm/s. The average velocity of the exhaust gases is 1500 ft/s relative to the engine. If the engine exhaust effective cross section area is 3.5 ftl, estimate the density of the exhaust gases in Ibm/ft3.
For
I~ I exhoIJ.S-I ~~==~~-~-~-~-~!OS6
conh-o I
sfeody flow
VtJlume
or
Ij A3 ~ ::: Thus ~
,
-
111,
m-tfYI I
+- m..2,
2.
65 ~
+-
O.
CO
!?
(3. 5 ~+2) (/500
5'- 10
fJ )
(J"f
5.IL 5.12
Air at standard atmospheric conditions is drawn into a compressor at the steady rate of 30 m3 /min.The compressor pressure ratio, PexiJ Pinle!' is 10 to 1. Through the compressor pi pn remains constant with n = 1.4. If the average velocity in the compressor discharge pipe is not to exceed 30 mis, calculate the minimum discharge pipe diameter required.
seclion(t) (inlef)
, --'r--------
•
I
----.I,
secl/on(Z)
I
I c'mf~1 (eX)f),: , VtJ!u/tJe. ~i-----1I~--7
compressor
FoY' sfeady , . m,.... =f}?I
flow
-
;:
or -V f!A :2 ~ 2-
50
d2.
di.tc/1drj€. pipe
(L I
Q, :=
-1r ~ ~
,
-
(;' yo;
d 2 ::: 0.004 m
5, / I
30 PlII'/ ~
-
/0
fil1ally
:3
--'- (
~
)
30!j-
60~
min
5·/3 Two rivers merge to form a larger river as shown in Fig. PS.13. At a location downstream from the junction (before the two streams completely merge), the nonuniform velocity profile is as shown. Determine the value of V.
fhe..
U$e IYJ
fne
~
CI/YJ1Y?J1 VO/vln1C 5h~wn s ke I-c-t-t ab(}Ve.. {tie nole
~
((/Y)s~(Vtvf/on
gelmI
+VYl 2
oA v
{
I
I
t
fha!
brol
of masS' fY/YJl1pJe
.
_ m
wit1J1~
-
3 -
Y11
().fv
t-
fA0.1 II0.2 V
fA z. ~2 ..
v
tv1
vve..
v
+ f A\I
V
=(50ftJ(3ff)(3f! t(iM! {Slf)ftft) ('Yo.ft){6ff)(o, '6)t (7()ff}{i H)
v
~
3·63
=-
ti
5
5-/2
5./tt Section (1)
5.14 Oil having a specific gravity of 0.9 is pumped as illustrated in Fig. P5.14 with a water jet pump (see Vi(ko V3.6). The water volume flowrate is 2 m3/s. The water and oil mixture has an average specific gravity of 0.95. Calculate the rate, in m3/s, at which the pump moves oil.
~
Section (2)
Water and oil mix (SG = 0.95)
t
Oil (SG = 0.9)
{or sfeadJ flow
FIGURE P5.1tf
.
m+m , A.
Or ~~ +
A/50 I
t Qz. =
Since
Q"
Comb/nt'nJ
I? Q,
(I)
fhe wakv and oil /11ay be. Qz
-f
Ij Q3
=
£9.$. /
a/l7d
z
we
If ({), f
(jef
0( 2
)
or
SGzQ,
= 5G J
(Of
f
Qz)
Qnd Q'2.
-
Q,
(
1- 5G3 )
SG 3 - 5G 2-
Thus
~
::
(1- ;3)( / _ 0.95) O.9S -
Ii, clJrnlre»/61e.
(2)
Q3
+ /} Qz
QI +
cOh.$t'de red
:::
2.00
:!!. J s
0,90
5- /3
5·/~
I
Air at standard conditions enters the compressor shown in Fig. P5.15 at a rate of 10 ft 3 / s. It leaves the tank through a 1.2-in.-diameter pipe with a density of 0.0035 3 slugs/ft and a uniform speed of 700 ft/s. (a) Determine the rate (slugs/s) at which the mass of air in the tank is increasing or decreasing. (b) Determine the average time rate of change of air density within the tank .
- --
.,.-----
Tank volume = 20 ft3
-~\ 1.2
In.
! _
700 ftls
- - - - -r b.0035
W; ff"n the hyo/< ~YJ //nc5. c (/y}se. '(VCi -h'dY) of m~>s prj ~/-e we !}ef
use the.. CtJrl-lrvl (oj ~
D MJ'(!
+ne _
vtJ/umf:.
-1"7 [ I
,
_~
Dt-
f)iJ1WJ = 0.00'15£ [)t
(h)
oMs-ys Pt-
So
~ Of
-
slugs!ft3
O{f
~/}CJ..-L)
01At-
c,~
SIll!!
'" "'Yea.! J'!}
~r~ ) = V$,s
Ot :::
A V Out
(Jur
-
s
0.00 LfS{
r OLAi
In
r!J.J J"
De Dt
-
:::
o. ooLf56
6.00'fS£ 20 ff'?
~>fI
5-/'1-
s!t{.J .s
-
~/Lf!J
.s -tf 2·ZPXI0 51'1J
-
-
I'f~
>
5.16
An appropriate turbulent pipe flow velocity profile is
V =
lie
'rn
(R ;
i
where lie = centerline velocity, , = local radius, R = pipe radius, and i = unit vector along pipe centerline. Determine the ratio of average velocity, u, to centerline velocity lie for (a) 11 = 4; (b) n = 6; (c) n =8; (d) n = 10
For'"
Ii?
any =
sec lion
CYOSS
,..oA u
-
cross seclion ayec<
area
J,P v. ndA A
A/50 ~
.J..
~A
.A
V. n ; V. l = uc for
u.
lIni~rm/y dislribufed
a
Io
f<
u (RC
u
Uc
t:)
R /
*
dA == 2rrr dr densify / ~ over Cltet/ A
Z1rrdr
-
and -
(R,/)
11
R
-
Z
n
..1-
f f -; )(~)d(;) o
_
fA. lie
if-
0.71 f
6
0.79/
10
0.866
5-/5
-
2 n7. 2n2.+
3n
+
I
5,/7
5.17 The velocity and temperature profiles for one circular cross section in laminar pipe flow of air with heat transfer are
where the unit vector and
The
i is
.
m
=/t
-l.
=
Te [ 1 +
i (;r - ~ (;)]
The subscript c refers to centerline value. r = local radius, R = pipe radius and T = local temperature. Show how you would evaluate the mass flowrate through this cross section area.
along the pIpe aXIs,
15
f/OWYfJ/e
1'h~~J
T
/'
v·n dA
A
/?;r
acft;y as
(Ur
tin
ideq/ !ftlJ'
P f RJ IIlr
f~y
dA ThU5 fY1
Ciy~/qj"
a..,
---- fir
217'rdr /?
()
and h1
oJ'ea.
cr()S5, feelloY!
ror
[I - f; )2] Zfr r
U
dr
C
II,;,
a ul1ikm!J til;h--i6u Ie rI
=21ft~ ~i,-r:::
J ~
fJ [
I<
'1
[I - (;) ] Iff (I /--
sIalic
ieft)
;(;)7
If we
Ii
~
fhen O.7(!)/
=
5-/6
pYe5SLlye
5 .I~ * To measure the mass flowrate of air through a 6-in.-inside diameter pipe, local velocity data are collected at different radii from the pipe axis (see Table). Determine the mass flowrate corresponding to the data listed below.
r (in.)
Axial Velocity (ft/s) 30 29.71 29.39 29.06 28.70 28.31 27.89 27.42 26.90 26.32 25.64 24.84 23.84 22.50 20.38 18.45 16.71 14.66 0
o 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 2.9 2.95 2.98 3.00
The
W1 a>
In :::
> flowvlJ.k
I
calCM.LaIe.d. w/tt.
Is
R
;OfA
21ir-dr
:
o
R :: p=
0.00238
U:: 1~CtJ.{
slu9 f+3
ax.la! ve.{dci+y ifl ff S'
r :: I oCd. I yadiu.s ,;' j". and
f~
11 rd r
/5 (. va./u.a. leo! IJUfltU;c.a.tI~ w; fI1 fk ~rJIJ"d.().1 'rule
()
wi/t1. une1tA.M l'flJblern
IS
J;"~v().ls.
II~ led
On
Thf!-
the
COY/ltp/A,~
Y1ex. t-
pa.je. .
(con't) 5-/7
PW)(I Y a/V\.
u.s~d
ro
Jollie
/his
Ii. J~
I
(coi'1'f)
100 CL:-:; 110 PRINT " t::.t -* ;t:.:t *:t: * t: -* :j:: *-:t)< :*--:t:>I::t: .i'- :t:i t* :t,;r:t :t.* .~.- i-*;I; i: :t·t *_:.t i, t" 1.2' 0 F)F~ I 17r "of.-;f: Thit:. program c:omputes the ma:=;;s flow rac:e ;t,,~" 130 F'F~II~T ,,** for problem 5,16 using the trapezoidal rule it" r. FPIKT ")::+: applied to unequal intervah:.. *t" 150 PF'INT ":t.***.********** i:***:t;*:*****:t** ***:t *i~ t.:;j::t::t.*******:Lt.:t" 160 PF:IIH
*
*
-1
*** **
*
**
1\
1.~U
*
170 DIN U(19),
180 190 200 210
R(19)
' 'Initialize the variables N = 19 RHO = .00238
220 P I = 4!
* ATN ( l! )
230 FOR I = 1 TO N 240 READ R < I ), U ( I ) 250 R(I) = R(I) / 12! 260 NEXT I 270 DATA 0,0, 30.00, 280 DATA 0.8, 28.70, 290 DA~A 1,6, 26.90, 300 DATA 2.4, 23.84, 310 DA~A 2,95,16.71,
0.2, 1.0,
29.71, 28.31,
I.e, 26.32, 2.6, 22.50, 2.ge,14.66,
O. 4, 1. Co, '"' 2. 0, Co. " 8J
29. 3:;, ~ 27. 89, ?h 64, 20. 38,
.:::)
00. 00
. 0,
"- ...J •
O. 6,
.... 4, 1
?
'-'
. '-', ':I
2. 9,
29. 06
27. 42 24. e-416. 71
'~::'.!=,rnpute integral using trapezoidal rule 340 FOR I = 2 TO N 350 SUM = SUM~(U(I-l)*R(I-l)+U(I)*R(I»*{R(I)-R(I-l)/2! :~:3C
·:,6;) NEXT
I
:::',,(,0 Jl1DOT 3130 '
=
F~HO
* 2! * PI * SUM
3g0 'Print the results 400 PRINT 410 PRINT USING "The mass flow rate is ##.#### slugs/s";
************************************************* ** This program computes the mass flow rate **
** **
for pr oblem 5. 18 usi ug the trapezo idal rule applied to unequal intervals.
** **
*************************************************
The mass flow rate is
0.0114 slugs/s
5-/8
MDOT
5·/9 As shown in Fig. P5.19, at the entrance to a 3-ft-wide channel the velocity distribution is uniform with a velocity V. Further downstream the velocity profile is given by u = 4\' 2l. where u is in ft/s and y is in ft. Determine the value ~f V.
Use the U/YJ~/ V~/ume if, e s k-elvh a hwe . ~
~ CIfYJ.J'-ervt( -h'or-- of
CL
I
'1 AI
=
Q1..
=
f
LA
V
-
4
3{~.7§)
/'I?ClSS
I
pYj'y}c/pJe
ff
flly - 2y':J b dy
dfJ
A2.
V (iJ.7Sff) b
by /he hY1J/<.eYl //YJe.J
iJ101/cttkd
(J
y
-
fi!2_
/ If
2y311>= 3
-
/.78'
Ft oS
/)
3 fb 11 s :3
-
J/J
5.20 Section (2)
5.20
Flow of a viscous fluid over a flat plate surface results in the development of a region of reduced velocity adjacent to the wetted surface as depicted in Fig. P5.20. This region of reduced flow is called a boundary layer. At the leading edge of the plate, the velocity profile may be considered uniformly distributed with a value U. All along the outer edge of the boundary layer, the fluid velocity component parallel to the plate surface is also U. If the x direction velocity profile at section (2) is
Section (1)
\
~u
u
Outer edge
of boundary layer
·1 FIGURE PS.20
develop an expression for the volume flowrate through the edge of the boundary layer from the leading edge to a location downstream at x where the boundary layer thickness is b.
From +he
Conservt) lioY!
II()w fh J"()u.1h rhe
of-
con/nJ /
"""ttSS
prlnc.iple.
vo I t.l ",., C
sitown
have.
~
.
/1'7
-::;..
6/
2
- 1~
-.l
'I.
\/. n
dA
A2
::::
Ovtd
.
tn
WIdth
()f ~c
p/afe
fhus
Q
t:
61
1 7JJ. b 8
5-Z0
C<.ppl/ed fr;
,;., fhc
fk.
/igul'e
5.22. 5.22 How long would it take to fill a cylindrical shaped swimming pool having a diameter ofIf) m to a depth of 1.5 m with water from a garden hose if the fiowrate is 1.0 liter/s?
defu.rm;IIJ c.onfrt; I V() il,,( fYt.e
From appl/cah'on
Iv fhe.
of fhe
volume
COrllro1
COl'7hlY-J
'ny
CJf
J?!1.tlS$"
wak-r only
Cf S
principle $hown ,;"
have
we
+
c,onservo,h'oVi
r V. n dA J(' -l
/\
==
0
cs
FOr Incontpres.si6le flow
Thus -t
or
2-
trJ) fa
==
11 (IDYl") 1. (;.5"",) (lOCO
::=----if Q.
tf
(1. 0
~)
I ;e~) (?600tf )
1 5"- 2. J
5.2.3
J
5.23 The Hoover Dam backs up the Colorado River and creates Lake Mead, which is approximately 115 miles long and has a surface area of approximately 225 square miles. (See \jdl'o \'2.3.) If during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and the outflow from the dam is 8,000 cfs, how many feet per 24-hour day will the lake level rise?
Fa r the
cont ('0 I volt/lYle .shoWfJ:
min - moul == ,
k f ~ d-1I cVwafer
or since m:::: pQ. )
Qin - Qouf -
Thvs,
d.h"
err
k (II/ok. h) '" "Joke 1b. Q.ut - Qin " ('1s,ooo: 8, odo) .fjJ z fI/lJ,t.e.
22S mi (52.9 0
{t.)
= 5. 90X/O i .if N
= .s. 'lOX /O-61f- (3,; pO()ir ) ( :ttf.lf;,) :::: 0.5/0 idy
5-22.
de foYdII n!J.
5.2 Lf
Stonn sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 1500 ft 2. What capacity (gal/min) pump would you rent -- - - J to (a) keep the water accumulated in your basement at a constant / /1 level until the stonn sewer is blocked off, (b) reduce the water / ' accumulation in your basement at a rate of 3 in./hr even while / / ) _ _ _ _ the backup problem exists? J - - -/- -
I
I /
/ /
I
-
I
Con serva-hfJYI
01- maS5 t-
or
cs
Con5ft:ln1
dh
F;y
Q.
d-t
A
II?
pari ct (J 6u.f
ro
:;:;
- A d-tdh
and
/ I _ _ .t. / _ -:.
-
-
-,
Cji flow
h
/
"~ f/~oY ayea ~l/
/
-=
A
In
fr;
aYlc( ar'ea, (A)
(I)
~O
leads fr;
Qin
eva/utile ()Ii,)
Q,;"
_
-- 0
c:leJ1Sily
QDwf
Etj. /
.I
n ciA
flt(id -r
/1
priJ1c;,Ie. (Etg. 5.17) /eaC/s
1,;O~'
,
'
u;n!l1/J"S fire WtI/ey Skef.ch ClbJve}) The
COn/nJI baS"ewtenf floor- (.see
fhe
OIlf!Y
ClJnfriin 5 water
- - - - - -
_____
How nut Vd/('(me fhaf
For a aefwmil1j
r
"t~t~
1
l/Jrrfl71/ ll olume fhaf
~
use
fl{ /
= (/500 f/') ( /
wIth
Q~wf =:0.
/h) (--L ) hr
::;
1Ju,s" 125
12 Ih.
f+1
-
hr-
Ff
Qokf
::;
(125 fl)(7. '18 :J~ )(',-!-- .) "' hy {f~
~tJ ~
/5:6
JE min
hI"
FoY
1
parf boJ
E~./
yields
, !;
j
I
5.2.5
. 5.25
A hypodermic syringe (see Fig. P5.2?) is used to apply a vaccine. If the plunger is moved forward at the steady rate of 20 mml s and if vaccine leaks pass the plunger at 0.1 of the volume fiowrate out the needle opening, calculate the average velocity of the needle exit flow. The inside diameters of the syringe and the needle are 20 mm and 0.7 mm.
a delDr"'/~
USing w~
£1'
tJ61r:lln (see
-;0 A, ~
3
+ ;0 Q2.
Since ;;:: C/)nJ/rJnf) ~
obm/~
1.1
~~
~
£1.
=
A, ~
or V:z..
::
(;, ) -~ A.1 1./
and
,,{ of'
FIGURE
LIP/lime
( Etj. 5'.17)
!JI'/;'c.iple.
/J1(1 Sf
C()n/YrJ/
I
ps.25
QJ1d fhe
COnservClI/tJn () f
In
€x.ohlple, S. 5';
as ot/II/ned
Ex Qmple s: g)
I' Q leaJ:.
1:
(;)
0
Qleak.. :. o. / Q .... 1
-(1) - d! -;;I?
1
=
(z 0 n,,,,, Is ) ( d. 7 ~,., ) .. (I. I ) (tot}o :~)
( Z 0 yrJl'k)
I ,
V;
::
IJf.
& -m 5
J
5.271 5.17 It takes you 1 min to fill your car's fuel tank with 13.5 gallons of gasoline. What is the approximate average velcoity of the gasoline leaving the nozzle at this pump?
v
-
A
hOJ31e
f10JJ/e
-
~
2.
Clhtl
So
::
AnOJ31e.
VI1P}3 ie
7!.!h()'J1/~
,..
If == {I'~. 5' )(JI)(12 )
vn()1Jle ==
2.
(7T~!r/L(7.'l-8)(bd) 17. if 1"1S
.. 5-25
5,2.6 A gas flows steadily through a duct of varying cross section area. If the gas density is assumed to be uniformly distributed at any cross section, show that the conservation of mass principle leads to
+ dV + dA = V A
dp p
0
where p = gas density, V = average speed of gas, and A = cross section area.
Foy a
sfeady,l
pYiJ1tift1!
;aAV
/elJd.$
=
oJl1e - dIMensioA(J/
-I7J
£~. 5./2
/i~~ fhe
CIJnSer-va/ion
I'dA V
-0
01
PI(J'>]
or
Con sfa",d
7hus
cl(fAV)
:::0
;aA d 'V T d{J AV + by ",A V we P/v/d'1f ~. /
Or
~ + dV + dA ;<1
V
-A -
5"'-
obfl1/~
0
26
(I)
5.2.9 A IO-mm diameter jet of water is deflected by a homogeneous rectangular -block (15 mm by 200 mm by 100 mm) that weighs 6 N as shown in Video V5.4 and Fig. P5; 2.9 Determine the minimum volume nowrate needed to tip the block.
bloc~
From fhe free body diaqram
of fhe block when if is ready
10 lip Z i110 =: oJ Rx JRx ::: WIII' where Rx ()r
is
the fore iha+ the wafer pllts on fhe block.
Thlls
J
(0..0/5
)
W/W - 6N ~", ~O,90N
Rx -== tRx
~
o.OSom
For fhe conff'o/ vollJme shown Ihe X"-componerr! of the m()/l/8I1fllttl e qv" lion
Sup V'n dlJ cs
becomes
c
Z rx,
5.JO Water enters the horizontal, circular cross-sectional, sudden contraction nozzle sketched in Fig. P5.30 at section (I) with a uniformly distributed velocity of 25 ftls and a pressure of 75 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determine the axial component of the anchoring force required to hold the contraction in place.
r::f:9 ,.fhe
-fh;s plDh/em
as ~//
hOJ}/e
ac,f/Yl9
.foyce s In
fhe
()V1d
()YJ
/n
+he
no!-
d/"ecfid~
f~rA+
sh()wYl.
cOMpaneYlf
fo fhe.. flow
th~ sl::ekh
fhY()U111
4M
ins!-zrvtl 6~~en sechiJns The hOY'i}lhfal
above.
ihe.. CJfWl()S pheriL foy-,eJ
AppliCA.htJr.
of +I1e
/ /jlfeat
of -fhe.
CAY/eel
hOYI]tJYlftJ.1
}?IUYkeYlfu~
n1t:!y
m(t.4" - ~ 1):= Or
I'YIQ-s.r
er~n~
( 1:::).5. /2) we
=.
1/
F. A
~
p'A, - ~
(I)
tJbltt/~
-Ii 1I"l-
/',11, -f{A1 +- tYJ 6,,,-U,) = f, 1"-1-/)' -
rr;;!l ):!(~.j_ 'i {II'
x-
6e~S~rI".s ;t
F and
b~
f
fhis ~J11nJ1 v()luYHe yieLds
Frorvt iN... tfrt'lSf/rv~I/,,1') af m -- fllA. AJ =to 41 A,.
f1- (I)
tJ'r
(JIA.
eqtJ.a.fion (Gr- >.22)
P.A -1=A - r2 vA l 'I
1bu.~
V()/Uftllf!.
of the UJn-/-nJ1 vaJu.~ a,.t shown
coll1fey,fJ
sk.e fCh _ Note
aye
incll1de. I;' the C();,~/
Me wpf~1" al
as
(/)al"ld(Z)cu Ihdicaf~
we
'I
~
2.
r;, 1£f}1. -flu, TI'p, (u 2 -fA,) ..., If
016- f'llf!,uznIz5 £t ) Jr(3,:..j IO() C -ft ' / l~ .s.-l tj"" (N'f ,;.,.?f,.~
~:=~ Ih
5-28
ft- 25ft
r Ib.~1 SI"1.
1
5.3 I A nozzle is attached to a vertical pipe and discharges water into the atmosphere as shown in Fig. PS.31. When the discharge is 0.1 m 3 j s, the gage pressure at the flange is 40 kPa. Determine the vertical component of the anchoring force required to hold the nozzle in place. The nozzle has a weight of 200 N, and the volume of water in the nozzle is 0.012 m 3• Is the anchoring force directed upward or downward?
to.lom •
3/S
FIGURE PS.3i
(I)
(2)
~
Cr:J')1
Q)..
serl/(J..h·/JI"\
~ Q, (1,
A').
tJI "" a.rJ
~e D~I"Jn
~. ':' ~DON ~ 'ZOON -ill.6 N
-
0 N -:
'f82 N
downwttrCf
5.325.3 Z
Determine the magnitude and direction of the x and y components of the anchoring force 0 required to hold in place the horizontal 180 elbow and nozzle combination shown in Fig. P5.31.. Also determine the magnitude and direction of the x and y components of the reaction force exerted by the 1800 elbow and nozzle on the flowing water.
Section (2)
x
12 in.
FJGUI~E 1>5.32
PI
= 15 psi
VI = 5 ftls
,
Fo~ defet'm;~/''' fhe
a.
fone
Coi11J..D I
fOY'ces in vIJ/ved
(z.)
5htJWh
QI't.
y dl;ec.fi~Y}
is
used.
In
the
c.()mptJn~n+ of -file
y cI/recl-;on
c(J¥Vtpf)nenff
fhr;t f con llI;nf ';'he elbow.l
1/f)/uJne
S'ecfioYls (/ JAnti
/:JelweeYl
the
and
X
1'10}j/e
pnd MIt/fer
Th
C oJ'dnJ/ v"/l(l11e a;1(1 f'/'e
s/t:.~fci1.
above. /tj?p/,"cern'Dr, 01
lJ~eclY YYJ()YJ!Iel"t.ftA~ quafltJJI)
(E1' ~. 22) leads -/-t;
r:
= 0
1-Y 01 -fh~
Ap,IIc.(J.+iDVI e~ fA a A'oYl - LAI
YIelds
f
U{AI
-
U2
f
A
M1.:2
From fA e C(n,.,serllaf,'on
rn = !Ju/ A,
&t.
1hUj
CIJ~p~neyd 01- +k linea)! mOWlei'lful'n
Xf:kt'ecf/()n
I
:;
of
fJlA2.
-:::
WUlSS
PI AI
-
FA + f.. A'2.. ~ X. 2
e1 ud-f,'on
A '2,
(Z)
May be eXl'fej~ed (N~
- P'fA (lA, of
U.) ::
p, A, - ~," + f,A2
~V\d
F :; f~, A/w,-tIJ.,) f P, A, t P. A2 =t u , : o,'(IA,tfA.) +p. A/x Als6 ~
II)
9. 2
ThUS
F
AX ,
(con't) 5-31
5.32.
(con'-/; )
r
(is lJ.
) 11'
(J2 ,•. J if
{n.'l.
Ftly
,
defwMJY1;11j fhe x aYlq y cOMf~nenlJ of
fhe
yetlcii())'J
-hY(;e
v() IUMe fhaf ClJytfa,i-u (1;1Iy the. Wit k,y- between sec/lam (I) fNt;td (A.,) ;f us~d. A p"lklAh'6-h of fhe y dlrecht)Y} CiJM-pIJt1(Jnf tit
of
COn fro/
fhe.
hhe4Y
VUOJ1-f~;, fuWl
e&JUtX.·fion
yieLdJ
I?., := 0
AfP /;caf;'Oh hi} () /)If
of fhe
e k fuJ'V/
fYJ ua f/(J i1
X
d/yecf/ol'J
leadf
cOl"l-'lptJnenl
of -the. II;'; ea v
fr;
Or
=-
/i9~
/b Section (2)
x
Section (1)
5-32-
5·33
1
Water flows as two free jets from the tee attached to the pipe shown in Fig. P5.33. The exit speed is IS m/s. If viscous effects and gravity are negligible, determine the x and y components of the force that the pipe exerts on the tee.
Use fne
Lo)'lfyl)1 vollAYhe.. s-h()wn.
Foy fhe
X - c.offlf() Y1 enf
of fhe -force. eX~l'kcl by +he
+he fee we. lAS'~ fhe x- c.oYhplJneni i'hO m-e. nItA YVI .€1/A a.ll(Jy). OYJ
pIpe.
r:;f the l/neCir
If A.J - ~f,A,""fl3)+ f
- ~(' l{ AJ -I- ~f~1J3 :: f,A J -
"().~r.;t..)Al- (-;;. ~)43- ~-,43) -J- ~ =p A I
I
-t-
F
/1) l'
)(
9t1t:je
To 1e.f. '1 we LASe cOYlservah'rm 61, = Q 2 +- ~3 A, ~ -= liz V2.
6y
v.::;
50 To
AI
eJ f-/.-necle.
mass
11"1 V,3 A~ IIJ = (0, 3M2.) (IS" m1 ) +- (0. S "'')(It;" m/.s) :::
1-
A.l. V:z.. +
I
01-
I
1'Yl1
/~~
we.
f/;fjQff!
- 72 ()()() N:.
/2
l-1.{e
8~Y'I1(hll/i J' ~~6~ ~ f/(IW
belwr£." (1)4t101(2..)
FJ(
F":: 72/~OON ~ .x Fw /1u.. .Y ~!~ ff I1u ~ exekfecJ /:;f hr-t-, PiPe Oh flu. lee, Y (IYYtMc"f Itl- h..L !;~e6N" hlnntA11w.-" er ah ,., ..fe, 9~f So
\.1',2.
~ II). ,4 2
::
5
(JSJXq'f9!:!)(l5;)(D.1"'~) ~ {,~'fOoN 5'-33
t "Fj
,ve
L).5~ #V
(,mltya,1
(r)
- .......~
.j. 10 psi
... --
5.3'1-
Water flows through a horizontal bend and discharges into the atmosphere as shown in Fig. P5.3'1. When the pressure gage reads 10 psi, the resultant x-direction anchoring force, FAx' in the horizontal plane required to hold the bend in place is shown on the figure. Determine the flowrate through the bend and the y direction anchoring force, FA\" required to hold the bend in pJace. The flow is not frictionless.
Q
P, A
=? ~
I
Area
•
"
I
~
Water
!
~l-
\i=.======::::'1
=0.2 ft 2 Area
= O. 1 ft 2
VOt..cme.
I.
"
(2.)'if:;(
~
FA, = 1440 Ib
" FA,=?
/~ •
FIGURE P5.3tj-
A uh1 .fl..D/ voluMe!. -f1.taf 'O,,~iJ1S -H,e. 1oe.V1d OlH& -tkt wakr wi+k;Y\ the loeHd be:fwee.n "S«'tlOl'1oS (I) a~d (2.) as s/"OI.\JI1 J~ -H1~ 51<.e.1-'/'" aho~ ''S t> f
Applic~no~
uS'-ea.
-HI Co Ii rJt.<1 r
- u, p Q
,.t'\ 0 i)\ en
tu,..,
01-
eti.. tAa:n 011
-0
v. cos CfSo a. -=. I
I/o
u ie Ids J
- F'Ax.+ rDA I I
f)
e.~.
or
J
-
~
al-1d
A,
-QAz.
Va. :
loe'b~(.s 1-
0
~-a.r
Qrc:OsLfS
AI
AL
..
+Cy po.r+ Co.)
Q-::
5-31.f
~I AI -
0
~Qge
,1i)J1A1. cos '1-5
w',H" ~,
x - direc+ion
file.
F
Ax
0
~r+ (h)
ft,v
, i rl eo. y
•
J=
AY
~ ~/n q.,
y'" dtre.c+;oll\
~ IA.I:f. -He",
IA-\ 0 MtLl1 fu WI
~
1{e.
We ltS't.
C>
C!.O'M.poYlf!"1f
of fhe
-ft, qe +
f~ - A,. R.
Sin
'-+5 • fQ.
()f '1..
t=AY
Q :.
a.~a
FAy --
0
os I'VI if S
A2.
f
1.
(?." I +t') . 'i-S" • (I. 9'1 ~ oS I '"
(O,Ol
f+")
S~)~ Ib.~'" $I ""'1.
-Ft )" 'l'i - Ib
-
Thrust, vector control is a new technique that can be u~ed to greatly :mprove t,he maneuverability of military fighter
aJrc:aft. It consIsts of USIng a set of vanes in the exit of a jet engIne ~o deflect, the, exhaust gases as shown in Fig, PS,3S, (a) DetermIne the pI,tchIng moment (the moment tending to rotate the nose of t?~ aJrc,raf: up) about the aircraft's mass center (cg) for the condItIOns IndIcated in the figure, (b) By how much is the thrus~ (force along the centerline of the aircraft) reduced for :he case IndIcated compared to normal flight when the exhaust IS parallel to the centerline?
ro.r petrI (0,) e7~1/'tIYJ
CA-ff'J
WL
Vane
- -- i" - -Pj';= !lim !li =
out
0-
-
-_ .....
= 16 slugs/s
the.. (p/'Yfp()nenf of file J')?()hlt!nf-of-tJ1(JMen~'"
;'r pe."pend,'udar -/tJ the plane of i1Je ske}~h of fhe alrcraff -k the. c()nfe.",ff 01 fhe. cvn-!n,/ Vo/",~e .show,? -k 9f!i fl,af
i olAf V
()c..
(2 0
f
oFf) (/~OO /
mf - j
>/;' ()
~/;' 8' (i6
fj )
SILl,.
O/A
. V. h,.
In '''I
=
pi fc~ ),14 J'))otne"f
II?
./
fr.£j _ 1,.. p(){) fJJf'fr)
J
ff-
"rrd.""j ", •..e~ f
I sluy.lt /J.
/1,. $2..
bh IStJ() _ 1000
.
1.
-:: pilej'J'y
fl. /b
I"
s~
~P1~hl
~ p~r(b) ~ apply !he /U)Y/3()YI/R1 Cdn7{'(jY)enf t)/ /he I/YJeoY' yYJ 0 'I'YIeYl
Iu ~
g
Il 1A.t:c. f/vYl
-ft;
~
(/m ~Ir
t)f
htL
U/n ~/
V()/U hi ~
T1wus+ .fa
(/~(}O
if)(UH/-W '1(6 1l-
J
/
~//,(i'
If
16 . .5'
=
5-36
)
Iv 9d
5,,36 The thrust developed to propel the jet ski shown in Video V9.7 and Fig. PS:3G is a result of water pumped through the vehicle and exiting as a high-speed water jet. For the conditions shown in the figure, what flowrate is needed to produce a 300 lb thrust? Assume the inlet and outlet jets of water are free jets.
• FIG U REP 5 . '36
For the cordrol volume ii1d/cafed. the x- componenf of ine fJ10menlllm
V2.
erlJ~h()n
Sup g'n dlJ
(1.)
: : ~ Fx becomes
cs (~ C o.s 3 011 ) f
(- lit )III of Vz P(+V2. L42 ::: Rx
where
have asslJtned ihai
We
I:: 0 on the ell/ire control s/Jrl4ce
and fhai -fhe ext'l/fJ9 wafer jet is hori~(Jnlal. W/fh m::: eli, ~ == fA,. V2 £1. (I) beaJlll6S Rx -; m(V2 - ~ cos e) =- f Vt AI (v" - \1 cos.30")
(I)
A/so) A, ~{ ; A2. ~ so fhal \I. AI ~ V s= - 112.
z
•
=
*
7- S in." (3,5 in.) 2-
,I
v, =
\I
(2)
2.60 v,
By comb,'nin9 frs. (/) 4nd tJ.) ;. Rx ~ pV/'A, (2,60-cos30·) or
: : r .Jo o,!}; l ~ l(l. 9¥ ":~)(iJrl') (2.60-co.s30')J
Thvs) Q ::: AI V,
=:
~/:~ft~) (:;'2.7 ¥);: 3, 9'f-¥ 5-37
Jfz.
= 22 7 11 '.s
5.:37 5.37
Water is sprayed radially outward over 180 as indicated in Fig. PS .3 The jet sheet is in the horizontal plane. If the jet velocity at the 0
nozzle exit iszo ft ls, determine the direction and magnitude of the resultant horizontal anchoring
H' _"
force required to hold the nozzle in place.
FIGURE P5.37
The Lo~1 volu,.,e includes ik ""'3/e anti waler i:Jdwren secfionJ (I) a..rI (Z) 10""'''1..,., ~a6'"" ·i el"s
o.c
or
F
A,y ()
F
~,y
:: 0
ApI'IiCAnM ()f fhe eg uo.lio'1 le.df n, F
A, x
or
and
'1-3 16
5.3~
-t--~
5.3B
A circular plate having a diameter of 300 mm is held perpendicular to an axisymmetric horizontal jet of air having a velocity of 40 mls and a diameter of 80 mm as shown in Fig. psS8. A hole at the center of the plate results in a discharge jet of air having a velocity of 40 mls and a diameter of 20 mm. Determine the horizontal component of force required to hold the plate stationary.
The U)/J!Yo1
Plate
';;+""':"":;::L.:!..JJ ~
FIGURE
40 m/s
PS.3E
fhe p/ak l./JIui {low,;' cor a.> Inri/caler! ab"ve - Appliudion of I-h-e. hOfljtJJI1 f a/ t.JY X
lIoluYHe Cl)nfa/n 5
Ih -the f/<.efdt d,;ecJi o/l1 cOMpoYlenf - u, f uIAI +
or
( I
f
A
IAl 2 . I.L2 -
of -fi1~
//Y/ear
mOMen-lum ~ua.f;~ yIelds
Q
l
om
5. ?=I A sheet of water of unjfonn thickness (h = m) fl ows from the device shown in Fig. p5.39. The waler enters venicall)' through the inlet pipe and exits horizontally with a speed that varies linearly from 0 to iO m/s along the 0.2 m length of the slit. Detennine the y component of ancho ri ng force
F AY
necessary to hold this device stationary.
~
--
~ ;I~ - - ... I , ,~
-- - -
...
",,""
• FIGURE P53~
A ",~t",1 volun\e. tMt ",n+"'i"s he. bOl( poyf,,,... ot l1,c. dev;,e G< o;s ,hown In -Ine ~ke.itJ" above. 'I ' u,~ , ArpliC
F = Av
I
1)
fI
~V , r. dll ~
5t;+
.
D \
'
,,'
0
T~<. var' a.+:o" of v- .. ;#, x
..
o.t.
f \ (<;O)()\ ilx
=
o
or FAY
:(q~~~
hd ...
y-so
-
s-~o
li"'e
5.40
change in the direction normal to the paper . Calculate the drag force (reaction force in x direction) exerted on the air by the body per un it length normal to the plane of the sketch .
The results o f a wind tu nnel test to de-
termine the drag on a body (see Fig. P5.40) are summarized bel ow . The upstream [section (l)J velocity is uniform at 100 ft /s. The static press ures are given by PI = PI = 14.7 psia . The downstream velocity distribution which is symmetrical about the centerlin e is given by
u
~
100 - 30 ( 1
u
~
100
-if)
- 100 IUs
Vz
l y l s3 ft l yl> 3ft
1:'-C::3 =::\
3 It I
Section 11
where u is the velocity in ft l s and y is the distance on either side of the centerline in fee t (see Fig. PS. 42). A ssume that the body shape does not
-- -
~
- - - --
LutrlrDJ..til..",. FIGURE PS.40
Section (2
i-
~
conm/;' i"j Plr Mfy PS Shown In fAe ry"ye if lIfed , AppliCAfion d f/,. X dlftcl-;O" componenf of Ihe linea, ""me"fuM
The
(017 &.1
e.
volume.
iead!
*
3 ff
-V;r1!,A,+
-Rx
ZfufUdy o 3 II
Zf
=
I
f)]
[100 - 30(1-
~ fhe c.#hscrvlAh'o,. !Applie d ,,,!weel) sed/onf (I) a"d(z)-as loll"",
To de k.,.m l ne.
fh e di;1tJ",u h
(I)
dy
01
.-.'lt1H
l!flA.tA.h;"'" is
3ft
ph -u;
=z
fJ
d'j
()
Thuf
3fi
h
; ~ [fOO - 30(1- })7 "1 ,
or
J u.
2.
h
0
". , fI')
=
( z ) {2 55 :;-
5.1 fl
(100 !!)(I il) Then
fr.,., lj'
J
.
R"r.m731J1"JJ )(too fJ) {5_1f+ill/, x
-II' /
(-'1"3- if
)(i
ff )
s'
-
l
(o,002JfJ/"!,) ( 7Z1, if00 ffj )
17, 1 Ib
P"'r
[!') Ii I.
sl.
{ ~;I.f.1' (!
.'
)
ff of Je~ nor"'41 fo fhc pl.ne of Ihe skeIGh
s- 'II
So 4/ 5.41 The hydraulic dredge shown in Fig. P5.41 is used to dredge sand from a river bottom. Esti.mate the thrust needed from the propeller to hold the boat statlOnary. Assume the specific gravity of the sand/water mixture is SG = 1.2.
• FIGURE PS.41
lJ.rilJj ~ (,1Ih~1 vptut11e fh4'vl>? by ~ hi?;~ I,'v,e In /t..L- s/::.elct, a6;ve W~ lASe fht., hoy1foJ?W oY X CllwtftIYJenf of ~ IlfJea-r /YlrM1enft..vm e~lA.P.hin" ~ gel~ =~/J V V ::: f. (sq) .2.
2 .2. :2. X
Hz 0
2.
"2
7I5!2 l.f
V V .2.
2-
CO.J
.fa fJ
whe!re ~eefi~ / is' where IlfIW en/t,yf Ik C-tM-h-s/ vp/ume verh·ca.llj tA.i?d secf/(/)1 2 iJ v.;),eye flctw J~ve.s- k ~I v"kn,e. aT h'1 t:l/}Jle of .?() 0 ~ ~ hor-'Jtr;lhl oIiYec-h~. N~1e That Me'l'~ is" no fu·y-t]tI>?kl d.io/'eC/h~ //Y1e~ #thne;,1u~ Flow a1 5'€c-fiQn I .
F ::: 1/.9'1- s/uffl(J,l1) ><
\"
fi$ /
F.. = {6SIJ /6 ){
7T (.2.
3..)
Iffro f! )(so :~) tlJs70o/
f-.s
(
f/-:
$lw!J
s2-
5.42.
Water flows vertically upward in a circular cross section pipe as shown in Fig. P5.4Z. At section (1), the velocity profile over the cross section area is uniform. At section (2), the velocity profile is V =
We
(R ; ryn k
Section (2)
where V = local velocity vector, We = centerline velocity in the axial direction, R = pipe radius, and r = radius from pipe axis. Develop an expression for the fluid pressure drop that occurs between sections (1) and (2).
h
Section (1)
FIGURE PS.4 Z.
The anCl!ysh' for fhls pmb/em IS fin;i/ar 7z, Ihe one of Exa~e. 5:/? The (onfml volUMe confaln5 fhe fiuid only belw~eYJ -sechorJs(/)tJnd{z.) as Ind/cllIter/ in the .ske·h;h. Apf//cafion ~ fhe verI/col Or ~ jt'neor mohfen~m
of fhe
cOl'Ylf"nenf
leads ~
qua/itJYI
R
j{JJ.,.. f
+
P,- P. The
=
W
z
21"1 rdr
o
~~
::: p" A - Ifr R
_pw,-" +
e~tJ4.fJoY1
(,(s -/O/IOVJ5
f~ AI :::
be obfa,y,yJ
Jf tU~ (,
r..2.2.. A - Ww
.l.. Z.
]r f"
f.~1f [{we ('~~I)
we;~/d of fhe I/IKJ/w I"" -f-he confYDI volu,n,e
w~ = 9fAh The value of w £ may
1-
dr -t
may
(I)
6e exp,/(;ffed as
frOi'VI fhL col'lseyvalion of ma.S'.f
l 7
2rrrdr
or (2)
10 evalu.afe t>(
= R-r R.
fhey, dt>(
= -dr R..
(If )
s,lf2.
I (con'f) and
~
~
-!.
('7/;"7 ydr
-= - /
0
1..
C{
7(1_1><) R
1
riC<'
~s. 2. a",«1 5" we () 1::,.1-... /"
C.omb/n,'.",
= "19 €.E- tvJ
u.J c
Th u. 5
frlJWJ
~-P.. 2.
&J. / R
(6)
2
;::;:~-fWI+ lTRl.
fo ev()/t,ux/-e 1-he.
Thus
R
2
= 1'1'-/ -'19
[(;;r/rd~ o
R
2.-
and ~. 6 beCIJmes
or-
/Vofe
fhaf
a very a
In
$"mtl//
c,ht11?ge
t"n
[tJntrAsf
Iv
-fhe
restllf- Of Exttl?'lp/e
porf/on of' -the fh~
mtJmen-hon
pyessuYf!
//ow
drtJj7 if
s: 13,
only
clue fo
befWeeh seci/onf /
£#3
I average velocity, u, with the axial direction momentum flowrate calculated with the nonuniform velocity distribution taken into account.
5.43
In a laminar pipe flow that is fully developed, the axial velocity profile is parabolic, that is,
,,
I , as is illustrated in Fig. P5.43. Compare the axial direction momentum flowrate calculated with the
\j
"I
I
!, ...........
( 'u ......
,
I
I \ \
-
~uc ~~ .~
\JI
FIGURE PS.43
The axial d,;ecfitJl1 fr/orwen-furn IIowrafe. btlsed pyolde wifh u =: U /s
M~
::::
x Uf/ifOYIJ1
ufu A =
J
1he axial direcfion
M~ J
nOI1-
ju
= r
/Jrl"1'rR1. rlow~afe
ff/()l1teJ1TUh1
IJl7ifr;rm
UZ'1l
hi,-
a un/101m velocity
{
fltlra btJl/e-.t<. ve locify pr~fi/e
un; fOYM
OJI/
"
(1
if
ba!Jed I
the
fjOfJ-
2-
0r;)J(I)ff) ~
u~ 271' R'
0
On
L
MFx nol'} J
(Jl)i faYn?
To ob-fa l"n a
ye/~·lionfhip
befween
con serva/-l'on of mass e~lt.ahdYt fU 1TR'
~
f hR' Uc;
U tl5
fo/I~ws
[rt-({Jj(;) d(:)
- -- -lAc, Z
U
Mi=;( nOn-
MAti u ~ vve use -fhe
-
-3if
J
U"ifm",
5-'1-5
~ 5.44
For the pipe (6-in.-inside diameter) air flow data of Problem 5./8, calculate the rate of flow of axial direction momentum. How large would the error be if the average axial velocity were used to calculate axial direction momentum flow?
F.) :.
tJ. 002'31'
> (1.(9
H:1
I
/OCA./ ve/()ci/y
Vi:::
r ::. local
J
s: /6
)
"3 in.
r~J~ 1-r CIt, -,,/
/·""-"1'''' .... 1""_,,"'/
e va / u.tA.f'Ct;If 1_.-1
1$
VI
by
• / nbtl11eY,CI(
. J. L l¥lfCfY~n~
~
f-/5!ny
frate.fdic/Ift/
ft,c.
Lt/nI.(J1A.
k¥
//lex f
pt1Jje.
Etj. /
~f (~ Table 61' /'r1Jb!eu,.,
radius; in. ( .fr~ Table. of I'mb~ s./B)
I? ': pipe YtJdiu.,J ~.Jtn<-
I
PYlJfYd-Wl
lA5ed.
fw
The ve.$ L( If of
is MFx -:: 0.29'1
MFx
wl-te>/e
v =
( MFx -
you/e
MF) /f7p
/VI~
)(
WI·!t-t
ftJ/s
4neg l1 a.)
,;'/e-yVIJ.!;.7}tG
)7v. y,o/f..
»'1 I,.v",e,y;'c.czJ
/S 1/J1e'/ 6n /'he, t'J1leqvahon aJ1d s~11A h~J7
of
conti: )
100 CLS 110 PE I NT ":t:::f. *;t:,.t* *****:~t:**;,j *.:j *::t~.* **-*:** ** ** i:t J>I':,tt.*:.;;f *t::t :tJt* ::1'::t. *,. 120 PE I KT .. i.* This program computes the axial-direct ion ;j;;'." 1::".0 PR I NT "** momentum f lovr rate for problem 5. 44; usi ng :i';:>i:" 140 FRnn tI.t:* the trapezoidal rule appl ied to unequal **" 150 PRINT "** intervals. til! 160 FR I NT "t:i:*::t::t*******::t;j.*****:t:*********i.*;j~;;j.**********:l:>\::**" 170 PRINT 180 DIM U ( 9 ) ,
,
R (19)
190 200 'Initialize the variables 210 1{ == 220 RHO
19
=
.00238 PI = 4! )\: ATN (1! 240 FOR I = 1 TO N 2:~,0
)
250 READ R
2~'.')O
t-'. "./
~
'
-)
j
,
2fJ. 71,
0.4,
29.39,
0.6,
1
27.69, 25.64,
1. 4,
':)
2e.31, 26.32,
'2. (),
22. ::.0,
,--,.
-l..
s~
'-'"
p, '-
t
,
.3.0,
250 FOR I
=
00.00
2 TO N
'! (; ~~E)~'T 1 :. ,:~. (:~'.:::: X. = RHO ,;
~::
,-
::;.
,
'11~~tJt~t*~****1***tt**~~*******~*:t:****~ ~*
Thi~
program computes the axial-direction momentum flow rate for problem 5.44 using tt the trapezoidal rule applied to unequal
~*
** ** **
*~*****************************************t*** The axial-direction momentum flow rate is 0.284 slug-ft/s2
5' -If?
27.42 24.. =,4 16.71
Consider unsteady flow in the constant diameter, horiZontal pipe shown in Fig. P5.45. The velocity is uniform thro\!:ghout the entire pipe, but it is a function of time: V = u(t) i. Use the x component of the unsteady momentum equation to determine the pressure difference PI - P2' Discuss how this result is related to F, = mat"
II (I)
," I I
:
.: I
---------------t I
p = density
--
(1)
U5/11}
the C-tJn-/rD1
VIJ/umf(.
-------
:
DI
--x
I
(2)
I~ fhe 5kel-ch and etffJ/y,'IJ fhe
shpwYl.
f of /he unstead) //YletfY' h/()n1eY)..J-.q/'Yl 8fl,fetl)~ -fo; !1Je Contentr of ft;/ s c.v fN~ ge l-
x-
U)t?1(J()t'J eYl
~ JflA.dil ")-t
f
CV
f rIA ~. ;:. d
1/
2
-
rx
(J
or
d (f~ 7r pl., df: 'f
)
r"fA 2.A I : :. I' u7.2./1"
CI~ ~t.un;'':}
~ = tt,
I
{Y/c/,tmlesJ
5-48
(). t -e I/f/Yy
f;fJW
s .1./-6 SAt)
The propeller on a swamp boat produces a jet of
It;r having a diameter of 3 ft as illustrated in Fig. P5.46. The ambient air temperature is 80 of, and the axial velocity of the flow is 85 ftls relative to the boat. What propulsive forces are produced by the propeller when the boat is stationary and when the boat moves forward with a constant velocity of 20 ftls?
3 It diameter
.. FIGURE PS.46
F;.., ~ sla..J.Iona Y!1 btJa-t +he hoyi]ontlA J Uwtl'~ tJ/ fhe h~e(),Y y11trme;,fu ...... e~/,4aJ"/m ().PI'/led -It; /tu l6?~ ,,{ IItJ cmM VD/Un1e S htJWn Ii-; ~ ~":. ei'(,h A,J,cve yields
FJ F
w, <.~ - VI)::.bA VV V2 r 2 ~ :z.
::
+"'YUS-t-
=
fj,YUf-i-
\.Iff,
2.
/
51nce
V.
/1,
V« I
V2
in ~ )
~
'2
/LA .I'):i~ (I'I.7rn2..1/lf¥.f~...J 'TT'{JfI)('I5.f-1 (
'I-
R-r
If /j~.). ibpt.{f )
15-3.3 -fl. I: )(5'1/)'7<) /~. 'R
\ ..
(-
,J,.S:l./
= //7 Ii;
F +),.,.uff
ft?Y
fhe lo()tAf Wf()vi, fd.,tNf,VYr/. J/V}'rt.. ~ ffuJ 0/
.5~e.
(.~ftb-{ \Id/u~e fhfIWn )n
h~ IttL ~/a,li~ ~
aINJV~ is u.red)
the S-kelc.J/ W
If,{; -the
is h~ /;"fJI¥~'
~ hIJV/]dvrk/ ~ ()/ fj,~ /J~eA¥ Yh(fWJe;,~
e~ ~
F.-IhYIH..}- ::
gel1'1/W (
I
2.
Fw tIu- yn()v;~ h~
F
V(,ltrt,:,'1
2tJ
';hY'Laf
=
-
8,/.2
/6
w) := I ~II~ W (w - w) :: R..E-7?-~:lV\{(~-W) r ,
~ ~ gS-
ft
2
.'1
tw\,,(
<.j
I
V\,j
=20ft
anti .>0
5". if 71 A free jet of fluid strikes a wedge as shown in Fig. P5.47. Of the total flow, a portion is deflected 30°; the remainder is not deflected. The horizontal and vertical components of force needed to hold the wedge stationary are FH and Fv, respectively. Gravity is negligible, and the fluid speed remains constant. Determine the force ratio. FHI Fv'
The h()rjt1nfa / QYJd. ve-y-h'Ul/ (tJn..,fJOneYJh of /he lInea ~ m 0 me n fu ffJ ~ M 17(; n cut OfP //ed f1J -I'hi. [ffn /evrh of ft,e CPYJ~I vO/l.Il'Yle ~hdwn fu fief - V, f' V, A +- ~ f ~ A:J. t- ~ CbS ?O ~ ~ AJ j
~ f/~:?O·~~ AJ )-mweve¥"' v/ -= v2.. -- II:I
V It'
A3
~
_
-t A3 flY)
V
(2) SeJ
eo>. (l)a n d{2) beUfrYJe P
II,) -- - Fit
~(). -
&Of.
AJ ::;.
(/ )
FI/
--
('3)
A2. +Iq t()$" $0 "-11,
C(}YIS( Y'VCt
G( I
....
30 tJ ~
Fv ~
~
~ ~
-
V.2.".o (Ill
: -
h~
~.2.
$ /;"
3 () 0
of
yY1
cus we.. ge f
t- blJ
or anPl
A J:::' Az. r /lJ
~b i:"/~
€,
f .
(?) aAt-d (tf) twt-- qe f
.4, -I- /;7 CQ.J
JIJ 0
-
Ill.. - //1
fJ 3 {COS?OO
-~------,
113
The ne'jCihve ul
CiS
>,1;' JO
5/tjY1
5" how}?
I;'
A7 »,-, 3IJ
i)
Indht:l/el'
the
0
0.27
-
~f ~ if ~wn r-~~ /Mn
.Jl:..eM~
5-50
-I _
S.lJ8
Water flows from a two-dimensional open channel and is diverted by an inclined plate as illustrated in Fig. P5. When the velocity at section (1) is 10 ft/s, wh~t horizontal force (per unit width) is required to hold the plate in position? At section (1) the pressure distribution is hydrostatic, and the fluid acts as a free jet at section (2). Neglect friction.
.......""'" l.Oft •
-
2.13
1b
5- 51
FIG U REP 5 .'f~
5.5'0 5.50
A vertical. circular cross section jet of air strikes a conical deflector as indicated in Fig. P5.50. A vertical anchoring force of 0.1 N is required to hold the deflector in the place. Determine the mass (kg) of the deflector. The magnitude of velocity of the air remains constant.
conhol 'Jolu;Y\e
Sec·tiM v
=
(I )
30 m/s
FlGURE PS.50
deteYmlne. fh~ Y>'fas.5 of fhe UJn/c,(,1/ de.fiecJ'()'r we use fJ,e.
To
SIP.lionaY"y /
/tp,P//CIIl h~n
fA bdlle .
the
LfI)1/n;/
v:
Ii! (-
of
In tJl'H en ~"'"
/1';-,e.tJ. yo
Ih/$
d~fuy»1/~
)1dl"} -
t-;.
VtJ/Unl~
v~h'clA /
fite
~hdWJ1. In l1te sl:.e kJ"
a'/yec ndn
C/)}IJI?pr;nenf
of
t!j tlal/o¥! ( c 1· 5'. 2. 2) fp -/he ChlIeH Ir (J f
y/e/JJ
VtJ/Ul"He
of
c.unfnJ/
UJS 30
0) ::: - 'A
-W
Cone
Or
Wc~nt'
~l1ne J - M (~- ~ UJf f()~
-
Hd/N"€,vevl!;'
::
~
::
711), If
tint:{
A;
E~. /
Thll)
).
Ct2J1t
be
-= f
m
t.tJne..
O
frO,").
as
I{ (~- ~ CbS 10 D) -
CJy frJ
':=
ClJne.
f2? A-J) /'I'fJ
ancJ IYI
CMe
::-
1).108
~ .9
fj'
*1
5-52
- ; :;:
f~~(/{-J{CPJJO~-~
(;)
5:51
I
Water flows from a large tank into a dish as shown in Fig. PS.SI. (a) If at the instant shown the tank and the water in it weigh WI Ib, what is the tension, T I , in the cable supporting the tank? (b) If at the instant shown the dish and the water in it weigh W2 Ib, what is the force, F2 , needed to support the dish?
Pifr ~ ~) we ~j?I'I
!he Y~rh'c.a" J
c~mptJneYlt of
the linef/ Y m(JI'l'1eh htlM ~/Aa~~ Iv /he lWJ1en1J of tdnfrtrf vr;/tlme A) C~;J -Iv Je-I
(j) Td qef /,¥flue 01 ~ill
Of!",!
/Ne
!,exnfJu/Jj f erJ.tCk~~ -h; ~ fttMI ff(/WJ .'f'N. (yee IN&< -kv J;' /hL -1?tY11<, -k ~ -Ia~J, ewi Ie f fr; gef Vu",!
Vz9 "A
0-
Then ~
- (Z'
't
=
VCz) (J 'Ii )O() Z.2-
fl.)
i,.(1) we. gef
f:)(''1,{ i/f J{z., If) If
~/u).
/
ZS· f
-
~!
StA-YhttL
ltv.
f!
2
7T /
it)
/
= T, - W,
/-1
16 . .5'"
ve YHca /
pi
;; ere- f 1tV~
~n-fr::>
M
C~ V1';.fv
cV8 we 1A5-e
II')
c~k.
~;-,~ =
~
C\I$
+v '~e su¥/-e,u fj (hlJ
-t-ho ) -
G CO • (2)
.fr-v.-..
37. (,
One)
qef
0-1-
I#a.k- I;' di.fh Iv
,e
(i,,){IO 1-1 rl2,-H)
-:
Z(?2. z.
j
,,;,/.,., we L{S' fn,.,...
(
b
~l"no""IJ; 's €-1l-tWt$"/:rn be-Iwee-n free furtee of
c~
50
CV B
~ ~- IN'z
/;"-/0
Vz.
CVMj'I{/';JC/h./ ()(
t
~.J..,., a! ""'
~ 7(;f
';",;'", c~
W '"l..
+
14· 7 Ib_ 5"-53
J 7-6
P
,.f f ~I-!..f
';'6..
..fa..,/c..
1.J - r;. - W2-
tf)/J.lj'l ~ YzS". if ~ ) 1Tt· 1 ff/(I /~ C' -frJ Jl" .s ) 'I '>/v9 . .ff
5
F2.. ;:"
~
j
S.52
I
t
5.52 Air flows into the atmosphere from a nozzle and strikes a vertical plate as shown in Fig. PS.S2. A horizontal force of I Z N is required to hold the plate in place. Determine the reading on the pressure gage. Assume the flow to be incompressible and frictionless.
~:~= (oz,).1
•
(
-=-.- ..... •
.(/)
V=.==:f= \===::::7
To de~"'M·It'\e. +ke. sk+ic:. Cone:; i dey (1) -tz, (2) .
£:
Area
I ':
C.o~Io:V\;V\~
i
-= r!.~- +
-t
Q'2.,
,c
sfz:\..t;OY\ (I)
"",f! .f(rs+
1..
_Vl
"Z
VI aV'(;.l Vt
of
((.t!V\Se.No..1-tDy\
Q
.:::f 0 5Q
,+tw.+
W'<.. V\O~
= 0.003 m 2, I {3}7'"
fi..\c. frtC'Ho\f\less (}.V\o. i~C.cmpv~ssi bte. flow of air ~ The BernolA I"I e~(.A.a. +1 0..-. fv.r ~'l S Ho",", "s
+- V~ ~
~
9N
FIG U REP 5.52.
pr~SSLlrE.. at
'f'Ie.
4-
I~; ~
Area = 0.01 m 2
•
I
\iYlked b~ ~e ~oV1+iyu"i+y
(lV-I{
~oss) e~~.a:tioV\
~r
'\ V I
E,\s. l Gtt'\O
(~~ V'-)L
'l.
vI"-
okl..,ir"
'3.
_
A2. V2-
::
-V'-2.
\iY\oW\eyt~~ eolA.A,,"c,,", ~'f" +h~ -fltJ\N ~ \.2) ~(3). FISY -tL-vt C,.6Y\-tn1 "o\M.me. s~ft.-V\~J o..b:>ve +he l;~e.ay )"Y\ot"'\ewtullV' py'\Y'lc.·I~Ie. ':j\e\ds
W<.
use +i1c
\;Y1eav
(
12. N
or
57
-F"
'5
((!on' + )
5:52.
1 (um'+) NdW, w;#-1
P. " \
9· 3
'l.
l~
S--55
5.SLf
I
ex ~~~~1' ,-------"
5.54Two water jets of equal size and speed strike each other as shown in Fig. P5.5~. Determine the speed, V, and direction, 0, of the resulting combined jet. Gravity is negligible.
'/
/.
V2
=10ftlS--':b Io.lft
,/"
T
/.'" 1 r
------~ 90 ,11 0
I I I 0.1 ft ,
,I-
t'
Vj=lOft/s
•
FIG U REP 5 . stj ,
the. c.orrty~l vch.u)\e skow~ lrt -\i1 t. s«.ef-"h aloove +he. l&.rea. ( mOM eyd~M e-~ ~:tjo'" for +he x o.Vld y dl~c+tot\.s are I for 1l-t e. X d l fe.C:f,'oV\
For
=0
- V7. ~ V2. A'l. + (V CfJS e ) ~ VA
y dLr~c:hoV\ - VI ~ VI A ,-+-(V "SlV'l. e) -e Vf\ :'
£1t1ct
~y ~.:
C
Also .for coV\<;erva+1l>v\ of vY\O.ss we. PIV( AI -t pV1. "'1. - ~V~ = c
~\le
E~s. I ar\d 2 We.. ~e+
F-roFV\ :l-
V'l. ~I..
CJ)
='
\J~ AI
~o e
~
e
-= c.O(1 V:A-z,
\.1. '41
N bWI
U) M
= O)t- ~
se
= cot_I~( 10 !;) 1r
- V~ A, 1'" V os in
(~t~
'10
II.
n I
~ \ r\ i n ~
1.
(
E~ s.
e
\
'"
0 f~) 1t' (~l.ft)
I). C1.t\O
(V I A I + "2
'3 IN t
AIJ -= 0
V ::
V .::
v= 5'-56
'i
q€.+
-=
v
S. 55 5. S£ Assuming frictionless, incompressible, one-dimensional flow of water through the horizontal tee connection sketched in Fig. P5. s5 , estimate values of the x and y components of the force exerted by the tee Jon the water. Each pipe has an inside diameter of 1 m.
z Section (2)
FIGURE
We
COJl'l
eCjtAa-fitTYI
U$e
'feacf7'oYl
-fhe..
cor.+rol
=
6 m/s
=
200 kPa
P5.55
fhe x and y CCIYVL(ltJnel1'/-r of Ik I/neay YY1()YY1eYl-ft"YVI
(&J.
-fhe.
V1
Pl
5.
22)
To
defen-'Yfine
exeyfed. by -!-he
foyce vo tUYJ1e
fhe
x and Y ct:JYVl{J()YJenls ()f
Fw f.22 leAds
wa.fer Oh -the.. fee.
coYtfA/~jn.J wafey I~ the, fee, E$'
(2)
1/
V3
:::
ll.1
1T!J.~
Li' (con 'I: )
$'-57
5'". 55
COn
If )
S,5h 5. Water is added to the tank shown in Fig. P5. 56 through a vertical pipe to maintain a constant (water) level. The tank is placed on a horizontal plane which has a frictionless surface. Determine the horizontal force, F, required to hold the tank stationary. Neglect all losses.
i
l'
-'-+)C ,~ _C0!lst~nt. water level 1 -:-:~--~-:~i-
!
(2.) ..
Jet area
= 1250 mm 2
'-.+1. m '. ',lm
F~
Frictionless surface i
I
-~--~~:;::;:~-';:-
..i
Jet area
= 625 mm 2
I
! /'
•
,
FIG U REP 5 .
5.6
A?p\y'It"~ -the. x- dlV~,tioil' c.oMPOV\e.I\+ of -the l jr\eAr W\OWlCY\-tuM -e~(AClfi ov) to the. U>~~w\s 0+ +""- (.O~~l vo luM(. Sk.tt.~J dloove vJ~ ~e..f ,
V,~VI~'- V2..~Y'l-A"l. -:: F
(I)
8err\OIA\l','~ ~uo.fjon ~ d.~~c.r~1oe.. -the .ffidio~l-ess -flow frbM -tltC'- c.o\'\~-h\wt WG\kY" SU1~'E:. level +0 +nCo .flow l~o..Vl~~ s.m,1io~cz, (l) and (z) VJe... alo~in USlh'3
V?. ";
[z ~ h"-
(2.)
I/I=~ ~1o;v\ll'l'3 ~5. l) 1. dVld. '3
F
= ~C3 ~, fA. -
or
F= D~
we qet
L~h2.~ A2.
o.t
5.57
I I 1.L-_ _.,....llL.
.........
(/f'0
Water flows steadily into and out of a tank that sits on frictionless wheels as shown in Fig. PS.S7. Determine the diameter D so that the tank remains motionless if F = O.
'"I
I
, 1
I
I
'~ '~;. :.o. ~"; ;" '~3-)
AffJ/Y1nJ fhe hfFfljQVlIa/ l{fVl1p(}ntVlf pI f1tl I/Ylea4 Wlornen..f.um -e,u&&-I/f/h [tin ~f.s
fn -hu. Vd tUY'YJe
0f
!Itt.
---if
(ltD
(gy,.fn; /
we qef:
s h(JWYI I;' the
fcs VtV.;dll :
Vd~= V pI-- -I-VdL. J
2.
I
A'}t:un / fl»c.e.
= ~ -= VlJh we gef
Vz.
V;cl7.. = ~p7,..f ~ ti1.
~o/c.IVJj ~f ~$. If
V3 L.. V;
(I)
CfAYlJI'Io-f
e~.~)
e8 . (I )
If
(2)
~ (z)1-or k
eS' (/) e~. (J.)
'-t4t.-
I..t
CaJIIYl,f
et. (j ) ~vo. k
~. lIs-;-~ L?. _
~
V'S ~ "'"
Vj -;
(7-)
~[)=O h wt..u..!' f
vzg h
w<.
~kvk.
I:x 5iJ.HrteJ
be 5df,j.fl~
CtU-t
C. (:( 1'1
F
~
sall.rh'ecl
be $t$--H.r~e,j
~!he,{ w:/t. j) -;; 0 ~ ~-hj/je"l J/l/'tt. p:: 0 ,$a.
k.L.Je f
::: l-j
5'-60
Jf)
fhal
of each is atmospheric. and the flow is incompressible. The ..' " contents of each device is not known. When released. which TI he four de~lces shown In .Flg. P5.5~ res.t on fnlctlOn. devices will move to the right and which to the left? Explain. Iess w hee s, are restncted to move In the x directIOn on y anc are initially held stationary. The pressure at the inlets and outlets (I) ,-- - - ~ ~I .
horijoYlfal
We (),pply jt.e.
/urn
/he
eg/,f(.f IltJn f,
I I
,
(Z),~~.__::=;:_-=::J'~~L-'
c()mPOYlenf. of f1.,e II~ e~ ~ Y>1 Om e,..,
~
\
. -'
I
(oj
~f fl,e. CdYlh-;;/ I/IJluYlfe
("OJ1kj1f$
(bfl);:'en liYles) aY'ld delel"n1J;'e the. 5e n$e of fll e
.,
;If F.
Is
Shown I~
tll'l cJ"
fhe
/n
J
~;t1
..fo.taL ~.
d"~cJ-i()n
(cJ
(eI)
the .fke~l,es, nulion
will be. f.o {he /e fl. .If ~ if
/n
~ d{Y€.cf/(Jn
riJhf . Fr,-y-
()flPIJS;
Ie
{-Jaaf shown.! the. Y??o~;'Y)
fr;
.J..F FA:O.; there- IS
1'10
)'(J'rJ
jonI-,. /
is'
n
fhe
l?1(rhOn.
ske tel, (a.)
- v, ~v, A
I
~ ~ ~ Al.
-
5(ncf:.
~
PlN s/r:.e
kh (b)
=
~
fo f1-,e lei f /
is
mIJh'on
/J -/tJ
fhe.
YO;J},!.
-~f'~,A,+ ~f~A2.=:F ().nol {y()n/) Ulnset"vah'trn of YY'Ja>J
fJ V, A, :::"p ~ 4 l.. ~I'ki
V '> V
)"tnee
2.;
I
rhe"
Fov s~~ ~) (nD~:
FLI
If
n
-/7J 11.e /eff
flow i.r ,;".ft, c va:f
a'JPI 1"I1(J~~",JIoIl.e~;flu.. ,/
(t )
-l1p V, A = ~ I
a~d ~ ;.1 h
ft,e lelf-
f,
nro hon /s
h lJ,e
njAf.
FIN' 5Ke. f-c~ (d) -v,p~~ + and ~
~ p~,.42.
UJYI.JerVtA 170 n
,.0 V, 4J ~ f'V2, {And ~ <.
so
= F;; of ~a.fS
A2.
"2.
~ if -10 fie r/!/'t-
lAne!
~Pf.~r'J /$ /r; the Jeff.
5-61
~. 5'1
I
Water d ischarges into the atmosphere th rough the dcvice shown in Fig. PS.59. Determine the x component of force al the flange required to hold the device in place. Neglect the effect of gravity and frictio n.
To
30
fl~,~
______
,,,
,,,,,,,/~k fht ;(-d,.-
{o".
"fU'Ye,) to hold f/,e device In flac¢ / the x- d/.-edio'1 CO>'>1I'OYI,,,1- of the 1,'.,,,,1" mommlu.., egl/ali."" is fAJed 0" h.e (ft,k-.. iJ ",-,clo.,,'''3
fj,e
+0 .I.fa;n;
skt-kJ.,
-V,,,,VA -VflV4 \ 112.2.1. To
dekTI'M/ne
7-D
./,fa ,';., :
VJ
I
.1A
+ V,, cos'~'f' V3 A3 = - F.1/ +f." 1/ CIJn5€rvAT7 J, • ..,
LJ
Jn ~
Q.,=>;<" T Q,} ~ II, = Ii A, f
>J A J
(20 ~f;:O. 8r~') ~f(J~t)(f/'ff'f')t- v,. (b.3h') f.
V3 =S ff S
fY7f'r->
E3 ' / we
,e!-
_ ('20 V)(I.U it}(2O ¥j{d.Zff')
FA
-=).'1 '+0
Ib
1>-62
(i )
S.60 5. 60
A vertical jet of water leaves a nozzle at
a speed of 10 mls and a diameter of 20 mm. It suspends a plate having a mass of 1.S kg as indicated in Fig. PS,bO . What is the vertical distance h?
FIGURE
fhe ve~H(.eI1
To de/eYYJlllne.
fhe. II~eIlY
cowtpbY1enf
tJl
wafe r
fhe
I;"
- R~
-;()O
r
.7
con-fn,/
h
cI/sfttnce
I"n()YHehfu".,.,
P5.60·
apply fk verhl.al tilfecho""!
we
e'6u().h·~h (61' >.2Z) fD ft,e
shown 1i1 fhe .s-kefct a.bove.. Thlls,
voluY11t2-
-ifwale,.. -- - VI f AI VJ
== -;0
J.
l{ ~ ~
2-
(J)
tt
7Ae verfico,/ reaciion force of lhe ;/ole on Jhe kI~/er is efu,o./ in 1n49nilu de 10 lite l,fKiIighl of Ik pJII.le) ()r
RJ : :" 9 !11p /a l e
=
('1. $1 !f) ( I. 5
~J:: I if. 7 N
A /050./ !he. wBiyh f of M£ wQ.ler willl/n I/J£ cfl)1IYo l volt/me J fJ.9 ~41e.r J ;.s 17~1 lig iiJle I and fAe /?It.Z.fS flow role is
m jJ ~ l{ =:
== ;; Ao Yo
-:: ( 999 ~ ) f
(/J. o:J.
v,::
N
hi )'- (
10 ; ) ;:: 3, /3
l!f
Thull Elf / becames - /~. 7
II ::: -
~
m
or
,
11f-. 7
- If. 70
3,/3.1t1s-
From lite Berl1lJtI// £tjpg!itJn (Er;.3.7) ~ + i. fl v,,:l+ J1zo
::::
-f1 .J. 1:(/ 'v; ~ 0' ~
I
Wi!
1fU2~ ifJl1~
/'0 -::It :. 0
1#
0
J
2,
:::,.1
h
l' ~~?
or .Ji/lce L If
h9ve
wAe.r8 20
T/;I/~
tp
=:21 (~,. /I 1) I
0
-
I
I =::"2. (q.81!f{J
(:z.. :z..) m ~ /0 -- JI.. 70 ~
=:: 3. q 7
/J'l
..-
5.6'1 Exhaust (assumed to have the properties of standard air) leaves the 4-ft diameter chimney shown in Video VS.3 and Fig. PS." ,."ith a speed of 6 ftls. Because of the wind, after a few diameters downstream the exhaust flows in a horizontal direction with the speed of the wind, IS ftls. Determine the horizontal component of the force that the blowing wind puts on the exhaust gases.
15 Itls
• FIG U REP 5.61
For -the
control volume indic4fed fhe x-Gomponen!
()r the
momenlvm eCI'Jafi~f7
s:u pV·/1 dll
= ~ Fx
become.s
cs
~
p Vz- Ill-
where Rx ".s the nef hOl'i'l.()IJio./ force fha1 -fhe winrh pvis 0" -foe exhaust 9a.ses. ===
Rx
J
ThtJs
Rx ~ m:;. V~
m2 ::: e!9" \4 == ~ A, V, (i. e. m,:::: '+',. ) m~::: (0,00238 sJ?-)I1J (lfHt Jr 611) ::: 0, /79 ~~~
J
or
Whel'B
He nCB.!
Rx '" O.1793J?(Jsf}),: 2,6'1 0f.!1:= 2.tQ/iJ
S-6tf
5· 62-
I
5. 62. Air discharges from a 2-in.-diameter nozzle and strikes a curved vane, which is in a vertical plane as shown in Fig. P5. b2.. A stagnation tube connected to a water V-tube manometer is located in the free air jet. Determine the horizontal component of the force that the air jet exerts on the vane. Neglect the weight of the air and all friction.
•
P5.~~
+hat
Note.
we 'Iqnoyc. -\-l,<. ef~+ of o~ost>h(,ri c.. Pf.f'SSUfC On ~h(. \lal~~ ;n OUf SOlu1iOVl below a~ck lASe. ~~e. pye~SLtfe.\. I\s 'In~\c~6 in
R
0+
FIGURE
)(
L '
Ex.~mp\e. 5, 10 J '#\(.
.
~~Sphc":l to ?y~ ~~uye -rvrc.e. MOJ-j r\ee.d c()ns I"~r~+ll>"" whe", id.t~ti~iY'l~ rto.L+ioV\ +orc.e!,. For ~e Qty .f'low·It\~ ihY¢~~h+he C-Ot\+rol votu,,",e., ~ ~~ a\oove) +he. X - d 'f~C -He>'" COMpot'le.\I\i of +he. I', neo.'(' M()mc.r'\1u~ ~(AO.,fi on 'I~ - VI P
,vI AI
V'l. eos
-
\Qlf
I\fP\/\~tioY' 0-(
(,) -tv
('2..)
0~.
30
V2,.Al..
Q\r
= - R;(.
(I)
8eroololll\ IS ~W)..p.oY\ 11r t-he. flow fn>M
y'le\c:is
V2, =VI Ihe n ) +ron'l +he. C()",c;erV'a.-i1on of Mo.S~ t>Y'Ylc.iple (~)
A,V, :: AI...V't.
m
e..rlAAfioY\ ~dl~ -+0 oJ::,~~1'\ +he .fDlloWl·t'\~ e,~+lOY\ fz,y ~ S~~~(7Y\ -hAbe decele.~~Oh
We
lAse
~
1
+
~ (l\r
For +he
t3c.rnotAll',
VI
:::
(~)
'2..
w\.t-h. +h-c.
V\f\o."'o me.kr'" I
Pt\~ +
h fY'~ t'\OCS' W
Q.
\.e,y
-
h
0'.
1Ytt;tr\g
o.lr
-
e.\IAct1io,""
1:s+o. ~
(s)
( G,)
( eDV)'i )
5·62..
I
(COn't)
Rx. ==
~x= ThiS
2.9f, \b
--
is the .f-ov'ce -eKe..rkd. ~~ -\hc. vane
--the. .f.lovl,~ ~~V'.
\?~ th.t. ·tlc,w 1~'1 <:\\1 txer+s on 1i1e VCl.n~ ~ ~~ ~y\ ~7 \I'\\-Mo\4. b~ Of posi k iYl d ;~c:nCTY\ 00 -Ihe rl1h-f)
The. .fu.ru2. -ex.uktl \
CY\
5"- 66
5.6S 5.6 S
A 3-in.-diameter horizontal jet of water strikes a flat plate as indicated in Fig. P5,65. Determine the jet velocity if a lO-lb horizontal force is required to: (a) hold the plate stationary; (b) allow the plate to move at a constant speed of 10 ft/s to the right.
(al
(bl
FIGURE
The COI1/-rr;/ volume
In The
shoWn
the
htJYlj()nft?1
tyJ () Me yt
/u,yy,
x- dlf'ec.iltJY1
OY
e~ ua/j(7}/l -
is used. The
skefch
sfah'oYlPYJ plale CAfe if. cPI1J'ldeYea
"5.65
IIpfllicaltan ~I
fi;ff.
cOYJ/If1~nent
of
fj,e /;i1eor
y / e/dj
-~
J
X
Oy lA,
Thus
()nd
f.,(,
plate
the
tit
fP.eed r
I
V= 10 :s ff /
-the lin eery hI~hZehluh-t erUIl/;OY'j
x - dIYec,fioY]
or and -==
/)10
5- 67
ZOo Z
====
ff J'
VI;'" platt!
5,66 A Pelton wheel vane directs a horizontal. circular cross-sectional jet of water symmetrically as indicated in Fig. PS.66 and Vidtm V5.4. The jet leaves the nozzle with a velocity of 100 ft/s. Determine the x direction component of anchoring force required to (a) hold the vane stationary. (b) confine the speed of the vane to a value of 10 ft/s to the right. The fluid speed magnitude remains constant along the vane surface. D
=1
.,; m.
(a)
(b)
FIGURE P5 ..6.6 (A)
To
deferm,nG -fJ,e. )(- dl;ecli~n cpl'H{J4J1ehf of qJ1c.h~Y'I~ ~e !7!fUI'i-etl iP
hlJ/d fhe
e,
tU?IJV~
skwn
S"1ei1/onlJry JIVe use file S'lRh'dJ1~'y on"/rrJ/ v~/JtM(!
V4J1{!
tlJtd th~ x- d,yeCndn
(£1' Ii· 2"2 )
MA:n#1I1
ct?m;;~l1ehf of- the //heo.- m,Jlfle"lu"",
l"hu~
•
~ :: m(tft ~ I.IJS 'Is ') :::j7AJ ~ ('1 t~ ~Jljl"1 :: I' ~p/I{~ of ~ t;.5'1~'
F. :: Iii Ib
" de!e,YYn,;'e
(6) 70
fhe
X'- d,;e&h'Oh
11J U)wh'ne .fife. Vane.. ~
life
&t
Ii
I
of"
W CIJ$ Y.r D) l.
&
/
JI16-k 11tA-1-
/'1-1 'I,
s- 68
fr";"
fo
ftt~ Y/1hf WJ'fh
()f
tHe. I/;r.u.r
clh1fy,1 vO/~Ylte (£1' $".21).
(vv J
or /lJ1cJ,fH'~ ~e ",(HIed
Chlsnrl1f s/eed of It)
Ynov/nf/
V()/ume.
f YtlJ1.rlitn·"?j
,: ::. ~ A, ltV
We
()l
me x- d/ved/on comp4J1od
and
IrN-
Clhrlrol
fr;
CIJ"VJ()nel'1f
p
7r~). W
(it'
t{
IRe Yl9"'f w-e
$,ud pf /0 ~f"
MOJll1.tJ1/ztm
ejlo"h',,,
lJut~"
/W
' ('
I
t W 1.
U.f ¥.f D) /
{t}
'i.h7
I
5.67 How much power is transferred to the moving vane of Problem 5.66?
P"wer
~
EA V
)
where frolf} Prohlem
( l/.fb \b ) ( ID PDWer :::
s. 66 Fa -::. II/bib
ft ")
t;S'o ~+.lb \ ( ~. hp J
':2
2.6S hp
~-
69
n
5.68
Water enters a rotating lawn sprinkler through its base at the steady rate of 16 gal/min as shown in Fig. P5.68'. The exit cross section area of each of the two nozzles is 0.04 in. 2 and the flow leaving each nozzle is tangential. The radius from the axis of rotation to the centerline of each nozzle is 8 in. (a) Determine the resisting torque required to hold the sprinkler head stationary. (b) Determine the resisting torque associated with the sprinkler rotating with a constant speed of 500 rev/min. (c) Determine the angular velocity of the sprinkler if no resisting torque is applied.
! r ij. /NOZZle exit . --t;--/,.,..----=
8
area
/ /--
I
---
=
2
O.04m.
.......
1/
'\ ~ ~
V
~~ "
.//j
...-///
--- -- -::-:--t(sio.fit1nd. YJ t.........
@
ClJtrfl7Jl
IItJlu~e.
=16 gal/min
--- -
I
~
FIGURE PS.,8
Th;s if f/~;I"y -It> GK'amfJle 5: 17. (a) To dekv»tlne fhe ~5"ir/Jy,f 1t;Yi Jt~ r~tlli~d fz> ;u4d M~ 5jJn'hk/erhead sfafl~l1t:1'Y we U$f:.- The. m"JIIt.~f - tJl-;;noJ1teJllfu"" ~f"Iue
etlA.alidn {EI' ~. 50}. i'Jul5,. (;)
= 6'/./7 t! oS
-r.$/uil-I : -
z.96 ff.16
dehYm,ne the r~5iJh'~1 -I-oY'ltl~ a>$dd~f.ed w/ft,.. tt spYIVJkiet' ~ttd of 5"t1~ ye.v we. a.5e c1· / attll;'. fhn.veVey I (/ot'i fA ro/7tI/t/YJ Wf.. /,ave.
(b) T()
ml;')
(Z) FI7Y W ,.. Wl..
We c
use bL
2 AI1()J}~ exi,,"
-
(16 !~a(
) (Ill'! ;;~~)
(2){ ". ()'f (GLJI1'
,;"....)
f ) 5',70
(7. '1R
9,4 ,
F/~
I
-= ~ ~.L )
....,;.,
{, '1./7 ff. J
S:6B
I
(cOJ1'f)
r~y
Z{
/,Ale ('(fe
= (9 In.) (soo ~ ) (i7r ftt ) (6tJ ~ )
(;2 I;") off
£1' 2. (,N~ htMe.
Th{J J with
:: 6'1./7 If - "llf.91 ff =T .J
V.
~2.
C/Jlfti
;Jfm
W~
E't. /
wilt,
e-z
-
(~'fK
,$/11llf
Y8 In.j{zf. zt t) (/ iJ!..
f.."/ lJcin . / •
I) (bIJ ~)
9tf rf2
alo'U/
ff-
.J"
tJ6"kt,;"
- (to 9'1 f.!!J.!)(I~
T
2f/- 2{
him
sll4,.
/12. (
i" ' )
fI-
)
.sa..
H
/. 3~ f-l. /h
CC) To de/e",m/YJe fAt.. ~n.I~/ar veto~/ fy ~f- I1rL >jJ}'/nlder /f ~ Y~/>h~ f()y~ tl.e fl;
p-p/I/ed we.
!s
~/Y1tA-h(;YI ()f CtjS. / aJlut 2
() bID. Ii?
tf
::W
w
=
)..
2.-
~
W
2
=:.
(6'1.17 ~/)(I Z
yo.....
7h e
the
U.${!
Yl; ftJy
":::.
(S,;'.)
~ 1MU.I IV.I i.r
N -::::(?I.?
if)
YQd)
s
fI, Ci .J
(b()~)
/(277
~d)
=
rev
5-7/
920
~
-======A=,=''''
9~·J
ypd
s-
5.69 S.t19 Five liters/s of water enters the rotor shown in Video V5.5 and Fig. P5.69 along the axis of rotation. The crosssectional area of each of the three nozzle exits normal to the relative velocity is 18 mm 2• How large is the resisting torque required to hold the rotor stationary? How fast will the rotor spin steadily if the resisting torque is reduced to zero and (a) () = 0°, (b) () = 30°, (c) () = 60°?
II FIGURE P5.69
7b de/erm/l'Je
fh~ flJ~u~
we use, -10 obfa/~
~oJ+tenf -
fhe.
~ y
;- Shlll'f --
We
nl)f~
J1I1
::
~uI-
V
duf-
"'~UI"ed -/zJ
}u,/d ..fhe. YrJl»y fiP./7i;)Ylaq ,,1-- m()JIH~Iu;+1 rorf"G I!.j;UIJtIJ'un (1:1' s: So) (I)
t115!)
fHaT
p t:2.
3A""'"1le eXl!
£1.$.
I J Z. aJlld 3
r' Q 3 To
riekY"w,ine
oSha! f
iDrfJ/lfe
e~J,(ah'oV1
nofe
Y)1
(it)
~t co.> 6'
AI'I6JJle
exit
fhe
Y7Jtrrr
fAJ19/,1Itt y
Wl.
~!IJ In
use.
( €'j. ~. So)
~hah-:: W-e
2
we
fhe
+t> "hm,n.l
~"f (tv 1)1A.7
CbS
veltJvify
it,)$
e
t1>s()c/tJtf~d wifh
m()~eJ?f- of - fflomel'1luW'l -/Dt"ffJe
l7~e
qt.tf- )
w /t?t hJIRhol'7) (5)
tha.f
(7)
(cOYl'f) 5-72
3ero
5.69
I
(c~n'f)
r;half :::
I
s"4.11 Fr~ £1"
w
= Zt:Jo
AI.
not
w~ obJ,un ~
fl.t:?H::
tJ
If I~/ (OJf 3()")(/PI?P ~),.
=
-= IbtJ
3(18~~L-){ IPI}~ ~)((J.S"..)
e ==
(c.) F7Jy
r~4fi
&
60 d wt use (9ff!1) (s ~3
£" 'I
Iv ge.f ~
~~y~ (3) (IYn-.n,~)
py
III N.~
~ t'j. f W
~
we 16hun -fp.., ~
(S
.$
~/(t).'-'")(Co! to') {tPtJtI : , . : /(f1'/;" )
(/IJIJIJ
rJhaff ::
--
YAel
~)(UJ6tJ~
L
=- ()
'firr (IPPd
~
2-
(3J(13ht,.,~J(lOPcJ ~) (d~~;)
;-73
f>
b1 j1.
5.7 J A water turbine wheel rotates at the rate of 50 rpm in the direction shown in Fig. P5.7J. The inner radius, T 2 , of the blade row is 2 ft, and the outer radius, r I , is 4 ft. The absolute velocity vector at the turbine rotor entrance makes an angle of 20° with the tangential direction. The inlet blade angle is 60° relative to the tangential direction. The blade outlet angle is 120°. The flowrate is 20 ft3 / s. For the flow tangent to the rotor blade surface at inlet and outlet, determine an appropriate constant blade height, b, and the corresponding power available at the rotor shaft.
f-
Section (1)
since
•
Q. 211;-; b ~,I +hen the blade /'el,kl-, b I
Section (2)
FIGURE PS.7!
II::
a
b
:Iffr;
The
1.5
51101f
OJ
~I
•
P()WeY'"
vv:.~tlH
) 15"
lief (NIt
powef" ~uah'tJn (£1.5'·53) . •
-vtt;.haff .::.
(Ina
nef (Jt.ri the Use
1/
of 11+ 'I
.,
wi It,
0,.
l{ V~ 2. d el'end.s
0"
wire/her
Ve 2 /s- opposi Ie. 7-0 or i" +he. 5"ante d/rechon as V; respecfively. / To defermine the value of ~,I we use. the velocify fYiong/( af se c. Non (I). Thus we hallf: I
V.R,I V, Wi~ fhe ve/~ci1y /Y/(Jn9Ie
~,I
hn
=
2()(J
V.R~ I fAn 60
we.. have
+ V, 0
However
V;
-
r;w
(con't ) 5-74
5".7/_1
(c()n't)
thus fq.]
V
leads fr;
rOd) (if ff) (5() rpm) (/-271' ;;;
r,w
::
:::
/ -L _ I ') ( hn 2.0~ fDn 60,/
R,/
tJbh'n (to #1)
tV i ff" ~9 ,I we
~
b
211"('1/1 )(9.65 1
=
9")
fhe blade lIe/ocifl'es 1i1 E~.
For
7J = r, (J.)
('Zff) (5 0
-
~
we
(Ve, 2. ntrt h
VB 2
::
I
_".
6 ,1
T
mh 2(}
0
COnJirucf
t1
ff of' ./
salle)
~ Z fan ~() 0
V
8,2.
-
(It )
t{
cOf/servaHon of Mass
:
~ 2~. ~ l
The se.cfiol? (2.) ve/(h:,fltj fy/1I1'10/(: ske.~kd 6elfMI
I
R,2.
ti
s
HI/I?
L{ yeahie -fhaf
We
t'p~)
fhe, velrJc/ Iy iYj(Jnv1e af secf/(}n (I) -h; ~blzi,;'
I
V.~ 2
.s
60L
~I fz:tn 20
,::'
I
'II.:
= 20.9'1 If
"'11" ( 2-71' ~) --------------~'~~~---== ~.~7
= J: w =
Foy ~I J we we
From
we ref
(60 .s: )
1.l:z.
(AYld
if
('I Tn (so YPWt) ( 2TT ~)
==
I
F-or-
2
tJ. O~25
V.
A,:::
/(, I A
2
(con'f) 5-75
wi Hz
50
£~.
V~ 2. -= Finally
'T we obmln
{ltf.3
€r.
w/111
virshall-
~f)1-rJJ1 30 2
~ I!,ti'l (,"
net ou.t
0 _
/&.'17
ft
== 0.673
we (}b~/~ sluff\fi.o if) )!(zfJ.71f £1 )1;.652"£rl-f{O.'f7 fJ:)'/O.6 7JB)}/i !it- ) 1-+1/, s If co S /(4. S/ s I· s"(j{ sllI1.ft J~
Or
W s)1tI.H net
":
'2.. /8 X I() 'I ft. II:;
-s
4ltf
a~c{
-Wsna.If nef"-o/,tf
;f
-
1.. iN X II) s~o
'I
If· 110 s
-::: '39. 6 hp
ff./b s. hI'
5-76
5.72
An incompressible fluid flows outward through a blower as indicated in Fig. PS.72. The shaft torque involved, T'haft' is estimated with the following relationship: Tshafl
= t1U2 VO.2
where t11 = mass flowrate through the blower, = outer radius of blower, and V O•2 = tangential component of absolute fluid velocity leaving the blower. State the flow conditions that make this formula valid.
'2
.....-
_...
FIGURE PS.1Z
I valId
S-i1a It
-
J1I1a..y 6e
c(JY14fJ~neJ1f
of
i11~
~)2
Ide",.l/h·ed
d.
e.
V
0,1 no
6y
Ee. s· 'f2..
a.. s tah'oJ1.IJl;':.Y and ". slea.dJ - I;' c. ne&I'j/6Ie
(;)
- flit! -
UJ>?1ffJl'I
~&J' I
co hd i -n (n,,$
Thast!
with -IJ,e /Ix/a I ar~
h~J? - de/w~/~ CQ'y1~1 volume
MeQI'I
She4Y sIY~s.! 1oY9UL
sftt!>~e~
d /s fy/huh'on
(see .rtelc~ a6uv&)
,clow
::: 0
f. u J1 / flJy~
"J
of t,.;, I
~77
2
w;~ ve>Iee-f ~ tlx/s IJ{ ttJ/r../JOh
5.73
V R2 = 30 ftls
5.73 The radial component of velocity of water leaving the centrifugal pump sketched in Fig. PS.73 is 30 ft/s. The magnitude of the absolute velocity at the pump exit is 60 ft/s. The fluid enters the pump rotor radially. Calculate the shaft work required per unit mass flowing through the pump.
•
aMt! 11011- d~/PYMlry
The stah'o}fcuy Sk~l-ch
lIni!
./
bIt/de
1""1_
r,
$ntlrt
Id~
V9)
2.
~".ud I tJ;. )
"l-
cd;r
ThuJ I
.frn.n
= (v:-
s: S"~.
shlJlf wOy/<.. fer
TJII~.J
(I)
be ()6M,ned
CAli
I
(lS
..fiJI/(}w>,
~/;'
)fTT ~)/l)::: yeV (~O...E.-
k
dbl,J1net/
~
The. fangeJ1hd ve/tJti/-yJ ~
~2
£4. II
(he
2-
::: C W ::;; (0.5 ff )(2.000 y!!'
V
ct)J1irrJI vt)/tlYJt~ shown /~ -IJ,e
/0 ttebYmln-R C~"'" l«,. U~I ""'L
I""""
= 1J
w.$/'tlfl7}te
used.
i.s
abtJve
Wlt/5J
FIGURE P5.73
(/J
s
fp/;;/.,;~
v~ ... )i [(6off/-(30~t/l ~ ='
JD5 If
52
{f s
-
£$' I
=
(105 :1)( ~2
t;lf60
ff·/6
s/uJ
~78
5.74
A fan (see Fig. P5.74-) has a bladed rotor of 12-in.-outside diameter and 5-in.-inside diameter and runs at 1725 rpm. The width of each rotor blade is 1 in. from blade inlet to outlet. The volume fiowrate is steady at 230 fe/min and the absolute velocity of the air at blade inlet, VI' is purely radial. The blade discharge angle is 30° measured with respect to the tangential direction at the outside diameter of the rotor. (a) What would be a reasonable blade inlet angle (measured with respect to the tangential direction at the inside diameter of the rotor)? (b) Find the power required to run the fan.
The ~lah(;Ylary and skel-ch above. tJ~1/e
/>
is UJed.
" ....
....
,c
.;;:--._--FIGURE PS.74
delrrrm/~J confn;/ VtJ!ui11e de~n1/h-e.
10
I
I
1pin.--J I-
.fjzDWYl In
-the
a retrst)JlUf61e. blt:lde
6/adl. $htJl1/d ft,e /-t;;n~Mf fo
ve/afive ve/ocJfy ttl lite Iy,/~f. JkekhuJ belttw.
The.. Inlef ve/tJcl/-y /Y/an!/e
we
aSSume
fhat
WI
(;)
Now Q
::.
-
A,
Q
= 35./
I
(12.
wifh E~. /
In)
ff fI-
e = +aM -t{J§· I J) 7 = ,
(37-"
f!
S
= (2.5 11].) .
== YoJ
7hu;
I I
the
Ih/ef
-Ihe
nOn -
"
51 ) ]
'f 3
0
(Con'c)
5: 7/f
I
(con It )
.
The power ye~~"y.ed) ~J,tlll) may be tJb-blird wi/tJ Thus
&1-
.
(2)
W:::mVV fha/l-
VV1IiH -!Iow-ra-!e /
th< 1-
'"
jl Q
~ "L
z.
1..
rl17-J
='
~bIt;l~i
YYlay k
a$ hl/lIWf_
(2-31 X 10-3 S/"ff (z"i0 tf Y -'- );; q.12 xmJ~ if to.!... J
J
Also
5".53.
/
fIIlh).
S
111,/)
u.. : ~w '"
(6 In. ) (172 ~ (/2
J£f J{27T fl! ) '" 90."3
In.) (6tJ .ff-: )
H
t!. .s
/Hln
The value. of ~I ~ may be. o6-/r:(Jnet/ by cOh-rld'Ylj .fh~ lIe/«:i7 friPYJ1/e -fpy -Ihe. flow letW'l1j -lite j?J~ af 5ec;hdn(~_ The ye!/ftfNc velocily at /he Yl)~ exjf- i> UMsJdereti -IT; be. IoWf fld -Iv the-- hhde fheY'e _ !he yafw exi f flrJ/N vtl ()~i fj -IY/aY/le is
skel-d1eJ
belfIN.
-
5.7S
An axial flow gasoline pump (see Fig. P5.75) consists of a rotating row of blades (rotor) followed downstream by a stationary row of blades (stator). The gasoline enters the rotor axially (without any angular momentum) with an absolute velocity of 3 m/s. The rotor blade inlet and exit angles are 60° and 45° from the axial direction. The pump annulus passage cross section area is constant. Consider the flow as being tangent to the blades involved. Sketch velocity triangles for flow just upstream and downstream of the rotor and just downstream of the stator where the flow is axial. How much energy is added to each kilogram of gasoline?
c~m+-ml
~5C
VIJ /Utl1e I
0
60
I
j£
I
I
I mean radius blade sections I
I
:7U:~coYlfro l
J
I I
I
I l
VDlume
)
FIGURE P5.7S
lhe veioc-i-f'l l'IiI1YJj/e." f;;r {low jusf (Jff~tM11 ~ /he, y,,1-tJy Is belaJ
SKefrfrled
fhe arithmefic mean Yael/uS.
lOY
WI
,- s-
V-"3rn
With
W
'he TriorJj!e., :;
J
V, cos 60
We ::.
6
con~(ude fhai
(3 ;:-) Co> 66"
ay,d
=
6
m s 5,2
v;= 5"- 81
5.75
-coy/Ii)
The ve-I()ciiy -/-vldYJg/e I()r (I(}w jllsf c/()/)JJ1.JIYetlM ske hhed
b~/(NJ f()y
t'i7UJn1fJm5ib/~
Thus
flow
lor relafive
pf fhe
Y()1-or is
-the tJyifhmelic, JlY/e tlYi Yadws, ~ ~)2.:::' ~.
f:or
/J1eon ratift.l.f t~
lIow
faYJgeJllf fr; fhe blade ve(rJcify -Iv/angle.- .>kefd7ed belM.
Wt
l{=Z{. obkiln iAe
W2.
~lI With fhe fY/an,/e we undude thaI-
Vn., ~~
= U~ -
W@JZ -:; TJ2 -
v:J21anl/§D:::
5"- 82
5,2
m_ h s
l-
m) -M~ '1/= $
2.2 ?Z' S
5.7~ 5.7" A sketch of the arithmetic mean radius blade sections of an axial-flow water turbine stage is shown in Fig. P5.76. The rotor speed is 1000 rpm. (a) Sketch and label velocity triangles for the flow entering and leaving the rotor row. Use V for absolute velocity, W for relative velocity, and U for blade velocity. Assume flow enters and leaves each blade row at the blade angles shown. (b) Calculate the work per unit mass delivered at the shaft.
J ~76~Ufe /
/
~
'A' I
/ 45¥' , I
I
FIGURE PS.7b
J
I \/oll.(jIr\e (0,.+11)1
I
I I
\lOhl~e
Blade sectIons
arith~etic
at the mean radius
17.5
0
I
The ve/oci.fy -/Y-/dr},f/es foy fhe.. Tk>w enleYl?J and fhe flow leav,h9 fh e
vofrJr
cd -!he dJ'lfhmelic.-
Yow
metlh Ytlc!tu.J (1ye JKelched
below.
1J2.
ari~lJtefic
Ai fhe
V:: t{ I
m~a;1
rw
::
f\!I
=
vCid/uJ)
the bltlde ve/oclfy ) ~ is
(fD ,;".)
I DIJO
re.",
nt/~
(12 In.) ('02-: ) If
(Z'Tt' Yev ~)
:: ?:? '3 II s
)lHlh
W;t4,
th~
~>in
7()fJ
V, (;() 5 70 W
"
I §irl LfS II
\tV I
fYl'aYl!/Ie.- f,Y fh~ flow l nHn-IYtJ the Y7JIuy Wl- cmclwle fIdf
vel~cify
-= VgI
(I)
I
= ~I
(2..)
= V~ ,- u
('3 )
Co> '1-5" ;:
'Ix,
I
(Con
'e )
5"'- 83
('I)
(con '7; )
5. 76
Fro}1l1 fhe rtlfio tJ{ ~5 . 3 /AMI if we. Obf71'~
V.X,, when um6/necl wift;. Elf.
which
(J
=
1af'llfS
I
~d 2.
y/e/ds
vI U)S 70"
or
u
=
[5/i1 70
0 _
=- ~'7.fofl.s
TheYl
\.j' S1i17i'
VBt
::
V
= \tj
= (f7' ~/)
=- SZ,3 If
'S,;" 70 ()
.>
I
~/
fa7.b ~+)
7(/ -=
0
(()5
70
:-
z"
'1
-Sfl-
I
()nd
WI
(.f)$
Vx , I
=
cos fs D
-
(~'1, q ~) U'Js If r
::
()
12, '-I
!! s
W/th fl1~ ve/ocify -/-Y/OYl9/e fllY fhe fl()w leavlnJ fhe r-olrH conclude ihaf
we.
(5)
.
\I
v~ l
-
-
V2. - W2- S-1Y1 ¥s
()
I
\{ $,;' «z.
Vz. (,0$
O{z.
= _ Zq, Cf {lS
(conI t )
G~. 5
{yom
ThUJ
= '12. 'I f-f .$
oJIId fr-uwt ~.
fLJ
::: U _ W ~1Y, Iff; 0
V (},'2-
")..
.x; aJ1d
= -fan-I
mass
22.l{-!J
4:.
.J
y/e/dJ
aMd!
I '-) -=
fru-,
Vx, ~
-!
(2.2.4
( Z'i, 7
it )J
=
37
0
~+; J
-IYl7Y'n E,/ . 7
v:z. We
(Vt9
f! _(ttl,'! f.f) ~/~lf) S $
valio or £15. 7
The
= 52.3
2.
CLJn
Ve, z U5~
Etg. S:S'I
deJ/veyed W
5htl.fl-
().f
::::
-
WdYf,.
per u~;f
/he s-ha.ff. 71zt-1I
TJV J
:f
C'a/ClAltlk fhe
fz;
(5"2, 3
~I
+ 7J
~
V fJ. I
Z
ft)(n.3 ~+) f (5"z.3 f f)(z~. '/ fI-~ !.L ) s1Jl/t flUj. (f 5
J )
5).
::: _ :3130
fl. Ib 5/uy
5- 85
5.77 Sketch the velocity triangles for the flows entering and leaving the rOlor of the turbine·type flow meter shown in Fig. PS.77. Show how rotor angular velocity is proponional to average fluid velocity.
..
,
'"
• FIG U RE P 5 . 77 (CoutICsy or EG&G Flow Tech. nology, Inc.)
F"" '" 5edjt7l-> of {he /uYbj'lG Mo.de. af r", w i /I, '" vdQc.i/y V = rrAj. 1k ve/(Jei!;t Irl"'''9/e ! hltJ.y be fkl.heo( as ShtM>'> .
W, V,
v v
~2. J
V
/.1s'''J
IE$. 5". '50
IN·
'lei
- VX 2.-+-1... - r W J
50
cu fJ-86
5".78
S.7f:3 By using velocity triangles for flow upstream (1) and downstream (2) of a turbomachine rotor, prove that the shaft work in per unit mass flowing through the rotor is Wshaft net in
V~ -
=
Vi
+ U~
Uf
-
2
+
Wr -
where V = absolute flow velocity magnitude, W flow velocity magnitude, and U = blade speed.
Any
or veJo(ify fnaJ19/e.- fw
.fef
wtJlAld 9/Jle.. fh~
YOw
~
=
relative
,rIM fhYI/lA;/'
I'e~u.lt. We
Jdme
IA.Sc
t{
/z,tvbt'YJltPch/fe
~ l;r/dnt/es Df 10,. ?S. 77.
V
~2
Fr()/'YI
,'rile! Ilow
fhe
fYl'aYJ9/e w4Z qef
velocify
l."l.
':2.
V.KII
'=>.
V -)/ V
(;)
v, I D
I
tlnd 1.
': W J - (~~ I +u) I
c~(, /Y)J nJ ifI ~I
I
-
V
y
'2
~ 2..
vl
_
'1.
- 2 Uv -U ~ I I e I
\AI - V
).
(2)
\
c.vc ohfaly) _
til
I
I
Z
-fh t!.
FrO)IVI
'2..
'l.
:
z
t:IM-d
£Cjr. /
2.
outJe f
flow
v~/pt;fy
fy/an9/e.
we- gef
'l :::
('f)
V .2.
and
(s) /
(COy/It) 5- 81
ft)/Py
? 78
I
. COY) 'f)
C(YI/yt /n'n /Y/J
Etj~.
Ull ~ 87..
'I
tA nd
the
T
:;?.
(,)
l{
Z
I
;t:-()y
"l-
lj'-W""2-
-
we. b6h/~
S'
~ef
of veltJcify
-
uJ
sh4f+
U!/ I ~I
f
-Iv,'a 119 /es (7)
~~~ I
he! /n
Corn b/YJ ,fij
~1s.
3/ h ClL-vtd 7
V 2_ V W5htff fhel Ii-?
:::
;).
/
ok-/-,; I;'
(AJ(.. l.
2-
2-
2.
+W-t V-U :a.. I ~
2
5"-88
W"22-
5.70/ 5.71*
Upstream of Rotor
Summarized below are air flow data for flow across a low-speed axial flow fan. Calculate the change in rate of flow of axial direction angular momentum across this rotor and evaluate the shaft power input involved. The inner and outer radii of the fan annulus are 142 and 203 mm. The rotor speed is 2400 rpm.
1ItL
Radius (mm)
Axial Velocity (m/s)
Absolute Tangential Velocity (m/s)
Axial Velocity (m/s)
Absolute Tangential Velocity (m/s)
142 148 169 173 185 197 203
0 32.03 32.03 32.04 32.03 31.09 0
0 0 0 0 0 0 0
0 32.28 32.37 31.78 31.50 29.64 0
0 12.64 12.24 11.91 11.35 11.66 0
lit
chA.llje
across
Downstream of Rotor
the.
~FAMx
ov
(/ )
f}Jhey'~
r
~ nnu II,( S
t»td (f)
(,
~ anti ~
aye local
ClAd 5c,c+io~
~
OVId.
~2
.
lIx'/L
As
V~
I
. Ih1/e¥
()nd
y-a.dii af secfil)n
t1tt. /ey
rad'-;
(2) dll.lll1s/tla-11-! ()f
1aJ1
y,hy
(;) I.If$~ of ~ ~iur ()Yf!.,
local aj,5tJluk k"le)1h4! vdlJcJly
al
secftdJ7J(2)IJIUiU)
I
aMd VX, J
5Wj1t!,JktJ.
7:Jlu/ff
bj
a-x.i(J.1 vt-/()( i ffes
a~ (1)(4/
aJ
Sec piJn.1 (2)
a-Itd (I)
t~. 5· '15
(2)
~.tJ FAMJ(
Gj. 2- IS eva {ua./(tA nUYY/fJY,"{,Il,/l witr... a. U/mlufw fYP
and
.
~h((ll) ;f ev~ 4-takD/ w'-f(., Gj. 5. '17. ThUi.)
.
WrA4ff- =- ~Illf :3
;' s
.eN1t tlA.ojuI
fA.)
( 3) 6~
IIu.. ~ujt,./ tJr;r;Y~ I/)/e~ on bu
(um If ) 5"-
8'1
(Con 't)
100 110 120 130 140 150 160 170 180 190 200 210 220
230 240 250 260 270 2130 29U 300 310 320 :3~~
0
340 ,3'50
*
PR HIT ., :+. tt*******:t;*.******************:*********:*********:.t:" PRINT ":rt This program computes the change in rate of *:t" PRINT ">1* axial-direction angular momentum and power**" PRINT input for problem 5.79 using the trapezoidal **" PRINT "** rule applied to unequal intervals. **" PR I NT "i' t::!.~~.t*****~~*********:t:*.*:t:* :j;~;~;~;**:t,*t*******i:*;t:*****i:" PRINT DIM UXU (19), UTU (19), UXD (9), UTD (9), R <19 ) ' , Initialize the variables N = 7 RHO = .L • .:::.. PI = 4! * ATNO! ) RPM = 2400! FOR I = 1 TO N PEAD R (1) , UXU (I) , UTU (I) , UXD (I) , UTD ( I ) RO) = R (1) ,/ 1000! NEXT I DATA 142.0, 00. 00, 00.00, 00,00, 00,00 DATA 143.0, 32,0:::-" 00.00, 32,28, 12.64 DATA 169.0, 32.03, 00,00, 32.37, 12.24 DATA 173.0, 32.04, 00.00, 31. 7,13, 11. 91 DATA lot=: 0, 32.03, 00.00, 31. 50, 11.35 DATA 197. 0, 31. 09, 00.00, 29.64, 11.66 DATA 2():3.0, 00.00, 00.00, 00.00, 00.00
"**
1
")0 ,.~
~) -~ ,
'Compute integral u:sing trapezoidal rule 380 .:3UJ{U :;:;: O!
~wo
SU]lfD = O! 400 FOR I = 2
:3(~' (I
TO N
410 TEMPU=UTU(IJ:t:UXU(I)*R(I)A2+UTU(I-1)*UXU(I-l)tRCI-l)A2
420 TEMPD=UTD(I).tUXDrI)*RCI) 2+UTD(I-l)*UXDCI-1):t.R(I-1)-2 430 BUNU BUMU + TEMPU (RCI) ReI - 1» / 2! 44(1 SUMD = ,sT):MD + TEMPD (R(I) - R(I - 1) / 2! ft
* *
450 NEXT I 460 MFXU - RHO
* *
*
2! PI 470 }{:FXD = RHO 2 ! t PI 480 POWER =:: 01FXD - MFXU) 490 ' 500 'Print the results
* SUMU * SUMD * 2! .t
PI
* RPM
/
(60!
* 1000!)
510 PRINT
520 PRINT USING "The shaft torque is ##.## N-m"; MFXD - MFXU 5:30 PRINT USING "The power input is ##.## KWH; POWER
**************************************************
** ** **
This program computes the change in rate of ** axial-dire . ::tioYl angular momentum and power )j~* input for problem 5.7~ using the trapezoidal ** rule applied to unequal intervals. ** ********.t*****~***.t*******************:t:*******.t***
**
i
IL.. The The
shaft torque is 4.79 N-m pm',er input is 1.20 KW
5"-'10
S.l:{() Air enters a radial blower with zero angular momentum. It leaves with an absolute tangential velocity, VB' of 200 ftls. The rotor blade speed at rotor exit is 170 ft/ s. If the stagnation pressure rise across the rotor is 0.4 psi, calculate the loss of available energy across the rotor and the rotor efficiency.
10
+
!he.
~haff w()y/::...
nef
~ 11"1
lNSh4 1:1-
j
tlt:'f ,},
wcJvl,- e~ lA.a:f,'1IY\
MCJYI1eJ1fuYYl
V
v;,uf-
:=
uJS;'ulH net-
~
CAh
In
be ()6~/~e?l wilt-..- the" V/.1tA11eVlf-of-
(. E~.
s·r-
(Z)
dkr
In
c~ ndnY')J'""J loss
tr
J·
-
/~df -ID
14-J,tt/2 ~ I';. "
-
Po
"
auf
r tr
i71A-f-
;0
loss -=
(lJ1,cI
loss
1t9tJo
rf.
(I)
It
5"- 9/
V
t'1,t,J
5.81
Water enters a pump impeller radially. It leaves the impeller with a tangential component of absolute velocity of 10 m/s. The impeller exit diameter is 60 mm and the impeller speed is 1800 rpm. If the stagnation pressure rise across the impeller is 45 kPa, determine the loss of available energy across the impeller and the hydraulic efficiency of the pump.
The
antl/y.fi$ of
~. Z 7
EXtfI,nf'le
/5
10 S()/V)~
a/pi/cable
#til
!.JS1i1J €~. ~ &f Exq"".pJe 501.7 we
py()bleVl1'
uv&;2 -
tJb/aln actual ir)'fp/ Py($$U~ Yise ({&-Yr;$f I),.~ fk""
2
H()lIIe 1If/r"J
~
:::
t;.UJ
=
(60
yY'''''
(1900 ~) (17T
)
(2)(!()()O ~ )
,.,.4 ) Yw
= 5.66 ~ .5
(bo.!-. ) }'?'lIn
Thus /0$5
::
(5.66
J)(!Os'jf~,~", ~ s2-
:::
/055
Fronz
&1' 5'
/ /. 6
of
(
5)(10
/II. ~
7
EX~k
{)C~aJ foia/ ;::
t ' ;;i)(f1f~~)
f1"t'eJj(Jf(
to
5. Z 7
we
ti.re aCYlJJ
tJbM,;" I~pe/I""
AI
/
5.82. Water enters an axial-flow turbine rotor with an absolute velocity tangential component, VII' of 15 ft/s. The corresponding blade velocity, U, is 50 ft/s. The water leaves the rotor blade row with no angular momentum. If the stagnation pressure drop across the turbine is 12 psi, determine the hydraulic efficiency of the turbine.
To
defeo'J1lne
ocfua/ //II()r/<..
achttl-I
momenf -
f),e, fuy/:;/ne
We
use
acfua/ \Nod~- ou.f
'1 The.
ef{/c/ency of
fhe
our
o('vf;
Wf.!'rK
(I)
loss
of
w,ft,af{
}
Ir
fief ouf Wt7Yf:.
of - rnomenfuYl'l
v(f J~
(2 )
I
To
fhe
rofvv
use
fhe
eYJ~ j
e~ lAcL/-;'tJy, (61' 5:?2) ~ ~6hln ~
L
loss =
eneYO'r ac.Y()sr the
of a vtf.f/la6/e..
I ()5)
-I-
~;..
-
~Uf
~ n~,Juf
-I-
9(~/" h~~o + ~h(1fl-
2
p.-I! f ", I 411, IJu. J1
V
Ii?
t/6j I;..
V·In tI~ II?.
7.J V . 1III ~
= /055
In
a.nd
5-'13
1-
l'Jef ,;.,
(3)
5.$13
j .
5.83 An inward flow radial turbine (see Fig. P5.83) involves a nozzle angle, 0:'1' of 60° and an inlet rotor tip speed, U l' of 30 ft/ s. The ratio of rotor inlet to outlet diameters is 2.0. The radial component of velocity remains constant at 20 ft/s through the rotor and the flow leaving the rotor at section (2) is without angular momentum. If the flowing fluid is water and the stagnation pressure drop across the rotor is 16 psi, determine the loss of available energy across the rotor and the hydraulic efficiency involved.
.ff3ctd CJ)...t.. I \IOfct,ae
---:o-:~~-~ ~.} ~,~ ~
I
:\ ,,~'-1IiIIg1!
,----•
FIGURE PS.83
analyJf.~ //ke fhe One of ~)(tlmf/e 5.Z8 IJ/(juld be liff~flyjok fw 50/V;'!j fhis flYbh/eJ'J1. ~j"ce t:4 fur6ine is Inwlv~d lit fh/s
An
pJ1)b1eJ1t'1.1
we Cdn
and fnm" ~.
wshaff :: - W$half nd i" hel (Jut c.())1c/ude that s itlfn ali ()'"
/055
pf"e.s.5'I'1~ dfPl' acyO~s
I
()f
EXtJI")flJe ~ U
yofuy
r'
=
If) $5
10
-=.:
51?A(P1~Ht1YI
pt(.s.sU'l(
V 1/~ I J
dY'Pf o.CrtJSf YrJIuy
(r)
;0
c/elwTJ'1J~e"
ft;y fn(..
-
-fh(.
I/(JW '€~/ny
hvm fhe velOcity
we eX'tlm/~e file.- ve/dclfy fy/ol1j/e fhe, ~ fhal- iJ JJ::.ekhuJ k/vw
V()/{"(e
fr;'tll1,fJe
o-f ~ I
we Obltil!1
From
ob1-al;'
we,
(;6 f:. ~ )(/'I'Il,': )
It/55
(t 1'1 f!:J $) {f"3 =
fh ,,,
I'I!J
tv$htlff
f
sla?l'u/jJ'fFYl piU.> u re
J655
;0
hel ()uf Oy
In
ofhe,y wWti.J /
/n
re.Jt.l/ir
7hl/ 5
J'ju/~1
lite l/VlI'Yk
me aA 1~:fiA.1
&f
Slayifa.'6'on preIS u Y!.
alu:/
e ff;c-ieYlC;Y
UJShtlff l1er OlAf
=
dJ0/J acrou ihe. roJuy
/0$.5 If
dyojO
I/f dt/tJ/'/a6/e
PlU'~.r
~er,/
-Ihe.
.
n:drJy
5.8 if An inward flow radial turbine (see Fig. P5.83) involves a nozzle angle, 0: 1, of 60° and an inlet rotor tip speed of 30 ft/ s. The ratio of rotor inlet to outlet diameters is 2.0. The radial component of velocity remains constant at 20 ft/ s through the rotor and the flow leaving the rotor at section (2) is without angular momentum. If the flowing fluid is air and the static pressure drop across the rotor is 0.0 I psi, determine the loss of available energy across the rotor and the rotor aerodynamic efficiency.
P,-~
105$
/he WOr/c.
slJa{f
wor1..,
CEi.
e~~~
wS/-'Q.ff_
=
l1ef ,;,
ol1d
C~61""/~ /t7$5
-
WSJulf~ ,,(!f ,;,
5".5'1/).
-=
-VI/.: I ~I
J?~ t
I t:tA-fd z
~s.
P. - P-a. /l
r
W: ha (.1-
v,2_
DI1..!-
y/elds ~~
2.
-
~I,
~I ,
(Coy/f) 5- 96
U\I I
~
I
con'f)
With the velocifJ
~
::
/-Y/an 9 Ie.. /life
~)
(-Zo
I/O
==
.fls
and
v~ :: 5iYl c e
V
0
('fo -H)
-::
SIn
I
the
flow /eav/~ fhe YD/w is
V. =- l/ ~
~
SIn b()
I
R,L
£1.3
lOSS Ie
-=
=
S
~()
0
::
Yael/a/
20 f+ S
we-
tJ6~/;"
(b, 0 I I". /!!-.) (I'1/'1'1 f+~ i!J. ...)
(1a1X10" i.J.1j ') ffi/
fl().~
lhe
t.- {{/c/ency
1 ::
may he
(lclvtai woY/L
aclvtal wov/£.oy
o!?h:lIi1t:.d
Qui
o-wt + / ()S$
3'f.~Lf it
S
wilt,
I
theY!
S·es I 5.S5 How much available energy is lost during the process shown in Video V5.7?
All fr;,
of tA~
pore"IitJ.1 eneY9!!
of Ihe
~y
~
/0;1-
I;'
the. bo7ldh?_
I'YJOI/ly
~
fhe
What is the size of the head loss that is needed to raise the temperature of water by 1°F?
in
c(r~lAl - r) })J
Ib~ . It lb.
h = 77f -= If L
S"-qq
\;j7f
f.f.
Ik \
57..) . Bn.)
5.B1 S.K7 A 100-ft-wide river with a flowrate of 2400 ft 3/s flows over a rock pile as shown in Fig. PS.87. Determine the direction of flow and the head loss associated with the flow across the rock pile.
IIlI FIGURE P5.87
To dele.rrn,;'e Hte d/'~.c~ of ~ we will
Q$.rLinte p.
dJ~cIJ8n
:/
u.r~ -fivL
eA?~ e~~ (Gfo 5":tt'l) aJ'u;/ C'QlclAlQk the. > /bS,J. .Ir I/t.L /vaal 11)5$ is ?()fi~·ve '" OUy aSJ'tlmec/ c4~o)"
head
I..f fJ.,e hR.atl /lJf~ if J?ryal,~ whjlh iJ J1td fJAyfic~/0 ;tJ()ss;ble.,J OrA¥' a.!stJ""et/ d/~c.-hlrn (JI }Iqw iJ w~. t'f
ft~ if ~CI:
50.;
tlSfllm/~
POI'" r (i)
-Iv
is ~ Y&Af -k lei; (/Y' ~ po/ni- (2.) I~ !he .5'kehl, I1bcv<./ we gef {he
HdlN
.JJ" (J
fA sin}
N(IIN
(Jl'ld
~ = (:Zlf(}() fr3) \0 A,.(Lj fi) (in: ff)
V2,
-:::
~ -::: ('].lf~o
AL
ft )
c
G f+
r
12
(2/-1- :(100 ff)
fr
S l..
So
_ (If+)
~_/oo
If wDrL
5". S8 5J:{S If a t-hp motor is required by a ventilating fan to produce a 24-in. stream of air having a velocity of 40 ftls as shown in Fig. P5.88, estimate (a) the efficiency of the fan and (b) the thrust of the supporting member on the conduit enclosing the fan.
40 ftls
--+-
1m FIGURE PS.8S
(0.)
The s~/uf;d"h -IZJ fJ,if fM] df ht.t poblew, iJ It"k.e b:dMJ'1e s: ),'1. We. lAse-
7 = ~~
fr,
ca;/Ctf.l4'/~'
fJ,e.
We use fl,e ~-h-t VI//ume f.oll~5"
~
fJ 8fA f
fUse}
IPs-h,r/!. e .f17·c /eYlc).
rd41
~ e~a,,;,rnr11.5;S2.)-Pr f/(IW -fA~Uj'h the
fJ.elel,ed abow.. 10 c,/A/cl1la,!e fh<.
+ '/,.2 -I- a c 2" .I l. !{ ::
/"fS
-
R a-rJ
~ :: fA~~
-=-
!i. + ~~ + 9r
r'
2-..z -:
':l-
~J
\.j::
j ~
~clJ. V;z... I?T 'I
=L
So
~
101
of I
0)
los s
WfJ,tJlft nef in
wfJ,~1 ~f~
iPJS II!
==
~ m
5.88
I
(c.orj'+)
II. /6
'Itt- 1-/./6 _ 2 i/.?
/0;5 -
'f'f ~~ _ 1f!,2 fU 16_ Ihl""
7
-I'././b
1~Yt?
Ibn-,
50
= /9.2
-
~·~b
¥Y ..p.~ 1/'/!tO
Frrr (/:;) We uSe fhe hOY)Jdn/a} ctJYl'IptJnehf of fhe //YJea yo YY)Qyyle,n ~ etjvtt:~,;h~ ~ e valuak /I....L anch()Y);'J fWa reJu ,.;e 4 Iz; h()/d ih( 1ztv? In place
FAX h-~
= Vii? 2.
pf).;{f (fA.,)
n:, .
=
2.
L
1T dv f
L{ =
m
=
9.t!-1 Ibm
F
-
('10 ft )~ 'II
fo Ax
,
s
'f')
(~2.2 (k.f+ ) . Ih.
$').
5-/02
= / 1.7 II;
-
U! ___ .___________ C!:)~.~ Air l
Air flows past an object in a pipe of 2-m diameter and exits as a free jet as shown in Fig. P5.89. The velocity and pressure upstream are uniform at 10 m/s and 50 N/m 2 , respectively. At the pipe exit the velocity is nonuniform as indicated. The shear stress along the pipe wall is negligible. (a) Determine the head loss associated with a particle as it flows from the uniform velocity upstream ofthe object to a location in the wake at the exit plane of the pipe. (b) Determine the force that the air puts on the object.
(rA) T(')
ene49J
/O$}
'1-
()'"r
e;...,~
hL
::
1
•-
:
, , - ....... "
"I(K ,t
,-"..t
II
I
':~W~ke ~ ,
2m-dia.
'_
.
T
1
l
l-m,dia
~4 m/s
l 1 '«
•r-:::: __ ~ ___ - - - -- -_-J
I' = 50 N/m2
rlJfle~eel h,
fii';~;gv/ .-1'~ = Tr.;i9V, =
.
'\
12m/s
Exit/"
V= 10 m/s
flu/d po~/i"k (}{s;1 !loWJ -10 ~ I()u"h~n I~ lite VII()*e ,,-I- (2.) we (J.fJply fht.. e$uaA'on (€f' ~. 8l/) -fr; thaI- pay/7'de floIN' 10 'Ie!:
de/-fH--m/(Je fhe
f,--qn,a (I)
J,L
~
I_
1_
1
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f,
'I
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+ -~ 2J
(5"q ~~)
('7-~)
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(10 ; ) 2
h -
(t)
L
'1
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+
el
(If
1)
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~(9,'1 ;'~)
(9.i';i)
ffVA)
+
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(coY'/f) 5"-103
-
--
$'. '15
n7
avu::J
1/ r'lx
=_lIt)
IV
5-IOif
5.90 Oil (SG = 0.9) flows downward through a vertical pipe contraction as shown in Fig. P5.90. If the mercury manometer reading, h, is 100 mm, determine the volume flowrate for frictionless flow. Is the actual flowrate more or l~ss than the frictionless value? Explain.
~- f .. ;0
_ h(
- 9
$G/fJ
(tf )
5(;011
COMb/I'JJ~
v
:;I.
Oy
J _ (' OD
",m) 't
']()O PtM
a J1d
.fnJn"
~f' I
ha ~
wC
1Y (Oo/",,) 1- (S.
7-9;') :.
I-I(Jw rill<-
!he /05.5
WO lAId.
wdl-1.IJ
o¥z ~ 3 s
~
AcAv.~ /
O.
j,!.
Ie.>.! fhol'!
fhe.. {y/c,!-/"I',1esJ \Iv Iwc /.e-l.4MJe-
be.. frt..a ky- livtl'l fhe. }e-nJ a""unl
5-105
t-t-;.eP(
a/'ov( .
5.91 . A~ incompressible liquid flows steadily along the pipe shown 10 FIg. P5.91. Detennine the direction of flow and the head loss over the 6-m length of pipe .
•
FIGURE P5.91
A.sSllme
f/()'-'V /Y(Jm
(I)
efut((-,'t:)n (£1- S:BI./) h c(}n~1 volume .>howl1:
f{
a I'td 1,1.5 e
(2)
gef hY- 11te,
fhe.
enedj'J
CQnfe-l?h of fJ,e.
f, +
-r}. -
-
1-0
"I
0
145
h
~
-- p, _ (f
~
(f
+
Z - 'l:I
J.
-::.
'3
5_106
tNt
-
I.
0 Yn
_
I, S"."
=
tJ. 5 ~
5.'1Z
I
c
5.92. A siphon is used to draw water at 70°F from a large container as indicated in Fig. PS.9Z. The inside diameter of the siphon line is 1 in. and the pipe centerline rises 3 ft above the essentially constant water level in the tank. Show that by varying the length of the siphon below the water level, h, the rate of flow through the siphon can be changed. Assuming frictionless flow, determine the maximum flowrate possible through the siphon. The limiting condition is the occurrence of cavitation in the siphon. Will the actual maximum flow be more or less than the frictionless value? Explain.
FIGURE PS.92..
Q
To A
::: Aa
01,1-0 I~
B
anrJ
j: or
'1
Vs
-=
(I)
frY8 VB' Lf
apply -I-he ene,,) e~lAah'()n t'n fh e skefrh above -Iv ()bftA/n VB'
we
v.'
..P.. 1- 9Z~ ==
jOJO rh ~ 1-
2",0
.!.. 2
f- !&/I
0
lf "leI in
2
~ ::; 9{~ -~e) - los} 1
and
(3)
('f)
or
Vc
==
Z (q, 11
!!:!) (- 3 ff )f030'/s ",,) ~ii
s..
(c.on't) ~-I07
1- (101000 /'I n,-z
'"
-122 B N ) lh~
(999. 7 ~)(/ ~ )
,.,J 1t,,;:,
::;
9.0'18
!?!
s
b. q1...
(con If )
= Ac ~
Q
z.
~c ~
=
we have. hv Ik-
if
fitJW~~
mtLX/Y'1/Um 1.
Q. '::
-fhyoVfJA ih~ 5//ho Y1 ;
2-
'!YO /Y1.)
(f).1()ft
If (Iif'! Ii,.
7.)
In) (tJ,Olf8 P1') fi
S
";
~.)8.kIIJ-:J ~3 s
7'P'
With
E1f. J
/ovJ-e,.r Oi/IvtJ
flowr~1e
tlJ'lrj
¥ we
Co~clkde
Mal Pny los!
WOWd pt.f Ir;
va//l..e of V /11 Ik SJf;hIJYJ OYld I'1w"f maJ..e. fJ"e max/mum f/owmk wi#! fy/chon /e.fs fft4l'1 IJ,e ~axl"mlJl'VJ
the.
wi/11ou t
{Y/cfi't1n.
5-/08
5.93
A water siphon having a constant inside diameter of 3 in. is arranged as shown in Fig. P5. c;g. If the friction loss between A and B is O.5V2/2, where V is the velocity of flow in the siphon, determine the flowrate involved.
4ft
i
12 ft
4ft
1 3 in.
FIGURE 1)5.93
10
c/elel'hJlne -fhe /Iowrak I Q/
Q .: A V =
we
Use
1t11. V
(/ )
J/
obfo/~
To
po/~ Ir
w~
V
A anti 0
apply
eneyfY e~lAAj/on (E~. '.i2.)
IAL J/aJd
I;'
/fOr v/+ F,
fhe
a
;4:4:fJfl
&0 lie. • 7-
1J-..M S
6elweeYl
I
~hoff
los.!
nef- ,-.,
tJr
~
J -Ill
-
().
'8 V
2
].
v= OVId
9(~A - ~II) 0, q
-
=
o.q
wifl1 £Ij. I '2
~
-=
1Y ('3/n .) ~ (I'll/-
{I 0.1 ;f)
'!:....)
::::
O.l3 {fl J
ff'~
£"-Ioq
f6. 9 ff s-
5,95 J
r
5.95
Water flows through a vertical pipe as is indicated in Fig. P5.95. Is the flow up or down in the pipe? Explain.
r(AA
'-"-i'~
•
I
I
I
I
I
I I
I
I
: : 1.:::r H
~ h !:a!i;;;;
"""""
Mercury
I
FIGURE P5.95
~onIY()1
TheFov
5hown
vtJ/ume
Sf~t2.d'tl ;"nCQM.fJlessible
we
~
Dbfelin
2
Ps
1-
r'
~
~
f
2:'
~r
C~.5eYII(lL·l(}}'1
Tnu s A
~
= V8
fi..urn loss
~,. I
= q H
8
J./tJWever
fhe
~;;;
[JJhic--h is
skefc-h ah()V~ i.! uset/.
fhe
11'1
flow
dowJ'!wa,rd
ft. s.n fI
t)f
~
1
F;:; + ~ 1-
to
2
Wlfi>$
we
g ~A
B
wh; ch
los S
-
A
(onv/llfde
8
fhaf
+
ej1Aah'un (5 ee
f?1til'JtJmefer
Sec-lion 2.6)
y/e/ds
- g[ h(l- S't,)-H} a.
ne$tlfive. ~t(al1lif! .5/nce
SG
~ /3.' . /I ne.Jt:i.f/ve
lij So -fh~
j() S ~ is rJ(}f fJhys i Cd /17 pt1ssJ6/e vpwllyd ~ 8 -k A. Fur ~wtJ/rd
leads
.fnw.. (AJ fD (8)
flow /'nMsf be
fldW fAe
tlbt1 ve anti;SiJ
-Iv
It; 5 SA =!J h , of
p()Ji five
( 5" GHi lIlY/d
1)
fh~Yc 1rH--t.
5'- 110
pAj'si CA 1ft p"ea.5tJn.a6/e.
5.96
A fire hose nozzle is designed to deliver water that will rise If 0 m vertically. Calculate the stagnation pressure required at the nozzle inlet if: (a) no loss is assumed; (b) a loss of 30 N'm/kg is assumed.
1b dele¥mlhe fhe- rlaiJllan't7n ptt:sS~ a f ~ MJJk Ih/ef we. a SJ ul'J1e ilia! fhe. 5ktfMt:l/;'gy, ples.fUle aJ- htL ;1()]}/e exif if & Same af fl"e. >1a5 Y141ftfr' tY'tJJute a/ fh~ nOJ]k ;h/ef Clnd we apply fhe eJ'% ~J e,.,.".alJPYJ (f,. S:i'l) -/7; Ik -f!tJw ~ h )203Jlc ex/t iv t1u.. i1It~o(;muYJ-t e/~ vt;"nift" of IN. w~ ~IAN -,0 ~el ~ (a)
~
;110
p
:=
0
(b)
Fw
?f l1 ~
-.:=.
/()SS
+ 17{I 0$$
IO~J ., ~~. /
-
30
tv.
-Iv
/eods
(9. '(D n,11< AIl'l
(J "" )
-=-
"?q1-
-kN =S ",-z.
1'1
*7
(I )
)
.)
&t.!
s:.lll
y;'eldf
"2
-k.. PC<..
5.17
For the 1800 elbow and nozzle flow shown in Fig. P5. q7, determine the loss in available energy from section (1) to section (2). How much additional available energy is lost from section (2) to where the water comes to rest?
.j:'"
6 in.
- - ...
(I
"\
rrt,., I V6h.tme
:YL I
12 in.
x
I
J = 15 psi = 5 fUs ~ u.---l.~----,~~
Pl
Vl
Section (1)
FIGURE P5. q7
so/v/ny fhe II's f fClyf of fhi.J jlT()bleYn,l the [(};rfrol 1/()/tI~e &h(}wn t'n fhe sl::eh;h ahtJlle 15 lIser-1.. To defe1"nriYJe the /1J~5
fOr
acct!n1.p()lIYI7 CiS'
flow fYl'/YJ1 sech'P>1
I it;
Sec/ion 2.
E9'
5".
71
Cdn
6e UJfd
.jfJ//OW5.
-
~-
P-z. '2
;0
x -y
5',';'ce
C.()oyd,;'",ks
hot,'},,, fa /
15
Fr{frlA
fh (!
2, - ~2.
a",eI
of
C-I)-1se rva,n,,1'l
o· A/so.; ?Z'::' ~h.,::
:
JI'Ylflf!
PJ'/;u./ple vv~
0 ?'si.
(..QYIc/lAtie Iha f
v,f;:j
112. :::
~
-(iJJ 2.
ThU5
( loSJ:;..
{(I
::
Z
%
oY I
/055
::; ;l.
--
For
second
fhe
5/U! ptWT- tJ{
flu/d f",yh'c.le. ~ Jec -hOYl 2.
fo
jtJss 2.
Nofe.
we c~n5;de" .Jhe Ilow 01 p.. .$/nk o~ mf;, 4'£8- ~ 79 /f!Atis -1-0
th/S' ,roh/en.,
tt
T1ta:f
7hllJ
5"- 112
t:f
1g
5•. An automobile engine will work best when the back pressure at the exhaust manifold, engine block interface is minimized. Show how reduction of losses in the exhaust manifold, piping, and muffler will also reduce the back pressure. How could losses in the exhaust system be reduced? What primarily limits the minimization of exhaust system losses?
We. an1y -f~ e)1~J ~~'I/Y) ( Eb · ~. I)) h f/1L lJo~ ~ Me. en7111e bloc/<. I ex-haus t YJ14J1Jift;ld I~/er-kta fp IN- ex.l1auff J!lJ~ eXif fo get
i;"r; :::; ~ Wifh E$. J
-r ;;( toss )
f
t.t f-
Y(,ducfJ'oYl
lIVe
5y>j-&w1. resulh b().cJe.- p1ers[JYe. .
In
a..
ItJlN'er
ftt(..
LtJsse J In
In
race
UlYJ I/"W
CiJrs.
fhe e.x. k YL f
fe r-
and
by
((JIl hjUY'":J !he ex hatlJ f
(Ina
~I'YlJjJYiak
CO.rff
fh" J
UI Wt(1t)}1 ~ n
()J
l11e enf'he.
C(J)A-IcI 6~
.J)'f /4",
M/,(ffler
-exfJa/,o-J.
/~
/f
.fue/' ILJ ()f~
/he
done
noise Qi1d emiJ.r/dns le!ps 1~/i(/)1 //m;fs with h /"hiJ ,k/Vlt:t of /~.fl ~cf,'()1'J can OCC/..IIY"
In
ve..T",nyefr/ el'rh
ike
los .J
OJlld
P;;,
!he
/)1
However,
10 con venh'onaJ veh/cler.
OJ'fd
of
of loss
exh~.ff
mVt7Y
re.tiu.ced by e l/wlI·Y14. rh' !J Caj-~lyh'c
valtle
(I )
area
S~P1.C
IOSI
.f ffJen,
clis~"6ul"()nf.
rec!.MciJOn P/{J'ny
can
aflo
6(("("('v
wilh few 6eHd.s
Howetl~,
of~ /et:ulJ Iv behci.r aJ?d t-UYJ'}.J" I" fhe PIPI"
/;mif the
exhxd of ofJf/"Yl/j/~
5 ... / /.3
C/J"eC(
diJIn·bu//d)'Js.
· _£. s~c--fj,n. ('2.)
5. 91 Water flows vertically upward in a circular cross section pipe. At section (1), the velocity profile over the cross section are·a is uniform. At section (2), the velocity profile is R V = We ( -Rk
I
,)117 .
I I
where V = local velocity vector, We = centerline velocity in the axial direction, R = pipe inside radius, and, , = radius from pipe axis. Develop an expression for the loss in available energy between sections (1) and (2).
I
I I
-f ~sedj~I'\(i)
fl 6 !AI
Fay- deI-e¥WJ/~;~ /~Sf we fld'WS
(jYli-/Qym
loss ~
€g. S.i7-
J
~-!{ -,/J
:::-
UfrI S eyv 4.
h fly,
of
M 110YJ-
ThuS' / -2
+
-tlte, eneYJJ eg/l'lafh'on
use.
0(, ~
J.'nti. sf
-
-:to
t(2.
~
-t
Z (tj. S.l3)
(;
J(~/- ~) I,tk
)
kve
II Is () ha lie /
/
S/nce
we.
ve/o( ilJ
-fhc..
~1
sollie ~1' /
Yields
-
fw
p>ob/~ 5./25 (c
_"2-
~-~
CJ· O~
jO
V;
1-!} ( ~ I - :Z"2. )
2.
5, J(Jo Discuss the causes of loss of available energy in a fluid flow.
50me
cc("uses of loss of Q.Vd.lllA.b/,e; fYVIf!H9't
In
a fluid fJ(jW
,. fj,i c 17'01'1
z.
heal trCln.!/w acft)ss
'3. /law
I1.CYOSJ' fA
secho..t2) i I tvYIoI db/a'n
pyt)-h'le. af .rec/'oJ'l(I) /$ un/Iov"",. /If
S'.3£ (sec. sol£( nOn
/.00
loss 5. 100
'
c:r: ::: J.-
7J.u... s-'
I
== /.0
C('
t1
km,-t-rt1Iure. d/fM8Ha.,
S ho ck.,
5- fiJI-
»
i
s: 10 I
[)5/~j f1,£ C-fhlm / v(}/umt. ,$41wn b hrD~~ I ;ne.s we '~l'Jfl/Y ~e aXJ~1
t &Inri
11. (y'r n1tI /
(. Q M!(Jne YI
Z ~ :: PY1d
So
h tJI.-/he I,'", ea yo
IJ
$i;'ce.
i I
3m
,I
.:1
_.:
- ---::
n-, ()me H Iu rVJ
t1,e~ i.l
no
t
0.1 m
Consider the flow shown in Fig. PS.91. If the flowing fluid is water, determine the axial (along the pipe) and normal (perpendicular to the pipe) components of force that the pipe puts on the fluid in the 6-m section shown.
c'
e-rlA..Q; h fn.. -Ir;
mflY"le,.!u"",
get-:
flow I~ !Itt 1"1(/~nfA/
.
rlJlech;..,
2 F ::. 0 ,fl;"c,e the flow i.s 4.$suI"'JeeJ. f~l/tf de",eUJp~P/ a ~CI the A he f- ~Y'1(JIM-lf- (Jf Qx/~I J.,,'~cf/(J n ~'r11enlf.1lN1 {Io IN 0{,.#1 of & CV if ~e~
I?N -
Wcos 8 ': 0
~: Wco.rB
OV
W = n! " "4 A.R = )'
2
~d '1 - f!. IX1/ ~ l'-- (17.'",)"'&..'.';= '162 tJ
_lor
,
/'1.;-
fJ :. S I';, ';::.
f
f!W- ~ 4X-/~J cJ ;~th~ ~ Al. + RA of W>';'
R}q :: r,DI} - r-zD Al.
(7-
~ II,
1- WSJ;'
(J
I
~ the-
f{'= fJ,us Pi.".,c/
,..,1( ~omefe,y
~ad/~.J
~r;(
f: =- d' h)
>h)...
I1-P2, = ~ (h;- h2
~
~N
R~
=
00;
=If:-f1.)11 ;
.f- W>I:' t9
)
= t/(h ,-h )f) -(Wfln ~ =r.cx1o':' )(7. 0 ","2
=0
Irt.SO)
o.s:Jtrt·I"'J~ (r62 N
32 N
5- }/5
j(.f;;'j9-5°j
5./02 5.102 Water flows steadily down the inclined pipe as indicated in Fig. P5.102. Determine the following: (a) The difference in pressure PI - P2' (b) The loss per unit mass between sections (1) and (2). (c) The net axial force exerted by the pipe wall on the flowing water between sections (1) and (2).
w/UfYte
t
6 in.
* (fA)
The.
d;ffeyeYlce
pr'eSSuf'e.; ~-
In
WlanOYHeter (see Jecf/oVl ~. 6)
p. - Por-
H2 o
2.
D 1/ -
P.
f(s it) S/~
:. - 6'
:=
2..
-42.4 \'"
l"
3(JD
Mercury
P2,
J
J')1~y be. obfa/~ed fY~ -the.
e~lAll.fiol1
with tAe fiLtid slllt:Nc.J 1-
(6 in.) ]
(/,;Z};)
+~
~~
( 6 I~
.)
(,;). f~)
Ik)fSff )Si"J11,o-f (0.50)j7+ ~3.6)(62..'f'&)((J.5.ff)= 2]7'1,
f.I i
f-l'J
-f+3
and
=
P,- P,
2~7.lJ2.
(h) The
IOSf
wlft,
EZ'
I
f+' (lw ~~
I).
per
".,if ,"7,;)
~. 71.
':
/.65 f
be!wee h
Thu5
R -= - 77. 2 110 )(
5'"- II 6
S /'
Sedi."I (J)
Ii mI (2)
J~1'4
"'''1 be 06i
s: 103 y
Water flows through a 2-ft-diameter pipe arranged horizontally in a circular arc as shown in Fig. P5.IO 3. If the pipe discharges to the atmosphere (p = 14.7 psia), determine the x and y components of the resultant force exerted by the water on the piping between sections (I) and (2). The .stea?y flowrate is 3000 ft 3 /min. The loss in pressure due to flUId fflction between sections (I) and (2) is 25 psi.
5.103
Section (2)
Section (1)
FIGURE P5.103
TO defeYYJlJJne
"y
ihe. w().fe Y
fhe X 4nd Y On
fhe. p/YJJ~
c(}ht{Jonenh
be~eYJ
of lite teru/fan"; -£1'C~ exerlep/
sechdn (I) and (2.) we lAS e the
01 -lite. I,j"eay mOYnenfu"1 e$uaf/OY/ (Etj £Z2) . fi';r fhe. CtJnlrtJl vo It/me Con-I-~,'ni~!he w(:('/ey I'? the p;r klwtey, secf/~YJ (,) 4nd (2) / £1' 22 le~d.f fo
X
and.
y
C()yY!po}1e HI J
~
R><-=-~It-V;fQ. =-RlI~ -!{f1~ and ¥ Ry -= r:....4,4 7- ~ f{}
7he resu /!-aMi IoYCe pipe ('h1 -jh~ vvCt~.
mayni ludt hi-fi
To de{oymln~ ~
Also
= ~ (loSS)
~ ~
()
A,
Elf. / C'R1t1d 2. att exeykd by-/JrG Yl.!tdktYll fir.u. of wah" lin p/,e if ~/ltd In
c.ompo)"Jenl:r l/-t~
(2)
In
o,,,aldr:. ;'" dl;~cf/oYl. PI we use. the e 11e,.-~y ~ua lIun) Efj. ~ Kl.
711.UJ/
':. :25 fS i
::
().nd
Rx (J..J.1d
watw
=_11.,850
the.
110
x dlt'~c-6~n GtNYtptJYlenf of ft,e,
On -/1t.t
p/"e bd~eey>
for-c.e.
se..ch'~J (/) tlMrJ (2.)
(con'f) 5- 1/7
exe,!ed &7 ~
Js +/2,850 110.
.oS
I
,J 03
(GOII '
t)
t;~. 2
{)JiM
vvc. ob-/-4t"n
~ = (1).92 Y
the
CU7Ij
fwee ( I)
y(2)
r!!!1!)
s .
ffl /
d J;let;H~Y)
eXf!/¥kd
aY! d
If) (ql{.
lJooOft)t ~ .) J ~ 15'10//; (~ no.. ~/"J' : :)(lo ~
UJmpoYJenf
of
In-e-
b'l the.
walw ~Vl the... pIpe
t5
IS ifd lb.
~
-//8
mil?
be~ sechdn.J
s. /041 S.lOLf When fluid flows through an abrupt expansion as indicated in Fig. P5.10~, the loss in available energy across the expansion, loss eXl is often expressed as
= (1
loss ex
~~
A)Z VZ
_....!.
_I
Az
2
I\ \I
II / \ ;'
'.1 S:ctiln
(2)
FIGURE PS.I0if
l
(I)
:::
I
sec -hoY!
(I)
-Iv se c. fr()yt
('G)
A AI - P.. A). ~ - ~ f AI l-j
f
()f f/'e /;~etlY" mIJYhel?/z"""
Oi'VtptJne n f
C.
c~ IAA-HtJ Y) (~~. 5· Z 2.) W ~e f!iA.,'d
Rx
(1)
V
\
we. olofrA/n ,.
fr-gyv,
Sectiln
10 c.afi ()(I ~ f .u.dio"., (I)
~-P.. + \1-11 -~ ex 2;0 I d,~echfJ n APply/)'},J f~e ax/a.
/OS5
II I
\
e'tIA~h'()Yl (r=IJ· ).!Z) fo the fl()w ~ Jech'qn(J)
fhe ene,.,'1
to secH()Y/ (2)
{j
-()
(1t+(t'",a~
where Al = cross section area upstream of expansion, Az = cross section area downstream of expansion, and VI = velocity of flow upstream of expansion. Derive this relationship.
A pply/nJ
I I
f ,'~
~_ _~.
'n
ctP'l.frv~ eo!
010 fa I~
we.
+
fhe. UJh-h--D1 voluYl1t (2 )
l( fA l. 112
NOw, ;f we.. consider sec.+io'VI (J) as tJCCUYY'lry af f;,e evrd of the. 5ma Iler di(lW/e~ p/pe (-the be91;'J1J~ of fhe /(A'rger d/(ll'11eiw pipe) uS 1~d.iCA.1-ed In -IAe.. skid. above I I fnll
£r
C~tJn.J/oy, /oss
fhe
yields
Rx + P, A 2. Nofe
th~f-
sec.flo n (I)
wife,
ES' 2. 1oe~.J
- - ~ t' AV, -t- ~ I' ..42 11:1.
P'1. A"
-
~nd
ptJsih'tJY/ed
(3)
at- /1,e, QJ?d of
fhe
Jma.lleY'
d/4me/e¥ flfe~ .t; ().dJ ovev- ar-e~ .A2.. Also, b(!CtUt.Je of fj,e jef HdW fy~ fl1e.. S"Ynalle~ d/a.~t-eY' pl;"e- Info f),e /orfe-r ft;
olher
ft,e.
Va./La.
p/,o-e) the
d/(lAeff!Al"
~ - P"2.
-;0
::
ie"-m.sV ~- V 2.
I
1;-,
~
of
;fx
E
/AI/II 11t~ f
~
fpe
sYna/1
eJ?4k,jh (.PwIf'4~tA
can dY()jO Rx' F~
(f)
A, A2,
L.UYYl h/YJJ.,
joss
eo(
5"-1;9
E;.3
FrI/WJ
CtJY?.Jervah'p n of
}'YJIIISJ
(&I.
s: /i )
we have
II:l. = I/.J _ II, A.".
/OJSex
> ill nA
f1S.
~h/1'117
~ (A~/ ~ 11)
=
,er
we
6
f
~_~_A_~~)
1
_l{_'2.__ 2
t:Jr
/OH e)(
~
~
f · 2((21 ) til.
Z
2
A, +-1 liz.
it,')}
'2..
(). Jll()
~{(- A~.)
/oss~x . '=
2-
"2,
5./05
5.105 Near the downstream end of a river spillway, a hydraulic jump often forms, as illustrated in Fig. P5.1 05 and Viu,'() V10.5. The velocity of the channel flow is reduced abruptly across the jump. Using the conservation of mass and linear momentum principles, derive the following expression for h2,
h2 = -
section {I
i (i J +
+ 2VgIhJ
The loss of available energy across the jump can also be determined if energy conservation is considered. Derive the loss expression
A'I'//ca;f/oi'/ etjt<~h'(JYI
of
!'he.
h()~Jjoj!J-/r,r1
FIGURE (15.105
cOWLl'lJnenf
of
fhe.
//neay hllJn,el'1iuw.
In fhe.. (dYl~1 Vo!un-tf! -hw. fo / -&- Un;f w/dlh ~f' FIII'W/
(Ef' >.22.) -fo the. wale,....
see-tim U) Iv se ch'(JY7 (2) /ea.dr - RX
}
+
h2.
'0_, _ ~
j
r; :h~
= - l{l'h, ~ + I{
J:)
r
h:L ~
(I ) we
dYf)p
Rx
d lola In
(:;;.)
(con'-f ) 5".. /2-0
5105
(COy)
I
t) / and _
2.
2 V
we ololtJ I;'
/; ] I
- 9( (&)-1 hi
2.
(
!:~) + h,
(
h~) 11,
__,
Zv'l.
9"',
=
0
2.
Jtll'YII
/IS!
::
g 'I h"2. hJ
5=-/21
(3)
5".JOb 5.1 % Two water jets collide and form one homogeneous jet as shown in Fig. pS.I06. (a) Determine the speed, V, and direction, 0, of the combined jet. (b) Determine the head loss for a fluid particle flowing from (1) to (3), from (2) to (3). Gravity is negligible.
1'1 = 4 m/s
•
FIG U REP 5.106
f:Dr +he wakr .flOWing throll9h the. tOntYl)\ volume -sk(.ttY!!td o.ioov€, the x. - tAY\cA 'i- dlt(.dior'\ GOYY\pof\et\n ()l -\1,e. lir'\u'f rY\OMe",h,tM ~cAo.tiOl-\ ate - ~~ ~ V2. A2. -t
"1> CfJS (8 ~\J3 Af
=0
(I)
Q..V\O
- V, ~VI AI + v~ Sin e fV 1 A3= C H-oy)\ -the. <..Ons-enb-hOt'\ of ~"S s p.,.i nc..~p\e we. qd" - ~\)J A I - ~V2.A"l. + PV3 A-?, -= 0 COY'\~"I)'\',,,a f=a..s. l aVId 2 We.. Obktl.-, r.' ICJ. 'l. JP d ' ) ~ITYI) +o.V\ e = v;- A = V I (~ ~ '\ "1.
l
I
-:7r7A V'). '1
\-
~
V ndl.
~
"
:. -----:...-- -= 0 ,';os b ( I' ~ '2.. 10 i-) 'IT~l.m) '1
So
NOW) c.ofYI~;n;n'3 f:~s.
-
\J~~A2. t
V;Cos
l
o..nc:l:; we. "Ie.t
e (~\j,Al T f
V'J.
"2.)
