., =
o + 2p(v.,>~[z- !(zn
(7 .10-1 O)
This result indicates that the pressure increases continuously from (p ) 0 at the entrance ofthe channel to (p ) 0 p ( v., )~ at the end ofthe channel. Differentialmacroscopic balances present no serious difficulties provided we can recognize when they must be used.
+
PROBLEMS o 7-1.
Make use of the continuity equation and the material derivative to show that the stress equations of motion given by Eq. 7.2-3 are identical to those given by Eq. 4.4-13.
7-2. If
f
(7.10-4)
v · ndA = O
A.,(t)
and the flow is incompressible, show that the macroscopic mechanical energy balance may be written as
divide by A llx to obtain
(p)Jz+&z (p)J.,
A
2p(v.,)~( 1 _ ~) = d(p)
-p(v!>A. + p(v!>Aiz+&z =
Qo
Use of Eq. 7.10-8 allows us to write Eq. 7.10-7 in the form
(7.10-2) so that the flat velocity profile will be neglected. We now wish to determine the pressure as a function of x, and we therefore need to develop a differential-macroscopic balance in the x-direction. We do so by applying Eq. 7.2-11 to the control volume shown in Fig. 7.10-2. Neglecting viscous effects and dotting Eq. 7.2-11 with i, we get
=
(7.10-5) m the y- and z-
~
fG
pv
)
dV
+
JG J
pv
2
)
(v - w). n dA
A , (l)
fa(t)
=
(7.10-6)
2
J
V •
t(n) dA -
A , (t)
prpV • n dA
+ W-
Év
d"(l)
7-3. A hydraulic brake that consists of a cylindrical ram displacing fluid from a slightly larger cylinder is shown in Fig. 7-3. The velocity of the ram is specified as u0 and we wish to know the force F required to maintain this =-
d(p)
(7.10-7)
dx
motion. Neglect viscous effects and use both the momentum and the mechanical energy balances to obtain two answers. State carefully the assumptions that must be made in order to obtain a solution. Ans: The mechanical energy balance gives
(7.10-8) F
=
pu~( 1T Di/4) 2[(D 1 / D0) 2 - 1F
276
Macroscopic Balances-lnertial Effects
Chap. 7
Problems
Fi Flg. 7-l. The hydraulic ram.
7-4. A jet of water issues from a 1-in. I.D. nozzle at a velocity of 30 ft/sec and strikes a flat plate moving away from the jet at 17 ft/sec, as illustrated in Fig. 7-4. What force does the water jet exert on the plate?
7-6. A very simple device for pumping 7-6. The fluid to be pumped is (instead of the force supplied b sec 0 ndary fluid issuing through th place vía viscous effects, and the
"1''
17ft/sec
l
Gravity
Fig. 7-4
7-5. Water issues from a pipe imbedded in a concrete wall as shown in Fig. 7-5. lfthe velocity leaving the pipe is 20 ft/sec and the cross-sectional area is 1 in. 2, what are the components of the resultant force exerted on the pipe by the wall? Assume that the pressure drop from the wall to the end of the pipe is negligible. Ans: F.,= 0.72Ib1 , F11
=
2.70 lb1
Fig
energy balance cannot be neglectec be neglected if the flow is turbule pump can be obtained with the mo1 the pressure rise between points 1 nozzle area A 0 while maintainin¡ Assume that the secondary fluid is
7-7. The Egyptians reportedly used wa 7-7. The radius r 0 of the circular the bottom of the bowl. Determin' quired if the depth of the liquid is
F
Fig. 7-5 hydraulic ram.
I.D. nozzle at a velocity of 30 ft/sec and the jet at 17 ft/sec, as illustrated in Fig. exert on the plate?
• 7-6. A very simple device for pumping fluids is the ejector pump illustrated in Fig. 7-6. The fluid to be pumped is called the primary fluid, and momentum (instead of the force supplied by mechanical pumps) is supplied by the secondary fluid issuing through the nozzle. The momentum exchange takes place vía viscous effects, and the viscous dissipation term in the mechanical "f' 01
r
----
Area= A
:
--1--
17ft/sec
l
Grcvity
7-4
in a concrete wall as shown in Fig. 7-5. ft/sec and the cross-sectional area is 1 in. 2, force exerted on the pipe by the from the wall to the end of the pipe is
Fig. 7-6
energy balance cannot be neglected. However, the viscous surface forces can be neglected if the flow is turbulent and a satisfactory design of an ejector pump can be obtained with the momentum balance. Derive an expression for the pressure rise between points 1 and 2, and note the effect of reducing the nozzle area A 0 while maintaining the secondary flow rate Q0 constant. Assume that the secondary fluid is the same as the primary fluid. • 7-7. The Egyptians reportedly u sed water clocks similar to that illustrated in Fig. 7-7. The radius r 0 of the circular bowl is a function of z, the distance from the bottom of the bowl. Determine the functional dependence of r0 on z required if the depth of the liquid is to be a linear function of time.
Macroscopic Balances-lnertial Effects
278
Chap. 7
Problems
7-8. Ifthe valve on the pipe shown in Fi elbow will decrease suddenly. The 1 for very short times when the velo effects are negligible. Determine p Hint: Two scalar components of ti Ans: p*
=
p0 + pgL 1 L 2/(L 1 + L 2)
7-9. A standard method for determinin of measuring the pressure differenc Fig. 7-9. Derive an expression for
1"""""'""'-'-'-~"'"'~" Areo
= A1
'
Flg. 7-7
Fi
í
y
Lx
L1
and 'the pressure difference measu the relative merits of the momentu is much superior to the other for t
7-10. What is the resultant force at the C< the pipe in Fig. 7-10, because ofth velocity in the 2-in. l. D. section is 7 lbm, and the interior volume is : 7-11. The device illustrated in Fig. 7-11 mine the resultant horizontal forc leaving fluid streams. Given: PA PB
p•
1 1
-f---
gauge pressure at A gauge pressure at B gauge pressure at e L2 Flg. 7·8
=
15 psig 20 psig
=
o
=
The average velocity atA is 10 ft/s• Ans: Fx
=
-28 lb1
7-8. Ifthe valve on the pipe shown in Fig. 7-8 is opened rapidly, the pressure at the elbow will decrease suddenly. The problem is a complex one but can be solved for very short times when the velocity may be set equal to zero and viscous etfects are negligible. Determine p* for short times after the valve is opened. Hint : Two scalar components of the momentum equation must be used. Ans: p* = p0 + pgL 1 L 2 /(L 1 + L 2) t 7-9. A standard method for determining turbulent flow rates in conduits consists
of measuring the pressure ditference across a flow nozzle such as that shown in Fig. 7-9. Derive an expression for the volumetric flow Q in terms of A 1 , A2 ,
Areo = A1
A 2 =oreo
T
7-7
Fig. 7-9
and ·the pressure ditference measured by the manometer. Consider carefully the relative merits of the momentum and mechanical energy balances, for one is much superior to the other for this particular analysis.
L,
E--- -
L2
- --
7-1 O. What is the resultant force at the connection between the nozzle assemb1y and the pipe in Fig. 7-1 O, beca use of the water issuing from the jet? The average velocity in the 2-in. l. D. section is 1Oft/sec, the mass of the nozzle assembly is 7 lbm, and the interior volume is 38 in.3
- --
-.-¡
7-11. The device illustrated in Fig. 7-11 is used to mix two miscible fluids. Determine the resultant horizontal force on this device owing to the entering and Ieaving fluid streams. Given: PA = 70 lbm/ft 3 PB = 59 lbm/ft 3 gauge pressure at A = 15 psig gauge pressure at B = 20 psig gauge pressure at e = o The average velocity atA is 10 ft/sec; at B, it is 15 ft/sec. Ans: Fx = -28 lb1
280
Macroscopic Balances-lnertial Effecu
Chap. 7
Problems
L. Grovity
Flg. 7-10
rate is the same in both cases.) E funétion of O, and assume that the m over the surface of the exit as indi Fig. 7-13b for assumption (b). momentum flux (a) Ans: momentum flux (b)
=
1 ~l1 (1 -
2
8 0=1in. Flg. 7-11
7-12. Water is being pumped into a tank as shown in Fig. 7-12 ata rate of 8 ft 3/sec. lfthe diameter ofthe pipe is 6 in. and the inside diameter ofthe tank is 30 in., what is the force required to hold the tank in position for h equal 24 in .. excluding the gravitational force acting on the tank itself? 7-13. In solving macroscopic balance problems, we always attempt to place the suñace of the control vo1ume so that the streamlines are parallel and the momentum flux is easily determined. As an example of the error that may occur if the streamlines are not parallel, compute the momentum flux at the exit of a circular diverging section assuming (a) the stream1ines are straight lines converging at point O, and (b) the streamlines are parallel. (Note: the magnitude ofthe ve1ocity vector must be adjusted so that the volumetric flow
(b)
Fig.
<
lcrclScc,Dic Balances-lnertial Effects
Chap. 7
281
Problems
L.
t h(t)
-t
12in.
~~~--~=---~_l Fig. 7-12 7-10
rate is the same in both cases.) Express the ratio of momentum fluxes as a function of O, and assume that the magnitude of the velocity vector is constant over the surface of the exit as indicated in Fig. 7-13a for assumption (a) and Fig. 7-13b for assumption (b). Ans:
momentum flux (a) momentum flux (b)
-l¡ - J
1 cos2 O In (secO) 2 (1 - cos 0) 2
= -
0=2in.
7-11
shown in Fig. 7-12 ata rate of 8 ft 3/sec. the inside diameter of the tank is 30 in., the tank in position for h equal 24 in., on the tank itself? we always attempt to place the the streamlines are parallel and the As an example of the error that may compute the momentum flux at the assuming (a) the streamlines are straight the streamlines are parallel. (Note: the be adjusted so that the volumetric flow
"o"~-:_--1!__----.................... ,,
Fig. 7-ll
282
Macroscopic Balances-lnertial Effects
Chap. 7
• 7-14. Fluid distributing systems often consist of an array of perforated pipes providing discrete distribution at any number of points. A single perforation in a pipe is shown in Fig. 7-14 . lf the diameter of the hole is smaller than the pipe-wall thickness, the jet is very nearly perpendicular to the main stream. Assuming that the volumetric ftow rate through the perforation can be expressed ast Q 0 = A 0 v 2(p 1 - p 0 ) / p, where Q 0 is the volumetric flow rate through the hole, A 0 is the area of the hole, and p 0 is the ambient pressure, use the momentum and mechanical energy balances to obtain an expression for the pressure ditference, p 1 - p 2 • Neglect viscous etfects and explain wh y you obtain two ditferent answers .
Problems
Air at 8 psig
8ft _::::;>
Air at 11 psig
---.,.:e.-
F
1 in.
Fig. 7-14
v0 = 25 ftlsec 7-15 . Calculate the discharge rate from the upper reservoir to the lower reservoir in Fig. 7-15. Take the fluid to be water and neglect viscous etfects.
Ans: Q
=
-.-t_______,
0.18 ft 3 /sec.
7-16. Does a converging nozzle on a garden hose place the hose in tension or compression at the junction between the hose and the nozzle? Don 't guess-analyze. • 7-17. The inclined flat plate in Fig. 7-17 is moving toward the plane jet at 10 ft/sec, and the jet velocity (relative toa fixed reference frame) is 25 ft¡sec . lf the jet is 1 in. thick, determine the force per unit width exerted on the plate by the stream of water.
Ans: fx
=
F
-106 lb11ft
• 7-18. Two miscible turbulent streams of densities p1 and p2 are flowing in a wide, rectangular duct separated by a thin plate. Neglecting viscous effects, use the momentum balance to calculate the pressure change in the mixing region in terms of (v.)¡, (v. )2 , h1 , and h2 . Assume the pressure is uniform across the channel and the velocity profiles of the unmixed and completely mixed streams are flat.
t This is an approximate version of a formula to
be derived in Chap. 8.
Fl1. 7-18. Mixing
Balances-lnertial Effects
Chap. 7
of an array of perforated pipes number of points. A single perforation diameter of the hole is smaller than the nearly perpendicular to the main stream. rate through the perforation can be p, where Q0 is the volumetric flow rate hole, and p0 is the ambient pressure, use rgy balances to obtain an expression for eglect viscous effects and explain why you
283
Problems
Air ot 8 psig
8ft __::::>
Nozzle diometer
Air ot 11 psig
= 2in
Fig. 7-15
1 in.
v0 = 25 ft/sec the upper reservoir to the lower reservoir and neglect viscous effects.
--~-------------
hose place the hose in tension or compresand the nozzle? Don't guess-analyze. is moving toward the plane jet at 10 ft/sec, reference frame) is 25 n¡sec. If the jet is it width exerted on the plate by the stream Fig. 7-17
f densities p1 and p2 are ftowing in a wide, in plate. Neglecting viscous effects, use the he pressure change in the mixing region in ~ssume the pressure is uniform across the ~ of the unmixed and completely mixed
~rmula
to be derived in Chap. 8.
Flg. 7-18. Mixing of two turbulent streams.
284
Macroscopic Balances-lnertial Effects
Chap. 7
7-19. A laminar jet ofwater issues from a horizontal nozzle as shown in Fig. 7-19. The velocity profile of the jet as it emerges from the nozzle is parabolic, but the profile becomes flat sorne distance from the nozzle owing to viscous effects. Neglect the effect of gravity and the ambient air and derive an expression for the final area A 1 . Ans: A 1/A 0
Macroscopic j Visco
=! Nozzle oreo, A 0
--------Fluid jet
orea, A, Fig. 7-19. The laminar jet.
7-20. Use the mechanical energy balance (rather than Bemoulli's equation) to show that v1 = v2 = v3 for the flow shown in Fig. 7.8-4. 7-21. Solve the problem of the moving water scoop illustrated in Fig. 7.9-3 by using Eq. 7.9-8 instead of Eq. 7.2-10. 7-22. Derive Eq. 7.10-6 by integrating the x-direction stress equation of motion over the y,z-surface. Neglect viscous effects.
In Chap. 6, we presented a qualit showed that the time-averaged stress as the instantaneous stress equations equations could not be used to det because the dependence of the turb velocity is unknown. Empirical exp length equation, may be used to detc cannot be applied to any arbitrary gc In solving turbulent flow problem: are important, we will always require of this chapter is the formulation e experimental data and the applicatior viscous effects must be considered.
*8.1
Friction Factors-Deflnit
In examining the momentum at 7.2-10 and 7.3-27), we see that knowJ,
Jten> dA
a
""·(t)
2
Chap. 7
horizontal nozzle as shown in Fig. 7-19. from the nozzle is parabolic, but from the nozzle owing to viscous and the ambient air and derive an
Macroscopic Balances: Viscous Effects
8
than Bernoulli's equation) to show in Fig. 7.8-4. scoop illustrated in Fig. 7.9-3 by x-direction stress equation of motion effects.
In Chap. 6, we presented a qualitative description of turbulent ftow and showed that the time-averaged stress equations of motion too k the same forro as the instantaneous stress equations of motion. However, the time-averaged equations could not be used to determine time-averaged velocity profiles, because the dependence of the turbulent stress i
*8.1
Friction Factors-Definition
In examining the momentum and mechanical energy balances (Eqs. 7.2-10 and 7.3-27), we see that knowledge of the integrals
f
t
and
f
<1>
?'".(t)
285
dV
286
Macroscopic Balances-Viscous Effects
Chap. 8
is required if we are to obtain satisfactory solutions to these equations. In the previous chapter, viscous effects were neglected and these integrals were easily evaluated. While reasonable results were obtained for a variety of problems, the flows investigated were restricted to cases for which we knew a priori that inertial effects predominated. In accounting for viscous effects, we must relate the surface stress integral and the dissipation integral to experimentally determined pressure drops and flow rates. A flowing fluid will exert a force on the solid surfaces that it contacts. In this chapter we are primarily interested in the component of this force in the direction of the mean flow, and it will be helpful to define a unit vector A which points in the direction of the mean flow (8.1-1) Before defining the drag or friction force that the fluid exerts on the solid, Eq. 7.2-10 will be rewritten in a form which is especially convenient for interpreting experimental data. Substituting Eqs. 7.3-7, 7.8-15, and 7.8-20 into Eq. 7.2-10, and using the divergence theorem for a scalar, we get
~~ Jpv dV + Jpv(v -- w) • n dA A,(t)
-;'"o(t)
J{-n[(p -
=
(8.1-2)
p0 )
+
pcp]
Sec. 8.1
Friction Factors-Definition
For convenience, the total drag force force
Frorm =A
A,+A,
Ftrlctlon
!=
~~ Jpv dV +Jpv(v- w) · n dA= J{-n[(p- p + pcp] + n · T} dA 0)
+
A,(t)
J{
-n[(p - p0)
+ pe/>]} dA +
A,+A,(t)
J
(8.1-3)
n·
T
dA
J
= ). ·
n[(p - p 0)
A,+A,(t)
+ pcp] dA -
A
.J
where
A* KE*
=
a characteristic a
= a characteristic 1(
There are many important processe! conduits of constant cross-section2 teristic kinetic energy per unit vol given by KE*
A*=
A,+A,(t)
We now define the drag force F D as the force which the fluid exerts on the so/id in the A-direction.t
Fn
·f
The drag force contains the negati~ beca use it is defined as a force the ftu Eq. 8.1-3 represent the force whicl Because ofthe way in which the drag n in Eqs. 8.1-4 through 8.1-6 is direc The separation of the drag force made because the former depends rr proportional to pu~, while the latter proportional to ¡.tu 0 • The word "forn of the drag force is greatly influencc while the word "friction" indica tes th primarily on the area of the so lid s The drag force for flow in clos correlated in terms of a dimensionle
+ n · T} dA
provided the density is constant. It should be kept in mind that for turbulent flow, T represents i + i(t>, and p represents p. Splitting the area integral on the right-hand side of Eq. 8.1-2 into the area of entrances and exits A.(t) and the area of solid surfaces (both fixed and moving) A, + A,(t), we get
A,(t)
= -A
A,+
.llfo(t)
-;'"G(t)
·J
t'
Because A will be orthogonal to th so lid surfaces, F n is given by
Fn= n • T dA
(8.1-4)
A,+A,(t)
t In the design of airfoils, an engineer is naturally interested in a lift force, the force the fluid exerts on the solid in a direction perpendicular to A..
for conduits having a constant eros only for mathematically smooth co for real conduits.
Balances-Viscous Effects
Chap. 8
solutions to these equations. In neglected and these integrals were were obtained for a variety of r.•otru·t .. n to cases for which we knew In accounting for viscous effects, and the dissipation integral to and flow rates. the solid surfaces that it contacts. In in the component of this force in the be helpful to define a unit vector A flow (8.1-1) that the fluid exerts on the solid, which is especially convenient for Eqs. 7.3-7, 7.8-15, and 7.8-20 theorem for a scalar, we get w) • n dA
(8.1-2)
pcp]
Sec. 8.1
Friction Factors-Definition
For convenience, the total drag force is split into a form force and a friction force Frorm
=A
J ·f
n[(p- p0 )
+ pcp] dA
Frrlction =
-A
(8.1-6)
n • 't' dA
A,+A,(t)
The drag force contains the negative of the Iast two terms in Eq. 8.1-3, beca use it is defi.ned as a force the fluid exerts on the so/id, while the terms in Eq. 8.1-3 represent the force which the surroundings exert on the fluid. Because ofthe way in which the drag force has been defi.ned, the normal vector n in Eqs. 8.1-4 through 8.1-6 is directed from the fluid into the solid. The separation of the drag force into a form force and a friction force is made because the former depends mainly on inertial effects and is roughly proportional to pu~, while the latter depends on viscous effects and is roughly proportional to ¡.¿u0 • The word "form" results from the fact that this portion of the drag force is greatly influenced by the geometry of the solid surface, while the word "friction" indica tes that this portion of the drag force depends primarily on the area of the solid surface. The drag force for flow in closed conduits is generally represented and correlated in terms of a dimensionless friction factor f defi.ned by
+ n • 't'} dA
f= ___!j¿_
(8.1-7)
A*KE*
be kept in mind that for turbulent p. Splitting the area integral on area of entrances and exits A.(t) and moving) A. + A.(t), we get {-n[(p- p0)
+ pcp] + n · 't'} dA
(t)
dA+
f
(8.1-3) D•'t'dA
(8.1-5)
A,+A,(t)
where A* KE*
=
a characteristic area
= a characteristic kinetic energy per unit volume
There are many important processes dealing with the flow of fluids through conduits of constant cross-sectional area. For such conduits the characteristic kinetic energy per unit volume and characteristic area are generally given by (8.1-8) KE* = tp(z\) 2 A*
= t wetted surface
(8.1-9)
A,+A,(t)
force which the fluid exerts on the
f
] dA - A • n • 't' dA
Fn
(8.1-4)
A,+A,(t)
interested in a lift force, the force pc:nlncuJtai to A.
Because A will be orthogonal to the outwardly directed unit normal at the so lid surfaces, Fn is given by
= -A
-J n •
't'
dA
(8.1-10)
A,
for conduits having a constant cross-sectiona1 area. Equation 8.1-1 O is valid only for mathematically smooth conduits and thus is only an approximation for real conduits.
288
Macroscopic Balances-Viscous Effects
Chap. 8
Frl~ion
Sec. 8.1
Factors-Definltion
Assuming that the stress at the wall · the numerator of Eq. 8.1-12 as Z=O
-A
Z=L
.f
n • 'T dA= -A;
A,
The wall shear stress
---,\
D
7'0
was previous
and we may use this definition to ex)
!= This result simply indicates that the dimensionless wall shear stress. The ·momentum balance Fig. 8.1-1. Flow in a circular tube.
Asan example, we will apply these ideas to flow in the circular tu be shown in Fig. 8.1-1. The characteristic area is given by (8.1-11)
A*= !rrDL
and substitution of Eqs. 8.1-8, 8.1-1 O, and 8.1-11 into Eq. 8.1-7 allows us to express the friction factor as -A
F
f = __ D_ = _ A*KE*
J
f{
-n[(p- Po)+ p
f
(!rrDL)(tp(v.i)
1}
(8.1 13a)
= nr = n 2 = n 8 = O at the tube wall
(8.1-13b)
n3 = n. =O
(8.1-13c)
A1 = Ar =O
(8.1-14a)
A2 = Ae =O
(8.1-14b)
A3 = A• = 1
(8.1-14c)
n[(p
Here, we have assumed that the v identica1, so that the momentum fl1 Forming the scalar product with A a Eq. 8.1-18 to FD =A
·f
""'·
and A has on1y one nonzero component.
+ n ·'
+ ""'·
(8.1-12)
For a mathematically smooth circular tube, there is only one nonzero component of the outwardly directed normal at the so lid surface, n1
O=
.4-¡
n • 'T dA
__:A~•!....,-_ _
It will be helpful at this point to a shown in Fig. 8.1-1. A rather care presented here, and future discussiom that these points are understood. F 8.1-1, the momentum balance reduc~
-n[(p- Po)
Dividing by A*KE*, we obtain th factor:
ic Balances-Viscous Effects
Chap. 8
Sec. 8.1
289
Frl~lon Factors-Definition
Assuming that the stress at the wall is independent of () and z, we may write the numerator of Eq. 8.1-12 as
-A
Z=L
.f
(8.1-15)
n · 't" dA= -Ai(n;T;i)TTDL= -TrzTTDL
A,
The wall shear stress
T0
was previously defined in Sec. 6.5 as at
r
=
(8.1-16)
r0
and we may use this definition to express the friction factor as
4r0
f =
(8.1-17)
tp(!\)2
This result simply indicates that the friction factor may be interpreted as a dimensionless wall shear stress. The momentum balance a circular tube.
to flow in the circular tu be shown given by (8.1-11)
It will be helpful at this point to apply the momentum balance to the flow shown in Fig. 8.1-1. A rather careful analysis of the stress terms will be presented here, and future discussions of the momentum balance will presume that these points are understood. For the control volume indicated in Fig. 8.1-1, the momentum balance reduces to
8.1-11 into Eq. 8.1-7 allows us to
O=
f
{-n[(p- p0)
+ ptf>] + D•'t"} dA
f-
+
(8.1-12)
n[(p- p0)
+ ptf>] dA+
J
n • 't"dA
(8.1-18)
A,
tube, there is only one nonzero at the solid surface, (8.1 13a)
Here, we have assumed that the velocity profiles at points 1 and 2 are identical, so that the momentum flux term in Eq. 8.1-3 is identically zero. Forming the scalar product with A and using the definition of Fn, we reduce Eq. 8.1-18 to
(8.1-13b) (8.1-13c)
Fn
=A
.J
-n[(p- p0)
+
ptf>] dA
+A
.J
n • 't" dA
(8.1-19)
A.
(8.1-14a) (8.1-14b) (8.1-14c)
Dividing by A* KE*, we obtain the following expression for the friction factor:
f =
i(
&'1
-
&'2) -+-
[A
J
n1 • 't" dA
Aentrance
J
+A
n 2 • 't" dA
.dexlt
J1A *KE* (8.1-20)
Macroscopic Balances-Viscous Effecu
290
Chap. 8
Friction Factors-Definition
and Eq. 8.1-20 may be written as
In obtaining this result we have used the definition
fJJ = ((p) - Po)
Sec. 8.1
+ p(cp)
(8.1-21)
f
D =
L ( &\
!-p(t\)2
and the area integral over the entrance and exit has been represented as two separate integrals. The stre~s terms are best treated in terms of index notation,
- !/.J -
1 A* Kl
If the structure of the turbulence does 1 stress terms will cancel and we are left
(8.1-22) where
Thus, the friction factor may also be in drop.
Remembering that the turbulent stress was given in Sec. 6.2 as -(t)
T;;
=
1
Dimensional analysis for the
1
-pV;V;
and the time-averaged viscous stress is given by f;;
Before examining the experimental of dimensional analysis to determine friction factor. Writing out Eq. 8.1-12
= !!:.(ov; + ov;) 2 OX;
OX;
we may write Eq. 8.1-22 as
A· (n ·
frictio~
~) = A;n;~(~:: + ~;) - pv;v~J
sr(-~~
(8.1-23)
oo
J=---
We now wish to evaluate this quantity at the entrance and exit where n has only one nonzero component n1 = nr =O } n2
=
n8
=
O
(8.1-24a) at the entrance and exit
n3 = n. = ±1
(t1TD1
Forming the dimensionless variables
Z=
(8.1-24b) (8.1-24c)
R=
Carrying out the summations indicated in Eq. 8.1-23, and making use of Eqs. 8.1-14 and 8.1-24, we get
~ (n · ~) = ± (~ oz av.
-~ - pv.v.
(8.1-25)
Ü= •
In obtaining this result, we have assumed that the velocity field is described by
Nne=
1\ •
Vr
=
Ve
v.
= O,
v;, v~, v;
-=1=-
1
-::j=.
)
O
allows us to express Eq. 8.1-29 as
O
The time-averaged velocity in the z-direction will be independent of z for all points downstream of the entrance region (i.e., in the region of onedimensional flow); thus,
ov. =o oz
(8.1-26)
!=
(~)(~)(Ll { o
This result indicates immediately that Reynolds number, but we must know
Balances-Viscous Efrects
Chap. 8
Sec. 8.1
291
Friction Factors-Definition
and Eq. 8.1-20 may be written as (8.1-21) exit has been represented as two
f = D(/JlJ L vr¡-
OJl)
vr2
1 (< -;-;) ( -;-;))7TD A*KE* pv.v. 1 - pv.v. 2
4
-
2
(8.1-27)
If the structure of the turbulence does not change from 1 to 2, the turbulent stress terms will cancel and we are left with the result
of index notation, (8.1-28)
(8.1-22)
Thus, the friction factor may also be interpreted as a dimensionless pressure drop.
given in Sec. 6.2 as
Dimensional analysis for the friction factor
Before examining the experimental values off, it will be wise to make use of dimensional analysis to determine what parameters will infiuence the friction factor. Writing out Eq. 8.1-12 gives
+ :;) -
pv;v;J
(8.1-23) (8.1-29)
the entrance and exit where n has (8.1-24a) entrance and exit
Forming the dimensionless variables
Z=_::
(8.1-24b)
D
(8.1-24c)
in Eq. 8.1-23, and making use of
R=.!:.
-,') -ov. - pv.v.
ü =!!L
D
(8.1-25)
•
the velocity field is described by
_ p(v.)D NRe---
# allows us to express Eq. 8.1-29 as will be independent of z for all (i.e., in the region of one(8.1-26)
2
1 ) JL/DJ "(- oܕ) f= (~)(!l.)(7T
L
NRe
o
o
oR
R d8 dZ
(8.1-30)
R=l/2
This result indica tes immediately that f depends u pon the ratio L/ D and the Reynolds number, but we must know upon what parameters oü.foR depends
292
Macroscopic Balances-Viscous Effects
Chap. 8
to complete the investigation. The continuity equation and the NavierStokes equations in dimensionless form are given by
V·U=O
au +u. vu = ae
(8.1-31)
-VfJJ
+
- 1- V' 2U
(8.1-32)
NRe
Sec. 8.2
Friction Factors-Experimental
lf the velocity profile is fully develope be independent of Z. lntegration wit yield the factor L/ D, which just cano entrance length for turbulent ftow is number and is on the order of 50 tub we may neglect the effect of the entra
The boundary conditions for this ftow may be expressed in both dimensional and dimensionless form as follows:
B. C. 1:
V=
0,
f=f(NRe)
*8.2
Friction Factors: Experi
(8.1-33)
U=O, B.C. 2:
V
=
V1(r,
U
=
U 1(R, (), e),
Z=0
(), t),
B.C. 3:
(8.1-34)
Z=O
z=L L
(8.1-35)
Z=D B.C. 4:
p = Pl = fJJ¡,
[JJ
pcp¡,
z= O
(8.1-36)
z=o
Although we cannot solve Eqs. 8.1-31 and 8.1-32 for U and fJJ, we can state that these two dependent variables are functions of the independent variables (R, (), Z, e), the parameters in the differential equations (NRe), and the parameters in the boundary conditions (L/ D). Thus, we can write
U=
u(R, (), Z, e, NRe• ~)
(8.1-37a)
=
[JJ(R, (), Z, e, N Re• ~)
(8.1-37b)
[JJ
Ifthe time-average ftow is steady, Ü will be independent ofthe dimensionless time e, and the gradient at the wall may be expressed as
oü. 1 = oR R=l / 2
function of
(e, Z, NRe• .!:) D
(8.1-38)
In Eq. 8.1-30, the () and Z dependence will be eliminated by integration, and the functional dependence of the friction factor is (8.1-39)
If the friction factor is measured u values of L/ D, we soon discover that represented by a single plot off vers performed the first experimental ~tu we immediately wonder, "Where dtd As is often the case with a mathematic the boundary conditions. In Eq. 8.1 zero at r = D/2, and while this seems true only for a mathematically smootl written
B.C. 1':
V=
O,
r
where the roughness function e(O, z) l that describes the roughness of the co magnitude of e is of the order of 0.001 "small effect," but for turbulent ftow i do not lead to small effects." This ex~ it is a classic example of a plausib erroneous conclusion. The early work by Darcy and other experimental study of the friction facto tubes. The rough tubes were obtaine< possible with sand of a definite grain si Nikuradse obtained a roughness ce encountered in commercially availab plotted as the friction factor versus th results are characterized by the relati•
1. H. P. G. Darcy; for an account, see 1 (New York: Dover Publications, Inc., 1963), 2. J. Nikuradse, VDI-Forschungsh. 361, A original paper is available as NACA Tech. M
Chap. 8
and the Navier(8.1-31) (8.1-32)
Sec. 8.2
293
Friction Factors-Experimental
lf the velocity profile is fully developed at Z = O, the velocity gradient will be independent of Z. Integration with respect to Z in Eq. 8.1-30 will then yield the factor L/ D, which just cancels the multiplying factor, D/ L. The entrance length for turbulent flow is nearly independent of the Reynolds number and is on the order of 50 tu be diameters. For large values of L/ D, we may neglect the effect of the entrance region and write
be expressed in both dimensional
F or large L/ D
*8.2
(8.1-40)
Friction Factors: Experimental
(8.1-33)
z=O Z=O Z=
(8.1-34)
L L
Z=-
(8.1-35)
D
z=O
(8.1-36)
8.1-32 for U and ~. we can state ~mct1011s of the independent variables equations (NRe), and the D). Thus, we can write (8.1-37a)
(8.1-37b) independent of the dimensionless be expressed as
(o. z, NRe• ~)
(8.1-38)
be eliminated by integration, and factor is (8.1-39)
If the f¡iction factor is measured using a variety of pipes all having large values of L/ D, we soon discover that the experimental values off cannot be represented by a single plot off versus N Re as Eq. 8.1-40 implies. Darcy 1 performed the first experimental studies leading to this conclusion, and we immediately wonder, "Where did our dimensional analysis go wrong ?" As is often the case with a mathematical analysis, the trouble líes with one of the boundary conditions. In Eq. 8.1-33 we specified that the velocity was zero at r = D/2, and while this seems like quite a reasonable statement, it is true only for a mathematically smooth tube. In actual fact, we should have written B.C. 1':
V=
0,
r
=
lD + e(O, z)
(8.2-1)
where the roughness function e(O, z) is sorne unknown function of O and z that describes the roughness of the conduit. For most commercial pipes the magnitude of e is of the order of 0.001 of the pipe diameter. Surely this is a "small effect," but for turbulent flow in pipes we will find that "small causes do not lead to small effects." This example should be remembered well, for it is a classic example of a plausible intuitive hypothesis leading to an erroneous conclusion. The early work by Darcy and others led Nikuradse 2 to carry out a detailed experimental study of the friction factor for smooth and artificially roughened tubes. The rough tubes were obtained by covering the surface as tightly as possible with sand of a definite grain size glued onto the wall. In this manner, Nikuradse obtained a roughness considerably more uniform than that encountered in· commercially available pipe. The experimental results are plotted as the friction factor versus the Reynolds number in Fig. 8.2-1. The results are characterized by the relative roughness parameter, ef D, where e l. H. P. G. Darcy; for an account, see H. Rouse and S. Ince, History of Hydraulics (New York: Dover Publications, Inc., 1963), p. 170. 2. J. Nikuradse, VDI-Forschungsh. 361, Ausgabe B, Band 4, 1933. A translation of the original paper is available as NACA Tech. Mem. 1292, 1950.
294
Macroscopic Balances-Viscous Effects
12xl0-~
1 9 8
4
1
•
1~ \
c/0=9 . 86xl0~4
r. '
o
= 1.98xlo- 3 • 3 . 97xlo- 3 6
o
•
=8.34xl0·3 = 1.63xlo- 2 3.34xlo- 2,
.
X
l}/0 =
Chap. 8
~-8~ JJ .-= 4
(commerciolly rough)_
Sec. 8.2
Friction Factors-Experimental
This equation is an indication that the to affect severely the parabolic veloc' pipes (say, efD > 0.10), we could ce laminar friction factor; however, this e If we use Eq. 8.1-17, in conjunction wi stress, we obtain 8J.L (v.)
To=--
Eq. 5.6-~ ~~ ~
3
1\
2.5
\
D
_... _. o;li
Eq. 8.2 /8 -. . loo...
2.0xlo-2
r'"'
!"" llfti!ll
.... fll
~rx:: .5,~~~~--~~~~--~~~~,~~~m-.~~~Krr~~-~~
indicating that the drag force is propo and the velocity. Under these conditio the friction force; hence,
1
FD
· 2 ~~~--+---~~~--+-~~~~~-.7 8*.2~-~9~-~i~~a~~~
1.0 x lo-2LW,..L---J-__-!---J-!--L--+--+-+~~~~:=;:=f:¿:L.:~ 4 6 8103 2 4 6 8 104 2 4 6 8 105 2 4 6 8 106 2
=
Frriction•
Critica! region
NflA>=u0 0/v
Fig. 8.2-1. Friction factor versus Reynolds number for sand-roughened tu bes.
is the height of the sand grain. Because all the tu bes were roughened in the same manner-i.e., the sand grains were placed as close together as possiblethe single parameter ef D was sufficient to characterize the roughness. The student should give sorne serious thought to these results, for they are a clear indication that we can easily perform an apparently rigorous analysis, which in fact neglects important effects. lt is quite common in engineering analysis to treat surfaces as mathematically smooth and impose boundary conditions of the type given by boundary condition l. Generally, such an approach is satisfactory (as it is for laminar flow); however, for turbulent flow in tubes, a small effect such as wall roughness can lead to a striking effect in the pressure drop-flow rate relationship. In the absence of experimental studies, the average investigator is not likely to formulate the boundary condition as indicated by boundary condition 1'; thus, we must be constantly on the alert for flaws and limitations in a mathematical analysis. Several aspects of this friction factor plot must be discussed before we continue with the results for commercial pipes. Laminar region
We note first that all the data in the laminar flow region fall on a single line, the equation for which was derived in Chap. 5,
f =
64
N Re
(8.2-2)
Pipe roughness is not a factor number, and the transition to turb critica! Reynolds number may depen the pipe, upstream conditions such as spurious disturbances such as buildin there is a lower bound for the criti however, recent studies 3 have shown t to Reynolds numbers of 2 X 104 by flow free of disturbances. In practic disturbance and the transition to tur region between NRe = 2100 and NRe · it is here that the transition is comple and in the critica! region the flow turbulent regimes. Transition and rough-pipe regions
For smooth tubes there is only o the two previously mentioned-i.e., flow existing for NRe > 4 X 103 • H the flow continues to change as the R1 the relative roughness is less than O. smooth pipe curve for a region in w 3. R. J. Leite, J. Fluid Mech., 1959, 5:81
ic Balances-Viscous Effects
Chap. 8
Sec. 8.2
295
Friction Factors-Experimental
This equation is an indication that the values of ef D were never large enough to affect severely the parabolic velocity profile. By using extremely rough pipes (say, ef D > 0.10), we could certainly find an effect of ef D on the laminar friction factor; however, this case is not of great practica! importance. If we use Eq. 8.1-17, in conjunction with Eq. 8.2-2, to express the wall shear stress, we obtain 8JJ. (v.) To= - - D
for laminar fiow
(8.2-3)
indicating that the drag force is proportional to the product of the viscosity and the velocity. Under these conditions, the drag force results entirely from the friction force; hence, FD 2
=
Ffriction•
for laminar flow
(8.2-4)
Critica! region
=u0 0/v number for sand-roughened
all the tubes were roughened in the placed as close together as possibleto characterize the roughness. thought to these results, for they perform an apparently rigorous effects. lt is quite common in mathematically smooth and impose by boundary condition l. Generally, is for laminar flow); however, for as wall roughness can lead to a rate relationship. In the absence of is not likely to formulate the condition 1'; thus, we must be ns in a mathematical analysis. must be discussed before we
laminar flow region fall on a single in Chap. 5, 64 (8.2-2)
Pipe roughness is not a factor in determining the critica! Reynolds number, and the transition to turbulent flow starts at NRe = 2100. The critica! Reynolds number may depend strongly on the inlet conditions to the pipe, upstream conditions such as valves and bends, and the presence of spurious disturbances such as building vibrations. lt is well established that there is a lower bound for the critica! Reynolds number of about 2100; however, recent studies 3 have shown that laminar flow can be máintained up to Reynolds numbers of 2 x 104 by taking extreme care to ~eep the inlet flow free of disturbances. In practica! cases, there are numerous sources of disturbance and the transition to turbulent flow starts at Nfi.e = 2100. The region between NRe = 2100 and NRe = 4000 is called the critica! region and it is here that the transition is completed. The transition is ,not a sharp one, and in the critica! region the flow alternates between the laminar and turbulent regimes. Transition and rough-pipe regions
For smooth tubes there is only one more region of flow in addition to the two previously mentioned-i.e., the region of fully developed turbulent flow existing for NRe > 4 x 103 . However, for rough pipes the nature of the flow continues to change as the Reynolds number is increased. Provided the relative roughness is less than 0.01, the friction factor curve follows the smooth pipe curve for a region in which the rough pipe could be considered 3. R. J. Leite, J. Fluid Mech., 1959, 5:81.
296
Macroscopic Balances-Viscous Effects
Chap. 8
hydraulica/ly smooth. As the Reynolds number is increased, each curve eventually departs from the smooth pipe curve, progresses through a transition region, and finally reaches a constant value depending only on the relative roughness. This latter region will be called the rough-pipe region offlow, and the zone between the smooth-pipe and rough-pipe regions will be called the transition region.
Sec. 8.2
Friction Factors-Experimental
numbers, the major source of resista turbulent eddies with the wall rough compared to the form drag. In the 11 factor is constant, and Eq. 8.1-17 i1 proportional to the density times the T0
Low Reynolds number
=
(~) p(v.) ~n 2
Comparing Eqs. 8.2-3 and 8.2-5, we s momentum transport exist in pipe flo the other of inertial effects. Also, it i. the formula developed in Chap. 7 fo1 which indicated that force exert {on the pla
where v and A are the velocity and parison,. we might interpret Eq. 8.2force per unit exerted on the { protrusions
There is obviously sorne similarity be Commercial pipes
Flg. 8.2-2. Qualitative deséription of wall effects in turbulent ftow.
We can explain the existence of the transition and rough-pipe regions only in a qualitative manner; however, the explanation will help to explain the nature of turbulent pipe flow and should be valuable for this reason alone. At low Reynolds numbers the laminar sublayer is generally thicker than the average roughness e, and the wall shear stress To consists primarily of viscous stresses. As the Reyno!ds number is increased, the energy of the turbulent eddies increases, and they penetrate more closely to the wall. This process causes the thickness of the laminar sublayer to decrease, and the irregularities of the tube wall begin to protrude through the laminar sublayer, as Fig. 8.2-2 illustrates. When the turbulent eddies come into contact with these protrusions, the interaction of the fluid and solid changes. At high Reynolds
Pipes and tubes used in engineer smooth nor be considered rough in roughened pipes. The roughness in < it subject to direct measurement. H1 to any given pipe if the friction fac Moody 4 has made an extensive inve~ cial pipes, the results of which appe< widely used in engineering design Values of the relative roughness are pipe. When using these charts, an engi of e/ D are only approximate and th relative roughness for any given typ< probable variation injfor smooth t1 ± 10 per cent is to be expected for co 4. L. F. Moody, "Friction Factors for
Chap. 8
number is increased, each curve curve, progresses through a transivalue depending only on the relative the rough-pipe region offlow, and · regions will be called the
Sec. 8.2
297
Friction Factors-Experimental
numbers, the major source of resistance to flow is the interaction of the turbulent eddies with the wall roughness and friction drag becomes small compared to the form drag. In the rough-pipe region of flow, the friction factor is constant, and Eq. 8.1-17 indica tes that the wall shear stress is proportional to the density times the velocity squared. _
7
o-
(/_)
8
(
P v.
)2
In the rough-pipe region of turbulent flow
(8.2-5)
Comparing Eqs. 8.2-3 and 8.2-5, we see that two very distinct mechanisms of momentum transport exist in pipe flow-one a result of viscous effects and the other of inertial effects. Also, it is of interest to compare Eq. 8.2-5 with the formula developed in Chap. 7 for the force exerted by a jet on a plate, which indicated that force exerted} {on the plate
=
v2 A
(8.2-6)
P
where v and A are the velocity and area of the jet, respectively. For comparison, we might interpret Eq. 8.2-5 as force per unit area} exerted .on the {protrus10ns
(f)
= -
p(v.)
2
(8.2-7)
8
There is obviously sorne similarity between the two flows. Commercial pipes
in
and rough-pipe regions only will help to explain the be valuable for this reason alone. is generally thicker than the -r0 consists primarily of viscous the energy of the turbulent closely to the wall. This process to decrease, and the irregularities the laminar sublayer, as Fig. come into contact with these solid changes. At high Reynolds
l'lJJiamtuu•u
Pipes and tubes used in engineering practice can neither be regarded as smooth nor be considered rough in the same sense as Nikuradse's sandroughened pipes. The roughness in commercial pipes is not uniform, nor is it subject to direct measurement. However, a value of ef D can be assigned to any given pipe if the friction factor is known in the rough-pipe region. Moody 4 has made an extensive investigation of friction factors for commercial pipes, the results of which appear in Fig. 8.2-3. This particular chart is widely used in engineering design and is often called the Moody chart. Values of the relative roughness are given in Fig. 8.2-4 for various types of pipe. When using these charts, an engineer must keep in mind that the values of ef D are only approximate and there may be significant variations in the relative roughness for ány given type of pipe. Moody has indicated that the probable variation inJfor smooth tubing is ±5 per cent, anda variation of ± 1Oper cent is to be expected for commercial steel pipe. Corrosion may also 4. L. F. Moody, "Friction Factors for Pipe Flow," Trans. ASME, 1944, 66:671.
298
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.2
Friction Factors-Experimental Pipe diamete
04 06 0 .2 0.3 0.5· 0.81
1\
Riveted steel
1/
\,)
1"'-111. 11 ro. Concrete
"
N
~
l"i
;::.
"
Woo d~ 1'\stave
·~
-;;
·¡:¡
~ ~ 8 •
a:
<:
.... Q)
.o
E ::>
e
en
'O
oe
>.
Q)
a::
o
...o .S "'...o t;
~
= o
]
,.;
,:.
•..
¡¡:
Pipe diomet1
Fig. 8.2-4. Relative roughr
J
JO¡:lO~
UO!¡:lDJ.:::I
cause large variations in ef D with tir analyzing pipelines which have been in time. There are two useful empirical form
5. C. F. Colebrook and C. M. White,"" Pipes with Age," J. /nst. Civ. Engrs. (London)
Balances-Viscous Effects
Chap. 8
Sec. 8.2
299
Friction Factors-Experimental Pipe dio meter in feet, O 5
0.1
0.4 0.6 0 .2 0.3 0.5 0.81
2
3456810
25 20
o.o 1a§gga!ffi~lnm~nmtEf1Ei±tlno.o7
0.04k: 0.03 ro. 0 .02
0 .06
[\
0 .05
40 60 100
300
Pipe diomeler in inches, O
Fig. 8.2-4. Relative roughness for commercial pipes.
cause large variations in ef D with time,5 and we must be cautious when analyzing pipelines which have been in operation for an extended period of time. There are two useful empirical formulas giving the friction factor in terms 5. C. F. Colebrook and C. M. White, "The Reduction of the Carrying Capacity of Pipes with Age," J. Inst. Civ. Engrs. (London), 1937, 7:99.
300
Macroscopic Balances-Viscous Effects
Chap. 8
of the Reynolds number for smooth tu bes. For Reynolds numbers less than 105 , Blasius6 gives the following equation
f =
0.316Nfi~ 4
For smooth tubes
(8.2-8)
Sec. 8.2
Reynolds number can then be determi1 assumption for f may be examined. If second trial must be made. In the follo type of calculation.
This equation is plotted in Fig. 8.2-1 and is in excellent agreement with Nikuradse's data for smooth tu bes provided N Re :::;:; 105. For higher Reynolds numbers, the Prandtl equation7 is useful:
JJ1 = 2.0 log (Jj N Re) -
0.8
For smooth tubes
(8.2-9)
This equation is also plotted in Fig. 8.2-1, and shows good agreement with the experimental data. The derivation of Eq. 8.2-9 is semitheoretical and should give satisfactory results for arbitrarily high Reynolds numbers. The Prandtl equation requires a trial-and-error solution to determine J; however, if the Blasius equation is used to obtain the first estímate, the final value can generally be determined with only one additional calculation. A useful equation for determining the friction factor in the transition and rough-pipe regions has been developed by Colebrook. 8 2.51 ) JJ = -2.0 log (e/DD + NRe,J] 1
For transition and rough-pipe regions
Friction Factors-Experimental
Determination of the flow rate for a pressure drop
Let us determine the volumetric fic 750-ft length of 4-in. diameter (nominal drop of 23.5lbcfin. 2 We shall assume d tional effects need not be considered. 1 D
=
L= 750ft
= 0.95 centi¡ p = 62.4 lbm/f
p.
t:J.p
= 23.5 lbr/in
.E_=
6. H. Blasius, "The Law of Similarity for Frictional Processes in Fluids," Forsch. Arb. lngr.-Wesen (Berlin, 1913), 131:361. 7. L. Prandtl, "The Mechanics of Viscous Fluids," in Aerodynamic Theory, F. W. Durand, ed. (Berlin: Springer-Verlag, 1935), 3:143. 8. C. F. Colebrook, "Turbulent Flow in Pipes, with Particular Reference to the Transition Region between the Smooth and Rough-Pipe Laws," J. lnst. Civ. Engrs. (London), 1938, 11:133.
0.0004
D
(8.2-10)
This empírica! relationship is, in fact, the basis for the curves shown in Fig. 8.2-3. lt requires a trial-and-error solution for f; however, such procedures are straightforward and this equation may prove useful if a digital computer is used for pipe-sizing calculations. Problems dealing with turbulent flow in pipes generally fall into two categories: determination of the pressure drop, given the flow rate, the physical properties of the fluid, and the geometry of the system (i.e., the length, diameter and relative roughness of the pipe); or determination of the flow rate, given the pressure drop, the physical properties of the fluid, and the geometry of the system. The first case is straightforward because the Reynolds number can be calculated and the friction factor determined immediately. The second type of problem requires that an initial guess be made for f, allowing us to calculate the average velocity by Eq. 8.1-28. The
4.03 in. (a
We may rearrange Eq. 8.1-28 to yield
and the Reynolds number becomes _ p(v,)D _ N Rep. For this example, we obtain 2
N
1 { [(2X23.5 lbr/in. ) Re - ,Jj (0.95 cen
--
x [ ( centipoise-ft se1
0.672 X 10-3 ]1>¡ or
4.
