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Chemical and Engineering Thermodynamics 3rd Ed. by Sandler (www.solutionmanual.net)

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Prévia do material em texto

v
Preface
This manual contains more or less complete solutions for every problem in the
book. Should you find errors in any of the solutions, please bring them to my attention.
Over the years, I have tried to enrich my lectures by including historical
information on the significant developments in thermodynamics, and biographical
sketches of the people involved. The multivolume Dictionary of Scientific Biography,
edited by Charles C. Gillispie and published by C. Scribners, New York, has been
especially useful for obtaining biographical and, to some extent, historical information.
[For example, the entry on Anders Celsius points out that he chose the zero of his
temperature scale to be the boiling point of water, and 100 to be the freezing point.
Also, the intense rivalry between the English and German scientific communities for
credit for developing thermodynamics is discussed in the biographies of J.R. Mayer, J. P.
Joule, R. Clausius (who introduced the word entropy) and others.] Other sources of
biographical information include various encyclopedias, Asimov’s Biographical
Encyclopedia of Science and Technology by I. Asimov, published by Doubleday & Co.,
(N.Y., 1972), and, to a lesser extent, Nobel Prize Winners in Physics 1901-1951, by
N.H. deV. Heathcote, published by H. Schuman, N.Y.
Historical information is usually best gotten from reading the original literature.
Many of the important papers have been reproduced, with some commentary, in a series
of books entitled “Benchmark Papers on Energy” distributed by Halsted Press, a division
of John Wiley and Sons, N.Y. Of particular interest are:
Volume 1, Energy: Historical Development of the Concept, by R. Bruce Lindsay.
Volume 2, Applications of Energy, 19th Century, by R. Bruce Lindsay.
Volume 5, The Second Law of Thermodynamics, by J. Kestin and
Volume 6, Irreversible Processes, also by J. Kestin.
The first volume was published in 1975, the remainder in 1976.
vi
Other useful sources of historical information are “The Early Development of the
Concepts of Temperature and Heat: The Rise and Decline of the Caloric Theory” by D.
Roller in Volume 1 of Harvard Case Histories in Experimental Science edited by J.B.
Conant and published by Harvard University Press in 1957; articles in Physics Today,
such as “A Sketch for a History of Early Thermodynamics” by E. Mendoza (February,
1961, p.32), “Carnot’s Contribution to Thermodynamics” by M.J. Klein (August, 1974,
p. 23); articles in Scientific American; and various books on the history of science. Of
special interest is the book The Second Law by P.W. Atkins published by Scientific
American Books, W.H. Freeman and Company (New York, 1984) which contains a very
extensive discussion of the entropy, the second law of thermodynamics, chaos and
symmetry.
I also use several simple classroom demonstrations in my thermodynamics courses.
For example, we have used a simple constant-volume ideal gas thermometer, and an
instrumented vapor compression refrigeration cycle (heat pump or air conditioner) that
can brought into the classroom. To demonstrate the pressure dependence of the melting
point of ice, I do a simple regelation experiment using a cylinder of ice (produced by
freezing water in a test tube), and a 0.005 inch diameter wire, both ends of which are
tied to the same 500 gram weight. (The wire, when placed across the supported cylinder
of ice, will cut through it in about 5 minutes, though by refreezing or regelation, the ice
cylinder remains intact.—This experiment also provides an opportunity to discuss the
movement of glaciers.) Scientific toys, such as “Love Meters” and drinking “Happy
Birds”, available at novelty shops, have been used to illustrate how one can make
practical use of the temperature dependence of the vapor pressure. I also use some
professionally prepared teaching aids, such as the three-dimensional phase diagrams for
carbon dioxide and water, that are available from laboratory equipment distributors.
Despite these diversions, the courses I teach are quite problem oriented. My
objective has been to provide a clear exposition of the principles of thermodynamics, and
then to reinforce these fundamentals by requiring the student to consider a great
diversity of the applications. My approach to teaching thermodynamics is, perhaps,
similar to the view of John Tyndall expressed in the quotation
“It is thus that I should like to teach you all things; showing you the way to
profitable exertion, but leaving the exertion to you—more anxious to bring out
your manliness in the presence of difficulty than to make your way smooth by
toning the difficulties down.”