( c..on'i )
==-
0
(2) (~)
(4 )
(C;)
(6)
anc! felk
of
(05$ :;::
22
S"
S
~#"
123
5".1 D7 5.1 07 The pumper truck shown in Fig. P5.107 is to deliver 1.5 ft 3/s to a maximum elevation of 60 ft above the hydrant. The pressure at the 4-in. diameter outlet of the hydrant is 1U psi. If head losses are negligibly small. detennine the power that the pump must add to the water.
Hydrant
II FIGURE P5.107
hs
= ~."3 f"l-
.
W.rhtlf.f her in
s- /2/f
5:}08 S.lOS What is the maximum possible power output of the hydroelectric turbine shown in Fig. P5.108?
6 mls Turbine
III FIGURE P5.108
lOr IIDw .(;.()fY1 sech'()n(t)-h; sechol'tfz.), yt.5.i2. y/e/cls
J.;.
:2 f- 92}. : !5 + '12
r.-
5/nce
P/ = I?2.
= /!,.--L r,vn,) 9 (i, -
'2..
~ 2..) - y~ 2
, MW
•
W;i1alf IJe{
ou f
llIaXihluhl
and ,
-W5 haH hef bfA..f
. W;haff hef ()IA./mil. x/ ,." £t JNI
~
W~hq{t
t
1'1(1 I;"
wfhq( r =-wSiJqH J1ef in
7htl5
+- Q
JO.,./I f ~
sltt,1f
lief ou.f ~/Jxirnl.t"'"
=
nel
- /t;'sf
.,d
-
loss
(; )
5.109
Estimate the power in hp needed to drive the main
pu~p of the.l~ge-scale water tunnel shown in Fig. P5.109. The
design condition head loss is specified as 14 ft of water for a flowrate of 4900 ft 3ts. Test section
II FIGURE P5.109
-hi; VJ
7he
Sf) It,{
of
t;x~fJ/e.
(Jf
fhif prob IeI't1 is fimi /a r ~ !he Ohe
i'. 6. L00I'I~ Cn;of.f I"et//jan t7F-
~ &iy 5ame, cl7JSS e. f (,url/tr;,)
-S"t;Jcn"rfr,
GS'
y?
~ B If
~ E8· 5: t> S
=J,f r C< . W
S"hq/f
nef
the.
we. UYlc!ude
hs = hL.. The
p.rtJunc/ iJ,~.
I;'
5-125
wafer tul1l?e/
-iunY)e/ bacl WI;'j
~fhe.
-It>
1/,e
ener,fJ
5, I/O
P.1. Q
Section (1)
5.11 0 Water is supplied at ISO ft3 / sand 60 psi to a hydraulic turbine through a 3-ft-inside diameter inlet pipe as indicated in Fig. PS.ll'"Q. The turbine discharge pipe has a 4-ft-inside diameter. The static pressure at section (2), 10 ft below the turbine inlet, is 10 in. Hg vacuum. If the turbine develops 2S00 hp, determine the rate of loss of available energy between sections (1) and (2).
= 60 psi_ = 150 ft 3 /s
VI = 3 ft
...,...~~
10 ft
L
P2
=
10 in. Hg ·vacuum
D2
=
4 ft
~
(I )
= (z /. 22
If) s
2-
(3 If) (Lf 1+1
FrlJW1 Gf. I
power I()s! (
-t'
5'50 f+. /10 ) S.hr
p2.1-. f! )~(J f1- J(Islu!.110 If _)\ s~
sa.
- 2500hp pllWe¥ t()ss":
30 I hp
5"-/26
:: II. '1'1
t! .5
5, /1/
5.111
A steam turbine receives steam having a static pressure, PI, of 400 psia, an enthalpy, hI, of 1407 Btu/lbm, and a velocity, VI' of 100 ft/s. The steam leaves the turbine as a mixture of vapor and liquid having an enthalpy, h2' of 1098 Btul Ibm, a pressure, P2' of 2 psia, and a velocity, V 2 , of 200 ft/s. If the flow through the turbine is essentially adiabatic and the change in elevation of the steam is negligible, calculate: (a) the actual work output per unit mass of steam; (b) the efficiency of the turbine if the ideal work output is 467 Btu/lbm.
(tl)
This
pmblem
If
fimi/ay
~
Ex.am/,/e
5:21.
=
2.
/'1-07 ~ _ /Ofl KIw
11:
/,,~
308
=
f
Ib~
l.
(loa fj ) 2,
(200
('2.').
f! )
Ibm. ft)(771 II, .
s 2.
f.f.,j,) flhA
'!i::-
=/hm
(b) /I reaS()fllJ,hle it;
Jd&.ll
e,(!-/ciency
15
out-puf
() y
I/V'lJ rk
J()B 8!!3
_ _ _/~"I'f'>;..-
'It? 8&
X
100
/he
rtlfit)
- 6610
Jb;..
5-17-7
of ac&aJ worJ. ()ulpwf
5//Z
I 5.11 '2. A centrifugal air compressor stage operates between an inlet stagnation pressure of 14.7 psi a and an exit stagnation pressure of 60 psia. The inlet stagnation temperature is 80 oF. If the loss of total pressure through the compressor stage associated with irreversible flow phenomena is 10 psi, calculate the actual and ideal stagnation temperature rise through the compressor. Calculate the ratio of ideal to actual temperature rise to obtain efficiency.
We fin
-II, 0..1- -the air compressdr
aS5ume ideal
()//Id thus Or
Compl"e~$;"Y1
~cc.oyd/~
/5e//l /Y~pic..
an
IJ1tJsf
IMVO/VC-
cdo/d/Ylate.;;
()PI'ecw ft-r(..
a5
/0 jJf i
fl"/c fitlnleSf
Eg. 5".101) if is
+0
£9'
Wi fh
pYOC. e 5S.
aclu~ I
fhaf
pyoces~ is
ad"abpn'c.
5". It)!
idt!/A I al'ltl
l~dicak
J~$
I~
-the.
In S
tJdia6afkal/y. IAl1d (;l.d/abalk
ct/J15fdM.f e;?1YOfl1 c.I)JI'!
(//ude
pY()ceSJ witt, rl"/c.h'tJYJ femfJe,yr,tMl
-141:/nJpy
/Ae-fJ.t.a1
C~~I/~$,ti()'yI PYPC.ti!CS
~I
6eltIW. A/~o shMn
~f"1al"()Yl
70psitl.-
tJJ1
IlL'
re.J
we p.,1$t)
CVWt/J;IC$$/()YJ
aM en~y In~YCIl.>e.
fhf!.,
Opera
pY~S$ uYt!
::
,Po'
dlA~
'::
~
fYic,IJ()J'I.
___...,.~_~;......~_o !f:JA. ~ ~ 2, tfchtJ ~2, ideAl
,,
"2 4cl-wJ.I
-----.,01'--...-.--7; 2. J
I I
S--1z'8
I
id~(J.1
i5
5:/12-
I (lIJ,,' t )
5. //'f
5.11q* Total head-rise values measured for air flowing across a fan are listed below as a function of volume flowrate. Total Head Rise (mm H 20)
Q (m 3 /s)
o
79 79
0.14 0.28 0.42 0.57 0.71 0.85
76 67 65 70
76
0.99
79
1.13 1.27
75 64
Determine the flowrate that will result when this fan is connected to a piping system whose loss in total head is described by loss = KLQ2 when: (a) KL = 49 mm H20/(ml/s)2; (b) KL = 91 mm H201 (ml/s)2; (c) KL = 140 mm H20/(m 3/s)2.
7he
a.. -IaJl1.
of !ttL Ufrn6/i?af/tJYI of f)fJll1J
fan
sys~
pump
Oy
S'ls~
/t/H
qe/~)IJIJI'y]ed
IS
head y/u. vs
I/olul'l'le.
pump
by /J1t.. V/J(Uhle..
.f/(jyo/ yak...
an4 IL uJ1l1eckd
/nwsecl-/",., fl6NY'tLk-
uo've.
7~
of
/h.e
utrv{.. ~l1.tJI
I1te dehm/J1e I1u
!It.t.. ~ of !tz,/,J 'p~b~ /{ ~nYleck~ f?J Il1.L ft,l"e~ [( aJ(b) ~(c)] p/P)i?y sys~s / Ik. )j.,/eysec-f/Ph.J <:J f fkt. III 111'2} .sy5' ~ / tJ S S vs. b( W YII'f..J () i1 ~ flu. fan -roW hc~d yise... v.s. ~ UuYlle fil- weye de/wY11iJ1Cd flO'Wya.k
Y'esu/~
tis·
tN
II~wy-ak
Wh~
wirh I1u U/V'J-tfJlA./e¥ pn;rY~ liJk,l On ~ ~?/IIII'W)~ ftifeJ. A p~/yn()mia I /etl,rf fjuwye.s 4lyl/e. f/I ()f ~ /tJ..6ula..k.d cL~ /5 used· 1k- jY1/e-rsech'()"¥I pd Jj;,f> wey~ dejeyJl"1/~ee:l 1# i ttt
I1v..
Ne.w+oYJ - I(O<.f/h JdYI
(UJn't) ~-/30
f- e. c.-~ ;'; 11/ e .
5.//1
100 110 120 130 140 150 160 170 180 190 200 210 220 230
240 250
CLS PR ~NT "****************.***********t;************************ :t*" PRINT "*:t, This program determines the intersection of the PRINT "** head loss and head rise curves for problem 5. 11lf *:r." PRINT "** A least square fit polynomial of the form: :t.*" PRINT "** Y = dO + d1*x + d2*x 2 + d3*x~3 + ... t:*" PRINT ":t:.t: is used to describe the head rise dat.a. **" PRINT "*****************)!'************************************" PRINT DIM B(21), DC2l), S(21), XClO1) , WOOl), YOOl), FOOl) DIM ERRF (01), PJ (101), PJMl (101), YBAR ClO1) ' 'intialize t.he variables NTERMS = 8: NTERMSAVE = NTERMS NPOINT = 10 INPUT "Enter the head loss coefficient."; KL PRINT FOR I = 1 TO NPOINT READ XC!), YO)
**"
A
260 270 280 290 W(n = 1 300 F(l) = YO) 310 NEXT I 320 DATA 0.00, 79.0, 0.14, 79.0, 0.28, 76.0, 0.42, 67.0 330 DATA 0.5 7 , 65.0, 0.71, 70.0, 0.85, 76.0, 0.99, 79.0 340 DATA 1.13,75.0,1.27,64.0 350 PRINT "The polynomial fit to t.he head ris:;e dat.a is of ordey-"; .360 PRINT U'::~ING "##"j NTERMS - 1 370 ' 380 'determine the polynomial coefficients 390
400 410 420 430 440 450 4()O 470
PRINT "The coefficients of the polynomial are:" FOR I = 1 TO NPOINT FCI) = FCI) - D(NTERMS + 1) * XCI) - (NTERMS) IfEXT I
FOR J
=
1 TO NTERMS
BU) = 0 DO E~
)
(J )
=
0
= =
0
NEXT J
480 C(U
0
490 FOR I = 1 TO NPOINT D (1) + F (1) 500 D (1) * W(l) ~,10 EO) = B (1) + X (1) .* W (1) 520 S(1) = S Cl) + WO) 5::::0 NEXT I 540 DO) = DO) / S(1) 550 FOR I = 1 TO NPOINT 560 ERRF(l) = Fer) - D(l) 570 NEXT I 580 IF NTERMS = 1 THEN GOTO 850 590 B(l) = B(l) / S(1) 600 FOR I = 1 TO NPOINT 6 1 () F! J M 1 ( I)
=
1
620 PJ(l) = X
(un 't) ~-.I31
(~n't)
640 650 660 670 680 690 700 710 720 730 740 7e:-.;0 760 770 780 790 800 810 820 8,30
840 850 860 370
e.r:,o 890
900
= =
FOR J FOR I
2 TO NTERMS
1 TO NPOINT
*'
P = PJ(l)
WO)
D(J) = DeJ) + ERRF(I) * P P = P P]CI) B(J) = B(J) + X(I) * P seJ) = S(J) + P NEXT I D(J) = D(J) / S(J) FOR I = 1 TO NPOINT ERRFCI) = ERRF(I) - D(J) pel) NEXT I IF J = NTERMS THEN GOTO 850 B(J) = B(J) / S(J) C(J) = se]) / S(J - 1) FOR I =- 1 TO NPOINT P = PJel) PJO) = (XCI> - B(J» PJ(1) - C(J) PJM1(l) PJM1(I) = P NEXT I NEXT J PRINT USING" d# = +#.####-"~~"; NTERMS - 1; D(NTERMS) NTERMS = NTERMS - 1 IF NTERMS ) 0 THEN GO TO 400 ' 'determine the intersection using the 'Newton-Raphson method
*
*
*
*
910 qNF = 1! ~)20
QN
930 F
=
==
QNP O!
940 FP =- ot 950 FOR I = 1 TO NTERMSAVE STEP 1 ,. 1) (} 960 F == F + D .0001) THEN GOTO 920
*
*
*
*
ft
*
1050 F = O! 1060 FOR I 1 TO NTERMSAVE STEP 1 1070 F = F + D ( I) QN (I - 1)
=
H)80 NEXT I 1090 1100 1110 1120
*
ft
PRINT ###.##"; KL PRINT USING "Head loss coefficient: PRINT USING "Volume flow rate ----: ##.### m~3/s"; QN PRINT USING "Operating head ------. ### rom of H20"; F
(COn 'f 5"-1.32
)
5./1'-1
(Con't)
***************************************************~**
** **: ** ** **
This program determines the intersection of the ** head loss and head rise curves for problem 5. 11'1 ~:* A least square fit polynomial of the form: ** y = dO + dl*x + d2*x-2 + d3*x-3 + ... ** is used to describe the head rise data. ** ******************************************************
(e)
Enter the head loss coefficient? 140. The polynomial fit to the head rise data is of order 7 coefficients of the polynomial are: d7 = -1.7369E+03 d6 = +8.2623E+03 d5 - -1. 5353E+04 d4 = +1. 3788E+04 d2 = -5. 9543E+03 d? = +1.0551E+03 dl =- -6. 2329E+Ol dO = +7. 8983E+Ol
~he
Head loss coefficient: Volume flow rate ----: Operating head ------.
140.00 Q.705 m"3/s 70 rum of H20
****************************************************** ** This program determines the intersection of the ** head loss and head rise curves for problem 5.114 ** ** A least square fit polynomial of the form: ** *t y = dO + d1*x + d2*x-2 + d3*x~3 + .. , ** is used to describe the head rise data. **
**
**
**~*******************t*******************************
(b)
Enter the head loss coefficient? 91. The polynomial fit to the head rise data is of order 7 The coefficients of the polynomial are: d7 = -1. 7369E+03 d6 - +e..2623E+03 d5 = --·1 . 5353E+ 04 d4 = + 1. '3788E+04 -' r, c.':" - -5. 9543E+03 C:2 = +1. 0551E+03 dl = -6. 2329E+01 dO -. +7. 8983E+Ol Head loss coefficient: Volume flow rate ----; Operating head
91. 00 0.928 m-S/s 78 rom of H20
(COJ'7't ) ~-13.3
(Con If )
****************************************************** ** This program determines the intersection of the ** ** head loss and head rise curves for problem 5.11Q ** A least square fit polynomial of the form: ** ** y = dO + dl*x + d2tx-2 + d3*x 3 + ... ** ** is used to describe the head rise data. ** *****~************************************************
**
A
(al
Enter the head loss coefficient? 49. The polynomial fit to the head rise data is of order 7 The coefficients of the polynomial are: d7
=
d6
= =
-1. 73~59E+03 -+8. 2623E+03
-1.5353E+04 - +1.3788E+04 d3 = --5. 9543E+03 (i~~ = +1. 0551E+03 ell -- -6. 2329E+Ol dO = +7. 8983E+Ol
d5 d4·
loss coefficient: Volume flow rate ----:
Head
49.00 1.203 m~3/s 71 mm of H20
~ -/31/-
5'.115" ~ lJ)
Water is pumped from the tank shown in Fig. J =~ -~1~ PS .1ISa. The head loss is known to be 1.2 V2 /2g, where V is I the average velocity in the pipe. According to the pump man-: 6m ufacturer, the relationship between the pump head and the ftow- J rate is as shown in Fig. PS .IISb: h" = 20 - 2000 Q2, where, hi} is in meters and Q is in m'/s. Determine the ftowrate, Q. I
I I I
L
-
'FrN-!'he- C/Jnl-m J
Vd /41"t1e. f
:
-
2.)
I
.- - -
- - Pump
b2.
''',
(
,-? - .
1_
VIe wanf Iv 1:.'11"" !he ftowrak
J
(a)
-, i.......t
0.07 m
/"\,
Ii"
= 20-2000Q2
'.
oO':-------::--':-:----.i 0.05 Q, m3/ s (h)
hfJ 1/11Y7 )
a/p/,·u/f.htJn ()f /he.. fYherJ:; ej1AfA.hlgy. (G~, ).gl() y/ei£i.r,
£+ V~ (( 2J
f- 2-z
2-
= 7.I- ~
0
6':2J
~I 1- h; 1-
(z) ~
nd
hs
= hp
:::
20 - 20a?
Since ~::: V2. A1. w<-
h
'1
().J/}~
..L
2.J
=:
!..:.!:
Z9
(£A )
U¥nb:n'j
(!E.) Az
a
'Z.
h~ ..fn,."
"l,.
~.L (/~(])~d. ('1)
I/Jt!
).
=~
-I- -t 2.
-J-
2
~-
2,.000
I
S--/3S
et
z
,.
0,10
5.116 Water flows by gravity from one lake to another as sketched in Fig. PS.116 at the steady rate of 80 gpm. What is the loss in available energy associated with this flow? If this same amount of loss is associated with pumping the fluid from the lower lake to the higher one at the same flowrate, estimate the amount of pumping power required.
FIGURE PS.116
or
1/1 tJ
~-------:::::---T--------------------------------I
: 5.1/ 7
5.11 -, A !-hp motor is required by an air ventilating fan to produce a 24-in.-diameter stream of air having a uniform speed of 40 ft/s. Determine the aerodynamic efficiency of the fan.
The Qeyt)djJ'1dnt/c.
1
~
=
"dept
or
efFic./e"yzc,!/
fhe
fan" '(
yejl4lrld acl"tA1 Pd1Alev Ye.IU,;eJ/
IS
The. ;dr.al ~h(iff I'0JA/er rtjw;&:/" vV;Cleal J i.1 Wifhou.f loss tl.CYDS$ fire fal1. Thus •
•
lN7,.1D_1 :: 4r
/5
?~wey
fAcill1."'/ .rhflff p"wer resu/yM.) WAc.fl.(.::
/ tlFr:Jf.
I
I'M
2.
~uf
2
1.
::
'2.
()~/;,ed ~ ES' ,.32 hY /low
v3
I'A,ut V'ilT 2.~uf = -I' ?r..!!ou+ ~ut = 'f l. J
VV;ded :: tl. '135 hr
~-136
O.7S hI' .
5".118 Aerator column
5.119
Water is pumped from a tank, point 0), to the top of a water plant aerator, point (2), as shown in Video VS.S and Fig. P5.118nt a rate of 3.0 ft 3/s. (a) Determine the power that the pump adds to the water if the head loss from (1) to (2) where V2 = a is 4 ft. (b) Determine the head loss from (2) to the bottom of the aerator column, point (3), if the average velocity at (3) is V3 = 2 ftls.
• FIGURE P5.118
(a)
The enerqyeCfvof/on {rom {/J to (:;.) ~ + JZ + z +h - h ~ P:L + li2. of ~ ~ 2j I P 'J. 7 z1 J.
w,-fh
A ::: ~ :: 11 ~ ~ ::: 0 9/ves hp == hi- +z2.- Z, ::: iff! f(;O+3)ff-slf
*
Thvs , the pump power ;'s
~
==
't (J h.s
~ 62.'1-
==
12ff
J
(3!f ) (JJ-H) :: 22"'15
tt; Ib (551~)
= '/-.08 hp
5"- /37
0
S
~.IJq
t
(I)
(:=--"-r~"~~~"--:::--
The turbine shown in Fig. PS.119 develops 100 hp when the flowrate of water is 20 fto/ s. If all losses are negligible, I determi~e (a) the elevation h, (b) the pressure difference across I the turbme. and (c) the flowrate expected if the turbine were J removed.
1 h
/(/ (,V,. L
J
1
P3
0
I 1
/
~ .' ______ 12..in·_ (-aJ,' -
-
.::
-
T
~---=.l..
-
-
l
(
1'4
0
::--
/
'.' - . . . (2) ~~ i~ \
:. J
~
I
CVB
-,' - - - -:I' -
-
(/)
-
2 S-·5 li..J
(C)
[;"-/38
-/ Free jet
5. /zo
J
t -7
5.120 A liquid enters a fluid machine at sections (1) and (2) and leaves at section (3) as shown in Fig. P5 .12.0, The density of the fluid is constant at 2 slugs/ft3 • All of the flow occurs in a horizontal plane and is frictionless and adiabatic. For the above-mentioned and additional conditions indicated in Fig. 5.120, determine the amount of shaft power involved.
Section -..r;--Section
&j5.
= =
(1)
80 psia 15 ftls 30 in. 2
P3
= 14.7
V3
=
A3
= 5 in.2
5'.64 J S. 65 and S.7' lead
+ ~2) _ n1 2. /
I
nef- in
FIGURE PS.12D
-h>
(fl.fO + iI{ ) + 2
(I)
)
S/nce •
V
rn I UJ
-
. v . v YYJ U).. - YY13 ()3 2
psia 45 ftls
(JtI/abah'~ flaw +hyoUj" -/-his IlbO(J WlllCh/ne
rAJIlri.
virS'htlff :: m3 (!l ;0
=
L (3)--1 .6~
1..------.
Section (2)
Pl V1 Al
Fi:J~ fhe iricfiOY1/tSJ
Pl = 50 psia V 2 = 35 ftls
=
( .
.) v
m1- + m3
.
v
+ n,3
IIf secfioYl (3)
~-13'1
.
'"
V)
. (.,
~ 0;: Yn2 lA,-iA 2.
m.l. U).- n1J
(u-u ) :: I
'2.
0
5./21
Section (2)
5.121 Water is to be moved from one large reservoir to another at a higher elevation as indicated in Fig. P5.121. The loss in available energy associated with 2.5 ft 3/s being pumped from sections (1) to (2) is 61 V 2 /2 where V is the average velocity of water in the 8-in.-inside diameter piping involved. Determine the amount of shaft power required. HGURE PS.121
.(y.IIYn Sech'oh (t) -Iv secft'ol'J (2) E~. 'i.iz leads
flow
loss],: ;at; [!J(~-~) the
FrtJVJ.-t
\I
:::
v()/ul'He
bl
If
-=
flowrale ~
;;:-i/-Lf
=
w-e (2.5
obhl'n
{t7)
11- /!.i:!... ¥ ( 12 117. fT
.
W
s-liIlH ner i"
=
£-1'1-0
f
)z
7.162
fJ:. .J
",
fj
(I)
~.12
z..
Oil (SC = 0.88) flows in an inclined pipe at a rate of S ft3/ S as shown in Fig. PS.l22. If the differential reading in the mercury manometer is 3 ft, calculate the power that the pump supplies to the oil if head losses are negligible.
(2) 6 in.
H
--+ I
12 in.
3ft
L -.
S" IIi' ~ =
.r 1-
; ( : ft)
~-Jlf/
--
_"':::t . . ,::r·,.~r
1'1-
ZS'·5 J
5.124
I
5.12'-1 The velocity profile in a turbulent pipe flow may be approximated with the expression
E.
=
(~)lln R
Uc
where u = local velocity in the axial direction, U c = centerline velocity in the axial direction, R = pipe inner radius from pipe axis, r = local radius from pipe axis, and n = constant. Determine the kinetic energy coefficient, a, for: (a) n = 5; (b) n = 6; (c) n = 7; (d) n = 8; (e) n = 9; (f) n = 10.
ror -the kinetic. eneryy ctJelhci~l1~ ~ we ma, use Efj. s: 3, . 711/,'($,
rx
= ;:
I<
1
1f'U
27r;-r)y
= 2
1/tlJ(f) ,,~) fr= 2 c!, f- i)rf)d(f) 1
.L
I
U
b
-3
fA
-3
{.(
(Z)
(~)
(5)
IX :::
(C)
/.08
For n = 7
(d);=w n=
r
IJ( == 1·()5
(e)
I=tJv. n::: q 0(' :::
/.0"1
(f) ~n=l()
tr
::::"/·~J
5"-/.'1-2
5.125 A small fan moves air at a mass flowrate of 0.004 lbm/s. Upstream of the· fan, the pipe diameter is 2.5 in., the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, (11, is equal to 2.0. Downstream of the fan, the pipe diameter is 1 in., the flow is turbulent, the velocity profile is quite flat, and the kinetic energy coefficient, (12' is equal to 1.08. If the rise in static pressure across the fan is 0.015 psi and the fan shaft draws 0.00024 hp, compare the value of loss calculated: (a) assuming uniform velocity distributions; (b) considering actual velocity distributions.
(Jml
loss
:::
3.36
fl-· Ib
5-/Lf3
5.126
Force from a Jet of Air Deflecled by a Flat Plate
Objective: A jet of a fluid sU'iking a flat plate as shown in Fig. P5.126 exerts a force on the plale. It is the equal and opposite force of the plate on the fluid that causes the fluid momentum change that accompanies such a flow. The purpose of this experiment is to compare the theoretical force on the plate with the experimentally measured force . Equipment: Air source with an adjustable flowratc and a flow meter; nozzle to produce a unifonn ai r jet; balance beam with an attached flat plate; weights; barometer; thennometer.
Experimental Procedure: Adjust the coumer weight so that the beam is level when there is no mass, m, on the beam and no flow through the nozzle. Measure the diameter. d, of the nozzle outlet. Record the barometer reading, H oam , in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Place a known mass, m, on the flat plate and adjust the fan speed control to produce the necessary flowrate, Q, to make the balance beam level again. The flowrate is related to the flow meter manometer reading. II, by the equation Q = 0.358 1Il/1, where Q is in f(l/s and h is in inches of water. Repeat Ihe measun:ments for various masses on the plate. Calculations: For each flowrale, Q, calculate Ihe weight, W = mg, needed to balance the beam and use the continuity equation, Q "" VA, to determine the velocity, V, at the nozzle exit. Use the momentum equation for this problem, W = pV 2A, to determine the theoretical relationship between velocity and weight. Graph: Plot the experimentally measured force on the plate, W, as ordinates and air speed, V, as abscissas. Results:
On Ihe same graph. plot the theoretical force as a function of air speed.
Data: To proceed, print this page for reference when you work the problem and dick ,,~/'(! to bring up an EXCEL page with Ihe data for this problem.
II FIGURE PS .126
(C!on't) 5- I'f'f
(COJ1'-t )
Solution for Problem 5.126: Force from a Jet of Air Deflected by a Flat Plate
d, in. 1.174
Hatm , in. Hg 29.25
m, kg 0.010 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 0.150 0.200 0.250
h, in. 0.54 1.08 1.52 2.18 2.72 3.25 3.81 4.32 4.92 5.46 8.13 10.85 13.72
T, deg F Q = 0.358 h"0.5, with Q in cfs and h in inches of water 70
Q, ft"3/s 0.263 0.372 0.441 0.529 0.590 0.645 0.699 0.744 0.794 0.837 1.021 1.179 1.326
Experimental V,ftls m, slug 35.0 0.00069 49.5 0.00137 58.7 0.00206 70.3 0.00274 78.5 0.00343 85.8 0.00411 92.9 0.00480 98.9 0.00548 105.6 0.00617 111.2 0.00685 135.7 0.01028 156.8 0.01370 176.3 0.01713
Experimental: V = Q/Awhere 2
A = nd /4 W=mg
=n*(1.174/12 ft)"2/4 = 7.52E-3 ft"2
Theoretical: 2
W = pV Awhere p
= Patm/RT with
Patm =YHg*H atm = 847 Ib/ftJ\3*(29.25/12 ft) R = 1716 ft Ib/slug deg R T =70 + 460 =530 deg R Thus, p
=2065 Ib/ftJ\2
=0.00227 slug/ft"3
(con't) 5 - /JfS
W,lb 0.022 0.044 0.066 0.088 0.110 0.132 0.154 0.177 0.199 0.221 0.331 0.441 0.552
Theoretical W,lb 0.021 0.042 0.059 0.084 0.105 0.126 0.147 0.167 0.190 0.211 0.315 0.420 0.531
s: /2-' (Con' -t )
Problem 5.126 Weight, W, vs Velocity, V
0.6
O.5
i
.
--.--~-- -------~--~---~-.-------~-,-
•
I
~--I
II ,
-----~--i
0.4
!
!
•
..0
~ 0.3
I •
i
0.2
--------------~---~--~-_r_-----------
--1I
J
0.1
,
0.0 +,- - - - - r - - - _ r - - - - - - - - , - - - - - - - - - j
o
50
100
150
V, fUs
5"-/'16
200
Experimental
1 - Theoretical
5 . 12/ 5.127
Pressure Distribution on a Flat Plate Due to the Deflection of an Air Jet
Objective : In order to defl ect a jet of air as shown in Fig. P5.127, the fl at plate must push against the air with a sufficient force to change the momentum of the air. This causes an in· crease in pressure on the plate. The purpose of this experiment is to measure the pressure distribution on the plate and to compare the resultant pressure force to that needed, accord-
ing to the momentum equation, to deflect the air.
Equipment: Air supply with a flow meter; noule 10 produce a unifonn jet of air; circular flat plate with static pressure laps at various radial locations; manometer; barometer; thermometer. Experimental Procedure: Measure the diameters of the plate, D, and the nozzle exit, d, and the radial locations, r, of the various static pressure taps on the plate. Carefully center the plate over the noull! exit and adjust the air f1 owrate, Q, to the desired conSlaIll value. Record the static pressure tap manometer readings, h, at various radial locations, r, from the centcr of the plate. Record the barometer reading, HOlm' in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. C alculatio ns : Usc the manometer readings, h, to delennine the pressure on the plate as a function of location, r. That is, calculate p = "Ym h, where "Ym is the specific weight of the manometer fluid. G raph :
Plot pressure, p, as ordinates and radial location, r, as abscissas.
Results: Use the experi mentally detennined pressure distribution to detennine the net pressure force, F, that the air jet puts on the plate. That is, numerically or graphically iIllegrate the pressure data to obtain a value for F = P dA = f p (217"r dr), where the limits of the integration are over the entire plate, from r = 0 to r "" D/2. Compare this force obtained from the pressure measurements to that obtained from the momentum equation for this now, F "" pylA, where V and A are the velocity and area of the jet, respectively.
I
Data: To proceed, print this page for reference when you work the problem and did..' "('re to bring up an EXCEL page with the data for this problem.
II F IGURE P5.127
(c on 't )
57 /2.7 I (c()n 't )
Solution for Problem 5.127: Pressure Distribution on a Flat Plate due to the Deflection of an Air Jet
D, in. 8.0
d, in. 1.174
r, in. 0.00 0.39 0.79 1.24 1.59 2.04 2.41 2.85 3.23 3.67
h, in. 6.62 5.92 3.04 0.55 0.19 0.13 0.09 0.05 0.03 0.00
Halm , in. Hg 29.25
T, deg F 77
p, Ib/ftA2 34.42 30.78 15.81 2.86 0.99 0.68 0.47 0.26 0.16 0.00
Q. ftA3/s 1.41
p, Ib/in."2 p*r, Ib/in. 0.2391 0.0000 0.2138 0.0834 0.1098 0.0867 0.0199 0.0246 0.0069 0.0109 0.0047 0.0096 0.0033 0.0078 0.0018 0.0051 0.0011 0.0035 0.0000 0.0000
P = YH2o*h P
=Palm/RT where Palm =YHg*H alm = 847 Ib/ftA3*(29.25/12 ft) = 20651b/ftA2 R = 1716 ft Ib/slug deg R T = 77 + 460 537 deg R
=
Th us, P
=0.00224 slug/ftA3
Using the trapezoidal rule for integration Fexp
=2n*0.5*L:J
TO
9[(prj +pr;+1 )*(ri+1 - rj)] = 2n*0.5*0.189 = 0.594 Ib
Theory: 2
F = pV A where
= =
2
= =
=
A nd /4 n*(1.17 4/12 ft)"2/4 0.00752 ftA2 V Q/A (1.41 ftA3/s)/(0.00752 ft"2) = 188 ftls Thus, Flh 0.00224 slug/ftA3*(188 ftls)"2*(0.00752 ftA2) = 0.595 Ib
=
5-IJf8
1 2 3 4 5 6 7 8 9
pr;+pr;+1 0.0834 0.1701 0.1114 0.0355 0.0205 0.0174 0.0130 0.0086 0.0035
r;+1 - rj 0.39 0.40 0.45 0.35 0.45 0.37 0.44 0.38 0.44
>'127
Problem 5.127 Pressure, p, vs Radial Location, r
40 35
.-.--~----~------~~--
30
.----~.. -------------------~-
--- --_. __ ..-.- -
------_._---------j ------- ----- -
-----------..j
I
25
- - - - -- --------.. . . ------------------------.-.------- --- ---i
. c 20
---'\-------~---~--'------~~I
ci: 15
--~~
10
------4
5
----1
N
<
I
~
! --.- Experimental
i
I i
i
i
0 1
0
3
2
4
r, in.
------
--
Problem 5.127 Pressure Times Distance, p*r, vs Radial Location, r
0.10 0.08 c: ::: 0.06 :£
1--.- Experimental 1
..
{ 0.04
-\-----~--~.---.~--- -
0.02 0.00 0
1
3
2 r, in.
5" - IJfCf
4
5.129 5.128
Force from a Jet of Water Deflected by a Vane
Objective:
A jet of a fluid striking a vane as shown in Fig. PS.128 exerts a force on the vane. It is the equal and opposite force of the vane on the fluid that causes the fluid momentum change that accompanies such a flow. The purpose of this experiment is to compare the theoretical force on the vane with the experimentally measured force.
Equipment:
Water source; nozzle to produce a uniform jet of water; vanes to deflect the water jet; weigh tank to collect a known amount of water in a measured time period; stop watch; force balance system.
Experimental Procedure:
Measure the outlet diameter, d, of the nozzle. Fasten the 8 = 90 degree vane to its support and adjust the balance spring to give a zero reading when there is no weight, W, on the platform and no flow through the nozzle. Place a known mass, m, on the platform and adjust the control valve on the pump to provide the necessary flowrate from the nozzle to return the platform to a zero reading. Determine the flow rate by collecting a known weight of water, WwaleP in the weigh tank during a measured amount of time, t. Repeat the measurements for various masses, m. Repeat the experiment using a 8 = 180 degree vane.
Calculations: For each data set, determine the weight, W = mg, on the platform and the volume flowrate, Q = Wwate'/( '}'t), through the nozzle. Determine the exit velocity from the nozzle, V, by using Q = VA. Use the momentum equation to determine the theoretical weight that can be supported by the water jet as a function of V and 8. Graph: For each vane, plot the experimentally determined weight, W, as ordinates and the water velocity, V, as abscissas. Results:
On the same graph plot the theoretical weight as a function of velocity for each
vane.
Data:
To proceed, print this page for reference when you work the problem and dick herl' to bring up an EXCEL page with the data for this problem.
~
FIGURE PS.123
(con't) 5-
ISO
5:123 I (COfl't)
Solution for Problem 5.128: Force from a Jet of Water Deflected by a Vane
d, in. 0.40
t, s
m, slug
Experimental W,lb Q, ftA3/s
Data for 8 = 90 deg: 0.02 7.71 0.07 8.66 0.17 8.87 0.12 8.92 0.22 9.66
29.8 18.2 10.1 12.6 10.6
0.0014 0.0048 0.0116 0.0082 0.0151
0.044 0.154 0.375 0.265 0.485
Data for 8 = 180 deg: 0.05 6.81 0.10 9.02 0.20 8.84 0.25 7.88 0.30 8.86 0.35 7.97 0.40 6.37
24.5 20.8 13.2 10.9 11.1 9.5 7.6
0.0034 0.0069 0.0137 0.0171 0.0206 0.0240 0.0274
0.110 0.221 0.441 0.552 0.662 0.772 0.883
m, kg
Wwater,lb
W=mg Q = Wwate/(y*t) V = Q/Awhere 2 A = nd /4 = n*(0.40/12 ft)A2/4 = 0.000873 ftA2 Theoretical: W = pV2A for 8 = 90 deg and W = 2pV2A for 8 = 180 deg
(Con'~)
5-/5/
V, ftIs
Theoretical W,lb
0.0041 0.0076 0.0141 0.0113 0.0146
4.7 8.7 16.1 13.0 16.7
0.038 0.129 0.440 0.286 0.474
0.0045 0.0069 0.0107 0.0116 0.0128 0.0134 0.0134
5.1 8.0 12.3 13.3 14.7 15.4 15.4
0.088 0.215 0.512 0.597 0.727 0.803 0.802
CDn't)
5:/28'
Problem 5.128 Weight, W, vs Velocity, V
1.0
-,-----~------------1
0.9 0.8
---
------~~-----------------
-
-----r;~-------
0.7 -0.6 ~ 0.5
•
-.#-+--
I
----..rr-----~-~-~__J
-~-~~---~~cF----~~---'----:~---j ~--.-----.--
0.1
o. 0
•
Experimental, 180 deg
Theoretical, 180 deg
0.4
0.2
Experimental, 90 deg
Theoretical, 90 deg
..c
0.3
•
!
----- --------1
J-....-I!~=--..,----r-------t------j
o
5
10
15
V, ftls
5""-1$2
20
Force of a Flowing Fluid on a Pipe Elbow
5.129
Objective: When a fluid flows through an elbow in a pipe system as shown in Fig. P5.129, the fluid's momentum is changed as the fluid changes direction. Thus, the elbow must put a force on the fluid. Similarly, there must be an external force on the elbow to keep it in place. The purpose of this experiment is to compare the theoretical vertical component of force needed to hold an elbow in place with the experimentally measured force. Equipment: Variable speed fan; Pitot static tube; air speed indicator; air duct and 90degree elbow; scale; barometer; thermometer. Experimental Procedure:
Measure the diameter, d, of the air duct and adjust the scale to read zero when the elbow rests on it and there is no flow through it. Note that the duct is connected to the fan outlet by a pivot mechanism that is essentially friction free. Record the barometer reading, H atm, in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Adjust the variable speed fan to give the desired flowrate. Record the velocity, V, in the pipe as given by the Pitot static tube which is connected to an air speed indicator that reads directly in feet per minute. Record the force, F, indicated on the scale at this air speed. Repeat the measurements for various air speeds. Obtain data for two types of elbows: (1) a long radius elbow and (2) a mitered elbow (see Figs. 8.30 and 8.31).
Calculations: For a given air speed, V, use the momentum equation to calculate the theoretical vertical force, F = pV 2A, needed to hold the elbow stationary. Graph:
Plot the experimentally measured force, F, as ordinates and the air speed, V, as
abscissas.
Results:
On the same graph, plot the theoretical force as a function of air speed.
Data:
To proceed, print this page for reference when you work the problem and click /tat' to bring up an EXCEL page with the data for this problem.
Centrifugal fan
II FIGURE PS.129
5"-/53
s: /2 9
I (COli 'f
)
Solution for Problem 5.129: Force of a Flowing Fluid on a Pipe Elbow
d, in. 8.0
V, ftImin
Hatm , in. Hg 29.07
T, deg F 73
Experiment V, ftIs F,lb
Long Radius Elbow Data 0 0 1200 0.38 1420 0.51 0.79 1800 2160 1.05 2440 1.38 1.65 2700 2900 1.91 2.19 3100 3520 2.83 3.12 3750 3950 3.38
0.0 20.0 23.7 30.0 36.0 40.7 45.0 48.3 51.7 58.7 62.5 65.8
Mitered Elbow Data 1400 0.30 1780 0.55 0.74 2000 1.12 2300 1.44 2630 1.72 2900 3150 2.06 3360 2.38 3550 2.62 2.74 3620
23.3 29.7 33.3 38.3 43.8 48.3 52.5 56.0 59.2 60.3
Theory V, ftIs Fth,lb 0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 65.0
P = Patm/RT where Patm = YHg*H atm = 847 Ib/W3*(29.07/12ft) R = 1716 ft Ib/slug deg R T = 73 + 460 = 533 deg R Thus, P = 0.00224 slug/W3 A = rrd 1l 2/4 = rr*(8/12)1I2/4 = 0.349 ftll2
(CtJl1'-/: ) 5"- 15"1-
0 0.02 0.08 0.18 0.31 0.49 0.70 0.96 1.25 1.58 1.95 2.36 2.81 3.30
=2052 Ib/W2
Problem 5.129 Force, F, vs Velocity, V 4.0
------~,---,------
3.5 3,0 -
Theoretical
2.5
.c
•
-_ 2.0 LL
----.. . --- ··---------1 I
1.5 1.0 0.5
O. a -t-..e::::.---,------'""1r-----i------l
a
20
40
60
V, ftls
5"-/55
80
Experimental: Long radius elbow A Experimental: Mitered elbow
'-./ J
6.1
The velocity in a certain two-dimensional flow field is given by the equation V
= 2xti
- 2ytj
where the velocity is in ft/s when x, y, and tare in feet and seconds, respectively. Determine expressions for the local and convective components of acceleration in the x and y directions. What is the magnitude and direction of the velocity and the acceleration at the point x = y = 2 ft at the time t = 01
FroM
ft,.,. II~ I"c; 1-:; ,)
eJGj>rt'$S 14J"
5:/nce
a"
(,{ Ju. +1/' Pll. !);< ~
:
(Ci!Jnv)
a~
pi/"
::
Tt:+
r- Z!:Jt)(o)
,+xl:2.
S/mi Jell" /'J)
= (Z.X'.1;) (2.t) -t
v-q
u ~-t
t9;(
J~
CO'1 PI
1.!! =- = at-
a;
(local)
a.,
(C~nj/.) .:
2!:J
u~t-1r~ d !J
l);<.
-
(2xt)(o)
-+- (- 2}jt)(-l-t-)
= If lj t-2.. At
x.-:.1.j=2ft U = 2.
t~o
Clnd (2) (0 )
=0
v- : -
2.
(2 ) ((}) .:: 0
~
So
-that
V==o
a~
=
Cl = j
a: =
.2;<. + il-xt'2
-2J +
'f.!Jt
=
2 (r.) + _ ,2
2- .:
41: - LfJ +-tis"
(2) -t
tt (2) (0) =
t
(2) (0) :::
Lf .ftls z. - If R Is 2.
6.2
Repeat Problem 6.1 if the flow field is described by the equation
V
= 3(x 2 - y2)i - 6xyj
where the velocity is in ft/s when x and yare in feet.
and
a;( (tDnv)
= [.(.~~
+r~ -:: 3{;tl:.!:J2.)(b,t)-t (-6~.!J)r-b.!J)
= / g(X. 3 +
qnd
a..'J (/OCtJ/):: ~; -
a.j (~nJl): ==
U
= 3 ~
2.)
0
ff r rf T
Je(.x 2 ':J
At K=!1 = / ft (/hd fA.
XJ
[O/-ut]
==
3 (i 2_ !J "l. )
(-6.!f) -t
(-h)(!J) (-, I.. )
~ !i 3 )
i:~o
v- :: -b (,)(i) = - Ie
=0
A
V= -&,j
Qnd
a i = /3 (x 3+x :/,)
= 18> [(0 3 ..,. (/)(t)~
::
a!J = IS( x~-+y3)
= 18[{/)2{1)
= 3{, -ftJs~
,,-z
3 -t{n ]
gtlt/.s·
6 ••~ tion
The velocity in a certain flow field is given by the equa-
v
=
xi + X2zJ + Yzk
Detennine the expressions for the three rectangular components of acceleration.
From e)(.frt'j>/~Y1 /t,1" ve /()'l"t!1)
fA. -::
><
Sinc.e a.~
::
a~
::
-!hen
0+ (
><
)(1) ,. (i~)( 0
)
+~r:)(O)
X
--
5t'mt/arl'j./ a - olr :;- t>t + and
a. ~
0 +
~w +
fA..
~ -J-
o 1-
(
X
xLi'·
T
:1 r
p~
ox
'lr!J!F 1" 8~
W'
Jur
J~
) (0) + (X2~) (~) + 6~) (fj ) 2-
6.4
The three components of velocity in a flow field are given by u = x 2 + y2 + Z2 U
=
xy + yz +
W
=
-3xz - z 2 /2 + 4
Z2
(a) Determine the volumetric dilatation rate, and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field?
=
V()/I.{me.fr/c. d,/t
(t:L)
Thus) {"Y fle/()c,fy I/()
This
tdJ/'rJjJtt7l'1fl'll-s
d;/a.ta.t,b"
I"tn~ In Co
rate -:
Volume. e>,f. a.. +Itlid elemfl1i
/I)Cltt'OIJ to (j,)
Jir
+ Jy T
f/ien
(x.,. r)
-t- (- 3 x.
-7:)
chtllfl/e I;' (I~ it rn~lIeJ -hm
-the
2.t. -r
ih~1: 1heyt! I~
/nd,ctl+es
YfSU/t.
J(.I. ;a;:
nD
dJl1e.
(lI1IJ1he r.
F'IPm .Gss. 'lIven: W
I
~
W;(.
CU!f
:: :i
( ;; - ~;) =;
.
/s
1:
(1;-- ::) = -1 [0 - (y-tZr)J " - (!J: H) ~):: [2i - (-3z.)J1 = sz~ -- - ( ~ I
= :l I
;;..
oi:.
_
.!...
p;(;L
- r-; +7:) i SI'nt.e
(!f - 2!J) = -
~
/.5 nat
W
nCJt
I
)fr~
+
S-fo 2.
~
J
elleffjwheve -the fl{)w field
~Y'oia-t';'1111 J. N ~ .
.=
0
6.5 Detennine an expression for the vorticity of the flow field described by v = - xy3 i + y4j Is the flow irrotational?
FY'CJm C( n
cl
e;<"fr"t~s,i;n -h,Y ve/tJci-l-!j) tv ;' rA : 1- (~_ W;e. W 11
~ ~
W = ~
/i
Iolll)ws
Jy
:2..
(
~
-
tl
=-
X!J3 J tr= !:J'f)
or
JJr)
( £t
~. /3)
~~)
(po.
6.
1- (Jtr_~) ;L
ax.
CJ~l )
r+
(0)
J
j + (f. ~f1") k
.-.l.
S/nce /~
:f hot.
,'.5
not jer()
/;/,()t4",6'(,I1I1I,
evef'fjwhere -f;,e
N tJ.
11f)
(EZ·k./2.)
0 !:j
inA-t
- z [CO)
ond w-= o}
t:.. ~
I
6.6 A one-dimensional flow is described by the velocity field
u v
ay + by2 w= 0
= =
where a and b are constants. Is the flow irrotational? For what combination of constants (if any) will the rate of angular deformation as given by Eq. 6.18 be zero?
i=tP1'
/~rottl.~/6,~q/ ./-/~w-
---"'
W
'=0
,
tJ1r) _ -0 of:
Thu.5) c;)
is
/5
no i
n~i
7'fY't)
irr~l:-tL~itPn(J/,
Since.. (~m
~//~w.s
N tJ
.
J!;j
a + zbfj
Thus) 1her-e tire no va. /1{ ts et(,{q/ -h> /~r()) -uud. wil/
:to
f'/tJu/"
.fr:,,, hie.. IIt/I)(./-/:y c/iIfn'hlli.;()J1 7 1J1fJ1 1114.1:
i=~f
fife
(If/Pi
I:i$. t,.18) • oU- JL.4
~:: ~ t-
(I:
ellfr!1w/'ere
N~ne.
tJ/
t4 1111(/
rlv~
a-• =-0
6
( eJt~~fi:
.r;,r
4//
k11t I)a/wt'.s
6.7 For incompressible fluids the volumetric dilatation rate must be zero; that is, V . V = O. For what combination of constants a, b, c, and e can the velocity components
u v
= ax + by =
ex + ey
w=O be used to describe an incompressible flow field?
For-
CI n
ine(f)mp tess //'Je
Ju. ~;(
lv/In
-the
Jtr -t-
Tj""
Ve /C;c./-/!1
~1Jt-o
o~-
d,.!J'.f,.; bl('t;{)~ r I i/el1
~tr
;'1 =e
a. +e =
/I,,/t/)
0
oW- _ -
~t
-0
6.8
An incompressible viscous fluid is placed between two large parallel plates as shown in Fig. P6.8. The bottom plate is fixed and the upper plate moves with a constant velocity, U. For these conditions the velocity distribution between the plates is linear, and can be expressed as
u =
u Moving plate
ux.b
Fixed plate
FIGURE P6.S
Determine: (a) the volumetric dilatation rate, (b) the rotation vector, (c) the vorticity, and (d) the rate of angular deformation.
(a)
(/e>Jtlmett/c
(h)
,t:;,,.
d//a.. t~ tt4?JI
(/e/~c./ry
tv~ = .J.:2-
CC)
cd. )
~
w
~
~
-S ::
24) •
"0
::
otr
(' =
-[rb
cW41:.
.=.O
--
I
'11 11 1:11
(~_ dU ) = oJ< 0 'J
j
- -2b
-zbV
.-'\
V ~
=
U -{'"
.b
cJu
-r p;<. J!:J
Thus)
~i"
()fj
'" Wi! -/;..
UJ~
Thus;
te ~ ~ +
d,'.s I,; hI.( f,'eJ11
~
4ntl
f'"11
(~t b,le)
6.9 A viscous fluid is contained in the space between concentric cylinders. The inner wall is fixed, and the outer wall rotates with an angular velocity w. (See Fig. P6.9a and Video V6.1.) Assume that the velocity distribution in the gap is linear as illustrated in Fig. P6.9b. For the small rectangular element shown in Fig. P6.9b, determine the rate of change of the right angle 'Y due to the fluid motion. Express your answer in terms of ro, ri' and w.
(a)
• FIGURE PS8
Foy the. lInear d,.sfy/buf,o"n t<...: -
r; co !J ~-Y"','
.5c
thai:
d"!. - _ V;w J!j Yo-Y;; alllf
S/"te
r= 0
The nejal-i II~ rfjlJi- anjle
sifn
(b)
~./O
I fl.l!) Some velocity measurements in a three-dimensional incompressible flow field indicate that u = 6xy2 and v = - 4 y2 Z. There is some conflicting data for the velocity component in the z direction. One set of data indicates that w = 4 yz 2 and the other set indicates that w = 4 yz 2 - 6 y 2Z . Which set do you think is correct? Explain.
To
;5A..t,54 -In e ~"-t1;1I,"1!1 e ~ u.~.fItD~ Ju air u..,. 75
dU :; ts,!J 2-
J"lnce -tlIfl1
~w-
1"
Tt.
~fJ
b3' 0 ) fS,:; '2._ 9fJ % + oWJt-
Thus)
2t,U~i:
Ff "A-f", iJ
CZ)
-
(!fin
w-= :5et ~f
c.
be ink Jra-kd w/fh Y~S~ct h :::
~r
::'0
8'fJr -~d2.
fdw- J 8!1 The
Jzr= - 3'1 i
II i1 II
~x
IttJtn
(/)
.= 0
'ffj1: 2 _
dA-bo.
1: d:J -
":J 2.-J:
(tvJ'-n-,
f '!J1.d ~ + +
(;I.;
1-
'(Yo) ~)
kt,'J) =0)
w- ::: 4/J:r 2. - ~!:J l.:z woui tA
tlflfllY
-10 he fne ~ff'ec,t se i:: .
('-10
i-c
i
()j,flJl)'
!1 )
Z)
b. J I
I 6.11 The velocity components of an incompressible, twodimensional velocity field are given by the equations
u = 2xy
v
= x2
_
y2
Show that the flow is irrotational and satisfies conservation of mass.
II
-the
t:.UJo-d/mfnSIr;I111/
.{jew
UJ =".J... (~V-
:e
F;; r
Ct 11
1J1e.
c/
ve /()Cit,
71te
~;<
,;J.
15
_ 17Ju.)= ':J
d/~ fri /:)/1'1-1';/1
0
f' ven.l
IS
-r; Sllt;.s!::t ttP;1.serJ/ail4J1f "I eJu. Jtr_ ~-;.
-then
1~~()I::a..be,ntI/)
"t"
P!J-()
mass
I
'./2..
I 6.12 For each of the following stream functions, with units of m2/s, determine the magnitude and the angle the velocity vector makes with the x-axis at x = I m, y = 2 m. Locate any stagnation points in the flow field. (a)
'" = X)'
(b)
'"
= -2x 2 + Y ~f ~~ ~fY'e/lrn fl.4 Hc..:hiP J1 J tr-= -
l{'::
dt ~,
=x
.
Ai: x::: l.-m) 'f ~
J-,.",,) I
~ d'X
V-=-~ ::-~
i
~~
-4/1IJuJ.J ~.f::
Th~s I
-~ 5Jn!c '" =0
L
C"a n 9 ":
ClI1"
",I:.){:: C>
(!)~C.IAY.s
at:.
u. -= 0 cf -=
J hM j
p()/n I: (j,)
>Co
X
-
v:. 0
z.
a--t
~::o /
tt
j+a9n4f../~;'
=- ':J .::. ().
Ft'y
TtJ
A 1:- )(. =
jlWl}
I:J;::
1r.:: 2-/I't1 I
~ ~ '!-x D~
it ~11f)W5 fhA-/.
tA.:: I
~
(.f"d
1r= ~
0/
'-. , '3
I 6.U The stream function for a certain incompressible flow field is '" = lOy
+ e-Y sin x
Is this an irrotational flow field? Justify your answer with the necessary calculations.
~'r 1'h~ /f"w -tD hfC
W ~ = -L 2. alt~
/z,r
f;,~
l;r()-ttLt:lo~a I
~=
=-
11"":: -
OlA.
-
~j
~()
J
e-~/';~
/0 -
=e
W~ ::' 11Wl;:O
J tf :: - e 7i
-':1
--/J1a;t
Jln"e
1../2.)
(~ - d!?~)-o ~/.
-:1
)
l:g.
S.f:rerun .f.U)1c. t.,';rl (J I'veI'!
tA.-
ThUj
(S4'e
I
.
~s ;( -~ . alr_ - - e SIJIJ
.sIn ~
~
~;(.
( e-~jJ~ -r;1J':s
JS
-!j
l-
e s int.
)
='0
an J'rroi:(L+r()~t:t1 flow -I,eid.
Yes.
~./'f
I 6.14 The stream function for an incompressible, twodimensional flow field is
IjJ = ay2 - bx where a and b are constants. Is this an irrotational flow? Explain.
Ql1d
#r the slrellln IA.::
()If ~
u-::: - ij: ~;(.
-
=b
Thus;
S / n t. e
w~
(t{n/-es~
G
1:- 0 a =0) .
~1'I'£~J1 CL
j
.
7 HI(J1 J
lO./5
I 6.15 The velocity components for an incompressible, plane flow are Vr = A,-l + Br- 2 cos f) Vo = B,-2 sin 0
where A and B are constants. Determine the corresponding stream function.
Fr~m
1he de/;;'/i,blt of th~ StYi'II/11 Iwnc,6/~n J ..,,... _ ..!.. dip .,,... :: _ _dip v~ -
So
r
~
+OY' fhe velocity I cJl; -I ;:. Ft9 :- A- r T B
tha.t.
- /3}-
;
;r;,.
Ii? Ie 9n1fe
c/Js'ft;'I7J,(.it~1/ fll/en, -2
I:'j'())
~s
I-
-2.
J~
Or
Vt;
S;'11
Jd¥ J{A+ ::
(/)
( 2.)
&
to e
r-eSf~t-t:
w/-ht
e
8 }--/ e.~ s
If:: Thw5)
.fo
=-
-
fa
j--2 sil1 l!} d/'
13 j- - 'sin CI
.sa,tIS.f.!1
both
+
.f. (r)
+
+0
I'.. fB)
~ (t) )
1-
./;"gs.
~b-b,/~
e ) da
S/~/Jar01 /nte'lrllte E~,l Z} u/rlh Y'l'~fec-i
fd if
to
(.3) 1(l1d/lf)
t.f:: Ae+ /3~-I.5/I1C1 +C where C ~
an
Clrb /trA r.!:J
c~n.5itll1 t.
I
t,.I'
I
6.1 G,
For a certain two-dimensional flow field
u v
=0 V
=
(a) What are the corresponding radial and tangential velocity components? (b) Determine the corresponding stream function expressed in Cartesian coordinates and in cylindrical polar coordinates.
(()..)
A i-
an arh i fyay'J pfJ,ni P
(.5~e ti'llAY'e) Vr=V~/na
ve = V ~s Cb)
B
Since. u=
Jif d':J
/t
v-= -~:::v oX
=0
f" I low thai fII
wher-e.
C
IIlsrJ) tui"ht
.
is
n()t
15
~ f/,fnc I:ltJ~
l)
f.
Lf=-Vx: -fC lin
.x.:::
'1=
Ii r
Y
b / IrA 1"':1 UJ1siol1i .
t.t!;S
B
- V f"UJS& fC
I;j
y
6.17
Make use of the control volume shown in Fig. P6.17 to derive the continuity equation in cylindrical coordinates (Eq. 6.33 in text). Volume element has thickness dz
8 x
z
FIGURE P6.17
~
f
fo'¥ +
f-V·:' 0',4
J~ f ~ t/-tr
-
"......
Cp-
f~y,~ dA -
(I)
net
i"ll.te
oS /'(Y'ffu.es
Cs
o! mNSS ()ui//()w fhr()II9h cf
~nff"()/
( f V/. r FrPR1
h'lwn
4
r, . .,. '4.f3 f)y
2
v,.
J.I' of tk) .". d(; d1:
q.,JI'
J/()/wme
d::1- t) (I- t )de d~ or
f ~J;h t.
Ne t: rLtc 01 m~~s ()U-tf/DW /n r- dJ';-e,..i/(J~ ( /J r,:
&,,)'1)
cs
CV
Clnd
( .E'S,
"'0
= .
- {t ~ - ~ r)(r- t)dGdl C 2.)
~,/7
(con i)
FrfJl11 !t9~"{ at- ttfht.' ~/
Net Y'a I:e ~utfl"/N
(;~
r
,;.,
/nASS
t) - d/ree:t";JJ
~ ~e)
drd
=
r
(fife - dfife dB)cJl'"dr:
~~~____
-fv;; - ; ~~ ~~ )drdilaI'/J{; ely d~d~
09
~
( 3)
(J$
hitlYe
I'lfPl'n
!Ve.t
f"11
te (;/
{)u-t-R~u;
r~
/IJV"; (I
at: rl9lJt:
In
()f"i-
Ji
y
tnQS.J
r - el;~c....j.ldH
~) rd&df 2-
- (/1'j. - of"i dl:- ) rd~ til' )2: ;l, (;t~ ydrd{)d~
('I)
(:)r
5u"sl-/-j.j,tI-'~1j
~f
Gjs. to
¥t r dydf)Jr -r Jr:r -+
a/v[;
,"nl-o £1. "./9 'j;eIJs
inY'/'< (fl.)
f- eli"
drdf)d2
)9
11'6 dt +
+! v;.
0 f~ r dy-dedr : 0 J~
or
De.
wr/ He 1'1
dY'd&d:&
,q .s
%.,. -/: fi (r~ lIj.) + -f: ;j~Ilj;) .. ,,-/ g
~,/8
6.18
It is proposed that a two-dimensional, incompressible flow field be described by the velocity components u
= Ay
v = Bx
where A and B are both positive constants. (a) Will the continuity equation be satisfied? (b) Is the flow irrotational? (c) Determine the equation for the streamlines and show a sketch of the streamline that passes through the origin. Indicate the direction of flow along this streamline.
(u)
10 :5otisry the Ct'JI1III1U/1y Sin ee
)
J u...,. d II"" =-0 p;( J';J /1:;r 1J1e ve IDtif.y
e~uai'';J1
c/''s-';''''/b,,/-ti;/J 1i1I~11
q,x =-
&..!:!. -0 t>,K. -
tJ~
0
~l1hiJl,/t1 e.! ",a. &;'J1 ls 5 a t;/sfietl. Yes, (b) :t.n ~I"der .ft,,.. the .fl(!)w +D be 1~f'~t:flb~i1111 wz.-=Oj
the
w~~l(t-~~) I ;L
(I3-A)
Thus; f/~w will ()fJ/Y be ;rr{)l:atl()~t:// ,'f (!)
4/~n9
(~%-b,/Z)
~ =A d r:1
tf-'/3 a;(. -
wi =
IAJheY'{!
,4=-8.
s.fre/lI'l1/;;'e
a.
dy _ (/ d~
-
fA.
S" that for fhe tle/Deit!! dl;fn'bu!tPH
ttl tie" (/11"
cI!I =
B x.
dZ
fi!:1
1htre fe,re
'1 c/!J
slope::
=-J)(. d x
.t:11.ferrlll/!:)h 'fields t./ 2 -
J
-
/.3 -;r
,,(,z+ C
r1
(p.jq
I 6. I q
In a certain steady, two-dimensional flow field the fluid density varies linearly with respect to the coordinate x; that is, p = Ax where A i~ a constant. If the x component of velocity u i& given by the equation u = y, determine aq expression for u.
For
p..
vCll"it:l/;/e.
del1~;.J.'1 f/~t.V)
J fcou.) ~ J (fv-) ~x.
J!:f
(t<..:: C4~)(J) /t;/llws -thllt
11-
Th
= AX!J
otlA.J . Ay
(As)
Jt'fV-J :: -AfJ
( I )
t1b
In Ie 'I rille
eg .(;) wi fh
jdrtV-) With
!
r~sl'ec.1:
tD
= -jAjd'J
=A-x
v=
v- _- - ~ z.
+
f{)I.)
:2.,;(,
where
~-2..0
!f 10 ~61rt/"
+
fIx)
(p, 2,0
I y, ft
6.20
In a two-dimensional, incompressible flow field, the x component of velocity is given by thl" equation II = 2x. (a) Determine the corresponding equation for the y component of velocity if v = 0 along the x axis. (b) For this flow field what is the magnitude of the average velocity of the fluid crossing the surface OA of Fig. P6.2D! Assume that the velocities are in ft/s when x and yare in feet.
1.0
~
________ A
CPoA
:
\
1-7 I I
0'
:8 ~8 I
.f ~DG
x, ft
1.0
FIGURE P6.20
ro
(a)
s~.J/s~ the t.D"tl;'U~.f!J
dlA
I
t
un/t
(i.,
tnJcJ:.lllSS
=1ft)
Jtr_
Jb
'?X -r
SInce.
(con~'der
-Rg U4 b()~
-0
'\
~~
=2-
~/I"ULS
thft-l. JLr=_Z
(I,)
t?!1 rn.f.e11'~6p;' "I
E"1,fJ)
1r= -
rf
~/tP"9
'2r=O
With
resrc.-t. to :t
2:; +
f{~)
1/= (h)
1;, ~al';;4
(j)AI3
/f/0I19
Thus)
t!)13
=
= ~ (,) = 2
U
AA8 = (.2..
'1/"
=0
Sa
~,tI = {/J;/3
Y;v
=
So thd
=0
:2..:J
%,4 :: 0;.~ - 4113 U
/-(;1.)
~f m~s.3
C4?11.st!YIIA-k"o'f
tt/()n'j A/J
7hen
(f/:'O)
X-AxIS
'j/eJdJ
::
{Jt;1J
a f"elt.
()A
(,,~~ !t7t1~) oft-
S
1h4.. t.
.50
ItIs ) (/ ,ct)(/ ,ct) = ih~t 2
(/)Ot3
= CJ ,
-Ft 3
-5 =
2
+-1: 3 s
v:r
+t'J.
-
...
/. If. / +-1: S
cD. ;2/
I 6.21 The radial velocity component in an incompressible, two-dimensional flow field (v: = 0) is
vr
=
2r + 3r 2 sin ()
Detennine the corresponding tangential velocity component,
vo' required to satisfy conservation of mass.
J.. r S/nce
~ (rtlj,. ) ~
r
/ J~ -r -;. gB +
-v; ::.tJ ) oU{; :
.
Pllc/
de
wIth
r
l"f fr;//ows
v:.r ::'
("I)
:<. r
+
2
31"3 05/ 11
e
1htt t
~ (r 'J/j,.) -
Thu~
Tr E$. (j) ~
-
v:
E$Ufli:I4?I1(l)
be,-",-"es
(II r
=- -
qr 2:S/11 e)
Ctin be ;"'-teff'tLieq'
Jd vg
= -f
VB::' !( r)
f
I";
an
with r~.5rc.t
(If I-' f- qrz. si" e) dB r
tfre - 9 ,...zcos ~ + .; (y.) Hncle ter/rJ,npd
'-.2 Z.
Ittl1ctle;'H
~/ Y".
6.22
y, m
The stream function for an incompressible flow field is given by the equation IjI =
3x2y - y3
1.0 B
where the stream function has the units of m2/s with x and y in meters. (a) Sketch the streamline(s) passing through the origin. (b) Determine the rate of flow across the straight path AB shown . in Fig. P6.2].
A x, m
1.0
FIGURE P6.22
If.: 0 (a..)
fJ/ 6Pns/:t?lJ I: If ore str(1/1If}/Ii7es. tf: 3;( 2; - 1 ~ tht! slre/lllllJ;'e
Li~,s
FOr-
fX'ssinf
has
thrDllfA vpl,,~
&t.
e$tlat-ldJl1 .{()Y
1he
()f' 1
9
In
( b)
~fel:c"h
Tfltl~
-the
If=o
fHl!.. S-f"flll11/iIlPS ihmllfk
-----7(----"':""-":"'-'
() :: ax z!J _ 'j3
zv3x
()f these sfrellm/Jl1e.s 15
4?=Y8-~ ,4 t 13 .x. =
lfs ::
cf=o,
(~:=~ !I=())
I~
j=
4-
1h~ ()I'if,n
(»)
':1:: I MI
sJU)Wi1
il1 1Jfe ;;jllr~,
.so tha..t
B(o) ~/) - (1)3
= -
I Mf~/s (per tlnd WJdtA)
Ai 3(1)'(0) - (0)
3 --
6
Ihus) ne1al-l~(' sI9n l~d,C4k.s inll';' the Iltjw /s IT()I'YJ rl ,h t .f.o Jett QS We /fPtJK ./rpm A Ie E
The.
t-23
X
I
~.23
6.23
The streamlines in a certain incompressible, two-dimensional flow field are all concentric circles so that v, = O. Determine the stream function for(a) Vo = Ar and for (b) Vo = A,-l, where A is a constant.
Prd)m
0/
"the. de-h/1;fltPll
.J..a
.5.fr~(Jm knc.t'()'11 )
-the
v-:(;;
~= y Je
/hilt tt)/f1t and 1l1prej,,,.e
ht
Jr
1/.~-t
;f '/:'//t)tuS
So
(A)
-= - ~
~
~
:::'0
= f(r)
~ = A;-
dlf :-Ar
( I )
Jr
t:.
EIl1) with 1'I'.s/,ct
l-
fd", = -jArdY 'f = JIoweVf') .sill~e
(jl is
-
A:
1/6t
t.f = Where.
C
I~
Cb) SJfn"/tlY'i':J)
tin
-k.,.
fdtfor
A.
+ ~ (~)
'2.
htl1C.i./()1I lJl &1 It-
t!f2. f C
Qrbifral''1
Uf1sMl"rl:,
~ = A- y-I == -
fA-r-'dr
tf = - A 1/1
Y
+C
('-24
,t; //"U/.! thL t
6.24-*
The stream function for an incompressible, two-dimensional flow field is 'II = 3x 2y + y For this flow field plot several streamlines.
The 12.~ ua.:tltJlJ Icy a ~.frellm/il1~ ,:S -h,fll1d'/''1 ~fh~ tjJ =~lJsizJl/i
/n
ffJe
ejua.f::lon -Idr 1JIe offlllllll -func.t:j~l1. Thusl lew -t11e
9/11el1 ~fre/l111
./uHtittJ#
Lf = / I:: ~ 11t)u)J
inai
t1Je.
'j= kJh-ere
VIIY'I()/,LS
XI
t/J 1 + 3x"2
.
loS
.
Ct;IJ,st:""t 11r,III'.s erNI be Q~.sJ1 ned Iv If-;
fbm,'/.!t "f ,jrrf'lIm//lJes.. I-J- Py",NIn? '1 eoord/n;ks ,,{ 1/'4Y'1(;~ Sfreqm/;d/ZS "?;//f)II/S
+0 t)hto/" -rhe
4
3i y +-b eStJd..tl(f)JJ c I
tl
{" yo
cq Jc" 14.
1:, ~
-':'00 cls
110 120 130 150 160 162 165 170 180 190 200 210
print "****************************************************" print "** This program calculates the X,y poin+.s for **" print "** various streamlines **" print "****************************************************" print dim y(4,) print "x y(Psi=l) y(Psi=2) y(Psi==3) y(Psi=4-)" for x=-10 to 10 for psi=l to 4y(psi)=psi/(1+3*x 2) next psi print using "###.# ###.#### ###.#### ###.#### ###.####";x,y(1) R
2) ,y(3) ,y(4,)
220 next x
( C&>I'/t )
,y(
res", Its
Tabu/alea
..:51-f"ell/l1I1~~
if = /;
-h'l'
4J1'~ 71?-t'11
21
~ If
ol1d a... fltrt .sh"uJ/~" the
..6tluw.
~***************************************************
ylPsi=l) 0.0132 0.0162 0.020L,. 0.0265 0.0357 0.0506 0.0769 0.1290 0.2500
0.2:;,00 0.1290 0.0769 0.0506 0.0357 0.0265
0.0530
r)
0.020~
0.0~08
:'
0.0162 0.0132
0.0324 0.0263
0.0~86
O.06~8
0.0395
0.0526
-2.:'-
-2.0 -1. 5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5
:'. a Lj,. '-
~.
0.571~
1.1~29
1. 0000
2.0000 1.1429 0.5000 0.2581 0.1538 0.1013
0.571~
3.5 !
y(Psi=2) 0.0263 0.032L,. 0.0L,.08 0.0530 0.0714 0.1013 0.1538 0.2581 0.5000
y(Psi=3) 0.0395 0.0L,.86 0.0612 0.079:) 0.1071 0.1519 0.2308 0.3871 0.7500 1. 71~3 3.0000 1.71L,.3 0.7500 0.3871 0.2308 0.1:)19 0.1071 0.0795 0.0612
:7:
-5.0 -L,..5 -4.0 -3.5 -:3.0
:).0
0.071~
y(Psi=L,.\ 0.0526 0.06L,.8 0.0816 0.1060 0.1~29
0.202:) 0.3077 0.5161 1.0000 2.2857 ~.OOOO
2.2857 1.0000 0.5161 0.3077 0.2025 0.1~29
0.1060 0.0816
5~----------~----------------------------------~ Legend .~--E)Ps1-1
4
Ps1-2
"
6
•
• Ps1-3
~---+I
Ps1-4
1
o
-1L-~
-6
__
~~~
-4
____
~
__
-2
~
__
~~~
0
x
__
~~~
2
____
~
4
______
~
6
6.25* The stream function for an incompressible, two-dimensional flow field is 'I' = 2r l sin 38 For this flow field plot several streamlines for
o ~ (J ~ n13.
Tile e$u ..b~n ;;" . 1' 111e sfr~"m Alndlc" . TflIIs~ 'lIven sfrfl1lJ1 Itt/utt;,,,
2;-30$';" 3~ file fft"ai:IOI1 cf ... . strellm/lli.,
If=
It
1<> //(lI
r=//f
(Z
where VPrt"ll.5
t"",f,,,,i
1-0 obb"j., a ibm//!!
",.!J
()r
~i. 3 B
VII/lies
IJ
)~
be I/J5/fjlled-lt, shwm/Jites. ;1- ~r()Jrqm ,I"()I" CQ/1
epic/daMI)' f"lJe ~"rp;,~ks (t./;'"re JC=yc~se 'j= /-5 /"19) o-f yarlt'UJ sftepm//II~ ~//ows .
"",/
100 cls 110 print " ~* '~***~*********************************************', 1 20 pri nt " -..>1' This program ca lcu lates the x .y po ints for **" 1 30 print "** various streamlines **" 150 print "~****** *** **************************************** ** " 160 print 162 dim psi(4). , :C4J,y(4) 161.1 print" Psi"'1 Psi =5 Psi=10 P si;20 " 165 prin~" x y x y x y x y" 166 pi=l.I*at n( l.O) 1 67 date 1 . 5.1 0 .2 0 168 fo!:' i=1 t o 4 1 69 rea d psi(i) 1 70 n ext i 1 75 fo r theta=pi / l 80 t o 59 *pi/ l SO step Pi/45 180 for i""1 to J,. 18 2 r"'(psi(il/(2*sin(3 *thetaJ))-(1/3) 185 x ( i J"' r* cos (theta J 190 y Ci)=r*sinltheta l 200 n ext i 210 print u si ns " 1t.UIt M.UM II.UM *,UII II.UIt it,UM II.UII x( 1 ) .y( 1) ,x C2) , y(2) , x (3 ) ,y ( 3 ) .x ( 4 ) ,y(t... ) 220 next theta
It.UIt" :
( C{)t1 '-L
)
Tab,,/a.f(d Y-l'.sllih The ~ fl'"ellm/J;je,.s
~r If ::: I; S; /~ 2CJ Cire 9 1tifJJ bellJw.
~***************************************************
** This program cal culates the x, y point~o foy *)f' ** various streamlines ** **************************************************** Psi=l y x 2.122 0.037 1.2/,),1 0.109 1.020 0.162 0.902 0.208 0.826 0.252 0.296 0.770 0.728 0.339 0.385 0.695 0./,),34 0.668 0.649 0.489 0.635 0.552 0.630 0.630 0.734 0.638 0.672 0.892 1. 235 0.802
Psi=10 y x /,),.571 0.080 O.23/')' 2.673 2.197 0.348 1.9/,),4 0.4/,),9 1. 779 0.544 0.637 1.659 1. 568 0.731 0.829 1.496 0.935 1. 440 1.397 1.053 1.190 1. 368 1. 357 1.357 1. 374 1. 580 1.923 1. 449 1.728 2.662
Psi=5 y x 0.063 3.628 0.186 2.122 0.276 1. 744 1.543 0.356 1. 412 0.432 1.317 0.505 1.244 0.580 1.188 0.658 1.143 0.742 1.109 0.836 1. 086 0.944 1.077 1. 077 1.090 1.254 1.150 1. 526 1. 372 2.112
Psi=20 x
5.759 3.368 2.769 2.450 2.241 2.090 1. 975 1.885 1.814 1.760 1. 724 1.710 1. 731 1.825 2.178
y
0.101 0.29:: 0.439 0.566 0.685 0.802 0.921 1.04:) 1.178 1.327 1.499 1.710 1.991 2.422 3.353
5 0 0 ~
4
0
-1~~~~
o
Legend oPsi=1 a Ps i=5 l>Psi=10 e> Ps 1=20
__~~~__~~~~__~~~~__~~~__~~~~__~~~ 4 1 5 6 2 3 x
6.2~
y
A two-dimensional flow field for a nonviscous, incompressible fluid is described by the velocity components
B(O,})'
u = Vo + 2y v =0 where Vo is a constant. If the pressure at the origin (Fig. P6.2') is Po, determine an expression for the pressure at (a) point A, and (b) point B. Explain clearly how you obtained your answer. Assume the units are consistent and body forces may be neglected.
Chec.K ~
;U.
;'{
W
,?'/()uJ
::.J..
2:
J-
A(1,O)
Po
x
FIGURE
P6.2~
is Ir'r~t(J.:I:r()~A/. 5 1Y1 c'e
(!.Jt' _ d ~;(.
LI
)
~
(
/3"'.~.I2.) b
an d lor fne rive" (le/~ctf'J c!tSfrlbU.bl()#.J ~::::.{) 1111# j~ =Z I ; f fr; II ~W.J thfi I- tU~:j:. O. Sinee I/~w i.5 nat / 'rr()ta it{;;14/ CLJl1noi ().ffJ/'1 the. i3ernou.lli .f!gU(:I:I~1I bet.Wtel1 an3 fw{) poi n 1s
In
the.
(a)5/I1~e 5.a me.
f/()w
lie/d.
1/=0, .5
1"he
frelJln //"e..
..po ;r
"t-
~Y'19;11
and
p()/ni A t'/Y'e
{)IJ
Th us)
~ 2,
= -P14
:L~
(I)
4'
At the Dr;'r1n f.r~m E~.
0)
~~~
(.b) ~/ ~ t
15 Ij /Jot ~11 St'lh1e ~rrY?II/'11j/lle as tJJ"/9In ~o aff)::; Ber/')tJul// e!"lLt'~# betw~el1 13 pnd tJ. 10 Use the '1- wm~J1ent ~ Gu.lfY'; .e~tlA,t:'tJl1'): L) ~ -!...! = -
r- (f!J
; ':J
the
r
[JIl" . ,. pi
1fJ =0
)
=0
c,- 2'1
pX +
; 'J
l'
) t
J
, ~,l7
J b.Zl
In a certain two-dimensional flow field the velocity is constant with components u = -4 ft/s and u = - 2 l'tis. Determine the corresponding stream function and velocity potential for this flow field. Sketch the equipotential line 4J = 0 which passes through the origin of the coordinate system.
*
k= "" -t11e lie /"Clf-rt
~
t~I7If'J"flfh J/vell 0)
=-i{-
p':;
~
_
{2
'7
J
J;< - .(,..
£'1' (;) /AI;rh Yt'5f~Gt ti, b +0 OhtrUh
jd If
::-
f- 1/ d!1
If =- If ~
fd'l .: I =
If
Th U5) -h
Sa
I: I'S 1'1
/;ith
If,: wher'e
C
IS
till
+; (x)
-(-
iI/feJr4i:e e1. (Z)
~ (,-)
7-
/AJ i fh
Z d;l.
Z X. -r
res/ni. .Jo)( .j." "b1711 h
+ .{ (!J)
I&. (j )
£s s, (3) a 2;(
( .3 )
11 d
(If)
-Lf!J -f-C
flrbi 1rtJY.!J ~51:t;I1i.
tf(J)H1 1he deli'''t'.f'47;' 0./ the lIe/()cif!J ffJ-/;tlfhQj fA..
00
'=
@j
v=
~~
fh~t /r;, the lIe/f)CI'nj
E-i J;<.
=: -
II p~
Mf11POflfttt.s
'lIve fI (5")
4-
!.-4> = - z
(~)
J':;
{,-30
t..27
In.f-e frio k
E ~ . (5 )
(,IJ
/tn
'('eoSfe t
fd4 = f- ~dx ~ ::- - It-x t
k+efYllk 13;,
(It,)
'Sa f:/'5.f!:t
0
/,f//I it
~(!J)
+
w/h re5fee..i b J -k aj,.j",;,
9:: - 2.)
Thus) -k
.f,;
!ar!J)
fdp = J-z d:; OY'
x
f.j.y,
b()P,
-r
-t
1Ij.(~)
4('1.)
E$ S, (7 J
tlJtti {a)
¢=-/ft.-Zjt-C t.Jhfte
C l:S
(1;1
aY'blfr'4Y; ~slo!1i.
jJl1C.e, ihe .e!lIlfp,LPHf,q/ 11I1e) rI~a) flfs.se.5 1hr()1I9h Pte ()r"91i1 ()C:.y=o)) ihel1 C.:::. 0 ;'it E"f,fr) So 7hlff -file e1litt1n~H ()f 71t( ¢;:::o egfJ'Jeftlff'4/ /111' IS 2
fJ
=: -
'Ix.
~=-,Z)(
,,-31
( 7)
I
C:..Zg
6. 2. ~
The velocity potential for a given twodimensional flow field is
¢ = (!)x 3
5X."2
-
Show that the continuity equation is satisfied and determine the corresponding stream function.
to sai:ls/!1
~"-t/I1,,,if!:1 ~ 0UIL t/(~i1 )
fhe
~ fi,y
~ =0
-r
-t;,e- ?/~t'1'I
tle/~c,'..fy f'ctel1tJdil.;
u= ~
= (3)({))C2.- S !JJ..
fA. ='
,
and
~
P!J
I/'}I-e!rl/I'h~
::-
Sx 2-5!:J2
IOx.~
-
Since
~
S';{2. _ S"!:J l.
.
re.s'pe~i.f()
With
//11- : if
5";( "--
~
9Jlle.J
5'!/ ) d!1
~ =- S (x1 - }J) + !, ()f )
51m l 'Jar l, ) ...,,.. - _ JJ v ~X til'ltl
Inff,rAitl);'
== -
wi"tl,
If::: 5';( 2!:J
10 sa-t-';/!;
lOX w ..J
re.s;evt to 7-
0)
-f;z. (!J)
both Ets, OJ qn~
(2),)
tf -- S x 2!J - f. :J 3 -f C
x.
q J i/e..s ( <)
6. z'i
Determine the stream function corresponding to the velocity potential
¢
=
x3
3xi
-
Sketch the streamline If! = 0, which passes through the origin.
= q.J: :: ~ ¢; =- 3X
fA..
ox
J'j
Zn let "" it!
tV I
n, (,PS/~C.:I;
!d fr I.f :
..5!;"';/lIr0
pi
fix!!
tJb';"1;-;
dJ
Iex!:J
Y'eJ/Pc,t 1:.
with
jd If ::-
-t()
~ ~!j 'J.J. = -
ij:
1~I-e'lrAfllJ',
3!:J ~
-h !1
3;1.2- 3fj 1..)
J
1/.- -
and
2_
x.
f1/~'/ds
d;( (2.)
If -= 3x 2!/ t !;.fjJ
To
'stl.ti.sfJ
h(!)"/h
GiS'
(I) 11114 (Z)
Lf:- 3 x"J. y _ :/3 + wheY'e
C t.s
passes
1hr~tlfh
4h
{'
et)lIsi7Il1t. '51;'C~ -the ~+re4m/;l1e 0=0 (x.:=())y::o) t"t 1:,/I~tUS th~t- C=o qi?~
tlY'bilrlfll"fJ
1J1e.
Drlflit
tf=
3;(':; -!:f
"3
The. -e~uQ,bol1 0/ the stream line p(J..sS/n~ tnrotA.:Jh the ()Y'lgl;' I~ /o/.{l1d b'.t 5f!H;'n:; If=o IH £~.r3) -1-0
(3)
r=c
ij/eltl x
,4 ske icJ, 4/'e
fJ=!-V3x ~I ilJe If =0
~hotVl1
IH
sfrellm);I/(;S
-the lijwye.. b-33
630 tion
A certain flow field is described by the stream func-
= A (J + B r sin (J
r/J
where A and B are positive constants. Determine the corresponding velocity potential and locate any stagnation points in this flow field.
.er + IS ~S e
1h - .!. Jtf - d ¢ r- r;n;--;n:,-
:t"'nkJrai:e. Win, rf'.s~",f ~ ~ ~ ~bl-olj,
fl ~ f 1-t 8 ~.s(?) dr
Or
tf ~
A /11 r
t B yo
S/mj'JAr/~ ) ;
Tr -
VI? -
(
aH~
f; ~ =- ) 8 tj:; /3 J~a,i-'s.f!j
To
tf : -
~ (~)
-t-
r- ~in l} d ()
~s~
+ t; 0-)
(If )
bO in E JS. {2) tt H PI. ;.J. / n r + B r- et;s e
fA)hey~ C /~
(2)
= - B 51 n8
~
;.
~.s ~
arb/.fr~Y.!:J
tin
{If}
+C
~nsi:(JI1I-.
51:a.'I14./:"18/~ ~/;,-h 6CC.~~ U)"er~ Vj.:;() al1ll ~.:o . PrtP/11 ~r. (3) ~=-o IJ..-t: G:o a~",< () =-rr. t=r~1?1 ~!. So
(/)
wi'fh 9=0
1). = -$
thlJ,i
t:(t'
b~Th
fie!},,+-, v~
Ii- -t
7J;. :
t9:; 7T
-thAt
r
h>r'
=:
A
-"8 .
/1
/'1TJ£l.}ev~v.;
•
~/11"!
p",joh,i.e, ~".$'frlYJ -Is "n1l~ Yes,,/-/:. il1d,ca.1-e.s va/"e ~I' r 1AJ/11i.J, /oS I'}()t de fined.
17f,. ~o
0
1- /j
=1
+
e
ry.. .: : 0
sIT;..111 Ai-T~JI
f¥'llit o (!C UY$ a,t; ,A.
e==-lT
aHA
r="Jj
A-
411d []
a..
~,
31 6.3 r
It is known that the velocity distribution for two-dimensional flow of a viscous fluid between wide parallel plates (Fig. P6.3 \) is parabolic; that is
FIGURE P6.31
with v = O. Determinoe, if possible;o'the corresponding stream function and velocity potential. -
To d e-t-ernll;,e
-
--
-
let
-!he sireQI>1 hnc/.JoJ1
~=~ = [{ [I-(t)~] an'; 111/efrg.fe
fdY! :: If = 7/= -
YIZSf~C.1:
WIn,
1-0
f~[,-{t)jd,
3~:
0; [ :J -
J'f =a
lX
) I.f
[{!1 where C /s
.fv.!:f
]
Ij
f
J; (;( )
net
I'UJ1c. t-/O~ /t).{ X
A.
f
/t)
-/hit t
[1- ; (-/)Y+ c
an ar/n.ftor!:1 6!JJ1stal1t.
Tc de.termJYJe. ihe /I~/tXrl"!:J ,()ifJ1-/;/~/ let: fA. = ~f 7{ [I - ({) 6)
::
I~ief"..tlie
wi f1t
PI J
Y'e~fect
=
1 == ~,..or/; v - d!J -
.x.
tlnd~.
desct"lues
7:{ [, -
-/;0
1:0 ob-l:(lil1
x
(%f-) d;<
L{
[;< - £i)6;e] + i. t )
0
-
_.2D(X!1
+ Jo-fz(.!J}
dJ
h'J.
Thus) "there 15 not q fl1/j -'c/otd (" 17te .fl()w
====================~
6-35
ve/oc./"ft1 ,Poi:.ent/q/ 17111"/: IS
/J(Ji
l'rrc;ill..-btJnql) .
6.32, is
The velocity potential for a certain inviscid flow field
cP = - (3.ry -
l)
2
where cP has the units of ft / s when x and yare in feet. Determine the pressure difference (in psi) between the points (1, 2) and (4, 4), where the coordinates are in feet, if the fluid is water and elevation changes are negligible.
1n~ 111)4.1 -/ie /4 1..5 de.sC!·j but by A. I/e/{J"-fr; pot:ellt';'/ -fhe /I'H~/'r ro &a. t-II)H (J 1 01'1 d '111 ~ 13~".n~"//i -eJ/,la..t,tD~ ~11 be appl,;d beiweei1
5/;' Cf!.
is
tin (j
lit", s"
.fwo pDlni:s.
3+ JfA..':: P;t. 0 t/J
At
)(.= /
-I-i
I
= -
'V'"r :: I
A- t
~
{, )(2.) =
-/2.f:t:
s
3 (I) 2+ 3 (t) '3..:. '1~ S
V;"2..: 4/'''' 'Pi 2.::: (-/2
thll.. i::
x =- ¥ ft
1h4.i
OJ
: -" x. ..J'"
J
1) z.. T('f +: ) =
'J:::' if ft
U2,= -b{If){Lf)
So
=
'J = 2 /-i:
I
U
00
Yt'2.
2..#-
=- -
ft:.
£:# oS
1-i ::
-3(tf)4-r 3{JI)'l.=o
~ 2..:"
(_
f~ ~) ~
2..
2- Z£"(7) 2.
(p.33
6.33
Consider the incompressible, two-dimensional flow of a nonviscous fluid between the boundaries shown in Fig. P6.33. The velocity potential for this flow field is
y'
\
(a) Determine the corresponding stream function. (b) What is the relationship between the
discharge. q. (per unit width normal to plane of paper) passing between the walls and the coordinates Xf' Yf of any point on the curved wall? Neglect body forces.
"r /
1/;=0
z x
FIGURE P6.33
(a.)
To qet.errm·"e tf
1~t-e1rll k
(#
fd tf = p.)(
/h re.sfu,.1: .;.. !J
.;.. ob-blIH
d!:J
'f:: Z.x!J + "Ix) J If _ '1r~ - ~ -
SIMi/III" /'1.1 So
Jd
thAt
If To
6"th
sa./;i.si!;t
if:==
~$
o!J
C.b)
The
the
~
IS
2..x.!J + h l j )
E$s. (/)
tln
tll1e{
dischaY'~e)
(2 I
'1111/ 11..)
C
ArlJilrl'lf'fj
if =-
-;: - Z ':J
!Z!:J dx
tf=:LXj+
where C :: 0
(I)
ttf)II-f-/:qHt.
S,nt.e tf=O
b)
pqssin9 1hrf)U94 tll1fj stlr~ce +WD UlQlls) stich as .48 (see nfllre) I /:S
F1
.l3))
~ =0
~
!f=O (3)
Z)C'.:J
t~
a/"nJ
- 911
6.3 Lf The stream function for a two-dimensional, nonviscous, incompressible flow field is given by the expression
!/J
= -2(x - y)
where the stream function has the units of ft 2 / s with x and y in . feet. (a) Is the continuity equation satisfied? (b) Is the flow field irrotational? If so, determine the corresponding velocity potential. (c) Determine the pressure gradient in the horizontal x direction at the point x = 2 ft, Y = 2 ft.
(a..>
-r; sa.t/s 4
,.;ne
i,," fl1
C{)n II
J
fA
F;;y
iJJe
:5l-rellh1
u=
--
qJf P!j
a.J!' =- 0
-t-
p.JC.
eSti a.-/;I rd I'J)
Ii !1
knc.-tI4Jn
.
91,;~n)
fr s
2-
v-=
-l!t aX.
=2 .It:
s
So
anc/
~111"/~tJ/';'!1 elulJ.i/~J1
7Jte
( fl/ofe: 0J,e" ~ //(')w .field /5 -the ~l'1il;'u/-f!J egua.iliJ;t /,$ (h)
1.5
's4tls/;ed.
dt:-hht>d h!:J (lIWtHIS
A
Yes. .sfretlm IUnc.t:I()·J!
/del7l-lc~,;1.:t .saill;·//ed.)
S/~{~
~fA
(/11"
-
alj
; t-
;;tr
'='0
4//aws
a;<. .::
04
~;<.
'l1'.fefr4t:,olJ
Clnd
CJJ = a
1hat
tA-=
CJ
~
=
the .(/()(,U -hidcl
Clnd
15
2-
!f / idcb
¢ =2{x + !J)+ C CJhere (c)
Wi1n anll
a.i:
C /s
an "Y'bl.fran;j C/Pl?si:"d.
.x - a;(t'~ - ~ -= /J (IA. lJ;<: (~
the
X=
2.
~t:)
,
=2.+-1;
h~Y;1/)11"~ /) d~ = 0 0'"
T
a nv(
J
v- ~ )
J';('''~
~f = ~
[z
~r (0)
11
T-
2 ~t- (0)J
iY'r~ia-ti()1f4/.
Yes,
=
c;,. 35"
I
6.35
In a certain steady, two-dimensional flow field the fluid may be assumed to be ideal and the weight of the fluid (specific weight = 50 Ib/ft 3 ) is the only body force. The x component of velocity is known to be u = 6x which gives the velocity in ftls when x is measured in feet, and the y component of velocity is known to be a function of only y. The y axis is vertical, and
( a.)
at the origin the velocity is zero. (8) Determine the y component of velocity so that the continuity equation is satisfied. (b) Can the difference in pressures between the points x = 1 ft, y = 1 ft and x = 1 ft, y = 4 ft be determined from the Bernoulli equation? If so, determine the value in Ib/ft2. If not, explain why not.
To sa iisly the (!t),rl:,~tljf!J eglla 1:U)# ) Ju. ocr_t:) ~x T
~ -
17 u ::: ~ ,,;(.
it
J~
= -t,
;v-
bi!
(!q11
71111 i
/J,.J.efrAfe,f w~1h re.ste~t
1/= -
Since
.t~ / /{)141.$
t:~
!J -Ie :J"eJ~
~J + ~ (;()
11" IS I7tJt a. HtIfCi:.If!i"
,,(
x
Rnd
JJ
~er()
at The.
OrJif/ H )
-rr=-6 ':J I3frn"u.JJi Rgu4f/oJt Ct/n he QPpJ/~~ hel:wten 4"!J +Wo POln+':' /f -I-h e ·f--/"w l.s J yo /"'(') i It i:t011" J. :s JJ''lCe..
,.b) The
t.v=J.(~V-_~) i-
:2..
Jx
" 'j
(Eq.h,JZ) ()
( 2)
6.3~
The velocity potential for a certain inviscid, incompressible flow field is given by the equation
4J =
2x 2y -
(ih,3
where 4J has the units of m 2 js when x and yare in meters. Detennine the pressure at the point x = 2 m, y = 2 m if the pressure at x = 1 m, y = 1 m is 200 kPa. Elevation changes can be neglected and the fluid is water.
1h~
SinCe
,fJ{)w
.p, -J-
i..s +
;Y'() -64. 61f)ila J) .fJ2. .,. \{.2.. Yj2. - :: -r '-} .2.; I
)'~it11 II; J
(Pith
~c$
~::
Ai
fDI;' f
/
u.. I -- If CI n
cI
V
2..::-
/
f()/~ t
Ie t (1)(1)
('I
)(. =. I '::
~ ),.
h7I
~()
tf~ S
"
/1112..
::
$ '&
:2
IA :: J.f (z)(z.) ~
-s . . .
,uA"Z.
(/" p)~ -
::.
&'0. I ~
Pa..
7.
Z;{ -2!J
!J =. 11m v-:I :- ~ (I) 3. -
(:oul
'J J ve J1
J
'Z.
tn,i :2.. (J)z
=
0
6.37 (a) Determine the velocity potential and the stream function for a steady, uniform, incompressible, inviscid, twodimensional flow that makes an angle of 300 with the horizontal x-axis. (b) Detennine an expression for the pressure gradient in the vertical y direction. What is the physical interpretation of this result?
Pr'l?7
(ct,)
1::$5_ (,..80 anol /'.8/
tf ::~r
(iI/Pi
C/ (.>< t4sol.
-+-:J
:s I;'
(!='J. ~, 36)
01..)
ot.::: 30"
tf = 0- (;< OJS1o" + ~ S/11300)
~ V-(~,8'b;( + ~.5'h~!1)
Sin" /111' /!j tf = v (!f U;sd - x. sin~) I
t/ h
a.
fpr 0(.:; ;-" c
If '" SJ~'{'
(bJ
(Ef. ,,80
v ('J ~j JiJ'- ><
U. :: l:; cfi ~
,
J /"
1,r..::
t'lt11l
:11/) .. l/
(~. ffU,J -
O. S()(),( )
~
~'::J
If ~11~tI)J 1nd 1A.::~.f"Vp:;,.f)I?J
(
anI!
itt e
1) t:1
if ~
-
hi I in
-R. ~ tl.1L -tJix
I
~ "the
£.1 -:: I[) (0dtIT+ u ~Jr -r 4~.x
J~
1r:='
&;115.fz1l,
t
Clud
IIfr1-/CA. J
V-
.!J - d I y~dtolt
e.r + JAr E) P!1 dr
(£='9, '.Slb) 6
#1):: - ~
*~ -tJ
or
/fJJ~
t=IA fev
V-:::. ~. 5"00 r;-
Cll'ttl
'r'f1jI.(JI::
/Yld,C4.J.e.s jh4/- 1he pres.suve dJshi hU.J-,D~ 1.5
h'1d"~s-taI-(G ,Th/~ I~ nDi ~ ~l1r/,Y/$ln, Vt'.5u/-I: .sInce. -tJ, ~ ~ frn~/,( II; e$ U¢I-/~11 Inti, c.al-e..s ~ f II- 1Ittrt.. loS no ehfll1'1e. 111 Vc. J~'-/:;'.!::J 1h~ eltlll-1pt' 111 pv-e.fS"Y~ Ii Slmpl 'J ~jtl e ..fa */he. We 19h f of fh(. ,fluId.; L: Po) a..
h fj d rt>.sta..J.ti.
v4 y, a. 1-tt> n'
6. '38 The streamlines for an incompressible, inviscid, two-dimensional flow field are all concentric circles and the velocity varies directly with the distance from the common center of the streamlines; that is
=
Kr where K is a constant. (a) For this rotational flow determine, if possible, the stream function. (b) Can the pressure difference between the origin and any other point be determined from the Bernoulli equation? Explain. Vo
(a)
~=
En.f-e? ra te
_ J If
=k
17","
y-
1£,.lI} w/1h
re.sfec.t.
/dt{; - for :-
"y-
if Sinct:.
(j,)
The
- ;z.
_ ( ~ If
11 - ? u;
/t Iol/()ws 1ha:t
wher-e
1<: r"
::
C Ilew
e~ U~&/DI1
.
t.p
Js
ntJ1..
- J<;-6.
an
IS
-r fIe)
=0 J.5
~=
dr
a.
fu 11 C. t /tJJ1 ()I a
a;,d 1heye~ye.
+c
"" +
arb " va. V:J
l"()ta..tl()~IIJ
ohd
Cton S1:411 t .
Therefore
fne
13erl1t/uJJi
o..pp);'ed betWteh --the.. oY/~/i1 a h d q 11 ~ pOInt J SI h ce These fo/nb QV€ }1{Yt 6)J1 The Same ~fr"etlm //l1e. Nt). (I?e.fer 1-0 d'SCtI$S'DI1 t'fS.sOCICl·l:t-tI tu/fh der/~IJ.I:I~·h olE; Z. it>. ~7. ) CA nt10t
be
I
'.31
y
6. 39
The velocity potential ¢ = -k(x 2 - y2) (k = constant) may be used to represent the flow against an infinite plane boundary as illustrated in Fig. P6.~t For flow in the vicinity of a stagnation point it is frequently assumed that the pressure gradient along the surface is of the form
ap
ax
= Ax
/
.. _._----
---_.-
X
FIGURE P6. ~9
where A is a constant. Use the given velocity' potential to show that this is true.
Fe>y
the
II
e /() c. /-J.!J poi:en.f,Q J 7' ve fJ ~cP
LA.. ==
a;.
11"::Clnc!
tne.
S~JtJA.ft~11
;:';y
'this
-6fea , ':J.-
o~ =-~b
pt!),~i
() CCtlrs
dlmt/l.s/~II/
1-kJO -
~ a~ =-t (IA, 1))<
(lIon?
SUyfac.e.
-the.
?l. :
/JU
1J.x. t-;..t)m
E1. U )
I
u ': -
(Z)
2. '" !:J
-ij. Qnq
u)
- -2.~)!.
1"
at the
...
~r'.j In
/It;w
r~)
(£:. ~>/o.)
tJ !:J
(:;=.o)
v-= ()
So
~
dX.
Z--k:X.
au.: -z/{ p;(.
qnP(
Gg . (3)
beu;me.s
~ :: PX
where
fi) (- Z
I
Ie)/ ) r- z~) - IfJe G,x
1J,a. i (..3 )
'.'fO
Water flows through a two-dimensional diffuser having a 20° expansion angle as shown in Fig. P6.40. Assume that the flow in the diffuser can be treated as a radial flow emanating from a source at the origin O. (a) If the velocity at the entrance is 20 m/ s, determine an expression for the pressure gradient along the diffuser walls. (b) What is the pressure rise between the entrance and exit?
\
2m
~\
~-----
o
,.
~
-~~
:~ran~
-"",,~,
Flow
7m
.
~/Exlt
1
(Sf'e
11 =2.0 ~~ 1??'1
Pr~i'Yl
-=
2.
5(>
rr y. 7rj,.::
2. rr
:50
'\
)
"/hILi:
("2. hH)
(2t!;
?)
/he. /3fYI/~ul/,' ~ lJ.4.;/:J()·}1
-p -+ it 1"';-"2.. =
JJnc.e
Ta'ie ,. J
~(Pl1slr1l1t:.
~ = -~ v;. ~ tJ;.
"y
., ,..,.,.", VI'":;
~1T'Y
(I)
~r
I -1-1.
Inel1
0
v;.
~ Y'
==
A
6.4 I An idealfliilcffiows betweenthe-inciined walls of a two-dimensional channel into a sink located at the origin (Fig. P6.41). The velocity potential for this flow field is
"
m ¢ = -In'
2n where m is a constant. (8) Determine the corresponding stream function. Note that the value of the stream function along the wall OA is zero. (b) Determine the equation of the streamline passing through the point B, located at x = 1, Y
x
o FIGURE P6.4\
= 4.
(et)
r
( I )
n fe~ t"rt .J.e.
~g ./1)
fe/if t>Y'
w/fh
5/ nee. Lf
n()i
is
/t11~
~6io/11
+ ~ f;")
qj
= 0
oB
(1. )
tu nc.+;~/1 01
A
Ir,
~
:l7T
-27T I -~ tPi'"
~= ,
f ~ it;
=
if=
to
r-e.s/ec.t:
r-
eS .f'l) bectJl1Us
So
~C if:: ~ .:;,,,.
C
where -So
~
~a-t
C
qh" ( .b)
At f11e
I/Idw€
of tf if ~ fI1e
Ie.'
It tlY'e
CtlIJ
be
(~~
/1"n
It .3
Possil'1f t?'n
-- ) I
({) .:- /. :3 3 r d d . J:;.I!) /YJ ~ g .t 3 )
eg(J~biJ,J1~.f
seefl
.
-t;,Y~w'lh ih/5 ?(),hi
(1.33 - ~ ).:211 =
(..3 )
&:.
Pt II. i:
So
(). 0 "SO 11'11
IS
~~
;;'1'
n?1
I-tJ n ~:o;-
8
!p:=.o
= -C.
/jJ.:
4,nd there~re
(fj/)
~I1SiP"t . If /~&)
~
J,j
/)?1
~1fS"olYl1 .
1he. sirellll1hlJe. plIsslnJ 1hrtJL(jh 13
(~~
/rom '&1.
D.
1.5
-
(3)
i) 1htl.t fAt!
,;Irflllf1 lilJfS
II/I ~"""'1I9I1t };heJ f'k~.sl'n~ 1hrlJtI&11 -!he "yiglit. )
6.42.
It is suggested that the velocity potential for the flow of an incompressible, nonviscous, two-dimensional flow along the wall shown in Fig. P6.4Z is ¢ = r~/3 cos !8
Is this a suitable velocity potential for flow along the wall? Explain.
suifaJ,ie ~ -fiJe C/!)rre.sp()ndln~ tfJ must hAVe. I/o/we fI/on~ the 1U1/1/ (s/nce -the tvll/I tnust tt:)rl"'''~}'~l1d --h ~ d¢ - If- ~ If Vj --...Lr JB =- ;r;. - "3 r ~~.s"3 &
:tl this ;5
et
:1:nk1t'Aie I:iS.tJ) w/tIJ
.!j f-"~ c..S
If:: !- +£/.8
,JIb
a
sfr~A/IJ/i"e). (I)
1:a B 10 ohtolH
reSfect
jd if:: f
et u;IIsT:t!l1i
~B
:
e
T
~ tr)
(2)
:: - 3~ r ~. o5J :t:. e .3 I,
/I
J:j
if =
If
r -; .5;'1'3
: S +
+;. (8)
(3)
both E'Is, (2) an4 (.J) If:: f 1f~ SIn : B + {
where
C
is
an (/rb;~rlJr!1
e()nsbtllt.
A-JotJ? one se~b~n ~I fJu! wl//I; t) =0) Qnd tj;::: C. A-/()nf 1ne other set..-bo'n e::: a nil tp C. Th us) tp hils ~ c.~ n 5 1::11 n i va/we. a/f)!'''} ihe WtI/J and the 1/vel1 j)e/~c.;-h; t~)'fe I'rh q, j Ct/n b.e used -k y~tYese" t f~~ a/tJ"1 the Will/. Yes.
V
=
ft" if.3
I
6.43 As illustrated in Fig. P6.43 a tornado can be approximated by a free vortex of strength r for r > Ret where R, is the radius of the core. Velocity measurements at points A and B indicate that VA = 125 ft/s and VB = 60 ft/s. Determine the distance from point A to the center of the tornado. Why can the free vortex model not be used to approximate the tornado throughout the flow field (r 2: O)?
\
\
•
Thus) (:011(
fA
ILl:
a.-t
Th ere /r:;re
)
13
)
/zt5"
~:
(Po
fiJ
Go
J
/ ZS'
.
~/nce
/1-
~:
J 2.5"
'A : :
fA. vorte.(.
1nrou'lhoJl.t
t/e/o~ifrt
.5
)
H
-:r )
\
""......
". . . - ---
---
} /
-,.,.,....
./
fA
t R
x
/ I ,"".. I //~
FIGURE P6.43
Sf)
ina i
Ie =/zS' fA
..50
that:
/G::
60;-/3
'B
"k - fA =/{)() l-i
f~//t9IPJ ~t
7h e hee
.::.
it
\
the
I> 0
~ '11.3
(/~O + fA )
It
C/ln/Joi b~ tlsed .j." CljJpro,( i"'41:.e t<. /:or)'UIII'o /-ltJUJ .fle/d Since at:- r=o The
b~~me5
/;'f;';,/fe.
6.44 The velocity distribution in a horizontal, two-dimensional bend through which an ideal fluid flows can be approximated with a free vortex as shown in Fig. P6.44-. Show how the discharge (per unit width normal to plane of paper) through the channel can be expressed as
q =
cJ¥(1
where Ap = PB - PA' Determine the value of the constant C for the bend dimensions given.
b
= 0.5 m = 0.9 m /
•
~A
1<
= a:.
~S:
fh e. 13fl'o/)~f,t / It; eba fi" -6u:;'n 1~ .,. ~; = P13 +~
2..~
±t
k
7: ~
-ve.e 2-J.
d-
~
FIGURE PS.44
(~A" - 7Ijg8'- )
= ;;; I("l. ( (J.I,.
=
or
1<.
Thus
)
7-
::
.111 .J!. tt.
r"l. -1t (~ -f: )1-
tJp=
Of'"
t Clncl
. /,Q/th
--
(.tL
1he~eh"~
t
c=
::
.b
C
a..
rY
V2 jn
r~
( ;~ - i~) -k
;;; - i~
('
')
%
VTh
J,
~
--
iff I' tJ,'1
(),S'
V(ft; . .), - -
I
bl.
hI'f hwt
=/
{o, ~",,/"
b-'f~
~,S"oo m'I
6.45" When water discharges from a tank through an opening in its bottom, a vortex may form with a curved surface profile as shown in Fig. P6.4S' and Video V6.2. Assume that the velocity distribution in the vortex is the same as that for a free vortex. At the same time the water is being discharged from !he tanlc ~t point A it is desired to discharge a small quantity of water through the pipe B. As the discharge through A is increased, the strength of the vortex, as indicated by its circulation, is increased. Determine the maximum strength that the vortex can have in order that no air is sucked in at B. Express your answer in terms of the circulation. Assume that the fluid level in the tank at a large distance from the opening at A remains constant and viscous effects are negligible.
• FIG U REP 6 .4-S
,. b ) (14-
8.,,-l.r'Z..J A-/r w/Jj
be.
_ J/ I sUCJC.et
/ hws)
Ir I =
/01
-s
....L
/I1r;~
P'fe.
~ J h en "'"
7.. ~
J::...
= - / -r-r: f' 1
.fr, ".
;- =- 2
it .
__
~B
6 •. /fb The streamlines in a particular two-dimensional flow field are all concentric circles, as shown in Fig. P6.1f{.. The velocity is given by the equation VII = wr where w is the angular velocity of the rotating mass of fluid. Determine the circulation around .thep~~!!.~_~CD~ ____..._.....
r=
Pri- d-; - f ~ f
FIGURE P6.Lf:b
ABeD
j,
df)
A-13
Sln~e
7/j. =0 (7::
+
-v;.
+f~ adt;
dl-
fezi/' tv
~I
~= CUr
J
d~
+
+0
Iv;. lJA
CD
/3,
4ntl
+
E~. U)
fe, t<)
~L
be OJ) m e.s
a.. l. c/ e r{)
fli'
(I)
t6.~7-
4 fils
6.47 Water flows over a flat surface at 4 ftls as shown in Fig. P6.47. A pump draws off water through a narrow slit at a volume rate of 0.1 fe /s per foot length of the slit. Assume that the fluid is incompressible and inviscid and can be represented by the combination of a uniform flow and a sink. Locate the stagnation point on the wall (point A) and determine the equation for the stagnation streamline. How far above the surface, H, must the fluid be so that it does not get sucked into the slit?
~
------------
..--------~....----t
H
•
-
t-I$~
0.1 ttl/s (per foot of length of slit)
• FIGURE PS.47
t.f= t.f
Thu.5; Cind
If:sl~k.
7-
"'I)I ;;,f"In
Hew
JIp
?If :' '"F dCJ I
V
- Ur SIj,e-
CfJsB-
!!!e
0)
2.7/
m1
(2..)
2rr~
- ~tp - -US/~e 1/;= I) ~ Y" -
1I-/c)fJ?
the
~:: 0 )
willi
wheY'{: 7/j. =0
J
.so
1111 t
the s.fa9IJAiltf)H
ClI1r/
p"J/1i
(jcc('{I"S
Beg. t'/.)
..f~m
-
-
/YY1
t; ::
.s
Ql1d
-bo
the
.Jta91J~t,ol} fo/ni
1he f"lfl1t
cl
pn the
IS
S//t. ( CfPl1 '-L
('-5/·
)
teJlII/
t!),
007'f~
+t.
t..47
I The IS
( COl?
'i )
c/ If Itt the. i.e,..o (E"'g.1) :so thAi (/Q/&1e
sfr~t1mhi1~
Sinle can
b£
si:fltll1llitOJ1 f~,~i (r = ~.t)()7~~ It) e :'0°) the- e~t.iai:'4)H ~/ the cjk'1I1'1t/to~
i,s
y=- rSln . (; W/I'I ffe 11
1he.
.eJ
Ii it.
01
t/~1J
the. ~*,9/1I1-bP" sf"'~J1",J/1I e
as
!:J=
:::Va
the. ;5i-dfn~I:'~11 sfrellml1l1e w;11 luJi b.( sucked 1~+lJ .:5/' 't. Tlte tn4;(imlAm c/,~tt1h,e.1 HJ ~y -Inc sl-otf/llhd,,,, s-trel/mJIi1e t;~Ctlr..s as c9 ~ 7T :s 0 ~ t P/u/e/
aboVe
tJ.2
H=
tt2. s
Z (If '{~)
--
(Noi~: AI/ th~ ./It//d be/f)W -tne s-lA9nl( ttDI1 s+f"ellml,;'e must 111r()tl 9h the. .s//t. Thus) f,."m t4J#S~rJ/4t'4JH ~I mASs /-IV =- ,f-/()W
,,iii)
~, ~ 250
'f~ .$
wh ic..A
e-hec.k:.s
WI'fh
71te.
~
/;i
.ft
tll1swer a J,~lIe • )
'-52
"'.!.f8
Consider two sources having equal strengths located along the x axis at x = 0 and x = 2 m, and a sink located on the y axis at y = 2 m. Determine the magnitude and direction of the fluid velocity at x = 5 m and y = 0 due to this combination if the flowrate from each of the sources is 0.5 m 3 / s per m and the flowrate into the sink is 1.0 m 3 /s per m.
':1
At pel~i. Pr IlI~H1 The )C.-tH.U A.~ Th( V e.\oc..; +-I:U dl.l" -h Th~ -two e:t~~
11te
!:I1~\c..
a.v(
j.,,::
SMA
SOI4 VC.eJ
,G,llow.s:
a.s
i
SOuVce
{VY1 } A=
::
/#tl.
FoY' The
~
fYV\
Th~s,
(S' IW\ -l""" J
1.0 ~ A=
2iT
n~
a.-i A LA..
y-=
W hev(
=
~'(2./M) "l.;{s3l-
=
~ 2~ ;;.,.
2..rY"
-i'hD.lt.
(V-y )
?rr
i)\ 1<-
(1r~ ) A -= ;)0
=
(r- z"",)
2lT
0,5 "5
-=
'rv:'
"I. ':
C>.OZq~: ~
1M
ht>,(,I'~t>",~J ve\oc.;~ ~VY\po)'\e . . tl
0.6\
S"'i
~+
0.0"2.(05
~
~,
(A.j
(0.
\.~
O"2.q~ ') ~
=- O.O\\f'i~ o.VI~
-tne
ve v+ic~1 ve \Oc.~+1j ~YY\fo~e,,",i) 11", LS
v= T~e
V€.\ 0 c. \
+, VA:
2.
V2q
(I..t
(O.02.qt)C!:! = O.Ol\O~.. ~ S A \~ nevety"~
f Lt~-t1rl = ~ ( o.
0 \ 4-q
~'; 1.+
, (0. 0 II 0
~) 2-
O. olg5
~ ~Lf-c
y
6. ~ 9
The velocity potential for a spiral vortex flow is given by ¢ = (f/2n) (j - (rn/2n) In r. where f and In areconstants. Show that the angle. a, between the velocity vector and the radial direction is constant throughout the flow field (see Fig. P6.~n. ::c
FIc;URE P6.1.f ~
HW
me
p~ 1::e11 til:! I
I/e /0'/"::1
'1=-
~d
~
:'
~
~ .".;-
~I-
V·e y = I VI
Sln"ce
--100
oord
~
V :-
'II IIf H J
A
-1
r;. er
:
r 0&or/> : I
f7 2.7rr
CtJscl.
+ ~
"'\
eB
~
-then Cbs~:
"" V· eJ-
IVI
=
7/;.
Y1/j.1.
-I-
VB
1. 1
=
(..i2 )
1-
.2..7T,...
/+
( _ ; , . . . ) 1.
Th US) a..
;;, .,.
Ct; J1 st:a n
a.
t.
rand
angle
01.
.
IS
~.5"O
I 6.50
For a free vortex (see Vidt'tI \'(1.2) determine an ex-
pression for the pressure gradient (a) along a streamline, and (b) normal to a streamline. Assume the streamline is in a hor· izontal plane. and express your answer in terms of the circu-
lation.
1r. " .L &lIP =
r
r
~B
fne .fy~~ 114r&" refre,fnH ,ql1
5,;',,,, ..f-/d4 ,
-me
I3fY'm~II;
1: + X.:
's (a.)
0
t
';r"td'M"/
Ilow
f?l''(L-t'o;, i:
= i!J11ls7:t1l1 f:.
(j )
Jz...~ /lei,#( ht.'!:w,-el1 an, -I:wo poi".b. ,4/,1/1 u .ffru",I,Atf; (r-=- wnsf""i), ~ )j ~"sfq.1all'" 1).=0 $0 -tIIaf from £'3.(11 w,111
:z.
i!,p/ls/-tt"f-
.-In .. ,
press""
£..e ': ot9-
,J C_:;h:lltf) I..'.e.)
0
"lretlml,i1~
wi'" 10- =0
~ 2. + i: -= ~"s.Jq"", 2-~
.5.
(,-S5"
({nil t-= cpns1:J,.f
~.s/
6.51
Potential flow against a flat plate (Fig. P6.51 a) can be described with the stream function '1/
the stagnation point at O. By adding a source of strength, m, at 0, stagnation point flow against a flat plate with a "bump" is obtained as illustrated in Fig. P6.51 b. Determine the relationship between the bump height, h, the constant, A; and the source strength. m.
= Axy
where A is a constant. This type of flow is commonly called a "stagnation point" flow since it can be used to describe the flow in the vicinity of
y
y
[
0 /'
X
x
(a)
Source (b)
-fiGURE PGSI
lj;= A-x ~ ..,.. For
1J,e. bump
the
/h1 ITt;; -- A;z..
, J if V;=?ae = A-r (JIn'
J/f v::: e -
or-
Th e lOIn /;) B:
v;.. =0
~ /)''Ue
-
r)
28 +
(!!J ~7T
B
po,'"i will "cct/y Ai f-=-P..). FDy -the ~'L1e H o t.retlm Iu YlC- tl (11)
~t:(l.1nlt tlOh
(B= ~)
.>G= 0) ~=/,.
}--lS/11
AI-
~~ z.!I :)111
oJ.
0.
i:
(' 1 )
2.8
r- =- t,) w/// be
v;;.::: a
qn
:z.1T'r
A
.:;-t:a1I1tft f/~;,
-htJ~ jJo/n t:.
o ::: A h C"$ 7T +
Th us)
pOlni:
Ir
Ir!Jm £'1.
(J )
'.52
I 6.5Z The combination of a unifonn flow and a source can be used to describe flow around a streamlined body called a half-body. (See Video V6.3.) Assume that a certain body has the shape of a half-body with a thickness of 0.5 m. If this body is placed in an air stream moving at 15 mis, what source strength is required to simulate flow around the body?
2rrb
b= EZ·{,·9'1
b= wheye
rn
Sf-renJth) (JHd
15
iheve-/Cye
(/51f)(t',:;) 7.SCJ
/h12.
-S
(P.5'3
6.53
A body having the general shape of a half-body is placed in a stream of fluid. At a great distance upstream the velocity is U as shown in Fig. P6.S3. Show how a measurement of the dif· ferential pressure between the stagnation point and point A can be used to predict the free-stream velocity, U. Express the pressure differential in terms of U and fluid density. Neglect body forces and assume that the fluid is nonviscous and incompressible.
y
--u
FIGURE P6.S3
!3ernrJulh' esuQ,.fl{)~ bei:ween A- ~ ()iPTxlI ;,
1;1:a.J = ~ It Is f»
\I.: 2. It
='
V
2. ( /
~
!=.>' ) ;:2-
e,,, s e +
(~'t.
P()/~t
JT
5u. /:'51:: i tLl. t/(;11
IJ
b(rr-I) SliJ
:::-
/rtJl11
7Tb 2.
( :l )
7T
f [;b . (2) ,rl.f-c)
-F]b.j =
Thus)
r-:z..
t
2-
.:
~ :2.: V1n~yt krt!
7h1J.
So
2-
t:= A
-'hb
(E'g . tt..J()" )
e
05,-;.,
e.:.
A
'./01 )
If
b (7T-B) j-=
Clnd
+ 2.-
(I)
A
ql1d
At
~t ~ 2.
7"
7. ( /
+ 0 +
E j. {,..
10 J
:flf feb
=iz.)
J;g . (/)
fA
+
it
T/
7.. ( /
+ :;... ) -
~
t
r tJ.7p3
V '-
6.54 One end of a pond has a shoreline that resembles a half-body as shown in Fig. P6.S4. A vertical porous pipe is located near the end of the pond so that water can be pumped out. When water is pumped at the rate of 0.08 m 3/s through a 3-m-long pipe, what will be the velocity at point A? Hint: Consider the flow inside a half-body. (See Video V6.3.)
15
•
m----J
FIGURE P6.sLf
lud.f -Io(!)d~ )
tf -= V- r-SJh B -that.
.::50
~~
VB =- JIQIIP(
dfj; '1/.'= 1.. I" ae
-
l-
Thus 41: /
p~J;'
t A
:::
V
:::
(l2g.
"f"
e.::t:J
97)
.51j"B
77 &;S e
I
(p,
.,. .
~
:tvJ.-
/-.:: IS'_
)
q,,~
l..-B .:: 0 1/:.= r ~: V+
!=or
4\
Sou rc e
AN(
( /)
~TT(JS)
f/()w ra te ~ /-reng'/h
/OI'JJ f Ipe ) the.
J~
(I:g. '.99)
then V= Z1T (S,,")
{g,37 )( 10
-¥
;IJ4f
S
+
--
(P.37
-It M4 xl/)
7
6•.5.5*
oj, - TbU
For the half-body described in Section 6.6.1 show on a plot how the magnitude of the velocity on the surface, V., varies as a function of the distance, s (measured along the surface), from the stagnation point. Use the dimensionless variables V,I U and sl b where U and b are defined in Fig. 6.24.
On 1JJ~ Stlr/ttce,
6/
the hflJf-b()dy .b (7r-e)
/-=
d5 X ::.
wIth
r
1((d;<)
=:
t~s
e
S/h 2.-1
{d.J).1
ol1d
!}
!1 =
y- (- Sill b) db db = V' (~se) dB
-t
d;(.:
ClitP
5,;, 8-. It
Y'
of-
~1/(JuJ.s ihpt.
til" S/nB d~ C(),s (}
there~ye
ds ::
"V Let
V r '2.(d e)
sib
+- (d yo )
l.
,
.
~
r2-f-
(~)" de
qncl
y1' :
1/6
ds -:: .::s~=
2.
So
7hA. t
,
ds;;' r-r~m
E Z.
=
V{rI'F+(UjZ 5111
-de 'f}u..
de
(I)
19
( 2. )
c.. /fJO
dr~_
Th IA S)
~
( rr- B)
C6'J
s+
IS
"J lit 11
by
//;;1<)'"1- {~/- de 1T
O~cP='7T.
r
S/n'17
tire / e12jfh
.5~=
e
(3)
(e&r/t)
The. ve/oc; ~ .I ~ I (')11 -!he. .s",.fp(~ ~f. th~ hal/-b~J!1 CIt H WI"'I'Hl'11 111 11t~ ft,rm (!)b#//;"" #Om £'1. /()/
".
V~=
Thus I ~Y' S""
a..
.frpl'II
-Vtis
=-
D
-r 2. I!fJs!, +
7Jv't'fJ e )
el. (3 ) J
j---110
".-1' 0111
tlif II
I ] [p,.) l-
~
he ~6kllle~ fr~",
V #- fj..Ptn
Ef
.flr).
b~
tf. t. /~O)
,4- fY"()9f'(/ffI /t,y-
CtlJ~lAla.J-/nJ
V* tls a iul'1c.t.Jl)it fJ/ S'" f,;//()u}s. (Not.~:.I;,
1}1e pYf),rlllfl
V""
1.5
c/e~J9n.kd
4.5
V' 41111 S'" QS
S.)
100 c:1""
110 120 130 150 160 170 171 180 190 200 210 220 222 22~
226 228 230 232 236 238 240 250 260 270 280 290 300 310 320 330 340 350 360
print "*******************************************************" print "** This program calculates the velocity distribution **" print "** over the arc-length of a half body **" print "*******************************************************11 print dim th(18),r(18),s(18),intgd(18),v(18) pi=~.O*atn(1.0)
n=18 dth=pi/18 s(l)=O. for i=l to n th(i)=pi-(i-1)*dth if i>l then goto 230 r(i)=1. drdth=O. goto 236 r(i)=(pi-th(i))/sin(th(i)) drdth=-(sin(th(i))+(pi-th(i))*cos(th(i)
))/sin(th(i))~2
intgd(i)=(r(i)~2+drdth~2)~0.5
v(i)=(1+2*cos(th(i) )/r(i)+1/r(i)~2)~.5 next i for i=2 to n sum=(intgd(1)+intgd(i))/2 iml=i-l for j=2 to iml sum=sum+intgd(j) next j s(i)=dth*sum next i print" Theta Arc-length Velocity" for i=l to n print using" ###.# ###.#### ###.####";180/pi*th(i),s(i),v(i) next i
(con't )
(~I1t.)
to.55 W'-I
TObU/ofed
da-ta..
(Jlla
A.
~f
fie-/:.
-the cf4-itA. tire
gnifH be/tJl().
******************************************************~
** This program calculate::-. the velocity distribution ** ** over the arc-length of a half body ** ******************************************************* ..s
Arc-length b 0.0000 ) 0.1751 0.3527 0.5352 0.7255 0.9269 1.1437 1.3811 1.6464 1. 9495 2.3052 2.7366 3.2814: 4,.0079 5.0539 6.74,87 10.14,19 21. 54,87
Theta 180.0 170.0 160.0 150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10.0
Velocit.'}T V5 0.0000 ) U 0.1739 0.3444 0.5078 0.6611 0.8013 0.9257 1. 0322 1. 1192 1.1854 1.2306 1.254:7 1.2588 1. 24,42 1. 2134 1.1693 1.1159 1.0577
1.5
....... ::>
~
r 1.01-
>-
~
...c U
0
.,
~
>
0.5 h
Gl Gl III ~
c::
0
...c Gl
c::
III
E ...c
0.0
0
-0.5~
o
____
~
____
I~
5
__
~~~
__
~~
10
__
~~
__
~~~
____
15
Dimensionless erc-length.
____
~i
20
sIb
~
____
~
25
~
".5''''
Consider a uniform flow with velocity V in the positive x-direction combined with two free vortices of equal strength located along the y-axis. Let one vortex located at y = a be a clockwise vortex (1jJ = K In r) and the other at y = - a be a counterclockwise vortex, where K is a positive constant. It can be shown by plotting streamlines that for Val K < 2 the streamline IjJ = 0 forms a closed contour, as shown in Fig. P6.56. Thus, this combination can be used to represent flow around a family of bodies (called Kelvin ovals). Show, with the aid of a graph, how the dimensionless height, HI a, varies with the parameter ValK in the range 0.3 < ValK < 1.75.
y
U
----
H
----L-l----t:--",.L1---x
--
--
_7/
(I)
e ,&.
yk: H~. tttUP..f,f)fl
(/) Illul
(.!j -t j)
2-
(I) ( 2.)
(-* -I)). :Jlec;';;~1fJI va/1ft ~I Va./k 1=1.L2) ~11 1)(: S'tJ/tleA b'1 ~ +r/~j tfnll err~r :)o/(.('h{;11 /0 t'6-klfl HI),. 5 IPme -itJ/;ulak4 till/ties Alit( the t;,y
A
~6?rre~ff)I1I"i1j'
tjl"R,lJ1J tJre
$hpWIJ
bel/9/.().
3
2.5
Ua/K
H/a
0.30 0.50 0.75 1.00 1.50 1.75
2.65 2.09 1.74 1.54 1.32 1.25
2
H/a
"1
0.5
.-.- - --
---- -- ---
-- -
---
-- -
- -.. __ .__________ _
-----------1 - --
-------1-
O,------~I------~I------~I-____~ o
0.5
1
Ua/K
15
2
ro.57
I 6.57 A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft 2 /s, and separated by a distance of 12 ft along the x axis, with a uniform velocity of 10 ftls (in the positive x direction). Determine the length and thickness of the oval.
).
::
a...
[1r~~
+ I
J~
(Et b./07)
~ [( ~ ) • _ I ] tt< ~ [2. CT;':") : ]
{ a..
I/.,
::
( E'b.
/3, /
Th e fh /ck.neS5/ +rla'/ '111'"
w;'th
2..l)
AS$"fme
fl'"Y'fJ)"'.
rl9 II t
Ctln
A4nd Side
O.
C!J. ZSo
t).2'Z
~.
o.
25"2
al1c/
~a., ~
r (!>.l~) fJ 2.
2~9
0.25"'1
2S"I.
0.25"0
O. 25"3
Thus)
Ii
be de term/ned frf!)fYJ Fg. '.IOC, .h:J I/Q/"Ie ~.... -/"/It.. Cfl1d CtPmlflY'e ~f E$. to. It) ~. (See -& J,1e. J,(, IDw. )
~ [(4)' - IJfa h
-:.
~
use
D.ZS3
-thickness
~. /0' )
- 2J.
= 2. (".ft)(~. 25"3)
-
3. a'l- -fi:
6.58*
Make use of Eqs. 6.107 and 6.109 to construct a table showing how (ila, hla, and rJh for Rankine ovals depend on the parameter n Val m. Plot h versus n Val m and describe how this plot could be used to obtain the required values of m and a for a Rankine oval having a specific value of rand h when placed in a uniform fluid stream of velocity, V.
rl
cauf
w h{)Y~
Itl1fl1t ()f. the hody I~ Z), tlnd the W/dth IS z~. /r;r tt J /vtl1 /l'p/ue e>f 'IT 7/tt, //1'11, F'f. (P. /07 Cfln b~ Jt!)jtJed /r,y .i/a / dlU{ Ef. ,./tJ/f CIII1 be .s~/i/fd ("SJ~.f 4n ~ ceY'l-ft:Iff)d jJYtJcetlu~) kr ~/~. The Yllt-/o ~/-i Mn 1Jtel1 be dekrmll1f4.
the
jJr011"1I1')1 iDy Ck/~"lgl/111 t/a) of 7T V tJ.-/11?1 ~//()IVS.
Ii-
-A/a.,
"Nd
l/J
as
t1
htnc-I:I()~
:00 c15 2.. 10 pc:int, "* *~: *:+ * *** ** * * * * ** ** * * ** ** * * * ** ** *** *:j: * * *:t. * * * * *:t: * * ~,: I' :L2'J print. "** This program calculates l/a, h/a, and l/h as a **" 130 p:r:'int "** f1.lnct,ie,n of pi*U*a/m fOT Rankine ovals :tic" 150 print. II ** *** **** **** * ********** * ** *>l:******ll: **:k ****:**:)1:0+ *' *:t:** * *''' 160 print 162 print "pi*U*a/m l/a h/a 1/h" 168 data 10.0.5.0,1.0,0.5.0.1.0.05,0.01 170 for i=l to 7 172 st.art.=O. 001 175 read a 180 la=(1!a+1)~.5 190 for has=start to 10.0 step 0.0001 210 ha=O.5*(has 2-1)*tan(2*a*has) 220 if abs(1-has/ha)O then goto 230 222 next has 230 lh=la/ha 250 print using "##.#### ##.#### ##.#### ##.####";a,la,ha,lh 255 start=ha 260 next i
*
*
A
*
** '"
Tabu/pted dabJ.. a"d g/l/fl1
I(
pi,,-/:.
()f-
lit. as
a. .ffl11~t/~n Df 7T'Tr~/m1 tire
be/otV.
r****************************************************** **' This program calculates lla, h/a. and l/h as a **
**
1'* function of pi*U*a/m for Rankine ovals
******************************************************* !:i:tU*a/m 10.0000 5.0000 1.0000 0.5000 0.1000 0.0500 0.0100 10
For
-2
tA.
10
Rfll1KIIJe
f!6;U/d 6~
(2.)
(....3)
h/a 0.11,,27 0.2632 0.8601" 1.301,,2 3.1022 1".1,,227 9.9538
l/h 7.3,*83 1".1623 1.61,,37 1.3281 1.0691 1.0362 1.0096
1
10
(J)
l/a 1.01,,88 1. 0951" 1.411,,2 1. 7321 3.3166 1".5826 10.01,,99
O{/4!
~//OWftl It;
-1
iAJl1H
1.
tinA
det:.fJl'InI!1e.
/7;.". a. 9Jvfl1 1.1.Jr. from -nte. gl"aph.
i.
:spec.;!J(~d the IoJJf)wJn~ s.J..eps
/Yn and 4..:
d~i:.f~/7J'I1·e 7J,e reSIJIN'd value t:Jf 7l[r1Z)m
tI~/I1.1 111/~ J/p/ue ~f 7rD"~ //1'11 ~h/cLtla;i:e 1~ +-1'{)1I1 Eg. ,. /" 7. W/1h 'the. VII/we of .ellt. c1eiermJ~e~1 tlnd ,R. :Jj>f'C,'tl p'd) dei:.frI11Ji1e
the '!I4/we !)f a.. LIf) 0;171 7T77a/,m Clnd a.. de tf}l'm,He'!.; /.5
-1r110WJ1/ Q 11 d
the
/r;r ~ jlVfJ1 1/ the
/,It}
lue
I/p/U(!
0-1
0
f t?11
u1m /s f;.xe4.
-
6.5"'f
Assume that the flow around the long circular cylinder of Fig. P6.59 is nonviscous and incompressible. Two pressures, PI and P2, are measured on the surface of the cylinder, as illustrated. It is proposed that the free-stream velocity. U. can be related to the pressure difference /)./) = fJl - fJ2 by the equation
u
-
u~c~ where p is the fluid density. Determine the value of the constant C. Neglect body forces.
y
FIGURE P6.59
si:a.5n4. tlon =
-P'2. + ~ 1.,
7
(I)
~
r:,2:: .2..
-J: ;0 So
u:
)ffy~.
1hey~~t"e
c=
V-f
(3 U:t) = -j-!, l!
«r;¥ ·
<.
6.60
An ideal fluid flows past an infinitely long semicircular "hump" located along a plane boundary as shown in Fig. P6.60. Far from the hump the velocity field is uniform, and the pressure is Po. (a) Determin~expressions for the max-
r
imum and minimum values of the pressure along the hump, and indicate where these points are located. Express your answer in terms of p, U, and Po. (b) If the solid surface is the'll = 0 streamline, determine the equation of the streamline passing through the point () = n/2, r = 2a.
( ~)
FIGURE P6.60
ine suv-1ace cf' ihe hump)
()1'1
1;. = 1:," ;/ l/ The. (/11
max.IYnU/?? pY"e.sst-ft"'e
4.-t 1'hese..
d
. .
rn, n I rn III rn A
t
t71{5
2
(I -
J{.
si.'-a)
0hef'e
Dec.fAY'S
$1;" 8::0)
~Y"
a..-t
t9.:q 71;
f~;n-l=s
1; (rm ~IJ(.)
~
10 +
~! V'-
(ILt
t9 =(J
0-#/
7T )
f~t5S u Y'~
!Dln t
1::.s (mlln ) (.b)
-----
V, Po
;:;r un/form f/r;w
If:: -
In
V
(4.-1: e:::
:1h( J1e,14h~e
..JC-
r)
d/t'Pc.b()~ )
r (I - f:~~).r ~ a I
d;5C"SS/~J11 as.s(!)~,ate'" w,·f;.,
the
-a..)SII,B .
~-v-r(l- _r'ttl
dey/vA-flO;' ~I £g. ~.JI:;').
~----
6.61 Water flows around a 6-ft diameter bridge pier with a velocity of 12 ftls. Estimate the force (per unit length) that the water exerts on the pier. Assume that the flow can be approximated as an ideal fluid flow around the front half of the cylinder, but due to flow separation (see Video V6.4), the average pressure on the rear half is constant and approximately equal to 1/2 the pressure at point A (see Fig. P6.61).
u= 12ftls
~
•
-----
.. FIGURE P6.61
hJ.
'.2.~
t"t
!-o11"lJ/!
~I?
C{
.sec..i::/~J#J
(~twei'h 6)=0
of
tA
Clrcw/(lY
clj/lnt/fY'
DrP.J ~
-t:he drttj
7nA.-t
,9=""J
C/J;'Jj(
7 1 ,,1('110<."'1
t.3
~ =
~JU4.i:-/~'11
17te.
-1 t ep~t9
",dB
C)
-k>l'ce
;::;1" 1;,(:,
me
6)n
.j;.P4-t.
(
/::0
t1t{e
f.s = ~ d.l1P<
sinCe
V-J(..
fh~ //PUlIJ,j
~ =tJ.
SrrnlJ'Jel-rlj
tPl1/~
(Ire.
;:/,,/d
II1.fert's/-PII 7r
1-;1 =
i
2.
f -1; '11lz
'IT
i1T5/~ 'Z-f) ~~l7dB 1!/z-
7T
-
1lt~
I",
OJ
6:.. lib
rEt.
.51;' 2-!) )
-brce ci(.(e
6,J16)
6-0
h~hn 5!.f))
/
1/2.
(J -
17r/1.- "
t-t;.sBdB == :5/"P
'1f/,
Chid
-
If
I'f!t 1b'=6. 7hus
(VIII
h/of!
e;:(
rrt)/n
;! u ~ (; -
-t
j; ~s ~ ~ d t9 1T/z
r::;.::- - z Dill!.
hq/f ~ ~e c.ylln/fr
't .5/~ 1.f) ) cpse v... de
-I
jf~3f}lIT
7T~
.
('-70
.: -j
( Z)
~. ' I
I
{(!or;'i ) It
f,;1/()IIIS
+r()/?1 i:?! .(2.) -ina t
t:XI ~ lV()f~
_
(-.,0.3U-za.
-1J",f 1h~ neja.I:JI/~
slid /nciJcJi:-er %,.+ 1I1~ w«lry J~ a c&all.!/ 'P4IFn.:; DI1 1J!e. c..rlOftlPy (,frpl1t h4/f) I'; tHe upst:re/lm c/1J'f'c:bf)I"'I. I-I-e>wevt~ (,Jhe It the -e .f.fe~1:.. of- t'lte Y'eIiY holf' of the ClfIJIfAel'" 6 tAJcPI1 Inib aC~ilJ1i (lit ct YPtd Flu,'d) -then fu,'ll be A net drtlj in til, dlYer;hD/f oI-f/lJlJJ. It
The prtS~I(JI'~ A.I: a/,(&(
1ht
-I:Dp D/ fh.f (!~kndfY' (pt>J~t ,40) /J
P.s =: ~ + 1; 7)"2. (J - Lf -S/I1'l.{;) WI n. {) = 71'Iz.. -PI, :: -A - ; f V' ~
SInce. A-o ~ -
3
PA- :; - ""i! if
lJi)fo( 1),/-r ~n~
1ne.
SC f'n .. /;.
~t
t- ::: .1 Ii< 7.. If
f:;.
~ >< Z
f
{ig . •. IJ~
1-
/J@Jllt/iI~ P(~.5S{,4Y.f
/:; : -
w,·l/ '1JJt..
4.
prf>Jt'"ft.d a VE'~ = -
'1.
~
'Z.
1/ (2.a.){I) :: 1: fD u-
= ~I = _
+ ~lp. V-~ + 3
- ffrJ').~ [,.u I ~
fl'/tK ""
-In e datA fJ I ~e 11 )
t-; =7; (J.1~ ~: )(t2 ;~) 2(3 h)
fo-11
=
D()..s/~f,J~ ~ ~ ;(.. 1.. a.. (I)
/
~L
6.62*
Consider the steady potential flow around the circular cylinder shown in Fig. 6.26. Show on a plot the variation of the magnitude 'Of the dimensionless fluid velocity, VI V, along the positive y axis. At what distance, yla (along the y axis), is the velocity within 1 % of the free-stream velocity?
-:~ I.J./on, D
f 1I1e.
-the
!1-I1;(JS
~e/I!)c.;ff1)
V,
1I'j..=O So th4t The m4'}fJrl-~de. i...J e!/IAI -fa /118/' J I;;Ut!..
- V- (/ + ;-:)
(Pt. ~.JIG')
SIn b
(a=¥:) r;::f;j)
v
v
/+
/+
/ -(-1:)~
100 110 print "* * * ***-*- * *- *- * * * * *-* * * *- ** ** * ** * *' *:t: *- *- *- *- *- *- ** * * * * * * * * * * *:1''' 120 print "** This program calculates t.he velocit.:,r profile **n 130 print fI*:j( on the +y-axis for flow ar()und a cylinder **" 1~0 print. !! **:t:or. *:0: ** * * ** * * * *- * * ** *)\: ** *:I::;j: * *- * *;~ * *- ** * *- * ** * *)1: *;1: *:t." 150 print 155 print " y/a V/U" 160 for ya=1.0 to 10.0 170 u=1+1/ya~2 180 print using "##.## #.####";ya,u 190 next ya
*
*
( UJ:'J
t. -72.
t)
(C4'JI'/t)
n tI a. fl"t ().f /he data. fr"l?1 fh'J~ ye.su/-h 1h4i ",..
da,ipS(Jfl1
(j
'1 > ICJ
a.
~*************************************************
**'
Thi:: program calculat.es the velocits profile
**
** on the +y-axis for flow around a cylinder ** ************************************************** y/a 1. 00 2.00 3.00 4:.00 :).00 E,.OO 7.00 8.00 9.00 10.00
V/U 2.0000 1.2500 1.1111 1.0625 1.04:00 1.0278 1.0204: 1.0156 1. 0123 1. 0100
2.0~--------------------------------------------------------,
1.8
1.6
1.4
1.2
1
3
7
5
y/a
9
11
u
6.63
The velocity potential for a cylinder (Fig. P6.63) rotating in a uniform stream of fluid is
¢
= Ur
(1 + ;:)
y
~
cos 0 + ;n ()
x
where r is the circulation. For what value of the circulation will the stagnation point be located at: (a) point A, (b) point B?
FIGURE P6.63
(1
(a.)
(.I:, )
II I /"f' r;
•
~/l1t
/3.)
e.S~j -('::
371' 2..
¥7TlIa
J
S;'n
~
6. b Lf
A fixed circular cylinder of infinite length is placed in a steady, uniform stream of an incompressible, nonviscous fluid. Assume that the flow is irrotational. Prove that the drag on the cylinder is zero. Neglect body forces.
D"fl~
f. 1's
:J.rr
: F; : -
eeS(J
~ de
fs :: ~ + -it [/2(1 -
-
f. f
tf
S/~~f))
[
27T
Css t7 dB
2.7r S I j,
=
~ ~ ~s e dB ~
. 3
s,~
J2. (;} o
C>
if f.o Jj DW5 ih(l. -t.
Drlt j = 0
lT
=0
( Eg. ,.
"I, )
-
-
_.
6. ~S
Repeat Problem 6.~4 for a rotating cylinder for which the stream function and velocity potential are given by Eqs. 6.119 and 6.120, respectively. Verify that the lift is not zero and can be expressed by Eq. 6.124.
27T
t = -fo +
I
t
F;,. " - [
Dr-tl') '"
CfJSI9 a. dt9
1.2.('
~ITI
( / - if
e
de '"
sin
17].
7r~ TJ
=0
u
2'Jr
1. :;I·M"'~
I
2f1s/nB
lrr
,ll
C!JS
. ~~ 5111 l7 of-
• 3
Cilsede- "
z1/
Ct)58 51;' 9 de-
=
o
d: +0 II ows fh- -l: DY'a.~
=0
J
2T
~ 3
0
'="0
,.17"
5/·~e de =
[
J~:iH
3
e de .-
£)
i s/~"e "jJ
de "
J-t !cIJoW5 1h().-t.
~·-71
6." ~
A source of strength m is located a dis~ tance etrom a vertical solid wall as shown in Fig. P6.G,b. The velocity potential for this incompressible, irrotational flow is given by
m
e)2 + y2]
+ In[(x + e)2 + y2]} (a) Show that there is no flow through the wall. (b) Determine the velocity distribution along the wall. (c) Determine the pressure distribution along the wall, assuming p = po far from the source. Neglect the effect of the fluid weight on the pressure.
k
(A)
S/~Ce. )
QYltA.
~e-i FIGURE P6.l,.~
= ~rJ ~x.
Ix 1., [ ex -;./"+ ~.]
::
U:
Mt
-Jf7r
)
(~_). ) ~ -t ~
~ f. n [l;<. 1-1 ) ~ j ~J =
It 10 /J()WS 1tll. t
2 ( x-~
[2. ex -.e )
2. ()(11.)
ex -rQ ) 2
(X-).)2 rfJ 'l.
"J..
+
-
';l.
+-.!:J'l..
C;{ -t..e )
(x+~ ) 2.+ 'j 2-
J o
Thus theye )
(b)
Tn!
l/eJ~c~+j
I
IS
nD i/tJw ihr-ough -the o/f!)IJJ 1r=
Cln d
U)I'th
7/=
the. t1?1
'fTr
Slhce
IJ •
u.. =o. A-l.so
~tfi
dlj
.
(il v'f? '" [
V.w- :. 7T
/AJa 1/)
Wa
tie 1t:)'/~
4 ~ ( x-}) '2-r!:J 2-
( c.~J1't ) '-7B
+
ftJ
leI''':'' ~ /
z~
(x+J)l. + ~ ~
J
(
/)
( 2)
( ~)
!=; y
Ir~m ih(!.
-PO _ -Pur 7 - 7 whey-e.
1;) =f;;
Sou Y'C e.l
+
Qh
d
Th us J
V; 2.d
fw- l".s the fyeS5Uye ai the
PJ,r:: ~ -
V ~ ().
-1! VIu-
'J.
1.J4/})
.s~
1h4i
6.t. 7
A long porous pipe runs parallel to a horizontal plane surface as shown in Fig. P6'(.1. The longitudinal axis of the pipe is perpendicular to the plane of the paper. Water flows radially from the pipe at a rate of 0.5 IT ft 3 /s per foot of pipe. Determine the difference in pressure (in lb/ ft2) between point B and point A. The flow from the pipe may be approximated by a two-dimensional source. Hint: To develop the stream function or velocity potential for this type of flow, place (symmetrically) another equal source on the other side of the wall. With this combination there is no flow across the x-axis, and this axis can be replaced with a solid boundary. This technique is called the method of images.
sou.,.,e )
/s
rnea.suJI'ed
.sh~wn
/n r?.:
C/ lui
~r
1"''2-
-fne SOl.{r,e. W,"th
;".I!)fYJ
()
~
~
X"2.+ {'j_3)2. / ()Wtt"
--the.
C!t9mblnt'c/ 2
+ (!j-3
(~-:3) 2.]
+
)~J . . in [x'+ 6+3)'J]
=
2 iYl [ )( "2..-t-l ~ +3 )1-] = ;;;<
2.>< )(2
-+ (f;j-3)
Vw-
= u
=0, =
( Con
1/
=0
)(2.+
n;; (-X-'2.
(:;'1"3) 1-
and
lf
--r-i..q- )
't )
l-
2)(
/0//"w.s -thai
WII/I) '1
sou rre )
S(U(Y"ces
()X
[ ;("Z
the. C.tJOJl'dIl1~/e
fi9ure
l' = ~ f..en [x u=
/1:.
2n
t'2.:
so -inA. t
and
~ .".".
At
f()Jn
t
A)
x.:.
VW-A
V'/(r/3
irtJm
the 1>13
t;ta. [
Lf
/'YY1
=-
0.57'
-¥2.)
J-_~
(~Pt.)
(If- -Pi:) '1. + 9 -PI:.
Jf- 7T
.2 5"
2.
:ft S
an d
)
=0
/3ern()ulj i ~
Vt.crB::
-+ -
5
wJ1'h
find
O. SIT
=
x. =0
At- p{)/n t B)
It)
Jj.
:L~
Or
~-~ = ::
J..L 2
I
.2.
d VWA Jb)
(~2.IfH3 :L (32. 2 ~ )
(3.
fE)
25'S
.2-
-
O. OOG> 2t> psf
6.68 At a certain point at the beach, the coast line makes a right angle bend as shown in Fig. 6.68a. The flow of salt water in this bend can be approximated by the potential flow of an incompressible fluid in a right angle comer. (a) Show that the stream function for this flow is I/J = A r2 sin 28, where A is a positive constant. (b) A fresh water reservoir is located in the comer. The salt water is to be kept away from the reservoir to avoid any possible seepage of salt water into the fresh water (Fig. 6.68b). The fresh water source can be approximated as a line source having a strength m, where m is the volume rate of flow (per unit length) emanating from the source. Detennine m if the salt water is not to get closer than a distance L to the corner. Hint: Find the value of m (in tenns of A and L) so that a stagnation point occurs at y = L. (c) The streamline passing through the stagnation point would represent the line dividing the fresh water from the salt water. Plot this streamline.
(a)
1n~ 1lve~
f;r
~
a/tin,
y
source (b)
(a)
• FIGURE P6.68
$f-rtam fun(.:f/~A)
= A- r 2$;"1 '2 e9~,(
if=-o
().=()
Thus) -f/,(, ('A~~ G=o
B-=-rr/2,
lP=o ,
replac.eeA lv/Tn. a. "5&/;" b~Jllu'"y't 4/(!)A1 4Jh,;;'h the Sfrefll?f -Iunc.-tItP~ r»1tJJr bt ~n$hnf. 7}J/5 b"u.;;tdf(Y"~ .j;;rmJ ~ Y"lint tlJllfJe. Allit 1'nff'e,/rJle Ink! sf-reAm ful1(...I-,D /I Cfln b" U$~~ fD yefr~.sfnf ft4~ JJ4' ~ YO/ill! qrtfle ~rl')er. (J:;)
(JHP.
B,. 17'/2.
.
5'/1'l'~
11::::'
r
a..b
1r
t9:: -rr/'l,.. 1;-: :: 2 A
r
=- ZA-r
r Uls7T
Zvr 1')?1
To J~ I;-
Cr-t (L ~fL
"lrr
Ct
~ 1-1t7"lVbf)H
= 11';.1 (~tp" 't
)
b-B2..
(!,(JI'I
be
( CfJl1't) _
/n1
- -2.7rL
2. hL
tjJ= A-r2.s/~'2.B -t CI n a.
g ()
rrn =: If rrA-L"l.
IA-Jl' th
ljJ= ,A-;-2Sln 28 T lAL!'e The
~4/ue "l /jJ 41- the. SJ.afnaf/~~ ~In·t (r=L) G::: tr/z ) J~
'&/ A L2 = ,4L Th'-lJ)
$1'"
2-
7T
'1-
2.41..'"("1£)
1T
-the €!"UviIDI' .;,,,. 1he ~fre(unJltj.e, pa~sln,
1n~ sh911aA-/~i4 p~/;'f Ij
,4L '2.11
f.,..
r
I
-
-
f-
_
L
-
-
~Y'" pj~H-;'l1j ,
X. t:lnlA
= r-
A- rs/n 2B
7T-28 St'11 2.8
V
t-
2.A L2 e
,
let.
I
~ pjr;z
J?~.cn
=:
CDS
().f-
a In~ d/I/;d,ltj .sfY"e4mJ/~
I~ sh() wn ~1'1 17u. ~J1DW/~~
;r~m
pa.,j e.
in v"~UfJ.t
Theta(deg) Theta(rad)
10 20 30 40 50 60 70 80 90
0.175 0.349 0.524 0.698 0.873 1.047 1.222 1.396 1.571
~ .~~
.
I
0.80 "» 0.60 0.40 0.20
!
IStreamline I .-;:::::::,:==..,..----,-!, -----, j
t----F--~_j~~~~~----~---ti---j
.f-----+_!,------=i~-_--±r--.---_--+-----r--------j
t----t------j------t---=~--iiiiiiiOO;~:.:__1 +------l-----I----_!__--__+----+__--_l +------l-----I~--_!__--__+----+__--_l
.00
Source
y' 0.496 0.667 0.778 0.856 0.912 0.952 0.979 0.995 1.000
x' 2.814 1.832 1.347 1.020 0.765 0.550 0.356 0.175 0.000
-.----.,.-:'- - - - - , r -
I~O.OO
I
r/L
2.857 1.950 1.555 1.331 1.191 1.100 1.042 1.010 1.000
0.50
1.00
1.50 X'
2.00
2.50
3,00
iL_______________________________________
J'
~. bet
I 6. ,~
The two-dimensional velocity field for an incompressible, Newtonian fluid is described by the relationship V = (12xy2 - 6x 3)1 + (18x 2y - 4y 3)j
where the velocity has units of mls when x and yare in meters. Determine the stresses a:m ay)', and Tn at the point x = 0.5 m, y = 1.0 m if pressu~e at this point is 6 kPa and the fluid is glycerin at 20 °e. Show these stresses on a sketch.