NRe=-
Now we need only find the value ofjJ equation and the relationship between The mínimum value off in this case
Balances-Viscous Effects
Chap. 8
For Reynolds numbers less than
(8.2-8)
Sec. 8.2
Reynolds number can then be determined, and the accuracy of the initial assumption for f may be examined. If the two values differ significantly, a second tria! must be made. In the following example, we consider this type type of calculation.
and is in excellent agreement with N Re ::;; 105. For higher Reynolds
For smooth tubes
(8.2-9)
2-1, and shows good agreement with of Eq. 8.2-9 is semitheoretical and "ly high Reynolds numbers. The solution to determine f; however, n the first estímate, the final value one additional calculation. friction factor in the transition and by Colebrook.8 2.51 )
NReJJ
For transition and rough-pipe regions
301
Frlctlon Factors-Experimental
Determination of the flow rate for a given pressure drop Let us determine the volumetric flow rate of water at 75°F through a 750-ft length of 4-in. diameter (nominal) commercial steel pipe for a pressure drop of 23.5lbr/in. 2 We shall assume the pipe is horizontal so that gravitational effects need not be considered. lt is given that
D = 4.03 in. (actual diameter) L= 750ft p
= 0.95 centipoise
p = 62.4 lbm/ft 3 l:lp = 23.5 lbrfin. 2 !_
D
(8.2-10)
= 0.0004
We may rearrange Eq. 8.1-28 to yield the basis for the curves shown in solution for f; however, such promay prove useful if a digital
1 (21:1p D) Jj ---¡;L
112
(vz) =
and the Reynolds number becomes in pipes generally fall into two drop, given the flow rate, the the geometry of the system (i.e., of the pipe); or determination the physical properties of the fluid, rst case is straightforward because and the friction factor deterproblem requires that an initial guess averagevelocity by Eq. 8.1-28. The Frictional Processes in Fluids," Forsch. Fluids," in Aerodynamic Theory, F. W.
N
- p(v.) D - _!._ (21:lpp Da)I/2 Re-
p
-Ji
p2L
For this example, we obtain 3
N Re=~ { [(2X23.5 lbr/in.~62.4lbm/ft X4.03 in.)
-J¡
3 ]
(0.95 centipoise) 2(750 ft)
(__.!!__) J}
2
x [ ( centipoise-ft sec ) (32.2 lbmft) 0.672 X 10-3 lbm lbr sec2 12 in.
1 2 '
or
_ 4.13 N Re-
X
Ji
104
43.
with Particular Reference to the TransiLaws," J. Jnst. Civ. Engrs. (London),
Now we need only find the value ofjfor ejD = 0.0004 that will satisfy this equation and the relationship between f and N Re given in the Moody chart. The mínimum value off in this case is 0.0 16, which yields a value for the
Macroscopic Balances-Viscous Effects
302
Chap. 8
Reynolds number of N
- 4.13 X 104 - 3 26 105 .Jo.016 - . x
Re -
Examination of the friction factor chart for s/ D 3.26 x 105 gives != 0.0178
=
0.0004 and N Re
=
Sec. 8.2
Friction Factors-Experimental
pipe sizes is available in handbooks, 9 of facing page) will be cited here for a The variation in inside diameter is s designing any real system. Throughout dimensions of any system are assumed t~ Flow in closed conduits of noncircul
which in turn yields the second approximation for the Reynolds number, 4.13 X 104 = 3.08 X 105 .Jo.0178 Returning to the chart, we see that a Reynolds number of 3.08 x 105 gives a friction factor of 0.0178 for this particular relative roughness. Having established the Reynolds number, we may now determine the volumetric flow rate, Q. NRe
=
2
(7T~ )(v.) = (NR~:D/l)
Q=
= [(3.08 x 105)(3.14)(4.03 in.)(0.95 centipoise)J (4)(62.4Ihm/fe) 3
X
=
0.672 X 10- lbm ) ( ft )] [( centipoise-ft sec 12 in.
0.83 fejsec
Nominal pipe diameters
Closed conduits of noncircular although they have received much les laminar flow, each noncircular conduit problem, for the result from one case e and Katz11 list severa! solutions of t dimensional laminar flow in noncircula Bateman12 give a more thorough discu severa! solutions. Since an engineer i. pressure drop-flow rate relationship, t suggested ·by Gaydon and NuttaP 3 is of odd-shaped conduits. The method re effort to determine the pressure drop-f per cent. Turbulent flow has been studied by rectangular and circular notched condUJ in Fig. 8.2-5. For each conduit, the fr·
J=At
It is unfortunate but true that the nominal pipe diameter is not the actual pipe diameter; we must take this into account when specifying pipe diameters. In the previous example, the actual diameter of a 4-in. commercial steel pipe was 4.03 in. The difference between the nominal and actual diameters depends upon the diameter and the so-called schedule number of the pipe. The latter is simply a measure of the wall thickness and is therefore an indication of the pressure that the pipe can safely withstand. Information on Nominal Pipe Size, in.
Outside Diameter, in.
6
6.62
Schedule No.
SS lOS 40ST,40S
soxs, sos 120 160
XX
Inside Diameter, in. 6.41 6.36 6.06 5.76 5.50 5.19 4.90
where A* and .KE* are given by Eqs number is given by 4 NRe=9. Chemical Engineers' Handbook, 4th ed. lnc., 1963), Sec. 6. 10. L. S. Marks, ed. Mechanical Engineers Book Company, Inc., 1958). 11. J. G. Knudsen and D. L. Katz, Flu1 McGraw-Hill Book Company, Inc., 1958), Cl 12. H. L. Dryden, F. D. Murnaghan, a Dover Publications, Inc., 1956), Chap. 2. 13. F. A. Gaydon and H. Nuttal, "Viscou Cross Sections," Trans. ASME Ser. E 81, 19 14. L. Schiller, Z. Angew. Math. Mech., 1 15. J. Nikuradse, Ingr.-Arch., 1930, 1:3 16. L. Prandtl, Proc. lntern. Congr. Appl.
Sec. 8.2
ef D =
0.0004 and N Re =
303
Friction Factors-Experimental
pipe sizes is available in handbooks, 9 - 10 and only a single example (see foot of facing page) will be cited here for a 6-in. diameter steel pipe. The variation in inside diameter is significant and must be considered in designing any real system. Throughout the remainder of this text the nominal dimensions of any system are assumed to be identical to the actual dimensions. Flow in closed conduits of noncircular cross section
v,. .... ~.UV'U 4
l0
= 3.08
for the Reynolds number, X
105
Reynolds number of 3.08 x 105 gives particular relative roughness. Having may now determine the volumetric
Closed conduits of noncircular cross section are occasionally used, although they have received much Iess attention than circular tubes. For laminar flow, each noncircular conduit must be Iooked upon as a separate problem, for the result from one case cannot be extended to others. Knudsen and Katz11 Iist severa! solutions of the Navier-Stokes equations for onedimensional laminar flow in noncircular conduits. Dryden, Murnaghan, and Bateman12 give a more thorough discussion of these problems and present severa! solutions. Since an engineer is generally concerned with only the pressure drop-flow rate relationship, the approximate method of solution suggested by Gaydon and NuttaP 3 is of great value in obtaining solutions for odd-shaped conduits. The method requires only a small computational effort to determine the pressure drop-flow rate relationship to within a few per cent. Turbulent flow has been studied by severa! investigators14- 16 for triangular, rectangular and circular notched conduits; the experimental results are shown in Fig. 8.2-5. For each conduit, the friction factor is defined by
FD f= A*KE* nominal pipe diameter is not the actual specifying pipe diameters. ,.,uu....,... of a 4-in. commercial steel pipe nominal and actual diameters so-called schedule number of the pipe. wall thickness and is therefore an indican safely withstand. Information on l""~·v .. •cu when
Schedule No.
lnside Diameter, in.
5S lOS 40ST,40S 80XS, 80S 120
6.41 6.36
160 XX
6.06 5.76 5.50 5.19 4.90
(8.2-11)
where A* and KE* are given by Eqs. 8.1-8 and 8.1-9, and the Reynolds number is given by _ 4p(v,)Rh N Re-
(8.2-12)
¡;, 9. Chemical Engineers' Handbook, 4th ed. (New York: McGraw-Hill Book Company, lnc., 1963), Sec. 6. 10. L. S. Marks, ed. Mechanical Engineers Handbook, 6th ed. (New York: McGraw-Hill Book Company, Inc., 1958). 11. J. G. Knudsen and D. L. Katz, Fluid Dynamics and Heat Transfer (New York: McGraw-Hill Book Company, lnc., 1958), Chap. 4. 12. H. L. Dryden, F. D. Murnaghan, and H. Bateman, Hydrodynamics (New York: Dover Publications, lnc., 1956), Chap. 2. 13. F. A. Gaydon and H. Nuttal, "Viscous Flow Through Tubes of Multiply Connected Cross Sections," Trans. ASME Ser. E 81, 1959, 4:573. 14. L. Schiller, Z. Angew. Math. Mech., 1923, 3:2. 15. J. Nikuradse, Ingr.-Arch., 1930, 1 :306. 16. L. Prandtl, Proc. Jntern. Congr. Appl. Mech., Zürich, 1927.
304
Macroscopic Balances-Viscous Effects
'
0. 1
""
O. 1 0. 08
["-.
O. 06
1"" O. 01
O. 01 O. 008
r
~
lb..
,_'
o N1kuradse . Re f. 15
O. 004
O. 001
'<
.....
•
Schl11er , Ref . 14 1
4.0
_)---¡
1
2.0
Eq 8. 1-8
~ 'So ~o· ·o
~ 1--:-:::-.,_ equ 11 atera¡ J \
Co
./'
-o-_ ~-1f
kC!i~ ~~!~
""" ~"~''''"'•'lJt~
O. 006
10.0 8.0 6.0
J 6-38 Eq . 5.
"k~ " "'\ ~
Chap. 8
1
oy
1 r.... ~!r le._1
-o-:. :!!_tched
--.,__In.
"'
r-<
1"--
~
! f
~ r-<
1--
"
t'-....: ~ ~"""-
~¡...
1.0 0.8 o. 6
r--
o. 4
o. 2
o.1
10
2
4
6 81Q2
2
4
6
11103
2
Fig. 8.2-6. Drag coefficien
1 1
1
The characteristic area A* for imme projected area in the direction of fio Flg. 8.2-5. Friction factors for noncircular conduits.
A*-
where Rh is the hydraulic radius. The results illustrated in Fig. 8.2-5 indicate excellent agreement with the Blasius equation; in addition, the values in the laminar fl.ow region fall on the line given by
TTD2
4 '
A*= DL,
The characteristic kinetic energy is ta velocity far removed from the imme and 8.2-15, the drag force acting on
(8.2-13)
FD=eD(
These results must not be construed as verification of the friction factor chart for all shapes of closed conduits, for it is well known that laminar ftows may deviate significantly from Eq. 8.2-13. However, we may conclude that if the shape of the conduit is not far removed from circular the friction factor chart can be used to obtain :.atisfactory results. Drag coefficients for spheres and cylinders
The friction factor for solid bodies immersed in a flowing fluid is traditionally referred to as a dimensionless drag coefficient and defined by
e D-
__fJ¿_
A*KE*
(8.2-14)
and the drag force acting on a cylindl
FD= eD
Experimental values of eD for spher Dimensional analysis would again 1 function of Reynolds number and reJa has not been studied extensively bec with reducing the drag on immersed 1 of accomplishing this end.t
t This is not always the case, for somet drag. This phenomenon, along with the c detail in Chap. 11.
10.0 8.0 6.0
Chap. 8
4.0 2.0 Co
1.0 0 .8 0.6
"'
~
' ""'
~"
....
Circular cylinder
"
0.4
Sphere
_....,
t---
0 .2
o.1
10
\.11 2
4
6 81Q2
2
4
6 8¡Q3
2
4
6 8
104
2
4
6 8
105
2
4
6 8¡Q6
NRe
Fig. 8.2-6. Drag coefficients for spheres and cylindl:rs. 6
8
The characteristic area A* for immersed bodies is generally taken as the projected area in the direction of ftow; thus, for noncircular conduits.
A* =
results illustrated in Fig. 8.2-5 indicate · ; in addition, the values in the 64
(8.2-13)
verification of the friction factor chart is well known that laminar ftows may However, we may conclude that if the from circular the friction factor chart
2
D
for a sphere
(8.2-15a)
A* = DL,
for a cylinder
(8.2-15b)
7T
4 '
The characteristic kinetic energy is taken to be !pu;,, where U 00 is the fluid velocity far removed from the immersed body. On the basis of Eqs. 8.2-14 and 8.2-15, the drag force acting on a sphere is (8.2-16) and the drag force acting on a cylinder of length L and diameter D is (8.2-17)
immersed in a flowing fluid is tradidrag coefficient and defined by
Experimental values of eD for spheres and cylinders are shown in Fig. 8.2-6. Dimensional analysis would again lead us to the conclusion that en is a function of Reynolds number and relative roughness. The effect of roughness has not been studied extensively beca use engineers are generally concerned with reducing the drag on immersed bodies; keeping them smooth is a means of accomplishing this end.t
(8.2-14)
t This is not always the case, for sometimes increasing the roughness can decrease the drag. This phenomenon, along with the curves shown in Fig. 8.2-6, will be discussed in detail in Chap. 11. 305
306
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.3
Pipeline Systems
For Reynolds numbers less than 1.0, the inertial terms in the NavierStokes equation can be neglected under certain circumstances, which leads to a linear set of equations first solved by StokesY The drag coefficient for this condition is - 24 (8.2-18) en-
T
NRe
and the drag force acting on the sphere is given by
FD =
31TflU 00 D
2 00'
(8.2-19)
Equation 8.2-19 is known as Stokes law and has been used extensively by engineers to describe the motion of solid spheres moving through gases and Iiquids. The discussion here has been quite brief; however, a detailed treatment of flow around immersed bodies is given in Chap. 11 .
l_ 2'
8.3
Pipeline Systems
One ofthe standard problems that an engineer may encounter is the design of a piping system such as the one illustrated in Fig. 8.3-1. In the design of a new system, the flow rate and the physical properties of the fluid are generally given, and the engineer must determine the pipe size and the power requirement for the pump. The final solution will naturally be subject to certain economic constraints beyond the scope of this text. The main economic problem hinges on the fact that the cost of pumps and pipes depend on the size and that the final design of the system should satisfy the physical conditions at a minimum cost. The mechanical energy balance is the most suitable macroscopic balance w use in solving pipeline problems, because the troublesome viscous effects are nicely lumped into a positive dissipation term. If we restrict ourselves to steady flow, fixed control surfaces at entrances and exits, and incompressible fluids, the mechanical energy balance (Eq. 7.3-27) takes the form
f
Net outftow of kinetic energy
2
f -f
(lpv )v • n dA=
T
20'--
Pump flg. 8.3-1. 1
If we now take the positive z-coord gravity vector, the potential energy fl
In addition, we will consider control entrances and exits is normal to the e
f V·
t(n)
dA
A.
prf>v·ndA
A.
dA=
A.
Rate of work done on the system at the entrances and exits Rate of work done by gravitational forces, or the net outftow of potential energy
V • t
(8.3-1)
Substituting Eqs. 8.3-2 and 8.3-3 intt volume with one en trance (designa te< we obtain
Rate of work done on the
system by moving solid surfaces
-É,
Rate of viscous dissipation to interna! energy
17. G. G. Stokes; for an account, see H. Lamb, Hydrodynamics, 6th ed. (New York: Dover Publications, Inc., 1945), pp. 597-604.
=
-p
If the velocity profiles are nearly fla
Chap. 8
Sec. 8.3
307
Pipeline Systems
1.0, the inertial terms in the Naviercertain circumstances, which leads to a StokesP The drag coefficient for this
T
(8.2-18)
200'
(8.2-19) law and has been used extensively by spheres moving through gases and brief; however, a detailed treatment of in Chap. 11.
j_ 2'
engineer may encounter is the design in Fig. 8.3-1. In the design of a properties of the fluid are generally the pipe size and the power requirewill naturally be subject to certain of this text. The main economic of pumps and pipes depend on the should satisfy the physical conmost suitable macroscopic balance the troublesome viscous effects term. If we restrict ourselves to trances and exits, and incompressible 7.3-27) takes the form
20~
T
Pump Flg. 8.3-1. Pipeline system.
If we now take the positive z-coordinate to be oppositely directed to the gravity vector, the potential energy function, cp, takes the form
cp = gz .
(8.3-2)
In addition, we will consider control volumes such that the velocity at the entrances and exits is normal to the control surface; thus,
J
V•
t(n)
dA = -
J
(8.3-3)
pv • 'n dA
Rate of work done on the systetn at the entrances and exits
dA
Rate of work done by gravitational forces, or the net outftow of
potential energy
(8.3-1)
Substituting Eqs. 8.3-2 and 8.3-3 into Eq. 8.3-1, and considering a control volume with one entrance (designated by 1) and one exit (designated by 2), we obtain
Rate of work done on the
system by moving solid surfaces Rate of viscous dissipation to internal energy
lp(v3 )zAz - lp(v 3 )¡A¡
+ pgz (v)zAz 2
pgz1 (v)¡A¡
= -pz(v)zA 2 + p 1 (v) 1A1
+ W- E"
(8.3-4)
Hydrodynamics, 6th ed. (New York:
If the velocity profiles are nearly flat, we may use the macroscopic mass
308
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.3
Pipeline Systems
a valve, a pump, etc. Writing Eq. 8.3
balance (8.3-5)
(v? + ]!_ + z} - {(v)2+ {2g pg 2 2g
to reduce Eq. 8.3-4 to (8.3-6)
(v)2 + ]!_ + z} - {(v)2 + { 2g pg 3 2g
Each term in Eq. 8.3-6 has the units of length, or may be converted to units of length by multiplying g., if English units are used. Thus, the tcrm
---------------(v)2 + .E_ + z} - {(v)2 + 1?.
(v): + P2 + z2} {_!_ 2g . pg
{_!_ (v)~ + Pt + Z¡} = W - E,
-
2g
pg
pgQ
pgQ
{ 2g
or _P_ pg pgfg.
}!_
is often called the head or pressure head, the designation resulting from the fact that any pressure can be related to a column of static Iiquid by the equation, }!_
pg
=
h,
for a static fluid
and the fact that the height of a static liquid above a datum plane is traditionally referred to as the liquid head. The combination of terms, pf pg + z, is called the piezometric head, and (v) 2/2g is referred to as the velocity head. In keeping with this traditional nomenclature, the dissipation term E, divided by pgQ is called the head /oss, and is generally separated into two terms, h1 and h,.. We will use h1 to refer to frictional head losses in straight sections of conduits having a uniform cross section, and hm to refer to head losses occurring at sudden expansions, gate valves, etc. These losses are often called minor losses beca use the major source of viscous dissipation is in the comparatively long sections of straight conduit. The change in head w¡ pgQ owing to the power supplied by a pump or the power delivered to a turbine, is designated as Htl), the subscript w indicating that this term refers to the rate of work done on or by the fluid. The term H will be positive for pumps that supply energy to the fluid and negative for turbines that extract energy from the fluid. Using this nomenclature and simplifying the subscripts for the entrance and exit, we may rewrite Eq. 8.3-6 in the form ti)
+.E_+ z} - {(v) + J!.. + z} = Htl)- h {(v,)' 2g pg 2g pg 2
1
2
-
hm
(8.3-7)
1
Before discussing the head loss terms in detail, it will be helpful to indicate the method of application of Eq. 8.3-7. Let us consider a pipeline to be split up into N - 1 separate sections, each of which contains either a straight section of conduit or one of the numerous special elements such as an elbow,
pg
2g
N-1
P8
(v)2 + ]!_ + z} - {(v)2 + 1?. { 2g pg N 2g p
Summing all these equations, we see remain on the left-hand side, while th the head loss and HID terms.
(v)2 + ]!_ { 2g pg
+ z} N
- {(v)2 + ]!_ . 2g pg
Using !i to indicate differences bet~ Changc in vclocity hcad
!1{~~2} +
Changc in pressurc hcad
!ip pg
Changc in elcvation
+
!iz
=:
1
At this point, the student should begi the derivation of the Navier-Stokes energy equation, and the macrosco¡ each was an important, logical step the complexity of flow through a pip analyze such a system with the ease result. With the aid of experimental coefficients we can use Eq. 8.3-9 to de: system. To determine the friction head 1< scopic momentum and mechanical shown in Fig. 8.1-1. The momentu11
Sec. 8.3
309
Pipeline Systems
a valve, a pump, etc. Writing Eq. 8.3-7 for each section of the system gives (8.3-5)
W + -P1pg + Z¡} = -pgQ
- +-p + Z }
(v)2 { 2g
E (8.3-6) -"
the designation resulting from the to a column of static liquid by the a static ftuid
2 (v) { 2g
J!... + z} = pg
1
Hw- h1
-
hm
(8.3-7)
in detail, it will be helpful to indicate Let us consider a pipeline to be split of which contains either a straight special elements such asan elbow,
pg
2g
{(v)2 -
-
2g
3
N-1
- + -p + Z }
(v)2 { 2g
pg
+ -p + Z } = pg
{Hw- h,- hm}1
1
+ -p + Z } = {Hw- h,- hm}
2
pg
2
- {(vJ2 + 1!... + z} = {Hw- h,- hm}N- 2 2g pg N-2 -
{(v)2 -
2g
N
+ -p + Z } pg
N-1
= { H w - h, - hm } N-1
Summing all these equations, we see that only the first and the Nth terms remain on the left-hand side, while the right-hand side consists of the sum of the head loss and Hw terms. 2
liquid above a datum plane is tradiThe combination of terms, p/ pg + z, is referred to as the velocity head. the dissipation term E. and is generally separated into two to frictional head losses in straight cross section, and hm to refer to head gate valves, etc. These losses are source of viscous dissipation is in conduit. The change in head a pump or the power delivered to a w indicating that this term refers The term Hw will be positive for and negative for turbines that extract ~nc:Iature and simplifying the subscripts Eq. 8.3-6 in the forro
pg
+ 1!... + z}
{(v)2 -
-
2
- +-p + Z }
(v)2 { 2g
pgQ
of length, or may be converted to units units are used. Thus, the tcrm
pg
{(v) 2g
2
+ 1!... + z} pg
N
_ {(v) 2g
+ 1!... + z} =! Hw-! (h 1 + hm) pg
1
N
(8.3-8)
N
Using d to indicate differences between the outlet and the inlet, we obtain Change in velocity head
d{~~2} +
Change in pressure head
dp pg
Change in elcvation
+
dz = !Hw N
- !h, N
- !hm N
Sum oí head losaea or gains íor pumpa and turbines Sum oí losaea íor allstraighl sectiona
(8.3-9)
Sum oC!osaea for all valuea, ftuings, etc.
At this point, the student should begin to appreciate all the effort involved in the derivation of the Navier-Stokes equations, the differential mechanical energy equation, and the macroscopic mechanical energy balance, because each was an important, logical step in arriving at Eq. 8.3-9. If we consider the complexity of flow through a pipeline, it is an achievement to be able to analyze such a system with the ease and simplicity suggested by the above result. With the id of experimentally determined friction factors and loss coefficients we can use Eq. 8.3-9 to design a new pipeline or analyze an existing system. To determine the friction head loss, h1, we must apply both the macroscopic momentum and mechanical energy balances to the control volume shown in Fig. 8.1-1. The momentum balance is given by Eq. 8.1-28, which
310
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.3
Pipeline Systems
condense the expression for hm to t
we may rearrange to get
z} {_E_+ pg 1
{L + z} pg
=!(L) (v)2
(8.3-10)
2g
D
2
where
where K is a function of the Reynold that enter into the boundary conditi
(v) = (v.)1 = (v.)2 Substitution of this result into Eq. 8.3-7, Hw and hm being zero, gives
The sudden expansion
h1
=!(L) (v)2
for circular tubes
D 2g'
(8.3-11)
Darcy-Weisbach 18
This result is called the formula. For noncircular conduits, we must remember that the term L/ D resulted from
i wetted surface cross-sectional area
!7T DL 7TD
The sudden expansion in a pipe! which we can compute the head loss momentum balance predicted the f
L
= - -2 = -
D
4
Making use of the definition of the hydraulic radius Rh> we have a more general form of Eq. 8.3-11,
_ (_.!:._) (v)
h, -
f
4R~~,
for cir~ular and noncircular condUits.
2
2g
'
(
= m
É,
=
pgQ
f
ci> dV
"Y ___
pg(v)A
f
Vv: d dV _.:...."~'____
2¡-t
=
(8.3-13)
pg( v)A
Using R~~, as a characteristic length and (v) as the characteristic velocity, we may put the volume integral in dimensionless form to obtaint 2
hm = (v) {.....::!__ 2g p(v)A
f "~'
VU: D dV)
(8.3-14)
¡-tR~~,
Since
p (v )A/¡-tR~~,
i
_ 8·3 12)
Equation 8.3-12 and the friction factor-Reynolds number chart provide us with sufficient information to determine the friction head losses in a pipeline. The minor losses must also be determined experimentally and tabulated for all the different types of fittings. The equation for hm for a Newtonian fluid is
h
o,
Fig. 8.3-2. S
thus, we may make use of both balance to determine hm. The dime
~p = 2 [ ( ~J
- J( 1
and
~p =
[(Dl)4 - t] + ~ D 7TD
1 p(
2
Equating the two results gives an
=
h m
e~
4Ev 7T D~pg( v.)
and the predicted loss coefficient, K
is directly proportional to the Reynolds number, we may
t Here the volume dV is actually dimensionless so that the term enclosed by braces { } is dimensionless. 18. H. Darcy, "Experimental Researches on the Flow of Water in Pipes," Compt. Rend., 1854, 38:1109.
The result is in error by about 20 I Reynolds numbers.
Chap. 8
Sec. 8.3
Pipeline Systems
311
condense the expression for hm to the traditional form,
+ z} =!(.!:_) (v)2 D 2g
2
h =K (v)2 m 2g
(8.3-10)
(8.3-15)
where K is a function of the Reynolds number and any dimensionless groups that enter into the boundary conditions. , Hw and hm being zero, gives The sudden expansion for circular tubes
(8.3-11)
18
formula. For noncircular conLf D resulted from
The sudden expansion in a pipeline, shown in Fig. S.3-2, is one case for which we can compute the head loss hm directly. In Sec. 7.5 we found that the momentum balance predicted the pressure rise with reasonable accuracy;
i1rDL L =--=-
~t~~' ?b==---..:
radius R,., we have a more
Dz
01
for circular and noncircular conduits.
Reynolds number chart provide us the friction head losses in a pipeline. ~rnunt:o experimentally and tabulated equation for hm for a Newtonian
f
!6==---= l
)
(8.3-12
¡
Fig. 8.3-2. Sudden expansion.
thus, we may make use of both the momentum and mechanical energy balance to determine hm. The dimensionless pressure difference is given by Momentum balance (Eq. 7.5-13)
2,u Vv:d dV "Y
(8.3-13)
pg(v)A
' (v) as the characteristic velocity, we form to obtaint (8.3-14)
1
and Mechanical energy balance (Eq. 7.5-31)
Equating the two results gives an expression for
=
h m
4Ev 7T
Dipg(v.)
= [1 _
Evf pgQ,
(D1) D2
2
2 ]
Mi 2g
(8.3-16)
and the predicted loss coefficient, K, is given as to the Reynolds number, we may so that the term enclosed by braces { }
(8.3-17)
on the Flow of Water in Pipes," Compt.
The result is in error by about 20 per cent and may be used only for large Reynolds numbers.
Sec. 8.3 Macroscopic Balances-Viscous Effects
312
When a pipe discharges into a large tank ora reservoir, loss is 1 velocity head, h
(v)2
m
K=
1, and the
2g
The diffuser
If a gradual enlargement or diffuser is used to accomplish the transition from a smaller pipe toa larger·one, the loss can be reduced significantly over that obtained for a sudden expansion. The experimentally determined coefficient Ct is shown in Fig. 8.3-3. These results were obtained by Gibson 19- 2o for water at high Reynolds numbers, and the friction and minor losses were represented as f
+ h m = ef
It is convenient to express the head 1 cient K may be represented as
K=C{
(8.3-18)
= -
This amount is termed the exit loss and physically represents the complete dissipation of kinetic energy to interna! energy by viscous forces.
h
Pipeline Systems
Chap. 8
[1 - (D1)2]2 (v¡)2 D2 2g
(8.3-19)
and Ct should tend toward 1.0 as () decreases indicating a decrease in the streamlined. For values of () equal te the minor loss are of comparable roa that the form of Eq. 8.3-19 requires in brackets tends to zero while the f1 Generally the losses in sudden el vortex formation that occurs when ft be supplied to maintain the vortex m and this energy dissipation shows u visualize an increase in vortex moti therefore expect Ct to increase witl larger than 1.0 present a puzzling s vortex inotion than that occurring i increasing values of Ct with dec1 reasonable. Gibson suggested a pos but he based his arguments on the a (see Birkhoff intuitive hypothesis II values of B. It would be difficult to previously mentioned movie "Fiow Aside from providing an excellent d this movie drives home the point tt topology before we can construct a
1
11'12
1T
The reducer
()
The loss in head at a gradual re< that for a diffuser. The difference n and hence vortex formation, does 1 accounted for by the friction loss h1 The sudden contraction
The sudden contraction illustratt reducer, for the abrupt change in Fl1. 8.3-l. Gradual expansion or diffuser.
19. A. H. Gibson, Proc. Roy. Soc. (London) Ser. A, 1910, 83 :366. 20. A. H. Gibson, Engineering, 1912,93 :205.
t lt is a plausible intuitive hypothesis 1
t Prepared by Professor S. J. Kline of tional Services, Inc., 47 Galen St., Watert•
Sec. 8.3 ic Balances-Viscous Effects
Pipeline Systems
Chap. 8
313
tank ora reservoir, K= 1, and the
It is convenient to express the head loss in this forro, because the loss coefficient K may be represented as
(8.3-18)
(8.3-20)
physically represents the complete energy by viscous forces.
and C¡ should tend toward 1.0 as () -+ TT. As () -+ O, the coefficient Ct first decreases indicating a decrease in the minor losses as the system beco mes more streamlined. For values of Oequal to approximately 7°, the friction loss and the minor loss are of comparable magnitude and C¡ begins to increase. Note that the forro of Eq. 8.3-19 requires the C¡-+ oo as O-+ O, because the term in brackets tends to zero while the friction loss remains finite. Generally the losses in sudden expansions are explained in terms of the vortex formation that occurs when flow separation takes place. Energy must be supplied to maintain the vortex motion against the action ofviscous forces, and this energy dissipation shows up in the term, hm. It is not difficult to visualize an increase in vortex motion as the angle O is increased,t and we therefore expect C¡ to increase with increasing O. However, values of C¡ larger than 1.0 present a puzzling situation, for it indicates a more intense vortex motion than that occurring in a sudden expansion. furthermore, the increasing values of C¡ with decreasing values of A 2/A 1 do not seem reasonable. Gibson suggested a possible explanation for this phenomenon, but he based bis arguments on the assumption that the flow field is symmetric (see Birkhoff intuitive hypothesis III, Sec. 1.3), and it is not, for intermedia te values of O. It would be difficult to describe the flow phenomenon, and the previously mentioned movie "Flow Visualization" is strongly recommended.:J: Aside from providing an excellent description of the flow field in a diffuser, this movie drives borne the point that we must have sorne idea of the flow topology before we can construct a meaningful analysis.
is used to accomplish the transition Ioss can be reduced significantly over The experimentally determined coeffiresults were obtained by Gibson 19- 20 the friction and minor Iosses were
_ (D1)2]2 (v1)2 D2
2g
1=2.25
1T
8
"2n
p¡pansion or diffuser.
n) Ser. A, 1910, 83:366. 05.
(8.3-19)
The reducer
The Ioss in head at a gradual reduction or reducer differs markedly from that for a diffuser. The difference results from the fact that flow separation, and hence vortex formation, does not occur. The loss in a reducer can be accounted for by the friction loss h1, and we need not include the term, hm. The sudden contraction
The sudden contraction illustrated in Fig. 8.3-4 is very different from the reducer, for the abrupt change in cross-sectional area does Iead to vortex
t It is a plausible intuitive hypothesis regarding the flow topology. t Prepared by Professor S. J. Kline of Stanford University and distributed by Educational Services, Inc., 47 Galen St., Watertown 72, Mass.
314
Macroscopic Balances-Viscous Effects
Chap. 8 Sec. 8.3
Pipeline Systems
-- - ---------=~------------~-
hm Square-edged
Jb
~
~
I-~--r
Well rounded
Reentrant
Table 8.3-1 Loss
CoEFFICIENTS FOR A SUDDEN CONTRACTION
~~ ~ ~ ~ ~ ~ 1~ 1~ 1~ 1~ 1 ~ 1 ~ 1 0
A 2 /A 1
0.1
0.2
1
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
diameter (D 2) downstream from the contraction. This mínimum area is called the vena contracta. Further downstream, viscous effects cause the jet to spread and fill the conduit. Obviously the loss in a contraction is severely influenced by the geometry of the contraction, for the geometry controls the degree of vortex motion and hence energy dissipation. Three pipeline entrances from a large reservoir are illustrated in Fig. 8.3-5 along with approximate values for the head loss. The loss coefficient for a square-edged entrance is generally taken as K ~ 0.5, while the coefficient for a well-rounded entrance is K ~ 0.04 and the losses can be neglected. In this latter case, as in the gradual reduction in a pipeline, streamlining prevents the occurrence of a vena contracta and the accompanying vortex formation. The re-entrant opening is the extreme case, giving 21. J. Weisbach, Die Experimental Hydraulik (Freiberg, Germany : J. S. Englehardt, 1855), p. 133.
g
~
Fig. 8.3-4. Head loss in a sudden contraction.
formation. The loss coefficient K determined by Weisbach 21 is tabulated in Table 8.3-1. An important point to notice in Fig. 8.3-4 is that the major portion of the loss takes place downstream from the abrupt change in cross-sectional area. The jet formed by the sudden contraction attains a minimum area about one
~
~ ~
a minimum cross-sectional area ~ degree of vortex motion, which, of co In practice, these minor head however, we can learn from these reduce the vortex motion and t relatively unimportant in pipeline p. importance in the study of flow arm Numerous types of fittings may l for st:veral appear in Table 8.3-2. T been stlJ<:lied as thoroughly as the there may be considerable variatio 8.3-2 and the actual value for a give Regarding the actual desi!¡n of existing pipeline, the engineer may The essential elements of the proble (a) geometry of the system (pipe entrances and exits); (b) pumping or power requirem (e) fluid properties (fl, p) and flo (d) the economic constraint.
Balances-Viscous Effects
Chap. 8 Sec. 8.3
Pipeline Systems
315
Squore-edged
Well rounded
by Weisbach 21 is tabulated in .3-4 is that the major portion of the change in cross-sectional area. attains a mínimum area about one
0.10
0.06
0.02
o
0.8
0.9
l.O
This mínimum area is viscous effects cause the jet infl.uenced by the geometry the degree of vortex motion entrances from a large reservoir ~proxi·mate values for the head loss. is generally taken as K~ 0.5, is K~ 0.04 and the losses the gradual reduction in a pipeline, a vena contracta and the accomopening is the extreme case, giving (Freiberg, Germany: J. S. Englehardt,
Reentront
(v)~ hm ~ 0 .8 Zg ·
Fig. 8.3-5. Losses at pipeline entrances.
a mínimum cross-sectional area for the vena contracta and a maximum degree of vortex motion, which, of course, gives rise to a maximum head loss. In practice, these minor head losses do not play an important role; however, we can learn from these examples that streamlining may greatly reduce the vortex motion and the accompanying energy losses. While relatively unimportant in pipeline problems, these ideas will be of primary importance in the study of fl.ow around immersed bodies. Numerous types of fittings may be present in a pipeline; loss coefficients for s~::veral appear in Table 8.3-2. The head losses for these fittings have not been st11ciied as thoroughly as the simple expansion and contraction, and there may be considerable variation between the value of K listed in Table 8.3-2 and the actual value for a given piece of equipment. Regarding the actual desig,n of a new pipeline, or the analysis of an existing pipeline, the engineer may be confronted with severa! possibilities. The essential elements of the problem are: (a) geometry of the system (pipe length and diameter, valves and fittings, entrances and exits); (b) pumping or power requirements; (e) fluid properties (p,, p) and fl.ow rates; (d) the economic constraint.
Macroscopic Balances-Viscous Effects
316
Chap. 8
COEFFICIENTS FOR FITIINGS AND
Type of Fitting or Valve
VAL VES"
Loss Coefficient, K
45° ell standard long radius
0.35 0.20
90° ell standard long radius square or miter 180° bend, close return Tee, standard along run, branch blanked off used as ell, entering run used as ell, entering branch branching fiow Coupling Union Gate valve open topen topen topen Diaphragm valve open topen topen topen Globe valve open topen
Pipeline Systems
Power requirements for a pipeline
Table 8.3-2 Loss
Sec. 8.3
0.75 0.45 1.3 1.5
In this example, we wish to det system shown in Fig. 8.3-1 if it is t galjmin of oil. The following are the
Q = 100 galjmin (volumetric flo p = 60 lbm/ft3 (density) ¡.t = 10 centipoise (viscosity) D = 4 in. (pipe diameter)
The gate valve is fully open, and the smooth. The average velocity in the ~
0.4 1.3 1.5 0.04 0.04 0.20 0.90 4.5 24.0
2.3 2.6 4.3 21.0
= 184fj
and the Reynolds number is N Re= p(v) D ¡.t
=
{(60 lbm/ft')(1840 (10 cen 4
x { (4.78 x 10 centipoise) lbr-secjft2
Under these conditions the flow is tu formulas discussed in this section. 1 diagram yields a friction factor of
6.4
!=
9.5
From the macroscopic mechanical en a From Chemlcal Englneers' Handbook, 4th ed. (New York: McGraw-Hill Book Company, lnc., 1963), Sec. S, p. 33.
The route to a solution simply requires the application of the macroscopic mass and mechanical energy balances, along with experimental values of the various head loss coefficients and friction factors. In the analysis of existing systems, the economic constraint is often of little or no concern and the engineer needs to determine either (b) or (e). In the design of new systeins, (e) is usually given and the economic constraint requires that the job be done at a "mínimum cost." This generally involves a multitude of factors and makes the application of the economic constraint the most difficult part of the problem. However, the physics of the design are adequately taken care of by Eq. 8.3-9, which is all we concern ourselves with in this text.
~{~~2} =o
initial ar
~p =0
The pr entrance
pg ~z
=200ft
L h1 = =
f
2
(v)
D 2g
LL
(0.033) (1840 (4 in.) (2X3
= 3.11 ft
Balances-Viscous Effects
Chap. 8
8.3-2
Sec. 8.3
Pipeline Systems
317
Power requirements for a pipeline system
FITIINGS ANO VALVES0
Loss Coefficient, K
0.35 0.20 0.75 0.45 1.3 1.5
In this example, we wish to determine the power requirements for the system shown in Fig. 8.3-1 if it is to handle a maximum flow rate of 100 galfmin of oil. The following are the specifications:
Q = 100 gal/min (volumetric flow rate) = 60 lbm/ft3 (density) p. = 10 centipoise (viscosity) D = 4 in. (pipe diameter) p
The gate valve is fully open, and the pipe is to be considered hydraulically smooth. The average velocity in the pipe is
0.4 1.3 1.5 0.04 0.04 0.20 0.90 4.5 24.0 2.3 2.6 4.3 21.0
(v) = 4Q = 4(100 gal/min) (231 inf
7TD 2
(3.14)(4in.) 2
gal
= 1840 in.fmin and the Reynolds number is N Re= p(v) D = {(60 lbm/ft~(1840 in./min)(4 in.)} p. (10 centipoise)
(_Jf_) (60minsec) ( 32lbrlbmsecft
4
X {
2
(4. 78 x 10 centipoise) lbr-secjft 2 144 in. 2
) }
= 7600
Under these conditions the flow is turbulent, and we may use the head loss formulas discussed in this section. The smooth pipe curve on the Moody diagram yields a friction factor of
6.4 9.5
!= 0.033 From the macroscopic mechanical energy balance given by Eq. 8.3-9 we note
ed. (New York : McGraw-Hill Book
the application of the macroscopic with experimental values of the factors. In the analysis of existing of little or no concern and the (e). In the design of new systeins, · requires that the job be done involves a multitude of factors and constraint the most difficult part of the design are adequately taken ourselves with in this text.
~{~~2} =o
initial and final velocities are zero
~p =0
The pressure in the reservoirs at the entrance and exit Ievels is a constan!
pg ~z
=200ft
2, h1 = f
(v? 2, L
D 2g 2
ft) = (0.033) (1840 in./min) ( 310 (4 in.) (2)(32.2 ftjsec 2) = 3.11 ft
Chap. 8
Macroscopic Balances-Viscous Effects
318
Sec. 8.3
Pipeline Systems
The su m of the minor losses takes a particularly simple form beca use the pipe diameter is constant. Gate valve
Elbows 2
,L hm =
(v? 2g
Entrance loss
+ 0.2 +
,.L K= (v) {2(0.75) 2g
0.5
Exit loss
+
1.0}
2
= (1840 in./min) ( 3 .2 ) = 0 .32 ft 2
(2)(32.2 ft/sec ) As often happens, the minor losses are quite small, and, in fact, disregarding these losses would have caused an error of only 10% in the viscous losses. Substitution of these results into Eq. 8.3-9 yields
L Hw =200ft + 3.11 ft + 0.32 ft = 203ft Solving for the power required,
W, gives
w= =
(203ft) (pgQ) 2,720 lbt ft/sec
or
W = 5.0hp Hence the pump must be capable of delivering 5.0 hp to the fluid if the maximum ftow rate is to be obtained. In studying this example, the student should stop and give sorne thought to the ideas behind it. We might list them as follows: l. the /aws of mechanics leading to the Navier-Stokes equations and the mechanical energy equation; 2. the principie of conservation ofmass; 3. kinematics-the Reynolds transport theorem leading to the NavierStokes equations; 4. the concept of stress; 5. mathematics-the general transport theorem leading to the macroscopic balances. The solution of pipeline problems by Eq. 8.3-9 is obviously a straightforward problem; however, understanding what one is doing is by no means a simple matter. Pipeline networks Very often piping systems may be much more complex than the one illustrated in Fig. 8.3-1, and the ftow from a given outlet may come from severa! sources. Such a system is illustrated in Fig. 8.3-6, the design of which
Fig. 8.3-6.
would require specification of either a given ftow rate at the single outlet. conditions must be satisfied.
l. The sum of the pressure dron the pressure is a continuous, , 2. The mass balance must be sa1 3. The mechanical energy balan(
In single pipelines, only condition 3 represent obvious physical concepts networks is simply one of algebraic The problem is solved by first sp en ter and lea ve the system and of tht values are chosen so that the mas assumed ftow distribution is used to system that generally does not satis calculation, adjustments are made conditions are satisfied. An efficie1 described by Venard, 23 and we shall r
22. H. Cross, "Analysis of Flow in 1 11/inois Eng. Expt. Sta. Bu/l., 286, 1936. 23. J. K. Venard, "One-Dimensional Streeter, ed. (New York: McGraw-Hill Bo
ic Balances-Viscous Effects
Chap. 8
Sec. 8.3
319
Pipeline Systems
!y simple form because the pipe Gate
valve
Entrance
loss
Exit
loss
1.0} uite small, and, in fact, disregarding of only 1O% in the viscous losses.
Fig. 8.3-6. Pipeline network.
5.0 hp to the fluid if the maxishould stop and give sorne thought as follows : Navier-Stokes equations and the
theorem leading to the Navier-
theorem leading to the macro8.3-9 is obviously a straightforward is doing is by no means a simple
much more complex than the one a given outlet may come from in Fig. 8.3-6, the design of which
would require specification of either the pipe or pump size required to provide a given flow rate at the single outlet. In pipeline networks, the following three conditions must be satisfied. l. The sum of the pressure drops around each circuit must be zero-i.e., the pressure is a continuous, single-valued function. 2. The mass balance must be satisfied at each junction. 3. The mechanical energy balance must be satisfied for each section.
In single pipelines, o ni y condition 3 has to be considered; however, 1 and 2 represent obvious physical concepts, and the problem of designing pipeline networks is simply one of algebraic complexity. The problem is solved by first specifying the flow rates of all streams that en ter and leave the system and of the streams within the system. These initial values are chosen so that the mass balance requirement is satisfied. The assumed flow distribution is used to determine the pressure everywhere in the system that generally does not satisfy condition l. On the basis of the first calculation, adjustments are made and the process repeated until all the conditions are satisfied. An efficient method, worked out by Cross, 22 is described by Venard, 23 and we shall not discuss this subject further in this text. 22. H. Cross, "Analysis of Flow in Networks of Conduits or Conductors," Univ. 11/inois Eng. Expt. Sta. Bu/l., 286, 1936. 23. J. K . Venard, "One-Dimensional Flow," Handbook of Fluid Mechanics, V. L. Streeter, ed. (New York: McGraw-Hill Book Company, lnc., 1961).
320
Macroscopic Balances--Viscous Effecu
Chap. 8
In industrial practice, pipeline networks can become exceedingly complex, and digital computers are required to perform the analysis. Ingels and Powers 24 have presented a useful method of digital computation.
8.4
Sec. 8.4
Unsteady Flow in Closed Conduiu
The boundary conditions for this proll B.C. 1 :
v. = O,
B.C. 2:
v. is finite,
B.C. 3:
v.
Unsteady Flow in Closed Conduits B.C.4:
In general, engineers would like to deal with processes operating at or near steady state conditions. Unstable or oscillating systems are usually undesirable, and considerable effort is made to control processes at sorne steady state. Unsteady state conditions naturally occur during the start-up and shut-down of any process, or if the steady-state operating conditions are changed. Very often it is important to know the time required to reach the
t
= O, for
=o, r =ro op (Pi -=---
oz
The solution of this equation is beyon a knowledge of partial differential equ by Szymanski, 25 and it may be express (v.) (v.)oo
=
1 - (0.963e-0· 3681
where (v.) 00 is the average velocity for t of an infinite series of exponential ter for satisfactory results. The dimensio .
.
t
0 =-
~---------L------•~1
h
___ _l____
and/00 is the friction factor at t
=
oo.
Joo = p=p,
If the final fl.ow is turbulent, the involving an initial transition in the la the turbulent region. lf the final Reync will be turbulent during most of the tr approximate solution by the mechani< If the liquid depth h undergoes a r the pressure drop in the tube may be
Fig. 8.4-1 . Unsteady flow in a pipeline.
steady state, and we will apply the mechanical energy balance to two transient fl.ows to illustrate the important steps in the analysis. Figure 8.4-1 illustrates a supply and pipeline arrangement for which we wish to determine the time required for the ftow to reach the steady state value. lf the ftow is always laminar, and the entrance length is small compared to the tube length, the differential equations of motion reduce to (8.4-1) 24. D. M. Ingels and J. E. Powers, "Analysis of Pipeline Networks," Chem. Eng. Prog., 1964, 60:65.
flp pg
=Jo
where
tlp pg
Taking the tube as the control volum
~dt J(tpv
2
)
dV =
-r
25. P. Szymanski, "Sorne Exact Solutim Cylindrical Tube," J. Math. Pures Appl. Ser.
Macroscopic Balances-Viscous Efrects
Chap. 8
can become exceedingly complex, to perform the analysis. Ingels and method of digital computation.
Sec. 8.4
The boundary conditions for this problem are B.C. 1:
v. = O,
B.C.2:
v. is finite,
B.C. 3:
v. = 0, _
t = O, for r
=
Op = _
B.C. 4: to deal with processes operating at or near or oscillating systems are usually undesirto control processes at sorne steady naturally occur during the start-up and steady-state operating conditions are to know the time required to reach the
321
Unsteady Flow in Closed Conduits
oz
O :==:::: r
for O~
r
:==::::
:==::::
r0
r0
r0 _,_,_(P-=--i----"'P-"'-o)
t
L
>o
The solution of this equation is beyond the scope of this text, for it requires a knowledge of partial differential equations. The result has been presented by Szymanski, 25 and it may be expressed as (v.) (v.)oo
= 1- (0.963e-0·3691 oo
+ 0.036e-1. 9091 oc + · · ·)
(8.4-2)
where (v.) 00 is the average velocity for t = oo. The complete solution consists of an infinite series of exponential terms, but only the first two are necessary for satisfactory results. The dimensionless time 0 is defined as
0
~---------L--------~~1
and loo is the friction factor at t =
=
t(v.) 00 D oo.
(8.4-3)
foo=~
(8.4-4)
NRe,oo
If the final flow is turbulent, the transition will be a complex process involving an initial transition in the laminar region and a final transition in the turbulent region. Ifthe final Reynolds number is large (say, 106), the flow will be turbulent during most of the transition period, and we may obtain an approximate solution by the mechanical energy balance. If the liquid depth h undergoes a negligible change during the transition, the pressure drop in the tube may be expressed as
~p = f oo (
~'-"''all .•..,a,
energy balance to two transient in the analysis. Figure 8.4-1 illustrates which we wish to determine the time state value. the entrance length is small compared of motion reduce to · (8.4-1)
pg
(8.4-5)
where
~p = h pg Taking the tube as the control volume, the mechanical energy balance gives
~~ f (!pv -y
of Pipeline Networks," Chem. Eng. Prog.,
L) (v.);, 2g
D
2
)
dV =
f
V•
t(n)
dA - E,
(8.4-6)
A,
25. P. Szymanski, "Sorne Exact Solutions of the Equations of Motion for Flow in a Cylindrical Tube," J. Math. Pures Appl. Ser. 9, 1932, 11:67.
