Solution Manual for Introduction to Chemical Engineering Thermodynamics 9th Edition – Smith, Van Ness, Abbott

SOLUTION MANUAL




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,1) Introduction
2) The First Law and Other Basic Concepts
3) Volumetric Properties of Pure Fluids
4) Heat Effects
5) The Second Law of Thermodynamics
6) Thermodynamic Properties of Fluids
7) Applications of Thermodynamics to Flow Processes
8) Production of Power from Heat
9) Refrigeration and Liquefaction
10) The Framework of Solution Thermodynamics
11) Mixing Processes
12) Phase Equilibrium: Introduction
13) Thermodynamic Formulations for Vapor/Liquid Equilibrium
14) Chemical-Reaction Equilibria
15) Topics in Phase Equilibria
16) Thermodynamic Analysis of Processes




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, Chapter 1 - Section A - Mathcad Solutions
16.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve
this equation by setting t(F) = t(C).

Guess solution: t
0 Given t = 1.8t 32
Find(t)  40 Ans.

F
16.5 By definition: P = F = mass g Note: Pressures are in
A gauge pressure.

2
P 3000bar D 4mm A  A  12.566 mm
2

D4

m F mass 384.4 kg
F P A g 9.807 mass Ans.
2 g
s


F
16.6 By definition: P = F = mass g
A

2
D 2
P 3000atm D 0.17in A
4

ft F mass 1000.7 lbm
F P A g 32.174 mass Ans.
2 g
sec



16.7 Pabs = g h Patm

gm m
13.535 g 9.832 h 56.38cm
3 2
cm
s

Patm 101.78kPa Pabs g h Patm Pabs 176.808 kPa Ans.


3

, gm m
1.10 Assume the following: 13.5 g 9.8
3 2
cm s
P
P h Ans.
400ba h 302.3 m
r g

1.11 The force on a spring is described by: F = Ks x where Ks is the spring constant.
First calculate K based on the earth measurement then gMars based on spring
measurement on Mars.
On Earth:
m
F = mass g = mass g 9.81 x 1.08cm
K x 0.40kg 2
s
F N
F F 3.924 N Ks 363.333
Ks
mass g m
x
On Mars:

x 0.40cm FMars 3
FMars 4 10 mK
FMars K x
mK Ans.
gMars gMars 0.01
mass kg


d M d M P
1.12 Given: P= and:
P
Substituting: P=
g dz = R g dz R T
T

P Den zDenve
1 r M g
Separating variables and integrating: ver
dz
dP =
P R T
Psea 0

PDenver M
zDenver
After integrating: ln = g
Psea R T


Taking the exponential of both sides M g
z
2

, Psea R Denver
and rearranging: PDenver T
= e
gm m
Psea 1atm M 29 g 9.8 2
mol
s




3

, 3
cm
R T (10 273.15)K zDenver 
atm
82.06
mol K 1mi
M
g zDenver 0.194
R
T

M z
g
PDenver Psea e R T PDenver 0.823 Ans.
atm
Denver



PDenver 0.834 bar Ans.


1.13 The same proportionality applies as in Pb. 1.11.
ft ft
gearth gmoon lmoon 18.76
2
32.186 s 5.32
2
geart s
h
learth learth 113.498
gmoon
lmoon


M Ans.
learth lbm M 113.498 lbm

wmoon wmoon 18.767 lbf Ans.
M gmoon


5.00dollar hr 0.1dollar hr
1.14 costbulb s 1000hr 1 da costelec s 1 da 70W
0 y kW hr 0 y
dollars dollars
costbulb 18.262 costelec 25.567
yr yr
dollars
costtotal costbulb costtotal yr Ans.
costelec 43.829

1.15 ft
D 1.25ft mass 250lbm g 32.169
2
4 s

, 2
Patm 30.12in_Hg A A 1.227 ft
2
D
3
(a) F Patm A F 2.8642  lbf
4 Ans.
10
mass g

F Pabs 16.208 psia
Ans.
(b) Pabs
A
3
(c) l Work F l Work 4.8691 Ans.
10
1.7ft ft Plbf
E 424.9
PE l Ans.
ft lbf
mass g

1.16 D mass 150kg m
0.47m g 9.813
2
s2
Patm 101.57kPa A
2 A 0.173 m
D
4 Ans.
(a) F Patm A F 1.909  4 N
mass g 10

F Ans.
(b) Pabs Pabs 110.054
kPa A

(c) l Work Work 15.848 Ans.
0.83m F l kJ
EP 1.222
EP l kJ Ans.
mass g
m
1.18 mass 1250kg u 40
s
EK 1 EK 1000 Ans.
2 kJ
mass u
2
Work Work 1000 Ans.
EK kJ


mass h
1.19 Wdot = g 0.91 0.92
time
5

, m
Wdot g 9.8 h 50m
200W 2
s




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, Wdot kg
mdot g h 0.91 0 mdot 0.488 Ans.
.92 s
1.22 25.00
a) cost_coal  ton
1
MJ cost_coal 0.95 GJ
29
kg
2.00 1
cost_gasoline 14.28 GJ
gal
cost_gasoline 
GJ
37
3
m
0.1000 1
cost_electricity  cost_electricity 27.778 GJ
kWh
r
b) The electr ical energy can directly be conver ted to other forms of energy
whereas the coal and gasoline would typically need to be conver ted to
heat and then into some other form of energy before being useful.

The obvious advantage of coal is that it is cheap if it is used as a
heat sour ce. Otherwise it is messy to handle and bulky for tranport
and storage.

Gasoline is an important transportation fuel. It is more convenient to
transport and store than coal. It can be used to generate electr icity by
burning it but the efficiency is limited. However , fuel cells are currently
being developed which will allow for the conver sion of gasoline to electr icity
by chemical means, a more efficient pr ocess.

Electr icity has the most uses though it is expensive. It is easy to transport
but expensive to store. As a transportation fuel it is clean but batteries to
store it on-board have limited capacity and are heavy.




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