An Instructor’s Solutions Manual to Accompany Fundamentals of Chemical Engineering Thermodynamics

751 Pages·2015·8.407 MB·English

by  Kevin D. Dahm & Donald P. Visco

An Instructor’s Solutions Manual to Accompany

FUNDAMENTALS OF
CHEMICAL ENGINEERING
THERMODYNAMICS
KEVIN D. DAHM AND DONALD P. VISCO, JR.
© 2015 Cengage Learning ISBN-13: 978-1-133-59845-9
ISBN-10: 1-133-59845-5
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1 2 3 4 5 6 7 16 15 14 13
I ' S
NSTRUCTOR S OLUTIONS
M
ANUAL
A
TO CCOMPANY

FUNDAMENTALS OF
CHEMICAL
ENGINEERING
THERMODYNAMICS

FIRST EDITION

KEVIN D. DAHM AND DONALD P. VISCO, JR.
CONTENTS
Chapter Page
1. Introduction ……………………………………………………………... 1
2. The Physical Properties of Pure Compounds …………………………… 19
3. Material and Energy Balances ………………………………………….. 55
4. Entropy ……………………………………………………………… …. 119
5. Thermodynamic Processes and Cycles …………………………………. 185
6. Thermodynamic Models of Real, Pure Compounds ……………………. 253
7. Equations of State ………………………………………………………. 321
8. Modeling Phase Equilibrium for Pure Compounds ………………………395
9. An Introduction to Mixtures ……………………………………………. 437
10. Vapor-Liquid Equilibrium ……………………………………………… 475
11. Theories and Models for ………………………………………………… 509
Vapor–Liquid Equilibrium of Mixtures:
Modified Raoult’s Law Approaches
12. Theories and Models for ………………………………………………… 561
Vapor–Liquid Equilibrium of Mixtures:
Using Equations of State
13. Liquid-Liquid, Vapor-Liquid-Liquid, …………………………………… 623
and Solid-Liquid Equilibrium
14. Fundamentals of Chemical Reaction Equilibrium ………………………. 651
15. Synthesis of Thermodynamic Principles ………………………………… 689

Chapter 1: Introduction
1-16) The value g = 9.81 m/s2 is specific to the force of gravity on the surface of the
earth. The universal formula for the force of gravitational attraction is:

(cid:5)(cid:6)(cid:5)(cid:7)
(cid:1) = G (cid:7)
Where m and m are the masses of the two(cid:8) objects, r is the distance between the
1 2
centers of the two objects, and G is the universal gravitation constant,
G = 6.674 × 10-11 N(m/kg)2.
A) Research the diameters and masses of the Earth and Jupiter.
B) Demonstrate that F = m(9.81 m/s2) is a valid relationship on the surface of the
earth.
C) Determine the force of gravity acting on a 1000 kg satellite that is 2000 miles
above the surface of the Earth.
D) One of the authors of this book has a mass of 200 lb . If he was on the surface
m
of Jupiter, what gravitational force in lb would be acting on him?
f

Solution:
A) Measurements obtained from different sources will vary slightly.
D ~ 12,742 km D ~ 142,000 km
Earth Jupiter
Mass = 5.97 × 1024kg Mass = 1.90 × 1027kg
earth jupiter

B) Mass = 5.97 × 1024kg Radius = 6.371 × 106 meters
earth Earth
(cid:1) /0 1
(cid:9)(cid:10)(cid:9)(cid:11) (cid:28)(cid:6)(cid:6)(cid:29)(cid:30)(cid:11) "#.$%×(cid:6)&(cid:11)'(cid:31) ( (cid:20)234(cid:11)!
(cid:1) = G (cid:12)(cid:11) (cid:1) = m(cid:14)(cid:15)(cid:16)(cid:17)(cid:18)(cid:19)(cid:20)6.674 × 10 (cid:31) (cid:11) !(*.+%×(cid:6)&, (cid:30))(cid:11).((cid:6) (cid:29))5
(cid:1)
7
6 = 7(8.9:;<=>)
C) 2000 miles = 3218.68 km = 3218680 m

(cid:11) (cid:11)'
(cid:28)(cid:6)(cid:6)(cid:29)(cid:30) (#.$%×(cid:6)& AB)
(cid:1) = 1000kg(cid:20)6.674 × 10 (cid:31) (cid:11) !(*+%(cid:6)&&&(cid:9)C+(cid:7)(cid:6)D*D&(cid:9) )(cid:11) = EFE9 G

1
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Chapter 1: Introduction
D) Radius = 66854000 m Mass = 1.898 × 1027 kg 200lb = 90.7 kg
Jupiter Jupiter m

kg m
(cid:7) (cid:7)% K (cid:7)O
(cid:28)(cid:6)(cid:6)Nm (1.90×10 kg) sec
(cid:1) = 90.7 kg.6.674 × 10 (cid:7) 5 % (cid:7)J P = >>9: G
kg (7.10×10 m) (1 N)

1-17) A gas at T=300 K and P=1 bar is contained in a rigid, rectangular vessel that is 2
meters long, 1 meter wide and 1 meter deep. How much force does the gas exert on
the walls of the container?

Solution:
1 Bar = 100,000 Pa

AreaTUVWX(cid:18)(cid:17) = (2×W×H)+(2×W×L)+(2×H×L)

Force = Pressure × Area
N
(cid:20) (cid:7)!
m
Force = (100000Pa)(2×1m×1m+2×1m×2m+2×1m×2m)b c
Pa
e
Force = :×:d G

1-18) A car weighs 3000 lb , and is travelling 60 mph when it has to make an emergency
m
stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
B) Assuming the rate of deceleration is constant, how much distance is covered
before the car comes to a stop?

