CHAPTER TWO 3 wk
7d
24 h 3600 s 1000 ms
= 18144 . × 10 9 ms 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h
2.1 (a)
(c)
2.2 (a)
554 m 4 1d 1h d ⋅ kg 24 h 60 min
1 kg 108 cm 4 = 3.85 × 10 4 cm 4 / min⋅ g 1000 g 1 m 4
1 m 1 h 760 mi = 340 m / s h 0.0006214 mi 3600 s 1 m3 = 57.5 lb m / ft 3 35.3145 ft 3
(b)
921 kg 2.20462 lb m m3 1 kg
(c)
5.37 × 10 3 kJ 1 min 1000 J min 60 s 1 kJ
1.34 × 10 -3 hp = 119.93 hp ⇒ 120 hp 1 J/s
2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = . × 10 6 ≈ 5 million balls = 518 ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr
3600 s 1.86 × 10 5 mi
1d
1 h
1
s
3.2808 ft 0.0006214 mi
1 step = 7 × 1016 steps 2 ft
2.5 Distance from the earth to the moon = 238857 miles 238857 mi
1
m
0.0006214 mi
1 report 0.001 m
= 4 × 1011 reports
2.6
19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost
Total Cost
American
European
= $14,500 +
= $21,700 +
$1.25 1 gal gal 28 mi
x (mi)
= 14,500 + 0.04464 x
$1.25 1 gal x (mi) = 21,700 + 0.02796 x gal 44.7 mi
Equate the two costs ⇒ x = 4.3 × 10 5 miles
2-1
2.7 5320 imp. gal
106 cm3
14 h 365 d
plane ⋅ h
1 d
1 yr
0.965 g
220.83 imp. gal
1
cm
3
1 kg
1 tonne
1000 g
1000 kg
tonne kerosene plane ⋅ yr
= 1.188 × 105
4.02 × 109 tonne crude oil 1 tonne kerosene
plane ⋅ yr
7 tonne crude oil 1.188 × 10 tonne kerosene 5
yr
= 4834 planes ⇒ 5000 planes
2.8 (a) (b) (c)
2.9
2.10 2.11
32.1714 ft / s 2
25.0 lb m
1
lb f
32.1714 lb m ⋅ ft / s 2 25 N
1 kg ⋅ m/s 2
1 9.8066 m/s 2
10 ton
1N
1 lb m 5 × 10
-4
50 × 15 × 2 m 3
500 lb m
= 2.5493 kg ⇒ 2.5 kg
980.66 cm / s 2
1000 g ton
= 25.0 lb f
1 g ⋅ cm / s 2
2.20462 lb m
35.3145 ft 3 1 m3
85.3 lb m 1 ft 3
1
1 m3
kg
2.20462 lb m
1 dyne
11.5 kg
= 9 × 10 9 dynes
32.174 ft 1 lb f = 4.5 × 10 6 lb f 2 2 1 s 32.174 lb m / ft ⋅ s
≈ 5 × 10 2
FG 1 IJ FG 1 IJ ≈ 25 m H 2 K H 10K
3
(a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2
ρc
ρfh
(30 cm − 14.1 cm)(100 . g / cm 3 ) ρc = = = 0.53 g / cm 3 H 30 cm ρ H (30 cm)(0.53 g / cm 3 ) (b) ρ f = c = = 171 . g / cm 3 (30 cm - 20.7 cm) h
2.12
Vs =
πR 2 H 3
⇒ Vf =
; Vf =
3
−
πr 2 h 3
;
πh Rh
2
2
f
H H−
h3 H2
3
2
= ρs
H3 = ρs H 3 − h3
h
3
r
2
2
ρf h
R r R = ⇒r = h H h H
FG IJ = πR FG H − h IJ − 3 3 H HK 3 H H K πR F πR H h I ⇒ρ =ρ H− G J 3 H 3 H K
πR H 2
ρ f V f = ρ sVs ⇒ ρ f = ρs
πR 2 H
