Chapter 7 Entropy
Review Problems
7-137 It is to be shown that the difference between the steady-flow and boundary works is the flow energy.
Analysis The total differential of flow energy Pv can be expressed as

d Pv  Pdv  vdP   wb   w flow   wb  w flow

Therefore, the difference between the reversible steady-flow work and the reversible boundary work is the
flow energy.
7-138E An insulated rigid can initially contains R-134a at a specified state. A crack develops, and
refrigerant escapes slowly. The final mass in the can is to be determined when the pressure inside drops to a
specified value.
Assumptions 1 The can is well-insulated and thus heat transfer is negligible. 2 The refrigerant that remains
in the can underwent a reversible adiabatic process.
Analysis Noting that for a reversible adiabatic (i.e., isentropic) process, s1 = s2, the properties of the
refrigerant in the can are
P1  120 psia 
R-134
 s1  s f @80 F  0.0774 Btu/lbm  R
120 psia
T1  80  F

80F
s 2  s f 0.0774  0.0364
x2 

 0.2222
P2  30 psia 
s fg
0.2209  0.0364

s 2  s1
 v  v  x v  0.01209  0.2222 1.5408  0.01209   0.3518 ft 3 /lbm
2
f
2 fg
Thus the final mass of the refrigerant in the can is
m
1.2 ft 3
V

 3.411 lbm
v1 0.3518 ft 3 /lbm
7-109
Leak
Chapter 7 Entropy
7-139 An insulated rigid tank is connected to a piston-cylinder device with zero clearance that is maintained
at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into the cylinder.
The final temperatures in the tank and the cylinder are to be determined.
Assumptions 1 Both the tank and cylinder are well-insulated and thus heat transfer is negligible. 2 The
water that remains in the tank underwent a reversible adiabatic process. 3 The thermal energy stored in the
tank and cylinder themselves is negligible. 4 The system is stationary and thus kinetic and potential energy
changes are negligible.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s 2 = s1. From the
steam tables,
v  v g @500kPa  0.3749 m 3 /kg
P1  500 kPa  1
 u  u g @500kPa  2561.2 kJ/kg
sat.vapor  1
s1  s g @500kPa  6.8213 kJ/kg  K
T2, A  Tsat@150kPa  111.37 C
s 2, A  s f 6.8213  1.4336
P2  150 kPa 
x 2, A 

 0.9306

s fg
5.7897
s 2  s1

sat.mixture  v 2, A  v f  x 2, A v fg  0.001053  (0.9306 )(1.1593  0.001053 )  1.079 m 3 /kg
u 2, A  u f  x 2, A u fg  466.94  (0.9306 )( 2052.7 kJ/kg )  2377.2 kJ/kg
The initial and the final masses in tank A are
m1, A 
VA
VA
0.4m 3
0.4m 3


1.067
kg
and
m


 0.371kg
2, A
v1, A 0.3749 m 3 /kg
v 2, A 1.079 m 3 /kg
Thus,
m 2, B  m1, A  m 2, B  1.067  0.371  0.696 kg
(b) The boundary work done during this process is
Wb,out 
 PdV  P V
2
B
1
2, B

 0  PB m 2, B v 2, B
Taking the contents of both the tank and the cylinder to be the system,
the energy balance for this closed system can be expressed as
E  E out

E system
in




Net energy transfer
by heat, work, and mass
Change in internal,kinetic,
potential,etc. energies
 Wb,out  U  U  A  U B
Sat.
vapor
500 kPa
0.4 m3
150 kPa
Wb,out  U  A  U B  0
or,
PB m2, B v 2, B  m2 u 2  m1u1  A  m2 u 2 B  0
m2, B h2, B  m2 u 2  m1u1  A  0
Thus,
h2 , B 
m1u1  m2 u 2  A 1.067 2561 .2  0.3712377 .2