Solution:
A)
(cid:6)lV #(cid:7)D&W(cid:19) W(cid:19)
Force = mass×acceleration 60mph(cid:20)+*&&m(cid:17)(cid:18)!(cid:20)(cid:6)(cid:30)no(cid:17)! = 88m(cid:17)(cid:18)

velocityWnsXo −velocitynsn(cid:19)nXo
Acceleration =
time
ft
88 −0
sec
a =
5 sec

ft
a = 17.6 (cid:7)
sec

2
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Chapter 1: Introduction

ft
Force = 17.6 (cid:7) ×3000lb(cid:30)
sec

z{7|}
F orce = y>ydd >
;<=
B)
(cid:11)
X(cid:18)(cid:18)(cid:17)o(cid:17)VX(cid:19)n(cid:14)s∗(cid:19)n(cid:30)(cid:17)
PositionWnsXo = (cid:7) +velocitynsn(cid:19)Xo ×time+positionns(cid:19)nXo
(cid:127)€ (cid:11)
(cid:20)(cid:28)(cid:6)%.*234(cid:11)!(#m(cid:17)(cid:18)) W(cid:19)
PositionWnsXo = (cid:7) +(cid:20)88m(cid:17)(cid:18) !(5sec)+0

PositionWnsXo = >>d |}

1-19) Solar panels are installed on a rectangular flat roof. The roof is 15 feet by 30 feet,
and the mass of the panels and framing is 900 lb .
m
A) Assuming the weight of the panels is evenly distributed over the roof, how much
pressure does the solar panel array place on the roof?
B) The density of fallen snow varies; here assume its ~30% of the density of liquid
water. Estimate the total pressure on the roof if 4 inches of snow fall on top of the
solar panels.

Solution:
A)
(cid:129)(cid:14)V(cid:18)(cid:17)
Pressure = ‚V(cid:17)X Area = Length×Width Force = Mass×Acceleration
ft
(900lb(cid:30))(cid:20)32.2 (cid:7)! 1lbW
sec
Pressure = b c
(15ft)(30ft) lb(cid:30)ft
32.2 (cid:7)
sec

z{|
P ressure = > >
|}
B)
o(cid:15)1 o(cid:15)1
Densityms(cid:14)† = (30%)63.3 W(cid:19)ˆ = 19.0 W(cid:19)ˆ

Forcems(cid:14)† = Volumems(cid:14)† ×Densityms(cid:14)† ×gravity

Forcems(cid:14)† +ForceŠXs(cid:17)om
Pressure =
Area
4inches = 0.333ft

3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Introduction
Pressure
ft lb(cid:30) ft
.(900lb(cid:30))(cid:20)32.2 (cid:7)!5+"(15ft)(30ft)(0.333ft)((cid:20)19.0 +!(cid:20)32.2 (cid:7)!
sec ft sec 1lbW
= b c
(15ft)(30ft) lb(cid:30)ft
32.2 (cid:7)
sec

z{|
P ressure = 9.FF >
|}
1-20) A box has a mass of 20 kg, and a building has a height of 15 meters.
A) Find the force of gravity acting on the box.
B) Find the work required to lift the box from the ground to the roof of the
building.
C) Find the potential energy of the box when it is on the roof of the building.
D) If the box is dropped off the roof of the building, find the kinetic energy and
velocity of the box when it hits the ground.

Solution:
A)
(cid:30)
Force = (20kg)(cid:20)9.81m(cid:17)(cid:18)(cid:11)! = :8e.> G
B)

Work = Force×Distance

Work = (196.2N)(15m) = >8EFG7 = >8EF ‹
C)
Potential Energy = Mass×Height×Gravity
m
Potential Energy = (20kg)(15m)(cid:20)9.81 (cid:7)! = >8EFG7 = >8EF ‹
sec
D) We know that Energy is conserved, so if the box is dropped from a height of 15
meters, its Kinetic Energy at Height = 0 meters will be the same as its potential energy at
Height = 15 meters.

Kinetic Energy = >8EFG7
To find the object’s velocity as it hits the ground:

(cid:6) (cid:7)
K inetic Energy = (cid:20)(cid:7)!Mass×Velocity

(cid:6) (cid:7)
2943Nm = (cid:20)(cid:7)!(20kg)×V
kg m (cid:7)
(cid:7)
Nm sec m (cid:7)
K294.3 Ob c = 294.3. (cid:7)5 = V
kg 1N sec

4
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Chapter 1: Introduction

7
V = :Ž.:e
;<=
1-21) 100 kg of steam is enclosed in a piston-cylinder device, initially at 300°C and 5
bar. It expands and cools to 200°C and 1 bar.
A) What is the change in internal energy of the steam in this process?
B) If the external pressure is constant at 1 bar, how much work was done by the
steam on the surroundings?
C) Research and briefly describe at least two examples of machines, either historical
or currently in use, that harness the energy in steam and convert it into work. Any
form of work is acceptable; you needn’t confine your research to expansion work
(which was examined in parts A and B).

Initial State Final State

T = 300°C T = 200°C
1 2

P = 5 bar P = 1 bar
1 2

Solution:
Define the gas as the system.
A) From steam tables:

kJ
(cid:143)(cid:144)(cid:6) = 2803.2
kg

kJ
(cid:143)(cid:144)(cid:7) = 2658.2
kg
So
(cid:31)–
Δ(cid:143) = “"(cid:143)(cid:144)(cid:7) −(cid:143)(cid:144)(cid:6)( = (100 ”•)(cid:20)2658.2−2803.2 (cid:31) ! = −:E,ydd ˜‹

5
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