H
H
2
ρs
s
R
1
1−
FG h IJ H HK
2-2
3
ρf
2.13
Say h( m) = depth of liquid
y y= 1 dA y=y=1––1+h h xx
⇒ A(m 2 )
1− y
dA = dy ⋅
∫
h
x = 1– y 2 y= –1
dA
2
−1+ h
( )=2 ∫
dx = 2 1 − y dy ⇒ A m 2
2
1 − y 2 dy
−1
− 1− y 2
⇓
1m
Table of integrals or trigonometric substitution
π 2 A m 2 = y 1 − y 2 + sin −1 y ⎤⎥ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) + ⎦ −1 2 h −1
( )
b g
W N =
4 m × A( m 2 ) 0.879 g 10 6 cm 2 cm
3
1m
E Substitute for A L W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g N 4
2.14
3
1 kg
9.81 N
3
kg N
10 g
= 3.45 × 10 4 A
g g0
2
b g π2 OPQ
+ sin −1 h − 1 +
1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m 1 1 poundal = 1 lb m ⋅ ft / s 2 = lb f 32.174 (a) (i) On the earth: 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 5.63 × 10 3 poundals 2 s 1 lb m ⋅ ft / s 2 (ii) On the moon 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 938 poundals 2 6 s 1 lb m ⋅ ft / s 2 (b) F = ma ⇒ a = F / m =
355 poundals 25.0 slugs
1 lb m ⋅ ft / s 2 1 poundal
= 0.135 m / s 2
2-3
1 slug 32.174 lb m
1m 3.2808 ft
2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 )
⇒
FG 1IJ = 5.3623 bung ⋅ ft / s H 6K
2
1 fern 5.3623 bung ⋅ ft / s 2
3 bung 32.174 ft 1 fern = 3 fern 2 6 s 5.3623 bung ⋅ ft / s 2 On the earth: W = (3)( 32.174) / 5.3623 = 18 fern
(b) On the moon: W =
2.16 (a) ≈ (3)(9) = 27
(b)
(2.7)(8.632) = 23 (c) ≈ 2 + 125 = 127
(d)
2.365 + 125.2 = 127.5 2.17 R ≈
4.0 × 10−4 ≈ 1× 10−5 40 (3.600 ×10−4 ) / 45 = 8.0 × 10−6 ≈
≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4 4.753 × 10 4 − 9 × 10 2 = 5 × 10 4
(7 ×10−1 )(3 × 105 )(6)(5 × 104 ) ≈ 42 × 102 ≈ 4 × 103 (Any digit in range 2-6 is acceptable) (3)(5 × 106 )
Rexact = 3812.5 ⇒ 3810 ⇒ 3.81× 103 2.18 (a) A: R = 731 . − 72.4 = 0.7 o C X=
. + 72.6 + 72.8 + 73.0 72.4 + 731 = 72.8 o C 5
s=
(72.4 − 72.8) 2 + (731 . − 72.8) 2 + (72.6 − 72.8) 2 + (72.8 − 72.8) 2 + (73.0 − 72.8) 2 5−1
= 0.3o C B: R = 1031 . − 97.3 = 58 . oC X=
97.3 + 1014 . + 98.7 + 1031 . + 100.4 = 100.2 o C 5
s=
(97.3 − 100.2) 2 + (1014 . − 100.2) 2 + (98.7 − 100.2) 2 + (1031 . − 100.2) 2 + (100.4 − 100.2) 2 5−1
= 2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-4
2.19 (a)
12
X=
∑X
12
i
i =1
C min=
= 73.5 s= 12 = X − 2 s = 73.5 − 2(1.2) = 711 .
∑ ( X − 735. )
2
i =1
= 12 .