 2659 .3kJ / kg
m 2, B
0.696
At 150 kPa, hf = 467.11 and hg = 2693.6 kJ/kg. Thus at the final state, the cylinder will contain a saturated
liquid-vapor mixture since hf < h2 < hg. Therefore,
T2 , B  Tsat @ 150 kPa  111.37 C
7-110
Chapter 7 Entropy
7-140 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in the
room and the entropy change during this process are to be determined.
Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are
constant at room temperature. 3 The system is stationary and thus the kinetic and potential energy changes
are zero. 4 There are no work interactions involved.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heat of water at room
temperature is C = 4.18 kJ/kgC (Table A-3). For air is Cv = 0.718 kJ/kgC at room temperature.
Analysis The volume and the mass of the air in the room are
4m  5m  6m
V = 4x5x6 = 120 m³
PV
100 kPa  120 m 3
m air  1 1 
 141.7 kg
RT1
0.2870 kPa  m 3 /kg  K 295 K 




ROOM
22C
100 kPa
Taking the contents of the room, including the water, as our system,
the energy balance can be written as
E  E out
in

Net energy transfer
by heat, work, and mass
or

Heat
 0  U  U water  U air
E system



Water
80C
Change in internal,kinetic,
potential,etc. energies
mC T2  T1 water  mC v T2  T1 air  0
Substituting,
1000 kg 4.18 kJ/kg CT f




 80  C  141.7 kg  0.718 kJ/kg C T f  22  C  0
It gives the final equilibrium temperature in the room to be
Tf = 78.6C
(b) Considering that the system is well-insulated and no mass is entering and leaving, the total entropy
change during this process is the sum of the entropy changes of water and the room air,
S total  S gen  S air  S water
where
S air  mC v ln
S water  mC ln
T2
V
 mR ln 2
T1
V1
0
 141.7 kg 0.718 kJ/kg  K ln
351.6 K
 17.86 kJ/K
295 K
T2
351.6 K
 1000 kg 4.18 kJ/kg  K ln
 16.61 kJ/K
T1
353 K
Substituting, the total entropy change is determined to be
Stotal = 17.86 - 16.61 = 1.25 kJ/K
7-111
Chapter 7 Entropy
7-141E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a
specified temperature and pressure. The entropy changes of the helium and the surroundings are to be
determined, and it is to be assessed if the process is reversible, irreversible, or impossible.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the
kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is
negligible. 4 The compression or expansion process is quasi-equilibrium.
Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R (Table A-1E). The
specific heats of helium are Cv = 0.753 and Cv = 1.25 Btu/lbm.R (Table A-2E).
Analysis (a) The mass of helium is
PV
25psia  15ft 3
m 1 1 
 0.264 lbm
RT1
2.6805 psia  ft 3 /lbm  R 530 R 




HELIUM
15 ft3
PVn = const
Then the entropy change of helium becomes
S sys  S helium

T
P 
 m C p ,ave ln 2  R ln 2 
T1
P1 


760 R
70 psia 
 (0.264 lbm) (1.25 Btu/lbm  R )ln
 (0.4961 Btu/lbm  R )ln
  0.016Btu/R
530 R
25 psia 

(b) The exponent n and the boundary work for this polytropic process are determined to be
P1V1 P2V2
T P
760 R 25psia  15ft 3  7.682 ft 3


 V2  2 1 V1 
530 R 70 psia 
T1
T2
T1 P2

n

 P  V 
 70   15 


  2    1  
    
 n  1.539
 
 25   7.682 
 P1   V2 
Then the boundary work for this polytropic process can be determined from
2
P V  P1V1
mR T2  T1 
Wb,in   P dV   2 2

1
1 n
1 n
P2V2n
n
P1V1n


0.264 lbm 0.4961 Btu/lbm  R 760  530 R  55.9 Btu
1  1.539
We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heat
transfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as
E  E out

E system
in




Net energy transfer
by heat, work, and mass
Change in internal,kinetic,
potential,etc. energies
 Qout  Wb,in  U  m(u 2  u1 )
 Qout  m(u 2  u1 )  Wb,in
Qout  Wb,in  mC v (T2  T1 )
Substituting,
Qout  55.9 Btu  0.264 lbm 0.753 Btu/lbm  R 760  530 R  10.2 Btu
Noting that the surroundings undergo a reversible isothermal process, its entropy change becomes
Ssurr 
Qsurr ,in
Tsurr

