8-2
Entropy and the Increase of Entropy Principle
example, receives more heat than it rejects during a cycle.
8-4C That integral should be performed along a reversible path to determine the entropy change.
Otherwise, the entropy must increase since there are no offsetting entropy changes associated with reservoirs exchanging
heat with the system.
single inlet and outlet is true only if the process is also reversible. Since no real process is reversible, there will be an
entropy increase in the fluid during the adiabatic process in devices such as pumps, compressors, and turbines.
8-3
8-12C Always.
8-17C Greater than.
8-19E The source and sink temperatures and the entropy change of the sink for a completely reversible heat engine are
given. The entropy decrease of the source and the amount of heat transfer from the source are to be determined.
Assumptions The heat engine operates steadily.
8-4
8-20 Air is compressed steadily by a compressor. The air temperature is maintained constant by heat rejection to the
surroundings. The rate of entropy change of air 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
negligible. 3 Air is an ideal gas. 4 The process involves no internal irreversibilities such as friction, and thus it is an
isothermal, internally reversible process.
outin
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by
nsferenergy tranet of Rate
outin 0
QW
EE
EEE
Therefore,
kW 15
inout WQ
Noting that the process is assumed to be an isothermal and internally
reversible process, the rate of entropy change of air is determined to be
kW/K 0.0503 K 298
kW 15
sys
airout,
air T
Q
S
8-21 Heat is transferred directly from an energy-source reservoir to an energy-sink. The entropy change of the two
reservoirs is to be calculated and it is to be determined if the increase of entropy principle is satisfied.
Assumptions The reservoirs operate steadily.
AIR
T = const.
P2
P1
Q
·
15 kW
600 K
8-5
8-22 It is assumed that heat is transferred from a cold reservoir to the hot reservoir contrary to the Clausius statement of the
second law. It is to be proven that this violates the increase in entropy principle.
Assumptions The reservoirs operate steadily.
Analysis According to the definition of the entropy, the entropy change of the
kJ/K 0.0833 1667.008333.0
total LH SSS
8-23 A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to
be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.
Assumptions The heat pump operates steadily.
Analysis Since the heat pump is completely reversible, the combination of the coefficient
kW 17.17
17.47
kW 300
COP revHP,
innet, H
Q
W
7°C
24°C
8-6
8-24 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropy change of the
working fluid, the entropy change of the source, and the total entropy change during this process are to be determined.
Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is given by
SQ
T
Noting that heat transferred from the source is equal to the heat transferred to the working fluid, the entropy changes of the
fluid and of the source become
kJ/K1.337
K 673
kJ 900
fluid
fluidin,
fluid
fluid
fluid T
Q
T
Q
S
(b)
kJ/K1.337
K 673
kJ 900
source
source out,
source
source
source T
Q
T
Q
S
sourcefluidtotalgen SSSS
Source
400C
900 kJ
8-8
8-26E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink. The entropy change of the
working fluid is given. The amount of heat transfer, the entropy change of the sink, and the total entropy change during the
process are to be determined.
8-29E A piston-cylinder device that is filled with water is heated. The total entropy change is to be determined.
/lbmft 25.1
lbm 2
ft 5.2 3
3
1
1 m
V
v
8-10
8-31 The radiator of a steam heating system is initially filled with superheated steam. The valves are closed, and steam is
allowed to cool until the temperature drops to a specified value by transferring heat to the room. The entropy change of the
steam during this process is to be determined.
Analysis From the steam tables (Tables A-4 through A-6),
001008.095986.0
C04
KkJ/kg 7.2810
/kgm 95986.0
C015
kPa 200
2
2
1
3
1
1
1
f
T
s
T
P
vv
v
H2O
200 kPa
150C
8-12
8-34E R-134a is compressed in a compressor during which the entropy remains constant. The final temperature and
enthalpy change are to be determined.
Analysis The initial state is saturated vapor and the properties are (Table A-11E)
RBtu/lbm 22477.0
Btu/lbm 98.103
F6 @ 1
F6 @ 1
g
g
ss
hh
The final state is superheated vapor and the properties are (Table A-13E)
Btu/lbm 52.114
RBtu/lbm 22477.0
psia 80
2
2
12
2
h
T
ss
PF75.7
The change in the enthalpy across the compressor is then
Btu/lbm 10.5 98.10352.114
12 hhh
8-35 Water vapor is expanded in a turbine during which the entropy remains constant. The enthalpy difference is to be
determined.
Analysis The initial state is superheated vapor and thus
KkJ/kg 5432.6
kJ/kg 3.3178
C400
MPa 6
1
1
1
1
s
h
T
P
The entropy is constant during the process. The final state is a mixture since the
entropy is between sf and sg for 100 kPa. The properties at this state are (Table A-5)
kJ/kg 9.2370)5.2257)(8653.0(51.417
8653.0
KkJ/kg 0562.6
KkJ/kg )3028.15432.6(
22
2
2
fgf
fg
f
hxhh
s
ss
x
The change in the enthalpy across the turbine is then
kJ/kg 807.4 3.31789.2370
12 hhh
T
s
2
1
T
s
2
1
8-16
8-40 R-134a undergoes an isothermal process in a closed system. The work and heat transfer are to be determined.
Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work
interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis The energy balance for this system can be
)( 12outin
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
uumUQW
EEE
2
1
R-134a
320 kPa
T1 =T2 =40C
8-20
8-44 The heat transfer during the process shown in the figure is to be determined.
Assumptions The process is reversible.
Analysis No heat is transferred during the process 2-3 since the
2
1
600
3
kJ/kg 471
2
8-45E The heat transfer during the process shown in the figure is to be determined.
Assumptions The process is reversible.
Analysis Heat transfer is equal to the sum of the areas under the process 1-2 and 2-3.
1
3
R.0)Btu/lbm1(3.0
2
)R460(360)R460(55
)()(
223212
21
3
2
2
1
12 ssTss
TT
TdsTdsq
T
(C)
s (kJ/kg∙K)
0.3
1.0
T
(F)
360
2
3