12 − 1
C max= = X + 2 s = 735 . + 2(12 . ) = 75.9
(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness 2.20 (a), (b) 1 2 (a) Run 134 131 X Mean(X) 131.9 Stdev(X) 2.2 127.5 Min 136.4 Max (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14
X 128 131 133 130 133 129 133 135 137 133 136 138 135 139
Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5
3 129
Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9
4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133
Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4
140 138 136 134 132 130 128 126 0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q ' =
2.36 × 10−4 kg ⋅ m 2
(b) Q 'approximate ≈
h
2.20462 lb 3.28082 ft 2 m2
kg
1
h
3600 s
(2 × 10−4 )(2)(9) ≈ 12 × 10( −4−3) ≈ 1.2 × 10−6 lb ⋅ ft 2 / s 3 × 103
Q 'exact =1.56 × 10−6 lb ⋅ ft 2 / s = 0.00000156 lb ⋅ ft 2 / s
2-5
2.22 N Pr = N Pr ≈
Cpμ
=
k
0.583 J / g ⋅ o C
1936 lb m
0.286 W / m ⋅ C −1
1 h 3.2808 ft
ft ⋅ h
o
3600 s
1000 g
m 2.20462 lb m
(6 × 10 )(2 × 10 )(3 × 10 ) 3 × 10 ≈ ≈ 15 . × 10 3 . The calculator solution is 163 . × 10 3 −1 3 2 (3 × 10 )(4 × 10 )(2) 3
3
3
2.23 Re = Re ≈
2.24 (a)
Duρ
μ
=
0.48 ft
1
m
2.067 in
1 m
1 kg 10 6 cm 3
0.805 g
s 3.2808 ft 0.43 × 10 −3 kg / m ⋅ s 39.37 in
cm 3
1000 g
1 m3
(5 × 10 −1 )(2)(8 × 10 −1 )(10 6 ) 5 × 101− ( −3) ≈ ≈ 2 × 10 4 ⇒ the flow is turbulent 3 (3)(4 × 10)(10 3 )(4 × 10 −4 )
kg d p y D
1/ 3
⎛ μ ⎞ = 2.00 + 0.600 ⎜ ⎟ ⎝ ρD ⎠
⎛ d p uρ ⎞ ⎜ ⎟ ⎝ μ ⎠
1/ 2
1/ 3
⎡ ⎤ 1.00 × 10−5 N ⋅ s/m 2 = 2.00 + 0.600 ⎢ ⎥ −5 3 2 ⎣ (1.00 kg/m )(1.00 × 10 m / s) ⎦ k g (0.00500 m)(0.100) = 44.426 ⇒ = 44.426 ⇒ k g 1.00 × 10−5 m 2 / s
1/ 2
⎡ (0.00500 m)(10.0 m/s)(1.00 kg/m3 ) ⎤ ⎢ ⎥ (1.00 × 10−5 N ⋅ s/m 2 ) ⎣ ⎦ = 0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m)
y
0.005 0.010 0.005 0.005 0.005
0.1 0.1 0.1 0.1 0.1
D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s) 1.00E-05 1.00E-05 1 10 1.00E-05 1.00E-05 1 10 2.00E-05 1.00E-05 1 10 1.00E-05 2.00E-05 1 10 1.00E-05 1.00E-05 1 20
kg 0.889 0.620 1.427 0.796 1.240
2.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2
200 crystals 0.050 in 25.4 mm 10 crystals 0.050 2 in 2 − min ⋅ mm in min ⋅ mm 2 238 crystals 1 min = 238 crystals / min ⇒ = 4.0 crystals / s 60 s min
(25.4) 2 mm 2 in 2
(b) r =
b g
(c) D mm =
b g
D ′ in
FG H
IJ K
crystals 60 s 25.4 mm crystals = 25.4 D ′ ; r = r′ = 60r ′ s 1 min min 1 in
b
g b
⇒ 60r ′ = 200 25.4 D ′ − 10 25.4 D ′
g
2
2-6
b g
⇒ r ′ = 84.7 D ′ − 108 D ′
2
2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f ⎡8.27 × 10−7 in 2 9 × 106 N 14.696 lbf / in 2 ⎤ (b) ρ = (70.5 lb m / ft 3 )exp ⎢ ⎥ lbf m 2 1.01325 × 105 N/m 2 ⎥⎦ ⎢⎣ 70.57 lb m 35.3145 ft 3 1 m3 1000 g = = 1.13 g/cm3 3 3 6 3 ft m 10 cm 2.20462 lbm
(c)
F lb IJ = ρ ′ g ρG H ft K cm F lb IJ = P' N PG H in K m m 3
3
f 2
1 lb m
28,317 cm 3
453.593 g
1 ft 3
0.2248 lb f
12
2
m2
39.37 2 in 2
1N
d
= 62.43ρ ′ = 145 . × 10 −4 P '
id
i
d
i
⇒ 62.43ρ ′ = 70.5 exp 8.27 × 10 −7 1.45 × 10 −4 P ' ⇒ ρ ′ = 113 . exp 120 . × 10 −10 P '