(j"
:: -
(T,
-
>t){
~~ ~
-F -r
ih~
dU
-:: ~x
12!J
'2
2)tr)
(OU
( ~O.
~J<
ay'"
d;sf~ibui/~J1) With 4
-/~;(
:::
'Z.
12 (/'0) -Ie
X::.t1),S"",. ~
j::1.0M-1: /
I
,:1.if (~,S)(I,D) ::
1;)..6
S
~=
36X::;:
3b (~,S)(I,o)
If,
s
~ ()"' ==
/dX -IZ!:f
-
(JI1J
~.I'Z.£'d)
(b,s) ::: 7.StJ -s
~.:
d!:J
b,lz5'a. )
( '=g. b, J 15' 1.,)
O~
lIe/t!>';+:;
9/(,/(11
( Eg.
,,(
-p-r~ ~()'"
txJ ~ f Fo.,.
2.;« ~l.!:
I
(j
~)(
C)
alj
2
2
Thus) .f.o~
and /"-:: /,5'0 +
cr'J~
::
tj = (1.5'D
L
X
2
~:
(I, 5"' () ~) ( 7, ~() f
)
) :: - S. r8 ..Ie fa.
6.70 Typical inviscid flow solutions for flow around bodies indicate that the fluid flows smoothly around the body, even for blunt bodies as shown in Video V6.-1. However, experience reveals that due to the presence of viscosity, the main flow may actually separate from the body creating a wake behind the body. As discussed in a later section (Section 9.2.6), whether or not separation takes place depends on the pressure gradient along the surface of the body, as calculated by inviscid flow theory. If the pressure decreases in the direction of flow (a favorable pressure gradient), no separation will occur. However, if the pressure increases in the direction of flow (an adverse pressure gradient), separation may occur. For the circular cylinder of Fig. P6.70 placed in a uniform stream with velocity, U, determine an expression for the pressure gradient in the direction flow on the surface of the cylinder. For what range of values for the angle will an adverse pressure gradient occur?
---
-u
•
e
FIG U REP 6 .J 0
e
rYP/71
Eeg .
~. /I ro
11 = -to + i
fVl(i -
if
;1'ln1.B)
( I)
Since
el/J
p?Slh;e., I
Hils
/17
adverse fr~.s>nre J rdQle",,-t
t3lljPB)
/r
1J;e rAl1fe
-!oj/Pills
tJf
2.
hom
rt'cJ
#r
Ei .f/) 1/11
tJdJlerse
p('e.sst/~ 1I'IU//I'I1/. Th/J 1"t/f1/fe e87rrRspl1ti.I -Jo Y'etlr half "I 171e C!f/ltlder.
'.71
I 6.7'
For a two-dimensional incompressible flow in the x-y plane show that the z component of the vorticity, C, varies in accordance with the equation
DC Dt
=
v V'2[. -<
What is the physical interpretation of this equation for a nonviscous fluid? Hint: This vorticity transport equation can be derived from the Navier-Stokes equations by differentiating and eliminating the pressure between Eqs. 6.127a and 6.127b.
~~. 1../27b (1.)
f ,
b, fleyel1tJ~te V'.esfe,:i:
fc) )( )
ax;J (JV'it
-t
Eg.") W,.th Ye'f eci +0 l:J cliol E ~ . tZ) wi-h a""pA $Ubtl"'U.t, G:fJI) .fr fJ " , E~.lZ) ,,~-I:'IIH
*
Jtr) - a!1d (JU ~"t
dtr
l.(.
~x .,. V ~!J
~ [ d (~ 1. tr ]X. ~X'L
"1
'8~ cJe//","/-,()i1
~•
V-)
~-z..
T
+
u
dIA. ax
d ( ~ 't (,.(
~
-
+
o}('L
-t
v-
& '""i
JI.(.)
v- d ~
d'1. '-<.
d':J'L
)J
-
(3)
(s~e li'~. '.17) ;;I..L
~v-
f~:~-r-; Re-wr;te
Et·(.3}
~
r;j,-/;g/h
d ( ~v- d L< (~tr_~)+ u. 'fX ~x- d!:1 Pfj
- ,X C>
~t:
7[ at· ~7..
(o~
dU
ax. - J;
)
+
b-87
)
-+
e'"
ol. (~_J'<)J oJ< t3 'J
~!J 'I.
(j,,)
a>< - ~
( If)
(P, 7/
1
.
-fer-rn :5/nce each it iol/()ws Iha.t
d~ tit
-
The I eli
D f~
+
f:J
afr
tA.
-Dt
de f"iva tille
D ()
IS
t
lor
d'f. )
+~
(S)
J.!:J 2.-
1hf!
f'slJe' G!.lf.S)
mtJfenid
Dt
s; eJ
Th e rl ~ h 1:; ha J1 d
.
.
(If)
expyessed a. s
be
CtV1
~teyab:;r
the
t.J here
-
~ Z.(5)
of
Erg.
In
h.(~ ;0 J.x2..
~ fiL = J!}
-r7.J'"
a"><.
icJe
pqyen thR SI'S
01
0
t.
E ~ . (5)
I
~h
b-e
ex.pYessed as
-v -V =)L~
where
Fo y
a.
V2. ~ So
thai
nfz :: Dt
-V
1'Jt:n{//~C.()u..s
.{/uid
~"11 b~
E''!. tG")
tdr/fltH I¥~
l7~
.z
J
-zj=a J
QJ1c/
111
Thu Cq.s-t!
D f1:-
Th us ) -{O.,..
a..
Df:
fwo- ttllmpI151OA4J
lIon I//SC/J(J.s ,f-/"I(j I
·/-Iu,'d j)ru·t,·cJe is -,.fl"o.
-=0
-In e.
c.hal1/1€
as t't
f?1 ()
lies
+/~w
/H
the.
I){
On
Ji1~ompt~.s5aJ/p
J
v();-iJ~/r':1 t>f "-
ih Y'~"9h the..
!..jaw It'eld
~.72.
I 6.72
The velocity of a fluid particle moving along a horizontal streamline that coincides with the x axis in a plane, two-dimensional incompressible flow field was experimentally found to be described by the equation u = x 2• Along this streamline determine an expression for: (a) the rate of change of the v-component of velocity with respect to y; (b) the acceleration of the particle; and (c) the pressure gradient in the x direction. The fluid is Newtonian.
(~) From
C4Jni./~ul'-f!:t e8ua.tl~~)
-the
~V'"'-o ~ -
c;u
ox.
.so tn~ t:
tv
"1'
in
u. = )(
=
?;tr _~ ~:: oJ<
A/so) ~g .II)
Cal-(
theV'ekl"t.
f
X-fiX/J
is
(;<)::: 0
Jo
'1r
(b)
(/)
wit;, res.ft>ct 1:.0!:J +0
OJ,-i-4/~
J- z~ d!1
=-
1r::: -
the
-;LX
btl ;".f.@9Y'4ftM
Jel 11'" SIHce
:z..
=
z)(:; + 0.
f-l)l.)
.s/-Ytlll'l1/Jhe) r=o 4/"11, "this
-tit. t: -zx';j
o;tiJ
(J/I1~
'.73
I
6. 13 Two horizontal, infinite, parallel plates are spaced a distance b apart. A viscous liquid is contained between the plates. The bottom plate is fixed and the upper plate moves parallel to the bottom plate with a velocity U. Because of the no-slip boundary condition (see Video V6.5), the liquid motion is caused by the liquid being dragged along by the moving boundary. There is no pressure gradient in the direction of flow. Note that this is a so-called simple Couetle flow discussed in Section 6.9.2. (a) Start with the NavierStokes equations and determine the velocity distribution between the plates. (b) Determine an expression for the flow rate passing between the plates (for a unit width). Express your answer in terms of band U.
T"
Z
I
t
1/ >-
Ii' I
t
I
,
It
Z
b
1
~
77 7
7 I
7 I I I n-X
J:/ )(~d
(a.. )
7htl~ ~r ~.,f'O f'''.5..fIlY~ ~YdQ/;'''t )2.k,
7JJz ..50
At
1hd 'j=D
at: ::J =-~
U::: C, fA. ='0
c;::
-0
j -t Cz.
and ,.f -h//t)/.IJS 1h~t
V
u.-= V and
TheYekre J
(1:,)
_
¥!J
fA..
f
b
u (I) d:J
0
where
C/ ==];
'6
=
V
J;
f~':J d::; 0
-
E~f -j, 2.
0
is 111 e. flDwra te per unit- WIdth
Vb 2..
pJ ..-le.
r;.71f
I 6.74 Oil (SAE 30) at 15.6 °C flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the channel is 20 kPa/m, and the distance between the plates is 4mm. The flow is laminar. Determine: (a) the volume rate of flow (per meter of width), (b) the magnitude and direction of the shearing stress acting on the bottom plate, and (c) the velocity along the centerline of the channel.
_
(k)
1-~)( Sln'~
I (tJ/,i. ~
lA.. :- -
I
').)A
all1d
;f
~tr) -r
(Eg.
~;<.
b.J2!i"J.)
op aX
-.......
v-=o iO//f!)wS thai
-~k a ':1 -
eE (Z!1 )
I
~ ~x
Ot..r
a.x
-=0
and -theY'e kwe ~ ('1)
Tj,x. ::
At- the boll-om pia ie) '1 =-,,£. ) (/11 d S il1 C(! ~ ~x. ......:\.. =: 61' (-R,):: (ZO;
::
- ¥J
/WI
(C )
-If
(2.81
xlO
~
~)
(Z)(Z)( IO-3M-{ ')
- D.
105
~
fD·75
I 6.75 Two fixed, horizontal, parallel plates are spaced 0.2 in. apart. A viscous liquid (J-L = 8 x 1O-3 lb· s/fr-, SG = 0.9) flows between the plates with a mean velocity of 0.7 ft/s. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel?
Tn r.(5)
~ i ..
= 1. If 2 ~z. p.,,..It::
'3 ::: 2:
(P-b) 0.7 5 =
I.05
tt S
6.710 A layer of viscous liquid of constant thickness (no velocity perpendicular to plate) flows steadily down an infinite, inclined plane. Determine, by means of the Navier-Stokes equations, the relationship between the thickness of the layer and the discharge per unit width. The flow is laminar, and assume air resistance is negligible so that the shearing stress at the free surface is zero.
3)1. = 3 w/th 7/':0
the J
UJoyd'l14.t~
t..cr::O
froM -the
J
and
~'1s/-em
#()fl1
Sh"Wh
fJ.x.
+
d...
,it the hjUye
ox. =0 Thus) lilli/lty -S.f.tJke.s ~!UlLtl()H~ (Et. '.I 2 7a..))
eLJl1tinuif!J e.tJtl4:i:IOH ~
1l1e
)(.- ~m'p'Mfl1t of the
() = _ i j
S/11
1'1}
Slit
(f
()
eJ/
+j
dZu
d!J
(I)
2-
A-/~o) s Jj.,{~ there /S 4(. free slwface there CtiIfMt. be a. pre.ssf.lY-e,. jYrJdlent Ih the X-dlY'ecf/~H So -h.c.i: E..e -= 0 Qut/ E'I. (I) L·.1L, axCqh
De.
wrt 7T~11
I/lS
( Z)
6:,-1'3
(P.77 Direction of flow
6.77
A viscous, incompressible fluid flows be-; tween the two infinite, vertical, parallel plates of Fig. P6.11. Determine, by use of the NavierStokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform.
~h+h~ FIGURE P6.17
With the ~f'dtn4~e .j,/.stem ~hf)UJn t<. = 0) W- =0 C(lId .frt!Jm 111e UJni/nu/f!:1 (Aa.I::/~J1 ; : ; = o· Th liS) rr"m 11te !/- UJmjJDJ1bl'"i of -the Nllllley-SlDk~.s -ejtlal-ID;"~ (EI· ~.1Z.1h)) tul1Jt ~:J -= -J)
ez
6
=
Jp
-h -f3
+;«-
cJ 2 tr
(I)
d;<'2.
S/I1C~ 'fhe ,"~~stlye is I'}()t a h(J1(:.-b~n be wl"iHeJ1 IJs d 2 v- _ .p
01 .x.) £'1'
1/)
C/l11
-tlfi'L -/--
(t.Jhe~e
P =
# +;03)
dt.r:: d.JL
s'f/7J/nelr~
J:;1!)f)1
of
1£,.
Thu5
)
(2)
~if1/d.s
w/1'J-,
v
~r =c
/nff?rp/-en P X T C,
ol1d
r o.t:
x=
(J
.j~
(me~Yl veJocj~ ) 9ivel1
V= 2:h '!hai
~ '4!:J -
3
/'-
( Z)
r..78 6.78
A fluid of density p flows steadily downward between the two vertical infinite, parallel
plates shown in the figurefor Problem 6:17. The flow is fully developed and laminar. Make use of the Navier-Stokes equation to determine the relationship between the discharge and the other parameters involved, for the case in which the change in pressure along the channel is zero.
Problem
See.
'2.
Z=
!
where
J5
.f = P':J !.e
~
Thus) 3~g 2.. J
~ +/J - CJp =
TY
!=b~
~
':;
p-l3
3 ~ the ci'scha"ge pel" "1I11 t.vidth
+/,q.
"to
to. 8 3
-3
2.
-It.
11. -fJ
=0
(lIo/:e: Jj,(! ne!a·h;,~ SI'!), iJ1dlC4ks -fh4t 1Ite d'l"e"tJ(/n ~( IIf)W l?1us t: b.fl
d"wnw~lI·eI
.f.o
create.
a IeY'O fY't'ssl.J~~ jradJlfrlt.)
6.79 Due to the no-slip condition, as a solid is pulled out of a viscous liquid some of the liquid is also pulled along as described in Example 6.9 and shown in Videu V6.S. Based on the results given in Example 6.9, show on a dimensionless plot the velocity distribution in the fluid film (vIVo vs. x/h) when the average film velocity, V, is 10% of the belt velocity, Vo.
l' VeJ1 b'J the ~tu4:h~)I
en
( 3)
(Lf' )
r",.,,,, tlnll
eZ.
e1.(3)
(J J:..~
~VD
(If)
~''U1
V. ~ J.
~
wr, HfH
W
3S
2.7 a.S
2·7 [~ ) +
(f. )
1_
J
A pia+- ~ f. -tn(! I/~ /Oc./-hJ c/I'flt-vi 6/.(.J-/~·1'I xlh
o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
vNo 1.000 0.744 0.514 0.312
I
1.500 ,---,----,---r--...,.--,----,-..,---,----.----. ! i
I
1.000
0.136 -0.013 -0.134 -0.229
f'...
<>0 0.500
0.000
l_
~
~
i
I
!
!
I
lr-....t-..,. i
-1--+--+-+--~
___~"...__-I---t--4----l -I~~:...-;Lj.
r
-0.500 -I--......!-I--L._.L...--L-~!----l_..l.---L---L--I
-0.296 -0.337 -0.350
~ Calculated from
~.,........,.-...,....."....--,
Eq. (5)
a
0.5 x1h
~-~~------------'
G,.FO
I
6.80
An incompressible, viscous fluid IS placed between horizontal, infinite, parallel plates as is shown in Fig. P6.80. The two plates move in opposite directions with constant velocities, VI and V 2 , as shown. The pressure gradient in the x direction is zero and the only body force is due to the fluid weight. Use the Navier-Stokes equations to derive an expression for the velocity distribution between the plates. Assume laminar flow.
FIGURE P6.80
c~l1di .J-IDt1~)
:steel h iod
RPr fne
171tl.t the.. X.-6Pmf~l1p"t ( EZ' ~. /2'71<.) reduces -1::0
In.f-e1rll 6,p;' t!J.f-
Gr
0)
J.
F;; r
'J
=.
b)
0
fA.. '::
V,
tit:.
p,l(.
£)
I
1:111"
! :: 0
J(.,)
e'l"lttJol1S
u)
J
of ("2.
( 2.)
u.. =--L{
c
w-:: 0 ,)
7Ie/cis
(I)
U:- ~
:J=
.:.
=")
01 the Natlle~ - S-iDkfJ
Sa
d zf.,( d!J 2.
v
=- - V ~ ~o
1ha t
V;=C,h-~
c= I
u; -t u;. ) u..= ( b
t;J
-
V
"
6.81
Two immiscible, incompressible, viscous fluids having the same densities but different viscosities are contained between two infinite, horizontal, parallel plates (Fig. P6.E,J . The bottom plate is fixed and the upper plate moves with a constant velocity U. Determine the velocity at the interface. Express your answer in terms of V, 111 ~ and 112' The motion of the fluid is caused entirely by the movement of the upper plate; that is, there is no pressure gradient in the x direction. The fluid velocity and shearing stress is continuous across the interface between the two fluids. Assume laminar flow.
u
Fixed
plate
FIGURE P6.81
the. spec/hed C~lId;'tloIf.SJ 7/::: 0 ; w-:::0.l if. :'0j QHIiI Ix. =oJ -Se> '/hat the x- ~mf()nMt of th~ IVallier-5fDkes -esaa:l:loHs (E?fJ. ',/Z711) -hI- e/they fht! up~Y tJr IOIVIY laffer vedtlces t.o /=r;r
dzu
-
I.nk~rA:ti"lt
~I
E S.
d!J
(/)
':0
(I)
J,.
fj/e/ds
u=A-~ -f8 fA.)
tJt~
h ;c h '1 Jiles
r",
the
lIe/f)c/-ht dt.sfn~ktIDN
uppey /4. reY' at-
'j:= 2 -'.) '"
I
it
= Z7
e/fhev
.so
J alJ"'Y".
/ha.:t
B -: V - A, (2.1,) I
her~ the S'IbSC. "'pi POI' the lowe#' latter-
J
W
re fers 10 11te U.f'f~v la.Yfy', CIt fj::oJ u=o ..so thai
=0
B2-
(..)her~ the. '5L1bu,JI'lpt U I -::
tlnu
fA 2.
AI:
'J='{
I
AI
Jl'eier.s +0 the
Z
( ':J - 2
U, ::: U2
A:z - -A I
JaY-iV'. Thus)
t..) +- 7J
~
= A'2,
Jo
AI (~-Z.h) + U
",...
JO£Oty
+
.::
1ha-l:
AI. --{
u
~
(c'()n
't )
(z )
( CI)I1'i )
S/~(e
he
1h(. lIeJ()C"-ht chs'frJ'blll:lDh IS /'~elll' '11 e~c.h Jtl~er 5he~Y'/n,
~fYi.s.s
tr)::
L':Jx
Is
.u. .i!::
( J u. .,. Ii J ~)x
"f-..
-jA.
/-
d!J
~l1si:ql1i 111 r()l.(fIHt)I(,f. eac.h lA-iffY'. hr the "'fpl'Y J~'ttr
L; ~ /-1 A, a J1 eI
laf'
the
)tJWdy
/tl'1fY'
Li
":!
jJ,. A
",... £., =-
2.
~ £.z.
Co,.J.,a -I: lit
v
PI A, -=)tz. A
I-
ft, -
liz. -
l:!-~
~,
u
Fixed plate
6.82 The viscous, incomp.r0:Jslb1e flow between the parallel plates shown in Fig. P6.8~ is caused by both the motion of the bottom plate and a pressure gradient, iJp/iJx. As noted in Section 6.9.2, an important dimensionless parameter for this type of problem is P = -(b 212 p.U) (iJp/iJx) where p. is the fluid viscosity. Make a plot of the dimensionless velocity distribution (similar to that shown in Fig. 6.31b) for P = 3. For this case where does the maximum velocity occur?
u • FIGURE P6.82
I
U ': ~
At
t.< :'
V
C, :: -
(dP) 2 ~ ':f t Jo
"'1}t1A..i:
..../'I , 'J Cz. ::-
-t
~L
v:
A -I:
.J. (lj") b _ ~
/)..1"
~;< /
b
(~ ) (!J "- b:;)
7
(1 - ~)
[,r
i.~ (~)(Y r; - ttr ,;( b )(!LI) J,
'::J
u -
(I)
t I
b
,
P=-iih"'-rr ('?i') ~
SII1C<)
1:1 . (I)
Can
b<
wr/ffel'1 as.
u. ... - -p(t)(l-)-f7 IJ-
p/oi 6/ -this
/$
sh()wn.
On
VI! / ~(..;
~
];
t
I
( 2.)
+rt chsf(; b/.( -tIC ';" loy P =- 3
-h1t!. .ft, / /() u) I n' ,
, -100
p4.J
y/b
u/U 1 1.17
0.1
1.28
0.2
1.33
0.3 0.4
o
1.32 1.25 1.12 0.93
,
0.68 0.37
o
1.2
..,--.,--,...---,-\---,..--,----r--,--;-\-,--,-I----r--,---'I,--;-1
1 0.8
t:tl~~j;~t:r:[' JI=JI=JI=I!=rI=L ~i I! I i
- ,
1
i
]
0.5 0.6 0.7 0.8
It~:~1 I! ~ O. 6 +---+-+-+----+----+-----j---I----l----tl---i~..... ~d~-----r--T!-----j
0.9 1
0.2 +-+----+--+--+----+--+---+---t----t--t__-t-~--r-::~-,r--:-I--j
0.4 +-+--+-+-+--+-+-+--1--+,-+-----+----i--1'l}l-tI, --j
, 1
0-l--...I....-...I....-.L.....-.L.....-j---!l.--J----1--L---'-=:;.l..---,----,---,--;
o
Calculated from Eq. (2) with P =3.
To
u/U L-____________________________________ i
de. krrrlJ~.c.
whp,,<
c/,/ft.r-tl1tI4·+-t!..
d .(1A-1v)
~'J ClntA.
WJ
1-;,
E'!. (2)
)114)(1I1U011
Cll1d
.sei
V€ )0(, I ~
J,a
=_3 ~
.= 0
[i (2 i-1ij - ±= ~
.. 1
"I""
.b
3
&'-/0/
DC'" C..O"
eS~AJ ~ 'ty~.
0[1. (J. ) - 1] - J.b
~-..L
dey
-InlVr
fh{
P= 3
c{(fA./v ) So
1.5
1
0.5
Thus"
6,83 A viscous fluid (specific weight = SO Ib/ft3; viscosity = 0.03 lb . s/ft2) is contained between two infinite, horizontal parallel plates as shown in Fig. P6.S3. The fluid moves between the plates under the action of a pressure gradient, and the upper plate moves with a velocity U while the bottom plate is fixed. A V-tube manometer connected between two points along the bottom indicates a differential reading of 0.1 in. If the upper plate moves with a velocity of 0.02 ft/s, at what distance from the bottom plate does the maximum velocity in the gap between the two plates occur? Assume laminar flow.
u= 0.02 ftls
•
'~~_. m_BB~ ~ .1= •
Fixed plate
r = 100 Iblff • FIGURE P6.83
MCI;(llnt.f/YI
at cllS-h}fce ~ntf tJl1t~e
ve/oc,1t:t ltI/// 4'JCCllr
Thus)
( J)
For rnanofYItte.,.
(St~ h~lJrt 1:0 r"IhtJ)
~+ ~f Ah -
'(1 f .An
t, - 1'2. ~ ('0}t - ~
= (J
()O
it
= f1.
) Ah gD
11
l:! ) (
6. J
.ft'3
I'
111.) -O.I~7-~ _ Ih .ft.
2.~.
+t-
A-Is 0 J
JP
ox.
::
.p, -R'l.
-
~
=
O. /(P7 (
(01')1. 12. ,'\01.
n
1'-'
4=t;2.
)
lb = (). ~3LJ H3
I. 0 I~'
+
12. ,'..,.
~
Z
Shaft
6 •. S1+A vertical shaft passes through a bearing and is lubricated with an oil having a viscosity of 0.2 N·s/m 2 as shown in Fig. P6. g'r. Assume that the flow characteristics in the gap between the shaft and bearing are the smne as those for laminar flow between infinite parallel plates with zero pressure gradient in the direction of flow. Estimate the torque required to overcome viscous resistance when the shaft is turning at 80 rev Imin.
75 mm
~"'i"
•
FIG U REP 6.84-
dF== "t'dA
The -bOY~ ue ~)\I\
.
1.5
'yce d r ac.+'~j d~++eyet,.hal ~vea.) dA-=- vi.Rc1e J
().
C$e E'
+0
due.
+1;
ijll'.iC!
cI~ ~ y. .J
().Iheve
t
IS
l
tne.
Il.t
Tr.·2dB L
VI·'~t.)
d F :: y.l. 'J. r.1 d8 ~ neAyil1j styes s.
i
Thu S J
1oJT"
d8
=
:1.11"
u= lJ.:t. b
Th us) from
=
E Z . II J
J= 2iTr/"r;«-¥)f. = = ),11 (0. 015 {0.2
Y
m1
0.355 N·
rm
r/ t ;
(I )
6.85 A viscous fluid is contained between two long concentric cylinders. The geometry of the system is such that the flow between the cylinders is approximately the same as the laminar flow between two infinite parallel plates. (a) Determine an expression for the torque required to rotate the outer cylinder with an angular velocity w. The inner cylinder is fixed. Express your answer in terms of the geometry of the system, the viscosity of the fluid, and the angular velocity. (b) For a small rectangular element located at the fixed wall determine an expression for the rate of angular deformation of this element. (See Video V6.1 and Fig. P6.9.)
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t .... sheanng s+ress (ill)
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6. 8~* Oil (SAE 30) flows between parallel plates spaced 5 mm apart. The bottom plate is fixed but the upper plate moves with a velocity of 0.2 mls in the positive x direction. The pressure gradient is 60 kPa/m, and is negative. Compute the velocity at various points across the channel and show the results on a plot. Assume laminar flow.
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cis print ,,******* *** **** ******** ** **** ***:+ *************** :t-* * II print II ** This program cal culates t.he veloci t.y prof il e * i l l print ,,** for Couette flow **" print "**************************************************" print print." y U(y)" for y=O.O to 0.0051 step 0.0005 u=40*y+78900*(0.005*y-y~2)
print using "#.#### next "}t
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4Hd A.
p/tJi ()f- the dlLia. a;e
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*************************************************w ** This program calculates the velocity profile ** ** for Couette flow ** ************************************************** y
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050
u(y) 0.0000 0.1975 0.3556 0.4742 0.5534 0.5931 0.5934 0.5542 0.4756 0.3575 0.2000
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6•.~ 7
Consider a steady, laminar Bow through a straight horizontal tube having the constant elliptical cross section given by the equation:
x2 y2 a 2 + b2 = 1 The streamlines are all straight and parallel. Investigate the possibility of using an equation for the z component of velocity of the form
X2
y2)
w=A (1 - - -2 a2 b as an exact solution to this problem. With this velocity distribution what is the relationship between the pressure gradient along the tube and the volume fiowrate through the tube? Fr(Pin
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A fluid is initially at rest between two horizontal, infinite, parallel plates. A constant pressure gradient in a direction parallel to the plates is suddenly applied and the fluid starts to move. Determine the appropriate differential equation(s), initial condition, and boundary conditions that govern this type of flow. You need not solve the equation(s).
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6.89 [t is known that the velocity distribution for steady, laminar now in circular tubes (ei ther horizontal or vertical) is parabolic. (See Vidl'Cl "\'6.6 .) Consider a IO-mm diameter horizontal tube through which ethyl alcohol is flowi ng with a steady mean vclocity 0.15 mls. (a) Would you expect the vcloci ty distribution to be parabolic in this ease? Explain. (b) What is the pressure drop per unit length along the tube?
(a)
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= g
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6.90 A simple flow system to be used for steady flow tests consists of a constant head tank connected to a length of 4-mm-diameter tubing as shown in Fig. P6.90. The liquid has a viscosity of 0.015 N . s/m 2, a density of 1200 kg/m3 , and discharges into the atmosphere with a mean velocity of 2 m/s. (a) Verify that the flow will be laminar. (b) The flow is fully developed in the last 3 m of the tube. What is the pressure at the pressure gage? (c) What is the magnitude of the wall shearing stress, Tn' in the fully developed region?
~
_ _ _ _IIIIIiiI_m
Diameter = 4 mm
\.
3m---I.!
• FIGURE P6.90
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6.91 A highly viscous Newtonian liquid (p = 1,300 kglm 3 ; J.L = 6·0 N . s/m2) is contained in a long, vertical, 150-mm diameter tube. Initially the liquid is at rest but when a valve at the bottom of the tube is opened flow commences. Although the flow is slowly changing with time, at any instant the velocity distribution is parabolic, that is, the flow is quasi-steady. (See Vidt-o V6.6.) Some measurements show that the average velocity, V, is changing in accordance with the equation V = 0.1 t, with V in mls when t is in seconds. (a) Show on a plot the velocity distribution (v: vs. r) at t = 2 s, where v: is the velocity and r is the radius from the center of the tube. (b) Verify that the flow is laminar at this instant.
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(b)
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6. ~ 2
(a) Show that for Poiseuille flow in a tube of radius R the magnitude of the wall shearing stress, TTl> can be obtained from the relationship
I(Trz)wall!
=
4JlQ nR3
for a Newtonian fluid of viscosity Jl. The volume rate of flow is Q. (b) Determine the magnitude of the wall shearing stress for a fluid having a viscosity of 0.003 N·s/m 2 flowing with an average velocity of 100 mmls in a 2-mm-diameter tube.
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6. 'J3
An incompressible, Newtonian fluid flows steadily between two infinitely long, concentric cylinders as shown in Fig. P6.~3 . The outer cylinder is fixed, but the inner cylinder moves with a longitudinal velocity Vo as shown. For what value of Vo will the drag on the inner cylinder be zero? Assume that the flow is laminar, axisymmetric, and fully developed.
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An infinitely long, solid, vertical cylinder ot radius R is located in an infinite mass of an incompressible fluid. Start with the NavierStokes equation in the () direction and derive an expression for the velocity distribution for the steady flow case in which the cylinder is rotating about a fixed axis with a constant angular velocity w. You need not consider body forces. Assume that the flow is axisymmetric and the fluid is at rest at infinity.
v;. =0) -z.; =0;
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6. qs
A viscous fluid is contained between two infinitely long vertical concentric cylinders. The outer cylinder has a radius and rotates with an angular velocity w. The inner cylinder is fixed and has a radius ' i ' Make use of the Navier-Stokes equations to obtain an exact solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position () within gap) and that there are no velocity components other than the tangential component. The only body force is the weight.
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6.96 For flow between concentric cylinders, with the outer cylinder rotating at an angular velocity w and the inner cylinder fixed, it is commonly assumed that the tangential velocity (ve) distribution in the gap between the cylinders is linear. Based on the exact solution to this problem (see Problem 6.95) the velocity distribution in the gap is not linear. For an outer cylinder with radius ro = 2.00 in. and an inner cylinder with radius rj = 1.80 in., show, with the aid of a plot, how the dimensionless velocity distribution, ve/row, varies with the dimensionless radial position, r/ro' for the exact and approximate solutions.
(I)
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(see PrIJf,Jem
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(1- !.:':) ...." 1/ MAt;; := 2... Ot)
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",/r.w 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
Exact Yo/row 0.000 0.131 0.260 0.387 0.512 0.637 0.759 0.880 1.000
r/ro 0.900 0.913 0.925 0.938 0.950 0.963 0.975 0.988 1.000
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().97 A viscous liquid (p. = 0.0121b . s/ft2, p = 1.79 slugs/fe) flows through the annular space between two horizontal, fixed, concentric cylinders. If the radius of the inner cylinder is 1.5 in. and the radius of the outer cylinder is 2.5 in., what is the pressure drop along the axis of the annulus per foot when the volume flowrate is 0.14 fe Is?
Check
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Plot the velocity profile for the fluid flowing in the annular space described in Problem P6.Q7. Determine from the plot the radius at which the maximum velocity occurs and compare with the value predicted from Eq. 6.157.
ch~ fy; btl. tlt)J1
The veloc/f!J
.
the annulus
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f.ti»[;< ~- ~ or r -~ - -!.
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tt
r·:: J.
o. 016 I~ ,s/~i 2. I
If fo//()ttlS "1hA i
~= -
(:u.'I i!:3) if
(d. /)/1. ~ ) "-L
wh(JY~
~
1,5 1'2.
ft
-L..
:00 110 120 130 1~0
1--=
l
~
as
~,ft 12.
~
)
J-
JH
.,Ltll1c. t:1(/~ af
J
1//1 tI
Y'()
= 2. S' I'".
(¥.. ft)':.M~)ln
I:z.
w;th
.fils
111
c~/Ctl /~t-/ng
rr~ (Yft):
I, 5" /11.
2..~ 1.5
n
ft. A r
.
/"
~J !;{.ft
fY'c9rtl!n .loy
the.
Y'An'l~
k//()WS.
cls print "**************************************************" print "** This program calculates the velocit.y profile **" print "** for flow in an annulus **" print. "* * ** ** * ** * * * * * ** **** **** *** * * * ** * ** ** * * * >I: *** *** * * *" print print II r (ft) v (ft/s)" for r=1.S/12 to 2.501/12 step 0.1/18
150 155 160 170 v=-~~~*(r-2-0.0~3403-0.05~38*log(r/0.208333)) 180 print. using "##. ### ##. ###"; r, v 190 next r
I
( Ct?JI)
i )
'--/22\
'.1.&
(eO!?!)
.110.
Tahu}qted da.ttL ~l1d FrdJrn 1hp.5e dat:tt. 1--1; oCCuvs a.t-
.
JS
?/tJ t a.f the d" ia. ()r~ f/IJI(Jn beJ!)f.(). speJ1 1hllt -lite. f'n"XllYlttm lJeJoc.I'i!:J ~.
/"5 ft
************************************************** ** This program calculates t.he velocity profile ** ** for flow in an annulus ** ************************************************** I
r (ft) 0.125 0.131 0.136 0.14,2 0.147 0.153 0.158 0.164
(
v (ft/s) 0.000 0.l±19 0.768 1.048 1. 265 1. 419 1. 514 1. 552
'~I) t)
0.169 0.175 0.181 0.l86 0.192 0.197 0.203 0.208
1.53l± 1.l±64 1.3lj.l l.169 0.947 0.678 0.362 0.000
0.23~------------------------------------------------------~
0.21
11111/1''''
,/
~
0.19
...::. 0.17 L
Hat/mum
ve1ot,tlj OC('LO'~ a.t r= O./("SPt
0.15
0.13
O.11L-~
0.0
__
~
__
~
__
~
__
~~
__
~
__
~
______
1.0
0.5 v
(ft/s)
/;.-/'2-3
~~
__
~
__
1.5
~
______
~~
2.0
6. q,,)4c As is shown by Eq. 6.150 the pressure
zlC R(z)IRo
gradient for laminar flow through a tube of constant radius is given by the expression: iJp
r r· 1.00
1
0.73
02
1
. 0.67
03
1
. 0.65
0
1
.4 0.67
r· r· 5
0.80
6
0.80
811Q
For a tube whose radius is changing very gradually, such as the one illustrated in Fig. P6. qq , it is expected that this equation can be used to approximate the pressure change along the tube if the actual radius, R(z), is used at each cross section. The following measurements were obtained along a particular tube.
.,
f.
K!4J ft [R(rT'tdi = ;e / Ro
I
I
!Jf>= g;;C/J).'jrR")-'f dr.-#. 7rR() ¥
(4)!),sitJrli
0
ra.d;us t:ube. (oJt'e bS. 6./S/)) ~A Cp}. 1r'F'() If
So
tha.i LI P (/1()nl/()J ~rtn -tube)_ fj f
This
it?te'jrtlJ
SIft1PjON
(tln,term I:llbe)
Con be.
Cll1d
r·
9
0.77
/"
i
(fhe prl'.sslIY'e dY'tJ,) ; t
a" II Ie ""
. 0.73
/'//////.
,.
FIGURE P6. '1 Cf
r/J.
08
1
11.0 1.00
Compare the pressure drop over the length efor this nonuniform tube with one having the constant radius Ro. Hint: To solve this problem you will need to numerically integrate the equation for the pressure gradient given above.
-=--
;l.Jt..:;:
07 . 0.71
1
etltl/uated numfY'Jct:/l&
-the da1;tt,
~/veJ1.
USII1.fj
(~n't ) ~*************************************************
** This program perf01.1l1S numerical integration *>1' ** over a set a set of an odd number of equally ** ** spaced point,s using Simpson s Rule ** ************************************************** I
Enter number of data points: 11 Enter data points (X , Y) Ncte: X/V t¥" and '(,.., ? 0.0,1.00
(RJt..)
_¥
? 0.1,3.52
7 O.2.~.96 ? 0.3,5.60 7 O.4~4.96 ? 0.5,2.'-1:'-1: ? 0.6.2.'-1:'-1: ? 0.7.3.9'-1: ? 0.8.3.52 ? 0.9,2.8'-1: ? 1.0,1.00
The approximate value of the integral
1S:
LJ f (non IIn'#'rm -f"b~) = 3,57
Ap
(un,ftJrh1
i::uj,~)
+3.5707E+00
6,. /00 6.100
Show how Eq. 6.155 is obtained.
;:;y -/It)tt)
lit
~11
Ql1 l1tt/ tl S)
r=
y. (".
t;'"
ft<
1i=D a.t
I) ..::
t:ll1a
~oJ IIlitf
L
~r-
ttl:
r==
/rtJl'h
Er.
~.11/-7
1:
Thll~
(l/?) '1;2. r tJz.
..1 (le
tf)A.
1': .:: c>
) r:.{. d i: (!,
2
C;
'z.
-L I~)
C .:: I
- ~ L~£
t; l'
Cl.
el 111 Ii 1" Cz.
t-
tI)1~
In
t:l11 pt
Y:~
hnlle
tJt!
(z. fJ -
~.
.2.) (I)
111 (~. ) l.2..
~ -~
/n
f;,
)
(z)
In (~. )
'l:l. / 't1.:, .-,,{)
yo ]
/n(~.) n ~
I~
~. I()I
I 6.101 A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pipe, by how much does the presence of the wire reduce the ftowrate if (a) diD = 0.1; (b) diD = 0.01?
The
(I)
(z.. )
//)c)
4 tk,d:.
~
';;'y
cp =
=0
711(; 'r I::.p
~
0:;yre.rjJt)J1ds
w/re)
( h0
~
?r';uJ. PO/Jet/Illes Ltw
= o. I J E 1 ,c if =- 7T~ If. AI'
$)A 1. Th us) ~'y 1ne stirn e
l)
(Eg.
'7 /(/~5
[I _
6. / b'1) . '2..
(tJ.l) '1-+
[t -
CO. J)
91
(n (o. J) ) -f/Ol()r~b:. l:.s reduced
f:.,p +he
I"
(J:J)
r(o tedttcl:t~1'1 t?.=- (; - ~. S7¥).>C )tJfJ St ~ a(ty~ .,t;;v 1; -=-~. () I E1. 91 (2.)
cp; rid
::=
'f2.
~I' [ , _ (/).01) 1',... V- (~. 01)9 ~7=-
reductIon
J 11
/11
6.';7'f
b::J
b%
tJ-l'.j
y.
g;uJ.
t:>t
==
(0,0/)
tjJ:.- (; - (j.7 F3) oX
D.]S3
j
/00 :-
:2. 1,7 0/0
;tit; ie 1Hlrl- t;,e lYeJfH(€ of e t/P 11 ~ t/f V1 S/'hfl/I un Ie Q!()jIf f -tlt~ kJe Cef1ffrJJ~e h II~ t:i ~';II; tC~Jllt e~c.-t t!Jn tke f/otJmde b ~/ 2.7
7. / 7.1 The Reynolds number, pVD/I1-, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for water (at 70 0c) flowing at a velocity of 2 m/ s through a 2-in.-diameter pipe.
Re'111 ()/tis num bit' =
fJ-VD
(FL- If r 2 )(I. T-)(L)
-
~
FL- 2
. /=;;;-- wa..t:er at LJ
r
=
Cj77.
g ..k,.
/YYI~
10
tJc)
~
=
(Ta b/e
r
(M L- 3 )(LT-,)(LJ ML
_1,-' ;V.s -
/m 2
13. 2 ,,, AppendlJ( B) .
7-1
.
=
Qnd
rOL~T~
M~LD
TD
7.2
I 7.2 What are the dimensions of density, pressure, specific weight, surface tension, and dynamic viscosity in (a) the FLT system, and (b) the MLT system? Compare your results with those given in Table 1.1 in Chapter 1.
=
. p = fr~S5UYe:" .../arce. are a.. .::: tv(liltj ~ = spec; lid. we/rift .:: Wlllt vi/tune L.~
-2-
HL T ...:.
.£ 1-
3
L..3
-
MLT L
-2
-.
/-1 T 1- Z
'
;1.1
T2..
/. -2.';;L_2.. . ()/ILT T -I
-
(a)
1ft
.
t:: 1'::' &; r:T=
;"
-.
the. FtT ~'t.skmJ J= L -If T t=L-
(j,)
p;
ft1L-'
r-
2 ~= ML-
FL- 3
0-::-
J
FL- 2
!VJ L r .5Lf.sffm)
the.
3 I~ HL-
2-
2
FL-
/11
/-=
T
7-2
M T -
2
r- 2 2.
ML-'T- I
-
-LTM
7,3
1 7.3 For the flow of a thin film of a liquid with a depth h and a free surface. two im,E2rtant dimensionless parameters are the Froude number. V/V gh. and the Weber number, pV 2h/cr. Determine the value of these two parameters for glycerin (at 20°C) flowing with a velocity of 0.7 mls at a depth of 3 mm.
V
'f;h
. --
f- V 2.h -
/WI
6,7 "5
j (r.81 Cf-z)(O,OtJ31M) (/ "2 fDo
0-
--
i
If. 08
Je !!-a )(CJ. 7 ~)~ ~. ~t)3 ,
trn )
.33 )I.)D-,2. - "
-
2 9. 3
tm
7,Lf
I
7. Lt
The Mach number for a body moving through a fluid with velocity V is defined as V / c, where c is the speed of sound in the fluid. This dimensionless parameter is usually considered to be important in fluid dynamics problems when its value exceeds 0.3. What would be the velocity of a body at a Mach number of 0.3 if the fluid is: (a) air at standard atmospheric pressure and 20 °e, and (b) water at the same temperature and pressure?
V :: (),3
(t:?-)
c
!=or So
c.b )
tllr
tit-
20De
V-=
tJ.
111 a. t
Pol"
WA-ter .:s () 1;Iat
J
c
=3Lf.3. 3
3 (3'f3. 3
at to DC
I
T)
S
::
/03
c :: /'1-8'/
V = ~, 3 (/'f~ / c;t) 7-3
( Tq6J~
!!!1
t!!1. S
e. if
J;'
//ppfnix /3 )
-
I'm
S
(Ta.6/e 8.2
'flfLf ff!1. s
.
111
/t,pfrld/~ 8)
7.S
I 7.5 At a sudden contraction in a pipe the diameter changes from D j to D 2 • The pressure drop, !::.p, which develops across the contraction is a function of DJ and D 2 , as well as the velocity, V, in the larger pipe, and the fluid density, p, and viscosity, /1. Use D J , V, and /1 as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?
? -tlJe~rtm) &,- 3 = ~
n&m
the. pt.'
l1)
~ t/Md?
dlmfl1"M/ess
fClYl/metfYS
== FC
2
r
reg "in''''' Use
as refe~/;/n1 variables. Tl1us) a:
b
C
1Tj = iJp q V ~ (;= L- '-) (L ) a- (L T -) h ( PL -J. T) c~ j: ~ L"T
0
I-rC=O
-2
+~,../:,-:2c=o
- j, + C
1t
~ jJ"w.s
ih" i
a. :£ / ~ .h:: -
='0
J,) C:: -/ ) 1/11 PI
1J1fY'~ f"r~
/JI. /),
TT.=
t::..
V/,
I
{, heck. d"';)fn.sltJfJs u~/n, 1/4 L 7 ~'Iskl11 : IJ.t DJ ~ (ML-1r-Z)(L)
·v)<-'
( L I-i) ( HL-JT-)
'11),
11;::
((,
V;a .b
c
L (L)Q. (LT-jb (FL- r) c.: rOL~rf) 2
'i'"
c=o ~ +b -).C=o
-.b + c
(.=-1
I
7r.. 2
':0
h=c»
=
c=o
Dz. D, 7--'1
.:...
M"LoT o
.'. ok.
7. s
I
( CC)/I t: )
7Tz
1:S
'{?,y
7T3.'
Obv/~usJ'1
cilmfnslol'I/ess.
((.
b
I-
C
7T3~ /,£1 V (F L - 1fT 2) ( L) Q. ( L7 - ') b ( F 1.. - 2. T) C~ poL~ T I+G
C
=-0
-If +a -r j, -2C. =0
:l-b -t-
11
~J/f)tIIS thai
et
C
(Ie"
= 0
= I) h = I; c =- -/
( .
~
-
1h-eJ'e{o~e.
f/HP(
77:= t-D,V 3 ~ . . Chec,/c JlmpnSIIJIJS us/nJ fv1 1.. T .5lfskm .' 4/), 11 (Mi.- 3 )(L.)(LT- ' ) ML- ' T-/
-d
T)
MDI/TD
., tJ/t..,
Th US)
~Yrhn~ /f!;J -rr
wheY't Vs
/s
.5
,j .sh"u Id
7r
~_
2..
V 7j. D, - ~ ~ D2 1he ve/l't:.;t!J I ~ the
Vc
lis
eg U(J..;tt"·J'} ) jtn4
J/ey pi fe.. Since
=(!j )2. V /)z.
(lot inrie(Jf}1deni of ~ I P2) I1J1A J/ til1d 1heYe~tt! no t be /11c,luded. Q~ tJYI /l1depeY1deKi lJav,able.
7-5
7.6 Water sloshes back and forth in a tank as shown in Fig. P7 .6. The frequency of sloshing, w, is assumed to be a function of the acceleration of gravity, g, the average depth of the water, h, and the length of the tank, Develop a suitable set of dimensionless parameters for this problem using g and as repeating variables.
e.
e
• FIGURE P7.6
w=. T- ' 1/,t. p,' +11 eo rem) 4 - 2 -=. Z d J'me 11's Ibn less paY"'!'€t.eYs Y'e!",'red. Use 3- ann L a.s Y'.ep~(J.·h;'~
FY-"rn
vaY'ltJ/JJe.s) Thll5)
.b
"'" = W? a..1 (T"'I)(L T-2)Q.(L)b::!; LO T
6
and
(,f;,r L)
z+
h)/~ws
b = y'z
Chec..k
I
(r;,~ r) a"d 1here/r,ye
, Dk
• ¥
Fo yo "TTl.:
J
- 'l tL '::'0
)(.. b L() T
q IJ
7r'}.. := ?1 d L (L T-t. ) a:.(L.).b == / + a+b = t;
tJ
7.7
I 7.7
It is desired to determine the wave height when wind blows across a lake. The wave height, H, is assumed to be a function of the wind speed, V, the water density, p, the air denand sity, Po' the water depth, d, the distance from the shore, the acceleration of gravity, g, as shown in Fig. P7.7. Use d, V, and p as repeating variables to determine a suitable set of pi terms that could be used to describe this problem.
e,
___..~+_H_··_··_·~t_____________~_·~~~~_V_.~ '1
#= L F;."111
f -=
V'= LT- 1
FL -"r
fa.:
7- 3
the 1/ fh et>YftJ1 J
d J ~ 411" I
2
FL-
= Lf
If
rl-
(L ) ( L J(;.
a
L
f/ .ferms ret(; u/rpti.
4.s fete4.-/;;'lIj VP rt'a6/es
tT, = II d
d::'
/I b;
, Thus)
C
(L T -I) b (Ft - 'f T ~)
C
=
F ()L" TO ( hI" F)
~=o
I
+
a... rb - 'Ie
- b
(~r L ) (.{".,. T)
=-0
-f"Z.C =0
Cl=-/
oJ
Use
1,=0 J C=6 I
anti
ThftelCte
1r~J:t. d
I
oj, '/J~{lSi.:J tTl.:::
dJMtns/on/ps...5.
fa. a4. tr b t
C
[FL -'I r7) tl.. ) fA. (LT-1 ) b (FL -IfT~) t: = I
-f C
- If -t- a. -t-b - 4- c 2-b-r2C
=0
=
0
==0
((!6)n 'i)
7-7
POL TO 6
(~r' r ) (full' '-) (,fr,'(" T)
7. 7
til'll'
I
( c~I1'i)
e::{S
J;r
7T/) -rr
a
=-I) b= 0)
-
cr
1';3 -
/;,. 7T'f:
11"'f:;-
C=o
,50
That.
).
d da. f
c
J, / '
{LT--z)(L)4.(LT-'jh (pL- lf r2-)c=. fDL()T" (.for t) (-hf' L)
C'.::. 0
/ -r
A. ~h
-
-2. -b
t.t- c.
=0
(f;r
"1"2(.-=0
C ::
0 )
,, I
H _ ).. d - r
(Ia.
f')
ft
d..)
7-3
~) V'"
01<
T)
:-:--:s--:::;:S-:-:-~-::-::::-
t H
Water flows over a dam as illustrated in Fig. P7.~ . Assume the flowrate, q, per unit length along the dam depends on the head, H, width, b, acceleration of gravity, g, fluid density, p, and fluid viscosity, J.L. Develop a suitable set of dimensionless parameters for this problem using b, g, and p as repeating variables.
7.S
1
f---b-l
'f
~ L 2. , - '
FrtJn?
bJ 31 and
Jf:: L
j,..:.
I
relea-if h:J
as
FIGURE
Pi -'111-
1,-3= 3 pi tfrmJ regUJY'f'd.
-the IJL the()rem Vlnd
;= L ,-2 1=
L
•
vt/n 'af;/es
77f= ~ b aJ~(c (L Z T -) (L ) et ( L T - 2.) b (F L - lIt
:2.)
Co .-:
a... -t-b -LfC
- / - z.b
-I-L C
f=- FL-2.r Use..
Th(Js~
pOL TO 0
(.for ;:)
c=-O 2 -f
P7.~
(:ky L) (lor T)
=0 '::0
'J
'p3~ J ~ C heck
d/~en~/~M
us,;' j
1'-1 L T .5't:,f.em.' ,
77;.-- JI 6
t:£
/,
j
!
C
(L)(L) ~ (LT-z).J(,t=t-'fT-a.j c. ~ FOL C=o I-t- ~ rb -tfC =0
- 2-b -rl. C
=0
a =-1) b ::()) 11:2.. 1'.5
() b 1/ J ~ ~.50
= b1-1
dt'm !14.5IPI') Jess
7-Cf
oro (J::,r r)
thy (.ky
L) T)
()/<
7.8
J
(~I1'i )
a.
"
113=)<- b j f (FL -2. T) (J-) (J- T-j /, (;=-L -If-T 2) c --= pOLO TO Co
#I.
/
-). 1"
t{
-I-C
(,k,- F)
=0
(J;y L)
-t.b -tfc-=o
/ - 2b-r 2 c.
:t.t- 10 IltP4IJ thai
a
n= 3
o//rnel1~/i;f1J
f:,9Iz
= - ~)
if 'lz..f
~
= - 1z .)
c= -J
L
I/Iz
u5iny
?:
(oCy T)
=0
~
-
I
/'1 ML T
s"'f~.J-em.'
(I1L-'r- ' )
(L)3/z(I..T-V'IZ (M[3)
-
11
()L.I tJ)-o
,
" ())(
7. 'f I 7. 'l
The pressure rise , !:J.p, across a pump can be expressed as tJ.p = f (D, p, ro, Q) where D is the impeller diameter. p the fluid den· sity, (j) the rotational speed , and Q the flowrate . Determine a suitable set of dimensionless param-
eters.
-2
II .fellows 1h~t
= -2) b' -/ , _ 4p
= -2
)
. . Tli- D'(o4J' CheclL. dl mel15l()11S IlSin, NI-T slfskm
.-
tZ
t1p
D,/,tcJ'
C.
qnt/
111trehte
Mf. -'T-'"
-
.'. bl<.
(L) Z (J..,J:')(r~·
n; = rP D"'/4J c
(L 3r)(L.),''(;:r¥r)b(rJ C: rLOT"
I
31''' -
.b=o
(;;,~ ,t)
~b =0
(f"y '-) (Tor T)
-1+2/'-C.=0
:U f",11.ws 1hai
a. =-3I b = 0, c" -/ ,
- cp
. Checir
4n4 1h'",/(,re
1r;z -
dimenSIOns
D34J
1151;', Mt..T ~'fsJ..em
CD D0p,)
_
"
L"T-' _ == NDLDFo (L)3 (r-')
7-1/
.: OK.
7. /0
I The drag, ~, on a washer shaped plate placed normal to a stream of fluid can be expressed as
7. 10
ill = f(d" dh V, fl, p)
where d, is the outer diameter, d2 the inner diameter, V the fluid velocity, f1 the fluid viscosity, and p the fluid density. Some experiments are to be performed in a wind tunnel to determine the drag. What dimensionless parameters would you use to organize these data?
~ == F rrlJ)tn
d,)
"-,
-= L
dz. :d: L V ' : : L T,-~
tJte. p/ fhet7r'rm J
V;
tlno
I
(4
Al
pV
d,
tl
j/
F"L ~ Tfj
/-t-C=O
(/;,,,.. F)
a.+h -Ifc=o
(-ICY' L..)
-.b-+2.C=o
11: r;,//PI(I$ ihll i
(.fey
/<..=-2.; },=-2, C ='-/ ;
ChecK ql mfl1SI;'ns
NtT ~lfs-t-em :
::::::
MLT (L) 2. (LT-') (I1L-3)
712 =
-2
~
d;L d, V
b
!
c
(L )(L) tZ (LT -') b lfL - 1f T 2) C1
/-t-a
1here'ye
0/ VI'
"'SlfJ.!)
~ d/' VZ;
tJl1t/
T)
~
1T:~ I
71..' .z.
FL _aT
V4ntl"/e~. Thus)
"C' t (L r-j b(FL -'trz) c: =
rr;.:
)A =-
fi Hl"l71s Y',!~,;'ed. lise
~ft'a,6/11~
""
(F) (L)
=3
1
=0
+ b - 1Jc. =0 .b +:J..c ':;0
-
7-12
c
=
-
NIJttJTO
...
()I<.
7. 10
I
.zt
-Iol/e;lVs 1h1l t
t:l. :: -
I; .b = ole:: 0
I
tI/I t/
1heY'e /:;"e.
17:=~ 2. d, tJh Icn
fOr
til tnf'I1.5I(J/Jlpss
ObV/()VS/;
IS
.
7?3:
- J. -t
t(
IrC .::a
(~r )=")
-t./J -ifc :: 0
(.j;,r L)
/ -.b -+ 2.c
=lJ
("'yo T)
11:= .3
.:
d, Vj1
~k
II)
S IhC(!;
I-V d
I~
l
a.
I"-
(Re!;//J"lds num bt!}")
ex-tressed
(J
I
sirJl1d£trtl
E 3.
dl ~pn~/~l1less PtlrqmebY'
AJould
(I)
rnOYe
Comm t)l11':1
be.
5
(Z)
As h,y til? tI (Z)
AS
qre
cfJmfl1,sIOl1et
J
/Jllq
!t1Si.s
.fJgu/llalen t" .
?-/3
Ij
~f1C'erl1pd I
,=;s.
(J)
7. 1/ I 7.11 Under certain conditions, wind blowing past a rectan- . gular speed limit sign can cause the sign to oscillate with a frequency w. (See Fig. P7.11 and Video VY.6.) Assume that w is a function of the sign width, h, sign height, h, wind velocity, V, air density, p, and an elastic constant, k, for the supporting pole. The constant, k, has dimensions of FL. Develop a suitable set of pi terms for this problem.
• FIGURE P7.11
1
tv: r- I b:.L -4t=L V== LT- f= r-L-'tr 7. .,k';FL PY'611?1 1'nt. p/ 1I1eorem '-J = 3 pi hn?lj YeZUlyptJ/. Use b.J ~ tin,!;; 11.5 repea.-t,Hfj INIYlable...s. Thu;~ tJ; ::- tv j, a. V f; C
(T -I) (L)tO. (L r-jh(rL "''fT~ c: Pi- "TtJ a
(tor r) (/:,y L) (Joy T)
C=:.o -t J, - i.fC 0
=
- I - j,
i"lC.
a: I
t t /-oIJIf}IJI..$ #fa
J
=0
j,:. -I) C:. 0 ) CI If a 1htre ~re.
wb
1Ti= V C he("k
c/"YlfI/S/()I1.5: fA) b
V-'::
(
(
T- 1) L) . . :. . L() TO
7T; - h j,Ow
v b/
C
(L) (L)(J. (LT-1)b (FL-'f (! =0
J -t
,: 01<..
(L,J)
a. +b- ¥c. =0 - .b "i"Zc.. =0
7- 1'f
r2-)(.=. pOLO TO (~y F) (~Y'
L)
(,(oJ" T)
7 1/
I
(tt!)l1t)
/=;r TT3: 1T3::
~ J/~' Vb f
(r-L)( L)Q.. { L7-0
h (
c:.
FL-'I
r ).) C = toL T 0
iJ
/-I-C!. = 0 l-t a...,. b - Jfc.. = a
-1 -r Zc.
:Ii
~/jf)Wj
1;J"i
173
=0
a -= - 3)
.b: -l) c:::- - I) ~J1A 1here (ore.
-k
= .b'V'1
Ch~CK r/;mel'l~PII.s ~ Slh..J ML T .s'l~klt1f I'
-4<
.
),3V'/~
=
HL'1. T -
Z
(L')(LT-,)2(HL-3)
7-1S-
o
- fi1 ~ L ~ T :. Dk.
7.12-
I 1.12. The flowrate, Q, of water in an open channel is assumed to be a function ofthe cross-sectional area of the channel, A, the height of the roughness of the channel surface, c, the acceleration of gravity, g, and the slope, So' of the hill on which the channel sits. Put this relationship into dimensionless form.
(p::' L3 T- 1 FYO/YI
the ?'" 1he~ye WI
,4 tlnli
S- -"2:. '3 P/ ffrrns Y'e 0/J/~r". US~
I
9 as rffJ(Jul;'~ Vt(v/ab Je..s 11j = ([J ,44J 6
Thus)
(i ~ r-I)(I..") (L r-1.) "=- L() TO I(.
E+Z4-tb=c =D
- I - 2 J,
a. :' - 5"/'f ~
-
b:: - I/z. 1 a,,1 1Ite"'~ fo,re.
J
--=-.(/)......""",,==-
A5"/'fff
1-
Check .: 01<
I -t2.&L-tb
1J,p f
a ':=
- 2..b -
=
=
1/4,)
0
(toy
D
(rfr~ T)
b = (:)) Ii 11 ~
1'ltfre
f;,1I'e
1T.2-- - e IJ
t>bv/~I4.5J:J -SC)
Jj
r.A
d,me"s/t!JJ1/eS'J. Th( 17l1vd f,' ~m dun tl1SJ~IIJ fSS. 7ft us)
7-/(II
Ij
L)
7.
/3
1
I.U Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface as shown in Fig. P7.!3. (See Video V1.S.) The maximum thickness, h, of a square of material that can be supported is assumed to be a function of the length of the side of the square, e, the density of the material, p, the acceleration of gravity, g, and the surface tension of the liquid, 0'. Develop a suitable set of dimensionless parameters for this problem. I'
{'
-I
~);fit~~8 i{Fm>4');~I~1
T ~=f(.J.)
-l=' L l='L ;= Fy~rn the
t)
ffJ
p/
(lI1d
I
..,0-)
1)
()
(/')
d=
1=I--'t,2-
1he~Y'em)
Lr- 1
b-- B ::. Z fi
.f.fr/Y1J
Y'R11J1rnl...
(Is retea,6;~~ vl/f'iflbJe..s. Thus)
6;
1Ti -:: ),. .1 a.1-
C
(t.. )(L)~(l..T-~) b(FL- If
r2.)
C! ::: '-t-~-+,b-,+c..
c_'
rfJLD(-
0
=0
- 2.b +2.. c.. .:= ()
a=-')
e:&J,
j:.o)
(/1111
171tJrt"re
-It flt=7
wAlch /5 ~j,JlJ~II.sJ!:1 cit in f,,~i!JIJ!1 s S . /=by 7Tz....'
~
c:.
IJ
rr t d !
7T"J.- =
(FL-')(L) 4..( L T--~ b (FL-'fr 2.) Co = / -+ C
:=
-/ +tJ... -rb - 'Ie _ 2.~ -r '2C
a:-z. )
j='tiO
I
0 =0
=0
l,-:-/
)
c.=-/ ) t/htl there{;r<.
rr j.~t If'!'J;'~
/vtt.. 7 0-
J, 2#f
-f:=
'
~'Is/e1l1 ( fo1 7- "L.. )
f
= (L 7.)0. 1- ){ML- ~ ) Z
cf (t-f-it
)
7-17
.-:. M ~L 0 T
P
..
7.Jlf
7.J 4 As shown in Fig. P7.14 and Video V504, a jet of liquid directed against a block can tip over the block. Assume that the velocity, V, needed to tip over the block is a function of the fluid density, p, the diameter of the jet, D, the weight of the block, 'lV, the width of the block, b, and the distance, d, between the jet and the bottom of the block. (a) Determine a set of dimensionless parameters for this problem. Form the dimensionless parameters by inspection. (b) Use the momentum equation to detennine an equation for V in terms of the other variables. (c) Compare the results of parts (a) and (b).
•
V = -f
(t)
V; t.. ~-I f =-
r- L
(a)
j';S?eVr/DI1
US/IfJ
71.,....
h:,y 11,
Y.l
J~~
11".z-
=:
lor 17j
'-3 :
b,,"In
3
(tDI1Hlil/;'..1
-krrns V)
b d: L
cJ -=
L
r ~tIJJY'ed .
(L 7-')(/...) ({ r:~~T~)
"," rot!'
1
= b
d
1T~ tll1'{
D tT) tire oJ:, JlI ~u.s 1':7 c/; men~/lJl1less .
{.hI t=;r Irnpend/n1 flPplnj aY~I(;1f' L MIJ =e> 50
p,'
=f
cJ
rr,3:: til1lt
J
t:p,J
I1LT:
Vb
aha
D:: L
11i = V D f:G '~
Check
H,r
~) b; d )
-1f T -a.
1nt. ?/ 1htJ)rem
p;.1)f11
B/f
DJ
FIGURE P7.14
171(1,,.1:
0
h
t= cl
= t:tJ (-;.)
(I )
o
7- 18
7. I tf I
( 6PJ1 ~ )
f:y-IP""
C9
!f
u ;.;.
f 7h H~j $;0
.
m",m~l1ftlm ~"~Jr/eya,t/~ t{51;'~ tJ,,~ CV
dA =-
sh(Jwlt
L F"
V2,A = F
I='1. {J.}
/rPTn
(;V2.A)(d) = ~ ({) i114, t
V = V~ (j,)
•
2.f,4 d
" nil
,4 = . ,.,fir . DZ
/A/11h
11= \/21v b r
(
V 1?"f . £2.)
1
part Ca.) ... I re;;;-'
(') Front
{ 2. }
7Tf'd-D-
V~
Clll1
b£
A '1
(id) E.) D
IN yo I
ifr 11
as £.3)
.t:f h//oltJs b'f
&JlI1flJr'Jn.J
E ,/s.
(2)
tfn,;;{ /.))
1VJtJ.t
f (~) ~ ) = (~)(-ff)' So
1/t..,t
1 (~)
~)
/.s
7- 1'1
actua
":1 inrJetflf("J~nt
tJ
f
-i .
7. /s 7.15 A viscous fluid is poured onto a horizontal plate as shown in Fig. P7. r5". Assume that the time, t, required for the fluid to flow a certain distance, d, along the plate is a function of the volume of fluid poured, ¥, acceleration of gravity, g. fluid density, P. and fluid viscosity. fL. Determine an appropriate set of pi terms to describe this process. Form the pi terms by inspection. /
I---d--j •
FIG U REP 7 • 1%S
;.= J...T- !;;: FL.- r2. !- ~ FL-2. r lf
z
i:-= T t=rpm -the
B:;
pt,' 7he"rem
/n.5f~dtt3;j / f;r
70
6, - 3 = 3 pi t.errns
(('~11i:4Iitll1.7 -t) : II,
1T;=iv'J
rejVI"retJ..
_
(T)(I-T-7L)
CL ) ~.z.
7-20
L
==
pOLdr o
,',
~/:.
I
7. Ie,
7.It;
Assume that the drag, ID, on an aircraft flying at supersonic speeds is a function of its velocity, V, fluid density, p, speed of sound, c, and a series oflengths, ell ... , ei , which describe the geometry of the aircraft. Develop a set of pi terms that could be used to investigate experimentally how the drag is affected by the various factors listed. Form the pi terms by inspection.
oJ.:: f ('0 /.1 V::;L. 7-1
FL-If T 'Z
==
)1) ., " Y/., ) Co
== LT-'
("/--fL)-3 = /+i pi ffl"tns Y"f'J"/re~ where IS the ntlmhtr of /en91h itrmJ (t'= ~ ; 3) etc. ).
r-rtJtn L'
I
C)
the pi thetJI'etn)
Blj /J1Sfectl(;H) fer
1T. I -
Check. us/nj
7T; (Ct)J1fq/111)1~ otJ) .,(J -=F If I V2..J,/, - (FI-- Tl.)t1.r-'j 2.fLj 2-
MLT:
~
Fbf ~
/i;r
rM/.-
(c~nl-(jin';'j
7Ti. Cll1d
Mt..T
both tI / /
(lYe
() the,.
=
VCo
these
Thus
!=~L 0 TO
_2.
3) ( LT- I ) z. (2) ~
t)/<..
c) . Dr
~ C
abl/lows/'1 dllnens/on Jess. ft'
.feyms c,tPl1 kI/~/iJ!J 7T.' =- t,' t
til1d
::::
tefm5
J.L'
J/~
/~I1()/III;'-' 1he
1;
J
wheyt:.
7- 21
CI re
obJl,otl~ Iy dim ens 1011 Jess.
7, J7
I
7.11
When a fluid flows slowly past a vertical plate of height h and width b (see Fig. P7 .17), pressure develops on the face of the plate. Assume that the pressure, p, at the midpoint of the plate is a function of plate height and width, the approach velocity. V, and the fluid viscosity, /.L. Make use of dimensional anaJysis to determine how the pressure, p, will change when the fluid velocity, V, is doubled.
Plate Width
= Ii
1'= .f (~) b) tI))<--)
t=
~c2.
J..':: L
l!:: Lr- ~= /=C' T I
b -= L
FrfJfJ? the pt' 1heOYfm 5- 3 .::. z. fi' +trl11J 8'1 1;"'pec,t:/~:/II.1 h,,.. m (dd?11 1-4,iuh, -;):
1Ji = ~-t. ~ V)A--
Check.
z
(FL- )(L) .:.. Ct.. (rl--:Z r)
,I )
IA$/~.! fit f.., T :
1'-4.. _
fHL
_lr--a.)(L) -.: (L 7-1 ) ( Mt.-',-')
V)<-- ff; r ~
(CtP,rltJ/'1I;'J
frJ-
=i
whIch JJ obvifJ('(sb
M"L()T
tJ
, CJK..
b) :
dlm·e~~~oll/tS.5.
Tn u~,
ti~ef({) V;«So 0)
Fr~m 13"1.
if ,f:,;I" WJ 1),a1: -kr A. ?J v(~ re&flefr; and VI~t~5Ifr/ if 1At! v(»eI(I+'1/~ Ii dou~/ed -t1te.. l...rlS5l1fe) wal he dCtlb/fd. tJ)
r:
7;3 1 7.1 ~ The pressure drop, Ap, along a straight pipe of diameter D has been experimentally studied, and it is observed that for laminar flow of a given fluid and pipe, the pressure drop varies directly with the distance, between pressure taps. Assume that Ap is a function of D and e, the velocity, V, and the fluid viscosity, fl.. Use dimensional analysis to deduce how the pressure drop varies with pipe diameter.
e,
fJ I'
At=-
I=L-'J..
=-
f. (/)/ J) ~)t)
D-'
L
}-' L
Frt)m 1he p/ fhet)~pm
1-;
V.; LT- '
FL-'-r
s- 3 =.l pI.' 17rms ye$tlIYed. Bt{ II1:5fec.l-u;H) -foy -n; (CtP)1.Jqil1lhj A.p): J
7r= /l.f>J) I /' V
ChecK.
uSln~ ML T :
~
Af b !'" Y
FDy
-n;
::!:
2
(I=L- )(L) --.: (FL- 2 T) (LT-')
. 11 6 LO TD
(I,/}L""7- 2)(L)
(CtPnltlil),n,')'
i
Oblli()usJlj dlmeY/51()J1less.
LJfb ~V
Thus;
=1(1. )
(I)
D
1Jte pY'()/'iem) LJ p oC 1 be t).f. the ~rl'Y1
shfrll1fJ1t af
rn us i
)'IV I{
IS
c50me
1J14i
D tons/:QJ1t.
-D'I 2. a.
So
1..
4f D == k
for
.: 01<.
(Mt..-'r-')(LT-')
1Tz =
whfll'e
rOL~To
!lIVen lteloci~. 7-23
It
'thus JcJ/ow~ 1hot
I
7.11
I. [q The viscosity, /-L, of a liquid can be measured by de. termining the time, t, it takes for a sphere of diameter, d, to in a vertical cylinder of settle slowly through a distance, diameter, D, containing the liquid (see Fig. P7 .19). Assume that
/Cylinder
e,
t
= fee, d, D,
{
t
y
1ht f/
/-L, ~'Y)
, - 3::,
1heOYflf1)
i:
,-
Che,k.
{ ,) (f:t..- a) (L)
L1a d. - . ,)A.
MI-T: -t Ar d
USlhj
-!
( FL -2. T )
(T ) (N C 2T-2.~IL) ~
~
f;, r
11;.
(CSJI11rti/11;'j
J
3
7Ti ( tlPn hi11 IIi.J 1: ) : 1T: -
,>-
Sphere
where ~'Y is the difference in specific weights between the sphere and the liquid. Use dimensional analysis to show how t is related to /-L, and describe how such an apparatus might be used to measure viscosity.
rYllfl1
\~~~
( "-1. L-'T- ' )
D).' 11'1.
=~
dlmf"si~I1J~5s
"bVIDUS).!1
(~fPn ft/ IJlIH .J
J.) :
L
1T.3-- t:i()6111~U6/~
Thus,
dlmtl1s/o,,/es.5.
-I::lJoa_- rf,(O ft) r d.;r d..
hied tjeemef,. ~
If.
-/:Llad.= C
C
~
J:s
q
e.,11 :, -/-aM:i )
~r
)A- == ~
!J /I
0..
-f
:::
{!,
Mt
~"s.f-a",.. + ~ 1" a f.1 ~ eJ.
3 e()'l1 e fr.!:J.
Tit ~sI .fr,.".
)A. = ~ Dr-/: b'J ~a/,'b¥'aJ'f)')( w,1lt ~/Ul" ,/ ~iI~lJ)n Ih5eP5rh. {P, Th C, ~ h,w II fh( VIS·COS,'-#, ~ I D 1/1(r fl tI' CD Ok-t. hL dt-/:e YlrJt /I'e '" ih Y'OtJj h 4 IYl eQs II. llem P4 ~I- 1'1u. -f.IMt! -t: Ii! t.pnjuncf,f)'x L
Ilt.e e,1I,s.f.II/1-t (, talt
be
at.termlt1'(,p{
7- 24-
A
7.2.0
I
1.40
A cylinder with a diameter, D, floats upright in a liquid as shown in Fig. P7.20. When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequency, w. Assume that this frequency is a function of the diameter, D, the mass of the cylinder, m, and the specific weight, y, of the liquid. Determine, with the aid of dimensional analysis, how the frequency is related to these variables. If the mass of the cylinder were increased, would the frequency increase or decrease?
.
w::.T Fr~'"
8'J
-1
C
/Cylinder diameter
i'
D=L
1/1e p/ 1heD~~W1, 4 -3 = I;' 'f'ec.;f-rD~tJ .
71:= I Check.. ,: OJ(
S/~Ce-
where
-there
C
. ",,1, /
JJ
p/ -/-en"" ;~ P~IIDWS
W -D ~:: J'" Is
it.
~"s1-aHf .
1h14:t:
C Thus)
tv= CJ)
vr
Fr~m p,i$ reSHli /f ~1/~klJ tnoi:. /f tv w/Jl decrease.
7-2~
tn1
,j
incy-ease"
= f)
7.:lJ/f:
I
*7.21 The pressure drop, ap, over a certain length of horizontal pipe is assumed to be a function of the velocity, V, of the fluid in the pipe, the pipe diameter, D, and the fluid density and viscosity, p and p,. (a) Show that this flow can be described in dimensionless form as a "pressure coefficient," Cp = ap/(0.5 p Vl) that depends on the Reynolds number, Re = p VD/ p,. (b) The following data were obtained in an experiment involving a fluid with p = 2 slugs/ft3, P, = 2 X 1O- 3 1b· s/ft2, and D = 0.1 ft. Plot a dimensionless graph and use a power law equation to determine the functional relationship between the pressure coefficient and the Reynolds number.
V, ftls ------j------_._._.__ . _-_. 192 3 11 704 17 1088
1280
20
(c) What are the limitations on the applicability of your equation obtained in part (b)?
A
(a. )
I.Jp=
FY'1'1I
r +- (~
DJ
=:
Y=LT-
PL-1.
f; f< ) D=L
I
p/ theDre m J 5"- 3 ::
fh(
8~ ';,spec.. .hc if .fc>r 7T,.)
-rr.I = Che (.k
U'SIII'j
-==
A. P
/,v:z.
11 LT ~'I.5.J-er11 : Ap ML-1r-Z.
(Hc3) (LT-j
IV,," ==
TT2 us,;"
fJ VD
=~
:. -
(PL- lf ,2.)(LT-1){Lj
-
( FL.-Z.
_
T )
MLT S'lsiem :
I-VP::: (ML-a) (LT- (L).= H"L/)T~ I
,#--
Thus)
l-
Ok..
( /vi
L- '
7-
)
:. OJ::.
1 )
J1p
~
"
-IV i..s
2.
aJit
o.G"" Ct:ln b~
AL
unk.hOUJH
inc.-Iude d
_ .1 aStV2. - 't' Thw,S/ (,uhel"t. Cp ntlt11 bey,
.{u/lfc.;hoi1 1 a ..{(J.Gl::or 111 Tr, (/1 deS/re,) $0
(~V.b)
~.
Lp - c)(R~) ,..; tnt. pre~sur.e ~eff,c.I;' t (Co c>n !t.)
tll1d
Re 1lr-e.
Re~"d/J..s
7. 2 I
I
( CDn 'z. )
(b)
inl!. da.-/::a '7 J lie n )
fA $1';1
c;.
_ _.d~p---:----:-_ = 6,SfV'- - (6.5){2. V' .6. P
eli-II(
t€ ::
1;,r)
=-
/:.YJ:>
2
-
(~~)( ) r / (", ok 3Y (D. -M:..i
_
(DD
V
1...>< 10-.1 I~'),..
e)ATa~U/4-k-~
/01" Cp
valtleJ
1J,e 4 .. t-a
tire
v, ftls
Ll p, psf
3 11 17 20
192 704 1090 1280
Show H
C/J!fP.
Re
be /,u; .
Re 300 1100 1700 2000
Cp 21.3 5.82 3.77 3.20
.....
25.0 -r---------~---__. • Cp = 638 Re- 1.00 20.0 f - - - - - ' \ - \ - - - - ' - - - - - - - -..:g 0. 15.0 +---~'-------------l ~ () 10.0 f----~ .........~~-.. -. .. - - - - - - - - - - 1 cQ)
;g ~
5.0 1 - - - - ~----=----iiOiiiiiii,;~~---.... 0.0 +----,---...,.-----,.---...,.-----1
£
o
500
1000
1500
2000
2500
Reynolds number, Re
{hlL
?() u.Jt V /t(w
Y'e
c~ (C)
BII.sep
"11
1he
1nL ern;/y./ca I a pp/" c..
=
[1)
Re
Ust-d
y-e/ ... ·/-,~II.sh, i')
300
1h...
1'h~ rille" aa..f:a. ~ £$. (j)..) WI)u./~ (!)"J'J b4! 411,{
Re If 1101"$
< Re. 5...
-me
,j
~3'8
tI~rJ46Jej
"I... ,.,,-
ND1:.e.: 41-thPI(tlh
/d,o '"Sh, f
h f.I {'II ~/Y
rA 11'i&.
200D
efu£t/~11 I71JfhC
he.. VI//,&I ""b/de -fh/J rtJJ19€1 re1/,(!-h .s-hflu/I( /lot- I:u.. ex. t:/'"#..fJdlal:e' loe,~"1f 1he rlll'lfe ()f d~ta used.
?-1.7
7.22 The height, h, that a liquid will rise in a capillary tube is a function of the tube diameter, D, the specific weight of the liquid, y, and the surface tension, (J. Perform a dimensional analysis using both the FLT and MLT systems for basic dimensions. Note: The results should obviously be the same regardless of the system of dimensions used. If your analysis indicates otherwise, go back and check your work giving particular attention to the required number of reference dimensions.
Us/;"
P/... T ~fjstem : -= L .D == L
~= PL- 3 CT= /=t-FrtJm The P": the()rem I 1./--2 = .z pL'tertnJ 8'1 InSfec..ttDII) for iTj (~ni4i/1liJ, -l) :
-i
I
11i=:f /.5
OJ,V'Dt/S!y dt'/nfnS/onJess.
;r
( cDnt:olnln,
tlnd
rr) : FL -/
{= ~ (a-a;.. ) USlfl'~ ULT ~'fJkm .-1,. L D-' L
=
/.J/tJ,o~,), '!here
~= ML- 2 ;-2.
a.fftll Y'5
f() be
0-=
M
r-2.
dl;'fI1S/~~)
.3 reierel1((!..
tJJ1/Y
,t reierFJ1ce dlmeflslol1-5 dYe act",ol/!I Yt'$tlIYfd ( Land M T-z) 1-0 de5cr/~e 'the i/qnqbJes, 8,/ inspec.i't!)"n,l hr 77; Csfle 41"ye)
CU1C,(
~Y'
tt=:l. I .D
7Tz
(~ittIHI"" ~=
0OJ)2.
d- (JJ1t1 (j) .' =
/VI T - 2. _ (ML- Z T-z."j(L)
7-28
_ l
H DL ~ TD
7:2g
I 7.23 The speed of sound in a gas, c, is a function of the gas pressure, p, and density, p. Determine, with the aid of dimensional analysis, how the velocity is related to the pressure and density. Be careful when you decide on how many reference dimensions are required.
(?;f)
c= f'
fJ:l: FL-'2.
/flf/J()tl9i1 1he~ ~'p'"r.;
to k
reference c/Jmfl1~II)IJS
3
(wjJlch would I~c!tcat~ thac rhl'r~ tlre ilo 011/'1
2
(LT- 1
re/erel1ce. d,rn.fI1SII)II.s
aciU4//'1 resulted since
t=
(rt.- )(LT
.'
~ '..0 1T. = ~
Che~k
U51J1J
_,/2-
(1-T
_I) 2 r~ F
(LT- 1 )2.(M/..ML-'7- 2
onJIj /
fL' term)
~= C 1>
a
CtJHstani.
c=
c=
where.
C, JS
a.
U
-2,1
-,)-~
L / ( L T d : riJL t) TO
PI..-'2.
N L T.'
where C J~
aYe
3 -2 = / Pl' term re~ulred. ==
-p
J
FJ...- 2 )
(/11&/
2
Thu~ fr~m 1Jte pl' 1Jzel)~m)
8'1 In~pecl::l{;11
foss/Me. pi I:eyrns);
Clff
CliI1S,j;qHi
(4= rC ) .
3
):
MDLtJT
D
, OJ<
7. 2. 4 H r
-'I-
I. 2.L/ The pressure rise, t1p = P2 - PI' across the abrupt expansion of Fig. P7.24 through which a liquid is flowing can be expressed as / PI
o
where A 1 and A2 are the upstream and downstream cross-sectional areas, respectively, p is the fluid density, and VI is the upstream velocity. Some experimental data obtained with A2 = 3 1.25 ft2, VI = 5.00 ft/s. and using water with p = 1.94 slugs/ft are given in the following table: AI (ft 2 )
\
0.10
I
0.25
\
t1p (lb/ft 2 )
I
3.25
I
7.85
1
0.37 10.3
\
I
I
0.52
I
0.61
I
11.6
Vi ........
Area = Al
I
12.3
Area = A2
Plot the results of these tests using suitable dimensionless parameters. With the aid of a standard curve fitting program determine a general equation for t1p and use this equation to predict t1p for water flowing through an abrupt expansion with an area ratio A 1I A2 = 0.35 at a velocity VI = 3.75 ftl s.
tJf=
FL-
p;.~m iJ,e
2
A;L::: L
AJ == L1.
p"
f::'
1.
fJ1e~l'(tn)
s- 3 =
2.
AE _
ChecK
.
USJI1!j
MLT: ~
I~~
-
{;y 7T2- (tlllftli4lrlj A-,
M L-/ T-2. tiff"
Az
tI~/;'j
-the daia.
-
J) 1-
M~L~TO
):
J.
1JI/el11
,+
~/I()JI/.5
tha.i
LlPR~2.
o.ofp7
0,",2
0.212
o.2J~
A,/Al.
O.ORO
O.2()o
().lq~
6.~lb O.4-F~
( r!41/ i)
V, -= L7- 1
-.
~;
I r,
"2.
.'
Fi..-2. (FL-IfTl.)(Lr-,)l-
(1-11..-3 )(L r-
r
ffrmJ '("e~tI/¥'et1.
pi
B~ /,,~pec.tlbJl /;;r ~ (t:£;n.fail1lhg fl f)
7r.= I Iv;:t
FL-'1
0.2.5'4-
.
~I<.
0.3 . 0.25
I
I
·····--i------~----I
.,- 0.2.
I
____
IV
0.1
--f-.- · ~I-.-
..
·7------------------------------- -
i:J... 0.15 .
0.05
.__ /
i
,-----
I--I-------~------- ------------L------------
I
'
o -f------!------!,f----+,----+-----I, o 0.5 0.2 0.3 0.4 0.1 Pi2
The
(!urlle
7'V1J11
7)r~
dl't:ltvn
1"'" 111 e ~ ylt ph
.e~IiIt.I:I"1f
~ / J0 /,4/ ) 1. + I. 0 7 ( 4'z. ) fl-j ~ - - . {:4z.
7- 31
-
0, 0 103
7.2.5 7.2;-
A liquid flows with a velocity V through a hole in the side of a large tank. Assume that V = f(h, g, p, 0")
-A =L
Fr~m 1/1e pt' the()/"em I
B'1 iYJS;eCt./~H
3.13
4.43
5.42
6.25
7.00
h (m)
0.50
1.00
1.50
2.00
2.50
Plot these data by using appropriate dimensionless variables. Could any of the original variables have been omitted?
where h is the depth of fluid above the hole, g is the acceleration of gravity, p the fluid density, and 0" the surface tension. The following data were obtained by changing h and measuring V, with a fluid having a density = 103 kg/m3 and surface tension = 0.074 N/m.
V= LT-'
V (m/s)
t=
;.; LT-2.
=2
p" .feJl'/'J'1S lOr 7T; (~/,-t(Jlnl.l1j V): S- 3
.Y- _
f"f'ifllY'fd.
/. r- I
77f = '{if {t- T -~ 'I,. (L) TT2. (eon /rJ;'lInfj t IIlfd rr) .'
/=r;r
/-3-1.2. ==
IT: :::z.
Check
~
-
1. 0 TO 1-
(Ft..- If r2.)(/.T-'1.)(L)
(/
tAStnj
0":: /="L- 1
FL-"'r'Z-
..:. rOL o TO
r-L-/
IML- 3j(LT-') (L J_2. ==
/l1"LD TO
M 7-2.
v V3..A .. ;::;,,. fit€.
data..
f#*z.lrr Vl'fih ~
-v
3."3 J )(.104-
-the.
13.3 )(10 t'
2.'1.3 X 10 If
I. '1-1
/. If I
/. 'f I
S3.0
x 10;
I. ifl
J' 2. 'I x If)
'I'
/.Cf/
I· __ .-•• j
-: -j
-
..
:~~=J.:
.
I.P~--+-~~~~~~~~~-+~~~~~~~~~~
rjT
The 9rr~ph
~J~en:
411d
vtlnil6/es
+Able shDw- thaI: VIVjh ~ anti
rr
~,,/d hI/lie
I{J
l~clepfndfHi elf fJ#J.)./rr. Thus"
hfen ~mdft-c/.
7 2~ 1 7. 2 ~ The time, t, it takes to pour a certain volume of liquid from a cylindrical container depends on several factors, including the viscosity of the liquid. (See Video V 1.1.) Assume that for very viscous liquids the time it takes to pour out 2/3 of the initial volume depends on the initial liquid depth, the cylinder diameter, D, the liquid viscosity, J1., and the liquid specific weight, )'. The data shown in the following table were obtained in the laboratory. For these tests = 45 mm, D = 67 mm, and), = 9.60 kN/m3 . (a) Perform a dimensional analysis and based on the data given, determine if variables used for this problem appear to be correct. Explain how you arrived at your answer. (b) If possible, determine an equation relating the pouring time and viscosity for the cylinder and liquids used in these tests. If it is not possible, indicate what additional information is needed.
e,
e
t(s)
II
17
39
61
107
IS
23
53
83
145
((1) ~ jA) ~) w2 ])=L ?-= FL r
i
=
J..=L FYtJrn -the
8 0/
pt' meoYern
-/'0;- 7J;
Jh~Fec..--/-tt:;"/1)
77: = /
Check
t- tr [)
D
(!=L-~
)A--
M LT
{,/SII1,j
5-3 == 2 p/ ffrms Y'efftl.,yed. ( CCJI1.J-t/;l1ln!J 1:) (T)(FL- 3 )(L) . pOLoT
+;r-D
~ 'I ~1-e1l1 : (T)( ML-2.
=-
~
.
Wh/ch
Is
C)
dt1.,~
h~m
(M
r-2)(L) _ Me. L TO L-' /-1) 0
b v /o{.{ 5 ~ dlmeJ1 j/lJnless . Th"s) t-tf D - cf (; ) )<-
For -th(!
r)
.J/v(n
.1--
75-
Eg. (/)
I.j.S-/TIl /iH
'7/fHhf1
(/ )
== o. Is, 72 ( a e£)115-kl 11+
(!()IIS I-oll-t ) .
IT-
10 1/005
the dt'l.:ta. 9 I v~ Y1
i:lr.D
?= 7- 33
:
7.2Co
J ?i77 Since ~
J~
? 70
871f
-e.~sen1:.IQ.IJ.!J
1ne eJt-peYJin en-ba. J da.,i-a.. a.. p pea y b be a6)Yrec.i:.
375
g 72
CtPns1::ant ~lIey 1he. Y'an~e ofthe VtlYla6/es 1/6f4 ~Y' the pYtJ/;/em
Lb) The
So
i: = WI fiJ
NtJ-I:.~
.for :L)B
-I:- I~
Tha:i
1.31. h
seuMds
f.
JJ t'n
tJl1''f...J
tJ{ lV,s//Wt~
r~s.f-Yl~~d ~zud/~JI /j 0'111 1/a.I/d ~;0:::~. (, 7t,; [) = ~ 7/111-*1) and 6- ~ 9: /pt}4eN//fn 3 Uil
1ha.t
t>.f-
1J1e
f)US
J/JI'+U!J' /
I/o
iwme be,n.J pour@d.
7- 3 '-f
n
7,.;2.7
I
7.27
The pressure drop per unit length, Apr, for the flow of blood through a horizontal small diameter tube is a function of the volume rate of flow, Q, the diameter, D, and the blood viscosity, jl. For a series of tests in which d = 2 mm, and jl = 0.004 N's/m 2 , the following data were obtained, where the Ap listed was measured over, the length, e= 300 mm.
Ap (N/m 2)
Q (m 3 /s) 3.6 x 10- 6 4.9 X 10- 6 6.3 X 10- 6 7.9 X 10- 6 9.8 X 10- 6
1.1 1.5 1.9 2.4 3.0
X X X X X
104 104 104 104 104
Perform a dimensional analysis for this problem, and make use of the data given to determine a general relationship between Apr and Q (one that is valid for other values of D, ~ and 11).
/' =- FL. -2.r Fr()tn the ft' fhe"Yftn
J
/.f - 3 = / pi tfym
81f In;pec.tlon ~ ~ LlAo D' _ I ~ cP
res
lilyed.
(Fr 3) (L)
Y
(PL-'Z r) ( L3 7-')
MLT:
.Ll~ Dif ~Q
(ML-1T-Z)(L)'f
::. .' OJ<.
(M'- -lr-I)(Lsr-')
S/nce -there IS
wheve
= /,33 x a J1 d fnere ft, re
U&ln,9
the. da.&
IH
L1i1 Dlf
/0
'fo." ~, 7
40. /
itO. If
'fo.7
-~ If)
.6p
Q
"7.28 As shown in Fig. 2.26, Fig. P7.28, and Video \'2.7, a rectangular barge floats in a stable configuration provided the distance between the center of gravity, CG, of the object (boat and load) and the center of buoyancy, C, is less than a certain amount, H. If this distance is greater than H the boat will tip over. Assume H is a function of the boat's width, b, length, and draft, h. (a) Put this relationship into dimensionless form. (b) The results of a set of experiments with a model barge with a width of 1.0 m is shown in the table. Plot this data in dimensionless form and determine a power-law equation relating the dimensionless parameters.
e,
e,m
h,m
H,m
2.0 4.0 2.0 4.0 2.0 4.0
0.10 0.10 0.20 0.20 0.35 0.35
0.833 0.833 0.417 0.417 0.238 0.238
iii FIGURE P7.28
ft= i(b;J-)~) (r"l?1 the fL' 1h eoyrm) '-/-/ / I? .$pe c. f,o n "
=~
p,'
krl71s rP~l4;ye,f. 91;1
1=~(i)4)
A/I
~.f. 1h~ p/
krms tjye obv/ousl!:! dJ mens/on/esS. (/:,) /="0". the dL.J:.a fivel1 1-4bltl"ktl va/U(ls for H-/b) l1/b J ~t1~ a y~ ShbtJIlt btl ow. I h/b 0.10 0.10 0.20 0.20 0.35 0.35
fib 2.0 4.0 2.0 4.0 2.0 4.0
H/b 0.833 0.833 0.417 0.417 0.238 0.238
,----------------~--------
..c
--
I
0.8 0.6 0.4 0.2
I
.{Ib -
H/b = 0.0833 (h/br 1OO
"-
"
--
--
o
I
0.00
I
0.10
0.20
i
0.30
0.40
h/b
~------------------------
All
J~sf"cA·,()'1I
61-
1He.H d"i-A. reI/eli 15 1H1l.-t "~t iI-ep~nd fJ~ "l./b; (.f',,) 111-t "Sl1l?1f Ua luE' Is obl-ttll:'~,( ICY' cI,'Flerf'''' t (/a {wtJ$ IJ I- .l.
H-/b d"e.s Df. H-/b
Ii, . Thus,
41'1 d
.e$
4{)t1f
t;,~ plDt 01 ihe da.ttl.)
u,4,. .f/D YJ
1+
('-Pc.)- I."
h = 0.0933 1:
lAS
0
U1~
0...
pllwev- law-
7.2,Q A fluid flows through the horizontal curved pipe of Fig. P7."2.Q with a velocity V. The pressure drop, !1p, between the entrance and the exit to the bend is thought to be a function of the velocity, bend radius, R, pipe diameter, D, and fluid density, p. The data shown in the following table were obtained in the laboratory. For these tests p = 2.0 slugs/ft3 , R = 0.5 ft, and D = 0.1 ft. Perform a dimensional analysis and based on the data given, determine if the variables used for this problem appear to be correct. Explain how you arrived at your answer. :;£;::;ft2 )
v - >L---_I--.
-
I, ~:~ I ~:~ ! !:~ I ~:; 5 - 3 = 2. I"
130/ ItJ$;e~l1)
(C4'nt:III#lny
.(;;y
-u;
..6p _ USII1?
1::('1"/71 S
ret"'/r(#~;
tip) :
(Fl-
Z)
(~L -IfT1.) (LT-)
1Tj = ,,4 V 2. C,heck..
FIG U REP 71. 2 ~
•
Prt9m 1Jte pi thepl'em)
~-,,--""'---
Z
NI..T s'fs/-em . ~ f'
_
( M'- - I T -I. ) (1'11..-') (I.. T-tj
Hr 1Tz. (~n£-~iH,il7 R
~/,
" Z
PI1P /) ) D
1T;..=R which
iJ
~bv;'o{,/f;,/~
ThtlS)
diM fI1SIOII/P.5,S_
/jQ -
;? j/ 2.
-
~
'f'
(-#) r-
(I)
D _ (). /.11:: _ /
--s
;; t.
(j)
tv J'"'h-1 ~
~1?.s-l:-t1l1i .
~
CPH St:qH
~«/eve~
/0 /J"WJ
Thu51
-IrzJ
/ 1:';- f,:; 1/() f().S 4 pi·, -the
jJYD/,ltm tJlre
n~i
&f'f'ec.-F· I
7-37
J.
#1" the
C4? 1'1~ brJt i
'/he
i
C,rls&:tn#
7. 30
I 1.3 ()
The water flowrate, Q, in an open rectangular channel can be measured by placing a plate across the channel as shown in Fig. P7.30. This type of a device is called a weir. The height of the water, H, above the weir crest is referred to as the head and can be used to determine the flowrate through the channel. Assume that Q is a function of the head, H, the channel width, b, and the acceleration of gravity, g. Determine a suitable set of dimensionless variables for this problem.
I-
-I
b
V
l
Weir plate
+(
Q :: H; b) ~) h:i L d = i 7- 2 p/
the~yefYll
'1- 2
p/ +(ftnj In 'spe d-/~/f .J:,y- 10 (t...~ntot'J11 n, 4'):
4J
'::: 2.
:=.
HSIz J IJz. tr;z.
= -Af+
7- 38
ri'~UJypJ.
L3 T- 1
(L ) 5""/z. (L T-z)
712.. ( CAIt HI#/l19 /;)
T H
1)2-
7.
3/
I J.51
From theoretical considerations it is known that for the weir described in Problem 7.30 the ftowrate, Q, must be directly proportional to the channel width, b. In some laboratory tests it was determined that if b = 3 ft and H = 4 in., then Q = 1.96 ft 3 / s. Based on these limited data, determine a general equation for the ftowrate over this type of weir.
7.30)
(j)
-
H5"/z..
where
C
J:.s
tt
J 'Iz.
(/)
- c (J;) Th (,lSI -ft,y he dd4. 9' Ilf)1
CtPn s ftvtt f.
/. 91, if3
-
c 1h e
({J
Jen&Y'4 / egIi a:t:zPI1 == 6. 5"'1t b
Y&ft3
]
J.5
.
7.32
l 7.31 SAE 30 oil at 60 OF is pumped through a 3-ft-diameter pipeline at a rate of 6400 gal/min. A model of this pipeline is to be designed using a 3-in.-diameter pipe and water at 60 OF as the working fluid. To maintain Reynolds number similarity between these two systems, what fluid velocity will be required in the model?
Pol" Re~I'J~/ds num btl' s/m;/tlr/f!:J;I VhH D.-m
:::
VD
Or ~-
(I)
and I '-I. 3 I-t 3
S
+hen
Thus/
v= ff-()m
\I. m1
r
- 2.02
(sft) z.
ft s
£r. lJ )
= (1.2/ (
5 X- /0-
If. 5" x /0 - 3
{t; (3 ftl (2.02 £t)
if.) ( ?z. ft)
7-t.io
oS
-
-z.
~.52 ;C/o
7·33
I 7.3 3 Glycerin at 20°C flows with a velocity of 4 m/ s through a 30-mm-diameter tube. A model of this system is to be developed using standard air as the model fluid. The air velocity is to be 2 m/ s. What tube diameter is required for the model if dynamic similarity is to be maintained between model and prototype?
FOr d'lntlliJ/c.. sim i /tTr/fy / i'he Reyno/cis numblY' rnu.st. be
the
S tim e
aJ.1d prototype.. Th us J
(;,r mode I v~
Dtm
~ So
Vb
-
v
thai
DhY\
-- V-
--
~
D -
V
~
CJ.
73IP ~/D
(I, '16 XI ()-!J'"-;. 2. ) (tf f1 ) ( t. / 'I X /0" 3!l!1. 2.) (z ~-) .s s
.. 3 I"WJ
..
7-Lf!
O. 7 3 fe,
/'WI I'm
(O,/)3~/1f1)
7.3Lf
I 7.3 if The drag characteristics of a torpedo are to be studied in a water tunnel using a 1:5 scale model. The tunnel operates with freshwater at 20 °e, whereas the prototype torpedo is to be used in seawater at 15.6 0c. To correctly simulate the behavior of the prototype moving with a velocity of 30 mis, what velocity is required in the water tunnel?
FOr d'lnQ"',"6 s/m; /tlr/ ty I the Re'l,,~/tlJ number must be fhe. SeNne hI'" m~r/e/ ClI1c1 prtJioiype. Thus J
'i!l
Vmr D":} _ ~
1/
~=-V411D /)?1
5 t'nce) ~ -V
(Warey @
V
~
~()oc ):::
V /.
()Olf
;(./o-ID 11)12.6
(SeplVaier @ 15:boC ):: /./7~/()""6-111~
(r;6/e /.~)J lind
.D/])/M =5 I;t follows thai
_, ,.",,2.) ( V::: /. 00 II- x. I/) ""5 M
(/.
17 X IIJ -6 ' ; 2. )
(7db/e B. 2) )
/2 Cf
7.35
I 7.J5 The design of a river model is to be based on Froude number similarity. and a river depth of 3 m is to correspond to a model depth of 50 mm. Under these conditions what is the prototype velocity corresponding to a model velocity of 1.2 m/s?
Pi:;y.
I=;."ude
17/,(1n j,pr
V.;m
V!/WI ctw.
Wheve
s I'm ~ J~ Y'if!J )
=
V
V#- tJ..
. the fflAIC;
d
I~
v: I/f{ flirt d4H Cll1d
w,'1h
V=
! :: j'WI
Vt
~
depth .
7h (.{ S
J
~
= ~(. ~) = 0. O.5() Z 5 Mt
7-'13
•
1.30~
7.3(0
I 7.3 ~ For a certain fluid flow problem it is known that both the Froude number and the Weber number are important dimensionless parameters. If the problem is to be studied by using a 1: 15 scale model, detennine the required surface tension scale if the density scale is equal to 1. The model and prototype operate in the same gravitational field.
t=br
dlfnqm/e- S;lru'/f/I";'+!J)
Vim
V
::"
VjJ-
V?rm )., ay,d
fm, V'; )~ 0;",
To saflsly
F;.t:JlJ de
_ -V V~
et..,d
=
f- V'-J· cr
HumhtY' .s/~/'Jlly/-J~
f¥
inere. -h,re !Dr Webev nlAmher a;, :: ~ (~)~ & ()
Thus) w'rth JIWI
(J
/1. ~
I/I~
( w~Th
a= J~ ))
.5 '/ mJ'lt:? Y/-f!J
" ~(-r)f=~(;r
V
f
and
~ /f=
IJ
7. 37
I 7.37 The fluid dynamic characteristics of an airplane flying at 240 mph at 10,000 ftare to be investigated with the aid of a 1:20 scale model. If the model tests are to be performed in a wind tunnel using standard air, what is tht" required air velocity in the wind tunnel? Is this a reaHstic vefocity?
For d ~J14m"~ sim; 14;i-lJ,I -the. ReYfJ()/d5 nUn? hiY' rn us.f ~ the. SaMe. #>Y' model tlnd pY'~ioi'J.pe. ThUS)
(/)
jJJ1Cf)
/-~ 3. 5'3Lf.x JO-
A = 3.711- x JD-7 ""
),,/1,,"
'h,1I<
-
No
I
=1..0 ,+
3750
J
7
~ ftl.
)
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I
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)
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I s ut..s h J
~ = 2, 3 r~)f) -3 slui'
+i'
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(TaMe C.I) (~kJe 1.7)
-thai
mph
If / s not 4
rea//stti, ve/()c/f::; - muc.h
7-'1s
fDa h/811 .
7.38
If an airplane travels at a speed of 1120 km/hr at an altitude of 15 km, what is the required speed at an altitude of~ km to satisfy Mach number similarity? Assume the air properties correspond to those for the U.S. standard atmosphere.
( V) Th e
~ fRfcl
C
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(13 ~, /' Zo)
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s" ~o
°C
+ 273.16
-
J./&',7 k
(Table (,2)
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8 ~~
(4'1.
a,
t.
C
3b.9,/ t:JC f ,;J.73,)~ I~'*1Wf oj I-,I",dt!.
=l/-r;-.11-0-)(-2.-g~-.ri-.d---)-(:;.-/,-,7-k-) =
/;-6."", Ci 11 d
(I)
/ ~ ~"'" olh I-ude I
T= Thus)
(V)
-
I<.:j, ~())
til y.)
T= aHti
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() f:
if
-Ie#- '/<
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r------------------------C(j ~ =. . 1(/'tfo) (21,.1 ~J' ) (13&,21<) :: J 1Il; ./< ()~.,.,
From £1' (J) ~.h~
308 ~ s
~
308=( ~ ) 1120
=
.2.'15"~ 5
1170
~ hr
(I
{M( ) h ...
7.t;o
I 7.40
The lift and drag developed on a hydrofoil are to be detennined through wind tunnel tests using standard air. If full scale tests are to be run, what is the required wind tunnel velocity corresponding to a hydrofoil velocity in seawater of 15 mph? Assume Reynolds number similarity is required.
For ?e'fn()/dJ number s;m "Jllr/f~ Vh)1 .i~
=-
~
J.
IAJhere
J
rt 7/
I~ Some cnoY'pcten'Sf,c, I eYJJth
of the
hfjdf'Dfoi /.
Thus)
~= -u" V
m1
a Ilt/
tv
/"ht
..,e /11tn =I -z4,
7:l Y
1.
),.",
V
(.,cull scale kst) _ fl,5"7 x /0-If !j'1..)
-
(/, z' x
7-"17
10-5" .ft'&.)
.s
(/5 mph)
7.'1/ 7.41 A 1/50 scale model is to be used in a towing tank to study the water motion near the bottom of a shallow channel as a large barge passes over. (See Vidl'O V7.7.) Assume that the model is operated in accordance with the Froude number criteria for dynamic similitude. The prototype barge moves at a typical speed of 15 knots. (a) At what speed (in ftls) should the model be towed? (b) Near the bo.ttom of the model channel a small particle is found to move 0.15 ft in one second so that the fluid velocity at that point is approximately 0.15 ftls. Determine the velocity at the corresponding point in the prototype channel.
Ca)
(1)
:51 ~ C-e. f-rom
t;;~. (I)
V"" -V
S 0 ~a.-i:-
V=
V1~
-1
= V~O ~
'/sO (0.15 ~) =
/.
O~ V
7. ¥2
J 7.4'Z.
AI: 40 scale model of a ship is to be tested in a towing tank. Determine the required kinematic viscosity of the model fluid so that both the Reynolds number and the Froude number are the same for model and prototype. Assume the prototype fluid to be seawater at 60 oF. Could any of the liquids with viscosities given in Fig. B.2 in Appendix B be used as the model fluid?
As
dISCIISS(?c/
/n Sec./;/()11 7. e. 3 if) rnQJ~ tal h both ReynoJd.J
nurn ber- 411d Fr""r/e number SlInllq n'7-;; -V ].
;
: (AJ.) ~ 3
) --;: ( /, /7 ( -!.. Jro If. 6 J.
lues tJf y
!VI).
The
aYe
a/I much
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?-'fr
- if JD
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I, Z't/lds tf It/t'n /It Ine. re~ lJ t'rer) va/ue. the
Fi9. B, 2
7. '-13
I
1. Lf.3
A solid block in the shape of a cube rests partially submerged on the bottom of a river as shown in Fig. P7.43. The drag, q]j, on the block depends on the river depth, d, the block dimension, h, the stream velocity, V, the fluid density, p, and the acceleration of gravity, g. (a) Perform a dimensional analysis for this problem. (b) The drag is to be determined from a model study using a length scale of 1/5. What model velocity should be used to predict the drag on the prototype located in a river with a velocity of 9 ft/s? Water is to be used for the model fluid. Determine the expected prototype drag in terms of the model drag.
r1J =.f. ( d.) ~)
(a..)
V)
fJ
J)
i) == f d= L -l:: L V= )..7- 1= t=L-lf r /=1""111 the fi -fh-e~Y.flY1) ~-3 = 3 p/ 1-~"I?1~ r~1"I;p~, 1
'1-
()..
(b)
d"n·.eI1$/~111J1 an4/151J
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and
1 (~) ~)
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=
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prDf IJ -t:!:/ pe ::-
.JL V8~
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v:1IfJ--
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7V
= Yfr9~
The fr'~dl~-I-JDn elua.-t1fPn J".$ Jj = ~m1 fV1. hl !'WI V,,; h~
7- so
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-r:
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7. /1.11-
rtit)
7.44 The drag on a 2-m-diameter satellite dish due to an 80 km/hr wind is to be determined through a wind tunnel test using a geometrically similar O.4-m-diameter model dish. Assume standard air for both model and prototype. (a) At what air speed should the model test be run? (b) With all similarity conditions satisfied, the measured drag on the model was determined to be 170 N. What is the predicted drag on the prototype dish?
E~. 7.19) Re'lJ1fJ/ris number
I=rtJhI
~ lJ....
v .... whey!.
D
I.J
slm; 14r1Tfj
.
IS
Y'eglJiYfl'i. Thus,
VD
-
-
-z/
the dt'sh clJtlrneteY".
It
~//"WJ 1f14t
~:~.12. V 7/ ~ 1m
and
w/n, ~
/v = / ~".,., = ( ~.2. J.j IYYI )/~~ ~-m) = ( ~ h Y' /WI
(g~ ~~
hI"
)2. 17D tV
(tVote Th_+ I!) :.,/j1Yr! Ii-, -IJ,ti.s pJ'tJb/eWl) 5/~(e I....fJm The. CJJtI(ltf,,,'yj of Re'lI'1()Jds nwm My S/ffll·/tll',.-ffj) v'll v,; = D; ID~ ThiS IS nrt
rrue
111
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7-S/
7. '15"
I 7.45
The pressure drop between the entrance and exit of a 150-mm-diameter 90° elbow, through which ethyl alcohol at 20°C is flowing, is to be determined with a geometrically similar model. The velocity of the alcohol is 5 m/s. The model fluid is to be water at 20°C, and the model velocity is limited to 10 m/s. (8) What is the required diameter of the model elbow to maintain dynamic similarity? (b) A measured pressure drop of2.0 kPa in the model will correspond to what prototype value?
Foy //()W In a. c/(Jsed C6ndtJ/f; De,PtfldfHt pi
I-e,m
=
f (-i', ;
T-;r 1J,/,J fJIlY/-lcu/lJr Py()b/fn? the deppndent
7! )
J
(Eg. 7. 10)
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1J.fJ
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r
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r
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7-53
7¥6
I 7.4 ~
For a certain model study involving a 1: 5 'scale model it is known that Froude number similarity must be maintained. The possibility of cavitation is also to be investigated, and it is assumed that the cavitation number must be the same for model and prototype. The prototype fluid is water at 30°C, and the model fluid is water at 70 0c. If the prototype operates at an ambient pressure of 101 kPa (abs), what is the required ambient pressure for the model system?
for naurie. number ~/m;J(JI'I+'J)
V-*,
-:
VJfW\ "i~ $0 -hrAt
V
ViT
(WIt;.. j:'Mt)
V; V!;
OJ
=
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(-p". - ?v- ttf'l = (1-;. - p~ )
-t(hI\ V""l. It
.f.o J/ow-s ~t
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=
if V4.
& ~
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J.
'""1. ll) (-fr - P... L... =
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use
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(AJr.,. -,lJv
)
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(a/'o5)
7-51(-
(Z )
7.0/-7 7.47 As shown in Fig. P7.47, a thin, flat plate containing a series of holes is to be placed in a pipe to filter out any particles in the liquid flowing through the pipe. There is some concern about the large pressure drop that may develop across the plate, and it is proposed to study this problem with a geometrically similar model. The following data apply.
•
Prototype d-hole diameter = 1.0 mm D-pipe diameter = 50 mm ,u-viscosity = 0.002 N . s/m 2 p-density = 1000 kglm 3 V-velocity = 0.1 m1s to 2 m1s
Model
d=?
D = 10mm ,u = 0.002 N . s/m2 p = 1000 kg/m3
V=?
(a) Assuming that the pressure drop, Ap, depends on the variables listed above, use dimensional analysis to develop a suitable set of dimensionless parameters for this problem. (b) Determine values for the model indicated in the list above with a question mark. What will be the pressure drop scale, APm/Ap?
•
fa)
Llp=/{d;D)t;f;V) J.p=- p L-2. do:: L D=- L f<:: FC2.T Fr~rJ1 the. pi theoJIPm, 6,- 3= 3 pi dlinen.5li)l14/ ~1')41'f.si.s ';jltld.s
~
fJV:z.
(b)
FIGURE P7.47
+--e t-rnS
t= FL-'tT Y'e~UI rrd;
'2.
V= L T-
tI ncJ
I
a..
= r~ (.!L fX!» D) ~
Foy ~/~,I~Y;+!J)
cl"...
cI D~:: 15 w in 1n ~ fA a,./:a
cJ ".,... :
0= 0
O. lOO""'hH
d
~ } J0 wS
::
~-
ThtA. i.
V
D.50Dlf fr~ ID.O~
7. 'IS , 7.4 8 At a large fish hatchery the fish are reared in open, water-filled tanks. Each tank is approximately square in shape with curved comers, and the walls are smooth. To create motion in the tanks, water is supplied through a pipe at the edge of the tank. The water is drained from the tank through an opening at the center. (See Video V7.3.) A model with a length scale of I: 13 is to be used to determine the velocity, V, at various locations within the tank. Assume that V = f (e, ej , p, j.L, g, Q) where eis some characteristic length such as the tank width, j represents a series of other pertinent lengths, such as inlet pipe diameter, fluid depth, etc., p is the fluid density, J.L is the fluid viscosity, g is the acceler-
e
pi thf'ore", dl/neI151;P/14/ anll/';Jsi..J Fr()h7
'f), e..
- ~
q Thus
I
fj/~/d.s
T
Q z. ) }.:r:J )
1h~ ~irn;Jtlf,'f!1 r-egulrernellfs
)/.'/YYI :. ~.
J-m. Clhd
7 - 3 ~ If
I
(1.1.:
V.12. _
ation' of gravity, and Q is the discharge through the tank. (a) Determine a suitable set of dimensionless parameters for this problem and the prediction equation for the velocity. If water is to be used for the model, can all of the similarity requirements be satisfied? Explain and support your answer with the necessary calculations. (b) If the flowrate into the full-sized tank is 250 gpm, determine the required value for the model discharge assuming Froude number similarity. What model depth will correspond to a depth of 32 in. in the full-sized tank?
are
CD 2.
(J)'ht2.::.
1-; d-m J. :>J
"
pye dt C.ht)J, etu.~/;iC) H 'n2-
V;::.
_
Q
I.J
V;111'\ oj 1M2.
-
QM-i J:r()(n -fhe Jllst sirn//4ra-~ re2t1/remetit (j)nn ==
Q
0
-e
~ )~ ft??1)1M1 I-
017;"
0
~ J.
Ir~pt 1he cSec.,r1JJ1d SI'm',/4rl '-J-!1 Y-l''jtl l remeJ1.i With
0"" Cp
= (b .J-
Fz.
5//?ce --these iwo yeg u l'remett b aY~ / 11 et)J1rJ/~ t I'tIoJ/~ws /n(i= 1he s;'m;/4f,'-!:J V-e~tllremell+S CtU1J1ot b-e 5~'= 15h e'd. 1/()
(COI11: ) 7-510
Ge (!)me t Y"J C.
SI
in / I (j rl f!t r-ez fA I YeJ .1 ""'" :: !:.i. I.~
j.'#H _
r
() I'"
SO
-fn~1:
). ,R.~ j..
-t;,f).-t
_.-L J3
a/I /eY1;1hs sca.Je as 1Jte lel1f th S~tJ.Je
(d efth )17J(}del ::: ( =
IF )(depTh ~y.-b,-t!Jpe (/~) (:12 in.) = 2 ¥-b I
7- 5 7
In.
.
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7.49 The pressure rise, Ap, across a blast wave, as shown in Fig. P7.49 and Video V11.5, is assumed to be a function of the amount of energy released in the explosion, E, the air density, p, the speed of sound, c, and the distance from the blast, d. (a) Put this relationship in dimensionless fonn. (b) Consider two blasts: the prototype blast with energy release E and a model blast with 11l000th the energy release (Em = 0.001 E). At what distance from the model blast will the pressure rise be the same as that at a distance of 1 mile from the prototype blast?
/J
(a)
f
fj ==
P= .f ( E') f)
Fyt)/I1 the p/ theorem a
I
Ii FIGURE P7.49
c)
d)
f:L-'I T ~
S - 3:: 2 PI
d /me"s/tJI'/4 I qnal'tSls !tIe Ids
..1P
= CP
jJc2.. (j,)
f ==
E';' F L
F L- 2.
f--d----!
iiI" ~/~aanry
I
=
E"... ~h)I C~ d;, /4I/'fh
~::.! J
d
E c.. "Z.d 3
f c~ =- C I 1:
IE : : ~. I 0"., = (c,. OtJ
cI ::
tfll1 d
3
d 11M
If
'=~ cl 3
J:::
;n,
E".,
(E) /,c2.d 3
()t>
I )
(I ;tHi )
3
= 0. /OtJlYtZL,
Ju/'th 1'hi.5 ~/ml'J~rl'+~ re~/n'l"ernfl1t p/ed/~fro'"
.I2Jua tliJ'i I~ .Ll -Pht1 =- Ap fJC'" f~ c,,:
-
-th~re
hre. LJf~= L1
d
P
~.:: ~, 100 11111.:
?-S8
7.
So
I
7, !;-/)
The drag, qv, on a sphere located in a pipe through which a fluid is flowing is to be determined experimentally (see Fig. P7.50). Assume that the drag is a function of the sphere diameter, d, the pipe diameter, D, the fluid velocity, V, and the fluid density, p. (a) What dimensionless parameters would you use for this problem? (b) Some experiments using water indicate that for d = 0.2 in., D = 0.5 in., and V = 2 ft/s, the drag is 1.5 X 10- 3 lb. If possible, estimate the drag on a sphere located in a 2-ft-diameter pipe through which water is flowing with a velocity of 6 ft/s. The sphere diameter is such that geometric similarity is maintained. If it is not possible, explain why not.
the fi fhfl)Yetn) 5"-~: 2 dll'n el1 ~;'tJl1d / 411 ~ / '1 " j (11 ~ / d,J
Wf)h1
--
-v . Sphere·
The s/;rn'/",./f.7
rf~"ll'emt'lI-1:
/J
d'm _ d
- f)-*1 - D O. "2 I)", (). 5"' /n.
d =
0,
=
e ~-I:.
Z
-I./:
(I"e'jlllf('d dltfhleffr).
7 hus, the P f!'dlc.tle/n
E
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1
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,&
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I
7.
SI
1
7.51 Flow patterns that develop as winds blow past a vehicle, such as a train, are often studied in low-speed environmental (meteorological) wind tunnels. (See Video V7.S.) Typically, the air velocities in these tunnels are in the range of 0.1 mls to 30 mls. Consider a cross wind blowing past a train locomotive. Assume that the local wind velocity, Y, is a function of the approaching wind velocity (at some distance from the locomotive), U, the locomotive length, height, h, and width, b, the air density, p, and the air viscosity, J-L. (a) Establish the similarity requirements and prediction equation for a model to be used in the wind tunnel to study the air velocity, Y, around the locomotive. (b) If the model is to be used for cross winds gusting to U = 25 mis, explain why it is not practical to maintain Reynolds number similarity for a typical length scale 1:50.
e,
(0..)
V -: .p (~ ,1.1 J,; b) ~ jJ-) -It 1. V==LT- 1 V= LT- 1 J.:' L J" -= L J, == L f= FL T From 1he. pi. tneoV'em 7 - 3 .:: '+ fl' -i-noms Y'egu'-y~tI. I d /meYls/~yJa I aM a I ~ ~J:S ' !j' eIds
11 fV'
=' r-L-1. r
Q rt pC,
r-
A-
0/)
~ 4 ( ~ ) .t ) <:
T~ us The ~jl1l" 14y,:f'j reg UI Y'fm ~11.+.s aye. ;.;' _ j. h~ .! t:'tK 11M! Tl~:= t- h l/ TIWf -;,., --It,.. h ~jG==..,..====-)4 __
=
T~e. pred;".J-loJ.. €ZJ)..~-/'Ii>~ V - VM\
1".5
V=-1Z.
( 1) 5/~ce
1he den,j~
ant< VljC"SI~
at- tHe Ci/'r f/OWIHfj 4rouI1d
The
1rg/,., Cind -the (Ur In 'hH Wlj1'd tunnel kJ~u/~ b< pYac.1-,"c.4 l1!t -tn~ same (I'WJ~! )),*~p..)) /t +olloWJ +rom tn"e IIHi 5Jm/lt(Y/+; vezu/yemen:i.. CWn,cl-1 ,J --f1.u ~ey"oldj nurn},.er)
111 ~t
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nu~) w/-fh V::- 2. S /1'n /s Th/~ Ca J1
-ttA4of
/eJ1j1n Scale. of
1:50
C/;1(:/
w;1h
~ ::- (sO) (Z§mf/s) = ~ Z§~ ~/S
ye'SlJlrf'd mfJael t/e/~;-f!J ,,5 rnuch be Ct. c h ~ ; v -e a' / h '171 e !AI ",i d
'1heve-hY(' ,"/l1ul111JtI""
+t90
e::(
::('-A ) U
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h 19 h .
hl9he.r lhal1 i:. I,.( 11;-1e. / Ci n d /Yr1l1lrl';"/~ Re'tho/d..s
v~J&II;eel rn od(l./ v-e!oc'+Ij I
1.5
7.52
I 1.5l
An orifice flowmeter uses a pressure drop measurement to determine the flow rate through a pipe. A particular orifice flowmeter, when tested in the laboratory, yielded a pressure drop of 8 psi for a flow of 2.9 fe /s through a 6-in. pipe. For a geometrically similar system using the same fluid with a 24-in. pipe, what is the required flow if similarity between the two systems is to be maintained? What is the corresponding pressure drop?
cp = + ( 4p)
A-5sllm~ Wh rre.'
d) D) f;;')
Q'" /iDlUya,te.:' L~ r-') LJ p "'" pressure dr~?::: t= C 2) d ~rl Ii (~ c/t'lfIhlef;,r-': L) D"", I'lpe c/,itlnetfr'= L ) !N .r/~/d df"~i-I:J:::: FL -'lr",,; I/Iul )AN +lwlO vis~sH!:J =- FL -2 T. Ao
fr~f}1 1/Ie p/ the~re",,) ~ - 3 ::: 8 p/ -trttrlS re$uiyed d;'mt'n5/~~lIj IIl1aJ'1~/.s f1it/JJ
t.2.
)lD
=
t (Pi
I
tlHa
tt
!Jl D't )
/)J)<2.
-p""", d
d
D
.Ll f>.,...
D': !
-'"1
=
D-w. _ - D
~ P D'"t
?
),=)1.", I
!
tJ t
=/~
=(iff) ~ AI.. - ff;.fri,!si):
~ CJ~
/~ Dm,
o. 5'p~ psi
~ ~ ;fA-.D
Q=(*"f~ j{)(?~ =(i)4J"" =- (If )(2. f .fP) =- p. tIt
7.53 During a storm, a snow drift is formed behind some bushes as shown in Fig. P7.53 and Video \'9.4. Assume that the height of the drift, h, is a function of the number of inches of snow deposited by the storm, d, the height of the bush. H. the width of the bush. b, the wind speed, V, the acceleration of gravity, g, the air density, p. the specific weight of the snow, 'Ys' and the porosity of the bush, TJ. Note that porosity is defined as percent open area of the bush. (a) Determine a suitable set of dimensionless variables for this problem. (b) A storm with 30 mph winds deposits 16 in. of snow having a specific weight of 5.0 Ib/ft3• A half-sized scale model bush is to be used to investigate the drifting behind the bush. If the air density is the same for the model and the storm, determine the required specific weight of the model snow, the required wind speed for the model, and the number of inches of model snow to be deposited.
~
:: +( d) H
-It -= L
d= L
ra)
t ==
~
/+~
L
4/so)
-
Vhh
Vd,,"~ So
'h1J
fAltr"
g~= ~
V11f = a'ncl
=
d,1t" Ji-1t'J'I
ri
FIGURE P7.S3
1=
=t.
V -' L T- 1
d=
L T-2.
rDL~TO
1'/ .J.trnu reZ~/;e~1
tlH&./
'jle'J.s
dna/'1siJ
~ = ~(1;)"~)
1
Dritt
.~!
h=L
FV'tJ/11 1h4! pi. 1'heare"",) Cf-3 a. dlme'hSltJnaf
.
3-) f) IS) 'Z )
ir".$~ FL- 3
FC'f-T "2.
&
•
bJ ~
J
Bush
9s / rJi,) '1)
V
V"g.H #1tI4 / )J. :::1.'l-
t:{ut;!
V
= V(d)"'
d
-
7
(3(),-,..ph)
:=
J./.
;1.~pJ,
£t ~
d~:(W)
({ ) (I ~ in.)
::
~. 00 ~n.
7. S ~
As iIlu strated in Video V7.2, model s are commonly used to study the dispersion of a gaseous pollutant from an exhaust stack located near a building complex. Similarity requirements for the pollutant source involve the following independent variables: the stack gas speed, V. the wind speed, U, the density of the atmospheric air, p, the difference in densities between the air and the stack gas, p - PJ' the acceleration of gravity, g, the kinematic viscosity of the stack gas, II ... and the stack diameter, D. (a) Based on these variables, determine a suitable set of similarity requirements for modeling the pollutant source. (b) For this type of model a typical length scale might be 1:200. If the same fluids were used in model and prototype, would the similarity requirements be satisfied? Explain and support your answer with the necessary calculations.
(a.) 5/" Ce.
v=
L i-I
if ~
1.. , - 2.
EJ'; L T -I ~:!: 1'/'7 -I
;;=
FL -1f T
D= L)
f-f.s ::i F C'+T4
2
~ //t)UlJ. ~r~/7t the pi.
f
I
--thAi 7-3 Jf- j:>;' ffrrn; 4r~ Yi'p/,I/ypd. CinA/1~/.s fjlf,ld...s .:::!...) ~) 2:2.) Clnd ~ I r . L ~L. ~ j. 0. J • I I' .
/+
-theoytm
se t
r- pI ~ ~ II I:)
Z7,m
(};) ~r
~rrn5.
I/",.
ret"': rt "'1'1'71.
'In~
511111/111'1 r!:l
D~ = V.D
Vitti 2
~M1"11
Y
::: -D",.,., D
rJ"'~
/
Z()O
10:>
and
V£1rtt =- Ys 1ne
Un, _ -VS""
D
V
D/m ..
- -I/s
:=
J
retJIJll'emfl17::l Z
=-
Y;:-DItIf
tfs
a..
~5j,;)t! J.I
toOl/I«
1_.
0::: :
0-$ ),. , _(I-A)
L
dD
~
~eC&>I1"
Zoo
dU':'fI1SIt)n~/
- I'
j/~/'/lIrl'-h:t
(see abt)ve)
ThJS re5t-t/J:: ~/7.f/I;;1-J W / 1h 1I11t -t In> m -j;, f. S ect:J;f,( :fI till '!" y/f '1 re% Ui re f'Yl erll:/ lind 1hfrej,ye. 17I~ j/hn,')lIr,'ft:J re~ U I remen+S (!OJ1f/pC be 5~.f:./jj,e4 UHt/er 1h'f .s+A.fe4 &;n drf-Ion 5 ,
Iv'tJ.
7.:rt 7.55
The drag on a small, completely submerged solid body having a characteristic length of 2.5 mm and moving with a velocity of 10 m/s through water is to be determined with the aid of a model. The length scale is to be 50, which indicates that the model is to be larger than the prototype. Investigate the possibility of using
t
either an un pressurized wind tunnel or a water tunnel for this study. Determine the required velocity in both the wind and water tunnels, and the relationship between the model drag and the prototype drag for both systems. Would either type of test facility be suitable for this study?
As
c/ehl"I'J.J.frll~rd IJIl Pg. 7./fl fo,y .f/~w (J rlJul1d /mmfrst"d b()cJ/~, Re'Jn~/d.s numbty s/m; /"1"1 =4 /J f".fl tllY'e" J() ihd ~j.-, -tiM!
-_ -v~ Y
~At'! = ~1 Y 7/ J._ ./-esls IIYf YUh In unpr~.S'.slo,,/'t4 wlnt-1 -hlill1tl I 1kel1 ~ (sfqildllJl'd (Jlr) .:: /. 'fb X. 16-f,-/WIYs I ~J1d 1J (~pttY'): /. /2 XIO-'l'fHl/s
J:f
fYlodel
So 1hlrl:
Vm. =
(/.I.,,~ X
-
VIWI:: ()
~
J~
l.
a!!. ) (/ ) / (1.12;( ID-~ ~ .... l~ (IO /()-!J
r.{ mode! fest-s t:(re
SIY1C.e
/1'101
~) = 2,1. I~ (+., wIHII 1,,..,/) water t.unnel WIth. -14 =~ ihel'1
Irl
)(l() )(10 7)
Y'iIJ5~illlbJe J~
both
or 111 e. Wa fe r .ftt JlIl e/ Co u /d ?v/I;.
J
'1etPl11efl'lc' 411t{
I
d'll1l1lfJl~
dJ V';R '-
_
=
~. 200
CAstS J
b~
~I/~U'.s
-
~ ~J. Jt1If2-
~:: ..L.J:..2. 1"~
rJJ
hi"
~
v,..."
(~~
e;1I1er 1he.
wSe d . .s-;hllil/f/-f~1 /t
~"."
T
J..ff41.
2.13
0, Lf LUI-
w.. fe#' funnel) Wmd
thllt
+ulInel
7. '5b The drag characteristics for a newly designed automobile having a maximum characteristic length of 20 ft are to be determined through a model study. The characteristics at both low speed (approximately 20 mph) and high speed (90 mph) are of interest. For a series of projected model tests an unpressurized wind tunnel that will accommodate a model with a maximum characteristic length of 4 ft is to be used. Determine the range of air velocities that would be required for the wind tunnel if Reynolds number similarity is desired. Are the velocities suitable? Explain.
,Rel(IIt>ld~ lJumb.tr s;ml'/4f11!J;
!1" V
IIPI
~~ = J- V..R
r
)A~
V'/m SlfUe.
=-7;t t
the. W1I1t1 ..fulllfel
aP!r(Jxirn~telt; -I1te feciuce.; -f..o tll1c{
#~
IRe
1,$
St1I?1t.
dll~
~ ",.,
V
OJ
Uh/!,ftSSII}'I1ft/
~I'"
the
medel 1"l1d
1/1
r pr~Pfrh'e.s
p~/()f!lpe.
TI1/1SI
w//J be €'$. e/)
?llIe""
:: (20
It) J!
=
5" V
( tf -It)
Ttterefr;rf) al IoU) Sfted /I""., :: S (ZtJ mph) =- /o~ mph aHa af 11t9h 6,~ed
~ :: S (q~ IWIP),) :
00
fhg-f the
mode / vel!)c
in,
Y'~11ge
tfb()
/s
trrtp h /Oornp;' fo 'f.>7J mph
1111,/1 vel()clfzt IH the WJ~tI 1-uJ1fJel e(f)/)1pr~ss~b/)'+!J ,,,f the Qir W()u! tI . .6ft1r,t -to be~me 411 lin'(JY'k~t titclr;r/ 14JherftlS
,41 i/Je
Cbmpressllllllf,; l.r /Jot: /n7'p~yfy,l1t -fi,y 1fte pYbfoftjpe. nlt/S, ine ht'ql?er l/e/~cii-!1 y;oe!tl1retl ~r 1ke ff1()tie/ wf)IIld /Jot be $UJ14j,/e. "v~.
7. 57
I 7.5 7 If the un pressurized wind tunnel of Problem 7. 5~ were replaced with a tunnel in which the air can be pressurized isothermally to 8 atm (abs), what range of air velocities would be required to maintain Reynolds number similarity for the same prototype velocities given in Problem 7. 'S{,,? For the pressurized tunnel the maximum characteristic model length that can be accommodated is 2 ft, whereas the maximum characteristic prototype length remains at 20 ft.
Re'fIJ(>Ic/s n IIm};!y SI'm; /IInfy .;
~ J":!. ::
11m
)A~
~ ::~ Ib1
t-E ~
& .l. ~ j..,..
fnt'!
V
--;:::I't€ T
/=PI'" lin Ide,,} 9t:1s )
/1 )
I til'ld
.{;,y ;'s~thfrf)J41 t!.f)/npf"e5S/~H
:1=. =~11 S I:n i1 t f'
~=J:.. fJ
tl'fYJ
L-£
tt7'1 - ft'l?'1 ( ass tJml;' 7 )J1YI -:::.; ) I)
-
.:f...J;.. V
~ - 1'1f'OI:4
'l,'6-mf!)~phfr'~ pr-e.sswY'<' (PY't'SStll"e at: tJ)""h PY'oto tljpq and ~ Is !f?.5SUY'f t:>f ~jJf't'ss('d fli}- In the Wll1d -ftt11t1el.
where. p /.s 0{f'Y"Ilf-eS ) J
ror tm =af>
~
==
(.J...)(zo It) V=-
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Thus J 4t
/fJl.J
f
( z .ft)
'If
hlfjh
V
.sIted ~ ::. I, Z ~ (2~
CH1t1
/.2;-
mph)
= Z 6' /W1 ph
s?~ed
V"",:: /. "25"" (90/J11ph) ::- 112,5"tmph
There ~rel
1)1(.
dis mph ..fr:,
r'e~(/Irett tnfJdel ve/f)cl1:J rnl1ge /~ //2.5"trnph.
?s~
I 7. Sg
The drag characteristics of an airplane are to be determined by model tests in a wind tunnel operated at an absolute pressure of 1300 kPa. If the prototype is to cruise in standard air at 385 km/hr, and the corresponding speed of the model is not to differ by more than 20% from this (so that compressibility effects may be ignored), what range of length scales may be used if Reynolds number similarity is to be maintained? Assume the viscosity of air is unaffected by pressure, and the temperature of the air in the tunnel is equal to the temperature of the air in which the airplane will fly.
slm; 1f/f'lf!1 ) ~ Y:" ,R~ ~ I- v.1
l1/,fhJl1ty
So
,)1-i11
th,,1:
i,,"
): fir
fin
Icle~J "as)
~ ~
1:.
(f)
~~~
? =;,e~
~-
5 ~ . (J)
t;y the
CttI1
IJnd w/1H
~s.f.tI/Ji femppnr/uYe.
J
w/fH
1m,
he 411'/#e 11 tI.s
);fIIf _ 1;)
V
)
~
- t..
dll~ '1 J ;;el1
J.4f4 -
al1d
k
:E.. = e,,,.s fa 11 t f -p _ -4-
or (fntl
/-
(tv i1H )Atttt :::)A
)
f;o/~fa.)
V k Pa. ) ~
T- O500 ~ = (Ii: (). z) VI /7 .fr; /loliJ5. 7hfJi j~
(/oJ-IGfl )
-1- - (/ 3()().k
fa. )
J ( / j; (), 2)
CJ, Ot,/f 7 to t). 0'17 I ,
7. SCf
I 1 Stf
Wind blowing past a flag causes it to "flutter in the breeze." The frequency of this fluttering, w, is assumed to be a function of the wind speed, V, the air density, p, the acceleration of gravity, g, the length of the flag, .e, and the "area density," PA' (with dimensions of ML -2) of the flag material. It is desired to predict the flutter frequency of a large = 40 ft flag in a V = 30 ft/s wind. To do this a model flag with = 4 ft is to be tested in a wind tunnel. (a) Determine the required area density of the model flag material if the large flag has PA = 0.006 slugs/ft2. (b) What wind tunnel velocity is required for testing the model? (c) If the model flag flutters at 6 Hz, predict the frequency for the large flag.
e
W= 7- 1
V= L7-1
e
1= MC 3 !=' /...T-1.
fA':: NL-2.
l;;'L
j:Y'~f)1 the. ?/ 1he~rem) ~-3 = 3 f/ +erl11s re$tIlYfJ.} t1l1d a. d/ln e t1 ~/PII~/ tin ,,/ '1'IS tt I.e I"~
W
Vj
Slhl//Ilf'I1-Y
= cf (~
)
J
~~
~ .J/t'n
l V8.R.
Vtrn Vdr;r,)~
gl'm = d \!",,:: (t)
W,1"h
fhe simi I~ r/1-!:I
e!tl.Q.':t:J~M
j5
fn~-/;
V
::
.
reI U remPt11=s s~.:t = tu." V~
V-f W - ViLv
jo
Vf
!~
J
J
V~ J-
,
I
~+I pi 7H('
pr"ecl,'c t./~i1
aht1
4!
~
(~ II,) = I. ~ 0 = r.!t:E ~o .f-.f 1
f-l)
I,
(Po
I 7. fDO River models are used to study many different types of flow situations. (See, for example, Video V7.6.) A certain small river has an average width and depth of 60 ft and 4 ft, respectively, and carries water at a flowrate of 700 ft3jS. A model is to be designed based on Froude number similarity so that the discharge scale is 11250. At what depth and flow rate would the model operate?
fi;y
P.rpud~
w herll 1.
IJ
n 1-1111 J,~ Y 5/;"'I'IlIrJf.!l V/??1 = Vi). -, Vj~ ~~ SOn1t c,naY/lc. t-en s' 1-, C 1('HJt;, )
£mt.=. (). ah to{
for-
t
model
/1117],
If f-t
the.
IJ0
().. prof:" -I:!f~ dep1lt t;f
e" rr-t's/o/1 d lit'!
I1I1K
dep1J1 (s
J.t'm ::: ((J.II 0) (If H)
== O.If-'f{).f-t
! ""' ;:: j
7. ,/
I 7. r..l As winds blow past buildings, complex flow patterns can develop due to various factors such as flow separation and interactions between adjacent buildings. (See Video V7.4.) Assume that the local gage pressure, p, at a particular location on a building is a function of the air density, p, the wind speed, V, some characteristic length, and all other pertinent lengths, i • needed to characterize the geometry of the building or building complex. (a) Determine a suitable set of dimensionless parameters that can be used to study the pressure distribution. (b) An eight-story building that is 100 ft tall is to be modeled in a wind tunnel. If a length scale of 1:300 is to be used, how tall should the model building be? (c) How will a measured pressure in the model be related to the corresponding prototype pressure? Assume the same air density in model and prototype. Based on the assumed variables. does the model wind speed have to be equal to the prototype wind speed? Explain.
e,
e
(a.J
1>= rL-· FY~fl1 -rj,~ P"
f
=F L-'fr
V -= L T - ( 1 :.- L Y,,' :: 1 1]1eDref"YI/ s- 3= 2 -p,: kY"l71s ve~",,'red, Ql1d dl n, -en SI;I'1IC./ Ci 11 4. J 1 S 1:S 'J J ~ I OJ 2.
c..
I'~Z = ~ (f: ) georne fy/c, 'S1'm,'l a rj'+j
j.~
_
-
ti,m
J.1)rt ..
T -
1 ).,:
J- ~'hff J.i
{"IIDWS ThAt 4/1 P(Y't/~fl4t lenfth~ aY( Scaled le~ffh Sca/(. f.".,. / j., Thus, WITh j;ttl I;. = \/aoo
t"f-
l'mode.1
h-eicd\,t -= oJ
.ill.Ji::: O.3~3 '300
+t
f'Wt =f
( 1 -P~.lf~) f~
Wd1t 1Jr( ~e1: of. '1' ~e r1 Va r/a hieJ 1hfY(!' IS no ve8~;ref1l~t foil" f'he V~JDC.·I~ sca./t!; VlYI4jV) 5fJ fk~ rnod~\ WI~e.{ -speed. dDe~ npt hail~ -to be .e~u~ ilt... tJu prI)-/:oi y pre. WU1~ opee&. No .
7- 70
7. ~2 j
n
7.62
A scale model is to be used in a towing tank to determine the drag on the hull of a ship. The model is operated in accordance with the Froude number criteria for dynamic similitude. The prototype ship is designed to cruise at 18 knots. At what velocity (in mls) should the model be towed? Under these conditions what will be the ratio of the prototype drag to the model dra~? Assume the water in the towing tank to have the same properties as those for the prototype and that shear drag is negligible.
Foy FR.oude n tim blr- s·lm~ jar; +!;j )
Qnp(
IN ,Tn
w,·th
Ym, = V!hO\ ~".,
vF
~:Vfv--
Vi
?=-im,
Frowde
lIt1mber ~im i /4 f ,+, ptJ :: ~I)M fy'~J.~
J,j
V
~)-- /. s/ a!!. s
I/ /~ ~lIdfs )1 (0, S/J.f.1{- ",,~t
tlltd
9tfPllleffli
SI ",dan
f'j I 1hel1
~ ~loJ:
-the dra.9.
~=.t.
otJ~
5Jiu:e.
I =/fWI
and
4:
(t)
fm,
V/Vmt::"
(fJ
J
V)/-f
M4 I
,·t
= (J) ( 5"0)
M'1
7-7/
~JJ()/JJj thai
3 '"
/.
~
2 X 105'
7.64 Assume that the wall shear stress, 'Tw, created when a fluid flows through a pipe (see Fig. P7.64a) depends on the pipe diameter, D, the flowrate, Q, the fluid density, p, and the kinematic viscosity, v. Some model tests run in a laboratory using water in a O.2-ft-diameter pipe yield the 'Tw vs. Q data shown in Fig. 7.64b. Perform a dimensional analysis and use the model data to predict the wall shear stress in a O.3-ft-diameter pipe through which water flows at the rate of 1.5 fefs.
(al
'::: 0.7 J3 0.6 0.5
J
;, ,
,
----1-.-__._+._.__ -L_. __.
o
0.5
1
1.5
Flowrate. Q. ft\/s (bl
LLu- -= lw--=-
FL-'-
FY'{)f)1
1he
+(D
J
D= L pi
ct c/, menS,bJ'l4 I
4>:::
3
L
-V) ,-'
=
tr. eOytM) 5- 3 2. fUt allj'SI'..r
;cp~~ SI
cr ) f)
•
in // fI r,'+~ w:>~
==
fjl 'e' d..s
1 ( !J )
FIGURE P7.64
2
7. ~f)
7.65" The pressure rise, f!.p, across a centrifugal pump of a given shape (see Fig. P7.6Sa) can be expressed as f!.p
= I(D,
8r-------------------------~ Model data
w, p, Q)
Wm = 40!l" rad/s
where D is the impeller diameter, w the angular velocity of the impeller, p the fluid density, and Q the volume rate of flow through the pump. A model pump having a diameter of 8 in. is tested in the laboratory using water. When operated at an angular velocity of 407T rad/ s the model pressure rise as a function of Q is shown in Fig. P7.6S'b. Use this curve to predict the pressure rise across a geometrically similar pump (prototype) for a prototype flowrate of 6 ft 3 /s. The prototype has a diameter of 12 in. and operates at an angular velocity of 607T rad/s. The prototype fluid is also water.
Dm
= 8 in.
6!---------··~-------~=
°O~----~----~~~~~----~ 0.5 1.0 1.5 2.0 Qm (ft 3 /s) (b)
Q
•
FIG U REP 7.65
Centrifugal pump
Jp = j ( D/~.) f.J
(0)
tJ t ~ f '-- 2.h-If)tn
D -' L
.
(",,).::: T
1=
-/
a)
FL-'tTl.
t:-he PI.: 1:hetJytlYl/ S-~: .2 Pi t!'rrns
t:ift~/'fsi5
&' e/ tis
LIe Itt) 11>
ret II/yeti) 11114
t4
l-
Ye! VI r-emt'lt t ~~
;;;-£3 '*"
Q'= L 3 T- J
I..J
cp
=:
tV
D3
~
a;ui jpy 1;,e dd£ ;,/iJev
tj),., =-
(folT tf') ( flit,):3 ( I. ( ~7T ~d) /2./n. ..5
Fr()1J1 tHe 1ntlJ,
(r;,.
P7.bS';)
LJp
.t1 /:m
.pP) _ I.
/'1 {!J
.5
:: .s:S'o ?.5l.: ~y til =/. / 'i ;tJ.
::
1-73
fit liS)
7.~ (P
I 7. "
Jt<
Jtr=o PIj
J;( r
;0
(~-t U ~ .,. 7r ~)O, - ~ 1")A (~~ 1" ~ ;~ )
If)
(Jrr + t< _ P~
~"
4s In(/ic.a/-PII
111
I
k
U- -r
fA 11= __
=~
X
f
~
/
(~-r JY} (e~. 7. 37) 0,( l. ~:; 1.) ,
-p "..; 1:. 1;,
V
V X
-riA-
- , 0 II
Sec.·ltoN 7.10 let ?i 14:. :JC
"""= ..!:!:.. .;.
viJ!:) = - a~ d ':1 !1
(liS. ? 36)
_1:1
..J....L
v.J
C7"-=
t-~
1r((i1.sk~IYM.j./~'n..s t4h be mille as /c/JClJIs: J(Vu.~) Jx'" V Ju-iax. ~;(, /4. d x. :: --;. J x'-tIr
Th e
/llit/,,'US
1f11,{
au. _ j'/ln ; /(11" & )
J v- _ ~ J I/#-
;2 -.).
,4/so)
J "2 fA.
_
p,(2. -
I;<*"
~ .:
v d (0 U -It) ~It -: Y-X U.,. a;.. a~ 12.
Clnc/ jim/iff r/~/
J 7- V- _ tJx2. -
J:: !..!!:. f!.
J'j.J. Jyf.
1/ J 2.v-¥ ~L
F
27.(A. _
p;2. -
1/
0'1.", i'.
Ji J!j'loZ-
fEy 17te local acce/fy~-b~·If.J dUo J rvu4') Jtl. = V Ju" rJt:: Jt--f.. it ~ Jt+ til/it ~ /m ,'k,yo)'11
J U- _ V J
n -~
]/-J.
;-c~
0 1.1" _ 1:: lY v.
; ; - ~ o!J+
J Z U I~x1tZ-
7. ~ ~
I ( ~11 Z ) roY' -the. pY'e.ssure I-erm.s,) ij::, _ ~ Pa f;J.. d.JG ~ :: J.x. ~x. aX. t/hd s/m;/Ilr/y)
li:.:: A li:..jJ. o!J
j.
d!J*
5Ub.'Jf,i-uf,il1 eJ/ 1h~ I/tlrI()~ ftrms J -(xpresspo/ In Hrms ill tnt! dlln~I1.5;{)I?J-es.r vtJY'lobJe.s, Crill h< IIltlde I~t-~ the (!)Y'I'gJ;'ql d, fleVfl1h4/ eglAa.t:I()AS (£I S . 7. 3~~ 7.3~,; ~1'1'" 7. 37) -1-0 tjie/ tI Ef~. 7· 38) 7· ~'J 1/(111{ 7. 'fo. To o/'.ftt"j the hh~ / !orin for ES5. 7. 'fJ tll-ul 7. 'f2 dlV, de each krm
hy
IvYi.
r
7.'-7
I
I. Ie. 7 A viscous fluid is contained between wide, parallel plates spaced a distance h apart as shown in Fig. P7.67. The upper plate is fixed, and the bottom plate oscillates harmonically with a velocity amplitude U and frequency w. The differential equation for the velocity distribution between the plates is
au au p at = J.L ay2
(
f
--....11
h
2
(
where u is the velocity, t is time, and p and J.L are fluid density and viscosity, respectively. Rewrite this equation in a suitable nondimensional form using h, U, and was reference parameters.
) Fixed pi ate
I' l
x
•
II
= UCOS
u. -V I
Le i:
TJ' v Wou* -
0-(;'*
~= O!j
02U _ ~t-
7-7~
WI
I
7. ~8 7. ~8
The deflection of the cantilever beam of Fig. P7.bi is governed by the differential equation
d 2y EI dx 2
P(x - C)
=
where E is the modulus of elasticity and I is the moment of inertia of the beam cross section. The boundary conditions are y = 0 at x = 0 and dy / dx = 0 at x = O. (a) Rewrite the equation and boundary conditions in dimensionless form using the beam length, r~ as the reference length. (b) Based on the results of part (aYwhaiire the similarity requirements and the prediction equa': rion for a model to predict deflections? (
(,L
J
Le i.
J1- = 1;
-
Co( iii&{
rJz!:1
-d.x,
= l..
;(.14 =
dt (J. b~ ) d )(. ~ _ d )(. .,.. -d;<. -
did d~
and
-
FIGURE P7.1D8
.so 7"ha.i
)(.
X
J. dt1~ dt,JJr
E. (!li*) !.£ '" =...L dx )..
J;e"* dx.l/-
(-i )
::
~ d)(,~
rJ'!f 'I-
dx" z. hec.Dme..s
OV'
(.b )
re~ UI veme,.:tr X,w. _
et Y'f. ~
J .... - )...
-
7-77
I
7. fo'f 7. fo9
A liquid is contained in a pipe that is closed at one end as shown in Fig. P7."~ . Initially the liquid is at rest, but if the end is suddenly opened the liquid starts to move. Assume the pressure PI remains constant. The differential equation that describes the resulting motion of the liquid is
where Uz is the velocity at any radial location, r, and t is time. Rewrite this equation in dimensionless form using the liquid density, p, the viscosity, jl, and the pipe radius, R, as reference parameters.
rlf =..t R
)
t1-:
± , ) tll1t1
initia!l~
_P_l_ _ _ _ _--1-_-----. End sed
r
auz = PI + (a 2uz + ! auz) P at e JL ar2 r ar
Le t
' _
rLz
~vz
t R T
,/ V
1-1.- - - 1 - - - - 1 , 1
~*= ~ ~ V
Whe~~
r
-?~me ~m'JI~lftt/1'J.S of the p4rtimekYS /)/-I QM~ R hal/ll1? ihe dlmeI1SI()~ ~f t.lme) tJn~ V I~ J~mt Ct;mlJ/ntt/~~ cf the sqme p4fllmeter.s hHtllht 1?,e d/rnfrlsl!;/LS t>f tt., tle/()c/7Tj. Let: 1: = /-,R?=. (pl.-If r"L) (L) ~ /" FL-2. T
IS
r
alit!
JA. V= ...£~
fJ R.
(
-2 T I=i. F/..-'f r2.) (L)
d
~ L T-
W,1h these d/~ei1S;~l1less IIlJyltd,ies .a~: a(v~~ ~~:- V J~" (...!..): Jt-l-
dt
at
W
1:
I
ff)!}!:. ) J~ "_ ~,R. (1£2. ~t:1ft -
(e)~ dV;"" fJ
1<3 Jtf'
~ = ~ (V'Zi-tt-) or·: v ~-I- fJ.): (~ )(!.) p~-Jt .:- A Jtfr+ ar ar -to. dfJ}-- i- ( i< f> ~ fo Jyo$. (';€ ,. 0 Y' "" a7tJi: _~ ~ (JVjI)k! = ~ d 21it~ /1) :: ~ JZ1Jj~ ayo 1',e2 ,It#- --r;;:-. J r I'Rz ~ rj.2. (~ ;o;e.3 () r1' z. l.
-
The ()l"IflJ14J d,'.fkrfntlq/ e$tlflf:to'n
[; ~r:f.] ~;: =
:. +
ClJM
n~w b~ ex.p'l'tSSfd aoS
~ (f;e3)J(~';:: +?~)
7-78
I
7. 70 7.70
An incompressible fluid is contained between two infinite parallel plates as illustrated in Fig. P7.70. Under the influence of a harmonically varying pressure gradient in the x direction, the fluid oscillates harmonically with a frequency w. The differential equation describing the fluid motion is
au
p -
at
=
h -f-
YL~u x
h
a2u
x cos wt + f.l -ay2
FIGURE P7.70
where X is the amplitude of the pressure gradient. Express this equation in non dimensional form using hand w as reference parameters.
f
~j(=
Let
I
ti-:. 4J i
dr
~: ~
(J.tu '" -I-~j.._,; 'C""' t3t
&IA. _ ~
fA fA) ().:"'") J.!1" ~
;!:J -
dZt-< Jj
l.
::
J :;-11-
t.U
~ -
,
c/l111
i
tu
,,+::: {~
~"" (w) J t:1c
-I, tv ~ ~ djf.
"*)lit
d ( Pu -'Ie. ; !J f. a!:1~ J!:J.
X
UJ
Ct(11
~~
J!:1'*'
LL) = tU
0J~ 2.. {h
The OY'J9/f1lt! d,/nY.fJ1ttal eg IAlJ.tJtPH
[;-AtJj ~~: ~
J r It
(J) =-
UJ () Zu 1'.
1h4.i '
IttU'-~.f.
=
~
So
-::t:
dZu -11 b !J;' z.
naw b.e -e.x.pytssed
c:~s tt' + ~-i!] ~>.~z
a~
7.7/
I
7.7 I A viscous fluid flows through a vertical. square channel as shown in Fig. P7.71. The velocity w can be expressed as
aplaz)
w = f(x, y. b, J.L. Y. V,
where J.L is the fluid viscosity, y the fluid specific weight, V the mean velocity, and ap Iaz the pressure gradient in the z direction. (a) Use dimensional analysis to find a suitable set of dimensionless variables and parameters for this problem. (b) The differential equation governing the fluid motion for this problem is
I-b--I
(a-+-~ w a2 w) 2 ax ay 2
ap= -y+J.L az
Write this equation in a suitable dimensionless form. and show that the similarity requirements obtained from this analysis are the same as those resulting from the dimensional analysis of part (a).