Which appeared in The Forms of Water, published by D. Appleton (New York, 1872).
Solutions to Chemical and Engineering Thermodynamics, 3e vii
Finally, I usually conclude a course in thermodynamics with the following quotation 
by Albert Einstein:
“A theory is more impressive the greater the simplicity of its premises is, the
more different kinds of things it relates, and the more extended its area of
applicability. Therefore, the deep impression classical thermodynamics made
upon me. It is the only physical theory of universal content which, within the
framework of the applicability of its basic concepts, I am convinced will never
by overthrown.”
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0 5
 
 3.11 (a) This is a Joule-Thomson expansion 
⇒ =( ) = = °( ) ≈ = °( )
=
$ ? $ . $
.
H T H T H T70 bar, 10133 400 1 400
32782
 bar, C bar, C
 kJ kg
and T = °447 C , $ .S = 6619 kJ kg K
(b) If turbine is adiabatic and reversible &Sgen = 0c h , then $ $ .S Sout in kJ kg K= = 6619 and P = 1013.
bar. This suggests that a two-phase mixture is leaving the turbine
Let fraction vapor
 kJ kg K
kJ kg K
V
L
x
S
S
=
=
=
$
.
$
.
7 3594
13026 
Then x x7 3594 1 13026 6619. . .( ) + −( )( ) = kJ kg K or x = 08778. . Therefore the enthalpy of
fluid leaving turbine is
$
. . . . .
$ $
H
H H
= × + −( ) × =
( ) ( )
08788 26755 1 08778 417 46 2399 6
V L
 sat’d, 1 bar sat’d, 1 bar
kJ
kg
Energy balance
0 = + +& $ & $M H M Hin in out out &Q
0 + −&W Ps
dV
dt
0
but & &M Min out= −
⇒ − = − =
&
& . . .
W
M
s
in
kJ
kg
3278 2 2399 6 8786
(c) Saturated vapor at 1 bar
$
. ; $ .
& . . .
% .
.
.
&
& . . .
S H
W
M
S
M
s
= =
− = − =
( ) = × =
= − =
7 3594 26755
32782 26755 602 7
602 7 100
8786
686%
7 3594 6619 0740 
 kJ kg K kJ kg
 kJ kg
Efficiency 
kJ Kh
in Actual
gen
in
(d) 0
0
0
1 2 2 1
1 1 2
1 1 2
= + ⇒ = −
= − + + −
= − + +
& & & &
& $ $ & &
& $ $
&
&
M M M M
M H H W Q P dV
dt
M S S Q
T
S
sc h
c h gen
Simplifications to balance equations
&Sgen = 0 (for maximum work); P
dV
dt
= 0 (constant volume)
& &Q
T
Q
T
=
0
 where T0 25= °C (all heat transfer at ambient temperature)
W
Q
Water
1 bar
25° C
Steam
70 bar
447° C
$
.H sat'd liq, 25 C kJ
kg
°( ) = 10489 ; $ .S T sat'd liq, 25 C kJ
kg K
= °( ) = 03674
&
&
$ $Q
M
T S S= −0 2 1c h ; − = − + − = − − −
&
&
$ $ $ $ $ $ $ $
max
W
M
H H T S S H T S H T Ss 1 2 0 2 1 1 0 1 2 0 2c h c h c h
−
= − × − − ×
= + =
&
& . . . . . .
. . .
max
W
M
s 3278 2 29815 6619 10489 29815 03674
1304 75 4 65 1309 4 kJ kg
3.12 Take that portion of the methane initially in the tank that is also in the tank finally to be in the
system. This system is isentropic S Sf i= .