322
Macroscopic Balances-Viscous Effects
Chap. 8
Assuming the velocity profile to be flat (it certainly won't be during the early stages), and representing the stress vector at the entrance and exit as t(n)
=
-np
(8.4-7)
we find
Sec. 8.4
Unsteady Flow in Closed Condut
Both Eqs. 8.4-14 and 8.4-2 indicate t the time becomes infinite; however, f~ say that steady state has been reache~ of the final value. The time is therefo
11.8
Cl
~ {tp(v,) AL} = 2
dt
(v,) ~pA -
Ev
(8.4-8)
1:/gg%
(f) g
d(v,) dt
= ~P
Ev
_
pg
pgQ
1:/gg%
(8.4-9)
=f
Ev = h, = pgQ
!( L) (v,) D
foo
5.3
(8.4-11)
2f 1 ~;2 =Joof d0 +e
(8.4-12)
z
The approximation given by Eq. 8.4-11 will not be valid at short times when the flow is laminar; on the other hand, if the major portion of the transient time is associated with the rough-pipe region of flow, the approximation might be quite satisfactory. Integration of Eq. 8.4-12, and application of the boundary condition
U,= O,
0 =0
t99% = o
--j
(v,)oo
To g·a in sorne familiarity with the could say for turbulent flowt
foo = (
D =(
(v,) = (
Under these conditions, the time wo1 t99%
=
O[(0.00:(5.
= 0[1.77 n
We separate variables and integrate to obtain
B.C. 1:
(v,)oo
0
Solving this equation would be a difficult task beca use of the complex variation of the friction factor f with the Reynolds number. A numerical solution would be straightforward (and might make an interesting class project), but to obtain an analytic solution we will assume
1
'
11.8JJ
2g
(8.4-iO)
>:::!
00
= --
t99%
and if we substitute Eq. 8.4-5 into Eq. 8.4-9 and use the dimensionless time 0, we can write Eq. 8.4-9 in the form
j_
'
It is difficult to comment on these two time for two distinct types of flow.
If we remember that 2
00
5.3
Cl
Carrying out the differentiation and dividing by p(v, )Ag, we have
= f
From this order of magnitude calcul~ becomes significant for large diamet¡ that this analysis assumed that the le the entrance length and is therefore 1
Oscillating systems-the U-tube r
(8.4-13)
Various fluid systems are subject latory motion. Actual processes ma) analogue or digital computers may be even approximately. The purpose ol gain sorne insight into the import
(8.4-14)
t Read Ü(0.005) as "on the order of rm
yield the final solution
Balances-Viscous Effects
Chap. 8
(it certainly won't be during the early at the entrance and exit as (8.4-7)
=
(v.) 6.pA -
Ev
(8.4-8)
viding by p(v. )Ag, we have 6.p _
pg
Ev pgQ
(8.4-9)
Sec. 8.4
Both Eqs. 8.4-14 and 8.4-2 indicate that the final velocity is not reached until the time beco mes infinite; however, for practical purposes it is satisfactory to say that steady state has been reached when the velocity is within 1 per cent of the final value. The time is therefore given as 1 8 1.
laminar flow
5.3 0 99% = foo '
turbulent flow
0
99%
= foo '
It is diffi.cult to comment on these two values, for they represent the transient time for two distinct types of flow. The real times for the two cases are
!(.!:.)D (v.)2 2g
t 99 % t
8.4-9 and use the dimensionless time
(8.4-ÍO)
323
Unsteady Flow in Closed Conduits
11.8D
= ( -J.) , v.
00
(
vz
)
00
"!> n~ b
re
00
5.3D ----
99% -
laminar
J. '
turbulent
00
To gain sorne familiarity with the orders of magnitude of these times, we could say for turbulent flowt
/oo = O(0.005) task beca use of the complex varianumber. A numerical solution make an interesting class project),
D (v.)
= 0 (l ft) = O(1 ftjsec)
Under these conditions, the time would be (8.4-11)
199
%
=
= (8.4-12) will not be valid at short times when if the major portion of the transient region of flow, the approximation of Eq. 8.4-12, and application of the
0 =0
(8.4-13)
(8.4-14)
o[
(5.3)(1 ft) min (0.005)(1 ftjsec) 60 sec
J
0[1.77 min]
From this order of magnitude calculation we see that the transient time only becomes significant for large diameter tubes and low flow rates. Remember that this analysis assumed that the Iength of the pipe was long compared to the entrance length and is therefore not valid for short pipes. Oscillating systems-the U-tube manometer Various fluid systems are subject to upsets which may give rise to oscillatory motion. Actual processes may become extremely complex, and large analogue or digital computers may be required to determine the fluid motion even approximately. The purpose of studying the U-tube manometer is to gain sorne insight into the important aspects of oscillatory systems. In
t Read 0(0.005) as "on the order of magnitude of 0.005."
324
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.4
Unsteady Flow in Closed Conduits
addition, this example is of sorne interest because it can easily be studied experimentally, thus introducing the student to the errors incurred in an approximate solution. The U-tube manometer to be analyzed is shown in Fig. 8.4-2. Once again, the differential equations describing the motion are too complex to be solved, and the mechanical energy balance is required. We start with the complete equation and note that W is zero to obtain
For the moving control volume i'a(t) that the gas-liquid interface is horizon 8.4-2. Actually, it is not because the interface will be as shown in the inset. Considering Eq. 8.4-15, we note th1
~t J(!pv
and the convective transport term on energy balance is identically zero. If vi! are neglected, we may assume that
2 )
J
dV +
i'"0 (t)
2
(!pv )(v - w) • n dA
A,(t)
=
J
V•
t(n)
dA-
A,< O
J
pgzv • n dA -
Ev
(8.4-15)
.91' aW
'
__("' Gas pressures suddenly equilibrated
( - - - - -- - -
V=W
V•
~n> hert =
and the first term on the right-hand assumption is open to question, becaus depends upon whether it is advancing solve the detailed problem associated v assume it is ftat and make the assump
T ho
_________________ l __ _
Final position of manometer fluid
1 - - - i + - - lnitial pasition of manometer fluid
1
z=O
¡
If the flow in the manometer were t 8.4-17 could be replaced by average term could be expressed as in the pr ftow in a manometer is likely to be la mate analysis for laminar flow. Assur one-dimensional, we write v(r, t)
Manometer liquid
Advancing liquid
"''""' '"'"'"" 11111 Liquid
Receding liquid
u
Fig. 8.4-2. U-tube manometer.
Liquid
= 2(v)l
Here, v(r, t) represents the component Expressing the transient velocity profil is a common technique (and one whi obtaining approximate solutions to eng method depends on the variations witl velocity profile is never far from par we shall explore the validity of this as! The average velocity (v) in Eq. 8.4 defined as positive when the flow is fr< V•
n heft
=
and the position of the two interfaces Z
= -h heft•
ic Balances-Viscous Effects
Chap. 8
because it can easily be studied to the errors incurred in an
Sec. 8.-4
For the moving control volume "Ya(t) we ·choose the fluid itself and assume that the gas-liquid interface is horizontal and well defined as shown in Fig. 8.4-2. Actually, it is not because the fluid adheres to the tube wall, and the interface will be as shown in the inset. Considering Eq. 8.4-15, we note that
f
~a(t)
on
V=W
dA-
325
Unsteady Flow in Closed Conduits
and the convective transport term on the left-hand side of the mechanical oenergy balance is identically zero. If viscous effects at the gas-liquid interface .are neglected, we may assume that pgzv • n dA-
E" (8.4-15)
""·(t) Gas pressures suddenly equilibrated
T ho
______ l __ _
¡
z=O
J of
V•
t.:n)
u L iquid
=
-V •
t
lrlght
(8.4-16)
and the first term on the right-hand side of Eq. 8.4-15 is also zero. This assumption is open to question, beca use the nature of the interface obviously depends upon whether it is advancing or receding. However, we must either solve the detailed problem associated with the moving interface, or we must assume it is flat and make the assumption indicated by Eq. 8.4-16 to obtain
~f (!pv
2
)
dV
= -
f
pgzv • n
dA - E,
(8.4-17)
Jll'G(t)
'j'"G(t)
If the flow in the manometer were turbulent, the various ve1ocities in Eq. 8.4-17 could be replaced by average velocities and the energy dissipation term could be expressed as in the previous example. However, oscillating flow in a manometer is likely to be laminar, and we must devise an approximate analysis for laminar flow. Assuming the flow to be quasi-steady and one-dimensional, we write v(r, t)
Receding liquid
hert
(~)]
= 2(v)[ 1 -
(8.4-18)
Here, v(r, t) represents the component of the velocity vector along the tube. Expressing the transient velocity profile in terms of the steady state solution is a common technique (and one which should be used with caution) for obtaining approximate solutions to engineering problems. The success ofthis method depends on the variations with time being small enough so that the velocity profile is never far from parabolic. After obtaining the solution, we shall explore the validity of this assumption. The average velocity (v) in Eq. 8.4-18 is a function of time, and will be defined as positive when the flow is from left to right. Since V•
n hen
=
-V •
n lrlght
(8.4-19)
and the position of the two interfaces is given by, Z
= -h hert>
z=
+h lrtght
(8.4-20)
326
Macroscopic Balances-Viscous Effects
Chap. 8
Sec. 8.4
Unsteady Fl 0 w in Closed Conduits
time 0 as,
We may simplify the area integral fn Eq. 8.4-17 to obtain
~1(4p7TL(v)2f" [1 - (~rJ r dr) = - 2pgh(v)7Tr~ - J
h ho
H=-
2
<1> dV r-.(t) .
o
(8.4-21)
0 =t
From Table 7.3-1, we find the dissipation function to be <1>
= 2~-'[av~ + or
(!r ovao + ~)r + (av.) oz 2
+ ~-'[ (::r) + (:~·)
r
2
]
r
+ ~-'[;(:;·) +
(::8)
r
= v, and <1> reduces to
16J,L(v)2 (~)
2
=
where
In terms of the dimensionless variables,
In this case, there is only one component ofvelocity, v. 4>
d H + 2{3 dH d02 dG
2
8
+ ~-'[' :,(;) + ~(:;r)
and Eq. 8.4-24 reduces to
(8.4-22)
Substituting this result into Eq. 8.4-21 and carrying out the integration, we get (8.4-23)
B.C. 1':
H= 1,
B.C. 2':
dH =O d0 '
The solution of this second-order, ordin by assuming solutions of the form
H= which when substituted into Eq. 8.4-26 eme(m2
Because the time rate of change of height of the gas-liquid interface is equal to the average velocity,
dh - = (v) dt
2
r~ dt
2L
(8.4-24)
=
B.C. 2:
dh dt
h0 ,
=0
m1
= -({3 +
m2
=
-({3-
Provided m 1 and m2 are distinct (i.e., {3
This equation is to be solved subject to the boundary conditions h
+ 2{3m ·
Thus, m has two possible values, given
dh+ (6v)dh + (3g)h = 0 B.C. 1:
This equation will be satisfied, and E differential equation, if m is chosen so t m2
We may rearrange Eq. 8.4-23 to obtain
dt 2
+ 2{3¡
t=O
(8.4-25a)
=o
(8.4-25b)
H
=
C¡emle
However, if the roots of Eq. 8.4-30 are ,
t
It will be helpful to put the differential equation and the boundary conditions in dimensionless form. To do so, we define a dimensionless distance H and
H
=
(C1
+
26. L. R. Ford, Differential Equations (N Inc., 1933), p. 70.
Chap. 8
Sec. 8.4
time
. 8.4-17 to obtain
327
Unsteady Flpw in Closed Conduits
e as, h
H=-
-2pgh(v)1Tr~- J<1> dV
=
ho
(8.4-21)
E>=t
7".(1)
!3i
~-u
and Eq. 8.4-24 reduces to
+
(~:·n
d2H
)r
+ ~t[;(~~·) + (~:8)
r
d82
+ 2{3 dH + H = O
(8.4-26)
dE>
where
fJ =
f2L r~ ~Ji
3v
In terms of the dimensionless variables, the boundary conditions are ofvelocity, v.
= v, and <1> reduces to (8.4-22)
and carrying out the integration, we
B.C. 1':
H= 1,
e =O
(8.4-27a)
B.C. 2::
dH =O dE> ,
8=0
(8.4-27b)
The solution of this second-order, ordinary, differential equation is obtained by assuming solutions of the form
H= eme (8.4-23) of the gas-liquid interface is equal
(8.4-28)
which when substituted into Eq. 8.4-26 gives
em 9 (m 2
+ 2flm + 1) = O
(8.4-29)
This equation will be satisfied, and Eq. 8.4-28 will be a solution of the differential equation, if m is chosen so that
+ 2flm + 1 = O
m2
(8.4-30)
Thus, m has two possible values, given by m1
(8.4-24)
m2 = -({3Provided m1 and m 2 are distinct (i.e.,
the boundary conditions
=o
(8.4-25a)
t =o
(8.4-25b)
t
= -({3 + .Jf32 - 1)
H
=
.Jf3
fJ =1=
C1 em1 e
2
-
1)
(8.4-31a) (8.4-31b)
1), the general solution is
+ C emsE> 2
{8.4-32)
However, if the roots of Eq. 8.4-30 are not distinct, the solution is 26
and the boundary conditions a dimensionless distance H and
H
=
(C1
+ C2 8)e-e
(8.4-33)
26. L. R. Ford, Differential Equations (New York: McGraw-Hill Book Company, lnc., 1933), p. 70.
328
Macroscopic Balances-Viscous Effects
Chap. 8 Sec. 8.4
Unsteady Flow in Closed Conduits
The solution is again given by complex function containing bo part independently satisfies the conditions, and takes the forro
This result indica tes that H takes which decrease exponentially wi damped." -i.OL------------:8::--------
-
Fig. 8.4-3. Position of the gas-liquid interface for an oscillating manometer.
The motion of the fluid in the manometer may have three distinct modes depending on the value of {3. The three types of motion are illustrated graphically in Fig. 8.4-3. We may discuss them as follows .
l. {3 > l. For this case, both m1 and m 2 are real negative numbers, and the constants of integration in Eq. 8.4-27 are readily determined to yield (m 2emt9 _ m1em28) H=~::.:.-----2..::...~ (8.4-34) (m 2 - m1) This type of motion is called "over damped," and the fluid approaches its equilibrium position exponentially. 2. {3 = l. Here, the roots are indistinct, and the solution is obtained by application of the boundary conditions to Eq. 8.4-33, yielding
H
=
(1
+ 0)e-9
(8.4-35)
The fluid motion is similar to the previous case in that the equilibrium position is approached exponentially. However, this solution represents the most rapid approach to equilibrium without "overshooting" the equilibrium position. Such motion is called "critically damped." 3. {3 < l. Under these conditions, the roots become complex, and we must take special care in obtaining a solution. We write the roots as
{3<1
(8.4-36)
Without the preceding analysis, we could occur in a mano meter. If viscous that the fluid should slowly and steadil But if inertial effects predominate, we behave somewhat like a pendulum, os tion. Since {3 is, in effect, the ratio o dependence of the motion on {3 is just though the analysis agrees with our int question, "U nder what conditions can We can answer this question, in part, analysis.
Order of magnitude analysis
Two major assumptions were impo
l. The gas-liquid interface was flat. 2. The velocity profile was parabol
The first assumption is best investigat1 the second assumption can be discuss1 away from the interface and the curve1 flow will be one-dimensional, and the t
av --+ éJp P< éJz
p-=
éJt
From previous work we know that if compared to the viscous term, the velo Making use of Eq. 8.4-18, we can p tude analysis using the solution to Eq. 8
ic Balances-Viscous Effecu
Chap. 8 Sec. 8.4
Unsteady Flow in Closed Conduiu
329
The solution is again given by Eq. 8.4-32; however, the result is a complex function containing both real and imaginary parts. The real part independently satisfies the differential equation and the boundary conditions, and takes the form
fJ
o
-
gas-liquid interface for an
may have three distinct modes types of motion are illustrated them as follows. m2 are real negative numbers, and Eq. 8.4-27 are readily determined to
Without the preceding analysis, we might easily guess that such motions could occur in a mano meter. If viscous effects predominate, it seems natural that the fluid should slowly and steadily approach the equilibrium position. But if inertial effects predominate, we would certainly expect the system to behave somewhat like a pendulum, oscillating around the equilibrium position. Since fJ is, in effect, the ratio of viscous forces to inertial forces, the dependence of the motion on fJ is just what we would expect to find. Even though the analysis agrees with our intuition, we should still ask the important question, "Under what conditions can we expect this result to be accurate?" We can answer this question, in part, by performing an order of magnitude analysis. Order of magnitude analysis
(8.4-34) Two major assumptions were imposed on this analysis.
and the solution is obtained by to Eq. 8.4-33, yielding 0)e- 9
(8.4-35)
previous case in that the equilibrium y. However, this solution repreequilibrium without "overshooting" · is called "critically damped." the roots become complex, and we a solution. We wriie the roots as
fJ
(8.4-36)
l. The gas-liquid interface was fiat. 2. The velocity profile was parabolic-i.e., quasi-steady fiow. The first assumption is best investigated experimentally, but the validity of the second assumption can be discussed mathematically. At sorne distance away from the interface and the curved portion of the manometer tu be, the fiow will be one-dimensional, and the equations of motion reduce to
av ot =
P-
a (rav.) - -op + pg. + p,oz
or or
(8.4-38)
From previous work we know that if the local acceleration term is small compared to the viscous term, the velocity profile will be parabolic. Making use of Eq. 8.4-18, we can perform the following order of magnitude analysis using the solution to Eq. 8.4-26 to estímate the magnitude ofthe
330
Macroscopic Balances-Viscous Effects
Chap. 8
terms in Eq. 8.4-38. Attacking the local acceleration term first, we writet
ov =o( P o
Flow Rate Measurement
(8.4-39)
Here we see that the inequality given b but will be satisfied for all dimensionl damped flows we may expect reasonabl) experiment, except for very short times
(8.4-40)
8.5
2
P ot
Sec. 8.5
The viscous terms are expressed as
Our requirement that the local acceleration term be small compared to the viscous term immediately leads us to the inequality, 1
~~ ~ ~ ~~ 1
1
(8.4-41)
1
or, in dimensionless form,
Flow Rate Measurement
There are severa! common devices meter, Pitot tube, and the sharp-crested velocities. It is of interest to analyze engineering practice, and since these experimentally, we have an opportunity
(8.4-42) If the system is under damped ({3 < 1), the first derivative dHfd0 takes on zero values; at these times the inequality can never be satisfied. For the under damped case, we can expect only qualitative agreement between the derived result and a real manometer. For the critically damped case our solution indicates that
d2H = -(1- El)e-9, d8 2
dH = -ee-9 dEl
Thus,
~~~~ ~ ~~~1· ~~~~ ~ ~~~1·
for for
Venturi meter
The Venturi meter, illustrated in Fig constriction that accelerates the flow pressure; this process is followed by a loss to be minimized. The pressure di related to the flow rate by the appli This presents a situation for which mentum balance because the pressure
e -o
1 1
e- 00
and we can again expect only qualitative agreement between theory and experiment. This situation is left rather wide open, as it should be. If the agreement between theory and experiment were poor, we should not be surprised. Neither should we be surprised if the agreement were good, for it would only mean that significant deviations from the parabolic profile did not greatly alter the value of the viscous dissipation term. The analysis for the over damped flow ({3 > 1) is more complex than the other two cases; however, if {3 ~ 1 we may write
...•
"1" 1
1
"
-
Z4--------+---l 1
L--/.
...:
z2 _____ ----- -----
z1---------
t The symbol O should be read as "the order of magnitude of." This type of analysis will be discussed in detail in Sec. 11.3.
Fil. 8.5-1. Ver
ic Balances-Viscous Effects
Chap. 8
acceleration term first, we writet
¡.t(v)) ( r~
= O(/!:._ dh) r~ dt
dh dt
(8.4-39)
(8.4-40)
8.5
(8.4-41)
1
331
Flow Rate Measurement
Here we see that the inequality given by Eq. 8.4-42 is not satisfied for 0--+- O, but will be satisfied for all dimensionless times greater than {3-1 • For over damped flows we may expect reasonably good agreement between theory and experiment, except for very short times.
term be small compared to the the inequality, '}/ 1 r~
Sec. 8.5
Flow Rate Measurement
There are severa! common devices-such as the orífice meter, Venturi meter, Pitot tube, and the sharp-crested wier-used to measure flow rates and velocities. lt is of interest to analyze these systems as an introduction to engineering practice, and since these metering devices have been studied experimentally, we have an opportunity to compare analysis with experiment.
(8.4-42)
< 1), the first derivative dHfd0 takes can never be satisfied. For the qualitative agreement between the so1ution indicates that
-dH = -0e-9 d0
for
Venturi meter
The Venturi meter, illustrated in Fig. 8.5-1, consists of a relatively abrupt constriction that accelerates the flow and produces a decrease in the fluid pressure; this process is followed by a gradual expansion allowing the head loss to be minimized. The pressure difference between points 1 and 2 can be related to the flow rate by the application of the macroscopic balances. This presents a situation for which we should definitely not use the momentum balance because the pressure (and thus the surface force) varies
0--+- O "1"
for 0 --+- oo
"2" 1 1 1 1
agreement between theory and wide open, as it should be. lf the were poor, we should not be surif the agreement were good, for it from the parabolic profile did s dissipation term. flow (/3 > 1) is more complex than 1 we may write
Control volume
--,
---:-'--- - - - - - - -
21---------
1
Gravity
of magnitude of." This type of analysis
Z=O----------Flg. 8.5-1. Venturi meter.
Macroscopic Balances-Viscous Effects
332
Chap. 8
from 1 to 2 in an unspecified manner.t The mechanical energy balance is well suited to this type of problem, and for steady, incompressible flow, the various terms in Eq. 7.3-31 are
f
lpv2y • n dA
=
-lp(v
3
) 1 A1
+ lp(v
3
) 2A2
Sec. 8.5
Flow Rate Measurement
include an estimate of the effect of t pressure; however, we shall neglect this by the centerline pressure to obtain
(8.5-1)
Q=A2a
(8.5-2)
If this formula is expressed in terms o z 2 - z1 yields
.ti.,
f
V •
t
l Q =A 2 J2(pm p[l
.ti.,
Jpcf>v • n dA= -p(
+ p(cf¡v) 2A2
1
(8.5-3)
..~~..
For the present, we shall assume that viscous effects are negligible and write
E" =
o
(8.5-4)
Substituting Eqs. 8.5-1 through 8.5-4 into Eq. 7.3-31, and assuming flat velocity profiles, we find lp[(v)~A 2 -
=
(v)~A 1 ]
(v)1(p) 1A1
-
(v)a(p) 2A2
-p[(cf¡)2(v)2A2- (cf¡)¡(V) 1A1 ]
(8.5-5)
Application of the macroscopic mass balance (8.5-6)
(v) 2A 2 = (v) 1A 1
allows Eq. 8.5-5 to be simplified to (v)~ - (v)~
2
= p [(p + pcf>)t
- (p
+ pcf>)2]
(8.5-7)
where Pm is the density of the manome Experimental studies indicate that accurate for engineering practice, anda to give
Q = CaA2
A plot of Ca versus Reynolds numbe~ meter with (A 2/A 1) = 0.25. We see tha sis is good because Ca is nearly unity. we should not expect the analysis to ho velocity profiles are no longer flat. A profile into aC<:ount, can lead to the rapidly as laminar flow is approached; l exercise for the student. 1.00
The mass balance may be used to eliminate (v) 1 and obtain an expression for the volumetric flow rate
Q=
J
(v) A = A 2 2 2
2 [(p
+ pcf>)
1 -
(p
+ pcf>)2]
p[l - (A2/At)2]
(8.5-8)
Now, 4> is a linear function in z; therefore, the average value is the centerline value, and (8.5-9) However, the pressure at point 2 is not a linear function of z because the streamlines are curved,t and strictly speaking we cannot replace the average pressure by the pressure at the centerline. A more detailed analysis might
t Does this suggest a differential-macroscopic balance as a possibility? ! See Sec. 7.4, Eq. 7.4-21.
{2(;;: .J~
/'
0.90
/
V 0.80
1/
2
4 681Ql 2
4
NRe=
Fl1. 8.5-:Z. Venturi ~
Chap. 8
The mechanical energy balance is well for steady, incompressible flow, the
Sec. 8.5
include an estimate of the effect of the curved streamlines on the average pressure; however, we shall neglect this effect and replace the average pressure by the centerline pressure to obtain
Q- A
(8.5-1)
(8.5-2)
333
Flow Rate Measurement
2
J
2(p¡ - P2) p[1 - (A 2/A1) 2]
(8.5-10)
If this formula is expressed in terms of the height of the manometer fluid, z 2 - z1 yields (8.5-11)
(8.5-3) viscous effects are negligible and write (8.5-4)
where Pm is the density of the manometer fluid. Experimental studies indicate that Eq. 8.5-11 is not always sufficiently accurate for engineering practice, and a discharge coefficient e, is introduced to give (8.5-12)
into Eq. 7.3-31, and assuming flat
- (vMp)2A2 - (t/>)1(v)1A 1 ]
(8.5-5)
(8.5-6)
A plot of e, versus Reynolds number is shown in Fig. 8.5-2 for a Venturi meter with (A 2/A 1) = 0.25. We see that at high Reynolds numbers the analysis is good because e, is nearly unity. For Reynolds numbers less than 2300, we should not expect the analysis to hold because the flow is laminar and the velocity profiles are no longer flat. A more carefulanalysis, takingthevelocity profile ·into account, can lead to the conclusion that e, should decrease rapidly as laminar flow is approached; however this problem will be left as an exercise for the student.
(8.5-7)
1.00
~v
(v)1 and obtain an expression for
0.90
(8.5-8) the average value is the centerline (8.5-9) ~~"""'F.
we cannot replace the average A more detailed analysis might
balance as a possibility?
/
eti 0 .80
V
~
lo-"
V
V
V
2
4 6 8 103 2
4 6 810" 2
4 6 8 1dl
Nrt.= (v) 1D1/V Fl¡. 8.5·1. Venturi discharge coefficient.
334
Macroscopic Balances-v 1scous Effects
Chap. 8
Sec. 8.5
Flow Rate Measurement
Orífice meter
The sharp-edged orífice shown in Fig. 8.5-3 is another device for accelerating the flow and obtaining a measurable pressure difference related to the flow rate. The orífice is often used as a metering device in preference to a Venturi meter because of its versatility and low cost of construction and installation. Because the area of the jet at point 3 (the vena contracta) is
t---.... !'--0 .80
!'--t---.... t-. t--r--..,
Control volume
"1''
"2"
1 1 1 1
0.60 2
r--- ~J
... r--..t- ¡... t--r--.., t-11 ... l-..
4
6 8
4
10
2
4 6
Fig. 8.5-4. Discharge coefficien '
Our analysis would indicate that
Ca.=~
'V [1- (
Flg. 8.5-3. Sharp-edged orifice.
unknown, the solution to this flow problem will not be as accurate as that for the Venturi meter. Calibration of orífice meters is therefore a necessity, and, in general, they are considered less reliable than Venturi meters. If we express the area of the jet at the vena contracta as (8.5-13)
where A2 is the area of the orífice and Ce is the contraction coefficient, the flow rate is given by _ CA
Q-
e
J
2
2(p¡ - Pa) p[1 - (CeA2/A1) 2]
(8.5-14)
Here we have applied the mechanical energy balance, neglected viscous effects, and replaced average pressures with centerline pressures. Because the contraction coefficient is a function of the ratio of areas A2/ A1 , the equation is generally written in terms of a discharge coefficient
Q = Ca.A 2~ 2(p 1 - p3 )/p
(8.5-15)
Because Ce changes with A2/A 1 it is ratl with experimental results. However, as coefficient should equal the contraction e simplified. Figure 8.5-4 shows sorne exp1 sharp-edged orífice. For small values o approximately 0.6, a reasonable value again the analysis is only satisfactory at Flow nozzle
A flow nozzle, such as that illustrat difficulty of an unknown contraction cot parallel as the fluid emerges from the nm is applied as before to yield
Sec. 8.5
Flow Rate Measurement
335
¡-..... r-...
8.5-3 is another device for acceleratpressure difference related to the a metering device in preference to a and Iow cost of construction and jet at point 3 (the vena contracta) is
0 .80
0.70
1·~
1'-.
~
0 .60
¡... ..............
r--
0 .40
........
0.60 2
...... t- """'¡... ..._¡........
0.20 0.10 0.05
......
4 6 8104
2
4 6 8105
2
4
6
81o6
2
NR• =
Our analysis would indicate that (8.5-16)
meters is therefore a necessity, and, than Venturi meters. (8.5-13)
c. is the contraction
Because c. changes with A 2/A 1 it is rather difficult to compare our analysis with experimental results. However, as A 2/A 1 becomes small, the discharge coefficient should equal the contraction coefficient, and matters are somewhat simplified. Figure 8.5-4 shows sorne experimental values of C,¡.for a standard sharp-edged orífice. For small values of A 2/AI> the discharge coefficient is approximately 0.6, a reasonable value of the contraction coeffi.cient. Once again the analysis is only satisfactory at high Reynolds numbers.
coefficient, the Flow nozzle
(8.5-14) energy balance, neglected viscous with centerline pressures. Because the the ratio of areas A 2/A 1 , the equation coefficient (8.5-15)
A flow nozzle, such as that illustrated in Fig. 8.5-5, does not offer the difficulty of an unknown contraction coeffi.cient, because the streamlines are parallel as the fluid emerges from the nozzle. The mechanical energy balance is applied as before to yield
(8.5-17)
Macroscopic Balances-Viscous Effects
336
"2"
"1"
Chap. 8
Sec. 8.5
Flow Rate Measurement
The experimental values of Cd shown is in error by about 10 per cent or less, area to the duct area, A 2/A 1 • As the n erated more severely, and inertial effec the Reynolds number is greater tha1 viscous dissipation begins to show i decreases.
1 1
Pitot-static tube
The pitot-static tube shown in Fi velocities ata point; it may be used t pro file is flat or if the en tire profile is m at point 2 which is the stagnation poin holes are drilled around the peripher_)l static pressure. The pressure at point 2 difference between these two pressure and relatéd to the fluid velocity. In thi the pressure along two streamlines rat
Fig. 8.5-5. Flow nozzle.
1.00
A)A,Lb1.~
í1
1
11 1 1_1 1 -0.50
~ """"
A2/A1 A2l;A 11=0.65
~
u
~~
0 .90 f-A 2 !A 1 = 0 .30
0.80
2
Th 4
6
8104 2
4
6
8105 2
4
6
81o6 2
l_
NRe =
Fig. 8.5-7. Pit
Sec. 8.5
Flow Rate Measurement
337
The experimental values of Cá shown in Fig. 8.5-6 indicate that the analysis is in error by about 1Oper cent or less, depending u pon the ratio of the nozzle area to the duct area, A 2/A 1 • As the ratio becomes smaller the fluid is accelerated more severely, and inertial effects very definitely predominate, provided the Reynolds number is greater than 105 • For lower Reynolds numbers, viscous dissipation begins to show its effect and the discharge coefficient decreases. Pitot-static tube The pitot-static tube shown in Fig. 8.5-7 is a device for measuring fluid velocities ata point; it may be used to determine the flow rate if the velocity profile is flat or if the en tire profile is meas u red. The tu be has a small opening at point 2 which is the stagnation point, i.e., the velocity is zero. A series of holes are drilled around the periphery of the tube at point 3 to measure the static pressure. The pressure at point 2 is called the dynamic pressure and the difference between these two pressures may be measured by the manometer and related to the fluid velocity. In this particular system, we wish to compute the pressure along two streamlines rather than over a control surface, and we Flow nozzle. Boundory loyer (thickness is exoggeroted)
"1"
"2"
·~========::::-
~.
4 6
e105
2
4 6
e o6 1
Th 2
1__
=
for a flow nozzle.
Fig. 8.5-7. Pitot-static tube.
338
Macroscopic Balances-Viscous Effects
Chap. 8
shall use Bernoulli's equation. Application of Eq. 7.4-17 to the streamline between 1 and 2 yieldst PI
Since v2
=
+ !pv~ + pgzi =
P2
+ tpv~ + pgz2
(8.5-18)
O, the pressure at point 2 is P2
= PI
+ !pv~ + pg(zi -
Z2)
(8.5-19)
Sec. 8.5
Flow Rate Measurement
Sharp-crested weir
Flow rates in open channels may the channel over which the fluid m monly used weirs, but we will consi Fig. 8.5-8. It represents a rather e sence of experience and knowledge,
Along the streamline from points 1 to 3 we write PI
+ fpv~ + pgzi =
Pa
+ !pv~ + pgza
(8.5-20)
To continue this analysis, we must know something about the flow field around the tube. We need not guess the flow topology, for experimental and theoretical studies have been carried out 27 indicating there is a thin region (called the boundary !ayer) around the tube where the velocity changes rapidly from the free stream value to zero. At high Reynolds numbers, the boundary 1ayer thickness is small and the pressure difference across it is negligible. Under these conditions, the pressure at point 3 is very nearly equal to the pressure at the static hole at the surface of the Pitot-tube. We take the streamline from points 1 to 3 to be outside the boundary !ayer; thus, (8.5-21) and the pressure at the static hole is given by Pa = PI
+ pg(zi -
Za)
(8.5-22)
Substituting Eq. 8.5-19 into Eq. 8.5-22 and solving for the velocity, we get vi
= J1.[(P2 - Pa) + pg(z2 - za)J .
(8.5-23)
//7//////§//ff//,ij¡
Fig. 8.5-8. S
p
As in the other systems studied in this section, the analysis is not exact and Pitot-static tu bes must be calibrated in terms of the equation, vi=
e J2(p2;
Pa)
(8.5-24)
The gravitation terms have been dropped because they are always negligible. If a Pitot-static tube is properly designed, thf" coefficient e can be very close to unity. The implication is not that the analysis is exact; it simply indica tes that severa1 errors have cancelled. t Remember that Bernoulli's equation neglects viscous effects. 27. H. Schlichting, Boundary Layer Theory, 4th ed. (New York: McGraw-Hill Book Company, Inc., 1955).
attack the problem. Both the mon would seem to be suitable tools fo equation and lea ve the application o Along any streamline we may w p
+ !pv
provided viscous effects are negligi the thickness of the jet at the weir the weir will be small. U nder th specified by applying Eq. 8.5-25 at
ic Balances-Viscous Effects
Chap. 8
tion of Eq. 7.4-17 to the streamline (8.5-18)
(8.5-19)
Pa
+ tpvi + pgza
Sec. 8.5
Flow Rate Measurement
339
Sharp-crested weir
Flow rates in open channels may be measured by a weir, an obstruction in the channel over which the fluid must ftow. There are severa! types of commonly used weirs, but we will consider only the sharp-crested weir shown in Fig. 8.5-8. It represents a rather complicated ftow problem, and in the absence of experience and knowledge, it is not at all obvious how we should
(8.5-20)
know something about the ftow field flow topology, for experimental and out27 indicating there is a thin region the tube where the velocity changes zero. At high Reynolds numbers, the the pressure difference across it is pressure at point 3 is very nearly at the surface of the Pitot-tube. We to be outside the boundary !ayer; thus, (8.5-21)
(8.5-22) and solving for the velocity, we get
(8.5-23) section, the analysis is not exact and terms of the equation, (8.5-24) because they are always negligible. thP coeffi.cient C can be very close analysis is exact; it simply indicates viscous effects. 4th ed. (New York : McGraw-Hill Book
Fig. 8.5-8. Sharp-crested weir.
attack the problem. Both the momentum and mechanical energy balance would seem to be suitable tools for the analysis, but we will use Bernoulli's equation and Ieave the application ofthe macroscopic balances asan exercise. Along any streamline we may write p
+ tpv 2 + pgz =c.
(8.5-25)
provided viscous effects are negligible. If the channel is deep compared to the thickness of the jet at the weir (H ';J> h), the velocity far upstream from the weir will be small. Under these conditions, the constant c. may be specified by applying Eq. 8.5-25 at sorne point far upstream from the weir.
C, = p
+ pgz
(8.5-26)
Macroscopic Balances-Viscous Effects
340
Chap. 8
Since the velocity in the z-direction is negligible there, the pressure is just the hydrostatic pressure, (8.5-27) P =Po pg(H - z)
+
and we see that every streamline has the same constant given by (8.5-28)
C, =Po+ pgH We may now write Eq. 8.5-25 as
p
+ l pv + pgz =Po + 2
pgH
= ..}2g(H- z)
PRO
(8.5-30)
The volumetric flow rate over the weir is given by z ~ L+h
f
Q= b
v ·i
dz
While experiments must be carried 01 this analysis does have value in that i rate on (H - L). In Chap. 9, we w show how practica! flow rate measuri theory, thus providing a more reliab The results presented in this sectio on fluid metering 28- 30 to obtain both and a thorough discussion of the m tions.
(8.5-29)
At the crest of the weir the fluid pressure is unknown; however, the pressure at both the top and bottom of the jet is atmospheric, and the pressure within the liquid should not be too far removed from this value. Applying Eq. 8.5-29 to the streamlines at the crest of the weir and setting p = p 0 , we find
v
Problems
(8.5-31)
z= L
where bis the width ofthe weir. While Eq. 8.5-30 may be a satisfactory expression for the magnitude of the velocity vector, it does not provide us with the x-component of the velocity vector. If we make the approximation (8.5-32)
8-1. To determine the pressure drop-fl<
bundle used in heat transfer e configurations similar to that shown in Fig. 8-1. The tubes are arranged in a triangular pattern with the mean flow directed along the diagonal of the square. Describe how you would interpret the experimental results obtained i such a system-i.e., how would you define the friction factor, what parameters would f be a function of, etc.? Neglect wall effects and assume the tubes are infinitel)i long. 8-2. Derive an integral expression for tube shown in Fig. 8-2. Express
Eq. 8.5-30 may be substituted into Eq. 8.5-31 and the integration carried out to yield (8.5-33) Q = fb..}2g{[H- L]3 i 2 - [H- (L + h)] 3i 2} In practice, the height of the weir L is known, and the depth far upstream H is measured with a depth gauge ; however, the position of the free surface at the weir remains asan unknown. Traditionally, we assume H- (L
+ h)
~
H- L
Q
EJ A1
---
-----.:
(8.5-34)
Flg. S.l. Flow iJ
and introduce a discharge coefficient to obtain (8.5-35) We expect the discharge coefficient to be less than unity, and experiments indicate that (8.5-36)
28. " Flowmeter Computational Handt lication, New York, 1961. 29. " Fluid Meters : Their Theory and New York, 1959. 30. H. W. King and E. F. Brater, Ham Book Company, Inc., 1963).
Balances-Viscous Effects
Chap. 8
negligible there, the pressure is just the (8.5-27)
pg(H- z) same constant given by
(8.5-28)
=Po+ pgH
Problems
341
While experiments must be carried out to determine the discharge coefficient, this analysis does ha ve value in that it predicts the correct dependence of flow rate on (H- L). In Chap. 9, we will analyze the broad-crested weir, and show how practica! flow rate measuring devices can be altered to fit available theory, thus providing a more reliable method of measurement. The results presented in this section are brief; we must refer to handbooks on fluid metering 28 - 30 to obtain both accurate values for discharge coefficients and a thorough discussion of the many metering devices and their applications.
(8.5-29)
is unknown; however, the pressure atmospheric, and the pressure within from this value. Applying Eq. the weir and setting p = p 0 , we find (8.5-30)
(8.5-31) Eq. 8.5-30 may be a satisfactory expresvector, it does not provide us with the we make the approximation (8.5-32) . 8.5-31 and the integration carried out
PROBLEMS 8-1. To determine the pressure drop-flow rate relationship for flow through a tu be bundle used in heat transfer equipment, experiments are performed on configurations similar to that Tube diometer is O shown in Fig. 8-1. The tubes are arranged in a triangular pattern ¿/'¿/'##/$#$!'#/#/& with the mean flow directed along o o o o the diagonal of the square. Describe how you would interpret the Mean o o o experimental results obtained in - - flow such a system-i.e., how would o o o o you define the friction factor, what o o o o o parameters would f be a function of, etc.? Neglect wall effects and //7//////////////////////////// assume the tubes are infinitely Flg. 8-1. Flow through a tube bundle. long. 8-2. Derive an integral expression for the pressure drop in the expanding circular tu be shown in Fig. 8-2. Express !lp as a function of Q, A1 , A2, and f The r
- [H- (L
+ h)]al2}
(8.5-33)
, 1
1
known, and the depth far upstream H the position of the free surface at iiuium1a11 , we assume (8.5-34)
1 1
1 1 1
L,
1
A~~-----------------------'~,,~ Az
Flg. 8-l. Flow in an expanding tube.
_ L)312
(8.5-35)
be less than unity, and experiments
(8.5-36)
28. "Flowmeter Computational Handbook," Am. Soc. Mech. Engrs. Research Pub/ication, New York, 1961.
29. "Fluid Meters: Their Theory and Application," Am. Soc. Mech. Engrs. Paper, New York, 1959. 30. H. W. King andE. F. Brater, Handbook of Hydraulics (New York: McGraw-Hill Book Company, lnc., 1963).
342
Macroscopic Balances-Viscous Effects
cross-sectional area is given by A = A1
+ (A 2
-
Chap. 8
Problems
)(i)
A1
and yo u are to assume that f is constant and the flow is turbulent. 8-3. Water is pumped through a 1-in. diameter pipe 100ft long at arate of 120 in. 3 /sec. Neglecting entrance effects, calculate the pressure drop for a relative roughness ej D of 0.002. 8-4. Calculate the pressure drop for the conditions in Prob. 8-3 if the pipe is replaced by a square duct, 1 in. on a side. 8-5. Crop-dusting of the artichoke fields around Castroville, California, is traditionally done by pilots flying at an altitude of 17 ft. If the particles of insecticide are approximately spherical with an average diameter of 10- 2 cm, anda density of 127lbm/ft, how long does it take them to fall to the ground? If the nearest population center is 1 mi from the fields, what is the maximum wind velocity (directed from the field toward the town) that can be tolerated before the dusting operation must be stopped? 8-6. The flow rate through a flow nozzle was given in Sec. 8.5 by 2(p¡ - P2)
Pipe diometer =0 0 Surge tonk diometer =fJ
Flg. 8-7. Use of a surge t
an approximate solution for the t~nk h'(t) may be obtained. Det maximum damping, and comme Hint: Assume a solution for h'(t h'(t)
where the discharge coefficient Cd had to be determined experimentally. Repeat the derivation retaining the viscous dissipation term in the analysis. Explain what experiments you would perform, and how you would correlate the data to obtain a satisfactory method of predicting flow rates by a flow nozzle. The assumption of flat velocity pro files is a reasonably good one and should be incorporated in your analysis. 8-7. The pump illustrated in Fig. 8-7 has an output given by Q = Q0
If aQ ~ Qo
the friction factor may be taken as a constan t. In the absence of a surge tank, the fluctuations occurring at the process are identical to those at the pump and the process is therefore difficult to control. You are asked to design a surge tank which will reduce the amplitude of the fluctuations by a factor of 10. Set up the problem in a general fashion, locating a surge tank of diameter D1 at a distance rxL (O : : :; rx ~ 1) from the pum p. Neglect minor losses and inertial effects. An exact solution is difficult to obtain; however, if the depth in the surge tank is represented by a steady term and a time-dependent term =
8-8. Apply the momentum and mecha for the flow rate over the shar fully list the assumptions that mu
+ aQ sin wt
where Q 0 is the time averaged flow rate, and aQ represents the amplitude of the volumetric flow rate variations . The frequency of these variations is w.
h
and make use of the approximat1
h0
+ h'(t)
where h'(t)
~
h0
8-9. Determine the water fiow rate fa the siphon shown in Fig. 8-9. Th inlet to the siphon is placed 10 1 below the water leve! and the ow let is even with the bottom of th tank. The pipe diameter is 1 ft, th totallength of the piping is 450 f¡ and the relative roughness is 0.0~ The fittings in the lineare standar 90° elbows. Assume that the levf of the water in the tank i constan t.
Balances-Viscous Effects
Chap. 8
Problems
343
pipe 100ft long at a rate of 120 calculate the pressure drop for a relative conditions in Prob. 8-3 if the pipe is si de. around Castroville, California, is traaltitude of 17 ft. If the particles of with an average diameter of w- 2 cm, does it take them to fall to the ground? · from the fields, what is the maximum toward the town) that can be tolerated stopped? was given in Sec. 8.5 by 2(p¡ - P2)
p[1 - (A 2/A 1) 2]
had to be determined experimentally. viscous dissipation term in the analysis. perform, and how you would correlate of predicting flow rates by a ftow profiles is a reasonably good one and
r
-a)L.---..l•l
and the ftow is turbulent.
\\ ;f'0 Pipe diometer = 0 0 Surge tonk diometer
=0 1
Flg. 8-7. Use of a surge tank to damp pump functions.
an approximate solution for the amplitude of the fluctuations in the surge tank h'(t) may be obtained. Determine the value of oc that will provide the maximum damping, and comment on the effect of increasing the ratio D 1/ D 0 • Hint: Assume a solution for h'(t) of the form h'(t) =a,. sin (wt - 8)
and make use of the approximation
y¡,
=
y-¡;0
(t + ~ h'(t)) ho
8-8. Apply the momentum and mechanical energy balances to derive an expression for the flow rate over the sharp-crested weir discussed in Sec. 8.5. Carefully list the assumptions that must be made to obtain a solution in each case.
+ aQ sin wt and aQ represents the amplitude of The frequency of these variations is w.
In the absence of a surge tan k, are identical to those at the pump to control. You are asked to design a of the ftuctuations by a factor ~r locating a surge tank of diameter the pump. Neglect minor losses and to obtain; however, if the depth in term and a time-dependent term
8-9. Determine the water flow rate for the siphon shown in Fig. 8-9. The inlet to the siphon is placed 1O ft below the water level and the outlet is even with the bottom of the tank. The pipe diameter is 1 ft, the totallength of the piping is 450 ft, and the relative roughness is 0.02. The fittings in the lineare standard 90° elbows. Assume that the level of the water in the tank is constan t. Flg. 8-9. Flow in a siphon.
Macroscopic Balances-Viscous Effects
Chap. 8
Problems
8-10. Explain how you might estimate the totalloss owing to the presencc of a flow nozzle in a pipeline. Hint: Apply both the momentum and mechanical energy balances to a suitable control volume. 8-11. Determine the cros~-sectional area of the jet illustrated in Fig. 8-11 as a function of z. The area of the rounded orifice is A 0 , and you are to assurne that the velocity profile everywhere in the jet is flat. Treat the flow as quasi-steady. Neglect viscous effects and sol ve the problem using both Bernoulli's equation and the momentum balance.
r
p = 68.5 lbm/ft J.L
Ll.t:z::z::~:::z::¡~¡____z =o
¡
:
1
I¡J,II:J.IIr.t.liJ.II~11r.t.IU,II¡.¡,I.J • 1 2
A1 =0.10 ft Flow
!
Sudden contr 6in. I.D. to
5' .L_
Seo wot
Flg. 8-13. Flo
"2"
1
'J .
centipoise
h
8-12. Water at 70°F flows through a Venturi meter as shown in Fig. 8-12. Cavitation occurs when the local pressure falls below the vapor pressure of the liquid, allowing vapor bubbles to form. The vapor Flg. 8-11. Flow through a rounded orifice in a tank. pressure of water at 70°F is 0.36 psia. Calculate the maximum flow rate that this meter can handle without cavitation. "1"
= 1.2
3
:
~~~¡¡wwuu~
1
! i
1
~~·~~
1
i
A2=0.025 ft
/1 (/ti(/! 1111111.1..
2
_......·.:.r:, _ _ 1/T~--
"~mnmm~~
Fig. 8-12. Flow through a Venturi meter.
8-13. Figure 8-13 shows a pipeline system for pumping sea water into a reservoir. The Venturi meter has a discharge coefficient of 0.96 anda throat area onehalf that of the pipe. The pipes are 4-in. and 6-in. I.D. commercial steel, and the manometer fluid is mercury. Determine the horsepower requirement of the pump and the volumetric flow rate. The density of sea water may be taken as 68.5 lbm/ft 3 and the viscosity as 1.2 centipoise. 8-14. A i-in. plastic sphere (p = 38 lbm/ft 3) is released from the bottom of a deep lake. What is the terminal velocity? 8-15. Calculate the horsepower delivered by the pump to maintain a steady flow ofwater in the system shown in Fig. 8-15. All piping is 6-in. I.D. commercia1
Flg. 8-15. Flo·
Balances-Viscous Effecu
Chap. 8
Problems
345
totalloss owing to the presencc of a flow the momentum and mechanical energy
r
p =68 .5 1bm/fl 3 JL
= 1.2
centipoise
Pump
h
L~p::z:ZJ-z=o
Sudden controction from 6in. I.D. lo 4in . I.D.
Flg. 8-11. Flow through a rounded orifice
in a tank.
5' .L...
Seo water Flg. 8-13. Flow in a pipeline.
for pumping sea water into a reservoir. eoeffic.ient of 0.96 and a throat area oneand 6-in. I.D. commercial steel, and the horsepower requirement of rate. The density of sea water may be as 1.2 centipoise. is released from the bottom of a deep the pump to maintain a steady flow
5. AII piping is 6-in . I.D. commercial
Flg. 8-15. Flow in a pipeline.
Macroscopic Balances-Viscous Effects
346
Chap. 8
steel, and all fittings are 90° standard elbows. Assume a value of 0.7 for the contraction coefficient of the 4-in. diameter orifice in the si de of the tan k. 8-16. Electrical transmission towers are placed at 1000-ft intervals, and i-in. diameter cables are strung between them. Determine the drag force on a single cable between two towers if the wind velocity is 60 mph. 8-17. When 15 ft 3 /sec ofwater flow in a 12-in. I.D. pipeline, 85 hp are lost in viscous dissipation for every 1000 ft of pipe. For these conditions, calcula te the head loss, friction factor, and shear stress at the pipe wall. 8-18. Calculate the flow rate of water from the tank illustrated in Fig. 8-18.
l
Water
1 in. 1.0. smooth tubing
lOOft
k----------250ft---------·~! Fig. 8-18
8-19. Calculate the smallest flow rate required to keep the pipeline shown in Fig. 8-19 running full. The pipe is 1-in. I.D. commercial steel. Neglect minor losses and keep in mind the phenomenon of cavitation.
r-30'----j
t
50'
t
t
Water at 70° F
LL
l
100'
j
\
)._ Pump
Flg. 8-19. Flow in a pipeline.