(tL)
w=
w-= { ( LT-
H()t'Y1
1
;(=
J=- L
L
•
if. )
!i; JJ.;)<.1 0.1 V"
J<.)
?
b=' L
==
FIG U REP 7 . 71
cY
P1..-;. T
=PL-
1he p(.: 1hel)i'e~ ?-3 == S J'i i-f>rms ye~IJly.etL
tll1t1lf1s/~
I
3
tlnd
1 .?E...:.. c::- - 3 V-:'L-- I Jt: - I L A.
y/e/Pi..s ~
(I)
j/
(.b)
dw_ ~X
~;<..
-
V ];
oX
t
So
V~1f J: h<.+
d /dWi-) h(" _ Ot:l- (
~;< 1i:-
~~tv
5, hit 'IIIY Iy .-'
lhe.
x*= ~ ) (jJl.=
d(VW~)())t.Jj.=
JZW- .=
gXl-
r)
w~=
Let
JX V
);; ~ = T...
&1'/'1);,,,) d;l{eyel1-6iJ;
Ul. )c
hZ
-=
~2W-J/c.
J .'}f:. 2-
efud,oJl
can nov b~ -f'xjJressp,<
-J- + tW](~1' h2. ~;<,JJ. 2-
qp = _
)tV Ji!:
-
(rbL. -r
I'V
+
(» 2 W :lc -r gx"'z.
7-80
as
~~) g~*' 1.-
~"2~) g~+2-
(2)
7.71
1
((1;/1't )
E1' (2)
/~titcRk
thai
kr+-=
== 'r
.if
j,
('/;* /\
-¥
~)
cr j, 2.
)All)
.b.2. J P
~V
n
) (.3)
d~~.s t1D~ all'eAr' i:v mtt.1-cM ih~ eJlldf~" "j,Ioo',l1erl b1 tltmfl1S'M~( tl"""{'1~Jj (E'j,. I)) 1t.< If/ 1'h()1I! h
las I:.
1'hu,
r.e';WJ-C
PI
+fYI11S
1-Wo
/11
£7 .t /)
(-1.rlz: q)(~J/ 11Iz.2.) :50
ttl! d
the Ii re
£,
-/7ta.. f
-r
1h/~
=
(
Y~>u./f
~/'ml'/C{y/~ -the
1
.! ))
e.11J1
!4,z
b~ ~m blAe4 ftJ y/~/ tI
= ~ 2lf
~ V Jt: II b.se b~ w r/ 1feJ1 a.s
f1 ?r/; ~ b"2... i l ) X b) b.J ~ V ) ,.ft y ~t
-lite $time IJS Y'eJtI;Yempl')-/s I~d,cq)-e~
sam e .
IS
( If)
/:;7
£9 J3).
£ t.J.
7htl5~
(3) 4-M.P( (If)
7.72. 7.72
Flow from a Tank
Objective:
When the drain hole in the bottom of the tank shown in Fig. P7.72 is opened, the liquid will drain out at a rate which is a function of many parameters. The purpose of this experiment is to measure the liquid depth, h, as a function of time, t, for two geometrically similar tanks and to learn how dimensional analysis can be of use in situations such as this.
Equipment:
Two geometrically similar cylindrical tanks; stop watch; thermometer; ruler.
Experimental Procedure:
Make appropriate measurements to show that the two tanks are geometrically similar. That is, show that the large tank is twice the size of the small tank (twice the height; twice the diameter; twice the hole diameter in the bottom). Fill the large tank with cold water of a known temperature, T, and determine the water depth, h, in the tank as a function of time, t, after the drain hole is opened. Thus, obtain h = h(t). Note that t ranges from t = 0 when h = H (where H is the initial depth of the water), to t = tfinal then the tank is completely drained (h = 0). Repeat the measurements using the small tank with the same temperature water. To ensure geometric similarity, the initial water level in the small tank must be one-half of what it was in the large tank. Repeat the experiment for each tank with hot water. Thus you will have a total of four sets of h(t) data.
Calculations: Assume that the depth, h, of water in the tank is a function of its initial depth, H, the diameter of the tank, D, the diameter of the drain hole in the bottom of the tank, d, the time, t, after the drain is opened, the acceleration of gravity, g, and the fluid density, p, and viscosity, J.L. Develop a suitable set of dimensionless parameters for this problem using H, g, and p as repeating variables. Use t as the dependent parameter. For each of the four conditions tested, calculate the dimensionless time, tgl/2/Hl/2, as a function of the dimensionless depth, h/H. Graph:
On a single graph, plot the depth, h, as ordinates and time, t, as abscissas for each of the four sets of data.
Results: On another graph, plot the dimensionless water depth, h/H, as a function of dimensionless time, tg l / 2/H 1/ 2, for each of the four sets of data. Based on your results, comment on the importance of density and viscosity for your experiment and on the usefulness of dimensional analysis. To proceed, print this page for reference when you work the problem and click her~> to bring up an EXCEL page with the data for this problem.
Data:
r H
L iii! FIGURE P7.72
7-- 82-
7.72.
I
Solution for Problem 7.72: Flow from a Tank
H for big tank, in. 16.0
H for small tank, in. 8.0
h, in.
t, s
Big Tank with T 16.0 12.0 8.0 4.0 0.0
=57 deg C
Big Tank with T 16.0 12.0 8.0 4.0 0.0
=20 deg C
tg 1/2/H1/2
h/H
0.0 9.2 20.0 33.8 57.0
0.0 45.2 98.3 166.1 280.1
1.000 0.750 0.500 0.250 0.000
0.0 9.0 20.3 33.0 57.2
0.0 44.2 99.a 162.2 281.1
1.000 0.750 0.500 0.250 0.000
0.0 21.5 66.0 126.5 209.2 287.7
1.000 0.875 0.625 0.375 0.125 0.000
0.0 20.8 69.5 125.8 225.9 298.8
1.000 0.875 0.625 0.375 0.125 0.000
Small Tank with T 8.0 7.0 5.0 3.0 1.0 0.0
=57 deg C
Small Tank with T 8.0 7.0 5.0 3.0 1.0 0.0
= 20 deg C
0.0 3.1 9.5 18.2 30.1 41.4
0.0 3.0 10.0 18.1 32.5 43.0
7-83
(
~,,'t
)
Problem 7.22 Water depth, h, vs time, t
t:
..c:
18 16 14 12 10 8
-------~--~-==-----=~J
.
~ -~ ~-··-I
Big tank, T
=20 deg C
X
I
Small tank, T
=20 deg C
-------~:------~
4
2 0
•
--.-. Small tank, T = 57 deg C
.
6
Big tank, T = 57 deg C
-
+----~-..:----~~--_r--.---1
i
20
0
40
60
80
t, S
Problem 7.72 Dimensionless Depth, h/H, vs Dimensionless Time, t*(g/H)"O.5
1.20 - , - - . - - - - - - - - - - - - ,
I
1.00
: I ----·-·-·~-------------~~----·--I
I
--I
0.80 J:
:c
_I I
0.60
i
0.40
0.20
----------'--
I
I I
-l-----+--~,.,----+-----l
I
0.00 +----+-----1--__'*----1
o
300
100
t*(g/H)"O.5
?-9'1
400
•
Big tank, T = 57 deg C
•
Big tank, T = 20 deg C
- .. X
Small tank, T
= 57 deg C
Small tank, T = 20 deg C
7.73
Vortex Shedding from a Circular Cylinder
Objective:
Under certain conditions, the flow of fluid past a circular cylinder will produce a Karman vortex street behind the cylinder. As shown in Fig. P7.73, this vortex street consists of a set of vortices (swirls) that are shed alternately from opposite sides of the cylinder and then swept downstream with the fluid. The purpose of this experiment is to determine the shedding frequency, w cycles (vortices) per second, of these vortices as a function of the Reynolds number, Re, and to compare the measured results with published data.
Equipment:
Water channel with an adjustable flowrate; flow meter; set of four different diameter cylinders; dye injection system; stopwatch.
Experimental Procedure:
Insert a cylinder of diameter D into the holder on the bottom of the water channel. Adjust the control valve and the downstream gate on the channel to produce the desired flowrate, Q, and velocity, V. Make sure that the flow-straightening screens (not shown in the figure) are in place to reduce unwanted turbulence in the flowing water. Measure the width, b, of the channel and the depth, y, of the water in the channel so that the water velocity in the channel, V = Q/(by), can be determined. Carefully adjust the control valve on the dye injection system to inject a thin stream of dye slightly upstream of the cylinder. By viewing down onto the top of the water channel, observe the vortex shedding and measure the time, t, that it takes for N vortices to be shed from the cylinder. For a given velocity, repeat the experiment for different diameter cylinders. Repeat the experiment using different velocities. Measure the water temperature so that the viscosity can be looked up in Table B.l.
Calculations: For each of your data sets calculate the vortex shedding frequency, w = Nit, which is expressed as vortices (or cycles) per second. Also calculate the dimensionless frequency called the Strouhl number, St = wDIV, and the Reynolds number, Re = pVDI/-L. Graph:
On a single graph, plot the vortex shedding frequency, w, as ordinates and the water velocity, V, as abscissas for each of the four cylinders you tested. On another graph, plot the Strouhl number as ordinates and the Reynolds number as abscissas for each of the four sets of data.
11&
S/de
IIJeW
FIGURE P7.73
I
7.73
Results:
On your Strouhl number verses Reynolds number graph, plot the results taken from the literature and shown in the following table. St
Re
0 0.16 0.18 0.19 0.20 0.21 0.21 0.21
<50 100 150 200 300 400 600 800
Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem.
Solution for Problem 7.73: Vortex Shedding from a Circular Cylinder
T, deg F 70
b, ft 0.50
Q, ft 3/s 0.036 0.036 0.036 0.036
y, ft 0.82 0.82 0.82 0.82
D,ft 0.0202 0.0314 0.0421 0.0518
N 10.0 10.0 10.0 10.0
t, s 13.2 19.9 24.5 30.1
0.062 0.062 0.062 0.062
0.79 0.79 0.79 0.79
0.0202 0.0314 0.0421 0.0518
10.0 10.0 10.0 10.0
0.029 0.029 0.029 0.029
0.86 0.86 0.86 0.86
0.0202 0.0314 0.0421 0.0518
0.018 0.018 0.018 0.018
0.92 0.92 0.92 0.92
0.0202 0.0314 0.0421 0.0518
A
0.758 0.503 0.408 0.332
V, ftls 0.0878 0.0878 0.0878 0.0878
Re 169 263 352 433
8t 0.174 0.180 0.196 0.196
6.3 9.6 12.5 15.1
1.587 1.042 0.800 0.662
0.1570 0.1570 0.1570 0.1570
302 469 629 774
0.204 0.208 0.215 0.219
10.0 10.0 10.0 10.0
19.2 28.2 33.1 36.7
0.521 0.355 0.302 0.272
0.0674 0.0674 0.0674 0.0674
130 202 270 333
0.156 0.165 0.189 0.209
10.0 10.0 10.0 10.0
31.2 41.3 52.2 65.3
0.321 0.242 0.192 0.153
0.0391 0.0391 0.0391 0.0391
75 117 157 193
0.165 0.194 0.206 0.203
00,
cycles/s
= Nit V =Q/(by) 00
8t
=ooDN and Re =DV/v, where
v = 1.052E-5 ft A 2/s
7-8'
Data from Literature Re 8t 0.00 50 0.16 100 0.18 150 0.19 200 0.20 300 400 0.21 0.21 600 0.21 800
7.73
( C
Problem 7.73 Shedding Frequency, ro, vs Velocity, V
1.S
--r-------;-----~-----------,
1.6
1.4
.. -
..__._-----_. __._-.. _ - - - - - - _ . - - - - - - - - - - / -
1.2
--I
~ 1.0
----~~--___:;iI""""'-------J
,
~
u ~
O.S
+-~---------~-+--------;;;,;L------t-:...
8
0.6
+-------'-----/--~-___;;;;O..=-_::7'~---.~
0.4
- - - - - - - - - - - - - ----II
------lI
0.2
~~-=----
-+-0 =0.0202 ft _ 0 =0.0314 ft """"-0 = 0.0421 ft ___ 0 =0.0518 ft
.. ----.-------.. ---- ---- ------
0.0 - f - - - - - - ; - - - - - - i - - - - - - , - - - - - - j 0.10 0.15 0.20 0.00 0.05
V, ftls
Problem 7.73 Strouhl Number, St, vs Reynolds Number, Re
0.25
0.20
-
0.15
. _._.__._---_._---,---_._----------\I
0.10
I ---1
! j
en
I
0.05
~-~-,--_4--~--~~
! •
I t
0.00 0
200
400
600 Re
7-87
SOO
Experimental
I, I
1_ Data from literature I
1000
! I
I
7.7'/-
7.74
Head Loss across a Valve
Objective:
A valve in a pipeline like that shown in Fig. P7.74 acts like a variable resistor in an electrical circuit. The amount of resistance or head loss across a valve depends on the amount that the valve is open. The purpose of this experiment is to determine the head loss characteristics of a valve by measuring the pressure drop, 6.p, across the valve as a function of f1owrate, Q, and to learn how dimensional analysis can be of use in situations such as this.
Equipment:
Air supply with flow meter; valve connected to a pipe; manometer connected to a static pressure tap upstream of the valve; barometer; thermometer.
Experimental Procedure:
Measure the pipe diameter, D. Record the barometer reading, H.tm , in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Completely close the valve and then open it N turns from its closed position. Adjust the air supply to provide the desired f1owrate, Q, of air through the valve. Record the manometer reading, h, so that the pressure drop, Ap, across the valve can be determined. Repeat the measurements for various f1owrates. Repeat the experiment for various valve settings, N, ranging from barely open to wide open. For each data set calculate the average velocity in the pipe, V = Q/A, where is the pipe area. Also calculate the pressure drop across the valve, 6.p = T'mh, where T'm is the specific weight of the manometer fluid. For each data set also calculate the loss coefficient, K L, where the head loss is given by hL = 6.p/y = KL V2/2g and T' is the specific weight of the flowing air.
Calculations:
A
= 7TD 2/4
Graph: On a single graph, plot the pressure drop, 6.p, as ordinates and the f1owrate, Q, as abscissas for each of the valve settings, N, tested. Results: On another graph, plot the loss coefficient, K L , as a function of valve setting, N, for all of the data sets. Data:
To proceed, print this page for reference when you work the problem and click II/'re to bring up an EXCEL page with the data for this problem.
T 1 h
Water --""--',,
,-:--
Free jet
II FIGURE P7.74
Solution for Problem 7.74: Head Loss across a Valve
N
D, in. 0.81
Hatm , in. Hg 28.7
h, in.
Q, ft"3/s
0.235 0.195 0.169
0.479 0.386 0.341 0.289 0.214
V,ftis
N
KL
47.8 33.8 26.2
65.7 54.5 47.2
2 2 2
9.95 10.21 10.54
48.9 32.9 26.1 18.8 10.0
133.9 107.9 95.3 80.8 59.8
3 3 3 3 3
2.45 2.54 2.57 2.59 2.50
48.6 39.8 31.3 22.5 16.8 13.6 9.6 5.1
231.1 214.3 193.1 161.5 140.8 127.4 109.3 79.1
4 4 4 4 4 4 4 4
0.816 0.777 0.752 0.772 0.762 0.752 0.723 0.731
15.8 12.3 9.3 7.2 5.0 3.3
250.7 223.3 195.9 172.7 144.5 119.0
5
0.225 0.222 0.218 0.217 0.217 0.211
=4 Turns Open Data 0.827 0.767 0.691 0.578 0.504 0.456 0.391 0.283
9.35 7.65 6.01 4.32 3.24 2.62 1.85 0.98 N
Ib/ft"2
=3 Turns Open Data 9.40 6.33 5.01 3.62 1.92
N
~p,
=2 Turns Open Data 9.20 6.50 5.04
N
T, deg F 70
=5 Turns Open Data 0.897 0.799 0.701 0.618 0.517 0.426
3.03 2.37 1.79 1.39 0.97 0.64 6.p
=YH20 *h
=6.p/(pV2/2) where V =Q/A =Q/(n*D2/4) KL
and p = Patm/RT where Patm YHg*H atm
=
R T Thus, p
=847 Ib/ft"3*(28.7/12 ft) =20261b/ft"2
=1716 ft Ib/slug deg R =70 + 460 = 530 deg R
=0.00223 slug/ft"3
5 5 5 5 5
Problem 7.74 Pressure Drop, ~p, vs Flowrate, Q
60 50 40
N
--+-N=2 ___ N = 3
<
~
.c 30
--------~-
c:
20
..
10 +-----__
~
-.-N = 4 --e-N=5
~--~~----~~~4----~
0 0
0.2
0.4
0.6
0.8
Q, ft 3/s A
Problem 7.74 Loss Coefficient, KL , vs Number of Turns Open, N
,----------------.
12 10
----
8 ...J
~
I--+-N=2
I.N=3 I&N=4
6
,·_------1I
4 2 0
0
. 2
3
N
4
5
I---e-N=5
7.75 7.75
Calibration of a Rotameter
Objective: The flowrate, Q, through a rotameter can be determined from the scale reading, SR, which indicates the vertical position of the float within the tapered tube of the rotameter as shown in Fig. P7.7S. Clearly, for a given scale reading, the flowrate depends on the density of the flowing fluid. The purpose of this experiment is to calibrate a rotameter so that it can be used for both water and air.
Equipment:
Rotameter, air supply with a calibrated flow meter, water supply, weighing scale, stop watch, thermometer, barometer.
Experimental Procedure:
Connect the rotameter to the water supply and adjust the flowrate, Q, to the desired value. Record the scale reading, SR, on the rotameter and measure the flowrate by collecting a given weight, W, of water that passes through the rotameter in a given time, t. Repeat for several flow rates. Connect the rotameter to the air supply and adjust the flowrate to the desired value as indicated by the flow meter. Record the scale reading on the rotameter. Repeat for several flowrates. Record the barometer reading, Halm, in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law.
Calculations: For the water portion of the experiment, use the weight, W, and time, t, data to determine the volumetric flowrate, Q = Wht. The equilibrium position of the float is a result of a balance between the fluid drag force on the float, the weight of the float, and the buoyant force on the float. Thus, a typical dimensionless flowrate can be written as Q/[ d(p/Vg(PI - p))l/2], where d is the diameter of the float, V is the volume of the float, g is the acceleration of gravity, P is the fluid density, and PI is the float density. Determine this dimensionless flowrate for each condition tested. On a single graph, plot the flowrate, Q, as ordinates and scale reading, SR, as abscissas for both the water and air data.
Graph:
Results:
On another graph, plot the dimensionless flowrate as a function of scale reading for both the water and air data. Note that the scale reading is a percent of full scale and, hence, is a dimensionless quantity. Based on your results, comment on the usefulness of dimensional analysis.
Data:
To proceed, print this page for reference when you work the problem and click hue to bring up an EXCEL page with the data for this problem.
t Scale reading
Float
o ~
FiGURE P7.75
7.75 I
Solution for Problem 7.75: Calibration of a Rotameter
d, in. 1.40
V, in. A3 1.50
PI, slug/W3 15.1
Hatm , in. 29.05
T, deg F 78
Air Flow Data SR 14.6 21.5 28.1 33.6 39.2 44.8 50.2 55.9 63.1 68.6 73.5 76.2
(Q/d)[p/(Vg(PrP ))]1/2 0.142 0.200 0.257 0.305 0.351 0.400 0.444 0.496 0.552 0.605 0.653 0.671
Q, W3/s 0.229 0.321 0.413 0.491 0.564 0.644 0.714 0.798 0.888 0.973 1.05 1.08
Water Flow Data SR 13.1 18.5 24.2 28.2 37.1 45.7 52.6
W,lb 6.52 8.01 7.02 7.81 8.20 9.21 8.19
t, s 19.9 17.7 10.4 10.1 8.4 7.5 5.7
Q, W3/s 0.0053 0.0073 0.0108 0.0124 0.0156 0.0197 0.0230
(Q/d)[p/(Vg(Prp))]1/2 0.103 0.143 0.213 0.244 0.308 0.387 0.453
P =Patm/ RT where Patm = YHg*H atm = 847Ib/W3*(29.05/12 ft) = 2050 IblftA2 R = 1716 ft Ib/slug deg R T = 78 + 460 = 538 deg R Thus, P =0.00222 slug/ft A3
7-'12
7. 75"
Problem 7.75 Flowrate, Q, vs Scale Reading, SR
1
.!! M <
I-+-Air
0.1
-=a
i--waterll
0.01
0.001 100
10 SR
Problem 7.75 Dimensionless Flowrate vs Scale Reading
!::! ~
~
rc
--..s -
0.8
-r--~-----~---~--~'----i
0.7
-------------\
0.6
---JIP------ - ---~---1
0.5
,-~----~-----jI
Q.
i
--;---~--------I
Cl
> 0.4 :::::0.3
-I-~~__'_~~~C---~~_+_~~~-----:
"t:I
0
) '~_W""------------'-'------~-_j
0.2 0.1 0.0
,i '--~--------------~-i
~------
-t----,.-----..,----;-----t-----\
o
I
20
60
40 SR
1-93
80
100
-+-Air
--Water
a, I
I 8.1 Rainwater runoff from a parking lot flows through a 3-ft-diameter pipe, completely filling it. Whether flow in a pipe is laminar or turbulent depends on the value of the Reynolds number. (See Video V8.1.) Would you expect the flow to be laminar or turbulent? Support your answer with appropriate calculations.
Re == ~ = ~D
If Re >t/-ooo fhe flow
is
iI/rbI/lenT, The
correspondinq velocify is
5 .fi~ Re 11 - ('1-000)( 1."J XIO- oS ) _ 0 0161 Ii V- D 3 ft -, Mo.sf likely Ihe veloc/fy will be greQter Ihan this oS
J
furbfJ/en! flruv.
t:8'J
I
8,3
8.3 The flow of water in a 3-mm-diameter pipe is to remain laminar. Plot a graph of the maximum fiowrate allowed as a function of temperature for 0 < T < 100°C.
VD - Q -For laminar flow Re:= -V ~ 2.100 J where V-7f Thu.s} fhe maximum ~ is gil/en by Re : : (fk)D = ifQ =2./00 or Q= ~/OO 71' 1/ /) 11
71'11 D
J
if
or Q == 20/00 :(O,003m)p = 7!9S
f/
J.l.f1.
ll'D2
J
where 1/-=::~
Qlld
r;~¥
WHh valves of 1/ {rom TabJe B,z we Obtain T, deg C
o 20 40 60 80 100
v, m /s a, m /s 1.79E-06 8.86E-06 1.00E-06 4.9SE-06 6.S8E-07 3.26E-06 4.7SE-07 2.3SE-06 3.6SE-07 1.81E-06 2.90E-07 1.44E-06 2
3
Flowrate vs Temperature 1.E-OS - , - - - - , - - - - - - - - - - - - - - ,
E
:::~~: +~-"'----"~-----------~---~--~--'~----~J
ci
4.E-06
M$1
i
-------
--- ---- -
------
-.--------.-.-.
2.E-06
---
-
--
--
O.E+OO 20
40
60
T, deg C
8-1
.
- -- -
j
j
-1 1
,
I
0
--
80
100
8.9
I SA Air at 100 of flows at standard atmospheric pressure in a pipe at a rate of 0.08 Ibis. Determine the maximum diameter allowed if the flow is to be turbulent.
fY!inimllhl Re:::
or wHh
V = Ii)
7f
e~D
= ¥Q
11' D2.
}
for tvrbvlent flow is
Re::
(1161 )D
e
fi"@ )l
Re :::/f000.
~p Q rr'p D
;:::
::: '1-000
lienee}
Q-== lfO~O:J.i D
(I)
Given oQ::; O.OB~ J where O'::'}p and p:: fr ThIJ,s, .& == OJ!-. 7 x/Jflf fir.) = 0 002.20 s/()q.s p (/716s/IJ,. H'ib )('l-to+ 100)'P, . ft.3 fiR so fhat _
Q-
.u3 ::: /. /3 .JL
Jh
O. 08 S
(..32. 2fi)(000220S/UlJ,S) . s~· ft:S
s
LJ h Ji':= 3.'1# 1./0-7 Ib'S ( nence wif fl20 see Table B~3)J £rt. (I) 9ives J
(O,OO2.20~) (/. 13 ~) D- qOOO 11' f'- - iJooo"" (3-<>'f'l XI 0- 7 '~t) _
if
eQ
_
fi
If
2.27
8.5
I
8.5 Carbon dioxide at 20°C and a pressure of 550 kPa (abs) flows in a pipe at a rate of 0.04 N/s. Determine the maximum diameter allowed if the flow is to be turbulent.
For turbulent flow J Re or _ Jfe Q D _ f.e(i _
= fJ~D > 'fOOO
J
where Q =: VII ::
Re - 1l}i D2. - 1l'fJ'D - ~ooo Thvs, _ II- eQ II D - 11-00011'1'- J where ifG = 0.04-:s and j1. =/.'1Hencs J
_
D-
I ) (0.0'1 JL oS) ( 9.S/f..) 4000"" (J.Jf7 X10-5 !f;;f) Lf.
= 0.0883 m
8-3
*
D2.V
5
X/O-
'fn~ (Table I.B)
e.6
8.6
It takes 20 seconds for 0.5 cubic inch of water to flow through the 0.046-in. diameter tube of the capillary tube viscometer shown in Videu V1.3 and Fig. PS.6. Is the flow in the tube laminar or turbulent? Explain.
If Re = Y,9 < ')../00 the {fow /.$ /t~rnina~ .oS where 'O,5/f'J. / ( 12;f1./fl)3 20..$ V:: R :: 1//1:_ A JI D2. ~i~JI,E. f4)'If
'f (
f+
or
V=I. 25 :s {2Thvs wifh 11:::. /,1../ X lOS ~ (see Table I.s) J
Re --
!1 ( °iO:AU ) /,2.1 x/f.5 re
1.25
= 396 -< 2./00
oS
The flow is laminar.
8-'1-
• FIGURE
P8.~
8.7
I
'1,.7 To cool a given room it is necessary to supply 5 fe /s of air through an 8-in.-diameter pipe. Approximately how long is the entrance length in this pipe?
V:: % =
;
(!IeffY
= 1'1.3 !f
Thus, wilh
/If.3~ -/{. (~fJ.~tf) =6 OJ 700 Re ;;: 7VD ;;: 1.57x I() .!j-
if (see 74Me I. 7)
t/=J. 57 XI 0-'1
> if 000
J.L
1/
Hence}
~ ~ '-A 'I Re Y6J
or
~ ~ 'fJI (6q 700)~(ft) ::: 18~ If ff
e-.5
.
f hI I
so me 1/0'1/ /s IJr /) en .
8,8
I
8. 8
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 1.85 Ib/ft2. Determine the pressure gradient, ap/ax, where x is in the flow direction, if the pipe is (a) horizontal, (b) vertical with flow up, or (c) vertical with flow down.
-Ii
AfJ sine == ')..r{ Thus) willi 7".: 7; at r:: -#: tlnd ~:: --f fA,s bec()mes In genel"o/,
J/- 'IW W == - D -
M
It'. (J
Sih
f)
a) For a hori'ionla/ pipe ()~o
~
=_
b)
JfiW .. _ .II- (J,BS -#-2.) = _ Z ¥o ~ D
1'X
I fl
'
For verlica! flow up
e::: qOO
'fj
tI- (1, 65
Ul.. __ dX
-
_}to
U
__
-
I It
jJ
H
6"
LL
16
- ~,7' 1/3 == -
69.
and c) For verfiCt~! flow down e :::_qoD ail.. __ Lf'lW + l'I == _ 'f a. as ~) + 62 tf JA. = 55 1X - D O l ff . ff3
8-6
•8
0 I
Ib
fi3
.&
fl3
8 ,9
I
8.9 The pressure drop needed to force water through a horizontal l-in.-diameter pipe is 0.60 psi for every 12-ft length of pipe. Determine the shear stress on the pipe wall. Determine the shear stress at distances 0.3 and 0.5 in. away from the pipe wall.
For
hori:zonta I pipe
a
Thus,
_
'( -
1
=
b
~
(O.6X/Jf~.pp) _ 'J.(J').. If)
~rc- or
..!k.
- 3,6
r
'(=
-f ~ ,.,
FfL
J
where r it
Hence]
/b 'w :: 3.6 (o,S) 7Z =- 0./5 W
'J-
and wifh
r=(o.s-O.3)in. =O.2in' J '( == S.6 ( 9,~ ) :: 0.06
ft2.
Finally J w/fh r
8./°1
=:
(O,E -O,S)'17.
::<
0
in.
r =- 0
KHI Repeat Problem 8.9 if the pipe is on a 20° hill. Is the flow up or down the hill? Explain.
¥. 2.;:
For a pipe ()/J a hilI = + 0 sine J where e == ± zoo /lsslJfJ'Je fhe flow /s uphd/ ; e·= +20° Ih -rL ~ LJ ~ . 31 '>- J.. (05 )[0.6 xllPI- rp. lb· ;u!7 I flUS, I : : 2. L7 - j smf)j or Iw:: 2 /rfl I~ If -t2·'fff3SIIJ J Or
Ih fw ::: - 0.29£ H2o
Since we fnv.s1 have
?W >0) fhe flow IJ7vstnQt
be uphill.
Assume fhe flow is riownh,11: f:)= -Zo' .JA -rL L [~ . ,7 r [0.11 XIII-If FlJ. I nils J ' { r::: 2. :£ - i sJnf)j or '(:: 2: 12. If lienee, wilh
=1'1. 3 r ~
r =E -
l#iJh r = (0.5 - 0.3) in. :: 0.2. in.) (= /Jf.3 (~;) = 0,238
#i
r
=(0.5-0.S)i/J. ::0
J
'(
=0
6J
where r ~ ff . The
rjow is downhill
T,
fw = /~3(1;t) == 0,596 ~
Wilh
J
Ih·
+t2,/ffj3 stn20
I
e,ll
8.11
Water flows in a constant diameter pipe with the following conditions measured: At section (a) po = 32.4 psi and Za = 56.8 ft; at section (b) Pb = 29.7 psi and Zb = 68.2 ft. Is the flow from (a) to (b) or from (b) to (a)? Explain.
the {Jow is uphill. Thus) or wilh JIg = ~ ) iJSSfJme
h
'L
=
Ib +:z - I'i> - Z '1
a
4'
b
=
(b) (a)
Ij- + ¥i +~4 = 'I- t ¥i +Zb +0. .
(32.1 psi -2rt. 7psi)(;IfIff!:) +56 eff -682ft 6ZJf
or
hi =-5./7ff
!A.
~
'.
which is impossible. ThllsJ fheflowjsdownhi/l,frorn(bHo(Q).
8-8
x. J:2
Water flows downhill through a 3-in.-diameter steel pipe. The slope of the hill is such that for each mile (5280 ft) of horizontal distance. the change in elevation is ft. Determine the maximum value of if the flow is to remain laminar and the pressure all along the pipe is constant.
az
az
or
.tlZ ~.J silJfJ ~ - (s~eoff))( (-I,f¥- K /()-s) :::: O. / ()2.
8-q
If
8. /.3
I
8. 13 Some fluids behave as a non-Newtonian power-law fluid characterized by r = - C( dill dr)", where n = 1, 3, 5, and so on, and C is a constant. (If n = 1, the fluid is the customary Newtonian fluid.) For flow in a round pipe of a diameter D, integrate the force balance equation (Eq. 8.3) to obtain the velocity profile
u(r)
=
-n
(n + 1)
a) (12ee
1111 [ r(II"'1)
11
_
(D)lll-lI nJ _
2
i.f == 2-l so thai with (=:;-c(1fr)n Af=-2f(djflor #;:=-(/fp)* r-h ~
For any FlUid or
we ob14in
-L
-s du :::!(-2ti)j r*dr lL=-(2~1t(n~J)
which
iIJfe9rafes foqive
r(lW.) t-C, , where
0 is a conAanl.
.The fluid sficks to fhe pipe so thaf {). = 0 at r =: ~ • Hence 1 from Eq. (I)
,
(Ll)1;
C, = 2C~
n ( D (n,t') (ntJ)
so fhot
2" J
-L.
--'1- (- 2cL Af tJ. ;:: (n+J)
)n f- r ~ -f- (lL) flWlJ 2.
.. NOTe: Since we tire considering onl,y oJd inleger valves lor n we ct/IJ vse fhe felt;t ffJqt it" (1Ir)n= - KJ where 1< > oJ TPe/J ~::- K~ sO fhal
¥r
<0.
8-10
(/)
8.IJf'"
I
8.14*
For the flow discussed in Problem 8.1S. plot the dimensionless velocity profile u/V<, where Vc is the centerline velocity (at r = 0), as a function of the dimensionless radial coordinate r I (D / 2), where D is the pipe diameter. Consider values of n = 1, 3, 5, and 7.
From Problem 8./3, Ii. (r)
Let
:=
-IL (
(n+1J
Ll{J
2 ic;
)h fr(lJril) + (JJ..)(CW)] 2.
Yc = (). (r=o) ,
Nofe: For ?-
==
or
u.
(2£1 Jii (~ J-n
(2)
c(M)n wdh djf <0 and!) on odd inleger
( >0 J we tnllsf
By dividing
Yc;:: (~~I)
(I)
(nil)
I
to hove Vc >0 as if musT.
hove C~o. ThlJs, from £q. (2.) J
J
£~. (J) by £1.. (1.) we obiQin
[r](CW)
Vc = I - (~)
r
This resuH i.s ploHed below for n = J) 3J 5; t1nd~ wtfh
An EXCEL
pr0'll'am
0 ~ (~) ~ I.
was vsed 10 do the calev/41ion.r alkiploif/fl9.
r/(O/2) 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
n= 1
n=3
n=5
n=7
uNc
uNc
uNc
uNc
1 0.998 0.990 0.978 0.960 0.938 0.910 0.878 0.840 0.798 0.750 0.698 0.640 0.578 0.510 0.438 0.360 0.278 0.190 0.097 0.000
1 0.982 0.954 0.920 0.883 0.843 0.799 0.753 0.705 0.655 0.603 0.549 0.494 0.437 0.378 0.319 0.257 0.195 0.131 0.066 0.000
1 0.973 0.937 0.897 0.855 0.811 0.764 0.716 0.667 0.616 0.565 0.512 0.458 0.404 0.348 0.292 0.235 0.177 0.119 0.060 0.000
1 0.967 0.928 0.886 0.841 0.795 0.747 0.699 0.649 0.599 0.547 0.495 0.442 0.389 0.335 0.280 0.225 0.170 0.113 0.057 0.000
( con 'f) 9-11
r/(D/2) vs uNc
1 0.8
-a 0.6
--n=1 ---n=3 ··--·n=5
0.4
---·n = 7
1::
0.2
a
---
-~~----~------~~
+-----~------r-----~----_+----~
a
0.2
0.4
0.6
8-/2
0.8
1
f.IS
I K J5 A fluid of density p = 1000 kglm 3 and viscosity J.L = 0.30 N • s/m 2 flows steadily down a vertical O.IO-mdiameter pipe and exits as a free jet from the lower end. Determine the maximum' pressure allowed in the pipe at a location 10m above the pipe exit if the flow is to be laminar.
Jom O.lm
-Re ::: 2100 Th(}J'.; 2/00 :::
for maxilf/flm pres.rfJl'e.
epoVlJ
k
_ 1000~ Y (O,/I1}) 030 •
#:!.
PJ~
or
V= 6.30lJ Buf for Jam/par flowJ
V::
ThlJ~ 6,30 So
(~e -!'J sine) /)2. .3:z.,P J
f ::
(LlP -
I
t:::PIO::::
~I/O.j. {/()m)silJt-P()'))( ()'/m):J. ..32 (O.3() ~) C10m) ",
fhal
Af ~ - 3, 7, XI 0
where lJ eo/1m J ~/()IJI
Jf :::
If 11/'-
-.3 7. t
~-13
kfa,
I
4IJd
&:: -po'
JI
~1/olW.3
e./6 ~.16 Water is pumped steadily from one large, open tank to another at the same elevation as shown in Fig. PB.16. Determine the maximum power the pump can add to the water if the flow is to remain laminar.
Length = 100 ft Diameter = 0.1 ft
II FIGURE PS.16
v
8./7
I 8.17 Glycerin at 20°C flows upward in a vertical 75-mm-diameter pipe with a centerline ve; locity of 1.0 m/s. Determine the head loss and pressure drop in a lO-m length of the pipe.
P=I2.60!i tn 3 Il=/. 50 N·S r' m'J..
For /qmillfJr Flow in fJ pipe, V=- overQge velocify = f ~qX :: i{J~):: 0.5.1f Thus
J =/Om
k
o '_ .e VD _ (J2607,?)(O.S1j) (0. 07£ In) = 31.5 < 2100
ne -
p
I 50 N·s • m2.
-
The flow i.s /(Jlflil}tJr so ih4f
V=
32~1
ThlJ,s J .ALl r
h
(AfJ-K'I.sinB)J)Z
== 3.2 JI.!V
D
-I-
were
J
~i
e=no
PI = {J2. +AfJ
hL == 4f -1. A
.: ......
ej) 2,
D-=O.07Sm
I
l!l.
L.<:::
;:.32 1.50'/ii2. lom),o,s.s) + ('I.8/~){/').60~)(!Otn) (0.07.5m)2.· III
= 1.66 x/Os -J,'J..
/I/S(), 2. t; +z, + ~ :::
~
8
7
N•.5)(
(
\--
"F +Z2.
~2
f ;;
this 9 ille.s 05
1,.
/,61)(/0 = (9. 8111;)(1260
+hi..
J
J
or ~f::: /66
or wNh ~ -= ~ J
:a) -/0",
8-/5
=
3, ¥3 m
kPQ 22.-
"Z/;::J, and
8./8
I 8./8
A fluid flows through a horizontal O.l-in.-diameter pipe. When the Reynolds number is 1500, the head loss over a 20-ft length of the pipe is 6.4 ft. Determine the fluid velocity.
hi. -_.l::C.: f D 'J-, where since Re::: /500 <: ZlfJO fhe Flow J
ioS
f;: 6~/Re =t'l//.5o o ;::;
ThV$
J
6, If fI :: 0,01/-2.7
or
V :::
8,19
/4111 ina,. ,
2,01
O,O~27
.so thai
;LO ·Ff
(0,/ //'}./4)
J:l oS
I X.J q A viscous fluid flows in a 0.1 O-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.9 m/ s. If the flow is laminar, detennine the centerline velocity and the flowrate.
For /omif/or flow in a pipe u(r) =
Vc[I-(-¥-lJ J where
D=O./m aIJd IJ.:::O.8~
r ==
-rJ.. I ,,{lSI
2.
hi _ 1/
O.B S -
[1_ (2 (o.o3Sm))]
Vc -
so t/Jai Q -= 11/'V :::
\.
O./Om
*D'J.
rO.5 'Ie)
~
or
II -
°1m - O. O/2m == O.03Sm ~
I Sqf!L
Vc - "
s
fro. 1mi' (0.5)(1.89
g-/6
cd
1') ==
7. 'fZx 10-
3
.If.
8.20
J
I--I~-
Oil (specific weight = 8900 N/m\ viscosity = 0.10 N·s/m 2 ) flows through a horizontal 23-mm-diameter tube as shown in Fig. P8.20. A differential U-tube manometer is used to measure the pressure drop along the tube. Detrmine the range of values for h for laminar flow.
8.2.0
tor
Iamii'J4r
flow
t~ h
flJe milJiI!JvlIJ
e.:f- 5 )./00 h is h::: 0 (no
the maximvm h is for Re::: ~/oo. Hence, v(o•O').3m) ( a'lOONlm') 9.8/ IfJ /.s2. 2100 ::
HenC8J A
Dr A
P -:: It -f:J. -: :
fl :::
0.5/11
0.0305 0.0'-3 HI
30J -¥OD "y1m2-
From m4nomeler equal ions: " of 0 (H of h) - S G ~Hzo h -
t H::: f:;. J or
Ap:: {J,-/2.::: (SG ~2.0 -r)h Thvs = h
J.
3 OJ If. 00 N /m"
(7 (9 8ooN/m3) - 8900/1//11 3 )
Hence
o ~h ~ 0.50Q!YI
8-/7
-!
~•.-Oi_1_ _ _ _f..L-23~jn_m__,. (2) ~""12
Re ~ 2./00; or
e: : r~. Thv~
where
- - 0.5 m
::: O.50qm
1/(11) and
8.2/
J S.21 A fluid flows in a smooth pipe with a Reynolds number of 6000. By what percent would the head loss be reduced if the flow could be maintained as laminar flow rather than the expected turbulent flow?
For eilher laminar or furhu/eni flow
1;£
hL ~ f D Z9 .- Thvs J w/lh Ihe same ~"/), 4nd 9
hL /(J1b
_
{I,m
"hL -IVI'b
-
+Ivrh
If the flow is laminar If the flow is furbf)lenf with Re :::to()() (lAd "* :::(), then fro", fhe ltlooJ)' charf (Fi,. 8,2()) ~vrb:: 0,035 Thvs
J
h
_ L /fJ'"
hl.-I
lJrh
::: O. 0101 ::. O•. JI 16 0.035
The heqd/oss would be redvced Dy (hLturh -hL/41l1 )/h/.tU,.~ == /-0.'186
If:!
O,S/If,
or
5/,11-70
8.22
I SG
=
0.87_[1
el). :::==--='--'-
-
Oil of SG = 0.87 and a kinematic viscosity v = 2.2 x JO - 4 m:! / s flows through the vertical pipe shown in Fig. PB.22. at a rate of 4 X 10 - 4 m 3 / s. Determine the manometer reading. S.22
--
h.
P. c::4 m
-
20 mm
FIGURE PS.22.
SG = 1.3
(I)
(2.)
From manomeTer consideralions
p, +ahI - %h + ~ h2. :: f2.
ThusJ
where ~:: SGtrlHJ.O:: 1.3(fl.811$)=12.7/f ~ and h =h- h:J.+jJ or h,.+h, :: hfi
1
I,-P,. = Ap ::;-'O(h 1 +h) +rm h :: (om- o)h - 01. Combine Eqs.
(~) and (3) fg
qive
Lf3.7 ~ = (l2.7J1.-8.53)1$ h - (8.5.3 ~)(1-IIJ)
or
h::
18.5
m
(3)
8.23
J
SG = 0.87_[1
-
(2}.::=~_
8.23
Determine the manometer reading, h, for Problem 8.22 if the flow is up rather than down the pipe. Note: The manometer reading will be reversed.
-
-
20 mm
ics4m
- r-
FIGURE PS.23
1h,
~Cr' tQ
..
'SG = 1.3
• 2rm (I)
(z)
considerafion.s III - oh, -I- d;n h - rhz =- III J where Om == s~ ~,.o =/.3(p.er ~)=/2.71f ~
Fr()m /flan()mefer
and 172 =i+h-h, or h2-+ hi
=J+h
Th(Js,
(1,-1,.
-==
Af == ~(h2.+h,)- ~mh =
-(d;,,- a)h +d'i
(3)
Combine Ef(s. (2) and (3) if) 9ive /I/.q
-
~ =-(J2.7'!-8.~3)~h +8.S.3~ (lfllJ)
or
{2.)e ' - -
h=-/8.5m
1 J 'S.5m
Nofe: Since h
8-20
-;:::=(I).
~
8.24 For Problem 8.22, what fiowrate (magnitude and direction) will cause h = O?
Prob. 8,22, AI=- (%-(f)h-(f'l Thvs) with h =0) Af::: - (fi af}d From
71' (-~J +rJ,)/)1f /2.8 j/.i
Q=
=0 =
Note fhqf ~fJ :J:0J but Q=O .since Af+rJ:::o 8,25 8.25 The kinetic energy coefficient, a, is defined in Eq. 5.86. Show that its value for a powerlaw turbulent velocity profile (Eq. 8.31) is given by a = (n + 1)3(2n + 1)3/[4n4(1l + 3)(2n + 3)].
From Err,S, 86)
_ eJAVu V3rJIJ
0( -
k
2
dll
where V== overtJ'je velot;ify, /} =7TR/
P
v- (n+l)(zn+/J :l \/
Fi £; I 8 if U = Vc - ..t:..Jn R • rom xample . Th vs wilh dl/ = 2. 71' r d,. QlJd -
1/ [/
I
i-3d. = : v/'
-
R
where where y::: ~ ,
IX
J
2.n
s
~
JaJ,j/l =2 71'S~tr) -j ]"I'r dr = 271' R"V,/S[I - y]ydy y~o
r=O
S
y=o
(0)
}
Let x::: J - Y so thai y = I-x and dy =-dX J Hence 1.3 0 ~ J.::n [1_y]=11y dy =-Jx~(J-X)dx =J(X J
Yc
x~1
n
- n+-3 X
3
-X7f
(I)
+/
)dx
0
~
n
- 2n+3
X 2n+31 n
X=I
x=o (2.)
(ntll (1.fJ+l'1
'l-n+' (n+3) (2I)f3)
8.26 ~.Z6 As shown in Video V8.3 and Fig. P8.26, the velocity profile for laminar flow in a pipe is quite different from that for 1.0l
Turbulent with Re = 10,000
.!!... = 11 Vc
- .!..1115 R
0.5
_ _ _ _ _ _ _ -1- _ _ _ _ _ _ _
a
0.5
1.0
/I
V; •
FIG U REP 8.2.6
For lam/nor or furbv/enl (low Q ~ fJV::: 1TR'J-V:: fUdl] =[U(~llrdr)
R
J
=27Tf urdr r::O
a) LQtrJinor flow: R
7TR"V,. 271Vc, Jr[l-(i /}:lr Thus) V==t\{
i
b)
==
i
='
2-
2.
==
271'\f~- ~]
2-
== 71'4 Yc
For U::: V::: ~ fhe ervttfiOI1 for
0
I - (~)"" or (f)"'" t Thus,!' "'Vr R I
TlJrb {Jlen! fJow
7TR 1 V: 27TVc
R
-% r;
O.707R
Vi
I
Sr[l-ftolr
==
9;ves
27TR2.'Vc J(.f)[J-(f)] 5d(f)
o
0
LeT y:= I - (f) .so -fha-f (f) :: /- y and d. (f) ;: - Jy ThlJ's >:=0 I 7T
R"" V '" 271 R2.Vc
f
(I-y) y I(s (-tiy)
_/
::: 21lR'"
y- [.5
V=
21!R2.\f
S
U:::
t.s
2.
v':: 1* the or
.f(YIi- - yVs) 1)1 0
Vc '6 - It J = 271R Vc
if Vc hr ~ := #- '" [1- f
Or
:=
f '"
(').C) 7&
e'!vQfitJn fof'
O.7SO so
fh4"/
t'
~ 9ives ==
O.7S0
R
8.27
I
8,27 Water at 80 of flows in a 6-in.-diameter pipe with a ftowrate of 2.0 cfs. What is the approximate velocity at a distance 2.0 in. away from the wall? Determine the centerline velocity.
V=
R
II =
*(.f2.
PI
2.0"
VD
fO" ::; /0.2 s so fhaf Re::; r -
::
(JO.2.{i)
(Aft)
if
9.2-6 )\/0-6
.J..
The flow is /()rbu/enf with ~::: (J - t )n) where n';; 8.3 (see
-E
=S.5/ X/rr
Fiy. S.I )
Thvs, (see Example B. If) V Vc
::=
:J.n'(n+l)('-fJtJ)
or _ IO.2~ Ve -
/llso)
O.BJt.2.
7-(8.3)':::: (8.3.1-/)(2X8.3+/) : O.8Jf2
ff = /2./ :s
af r= 3ifJ.-2.0in. =/.Oil'),)
.L
-L 8.3
fi.(J-;~~~)= .
u::::Vc(I-f)fI:/:2.1 s
J/.S#
8.28 8.26 During a heavy rainstonn, water from a parking lot completely fills an 18-in.-diameter, smooth, concrete stonn sewer. If the flowrate is 10 ft 3 /s, detennine the pressure drop in a 100-ft horizontal section of the pipe. Repeat the problem if there is a 2-ft change in elevation of the pipe per 100 ft of its length.
(b) W/{h
f/fJW Vfh/I/ ~ - Z/ :
2
ff so fhat
£jf :: (62ll-!t3)(t.H)( /tf.~~:'''1 +O. 266/~1. (c)
:::
/'/3 psi
Wi-Ih flow downht11 .F2. - Z, :- 2ft s() fhQf .lJ{J:;:: (62·~ffo)(-1.fI) (II{.'J~?-)+ 0.266 *'=
:=: -
0.10/ psi
8.zq r 8.2 9
Carbon dioxide at a temperature of 0 °C and a pressure of 600 kPa (abs) flows through a horizontal 40-mm-diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m 2 per lO-m length of pipe.
8.30
I 8.3:0 Water flows through a 6-in.-diameter horizontal pipe at a rate of 2.0 cfs and a pressure drop of 4.2 psi per 100 ft of pipe. Detennine the friction factor.
= 0.0300
8-2.5
8.31
I -T
8.31 Air flows through the 0.108-in.-diameter, 24-in.-long tube shown in Fig. P8.3J. Detennine the friction factor if the flowrate is Q = 0.00191 cfs when h = 1.70 in. Compare your results with the expression f = 64/Re. Is the flow laminar or turbulent?
h
7,-
./"77;
(I) Air
•
•
..LJ.
i'
\/,2
""...!.L + Z
2p
where
I
FIGURE
Water
I I
I
_r---
L.L
.l..
i
0.108 in.
t
'Z,:: 2'2
fJ2. =: 0
;
~::O
J
and
V::: v.2. =
~
!i
O,OOlql 8
f ( 0'11
A =
oS
ff)2.
= 30.0
{i
Hence J Eq. OJ becomes
fJl = -fpV2(1 + f-Z-) or 8. e'l=;. (0.002.38 S~~)
*-
[If f (-o.~o~~Jl30.0~)"
Also Re = .p! or R _ (~Ii) (30,0 y) e -
1.57)(
IO~~ ~2.
Note~
47-e ::: 1'1
6'1-
1720
:: 0.0372
(2)
(I)
f, = Oq~o h = (6 2 .1f.!ffi) (ft FI) =8. 8'1 ~ A~~
V
P8.~1
~2. £ V2. 7'r~? J2. + ~ + Z2. + f 75 2p -
=
--I
-
24 in.
or
f =0.03 2.~
= /72-0< '2. 100.J -fhe fl()w is lamifJar
8.33 J
8...33 Detennine the thickness of the viscous sublayer in a smooth 8-in.-diameter pipe if the Reynolds number is 25,000.
51/
f
O.s = U. we obfain
J
IN
heref~2.. =- ('fMt)lf e and
ea
fw =
Thvs, .51/ .511D 5 6 = ""\ff' = 1Jr V'8 V
and
u. 4 ==
yJ
r _
Va vD ) or °
05 -
'>-
£w:: I
D1 'I•
f1.
Si~ce LlfJ == D "i"pV I
V
SD D
1tr
l1e Ye
From Fi9. 8. 20} for a smooth pipe with Re:: 2.5)( IO~ f == 0,02'1 Thus) from £,{.tl)
6
~
=
sf§' (R.~-I) = O.OO2.Jf3 fI
2,5XIO'"
Vo.
~
O")./f
8-27
(I)
B.3#-
I 8.34
Water at 60 OF flows through a 6-in.-diameter pipe with an average velocity of 15 ft/s. Approximately what is the height of the largest roughness element allowed if this pipe is to be classified as smooth?
5;
Lei h = rov9hness heiqhf. Thos} h =4 J where O:S::: with u~ =( "; )~ and?;::: e . Since AI' -== f-! ieV:l. we obflJin '>- - pfV2 --IT V (w - 8 or U., -Va
%i
For a smoofh pipe wilh Re = from h9' 8,2 0 f ::: 0.0125
t·
ThV.5, U· = (- O.; IZ5 (/s!j) or - . 5.fi.z,) 6' = .£it. = .5 (/,2/X/O oS :s
U·
o.sq,3
~
ift:::
{/s fj. ){j.:;~ = 6.19 x/Os we obtain //J./X/O -
.s
='
-=
O. 593 ~ /'02x/ri~ ff
8-28
8.3.5
J
8.35 A 70-ft-long, 0.5-in.-diameter hose with a roughness of e = 0.0009 ft is fastened to a water faucet where the pressure is PI. Detennine P I if there is no nozzle attached and the average velocity in the hose is 6 ft/ s. Neglect minor losses and elevation changes.
D=0.5 in.
V
~~~:------~~~(~ (I)
.£ = 70ft
0) 0.000'1 If = 2.16 X/o-z.
(~fJ)
= 2.07 X/O¥ we obfain
Hence} from
fJ, '"
f= 0.052.
Eft. (j)
(0.052) (;;'ff) /2.
ff
~
{1.91f
~"1f ) (6#/' '" 3050 ~ ~ 21.2.ps;
8- 2. 'I
8.36
I 8.36
Repeat Problem 8.35if there is a nozzle of diameter 0.25 in. attached to the end of the hose.
Jff+* +II = Pr +fi +Z2. +(
Iii
I
v
: D=o.5 in.
a-+1. (I)
=70 ff
~ (~)
D =: 0.').5 in . .2.
where Z6h I V,=V;61f, fz=O and V,. ::: Y!& =V.I (.Q..) -'6 f1 ) (z/"::: 2. '1-11 lil D~ \~ s .s
Thus, It =(V/·+{-t V:L)~ =-t p(v,.:/. +fl V'") r· 'Ih £. o.oooqff -2rrom Fi,- 8.2·0 WI D = (0.5 ft) =2.1& x/a and
Vb
Re=-= 1/
(6 f1 )(0.s 0) oS
-rr
5
/.2/ x10-
!F
~
=2.071
~
weobfaif)
(I)
(:::O.052
Hence) from £'1. flJ
1', := i
(I.9'1
S~~$)«2.'f fj-/ +(0.052) (~:) (6!1f)
=36 oq I$. '" 2.5.1.j}.
Note: To lI'Jainfain the same flowrale wdh fhe noy-zle aHocherJ QS compared fo fhat wlthouj fhe nO"i2/e (see Probe 8.35) the pressure mvsf be increased from 2/.2 psi 10 25.1ps,:
~-30
8.37 "J iteration procedure to obtain f. Plot a graph of the percent difference in f as given by this equation and the original Colebrook equation for Reynolds numbers in the range of validity of the above equation, with e/D = 10- 4 •
8.37*
The following equation is sometimes used in place of the Colebrook equation (Eq. 8.35): 1.325 = [In[(e/3.7D) + (5.74/Re o.9 )]F
f
for 10- 6 < e/ D < 10- 2 and 5000 < Re < 10+ 8 (Ref. 22, pg. 220). An advantage of this equation is that given Re and e/ D, it does not require an
Lei ..of == fapp-f
J
fapp
fap,:- approximafe
where
reSfJ/f
ohI4/iJe.d
(rum
/.:3 25
..J..... = -2.0 log[~ +M. 1(f 3.7/) Re1(f
Thvs with J
fa
t
/.3'-5
[In(2, 10X/0 -t
pp
and
W
5
= -2.0 loq
For
these become
== IO-1f
=.
3
Sx/O ~
where
Fapp
[2.
J
.s.7~A )11
Reo ~
7()x/O-.s -I-
t:Vr]
(.1)
f ca Icv Iate an"I p/1 07 IOOA:: ana f are obtained from £qs. (I) and (2.)
Re
~ 10
8
J00 ( fqpp - {)
J
Proqram P8·fI=.37 shown be/ok! :'00 120 130 1i!-0 150 160 170 180 190 200 210 220 230 2i!-0 250
2-
Wq-s
used f()r
f
J
!he cQ/cu/afions.
cis print "**************************************************" print ,,** This program calculates the difference **" print "** between the friction factor given by the **" print "** Colebrook equation and that given by the **11 print "** approximate formula provided. The Cole**" print "** brook result is determined by an iterative **" print "** routine. **" print "**************************************************" rr = lE-i!Re = 2500 faprox print " R e f f - faprox, %" for i = 1 to 16 Re = Re*2 faprox = 1.325/(log(rr/3.7 +
8-3/
8.37· I (con't) 260 fp = faprox 270 goto 290 280 fp = f 290 f = 1/(-2.0*log(rr/3.7 + 2.51/(Re*fp~0.5))/log(10) )~2 300 if abs(l - f/fp) > 0.000001 then goto 280 310 diff = ((f - faprox)/f)*100 320 J,?rint using "#.###~~~~ #.###### #.###### +#.###~~~~II;Re,f,faprox, diff ~30 next i ************************************************** ** This program calculates the difference ** ** between the friction factor given by the ** ** Colebrook equation and that given by the ** The Cole** ** approximate formula provided. ** brook result is determined by an iterative ** ** routine. ** ************************************************** Re f faprox f - faprox, % 5.000E+03 0.037505 0.037961 -1.216E+00 1.000E+0~ 0.031037 0.031138 -3.233E-01 2.000E+0~ 0.026101 0.02605~ +1.809E-01 ~.OOOE+O~ 0.022286 0.022196 +~.017E-01 8.000E+0~ 0.019319 0.0192~1 +~.0~7E-01 1.600E+05 0.017026 0.016985 +2.397E-01 3.200E+05 0.015290 0.015295 -3.227E-02 6.~00E+05 0.01~032 0.01~077 -3.176E-01 1.280E+06 0.013179 0.0132~6 -5.091E-01 2.560E+06 0.0126~3 0.012713 -5.513E-01 5.120E+06 0.012332 0.012391 -~.7~8E-01 1.02~E+07 0.012162 0.01220~ -3.~99E-01 2.0~8E+07 0.012072 0.012100 -2.298E-01 ~.096E+07 0.012027 0.0120~3 -1.362E-01 8.192E+07 0.012003 0.012012 -7.108E-02 1.638E+08 0.011992 0.011995 -2.875E-02
(f - fapprox)1f vs Re
0.6 0.4 0.2 0 0~ -0.2 >< 0 Q. -0.4
-
!
t
!
J
/
~
r\
V
Co
III
'to'to-
~
~
-0.6 -O.S -1 -1.2 -1.4 1.E+03
1.E+04
1.E+05
1.E+06
Re
8-32.
...
V
j,;"
..-.
l-<
1.E+07
1.E+OS
1.E+09
8,38 T 6.38 Water flows at a rate of 10 gallons per minute in a new horizontal O.75-in.-diameter galvanized iron pipe. Determine the pressure gradient, 6.p / e, along the pipe.
Q
:=
/0
4 ~ (JmJE..) (7. 3/ in}) ( 19 /--"3) -::: 0.0223 mm 60S / fa I 172' /IJ.
Thvs - Ji _
V-
~
0.02.2.3 s
J
II -
_
-r;:-H )2. -
:tI. (0.75 J{-
7.')..
7
It .s
.tt ;s
Now, for a horizon/IAI pipe
Af::
f -t t pV2.
Re :: til:: -If
y
where si"ce
7.27J}
(~2.fl) S f+
1.2/1.10--:s
::: 3. 76}( /0
~
and. S::. ::: O.-Ooosft ::: O.Oog D
e;-".s II)
if follolNS from Fi'l' e.20
fhal
f:: 0.037
ThvSJ 2-
E£:;; 0.0.37 (J.'I'fs/f/9S Ifl.3) (7.27(I/s) :: 3 o. ¥ 1 (f!;?f fl) (2.)
II; ( I-N~ 1{3 --'11'1 in.~)
= 0.2/1 psi/II
~-33
8.
if!
I 8.41 Air at standard temperature and pressure flows through a l-in.-diameter galvanized iron pipe with an average velocity of 8 ft/s. What length of pipe produces a head loss equivalent to (a) a flanged 90° elbow, (b) a wide-open angle valve, or (c) a sharp-edged entrance?
.L D VD ( 8!sf )( I ~ H) .3 where 'lid" ne:: --:n-::: -"".f:f! = Lf.25x If) y I.S7XIO
t.e ::: -KLffD' £.
0.0005 (f =(//12. H) :: O.~06
75 Thus
J
t.ef ::
.s
V= :!l~2 Thus,
hi =f-!s!i ' where
~
2.
8 {(Jm)Q. hL- .".2.(q.g/~)D5 artWn
IUbil1tJ &~ or
Nofe: The
lhil'JillJU/fJ
hL= fl ~
(::r/= ~~~'-
h'J. ::O.082.6r;s frt , UIhere hJ."'In,D,vIllJQ"'s /tI3
or
Re:: y# = 1T~~
f)lso}
.I
.
10-' m.1/ s . In your solution obtain the friction factor from the Colebrook formula.
Water at 40°C flows through drawn tubings with diameters of 0.025, 0.050, or 0.075 m. Plot the head loss in each meter length of pipe for flow rates between 5 x 10- 4 m3 /s and 50 x
For
TnlJS WI1'1
(see Table 8./) we oblain f:: 0, 0'15 Fi 9 ·8.20) kJ.. (/2- f-l) =1.85ZKi. or 0) 90' elbo'l} : k;. :: 0.3 or i~(J::: o.S56Fl oO-¥S (, . b) qlohe valve: K:: 2. or ie'l =3. 70 If c) sharp en/ral)ce: KJ.::: 0,5 Of is'f,=o.q2.6If
8.4-211 18.42*
orwdhi=lm,
(
-L.
I
-B'::
",m ~ 1.5XIO- m -6 I.S~ 10 ) where D ~m
=
-¥ Q
0.0015 co
f?e
(I)
6
1f(6.S8XIf)-7~) D OCCbrS for @1II1i'J
Relhin =/Jllfx/o" ~.~~~If = /2/100 rfln98 01 pqrameler.s considered. Hence, from Ef. (8.35)
J
or fie =1.9'1-x/06
(2.)
2
(3)
.
and D/lJllx. Thus,
The flow is
7urbv/cnt over the
J- == -2,0Ioq[.-f- + 2.. 5 /] or with £.:: /.5 x/o-'m Vf ,3.7D fie 'ff -L = -2 0
1ff ' Thvs for
100 [If.o.sx/o-7 + J
D
2.5 J 1
Re
vr]
~ .so x10-'1f-3 and w/lh D=: o. 0 2.5 rn O. OSO m J or 0.075 m de fermine Re frofll £1. (3); f from £'1' ('S'~ J
.5 xIO-~ ~ :s. Q
(fl.)
I
J
B-.3Lf
8,1f2"
j (conll)
and hI...
frOfh El(.O). These resvlls are calculafed and plotted below. (h. -::hJ.. (QJ D), See Pro 9ram P8# '1-2 sh()lIIn be/ow.
100 110 120 130 1~0
150 160 170 180 200 210 220 230 2~0
250 260 270 280 285 290 300 310 320 330 340 350 360 370
cIs open "prn" for output as #1 print#l, "**********************************************" print#l, "** This program calculates the head loss **" print#l, "** as a function of flowrate and pipe dia- **" print#l, "** meter, using an iterative scheme to **" print#l, "** determine the friction factor from the **" print#l, "** Colebrook foxTrlula. **" print#l, "**********************************************" D = 0 for i = 1 to 3 D = D + 0.025 rr = 1.5E-6/D print#l, " " print#l, using "For D = #.###:!t m with e/D = #.##:!t~~~~l!;D,rr print#l, " Q. m3/s Re f hL. m" Q = 0 for j = 1 to 10 f = 0.02 Q = Q + 5.0E-4 Re = 1.94E+6*Q/D fp = f f = 1/(-2.0*10g(rr/3.7+2.51/(Re*fp~0.5) )/log(10) )~2 if abs(l - f/fp) > 0.0001 then goto 310 h = 0.0826*f*Q~2/D~5 print#l, using II #.#:!t#~~~~ #.:!t##~~~~ #.#### #.##:!t~~~~II;Q,Re,f,h next. J next i
********************************************** ** This program calculates the head loss ** ** as a function of flowrate and pipe dia- ** ** meter, using an iterative scheme to ** ** determine the friction factor from the ** ** Colebrook formula. ** ********************************************** For D = 0.0250 m with e/D = 6.000E-0:, Q, m3/s 5.000E-04 1.000E-03 1.500E-03 2.000E-03 2.500E-03 3.000E-03 3.500E-03 4.000E-03 4.500E-03 5.000E-03
Re 3.880E+04 7.760E+0~
1.164E+05 1.552E+05 1.940E+05 2.328E+05 2.716E+05 3.104E+05 3.492E+05 3.880E+05
f 0.0223 0.0193 0.0178 0.0169 0.0162 0.0157 0.0153 0.0150 0.0147 0.0145
8-35
hL, m 4.718E-02 1.629E-01 3.384E-Ol 5.702E-01 8.563E-Ol 1.195E+00 1.586E+00 2.028E+OO 2.520E+00 3.062E+OO
8. ifZ"
= 0.0500 m with e/D = 3.000E-05 Q, m3/s Re f hL. m 5.000E-0~ 1.9~OE+0~ 0.0261 1.727E-03 1.000E-03 3.880E+0~ 0.0222 5.873E-03 1.500E-03 5.820E+0~ 0.0203 1.208E-02 0.0191 2.021E-02 2.000E-03 7.760E+0~ 2.500E-03 9.700E+0~ 0.0183 3.017E-02 3.000E-03 1.16~E+05 0.0176 ~.189E-02 3.500E-03 1.358E+05 0.0171 5.532E-02 ~.000E-03 1.552E+05 0.0167 7.0~2E-02 ~.500E-03 1.7~6E+05 0.0163 8.717E-02 5.000E-03 1.9~OE+05 0.0160 1.055E-01
For D
= 0.0750 m with e/D = 2.000E-05 f hL. m Q. m3/s Re 0.0289 2.516E-0~ 5.000E-0~ 1.293E+0~ 0.02~~ 8.~83E-0~ 1.000E-03 2.587E+0~ 0.0222 1.738E-03 1.500E-03 3.880E+0~ 2.000E-03 5.173E+0~ 0.0208 2.897E-03 0.0198 ~.313E-03 6.~67E+O~ 2.500E-03 0.0191 5.975E-03 3.000E-03 7.760E+0~ 9.053E+0~ 0.0185 7.876E-03 3.500E-03 ~.OOOE-03 1.035E+05 0.0180 1.001E-02 0.0176 1.237E-02 1.16~E+05 4:.500E-03 0.0172 1.~95E-02 5.000E-03 1. 293E+05
For D
PROBLEM PSIlI"f-2.
3
.J 1:
2
en
Ul
o
.J
C
1
oc(
III
D=O,0.50m
J:
o D= 0.075 m -1~__~____~______~__~____________________________~~
o
1 2 3
FLOW RATE.
No Ie fhe
sfl'on9
Q
(M3/S)
dependence of hI. on D.
g-36
5 3 ''10-
g,
if]
J 8.43 Air at standard temperature and pressure flows at a rate of 7.0 cfs through a horizontal, galvanized iron duct that has a rectangular cross-sectional shape of 12 in. by 6 in. Estimate the pressure drop per 200 ft of duct.
For a horjzontq/ dvcf Af::: 'thL = f or _ .
7
#3
V- (J~in.)(6in,)( I n~ 2.)
'fh /If'fIl1, WI D .= !td = #- (O.5f/1.) P
?J
fi
-I.If.o
J
sand Re" -
(/)
Vf)h T
= 0.66711
(2+/)fI
Thus, .f:/.) I. Re = (I'/-. 0 [O. 667ffJ. = !)
oS
h
_
i "ipV2._where V=~
&.9SXIOif
/.57 XIO'f.p.2. E.
Also J for 90 /I/QlJiled /ron E~ 0 .. 0005 fI) or 25h::: From ~'9' 8.20 we ohtain f= 0.02.2.7 Thus) from q. (/) w,fh i =20()/fJ
AP = (0. 022.7)~:6~ f
(2..38XIO-.3
O,QOQStt
0.6671/
s~!!J) (III-. o!f) 2. ==
8-37
I.Sq
:: 0.000750
Ita
=0.0 I/O
pSI
8,JfJf ~.44 Water flows at a rate of 2.0 fels in an old, rusty 6-in.diameter pipe that has a relative roughness of 0.010. It is proposed that by inserting a smooth plastic liner with an inside diameter of 5 in. into the old pipe as shown in Fig. PS.44, the pressure drop per mile can be reduced. Is it true that the lined pipe can carry the required 2.0 ft 3/s at a lower pressure drop than in the old pipe? Support your answer with appropriate calculations.
Old
New
ill FIGURE PS.44
yes", fhe /Jew 11;& n4S 41()wer At
=
8-38
8,Jf6 Flow reducer washer
8.:46
To conserve water and energy. a "flow reducer" is installed in the shower head as shown in Fig. P8A6. If the pressure at point (1) remains constant and all losses except for that in the "flow reducer" are neglected, determine the value of the loss coefficient (based on the velocity in the pipe) of the "flow reducer" if its presence is to reduce the flow rate by a factor of 2. Neglect gravity.
50 holes of diameter 0.05 in.
FIGURE P8.46
8-3q
j
8.lf7 SA7 Water flows at a rate of 0.040 m3 js in a 0.12-mdiameter pipe that contains a sudden contraction to a 0.06-mdiameter pipe. Determine the pressure drop across the contraction section. How much of this pressure difference is due to losses and how much is due to kinetic energy changes?
D, =0.12m
~ ~* =;(00;;),,'I• m
Thus) wifh
• (0
~:lo Vl + l/ \.{2. J.. = !If + :lj- +%2. nL :;.'} wI/ere
and",3
:: 3.5'f-P-
v..
3 61=0,O'ff!!.. Sf
_ z, -~;.. "::t
=! =~O'f'T f O.o6m If.
'(2.)
--
m3
J
D:z. = 0.06 f17 (
( v,:z. + ¥LJJ +2j z,
1
(
-t ==(~~)2.-( ~.~:: )2= 0.25 we Ob/flin
( )
I
= IJM~
from
Fie;. 8.30
KJ.= O. 'fO Hence, from £~, 0) {J,- (J:z =
or
-t f{KLIS."-+ .l
fJl-/~::: 3Q.7 X /O
N
~'-- V, S] =f(999 ~)[o. '10 (11f.1 :f"+ (11f.Iq.l-(3.5'ff~ -3
N
fi2- + 93,OX/u- fii :: 133 kPQ
This represents a 3 q. 7 /rPo drop from losses Qnd a 93,0 kPa drojJ rif)e fo on incretlse in kinelic ener9,.
8-'10
8. Jfq
I
8•.4Q At time t = 0 the level of water in tank A shown in Fig. P8AQ is 2 ft above that in tank B. Plot the elevation of the water in tank A as a function of time until the free surfaces in both tanks are at the same elevation. Assume quasisteady conditions-that is. the steady pipe flow equations are assumed valid at any time, even though the flowrate does change (slowly) in time. Neglect minor losses. Note: Verify and use the fact that the flow is laminar.
t 2 ft at t !
-3ft-
-::=:=::------ ----::::=::=
--:j:j:::::::---';;';;-::;::::-'~---25
0 ft-
~__ B ____~~~~c=====~
FIGURE P8.4Q
2~/- ho = 1f!: -i Ii = 3;'$1 V, or bl uS;/)9 Ef#!3) -h0 = _(~)2. 32ft J dz/ D 'I D'J. at Lef FE so fhed ~ = ,
z,-iJ:-
' Dr)2 3 2jJi dF
2. r = - (D
r D'J.
__
"'-o.l-in.-diameter, gal. iron
Thus, Er.s, (J.) and (If) 9ive
2Z
i-3ft-
11- and £q.
(fF
(con)/J 8-1.f{
(05)
bec()mes
~AL-~
or Thvs
0(
dE + F== 0 J where ex. =
(ff
fdF
J
.r
()( F = -j dt
f/ence,
t. .
F =C e-(~)
or
0(
16 # I. (Dr lD2. D
In F =- t + C)
)2.
where C:::consTfJlJf
. ho C -(tlo<) 'lh lh . .J. I _J-/T::: e WI r. /, e Inn/tJ CO'KJI1/~n z/ =ho when t =0 J or C ::: ~ Thu.sJ z I -.!:k == l1.!. e- O/tX) 2. :2nof
TII
or
2/::::
IS} Z/ -
/Vole; As l:'-"OlJ J Zl-+~
if-[J+e-(i/o()]
For the cond/liof/.S 9iven} ho:: 2 ff tJnd
~)(ZSN)(~N )2. 5 «== {62.'f~)(~r1r (WD = 2.801
Hence~ Z I -- J
5
t
+e -(;..flX/O:S)
)
This resu/f is ploNed
h Zt.tV ff and were
1
r; ,..,. .s
be/ow. (Nofe: lim z, = Iff) l"'Ob
0.5 +--+--------1f-------I---+--t----t---t-----r---t----. O+---~--~~--_+--~--~--~--+_--~~
o
20qooo
40QOOO
60QOOO
t, sec
BOqOOO
1OOQOOO J
8.50~
f
8.50* Repeat Problem 8.4Q if the pipe diameter is changed to 0.1 ft rather than 0.1 in. Note: The flow may not be laminar for this case.
2 ft at t
=
0
l O.l-ft-diameter, gal. iron
FIGURE P8.·4Q (I)
(2.)
(0. 25Lflj)2.
(
or Zt =(i+O.00382)fI
it)
2 32.:z. S.1.
£::::
75
{f
0.005
o.lff
:; o. os
(see Table B.O
(con'/J 8-tf3
Hence,
vr I
= -2.0
J09 [ 1.35·xIO-3 T
_~
vYf
3.0'f-X}O
]
J
where V-v 11 .s
(5)
£r. (ff)
for Z, $2(f)J siarfin9 wdh in/fiat cOl1ri;lion Zt::: 2 ff 41 t::: o. Ohfa/n f {which is a {()ncl/oll of i becavse V= - t'/oo ~ (Ef. (3)) ,:r Q runc hon () t ) from £'1' (s). ProqrQm P8NEO shown he/ow WQS /}Sed 10 oldfll" fhe results.
Solve
(t:e. ifJfe9rtlle)
r
100 110 120 130 1""0 150 160 170 200 210 220 230 2""0 250 260 270 280 300 310 330 3""0 350 360 380 385 390 ""00 '*05 ""10 ""20
cls open "prn" for output as #1 pr in t # 1, " * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * " print#l, "** This program calculates the water depth **" print#l, "** as a function of time. The friction **" print#l, "** factor is obtained by itteration from **" print#l, "** the Colebrook formula. **" print#l, "***********************************************" print#l, " " t = 0 VP = 0 rr = 0.005 dz = 0.05 print#l, " z, ft dz/dt, ft/s f t, sIt for i = 1 to 20 z = 2 - (i-1)*dz f = 0.02 dzdt = -7.98E-""*(z - 1)AO.5/f AO.5 V = -900*dzdt Re = 8.26E+3*V fp = f A f = 1/(-2.0*log(rr/3.7 + 2.51/(Re*fp O.5))/log(10) )A2 if abs(l - f/fp) > 0.001 then goto 3""0 if abs(l - VP/V) < 0.01 then goto ""00 VP = V goto 300 t = t - dz/dzdt zn = z - dz print#l, using" #.#### +##.##### #.#### +#.##AAAA";zn,dzdt,f,t. next. i
8.so t I (con'l) *********************************************** ** This program calculates the water depth ** ** as a function of time. The friction ** ** factor is obtained by itteration from ** ** the Colebrook formula. ** *********************************************** '-' , ft '" 1.9500 1.9000 1.8500 1.8000 1.7500 1.7000 1.6500 1. 6000 1. 5500 1.5000 1. 4500 1.4000 1.3500 1.3000 1. 2500 1.2000 1.1500 1.1000 1.0500 1.0000
dz/dt, ft/s f -0.00440 0.0329 -0.00429 0.0329 -0.00417 0.0330 -0.00405 0.0331 -0.00392 0.0332 -0.00379 0.0333 -0.00366 0.0334 -0.00352 0.0335 -0.00337 0.0336 -0.00322 0.0337 -0.00307 0.0339 -0.00290 0.0341 -0.00273 0.0343 -0.00254 0.0345 -0.00234, 0.0348 -0.00213 0.0352 -0.00189 0.0358 -0.00162 0.0365 -0.00130 0.0377 -0.00089 0.0402
t, s +1.14E+01 +2.30E+01 +3.50E+01 +4.74E+01 +6.01E+01 +7.33E+01 +8.70E+01 +1.01E+02 +1.16E+02 +1.32E+02 +1.48E+02 +1.65E+02 +1.83E+02 +2.03E+02 +2.24,E+02 +2.4,8E+02 +2.74E+02 +3.05E+02 +3.44E+02 +4.00E+02
2.0 1.9 1.8 1.7
;: ~
1.6 1.5
N
1.4 1.3 1.2 1.1 1.0
o
100
200 t, sec
8-t;-5
---------
300
400
8.5/
8 .. 5/ As shown in Fig. P8.5/ , water flows from one tank to another through a short pipe whose length is n times the pipe diameter. Head losses occur in the pipe and at the entrance and exit. (See Video VS.4.) Determine the maximum value of 11 if the major loss is to be no more than 10% of the minor loss and the friction factor is 0.02.
• FIGURE P8.51
If hLhJ4Jor .= lo'ZhL..~ ) then mlfJor 10
f
2., - '£ 1\ 1jv2.
1 v'" D
2: KL
or
1- _ £KJ. D - 10
f
KIJ. en 1rf'4nce + k:'L ex/I = O. e +I :: I. e -rhvs wifh f:: 0.02 and 1: n /) £r. {/J hecomes where
=:
j
-nD D
-
/.8 /0 (0,02..)
Of'
n - q --
(I)
8.52
J
8.S 2. Gasoline flows in a smooth pipe of 40mm diameter at a rate of 0.001 m3/ s. If it were possible to prevent turbulence from occurring, what would be the ratio of the head loss for the actual turbulent flow compared to that if it were laminar flow?
Lei ( )/ denote fhe turbulent flo/IJ and { ~ the - /, r 1. yt Thv.5J hif - ft D:1-9 olld h/.J ::: ~ D 21 where V= I; _ 1/ _ Q _ O.OO/~3 m VI: - VJ - 7f - 17'( )2. = 0, 79{-:s
/lJ/fJlfJor flolll, (I)
7j O.()lfm
f =6eo~ Qnd J1. =3./X/O- 1I ~~ Re = eVD = (680~)(O.796~)(o.O'fm) = 6. 98X/O'f
Fro", Table
I. 6
So
thaf
.3 •/ x/0-'1- !!.:!. 1112-
p.
ft::
lielJce} ,from h9. 8.20) for a smoolh pipe 0.0/92 while for lominar flow £'1 = 4! ~ "6 '16~ ~ = 9.lb X / O- 1f "e • 8x/O Thus) from Eft fl) hl.t _ -
hLJ
-
-
ft
~
_ -
O.Olq:;.
9.1t XIO-
'I-
=2/.0
8.53
l
8.53
A 3-ft-diameter duct is used to carry ventilating air into a vehicular tunnel at a rate of 9000 ft3/ min. Tests show that the pressure drop is 1.5 in. of water per 1500 ft of duct. What is the approximate size of the equivalent roughness of the surface of the duct?
(/)
8.54-1 Natural gas (p = 0.0044 slugs/ft 3 and v = 5.2 X 10- 5 ft2/s) is pumped through a horizontal 6-in.-diameter cast-iron pipe at a rate of 800 lb/hr. If the pressure at section (1) is 50 psi (abs), determine the pressure at section (2) 8 mi
8.54
downstream if the flow is assumed incompressible. Is the incompressible assumption reasonable? Explain.
lr 11 +~ +~;::- '1+¥j+Z3. +fD~
where Z/~~Z and ~=-~ Ib 0 2:1.2 i.!L 'tQ==800 J!;r \3600s LIIu:-)=O.Z2Z-- orQ= '.s ;s ~ (32.2~)('f.'hlo-3.f!J!1) 2.
2.
J
/II.so}
Thvs V= !i.. == }
II
Wlfh £ 75::: j)
7/2.
Re
c:
1.57 !f3 = 800 li {f3 .1l(.k.-)2. ..s ~. ,,-if lt l6 ) e'L~ = Vll D = (8.00$ )-~7f). = 7.69 x/o'l- and (Fr()/J'J Tahle 8, t)
o.oooasff
;:
1J -
TI
5.2X/O
.r-
(6/12ff)
'I'
-,:Qool7 weOl) tlln
f == o.o2lf5 oS
rJ...L D V2. = 50 Ib - (O.02~5) D 2 r Ii?-
::: SOfJsi
qives (J
f
2.02psi SOfJS;'
t' r', I "vs rro/)'J t='!,(/./
..,-1.. I
J
(Smi)(S).8ofj,.) La,. IfXI0.3~)(8.00!:t)2 o.s ff 2. C • 1f3 oS
-:;.q I ~ = (so -1...o2)psi or
Nole: pj-~ - I, -Since
fl3 =/,57s
= 0 .OLLOL1. a , . T.J
-A.;::- ~8. 0 Lf..O
psi
7. cL/lnoe in pressure. n "/
==pRTJ w/fh T essenlia/ly co,;s/Q/Jf.J IJ small choIJge ill I SIIJ4// ChIJb96 il) p. TiJtI.t Ihe /l')cfJlIJ,ore.ss//;/e 4.IsfJPJ,tiofJ is valid.
8.55· J
8.55*
Water flows in a 20-mm-diameter galvanized iron pipe with average velocities between 0.01 and 10.0 m/s. Plot the head loss per meter of pipe length over this velocity range .
h f .l.D '),.9v~ z::
J
or
wilh i
hi ftt.:,.",) 2~~8/~ ;=
I//s o} Re ::: X,f.: r
J
=/m qnd [) == O.02. 0 m
hi = 2.SS f V2.
Of'
V (0._0:117; /.I2X/0
*::
or
4L }
(I)
Re =1.79 x/Oil- V
(2.)
FOr this pipe) O~':::"III = 7.5 )(/0- 3 (see Table 8. I ) fhe Colebrook /ortrJlJ/o hecomes (£f{. 8. 3S) 2.£ ] vr : : -2.0 °9[£, i% + Revt
.J...
~
or
r w:: -2.0 logJ!.03XIO I
-3
$0
Ihat
2. 5 ]
+ Pte 1fT
Nole: II Re ~ 2100 the (low is lominqr qnd f::: ~ If I?e ~ '1000 Ihe flow is f{)rhlJlenl and f is ohftlined from $"r. t:J). For 2/00< fie < 'fOoo rI;..s ngf clear which value of f 10 f)se, For simplici& a.s.svllJe lamiIJtJr flow is tr)oiIJlainerl up fo Re ='10()(). From £0,](:J.) V= I.7¥~oo oS' == 0.22.3 ~ when /?e -:: '1000. ,) x/a Thil~ {or 0.0/ ~ V~ O.223~ obf4in flo a/ld
(.3)
(¢)
r frollt £tts.(1.)PlJdC'I)
and hL from £1'(/)' FOr O.~23< v~ /0-1 06111il) Re and f fro/l/ Ef.r. (2.) and ('I). The lIalue.s of h. 4re CQ/c()/Qfed and plolled he/ow (see pro9l'fJ/YJ P8#55). 100 110 120 130 14,0 150 160 170 200 210 220 230 300 310 320 330 34,0 350 360 4,00 4,10 4,20
cls open "prn" for output as #1 print.# 1, II **************************** *********** ********* If print#l, "** This program calculates the head loss in **" print.#l, "** pipe as a function of velocity. The **" print#l, "** friction factor is obtained by the Cole- **" print#l, "** formula. **" print# 1, "********************************* *************:** ff rr == 0.0075 V = 0.005 print#=l, If " print#l, " VI m/s Re f hL, m" V == 2*V Re = 1.79E+4,*V f
= 64,/Re
if Re < 4,000 then goto 4,00 fp == f f == 1/(-2.0*log(rr/3.7+2.51/(Re*fp~0.5) )/log(10) )A2 if abs(l - f/fp) > 0.0001 then goto 34,0 h == 2.55*f*V~2 #. ### A~. #.### print#l, using" ###.### #.###A~A~ if V < 10 then goto 300
(con'l)
A
~";
V, Re, f, h
************************************************ ** This program calculates the head loss in ** ** pipe as a function of velocity. The **
**
**
friction factor is obtained by the Coleformula.
**
**
************************************************
I
1.E+OO
~~§ill~!~II~~)lfIJII /
E
1.E-01
§'III~tml/~mll
.:. .c
I)
I"
1.E-04 .4,::1"_L......I...J..~JJ.l.I-.....J......J....J...I...J..j.""'""+_................................."'"'i 0.01
1
0.1
V, m/s
8- SO
10
- - Turbulent flow - - ~ Laminar flow
8.56
I
8.56
A fluid flows through a smooth horizontal 2-m-long tube of diameter 2 mm with an average velocity of 2.1 m/s. Determine the head loss and the pressure drop if the fluid is (a) air, (b) water, or (c) mercury.
R,;c
, where hi -: : {7i"1 J or hi ==22S{ m
tt)
b)
l.¥6x/o-s
air wafer
287 lQl11in.r 61/- 0.2.23 1{e= 6 1./210.'0- 3150 ..,turbfJlem 0.0'1-0'1-
c) mere-vry
8.57
I
1./05 )./0-
7
36,5"00
furbulent
0.021.0
ro
~.
2
Z, == &2 J and
~ =:-V2. 0)
602-
9.09
/2.0 q 800
B,9/X/o'"
'1:95
/33,000
6.58Xl05
8.57
Air at standard temperature and pressure flows through a horizontal 2 ft by 1.3 ft rectangular galvanized iron duct with a flowrate of 8.2 cfs. Determine the pressure drop in inches of water per 200-ft length of duct.
Jr"
~ +1-1 +&,
_ (J~
~2.
T + ~I +Zz. ~ f
-
.l ~2 lJ
h
Z-1 ' where z/= z~ (lI1d ~ =~
Also D:;!M.. = ~(2f1)(1.3fl) =I. 576 ff
and)
V=
11 (J)
1f
P
=
3
1£
z[:Zf/ +1.3f.1.]
8.20$ =3/511 (2f1)(1.3t/) . S
Th tJsJ Ir1'. '" f ~ f pV', where {or 9o/r/Qniled iron Eo = 0.0005 {f (robleS. I) Hence, -t..n '" f~~:O:/' o. f)00317 and Reh~ '{.P,I = (I.S7tffJ!:;:/J"'31600 ' , /.S7X/O.$
'"
J
so from Fit). a:to, f=0,01-5
ThusJ ft -f:J.. ==lO.02.S / )(2.00 If , I ( -3 §!J!s) I 11)2 . Jb 1.57&F1J2: 2.38X/O ~ ,3,/S.$ =O,037'ffP or wdh fJ,-f:z. :::: ON;.o hJ
h=
f1-;ta ;,f)
.IJL
=
0,::':'1-/12.
==
6.00XIO-~ff = 0.00720/17.
f1~
~-51
of wafer
8.58
I
8.58 Air flows through a rectangular galvanized iron duct of size 0.30 m by 0.15 m at a rate of 0.068 m3 /s. Determine the head loss in 12 m of this duct.
h : : r1 r L
Dh
Clnd
~(Q
V
where D.
2.7
J
~
m~
=
::=.!J
O.068-r ::: /5/1l1.. (0.3m)(O,15ni) . oS
7f ond from Table a.1 J i:. ::= o,lsx/o-.3m = 7.S x/o-~ Dh
0::
4(O.3m)(o./sln)
::=
2.[a.3m to.ISm]
r-
I//s
0.2 m
. JIL D = Vl>h.::: (I. 5I oS) (o.1.m) oJ neh T /.Jf6 X/O-.$".oe
::=
20700 J
.s
lienee.! from
0.21'11
Fi,. 8,~O
F=O,027
so fhat
hL : : (0.027)(12.111 ., O.2m) 6.5 q
I
{/.5/f-i
= O./88m
2(q.a/~
8. S9
Air at standard conditions flows through a horizontall ft by 1.5 ft rectangular wooden duct at a rate of 5000 ft3/ min. Determine the head loss, pressure drop, and power supplied by the fan to overcome the flow resistance in 500 ft of the duct.
r hL -f1. Dh 2.9and
J
!tf1 _
fO(I.SrI) _ Dh == P - 2[1Ft t/.5ff] - /.2
II/so} neh::z D
VDh 11
£ ~ 0,0006 f-I ~ 0.0018 ff Dh
=
here V-.!i - (5000fln)(-1!W-) = 55.6 lisT - II (J ff) (1.5 fI)
141
/.2.1-1
{l (I
ff
,
(/.2. f-I) n .5 = (£5.6!j) ,.t.l~ = 7.2.5 X/O 1.57 x 10- .lj-
and
f,..,,'IIJ
Tt
v.
-r..4"~ 8,/ '~O't7
to 0,,·003 If. Use al') ''avera1c'J £.::= 0.001 8 fI so -!hat i r 8"0 f =0.0015 IIflIJ-SJ Trom rL''9"~ ::: o.o'-zJ or ff
EOO ff _, (5S.61Jf
h'i.= (0,022) ( /,2(f ) .2(3.2.2£) = 'f~O For this horitonial pipe IJ!- +!i +z,:: 11 +If +Z:;. f hL} and V,;: v,. . 2
where :Z,;: 2:;.
Thus} fl-P:J.::: O'hL :::::(7. 65X/O-2.. tJp)(l/-'f()f/) == 33.7 ~~%.;: O.23'1-pS i P =rQ hL ,: 6( (,,-f:t ) ,: (sooo !/:~~;~b) (33. 7 It> ) =(2810 ~1r55~ ~ j] or p =- s.// hI'
8-52
8.60
I
8. 60
When the valve is closed the pressure throughout the horizontal pipe shown in Fig. P8.60 is 400 kPa, and the water level in the dosed surge chamber is h = 0.4 m. If the valve is fully opened and the pressure at point (1) remains 400 kPa, determine the new level of the water in the surge chamber. Assume the friction factor is f = 0.02 and the fittings are threaded fittings.
Closed Angle valve
5?
f,....-- Surge chamber
-.0.5 m T
! 1
D
~ Tee
=
0.02 m
~
~C:::r=:::::::::::=::::!~.J=======~====::{ (1) (2) C3)· - - - 8 m' -t - - - I . \VfH!\-'- 5 m - - f ,+
•
.I
I
FIGURE P8.60
'f+¥i+ z,::; 1!j-+1i +z.z+(fifZKL)¥i wherez,=Z:l.J ~ I. V" Thus) If = ~ f(flS +L ~)'1 wilh v:: ~ . J
~::~':l:::'O
2
Wi.Jh
~:: 2 for 4n an9/e
valve ond
we obtain '1-00
Jsjf.
==
9.80.1$
or
I 2 ('1.81!jsJ
K::: o. q
for The fee (see Table 8.2.)
[I +(O.02)(-(B+S)/TJ \ + 2.+0.91 V2. j
O.021h -;
V== 6,88%
Tho.!) fJ.3 is de1ermined from
4s y;J r Jf+i'gv,~ +2,::: "'f ').1 +Z.3 +(f D +~ K):z.g -I-
J
where 2,==Z.3 ond ~=O
f)Js o} ~ =V HenceJ
'1 == .t;
(J +fl +£ Kt.)¥j J where j,== em and K::o Thus) .! y-'J. l 2(13 = II -(J +f D )21°':: f,-{/+f"5)te V O == '-foo kP - (I +-(O~02.)(-a.~:m )~ (qqq~) m (6. 89lJl/' = 'fO /rP-2.13X,cf.!\ m -f
orfJ;a::: /e7 kp
Thus} Q(1ri
LJ~ = '1-00 kRJ
,-.
;;.3 : : 187 /{pq
M:::: /J74.s.s of air ill surge chamber -::: f ¥ ;:: ~Ql)sfalJt, where ¥ '= II (o.5m-h) ond fJ::: pl?~or p=if Thus, wdh ( )c rienoli1J9 fhe closed valve condition)
Pc ~ ==fo"Va J or 41J(O.5 -0.'1-)
Or
"as I-V
with the I/q/ve closed wheh h:::: O·lf-m wilh the valve open qnri h =:hD
==
n c
.....
O.5m
~
.
I'IJlr
~
A
~ (a)
RII~ /I(o.5-h,) IIssume ~ ~To D xk
where pc:::: 'foo/r~-ohc ='I ool
g-53
8.61
8.61 What horsepower is added to water to pump it vertically through a 200-ft-long, I.O-in.-diameter drawn tubing at a rate of 0.060 ft 3 / s if the pressures at the inlet and outlet are the same?
.
('J.)
0)
8.62 8.62 Water flows from a lake as is shown in Fig. P8. 62 at a rate of 4.0 cfs. Is the device inside the building a pump or a turbine? Explain and determine the horsepower of the device. Neglect all minor losses and assume the friction factor is 0.025.
(Il
525 ft
8.63 8.63
525 ft
Repeat Problem 8.62. if the flowrate is
1.0 cfs. 495 ft
300-fi-long. O.4-fl-diameter
8-55
8.6 If 8.64 At a ski resort water at 40 of is pumped through a 3-in.-diameter, 2000-ft-Iong steel pipe from a. pond at an elevation of 4286 ft to a snow-making machine at an elevation of 4623 ft at a rate of 0.26 ft 3 /s. If it is necessary to maintain a pressure of 180 psi at the snow-making machine, detennine the horsepower added to the water by the pump. Neglect minor losses.
8-56
8,65 I
8.65
Water flows through the screen in the pipe shown in Fig. P8.65 as indicated. Determine the loss coefficient for the screen.
t1
-.-
6 in.
t ...l.--il-
~-S7
", "'SG
= 3.2
8,66 8.66 Water flows steadily through the 0.75-in. diameter galvanized iron pipe system shown in Video VS.6 and Fig. PS.66 at a rate of 0.020 cfs. Your boss suggests that friction losses in the straight pipe sections are negligible compared to losses in the threaded elbows and fittings of the system. Do you agree or disagree with your boss? Support your answer with appropriate calculations.
6 in. length
elbows
4 in. length
Closed ball
J.
v2.
Major loss ~ { If q. where j::(6+6+'t+j)in. ==/7//I.~ D:::O,7S/n.
an rJ.
V= ~:. 1f Th{)sJ wi1h
~
0, 0.2 (0. 7S//;l.)
e
f-l
''f.1, :. 6,$'2 :s
H ( O.7S' r ) == 6,5)'--;s ~it P J. 2/ x. /0-.5
R =~
FIG U REP 8 .6~.
•
P
::: 3.37 X10'1- and
ooos ff = 8x/o-3 ( see ..,.. hI) 11' ( C". 8 ) (0.7,s;4) fa Ie 8,1 we OD70ln see T'9. .'2.0 -rr .1 JC' _ /7 ill. JC' V'f :::; 0.038 .so fhai f D~? - 0.038 0.7.5/11. ~p = O.·~61 J:j
t. D::
0,
Al.so)
Minor loss
=
r:
V"
V2. KJ..'! : : [2 (J .s), + 2 of 0,)05' ] Zj ::: 5,1.5' ~ o /iee / / ~. 2- ::0,6'f 90 elbow rcdvcer wHo'L A:1. ( !!'O/~) V"
7f;
-rhvsJ from major loss mino!' loss
£'(.s. (I)
-
fi
s,/s ~
=:"
(see
and (j.),. :
o. 861
(I)
O.iSln·
h1'
8.26)
== 0./67 :: It,? %'
:J-f
Prob(),bly dl.saZ/'Be wll/J boss hec4v.re pipe frict/(J1l 17% of other losses.
8-58
/S
ab()!)!
(:l.)
8.67
I X.C) 7 Because of a worn-out washer in a kitchen sink faucet. water drips at a steady rate even through the faucet is "turned off." Readings from a water meter of the type shown in Yidl'O 'S.7 indicate that during a one-week time period when the homeowners were away. 200 gallons of water dripped from the faucet. (a) If the pressure within the 0.50-in-diameter pipe is 50 psi. determine the loss coefficient for the leaky faucet. (b) What length of the pipe would be needed to produce a head loss equivalent to the leaky faucet?
(a)
f/ence;
)(IIf'fVIi::: il/) (50 ffi.. 10." 62.'1 ~
or
K'L. :::
7. 01 x/O
(O,032.1fUl"
IC
2. (.32,2.
i
/t)-
I
(b) (0, O.3:J. If
#-)ffl·(W-li)
-5
I,l/XIO
_
S'
g-sq
;' //2 . : : 2-/()()
Sf)
.
.
the flow Is lam/liar
8.68 I S.6X Assume a car's exhaust system can be approximated as 14 ft of 0.1 25-ft-diameter cast-iron pipe with the equivalent of six 90° flanged elbows and a muffler. (See Vilko VB.S.) The muffler acts as a resistor with a loss coefficient of KL = 8.5. Determine the pressure at the beginning of the exhaust system if the flowrate is 0.10 cfs, the temperature is 250 of, and the exhaust has the same properties as air.
d'
0)
8-60
8.6Q
I
8.69 Air is to flow through a smooth horizontal rectangular duct at a rate of 100 m 3 /s with a pressure drop of not more than 40 mm of water per 50 m of duct. If the aspect ratio (width to height) is 3 to 1, determine the size of the duct.
(I)
(2.)
(3)
(If)
Reh Triol and error so/lJfi()n of Efts. (2.~ (3)) (If) for
/lss{)me f ~ 0,02 Reh:::
Sf)
fJ Re and V: J
thai o. 01- ~ o. 017:J.h or h :::/. 03/IJ. ;:rom Ef. (:;~ =3.32 x/ot which fro/fJ h1' a2.o~ive.s F=O.009t :10.02 S
.3·t:-;,q6
Assume f = 0.0096 which ,ives h = 0, 890m, Thll~ Reh :: 3. etr-xlr! or f;:; 0,0093 r/ 0,009t /lsslJme
f
0:;:
0,0093 ) or
h:::: O.8fPfm. 7IJIIS.) ReI..n = 3.87x/06 of f::: 0.009..3 J
which ofjrees ~Vllh fhe (JsslJmed va/uB. Thus, Ihe dflci is
h:: 0.88'1 m /;,
8-61
3h::: 2. £5 m iIJ
s/j.e.
8,70
8.70 Repeat Problem 3.14- if all head losses are included. The pipes are I-in. copper pipes with regular flanged fittings. The faucets are globe valves.
•
Bill from 74h/e 8.1,
-5-:: fi
olJd .s ;"ce'
FIGURE P3.1"'"
.5x/o-'ff -5 --:-:---:--- = 6.0 x/ 0
(JU)
R =.lJJL::: (ik. f.I) (2.0$ ) =/.,.38 X 105 if {olio ws from e l' /.2.1 x 10-5 IF"
hr. 8. 20 IhaT
f= 0.0/65 Helice.! from
£y- {/)J
If:::
7f1.,. 6. ')../ (8/1-(0.0/0$) +11)1-1 = 83,9 II
11 : : (83 • 9 (-I) (t2.'f~)( Jlf'!!:'~ )=3~, ¥-,.ri Now; O.fsllnJe I, I'ema/IIS ihe S4me re14rdle.s-s or
This is e.s.relrlra/f irlle if fhe Sf)I'I/Y line /e4dilJf IP /01'98 compared 10 the fJj;es IIJ Ihe h{)lIce.
wh/c.h fallcel is ojJen.
the pifJes ;.s re/41iflefy
S.70
1(con/f) (0) Open h4se/lJenf favcef:
v,:z.
Ij!- -I-;:j AJ ::
(f".J
0J
~2.
.A:J
of
%, ::
if +~ +Z.3 +(f D t z: KJ.)zi
2-
).
v,:- ~ :: V Ie-::: 93.9ft i::: 511 .)
z:: K :: ~o'
t
KL fjlohe :::
sl'l1w
~
.J
(J0
(),3
+10
:;r
)
IJl1d
/0.3
V6/1/8
Thvs
V2.
J
83.9f/"
-sfl +( f
tZJ J-/o,3);.(31.2¥5
Of'
2.
5730 = (60 f f /0. 3)V1 £
II/so
1):::
whel'e v~ 1j
OJ
-5
6.0 x 10
n DV "e::: T =
ana (f,:fI) V
1 2/1./0-5 it
•
=
(I
3.
L t. 89 xlD V wnere v~:s
(2)
sl.
TrifJ/ IJlJd erl'or .so/t/tion: IIssume f:: IJ, OldS as be/()f!e. Thvs from Ef, II) j
V=
,..-----.5 730) (60)(0.0/&S. +/0.3
= 22.
s!f
QIJJ
frolJ1 £o.(~) J
Re =6,8'1 x/rl (22.~) :: /..5.5 x/oS so fhqf from Fit;. fl. 20; f:: o. aIDS, This afre8s with tlJ~ (J.r~vmed va/de ~ Hence V= 22. ~ {i j
I/s 4bove (excepl /,elweel1 pOilJ7.s (I) IJlJd (If)J
(h) Open 2nd floo! fflvcel: Zi' -f
(f'£ +L /0
r; '" f
and
£ KI. ::: ~/ee +~ 90'
+ ~ 91,;8
elboll/
'V'
1!
wheN Zi' '" If/II, 1 "/fI~ 1f '" 8 3 .9't
= 0.2 +0.3'" /0
~ /(},E
vallie
Heno eJ 19+({(i!) +/0.5)2(31..2.)
=83.9)
or
/f180==(2.29{t/o,s)V
where VNli
Fr()1J1 £r. (3)j V::/Z()# ,so {~omE?(:;..) s TlJVS.l (rom Fi'9, 8,2-0 w//h f:- b'() X/()-~ f= 0,d17o$
/JsstJI1ISf==O.O/7S.
Re -:: /1/7XjO-'. whic;~ 41/'ees Wtlh
lIence.l
the
(JofS()lIIerJ
V=17. o!J. 8-63
ralfle.
2.
(i)
8.71
I
r--I'- - 1 8
in'-~'I(J)
A------------~.~.---Q
("+-----------..,.,.
8.71
Water at 40 of flows through the coils of the heat exchanger as shown in Fig. P8.71 at a rate of 0.9 gal/min. Determine the pressure drop between the inlet and outlet of the horizontal device.
(
J
(
Threaded 180 0 return bend
) (
•
(2)
•
' " O.5-in. copper pipe (drawn tubing)
lienee) {rom Fif}.
8.;1.0
f= 0,0#/ fllld
from £f· (/)
It -flo = ( O.OJfl (
:;t.f/) +/0.5) (l ) {/.9~ .sir'P )(!.lf7!t /" 1"-
of
f, -f2. = 96.B~ = O.32SP$;
8-6'-1-
8.72
I Threaded
elbow \
1"--30 ft
(2) '\
)h--.....,~--!
lot
ft \3-in.-diameter," 8.72 Water at 40 of is pumped from a lake as shown in Fig. \7.(1} -I f = 0.02 Pump PB.72.. What is the maximum flowrate possible without cavita- ~~=~ ++b=== tion occurring? - -- --:-::=:-::;.- ::;:::~--------:-:~:::::;
20 ft
l
FIGURE PS.72.
if +~2fZ = If I
1.
y."~
~ -I- Z2 -I-(f-o +r; K)~? ' where 2.
-I-
2, ~OJ
Z~'" Illfl,
(I)
/J;:; 1'l.7~(abs) ~ =0) V2 =V CI/Jd fro/h fahle 8./ f3. =0.12./7 {ffilabs) =/7.52 ~ Thus} with the 9illen f=O.02 we obtai;' from£f{.(I) J
J
.& ) I If in"!') ( /1f.7 in'-. cI'f 1fi 62 ~ ./1L •
JIL
- 17.52 pt2.
=-
loff
ffl
+(O.02( SOH) + I + 1.5 + o~e) V A ff 2(32.2,tp, 2-
1'-
where we have (Jsed K;.::: 0, e for the enff'tfnceJ K;.:: /.5 for (see 09.8,,.2- tlnd Tahle 8.2.)
S !
the 90' elbOW
Thus) V= 1,/-,116fj so -fhal Qc:llv=*(lifl)2.(JJf.If6~) ~ 0.7/0
8-65
if
Q,., 0.010 cfs
8.73 Nozzle tip diameter,., 0.30 in.
8.73 The i-in.-diameter hose shown in Fig. PS. 73 can withstand a maximum pressure of 200 psi without rupturing. Determine the maximum length, e, allowed if the friction factor is 0.022 and the flowrate is 0.010 cfs. Neglect minor losses.
~ + ~2. +~I _
:=
If +if +Z;. + rl ¥;
_.!!. _
fo - 0 J ~ - II, Thvs klilh J
62.'1-14"3
1:::
¥
0.01
(I)~I
L-j
LL_~=:[p~um./p
where
J
F+
f (~fI")
1=0.022
(2oolJi!-)(I'f'f~) + or
Water
= 7. 33 s)
fIlJ'
A
;".~
D ,., 0.50.
~
10ft
j
FIGURE P8.73
~ ::oJ $;z.:: /off, f1t =:;'OfJjJSl: _ g _ 0. 01#3 _ fJ.
~ - 112. - f{ ~ ft)~ - 20.'1-
oS
£". (/J hecomes (vsin9 V=VJ) f7.33Jj),- _
(2.0.lf.fj)').
2(3:1..2%) -
2(3.2.2~)
i
,*,/0f/+O.022( 0.5
)
(7.33 !})2..
2(.
rrfl
31.
,.Ii ) S:J.
1012. ff
q Nozzle tip diameter = 0.30 in.
8.74 The hose shown in Fig. PS.73 will collapse if the pressure within it is lower than 10 psi below atmospheric pressure. Determine the maximum length, L, allowed if the friction factor
is 0.015 aod the flowra!e is 0.010 cfs. Neglect miDorlosses.
-If + vL' +21 = if +¥i +Z.1 +{iii, ::z.
.:~
0 50 in
D
Water-
=.
2
where
1/
3
It
Pump 0:::
0,
Vi::
I;'
_
6i
0.01
_
J
(-1Of!.){tIflf·j~) +(1 f{O.OI.5J( L )) (7.33 ~)2. 62.¥ iJh t ..rg FI 2. (32.2. :g) l'k S J
TO
or
L = 8'1.0 fl
=
0.010 cfs
~. 4{r' .
.
ne;," . L~~'
<
••
10ft
j
FIGURE P8.73 OJ :2, :: 311J
f1 _ Z2.:::: 0 J fJ~;::; -I01h!- J olJd ~ - If :;: 1l (o.s ~'). - 7.33 s - V ,. If n:FI Thvs wilh r:: 0.0/s £it. (I) becomes '"I
3rf '"
(I)
(I)
8.7S ~\'75
The pump shown in Fig. PS.75 deli'Vers·a. hF.mi of 250 ft to the water. Determine the power that the pump adds. to the water. The difference in elevation of the two ponds is 200 ft.
v =
Pipe length 500 ft Pipe diameter = 0.75 ft Pipe roughne55 0
=
• FIGURE P8.75
or
~
(667 f +/2.8) V /llso Re== iV/J J
).t
::3],1-0
= (1.9~~)V(O.7$tl) 2.3'1-X/()-s
I~~
or , Re :::; 6.22 x/o'f V
(l)
and {rom Fi9- B.2tJ:
f
8-66
8,76 8.76 As shown in Video V8.6 and Fig. P8.76, water "bubbles up" 3 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length of the O.75-in.-diameter galvanized iron pipe between point (1) and the exit is 21 inches. Determine the pressure needed at point (1) to produce this flow.
•
~ +~2 +Z -h ~
2j
l:l
where 2, = 0 J ( I)
==
+ ~2. +Z2-
bIt
12. -::0)
FIG U REP B . 76
~j
~ ~o
Thlls)
£!r :;: Z:z. +h~:z. ~ - f? Where V,::: ~ ~ V Wilh no head loss from (3) to (2) and 12. :::f,g ::
-q: +1,3 '" ~:z.
Z ..
J
~
;::-0 lUll.
or ~ =:{~j(Z2. -Z3) =.y2(32.7.!t~) (
oblaill
;%:f.I) '" J{;OIg.
Thvs ft (0.1S If Re== Y.lJ.:= '9J2. = 'I,Ols _-rr -, = 2. 07 XIO'lJ
11
1/
and.
£ ::::: 0.00005
ff
(~)fi
D
f ==
0,039
Also) hL = Henc~
*'"
J
1.2./XIO 5 .s "
=0.008
:z. 1 V f D -rj 1-
(see /o,h/e /0, I) J so fh4f (see F;9' F/.2lJ)
r: ~ ~
where l: KJ. :: 3 (I.s)
2-
£0.1 (I) hecomes
2.
z.. +[ft +~ k.t] ~ ~
=
1.73
i1
where V,
0.75
"v
32..2.
tf
ihlls I, ~ (62.'f~ )(I. 73 ft) J
:=
108
¥. S
2-
~ ~ -/~ fI +[O.03 Q :;"II~./n.' +'1:S-I] ~'l:OIY.X) 2 ~
=::
1f-2- ~ 8-67
0.7.50
ps/
52-
=(0.583 +1.I'I7)ff
8.71 X. i7 The pressure at section (2) shown in Fig. PS.77 is not to fall below 60 psi when the flow rate from the tank varies from o to 1.0 cfs and the branch line is shut off. Determine the minimum height, it, of the water tank under the assumption that (a) minor losses are negligible, (b) minor losses are not negligible.
FIGURE P8.77 All pipe is 6-in.-diameter plastic (tID == 0). flanged fittings /';
6 ft
1
••
l-:i~~ [b
e ~~::!:@iWW"i .1,
900
90° elbows
.ift-t-f +~I == £t+-ii +Z2. +(f-t t&K)¥j. ond Z2,==o /6 +h = I}
ThvsJ wilh
J
where It=:~
~:1 ~i~~
ft~
~=OJ z,=16f11hJ
V::~
+¥; +(fl +&I
• =::
I J' Vj -
-
(J/Jd ~
rr~(1Im
in~ h == -1611 + (6°7?)(~f1i=)
lienee)
IJ..
62. 'fw:~ ,,-
or
+(1 +fr
7[ -
*
oS
r~)"
h+6+60 0 +9( 0 )
7f
==
fJ..i
-
J
S. 00
.s
it
2-
+ Elf;.) (-'.09 oS)
2.(32.2 f,,)
h=/~2..5+(I+f(JS~~5+h)tLK)(o.'I-(}2.) fI) where h"'f/ With -! =0 (Jlld Re == (s.o9#)ff:/J) == 2.10X/O s we ohio/I)
YI- :::
f= 0.015.5 0)
(see
I.ZIXIO
oS
hl-- 8.20)
Ne9/ect minDr lo.sses (£ K=:O): From EfJI) h=/Z2~&-( / +(O.OIG!» (15:'~ +h)) (O.l}02) or h = IJf3 ff
b) Inclvde m/nor losses: £ If;. == Jf;.enlrtJlJce +/5 ~e/IJ(Jw + ~/ee
+/5 (0,3) + 0.2 =S. 2 (see TobIe 8:2.J q,s.rUfIIC f/1J1J9Cd
::: 0.5
fif/iIJ9 5 )
Thus} from Ef. (JJ
h :::122~5+(J +(0.0/55)( /~o:th) +s.2.) (0.'1-02.) or
h = I#-6 ft
Note: For fhis cqse
minor losses ar~ no! VerI i"'floridlJl. 8-68
OJ
8,78
I FIGURE PS.71
S.78 Repeat Problem 8.77 with the assumption that the branch line is open so that half of the flow from the tank goes into the branch, and half continues in the main line.
b) Inc/tide minor /().Sses : £1<;. q =: ~L emr4nce ,J + /.5 ~'l.i!ICJ(IW Md '& IV..Lo -::: f(;'l.fee::: 0,2 From £f,t:;..) IL_
= 0.5 t Is (0.3)
All pipe is 6-in.-diameter plastic (flD =: 0). flanged fittings
=.s. 0 (see Table
8,2,'"
~~~)
(/SSV/He ",
t +h)+ 5. O)(O.tl-Ol) +(1800(0.01705) +0.2) (0./0/) h:=12-2..5+(/ +(o.0Io5S)('600.5
or
h = 137ft Nole : For fhis cqse milJor losses are nol veri imporfllnl.
8,79 3.43 A smooth plastic, lO-m-Iong garden hose with an in. . side diameter of 20 mm is used to drain a wading pool as is Repeat Problem 3.43 If head losses are mcluded. shown in Fig. P3.43. If viscous effects are neglected, what is the ftowrate from the pool? 0.2 m
0.23 m
&~ +1i ~ + Z'3- +(f J..D 4- &~II. ) :t '1 where It 0=/,. #0.1 ~ =0, 22. =0 J ~ a 'f3th
f.t:.? + Jr ').1 +2
(I)
::::
I
=
dnd liz. V
lIS
ThIl$J
wifh
O. If3 IIJ
or
~
£~.
oJ
V"
.,:5 ( I + f
2 (9.81
8. 'fJf =
Also
~tI,,1 ::: 0.9
:::
£e. (/) beco/l11J.r /0/11
o.02.m+ O. 8)
,.. (/,8 +500 f ) V where J
Re::: PY. ::: (o~ O~m) V
1,/:J.x/(j61£
V
V~.p-
= 1.7Q
X/O/!-
(2.)
V
(3)
Trial and error so/vtion: f} sSlJlhe
f : : o. 02.
r
8./f~
V --
or {rpm £
r. (,.)
]~ - 0 811-6!!L (1.8+500(0.02.)) -, s
so from
£r. (3) Re::: 1.7Q x/o' (0,8'1-6) J
::: 1.5/ XjO 'l
Wilh this Ae valve and f:::o we obtain {rom the Moody charI (Fi9. 8,:JO)) f;:: 0.02. 7 wh/ch is /Jot fhe assumed 'la/vB. Tht/s fry 49aiIJ. /lsSI/I»8 {=-O.027 or fro", £tts, (z)ana (3~ V= O.1Lf2. If 4nd Re:;: 1.33 x/o~ 71JtI~ {rpm chari f::: 0.028 ::J: f). 0'). 7
lI.ssvme f::: ond from
W~i(;A 9i t/es V~O,731 1JL) ((eo; /.3/ XI{JJ~ Mu()dy charf) f ~ 0, 028.1 f/;e t/s.flllYJed '14ft/f).
0.02.8
the
fhe h10()o/
Hence V:::O.731 'f ~nr1 hi q ,I} V == 7j11' (O.OZOfh)2 (O.731 T) -;:: J
t=:
8-70
-If f1I
2.30X 10
S
3
8.80 ~.so The exhaust from your car's engine flows through a complex pipe system as shown in Fig. P8.80 and \ith~(j \ K5. Assume that the pressure drop through this system is ApI when the engine is idling at 1000 rpm at a stop sign. Estimate the pressure drop (in terms of API) with the engine at 3000 rpm , when you are driving on the highway. List all assumptions you Intake manifold made to arrive at your answer. II FIG U REP 8 . 8 0
For sfeady
and ~ c ~ -12- we o~-J4i/J
:E,::: Z.z ilf =: It
~I;:;
Sf)
f1741 will, ~
== [
J..
V1
f D +,,] iI
0'4 : : 0 (f])J +J()"ijV2. = i f V (f 1 +K ) 1.
J)
lIenee
J
A~()f)()
A
(2.)
{fowl
If t i!, +fi -4. '" ~ H2o t ~. !lssvm8 Alld
Exhau
I/o Of)
/JsJ'pme
_
-
t BoDo ~0~0
f +I
f31t1Q
-I-
eooo ::: ~(J(J()
anJ ~()()O::: f.]()"(1 ((:e. f ilJcJ~endelll of /(e )
1\, ( .f ~
rnvs
4b:" _( 'vj ... \""
Ll{JIOOO
-
~ooo j
Re
Buf V-=:: ~ where Qis a.r.rtll)1eri pl'OjJol'fi()lIdJ/ If) ef)?i/J~ r;;Pl.
T/JP,1 is
~O()o;:;.J ~otJt)
~fJOOO ~ (3)1... ~
~f/Of)O
Sf)
fhai
q =
8-7/
8.8/
I
s.~ 1 Water flows from a large open tank, through a 50-ft-long, O.lO-ft-diameter pipe and exits with a velocity of 5 ftls when the water level in the tank is 10ft above the pipe exit. The sum of the minor loss coefficients for the pipe system is 12. Determine the new water level needed in the tank if the velocity is to remain 5 ftls when 20 ft of the pipe is removed (Le., when the length is reduced to 30 ft). The minor loss coefficients remain the same.
(lJ -u -:;:
, I
I
f
I
I
:J
V=Vz. ..... 1
r
, h
LV2 (~
Of'
h::: 8.02-ff
8-71-
x.Sl Water is to flow at a rate of 3.5 ft3/S in a horizontal aluminum pipe (8 = 5 X 10-6 ft). The inlet and outlet pressures are 65 psi and 30 psi, respectively, and the pipe length is 500 ft. Determine the diameter of this water pipe.
I!L
~r.
r +L/ +iI ==
*'"
where V= Thv~
f.
1
/'-
-& ;
;.1~:a.+f1D ~
~/)~
:::{-t Ii
*
~
3,5
=
~::: Z:J. V, ::: ~ :: V
'1
+22 +
v
;
J
where D~(f, V~ fI/.s
Ih I jn~ (6S - 30) 7i~ lllflf
~
6 ~ /1>
Or
2.
which simplifies to
D =/./38 {liS AlsoJJ from Table 8,/ J
1fi ) _ f -
1fS
§OfL J)
'1-.111 ) 2'Jif 2(32-.2) (
-, x/
(I)
.£- ;:: -!D !..
(MfA
::; e.,VD ;:: 1.9# (~)D ()r Re::: 3.70X/OS Re -p ')... 3/f 't 10-'& D Tri4/ and erl'fJl'so/vfion : -¥ IIAK/lQWns (~ j; Re r); 're9 v4Iio/)s ((/)J(~ J
(.J),.41Jt:A /)food, Ch4rf (F;1~ 8.').0))
Assume
f -== 0.02 so from £'1_ (I) D::: O.S'-o rf. TIll/sol frolf} Efs. (l)alJelliJ Re ~ 7,// x/osand i ~ '1.1 x/P-~ so froll! 0,.8'],()J f: D. 012.8 r/;().02.
AssUl1}e
f .. 0.0/:2.8 whir;h 9iv()S qnd
-t::: I,/x/(}-~
Q9rees TlJv~
/)
=:0. If 71 II)
TI)(/~ from
w,-Ih the tJsslII1Jed
/) ~ 0.'1-71 f1
Ft'?
vallie.
Re ~ 7. 77X/O~
8.2-~ f:: 0.0128 1V1}J~h
8.83
J
8.&3
Water flows downward through a vertical smooth pipe. When the flowrate is 0.5 ft 3 / s there is no change in pressure along the pipe. Determine the diameter of the pipe.
'1
r=
m V
~ 1
(I)
( :J.)
(3)
Triol find er~/' so/ufioh : .3 UlJtlJlJWIJS (D~ Re
J
f) and 3 ef/(/41i()lJs ((~t3~ ~lJd
Fi9. 8.20)
(;:0.02. so from £f(. (~ /)~o,/t'fl (llld frum ftt.(3).) Re::: 3.18 x lOs. T/Jvs~ from Fii. 8,").0.1 f:::. O.O/If :I O. 02
Assume
Assume f:~ o. O//f .so fh4f D;: o./ssfl and !Ie :::3.1f2 XIO~ 7hvs" from
Fi9' 8.~~
Assvmea Vo./1I8. Thus D:: 0./55 fI J
f-O.llf
which
cheo~.I
wilh fhe
8.8/f 8 .. 8'" As shown in Fig. P8.8f, a standard household water meter is incorporated into a lawn irrigation system to measure the volume of water applied to the lawn. Note that these meters measure volume, not volume flowrate. (See Video V8.7.) With an upstream pressure of PI = 50 psi the meter registered that 120 ft3 of water was delivered to the lawn during an "on" cycle. Estimate the upstream pressure, PI' needed if it is desired to have 150 ft 3 delivered during an "on" cycle. List any assumptions needed to arrive at youl answer.
• FIG U REP 8 .84
The ener9Y €9tJalion {Of' fhis flow i.! ~
~
z
~~lf+z,-[fJ+L~]~ ~*~~.Ll:l where Z, =:: ~:J.
/hvs
from
J
f:L:;-o
£,. (I)
1'1 frV,'- [ft
~ ~ ~ and
V2. =:
4; ~
2-
r ~ +(4&) -/]
+
J
Bof Q::: II, ~ :::
J
(I)
¥
where VIS fh~ VO/1I1110 01 kldler so/II/od dvrifl9 a/) '1ln" cycle and t is fhe /elJ9fh ollhe cyc/~. J
9iven sysfem 2: k;. is indepenJofJl 01 Q. S'/JI,lt1l'ly./ fot' Re pipe flow) f is ifJde{JendcnJ of Re ( or ~). Thv~ [f1 +Z kJ. of(! )2- /J is ~on.sf(J"IJ liJdependf}/J! uf Q. Hence, from £'1' (;')J if fhe len,fh of the cycle is cOlJsianf Fof'
tl
l/Jif/9
J
{JI
t,ft = ~ pv,~t
It )17.011
3
2"
P'" ~:z.o
or
f,) :;; 1.£6311;) 150
=:&
=[ v,),r.
l V,
)/;10
]~= r~E' ):2.;0 (.!flo t:: I. 5"63 \"
I UJ
/.SI3( sopsi) ~ 78./ fJSi
/1.0
8-7.5
8.85 X.SS When water flows from the tank shown in Fig. P8.85, the water depth in the tank as a function of time is as indicated. Determine the cross-sectional area of the tank. The total length of the 0.60-in.-diameter pipe is 20 ft, and the friction factor is 0.03. The loss coefficients are: 0.50 for the entrance, 1.5 for each elbow. and 10 for the valve.
--Z-................-.......,............"....,.,
100
200 1.5
!Ii FiGURE
8-7t
PS.S5
300
8.96 KS6 Water flows through a 2-in.-diameter pipe with a velocity of 15 ftls as shown in Fig. PS.S6. The relative roughness of the pipe is 0.004, and the loss coefficient for the exit is 1.0. Determine the height, h, to which the water rises in the piezometer tube.
Open
I h
15 ftls
(I)t-I.- - 8 f t - - - - i •
OJ
h = /6.S If
FIGURE PS.S6
8.e7 H.H7 Water flows from a large tank that sits on frictionless ~ - wheels as shown in Fig. P8.87. The pipe has a diameter of 0.50 m and a roughness of 9.2 X 10-5 m. The loss coefficient for the filter is 8; other minor losses are negligible. The tank and the first 50-m section of the pipe are bolted to the last 75-m section of the pipe which is clamped firmly to the floor. Determine the tension in the bolts.
-
I) - - -
1\1 FIGURE
P8.87
{OJ ~--.-----' (~
CfJntro/ YDJlltne
or (I)
v-
/9,9 q f ~50f
-V IIlso Re:: ~ ::: or
(2.)
Re == ~.~t x/os V and
s:: D
rnvs
f \
5
9.2 XIO m ::: l.i/fXli'f O.ShI
from
Ihe AI(jot1j cnarl:
~ *~J.Jlfxlril' Re (con 'f)
h boH
8.87
I(c,on'tJ T/'/0/4fJd error srJ/vliof) :
Assume f:: 0.0:;' ..£l V::: s.2-?;- !E- Re := 2. ~X/06!:!!.. I::: 0, o/~ :I (),()2flSSPIlI9 f:: 0.01'f'-!!'!' V::.s: 60qL .!!:!- Rf) ~ 2.SX/O I (.1). {::: O. f)/~ 7i7~
V== SIIO.fL
/I/so ~
f;
~z
+:i!.3
+i!;
where
Z ~::: ~~
1'.3_ '7 or
=:
h
J
'LJ-2.
-4
J
~~
-2. :=
V:z..:::'~ IJn~
f!; ~ Zz +~ f,.. ~ ()
Sf)
fl1al
-;: (f b. -I- k;)J[.1 ;-(0.0/.,. (?E!L)-I8) /)
ll.
z,!
f). SAl
(S.6o.fl)2 == 1£,/1l'J 2 (9.11i!!..)
{1== p,aox/o3# Iii (/~./fII):;; I.se X/O.s-AI AJi
TIJ v~ from Erp (0): FboH -;: (iJ 113 +PA.l ~ 2=p.~x/o$f.):; (O.ShI) 2- +(9P9 tJ)1j-(o.,smll-S'.IIJ11-J 1. ::
3./(JXIO~N + 6,ISX/O.Jjr/
8-79
,r2.
8.88
8.88
Water flows through two sections of the vertical pipe shown in Fig. P8.8 B . The bellows connection cannot support any force in the vertical direction. The O.4-ft-diameter pipe weighs 0.2 lb/ft and the friction factor is assumed to be 0.02. At what velocity will the force, F, required to hold the pipe be zero?
Pipe weighs
0.20 Ib/tt
D = 0.40 ft
FIGURE PS.BS
From Ihe m'OmenTum e9uafion appn'ed 10 fhe
con/rol volume indicofed
I, II, - ~:l0 -
~i,!ce Thvs .;:; WH~o+WPif' = '0 ill, + I (:t=) J 1/ AI II, Wpipe
=rh (V,. - ~) : : 0
or~ == 0/ + (O.20 lt)P :: 0/ +1.59 i where ~ ( 0.Jffl)2
I!'j
J
VI:: Vl f)
NiA
1// n )
1"'//
,t;/soJ
4t- +¥g2 HI 't + .r:;1 U;J. +f /; li =
\1 = l{ ~ V, ThlJs)
f)
TI
J
where 12::0,
2, = 0, (Jnd 2,-:::J = };tZ +fl..LfV~ Q
2.
D 2..
J
or when comhined wifh the ()br;Ve force DC/lance resvll 11 : : oj+f-t ip V2. = 0'1 +1.591 i
TL I· IfJOiIS,
fDVV2.
-Tii
==/.59
or
v--
_I 2D(/.S9) -
Y
Nofe: Til is answer is indepenrJelll
8-80
pf
-
2(O.JI.)(/.S9) ==5.7'3
(1.9'1)(0.02.)
of the pipe /enqth J 1.
=
fioS
8.8q
I
\7(1) t ------:-:--:-----
60-mm-diameter, 30-m-long pipe; I.. f=/ 0.016 _ _ _rJ~
8.Sg
The pump shown in Fig. PS.89 adds 25 kW to the water and causes a flowrate of 0.04 m3 /s. Determine the flow- 40-mm-diameter no~zle ~. rate expected if the pump is removed from the system. Assume / f = 0.016 for either case and neglect minor losses. ----~===========:t,~ " £ " - L - - _ . - - l
J .-
(2.)
Pump
FIGURE P8.S9
8-81
8.90
I
8.90
A certain process requires 2.3 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90° elbows, determine the pipe diameter. Elevation differences are negligible.
+ 29v.'" +
o::z
~I
~2. £ V'" . +~+l:t +(fD+r~)~ J where~==30fJs, J I, ~ 60 fJ si, ~=O It == V= !l.:: 2.3 fJ ::: ~ £i wifh DNff J It'" f j . ! r D'" D2. S J
b.
.fZ/:: ;
= 7.
2.
If
Thvs,
l1-f2. ::: (f~ +L")fp V2-
orr60 -30) ~ (PI-If #i)=(ff{( 2~Oft) + 6(I.S) +0.5) (~lIf-1jf·(t.)(1.9'f~) where we hove used 2: f( :: 6 KelhtJw +"elllrlJhce = 6 (I.s) -1-0.5
Thus,
Jl.tJ, Lf :: (J + /~Ol)
A/so,
iff
(I)
_ VlJ _ ('i!) D :: 2.9..3!f ~ D .5' I Re - y - 11 1.2IXIO-!;tpD' or Ile :: 2.Jf1.X/O IS
and from TahIr: 8.1 e o.ooo.sN
75 == Finol/y) from
D
Fi9' 8.7-0'·
Trial and error so/vfion of Eqs. (I ~ (:J.); (JJ, and ('I) for fJ DI ! J and Re.
f
D Normally if is e4siesf fo 9uess a va/V6 off, cQ/cv/afe ~ efc, In this cose (becoIJSe of minor losse), Eq,{/) is no! eosy to IJse in fhis fashion, TIlus) Qsstme DJ calcfJlafe r (Ef{. (/))J lie (£", (2.)) J ond f (£f. (.1)). Look up f;1) Fi7~ 8.20 (cr· (~)) IJlJd c()",pare wifh fhal fro", Eft· (I), /issII",e D=o./ffi. T/;/IS, f=O.00557, Re:6.05XlosJ f;;:o.oo/:l.S or from Fi9' 8.:Z 0 f:: 0.021 -:f: 0.00557 Assume D=O.5f-l j f=o.OSSIJ
Re =II;8/f/
Assume D:: D. 'is ff.J F:: 0.02 -s'3 J fie:: $,38"'0$J
.g-:::O.OO/orf=O.0203¢O.OSSI -DE
= 0.0011/ or f == 0.02.05 :to.Ol/l-.3
Re =,s..sox '0: !-::: D. 0 0//'1-0/'" f=O.0205t:O.OlQ7 INfer e/J ofJ9h trifJ/5 ()hloifJ D= O. If'/-2 fi Nofe ; If H9. 8.20 (Ef - ('I)) ;0$ replaced hJ1he Colehrook erll4lio/J fhis prOblem cOllld be solved efJ.II'Ij will; a cfIJ'J,PtJler.. /J~siJme
D= O.llifff;
f::: O. 0197 J
8.Q'
The turbine shown in Fig. P8,QI develops 400 kW. Determine the ftowrate if (a) head losses are negligible or (b) head loss due to friction in the pipe is considered. Assume f = 0.02. Note: There may be more than one solution or there may be no solution to this problem.
120 m of 0.30-m-diameter cast iron pipe
FIGURE P8.'U
'J!-+¥J fZ, ::: iff ri +Z2- +ft¥; -l-~ ThusI , z=
Z 2. ::: 0
Z, - 2j
Thu,s,
''7'
J
lr
XI
where
(/)
""
hr
3
L:=
¥oo x/o
N.m
-:s
~om=2(9.81:;") +
.52..0
~
or
52.0
(q.80 x/rr P.3)!j(Jm/ ~
rQ
\/2.
V2
z,c20tnJ
"]..9- +{ D 2.9- +1-. ,,.,.
a)#eq/ecf head losses (f==o):
_ vl + l.
I,;:~ =0;
where
I
,,3 Y2.
\I
-392v,.+/020=
;: T
m
0
(2.)
De/ermine Ihe roofs of fh,-s cuble erVfJ lion. Lei v,.3-392~ f/O')..O=F
kIanf F:::0. Note Ihol ~ = 3V/'-392 so fhaf ~ :::0 aI ~;: r//.'1l}A/so} F::: 1020 when ~ =0 J F::; ~ 007 wheh V,.::: -II. ¥ J and r = - /9t7 whe/) ~::: I/.Jf. IJs ~- ()b J r F lis ~ - - 00; F - -ctJ. This liJformaliof) Indicales Ihere 4re IUlo posilive real rools (.see Ihe fi9ure). The nefolive rool hlJs no physical meQlJifJ9' So/ulion of £'1' (1-) 9ives -1967 .--Vi = 2. 6s ~ Or ~::: 18.3 ~ Thlls, Q -:: 1i:J. liz. := (J tnt V;. -0().
or
~
*
3
3
Q= 2.08!f or Q=/~Jf~
b) Inclvde heQd loss (f ;:0.02): From E'(. II) V = o~o", =(lfO.o2/:/~om )(JJ.J)~) Vl + 52.0 m n~
0.3 m
3
~
2 (q, 81-$.)
-O.398~ +1.03'1=0
'h,t2 = ~ (.~)2.= ~ (~.~m) ::::.//./~
V~
Lei G= ~
3
-O.3?B'Vz +1. 03/f j
de1ermiIJ8
v,. fhot 9ives 6=0. lis above) G-±ooa,s v,.-tooj ~~ ::::3~2..-0.3?8=O for Vi =:J:. O. 36Jfj G=/. 03'1- for v,. :::0) an d G:: 1.13 for ~.::' - O.36oS' I
of V,,== +0,36'1: Thus, fhe qr4fJh Df G looks QS shown, A cuhic e'lIJClfiol} has al most two min. or max. IJs ,shown, there is no pos/live real roof, The Flow clJlJn'" occtJr (mvsf have P<'f(}{)~W)
G=O.Q37
G
-0.361f
8.92. I K.n
A fan is to produce a constant air speed of 40 m/s throughout the pipe loop shown in Fig. PS.92. The 3-m-diameter pipes are smooth. and each of the four 90-degree elbows has a loss coefficient of 0.30. Determine the power that the fan adds to the air.
-- --
(;1: f--
.. /
20 m
i'"
v= 40 mls
1
I~
~
D=3m-
10 m
~J ~I
(1).::{21
--
_\(Fan
'i'-
ii2 FIGURE P 8.92
~
:.-
8.9.3 , KL KL
g.~3
elbow
exit
=.0
= 1.5
.V
Water is circulated from a large tank, through a filter, and back to the tank as shown in Fig. PS.93. The power added to the water by the pump is 200 ft·lb/s. Determine the ftowrate through the filter.
~2.
V2
1- + +# +h If ~ +41 +( f l +f 1V.,.)1j 2.
p '"
Z;
(I)
Zz.
where
,,-:::fJ,.
~ =- ~ =-0"
J
AJsoJ ~::: r() hp
Dr
f1.:1!
200 ,s:
h -
and Z,:: 23-
p - 62.,~ (¥(O.lfO"2)V
ThusJ £0,T (I) '1-08 V
or
=
-
1/-08
- V
becfltnts
( 200 ff
0,111
2
I. )) V f +,0.8+05(1.£) -I- /2 +6 +/ 2(32.2~}
¢'
13./.3
3
V ~ (f +0.01365)
1:: e
VD =-
-p
(2)
1.1",,5wrv!1){O./If) 2 a~ x/o& •
Iifll."is.
or Re::: B290V
Tri.1 and error so/vl/o/J: flssum~ r::: 0.0'1. From £,. (2) J V::: 6.26 # ; from Ft. (3~ Re :: 5. '2. 0 x/0 ~. Thv~ from Fi~. e. 20.1 f:::: O. 039 =I: 0.0/1 /Jssllme (::: O,o3Q J or
V:: 6. 2 9.if.
IJnd
Re ::: S.21 X/O~ and f -: : O.O:Jf
(Check)
Tf,v~ Q:::/I V::: "'(O.lf/l(6.29~) = o.O.y.9~-!f3
(3)
8.94
Water is to be moved from a large, closed tank in which the air pressure is 20 psi into a large, open tank through 2000 ft of smooth pipe at the rate of 3 ft 3 Is. The fluid level in the open tank is 150 ft below that in the closed tank. Determine the required diameter of the pipe. Neglect minor losses. . (IJ
T 1501+ ~
(2.)
1=2000H
(I)
(2.)
_ P(~)D _ -
P
-
1.9~ (3.82)
2.3¥'tIO- s
/)
or Re
==
3./~'JIO~
I
Trial lind error solufio!) : IIssume f=O.02 S~ from Ey. (1.)" /):; O.S'fo II lind from £".(3) Re =:.£.87X/O S , Thv~ fr(J/II F". 8.UJ(w/lh -i-.=:o) f::o.ol.3 :;0.02. !JssulIJe f =:0.0/3 wh/cA 'lives lJ ~ O. 'f9.5 II J f(e:: 6.. 1fQ Assume f
Thvs
J
:O.OI2.S,
~o D=O. ~91f-lJ
D ::: O. '1-9/ (I
g-86
lOs
J
QlJd f::o.ol2..5
He ~6Jf~,(/OsJ f ~ O.O/'-S (checJu)
8.,95
I
8..95
Rainwater flows through the galvanized iron downspout shown in Fig. P8.'15 at a rate of 0.006 m3 /s. Determine the size of the downspout cross section if it is a rectangle with an aspect ratio of 1.7 to 1 and it is completely filled with water. Neglect the velocity of the water in the gutter at the free surface and the head loss associated with the elbow.
70 mm
L
T~. ~. Ifm
LI"li'~l\W#Hi$l'W""''> ~ \.
3m---l·1
8.Q6
*1 8.96
*
Repeat Problem 8.QS spout is circular.
if the down-
(2.)
(3)
(5)
(6)
f=
f
-2.0
2
JO'J[¥."tX/~-6+ 3.6
W J
(7)
O-9D]
Solve £rs. (t) and (7) iferqlive/y. S fflrl "11Th Q.ss/Jhlod vfllues D;;: 0./ J f= 0.02. From El(. (I) o61qill Q /Jew D valve WI/h Mis new D cQ/cvlole 0 /Jew f vqlve from fro (7). Re,PfJ4t suc.h CQ/cp/otir¥ls 0
unfit the
nih lind
0
IJ_/.s f I/IJ/ves
J < 0.001 ) 1- i!L fn- I
s4iisr, The CfJIJV8ry8nce
and II - ~
~I
8-88
}
crderiQn
Pro9ram P8# 96 .BIJS show/) below kills used 10 solve £ro5. (6) QlJd (7) 405 indictied above 10 9/~ve D=o. OJf~Sflla/lJ
f= 100 110 120 130 1l±0 170 180 190 200 210 220 230 240 250 260 270 280
0.02.78
cIs print "*****************************************************" print "** This program determines the friction factor, f, **" print "** and the diameter, D, solving iteratively **" print "** Colebrook's equat.ion **" print "*****************************************************II print f=0.D2 d=O.l dp=d fp=f d=( (f+0.1429*dp)/195600)-0.2 f=1/(-2.0*log(0.OOOOl±05!d+O.000368*d/fp~.5)/log(10))-2
if abs(l-f/fp)O.OOl or abs(l-d/dp)O.OOl then goto 210 print. print uSlng "The friction factor is f = +#.####----";f print. uSlng " The diameter is D = +#.####----";d
****************************************************~
**
This program determines the friction factor, f, ** ** and the dia~meter, D, solving iteratively ** ** Colebrook's equation ** ***************************************************** The friction factor is f = +2.7842E-02 The diameter is D = +4.l±518E-02
g,q 7
1
8.97 Air, assumed incompressible, flows through the two pipes shown in Fig. P8.97. Determine the flowrate if minor losses are neglected and the friction factor in each pipe is 0.015. Determine the flowrate if the 0.5-in.-diameter pipe were replaced by a l-in.-diameter pipe. Comment on the assumption of incompressibility.
p = 0.5 psi
I
~FI. (0)
1 in.
~
m
t 20 ft
0.50 in.
!
(2.)
(.!)
'II .1.__t 20 ft- ~,
FIGURE P8.97
{?a ~ fJ.j_ ::
lJo
l'i
/~. 7p.Ji . ;:: 0.967 (0.5 f/1f.7)PSI 3-QO
The floJV is neflr/j JiJcompressi!JJe.
8,178
*1
T\iD
p = 0.5 psi
= 150°F
*8.98 Repeat Problem 8.97 if the pipes are galvanized iron and the friction factors are not known a priori.
(0)
1 in.
*
0.50 in.
t
(I)
(2.)
(.l)
I..--_ t 201:._........-..1-20 t \lJ It-- J. ~
FIGURE P8.Qg
(2.)
(3) (If)
I
I
(s)
and and
ReI == ~DI 1/
J
11 == ..l! =
fo
Hence} Re 1
ano
Re2. == ~ V~ J where fr()m 'able 8,3 1(.4 ·7 ~ '1.181./0 {fl. = 2.00X/O·¥ .I-L['.12.
O.0020q ~
== (O,2.sV,J I
(-k fl.) =/O/f V,
(
.s
I
S For lurhulenf Flow £'(.
=2 08 ~
8.35
(7)
vrI =-2.0 Jog [ .3~
9iV~s
£
i-
2 5/ ] R~yr
(8)
comhinifJ9 £fs. ('1-) fhrough (8) we obtajn
I [ -3 W,I ==-2.0/09 I, 62xl0 +
w: =-2.0 Jog[-3 3,2.lf.x/o
and,
~
2. OOX /0-11-]£
o == V2. Vf H) nez 2 .OOx/o·if f:J:
By
.s
ffS
t
',1111
V2.1{f;
2
I 1-1 X 10.
'4 W;
]
(c,onJl)
(q)
(1 0)
B.98
f
,
(conI!.)
fhe
Solve E"q,s.(.3)) (9)J Qnd (jo) {or
If D, == D2 then J
Re =Re2 =
~~~
V:z. D:J. = 1/
I
Thlls, Ef. (I) hecomes
J
~
::
-t; == i2. == 0.006
-& since
V:t (7i:ft~
2.00)(/0-"']£ ..s
h.J and Ih. (see below).
unknf)w/Js ~J
J
qnd
= '1-16 ~
tPo =ie ~~ [~(el;:~) +IJ or
(o.s1::,.)(JJflf ~ ) :: i (0.00209
¥f; ) \{:l. [f2. (1
0 ;:)
+ I]
Hence 6, eq x 109 = ~~ [ '180 -& +1] l
Also) frohl El(. (8) .-L =-2.0 lo~ ~.62x/0-3 + vr;. L Solve £'(s.
Nole:
(II)
(II)
w:
3
6,03)(/0-
V:t
]
(/2)
ond (/7-) for &. tJnd ~ (see he/f)w)
S/nce {J:::: pRT
il fo/lolAls +hal
.fi._(,ln)_hJi. 1' fo - ( A) -]10 73 If we (J.S.SVtrJe 7; =70 (/'prohfJbly will nf)f he RTo buf rt should be Q reasof)oble apprOXillJaTi()n) fhen ~.!I :;: \ 0
IJt- = (0.5Ilf.tllf.7 7 fJ~i psi
:::: OJI67
, ()
Pro9rIJIiJ P8# qg .shaWl) be/ow
'lidoS
The flow is neqrly incompressible.
used fo Dbl';1J the f()//f)lIIiIJ9
resfJ/ls : 0)
W,-Ih fhe fwo different diall1efer pifJes ; Q:::
b) With 100 110 120 130 140 170 180 190 200 210 220
fhe sill'lle pipe:
J
0, 07~6
.f
Q=: 0.339 #3
cls print 11 ********** ****************** ********************* *** * f1 print "** This program determines the frict.ion factors, **" print "** f1 and f2, and the velocity V, solving **f1 print "** iteratively Colebrook's equation **" print "*****************************************************" print f1=0.002 f2=0.002 f1p=f1 f2p=f2
230 240 245 250 260 265 270 275 280 290 300 380 385 390 400 420 430 445 450 470 480 490 500
v=(68900/(15*fl+480*f2+1))~0.5
fl=1/(-2.0*log(O.00162+0.0241/(v*flp-.5) )/log(10))-2 f2=1/(-2.0*log(0.00324+0.0121/(v*f2p-.5) )/log(10) )-2 if abs(l-fl/flpl)O.OOl or abs(1-f2/f2p»0.OOl then goto 210 print print "For the CB2'.e of unequal diameter pipes: II print. using "Tbe friction factors are fl = +#.####'·~'·-I!;f1 print using" and f2 = +#.####----If;f2 print. using" The velocity is V = +#.lHt##··--- ft/S";V Q = 3.14159*(0.5/12)-2*v/4 print using" The flowrate is Q = +#.####---- ft3/slf;Q print print. print "For the case of equal diameter pipes:" f2=0.002 f2p=f2 v= (689001 (480*f2+1) ) -0.5 f2=1/(-2.0*log(0.00162+0.00603/(v*f2p-0.5) )/log(10) )-2 if abs(1-f2/f2p»O.001 then goto 420 print using" The friction factor is f2 = +#.####----";f2 print using" The velocity 1S V - +#.####---" ft/s";v Q=3.14159*(1/12)-2*v/4 The flowrate is Q = +#.###---- ft3/s";Q print using If
k**************************************************** This program determines the friction factors. ** ** f1 and f2. and the velocity V, solving ** ** iteratively Colebrook!s equation ** ***************************************************** ;j:*
For the case of unequal diameter pipes: +4.2508E-02 The friction factors are f1 and f2 = +4.4593E-02 The velocity is V = +5.4682E+01 ft/s The flowrate is Q = +7.4561E-02 ft3/s For the case of equal diameter pipes: The friction factor is f2 = +3.5069E-02 The velocity is V = +6.2160E+01 ft/s The flowrate is
Q
=
+3. 390E-Ol ft.3/s
9.100
I
Elevation
=
15 m
_~(A>
Diameter of each pipe
----;!--~-
=
0.10 m
·A
1..0.0..._
8.100 With the valve closed, water flows from tank A to tank B as shown in Fig. P8.100. What is the flowrate into tank B when the valve is opened to allow water to flow into tank C also? Neglect all minor losses and assume that the friction factor is 0.02 for all pipes.
(/)
FIGURE P8.100
Q, == C¥2 +Q3 where QI' = 11£ V:. == f 4~~' ThvsJ since /)J:: /)2:: /).3 iI fo/lows fhol
J
i::
~2Jl
~ == l{ + ~
(I)
Also, for flvid flowin9 from II loB, &. +Yl +:!i +Ze tf,I 1L;!L.2 ~ 21 +z~ =& I' 21 D, + t2 j. D,.
2.,
v:
'J.~ J
where Tn LJ,q: y~ A78 == 0,
~ == ~ =0) ZA ::/5 ~2.rJnd Ze==O' Thus) zA = ,,1L.:iL +t:t !.~.2. or I D, ~, D:t 2j ~ 15m == . (0,02) [(801'h)~3.+(ifO/1})~2] (0.1/11) (2)(1.8/ $.)
OJ
Hence, J8,Jf == \,12- + 0,5 ~2. J
l'l
N
~
(.3)
SiIYJ'/rJl'/y J lor fluid {/owin9 from IJ fo C, tfJ/J +~ +Z. T 21 /i
:::
f&. +z.c +£.! }IL.:J. + £ h li2. where LlLJ ::AJC:::(), r +~ 2? I D, 2.1 3 /).3 ~1 J Vb yc
~r VccO 219 =/.5m, IJnd Zc= 0 J j ~2. b ~2. ThIJS, :Z/J == ~ + ~ '"9
£ i1 b
, r s (J J ( ) r I r 12. V: _£ b vI By comptJrth9 k.r . J., ana ~ we Tina T2 D; 1:j - .3 D3 29 or since &=~ and 4.::: D~ J ~ 142. =~ }6.2 Thvs) ~o ~2.= 7..5~.2 or ~ ==1.369 ~ Solve £9s. (IJJ (3); IJnd (S) for 11 J ~ find II.J • J
From £fl' OJ and (s); fB.if
~ ==
I, 3t9 ~ + ~ == 2,369 v.s. and from Ef . (3)
=(2.369~)'). +().5 (1.369'4)2
and ~ =2.1-9 f + /. 67& ~~
Vz
==
*"
r
or
'V.J = I. t7t
= If. 00 ~ 3
(0, 1mi' (2.2.9.p) ::: 0.0/80 ':
'!-
J
~ =1.319(l.t71f)
=2. 7- 9 11-
(~)
(sJ
8.101
-I
Elevation = 15 m
F-'¥~d (A)
Diameter of each pipe = 0.10 m Elevations = 0
,'::A
*8.101 Repeat Problem 8.100 if the friction factors are not known, but the pipes are steel pipes.
' - - ---:-
LBO c-++-"\" m+;:'~
CS)
40
ee)
-J/)
E=75=m::;:;;(3~)~~"f'Y>l-:~.:::"'1_~C FIGURE P8.100
Q/ = Q2. +Q3 where Q/ ::;At'l{' :: f4·~Vt· /::~J./3 Thus, silJce D,:: D2 :: D.3 if fo/lows fhot I
ro
~=~f~
Ais 0) for
f
+
¥i
fluid fJowin9 from It to B t ~/I = '1 t +28 t
¥$
\?i:: '4 =0
J
where IA ::14 :: 0,
2 11 ::::/Sml a!)d.&o:::O
Thvs} Z/I :: r; J
/5", =
t! ¥i + (;I. t ~i J
J.
D'
2
I.
~ -I- -& f
I 2,
:2
J.
li 1-1 } or
(2.)
[f,f~.~:nw+M~::)'6·]2(q.~/~)
or 0.368 == ~ ~2 + 0.5 f:J. ~2.
(3)
SinJl/fJr/y, for flvid flowing from 1110 C ik v,..,z _ tPc_ + ~2.+ =z r la .M: + r iJ~:l L ?1' +~ +Z/i - T 2.1 ~c 4- TJ D,'" ~ D; '2-1') wnere ~ =Vc ::: oJ ~A= IS 111) lI!)d Zc==(} IL f, 11. ~2. r J~ Y.j2 I nus J ZIJ = I D iii + TJ 7i :z.! I ~ :3, Z J
By cu"'parill9 EfS' or since D2 D~ II::
fz i ~ = &
(2) afJd (If) we find
(111::fJc::
J
f2 J2. Vzl.::: -& i~ \&2. Thlls} "f()~ V/" ::: 75 ~ ~z or ~ = I. 3&9(
J::
-2.. oJo 9[! + 2..SI] 3.7
Re
£:: O. OJ.f.-S /1')hJ so fhol for each pipeJ VlJ II/so} Re = V or for i::~2}3
Re. = V~'Di = l
tI
E
7S
~. (O,/m) = 8.93X/O"" 'i'
/.12 x/o-I>f
I
(con't)
vr
('1-)
-! ¥i
.
35 From £q,8./ {,. J
0,
t) \~3 ~
where from TolJle
= o.o~.s mill = 11:5 XIO~ '00111111
•
(S)
8.1
(6)
.I
Solve 6 ~J
e'll/(JtiolJs
& & I
I
for b
~ J ~J ~.