10.2 Btu
 0.019 Btu / R
530 R
(c) Noting that the system+surroundings combination can be treated as an isolated system,
S total  S sys  S surr  0.016  0.019  0.003Btu/R  0
Therefore, the process is irreversible.
7-112
Q
Chapter 7 Entropy
7-142 Air is compressed steadily by a compressor from a specified state to a specified pressure. The
minimum power input required is to be determined for the cases of adiabatic and isothermal operation.
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 with variable specific heats. 4 The process is reversible
since the work input to the compressor will be minimum when the compression process is reversible.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis (a) For the adiabatic case, the process will be reversible and adiabatic (i.e., isentropic),
thus the isentropic relations are applicable.
T1  290 K 
 Pr1  1.2311 and h1  290.16 kJ/kg
2
2
and
Pr2 
T2  503.3 K
P2
700 kPa
Pr1 
(1.2311)  8.6177 
h2  506.45 kJ/kg
P1
100 kPa
The energy balance for the compressor, which is a steady-flow system,
can be expressed in the rate form as
E  E out
in

E system0 (steady)




Rate of net energy transfer
by heat, work, and mass
AIR
Rev.
1
0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
 h1  m
 h2  W in  m
 (h2  h1 )
W in  m
Substituting, the power input to the compressor is determined to be
W in  (5/60 kg/s )(506.45  290.16 )kJ/kg  18.0 kW
(b) In the case of the reversible isothermal process, the steady-flow energy balance becomes
 h1  Q out  m
 h2  W in  Q out  m
 (h2  h1 ) 0  Q out
E in  E out  W in  m
since h = h(T) for ideal gases, and thus the enthalpy change in this case is zero. Also, for a reversible
isothermal process,
Q  m T s  s   m T s  s 
out
1
2
2
1
where

s 2  s1  s 2  s1

0
 R ln
P2
P
700 kPa
  R ln 2  0.287 kJ/kg  K ln
 0.5585 kJ/kg  K
P1
P1
100 kPa
Substituting, the power input for the reversible isothermal case becomes
W in  (5/60 kg/s )(290 K)(0.5585 kJ/kg  K)  13.5 kW
7-113
T = const
1
Chapter 7 Entropy
7-143 Air is compressed in a two-stage ideal compressor with intercooling. For a specified mass flow rate
of air, the power input to the compressor is to be determined, and it is to be compared to the power input to
a single-stage compressor. 
Assumptions 1 The compressor operates steadily. 2 Kinetic and potential energies are negligible. 3 The
compression process is reversible adiabatic, and thus isentropic. 4 Air is an ideal gas with constant specific
heats at room temperature.
Properties The gas constant of air is R =0.287 kPa.m3/kg.K (Table A-1). The specific heat ratio of air is k =
1.4 (Table A-2)
Analysis The intermediate pressure between the two stages is
100 kPa
27C
100 kPa 900 kPa   300 kPa
Px  P1 P2 
900 kPa
The compressor work across each stage is the same, thus total
compressor work is twice the compression work for a single stage:




kRT1
Px P1 k 1 / k  1
k 1
0.4/1.4

1.4 0.287 kJ/kg  K 300 K    300 kPa 
2
 1


 100 kPa

1.4  1



 222.2 kJ/kg
wcomp,in  2 wcomp,in, I  2
Stage I
W
Stage II
27C
Heat
and
W in  m wcomp,in  0.02 kg/s 222.2 kJ/kg   4.44kW
The work input to a single-stage compressor operating between the same pressure limits would be
wcomp, in
and

kRT1
P2 P1 k 1 / k  1  1.4 0.287 kJ/kg  K 300 K    900 kPa

k 1
1.4  1
100 kPa






0.4/1.4

 1  263.2 kJ/kg


W in  m wcomp,in  0.02 kg/s 263.2 kJ/kg   5.26 kW
Discussion Note that the power consumption of the compressor decreases significantly by using 2-stage
compression with intercooling.
7-114
Chapter 7 Entropy
7-144 A three-stage compressor with two stages of intercooling is considered. The two intermediate
pressures that will minimize the work input are to be determined in terms of the inlet and exit pressures.
Analysis The work input to this three-stage compressor with intermediate pressures Px and Py and two
intercoolers can be expressed as
wcomp  wcomp,I  wcomp, II  wcomp,III