Problems
8-20. Calculate the flow rate through tht submerged orifice in Fig. 8-20 The orifice diameter is 2 in. Esti mate the discharge coefficient.
8-21. Derive an expression for the vol· umetric flow rate in a Ventu meter (see Sec. 8.5), assuming th velocity profile is always paraboli Compare your result with th experimental values of the dis charge coefficient shown in Fig 8.5-2.
ic Balances-Viscous Effects
Chap. 8
elbows. Assume a value of 0.7 for the orífice in the side ofthe tank. at 1000-ft intervals, and i-in. diamDetermine the drag force on a single velocity is 60 mph. .I.D. pipeline, 85 hp are lost in viscous For these conditions, calculate the head at the pipe wall. the tank illustrated in Fig. 8-18.
1 in. /.D. smooth tubing
250ft _ _ _ _.., 8-18
to keep the pipeline shown in Fig. I.D. commercial steel. Neglect minor of cavitation.
t 50'
t
r-30'4
l
lOO'
Problems
8-20. Calcula te the flow rate through the submerged orífice in Fig. 8-20. The orífice diameter is 2 in. Estímate the discharge coefficient.
8-21. Derive an expression for the volumetric flow rate in a Venturi meter (see Sec. 8.5), assuming the velocity pro file is always parabolic. Compare your result with the experimental values of the discharge coefficient shown in Fig. 8.5-2.
347
Air ot 4 psig
Atm. press.
5'
j__ Fig. 8-20. Flow
orifice.
through a subrnerged
Sec. 9.1
Open Channel Flow
9
Uniform Flow
the curvatu re of the free surfacet is sm and rapidly varied ftows, where the cu In the study of ftow in closed con~ fiow rate, pressure drop, and geometry will seek a relationship between the geometry [Q, h(x), b(x), 'l)(x)]. We sha terms of the macroscopic mass, mome and then go on to consider the flow ir This willlead us to equations defining shall analyze rapidly varied ftows, suc
9.1
Uniform Flow
Consider the ftow in a rectangula sluice gate, such as that illustrated in gate, the ftow will be nonuniform ar
The study of open channel flow is primarily directed toward applications in irrigation, ftood control, and water suppty. The analysis of ftow in open channels is more complex than the analysis of flow in closed conduits for two reasons : the existen ce of a free surface complicates the analysis; and the actual channels encountered in practice are comparatively complex. A large portian of the ftow in closed conduits is restricted to pipe flow, where the diameter and relative roughness describe the geometry fairly well. However, in dealing with flow in open channels we may encounter anything from a carefully constructed aquaduct to a grass-lined irrigation ditch. The flow characteristics of the former are fairly well known; however, those of the latter must be studied experimentally for every individual case if accurate results are desired. The study of open channel ftow could easily be the subject of an entire course, and in a single chapter we can only hope to comment on the most important aspects. These ftows may be classified as either steady or unsteady, uniform or nonuniform. With the exception of the solitary shallow-water wave studied in Sec. 9.3, al! the ftows investigated in this chapter will be steady. A uniform ftow is one for which the fluid depth h above the channel bed is constant. Nonuniform flows can be further classified into gradually varied flows, where 348
_j~___ ---·-Fig. 9.1·1. Formati
distance downstream, a balance .bet will be achieved and the flow will be a representative rectangular channel indicate a rather curious situation. B wetted surface is the only force opp leads us to expect the maximum velo
t The term "free surface" denotes an that the tangential stress is zero and the no
Sec. 9.1
9
Uniform Flow
349
the curvature of the free surfacet is small compared to the depth of the fluid, and rapidly varied flows, where the curvature is comparable to the fluid depth. In the study of flow in closed conduits, we sought equations relating the flow rate, pressure drop, and geometry (Q, !l.p, L, r0 ), while in this chapter we will seek a relationship between the flow rate, fluid depth, and channel geometry [Q, h(x), b(x), r¡(x)]. We shall first analyze a simple uniform flow in terms of the macroscopic mass, momentum, and mechanical energy balances, and then go on to consider the flow in a channel with a changing geometry. This willlead us to equations defining gradually varied flows, after which we shall analyze rapidly varied flows, such as the hydraulic jump.
9.1
Uniform Flow
Consider the flow in a rectangular channel formed downstream from a sluice gate, such as that illustrated in Fig. 9.1-1. Just downstream from the gate, the flow will be nonuniform and rapidly varying; ho~ever, at sorne
· y directed toward applications supply. The analysis of flow in open · of flow in closed conduits for two complicates the analysis; and the are comparatively complex. A large is restricted to pipe flow, where the the geometry fairly well. However~ we may encounter anything from a lined irrigation ditch. The flow well known; however, those of the for every individual case if accurate annel flow could easily be the subject r we can only hope to comment on ther steady or unsteady, uniform or solitary shallow-water wave studied · chapter will be steady. A uniform abo ve the channel bed is constan t. into gradually varied flows, where
_j)___ ----------------------Fig. 9.1-1. Formation of a uniform flow.
distance downstream, a balance .between gravitational and viscous forces will be achieved and the flow will become uniform. The velocity profile for a representative rectangular channel is shown in Fig. 9.1-2, and the curves indicate a rather curious situation. Because the drag force occurring at the wetted surface is the only force opposing the gravitational force, intuition leads us to expect the maximum velocity to occur at the centerline on the free
t The term "free surface" denotes an air-water interface, and has the characteristic that the tangential stress is zero and the normal stress is - np 0 •
Open Channel Flow
350
Chap. 9
M ximurri velocity
Sec. 9.1
Uniform Flow
Time-averaging of all the equation. because the open channel fiows of p turbulent. The general macroscopic 1
:J
pv dV
?'"a (t)
z h
+
J
pv(v - w) •
A, (t)
can be dotted with i and applied to t
Because the fiow is uniform, (v!)2
Unes of constont velocity ( Vx)
and Eq. 9.1-6 reduces toa balance o Fl¡. 9.1-l. Velocity profiles in a rectangular channel.
O = pgzAL surface; however, the maximum velocity is located below the free surface. This anomaly appears to result from the secondary flow (indicated in Fig. 9.1-2), which carries the low velocity (vz) fluid upward along the sides and then out over the free surface.
Mass balance
dt
f +f p
dV
p(v - w) • n dA
A,
?'"4 (t)
.
=
O
Assuming that the shear stress at th should be for uniform fiow), Eq. 9.1
(9.1-1)
Defining an average wall shear
stres~
l '~'o=--
J
v·ndA =O
S,
we obtain
O= pg sin (9.1-2)
A,
Application of Eq. 9.1-2 to the control volume illustrated in Fig. 9.1-1 yields the obvious result (9.1-3)
= A1 =
d
(pg)l
and simplifying it for a fixed control volume and incompressible fiow, we obtain
Since A2
f [-
where S is 'the wetted perimeter, b -1 that
Starting with the general macroscopic mass balance,
f!..
+
bh, we obtain
(9.1-4)
Following the procedure used in tC arrange this equation in dimensionle Dimensionless wall shear stress
t The use of the net stress vector is a inasmuch as t 1:, is zero at a free surface.
Open Channel Flow
Chap. 9
Sec. 9.1
Uniform Flow
351
Time-averaging of all the equations used in this chapter is understood, because the open channel flows of practica! importance are almost always turbulent. The general macroscopic momentum balancet
:J
pv dV
7"• (t)
z ¡¡
+
J
pv(v- w) • n dA
=
A. (t)
J
+ t~>
7" • (t)
.9/• (t)
pg dV
J
(9.1-5)
dA
can be dotted with i and applied to the fixed control volume to yield p(v! )2 A - p(v! )1 A
=
pg.,AL
f
+ t~> • i dA
(9.1-6)
.9/
Because the flow is uniform,
(v.~:.. )
(9.1-7) and Eq. 9.1-6 reduces to a balance of surface forces and body forces,
a rectangular channel.
O = pg.,AL located below the free surface. This (indicated in Fig. 9.1-2), along the sides and then out
+ f [-i · npg + i · (n • T)] dA
(9.1-8)
.9/
Assuming that the shear stress at the solid walls is independent of x (as it should be for uniform flow), Eq. 9.1-8 becomes (9.1-9) S
where S is ·the wetted perimeter, b that
+ 2h.
(pg)]
(9.1-1)
=
Once again uniform flow requires
(pg)2
(9.1-10)
Defining an average wall shear stress as (9.1-11)
and incompressible flow, we we obtain
=0
(9.1-2)
O= pg sin O AL-
-r0 SL
(9.1-12)
Following the procedure used in the analysis of closed conduits flow, we arrange this equation in dimensionless form to obtain illustrated in Fig. 9.1-1 yields
Dimensionless wall shear stress
Dimensionless body force
(9.1-3) (9.1-13) (9.1-4)
t The use of the net stress vector is a convenience in analyzing open channel ftow inasmuch as t(:) is zero at a free surface.
Open Channel Flow
Chap. 9
where the hydraulic radius is given by R~
cross-sectional area
bh
wetted perimeter
S
= ----:---:---
n2
116 R~13
Uniform Flow
Flow rate-channel depth calculati uniform flow
Severa! empírica! expressions exist for the friction factor, f The most popular is the work of Manning, 1 which we may write in the form
f =
Sec. 9.1
If the geometry and fluid depth fo of the flow rate is straightforward. If and solve for (v:z), we get
(9.1-14)
The coefficient n is a measure of the roughness, and values for severa! types of surfaces are Iisted in Table 9.1-1. Since f is dimensionless, we must be Table 9.1-1 T ABLE OF ROUGHNESS COEFFICIENTS na Type of Channel
8g sin()
bh
Q=
R:'
116n 2
n, ft' /6
If the volumetric flow rate and channe
Artificial Channels of Uniform Cross Section Sirles and bottorn Iined with well-planed tirnber evenly laid . . . . . . . . . . . . . . . . . . . . Neat cernen! plaster, srnoothest pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cernen! plaster (3 cernen! to 1 sand), srnooth iron pipes . . . . . . . . . . . . . . . . . . . . . . Unplaned tirnber evenly laid, ordinary iron pipes .. . .... .. . .. ........ ... .. .... Ashlar rnasonry, best brickwork, well-laid sewer pipeb . . . . . . . . . . . . . . . . . . . . . . . . Average brickwork, foul planks, foul iron pipes, ordinary sewer pipes after average uneven settlernent and average fouling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Good rubble rnasonry, concrete laid in rough forrns, poor brickwork, heavily incrusted iron pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
From the channel geometry and fluil volumetric flow rate. For a rectangul
0.009 0.010 0.011 0.012 0.013
is to be determined, we use a trialchannel; we multiply Eq. 9.1-15 by bJ R~¡=
and rearrange to obtain
0.015
h=![(2h+b
0.017
Channels Subject to Nonuniforrnity of Cross Section
If we assume a value of h, we can th value of h and test the validity of th1
Excellent clean canals in firrn grave!, of fairly uniforrn section; rough rubble, "drying paving" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.020 Ordinary earth canals and rivers in good order, free frorn large stones and heavy weeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.025 Canals and rivers with rnany stones and weeds . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.03-0.04
yield close agreement between the a¡ fluid depth is determined. More con channel, Iead to a more tedious tria! is straightforward.
a Data fro m L. Marks and T. Baumeister, eds. , Marks' Mechanica/ Englneers' Handbook . 6th ed. (New York : McGraw-Hill Book Company.Inc. 1958). pp. 3- 81. A more tho rough list of values.is given in Ref. 2. b This last value should be used for the previous categories, if doubt exists asto the exccllencc of construction and the maintenance free from slime, rust, or other growths and deposits.
careful to remember that n has units of ftl/ 6 • Note that the value off given by Eq. 9.1-14 is independent of the Reynolds number, indicating that it is only valid for the " rough-pipe" region of flow where inertial effects predominate. A detailed discussion of the drag force is given by Chow.2 l. R. Manning, "On the Flow of Water in Open Channels and Pipes," Trans. Inst. Civ. Engrs. Jreland, 1891,20 :161. 2. V. T. Chow, Open Channel Hy drau/ics (New York : McGraw-Hill Book Cornpany, Inc., 1959), Chap. l.
1
Mechanical energy balance
If the flow is incompressible and may use the results of Prob. 7-2 to 1
:t fr.w(!pv~
dV
+j
= f V • ttn) dA A.,(t)
A
-
Open Channel Flow
Chap. 9
Sec. 9.1
353
Uniform Flow
Flow rate-channel depth calculations for uniform flow
bh
S
for the friction factor, f The most we may write in the form 2
16~ R~ 13
If the geometry and fluid depth for a channel are given, the determination ofthe flow rate is straightforward. Ifwe substitute Eq. 9.1-14 into Eq. 9.1-13 and solve for (v.,), we get
(9.1-14)
uuJ~nn.c:::~s.
Since f
and values for severa! types is dimensionless, we must be
8g sin
··· ··b ····· ··· ······· ····· ·· ·. .
····· ·· ·· ···· ·· ······ ·· ··· ... .
0.015
forms, poor brickwork, heavily
uniform section; rough rubble, 0.020 , free from large stones and heavy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.025 . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.03-0.04 Mechan/cal Englneers· Handbook . 6th ed. (New York : thorough list of values.is gi ven in Ref. 2. if doubt exists asto the excellence of construction and deposits.
ft 116 . Note that the value off given number, indicating that it is of flow where inertial effects predrag force is given by Chow. 2 Open Channels and Pipes," Trans. Inst. Civ. (New York : McGraw-Hill Book Company,
(9. 1-15)
eR:'
3
rectangular channel
(9.1-16)
If the volumetric flow rate and channel geometry are given and the fluid depth is to be determined, we use a trial-and-error solution. For a rectangular channel, we multiply Eq. 9.1-15 by bh, express the hydraulic radius as
and rearrange to obtain
1[
h = - (2h
b
0.017
······· ·· ···· · ··· ···· ... o·· · ···
413
116n2
116n 2
n, ft 1 i 6
pipe .. .. . ..... .. . . ... . . .. . . . ordinary sewer pipes after average
eR 11
From the channel geometry and fluid depth, we may readily compute the volumetric flow rate. For a rectangular channel,
Q = bh
0.009 0.010 0.011 0.012 0.013
8g sin
+ b) (116Q2n2)3/4]2/5 8g sin e
(9.1-17)
If we assume a value of h, we can then use Eq. 9.1-17 to calculate a second value of h and test 'the validity of the assumption. A few trials will quickly yield close agreement between the assumed and calculated values, and the fluid depth is determined. More complex geometries, such as a trapezoidal channel, Iead to a more tedious trial-and-error calculation, but the method is straightforward.
Mechanical energy balance If the flow is incompressible and there are no solid moving surfaces, we may use the results of Prob. 7-2 to writc the mechanical energy balance as
:t J(ipv~ r .(t)
=
dV
+ J(!pv )(v 2
w) • n dA
.A:,(t)
Jv • t7n) dA - Jpcf>v • n dA A , (t)
(9.1-18) É,
Open Channel Flow
354
Chap. 9
Sec. 9.2
Gradually Varied Flow
For the case under consideration, Eq. 9.1-18 quickly reduces to
J(tpv )v · n dA = - J 2
A,
n dA -
PuV •
J pfv • n dA -
9~2
Ev
Gradually Varied Flow
(9.1-19)
A,
Here we have used the arguments presented in Sec. 8.1 to reduce the stress on the entrance and exit to a simple pressure term, -np 9 • Writing the gravitaticnal potential as (9.1-20) cp = g(1](X) Z COS (:1]
In this section, the macroscopic ba illustrated in Fig. 9.2-1. The width an< to a variable fluid depth h(x).t The v shown in Fig. 9.2-1 areexaggerated, bec
+
and expressing the gauge pressure as
Po
=
(9.1-21)
pg[(h - z) cos (:1]
we may combine the pressure and gravitational terms in Eq. 9.1-19 to obtain
-J A,
PuV •
n dA -
J pfv • n dA = - pg J(17 + h cos (:l)v • n dA
(9.1-22)
A,
A,
Application of Eqs. 9.1-19 and 9.1-22 to the control volume indicated in Fig. 9.1-1 gives
tp(v!)2A2 - lp(v!)IAI = -pg[(1J 2 + h 2 cos (:i)(v.,) 2 A2 Inasmuch as the flow is uniform,
-
(17 1
+h
1
cos (:i)(v.,) 1 A1]
(9.1-23)
(v!)2A 2 = (v!)¡A 1 (puv.,)2A2 = (p"v.,)¡A 1
= (zv.,) 1 A 1 (v.,)2 A 2 = (v.,) 1 A 1
mnmmmmmlnm,
The mechanical energy balance reduces to
lx=x
1
1J lx=x,] (v.,)A -
-
=uLM«UU«m~ b(x)
(zv.,) 2 A 2
O= pg[17
dr¡ ton (}=- dx
Ev
(9.1-24)
Dividing by pgQ and noting that
1J '"'="'• we can rewrite Eq. 9.1-24 as
-
1J
'"'="'• =
L sin
(:1
= h1
(9.1-25)
L sin (:1 (9.1-26)
Thus, the viscous dissipation is just equal to the decrease in potentia/ energy for uniform flow. The head loss can be expressed in more familiar terms by substituting sin (:1 from Eq. 9.1-13 to obtain (9.1-27) However, the physical significance of the viscous dissipation is best expressed by Eq. 9.1-26.
~rrTm77777,i7771i77777i~
Fig. 9.2-1. Grad
only if the geometry is changing slowl) channel must be small compared to the of the channel bed must be small comJ fore treat the flow as one-dimensional, 1
t The phrase, "variable fluid depth" is th balances are in order.
Open Channel Flow
Chap. 9
Sec. 9.2
355
Gradually Varied Flow
9.1-18 quickly reduces to · n dA-
Jpcpv •
n dA-
Ev
9-.2
Gradually Varied Flow
(9.1-19)
nted in Sec. 8.1 to reduce the stress on term, -np 9 • Writing the gravita-
+ z cosO]
(9.1-20)
- z) cosO]
(9.1-21)
In this section, the macroscopic balances will be applied to the channel illustrated in Fig. 9.2-1. The width and slope are allowed to vary giving rise to a variable fluid depth h(x).t The variations in channel slope and width shown in Fig. 9.2-1 are exaggerated, beca use the analysis presented here is valid
-
tational terms in Eq. 9.1-19 to obtain -pg
J(r¡ +
h cos O)v · n dA
(9.1-22)
A,
to the control volume indicated in
(v~) 1 A 1
(pgv.,) 1 A 1
b(x)
Plan view
(zv.,) 1A1
(v.,)¡A 1
(9.1-24)
(9.1-25) Cross section
(9.1-26)
to the decrease in potentia/ energy expressed in more familiar terms by
(9.1-27)
viscous dissipation is best expressed
wm7?7777J7777J"7777i77/~ Fig. 9.2-1. Gradually varied flow.
only if the geometry is changing slowly-i.e., the curvature of the walls of the channel must be small compared to the width ofthe channel, and the curvature of the channel bed must be small compared to the fluid depth. We can therefore treat the flow as one-dimensional, even though the streamlines are curved.
t The phrase, "variable fluid depth" is the clue indicating that differential-macroscopic balances are in order.
Open Channel Flow
356
Chap. 9
Sec. 9.2
Gradually Varied Flow
and Eq. 9.2-1 reduces to (v.,)bh
(al
!"'+t."' -
Dividing by D..x and taking the limit D. scopic mass balance, d dx ((v.,
h(x) Side view
Momentum balance
For steady flow, we may dot Eq. 9
Jpv.,v • n dA
Free surfoce al x
=
A,
Starting with the momentum flux term tial control volume to obtain
Perimeler al x
Jpv.,v • n dA = p(v!
Perimeler al x+llx
(b)
f_ p 1"
A,
The body force term yields Flg. 9.2-2. Control volume.
pg.,dV =
-r
A more detailed version of the control volume to be used is shown in Fig. 9.2-2. It is formed by two parallel surfaces at x and x + D..x perpendicular to the x-coordinate.
The area integral of the stress vector
f
i · t<:> dA
~
=-
f
i ·1
~
Directing our attention to the pressur
Mass balance
l. Pu = O, on the free surface; 2. i . n = O, on the bottom of the
Application of Eq. 9.1-2 to this control volume yields
f
f
V•
n dA
A,(z)
where the cross-sectional
=
f
v • n dA
1.,
A,(z)
ar~a
f
+
v • n dA
I.,+A., =
0
(9.2-1)
A,(z)
=
(9.2-2)
bh
-l db , on the sides of dx
Under these conditions, we may writt
of flow, A.(x), may be expressed as A.(x)
3. i. n =
Ji •DPu dA = f Pu dA LA., - j ~
A
A,(z)
Note that v • n must be zero at the free surface for the flow to be steady. Assuming that the flow is one-dimensional, we may write V•D
= ±v.,
At the entrance or exit of the control volurne
(9.2-3)
-
(~
Open Channel Flow
Chap. 9
Sec. 9.2
357
Gradually Varied Flow
and Eq. 9.2-1 reduces to (v,.)bh
Dividing by t:u and taking the limit scopic mass balance,
1
1 1 1 1
1
1 1 1
!"'+t.'" -
d
Side view
-
dx
(v,.)bh
~x
j,. =
(9.2-4)
O
---.. O, we find the differential macro-
((v,.)bh)
= O
(9.2-5)
Momentum balance
For steady flow, we may dot Eq. 9.1-5 with i to obtain
Jpv,.v • n dA = f_ pg,. dV +Ji· t7n> dA -r
A,
(9.2-6)
.!#
Starting with the momentum flux term, we apply this equation to the differential control volume to obtain
J
pv,.v • n dA = p(v!)bh
lzHz -
p(v!)bh
¡,.
(9.2-7)
A,
The body force term yields
J
pg,. dV
= pg,.bh ~x
(9.2-8)
-r
volume to be used is shown in Fig. at x and x + ~x perpendicular to
The area integral of the stress vector may be expressed as
Ji ·
t<:> dA =
.!#
-Ji·
np 11 dA
.!#
+Ji ·
(9.2-9)
(n • 't') dA
.!#
Directing our attention to the pressure term, we note the following: volume yields
(9.2-1)
l. p11 =O, on the free surface; 2. i • n = O, on the bottom of the channel; . db . . db 3. 1 • n = -i-d , on the s1des of the channel, prov1ded~
X
~l.
Under these conditions, we may write may be expressed as
(9.2-2)
Ji •
Dp 11 dA
.!#
J
f
A,(z)
A,(z)
= Pu dA L&z -
P11 dA
L
Pressure forces on the entrance and exit of thc control volume
for the flow to be steady. we may write At the entrance or exit of the control volume
(9.2-3)
(9.2-10) Pressure forces on the sides or the channel
358
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varied Flow
lt should be noted that the last term in Eq. 9.2-10 is easily overlooked, for our attention is generally directed to the forces at the entrance and exit of the control volume. If we further assume that pg is independent of position across the channel, Eq. 9.2-10 takes the form
and applying this result to Eq. 9.2-17, 1ess form of the differential-macroscop
-J
For practica! purposes, gradually vari () is Iess than 10°, and under these con
h(x)
h(x)
f i · npg dA =f pgb dz ""'
"""
pgb dz
1
x+.ix
o
h(x)
1
~xf pg dz
db dx
-
x
o
(9.2-11)
2
dh[g• _ (v.,) ] dx g gh
The viscous shear stress term is easily handled by the definition of 'To given in Eq. 9.1-11
f
i · (n • -r) dA
= -S ~x To
g
¡
p(v!) bh
h(x)
-
o
LHx - f
dh (1 dx
= pg.,bh ~X
pgb dz
l., -
o ~x
R!
t1
g
which Ieads to
h(x)
f pgb dz
Dividing by
lx
cos
o
Collecting the various'terms in the momentum balance yields lx+dx -
=
-g,. = sm fJ
(9.2-12)
J4
p(v~) bh
gh
~
o
(v,.Y
_
and taking the Iimit
e:) ~X ~x ---+
h(x)
f pg dz - S
~X 'To
(9.2-13)
o
We have replaced (v.,) 2 /gh with the Fr the surface profile depends on whether for which NFr = 1 are called critical.fiJ .ftow and NFr < 1 a subcritical jlow.
O, we get
h(x)
p .E_ ((v~)bh) = pg,.bh - !!._ f pgb dz dx dx
)
f
h(x)
+ (db) dx
o
Mechanical energy balance
pg dz -
T0
S
(9.2-14)
o
Starting with Eq. 9.1-19, we expre
Expressing the gauge pressure as
pg = pg. [h(x) - z]
2
2 )
h db + pg.-2 dx
T 0S
(9.2-17)
2
(tpv )v • n dA
= iP<
A.
Using Eq. 9.1-20 to express
e/>
pg
and wri
= pg[(h
we can combine the pressure and gra'
- f pgv•ndA- f pcf>v·ndA
Here we have made use of the fact that
Q = (v,. )bh = constant
(9.2-18)
Differentiating Eq. 9.2-18 or Eq. 9.2-5 to obtain
bh d(v,.) + (v,.)b dh dx dx
f
(9.2-15)
carrying out the indicated integration and differentiation, and using the approximationt (9.2-16) we get d(v,.) d (bh p(v,.)bh - = pg,.bh - pg.dx dx 2
NFr)- NFr(~)b
+ (v,.)h db = dx
O
(9.2-19)
t Chow (Ref. 2, p. 28) indicates that Eq. 9.2-16 may be in error by as muchas 20 per cent; hence, any study of real systems must incorporate the necessary correction factor.
A.
=
A0
Applying this result to the differenti collecting the terms in Eq. 9.1-19, we
Open Channel Flow
Chap. 9
Eq. 9.2-10 is easily overlooked, for our forces at the entrance and exit of the
Sec. 9.2
Gradually Varied Flow
359
and applying this result to Eq. 9.2-17, we can obtain the following dimensionless form of the differential-macroscopic momentum balance, dh[g• _ (v.,?J dx g gh
(9.2-11)
_ (v.,) (~) db =
2
gh
b dx
g., _ T 0 S g pgbh
(9.2-20)
For practica! purposes, gradually varied fiows are restricted to cases where O is less than 10°, and under these conditions g. =coso~ 1 g
· uD -g., = sm
(9.2-12)
g
~
tan
= - (dr¡) dx
{)
which leads to balance yields dh (1 - NFr)dx h(x)
)
tlxf Pu dz - S tlx To
(9.2-13)
-o, we get dz
+ (~~)
NFr(~)db = b dx
(dr¡ dx
+
(9.2-21)
ToS )
pgbh
We have replaced (v.,)2/gh with the Froude number, so that it becomes clear the surface profile depends on whether NFr is greater or less than one. Flows for which NFr = 1 are called critica/ fiows; NFr > 1 represents a supercritica/ jlow and NFr < 1 a subcritica/ fiow.
h(x)
f
Pu dz -
ToS
(9.2-14)
o
Mechanical energy balance
Starting with Eq. 9.1-19, we express the kinetic energy flux termas
J
(9.2-15) and differentiation, and using the (9.2-16)
2
(ipv )v • n dA = ip(v!>AI
"'
Using Eq. 9.1-20 to express 4> and writing the gauge pressure as
Pu (9.2-17)
(9.2-22)
- ip(v!>AI x+4x
A,
=
pg[(h - z) cos O]
(9.2-23)
we can combine the pressure and gravitational terms in Eq. 9.1-19
-JPuV • n dA - Jptf>v • n dA = - pgJ(r¡ + h cos 6)v • n dA (9.2-18)
A,
A,
(9.2-24)
A,
Applying this result to the differential volume element in Fig. 9.2-2, and collecting the terms in Eq. 9.1-19, we get db + (v.,)h=O dx
(9.2-19)
.2·16 may be in error by as muchas 20 per the necessary correction factor.
iP(
- (v;)AI ) = - pg[(r¡ x
+ h cos O)(v.,)AI x+4x
- (r¡
+ h cos O)(v.,)AIJ
-
Ev
(9.2-25)
Open Channel Flow
360
Chap. 9
Dividing by !::u and taking the limit ó.x -- O, we have the differential macroscopic mechanical energy balance
iP .!!._ (
dx
+ h cos e)(v.,)A] - e,
(É")
Gradually Varied Flow
to eliminate the velocity gives
(9.2-26)
where • 1'1me,=
Sec. 9.2
(9.2-27)
Az-o Ó.x
Assurning that (9.2-28)
For a given volumetric flow rate and s h. Exluding negative and complex valu values of h or none. The solutions are for each value of Q2/2gb 2 there is a mil is no solution for h. By inspection of comes Iarge the fluid depth may tend t
and using Eqs. 9.2-18 and 9.2-19, we can rearrange Eq. 9.2-26 to obtain
!!_ [ (v.,)2 dx
2g
+ h] = -
(d1] dx
+ ~)
(9.2-29)
pgQ
6
providing e is less than 10°. Referring to Eq. 9.1-26, we see that the viscous dissipation per unit length e, for uniform ftow is given by
~=~= pgQ
sin
e=
- d7]
pgQL
for uniform flow
(9.2-30)
dx
4
lf the viscous dissipation for gradually varied flow is very nearly the same as for uniform flow, we may expect the right-hand side of Eq. 9.2-29 to be small. Integration of Eq. 9.2-29 gives 2
(v.,) 2g
+h=
-J
5
(d1] dx
+ ~) dx + constant
h
(9.2-31)
pgQ
which is generally written
(v.l
--
2g
+ h-E Bp
(9.2-32)
E 8P is known as the specific energy, and it should be very nearly constant for gradually varied flow, although the changes in the geometry and roughness of the channel will certainly lead to small variations. Ifthe flow rate and channel width are given and values for ,.o are available, either Eqs. 9.2-21 and 9.2-18 or Eqs. 9.2-29 and 9.2-18 can be used to solve for the two unknowns, (v.,) and h. In general, the process requires numerical integration, although analytic methods may be used to obtain approximate solutions for simple cases. The mechanical energy balance is not as powerful a too! for solving open channel ftow problems as it was for solving closed conduit problems; however, it can be used to good advantage to illustrate sorne important characteristics of open channel ftow. Combination of Eqs. 9.2-18 and 9.2-32
Fig. 9.2-3. Fluid depth as a
We may determine the value of h fe the derivative dEsp/dh equal to zero a1 dEsp = 1 _ dh
This condition corresponds to what i energy, fluid depth, and velocity all a:
Open Channel Flow --+-
[('17
Chap. 9
O, we have the differential macro-
+ h cos O)(v.,)A] - e,
Sec. 9.2
Gradually Varied Flow
361
to eliminate the velocity gives
º2
- -2 2
(9.2-26)
2gb h
+ h = E sp
(9.2-33)
For a given volumetric flow rate and specific energy, Eq. 9.2-33 is a cubic in
(!:)
(9.2-27)
(9.2-28) rearrange Eq. 9.2-26 to obtain
h. Exluding negative and complex values of h, this equation yields either two
values of h or none. The solutions are shown in Fig. 9.2-3, where we see that for each value of Q2/2gb 2 there is a mínimum value of Esv below which there is no solution for h. By inspection of Eq. 9.2-33, we can see that as E sp becomes large the fluid depth may tend toward either zero or Ew
(9.2-29) Eq. 9.1-26, we see that the viscous flow is given by
= - d17
for uniform flow
(9.2-30)
dx
flow is very nearly the same as side ofEq. 9.2-29 to be small.
riL-Jitar:tu
~) pgQ
dx
+ constant
h
(9.2-31)
(9.2-32) should be very nearly constant for in the geometry and roughness of
Esp
Fig. 9.2-3. Fluid depth as a function of specific energy.
given and values for r 0 are available, and 9.2-18 can be used to solve the process requires numerical be used to obtain approximate as powerful a too! for solving open solving closed conduit problems; to illustrate sorne important · of Eqs. 9.2-18 and 9.2-32
We may determine the value of h for the mínimum value of E sp by setting the derivative dE5 vfdh egua! to zero and solving for h. (9.2-34) This condition corresponds to what is called critica! flow, and the specific energy, fluid depth, and velocity all assume their critica! values. Expressing
362
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varied Flow
Eq. 9.2-34 in terms of the velocity, we find the critica! depth h. to be h = (v.,)~ e
(9.2-35)
g
----
and the critica! velocity is (9.2-36)
Substitution of these results into Eq. 9.2-33 indicates that the critica! specific energy is always three halves the critica! depth (9.2-37)
In terms of the Froude number, we note that Eq. 9.2-35 gives Flg. 9.2-4. Establishm
for critica! flow
(9.2-38)
There are perhaps two reasons why the condition NFr = 1 is designated as the critica! ftow. Ifwe remember that Eqs. 9.2-29, 9.2-30, and 9.2-31 indicated the specific energy was only approximately constant, we can certainly expect small variations in Esp from changes in geometry and channel roughness. At the critica! condition, the slope of h versus E 8 P is infinite; hence, small variations in E 8 P lead to large variations in h and wave motion is likely to occur. For this reason, designers try to avoid such ftows. A second reason for designating NFr = 1 as the critica! ftow is that it separa tes two fairly distinct ftow regimes. When NFr < 1, the fiow is subcritica/ or tranquil and disturbances can travel upstream; thus, a change in downstream conditions can alter the ftow upstream. When NFr > 1, the ftow is supercritica/ or rapid, and downstream disturbances are not propagated upstream. The speed at which a disturbance is propagated will be discussed in Sec. 9.3.
As an example of the application of the equations developed in this section, we will investigate the gradually varied fiow indicated in Fig. 9.2-4 to determine the length required for establishing uniform ftow. This example will also serve to introduce the student to the variety of surface profiles he may encounter for any particular fiow system. We start with the differential momentum balance, expressing the Froude number in terms of the volumetric fiowrate and restricting the analysis to a rectangular channel of constant width.
2 3
r0
= 14.5p
If we further restrict the analysis to ver¡ the approximations
Under these conditions, we can combir
dh(1- ~)=sil 2 3 dx
gb h
We have replaced -dr¡fdx with sin O. T simplified somewhat by noting that
Establishment of uniform flow
dh(l~)- -(dr¡ + ~) dx gb h dx pgbh
Assuming that the Manning formula ftows, we can combine Eqs. 9.1-13 and! of the volumetric ftow rate to obtain
º2 (gbº2h
gb 2 =
2
because Q 2 fgb 2h~ is the Froude number • to one. Equation 9.2-41 can now be an
~~[1- (~n =sin o
If we designate hnt as the fluid depth f (9.2-39)
t The subscript n is used because the de¡ called the normal depth.
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varied Flow
363
find the critica! depth he to be (v.,)~
(9.2-35)
g
-- --
(9.2-36)
indicates that the critica! specific depth (9.2-37)
that Eq. 9.2-35 gives Fig. 9.2--4. Establishment of uniform ftow.
(9.2-38)
the condition NFr = 1 is designated as s. 9.2-29, 9.2-30, and 9.2-31 indicated constant, we can certainly expect geometry and channel roughness. At versus Esp is infinite; hence, small in h and wave motion is likely to avoid such flows. = l as the critical flow is that it When NFr < l, the flow is subtravel upstream; thus, a change in flow upstream. When NFr > 1, the disturbances are not propagated is propagated will be discussed in
Assuming that the Manning formula can be applied to gradually varied flows, we can combine Eqs. 9.1-13 and 9.1-14 and express the result in terms of the volumetric flow rate to obtain To
=
Q2n2
(9.2-40)
14.5p 2 2 1/ 3 b h R,.
If we further restrict the analysis to very wide channels, b the approximations
> h, we can make
R,.~h s~b
Under these conditions, we can combine Eqs. 9.2-39 and 9.2-40 to find
dh( 1 -
-
dx
Q2 )
~
gb h
. Q2n2 = sm () - 14.5 2 1013
gb h
(9.2-41)
We have replaced -dr¡/dx with sin O. The left-hand side of this result may be simplified somewhat by noting that of the equations developed in this varied flow indicated in Fig. 9.2-4 · uniform flow. This example to the variety of surface profiles he system. balance, expressing the Froude and restricting the analysis to a
-(dr¡ dx
+ ~) pgbh
Q2 -
gb2 -
(L)ha- ha gb2h~
e -
e
(9.2-42)
because Q 2jgb 2h~ is the Froude number for critica! flow and is therefore equal to one. Equation 9.2-41 can now be arranged in the form
dh[1(he)3] =sin ()(1- 14.5Q2n2 ) dx h gb h 1 sin () 2 10 3
(9.2-43)
If we designate hnt as the fluid depth for uniform flow (h-+ hn as x-+ co), (9.2-39)
t The subscript n is used because the depth under uniform ftow conditions is often called the normal depth.
364
Open Channel Flow
Chap. 9
Sec. 9.2
we may rearrange Eq. 9.1-16 to obtain (9.2-44) Our original differential momentum balance may now be written in compact forro dh dx [ 1 -
(hhe)3] = sin e[1 - (hhn)10/3]
(9.2-45)
which is to be sol ved subject to a boundary condition of the type B.C. 1:
x=O
(9.2-46) Critic
Surface profiles
Before going on to obtain an approximate solution to Eq. 9.2-45, we will discuss the possible surface profiles this equation may describe. Our purpose is not to acquaint the student with the nine specific surface profiles associated with the flow downstream from a sluice gate, but rather to drive borne the point that flows in open channels can assume many possible forros. In his discussion of gradually varied flow, Chow (Ref. 2, p. 222) lists severa! dozen commonly encountered surface profiles. Thus, in analyzing flow in open channels it is extremely important to have knowledge of experimentally observed surface profiles, for it is especially difficult to "guess the flow topology" if the various possibilities are unknown. For a given volumetric flow in a given channel, it should be intuitively obvious that the normal depth hn will decrease with increasing slope. This idea is expressed quantitatively in Eq. 9.2-44 and leads us to classify channels as mild, i.e., having a slope small enough so that the normal depth is greater than the critica! depth, (hn > he), critica/ (hn = he), and steep (hn < he). These conditions are illustrated in Fig. 9.2-5, and the flow at the critica! slope is wavy or undular in agreement with our previous comment on the instability of critica! flow. We now wish to examine Eq. 9.2-45 for mild, critica!, and steep channels downstream and upstream from a sluice gate. For mi!d channels, hn > he, and we must consider three possibilities in the gradually varied flow region. Ml. h > hn >he, subcritical flow; M2. hn > h > he, subcritical flow; M3. hn > he > h, supercritical flow.
Steep
Fl¡. 9.2-5. Classifi
Examining Eq. 9.2-45, we find the foil Ml. dhfdx M2. dhfdx M3. dhfdx
> O, backwater curve; < O, drawdown curve; > O, backwater curve.
The condition that h increases with in called a backwater curve, while a dect The three profiles for mild channels are approaches the critica} depth, the profil as Eq. 9.2-45 indicates, the flow cann< The Ml backwater curve is the typc dam (or the sluice gate shown in Fig curve is important to reservoir desig
Open Channel Flow
Chap. 9
Sec. 9.2
365
Gradually Varied Flow
---
(9.2-44) nce may now be written in compact
(9.2-45) condition of the type
x=O
(9.2-46)
solution to Eq. 9.2-45, we will equation may describe. Our purpose specific surface profiles associated gate, but rather to drive home the assume many possible forms. In his (Ref. 2, p. 222) lists severa! dozen Thus, in analyzing flow in open have knowledge of experimentally difficult to "guess the flow unknown. channel, it should be intuitively decrease with increasing slope. This and Ieads us to classify channels so that the normal depth is greater (hn = he), and steep (hn < he). and the flow at the critica! slope previous comment on the instability Eq. 9.2-45 for mild, critica!, and from a sluice gate. consider three possibilities in the
VAIIllla'""
........ ....
.........
Steep
.............
........ .............................
Fl1. 9.1-5. Classification of channels.
Examining Eq. 9.2-45, we find the following restrictions on dhfdx: Ml. dhfdx M2. dhfdx M3. dhfdx
> O, backwater curve; < O, drawdown curve; > O, backwater curve.
The condition that h increases with increasing distance down the channel is called a backwater curve, while a decrease in h is called a drawdown curve. The three profiles for mild channels are illustrated in Fig. 9.2-6. As the depth approaches the critica! depth, the profile is drawn with a dashed line, because, as Eq. 9.2-45 indicates, the flow cannot be gradually varied as h --+-he. The Ml backwater curve is the type of profile that exists upstream from a dam (or the sluice gate shown in Fig. 9.2-4). Knowledge of the backwater curve is important to reservoir design, because the profile líes above the
366
Open Channel Flow
---------~---- ~--
MI
-..._....._
-------~
----
-- ----
---------------__------- ---.
__
-_-:~
Chap. 9
......
--
............_.............._
Sec. 9.2
Gradually Varied Flow
The C 1 and C2 backwater curves o< may exhibit hydraulic jump. The C3 ba from a sluice gate, and a mild, or undull supercritical to critica! flow.
......__---- , - - - - '
~ --·- -----... --<:......__
~ --......
Flg. 9.2-6. Surface profiles for mild channels.
normal depth of the channel and the flooding behind a dam may extend beyond a plane established by the depth of the water in the reservoir. The M3 backwater curve occurs downstream from a sluice gate. As the critica! depth is approached a sudden transition from supercritical to subcritica! flow takes place. This transition is known as a hydraulic jump. The M2 drawdown curve does not occur for the configuration shown in Fig. 9.2-4. However, it can occur when a change in channel slope from mild to critica! takes place. For critica! channels, h.. = he, and again we have three possibilities in the gradually varied flow region. Cl. h >h .. = he, subcritical ftow; C2. h11 = he = h, critica! ftow; C3. h 11 = he > h, supercritical ftow. Examination of Eq. 9.2-45 for this channel indicates: Cl. dhfdx >O, backwater curve; C2. dhfdx > O,t backwater curve; C3. dhfdx > O, backwater curve. The three profiles for critica! channels are illustrated in Fig. 9.2-7. t This result is easily obtained by application of I'Hospital's rule to Eq. 9.2-45; however, the actual existence of such a profile is open to question.
__
------- ---~, , -
Fig. 9.2-7. Surface profilc
F or steep channels, he > h11 , and th SI. h > he> h.., subcritical flow; S2. he > h > h.., supercritical flow; S3. he > h 11 > h, supercritical flow. These conditions lead to the following SI. dhfdx > O, backwater curvt:; S2. dhfdx < O, drawdown curve; S3. dh fdx > O, backwater curve.
These surface profiles are illustrated in The Sl backwater curve is the type t The S2 and S3 profiles can occur downs1 being similar to that illustrated in Fig.
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varied Flow
367
The Cl and C2 backwater curves occur upstream from a dam, and both may exhibit hydraulic jump. The C3 backwater curve may exist downstream from a sluice gate, and a mild, or undular, jump occurs at the transition from supercritical to critica! flow.
for mild channels.
flooding behind a dam may extend of the water in the reservoir. from a sluice gate. As the transition from supercritical to subís known as a hydraulic jump. occur for the configuration shown in a change in channel slope from mild again we have three possibilities in the
Fig. 9.2-7. Surface profiles for critica! channels.
For steep channels, he> hm and the possible types of flow follow: SI. h >he > hm subcritical flow; S2. he > h > hm supercritical flow; S3. he > hn > h, supercritical flow. These conditions Iead to the following surface profiles:
indicates:
are illustrated in Fig. 9.2-7. ofi'Hospital's rule to Eq. 9.2-45; howto question.
SI. dhfdx > O, backwater curvt:; S2. dhfdx < O, drawdown curve; S3. dhfdx > O, backwater curve. These surface profiles are illustrated in Fig. 9.2-8. The Sl backwater curve is the type that may occur upstream from a dam. The S2 and S3 profiles can occur downstream from a sluice gate, the S2 profile being similar to that illustrated in Fig. 9.2-5.
368
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varled Flow
allows us to write Eq. 9.2-45 as
To extractan analytic solution from mation ][10
If t < H < 2, this approximation w however, it is not excessive when coro in the use ofEqs. 9.2-16 and 9.2-28 a factor. Simplifying Eq. 9.2-47, separatin
where
e is the constant of integrat
B.C. 1':
H=H0
The integral in Eq. 9.2-49 is available condition 1', is
Fig. 9.2-8. Surface profi.les for steep channels.
Once again, the point to be made in this discussion is that a variety of surface profiles are possible for a given geometrical configuration and it is necessary to have sorne idea what to expect before plunging into an analysis. Returning now to Eq. 9.2-45, we wish to extractan approximate solution for the approach to uniform flow downstream from a sluice gate. The fl.ow at the sluice gate cannot be classified as gradually varied; however, at a short distance downstream fwm the gate the variations in fluid depth are more gradual and Eq. 9.2-45 is satisfactory. Defining the dimensionless quantities H
= -h ,
dimensionless fluid depth
hn
X = x sin O hn
dimensionless length
'
fJl = he , ratio of critical to normal depth hn
+ j_ [tan_1 (2H + 1)
J3
J3
F or practica! purposes, we can assu when h is within l per cent of h.,. or ll required to achieve uniform ftow · designated by L,.
Entrance lengths
For mild channels, only the M3 gate, and a hydraulic jump occurs ir type, 9l < 1, and the maximum entr because the term in braces ({ }) is al
3. H. B. Dwight, Tables of Integrals a. Macmillan Company, 1961), p. 42.
Open Channel Flow
Chap. 9
Sec. 9.2
Gradually Varied Flow
369
allows us to write Eq. 9.2-45 as 3
dH( 1 - Ha [Jt ) dX =
(
1 ) 1 - Hto¡a
(9.2-47)
To extractan analytic solution from this equation we must make the approximation ¡pota~ na (9.2-48) If i < H < 2, this approximation will lead to an error of about 25 per cent; however, it is not excessive when compared to the percentage of error incurred in the use of Eqs. 9.2-16 and 9.2-28 and the Manning formula for the friction factor. Simplifying Eq. 9.2-47, separating variables, and integrating, we get
f(~3-=-~ ) 3
where
dH = X
e is the constant of integration to
B.C. 1':
+e
(9.2-49)
be determined by application of
X= O
(9.2-50)
The integral in Eq. 9.2-49 is available, 3 and the result, after applying boundary condition 1', is for steep channels.
H2)
Ho)2]
(H- Ho) +(&la- 1){In [( 1 + H + (16 1 + H 0 + H~ 1 - H
(9.2-51) in this discussion is that a variety of geometrical configuration and it is before plunging into an analysis. sh to extractan approximate solution m from a sluice gate. The flow gradually varied; however, at a short variations in fluid depth are more Defining the dimensionless quantities
For practica! purposes, we can assume that uniform flow has been reached when h is within 1 per cent of h .. or H is within 1 per cent of unity. The length required to achieve uniform flow will be caiied the entrance length and designated by L •.
Entrance lengths For mild channels, only the M3 profile occurs downstream from a sluice gate, anda hydraulic jump occurs in the neighborhood of H = 81. For this type, ,tjt < 1, and the maximum entrance length will be obtained for 91--+ 1 because the term in braces ({}) is always positive. Letting 81 = 1 to obtain
to normal depth
3. H. B. Dwight, Tables of Integrals and Other Mathematica/ Data (New York: The Macmillan Company, 1961), p. 42.
Open Channel Flow
370
Chap. 9
Sec. 9.3
The Solitary Wave
an upper bound gives
L = (0.99- H 0)h., sin 8 • and an order of magnitude estimate of the entrance length is
(9.2-52)
L.=o(~) sm O
(9.2-53)
Since h., is generally on the order of feet, and sin O may be 0.01 or less, the entrance length is on the order of 100ft or more for mild channels. For critical channels, fJt = l and the entrance length is given by Eq. 9.2-52. For steep channds, we see from Fig. 9.2-8 that there are two possible profiles leading to a uniform flow.: S2. fJt S3. fJt
~
H ~ 1;
~
1
~
H.
For these two limiting cases we will take H 0 ~ 1 and H 0 obtain; S2. L.= (h.,fsin O) [1.44fft3 S3. L. = (h.,fsin O) [2.02fft3
+ (1.01 + 0.99].
- H 0)];
< 1, respectively, to (9.2-54a) (9.2-54b)
In bath cases, the en trance length can be hundreds of feet long, and it should be apparent that uniform flow is the exception and varied flow the rule for most practical cases.
9.3 The Solitary Wave To understand sorne ofthe phenomena encountered in open channel flow, we must determine the speed at which a disturbance to the surface profile moves. We first consider a disturbance caused by a movable wall which boun4s an initially quiescent body of water. This process is illustrated in Fig. 9.3-1, and a sketch of the control volume to be u sed in the analysis is shown in Fig. 9.3-2. We assume that the wave profile is preserved and choose a differential control volume that moves with sorne specific portion ofthe wave. Note that the control volume is not a material volume, for it is the profile which moves, not the water. In this respect, Watts' has written that: "Buddhism has frequently compared the course of time to the apparent motion of a wave, wherein the actual water only moves up and down, creating the illusion of a 'piece' of water moving over the surface." 4. A. W. Watts, The Way ofZen (New York: Pantheon Books, Inc., 1957).
Flc. 9.3-1. Gener:
r------- - - flh
ho
h(x}
~"""'"~'' z
Lx
Flc. 9.3-2. Moving control vol
Open Channel Flow
Chap. 9
Sec. 9.3
371
The Solitary Wave
---
(9.2-52)
(9.2-53)
' and sin () may be 0.01 or Iess, the or more for mild channels. the entrance Iength is given by Eq. 9.2-8 that there are two possible
H0
~
1 and H0
~
1, respectively, to (9.2-54a) (9.2-54b)
hundreds of feet long, and it should and varied flow the rule for
Flg. 9.3-J. Generation of a solitary wave.
encountered in open channel flow, a disturbance to the surface profile caused by a movable wall which . This process is illustrated in Fig. to be used in the analysis is shown profile is preserved and choose a sorne specific portion ofthe wave. material volume, for it is the profile Watts' has written that:
Control volume
r----------flh
1----c ho
h(x)
the course of time to the apparent water only moves up and down, moving over the surface." : Panthcon Books, Inc., 1957).