From £'(. (s))
~ =' 0.73 0(-
;
£rs, O)J (,g) J (&) /6),(7)) (Jnd (8) for
(J/)KIJoMlIJS;
Ir/a/
(7)(~
olJd
error so/pi/on
tiS
fo//pw.s:
t; t- ~,which when combined "lith
EC(. (J) g/ves
~ = [I to. 730(~)-'i] ~
Thlls by com bini/}! £rts
t;;r
J
0.368
or
~
(q)
I
(3)
='f~ [I +O.730(
-f [I
2.-
and (q) we obfqin to.S
0.368
~ +0.730(4; Y']2.+ 0.5 f2-
fI/s~ from £'1'
fz ] v,,2 ]
~
(10)
(I)J (II)
So/vfion method !o)GlJess valves of "J fl.; and & (1I9ooJ siQrfiIJ9 valve IS fne /Q/'7e Re valvB fDr i:::9.sx/o¥-, or "=-&=£-=o.OI7~ h) Ca/clJ/ale ~) ~J alJd Va fro/IJ £ys. ('I); (/0)1 dl/d (1)/ c) Ct,jc(}/4fe F2 J qnd~ frmEts. (6~ (7~ (8) j d) Compare the new ~. IV/lh fhe pre vious ones j e) If IIPi 9t1oc/ elloVfA tl9reellleflfJ refJeal wt"-Ih the
t, J
new ~. as fhe 9()ess~ Program P8#/OI .shown below
Wf}-S
used ..fo ca/c()/aie the fO//OIViIJrI
resulfs:
QJ::
m3
0.03.3/ S
J
Q2:::
1»3
0.0193 T
J
l(j:::
m3
O.0/38T
8./01 100 110 120 130 1l;0
150 160 170 200 210 220 230 2/,00 250 260 270 300 305 310 320 330 3/,00 350 360 370 380 390 /,000 410 iJ20
II
I
(conJi)
cls open "prn" for output as #1 print fI************************************************" print "** This program calculates the flowrates in **" print "** the three pipes using the Colebrook form- **" print 11** ula to determine the friction factors. **" print H** An i terat.ion scheme is used. **" print "****** *********************** ***** ******** ****** II dim f(3), fp(3). V(3). VP(3). Re(3) for i = 1 to 3 f(i) = 0.017 VP(i) = 0 next i rr = /,o.5E-/,o print II II print "pipe no. Re f V. m/s Q. m3/s" del=O V(2)=(0.368/(f(1)*(1+0.730*(f(2)/f(3) ) ft O.5) ft 2+0.5*f(2)) ) O.5 V(1)=(1+0.730*(f(2)/f(3) ) O.5)*V(2) V(3)=V(1)-V(2) for i = 1 to 3 fp(i)=f(i) Re(i)=8.93E+/,o*V(i) if Re(i)(2100 then goto /,000 f(i)=1/(-2.0*log(rr/3.7+2.51/(Re(i)*fp(i) ft O.5) )/log(10) )ft2 if abs(l-fp(i)/f(i) )0.001 then goto 3/,00 go to /,010 f(i)=6/,o/Re(i) del=del+abs(l-VP(i)/V(i)) next i if del
ft
500 510 520 530 540 600 610 Q=(3.14159*0.1~2/4)*V(i) 620 print using" ## #.##ftftftft 630 next. i
#.####
##.###
#.##~ftft~";i.Re(i),f(i)~
(V( i) ,Q
************************************************ ** This program calculat.es the flowrates il"} ** ** the three pipes using the Colebrook form- ** ** ula to determine the friction factors. ** ** An iteration scheme is used. ** ************************************************ pipe no. 1 2
3
Re 3.76E+05 2.19E+05 1.57E+05
f
0.0176 0.018/,0 0.0190
V, m/s /,0.211 2./,051 1. 760
Q, m3/s 3.31E-02 1. 93E-02 1.38E-02
8.102 8.102. The three water-filled tanks shown in Fig. P8.102 are connected by pipes as indicated. If minor losses are neglected, determine the flowrate in each pipe.
D
=
f
=
f = FIGURE P8.102
Assume fhe fluid flow.s from IJ 10 Band /110 c. TIJl/sJ or f (o./m)" ~ ::: f(O.OBhJ/v,. +*(O.0811ii'~ Thus, ~ = O,6Jf ~ rO.6~ ~
()I:
0.08 m 400 m 0.020
Q,. +Q3 (I)
For fluid flowiIJ9 from /I foB with fA: /,8 ~O and ~ ~ ~ :::-0) Z = Z + t i, .:it + ~ .&. Yl' /I 8 I DI 2.1 2. D~ '1 or
60 m - 2 0 111 = (0.. 0105)
Hence
V/2.
(200/1)) O,/m
2(fl.8J.ff's)
+(
)(200h) )
V/
o,oem Z(q.811f,J
0.02J)
J
LfO =/,529 ~2. + 2.55 ~2. Simdqr/y, for fluid f/owinfJ from II fa C .wilh fJA ::fc =0 Z1J =2 C +~k.it + f3 b. YJ.'I D, 2., D3 ,., or
'-
am)
/f.0()
IIJ)
(2.)
C/lJcI
~::: ~ ~OJ
V;
60", :: (0, 01.5)(-2 o V, +(0.020)( o.oem ( Il!.) 2. 9. 810$2. O,lm 2.(fl.B/~')
Hence 60 =- /,,529 V/ + 5.10 ~2 Solve Ers. (/)J(z)lond(.3) J
(3)
for 11,) ~14I1dl(g. From£fs.(J)and(3): 60= 1..5'29 (o.tJTi(v,. + ~/ +.5./0 ~:ZJ or 95.8 = (V~ +l4l + B./If ~2.
Su6lr(Jcf £'1' (2.) from Ef. (3) 60-'1-0
:
= .s,/0~:Z + 2.55 \,42 or ~
=12 ~'- - 7.8'1- I
(If)
(5)
Thvs from £I(s. (If) fJnd (s): 8,//f 1/./ +(j2. vi _7.ei +'V.3 )2.- 95.8 :::- 0 J
This
C4fJ
be sillJplified fo
I ~ I 2 ~ l' .2. ~ -7. 81f
rearrange
10 qive ~ If -
-I/./Jf ~
2-
Square both sirles and /9,63 ~2 + Q2. 5 =0 which C4n he solved
:= 103.6
by fhe qlltJdr4/ic formfJla fo give ~2.= Iq.63:±JI9,~":'-tf(q2..5)' ;: /1.77 or or ~
== 2.BO.!f
7.86
Thvs ~::3.4l3f
(6)
8./02'
(con'i)
Nofe: The value ~=9.'f3: is not a solufion af E,(s. (I), (2.), (lnd (3).
the orifjino/ e'lufJlions Wah this' valve fhe rl9hf hand siJe of EfJ6)
J
is neqaliv8 (i.e, /0.3.6 - II./If ~3.:; 103.6- //.IIf (3.11-3)2. = - 2.'1;5), lis seen from the lefl hand side of E".(6)J -fhi.s etlAIl"; be. This ex/ro root WQS inTroduced by squariIJ9 Eq. (i),
Thus} ~= fJ3~=f(O.oBm)'"(2.fJO!f) lJ/so from Eq, (.3) ~
;:;0,0IJfI.;-3
J
6o=I,S:J.9V/'+S.JO(2.8 0 )" or ~.::: 3,62f} 1l 2. ",3 or Q/:: AI ~ = ;yo (O.JOm) (.3.62 f) = O.02.B'f-:s
olJd from £'1' (0: 3.62:: 0.6'/- ~ +0.6'1(2.80) W
or Vt. =2.86f
Q2. =fiz ~ ::::.1J (0.08m/ (~,86 ~)
3
=O.0/'1-3.!J
8./03 K.IIJ3 Water is pumped from a lake, into a large pressurized tank, and out through two pipes as shown in Fig. PS.I03. The pump head is hp = 45 + 27.5Q - 54Q2, where hp is in feet and Q (the total flowrate through the pump) is in fe/so Minor losses and gravity are negligible, and the friction factor in each pipe is 0.02. Determine the flowrates through each of the , pipes, QI' and Q2'
Free jets
= 5 in.
Q
f----1200 It - - - - I . j (2.)
II FIGURE P8.103
(I)
/ +0.02. (1ooH/(6112.1-I))
= O. 703 'vi
---:'---:ll~/-'-----:~)
/ +0.02 (/2fJO rT/ (SII1. If)
T/]vsJ Q:z.
- - ==
Az.::: Vi
~
f
0
r'" _,., LL nD!":) I.f:z. - v. TU u uri .s 6
J./. -'-
{naT
II, V, (12#) "" (fl of c(:z. ::: Q, +O. '1-11 f¥1 ::: /, '1-11 fYl
({ I
Q=
A/.t0 wi-lh no
If
ff,fnt (0. 703V,) 6
J
of
/O.r.t8s
(rr;m Ihe lake 10 the f41J~ S~ 11141 wf!h Ie ::l) al/d
fi +hp = It +ir f?
(a)
h, =::
('1-)
hfJ,.., :: JZ[/+£l] I /)J
W/JldJ
CAli
~
'"
<'
w/I/J -4 (I) ff) five
De r;()/'IIb//lerJ
WlW6
Y. ~ ::"~
l, :: I
.1Z '"
(~ • ,,.11)
"'..5. o91/i
hp::: 'l-S J-2.7.S' -S'f~:z. so fhal Elf (f:) ber;ollle.r '$'~ +:1.7.£ ( /.IfN~/) - ~If. (I. 'PI (t, / = :r;:.~([! .fQ. 02. llj ] and
or
I~
/31r.r/·-~(}.9QI -~Sr:O wht'clJ /;4,S the so/v/;'61l 1:>/ '-r,
=
I/-o.9.t
ii'l().f)"+'f(~)('3,j (/3 J)
-
0 76 •
3 S
2
II Iso
J
t¥2.. :::
O.'MJ Cl
::: O,'HJ((),7P31j3) :::: 0.37.2. !J3 8-100
J
II I
\ or
-
br.
,J.' ,/.
O.7V '(). W{I/r;II
h
'tiS IJO
phYJ"/&4/ II1 Caf}iIJf)
2
BJOIf
J X.IO-l A 2-in.-diameter orifice plate is inserted in a 3-in.diameter pipe. If the water flowrate through the pipe is 0.90 cfs, determine the pressure difference indicated by a manometer attached to the flow meter.
Q _
to' -
~-IOJ
.-.(
1L _ 2. in.
t
~=3jn.
=ltf.26
1..1. d==2in. IT
Ji3
_ 2
D - 3 in. -"3 J
fJ 0--lld If V~1l~ = O.7-!f If D f (f,. rf)~
= 2.'1Sx/o6
v ,
Ii ~
Q:: O. 70 s
J
and
8./0S
I
8.105 Air to ventilate an underground mine flows through a large 2-m-diameter pipe. A crude flowrate meter is constructed by placing a sheet metal" washer" between two sections of the pipe. Estimate the flowrate if the hole in the sheet metal has a diameter of 1.6 m and the pressure difference across the sheet metal is 8.0 mm of water.
or Q == 29.5 Co
-¥
0)
DV
= Ais OJ Re : : -:;;"
and d
~; -D =
V
(2m)
---=-----";"----=----rv
/./fb X /O- S 1!L.s
1.. 6m
2.0 hi.
or
(2.)
:::0.8
and errol' so/tJlion: Assume G,;::: 0.6/ so fhQf
Trial
m v- 7fQ -- IlJao-:s ( . )2. 3
Ll
nence
J
-
JI.
Froth £q'(~)J
2.01l'J.
3
from Eq.O), Q :::29.5 (0.61):: /8.0..!p-
::. 5 ..73 ./!1.. oS
Re -:: 1.37 x/Os (5. 73) ::' Z 85 x/OS
This Re and @ 9ive Co::: 0,6/ (see Fi1. 8.'1-.1) wh/ch "1rees w/fh fhe ps.!vmed valve. m3
ThfJS; Q-:: /8.0 r !
8-102
9./06
T 8.106 Gasoline flows through a 35-mm-diameter pipe at a rate of 0.0032 m 3 /s. Detennine the pressure drop across a flow nozzle placed in the line if the nozzle diameter is 20 mm.
-cn nn Ll /2(-p,-f>l)' p(J-~"')
1.'1 '¥ -
h
,were
t:J~.d.D = 35mm 20 mm
t"
= 057/
.
7
IJn = lld Jf
-From To ble /. 6 \0.:: 680 ~ and }l =3./ x/0'1- ~':' m 3 ThuS Re : : ejD where V=~ == O,00321F = 3.33 111 fh f ~D .ff (O,o3smt s so OR = (68of!&)(3.33f)(o.03 5 m) = 2.56 x/Os e 3.1 x IO-1f'!!.-:f m 1
J
lien eel from Fi9- 8.4.3 J Froth E'{. (J) O. 0032 or
I
I
;,if
~ =(O.986).1j (o.02.0m)" (.690""-1
11 -fJ,- =
8./07
en == OJI86
'''-
3.2'fX/O¥.g.'1. =
8.107
32JfkPa
Air at 200 of and 60 psia flows in a 4in.-diameter pipe at a rate ofO.S21b/s. Determine the pressure at the 2-in-diameter throat of a Venturi meter placed in the pipe.
8-/ 0 3
-fo) If)
J -0.57/
2
(I)
8. /OPJ
..! h:: 7.3 mm H,.o
8.108
A 50-mm-diameter nozzle is installed at the end of a 80-mm-diameter pipe through which air flows. A manometer attached to the static pressure tap just upstream from the nozzle indicates a pressure of 7.3 mm of water. Determine the ftowrate.
Q
J::
C.
2. (PI-e;).L
11
pO _~If)
n nn
I
f
v
50mm
Wnere ~ = 80mm = 0.:625
an
d
1,-/2 == 0H oh =(q800~ )(7.3XIO- 3m) ==7/.5J:{,"
Thus, with lin;:: If d~" Q=C. 1l (O. 050ml" ~ (7/.5~) n
m
(/IZ3~)(J-0.62.SII)
If
or 3 Q:: 0.0230 en IJssiJ/1'Je Cn =0.97 so that Q= O.0223J} and V=JL = O.0223-f-3 :: 'f.'1'f/ll fD'" 1J(O,o8m)'" oS or
~
Re :: JLQ = (If.lI/f oS )(O.o8m) !.lft x10-.5 !:~
'Y
Cn = 0.963
:/:0.97
:;
~,Jf3Xlo'l Wilh fhe Re and a we oblain
.
\"'
(fhe ass{)med value) (See Flq.8.Jf3) 3 0.963 So fhai Q = 0.0230 (0.963):; O.022/.if
Thus J assume en:: J JlL3 ana V= 0.02.2/ = If: 'fO J!l $
f(O.OBml S I./'th Re == (Jf.Jfo~)(0.oBm2 ~ b1 ' C. C'LneCK'- Cn . WI /.~6xJO-.s.!§3 = 2./fIX/O we 0 10111 n -;::: 0.963 from Fi9- 8.11-3 (checks) TL ~ m3 I nUs, l\';;:: O. 02. 2/ T L
8-10'1-
8./09
J
.... .."rQ/r : --1h =3.lff .
,
8.109
A 2.S-in.-diameter nozzle meter is installed in a 3.8-in.-diameter pipe that carries water at 160 of. If the inverted air-water U-tube manometer used to measure the pressure difference across the meter indicates a reading of 3.1 ft, determine the flowrate.
Q=Cnlln
2({1-PtJ.)
C ,rom
8.J:
e{l - ~'1)
7:ll IQole
VIf.
() !
where ~=Ji D
J
P=I.896
slv9:!, fl3
(I) \...
- ..} D=3.Bm.·
= 2.5(n, 3.8 m,
+
• d=2 5J'n
~f'
=0.658
-& Jb's
JjI-=8. 32X / O -fP. sofha
,
(I)
t
Re:= £VD = (J.eQ641¥) v(flfI) 8.32 x/o-
ft-
6
Jti!
Re :: 7. 2.2x10 ~ V where V. . 1i f)/SOI wilh (} =:: 1f D2 V £1. (JJ bec()mes (usi"9 {1-(J:;.:;: 0' h): .1l(~ fI)2. V= C .1l(3:§.ff)~[ 2 (32.2.~)(1.8q6 ¥;f)(3.lf-l) J~ or
J
If
or
n
/2
If
0$
/2
(1.8Q6
o$il105) (1-O.658/f)
V::. 6.78 en
(3)
Trial and error solo/ion ().Sif}9 Fi9' 8,/f3 for en -:: Gil (Re ~:::a6s8) ; lissome =0.Q9 From £1,(3) V= 6.78 (0.99):: 6.11ij I
en
From Ef(. (2) Re;: 7.?..2.XIOII-(6.7/~) = 'f.8/f"t/o s which from F'9. 8. tt7 9ives =- 0.99 (checks wifh fI.uumed vqllJe)
en
Thf)~J
(2.)
V=6.71# Ql1d Q=~llVr:::f(~"lft)2(6.7/If):: O.5281j-3
8-/ 05
8,110
J 8.110
Water flows through the Venturi meter shown in Fig. P8.110. The specific gravity of the manometer fluid is 1.52. Determine the flowrate.
*.r 6
(2.)! .
(I) e
• 3
~
1~_:2
Q
'\
~
In.
I
1---------1/
in.
~tm
t
'SO = 1.52
FIGURE P8.110
h ~ == JL == 63 in.in, -- 0. 5 Q::: Cv IIT Z({1-"fo.)· p(J-(3'1-) , were \" D Also, f' +'II = f~ +h'(l-h) +~(SG)h or f,-I2.:= o(SG-J)h = P9 (SG-I)h Hence
I
/) - G II
L\' -
v
T
/2 eg (S6-nh
p(J- ~If)
,
or
[ 2(3:;..2fA)(J'E2-J)(lifl)]~ Q= Cv l£(~f.ll If /2. [ (1- 0.5 11 ) Thus, Q= 0, I/qB Cv Assume Cv ';:. 0,9£1 so tha+
Hencs
1i.~
~H7~
J@
V;;:"2. IfD
:::;
1l (6 )1. 'I- jiH
:::; 0.596
li oS
so
Q;:. 0.1198 (O.98)=O.1J71j3
thaI
Re = 'tQ == (O.5q6!f)(~FI) :::; 2.'1-6 X/O'ls 'Y 1.2/XIO #From Fi9' 8.'15 cd this Re Cv~O.96 :/=0.98.1 fhe assumed value. J
HehceJ ass ume Cv = 0.96 J
or
Q'" O. /I fl8 (0.96) '" 0.11 s1f3 and
V=':('1)'"
0.586
y
If 1'1.
6
( There {o,.°e Re -- 0.5S6 nJ -- 2 . Lf2.XIOIf so /. 2J x 10-.£ Cv ~ 0.96 Checks 'IIdh l).r-svlfJed valve. J
Hence)
::
Q::: 0, I IS ¥3
8-/06
Inr.rI hrom R'9, 8 ' ¥s J
B.II/
I
3 ~ *6 'tPl .1,----_---.J/ ;0
(l,)
')
~,." ~
X. j II If the fluid flowing in Problem 8.110 were air, what would the flowrate be? Would compressibility effects be important? Explain.
1. ~_:2 in. ~ tm
'sa
t
= 1.52
FIGURE P8.110
or
Q:= 5. elf Cv As-s(}me Cv =0.96, Hence. D _
FIe - -
v
1:1
or Q= .s. 61 'S so thai V=!fD2
('.9- )
(l8.6o$) p.fl -'l-Rl I,S7x/O :s
VlJ -
.fl.'!
~
fl3
S.61
=
f (f,.Jfl oS
fi.
-= 28,6 s
Firom r/f]r fh SA'S is qive.s
- CJ it -7,1/ X/O
C\l~O.975 or Q=5.8Jf(O.97S)== .5.69!!.3
Nofe: Wah v =
"* = :/1-)". 'I1'-
"'
29.0
Re = 29.0(1£) = 9 2IfX/O"" /,57)(/0-'1 "
f/-
we oh14in
T/J/)SJ from Flo. 8.'»'5 c.v~ O.97SJ 7 which 49rees wdh /he Qss()/lled
valve.
g-/07
(:2.)
8./12
I
8.112 Water flows through the orifice meter shown in Fig. P8.112 at a rate of 0.10 cfs. If d = 0.1 ft. determine the value of h.
e=*
II/so, -lL _ 0.10#3 _ it V- .g/)2. - 14 (Ii. Hi - /f. 58.s ==
Y!2 == (Jf.58fJ)(k~t) s /.2.lx/f
1/
~
FIGURE PS.1l2
Q='CoA.y ~t':p{ J where
Re
h
='
so
~/; =0.6, f1t -/~ 4h =e~;'
(f)
thor
= 6.3/ XIO~ Hence) from Fi.·9.8Jfl) Co::::O.6/6
¥
Therefore} from E1. {/J!
¥(0. Iff?"
O.IO.[j-3 =(0.616)
2 P(32.21iJ h or
pu- 0.6/1-)
h ==5.. 771; TJ
9,/13 8.113 Water flows through the orifice meter shown in Fig. P8.112 at a rate of 0.10 cfs. If h = 3.8 ft. determine the value of d.
r L1 Q== V()11o
e(J - (J~)
2{-(J,-(Jz.)
and, V= Si... = .:JlD2. ~
J
h
a tl r:&
were \:' 15
= _d_ . I~ ft
-
-
It''h =: f'J h
fJl f:J., - 0 .
J
O.IO~ = ~.58 It R :: ~::: ('!.ssY ~(Aff) (.!. f~)2. e ".. I 21 XI 0 ..s1L
1l
#
I~ ~
S
)
•
Trial and error sO/lJfion: From Ef· O) I x. o..s ,if == c.0 11. (d ff)'2.f 2 e(32 .zYa)(3.BfI 2. or QOOB/If= If f ( /- ~If) •. {
)J
(j)
== 6.3/ X10'f
s
C
Q
d2.. ) Whe1'e d~ NI
J _ ~'f
Ct.)
/Jssvme ~= O.6} or d= I~~ =: ~ (0.6)== 0, I off. rhv~ from £rt.(:t-) Co = 0,7£9. !loweverJ (rom Fi9,8.91 for fhis He and ~J Co=O.6IS rFO.7S9 Assume ~ =0, 65, or d == -h (0.65) = 0, J08 (-I. From £". (;.) Co = 0.6 33. Fro", F;,- e.III J Co = 0.618 :# 0.633. . A,ssume ~ =0.67) or d== 12. (o.67) ::: 0.112 fl. From £f' (,,) Co = 0, SBo From Fi9' 8.'fl J Co .::0,6/9 :;:0.S90 ThlJs, d~ 0,109 (1 8-108
e.II'I-
I
I
~
K 114 Water flows through the OIifice meter shown in Fig. PS.1 12 such that h = 1.6 ft with d = 1.5 in. Detennine the flowrate.
Ll Q =c.ono
2 (f1-(h.) p(J-~'1)
J
h
were
h =1.6 (f
FIGURE PS.1l2
d }oth p;: 15d = 2./.5ill. in. :;:; 0.75 an /1-1:1.-:: 0
;:
e1h
Thvs, it. ~ =c 1l(1,5 )2.[ 2 e(32.2~){J.6H)] Q o I/- IT ff f ( /- 0.7£11") or Q=O./S/ CO Also Re::: ~
. 1Z. = 1,'J.IX/fT V(7i.f1) ;:: 1.3BJ
()ssflmed VtI/lJe.
lienee,
r;.:: 0,0936 fJ! s
8-IOQ
Th/lS V:::~S.B (0.0936)
8.'1-/ J Co =O.62; Me
(I)
(2.)
8. lIS S.I 15 The scale reading on the rotameter shown in Fig. P8.115 and Video VS.6 (also see Fig. 8.46) is directly proportional to the volumetric flowrate. With a scale reading of 2.6 the water bubbles up approximately 3 in. How far will it bubble up if the scale reading is 5.0?
t ~
I!:L +Z -tYL7.- h : : jt I 2{f rl where
I, ::I,. ~IJ J
(0
FIGURE PS.115
1J2.~.J.Z +~ z. 2.?
~
Z, ::: OJ ~ ~()~. s() thai
.
w/lh no /o.r.res
(hI. ~(J~
It/2.
"£j
::=
For
the f'Of4meter r¥ ::: KI(SR
Zz
KtSR
/)
~::-ft~T
Al-(z!-)
,.
Jil/iJiA~
(s.o/-
(2.&)7-
==
so
th4 i whell
c()lfJ/;i/J~J w/IA Elf. (/~
K2.(l./~'- _ .l
,l<2.(SB)2. :::£
By
reat.iilkl (,fAd
J{ is a CQIJ.S141J1.
Thv.r J
WhC/'9 SR ~sca/e
or
1; (~,)
- (r;. fI)
11J6S8 fWfJ e(I'/4ti()ll~
-f--
(r;.ff)
or f;
== O.92s"
8-110
If = //'/117.
8.116 I
8.116
Friction Factor for Laminar and Transitional Pipe Flow
Theoretically, the friction factor, f, for laminar pipe flow is given by where the Reynolds number, Re = pVDIIL, is based on the average velocity, V, within the pipe and the pipe diameter, D. Also, the flow is normally laminar for Re < 2100. The purpose of this experiment is to use the device shown in Fig. PS.116 to investigate these two properties.
Objective:
f = 641 Re,
Equipment:
Small diameter metal tubes (pipes), air supply with flow regulator, rotameter flow meter, manometer.
Experimental Procedure: Attach a tube of length L and diameter D to the plenum. Adjust the flow regulator to obtain the desired flowrate as measured by the rotameter. Record the manometer reading, h, so that the pressure difference between the plenum (tank) and the free jet at the end of the tube can be determined. Repeat for several different flowrates and tube diameters. Record the barometer reading, H bar, in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Calculations: For each of the data sets determine the pressure difference, Ap = 'Ymh, between the plenum pressure and the free jet pressure. Here "'1m is the specific weight of the manometer fluid. Use the energy equation, Eq. 5.84, to determine the friction factor, f Assume the loss coefficient for the pipe entrance is KL = 0.8. Also calculate the Reynolds number, Re, for each data set. Graph:
On a log-log graph, plot the experimentally determined friction factor,/' as ordinates and the Reynolds number, Re, as abscissas.
Results:
On the same graph, plot the theoretical friction factor for laminar flow,
f = 64/Re, as a function of the Reynolds number. Based on the experimental data, determine the maximum value of the Reynolds number for which the flow in these pipes is laminar.
Data:
To proceed, print this page for reference when you work the problem and elide hen' to bring up an EXCEL page with the data for this problem.
II FIGURE PS.116
e-/II
8./16
I (CDIJ'f) Solution for Problem 8.116: Friction Factor for Laminar and Transitional Pipe Flow
D
L, in. 24
Hatm , in. Hg 28.9
T, deg F 73
h, in.
a, ml/min
a, cfs
v, fps
Re
f
0.003887 0.003652 0.003534 0.003328 0.003033 0.002945 0.002863 0.002709 0.002651 0.002179 0.001708 0.001060 0.000648
61.11 57.40 55.55 52.31 47.68 46.29 45.00 42.59 41.66 34.26 26.85 16.67 10.18
3202 3008 2911 2741 2499 2426 2358 2232 2183 1795 1407 873 534
0.0341 0.0349 0.0345 0.0344 0.0349 0.0339 0.0325 0.0322 0.0307 0.0338 0.0461 0.0758 0.1194
0.000330 0.000280 0.000250 0.000186 0.000130 0.000097
28.58 24.24 21.69 16.08 11.28 8.42
638 541 484 359 252 188
0.1007 0.1134 0.1311 0.1785 0.2348 0.3256
0.000545 0.000401 0.000312 0.000191 0.000112
25.17 18.50 14.42 8.84 5.17
770 566 441 270 158
0.0838 0.1140 0.1431 0.2270 0.3893
=0.108 in. Data 7.5 6.75 6.26 5.54 4.66 4.29 3.92 3.48 3.21 2.34 1.86 1.11 0.63
D
6600 6200 6000 5650 5150 5000 4860 4600 4500 3700 2900 1800 1100
100 2100
0.6400 0.0305
=0.046 in. Data 9.52 7.68 7.08 5.26 3.39 2.61
D
Theoretical f Re
560 475 425 315 221 165
=0.063 in. Data 4.58 3.32 2.51 1.48 0.86
925 680 530 325 190
P = Patm /RT where
=YH20*H atm =847 Ib/ftJl3*(28.9/12 ft) = 2040 Ib/ftJl2 =1716 ft Ib/slug deg R =73 + 460 =533 deg R
Patm
R T
= =
Thus, P 0.00223 slug/ftJl3 and Y Also, ).l 3.83E-7 Ib s/ft Jl 2 Theoretical for laminar flow: f
/!p/y
=p*g =0.0718 Ib/ftJl3
=64/Re =64/(pDV/).l)
=(fUD + KL + 1)(VJl 2/2g) where KL = entrance loss coefficient =0.8 and V =a/(nDJl2/4)
~-//2-
8.//6
Problem 8.116 Friction Factor, f, vs Reynolds Number, Re
1.00
~:-=:- -=~i:-:==
f!
!
i : :~
-+---~~-~~: . . --.-~.-~--'-----'-----lI
0.10 - -.-----.--
.------.. - .. -
--
------~---'b.-"-----
--------~:r-
.. ---------------.;......-~-
.. - .. --------,--~--'--'
-------~------_r_~--+--i---""'---.
:' I , -- --------4------+--'--+--\-------"~__.;-~->___+-+--+--+
=0.108 in.
•
Experimental, D
•
Experimental, D = 0.046 in.
• Experimental, D = 0.063 in. --Theoretical, laminar
0.01
-l-----------'--"-r------~---~""___1
100
10,000
1,000
Re
8-/1.3
8,117
8.117
Calibration of an Orifice Meter and a Venturi Meter
Objective:
Because of various real-world, nonideal conditions, neither orifice meters nor Venturi meters operate exactly as predicted by a simple theoretical analysis. The purpose of this experiment is to use the device shown in Fig. P8.117 to calibrate an orifice meter and a Venturi meter.
Equipment:
Water tank with sight gage, pump, Venturi meter, orifice meter, manometers.
Experimental Procedure:
Determine the pipe diameter, D, and the throat diameter, d, for the flow meters. Note that each meter has the same values of D and d. Make sure that the tubes connecting the manometers to the flow meters do not contain any unwanted air bubbles. This can be verified by noting that the manometer readings, hy, and ho, are zero when the system is full of water and the flowrate, Q, is zero. Tum on the pump and adjust the valve to give the desired flowrate. Record the time, t, it takes for a given volume, V, of water to be pumped from the tank. The volume can be determined from using the sight gage on the tank. At this flowrate record the manometer readings. Repeat for several different flowrates.
Calculations: For each data set determine the volumetric flowrate, Q = VIt, and the pressure differences across each meter, IIp = 'Ymh, where "1m is the specific weight of the manometer fluid. Use the flow meter equations (see Section 8.6.1) to determine the orifice discharge coefficient, Co, and the Venturi discharge coefficient, Cy, for these meters. Graph:
On a log-log graph, plot flowrate, Q, as ordinates and pressure difference, IIp, as
abscissas.
Result:
On the same graph, plot the ideal flowrate, pressure difference.
Qideal
(see Eq. 8.37), as a function of
Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem .
.. FIGURE P8.117
8-/ltf
8.//7
T(con)" )
Solution for Problem 8.117: Calibration of an Orifice Meter and a Venturi Meter d, in. 0.625
0, in. 1.025
V, gallons
t, s 27.0 13.2 34.2 16.6 12.0 11.7 15.4 25.1 20.4 17.3 15.7
ho, in. 9.3 37.1 5.5 23.9 43.2 51.3 27.9 10.1 14.7 21.4 26.7
hVI in. 3.8 14.5 1.9 10.1 18.1 21.7 11.2 4.2 6.2 8.7 11.2
2.00 L1PoI Ib/W2 L1PvI Ib/ft"2 48.4 19.8 192.9 75.4 28.6 9.9 124.3 52.5 224.6 94.1 266.8 112.8 145.1 58.2 52.5 21.8 76.4 32.2 111.3 45.2 138.8 58.2
3
A, ft /s 0.0099 0.0203 0.0078 0.0161 0.0223 0.0229 0.0174 0.0107 0.0131 0.0155 0.0170
Average discharge coefficient:
a = V gal/t s x (231
Co 0.611 0.626 0.627 0.620 0.638 0.600 0.618 0.631 0.643 0.629 0.620
Cv 0.956 1.001 1.067 0.953 0.985 0.923 0.976 0.978 0.990 0.986 0.957
0.624 orifice
0.979 venturi
Ideal C=1 L1P, Ib/ftA2
in."3/gal)x(1 ft"3/1728 in."3)
L1P = YH2o*h = 62.4 Ib/ftA3 *h ft Ov =A2/[1 - (A2/A 1)"2]"0.5*C v*(2*g*L1p)YH2o)"0.5 and 0 0 =A2/[1 - (A2/A1 )"2]"0.5*C o*(2*g*L1PoIYH20)"0.5 where A1 = 1t 0"2/4 = 1t (1.025/12 ft)"2/4 = 0.00573 ft"2 and A2 = 1t d"2/4 = 1t (0.625/12 ft)"2/4 = 0.00213 ftA2
Problem 8.117 Flow Rate, Q, vs Pressure Difference,
i i i....,.
J•• :
,
~p
:'
• • I
:
i :: i
iii! i
I
!
I
!
I '. i I
0.001 +-_--'-_..!...i---L:.....I-~'. !. .:
: Iii I:
i i!
!---'----.!..--'---'-...........-.W....{
'-tl
1000
100
10 ~PI
Ib/ftA2
8-/IS
Experimental, orifice Experimental, venturi Theory, C
=1
18.0 75.5 11.2 47.7 91.4 96.1 55.5 20.9 31.6 44.0 53.4
8.//8
8.118
Flow from a Tank through a Pipe System
Objective:
The rate of flow of water from a tank is a function of the pipe system used to drain the tank. The purpose of this experiment is to use a pipe system as shown in Fig. P8.118 to investigate the importance of major and minor head losses in a typical pipe flow situation.
Equipment:
Water tank; various lengths of galvanized iron pipe; various threaded pipe fittings (valves, elbows, etc.); pipe wrenches; stop watch; thermometer.
Experimental Procedure: Use the pipe segments and pipe fittings to construct a suitable pipeline through which the tank water may flow into a floor drain. Measure the pipe diameter, D, and the various pipe lengths and note the various valves and fittings used. Measure the elevation difference, H, between the bottom of the tank and the outlet of the pipe. Also determine the cross-sectional area of the tank, A tank • Fill the tank with water and record the water temperature, T. With the pipeline valve wide open, measure the water depth, h, in the tank as a function of time, t, as the tank drains. Calculations: Calculate the experimentally determined flowrate, Qex' from the tank as Qex = -AU!nk dh/dt, where the time rate of change of water depth, dh/dt, is obtained from the slope of the h versus t graph. Select a typical water depth, hI> for this calculation. Graph:
Plot the water depth, h, in the tank as ordinates and time, t, as abscissas.
Results: For the pipe system used in this experiment, use the energy equation to calculate the theoretical flowrate, Qth, based on three different assumptions. Use the same typical water depth, hI> for the theoretical calculations as was used in determining Qex. First, calculate Qth under the assumption that aJ1losses are negligible. Second, calculate Qth if only major losses (pipe friction) are important. Third, calculate Qth if both major and minor losses are important. Data:
To proceed, print this page for reference when you work the problem and clu:k hen' to bring up an EXCEL page with the data for this problem.
D
Floor
.. FIGURE P8.118
8-1/6
Solution for Problem 8.118: Flow from a Tank Through a Pipe System
The pipe is galvanized iron with threaded fittings. The system contains: one sharp edged entrance one fully open globe valve two 45-deg elbows four 90-deg elbows
D, in. 0.595
A tank , ft"2 0.654
h, ft 1.00 0.90 0.80 0.70 0.60 0.50 0.40
t, s 0 13 26 40 54 67 81
H,ft 1.00
Total pipe length, in. 135
T, deg F 71
Experimental: Q ex = -(dh/dt)*Atank = -(0.0074 ftls)*(0.654 ft"2) = 0.00484 ft"3/s Theoretical with no losses: Qth = V 2*A2, where when h = 0.90 ft V 2 = (2g*(h + H))"0.5 = (2*32.2*(0.9 +1.0))"0.5 = 11.06 ftls and with A2 = rcD"2/4 = rc*(0.595/12 ft)"2/4 = 0.00193 ft"2 Qth = 0.00193 ft"2*(11.06 ftls) = 0.0213 ft"3/s Theoretical with major losses: Qth = V 2*A2, where the energy equation gives h + H = V//2g(1 + fLID), where again use h = 0.90 ft and f is a function of Re and £ID Thus, with h = 0.90 ft, 1.9 = (V//64.4)*(1 + f*135/0.595), or 122.4 =V/*(1 + 227f) Re = V2D/v = V2*(0.595/12 ft)/(1.04E-5 ft"2/s) = 4768*V2 and £ID =0.0005 ftI(0.595/12 ft) = 0.0101 Trial and error solution: Guess f, solve for V 2, calculate Re, obtain new f from Moody chart The solution is: f
=0.041, V 2 = 3.44 ftls, Re = 16,430
Qth = 0.00193 ft"2*(3.44 ftls) = 0.00664 ft"3/s Theoretical with major and minor losses: The energy equation gives h + H = (1 + fLID +L.KdV//2g where L.KL = 0.5 + 10 + 2*0.4 + 4*1.5 = 17.3
=
Thus, with h 0.9 ft 1.9 = (V//64.4)*(17.3 + f*135/0.595), or 122.4 = V/*(17.3 + 227f) Trial and error solution gives: f
=0.42, V2 = 2.14 ftls, Re =10,200
Q th = 0.00193 ft"2*(2.14 ftls) = 0.00413 ft"3/s
(con J-/) 8-117
~-------------------------------------------------
I
Problem 8.118 Water Depth, h, vs Time,
S
1.2 1.0
-6c-----+----~----_r___----~-1
0.8
I
~ 0.6
-------~~~~----1
.c:
I
0.4
I
I i .
h = -q.0074t if 0.9965! 0.2
!
0.0 0
20
40
60
t, S
8-//8
80
100
Experimental I I I-Linear h vs t II I •
8.//9
8.119
Flow of Water Pumped from a Tank and through a Pipe System
Objective:
The rate of flow of water pumped from a tank is a function of the pump properties and of the pipe system used. The purpose of this experiment is to use a pump and pipe system as shown schematically in Fig. PB.119 to investigate the rate at which the water is pumped from the tank.
Equipment:
Water tank; centrifugal pump; various lengths of galvanized iron pipe; various threaded pipe fittings (valves, elbows, unions, etc.); pipe wrenches; stop watch; thermometer.
Experimental Procedure: Use the pipe segments and pipe fittings to construct a suitable pipeline through which the tank water may be pumped into a sink. Measure the pipe diameter, D, and the various pipe lengths and note the various valves and fittings used. Measure the elevation difference, H, between the bottom of the tank and the outlet of the pipe. Also determine the cross-sectional area of the tank, Atank • Fill the tank with water and record the water temperature, T. With the pipeline valves wide open, measure the water depth, h, in the tank as a function of time, t, as water is pumped from the tank. Calculations: Calculate the experimentally determined flowrate, Qex' from the tank as Qex = -A tank dhldt, where the time rate of change of water depth, dhldt, is obtained from the slope of the h versus t graph. Graph:
Plot the water depth, h, in the tank as ordinates and time, t, as abscissas.
Results: For the pipe system used in this experiment, use the energy equation to calculate the pump head, hp, needed to in order to produce a given flowrate, Q. For these calculations include all major and minor losses in the pipe system. Plot the system curve (i.e., pump head as ordinates and flowrate as abscissas) based on the results of these calculations. On the same graph, plot the pump curve (i.e., hp as a function of Q) as supplied by the pump manufacturer. For the pump used this curve is given by hp
= -2.44
X 105 Q2
+ 51.0 Q -
12.5
where Q is in fe Is and hp is in ft. From the intersection of the system curve and the pump curve, determine the theoretical flowrate that the pump should provide for the pipe system used.
Data:
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•
FIGURE PB.119
Solution for Problem 8.119: Flowrate of Water Pumped from a Tank and Through a Pipe System
The pipe is galvanized iron with threaded fittings. The system contains: one sharp entrance eight 90-deg elbows two 45-deg elbows two globe valves one union
0, in. 0.625
A tank , ft"2 0.647
H,ft 3.50
Total pipe length, in. T, deg F 62 242 Pump equation hp, ft 12.50 12.31 11.63 10.46 8.80 6.66 4.02 0.90
t, s h, in. 25 0 7.6 24 16.1 23 25.2 22 32.3 21 40.8 20 48.9 19 57.7 18 65.7 17 74.9 16 82.7 15 Experimental: Qex -Atank*(dh/dt) where from the graph, dh/dt
=
Q, ft"3/s 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007
V,ft/s 0.00 0.47 0.94 1.41 1.88 2.35 2.81 3.28
System equation Re f
o 2070 4140 6210 8281 10351 12421 14491
=-0.1204 in.ls
Thus, Qex = -(0.647 ft"2)*( -0.1204/12 ft/s) = 0.00669 ft"3/s Theoretical: The energy equation gives h +hp - hL = H +V2/2g, where 2 hL = (fLlO + LKd*V2/2g = (f"(242 in.lO.625 in.) + 0.5 + 8*1.5 + 2*0.4 + 2*10 + 0.08)*V /2g 2 = (387*f + 33.4)*V2/(2*32.2) = (6.01*f + 0.519)*V
=
=
Thus, with h 18 in. 1.5 ft, 2 2 hp H - h + hL + V 2/2g 3.5 - 1.5 + (6.01*f + 0.519)*V + V /(64.4)
=
=
or 2 hp = 2.0 + ( 6.01*f + 0.535)*V But V Q/A Q/(rc0 2/4) Q/(rc*(0.625/12 ft)"2/4) 469*Q Thus, the system equation is hp = 2.0 + ( 6.01 *f + 0.535)*(469*Q)2 = 2.0 + (1.32E+6*f + 1.18E+5)*Q2
=
=
=
=
Also, obtain f from the Moody chart with Re VO/v V*(0.625/12 ft)/(1.18E-5 ft"2/s) 4414*V c/O = 0.0005 ft/(0.625/12 ft) = 0.0096 From the graph, the pump and system equations intersect at Qth = 0.0051 ft"3/s
=
=
=
( ('0,,'1)
8-120
hp, ft
2 0.0309 0.0490 0.0470 0.0450 0.0430 0.0425 0.0420
2.16 2.73 3.62 4.84 6.37 8.27 10.50
8./19
Problem 8.119 Water Depth, h, vs Time, t 30 25 20
C
.£
•
15
Experimental
-Linear h vs t
10 h=-O.1204tf+-24.9 ; 5 0 40
20
0
60
t,
80
100
S
Problem 8.119 Pump Head, hp, vs Flowrate, Q
i
•
L'!
'
8 +--l--~I--i'-+-+~.f-.----'Il-+-¥-\.·-+--+-crr-+--'---+ I
\. i ' /
-+- Pump curve -+- System curve
..Jf' I
!
I
!
;
1
I
i
I !
!
i
_\l
!
"'1
o +---J.--'--i--'--~---r_,..J,J_-,--r--'--........- f 1.
0.000
0.002
0.004
0.006
Q, ftJ\3/s
8 -/"-1
0.008
a, }20
I
8.120
Pressure Distribution in the Entrance Region of a Pipe
Objective:
The pressure distribution in the entrance region of a pipe is different than that in the fully developed portion of the pipe. The purpose of this experiment is to use an apparatus, as shown in Fig. P8.l20, to detennine the pressure distribution and the head loss in the pipe entrance region.
Equipment:
Air supply with flow meter, pipe with static pressure taps, manometer, ruler, barometer, thennometer.
Experimental Procedure: Measure the diameter, D, and length, L, of the pipe and the distance, x, from the pipe inlet to the various static pressure taps. Adjust the flowrate, Q, to the desired value. Record the manometer readings, h, at the various distances from the pipe entrance. Record the barometer reading, HbaI> in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Calculations: Detennine the average velocity, V = Q/A, in the pipe and the pressure P = 'Ym h at the various locations, x, along the pipe. Here 'Ym is the specific weight of the manometer fluid. Graph:
Plot the pressure, p, within the pipe as ordinates and the axial location, x, as
abscissas.
RESULT: Use the graph to detennine the entrance length, Le , for the pipe. This can be done by noting the approximate location at which the pressure distribution becomes linear with distance along the pipe (i.e., where dp/dx becomes constant). Use the experimental data to detennine the friction factor for fully developed flow in this pipe. Also detennine the entrance loss coefficient, KLent • Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem.
r
h
~
iIii FIGURE P8.120
~-I2.2.
Solution for Problem 8.120: Pressure Distribution in the Entrance Region of a Pipe
0, in.
L, in.
Q, ft"3/s
H atm , in. Hg
0.74
50
0.481
29.7
x, in. 0 1 2 4 6 10 15 20 30 40 50
h, in. 9.98 7.21 6.61 6.19 5.82 5.15 4.23 3.64 2.28 1.09 0
T, deg F 75
p, Ib/ft"2 51.9 37.5 34.4 32.2 30.3 26.8 22.0 18.9 11.9 5.7 0.0
P = Patm/ RT where Patm = YHg*H atm = 847Ib/ft"3*(29.7/12 ft) = 2096Ib/ft"2 R = 1716 ft Ib/slug deg R T = 75 + 460 = 535 deg R Thus, P = 0.00228 slug/ft"3
v = Q/A =(0.481
ft"3/s)/(n*(O.74/12 ft)1\2/4) = 161 ftls
p = YH2o*h From the graph, the p vs x results are linear after (approximately) x
=15 in.
Thus, Le
=15 in.
For the fully developed flow portion, dp/dx = -fpVI\2/20 and from the graph dp/dx = -0.635 (Ib/ft1\2)/in. Thus, f = 0.635 (I b/ft1\2)/in. *2*0. 74 in./(0.00228 slugs/ft1\3*(161 ftls)1\2) = 0.0159 From the entrance to the exit of the pipe Pent = (KL + fUO)pVI\2/2 Thus,
KL = 2Pent/(pVI\2) -fUO= 2*51.9Ib/ft1\2/(O.00228 slugs/ft1\3*(161 ftls)1\2) - 0.0159*50in./0.74 in. =0.'682
Results: Le
= 15 in.; f =0.0159, and KL = 0.682.
~-1').3
Problem 8.120 Pressure, p, vs Axial Location, x
60 50
-j~------:-----~-------1
40
••
N
<
~
~
. c 30 --+c.
•
20
-
10 0 0
20
40 x, in.
L~
__
8-/1-'1-
60
•
Experimental
t:.
fully developed flow Linear (fully developed flow)
8.I2.J
I 8.121
Power Loss in a Coiled Pipe
Objective:
The amount of power, P, dissipated in a pipe depends on the head loss, hL' and the flowrate, Q. The purpose of this experiment is to use an apparatus as shown in Fig. PS.12l to determine the power loss in a coiled pipe and to determine how the coiling of the pipe affects the power loss.
Equipment:
Air supply with a flow meter; flexible pipe that can be used either as a straight pipe or formed into a coil; manometer; barometer; thermometer.
Experimental Procedure: Straighten the pipe and fasten it to the air supply exit. Measure the diameter, D, and length, L, of the pipe. Adjust the flowrate, Q, to the desired value and determine the manometer reading, h. Repeat the measurements for various flowrates. Form the pipe into a coil of diameter d and repeat the flowrate-pressure measurements. Record the barometer reading, H bar , in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Calculations: Use the manometer data to determine the pressure drop, IIp = "Imh, and head loss, hL = IIp/y, as a function of flowrate, Q, for both the straight and coiled pipes. Here "1m is the specific weight of the manometer fluid and "I is the specific weight of the flowing air. Also calculate the power loss, P = "I QhL' for both the straight and coiled pipes. Graph:
Plot head loss, hL' as ordinates and flowrate, Q, as abscissas.
Results:
On a log-log graph, plot the power loss, P, as a function of flowrate for both the straight and coiled pipes. Determine the best-fit straight lines through the data.
Data:
To proceed, print this page for reference when you work the problem and click /rae to bring up an EXCEL page with the data for this problem.
1 h k - Manometer
- T
Air supply
1m
r
'I
...."
-
'
~)Ol ed pipe
-
--
Free jet
•
FIGURE PS.121
8-/25
Solution for Problem 8.121: Power Loss in a Coiled Pipe
D, in. 1.44
L,ft 18
h, in.
Q, ft"3/s
Hatm , in. Hg 29.9
T, deg F 80
~p,
Ib/ft"2
hL' ft
P, hp
Straight Pipe Data (d = infinity) 10 1.19 8 1.06 6 0.913 4 0.731 2 0.505
52.0 41.6 31.2 20.8 10.4
709 568 426 284 142
0.1125 0.0802 0.0518 0.0276 0.0095
Coiled Pipe Data (d = 8 in.) 10 0.835 8 0.745 6 0.641 0.517 4 2 0.357
52.0 41.6 31.2 20.8 10.4
709 568 426 284 142
0.0789 0.0563 0.0364 0.0196 0.0068
~p = YH20h where YH20 = 62.4 Ib/ft"3
hL =
~p/y
where Y =gp
P = Patm/RT where Palm = YHg'"H atm = 847Ib/ft"3'"(29.9/12 ft) = 2110 Ib/ft"2 R = 1716 ft Ib/slug deg R T = 80 +460 =540 deg R Thus, P = 0.00228 slug/ft"3 and Y = 0.0733 Ib/ft"3
P = (yQhJft Ib/s'"(1 hp/550 ft IbIs)
e-I2.6
8.11-/
Problem 8.121 Head Loss, h L• vs Flowrate, Q
.
800 700 600 500 ~ 400 .s::. 300 200 100
,,
•
L
f
)
1/
/
~Straight
/ / -/
pipe
_Coiled pipe
rf ('
a
a
0.5
1.5
1 A
Q, ft 3/s
Problem 8.121 Power, P, vs Flowrate, Q 1.000 -,.------~-----...,
•
Straight Pipe
•
Coiled Pipe
0.100 +------>;Q;;----t1--------; (coiled)
c..
.r:.
-
0:
P = 0.0679Q2.87 (straight)
0.010 +-----,...--1---------;
0.001
+-------+------.. . .
0.1
10 A
Q, ft 3/s
~-/27
Power (Straight Pipe)
- - - Power (Coiled Pipe)
q,/ p
=
-0.25 pU2
9.1
Assume that water flowing past the equilateral triangular bar shown in Fig. P9.1 produces the pressure distributions indicated. Determine the lift and drag on the bar and the corresponding lift and drag coefficients (based on frontal area). Neglect shear forces.
U
= 5 ft/s
~
p
= 0.5
pU2
b = length =
4- ft
FIGURE P9.1
olJ = {fJ cose dll + f ~ sin e JII, where ~ == 0 Thu.s, rIY= {fJ ~os e dll +J f cose dA +Sf cos e J/I I
::: 2
or
JfJ
2.
cos
60° dll
1
J
.3
- fJ dll == 2 (o.S~lI:z.) cos 600 iIJ 3 -(-0.25 f {j'-) i b
pU" ib .so fhat JJ= O.S (I.f/I} .s~~.s )(5 ~+ /"(o.lft)(LfftJ otJ=
0,5
Becavse of syrnmefry of fhe object} ;t::: ~ A/so} from El(. {/J
C - ~ - ~sfY1b = ;,00 D - i p7J'-A - f pY2.Jb and .since ;t::: 0
cIC"L = -f pTlIJ
0
=-
(I)
== 9.70
/b
9.2
r
f' =
9.2
Fluid flows past the two-dimensional bar shown in Fig. P9.2. The pressures on the ends of the bar are as shown, and the average shear stress on the top and bottom of the bar is Tavg. Assume that the drag due to pressure is equal to the drag due to viscous effects. (a) Determine Tav a in terms of the dynamic pressure, pU 2 /2. (b) Determine the d~ag coefficient for this object.
a)
elf:;:: {ric/ion
u
p= +pu2 ~I
..
Width = b
drfA1 = 2 7;'19 (/oh b)
I.
-0.2(+pU 2)
-- ;/ r avg
r avg
1011
i
.,
== 20 hb ~1I1
anrt
cJ,. =pf'e.$.Svre Jr41:::: -I: p yz.( hh) -(--I; pU'"(O.2)) (hh) :: 1.2 (1. py2..)(bh) TlJv~ if 20
4 ~ ~f
hh faV!
= 1.2
fhsA I y1 (bh) '£ P
or
tv, : : 0.06 (fp7J~) h) t#= 4 +4 ~ C ~ pu2.lJ:: C tp1J2. bh ll
Thvs
ll
~
2,0/76 or
fall,
+/.2
20 (0.06) (fpU
ThvSj eLl::::
2. if 0
2
(ipy2) (bh) )
+1.2(ifU~)
=CD dpU" hh -eel)
(-feV:I.)
9..3'"
I
9.3* The pressure distribution on the 1-mdiameter circular disk Fig. P9.3 is given in the table below. Determine the drag on the disk.
\
Cll L(2.)
......
\~
~-
FIGURE P9.3 D
of} ==
r::::r
f frill - f f I
rilJ =
2.
J f (271' r dr) -11 : D2. si/Jce dlt:: 27Trdr r=o J
ThlJS J 0.5 m of) == 271' { f r dr - (-Sif,.)fOmJ.)
0 •.5
:=
o
fr
J
kN/m
o
o
0.05
O.")../tf
0./0
0.'106
0.15
0.558
0.20 0.25
0.62-0
0.30 0.35
0.71/ 0.662-
O.JfO O.¥S
O.S6~
0.50
0.000
/lSln9
r
dr
oJ-
3.93 kN
where rLjAJ ~ r.-v In /11} fOUowiIJ9 In/of/ranJ: 0
Evu/lJale fhe /nle9ral numerically tlsiR9 the
By
Z7TS f
o
0 .. 333 pr09rlJH1
4.34 4.28 4.06 3.72 3.10 2.78 2.37 1.89 1.41 0.74 0.0 0.5
0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
O.6QS
fhe
p (kN/m2)
r (m)
SIMPSON. OilS we ohlain
~*************************************************
** This program performs numerical integration
**
S{Jr dt' = O.2LJ/ 0
·t"
over a set a set of an odd number of equally ** ** spaced point.s using Simpson s Rule :t::t ************************************************** f
Enter number of data points: 12 Enter data points (X . y) ? O.OO.O.OOCI ,. . 0.4-5.0.333 '; 0.05.0.:14
r)
0.50.0.000
? 0.10.0.4-0E '?
O.15~O_55P
? O.20.0.62C ? 0.25,0.695
?
O.30~O.711
?
O~35.0.662
·~I
O.4:0.D.564
The approximate value of the integral
q-3
15:
+2.4-083E-Ol
9.11-
I br:Widfh
9.4
The pressure distribution on a cylinder is approximated by the two straight line segments shown in Fig. P9.4. Determine the drag coefficient for the cylinder. Neglect shear forces.
o
f--
_pu 2
8cO
\:
:
\
I I I
I \ I
______ '\1
Sf- cos() (brJJe fJ ::: -f U"if 1f
1f'
cosf) df)~
-pU 2.(J C(Js(jd9::: ~pV2.sine I
:: 7Th
~
1V'2
and. ~
f)::()
e r=: f)
e
V 2.
~
{ fJ CQS() tie =TPV'f [I -Ie] c()sede:: i-fV2.[siIl 9 0 0
(JJ
-*
~ (cos9 f9sin8)}
=t pU2.[I-~ (f) -(-1;)] :: t('u~[ ~ -2] Th/l~ fr()111
tb=2br
Frs. (I), (2), IJlld
Ii c()s8 de
(J)
0
Thus)
Sf
e deg
180 ' I I I
fJ :: - PyZ for of ~9~" find fJ =: -#:(,172.[1-#9] f(JrO~8~ f (,:e. fJ::: 1fU2. if 1T
I
90
"I \ I ,I
.".
,fJ-fpC()S8dlJ:: fpcosB (brd9) =2. where
I
<\
P
8 'I!: 2,,,,,
eJr
~
=2
(3)
hr [if u2.{fr -2) +p U ] =ipv'('-;,hr) Z
()
so thai
9-If
0 (.1)
Linear distribution
9.5 Repeat Problem 9.1 if the object is a cone (made by rotating the equilateral triangle about the horizontal axis through its tip) rather than a triangular bar.
u = 5 ft/s 0.1 ft p= 0.5 pU2
b = length
•
By symmelry the fifl
IS 2ero.
FIGURE
= 2 ft
P9.1
Thvs; tf. ::0 and CL ::0
X :::O.11iJ
A/so ,jJ = r/{ron+ + 'tear where
4ronf :: SfJ
fJ/
and
dll and dlJ;::; ~rrrdx
fJ ==o.sf U2. ( J -lOX)
Thv-S.J wilh
i.e.
C()So
==
o,spU2.
xc:O
fli :: 0
J
we hove
x~~/
4ronl '" or
f
0/ •
~J
cole) dx =27rCOSR.60'(0.5 pT72.)f(X-IO)t)dx
0.5 PU2. (J-/OX) 271(X
X=O
0,'
~ronf '" 27r(0.S/(o.spU")[ 4 ~xsl,o" 4runf ='0.00130q eU" 2
_
Also) 4ear':::: _(o.1.SpTJ2.)
.so thaI
o
*
(0.1)' =0,00/96
eU2
o:9=4ronl +tiJ,.c4r -::: (0.00130'1 -1-0.00/9/) p7J2.=O.00327 f7J2. -so w/fh
p =/,9'1
s;;rs
and U:: 5
tx9::= 0,00327 (/,?/fJ(5)'- :: 0./s9 II;
9-5
#
J
Q.6
I 9.(; A 17-ft-long kayak moves with a speed of 5 ftl s (see Video V9.2). Would a boundary layer type flow be developed along the sides of the boat? Explain.
Re
=
Pf-
J
SO
w/.Jh
Re-
wdh 1:::/7ff Ql1d 5
1/= j.Z/x 10-
1f2
TJ:: we
5!1
and 00'F w4fer
halle
(J7ff)(Sfj.) - 7.02Xj06
- I.Z/x/o5
if -
Since Re ~ }OOO is offen assumed +0 be the lower Jim/I for bOlJndary layer fype flow J if is clear thai a b()undary loyer wov/d develop l1/on9 the sides of the kayak. ~.
9.7
,
9,7 Typical values of the Reynolds number for various animals moving through air or water are listed below. For which cases is inertia of the fluid important? For which cases do viscous effects dominate? For which cases would the flow be laminar; turbulent? Explain.
Animal (a) (b) (c) (d) (e)
large whale flying duck large dragonfly invertebrate larva bacterium
Speed
Re
10 mls 20 mls
300.000.000 300.000 30.000 0.3 0.00003
7 mls 1 mmls 0.01 mmls
IlJerlia important if Re ~ / (i.e whale, d{}ck dra9()nfl;) I
VisctJl4 effecTs arJlfJin41e if Re~1 (l.e larva I haefer/vliI) BOlJnrilJfl), layer flow becomes
lurhllleni
for
Re
()/J
-Ibe
order of lOS' 10 /o~ (t',e, whale ana psrh4ps Ihe dlJ&~) The flow w()u/d be /amilJQr for fhe dr4?OllfifJ /1J1'1I1J, 4nd haGTer/llh1 anJ perhQp5 Me duck.
9-7
'l. 9
I 9.9
Approximately how fast can the wind blow past a O.25-in.-diameter twig if viscous effects are to be of importance throughout the entire flow field (i.e., Re < 1)? Explain. Repeat for a O.004-in.-diameter hair and a 6-ft-diameter smokestack.
UD
<
b
if viscous e ffecfs are fhrouqhoul fhe flow.
If For sfandard air 1/ -= /. 5 7 x/o- 1-2
Thu-s}
U<
-If
1.57XIO
D
object
J
where D l.s fhe diamefer in feef.
DJ ff
yll J S
fwi']
2,OBXI02.
7, S'I- X/0.3
hair
3.331-10'"
O.if7/
smokesfack
6
2.62. x/os
9-e
fo be ImporlQnf
9./0
I 9.10
A viscous fluid flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm. Determine the boundary layer thickness at distances of 0.20, 2.0. and 20 m from the leading edge. Assume laminar flow.
For laminor flow 0::: c1(i J where C i.s Q cOllslQn7. Thus) - L _ 12XI0.3m where X~m J o~tn C - ...Iv' - {i:3m = 0.0105 or 6 = 0.0105 rX
x) m
'1,//
rx
1.3m
6J
m
6) mm
0.2.
0.00'-170
If. 70
2.0
O.OI'fB
/'f. 8
20.0
O.OJf70
Jf7.0
I
9 •.17 If the upstream velocity of the flow in Problem 9.10 is U = 1.5 mis, determine the kinematic viscosity of the fluid .
For /tlmifJfJr flow
Thus, 1/ =
m
. m 6 = sVT ) or 3
'2.
(I.5-r)(J2 Xlo-m) 2-5 (/,3m)
=
6
j/
-6m2-
.6SX/0
S
TJJ2. = :;.s X
9./2
I 9.12 Water flows past a flat plate with an upstream velocity of U = 0.02 m/ s. Determine the water velocity a distance of J0 mm from the plate at distances of x = J.5 m and x = 15 m from the leading edge.
the U = u f r,,) From
BJasivs sO/III/on For bovndary laygr
I
17 =
yllx
Vq/ves of f(J7) are "iI/en in
I.
Ux
(O.02./f) (J5m) = J./2.x/tJA ~
crifical Rexcr = 5)(/0
X,: /. 5 m
5 J
andy::
n -(lox 0-.3
'f, -
'/
/'n
)
TabJ~ 9.1.
5
==
is Jess fhan fhe the boundarv/ Jailer {/Pli/ is labJilJar. F
2..68 x/O
if fol//ws thai
3 /ox/0- m We obfain:
O. 02 ~ (J./2X/f 6..!!l!) (I.s m)
- IOql -.
oS
Linear /nferpo/afion from Tahle q,/ 9ives:
f
I
= 0.26'1-7
-I-
(O.3Q38 - O.2.6'1?l (J.2 -0.8) (/.Oq/ -0.8) :::: O.3EQ
lIence, U, == [j ('(il)
SihJilarly ot J
=(O.oZ1f) (0.3 5 9);:; Xl. :::
-3)
I
T!z ::llox/O rn
0.00718.t; 3
15m and y:: /ox/f m we ohfain: .JJ1.
0.02,..s
(I.I2.Xlo-6:J!f) (Js/h) s
= 0.3'15
L/I)etlr ;nferpo/afi()n {rohl Table 9.1 q/ves :
f 1=
0.0
+
(0./328 -0.0) (0.8 -O,¥)
(0.3'1.5 - 0.0)
lienee, tlz ~
on a flal plate}
where YI J fhe simi/drily variable) I'~
J
SInce Rex = ~
/II
flOk!
u frlJ2,)
=(0.02~)(0.II'fS)=
O,002.2Q.tp
'1-/0
-= 0,
/IJfS
9. /3"
T
"'9,13 A Pitot tube connected to a water-filled U-tube manometer is used to measure the total pressure within a boundary layer. Based on the data given in the table below, determine the boundary layer thickness, 8, the displacement thickness, 8*, and the momentum thickness, 8.
eflJaliofJ) ~/Ih
From fhe Bernoulli
fa if' .c::<
PH,20
H follow.s thai
PI + -t fQ,r v,2. == 12 + i eq,-,. ~'J.J where V, -= {j, I ~ ~ 0 J fl,::: 0 and '2.::: 'tNJ.o h Thus}
U =
I¥ 2
J
~zoh =
I. II 2l9BOO pjJ)h m
/.23~
Pair
or
m
11L
/26.2. Vh
tJ..=
I
where h-m} tJ~s
J
y (mm)
h (mm)
o
o
2.1 4.3 6.4
10.6 21.1 25.6 32.5 36.9 39.4
10.7 15.0 19.3 23.6 26.8 29.3 32.7
40.5 41.0 41.0 41.0
For Y> 26.8 mf11 we see that h = '11.0/11111
Thvs U = /~ 6.Z/(O.OJfI/ = J
For y::: 23.6mm J or
U
V;;;
6~
Thvs J
The
25.55.p.
U::: /2-6.2 /(O.OJl.05)
== 25. '1-0 .p-
25,.1/<) 2.5 ..5.5
= O. 99If
23.6
mm
displacemeni fhickness J
l =S(1-11- )dy
6~ is
oD
~ = U
/='
o /2. 6. 21/h 25 . .5.5
0,0268
or Since
= '1-. 9~''i( ,n
IhIS. hecome.s
In
{(I-'f.9!fl/h)dy y=o
Numerical infeqrafion of fhe tabulafed dafa qJves (See nex+ PtJge.)
9-1/
6-- ~/8X/O-3 m
IJ/so} the momenlvm f!J
is
fhicknes.s) (iJJ aOU8m
~
f
= V(J - -V-) Iy '" 4i. 9/f ~ 1(h (J -1f-.9'f fh) dy . o
y~o
Nt/merical infe~rlJlion of the fabulafed JatfA gives @::: Use proqrafYI T!(IIPEIOlio
Ym J
2.23xI0-
;rde9rqle the Infe9rands fo611/afed be/olJl:
1h (/-If,9ifYii')
( 1- 'f.?'ffh) I
0
0.1f9/ 0.282 0.210 Q./Oq
0.0506
0 0.002.1
0.00'1-3
0.006'10.0107 0.0/50 O.OIQ,3 0.0236 0.02.68
0.02.93 0.0327
O. O~/O 0.0335 0.01Q7 0.0098/ 0. 0 0386 0.001/8
0,0511
O.OI9Jf 0.0058'1
0 0
0 0
0
0
****************************************************~
** This program performs numerical integration ** ** over a set of points using the Trapezoidal Rule ** ***************************************************** Enter number of data points: 9 Enter data points (X , y) ? 0,1
? 040021,0.4:91 ? 0.00f,j,3,O.282 '? 0.0064:-.0.210 ? 0.0107,0.109
'? 0.0150,0.0511, ? 0.0193,0.019f,j, ? 0.0236,0.00584:? 0.0268,0 The approximate value of the integral
Th"s~
0.0268
S(/-Jf.9Jftfh)dy==o.OOlfI8
o
(con't) '1-12
-
.
· l" c '"
+f,j,.1777E-03
3
m
****************************************************~:
** This program performs numerical integration ** ** over a set of points using the Trapezoidal Rule ** ***************************************************** Enter number of data points: 9 Enter data points (X , y) ? 0,0 '? 0.0021,0.0506 ? 0.0043,0.0410 ? 0.0064,0.0335 ? 0.0107,0.0197 ? 0.0150,0.00981 ? 0.0193,0.00386 ? 0.0236,0.00118 0.0268,0
"
The approximate value of the integral is: +4.5206E-04
0.0'-'
Thv.s or
~
J
S Vh (J-q..qJf-Yh),y = ~.51.X/o"fI. o -3 = Jf.9~ x ~,S2 '1.10- ¥ =2..23 x/O In
~-I3
9.14 Because of the velocity deficit, U - /l, in the boundary layer, the streamlines for flow past a flat plate are not exactly parallel to the plate. This deviation can be determined by use of the displacement thickness, 15*. For air blowing past the flat plate shown in Fig. P9.14, plot the streamline A-B that passes through the edge of the boundary layer (y = 15 8 at x = C) at point B. That is. plot y = y(x) for streamline A-B. Assume laminar boundary layer flow.
Rej:::: T:::: U£ . 5 mee
Streamline A-B
u=
--
1 m/s
-+-
A -
-----" Edge of bOlJndary layer
-
f = 4
/_
m------H.!
FIGURE P9.14
= 2.7Jf X /0 s <.::
(If) ( Jf In) /.Jft X/O-OS..f!!J-
5 X 10 5
J
fh e bounat/ry J
5
layer .flow remain.5 laminar alonq fhe enlire p'lafe. flelJce,
6=
sj ~.
or
68 ;5 [
u'm~o;-1')(tf"')] ~ O.0382-m
The flowrQle aJrried hy fhe acfual Dovndtlry loyer is by de{'l/dion ef/val 10 thaf carried by a vniffJrm velOCity w/lh Me plare d/sp/qced. by an aRlotJnf 6~ Since there is no flow fhrouf/h the plate or sfreamline II-B J
u
u
~
i ,----""
6
-1--(I)
QA :::QB or UYI/::= (68 -6;) U where 6'= 1. 721 or(. [ (/.'l-6XJ(r.5:fflj{lfm)]~ 0 0 °8=/·72.1 IT J=' J315m J
1';; ·
~
Thus, y,q == 68
-
dB
,
-= O. 03B2-m -
O.O/315m :: 0.025/ m
S[": Y)~ Y-~7 _ L~
lienee, for ooy x-lOCI/lion A QA == Q or [/>A::: U(y -6 ) or y :::: IA + 6.J= ~ t 1.72/ /
v:
;;:. O.OUI
mt
-0.04
[(/.~x/o
-:s) X m
~"""r~-r-r""""7/'r'7""':)r-r-Tj"7")/"rrr
= 0.025/
/.If
0.035 E
]i
I
_5m'J..
1.7').1
: sfream//ne
+ 6.58 X/031x' m
where X""m
_._-, ------.---.--1
--------1
--.---------~
--~----
o~o~;
~ o~o~~ =~-=~-------- --:-----=-~-==~-~-===~= ~ O. 01 ---------- - ----- -----{I ---------------~-----------~----------------~---.
i
0.005
- l
Or--------r~------~------~------~
o
1
x, m
9-/Jf
I
2
3
4
9.15 Air enters a square duct through a I-ft opening as is shown in Fig. P9.15. Because the boundary layer displacement thickness increases in the direction of flow, it is necessary to increase the cross-sectional size of the duct if a constant U = :2 ftls velocity is to be maintained outside the boundary layer. Plot a graph ofthe duct size. d, as a function of x for 0 :s x :s 10 ft if U is to remain constant. Assume laminar flow.
FIGURE P9.15
For incompressihle flow Q0 == (}(x) where Qo::: flolVrafe info !he dud n3 :::: (JAo ::: (2. fJ) ( J f/1) ::: 2 -:s and Q(x):: [J/II where fJ:: (d - 2. 6')2. /s Ihe effective areq of Ihe dlJ~f (1J//owiIJIj for fhe decrefl.sed f/owrale if) fhe bOllndary layer).
Thus, Qo :::: U(d -26·)2 or d = / fI f where
f1/X'
6' = /. 721V-1!-
Hence, from
d=
= I. 71.1
[r.
2t5*)
(0
J ~ ~1;.'" )X -'1.f:J..l.
1.57
0:
0.0/52
fX {~ where xNf/
Ef. (I)
/ + 0.030'1
rx ff
For example) d:: / IIIJI x:::O ond d::: l.oQ6 fl of x ~ /0 ff. d vs x
1.10 1.08 1.06
---.--------.~~-
~ 1.04
't:J
1.02 1.00 0.98
-I
---;---------r-------,----,------j
-1-,
o
2
4
6 x, ft
9-/S
8
10
9.16
I 9.£6 A smooth fiat plate of length e = 6 m and width b = 4 m is placed in water with an upstream velocity of U = 0.5 m/s. Detennine the boundary layer thickness and the wall shear stress at the center and the trailing edge of the plate. Assume a laminar boundary layer. 2.
6::: 5/1/TJX
•==-5 (/./2.x/o-t) X
-,
== 7.¥8 x/o-
0.5 oS
and
3
fi)"J?
O./2Lf.
N m'2.
J
Thvs J af X=3m
here
W
=6 m
X~m
6=7,lf8X/(j.31(3::: D.O/30m ">- _ 'IV' -
while of X
Vx m
_ 0 07/6 J!. f3' ' In'-
0,/,.1/
o= 7,118 X/031(6' 0./2'1-
=
O· 0183 m N
?w:: 16' = 0.0506 m"
'1-16
J
where x~m
(999.15,;3) (1.12. XIO- 3 !!.;f,.) X
3
fw =0.332 U ~v-ef· =0.332. (o,s1}-)Y2.
- rx
3
.
9./7 9.17
An atmospheric boundary layer is formed when the wind blows over the earth's surface. Typically, such velocity profiles can be written as a power law: u = ay", where the constants a and n depend on the roughness of the terrain. As is indicated in Fig. P9.17, typical values are 11 = 0.40 for urban areas, n = 0.28 for woodland or suburban areas, and n = 0.16 for flat open country (Ref. 23). (a) If the velocity is 20 ftls at the bottom of the sail on your boat (y = 4 ft), what is the velocity at the top of the mast (y = 30 ft)? (b) If the average velocity is 10 mph on the tenth floor of an urban building, what is the average velocity on the sixtieth floor?
( '"n) U -- C yO./6
Th vs}
tJ2. U,
(b)
~
o.~
lJ. = C Y
ThlJs
,
J
U
J
FIGURE P9.17
h e 'IS a cons/tin J f were O 6 •J ff
(V)
=r ~
or Uz.;:
-
where C is
;; =
>.
t1
(30 If )0.16 lIff
== 27.6
1i S1
consTanf
(Y; )0. tlo ~
20 S
Or U2. = lomph
q-/7
(
60 )o.l!== 2.0.s mfJh
/0
9.18 A 30-story office building (each story is 12 ft tall) is .built in a suburban industrial park. Plot the dynamic pressure, 2 pu /2, as a function of elevation if the wind blows at hurricane strength (75 mph) at the top of the building. Use the atmospheric boundary layer infonnation of Problem 9.11.
From F/q. P 9./7 fhe houndary layer ve10cdy prof/Ie is (Jlven by 0.28 C O. '2.9 h U Y ) Or u.:: Y J were C is a consf4nf. /'V
Thus) .JL ()./ or
::::(L)0.26 X
028
U= /JO(3rO) . fj.
where y ~ ff a
Hence}
o
t pu~.= i(2.38X,03~¥4)f,o( ~~or;r or
Th/.s
/.5
o
X= 301..12
=360ff
o o n o
,,--7J~~-T-'i""7--;~rr'7-"
0.5 6
-t. f u'J.= /'1: tf (k) fr:L
n
J
where Y'" If
plolfed in the fi9l1re below.
400~----------~------------~-----------,
350 3 0 0 - ------~--- --------------- --- ----250 --------------------
=~ 200 >-
150 100
-------------------"-~-~------
50
-------------
~~-------------------- - - - - - -
o +-~~--------~------------r_----------~ o
10
5
,/-18
15
9.19 The typical shape of small cumulous clouds is as indicated in Fig . P9.19. Based on boundary layer ideas, explain why it is clear that the wind is blowing from right to left as indicated.
FIGURE P9./9
lis indicafed in
h~. P9./7) beCIJilSe
of the afmosphmc bOtJndary
layer the velocily of fne wind 'IM.rally increases wiM ajf;/pde. ThlAf, fhe fop porlioM of a cloud fravels Fosfer Ihun ifs base - Ihe cloi/ds fend fo "lip " fOlllard fhe direclion of the wind. Tha! is J Ihe wind is from ri9M fo leff.
9.2.0
I 9.20
Show that by writing the velocity in terms of the similarity variable '7 and the function f(,}) the momentum equation for boundary layer flow on a flat plate (Eq. 9.9) can be written as the ordinary differential equation given by Eq. 9.14.
The 90verninq
e'lvtJl/ot}s
tin d
dl/. +- U -0 IX ~y -
U
CI/'Ie
(I)
#- + v ~ = r 41t
COIJsirler
and;; = Thvs J JJ7
/).=
(2)
r; f(n) and
(lxi'-y
v=(%:%)~(1 r'-f)
--% .1. II rx = -2., ~ fV- yx =- 2. X
,so
fhaJ
JJ'l
where (
/= 1"
(2.1)
1fiJ'-k
ond Ty = rJi X :z
(.1)
"
~= !x (Uf') =utf = U~¥X =-i-¥i!f'l ~
and
(If)
~
.;,
If ::; (r;})2it (17 f '- f) = (fJJ~ (" f" i- fl - f')f9- =(¥ff )'-'1 f"Yf x-~ °fr :: i Jf 1'/ {'I (5) ~y
-rh{)~
by tJsi/J9 £1s. ('I-)tJIJd (s) we see fhai .Er rl) is .sqj;Slied fol'
any funcfion (tif).
II/so, ~ = 1#and
.k
#- =In (ur') [Yf £li] :: (rJ'/) ~ f"
a'"U = ('.1.)Js. .Lf.1 =( U3)~ vx ~y ~x
1"y2
Thus) by
lISiIJ9
Efs.
(2./)) (t))
(6)
fill ry u/ ::: J.t:. fill 1/X
(7)
fJlJd (7) w/lh £r. (2.) we ohltli/)
(U f') (~-¥ J? {") +( 1f¥)Y,. (1t f '- f) (!i)~ [" =
vll f'"
which simplifies 10 : 2. fill -
ff"=o
From El(. (2.1) the bovndary conditiofl.S at
u.:: 0 = U fco)
y=O (i.e. 1=0) beC()me aIJd V =0 =(!!f)~ (0 (to) - fro))
That is J fro):::o and f'roJlllo
Similarly J usI y -II' 0() (l.e" £1,(2..1)
f- /
as
l? -.. otJ)
".-.00.
we re'lvlre
1).'-'
u.
r/Jv~ from
9.21
"I
9.21*
Integrate the Blasius equation (Eq. 9.14) numerically to determine the boundary layer profile for laminar flow past a flat plate. Compare your results with those of Table 9.1.
Solve the fo/lowin9 third ()rder differenfiaJ ef/llafion by a nvmerical in fe9fafion feGhniqve: III
Z f +ff
1/
with boundary condil/oHS f:: f'=o at '1=0 and f'-I as 7'/- 00 (( )/: Wrile this third order ef/IJtJfion (/.f 3 lIi'sf order e'ltlalio/lS (,l/Jd /J.se a RVfJ94 - kuffa IIU/fl81','C4/ techniqfJe TO i/Jf~rale fhem. 7h1/~ /e f >1 == f J X ~ f I::. Y.z J ~ I ::: f ": ~ dlUI }) / = {"/::: - t f{'/:: -i ~ 13 Thai is: -::0
In )
J
/
}j=~
X./::
~
ond
~/=_}jY3h·
These can De tlfJl'roximqted q.s A)j = ~ AI?
.A~ = ~ Ai? J find A)j :: (- J1 Y.J/2.)Ar; start wilh Yt ~ y,.=O af 1;::0. ;?sstlllle ~ == c af i; =() (w/;el'e C is s()me 9ivell cIJI)J'rtJllf) IJlJd',hfe!r*le ff) J;:: IX) II by yt.:: X (0) +2; AX··LJI; J
I
;;
If Y2 (00) :f / (t:e j f't()()) of I) tlt.(jlJ.jt the /la/tie 01 c (t:o; f'lo)) a/Jd fry Q9a/n. Tile IwofJlJifJl bOllllldl')' Jl4/ve jJ/'IJb/e/fJ (~;e) f(o)=:rro)~ 0 and f 't¢) == I) is sp/ved by l/er41/()/J fiG IJIJ //I/iial Jla/()e problem
r,:e)
fto) =: ('to):/J) fto) ~ c).
Pro9ram P8#J.' show/J below was vsed for fhe c4/c()/atiI}As. rAe f/n4/vallle 01 C :: 0.332 and f/Je Jle/()c/"fy 'prollie J U -:: V f'r}J), rJ9ree very well w/lh fhe sfQfJJfJrJ valves ~;velJ in Tab /e 9.1 lOO 110 120 130 14:0 150 160 170
cls open "prn ll for output as #1 print "*** ******* ****** ******* *** ****,..******* ******** *****" print H** This program integrates the boundary layer **" print "** equation (Blasius equation) for a flat. plat.e **" print "** using a Eunga-Kutta type routine. The user **" print "** must specify an initial value of f'! (0) :=;,0 **" print "** the boundary condition 'at infinity' (f!! '= **"
9. 21 41 (con't) 180 190 200 210 300 310 320 330 335 336 340 350 360 370 380 385 390 395 400 410 420 430 435 440 445 450 460
print "** 1) 1S satisfied. **" print "***************************************************" print " " print If " print "Input a value for f" (0)" input c print "Input stepsize and number of stepslf input dx, n print "Input how often to print output (number of steps)" input. nn print" eta f f! f! '" yl = 0 y2 = 0 y3 = c x = 0 m = 0 for 1 = 1 to n m = m +1 x = x + dx yl = yl + y2*dx y2 = y2 + y3*dx y3 = y3 - (y1*y3/2)*dx if m ( nn goto 450 print using" ##.#### +#.##~~~~ +#.##~~~~ +#.##~~~~";x,yl,y2,y3 m = 0 next. i got.o 210 t************************************************** This program integrates the boundary layer ** ** equation (Blasius equation) for a flat plate ** ** using a Runga-Kutta type routine. The user *)f' ** must specify an initial value of f" (0) so ** ** the boundary condition 'at infinity' (f' I '= ** ** 1) is satisfied. ** ***************************************************
**
=~pu~ a value for f" (0) ? 0.332 Input stepsize and number of steps ? 0.01 700 Input how often to print output (number of steps) ? 50 f' , f' f eta O. :,000 +4.07E-02 +1.66E-01 +3.31E-Ol 1. 0000 +1. 64E-Ol +3.30E-Ol +3.23E-Ol 1.5000 +3.68E-01 +4.87E-01 +3.03E-01 e~ 2.0000 +6.47E-Ol +6.30E-Ol +2.67E-01 2.5000 +9.93E-01 +7.52E-01 +2.17E-01 ~ r3.0000 +1.39E+00 +8.47E-Ol +1.61E-01 6~ 3.5000 +1. 83E+OO +9.14E-01 +1.07E-01 4.0000 +2.30E+00 +9.56E-Ol +6.38E-02 Lfr4.5000 +2.79E+00 +9.80E-Ol +3.36E-02 5.0000 +3.28E+00 +9.92E-Ol +1.56E-02 2. 5.5000 +3.78E+00 +9.97E-01 +6.41E-03 6.0000 +4.28E+00 +9.99E-01 +2.32E-03 6.5001 +4.78E+00 +1.00E+00 +7.36E-04 o 7.0001 +5.28E+00 +1.00E+00 +2.06E-04
1
/
ov.
.7
0.5
If'
I
9.22 An airplane flies at a speed of 400 mph at an altitude of 10,000 ft. If the boundary layers on the wing surfaces behave as those on a flat plate, estimate the extent of laminar boundary layer flow along the wing. Assume a transitional Reynolds number of Rexcr = 5 x lOs. If the airplane maintains its 400-mph speed but descends to sea level elevation, will the portion of the wing covered by a laminar boundary layer increase or decrease compared with its value at 10,000 ft? Explain.
h U=/fOO h( Ihr )(S2BOff) =587.£i ReXcr = UXcr 11 ,were mp 3600.5 mi Ib .s -1 '5 T II p - 3.S3#-XIO -w:(Jf} d f rom I aD e C. I) 11 - f -3 §.l~ /,7S6'X/O
lienee) wifh
RexC,. = 5X/O
ft3
= 2.01 X/O-¥ 1j-2.
S }
X ':: 1/ Rexcr _ (2.0/ x/oJ/. fj2.) (S x IDS) :: 0./71 If cr V 587 i! s
IN sea- level: (b)
ReXcr ==
Xcr
VlI
J
Hence,
where
u
and
1/ == 1.57 x /0" 1'/2.
== 'fOO mph (
J hr
3600S
) (5 2 9,0 ml
Hl 5 - 11 Rear _ (1,57 X/0 -::s )(5 xJO ) _ 0 I
Xcr -
ft) :: 587 i± S
-J/.
Tr
v
-
587
ft
-:s
-
'lJ'[
fl
,..::1....
,
===
The laminar boundary layer occtJPt"es the ft'rsf o.13Jfff of Ihe winq at seQ Jeve I and (from par! (a) rrbove) fhe f/rsf O. 171 if af an alfilude of / ~ 000 ff. This is dve mainly f" fhe lower dellsilt ( iarger kinematic. viscosi1 y ). The d),!7amic visco.sifies are approximaiely fhe same.
9,2/f I 9.24 A laminar boundary layer velocity profile is approximated by u/U = [2 - (y/o)](y/o) for y:s; 0, and u = U for y > o. (a) Show that this profile satisfies the appropriate boundary conditions. (b) Use the momentum integral equation to determine the boundary layer thickness, 0 = o(x).
ra)
t::: 9 (1) #:1
TlJvs
J
==
2Y _12. where Y:: yld
-=
0
-#-/ = 2 - j
as if mllst J
y~J
y=O
/1 /J1 wI.
4.$
rr = U[ f -~]
A/so,
J,
-:; / 01' IJ -:: lJ ()
so 1114t a1J~ 77[f y~J
I y:::J
r] ~O
(b) From the momon/vI'» infer;rlAl (j1fJafiof), I
o=
2-i~,P)(, where c, ~ (, (l-fo)JY and c~ ~
d
1r) 1:: 0
T/;t/s)
I
c, '" ((:lY-Y') (J-zY+Y'")JY o
0
--I-S-I-/--#:3 ~
and CL
=:
=~ /-5'
(2 -2yj
Ire 0
so -fhat
J=
/
fie f/c S;
= !(2Y-Sy1+ lfy 3 -Y'l)dY
2
~2.) 1/ X '"' / 3O;X
r:s- 7l w/fh Rex::::
<5 _ -{.30 -X - VRex
1ff
_
S.'hP
-
1/ f/.e)(
J
9-2'1-
9. z-s 9.25
I
y
A laminar boundary layer velocity profile is approximated by the two straight-line segments indicated in Fig. P9.25. Use the momentum integral equation to determine the boundary layer thickness, t5 = t5(x), and wall shear stress, T" = T,,(X). Compare these results with those in Table 9.2.
0
/1
71
012
/l
[7 0
I I
I
I I
!
I
2U 3
U
I
I
FIGURE P9.25
Compare these results To fhose in Table q.2..
,U
9,261t
I
ylb
9.26*
An assumed dimensionless laminar boundary layer profile for flow past a flat plate is given in the table below. Use the momentum integral equation to determine b = b(x). Compare your result with the exact Blasius solution result (see Table 9.2).
From f
o
=
the
momentum infe 9/'al
2. 1/X J ve, ' Cz.
J
eqllof/on
where C = 1#3-} d ~ dt
and I C, -:: S9( }-~)dY with o
1=0
1f ::~(Y)and y=f
The I/Q/oe of Cz. ClJn he appr(Jximafed tiS C2. ~
and Ihe
VQ/IJe
of ~
C(JII
ulU
o
o
0.080 0.16 0.24 0.32 0.40 0.48 0.56 0.64 0.72 0.80 0.88 0.96 1.00
0.133 0.265 0.394 0.517 0.630 0.729 0.811 0.876 0.923 0.956 0.976 0.988 1.000
A(~) A
(f)
0./.13
-= 0.080::; I. 66 :t- cO
~
be oblo/ned from IJPhJ81'1&a1
infe9r41ion (pro9rtlfn Tf(flPElO])
Y o 0.8 0./6 O.ZIf 0.32O.IfO
O.Jf8
0.56 o.61f 0.7,.
().eo 0.88 0.91 / .00
,#(1-9) o
fI.s ilJdictried helow;
0.115 0./'.5
0.2.39
C, -=
0.2.50 0,233
o
0.07/ O. 0If2 0.02.3 0.0/2-
so thaf
I
(f.66) J1: - 5 03 (11 X)1: 6 =[- 21/X U (0./31) -. u
0./98 0./53
O.joq
J~(J-VJdY ~ 0,/3/ I
or
x<5 -_ -V 5.03 Re;<
h
J
eV}.
were Rex::: p
Nole: The Blasius sollliion h•.s ~ nol s .03
o
1~***************************************************
** This program performs numerical integration ** ** over a set of points using the Trapezoidal Rule ** ***************************************************** Enter number of data points: 1'-1, Enter dat.a points (X , y) ? 0.56,0.153 ? 0.00,0.000 ? 0.6'-1,,0.109 ? 0.08,0.115 ? 0.72,0.071 '? 0.16,0.195 ? 0.80,0.0'-1,2 ? 0.2'-1,,0.239 ? 0.88,0.023 ? 0.32,0.250 ? 0.96,0.012 0.40,0.233 ? ? ? 1.00,0.000 0.'-1,8,0.198 The approximate value of the integral is: +1.3096E-Ol
9.27*
I
9.2.7* For a fluid of specific gravity SG = 0.86 flowing past a flat plate with an upstream velocity of U = 5 mis, the wall shear stress on a flat plate was determined to be as indicated in the table below. Use the momentum integral equation to determine the boundary layer momentum thickness, 8 = 8(x). Assume 8 = 0 at the leading edge, x = O.
S /nce 7W = f TJ:l. ~ ;t fol1owtS that d@ =
e?WU
2
d
)(
which can be irdeqrorec1 fo give (t}s//Jq @=O af x =0) X x (j) '"
f
~~2 Iw dx =(o.e6)(1O~~)(5P-)'2. ~ Iw dx
or (j)
-==
-5 (
if. 65 x10
x
J fw dx
J
where
(f),.., m J x ~ rn J
N
and fw..v ;,;.
(I)
o
For 0 ~ x 6 2.0 m, iIJfe9r4fe E'{. (I) fo defermine @os a {()ncl/on of x. To do soJ we need fhe va ItJe of 7,; af x=: 0) whic;h is !Jof 9itlell in the fable. Theorel/cally J fw ~ 40 aT the /eadif)}. For our pvr,os8~ based On the exfrapolaTed curve be/olllJ assume ~ =: 22 -ffi at X:: 0 'tw VS X
x (m)
o
25
0.2 0.4 0.6 0.8
i
20
.
-
'"E 15
-
z
i 10
~
Or---~----~----~--~I 0.0
0.5
1.0
1.5
1.0 1.2 1.4 1.6 1.8
2.0
2.0
x,m
Pr09r4rn P91127 showl1below was {}Sed for fhe calcvlafio/Js.
(con'l)
13.4 9.25 7.68 6.51 5.89 6.57 6.75 6.23 5.92 5.26
9.27,1. I 100 110 120 130 140 150 160 170 200 210 220 230 240 250 260 270 280 290 300 310 3 20 330 400 410
(con 't)
cls open "prn" for output as #1 print# 1, "******** * * ******************** ******************** print#l, "** This program calculates the momentum **" print#l, "** boundary layer thickness as a function of **" print#l, "** x from the given wall shear streSfj distri- **" print.#l. n** but.ion. **n print#l, H**************************************************" dim t.au (11) tau(1)=22.0 : tau(2l=13.4 : tau(3)=9.25 : tau(4)=7.68 tau(S'=6.51 : tau(6)=5.89 : tau(7)=6.57 : tau(S)=6.75 tau(9)=6.23 : tau(10)=5.92 : tau(11)=5.26 print#l. " If print#l, 11 x, m momentum thickness, m" for l = 1 to 11 x = 0.2*(i-1) if l = 1 then goto 400 intgrl = 0 for j = 1 to i-I intgrl = intgrl + 0.5*0.2*(tau(j+1) + tau(j» next. j theta = 4.65E-5*intgrl print#l, using" ##.### #.*##~~~A";x.t.heta next l tf
~*************************************************
** ** **
This program calculates the momentum ** boundary layer thickness as a function of ** x from the given wall shear stress distri- ** ** bution. ** ************************************************** x. m 0.000 0.200 0.400 0.600 0.800 1. 000 1.200 1.400 1.600 1.800 2.000
momentum thickness, m O.OOOE+OO 1.646E-04 2.699E-04 3.487E-04 4.146E-04 4.723E-04 5.302E-04 5.922E-04 6.525E-04 7.090E-04 7.610E-04
I
0.0008
~I
0.0006
E ~
0.0004 0.0002 0 0.0
1.0
0.5
x,m
1.5
2.0 I
I I
~------------------------------------------------------------I
'1-28
9.Z8 9.28 The square flat plate shown in Fig. P9.28a is cut into four equal-sized prices and arranged as shown in Fig. P9.28b. Determine the ratio of the drag on the original plate [case (a)] to the drag on the plates in the configuration shown in (b). Assume laminar boundary flow. Explain your answer physically.
...!!..
T e
1 1 - - - - - - - 1.. T
• • • • •~e/4' 1
.. FIGURE P9.28
By comparinq £'ts . CO and (:J.) we see fhaf
4a = In
2. 0
4h
cose (b) fhe bOIJndary ItJyer on fhe reor plale i.s fhicAel' Iholl on fhe fronl plate. lIellce fhe shear sfre.tS is /e.ss I)" fhe r{](Jf pfafe fhall if is on fhal p/afe In confl9vrafion (a)) 9,vin, /es.s dray for c4!e (b) fhan for cqse CO)J eve" fholJ9h the fola/ areas are the Sqme.
9.ZQ
I
9.29
A plate is oriented parallel to the free stream as is indicated in Fig. 9.2.9. If the boundary layer flow is laminar, determine the ratio of the drag for case (aHa that for case(bl. Explain your answer physically.
u
PLATE
~
(a)
I(b)
FIGURE P9.29
For case (a): JCI.
_..L T/2.C 2- P Df
c.'Df -- V1.329 - 1.3)..8 Ret Yu J,'
h
/I
were
I
Y-hvs J
'1a =feU" ~('fJ')
For case (6) t PU"CDf /I
4,,:=
where
PI
an
d
IJ
02-
==
Lf"
V :=
2.56
pU% y:v;%
c.... = (ffijJJ U(tflJ
(I)
and ,4:'f12
UT
Thus, oUfb
-
11
== ...L a U2. 2.\
1.32.eW ('fill.)
y4{fi
i
:,t
= ...L (2 56 OU~ 1fV f)~) 2.. \ ~
(2.)
From E~s. (J) and (2.) we see thai
4a cftfb
=
Z =
The shear sfress decrease.s wdh disfance from thrl /eadifl9
erlfje
of fhe plafe (i.e.} fhe fhickenifl9 of the bOlJnriary/tJ!er). ThIJS.I eve/J Ihofl9h fhe plate !Area /J the same for case (aJ or (b) J the QVfJra9fJ shear ,stress (and the drat;) is 9realer for ca.se (a),
fl. 30
I 9.30
If the drag on one side of a flat plate parallel to the upstream flow is S) when the upstream velocity is U, what will the drag be when the upstream velocity is 2U; or U/2? Assume laminar flow.
C ::: 1,328 Vf
and
9-31
J~i
I
9.3/
I
9.31
Air flows past a parabolic-shaped flat plate oriented parallel to the free stream shown in Fig. P9.31. Integrate the wall shear stress over the plate to determine the friction drag on one side of the plate. Assume laminar flow.
Treat each .strip of thickness dy and /e/J9lh i=i(y) liS a SRJf/I/I f/Q/ plate wilh rlr49 dtb where for lal1linllr flow ,,jj= CD, tpU JA with rill ~ I tly olla CD1 " 2
Thlls
J
0/ t eU I rly =
/.32 B
db", V
2
O.66-!t-
1r::::: 3..i ff'f' U
~
=
1.328
ret!
1/h
rt tiy
But 1:::: If _y2 so fh41 +2-
,/) :::: JrrlrB Jr0.66'1- ~p' U VJf-y~· dy =:
3/2
/Vole: The un/Is ()/J the inTftJftI/ are HlIa (i.e. 2 rtd: L ~/.a)
9-32
9.32
I 9.32
It is often assumed that "sharp objects can cut through the air better than blunt ones." Based on this assumption, the drag on the object shown in Fig. P9.32 should be less when the wind blows from right to left than when it blows from left to right. Experiments show that the opposite is true. Explain.
'.
U'?
FIGURE P9.32.
!l si9IJiliconf porfion of fhe drq9 on an ohject cqn he from fhe relaliflely low pressure developed in Ihe wake re9ion hehind Ihe objecf. By maxin9 fhe obJeci sfrean,!;/Jed (i. e./ (luI#! fro/)} left fo ri9ht) nol rl9hf 10 Iell /f) Ille ahove !J9f)re) botJndary loyer separation /s avoided I.IlJd (J relafively fhill woke lVill; I()w drtJ9 is O/)"/tlifled. Whefher fhe front of fhe obJec1 is ~h41'j) or ''/;/u"r does noll.llfecf the cOl)lr/{;(}/iofJ to fhe dra9 frulIJ the frunf pal'7 of Ihe body -of /eqsf not CIS mvclJ CI fhe widfh of fhe wake affecfs fhe dr4g. /I
'1-33
9,33 9.33 Two small holes are drilled opposite each other in a circular cylinder as shown in Fig. P9.33. Thus, when air flows past the cylinder, air wil1 circulate through the interior of the cylinder at a rate of Q = K (Pi - P2). where the constant K depends on the geometry of the passage connecting the two holes. It is assumed that the flow around the cylinder is not affected by either the presence of the two holes or the small flow rate through the passage. Let Qo denote the flowrate when = o. Plot a graph of Q/Qo as a function of for 0 s; s; TT/2 if (a) the flow is inviscid, and (b) if the boundary layer on the cylinder is turbulent (see Fig. 9.17c for pressure data).
e
.
...
-~
....~---
~
...... ~...... ~-.~...........................~.,.
•
(0)
u
-'
;?'
",'"
~,.,
(0)
e
e
... ...,... .......
FIGURE P9.33
For invi.rciJ {Jow : fJ-lJf) C ::: r T' == / - 'I-.si!) 2.(} 'f tpu"
ThtJs) Q -= K (II - f:J. ) ::: K[r" -fo) -(f,- -fo) ]
=K[ -tpU'-( I-Jfsl,,2.(}) - t.(JU2.(J-tf.sin'J.(e+1r))] nul si,/O e sirl" (f) +71') Hence J Q:: 0 fir if/viscid flow " /Vote! !his is 10 he eXfecred becafAre
01 the .rYlI/mell'l~/
pl'8.sS fJrB
(6) For a ftJrbtllenT hotlndary
dislr/blllioll.
layer =
Q ::: K ((1, -(12) ::: K[( P, -Po) - (/2. - fo) ] ::: t pllzK [ef , - c,:z.] where efl is for ()
CllJd
c,:J.
is fol' 18o-e dey,
Ohta,n ~ 141a fr()m [;1- 9.17 Noi e; C :;; / fur ().:: () 4/Jd C, z~,1f fur B::: / It) Ie!
f TfuI.r Q0 ,so fhaf J
.R:: Qo
== Q
I ::; if 717.K[ / - (-0, t,t)]
e~o
teu'k[cp,-ce:a.] _ I. If (1: pTf'''K)
CPI -Cf2.
-
The resvlls are f4/;vlaled
t
:: /. If ( l!J1-K)
I,If 4J'1d
(c ()n'-I)
;Ioiletl be;PW.
(con If)
9,3,3
e,deg
180 - e, deg
cp1
c p2
0/0 0
0 15 30 45 60 75 90
180 165 150 135 120 105 90
1.00 0.70 0.00 -0.90 -1.70 -2.10 -1.90
-0.40 -0.40 -0.42 -0.43 -0.45 -1.30 -1.90
1.00 0.79 0.30 -0.34 -0.89 -0.57 0.00
0/0 0 vs 8
1.0 0.8
i
0.6
_____ 1
i
0.4
0
a a
0.2 0.0
--~--------.---~--~~.-:-~~~-~-
-0.2----------~~~----:-~~-~-
-0.4
--~.
---------------
- 0 . 6 - . ------ ----~--------~~-0.8
-
-------,-------------
-1.0 + - - - - - - , - - - - - - - r - - - - - j - - - - - - r - - - - - . - - - - - i 90 30 45 60 75 o 15 8, deg
9- 3S
1
9.3'1-
9.34 Water flows past a triangular flat plate oriented parallel to the free stream as shown in Fig. P9.34. Integrate the wall shear stress over the plate to determine the friction drag on one side of the plate. Assume laminar boundary layer flow.
U =0.2 m/s --....
•
.b ~ {7W dl/ where Iw::: 0.332. V Thvs ' " '" 10 • 33
3
till
(2)
f f
(2)
IY
dA x=o
rf.f}x
y=o
f D'~'YiX
x=o
:: 0.332
ftf.
x==o.S y=O,s-x
XeO
=0.332. U /7- P"f;;-
FIGURE P9.34
*
.2//4. Ve,u.~ {
=0.332 7l~Vpp
34
l.Om
dx O,S
7l4.VfjA (2.J[o.~('-)xt - ix~i o
= O. 66'f(O.2l})3/,.
999!,Ml.fU/o3 ~.;) [Vo.S - y. (0.5)3;" ]
or
JJ == 0, 02.96 N
')-36
9,35 9..35 A three-bladed helicopter blade rotates at 200 rpm. If each blade is 12 ft long and 1.5 ft wide, estimate the torque needed to overcome the friction on the blades if they act as flat plates.
Let dill = fOrYIlJe {rom fhe dr49 on area elemenf dll or 2
dlll= (4op +cbbcilom ) Y==z(tPU CDf dll) y where [J = t»y
4/Jd fof'
1.328
!/~
CDr =
laminar floW *
with ReD = UJ
Rei
0
.,lI
Thvs) M = Dy2 d,,; \ or WI'fh
1.329
(llL~ 11
U=Q)Y
el/Y/:::
1.~2e
l~
p tJ)~"3. J ..k~ V ~~ Ys.~ tiy
=(1.326)(2
or
Y1 dy = 1.328 PU~l~ 11 ~ Y dr
~
3 ,h(.200 rc..v )( Imifl)(2.lI'rad)l(~ 5F/)-!'(1 S7Xlo¥1il.)i. ~ J ,38XIO- ~ ~~ 'L\ min 60.s I rev ~ . , .s 1'"'11 5~ 2.
d/YJ = O.OJ25 Y
dy ff·lb
J
where Y fI N
Thus) fhe net f()r9ve for the fhree hlades
!YJ = 3 Sd/J'J == 3 (0.0 /2E
~
).f y-t dy
~
== 3(0.012.5 )(
-f ) (12.)~
or
M=
6'1-.1 1f,IIJ
Nofe: The forf{ve covld be greate,. ,'f Ihe bOfJi'Jdal'y layer is fvrhlJ/enl.
* III fhe and.
lip y ==,~ fl so fhaf TJ ==
Re
JEt
=t ~:op )(2-11 f!tv)(J2 fl)~ r.eV
(J)
Y
2.5/
1)
tnln
- 25/ -s ) (1.£11) 6 ,L' h 1 JL JL ~tip - /,57X/O-'I- f: l. =2.'I-OX/O , 'Alnlc is 9re aTer rnan rne oS
ori/ical valve. Thl/~ the bOlJndIJry is t(Jrbf}/ent al the lip and laminar at the huh. FbrsiIiJP//t:.Ity ass()me a laminar b()undary layer fhr()v,h()v1. 9-37
9. .36
I 9.36 A ceiling fan consists of five blades of 0.80-m length and 0.1 O-m width which rotate at 100 rpm. Estimate the torque needed to overcome the friction on the blades if they act as flat plates.
Lei dllJ = 1orq{)e fromfhe drfJ9 dA of fhe hlade
01}
elemenf
dA
+ _ J.. 2. ) 2 d/IJ - (oV.top 4oHom) Y - (2. pUC/)1 d~ Y where {} =tily and tJJ=/oo..£!).. ( liYlilJ)( 2-'!trod) mm 60S I rev or _
A
or(j.}::: 10, '17
E -J ,,~ U ... ..... dy
IP (
Y
::J.
dF
-flO
.1
Ji =O,lm T
0' 'J
L== 0.8m
The fr/()ximvPJ R~ will ()cc()r at t:inf OJ where y:;: L ();
Re
U R. = wLR
==
:1/ ThlJs ,
1/
= (10.'1-7 CfL)(O.8m) (a.lm) 1.1}6 xI 0-.5 .Di!
1/
= S.7~X/OJI.
s
oS
af a/I poinfs on fhe blade Rex < Rexcr = C = 1.329 = 1. 3 )'sW
fhe f/uw is JlJlfJil1or.
SXIO Clnd
if ui
Df ~ ReJ, I so fhrrl from Elf. CI)
d/YJ =
r
r;2
'HfJ!'W (i dy) y
== 1.32.8
d/Yl::: /.328 f!JJ~ Yvi' y"% dy = /.31.8
or
(1.Z3~) (10.'1-7 o/l~-g
3~
f U 2 Vll t ydy ,hv/ Wit/; Tf::t(Jy ,
s ,'.¥,{x/o- .()( O./IIIt"
/4Jy
~
d/l1 =0,0669 Y dy N'm J where YAJ'" ThtJs the nef fOfYjve O/) the four hlades is 2.
0.8
J
S
M= 5 d#J or
s 0.0669 Y ~oy S y=O
== 5
o.aM
/'1:::; 0,0 Jf38 N'/IJ
9-38
( 2) y k-l 2-
5 (0.06t9) "7
0
9.37
9.37 As shown in Video V9.2 and Fig. P9.37a, a kayak is a relatively streamlined object. As a first approximation in calculating the drag on a kayak, assume that the kayak acts as if it were a smooth flat plate 17 ft long and 2 ft wide. Determine the drag as a function of speed and compare your results with the measured values given in Fig. P9.37b. Comment on reasons why the two sets of values may differ.
fled plate cIf~ fpU'-CD,II where
For
a
JJ ::
/711(2 ff) ::: 3/f ff:;' tlnd IJ1 is a {tinct/of) of
c
Thvs
Re
J::;-
J7 N TJ it I. 2/Xlo5 ft ~
:e Consider I ~ rJ!E
(I)
Ret:::
11J
= /JI-())( /0 6 7J
8ft
(2-)
6
or /,ifOX/O ~Re.J ~/.12xJ07 From Fi9' 9,/5 we see thaf in this ReJ rtJ"ge fhe houndary layer flow i.s in fhe fransitional r0/19c. Thlls from To.h/e 1.3 GDI :; O.lf.5.5/(/09 Rel)2.&8 - /700/Rel By cOlnbt'nin, Ef(s. (0.1 (2); anti(3); J) :::: -! ( I. q'fs.J.~9f )*U 'J.CDf (3'1-f/J.) or J
J
rIJ= 33.0 V" [0.",.5"5/(101 (1.'10I./iU»)'~1700/(/.'fOX/o617)J The reslJll.r from fhi,s erilarion tire ploffed he 10 w, 7l, ff/s rf}J Ib 8~----~----~----~----_.
61------+-----~-----+---+--j
dvalves
rYIf>Q5
~ Ib
I 2 .3 if S
o.oQe6
6
.3.L/-3
7
41---------+------r----~~--~
8
..---.- theory (E . ('I)
2~----+-----~~--_+~~~
• 2
4 Kayak speed U. ftfs
6
8
FIG U REP 9 .37
O,/fIO
o.9oq 1.58 2.'1-2-
'I-.sq ,5,90
(3)
(II)
9.38
I 9.38 A sphere of diameter D and density Ps falls at a steady rate through a liquid of density p and viscosity J.L. If the Reynolds number, Re = pDU / J.L, is less than 1, show that the viscosity can be determined from J.L = gD2(ps - p)/18 U.
"'- diameTer D
For .steady flow or of) + FE == W
J
E F; ::0 where
and
Ii:: bVOytJRt force = f~ 11 = f1( f )77'( ~ t W :: wei,hf = Ps g V ::: flJ (4) Tf(~ IJ == drtJ7 ::: CD f f '% D2) or siIJce rf} = 37l'Dll,P
Thvs} 31l'DfJl-" +
f9(.j-)1!(-B:)3 = fs9(l)7T(~)3
which
be reorf'IJngecJ fo ,ive
f
=
CC/II
g D2.(e.s-p) IBTl
tu
t
Re< I
I
9. 3 q
9.39 Determine the drag on a small circular disk of 0.01 ft diameter moving 0.01 ftls through oil with a specific gravity of 0.87 and a viscosity 10,000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
e~ (o.e7)(J.9~-§.~~)::
o/}= CD ~ pU:J/1 J where
p=IO')tHz,O::: Il1(z.311XIO s
and Thvs
0
ne
J
so
::
UD:: eIli)
P
11
/'688
~~
1:;~)-:::o.:z.3~
0.23/f-w:
is valid. FOr fhe df.sx normal fo fhe flow) 2~: = 2.O.~ -.# = 2~300 so thai frolll £~Itl) 7.2./ XIO /} = 28,300(t )(1. t8B S~} )(O,Olfj)'- f (O.OlfJ)2. ~ I. sax/o-If Ib
r;:::
If Ihe disk i.s PQraJ/e/ fo the flow, ~::: 1~~6 so thai JJ"OI'1IIQ/
9. Jf()
c.
( 13,6)
Dpar41/81 = 7[; = 0.667 CD"or/YIQ/ ( 2iiel/)
a
33.3
7. reduction
J
, 9.40
For small Reynolds number flows the drag coefficient of an object is given by a constant divided by the Reynolds number (see Table 9.4). Thus, as the Reynolds number tends to zero, the drag coefficient becomes infinitely large. Does this mean that for small velocities (hence, small Reynolds numbers) the drag is very large? Explain.
or
For a 9iven ohjecf CD ~ ~e (where the 1/4/118 C depends on the shape of the ()bject) provided Re ~ I, TIllIS" (loS J
Re -
ca.
OJ- CD -
lIowev(J~ ~
..L
C
L
~
= CD 2. pU /I ;: (e!D.) ~ eU II"'" C/ TIJqt iS as lJ--';; (t:e. Re-O)I Ihen 0#- U TAvs dfJes C/) l')Je4A fhaT ~-o f No. 2.
rU
J
J
It;
O.&88,#!fp)(O.Oll)(o,OJf-lJ _ 7."'/ I -If«/ Ih's .~ x 0
=
tho! the low Re d41Q of TaMe ?,1f
~ pqral/el ;;::.
(I)
-Ot)
9-'11
f'r
9. 'l-f 9.41 Compare the rise velocity of an i-in.diameter air bubble in water to the fall velocity of an Hn.-diameter water drop in air. Assume each to behave as a solid sphere.
(a) air bubble in wafer: For sfeady rise
L:r;
D== O./25in.
diQ. r;
0
or
Fe =
W +tb' } where ct=JrQ9 = CD {pif2.f
W= we'jhf ~ ~ir -V Oaifl ~71' ( :=;
~J
Fa == buoyanf
force
=
CJ p
Hence J or
U=
i17' (Jt )3 == CD f
D3
=
fY2.
f
=/
D2. or U
¥ ) where CD -= C/) (Re) and Re:::
0.669
_rr-' yeD
From Fig.
g)3
~o !fJl(T) if f()//o'Us fhal W<'< F8
dN;LO -V
However} since rair.eC.<. ~,.O or
F8 =d7
J):1.
(0115)
or
-Tc H-5tJ~ U Re = I,l/XIO -:t-
9,ll :
I/- D g 3 CD
·=
4 (!2!.f¥)f1 (31, 3 CD
u.v/)
(J)
= 861 U
(2.)
v:
Tria I and error solufion for fI.s.svm~ CD U {r()tn E'{. (OJ Re from Ef. {:J.)j check CD from Elf, (3).1 the 9raph,
Re
IIS.sUII/8
Assume
j
obtai!) (3)
Cb = 1--.-. Y;O.66,1---fI- Re =576 - - ClJ =0.5:1: J CD:fJ,S--U-=O/I/ftl -- Re :: '615 - CD = 0,5 (checks) Thv$J U= O.9¥-6
.g.
(b) wafer drop in air: Since ~il'« (fH:I.() ) F8 <:<: W Thus J W=~ J or 4. 0 !If-(l}? = cD u2.fIJ2.
_/4
orU
D olt.o I S feD
fUso}
Re =
J
"11
(
J.
/I (~R)(6Z,'/-") ]~ /9,1 Jj. 3(2.38)(IO-3~~j} )CD ~
L
'17..
(Jf)
VC;
0 /').5
UD =
te
2.
f!)
1.5 7xlO
~
'flf
= 66.3 U
all'
t
U
(5)
Trial and error solution of £'1s. (If), (&)) and 9raph 13) : Il.s.sume CD:: O•.s ---. U -= 27. f) fj - Re ::: /790 - CD:::; 0, if "1= o,s Assume C/):: O,tf - U::: 30.2 Re =2000- C/)::: OJ,l (checks)
P. -
Th
Nofe; Becavse of fhe 9raph (Fij. Po 2.1) the tiS J answers are nof ()ccvrafe to fhree sI91J1//canl N9l1res •
[J = 30,2 ff .s
9.42. A 38.1-mm-diameter, O.024S-N table tennis ball is released from the bottom of a swimming pool. With what velocity does it rise to the surface? Assume it has reached its terminal velocity.
tU dia, D=-, For sfeady rise 21&=0
tFB
(
or
~W
Fa = W+,/J) where /J -:; drag =CD i fu2.f D2. VI
~
=:
water f'jf
TJ1
weigh! =- O. 02't5 N
:::: bl/oyanf force = oJ{-
}
38./mm
='t if371 (~ )2
or CD y2 = o~ 1/-.55 J where U~ ~
(J)
A/so, Re = f{P or
Re --
U(O.038Im)
J./2X/o- 6J1J.! ..s
Fino//y, from
=
3,1./-0 x/O
{l
'(/ J
where
V
rn
~-:s
(2.)
Fi~. '1.21; Gill "--v--
(3)
Re Trial tlnd error so/ufion : /I.s.sume CD j obiain TJ from Ef. 0)) Re from E'l~(J.) j check C/) from £1' tJ) the 9r4ph., J
A.ssf)m~ CD =0.5 -.-. U = O.Q599 Il.s.sume
CD =OJ; -
ThlJs, V =
U=
I. 06
~
---..
Re
= 3.2JfX/O'"-
Cl)
=O,If '* 0,5
Re::: 3.62.1./ ol.f-.-. CD =0. ~ (checks)
1.06.p
Nofe : 8eco/J.se of the qraph (Fig, 9.2.1) the answer,s are nof accllrate fo three sI9ni{icanf fi9lJres.
9.44 A hot air" balloon roughly spherical in shape has a volume of 70,000 ft3 and a weight of 500 Ib (including passengers, basket, balloon fabric. etc.). If the outside air temperature is 80 of and the temperature within the balloon is 165 of, estimate the rate at which it will rise under steady state conditions if the atmospheric pressure is 14.7 psi.
For .sfeady rise ~ F;
where
or
=0)
Fs ;:: W+ /! volIJme 11
2
o!) ::: drag ::: CD t. f U : D2. ~ :: 6110yanl force == 0' ¥ and.
W= lofal welJhf =.500 Ih + O;n .y
!Vow
D=
\
-13RT
0= p~ ::: and
0; =
.1i:.
In
R"Tin
~
")2-
Jb ) {
(
I'f,7-w. I). In· = 0,002-29 slllfM {!715 s ,,; .• )('16oteo)OR fI.:I (0.00229 .stjf)(32~2~.)::: O. 0736 Jb • 2. Ii == (11f. 7 7ii?-){t2.%) (32.2 .s~) _ L Ib - 0.0036 N.i
¥p
(1715~!;~R)('f.60+/65)'R
/'Iole: Since
Thus wifh 11::: 7xIO'lfl3: ~~f)3 , or D=
51.1
*
i~ lJeal'/}, Me same
ft we obtaIn
JJ = CD iro. 002- 29) U2. (51., )'1. ;:: 2.36 CtJ rJ2. /b) where U~.g. Ills 0) W;:: Soo Ib +(0,0636 PJ..i)( 7~ 000 fi3) ano rr-
fJlso J Re ... or
Re :::
ond
51./
!II-
+ 2.36 Co TJ
ff U '
QJ
::: 9952/1>
Tht/sJ Fa::: W+rf} 9ive.s 2-
2.
or CD TJ ::: Blf.7
.'
(J)
.s
=3.2.SX/O U J,S7X/O ~ , I~ ~ from [;7- 9.21 CD~ -~.fi!"
(:;.)
(3)
Re
£i.
Tr;al 4nd error solufion: IIssv!1J8 CDj ohrQ/t} lJ.s..stJme
CD::: 0 •.5 -+
ill~
o()fsir/e.
~;:::(O.(J736f!P)(7o"OOOff.3) =SIS2Ib 5/52 /b ::: 1'-952 Ih
fhe ba//oon i.s
open at the hoftoll1 J fhe pres.spre within fhe Da//ODn
(/ from OJ) Re from Ef{.t:J.); check C/) from £,.(3)J the qraph. U::: 13.0 #- ~ Re :::1.J..23 x/o 6 ---. Cb ::: o.2.¥;t O~S
lJ.sst/me CD =0.2.'1-' V ::/8.8 ft
-
Assume CD:: D.30~ U::::/6.8§. ~
Re = 6.11
X/06 -
Re ;5.'16 x/06 ~
CD :=0.30:/=
O.2~
~ :: 0.30 (checks)
9. If5
9.45 A 500-N cube of specific gravity SG = 1.8 falls through water at a constant speed U. Determine U if the cube faBs (a) as oriented in Fig. P9.45a, (b) as oriented in Fig. P9.45b.
(a)
•
(b)
teD
FIG U REP 9.45
For steaay fa II) :£ F;;; ma:::O
or
W == oI! +Fa, where W::: wel9hf :::500 N
(I)
Fa::
buoyanl force::: 0 D3 of}::: f f UCD A -:; dr4fj
Qfld
But
W~ 0; D3::;
SG
3
0D ,
3
.3
or 500 N ::: I. fI ('I, eox 10 -j.) D
ThvsJ D == 0.30.5 m Sf) that from £~. {/J .5ooN:: i(qqq~)UCD (O.305m)% +('l80 X/0 3!3) (o,305m)3 or U :z. CD ::: '1-.78 where V,.. l}(a) for case fa) CD::: 0.80 Hence.l
(see
F/". q.2Cf)
U:: ( ~:;~ )1/,. ~ 2.Jfif if-
(b) For case (b) CD:::: /.0£
Hence)
U:::
'1-.78 ( 1.0.5
)"2. ~
WI
2.137"
Q-'f5
9.46 The 5 X 10-6 kg dandelion seed shown in Fig. P9.46 settles through the air with a constant speed of 0.15 m/s. Determine the drag coefficient for this object.
•
FIGURE P9.46
For sfeady faJ/if19 41 tl cQl/stani .fl'eed,
rfj c VI or
/IJ?-::
C/) i p7J2.IJ
ThIJs;
sx/O-6kj (Q.II-fi) ~~ ({)(1.23i1a)(O.lst)1. f(O,OlflJl)2.
9,Jl7
9.41 A 22 in. by 34 in. speed limit sign is supported on a 3-in. wide, 5-ft-Iong pole. Estimate the bending moment in the pole at ground level when a 30-mph wind blows against the sign. (See Video V9.6.) List any assumptions used in your calculations.
For e9va//br/tltfJ I LJ IY/q (YJB
=
2.,.5'
fl a9f'
of
I---l-.....
T
::: 0 or
(s + H)ff
4)
where
4 .: dra1 on the pole and d& :: dray on 1Hf sir/? From Fic/.
9.28
wifli ilD.: 0,/ for the si7rJ
J
(/)
3S1 i!,p~ D
1
CDs::: /, q
From ~7' 9./9 if the po.rf aots as tf srvare rod ....., 1-1 with s hal'p cornel'S cnp == 2.2. Th/J~ wilh 17== 3 0nJl'h ;:: 'f'f tt t:l ::
~
-J-DT!:z.C ~
r
'j)
~
17 tn.
I
,,==...L '0 001-3~ §,/vf;!..) (tf'fii)2(19) (~).. (9Jf)ff2.) -22 III s I'1s 2. \ f{3 ..r' / ~ -¥- . I/O I
,
and
~f =1: W2.C/)P lip =' i(o.oo:m ;'1;-) ('f'ff1. t(2..2.) ( ~ (S)ft) " 6.3'f/l, -rhtJs; from Eq.,(;) ~
MB :::; ~.sff
(b.31f/b) +(s+!J)ff{22,716):::: /62
9-'1-7
ff-Ib
9.48
Determine the moment needed at the base of ZO-m-tall, 0.12-m-diameter flag pole to keep it in place in a 20 m/s wind.
FOr ervi/i Driom I
Jj=c. •
Since
I>
thai
Thus,
i y2.iD P UD Re =-:r::=
fYJ =
l,jJ
(2.0'1- )(0.12 m) -5
/.lf6 X IO
2.
!}
.s
==!.6'fx/0
J
CD == /.2
&0=
/.2
(f )(I.~ ~3)(ZO'f)2.(2()11I)(O.I2.IIJ):;; 70aN
Hence J from £r· (/)
M:;;
(I)
where
~ ( 70 8N)::: ?,080;Vi?} 2.
9,1ft(
I ~}.49 Repeat Problem 9.48 if a 2-m by 2.5-m flag is attached to the top of the pole. See Fig. 9.30 for drag coefficient data for flags.
=4tbi
For eqtJi/ibritJm, /YJ +(1, - ~ )~2. where ~ = 2 0 m) 12.. = 2.5 mJ (Inri D2 ::: 2m. From the so/v-lion +0 Problelh q, Jf8 ) ~I ~::
IJ/so}
/}2.., =CD
t pTJ 12. D2 2.
.:.11 9.30 WllfJ
) where from FIg.
we obfoin ClJ = 0,08.
(I)
(2.)
n12. == T2.5 :;: 2-
Thus}
082 == o.o8(1)(I.1-3~)(20':)~(2IEhJ)(21T1) = 98,'111 By combinifJ9 £'1s . (/), (2.)) and (3) we ohfain /'1:: 7,080 /V.", +(20m - 1m) (98,';' N) = B)qSO N'm
/.2S'
9,SO 9.50 During a flood, a 30-ft-talJ, lS-ft-wide tree is uprooted, swept downstream, and lodged against a bridge pillar as shown in Fig. P9.S0 and Vjdl~O V7.(j. Estimate the force that the tree puts on the bridge pillar. Assume the tree is half submerged and the river is flowing at 8 ft/s. See Fig. 9.30 for drag coefficient data. '1'
8 !tis
.--+-
.. FIGURE P9.50
Force of free on bridge ::: dr(J1 on free c r/J: c" fell'lJ whe re lJ== 8 H&/lS,Sl)me fhe free is .rha;ed as a/} e//ifue anti -t I~ /n fhe 1114/91', 7htJsJ A :;; (dJ (30 It) (/ S fl) :: /771-12. FrDfJ1 Ft9' 9,3(} tis the wind pas I a free ;f)cr~a.re~ I/Jc clrllf ctJe!l;c/~/J1 decre.'IIsBS (fhe leaves "(Did /;P,(;X 1I.r.rv",e //;e.rafJJe Inif)! l1o//BIJ.r III
*
P
).
wafet
411d
(/.re
c;::; 0.1..0,
7h{).JJ h rx9'::: o,,,o(f)(/,9~.!;')(eP/· (!77f1:L)
9-£0
-::: 2Z00//;
9.s/
I
9.51
If for a given vehicle it takes ZO hp to overcome aerodynamic drag while being driven at 65 mph, estimate the horsepower required at "15 mph.
9-SI
9..52.
I
9.52 Two bicycle racers ride 30 km/hr through still air. By what percentage is the power required to overcome aerodynamic drag for the second cyclist reduced if she drafts closely behind the first cyclist rather than riding alongside her? Neglect any forces other than aerodynamic drag. (See Fig. 9.30).
~D
==
power when nof dl'offing
==
Volta :: U CDJ/IJ f py"A = q,ND ipV~
dnd
Po
3
dl'oflilJfj :: CD -ipu A
= power when
From Fi9' 9. 30 Thus,
CD
WI)
== 0,88
-
D
Q/Jd
q,D== 0,50
0.88 -O,5Q. 0,88
= 0,'1.3)",) i.e',) a
1,L3,27, decrease
9.S'f
I 9.54 On a day without any wind, your car consumes x gallons of gasoline when you drive at a constant speed, U, from point A to point B and back to point A. Assume that you repeat the journey, driving at the same speed, on another day when there is a steady wind blowing from B to A. Would you expect your fuel consumption to be less than, equal to, or greater than x gallons for this windy round-trip? Support your answer with appropriate analysis.
1rlp wHh fhe lafrJer pOW81' lOST aV8 f() qerodYI)/Jmic drdf willllJ'e fhe mO.It 94.f. L ef ( ), mean winJu tlnti ( )2. mean //wine/II,
np
No wino!: 4 :: c/) -t pu"'l; fof' ~ofh 11.-. B
0)
41ld
B~II
).
B
Thtl~
-;; ~ p()wer ::: 1J~ :;: 1:p7l}[)/l
(2)
Wind
(7Jw -'=
t
,j~ ::: CIJ f
wilJd speed;
ttl f7lwfll
/)2. == Col) 1. P(ll"'~J2.fi TlJIJ.s
as.sV/IIe ~<:7J):
for /1-+8
B~II
fDt'
~
11 ~ tp(U+1lw) 7Jcj)1I I
~
:: i
fot' I}~B
(J (rJ-7Jwt
7J ~ IJ for 8-,./1 £ ner'lf f).sed ::: ~( where ,,::: filJ/B III 1Q froll/ /I~ B PI' !~Il Thvs E, == 2- (te 7J3c/JIJ)i J
(Jfld
£2. ==
..1.. 2
f (77-171) 2. UC/) Il t
.L
-I- 1.
p( TJ -Uw)2. UCb/l rI ::
t. e.
mf)/,fi fvel needeJ whelJ Windy
q-53
-
I
/ + (71w /7J)
:z. <: I
9..5.5 9.55 A 2S-ton (SO,OOO-lb) truck coasts down a steep 7% mountain grade without brakes, as shown in Fig. P9.SS. The truck's ultimate steady-state speed, V, is determined by a balance between weight, rolling resistance, and aerodynamic drag. Assume that the rolling resistance for a truck on concrete is 1.2 % of the weight and the drag coefficient is 0.96 for a truck without an air deflector, but 0.70 if it has an air deflector (see Fig. P9.S6 and Video VIJ.S). Determine V for these two situations.
v --TrUCk width
•
For cOlJs/ant .rp~eJJ
&
FIGURE P9.55
Fx : mtlx ::: 0
!)
Of'
Wsi»e::: a9 -/- F where (J == Qrc fan (i/o) :::
'I.
~00 fA (sin 'f.oode,) =t
or 3 'f9a 16
(aJ
1.
= O,/~3 UcJ)
(0.001-38
J
c/j:::
O,7() J
:./;~ ) lJ'-C/) (/2 N1%fl) +0. 0/2 (5'~ 000 /b)
+600 Ii
1 r C/) ::. 0.9 t fhen II =/If£!-
(b) If
) tiJ= t f 712. c/) /I
J/.. 00 tie,
Thvs £0,
then U:::
= 10 ft
/70 ji
=:-
qi//meA
= /lbPlpn
9. .56 9.56 As shown in Video V9.S and Fig. P9.56, the aerodynamic drag on a truck can be reduced by the use of appropriate air deflectors. A reduction in drag coefficient from CD = 0.96 to CD = 0.70 corresponds to a reduction of how many horsepoer needed at a highway speed of 65 mph?
(a) CD
=0.70
(b) CD
•
-p ::: poWer
==
gO'
FIGURE P9.56
where
2-
rIJ == f p U CD A Th us) tJ. P ;;: reJl1cliolJ if) power ==1b-~ ::: t p7J 3 A[CDb
W;I h 71::: 65' mph :::
-
C~a]
95 . 3 .ft;sJ
b'P ;::. i (O,00238!.}:J-!.) (9S.3.fi)3(JOff)(Jlf/) [0.96 -0.70] ;::. 32;; 100
¥ ( £i-ohb) : : .58. if hp oS'
9-S5
=0.96
9.57 9.57 The structure shown in Fig. P9.57 consists of three; cylindrical support posts to which an elliptical flat-plate sign is attached. Estimate the drag on the structure when a 50 mph wind blows against it.
where If we a..r.r(/llJe the s;911 is (Jf) el/~eJ
/l:: f
9.0 0 1-12.:: 0, aft (;0511) ::: /2.,0 Ifl. /lAd
IJ'J. ~
IIJ
(;off) (5f1) ~ 39.3 fl2.
At.;:
0.6 ff (15ft):::
l(f(lsfI)
== IS; off)"
From Fif. fl, z? J f()r tI Ihin disc ~/::: /,/ For -fhe cylindricaJ po.ri obl4/n Cb fr()m Fif - 9.'J-/ Cis ~ (7J :::S()IHIA::: 7.3.3.f!.)
Re ~ ~ :: 2.
l'
13.3
~!OI6~) 1p.
== 2.8X/().S' - - + C
/)~
/,S7X/() i'
::: 0.6
S inl/lar/YJ
Re3 ::;: .3 7x/O.5s~ C.o.J ~ 0. s 1
{(elf =:Jf.7X/o -
Cb~
ThvsJ from £f. (/):
rff =i-(O.O()2.33 ~:f.t ::=
=: ().'J-S
)
(73.3P.tf!./ (39.3 ft) +O. Irp.ol-!j +O.S(!1. ff2)+o.2-S (lsf:,l2)]
3 7~ /iJ
9-56
9.SY As shown in Video V9.5 and Fig. P9.S9, a vertical wind tunnel can be used for skydiving practice. Estimate the vertical wind speed needed if a ISO-lb person is to be able to "float" motionless when the person (a) curls up as in a crouching position or (b) lies flat. See Fig. 9.30 for appropriate drag coefficient data.
For eC((J///hritJm cond/Ii of},s
W=rIJ = CD t f TJ2.A fls.sume
• FIG U REP 9.59
w tv
W::: /60 Ih and C/)/}:: q fl:/. (see Fi'9. 9.30)
ThlJ,s} /601b == (f)(O.00238S1pS) U'l-(9f1'-) where U..... or ff)( S28of+ Iml )(36 0 0..$) U-( - 122 5 I hn :::83.2",1'h
fj
Nofe : If fhe skydiver /~flr/erJ lip il1fo a haIr, fhen Call?:. (see
Fi9 .9.30) and
(J=
/s8mph
9-S7
2.,slf
9. 6 o'It
u
9.60*
The helium-filled balloon shown in Fig. P9.60 is to be used as a wind speed indicator. The specific weight of the helium is y = 0.01lIb/ft3 , the weight of the balloon material is 0.20 lb, and the weight of the anchoring cable is negligible. Plot a graph of f) as a function of U for 1 sUs 50 mph. Would this be an effective device over the range of U indicated? Explain.
FIGURE P9.60
For- fhe balloon fo remain stationary 2: Fx ~ 0 and 2: Fy = 0 rb Thlls) IJ = T cos e or T == c()..s e
~
W + T silJ e +k1Je which comhine 10 qiV8
I4nrJ.
Fa .: : W t
c!J
~e
==
. I/J. = IenSlon
CQ
ble
ftJ178 + ~e
Sui W.= 0,2 /6 ) and
pJP
==
(I)
F8 =eq"V =(7. 6£ x/ti2.~) ¥(-f H)3:::: o.32o'llb
ONe V = (0.0 II ~) 'i:{l (1: H)3:: 0.01,16/ /6
ThLJs) £f/I (I) becomes o.320'l1i; = 0,2 /b
or of! fan
+,j} f4/JB + O.oi.jt/ II:,
e :: O. 07l(-3 /6
(2)
(I/-)
100 110 120 130 1'"'0 150 160 170
cis open "prn" for output as #1 print#l, "***************************************************" print#l, "** This program calculates the angle of the **" print#l, "** cable as a function of the veloci t.y of the **" print#l, "** air. Values of the drag coefficient are **" print#l, "** obtained from Fig. 9.23 as a function of **" print#l, "** Reynolds number. **'1
(con'!:.) Q-S8
(con 'I;)
9.60' 180 190 200 210 220 230 240 250 260 270 280 290 300 310
printfl:l, "***************************************************" printfl:l, " II pi = 4*atn(1) printfl:l," U, mph CD theta, deg" for i = 1 to 10 print " " input "For a velocity of (in mph) U=",U Re = 1.27E4*(88*U/60) print using "the Reynolds number is Re=#.###~~~'-II;Re input "Enter the drag coefficient: CD=".CD theta = (atn(19.9/(CD*(88*U/60)~2)))*180/pi print using "For U = fl:fl:#.#fl: mph theta = ##.###fl: deg";U,theta print#l, using II ####.# ##.### #fl:#.##fl:";U,CD,theta next i , *************************************************** ** This program calculates the angle of the ** ** cable as a function of the velocity of the ** ** air. Values of the drag coefficient are ** ** obtained from Fig. 9.23 as a function of ** ** Reynolds number. ** ***************************************************
I
theta, deg CD U. mph 87.524 0.400 1.0 79.707 2.0 0.420 0.540 34.421 5.0 10.0 0.550 9.548 7.102 15.0 0.330 20.0 0.100 13.022 25.0 0.080 10.482 6.516 30.0 0.090 2.759 40.0 0.120 50.0 0.160 1.325 I --------------------------------------------------,
i
e vs U
I i!
90
1\
80
\
70
1
60+--\~--~~------+-------~------~------~
~ 50+--4-\----~------+_------~------~------~ ! 40+---\~--~------_+------_4--------~----~ 30+---~\--~-------+------~--------~----~ 20~-~\~--~----~--~--~
\
./
""--
10+-------~----~~--~~_4--------~----~
~
./
"""'-I---
O~--~----~--~~==~==~ o
10
30
20
40
U,mph
50
lNofe :oBe:cause of the sudden chon98 in C/) when Ihe
boundar! layer DecrJfII9S furhu/slll (at ahout /S mph), the e vs U eUl've i.s hi9hly non - linear. 1/J loe" for So/l# valves of e there is more fhfJA one po-ss/bie va/fie 01 Y. If would /Jot wo"k well rJ,f a INiAd speeJ indica/of'
q-sq
,n Ih;s f'4IJfR.
9.61 A 2-in.-diameter cork sphere (specific weight = 13 Ib/ft 3 ) is attached to the bottom of a river with a thin cable, as is illustrated in Fig. P9.61. If the sphere has a drag coefficient of 0.5, determine the river velocity. Both the drag on the cable and its weight are negligible.
9.61
60°
r: Fx =
or
0
t.#-::: TC(Js6o'
(I)
and
ZFy::: 0 or Fa - W= Tslnl o'
(1.)
T
Since
'60·
11 ::Vo/Vl1Js fJf cork== -!llr3 = /71'( Ji fl)3:;
2.1-2 t/o-.1 f+
4.10 V - ¥:fJl'lt. V :: 7 si/l bil (62../f -13)#" (j..'l-2 x/oTlJvs~
T:::
0./3
,-
J
J
3
if f()llows frf)/II Ff. (2.) thai
or
JfI'
t:: . , -
si",rI
aIh
From £'1' (I) I ~::: CD feu%1J : ; 7 c()Sto() where
Thvs
1/
J
D ~
U=[ 2. T (;(J.s6~ ] =[ CD
e7r r
fJ
:2 (O.l3sIIJ)(;t7S6o ] 0.5 (1,'I1f ~:¥) 1f( ;tft)'I.
Q-60
k 2_
/I'lI!1Tr
ft
- 2 .S.5-:s
2
9.12. 9.62 Two smooth spheres are attached to a thin rod that is free to rotate in the horizontal plane about point 0 as shown in Fig. P9.62. The rod is held stationary until the air speed reaches 50 ftjs. Which direction will the rod rotate (clockwise or counterclockwise) when the holding force is released? Explain your answer.
Lef 4 and
.. /O.7-ft diameter
4
del/ote fhe dra9 forces on fhe fell and ri9h f ~pher8.s respeclively. If 4 >~ fhe rod will To/ale covllier CIDG~t;ise. _ I 7Js
:r~
J
4. - £;J. f
UD.
#OIi0
O.0023S
e.. ;:= 1..#
Re:::
and. De
!Tj
=::
D UDR
l-
R
CJ)L
D8p. .....tJ
__
Fif-
11
C/)L -¥-
4 >4
fii
"1 ~
,.;!
B.21 Or 8.25 J
2.
v/..
1$
0.002-38 -!fp(SO~) (O.7f./)
alld ~R::: 0.5
CDR Lf DR == .1l 1\2
Since
_
3.1'1-1./0
=0.06
(17/..
Sf)
thai (J)
sllJ ~(50#)(I,Sf/)
3.7'1'/./0
fJ'
Thv~ frrJlII
:z
wlJere CL) =:: CJ) (Re J.
IJA. =:: CbJ. IIJ. CDt(
;(
e
~ and a&J? ~ CJ)R t 71 IIR
2.
~
1u
5
=:/1.771./0
.5 ::; 2,23X 10
for smooth spheres
$0
thai
Eq.f/) 'lives
2.
:
0.5 (O.7{~) O.06(J.5f1)1 == /.81
Ihe r()d will rotafe counter clockwise.
/lole; II11hOUfh Ihe rip,1 sphere /s S llJaller fhan Ihe Ielf, If has more dra'j hSCfJlI..re il n4S a It/lYe rJr41 coefficielJf (/al1li/JfJr boundary layer, wiJe waKe). The /4r,9 sphere has a smaller drtJ1 coefficient (lurbulenl bounJ4ry layer J narl'pw waf.e).
9-61
Q.63
I 9.63
A radio antenna on a car consists of a circular cylinder t in. in diameter and 4 ft long. Determine the bending moment at the base of the antenna if the car is driven 55 mph through still air.
FOr e9ut//nrivm J g /YJ():: 0 J or fYIo = lrlJ
,dia- D V=5 S mph - - ff
==80.7 -
ltd} 1
2.=2ff
oS
where of) = CD fpV2.f1 (J) . Re -- Jll. - (80,7#)-I/o VB - ,I 07 X/0'1) 5 /l7ce .y ft'-ft -
c:;,slRx
:sit follows from Fi9. q. 2/ fhol C~ ~ I. 3 lienee 01} = /..3 (t )(0.002.38 =!.~~5 ) (BO. 71j)'J. (if (1) ( ~ fI)
/Yl()
/,57 x 10
Thus) Mo = (2 fI) (0.8'10£) = /.68 Ff •Jb
9. 65
Estimate the wind force on your hand when you hold it out of your car window while driving 55 mph. Repeat your calculations if you were to hold your hand out of the window of an airplane flying 550 mph.
b= CDterf2./I
J
where
U=(ssmph)(
Ii
~~m;h)
Ass/)f/)c your hand 1'.5 'fin_ by bill, i/) si~e a t~in disc wifh CD ~ 1.1 (see Pi-g.9.zq).
=
80.7!t
dnd acl.s like
Thvs) rIJ=(J,J)(~)(O,00238)(eo.7fj)').(11f1)(&N) == /, 1f2 Ih If your hand is norma} 10 the fhe lif! force is zero.
For U =550 mph == e07 Y (i,e') a /0 fold increase in rJ) fhe 2 draq will increase by (J factor of 100 (i.e~ 08- U )J or JJ= /'f2/b
Note: We have assumed fh4t CL) is not a funcl/on of v. That /sJ if is /Jof a fvn&lion of either Re:::: ~ or /Y/fJ -::
*" .
9-62
?, 66 It I
9.66* Let 0>0 be the power required to fly a particular airplane at 500 mph at sea-level conditions. Plot a graph of the ratio 0>/0>0' where 0> is the power required at a speed of U, for 500 mph =:; U =:; 3000 mph at altitudes of sea level, 10,000 ft, 20,000 ft, and 30,000 ft. Assume that the drag - coefficient for the aircraft behaves similarly to that of the sharp-pointed ogive indicated in Fig.
9.2f.
3
-p == Urb ::; CD i
pV fJ
'P
t eV3C
D
ro
2. 00
Do
fhaf -::;:}:;::; ..L eJ1!C A -
.sO
(88 ft )
Nodw1 Yo =(s 00 mph) (6()m~h) = 733~ an /Ylo = where C k RTi so fhat Co -::
¥-
or
=1 Ii.
J&. = 1//7 7 33 ~ 11
M = 1'lfJo Co
Hence J from Fi9.
.
eno = 0,2
9,ZtfJ
so
fils 0,
or
Ma =
3
e (-JL) PoC Va
fhal £". (f)
(I)
D Do
/1. If (17/6/it '~~R ) ('1-60+ 5'1) oR
i
9
hecOf!J8.5
(- U 733
J
where Q -
C
-p _ -6 C V.3 h if )3 or 70 - 5,33 x/o r D were ffS. U,., Ii and CD =Cb(fYla) from h~. 9.2'1- CDI Fe~ s;:r(3J
e CD
1> _
% - (0.002 38 sllJ~)(O.2) M-
-
"-/ = / / /71f
== 0 656
.s
A
~
U Vf.JI. (1716~J~~t/l) (Jf6() t T) 'R' U
where TJN fj and T'" 'F
I
,; 2 Lf 00 ('1-60 +T)
J
Thos J for fhe givel) a/fi1ude obfain Tand p in Table C.I. Se lee f bOO mph ~ U~ 3,()()O mph (i.e~ 73.3!f ~ U:: 'I-~oo 1j )J determine /!Ja from £'1,,('1) J C[) from Ef(J3J (the graph)) and ~ frblll Eq. (~J. The reslI/fs are ploHeri be/olll. 0)
WI
IN /~OO() ffJ p::: 1.76 x/o- 3 Thus, i:,::; 9,98 x/o9 Cb V3 (J
J
Tt
T=
:2 3. Lj.
"F
Ma= 10eo 3 b) 11-1 2~oooflJ P=/.27 x/o- S/7!- 7= -/2.3 'F J
Thvs
:e
-9
~ :: 6.77 X /0
Nf _ I'IQ -
Cb
U
{:l.)
3
U IOIf()
9-63
(If)
9.66 4
I
(conll)
and
3~OOO ff Thus ~:::
c) IN
J
M-
I'-I(J.-
Proqratn P91166 WQS f4keIJ fro", 100 110 120 130 ll,o.O
150 160 170 200 210 220 230 240 250 260 270 280 300 310 320 330 3l,o.O 350 360 370 380 390 ~oo
410 ~20
430 44·0 450 460 465 470 l,o.80 l,o.90
e~ 8.9/
J
x/o-If.
sf:¥-
J
T =- tf7.8 fir
'I,7S X/O- 9CD y3
U
q'ls
s/70Wh
belolV Wqs flsed for the CQ/cfJ/4Iiol)s. Cb vs (;14 rit/fa
Fi9J ? 2'1-.
cIs open "prn" for output as #1 dim M(17), C(17), A(3), AA(3) print#l, "*************************************************" print.#l, "** This program calculates the power rat.io **" print#l, "** at different altitudes of flight for **" print#l, "** flight speeds from 500 to 3000 mph. **" print#l, "*************************************************" C(1)=0.19 : C(2)=0.23 : C(3)=0.l,o.2 : C(l,o.)=0.55 : C(5)=0.52 C(6)=0.49 : C(7)=0.46 ! C(8)=0.43 : C(9)=0.40 : C(10)=0.38 C(11)=0.37 : C(12)=0.36 : C(13)=0.35 : C(ll,o.)=0.35 C(15)=0.34 : C(16)=0.34 : C(17)=0.34 A(1)=1080 : A(2)=10l,o.O : A(3)=995 AA(1)=9.38E-9 : AA(2)=6.77E-9 : AA(3)=4.75E-9 for i = 1 to 17 M(i) = 0.25*(i+l) next i for i = 1 to 3 z = 10000*i print#l, " " print#l, using "For an altitude of z = ######. ftl!;z print.#l. " U, mph Ma CD P/Po" U = 0 for j = 1 to 6 B = 0 U = U + 500/60*88 Ma = U/A(i) for k = 1 to 17 if B = 1 then goto 450 if Ma > M(k) then goto l,o.50 CD = C(k-1) + (C(k) -C(k-l) )*(Ma - M(k-l))/(M(k) - M(k-l)) B
=
1
next k ratio = AA(i)*CD*U 3 UM = U*60/88 print#l, using" #####.# next. j next i ft
#.####
#.####
(con II;) 9-6'1
#.###ft"ftft!l;UM,Ma,CD,ratio
?# 6 6 4~
(con'-lJ ~~~*********************************************~
**
**
This program calculates the power ratio at different altitudes of flight for flight speeds from 500 to 3000 mph.
**
**
** ** ************************************************* For an altitude of z = 10000 ft U, mph Ma CD PIPo 500.0 0.6790 0.2186 8.088E-Ol 1000.0 1.3580 0.5370 1.589E+Ol 1500.0 2.0370 0./,r556 /,r.550E+Ol 2000.0 2.7160 0.3827 9.061E+Ol 2500.0 3.3951 0.35/,r2 1. 638E+02 3000.0 /,r.07/,rl 0.3/,r00 2.717E+02 For an altitude of z = 20000 ft U, mph Ma CD PIPo 500.0 0.7051 0.2228 5.9/,r9E-Ol 1000.0 1./,rl03 0.5308 1. 13/,rE+Ol 1500.0 2.115/,r 0./,r/,r62 3.216E+Ol 2000.0 2.8205 0.3772 6./,r/,r5E+Ol 2500.0 3.5256 0.3500 1.168E+02 3000.0 /,r.2308 0.3/,r00 1.961E+02 For an altitude of z = 30000 ft U, mph Ma PIPo CD 500.0 0.7370 0.2279 /,r.270E-Ol 1000.0 1./,r7/,r0 0.5231 7.839E+00 1500.0 2.2111 0./,r3/,r7 2.198E+Ol 2000.0 2.9/,r81 0.3721 /,r./,r61E+Ol 2500.0 3.6851 0.3500 8.196E+Ol 3000.0 /,r./,r221 0.3/,r00 1.376E+02
100
- - z =10,000 ft
o
a.. it
- - - z = 20,000 ft - - - - - z =30,000 ft
f------
100
---~,--
.
10000
1000
U, mph
9-65
9,~7
J
1).67 A 0.50-m-diameter meteor streaks through the earth's atmosphere with a speed of 1800 mls at an altitude of 20,000 m where the air density is 9 X 10- 2 kg/m 3 and the speed of sound is 300 m/s. The specific gravity of the meteor is 7.65. Use the data in Fig. 9.24 to determine the rate at which the meteor is decelerating.
z:: r::: ma
or
tb= mq
TIJvs,
1: pref) f 1/- == fm .; 1/(.g)3 fA where If. (/) 3 [ l!. 1 'I- (0.5111)3 m ~ Pm -:3 11 ~) ::: 7,65 (1 000 /113) J r 71 --r == IIlso U /fIa ::: C" = j
9-66
Phi ~ SG PN~o I.
£0/ ~9
9.68 9.68 A 30-ft-tall tower is constructed of equal I-ft segments as is indicated in Fig. P9.68. Each of the four sides is similar. Estimate the drag on the tower when a 75-mph wind blows against it.
flss llme no inferference be/ween
4~===~~ .... ....
fu
the
fronf and back portions of the lower. A/so, neq/Bd fhe cJrfHI on the sider of
2 in.
FIGURE P9.68
(I)
L t
t
Thus) from
£r.. (t)
,j} = 30(l. (O.OO2.3FJ s~~s )(110 ~l)[(J.q8)(2.)(I ff)(-kH)+(18)(kH)( t
et~~8V2 rt) (1.82.) (z){/f/)(k H) +(1. rO(J. f+)( 8 t8/~812 r+)J
or ,j} = BSq Ib
9-67
9.69 9.()9 A 2-in.-diameter sphere weighing 0.141b is suspended by the jet of air shown in Fig. P9.69 and Vidl:o V3.1. The drag coefficient for the sphere is 0.5. Determine the reading on the pressure gage if friction and gravity effects can be neglected for the flow between the pressure gage and the nozzle exit.
Area = 0.3 ft2
t Air Area = 0.6 ft2
9-68
Q.70 9.70 The United Nations Building in New York is approximately 87.5-m wide and I 54-m tall. (a) Determine the drag on this building if the drag coefficient is 1.3 and the wind speed is a uniform 20 m/s. (b) Repeat your calculations if the velocity profile against the building is a typical profile for an urban area (see Problem 9.17) and the wind speed halfway up the building is 20 m/s.
(C/)
of) == CD f f v'lJ ==
/.3 (i
T h
)(1.23 ~)(2of t(15'1m)(87.SIfJ)
or
JJ ==
/1,3/ x /0
6
N=
'1-.3/ MN
y
1
h==l.5ifm
(b)
For ~n vrDQ/J areo, U;::. C yo.1.}
Thv.s J wdh u =2.0 lJ- af y -:: -b: : 77fJJ we obfain C
= :,.11-
== 3.52.
J
or
f).
== 3.52
The 10141 drtl9 is
f
yO.4
wilh U 1} J Y'" In IV
y= /s'!-
cb = deD =,fCD ip u. dlJ = -freD {(3.s2. 2
yO.If/- (87.5) dy
y::o
or
15~
.() =f(J.23) (1.3) (3. Sz/(87. S)
S
Thvs)
0
/,8dy = 967C~
rf} == 'f. /7 MN
9-61
) (f5'1r=
'/;/7
~/06 N
9.72 When the 0.9-lb box kite shown in Fig. P9.72 is flown in a 20 ft/s wind, the tension in the string, which is at a 30° angle relative to the ground. is 3.0 lb. (a) Determine the lift and drag coefficients for the kite based on the frontal area of 6.0 ft2. (b) If the wind speed increased to 30 ft/s. would the kite rise or fall? That is, would the 30° angle shown in the figure increase or decrease? Assume the lift and drag coefficients remain the same. Support your answer with appropriate calculations.
u = 20 ftls
•
/r':31b
•
(a)
z:: Fx
::: If) Ox =: 0
0'"
t
FIGURE P9.72
~::: Tcos30fJ :::(316JC()s30o :::: 2.IOI/;
Thfl.fJ
e/) ==
tiJ t('U2-A
-r=3Ih W= o.911J
of' e/):::
O,i)/(J
/llso;
zry =mtly=:O !hils;
t
C.l. ==ff TJ'''I1 =~
or 2Jfo/6
or
G. :: 0.8'1-0
0) (2.)
9-70
9.73
I
9.73 A regulation football is 6.78 in. in diameter and weighs 0.91 lb. If its drag coefficient is CD = 0.2, detennine its deceleration if it has a speed of 20 ftls at the top of its trajectory.
0.0238 Ii> 0,0283 S/fJr;s
::: O.Sil-1
ff
~ oS
9-71
fl. 7'19.74 Explain how the drag on a given smokestack could be the same in a 2 mph wind as in a 4 mph wind. Assume the values of p and J..L are the same for each case.
/)= CD
LeI (
i
z
f U'J./J :::: CD £pU 1l'Dt
)1 denofe conJilions
with lJ= Jl}-
with U= 2. !p7/;lIs with p, =f:l 1 fo hove /J;::: ciA We hove ' u,2 rrDLA=V[)z2f2. r J U1.2. lTDLIJ or CDJl1 ~:::CD:;. Uz2CD/2ft and ( )1. J
I
Thai i.sJ CD1 :::: LfC/)2
where CD, and C/)2- ore fvncfions of Re == ~D tJs shown in Fi9vre 9,21 (a), and ~ = Uz if foJ/oU/s fhat
Since 11 =uRei = 0,5 Re 2.