(a) The ideal gas solution
S S T T
P
P
N PV
RT
N PV
RT
N
P V
RT
N N N
f i f i
f
i
R C
i
i
i
f
f
f
f i
p
= ⇒ =
F
HG
I
KJ =
F
H
I
K =
⇒ = = = =
= − = −
300 35
70
1502
1964 6 mol; 1962
17684
8 314 36
.
.
. .
.
.
 K
= mol
 mol∆
(b) Using Figure 2.4-2.
70 bar ≈ 7 MPa, T = 300 K $ . $S Si f= =505 kJ kg K
$
. ,
.
.
.V m
N
i i
i
= = =
=
×
=
00195 07
00195
3590 kg.
35.90 kg
1282
m
kg
 so that m
m
kg
1000 g
kg
28 g
mol
 mol
3 3
3
At 3.5 bar = 0.35 MPa and $ .S Tf = ⇒ ≈505 138 K. kJ kg K Also,
$
. ,
.
.
.
.
.
V m
N
f f
f
= = =
=
×
=
0192 07
0192
3646 kg.
3646 kg
1302
m
kg
 so that m
m
kg
1000 g
kg
28 g
mol
 mol
3 3
3
∆N N Nf i= − = − = −1302 1282 11518 mol. .
3.13 dS C dT
T
R dV
V
= + eqn. (3.4-1)
∆S a R bT cT dT e
T
dT
T
R dV
V
= −( ) + + + +LNM
O
QP + zz 2 3 2
so that
S T V S T V a R T
T
b T T c T T
d T T e T T R V
V
2 2 1 1
2
1
2 1 2
2
1
2
2
3
1
3
2
2
1
2 2
1
2
3 2
, , ln
ln
 a f a f a f c h
c h c h
− = −( ) + − + −
+ − − − +− −
Now using
PV RT V
V
T
T
P
P
S T P S T P a T
T
b T T c T T
d T T e T T R P
P
= ⇒ = ⋅ ⇒
− = + − + −
+ − − − −− −
2
1
2
1
1
2
2 2 1 1
2
1
2 1 2
2
1
2
2
3
1
3
2
2
1
2 2
1
2
3 2
, , ln
ln
 a f a f a f c h
c h c h
Finally, eliminating T2 using T T P V PV2 1 2 2 1 1= yields
S P V S P V a P V
PV
b
R
P V PV
c
R
P V PV
d
R
P V PV
eR P V PV R P
P
2 2 1 1
2 2
1 1
2 2 1 1
2 2 2
2
1 1
2
3 2 2
3
1 1
3
2
2 2
2
1 1
2 2
1
2
3
2
, , ln
ln
a f a f a f
a f a f
a f a f
d i d i
− =
F
HG
I
KJ + −
+ −
+ −
− − −
− −
3.14 System: contents of valve (steady-state, adiabatic, constant volume system)
Mass balance 0 1 2= +& &N N
Energy balance 0 1 1 2 2= + +& &N H N H &Q
0 +Ws
0
−P dV
dt
0
⇒ =H H1 2
Entropy balance 0 1 1 2 2= + + +& & &N S N S Sgen
&Q
T
0
⇒ = − =∆S S S
S
N2 1
&
&
gen
(a) Using the Mollier Diagram for steam (Fig. 2.4-1a) or the Steam Tables
T
P
P
H
T
S
1
1
2
2
2
2
600 K
35
7
30453
293
7 277
=
=
=
=
⇒
≈ °
= bar
 bar
 J g
C
 J g K$ . $ .
$ $
.H H1 2 3045 3= = J g . Thus $ .S1 65598 = J g K ; Texit C= °293
∆ $ $ $ .S S S= − =2 1 0717 J g K
(b) For the ideal gas, H H T T1 2 1 2 600 K= ⇒ = =
∆
∆
S S T P S T P C T
T
R P
P
R P
P
S
= − = −
= − = ⇒
=
2 2 1 1
2
1
2
1
2
1
1338 
0743
, , ln ln
ln .
