Chapter 7 Entropy
Review Problems (cont.)
7-154 The validity of the Clausius inequality is to be demonstrated using a reversible and an irreversible
heat engine operating between the same temperature limits.
Analysis Consider two heat engines, one reversible and one irreversible, both operating between a hightemperature reservoir at TH and a low-temperature reservoir at TL. Both heat engines receive the same
amount of heat, QH. The reversible heat engine rejects heat in the amount of QL, and the irreversible one in
the amount of QL, irrev = QL + Qdiff, where Qdiff is a positive quantity since the irreversible heat engine
produces less work. Noting that QH and QL are transferred at constant temperatures of TH and TL,
respectively, the cyclic integral of Q/T for the reversible and irreversible heat engine cycles become
Q 
 
rev
  T

 QH
TH


 QL
TL

1
TH
Q
H

1
TL
Q
L

QH QL

0
TH TL
since (QH/TH) = (QL/TL) for reversible cycles. Also,
QL,irrev QH QL Qdiff
Qdiff
Q
Q 




0
  H 
TL
TH TL
TL
TL
irr TH
  T
Q 
 0.

  T
since Qdiff is a positive quantity. Thus,
TH
·
QH
QH
·
·
Wnet, rev
Rev
HE
·
Wnet, irrev
Irrev
HE
·
QL
QL, irrev
TL
7-126
Chapter 7 Entropy
7-155 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount
of heat transfer through the glass and the amount of entropy generation within the glass in 5 h are to be
determined
Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant
at the specified values. 2 Thermal properties of the glass are constant.
Analysis The amount of heat transfer over a period of 5 h is
Q  Q cond t  (3.2 kJ/s)(5  3600 s)  57,600 kJ
Glass
We take the glass to be the system, which is a closed system. Under steady conditions,
the rate form of the entropy balance for the glass simplifies to
S in  S out


Rate of net entropy transfer
by heat and mass

S gen

Rate of entropy
generation
0
 S system  0

Rate of change
of entropy
Q in
Q
 out  S gen,glass  0
Tb,in Tb,out
3200 W 3200 W 

 S gen,wall  0  S gen,glass  0.287 W/K
283 K
276 K
Then the amount of entropy generation over a period of 5 h becomes
S gen,glass  Sgen,glasst  (0.287 W/K)(5  3600 s)  5160 J/K
7-127
10C
3C
Chapter 7 Entropy
7-156 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened
and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses
heat to the surroundings. The final temperature in each tank and the entropy generated during this process
are to be determined.
Assumptions 1 Tank A is insulated, and thus heat transfer is negligible. 2 The water that remains in tank A
undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible.
4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no work
interactions.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the
steam tables (Tables A-4 through A-6),
Tank A :
P1  400 kPa 

x1  0.8

v1, A  v f  x1v fg  0.001084  0.80.4625  0.001084   0.3702 m 3 /kg
u1, A  u f  x1u fg  604 .31  0.81949 .3  2163.75 kJ/kg
s1, A  s f  x1 s fg  1.7766  0.85.1193   5.8720 kJ/kg  K
P1  300 kPa 

s 2  s1

sat.mixture  v
x 2, A
2, A
T2, A  Tsat@300kPa  133.55 C
s 2, A  s f
5.8720  1.6718


 0.790
s fg
5.3201
 v f  x 2, A v fg  0.001073  0.790 0.6058  0.001073   0.479 m 3 /kg
u 2, A  u f  x 2, A u fg  561 .15  0.790 1982.4 kJ/kg   2127.2 kJ/kg
Tank B :
v  1.1988 m 3 /kg
P1  200 kPa  1, B
 u1, B  2731.2 kJ/kg
T1  250  C 
s  7.7086 kJ/kg  K
600 kJ
1, B
A
The initial and the final masses in tank A are
m1, A
V = 0.2
steam
P = 400 kPa
x = 0.8
V
0.2 m 3
 A 
 0.540 kg
v1, A 0.3702 m 3 / kg
and
m2 , A 

m3
B
m = 3 kg
steam
T = 250C
P = 200 kPa
VA
0.2 m 3

 0.418 kg
v2, A 0.479 m 3 / kg
Thus, 0.540 - 0.418 = 0.122 kg of mass flows into tank B. Then,
m2, B  m1, B  0122
.
 3  0122
.
 3.122 kg
The final specific volume of steam in tank B is determined from
v 2, B 


m v  3kg  1.1988 m 3 /kg
VB
 1 1 B 
 1.152 m 3 /kg
m 2, B
m 2, B
3.122 m 3
We take the entire contents of both tanks as the system, which is a closed system. The energy balance for
this stationary closed system can be expressed as
E  E out
in

Net energy transfer
by heat, work, and mass

E system



Change in internal,kinetic,
potential,etc. energies
 Qout  U  (U ) A  (U ) B
(since W  KE = PE = 0)
 Qout  m 2 u 2  m1u1  A  m 2 u 2  m1u1  B
Substituting,
 600  0.418 2127 ..2  0.540 2163 .8 3.122 u 2, B  32731 .2
u 2, B  2521.7 kJ/kg

7-128

Chapter 7 Entropy
Thus,
v 2, B  1.152 m 3 /kg 
T2, B  113 C

u 2, B  2521.7 kJ/kg  s 2, B  7.225 kJ/kg  K
(b) The total entropy generation during this process is determined by applying the entropy balance on an
extended system that includes both tanks and their immediate surroundings so that the boundary temperature
of the extended system is the temperature of the surroundings at all times. It gives
S in  S out


Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
 S gen  S A  S B
Tb, surr
Rearranging and substituting, the total entropy generated during this process is determined to be
Qout
Q
 m 2 s 2  m1 s1  A  m 2 s 2  m1 s1  B  out
Tb, surr
Tb, surr
600 kJ
 0.418 5.872   0.540 5.872  3.122 7.225   37.7086 
273 K
 0.912kJ/K
S gen  S A  S B 
7-129
Chapter 7 Entropy
7-157 Heat is transferred steadily to boiling water in a pan through its bottom. The rate of entropy
generation within the bottom plate is to be determined.
Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant at
the specified values.
Analysis We take the bottom of the pan to be the system,
which is a closed system. Under steady conditions, the rate
form of the entropy balance for this system can be expressed as
S in  S out


Rate of net entropy transfer
by heat and mass

S gen

Rate of entropy
generation
0
 S system  0

104C
Rate of change
of entropy
500 W
Q in
Q
 out  S gen,system  0
Tb ,in Tb ,out
105C
800 W 800 W 

 S gen,system  0  S gen,system  0.0056 W/K
378 K 377 K
Discussion Note that there is a small temperature drop across the bottom of the pan, and thus a small
amount of entropy generation.
7-130
Chapter 7 Entropy
7-158 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise
the water temperature to a specified temperature and the entropy generated during this process are to be
determined.
Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in
the container itself and the heater is negligible. 3 Heat loss from the container is negligible.
Properties The specific heat of water at room temperature is C = 4.18 kJ/kg·°C (Table A-3).
Analysis Taking the water in the container as the system, which is a closed system, the energy balance can
be expressed as
E  Eout