 




nRT1
nRT1
nRT1
1  Px P1 n 1 / n 
1  Py Px n 1 / n 
1  Px P1 n 1 / n
n 1
n 1
n 1
nRT1

1  Px P1 n 1 / n  1  Py Px n 1 / n  1  Px P1 n 1 / n
n 1
nRT1

3  Px P1 n 1 / n  Py Px n 1 / n  Px P1 n 1 / n
n 1
The Px and Py values that will minimize the work input are obtained by taking the partial differential of w
with respect to Px and Py, and setting them equal to zero:






 Py

 P
 x




w
n 1 1

0
 
 Py
n  Px


n 1
1
n
w
n  1  1  Px
 
0
 
 Px
n  P1  P1


n 1
1
n

n  1  1

n  Py
 Px

 Py





n 1 1


n  P2
 Py


 P2




n 1
1
n

n 1
1
n
0
0
Simplifying,

1  Px

P1  P1



 Py

P
 x




1
Px
1
n

1
n

1

Py
 Px

 Py





1

P2
 Py

P
 2




2 n 1
n

2 n 1
n
1 P

 n  1
P1  Px



1  Px 

 Pn  P 

y  y 
1 2 n
1  Px 
1  Py 
 

Pxn  Py  P2n  P2 
which yield
 
 P P 
Px2  P1 Px P2 
 Px  P12 P2
Py2  P2 P1 Py 
 Py
13
2 13
1 2
7-115
1 2 n


 Px21 n   P1 Py
1n

 Py21 n   Px P2 1 n
Chapter 7 Entropy
7-145 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Some
steam is extracted at the end of the first stage. The power output of the turbine is to be determined for the
cases of 100% and 88% isentropic efficiencies.
Assumptions 1 This is a steady-flow process since there is no change
with time. 2 Kinetic and potential energy changes are negligible. 3 The
turbine is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
P1  7MPa  h1  3410.3 kJ/kg

T1  500  C  s1  6.7975 kJ/kg  K
P2  1MPa 
 h2  2879 .4 kJ / kg
s 2  s1

7 MPa
500C
STEAM
13.5 kg/s
STEAM
15 kg/s
I
II
1 MPa
s 3s  s f
6.7975  1.0910

 0.8775
P3  50 kPa  x 3s 
s fg
6.5029

s 3  s1
 h  h  x h  340 .49  0.8775 2305 .4   2363.5 kJ/kg
3s
3s fg
f
50 kPa
10%
90%
Analysis (a) The mass flow rate through the second stage is
 3  0.9m 1  0.9 15kg/s   13.5 kg/s
m
We take the entire turbine, including the connection part between the two stages, as the system, which is a
control volume since mass crosses the boundary. Noting that one fluid stream enters the turbine and two
fluid streams leave, the energy balance for this steady-flow system can be expressed in the rate form as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)



0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m 1h1  (m 1  m 3 )h2  Wout  m 3h3
Wout  m 1h1  (m 1  m 3 )h2  m 3h3
 m 1 (h1  h2 )  m 3 (h2  h3 )
Substituting, the power output of the turbine is
W out  15 kg/s 3410.3  2879.4 kJ/kg  13.5 kg 2879.4  2363.5 kJ/kg
 14,930 kW
(b) If the turbine has an adiabatic efficiency of 88%, then the power output becomes
W a  T W s  0.88 14,928 kW   13,140 kW
7-116
Chapter 7 Entropy
7-146 Steam expands in an 84% efficient two-stage adiabatic turbine from a specified state to a specified
pressure. Steam is reheated between the stages. For a given power output, the mass flow rate of steam
through 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 negligible. 3 The turbine is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
Heat
P1  8 MPa  h1  3398.3 kJ/kg

T1  500  C  s1  6.7240 kJ/kg  K
2 MPa
P2 s  2 MPa 
 h2 s  3000 .3 kJ/kg
s 2 s  s1

Stage I
2 MPa
500C
Stage II
P3  2 MPa  h3  3467.6 kJ/kg

T3  500  C  s 3  7.4317 kJ/kg  K
8 MPa
500C
P4 s  100 kPa 
 h4 s  2704.2 kJ/kg
s 4s  s3