$
.
a f a f p
J mol K
 J mol K
3.15 From the Steam Tables
At 200oC, 
P
V V
U U
H H
S S
H S
L V
L V
L L
L V
=
= =
= =
= =
= ⋅ = ⋅
= = ⋅
15538 MPa
0001157 012736 m
85065 25953
852 45 27932
2 3309 64323
19407 41014
.
$
. / $ . /
$
. / $ . /
$
. / $ . /
$
. / $. /
$
.
$
.
 m kg kg 
 kJ kg kJ kg
 kJ kg kJ kg
 kJ kg K kJ kg K
 kJ / kg kJ / kg K
3 3
vap vap∆ ∆
(a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties
of the steam and water will be
At 150oC, 
P
V V
U U
H H
S S
H S
L V
L V
L V
L V
=
= =
= =
= =
= ⋅ = ⋅
= = ⋅
04578 MPa
0001091 03928 m
63168 kJ 25595
632 20 kJ 27465
18418 kJ 68379
2114 3 4 9960 kJ
.
$
. / $ . /
$
. / $ . /
$
. / $ . /
$
. / $ . /
$
. / $ . /
 m kg kg 
kg kJ kg
kg kJ kg
kg K kJ kg K
 kJ kg kg K
3 3
vap vap∆ ∆
(b) For simplicity of calculations, assume 1 m3 volume of tank.
Then
 
Mass steam initially = 0.8 m
0.12736 m kg
 kg
Mass water initially = 0.2 m
0.001157 m kg
Weight fraction of steam initially = 6.2814
179.14
Weight fraction of water initially = 6.2814
179.14
3
3
3
3
/
.
/
.
.
.
=
=
=
=
62814
17286 kg
003506
096494
The mass, energy and entropy balances on the liquid in the tank (open system) at any time
yields
dM
dt
M dM U
dt
M H dM S
dt
M S
M dU
dt
U dM
dt
M H H dM
dt
M dU
dt
dM
dt
H U
L
L
L L
L V
L L
L V
L
L
L
L
L V V
L
L
L L
V L
= = =
+ = =
= −
& ;
$
& $
$
& $
$
$ & $ $
$
$ $
 ; and 
or 
c h
Also, in a similar fashion, from the entropy balance be obtain
M dS
dt
dM
dt
S S dM
dt
SL
L L
V L
L$
$ $ $
= − =c h ∆ vap
There are now several ways to proceed. The most correct is to use the steam tables, and to use
either the energy balance or the entropy balance and do the integrals numerically (since the
internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature.
This is the method we will use first. Then a simpler method will be considered.
Using the energy balance, we have
dM
M
dU
H U
M M
M
U U
H U
M M U U
H U
L
L
L
V L
i
L
i
L
i
L
i
L
i
L
i
V
i
L i
L
i
L i
L
i
L
i
V
i
L
=
−
−
=
−
−
= +
−
−
F
HG
I
KJ
+ +
+
+
$
$ $ ,
$ $
$ $
$ $
$ $
 or replacing the derivatives by finite differences
 or finally 1 1 1 11
So we can start with the known initial mass of water, then using the Steam Tables and the data
at every 5oC do a finite difference calculation to obtain the results below.
i T (oC) $UiL (kJ/kg K) $HiV (kJ/kg K) MiL (kg)
1 200 850.65 2793.2 172.86
2 195 828.37 2790.0 170.88
3 190 806.19 2786.4 168.95
4 185 784.10 2782.4 167.06
5 180 762.09 2778.2 165.22
6 175 740.17 2773.6 163.42
7 170 718.33 2768.7 161.67
8 165 696.56 2763.5 159.95
9 160 674.87 2758.1 158.27
10 155 653.24 2752.4 156.63
11 150 631.68 2746.5 155.02
So the final total mass of water is 155.02 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3
leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, the
final volume fraction of water is 16.91%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
For comparison, using the entropy balance, we have
dM
M
dS
S S
M M
M
S S
S
M M S S
S
L
L
L
V L
i
L
i
L
i
L
i
L
i
L
i
i
L
i
L i
L
i
L
i
=
−
−
=
−
= +
−F
HG
I
KJ
+ +
+
+
$
$ $ ,
$ $
$
$ $
$
 or replacing the derivatives by finite differences
 of finally 
vap vap
1 1
1
11
∆ ∆
So again we can start with the known initial mass of water, then using the Steam Tables and
the data at every 5oC do a finite difference calculation to obtain the results below.