Flg. 9.3-1. Moving control volume for analysis of the solitary wave.
372
Open Channel Flow
Chap. 9
Sec. 9.3
The Solltary Wave
Neglecting viscous effects and app, in Fig. 9.3-2, we bave
Mass balance
d
We begin the analysis with the macroscopic mass balance
~t Jp dV +Jp{v 7'"0 (t)
p dx
(9.3-1)
w) • n dA = O
Expressing the gauge pressure ast
A,(t)
pg=
and assume that w is a constant vector such that
w.,=c,
and carrying out the integration it (9.3-2)
w11 = w. =O
The mass in the control volume is a constant so that Eq. 9.3-1 immediately reduces to
J
p(v - w) • n dA
I.,+Az
+
A,W
J
p{v - w) • n dA
1
=O
d [(v.,(v., dx
p-
Assuming that the velocity profil
(9.3-3)
((v.,)-e)
A,W
Here we have used the fact that
Use of Eq. 9.3-7 to eliminate d(v.,
(v - w) • n
= O
h(z)
(v.,- e) dz
o
1
z+Az
-J
ll(z)
(v., - e) dz
o
1
.,
=O
(9.3-5)
d
dx [((v.,) - c)h] =O
(9.3-6)
+ (v.,) dh =
(9.3-7)
which we may also write, dx
e dh
dx
Momentum balance
The momentum of the fluid in the control volume will be a constant, so the x-component of Eq. 9.1-5 becomes
J pv.,(v- w) • n dA= Ji· [-np., + n ·~]dA ..t,(t)
.w
JI(
or
e=
Provided the amplitude of the wa depth !1h ~ h0, the fluid velocity (v.,) ~e, and Eq. 9.3-14 reduces
e
where the density has been removed because the flow is incompressible. Dividing by !1x, expressing the integrals in terms of average velocities, and taking the limit !1x --+ O, we get ·
h d(v.,) dx
(e-
(9.3-4)
at the free surface
Note that Eq. 9.3-4 does not imply that the control volume moves with the fluid (w = v), it simply means that the control surface always coincides with the free surface. Evaluation of the terms in Eq. 9.3-3 gives
f
[(v.,(v.,-
(9.3-8)
The result of our analysis app tions-i.e., the wave profile is pt speed of a particular point on tt constant. However, our analysis of waves m channels, for observ faster than the leading or trailing in front of the crest. However, if over the wave, and a reasonable :
t This equation assumes that the hJ the variations caused by fluid motion i t There is sorne tradition associate• time rate of change of position of the w word speed for v = .,¡;:-;: However, misinterpret. 5. J. J. Stoker, "The Formation o 1 :l.
Open Channel Flow
Chap. 9
Sec. 9.3
373
The Solltary Wave
Neglecting viscous effects and applying this equation to the control volume in Fig. 9.3-2, we bave /0(.,)
ros,:oo:tc mass balance
p ddx [(v.,(v., - e))h] = - ddx (9.3-1)
(9.3-9)
o
Expressing the gauge pressure ast
pg = (9.3-2)
Jpg dz
(9.3-10)
pg[h(x) - z]
and carrying out the integration in Eq. 9.3-9, we find
p ~ [(v.,(v.,- e))h] dx
stant so that Eq. 9.3-1 immediately
= -pgh dh
dx
(9.3-11)
Assuming that the velocity profile is flat, and using Eq. 9.3-6, we get (9.3-3) ((v.,)- e)h d(v.,)
= -gh dh
dx
dx
(9.3-12)
Use of Eq. 9.3-7 to eliminate d(v.,)fdx leads us to (9.3-4) the control volume moves with the surface always coincides with in Eq. 9.3-3 gives (9.3-5)
( e-
(v.,) ) 2
e= (v.,) ±
(9.3-6)
(9.3-7)
(9.3-13)
../gh(x)
(9.3-14)
Provided the amplitude of the wave is small compared to the quiescent fluid depth !l.h ~ h0 , the fluid velocity will be small compared to the wave speed~ (v.,) ~e, and Eq. 9.3-14 reduces to
e= the flow is incompressible. in terms of average velocities, and
= gh(x)
or
±../gh(x)
The result of our analysis apparently contradicts one of the first assumptions-i.e., the wave profile is preserved- for Eq. 9.3-15 indicates that the speed of a particular point on the profile depends upon h(x), which is not constant. However, our analysis does predict qualitatively the real behavior of waves m channels, for observation5 indicates that the wave crest travels faster than the leading or trailing edge and that the wave tends to get steeper in front of the crest. However, if !l.h ~ h0 , the speed will change only slightly over the wave, and a reasonable approximation for small amplitude waves is
e=±~ ntrol volume will be a constant, so
• [-np,
b
+ n •-r) dA
(9.3-8)
(9.3-15)
(9.3-16)
t This equation assumes that the hydrostatic pressure variations are large compared to the variations caused by fluid motion in the z-direction. t There is sorne tradition associated with the use of the word ce/erity to describe the time rate of change of position of the wave profile, reserving the word velocity for v and the word speed for v = VV:V. However, the term wave speed or wave velocity is difficult to misinterpret. 5. J. J. Stoker, "The Formation of Breakers and Bores," Comm. Appl. Math., 1948 1 :l.
37-4
Open Channel Flow
Chap. 9
We must clarify the assumption associated with Eq. 9.3-10. Expressing the pressure as hydrostatic is satisfactory only if pv!fh0 is negligible compared to pg (inertial effects are therefore small compared to gravitational effects). A careful study of wave motion• indicates that this assumption is satisfied when the wave length A. is long compared to ~he fluid depth. Such waves are often called shallow-water waves, and the wave speed is given correctly by Eq. 9.3-16. Waves for which the wave length is less than the fluid depth are called gravity waves, and their speed is given by
e= ±
/gi,
-J2;
gravity waves (A< h,}
Sec. 9.3
The Solltary Wave
.fih-
(9.3-17)
We see that gravity waves move ata slower velocity than shallow-water waves. If the wave length is very short-say, less than 1 cm-the wave velocity is governed by surface tension effects and these waves are called ripp/es. Their wave speed is given by e --
±
!5'"pA.(f '
ripples (A < 1 cm}
(9.3-18)
Both Eqs. 9.3-17 and 9.3-18 are restricted to the condition that the wave amplitude is small compared to the wave length, Ah < A.; thus, the results are applicable only for small disturbances. An important point to be cleared up before we discuss the result derived for shallow-water waves is the sign ± associated with the wave speed. There is nothing in the analysis which tells us whether the wave generated in Fig. 9.3-1 will travel to the left or to the right, but there is little doubt in our minds that it will move to the right. One might ask, "If our analysis is correct, why shouldn't it predict the direction in which the wave moves ?" The answer is that one must investigate carefully the actual generation phenomena in order to predict the direction of wave propagation. Stoker7 has presented a thorough treatment of this problem indicating that a disturbance always propagates into the "region of quiet," i.e., the undisturbed region. In our problem the undisturbed region is to the right of the movable wall; hence, the wave moves to the right. An analysis of wave propagation in a flowing channel is similar to the development presented here and Eq. 9.3-14 is again obtained with the velocity (v,.) being the normal flow ve1ocity. It is thus indicated that a disturbance plus the stream may propagate downstream at a velocity equal to velocity, and upstream ata velocity equal to minus the stream velocity. This situation is illustrated in Fig. 9.3-3 where a disturbance has been created by momentarily immersing a gate into the stream.
Jih
Jih
6. L. M. Milne-Thomson, Theoretica/ Hydrodynamics, 4th ed. (New York: The Macmillan Company, 1960}, Chap. 14. 7. 1. 1. Stoker, Water Waves (New York: Interscience Publishers, lnc., 1957), p. 303.
Fig. 9.3-3. Propagati
The interesting case to consi is given by Eq. 9.2-36 as Thus, for critical flow a disturb nel; for subcritical flow (such a• may propagate upstream; and ~ carried downstream. Note tha~ waves because they have a small may propagate upstream in a e¡
Thus, if the fluid depth is Ü(fe Such small waves are rapidly that shallow-water waves, grav· in supercritical flow.
Open Channel Flow
Chap. 9
Sec. 9.3
375
The Solitary Wave
with Eq. 9.3-10. Expressing the if pv!fho is negligible compared to to gravitational effects). A this assumption is satisfied when fluid depth. Such waves are often . speed is given correctly by Eq. 1s less than the fluid depth are called
.¡g¡; -
waves (A < h1)
(9.3-17)
velocity than shallow-water waves. than 1 cm-the wave velocity is waves are called ripples. Their
(9.3-18)
to the condition that the wave length, M ~ A; thus, the results
we discuss the result derived lociatc:d with the wave speed. There the wave generated in Fig. there is little doubt in our minds "If our analysis is correct, why the wave moves ?" The answer is generation phenomena in order Stoker7 has presented a thorthat a disturbance always proparegion. In our problem movable wall; hence, the wave flowing channel is similar to the is again obtained with the velocity thus indicated that a disturbance equal to plus the stream to minus the stream velocity. where a disturbance has been into the stream.
Jih
Jih
4th ed. (New York: The ~crscien
Publishers, lnc., 1957), p. 303.
Fig. 9.3-3. Propagation of a disturbance in a mild channel.
The interesting case to consider is that of critical flow where the velocity is given by Eq. 9.2-36 as (v~) =
Jih
(9.3-19)
Thus, for critica! flow a disturbance may remain fixed at a point in the channel; for subcritical flow (such as that illustrated in Fig. 9.3-3), a disturbance may propagate upstream; and for supercritical flow all s~pall disturbances are carried downstream. Note that this latter statement also applies to gravity waves beca use they ha ve a smaller velocity than shallow-water waves. Ripples may propagate upstream in a critica! channel, provided ,
27T(]
,.<-
(9.3-20)
pgh
Thus, if the fluid depth is O(feet), the wave Iength must be about I0- 2 cm. Such small waves are rapidly dissipated by viscous effects, and we conclude that shallow-water waves, gravity waves, and ripples are carried downstream in supercritical flow.
376
Open Channel Flow
Chap. 9
In effect, we have airead y come to this conclusion in our analysis of the ftow profiles upstream ofthe weir in Fig. 9.2-4; however, it will be worthwhile to consider the matter further. In Fig. 9.3-4 the surface profiles upstream of a dam have been sketched for mild and steep channels. In each case, the dam causes a dist'urbance in the ftow. For the mild channel, the ftow is subcritical
Flow Over Bumps, Crests, a
9.4 Flow Over Bumps, Cr
The presentation in this sectio to examine the broad-crested wei Flow over a bump
Bockwoter curve
-.,.....___
Sec. 9.-4
-------
. ~----
.___
~-
-
We wish to examine the fto whether the surface will hump o channel. Intuition would natural appear but this assumption is no are no variations in channel widt
dh -(1-
dx
NFI
Upstream and downstream from
Under these conditions, Critico! depth of .. flow without dom Flow cond1t1on controlled ot the ~ 1 upstreom end 1 1
Hydroulic jump
Steep or su
1
. Percntico¡
dr¡ ( dx
The difficult question to answer i. bump ?" The behavior of dr¡fdx ü decreases on the upstream side of side. Our comments regarding T and for a start we shall assume th over the bump. This statement is forms as illustrated in Fig. 9.4-1
slope
dr¡ +.ToS ) ( dx pgbh Fig. 9.3 .....
Su~face
( dr¡ + 'ToS ) dx pgbh
profiles upstream from a dam.
and the disturbance is propagated upstream forming the smooth backwater curve. In the steep channel, small disturbances are carried downstream; however, the dam presents a finite disturbance and a hydraulic jump is formed which propagates upstream at a velocity just equal to (vz>· The characteristics of hydraulic jumps will be treated in Sec. 9.5.
<
We now consider two cases: l. 2.
NFr NFr
> <
1, supercritical ftow 1, subcritical ftow.
t This statement assumes that we e channel.
Open Channel Flow
Chap. 9
conclusion in our analysis of the ; however, it will be worthwhile the surface pro files upstream of a channels. In each case, the dam channel, the flow is subcrítical
Sec. 9.<4
9.4
Flow Over Bumps, Crests, and Weirs
377
Flow Over Bumps, Crests, and Weirs
The presentation in this section is mainly qualitative, the objective being to examine the broad-crested weir and describe how it works.
Flow over a bump We wish to examine the flow illustrated in Fig. 9.4-1 and determine whether the surface will hump or dip when passing over the bump in the channel. Intuition would naturally lead u's to belíeve that a hump would appear but this assumption is not necessarily correct. We assume that there are no variations in channel wídth so that Eq. 9.2-21 reduces to
dh (1 - NFr) = - (dr¡ dx dx
+
ToS ) pgbh
(9.4-1)
Upstream and downstream from the bump the flow is uniform,t and
dh dx
=o
(9.4-2)
Under these conditíons,
dr¡) ( dx
=-
( r0S ) pgbh
(9.4-3)
The difficult question to answer is, "What happens to these quantities at the bump ?" The behavíor of dr¡fdx is defined by the geometry; its absolute value decreases on the upstream side of the bump and in creases on the downstream side. Our comments regarding r 0 and h can only be vague generalizations, and for a start we shall assume that nei.ther h nor r 0 changes as the fluid flows over the bump. This statement is consistent with the assumption that a hump forms as illustrated in Fíg. 9.4-1. Based on this assumption, we may wríte
dr¡ ( dx ( dr¡ dx forming the smooth backwater are carried downstream; and a hydraulic jump is just equal to (v,.). The in Sec. 9.5.
+
ToS ) pgbh
> O,
+ 'roS ) < O, pgbh
on the upstream side
(9.4-4a)
on the downstream side
(9.4-4b)
We now consider two cases:
l. NFr 2. NFr
> 1, supercritical ftow; < 1, subcritical flow.
t This statement assumes that we do not have supercritical flow on a mild or critica) channel.
Open Channel Flow
378
L
Chap. 9
Sec. 9.'1
Flow Over Bumps, Crests, an
Supercriticol flow
-------------------1~:..-------~bcriticol
Flow over a crest flow
We now wish to examine the ¡:¡ in Fig. 9.4-2. Water flows from a reservoir formed by an adjustable take a chronological approach.
~
Z=H
Z=L----------Flg. 9.4-1. flow over a bump.
For case 1, Eqs. 9.4-1 and 9.4-4 indicate that dh
dx
>o
'
dh <0
dx
'
on the upstream side
(9.4-Sa)
on the downstream side
(9.4-Sb) Fig. 9.4-l. Flow
Thus, for supercritical flow the profile rises up over the bump, and our assumptions regarding h and To in the neighborhood of the bump may be correct. For case 2, Eqs. 9.4-1 and 9.4-4 indicate that dh
dx dh
'
.
-> 0 dx '
on the upstream side
(9.4-6a)
on the downstream side
(9.4-6b)
Thus, for subcritical flow our analysis indicates that the profile dips down over the bump as indicated by the dashed curve in Fig. 9.4-1. This conclusion, of course, violates our assumption that h and Toare constant in the neighborhood of the jump. However, if h decreases and -r 0 increases owing to the increased velocity, the inequalities given by Eqs. 9.4-4 still hold, and we conclude that the profile rises over a bump when the flow is supercritical and dips when the ftow is subcritical. We must note· that this discussion is very qualitative in nature, and the behavior of the surface profile in the neighborhood of a bump is best understood in terms of experimental observation. The analysis is put in its proper place if we simply state that it does not contradict the experimental observation.
Let us consider first the case ftow occurs and the..water in the 1 so that a small fiow takes place, t profile will occur at the crest. F increase the ftow rate until the crit1 ftow has been subcritical and c1 However, once critica! flow is rea propagated upstream, and further out the downstream portion of th' tion is illustrated in Fig. 9.4-3, wh the crest, the critica! fiow at the c1 of the crest. The broad-chested weir
A broad-crested weir is shown as ftow rate measuring devices, a viously discussed sharp-crested v analysis of this ftow is identical t1 where the application of Bernoull
p + lpv2
•
Open Channel Flow
Chap. 9
Sec. 9.4
379
Flow Over Bumps, Crests, and Weirs
Flow over a crest We now wish to examine the problem of flow over a crest, as illustrated in Fig. 9.4-2. Water flows from a reservoir over the crest and into a second reservoir formed by an adjustable gate. In examining this problem, we shall take a chronological approach. Z=H----------------~L_-------------_
Z=L----------- h
(9.4-Sa)
(9.4-Sb) Fig. 9.4-2. Flow over a crest in a reservoir.
up over the bump, and our ~~lJLuuiul;ouu of the bump may be
(9.4-6a)
wnstream side
(9.4-6b)
that the profile dips down over Fig. 9.4-1. This conclusion, of are constant in the neighborhood increases owing to the increased still hold, and we conclude that supercritical and dips when the discussion is very qualitative in in the neighborhood of a bump The analysis is put in not contradict the experimental
Let us consider first the case where the control gate is raised so that no flow occurs and the...water in the reservoir is quiescent. If we Jower the gate so that a small flow takes place, the flow will be subcritical anda dip in the profile will occur at the crest. Further Iowering of the control gate will increase the flow rate until the critica! velocity is reached. Up to this time the flow has been subcritical and controlled by the downstream conditions. However, once critica{ flow is reached small disturbances can no longer be propagated upstream, and further lowering of the control gate simply washes out the downstream portion of the dip in the pro file at the crest. This situation is illustrated in Fig. 9.4-3, which shows the subcritical flow upstream of the crest, the critica{ flow at the crest, and the supercritical flow downstream of the crest. The broad-chested weir A broad-crested weir is shown in Fig. 9.4-4. Broad-crested weirs are used as flow rate measuring devices, and have a certain advantage over the previously discussed sharp-crested weir (Sec. 8.5). The initial portion of the analysis of this flow is identical to that presented for the sharp-crested weir where the application of Bernoulli's equation to any streamline gave us
p + lpv 2 + pgz = p0
+ pgH
(9.4-7)
380
Open Channel Flow
Chap. 9
Subcriticol flow
Sec. 9.4
Flow Over Bumps, Crests, and
The broad-crested weir has its grea reached the critica! value. Under ti
Critico! flow
V . e
and Eq. 9.4-9 is written as
J2i
Ve=
Supercriticol flow
We may now eliminate h., solve for rate,
Q=b~
'
Flg. 9.4-3. Flow over a crest in a reservoir.
Over the central portion of the weir, the streamlines are straight and the pressure is hydrostatic, (9.4-8) p =Po pg(he L - z)
+
+
This result ts similar in form to t however, in deriving this equation coefficieut to correct the assumpti been able to specify the fluid depth The results obtained in this se flows and negligible viscous effects. should be regarded only as estímate
and the velocity in Eq. 9.4-7 is given byt
v
= J2g .Jcn- L)- h
Calculatiorl of flow rate (9.4-9)
This equation could be used to determine the volumetric ftow rate over a broad-crested weir if JI and h were measured with depth gauges. Because the streamlines are straight, the volumetric ftow rate per unit width is
q = h.fii
.Jcn- L)- h
(9.4-10)
For a value of 11 L-= 0.525 unit width, and the fluid depth on By Eq. 9.4 -13 the volumetric fio
q = .Jg'[i(H- L))312 = or
q=l. Because the flow on the weir will b
q =vA=
Jih.
and. sol ve for he:
he= i(H
Experiments described by Doering conditions, q= 1
he= O Flg. 9........ Broad-crested weir.
t Remembcr that Bernoulli's equation neglects viscous effects; therefore, it is not unreasonable that Eq. 9.4-9 indicates a ftat velocity profile.
Thus, the actual flow rate is within the measured fluid depth on the wei value. In view of the simplificatio between measured and calculated f1
8. H. A. Doeringsfeld and C. L. Bark
Open Channel Flow
Chap. 9
Sec. 9.4
Flow Over Bumps, Crests, and Weirs
381
The broad-crested weir has its greatest utility when the flow at the weir has reached the critica! value. Under these conditions, the velocity is given by
.fihc
(9.4-11)
J2i ~(H- L)- h.
(9.4-12)
v.
=
and Eq . 9.4-9 is written as
v. = ulic jump
We may now eliminate h., solve for rate,
v.,
and obtain the total volumetric flow (9.4-13)
L- z)
(9.4-8)
This result IS similar in form to that obtained for the sharp-crested weir; however, in deriving this equation we have not had to introduce a discharge coefficieut to correct the assumption of a flat velocity profile, and we have been able to specify the fluid depth at the weir. The results obtained in this section are restricted to gradually varied flows and negligible viscous effects. Calculations made using these equations should be regarded only as estimates. Calculatiorl of flow rate
(9.4-9)
the volumetric flow rate over a with depth gauges. Because rate per unit width is L)- h
(9.4-10)
For a value of H - L = 0.525 ft, determine the volumetric flow rate per unit width, and the fluid depth on a broad-crested weir. By Eq. 9.4-13 the volumetric flow rate per unit width is given by
q = .Ji[i(H- L)]312 = ~32.2 ft/sec 2[(i)(0.525 ft)] 312 or
q = 1.17 ft 3/ft-sec Because the flow on the weir will be critica!, we may express q as
q =vA= Jg'h;h. = }.g[i(H- L)]a12 and. sol ve for h.: h. = i(H - L)
=
0.35 ft
Experiments described by Doeringsfeld and BarkerS indicate that for these conditions, q = 1.08 ft 3/ft-sec
h.= 0.28 ft weir.
viscous effects; therefore, it is not profile.
Thus, the actual flow rate is within 8 per cent of the calculated value, while the measured fluid depth on the weir is 20 per cent smaller than the calculated value. In view of the simplifications made in the analyses, the agreement between measured and calculated flow rates is remarkable. 8. H. A. Doeringsfeld and C. L. Barker, Trans. A.S.C.E., 1941, 106:934-46.
Open Channel Flow
382
9.5
Chap. 9
Sec. 9.5
Hydraulic Jump
after the jump. Expressing the left-h and evaluating the right-hand side, '
Hydraulic Jump
A hydraulic jump occurs at a transition from supercritical to subcritical flow. This transition may be brought about by a change in the slope of the channel, such as the jump illustrated in Fig. 9.4-3, or it may occur when a supercritical flow is formed on a rnild channel by a sluice gate. This situation is illustrated in Fig. 9.2-4. In analyzing the hydraulic jump, we will consider the flow shown in Fig. 9.5-1. As shown, it consists of a supercritical flow coming off a steep channel onto a mild channel. On the mild channel, the depth gradually increases and then undergoes a sudden change, which is called an hydraulic jump. We will, in the course of this analysis, show that the jump may be lo<-ated as indicated here or up on the steep portien of the channel.
p(v!)3h3
-
p(v!)
Assuming flat velocity profiles and we have
There are two possible solutions to
h2 =
ort
Groduoliy voried flow
Supercriticol flow
Here we have excluded the negativ is not difficult to eliminate Eq. 9.5-~ 9.2-21. For a supercritical flow on Eq. 9.2-21 will be negative, as will that dhfdx is positive and Eq. 9.5-6 Recognizing that q 2/gh~ is the Ii jump, Eq. 9.5-7 reduces to
ha=-~2
Fl¡. 9.5-1. Hydraulic jump.
lnasmuch as the hydraulic jump is highly turbulent and energy losses are large, the momentum balance provides the only route to a direct solution. Assuming steady flow and neglecting viscous and gravitational effects, Eq. 9.1-5 reduces to
f
pVzV •
n dA = -
A,
f
i · np, dA
(9.5-1)
Ñ
when dotted with the unit vector i. The assumption of nearly straight stream1ines before and after the jump allows us to express the gauge pressure as p, = pg(h 2 - z), (9.5-2) p, = O, h 2 < z :-: ; ; h 3 before the jump, and
p,
=
pg(h 3
-
z),
O :::;;z :::;;h 3
(9.5-3)
The nature of the jump depends ve1 approaching flow; Fig. 9.5-2 illus1 brief description of these different t:
= 1 to 3. Standing wavc undular jump. The surface is 2. N Fr = 3 to 6. Rollers develc downstream surface remains 1 3. NFr = 6 to 21. The high-Sf unstable jet producing Iarge can travel for miles and cause 4. NFr = 21 to 80. The action ol and position are Ieast sensith dissipation:t: ranges from 45 t• l.
NFr
t Work done by the Bureau of Red between Eq. 9.5-7 and experimental obser· t The major effect of an hydraulic jum
Open Channel Flow
Chap. 9
Sec. 9.5
Hydraulic Jump
383
after the jump. Expressing the left-hand side ofEq. 9.5-1 in terms ofaverages, and evaluating the right-hand side, we get from supercritical to subcritical by a change in the slope of the Fig. 9.4-3, or it may occur when a by a sluice gate. This situation consider the flow shown in Fig. flow coming off a steep channel the depth gradually increases and is called an hydraulic jump. We that the jump may be located c1s of the channel.
(9.5-4) Assuming flat velocity profi1es and applying the macroscopic mass balance, we have (h2 - ha)[(h2
r-------~-~-~-~~~~~
1 1 1
J=
h2 =ha
O
(9.5-5)
(9.5-6)
ort a
1
2q2 gh2h3
There are two possible solutions to Eq. 9.5-5; either
h
flow
+ ha) -
= _ h2 + Jh~ 2
4
+ (.!L) 2h2 gh~
2
(9.5-7)
Here we have excluded the negative root for ha as physically impossible. It is not difficult to eliminate Eq. 9.5-6 as a possible solution by examining Eq. 9.2-21. For a supercritical flow on a mild channel, the right-hand side of Eq. 9.2-21 will be negative, as will be the term (l - N~·r). Thus, we conclude that dhfdx is positive and Eq. 9.5-6 is not a possible solution. Recognizing that q 2/gh~ is the Froude number for the flow prior to the jump, Eq. 9.5-7 reduces to (9.5-8)
turbulent and energy losses are only route to a direct solution. and gravitational effects,
(9.5-1) assumption of nearly straight us to express the gauge pressure (9.5-2)
(9.5-3)
The nature of the jump depends very strongly on the Froude number of the approaching flow; Fig. 9.5-2 illustrates the four main types of jump. A brief description of these different types of jump is helpful. l. NFr = 1 to 3. Standing waves are formed and the jump is called an undular jump. The surface is smooth. 2. NFr = 3 to 6. Rollers develop on the surface of the jump, but the downstream surface remains smooth. The jump is called a weak jump. 3. NFr = 6 to 21. The high-speed flow entering the jump forms an unstable jet producing large waves of irregular period. These waves can travel for miles and cause serious damage. 4. N~-r = 21 to 80. The action ofthis jump is well balanced and the action and position are least sensitive to the downstream depth. The energy dissipationt ranges from 45 to 70 per cent.
t Work done by the Bureau of Reclamation (Ref. 9) indicates excellent agreement between Eq. 9.5-7 and experimental observations. t The majar etfect of an hydraulic jump is the dissipation of kinetic energy.
Open Channel Flow
38-4
Chap. 9
Sec. 9.5
Hydraulic Jump
Undular jump
From Eq. 9.1-17, with b
~h,
we o
h¡ = ha=
~-:, ....,..........- - - - " ' -
_;
::::__--
Weok iump
NFr=3-6
Knowing ha and the flow rate, we fluid depth, h2 • h2-
Obviously, the jump from 2.04 to 2.6 the Froude number just upstream o
N F.,
Steody jump
\~~~~\'\'\~~""""'0-.. Fig. 9.5-2. Various types of hydraulic jumps.
Under these conditions, we have a1 in Fig. 9.5-2. The velocity just upst may think of the standing wave wave velocity of -9.8 ft/sec. To determine the location of th at the start of the mild channel is t 0.45 ft. This statement assumes tha to a gradually varied flow that do upstream. Application of Eq. 9.2beginning of the mild channel to t If we decrease the slope of t we obtain the conditions, hl =
h2'
ha' 5. NFr ¿ 80. The jump action is rough, and energy dissipation may be as high as 85 per cent. Undesirable waves are formed which travel downstream. lt i~ important to ha ve sorne idea how to loca te the pósition of a jump; to outhne the method, we shall study a specific example for the flow shown in Fig. 9.5- 1. Location of the hydraulic jump
Let us consider the problem of locating the position of the hydraulic jump shown in Fig. 9.5-1 for the following conditions:
(a) (b) (e) (d) (e)
01 = 30°; 02 = 5'; q = 20 fta/ft-sec; channe1 width ~fluid depth; n = 0.014 ft11 6 •
Obviously the jump is a stronger upstream from the jump (NFr = 2 this case, Eq. 9.2-41 indicates tha beginning of the mild channel. FUI will strengthen the hydr11ulic jump a the start of the mild channel. Whe1 jump will occur at the beginning of channel is decreased even more, the as Fig. 9.5-3 illustrate~. The jump under these conditions, because it n both the steep and mild channels. fairly well established. A detailed carried out by the Bureau of Recl~1r
9. "Hydraulic Design of Stilling Basi1 graph No . 25, Hydraulics 1 aboratory f Colorado. 1958.
Open Channel Flow
Chap. 9
Sec. 9.5
Hydraulic Jump
385
Undular ·ump
From Eq. 9.1-17, with b ';t>h, we obtain the two uniform fluid depths hl = 0.45 ft ha= 2.62 ft Knowing ha and the flow rate, we may rearrange Eq. 9.5-7 and solve for the fluid depth, h 2 • h2 = 2.04 ft Obviously, the jump from 2.04 to 2.62 ft is nota strong one, and ifwe calculate the Froude number just upstream of the jump we find NF.r
Steody jump
Under these conditions, we have an undular jump such as the one illustrated in Fig. 9.5-2. The velocity just upstream from the jump is 9.8 ft/sec, and we may think of the standing wave f9rming the undular jump as having a wave velocity of -9.8 ft/sec. To determine the location of the jump, we first note that the fluid depth at the start of the mild channel is the same as that in the steep channel, i.e., 0.45 ft. This statement assumes that the change in channel slope gives rise to a gradually varied flow that does not allow disturbances to propagate upstream. Application of Eq. 9.2-51 indicates that the distance from the beginning of the mild channel to the hydraulic jump is 700 ft. If we decrease the slope of the mild channel from () = 5' to () = 3', we obtain the conditions, h¡ = 0.45 ft
h2 and energy dissipation may be waves are formed which travel to loca te the pósition of a jump; example for the flow shown
the position of the hydraulic conditions:
= 1.46
=
1.71 ft
ha= 3.06 ft
Obviously the jump is a stronger one, although the Froude number just upstream from the jump (NFr = 2.5) still indicates an undular jump. For this case, Eq. 9.2-41 indicates that the jump takes place 560 ft from the beginning of the mild channel. Further decreases in the mild channel slope will strengthen the hydraulic jump and move the position of the jump toward the start of the mild channel. When the calculated value of h 2 is 0.45 ft, the jump will occur at the beginning of the mild channel. lf the slope of the mild channel is decreased even more, the jump will take place in the steep channel, as Fig. 9.5-3 illustrates. The jump is a good deal more difficult to analyze under these conditions, because it may take place over a region occupied by both the steep and mild channels. In any event, the location of the jump is fairly well established. A detailed experimental study of hydraulic jumps carried out by the Bureau of Rechrnation 9 has provided accurate design data. 9. "Hydraulic Design of Stilling Basins ancl f'nergy Dissipators," El{![ineeril{l{ Monograph No . 25. Hydraulics Laboratory Branch, U S Rureau of Reclamation, Denver, Colorado. 1958.
386
Open Channel Flow
Chap. 9
Bz Flg. 9.5-3. Hydraulic jump.
Problems
9-3. Determine the slope of the mild the hydraulic jump takes place at Ans: 82
=
0.5 x 10-' radians
•
PRO~LEMS
9-4. Apply the momentum balance to t the wave speed in a flowing strea1
9-1. Apply the same type of analysis used in examining the motion of a solitary wave to deduce the surface profile for the depression wave shown in Fig. 9-1. Remember that the disturbance will propagate into the undisturbed region. Comment on the shape of the disturbance as time progresses.
9-5. A uniform flow of water takes J with evenly laid unplaned timbe horizontal is 6•, the channel is 10 the velocity, flow rate, and Frou
9-6. The surface roughness and slope by design considerations; howeve1 ratio of the fluid depth to chan area of flow is 100 ft 2, that will
9-7. How deep will the water flow in channe1 if the slope is 0.001 and 1 mine the wall shear stress, T0 , an1
Ans: h = 1.3 ft, -ro= 0.065 lbtl!
9-8. Determine the volumetric flow n flow depth is 3 ft and the slope il
Flg. 9-1 . Creation of a depression wave.
9-2. The positive surge, or elevation wave, shown in Fig. 9-2 may be formed in a riverbed by a cloudburst, or in an estuary by a rapidly advancing tidal front. In the laboratory, it can be formed by sudden changes in the position of a sluice gate. The positive surge is often called a moving hydraulic jump. Analyze the flow shown in Fig. 9-2 to obtain an expression for the velocity of the surge in terms of h1 and h2 •
9-9. U se the mechanical energy balane the sluice gate shown in Fig. 9-9 at the gate, h2 • Because the stre traction or discharge coefficient clearly why this is so). Show h depth h3 could be used to deterrr
9-10. The volumetric flow rate in a 1C mínimum specific energy possible the critica! veloeity?
Ans: E 1 P = 4.6 ft, he= 3.0 ft,
1
9-11. Rederive the differential-macro corporating the effects of a con
Open Channel Flow
Chap. 9
387
Problems
Fig. 9-2. The positive surge.
9-3. Determine the slope of the mild channel in the example in Sec. 9.5 such that the hydraulic jurnp takes place at the beginning of the mild channel.
Ans:
82 =
0.5
X
w-' radians
9-4. Apply the momentum balance' to the flow illustrated in Fig. 9.3-3 to determine the wave speed in a flowing stream. in examining the motion of a solitary depression wave shown in Fig. 9-1. ,.u ....at:au: into the undisturbed region. as time progresses.
9-5. A uniform flow of water takes place in a rectangular channel constructed with evenly laid unplaned timber. If the angle between the channel and the horizontal is 6•, the channel is 10ft wide, and the water is 2ft deep, compute the velocity, flow rate, and Froude number. 9-6. The surface roughness and slope (i.e., sin 8) of a rectangular channel are fixed by design considerations; however, the width and depth are not. Calculate the ratio of the fluid depth to channel width, subject to the restriction that the area of flow is 100 ft 2 , that will provide a maximum discharge. 9-7. How deep will the water flow in a 15-ft wide, smooth, concrete, rectangular channel if the slope is 0.001 and the volumetric flow rate is 87 ft 3/sec. Determine the wall shear stress, T 0 , and the Froude number.
Ans: h = 1.3 ft,
T'O
= 0.065 lbtlft 2 ,
NFr
= 0.48
9-8. Determine the volurnetric flow rate in a 10-ft diameter sewer if the uniforrn flow depth is 3 ft and the slope is 0.0002. 9-9. Use the mechanical energy balance to derive an expression for the flow through the si u ice gate shown in Fig. 9-9 in terms of the fluid depth h1 and the depth at the gate, h2 • Because the streamlines are not parallel at the gate, a contraction or discharge coefficient must be included in the analysis (indicate clearly why this is so). Show how experimental measurements of the fluid depth h3 could be used to determine the contraction coefficient. in Fig. 9-2 may be formed in a by a rapidly advancing tidal front. sudden changes in the position of a called a moving hydraulic jurnp. · an expression fo r the velocity
9-10. The volumetric flow rate in a 10-ft wide channel is 300 ft 3/sec. What is the mínimum specific energy possible for this flow? What is the critica! depth and the critica! veloeity?
Ans: E 1 p = 4.6 ft, he = 3.0 ft, v 2 = 10.0 ft/sec 9-11. Rederive the differential-macroscopic mass and momentum balances incorporating the effects of a constant seepage into the channel. Express the
388
Open Channel Flow
Chap. 9
Problems
9-15. Apply the momentum balance to an to obtain a relationship between the the bottom of the channel. 9-16. Derive an expression for the horizo in Fig. 9-9.
h,
L~~k-
9-17. Water in a mild channe1 flows ata from a holding basin to an irrigat adjustment in a sluice gate at the wave, how long will it take for this system?
Fig. 9-9. Fiow under a sluke gate.
9-18. If the amplitude of the wave in Prol estímate the distance it will travel b the initially symmetric wave profile
volumetric flow rate into the channel as cubic feet per second per unit length of channel, and assume that the fluid enters the channel normal to the channel walls. 9-12. Perform a numerical integration of Eq. 9.2-47 and compare the results with Eq. 9.2-51 for the following thn:e cases: l. !JI
3.0,
=
H0
=
H 0 = 0.5
3. !JI
H 0 = 0.1
0.5,
1
6ft
~!
______
2.0
2. !JI = 3.0, =
Ans: 10.4 min
9-13. Water, ata velocity of 3 ft/sec and a depth of 2 ft, approaches a smooth rise in a channel as shown in Fig. 9-13. Estima te the depth of the stream after the 3 in. rise.
Ans: h
=
1.7 ft
Fig. 9-19. Flow thr
Fig. 9-13. flow over a rise in a channel.
9-14. Apply the mechanical energy balance, in conjunction with the momentum balance, to an hydraulic jump in a rectangular channel to derive an expression for the energy dissipation. Express the Ioss as a fraction of the velocity head v2/2g and Froude number preceding the jump. Examine the derived expression for NFr = 1 and NFr-+ co.
9-19. A rectangular channel 6ft wide hai metric flow rate of 60 ft3/sec. A C< trated in Fig. 9-19, and it may be as changing the surface profile at the varied on a mild channel. (a) Determine the minimum value remain undisturbed. (b) Describe the surface profile u¡ striction for values of b0 greate (e) If b0 is decreased further than th surface both upstream and dov.
Open Channel Flow
Chap. 9
Problems
389
9-15. Apply the momentum balance toan hydraulic jump in a 90° V-shaped channel to obtain a relationship between the two fluid depths, h1 and h2 , measured from the bottom of the channel. 9-16. Derive an expression for the horizontal force acting on the sluice gate shown in Fig. 9-9. 9-17. Water in a mild channel flows ata velocity of 0.8 ft/sec and ata depth of 8ft from a holding basin to an irrigation distributing system 2 mi away. If an adjustment in a sluice gate at the holding basin produces a shallow-water wave, how long will it take for this disturbance to be felt at the distributing system? Ans: 10.4 min
cubic feet per second per unit length the channel normal to the channel
9-18. lf the amplitude of the wave in Prob. 9-17 is 6 in. and the wave length is 40ft, estímate the distance it will travel before a pronounced skewness is formed in the initially symmetric wave profile.
. 9.2-47 and compare the results with
of 2 ft, approaches a smooth rise the depth of the stream after the
---- -Fig. 9-19. Flow through a constriction.
in.
1 in conjunction with the momentum ,--··-··,...- ..~· channel to derive an exthe loss as a fraction of the preceding the jump. Examine the
9-19. A rectangular channel 6ft wide has a uniform flow depth of 2ft anda volumetric flow rate of 60 ft 3/sec. A constriction occurs in the channel as illustrated in Fig. 9-19, and it may be assumed that inertial effects predominate in changing the surface profile at the constriction. Treat the flow as gradually varied on a mild channel. (a) Determine the mínimum value of b0 for which the upstrearn depth will remain undisturbed. (b) Describe the surface profile upstrearn and downstrearn frorn the constriction for values of b0 greater than, or equal to, the rninirnurn value. (e) If b0 is decreased further than the mínimum value, what will happen to the surface both upstream and downstrearn of the constriction?
Sec. 10.1
The Governing Equations for Co
Now, however, our problem has been density as a function of temperature an
p = p(l
Compressible Flow
10
Thus, we find temperature as a new va1 introduced, which is, of course, the tot2 Conservation of energy The fundamental postulate of con words as the time rate of ) change of interna! _ and kinetic energy { ( of a body
the rate at which heat is transferre to the body (energy transfer through a rigid, diathermal wall)
+ Up to this point we have treated only incompressible ftows, or ftows for which we could assume that the density p was a constant. Our purpose here will be to investigate the simplest type of compressible ftow-i.e., onedimensional flow in closed conduits. This study will provide an introduction to the use of the principie of conservation of energy in fluid mechanics and to the phenomenon of shock waves. In presenting this material, we assume the student has had, or is presently taking, a course in thermodynamics. Currently, the most important aspect of compressible ftow is the supersonic motion of aircraft and rnissiles; however, this subject requires a more detailed study than we have time for in an introductory text. For a more comprehensive treatment, the student is referred to the two-volume work of Shapiro. 1
The reference to heat as energy transfe~ work as energy transfer through a mova that heat and work are simply two di actually deal with "walls" or surfaces mechanisms. lf we integrate Eq. 10.1-3 with res1 written as fl(U + KE) where
= = Q=
U KE
W
10.1 The Governing Equations for Compressible Flow In previous chapters the implied equation of state has been p
=
constant
(10.1-1)
l. A. H. Shapiro, The Dynamics andThermodynamics ofCompressib/e Flow, Vols. 1 and ·
2 (New York: The Ronald Press Company, 1953).
390
the rate work is < { the body forces su
=
internal energy of the bo kinetic energy of the bo heat transferred to the b work done by the body (
Equation 10.1-4 represents the first la closed systems (i.e., matefial volumes). statement of the principie of conserva usually encountered in thermodynami when the surroundings do work on th between the body and the surrounding! natural to retain the sllme sign conve energy transferred to the body is a ~ the transfer is accomplished by heat o t Correctly speaking, Eqs. 10.1-4 and 1 the first law of thermodynamics being that
Sec. 10.1
The Governing Equations for Compressible Flow
391
Now, however, our problem has been expanded, and we must represent the density as a function of temperature and pressure. p
10
=
p(p, T)
(10.1-2)
Thus, we find temperature as a new variable, and another equation must be introduced, which is, of course, the total energy equation. Conservation of energy The fundamental postulate of conservation of energy may be stated in words as the rate at which ) the time rate of heat is transferred change of interna! _ to the body (energy transfer { and kinetic energy} ( through a rigid, of a body diathermal wall)
+ incompressible flows, or flows for p was a constant. Our purpose here of compressible flow-i.e., onestudy will provide an introduction of energy in fluid mechanics . In presenting this material, we taking, a course in thermodynamics. of compressible flow is the super' this subject requires a more an introductory text. For a more referred to the two-volume work of
(10.1-1) ofCompressible Flow, Vols. 1 and
(10.1-3)
the rate at which } work is done on { the body by body forces such as gravity
The reference to heat as energy transfer through a rigid, diathermal wall, and work as energy transfer through a movable, adiabatic wall is given ~9 emphasize that heat and work are simply two different modes of energy transfer. We actually deal with "walls" or surfaces which may transfer energy by both mechanisms . If we integrate Eq. 10.1-3 with respect to time, we obtain a result often written as (10.1-4) d(U KE) = Q - W
+
where
= interna! energy of the body = kinetic energy of the body Q = heat transferred to the body (by the surroundings)
U KE
W
of state has been
+
(the rate at which sur-¡ face work is done on the body (energy transfer through a movable adiabatic wall)
=
work done by the body (on the surroundings)
Equation 10.1-4 represents the first law of thermodynamicst as it applies to c/osed systems (i.e., material volumes). The only difference between the word statement of the principie of conservation of energy given here and the form usually encountered in thermodynamics texts is that work is taken as positive when the surroundings do work on the body. Since energy can be exchanged between the body and the surroundings in two ways (heat and work), it seems natural to retain the slime sign convention for both these mechanisms. Thus energy transferred to the body is a positive quantity regardless of whether the transfer is accomplished by heat or work. t Correctly speaking, Eqs. 10.1-4 and 10.1-3 are definitions of the interna) energy; the first /aw of thermodynamics being that the interna) energy is a state function.
Compressible Flow
392
Chap. 10
(b)
~~ f p(e + !v2) dV = f,.(t)
- f
(e)
q • n dA
(d)
+ Jt
S'l,.(t)
.r.f,.(t)
pg • VdV
The Governing Equations for Co
Applying the divergence theorem to th us to put all the terms under one inte equation.
The mathematical statement of Eq. 10.1-3 is
(a)
Sec. 10.1
(10.1-5)
f,.(t)
Expressing gas -
where e = interna! energy per unit mass q = heat flux vector
Vr/J, we may write
pg·v = -
The term (a) in this equation represents the time rate of change of internal and kinetic energy of the material volume "Y m(t)-i.e., the body. Term (b) represents the heat transferred to the body. The negative sign results from the fact that if the heat flux vector is pointing inward toward the body, the quantity q • n will be negative. Thus the negative sign in front of the integral is required. In terms of the symbolst used in Eq. 10.1-4,
Q = - fq·ndA
Remembering the formula for the mat
+ PE) =
(10.1-7)
Q - W'
!!_fps dV =fp DS dV Dt Dt ,.(t)
r
t
~/e+iv2)dV=-f q·ndA+fpg·vdV+f(n·T)·vdA .of,.(t)
f,.(t)
J1J(,.(t)
(10.1-8)
t As usual, the dot over Q indicates a time derivative.
Time rate of ehange of lntemal, klnetie and potential energy per unlt volume
p !!._(e
Dt
~
11
u
+ tv2 + r/J) =
We will refer to this resultas the total e show how the thermal energy equation mechanica/ energy equation from Eq. In the analysis of compressible flows useful form ofthe principie of conserva tia to compressible flow, we will only use and Eq. 10.1-13 must be integrated. presence of the density p outside the thfs difficulty before proceeding. Letti unit mass,
Ti
Eq. 10.1-5 may be written as
;r,.(t)
Substitution of this result into Eq. equation in terms of interna!, kinetic,
tJ
,.
and express the stress vector as
f p
ot
pg•v=-
Here W' represents only the work done by surface forces. If we make use of the special form of the Reynolds transport theorem derived in Chap. 3,
r
Dt
(10.1-6)
Term (e) in Eq. 10.1-5 represents the rate at which work is done on the body by surface forces. Term (d) in Eq. 10.1-5 represents the rate at which work is done by the force of gravity. We will show shortly that this term can be represented as the material derivative of the potential energy of the body, thereby allowing us to include it on the Ieft-hand side of Eq. 10.1-5: This form of the energy equation may ha ve been encountered in thermodynamics texts where it would be written as
KE
aq,
Since the gravitational potential q, is ind may combine Eqs. 10.1-11 and 10.1-10·
d,.(t)
~(U+
vr¡,
-=--
and expanding the material derivative i
an + pv·VO = -
p-
ot
Compressible Flow
Chap. 10
(d)
+Jtcnl • VdA+ .o/ ,.(t)
J
pg • VdV
The Governing Equations for Compresslble Flow
393
Applying the divergence theorem to the second and fourth integrals allows us to put all the terms under one integral sign and extract the differential equation.
10.1-3 is (e)
Sec. 10.1
(10.1-5)
(10.1-9)
j/" ,.(t)
Expressing gas -Ve/>, we may write pg • v = - pv • V e/> the time rate of change of internal "Y mCt)-i.e., the body. Term (b) body. The negative sign results from · inward toward the body, the negative sign in front of the integral in Eq. 10.1-4,
(10.1-10)
Remembering the formula for the material derivative, we write
Dc/>=ocf>+v·Vcf> Dt ot
(10.1-11)
Since the gravitational potential e/> is independent of time, ocf>fot = O and we may combine Eqs. 10.1-11 and 10.1-lO·to obtain
(10.1-6)
De/> Dt
pg•v= - p at which work is done on the body 1-5 represents the rate at which work show shortly that this term can be of the potential energy of the body, Jeft-hand side of Eq. 10.1-5~ This been encountered in thermodynamics (10.1-7) by surface forces. of the Reynolds transport theorem p DS dV
Dt
f
pg · v dV
+
Substitution of this result into Eq. 10.1-9 yields the differential energy equation in terms of internal, kinetic, and potential energy. Time ratc of changc or intcrnal. klnctic and potcntial cncrgy pcr unit volume
D 2 p - (e+ tv
Dt
Ratc of hcat transfcr pcr unlt volume
Ratc of surfacc work pcr unlt volume
+e/>)= -V· q +V· (T • v)
(10.1-13)
We will refer to this resultas the total energy equation. In Sec. 10.2, we will show how the thermal energy equation may be derived by subtracting the mechanica/ energy equation from Eq. 10.1-13. In the analysis of compressible flows, the total energy equation is the most useful forro of the principie of conservation of energy. In this brief introduction to compressible flow, we will only use the macroscopic total energy balance, and Eq. 10.1-13 must be integrated. The integration is complicated by the presence of the density p outside the material derivative; we must remedy thi.'s difficulty before proceeding. Letting O represent the total energy per unit mass, (10.1-14) n =e+ tv2 cP
+
f
(n • T) • v dA (10.1-8)
derivative.
(10.1-12)
and expanding the material derivative in Eq. 10.1-13, we get
an
p - + pv ·Vil= -V· q +V· (T • v)
ot
(10.1-15)
Chap. 10
Sec. 10.2
The Thermal Ener¡y Equation ant
Noting that n [(opfot) +V· pv] is zero by the continuity equation, we may add this term to the left-hand side of Eq. 10.1-15 to obtain a form suitable for integration.
Adding
a~d subtracting p/ p from O yic
Compressible Flow
394
(10.1-16)
dt Carrying out the integration over the arbitrary moving volume "Ya(t), and applying the divergence theorem to the second, third, and fourth integrals, we get
+ J(pvO) • n dA=
(pO) dV
1'"0 (!)
d
0
-
f d
4
(l)
(10.1-17)
do(l)
Applying the general transport theorem, and noting that the convective energy flux term will be zero except over the area A.(t), we obtain the macroscopic total energy balance
~t
f
pO dV
+
f
,V.(t)
f
pO(v- w) • n dA=- q • n dA+ Jt
.A:,(t)
.¡¡f.(t)
(10.1-18)
.¡¡f.(t)
Splitting the rate of work term into area integralst over A,(t) and A.(t) allows us to write
~~
f
pO dV
1'"0 (!)