Esystem
in




Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
Water
40 kg
We,in  (U ) water
W e,in t  mC (T2  T1 ) water
Substituting,
Heater
(800 J/s)t = (40 kg)(4180 J/kg·C)(80 - 20)C
Solving for t gives
t = 12,540 s = 209 min = 3.48 h
Again we take the water in the tank to be the system. Noting that no heat or mass crosses the
boundaries of this system and the energy and entropy contents of the heater are negligible, the entropy
balance for it can be expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
0  S gen  S water
Therefore, the entropy generated during this process is
S gen  S water  mC ln
T2
353 K
 40 kg 4.18 kJ/kg  K ln
 31.15 kJ/K
T1
293 K
7-131
Chapter 7 Entropy
7-159 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate.
The rate of entropy generation in the surrounding air due to this heat transfer are to be determined.
Assumptions Steady operating conditions exist.
Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closed
system.. The system extends from the outer surface of the pipe to a distance at which the temperature drops
to the surroundings temperature. In steady operation, the rate form of the entropy balance for this system
can be expressed as
S in  S out


Rate of net entropy transfer
by heat and mass

S gen

Rate of entropy
generation
0
 S system  0

80C
Rate of change
of entropy
Q in
Q
 out  S gen,system  0
Tb,in Tb,out
Q
Air, 5C
2200 W 2200 W 

 S gen,system  0  S gen,system  1.68 W/K
353 K
278 K
7-132
Chapter 7 Entropy
7-160 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratio
of the mass flow rates of the extracted steam to the feedwater and entropy generation per unit mass of
feedwater 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 Heat loss from the device to the surroundings is negligible.
Properties The properties of steam and feedwater are (Tables A-4 through A-6)
1
P1  1 MPa  h1  2827.9 kJ/kg

T1  200  C  s1  6.6940 kJ/kg  K
Steam
from
turbine
1 MPa
200C
Feedwater
h2  h f @1 MPa  762.81 kJ/kg
P2  1MPa 
 s 2  s f @1 MPa  2.1387 kJ/kg  K
sat.liquid 
T  179.91  C
3
2.5
MPa
50C
2
P3  2.5 MPa  h3  h f @50 C  209.33 kJ/kg
s  s
 0.7038 kJ/kg  K
 3
T3  50  C
f @ 50 C
4
 h4  h f @170 C  719.21 kJ/kg

T4  T2  10 C  170 C  s 4  s f @170 C  2.0419 kJ/kg  K
P4  2.5 MPa


2
sat. liquid
Analysis (a) We take the heat exchanger as the system, which is a control volume. The mass
and energy balances for this steady-flow system can be expressed in the rate form as follows:
Mass balance (for each fluid stream):
 in  m
 out  m
 system0
m
(steady)
 in  m
 out  m
1  m
2  m
 s and
0  m
3  m
4  m
 fw
m
Energy balance (for the heat exchanger):
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
 1h1  m
 3h3  m
 2 h2  m
 4 h4 (since Q  W  ke  pe  0)
m
Combining the two,
 s h2  h1   m
 fw h3  h4 
m
 fw and substituting,
Dividing by m
m s
h  h3
719.2  209.33 kJ/kg  0.247
 4


m fw h1  h2 2827.9  762.81kJ/kg
(b) The total entropy change (or entropy generation) during this process per unit mass of feedwater can be
determined from an entropy balance expressed in the rate form as
S in  S out
 S gen
 S system0  0




Rate of net entropy transfer
by heat and mass
Rate of entropy
generation
Rate of change
of entropy
m 1 s1  m 2 s 2  m 3 s 3  m 4 s 4  S gen  0
m s ( s1  s 2 )  m fw ( s 3  s 4 )  S gen  0
S gen
m fw

m s
s 2  s1   s 4  s3   0.247 2.1387  6.694   2.0419  0.7038   0.213kW/ K  kgfw 
m fw
7-133
Chapter 7 Entropy
7-161 Problem 7-160 is reconsidered. The effect of the state of the steam at the inlet to the
feedwater heater is to be investigated. The entropy of the extraction steam is assumed to be
constant at the value for 1 MPa, 200°C, and the extraction steam pressure is to be varied from 1
MPa to 100 kPa. Both the ratio of the mass flow rates of the extracted steam and the feedwater
heater and the total entropy change for this process per unit mass of the feedwater are to be
plotted as functions of the extraction pressure.
"Knowns:"
WorkFluid$ = 'Steam'
P[3] = 1000"[kPa]" "place {} around P[3] and T[3] equations to solve the table"
T[3] = 200"[C]"
P[4] = P[3]"[kPa]"
x[4]=0
T[4]=temperature(WorkFluid$,P=P[4],x=x[4])"[C]"
P[1] = 2500 "[kPa]"
T[1] = 50"[C]"
P[2] = 2500 "[kPa]"
T[2] = T[4] - 10"[C]"
"Since we don't know the mass flow rates and we want to determine the ratio of mass flow
rate of the extracted steam and the feedwater, we can assume the mass flow rate of the
feedwater is 1 kg/s without loss of generality. We write the conservation of energy."
"Conservation of mass for the steam extracted from the turbine: "
m_dot_steam[3]= m_dot_steam[4]
"Conservation of mass for the condensate flowing through the feedwater heater:"
m_dot_fw[1] = 1 "[kg/s]"
m_dot_fw[2]= m_dot_fw[1]"[kg/s]"
"Conservation of Energy - SSSF energy balance for the feedwater heater -- neglecting the
change in potential energy, no heat transfer, no work:"
h[3]=enthalpy(WorkFluid$,P=P[3],T=T[3])"[kJ/kg]"
"To solve the table, place {} around s[3] and remove them from the 2nd and 3rd equations"
s[3]=entropy(WorkFluid$,P=P[3],T=T[3])"[kJ/kg-K]"
{s[3] =6.693"[kJ/kg-K]" "This s[3] is for the initial T[3], P[3]"
T[3]=temperature(WorkFluid$,P=P[3],s=s[3]) "Use this equation for T[3] only when s[3] is
given."}
h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])"[kJ/kg]"
s[4]=entropy(WorkFluid$,P=P[4],x=x[4])"[kJ/kg-K]"
h[1]=enthalpy(WorkFluid$,P=P[1],T=T[1])"[kJ/kg]"
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])"[kJ/kg-K]"
h[2]=enthalpy(WorkFluid$,P=P[2],T=T[2])"[kJ/kg]"
s[2]=entropy(WorkFluid$,P=P[2],T=T[2])"[kJ/kg-K]"
"For the feedwater heater:"
E_dot_in = E_dot_out"[kW]"
E_dot_in = m_dot_steam[3]*h[3] +m_dot_fw[1]*h[1] "[kW]"
E_dot_out= m_dot_steam[4]*h[4] + m_dot_fw[2]*h[2] "[kW]"
m_ratio = m_dot_steam[3]/ m_dot_fw[1]
"Second Law analysis:"
S_dot_in - S_dot_out + S_dot_gen = DELTAS_dot_sys"[kW/K]"
DELTAS_dot_sys = 0 "[KW/K]" "steady-flow result"
S_dot_in = m_dot_steam[3]*s[3] +m_dot_fw[1]*s[1] "[kW/K]"
S_dot_out= m_dot_steam[4]*s[4] + m_dot_fw[2]*s[2] "[kW/K]"
S_gen_PerUnitMassFWH = S_dot_gen/m_dot_fw[1]"[kJ/kg_fw-K]"
7-134
Chapter 7 Entropy
m
S
0.2191
0.2222
0.2255
0.2287
0.2318
0.2349
0.2378
0.2407
0.2435
0.2462
[kJ/kg-K]
0.1799
0.1838
0.188
0.1921
0.196
0.2
0.2038
0.2076
0.2113
0.215
P
[kPa]
732
760
790
820
850
880
910
940
970
1000
0.215
P3 = 1000 kPa
Sgen,PerUnitMassFWH [kJ/kgfw-K]
0.21
0.205
0.2
For P3 < 732 kPa
0.195
0.19
Sgen < 0
0.185
0.18
0.175
0.215
P3 = 732 kPa
0.22
0.225
0.23
0.235
mratio
7-135
0.24
0.245
0.25
Chapter 7 Entropy
7-162E A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and is
charged until the tank contains saturated liquid at a specified pressure. The mass of R-134a that entered the
tank, the heat transfer with the surroundings at 120F, and the entropy generated during this process are to
be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the
process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains
constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The
direction of heat transfer is to the tank (will be verified).
Properties The properties of R-134a are (Tables A-11 through A-13)
v1  v g @120psia  0.3941ft 3 /lbm
P1  120 psia 
 u1  u g @120psia  105.06 Btu/lbm
sat.vapor 
s1  s g @120psia  0.2165 Btu/lbm  R
v 2  v f @140psia  0.01386 ft 3 /lbm
P2  140 psia 
 u 2  u f @140psia  44.07 Btu/lbm
sat.liquid 
s 2  s f @140psia  0.0902 Btu/lbm  R
R-134a
160 psia
80F
R-134a
120F
3 ft3
Q
Pi  160 psia  hi  h f @ 80F  37.27 Btu/lbm