i T (oC) $SiL (kJ/kg K) $SiL (kJ/kg K) MiL (kg)
1 200 2.3309 6.4323 172.86
2 195 2.2835 6.4698 170.86
3 190 2.2359 6.5079 168.92
4 185 2.1879 6.5465 167.02
5 180 2.1396 6.5857 165.17
6 175 2.0909 6.6256 163.36
7 170 2.0419 6.6663 161.60
8 165 1.9925 6.7078 159.87
9 160 1.9427 6.7502 158.18
10 155 1.8925 6.7935 156.53
11 150 1.8418 6.8379 154.91
So the final total mass of water is 154.91 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3
leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, the
final volume fraction of water is 16.90%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
These results are similar to that from the energy balance. The differences are the result of
round off errors in the simple finite difference calculation scheme used here (i.e., more
complicated predictor-corrector methods would yield more accurate results.).
A simpler method of doing the calculation, avoiding numerical integration, is to assume that
the heat capacity and change on vaporization of liquid water are independent of temperature.
Since liquid water is a condensed phase and the pressure change is small, we can make the
following assumptions
$ $ $ $ $
$ $
;
$
U H H H H
dU
dt
dH
dt
C dT
dt
dS
dt
C
T
dT
dt
L L V L
L L
L
L L L L
≈ − =
≈ ≈ ≈
 and 
 and 
vap
P
P
∆
With these substitutions and approximations, we obtain from the energy balance
M dU
dt
dM
dt
H U M dH
dt
dM
dt
H
M C dT
dt
dM
dt
H
C H
C
H
dT
dt M
dM
dt
C
H
M
M
L
L L
V L L
L L
L L
L
L
L
L
L
L
f
L
i
L
$
$ $
$
$
$
$
$
$ ln
= − → =
=
=
−( ) = FHG
I
KJ
c h 
Now using an average value of and over the temperature range we obtain
 or
vap
P
vap
P
vap
P
vap
P
vap
∆
∆
∆
∆
∆
1
150 200
and from the entropy balance
M dS
dt
dM
dt
S M C
T
dT
dt
dM
dt
S
C S
C
T S
dT
dt M
dM
dt
C
S
M
M
L
L L
L
L L
L
L
L
L
L
f
L
i
L
$
$ $
$
$
$
ln .
.
ln
= → =
=
+
+
F
H
I
K =
F
HG
I
KJ
∆ ∆
∆
∆
∆
vap P vap
P
vap
P
vap
P
vap
 
Now using an average value of and over the temperature range we obtain
 or
1
150 27315
200 27315
From the Steam Table data listed above, we obtain the following estimates:
C U T U T
C T
T
S T S T
C S T S T
L
L
L
P
o o
o o
P
P
o o
C C
C - C
kJ
kg K
or using the ln mean value (more appropriate for the entropy calculation) based on
C C kJ
kg K
=
= − =
=
−
=
⋅
F
HG
I
KJ = −
=
= − =
+
+
FH IK
=
−
FH IK
=
⋅
$ ( ) $ ( ) . .
.
ln $ $
$( ) $( )
ln .
.
. .
ln .
.
.
200 150
200 150
852 45 632 20
50
4 405
200 150
200 27315
150 27315
2 3309 18418
47315
42315
4 3793
2
1
2 1a f a f
Also, obtaining average values of the property changes on vaporization, yields
∆ ∆ ∆
∆ ∆ ∆
$ $ $
. . .
$ $ $
. . .
H H T H T
S S T S T
vap vap o vap o
vap vap o vap o
C C kJ
kg
C C kJ
kg K
= × = + = = × + =
= × = + = = × + =
⋅
1
2
150 200 1
2

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