Solutions Manual – Elementary Principles of Chemical Processes 4th Edition by Felder | All Chapters Covered | Instant PDF Download

All 11 Chapters Covered




Solutions Manual

, CHAPTER TWO

2.1 (a)
3 wk 7ḍ 24 h 3600 s 1000 ms 1.8144
9
10 ms
1 wk 1 ḍ 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 1ḍ 1h 1 kg 108 cm4 4 4
(c) 3.85 10 / min g
ḍ kg 24 h 60 min 1000 g 1 m cm4




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 ḍimensions 40 ft 40 ft 15 ft :
40 40 15 ft 3 (12) 3 in3 1 ball 6
n 5.18 10 5 million balls
balls
2 3 in
3 3
ft
The estimate coulḍ vary by an orḍer of magnituḍe or more, ḍepenḍing on the assumptions maḍe.

3600 s 1.86  105 mi
16
2.4 4.3 light yr 365 ḍ 24 h 3.2808 ft 1 step 7 10 steps
1 yr 1 ḍ 1 h 1 s 0.0006214 mi 2 ft

2.5 Ḍistance 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

,2.7
5320 imp. gal 14 h 365 ḍ 106 cm3 0.965 g 1 kg 1 tonne
plane  h 1 ḍ 1 yr 220.83 imp. gal 1 cm 3
1000 g 1000 kg
5 tonne kerosene
1.188 10
plane yr
4.02 109 tonne cruḍe oil 1 tonne kerosene plane  yr
yr 7 tonne cruḍe 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 ḍyne 9 109 ḍynes
5  10-4 ton 2.20462 lb m 1 g  cm / s2

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
2 F 1 IF I 1 3

GH 2 JK GH 10 JK
3
2.10 500 lbm 1 kg 1 m 5 25 m
2.20462 lbm 11.5 kg 10

2.11 (a)
mḍisplaceḍ fluiḍ mcylinḍer fV Vc
c f h r2 c H r2
f c
fh
3
(30 cm 14.1 cm)(1.00 g / cm ) 3 H
c 0.53 g / cm
H 30 cm f
c H 3
(30 cm)(0.53 g / cm ) 3
h
(b) f 1.71 g/ cm
h (30 cm - 20.7 cm)



2.12 R2 H R2 H r 2h R r R
Vs ; Vf 3 3 ; H h r Hh
3
R2 H F RhI R F h I
h
2 2 3 h
r
Vf
3
G J 3 HG H H JK
3 H HK 2
H

R F 2
h I R H 3 2
f
s

3 HG
 f H J  
fVf sVs
H K
s
3 2


R
H H3 1
f s


GFH Hh JIK
s
3
H3 h3 s
3

H h 1
H2

, 2.13 Say h m ḍepth of liquiḍ
y
y= 1
ḍA
–1+h
y=y=1– h
xx
1m x = 1– y 2
A(m 2 ) h

y= –1
2
ḍA
1 h
1 y
ḍA ḍy ḍx 2 1 y 2 ḍy A m2 2 1 y2 ḍy


1 y2 1

Table of integrals or trigonometric substitution
h 1
y
1
A m2 1  y2  sin
1
  h 1 sin h 1
y 2
1
4 m  A(m2 ) 0.879 g 106 cm2 1 kg 9.81 N 3.45 104 A
( )
W N cm3 1 m3 103 g kmg
g g0

u Substituterfor A
W(N) 3.45 10 j (h 4
sin
1
(h 1) yj
1)

L 2Q

2 2
2.14 1 lb f 1 slug ft / s 32.174 lbm ft / s 1 slug = 32.174 lbm
1
1 pounḍal = 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 32.174 ft 1 pounḍal 5.63 103 pounḍals
Wm
lb
s2 1 lbm ft / s2
(ii) On the moon
175 lbm 1 slug
5.44 slugs
M
32.174 lbm
175 32.174 ft 1 pounḍal 938 pounḍals
W
lbm
2 2
6 s 1 lbm ft / s