Ti  80  F  s i  s f @ 80F  0.0774 Btu/lbm  R
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.
Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and
internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed
as
Mass balance:
min  mout  msystem  mi  m2  m1
Energy balance:
E  Eout
in


Net energy transfer
by heat, work, and mass

Esystem



Change in internal, kinetic,
potential, etc. energies
Qin  mi hi  m2 u2  m1u1 (since W  ke  pe  0)
The initial and the final masses in the tank are
V
3 ft 3

 7.61 lbm
v1 0.3941 ft 3 / lbm
V
3 ft 3
m2 

 216.45 lbm
v2 0.01386 ft 3 / lbm
m1 
Then from the mass balance,
mi  m2  m1  216.45  7.61  208.84 lbm
(b) The heat transfer during this process is determined from the energy balance to be
Qin  mi hi  m2 u 2  m1u1
 208.84 lbm 37.27 Btu/lbm   216.45 lbm 44.07 Btu/lbm   7.61 lbm 105.06 Btu/lbm 
 956Btu
7-136
Chapter 7 Entropy
(c) The entropy generated during this process is determined by applying the entropy balance on an extended
system that includes the tank and its immediate surroundings so that the boundary temperature of the
extended system is the temperature of the surroundings at all times. The entropy balance for it can be
expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
Qin
 mi s i  S gen  S tank  m 2 s 2  m1 s1
Tb,in
Therefore, the total entropy generated during this process is
S gen  mi s i  (m 2 s 2  m1 s1 ) 
Qin
Tb,in
 208 .84 0.0774   216 .45 0.0902   7.610.2165  
 0.0637Btu/R
7-137
956 Btu
580 R
Chapter 7 Entropy
7-163 It is to be shown that for thermal energy reservoirs, the entropy change relation S  mC ln(T2 / T1 )
reduces to S  Q / T as T2  T1 .
Analysis Consider a thermal energy reservoir of mass m, specific heat C, and initial temperature T 1. Now
heat, in the amount of Q, is transferred to this reservoir. The first law and the entropy change relations for
this reservoir can be written as
Q  mC T2  T1  
 mC 
and
S  mC ln
Q
T2  T1
T2
ln T2 / T1 
Q
T1
T2  T1
Q
Taking the limit as T2  T1 by applying the L'Hospital's rule,
Thermal energy
reservoir
m, C, T
1 / T1 Q
S  Q

1
T1
which is the desired result.
7-164 The inner and outer glasses of a double pane window are at specified temperatures. The rates of
entropy transfer through both sides of the window and the rate of entropy generation within the window are
to be determined.
Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constant
at the specified values.
Analysis The entropy flows associated with heat transfer through the left and right glasses are
S left 
Q left
Tleft
S right 

Q right
Tright
110 W
 0.378 W/K
291 K

110 W
 0.394 W/K
279 K
18C
We take the double pane window as the system, which is a closed system.
In steady operation, the rate form of the entropy balance for this system can be
expressed as
S in  S out


Rate of net entropy transfer
by heat and mass

S gen

Rate of entropy
generation
 S system  0

0
Rate of change
of entropy
Q in
Q
 out  S gen,system  0
Tb,in Tb ,out
110 W 110 W 

 S gen,system  0  S gen,system  0.016 W/K
291 K 279 K
7-138
Air
6C
·
Q
Chapter 7 Entropy
7-165 A well-insulated room is heated by a steam radiator, and the warm air is distributed by a fan. The
average temperature in the room after 30 min, the entropy changes of steam and air, and the entropy
generated during this process are to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The kinetic and
potential energy changes are negligible. 3 The air pressure in the room remains constant and thus the air
expands as it is heated, and some warm air escapes.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, Cp = 1.005 kJ/kg.K for air
at room temperature (Table A-2).
Analysis We first take the radiator as the system. This is a closed system since no mass enters or leaves.
The energy balance for this closed system can be expressed as
E  Eout
in


Net energy transfer
by heat, work, and mass
Esystem



Change in internal, kinetic,
potential, etc. energies
Qout  U  m(u2  u1 )
(since W  KE = PE = 0)
10C
4m  4m  5m
Qout  m(u1  u2 )
Using data from the steam tables (Tables A-4 through A-6),
some properties are determined to be
Steam
radiator
3
P1  200 kPa  v1  1.0803 m /kg
 u1  2654.4 kJ/kg
T1  200  C 
s1  7.5066 kJ/kg.K
v  0.001043 , v g  1.6940 m 3 /kg
P2  100 kPa  f
u  417 .46, u fg  2088.7 kJ/kg
v 2  v1   f
s f  1.3026 ,
s fg  6.0568 kJ/kg.K
x2 
v2  v f
v fg

1.0803  0.001043
 0.6375
1.6940  0.001043
u 2  u f  x 2 u fg  417.36  0.6375  2088.7  1748.9 kJ/kg
m
V1
0.015 m 3

 0.0139 kg
v1 1.0803 m 3 /kg
Substituting,
Qout = (0.0139 kg)( 2654.4 - 1749.0)kJ/kg = 12.6 kJ
The volume and the mass of the air in the room are V = 445 = 80 m³ and
m air 
P1V1
100 kPa (80 m 3 )

 98.5 kg
RT1 (0.2870 kPa  m 3 /kg  K)283 K 
The amount of fan work done in 30 min is
Wfan,in  W fan,in t  (0120
.
kJ / s)(30  60 s)  216 kJ
We now take the air in the room as the system. The energy balance for this closed system is expressed as
Ein  Eout  Esystem
Qin  W fan,in  Wb,out  U
Qin  Wfan,in  H  mC p (T2  T1 )
7-139
Chapter 7 Entropy
since the boundary work and U combine into H for a constant pressure expansion or compression
process.
Substituting,
(12.6 kJ) + (216 kJ) = (98.5 kg)(1.005 kJ/kgC)(T2 - 10)C
which yields
T2 = 12.3C
Therefore, the air temperature in the room rises from 10C to 12.3C in 30 mi.
(b) The entropy change of the steam is
S steam  ms 2  s1   0.0139 kg 5.1638  7.5066 kJ/kg  K  0.0326kJ/K
(c) Noting that air expands at constant pressure, the entropy change of the air in the room is
S air  mC p ln
T2
P
 mR ln 2
T1
P1
0
 98.5kg 1.005kJ/kg  K ln
285.3 K
 0.8013 kJ/K
283 K
We take the air in the room (including the steam radiator) as our system, which is a closed system. Noting
that no heat or mass crosses the boundaries of this system, the entropy balance for it can be expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
0  S gen  S steam  S air
Substituting, the entropy generated during this process is determined to be
S gen  S steam  S air  0.0326  0.8013  0.7687 kJ/K
7-140
Chapter 7 Entropy
7-166 The heating of a passive solar house at night is to be assisted by solar heated water. The length of
time that the electric heating system would run that night and the amount of entropy generated that night are
to be determined.
Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in
the glass containers themselves is negligible relative to the energy stored in water. 3 The house is
maintained at 22°C at all times.
Properties The density and specific heat of water at room temperature are  = 1 kg/L and C = 4.18
kJ/kg·°C (Table A-3).
Analysis The total mass of water is
50,000 kJ/h
mw  V  1kg/L50  20 L  1000 kg
Taking the contents of the house, including the water as our system,
the energy balance relation can be written as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