100 kPa
Analysis The power output of the actual turbine is given to be 80 MW.
Then the power output for the isentropic operation becomes
W s ,out  W a,out / T  (80,000 kW) / 0.84  95,240 kW
We take the entire turbine, excluding the reheat section, as the system, which is a control volume since mass
crosses the boundary. The energy balance for this steady-flow system in isentropic operation can be
expressed in the rate form as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)



0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m h1  m h3  m h2 s  m h4 s  W s,out
W s,out  m [( h1  h2 s )  (h3  h4 s )]
Substituting,
95,240 kJ/s  m (3398.3  3000.3) (3467.6  2704.2 )kJ/kg 
which gives
  82.0 kg/s
m
7-117
5 MW
Chapter 7 Entropy
7-147 Refrigerant-134a is compressed by a 0.5-kW adiabatic compressor from a specified state to another
specified state. The isentropic efficiency, the volume flow rate at the inlet, and the maximum flow rate at the
compressor inlet are 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 The device is adiabatic and thus heat transfer is negligible.
Properties From the R-134a tables (Tables A-11 through A-13)
v  0.14549 m 3 /kg
P1  140 kPa  1
 h1  243.40 kJ/kg
T1  10  C 
s1  0.9606 kJ/kg  K
2
P2  700 kPa 
 h 2  296.69 kJ/kg
T2  60  C 
0.5 kW
R-134a
P2  700 kPa 
 h2 s  278.06 kJ/kg
s 2 s  s1

·
V1
Analysis (a) The isentropic efficiency is determined from its definition,
1
h2s  h1 278.06  24340
.

 0.650  65.0%
h2a  h1 296.69  24340
.
C 
1  m
2  m
 . We take the actual compressor as the
(b) There is only one inlet and one exit, and thus m
system, which is a control volume. The energy balance for this steady-flow system can be expressed as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)



0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
W a,in  m h1  m h2 (since Q  ke  pe  0)
W a,in  m (h2  h1 )
Then the mass and volume flow rates of the refrigerant are determined to be
m 
W a ,in
h2 a  h1

0.5kJ/s
 0.0094 kg/s
296.69  243.40 kJ/kg


V1  m v1  0.0094 kg/s  0.14549 m 3 /kg  0.00137 m 3 /s  82L/min
(c) The volume flow rate will be a maximum when the process is isentropic, and it is determined similarly
from the steady-flow energy equation applied to the isentropic process. It gives
m max 
V1,max
W s ,in

0.5kJ/s
 0.0144 kg/s
278.06  243.40 kJ/kg
 m max v1  0.0144 kg/s 0.14549 m 3 /kg   0.00210 m 3 /s  126 L/min
h2 s  h1
Discussion Note that the raising the isentropic efficiency of the compressor to 100% would increase the
volumetric flow rate by more than 50%.
7-118
Chapter 7 Entropy
7-148E Helium is accelerated by a 94% efficient nozzle from a low velocity to 1000 ft/s. The pressure and
temperature at the nozzle inlet are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Helium is an ideal gas
with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus
heat transfer is negligible.
Properties The specific heat ratio of helium is k = 1.667. The constant pressure specific heat of helium is
1.25 Btu/lbm.R (Table A-2E).
Analysis We take nozzle as the system, which is a control volume since mass crosses the boundary. The
energy balance for this steady-flow system can be expressed in the rate form as
E  E out

E system0 (steady)
0
in




Rate of net energy transfer
by heat, work, and mass
Rate of change in internal,kinetic,
potential,etc. energies
1
E in  E out
  pe  0)
m (h1  V12 / 2)  m (h2 + V22 /2) (since Q  W
0  h2  h1 
V22  V12
V 2  V12

 0  C p ,ave T2  T1   2
2
2
Solving for T1 and substituting,
0
1000 ft/s2
 1Btu/lbm

21.25 Btu/lbm  R   25,037 ft 2 /s 2
From the isentropic efficiency relation,
T1  T2 a 
V22s  V12
2C p
 180  F 
N 

  196.0 F  656R


h2 a  h1 C P T2a  T1 

h2 s  h1 C P T2 s  T1 
or,
T2 s  T1  T2 a  T1  /  N  656  640  656  / 0.94   639 R
k 1 / k
T2 s  P2 
From the isentropic relation,
 