P ' = 9.00 × 10 6 N / m 2 ⇒ ρ ' = 113 . exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113 . g / cm 3 cm = 16.39V ' ; t bsg = 3600t ′b hr g d i V ' din i 28,317 1728 in ⇒ 16.39V ' = expb3600t ′ g ⇒ V ' = 0.06102 expb3600t ′ g
2.27 (a) V cm 3 =
3
3
3
(b) The t in the exponent has a coefficient of s-1. 2.28 (a) 3.00 mol / L, 2.00 min -1 (b) t = 0 ⇒ C = 3.00 exp[(-2.00)(0)] = 3.00 mol / L
t = 1 ⇒ C = 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 − 3.00 For t=0.6 min: (0.6 − 0) + 3.00 = 14 . mol / L Cint = 1− 0 Cexact = 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L For C=0.10 mol/L:
t int = t exact
1− 0 (010 . − 3.00) + 0 = 112 . min 0.406 − 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00
(c) 3.5 C exact vs. t
3 C (mol/L)
2.5 2
(t=0.6, C=1.4)
1.5 1
(t=1.12, C=0.10)
0.5 0 0
1
2
t (min)
2-7
p* =
2.29 (a)
(b)
60 − 20 (185 − 166.2) + 20 = 42 mm Hg 199.8 − 166.2
c
MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END OUTPUT DATA 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 # # 215.5 100.0 105.0 1.8 # # 215.0 98.7
2.30 (b) ln y = ln a + bx ⇒ y = ae bx b = (ln y 2 − ln y1 ) / ( x 2 − x1 ) = (ln 2 − ln 1) / (1 − 2) = −0.693 ln a = ln y − bx = ln 2 + 0.63(1) ⇒ a = 4.00 ⇒ y = 4.00e −0.693 x
(c) ln y = ln a + b ln x ⇒ y = ax b b = (ln y 2 − ln y1 ) / (ln x 2 − ln x1 ) = (ln 2 − ln 1) / (ln 1 − ln 2) = −1
ln a = ln y − b ln x = ln 2 − ( −1) ln(1) ⇒ a = 2 ⇒ y = 2 / x (d) ln( xy ) = ln a + b( y / x) ⇒ xy = aeby / x ⇒ y = (a / x)eby / x [can't get y = f ( x)] b = [ln( xy ) 2 − ln( xy )1 ]/[( y / x) 2 − ( y / x)1 ] = (ln 807.0 − ln 40.2) /(2.0 − 1.0) = 3 ln a = ln( xy ) − b( y / x) = ln 807.0 − 3ln(2.0) ⇒ a = 2 ⇒ xy = 2e3 y / x [can't solve explicitly for y ( x)]
2-8
2.30 (cont’d) (e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2) b ]1/ 2
b = [ln( y 2 / x ) 2 − ln( y 2 / x ) 1 ] / [ln( x − 2) 2 − ln( x − 2) 1 ] . ) = 4.33 = (ln 807.0 − ln 40.2) / (ln 2.0 − ln 10 ln a = ln( y 2 / x ) − b( x − 2) = ln 807.0 − 4.33 ln(2.0) ⇒ a = 40.2 ⇒ y 2 / x = 40.2( x − 2) 4.33 ⇒ y = 6.34 x 1/ 2 ( x − 2) 2.165 2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope = m, Intcpt = − n (c)
1 1 a 1 = + x ⇒ Plot vs. ln( y − 3) b b ln( y − 3)
x [rect. axes], slope =
a 1 , intercept = b b
(d)
1 1 = a ( x − 3) 3 ⇒ Plot vs. ( x − 3) 3 [rect. axes], slope = a , intercept = 0 2 2 ( y + 1) ( y + 1) OR 2 ln( y + 1) = − ln a − 3 ln( x − 3) Plot ln( y + 1) vs. ln( x − 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a ⇒ slope = − , intercept = − 2 2 (e) ln y = a x + b
Plot ln y vs.
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
(f) log10 ( xy ) = a ( x 2 + y 2 ) + b Plot log10 ( xy ) vs. ( x 2 + y 2 ) [rect.] ⇒ slope = a, intercept = b
(g)
x b x 1 vs. x 2 [rect.], slope = a , intercept = b = ax + ⇒ = ax 2 + b ⇒ Plot y x y y OR
b 1 1 b 1 1 vs. 2 [rect.] , slope = b, intercept = a = ax + ⇒ = a + 2 ⇒ Plot y x xy xy x x
2-9
2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0169 . ).