22C
water
Change in internal, kinetic,
potential, etc. energies
80C
We,in  Qout  U  (U ) water  (U ) air
 (U ) water
 mC(T2  T1 ) water
or,
W e,in t  Qout  [ mC(T2  T1 )]water
Substituting,
(15 kJ/s)t - (50,000 kJ/h)(10 h) = (1000 kg)(4.18 kJ/kg·C)(22 - 80)C
It gives
t = 17,170 s = 4.77 h
We take the house as the system, which is a closed system. The entropy generated during this
process is determined by applying the entropy balance on an extended system that includes the house and its
immediate surroundings so that the boundary temperature of the extended system is the temperature of the
surroundings at all times. The entropy balance for the extended system can be expressed as
S in  S out


Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
 S gen  S water  S air 0  S water
Tb,out
since the state of air in the house remains unchanged. Then the entropy generated during the 10-h period
that night is

Q

 out
 water Tsurr
295 K 500,000 kJ
 1000 kg 4.18 kJ/kg  K ln

353 K
276 K
 750  1811  1061 kJ/K
S gen  S water 
Qout 
T
  mC ln 2
Tb ,out 
T1
7-141
Chapter 7 Entropy
7-167E A steel container that is filled with hot water is allowed to cool to the ambient temperature. The
total entropy generated during this process is to be determined.
Assumptions 1 Both the water and the steel tank are incompressible substances with constant specific heats
at room temperature. 2 The system is stationary and thus the kinetic and potential energy changes are zero.
3 Specific heat of iron can be used for steel. 4 There are no work interactions involved.
Properties The specific heats of water and the iron at room temperature are Cp, water = 1.00 Btu/lbm.F
Cp, iron = 0.107 Btu/lbm.C (Table A-3E). The density of water at room temperature is 62.1 lbm/ft³.
and
Analysis The mass of the water is
mwater   V  (62.1 lbm / ft 3 )(15 ft 3 )  931.5 lbm
We take the steel container and the water in it as the system, which is a closed system. The energy balance
on the system can be expressed as
E  Eout
in


Net energy transfer
by heat, work, and mass
Esystem



Steel
Change in internal, kinetic,
potential, etc. energies
Qout  U  Ucontainer  Uwater
WATER
120F
 [ mC(T2  T1 )]container  [ mC(T2  T1 )]water
Q
70F
Substituting, the heat loss to the surrounding air is determined to be
Qout  [mC (T1  T2 )]container  [mC (T1  T2 )] water
 (75 lbm )(0.107 Btu/lbm   F)(120  70)  F  (931.5 lbm )(1.00 Btu/lbm   F)(120  70)  F
 46,976 Btu
We again take the container and the water In it as the system. The entropy generated during this
process is determined by applying the entropy balance on an extended system that includes the container
and its immediate surroundings so that the boundary temperature of the extended system is the temperature
of the surrounding air at all times. The entropy balance for the extended system can be expressed as
S in  S out


Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
 S gen  S container  S water
Tb,out
where
S container  mC ave ln
S water  mC ave ln
T2
530 R
 75 lbm 0.107 Btu/lbm  R ln
 0.72 Btu/R
T1
580 R
T2
530 R
 931.5 lbm 1.00 Btu/lbm  R ln
 83.98 Btu/R
T1
580 R
Therefore, the total entropy generated during this process is
S gen  S container  S water 
Qout
46,976 Btu
 0.72  83.98 
 3.93 Btu/R
Tb,out
70  460 R
7-142
Chapter 7 Entropy
7-168 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates,
the exit temperature of air and the rate of entropy generation are to be determined for the cases of an
insulated and uninsulated evaporator. 
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential
energy changes are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specific
heats at room temperature.
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat of
air at room temperature is Cp = 1.005 kJ/kg.K (Table A-2). The properties of R-134a at the inlet and the exit
states are (Tables A-11 through A-13)
P1  120 kPa  h1  h f  x1h fg  21.32  0.3  212.54  85.08 kJ/kg

x1  0.3
 s1  s f  x1 s fg  0.0879  0.3(0.9354  0.0879 )  0.3422 kJ/kg  K
T2  120 kPa  h2  h g @120 kPa  233 .86 kJ/kg

sat.vapor
 s 2  h g @120 kPa  0.9354 kJ/kg  K
Analysis (a) The mass flow rate of air is
P V
100 kPa  6m 3 /min
m air  3 3 
 6.97 kg/min
RT3
0.287 kPa  m 3 /kg  K 300 K 




We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances
for this steady-flow system can be expressed in the rate form as
Mass balance ( for each fluid stream):
 in  m
 out  m
 system0 (steady)  0 m
 in  m
 out  m
1  m
2 m
 air and m
3  m
4 m
R
m
Energy balance (for the entire heat exchanger):
E  E out
in



Rate of net energy transfer
by heat, work, and mass
Esystem0 (steady)

6 m3/min
0
R-134a
Rate of change in internal,kinetic,
potential,etc. energies
AIR
3
1
Ein  E out
2 kg/min
m 1h1  m 3h3  m 2h2  m 4h4 (since Q  W  ke  pe  0)
Combining the two,
Solving for T4,
 R h2  h1   m
 air h3  h4   m
 air C p T3  T4 
m
T4  T3 
2
4
sat. vapor
m R h2  h1 
m air C p
(2 kg/min)(23 3.86  85 .08) kJ/kg
 15.5C = 257.5 K
(6.97 kg/min)(1. 005 kJ/kg  K)
Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this
steady-flow system can be expressed as
S in  S out
 S gen
 S system0 (steady)




Substituting,
T4  27 C 
Rate of net entropy transfer
by heat and mass
Rate of entropy
generation
Rate of change
of entropy
m 1 s1  m 3 s 3  m 2 s 2  m 4 s 4  S gen  0 (since Q  0)
m R s1  m air s 3  m R s 2  m air s 4  S gen  0
or,
S gen  m R s 2  s1   m air s 4  s 3 
where
7-143
Chapter 7 Entropy
0
T
P
T
257.5 K
s 4  s 3  C p ln 4  R ln 4  C p ln 4  (1.005 kJ/kg  K) ln
 0.154 kJ/kg  K
T3
P3
T3
300 K
Substituting,
S gen  2 kg/min 0.9354 - 0.3422 kJ/kg  K   (6.97 kg/min)(0. 154 kJ/kg  K)
 0.113 kJ/min  K
 0.00188 kW/K
(b) When there is a heat gain from the surroundings at a rate of 30 kJ/min, the steady-flow energy equation
reduces to
Q in  m R h2  h1   m air C p T4  T3 
Solving for T4,
T4  T3 
Q in  m R h2  h1 
m air C p
(30 kJ / min)  (2 kg / min)(233.86  85.08) kJ / kg
 11.2 C
(6.97 kg / min)(1.005 kJ / kg  K)
The entropy generation in this case is determined by applying the entropy balance on an extended
system that includes the evaporator and its immediate surroundings so that the boundary temperature of the
extended system is the temperature of the surrounding air at all times. The entropy balance for the extended
system can be expressed as
Substituting,
T4  27 C +
S in  S out