T1  P1 
T
P1  P2  1
 T2 s



k / k 1
 656 R 

 14 psia 
 639 R 
1.667/0.667
7-119
 14.9 psia
HELIUM
N = 94%
2
Chapter 7 Entropy
7-149 [Also solved by EES on enclosed CD] An adiabatic compressor is powered by a direct-coupled steam
turbine, which also drives a generator. The net power delivered to the generator and the rate of entropy
generation are 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 The devices are adiabatic and thus heat transfer is negligible. 4 Air is an
ideal gas with variable specific heats.
Properties From the steam tables (Tables A-4 through 6) and air table (Table A-17),
T1  295 K 
 h1  295.17 kJ/kg , s1  1.68515 kJ/kg  K
T2  620 K 
 h1  628 .07 kJ/kg, s 2  2.44356 kJ/kg  K
P3  12.5 MPa  h3  3341.8 kJ/kg

T3  500  C  s 3  6.4618 kJ/kg  K
P4  10 kPa  h4  h f  x 4 h fg  191 .83  0.92 2392 .8  2393.2 kJ/kg

x 4  0.92  s 4  s f  x 4 s fg  0.6493  0.92 7.5009   7.5501 kJ/kg  K
 in  m
 out  m
 . We take either
Analysis There is only one inlet and one exit for either device, and thus m
the turbine or the compressor as the system, which is a control volume since mass crosses the boundary. The
energy balance for either steady-flow system can be expressed in the rate form as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)



0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
For the turbine and the compressor it becomes
1 MPa
620 K
Compressor:
 air h1  m
 air h2  W comp, in  m
 air (h2  h1 )
W comp, in  m
Turbine:
 steam h3  W turb, out  m
 steam h4  W turb, out  m
 steam (h3  h4 )
m
Substituting,
W comp, in  10 kg/s 628.07  295.17 kJ/kg  3329 kW
12.5 MPa
500C
Air
comp
0.98 kPa
295 K
10 kPa
W turb,out  25 kg/s 3341.8  2393.2 kJ/kg  23,715 kW
Therefore,
W
net,out
 W turb,out  W comp, in  23,715  3329  20,386 kW
Noting that the system is adiabatic, the total rate of entropy change (or generation) during this process is the
sum of the entropy changes of both fluids,
S
 m ( s  s )  m
(s  s )
gen
air
2
1
steam
4
3
where


P 
1000 kPa
m air s 2  s1   m  s 2  s1  Rln 2   10 kg/s  2.44356  1.68515  0.287 ln
P1 
98 kPa


m steam s 4  s 3   25 kg/s 7.5501  6.4618 kJ/kg  K  27.2 kW/K
Substituting, the total rate of entropy generation is determined to be
S gen,total  S gen,comp  S gen,turb  0.92  27 .2  28.12 kW/Κ
7-120
Steam
turbine

