978-0078027680 Chapter 8 Part 2

subject Type Homework Help
subject Pages 14
subject Words 5219
subject Authors John Cimbala, Robert Turner, Yunus Cengel

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8-22
8-47E Steam expands in an adiabatic turbine. The maximum amount of work that can be done by the turbine is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
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8-27
8-53 Steam enters a nozzle at a specified state and leaves at a specified pressure. The process is to be sketched on the T-s
diagram and the maximum outlet velocity is to be determined.
5)-A (Table
KkJ/kg 8902.5
kJ/kg 6.2784
1
kPa 6000
1
1
1
1
s
h
x
P
For the maximum velocity at the exit, the entropy will be constant
during the process. The exit state enthalpy is (Table A-6)
kJ/kg 5.24924.19858533.033.798
8533.0
3058.4
2159.28902.5
K kJ/kg 8902.5
kPa 1200
2
2
2
12
2
fgf
fg
f
xhhh
s
ss
x
ss
P
We take the nozzle as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream
enters and leaves the nozzle, the energy balance for this steady-flow system can be expressed in the rate form as
2
0)pe (since
22
0
2
1
2
2
21
2
2
2
2
1
1
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by
nsferenergy tranet of Rate
outin
VV
hh
QW
V
hm
V
hm
EE
EEE
  
Solving for the exit velocity and substituting,
 
m/s 764.3
5.0
22
2
5.0
21
2
12
2
1
2
2
21
kJ/kg 1
/sm 1000
kJ/kg )5.24926.2784(2m/s) 0()(2
2
hhVV
VV
hh
T
s
1
1200 kPa
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8-29
8-55 Heat is added to a pressure cooker that is maintained at a specified pressure. Work is also done on water. The
minimum entropy change of the thermal-energy reservoir supplying this heat is to be determined.
Assumptions 1 Only water vapor escapes through the pressure relief valve.
Analysis According to the conservation of mass principle,
CV
mm
dt
dm
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8-33
Entropy Change of Incompressible Substances
8-60 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the
total entropy change are to be determined.
Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-
insulated and thus there is no heat transfer.
Properties The density and specific heat of water at 25C are
= 997 kg/m3 and cp = 4.18 kJ/kg.C. The specific heat of
copper at 27C is cp = 0.386 kJ/kg.C (Table A-3).
Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closed system since no mass
crosses the system boundary during the process. The energy balance for this system can be expressed as
U
EEE
0
energies etc. potential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
or,
WATER
Copper
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8-35
8-62 A hot iron block is dropped into water in an insulated tank. The total entropy change during this process is to be
determined.
Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-
insulated and thus there is no heat transfer. 4 The water that evaporates, condenses back.
0.45 kJ/kg.C (Table A-3).
Analysis We take the entire contents of the tank, water + iron block, as the system. This is a closed system since no mass
U
EEE
0
energies etc. p otential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
C26.7
2
T
The entropy generated during this process is determined from
 
 
kJ/K 314.12
K 291
K 299.7
ln KkJ/kg 4.18kg 100ln
kJ/K 232.8
K 623
K 299.7
ln KkJ/kg 0.45kg 25ln
1
2
avgwater
1
2
avgiron
T
T
mcS
T
T
mcS
kJ/K 4.08314.12232.8
waterirontotalgen SSSS
WATER
18C
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8-36
8-63 An aluminum block is brought into contact with an iron block in an insulated enclosure. The final equilibrium
temperature and the total entropy change for this process are to be determined.
Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specific heats. 2 The
system is stationary and thus the kinetic and potential energies are negligible. 3 The system is well-insulated and thus there
is no heat transfer.
Properties The specific heat of aluminum at the anticipated average temperature of 400 K is cp = 0.949 kJ/kg.C. The
specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45 kJ/kg.C (Table A-3).
U
EEE
0
energies etc. p otential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
K 382
2
C109T
The total entropy change for this process is determined from
 
 
kJ/K .2212
K 413
K 382
ln KkJ/kg 0.949kg 30ln
kJ/K .4722
K 333
K 382
ln KkJ/kg 0.45kg 40ln
1
2
avgalum
1
2
avgiron
T
T
mcS
T
T
mcS
Thus,
kJ/K 0.251221.2472.2
alumirontotal SSS
Iron
40 kg
60C
Aluminum
30 kg
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8-37
8-64 Problem 8-63 is reconsidered. The effect of the mass of the iron block on the final equilibrium temperature and
the total entropy change for the process is to be studied. The mass of the iron is to vary from 10 to 100 kg. The equilibrium
temperature and the total entropy change are to be plotted as a function of iron mass.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
Stotal kJ/kg]
miron [kg]
T2 [C]
0.08547
0.1525
0.2066
0.2511
0.2883
0.32
0.3472
0.3709
0.3916
0.41
10
20
30
40
50
60
70
80
90
100
129.1
120.8
114.3
109
104.7
101.1
97.98
95.33
93.02
91
10 20 30 40 50 60 70 80 90 100
90
95
100
105
110
115
120
125
130
T2 [C]
10 20 30 40 50 60 70 80 90 100
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
miron [kg]
Stotal [kJ/K]
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8-40
8-70 For ideal gases, cp = cv + R and
12
2
11
22
PT
P
P
VVV
1
2
1
2
1
2
1
2
1
2
lnln
lnlnln
P
P
R
T
T
c
P
P
R
T
T
R
T
T
c
p
v
8-71 For an ideal gas, dh = cp dT and
v
= RT/P. From the second Tds relation,
P
dP
T
dT
T
dP
P
RT
T
dPc
T
dPv
T
dh
p
Integrating,
2
2
P
T

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