Rate of net entropy transfer
by heat and mass
S gen


0 (steady)
 S system

Rate of entropy
generation
Rate of change
of entropy
Qin
 m 1 s1  m 3 s 3  m 2 s 2  m 4 s 4  S gen  0
Tb,out
Qin
 m R s1  m air s 3  m R s 2  m air s 4  S gen  0
Tsurr
or
Q
 R s 2  s1   m
 air s 4  s 3   in
S gen  m
T0
0
T
P
261.8 K
 0137
.
kJ / kg  K
where s 4  s3  C p ln 4  R ln 4  (1.005 kJ / kg  K) ln
T3
P3
300 K
Substituting,
30kJ/min
S gen  2kg/min 0.9354  0.3422 kJ/kg  K  6.97 kg/min  0.137 kJ/kg  K  
305K
 0.13315 kJ/min  K
 0.00222 kW/K
Discussion Note that the rate of entropy generation in the second case is greater because of the
irreversibility associated with heat transfer between the evaporator and the surrounding air.
7-144
Chapter 7 Entropy
7-169 A room is to be heated by hot water contained in a tank placed in the room. The minimum initial
temperature of the water needed to meet the heating requirements of this room for a 24-h period and the
entropy generated are to be determined.
Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with
constant specific heats. 3 The energy stored in the container itself is negligible relative to the energy stored
in water. 4 The room is maintained at 20°C at all times. 5 The hot water is to meet the heating requirements
of this room for a 24-h period.
Properties The specific heat of water at room temperature is C = 4.18 kJ/kg·°C (Table A-3).
Analysis Heat loss from the room during a 24-h period is
Qloss = (10,000 kJ/h)(24 h) = 240,000 kJ
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
E system



  Qout  U  U water  U air 0
Change in internal,kinetic,
potential,etc. energies
10,000 kJ/h
or
-Qout = [mC(T2 - T1)]water
Substituting,
20C
-240,000 kJ = (1500 kg)(4.18 kJ/kg·C)(20 - T1)
It gives
water
T1 = 58.3C
where T1 is the temperature of the water when it is first brought into the room.
(b) We take the house as the system, which is a closed system. The entropy generated during this process is
determined by applying the entropy balance on an extended system that includes the house and its
immediate surroundings so that the boundary temperature of the extended system is the temperature of the
surroundings at all times. The entropy balance for the extended system can be expressed as
S in  S out


Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
 S gen  S water  S air 0  S water
Tb,out
since the state of air in the house (and thus its entropy) remains unchanged. Then the entropy generated
during the 24 h period becomes

Q

 out
 water Tsurr
293 K
240,000 kJ
 1500 kg 4.18 kJ/kg  K ln

331.3 K
278 K
 770 .3  863 .3  93.0 kJ/K
S gen  S water 
Qout 
T
  mC ln 2
Tb ,out 
T1
7-145
Chapter 7 Entropy
7-170 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other
side contains He gas at different states. The final equilibrium temperature in the cylinder and the entropy
generated are to be determined for the cases of the piston being fixed and moving freely.
Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the
container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible.
Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K, Cv =
0.743 kJ/kg·°C and Cp =1.039 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K, Cv = 3.1156 kJ/kg·°C, and
Cp = 5.1926 kJ/kg·°C for He (Tables A-1 and A-2)
Analysis The mass of each gas in the cylinder is
 
 PV 
500 kPa  1m 3
m N 2   1 1  
 4.77 kg
0.2968 kPa  m 3 /kg  K 353K 
 RT1  N 2

m He
 

N2
1 m3
500 kPa
80C
 P1V1 

500 kPa  1m 3



 0.808 kg
3
 
 RT1  He 2.0769 kPa  m /kg  K 298 K 


He
1 m3
500 kPa
25C
Taking the entire contents of the cylinder as our system, the 1st law relation can be written as
E  E out

E system
in




Net energy transfer
by heat, work, and mass
Change in internal,kinetic,
potential,etc. energies
0  U  U  N 2  U He
0  [mC v (T2  T1 )] N 2  [mC v (T2  T1 )] He
Substituting,
4.77 kg 0.743 kJ/kg  CT f
It gives




 80  C  0.808 kg  3.1156 kJ/kg   C T f  25  C  0
Tf = 57.2C
where Tf is the final equilibrium temperature in the cylinder.
The answer would be the same if the piston were not free to move since it would effect only
pressure, and not the specific heats.
(b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
0  S gen  S N 2  S He
But first we determine the final pressure in the cylinder:
4.77 kg
0.808 kg
m
m
N total  N N 2  N He       

 0.372 kmol
 M  N 2  M  He 28 kg/kmol 4kg/kmol
P2 


N total Ru T 0.372 kmol  8.314kPa  m 3 /kmol  K 330.2 K 

 510.6 kPa
Vtotal
2m 3
Then,

T
P 
S N 2  m C p ln 2  R ln 2 
T
P1  N
1

2

330.2 K
510.6 kPa 
 4.77 kg 1.039 kJ/kg  K ln
 0.2968 kJ/kg  K ln
  0.361 kJ/K
353 K
500 kPa 

7-146
Chapter 7 Entropy

T
P 
S He  m C p ln 2  R ln 2 
T1
P1  He


330.2 K
510.6 kPa 
 0.808 kg 5.1926 kJ/kg  K ln
 2.0769 kJ/kg  K ln
  0.395 kJ/K
298 K
500 kPa 

S gen  S N 2  S He  0.361  0.395  0.034kJ/K
If the piston were not free to move, we would still have T 2 = 330.2 K but the volume of each gas would
remain constant in this case:
0