±
From F;~, 9,1-1 tq) we cqn deTermine CD a value of Re2 such fhat Re, -: : 0. S Rez and CD1 =- Lf CD:;. j hence the drd9s
are e'ltJo/ eve"fhou9h the velocHies are /JIJerlla J. This occurs beCOIJ.J'8 of fhe s()dden drop in ClJ as the bo()ndafY Jqyer Decomes tlJf1hu/(}n1.
q-72
J.
CD2,.== 0. 25
• ____ 1__ _
9.77 Radius
= 0.845
in.
9.77
A strong wind can blow a golf ball off the tee by pivoting it about point I as shown in Fig. P9.77. Determine the wind speed necessary to do this.
- - Weight = 0.0992 Ib
When the ball is ahout 10 be blown from Ihe fee fhe free bOdy dia9rtlm is 4S showlI. flence.; hy .full/min, IfJD/lJ8111s ObO()1 (I): [J ~ ~ 0 or Wi == tJ r vs) (0. O'!f/2 /6) ( o. 2. oin.) = IJ (0.821 i".)
Tn
J
or
08= o. 0292/iJ
I
where
/Ie ~ fflf21Tr2
-ri. IfIUS)
D.B2/ in. .
Il
O. O'}.'I-~/O;:
CD 1...1. £(}fS) 2. (0.00138 If'
or
C/) U2. =: 130S where 1
V%r1°.89SIIJ) In i{!;.
2-
n' x,:::O.lOln.
~T1
u- P.
(I)
For a sphere4- C/) ==~(Re) (see Re == ~ p -
r=O.8IfsilJ.
[;'9- 9./..5) where
(2.)
(O.00233S'''9S/FlJJU(2.(().8~5)Il2.r.l) 3.1f7X/0 7 00',slfl3.)
01'
Re ~ 9tt rJ.; where V~!!-
(.1)
Trial and erN!' s()/fliion: IlSSf)hJ e C/) =: O. If oS 0 Intll frf)/IJ £1- (/)J lJ::: S l fro/l} £o.{,1) Re::: 9bl (~7./) 1 J
-::!
/!)
and
s. S 2 X/O ¥. TAvs fro/J? Try 41(J//;. J
Fl9' 9.],5, Cj; -;: O. 2.5 :f (), ~Il !Js.su/lJe Cb =: 0.1.2 so fh4t V::: 77,() #- tllJti Re:: 7,1f'rX/O~ T/JlIs.; Fro/IJ r;,. 9.~5J Cb -;:: 0.22 ChecKs. lienee;
U;:;
77,tJ
#-
9.78
I
9.78
An airplane tows a banner that is b = 0.8 m tall and e= 25 m long at a speed of 150 km/hr. If the drag coefficient based on the area bEis Co = 0.06, estimate the power required to tow the banner. Compare the drag force on the banner with that on a rigid nat plate of the same size. Which has the larger drag force and why'!
P = rbV where oft= CD ±fy'J/I w/I/; f}::: bi. Thus wilh CD=O.06tJnd lJ:::~.50~)( 3}0'::.5)( 'f~O: ):::~1.7.p J
J
fhis 9ive.s -p = (0,06) (± )(1, 2.3 ~)('fl. 7.!fl
(O.Bm)(-;.Sm) ;: 5
3, S
xl 0.3
W :: S 3,5 k W
For a r'9id fla-l plate P = rlJ7J = 2 CD 1fV~bi
(fhe f4c1or of fwo is needed hectlf)Se
the
drQ9 coefficjenf is bClSed QIJ file dra1 QIl ()ne
iI,'fh Re/J ~
Wj
side ()f V},
-u':=
.e"
FI9J 9./s a
va/tie
('l-1.7,f-)(25hJ) UI'
-$
k-rD X/O
of
m£ oS
Ihe pIa If) )
= 7, /~X/O 1
CD:::: o.O();).s
IJ·.f'
We ()oraln
rrom
for a smooTh pia/e.
Th()s) 'P ==
2- (0.002.5)( J..) (1.231;) (If!. 7-lJ!-i( o. 8m )(2-.5",) = ¥-.lft x IO.3W :: /f. f~ AW
For +he flal p/4le cQ.!'e the d1'4! is re/alillefJ sll/al/ becfJlJSe /1 IS doe elJf"f'e~ 10 sheqr(viscotJ,s) forces. /)V{) 10 fhe I'/flvller/1J9 of 1);61 banner, (/ 900d porl/o!} of /4 JrtJ1 (4!}ril;elJce power) is (J reS'()/1 01 preS-SlIf'8 forces . II /'.t /lui (/s slreomh"ned tiS a 1'I"tiel /hI pIale. h
9-7'1-
9.7Q
I 9.19
By appropriate streamlining the drag coefficient for an airplane is reduced by 12% while the frontal area remains the same. For the same power output, by what percentage is the flight speed increased?
p :;: g TJ
J
W
I;ere rfJ ::: CD
Lei (
)0 denole the
Thvs J
w/lh
eDt)
r;:: 11
feo Tl;;2.flo Yo
3.3 Tfo.C ao == u.s c[)~
...
==
f f 7Jit
ori,inlJ I conlijllrtlfion t:fnJ (
CDs i
)05
the slream/ied one.
we ohfa/n
f!s (!./I1.s Us or w/-/IJ t90 ~lJs ) po -:: f~
ThllsJ
[c.]~ T:::: cl>P. =[ Vo Ds C
tl
Do
C.
iJo
-O,/2CD(')
J3=I, k
0'135
h speed Increqse Note: 'P~ U3Cl> .so/hal JfJ==3V2.GbJU.fU3JCj). Thvs w/IJJ JfJ=O this give.s t5U = - f Jet) =- -0;2 = -+0.01/- == 4-% t:e'J a '1-.35
J
TJ
J
9.80
CD
I 9.BO The dirigible Akron had a length of 239 m and a maximum diameter of 40.2 m. Estimate the power required at its maximum speed of 135 km/hr if the drag coefficient based on frontal area is 0.060.
P:;alJU J where ob'=Cf)teU'll Thus) with [J:::(/3.5~)( 31 oh;s)( ',Okhlm) = 37.5.lfP = CD -tpU 3 f}
=(0.o60)(i)(J.2.3 !;;-3) (37.S!p)3Cf) (lfo.2tn/- == 2.'1- 7 X/06 or
p
3
= 2.1- 7 x 10
AW (/.3/f.~)= 3310 hp
9-7S
k~:2.
9.8/ I 9.81
Estimate the power needed to overcome the aerodynamic drag of a person who runs at a rate of 100 yds in lOs in still air. Repeat the calculations if the race is run into a 20-mph headwind; a 20-mph tailwind. Explain.
9-76
Q.83
I A fishnet consists of 0.10-in.-diameter 9•.83 strings tied into squares 4 in. per side. Estimate the force needed to tow a 15 ft by 30 ft section of this net through seawater at 5 ft/s.
o.lo-in.-dia. cylinder ifin.
/
Jfin·/.V ~/
/ /'.
I U~ecfjon 1/ / / / The net cQn be freafed as one lon9 .... 1-~--.!'-/~--t---t---1~ 0,/0 -/n. -diameTer .circular cylinder wlfh
ell = C; f [JUl./} J where U::: 5 #-
. Each
I fl' .sectirJn
Qr Ihe net
conTains 6 feel {}f sfri171 (donoi C()IJnt fhe edges fwice). ThPJ.; f/;e fola/ sfri/Jy lell,lh ;0$ 0fJlioximalely 1 =(6 ff2o) (/5 f-l)(30f/) = 2700 Ii f)lso, since p= /,99 s~¥ and 1/:::: /.26 x/o-.s ~2. (see Table 1,5)
Re:::: 0:::
Vb
rs
q,;. fI) 1,26 x10-.5 1J! 1J)(
= 33/0
ThlJs}
/J==(I,I)(1)(1.99%'1S)(sfj/{~ft)(2700ft):::: 6/61b
9-77
u 9.84
An iceberg floats with approximately, of its volume in the air as is shown in Fig. P9.84. IT the-wInd velocity is U and the water is stationary, estimate the speed at which the wind forces the iceberg through the water.
-
•
7 volume in air
11
---.~--------~-FIGURE P9.84
Lei ( ~ denofe fhe portion of the /cebel"fJ in fhe air dnd ( )w Ihal porlio/) in Me w4ler. Thv.s} ~::: and Vw::= f V J where 11:: vo Itlme 0 f fhe ice ber9
t-v
For sfeady mofion)
4::: 08w
J
where and
wlln
4 -= ~q i ~(U - rh')2.Aa A _
GVw-
C
.1. 0
11 Z Ll
'Dw2. \W Vb
nw
i1 :: speed ot fhe iceber9 (I)
9-78
9.85
I 9$5 A Piper Cub airplane has a gross weight of 1750 Ib, a cruising speed of 115 mph, and a wing area of 179 ft2. Determine the lift coefficient of this airplane for these conditions.
'1.86 ) A light aircraft with a wing area of200 ft 2 and a weight of 2000 Ib has a lift coefficient of 0040 and a drag coefficient of 0.05. Determine the power required to maintain level flight.
9.86
t~ W=2 0 00/b == ~ ifU~1l
For eqlJi/;hrium or
2000 / b
.sIll
2-
= (0.'100) f. (0.00 23 8 -;/f) U (zoo ff')
Hence)
TJ = 1'IS !i Also; p::: power =: It Tl) where rIJ= e/) i:pV'-;; =(O,OS)i(O,00238E~¥)(/'1$fil·(2.oofl2)~ 250/1; /Vofe; Thi.s valve of rb C()vld he ohlailJed frfJm ~ == ~ := f!:. :: (),1f() :: 8 fJr d9 ~ ~ : : ?:..,ooo//; ~ :;.50 II; /IV
(7./
c{)
o.oS.J
8
0
ThvsJ h£ ) :: P ~ 250/lJ (1'1-5 JJ s )::: 3.b3 x/O /f .!.:Ul. ~ ( sso/ {(.J1J -:r
9-79
65.9
hp
9,17 9.87 As shown in Video V9.9 and Fig. P9.87, a spoiler is used on race cars to produce a negative lift, thereby giving a better tractive force. The lift coefficient for the airfoil shown is CL = 1.1 and the coefficient of friction between the wheels and the pavement is 0.6. At a speed of 200 mph, by how much would use of the spoiler increase the maximum tractive force that could be generated between the wheels and ground? Assume the air speed past the spoiler equals the car speed and that the airfoil acts directly over the drive wheels.
Tracfive force
=
Fz ~ Ji ~
where j1. ~ coef{j'c,ieni rI {net/on Thf)~)
b
= spoiler length = 4 ft
JW w::
O. 6
fl A~ =ft i!.; where
LlF;. ,".r fhe increase in fracl/ve force d{)e fo the rdowlJwardJ //I'f. Hence will; lJ:::; 200 hJnh:: 293 fils .Af; :::; )
,-
J
t= ipU 2 CL IJ = i(o.OO').3e SK¥-)(2Q31j/·(;'/){J.5ff)(lfffJ-:= 67¥/b; and ~
S ::=
0,6 (67~/b)-:: ,!-os /6
9-30
9.98 9.88 The wings of old airplanes are often strengthened by the use of wires that provided cross-bracing as shown in Fig. P9.88.1f the drag coefficient for the wings was 0.020 (based on the planform area). determine the ratio of the drag from the wire bracing to that from the wings.
Speed: 70 mph Wing area: 148 ft 2 Wire: length = 160 ft diameter = 0.05 in .
•
FIG U R E P9.88
so -Ih'rf
andsillce
Re -- Jl!2 1/ --
88# )( -,om'" --n:f.I
(7olYlph) ( I. ~7 x IO-JI. J:L2.
0.0.5
)
:::: 2720.
oS
From Fi9'
9.21 }
w/lh Re =: 271-0 we obTain
Hence)
clJwire rx9lJ1ifJ9
-
(1.0)(0. 667f-lZ.) (0.02) (IJf8 f/2.) :: 0.225
9-e/
J
or 22.570
9.89
I 9.81) The jet engines on a Boeing 757 must develop a certain amount of power to propel the airplane through the air with a speed of 570 mph at a cruising altitude of 35,000 ft. By what percent must the power be increased if the same airplane were to maintain its 570 mph flight speed at sea level?
-fJ:::: p()wer :::,fJ7J
ir 1]3Cb 1/
=
Lef ( )0 and ( )3$ den()le cOllrl/II()M af .seo, lei/el Tnvs z?;:: 77;. .sf) fh41
&llJri
3.s;()()() ftrCfJflC!illof)
j
7j
-11.s- 16 ~s ~ ~
13s-
::::
..L
71.
3
c/)t)lio df.;s ~ C.o.s./~~ 2.
po
~\M
~
Sf)
0
or
__ 0.00238
w,Yh r()
~ ~ ~
0.0007.38
stqf
II A:, ~/lJ.5 411d
c/Jo
~~3') J
Vq/vOs from iaiJ/e cJ,
=8, 22
= 32'1.. tI-f
·
.I'D
e
h~~M
f/;ell
9.90
J
9.90 A wing generates a lift 5£ when moving through sealevel air with a velocity U. How fast must the wing move through the air at an altitude of 35,000 ft with the same lift coefficient if it is to generate the same lift?
)
~ /IJIJ )Ji - ~2.38)(/0-3:7I! ~eQ leve/- 7. 38 X/O"" 1/* q•• level
l{¥oofl
= .J. 80
r{eQ level
S
9.9/*
When air flows past the airfoil shown in Fig. P9.91 the velocity just outside the boundary layer, U, is as indicated. Estimate the lift coefficient for these conditions.
f--
I
r0.8
If shear forces are negligible ~=
Jf cos e dlJ
J
FIGURE P9.9/
where fhe + siqn is used on
IIlso,
fhe lower surface j - sI9n on upper surf(lce.
fJ = 1'0 +ipV2.- -feu.'-· consfcmf aim().spheric
Since
preSSllre
the ~
doe.s not
c.onlrib"fe fo the liFt, we sei f. ~O Thvs J fJ = fpy2.[J - (~tJ So fh4t
f.
u
-Ie -7.ZZz:z=-
J~a
f#f(,u [I-(tt] cos8dR 1
1.=
1.0
x c
However, dll~ j Js where j: will9 leI/II/II or cose dll:= cose ids = i rk
dA
~. '(~
~
Hence I
£=-fpu
2
j[t -(~):L]idx +ffV2.f[I-(~)2-J 1 dX
upper or
t :: f
f rl"l
lower
f[(%)2
- (%)~ ]dx
upper
lower
II/so; since G;. = f f' ~2A = i if fo/lows thaf I
CL
= j[ (-fff' .!.::O
c
upper
-(-~)2fOlller
] d(i-)
e~2tC
I
wnere
Thi.s
C: chord
len9th
Infe9ral is obfained by nVQJcriCQ/ infeqrafio/J of fhe d4ta 9ir1en in fheh9vre.
The fo//OW/tJ9 f4h/e of Jaf4 is oJ/a/ncd ; (con'£) 9-H'f
9.9/
"1
(conli)
-cx
[(~~:per - (~)~wer J
0
0
0.05
0.65 0.52 0.'1/
0./0 0./5
0.30
0.36 0.32 O. '2.q
0.35
0.2'/-
O.lfo 0.Jl.5
0.'2.0
0.20 0.2.5
D.SO 0 . .55 0.60
0.6..5 0.70
O.7E
0./8 D.17
O./S O./~
0,12. 0.1/ O.Oq 0. 0 7
0.80 0.85 0.90 0/1.5
0.01
/.00
0
0.05
0.03
By lJ,sinJ the pro9ram TRAP£~OI wHh the ahove valves of Ihe infcqrIJluJ we ohf4in ***************************************************** ** This program performs numerical integration ** ** over a set of points using the Trapezoidal Rule ** ***************************************************** Enter number of data points: 21 Enter data points (X , Yl ? 0,0 ? 0.55,0.15 ? 0.05,0.65 ? 0.60,0.11,t. ? 0.10,0.52 ? 0.65,0.12 7 0.15,0.lxl '? 0.70,0.11 '? 0.20,0.36 ? 0.75,0.09 ? 0.25,0.32 ? 0.80,0.07 ? 0.30,0.29 ? 0.85,0.05 ? 0.35,0.2lx ? 0.90,0.03 ? 0.I,t.O,0.20 ? 0.95,0.01 ? 0.1,t.5,0.18 ? 1.00,0 ? 0.50,0.17 The approximate value of the integral is: +2.0550E-Ol
Thus)
~:::: 0.206
9.93
9.93
A Boeing 747 aircraft weighing 580,000 lb when loaded with fuel and 100 passengers takes off with an airspeed of 140 mph. With the same configuration (i.e., angle of attack, flap settings, etc.) what is its takeoff speed if it is loaded with 372 passengers. Assume each passenger with luggage weighs 200 lb.
(I)
"£::: CL ifV~A -::: W Lei ( )'00 denofe conditions wifh /00 pdsset/ljers ond ( )372 wilh 372 pqs.renrer.s. ThuS; w,lh ~/OO'" ~37.2 J ~oo ;:: 11372. J (}nd 800:::: ('37.2 Ef · (I) 91ves
For
sfeady fli9ht
til 0 0
=
;/.37.2
Th/)~J
9,91f
U;:o -2.-
~7:Z
or
u u700 ~r58~ooo580, +(372 ==
~12:: /'f6 mph
9.94
Show that for unpowered night (for which the lift, drag, and weight forces are in equilibrium) the glide slope angle, is given hy tanH = CnIC t .•
e,
FOr steady L Fx
vn~wered
v~
fh9hf
=0 9lve.s
b= Wsil}8
== 0
t : : W&Dsf)
Te
tllld
~
Fy
91i1es
ThlJ s }
b_ W sinG
y-
-100) (2.00)J
000
'.312.
_.J. () 1 (JIJ J
Wt Os e -
Cb
fan e = Ct.
9-86
Ib
Ib]~
III TJ
J WI
_
'100-
JlfOlfl/J1J
9. q 5 J
If the lift coefficient for a Boeing 777 aircraft is 15 times greater than its drag coefficient, can it glide from an altitude of 30,000 ft to an airport 80 mi away if it loses power from its engines? Explain. (See Problem 9.94.)
9.'J5
I
From Proble In 9, q if J Hence, I or d=if;S)
J
/5
~ooorl d
= 8S.21J7/
lienee; the plane can q/ide 80m,:
9.96
I
9.96 On its final approach to the airport an airplane flies on a flight path that is 3.00 relative to the horizontal. What lift-to-drag ratio is needed if the airplane is to land with its engines idled
back to zero power? (See Problem 9.94.)
e=
y-.pr
-.;:::!J
/??~1~-;~J
From Prohlem 9.9 if J fan
~ _
CD C'-
c'9 -==/,/ CD
9-87
9.97
9.97 A sail plane with a lift-to-drag ratio of 25 flies with a speed of 50 mph. It maintains or increases its altitude by flying in thennals, columns of vertically rising air produced by buoyancy effects of nonunifonnly heated air. What vertical air speed is needed if the sail plane is to maintain a constant altitude?
WJ1h no vertical air h/o·l;on Ihe sailp/ane w!.IJ/d 91ide wilh a slope Qn9/e OJ where since L F =0 tIf = 1.11 sine (Jnd t = Wcos{;. Helice) .!It::: 1I/si,,() ~
or since cIJ-::: ffV C/) /I
t = 1: PY
4
and
Cl II if follow.s
~
Ihat tan B=
Wc~B
==
tane
-*.
Therefore in sf/I/ 4ir the s4i/p/one w()vld 10os8 alfilllrJe ctf a role ()f Usin()) where
t
fan) ( g~) : : fan' 1.sj:=2,29 ~ lienee, an upward wind 0 f(so "'ph) sin 2.29" =: 2,00 mph wI/I a/low hori2 o nlo/ fl'9hl.
() ==
9-ea
U
~
9Jl&
I 9.98 Over the years there has been a dramatic increase in the flight speed (U) and altitude (h), weight ('W), and wing loading ('WI A = weight divided by wing area) of aircraft. Use the data given in the table below to determine the lift coeffiv __ cient for each of the aircraft listed. Aircraft Year 'W,lb U, mph 'WIA, lblit 2 h, ft .................................................................................................................................................................. Wright Ayer Douglas DC-3 Douglas DC-6 Boeing 747
~
Ct. ::: .j f 7J'-,tJ
Tnt/oS
J
Wri9hf F7yer
1903 1935 1947 1970
-
750 25,000 105,000 800,000
35 180 315 570
W
1. f u2j}
e s/V1.$/{f3 J
3Jl
-.3
1.5 25.0 72.0 150.0
_
2
- p77 L
'" ffls
0 10,000 15,000 30,000
(.W)
rr
w}IJ.
/61(12-
ct..
X/o
SZ3
~.s
o.¥--&o
DC-.3
I. 7t 'tJO- 3
2t/f.
2. S". 0
O.'ff)9
DC-6
/,sox/o
3
'It2-
72-, ()
O. f.SI
7¥7
8.9/x/o-¥
83t
ISO
OJI-J2
2.
9-e9
9,99
J
9.99 The landing speed of an airplane such as the Space Shuttle is dependent on the air density. (See Video V9.1.) By what percent must the landing speed be increased on a day when the temperature is 110 deg F compared to a day when it is 50 deg F? Assume the atmospheric pressure remains constant.
('1-60 +/10)
(~60 +£0)
== 1.//76
'ThlJS)
IT110.,
== VI, // 76
lloo :. 1,0572 77;0~
9-Qo
or a 5.727. IlJcre4Se
9,/00
J '.100
Commercial airliners normally cruise at relatively high altitudes (30,000 to 35,000 ft). Discuss how flying at this high altitude (rather than 10,000 ft, for example) can save fuel costs.
FOr /eve I Fli9ht W:::aircrafl wei'lht = ~::: ~ 1pu'"jJ T/;/JsJ for given ~ C; J tI/ld f} Ihe dYIJ41J1/C pressure is COllst ani /ndependenf of a/h'fvde. Thalls .1. V 2) - .L 7/2.) or lJ - (-J14t1t1~)~ U 2.. P 10., 0001; - 2 e 4~OfJ() ff J :3~fJOO - f.3rJ,;OODj I~OOO J
Hence ~o~ OO() > q" O()O li/so,} s/I)ce flJe drill ;'S .I
,/)
'"
=C/).J:pll2.I/)
.3";,OOD
.1 0" "(JI
:a-
t# ~c/) -fpV /I II
=C/JfplJ'II)
I~QIO
slIJce
f()//()JII..$
fhal
tpT{:fJI(J =f:e~(}tJl -;
Hence fhe o/rcrarl can fly fosler af hif};f) a/l/ltJr/ed wllh file .same (JlJlotJlJf of dr49 ( 4~,oll ~~"()o) J
::
9-9/
9.102
9.102. Repeated controversy regarding the ability of a baseball to curve appeared in the literature for years. According to a test (Life, July 27, 1953) a baseball (assume the diameter is 2.9 in. and weight 5.25 oz) spinning 1400 rpm while traveling 43 mph was observed to follow a path with an 800-ft horizontal radius of curvature. Based on the data of Fig. 9.39 do you agree with this test result? Explain.
For steady IJ')olion a/oIJ! Ihe curved
path ZJ F,. = m Or or.2.
¥
£ = m J{ ::: # where V={lf3mphJ( :os~:h)=63.1~ Thu~ ( ~62S Ib)(6 3.1 ~),. _ ~-
1(::: 800
f-l
J
(32.2 iO(800 f+)
BUT)
- 0.0507 lb
t
~ = CL -J: fy2.fJ = ~ Py2.fD2. or _ 9t _ e (o.OSo71J,) L C - 7l'pTli!.D2- - 1l'(O.OO2.38-?'fI)(6.3.I{fi"( #(1/- =0.:2.33
(I)
From h.,. ?39 wHh D PL2. 7J
If
= (1
rev)( Jmin )(2.11 rad)(2-.9 U) -;;o.s Jrev n 2.. (63 . Ii} ) .
OOMfii
CL ~ O.OB which is less Ihal) IhB
:;
0.28/ We ohft:lin
t; =0.23.3
in
£,,_ (I).
lienee (J smorJlh sphere would not curve QS mvch perhaps a rotJ9h ball (i.e 'J (J/le w/fh seams) wov/d#
9-92
(IS
indicated, but
'I, /0.3
9.103
Boundary Layer on a Flat Plate
Objective:
A boundary layer is formed on a flat plate when air blows past the plate. The thickness, 8, of the boundary layer increases with distance, x, from the leading edge of the plate. The purpose of this experiment is to use an apparatus, as shown in Fig. P9.l 03, to measure the boundary layer thickness.
Equipment:
Wind tunnel; flat plate; boundary layer mouse consisting of ten Pitot tubes positioned at various heights, y, above the flat plate; inclined multiple manometer; measuring calipers; barometer, thermometer.
Experimental Procedure: Position the tips of the Pitot tubes of the boundary layer mouse a known distance, x, downstream from the leading edge of the plate. Use calipers to determine the distance, y, between each Pitot tube and the plate. Fasten the tubing from each Pi tot tube to the inclined multiple manometer and determine the angle of inclination, e, of the manometer board. Adjust the wind tunnel speed, U, to the desired value and record the manometer readings, L. Move the boundary layer mouse to a new distance, x, downstream from the leading edge of the plate and repeat the measurements. Record the barometer reading, Hbar, in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Calculations: For each distance, x, from the leading edge, use the manometer data to determine the air speed, u, as a function of distance, y, above the plate (see Eq. 3.13). That is, obtain u = u(y) at various x locations. Note that both the wind tunnel test section and the open end of the manometer tubes are at atmospheric pressure. Graph: Plot speed, u, as ordinates and distance from the plate, y, as abscissas for each location, x, tested. Results: Use the u = u(y) results to determine the approximate boundary layer thickness as a function of distance, 8 = 8(x). Plot a graph of boundary layer thickness as a function of distance from the leading edge. Note that the air flow within the wind tunnel is quite turbulent so that the measured boundary layer thickness is not expected to match the theoretical laminar boundary layer thickness given by the Blassius solution (see Eq. 9.15). Data:
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Boundary layer mouse /Pitot tubes
-.::::Flat plate
Water-E........~
•
9-9.3
FIGURE P9.103
Solution for Problem 9.103: Boundary Layer on a Flat Plate
8, deg 25
Halm , in. Hg 29.09
T, deg F 80
y, in.
L, in.
u, ftls
Data for x = 7.75 in. 0.020 0.20 0.035 0.35 0.044 0.48 0.060 0.70 0.096 0.95 0.110 1.06 1.21 0.138 0.178 1.44 0.230 1.70 1.85 0.270
19.9 26.3 30.8 37.2 43.4 45.8 48.9 53.4 58.0 60.5
Data for x 0.020 0.035 0.044 0.060 0.096 0.110 0.138 0.178 0.230 0.270
0.15 0.35 0.45 0.71 1.20 1.30 1.56 1.77 1.95 2.00
17.2 26.3 29.8 37.5 48.7 50.7 55.6 59.2 62.1 62.9
19.9 28.8 31.5 37.5 44.0 45.8 50.7 55.2 59.0 61.0
Data for x = 1.75 in. 0.020 0.20 0.035 0.50 0.044 0.68 0.060 0.90 0.096 1.51 0.110 1.70 0.138 1.90 0.178 1.95 0.230 2.00 0.270 2.00
19.9 31.5 36.7 42.2 54.7 58.0 61.3 62.1 62.9 62.9
Data for x 0.020 0.035 0.044 0.060 0.096 0.110 0.138 0.178 0.230 0.270
YH20, Ib/ft"3 62.4
=5.75 in. 0.20 0.42 0.50 0.71 0.98 1.06 1.30 1.54 1.76 1.88
L, in.
y, in.
=3.75 in.
2 pu /2 = YH20 *L sin8 where p = Palm/RT where Palm = YH20*H alm = 847Ib/ft"3*(29.09/12 ft) = 20531b/ft"2 R = 1716 ft Ib/slug deg R T = 80 + 460 = 540 deg R Thus, p = 0.00222 slug/ft"3 Approximate boundary layer thickness as obtained from the graph: x, in. 0, in. 1.75 0.15 3.75 0.20 5.75 0.27 7.75 0.30
9-9'1-
u, ftls
Problem 9.103 Velocity, U, vs Distance, y
c: >-
0.30
,-----,----~---,------,I
0.25
----II--+-~--~-I
0.20
---~-------~~-------_H'__iI_+_-----1
I
I
-
0.15
-I-------------i--------I+--f--4-----1 I
-+-x
=7.75 in.
__ x = 5.75 in.
=3.75 in. -e-x = 1.75 in. --.- x
I
0.10
j I
I
0.05
~~.-:;~-----------~----I
I
0.00
o
U,
80
60
40
20
ftls
Problem 9.103 Boundary Layer thickness, 8, vs Distance from Leading Edge, x
0.35
I
----"--~----+----~~___1
0.30 0.25
-
-
~~~~~~
------'---
c: 0.20 r-O~
I
Approximate boundary layer thickness
Best fit power-law curve I ; '---~_ _ _ _ _ _ _
0.15 0.10
•
--.---------~------ ---.------~-------_i
-
8 = OJ 12x°.48 ---~-----------------------------~
I
O. 05
+=----t-----~---t----;----__t__
0.00
+-----~-----r-----T----_,----~
o
2
4
x, in.
6
8
9-9.5"
10
---.J
9./otf.
9.104
Pressure Distribution on a Circular Cylinder
Objective: Viscous effect within the boundary layer on a circular cylinder cause boundary layer separation, thereby causing the pressure distribution on the rear half of the cylinder to be different than that on the front half. The purpose of this experiment is to use an apparatus, as shown in Fig. P9.104, to determine the pressure distribution on a circular cylinder. Equipment: Wind tunnel; circular cylinder with 18 static pressure taps arranged equally from the front to the back of the cylinder; inclined multiple manometer; barometer; thermometer. Experimental Procedure:
Mount the circular cylinder in the wind tunnel so that a static pressure tap points directly upstream. Measure the angle, /3, of the inclined manometer. Adjust the wind tunnel fan speed to give the desired free stream speed, U, in the test section. Attach the tubes from the static pressure taps to the mUltiple manometer and record the manometer readings, L, as a function of angular position, O. Record the barometer reading, H bar, in inches of mercury and the air temperature, T, so that the air density can be calculated by use of the perfect gas law. Use the data to determine the pressure coefficient, Cp = (p - Po)/(pU2/2), as a function of position, O. Here Po = 0 is the static pressure upstream of the cylinder in the free stream of the wind tunnel, and p = 'YmL sinf3 is the pressure on the surface of the cylinder.
Calculations:
Graph:
Plot the pressure coefficient, Cp' as ordinates and the angular location,
e,
as
abscissas.
Results:
On the same graph, plot the theoretical pressure coefficient, Cp obtained from ideal (inviscid) theory (see Section 6.6.3).
=1-
Data:
4 sin 20,
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Water
.. FIGURE P9.104
Solution for Problem 9.104: Pressure Distribution on a Circular Cylinder
~,
deg
Hatm , in. Hg 29.97
25
e,
deg
L, in. 1.2 1.1 0.7 0.1 -0.6 -1.6 -2.4 -3.1 -3.0 -2.7 -2.7 -2.6 -2.6 -2.6 -2.6 -2.6 -2.7 -2.7 -2.8
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
P = YH20*L
T, deg F 75
U, tUs 47.9 Experiment Cp p, Ib/ft"2 2.64 1.00 2.42 0.92 1.54 0.58 0.22 0.08 -1.32 -0.50 -3.52 -1.33 -5.27 -2.00 -6.81 -2.58 -6.59 -2.50 -5.93 -2.25 -5.93 -2.25 -5.71 -2.17 -5.71 -2.17 -5.71 -2.17 -5.71 -2.17 -5.71 -2.17 -5.93 -2.25 -5.93 -2.25 -6.15 -2.33
sin~
P = Patm /RT where Patm
R T
= YHg *H atm = 847 Ib/ft"3*(29.97/12 ft) = 2115 Ib/ft"2
= 1716 ft Ib/slug deg R
=75 + 460 =535 deg R
Thus, P = 0.00230 slug/ft"3
9-97
Theory
Cp 1.00 0.88 0.53 0.00 -0.65 -1.35 -2.00 -2.53 -2.88 -3.00 -2.88 -2.53 -2.00 -1.35 -0.65 0.00 0.53 0.88 1.00
Problem 9.104 Pressure Coefficient, Cp , vs Angle,
e
1.0 0.5 0.0 -0.5
•
Experimental
Co
u -1.0
-
-1.5 -2.0 -2.5 -3.0 0
30
60
90
e,deg
120 150 180
Theoretical (inviscid flow)
I
/0, I
10.1 Water flows at a depth of 2 ft in a lO-ft-wide channel. Determine the flow rate if the flow is critical.
/0.21
10.2
The flow rate per unit width in a wide channel is q = 2.3 m2 /s. Is the flow subcritical or supercritical if the depth is (a) 0.2 m, (b) O.Sm, or (c) 2.5 m?
v= -AQ = :tl :: -i-Y yb or
Fir --
m~
t
2..3-:9 .l!l. '
19.81 s" y
y, m
/0.3
so fho f Fr ::: ~ = if vg Y yfiY
~
=
h
0.73# S/2
Y
Fr
I
were y- m
flow fype
a)
0.2-
8.21
b)
0.8
1,03
supercrificQ/ supercrif ical
c)
2.5
0.186
slJhcri/icol
J 10.3
Water flows in a canal at a depth of 2.8 ft and a velocity of 5.3 ft/s. Will waves produced by throwing a stick into the canal travel both upstream and downstream, or will they aJl be washed downstream? Explain.
C=V~Y =j(32.2!J)(2.8ff)' ==9..5'O¥ ThlJs, with V::5.3 ~
c fhe wove CQI) frl1vel upsfream Cl9tJiIJSi fhe c{)rrent. Re/alive to fhe sfretJlh bOI1K fhe wave fravels fJps1re4m wi~h velocity t- V = 9.sofj -5.3 !+ = ¥.2 !/-.1f also frove/.s dOlllfJsfream. <
fhe flow i~ slIberifical. Waves travel both vp-rfreCfff} 4IJd downstream.
Nofe: Fr =
¥ :: :~!l = 0.558
j
/0-/
.
/o.Jf10.4 Consider waves made by dropping objects (one after another from a fixed location) into a stream of depth y that is moving with speed Vas shown in Fig. PlOA (see Video \'9.1). The circular wave crests that are produced travel with speed c = (gy)l/2 relative to the moving water. Thus, as the circular waves are washed downstream, their diameters increase and the center of each circle is fixed relative to the moving water. (a) Show that if the flow is supercritical, lines tangent to the waves generate a wedge of half-angle a/2 = arcsin(l/Fr), where Fr = V/(gy) 1/2 is the Froude number. (b) Discuss what happens to the wave pattern when the flow is subcritical, Fr < 1.
v
•
FIG U REP 1 0 . 4
(a) In a lime inferval of i slf)ce the ohject h,1 fhe wafer (allJ i11141i4ff)J fhe Wave) the cef)fer 0 {he wave has heel) swepl J()U'fNfreClfYJ a d,'.s14f)Ce Vi and fhe W4Ve has eXjlallJerJ. If) 6e a dis/af/ce ct frpm ifs eenfer. Tn/s i-r .showlI 1/J1/Je f/ytJre be/ow. !Vole fhal Vi >c-t if V>c (,·.e, Fr>/) .
r
J
.......
01'
(6)
-
--
./
ct vt
f::: arcsin (/ /Fr)
_ c
-1'
If Fr
/
I
-,,,ci
\ U '"
\ J
Vi /
Vi <. ct if
'-.
Fr
-
-
/0.5 10.5 Waves on the surface of a tank are observed to travel at a speed of 2 m/s. How fast would these waves travel if (a) the tank were in an elevator accelerating upward at a rate of 4 m/s 2, (b) the tank accelerates horizontally at a rate of9.81 m/s 2, (c) the tank were aboard the orbiting Space Shuttle. Explain.
Since c:;; 1{iy-' H follows That _ c2. (2 ~)~ = ::: O. ¥o 8 m Y 'I Q.81f;.
the fank aep+h ~
(a) If ihe rank accelerafe.s upward wifh acceleration 0. J Jhe effective acceLe.rcrl ion of qrav/fy is g ff -:: 9+a '= (9.8/ +tt)f! -= 13.81 fi. e
Thus C =::
J
V-ge-(r--'y'"
==
1('3.8/If,. ) (O./f()8m) ==
2,37f-
If the tank accelera1es hori2onla/& 'II/Iii acceleration aJ the effeciive acceleration is jefr ~ / i~ + a2. ':;: /9.8/2.+'1.8/": =/3.87.r;:..
(b)
Thus) ,...---------. c::: (13.87!f;.)(o,If08m) = 2.381} (c)
In orhif ~err -::0 (weI9hfless) .so
10-3
a
c:!!.....
10.6
I 10.6 In flowing from section (I) to section (2) along an open channel, the water depth decreases by a factor of two and the Froude number changes from a subcritical value of 0.5 to a supercritical value of 3.0. Determine the channel width at (2) if it is 12 ft wide at (I).
10,7
I
/~w,,,c,,st
10.7 Obser\'ations at a shallow sandy beach sho\\' that even though the waves several hundred yards out from the shore are not parallel to the beach, the waves often "break" on the beach nearly parallel to the shore as is indicated in Fig. PI O. 7. Explain this behavior based on the wave speed c = (gy)v:.
~Cll Ocean (2.).' . .- - \ ........--""H»-jH!>!>~C·7g'7 . .; . :....
M~
Bea~h'"
. .' •...
'.
FIGURE PIO.7
Since C =",,9Y il {OJ/Ollis fhaf c, > C2 becQlJse of the fact lhat YJ >)t . There fore a~ fhe waves move, fhat porfion in Ihe deeper water fends fo ~afch up wilh fhat porliDn closer fo shore in the shallower wafer. The wqve cresf Je/ld.s 10 become more nearly pIJra/le I fo Ihe shore line. The waves "break" on fhe shore 4S if The wind were blolllin9 normal 10 fhe shore. I
l
/I
ID-J.f
/0.8
I 10.8 Waves on the surface of a tank containing water are observed to move with a velocity of 1.8 m/s. If the water is replaced by mercury, with aU other conditions the same, determine the wave speed expected. Determine the wave speed if the tank were in a laboratory on the surface of a planet where the acceleration of gravity is 4 times that on earth.
Since C =(g i
if foJlows fhat the walle speed is independenf fhe f/vid density. Thus CH;J.o :: CH, ::: /.8!f on eqrlh.
of
J
.t
However) on 'he planet
-/
Cplanef -
or
'10.9
,
~/anefY
·-f
ge4rfh) (
gearfh
-
Cpt/l1Iet = ( Lf )~Cearfh
fjplanef/
=('f)~ (1.8!f) =
)~ -- ( gearlh 9PI(Jnet)~ .!z: (~eQrth y) 3.60
If
for wafer or mercury
10.9
Often when an earthquake shifts a segment of the ocean floor, a relatively small amplitude wave of \'ery long wavelength is produced. Such waves go unnoticed as they move across the open ocean: only when they approach the shore do they become dangerous (a tsunami or "tidal wa\'e"), Determine the wave speed if the wavelength. A. is 6000 ft and the ocean depth is 15,000 ft,
From
£'(,
/0,11-:
C =[~
or
fqnh(2.JX )]~ k
C = [ (.92.Z[l,.)(6oooff) 2""
f h(-2.17'(J5~oooft))~1 ~Qn
6000 fl
/0-5
~
/75
H s
la,to 10.10 A bicyclist rides through a 3-in. deep puddle of water as shown in Video V.lO.l and Fig. PIO.IO If the angle made by the V-shaped wave pattern 'procfuc-ed by the-1ront wheel is observed to be 40 deg, estimate the speed of the bike through the puddle. Hint: Make a sketch of the current location of the bike wheel relative to where it was !::..t seconds ago. Also indicate on this sketcr the current location of the wave that the wheel made !::"t seconds ago. Recall that the wave moves radially outward in all directions with speed c relative to the stationary water.
AI f ime t :: 0 the froni whee I wos af point (OJ. AI fhe Gvrrent lime t:; LJ. t Ihe wheel has traveled (J.
• FIG U REP 1 0.10
J
wave proallced 01 (0) when t =0 V
J
and is at point (j), Ai t:rne t ==A( a wave prodvced by the wheel when N Was of (0) wi If he a diS1ance C At from (0) as indicaf e d i/l-lhe fi9vre. Waves prodvced 01 variOf}.s /i,."e.s (from t:; 0 -/0 i =: At) hy the from wheel will form fA V-shaped Wo,ve as shown ill fhe sec()/Jd fJ'rttJfIe
dis/an ce
d::
V.~i
(proviJed V>c jsvperor,.fical
(/)
~
c
I I \ \
bike speeJ),
CA.i
or
c sinf
V= ThtJs
V=
where
C
~ fiY '" [ 32.2 ~ (~H)Ji:::-; 2..e~.p
J
2.f'ffj
sln
=:
8.30#
20rl
10-6
/0.1/
J IO.1l Water flows in a rectangular channel with a flowrate per unit width of q = 2.5 m2 Is. Plot the specific energy diagram for this flow. Detennine the two possible depths of flow if E = 2.5 m.
E - y + ~" 21 y" Thus, plot £ ::: y .,. 0.3/'1 y" 0.'-
Nole; Yc =(T) 01}
~
J
wAel'e £ my'" m ). N
((2 .sJ11!fJ.)~ ~ ::: 0. 86
=
IF
d Emili
3
_ 3
=3.' Yt: -
Q.8I s t.
0 111
2: (0. 86 0 m) ::: 1.29m yvs E
2.5
/)
1.5
/'
E
V
1/'/'/
~
1
// 0.5 0
/
~'"
2
,"
,,
," Ell
n~/.~
l-':: ~
0.8 ,0
t"-......
IL
0
0.5
1
1.5
2
2.5
3
E,m
0.3/9
For £ =2.5 m J Ef. (I) is 2.5::: Y + yi or y3 -2 . .5y2 +0.3/9 ::: 0 The roo/.s ff) fhi.s e'!volion ore y::: 2.'1-5 , 0.338 tJnd - 0.335 I
Thus,
y =2...#5 m or y ::: 0.386 IIJ
/0-1
/0./2. 10.12 Water flows radially outward on a horizontal round disk as is shown in Video VIO.6 and Fig. PlO.12. (a) Show that the specific energy can be written in terms of the flow rate, Q, the radial distance from the axis of symmetry, r, and the fluid ( 1 depth, y, as
Q)2
E=y+
27r1"
~(
2gy2
Thus,
Q)2.
£=y+ (21Tr
'----J" /
./
/r
\
(b) For a constant flowrate, sketch the specific energy diagram. Recall Fig. 10.7, but note that for the present case,. is a variable. Explain the important characteristics of your sketch. (c) Based on the results of Part (b), show that the water depth increases in the flow direction if the flow is subcritical, but that it decreases in the flow direction if the flow is supercritical.
(a) The specific elJergy i.s
/
'4 I
\
"
I--r-
/ /
r--~ •
FIGURE P10.12
E. ~ y + t-gJ.) where V:= ~ :::
Q
21lry
I 2fjY'" "-12.
(b)
Lei ~ ;;; 2":r so Ihaf E =Y f ~yt. ",hid, is Ihe slime tis For fINO d/mensi()(JQ/ Flow wifh ff s:: f hei119 replaced by I!pwever; for fWo dimensional flow 't is co/ul4nfj for radial flow 'i Is &l variable since r Varies. Bvf ~ vs YC/JrVes for cons/ani wovla look as shown below (Fi9~ /0,7). / ~ubc'i1;CQI
¥.
r
(e) Front fhe Bernoult· e'lvalion £, ;:: £2. or £ =- cons/QlJt for this
Y
//
/-.J'l.
flow.
,.
Consider suhcr/fic4/ flow -point II. For olAf/ow r incrB4se.s so Ih,,1 f decre4ses. Thv.s since £:: c.oMl.J the flow 90es
{f'()HJ
sfetfe ~ If) II~;
the depth increQsc.s. F()r sUb -
E
crilic.a/ in (10111 r decreases J ~ increases E, =E2, fhe flow 'lues fro/ll II, 10 11.1) and the depth decreases. For s upercr/IiC41 Flow f is true. 'ThIlSJ stlhcrifiCIII depfh oulfoll! increases fJ decreases ~ j or inflow decreases fr()fn B, fo 82 - decreasin, depl-h. depth oliff/ow Supel'cr/llca,/ inflow from B, fa 8.3increases J
increasifl9
depth.
/0-8
supercriliccl depth
incre4ses deplh decreases
/0,/3-
-
I
numbers. Plot the specific energy diagram for this flow. Repeat the problem for E = 1, 2. 3. and 4 ft.
10.13*
Water flows in a rectangular channel with a specific energy of E = 5 ft. If the flowrate per unit width is q = 30 ftl/s, determine the two possible flow depths and the corresponding Froude
(0)
(I)
(2)
y, fI
r;.
(s()bcrificalJ
Fr
YJ ff fsvpercrdical)
I
2 .3
no .solufion poss,"J,l
I
if .5
.J
if. 21
s
Note Ihai Emili ='2 Yt: - whers Yc::
(
J
Thlls , E. mlfJ
=-
2.26
0.6/2.
~ f) ~ =I(30 If:L)2] {! 1.
/,56
= 3, 03 ff
32.2
~s.s ft
!... (3. 0311)=
As shown on the 9rllph be/fI~J there are no posifive real roots of Ct{. (I) if £ < £mi,,::: '1.S5 rl. 7~-r--~~~--~~--~-r--~~--
. . j"
6+-~--~~~--~-+~+--+--~~~
/V
5+-~--~~~--~"'~/+--+--~~~~~ 1; ...... /
It: 4 -t--+---+---+-~~-A--+--+-----+--.J---I---I
.........
1,/ >f ;:.3.o3H
~3+-~--~~~-~~--+--+--~~~~~ /" \ EI7./~ = Krs f.I 2-t--+--,,~--~4--4~~_~___ --+~--~~~--~ 1
o
r--r-+---L-.I
./
~" II
o
1
2
3
4
5
6
E,ft
/o-q
7
8
9
10
11
/ D.llf \
10.14
Water flows in a rectangular channel at a rate of q = 20 cfs/ft. When a Pitot tube is placed in the stream, water in the tube rises to a level of 4.5 ft above the channel bottom. Determine the two possible flow depths in the channel. Illustrate this flow on a specific energy diagram.
= ~.S -y or y.3 -If.Sy2+ 6. 2 ! =0 I where y-{/
The rOOTs ()I this
Thv-s,
y .:
gfl'lt1IiOIJ
are y ~
9. I'fJ
I. If1-, lind - I. 0~
~./'l-II or y ;: 1.'1-2 If
(I)
/0-/ 0
/0./5 10.15 Water flows in a 5-ft-wide rectangular channel with a flowrate of Q = 30 f~ /s and an upstream depth of Yl = 2.5 ft as is shown in Fig. PIO.15. Determine the flow depth and the surface elevation at section (2).
FIGURE PIO.15
A 7 f
IJ~ ~2 t?~2.fZ, = -I ti1+Z2} where ~~I1.=:0J z,='t-=21I (30 !f) (2.£+)(5 fI)
-= Q =
V, tii
=
Thvs, (3 Ii)2. .s
2(32. 2 ~)
or
3
fJ. S
:' Q :::
J
(1. fi) 2. .. y~ oS
+ 2 ff =
3.2
!Vole: Fr::jL
If
•
Y1.
-.A.
- Y:z.
f 0.2 ff f ~
v;y;
=
•
wl"chhQsroofs Y1.='I.77'1- I O.632.
11IJd-o.632.
1
3!f
K32.2~J(2fnJ~
= O.37/f < I
fheh Fi;,>/. This cQlJnol he silJce there ,".s no "6vPlpl bel.,een (tJ anti (2.) of whi&n critical COfldihfJ"S CfllI Oa&Vr.
Thus}
30 i}.J
(5 If) Y1.
Z2=o.2rl+Y~J
2. (32.2'f{)
~ -1.9/fy,. to.5Sf/=O I
IJlJd ~ 7&
J
Y2.=O.63:J. J
=I, 77~ II (1I1d 22. =1.97'1 ff
10-1/
.
y
• " (I)
,
, • (2.) •
I
,
E
y,:: 2 ft
Y,2,:: I. 77'1-fI
/0,/6
10.16
Repeat Problem 10.15 if the upstream depth is Y I = 0.5 ft.
FIGURE PIO.16
Thvs}
~ :: 0.528
f1 olld :&3. =:: O. 72.8 ft
10-/2
/0.174
I
10.17*
Water flows O\-er the bump in the bottom of the rectangular channel shown in Fig. PIO.17 with a flowrate per unit width of q = 4 me/s. The channel bottom contour is given by ZB = O.2e- x :. where ZB and x are in meters. The water depth far upstream of the bump is Yl = 2 m. Plot a graph of the water depth. y = y(x), and the surface elevation, Z = z(x). for -4 m s; x s; 4 m. Assume one-dimensional flow.
Ii +i1'1.2. + -f +ijV2.+Z -x2. or Z = Y t 0.2 e J
nlls,
or
(
2
11!..)2. oS
J
(!l m)"
+ 2 m .::
-0.2 e-
FIGURE PIO.I7
,,'2-1h
y:S
2. (9, 81;l)
y3 _( 2.2.0
O.2e -x 2
where fJl o:::f ==0) ~::: ~:: 2m J 22.:: y +28 v,!t ~ Iff.., I a. If ,:::: " .:: :: 2!:r tlnQ V=.Jj.. :: -
2/ ::
TL I
ZB =
-+
Y
J
Y
Y+0.2 e- x:2
2(fI.&//fi) X2 )
yZ+ 0.8IS::0 where y-m
with - ~ ~ X~ Ifm /Vole: Fr: ~; - = 2 f-
Solve for y
y = 0"1-52. ""
'iy, [(f/.8/~){21n~Ji. ThiJs, the flow will remQin slJhcrilicQ/ I
/
1/
fhrof},holJf-lhe largesl roof of Er. (I) wdl be the correc.1 one. Use proJrarn C UBI C
and fhen z = y to.2 e- x
X}m "! 'f.O ;t .3.5 ± 3.0 t 2 ...r
± 2.0 !:. I.S !:. /.0 ~o.s
0
_xl.
- (2.20 -0.2
e )
fo solve For
~
for -'I RJ ~ x ~ If"" y
- 2.200
2.000
- 2/2;.00
2.000 2.000
- 2.200
yct) froth Ef. (fJ
/ .995
- 2.199 - 2,196 - 2,179
1,990
- 2./26
1.900
- 2. Olf'/- 2,000
/.71/0
/.961/ 1.72. 7
/0-/3
-x2. Jm
Z== Y+O.2 e 2.000 2.000 2.000 J .q,/s
1,99'1/,990 1.979 1.9Jf6 1.92.7
(/Qrpesf roof)
(cantO
/0.17"/
The ahove resolls are plotfed in
the 1ra,h De/ow, I
yvs
i I
X
2.00 1.95 1.90 E_ 1.85 >-
1.80 1.75
/i
V
/
/
V
L
~
1.70 0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
x,m
zvs x 2
I
1.99
L
1.98
L.----
t-
jV
1.97 E_ 1.96
/
N
1.95 1.94
/
J
/
I
1/
1.93 . - ' i
1.92 0.0
0.5
1.0
1.5
2.0 x,m
/0-1'1
2.5
3.0
3.5
4.0
/0.184
*10.18
Repeat Problem 10.17
jf
the upstream depth is
O. 4 m.
:n = O. 2 e _.\2 FIGURE PIO.IS
4'-+ 21'VJ2 +2, :: ~+'ij.V:J.. +zJ LJ
or
Z:z. :: y +0.2. e-
Thl/s,
(IOr/-)2. 2 (fl.811f)'l.
or
y3 _ (5.50 -
x2.
~;::
J
where 1t:::/~OJ
t::: '1-4 o.lim
~ ==--1-
V:::!f- : : 1-
::: lolJlJ dnd
2 {'lBlf)2
In
x )y2 + O. 815;;0 where ylWm 1
Solve for y wilh -7' ~ X ~ /,t m
Nofe: Fr
z,.= y+ZB
(if !f)2 + Y -1-0.2 e-x:J..
+ O,Jf, =
0.2. e-
2,::Y;:::0,Jfm J
2
IO.!!l ~
==
(J)
y
=5.0.5>/
Yj» [('l.8/~) (o,'Tmil~ Thvs the flolI! will remoin svpercrilical fhroll,hotJf --fhe shQ//esf posit'vB roof of I
J
f,(.O)
will he Ihe correcf one.
(joSe prOf/ram CVBIC roof) alld
X)rn -!. if. 0 t .3.5
±3.0 ± 2.5 ± 2.0 ± 1•.5 ± /.0 ± O.S 0
/
then
~::
10
solve fOry(x)frofIJ Fr.O) (..sma//esf positive
y +0,2 e-
- (5,SO-0.2
e- x
2-
K2.
for -tf.m ~ X ~ Jfm.
y
)
-5 . .500
0..',.000
-
O. 'fOOO O. 'fOOO
S.500 5 • .s Of) 5 . '1-99 S. '1-96
0.3998
2.
2;;
y -I-a2 e- X m J
O. 'fooo O. Jfooo O. JfOOO
O. I..fOO~ O. if. 036
-.s. 3/flf
0.3999 O.Jfo06 O. J/OZB 0, '1-063
O. Lj 76Lf 0: 5 6 21
- .S'. 300
0.11-082
0.6082
- S .479
- s. 'f2.6
/0-1.5
0.lf~/7
/o,/8~
I
(conI')
The ahove resv/ls are plolfed on the qrllfJh iJe/ow.
zvs x 0.65 . . . . . . - - - - - - . - - - - - - - - - - - - . . . . - - - - - - - .
/0-/6
VI
10.1 q
Water in a rectangular channel flows into a gradual contraction section as is indicated in Fig. PlO.lq. If the ftowrate is Q = 25 ft'/s and the upstream depth is YI = 2 ft, determine the downstream depth, Y2.
Im77r-fflTflZTA3 . 1 ~wkvvvm
~ bl'" 4 ft
b2
= 3 ft
~ V2
p:?2Z2ZZZZZ7azz?:
Top view
FIGURE PIO.,q
Side view
or
'h.3 -
2./5 ~2
+ /.077 =0 which htJs roofs Y1.:: I. 8 2 8J O.9~6J and -0.62.3 . I1
\1, Note: F0 ~ vi>' ~~ 3.~ s ]k~ ~YI
(32.2.$ )(2.fI)
Since Ihere is no
0.390..:.1
7-
relotive lIIinilllVIf/
area he/ween OJ and (2.) where critical fJow can occur if Fo/lows fhot Fr:z. ~ I also. Thlls II is nof pOSS/b/6 10 have Y:z.:: O.9¥t J
Thvs) Y1.::: 1.828 ff
/0-/7
y
(I)
10.2.0
10.20
Sketch the specific energy diagram for the flow of Problem 10./'1 and indicate its important characteristics. Note that qi =;6 q2'
FIGURE PIO.20
l:_-~J~~L~-_:~--- .
\']
~.
_________
(2)
,i
---'O.-'-'-'-~
Yl
12
-+\'2
t
>;//fTm?//7/7/7'/////1»//7//7/7////7// 'T/T/Jd// (1)
_ Q
!/-J =6. 25 -.sfl~
_ 25 Jff+
Cf,- T or
-
+
E- Y For or
fil.~
(6. 2 5.s )
2(j2,2~)y1.
+ 0,607 E- Y y:z. _
or
the b2- = 3 II chQ!1IJe) J
,/1.
!i.
= b1. =
fi2. 2-
E=
t
(8,3~)
Y 2(32.,2~J y2.
Noh; Yc =(and
.x
r)~
or E:= Y +
sofhaf Yc, ~ (
k
2
(2)
2. S .J:ti
3H
(I)
fi2. = B.33 .s
1,077
(JJ
y2.
f~
=(
17l=
(~~Z;
I.067R
k
=(~)3= (-(8;33~)2.):3= 1,2.92 f-l Ij
Cz.
A/s~ Emin
=-: Yc
32.2.
%
or Emilll
J
£/1IilJz.
=f(I.067{f)::: 1.600ft :=
f
(1.1-12 It)= 1.938f/
The specific ene"9Y ditJ9ro/fJ.S (EfS. (I) IJlJd (J.) tire p/offed he/w: . .
..
~- ~
j
1
I
I
~,
\I
~
ff
;
2.
---------r------
I '
• '
Jemf). ''') ~I~
/0-/8
10.21
10.21
Repeat Problem 10. Iq if the upstream depth is Yl = 0.5 ft. Assume that there are no losses between sections (1) and (2).
Top vie\\<
FIGURE PIO.2,' - - - -_ _ _......:01,....:...(2_)_ Y2 ~\'2 >7??>tW?7Tffff/?/T77ff/?l??Tff7ffffff/J/ t77?Tff/J/ (1)
or
3
2
(2)
'
Yz - 2.93 Y2 + /. 077 :: 0 which h,s roots Yz -:: 2. 79, 0.699, ond - o. SSE AI f . Fi v, == 12..S!! IVO e' r,::: = 3.12>/
Vii'
[(32.2{~J(O.5ff~~
fhere is no relafive milJilllvl1I arCQ between (JJ qnd (2) where cri/icfJi Flow COli occur if follows fhal Fr2 > I also. Thvs if is noT poss/6le 10
Since
J
hove Thus,
~ :::2.7 q-
(fhe s{)bcrifiCIJI roo/).
~ =O. 6?~ If
/o-/q
y
/
E
/0.7-1.
I 10.22. Water flows in a rectangular channel with a flowrate per unit width of q = 1.5 m 2 / s and a depth of 0.5 m at section (1). The head loss between sections (1) and (2) is 0.03 m. Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram. Is it possible to have a head loss of 0.06 m? Explain.
(I) 1.1
•
~
1.6
--,---...,.....----r--....,.----,--.,.--...,....-...,.....-.....,
1.4
-t---t---+---+-----j--+---+--~V~---t
"'/
~;
1 .0 -r---t---t----t--I-----:;; ...rl""/'---...--/---+-----f-----I ~
0.8
>.
..,. ......
-t---t---t----t--~~-/-A----I---'-h-.,9---I ..,.f'f' 1. ,/EJ. F O.q.
0.6 +---t---t---:,...,-F---f--~---'f-cv:-.A-----I-----1I-----t "" ... '" ""min= ).9Ia-':';;'r--.....
E; =0.9.9
0.4 -r------r--... --:. ...f----t---='---'-'--t--'--'---'--+----"""oo...._d_:-:::..-.:..--:..~--:...-_~--I 0.2
-
... '"
i-~...~--+--+---j--+---+--_+_-_I
1;'" '" I 0.0 ~--+--+---+--_+_--!---!---+--__!
0.0
0.2
0.4
0.6
0.8
1.0
E,m 10-20
1.2
~ •t
---~
Y' t 7777777777777?7
~:/ 1.2 -t---t---+---+-----j--+------::-..f"V'----F-----j----t E
t
(~)
1.4
1.6
Note: If hL .:: 0.06111 wilh £, =O.9SQm so thaf £2.::£, - 0,06, fhen £,. ==0. eqq III <: Emil') ThusJ if is not possible
to hOlle hL -:::o.o6 wifh fhe qivclJ ~ andYt
.
/0.2.3
I 10.23
Water flows in a horizontal rectangular channel with a flowrate per unit width of q = 10 ft2/s and a depth of 1.0 ft at the downstream section (2). The head loss between section (1) upstream and section (2) is 0.2 ft. Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram.
. , \1~ Y,
(1..)
f · ~-~
OJ
77)777777777 )777
yvs E
3.0
--r----~--....,------r-----r"-----,--,-I'~,'I""V"'"7'""-.....,
2.5
-i------+-----+-----+----+--~I'~--+--t---------I
2.0
,.' /:7'-i---t__-______I -i------+-----+-----+------:I'.-F----
1.5
-i------+-----+---~I'~--~.\~~---t---------I
1;"1"/
",,'/V
;:= ~
,,'
I/E,.~;IJ :: 2.1'
1','
~/E2. =. ,s51f
1,.,,,,,,,,,
>1.0
- i - - - - - - + - - - " ,..F----+---~--____f"I.... ______ ;;::::-:--+__-______I
~~~~----~
",'"
0.5
II
,
",'
l£ I =2..7sH
-i----~--__+_--__+----t--~r___------'---+__-______I
, , '" '"
1,.,'" 0.0 -¥---...f---+----+----+-----i-----+------1
o
0.5
1
2
1.5
E,ft
10-').'
2.5
3
3.5
10.'1..../1
I 10.2+
Water flows in a horizontal, rectangular channel with an initial depth of 2 ft and initial velocity of 12 ft/ s. Determine the depth downstream if losses are negligible. Note that there may be more than one solution. Repeat the problem if the initial depth remains the same, but the initial velocity is 6 ft/s.
t
~
pi t1'
~ -
-
~
/2-
~ =2 - - -~ 77'7117771711/717777J7//
V. =12 fp.s I
,
b;:: widfh
which
has 3 roofs; ()ne ne9alive (no phY.Jic41
rneanin9J1 one is ~ =: 2. ff (no change in depth), and ~ :::: 3,51 If (an increasei" deplh).
If ~::: 6 11 the/) I
v.: == ( 6 fj )(2.11)
.s)
Yz
2.
and £'1.'(/) becomes 2 + 6
2
- \/ +
2(32.2.) -
J'2
(
/2.
~ .ll:. Y1
)2-
2i_ .
2 (32,2.)
or
X. 3 -2.56 Y:z.2- +2-.2-'1- ~O
The p()s/live real roQ/s are ~ c: 2 fl J or ~ ::: /,38 ff (a dec.rease ill deplh)
10-2.2
~'''''''-''''TTh.'''·
/0.25
""''''I'''''''''''''''
10.25
A smooth transition section connects two rectangular channels as shown in Fig. PlO.2S. The channel width increases from 6.0 to 7.0 ft and the water surface elevation is the same in each channel. If the upstream depth of flow is 3.0 ft, determine 11, the amount the channel bed needs to be raised across the transition section to maintain the same surface elevation .
.L),
1/1.
Z Zj'
-l.k!- +..!L f
r
= l!f. . f
V2
J ~~ Z, 3.
J
here
WI
6ft
&",L",,,,~ FIGURE PIO.2S (I)
Top view
(2.)
S7
4 ~ Ll.. ::: 0 (J/Jd ~,::: 22, r~
•
Side view
'-'
lienee, 11, :::112. or (6 ff) (3 (t) =(7ff)(3f1-h)
=
or
h:::;
Nofe:
Md
0.'129 ff
~/:::; f, =
f
IJ/Jd
'/2.::; ~ :::: ~ < rt,
~2
~
£, : :; Ii +21 IJnd £2.::: ~ +"iff fhaf ~ - Ez : :; ~ - ~
Thv~ .since ~ c: ~
II fo/lows
The correspoIJdin9 specific eIJer9Y di(J9ralll is qs ille/kateri be/oJP=
y
//
/
~ - ~ - ~ - - - - - - / ' - • (J) /
~ -------"/./ / / /
fj.,
/ / LL....-_ _ _
1
~~,,~~,,~~~~
ThIlS, ~;::"~ or
*t
7ft
~---'--_
/0-23
i:2
/0.26
I
10.26
Water flows over a bump of height h = h(x) on the bottom of a wide rectangular channel as is indicated in Fig. P10.26. If energy losses are negligible, show that the slope of the water surface is given by dy/dx = -(dh/dx)/ [1 - (V2/gy)], where V = V(x) and y = y(x) are the local velocity and depth of flow. Comment on the sign (i.e., <0, = 0, or >0) of dy/ dx relative to the sign of dh/ dx.
For any fwo points on Ihe
f(x)
-.2.,=,",,"-=-,,-=-= __ ~ _ Yl
JT
= -_ __
t---
~~7/~/r~~~f ....
h(x)
FIGURE PIO.26
I
').~
(JX
~
~)
free s()rface:
¥i .,
:V't dh fit. ').9 Tx
.:--=-=-----:-r=-----c-=--
-{
-If!- +- if t &, :: ~ $.2 Thvs, JC + h +y == cOfJslonf so -I-
VI
where fJ, -:: 1.2 ::: () J Z, :: M) (u,d ~2 =h f Y2
tha+ hy diflerehlialillq
=0
(I)
Also, for cOl/servoliDA 01 h/qss
Vj >1 :: Vy or V~ t Y ~
;:;0
or ~:: -
f 1x
(Z)
Combine £'10$. (I)(I/Jd (2.);
V
~
(-J!. dy ) +dh y
dX
+ d.x -:::0 "dy _ d1 di J or OJ -
/lole: If Fr ==ffy
<:
I j . fhen
~ (/nd
-(df)
(/_(fy'))
¥X
hove the OpPosite sijn 9
If Fr > I J
have
fhe
the" dA and!U
SQllle
Tx
sifjlJ.
dX
V · dll - - -
-..-
:;:<0
~ -x
Fr
di>o
Fr>/
/0-')../1-
/0.1-7
j
10.27
Integrate the differential equation obtained in Problem 10.26 to determine the "drawdown" distance, e= e(x), indicated in Fig. P10.26. Comment on your results.
From Prohlem
or
\1. 2
y.3_(y,t 26
/0.26 :
1J'2
-h)Y:Zt(v~;
&
)::0
Obtain y== y(x) fro/}') £'1' (jJ and fhen or j:: YJ -h-y
(I)
1 =J(x)
from y,:: h +y +1
Ihon Inc Bern(){)//i efV4//o/) : wilh V,,~ so Ih4t
Nofe: £'1' (I) is r/fJlhill9 ",ore
y; +y, " 1;+
y+h
2;~2 +}j::: (¥J.2 ').~ y+h t
l
wnkh sim/Jlilie.s
/0-25
1
70
[r. (I),
/0.28 I
10.28 Determine the maximum depth in a 3-m-wide rectangular channel if the flow is to be supercritical with a flowrate of Q = 60 m3/s.
3
_ Q _ 601]V- 7f - (3m) y
A!.so,
_
-
20
y , where y:depfh
(Vl.) Fr '" YjY - f9. 81;) y]% '" V
Thvs wdh Fr:/ J
J
6.3Q
y3Az
y =(6.3q)~
/Vole: Fr decrt!Qses as y increases. == :).'1I1-/fJ
10-26
/ o. 30
I J 0.30 The following data are taken from measurements on a river: A = 200 ft2, P = 80 ft, and So = 0.015 ftlSO ft. Determine the average shear stress on the wetted perimeter of this channel.
rr-
rW
==
Y D
0
(\
nh~o
J
h were anri
D
"I, ==
pA
==
200//2.
ThvsJ ?W:::: 62.,'f#S (2,s()11) (0. o()030)
/0.31
::::
eo(+
;::
-"
2.$01"1-
O,O/fD~1ff:L
1
A viscous oil flows down a wide plate with a uniform depth of 8 mm and an average velocity of 50 mm/s. The plate is on a 3° hill and the specific gravity of the oil is 0.85. Determine the average shear stress between the oil and the plate.
10.31
t~
b :: plate widlh - . 77; 7 77 77; ,;;, 7777,/ ;; 7 ;;"1 y= Shim
lw == 't~ So
J
where 0:: O. fJS rH,.o = 0.95 (QgOOU;3)
= 8330 ~3
For a wide flal pIa/e) /)::: hy and p:: b so fhat Rh'::';:: y~ 8XIOA/SOl S(J:; sin 3 ' so fhaf
/0"'27
3 p/
10.32 The following data are obtained for a particular reach of the Provo River in Utah: A = 183 ft~. free-surface width = 55 ft, average depth = 3.3 ft. R" = 3.22 ft, V = 6.56 ft/s, length of reach = 116 ft. and elevation drop of reach = 1.04 ft. Determine the (a) average shear stress on the wetted perimeter, (b) the Manning coefficient, n, and (c) the Froude number of the flow. X" D
S
aJ 7;=Ol)h'o
}
h 5 0 = 116t/1.0Jf rf were
=0.00897
Thus} ?;, = (62.Jftfs)(3.2.2 ft )(O.OOgQ7) = I. BO~ b)
rt:::
Thvs} c)
~ A Rh2.;3 So %. :: /I V J where K::: l.if?
n
= I.lf9 R:/.3
V
Fr =y{yy j
s;:. ::
(t.Lf9) (3. ').2./"/s(0. OOeq7)~_
6.56
=r(32.2 ;;~6!f)]~ '" 3.3ft 05
2.
10-2-8
0.636
- 0.Oi.f69
< I (subcrifico{)
/0.33
I
10.33
Center board
r1
/
By what percent is the flowrate reduced
in the rectangular channel shown in Fig. PIO.33 because of the addition of the thin center board? All surfaces are of the same material.
bl2
\J ----:-----::-:----..::.~---:---
~-bI2---
W7m~~~~~
I.
b---+l·1
FIGURE PIO.33
Q~ ~ fJ Rh~ S/i Wi/houl fhe centerboard
(I)
(2.)
/0.3 if
I 1034 Water flows in an unfinished concrete channel at a rate of 30 m 3/s. What flowrate can be expected if the concrete were finished and the depth remains constant?
-* f} R;/3 S~
Q=
Lei Thus) since fJ u == Af ) QLl Q.f
==
()f denote finished j ( Rhu -:: Rhf and SOil -::: S"-F
)( fJ o.~ ti nu. U 11/1 Sou _ IJ
Ji. II
nf
.f
From Table
Rh~ sOf~
-
f
/0./
nu:::
!!L. nu
O.D/if
or
Qf ::.!1JL nf
n =
\fu
O.O/if
(3 0
l
l!L
0.0/2.s
)
J
nf
:: 0.012-
= 35.0.11l .s
)u denote vnfinished. J if fo/lows fhal
10.3.5 10.35 The great Kings River flume in Fresno County, California, was used from 1890 to 1923 to carry logs from an elevation of 4500 ft where trees were cut to an elevation of 300 ft at the railhead. The flume was 54 miles long, constructed of wood, and had a V-cross section as indicated in Fig. PlO.35. It is claimed that logs would travel the length of the flume in 15 hours. Do you agree with this claim? Provide appropriate calculations to support your answer.
• FIGURE P10.35
i ::: dis/alice ~
::=
1.
frQl/e/ ed ~ ~119
I: • TlJI/~
~ (S~ mi) (5l eo fll",,') (/s hr) (3o(JO S /hf')
== S, 2c9
1i.
t S De/ermine the t/veratj8 wafer velocily; ~ (Inri cPPlfaf'O lot
V ~ Rh1-~1f'fo ) where }(.::/. If?; 11:! t z::
_H. I
$0 7fl(J1
Also; ~
So ::
(J {il.)
,I w/lh 1I)j~' ~Of'
~ O,S f/~J p::: 2 II
Rh -- fi. f-l ~ -- 0 ,2 ~,.. f'.lT p -- o.~s#-
z
T:::
(~500 -300)
H
(5/fmi) (52 eo filmi)
= o. O/lf7
Thvs wtlh n::: o. 0/2 (see 74h/e /0. / J planed wood)J V= 0,0/2 /.119 (OI2.5)~ 1o.01lf7 = .6.97 Ii ~ J
Nole: V;s slif/1rl1y J4rger than 0"1.
10-30
TIJ/I~ the claim Clt/eal'S
curr(J(;t. Ves.
If) be
/0.36
I 1036 Water lows in a river with a speed of 3 ft/s. The river is a clean, straight natural channel, 400 ft wide with a nearly uniform 3-ft depth. Is the slope of this river greater than or less than the average slope of the Mississippi River which drops a distance of 1475 ft in its 2552-mi length? Support your answer with appropriate calculations.
(I)
V : : ~ Rh ~ rs;
J
where X =:/JI~
V=.3ff/s, Y=3ff b=: t;.ofJ1f.l l/::iJy:: /20()f/~ P:::b+2Y~'f(J6rl J
TiJfls D J II I1h := P::::
/2-()o/l I/-otff
IIfro (rom TaIJ/e J
..3
c
/J
1'1
=2./tTT
n ~O.03
/0//)
I. Jf9 ( '). , 9t) ~ 0,Q3
~ SrJ
rs:
11;41
"
Of'
So = O. 000858 ihB aVerIJ9B lJIIi.rs'ss;ili .s/ofe /:s
10-31
fl'fJIYI
/5."1- (I):
/0.37 6or---------------------~
10.37 At a particular location the cross section of the Columbia River is as indicated in Fig. P1037. If on a day without wind it takes 5 min to float 0.5 mi along the river, which drops 0.46 ft in that distance, determine the value of the Manning coefficient. n.
QO~--u40~O--~80~O~~1~20~O~~=-~· Width, ft
FIGURE PIO.37
c:
5280 if,.) v::: (0,5 (5/11mmi)( ,)(60mrn S)
.
[rom the qlven dalo)
==
8 8 £t
'.s .
From fhe lIlanning ettlJlJli()n,
V== nK Rh~ So ~
J
Q~H where ><~/Jf?, so= (I ~ ')(5 JJ) = 0.000/7"1o A o. ml 280 m i
)
IJlJd !lfJ ::::
!J and
ilppr()ximaie
p .
P fro", fhe fi9vre a.s
II ~ tby -;: -J: (I 7oo(1)(lflfff) :::: 3~¥()(){l2 and P~/8ooff
Thvs i?Jh~ 37.,JfO~/J2 =zQeff 1800
lienee J from £1- (I) 8. e ::
n- ( 1.~9
20. s)
J
;
26
(0.000/7#)
~
or n:: 0.0/69
IO~32
(J)
10.38 ,
10.38
If the free surface of the Columbia River shown in Fig. PlO.37were 20 ft above the bottom rather than 44 ft, as is indicated in the figure, how long would it take to float the O.5-mi stretch considered in Problem 1O.37? Assume the elevation change remains 0.46 ft.
Width, It
FIGURE PIO.31
Lei ( )20 denofe fhe 20 Hdepth and ( )#~ fhe Thvs,
X
213
'PI-
rf depth.
J.i
~o = n,.o Rh"o so~o and
$020 :::
S,,¥¥
HenceJ '" 0
V~Jf
=(
Rh20 )2.h
From
Rhl#l-
the Nivre
I/~'f ~ f by =i:(J700!+) (tf'ffl) :: 37 'fOOfl2
~If ~ /800 Ff
@d
~ ~ -E!......::( II.S f+) ::: Vito 20.8 f1 1
- (
L9.0 -
\.{,,,,) V:10
t #~ --
(I)
~
A:J.o';::
dby:: i
F;.o ~
/5S0 - 600~ /oooff
O.b7~ 5 milJ
0.67'1-
(/S50-Ifoo)(20)ff :: 1~5ooff
2.
d so fhal wilh j= Vi (Jnd 1,.u=:Pp; :O. .5nll;
- Z ¥2
-
J
~
10-33
.
111m
/0.39
I lO.3Q Rainwater runoff from a 200-ft by SOO-ft parking lot is to drain through a circular concrete pipe that is laid on a slope of 3 ft/mi. Determine the pipe diameter if it is to be full with a steady rainfall of I.S in./hr.
Q;: K n ARh7.;:!I So ~~ where
A=
*
D1-
= 0.000568
J
and
From Table /0./, n =0.0/2 AIso} ~ = Alof r J where r ~ rainf411 rale :::
Thus,
.
3~tl1 ~ :'~tt;~ (:q D'") ( 1lt)~ (0. 000568) ~ 0=
,
N
. 1.5
Jh
fr rl3
Q:: (').oOH)(sooff)(l. s fr)(Ii1nJ( 360;.s) ==3.lf7.!j-
Hence J from Elf' 0):
or
,
1.6Lfff
/O-.3Jf
(I)
/O.lf~
10.40 To prevent weeds from growing in a clean earthenlined canal, it is recommended that the velocity be no less than 2.5 ftl s. For the symmetrical canal shown in Fig. P10.40, determine the minimum slope needed.
V== nx Rh~ So"k where X ~/.'1'1 J
D II and nh ;:: l'
(I)
II :: t (If f.I +12. f+ ) (.3 H) :::: 2 If fll. Qn d p:::: rt ff +:2 (5 [.0 = /IT ff Thus) R = 2.'Tf/2 =171'1(1 h
II/- fl
•
From Table 10.1 J n = 0.022 _ /.1f9 (
2.s - 0.022 /.7/Jf)
"4
~
So
or
so
lhQf Ef.(J) qives (with V::
2.5
!J)
SD ~ 0.00066'+
10. Jf I 10.41 The smooth concrete-lined symmetrical channel shown in Video V10.3 and Fig. PI0.40 carries water from the silt-laden Colorado River. If the velocity must be 4.0 ftls to prevent the silt from settling out (and eventually clogging the channel), determine the minimum slope needed. FIGURE PIO.40
h K:: I. If? and Rh:: A p V::: 7fJ< Rh2h 5 V2. J Were fJ ::: t (if fI +12 fO (.1 ff) == 2 Jf f+ 2. an rJ P =if ff f 2. ( 5 f+) =Ilf ff Th liS, Rh = 21/H2o Ilfft = I. 7/Jf ff 0
From Table 10./) n::: 0.012 so fhal £'j.(J)qives (with V=If.fJ) 1.11-9 ~ If..O ;:: 0.0/2 0.7/J-f)
~
_
So or So -
10-35
0.000505
(I)
10.'1-'2.
10.4Z
The symmetrical channel shown in Fig. PlO.40 is dug in sandy loam soil with 11 = 0.020. For such surface material it is recommended that to prevent scouring of the surface the average velocity be no more than 1.75 ftls. Determine the maximum slope allowed. FIGURE PIO.40
x ~ So~ ) where V=rrR h
K=I.J.f9
A and Rh=P
(I)
A ;:: t (Lf ff t 12 H) (3 ff) :;:: Z If f f 2. and P == If fl + 2 (5 f I) -::; 1if ff a Thus) R :: 2lfff =I.7IJf{1 h IJfff ff 'Wilh n == 0.020 and V= 1.7.5 T fli.aJ qives
= 6:~:o (J. 711/' /~ S~
/.7.5
or So :: 0.0002 6q
10.'13 10.43 The flowrate in the clay-lined channel (n = 0.025) shown in Fig. PlO.43 is to be 300 ft 3 /s. To prevent erosion of the sides, the velocity must not exceed 5 ft/ s. For this maximum velocity, determine the width of the bottom, b, and the slope, So.
v== '*
FIGURE PIO.43
:
where A = f [b t ( b +i, +12)] y wilh ~ == h~:~o 3. '1-6 fl o =2 ff and J,. = 3 Thus) 5 11 = 300!f or b::. 27..3 ft s fEb +(b+.3.'l-6fff:ZU5J2fO J === J
ta::S K
~
%/,a
A/so} V: : n Rh So } where "1.==/.'19 and from Table /0./) n::. 0.025 From Efj.tO} A;:: i [2 (27. 3 to .,. 3. '1-6 ff +2 rtJ (2 ff) :.60.0ff2. Also}
2ff
P == b +J~ .,. J.u :: 27.3 ff +Sin- SOD Thus} Rh :: A.p = 60.0£12. :: I. 76 ff so 3lf.1 ff • ~
_1,11-9
5-
0.02..5
T
(1.76)
24
So
_
~ I
or So-
10-36
+
2. ff . LU 0 SIn ,oJ
= 3'1-.1. ff
fhaf ££1. (2.) becomes G
0.00331
(I)
10.Jf4- J 10.44 A trapezoidal channel with a bottom width of 3.0 m and sides with a slope of 2: 1 (horizontal: vertical) is lined with fine gravel (n = 0.020) and is to carry 10 m3 /s. Can this channel be built with a slope of So = 0.00010 if it is necessary to keep the velocity below 0.75 mls to prevent scouring of the bottom? Explain.
3
Determine V wilh ~== 10./f and So :: 0.00010, Q= ~
AR;hVS;; where /l= fy[3+(.3+lfY)]::: 2Y2.+ 3y Rh :: JJ wdh p:: 3 t 2 (1/5'y)
f/nd
J
Thvs,
--L (zy:L+3y)
J0:::
0.02
or
f
2. y
2]~ I r 3 y (o.oooJ)J4.
3 t 2. V5
Y
~
(2. y z.,.3y) :3 20 :: ( -v:r )~
wht"cn
3 +1 ;.s 'y
:J.y
2
he wrdlen
COli
( O.lI.f-3y-6.03 3fl1/Ey) ==0
QS
=F(y)
0)
Solve (hy frial and error) E'l' OJ for y: YJ m '2.20 2.2.2 2,2.Jf 2.26
F (y)
fhal y;:;
0.1 -
-O,'f.Sq -0.2.6'1 -0,077 0,1/7
From fhe 9raph 2. 25
fry) -
..
o
'118
J
that
V==
II :::
~
~2.20 2.2~"f 2.2.6 ~. I
I
2-
2. (2.2..5) + 3(2.2.5) ==
3
JOs
-
I
-0,6
m
l:: /6.9m2.
ihlls"
see
m
lIence from £'1' (0) so
(0)
::: 0,592
/6.9 IJJ 2-
f
V< 0.7.5.f so fhat scofJrin, will nul occur,
/0-37
10.4-5
J 10.45 Water flows in a 2-m-diameter finished concrete pipe so that it is completely full and the pressure is constant all along the pipe. If the slope is So = 0.005, determine the flowrate by using open-channel flow methods. Compare this result with that obtained by using pipe flow methods of Chapter 8.
For open channel flow
Q== ~ f} Rh~ S/~·
where x:::: J
J
A/soJ A=fD2 ==1j(zmt::::3.1'fm:J. and P::::"D=6.28m so thai R _.1i - 3 .IIf /17'- - 0 5 h - P - 6.2.8 m -
Hence
wifh n::::
J
I Q = 0.012.
For pipe flok!
. m
For finished concrete (see Table 10./)
0.012
(3.1'1-) (0.5)~(~ 0,005) 2.
3
::::
111 11,7 -:s
(
open channel )
with cons/ani pressure:
+f!t'
2,
t£Jt JL.2 +z, = b + Yl +Z:J. ~ ~ 21
where PI == f~ and It;' = ~ Thus, with z/ - 2:;. == J So •
(2.)
Z~=o
f So :::: f 1r D ').g
or
fV'-:::
2.tjDSo :: Z(9,8/~.)(2.m)(O.005)
F'rom Fig. B. 2.2 J (or smooth concrete ~ 6V AI 0 - Y.Q - V(2111) SOl
== 1.5 x/o-If(2.)
ne - 11 - 1.I2.XIO'm..2.. - 1.79x/o ,s
and from fhe Moody chari (Fi9' 8.23) : Solve £'15. (I), (2), tJlJd (3) for ~ VI Re : ils.sul!1e
f=
0,015
0'/96]~ = V ::::[ 0,01.5
so
fhqi from
E,.(J)
Re
3 61l!1.
or
'..s
Re.: I. 79x/O~(3,61):: 6J1-6x/06
.'
.
£f. t.1) (t1foldy c.hqrf) 0.196] J.2 V=[ 0,013 = .3.8 801 .s
Thu~ from
/lsslJme f c 0.013 or fhat Re ==/, 7?XI0 (3. 88) =6.'15 x/o 6 T/;/I~J from £1. (3) f:: 0.0/3 (check.s
f:::: $0
f
0.013 :I 0.015.
I
6
wHh fhe t/.t.svlIIsd vo/ve)
lience V=..3. 88'; ()r J
Q=!9 V.::: !(2.111)" (.J.88~) :' 12,2 1f3 10-38
3
(pipe flow) :::, II. 7~ (open c.hannel
flow)
IO·Jf7
I lO.4i Because of neglect, an irrigation canal has become weedy and the maximum flowrate possible is only 90% of the desired flowrate. Would removing the weeds, thus making the surface gravel, allow the canal to carry the desired flowrate? Support your answer with appropriate calculations.
Lei ( )", and (~ denofe weedy Qnd r;ravel co"difIPf)SJ re.sle6 //l/ely. TlJv.JJ (I) (1.)
X ~ 1r:::Qw-_ /1w!lw ~w ,s"", and
Q., =
f 1 Rny"4 ~
wJw./ wdh !he Ch4M9/ M/ J flU!« 111 J lIe/Jce) /; y diV iJ;f}~ /;1 Ff (/J :
£, (:;.)
Q
.-1-
J : : : !!L = !jf- ) WAel'B Qw
*~
ThtJs) ('J :: \.1(,
0. Or li
f",fJjf/
";"/;/0 /().j:
1 == /, 2
(). 03 0 I) ~£.&I
4 Evt ~w:: 0.9 lftJeslre rJ 1
•
taIV So
4,,/ R",
fH4+
CV, :: If 2 (0, q Q"esil'eti) = /, 0 ~ qriesif'84
YeS ;1 WOIIIJ worA. J
/0-39
J
(lnd -5;,,,, ~ s;"
/)w::: O.03();
I? : : f},().2.,s-
10·f8 I 10.48 An open channel of square cross section had a flowrate of 80 ft 3/s when first used. After extended use, the channel became half-filled with silt. Determine the flowrate for this silted condition. Assume the Manning coefficient is the same for all the surfaces.
LeI ( }n alJd ( ).r denoTe
(I)
10.49 A circular, finished concrete culvert is to carry a discharge of 50 ft 3/s on a slope of 0.00 I O. It is to flow not more than half-full. The culvert pipes are available from the manufacturer with diameters that are multiples of I ft. Determine the smallest suitable culvert diameter.
x 2..-1 k Q=n ARh 50"') where X= l.'f9 ) S/)= 0.001) anrJ(frollJ TabJe
/0.1)
n = 0.0/2-
For a circular pipe half fuJI /I:::
i D'"
J
p:::
f
D so that Rh =~ =
*
Thus, 50 ::;o~o~~ (*D'')(~)''h(o.ooJ)~) or D::.5.21ff To make sure if i.s not more thon holf (vII use fhe 6 ff diameler pipe.
/0,501 10.50
A rectangular unfinished concrete
channel of 28-ft-width is laid on a slope of 8 ft/mi. Determine the flow depth and Froude number of the flow if the flowrate is 400 fe/so
¥I
h 8U = 0.00Is/5 Qnd Q: : j(AR2~s~ n h 0 J were k= I.'f'l J So = .52BOff 1 from Table /0./ n = 0.01,/A 2By AIso) A = 28 Y and p= 2Y+2B SO fha f Rh :::: P ::: 2y +28 Thus)
'100 ==
or
/,lI" ( 2By O,O/If 2y+ZB
)~/3(2SY)(O• OO/5/S)~
5'/,9
Y O.5QJf = -....:......-(y t/lli/3
Hence
l
O.tf58(yt/if)-Y
5./.i 2.
0,5 ...
:::0 !fiFfy)
Trial and error soJufion fay Ffy)=:O From Ihe 9rQ ph r=O when y= 2.23
F \
o l---...L..-\/-4-....L-_....I I
2.2
Thvs y:: 2 .. 23 ff J
V--
Q
Jl-OO¥
If :: (2BfO{2.23ff) =
6
VI
Jt
.,I.s
so thof £i. 1.. 6. JfI .s Fir=V'lY =[(32.2¥a)(2.:l3ff)J~ = 0 .7.56
IO-LfI
. y:2.23
-0.5 I-
1_ -
2.3
Y
/O.S I
I
10.51 A 10-ft-wide rectangular channel is built to bypass a dam so that fish can swim upstream during their migration. During normal conditions when the water depth is 4 ft, the water velocity is 5 ft/s. Determine the velocity during a flood when the water depth is 8 ft.
( ~
L ef
and ( 1- denule f/Ol'nJa I
af/a
flood ctJlJd/fiol/s, f'cspecfillely.
ThvsJ 0)
n;;k
II
Vn :::
Clf)d
D 1,./3.fr;::
ySolI
f1hlJ
\/ )( R 2-4 Ir;:-' Vf
z::
-n;
hf v~{
where nn ~ nf
J
~on -::; S()I and
lin :: I (J If (if fIJ ::: ~ (JII~ J
11;:: 10 If (&f-IJ ::; fo til.
Pn = 10 If f 2 ('ffT) =Ii If ~ fJ ::: lofT +2(111) ;: 2.of-l Th VS Rn/}:;J
anJ
_ II
~r -
flence~
A t-::: j;ff112- :::
71"
d/v/Je
8()t/L :z.I If
Eq(:J.)
2..:J.:J. II
'" 3. on.;
by £q(jJ t() ()~/4in:
JfVn -_ (Rhl )~~ ( .3.oe# )~~ R 2:).~ff - 1.2/f pn
so
that
~ ;= I.;z.~ ~
::: I. 2'f (s ij)
::: 6,).2-#
to-If2-
/0.52
J 10.52 An engineer is to design a channel lined with planed wood to carry water at a ftowrate of 2 m3 Is on a slope of 10 m/800 m. The channel cross section can be either a 90° triangle or a rectangle with a cross section twice as wide as its depth. Which would require less wood and by what percent?
Qr: -H- AR;h S/i
(I)
Let ( )t denote fhe frianguJar cross -sec/ion ana ( ),.. denofe the rectangular- cross-section 3 10 Thvs 4Jr:= tVl =2.!J- J ~,.. =.s;,t = 800 and nr = nl So thai £q. (I) give.s 1-2Yr~
IT]
J
11 ArRh,..~!;: =III Rh: where Rh :: p
(2.)
J
Hence, iJ,.:= 2. Yr :z.
~
J
p,. = lfYr
S(7
Also)
Ai = i(2)i)X:= ~~ J
fhal Rhr := ~t = ~ Yr
'1 = 2 (1{iyt ) so tho! Rhl = 2~ .
Thus) from Ef . (').) : ' 2Yr 2 ( "2I)~ Yr ::::)1 ,. (J)% 2."12 M J or Yr = 0.707 Yt
The amotlfJf of wood i.s proporliolJfll fo fhe welled perimeter, P. Since
!l= P,.
21/i'Yt == zl'I'y;. if Yr q. ( 0.107 ) Yt
= /.00
the friqnq/e requires lhe same amount of wood as
'he recTangle
/0.£3 10.53 The two channels shown in Fig. PIO.53 are laid on the same slope and lined with the same material. When these channels are flowing half full, in which one will the flowrate be the greatest? Show any calculations needed to obtain your answer.
• FIGURE P10.53
0) (~)
/O-1./-'f
10.5 If
10.54
Water flows in a channel with an equilateral triangle cross section as shown' in Fig. PIO.54. For a given Manning coefficient, n, and channel slope, determine the depth that gives the maximum flowrate. FIGURE P]() ..5+
'---I
1 -- fan2.h60° b = 2lh-y) fan 60° /J _ ~.s -
There fore
K J ( 2,)[ J- h y - y:J.. Q -_ 7f fOI1~o' lhy-y 2(h + y)
For fhe dE
J
!rJQxi/1JVIfJ
.'%
]:3So~
-= 0 J which is e'{lJivQ/eni fo ( : l )51.3 == ( 2hy - Yo)~ Vpon cli!ferelJlitJlioIJ
{/owrale J
Or = OJ where F7y)
~
cos 60°
y+ hco.s60
and simp/ilicrrlion !his 9ill8S S ( Y+h cos 6otl) (h - y) - ( 2.h Y - y" ) =0 or tl +! y 2. +(.5 h cos 60() -.3 h ) y - 5 h2. C os 60 : : 0 which cqn he wrilfen as
e({-t:.- (-f) - 5 Hence}
=0
Y _ J;t '~I+-'1--(-8)-(5---"),
h -
Y
Sin600
/6
= -0.73/ or + 0.856
The ne9afive ro01 has no physical meQnin9'
Thus} y =O. 856 h
10 -'1-5
/0.55 ( 10.55 At what depth will 50 ft:1/ s of water flow in a 6-ftwide rectangular channel lined with rubble masonry set on a slope of I ft in 500 ft? Is a hydraulic jump possible under these conditions? Explain.
l .~~--j 6ft
n c: 0.02.5
Q:: /.onIf.1 ARh~ vs, where LJ /y 0 A _ 6r'n
~
0
1111::
J
anfJ.
n::
P -
"ly+6
(see Table.
0,01-5
10,/)
Thvs) 60
J
'-,.1
1.1f9
-c:
0,02.5
(6Y)[ 6y ] :2. y+ 6
'3(0
Whic.h becomes y5/3 := (2y+ 6 )'-/3 (O.9~g) The Irial and error solvfion
Y= 2.S3fl -rl... V - !l f/VS IJ = I
-
J
50
'
002.)~
70 this
ft~
6 (2.$"3) ffl:::
so thai V Fr ::~~
e9tJal;~1J
IS
3 · ')..9 f"TIS /.
3.29 ff4
v'J y
S/nce
Fr
< I if
/s
not poss/hle
/0-'1-6
10 have a hytlr4V//c .ifl~l.
10.56 10.56 .
I Water flows in
t~e s~mmetrical, unfinished concrete ~~21
~~~~~~~als c:,~f~/i~~~t~ ~~~~i~~~~e6n~tr:;::~;[u~~~ :;:~~
~':'
.'V==:<-:-o-: --,g'oo' -
of concrete needed to line each 1000 ft of the channel.
~
35 0 ' . _ - - , - _. • • ."
\ I.
-':--:.;::;:::-:-::.;:-:-=-=--~.
_
., •.••.• d
_
.•
.. I . • • . • •
,0
4-in.-thick
" ' ; " .;...
10ft
.1
concrete lining
FIGURE PIO.56
Q::: 7fX fJ RhfA3 So% where
from
%J
Table /0.1
n;: 0.01'1-
A= t(~ +/off)y = t (20{ff2.86Y)Y
~
A==(Jo + 1.¥Jy)y
'" =/011
A/SOl or
~-.A.-
;t~
~/off--lV
II
+).
12 = sin~sO
h - P - (joll 102/1.)
or
'i
(I)
R _ (JOt/.'f.3Y)Y
fQ~3.sfl :: I. 7¥ Y
h - (Jot3,lfBy)
Hence J with
x:: /.~p
£ 1{. (I) becomes
J20= I, lf9 (10 t/.IfJy)y O.O/If
or /tf,eqo =
IO-t/·JfJY,yJ~~(11-.2.) ~2 (jo+ 3. If By) 2000
r
(10 y +1. #-.3y2.) S
(JO+3.~8Y)2.
"
(
J
or /22
Solve (frial qnd error) for
SA (JO+.3/fSy) -(Joy+!'Jf.3yZ) 2=O=F(y)
F(y):::O
F
,v
/.60 1.6S
185
1.70
-III
200""
F
1.f2.S
}oo
ThIJs y -= 1.66'f fl J
v-= volume of concrefe per Jjoooff
yo;:. I. 66Jf ff
O~L-------'~~--~' 1.60 1.65 ~y /,70 - /00 I-
.:: (P + Lf ff) ( 'JOOO £1) ( ~ fI) where p= loft +2/,. :: /ofl +2. (I.7Jf)(I.tl'fff)== 15.8ft
Hence
J
11 =(15.8 {f +Lffl){IJOOO f·I)(~H) =
lo-'t7
6}600 {fJ(
k~~~) = 2-~'f yd 3
~
10.57
J 10.57
Determine the critical depth for a flow of 200 m'/s through a rectangular channel of 10m width. If the water flows 3.8 m deep, is the flow supercritical? Explain.
Q=
*
fJRh1:1 S/2. where for crilicQ/ flow
Thvs, wifh V"
*"
¥- : : V
or Y= 3.tf'f I'J7
Q.81
If The
i
y'
[!L3
~~;In""i
" 2/ ';
Fr =/ or V=
1
1iY
we hqve
~ fhen V:= f.~ =S.26.f- and Fir = V9Vy = [(q. 81S.:z.6 p,.)(3.8111 )]~ = 0.862 fJo w i.s slJbcrilicQ/.
y=3.8m J
lO-tf-8
10.58
I 10.58 Water flows in a rectangular, brick-lined aqueduct of width 1.2 m at a rate of 73,000 m 3 /day. Determine the water depth if the change in elevation over the 16-km length of this channel is 9.6 m.
l
~/3.~
X
Q == 1) f) Rh vSo where X= I,) n::' o.o/s (see TalJle
Q;"
14':l
",3
/0.1) J
J hr
DA
Thvs~
Ea.(!) I'
0,8J1.S ::
~
P
z::
-L (I.').y)
0.0/5
Yr
and O. BIfS
m3
s-
0.0006
1·2.Y_
= i"y+I.1..
becomes:
I /. Y. .] ~---------"'A
2.
';y+I,,.
_ [If" Y ]
O. ~3/ -
)
(9.6h1)/(/6x/o3m) :::
A ~ /. '). Y and Ilh
of
(f)
7~ 000 j,y ( ').I/-hr) ( 36 oos
Also) So =
t Y
VO,QOfJ6
~ '.3
:2ytl.2
IJ fria / and error soluiion
of
Y= 0.86/ hi
lo-~q
Ihi.s eCI'Ja/ion 9 j l)es
/0.59 J 10.5'1 A smooth steel water slide at an amusement park is of semicircular cross section with a diameter of 2.5 ft. The slide descends a vertical distance of 35 ft in its 420 ft length. If pumps supply water to the slide at a rate of 6 cfs, determine the depth of flow. Ne£!Ject the effects of the curves and bends of the slide.
r
1
2 .5 !f
.. ~f~ .'. . ~~~
:.::~.:\.~./~y
Q::: nk Ii Rh2A3 So"k
J
35 ff
where
k;::I.'f9 J SD:: J/.2oFt :: 0.0833 1
and from Table /0./ n ==
0.012
Also (see £xomple /0.5)
IJ == '8 (f) -smB) and R_
J
Thus, Q=
*So~ I
94 [ (e 8 (1/)"4
D
or 6 0 :: 1.1f? (O.0833)~ •
0.012.
lIenee
h-
O.2.q3f)~=(e-si"B)
where D::: 2.5 ff
D(9-sin8)
'1-8
J
-sinef4 j
e~
where
J
8~
8
JI
e r. N
Q
J
6 ,
s/:
[ (() -sillS) (lfl/.3 fj i l3
~
•
.
(2.5)
J
2.
D2.
Q= 6.0 te .s
0.02528
~]
:z..
-(e-sme )5 ==O=F(e)
Trit,j (Jnd error so/ufi()() for F(e) =0 eJ~d
F
.
1.50
O.02tf7
1.57
O.oolqS
1.60
-O.OJ3.5
0.02
F 0,01
~
~
J
0
Thvs,
f)=(1.57/f
1.50
-0.01
l-
-0.02.
I-
or since
y:;:. ~ (/- cos (!)) if follows fhol
=0.369 ff
/0-50
I
1.55
•
I
~60 •
rq,j)( J~or:?) = ,qO.2-
y =(¥ fI)( J- co.s( ~))
Ie::: 1.57¥ j
0
e
10.60 10.60 Two canals join to form a larger canal as shown in Video VI0.2 and Fig. PI O. 6Q. Each of the three rectangular canals is lined with the same material and has the same bottom slope. The water depth in each is to be 2 m. Determine the width of the merged canal, b. Explain physically (i.e., without using any equations) why it is expected that the width of the merged canal is less than the combined widths of the two original canals (Le., b < 4 m + 8 m = 12 m).
II FIGURE P10.60
Q.3 = Q/ + Q2
where
for &':::1,2,3
Q{' :: ~. A,. Rh~~ VSOl~ Thvs K fI R 1;j 1r;-' X ~ 1r;::-' }< 2.4 t~ n3 l3 h~ r50,3 :: "1i1. A1. Rh, r So,- + If; AI R Il, rSo, J
But
n,::: n2
and
::: n3
SOl::: SOl..
~ SO.3
.$0
(t)
thaf Eq. (J) become.s
A3 Rh32/3 :::!l2. Rh,.='-h -I- III Rhi2.h where
AI:::
2m (4111):::
A:z..:::
2m
(2)
8"l F( ~ (2f2+Jf).t8m J
(em):::/6tn'"
f.
:z.
-fhtat Rh, ~ r: : ; ::: 1m Fi::: (2-+2+B):#/:Lm So +ha1 Rh2. =~:: 12-111 //,,,," -=-/,333m so
;~
J
and. /).3 == 2 b /112-) f1, :: (~+2+b) ==(b+¥)m so
Thvs Er.(2) becomes
rhal
Rh3 ==
t : (!~9-)
J
!~J"/.3 (2b)L(b+~) == /6 (1.333) ~ .} or5
8 (J) ~ :::
27. 7'-
~A
b /3
= 8.63 (6+'+)
(3)
3
A tria I and error or e9/Jaiion so/vel' so/vrion 10 £ft- (3) 9/ves b = 10,6& m I {
Ihe two ori'lin4/ canals
mer'jed
If) form a/2m wide cafJQ~ fhe
wafer depfh flJov/d he less fhlJlJ 2m h eC4t1.se wdhovl the fwo w~/Is thele would be I~ frief/on force hold the wa1er b4CK. ThllsJ to lJIallllain +he 2m depfh W~f}1tJ.rl have b<'/:Lm.
g.3
1
fJ. 1=1=1 j..--8rn~1 ~/:J.m-
/0-51
10.61 ..
10.61 *
Water flows in the painted steel rectangular channel with rounded corners shown in Fig. PlO.61. The bottom siope is 1 ft/200 ft. Plot a graph of flowrate as a function of water depth. for 0 :s y :s 1 ft with corner radii of r = 0, 0.2, 0.4, 0.6, 0.8, and 1.0 ft.
*
ol 2ft----..
c_'=---=::=:=:::=;=---~----~--:=:-:-=:-:-=::::=:--y-=r_=::_=:-=-_-_=-_:_=::::d
J
Jff
= 0.005
200ft
,,.. .. - - 2. fI
Hence wi! h Rh :: J
1.1:: I.Jf9 \¥ 0.011f
A~3 - ' -
(b)
I
(0
p~'
or
Ef(s. (JJ, ('J.)J lind (3)
7.53
~
["Y-('1-1I)r+2.]J~
1 -.-i
-----t
.,.y-r (2.)
(3)
give ff3
r TOr
r~y:!JJ where r-fl,y,.ft/;",s
/O.S J
(If)
+
Assume Y ~ r: Thus, A = AJ +f}2. +A.3
From Example
(I)
005)~
[2Y-(2.-¥)r2.]
Q=
r
4nol
AsslJme y ~r : Thus) A=2(y-r) +r(2-Zr)+ i 7Tr2. or A=2Y-(2.-1[)r2. and p = 2 (y-r) +(').-Zr)+ llr or p:: 2y -(Jf-Tf)r +2
(Q)
I
FIGUnE "10.61
Q= fJ Rh sji, where X= I. Iff from 7Qb/e /0./ n::: o. O/~J So =
l
•
.
: r:z.> y
'IIilh
I
.
• (3) I
D=2r
A,+A3= (Zrf(9-sin8) where fJNrad and
e
cosl =
r;y
Hence) A = .f(e -sin8) of- (2-2.r)y
AIs 0 p= 1
2-2r
(5)
where
from Example /0.5) R+~ = (2i)9 = re Thus p:: 2- 2r tr8 = ~ +(9 -1.)r By combing £'1.5. (OJ (5)J and (6) We obtain: J
() - J.~9
q I 0.0111
or Q=7.53
A5.I.s~ I
(0 005)./i ,
['2.. -f (9 -sine) +(2-2.I")YJ5~ 2A
[2. +- (8-Z) r];3
for O~y6r where r",ff y,..fI
Q'"
(con/i) IO-.5~
fJ-3 and ~ =2 co£'( r; Y) "" ;ad J
(7)
10.61 .. ,
(&onJt)
For
Iff pl()f Q:::Q(y) from either ft{. (#oj or Eq. (7) for ()~ y~ I ft. Pro9ram P /o#~ I show" De/()W wtU used 10 calclJlate fhe .tall1jJle re.rf)//s .J'howfJ. r=OJo.2JO.I/-JO.6Jo.8Jand
:"00 cls 110 open "prn" for output as #1 120 print# 1, "************************ * ** *** ***** ** * **** *****" 130 print#l, "** This program calculates the flowrate as **" 1l,r0 print#l, "** a function of depth for various values **rt 150 print#l, "** of the radius of curvature of the **" 160 print#l, "** corners. **" 170 print#l, "***********************************************" 180 r = -0.2 190 pi = l,r*atn(l) 200 for i = 1 to 6 220 1.' = 1.' + 0.2 230 print#l, " " 2l,r0 print#l, using "With 1.' = ##.## ftrt;r 260 print#l, " y, ft Q, ft3/s" 280 y = -0.1 + 0.00001 290 for j = 1 to 11 295 y = y + ('.1 300 if y < = r then goto 500 320 0 = 7.53*(2*y-(2-pi/2)*r*r)ft(5/3)/(2*y-(l,r-pi)*r+2)ft(2/3) 3l,r0 got.o 600 500 th = 2*atn((r*r-(r-y) ft 2) ft O.5/(r-y)) 5200= 7.53*(r*r*(th-sin(th))/2+(2-2*rl*y)ft(S/3)/(2+(th-2)*r) ft(2/3l 600 print#l, using" ##.## #.###~ft~~";y,O 620 next j 6l,r0 next i *********************************************** ** This program calculates the flowrate as ** ** a function of depth for various values ** ** of the radius of curvature of the ** ** corners. ** *********************************************** T,tJith r = -0.00 ft 0, ft3/s y, ft 6.987E-08 0.00 3.0l,r5E-Ol 0.10 9.123E-Ol 0.20 1.700E+00 0.30 0.l,r0 2.613E+00 3.620E+00 0.50 l,r.699E+00 0.60 5.835E+00 0.70 7.017E+00 0.80 8.236E+00 0.90 9.l,r87E+00 1. 00
With r = y, ft 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1. 00
0.60 ft 0, ft3/s 2.806E-08 1. 79l,rE-Ol 6.220E-01 1.281E+00 2.122E+00 3.107E+00 l,r.198E+OO 5.357E+00 6.566E+00 7.815E+00 9.096E+00
10-53
With
1.'
ft 0.00 0.10 0.20 0.30 0.l,r0 0.50 0.60 0.70 0.80 0.90 1. 00
y,
1. 00 ft
0, ft3h:. 1.589E-I0 7.158E-02 3.112E-Ol 7.2l,rl,rE-Ol 1.305E+00 2.0l,rlE+00 2.918E+00 3.919E+00 5.022E+00 6.207E+00 7.l,r51E+00
Qvsy 10
/.
9
8 7
-
6
- - r = 0.00 ft
I/)
..,
t::
d
5
/
4
v", j;
,"
,,'
- - - r
- - - - - . r = 1.00 ft
3
2 1
.... ~ ...-:. -:--. .. "
0 0.0
~
...
0.2
......... . -0.4
0.6
i
I
I
y, ft .
I i
/O-EJ.f.
0.8
=0.60 ft
1.0
I
10.61.*
Water flows in the fiberglass (n = 0.014) triangular channel with a round bottom shown in Fig. PlO.62. The channel slope is 0.1 m/90 m. Plot a graph of flowrate as a function of water depth for 0 :0:; Y :0:; 0.50 m with bottom radii of r = 0,0.25, 0.50, 0.75, and 1.0 m.
~~-1 r
" 90
I
0
0.5 m
j
/
/V" FIGURE PIO.62.
(b) I/sslJme y~ 1,
A-= III +Al , where from £X(lIfJp/e 10.5 with e 1/ - sin 1I) _ (TT-2) r2. .~-------,F AI = (2r)2.(1l B 2. 2.~ I r::
Also, for the frapezoidal area iJ2. (see fiqvres):
1t2
If
= 1/2r +2[Yf(l'i-I)r]][y~(J-k)r] = [Y+(~-I)r][y-(J-*)rJ
ThIJs
J
11=[y+(~-J)r][y-(J-~)r] + (7T~2.)
r2
Also} P = 11 t ~J where R= "f r qnd ~ = 2 (YZ)[Y -(l-vi-)r] Thus, p: +212[Y-(J-y})r]
rr
10-55
(.3)
/0.624
r (con Ii) Thus, wifh Rh :: f:)
I
\{ = O.OJJf
or Q = 2,38
115/3
n
1-
I
Elf. (I) hecome.s (
p~ 0.00//J)
~
115/3
p~/3
(5)
Hence) for y ~ (I-;=} ) r calculafe 6i from £'1' (5) J lAIdh t1 and P from frs . (3) and (If), Th tls ; p/o-lQ~Q(Y) for O~y~O.51h wdhr=O 0.2050.50 O,7S/,Om .L I J J J
wI/ere
a)
b)
J
.S"/:
sin e):3 wil. h CQ.J.a() -::.:...rr - It i f y;!f (..L Q ::: 0.7SO r81a (e -e'-/3 1- y,J r and 5~ Q=2. 38 112, ~ with f) ::[Y +(/i-/)r][Y-(J- &)r] + (1T~2.) rand P . .3
p =fr +2y£[y-(J-~ )r] if y>(t-Vr)r
These restI/t.s are co/cfJla/cd and plolted below usifl9 PrOrrqm P/O.#62. 100 110 120 130 lLrO 150 160 190 195 200 210 220 225 230 21,00 250 260 270 2S0 300 320 31,00 360 1,000 1,020 500 510 520
cls open "prn" for output as #1 pr int.# 1, "****************************** ***********'* '* '* '* '*:t:" print#1, "** This program calculates the flowrate in **" prir.t#l. t!** a vee shaped open channel with a rounded Oli" print#1, "** bottom. '*:t!1 pr int.# 1, "* * ** * * ** * * * ** ** ** **** * ** ** * * ** ** ** * * * * * * * ** * * * *" pi = Lr*atn(1) r2 = 2"0.5 r = -0.25 for i = 1 to 5 r = r + 0.25 l)rint#1, " " print#l, using "l>Jith r = ##.## m";r print#1, " y, m Q, m3/s" y = -0.05 + 0.000001 for j = 1 to 11 y = y + 0.05 if y < = (1 - 1/rZ)*r then goto 1,000 A = (y+(3/r2-1)*r)*(y-(1-1/r2)*r) + (pi-2)*r*r/1± P = pi*r/2 + 2*r2*(Y - (1 - 1/r2)*r) Q = 2.3S*A~(5/3)/P~(2/3) goto 500 th = 2*atn( (r*r - (r - y)-2)~0.5/(r-y)) Q = 0.750*r-(S/3)*(th - sir.(th))~(5/3)/th-(2/3) print#1, using" ##.### #.###----";y,Q next j next i
/0-56
/O.62~
*********************************************** ** This program calculates the flowrate in ** ** a vee shaped open channel with a rounded ** ** bottom. *Jl' *********************************************** With r = y, m 0.000 O.OSO 0.100 0.lS0 0.200 0.2S0 0.300 0.3S0 0.4-00 0.4S0 0.500
0,6
0.00 m Q. m3/s 1. 194-E-16 4-.038E-042.564-E-03 7.559E-03 1.628E-02 2.952E-02 4-.800E-02 7.24-0E-02 1.034-E-Ol 1.4-1SE-Ol 1.874E-Ol
With r = y, m 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.4-00 0.4-S0 O.SOO
1.00 m Q. m3/s 3.4-36E-13 5.120E-03 2.263E-02 5.361E-02 9.839E-02 1.S69E-Ol 2.290E-Ol 3.138E-Ol 4.113E-Ol S.21SE-Ol 6.448E-Ol
---~~'--'-'-----
0,51-----
f O,Sm
t
0.'+
-----------
------------------------ --- ____ C_~~· ~~-~-"
0.3
0.2 i--------------------------------
0,'
------~f
O.5m
t 0
0
0.3
0.1 10-57
O.Jf
0,5
/0,63
10.63 The cross section of an ancient Roman aqueduct is drawn to scale in Fig. PlO.63. When it was new the channel was essentially rectangular and for a fiowrate of 100,000 m3 /day, the water depth was as indicated. Archeological evidence indicates that after many years of use, calcium carbonate deposits on the sides and bottom modified the shape to that shown in the figure. Estimate the fiowrate for the modified shape if the slope and surface roughness did not change. I ! I 0.2 0.4 0.6 0.8 1.0 m II FIGURE P10.63
surface
Original water surface
Calcium carbonate deposits
I
a
'. . :
.'
.
..
.'
where from measuremenls on the fi9/Jre
Ao'X
0, 6m (I. 25hJ) -:: 0,75,,/
and
Ao O,751'n2. 0 306 Rho = Pc ~/.25m +2.(0,6",)::' m
~O.6m
~
7nIJs from £'/_ (jJ: J
ori9inai
no (O.7S)(O,3()6) '".<5 1r:,SOD
_ )i
1.16 -
or
f. rs:
=
3.1f1
frlodified: Qm
J<
==
2./3 1'-;:-
n", flm Rhm
VSom
from fhe fi9vreJ Am -;: . I m ( /.2111/ O.3b1)
/.2-hJ
-;::: 0.75 m2.
and.
R 11m -
Alii"..
O.7S m'J.
:: O.2B3m
P,n"'" (0, SSm +/.Im + IhI)
o.ssm
modified
10-58
a.3m
10.61/·~------6m--------4
10.6'" The smooth concrete-lined channel shown in Fig. PlO.64 is built on a slope of 2 m/krn. Determine the flowrate if the depth is y = 1.5 m. 0.5
61=__> ,. . . . -
y'''' 3 m------1 -
••
0
••
,
•• ',,,,
••••
FIGURE PIO.64
_ L IlD~4S~ L _ _ 2m Q - n nnh 0 Wnere X-I) So- looom == 0,002.) and from TaMe jO./ n = 0.012Wdh Y =1,5m ,4= (.3m) (o.,sRJ) + f. (3m fom) (J,Ohl ) == 6/11'-
(J) .
J
J
ond
ok
p= /,Smf.3m fO.Emf (1'-+3 )2. m = 8.J6m ThlJs) ~ = ~ =B.~;; == 0.735 "', tJlld Ef. (I) 9ives 2
Q= 0.;12.
(6)(0. 735//3
(0.002)~
= 19.2-tf
/0.6.5 10.65
Determine the flow depth for the channel shown in Fig. PIO.64 if the flowrate is 15 m3 /s.
Q == ~ fJR;-1 SO~) where X=J So:= 3~:m J
A/so, fl:: 3y'" f
[3(y-O ..5)](Y-O.5) = ~
=0.003) and froh17aJJle/(J,/ 1I=0.01J..
y2..f-;' Y+ ~
p= yt3to,.5t L~(y-t)~+q(y-f)2.]k
tJnd
,....~3(_Y_-O_.5_).,-;.--r-_ _--. :Y-O.5 :+
2.
:: y +.3.5 of- '{iO' (y- 0.5)
~/6 y +/.92
::
T
L..01-.5_f1l_~
J and rr:: /S 1f-3 we obt4in
Hence J wifh Rh = 1
15 =v. 0/2 or 1'1
I
.5/3
2.
(/.5y fl.Ey +a37S)
(
Jf.//'yt/.l:J.
k
)"A (0,003)'" '3
0 I/-
2.01f(lf.16y+I.92)· -1.sy:J. -/•.sy-O.37S=O :;:F(y)
Tria/and error solufion for F(Y)~ 0
Y F 1.20 0.085.5 1.22 O.OOIfI /.2'1- --0,0786
0.1 F
0.05
o Thus) y ~ J.22m Nofe: Since y < 1.5m the wafer-o.os does nof conrad the Jef-f verrica/ wall -0.1 /o-Sq
+----'--_~---L-_
Y
/O.66~
z = 530 ft
10.66*
The cross section of a creek vaHey is shown in Fig. PlO.66. Plot a graph of ftowrate as a function of depth, y, for 0 :s y :s 10ft. The slope is 5 ftlmi.
30Y-120
yt1-6
Hence} Q:::: ,.lf9
0,03
or _
Q-
LfB.9 (
~
~
(8Y)[~ (O,OOOqIf7)~ yt/2J ySI.3
(Y-/f)5!z,
12.)2/.3 + 265
(
yt
t
I. 1/-9 (3 0 Y-I1-D) [
0.05
30 r-IJ.O] (o.OOO9'f7~ yt2.6·
.3
)'"A yt26 !I
for
'f~ y~effJ where Q~~
Y-8
y-l/
y ~ 8 ft : Q = Q, +4>2 tQ3+ QI/-
(c) For
with n,=o.o3, n2.=o.o5In.3=O.03.sJn~=O.15 eN Also, A, 8y 1J2. = 30(Y-9) A-i =90(Y- 8) , find 11/1- =50 (y-e) :=
I
J
gild
R,=e+etJ.f=2 o J
R= 3()tJf=3tf)
~=50t(y-8)= ytif2
fj=9 0 +(y-8)= y+82
/10
.so thol wilh Rn i =11,: and (c.onJ-t) fO-60
'
ond
(3)
y-s
/I. Rh~ S ~ == I.n'f!./I: RJt~/8 ( o. OOOqtj.7)1£:; Qt .:: Lntfj ttl ~ lilt
O.Oll-59
ThilS
[
n;
Q
I
_ O.Olf5q Q- 0.03
SA
51
(8)') ~ + O.(JIfsq [30(y-/f)] + o.o~sq 2.0'h
0.05
3'f""'3
0.035
fl., 'D.'th'l.h ~
QO(y -8)]
(y+82.),,13
3
+ O.0'l-5Q [.50(Y-B)]5/
or
(y tlf2.)'4
0./5
Q::. 6.6'1-y
~
.s~
(y-8)5/.3
(y- e)~
+2S.3(y-If) +2370 (y+BZt/3 + 208 (y+'I2.)~
ForY~&J
Q~1f
For o~ y~'O plol Q::Q(y) from £'(..5. (2.)J (3)J O{l (If')' ProqraPJ P/01166 sh()wlI be/ow wa.; used fo ca/cv/oie the res()/ls, 100 110 120 130 1'*0 150 160 170 180 190 200 210 230 2'*0 300 310 ,*00 '*10 500 510 520 600 700
cls open "prn" fOT output as #1 print#l, "*************************************************" print#l, "** This program calculates the flowrate in **" print#l, "** the creek at various depths of flow. **" print#l, "*************************************************" Y = -0.5 print#l, " " print#l, " y, ft Q, cfs" for i = 1 to 21 y = y +0.5 if y < = '* then goto 300 if y < = 8 then goto '*00 goto 500 Q = 30.8*y~(5/3)/(y + ,*)~(2/3) goto 600 Q = ,*8.9*y~(5/3)/(y + 12)~(2/3)+265*(y - ,*)~(5/3)/(y + 26)"(2/3) goto 600 a = 5/3 b = (y-8)~a Q=6.6,**y~a+25.3*(y-,*)~a+2370*b/(y+82)~(2/3)+208*b/(y+,*2)~(2/3)
print#l, using" ###.# next i
#.###~-~~";y,Q
************************************************* ** This program calculates the flowrate in ** ** the creek at various depths of flow. ** ************************************************* y, ft 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 ,*.0 ,*.5 5.0
Q, cfs O.OOOE+OO 3.559E+00 1.053E+Ol 1. 9,*3E+Ol 2.961E+Ol ,*.072E+Ol 5.252E+Ol 6.,*86E+Ol 7.761E+Ol 1.011E+02 1.350E+02
5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
(conJi) 10-61
1. 765E+02 2.2,*5E+02 2.781E+02 3.366E+02 3.997E+02 ,*.669E+02 5.872E+02 7.607E+02 9.755E+02 1.226E+03
('1-)
/O.66~
(conli)
The f lowrafe
os a funcfion of de plh ------
is pJo/fed be Jow.
---------i--~----<--~--,----~
- -
----r- -
1200~------~--------~--~----~-------4--------~
- -~---.------- ___._~ -.-.---~~----_+--------<. -------·-l~---. ~-
---- _~ _ ~===~E~~==---=~===3~~~~==--~--.
------
~J=~~_=~:=_=:=_~~-~~l -_~--_=_
, , --.------------+--- - - --.,------ ---~'-- ----------i----<---- -- .---- - - .-----. - - - - ____ J.________ ----".. -- --.---r-------------- --~:~-- ------- -_.
------ ------------i---~--~, ~---+--------+------~-- ~ -------~--+--.
----~~--==--== SOD
600
--:~-
==- -1--- =----=-=-1
--+-----;----
~------~--------_4--------_+----~--~~------+ -- - - ------ -1--------'---------1---- ----------
----------+-
------~----i----------
-
~oo~-------+--------~--~----+---~-+~---------+ -/ _--+_________ L _____ j ---
-
---
--"-------
0-
- - _ ••
----_.---t
------:- ------j
-~~-----+I-'~--I ,
t
200r--------+--~----+---~--~----~--4---~--~ "-----.
-f---~--
-----1------
10-62.
--- - - ---~- .. ----.-
10.67
I :.1·,1--'-----6 m - - - - - 4
l.d~:loo::"~-.
(j)(j)~,--..--l"''---®---l...
-,'-"-.,-., -
10.67 Repeat Problem 10.64 if the surfaces are smooth concrete as is indicated except for the diagonal surface, which is gravelly with n = 0.025.
i~~ 0.5
=.=. = ..:j
~ ·~t.=.=.=.. .,. . ,._m ~
/cref e
/
FIGURE PIO.64
II,Rh l :3 50"~ Q=Q,+Q2, = tijK"/:
+
)( f72.. LJ RZh ~ n;, h~ So where X=I J 50::: 0.002 J
1
nJ=:o.ozs,(lnd from TabJe jO./ n,.=o,OI2.
A/SOI II,:: t( I.Om)(3f11)== 1.50 hi"
J
P, =(1.0 2.+ 3.02.)~ == .3,/6m
,4, I.SOm" R = 3.16 hi = O.if.7S In A2 =(.3I17)(J,5m)= 'f.Sn/ J ~ =QSm +3117 + 1.5", = sm D
or Ilh
an rJ
=: -
J
D
112.
'I. Sin'"
or nh,,::: Ii = 5 In
flenc8J from
= 0,90 hi
Eq. (/):
I '"..3 Js. I Q = 0,02.5(1.50)(0.'1-7.5) (o.oo2J + 0.012. or 3
'"h
(#.5)(0.90)
(0.002)
~
Q= 17.3 fNote:
With al/.svrlflce.s
3
concrefe J
10-63
Q= 18.z.!f (see Problem
/O,&lf).
(fJ
t------2.5 m----l"
10.68*
Water flows through the storm sewer shown in Fig. PlO.68. The slope of the bottom is 2 m/400 m. Plot a graph of the flowrate as a function of depth for 0 ~ y ~ 1.7 m. On the same graph plot the flowrate expected if the entire surface were lined with material similar to that of a clay tile.
-- -
- - ------------ -
Rubble masonry
--:---:-:---: -:-:--.- --:---:----
T --'-l-----'«~.v-c,ay
tile
FIGURE PIO.68
(a) For 0 ~ y -==O.5m: The flow is Ihe SQl11e as fhal in 4 circiJlar pipe. Thf)sJ from EX4mple /0.5 wilh D= / fI1} X=/} and n = O.O/~ (7Q6/e /0.1); J< k D% (e- s inel.1.3 I ').111 1i (1)fV3 (8-sinell.3
Q= n So:z.
B ('ft~
or
=O.OIJl.
e2/.s
~
('l-oom) 8 (Jf?13
ei /
3
5~
.m! Q= 0 . 251 (f) -SinS) e2 / 3 S J and 8 = 2 cos-,(o.s-y)
where fj"'rad
(I)
O.S
(b) For
Y~ O,Sm:
LI4--- 2, 5m (2.)
y-O.5
X Q,= n;
D"~ n,11 nh, SO~
2' J
r ~-----,j.!m..l -----\..~m-l
'Ih n, = o.o/~}
W'
(J)
II,:: ¥(0.5tn)'-=o.39311J2., !?=7r(0.S)=/, 57m Sf) thaf R -==.AL =O.3Q3111:L =0.250 m hi P, 1.57m ThIJs} (;,= o.~IJI.(O.3q3)(O.2.50t.&(If~O )'1. :; O.787.!f3
Also,
.K
2~
li.
Q2 = n" fJ2. Rh2. So ) wilh A2.
=
(2,5m)
lienee, wilh
Q2.
'=
*}
(y-o.s) Rh2. =
n1
= 0.025 (see Toh/e /0./)
=2.5 Y -1.2.5 and 8. =2. (y-o. .5) T 2(~)=2Y
t
0,5
E'(. (;;..) hecolTle.s
~ ( _ )5/3 , (_2._)~ 002.5 2.5 Y 1.25 ( • 2y+4S)~ ~oo
Therefore)
Q::! 0.787 + /3.0
(2.)
_ -
(y_o.S)514 /3,0
(2ytO.S)%
s (Y-O.5)
~
(2. yt 0.5)'''/3
Plot Q= Q(y) for O:;y ~ 1.71'11
3
1f for y ~ O,Sm /)SJn9
(codi) /O-6Lf
Eqs.
(I)
and (3),
(3)
If fhe en/ire surfqce were lined with ",c,feria} with n, =n2 = o.o/if, Etfn. (I) would remain valid. The coefficielJl NJ3. 0/ in Eq. (.3) wovlJ become /3.0(0.025) = 23.2 For this case rO.D/Jf . • 1
Q =0.787
+ 23.2
(y~0.5l~ ~
(2Y+O.5) 1.3
l
-sm for y~ 0.5h1 3
Thi.s re.sulf i.s also plo-lted U.~. Q) frorn £r, {/J for o~ YS:O'S"I and Q from £'{. (If) for 0.5 4! y:5 1.7/11). See Proqram P /0#68 below. 100 cls 110 open "prn" for output as #1 120 print#l, "*************************************************" 130 print#l, "** This program calculates the flowrate in **" 140 print#l, "** the channel as a function of depth. **" 150 print#l, "*************************************************" 160 dim a(2) 170 a(1) = 13.0 180 a(2) = 23.2 190 for i = 1 to 2 200 print#l. " " 210 if i = 1 then goto 260 220 print#l. "With n = 0.014 for the entire channel" 230 goto 280 260 print#l, "With n = 0.025 for part of the chann8l" 280 y = -.J. + 0.00001 290 print#l, " y, m Q, m3/s" 300 fo~ j = 1 to 18 320 y = y + 0.1 340 if y < = 0.5 goto 500 360 Q = 0.787 + a(i)*(y - 0.5)~(5/3)/(2*y + 0.5)~(2/3) 380 goto 600 500 th = 2*atn(((0.5~2 - (0.5 - y)-2)-0.5)/(0.5 - y)) 5200= 0.251*(th - sin(th))-(5/3)/th-(2/3) 600 print#l, using" ##.# #.###~---";y,Q 610 next. j 620 next i ~************************************************
This program calculates_the flowrate in ** the channel as a function of depth. ** ************************************************* ** ~*
With n y,
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
m
=
0.025 for part of the channel Q, m3/s 7.552E-l1 3.293E-02 1.381E-01 3.089E-01 5.315E-01 7.870E-01 9.837E--01 1. 367E+OD 1.853E+00
10-65
(4)
/0,68·
(cadi)
0.9 1.0 1.1 1.2 1.3 loY, 1.5 1.6 1.7
2.y,07E+OO 3.010E+00 3.6L!9E+OO L!.315E+00 5.003E+00 5.708E+00 6.L!26E+00 7.157E+00 7.897E+00
With n = 0.01~ for the entire channel y, m Q. m3/s 0.0 7.552E-11 3.293E-02 0.1 1. 381E-01 0.2 3.089E-01 0.3 5.315E-01 O.L! 7.870E-01 0.5 0.6 1.138E+00 1.822E+00 0.7 0.8 2.689E+00 3.678E+00 0.9 L!.75L!E+00 1.0 5.89y,E+00 1.1 7.083E+00 1.2 8.310E+00 1.3 1.L! 9.568E+00 1.085E+01 1.5 1. 215E+01 1.6 1.7 1. 3L!8E+01
-
I
_ ...• _.... .1 ...
in:: O.Ol'!- for
10
_. _. __..____ ._._.__.________ ~e.l!flre . channeL . .
5
n= 0,025 for top porfion of channel . o,l/-
O,B
y, m
/0-66
/.6
10.6 q Determine the flowrate for the symmetrical channel shown in Fig. PlOAO if the bottom is smooth concrete and the sides are weedy. The bottom slope is So = 0.001.
K .1.A k. Q :::: QJ +
J
Hence wilh n,::: 0,0/2 and n,.:: 0.030 (see J
Table 10./)
we obfain:
Q= ~:~;2 (/2) (3/,4 (O,OOJ)""4-(2)~:~~ (6) {J.2),"/3 (0.00014 = IIq
/0-67
¥3
/0.70
I
10.70 Water in a rectangular painted steel channel of width b = 1 ft and depth)' is to flow at critical conditions, Fr = 1. Plot a graph of the critical slope, Son as a function of y for 0.05 ft :5 Y :5 5 ft. What is the maximum slope allowed if critical flow is not to occur regardless of the depth?
V=
I
I---
1<. zh Ji h X== J.lfq and from Table 10.1 7f Rh So J were
Also) Rh Thv.5
=J.: .,.01
Y )l~
J
IA'! (
and
Fr==m ==/
with ~
J
V=
/R
~ I
--~ .. I
n = O.O/if.
1fiY
[(2y+/)'fJ = 0.0028Jf Y
~
3
V32.2. Y
:=
o.O/Sf
EquafioIJ
(J)
is ploHed below. To de/ermine fhe mini/lJlJ/IJ critical oS lope
set daS;c == 0 .
zytl
That
dSoc _(J.) 0 ~ - 3 0.0028
-t
So:
Soc
or
is: -2h
If)[ (ZY+J)IJ] Y
f
3
if(2ytJ) (2)Y -(2 Y
y2
+J)lJj =0
Thus, y == so fhfff frol11 £C(. OJ S := 0,0028'1 [ (t,+ 1)'1 ]~ = 0.007.5 7 "6
oCmin
If 5 0
<0.00757
critical
fJoWl
cannot occur aiany dep-th.
The followinq volves are ohfained from Er· OJ. !Vole fhal Jim So
y_O
c
(2YtJ)I/o]~ Y = [ y~O
== O.0028tf Jill)
10-68
oa
and lim So =- 00 y_DO
C
1--/0_,7_°---1
(con V)
Soc
vs y
0.045 0.040 0.035
I
0.030 0 0
I
0.025
0.015
--~-~~'---'--~~-~~ ~ "~
0.010
~~-~--~ ~---"-l
Cf)
0.020
..
-~--.-
0.005
----------
0.000 0.0
1.0
2.0
3.0
4.0
--·-~--l
I
5.0
y, ft
SOc
0.0090 0.0088
vs y (expanded scale)
.,.----,....---,--....,-----r---....,-----~------__n
. ___________
~
___
t ~'_~~
__
___+__--~~--_--
____. _ _ _
-----0-----
0.0086 0.0084
g 0.0082 -
C/)
~ .~=~=~~-~~
0.0080
--
0.0078 0.0076 0.0074
---------;----~I;;;;;;;;--
+------r-----r------r-----.-------f 0.0
0.1
0.3
0.2 y, ft
/0-6 q
0.4
0.5
10.7'
1 y (i.e., Soc = CIY, where C I is a constant) and that as Y --') 0 the slope becomes proportional to ) Y -113 (.I.e., Soc = C2/Y 113 ,w here C· ~ IS a constant.
10.71
Water flows in a rectangular channel of width b and depth y with a Froude number of unity. The slope, Son of the channel needed to produce this critical flow is a function of y. Show that as y --') 00 the slope becomes proportional to
Show that the channel with an aspect ratio of b/y = 6 gives the minimum value of Sllr'
l4--b
.. f
r (/)
lis y -0)
(ZV b)" - f
SOc --.(- Xn9b'll.l)
To de/ermine dS oe dY or
so
Iha1
(J1)l Y -- c,. yY.i
fhe minimtlm Soc J calculafe
~~c =0
from
=(Xngb'l/.3 ) (1- J[ (2 ytY b)"'] -2.;3 ['1(2 Yfb) (2.)YY - (2ytb) 3
3
2.
3
(2y +b) [BY - (;;.y tb)];O
Thvs) y= ~
/0-70
£q.(J):
J!-
] == 0
10.72
I
10.72 Water flows in a rectangular channel with a bottom slope of 4.2 ft/mi and a head loss of 2.3 ft/mi. At a section where the depth is 5.8 ft and the average velocity 5.9 ftls, does the flow depth increase or decrease in the direction of flow? Explain.
J
her S = hL - 2.3ft s = '1-.2 ft w e f T - .5280# J 0 52-80H, and 1T= ~:::: 5.9 g. ::Q Jf32 (g y}li [(32.2.~)(5.8fl)J ~ .
Thus)
(2.3 -~.z) dl - I s;eo t =-0. OOOIf'lZ <: 0 The flow depthdecreQse.s in flokl direclittJ. dxx - O.lf32 There is less heQa JrJSS fhan chonle in e/evotiof) for fhis .suhcrifico/ f/uw. The flUid speeds up a/ld qefs shallower.
10.73
J
_'V
10.79
Water flows in the river shown in Fig. PIO.73 with a uniform bottom slope. The total head at each section is measured by using Pitot tubes as indicated. Determine the value of dy/dx at a location where the Froude number is 0.357.
, \'(3)
~(--l-):"-------------\' 1-1(4) ~ (~--------------- -- ~ J.
"-
~?77;77i~~~~"d/,J Zl = 620.1 ft Z3 = 628.3 ft x2 - Xl = 4100 ft
l
(.
(2)
Z2 = 618.7 ft 625.0 ft
Z4 =
FIGURE PIO.7S
dy
Sf
-so
(]X=: I-Fl'
J
where from/he fi'lvre S
f
or ,... = 8 os 10-'" d S .::: Zl-~:l = ..::J; • x an 0 R
Thus, dy
;rx=
10-11
=~ = 23 -i!~ J
X,-X:z.
(620,J-t/8.7){f
If/oof!
:::. (628.3-62.5.0)(+
¥/ooff
=3 .-,.li/X/OL{.
10. 7"1-
I !
10.7/f
I
Repeat Problem 10.73 if the Froude number is 2.75. 21 = 620.1 It 23 = 628.3 ft X2
-
X]
"2 = 618.7 It 625.0 It
Z4 =
= 4100ft
FIGURE PIO.73
ft:::: dx
5f - So 1- Frl
I
where from fhe fi9vre
-8 -If J$ or Sf - .05 x/o tina (J=
ThlJs, d = 1 dx
/0.7S
-it
2/-Z~
J
=
'f /-'1 _ _ 7. 07X 10- 5
8.051./0 -3. Ix 0
/-(2,75)2
I
:5:b ~r
10.75
Assume that the conditions given in Fig. PlO.73 are as indicated except that the value of Z4 is not known. Determine the value of Z4 if the flow is uniform depth.
Z
620.1 ft 23 = 628.3 It X2 - _'I = 4100 It 2]
=
t .
,/
~
z2
///////)'/7
= 618.7 It
FIGURE PIO.73
For uniform flow ~:::: "
Sf - S~
/-Fr
=0 or S~ J
= So
T
Thus,
- 2i i-Zz
Z3 - '2'1- _
J
or %'1 == 626.9
(
J
or z~ == 23 +Z2. - 2, = 628.3 +6/8.7-620./) ff
ff
10-12
I
10.76
10.76 A 2.0-ft standing wave is produced at the bottom of the rectangular channel in an amusement park water ride. If the water depth upstream of the wave is estimated to be 1.5 ft, determine how fast the boat is traveling when it passes through this standing wave (hydraulic jump) for its final "splash."
*: r-
-!'!.
d I +/1 +8 Fr,'" ] I.sfl
2.
Thus, Fr,:::
~ ::: Fr,
10. 77
V,
I
/,971 or since
#l = /.97
FiJ
= ~ f})'I
(32.2!t) (1.5 ff)
I 10.77
The water depths upstream and downstream of a hydraulic jump are 0.3 and 1.2 m, respectively. Determine the upstream velocity and the power dissipated if the channel is 50 m wide.
y2. .::: Y,
I. 2m
I
r. / / +8 fir, 2.']
0.3 m ==2"[1 +
if follows
that V,::
or
£~,Off
t
Y2.
I.S If =)1 , 7.f777 77 7777711t
or / 2.0 U +1.5 if) ::: ..J..[_I +jl +8 Fr, '" ] ~.
t
Fti =3./6 Thus,
(.3.16)[(Q.8J?) (O.3/h)]
~ == 5. t;2
.
sInce
!f
The power dissipafed is qiven by r -p:: lQ hr;.. J where !:E:.:: -I- F /' (I - (4- t) Yt /- ~ Y, 2. 12
0\ =(0.31>7)[1- ~.~: + (3~6)1(1-tt:::n] = Also, Q= A/~ :: Ytbl1:: (o.3m) (SO/H) (5.1f2!f-) Thll.s
:0:
O.SO'fm
S/.31}3
l
;1)::'(9.8 ~)(81.3.t;-3)(O.50¥fn):: ~Ol k~'m = '101 kIN
/0-73
/0.78 I 10.78 Under appropriate conditions, water flowing from a faucet, onto a flat plate, and over the edge of the plate can produce a circular hydraulic jump as shown in Fig. P10.78 and \'idl!i1 \'10.6. Consider a situation where a jump forms 3.0 in. from the center of the plate with depths upstream and downstream of the jump of 0.05. in. and 0.20 in., respectively. Determine the flowrate from the faucet.
.-0J"~
,~3 4/J~mp (2)
, / ' ,..,.... i
in.
I fI)i····· · ~.l:,: :; ; :,:,: ;:;: i: *: ;:;: : :.: : :;: : :+'t; : :; ; ; : :·.':H'· · pL ...•...
0.05 in. 0.20 in.
II FIGURE P10.78
For a hyriratJ/ic jump:
~
::=
i[-I +1{i+8Fr/· ] or ..
o. os in. == f £1 + Y1+N F;?-] 0.2.0111.
ThIJs
.so
fhai Frl ::: 3.1t
J
V, = 3,161.32.·:z.f{(o,os//;l.)11 ::: a/Jd
Q== AI ~ ~ 271R, Ii ~ ::
I.
/b!J. ff3
ff. f.I){qi~5 11) (/.It {t ) ::: 0.00759 s
271(
IO-7Jf
/0.79
I
.10.79 In order to have a hydraulic jump, the flow upstream 'of the jump must be supercritical. This implies that a wave made by a disturbance upstream of the jump cannot travel upstream; it gets washed downstream (see Video VI 0.(;). Show that for a hydraulic jump in a rectangular channel, the Froude number upstream, - Frl' and the Froude number downstream, Fr2' are related by -2 8 Frf Fro 2 - [(1 + 8 Frf)l/2 - 1]3
Plot Fr2 as a function of Frl and show that the flow downstream of a jump is subcritical. ;>77Y7777T777;;;
(I)
(z)
2.
2
Fr2
"
[I
8 Fr:
J
+8Fr/ -
y
(3)
Thi.s resuH iJ p/o/-/ed below.
J
,
3.5
I
T-------------------
3.0
FrJ
Fr2.
0.5 /
2.26
2 3
O.S'f7 0.1/10 O.33Q 0.296
if
5
2.5
~---
2.0
---.~
.!'
I
LL
1.5
- -- - -.--
~
--
-~-
-
-
.--
1.0 !
:
!
0.5 ---.
i
I
0.0 I
- I - - - - - - - , - - - - - r - - - . . , . . . . - - - - - , - - - - iI
0.0
1.0
Nole; To hQve a jump we mils! have FIf> I. FrOIl? the 9roph FI3. ~I if ~ '>/. NOTe; Fr/ = / rp"ve.s Frz =/. IJ/so (;QI1 show from £9' dFr2./rlFr, <0. Hence" Fr2<1 for Q JUMP,
2.0
3.0
4.0
5.0
Fr1 I
~----------------------------------~
/0-75
I
(3)
that
/O.BO
I 10.80
Water flows in a 2-ft-wide rectangular channel at a rate of 10 ft 3 /s. If the water depth downstream of a hydraulic jump is 2.5 ft, determine (a) the water depth upstream of the jump, (b) the upstream and downstream Froude numbers, and (c) the head loss across the jump.
i13
Q - 10 ;s \<2. --7[; - (2.5f1)(:;.ff)
(a)
or
Fr2, =:
2 0 Ii so - • S -
fhai Fr2. =
f:l. V:z. ~ = ~ .s (g.y,) ,. [(3;2..2~,.)(2.5ft)J~ \ I "
Obfain Fri from fhe reslJH of Problem
0.223
/0.7Q : ..
8 Fr,z or (0.223) 2r(J +8 Fr,2.)~_1]3- 8 F~/:z. [(JtBFr,2.)V2.-1]3 LI Triol and error solviion for f =0 ~ f
Fr: 2. ::
=0 == f( Fr,)
(I)
:2
f
Fr,
Fr: ::8.10/
5 -
8.0 -7,3q 8,/
- 0.26
8,2
7.25
o
Thvs, Fr," 8.10 =(v, )V~ IL3 g Y, II -
v, -
l,
-
I -
JO s (2. ff)}'I
Hence, Yt = 0,228 Hand
-s
, where
- 5 - ~
fh f
so "
\.1:: 0.;29
8./0=
I
8.0
V .1
I
8.2.
Fr I
f-
(ft)
(32.2. Y/)~
= 2.1.9#
(b) From pari (a) Ff) = 8./0} Fr2, == 0.223 (c)
A/so, hL ::: r; [I - 2l,
+
!if(J -(it)]
=O. 22B
ff[l- ~;;8 + (8~O)( J -( o-;~ey·)J
or
hL = 5./£ f+ Ir
Or could use ~:: f [-I +~ J + B Fi/] wilh Y:z. =2.sfl so fhat /1 :a. I I 2. 2. 5 +y. = ~-Vr-/-+8-Fi-f--:->2.' /Vow wdh Fr,2.:: J1.. = \CV (by,)) :: (JO/(:z.X)) I I I .) J 'J Yt 'J ~ 32.2 Yt
I
or
L. 2 _ 0.776
rr, -
v3 1/
5+}j
a::
y~ [I
J
we
~bf4in .k
-1-8(-°·;:6)]"
aMain X:L+ 2 ,5X
-0.62-/;:0
.By sqlJQri"9 both sirles which 9ive.s /0-76
ofJdsill1ph!JilJy we
)1.:0. 22 8 {las
ahove.
/0.
eI
I 10.81
A hydraulic jump at the base of a spillway of a dam is such that the depths upstream and downstream of the jump are 0.90 and 3.6 m, respectively (see Vidt'o V 10.5). If the spillway is 10m wide, what is the flowrate over the spillway?
~
t
/
-}'J:: O.9m
f
V~
Y2.'::: 3.6rn
777p 7 7777}7 / 7 /7
b = 100m
Y2. ='2I [ Y, -} I +v 1+8Fr,2,']
Hence) ~::: 3. 16
~=
J
J
or 3.6m o.qm
hur Fr, =
3.16 [(9.81 qi )(o.'I/11~Y2-:::
I
== 2:
(fJ ~Yi
9.3Q
[J -I +- 1+8Fi;£] so that
!}
ThuS
J
3
Q =IJ, V, : : by! U=( I o.Om)(o.QI7l)('I.3Ql}) = 8.tt,5 ~ 10.82 ] 10.82 Determine the head loss and power dissipated by the hydraulic jump of Problem 10.81.
hL= Y,[I-~ +1f?(J-(~t)] Hence}
I
where from~::: ;':~=i[-H-JI+8F"'{I.'J
Fr, = 3.16 so fhat
3.6/11 -#- (3.16 )~(I _(o.QI1l )?)J::: I 51 hL :::(0• q m)I 1- O/lm 2. 3.6m J ..
AIso) 'fJ::: rQ hi. Hence
J
where
v,::: (9Y)2. 0::: .J,.
[
m
(q.81:',.)(O.qm~,,%. (3.16)~ Q.3Q:
J
7'= (9.80~)UO.qf)J)(/Oom)(q.3q~)](J.51 m}= 12J500kN;111 ==
10-77
/2. J500kW
/0.8.3 10.8.3 Water flowing radially outward along a circular plate forms a circular hydraulic jump as is shown in Fig. PIO.83a. This is shown easily by holding a dinner plate under the faucet of the kitchen sink (see Video VI0.ll). (a) Sketch a typical specific energy diagram for this flow (see Problem 10.12) and locate points I, 2, 3, and 4 on the diagram. (b) Which of the water depth profiles shown in Fig. PlO.83b represents the actual situati~n? Explain.
-
_ _---'I
,
(1) (2) (3)
(4)
(b)
iii FIGURE
(a)
P10.83
From Problem /0./2 fhe. specific ener9Y dio'lram for this radial flow is
shown below. .~Uj,~~iJical
y
lncreas~flI' '
~, • • .. •
/
,
\ .s upe rcritica I • . G
/ . /fJcre4Sln9 m /increasin9 r
/
E" Up.slream of fhe jump fhe flow mv.sf be Slip ercr/f ica I so (I) and (2) are localed, Enerpy is consef'teJ- £, =E.2., The deplh decreases from (01rJ(21 1nthe jump enerqy decreases - £.3 ~£'I- < £~. The flow /s stJbcrdt"co/ dOlJlns-freo.m of fha jfJmp and fhe deplh jncreases. (See 4he G1hove qraphJ Th/)~
fhe flow is like fhe {o//owin,:
/O-7g
IO.Slf
I
10.84
I
Water flows in a wide finished concrete channel as is shown in Fig. PlO.84 such that a hydraulic jump occurs at the transition of the change in slope of the channel bottom. If the upstream Froude number and depth are 4.0 and 0.2 ft, respectively, determine the slopes upstream, SOl' and downstream, S02' of the jump to maintain uniform flows in those regions. The jump can be treated as a jump on a horizontal surface.
For uniform flow V==
= A:::
channel R
h
P
*
FIGURE PIO.8Jf
R;1.3 5,;1-',. where J
X-::;/·tf.9
and for a wide
~ Y since b»y h i
yb
2. y+
;
,y,Vi)!?, :: 'f
Upsfream of rhe jvmp Fi;:: (
so thai
v,::: (~Yi)~ ~ :: [(32,2 ~)(O.2ft)]J-i (If) =10.2 ~ Hence, frrJm ft{. (I): 0.0/2
S Y'2.
0,
J
or
.s - =' 0 0577 === 01 -
f[- )
Downs/ream of +he jump Y2. == +JI f8Fr,2.'] or y, Y2 =(i) (0.2 fn[-J+JI +8 (1f)2.' ] =I. 03~ ff Thvs J AI V,
==
1)2 ~
so -Ihat £". 0) 1/17 = :,'o~~
or 9ives
v,. =~ V, =(J.g3~~) (JO,2 ~) == 1.97l
(J.036{~ So~
_
I
~I..-=--=--b-------;---------'+f.J
From Table /0., n::: 0,0/2
10,2 ::: 1.'1-9 (o.2.)Zh
~
(I)
J
or Soz:::
IO-7q
0.00021./-0
10.8514-
I 10.95*
A rectangular channel of width b is to carry water at flowrates from 30 :s Q :s 600 cfs. The water depth upstream of the hydraulic jump that occurs (if one does occur) is to remain 1.5 ft for all cases. Plot the power dissipated in the jump as a function of ftowrate for channels of width b = 10, 20, 30, and 40 ft.
-po: rQhL
J
where hL " Yt [1- (~) + and
Also, C
10
,- -
rr, - (9 Y,>:)Y:& =
J
h \I,,-- II,Q -were
G-£J
[(32.2ft)(J.5fl)]~
Q
/.5 b
=O.OqSq
VI -
/
Yi =1.5 ff 774/ 717
,
Y:L ~
777~7771
f (J - (f,J)]
~ ::: ~ [- J +~ I+8 Fr~']
\I
t
Q
b
(J)
provided F"r,?;O
J
so
(2.)
fhQ7.1 r
LJ
nenceJ rrom
hL = (I.5J[J -(~) + (0. 00 if 60)( Z}'O-(f;. t)
J ff
J
£. 0)
tt·
where b"'flJ Q,.,if
(3)
and from E£(. (2.)
~ = i[-J+ (! +O.0736(~/")~J
(IJ)
For the given values of plof -p from
p= 62.ifQhL
ff:s'lb
for 30~ Q tr600
¥
(5)
Nole; If Fr,
b f+
QJ J -¥3
10
lOll 208
J
2.0 30 ~o
3/2-
¥16
Wilh h::: /~ 2.~:J~ or Jl.off calculate and plot -p from: 0) p::: 0 if ~<.QI b) 'P:: 62.# QhL ~ J where ohloin hI. from El(. ~ from £,/-(,,) if Q/~Q ~ 600ij-3
(.3)
wdh
The pro9ram and rcsulfs are 9iv81J he/ow. (See protjrdm P/o#85) (conI/: ) /0-80
(I)
/0.85·' ( con Ji ) 100 105 110 120 130 1l,t0 150 160 180 190 200 220 230 2l,t0 250 260 270 280 300 310 320 330 34-0 350 3 60 370
cis open "prn" for output as #1 print.# 1, H* * * * * * *** * * * * * ** * * * **** ** ****** * * ** *** * ** * ** * * * H print#l, "** This program calculates the power **" print#l, !!** dissipated by a hydraulic jump for **" print#l, H** various width channels and various **" print#l, H** flowrat.es. **" print#l, "***********************************************" b = 0 for i = 1 to l,t b = b + 10 print#l, " H print#l, using "With b = ###.##H;b 01 = 10.l,t*b print#l, using "If 0 < ###.# then P = 0 (no jump possible)";Ol print#l, " Q, cfs P, ft. ibis" Q = 50 for j = 1 to 11 0 = 0 + 50 if Q < 01 then goto 360 y2y1 = 0.5*(-1 + (1 + 0.0736*(0/b)~2)~0.5l h = 1.5*(1- y2y1 + 0.00l,t60*(Q/bl-2*(1 - y2y1-(-2l 1 1 P = 62.l,t*Q*h print#l, using H ####.## +#.###----";Q,P nezt. j next i
Sample OVlpVI! *********************************************** ** This program calculates the power ** ** dissipated by a hydraulic jump for ** ** various width channels and various ** ** flowrates. ** *********************************************** With b = 10.00 If 0 < 10l,t.0 then P = 0 (no jump possible) Q. cfs P. ft. ibis 150.00 +l,t.6l,tOE+02 200.00 +l,t.131E+03 +1.l,t32E+0i! 250.00 300.00 +3.i!27E+04. +6. 72l,tE+0l,t 350.00 l,t00.00 +1.165E+05 l,t50.00 +1.852E+05 500.00 +2.766E+05 +3.939E+05 550.00 +5. l,t0 l,tE+ 0 5 600.00 With b = 20.00 If 0 ( 208.0 then P = 0 (no jump possible) Q, cfs P, ft. ibis 250.00 +8.859E+01 300.00 +9.281E+02 350.00 +3.376E+03 l,tOO.OO +8.263E+03 l,t50.00 +1.6l,tlE+0l,t 500.00 +2.863E+0l,t 550.00 +l,t.57l,tE+04600 . 00 + 6 . 8 5 5 E + 0 l,t (con J/: )
10- 8/
/0, eSf.
