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Exam (elaborations)

Solutions Manual – Elementary Principles of Chemical Processes 3rd Edition by Felder

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Enhance your mastery of chemical engineering basics with the Solutions Manual for Elementary Principles of Chemical Processes, 3rd Edition by Felder, Rousseau, and Bullard. This manual offers fully solved, step-by-step answers to every end-of-chapter problem, making it an essential companion for chemical engineering students tackling mass balances, energy balances, and process design. Ideal for first-year undergraduates and students preparing for exams, this guide helps demystify complex topics like recycle/bypass streams, phase equilibria, psychrometrics, and chemical reaction engineering. What’s Inside: • Complete solutions for all end-of-chapter problems • Step-by-step mass and energy balance calculations • Diagrams, flowcharts, and conversion help • Clear formatting and process-based explanations • High-quality searchable PDF – ready for download Felder Solutions Manual, Chemical Engineering PDF, Mass Balance Solved, Energy Balance Problems, Felder 3rd Edition, Recycle Stream Help, Unit Conversions Guide, Process Design Basics, Thermodynamics Calculations, Process Flow Diagrams, Reaction Stoichiometry Solved, Psychrometrics Examples, First-Year ChemE, Chemical Process Problems, Chemical Stoichiometry Help, Process Control Calculations, Engineering Problem Solving, Felder Bullard Rousseau, Bypass Stream Solutions, Chemical Engineering Fundamentals

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Institution
Elementary Principles Of Chemical Processes
Course
Elementary Principles of Chemical Processes











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Institution
Elementary Principles of Chemical Processes
Course
Elementary Principles of Chemical Processes

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Uploaded on
November 20, 2025
Number of pages
586
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

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Elementary Principles of Chemical Processes
ST

– 3rd Edition
UV

SOLUTIONS
IA

MANUAL
_A
PP

Richard M. Felder
RO
Ronald W. Rousseau
VE
Complete Solutions Manual for Instructors and Students
D?
© Richard M. Felder & Ronald W. Rousseau


All rights reserved. Reproduction or distribution without permission is prohibited.
??

©MEDGEEK

, CONTENTS
ST
Notes to the Instructor iv

Section/Problem Concordance vi

Sample Assignment Schedule I ix
UV
Sample Assignment Schedule II x

Sample Responses to a Creativity Exercise xi
IA
Transparency Masters xvi

Compressibility charts xvii
Cox vapor pressure chart xxi
_A
Psychrometric chart – SI units xxii
Psychrometric chart – American engineering units xxiii
Enthalpy-concentration chart: H2SO4-H2O xxiv
Enthalpy-concentration chart: NH3-H2O xxv
PP
Problem Solutions

Chapter 2 2-1
Chapter 3 3-1
Chapter 4 4-1
RO
Chapter 5 5-1
Chapter 6 6-1
Chapter 7 7-1
Chapter 8 8-1
Chapter 9 9-1
Chapter 10 10-1
VE
Chapter 11 11-1
Chapter 12 (Case Study 1) 12-1
Chapter 13 (Case Study 2) 13-1
Chapter 14 (Case Study 3) 14-1
D?
??

iii

, CHAPTER TWO

3 wk 7d 24 h 3600 s 1000 ms
ST
2.1 (a) .
18144 u 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 Ÿ 26.0 mi / h
3.2808 ft 1 h
UV
554 m 4 1d 1h 1 kg 10 8 cm 4
(c) 3.85 u 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
IA
921 kg 2.20462 lb m 1 m3
(b) 57.5 lb m / ft 3
m3 1 kg 35.3145 ft 3
5.37 u 10 3 kJ 1 min 1000 J 1.34 u 10 -3 hp
(c) 119.93 hp Ÿ 120 hp
min 60 s 1 kJ 1 J/s
_A
2.3 Assume that a golf ball occupies the space equivalent to a 2 in u 2 in u 2 in cube. For a
classroom with dimensions 40 ft u 40 ft u 40 ft :
40 u 40 u 40 ft 3 (12) 3 in 3 1 ball
n balls 3 3 3
6.48 u 10 6 | 7 million balls
ft 2 in
PP
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 u 105 mi 3.2808 ft 1 step 7 u 1016 steps
1 yr 1 d 1 h 1 s 0.0006214 mi 2 ft
RO
2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1 m 1 report
4 u 1011 reports
0.0006214 mi 0.001 m
VE
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)
D?
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 Ÿ x 4.3 u 10 5 miles




2-1

, 2.7
5320 imp. gal 14 h 365 d 10 6 cm 3 0.965 g 1 kg 1 ton
plane ˜ h 1 d 1 yr 220.83 imp. gal 1 cm 3
1000 g 1000 kg
ST
ton kerosene
1188
. u 10 5
plane ˜ yr
4.02 u 10 9 ton crude oil 1 ton kerosene plane ˜ yr
yr 7 ton crude oil 1188
. u 10 ton kerosene 5
UV
4834 planes Ÿ 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
IA
25 N 1 1 kg ˜ m / s 2
(b) 2.55 kg Ÿ 2.6 kg
9.8066 m / s 2 1N
10 ton 1 lb m 1000 g 980.66 cm / s 2 1 dyne
(c) 9 u 10 9 dynes
5 u 10 -4 ton 1 g ˜ cm / s 2
_A
2.20462 lb m

50 u 15 u 2 m 3 35.3145 ft 3 85.3 lb m 32.174 ft 1 lb f
2.9 4.5 u 10 6 lb f
1 m3 1 ft 3 1 s 2
32.174 lb m / ft ˜ s 2

FG IJ FG 1 IJ | 25 m
PP
500 lb m 1 kg 1 m3 1
| 5 u 10 2
H K H 10K
3
2.10
2.20462 lb m 11.5 kg 2

2.11 (a)
mdisplaced fluid mcylinder Ÿ U f V f U cVc Ÿ U f hSr 2 U c HSr 2
RO
Uc
Ufh
(30 cm  14.1 cm)(100
. g / cm 3 )
Uc 0.53 g / cm 3 H
H 30 cm
Uf
U c H (30 cm)(0.53 g / cm 3 )
(b) U f . g / cm 3
171 h
h (30 cm - 20.7 cm)
VE
2.12 SR 2 H SR 2 H Sr 2 h R r R
Vs ; Vf  ; Ÿr h
3 3 3 H h H

Ÿ Vf
SR 2 H Sh Rh

FG IJ SR FG H  h IJ
2 2 3
h

H K 3H HK r
D?
2 H
3 3 H
SR F
2
h I SR H
3 2
U fVf U sVs Ÿ U f
3 H
G H J
H K
U
3 2 s Us
Uf

R
H H3 1
ŸUf Us Us Us
??
H
h3 H 3  h3 F hI
1 G J
3


H2 H HK

2-2

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