kJ/kg  K  0.92 kW/K

Chapter 7 Entropy
7-150 Problem 7-149 is reconsidered. The isentropic efficiencies for the compressor and turbine
are to be determined, and then the effect of varying the compressor efficiency over the range 0.6
to 0.8 and the turbine efficiency over the range 0.7 to 0.95 on the net work for the cycle and the
entropy generated for the process is to be investigated. The net work is to be plotted as a function
of the compressor efficiency for turbine efficiencies of 0.7, 0.8, and 0.9.
"Input Data"
m_dot_air = 10"kg/s" "air compressor (air) data"
T_air[1]=(295-273)"C"
"We will input temperature in C"
P_air[1]=98"kPa"
T_air[2]=(700-273) "C"
P_air[2]=1000"kPa"
m_dot_st=25"kg/s" "steam turbine (st) data"
T_st[1]=500"C"
P_st[1]=12500"kPa"
P_st[2]=10"kPa"
x_st[2]=0.92"quality"
"Compressor Analysis:"
"Conservation of mass for the compressor m_dot_air_in = m_dot_air_out =m_dot_air"
"Conservation of energy for the compressor is:"
E_dot_comp_in - E_dot_comp_out = DELTAE_dot_comp
DELTAE_dot_comp = 0
"Steady flow requirement"
E_dot_comp_in=m_dot_air*(enthalpy(air,T=T_air[1])) + W_dot_comp_in
E_dot_comp_out=m_dot_air*(enthalpy(air,T=T_air[2]))
"Compressor adiabatic efficiency:"
Eta_comp=W_dot_comp_in_isen/W_dot_comp_in
W_dot_comp_in_isen=m_dot_air*(enthalpy(air,T=T_air_isen[2])-enthalpy(air,T=T_air[1]))
s_air[1]=entropy(air,T=T_air[1],P=P_air[1])
s_air[2]=entropy(air,T=T_air[2],P=P_air[2])
s_air_isen[2]=entropy(air, T=T_air_isen[2],P=P_air[2])
s_air_isen[2]=s_air[1]
"Turbine Analysis:"
"Conservation of mass for the turbine m_dot_st_in = m_dot_st_out =m_dot_st"
"Conservation of energy for the turbine is:"
E_dot_turb_in - E_dot_turb_out = DELTAE_dot_turb
DELTAE_dot_turb = 0
"Steady flow requirement"
E_dot_turb_in=m_dot_st*h_st[1]
h_st[1]=enthalpy(steam,T=T_st[1], P=P_st[1])
E_dot_turb_out=m_dot_st*h_st[2]+W_dot_turb_out
h_st[2]=enthalpy(steam,P=P_st[2], x=x_st[2])
"Turbine adiabatic efficiency:"
Eta_turb=W_dot_turb_out/W_dot_turb_out_isen
W_dot_turb_out_isen=m_dot_st*(h_st[1]-h_st_isen[2])
s_st[1]=entropy(steam,T=T_st[1],P=P_st[1])
h_st_isen[2]=enthalpy(steam, P=P_st[2],s=s_st[1])
"Note: When Eta_turb is specified as an independent variable in
the Parametric Table, the iteration process may put the steam state 2 in the
superheat region, where the quality is undefined. Thus, s_st[2], T_st[2] are
calculated at P_st[2], h_st[2] and not P_st[2] and x_st[2]"
s_st[2]=entropy(steam,P=P_st[2],h=h_st[2])
T_st[2]=temperature(steam,P=P_st[2], h=h_st[2])
s_st_isen[2]=s_st[1]
"Net work done by the process:"
W_dot_net=W_dot_turb_out-W_dot_comp_in
7-121
Chapter 7 Entropy
"Entropy generation:"
"Since both the compressor and turbine are adiabatic, and thus there is no heat transfer
to the surroundings, the entropy generation for the two steady flow devices bec omes:"
S_dot_gen_comp=m_dot_air*( s_air[2]-s_air[1])
S_dot_gen_turb=m_dot_st*(s_st[2]-s_st[1])
S_dot_gen_total=S_dot_gen_comp+S_dot_gen_turb
"To generate the data for Plot Window 1, Comment out the line ' T_air[2]=(700 -273) C'
and select values for Eta_comp in the Parametric Table, then press F3 to solve the table.
EES then solves for the unknown value of T_air[2] for each Eta_comp."
"To generate the data for Plot Window 2, Comment out the two lines ' x_st[2]=0.92 quality '
and ' h_st[2]=enthalpy(steam,P=P_st[2], x=x_st[2]) ' and select values for Eta_turb in the
Parametric Table, then press F3 to solve the table. EES then solves for the h_st[2] for each
Eta_turb."
W
[kW]
20124
21745
23365
24985
26606