0.18 0.16 0.14 0.12 y
0.1 0.08 0.06 0.04 0.02 0 0
20
40
60
80
100
R
y=aR+b
U| V| W
. − 0.011 0169 = 2.11 × 10 −3 80 − 5 ⇒ y = 2.11 × 10 −3 R + 4.50 × 10 −4 −3 −4 b = 0.011 − 2.11 × 10 5 = 4.50 × 10
a=
ib g
d
ib g
d
(b) R = 43 ⇒ y = 2.11 × 10 −3 43 + 4.50 × 10 −4 = 0.092 kg H 2 O kg
b1200 kg hgb0.092 kg H O kgg = 110 kg H O h 2
2
2.33 (a) ln T = ln a + b ln φ ⇒ T = aφ b
b = (ln T2 − ln T1 ) / (ln φ 2 − ln φ 1 ) = (ln 120 − ln 210) / (ln 40 − ln 25) = −119 . ln a = ln T − b ln φ = ln 210 − ( −119 . ) ln(25) ⇒ a = 9677.6 ⇒ T = 9677.6φ −1.19
b
(b) T = 9677.6φ −1.19 ⇒ φ = 9677.6 / T
b
g
T = 85o C ⇒ φ = 9677.6 / 85
0.8403
b g T = 290 C ⇒ φ = b9677.6 / 290g T = 175o C ⇒ φ = 9677.6 / 175 o
g
0.8403
. L/s = 535
0.8403
= 29.1 L / s
0.8403
= 19.0 L / s
(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.
2-10
ln ((CA-CAe)/(CA0-CAe))
2.34 (a) Yes, because when ln[(C A − C Ae ) / (C A0 − C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0
50
100
150
200
0 -0.5 -1 -1.5 -2 t (m in)
Slope = -0.0093 ⇒ k = 9.3 × 10-3 min −1
(b) ln[(C A − C Ae ) /(C A0 − C Ae )] = − kt ⇒ C A = (C A0 − C Ae )e − kt + C Ae
C A = (0.1823 − 0.0495)e − (9.3×10 C =m /V ⇒ m =CV =
−3
)(120)
+ 0.0495 = 9.300 × 10-2 g/L
9.300 × 10-2 g 30.5 gal 28.317 L = 10.7 g L 7.4805 gal
2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 . × 10 −2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 . × 10−2 (c) V ( m3 ) = 100 . × 10 −3 exp(15 . × 10 −7 t 2 ) 2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k lnV 8.5
lnP
8 7.5 7 6.5 6 2.5
3
lnP = -1.573(lnV ) + 12.736
3.5
4
lnV
k = − slope = − ( −1573 . ) = 1573 . (dimensionless) Intercept = ln C = 12.736 ⇒ C = e12.736 = 3.40 × 105 mm Hg ⋅ cm4.719 G − GL 1 G −G G −G = ln K L + m ln C = ⇒ 0 = K L C m ⇒ ln 0 m G0 − G K L C G − GL G − GL ln (G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
3 ln(G 0-G)/(G-G L )
2.37 (a)
2 1 0 -1 3 .5
4
4 .5
5
ln C
2-11
5 .5
2.37 (cont’d)
m = slope = 2.483 (dimensionless) Intercept = ln K L = −10.045 ⇒ K L = 4.340 × 10 −5 ppm-2.483
G − 180 . × 10 −3 = 4.340 × 10 −5 (475) 2.483 ⇒ G = 1806 × 10 −3 . 3.00 × 10 −3 − G C=475 ppm is well beyond the range of the data.