+
f
pO(v - w) • n dA
=-
.A:,(t)
f
q • n dA
da(t)
+W+
f
t
.A:,(t)
(10.1-19)
Note that' the term W used throughout this text always refers to the rate of work done on the system by solid moving surfaces. In thermodynamics texts, this concept is often referred to as the useful work, while the work done on the system at the entrance and exits (the last term in Eq. 10.1-19) is referred to as the ftow work. While Eq. 10.1-19 is a completely general forro of the macroscopic total energy balance, we shall generally restrict ourselves in this chapter to fixed control surfaces at entrances and exits and steady state conditions. Under these conditions, Eq. 10.1-19 reduces to
f
pOv • n dA
.Á,
=-
f
q • n dA
.¡¡f
+W+
f
+ !v2 + tfo- ~)v · n d
p where the integral representing the ra been replaced by Q. As we have noted in several previ and exits may be approximated by ~·>
q • n dA+ Jt
(t)
h
Áo
=-;:¡ (pO) +V. (pvO) = -V· q +V· (T • v)
f ~~
Jp(
t • v dA
.Á,
for fixed control surfaces at entrances and exits and steady state conditions
(10.1-20) It will be to our advantage to express this result in terms of the enthalpy per unit mass h=·e+l?.
p
t This integral is zero ov~ the fixed solid surface, A,.
(19.1-21)
=.
Substitution of this expression for the 1 a cancellation of terms and we obtain
f
p(h
+ lv2 + 1/>)v
If we neglect variations of h, v1, and restrict oprselves to a system having write Eq. 10.1-24 as A(h
+ !v1 +
where m is the mass flow rate, and A exit conditions (where v • n is positive] Y· n is negative). With the mass, momentum, and en ready to analyze the propagation of a " and begin our investigation of compress we should tie up sorne loose ends re encountered in the derivation of the me section, therefore, we will derive the tl the viscous dissipation does indeed apJ1
10.2 The Thermal Energy Equa the Entropy Equation Our objective in this section is to p energy equations in proper perspectiv We start the analysis with the total en D
p -(e + lv') Dt
= -V· •
+ PI• Y+
Chap. 10
Sec. 10.2
by the continuity equation, we may Eq. 10.1-15 to obtain a form suitable
Adding
Compressible Flow
(10.1-16)
-V· q +V· (T · v)
arbitrary rnoving volume "Ya(t), and second, third, and fourth integrals,
-f
q • n dA+
.ofoW
Jt(n) •V dA .ofoW
q • n dA+
..,• (t)
Jt(n) •
V
dA
(10.1-18)
JI(. (t)
area integra1st over A,(t) and A,(t)
-f
q • n dA
..,.w
a~d subtracting p/ p from O yields
fp(
h
+ tV1 + cfo- E)v •n dA= Q + W + Jt1• 1 • vdA p
~
(10.1-22)
~
where the integral representing the rate of heat transfer to the system has been replaced by Q. As we have noted in severa! previous examples, the stress at entrances and exits may be approximated by
Substitution of this expression for the stress vector into Eq. 10.1-22leads to a cancellation of terrns and we obtain
J
p(h
+ lv1 + cfo)v • n dA = Q + W
(10.1-24)
..t. If we neglect variations of h, v1 , and cfo across the entrances and exits and restrict ourselves to a system having one entrance and one exit, we may write Eq. 10.1-24 as
(10.1-25)
+ W + f t(n) •V dA ..t.< t)
(10.1-19) this text always refers to the rate of rnoving surfaces. In thermodynamics the useful work, while the work done (the 1ast terrn in Eq. 10.1-19) is referred general forrn of the macroscopic total ourselves in this chapter to fixed and steady state conditions. Under to for fixed control surfaces at entr.ances and exits and steady state conditions
(10.1-20) this result in terrns of the enthalpy per
(19.1-21)
where m is the mass ftow rate, and ti indicates the difference between the exit conditions (where v • n is positive) and the entrance conditions (where v • n is negative). With the mass, momentum, and energy balances at our disposal, we are ready to analyze the propagation of a weak sonic disturbance (a sound wave) and begin our investigation of compressible ftows. However, before doing so, we should tie up sorne loose ends regarding the viscous dissipation terrn encountered in the derivation of the mechanical energy equation. In the next section, therefore, we will derive the therrnal energy equation to show that the viscous dissipation does indeed appear as thennal energy.
10.2 The Thermal Energy Equation and the Entropy Equation Our objective in this section is to put the total, therrnal, and mechanical energy equations in proper perspective, and to derive the entropy equation. We start the analysis with the total energy equation,
D Dt
p - (e surface, A •.
395
(10.1-23) (10.1-17)
and noting that the convective energy area A,(t), we obtain the macroscopic
-f
The Thennal Enercy Equation and the Entropy Equation
+ lv') = -V • q + PI• v +V· (T • v)
IOtaiCMr~JequalioD
(10.2-1)
Compressible Flow
396
Chap. 10
the stress equation of motion,
Dv = pg Dt
p-
+V·T
(10.2-2)
and the continuity equation
~: + V · (pv) =
O
(10.2-3)
Forming the scalar product of Eq. 10.2-2 with the velocity v yieldst p
!!_ (!v 2 ) = pg • v + v ·(V · T) Dt
(10.2-4)
The last term on the right-hand side may be expressed ast
v • (V · T)
=
V · (T · v) - Vv : T
mechanic:al energy equation (10.2-6) = pg. v + V · (T · v) - Vv : T Dt Subtracting this result from the total energy equation yields the thermal energy equation:
+ Vv : T
thermal energy equation
temperatures, or temperature profil studied in detail in courses in heat tr The total energy equation is gen temperature and kinetic energy are t course, for compressible flows. We must be careful to remember derived from the laws of mechanics, equations are two distinct equations. under special restrictions, can dege equation and the mechanical energy regarding the role of the fundament2 Entropy equation
p !!_ (!v2)
De = -V • q Dt
The Thermal Energy Equation a
(10.2-5)
Substitution of this expression into Eq. 10.2-4 yields the mechanical energy equation for compressible flow:
p-
Sec. 10.2
(
l 0.2-7)
In the next two sections, we wi entropy) processes. lt is therefore im so that we know what we mean by that the ·interna! energy may be sp?.ci
e= we may take the material derivative
De = Dt
(oe) os
D~
p
Dt
T = - pi+
De . = 1p Dt
"t'
p-
the last term on the right-hand side of Eq. 10.2-7 may be expressed as§
=
-pV · v +
Remembering now that the continuit) (10.2-8)
Our general form of the thermal energy equation for compressible flow is now written as
De
p-
= -V . q -
pV • V
+
§ See Eqs. 7.3-13 through 7.3-17.
we may write Eq. 10.2-12 as
(10.2-9)
Dt Thus, viscous dissipation always gives rise to an increase in the interna! energy and, therefore, the temperature of the fluid. The therma1 energy equation is most convenient to use when heat is being transferred to or from the system, and we wish to know inlet and outlet t See Eqs. 7.3-2, 7.3-3, and 7.3-4. t See Eqs. 7.3-9 through 7.3-12.
op
Multiplying this equation by p yields
If we write the total stress tensor as
Vv : T
+ (oe
De =Tp Dt
p-
Substitution of pDe/Dt from Eq. 10.2for the rate of change of entropy
Ds 1 p-=-(· Dt
t See Eq. 3.5-7.
T
Compressible Flow
Chap. 10
+V· T
(10.2-2)
• (pv) = O
(10.2-3)
2-2 with the velocity v yieldst •V
+ V. (V. T)
(10.2-4)
Sec. 10.2
The Thermal Energy Equation and the Entropy Equation
397
temperatures, or temperature profiles and heat fl.uxes. This equation is studied in detail in courses in heat transfer. The total energy equation is generally used when significant changes in temperature and kinetic energy are taking p1ace-the common situation, of course, for compressible fl.ows. We must be careful to remember that the mechanical energy equation is derived from the laws of mechanics, and that the total and thermal energy equations are two distinct equations. The fact that the total energy equation, under special restrictions, can degenerate into both the thermal energy equation and the mechanical energy equation, has led to sorne confusion regarding the role of the fundamental energy postulate in fluid mechanics.
be expressed ast
Entropy equation (10.2-5) . 10.2-4 yields the mechanical energy
mechanicalenergy equation
(10.2-6)
In the next two sections, we will be dealing with isentropic {constant entropy) processes. 1t is therefore important to derive the entropy equation so that we know what we mean by this term. Starting wi~h the statement that the interna! energy may be spP:cifi.ed in terms of the entropy and density e
energy equation yields the thermal :T
thermal cnergy equation
= e(s, p)
(10.2-10)
we may take the material derivative of e to obtain
De Dt
(10.2-7)
=
(oe) os
o
Ds Dt
+ (oe) op
s
Dp Dt
= T Ds Dt
+ .E. Dp p2 Dt
(10.2-11)
Multiplying this equation by p yields
p De= Ip Ds + P. Dp Dt Dt p Dt
Eq. 10.2-7 may be expressed as§ {10.2-8)
Reme~bering
now that the continuity equation may be written in the formt
Dp -+ pV·v=O Dt
equation for compressible flow is - pV •V
+ 11>
(10.2-12)
(10.2-13)
we may write Eq. 10.2-12 as (10.2-9)
De Ds = T p - - pV • V Dt Dt
p-
rise to an increase in the interna! of the fluid. convenient to use when heat is and we wish to know inlet and outlet
(10.2-14)
Substitution of pDef Dt from Eq. 10.2-9 and rearrangement give an expression for the rate of change of entropy
Ds 1 p - =-(-V. q Dt T
t Sec Eq. 3.5-7.
+ 11>)
(10.2-15)
398
Compressible Flow
Cflap. 10
Sec. 10.3
The Speed of ound
We now wish to determine the time rate of change of the entropy of a material volume. Rearranging the right-hand side of Eq. 10.2-15 and forming the integral, we obtain
f pD;dV=-f v.(~)dv+f (-q·~T +~)dv D 7'".,.
(10.2-16)
7'"...
Using the special form of the Reynolds transport theorem given by Eq. 3.5-12, and applying the divergence theorem to the first integral on the righthand side, we have an expression for the rate of change of entropy.
.!!..fpsdV=-f q·ndA+f (-q·.VT+~)dv Dt 7'".,.(l) ,,.
(10.2-17)
If a process is to be isentropic, the entropy of a body or of a material volume involved in the process must be constantt therefore,
J
ps dV= O, Dt 7'".,.
.!!_
for an isentropic process
(10.2-18)
This condition can be satisfied as follows: l. q • n =O on the surface, i.e., the process is adiabatic; 2. = O everywhere in the system, i.e., the process is "reversible"; 3. ~ = Oeverywhere in the system, i.e., "frictional" effects are negligible.
VT
In analyzing the speed of sound in Sec. 10.3 and flow in nozzles in Sec. 10.4, we will assume that the processes are isentropic even though viscous dissipation cannot be ídentically zero and the temperature gradients are definite1y finite. This assumption of isentropy greatly reduces the extent of the analysis while stil1 allowing us to investigate the key features of the flow.
10.3 The Speed of Sound We will determine the velocity ofsound by analyzing the density variations that occur when a small disturbance is created at the end of a tube, such as that shown in Fig. 10.3-1. This process is comparable to what might happen if the closed end of a cylindrical tube were hit with a mallet. If the flow were incompressible, the density would be p0 throughout the tube and the velocity would everywhere be equal to the velocity of the piston at the end of the tube. Since the flow is compressible, the density may increase above the value p,;
t
Strictly speaking this is a sufticient but not a necessary condition.
thus, the fluid far from the piston is · profiles are indicated above the tube are comparable to the surface pro shown in Fig. 9.3-1, and in what foil two different wave propagation phe We will analyze this problem Íl momentum balances, and because t the x-direction, we must make the x-direction. Figure. 10.3-2 illustrat~ used. We assume that the speed oft constant given by e; thus the contr
Mass balance Starting with the general macro
~tf pdV+ f .<1)
7'"
..t,(
Compressible Flow
Cl'lap. 10
Sec. 10.3
399
The Speed of ound
rate of change of the entropy of a side of Eq. 10.2-15 and
transport theorem given by Eq. to the first integral on the rightrate of change of entropy. ( -q. VT + ~) dV
..<'>
r•
(10.2-17)
r
of a body or of a material volume therefore,
r:::::::::::::::;
(10.2-18}
i.e., the process is "reversible" ; "frictional" effects are negligible. 10.3 and fiow in nozzles in Sec. 10.4, 'iserltr
by analyzing the density variations at the end of a tube, such as comparable to what might happen hit with a mallet. lf the fiow were lthr
Fi¡. IO.l-1. Generation of a sound wave.
thus, the fluid far from the piston is initially undisturbed. Illustrative density profiles are indicated above the tu be for each step of the process. The curves are comparable to the surface profiles for open channe~ wave formation shown in Fig. 9.3-1, and in what follows we shall see that the analysis ofthese two different wave propagation phenomena is similar. We will analyze this problem in terms of the macroscopic mass and momentum balances, and because the variable ofinterest p (or p) changes in the x-direction, we must make the macroscopic balances differential in the x-direction. Figure 10.3-2 illustrates the differential control volume to be used. We assume that the speed of the density variation (or sound wave) is a constant given by e; thus the control volume moves with the sound wave. Mass balance Starting with the general macroscopic mass balance,
~
J
J
pdV+ p(v- w)·ndA =O dt ~.
(10.3-1)
Compressible Flow
400
Chap. 10
Control volume
Sec. 10.3
The Speed of Sound
Momentum balance
We start with the general moment r-
,
~t
-W._ / ____ _ /___ L _ ~~ - - - - P o 1
1
Density profile
1
1
1
1
: 1 1 L
f
f
pv dV +
r.(l)
pv(v - w) • 1
A,
and note that the momentum of the the scalar product with i to obtain
'-~---e
f pv,.(v- w) ·n
1
___ J1
A,(l)
Remembering that viscous effects are Eq. 10.3-7 to the moving control voh
f pv.,(v., -
Fl¡. 10.3-2. Moving control volume for analysis of sound wáve.
e) dAI
.A.(I)
we note first that since the control volume moves with a specific portion of the wave having a fixed density, the first term is zero.
~ dt We see that i · w
f
(10.3-2)
pdV=O
7"'.(1)
= e, and the remaining area integral takes the form
f
pv.,(v.,-
A ,(1)
Again assuming flat velocity profile 6.x-+ O, we obtain d dx [pv.,(v., - o Using Eq. 10.3-5a allows us to simpl p(v.,-
(10.3-3) While the wave velocity e may be quite large, the actual fluid velocity v., will be small. Viscous effects may be neglected,t and the velocity profile may be assumed to be ftat. Under these conditions, we may divide Eq. 10.3-3 by 6.x and take the limit 6.x -+ O to obtain
o:+ Ao:
e)A(~
dJ
Taking the cross-sectional area A to eliminate dv.,fdx from Eq. 10.3-10 an
Solving for e gives
-d dx
[p(v.,-
e)A]
=
(10.3-4)
O
which may be rearranged in two convenient forms, (10.3-Sa)
p(v., - e)A = constant d
p-
dx
[(v"'- e)A]
t This is an experimental observation.
e= v,
and, provided v., ~ -Jdpfdp, our expr to
e= dp
+ (v.,- e)A=O dx
(10.3-Sb)
An analytical proof would be difficult.
Note that the analysis indicates th direction; however, we know intuiti1 as Fig. 10.3-2 indicates. A rigorow
Compressible Flow
Chap. JO Sec. 10.3
Control110lume
The Speed of Sound
401
Momentum balance
We start with the general momentum balance,
~t Jpv dV +Jpv(v r.m
----Po
w) • n dA
Jr.ct>pg
=
A.(l)
dV
+ Jten> dA
(10.3-6)
.w.co
and note that the momentum of the control volume is constant. We form the scalar product with i to obtain
J
pv,.(v- w) • n dA
=Ji •
A.(t)
ten> dA
(10.3-7)
(1) 0
Ñ
Remembering that viscous effects are being neglected and i · w = e, we apply Eq. 10.3-7 to the moving control volume in Fig. 10.3-2 and obtain for analysis of sound wave.
J
moves with a specific portion of term is zero. (10.3-2)
pv,.(v., - e) dA'
A.(l)
x+Ax
J
= -
pv,.(v., - e) dA'
A .(1)
.,
J
p dA'
A .(t)
x+A:t:
+ Jp dA' A . (1)
.,
(10.3-8) Again assuming flat velocity profiles, dividing by llx, and taking the limit llx-.. O, we obtain
~x [pv.,(v.,- e)A] =
-
area integral takes the form
e:)A
(10.3-9)
Using Eq. 10.3-5a allows us to simplify this result to (10.3-3) large, the actual fluid velocity vo: will t and the velocity profile may be we may divide Eq. 10.3-3 by llx
p(v.,-
e)A(~:"')
=
-(~:)A
(10.3-10)
Taking the cross-sectional area A to be constant, we may use Eq. 10.3-5b to eliminate dv.,/dx from Eq. 10.3-10 and obtain (10.3-11) Solving for e gives
(10.3-4)
(10.3-Sa) (10.3-Sb)
e= v., ±
p p
(10.3-12)
and, provided v., ~ ,Jdp/dp, our expression for the velocity of sound reduces to
e=±~
..J-;Jp
(10.3-13)
Note that the analysis indicates that the wave may propagate in either direction; however, we know intuitively that it will move from left to right as Fig. 10.3-2 indicates. A rigorous proof of this fact involves extensive
Compressible Flow
Chap. 10
Sec. 10.3
The Speed of Sound
analysis. Such a study has been carried out by Stoker2 indicating that a disturbance always propagates into the "region of quiet." If the process is adiabatic and reversible, the entropy, s, will be constant (i.e., the process is isentropic), and Eq. 10.3-13 is written as e
=m.=
Ta GAS CONSTANT
R
ANO
SOME
(J
Gas {the speed of sound}
(10.3-14)
For air at atmospheric pressure and normal temperatures, Eq. 10.3-14 indicates that the speed of sound is about 1100 ft/sec. Our definition of a "weak disturbance" will be that v., (in Eq. 10.3-12) is much smaller than the speed of sound. In the process under consideration, we may assume that the fluid velocity v., is equal to, or less than, the velocity of the piston; our analysis will therefore be correct provided the velocity of the piston is much less than the speed of sound. For a perfect gas, the pressure and density in an isentropic process are related by p constant (10.3.15) p7
-=
where y = ef}fe.,, the ratio of specific heats. The equation of state for a perfect gas is generally expressed as pV = nf!IT (10.3-16) where f!l is the universal gas constant and equal to 82.06 atm-cm3 /g-mole °K, and n is the number of moles. This form is generally encountered in thermodynamics texts where the quantity of material is specified in terms of the number of moles. In fluid mechanics, the mass is a more useful measure of the quantity of material contained in a system, and Eq. 10.3-16 is written in the form (10.3-17) pV= mRT where R = _ __:f!l_,_ _ molecular weight An even more convenient form results from dividing by V to yield p= pRT
This analysis has introduced us in the analysis of compressible fio the velocity of sound, compressibi velocities comparable to, or greate unusual situations where the co Therefore, the ratio M of the fi important measure of the compres after the physicist, Ernest Mach. 3
The common designation of the fi< of the Mach number are as follow1 M~1,
t
(10.3-18)
Here we must remember that R is different for every gas. By differentiating Eq. 10.3-15 with respect to p, using Eq. 10.3-18 to eliminate p, and substituting the result into Eq. 10.3-14, we can express the velocity of sound as
e= ../yRT
Hydrogen, H 1 Helium,He Nitrogen, N, Oxygen, 0 2 Carbon dioxide, C02 A ir Water vapor, H 20
(10.3-19)
Values of R and y for sorne common gases are listed in Table 10.3-1. 2. J. J. Stoker, Water Waves (New York: Interscience Publishers, Inc., 1957), p. 303.
1,
M> 1,
Because the velocity of sound in 1 mph), compressibility effects are 1 treatment of compressible fiows is
t Remember that the circulation ve) small compared to the speed of sound. 3. H. Rouse and S. Ince, History of 1 1963),
p. 195.
Compresslble Flow
Chap. 10
Sec. 10.3
The Speed of Sound
<403
2
out by Stoker indicating that a "region of quiet." the entropy, s, will be constant . 10.3-13 is written as
Table 10.3-1 GAS CONSTANT
R
ANO RATIO OF SPECIFIC HEATS
y
FOR
SOME COMMON GASES
y, standard conditions
Gas
(10.3-14) normal temperatures, Eq. 10.3-14 1100 ftfsec. Our definition of a Eq. 10.3-12) is much smaller than the we may assume that the the velocity of the piston; our the velocity of the piston is much
(10.3.15) of state for a (10.3-16) 3
equal to 82.06 atm-cm /g-mole °K, is generally encountered in thermomaterial is specified in terms of the the mass is a more useful measure of system, and Eq. 10.3-16 is written in (10.3-17)
Hydrogen, H 1 Helium, He Nitrogen, N 1 Oxygen, 0 2 Carbon dioxide, C01 A ir Water vapor, H 20
M=~
e
are listed in Table 10.3-1. : Interscience Publishers, Inc., 1957), p. 303.
1.33
(10.3-20)
The common designation of the flow regimes associated with various values of the Mach number are as follows: incompressible flow
t
1,
subsonic flow
M~
1,
transonic flow
M> 1,
(10.3-19)
1.41 1.67 1.40 1.39 1.29 1.40
This analysis has introduced us to the velocity of sound, a key parameter in the analysis of compressible flow. If the fluid velocity is much less than the velocity of sound, compressibility effects are usuallyt small, while fluid velocities comparable to, or greater than, the speed of sound lead to sorne unusual situations where the compressibility effects domínate the flow. Therefore, the ratio M of the fluid velocity to the speed of sound is an important measure of the compressibility effects; we call it the Mach number after the physicist, Ernest Mach. 3
M ~ 1,
for every gas. By differentiating 10.3-18 to elimina te p, and substituting the velocity of sound as
766.5 386.3 55.15 48.29 35.12 53.35 85.8
supersonic flow
Because the velocity of sound in liquids is approximately 5000 ft/sec (3400 mph), compressibility effects are rarely of importance in liquids, and the treatment of compressible flows is generally restricted to gases.
t Remember that the circulation velocities occurring in a heated pan of water are small compared to the speed of sound. 3. H. Rouse and S. Ince, History of Hydrau/ics (New York: Dover Pub1ications, Inc., 1963), p. 195.
Compressible Flow
Chap. 10
Sec. 10.4
lsentropic Nozzle Flow
the flow in a nozzle, the derived isentropic flows in ducts of variable important to note precisely where
Speed of sound In helium and air
We wish to compare the speed of sound in air and in helium at 72°F. For air, Eq. 10.3-19 gives Calr
= [
112
lbm- R lbr-sec = 1130 ftjsec Similarly, the velocity of sound in helium is calculated to be cae
r (x) 0
= 3330 ftjsec
Thus, the velocity of sound in helium is approximately three times as large as that in air. This difference gives a curious quality to the voice if one speaks with a lung full of helium.
10.4 lsentropic Nozzle Flow The simplest example of compressible flow is the isentropic flow from a large reservoir through a converging nozzle, such as that shown in Fig. 10.4-1. The assumption of constant entropy requires that the process be
Reservoir 1
Reservoir n
Pressure = p 0 Temperoture = Velocity ~O
Bock pressure = J&
r•
Fig. 10.4-2. Control volu
we will remove this restriction in We shall now discuss the form tH applied to the differential control vo Mass balance
Application of Eq. 10.3-1 to the d
-(p( dx
or
p(v., )A :
Flg. 10.4-1. lsentropic ftow through a nozzle.
adiabatic, and that q • VT and be negligib1e. The assumption of adiabatic flow is reasonable, and we may neglect viscous dissipation for flow in a nozzle without incurring much error. Because fairly large temperature changes take place in the nozzle, the assumption that q • VT is negligible must be based on the fact that the thermal conductivity of gases is small. Therefore, q
=
-kVT~::::i
O
Although our analysis in this section will be directed toward understanding
Momentum balance
Starting with Eq. 10.3-6, we for
f A.
pv!dA'
~t+&~t
-
f A.
t Here we have assumed that the dem the nozzle.
Compressible Flow
Chap. 10
Sec. 10.4
405
the flow in a nozzle, the derived equations will be generally applicable to isentropic flows in ducts of variable cross-sectional area. It will be especially important to note precisely where the isentropic restriction is imposed, for
sound in air and in helium at 72°F. (532 OR ) (32.2lbm-ft)] lb 2 r-sec
lsentropic Nozzle Flow
1 2 '
is calculated to be r
-'-----+--:----t--~f _ _ _ _ _
is approximately three times as large quality to the voice if one speaks
flow is the isentropic flow from a nozzle, such as that shown in Fig. requires that the process be
Rnervoir n Bock pressure = ¡¡,
x
Flg. 10.4-l. Control volume for analysis of nozzle flow.
we will remove this restriction in our analysis of shock waves in Sec. 10.5. We shall now discuss the form that the governing equations take when applied to the differential control volume shown in Fig. 10.4-2. Mass balance
Application of Eq. 10.3-1 to the differential control volume yieldst
!!..._ (p(v.,)A) = O p(v., )A
The assumption of adiabatic dissipation for flow in a nozzle large temperature changes take q. VTis negligible must be based on gases is small. Therefore,
=
constant
(10.4-1 b)
Momentum balance
Starting with Eq. 10.3-6, we form the scalar product with i to obtain
f A.
be directed toward understanding
(10.4-la)
dx
or
pv! dAj
., z+A
f
A.
=
pv! dAj "'
f .si
¡ . t<•> dA
(10.4-2)
t Here we have assumed that the density is constant at any given cross section of the nozzle.
Compressible Flow
406
Chap. 10
In evaluating the term on the right-hand side ofEq. 10.4-2, we must remember that the outwardly directed unit normal n from the control volume is given by at the entrance and exit i·n= ±1, at the walls of the nozzle i · n =sin O, where
1(
nv =tan- - dro) dx
_.,
f
p dA!.,Hz
+
f
A.,
p dA!.,+
(~:o) ~X
A.,
f
p ds
(10.4-4)
S
Substituting Eq. 10.4-4 into Eq. 10.4-2, dividing by ~x and taking the limit ~x -O, we obtain
t!.._ (p(v!)A) = - ci_ {(p)A) dx
dx
+ 27Tr0 (d'0)p
(10.4-5)
dx
Here p represents the mean pressure over the bounding circular strip of the control volume. If we now assume flat velocity profiles and a uniform pressure across the duct (i.e., (p ) = p = p), this result reduces to
ci_ (p(v.,)2A) = -A (dp) dx
(10.4-6)
h
+ t
(v2 )
The assumption that = (v.,) 2 is dr 0 jdx ~ l. Further restriction ofth is a linear function of temperature a
c'/)T + t
The constant may be evaluated by , (v.,) =O and T = T", which gives
c'/)T+
p(v.,) d(v.,) = _ (dp) dx
and
(v.,)• =
Equation 10.4-12 can be used to expr the local Mach number and the stag (10.4-7)
dx
T=
Energy balance We begin with the simplified form of the energy balance given by Eq. 10.1-24, and note that W = Q = Oto obtain p(h
+
tv 2
+
cfo)v • n dA = O
t
where r is called the stagnation tem is unusual and is done to clearly de~ and not the temperature at point pressure are the temperature and pr if it were isentropically brought to re it is often convenient to specify lo · temperature and pressure and the Ma important quantities. We now have at our disposal the along with the equation of state for sol ve for the four unknowns (T, p, f
dx
which we may simplify further by using Eq. 10.4-1 b,
f
Provided the ve1ocity profi1e is essen across the duct, Eq. 10.4-8 quick1y
h =e'/)
Inasmuch as we are neglecting viscous effects, the stress vector is given by t(a) = -np, and the right-hand side of Eq. 10.4-2 becomes i • t
lsentroplc Nozzle Flow
and Eq. 10.4-9 becomes
To keep the analysis reasonably simple, we must impose the restriction that dr 0 jdx ~ 1, so we may write . n n dro (10.4-3) sm v ~ tan v = - dx
f
S.C. 10.4
(10.4-1)
ro( 1
Three assumptions led to this resul1 perature; the equation of state for perfect gas law; and the flow is 011 isentropic flow has not yet been imp Since y > 1.0, Eq. 10.4-15 indic the velocity (and therefore the Mach of this conversion as one of thermal
Compressible Flow
Chap. 10
ofEq. 10.4-2, we must remember n from the control volume is given
S.C. 10.4
lsentroplc Nozzle Flow
Provided the velocity profile is essentially ftat and the enthalpy does not vary across the duct, Eq. 10.4-8 quickly reduces to
h
+
t(v.,)2
=
constant
The assumption that (v2 ) = (v.,)2 is reasonable in light of the restriction that dr0 fdx ~ l. Further restriction ofthe analysis to gases for which the enthalpy is a linear function of temperature allows us to write
walls of the nozzle
_ dr0 )
h = c'IJT + constant
dx
and Eq. 10.4-9 becomes
we must impose the restriction that
dr 0
(10.4-3)
=--
dx
the stress vector is given by . 10.4-2 becomes p dA!.,
+ (~:0 ) llx
J ds p
(10.4-4)
S
dividing by !lx and taking the limit
(10.4-5) the bounding circular strip of the velocity profiles and a uniform p), this result reduces to
-A(~:)
(10.4-6)
(10.4-7)
the energy balance given by Eq. obtain
)v · n dA= O
c'IJT
(10.4-10)
+ ! (v.,) 2 = constant
(10.4-11)
The constant may be evaluated by applying this result to reservoir ·1 where (v.,) =O and T = r', which gives (10.4-12) 0
where T is called the stagnation temperature. The use of "o" as a superscript is unusual and is done to clearly designate r as the stagnation temperature and not the temperature at point "o." The stagnation iemperature and pressure are the temperature and pressure that a fluid particle would attain if it were isentropically brought to rest. In the analysis of compressible ftows, it is often convenient to specify local conditions in terms of the stagnation temperature and pressure and the Mach number; thus r and p 0 are especially important quantities. We now have at our disposal the mass, momentum, and energy equations along with the equation of state for a perfect gas, p = pRT. We may now sol ve for the four unknowns (T, p, p, (v.,)). Noting that
yR y-1
(10.4-13)
= MlyRT
(10.4-14)
C'IJ=--
and (v.,)1
Equation 10.4-12 can be used to express the local temperature as a function of the local Mach number and the stagnation temperature.
Eq. 10.4-1b,
-e:)
(10.4-9)
(10.4-1)
T = To ( 1 + y
~
rl
1 M2
(10.4-15)
Three assumptions led to this result: enthalpy is a linear function of temperature; the equation of state for the gas can be approximated by the perfect gas law; and the ftow is one-dimensional. Thus, the restriction to isentropic ftow has not yet been imposed. Since y > 1.0, Eq. 10.4-15 indicates that the temperature decreases as the velocity (and therefore the Mach number) increases. Intuitively we think of this conversion as one of thermal energy to kinetic energy.
Compressible Flow
-408
Chap. 10
Our next step in the analysis requires the restriction of isentropic flows. By a straightforward analysis, we can show that the temperature and pressure in an isentropic process are related by _p_· = constant
lsentropic Nozzle Flow
Using the perfect gas law (Eq. 10.3of temperature and pressure, and rep
(10.4-16)
p ; >-1
we may write the mass flow rate as
and we may use this result in conjunction with Eq. 10.4-15 to obtain an expression for the pressure
p
Sec. lOA
= po(1 + y~ 1 M2r1-y
m= A~
(10.4-17)
Substitution of T and p from Eqs. lO This equation is only valid for isentropic flows. We will obtain an interesting result if we use the binomial theorem4 to expand the term in parentheses.
p =Po[¡+ _Y_ Y-1M2+ 1 1 -y 2
~y ~y~:
2
(y- 1) M4 + ...]
2!
4
(10.4-18)
Making use of Eq. 10.4-14 and simplifying the terms somewhat, we obtain
-
p- p
o[¡ - ~
(v,l +
2 RT
y(2y- 1 )M4 8
+ ...J
(10.4-19)
Application of the perfect gas law allows us to obtain a form somewhat reminiscent of Bernoulli's equation in the absence of gravitational effects
po[1 +
y( 2 y - l) M 4 8
=
p
+
+ ...]
! p < v., > 2
(10.4-20)
[ + ( ~ 1) JljY-1 1
y
M
So far, we have determined the temp terms of T 0 , p 0 , and M, and we nee flow completely. Substitution of Eq. result:
2
Let us imagine now that we have a free stream of gas at a pressure p moving at a velocity (v.,), which impinges on a Pitot tube such as that illustrated in Fig. 8.5-7. It is not a bad assumption that the flow along the stagnation streamline is isentropic ;· thus, the pressure at the nose of the Pilot 0 tube is the stagnation pressure, p 0 • If the flow were incompressible p would be given by p0 = p + !P (v.,)2 (10.4-21) However, Eq. 10.4-20 would certainly give a more accurate determination. This simple example provides us with an estima te of the effects of compressibility, for if M< t. Eqs. 10.4-20 and 10.4-21 will agree to within 1 per cent. Returning now to the analysis of the nozzle, we designate the mass flow rate as m and write Eq. 10.4-1 b as m = p(v., )A (10.4-22) 4. H. B. Dwight, Tables of Integrals and Other Mathematical Data (New York: The Macmillan Company, 1961), p. l.
Thus, T, p, m, and (v.,) are known a and p 0 are generally specified, the where in the duct to specify the flow , back pressure in reservoir Il is spe1 necessary to determine the tempera1 tributions. We may do so by assumin to calculate the Mach number in t determine the pressure distribution b~ at the exit of the nozzle is different we assume a new value of mand rep Rather than doing so, we will ex 0 specifying the pressure ratio pfp and m/ A throughout the nozzle. By worl unit area, mfA, we avoid specifying example we have assumed that the g~ is 300°K. The results are shown in F We want to think of the pressure pressure divided by the stagnation pre~ the. points on the curve represent the nozzle shown in Fig. 10.4-l. Let us ir
Compressible Flow
Chap. 10
the restriction of isentropic flows. show that the temperature and by
Sec. 10.4
lsentropic Nozzle Flow
409
Using the perfect gas law (Eq. 10.3-18) to express the density as a function of temperature and pressure, and representing the velocity as (10.4-23)
( 10.4-16) 10.4-15 to obtain an
we may write the mass flow rateas m=Ap
(10.4-17) flows. We will obtain an interesting to expand the term in parentheses.
Substitution of
r and p from Eqs. =
Apo -
Rro
1
M
(10.4-24)
10.4-15 and 10.4-17 gives us
. f*'( + - -
m
the terms somewhat, we obtain
fy viiT
y - 1 2)( y+l)¡ 2(1-y) M M 2
(10.4-25)
So far, we have determined the temperature, pressure, and mass fl.ow rate in 0 terms of ro, p , and M, and we need only specify the velocity to define the flow completely. Substitution of Eq. 10.4-15 into Eq. 10.4-23 gives us this result:
(10.4-19) us to obtain a forro somewhat absence of gravitational effects .. .]
[1 +_Y_ ( _2 1) M2
(10.4-20) ]1/Y-1
free stream of gas at a pressure p on a Pitot tube such as that illusthat the fl.ow along the the pressure at the nose of the Pilot fl.ow were incompressible p 0 would (10.4-21) give a more accurate determination. estima te of the effects of compressi4-21 will agree to within 1 per cent. nozzle, we designate the mass flow (10.4-22) Mathematical Data (New York: The
(10.4-26) Thus, r, p, m, and (v.,) are known as functions of p 0 , ro, and M. While ro 0 and p are generally specified, the Mach number must be determined everywhere in the duct to specify the flow completely. For practica! purposes, the back pressure in reservoir 11 is specified and a trial-and-error solution is necessary to determine the temperature, pressure, and Mach number distributions. We may do so by assuming a mass fl.ow rate and using Eq. 10.4-25 to calculate the Mach number in the duct. Once we know M, we may determine the pressure distribution by Eq. 10.4-17. Ifthe calculated pressure at the exit of the nozzle is different from the back pressure in reservoir 11, we assume a new value of m and repeat the calculation. Rather than doing so, we will examine the results of these equations by specifying the pressure ratio p/p 0 and determining the distribution of M and m/ A throughout the nozzle. By working in terms of the mass flow rate per unit area, m/A, we avoid specifying the geometry of the nozzle. For this example we have assumed that the gas is air and the stagnation temperature is 300°K. The results are shown in Fig. 10.4-3. We want to think of the pressure ratio p/p0 as being physically the back pressure divided by the stagnation pressure. When we loo k at it in this fashion, the points on the curve represent the values of M and mfA at the exit of the nozzle shown in Fig. 10.4-l. Let us imagine that the back pressure is initially
-410
Compressible Flow
Chap. 10
Sec. 10.4
lsentropic Nonle Flow
mlA
M
p/pO Fig. 10.4-3. Variation of M and rii/A for isentropic ftow.
equal to the stagnation pressure and that no flow takes place between the two reservoirs. In Fig. 10.4-3, then, p/p0 = 1 and M= m/ A =O. Now let us lower the pressure in reservoir 11, which according to Fig. 10.4-3, causes an increase in both the Mach number and mfA entirely in accord with our intuition. When the pressure ratio reaches a value of about 0.5, the mass flow rate per unit area reaches a maximum and begins to decrease. As the back pressure tends to zero (p/p 0 - 0), the flow rate tends to zero while the Mach number steadily increases to values greater than 10. This behavior is certainly not in accord with our intuition, and we had better reserve judgment on the validity of these curves. Experimental evidence is in order here, and the results indicate that mfA is constant at the maximum value even though the back pressure is continually decreased. The Mach number at the exit of the nozzle is also constant at a val1,1e of 1.0, and the flow is termed critica/. This situation is similar to the critica! flow we encountered in the study of open channel flow in that the flow rate is not controlled by the back pressure (downstream conditions) for sonic (critical) or supersonic (supercritical) flows. Figure 10.4-4 shows a Schlieren photograph of an air jet issuing from a convergent nozzle. The Mach number at the exit is 1.0 and the pressure at the exit is 105 times the back pressure. Under these conditions, there is a sudden expansion of the air stream as it lea ves the nozzle and the flow is no longer one-dimensional. lt has overexpanded and a shock wave has formed in the jet downstream from the nozzle exit. The performance curves for a convergent nozzle are shown in Fig. 10.4-5 and indicate the constant flow rate achieved for values of the pressure ratio less than the critica! value.
flg. 10.4-4. Flow issui Photograph Center, Va.
-----~--(al ~--(b) ~......--(e)
---(d) X
Flg. 10.<4-5. Performance
Compressible Flow
Chap. 10
Sec. 10.4
lsentropic Nozzle Flow
411
mlA /A
m/A for isentropic flow. no flow takes place between the = 1 and M= m/A =O. Now let according to Fig. 10.4-3, causes and m/A entirely in accord with our a value of about 0.5, the mass and begins to decrease. As the the ftow rate tends to zero while the greater than 10. This behavior is , and we had better reserve judgment and the results indicate that mjA though the back pressure is conat the exit of the nozzle is also is termed critica/. This situation is in the study of open channel flow by the back pressure (downstream · (supercritical) flows. of an air jet issuing from a at the exit is 1.0 and the pressure at Under these conditions, there is a leaves the nozzle and the flow is no and a shock wave has formed exit. The performance curves for a .4-5 and indicate the constant flow ratio less than the critica! value.
Fla. 10.4-4. Flow issuing from a converging nozzle. Photograph courtesy NASA Langley Research Center, Va.
--4---(o) ~+--(b)
-------(e)
m A
t------.
----(d) X
o
0.5 p/po
Fla. 10.4-5. Performance curves for a converging nozzle.
412
Compressible Flow
Chap. 10
Choke flow
The condition of maximum ftow rate observed in a converging nozzle is called choke jlow, and all engineers who work with fluids should recognize its existence as a fact of life. lt would be possible for an engineer, given a pressure ratio less than the critica! value, to "plug and chug" with the appropriate equations and calculate a mass flow rate for converging nozzle that is less than the value occurring at the choke flow condition. There are two other ways in which choke flow may occur. While these mechanisms are too complex for us to analyze in this limited treatment of compressible ftow, they deserve a comment. In a converging nozzle, we accelerated the flow by decreasing the area of the duct; however, there are other ways in which the flow can be accelerated provided the flow is compressible. Let us consider first the case of compressible, adiabatic flow in a long duct of constant cross-sectional area. The decrease in pressure gives rise to a decrease in the density, which in turn forces the velocity to increase. lf the pressure drop is suffi.ciently large, the velocity may reach sonic velocity at sorne point in the duct and choke flow will occur because the downstream conditions can no longer influence the flow rate. Choke flow may also be brought about by heating the gas. Such heating causes a decrease in density and an increase in velocity; thus, sonic velocity may be attained and a choked condition occurs. lt also follows that cooling the gas m:ay prevent the occurrence of choke flow. Converging-diverging nozzles
Sec. 10.4
lsentropic Nczzle Flow
As we begin our progress throug increases because of the decrease in because the decrease in area accelerat pressure ratio, the Mach number is l. are to continue along the performan mf A is going to decrease. In our pre intuition told us that points on the E were inadmissible. Experiments con choke flow was introduced. In this e imagining an increased cross-sectio decrease in m/ A, but what about the M continuing to increase while the are2 opposition to our intuition, but if we ¡ find the velocity does increase in t velocities are obtained. Subsonic and supersonic flow
The increasing velocity in the divet phenomenon which we need to analyz density decreasing faster than the area process in concrete terms. Starting carry out the differentiation and divi~
l(dp) + (v.,)1d.d
; dx
Differentiation of the energy balance 1 We now consider isentropic flow through the converging-diverging nozzle shown in Fig. 10.4-6. The equations derived at the beginning of this section apply to this system; however, in examining the performance curves in Fig. 10.4-3 we will proceed in a slightly different fashion. In this case, let us imagine that p 0 and PB are Reservoir O Pressure =p8 fixed, thus fixing the mass flow rate. Reservoir 1 Po T" The pressure ratio p/p0 decreases as we proceed through the nozzle from reservoir 1 to reservoir 11, and we ask ourselves the question, "Can we specify a value of A for each value of p/p0 such that both the performance curves in Fig. 10.4-3 and our Flr. 10.4-6. Converging-diverging nozzle. intuition ·are satisfied ?"
dh dx
+ (vz)
which we may simplify further if we dh
=
T
For isentropic fiows, we may write
(~~).Substitution into Eq. 10.4-28 yields a equation: (vz) (d(v"')) dx
s
Compressible Flow
Chap. 10
observed in a converging nozzle is work with fluids should recognize be possible for an engineer, given a to "plug and chug" with the approrate for converging nozzle that is flow condition. choke flow may occur. While these analyze in this limited treatment of In a converging nozzle, we area of the duct; however, there are accelerated provided the flow is
~m¡pressit>Ie,
adiabatic flow in a long decrease in pressure gives rise to forces the velocity to increase. If velocity may reach sonic velocity will occur because the downstream rate. by heating the gas. Such heating in velocity; thus, sonic velocity occurs. lt also follows that cooling flow.
Sec. 10.4
lsentropic Nczzle Flow
413
As we begin our progress through the nozzle, p/p0 decreases and m/A increases because of the decrease in area. The Mach number also increases because the decrease in area accelerates the flow. When we reach the critica! pressure ratio, the Mach number is l.O and m/ A is a maximum. Now, if we are to continue along the performance curve, the area A must increase if m/A is going to decrease. In our previous example of a converging nozzle, intuition told us that points on the performance curve for supersonic flow were inadmissible. Experiments confirmed our intuition, and the idea of choke flow was introduced. In this case intuition does not restrict us from imagining an increased cross-sectional area giving rise to the predicted decrease in m/ A, but what about the Mach number (and therefore the velocity) continuing to increase while the area increases? This process is surely in opposition to our intuition, but if we again turn to experimental evidence we find the velocity does increase in the diverging section and supersonic velocities are obtained. Subsonic and supersonic flow
The increasing velocity in the diverging portion of the nozzle is a curious phenomenon which we need to analyze further. Certainly it results from the density decreasing faster than the area increases, but we need to examine the process in concrete terms. Starting with the mass balance (Eq. 10.4-1a), we carry out the differentiation and divide by p
l(dp\ p dx}
+ _1 d(v.,) +!_dA= 0 (v.,) dx
A dx
(10.4-27)
Differentiation of the energy balance (Eq. 10.4-9) yields through the converging-diverging derived at the beginning of this examining the performance curves Fig. 10.4-3 we will proceed in a different fashion. In this case, us imagine that p0 and Pn are thus fixing the mass flow rate. pressure ratio p/p0 decreases as proceed through the nozzle from rp ~,,.rvm'r I to reservoir 11, and we ask the question, "Can we a value of A for each value 0 p/p such that both the perforcurves in Fig. 10.4-3 and our ·are satisfied ?"
dh dx
+ (v.,) d(v.,) = 0 dx
(10.4-28)
which we may simplify further if we use the thermodynamic relationship
dh
= Tds +! dp p
(10.4-29)
For isentropic flows, we may write
(~~).=~e~).
(10.4-30)
Substitution into Eq. 10.4-28 yields an interesting special form of the energy equation: (v.,) (d(v.,))
dx
s
=
_l(dp) p dx .
(10.4-31)
Compressible Flow
Chap. 10
This result is identical to the momentum equation (Eq. 10.4-7) ifthe entropy is constant. Thus, one-dimensional isentropic fiows may be solved by means of the mass and total energy equations, the momentum equation yielding no new information. Directing our attention now to Eq. 10.4-27, we specify that the fiow be isentropic and write the density gradient as,
(~:). = (:;).(~:).
(10.4-32)
and use Eq. 10.4-31 to eliminate the pressure gradient ~
~\ = -p(OP\ (v.,>(d(v.,)) ( dx}. op}. dx •
(10.4-33)
Substitution of this result into Eq. 10.4-27, and use of Eq. 10.3-14 to express (opfop). in terms of the speed of sound yield
(
- .!.2 + _1_)2 (v.,) d(v.,) + .!_dA = 0 c
(v.,)
dx
A dx
S.C. 10.4
lsentrople Nozzle Flow
incrbses with increasing cross-sectio cross-sectional area. We must keep i isentropic fiow that may not occur i ducts. The other fiow properties a changes from subsonic to supersonic; In studying the performance cu tinuously decreasing pressure ratio p that lead to a converging-diverging results for isentropic fiow from still that A(x) is speci.fied so that the no 10.4-6. If we specify the stagnation ~ derived equations to determine the 1 starting at reservoir 1 where p = p 0 p = PB· Such calculations lead to Throot
(10.4-34)
Expressing this result in terms of the Mach number, and rearranging, we obtain d(v.,) dA
= - [~ (1
-
M~J-1
(10.4-35)
(v.,)
From this result we see that if M < 1, the velocity increases if A decreasesi.e., the fiow is accelerated in a converging section and decelerated in a diverging section. However, if the fiow is supersonic, M > 1 and the velocity Table 10.4-1 SUBSONIC ANO SUPERSONIC FLOW IN CONVERGINO ANO DIVEROING CONDUITS
1. 1~
.• 0 1vergmg
Parameter
Subsonic
Supersonic
Subsonic
Supersonic
Velocity Pressure Temperature Density Entropy
increase decrease decrease decrease constant
decrease increase increase increase constant
decrease increase increase increase constant
increase decrease decrease decrease constant
o Fl1. 10.4-7. Pressure ratio in
first two curves, a and b, are for Iow always less than 1.0. Here, the nozzl being low where the velocity is high the diverging section. The pressure downstream cross-sectional area is s (a), (b), and (e) indicate increasing causing a Mach number of 1.0 to be gives the maximum flow rate for the and the solution for the pressure is
Compressible Flow
Chap. 10
equation (Eq. 10.4-7) if the entropy flows may be so1ved by means the momentum equation yielding no now to Eq. 10.4-27, we specify density gradient as, (10.4-32)
(10.4-33) , and use of Eq. 10.3-14 to express yield 1 dA
+--=0 Adx
S.C. 10.4
lsentropk Nozzle Flow
415
incrbses with increasing cross-sectional area and decreases with decreasing cross-sectional area. We must keep in mind that these conclusions hold for isentropic flow that may not occur in long reaches of converging-diverging ducts. The other flow properties also behave differently when the flow changes from subsonic to supersonic; these are indicated in Table 10.4-1. In studying the performance curves in Fig. 10.4-3, we specified a continuously decreasing pressure ratio p/p0 , and found that values of A(x) exist that lead to a converging-diverging nozzle. We now wish to look at the results for isentropic flow from still another point of view. Let us assume that A(x) is specijied so that the nozzle resembles the one shown in Fig. 10.4-6. If we specify the stagnation properties, A(x), and m, we can use the derived equations to determine the pressure at every point in the nozzle, starting at reservoir 1 where p = p 0 and continuing to reservoir 11 where p = p 8 • Such calculations lead to the curves shown in Fig. 10.4-7. The Throat
(10.4-34)
number, and rearranging, we
Ps=.q, (10.4-35)
Ps=Pc
velocity increases if A decreasessection and decelerated in a supersonic, M > 1 and the velocity
Forbidden region
r.l~
.• D 1vergmg
decrease increase increase increase constant
increase decrease decrease decrease constant
0
X
Fl1. 10.4-7. Pressure ratio in a converging-diverging nozzle.
first two curves, a and b, are for low flow rates where the Mach number is always less than 1.0. Here, the nozzle acts like a Venturi meter, the pressure being low where the velocity is high and a pressure recovery taking place in the diverging section. The pressure recovery is incomplete because the downstream cross-sectional area is smaller than the upstream area. Curves (a), (b), and (e) indicate increasing mass flow rates with the latter flow causing a Mach number of 1.0 to be reached at the throat. This condition gives the maximum flow rate for the nozzle at these stagnation conditions, and the solution for the pressure is double valued and may follow either
416
Compressible Flow
Chap. 10
Sec. 10.5
Shock Waves
curve (e) or curve (d). In actual practice, the pressure distribution at this flow rate would be determined by whether the back pressure were equal to
Peor Pd· But what of the "forbidden region" between Pe and Pa? Our analysis indicates that this condition simply cannot be reached, yet we know that experimentally we may adjust the back pressure to take on values between Pe and Pd· Once again, experimental evidence is in order, we find that if the back pressure is adjusted to sorne value between Pe and pd, a shock wave develops and there is a sudden transition from supersonic to subsonic flow accompanied by an increase in entropy. The pressure distributions for such flows are indicated in Fig. 10.4-8. This phenomenon is similar to the sudden Throot Flg. 10.5-1. Control of a shock wave.
increases. We will now reformulate equations for the control volume sho shock wave) without imposing the res that in the previous section the mo specify the flow; however, when we that the momentum equation is requt Mass balance
o
X
Since p(v.,)A is a constant, the m
Fig. 10.4-8. Formation of shock waves in a converging-diverging nozzle.
transition that occurs at a hydraulic jump in open channel flow. Note that these shock waves are advancing into the flowing stream at speeds greater than the speed of sound just as the hydraulic jump advances into a flowing stream at speeds greater than Jih-i.e., the velocity of propagation for a small disturbance.
or by Eq. 10.4-24 we may write
M (Y"_
Pt \) "R'T: Momentum balance
10.5
Shock Waves
In attacking this problem, we will make use of the experimental observation that shock waves occur over very thin regionst and show that a transition from supersonic to subsonic flow is permissible, provided the entropy t The thickness of a shock wave depends on the Mach number, temperature, and pressure and is on the order of 10- • in. (see Ref. 1, Vol. 1, p. 134).