S
[kW/K]
27.59
22.51
17.44
12.36
7.281

0.75
0.8
0.85
0.9
0.95
0.6665
0.6665
0.6665
0.6665
0.6665
Effect of Turbine Efficiency on Net Work and Entropy Generated
27000
30
comp = 0.6665
26000
25
20
24000
23000
15
22000
10
21000
20000
0.75
0.78
0.80
0.83
0.85
turb
7-122
0.88
0.90
0.93
5
0.95
Sgen,total [kW/K]
Wnet [kW]
25000
Chapter 7 Entropy
7-151 Two identical bodies at different temperatures are connected to each other through a heat engine. It
is to be shown that the final common temperature of the two bodies will be T f  T1T2 when the work
output of the heat engine is maximum.
Analysis For maximum power production, the entropy generation must be zero. Taking the source, the
sink, and the heat engine as our system, which is adiabatic, and noting that the entropy change for cyclic
devices is zero, the entropy generation for this system can be expressed as
S gen  S source  S engine0  S sink  0
mC ln
ln
Tf
T1
 ln
Tf
T2
Tf
T1
 0  mC ln
0
 ln
Tf Tf
T1 T2
Tf
T2
0
m, C
T1
0
 T f2  T1T2
and thus
QH
HE
T f  T1T2
W
QL
for maximum power production.
m, C
T2
7-123
Chapter 7 Entropy
7-152 The pressure in a hot water tank rises to 2 MPa, and the tank explodes. The explosion energy of the
water is to be determined, and expressed in terms of its TNT equivalence.
Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energy
changes are negligible. 3 Heat transfer with the surroundings during explosion is negligible.
Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6)
v1  v f @ 2MPa  0.001177 m 3 /kg
P1  2 MPa 
 u1  u f @ 2MPa  906.44 kJ/kg
Sat.liquid 
s1  s f @ 2MPa  2.4474 kJ/kg  K
P2  100 kPa  u f  417 .36, u fg  2088.7 kJ/kg

s 2  s1
 s f  1.3026 , s fg  6.0568 kJ/kg  K
x2 
s2  s f
s fg

Water
Tank
2 MPa
2.4474  1.3026
 0.189
6.0568
u 2  u f  x 2 u fg  417 .36  0.189 2088 .7   812.1 kJ/kg
Analysis We idealize the water tank as a closed system that undergoes a reversible adiabatic process with
negligible changes in kinetic and potential energies. The work done during this idealized process represents
the explosive energy of the tank, and is determined from the closed system energy balance to be
E  E out
in

Net energy transfer
by heat, work, and mass

E system



Change in internal,kinetic,
potential,etc. energies
 Wb,out  U  m(u 2  u1 )
E exp  Wb,out  mu1  u 2 
where
m
0.1m 3
V

 85.0 kg
v1 0.001177 m 3 /kg
Substituting,
Eexp  85.0 kg 906.44  812.1kJ/kg  8019 kJ
which is equivalent to
mTNT 
8019 kJ
 2.47 kg TNT
3250 kJ / kg
7-124
Chapter 7 Entropy
7-153 A 0.2-L canned drink explodes at a pressure of 1 MPa. The explosive energy of the drink is to be
determined, and expressed in terms of its TNT equivalence. 
Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energy
changes are negligible. 3 Heat transfer with the surroundings during explosion is negligible. 4 The drink can
be treated as pure water.
Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6)
v1  v f @1MPa  0.001127 m 3 /kg
P1  1 MPa 
 u1  u f @1MPa  761.68 kJ/kg
Comp.liquid 
s1  s f @1MPa  2.1387 kJ/kg  K
P2  100 kPa  u f  417 .36, u fg  2088.7 kJ/kg

s 2  s1
 s f  1.3026 , s fg  6.0568 kJ/kg  K
x2 
s2  s f
s fg

2.1387  1.3026
 0.138
6.0568
COLA
1 MPa
u 2  u f  x 2 u fg  417 .36  0.138 2088 .7   705.6 kJ/kg
Analysis We idealize the canned drink as a closed system that undergoes a reversible adiabatic process
with negligible changes in kinetic and potential energies. The work done during this idealized process
represents the explosive energy of the can, and is determined from the closed system energy balance to be
E  E out
in

Net energy transfer
by heat, work, and mass

E system



Change in internal,kinetic,
potential,etc. energies
 Wb,out  U  m(u 2  u1 )
E exp  Wb,out  mu1  u 2 
where
m
Substituting,
0.0002 m 3
V

 0.177 kg
v1 0.001127 m 3 /kg
Eexp  0.177 kg 761.68  705.6 kJ/kg  9.9 kJ
which is equivalent to
mTNT 
9.9 kJ
3250 kJ/kg
 0.00305 kgTNT
7-125