T
V
330.2 K
S N 2  m C v ln 2  R ln 2   4.77 kg 0.743 kJ/kg  K ln
 0.237 kJ/K


T
V
353
K
1
1

 N2
S He
0

T2
V2 
330.2 K

 m C v ln
 R ln
 0.808 kg 3.1156 kJ/kg  K ln
 0.258 kJ/K


T1
V1 
298 K

He
S gen  S N 2  S He  0.237  0.258  0.021 kJ/K
7-147
Chapter 7 Entropy
7-171 Problem 7-170 is reconsidered. The results for constant specific heats to those obtained
using variable specific heats are to be compared using built-in EES or other functions.
"Knowns:"
R_u=8.314"[kJ/kmol-K]"
V_N2[1]=1"[m^3]"
Cv_N2=0.743"[kJ/kg-K]" "From Table A-2(a) at 27C"
R_N2=0.2968"[kJ/kg-K]" "From Table A-2(a)"
T_N2[1]=80"[C]"
P_N2[1]=500"[kPa]"
Cp_N2=R_N2+Cv_N2"[kJ/kg-K]"
V_He[1]=1"[m^3]"
Cv_He=3.1156"[kJ/kg-K]" "From Table A-2(a) at 27C"
T_He[1]=25"[C]"
P_He[1]=500"[kPa]"
R_He=2.0769"[kJ/kg-K]" "From Table A-2(a)"
Cp_He=R_He+Cv_He"[kJ/kg-K]"
"Solution:"
"mass calculations:"
P_N2[1]*V_N2[1]=m_N2*R_N2*(T_N2[1]+273)
P_He[1]*V_He[1]=m_He*R_He*(T_He[1]+273)
"The entire cylinder is considered to be a closed system, allowing the piston to move."
"Conservation of Energy for the closed system:"
"E_in - E_out = DELTAE, we neglect DELTA KE and DELTA PE for the cylinder."
E_in - E_out = DELTAE
E_in =0"[kJ]"
E_out = 0"[kJ]"
"At the final equilibrium state, N2 and He will have a common temperature."
DELTAE= m_N2*Cv_N2*(T_2-T_N2[1])+m_He*Cv_He*(T_2-T_He[1])"[kJ]"
"Total volume of gases:"
V_total=V_N2[1]+V_He[1]"[m^3]"
MM_He = 4"[kg/kmol]"
MM_N2 = 28"[kg/kmol]"
N_total = m_He/MM_He+m_N2/MM_N2
"Final pressure at equilibrium:"
"Allowing the piston to move, the pressure on both sides is the same, P_2 is:"
P_2*V_total=N_total*R_u*(T_2+273)"[kPa]"
S_gen_PistonMoving = DELTAS_He_PM+DELTAS_N2_PM"[kJ/K]"
DELTAS_He_PM=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))R_He*ln(P_2/P_He[1]))"[kJ/K]"
DELTAS_N2_PM=m_N2*(Cp_N2*ln((T_2+273)/(T_N2[1]+273))R_N2*ln(P_2/P_N2[1]))"[kJ/K]"
"The final temperature of the system when the piston does not move will be the same as
when it does move. The volume of the gases remain constant and the entropy changes a re
given by:"
S_gen_PistNotMoving = DELTAS_He_PNM+DELTAS_N2_PNM"[kJ/K]"
DELTAS_He_PNM=m_He*(Cv_He*ln((T_2+273)/(T_He[1]+273)))"[kJ/K]"
DELTAS_N2_PNM=m_N2*(Cv_N2*ln((T_2+273)/(T_N2[1]+273)))"[kJ/K]"
"The following uses the EES functions for the nitrogen. Since helium is monatomic, we use
the constant specific heat approach to find its property changes."
E_in - E_out = DELTAE_VP
DELTAE_VP= m_N2*(INTENERGY(N2,T=T_2_VP)INTENERGY(N2,T=T_N2[1]))+m_He*Cv_He*(T_2_VP-T_He[1])"[kJ]"
7-148
Chapter 7 Entropy
"Final Pressure for moving piston:"
P_2_VP*V_total=N_total*R_u*(T_2_VP+273)"[kPa]"
S_gen_PistMoving_VP = DELTAS_He_PM_VP+DELTAS_N2_PM_VP"[kJ/K]"
DELTAS_N2_PM_VP=m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_VP)ENTROPY(N2,T=T_N2[1],P=P_N2[1]))"[kJ/K]"
DELTAS_He_PM_VP=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))R_He*ln(P_2/P_He[1]))"[kJ/K]"
"Final N2 Pressure for piston not moving."
P_2_N2_VP*V_N2[1]=m_N2*R_N2*(T_2_VP+273)
S_gen_PistNotMoving_VP = DELTAS_He_PNM_VP+DELTAS_N2_PNM_VP"[kJ/K]"
DELTAS_N2_PNM_VP = m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_N2_VP)ENTROPY(N2,T=T_N2[1],P=P_N2[1]))"[kJ/K]"
DELTAS_He_PNM_VP=m_He*(Cv_He*ln((T_2_VP+273)/(T_He[1]+273)))"[kJ/K]"
SOLUTION
Variables in Main
Cp_He=5.193 [kJ/kg-K]
Cp_N2=1.04 [kJ/kg-K]
Cv_He=3.116 [kJ/kg-K]
Cv_N2=0.743 [kJ/kg-K]
DELTAE=0 [kJ]
DELTAE_VP=0 [kJ]
DELTAS_He_PM=0.3931 [kJ/K]
DELTAS_He_PM_VP=0.3931 [kJ/K]
DELTAS_He_PNM=0.258 [kJ/K]
DELTAS_He_PNM_VP=0.2583 [kJ/K]
DELTAS_N2_PM=-0.363 [kJ/K]
DELTAS_N2_PM_VP=-0.3631 [kJ/K]
DELTAS_N2_PNM=-0.2371 [kJ/K]
DELTAS_N2_PNM_VP=-0.2372 [kJ/K]
E_in=0 [kJ]
E_out=0 [kJ]
MM_He=4 [kg/kmol]
MM_N2=28 [kg/kmol]
m_He=0.8079 [kg]
m_N2=4.772 [kg]
N_total=0.3724 [kmol]
P_2=511.1 [kPa]
P_2_N2_VP=467.7
P_2_VP=511.2
P_He[1]=500 [kPa]
P_N2[1]=500 [kPa]
R_He=2.077 [kJ/kg-K]
R_N2=0.2968 [kJ/kg-K]
R_u=8.314 [kJ/kmol-K]
S_gen_PistMoving_VP=0.02993 [kJ/K]
S_gen_PistNotMoving=0.02089 [kJ/K]
S_gen_PistNotMoving_VP=0.02106 [kJ/K]
S_gen_PistonMoving=0.03004 [kJ/K]
T_2=57.17 [C]
T_2_VP=57.2 [C]
T_He[1]=25 [C]
T_N2[1]=80 [C]
V_He[1]=1 [m^3]
V_N2[1]=1 [m^3]
V_total=2 [m^3]
7-149
Chapter 7 Entropy
7-172 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other
side contains He gas at different states. The final equilibrium temperature in the cylinder and the entropy
generated are to be determined for the cases of the piston being fixed and moving freely.
Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the
container itself, except the piston, is negligible. 3 The cylinder is well-insulated and thus heat transfer is
negligible. 4 Initially, the piston is at the average temperature of the two gases.
Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K, Cv =
0.743 kJ/kg·°C and Cp =1.039 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K, Cv = 3.1156 kJ/kg·°C, and
Cp = 5.1926 kJ/kg·°C for He (Tables A-1 and A-2). The specific heat of the copper at room temperature is
C = 0.386 kJ/kg·°C (Table A-3).
Analysis The mass of each gas in the cylinder is
 
N2
1 m3
500 kPa
80C
 PV 
500 kPa  1m
m N 2   1 1  
 4.77 kg
RT
0.2968 kPa  m 3 /kg  K 353 K 
 1  N2
3

 

 PV 
500 kPa  1m
m He   1 1  
 0.808 kg
3
RT
 1  He 2.0769 kPa  m /kg  K 353 K 
He
1 m3
500 kPa
25C
3


Copper
Taking the entire contents of the cylinder as our system, the 1st law relation can be written as
E  E out
in


Net energy transfer
by heat, work, and mass
E system



Change in internal,kinetic,
potential,etc. energies
0  U  U  N 2  U He  U Cu
0  [mC v (T2  T1 )] N 2  [mC v (T2  T1 )] He  [mC (T2  T1 )] Cu
where
T1, Cu = (80 + 25) / 2 = 52.5C
Substituting,
4.77 kg 0.743 kJ/kg  CT f  80  C  0.808 kg 3.1156 kJ/kg  CT f
 5.0 kg 0.386 kJ/kg   CT f  52.5  C  0

 25  C
It gives
Tf = 56.0C
where Tf is the final equilibrium temperature in the cylinder.
The answer would be the same if the piston were not free to move since it would effect only
pressure, and not the specific heats.
(b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
0  S gen  S N 2  S He  S piston
But first we determine the final pressure in the cylinder:
4.77 kg
0.808 kg
m
m
N total  N N 2  N He       

