Solutions Manual – Elementary Principles of Chemical Processes 4th Edition by Felder | PDF

ALL 11 CHAPTERS COVERED




SOLUTION MANUAL

, CHAPTER TWO

3 wk 7d 24 h 3600 s 1000 ms
2.1 (a) = 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
554 m 4 1d 1h 1 kg 108 cm 4
(c) = 3.85 × 10 4 cm 4 / 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
(b) = 57.5 lb m / ft 3
m3 1 kg 35.3145 ft 3
5.37 × 10 3 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 :
40 × 40 × 15 ft 3 (12) 3 in 3 1 ball
n balls = = 518
. × 10 6 ≈ 5 million balls
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 3600 s 1.86 × 10 5 mi 3.2808 ft 1 step = 7 × 1016 steps
1 yr 1d 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 + 0.04464 x
gal 28 mi

$1.25 1 gal x (mi)
Total Cost European = $21,700 + = 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 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 lb m 32.1714 ft / s 2 1 lb f
2.8 (a) = 25.0 lb f
32.1714 lb m ⋅ ft / s 2
25 N 1 1 kg ⋅ m/s 2
(b) = 2.5493 kg ⇒ 2.5 kg
9.8066 m/s 2 1N
10 ton 1 lb m 1000 g 980.66 cm / s 2 1 dyne
(c) = 9 × 10 9 dynes
5 × 10 -4
ton 2.20462 lb m 1 g ⋅ cm / s 2

50 × 15 × 2 m 3 35.3145 ft 3 85.3 lb m 32.174 ft 1 lb f
2.9 = 4.5 × 10 6 lb f
1 m3 1 ft 3 1 s 2
32.174 lb m / ft ⋅ s 2




500 lb m 1 kg 1 m3
≈ 5 × 10 2
FG 1 IJ FG 1 IJ ≈ 25 m
H 2 K H 10K
3
2.10
2.20462 lb m 11.5 kg

2.11 (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
H 30 cm
ρf
ρ H (30 cm)(0.53 g / cm 3 )
(b) ρ f = c = = 171. g / cm 3 h
h (30 cm - 20.7 cm)


2.12 πR 2 H πR 2 H πr 2 h R r R
Vs = ; Vf = − ; = ⇒r = h
3 3 3 H h H

⇒ Vf =
πR H2
−
FG IJ = πR FG H − h IJ
πh Rh
2 2 3
h

3 H HK
r
3 3 H H K 2 H

πR F h I πR H
2 3 2
ρ f V f = ρ sVs ⇒ρ
3 H G H− J
H K
f =ρ
3 2 s ρs
ρf


R
H H3 1
⇒ ρ f = ρs = ρs = ρs
H−
h3 H 3 − h3
1−
FG h IJ 3


H2 H HK

2-2

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

y= –1
dA
1− y 2
−1+ h
dA = dy ⋅ ∫ dx = 2 1 − y dy ⇒ A m
2
( )=2 ∫
2
1 − y 2 dy
− 1− y 2 −1


⇓ Table of integrals or trigonometric substitution
h −1
π
( )
A m 2 = y 1 − y 2 + sin −1 y ⎤⎥ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) +
⎦ −1
2

2
4 m × A( m 2 ) 0.879 g 10 6 cm 2 1 kg 9.81 N
b g
W N = cm 3
1m 3
10 g 3
kg
N
= 3.45 × 10 4 A
g g0

E Substitute for A
L
W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g
4 2
b g π2 OPQ
+ sin −1 h − 1 +
N
2.14 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
s 1 lb m ⋅ ft / s 2
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
6 s 1 lb m ⋅ ft / s 2
2




355 poundals 1 lb m ⋅ ft / s 2 1 slug 1m
(b) F = ma ⇒ a = F / m =
25.0 slugs 1 poundal 32.174 lb m 3.2808 ft
= 0.135 m / s 2




2-3