(con'/) With b = 30.00 If Q < 312.0 then P = 0 (no jump possible: Q, cfs P, ft.lb/s 350.00 +2.87~E+01 ~OO.OO +3.628E+02 ~50.00 +1.392E+03 500.00 +3.~9~E+03 550.00 +7.039E+03 GOO.OO +1.239E+0~ With b = ~O.OO If Q <: ~16.0 then P = 0 (no jump possible) Q, cfs P. ft.lb/s ~50.00 +1.129E+Ol 500.00 +1.772E+02 550.00 +7.201E+02 600.00 +1.856E+03
PvsQ 1.E+06
1.E+05
-
1.E+04
--b=10ft
rJj
.c
- - - b = 20 ft ..... b = 30 ft
.:=
Q..
1- . - . b = 40 ft.
1.E+03
1.E+02
1000 Q,ft/s
IO-~'}...
t
10.96
10.@6
Water flows in a rectangular channel at a depth of y = 1 ft and a velocity of V = 20 ft/s. When a gate is suddenly placed across the end of the channel, a wave (a moving hydraulic jump) travels upstream with velocity V w as is indicated in Fig. PlO.66. Determine V w' Note that this is an unsteady problem for a stationary observer. However, for an observer moving to the left with velocity V •. , the flow appears as a steady hydraulic jump.
- -----:---:-:::---- --:----
v=o
y
v~
FIGURE PIO.B6
FOr an ohserver moving 10 fhe lell with speed Vw the flolJl4ppe4rs as shown below.
Thlls ) freQ/ fhe flow CIS q Jump wiih H, v, (20+ Vw ) r ::; (1 y,)Ji = [(32,:Z¥)(Jf+)] ~ or
Fr, = 0./76 (20+Vw )
AI \Y,I = IJ2
III n so)
and
\I
V2
J
(I)
Y2 _ v, _ or V - V N
=
V.
(:J.)
W
2
~ =l[-J+/JfBJ:;-/] which
20~ v.,
2. 0 +Vw
when combined with £,(.s.(J)Qna (2.) become.s
~[-J+jJfB(O.I76i'(:lOfv.,l]
or
2(ZO+V...,)+ ~ or 2 ('f04- 3 Vw )
= Vw ( 1+(O.2'1-8)(ZO+Vwt)2.k 2 [
= Vw
2
I + (O.2.1f8)(Zo+Vw ) ]
J
which CQn be wrilfen as
O.2Jf8 VwLf +9.'12 'lw3
+91.2 Yw2. - 2JfO Yw-160 0 ::; 0 == f(VIAI ) Trial and error solution of £f(.(3) fo/' f(Vw)::o: Vw
.Jf
f(Vw )
'1-.20
-/87
~.2.S
-1'30 -72.2
J.}.30
'1-.'35 ¥.JfO
f
50
0
-/2.9
'1-7. 6
(3)
-50 -100
Thus, Vw.:: If. 36 {t
/0-83
#.2-
Vw
/0.87 a ......................
10.87 When water flows over a triangular weir as shown in Fig. PIO.87 and Vidl'O VIO.7, the cross-sectional shape of the water stream is clearly triangular in the plane of the weir (Section a-a). Farther downstream (Section b-b) the shape of the water stream is definitely not triangular. Explain why this is so. Hint: Consider the water velocity profile at Section' a-a.
a-a
bOb
II FIGURE P10,87
As d/sctJ.fJ'ed in Secfio/l /0. &. 2) fl;e speed ollhe wafer f/ow/1J9 over II weir is a function of h (see the ff?Vre ): U2.
::,j2?- (h +¥f )
ThlJ~ fvr () 7I'iaIJ9v/ar-
the wafer speed af
than
(3)
weir
is 9reafel'
af (If), The h"her -/he .speed) the faflflJefl " .fhe wafer ''.shoots ()vt t)r 1114/1r IIfJtier fhlJ+
lJ
the tic/ion of 11'tJlIltr The frtJjecfuf/e.r of {he wtJJfer.l fherefore al'e ~ .sh()wlI J
be/ow
I
...Yt
(If)
(~)
"-
V.3
"-
,
\
. -- - \"',
" .... '"\
"
'\
The rBsv/f is fA dis/of lion
ot fhe
ori91/Ja/ rrilJllflli4r crfN..r-secf/()1J
of fhe lIJlJJier sil'eam 4S .shfJwl1 if) lite videlJ,
JO-8tf
10,88
10.8e Water flows over a 5-ft-wide, rectangular sharpcrested weir that is P", = 4.5 ft tall. If the depth upstream is 5 ft, detennine the flowrate.
IO.eq lO.8Q
A rectangular sharp crested weir is used to measure the flowrate in a channel of width 10 ft. It is desired to have the channel flow depth be 6 ft when the ftowrate is 50 cfs. Determine the height, p ... , of the weir plate:
f Vii h H34 , where =: 0.61/ + 0.075 -H;
Q= Cw,. CWr
H-:::6ff -
Pw and
Thus}
Q= (0.611 +-O.07.5(6~PJAl)) (t)(2~)\ (6-Pw) or W 5oIf3 == (0.6/J +0.075 (6-;:'",)) (} )( 6¥''f~ )~ (/0 f+) (6- Pw ) Hence} [
B. J5 + (6-~)J P ( 6 - Pw)~ - /2.5 ;:: 0 w
e
F( p,# )
Trial and error so/vfion of £q. (J) for P(Pw) -::O!
Pw
F(pw )
'1-.5 if. 6
3.08
if. 7
O.OOqq -/,116
If.e
3
F
1150
o ~--~--~--~--~Thv.s J Pw = '1-.70 ff -I
/0-85
(I)
IO.qo 10. q 0 Water flows from a storage tank, over two triangular weirs, and into two irrigation channels as shown in Video VIO.7 and Fig. PIO.qO The head for each weir is 0.4 ft and the flowrate in the channel fed by the 90-degree V-notch weir is to be twice the flowrate in the other channel. Determine the angle for the second weir.
e
• FIGURE P10.QO
Q=Cwt k ton (!) 1/i.j H 4. S
(I)
where e, :: qoo J HI:: H2o :: O. 'fff QlJd QI;::: 2 61:2-
(2)
J
ThvsJ from C.~tl
From
Fit;. 10•25.1
= o.sQO £'(s.U) an
8 C'lit, Is
d
(2-) J
H 5/~ == C__ tan (6, -r ) ~r::::-: Y2.j If I
2,
8
1.
(f},. ) ~r::-:. y2-j
7$ TOil ~
H.~sA X 2
or 0,590 tan '1£" :: CW/ 2 iO/1 (~ )x 2
or C",t2. tan (~ ) -:: 0.295
(3)
Tr/a I and error so/vllon: 1I.r.svme &2::: 20°. From Fif. / tJ./6 CIII'I.z. :: 0,6u 1
ThIJsJ CWtz fa!} (~ )
=
o. 626 tan (10') :::
Rel'Bo.fed trie..r result in fhe 9r.aphiJelol,ll froln which We
82, ==.53 0 o.1f 0.3
CWt,. fan(~ )
0.2.
0.1
o
2,0
.30
ed
'1-0
Jf:Kj
10-86
e.2 wF 20° GonC/(Jde thai
0./10 == O,ZPS. ThllsJ
.50
6t)
/0.'1/
,
I
10." Water flows over a broad-crested weir that has a width of 4 m and a height of Pw = 1.5 m. The free-surface well upstream of the weir is at a height of 0.5 m above the surface of the weir. Detennine the flowrate in the channel and the minimum depth of the water Ilbove the weir block.
Q=Cwbbff(ff4H~J where Cwb ::
(I
0.65 t 1L)~ Pw
0.6£
:;:: (1+ o.5m)V2. l.S
Thus, Q=
(O,S63)(q.rn)(q.8/~)
~
2.
m
0,563
(23" ):3/2. (o.5m)3k
2==
Also, Ymin::: Yc =
1-H == (t;)(o.sm) == 0.333 rn
/0-87
10,92.
I 10.92 Determine the flowrate per unit width, q, over a broad-crested weir that is 3.0 m tall if the head, H, is 0.60 m.
Q
119 ( '32)31. H3~
Q 0 6.5 =1) = ( u
.iL
2.
J
where H:::O,6tn and Pw =3.0 m
o 1+ rw)'" Thvs, 0.6.5 .I!! ~ (2-)~ ()~ Cf= (lt0.E.)~ (q.8J s~) :3 O,6m
srn2.
O.Lf70
::::
3.0 lO.Q3 Water flows under a sluice gate in a channel of IO-ft width. If the upstream depth remains constant at 5 ft, plot a graph of flowrate as a function of the distance between the gate and channel bottom as the gate is slowly opened. Assume free outflow.
~-
f
Y,=sfi I
+
l..t
Q
______
777/777]' 'f 7 7777/
Q::fb= b~aJ2.gy,'J where YJ::5{f, b=Joft,o/JdCd is fromhij.lO.2'1. Thu~
Q::: ~
(loft) a [2(32.2ii~ )(5 rn]
~
~
0
00
0.5 1.0
10
0.6 0.58
1..5
3.33
2..0
2.5 2
QJ
rt
.5
2.,5
rn
Q,
0.55 0.53 O•.5J Q.50
k 2.
= j7q ~
a
1/
where a"" (f
J
¥
0 5/,9 9B.5 IJf2 183 2.21f
250
--
200
-----.----~----.------------
150-------------------
-----
------------ -------
",--
¢::
a
100
50 ------O-¥=----r----,------,-------,.--------;
0.0
0.5
1.5
1.0 a, ft
10-88
2.0
2.5
}O·IILf
I ~:~~::~--x~~
lO.Q4 Water flows over the rectangular sharp crested weir in a wide channel as shown in Fig. PIO.94. If the channel is lined with unfinished concrete with a bottom slope of 2 m/300 m, will it be possible to produce a hydraulic jump in the channel downstream of the weir? Explain.
t
Q;: CWI'
H=
U";f"m flow J'm:'
~J;t;:J;2~~'~;!N//' YI FIGURE PIO.Q4
So
=
2/300
"(29' b H.3/2- J where Cwr = 0,611 + 0, 075 (!) wilh
3m-2.2m =0.8117
and
!i,=2,2/'n
Thus) I Q= [0.611 +O.075(2~·~:)J (t) [2 (Q.8J ~)J~ b (O,8m)~ or
3
Q== ',3ifQ b ~ HenceJ
V. - ~
-~
A, - b>'J
I -
J
_
where b. . m I. 3 'f9 b
or
b y,
~:::: J.3'f9
For lJni forth flow ~ Q= nX fJ Rh:z.A:3 So:2) or
\'
Also) for a wide chann8/
1//= by!
VI;::
(I)
y,
J
X ~ ~ 7f Rh/ So J
and
- by, - v 'f b» /Iv Rhl--..&. P, -(2y,+b) -'I I Thus with n= O.O/If (see 74ble~/0.]) J
W
here X ~ /
J
R: 2 y, f b
.so
2m So::: 300m
that
I
Q = /.3lfQ b == o.d'If- (by,) (y,):3 (0. 00667)~ or
Yt -::
0, iflS 111
Hence J from II __ 1,3Jf9 VI I O.l/I.E
£r· (/) = 3,2S m. s
or
Fr/;::: /. 6 I
Since Fh
So
fh 4 f Fir. _- A.~ \I, _ 3.25 fL - r. vljr, L(Q.8IIf)(o.JfISfb)]r.I. /.L
I
>, if is possible +0 produce a jtJmp,
/o-gq
=0.00667
10. q.5
I 10._95
Water flows in a rectangular channel of width b = 20 ft at a rate of 100 ft'/s_ The flowrate is to be measured by using either a rectangular weir of height Pw = 4 ft or a triangular (8 = 90~) sharp crested weir. Determine the head, H, necessary. If measurement of the head is accurate to only ±0.04 ft, determine the accuracy of the measured flowrate expected for each of the weirs. Which weir would be the most accurate? Explain.
Recian9lJ/ar weir:
(0)
~= (0.611 +O.07S("*w))(1-)'YZi b H4. Thvs, 3
Q= [0.6/1
.,. 0.07.5t ~
or
Q=
With Qr:: /oo1jl this or 3.
3A
-.5.'1-0
/.25 1.30
-2,.53
.tl 3
where ~"'!f and H"'ff 9ive.s 0/135 =(0.6/1 +O.OI88H)H-%
- Iff/. 7 =0= F(H)
o
F(H):::O:
(I)
H=/,291-
I
I
1.2.0
1.1.5
VH 1.30
-2. -If -
O,~o
Thus H: : I. 2ql/-(f J
(b)
3
2 J
Trial and error so/ulion for H F(H) ,,20
where Pw=:/fff
)1 (} )[2 (32.2 ~2)J120ff) H4
J07(0.6J1 +O.O/8BH)H
( 32.S + H)H~
J
-6
Trian9lJ/ar weir:
Q= GlUt ~ fan (t) V29 H.5
4
,
:::
Cwt(~Xftm 1f5o)[2(32.2~)J~ H /2 S
or
~ n3 -¥o2e Cwt H:L I.f J where H,., ff and Cwf is from F,"9./0.25 For Q:::: looff) assume Cw+::: 0.58 so ThQt
Q=
~.28 (O,EB) H~ J or
H= ~39fl
Noh: : The ass{)med Cw+:::O.S8
ohecks (see A'l . 10.25) CaJclJJafe ~ forfPH,oo, ~()!o.o~J and #'00- 0 •0'1 from £f(s. (f) and (2): (Rectangular) HJ H QJ cf.s (Trian9u/ar) HI ft GJ cfs
"'rio:::
I. zStf J. 2.9lf J,33tf
95.3
H,oo
100 10'1.9
'v./iJh Hto.oifff if
Lt. 35 = ~. 39 ~.~3
QB.O 100 }02.5
is seen thai frian9{)/fJr weir is more acClJrale ((.e. smQller variafion inQ). IO-QO
(2.)
/0.96 10.% Water flows over a triangular weir as shown in Fig. PIO.96a and VidCt) VJO.1. II is proposed that in order to increase the flowrate. Q. for a given head. H. the triangular weir should be changed to a trapezoidal weir as shown in Fig. PIO.96b. (a) Derive ~,quation for the flowrate as a function of the head for the trapezoidal weir. Neglect the upstream velocity head and assume the weir coefficient is 0.60, independent
1/
of H. (b) Use the equation obtained in part (a) to show that
when b « H the trapezoidal weir functions as if it were a lriangular weir. Similarly. show that when b » H the trapezoidal weir functions as if it were a rectangular weir.
(, )
•
(b)
FIGURE P 1 0 .96
h~H
(
q=CJ U i dh Z
J
where Uz c 1fi.jh
h'O
ofld j = b +2(J/-h)
Thlls,
=(:;.J;.J,)-2h
H
Q;cJ 1{ijJi[(2Hfb)-2h]dh ",cfij
1/
1/
o
(1.iI+b)
S 1[hJh-c"Yij.(Z)I h /"dh o 3
3
= Cw 6j (2Htb) ~ /I 'f -Cw ~.
or (J)
(f,)
t
0
/ISh.
$
Q= CwH V2i b /I~ +.J Yij.!I ~] J Where Cw ~o, D Note: This e1v4/ilfl is s/!lf;(y Me SIIIII of q 101' a redo"9u1or weIr all" Q {Of' t/ /ri41l9U/rJl' weiI', T/7al Is Q'Cl =' Qp ~ Q... .
From fr. (J) Q='cwVij 1I~[ffb +J/I] TIllIs, if h« II, Q'Z C", Yij. /I "''f.£ /I] '" clll 1fij Is /I'%. W/;/C/; Is fhe e'1i1llii lfl for
tI
frI4f1~v/dr 41iel',
Si/llr1d~1 ~ il h»~ q ~ C,..1(1.i h f II~ whidl IS f/;e e'1wI'hii/l for a reciafl'1v/tlr wei/'.
10-'I{
10.97 10.91 A water-level regulator (not shown) maintains a depth of 2.0 m downstream from a 50-ft-wide drum gate as shown in Fig. Pl O. 97 . Plot a graph of flowrate. Q. as a function of water depth upstream of the gate. Yl' for 2.0":;; Yl ..:;; 5.0 m.
,j·"·"'·',j'T""·'·"'·
y,
1
j~=2m
a = 1m •
FIG U REP 10.97
Q::: b q. -::: b Ca a V~1 Yt
where a =/ m and b = 5011 (0, 301f8 ff) ::: IS,2m
I
Thus, Q= (ls,2fh) Cd, (/ m) V2 (fl81!ffi) ('h m) ,
== 67.3 Cd,
yy; %3
where y, m 'V
Obfain Cd from Fiq. 10. z q w/lh ~ :: 2. Y/J m
-Yt a
Cd
2.0
2.0
0
2.5
2.5
0.'12
Jfl/.7
3.0
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6/.8
3.5
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0.35
60
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2.6 2.B
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S
0 Q VSY1
90
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d
40
30 20 10 0 2
3
/0-92
4
5
6
•
Drowned outflow
•
Free outflow
10.'18
10.98
Calibration of a Triangular Weir
Objective: The flowrate over a weir is a function of the weir head. The purpose of this experiment is to use a device as shown in Fig. PlO.98 to calibrate a triangular weir and determine the relationship between flowrate, Q, and weir head, H. Equipment:
Water channel (flume) with a pump and a flow control valve; triangular weir; float; point gage; stop watch.
Experimental Procedure:
Measure the width, b, of the channel, the distance, PW' between the channel bottom and the bottom of the V-notch in the weir plate, and the angle, e, of the V-notch. Fasten the weir plate to the channel bottom, turn on the pump, and adjust the control valve to produce the desired flowrate, Q, over the weir. Use the point gage to measure the weir head, H. Insert the float into the water well upstream from the weir and measure the time, t, it takes for the float to travel a known distance, L. Repeat the measurements for various flowrates (i.e., various weir heads).
Calculations: For each set of data, determine the experimental flowrate as Q = VA, where V = Lit is the velocity of the float (assumed to be equal to the average velocity of the water upstream of the weir) and A
= b(Pw + H)
is the flow area upstream of the weir.
Graph: On log-log graph paper, plot flowrate, Q, as ordinates and weir head, H, as abscissas. Draw the best-fit line with a slope of 5/2 through the data. Results: Use the flowrate-weir head data to determine the triangular weir coefficient, CWI , for this weir (see Eq. 10.32). For this experiment, assume that the weir coefficient is a constant, independent of weir head. Data:
To proceed, print this page for reference when you work the problem and dick here to bring up an EXCEL page with the data for this problem.
II FIGURE P10.98
/O.Q8
(con'/) Solution for Problem 10.98: Calibration of a Triangluar Weir
e,
deg 90
b, in. 6.00
H,ft 0.231 0.224 0.211 0.192 0.176 0.156 0.136 0.106 0.091 0.088
t, s 8.2 8.5 10.7 12.5 16.5 19.5 27.1 48.2 62.9 68.1
Pw, in. 6.55
L,ft 1.50
Q, ft A 3/s
V, ftls 0.183 0.176 0.140 0.120 0.091 0.077 0.055 0.031 0.024 0.022
0.0711 0.0679 0.0530 0.0443 0.0328 0.0270 0.0189 0.0101 0.0076 0.0070
Q
=VA =V*b(Pw + H) where V =LIt
Q
=Cwt (8/15) tan(8/2) (2g) 1/2 H5/2 where from the graph
Q
= 2.76 H2 .5
Thus, Cwt = (15/8)*2.76/(2*32.2)1/2 = 0.645
Problem 10.98 Flowrate, Q, vs Head, H
0.10
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--I
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.
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1
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jO-9Jf
Experimental
'-------.:----=--
10.9'1
10.99
Calibration of a Rectangular Weir
Objective: The flowrate over a weir is a function of the weir head. The purpose of this experiment is to use a device as shown in Fig. PlO.99 to calibrate a rectangular weir and determine the relationship between flowrate. Q, and weir head, H. Equipment:
Water channel (flume) with a pump and a flow control valve; rectangular weir; float; point gage; stop watch.
Experimental Procedure:
Measure the width, b. of the channel and the distance, PW' between the channel bottom and the top of the weir plate. Fasten the weir plate to the channel bottom, turn on the pump, and adjust the control valve to produce the desired flowrate. Q, over the weir. Use the point gage to measure the weir head. H. Insert the float into the water well upstream from the weir and measure the time, t, it takes for the float to travel a known distance, L. Repeat the measurements for various flowrates (i.e .• various weir heads). For each set of data, determine the experimental flowrate as Q == VA. where V = Lit is the velocity of the float (assumed to be equal to the average velocity of the water upstream of the weir) and A == b(Pw + H) is the flow area upstream of the weir.
Calculations:
Graph: On log-log graph paper. plot flowrate, Q. as ordinates and weir head. H. as abscissas. Draw the best-fit line with a slope of 3/2 through the data. Results: Use the flowrate-weir head data to determine the rectangular weir coefficient. Cwn for this weir (see Eq. 10.30). For this experiment, assume that the weir coefficient is a constant, independent of weir head.
Data:
To proceed. print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem.
III FIGURE P10.99
/0-95
10,99
(con'/J Solution for Problem 10.99: Calibration of a Rectangular Weir
b, in. 6.00
Pw , in.
L, ft
6.00
1.40
H,ft
t, s 2.2 2.7 3.0 4.2 4.5 6.6 15.8 23.8 38.4
0.254 0.216 0.184 0.162 0.151 0.111 0.060 0.046 0.031
Q, ft"3/s 0.240 0.186 0.160 0.110 0.101 0.065 0.025 0.016 0.010
V, fUs 0.636 0.519 0.467 0.333 0.311 0.212 0.089 0.059 0.036
Q
=VA =V*b(Pw + H) where V =LIt
Q
= Cwr (2/3) (2g) 112 H3/2 b where from the graph
Q
= 1.79 H 1.5
Thus, Cwr
=(3/2)*1.79/(0.5*(2*32.2)1/2) =0.669
Problem 10.99 Flowrate, Q, vs Head, H
1.00
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.
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H,ft
/0-96
I •
Experimental
/0./00
10.100
Hydraulic Jump Depth Ratio
Objective: Under certain conditions, if the flow in a channel is supercritical a hydraulic jump will form. The purpose of this experiment is to use an apparatus as shown in Fig. PIO.IOO to determine the depth ratio, YZ/YI> across the hydraulic jump as a function of the Froude number upstream of the jump, Fri' Equipment:
Water channel (flume) with a pump and a flow control valve; sluice gate; point gage; adjustable tail gate.
Position the sluice gate so that the distance, a, between the bottom of the gate and the bottom of the channel is approximately I inch. Adjust the flow control valve to produce a flowrate that causes the water to back up to the desired depth, Yo, upstream of the sluice gate. Carefully adjust the angle, 0, of the tail gate so that a hydraulic jump forms at the desired location downstream from the sluice gate. Note that if 0 is too small, the jump will be washed downstream and disappear. If 0 is too large, the jump will migrate upstream and be swallowed by the sluice gate. With the jump in place, use the point gage to determine the depth upstream from the sluice gate, Yo, the depth just upstream from the jump, YI> and the depth downstream from the jump, Yz. Repeat the measurements for various flowrates (i.e., various Yo values).
Experimental Procedure:
Calculations: For each data set, use the Bernoulli and continuity equations between points (0) and (1) to determine the velocity, VI> and Froude number, Fr\ = VI/(gYI)I/2,just upstream from the jump (see Eq. 3.21). Also use the measured depths to determine the depth ratio, YZ/YI' across the jump. Graph:
Plot the depth ratio, Yz/Yl> as ordinates and Froude number, Frl, as abscissas.
Results:
On the same graph, plot the theoretical depth ratio as a function of Froude number (see Eq. 10.24).
Data:
To proceed, print this page for reference when you work the problem and click here to bring up an EXCEL page with the data for this problem.
III FIGURE P10.100
(con'/)
(con)')
/0./00
Solution for Problem 10.100: Hydraulic Jump Depth Ratio
Yo, ft 0.855 0.759 0.691 0.578 0.492 0.414 0.289 0.248
Y2, ft. 0.404 0.386 0.367 0.337 0.308 0.280 0.233 0.211
hft 0.055 0.055 0.055 0.055 0.055 0.055 0.055 0.055
V" ftls 7.19 6.75 6.42 5.83 5.34 4.85 3.95 3.62
Experimental Fr, 5.40 5.07 4.82 4.38 4.01 3.65 2.97 2.72
Theoretical Fr, yh, 1 1.00 2 2.37 3 3.77 4 5.18 5 6.59 6 8.00
Y2/Y' 7.35 7.02 6.67 6.13 5.60 5.09 4.24 3.84
For flow under a sluice gate: V, =[2g*(yo - y,)/(1 - (Y,/YO)2)],12
Theory:
Fr,
=V,/(gY1)112 Problem 10.100 Depth Ratio, Y2JY1, vs Froude Number, Fr1
8
------.-
-------------
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2
3
4
5
6
7
- - Theoretical
/0./01
10.101
Hydraulic Jump Head Loss
Objective: Under certain conditions, if the flow in a channel is supercritical a hydraulic jump will form. The purpose of this experiment is to use an apparatus as shown in Fig. PlO.lOl to determine the head loss ratio, hJYI> across the hydraulic jump as a function of the Froude number upstream of the jump, FrI' Equipment:
Water channel (flume) with a pump and a flow control valve; sluice gate; point gage; Pitot tubes; adjustable tail gate.
Position the sluice gate so that the distance, a, between the bottom of the gate and the bottom of the channel is approximately 1 inch. Adjust the flow control valve to produce a flowrate that causes the water to back up to the desired depth, Yo, upstream of the sluice gate. Carefully adjust the angle, e, of the tail gate so that a hydraulic jump forms at the desired location downstream from the sluice gate. Note that if e is too small, the jump will be washed downstream and disappear. If e is too large, the jump will migrate upstream and be swallowed by the sluice gate. With the jump in place, use the point gage to determine the depth upstream from the sluice gate, Yo, and the depth just upstream from the jump, YI' Also measure the head loss, hb as the difference in the water elevations in the piezometer tubes connected to the two Pitot tubes located upstream and downstream of the jump. Repeat the measurements for various flowrates (i.e., various Yo values).
Experimental Procedure:
Calculations: For each data set, use the Bernoulli and continuity equations between points (0) and (1) to determine the velocity, VI> and the Froude number, Fri ::::: VJi(gYl)IJ2, just upstream from the jump. Also calculate the dimensionless head loss, hJYl> for each data set. Graph: Plot the dimensionless head loss across the jump, hJYI> as ordinates and the Froude number, FrI, as abscissas. Results: On the same graph, plot the theoretical dimensionless head loss as a function of Froude number (see Eqs. 10.24 and 10.25). Data:
To proceed, print this page for reference when you work the problem and click ht're to bring up an EXCEL page with the data for this problem. Sluice gate
Point gage
1/1 FIGURE P10.101
Pitot tube
( con)t)
/o-qq
IO./O}
(con J/)
Solution for Problem 10.101: Hydraulic Jump Head Loss
Yo, ft 0.855 0.759 0.691 0.578 0.492 0.414 0.289 0.248
Y1, ft 0.055 0.055 0.055 0.055 0.055 0.055 0.055 0.055
Y2, ft.
OA04 0.386 0.367 0.337 0.308 0.280 0.233 0.211
Experimental V 1, ftls Fr1 7.19 5AO 6.75 5.07 6A2 4.82 5.83 4.38 5.34 4.01 4.85 3.65 3.95 2.97 3.62 2.72
hL, ft 0.364 0.313 0.271 0.201 0.152 0.117 0.058 0.042
Fr1 1 2 3 4 5 6
hL/Y1 6.62 5.69 4.93 3.65 2.76 2.13 1.05 0.76
Theoretical Y2 /Y1 1.00 2.37 3.77 5.18 6.59 8.00
hL/Y1 0.00 0.27 1.41 3.52 6.62 10.72
For flow under a sluice gate: )1(1 - (Y1 / yO) 2 )]1/2 V 1 -- [ 2g*Yo(- Y1
Theory:
hL/y 1 = 1 - (Y2/y 1) +Fr12[ 1 - (y 1/Y2)2]/2 where Y2/Y1 = [-1 + (1 + 8Fr/)112]/2
Problem 10.101 Dimensionless Head Loss, hL/Y1 vs Froude Number, Fr1
10 -
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10-/00
5
6
7
--+- Experimental I - - Theoretical
I
11.1 As demonstrated in Vidt'll \'11..1, fluid density differences in a flow may be seen with the help of a Schlieren optical system. Discuss what variables affect fluid density and how.
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,tdelll /
(/11
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e = Rr ..e.. ,$tJ
elf(I"I~J
pressure) ?) '/tJ/Y;aniJt?r c,f)(J/'n;
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11.1 Describe briefly how a Schlieren optical visualization system (Video" V 11.1 and Vll.1, also Fig. 11.4) works.
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.113 Are the flows shown in Videos V Il.l and VII ..:! compressible? Do they involve high-speed flow velocities? Discuss.
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1 11.5 Air flows steadily between two sections in a duct. At section (1), the temperature and pressure are TI = 80°C, PI = 301 kPa(abs), and at section (2), the temperature and pressure are T2 = 180°C, P2 = 181 kPa(abs). Calculate the (a) change in internal energy between sections (1) and (2), (b) change in enthalpy between sections (1) and (2). (c) change in density between sections (1) and (2), (d) change in entropy between sections (1) and (2). How would you estimate the loss of available energy between the two sections of this flow?
may be
(b) C1,11.9
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I 11.6
Helium is compressed isothermally from 121 kPa (abs) to 301 kPa (abs). Determine the entropy change associated with this process.
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I 11.7 Air at 14.7 psia and 70 of is compressed adiabatically by a centrifugal compressor to a pressure of 100 psia. What is the minimum temperature rise possible? Explain.
The
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YYlln I /'JII U Wi
fY/c li(}n less
wtJu /d (Jew" w / Ih
rise.
process /,4/hlch inv/)Ive.r
uns~nt
tf
-Flow. A Cc..lh'dlllj ft; -Ihe se.GQnd klw of fhe
enfyopi'
procesS' ~
Musf
Cdnnof
i';
illttsfvedef
IhCret1se. or
fhe
how
fheyJt!fodYJ1tJl'JlliCJ)
Y~1I11?
d~ cYetl.fe. .
The
en-/rt;I'Y /)y
E9' S. 101
d iOf YIIJII? f'Ke !rhed be/ow
p~sS' ~$ulfr I~ a. n-,I,,/mum 'kRl~K.
i5r3nfn:,pic
r~e.
~~
s Far fhe
isenfr-op/{.. pYocess J EI'
T (lut
- ~n
mi"i,.,/.(""
is- va/ld·
J /.211-
&=1
(P.. 7J"" t
= (£Jo'R)
f>,;"
()
T.
11-
()
6
'-i:..!
/fof)P$iA.)
(I¥. !114 7
_ '117 ~ - 530 ~
In
7iuAs I
.
lind
T/Juf min ;,..14""
)
c:U.iYl,!! 4n atllablfle-
CdI1.J'htn7
7-.s
Ifen~ic
I.:
1/7 "I(
fl.g
J
11.8 Methane is compressed adiabaticallyi from 100 kPa (abs), 25°C to 200 kPa (abs). Whati is the minimum compressor exit temperature pos-i sible? Explain. i
me
ctJmpressO'Y ex.t'f -fempeyt~lur-e
m/n/murn
ad/al,~tlc
elr/rtJp},
Cll1d
bY
Ir)cf,(jnks..s /JYlJces$
i$el'rlrr;plc Flaw.
fherm()tilrnqm/c.J.;
£9'
wtJuld occ..u~ tv/fJ, tin
whic4 invc/ves a unstzrnf
A cceJYdI~ -10 /he.. '5ec..()~d law of
5. /0/) -Ihe eh1-rojJ,Y m",.sf
/17c..yetlfe
or
adiabalic pr(Jce.5~ if- CQI/'1I'1Dt clecrel1(s~. The T-s d/tl1Ytlm .riefcAed 6elM ;I/t{sfraks htJw fAL iseh-iYPpic pn;cess resulf:> I;' D-- /()~r exi f fe?npeyt14"e, -/hQJ'l I1l7y
'Yemcl!'n
ctJ;;;-/lttl1f
aciut:d ad /a6a{;c
dllr1n.J
py()cess
Qn
j;elwe.e~
.fhe.
Sdl'J7e
/1ressuyes.
~v.t
T
S
C/mc.lutte ~ ~ .//.2'/ Ihal
Or
=
3?/1<....
11-7
/I.
q
I J 1.9 Air expands adiabatically through a turbine from a pressure and temperature of 180 psia. 1600 oR to a pressure of 14.7 psia. If the actual temperature change is 85% of the ideal temperature change. determine the actual temperature of the expanded air and the actual enthalpy and entropy differences across the turbine.
To
deret'm,;'e -th.e aclvtal fen;1pey~fwr'e. of I-l1e expal1atP/
ac~1
ei/lfttall''1
t:iJ;Jd
/D delent1lne -lite
{ii'sf
lhe ide,tI 7elNlperalul'e. ad ialott-n'c-
enn--ol'Y
idf'tJl fetnperalu~e cht:tJ19~
fYiV'h4Y1/~SS
IifHfIJ
The t:tchtetl Pl'lXefJ Ihwltl~s ~
/Alilh
T-
S
d;ffenHcef
c//&:~?r,1;a;,
iJ~}1 frt7pic
11uI;
6e/gw.
.r-kefrh
11 . 2 '/
~t(f iel~~1
5ince
~lJ.t
1cA..&/
T. ) = ~
(lcr(JSf
the
we nted
lite fllr6/ne.
turbine
eYpt:l¥15Ipn.
Jmallw ~walwt'e ct,tlJ1!~ as ;Iklh-tkd
t:l
s
(
IICl'tJ$f Me fuy6/ne
Chan!e
P,~
£1.
lJJ1tJ1
IAe. lurb/ne is tiss()c/ak, wi#' an
t:lCY()$>
t:lJlJd
air
O. 85 ( T
Duf
-
ffl
it:l~« 1
11-
7. )
g
11·10
I
11.10
An expression for the value of cp fori carbon dioxide as a function of temperature is ! = 9210 _ 3.71 x 1()6 8.02 x 108 p C T + P
where cp is in (ft . Ib)/(slug . OR) and where Tis in oR. Compare the change in enthalpy of carbon dioxide using the constant value of cp from Problem 11.2 with the change in enthalpy of carbon dioxide using the expression above, for T2 - TJ equal to (a) 10 oR; (b) 1000 oR and (c) 3000 OR. Set TJ = 540 oR.
'"
con5~1
v
c;" fhe cl1lflJ1,Je In e;r~~lpy.J h, - 1,,) ~y b~ eVdlfA~fepi will? Gf. I/. '}. rhus} {f;v
(h: -{):=
C
VtlY)'lY:J '"
So)
,,,,,sIMt .,
Fur
-e VtlluaW
wi th
v
I~
v
h, - hi"
(T~ - ~
)
6'{ I/. 3.
'"
Thus"
r~
J ('210 -
I
r;
(tJ.JhrY
9210
(rl.- T,) -
3.71 XIO'
'it
= 5'/-0 (lK
aAltd
J
k tI~) - ~.OlX IO~J.. ..!-) T ( 7: _
I
1-
T. I
~ = 55'0"R.
f~97 /'.;.. /6 ) $/"" .1(
aYld v
T"l.
T
Oy
2.
~.()Z X/Of) d r
"i.7(XIO '+
r;
7;
('h - h):::. ""YI't!,
v
cha)1~ 'M enlJta/pYJ h'J, -4, J h1~ Ix
the
c" dT
v
If
( n,,- hi) VtMY/~J c.,.
51) 000
fl. /1, SI"
(emil)
(sSO 01<_ 5''M of() :::: if 9,{}()O -li-.lb $iu.J
II. 10
I(con't)
and (.hv
2
V)
-I-,
J
.
VtlYtf/~
C"
(V
v
h - ~) Varytn.J . 2.
7
-::: 2 . / -, X 10 0
11-. /6
,
7 //.//
I
11.1 1 Why?
Does sound travel faster in the winter or summer?
firy.
ClIY"
~ ~ S"peL4
tJr
.f4Jw1d ~ EI' /1.3,
iJ
c -= Vf
Sf}
or
fla"l'?7eY'~
w/'el"l
r
p",p{
f1t.u.s //-10
C
If
j,~"),er:
1/·J2.
T
Il.12 Estimate how fast sound travels at an altitude of 250,000 ft above the surface of the earth.
~
eg. C:;
11.3b
VRTIrI,
~J;J ol-lit.,de ,,1 2S'O, ddtJ .f.J. ,) ~ Tlltb/~ C·I ~ ~r~~1I1e ~~ -Iv k - 81.77 6F 5D Af
C
= I f(?'6 fl. ~' ) ('160 -R_ ih) XI.l{-) Y~ siCA,. 'R It Ih) :: ( SII#-7.!!.. s').
1/-11
11.13
J 11.13
Determine the Mach number of a car moving in standard air at a speed of (a) 25 mph, (b) 55 mph, and (c) 100 mph.
raflD of ~Cd/ ve/()cify -10 Jleed D/.rIJtmri.
The M4ci1 nUY'1ber IS /he 711 us
Ma..
h
_ V
- -c
~ /t) J1d aypl a /y C ::
or
VRT*
=
_ (/117 If)
06()O L ) hr -
z 5'
I'1"PJ,
= O.03Z8
761.{, mph (h) FI) Y
V.::!i5 mf h
fVlA.
=
5~ n?/h
761.£ mph (C)Hrr
_
V = 25 WlI'~
MfA. .::
V
=
0.072 '2
~ 100 rnfh
Ma :
lao mph
/117 fl-
s
sj(~2fO:~)
C (a)Foy
0:
O· 131
76/. b Y"ph
1/-/2..
76 I. 6
mph
II. IS I 11.15 How would you estimate the distance between you and an approaching storm front involving lightning and thunder?
(J¥'J,e.
w()..!
ft; e$-h~"ie,
and f),ffYrJac,hu;'J fhe
1?/..{l'Wbey
/ i!t.thJj
fhe d/f~J'lce ioe-lwf'en YblA
Sffrm c./oud.$.J
tJ/
seco;ulJ
/
f
/
x) ;s -frJ C~UI!II
befwe.en 5eeJ;'q fj,e
fJ7
./
hear;, !-""ouler. U Pi'! ~'I'Yt?x"'ntf,(f(; value 01 lite, sl'eetl 01' .s1J"11~ IIf~
tlJ'It/
fj: (see
dJf/-r;.nct-.I}(
Table
B.3)
~
1/-/3
~
CdJi
PjJjJrtJx;~/e
/ /. /6 J
11.16
If a high-performance aircraft is able to cruise at a Mach number of 3.0 at an altitude of 80,000 ft, how fast is this in: (a) mph; (b) , ftls; (c) m/s?
(6) Wi~ Etg. /1, 'It,
V
== (Ma) c
eo, (Jot) 71- /n U· 5· siz:lYJdayo/ f?lrn",oS',Phere/ w~ ~vc ~ -fhe so lu fio 1'"1 0 f- p rob/.e;-n / /. / ~ ahd
af
C. ~
q7K
fr s
1hu.5
v ;::
C-~.o )('178 ~f ) ==
2.'130
t!.
=5
(cn lhel't =
(C)
2000 mph
Also 393 m s
II : :
1/-1 If
/ /.1'7
I
1 I, l7 . Compare values of the speed of sound in m/ s at 20°C in the following gases: (a) air, (b) carbon dioxide, (c) helium, (d) hydrogen, (e) methane. Give one example of why knowing this may be important.
To CalUAlak -fhc 5flted. of sdund ~tj. //.J(,.
lAse
C :.
111
tJfn /dea/ 91lS ~ can
Thus
tfRT~
'*
Wi!t1 va/ueJ of R aMti
{Y7Jm TaJ:,le /.7 we o6h/n
tl/ r
(AJ f,;,y
c
=
-
I()(JS In
.s
-
.5
(e) ~ C
me /lulYJ e
__
--------------------~---
(5iB.1.I.
-1J.9. K .
fJ I) (I
l' }.}.Pt
)
(7!13 K
~~
11..9 In
) ('.31)
)
(con't) 1/-/5
'f'f6
//·17 I ("on 't )
ie$-I-f fA
of Tuybtll'YJ« ch/Y)er
~Qf
fAlAt
If) w~ Y 'S"l'eer/ of flJLl he! if ftJmef,-WlCJ fo hl!A - MA.ch -l?tlm6er t:JP~Y'~fJ'f:Jn
be-II c.1, /'e vet?!
-Sl'~ed anti t,e~e }'Y\~chQI?/c" /
II. I tj
A,
w"fh les{
le.f'S" ,..fa lit)nil" I
sf~~.r;~J
PIY-1-S of fhe
-ferf
()1
fhe
rj.
I
J Lib If a person inhales helium and then talks, his or her voice sounds like "Donald Duck." Explain why this happens.
ThL 5r<-t~ n1He s ftw. 50u n dJ
of .s~j II? p.t.t ~ of
he/,',-
;!' "~(JY"; -/1,Yee
/I1AA'Id I;' a I r f Ilk if- d~f when 5fea./c.'j f
.I
11- /6
D
I/J...
pe-Yf ImJ
fJ"YVVtJh
VI)" u-
he//un-"I.
II. t q 11. Jq Explain how you could vary the Mach number but not the Reynolds number in air flow past a sphere. For a constant Reynolds number of 300,000, estimate how much the drag coefficient will increase as the Mach number is increased from 0.3 to 1.0.
Cans"del"ln~ ~'r
Mo., o.s
\II(.
can -f1
v
v c
MAo=
ide"l ~~S I
QS CAn
(I)
1he. Re.~r'l.o feh R~::: ~yd -== _
?~d RT)It
}.A
Loo\::.lt»q at ~\A&\.(.10tls. \ Q~ '2.. \iJtC. v(a~n 1WJ..t IN'<. C4n v(J.'l\j Mo. lA)~i\~ k.o\O·l~ Re. cd)\Sta.rf ~ va.v~{It'~ V A~d P OV\\~ v..J i +~ P V held c.Dl\c;i'lllf\.;' . SFro M +kc D)'(7Af h Ioeto w wcz. C61\c..lUGle +ko.-\- a,1- Re -:: ~ ~ to) d.ra.9 c.tdf;"'e.n1 'I~C)C.<;' fro~ O,Lt1 +z> 0.15 C(+ Ma.
the
~
,ACr.eo.<;eS
o. ~ hI, D . Ma
= 1.2
2.0 3.0
,,1.5
1.0~~~~~~~~:z~~~~~~
4.5 0.9----.-------
1.1
O.B - . - - - - . - - - - - - - - - - - - - ; - - - - 1 1.0 0.9 O. 7 ~;:':;";;;~::;-'.:.;.;"::; ... :::::::::::: ..--::;-:,.-:::;--=:;--;::==:;:::::;::::::::;1
0.6
r----1r----.--...E-
CD 0.51"'""'_ _
0.4
0.2--- .. 0.1 _.
. - - . _ - - " ' ...
__
..
---------
o:---"--_ _ _ _ _ _ _ _ _ _- - - l 2
3
4
5
6
7
Re x 10- 5
11- 17
B
9
FIGURE 11.2 The variation of the drag coefficient of a sphere with Reynolds number and Mach number. (Adapted from Fig. 1.8 in Ref. 1 of Chapter 9)
//·20
J 11.20 The flow of an ideal gas may be considered incompressible if the Mach number is less than 0.3. Determine the velocity level in ft/s and in mls for Ma = 0.3 in the following gases: (a) standard air; (b) hydrogen at 68 oF.
Frowr bCO' / /.46 we h.ave.
V = (Ma.) c when C/JrYt6;}u"d
{;Jp,ic/r.
V
wifj" ~~. II. '36 / leads
Iv
M~ VRT"k
=
v::
(I)
o. '3
A/so
v = (b) For A/so
(335
fJ) (0. 30'l-K ; )
hydrogen
D
af 68 F;
T:: S'21 d/(.
~
/02
!!!
=s
R =-;.'I-")(I()~
lhus wilh E1' /
ff·
l /:;
anel-lt=/''11
~ Tt2lJe /-7.
s""J . ~
= 1'290 !!
===
II/so
1/-18
J
11.21
11.21 At the seashore, you observe a highspeed aircraft moving overhead at an elevation of 10,000 ft. You hear the plane 8 s after it passes directly overhead. Using a nominal air temperature of 40 of, estimate the Mach number and speed of the aircraft.
x = V./;
Mp. =
SinK 1
V
-
(/ )
C
Also hllltJr
{om b;n/~ ~/n
- -r Vi c1 L
()(
( 2.)
/~d
r 5'1'"
::.
we
tJ6~/~
tJ(
t
C
{bStI(
z.
= /OCfh Then
=
IX ::
aM-cI
Ma,
~
- 121.7 •
0
o
2.0g
SIlT
Fuy-/hw
V =
23.7
(M~)
c
= (2.0!)( 1096 ~f)
If = 2z7° s
-
If s
If. 22
I 11.22. A schlieren photo of a bullet moving through air at 1:1-.7 psia and 68 of indicates a Mach cone angle of 28°. How fast was the bullet moving in: (a) mis, (b) ft/s, (c) mph?
W'-Ih
&ps. I/. 31 tl/luJ //·36
V
For-
~
SIn
tJ /r
frcrm
(b)
=
0("
:::
we
obk//1
VR-r-k Sin C\""
af
fable
Thuf
/
v
-: 2'-100
-=-
If
=s
(CA.) or -:: 732.
-
s
(c) and
//-2.0
ihSto#tlneOUs loeo.fion
/ J. 2.3
of lhe point 11.23 At a given instant of time, two of the pressure waves, each moving at the speed of sound, emitted by a point source moving with constant velocity in a fluid at rest are shown in Fig. P 11.2:J Determine the Mach number involved and indicate with a sketch the instantaneous location of the point source.
S()ur(e 0.1 m
1+----0.15
7he
Mach number
p(),~f
S~uyce
wi /I,
E'IMti
In V()/ve.r/
1/.3'1
w; fJ,
tlIs,flJc/ak4 I~
-fJ,e
~J
FIGURE Pll.2 j
Ine
= we O. OJ ""
(!).
p. dlm)(~./S"'" t- .£) I. :.
(d.
()/o/
/1+7
(J.IS"1?1 +- L
.R :::
0.1 ~ )
(f). ISm) ....
(~. ()f! m)
0.0,
J'J?
O.
l'I1onlJYI
skeM ab()l/(! i5
as $ho~YI be,/fIW.
Prd»? ..; h e.
m--~
!7- 9'1
= I." 7
L Of 0/67
/?'I
DT
8/1$
f1Je.
lIy tJ614111e4
IJ.2.if
11.2+
At a given instant of time, two of the pressure waves. each moving at the speed of sound, emitted by a point source moving with constant velocity in a fluid at rest are shown in Fig. Pll.21t. Determine the Mach number involved and indicate with a sketch the instantaneous location of the point source.
ifJslal/T4neIJUS I~on
~f
fhe pllilf Soulte 2 in.
10 ,
'
In.
sci
:C(i-tW/M)
-i
~ .:;J
Vt
FI GURE PU.24-
(I)
c t wlf,v-& J.I ~, -f'roYJ,t
fh e.
C (t --C~lIe)
0:;:
Sk.e-k~
~ In.
We hrv/~
aloove.
-::: C t
-
c tvJ"~ = /0 In. - cfwa~
ThIA~ , C ..J I. wave ()J1d
w,'ft" M~:::.
A/so
Ma
::.
--
/
£1'
I.
CJ In . - 2 ,'". ...... 8
/n .
SIn. SIn.
vt
ct
-=
vt
-
10,n .
~.h2S"
/lutf,
Vi
/0-025 )(/1) ,;.. )
':::\
6.2.S iI'],
/ /-22.
11·25'
J
11.2~ How much time in seconds will it take for the "bang" of a firecracker exploding to be heard after the blast from 200 yards away on a standard day (see ',ilh'o VI L5)?
we..
CCln
///7 .f.J.
s
We +he. s-peed (Jf sound t'n rlaYldarrf a/r .> ) fo fief the lime e.rh~afe SPtAJkfj ~ wilt,
t::
:s. -;: c
11·26
(2~() yJs ) ( (1117
3ft )
-=
f!)
0·5 s
J
j 1.26 Sound waves are very small amplitude pressure pulses that travel at the "speed of sound." Do very la:ge amplitude waves such as a blast wave caused by an explosIOn (see Vidl'O \- i 1.5) travel less than, equal to, or greater than the speed of sound? Explain.
The ~peed ,,{
,,$PUl1d IS
fhe speed al wh/c.h a V'J
pl'eSSUye dl.!fuy/,ance -frfAlle/.r Ii,YdUft.. f1u/J and it re?Ye.sent-..f fAe h'J/Y)/mtlPl .r;;-eecl f-lJiJ di.rlurbO)'Jce. Fin; Ie pl'~t!tlye c//.rlurbl'u')cef
infinire.>/md'/
a 01
fral/~ I
#.ft-e;- I-h~J-'J
hectlu.le {he /arr;ey f 1f.ffute diffe"ence. pcfs pf a 4y/veY" 61
fttf"/w
$t;tlnc/
m(}lIf:,»,)enf.
/1-23
waVeS'
/ 1.27
I
l1.27 Starting with the enthalpy fonn of the energy equation (Eq. 5.69) show that for isentropic flows. the stagnation temperature remains constant.
we have -I-
fiwv fhe
Fir/ ise.n-fr.6t;c.,
J{e,.( 1,..)] =-
en~J
P. hd (Qnel-:: O. S hfYlI/lH#)., v v '2. 1,0 =- h -I- ~
~ef
fl!.YYlIJ,'".r
,,,
of
W
s/, .. ff
/tef i"
tI/YJJ-Jm"f
en ttIf2t,.J if rJeH1'Jed t7 J
z
~o / ~ 1?e,jtj/.b/e cknfe In ele va. --hdn (ol=~ .;;, JQ.re;) a 17d ho .rA.tlft
IN'()'(k...1
V\j.t,(.ff)
//- 2'Jf
fheh
".2e
I 11.28
Explain how fluid pressure varies with cross section area change for the isentropic flow of an ideal gas when the flow is (a) subsonic; (b) supersonic.
Wi#'
"f
£"
chan~
In
help
With -the
dYeA.
dp
I' V').
:::
(b)
11.29
CA.}')
C~mmenf cJn
isen--h-o,o/c. f/(JW.
tin
FroJ'V1
haw 'Y~UYt. 13,.
VPYlCS
//.'/7 w~ t7b~/;"
( t)
ft
£1' I
raj Ftr; subson/c, fluw I
of A·
(A.e
c/ A
(l-tvla2) vJ"a~ges
II. l/ 7
If A
sU1~.rIJ 111 C f"ep..{es /
FttY su.perftJXic fldW,J
~. I
ro
I-/ A
cAaJ?!es of A.
SlAjlfeIir
-fhaf- Ch4;'J9~S of P p
Incretl.Jef
f/)/I~w
~nd v/c~ verst{.
thai cht1I?~J of p are "PfHf;ft
J;'crease.f /
P clec~a.Je.J and VIC~ versa..
1 11.2.9
For any ideal gas, prove that the slope of constant pressure lines on a temperature-entropy diagram is positive and that higher pressure lines are above lower pressure lines.
r:ri -
.fw..
tlYJd since v
d t,
hcwt.
(I)
Cp
h'e CtJYIclude -femfJeva.lurt. - en-lrupt
-fhdf the.. 5/ope of a Ct/J'Istanrd/atjY4fi1 1$ p"sif/ve·
pye.rJu~
line
~ fl' /1. ZLf we CUyzc!ut:!e -fhal
3= :: P, fDy ~n-y
~. //.7 is' VA-It'd l We
= I.
E1'/
t-UYfl,eY,
gel!
f
ds
a
an iderl/
::: C d T
and -thus dr Wif~
-r
(ii (1,) 1; )
iseh~/c PYlJce.ss
lowe yo pr-e.JSU1'(.
1;J'?~s
In
h1yher f'lesJtw(. lines fempeya:fu*_ - erd-Y17py c1;~!YAflfS. aJ1cX rhus
/I-ZS
t:J
re abpve
0)'1
//·30
I
Determine the critical press~re and temperature ratios for: (a) air; (b) carbon dioxide; (c) helium; (d) hydrogen; (e) methane; (f) nitrogen; (g) oxygen.
11.30
1he- crifjc..a/ ideal
preSSUye a y-e
90S
rafio
t::)s.
fyd'I¥!
I
Crih'ca/ tempem/uye raf/o -iOYoY'}
an.d The /I.~ /
and II·
63
..iL
p~
=
~
(2)* -Je .,.,
(I)
I
and
Ttl ~
(a)
'k +
2.
::-
rdY
(2) I
1<=
air)
fr.uw. Table
/·I./-tJ
/·7. Thus
,.'1<;
p*
~ (/~~I
Pa
) ' . . -1
=
,
6.5283
and
z ~
',lfo+1
(b) I7r; carbon
f
0.83'$1
~
d/()(/de ~3o
:=
F:,
k = /·30
I
/2- ),y;:; L~
-: :
/.'301"1
ftrm"
O· S'"ifS7
and T¥-
0.3676
=
(c)
m
hej,'u m
I
-k =/.(d,
p* -= I~)'~ Po ( /·6t+!
f. 66
-
O.l{-Kg I
()nd
T* _ ~
-
2-
=
0.75'19
( con't) 11-2.6
74b 1c 1·7.
Th(J~;
/ /. '3 0
I (con If)
a~d
r* ==
2
-
r;
PII+I
ThusJ
()nd
z
r;
0.1658
1·11+/
(I)
fm;. Tdb/e /. 7. 0.£28 '3
T
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~ (g)
:=
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2
/·'10+1
Fw tJxy'jen
P
P..
~
:;:.
I
(2
6. 83sj
::-
k ~ /. 'kJ ~ 7J.l.le /. 7. JA.u~ ~ ',110-1
):::
O· ~28J
/.'forJ
tJ
and T~
=
Z
O.KJJ'J
//-2.7
11.3 ,
Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross section area of the throat of the converging duct is 0.05 ft2. Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia; (b) 5 psia. Sketch temperature-entropy diagrams for situations (a) and (b). Verify results obtained with values from the appropriate qra.ph in Appendix D with calculations involving ideal gas equations.
Th/s problem is .!";rn;/ar -10 Exam'ple II· 5" Th e.- yy,t:l5 s fiOWVlff fe.. is' db fa IJ? e.d a f Ihe fhroAt w//h Gj. I;. Sit?
rn ": fth A fh ~h The ihr()af
To
Can b~ obhlJ1ePl
de.n!iiy
t1~ ~ /1 de(e,-in,),e.
r
I
r
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(~) M4 ]i':i 2
(2)
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2.
the.
with E'J' /l.btJ.
thnsaf
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n /A"...w
use.
INt.
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The .If fhen
fhen
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The l/eJocJty af fhe
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~"hed
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=}
4h T~
IS
15
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is
f/()W
(J
vr-R.-~-It(-
6/1/11;' etl
wi~
~f' /1·56'.
7;.
= / 1-
Thus
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(5)
(-k.; ')Mo.+~ (~n'f ) //- 2. 8
7JJ1I~
I/, j
(con If )
J
(a]
F~
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J
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+ (I'~ - ') (0. 7618 )
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we ObfaJ;'
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(k07)(IO-7;~;J{(J.05ff1J{g/)6.Z
=
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::
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value
Met
:=
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:::
Wi Ii.
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//Iliff,..
0.66'
1'1. 7 f1J iA.
The
finr
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{'J
=
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9
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we. rei ~
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) -: :; (0. 9
)(s/r~) :: lI6 7 ~
~
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11-2q
1/·31
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::
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) ( 2.'S.? Xlo -1 J/~)
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.: :
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klh ::
)(1019 ~) =
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)
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1/·31
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T
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/
Po T
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0
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pi
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1*
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Tl\'
/ " ' / t:.e s
11-31
1/·32
I 11.32
Helium at 68 OF and 14.7 psia in a large tank flows steadily and isentropically through a converging nozzle to a receiver pipe. The cross section area of the throat of the converging passage is 0.05 ft2. Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia; (b) 5 psia. Sketch temperature-entropy diagrams for situations (a) and (b).
]his p't?J6/eWJ is
The
The
- f.fh A-It..
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r
~
=
To cle-bym,'n t
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af- fAe #JYrlai w;1t,
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and Wi It,.
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With [1-/
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iN£'
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03
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t;,
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(aJ
(1:»
I---~~---- T
*
5
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II· 33
I 11.33 What is the static pressure to stagnation pressure ratio associated with the following motion in standard air: (a) a runner moving at the rate of 20 mph; (b) a cyclist moving at the rate of 40 mph; (c) Acar moving at the rate of 65 mph; (d) an airplane moving at the rate of 500 mph.
M~
we.
(/)
can
(11.5CJ) use
we
:: 1/17
ff .I
C}y
-=
ft!ta.,.
20 I'kl''''
-=
16/. ~ I'kPh
~.OZ62.
r
E.. Po ::"
L
(6) FOr V ~ 'f~ M tt;
I-tmph
'10 mph
76 I. h
O. () S 2 S
:::
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.e~
(C) ~
-=-
v::::
M~
-::;
[
J J+ ~17-(O.OS2.$/·
6 ~ n?'ph
C
Ii : :
=-
0, tJ i"S'I
( + ().
I "2-(0.08'5'+) v::: 5"OOtnph
(d) h /l1~
== 5ZJP rYlfJh
"3
tin.d. ~
=
[
I+O'~
S
., 7 . -""
o. ?9'frt
J
-::- t11. 65" 6
7('/.6 PIP/"
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o. q98
=-
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aJ1.d
7
3. t:;
til1d
.l (o.f/i6)'2.
7·5
J
/1-35
:::
~'7,+9
1/·34
T 11.34
The static pressure to stagnation pressure ratio at a point in an ideal gas flow field is measured with a Pi tot-static probe as being equal to 0.6. The stagnation temperature of the gas is 20°C. Determine the flow speed in m/s and Mach number if the gas is (a) air; (b) carbon dioxide; (c) hydrogen.
(a)
To
defeY"'lne -fh~ flow .s~~d
rhe
p;~S5UYe.
stan'-
the
fo s-htfntfJlitJn
f~mfJen:llure /
S-1-4jI'Za/li(;1'7
9/ven
M4Clt number hpvl1! ke;r j/ven
IJnd
7;
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we.
yah~ el1f~1'"
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I
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(JII'Jt:/ yea4 Fd vt/lue 0/ Mq. lkus
I/o/ue
In
wilt,
Fig. iJ./
MIA = 0.89
rdY"
M~
--
t;·19"
='
I ::.
F;9' /). I
fIves
1).16
7; p ~ tJI
/'tf,u s
T =-
(h)
Fol'
(r) 7;
1,;
~ (t).86
C4 r bIn
Thu5 we halle
r.
1ff~ -)-11. -j(t)
Ma =
I;
(}Y'
MA
-::: 0.11 J
--
(C,4nlf) //- 36
-
'fr( d.6) -;:;- ]e) _
I
_/
1·3-1
Z
-
I. '3 - I
wilh
//.3
con'-f )
-
_
~
CJ.8t¥
252.5k
lhen :;
11-37
1070!!2
=
oS
//·"35
I 11.35
The stagnation pressure and temperature of air flowing past a probe are 120 kPa (abs) and 100°C, respectively. The air pressure is 80 kPa (abs). Determine the air speed and Mach number considering the flow to be (a) incompressible; (b) compressible.
(to A$SlAJII1ln!) -fo
Iilw we
inCtJrnpYtHiblc
COJ1I'1e.ct
ihe.
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8eY'MIA/lis efuahiJn (Et.3.7)
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t:iYld
V =-/Z(Fo-P) I"D ~ =
aJIld
(I)
10
(2)
R7;
U/mb/YJl~
2
E"Is, / and 2
W~
Ob"IR/;'
[IZtJ 1tl'4(A6s)- 8(}~{~6F)](ze6,9 ~ V373~) *'/. K /-:: 267 ~ [i 20 fe~{"'b$)7 (/1!'i~," ) = f
number We need
Ma =
,er
sa.
v
-cv.
(~)
VRT1<
r T
= To
T :::.
337·S K..
W-t,
fA fe
V"(~-I) ~ 2 R R.
Ih. t:.- ~u.a:f,~n of mat/on (€ij. 11.91) It; (}~/;'
= ~73jt:.
_ (2/7 f )'2.(t.4_ 1)
(I.-!:!- )
2 (/."" ) (Z86.t1 N."" ) 'Ill k.
~r· !!! $"
1/.3£
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fI'+ )
Wi~ Etj. 3
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o6f?J,;'
2D 7 .s !!!.
-
p
_
0.67
Nil ::: ()·78
=-
Also
frlIWI
r
~ () 11 tl
::=
F /9. tJ.
f)·1
we
~ 4.c{
ali
ftt.u J T ::: ( 0 . ~1
) (3 7 ~ f:-) -
V -::. Ma?l RT -k
= (0.78J
132
Ie
0.725"
11.36
The stagnation pressure indicated by a Pitot tube mounted on an airplane in flight is 45 kPa (abs). If the aircraft is cruising in standard atmosphere at an altitude of 10,000 m, determine the speed and Mach number involved.
For
sfandard
/~ O()t) n7
P =
26· 5"tJ
tlfmtJsphere.
kIll. (abs)
dYld
T
-
P
=
223./k
lhu.J
Pa
zt. 5"tJ k~ (abs)
-
1).59
'1-5 VIIt (tiJ,s )
and -frmn
v=
P/g. 0./
we reacf
{/vlA)C
/I-~O
vve ~ef
-h-PYh Ta61e
C·2
If. 39
I
':'11 .JK An ideal gas enters subsonicaJly and flows isentropicaJly through a choked converging-diverging duct having a circular cross-section area A that varies with axial distance from the throat, x, according to the formula A
rhis is like S'il1c.e
Example //.8.
= 0.1 + xl
where A is in square feet and x is in feet. For this flow situation, sketch the side view of the duct and graph the variation of Mach number, static temperature to stagnation temperature ratio, TITo, and static pressure to stagnation pressure ratio, plpo, through the duct from x = -0.6 ft to x = +0.6 ft. Also show the possible fluid states at x = -0.6 ft, 0 ft, and +0.6 ft using temperature-entropy coordinates. Consider the gas as being helium (use 0.051 :5 Ma :5 5.193). Sketch on your pressure variation graph the nonisentropic paths that would occur with overand underexpanded duct exit flows (see Vilh:o V 11 A) and explain when they wilJ occur. When will isentropic supersonic duct exit flow occur?
A~ and
= (J.,
A
r X.
The
+- X l.
1r
we...
UiYl
(n
_ detet'n1me
tAye S ummol'J)ed
is.
r).uc,f
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=
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1"
r
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A If = 0./ f:1T ~ and
= I +-
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A"
Wi ft.. &t. The>e,
we...
2
.1
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Values
(2..)
CIA.J'1
(.(N"ell'ol1d/~ tv value! of k.
delfHfYllne
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A* !=rJy he" ~ YVI wt! en ~ pnJ~Y"~ MfA.. values w/fhiyr ~e rtlfl9(!" s,ecih~ci I;" -ihe pY{)blem sfAkJ¥/~.".f CJ bl-a /;" values of d (Gfj. fl· 71 )) X (£1' 2.) J T ('f.I/·~6) Aif 1;, ,E. (~1' 1/ Sq ) . Th e ~ e. VI) lues aft ~6ult4.k~ and f//"dpht.d on .(1,
p~,!es
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(c.~n 'I; ) 11- lflf
Over- and under-expanded duct exit flows will occur on approximate paths sketched on the magnified pressure variation graph below when the ambient pressure of the surroundings into which the duct is discharging is respectively greater than and less than the flowing fluid p-ressure at the duct exit.This illustrates how the flow adjusts to these pressure differences through oblique shock waves that involve irreversible and thus non-isentropic flows. When these two pressures are equal, the flow is "ideally expanded" and the flow into the immediate surroundings is nearly isentropic.
1.0
><.
(-ft)
II. 3'1 ." 11.] 9 An ideal gas enters supersonically and flows isentropically through the choked converging-diverging duct described in Problem 11.38. Graph the variation of Ma, TITo . and plpo from the entrance to the exit sections of the duct for helium (use 0.05 1 s: Ma:S: 5.193). Show the possible fluid states at x = - 0.6 ft, 0 fI, and +0.6 ft using temperature-entropy coordinates. Sketch on your pressure variation graph the non-
isentropic paths that would occur with over- and underexpanded duct exit flows (see \ ilku V.l l...l) and explain when they will occur. When will isentropic supersonic duel exit flow occur?
w
Th i! i! si,.,.i/aY' E)(ample // 9 · Thi! f,-ab le Wl involve! the dud of fydl.kM /1'39. IId~vey fht. flaw enf~Yf fv,,~ndY1;"'//y . We. CitY! usc value, fro"" the hb'es "f p,.,bk... II· 31 wi~ " l i Hle yeRYYtlnq~mthf to (/~U>{JI1i fur Ik ;vpersd nic enf&¥IYJ9 ('low.
he I"~..,
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1 . 1~5
1·4
O· 'l.D
2·771
o· ~o
J.6~7
'H 'I .•
4-441
7·4
5.113
/1. 0
1·'/0 2. 6 J
17.1 65>6 o.I~t.82
O . ~lV
0 . 10/07,
O.003f'J
/I.
/.oa
o.
(con't)
b
O .I ~.1J
O.So f .00
7.4J
a.
0 .14811 O·V1B08
O·041VI O· dNq,
0:1.0 O·YO O· 60 0. 10
lU,?.
0.'>188 O. V611'-, 0.2111 15
0.60
5 lAh5bn ic.
OHiO 0.223 0· 123 0.076 0.051
O,/D"7-
st.IG
.!..
C
Sol.hOn O. q1473 O. '8JB~ ~.'i'15'3 0."", 0'''9/¥
O. fYi/6 O .1f'8~
O· jr7'5 0·19522
o·,nlv
d
//,3q
(con'l)
/. t>
oL---------------------______~~======~ -/.00
-0.8
-D.&.
-08
-~.2
0
6.2
0.'1
Iuv
he//uJ'tl.
o.b
e,g
/.0
x (f+) Von'afi4n
of Mach
lIJuMhe~
/.0
D.E 0.'
I 0.4
0.2.
o-1,00 ~------------------------------------------~ -o.g -~.b -c. If - O. 2 0 a"2.. (). '+ 0.6 o.g /.0 x{ ff)
Va r iafion of
'SftJ. h'c-- ternperaiufl- 11 '> M5V1ttHOvt femper4.lrJlt r~ H6
//- 1f7
11·3Q
I (con't
)
Over- and under-expanded duct exit flows will occur on approximate paths sketched on the magnified pressure variation graph below when the ambient pressure of the surroundings into which the duct is discharging is respectively greater than and less than the flowing fluid pressure at the duct exit. This illustrates how the flow adjusts to these pressure differences through oblique shock waves that involve irreversible and thus non-isentropic flows. When these two pressures are equal, the flow is "ideally expanded" and the flow into the immediate surroundings is nearly isentropic.
x. (oft)
\·0
II-tfi
(con 'I) /.0 o.~
0.1,
1: ~
().Ij
0,1
__~,.0
o~======~~-------------------=~==~ -0.8 -0.6 -0, IJ - 0.2 0 fJ,'2O.Lf 0.6 0.1
-I.OD
x(f+)
Varjllt./ioYl of
5~n'-c.- preSSf).re. -fr; r!-P.?I1.4t-h'oY\ p~SSU't yaNo
fr;y hel iuW\
f: o.qt:/7K'I
P,
D.q------ - - - -
~.y
--
0.'3
r:. =0.00313 Po
0,1
o
--jL---
l . - -_ _ _ _ __
s
T-
5
I::; ~.IOIDZ
~
d/aCf
yaWl -{Py he //uW\.
1/-.'19
1/. LfO
I '<1 lAO Helium enters supersonically and flows isentropically through the choked converging-diverging duct described in Example 11.8. Compare the variation of Ma, TITo, and pip a for helium with the variation of these parameters for air throughout the duct. Use 0.163 ;s; Ma ;s; 3.221. Sketch on your pressure variation graph the nonisentropic paths that would occur with over- and underexpanded duct exit flows (see Viden V U A) and explain when they will occur. When will isentropic supersonic duct exit flow occur?
o6-kJ./~-ed
wi ~
T To
~ 11t1 .t:. Po
PYb9 r4.",.,
Mtte-L-t. numb.e.Y"! /iI1PtA. T .
.A ) A1f In
Values of
tI1e
)( • j($,) 0./
.ISe N71
is
X(rn)
£9. 5"
I Ma
IN
.fW
i It, It
he /itAm W~$
=/. 66. values
~ '3 ·22 J w~e.
user!
of
f
'I..
r;
ye (J. Yran t'jeol. -Iv y /e Icl )( . with. x !;;y al'; tJbtztIJ1eJ air M~
Po
I To
.E.. Fa (/,0369
~, 2.21
0.22' I
O. 02~g
2.80
-0.4 - 0.3
2.17~
O.Z~2.0 (J·?602.
0.01/14
2.ljg
0.0767
2. /L.f
O.Iq.g) O.21ltf4
/.76
'.37
0.72.7 J
tJSZ78
0·'188
/.0
0. 83~3
~S]23
/·37
0.7271
-0.2 - O· I
o
O. I 0.2(!J,J
'.85'9 ,·lIO Z
/.0
J.'-I() 2
~5
wi#,
- tJ.5
:;'.;2.0
of
O. J
-
Example 1/.8 The VQ-yianah ()f Ma" T anol ? ~ GKtUY//'le 11.8. h-eli:w, Po
whIch
fV/o...
0./, 3 ~ MfA.
Y'a.n9t:..
)( Vtere
w ift-t
(J.46t3 O. ,06, 0.7519
O. ,06'
/·95'8
O.46g3
Z.~z. 0
I
0.2. "1'1 0.1"18 '3 0.07h'1
O.?;39Q O.LI¥8,( tJ.s2/9 O.617S'
(J.0{04 0.102.7
()./8tfii;
1·76
o.617S
0.32.78 tJ.I'Jso
1.14
tJ.~219
0./027
(). 'f
2·778
0.3'02O.29z0
~.OLfI4
1·48
o. Lftfg'l
a.0604
tJ. ~
3.221
O. '2.2., I
O.oz38
2.~O
O·~SqLf
d.0369
O·6't83 ~. ~ If 39
0.69
0.'-17
O,Ql31 Q.QS77
(). (
(J.6~
2-
O.i66 't
(). 7.. 0.3
0.%0
O. '13 '11
0.116
O.~l.17
0.'32
o.'77'1?
0.4 0·5"
O.:2Z3 0./63
"",(, B1 ('j.'k31j
0.95
(?',9/3
0/11B3
0.23 O,f7
O.i8''15 0.991{3
'C;
(can't) /1-50
tJ.7Z7'I
().8~'
0·'13/)" 0.'1638
tJ.'l8ao
1/.40
(con If)
1.0
.
hellun.,~~
MtL 2.0
~
~
/""
.
air
'.0 o~--------~----------------------------------~ -D.S -o.'f -0.3 -cl'Z -0./ 0 0./ at..0.'3 D.4 a.S"
X(m)
/.0
o.i
I
7;, 0.(,
0.1./
O.2~-------------------------------------O. "2-tJ. 3 -0.1.{ -03 -0..2-- -0.1 0 ~ I
-os
______ ~ (). l(
X (hl)
/·0
O.i
0.2
o-o.!) ~--------------------------------------------~ _o.¥ -0,3 -a.2 -0. I 0 O. I a"2.x( rn ) ( c,()n I~ //-51
)
/ ,. '1-0 I
(c()n 't
)
Over- and under-expanded duct exit flows will occur on approximate paths sketched on the magnified pressure variation graph below when the ambient pressure of the surroundings into which the duct is discharging is respectively greater than and less than the flowing fluid pressure at the duct exit. This illustrates how the flow adjusts to these pressure differences through oblique shock waves that involve irreversible and thus non-isentropic flows. When these two pressures are equal, the flow is "ideally expanded" and the flow into the immediate surroundings is nearly isentropic.
\. ()
x(.ft)
//-52
) I· ifl "1 JA j Helium enters subsonically and flows isentropically through the converging-diverging duct of Example 11.8. Compare the values of Ma, TITo, and plpo for helium with those for air at several locations in the duct. Use 0.163 :s: Ma :s: 3.221. Sketch on your pressure variation graph the nonisentropic paths that would occur with over- and underexpanded duct exit flows (see Vith:,1 " II.") and explain when they will occur. When will isentropic supersonic duct exit flow occur?
The. vari(?fion
01
II
)
T
~J(-
06 fr:uned
Mach
with
-the
~I'
x
L
wift.,
IJGNTROP
ytll'1re
O. 163
r;6klnetJ
weY'e
~
Ma
hel,'u1'Yl
wtJS
~
PYOfyt:LVYI
ntAYltbeys '"
Inpuf. Values
(J/Yld
~
~
tv,'ft-t
Mel.
k = /.6€. l4/ues
~
?'221
~ye.
of
used as
witt..
O. I x =/:") 0·1 which is £fj. 5 of &)((';l 11'1;;/<- /1·8 tetlYYtV1gec/ fz, YIeld x. The., vayia.fjon tJf Md L t1nd.E.. w/ fit x fZ;y a/yo was ) T.. p. fhm, EXtffflp/e /I. 8 .
()6m/~ed
I)
helium !
x (WI) -0-5 -0."1 -0. :3
O./l> 3 O.z23 0·3/h
- (). 2-
0.'1'60
-0· 1
(J
o
P
TQ
.un
1.0
Pc
(J.~9/3
O.~7~3
0.l7
O.'j83t:t
0.'5"9
0.<)661
0.92/7 O.gl(3Cf o.('QI3
0.2'3 0.37.0.47
0·9'347 O. ~'6 q 0·751'1
0.'1'68/
supers~l7i c. 0.1
o. 60'~
/.'102 1.658
0.4653
0.2'&440.llf3;
o·?;
~.'32.()
0·3~02
(J.O]67
0·4 0·5
1.,. 778
0.2-
3.'2'2.1
Mo
O·2g2o 0.2.26 ,
0.69 1.0
T
T"
1:. ~
O.c(9Lf3 o.,)8'1S"
O·C]6'3Z
0.979'7
(7:=]315
O.98{)O
0.'/577
O. 'l5~b
O.l]/~1
0.727L/
(J.~33
0.5283
sokrh'on /·37
0.7271
0.3278
,.71,
(J.bOS
0.0'+1'1
2.14 2· 1ft
o. 'ttftl(
('-Ig50 0.1027 O.06r)'/
O.C03~
2.g0
0.'18'1'1
0.0%1
(J.~219
5ubscnic... 50/r~:HoYl 0·[
O.6~2
0.866'/
o.6Qs3
0.6'1
O. '7131
O.727J.f
0·2 0·3
O.q.~()
o.tH~7
~.
O·'jS77
0.851"
0.3t6
O.'J6H 1
().L{
0:2.2~
o·~83cr
o· 97.17
0·lf7
0.5
0.163
0.9913
1/-53
g-'f3,
0·32
o· 9799
O.9~¥J
0.1.'3
O.989~
f)'''31~ (}.~6 3g
O.978~
0.17
O. &)1:1'f3
o.ljgoo
(Con'f)
//.4' ,
i ,i I
I
-·--1--
_
j
~.2~------------------------------~------------------~ _os -0.1.{ -03 -0.).. -0. I 0 Q/ O.V fJ.3 0.5" X (I"'n )
O.i ·~O.b
P
p.
·-·o.~ I
-
:
0."2.
O·L---____________________________________________J -t1,S'"
-0;'3 I
-~~2.
-0.'
0
Ix(rn') 1/-5"1-
0.1
a '2-
(ctJn It )
11·'1-1 I (C()h't J
Over- and under-expanded duct exit flows will occur on approximate paths sketched on the magnified pressure variation graph below when the ambient pressure of the surroundings into which the duct is discharging is respectively greater than and less than the flowing fluid pressure at the duct exit. This illustrates how the flow adjusts to these pressure differences through oblique shock waves that involve irreversible and thus non-isentropic flows. When these two pressures are equal, the flow is "ideally expanded" and the flow into the immediate surroundings is nearly isentropic.
x.(.ft)
1/-55
/ /. LfZ 11.42*
Helium flows subsonically and isentropically through the converging-diverging duct of Example 11.S. Graph the variation of Ma. TlTo, and p/Po through the duct from x = -0.5 m to x = +0.5 m for p/Po = 0.99 at x = -0.5 m. Sketch the corresponding T - s diagram. Use 0.110 ::s Ma ::s 0.430.
Thi
is
S'
Foy
lA ~e
we..
he Hum
values
Extlt?1f/~ /1./0·
like IJ
of
T
.A~ )
+he..
To
yange.
chol
~
0·110
o.
based On A:I" F A
which is
va.lu.es
defc,ymJn~
of-
M.a...
f!,
= /A~(:~)
x
coyyespondlnJ +0
f
().nol
wi ft.t Ie: /.66 +0
ISEN7f(OP
pY09ratl1
E'fj.
fl.l(tA..
~
o·'f3o .
We...
ulclAlo..k- x. wire...
I
z
(I )
fJf
ex P. I'J'!f Ie
//·8.
S/nce the.. flow is har
+J,n~t
Af=
(IiA )-fl.l'o~f (.R 1 is ofo.!-A,;'eo( AI
wne-re
wi~
P'f'(xi'{'·aw,.
.lSfN1X~f
fw.
Aol' '-fYI't~(}o..f
Thus I
A*
=
O./m
,.
"L
O. Of. 8'
:::
Yl1
/.'17
al'fcJ..
~1'
==
X
wi ft1 f(r 2
x{Itt)
I
heL~U
!O.~68 (fo)
-
(2)
(J·t
W, ft... e.. ro,,¥a.~ IStNiROf wiftt
I
MA
A ~
I.
f. ~
~
,It=- /.6'
-0.5"0 -0·1./0
0./10 O./l.fq
3.8'3
O.~'2.7
O.'J'lIV
-O.SO
0.206
2.~{
O.']g62.
- 0.2. 0
0.28g 0.3&1
2.06
O.'f30
1·47
0.9734 0.95'13 0.9425
O.%§f::J O.93Y3
0'$81
1·6z.
0.95'13
O·2gg
Z.06
0.9734
-0·( 0
0 0·/0 0.200.30 0.'+0
0.1;0
15./6
"b2
O.CJ%O
0.<;900
si&de
a..
a.ggqo
0.206
'Z.~I
0·~g'6z
0·81,16 O. tj~c;o 0.9'11./3 (J.C}6?'
0.1'/-9 O· (/O
;.~3
0·9927
o.~8Ig
!;./ b
0.<]960
o.9~OO
(can 'f) //-56
b
c.
Ma.:: ().o/Jo.
/1.1./2-
(con 't ) 0.5
O.'"i
Ma.. 0·3
0.'2.
0./
- 0.5
-o.z.+
-O.l
- 0.3
-0. /
0
0./
0.7..
0·3
0.4
0.5
o. !
O. ?.
Q3
0.4
0.5
O. {
~.'].;
~.3
O./f
(j.S"
X(Yn)
1.00· O.9g
T
I~
0.% 0.9~
-D.5
- 0.4
-0.2
-0.3
-0. ,
0
X(M) /.00
0.95 P
P-
o O.')fJ
l!.~S
0.80 -0.5
- O./.f
-0.3
-0.1.-
_0.1
0
X (rn)
F :. 0.99
Po
,.00
-To7" =0.996
0.<)3
I 70
0.91. O.9~
f.. = tJ.
o.n
/1-57
/I. i.f 3
I 11.43
An ideal gas flows subsonically and isentropically through the converging-<.liverging duct described in Problem 11.39. Graph the variation of Ma, T/To, and p/Po from the entrance to the exit sections of the duct for (a) air; (b*) helium (use 0.047 :s Ma :s 0.722). The value of p/Po is 0.6708 at x = 0 ft. Sketch important states on a T - s diagram.
is like Example 1/.10· Since .f..::: O.~7()g at- x'=' 0 is
ThiS
qtettfe.r
Po
fY'ohlenn
11·3(J(a,) stJ/u-h'f)n)
50IlAh'oYl)
-the.
a/r aY'JP
dud
not
choked.
()ye
Ohfr:l/~ (o.)
Coyve.>ptJndiY'1j •
Foy
we
aIr
>!'=
= O.5"2gg fov air (sa..
F;.
fo,... hel/uYn (s(e ;:n>6/e""" I;'JO~)
and O.¥8fl
he/,itYVI I'ltJws fhvdlA9h the. c"ohver9Inj-d/veY9'ny
Fzrv
t
values of
vtllue s of
entey-
e..
Than
Ma;
• F=la., P.I
;
wift..d
./
And
()f dif!erenf values of x we ?
valuesPo () f l l
1
fiy.[).1
n,r
{IIJW
fhe
T or
If)
A
A = 1).1 If . ~
M~I
-Iv 9e.f
II~
(lIf
x::: 0
vcr/tie of .E..
d.t7ol
)
~
we gei A = /,oS A ~ = 0./
O.~'15
f-f 'Z:::
ff'1
/.05
A
A* The,
x'l.-fc./ ~
II~
UJYyfs pOYlcl/~
Fij.D.1
a~
-::
~'J. t- O.t
(lJ
0.,,'5 va /u~J
al.>o -fo 6/111a. k PI
of
~
IJ~
0 JI\
fhe
~n't
1/-58
Ylex f
)
)
Na.,;
ra.ge.
I ~
lOfA
.e.
Ii
Ir~
/1.43
(COtllt ) Witt. ~.I
F 1""01'\.\.
A
X (f-+)
Fi9'
D·I
A-I'
M",-
-1,0 -0. g
1/.6
0.05 O.og
O·Cf9
-0·6
1I.g 2.7
~.IZ
0.'19
O.'J~
0.2.2
O. tt'i
0.'j6'
/·S
~.44
()/jb
0.9.7 0.&6
7.8
-0.4
T
P
To
~
.,q
O.
O. 'j9
1.0 1·5
0.78
~.H
t1·LfI.l
2./
(J.2"1.
0·16 0.'19
O.':U·
O.h
4.2
0·/2
o. .,.,
0.4!f'i
C.g
7.8 11.6
0.08
0·19
o.'t~
a.o5
O'&]9
O.CJ'
0.8
0.6
MtL O.L1
O.L
-0.'1
-1.0
-D.b
-0.'1'
-D.'1.
P.2
0
(J.t(
0.6
Oog
,.0
0.&
,.
x(ff)
Van'af/oY!
01
M~ch
number
~
a;'y
/,00
O.~5
I
r;
o.lfo ~85
0·80 -/.0
-O.~
-4.~
-b.t..f
-0.2-
0
p. '2,
O·Lf
0.(:;
()
x{ff)
Vaviah'on fDy
of
slztfic femptya/urt. -hJ
sirA.'h~f,dn .f~J71p~Alure
41r
(con't) If-59
b
O·~7
/.0
0
a"
~·19
-0·2 0 0.20.4
I.D
S~te
rdno
c.
II. if 3
(Con/t) /. D.
r-----....'""'---
0.'1
E Po O.g
a6
~--------------------------------------------~ I. () ~. 6 tJ.8 -(J,g -D.6 -fJ.'{ -~.... 0 I.'], ().4 x(.f.f )
-/.0
Vayj~fi"Y'I
of
s172h'c. pyes.ruye
-hJ $-Ia,nlf,h'em pteS5ure y-af/o
/7Jy air
1: ::
0·'79
PD
1·00
! = 0·"''1 1D
0.95
r.
TD 0.90
0.85
4'<;0 ~------
,
1-5 (b) h¥ hel/IA~, W~ defCvrnll17e
aIr
va/u.~s
X;;
ral?ye.
JA~ (~71) -
Wf"ie"h is
blJ.s~rJ.
choked)
A"" 1-
1/"* =-
0"..
wit4 ~
==
/.6~ ~
~) L 1AJ1/1 P U;yy~>I'Imdl'nj -fD VIJ/I.(~.1 "f A~ 10 Pe, 0·17 ~ MIA ~ 0.7'2.2. We. CAIc.~/~ k X wd!.
/)f
wi tlti,.". the
.zS€N77et1P
PYO'jyaYl>1
lAse
0./
G'i'
A+hY'OtLf'
(I)
z.
"f PI7>J,lell1i'L.
ThlA.s
we
Atit nJAi
(r»n't) 1/-
60
ll·31j.
5 ince #ais
d.e~I'tI'I;V/e Il ~ w/J-{...
lIowi5 no!
(co n ' t: )
/ /. If.3
where
7iJus
(A14J.y()4.f
J
0./ ffl.:::
flK- :::
if l
0.0'1115
/.0135 ~. I
and
be c(} me. s x ~ j""'(O-..(J-9-3-IS-)-{-:-:-~-)---O-.-I I>ENTP..Of
Wit{"
Wi~ ~~. 2
All'
O,O.ti7 0.071
-1.0
- 0.8
r
A
M~
X (if)
o.m3
7.ttS'"
~."g3
c;.9959
0."S7 0.'86/
0.'719/ 0·96£3
o·'If 52-
0.3' 7'1
O.gS]'}..
(7. '708 0.'fJb7't
- 0.6
0·/15
4.'371
0·207
'Z.7f1n
- ()·2
~.tfiq
1.5IJlf'i J·0735
D.722 {}·lfl't
0.2 (j.Lt
,·sol(q 2. ,Q73
0.207
().~
O.{/~
D.
0.071
'·0
O.()47
.f oJ~~:z.
1'0
JJ.qq~lj
- f). if
0
(2)
0.'''52
().,
b
0.,65J
0·9161 O· ,,'57
4.'t378 7· ;5'1' n'f1f3'i
~k
o·n'!'
O.9H3
O·~95g
0"~'3
O·lj'rn.
C
o.g 0.6
Mcu 0.4 0.'2. 0
-/.0
-D.b
-d.g
-0·4
()
-0,1.-
0.1.-
(J·lI
(J.t,
0,£
,. D
X ( ff)
Va YIP" fioY'
t7f
MA.cV,
nt..o-."bt:IY f&v he //U;'YI
I.00r--- _ _ _~
-r-
0.95
T., Of}o
O'8$~-------------------------~--------------------------~ (J.(, (J.g /.0 _I.~ -0.8 -0.6 -0.'1 -0.2. 0 0.2 '·4
X (f+) va.,/~.h·Qn
of
s~h'c.
fem~(1.n.ye -fo slz<'jl'laf/t7Yl
(COI/1' t.
//- 6/
)
II. 'f3
(UJI1'i)
f
0.1
0 . / ,-O.g - - --0.6 --- - - - - - 0- - -().2. - - -0."1 - - -(J., - - -b.~ - - - ,.0 ' -1.0 -0.'1 -fJ.J, )( (.fI.. ) Va "/A.h'~n of smfle.
pye$,H(y~
h,
sf-a~/I'\f'. h'o" pYe.JSuye ra.h·tJ '" he.I/~~
f.. .. Po
~.8S
O.Ho~
I .. 0.'153210
0.80'--------
/1- 62
11.4Jf
An ideal gas contained in a large storage container at a constant temperature and pressure of 59 of and 25 psia is to be expanded isentropically through a duct to standard atmospheric discharge conditions. Describe in general terms the kind of duct required and determine the duct exit cross section area if the discharge mass fiowrate required is 1.0 lbm/s and the gas is (a) air; (b) carbon dioxide; (c) helium.
To
de leYJIYII;' e
air
t£jr
fhe ~;9.
)
~)(if
duc.f
. m
=
Aex.i-!Foy
-fhe
us~
(j)
ctensify I
Thu$
/;/
::
vve.
~)(it
ex; f- flokl O. / .
ex;! flow aveC(
US(!
Er //·60
or loy
]~~I)
I
/ I t-
we
/eXi/- )
(i!- ') Ma 2. .
e'Xd
2.
where.
~
(jlr
~xit = We.. obrn,;' Fig- D. /.
Ma.
RJ; Mq
(
;::I). D· /
valu .. of
exit
-frdWI
?exit
uSlny
€-1-
II- S1/ tJY
~
t:IIY/
7htlf
-r~1i;.1) J(t;)
~----------------------------
F;j. D. /
valuf:-
:: (M4eXi f-)
fA S
fA
(#-)
(5)
/uhc.-//()Yl of
,t;
o6-h l'n
lI~Xif
~'I)
J;
-
exit-
/VI 4 e)(i f-
lo
~
C
=
f/
(M~jt) /~ ~)f
(con't)
*
'------~---~~------~----------.. -.---------~
/1-63
(COIJ't- )
/J./fLf
whe re
fr.trm
TeX.i t
-
E~. /I. S"6 I OY
IPv
clIr"
~
F iq ·
P. I
_ _7;_ _ __ 1~-I)Ma 'l . exJf
I t-
( , 2.
~ air
(JY
7:~x.i f
~ (
--
r.TeXi!' Vel
Vexi f
D,I .fW A1t;.~w/'f) """.
"
thu!
anA
~ F'1'
lite
=
(M4eXl'l- )
(5)
(tlJ Fov air ~x.if
t;;
-
The
ILl· 7,os/tL
= tJ.5
j- _
I
2>" ,PS/A covyesjJona'IYIJ
-lex.i To
Q
val",es
tJ.86
~xil- = 0.63
~
Now
win,
£'1. :::
3
(25"(J$/a ) (1'I't (17/6
;.,~ 'l. )
f'f· 16 )( SI5 S" IUf. ";2.
(UlYJ'+ ) IJ~
6f.j-
(0. '3 -IZ-)
)
-J 2 ·7 S' )( ItJ .1'114. J .ff)
//.'f4
(C()J'7 ' f)
'P) (2.1, 11
Ct;n l/f!/rJIYlJ
XIP-l
dwc,./-
sl,; ) (q32H
would.
~I- )(nZ~)
5ufh'ce
= (). tJ / z If 2.
$//'(9
..fk eKil-
$/;'ce
fttJw i.s
S lIb?Sd>1 i C .
-
I
/ ·3 - /
/
-3
~Xif =
Wi Itt ~1'
$"
if. ()J' X 10
or !U'j -(./3
we fel-
~lC/f = (0.1123)
-:. 76
s: 7 /1-
-
.J
S' u.b.s())1/c .
(eon'c) /1-- 6S
!
II. 'fLf (C)
yi~/d 5
hJr he/,ulYJ ; Gj. 'I
r-----------------------
Ma e)(,j f
=0.3'1'1
-
/
/·66-1
~.05'6
-¥
x/a
f/U!
rf3
WI/-h ~1' 1/
veXjr
Wi ff,.
t
=
1:-). I
Aexif
w€- gef (0.81.1£1)
W'-
(I ~) 5 --------------------------------: (t.OS"6 )(10-ii
SI/,(~)(2lff'f
~3
A
C()l1VerjlrJj
I?tJl}/C
w;J/
d{)
fr1
51V)ce
3 Z. 2
O. 030? {flo
16 no) T~1
-Ik exit f/()w If f"b.xmic.
1/. L/-
11.45 An ideal gas is to flow isentropically from a large tank where the air is maintained at a temperature and pressure of 59 of and 80 psi a to standard atmospheric discharge conditions. Describe in general terms the kind of duct involved and determine the duct exit Mach number and velocity in ftls if the gas is (a) air; (b) methane; (c) helium.
-
10
the. dlAcJ
de/eywz/ne
E~.I/·s"'j
0,,-
Ildr
J
/
) we hlAm6ev / Ma~'f XI
Mach
Fi9' 0./ .
=
Merex,'r Ov f,v
.fv.,.
exit
lA~e
TAu>1
](~~)
(I )
IJIlr
/lA~
::
ex.it
Fi9- f).1 (]x/f
VA/tfc.
'if
~
Ibnc b-on
ve/pci.j,A '7" v,c:X'fI )
1-14,,,.if.
VeXif wheye
(7/
W~ IAf~
~)(.il-
(2 )
~
Y
R TcXif- -k
{"3J
1exit
1C~if
(5)
(a.) Ft;y CUr
-
Pexit
~
:;
==
()./f38
9" /sid.v.JA.d f1,u, > kvw. F /9. 0 '/1 ~
M4 )1.i/ ' ttJt.d "Tex/f" = 7;
:'
/·Z
tJ.b2
/1- '61
c.o-r r€$fol1d!n5 Vt7""~S are,
Con If)
1/. £/5
!he ~
wiftt. ftj. 5
/..
:::
ex-if
w; fh
CIP'II:A
we ohfa.tJ1.
(S?'1DI<)(~.61..
(17/6
-: (t."l)
Vexit
C-vn.c tude.
Erg. 3 we
rJ
-=
)
322. p.,.
fha f-
)(~22 ~1!.)(I.Lf) SI"1' 0.e ( I ~ ) -FI-./I;
~/Lilj.
•
fJ
C()nJleY7/~j - clive,Yf'J1j i;
:f/()W
n
yei:/~c1
; r
oJ)/e
!!:
=
ISSO
=
6ectUUe
ti J
M..t. ~xif
5'u(J(!'ySdn/c-.
(b) ~ Me..tiuJne
M()exit
:::
we. dblrA.,~ ~
€'1' I
~
= /.78
/ ,,')1-1 (!~. 7~s;a) Tj( gO~r/~
w~
Then wi Ih clJ- 'f
~x;t
gef
5' 19 III?
=
1 +3
(1.31 - )(1- 7
iN e.
l"fr,u;"
A c(/m)~/~ - dt'~71"'J fI(fW if (C) Fw
e-tiJ
.5(Al'ey~()nic-.
he//ui'YI
we
cJ61a,;'" ~ Gj. I
They,
/
-
with -
(VOy/lt )
//- 6e
26 'f. I "/?
/.71
1/. tf 5
Con't)
Ve'{ir = (1.71)
=
/J fft!W
577~ _ sif
C!Mllw,t'nJ - d/.)t:IY'iJ~J I
r
5'lA(JeySdY1;e, .
t1/)31/~
I)
ve5",yed Slnc~ the ex.jl-
II. '16 11.46
An ideal gas flows isentropically through a converging-diverging nozzle. At a section in the converging portion of the nozzle, AI = 0.1 m', PI = 600 kPa (abs) , TI = 20°C, and Mal = 0.6. For section (2) in the diverging part of the nozzle determine A,. p,. and T~ if Ma, = 3.0 and the gas is (a) air; (b) helium.
ro
lIN
(~~ )
-- =
II,
:::
A:z.
0)'
Alo we use Ecp. 1/.71
de-fe YIfII;' e
.
,
11f
(~ , )
r{1I..~ I
(~ ) , ~ (!~ f)
I t
M4;
2 Mtt,2
j",(*-
J
~+I :l.f*-I)
!(F;').
A,
A;.. -:;
I-
.L fo1q
PIt]. PI/
t;l!
j;~ [
A,
(:;)
a,y
rvY' r.)
0'1'
(/ )
(2)
( Fi 9,
10
dekH-rn,~
~ =
{
e.
we
(?-)
p~
= ~
(P.)
I
(3)
~
0"'-
frrr
airI
p.2- : : .
~
i(
f;:'9'
(F/9
To
defe.,m/,,-e
7:2- = T,
T~
(~)
-7;-
w~
=
0·
f),,/ vdlue.
~
Oy
.frH T~
[
7;
=-
T
,
! (fiq.
[, + I
l]
o.f p, P,.
f.... M
of..&
fw M.&o,)
I':, G'-g. 11.56
(A.s-e.
fT.) a,Y"
V6/U'
I
or
Iw' ti ';"/ (:'9.
(~~')M~,~] )
r-(~-I ) Mtj/" J 2-
f). I
Va llA.\!.
of
T~ f,.. To
11,,,, )
(Flq D.I
Va. I",,!
or
5 !w
M~)
(COn' f ) 11- 70
'"
('f)
0./. lhtls,
(5)
1
(~)
(~)
a/,. .. £'1-
p.()y
A2. : £9' 1./
1w.S' -Iv
2.
(0.//11')
:J
('1·3 (I. 2.
)
yt'elds
~ : [ 6o(J 1t.1i{~"r)7 aYl~
~f"
9 /lle s
r :: (293 fC..) :L
(0. ~6
)
{d.q~}
=
1/3
I<.
:=
11-71
O.2.S7n-?L.
//.!.f7 11.47 Upstream of the throat of an isentropic converging-diverging nozzle at section (1), VI = 150 mis, PI = 100 kPa (abs), and TI = 20°C. If the discharge flow is supersonic and the throat area is 0.1 m~, determine the mass ftowrate in, kg/s for the flow of (a) air; (b) methane; (c) he-. lium.
we.
fhe
de feYJI'JIIIYJe..
w/It,
Mach num6ey- at sec//on. (/)
(;) ror a/I of Ih"e .1tl~e.s ~
he cause
I
~ YI/wln3
fAe. Ma~s
//JvoIVu:/ ~tl6s(')n/c.
flow is
Illlwyak.
use
We
t1, t.. .flow
TJruS'..
slow.
/;/::.(!./y
is
i-l-
Ih e..
the. leaj//~
tl¥Ja.
6fj . 1/. '1tJ
~f
fhcd
is /t!ss 'lhdh
MQ,
-f),Ylx~.f HIl/AI
/J rJ?1l/.:.ed
-fo 0 61rJ l'n
(2)
ve/~ci fy / V~ we, lAse
V*~ VRT~-k To
o!om/;'
T
(3)
~ W~
use ff·
I/, 63.
Thus
J
(tf) ~r
.fur cur) T~ -=
~ (value
of'
T
~
1=,'9. O. J fiJv /I1a -=-
7"
To
de/e..ym,ne
~ ~y
=:
TD
r; [/
4/y
wt:..
use
r (: I )
€~
. 11-S6.
~/2 ]
, T,
(~nll::)
//-72
Thus"
11;"~
is sVl'erstJnic.
,;, =-,..
/.~
/.0 )
(5)
my
~OP'l't )
//,47
(q)
or ..foy
p¥ FaJ' Po ~ oy
fw ~
tllY)
=Po
(vAlue of
we.
use
~
j?'"
Po
~ /Vtt:(. =' I- 0 )
(I tJ)
ThMc;
E'I_ 1/-$'9 -
~
frI.
::::
P,
[
(~; I) /v1Q,L ]r-i
I f-
(II)
t[ ,;,.
?,
::.
(va.lue (If
.;;~
f: P,.
(a) '::w aIr
Mil,
':::;g_ P. /
we use
G$_ /
h
;;'9. 0, LJ
1
,tOy
Ma,)
!,-hi,;"
-
(). tt 372
2.'13 ( d_
'16
k...
::
]05
)
we.. tJ/:'hr I;'
/1- 73
K..
Po
/00 iPtl (ab$ )
-
7heJ'l wi fA
.
a.1
1t..1't:i (4f?S )
=
6~.
kJ'~{a"$)
8
~f- 3
(~t)\B
>(10
1
IJ ) /'I?'l-
':::
= fJ·
=.
N·n? )(25 ~l. I<.
fi' z BJ
t#itt.
I
M
5
/tIjJI{.(llI;s)] (0. 52 g2 8)
(236.9
H"na.t'J-
/I
Fi~.
of
It)
[1/5
;;:
;;>1
=
~.g7
6f.
pK
f~e ~e} f
w;tt..
we... ololA,;'
,.,.,3
kJ
if •
obI-. U'l
W"1!
3) (0./ Yh2.)(:1ltt
(6) F~,.. meihanG
we-
~f
".8'3
Gj.
lASe..
I
;)
:::;.
-
.J
db~;"
-fr;
050 ':)
03363
'( ~/I· 3 N. YI?
~.I<
Ilow is
2t.5 'k}
) ( 2.'i3 Ie) (,.'11
)
(I L )
a,.
c-tuk..ed
",f-
l!!.
+J,e
f3..
-::: tJ. 336']
~
:=.
(213
~)
cf
Lf
i#; ~
and..
+- (t.1~-1 )(0.J163)).] - Z~!·II<
[I
we. ohm 4;"
T ~ ::: (2.-93·/I<)C
2..
+
/.31 Th.
v,f.
F-~
£1'//
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D.JI
r
we..
.
ob!.,,..
;:::. [/()d kl(((4ks)] i It,
Elj. 'i
::: [i07. ~
25"S./ K.
I
I/X-
r
I
*7.
!!!)
f~
1.11
M(4bs)]( ~)/.7/-1 -::.
;O _ _-.::_~~~31:~
~
+ (t.1~ -') (tJ. 3361 / ] '.11-'
9e f
t '.31-t-1
1"heJ/7 WI ~ E-1· 3 we. ho,lIC ;y (5-g. J x 10 '3 1--_ _ _ _
-=
'!:..!!!.) (2.~C.II<){1.31) 1c.,. I:. (I ~
(Slg.)
-
)
!,.. )
)(253./~)
=
o. ~3S
(con/f) /1-
7Jf
5'3.5?
k
~J ;;; j
---------------------------~
/1. If 7
(con l-t ) £~.
r/yzally, witt,
vi?
2-
we ohm.,;"
~f)( (J.I hI·~)(~/!.6
== (0.tf3f
_
;')
",,3
_
C1.
aMrA vJ;tt,
If
~~
.s
= 2.9>.21<..
()6 m.,"n
Wf!.
T.r = (295".2. k /
it.~ _
0./'/92..
= (2-q31<) [ / + (1.~-I)(O.14.ti2)2]
7;
/8·1
_2_)
"Z I. 66 1-1
-=
222. k.
~ :: [ioo 1efl{{aI,~)] [ fq. o 'i
wiHt
and..
p.r::: Thc¥I
[ / ()
wi~
f1 if ~
1
.f, 6
1.9 fq, I'f( (4'S)
J(
- 2
)'.~-'
:::
/.6( tf
£,. B' 3
('If 7'1 X 10 ~)
-
0.107'7
(2 077 ~ ) r:,·Y)().lly
wilt.
(22J../<. ) ai·1< . €~.].. we 6/;'-IA II'?
I
/J.1
-
1/-75 .
-
s
11. Lf8
The flow blockage associated with the use of an intrusive probe can be important. Determine the percentage increase in section velocity corresponding to a 0.5% reduction in flow area due to probe blockage for air flow if the section area is 1.0 m:', Tll = 20°C, and the unblocked flow Mach numbers are (a) Ma = 0.2; (b) Ma = 0.8: (c) Ma = 1.5: (d) Ma = 3.0.
We
Iv
won-l
asceyll1/~
~l!JblD"J::.LC' To
de'lwvn/;'e V
fht.
()c/c.e.d..
U 11 hi
H;y
~Yl6I()C.~~
:::
(mb/ockeol
M~ M"
we
T
Ul"lb/f)c):,ed "I fI ,,/cui
aveA.. ve,t"c;1y / '{~bIDc..b') we ,/ R. T V U~ bI
k..
{lJ
DtJcc.J
lAse
~
J
I7; ~ Ma. Ullh/~d:£r/ rn.,.,. E~. /I.?G ~ MaIi"VN
7;; (
II) dc-~JII'1J;"e. I1u 6lodu..d
U5e
tu'&.
veLou'", ~
~/()c.kuI) ~
(z)
£A4..t.
~/()~J = ~tloc-lael ~R ~,tk,fk. Pw
M~ b/d(.k..v/
I'Ve..
we.
£f. / /.7/.
AJlochcI
t)Jt1d
dek¥~Jne
A~
,!1bldc,h.t/ ;'"~'jfA '''t!r
nr/p/ 4Atd ernrr·
A~
(5)
11-76
Iliff
(a)
~()nlt)
M~
Fiff
~ tJ. l
U ~f:I ~ c.k.ec/
w itt...
E~· I
~h61"t4.I
-=
46/ocJ:.e4
-
/1-;('
w;/4
/JJ1tX
MI(
2.tf()· 7
'=
k.
r----------------------(2gb.] ~ f~qO.71<)(I.'-IJ
(0.2..)
~.'"
ret
"',..)
('
1 f .;=
=(o.19SJ(2.'163S) = 2.9'1'7
,~J
olht/~
we
1d'$./l71
=~. 2 ()/
blluc.u(
£~~
W/I"tt
CI..,d J).S",
2
ha~
tv!-
e1 f . ~ (,fn~ 11.7/ +v
We us-e
E1S.
vV il't,
= (ZQ3 t:-) (~. 9rz()()
T II. Then
we. oh,*,;'
tfI'/6/~d~
s ~ ~f z,f"j:;
K
-=
Z
O/()., k
= "·l/i] %
-
(b) ~ M~ ==
o.g
~nb/~tk.wI men
obh,;' IN;/(,
V-/C
=
£,s:.2
11·5"6
D-ttP(
(zQ31c.)( ~.8g65Z.) =- 2.>"'13 ;:
£9' I we jef
tN)itt
~-------------------
O. 8
Vlutb1oU:uJ::::
We.
USc..
~'i.(.¥
A
a JIId..
(2gt.'!
CYltJ/. 1/.7/
f./.I?'\ )
~j.1<.
(zS9.!I<)(I.'1) (/
1'>'1 s~
r;~/8c1c.e4
-c:
lit¥: w; It,
£,,-
(0. 91S'")(/·03Izs) /I,
71
we
0
1·03]
hho;'
M~;/ockuJ ;;: ()·313
Wilt, E~. S
IV )
!J.?-
we
re"f
7i1"~ = I
293 k.. t-(t./f- /)(c1,g/~l "2.
=
(COn Ii
/1-
77
2S"3.8 K..
)
£,.
w; fh
'3 we have
V,
j, Il}t/
-;:: (0.113)
tlnd
(~/I)cJ:e4 - ~11h1(J(,~) x
1110
";>..
(2.b2.1
VLtnblD~ {Cj
l),~ n
258· 'I ;' ) (/tJtJ)
(?t"i.lf ;')
hr MA... =- ,. S- ,,r;~b/"~
~ s
M
wilt, Gjs.2
tJ);,Itl/;'
:: (2.93 K) (~,6g9ts)
ew.'{
-=
1/.5'6
2()2. /
k
cf. / We. gef . ~J'fblodu.(/ == (I' S-) ;-(234-b.-9-~-·-:-)-t-2-0).-.-,-)-(1-"-tj) wi It,
V
F<7. ,....
I
We. use.. EjS.lf and /1·7, -h 7-I.. ' A bl"cJ.erJ.
='
A-t ~Mci w/#t
1::7'
/1/
Iz!. -; .f"2-
(~. qr;S)(1.17, "2-) -= 1/. 71
(I
I. /7
w~ tJ"-k/~
ana
(~/othd
- IIUrt"/orkd)
(d) FoY ~ =- 3·0 Gn/:'/t/du-t!
Tht Yl
w;~
Gf.
/00
X
we.. ~~;"I;' '=-
"",lit,
Ets'2 e(hol II.S6
(2-1J 1<) (0.15 71lf) 1 W~
gef
(ct7I1'f)
/1 ... 78
=
/(JI/.6 I:.
)
=l.lf3 % ~
we.. {,(se
E1f. if
A61od<.t..4
-
(J n
d II. 7 /
(tJ Cf9S ) (Lf.23lf6)
-::
- If. '2.1 3
0
A~ Pl'7d
£'g.11.71
Wift,
Mil/
-:: 2. '19)
blocicul
E~. 5"
With.
we. ~bhl;'
we.
'let2'13 k..
T
b/()ckuI
I +- roll -
I)
(2. ct9S)),
2.
w!.
have.
Ond
('{)()d~~
-
~1'J61()Ci~.d)
~M 61odu/
X I () 0
-G
(6 fL!. 't If- - b/,-/. 9 ( 6 Ii(. 'j
;J
-;- )(i()b)
=_
f1, bit]
-
%
1/.50
11.50
For Fanno flow, prove that
dV V
-=
jk(Ma 1 /2)(dx/ D) 1 - Ma~
and in so doing show that when the flow is subsonic, friction accelerates the fluid, and when the flow is supersonic. friction decelerates the fluid.
5f~yft~
wifh
'*
.L ( / t-
d(v'")
M( 2)
2
~
( /+
(I)
11~1~ we j1lfve
€$.
d (/.111. MAl C OMb/niYIJ
d (M4).) Mal
V~
2
FrtJm
we. h~ve
E'j. /1. qs
=
)
r
vz.
(1c- , ) Mal.]
I -f
d(l/2.j
(2)
2 we tJbfa/Y1
G~,), , And Z
1i M/./'") d (v~) _ [It- (i!-I) M4"] d~ / 1/ 2 :2. v"
.L ( Ma 2- , .2
d{/.I1.j
)
f
:fA ft11l~)(:: 2.
t1
(3)
/)
-= -
and d (II l)
-=-
V 2..
f
M ,/. (
-k
M4: 1._ I )
dx f)
t-/oW(. II'tW' d(I/2.)
~s
:: :z.VdV
C(/Wl41/n,""..J
rf.!!
=
we ,ef
(,,)
/ - /111(''2 -fJ,~
~/c,f/ol'l
i~
1:11." 4Hd S
,ck (~"l)( ~}
II"
When
(5)
Ft""" is IA.C(
.f'vpeYJIJJlic
n,iG -//on
.s~6.s1J11; c
e/eya.lcs
~
'lite
(Ma> 1.0).,
de ce(eY4!u
(M({ < I. ())
-/114#1:1.
On I1tL
€f· 6
leads n,
#,e -/i",,/el.
//-80
1£,.
~
I et.ttis -h> dV = V
dtt.e.y
dv v
ha.Hd
= _
Q
"/- tI"d flutJ
,..,4elo1 fhe. neJ/
In
jl,ir
H~k/
Cl!ue
the values of static temperature, static pressure, stagnation temperature, stagnation pressure. and velocity at the inlet [section (1)] and exit [section (2)] of the constant area duct. Sketch a temperature-entropy diagram for this flow.
11.5/
Standard atmospheric air (To = 59 of, Pu = 14.7 psia) is drawn steadily through a frictionless and adiabatic converging nozzle into an adiabatic, constant cross section area duct. The duct is 10 ft long and has an inside diameter of 0.5 ft. The average friction factor for the duct may be estimated as being equal to 0.03. What is the maximum mass flowrate in slugs/s through the duct? For this maximum flowrate determine
7J,/s
51",aay 7z, E:xaYH/J!e
is
tlte
{'low ~ak
I'YttlX/""uJ'V/
(..tTn{/..""f
af'eJ\
ducf
ex./f ['see-f/on (2)]
thYfJH7h
choi:t!$
is
1.1).
As (..x-pllJ.';'uI II? £-xa""'I/~ 1/· /2/
//- 12.
due-I
fIt(..
and
f1tt
will
Math
lJr.e l'I'lax/munr,
wl,~n I/,~
C1C.c"u Y
ar
nUWlber liowrak
thl. c4tc.f
(}b/7l/~eP{
CAn be
wi 111 (f)
We
CUn4l4nf COlliS
Itt n f
duc.f
Arelflf
kv c.h.ok.ed
~
--~
i:=
CAn
7; V
-::
"=' t).
re.tA.()( va'/ues
()b tR.I
T"" ~ ~2:-)7 and
=V2
~,
=
0.6
=
f" _ (,,(~-.f) /
S -1+
()
Fifj'p, :2
Ct/n b~
"
belAIlIe
nO.Jl/e.. bu.f ple&r~RSe.J ft,'(ll,,(fi. IA~ 01 lYi~li()n. TAt/5.1 ~ 1'1. 7l'fia. .
(~.()3) (It; Ff)
. "-1.J) - £, )
from
i':f
.A/s01 Po is
=5/qD~_
flow
f" ( a~d
::: 1,,,1.
C.(}n ~rf
-the
II?
T1te ~ m{,~ hIJw J/"u
fhl'ol
the! 7; is ft,e. {low is a.4/alHlh·c VIOle..
~+
I COn
"
6~
o~
n
e,f
=
Mill)
witt,
v,
1;
];J vii' )
€,.
//.03
(;.:+1 /51'1'12- J
ple/("""" /I? t:d
=
" p~
7; ;.s
S/hCe.
~32 D,e
4~t(:(
:-
~, ptl- .
7he n
CtmJ /It '" l-
lJu,/ ./
~
wi""
(t 716 so
(con't)
//- 8/
/020
If" =
=.r
L( <6
/ /.51
I
(cOJ1't)
*' ) f (1-,(,
':S
b
r1t/ I -:::
~
~.
S7
:: 1·13
T'ftI I
:;:
0, 6
(~)
v~
;.
Po,
-L-
pr¥
1.86
0;
-= /. ').2.
4
fZrom C1'
2.
We.. 9e.f
T :: (I.13
) ('13)" f)12.)
_ 'fRYif)It!.
-===
I"
Witt. £,.1 V,
.=.
IN'"
(0.6
dbh.Jn )(1172.0
{f)::: J
::
~II
-
6/2
.J
Itt· 7 pii..
=
,.22
/.21.
" :::; (0.8
=
Ii
-
If. i' I's/~
:-
/.16-
//- 82.
'2-I'.s/a,
=
11-5'2.1 11.52 The upstream pressure of a Fanno flow venting to the atmosphere is increased until the flow chokes. What will happen to the flowrate when the upstream pressure is further increased?
fV Alf()
-
-
level
Ij'
UMcllA.~
-flbW
e., Iv
hILVwJ~
,1 + 0
flu.
7;
Ja-me
",jher ft,/Lf
fJv
At's/)
IS
f7V
7kuJ
If
':J -I1u. ~
1;'u-etJ-f
~-In-t ,.!,
he r'C.
vetil/it
tJ,e
s-e". iCJ I
~ I~ f
-h~ ~ f3'
d
j
Fa;'I?o
IJ 'ne./'
fW!,{".{'lI foe ~
//- 83
w~eYe
.}al'lre,
blA-f
,Y"
,'r,/e -I In
hlW
!-,1 4
~Lrt)
II.SG
tlfJ-a
41'1r1
tIu
we
!Yr SJ ~ 1')/ a cJ,okd J#nno /nCff-Ue (1/ .fIWl 4k Alto.
~ p~ df EX~/e 1/.11
a
Jn~~tI
,;, -ft,e-
axiAl IDUtti~ ()f Ike
fJne
}
if
ItJCAJ'/4Y1
QX iA/
/
-
-4 . . ani
Ma f
"YI"e
(AJ)1S~ /It f
I'
1
4/,
/!4f-
T
5"0
.-E f./} aIr;;;:;; :: - J'fT V""
Y
~ ~
.;;:., d,'{'~1 v,l/ueJ df
11.53
The duct in Problem 11.51 is shortened by 50%. The duct discharge pressure is maintained at the choked flow value determined in Problem 11.5/. Determine the change in mass flowrate through the duct associated with the 50% reduction in length. The average friction factor remains constant at a value of 0.03.
Jh/s is like Example
/1.13.
will
chee-K. OUr as~whlfJf,'on by CdH1f?""~
sf/II
c-htJ/(e
P~
wi th
~ p:tt" fire .flow is
ass umpf,'oY1
ca./cu./,:d-e In
p,;
.L f
t:11'1()ff1ey
Oy
QHd
t-ht?-
we..
mus -f
Fey
choked
vntlSf
st)/uh'on
iltfZ
de v/f'e
9uess 1hN"-
We
b~
mus-IHOWYA!e
pyo~/eYVI
01-
al1t7/11ey
w-e-
s-horknt&/ dud
vho I:.e d. For
Y'n.llde.
&IS
fhe,
/n
01/&1
II. S /.
F()y
.If noll
choked flow ~ 6xtl~/~ //./2
Gli1clulced flow,.
~fya,k9!/.
.flow'
:f' fI. -.,( ) P F;9' 0.2.
FrO>??
M~I
.:
T,
,If
::.
~
-=-
-If
V
P I
~
:-
o·
D
(d.S/'I) we
~etd
6" (I)
1./
(J,7
(2)
/.G
~I
Thus ;:: :::(0.75"
) (N. 7psilA.) ~ I I (con
;os1A., If)
//- at(-
fd
//.~3'
J
(con't)
;.0 5'ince
> ~ ; t. 3'1
~ ; •. 3{J I'I/tJ..
Itte
fl(}w
loA' -=
r;.
i.t
che1k.e4
4. f"
1'~/4..
tlJ.U.lnH!d.
be o6ht,J,ed wil4
Cdn
G1.1/.63 SInce
~
't CtJ)1.f/rnd.
_.r
I
lAIiftt. E1' I ~
/I'M
1A/e.
have
-
~ A:l, ~
=
~.Z.68
':
~
, 1'}1
The.
flit!
-
.f
vhtU1lj e.
,n
h1,u j
?~':- M,09x
!lflW YtItIt:
I"J
/00
'() If
The..
1'Y'·~Hs
J4/~
I'lowr..!c. i:J7
;;'cvt4l.ted
"r
sa)..
1/-.85
9. ~)f)
i;/lrel<'l.
f1....t
fub~
W4J
7h1l$...
11·5't
I l1.slf
If the same mass ftowrate of air obtained in Problem 11.51 is desired through the shortened duct of Problem 11.53, determine the back pressure, P2, required. Assume f remains constant at a value of 0.03.
rhis
.r/"",;'/t:{y -fo EKe:tht'pIB
i1
fl(}w~ale
due-I-
ach/el/ed
of
C;'felIHed f'yz,bl e,-.
Pmhle.,
m
/1.1
Prt.J6IeJ1l-l
11·~3", /Me.
.;."
P1"llb/~""" 11.,1 _ Tltu$,.frJy
II't
IMIlS$
is desil"l!c/ wi!A n,~ ~h/),.ft",-t(l/
11.51
I?eu
the stJ~e
Since
Acl.iel/-e,
'/17e.
fl,e. val'te ,,/- Mit,
vcdue. of M4.J.. t;tf~
Stlm(.
we. ~ ve.
lIS'!
D
F (J., -l,) £)
oy ".~ ff
./:)
W/fh
f(.lY-~'l.)
-=
().3
wt..
enfeJr
FiJ. Po
and r~~d
Z
lJ
0)
= 1.6 1h.e Vlt/ue of p '" ~bhJ;'ed /~ f ~ == 6 ·ttl (A.II\.;"
wi/1t
1:1_
I
P." =- ( ,.6"
PYDJ,/e-, 1/.53 iI
psit;..;.
we ,e! )(~. ~fi psi"-) ~
11- i6
(l-.
-
~$;a.
.rh'll va.I/Ii Sf) I
I;'
//.55
I 11.55
If the average friction factor of the duct of Example 11.12 is changed to (a) 0.01 or (b) 0.03. determine the maximum mass flowrate of air through the duct associated with each new friction factor and compare with the maximum mass flowrate value of Example 11.12.
f' =
(a) For
we have.
().Ol
F ( ~ 1'_.l., )
_
(1/.01) (7.", )
/)
c
0.2
(()./ rn)
fig. P.2 we ret2d
and
~.7
fv1~1
= /.
T,
rtf -VI
v~
0)
I
(2.. )
() ·73
~
FnIn-t 6 xantple
/1·/2
7,f :: 2 'f0 J<. and
= "3/0 ~
vAl 7hU5)
TI aYld
wifh
=(I. wif1,..
V, :
I
) (2. 'I() /<-)
I
f'j.2
(0,73
=
f>J ==
£,.
s
we 'jei
we
=
2& Lf k.
%fa,;'"
) (310 ~) $
= 22.6
c:!
S
0.72
(().72
= /.7 1/-
81
1l" S
Poy f;;
r(
we have
0.03
~ tl_l/)
=
((J./ ~ )
/)
Fig.
on
Ol1d
(2.,.,.. ) ::. (}.6
((J.03)
we-
P.2.
Yedd
-= 0.57
/V1 ~ I ?;
-= 1.13
rlf-
V,
= ().6
V.f'
1Au5
J
1; ~ (J./.3 )(z'ftJ t-) ==
=
27 /1<
'!?) -=
(0.6 )(3/0 /g6 ft'1 I S S
V
Fit]. [). I
Fro;'Yl
1, ::
read
We
~V"
Mtf, == ().S7
O. <;
P..
~J
1h1.A5I
R=
10
(0.2
de-k¥IM/ne
-k., =7 /.52-
max./mul"l? (ch()lc~d
The
.
WI
-
1·70
::
1.65
f"" O.t:J1
IY1
{ -=
0.02-
In f=(J.03
-Ii,
-
f
k, J
:::
1.5"'2. ~ ,.)
/1- g8
II. 50
I 11.5'6
If the length of the constant area duct of Example 11.12 is changed to (a) 1 m or (b) 3 m, and all other specifications in the problem statement remain the same, determine the maximum mass flowrate of air through the duct associated with each new length and compare with the maximum mass flowrate of Example 11.12.
FoY'
maXI"nUrn
12. - /', =-
raJ rOY
I'Iowrak- the I""
duct IS
we. have
tJ. / ".,
.J)
F ,...I' q
Fyc'f")-.
Ma, .::
[J" • .-
~. 7
T, = I. -T" _V,
I
(I)
n. v.73
':
II'"
(20)
Ex.~p/€
Frtn'YI
choked.
/1./2
T~ ':: J. '10 k tina
= 310 ~
1I:t< 1hus
I
~
~
wif'h
£'1' J we, tJb h ,;"
~ (I. I
and wili, £1,
)(:J.'10I<): 26iJK. 2
we 'Jef ) ('310 rn)
= 2. ,,6 S'!!!.
70 deieYm;¥Je f, we €nJw.
F i9. O·J
ltj::;
P,
~,
,
nus
~
(I
(0.73
s
= o./l. -
-
(0. 72-
(C On't )
/1-
8'1
wi#' Ma
J
=0.70
CLw;(
read
1/.5"6
I
(con'f)
f (1 ~ 1,) _ f ( 1.2 - £, )
-
J)
~
Fig.
NlfA J ::: O.
T,
II'" -
Thus
-r;
[J.2.
~7
J(2¥OJ<) :
(tJ.t
h1-
Fr~m
==
.
hi
le.(,: ;<,...,
= ::
.l")-~ =3.....
!!!
S
D· / we. Yead f,.". MIA,
.)-1=1"" l I
,
27/F:-
) (3/tJ !!!) = /3' $
m
J'YI
(d.1 ;If)
0.6
:;(/.13
V, -=
(O.tJ2)(3 W1 )
::. I. /3
r lf V,
/)
--
1·70
Ie,
-
.$
1.65
~
.J'
1.~'l1t9 J
//-.'10
=0.57
_ 1J.6
1/.57
I 11.51 The duct of Example 11.12 is lengthened by 50%. If the duct discharge pressure is set at a value of Pd = 46.2 kPa (abs), determine the mass ftowrate of air through the lengthened duct. The average friction factor for the duct remains constant at a value of 0.02.
:f (12. -.(, ) [)
=
Or
= d. ,
(O. tJ2 2(?n,2
,cu':.~)
-
:::
L
rr~
va(~(!
a.
9lAfSS
- • r'9'
r !..et'...l,)
,..,...,
v,
and
II-
•
a.
of MQ,.
V()/tle
£,_ I
H. Wi"",
MiS
wif;'
£)
Mt:i2. Ilhrl ger a. CdYl'tSI'OrJlil,y
of
o
ti)
--
f)
d'/hl
We
_ F(L'--t1 )
we
vnlue
Witt-.
I c"lc.ei/"fe
t/fel-"/.
of
f/ .(J!..~)
Vttltte of fa!~) of £) ~ value
w~ 9f!f ~
r/~. p.2-
p
M~J
()'ltd
MtAl.
Then
we.. t76Iai" Ii-rm,
':iys:
P./
with (2)
we
fie +-
VA/{,fe
&f
of
Pl. thai we
We 91Ae5f ~ value of .M~2.. w(. rea. d on F ,'3. D.2.
COn
=O.9S.
J)
J),en
with 6'1' I
we Oblal~
Fre'-L, ) =
(). 6 +
~.OO128
D
fw
fl.Jhie.h
Mtt
~ ""
j
FIJ-
-
0.2-
tJ· 57
( COn 't) //- 9/
tJ. (;
(b)¥tftlTt
fAlift" 11. = '11,2 kla(dhJ)~
CtJyye-'l'0nt::t/~
-h Ma2.=~''1S'
/1.57
I
(con 't )
[1-
Now wi ft"
R :::
[/0 I
2.
fhe aJf ll ""I'-h't1r&
cl
(~
= I 7:r. )
J
1)/ J
IJ Iso -from V~
_I (l.tJ6
I
1.96'
)
= 'I6,61t.At(ab.s)
/
k t.(I h ALte,.;.
tv1fA~:: ().f)5".
:: (tJ,81)
[/tJr~Ik(abs)]::: '31.kIQ{4bJ)
7;, ::: (0.1 'f ) ( 233J::-)
(
V,
)(-
1,
~I
T
'Ie f
~,o~ (tl6s-)] (0, B I
/1:;) P°
P =
'J
aHcI
Wt
(lose... el10UYh fo flu JI'v~11 c//5Ct,(1Yjt.. fll'tr$UYe
which is 7lt1l1
2
F
:; 2-7 I I<.
I
19. P.2
fur Ma, =
t7. s; 7
0.'/
=
anti f/1u5
V. :. (t). (JI J
(s~~ f,.)7r(O'/h/)2.(;S1 (286.? N.fJ'1 )(27/;:)('1)
*,.1::.
111•
-
/.55 -119 _
s
1/- 92
f!)
/1.58
1
11. 58 An ideal gas flows adiabatically with friction through a long, constant cross section area pipe. At upstream section (1), p, = 60 kPa (abs), T, = 60°C and V, = 200 m/s. At downstream section (2), T2 = 30°C. Determine P2, V2, and the stagnation pressure ratio Po) po.' if the gas is (a) air; (b) helium.
Ma 1 =
F Ij. p.2
T,
~
r"" ~
:::
?if ~II
;
= 0.55
:::
'Kif?.
/./ 'f (I)
19
... )
('~
/.25
p~
D
rrom the ftmper"fzo'e.- raf/()
r:lf::.
T,
~
/.1 Lf
,if
(~J3/C.) (/. 11f)
(z97. Jt:.)
Fiy. p.?.
(,.vi H,
7'l.
::;
/.01.{
!J:i.,. /
rtl/fd
-r (3)
&. :: /./
p~
(iifhd
(con/i)
(4)
1/. sa
I
(contf)
WiN,
~:: ~ ( P )F)( P,. (}nd
E.
Ii::
)
P, pi( I and '3 we obl.,;'
[60 f
-\(I/)
I
I
I.e;
AIs()
== 3'1. 7 1zf" (4bs)
wllh
I
~t. I
Pc, I
~/~ ..-- .
:;
) /_1 _1 = (J.g
(;
{ /.25'
~I I
-;
r ',-tally
(216.9 N.I?/ )(]()]fc.)(f.tf)
V.'J.. ::
"kf.k
(I
N)
ky.e
;:
?I'f ;'
s,.
(6) Hr/ heiJ~m)
The
Mach.
It? =/. 6{'
nUMbey
Itnd R
= "2.077
of- sec-han (/) is ::.
N.m
.f~ Ta61e /. B .
-/
(2.00 ': ) 0;:.
0./17
(5) T~
Or
(7)
(con'+ ) /I-9/f
I I. 58
I
(Ct) f'J If)
Wit? ftp - 5
obiaJ'n
we
T ~::: (331j()
:;: 2!i"3.2 K.
(I-j/~ )
Thu)
- -
T-z.
(3()3 ~)
r~
(Z~]. 2/:..)
T'j.
Wi~
-:::
1.ICf7
~
we...
t. /q7
T.I(-
M~2. = if (R-fL )
-
I (!!:)
!l : Jp.r
~
;:ft
Mal.
::: .l.
W;ff,.. !{ =
alid
MOl
i*
(1
.
]
l
r(*;')M&t{]
fl Z. )(1 +
{(ir' ~
~
~$_
f
I
r f I
(ft~,) ;
=,1-' ) [ (().SED3
Ma;)
(: )(
f$ )
~ tl'Yt4 8
/,Ale..
(i~)
J:
fe'!!:}jo.)gDl)"
?;:.:~';.
1Jj
2.
I
ha lie
(',/3
_ 18.¥ W'i(4 h5) -
Wi~
-F/Y'J t:i II'I ~
~::: M"'l..VR~'k.
r
077
1'0 1J..J.1<
)f303K)(t.66)
(I
~)
1<..9- ~
.r~
5<;3 !:!:!. -
5
1/- CIS
(8)
IffS
1+(!-'~-I)(~.5"g03/J
(O.>80~
f?;; [6o*Pq(~S)J L!-)(/.81))
~
J(/. 66-/
L (/. 19 7)
~ Jj1';::':/--'.-)ll.,,+,)L [;2- iT; '2.
7(f-)::: O. fig~ 3
¥) _I
2
/I. 5"9
I 11. 5"q
For the air flow of Problem 11.58, determine T, p, and V for the section halfway between sections (1) and (2).
If
is
secf/on (A)
halfway belween sec/70/'u (/) tlnd[2.) we
p/ac(ld
f (). ~J..,)
_ -f ('(~-.l2)
[) f/!~.fll)
and wifh
ha~
D
Pi,.,
we
(I) P~2.
D
(',IYld
Ye~P/ (O"l'"l"es,Pone/)Y:J
j)
values
(.2)
fl- (pas
where.
pJf
I/.
.
l1f
Ft'Ylctll'l)
A
witt,
() b..frL/rl ed
IS"
:= Ma.
"
70
ob-/Q;"Y1I!'/ Ii? t'lte so/u';'/()r} ()f PY7Jblem II. 58'
VR~
I(
F(,R~.I,) and
--
deferrn,;'~
we enk,. 1='7-
D
Ma.l
= a. S~
and
vead
aVId.
~O;..
:.
-= TheY!,.
D
0.01
ft. {
:f (./.1_1.,,):;
~_~_1 _ _-_0_._0/~)_ 2...
(C OI"1't) /1-96
with
So /14 '17'0.., of Prob/el'J? II. S!
0.9/
_ 0·7
D wi ~
1J·2
+
0.01
1/.51 1 (CO;1'+)
= ().36
f(j/-.l.A)
Wifh
f)
Ma
"
=
(),61
TA ;;; /./ r*
~ = /·7 Now
wi'M
r..
~
f"
2 we
obtai;'
-= (j./ ' )(2'1'1;:') ;
P-. :: (/. 7 A
)f-./. ? / £1-
v.II ;"
321 K.
(tJ, 63)
if
5 if
we
(Z~6J;
'1(~ (lIbs)
have /v."" )(32..3(1::.)(/''1-)
'kJ. I<
(I
2:!.- ) ;t.,.
~ .f~
_
2'27
~
--$
I
11.60 11. 60
An ideal gas flows adiabatically between two sections in a constant cross section area pipe. At upstream section (1), PO.I = 100 psia, To. I = 600 OR, and Mal = 0.5. At downstream section (2), the flow is choked. Determine the magnitude of the force per unit cross section area exerted by the inside wall of the pipe on the fluid between sections (1) and (2) if the gas is (a) air; (b) helium.
/ " CMSItI,,f 0_ pipe ~ / cc,,~1 voJuWte
r I
p,A ~l I I
(I)
(2) flow
The.
-f7, (2.)
s-/:;efch eel ~h~lIe is IAse4. Afl'lr/~ 1I1t. A'(./a,/
V" lume
fr-o/
(,011
fVfh'lIdJ)
cam.,()t1enT t7f fhe //net1y l'JI}()menful'Yl ~u.ah'oh (€f' 5. 22.) -Iv i11e col1Tel1fJ
of fhiJ (;1);11n;/ VOIUW1~
pipe wall
"11
I
Tt,us we (a.) Po;"
lAIr
T,
::-
-frff tlte -ffm:;(.... exe;led by lite.
the flu/ell 1<;( )
== p,A-~A
Rx
w~ ~f
-t
rn(~-v2.)
~ - p.,. + ;0, ~ (~-- V1. ) JIleetA
~) Pl.;
we.. enter
(I)
fJl) ~ aYJd ~. ,:; J' D. I with
CJ,7-;) Or
Tol J
tJlJd
f,
::
OJIJ.f
F:, .I
TAus
TJ = ((),95
and
535
t!
.f
/1- '18
1/.60
I
(Con 'f )
Af sechi;n (2.) the flow is c/1bked. 7J1.M<; we. use fhe '" s!z:.k. p/ fAe F-anno f//Jw/ Fig. 0·2 lilY s(.cH0f1(2). 6nk¥J~J Fif· /)·2 wi/1, Ma,=IJ.S we ,eAOl
and
v,
=
= (J,5'f
V*"
It; I, V,.
'fl1u.~
~:
£".
w/It,
f'/I)w
R.AJ(
;
~
''.>If
£~i6 if
2 - logo f!
(0.5 J.f)
1
S
~1Ie.
I#l.
=('1if ps/()..)(i'f'f Ij,.~) -(3't''ffJ.ri,,-)(I'Ilf
(b) For" hell~WI
(R=
/2/120 f-t·lb
£'J~' /1.~6 q""d //,Sf
-=
Mal
P,
='
f.
.fi.trw, fable /,7)
we. use
~·S QJlld o6trAin
/+
f·66 -')d.~l 2
and
=1·66
6~tJ d~
7;
/f(~t)Mq,'
1<.
an(/{
5/£,(,/. PI( /IV;J-!j..
1;,.2)
if"
-fi1-
1.6'
f- (~Y'W" 1t~ (IM,Si·i, f-{l;"}(M/1 ,.": I f-
aJ1t:A
1/-
qq
'MIS;...
/ /. b0
I
(CO /1'
t)
AI- secfioYl
(2)
fhe fl()w IS ch()ked. TJuA~/ we use. fhe '" ~-lA1e
of
f/tJW
fuy secfidYl (2). With EfjS. //./07 a.JIICi /1./01 tfnd
Fanno
MA I -= 0.5
we qef~/.
=
q pr/a.
12
~!(~) _-=Z-=--_ _
(P.5)
.L
'+{I·;:-0(~·sl . P 1{ = 36· q P s I~ =
/?
(~
:::;
~05() ff .J
N()w
w;tt.. G;. / we
hdll~
11- 100
=V
2.
11.6 I
An ideal gas enters [section (1)] a frictionless, constant flow cross section area duct with the following properties:
To
=
293 K
po
=
101 kPa (abs)
Mal
=
0.2 For Rayleigh flow, determine corresponding values of fluid temperature and entropy change for various levels of pressure and plot the Rayleigh helium. line if the gas is
Thi S if 5/11'1; lar
fr; lXt:lmple. 1/. IS.
pJDf ihe. Rayle/9h //ne
10
~s*uJ
for we. uJe E,.
1/.111
2-
p +
(,IlV) RT
::: ('onsfant ==
P
pI +
1-
(~V,) RT,
II)
P,
and Ell' //.7b S-~
Iv
('OI'l.J1ru"t
C()
('I'esf1(h?d/~
f( aylei?h
,r -
::: Cp
In
a
r;
-
R In p
(2)
r:
fable of vtllue~ of femperaturG (lnt:l/ t!nlnJl')' t.hllnJe press"r~
Ir> d;ffe.,e""f levels of
dow1JJfy~,~
t?-
flow.
10
de feKjNJ./;' e
f()
IJh /Q/YI
P,
P, we. lAnd
use flte 17
M4,
~
aJ1P-
E,.
11·59
-/rdm
,P"I
1() de fe?'~"";' e. 0 we. use r, tJlI7t7 fhen -Iv 1)6 -m,'n
74,
M~l
7;
ani £1;; -h-6W1
/1·56
I
~
r.. ==(}) 7;,
(~)
~I
I
We.
ob1rJ.,'n
~~~"oV- C4l'1rreu'-/
~
(5)
(C()n't) //- /0/
1/.61
I
(C()n't)
281.2. I::.
With ffj. 5 ~
v, =
we.
~Illlt.
('1 7.72 )(11) 1
flo }(O.}.)
(~)
/1- /02
11.61
(('0;'1
't )
P[~'IJJp.bs)]
T(/(:.)
S"-~~L)
90
583
3g33
KO
8B3.2
61tfB
11.,. /(.
7(J
IOq'2.
763 '-I
60
J2J2
'84- '10
~O
/236
fflZO
if5
12/5
q,oCf
'f0
, /72
qlh 5
35
/105
1135
25
903.3
g7g /
I~
€Jlo.'1
77CJ5
1100
leDO
T(t.) %0
7DO ;00
~o~-----------------------------------------------7N)O
~ooo
11-
}03
'l()()O
roo 0
1/.67-
I 11.6 Z- Standard atmospheric air [To = 288 K, Po = 101 kPa (abs)] is drawn steadily through an isentropic converging nozzle into a frictionless and diabatic (q = 500 kJ Ikg) constant cross section area duct. For maximum flow determine the values of static temperature, static pressure, stagnation temperature, stagnation pressure, and flow velocity at the inlet [section (1)] and exit [section (2)] of the constant area duct. Sketch a temperature-entropy diagram for this flow.
For
yYI(JX
Ray/e/9 A fl()w is c.hok.ed. fbY' I-he.
/rnlAYYI fiow/ f-he
isenfrup/~
nOJ3/e.
1.~I
~
T
0
-:;. p
P.,
~I
=
288;:'
;:.
/0 I itflfL (A~ ~)
"
10
defeyYJ7/-"e
fhe
we
need the
value of Ma"
~-/r;l.fic.-
af
5/-al-e
exif Rayleigh /;tJw I~/ef .I / deJcyYH'YJe M"t we... use
f/1~
70
nOJ}/e
hfJ 2. -
~(J)/
-
~ -=
c,o
TGp.
!l
-r
r =
(5"O~ (JIM ~) 4LJ
/
Or-::
~
~I
" 7;7. I
";;
0,37
c.h(Jked ZSs k.
- ~,
:
) -I- 29$1::
== 73£ k.
~)
P.f.1<
f/dw )
T
~2
:
~a
we
J
~f
== tJ."37
7g6 I<
we.-
eYlIe¥ -F;~. lJ.3 and
rene!
(I)
= 2./
~
T,
::
~4
-, 7,;4
~
I
T ~I I
With
7;,
::.
~2
rIM'!
Clnd nof/nJ fhat ~
~I
(7:
(2)
&.S'- .2....
~
/1- 104
/1.,,2-
I
{Coni!}
v, =
(}.2
Va.
~
=
1.lq
(4 )
forA I Wi#! E~. '-I
~(). I w;th
p'"
:::
I
-=
I./q
1Vl", I
P, ~/ I
we olofa,'n
~
100/effA(abs)
8q.· q fefa.. (().~s):;; 1:.
::
", "2-
/.1'1
we 'f'e.a. d fYtrM
0.31
r::ig.O./
= 0.94
'£. :::
0.93
(6)
7;"J
frs. ~ and 6
Witt,
~
::=
we, get
.)[ IOI1e..P~{~S)] :: C/'5
(O.9Lf
~ :. (0. rSfll.f ) (Z.SB t::..)
1huJ VJ ::. MfA,
r
R T, f<
ZBZ IZ
.:=
= (().JI)
?, 2./
II:({
=
.
v, ~. '2
{}.2
(ctJ1'1 't )
1/- /05
f
1/.62
(un't)
S-s ) .).
5.2.
-s) ': ct i l~
$2- -
~J':
f\ In
T. -
(,/
?~ P.I
/6 tJO .IJ. n-. ~!. k
7(1<) 60()
/
1100
p.; q5 I •
/
'"
"
kf4.(";$ )
~;'=2F£X z~o
/1- 106
/
~
/1.6 '3
11.63
An ideal gas enters a 0.5-ft inside diameter duct with PI = 20 psia, TI = 80 of, and VI = 200 ft/s. What frictionless heat addition rate in Btu/s is necessary for an exit gas temperature, T2 = 1500 OF? Determine P2, V 2, and Ma~ also. The gas is (a) air; (b) helium.
10
the f,eaf -/ran.Jfer yak we Use fhe
defel'm/ne
eneY9J qUaftbn
(fer 5'.69) -fo gef
Fo.".
• Clhef
=
n1a$$
.flow'ra fe
.
(I)
m
we. = P, Rr;
-r~'T
2-
1rJ) -
I
Lf
rAY1d
(2.)
V
J
r~J we.
U~e
£0. /1.56. 7hu~,P
/
~
To <7Y
use.
(3)
/ +(~2.-I) /vlt{2.
-!r;y air
-r _ {(Mo.)
-7; -
To def.eYm,ne
F/:J. P.
In
!?:: P, (fa. )(J wi fh 1:$_ //.
whel'e
-~p
It -::: If
or -fiN.
(1
(L/j
we. use.
If p,
I
ItA /23
)
(5) fp, /?~y/e;j),
It.
It!vf(). 1.
I~
P :: f (Mill) 111 flo. Foy eXir ve/()cify I '{::. Mtl2.
(1) ~ I
we use
VRT"l. k
We. dekYm,'rJe MaI wift.. V. /'vf a. ::. V, ::: I
flow
C,
(C ()n , f ) /1- /07
(q)
II. 63
I(con I! ) we determIne
Gl1d
r..
-
=
7;,.
Mal. wifh
(~~)(;
)
~
f~ . /I. lit -fot'
elY/of
T -:: ~
fri
/(ayle:Jh How/ nlllHe/y
T k ) M.
/ t-
k
(JO)
M~"
l'
(II )
02)
(a)
Foy
0/1'
we
defeYml;'e
Mal wilt,
£(r q.
1}1I'~
(ZotJ tf)
= o·/g
T, -:: f).
19
7;, I
T
~I
:::
= 0.99"
M", I = O.lt
Wdl,
Sr5~
T.
= 0./7
?
:: 2·3
we
read Qr
and
TAu>
PP... will,.
f..;. 7;..
-::;
G~. It) we obm/~
(I y60 'r< ) ~'f()
(0.
17
):=.~. 6 2-
til<
(C&n / t) 11- 108
/I.G?
I (COf'l't)
F~ h. ~
gef ~
we
0.62
Fig.
D· 3
~
Ma 2- ::: () .1/-0 GVld
~
Pa.
= 1."6
MaJ.. :::
Wirh
1?-
we. Yed.ci
0.'10
PIg.
On
£) ·1
0.97
=
7;;2J
Thus
/
1'1' -0.97
-
q.§"
we have
(J
~"]., ~ J
Then. wiM
Z020 DR.
R. ;: (2CJfJ/IA.)!~ ) (1.'16 (2.3
:1.
Wi th
8t.? t(
:=
(0. 'fp)
Wi I?t Gj~ 2 we .
=
pfJ&t.
we have (t7)t
ff. /6 ) (;q6~ 51u.9·
111
= 17
)
'~
(I .1!z..
:::
s~.fl)
,e f
~&I : '
;;.
(zOPJ/a. )(1'1'1
In. 'l)
7T (0,5 ft )). (ictJ
F/).
(1716
tJlZ)(f.l-f)
1'1. /6)(5~ 'I?-)
ft)
:: 0./22
~
(tf)
.slll~ !I?
11110< (NiH,
f2~ef '1
Gj.(
0.6*',;'
t,Je
: (tJ,/22. J'Iu,)(to()6 -flit) (Z02cJ~ 54) tJ/?
..r
I;,
.
(778
~Lf5~~)
fl.I") Bn..
( ConI f
) / /- loq
::
13'M 8~ J"
//.63
I
(c()/I'f)
(hJ!iJ-y- hellu'm J f<.:;;. /,66 And If W;Ht ~. '1 we h(lve (Z()tJ Mt{)
::
==
/;;'10/20
H.16 $/~. d"f
(t)
fY-~ Ta6/~ j.
-
.,
7.
a os97
Witt.
~ FtJ y
I
1v14,
::
"':
f f/! -/)M~ 1 ;
= ~ .() Sf( i.>
EI5. 6
(/f;'HO{ / /
If- /.66 -------------------: / I- (1.61 )(d. d5''1'i) l
(S'IIJ ~)~ t- (I. 6~ -
(J'tJ ()S99))-= ~'I~.' ()I! 1
,y/e/d 2,6o/~
lind
or (bId
Mq
:: d.1/6 2
We
use
added
!he JuhJ(h''/c
If"l1d
su,tH7rSd11/(
bY
s-, /t;
-
/IV: Ih
h~f
I
f~/uh~/
Mtil.. ... O.J/b >/},c.e
dddil,dn
.flow ~ a....
(Con If: ) /1- 110
we- cal?'?of .>'ubfdn/c
h~4.f is
be,:,
a.cce/e Yt:<~ ~ ctJV1dt'I/OJ.-a vf/JiYrta"".
/1.63
I
(Con 'i)
Wifh
=
J.1a l
aJ1ci wif{"
~~
£1. 6
We obi7A./~
bj.]
f-(i..6'-')(o.1I6 )2] 2 M.a.).. = O. //6 we have
:: (1960 4~) [
T. Wi~
0.//6
a-"".rJ
I
_1_r_/..-:'6:..-"___
;
;;
/fI~7 DR.
z. 60 Z
1+ (1.66)(tJ.U6/ IN i It,. Ef . 5" we
P.. ::
w; ty,
(
Gtj. g we
~
!Ali f?, {'J. 2 =
we-
19· 7 (ISla..
2.6tfll
Yv; V~
= (6.116)
m
-=
(Z() f.Jif). )/_1_ ) (2.602)
l.
---------------(/),,'f-?,O
Ff.11; 5/"9. DIl.
')ef
(20jJS/a.)(I'I'I
)(/q6~·R)(1.66) (
II,
/
f:.~~)
('1)
In
(1-
/1/
!!
=s
5IUff!)
7r (()·'5)#-(ZOO
Ql1et
-=- 737
jt) -::
-
.f
11.64-
Air enters a length of constant cross section area pipe with PI = 200 kPa (abs). T, = 500 K, and VI = 400 m/s. If 500 kJ/kg of energy is removed from the air by frictionless heat transfer between sections (1) and (2). determine P:T2 • and V2 • Sketch a temperature-entropy diagram for the flow between sections (1) and (2).
To
slrAfe of the aIr
fhe
defeYrnine
rech'on(z} we.- use fhe Iv c.a/culak ihe value of TC12. . ~5~
e~n{(ll/on (E~.5".61)
eheyCit, iI/
{)
Cff
:=
~ Ylet in
or~
111et-
::
I.,
~2.
We.
c"
()b~/~ v~/ue
M~ I
=
)
To,
~
of
/V(().., •
J
a
70,
+
~
-
-
~ nef ~
c,P
T,
w/'ic.r.
7;1 1
we
M
de fe.Ym/;1e
+
(I)
7;;, I
ve~tt
Ma I
fY.~
FI·9· /). / w/tt.
IV"~
~
:=
(2)
{R7;~
('/
we
abo
enfey
Pi9- P. 3
p,
t)11d.
tea.c/ volu es
of
wif"
p~
(~)
W//-h
thi!
~
de feYYYllhe
I
7;,.
Ill1d
P-,~ :;
(1/ it') f, P,
~
(~2- ) (~ ) '0
=:
v
wi~
2
('1 )
~
(oul
or
value
(5')
Ti
~ =(~~ )(:) v,
(6)
(con/t) //- //2
u6'+1
(con/f) ~
Gtj. 7,
We.. use M~,
_
gef
(lfo() 1)1) (z?{.] Mffl )(S~OI<)(i.lf2 1;.9- fr. 7 .!:!... ) t?,. l!!
(I
S?.
r,
-= O. It
~, I
p,
'=
t·/if
~ T,
~
= /.02
V, = o. '1 1I't:t
aVId
~,
,
fI~w
T. ~""
=-
wI ft.t
~
99 7;, I
= b. qq
~&(
((Of/It) //-1/3
M~
2-
.5. ::
;;
f).
IS
2.:3
~
7;. ;. d.1l 7;
aka
= IJ. () 7
V2,.
lie..
With 1-hese. yah'os
Gjs-.
LfJ ~ a wi 6
~
=: ( 2 .
3
antii fhose ral/os
,6ttu n
-1-0 )
~ ::: (C'!7
CPfYefl()ndJYy
r; / }[ ~
lao
l/Uai¥)]
)~ 1
_
fo Mq, = P.f7 We lAse
=.t!!..! ItJ/;. ("/;$) <63
k..
( I. f) Z
and
)~'f 1 No k Thi 1
That acetfh;/ I ~J if
no!
To~ ~ U~/I1'.J ~'" J
~I ~~
If.
E1'
w/M ::
resull-
Cf7n"f!cf
({sstJc/a/ed tvjlt,
!vftl
tJ :: 31 1?1 ) ('f ()!!2.) s -s
UJ/ny
11.13/-
I1rs7 Juwe-W/Y;, we..- deJ~WJ/~-e
ThuJ,
1/./3/.
;2.(iu-/) Mq/ ( / t-
(I
t-
~ Mq/) =
kMq/)1.
2 (I." rl )(0. [ / f-
ov ( C1JYl
'f)
/1- litt-
m;1i r(.{:!t:. tj if?
(I. if) (0. !13/ ]
II.
GLf
I
(cOn
'+ ) T,
Now
10, .I
I
T,
I
-------------
7;"
r;,
~
=
- (50PJt:-)
aS62¥
Now (IIJ; f4 E1-
We
~
) jc..
,,6~/h
~./776
w;ftJ
r;
::: el· 7'1 !<
A
/v1(f ... =
7hert
t)
f()iI f
Wi it... {'t. ~
TcI
we J"a ve
I
_(~(}~m
1,;
)
::::
(J.!6 21.f
O. !62'1
J
r;
~
I+(t,~-/) (tJ,F'i3)"l.
I
6$. I/. /U a.J-td Mil, = ().gq] Clnd Ma.z.. = ()./776
r (u-- k-) Ma ,]Z, : :
":: /:--.
/ f-
-k ~4,"Z-
[u
t- ,·if )(
/
~,rIj3)
+ {to If )(0. ~9'3l
1/-1/5
i
2-
==- /. 0
z,
/1.1(.
1~f
1/.{,4
J
(~f1(f::)
NOlA!
l;:::
h.a ve
E~ . ~ we
witt.,
(1)./666
J(--.!-)(5t;/)/C.)
==
8'1. /'1 k.
/.026
T;, :; &'/./1 K. if
(is
Fov
si2dv.-Ld
T-
OlAV
S-)'. l
J
:=
-
<
T:., = gr 7'11< ~~
~.