To obtain the momentum balan x 1 to x 2 to obtain
m( (v.,)2 -
Here we have assumed that the sho be neglected relative to variations in
Compressible flow
Chap. 10
Sec. 10.5
417
Shock Waves
the pressure distribution at this the back pressure were equal to " between Pe and pd? Our analysis ot be reached, yet we know that pressure to take on values between is in order, we find that if the between Pe and pd, a shock wave from supersonic to subsonic flow The pressure distributions for such phenomenon is similar to the sudden
Flg. 10.5-1. Control volume for the analysis of a shock wave.
increases. We will now reformulate the mass, momentum, and energy equations for the control volume shown in Fig. 10.5-1 (which contains the shock wave) without imposing the restriction of constant entropy. Remember that in the previous section the momentum equation was not required to specify the flow; however, when we allow the entropy to vary we will find that the momentum equation is required. Mass balance
Since p(v.)A is a constant, the mass balance simply reduces to in a converging-diverging nozzle.
in open channel flow. Note that the flowing stream at speeds greater ic jump advances into a flowing , the velocity of propagation for a
(10.5-1) or by Eq. 10.4-24 we may write P1
M
[Y'_
1\} R"J;
-
P2
M
Jy 2\} ~
(10.5-2)
Momentum balance
use of the experimental obserthin regionst and show that a transiis permissible, provided the entropy on the Mach number, temperature, and . 1, Vol. 1, p. 134).
To obtain the momentum balance, we simply integrate Eq. 10.4-6 from x 1 to x 2 to obtain (10.5-3) Here we have assumed that the shock is so thin that variations in A(x) may be neglected relative to variations in (v.,) and p .
<418
Compressible Flow
Chap. 10
Problems
Substitution of this result into Eq. 10.5
Energy balance We derived the energy balance in the previous section without the restriction of constant entropy, and we need only note that the temperature is given by Eq. 10.4-15; thus, the ratio of temperatures is 1 +y-1M2 2 1
T2 T¡
s2
-
s1 =(el>- iR) In
We may now use Eq. 10.5-4 to express Mach numbers, and we obtain
s2
(10.5-4)
-
s1
1+ -
= (cP- iR) ln (
1 +y-1M2 2 2
1+
Now, the second law ofthermodynamio process; therefore, Eq. 10.5-12 indicatt
Entropy change Starting with Eq. 10.2-10, we can quickly derive the thermodynamic relationship 1 (10.5-5) dh = Tds dp
M 2 =M¡,
or
+p
Rearranging and integrating ds between x 1 and x 2 gives 2
s2 -
S¡
=
2
2
f f f d~ ds =
1
(10.5-6)
d: -
1
1 p
Restricting our ana1ysis to fluids that may be treated as perfect gases, we may carry out the integration to obtain s2 - s1 =el> In
(i) - R In (~)
(10.5-7)
The ratio T 2/T1 is already available in Eq. 10.5-4, and we need only find f2fp 1 to determine the entropy change across the shock. It is easiest to do so by means of the mass balance and the perfect gas law. Rearranging Eq. 10.5-1 and neglecting the change in area from x 1 to x 2 , we obtain
M 2
Our analysis indicates that only transit are possible. The analysis does not in nor does it indica te that the transition si 1t only states that such a transition is increases. From the analysis of isentropic fl conditions (i.e., a back pressure in tH the possibility of an isentropic flow. might ask, "Why isn't there a smooth sonic to subsonic flow ?" The answer able at present. We can only state that flow occur abruptly.
PR.OB
(10.5-8)
10-l. Integrate Eq. 10.1-19 with respect the first law of thermodynamics meaning of each term.
Expression of the velocities in terms of the Mach number (Eq. 10.4-23) before and after the shock wave yields
10-2. Starting with Eq. 10.2-15, derive for a control volume, "Ya(t).
P2 P1
(v.,)t
-=--
(v.,)2
112 -P2 = M1(T1) p1 M2 T2
(10.5-9)
10-3. Derive Eq. 10.4-7 by integrating ti
cross-sectional area A(x)-i.e.,
f
and the perfect gas law may be used to express the pressure ratio as P2 p1
=
12 P2T2 = M1(T2) 1 p1T1 M2 T1
(10.5-10)
i · p [:;
A(z)
t
JdA
+ v • Vv
=
-
j
A(
Thus, we could have the situation M
Compressible Flow
Chap. 10
Problems
419
Substitution of this result into Eq. 10.5-7 and rearrangement give previous section without the need only note that the temperature of tempera tu res is
s2
-
s1
= (e~ - iR) In (~) - R In (~:)
We may now use Eq. 10.5-4 to express the temperature ratio in terms of the Mach numbers, and we obtain
(10.5-4)
s2- s1
=(e~- iR) In(¡+~ M:) -R In (~1) 1
derive the thermodynamic
(10.5-11)
+
r___! M2 2
(10.5-12)
2
2
Now, the second law ofthermodynamics tells us that lls 2 Ofor any adiabatic process; therefore, Eq. 10.5-12 indicates either
M 2 =M¡,
lls =O
M 2 < M1 ,
lls > O
or
(10.5-5) x 1 and x 2 gives
J -f 2
1
dh T
1
2
dp pT
(10.5-6)
be treated as perfect gases, we may
(10.5-7) Eq. 10.5-4, and we need only find the shock. lt is easiest to do so perfect gas law. Rearranging Eq. from x 1 to x 2 , we obtain
(10.5-8) 10.4-23)
Our analysis indicates that only transitions that decrease the Mach number are possible. The analysis does not indicate that such transitions do occur, nor does it indica te that the transition should be from supersonic to subsonic.t lt only states that such a transition is possible, and if it occurs, the entropy increases. From the analysis of isentropic fiows we know that there are certain conditions (i.e., a back pressure in the "forbidden region") which disallow the possibility of an isentropic fiow. With this information in hand, one might ask, "Why isn't there a smooth, nonisentropic transition from supersonic to subsonic fiow?" The answer to this question is apparently unavailable at present. We can only state that transitions from supersonic to subsonic fiow occur abruptly. PROBLEMS
10-l. Integrate Eq. 10.1-19 with respect to time to obtain the traditional form of the first law of thermodynamics for open systems. Carefully explain the meaning of each term. 10-2. Starting with Eq. 10.2-15, derive the general macroscopic entropy balance for a control volume, '"~'"a(t).
(10.5-9)
10-3. Derive Eq. 10.4-7 by integrating the x-direction equation of motion over the cross-sectional area A(x)-i.e.,
J
the pressure ratio as
i · p [:;
AW
(10.5-10)
t
+
JdA = - Ji ·
v • Vv
Vp dA
+
AW
Thus, we could have the situation M 1 > M 2 > 1.
J
i · pg dA
AW
+
p.
J
i · V 2v dA
AW
420
Compressible Flow
Chap. 10
Make the same simplifying assumptions as in Sec. 10.4, and use the Leibnitz rule (see Prob. 3-5) for interchanging integration and differentiation. 10-4. For a given value of T o, show that an upper limit exists for the velocity given by (v., )
<
..J~ y-=-} ~
10-5. A perfect gas at T1 and p¡ in a large reservoir flows isentropically through a convergent nozzle of area A2 into the atmosphere where the pressure is p 0 • Derive expressions for the mass flow rate mtaking into account the phenomenon of choke flow. (This type of calculation would be a key step in the design of a safety valve for a high-pressure chemical reactor.)
Flow Around lmmersed Bodies
10-6. If ata shock such as that shown in Fig. 10.5-1, M 1 = 3.0, T1 = 1000°R and p 1 = 0.2 atm, find the Mach number and pressure after the shock. Assume 1' = 1.5. 10-7. For isentropic flow in a divergent channel, the density decreases 14 per cent (p2 = 0.86p1 ) for a 10 per cent increase in the cross-sectional area (A 2 = 1.10 A1). Calculate the change in velocity between these two points. 10-8. A stream of air flows ata Mach number of 0.8. lf a Pitot tube is used to measure the velocity, compute the error incurred by the use of the incompressible form of Bernoulli's equation. 10-9. Air ftowing in a duct enters a nozzle at a velocity of 5000 ft/sec, a pressure of 100 psia, and a temperature of 800°F. The nozzle is designed to increase the pressure isentropically to 125 psia. If the mass ftow rate is 1O lbm/sec, what is the downstream cross sectional area of the nozzle? 10-1 O. Air is contained in a tan k at 200 psia and 7rF. lf the flow of a ir through the outlet valve is approximated as isentropic and one-dimensional, what is the maximum flow rate into the surrounding atmosphere? Take the crosssectional area at the valve as 1.0 in. 2 and the ambient pressure as 14.7 psia. What would the ftow rate be if the pressure in the tank were increased to 400 psia?
The objectives of this chapter are to ena associated with ftow around imm a rational explanation of the drag shown in Fig. 8.2-6 in Chap. 8. Bou thus giving sorne practice in setting U balances. We shall also use boundary performing an order of magnitude am
11.1
Descri ption of Flow
The study of ftow around imme subject to discuss with any degree of the fact that there are three distinct, inftuence the force exerted by the ftui the ftow past a thin airfoil at a zero a number. At high Reynolds numbers, except in a thin region close to the outside this thin region is governed a called irrotational. As the name impli
4
Compressible Flow
Chap. 10
as in Sec. 10.4, and use the Leibnitz integration and differentiation. an upper limit exists for the velocity
reservoir flows isentropically through a atmosphere where the pressure is p 0 • rate mtaking into account the phenomcalculation would be a key step in the chemical reactor.)
Flow Around Immersed Bodies
11
10.5-1, M 1 = 3.0, T1 = 1000•R and and pressure after the shock. Assume the density decreases 14 per cent in the cross-sectional area (A 2 = between these two points. of 0.8. If a Pitot tube is used to error incurred by the use of the incomat a velocity of 5000 ft/sec, a presof 800°F. The nozzle is designed to to 125 psia. lf the mass ftow rate is 10 sectional area of the nozzle? and n•F. If the flow of air through the and one-dimensional, what is the atmosphere? Take the crossand the ambient pressure as 14.7 psia. pressure in the tank were increased to
The objectives of this chapter are to introduce the student to the phenomena associated with ftow around immersed bodies and to provide him with a rational explanation of the drag coefficient-Reynolds number curves shown in Fig. 8.2-6 in Chap. 8. Boundary layer theory will be introduced, thus giving sorne practice in setting up and solving differential-macroscopic balances. We shall also use boundary layer theory to explore the method of performing an order of magnitude analysis.
11.1
Description of Flow
The study of ftow around immersed bodies is an enormously difficult subject to discuss with any degree of exactness. The difficulty results from the fact that there are three distinct, and fairly complex regions of ftow that inftuence the force exerted by the fluid on the body. Figure ll.l-1 illustrates the flow past a thin airfoil at a zero angle of inclination and a high Reynolds number. At high Reynolds numbers, inertial effects predominate everywhere except in a thin region close to the surface of the solid body. The ftow outside this thin region is governed almost entirely by inertial effects and is called irrotational. As the name implies, fluid elements in this region undergo -421
Flow Around lmmersed Bodies
Chap. 11
Sec. 11.2
The Suddenly Accelerated Flat Plac Fluid extends
y
-oo~--c=========~
t sO, the plote is stoti t >O, the veiocity of ti
Fl¡. 11.2-1. The suddenly acc lrrotationol
Fl¡. 11.1-1. Flow regimes around an immersed body.
translation but do not rotate; therefore, there is no shear deformation and viscous effects are negligible. A great deal is known about irrotational flow, because in the absence of viscous effects the equations of motion can be solved by fairly simple mathematical techniques. While the study of irrotational flow is well within the student's capabilities, it will be treated only · qualitatively here, and we will concentrate our attention on the boundary layer. The wake is extremely difficult to treat theoretically, and again we will restrict ourselves to qualitative comments regarding this region. Near the solid surface, viscous effects become important, and this flow region is called the boundary layer. The equations of motion here reduce to what are known as the Prandtl boundary layer equations. In the next section, we willlay the ground work for both the derivation and the solution of these equations.
the area of fluid mechanics. The systern Fig. 11.2-1. We start with the two-dimensiona1
. p-+v-+v-(ov~ OV~ ov.,) ot ~ ax ay ovtl ovtl ov") = - 1¡ p (- + V~- + V at ax ay , 11
11 -
and the continuity equation
OV~
ox
+~
é
Since the plate is infinite in the x-directi and p are not functions of x. The conl to
ovtl = oy
and application of the boundary condi1
11.2 The Suddenly Accelerated Flat Plate As an introduction to the analysis of boundary layers we will examine the fluid motion that results when an infinitely wide, infinitely long, flat p1ate is suddenly accelerated from rest to a constant velocity, u0 • We may rightly ask, "Why study a hypothetical flow that can never be realized in practice ?" The answer is that by performing the analysis we will gain intuition-i.e., we may improve our skill at guessing the flow topology. In addition, the equations we will solve are identical to those describing transient heat and mass transfer into a semi-infinite medium; thus, the analysis has value beyond
B.C. 1: V"= O, leads us to the conclusion that V11 is zero the equations of motion reduce to p
ov.,) (a¡ =¡
o=-(~~
We need solve only the first ofthese te
Flow Around lmmersed Bodies
Chap. 11
Sec. 11.2
-423
The Suddenly Accelerated Flat Plate
Fluid extends to y =oo y
-oo~--~c:::::::::::::::~:::I::::x::~:::Jr---~+oo t sO, the piote is stotionory t >O, the velocity of the plote is o constont, u0 Flg. 11.2-1. The suddenly accelerated flat plate.
the area of fluid mechanics. The system under consideration is illustrated in Fig. 11.2-1. . We start with the two-dimensional fonp. of the Navier-Stokes equations there is no shear deformation and is known about irrotational flow, the equations of motion can be techniques. While the study of ircapabilities, it will be treated only our attention on the boundary treat theoretically, and again we will regarding this region. become important, and this flow equations of motion here reduce to layer equations. In the next both the derivation and the solution
Flat Plate boundary layers we will examine the wide, infinitely long, flat plate is eonsta111t velocity, u0 • We may rightly can never be realized in practice ?" analysis we will gain intuition-i.e., the flow topology. In addition, the those describing transient heat and ; thus, the analysis has value beyond
ov., 11 ov.,) --'--+,u-+. op (o 2v.,2 02v.,) (otav., + v-+v"' ox oy - ax ox oy2 p(avti + v., ovti + vti ovti) = - op- pg + .u(éJ2v112+ éJ2v11) ot ox oy oy ox oy2 p--
(11.2-1)
(11.2-2)
and the continuity equation (11.2-3)
1,
Since the plate is infinite in the x-direction, we can argue logically that v.,, v and p are not functions of x. The continuity equation immediately reduces to
ovti =o oy
(11.2-4)
and application of the boundary condition
B.C. 1 :
1
v = O,
y=
O
(11.2-5)
11
leads us to the conclusion that v is zero everywhere. Under these conditions, the equations of motion reduce to
2 P(ov.,) =,u (o v.,) ot ol o=-(~~)- pg
(11.2-6) (11.2-7)
We need solve only the first ofthese to determine the velocity as a function
Flow Around lmmersed Bodies
Chap. 11
of y and t. Defining the dirnensionless velocity Uz as U
=
z
Vz
(11.2-8)
Uo
Sec. 11.2
The Suddenly Accelerated Flat Plat
At this point we can see that setting a sirnplifying our equation. Substituting setting a = 1, we get
we rnay write Eq. 11.2-6 in the forrn
(~~z)bytb-1 =
(o Uz) 2
oUZ ='V (11.2-9) ot ol which we rnust solve subject to the following boundary conditions B.C. 2: Uz =O, t =O, O ::;;y::;; oo (11.2-10)
= 1,
B.C. 3:
Uz
B.C. 4:
Uz =O,
t >O,
y= O
(11.2-11)
y= oo,
t ¿ O
(11.2-12)
The boundary condition B.C. 2 is often referred toas an initial condition since it specifies the velocity field at sorne initial time. The boundary condition B.C. 3 indicates that for times greater than zero the plate is rnoving at a constant velocity, thus an infinite acceleration is required. While this is physically irnpossible, it provides us with sorne insight as to how disturbances originating at salid surfaces are propogated into the surrounding fluid. Equation 11.2-9 is a partial differential equation. Until now such equations have been considered beyond the scope of this text; however, we can solve this one without expending too rnuch effort, and in the process we can becorne acquainted with a fairly general rnethod of solving partial differential equations. A solution rnay be obtained by rneans of Laplace transforrns, but here we will use a technique known as a similarity solution. Thus, we define a new variable 1J in terrns of powers of y and t, (11.2-13) where a and b are as yet undeterrnined constants, and we seek a solution of the forrn (11.2-14) Uz(y, 1) = Uz[1J(y, !)] If we find that Uz can, indeed, be represented in terrns of the single variable rJ, we will have been successful. lf we cannot represent Uz in terrns of 1], we rnust try other rnethods of solution. We note first that the two partial differentials in Eq. 11.2-9 rnay be written oUz ot
=
ot
oUz = d Uz (01]) 2
ol
2
drJ 2
2
2
ay
1] =
yt
Equation 11.2-16 reduces to
(:~x) = - ~ Putting the boundary conditions in B.C. 2':
Uz= O,
B.C. 3':
uz =
B.C.4' :
1, Uz= O,
Here we see that boundary conditions E terrns of the transformation of variables h We solve the differential equation t P=
-~
~
where
P=·
Separating variables and integrating gi1 lnP
=-
or
(dUz) 01J = (dUz) byatb-1 drJ
If we could sornehow elirninate y and would be reduced to solving an ord· accornplish this by setting b = -t, ther
(11.2-15a)
d11
au.,_ e
dr¡2 -
+ dUz (d 1J)
1
2
drJ
d Uz 1 b 2 =-(ayat) d1] 2
Forrnation of the indefinite integral giv
ol
+ dUz -[a(a- 1)y
(11.2-15b)
a-2 b
d1J
t]
Flow Around lmmersed Bodies
Chap. 11
velocity U., as (11.2-8)
Sec. 11.2
425
The Suddenly Accelerated Flat Plate
At this point we can see that setting a = 1 will eliminate one term, thus simplifying our equation. Substituting Eqs. 11.2-15 into Eq. 11.2-9, and setting a = 1, we get 2
dU"')bytb-t = ( dr¡
(11.2-9) boundary conditions
o :S:: y
:S::
00
y=O t¿O
(11.2-10) (11.2-11) (11.2-12)
11
(ddr¡U"')t 2
b
(11.2-16)
If we could somehow eliminate y and t from this equation, our problem would be reduced to solving an ordinary differential equation. We can accomplish this by setting b = -!, thereby defining r¡ as,
r¡ =
(11.2-17)
yr-t /2
Equation 11.2-16 reduces to 2
rred to as an initial condition since it time. The boundary condition B.C. 3 the plate is moving at a constant velocWhile this is physically impossible, how disturbances originating at solid fluid. equation. Until now such the scope of this text; however, we much effort, and in the process we general method of solving partial be obtained by means of Laplace · known as a similarity so/ution. of powers of y and t,
2
U.,)
dUx) = _ 2v(d ( dr¡ r¡ dr¡ 2
(11.2-18)
Putting the boundary conditions in terms of r¡, we find B.C. 3':
U.,=O, U.,= 1,
r¡ = 00 r¡=O
B.C.4':
U.,=O,
r¡ =
B.C. 2':
00
(11.2-19) (11.2-20) (11.2-21)
Here we see that boundary conditions B.C. 2' and B.C. 4' become identical in terms of the transformation of variables from y and t to 11· We solve the differential equation by first reducing the order to obtain p
(11.2-13)
= _
2v(dP) r¡ dr¡
(11.2-22)
where P=dU., dr¡
(11.2-14) L""~·~~·~-u
in terms of the single variable represent U., in terms of r¡, we
Separating variables and integrating gives
In P
¡trerentíals in Eq. 11.2-9 may be written
(~~"')
byatl>-
1
= -
2
!l._ 4v
+ C1
or (11.2-15a)
dU.,- C' e·· -·"14• d172 -
(11.2-23)
1
Formation of the indefinite integral gives
dU., -[a(a - 1)ya-2 tb ) dr¡
(11.2-15b)
f
~
u"' = e;
o
e-r"t 4 • d-r
+ c2
(11.2-24)
Flow Around lmmened Boclia
Chap. ••
Sec. ll.l
The Suddenly Accelerated Flat Plate
where -r is the dummy variable of integration. It will be convenient to define a new dummy variable Eby T
E=allowing us to write Eq. 11.2-24 as
---A~
#
~rv4.
u;l: =
c~J e-~~ dE+ C
2
u.
(11.2-25)
o
By boundary condition 3', C2 = 1, and application of boundary condition 2' leads to the result
o
(11.2-26)
Flg. 11.1-1. Velocity profil accelerated ftat plate.
The definite integral in Eq. 11.2-26 is known to be equa1 to expression for the velocity becomes
U:1:=
1 -2-
f
J;,o
.¡;;.¡2, and our Approximate solution
~¡v'4v
e-~ 2
dE
(11.2-27)
The latter term in this equation is known as the error function, abbreviated erf(r¡/../4v). Tabulated values of the error fl,lnction are available, 1 and our final expression is
U;l: =
1- erf ~
(11.2-28)
'Y 4vt
This result is plotted in Fig. 11.2-2. The key point of this analysis is that U,. reaches 99 per cent of the undisturbed value for y/../ 4vt = 1.8, and we can use this fact to define a "boundary layer" thickness () as
() = 3.6J;i
(11.2-29) For values of y larger than (), ·the velocity is, for all practica} purposes, zero. We may also interpret Eq. 11.2-29 by saying that the distance the disturbancc penetrates into the quiescent fluid is proportional to the square root of time t, and the kinematic viscosity, v. This result agrees with our intuition, which should tell us that a large viscosity leads toa rapid propagation of a disturbancc, and a zero value of the viscosity would not allow the disturbance to be fclt anywhere in the fluid. 1. M. Abramowitz and l. A. Stegun, eds., Handbook of Mathematical FunctiolfS (Washington, D.C.: National Bureau of Standards, 1964).
We will now solve this problem agai compare the results with the exact s~h "Why solve a problem wit~ an approxtm is already available-espectally when ~he any real situation ?" Aside fr?m keepmg that by comparing an approxtmate result sorne insight into the validity o~ th.e ¡ it provides an interesting apphcatton balance. The control volume for the approx1 l1.2-3. The upper boundary is located y-direction. We start w~th the complet and dot it with i to obtam
~~ f pvz dV +f pvz(v f
0
(1)
A,W
-1
We again impose the restriction that the v of x, and make the approximation (A) t
A. 1:
Flow Around lmmersed Bodia
Chap. 1 ,
Sec. 11.2
427
The Suddenly Accelerated Flat Plate
1---'....----i---- Exact solution t-----i - - - Approximate solution
(11.2-25) application of boundary condition 2'
o
o
(11.2-26)
Fls. 11.2-l. Velocity profiles for the suddenly accelerated flat plate.
known to be equal to
J;¡2, and our Approximate solution
(11.2-27) as the error function, abbreviated ft,mction are available, 1 and our (11.2-28) key point of this analysis is that U. value for y/../ 4vt = 1.8, and we ca: thickness ~ as (11.2-29) is, for all practica! purposes, zero. that the distance the disturbance to the square root of time t, agrees with our intuition, which a rapid propagation of a disturbance, not allow the disturbance to be felt
We will now solve this problem again, by an approximate method, and compare the results with the exact solution. The student might well ask, "Why solve a problem with an approximate method when the exact solution is already available--especially when the problem is somewhat removed from any real situation ?" Aside from keeping idle minds busy, the answer must be that by comparing an approximate result with the exact solution, we can gain sorne insight into the validity of the approximate method. In addition, it provides an interesting application of the macroscopic momentum balance. The control volume for the approximate solution is illustrated in Fig. 11.2-3. The upper boundary is located at y = ~(t) and is moving in the y-direction. We start with the complete macroscopic momentum balance and dot it with i to obtain
~~ Jpv., dV +Jpv.,(v f.w
J
i • t
(11.2-30)
-"'.w
We again impose the restriction that the velocity and pressure are independent of x, and make the approximation (A) that
Handbook of Mathematical FunctioiU 1964).
w) • n dA =
A.w
A. 1:
v.,
=o,
y ;;>8(t)
(11.2-31)
428
Flow Around lmmersed Bodies
Chap. JI
r:---------~ 6(f)
J -oo-1
1
1 ~
1 --;
1
1
1
1
The Suddenly Accelerated Flat Plate
We now make the assumption tha1 allowing us to express the velocity as
L.
1
Sec. 11.2
= v,
v.,(y, t)
A.2:
Note that this equation is comparable to
--•+00 7}
= .l
!5(
"----L ----.4
We further assume that the functional represented by a polynomial.
Flg. 11.2-l. Control volume for the suddenly accelerated ftat plate.
Of course, we know this approximation is very reasonable on the basis of the exact solution. A little thought will indicate that all the momentum flux terms are zero, and Eq. 11.2-30 reduces to L
L
dt ff pv., dy dx =f [r 6(1)
E_
o o
o
11 .,,
11=6(t)
r .,¡ Jdx 11
(11.2-32)
11=0
Substitution of Newton's law of viscosity for r 11., yields
v., = a + b(~)
A.3:
d JLJ d d JL{(ov.,). dt- o o pVz y X = ¡..t o -oy <>
-
1/=6(1)
(ov.,). }d oy v- o X
(11.2-33)
By approximation 1, we have specified v., = O for y ~ d(t). If we require the derivative to be a continuous function, then approximation 1 naturally leads us to the condition,t
(ov.,joy)
(!:"') =o,
v,.= O, ov.. _ o oy- , v,. = u
1
(11.2-34)
and Eq. 11.2-33 reduces to
(an intuit1
y= c5(t)
(continuit
y=O
(a bonafi
oyz- o, 0 Vz
-
y
=o
(from the
Equations 11.2-38 may be used to deter: obtain
6(1)
~ J pv., dy = dr o
y = d(t)
These three equations allow us to deten Eq. 11.2-37. We could obtain more derivatives of v,. to be zero at y = !5; h1 regarding the velocity profil~ ca~ be ?bu Examining Eq. 11.2-1, keepmg m nund we quickly conclude that 2
y= d(t)
ce
To determine the constants in the po . information regarding the velocity profil
0,
6 1
+
-¡..t
(ov.,) oy
(11.2-35) 11= 0
Here we have used the fact that v., is independent of x ; thus, the integration with respect to x can be dropped. This result states that, the force exerted by the plate on the fluid is equal to the time rate of change of the momentum of the fluid. tOne can also argue that all higher derivatives are zero at y
=
V =U
..
[1- ~(~·6. 2
o
Defining dimensionless variables as
v.,
U,.=- an Uo
we see that the velocity is given by
u.. = 1- !l
Flow Around lmmersed Bodies
Chap. 11
Sec. 11.2
We now make the assumption that the velocity profiles are similar, allowing us to express the velocity as
L.
v.,(y, t) =
A.2:
--,...+00
is very reasonable on the basis of the indicate that all the momentum flux to
-
T 11
.,¡
for
T 11.,
ov.,).
Jdx
(11.2-32)
11=0
yields
ay
11-o
x
(11.2-33)
v., = O for y ¿ !5(t). If we require the function, then approximation 1
y
= !5(t)
av.,) (ay
=
y !5(t)
(11.2-37)
A.3:
To determine the constants in the polynomial, we require sorne specific information regarding the velocity profile, which is given as
v.,=O,
y= !5(t)
(an intuitive approximation)
(11.2-38a)
ov.,_ o oy- ,
y= !5(t)
(continuity of stress)
(11.2-38b)
y=O
(a bonafide boundary condition)
(11.2-38c)
1
v., = u0 ,
(av.,). }d
- : ; 1/=d(t) -
(11.2-36)
We further assume that the functional dependence of v., on yjt5(t) can be represented by a polynomial.
the suddenly accelerated flat plate.
1/=d(t)
v.,[..L] !5(t)
Note that this equation is comparable to Eq. 11.2-14 if we designate r¡ as r¡
T 11 .,,
429
The Suddenly Accelerated Flat Plate
(11.2-34)
These three equations allow us to determine only three of the constants in Eq. 11.2-37. We could obtain more equations by requiring the higher deriva tives of v., to be zero at y = !5; however, a key piece of information regarding the velocity profile can be obtained from the differential equation. Examining Eq. 11.2-1, keeping in mind that p and v., are independent of x, we quickly conclude that
éJ2v., =O
ol
'
y=O
(from the differential equation)
(11.2-38d)
Equations 11.2-38 may be used to determine the constants a, b, e, and d to obtain
-¡.t-
(11.2-35)
(11.2-39)
11=0
~""'''~"""u''"' of x; thus, the integration
result states that, the force exerted time rate of change of the momentum
Defining dimensionless variables as
U - v., and . , - Uo
Y= .l ~
(11.2-40)
we see that the velocity is given by
U., = 1 -
! Y+ ! Y3
(11.2-41)
-430
Flow Around lmmersed Bodies
Chap. 11
S.C. 11.3
The Boundary Lay~r on a Flat Plate
Putting Eq. 11.2-35 into dimensionless form yields
~(oJlu., dY) dt o
=-
~(du., )
OdY
Ua;¡
(11.2-42) Y=O --"'
By substituting Eq. 11.2-41 and carrying out the differentiation and integration, we obtain a first-order differential equation for the boundary layer thickness.
do=(~) dt
_.... --"'
(11.2-43) --"'
Separating variables, integrating and imposing the boundary condition B.C. 1:
<5
=O,
t=O
--"'
(11.2-44)
we ha ve an expression for the boundary layer thickness as a function of time, <5
= 2.8J;i
(11.2-45)
While the boundary layer thickness predicted by the approximate solution is very different from the exact solution, the result is notas bad as it might seem. The comparison of velocity profiles is given in Fig. 11.2-2, and certainly the two profiles are in good agreement for values of y/#f less than 1.0. We conclude that this type of approximate analysis should yield reasonably satisfactory results and we are now ready to attack the more complex boundary layer flow illustrated in Fig. 11.3-1.
11.3 The Boundary Layer on a Flat Plate The flow we wish to consider in this section is illustrated in Fig. 11.3-1. lt consists of an infinite stream flowing past a thin flat plate of length L. The plate is infinitely wide and the flow is uniform at a velocity of U 00 • This flow has sorne of the same characteristics of the suddenly accelerated flat plate, especially if we view the problem as a flat plate moving at a velocity u 00 through an initially quiescent fluid . Under these conditions, the time t in the previous problem would be comparable to xfu 00 where x is measured from the leading edge of the plate. For this reason we may use Eq. 11.2-29 to estímate the boundary layer thickness as <5
~ 3.6-../vxfuoo
Fil. 11.3-1. Boundary layer
will be given and compared with an understood that the order of magnitude at analysis and has applications far beyond W e start with the two-dimensional continuity equation for steady, incomprc
v.,(ov.,) + v~~(ov.,) =-! ax oy P v.,(ovll) + v1 (ovll) = _! ax oy P ov., +OVIl ax oy
1
We use the free stream velocity and dimensionless variables
(11.3-1)
Our first objective in this section will be to perform an order of magnitude analysis of the equations of motion to obtain the Prandtl boundary layer equations. We will not solve the equations because. the mathematical techniques required are beyond the scope of this text ; however, the result
Vz
Uz=-, Uoo X
X=-, L
Flow Around lmmersed Bodies
Chap. 11
S.C. 11.3
The Boundary Lay~r on a Flat Plate
431
form yields
v(dU,.)
=-~
df
Y =O
(11.2-42)
out the differentiation and integraequation for the boundary !ayer (11.2-43) imposing the boundary condition t=O
(11.2-44)
layer thickness as a function of time, 2.8J;i
(11.2-45)
by the approximate solution is the result is notas bad as it might seem. given in Fig. 11.2-2, and certainly the values of y/#f less than 1.0. We analysis should yield reasonably ready to attack the more complex 11.3-1.
a Flat Plate
Fl¡. 11.3·1. Boundary layer fl.ow over a fl.at plate.
will be given and compared with an approximate analysis. It should be understood that the order of magnitude analysis is a general tool of engineering analysis and has applications far beyond the development presented here. We start with the two-dimensional Navier-Stokes equations and the continuity equation for steady, incompressible flow.
v.,(ov.,) + v (ov.,) = _ !(op) + v(o v., + 0 v.,) ox oy p ox ox ol v.,(ov + v (ov = _ !(op) + v(o v + ov ox oy p oy ox ol ov., +OVIl= o ax oy 2
· section is illustrated in Fig. 11.3-1. pasta thin flat plate of length L. The uniform at a velocity of uro . This flow of the suddenly accelerated flat plate, a flat plate moving at a velocity uro these conditions, the time t in the to x/uro where x is measured from reason we may use Eq. 11.2-29 to
11
2
2
11
)
11
11
(11.3-2)
2
11
)
2
11
)
(11.3-3) (11.3-4)
We use the free stream velocity and the length of the plate to form the dimensionless variables
(11.3-1) be to perform an order of magnitude to obtain the Prandtl boundary layer equations because. the mathematical of this text; however, the result
2
U= :z X
X=-,
L
P- Po P=--
V:z
Uro
'
pu!,
{)' = tJ(x)
L '
<432
Flow Around lmmersed Bodies
Chap. 11
Sec. 11.3
The Boundary layer on a Flat Plat
and write the origir.al equations as (1) (1)
(<5')
(~)
(1')
( 1) <5'2 3
(1) (<5')
( <5')
(<5') (1)
(~)
U.,lxo-AX
u au" + u au" = _ aP + _1_ (a u"2 , a u") "'ax "aY ay N Re,L ax ' aY 2 2
(1)
3
(11.3-7)
0
OU "') ( oX Xo
= UziXo+AX -
u¡ =UI +(X-X)au.,l +(X-Xo)2a2u.,l 2 + (X 3!
X o? aau., 1 aX 3 Xo
2!
+
...
Uzlxo-AX _
Provided the higher order terms are approximation is oU., 1 oX Xo
~ U.,lxo+
By adding Eqs. 11.3-11 and 11.3-12, w for the second derivative.
oU., 2
--~
oX
U .,lxo+AX -
2
From these two equations, we conc1ude for the derivatives may be expressed as
au., ax
(11.3-9)
To treat the estimate of a derivative, we need to expand the function in a Taylor series and develop sorne finite difference expressions. Expanding U., around the point X = X 0 gives Xo
•
2~X
azu.,
c5 o o
ax
1
-=·
0{~ l ff ~., dYdX] = 0(1) 0
3
Subtracting Eq. 11.3-12 from Eq. 11.3 deriva ti ve,
6
"'Xo
~x(~ 3
The generally accepted meaning of this expression is that U., is closer to 1.0 than it is to either 0.1 or 10.0. An order of magnitude is perhaps best looked upon as an estima te of the average value of a function over the region under consideration. Viewing it in this way, we might express Eq. 11.3-8 as
"'X
U.,lxo-
_~X3! (ooXU.,)
(11.3-6)
where the quantities in parentheses over each term indicate the order of magnitude of that term. Here we ha ve used what is called a length Reyno/ds number, noting this fact by giving the Reynolds number as N Re L· An order of magnitud e analysis of a complex set of differenÚal equations can be an extremely profitable undertaking. With this technique an engineer may often reduce an essentially unsolvable problem to one which still contains the pertinent elements of the real world yet is amenable to analysis. The technique requires skill, daring, and luck, and the reward may be great. We start with the x-component of the dimensionless velocity, U.,. Its value ranges between O at Y = O and 1 for Y ¿ <5', and we write the order of magnitude of U., as (11.3-8) u.,= 0(1)
U.,=
=
2
(1)
au., + au" = ax aY
3
~X (o U.,~ + 3! axa)
(11.3-5)
aX
oX 2
Letting X 0
=
-
t and ~X = t. we find
au., ax
Xo
oU 2
(11.3-10)
--"'oX2
Flow Around lmmersed Boclies
Chap. 11
Sec. 11.3
..33
The Boundary Layer on a Flat Plate
Letting (X- X0) equal +~X and -~X, respectively, gives (1')
U.,)
( 1)
2
2
1 ~X (o U.,) 1 U.,lx.+~X = U.,lx. + ~X (o oX x. + T! (JX2 x.
15'2
(11.3-5) (11.3-11)
(~)
(b')
+ _1_ (o U 2
oU 2
11
,
11)
Nlle,L oX2 T (Jy2
(11.3-6) (11.3-12) (11.3-7)
over each term indicate the order of used what is called a length Reynolds Reynolds number as N Re L· a complex set of differenÚal equations · With this technique an engineer problem to one which still real world yet is amenable to analysis. luck, and the reward may be great. dimensionless velocity, U.,. lts value Y¿ b', and we write the order of
Subtracting Eq. 11.3-12 from Eq. 11.3-11 yields an expression for the first derivative,
oU.,) = U.,lxoHX- U.,lxo-AX- .!~r(oau.,) 1 ( oX Xo 2~X 3! oX 3 Xo
+ higher order terins
Provided the higher order terms are not excessively large, a 'reasonable approximation is
oU., 1
Xo
2
0 U.,
+o oo
--~
U :elxo+AX - 2U :elxo ~x 2
+ U,,lxo-AX
( 11.3-15)
From these two equations, we conclude that the order of magnitude estimates for the derivatives may be expressed as
=
oU., _ 2
oX 2
Letting X 0
= ! and
~X
0(U.,) ~X
oX
O(U.,) ~X 2
-
= !, we find that
au., =
ax (11.3-10)
(11.3-14)
2~X
oX 2
(11.3-9)
+(X- Xo)2 iJ2U., 1 2! oX 2 Xo
1X o
U:eiXoHX - U:elxo-AX
oU.,
need to expand the function in a tlítlterence expressions. Expanding U.,
1
~
By adding Eqs. 11.3-11 and 11.3-12, we may also obtain an approximation for the second derivative.
expression is that U., is closer to 1.0
dYdx) ~ 0(1)
Xo
oX
(11.3-8)
looked u pon as an estima te of the under consideration. Viewing it as
(11.3-13)
(J2U., oX 2
0(1)
= 0(1)
(11.3-16a) (11.3-16b)
Flow Around lmmersed Bodles
In the second case, we are stretching our definition, for written ~XI
~X2
Chap. 11
could also be
= ! = O(lo-1)
S.C. 11.3
The Boundary Layer on a Flat
On ~the basis of the order of J expect these equations to be valid o express this restriction in terms of a
however, our final result would not be altered if we made this change. To determine the order of magnitud~ of U11 , we note first that
aull = _ au., = 0(1) oY ax UvJY=o = 0(1)
(11.3-18)
and we conclude that
u"= <5'[0(1)1 = 0(<5') The other order of magnitude estimates indicated in Eqs. 11.3-5 and 11.3-6 readily follow by similar techniques. In examining Eqs. 11.3-5 and 11.3-6, we see that the corresponding terms in Eq. 11.3-6 are of order of 6' less than those in Eq. 11.3-5. Thus, if 6' is about I0-2 , we can neglect the y-direction momentum equation and also (o 2 U.,joX2) relative to (o 2 U.,foY2), which allows us to simplify the equations tot
au., au., dP 1 (éJ2 U.,) u.,ax + u oY = - dX+N oY Re,L 11
2
au., aull = ax +ay
0
(11.3-19) Prandtl boundary !ayer equations (11.3-20)
These equations representan indeterminant set of equatio·ns, for we have three unknowns-U.,, U11 , and P-and only two equations. However, for this type of ftow we know that
dP =O dX '
or
NRe,a: ~
because xjL is always 1ess than one layer equations to be valid if the len¡ The solution of Eqs. 11.3-21 a1 possess at the moment; however, th difficulty and the result is compare
and by our order of magnitude analysis we conclude that variations of P throughout the boundary layer must be small. This statement suggests that dPjdX within the boundary layer must be nearly zero, and we are led at last to the Prandtl boundary layer equations for ftow pasta ftat plate:
1
1.0
~
" ,' 1
0.6
./,¡
0.4
0.2
11 " J
o
,..
.)
0.8
Y;;::: 6'
au., au., 1 (éJ2 U.,) u.,ax +u~~ aY =-¡:¡ayz ae,L
= ~L ~·3.6~(: U X 00
(11.3-17)
Using a finite difference formula in conjunction with Eq. 11.3-17, we find that
oU" = UviY-6' ay <5'
!5'
I'Eqs. 11, 3-21 ond 11.3-22
1.0
1
1 2.0
-
3.C
"'""Y.
FIJ. 11.3-l. Comparison of ex¡ profiles for ftow past a ftat plate.
(11.3-21) (11.3-22)
t This step relies very heavily on the plausible intuitive hypothesis that "small causa produce small effects."
The calculated and experimental v function of the single variable 1J = ) differential equations are transform The length Reynolds number for the 105 to 7.28 x 105, large enough tosa
Around lmmersed Bodles
definition, for
~X2
Chap. 11
could also be
S.C. 11.3
On Hhe basis of the order of magnitude analysis presented here, we expect these equations to be valid only if 15' :::;; 1Q-1• By Eq. 11.3-1, we may express this restriction in terms of a length Reynolds number as
altered if we made this change. of U11 , we note first that
15' =
~ ~· 3.6 ¡:;-(~) = ~ (~) L \) ;:::;: L JN Re,z L
(11.3-18)
set of equatiáns, for we have three equations. However, for this type
Nae,z :2::
we conclude that variations of P small. This statement suggests that be nearly zero, and we are led at last for ftow pasta ftat plate: 2
(3.6)1 X 10'
1.0
,.
)
0.8
9
V
~
(11.3-24)
,....a-·
JI/
¡;;r
0.6
u.
NRe,.r =~ V
' 1
0.4
+ 1.08xl0 5 <» 1.82xl0 5
)1
0.2
J
2
(11.3-23)
because x/Lis always less than one. Thus, we expect the Prandtl boundary layer equations to be valid ifthe length Reynolds number is greater than 10'. The solution of Eqs. 11.3-21 and 11.3-22 requires more skill than we possess at the moment; however, they can be solved without a great deal of difliculty and the result is compared with experimental data in Fig. 11.3-2.
Prandtl (11.3-19) boundary !ayer equations (11.3-20)
1 (ooYU)
10-2
or
indicated in Eqs. 11.3-5 In examining Eqs. 11.3-5 terms in Eq. 11.3-6 are of order of 15' is about IQ-2 , we can neglect the (o 2 U.,/oX2) relative to (o 2 U,fo Y2), tot
- Nae,L
:::;;
(11.3-17)
¡c~;J:um.¡uc::;.
- - - __.,
-435
The Boundary Layer on a Flat Plate
ll o
"
o 3.64xl0 5 • 5.46xl05 e 7.28xl05
t--Eqs. 11 . 3-21 ond 11.3-22
1.0
1
1
2.0
3.0 '/}:ay
4.0
5.0
6.0
7.0
¡u;; .¡-v;-
FIJ. 11.3-l. Comparison of experimental and theoretical velocity profiles for ftow
past a ftat plate.
(11.3-21) (11.3-22)
intuitive hypothesis that "small causa
The calculated and experimental values of the velocity are plotted as a function of the single variable r¡ = yV u 00 f'vx, which arises when the partial differential equations are transformed to an ordinary differential equation. The length Reynolds number for the experimental points ranges from 1.08 x 106 to 7.28 x 105, large enough to satisfy the restriction given by Eq. 11.3-24.
436
Flow Around lmmersed Bodies
Chap. JI
Sec. 11.3
The Boundary Layer on a Flat ~
We conclude that our order of magnitude analysis and simplification of the t-,o-dimensional Navier-Stokes equations have been successful. The result indicates that the boundary layer thickness is given by «5
=
4.9
r;; '.};¡:
(11.3-25)
which is not too different from the original estímate given by Eq. 11.3-1. An interesting comparison between theory and experiment is the measured and calculated drag coefficients shown in Fig. 11.3-3. It becomes obvious that theory and experiment are not in agreement for values of NRe,z less than lOS. 0.5
Fig. 11.3-
1
~
0.2
0.1
Co
-
o
"
D
~
0.05
o
(pv)v • n dA
o
5xl0 2
= 1
A,
~ ~.0
0.02 10,z
J
Experiment
Applying the x-component of this eq
.........
50
For a fixed control volume an takes the form
Theory
Momentum ftux atx + Ax
'
103
~ 2xl0 3
Momentu¡ at"
...............
5xl0 3
NRe,L
Flg. 11.3-3. Comparison of experimental and theoretical drag coefficients for ftow pasta ftat plate. (Experimental data are taken from NACA Tech. Mem. 1316, 1951.)
We ha ve made the following assumpti analysis that led to the Prandtl bou A. 1 : The viscous stress on the may be stated
Approximate solution
Although we have left the Prandtl boundary layer equations unsolved, we can analyze the ftow with an approximate solution. The control volume to be used for the differential-macroscopic momentum balance is shown in Fig. 11.3-4. In this analysis, we will start directly with the macroscopic momentum balance; however, we could integrate the Prandtl boundary layer equations to obtain the same differential macroscopic momentum balance. This problem will be left as an exercise for the student.
which is comparable to sp
a A. 2: The pressures at x and x contribute to the moment from the facts that (opfo outside the boundary laye
Flow Around lmmersed Bodies
Chap. 11
Sec. 11.3
437
The Boundary Layer on a Flat Plate
analysis and simplification of the have been successful. The result is given by (11.3-25) estimate given by Eq. 11.3-l. and experiment is the measured in Fig. 11.3-3. It becomes obvious agreement for values of NRe,z less
r.----------
--------~
L
Fig. ll.l...C. Control volume for approximate solution.
Theory
o
For a fixed control volume and steady ftow, the momenturn balance takes the forrn
J
Experiment
(pv)v • n dA
A,
=
J
J
~
~
pg dV +
(11.3-26)
t 1a 1 dA
Applying the x-cornponent of this equation to the control volurne yields Momentum flux atx + Llx
Momentum ftux
Momentum ftux
aty
atx
~a
Surface force aty =O
We have rnade the following assurnptions in accord with the order of magnitude analysis that led to the Prandtl boundary !ayer equations. A. 1 : The viscous stress on the x-surfaces is neglected. Thts assurnption rnay be stated r.,., <.{ r 11 .,
boundary layer equations unsolved, oxtmate solution. The control volume rnornenturn balance is shown in directly with the rnacroscopic integrate the Prandtl boundary differential rnacroscopic momenturn exercise for the student.
which is comparable to specifying that
02V
02V
ox
ol
-"'2 <.{ -"'
+
A. 2: The pressures at x and x tl.x are equal, and therefore do not contribute to the rnornenturn balance. This assurnption is derived frorn the facts that (opfoy) is negligible and that the pressure outside the boundary !ayer is constant.
•08
Flow Around lmmersed Bodies
A.
Chap. 11
3: The viscous stress at y = r5 is zero, which results from specifying that the velocity gradient (iJv,jiJy) must be continuous and is therefore zero at the edge of the boundary layer.
Dividing Eq. 11.3--27 by
~x
and taking the limit
~x-+
O, we obtain
6
t!_ dx
J pv: dy- pu~ (dr5) + dx
pU 00 V11 1
11~6
o
= -t-t
iJv., 1 iJy ~~~o
(11.3-28)
Sec. 11.3
The Boundary Layer on a Flat
Carrying out the integration and difl 39 (dr5~ 280 dx/ Separating variables, integrating, an
B.
c.
r5 =o,
1:
we get an expression for the bounda
where the shear stress has been represented in terms of the velocity gradient at y = O. Once again, we represent v., as a polynomial in {y/r5) (11.3-29) which is subject to the restrictions
v.,= O, V.,= U 00 , iJv., =O iJy '
y=O y=r5
(a bonafide boundary condition) (an intuitive approximation)
(11.3-30a) (11.3-30b)
y=r5
(continuity of stress)
{11.3-30c)
iJ2v., = O (from the differential equation) (113-30d) y=O iJl ' These conditions allow us to determine the four constants in Eq. 11.3-29, and the velocity is given by v., = u 00 {iY- !Y3) (11.3-31) where
Ó=
which is in excellent agreement witl velocity profiles for the exact and a 11.3-5, and the agreement is seen to agreement between the exact and a accelerated flat plate.
r5
= r5,
0.8
0.6
A~"'
Ux 0.4
0.2
the continuity equation may be integrated
v~~j
11~6
=
-f6(iJv.,) dy o
iJx
(11.3-32a)
3u"' (dr5) = -8dx
V
o
¿, ~
1..1
r'
/r'' 1.0
2.C ;
Fl1. 11.3-5. Velocity prc
and substitution of Eq. 11.3-31 for v., gives
v11 1y~6
1
1.0
Y=~ To determine v11 at y
1
(11.3-32b)
Substitution of Eqs. 11.3-31 and 11.3-32b into Eq. 11.3-28, and the changing of the variable of integration from y to Y, give (11.3-33)
To gain sorne insight into the nat1 the following example. For the flat F following conditions:
u"'= 100 mph p = 0.075 lbm/ft3 (e fl = 3.8 x I0-7 lbr· L = 6 in.
