Elementary Principles of Chemical Processes – Solutions Manual | Complete Step-by-Step Answers

CHAPTER TWO

2.1 (a)
3 wk 7d 24 h 3600 s 1000 ms = 1.8144  109 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
554 m4 1d 1h 1 kg 108 cm4 4 4
(c) = 3.85  10 cm / min g
d  kg 24 h 60 min 1000 g 1 m 4



760 mi 1 m 1 h
2.2 (a) = 340 m/ s
h 0.0006214 mi 3600 s
921 kg 2.20462 lb m 1 m3 3
(b) = 57.5 lbm / ft
m3 1 kg 35.3145 ft 3
3 -3
5.37  10 kJ 1 min 1000 J 1.34  10 hp
(c) = 119.93 hp  120 hp
min 60 s 1 kJ 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 in3 1 ball
n = = 5.18  106  5 million balls
balls 3 3
ft 3 2 in
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 3600 s 1.86  105 mi 3.2808 ft 1 step = 7  1016 steps
1 yr 1 d 1 h 1 s 0.0006214 mi 2 ft

2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1 m 1 report
= 4  1011 reports
0.0006214 mi 0.001 m

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.
$1.25 1 gal x (mi)
Total Cost American = $14,500 + = 14,500 + 0.04464x
gal 28 mi

$1.25 1 gal x (mi)
Total Cost European = $21,700 + = 21,700 + 0.02796x
gal 44.7 mi

Equate the two costs  x = 4.3  105 miles




2-1

,2.7
5320 imp. gal 14 h 365 d 106 cm3 0.965 g 1 kg 1 tonne
plane  h 1 d 1 yr 220.83 imp. gal 1 cm 3
1000 g 1000 kg
tonne kerosene
= 1.188 105
plane  yr
4.02 109 tonne crude oil 1 tonne kerosene plane  yr
yr 7 tonne crude oil 1.188 10 tonne kerosene 5


= 4834 planes  5000 planes


25.0 lbm 32.1714 ft / s2 1 lb f
2.8 (a) = 25.0 lb f
32.1714 lbm  ft / s2
25 N 1 1 kg  m/s2
(b) = 2.5493 kg  2.5 kg
9.8066 m/s2 1N

(c) 10 ton 1 lb m 1000 g 980.66 cm / s2 1 dyne = 9  109 dynes
5  10-4 ton 2.20462 lb m 1 g  cm / s
2



50  15  2 m3 35.3145 ft 3 85.3 lb m 32.174 ft 1 lb f 6
2.9 = 4.5  10 lb f
1 m3 1 ft 3 1 s2 32.174 lbm / ft  s2

1 kg 1 m3
F 1IF I 2 1
3
500 lbm
2.10  5  10 G J G J  25 m
2.20462 lbm 11.5 kg H 2 K H 10K
2.11 (a)
mdisplaced fluid = mcylinder   f V f = cVc   f hr 2 = c Hr 2
 h (30 cm − 14.1 cm)(1.00 g / cm3 ) c
 = f = = 3
H
c 0.53 g/ cm
H 30 cm
 c H (30 cm)(0.53 g / cm3 ) 3 f
(b)  f = = = 1.71 g/ cm h
h (30 cm - 20.7 cm)


2.12 R 2 H R 2 H r 2h R r R
Vs = 3 ; V f = 3 − 3 ; H = h  r = H h
R 2 H h Rh
2
F IR 2 F
h3 I h
 Vf = − G J = 3 HG H − H JK r
3 3 H HK 2 H
 = R F h I
2
R H 3 2



3 HG J=  3
 H− f
H K
f Vf sVs f 2 s s

R
H H3 1
  f = s = s = s
H 3 − h3
GFH hH JI K
3 3
H− h 1−
H2




2-2

,2.13 Say h(m) = depth of liquid
y
y= 1
dA
–1+h
y=y=1– h
 xx
1m x = 1– y 2
A(m 2 ) h

y= –1
2
dA
−1+h
1− y

dA = dy   ( )
dx = 2 1 − y2 dy  A m 2 = 2  1 − y2 dy
− 1− y2 −1


 Table of integrals or trigonometric substitution

( ) + sin−1 (h −1) +
h−1
A m 2 = y 1 − y2 + sin y = ( h −1)
−1
−1 2
4 m  A(m2 ) 0.879 g 106 cm2 1 kg 9.81 N
()
W N = cm3 1 m3 103 g kmg = 3.45  10 A
4

g g0

u Substituter for A
W(N) = 3.45  10 j (h − 1)
4
( )  yj
+ sin−1 h − 1 +

L 2Q


2.14 1 lb f = 1 slug  ft / s2 = 32.174 lbm  ft / s2  1 slug = 32.174 lbm
1
1 poundal = 1 lbm  ft / s2 = lb f
32.174
(a) (i) On the earth:
175 lbm 1 slug
M= = 5.44 slugs
32.174 lbm
175 lbm 32.174 ft 1 poundal = 5.63  103 poundals
W=
s 1 lbm  ft / s2
2

(ii) On the moon
175 lbm 1 slug = 5.44 slugs
M=
32.174 lbm
175 lbm 32.174 ft 1 poundal = 938 poundals
W=
6 s 1 lbm  ft / s2
2



355 poundals 1 lbm  ft / s2 1 slug 1m
(b) F = ma  a = F / m =
25.0 slugs 1 poundal 32.174 lbm 3.2808 ft
= 0.135 m / s 2




2-3

, 2.15 (a) F = ma  1 fern = (1 bung)(32.174 ft / s2 )
FG 1I J = 5.3623 bung  ft / s 2
H K6
1 fern

5.3623 bung  ft / s2
3 bung 32.174 ft 1 fern
(b) On the moon: W = = 3 fern
6 5.3623 bung  ft / s2
s2
On the earth: W = (3)(32.174) / 5.3623 = 18 fern

4.0 10−4 −5
2.16 (a)  (3)(9) = 27 (b)   110
40
(2.7)(8.632) = 23
(3.600 10−4 ) / 45 = 8.0 10−6
(c)  2 + 125 = 127 (d)  50  103 − 1  103  49  103  5  104
2.365 + 125.2 = 127.5 4.753  104 − 9  102 = 5  104
(7 10−1)(3105 )(6)(5104 ) 2 3
2.17 R   42 10  410 (Any digit in range 2-6 is acceptable)
(3)(5106 )
R = 3812.5  3810  3.81103
exact


2.18 (a)
A: R = 73.1 − 72.4 = 0.7o C
72.4 + 73.1 + 72.6 + 72.8 + 73.0
X= = 72.8o C
5
(72.4 − 72.8) 2 + (73.1 − 72.8) 2 + (72.6 − 72.8) 2 + (72.8 − 72.8) 2 + (73.0 − 72.8) 2
s=
5−1
= 0.3 C o


B: R = 103.1− 97.3 = 5.8o C
97.3 + 101.4 + 98.7 + 103.1+ 100.4
X= = 100.2o C
5
(97.3 − 100.2)2 + (101.4 − 100.2)2 + (98.7 − 100.2)2 + (103.1− 100.2)2 + (100.4 − 100.2)2
s=
5−1
= 2.3 C
o


(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.




2-4