(b) C = 475 ⇒
2.38 (a) For runs 2, 3 and 4: Z = aV b p c ⇒ ln Z = ln a + b lnV + c ln p
ln( 35 . ) = ln a + b ln(102 . ) + c ln(9.1)
b = 0.68 ⇒ c = −1.46
ln(2.58) = ln a + b ln(102 . ) + c ln(112 . )
a = 86.7 volts ⋅ kPa 1.46 / (L / s) 0.678
ln(3.72) = ln a + b ln(175 . ) + c ln(112 . )
. Slope=b, Intercept= ln a + c ln p (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV 2
lnZ
1.5 1 0.5 0 -1
-0.5
0
lnZ = 0.5199lnV + 1.0035
0.5
1
1.5
lnV
b = slope = 0.52
Intercept = lna + c ln P = 10035 . When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a + c lnV 2
lnZ
1.5 1 0.5 0 1.5
1.7
lnZ = -0.9972lnP + 3.4551
1.9
2.1
2.3
lnP
c = slope = −0.997 ⇒ 10 . Intercept = lna + b lnV = 3.4551
Z
Plot Z vs V b P c . Slope=a (no intercept) 7 6 5 4 3 2 1 0.05
Z = 31.096VbPc
0.1
0.15
0.2
Vb Pc
a = slope = 311 . volt ⋅ kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-12
2.39 (a)
sxy = sxx =
a= b=
∑x y
1 n
∑x
1 n
sx =
n
1 n
i i
= [(0.4)(0.3) + (2.1)(19 . ) + (31 . )( 3.2)] / 3 = 4.677
i =1 n
= (0.32 + 19 . 2 + 3.2 2 ) / 3 = 4.647
2 i
i =1 n
∑
xi = (0.3 + 1.9 + 3.2) / 3 = 18 . ; sy =
i =1
sxy − sx s y
b g
sxx − sx
2
=
sxx s y − sxy sx
b g
sxx − sx
2
1 n
n
∑y
i
= (0.4 + 2.1 + 31 . ) / 3 = 1867 .
i =1
4.677 − (18 . )(1.867) = 0.936 4.647 − (18 . )2 ( 4.647)(1867 . ) − (4.677)(18 . ) = 0.182 2 4.647 − (18 . )
=
. y = 0.936 x + 0182 (b) a =
sxy sxx
=
4.677 = 1.0065 ⇒ y = 1.0065x 4.647
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1 0 0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a
a = Intercept = 0.082 L / g
3 2.5 2 1.5 1 0.5 0
2 1.5 C
1/C
(b) b = slope = 0.477 L / g ⋅ h;
1 0.5 0
0
1
1/C = 0.4771t + 0.0823
2
3
4
5
6
1
t
C
2 C-fitted
3
4
5
t
(c) C = 1 / (a + bt ) ⇒ 1 / [0.082 + 0.477(0)] = 12.2 g / L
t = (1 / C − a ) / b = (1 / 0.01 − 0.082) / 0.477 = 209.5 h (d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2-13
2.41 (a) and (c)
y
10
1 0.1
1
10
100
x
(b) y = ax b ⇒ ln y = ln a + b ln x; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.1258 2
ln y
1.5 1 0.5
b = slope = 0.168
0 -1
0
1
2 ln x
3
4
5
Intercept = ln a = 11258 . ⇒ a = 3.08
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)
600
0
800
ln(1-Cp/Cao)
400
ln(1-Cp/Cao)
0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0.0062t
100
400
500
t
Lab 1
600
400
600
-4
-6 ln(1-Cp/Cao) = -0.0111t
t
Lab 2
k = 0.0111 s-1
800
0
0 ln(1-Cp/Cao)
ln(1-Cp/Cao)
200
300
-2
k = 0.0062 s-1
0
200
0
-2 -4
200
400
600
800
0 -2 -4
-6 ln(1-Cp/Cao)= -0.0064t
-6 ln(1-Cp/Cao) = -0.0063t t
t
Lab 3
k = 0.0063 s-1
Lab 4
k = 0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k = 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2-14
2.43 yi = axi ⇒ φ (a ) =
n
∑
di2
=
i =1
⇒a=
i =1
2.44
i
i =1
− axi
g
2
dφ ⇒ =0= da
∑ 2b y n
i
i =1
g
− axi xi ⇒
n
∑y x
i i
i =1
−a
n
∑x
2 i
i =1
n
n
∑
∑by n
yi xi /
∑x
2 i
i =1
DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X * 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a = 6.536, b = −4.206
2-15
=0
2.45 (a) E(cal/mol), D0 (cm2/s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0. (c) Intercept = ln D0 = -3.0151 ⇒ D0 = 0.05 cm2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = − E / R = -3666 K ⇒ E = (3666 K)(1.987 cal / mol ⋅ K) = 7284 cal / mol
-10.0 ln D
-11.0 -12.0 -13.0 -14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet T 347 374.2 396.2 420.7 447.7 471.2
D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05
1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0
lnD (1/T)*(lnD) -13.5 -0.03897 -12.9 -0.03447 -12.3 -0.03105 -11.7 -0.02775 -11.2 -0.02495 -10.8 -0.02296 2.47E-03 -12.1 -3.00E-02 6.16E-06 -3666 -3.0151
D0
7284
E
0.05
2-16
(1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06