Around lmmersed Bodies
Chap. 11
which results from specifying must be continuous and is boundary layer. the limit !l.x--+ O, we obtain
..,v,¡
v~6
= -p, ov., 1
oy
(11.3-28)
!1=0
Sec. 11.3
Carrying out the integration and differentiation yields 39 (di}) 3( V ) 280 dx = 215 u..,
(11.3-34)
Separating variables, integrating, and applying the boundary condition B.
c.
1:
15 =O,
X=
o
(11.3-35)
we get an expression for the boundary layer thickness
in terms of the velocity gradient a polynomial in (y/15) (11.3-29)
(11.3-30a) (11.3-30b)
IJ
(11.3-36)
...J ~
::.:::.0 ..:::
. /~
0.8
the four constants in Eq. 11.3-29,
~~
¿1'
0.6
u.
A
~....
~('
0.4
~"~"'
0.2
./
equation may be integrated (11.3-32a)
¡;; \};¡:
1.0
(113-30d)
(11.3-31)
= 4.64
which is in excellent agreement with that given by the exact solution. The velocity profiles for the exact and approximate solutions are shown in Fig. 11.3-5, and the agreement is seen to be excellent-far better, in fact, than the agreement between the exact and approximate solutions for the suddenly accelerated flat plate.
(11.3-30c)
- tP)
-439
The Boundary Layer on a Flat Plate
o
~
Exoct solution ---Approximote solution
~-
/
1.0
2.0 y VU(SJ/VX
30
4.0
5.0
FIJ. 11.3-5. Velocity profile for ftow past a ftat plate.
(:~)
(11.3-32b)
into Eq. 11.3-28, and the changing Y, give 5(dt})
S
3( V ) dx = - 215 """
To gain sorne insight into the nature of the boundary !ayer, let us consider the following example. For the flat plate shown in Fig. 11.3-1, we assume the following conditions:
u..,= 100 mph
= 0.075 lbm/ft3 (density of air) p, = 3.8 X I0-7 lbr-sec/ft2 (viscosity of air) p
(11.3-33)
L
=
6 in.
Flow Around lmmersed Bodies
Chap. 11
The length Reynolds number is 6 x 105, and the boundary layer thickness at x =Lis Ó!::::J(S) ¡:::::,¡
6in. ../6 X 105
Sec. 11.-4
Externa! Flows and Wakes
region, a wake region, and a stagnatio on the plate results almost entirely fron back surfaces of the plate. Under th proportional to the density times the
3.86 x 10- 2 in.
The point here is that the boundary layer thickness is in general quite small for conditions that might be encountered for air flowing past solid objects. Since the kinematic viscosity v of water is about a factor of 10 less than that of air, we can also expect boundary layers for the flow of water around solid objects to be very thin.
11 A Externa! Flows and Wakes We are now in a position to understand the nature of the boundary layer flow, and we must go on to the examination of the flow outside the boundary layer-i.e., the external flow-and the characteristics of the wake region. In the absence of a chapter on irrotational flow, we are forced to treat this problem in a qualitative manner. This will be done by giving a qualitative description of the flow about various solid bodies and examining the drag coefficients for these bodies. When a thin flat plate is placed parallel to the flow field, the drag exerted by the fluid is almost entirely from viscous effects, and we can use Eqs. 11.3-31 and 11.3-36 to calculate the drag coefficientt L
2b
e
J1-' (ov.,) oy
--..l!o_ _ _ __
D-
(2Lb)(}pu!,)
1.293
Fl1. 11.4-1. Flow past a plate at high Reynolds numbers.
dx
v=O
(11.4-1)
.JNRe,L with considerable accuracy, provided the boundary layer flow remains laminar and NRe L ~ 103 • However, if the plate is plaéed perpendicular to the flow, an entirely different situation, depicted in Fig. 11.4-1, is created. For this case, the flow may be divided into an irrotational flow
t The exact solution gives a coefficient of 1.328.
This situation is comparable to the je and the rough-pipe region of flow ( plate placed perpendicularly to the fn jected area and the kinetic energy pe
en-(1 --d
Both mathematicaJ2- 5 and experimen available for a flat plate placed pa These results are shown in Fig. 11.4 large difference in the drag coefficient For Reynolds numbers less than drag coefficients differ by less than intuition that the drag coefficient fo flow is the larger of the two. For Re effects predominate for the perpend' constant. The behavior of the parall coefficient ·continues to decrease with keep in mind that the drag force is 1 not increasing as fast as pu~. These n distinct types of drag forces to be ce there were two distinct mechanisms rough pipes. While the drag coeffi.cient for the
2. Y. H. Kno, "On the Flow of an Inco Moderate Reynolds Numbers," J. Math. Ph; 3. S. Tomotika and T. Aoi, "The Stea< Cylinder and a Flat Plate at Small Reynol 1953, 6:290.
4. B. A. Boley and M. B. Friedman, "Or of a Flat Plate," J. Aerospace Sci. 1959, 26:L 5. H. Schlichting, Boundary Layer Theor. lnc., 1955), Chap. 7. 6. Z. Janour, "Resistance of a Plate in NASA Tech. Note 1316, Nov., 1951. 7. S. Dhawan, "Direct Measurements of 8. S. F. Hoerner, Aerodynamic Drag (p New Jersey), Chap. 3.
Around lmmersed Bodies
Chap. 11
Sec. 11.4
441
Externa! Flows and Wakes
region, a wake region, anda stagnation region in front ofthe plate. The drag on the plate results almost entirely from the pressure difference on the front and back surfaces of the plate. Under these conditions, the drag is very nearly proportional to the density times the square of the free stream velocity. drag oc pu~ thickness is in general quite small for air flowing past solid objects. about a factor of 10 less than that for the flow of water around solid
This situation is comparable to the jet impinging on the flat plate (Sec. 7.8) and the rough-pipe region of flow (Sec. 8.2). The drag coefficient for a flat plate placed perpendicularly to the free stream is defined in terms of the projected area and the kinetic energy per unit volume of the free stream.
e D -
the nature of the boundary !ayer of the fiow outside the boundary of the wake region. In fiow, we are forced to treat this be done by giving a qualitative bodies and examining the drag fiat plate is placed parallel to the almost entirely from viscous effects, we can use Eqs. 11.3-31 and 1.3-36 to calculate the drag coeffiL
e
Jp, (ov.,) ay __
D-
2b
dx
J!.o_ _ _~~~o __
(2Lb)(!pu~)
1.293
.J NRe,L
(11.4-1)
considerable accuracy, provided boundary !ayer flow remains and NRe L ~ 103 • However, the plate is plaéed perpendicular to fiow, an entirely different situadepicted in Fig. 11.4-1, is For this case, the fiow may divided into an irrotational fiow
drag force (Lb)(!pu~)
(11.4-2)
Both mathematical 2- 5 and experimental 6- 8 values of the drag coefficient are available for a flat plate placed parallel and perpendicular to the flow. These results are shown in Fig. 11.4-2, and, as we would expect, there is a large difference in the drag coefficients for the two configura~ions. For Reynolds numbers less than one, viscous effects predominate and the drag coefficients differ by less than 50 per cent. lt is consistent with our intuition that the drag coefficient for the plate placed perpendicular to the flow is the larger of the two. For Reynolds numbers greater than 10, inertial effects predominate for the perpendicular plate and the drag coefficient is constant. The behavior of the parallel plate is quite different, and the drag coefficient 'continues to decrease with increasing Reynolds number. We must keep in mind that the drag force is not necessarily decreasing; it is simply not increasing as fast as pu!,. These results indicate clearly that there are two distinct types of drag forces to be considered for immersed bodies, just as there were two distinct mechanisms of drag to be considered for flow in rough pipes. While the drag coefficient for the perpendicular plate remains constant 2. Y. H. Kno, "On the Flow of an Incompressible Viscous Fluid Past a Flat Plate at Moderate Reynolds Numbers," J. Math. Phys. 1953, 32:83. 3. S. Tomotika and T. Aoi, "The Steady Flow of a Viscous Fluid Past an Elliptic Cylinder and a Flat Plate at Small Reynolds Numbers," Quart. J. Mech. Appl. Math., 1953, 6:290. 4. B. A. Boley and M. B. Friedman, "On the Viscous Flow Around the Leading Edge of a Flat Plate," J. Aerospace Sci. 1959, 26:453. 5. H. Schlichting, Boundary Layer Theory (New York: McGraw-Hill Book Company, Jnc., 1955), Chap. 7. 6. Z. Janour, "Resistance of a Plate in Parallel Flow at Low Reynolds Numbers," NASA Tech. Note 1316, Nov., 1951. 7. S. Dhawan, "Direct Measurements of Skin Friction," NACA Rept. 1121, 1953. 8. S. F. Hoerner, Aerodynamic Drag (pub. by author, 148 Busteed, Midland Park, New Jersey), Chap. 3.
Chap. 11
Flow Around lmmersed Bodies
'
5.0
1
:-... :-...~
!""
1'
2.0 1.0
."
1 1
1.1
~
1 1 1
1
1
Flot plote perpendicular to flow -
~~-
r~
0.50
["..
Co 0 .20
11
~
0.10
/
Flot plot& porollel to flow
Sec. 11.4
Externa! Flows and Wakes
erratic manner and then follows a s11 the curve for the 1ower Reyno1ds nur in the ftow, and in view of our knov turbu1ent ftow in smooth tubes, it st to the onset of turbu1ence. Both experiment and theory 6 ve 3 x 105 , and a schematic drawing o
K Tron
........
1.-¿...J
r-....
'
0.05
1-----Lominor____...,
......
't-..
0.02
~"""-~
!"'
0 .0 1 0.1 0.2
0 .5 1.0 2.0
5.0 10 20
50
lo2 2
5
5 104
10\3 2
NRe
Fl1. 11.4-2. Drag coefficients for thin, fiat plates.
for Reyno1ds numbers higher than those in Fig. 11.4-2, an important change in the drag characteristics of the paralle1 p1ate occurs. This change is illustrated in Fig. 11.4-3, where we see that the drag coefficient for a ftat p1ate behaves much like the friction factor for ftow in a pipe. At a Reyno1ds number of about 3 x 105, the drag coefficient suddently increases in an
Fl1. 11.4-4. Boundary ll
l{.r~
1
1
1 1 11
Q ~
1 T.
UUil~~IIVII
6
~
t-~~~~
~ ~~
eD
~~lí'
""" ~
N _., ~
mnor
/
t---.
-
Tur~ule1
-~
fl_ ~
-- ~
'r-.
'1'.
1(}3
l
NRa, L Fl1. 11.4-l. Drag coefficient for a fiat plate-the transition to turbulence.
As in the case of pipe ftow, the tra variety of factors. Schubauer and S intensity of turbu1ence resu1ted in a and by careful1y eliminating the free 1 number of 3 x 108 was reached. On the basis of our qualitative e p1aced both paralle1 and perpendicu1 three distinct types of drag forces: "fo acting on the front and back (or lel "friction force"; and a turbu1ent "fr these phenomena can be obtained o analysis and an extended survey of 1 description presented here does toucl in this chapter. . The next step in the process of sh
9. G. B. Schubauer and H. K. Skramst1 Transition on a Flat Plate." NACA Rlpt. 9
Sec. 11.4
Externa! Flows and Wakes
erratic manner and then follows a smooth curve decreasing less rapidly than the curve for the lower Reynolds numbers. Obviously a change has occurred in the flow, and in view of our knowledge of the transition from laminar to turbulent flow in smooth tubes, it seems reasonable to attribute the change to the onset of turbulence. Both experiment and theory 5 verify the onset of turbulence at N Re 40 ~ 3 x 105, and a schematic drawing of the flow field is shown in Fig. 11 ..4-4. Tronsition
for thin, ftat plates.
in Fig. 11.4-2, an important change plate occurs. This change is illusthe drag coefficient for a flat plate flow in a pipe. At a Reynolds oe1mc1·.em suddently increases in an Fl1. 11.4-4. Boundary layer formation on a ftat plate.
As in the case of pipe flow, the transition Reynolds number depends on a variety of factors. Schubauer and Skramstad8 found that an increase in the intensity of turbulence resulted in a transition at a lower Reynolds number, and by carefully eliminating the free stream turbulence, a transition Reynolds number of 3 x l ()8 was reached. On the basis of our qualitative examination of the flow past a flat plate placed both parallel and perpendicular toa uniform stream, we are aware of three distinct types of drag forces: "form force" from the difference in pressure acting on the front and back (or leading and trailing) surfaces; a laminar "friction force"; anda turbulent "friction force." A clear understanding of these phenomena can be obtained on1y with a more detailed mathematical analysis and an extended survey of experimental results; however, the brief description presented here does touch on the key points we wish to bring out in this chapter. . The next step in the process of sharpening our intuition is to examine the 9. G. B. Schubauer and H. K. Skramstad, "Laminar Boundary Layer Oscillations and Transition on a Flat Plate." NACA hpt. 909, 1949.
Flow Around lmmersed Bodies
--
5.0
Um
l]o
2 .0
r- 1- Flat plate
2 .0
e¡,
f-t
0.20
1
'
~
r-.
~QJD
Externa! Flows and Wakes
4.0
Cylinder
0 .50
Sec. 11.4
10.0 8 .0 6 .0
.,... t-¡..
1.0
0 .10
Chap. 11
11
1' ""¡..
Coi.O
0.8 0.6
0.4 0 .05 0.2
0 .02 0 .01 0 .1 0 .2
0.5 1.0 2.0
5.0
10
20
50
102 2
0 .1 10
1
11111 2
4
6 8102
2
4
1
6 8103
2
NRe Flg. 11.4-5. Drag coefficients for a cylinder anda ftat plate.
Flg. 11.4-6. Drag coefficient vers
cylinder.
drag coefficients for a cylinder and a flat plate shown in Fig. 11.4-5. At low Reynolds numbers, the drag coefficients are nearly the same, with the cylinder showing a somewhat higher value for CD· It is not at all obvious that the drag on these two rather different objects should be nearly identical in the low Reynolds number region. However, it appears that the degree of deformation suffered by the fluid as it flows past the plate and the cylinder is about the same, which leads to similar values for the drag coefijcient. The drag on the flat plate results entirely from form drag even at low Reynolds numbers, while the total drag on the cylinder at low Reynolds numbers is evenly divided between form and friction drag.3 From the results shown in Fig. 11.4-5, we conclude that the form drag for the cylinder must always be less than that for the flat plate, and that streanilining reduces the total drag at high Reynolds numbers. Recalling that the boundary !ayer on the flat plate become turbulent at a high Reynolds number yielding an abrupt increase in the drag coefficient, we might expect a similar change in the drag coefficient-Reynolds number curve for the cylinder. The complete curve is shown in Fig. 11.4-6, and, indeed, there is an abrupt change in the drag coefficient. But while the drag increases for the flat plate, the drag on the cylinder undergoes an abrupt decrease! Here it seems appropriate to quote Shapiro, 10 who, at this point in his classic 10. A. H. Shapiro "The Fluid Mechanics of Drag," distributed by Educational Services, Inc., 47 Galen St., Watertown 72, Mass. A written version of the experiments and commentary contained in the film is available in the paperback book, A. H. Shapiro, Shape and Flow- The Fluid Dynamicsof Drag(Garden City, New York : Anchor Books, Doubleday and Co. Inc., 1961).
movies on drag, said, "Well, this is curious phenomenon we must exa separation for both laminar and tur separation point when the boundary to the abrupt decrease in the drag e layer separation, we will present a b field outside both the wake and the t The externa! flow
Both analysis and experiment i boundary layer can be described r neglected entirely. For steady flow, t
pv • Vv =
and the fl.ow is termed irrotational. Ec of what is called a perfect fluid, or an fluid has a zero value of the viscos 11.4-3 may also be obtained by postul to any surface, requiring the stress te1
T=
and the stress equations of motion qt
ow Around lmmersed Bodies
Chap. 11
445
Externa! Flows and Wakes
Sec. 11.4
10.0 8.0
6.0 4.0
eno
2.0 ~ .......
~~
c.DI.O
1--
0.8 0.6
1\
0.4 0.2 0 .1 10
2
4
6 8102
2
4
6 8103
2
4
6 8
lo4
2
4
6 8~
2
4 6 81()6
NR• Fl¡. 11.4-6. Drag coefficient versus Reynolds number for a circular cylinder.
plate shown in Fig. 11.4-5. At low nearly the same, with the cylinder D· It is not at all obvious that the should be nearly identical in the Iow that the degree of deformation plate and the cylinder is about the drag coefl].cient. The drag on the even at low Reynolds numbers, low Reynolds numbers is evenly 3 From the results shown in Fig. for the cylinder must always be less · reduces the total drag at the flat plate become turbulent at a increase in the drag coefficient, we coefficient-Reynolds number curve shown in Fig. 11.4-6, and, indeed, .. 1 1,,1'-'ll"· But while the drag increases undergoes an abrupt decrease! who, at this point in his classic " distributed by Educational Services, of the experiments and commenbook, A. H. Shapiro, Shape and New York: Anchor Books, Doubleday
movies on drag, said, "Well, this is certainly unexpected." To explain this curious phenomenon we must examine the process of boundary !ayer separation for both laminar and turbulent flow, for it is the shifting of the separation point when the boundary layer becomes turbulent that gives rise to the abrupt decrease in the drag coefficient. Before considering boundary !ayer separation, we will present a brief, qualitative description of the flow field outside both the wake and the boundary !ayer. The externa! flow Both analysis and experiment indicate that the region outside the boundary !ayer can be described reasonably well if viscous effects are neglected entirely. For steady flow, the equations of motion reduce to pv • Vv = -Vp
+ pg
(11.4-3)
and the flow is termed irrotational. Equation 11.4-3 also describes the motion of what is called a perfect fluid, or an inviscid fluid. By definition, an inviscid fluid has a zero value of the viscosity coefficient; thus, fl = O. Equation 11.4-3 may also be obtained by postulating that the stress always acts normal to any surface, requiring the stress tensor to be given by
T= -pi and the stress equations of motion quickly reduce to Eq. 11.4-3.
(11.4-4)
Flow Around lmmersed Bodies
Chap. 11
Now, inviscid fluids do not exist in the real world. Yet there is much to be learned from the study of such fluids, just as we learned a great deal from the study of the suddenly accelerated, infinite, flat plate, even though it is a flow which cannot be realized in practice. Here, we will examine the solution of Eq. 11.4-3 for flow pasta cylinder, only to obtain a limiting condition for the form drag. The results of one possible solution ofEq. 11.4-3 for flow past a cylinder are shown in Fig. 11.4-7. The streamlines are shown in Fig. 11.4-7a and the pressure distribution over the surface of the cylinder in Fig. 11.4-7b. The analytic expression for the dimensionless pressure on the surface of the cylinder11 is
P - Pao = (1 - 4 sin 2 O) (11.4-5) !pu!, Remembering that an inviscid fluid exerts only a normal force, we may use Eq. 10.4-5 to express the drag coefficient as · CD =
Frorm =A (!pu!,XbD)
·f
21r
n(1 - 4 sin 2 O) dO
(11.4-6)
o
Inasmuch as A • n = -cos O, the drag coefficient is zero,
S.C. 11.4
Externa! Flows and Wakes
force on an immersed body, for expe bodies experienced a drag at high were presumably negligible. The two points of maximum p1 referred to as the forward (O = O) terminology is appropriate, because the fluid is inviscid and "slips" past rear stagnation pressure approachc sometimes called the pressure recove1 recovery is complete. We now have sorne knowledge of i.e., it is zero for the flow illustrated in description of inviscid flow somewha into the nature of the flow of real illustrated in Fig. 11.4-7 is just one o 11.4-3 subject to the boundary cond velocity on the surface of the cylinde Thete is a class of solutions for in as Helmholtz flows,t or free streamli shown in Fig. 11.4-8 for flow past a
b
CD = -
J(1- 4 sin O) cosO dO= O 2
(11.4-7)
o
This result, often known as D'Alembert's paradox,11 caused considerable consternation among early researchers in fluid mechanics. Even in the absence of viscous forces, it was felt that a streaming fluid would exert sorne 8='1TI2
(o)
Flg. 11.4-8. Helmho
-3 ~)
~
o
'11'/2
o
1T
FIJ. 11.4-7. Flow of an inviscid ftuid past a cylinder. 11. L. M. Milne-Thomson, Theoretical Hydrodynamics, 4th ed. (New York: The Macmillan Company, 1960), p. 157. 12. H. Rouse and S. Ince, History of Hydraulics (Ncw York: Dovcr Publications, Inc.,
1963), p. 102.
wake region was taken to be the sa pressure in the wake is taken to be s separation moves toward the rear < progressively smaller. When the prc forward stagnation pressure, the wa collapses to the continuous flow.
t
See Ref. 12, p. 201, and Ref. 11, Cha¡
Around lmmersed Bodies
Chap. 11
as we learned a great deal from the flat plate, even though it is a flow we will examine the solution of to obtain a limiting condition for solution of Eq. 11.4-3 for flow past streamlines are shown in Fig. the surface of the cylinder in Fig. ~rn~em¡íollllei¡s pressure on the surface 4 sin 2 O)
(11.4-5)
only a normal force, we may use
(11.4-6)
S.C. 11.4
+47
Externa! Flows and Wakes
force on an immersed body, for experience certainly indicated that immersed bodies experienced a drag at high Reynolds numbers where viscous effects were presumably negligible. The two points of maximum pressure shown in Fig. 11.4-7 are often referred to as the forward (O = O) and rear (O = 1r) stagnation points. The terminology is appropriate, because the velocity is zero at the points even if the fluid is inviscid and "slips" past solid surfaces. The degree to which the rear stagnation pressure approaches the forward stagnation pressure is sometimes called the pressure recovery. For the illustrated case, the pressure recovery is complete. We now have sorne knowledge of the limiting behavior of the form drag, i.e., it is zero forthe flow illustrated in Fig. 11.4-7. We can carry this qualitative description of inviscid flow somewhat further and perhaps gain more insight into the nature of the flow of real fluids past immersed bodies. The flow illustrated in Fig. 11.4-7 is just one of an infinite number of solutions to Eq. 11.4-3 subject to the boundary condition that the normal component of the velocity on the surface of the cylinder is zero. There is a class of solutions for inviscid flow past immersed bodies known as Helmholtz flows,t or free streamline flows. A possible flow of this type is shown in Fig. 11.4-8 for flow pasta cylinder. The pressure in the stagnant
(11.4-7) Helmholtz flow
paradox,11 caused considerable in fluid mechanics. Even in the a streaming fluid would exert sorne
,/'
\ _..,Continuous 1 "( inviscid / \ flow 1
--t.
\
\
,,_. ,
1
~
3
o
7T/2
,
8
(al
(b)
Flg. 11.4-8. Helmholtz flow past a cylinder.
(b)
O
fluid past a cylinder.
rr/2 0
(New York: Dover Publications, Inc.,
wake region was taken to be the same as the free stream pressure. If the pressure in the wake is taken to be sorne value larger than Poo• the point of separation moves toward the rear of the cylinder and the wake becomes progressively smaller. When the pressure in the wake is taken to be the forward stagnation pressure, the wake disappears and the Helmholtz flow collapses to the continuous flow. t
See Ref. 12, p. 201, and Ref. 11, Chap. 12.
448
Flow Around lmmersed Bodies
Chap. 11
Sec. 1 1.-4
Externa! Flows and Wakes
Inasmuch as the pressure distribution in the Helmholtz flow is not symmetric, a positive drag force results ; the flow is therefore in accord with our experience. In addition, the stagnant wake region also bears sorne resemblance to reality, although in actual fact it is unstable and a series of oscillating vortices are formed. Our treatment here has been extreme! y qualitative ; however, from the Helmholtz and continuous flow pressure distributions we may draw the following conclusion : When the .flow separa tes from the so/id surfa ce the pressure recovery is not complete and the form drag increases.
Boundary !ayer separation Seporotion point
When a real fluid flows pasta cylinder, a boundary layer forms, separating from the cylinder as shown in Fig. 11.4-9. A detailed sketch of the boundary !ayer separation is shown in Fig. 11.4-9, and a qualitative explanation of the Laminar flow at intermediate Reynolds numbers
Boundary !ayer
o "\
'\ \
1 1 1
\
\
/
-3
1
' ,_.., /
Continuous inviscid flow
o
1rl2
1T
{)
íbl
(al
Fig. 11.4-10. Separation of a lamín
which indicates that the sign (±) o whether the pressure gradient is positr of th,e boundary !ayer on a flat pla gradient does not vary across the bou opfox at the solid surface by examinin§ layer. Over the front surface of the cylind by Bernoulli's equation, then, the pr inertia and the pressure gradient are against the effects of viscosity. A decr a favorable pressure gradient, which remains attached to the sol id surface. '
Fig. 11.4-9. Flow of a viscous fluid pasta cylinder.
op. OX
separation phenomenon follows. We imagine a rectangular coordinate system located on the surface of the cylinder as shown in Fig. 11.4-1 O, and write the x-direction equation of motion for steady ftow and negligible gravitational effects. ov., p ( v., ox
+
Vu
ov., ) oy
= -
op OX
(ov ov., ) + f-t OX + oy 2
.,
2
2
2
(11.4-8)
At the solid surface, v, and vu are both zero and Eq. 11.4-8 is simplified to (11.4-9)
and Eq. ll.4-ll indicates that
a2v.,) < o (ol
This condition is illustrated by profile Let us now consider the possibilit: did not take place, and that the ft accurately described by the continuot For values of () > 1r/2, the pressure adverse pressure gradient wou\d be e: forces would thereby be acting against
Around lmmersed Bodies
Chap. JI
Sec. 11.4
449
Externa! Flows and Wakes
-ffi-
in the Helmholtz flow is not (b)
wake region also bears sorne fact it is unstable and a series of qualitative; however, from distributions we may draw the from the so/id surface the form drag increases.
Seporoted boundory loyer
Woke Seporotion point
r, a boundary layer forms, separating . A detailed sketch of the boundary and a qualitative explanation of the Laminar flow at intermediate RP.ynolds numbers
p-p,
__ a>
t pu!
~
o
\
1 1
'\
/
"'
\
-1
\
\
-2
/
-3
1
' ,_.., /
Continuous inviscid flow
o
rr/2
TT
()
(b)
Fig. 11.4-10. Separation of a laminar boundary !ayer from a cylinder.
which indicates that the sign (±) of the second derivative .depends upon whether the pressure gradient is positive or negative. Following the analysis of the boundary layer on a flat plate, we shall assume that the pressure gradient does not vary across the boundary layer; we can therefore determine opjox at the solid surface by examining the inviscid flow outside the boundary !ayer. Over the front surface of the cylinder, the velocity increases as x increases; by Bernoulli's equation, then, the pressure must be decreasing. Both fluid inertia and the pressure gradient are tending to propel the fluid forward against the effects of viscosity. A decreasing pressure leads to what is called a favorable pressure gradient, which gives rise to a boundary layer that remains attached to the solid surface. When the pressure gradient is favorable,
fluid past a cylinder.
op < o ox
imagine a rectangular coordinate as shown in Fig. 11.4-1 O, and for steady flow and negligible
é)v2"'
é)2v.,)
+.u-+( ox OX oy2 2
(11.4-8)
zero and Eq. 11.4-8 is simplified to (11.4-9)
and Eq. ll.4-11 indicates that
(oolv.,) 2
<
0
At the solid surface for a favorable pressure gradient
(ll.4-10)
This condition is illustrated by pro file (a) in Fig. 11.4-1 O. Let us now consider the possibility that separation and wake formation did not take place, and that the flow outside the boundary !ayer was accurately described by the continuous, inviscid flow shown in Fig. 11.4-7. For values of () > TT/2, the pressure would be increasing and a so-called adverse pressure gradient would be established. Both viscous and pressure forces would thereby be acting against the inertial force ofthe fluid, and there
<450
Flow Around lmmersed Boclles
Chap. 11
Sec. 11.-4
Externa! Flows and Wakes
is a possibility of a backflow occurring and a wake forming. The point at which the backflow starts is called the separation point. By Eq. 11.4-9 we know that an adverse pressure gradient leads to 2
(oolvlll)
>0
At the solid suñace for an adverse pressure gradient
(11.4-11)
Toe change in sign of the second derivative indicates an inflection point such as that shown in profile (e) in Fig. 11.4-10. This qualitative analysis indicates that the separation point should occur where the pressure is a minimum. The continuous inviscid flow solution would therefore indicate a separation point atO = 90°, although the actual separation point for laminar boundary layers13 is at O~ 80°. For all the approximations, the agreement is not bad. We shall conclude this discussion of flow around immersed bodies by examining all the flow regimes associated with uniform flow pasta cylinder. In practice, the engineer will encounter a variety of shapes of immersed bodies, and limiting our discussion to cylinders certainly does not do justice to the subject. However, the pertinent aspects of the flow around immersed bodies are contained in this one example, and it should be a satisfactory introduction to the subject. In Fig. 11.4-11, a series of flow fields is illustrated based on the observations and calculations of a number of investigators. 13- 17 At Reynolds numbers less than one there is no wake and the flow follows the curved surface of the cylinder in much the same manner as the inviscid continuous flow. At a Reynolds number of about 4, a visible wake appears. As the Reynolds number increases, the wake grows in size and the separation point moves forward. At a Reynolds of about 40, the flow tends to become unstable, and the two vortices behind the cylinder are alternately shed to form what is known as Kármán's vortex trail. This flow pattern is illustrated· in Fig. 11.4-12. The alternate shedding ofvortices produces periodic transverse forces on the cylinder that may cause severe damage if shedding frequency is comparable to the resonant frequency of the cylinder. The "singing" of electrical transmission wires in a high wind is a familiar example of this
Fl1. 11.4-11. Streamlines
phenomenon. The Kármán vortex trru range of from about 60 to 5000. At ' tends to become more turbulent in obscured. As the Reynolds number increases where a sudden decrease in the drag e~ of this section is to try to provide a phenomenon, and by now we have su From our examination of the bo that a transition to turbulence takes ¡:¡
13. A. S. Grove, F. H. Shair, E. E. Petersen, and A. Acrivos, "An Experimental Investigation of the Steady Separated Flow Past a Circular Cylinder," J. Fluid Mech., 1964, 19:60. 14. L. Prandtl and O. G. Tietjens, Applied Hydro and Aeromechanics, (New York: Dover Publications, Inc., 1957), p. 302. 15. l. Proudman and J. R. A. Pearson, "Expansions at Small Reynolds Numbers for Flow Pasta Sphere anda Circular Cylinder," J. Fluid Mech., 1957, 2:237. 16. A. Thom, "The Flow Past Circular Cylinders at Low Speeds," Proc. Roy. Soc. (London) Ser. A., 1933, 141:651. 17. D. J. Tritton, "Experiments on the Flow Pasta Circular Cylinder at Low Reynolds Numbers," J. Fluid Mech., 1959, 6:547.
FIJ. 11.4-12. The
Around lmmersed Bodles
Chap. 11
Sec. 11.-4
Externa! Flows and Wakes
451
and a wake forming. The point at
separation point. By Eq. 11.4-9 we leads to At the solid surface for an adverse pressure gradient
(11.4-11)
indicates an inflection point 11.4-10. This qualitative analysis occur where the pressure is a solution would therefore indicate a actual separation point for laminar the approximations, the agreement flow around immersed bodies by with uniform flow pasta cylinder. a variety of shapes of immersed certainly does not do justice of the flow around immersed and it should be a satisfactory is illustrated based on the obserof investigators. 1s-17 At Reynolds and the flow follows the curved manner as the inviscid continuous 4, a visible wake appears. As the in size and the separation point 40, the flow tends to become the cylinder are alternately shed to trail. This flow pattern is illustrated· vn•·tu•••c: produces periodic transverse damage if shedding frequency is of the cylinder. The "singing" of wind is a familiar example of this
Fil. 11.4-11. Streamlines for ftow pasta cylinder.
phenomenon. The Kármán vortex trail is observed in the Reynolds number range of from about 60 to 5000. At higher Reynolds numbers, the motion tends to become more turbulent in nature and the distinct vortices are obscured. As the Reynolds number increases further, it eventually reaches the point where a sudden decrease in the drag coefficient takes place. The main purpose of this section is to try to provide a rational explanation for this curious phenomenon, and by now we have sufficient knowledge to do so. From our examination of the boundary !ayer on a flat plate, we know that a transition to turbulence takes place for a length Reynolds number of
and A. Acrivos, "An Experimental a Circular Cylinder," J. Fluid Mech.,
~xp;msíons
r
at Small Reynolds Numbers for Fluid Mech., 1957, 2:237. at Low Speeds," Proc. Roy. Soc.
Past a Circular Cylinder at Low Reynolds FIJ. 11.4-12. The Kármán vortex trail.
452
Flow Around lmmersed Bodies
Chap. 11
Problems
about 3 x 105 • lf we are not cautious and apply the result for a flat plate to the cylinder, we may write
UooS) = 3 X 105 ( v transition
(11.4-12)
where s is the distance from the forward stagnation point to the point where the transition toa turbulent boundary layer takes place. If 01 is the angle at which the transition takes place, we may rewrite Eq. 11.4-14 as
(
uoo D) V
(_!!__) () 360°
=
3 X 105
(11.4-13)
t
or
Thus, if the Reynolds number is greater than 3 x 10 5, we might expect the turbulent boundary to existo ver a portion of the forward half of the cylinder. The fact that the sudden reduction in the drag coefficient occurs ata Reynolds number of about 3 x 105 certainly leads us to believe that this abrupt change is associated with the appearance of a turbulent boundary layer. Careful experimental studies of this phenomenon18 - 20 have clearly demonstrated that the appearance of the turbulent boundary layer causes the abrupt decrease in the drag coefficient, by the following mechanism. When the boundary layer becomes turbulent, the separation point moves to the rear of the cylinder, perhaps as much as 10°. This change reduces the area of the low pressure region, altering the pressure distribution around the cylinder so that the pressure at the rear stagnation point increases. Both these occurrences tend to reduce the form drag, and since it accounts for practically all the drag at these high Reynolds numbers, the drag abruptly decreases when the boundary layer becomes turbulent. The description of flow around immersed bodies presented in this chapter has been brief and very qualitative. However, we have touched upon the essential elements-laminar and turbulent boundary layers, externa! flows, and wakes. Extensive experimental studies have been conducted for a large variety of solid bodies, for both compressible and incompressible flow. None of these, however, will be discussed here, for to do so would add little to our knowledge of flow phenomena if we could not pursue a more rigorous mathematical attack. 18. See Ref. 14, p. 96. 19. A. Fage and J. H. Warsap, "The Effects of Turbulence and SurfaceRoughness on the Drag of a Circular Cylinder," Aero. Res. Council Rept. 1283, Gr. Brit., 1930. 20. A. Roshko, "Experiments on the Flow Past a Circular Cylinder at Very High Reynolds Numbers," J. Fluid Mech., 1961, 10:345.
PROB
11-1. Obtain an approximate solution fo Sec. 11.2) using a fourth order poi)
c2v
-"'=(
ay2
Compare the calculated boundary 1 exact solution. Ans: 6 = 3.65 y;¡
11-2. Integrate the Prandtl boundary laye expressing v., as a polynomial in Hint: Start by forming the integral d
av., v., ax +vil ( Jo and expressing v., as
11-3. Obtain an · approximate solution velocity as
~
Compare the calculated boundary those given by the solution of the Ans: 6
=
3.47 Vvxfuoo
11-4. Air at 72°F and 1 atm ftows past velocity of30 ft/sec. What is the bo the leading edge? 11-5. Consideran imrnersed body ofma an infinite, quiescent fluid at a ve! power required to move the bod characteristic area used in defining 11-6. The turbulent boundary layer t determined approximately if we ftow in smooth tubes. The velocit
Flow Around lmmersed Bodies
Chap. 11
453
Problems
apply the result for a fiat plate to
PROBLEMS (11.4-12) stagnation point to the point where !ayer takes place. If 01 is the angle at rewrite Eq. 11.4-14 as (11.4-13)
11-1. Obtain an approximate solution for the suddenly accelerated flat plate (see Sec. 11.2) using a fourth order polynomial and the additional condition
o2v
oy2,
= o,
y
=
o(r)
Compare the calculated boundary !ayer thickness with that obtained by the exact solution. Ans:
o = 3.65 Vrl
11-2. Integrate the Prandtl boundary !ayer equations between y = O and y = o(x), expressing v, as a polynomial in yfo(x), to derive an expression for o(x). Hint: Start by forming the integral than 3 x 105, we might expect the of the forward half of the cylinder. drag coefficient occurs ata Reynolds us to believe that this abrupt change boundary !ayer. phenomenon18 - 20 have clearly turbulent boundary !ayer causes by the following mechanism. the separation point moves to as 10°. This change reduces the the pressure distribution around the stagnation point increases. Both drag, and since it accounts for ...... v.u-> numbers, the drag abruptly turbulent. bodies presented in this chapter •nuJPv•"r, we have touched upon the boundary layers, externa! fiows, have been conducted for a large and incompressible fiow. None for to do so would add little to our could not pursue a more rigorous
of Turbulence and Surface Roughness on Rept. 1283, Gr. Brit., 1930. Past a Circular Cylinder at Very High
d
f ( :::z + V:z
o
d
Vy
:;:z) dy =
f :;:z
V
dy
o
and expressing v, as
11-3. Obtain an · approximate solution for flow past a flat plate, expressing the velocity as o s.; y s.; o
Compare the calculated boundary !ayer thickness and drag coefficient with those given by the solution of the Prandtl boundary layer equations. Ans: o = 3.47 Vvxfu 00
11-4. Air at n•F and 1 atm flows past a flat plate with a uniform free stream velocity of30 ft/sec. What is the boundary !ayer thickness ata point 5 in. from the leading edge? 11-5. Consideran immersed body of mass M (such as a submarine) moving through an infinite, quiescent fluid at a velocity w = :Au0 • Derive an expression for the power required to move the body in terms of eD and u0• Let A • be the characteristic area used in defining eD• 11-6. The turbulent boundary !ayer thickness on a smooth, flat plate can be determined approximately if we make use of experimental data for turbulent flow in smooth tubes. The velocity profile may be approximated by (see Sec. 6.3)
454
Flow Around lmmersed Bodies
Chap. 11
and the shear stress at the wall by
To =
(o
p (Üz )2
(see Sec. 8.2) thus leading to
lndex
To = ({)pu!o 1=
0.316c<5 VUoo
rl/4
Here we have replaced the average velocity in a tube with the free stream velocity, and the tu be diameter with 2<5. We can use the momentum equation to determine the turbulent boundary !ayer thickness. Consider carefully the boundary condition to be imposed on the derived differential equation for <5, Use these results to calculate a drag coefficient that may be compared with the experimental curve for CD shown in Fig. 11.4-3.
Acceleration, 82, 166 Adiabatic flow, 398 Ambient pressure in the momentum balance, 257 Angular momentum principie, 7 Apparent viscosity, 19, 68 Archimedes' principie, 53 Atmospheric pressure, 43 Averages area and volume, 5, 61, 109 time, 187 Fl¡. 11-7. Suddenly accelerated joumal bearing.
11-7. A bearing, shown in Fig. 11-7, is initially at rest. At time t = O it is suddenly accelerated to a constant angular velocity, w. If the viscosity of the oil in the bearing is 100 centipoise and the annular gap is 0.10 in., estimare the time required to obtain a steady velocity profile.
Barometers, 43 Bemoulli's equation, 230 Bingham model fluid, 19 Blasius, H., 300 Borda mouthpiece, 247 Boundary conditions, 42, 173 Boundary layer flow, 422, 430 Boundary layer thickness, 426, 431 Boundary layer separation, 448 Buoyancy force, 53 Cavitation, 22 Cavitation number, 22 Center of pressure, 49 Center of stress, Prob. 2-14 Olokc flow, 412
Flow Around lmmersed Bodies
Chap. 11
(see Sec. 8.2)
0.316Nil''
Index
velocity in a tube with the free stream 215. We can use themomentumequation !ayer thickness. Consider carefully the the derived differential equation for d. coefficient that may be compared with in Fig. ll.4-3.
Acceleration, 82, 166 Adiabatic ftow, 398 Ambient pressure in the momentum balance, 257 Angular momentum principie, 7 Apparent viscosity, 19, 68 Archimedes' principie, 53 Atmospheric pressure, 43 Averages area and volume, 5, 61, 109 time, 187
at rest. At time t = O it is suddenly w. lf the viscosity of the oil in the gap is 0.10 in., estimate the time pro file.
Barometers, 43 Bernoulli's equation, 230 Bingham model fluid, 19 Blasius, H., 300 Borda mouthpiece, 247 Boundary conditions, 42, 173 Boundary !ayer ftow, 422, 430 Boundary !ayer thickness, 426, 431 Boundary !ayer separation, 448 Buoyancy force, 53 Cavitation, 22 Cavitation number, 22 Center of pressure, 49 Center of stress, Prob. 2-14 Cbokc flow, 412
Coefficient of expansion, 13 Coefficient of viscosity, 14 Colebrook, C. F., 300 Compressibility, 13 Compressible ftow, 3, 390 Conservation of energy, 8, 391 Conservation of mass, 8, 92 Conservation of momentum, 8 Constitutive equations, 9, 128 Continuity equation, 93, 189 Continuity of stress, 42, 174 Continuity of velocity, 42, 176 Continuum, 1 Contraction coefficient, 334 Control surface, 33 Control volume, 33, 88 Convective acceleration, 82 Couette viscometer, 15, 177 Critica! ftow, 359 Critica! depth, 362 Critica! Reynolds Number, 5, 195 Critica! velocity, 362 Cylinder, ftow around, 305, 445 D'Alembert's paradox, 446 Darcy-Weishbach formula, 310 Deformation, 135 Density, 2 455
456 Derivatives (see Material derivative, Total derivative, and Partial derivative) Differential-macroscopic balance, 273, 355, 405, 436 Diffuser, 312 Dilatant fluid, 20, 68 Dimensional analysis, 158, 291 Discharge coefficient, 333 Divergence of a tensor, 123 Divergence of a vector, 84 Divergence theorem, 84 Drag coefficients, 304 Drag force, 286 Dummy index, 27 Dyad, 112 Eddy viscosity, 200 Energy equation, 391 Enthalpy, 394 Entrance length, 171, 369 Entrance Joss, 315 Entropy equation, 398 Equation of state, 9 Equations of fluid statics, 36 Equations of motion, 154, 193 First law of thermodynamics, 391 Flow meters, 331 Flow nozzle, 335 Free index convention, 27 Friction factors, 287, 293, 352 Friction Joss, 308 Froude number, 165, 359 Gas constant, 402 Gauge pressure, 45 Gauss' theorem, 84 Gibbs notation, 23 Gradient of a scalar, 40 Gradually varied flow, 355 Gravitational constant, 11 Gravitational potential function, 40 Hagen-Poiseuille flow, 61 Hardy Cross method, 319 Head loss, 308 Helmholtz flows, 447 Hydraulic jump, 382 Hydraulic radius, 159, 303, 310, 352 Hydrometer, 55 Hydrostatic stress, 36
lndex
Ideal gas Jaw, 9 lncompressible flow, 3 lndex notation, 26 lrrotational flow, 151, 445 lsentropic flow, 398, 404 Isotropy, 141 Kinematic viscosity, 16 Kinetic energy, 224 Kronecker delta, 130 Laminar boundary !ayer, 430 Laminar flow, 4, 56, 169, 294 Laminar sublayer, 197 Laws of Mechanics, 7 Linear momentum principie, 8, 36 Local acceleration, 82 Loss coefficient, 311, 316 Mach number, 403 Manning formula, 352 Manometer, 44, 323 Mass balance, 214 Material coordinates, 76 Material derivative, 78 Material position vector, 76 Material surface, 33 Material volume, 32 Mechanical energy balance, 228 Mechanical energy equation, 224, 396 Minor loss, 308 Mixing Jength, 202 Momentum balance, 218, 254 Moody chart, 298 Moving control volumes, 261 Moving reference frames, 154 Navier-Stokes equations, 154 Net stress vector, 257 Newton's law of viscosity, 16, 133, 139 Newton's second law of motion, 7 Noncircular r.onduits, 303 Non-Newtonian fluids, 16, 63 Normal vector, 35, 46 Nozzle flow, 247, 335 One-dimensional flow, 6 Order of magnitude analysis, 329 Orifice meter, 334 Oscillating systems, 323 Ostwald-de Wael model fluid, 20, 63
lndex
Partial derivative, 79 Path lines, 97 Perfect gas, 9, 402 Piezometric head, 308 Pipe flow: laminar, 57, 169, 294 turbulent, 196, 295 Pipeline networks, 318 Pitot-static tube, 337 Potential energy function, 41, 393 Power-law fluid, 20, 63 Prandtl boundary !ayer equations, 434 Prandtl mixing length theory, 200 Pressure: atmospheric, 43 gauge,45 vapor, 22 Pressure head, 308 Projected areas, 37, 53, 111 Pseudoplastic fluid, 20 Rate of strain, 14, 135, 147 Rate of strain tensor, 134, 138 Reducer, 313 Relative roughness, 293 Relative velocity, 156 Reynolds, 0., 4 Reynolds number, 5, 160 Reynolds stresses, 194 Reynolds transport theorem, 92 Rotation, 151 Roughness factor, 293 Schedule number, 302 Separation, 448 Sharp-crested weir, 339 Shear rate, 14 Shock waves, 416 Slip flow, 3 Solitary wave, 370 Sonic velocity, 3, 398 Spatial coordinates, 76 Spatial position vector, 76 Specific energy, 360 Specific gravity, 55 Speed of sound, 402 Stagnation pressure, 407 Stagnation temperature, 407 Steady flow, 6 Strain rate, 14 Streak lines, 97 Stream function, 103
lndex
Ideal gas law, 9 Incompressible flow, 3 Index notation, 26 Irrotational flow, 151, 445 n ...,ntrt)nir flow, 398, 404 lsotropy, 141 Kinematic viscosity, 16 Kinetic energy, 224 Kronecker delta, 130
Laminar boundary !ayer, 430 Laminar flow, 4, 56, 169, 294 1'-"''uu. mu sublayer, 197 of Mechanics, 7 momentum principie, 8, 36 acceleration, 82 coefficient, 311 , 316 Mach number, 403 Manning formula , 352 Manometer, 44, 323 Mass balance, 214 Material coordinates, 76 Material derivative, 78 Material position vector, 76 Material surface, 33 Material volume, 32 Mechanical energy balance, 228 Mechanical energy equation, 224, 396 Minor loss, 308 Mixing length, 202 Momentum balance, 218, 254 Moody chart, 298 control volumes, 261 reference frames, 154
One-dimensional flow, 6 der of magnitude analysis, 329 rifice meter, 334 cillating systems, 323 twald-de Wael model fluid, 20, 63
~
457
lndex
Partial derivative, 79 Path lines, 97 Perfect gas, 9, 402 Piezometric head, 308 Pipe flow: laminar, 57, 169, 294 turbulent, 196, 295 Pipeline networks, 318 Pitot-static tube, 337 Potential energy function, 41, 393 Power-law fluid, 20, 63 Prandtl boundary !ayer equations, 434 Prandtl mixing length theory, 200 Pressure: atmospheric, 43 gauge, 45 vapor, 22 Pressure head, 308 Projected areas, 37, 53, 111 Pseudoplastic fluid, 20 Rate of strain, 14, 135, 147 Rate of strain tensor, 134, 138 Reducer, 313 Relative roughness, 293 Relative velocity, 156 Reynolds, O., 4 Reynolds number, 5, 160 Reynolds stresses, 194 Reynolds transport theorem, 92 Rotation, 151 Roughness factor, 293 Schedule number, 302 Separation, 448 Sharp-crested weir, 339 Shear rate, 14 Shock waves, 416 Slip flow, 3 Solitary wave, 370 Sonic velocity, 3, 398 Spatial coordinates, 76 Spatial position vector, 76 Specific energy, 360 Specific gravity, 55 Speed of sound, 402 Stagnation pressure, 407 Stagnation temperature, 407 Steady flow, 6 Strain rate, 14 Streak lines, 97 Stream function, 103
Streamlines, 97 Stress equations of motion, 121, 131 , 219 Stress tensor, 112 Stress vector, 35, 108 Supercritical flow, 359 Surface profiles, 364 Surface tension, 21 Tangent vector, 49, 98, 230 Thermal energy equation, 396 Time a verages, 187 Time derivatives (see Material derivative, Total deriva ti ve, and Partial derivative) Torque, 7, 48, 179 Total derivative, 77 Transition region, 197 Transport theorem, 88 Turbulent core, 197 Turbulent flow, 4, 186 Turbulent stress tensor, 194 Turbulent velocity profile, 196, 203, 242 Two phase flow, 173 Uniform flow, 349 Uniformly accelerated flow, 166 Units, 10 Unit tensor, 129 Universal gas constant, 402 Unsteady flow, 6, 261, 320 U-tube manometer, 44, 323 Vapor pressure, 22 Vectors, 22 Velocity head, 308 Velocity of sound, 402 Vena contracta, 334 Venturi meter, 331 Viscosity, 14, 134 Viscous dissipation, 223, 396 Viscous stress tensor, 130 Vorticity tensor, 138, 152 Vorticity vector, 152 Wall shear stress, 205, 289, 351 Waves: gravity, 374 ripples, 374 shallow-water, 370 Wave speed, 373 Weir: broad-crested, 379 sharP-crested, 339