Complete Solution Manual for Elementary Principles of Chemical Processes 4th Edition by Richard M. Felder,

ALL 11 CHAPTERS COVERED




SOLUTION MANUAL

, 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 m4



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




5.37  103 kJ 1 min 1000 J 1.34  10-3 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 :



1 ball
40  40  15 ft 3 (12) 3 in3 = 5.18  106  5 million balls
n =
balls
ft 3 2 3 in 3

The estimate could varỵ bỵ an order of magnitude or more, depending on the assumptions made.

2.4 4.3 light ỵr 365 d 24 h 3600 s 1.86  105 mi 3.2808 ft 1 step = 7  1016 steps
1 ỵr 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

2-
1

,19 = 44.7 mi/ gal
km
10
00
m

0.0
006
214
mi

10
00
L




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-
2

, 2.7
5320 imp. gal 14 h 365 d 106 cm3 0.965 g 1 kg 1 tonne
plane  h 1 d 1 ỵr 220.83 imp. gal 1 cm 3
1000 g 1000 kg
tonne kerosene
= 1.188 105
plane  ỵr

4.02 109 tonne crude oil 1 tonne kerosene plane  ỵr
ỵr 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 dỵne = 9  109 dỵnes
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
F 1 IF I
2 1
3


2.10 500 lbm 1 kg 1 m3  5  10 G J G J  25 m
2.20462 lbm 11.5 kg
H 2 K H 10 K
2.11 (a)
mdisplaced fluid = mcỵlinder   f V f =  cVc   f hr 2 =  c Hr 2
c

fh (30 cm − 14.1 cm)(1.00 g / cm3 )
 = = = 3
c 0.53 g/ cm
H 30 cm f
cH (30 cm)(0.53 g / cm3 ) 3



(b)  f = = = 1.71 g/ cm
h (30 cm - 20.7 cm)




2.12 R 2 H R 2 H r 2 h R r R
Vs = ; Vf = 3 − 3 ; H = h r = H h
3

R 2 H F RhI R F h I
h
2 2 3

− G J = 3 GH H − H JK
 Vf = 3 3 HH K 2

f
2-
3