 0.372 kmol
M
M
28
kg/kmol
4
kg/kmol
  N 2   He
P2 


N total Ru T 0.372 kmol  8.314kPa  m 3 /kmol  K 329 K 

 508.8 kPa
Vtotal
2m 3
Then,
7-150
Chapter 7 Entropy

T
P 
S N 2  m C p ln 2  R ln 2 
T1
P1  N

2

329 K
508.8 kPa 
 4.77 kg 1.039 kJ/kg  K ln
 0.2968 kJ/kg  K ln
  0.374 kJ/K
353K
500 kPa 


T
P 
S He  m C p ln 2  R ln 2 
T1
P1  He


329 K
508.8 kPa 
 0.808 kg 5.1926 kJ/kg  K ln
 2.0769 kJ/kg  K ln
  0.386 kJ/K
298 K
500 kPa 


T 
329 K
S piston   mC ln 2 
 5kg 0.386 kJ/kg  K ln
 0.021kJ/K
T1  piston
325.5 K

S gen  S N 2  S He  S piston  0.374  0.386  0.021  0.033 kJ/K
If the piston were not free to move, we would still have T2 = 329 K but the volume of each gas would
remain constant in this case:
0


T
V
329 K
S N 2  m C v ln 2  R ln 2   4.77 kg 0.743kJ/kg  K ln
 0.250 kJ/K


T
V
353
K
1
1

 N2
0


T
V
329 K
S He  m C v ln 2  R ln 2   0.808 kg 3.1156 kJ/kg  K ln
 0.249 kJ/K


T1
V1 
298 K

He
S gen  S N 2  S He  S piston  0.250  0.249  0.021  0.020kJ/K
7-151
Chapter 7 Entropy
7-173 An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is
opened, and air is allowed to escape at constant temperature until the pressure inside drops to a specified
value. The amount of electrical work done during this process and the total entropy change are to be
determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the
process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air
remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat
transfer is negligible. 4 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R =0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A17)
Te  330 K 
 he  330.34 kJ/kg
T1  330 K 
 u1  235.61 kJ/kg
T2  330 K 
 u 2  235.61 kJ/kg
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting
that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal
energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as
min  mout  msystem
Mass balance:
E  Eout
in


Energy balance:


me  m1  m2
Esystem

 

Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
We,in  me he  m2 u2  m1u1 (since Q  ke  pe  0)
AIR
5 m3
500 kPa
57C
The initial and the final masses of air in the tank are



PV
200 kPa 5 m 


RT
0.287 kPa  m /kg  K 330 K   10.56 kg
m1 

P1V
500 kPa  5 m 3

 26.40 kg
RT1 0.287 kPa  m 3 /kg  K 330 K 
3
m2
2
3
2
Then from the mass and energy balances,
me  m1  m2  26.40  10.56  15.84 kg
We,in  me he  m2 u 2  m1u1
 15.84 kg 330.34 kJ/kg   10.56 kg 235.61 kJ/kg   26.40 kg 235.61 kJ/kg 
 1501 kJ
(b) The total entropy change, or the total entropy generation within the tank boundaries is determined from
an entropy balance on the tank expressed as
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
 me s e  S gen  S tank
or,
S gen  m e s e  S tank  m e s e  (m 2 s 2  m1 s1 )
 m1  m 2 s e  m 2 s 2  m1 s1   m 2 s 2  s e   m1 s1  s e 
Assuming a constant average pressure of (500 + 200)/2 = 350 kPa for the exit stream, the entropy changes
are determined to be
7-152
We
Chapter 7 Entropy
s 2  s e  C p ln
T2
Te
0
 R ln
P2
P
200 kPa
  R ln 2  0.287 kJ/kg  K ln
 0.1606 kJ/kg  K
Pe
Pe
350 kPa
T1 0
P
P
500 kPa
 R ln 2   R ln 1  0.287 kJ/kg  K ln
 0.1024 kJ/kg  K
Te
Pe
Pe
350 kPa
Therefore, the total entropy generated within the tank during this process is
s1  s e  C p ln
S gen  10.56 kg 0.1606 kJ/kg  K   26.40 kg  0.1024 kJ/kg  K   4.40 kJ/K
7-153
Chapter 7 Entropy
7-174 A 1- ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium
temperature in the tank and the entropy generation are to be determined.
Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank is
negligible. 3 There is no stirring by hand or a mechanical device (it will add energy).
Properties The specific heat of water at room temperature is C = 4.18 kJ/kg·°C, and the specific heat of ice
at about 0C is C = 2.11 kJ/kg·°C (Table A-3). The melting temperature and the heat of fusion of ice at 1
atm are 0C and 333.7 kJ/kg..
Analysis (a) We take the ice and the water as the system, and disregard any heat transfer between the
system and the surroundings. Then the energy balance for this process can be written as
E  Eout
in


Net energy transfer
by heat, work, and mass
Esystem



Change in internal, kinetic,
potential, etc. energies
0  U
0  U ice  Uwater
[mC (0  C  T1 ) solid  mh if  mC (T2 0  C) liquid ] ice  [mC (T2  T1 )] water  0
ice
-5C
80 kg
Substituting,
(80 kg){(2.11 kJ / kg C)[0  (-5)] C + 333.7 kJ / kg + (4.18 kJ / kg C)(T2  0) C}
(1000 kg)(4.18 kJ / kg C)(T2  20) C  0
It gives
WATER
T2 = 12.4C
1 ton
which is the final equilibrium temperature in the tank.
(b) We take the ice and the water as our system, which is a closed system. Considering that the tank is
well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed
as
S in  S out  S gen  S system






Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
0  S gen  S ice  S water
where

T 
285.4 K
S water   mC ln 2 
 1000 kg 4.18kJ/kg  K ln
 109.9 kJ/K
T1  water
293K

S ice  S solid  S melting  S liquid ice


T
   mC ln 2

T1


mhif


T


  mC ln 2
T1
 solid Tmelting 




 liquid  ice

273K 333.7 kJ/kg
285.4 K 

 80kg  2.11kJ/kg  K ln

 4.18kJ/kg  K ln
268 K
273K
273K 

 115.8 kJ/K
Then,
S gen  S water  S ice  109.9  115.8  5.9 kJ/K
7-154
Chapter 7 Entropy
7-175 An insulated cylinder initially contains a saturated liquid-vapor mixture of water at a specified
temperature. The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the
cylinder. The amount of ice added and the entropy generation are to be determined.
Assumptions 1 Thermal properties of the ice are constant. 2 The cylinder is well-insulated and thus heat
transfer is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy).
Properties The specific heat of ice at about 0C is C = 2.11 kJ/kg·°C (Table A-3). The melting temperature
and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ/kg.
Analysis (a) We take the contents of the cylinder (ice and saturated water) as our system, which is a closed
system. Noting that the temperature and thus the pressure remains constant during this phase change process
and thus Wb + U = H, the energy balance for this system can be written as
E  Eout
in


Net energy transfer
by heat, work, and mass
Esystem



Change in internal, kinetic,
potential, etc. energies
Wb,in  U  H  0
Hice  Hwater  0
[mC (0  C  T1 ) solid  mh if  mC (T2 0  C) liquid ] ice  [m(h2  h1 )] water  0
The properties of water at 100C are (Table A-4)
v f  0.001044, vg  1.6729 m 3 / kg
h f  419.04,
h fg  2257.0 kJ.kg
Then,
ice
-5C
v1  v f  x1v fg  0.001044  0.21.6729  0.001044   0.3354 m 3 /kg
WATER
0.02 m3
100C
h1  h f  x1 h fg  419.04  0.22257.0   870.4 kJ/kg
s1  s f  x1 s fg  1.3069  0.26.048   2.5165 kJ/kg  K
h2  h f @100 C  419.04 kJ/kg
s 2  s f @100 C  1.3069 kJ/kg  K
m steam 
V1
0.02 m 3