S"
7,030
~1=.L1vL
We
u.se
£7'
//.7t -fl? CA/t".J4 fc-
2 1<.9.1<.
200
100
o
~-----------------------------------------
1/- //6
~-~. TJrllJ,
11.65 -' 11. 6'5
Air flows through a constant cross section area pipe. At an upstream section (1), PI = 15 psia, TI = 530 oR, and VI = 200 ft/s. Downstream at section (2), P2 = 10 psia and T~ = 1760 oR. For this flow, determine the stagnation temperature and pressure ratios, To.21To.l and Po.]1PO.I' and the heat transfer per unit mass of air flowing between sections (1) and (2). Is the flow between sections (1) and (2) frictionless? Explain.
To
7;;2 , To,
pyeJ.fuYe
La e.
we
ro.i-io)
:da9na.noYl -femp€rafuYe. Cll1d the s1zl:JYlahon
fhe.
deferWI/;"e
t~'~(T, )(!i) Tz ~, ;' T"
OJ
(r.'·)(ff;., )(?)
(2)
:.
I
one(
fo-z.
=
I
~, /
p~
I.
P,
whe.,e
T 7;
-::
f(~,,)
=
r
(;/1'101
f' ~
70
(?1a)
For
mass
fhe
Fig. O. I
117
the. Mach.
defeYWI,;,e
Ma ::
.
111
J-1u;Hbey-
ed eoch. st-ch"OI'7 we. (.,fse
v
v :: C
Ve.IDC,' ~
at- s-ecfion
(2)"
~
J
t,v~ (lS!-
the. ~~h.Jenlafi;'Y) of
PY-/yZC.,f;a.1 -Iv ob~/;'
~~ = ~ ~ dy
~:. ~ ~ :: I P, ,.o~
Fo". Me
he~t
/-YOI1J/e¥
S-f.cllons (I) aMd {Z} we tJb~/n
CJlJel- = 111
)(!2-) ~
(~ / \ r;
pev UYI/t I17tl5S of dir /hWI-., 6elw~eY1 l1Se. IAe. el1eY'!i:; e$ U~. J1(,n ( E$. 5".6,) iv
h0 2. - ~~/I -=
We.. ~~fz:t/~
(COy/It)
1/-117
/I.
C-t;"
J
T~
(con 'f ) wl1(~tI,~y or" no! f;,e
asce.r-/Q/n
i r {Y/cndn/eSJ
(2.)
M~m~YlIun-,
~
~x
==
~U5~
Fiy} t we
fhe
ep-tmt;n (£~. S72.2.)
Rx = A( ~ - f2)
n
lA>~
f2.
of yj"
1-
air flow belwteYl secn~y)J (/) dht:/
ax/a. I
fz>
COm,bonel)f
(v, - ~)
A ~ {/v -~) ~ R-P1. "I
9'16
\ 5stJ ",(
ECi. 3 p
witt..
we
CAlculale
:=
With
J
::. ~.I f
)
7; -= 0,19
~I
e... -: :. (AYld
~I -few. Mf{"2::
-("). =
~.'97 0, ,/-f .)
0.96
~J2.
&. = 0.85
Po)
"J.,
(I-
Mt/J turd MtiZ . We. Jv:w~
=
M4
R7j
S
(zoo §t )
Ma. I =
Ii V; (~-~)
+
7hu~
\{ = (ISP.J/~ f76~ Ote) (2~~ j"") : N(fvV
I/,e //neerr
gef
€tj." -h detey"",,;'e ~.
/tl I'J/~
or
(con 't ) /1-118
~. I
B
().'f8
(7)
J/.65 ~perafu 'Ie
Thesf:.
+v
fye5~tJr
tH1d
ratio!
tr-ye UJed wi Itt
&;1. I &tI1A 2.
oh~/'y,
-
61'
5"
(6(){){ In
wi~
al7d
Rx
A
€J.
we.
d bf7../~
-F1. 16)( ~~ S '<) ( 3.112 -/) 5 1U.7. D/(
7 w~ hA.ve
(I~I'.J;t< _If) f1.JI'a.-)f'f'f
=
-f
I;', ~~
1.)
(l5"fJI'~) (Nt! ./fi')(2()O ttA I 71" -(f. Ih ) .>It
/<)1,
;=
2J)O
(£ f~ ~I<)
6,R
f.j;'
vv!.-
(.Qytdude.
fkq/
Me-
flow
/I-/lq
JJ
l10t
If)
fi- r'it f~ (~A Ib
3'12 16
A and
7.37 X If)' ff. /~ s(1tf
fr,'c{;'t/hlesJ.
'1-J2.
1/.66
11. 66
The Mach number and stagnation pressure of an ideal gas are 2.0 and 200 kPa (abs) just upstream of a normal shock. Determine the stagnation pressure loss across the shock for the following gases: (a) air; (b) helium. Comment on the effect of specific heat ratio. k, on shock loss.
W-e ~ hoc*.. I
F:.c:);
=
(/V"
(1- ~) 1;)(
x
(I)
I
To
fh~ J'1a~Y1a 1/~J1
defe YW//ne.
ThusI
PJ-eSJ(jy€
yan6
[j. 1/. /S"b'
1
1...
[(¥)Na:]~-{I r(~)~d:}I-~
((~)~: - (1<11 I~)IJ7{"!-;
(2)
L ~+I aIr"
Po) y
Pox I (a) FtIV tI!~(i;'~/.'I) we ha ~
~y
::
(J.72
TlnlS / wi tI1
E'j. I
J'v/P
x ::
2.()
./
~x "I
~x
-
(b) Ft¥
t;, Y
6';" I h )
=
§6"
=:=::..=.
~,att
At:We wi~ &J.2. (!.t6) f~ ~ :: [{I¥')(2.0)"2.]~·6W'[1 t(t~)(2.ot I-I." he//uYVt{~::1.66)we
~/JC ClYl d
t)
(fd4 k.~{tf6s) ](1- IJ.72
=
J
I
We
R, x OC¥bSJ
)=
0.7621
J[:l (/.~I)](2.~~2.. - /i.66-')1'~-1)
2l
with Et·
~
..
-
1:
1
,/
I
-=
1.6t f-I
( Util
we 9e-1 ["20"VP-.{aJ,.sJ] (1-
().7'7..1) ==
pcuff (a) OA-Id (b) we UMctYrda {)t...
ntNMA../ shtJck...
decrea...J't:...J 1/-/20
J2i..
M4 Ittt los> wilt,. aJil
I
kkl !i'a.5UI'e
iV/c~e
In 1<..
1/.6 7
11. 67
The stagnation pressure ratio across a nonna! shock in an ideal gas flow is 0.8. Detennine the Mach number of the flow entering the shock if the gas is air.
To
- flDW
d efe rrn I;'e
ClI i"
eMteY/~
>ltoU:...1 Max.l
fbl
pY..(!$st.u'e
-
Pc;y ~x
_J_
~y {,{vtd
)(
.I
we
= 0·8
read On F/g. D.Lf
/.83
1/-/7../
~Mkr
,,/ve.Y1 fl,e Flo. f). 'I .7
5 ~nq;f;;Y!
w/ft,
11. bB
Just upstream of a normal shock in an ideal gas flow, Ma = 3.0, T = 600 oR, and p = 30 psia. Determine values of Ma, To, T, Po- p_ and V downstream of the shock if the gas is (a) air; (b) helium.
Ma y
(I )
(2~ ) 1v1o.: _ I oy
.f;y ~/Y'
~-I •
W£
de fe n''''~lIrle
~y
= 7j, [
U5e
T
CJY
Fit}. D.l.} fi.,. fvl4; we.. use E~. /1.56.
Ii-
we tH"~
tlS 4
funchfM of A14.J('
ThlA.J~
f*;.I) M4/ ] ':,'9' lJ.1
(z)
~
To
("3)
or rClY
(5)
!} p)f.
1/-
122
as a:
Fi< hch~
of" /vIqx'
I /. ~~
I
(CO/1'-f)
(to FaY
we
tllY
reQd
~
Fig. p.Lj
-mv
Ma)(:: 3,0
= 0.4-75
Mit
'}
R = 10."3 .1
{n
PJ<
-=
'r ::
12
(10 )
fJ.96"
T
~y
~
£~.I w(: qef
= 16w of!!-
~
: (7...7 ) (loolJf<) and t/1us with. E'I./ tJ 7.
~y
::
Wilt, 6,. 7
E'j. 9
p..
~y
The Y1
we
=VO.3
~ aYlc{
Iy
==
6.9~
j, .frt ;"
=
J{3tJl'flP.)
3tJ11/'s/o..
y/e(p(.J
)(3tJ J?.fJf:,..)
= (12 wi#,
(J
Gi. b we
~ :: (0. q,75
0
:: '3~ j'75JCt..
-=
b IR,;'
)
-= 937
-==
(CDn'-t) 1/- /23
fl..5
II. C8
(con If)
~
= (J.t;z/
35"00 {f =J
/1- 12'1
II. 69
I I i):.e +he one sht;wn ,..., Vi cleo V. a.LI
11. ((/7 A total pressure probe"is inserted into a supersonic air flow. A shock wave fonns just upstream of the impact hole. The probe measures a total pressure of 500 kPa(abs). The stagnation temperature at the probe head is 500 K. The static pressure upstream of the shock is measured with a wall tap to be 100 kPa(abs). From these data, detennine the Mach number and velocity of the flow.
like We e h lev
/hiJ
iJ
GXdJ"J1p/e. 11.19· r::-;~. J). 'I w iff"
f'x
IO~ -lu'a..{tib~)
r-etJ..d
llJ1d
Ma We
O~ -kf~ ( 4(,,r)
5
fo,y ::
= =-1. 9
)(
de fer»lIn e.
the value of
~ wilt.
(I)
\{ :- Ma. x tfRT; k FC'fY ~
W~
7;.
flhce
~x:' we
~.7:;: 500 "-
~ve..
r
)(
OJ1.d
F,"g. lJ. I
o.S.g
=
~x One(
vea.d ~
-::(o.~6
wi~
v
J(
=
)
St)()/<:..
E'J. I
= zeto
I<.
we
,.?
/1-/2.5
~ Me{ 1( -:: I, q
($"ee videfJ v3·lf J 11.70 The Pi tot tube on a supersonic aircraftAcruising at an altitude of 30,000 ft senses a stagnation pressure of 12 psia. If the atmosphere is considered standard, determine the air speed and Mach number of the aircraft. A shock wave is present just upstream of the probe impact hole.
Thus/ ~y -::
/2.. pi}'"..
~
q.'57'!l ~1/~
witt,
tUta
F
fJ.
flt/!
vp!ue of
~I y
We
yettd
frt'YVl
Tx
O.tl
IlAtt = 1.25 'I
.=
nut)/
~
:::
!'v1~x V=R~ ~
-:
(17/6
f!J!! )(,+12.,2.'« {I. if)
rrl-tf. ' f(
( lIb
)
Slu,. £t
One!
sz
lIx :::
I 2 'fo
f+
-
..J
/ /- 12.6
1/.7, I 11.71
An aircraft cruises at a Mach number of 2.0 at an altitude of 15 km. Inlet air is decelerated to a Mach number of 0.4 at the engine compressor inlet. A normal shock occurs in the inlet diffuser upstream of the compressor inlet at a section where the Mach number is 1.2. For isentropic diffusion, except across the shock, and for standard atmosphere determine the stagnation temperature and pressure of the air entering the engine compressor.
The-- de-celevlfi-t/on process I;' !he- In/e..!- d/fhtser is ~fftl#fed I-D Ie ad/aha-ht SiV1CR we- (lYe- ~nJ'/dey/~!f i.seY)"lrlJ)'lie- d/ffUJ/~ eXGe,f- acyoss
To :
Shad::..
{he-
TNA ~
C!mr~1-
(I)
T
", diffuJ8Y /;"/el-
To
tht. d;f~5€1r l'nle 1-
4ete¥IIJ1Jne..
wi~
F ;9. p. I
1. =
r; Tahle
a~ti
fem,c¥~./l,l/rt
eYl1eY"
We
rea.d
~§
6,
IS" ~I'va
Af
Htt -:: :. 2.0
fia'jVlaf/oh
(2)
elevttnoY/
il?
-;1aJ1dClYc/
a/Yn()sjJp,~ye
fIVe veqc( ~
c. z.
T ='Thu r, wi H,
~t·£
C ::
&t f'.
2/~·5 I<
I t:lJIU:;{
To de fe-VWlI;'~ the..
2..-
we
'>'-laCjJ14-!J~n
b"u~
0
fye>S(J"c- cd
we use,
(con 'f ) //- /2-7
Itte.
CtJvv
I;'/~ t
/1·71
1 ((On'! ) We
hJ.
~
Fq,-!nAJtn-- ,;'Ie f
obla ,;"
f).
J
~ M~
dif!uJer 'hle!-
{ d,. (ftaf!-y IH It.f
~
Tttus
we
F/9 • .(J, I
?q,. rru re ~
:. 2,0.
~ve-
I;' lei
=
Fe/x
r: di Ruler ,~/ef
-=
0.13
/.0
J
tJl1ti ~
((/YYlfJ
1;'/~t'
~y
7tnl>1
w//t1
':=' /.
0
GJ- '3 We. ,,6-ki ,;,
t, um,I' inlef =['lJ, DOO:' (awi]{t. 0)(M'i2%)(i. ()) = 91.1.~'f,.{ttbfJ =91. #Q(ik) To
defer m/ne
e J11er
ftte
F Ij. D. /
~(J~ In/e..!-
~ Clm,
5 kt
wilt,
fyeJJuye
f,'< ,tV{ ~
egy..,p
, JhJd·
- 0.89
Inlef
Thl{),
1 ~/7l1
In J'lf
_
(o.8tf
/I-i28
at thL
C/frYIpYt-s.JdY
-= 0, I..f aI-it/(. retA-PI
I~/ef we
11·72
J 11.72. Determine, for the air flow through the frictionless and adiabatic converging-diverging duct of Example 11.8, the ratio of duct exit pressure to duct inlet stagnation pressure that \\ill result in a standing normal shock at: (a) x = + 0.1 m; (b) x = +0.2 m; (c) x = +0.4 m. How large is the stagnation pressure loss in each case?
ThiS is ~'lm;lar IZJ (a)Fc;y
c;-h:fndlY:J
a
Ex.t1m~/e 11·20.
5'hoci<:-
ntJYfYflit1
f~b/e
of EXample I/. ~ thar
Max =
/.37
Me
af
x -=- f
0./ rn
we nof~ {Y"h1
and
-Px = ()·3:7
'2
(I)
~)( 1
Fi9' D. '+1 for
FrtJrn
M~,
:=:
= /.37 ~ eJbTetI;'
M4)(
0, 75
(2)
Fig.
FYtJ1'J1 M~1
=
we. ftnrl
D.I
rz,y
0.75"
Ay ':' I. /
(3)
A* PtJr
02. /
=- r o. / AV-) is X
Tn
"
-Me
rnfio
of. ducf
:: 3.18
(con't)
ex i f
4Jr-e~
-IZJ local
(J!'eP-
con'+ )
/1.72
f'. -?" ~I
f:
=:.
~2.
~x
-e';!/ --
D
~x
r;,,)
(,-
= [lOt frl1J.(abs)]f-
P. i"j6
~)( ;I
(b) /7:;v tt. s-fal1e:1l~ th e.
1-t:l61e
sl1(:J(;k...
/7(JyYl'tn I
€'x al'T1l'le I/. I
of
Cl f
K:
Iha I
;; Ii: :: O. /1 ~K
Fr~
~i9. D.
MQ ::. f).,
'1) by M¢)(
~ t . ]{,
we
2
J
{v
_1_, = 0.33
~x
r i!J- o. I
FrtNn
,.
Mil -:::
w-e
f-/nd
fD,..
0.63
~ = /. 16 -= + 0.2""
Aot"
For
)C
~z. =
-
A~
;I
tJ./J")f"'-t
(o.S,..,) "Z.
I/'/A? ~ + (o.2m)
'2-
=
2.$"
(ctJn'f) 11- /30
(Jb!eiln
+- (). 2.;.., we
):=£.jl'a.
-
nf)1e fYlJ~
/ I.
72
1(UJ/J If)
0.2.0
Met,.:
and F;.
-Pz
::
-= ().
{~
~2. )
~7
1htO
~
~
=
F:, lJt~
loss
~I
D,:/
111
~z
-
(f J(P..,): (~97
-=
/?')( J
I
.JflfIf Hit
::
~x
no"
~)( -~)
::
preJ.Iure
5~ndlj
a
~J(
ht.6/~
of
Mtf
K
I?OY~4/
)
,
: [;orkl'q(ak;7{-tJ.?J
st,ock. tit
k :.f.o.ill¥1 we n~1e
ex.ample ~/.R r~-U:.ft-
=:;( • 'T~
& IF = O.,O( ~x
Pi,. O.Lf) {"y M~:: 2· Ifl we o/'frU;'
M~
=-
(J'~/S-
and
J;u
.2.::.
f;)C
= ~. 5" I
)
~
F/j. f).1
we ft'nd
Ma) ::: O. S I
Aft AJI(
=1·3 (con't) /1- 131
)~
17
~~
-
an4
~17/rn
== (J.t
If
t:)( (1- ~'t) I
(C) Ny
)f,9J
fmn the
11.72..,
I
(COn't)
()./rn'l. 7- ((J.S~) 1).1
A.. ~
A~
f
= 1·35
(o .I.f "",) '1
/11,. )/II!/) = (1.35")(1.3
(/;,
)
j(;'..,
1/,. =- ;. '6
W,' 'It,
n? "L-
').
we.
gef
==
I"Z
~ F/J. /)./
A"F
P,. 1:
~~ .I
F...
-
::
~/I2.
';;
nil"), P-z.. ~I I
=
]he I()ss
1:) =(0.,2J' .
1 =([ )( It, IX
In
~)( .I
.J-/ztf na oil/')
/,N~SJ" JI'~
If
/1- 132
)(1).5/)
= 1)·'!-7 ~
11.73
I 11.73
A normal shock is positioned in the diverging portion of a frictionless, adiabatic, converging-diverging air flow duct where the cross section area is 0.1 ft2 and the local Mach number is 2.0. Upstream of the shock. po = 200 psia and To = 1200 OR. If the duct exit area is 0.15 ft:. determine the exit area temperature and pressure and the duct mass flowrate.
ex/t -fe YY1peYtJl1u ye ~ 7i.., a n4
duct
fhe
We
o.,c
I/dll.{e..
FJj" ,D ~ /
&: = (:1.
It,.
tAl i
Pt..
MQ.j ';
I
Th~ Vtl /He
valtle d{ Mt;~ we /'If A'). wht'c), we
)/ 111/)
OJ
MC/ V
~
rJbrn./;'ed
o61r:t1;'ed
~
Thus
~
2.0.
F if!. F/~.
Fi9-
P. /
A~ =
/.2
Me
p)'1)b/~
w"
o· '-/
~ Max:: 2.. 0
we..
Yf2gol
slrtlemeYlI - /. S-
inu.;
A1.
A-f
D. /
f).if
A~
~
Fi1· lh
wi Itt EJ. I
= (I. 5) (i. 2
)
~~
vve = /.8
(con't) /1- /33
w~"th /he vt'I/ue
~ ,kH~lAIn v~/4e
'= ~. 5" '2>
CJi"ld ~
~
At<
va/~e of (::) is
The
(lHd
MC/2'
A, I( A*'
A"
of
file
~
gef
yt:V?OI
~
i::nolAIIYt.J
~
t>6-kun
o",SI/;" -fht:s~
~2.
7;4.I
F j 9 - lJ~ I
Cd",
fJJ'~.5JtI~
..frff- Ma) : 0·S3
or:
11.73
I
'+ )
(C~n
A,.
With
= /.g
AIF
Mql =
o.3~
-
=
T
=
10, 'J.
(2.)
t:lnd
~:z..
The..
T
va II.( e
p
~l.
'=
~)(
of
P
=-
=
(')"
Where
~y
lJuts
::
tJ.72
'- = (20() /'S ia. ) (~. 72
W//11
£, ~:: ~ £, s
we. ,,6111,;'
"3 QJ-1d S'
/7;.) ': '
(!2IJO'I<.) (0 .'7
,2.(7:;
Wi ft,
~ ~
~2.
s. 'I
~ 2.
:
nd 6' we
(5 ) ~2.. I
=
~
:=
)
A2. ilL.
(i'f~ J?.f'~ )((1.92
in =
O.g!
=
~
) : !.E. PJ1&t.
=
I:;;' I? 1.;
use.
~2
c2, ~
Mil
Ff.lb) o,e
(Itfo ".(2)
slu, .f
11- 131.f
P,. 1/~ AT,
2
,;, = (I 32ps/Q ) (1'I-If$.J't) r/~f.
/Idd
htf I/~
we.
tlYJc/
(1716
..e:
-
ft()wrak
I'??(;Sf
YY7
d
(6 )
) :::: 1'1"1 pJ;a.
~2
.
Ma'l-
y ~~ R
1/. 7Lf
I
11. 79
Supersonic air flow enters an adiabatic. constant cross section area (inside diameter = 1 ft) pipe 30 ft long with Mal = 3.0. The pipe friction factor is estimated to be 0.02. What ratio of pipe exit pressure to pipe inlet stagnation pressure would result in a normal shock wave standing at (a) x = 5 ft, or (b) x = 10 ft, where x is the. distance downstream from the pipe entrance?
This
IS
wifh
Ma:: 3.0
n
simi/Clr
Example. /1·21.
F'9. f). 2
we. enf(,y
I
f(£I'-_l,) __ _
(J.
Determine also the duct exit Mach number and sketch the temperature-entropy diagram for each situation.
IJ.nd '1ef
L:'~
;7 &-
D
We
tAat
nbfe
If
f (l~ll) =-
f (1 -1.)() + f ( (J( ... .(, )
-
{)
(~)
f)
qe f ~
Etj. 1 we.
WiH,
f(1~lx.)
Rx'" 1.., ~
F (1~1,)
:::
f( 1¥....1.)(} -:::
tJ.
S if
:f(1x- I ,) = f).5'z
f)
D
OJ
j)
D
_ (1J.1J2)(sffj
-(I {"of)
¥2
/)
V>litt,
.f(J.f":...I1() = tJ.¥2. [)
M")( Wi It,
Now
= 2· 5
MaJ( -= 2· 5
Mit ::
(j,52
wi tt,
M~7
'J
w~
= ~. 5 2
F;9· D. L/ (,fYl~ read
e nfer
we. 06 frL,,, ~ != i J. O. 2
~/f~ -~) J)
(cIJI1't)
//- /35
II. 7'1
T
(con/f)
we
qef .;-( ft'-l2,]:::
(J.
-/)
( / .f"f)
el1~/'rJ,/
tJYld
'f
Fiq.
P.2
{'(e1':-.iz )
wil1i
= ().'(
£)
Ma
::: 0.62
flfJ w
(SUbSf)h!C
-
2.-
_
)
t ~(j~£tJif;)(:7(£) Wi ~
!v1~2-
fz. -
= tJ.6 'L
we. tJJ:A I~ fr-lWt
(3)
M~:J;: O.~2 live
Wjfh
?"'9' (). 2-
/,7
?-
t7blat'n ~
Fig. 0.2
~ = 1.05
( if)
p.f
/11 a ::: ?.. £
Wi ft,
PJ
Q,,,,()
we '1ef
)(
Px
(2.)
Frvm P;9 ,P. 4
7
~
~b ItA /'n
Wr!..-
fn,.,..,
F iy. D. 2-
&. = ().3
p-l For M~
I
=
~. 0
we.. qet ~ F/J.
p. 2-
(~)
(C~n/t) /1- /36
I/. 7zf I
(con 't)
) =().213
-
( 'I)
,,
(>t) (I)
s we do J1~f have, vallAc.r of ~a.lufe- Or ptr!JSIAJe a YlyVJhey€ In fhe ftow / we.- Can skeld-I 1utdJ'~/Jve!J wha..f htuppenr "n T-J c-tJOYd/YJa.IeI. !he. T- S c//tJ..7Ytl-,-v)
5/nc e.
onlJ
wil! he.,
S/mi/ar
itJ /1t€-
/Jne
of
f:1!.
Gil.
21 (b) ~ Ind/o/lfed
above. (1;) I)/i/it E~. /
(//e.
."
f(1-1~)
gel- /rJy .f.x - ~J ~ ~52
_
::.
/0
ff
(().o~)(/~-rf)..::: 0.J2
(I rf)
= 0.32
Ma ~
With
=2
Max :: 2
Ma ::: 0·5' Y Na'AI wI fh
!/Itt!! == 0.5"
we ~bm;Y] ~
( co)'/f) //- /37
175.
D. 2
If. 7J.f-
I(coY) If ) f{J~1!J) ::
0.""2
p
SiJ!1ce
FU :'1)) =
f(t~J'1) _ 1'(-12 -£:;)
o
/)
we
qttf
f(l~l)) '::
0.,'2.
- ( o.o2.)(2oft)
,
(I -1'+)
f)
e h~/""
C/l1d
D
FI7,4J.2
win.
= 0.2.2-
-F(I!.P1.) p-
= O·"2.Z·
M't~ = (). e'I
W,/(,
M42, -:::
~
eC}
J; = ,.IL/ ptt' W;tt. 1v1IA., '" tJ.
§7
we.. (}t;-h,"n ~
F/j. O·?-
tJbl-,"n ~
Wf
~/J' P·2
Py¥:= 1.~6
(1 0 )
p
Witf"
Max:: 2
1; :: ,0"",
and
we
qef £.rm,
Fij. P·lf
If. 6
W~
()b~ in
(II )
1Yom F/7 ~
0.1.
Px :: 0.'-1
pI(
C~bi¥J/"1
Grp5.;).) 7)1;
't /().; /I (JWJl
dbtal"n
/2 we
) -::: ()·I6'
-
... ...(y) (2)
", \
T
./
..-
(x)
OJ
/1- 138
"
,
/I.
75
I 11. 75
Supersonic ideal gas flow enters an adiabatic, constant cross section area pipe (inside diameter = 0.1 m) with Mal = 2.0. The pipe friction factor is 0.02. If a standing normal shock is located right at the pipe exit, and the Mach number just upstream of the shock is 1.2, determine the length of the pipe if the gas is (a) air; (b) helium.
We
J1 ofe
f
tho. t
(11. -1,) =
(I)
j)
{)
D
Flq. p.2. ThUJ,
/)
Andw,'nJ Mal
we.. can
and Mal.
de,le~n1/ne
;-f!!:1,) a'1a f)
f (1~/,)
IJl1d
1:'1. I
with
we
f (12- 1, ) .
d61a/n
o (i/5t)
~
indWI?
we. cttn del-el"m/ne
f (t~l,) -
J2 - i,
().3
j)
and
f (~~12.) :: d. ()! T rhus" wi#! £8-
I
INC
~ve =
al1t1 /J Xl.
-l'/ :: (0.27
)(tJ./ nI ) o,t})..
c
/.3 5 .!'" -
( C(j 11'
t)
/1- 13q
~.
27
Wi Ih .f- and D
//. 7.> T
(C rJnit
)
(b) FfJY he/;~~ (It:: J.66 fY-nn ~61e ;'8) we havr.. with Gt· 2
f (1 ~f,)
=(...!.-
/)
J!'-(2.0? (2.f);'
1.66
T
('.66
TI]'" Lr{l.~'J2.0;' -, +(t.,;- 'Jz.o;'
! 2 ( J.l6)
= 0.213'
and
f!!l...R.) ~ /)
IL))rJ-{'~2n t r'J!!..!.}/~!I'!Fry,'2/ ~ (/.66 L J ! '2.(1.66) 1./ (1.2)
I t~.:IJ-~(J.l}J
,
Wi /fit E1' I
we
f(1 2 -~)::
1 'f ; ~
I
(JblrA I;'
0.21JI -
o.ozS()7
(C,/fl)(~.I~)
-=
0.'10/ m -
().()2
/1- I LlO
(!J·lf!
::
tJ.02tj07
1/.76
I 11. 76 Air enters a frictionless, constant cross section area duct with Mal = 2.0, To,l = 59 of, and pO.! = 14.7 psia. The air is decelerated by heating until a normal shock wave occurs where the local Mach number is 1.5. Downstream ofthe normal shock, the subsonic flow is accelerated with heating until it chokes at the duct exit. Determine the static temperature and pressure, the stagnation temperature and pressure, and the fluid velocity at the duct entrance, just upstream and downstream of the normal shock and at the duct exit. Sketch the temperature-entropy diagram for this flow.
Af- the du. c,f en1vance
7;;, ---
:;-qDF
r:,
/ '-/. 7 f r /a.
I
ai1d '="
-
I
sech'tJh (/ ) )
we h~ve
5/Q'R
I
Witn
M~
,
::: 2.1)
we. en/ey FJ9'
D./
AJI1P1
yeaol
(t) and
P, ~
~.I"3
(2.)
~, Thu~
w;#t
f1>.
" -= (0. So
/ and 2
we. 061a.li1
) ( SI'f 'I<) ~
zCJ /
-
and
:: (2.0)
/1- 1.If/
:.e
::: /660 If .=..r
II, 76J
(Cf)n
'f )
and
~d/x = f.',' (e /~J~f'oIX) ~I
,
t;~
Max =I·~
Mal:: 2.0 and
P(J>-
ItJ11
-
foa. I
fu)C -L-
('f)
we
year) ~
Fi9"
(15)
:; 0.7'1
-= 1./2
f;A I
,
Wi f{., these vlti-ios and (;11."3 a.wl. tf we, -= (~/qtJRY--1.
7:
F;
,)(
=
(t.S"
IL ::
f.S
we
2#1Jev
-
-=
0.61
aM
TheY!
Px
=(0.27
'Ix
-= Ma x VR7;./( -=(1.5)
)(11
~
fsir;.
FiJ- J)~l {;lM~ rl!fA.cI
Tox J
QYld..
Obtalh
) -: 518'7<
(ILf.7fSjl{)(~)((.12 )
M~ =
{).Ii tC,
) (().cJ/
'( /Y. 7q
OJ 'I
f).. 3
psi~ -= ~ fsit).
-
11-14-2.
Af secfl(J'yj (1) j()!f d()w;?s/YetJI/')1 of Fl,. /J.Lt- fUy M4.x ::: /.5
Me, shack. we obi,:,,;'
~
May,: (J.7
F; ::
2.5
~
T..
')' :: /·3 7;
~ :: I. 'j
-
V"
~
l! ~x ,
:=
~.ci '3
these rrrh'os ()'J1.d rAe hvl'Yllner} we. htJ.ve,
Wifh
Py :: (2.5
TY
af- Jec!/cNI(x) pYcvlduJ/~
)(3.tJ()fS /o..j -= 7·5I'si(J., J{lf/()(J/!..) ~ ~3~~
:::(/.$
V} ='
va.lue.> of fy()peYk~s
(ft{-10 I. q.
ft ) ~
~ 7 :: ((J. q3
73'1
f'
)(1/.0 fJ/~)
A/5~, f",;'c.e f1.,e flIJw across -lite
~lj :: ~x :;
:: If). ].I'si(;. hdY»tIl(
sA~cJ:..
/$
t:i.c//a{,p!/c/
5'99 tJ{(.
Af f1t.e- riM..cJ ex.;' f.l sechtm (2) W~ hrI. ~ I1tL .s u6.>c Yip f ,~" s /r:t Ie Ii? Pig. O· 1 5iJ?ce the How is choi:.ed -Ihere. Thus..fYrJm E~>. 5"" anr;( ~
T.(}tJ.. :::
we....
7;;,
~c,w.de
-:::. (5 If 01<.) (0. 7'1)
I
i
that
ar.4 _
(fl{. 7 ;OJ/A
(I.~)
6S7'I!. -:::.. ~ 'l-
?:
) =-
Cj. 'B !,sia
--
(COYI'f) //- 11.f3
II. 76
J
(COlt Y:
Wi~
)
Ma,:=' 2.0 W~ yead ..fiM.ft,w frm,
Ii. : :.
F19·
Po 3:
O·3~
Pit
V,
-
:: {.'!§"
~
TJ,Uf,
F;..
-
(/.9 Ips/a. )
( (7·1' (Z'lJ~) ~ = . (d.53
) "S:"
~. "31 fS/a..
:;;
-
S-~ "f<
P'Z..
72-
:::.
~
anPi
2.5"0
1 - -_ _ _......1-:--_ _ _....... 1__
o
1000
s-s I
/1- /Jl.J.f
(fl:Jl.ZO)oo SI"".-R
II. 77 and pressure, the stagnation temperature and pressure, and the fluid velocity at the duct entrance, just upstream and downstream of the normal shock and at the duct exit if the gas is (a) air or (b) helium. Sketch the temperature-entropy diagram for each flow.
11.77 An ideal gas enters a frictionless, constant cross section area duct with Ma = 2.5, To = 20 °e, and po = 101 kPa (abs). The gas is decelerated by heating until a normal shock occurs where the local Mach number is 1.3. Downstream of the shock, the subsonic flow is accelerated with heating until it exits with a Mach number of 0.9. Determine the static temperature
(a) ray all'" we. have ar -/1.e dUvT -lnIY-ancel sechtJn{/) Mtt J
~
~
:: 2o~ :: -;..''3 K-
I'
1; 7(}J
2·5
(lJ
==
)
and
P, :; ~!06 F;, I
'!Aws we- htAl/e T :: (d./.fLf
wi f?t Efj!. I and 2-
)(1-'13Jt:):: 13()/<
-
I
--TheY7 '{ :=
1014, !fo~~
= (1.5)
:::: ~7/ ~ 5
11+
(Lf)
11- If'S
11,77
Mt:\x = /. '3 -Tx -
lox
vvt- e n lev
:1
,]\
().75
)
aYJd fx = 0.36
ex I
1hu) I
0c =(tJ.7 5
J{JqS 1<)
tAhc( P>c,
TheYI
~
= ( (7'$6
)
=
2
Cfb IC.
[lf7.tf fd~(v./;?s)] :: !l.-!P~(~bs)
__- -
=fVt~x V~ ~ I<.
:=
(/.1)
(can't) /1- /
J.j.
6
//.77
I
(CI),,'I:J
Af 5eCn~n ('1) jl.l5f dpwJI},)/ream of Me. sluJci. we oblQ,;' ~ Fig. D. 'f .frN Mttx ~ /·3 Md y :: tJ·1Cf
!l = /·K p)( !Y. ==
/. 2-
7X
~ = /·5 V~
~
~.q8
=
~)( I
fhese
Witt,
}'1ktl() I
j
= (/.!
J A/s!),
) (2'15 k.) -=
::(0.1 ~)
0;::
['t 7. 'f MfaJ,f)] f/()w ().VY~.!J
the
"')1Y!ce
T()" I I
=?o.8 = k./k{ah5} ;;!!-
rtf-'f! .;.):: 2 q 9 !!!. (I.,) =- S
V,::
~1
.>echd;.(x) p1ev/pw/y
) [17. / illk[A-b5)]
r; :: (/. Z f
(If
we hav~
ciek4m/neri
R
Values of frrJ/fnh'c.J
Clnd
1;;, x
==
the,
'16. 'I M(~bs) ---
YJOYW1(J/
{hoc/<-;s ad/p.6an'c/
= 3qfi It:.
At- t1tt- dM. if exi f) se ~hlA, (2-) / we hMG (5)
~ = fJ (~ )(ft) Z:: Ty (7;.)( 72.)
{0J
~
Ty
7,jl-; 7;,., (~ )(7i,1 ~"L I
~Ij
=
10 I
III
(7)
(PDI«-)(t; 1.) _ p"
_
~~
~= ~ (~)(~) 1/0/
~"
)
Vy
(I)
I
F'oa. I
( COli
't )
/1-/'17
//.77
I
(con'.(-)
IIfJPy~py;ale
Fig. D.1
Y'lAn'os -/D use In ~r. fw May = tJ. 7~
§
#JrtM!h 9
Mt #J,fAIW
and
MuS I
2. ::
~
1-'3
.p,.
= I. /2
r;
::
I. () 2.
-
/, () 2
~
~
-T,.
~
~tt _ f). ttl" 7;~
I
t;-z.
--
~. f'l
J
?;;a. I
~/'1
~(),.
=
1.02
I
p.
~ :: /.01
t:a. I Vy
-V~ -_
0."
_V'),. :: O. 9 I v~
Waf.. tne.s f,
Y't4.:h05 and E$I. 5 tl,YVHfh. 1 we. ob~J;'
Ii::
[3()·1 k~ (rd,$)] (-'1·3
r;:
(3~11<) /
7; 1. I
--.l-) (/.02
('~02.
:l..
::'
(395' I<) / -L (
0.
'16
)(0. ~q
) (1.12 ) )
= 26.6 f<~(~f)
~ 35 I k _
) =
( C-DY7'f- )
/1- / 'f8
q.tJ7 K. _
-km
M42 =tJ, 'I.
con 'f )
/1· 77
f. ;:: ['f6.'f"*Pa.(tfJ,$)] / ~2
I{
('L16 ;' )(~,! I
:=
Fw
Sl:e~'/~
we
u;e
I)~ed vall1ef of 5-~.
- I? In ,P ~
eXAht.f Ie /
Sx - 51
::: (lOd'f
5 : 52.7
)(
r;
I
fuv-
s. -
d/~frtlht ~
T-f
I/L
= C.o 1/7 I
I
J
; ) ( tJ. 'il
I
S - 5',
fuf
.-!- ) (1.111
(lP2
J
L.) /n(;~ K)
~.I<
/~~ J
:r
~. K
5/Yn I e;t,y 1'1 J'
57 - SI =
~'36 L 7l.J. k.
p..ltl.d $'
~
-sI
:=
57tJ
L 1,.1::.
'100 (2) l')(Jy,"""
Sho'/<.
?a.O()
line 'f(~)
5k.l!
ft.'"
(nor ft;
S(a/e ) UJO
IOO~--------~----------~---------~---------~---------~I-------------o IDO ZOO 300 ~oo 500
5-5 J
I:F !~- ) ~.J:
11- lif9
11.771
-(cOn'fJ
(b) [by hell'urn (~: 1.66 and!<.:: 2()77 ~ (yOWl rah/e /.
Ma I
7;;
~ Z· 5 '=
20 "C :: 7- q31<
,I
'"~J ::: /0 I -k.l'a (a h $)
With Mil,::
2· 5
r.
I I
ThlA~
vJe,
-r;::
~se,
we
j-
i1ave.
1l.-1 IIA ). -''''4 "2 ,
wilt,
.
p:: (o.o5CJ9) I
O. $2.65
(to)
f~s. 10 aYLd. II
(0.3265) (2Cf3 J<.)
()nd
I
=-
Cj5.7 K
[1 0' iPa. (tllb$)J
=- 6.0 5 1e..f~(rAbs)
Then
(tZ)
To,
-'
::
O.7K7
(Con't) /1-/.50
J
/1.77 J(con't
Oy
ft" = /. q()5 I
F.a. I
;;,,. = 2.(1.6(,+1 )(1. '3) -
'1
I
I ~,.
-/;Q.
_
-
"J
Wi f?,
(! r 1.6,)
these
Tax -:.
yp.fios fJ.na
€tjs. 12-
-= [itJ'~I'(((j~s)]
(;--l-
we
r;,x =
1
::
t't6~ -1}).3)'l.
1*-1
[-It-(~) ""a; _
{IS}
-
ECJf: /4 tt.n Ii IS
~ :: (0.6'1 '2 )
(
'3"
0
~)
:::
Z 31 J< :::=
avtrJ
Px
0.6'12.
lL
,
w; +i-l
fl.u 5
St a-"A I J. S'1 -k Ijet
I
aha
~x 1
(I.
=
I
:'
36~ 1'-
)(J'()If) -= 55.' -kfr-t (ab$)
9~5
use &is.
7;
1;
=
1•
Mqx'" /.3
we. ge:1-
Y1d 13
(().7f7
1)(
Wi Itt
Q
(1..'131<)/-'-) (0,9671)
I
~
/.tJLf
I r(J.UJ(I.3/"
=(tJ, sZI J[ 55.,lefJ (a.b$)]
The;-.
v)( -= Mct x
4
if Rr;t
-=(1·'3)
:
,e·,
(20 77
-kfa.. (abs)
-= IMo !!!.. .= 5
(Con'.f ) /1- /s /
I/,
771 (c(Jn'f )
Af
Sec lion
(y) juS f
jt.f'lq ) Ii t§'o;, /I. lSI) II. 15"'1 and /tAQ)("l..
/VIp.y --
a/ltd Max:' /·3
/1·/5"6
+ (~.:J:t)
(1· 3J 'J.-t
(~)M4: -I
(':_1 )
:: do 7933
(2)(!./,iJ)(1.3/ _ I
~-I
-Px'r
of the shtJcK. we. ObftAln w//I, £1;'
dowY}siY'eam
(1.66-1 )
=
T
-rx1
::-
r; = (/, 2q 12'31 k)
.= ~
~
;::
(1ltd
;J =
1·'+«(3
f
~1
=
(tJ.
ld x = 36DK
7;'1 -:: I Af Me
I
(haf
~::: ~
y
_
exjf) ~ecfidyj {2.}
lIVe
(Pa.)(R ) P'f f:t.
httve (Ib) (17)
( Conic) /1- IS 2.
~" v, (~) (~ )
rzO)
E ~s. /6 il,YfflAyh.
A p P'f'P Py lale
ntfioJ
-h; we
In
E,r.
11./23 1
11.128/
//./31.1
wif?t
Ma'L::: o. &0.
and
//./3J
aJ1d
20
tJ¥e o61rt/~~pI
1/./2'1
fin,. l'1a,=:fJ.71J3
lhl/5 1
/ + /.6'
': /·/3'1
-
7#. -=
Ta..
t(/
t- /."
/ +-
~
/.06,)
)..
} (
o. fj 0 )
(1.66)(0.
]
:- /. ()
if3
fjlJ)"
Z (/.66 t 1)(fJ. 7'(33) 7.
[
1+
t'-;;-')(~' 7q33/ ]
>
-
[/ f- (/.66 )(O.7f3J)V J"l.-
- 1;1I
/1- 153
":.
0.'7671
1/. 77
I .
(C() n If J
~
= ((), 7933) f(1 f 1·66 }((),7i3~)
L-I r
Vet
v,. '"
(D_?O)
~
Wi~
[(1+1 6&)(0 '11) 0
0
=
T
= (2 ttl t: ) / -.!.-) (I.O'!J)
~
ffJl1(tJios)
= O.iJlfj
fhYhll'l4 2()
we tJb7a';'
)0.13'f):: 2-9·'1 kAt(IJ~.5)
P2.
[33.7
J
and 8~s. 16
Y"a.-h'f>I
~·!/f7
(t.66) (0. 77n)'J.}
, r{t.,,) (O,'b) ~
fheSl-
7-
7/.l-
1/~301
(/.o~5'
~
-::. )Jj2;::' -
T = ( 36 t1 !<) ~ ) (~fjq J'f J::: '37(7 k. ~z. (;"(71 =
1;). = [5't./ft,~(A~j)](;-!-r/·oO~) = l.tJ2/
I
V;z.. = (g~'f ~ )/-L )/b. fj1ttj S ( d.11 97 J( " Fuv $k.efe-h'h9 5 - 5,
!hut, f7v
ex~ple.,
s:JC -5I
= =
we V1e~ol valLteJ of s-~ . We use
In ?
o
~
(522~.2 )
/<)
/1'1 (21 1
~. ~
2327
*~(ab~)
== Cjd2 ~ .J
d/a.'jyltl'l-t
= c.p In L - R.
5)( - 5/ w
tt T-!'
£'$.2
=
lq5".7 I<
:r
it,. II-//
lOo I-
laO
f-
T(t<.) //J(}
(tJ
()
0
IODO
S' - 51
//- /Sif
I
I
ZDfJO
3d()()
(~;'I< )
/2.1 1.1.1 Water flows through a rotating sprinkler arm as shown in Fig. P12.1 and \,id('o \'12.2. Determine the flowrate if the angular velocity is 150 rpm. Friction is negligible. Is this a turbine or a pump?'What is the maximum angular velocity for this flowrate?
/ /
I
'"
/'
.-
,...-----
.......
.... '"-
/
",
\
I
\...~.-
.-
I
.-
0.3 in.
I
_ I 70'
'/
~\
(~\
T == m(r2 'Ve2. - fJ Vel) is no friction. I}/s~
\
fhere
::: 0 since ~I =0
""
I
I
I /
'-
",/
..............
.......
lienee} Ve:z.:::; 0 U2 .... (;}'2.
I
w =/S'D rpm
_---,..-
/
,.../
1-7in·-1 o
rev ( ? 71' rtNJ)( I min )(2 rl) ::: 150 min rev 60 s 12. TT
G
c
:: 1j.16!f •
ThvS
FIGURE
P12.1
1
~ s/"B :;: U2-
or
9.1" ~
~::: ~in 70'
::: 9.75 fi s
lienee) Q:: 2 W,. Al :: 2(1 (W (f)') (9.75!j) :=.
this
is
t:v
{J.oo9S8
if : :; if,3
-Iur!:J;ne. he.cause.
94 V"'in
ihe SfY,'~kley
fl}(JVe$
111
reS/HUlSe..
fA" d -flo -foyces. 5/Y7ce lYic -no"" if n~;'i,'/;/e tJ,~ n'JIlX;1I1uI'Y/ pn9J,f/4Y ve/()c,;1y .fIN tniJ f!IIWYHk if /he tJne U/Yfe.(/~I':J -Iv I5'OYI'YM
-ft;
-f I
ltV
(7Y
/2-0
rz..2
~----- .......
12.2 Water flows axially up the shaft and out through the two sprinkler arms as sketched in Fig. P12.1 and as shown in Video \ 12.2. With the help ofthe moment-of-momentum equation explain why, at a threshold amount of water flow, the sprinkler arms begin to rotate. What happens when the flowrate increases above this threshold amount?
!-7in.--j '~'....IfJ----rG..,----J;«· aa
FiGURE P12.1
fpY';y,kley- I r s I;"";/IIY' ~
ve/()C; Iy
fl,e.
f",A"J1e
fhe
...
IQ
(ll1e
fJF GX4m/,/e
s: /8.
sJ,OIN"" 1M
U/)IIIc.IMt:it. ~4 t
wlt.e ....e
t{
==
t;w
,e f
(/MIt" ;~,. "'J J Wit
-rSn41f
:= - y 2. (
S"d ./ wke,., ~
~II#
;}
tZ""tJ.
-.
(.,0
rdtw
,M" c,t{!4.Jt!S
T
.s-)'«If
Wz. SIYJ 7() I(~J
laYge el'tDaJh witt,.
-ihe, fpr,'",I
w:""
o
70 -
\
1$
-I/tJw ntlc-
witeYJ
/wZ >/~
G.)-:
tN~
t. l'D!.k.
~/~S
.fit -ril,e r
o +0
GAJ
J
is
h ()
/~ey
]tlrtJ
n~/'j"b!/ ~I/.J
z Yw
1;'~eJ
VA-Aft. of Wa-.J f.#t,
wifh I;'C~I!~.f. Wz. lit.,,»') 0:1 1'H4XJ~I.I~
fAts"
nal/e. I:;' -,I
"""'~/~""
Vet/we.
01 w.
/2.3 12. 3 The rotor shown in Fig. P 12a rotates with an angular velocity of 2000 rpm. Assume that the fluid enters in the radial direction and the relative velocity is tangent to the blades across the entire rotor. Is the device a pump or a turbine? Explain.
•
FIGURE
P12.-3
T= n? (r2, l4:z. - r Ve,) where V(iI =0
If ihe rOlor 2. 7T IJ
bJ v"
is of cons/ani
: : 21/ r2. b:z. ~:z.
or \I,r 2- .=.!Lv, r:;. r, and
UI :: n W
I:
TT J
hei9hf J fhen b, =b:z. and fhe conlilJ()ify e9lJat/on is
0,5-(/
0, q N
v:z.'= r2, W
SO
(20
ftS ) :::Jl1ii • s
fh af
TT -
V.2, -
r:;.u' r,
= 0,91-1 O.S ff
U.
I
Of'
U2. =/.e~ Hence Jhe fo/JowilJ9 veloeily frian9/es can be drawn. J
VI = 2'0
Vn . = 11,1
u,
If /s seen thai fhe rofor turns fhe flow info fhe direc1ion b/Qae motiof). Th/s is a pvmp.
/ )..-2
or the
/7.,'1 l' ..t At a given radial location, a 15 ftls wind against a ~~dmi11 (see Video \'12.1) results in t~e ~pstream (~~ an~ U2
downstream (2) velocity triangles shown In Fig ..P12.4. etc . an appropriate blade section at th~t radial locat~on and determine the energy transferred per umt mass of flUid.
i
_ _ ko _ _-+-!- _ ___
-
VI
= 15 ftls
= 20 ftls
60:(.
Gi FIGURE P12.4
We can il1 vtJ/ved of fhe
defer-a7lne
whethey
by comfdn~ The d/~c-lto#"J f.he r-o-hrY- blade Sect/on will, th~ veloci-ly ~ V. .If the fir f frrct. ()Yla
;f I;if
a iur&lnt:. iDrct!... 01'7 of the bla.de
d/yec f/OYl
fhe blade
ve/()cify
are
fhe lift
;nvolved.
I.f
orpo~ife.
d/~ecf/on.J/
a
a..fan
OY
iUYb/ne is ano{ hla.de Ve.ldCify are iYl
fhe
In
fayce fan
The
== -Ir:tJ1
/ Thu5}
-I
V; ==
V, fhe
dJyecli't;n
e;t
The d/yec;(ion (Jf fh~
~ the shape of lire /VIrJy
can
iangeYlt"" rhe relat/ve
a;yle / 4 frln- I (2..ofJ) = (15 ~f)
ei7feYl~ rel/lf/vc flow
4
saYJ'}.e
;..> In v(Jlv~.
be litferrfd blade secl/on s~el-(h ecj -to k a nd /~vi'?1 -fhe vofzry ~. ioYce.
/ i{'f
qxla/ flow fUYh#~cJ,/~e
fhr:
.I
~ows enleYl~
t4 53.1"
-mlor- b/a..dc Secn'tJns s-j~.efdtet/ below are t2fPYb/Jy/a!c OD
(con~f )
/'J..-E
/2 · II
(co n ' t
)
lifl fOrc~ aclinJ On each rohtr blade 5ec//()n is in fhe 5aYJ1e dlrec-h'OYl as" fhe. bftJde veloc/fl we. conclude fhaf +hi J fu. r hdn1. t1c-h/~e i.s a itJrb/ne.. The enertj'j ~Jeyy~d per- tJl1il mas> ;5 !he shaflvveJ'Y/::- PW- UI1/f Mt15s ./ W $'2Q J f I Nh;ch We 6::tn ck-hr-J'HIy;e f SInce fhe
wiTh
~tJ·
11.5.
lhas
.
(I)
v.6)2-
- W
W
:=W
5ih
:::
J
-
tI
1.
anti 2-
(,0
~
V 1-
V'2. -t- V"l. I
I
. 0 )ln60 -
20 -ff ..f
r
Ib rf
Sf"'J' 72 ;"
_
33,0
fl. Ib 51LIJ
tJJ
shafI-
-
~
- 33.0 f.J . I}: Sh
----~
(32.1. Ibf\'1) 5/1.4'
- 1,02
fl. tb I bfYl
j
12...5 U.5 Sketch how you would arrange four 3-in.-wide by 12in.-long thin but rigid strips of sheet metal on a hub to create a windmill like the one shown in Vidl'lI VI2.1. Discuss, with the help of velocity triangles, how you would arrange each blade on the hub and how you would orient your windmill in the wind.
~.V\d a'o~ -m_-lio~ dx.i~
>
V.
~~,
jtJ. -5
/2.6
J~.6 Sketched in Fig. P12.6 are the upstream [section (I)] and downstream [section (2)] velocity triangles at the arithmetic mean radius for flow through an axial-flow turbomachine rotor. The axial component of velocity is 50 ftj s at sections (I) and (2). (a) Label each velocity vector appropriately. Use V for absolute velocity, W for relative velocity, and U for blade velocity. (b) Are you dealing with a turbine or a fan? (c) Calculate the work per unit mass involved. (d) Sketch a reasonable blade section. Do you think the actual blade exit angle will need to be less or greater than 15°? Why?
~ IGURE
(a) See
P. /2.6
Fi9vre above.
= m(r2. Ve2 - n Vel) = th r;"e4Y1 (Ve,. - Vel) where Vel >0 and ~/':: 0 (see f'1IJre a hove)
(b) T
Tfws) T> o. (e) .w:shaf! ::
The macn/ne is
fk V8z -
VxL == Vx2 -= 50 fj f!'om the fitl'Jre jhal ! 0
=
fan.
--====-
~ Vel":: U(V0 2 - Vel)
SilJce
\I, cos /s
Q
J
where U= u,::: V:z.
if follows
u,
~o Ij
¥ V2. c~30· -:: so!J-
or ~:: S'I. fJ anti.
01'
V,2. =- 57.7!1-
so -Iha/
V9,==-V,
sin/So=-.5/.'l sin/,&()::::.-I'g.'f!i
and
VG2. -= \{ 5;" 3(/::: .21. V¥ AIso U=IVe d-f/Ve:J.{::: L/ 2. '2 -# J
Hence P{hQI./:::: L/7.. 2#( ~a.g J
(dH:;rom fh 8 Tl9tJre r-'
ftil)::: ()
g. - (-/3·lI
4 l/2.'i-H"J· 50 .J
or f:1= 'f8
ThtJs; the blade shape is as shown:
( con'f: ) /j. -
fj)) -
6
0
if: ==== 17(0
rJ"e ~c~1 J:,laele. 4nJ/e -It;
fA
el."'et/~
«
IS
(J
wil/ ne~d ~
flow ().~/t...
R
+J,~
ISD
f Iht "lade, e'Lil-.
&C¢I.lJe. Df ~'t{"'elQ'Y layer dtVeUpWleHf. th "lode, /
be. ksI 1k;.
01 Iltt/tI al'lJI-e. wilt be diffe.,eH'I In,., fj,e
bl,,,,/e ""'J/e. j.J! S$
-/uYYllly
(Jill
IMtI, .f'(,iYhas
'lJ,AYJ e1l-~elul w//I k
tJ cit ,',e Vt!e:I.
12-7
A.dwal!J
p .. 7
----- , ,, ,,
/ /
12.7 Shown in Fig. P12.7 is a toy "helicopter" powered by air escaping from a balloon. The ai r from the balloon flow s
S:::
,,
radially through each of the three propeller blades and out small nozzies at the li ps of the blades. The nozzles (along with the TOIali ng propeller blades) arc tilted at a small angle as indicated. Sketch the veloc ity triangle (i.e .. blade, absolute. and relat ive veloc ities) for the now from the nozzles. Explain why this lOy tends to ffio\'C upward. Is this a turbine? Pump?
I I I I
~
0
~ 0
,,
,, ,
I ,,
"
U
,
,
,
I
I I I / /
Balloon
•
FIG URE
P12 .1
If we assume Ihe helic.opfer is stQfionary, +hen/he blQde speed is w R in the horizon /al plane QS shown in the siae view below. The relalive velody, ki, is direc.led alon9 the mule, and the ahsolvle ve!uc,lYJ V= wf 17, is as indica/ea, 11
v The -roy
fe",:ff
-!.
""tlve I/pyVaffd be c".,ue fhe. {Io",,"
bloae r PlAs/" VI' 0" /,IqdeJ
On
/0
ihe.
Yr>fhle
a",b/e4f
the",.
/:"."
ai.
TJ-vL a i r '"
/2- 8
va.,. fk
fn.'Yr> fh.c bulloon f".v.c"-' /he
IC .fvJb/n l!. _ ffr,w~V(~/ tiS
0
f"""P '
11,..
bto.cle! ad
);2.8 A centrifugal water pump having an impeller diameter of 0.5 m operates at 900 rpm. The water enters the pump parallel to the pump shaft. If the exit blade angle, f32' (see Fig. 12.8) is 25°, determine the shaft power required to turn the impeller when the flow through the pump is 0.16 m'l s. The uniform blade height is 50 mm.
( E'p. /Z.ID) Tsh4R; :: 12.
tc.
I
tP ~ ~;.
(I)
D;, - Vt9 , (z )
~d ) Yev
~~.
f/t)w'ra:6e
(p=: ~2. ThuS) frpm
C(/1
P(
-...
i1-z
2.7T f2. /:;2,
(J 21l'.z
bz.
-
£::"7. (z) ~2- = (Z 3. , - Z. 6'f
11'1/111 F;.
b
C;~ 2
rtld S
tn1
5'"
.f.c, / /()WJ tha,i
JI lien; /-1:
1..5
:::
(e;,JI,
;:3 )
(2.1T)(t'. '2£,1#) (~. OSr1Yf )
Cot
2 S6 )
:::'
2.o~ t1H
s
~ = /f.2 ~
(I)
T,h4t '" ("1ff ~ )(0. If, ;-3) (C. Z&rH-1)( / u ;') ~ 7/" ~ N·«>t 50
..J
/2.·9
r
1~.9 Discuss the differences between a centrifugal pump and a positive displacement pump (see Vidco v 12 ..' for an example of a positive displacement pump impeller that looi(s like a centrifugal pump impeller),
A c eYJfr;10r1 'UYl'f~ ';"'I'~lIer ,wlttJ /wee.> Jt!ve/operl
"''1
surkeS'
So
(/Ilt/
/J"
ti
fluitl
do/YIj
I/';w,:'
jI1()VeS
II
~
v4luYJtt-e () f one
Dver /-)$
INftleol
-/'Iuid fI,~h.
A !'()S;/lve, difl/acemel'll fJU1'It1' ~J ~
"'-91l,;',r1- flte,
,,1I
IIY
_-1.
C~fUye.r
l'iuiel t).l1d ""'tIeS 1J;e el1-h~t.. v,,1"'"fe
"'CA.~11
*
~III() fl,er.
/2-10
•
/2.
./0
A centrifugal pump impeller is rotating at 1200 rpm in the direction shown in Fig. PI2.1O. The flow enters parallel to the axis of rotation and leaves at an angle of 30° to the radial direction. The absolute exit velocity, V2 , is 90 ft/s. (a) Draw the velocity triangle for the impeller exit flow. (b) Estimate the torque necessary to turn the impeller if the fluid density is 2.0 slugs/ft'. What will the impeller rotation speed become if the shaft breaks?
1
•
raJ
The
ex..Lt
Ve / ~ci
fitJw
9rp-p fu cttil!:J
+!f
P12.10
{Y;tlY'/jJe C411
(j1·d,c.a+fn
II S
FIGURE
in.J tL
belrJw)
-r Tr
-
v.Z -
1': 2
fA.)
tie ItX ,-for
= (o,s ft) In q119)e.-
t:a11;6'2.
=
J
12-1/
z. r
o
/2-.) 0
(
CAPI1
'i )
Th us) /r1J1It Pte t/e/oc/f!/ &I4I1j Je Vrz. ..... -
Wz
~c5
¥VIm
~2
CeffJ pJeie./'j
(.b)
Ft'tPh1
-F-k-
-s
8'O,()
W
Un d
-
1/..97)
tHe
kl1oWI1 I
v/ec/~e'd.
Fg.
r
12.
tv
Ishtlf.t::-
~
t
2.. Tl
;n1 =
=("<.0
VBI
/171
=0
V~.2
r;.
{/J
Vr .z.
r:J- /'2
~1; yZ1T)(().!>ft)(,~ ft) (90 ft)
&5
j~.
-:: 1ft). 8 s ~9.5 So
EZ. {/)
fY'~
-that
Is ha Ii:
=;-
(Lfo. ~ ~.s) (D,S ft)( 'to ;-) 5 I ~ 3 D
-
0
+t· l.b
fl 8
11'1
When the
.5h a fl 6rea k. r
IMpeller ~v-tn.fvl.I:d:;
.l
the -k>Y"tjJ;te bf?q;rneS
slop.r
6eco..use..
fhere is
sJU:j.{f-
-!-oY-fjfAe fo -fr;yce. if -/r:; y-ohfe. In -loY-rue. dr/ves !he.. i''hIpe//e-r and fh~
f1u./d.
On
ci r iv/11j
d rl'ves
!I,cyo hand ) {he I~pe lie.,,-. fA e-
0
Ii;?
/2-/2
a.
1 I 7-wrt?JY'Je.
Sef() and the no /tJnger a....
plA-l'?'IjI. ft,e, I"Y1pt:jley Y'10~.5
tit.
~e
I
.I
yY1()v/VI,y fll/f/d
12. I
J
[
Discuss the main simplifying assumptions associated with Eq. 12.13 and explain why actual head rise is always less than ideal head rise. Discuss how ideal head rise is head "added" to the fluid and actual head rise is head "gained" by the fluid.
fAa.! nb /~.s.J ()/ ava//a~Je.
a.fS "YI"IJ~
-fh YV>1j i, /-k
-flo w
Ik
a (., rps!,
the
ideal
'p1L1'nj"J
acroSS
7
the
{/owi
liu./d
. the
a..c.I",.1
I~
be/ny
/
IeSJ
12-13
fUn,~ 1;",De //er.
is
fhUJ ellAa I
I'/,l~;o I'"/;'v.J fhe
t,bwevw.l ~e ht!arl
t"an fhe
~/LtJ "rJ
Yea~j~~ ide" /
rl.Je
bl
/2. ) 2.. 12.12. A centrifugal radial water pump has the dimensions shown in Fig. P12.1~. The volume rate of flow is 0.25 ft 3 /s, and the absolute inlet velocity is directed radially outward. The angular velocity of the impeller is 960 rpm. The exit velocity as seen from a coordinate system attached to the impeller can be assumed to be tangent to the vane at its trailing edge. Calculate the power required to drive the pump.
0.75
•
FIGURE
in.~~~
P12.1L.
12. II
(i)
~:: I;. UJ To
obt;tfln
/ husI
//BZ
Ii~m
We.
S" ,it. ( s: Z "'itt:
) (
J
"se 1Ite ~
WI" ) r:
}I'~d)(
'1{p()-;;,~ \7...rr ~
ext. t
"'1
W6h .rb =U. 'i'f ~){t. Z5~~)(¥O '!)("fS~ J :-c) I () t 0
.f2t· /h
-:5 / () /0 f:1=;. Ib
) 2--/ 'I-
.ft
= If~, I 5
Ve fou-ht fot1ftr/e .sh{)w/"l be/ow. V!92.
Iii. (I)
I )
60 r:/~
/2.13 J Water is pumped with a centrifugal pump, and measurements made on the pump indicate that for a flow rate of 240 gpm the required input power is 6 hp. For a pump efficiency of 62%, what is the actual head rise of the water being pumped?
fr~m
Fs.
eZu. a;brP 11
12-.23
1h~ fum/->
eff; c~el1'c'7
,
u
if'cp ~a./SSo bhf ('I) (./;h,)(s-so)
J-Q (0. ~2){ C. hp)( S50 ;~;;
)
('Z.Lf ~~)[(ztf{) ~~)jc7.lf8 Ji~)((po ~,.~)J ~1.3
-fi
/2-15
12. / '-I
I
The performance characteristics of a certain centrifugal pump are determined from an experimental setup similar to that shown in Fig. 12.10. When the f10wrate of a liquid (SG = 0.9) through the pump is 120 gpm, the pressure gage at (1) indicates a vacuum of 95 mm of mercury and the pressure gage at (2) indicates a pressure of 80 kPa. The diameter of the pipe at the inlet is 110 mm and at the exit it is 55 mm. If Z2 - ZI = 0.5 m, what is the actual head rise across the pump? Explain how you would estimate the pump motor power requirement. 2...
+
:5 /n c e.,
G $. (J)
WI
d-
s
-f,;:3j. )
flo X
!3
+
//,Sm1
To e$-n"m,,/e- the plA~p
7::
0 Q.
M()-k"..
hp..
bhp (~§o)
Iv
)
fA.
( ~. 9) (f. g() ;<. 10 3)
J,.,~ =-
(/
=- Yz. A-z...
)~ 3 /J //J'I12. ) ~()i. JO?! ~ 1- (~.01S)(J33)( )D 3) ::.~
12.. =
~
;; (D. J1~tm) 2)1/ .4-J
Ifn tP<.
2-
(flO .J/~ ) ( t. 309 X /0 -
y,= I
7h US) ,f-r~1!1
-r-..z.. -.z:, -r
LJl., -1/,
Cfef
/2- 16
pOl#Cr
ye D1A1;~el?";'
UJ~ Ej.12.21
/2./5
The performance characteristics of a certain centrifugal pump having a 9-in.-diameter impeller and operating at 1750 rpm are determined using an experimental setup similar to that shown in Fig. 12.10. The following data were obtained duri ng a series of tests in which Z2 - Z I = 0, V:z = VI' and the fluid was water. Q(gpm)
120140160180
P2 - P I (psi)
/40.2 /40,1 138,1 136,2
Power input (hp)
1.58
1001120 33,S / 30,1
2,271 2,671 2,95
3,19
3.49
140 25,8 4,00
Based on these data, show or plot how the actual head rise, hi/' and the pump efficiency, 17, vary with the flowrate, What is the design flowrate for this pump?
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1/2./6 12.1 b It is sometimes useful to have flo - Q pump performance curves expressed in the form of an equation. Fit the flo - Q data given in Problem 12.' 5to an equation of the form flo = fl" - kQ2 and compare the values of flo determined from the equation with the experimentally determined values. (Hint: Plot flo versus Q2 and use the method of least squares to fit the data to the equation.)
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(f)n
ft.~1?f
da.-iL
-the
Problem
CAn b(' C!:1"'&l:-ed ant1f hrm etll"j/~
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Q (!-f/WI)
fable
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12. /5.)
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reStAffs qre ~bfr,I#ed.
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7 7. 3
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t.~= 'itf.5- d.oOl7b Ql. (/) wheye. Jru.- IS I~ .f1:. WI"-t-h 42 /11 gprm. A- p}"t Sh()W/Hj the ~m f~Y/s~n be:/;u.;eeJ4 71te -I'X-pev/null'l4./ dam. (inti -/Jte.. fr"e dlc..l-e~ Y'-esuJ-/:s (rr4J1I1
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I 12.17 In Example 12.3, how will the maximum height, ZI' that the pump can be located above the water surface change if (a) the water temperature is increased to 120 of, or (b) the fluid is changed from water to gasoline at 60 OF?
(0...)
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12.18
I 12./ g A centrifugal pump with a 7-in.-diameter impeller has the perfonnance characteristics shown in Fig. 12.12. The pump is used to pump water at 100 of, and the pump inlet is located 12 ft above the open water surface. When the ftowrate is 200 gpm the head loss between the water surface and the pump inlet is 6 ft of water. Would you expect cavitation in the pump to be a problem? Assume standard atmospheric pressure. Explain how you arrived at your answer.
- -l: I -
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pump
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12- 2..1
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19
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T (I)
12. let Water at 40°C is pumped from an open tank through 200 m of 50-mm-diameter smooth horizontal pipe as shown in Fig. P12.19 and discharges into the atmosphere with a velocity of 3 m/s. Minor losses are negligible. (a) If the efficiency of the pump is 70%, how much power is being supplied to the pump? (b) What is the NPSH A at the pump inlet? Neglect losses in the short section of pipe connecting the pump to the tank. Assume standard atmospheric pressure.
Sl
:-:--::-::ti-:-:-:--::-:::~-
3m
Diameter = 50 mmCZ)
1
PUMPj
•
(,~ N'"
r--\----.U FIGURE
Length = 200 m
P12.19
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Note
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+
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12.20 12.20 The centrifugal pump shown in Fig. P12.20 is not self-
(al
•
The jJllmp
head -1/()wl-a t.e tire ShC9wn
pUmp tlln ~cld
FIG U REP 1 2 . 20
charli?i:('r~s tJU -f.e,y
In T-iJ.
(bl
tJ..
1::Y?ICA,) c el1 t;n/ujtl!
rnt2~imtlH1 i1et(i. 1h~t -the t.)hel1 Q~o ((.:e.) at- si:P,.t- (,(fJ ft,,. extlll1;)e}
1l.11.
The
~(C.tI;.s when !hI] heil~ I~ 111 terms t?/ -/he .f/u/d 0 the '-UflJ,. )/ef/ed1nj /t)sses 411H the Ve!lOc.lry h~l:(d (/i"'« ~1I"/i~i:,i;11 eHee,,-h,) -fh~ pump CI/J1 lilt the ff"'I'd tt heljht H e$aa / to -the hellll added b'f the I"m,. ~k,le peri' f 1he liw/d J/; 1}u. pump iJ (I(.'r (L', e,) /Jot prlmpd) 1h~ heAd ",d';e~f 'S I" i::r'I-rtlS 6f H "I" In ~f ~IIY.. FoY' t='X41'11;Je/ 1'1- h«,:::~~ft 1ft e pUPlp eoll/d rllise. WA-b yo 1h~t h"h I'f j t 'J pnme,l. (Iille/! iv/hI WA-i:etJ..r.-f lJte ,oJlQJ,/) ,j /')bi primed (hl!ed k)17h "Jr.) 1JteH ?/Je pwtnj> CAIn t9n/y Y'rIlse w/l;t:.er up to ~ di.sftJ/'Jce,
)-f =- 3(; ft if-elite -/he
w/l--!::.ev /..VI-Ii
6(1;'
t".f:er !')lJi
= 30 Ii ~ef:-
lId-'/)
~ ()7~5 i4~) = (), O:;b 8 f.f:-
((;,z. if ~J)
1h~ pump.
12.2/ Owing to fouling of the pipe wall, the friction factor for the pipe of Example 12.4 increases from 0.02 to 0.03. Determine the new flowrate, assuming all other conditions remain the same. What is the pump efficiency at this new ftowrate? Explain how a line valve could be used to vary the ftowrate through the pipe of Example 12.4. Would it be better to place the valve upstream or downstream of the pump? Why?
(/)
V-=
t>v
w/11t
.::::
~= If) i- Ie.. Oil'f ~~//In/n ~ In
1 f::
4J
2
/0 -t 5.~{))()b-5
[~ (!a/Jnlll~
2-
(2. )
The l~if't"sec-iil!Jn of EgJZJ (the S!fsl-em ettuJlf!)~) WdH 7He f~rfr;rmtll1[e e. u rve .j;:,y the pump) CIS 5htPWi/ be /()W) /HC/JC~rl-eJ 1Jta t
1he hew
.flow i'd.i('
I~
cP =- /I.f tJo 1ii~ a~
and
-In l..:s
.f-/()tvYIt-te.
79.0%.
l.s
100'1----~--------r_----r__r---------
o ---!..----.--"---------'----.---- .-----, o
400
(GO Y) '.e )
800
1200
1600
F'lowrate. pI/min
/2-2..5
2000
2400
12.2.1
I
(Co;, 'i )
A
l,ne valve.
fo
1he
t:{c,f.r
tlf
CI()S'lny
Th,w.
val';'ab/~
CA
fhe
vI/I/~
fy/c..6iJnaJ ye,Ii.>lance is
efu/va/el'7-f ~ ~y
fy)cli~n
and 1?10VI'ny fh~ .$j'.J~ curve to fhe. lel'! J ~ Terse cn~ /he head W""//e A-!- 1/1 n t?,JJeyt;t li()I"7~/ poihl/},vdlv/hy less Iit;wY'p/rt fluln w/H, a m",.~ open valve-./e/Ii,.. This sy~~ C-t,Irve p~,e.
I/1l7ti
valve i.f sys~
if
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re WJOV)~ /Y/( ~~J?
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("pt>n/~ va Ii/e) .
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1/
would! te
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and C/MI/ ~fl"n ,PIJJ$/ble. pkc~~e,...,f 0-1 I'1vl va/ve.
12-26
w/)t,
---J.he
/fJlN
.fuel/ern
up.r~
/2.22...
I 12.2 2... A centrifugal pump having a head-capacity relationship given by the equation ha = 180 - 6.10 X 10 -4Q2, with ha in feet when Q is in gpm, is to be used with a system similar to that shown in Fig. 12.14. For: 2 - : 1 = 50 ft, what is the expected flowrate if the total length of constant-diameter pipe is 600 ft and the fluid is water? Assume the pipe diameter to be 4 in. and the friction factor to be equal to 0.02. Neglect all minor losses.
(I)
The
pllmp head·
lac; -i!1 re /a ht)~ sh, P
Ca.
1r.~::, /~ 0
- d:.. /0
The o;e('~tiI1J jJ{);,,,t w//I wheYe ItL = lip / 6r
Thk5.)
j
f 0 - 6.1 ~ X jlJ
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X
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a.t-
the
= 501- 3. b If X)6
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~
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12.23 J
(lJ
12.23
A centrifugal pump having a 6-in.-diameter impeller and the characteristics shown in Fig. 12.12 is to be used to pump gasoline through 4000 ft of commercial steel 3-in.-diameter pipe. The pipe connects two reservoirs having open surfaces at the same elevation. Determine the flowrate. Do you think this pump is a good choice? Explain.
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10
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P"mp
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e (froM
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fflNJt
2-
= Vb/v : :-
¥ ¢ (~t)/.;) (1T)(.i'/;z ft)('t.f XIO-'$J
£ - S. CJU{r t
f
(341. -It) (2..) (~2. 2ft.)
X ) ~ 5" f
eemlt1fVCl4! steel
A
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4.1: 11t~ f1pulr(J,~e. -!he. pllmp &,e~a.ie.s 6f Iton; wt:Juld A/pellY Ie ~~
nelfy
rype
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soo
::: 300
Bin. dia
10
10:
ID
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,
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•
I
+
!
/Z.21
T Determine the new flowrate for the system described in Problem 12.23 if the pipe diameter is increased from 3 in. to 4 in. Is this pump still a good choice? Explain.
D = tf/tz. ft y
(I/-o(!)a .ft.)
(t !-t) y.::
c:p
4J
:::
A $0
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(Z)(~2.2 ~)
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i1t iJ fjtlW rab. /p W (N ?'r%)/ 7ftIJ PII"'I
5/'; Ce.
, ' . I ----.. ·-~··.... --1---
the. i$"
110
P"l11f ef{/c)e~C.!7 /t"', ,e r- ec. /1o~d
12..-31
IS C
t,J';'/~
hot'~ .
/2.25 12.2 S A centrifugal pump having the characteristics shown in Example 12.4 is used to pump water between two large open tanks through 100 ft of 8-in.-diameter pipe. The pipeline contains 4 regular flanged 90 0 elbows, a check value, and a fully open globe value. Assume the friction factor f = 0.02 for the 100-ft section of pipe. Other minor losses are negligible. If the static head (difference in height of fluid surfaces in the two tanks) is 30 ft, what is the expected flowrate? Do you think this pump is a good choice? Explain.
j./P'p/IC4.·+'~;' ft;,;'tJ
(I) 411;(
(2.).J
jj +- ~2-
z..>
J'
r,th :. Iz.P
wJ111
tiJ1tA
-e"erj!:J .e.lua.-i,oh heiwt?el1 --!he +w()
o/' -/he
9/;"u f'
'
==-0
[If (~,~)
L 4..L :-
1hof f'I1ll1~r-
With
f
/I'.I
-= v..2.. =~ )
-r
/D T
Joss C(H!.fflC~e~fJ
A-
GZ.
6 Y
1J,~
cz,;
be CtPml'.5
5't5ffl'11
+
Z
1/= !E = an~
:zz. - i = 3() f-I:)
Clnt!
~
'? .Il t"- ~/~"
= -3D C41'1
E 1. (/) bec.tJl!}pj
J
~p =- '30 -Ft. + ~ Can be e)('/Y.fSSfd as
heat{ loss term
The.
.)
(I)
-Iv
i: +-
lL
~.O2.
tZ)
(/~() It)]
(fz,
"Ut:llY1e,t
Fe)
fY~1h
cp (.f~h)
V 2L{.3Z.Z{!)
La /ole
.P, '3. A/so;
L!) ~ ft)2,12...
-t 2..0'
[~(ft%)J
b{' wr,lflJ1
AS
-/..-1' = :h) t- Ul- XJ~-5 [t?(~~
2-
( -3' )
The /;'.fpr5ecj.J()~ "I- -fhof! s,!Stfm curve (E~. 3) bJJ~ 1YIe tLlI1l p . ell rlle).as .ShO('uh t:>1'7 th~
that
-f',lIre) If7c/J(4..feJ
~-=
/7'16 *"~ Since the e fl,c',eYlc,:/ 6t ih Jj -fJOk)rafe
/;
nellY
80
,P"k
e f.fti. "el1c!J; ~s sh~wl1 f)n-tHe. ff~t(Y'e -thiJ pump wfJ/lld b~
s(('H5~chl'ft .
•
I
20'
I I
a-a
400
1/ 1716Jltm -_·-------------1-____ ._
----------------800 1200
Flowrate. gal/min
/2..-32.
1600
2000
2400
/2·.).6 12.26 In a chemical processing plant a liquid is pumped from an open tank through a O.I-m-diameter vertical pipe into another open tank as shown in Fig. P 12.26(a). A valve is located in the pipe, and the minor loss coefficient for the valve as a function of the valve setting is shown in Fig. PI2.26 (b). The pump head-capacity relationship is given by the equation ha = 52.0 - 1.01 X 103 Q2 with ha in meters when Q is in m 3 Is. Assume the friction factor f = 0.02 for the pipe, and all minor losses, except for the valve, are negligible. The fluid levels in the two tanks can be assumed to remain constant. (a) Determine the flowrate with the valve wide open. (b) Determine the required valve setting (percent open) to reduce the flowrate by 50%.
T
3m
....-____ J..
30m
40~----------------~
-___00-_-_-_-_-__ -__-_-_-_-_:--)
(~:::-::;:jo:-:--::-:::f-:-:-=:
(a)
o
, ! I
o
20
60
80
(Closed)
100 (Open)
Percent valve setting (b)
•
t:?n'(
7h~
FIG U REP 12.26
111,'"/11
~:::-
11
/)
f'2,
=0 J
11::. If. =0
I
l-
I
-Ivf
he~4 iDs.s #f'm
_
(
vp/ Jlf
~)
fL
0fe",
J :::- 3~""'1
v= Ef.r 3)
(2 )
~
V'
= (kL -t
i-
kJ. ~
(~/"'rt-'t r; 1" P /2.
/.0
D )
2-d-
I:p . (Z) ( 3/)"" ) ]
(j)
33nn + [I. /)
if = !3""
elll1
V 2. /,;
2.
9/:;)
S~ tn",t t'.JJ1k
,,~ Ull'dr-f1f ~s trH )
Z,f.il"j...
.3
(1'j! )
be'Ui1I~..s
~f--
bf?c..&:iI?1e.I'
LL
[ /. 0 of' ~. 02 ( d./;Hf)
1-
12 B 0 (J)
Mt I
A.5
1/,,4.1>::~· 1M( I
= 33/m +
r, .=- 33
= 33 mI + r
k eJlfy~.sJ~""
ell"
L
~z. -
tin If
f'
Is.. 0 ](i26) [
S; 78>( If) 3
[
12-33
(j!
rr (tttI%)] 2.
c:!l)) "
{ S"j
12.20
s/n(~
Ihe film; -tfIJ(ti,,,h ,.;
1" = (/1'1'(
33
Er
St.
~ - /, ~( X (0 3[ Qf-r.3)J 2
~(/n
(6)
1 pa;ied
~ 3
(J;)
I::D de/;erf}1l14~ 1He IJDUJrtLie. Thus,;
.:2..
-t 57 7 8 X'1f) tP = S"z. () fd::-
(I:.)
3
I. () I
X JD
Q
2.
rm 3 &. t)SZf S
rf l;?e flow I"a-k ti if) be ctti 1;1 A~Jf :50 rf=- ~. ~ ;.-zr/z :: tJ. ~ 2 IPS" t11'/ 3h) -the. hel('/ b 11}?/.lJ r
r
J,j
~? = 5'2. t> - I,f)/ X I~
3 (
~.P2~S ~
.3)
2.
S!).brm Frpl'}1
Ft· (If) .5~. 6 m1
Prom
WIPt
XL
tlI1KI1[)WM
= d3r»7 + (.kL +1..0 )(S2.b) (~, P2~S' ~J) 2.
;:.;,. 12. Z'i (b)
-tnt.
J/4/Y(
w()"JA
obi:tJln -In 11. IeL
/2-3'f
be
/3% ~'pel?
t:o
/2.2.7 A centrifugal pump having an impeller diameter of I m is to be constructed so that it will supply a head rise of 200 m at a flowrate of 4.1 m3 / s of water when operating at a speed of 1200 rpm. To study the characteristics of this pump, a 1/5 scale, geometrically similar model operated at the same speed is to be tested in the laboratory. Determine the required model discharge and head rise. Assume both model and prototype operate with the same efficiency (and therefore the same flow coefficient).
F;'w
~;m;Jllrl:';~
+/()W
the 1YJ~c/e/ flll11f Inllst t)J>fya"ie at thii' ~e.pflcie;,t .,) t?'1. p /2.32. .) ..so thai:
St1l11e
(~3L : ( UJ~3)f wheye
tJ,e SU;'~CYI;t
? roiot!:J,ce.
re-HrJ to The
(/'M)
fnodel
and
(p)
-k
-the
Th U5)
Q,
=
~
tV,
rm
?vmr =~
::.
(D,m)J ~ DI'
tv/'.J
(I)
'
b".,. / Of'::"
Yb)
(;)3 (if: ) ~ 3)
2l ( h~) 2.(~ $11?'f
df> = ;
"'h1 .)
Vp
tv.m:='
D,.
tJfJ) ])/m
/2-35
ll.",
l'
/
j)P:= YS".J t1 J1 II
-t.~ = ltJo In.l
/2.2.g
I
! 2.2S Explain how Fig. 12.18 was constructed from t~st data. Why is this use of specific speed important? Illustrate with a specific example.
A variety of pump configurations like the ones shown in Fig. 12.18 were tested over a range of flow rates. Performance data like those shown in Fig. 12.17 were acquired. For each pump configuration, the operation at maximum efficiency was noted and the specific speed, N.s ' (Eq. 12.43) was calculated for that condition of flow. These specific speed values calculated at maximum efficiency operation were then used to distribute the different pump configurations as shown in Fig. 12.18. Specific speed is important because from desired design operational data ( UJ , Q, and ha. ) a specific speed value can be determined. With that value of specific speed and Fig. 12.18 the designer can decide what kind of pump configuration to use for maximum efficiency operation. F or example, at lower values of specific speed, a centrifugal pump is generally best. At higher values of specific speed, an axial-flow pump may be best. In between values of specific speed may suggest that a mixed-flow pump would serve most efficiently.
/1-36
12.29 Use the data given in Problem 12.15 and plot the dimensionless coefficients CH , C<;ih 7J versus C Q for this pump. Calculate a meaningful value of specific speed, discuss its usefulness, and compare the result with data of Fig. 12.18. trPI11
P/"/)/:; /em
Q (jpm)
.(;, /I~td '"j
1h~
/Zf/'5
2-0
Jfo
bo
dllt.1L
tVef"e
80
~b-b/;'~d
.'
j~o
/20
lifo
~~ eft)
1Z.~
12.;
87. '7
t~.S"
77,3
69.S
59.S"
~ (010)
2.'1·7
YI.l
'ff.i
s7.S"
'I.
(.o.,/-
S2.L
/.S-8
2·2.7
7.... (, 7
2.1~'"
;,. )CJ
J./f9
If, () 0
PD/()er Inpu.t(hp)
3
D= .J Jz..
(/!~3 ~~)
- Z. 88 X /~
-s
Q ("
!t) ;'1-
(if ft.):?
m)
lAtA. = (32.2 ~) ~Ik{.ft) eH = UJ}..Dl.
(/8'3.3 r:d)l(J~ k)2
= 1,70 X If3 C(J
=
1..
;L
(ft.)
WshAft. -- W~h6ft (hp) ( ~b~ ~:~; ) fUJ3 DS
(I.11f W:)(183J r;t!)2.~~~ftf
- /. qIf )( /he dak 2..0 Lf
/53/ -If
3,1)7 Y-JO
21·7
4-
•
W.s hAft Chf)
botle .' bO
?to
5. 76'(ID~.
Q
)D-
-3
jO
lot')
/./SX I~
I. 73 ..(lb-3 1. 3~t)D-3
~.IS7t
~.Nf3
'f. 'foxJb If]. 2-
-If
6.//f2J
J;J!i Ib- If s'7U J/)
'f9.1
57.7
( ~o" 't) /z- 37
-~
1'10
I2-D
-.3
-,3
Z.~J>x J(;
3./fI.XID
o./3J 7
0.1/ gil-
().Jo/f
t. 77 i/O
?7~f Jo
/,0.'1
5"2.6
-s--
h.J1xf~
tJ.3
11.o3XIO-J
12.2-1
(COYJ't)
W (ff-)
(bJ.. (9f""')
[ha(ff) ] "¥'t
frrr
50
6/.3%
N
>a IS'
.
In
wiftr,;'
F/J .
tA~
12 ./~
1;2.-38
12.3 0 A centrifugal pump provides a ftowrate of 500 gpm when operating at 1750 rpm against a 200-ft head. Determine the pump's ftowrate and developed head if the pump speed is increased to 3500 rpm.
h>Y
I{np
tt
~ll/el1
~ii. /.J
ptfm p "llIe n
the
efflci:
b!:J
E$s.
t:J/ /2.
t:t
iJ6
C-hl{n~e
4/11'(
/;-/
/2.37.
s~~d on
tp
ih/.t5)
cg 43. And
WI1J.t
tB
=
5"()OJpm/ ~
=-
;:; ... W.2.. Lv I
Lf'z - -
/(){)~
/J. fz.
/7~7) Y~mJ 4rlP'
/!')
_
1..(")
-
!ptn
Wz- l.
~IL =- z..~o ft. I
goo
It
toz =-
(
3 S"P"
rfM) /
(
/757)
rpm)
SS"oo rpm I
(51) 0
.J1m
)
-JheJ1
/2.31
I 12.3/ A centrifugal pump with a 12-in.-diameter impeller requires a power input of 60 hp when the fiowrate is 3200 gpm against a 60-ft head. The impeller is changed to one with a lOin. diameter. Determine the expected flowrate, head, and input power if the pump speed remains the same.
Ft>r Je(!)/)1e 'Y'/~a Ib SI ~//tll' PI.{ mp.s 6>,t::eva-t1l1' ai -!he same sfeeJ. 1ke el-tec:/; of- tt CMnge In IInp:z//PY' d//fme-ter" J.] t:jJIIRJIJ };, ~Is. /Z..39) 12.'1-0 /
/2·ft/·
ThIlS,) (j), -
tVL
=-
¢/
{j?z Hpl'Y1
Eg.
I)..
= (!i;2. ) 3 9/
'fo
Ja../
D, ==
~azSo
thflt
WJ1"k
2-
1)2.]..
Jz.~/::: 6~ -f'i-
-Itt{'Z : :-
(!l )2 '0,
t.~1
-
(If) ~ n. ) ~1-6 -Ii ) lZ J n·
(b ~.
. tV 5hllfiz.
W~jI4ItJ
IV
.JhtJ#;z..
= fl. 7 f-t
::-
60 h p
::
(B-) Sf,j PI .5h4i
/).. -J.f.[)
I
::- (/0/ ~ I z. Jn.
)!Iao hI') .: -
J. if.
JZ.lI_I)
J
np
12..32 12.3-2. Do the head-ftowrate data shown in Fig. 12.12 appear to follow the similarity laws as expressed by Eqs. 12.39 and 12.40? Explain.
f::i~. /2. /2
z/z"w ihe el+~c' 6f CAlIl1fJn,j hetltl- ./Nt()tA-te Cht:ly~cie1"/~hC:s. A-ccord;d.f
Ih e dalA In dlt-/mebr 6>/7 S/ff1J'JI/YI7fj
eA./YeSsfl'l
If/WS
tP, _ 4J2.
TIt US J ~~
-!he
E3.
/Z. 3'1
"nd
-
~
1..5
In CY'etl:;e....s
Z
-!he
(~t. 1l.3&J)
:,
jl1cyet1.s~d
ff&ffI
tiC C 6 rdlny
1:-0
c:P, __ (D;a )3 ~/
bJI? 10 7Ii?)
-to
1:tg. /2.. f-.o
,/),3
till/me t:ft
f/()WYI2-i:e
(f-~In t(
4.5
by
Impel/~Y
=:
PI
~/h.
.f",.
-k
71n.
/2.3'1
tfS
(~.) )~/ bJn.
t6 Jln.
= /.'>'1 cPt
no(
(frOJrt
t/n. if) !l I;'. )
5/';1/ /4rJ.J) fy"l11
E'j.
fl;p~ ~j~. ~v 7In.) t:iI'I C(
(fr~m
61n.
to gin.)
7h 145 6,. 41'1J !tt<.. =- l..S-D H j
4ft/meter' Ivl::- (8)
;/lIdl1
(S~e
IIn;e//Pfj tvh eY'e.
l"ln i I .5t(c.h h5. /~. JZ ~n
-!he
CR.
a()
(It) lJ here
"'1/OkJll1~ fa9~)
(/J =- IZ~ :I,m ~r
CtPrr-l'.spfPJ1r/Ir;,9 pr"etll,-+e/ POIJ1 i
= (;. S1 )
(/ 2
°1 lun)
It = (;. 3b) (2S0 Ii) 'til.
( CO~~)
/2.-
tfj
-=-
4lfd
1htt. b-1n. /,u"lA/P(
be
;2,
~2
(
~,/t)
40
80 Cotp.xily, &JUmin
/01' W
IP
8-/n. 4tAfJ1ebr /11'}~e I/ey
1J,e
u t v(
at:
be
q
=(2..37)
(IZt!) Jptn)
fr,4 : (;. 78) (2- 5""() ft) z.
P!9/;ifJ (13)
;:;~.
-/here 1:;,
)2./2.
#J!f)W lJo~ CUrv~
0,6-/:(11 'n
/InA (C)
1/ze .fh4. l£5
the
fred(c-k~ jJo/;'i,J
-/he
::- 2-8lf- #-pm1 - 'ftfS-~-t
ih( 6Prre.sjJ()l'/dl n3 C!urlle.s tI~m()/1strltt./l1,g -Ih~i. 1h~'1 do a.ppeAY hI; near
~/';'I'J4rt't!:J
/lftvS.
law.s
inl115 Ja -6e t:I. 1::-0 the he~d-,f/~tJr~~ ~ rre5poJ1dln.!}
right
-k
17u
.fo
fe.5' SI;"; ltlYl1-!1
ei CL~rd/l1!I
In
simp}:;
tin¢.
CltrJles
6'-1 n. dltfm'etfJr fWl11p 5 . I f /s cAetlv frfJJI11 1hu i j 1tr7er4//,!J ht?w -Ih(!. i'hrR€ ell,. lies
/2- ¥2.
The
6,-Irl. QI4rnef:eY' tA.
p k)4 yoIJi -i.e
ivy ~ h'9- /2. / 2-
~ye
re LtL-beP(.
7-/V].
'ik,d;
t/{J1t:1
/2,33 A centrifugal pump has the performance characteristics of the pump with the 6-in.-diameter impeller described in Fig. 12.12. What is the expected head gained if the speed of this pump is reduced to 2800 rpm while maintaining a flowrate equal to 200 gpm?
;:;'7.
Frpl'h
cp::
/2. / 2-
17~ :Jpt»1
( See
1t,1II"t
e ff.lc../enC!1
Pr
~r
~-I;'. rlltll11ebr 1m/tiler
the
o/erlLt/~;7 ~I:- -!.;,-~O rjM1.)
hi...=- 230 Ii. wh(n f!);f".~iJnj J peak.. e ff/~ ,eHc:; I;elow). Th U~) ,:f 17te ;lIm; /J st/I/ o;tY'/L-bd a.:t pt'ak:. 411;(
the
fA.) J -th
yeduted ~
SleeP{
2
Koo rpm
'1hfl1
-11"01+1
12.~~
( l?j. c5 c> -ftt.a.A
~L FrtJm
tt):z.
:-
tv,
(
=
23DO
d:f'tJo
rpm rp'"'
)r./70.J!~)
-
I~b ! PI'trl
k~. J2.'S7
l~J iii 2..so iktJ.. t
cPt
/2. if:;)
tv
=
2.
(l?t.
I
tvZ.1..
-it< :; (~2. ) ~f~I W, L
2.3~o rpm =- ( gs{)O rpm
y(2 (; h) .: 3
,-,- -, -,-_OJ I
O+--t--f--l--+.l"
a
~
00
lW
1~
200
C:lpxily. &.lUmin
12 -
13
2~
Ilt7
/2."37)
H
12.3Jf12.3'JIn a certain application a pump is required to deliver 5000 gpm against a 300-ft head when operating at 1200 rpm. What type of pump would you recommend?
,t:;r
(f==
6?e CI "fIC
5'000
sred
3 ptn) ~,(..:: ~OO /-f.. I
dnd
tV
=-
J2C)O
rpm
/
1.5
UJ (rpm)
Yq? (Ipm) ·
[I~ (.ft.)) 3/,/(;2.&0
YSt;~ Ipm (3 ~c It )31y-
I'Pfrl)
I
II g()
From Fij. 12. /t/ d 1h~ .:5!ec/ltc ~!.eed a. rtf.d/~J fhcu pClm; (Cf'nfn°1u9"/ !tI,I')1p) tu~tlJd P( rec.~()1meY1dfd.
/l.-/flf
the
12..35'
I 12.35
A certain axial-flow pump has a specific speed of 5.0. If the pump is expected to deliver 3000 gpm when operating against a 15-ft head, at what speed (rpm) should the pump be run?
Ns
=
tJ (Y'tldh)
reP (I-t%)
[~(kIs2) J/LL.ft:3
3
/'IWIth
(7.13 f/3) (bD ~t1 ) (S:o) [c?ZZ
=
Itjq
.s
)+eflce
-
I!fDD
rpm
;::)(I';-h~
3/'1-
/2,~~
I . A certain pump is known to have a capacity of 3 m 3 I s when operating at a speed of 60 radl s against a head of 20 m. Based on the information in Fig. 12.18, would you recommend a radial-flow, mixed-flow. or axial-flow pump?
Ivs = .;;,,.
0 :. ~o
Y"n' /s) Iv
=
.s
;t:i~.
t.;
Ycp ~o/S)
(Yilt/is)
&fm/s~) /,~ ?"'Jj3/cr
cp::- 3 ( ~o
YA 4
Is ) d:!
/WI)
r. 8/ /s; I'M
t1I1A
-the...
P" mp
'£::
h) V3 ~31.s '
[ (UI ""Is ~)(2~ "" t/~
/2, /~
/2- '16
1..5
ZCJ"Wt
.1.2..37
f Open
12.37 Fuel oil (sp. wt = 48.0 lb/ft3 , viscosity = 2.0 X 10- 5 lb·s/ft 2 ) is pumped through the piping system of Fig.
(Z)
-::B:'.;:.;:-:-~-::ar-:-:-:--:---~j
P12.3/with a velocity of 4.6 ft/s. The pressure 200 ft upstream
Exit
from the pump is 5 psi. Pipe losses downstream from the pump are negligible, but minor losses are not (minor Joss coefficients are given on the figure). (a) For a pipe diameter of 2 in. with a relative roughness e/D = 0.001, determine the head that must be added by the pump. (b) For a pump operating speed of 1750 rpm, what type of pump (radial-flow, mixed-flow, or axial-flow) would you recommend for this application?
5 psi :i:
rD = 2 in.
(l)
__ e
~r-{K...,..,'-
1.0)
-1
'r
v = 4.6 ft/s
1---200 It - - - 1
•
FIGURE
P12."3r
(I)
( 2.) Z
Th e head Il!>ss ..t:erJ11
T--RL
=:
(3Z.. 2.~ ,Sa.
)
CJ4n
[~ va/tI(
-+I.~-+
............
1.0 ~
eJb()w
e';(' ~
t
/;
. ~ ft:.J ) ( ( 1f6' BZ. Z 1$ If. ~ JiJ-.
tln f)(
tv;fh
MUS.J
-t...L
elb = ~, /) 0/ =
13.
i,
h
-f == tlhd
~.~ Z'f ~
4)(az. t7../ (/)
s
~71;(/()'f
(/Jhrl hI_ ,j.z~) . Ff.
(2.)
if::: 18. B -f-t
~= YIt =- ('It, ~ fT)~! ft)2 -=
or
hY
().It/o
rr = (~.I()~ f/3) (7 '/~ !It{ ) ~~,:,;,)
th/J
::1fecl'll c..
re C f!:)m 111 PI"? de d
.{;; yo
==
;)pf>etJ. a. radla/-flow Pump w()uJd b~ -fh d 4.'p)JJ I ~J,() ~ (s~e;:;j.. / Z.I a) .
/2- ¥7
/2·3'1
I
12.39 The axial-now pump shown in Fig. 12.19 is designed to move soon gal/min of water over a head rise of S ft of water. Estimate the motor power requirement and the VI VOl needed to achieve this 1l0wrate on a continuous hasis. Comment on any cautions associated with where the pump is placed vertically in the pipe.
E,. /2.Z~
rrdJ1lf IlYlti
hf!tJa Y/$e
fldWra1e, I"vtl/v~d. Thi$ i.r fJ,e WI;~i/'I'Ju~ P()wer Yejll/~etI '"
()clll~1Ie
f~ ~rlwWtaHce.
p ::
10
~ (jet file pDwe'l ef/l,'vltk",-I- -!rJ fhe.
7 (;(
hA.
e$n~Q-/~ fhe
~S~UI?I~ fJ,e
p(}VtkN
fhfl{f (1y "."fI'It,., {JdWev ~~ £(/Y~,"e,.,f"
~f-h(;/en':Y
1':'1-0
fhe
-
~/etif/e".
fhe
pf
5· / J,f
~ t. If lap
--
So
12.- tf8
n~d h
(J{ skill! IH ~
;ump pe¥'1w1lf1~~ce s"e~i-hi.cI.
-?P' (J."
CfMver.r;tJl'J
/Nt.
/7. ·31
I(COn't
)
7ht m~lIi?
is" ~ tio
,h-/Ah pi'pl!.
fJ,t:, C (J
ClZUI/on In plac/ny fJe,. pUPlf veY"h'c~/& I~
In
vi -1-4 -f,iln
$"0
a ~ay ~
In
pV4id
~1J//a(J.rt.. of' c.a V/*h~n
rAe pump. the.
pu 6/,le..J Ii, the pump e41!? -8'ryd~ pLlWI! 6/ael~ Ahtl ~I/,e~
wetted
JUf-k~S.
J,,e+WUJ'1
t-k #ef..
-Ihe ~n~.J
lip!,!!"
.J/'&r'c~ (I) tMtt:l-
e., uIl 'hm.1
,.!:S/f)
~ pu~p e)lt~n(-t! (2.) INe
qef ).
~ -P,.+ ?f
;;''1
+
?.l.
- P.
'l.
f-
V, ;-2- -I, 1 L
-
1.9
f
50
').
.&.:: 3. 'I
~, -
?.,. -
~
~
"
'6
fD
and
of
/#11~/W,f}e.-
-y
-
hJ,.
w-e m/n/WI'le
I
~/-~J.' To ()'hie~
n,. it we place the /Jump h"" verh"utl't I~ /he, I;' itJ:G
P,,,e..
!hi! w,-II -k~uJ -h ~
ellV; /anfJ..",
,f" Ik fh~
fumf
wk ic.IA
()(;GUrJ
kCl/YYle,
If:.S~
A
hif'" enllkjh
Wh~n
P,.
tlr4j11t
f1re
flw,id ..
12 - '19
Q1"1I1/tI"
h
aVA/a'
rdake1 ptesruYeJ
Jlo/lrP /,YeS,Itt."e
of
Il.Jf / A Pelton wheel turbine is illustrated in Fig. PI2A!. The radius to the line of action of the tangential reaction force on each vane is I ft. Each vane deflects fluid by an angle of 135 0 as indicated. Assume all of the flow occurs in a horizontal plane. Each of the four jets shown strikes a vane with a velocity of 100 ftl s and a stream diameter of I in. The magnitude of velocity of the jet remains constant along the vane surface. (a) How much torque is required to hold the wheel stationary? (b) How fast will the wheel rotate if shaft torque is negligible and what practical situation is simulated by this condition?
•
T= n mrm (a)
W/-!h
(u-V,)(J-cos~)
wheel staTionary
fhe
Tc:·Jfti; rm
~ (I-c().S~)
FIGURE
where no:: If.
U-:::O so
where ~
I. OS7 .s~f~
Thvs, T= - 'I ( /.0.51 !~f.s) (J fl) (/oo1}) (I-cos
Ef· (j)J whel) T=o.l the/}
Thvs,
V = CAJ rm = ~
01'
tv
:=
11..::: r",
J3S0) '= - 722
fI·lb
U=~
=-
/00!J.
/11
The ?/e- YO .Jvy~ ~ ca se- yep -reseYlis ~hq f+
(I)
fh4t
ni = pfJ V= (/ll'f :'~If) 1(1£ FIt (/00 g.) (b) FNm
P12.41
s ilvta-hb-n .
/;J.. -
so
100
raJ ( 60,S) ( S
=9SS rpm c<
hn k..e Y"l
min
I rev
2.11'
rad
)
12.42 Consider the Pelton wheel turbine illustrated in Figs. 12.24, 12.25, 12.26, and 12.27. This kind of turbine is used to drive the oscillating sprinkler shown in Video V 12.4. Explain how this kind of sprinkler is started, and sUbsequently operated at constant oscillating speed. What is the physical significance of the zero torque condition with the Pelton wheel rotating?
As
795 bel4w
()n PIJ!t!
s-nIJw¥I
E
12. >"0
,..; ~ (tJ -'1 )(1- (IJ~ t! ) of
"'4~{f wife. c~,,'j ~ /1 /J'y,u-r.
~.,{1' i~
Wh'fl n
jll~f. /lJHfey fj,al-1 Ik, yeSisf,', -hrrllfe rnJrI"rJ~d f~~ ~1'1"nAJe r ~ The
f'e/~ Wheel
'lite o.fci//(rhol'l
#te fI'Y/nlcl~r.
df
t"tJ~hfm /)NtJ S-IYi)?HkAr
tvn.fJa"J
vtt/ue of 0-
also
AlI?d
YOfrJ:Ht-s IJntl e/YIVtf A.f16v w~et!1
()scil'af/m b,9';'.1/
4"Y
a"'~ ~tl# Ttsul-h I~ ~
VAlue (){ h,
~h;
by
(J~~illa.-I;1n'I
IUVJ
#lit.!
YD!A·!i",. 5"1'('&::/
~Y;or/.
IF tJt! fJ,a.ff Cl/HJ1~c,h';J ~ (Jlc;I/t:f~~ S,ylnWw
-h
ft,~ It /-Iw, w/'eel IJY~4h duY/'Y o)lUntf,;,." ~
fJ,~
~pY/~k./ey
wil)
OSlillQ
(t.tl It!
n,?
al1d fl,e,
R/~
wJ,erl I#i il ru" 4:1 l4~.r~f r~htf.,.,Y)
~jJ€~"
~r'(/l17,j'j
-h
V
',--5/
= V,
.
12.43 A small Pelton wheel is used to power an oscillating lawn sprinkler as shown in Video VU.4 and Fig. P12.43. The arithmetic mean radius of the turbine is 1 in. and the exit angle of the blade is 135 degrees relative to the blade motion. Water is supplied through a single O.20-in. diameter nozzle at a speed of 50 ft/s. Detennine the flowrate, the maximum torque developed, and the maximum power developed by this turbine.
iI FIGURE P12.43
For the Pelion whee I .shown II -]I. 2.V,_1Z(-O'')..!1..f')2.( o H) I v, - If D, I - trI;J.. T S '1: Q-A or
Q~
3
O.O/oq
.ff
!n
From F,9. II, Zl &half == mrm ~ (I-co.rp)
0,;: 0.20 in.
V, :::50 f.l,~
max
and
~hQN
j
~@=13S~
= 0,2.5 mV/2..(1-C .5(1)
max
where
O
m:: eQ == /.9-1'- ~~ (o.OJOq If) '" 0.02 /) _ 0.0211
=!:?S (I~ f.I)(5 0 fjl(t-c:os 1.35') == 0.150 :!v:;;ft = 0, /50 fl, Ib
and
~haN :::
0, '-5 (0,02-1/
~:tM) (so 1}-/-( /- c os /3.5
m4>:
or:
~ha(f max
:::: 22.
S
H.Jj,
.:s-
I .5'So
hp
f!.:Jk. ::;
0, oif. 09 hp
..s
J2-S2
0 )
/2,JfLf A water turbine wheel rotates at the rate of 100 rpm in the direction shown in Fig. PI2.44. The inner radius, r2' of the blade row is I ft, and the outer radius, r j , is 2 ft. The absolute velocity vector at the turbine rotor entrance makes an angle of 20° with the tangential direction. The inlet blade angle is 60° relative to the tangential direction. The blade outlet angle is 120°. The flow rate is 10 ft 3 /s. For the flow tangent to the rotor blade surface at inlet and outlet, determine an appropriate constant blade height, b, and the corresponding power available at the rotor shaft. Is the shaft power greater or less than the power lost by the fluid? Explain.
ptMl&y- C/Alwlake-/ less Itta" Me rtJdweY
fhat #te .slr..ff
NDk.
b~/fM/)
W5nQ f f l•
.J
if
bl.l th~ -flu/ol bec~.J(! Jtlme- of ./ Section (1) the- PQWW //).$ f- by 1t,c, 1/1A1~ is clJJe • FIG U REP 12.44 h -flu /()( ~ 5haIJ. be4?'ln; /;;(;'17'oY) While Ih(... Ye.Jf is ~f,~e.'I'ed()-f hu 5i,(JH .. /OJf.
Section (2)
t
Q:::: 21T'/j b ~ cos 30(J where r; =: 10!f3 (mri" = 2 II II/s()J w/lh w::Ooo!Y.. )(llfliA)(2.7l'rarJ) --/0 'f7 rarl milJ ~().S rev , oS
;/ fo//ows that U, == r; /));:: (2f1)(IO.Jf7 q!) : : 20,9 U2.::: r,. W=(J II) (10,117 /0. '17!
!/
f!l.) -:
From the Law 01 Sines (see (jovre): or ~ :: 1/./2
~
b_ -
p.
T c1
7
so
and
~'."'~2'oo oJ "
fhal from Eq. fI) _ lofj3
2.1l'r, k!,co.s50o -
':'
_2_()_,9_fj~_ _ Sin (90'-ZOtJ-300)
2.1T(2f1)(II,IZj.) C(}.f30(J
=0.0826
Also) Ve, : : ~-I- WI, sih30 20.9 +/1.1'2 sin,30' l: :2.6.£ #Ulshf4(' == pQ (02. V9 2. - U, Ve,) where Q :: :L1l'r2. b Wz or ~ _ ~ _ 10 §-3 tJ
::::
2,-
2.Tl'r~bco.s30(J
-
(I)
u,::: 2.0,9
if
o;]d (1.)
CQs300
!i
Z7J(J11)(0.082.l>lf)cos30(J =22 . .2..r
~}.
From the L4W of Cosines (see fo/vre):
142.= (lO.1f7)2+(2.2..2.1"-2..(J0,lf7)(2.2.2)cos60() or ~ =lfl2!/ Thus) 142.::: IO.lf7 - 2.2.2. sin 30' -:: - 0, 63!) (lnd Fl(. (1J hecomes WshQ(t ::(1,9~ -!~fS) (lofj3) [001 Jf7f1) (-0. 63
Uz.:: /0,'1 7
V
3d
fl.) -(.20.9 #)(;z.t.s ¥)] : : - I. 08x 10'1 t!s:!9. ::: - /9.8 hp
1:2..- 53
12.1.f5 12. Lf5 A sketch of the arithmetic mean radius blade sections of an axial-flow water turbine stage is shown in Fig. P12HS. The rotor speed is 1500 rpm. (a) Sketch and label velocity triangles for the flow entering and leaving the rotor row. Use V for absolute velocity, W for relative velocity, and U for blade velocity. Assume flow enters and leaves each blade row at the blade angles shown. (b) Calculate the work per unit mass delivered at the shaft.
--~
t
1500 rpm
. / fia. tV
~
Blade sections at the arithmetic ~ mean radiUS
t
U
. /
45°V •
7450
FIG U REP 1 2 . '+5
(") U.I = 'i tAl an d U.2::: r.zlJ} were h ~.I = (15 00 .!!jt.)(J!!!j!!.) raJ):: 157 ~ '" min 60S (2"" rev .s so that wilh rm-=O's HI t..v
V, =(0.5 H)(JS7 !~J) ~ 78.5!t The inlei and extf velocdy irianq/e s
are as shown. Nofe: U, -= 02 (same rodilJs) and J')
U2. :::78.5
~ -= W2. (from conT/fwdj efn.
(b) ~Qff = ~
1/,;2 - ~ l4, =V(v~n. - Vel)
rrom fhe fi9/Jres: ~ cos 70() o/JrJ from Law of Sines
=
W,
0)
~ c ~s 'f,s0
_
78.s
sinJ..Oo - sin (70 0 -IfS tJ )
.1L.
so Tlla
f V. =
or
I
(63.S
#) C-f)S~sl1
cos 70·
11
:::
Or
W,
/3/. 3 Ji. . s
II
V9 ,=V;cos20 ::/ 23 $
AIso,;
Vel.::: V2.. - ~ sin~.5t1
= 78,~ f
- 63.5£1- sil'J'fS~= 33.~ 1/
Hence; from Eq. (I) JUShafl:::
{t2. = 702.0--:.. 78,S Ji[ 33.,{ # - /2.3 £i ] s
/;2- S If
=63.S!l
j 2.46 An inward flow radial turbine (see Fig. P12.46) involves a nozzle angle, ai' of 60° and an inlet rotor tip speed, VI' of 9 m/s. The ratio of rotor inlet to outlet diameters is 2.0. The radial component of velocity remains constant at 6 m/s through the rotor and the flow leaving the rotor at section (2) is without angular momentum. (a) If the flowing fluid is water and the stagnation pressure drop across the rotor is 110 kPa, determine the loss of available energy across the rotor and the efficiency involved. (b) If the flowing fluid is air and the static pressure drop across the rotor is 0.07 kPa, determine the loss of available energy across the rotor and the rotor efficiency.
U 1 = 9 m/s
t •
( a)
loss =
so fhai
Ioss
--
Po J - jJfJ2. ~
t
.ld.shaH
J
preSSfJre drop Qcross roi(¥' -=AI!s
qqq ~"S
( /I 0 x103-4'2.)
( 999
P12.4'-&
h were POI-fo2. :: s+a9f'J41i on
3 N I JOX/O m:l
-
FIGURE
- 0.8Lf9
.lj;3)
12- .53
'" . '-I'
( ot)
I
(Cd lI't )
IOSS = If) I f>- 1f)2.
-I- .J»:sht.tll
h
I
J
J
were 101 - 'fn.. ;:: sra9lJa1Jon pressure aro!
J
aero.; S
alJd
floif)f' :: A
JP;h411 ;:: liz Vel. -"0 VOl:: - ~ 14, since Vel. ~O
ThIlS) ~hafl:: - (q q-)( /1-
'f cos .30
0
)
::: -
Q3,S
f.."l.
II/so) .I:J
fJs
== II
-f,. +t p( V, 1:.. V~~)
=0.07 kPa + -}(i.23 ~)((f2 pi1- -(6'!'f) ( m
=(0.07 +O.066¥)AP4;:: O./3~'fkPr.f
T/Jus loss
~
=
3N 1 O,/31'1-X/O;;;;. _ Q3.£ ::: /Z if ~
(1123 ~)
and
..d
93.S .s~
=
13PfR) • 111~ (
1.23
O.8Lf3 == =
.!J1s
J~-S6
'!fa)
10
~
Is
12,47 For an air turbine of a dentist's drill like the one shown in Fig. E12.8 and Video VL!.5, calculate the average blade speed asociated with a rotational speed of 350,000 rpm. Estimate the air pressure needed to run this turbine.
1() esf,~.,.,,,/e, ..fhe..
air ,reI/lAiC I
p,,)
Ylt.ed~d ~
YUn
Mil
.Ju"'~;Yle ~ wt. esnMAk 1},a;! t1te nPJ?/e etif ve!oti,
;$ fJ.htnNi -Iw;c,o(.. VWnA ~
~'1~
At
IA!
ft,e, ~aI.t.
blAde
(JY
V = 2 U =
918
f'Vs
~1I>td.1~ MttcJ,.,. nu... 1..r I N) if ~;nrttlely
50 /
V _ '/8
Z
IIOt)
f-Ih
wilt. C
f1~
~J'n~"k"loJ. fi~,.J l/fJliJ:. J
(Vte (}.eJ ~
.
~
F,9· D. I
tN.. V~'uJ. 'f-
1$
.e.. --
d. /
~
/0 (I'1~
A.;
;v
f: ::: Z. -::: '
ffJ
1~
~d
r. {ffrrl~'hlt/liJ -h M= d.13
I'd
/01'
(). J
7If/If) :::: .:;t7 fJi~ /2-57
12AS A high-speed turbine used to power a dentist's drill is shown in Video V L!.S and Fig. E 12.8. With the conditions stated in Example 12.8, for every slug of air that passes through the turbine there is 310,000 ft ·lb of energy available at the shaft to drive the drill. One of the assumptions made to obtain this numerical result is that the tangential component of the ahsolute velocity out of the rotor is zero. Suppose this assumption were not true (but all other parameter values remain the same). Discuss how and why the value of 310,000 ft· lb/slug would change for these new conditions.
Ex~Je 12. I w(.
PYvw,
= -VVh f 1J~
W-,-,
$n~"'1-
fo
IN J, e,-1J"e",-
/;,~/.I
wfA41f
less ILwhl'nlJ J
btl.
\/f/;/1
:&..
17~
cfwll0 no! 3WD ,/ fl,e JIr d~,e.J1dJ' ~
Vt'll..
.
/#4f1
be
$MIVIlle~
{1,4""
iJ Iht.
IH ()v'
{..4.(e.
fhe dl'Hc..:h~Y) IJ~ rD~»~
I~.
If' /'/4. tie $ do
I~ E~~/G /2. r~
V
In.
,... -fJ,~ dl~c,-h;", If Ytlh:Ii""",1 !he !'If ~e
/es.! ,/ ~JIt" les! ww/c. iJ ex-!w.cferl ~ the, 1-{f/tN/~ 1lfA.-t',J. Jiu t the t:JPfJ~$;-Ie.;.$ /vt.L~ OVi
eG.~
(J.
hfJf
()Y'
f7/
I
ViJ)' i5
if
ha v-e.
wI.e '" +h_~
f1.t.
if
blt4.cle
If
(,LAd r.s fJ,~
tk
Ct4J-i.
wu,.,e ..;.."" I;'
'J
[X/l,IM,/(
1').-58
of tt,c fl..,; 0/
'2.3.
)
I
12·50
12.5 a A Pelton wheel has a diameter of 2 m and develops 500 kW when rotating 180 rpm. What is the average force of the water against the blades? If the turbine is operating at maximum efficiency, determine the speed of the water jet from the nozzle and the mass ftowrate.
VI.Sna L
'"
t
Thvs J
= T =f (JJ
3
Fw or 500 x10 !!.;.!!J. S
=(~In) r(JeO~)(! mi,,)(2.1l'rev raJ) ~ m,n 60s
F= 26 too N ==J===
Also) Wshatl :: pQ(J(U-~)(J-cos~)
Sf) fh41 af maxihllm efficiency
with ~::;/80o and V= ~ fhis qiVes W.shaN ::; \or.>1.f:J..'" (- 2-V,)( 2.) == Q2. Vt2. = Zn1 ~{L
e
0)
BvIII
VJ
= 2 rJ =2 I» 2. Jl = fA.) D= (!80J:!f. )(1.min) (311' raJ) (.2.m) = 3 7• b !!!.. mJfl &o.s rev s
2. (SOOXlrf
~)
{37.6 ';')2.
- 707 N·s tyJ
1~-5q
=707
kf/
s
/2.S I
12.5 I Water for a Pelton wheel turbine flows from the headwater and through the penstock as shown in Fig. P12.'?I . The effective friction factor for the penstock, control valves,and the like is 0.032 and the diameter of the jet is 0.20 m. Determine 'the maximum power.output. -
1 T
D=0.90m I = 1020 m
~h4f1 = pQU(U-V,)(f-C;()S@) or for rnaxilTJ()m power ~::./80oJ
U::-¥!-
Thvs)
=- p'-f:l\' 2:'\1,2-
\:J
Wshaft
max.
0)
:z.
:l
'1
By! f!.{ + ~ u. '" Ii.j~ +Z, +f I> ~ where p. '" " '" OJ z. ~ Q7,s1'J'l, &, ::::2.S011'J and Vo=-o '7
Hence
fi +f l !j.2.
J
"2 () ;::
or so
]X tI-
z, +
J
where Il ~ ::: /I V
d/ V, = -i D2.V • ThtIT_J is V=(rJ)2. -d- ~
=
(0.2 m)2 0.9 m V;
('J)
== O. OJf9'f
V, I
fhal £'1- (2.) hecomes:
97Sm'" UOm +
or ~ ;:: /lif.3
2(f.~;)[1 +O.032( 1:~:;yo.o~9¥f]
where
~~f
t;-
lienee J Q = II/V, ::: *(o.').hJtY//~3';) = 3.S~!f3
Thel'efore frofh £'1' tlJ: J
'NshQIf max.
=-(q9?.&i. )(3 st.!!t) IfJs I
oS
(11'1.3 2
iff)2 - 232 x/o6 !!:!!J. -:: 232.00 /rW . .s =J===
I).. - 60
/2..5'2..
I 12.52. Water to run a Pelton wheel is supplied by a penstock of length e and diameter D with a friction factor f. If the only losses associated with the flow in the penstock are due to pipe friction, shown that the maximum power output of the turbine occurs when the nozzle diameter, D 1, is given by Dl = D/(2fe/D)I/4,
I
WshaN = ptf U(u- ~) (J -Co~p) so the 1fJ4ximvIYI power ovlpui w/lh ~:: /BO O and Tf = ~. TlJvs, •
WshClN :::::
OCC()Y'.s
l-{2.
per i:
(0)
where
(J)
D,
1
v
~------
D
(I)
(2.) (3) ~
12--6/
/2.53
I 12.53 A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies diameter D, and friction the water to the wheel is of length factor f. Minor losses are negligible. Show that the power developed by the turbine is maximum when the velocity head at the nozzle exit is 2H/3. Note: The result of Problem 12.52 may be of use.
e,
(0)
~D J
fI
t
(I)
_ 3
-
~
~2.
ij.
or \1,2
2.1-
= til --
/2 -62.
12.5'1
J 12. {;J./. If there is negligible friction along the blades of a Pelton wheel, the relative speed remains constant as the fluid flows across the blades, and the maximum power output occurs when the blade speed is one-half the jet speed (see Eq. 12.52). Consider the case where friction is not negligible and the relative speed leaving the blade is some fraction, c, of the relative speed entering the blade. That is, W2 = cW I • Show that Eq. 12.52 is valid for this case also.
The inler and otJiJef veJoc/fy frianq/e.s tire as shown.
inlef
Thlls) VOl
= l0 and
V62 ==
oullef
U + W:z. cos ~
bvt W,,==cW', :::c(v,-V)
So
fhat
V8~: r;+c(~-u)c()S(1
Therefore J WslJatt ::: til
u[ u+ c("" - U) co.s~ - v,] ::: m[U( I-c cosp) - V, (I-c ==
For
maXilll{)hI lJowefJ
r
s h411 dW (1 U
m(/-cc~~)[2U-~];:;O
()r
::" 0
m(/- c co.s~)[u2._ lJ~]
rJr
u==-¥:
/2-- 63
CIS
~)]
12.
£.;1 12. SS
A hydraulic turbine operating at 180 rpm with a head
of \ 70 feet develops 20,000 horsepower. Estimate the power and speed if the turbine were to operate under a head 0(100 ft.
ru, =180rl'''' ho J
J
hr:z.
= /701/ , = / 170 II
,
!ls.s/Jllle fhe efliciellcy remQillS cons/ani; So wifh D, = D.. and ~ =v .. ; 170 (/80)"'-
~
100
all- or
aJ:z.= Z3't q'"
IIssume Ihe SQ"'e powercoefficie#: so wiih D, =D. and fl '" f.:
2~ 000
(t BOr 11. .56
=
WShg/i. (23'f)'
or
lIt.h.fl2
=
'f~ '{oo hI'
I
'-----
1-
..../
~~ ·1)~
12. 56 Drafl tubes as shown in Fig. PI2.% are oflC:n installed at the ex it of Kaplan and Francis turbines. Explain why such draft tubes are advantageous.
'@7
t1 .'
:~:f;W/
•,)
(
rc
\:;:
">on I,,,,
",\r,.,;w.,
c~~222a •
FIGURE
P12 .5t
i;J,1ho.n +he dra/f lobe there Ulollid be a relative/y 1ii9h steed. exd jel ( speed V, J f"Bssure 111=0). Wdh Me draff ItJj,e (which at;fs as a ddfwer ) the nif .rpee" is "'''vh smaller (I{,"O,/,1<#0). FrOI11 Bernou/k e?Jl4/10!J ,y Mlows fhal fJ, <0 (wdh ihe draN Me). lienee fheN is a Joryef head OII41/a6/e fo fbe fJll'bine. "ore ellcf'?! can be removed fro", fhe Iluid.
12.57 Turbines are to be designed to develop 30,000 horsepower while operating under a head of 70 ft and an angular velocity of 60 rpm. What type of turbines is best suited for this purpose? Estimate the ftowrate needed.
IJ. - 65
1.2.53
I
12.SS Show how you would estimate the relations~ip ~e: tween feature size and power production for a wmd turbine !Jke the one shown in Video V 12. 1.
+h~ ve/a-honrh'f k:lwtev, kluy(' $'3e Pi'll! f()wer--' p'r'rIdAAcfio", f'w t; vv/".,d ~.,b/ ~~ wt. u.$-e +he d;~YlJ;".."IeIJ pi 1eYWf! dl 12·29 &tNvJ 12·30 whh,t, ate. ~"/;CAbk ft.". ft, iIi'" u'" flus; ble flow. F(". {,";,/a(" .furl>:I'Itf a~'" '~A-h', (A 4; -I;",,)
10
(!sf/lh1l1tic
Err.
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w<- um1/' ivae
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ha I = hal..
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I 12. 5'1 Water at 400 psi is available to operate a turbine at 1750 rpm. What type of turbine would you suggest to use if the turbine should have an output of approximately 200 hp?
/1501[2iO'
:: If. 86
( Q2-3)S/I{-
wh ic h /s in fh e ralJr e
approprtale for an impulse ftlfhine.
/;;'-67
/2·60 I 12.6~) What do you think are the major unresolved nuid dynamics problems for gas turbine engines?
5~ ,n~qo/ (.¥YIrero!ve,) Fluid mect,QT1icr !'#.bI_ J
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pyn
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bl"pfe 10M, f"~J,,,,,,f
.fluid
12.6Jf The device shown in Fig. PI2.fif-is used to investigate the power produced by a Pelton wheel turbine. Water supplied al a constant flowrate issues from a nozzle and strikes the turbine buckets as indicated. The angular velocity, w, o f the turbine wheel is varied by adj usting the tension on the Prony brake spring. thereby varying the torque, T'''-fl' applied to the output shaft. Th is torque can be determined from the measured force, R, needed to keep the brake arm stationary as T"'./\ = Fe. where e is the moment arm of the brake force . Experimentally detcnnined values of wand R are shown in the following table. Use these results to plot a graph o f torque as a function of the angular velocity. On another graph plot the power output, Wollafl = TollaR w. as a funct ion of the angular velocity. On each of these graphs plot the theoretical curves for this turbine. assuming 100 percent efficiency. Compare the experimental and theoretical results and discuss some possible reasons for any differences between them.
•
FIG U REP 1 2. fo/f
w (rpm)
R (Ib)
o
2.47 1.91
360 450
1.84 1.69 1.55 1.17
600 700
940 1120
0.89
1480
0.16
(Q)EXflerimenlal: T- Rl =(O.sfl) R or T= o.sR rUb, where R-/h rai) rev) (.1m!/!.) (1;JI'rev and W·silo" -- r W T ( l
Or
W&h.f/ '"
Valves of
W, TJ
rhr((J-V)(I-('ls~)
Ji'
- O. 5'12 ~ - 53 7 11 - Zl(o.'IJ ,\2. • S 'I
m= pf¥= HeflC8,
TW H;'J, , wheN T-fllb, W~f'pHl
and ~h.rl are 9iVefl ill fh faDle and ,!/'aIA below.
Te
(bJ Thelrldico/: Q V.,-- 7r,
0./01/1
J
Jj;' f'I;
11
where I1ssvll18 ~=180', fld
lJ= 0./05 sips
(1.9'f S71/)(o.SJf:J. f
wdh U=wg e(;'..ff)(!~:;'·(,J ~oi) =0.OU2W#, (J}-rpll1
T= (0. /05 S~"9f)(F.:fI)[O.OU2W -53.7] or
T =
~
1.'11 ['f,88XI0 '"
-I]
rI·//;,
¥
where "'~rprn
(con 'I) 12-6q
(3)
J2,6lf
(crJII11)
IIlso)
~hQfl =
T(~f (I)) ::: O. lOll- 7 T
T tJ) =
tJ)
t~:Jj ) where T,., fl-Ib, o;-I',jt
Values of T and Ufhalf frtJdJ £rs. (3) and (fl.) are ploiled if) fhe 9r4/); helrJw~
e~periment T ) fl.1 b W.,LJI £:1;J! Snarr) s
o
1.235
0
360
0.9S.5
l/-5o
o/ltO
.36.0 '13.3
600 700
O.8'fS O. 77.5
9Jf(J
Iheory . -7: F/·/b -~h4If
J
0
l.lI-l /./6
1/-.3.8 5/.g 62..6
/./00
0.595
53.1 .5 6. 8 57.6
o.QQ7 0.92.8 0.763
1/20
O.I/IIS
52.2.
O.63Q
75./ 7S.0
IJfBO
0.080
12.'1-
0.392-
60.7
68.0
80
1.6
•
WshoU
J.~
T
H.Jh
/.2-
--~
60
"' JC
1.0
D.S
•
0,6
experiment ~ Jl.Q
"
" 2.0
0.40.2-
0
/000
w, rpm
/'2- 70
2000
0
~ .s
APPENDIX A Listing of Standard Programs
It -/
EXPFIT.BAS
100 cls 110 print "***************************************************11 120 print ,,** This program determines the least squares fit **" 130 print "** for a function of the form y = a * e ~ b*x **" 1'*0 print "***************************************************" 150 dim x(101l,y(101l,logy(101l,ybar(101l 160 print 170 input "Number of points: ",n 180 print "Input X, yl1 190 for i=l to n 200 input x(il,y(il 210logy(il=log(y(ill 220 next i 230 sx=O 250 sy=O 260 sxy=O 270 sxsq=O 280 for i=l to n 290 sx=sx+x(il 300 sy=sy+logy(il 310 sxy=sxy+x(i)*logy(i) 320 sxsq=sxsq+x(il~2 330 next i 3'*0 loga=(sxsq*sy-sxy*sx)/(n*sxsq-sx~2l 350 b=(n*sxy-sx*syl/(n*sxsq-sx~2l 360 a=exp(logal 370 print 380 print using "a = +*.*~Ht~~~~II;a 390 print using "b = +jt.***~~~~II;b '*00 print '*10 print" X Y Y(predictedl" 420 for i=l to n 430 ybar(il=a*exp(b*x(ill '*40 print using "+#.####~~~~ +*.*###~~~~ +#.####~~~~II;x(il,y(il,ybar(i) lj,.50 next i
A-2.
LINREG l.BAS
5 cis 10 print "***************************************************" 20 print "** This program determines the least squares fit **" 30 print "** for a fun c tion of the form y::;: b .. x **" ~O print "***************************************************" 4.5 print 50 dim x ( 1011,y(lOl),ybar(101) 60 input "Number of points : " ,n 70 print "Input X, yll 80 for i"'1 to n 90 input x (ll ,yCi ) 100 next i 110 5XY=O 12 0 5x3Q=O 130 for i=1 to n 14.0 5 Xy= s xy+x(il*y(il
150 160 170 180 190
sxsq=sxsq+x(i)~2
next i b=sxy /s xSq
print
print using "b ::;: 200 print 2 10 print 11 X
+* ....
220 for i"'1 to n 230 ybar(l ):b*x{i ) 2lj.Q print using " + •.•• U 250 next i
~~ A A
Y A ---
..
;b Y ( predicted)tl
+ •. UU" """
A-3
+1t.UU .. _ .... ";x(il.y(il.ybar{il
LINREG2.BAS
5 cis 10 print "***************************************************" 20 print "** This program determines the least squares fit **" 30 print "** for a function of the form y = a + b * X **" ~O print 1'***************************************************" 50 dim x(101).y(101),ybar(101) 55 print 60 input "Number of points: ",n 70 print "Input X. Y" 80 for i=l to n 90 input x(i),y(il 100 next i 101 sx=O 102 sy=O 110 sxy=O 120 sxsq=O 130 for i=l to n 131 sx=sx+x(i) 132 sy=sy+y(i) 1~0 sxy=sxy+x(i)*y(i) 150 sxsq=sxsq+X(i)A2 160 next i A 161 a=(sxsq*sy-sxy*sx)/(n*sxsq-sx 2) A 170 b=(n*sxy-sx*sy)/(n*sxsq-sx 2) 180 print 190 print using "a = +*.***AAAA";a 200 print using "b = +i.i*iAAAA";b 210 print 220 print" X Y Y(predicted)" 230 for i=l to n 2~0 ybar(i)=a+b*x(i) 250 print using "+~Lii**AAAA +~L****AAAA +*.i***AAAA";x(i) ,y(i) ,ybar(i) 260 next i
A-If
POLREG.BAS"
100 110 120 13 0
cIs print "***************************************************" print "** This program detennines the least squares fit **" print 11 ** for any o rder polynomial of the form: **" l~O print "** y :: dO + d1*x + d2*x"2 + d3*x"3 + •. . **" 150 print 11***************************************************" 160 print 170 dim b(21),d(211,s(21),x(lOll,y(lOl),f(101) 180 dim errf(101),pj( 1 01),pjml{lOll,ybar(101) 200 input "Enter number of terms in the polynomial: ",nterms 210 input "Enter number of data points: ",npoint 22 0 print:print "Enter data points (X • Y)" 230 for i:1 to npoint 2~O input x { i ) ,y(i )
250 d(i)=Q 260 f(i)"'y{i) 270 next i 280
print
290 print "The coefficients of the polynomial are:" 300 310 320 330 3lj.0 350 36 0 370 380 390 400 410 42 0 430 440 450 460 470 480 490
for i=1 to npoint f(i)=f(i)-d(nterms+1)*x(i)~(nterms)
next i for j=l to nterms b( j)=O d(j )=0 s{j)=O next j C(1)=0 for i=l to npoint d(1)=d(1)+f(il b(11=b(1)+x{i) s(1)=s(1)+1 next i d(1 ) =d(1)/s (1) for i=l to npoint errf(i)mf(i)-d(l) next i if nterms=l then goto 750 b(ll=b(l)/s(l)
,f-s
(con't ) POLREG.BAsa
500 510 520 530 5~0
550 560 570 580 590 600 610 620 630 6~0
650 660 670 680 690 700 710 720 730 7~0
750 760 770 780 790 800 810 820
for i=l to npoint pjm1(i)=1 pj(i)=x(i)-b(l) next i for j=2 to nterms for i=l to npoint p=pj(i) d(j)=d(j)+errf(i)*p p=p*pj(i) b(j)=b(j)+x(i)*p s(j)=s(j)+p next i d(j )=d(j )/s(j) for i=l to npoint errf(i)=errf(i)-d(j)*p(i) next i if j=nterms then goto 750 b(j)=b(j)/s(j) c(j)=s(j)/s(j-1) for i=l to npoint p=pj(i) pj(i)=(x(i)-b(j»*pj(i)-c(j)*pjm1(i) pjm1(i)=p next i next j print using "d# ::: +#.####AAAA";nterms-1,d(nterms) nterms=nterms-1 if nterms)O then goto 300 print print II X Y Y(predicted)" for i=l to npoint print using "+#.####AAAA +#.##1t#AAAA +4L####A~AA";x(i),y(i),y(i)-errf(i) next i
aThis program is based on an algorithm described in Conte, S.D. and de Boor, C., Elementary Numerical Analysis: An Algorithmic Approach, 3rd Ed., McGrawHill, New York, 1981, p. 2.59. '
POWER1.BAS
5 cls 10 print "***************************************************11 20 print "** This program determines the least squares fit **" 30 print "** for a function of the form y = a * x h b **" 4.0 print "***************************************************" 50 dim x(101),y(101),logx(101),logy(101),ybar(101) 55 print 60 input IINurnber of points: ",n 70 print:print "Input X, Y" 80 for i=l to n 90 input x(i),y(i) 98logx(i)=log(x(i)) 99 logy(i)=log(y(i» 100 next i 101 sx=O 102 sy=O 110 sxy=O 120 sxsq=O 130 for i=l to n 131 sx=sx+logx(i) 132 sy=sy+logy(i) 14.0 sxy=sxy+logx(i)*logy(i) 150 sxsq=sxsq+logx(i)h2 160 next i 161 loga=(sxsq*sy-sxy*sx)/(n*sxsq-sx h 2) 170 b=(n*sxy-sx*sy)/(n*sxsq-sx h 2) 175 a=exp (loga) 180 print 190 print using lIa = +#.###hhhAllja 200 print using lib = +#.###hhhhll;b 210 print 11 220 print Y(predicted)II X Y 230 for i=l to n 2~O ybar(i)=a*x(i)-b 250 print using 1I+#.####hhh- +#.####_h_h + # • # # # # h h h h " j X ( i ) , y ( i ) ,ybar ( i ) 260 next i
SIMPSON. BAS
100 cls 110 print "*",,,,,,,,,,*,,,,,,*,,,,,,,,,,,,** •• *,,,,,,*,,,**,,,,,,,,,**,,,,,,,,,,,,***********,,,,,,***"
120 print It** This program performs numerical inte gratio n **" 130 print ,,** over a set of an odd number of equally **" 1t..0 print "** spaced po ints using Simpson I s Rule "'." 150 print 11.**************.******",***********************",*."
16 0 print 170 dim x{101J,y(lOll 180 input "Enter number of data points:
", n
190 print "Enter data po ints (X • Y)" 200 for i 1 to n 3
210 input x (il, y (i) 220 next i 23 0 h : (x(nl-x ( l )/( n - l) 24 0 5"'0 250 for i=2 to 0-1 step 2 260 sEs+4*y(i)+2*y ( i+ll 2 7 0 next i 28 0 intgrl:h/3*(s+y(1)-y ( n» 290 print
300 print using "The approximate val ue o f the integral is: +*.UU ........ ";intgrl
TRAPEZO I.BA S
100 c ls 110 print
"*****************************************************"
120 print "** This program performs numerical integration **" 130 print u** over a set of points using the Trapezoidal Rule **" litO print u**** •• *******.*******************.*******************" 150 print 160 dim x(lOl),y(lOl) 170 input "Enter number of data points: ",n 180 print "Enter data points (X , y)" 190 for i=1 to n 200 input x(i),y(i) 210 next i 23 0 intgrl=O 2 ~ 0 for isl to n-l 250 intgrl=intgrl+O .5* (x(i+l' - x ( i)*(y(i ) +y(i+l » 2 60 next i 280 print 290 print using "The approximate value of the integral is: + •.•••• - -- - ";intgrl
A- -8
CO LEBROO.BAS
100 cIs 110 print
120 130 140 150
,,*.*** ••• *** •• ****.* •• ****.**.* •• * •• *.************* •• ****.11
print "** This program determines the fri ction factor. f. for **" print "** pipe floW' for the case of laminar or turbulent flow «." print "*'* (solving iteratively Colebrook's equation), given *. " print "** the Reynolds number and the relative roughness of **"
160 print ".* the pipe **" 17 0 print "* •• ** •• *.********.***.*.*.*.**** •• ****.****.****** •• *" 180 print 190 input "Enter Reyn olds number, Re = tItre
** ••
200 f"6/j./re 210 if re < 2100 then goto 260 220 input IIEnter relative roughness, rr = ". rr 230 fp · f 240 f~1/(-2.0*log(rr/3.7+2.5 1 /(re*fp~.S)l/log(lO»A2 250 if abs(l-f/£p»O.OOl then g o te 230
260 print 270 print us ing "The friction factor is f
;0
+1t . •• UAA ...... ";f
CUB IC.BAS
100 110 120 130 l(,j. O 150 160 170 18 0 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330 3(,j.0 350 360 370
cIs print print print print print input input input
"******************* ***************************************" It ** This prosram determines the real roots of a **" "** c ubic equation of the form x 3 + a*x " 2 + b*x + c '" 0 **" "************"'**"'*"' ****************************************" A
" " "
a • ",a b • " •b c • " ,c
'Check if the equation h as complex roots p"'(3*b-a'"'2)/3 q={2*a A3-9*a*b+27*cl/27 if q'"'2/(,j.+p"3/27{=0 then go to 250 print:print "The equation has complex root s ll:stop xO= - a/3+2/3 *( a"'2-3*bl"0 . 5 for i=1 to 20 x1~(2*xO "' 3+a*xO "' 2-cl/( 3*xO"'2+2*a*x O +b )
if abs (x1 / xO-1 )<0 .0001 then goto 310 xO "'x1 next i rn"'a+x1 n=b+a*x1+x1"'2 x2 z ( -m+(m ... 2-(,j.*n) ... . 5)/2 x3 -( -m -( m... 2-(,j.*n) ... . S )/ 2 print print "The roots of the cubic e quation are:" pri n t using "x1::: +*.UU ... ...... ... x2 .. +It . tUI x3=+It .• U, ........... ";xl.x2,x3 A
.......
'-
~o o
110 120 130 1'0 150 160
CLS
PRINT "**********************************************************" PRINT ,,** This program computes the one-d imensiona l Fanno or *:*,. PRINT
PRINT PRINT
"** Rayleigh flow functions f or a gas with constant **" "** s pecific heat and molec ular weight. (NOTE: k > 1 ) **" "**********************************************************"
170 ' Fann o flow f unctio ns 180 DEF FNFTTSTAR (K, MAl = (K + 1*) / (2* + (K - 11 ) 190 DEF FNFVVSTAR (K. MAl = SQR(FNFTTSTAR(K. MA l * MA
200 DEF FNFPPSTAR (K. MAl
=
I MA 220 DEF FNFLD (K. MA) = (11 K. MAl ~ 2) I (2 * K) 230 ~)))
*
2 ~O
260 270 280 290
MA
A
2) /
' Rayleigh flow functi o ns DEF DEF DEF DEF
=
FNRPPA (K, MAl = FNRTTA {K, MA l : FNRVVA (K, MAl = FNRTOTOA (K. MA l
+ UI
1* + (K - 11 )
(K
*
3 10 320 330 300 350 360 310 380 390 400 410 420 430 440 45 0 460 410 480 490 500 510 520 530
MA
*
~
MA
MA 2)
~
« K + 111 / (21
~
2)
~
2 /
2) + (K + 1.)
*
(2*
*
*
(K - 1
* LOG (FNFVVSTAR(
2*
( 11 + K) / (1# + K * MA • 2) (FNRPPA(K. MAl * MAl 2 FNRPPA(K. MAl * MA ~ 2 = 2* * FNRPPA(K, MA l 2 * MA
300 DEF FNRPOPOA (K, MA l = FNRPPA(K, MAl
I ( K - U) )
~
SQR (FNFTTSTAR{K. MA l) / MA
210 DEF FNFPOPOSTAR (K , MA l : ( 1* / FNFTTSTAR (K, MA l)
250 DEF FNRTMP (K. MA l
*
A
2
*
FNRTMP(K. MA ) /
FNRTMP(K, MAl / (K + 111)
, Get functions desired LOCATE 8 : PRINT ItProgram options 11 LOCATE 9: PRINT" (1 ) Fanno flow cal c ul a tions" LOCATE 10: PRINT" (2 ) Rayleigh f lo"," calcul ations" r,.OCATE 11: INPUT "Enter the number of the option desired: ". OPT IF (OPT <> 11 AND (OPT <> 2) THEN LOCATE 11: PRINT SPACE$(79): GOTO 360 '-- - Disp lay banner specif ing which fl ow calcul ation is being performed CLS
= 2 GOTO "80 PRINT 11**********************************************************" PRINT "** Computing the one-dimensional Fanno fl ow functions **" PRINT u** for a gas with constant specific heat and molecular **11 PRINT It** weight. (NOTE: k > 1) ** " PRINT "**********************************************************" GOTO S,.,O PRINT It*************************************************************" PRINT "** Computing the one-dimensional Rayleigh flow functi ons **" PRINT "** for a gas with constant s pecific heat and molecular **" PRINT "** weight. (NOTE: k > 1) **" PRINT "* **** ********** ***** ": ************ ******** ********************" IF OPT
5~O '--- Get the user specified specific heat rati o 550 LOCATE 7: INPUT "Enter the specific heat ratio. (k
(cont)
>
1 ): "
K
(K
(K
(Cf!)/')Z)
560 570 580 590 600 610 620 630 6~0
IF K (= 1 THEN GOTO 550 LOCATE 7: PRINT SPACE$(79): LOCATE 7 PRINT USING "The specific heat ratio is k=##.###"; K I
Get Mach number to solve for FOR I = 8 TO 16: PRINT SPACE$(79): NEXT I LOCATE 16: PRINT SPACE$(79): LOCATE 16 INPUT "Enter a Mach number to solve for (999 to quit): ", MA IF MA = 999 THEN END IF MA > 0 AND OPT = 2 THEN GOTO 710 IF MA > 0 AND OPT = 1 THEN GOTO 830 LOCATE 9: FOR I = 1 TO 6: PRINT SPACE$(79): NEXT I: LOCATE 1~ PRINT "Valid Mach number range: Ma > 0" GOTO 620 I
650 660 .670 680 690 700 I 710 I Solve Rayleigh flow functions for specified k and Ma 720 LOCATE 9: FOR I = 1 TO 6: PRINT SPACE$(79): NEXT I: LOCATE 9 730 PRINT USING" Ma #.####~~~~"; MA 7~0 PRINT USING" P/Pa = #.####~~~~"i FNRPPA(K, MA) 750 PRINT USING" T/Ta = #.####~~~~II; FNRTTA(K, MA) 760 PRINT USING" VIVa = #.####~A~A"; FNRVVA(K, MA) 770 PRINT USING "To/Toa = #.####~~~~II; FNRTOTOA(K, MA) 780 PRINT USING "Po/Poa = #.fi:fi:#fi:~~~~II; FNRPOPOA(K, MA) 790 I 800 I Loop back for another Mach number 810 GOTO 620 820 I 830 1 Solve Fanno flow functions for specified k and Ma 8~0 LOCATE 9: FOR I = 1 TO 6: PRINT SPACE$(79): NEXT I: LOCATE 9 850 PRINT USING" Ma = #.####~~~AII; MA 860 PRINT "f(l*-l)/D = "; 865 PRINT USING "#.####~~~~"; FNFLD(K, MA) 870 PRINT " T/T* = "; 875 PRINT USING "*.###*~~~~II; FNFTTSTAR(K, MA) 880 PRINT II V/V* = "; 885 PRINT USING "fi:.fi:###~~~~"; FNFVVSTAR(K, MA) 890 PRINT II P/P* = "; 895 PRINT USING "#.####~~~~II; FNFPPSTAR(K, MA) 900 PRINT II Po/Po,* = "; 905 PRINT USING "#.###*~~~~"; FNFPOPOSTAR(K, MA) 910 PRINT 920 ' 930 I Loop back for another "Mach number 9il-0 GO TO 620
It -II
ISENTROP.BAS
100 110 120 130 14,0 150 160 170 180 190 200 210
CLS PRINT "***********************************************************" PRINT "** This program computes the one-dimensional isentropic **" PRINT "** flow functions for a gas with constant specific heat **" PRINT "** and molecular weight. (NOTE: k > 1) **" PRINT 11***********************************************************11 ' I Isentropic flow functions DEF FNITTO (K, MA) = 2# / (2# + (K - 1~) * MA ~ 2) DEF FNIPPO (K, MA) = FNITTO(K, MA) ~ (K / (K - 1#) l DEF FNIRRO (K, MA) = FNITTO(K, MA) ~ (1# / (K - 1#)) DEF FNIAASTAR (K, MAl = (2# / ((K + 1#) * FNITTO(K, MA))) ~ ((K + 1#) / * (K - 1#))) / MA 220 I . 230 ' Get user specific heat ratio and display 24,0 LOCATE 7: INPUT IIEnter the specific heat ratio, (k > 1): ", K 250 IF K (= 1! THEN GOTO 2~0 260 LOCATE 7: PRINT SPACE$(79) 270 LOCATE 7: PRINT USING "The specific heat ratio is k=##.###"; K 280 r 290 ' Get Mach number to solve for 300 FOR I = 8 TO 15: PRINT SPACE$(79): NEXT I 310 LOCATE 15: PRINT SPACE$(79): LOCATE 15 320 INPUT IIEnter a Mach number to solve for (999 to quit): ", MA 330 IF MA = 999 THEN END 3~0 IF MA > 0 THEN GOTO 390 350 LOCATE 9: FOR I = 1 TO 5: PRINT SPACE$(79): NEXT I: LOCATE 13 360 PRINT "Valid Mach number range: Ma > 011 370 GOTO 310 380 390 ' Solve isentropic flow functions for specified k and Ma ~OO LOCATE 9: FOR I = 1 TO 5: PRINT SPACE$(79): NEXT I: LOCATE 9 ~10 TTO = FNITTO(K, MA) ~20 PPO = FNIPPO(K, MA) ~30 AASTAR = FNIAASTAR(K, MA) ~~o RRO = FNIRRO(K, MA) ~50
I
~60
r Display computed results 4,70 PRINT USING" Ma = #.####~~~~II; ~80 PRINT USING" T/To = #.####~~~~"; ~90 PRINT USING liP/PO = #.####~~~~II; 500 PRINT USING "RHO/RHOo = #.####~~~~"; 510 PRINT II A/A* = "i 515 PRINT USING "#.####~~~A";AASTAR 520 PRINT 530 ' 54,0 ' Loop back for another Mach number 550 GOTO 310
MA TTO PPO RRO
A--/;2.
(2#
SHOCK.BAS
100 CLS 110 PRINT "************************************************************" 120 PRINT H** This program computes the one-dimensional normal-shock **/1 130 PRINT 11** functions for a gas with constant specific heat and **11 14,0 PRINT 11** molecular weight. (NOTE: k ) 1) **11 150 PRINT "************************************************************" 160 I 170 I Normal-shock functions 180 DEF FNSTMP (K, MAX) = (2# * K * MAX ~ 2 / (K - 1#)) - 1# 190 DEF FNSMAY (K, MAX) = SQR«MAX ~ 2 + (2# / (K - 1#))) / FNSTMP(K, MAX)) 200 DEF FNSPYPX (K, MAX) = 2# K * MAX ~ 2 / (K + 1#) - (K - 1#) / (K + 1#) 210 DEF FNSVYVX (K, MAX) = (K + 1#) * MAX A 2 / «K - 1#) * MAX ~ 2 + 2#) 220 DEF FNSTYTX (K, MAX) = (1# + (K - 1#) * MAX ~ 2 / 2#) * FNSTMP(K, MAX) / «( K + 1#) MAX) A 2 / (2# (K - 1#))) 230 DEF FNSPOYPX (K, MAX) = «K + 1#) MAX ~ 2 / 2#) A (K / (K - 1#)) FNSPYPX (K, MAX) A (1# / (1# - K)) 24,0 DEF FNSPOYPOX (K, MAX) = «(K + 1#) * MAX ~ 2) / (2# + (K - 1#) * MAX ~ 2)) A (K / (K - 1#)) / FNSPYPX(K, MAX) A (1# / (K - 1#)) 250 I 260 I Get user specific heat ratio and display 270 LOCATE 7: INPUT "Enter the specif ic heat ratio, (k ) 1): ", K 280 IF K <= 1! THEN GOTO 270 290 LOCATE 7: PRINT SPACE$(79): LOCATE 7 300 PRINT'USING "The specific heat ratio is k=##.###"; K 310 I 320 I Get Mach number to solve for 330 FOR I = 8 TO 17: PRINT SPACE$(79): NEXT I 34,0 LOCATE 17: PRINT SPACE$(79): LOCATE 17 350 INPUT "Enter a Mach number to solve for (999 to quit): ", MAX 360 IF MAX = 999 THEN END 370 IF MAX )= 1! THEN GOTO ~20 380 LOCATE 9: FOR I = 1 TO 7: PRINT SPACE$(79): NEXT I 390 LOCATE 15: PRINT "Valid Mach number range: Ma,x)= 1" ~OO GOTO 3~0 4,10 I ~20 ' Solve normal-shock functions for specified k and Ma ~30 LOCATE 9: FOR I = 1 TO 7: PRINT SPACE$(79): NEXT I: LOCATE 9 ~~O MAY = FNSMAY(K, MAX) 4,50 PYPX FNSPYPX(K, MAX) 4,60 VYVX = FNSVYVX(K, MAX) ~70 TYTX = FNSTYTX(K, MAX) ~80 POYPOX = FNSPOYPOX(K, MAX) ~90 POYPX = FNSPOYPX(K, MAX) 500 ' 510 t Display computed results 520 PRINT USING" Max = #.####AAA~ft; MAX 530 PRINT USING II May = #.####~~~~II; MAY 5~0 PRINT USING" Py/Px = #.####AAAAft; PYPX 550 PRINT USING "RHOy/RHOx = #.####~~AAII; VYVX 560 PRINT USING" Ty/Tx = #.####A~AAII; TYTX 570 PRINT USING II Poy/Pox = #.####A~~~II; POYPOX 580 PRINT USING" Poy/Px = #.####AA~A"; POYPX 590 PRINT 600 I 610 I Loop back for another Mach number 620 GOTO 34,0
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