 0.0596 kg
v1 0.3354 m 3 /kg
Noting that T1, ice = -5C and T2 = 100C and substituting gives
m{(2.11 kJ/kg.K)[0-(-5)] + 333.7 kJ/kg + (4.18 kJ/kg·C)(100-0)C} + (0.0596 kg)(419.04 – 870.4) kJ/kg=
0
m = 0.0353 kg = 35.3 g ice
(b) We take the ice and the steam as our system, which is a closed system. Considering that the tank is
well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed
as
S in  S out  S gen  S system






Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
0  S gen  S ice  S steam
7-155
Chapter 7 Entropy
where
S steam  ms 2  s1   0.0596 kg 1.3069  2.5165 kJ/kg  K  0.0721 kJ/K

S ice  S solid  S melting  S liquid

ice

T
   mC ln 2

T1

mhig


T


  mC ln 2
T1
 solid Tmelting 




 liquid  ice

273 K 333.7 kJ/kg
373 K 

 0.0353 kg  2.11 kJ/kg  K ln

 4.18 kJ/kg  K ln
268 K
273K
273 K 

 0.0906 kJ/K
Then,
S gen  S steam  S ice  0.0721  0.0906  0.0185 kJ/K
7-156
Chapter 7 Entropy
7-176 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the
bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium
is established and the amount of entropy generated are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the
process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains
constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work
interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).
Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary.
Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and
internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed
as
Mass balance: min  mout  msystem  mi  m2 (since mout  minitial  0)
E  Eout
in


Energy balance:

Net energy transfer
by heat, work, and mass
Esystem



Change in internal, kinetic,
potential, etc. energies
Qin  mi hi  m2 u2 (since W  Eout  Einitial  ke  pe  0)
Combining the two balances:
Qin  m2 u 2  hi 
where
PV
100 kPa  0.005 m 3
m2  2 
 0.0060 kg
RT2
0.287 kPa  m 3 /kg  K 290 K 



10 kPa
17C

h  290.16 kJ/kg
Ti  T2  290 K   
 i
u 2  206.91 kJ/kg
Substituting,
Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ
5L
Evacuated
Table A -17

Qout = 0.5 kJ
Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we
reverse the direction.
The entropy generated during this process is determined by applying the entropy balance on an
extended system that includes the bottle and its immediate surroundings so that the boundary temperature of
the extended system is the temperature of the surroundings at all times. The entropy balance for it can be
expressed as
Sin  Sout


Net entropy transfer
by heat and mass
mi si 
 S gen  Ssystem




Entropy
generation
Change
in entropy
Qout
 Sgen  S tank  m2 s2  m1s10  m2 s2
Tb,in
Therefore, the total entropy generated during this process is
S gen  mi s i  m 2 s 2 
Qout
Q
Q
0.5 kJ
 m 2 s 2  s i 0  out  out 
 0.0017kJ/K
Tb,out
Tb,out Tsurr 290 K
7-157
Chapter 7 Entropy
7-177 (a) Water is heated from 16°C to 43°C by an electric resistance heater placed in the water pipe as it
flows through a showerhead steadily at a rate of 10 L/min. The electric power input to the heater and the
rate of entropy generation are to be determined. The reduction in power input and entropy generation as a
result of installing a 50% efficient regenerator are also to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point within the
system and thus mCV  0 and ECV  0 . 2 Water is an incompressible substance with constant specific
heats. 3 The kinetic and potential energy changes are negligible, ke  pe  0 . 4 Heat losses from the
pipe are negligible.
Properties The density of water is given to be  = 1 kg/L. The specific heat of water at room temperature is
C = 4.18 kJ/kg·°C (Table A-3).
Analysis We take the pipe as the system. This is a control volume since mass crosses the system boundary
1  m
2  m
 . Then the
during the process. We observe that there is only one inlet and one exit and thus m
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
 E in  E out
Rate of change in internal, kinetic,
potential, etc. energies
 1  mh
 2 (since ke  pe  0)
W e,in  mh
 (T2  T1 )
W e,in  m (h2  h1 )  mC
WATER
16C
43C
where
  V  1 kg/L10 L/min  10 kg/min
m
Substituting,
W e,in  10/60 kg/s  4.18 kJ/kg   C 43  16  C  18.8 kW


The rate of entropy generation in the heating section during this process is determined by applying
the entropy balance on the heating section. Noting that this is a steady-flow process and heat transfer from
the heating section is negligible,
S in  S out
 S gen
 S system0  0




Rate of net entropy transfer
by heat and mass
Rate of entropy
generation
Rate of change
of entropy
m s1  m s 2  S gen  0
S gen  m ( s 2  s1 )
Noting that water is an incompressible substance and substituting,
T
316 K
 C ln 2  10/60 kg/s 4.18 kJ/kg  K ln
S gen  m
 0.0622 kJ/K
T1
289 K
(b) The energy recovered by the heat exchanger is
Q saved  Q max  m C Tmax  Tmin 


 0.5 10/60 kg/s  4.18 kJ/kg  C 39  16  C
 8.0 kJ/s  8.0 kW
Therefore, 8.0 kW less energy is needed in this case, and the required electric power in this case reduces to
W in,new  W in,old  Q saved  18 .8  8.0  10.8 kW
Taking the cold water stream in the heat exchanger as our control volume (a steady-flow system), the
temperature at which the cold water leaves the heat exchanger and enters the electric resistance heating
section is determined to be
 (Tc, out  Tc, in )
Q  mC
Substituting,
7-158
Chapter 7 Entropy
8 kJ / s  (10 / 60 kg / s)(4.184 kJ / kg C)(Tc, out  16  C)
It yields
Tc, out  27.5  C  300.5 K
The rate of entropy generation in the heating section in this case is determined similarly to be
T
316 K
S gen  m C ln 2  10/60 kg/s 4.18 kJ/kg  K ln
 0.0351kJ/K
T1
300.5 K
Thus the reduction in the rate of entropy generation within the heating section is
S reduction  0.0622  0.0350  0.0272 kW/K
7-159
Chapter 7 Entropy
7-178 Using EES (or other) software, the work input to a multistage compressor is to be
determined for a given set of inlet and exit pressures for any number of stages. The pressure ratio
across each stage is assumed to be identical and the compression process to be polytropic. The
compressor work is to be tabulated and plotted against the number of stages for P1 = 100 kPa,
T1 = 17°C, P2 = 800 kPa, and n = 1.35 for air.
GAS$ = 'Air'
Nstage = 2 "number of stages of compression with intercooling, each having same pressure
ratio."
n=1.35
MM=MOLARMASS(GAS$)
R_u = 8.314"[kJ/kmol-K]"
R=R_u/MM"[kJ/kg-K]"
k=1.4
P1=100"[kPa]"
T1=17"[C]"
P2=800"[kPa]"
R_p = (P2/P1)^(1/Nstage)
W_dot_comp= Nstage*n*R*(T1+273)/(n-1)*((R_p)^((n-1)/n) - 1)"[kJ/kg]"
Nstage
1
2
3
4
5
6
7
8
9
10
W
[kJ/kg]
229.4
198.7
189.6
185.3
182.8
181.1
179.9
179
178.4
177.8
230
Wcomp [kJ/kg]
220
210
200
190
180
170
1
2
3
4
5
6
Nstage
7-179 … 7-182 Design and Essay Problems
7-160
7
8
9
10