7-1
Chapter 7
ENTROPY
Entropy and the Increase of Entropy Principle
7-1C Yes. Because we used the relation (QH/TH) = (QL/TL) in the proof, which is the defining relation of
absolute temperature.
7-2C No. The
∫ δ Q represents the net heat transfer during a cycle, which could be positive.
7-3C Yes.
7-4C No. A system may reject more (or less) heat than it receives during a cycle. The steam in a steam
power plant, for example, receives more heat than it rejects during a cycle.
7-5C No. A system may produce more (or less) work than it receives during a cycle. A steam power plant,
for example, produces more work than it receives during a cycle, the difference being the net work output.
7-6C The entropy change will be the same for both cases since entropy is a property and it has a fixed
value at a fixed state.
7-7C No. In general, that integral will have a different value for different processes. However, it will have
the same value for all reversible processes.
7-8C Yes.
7-9C That integral should be performed along a reversible path to determine the entropy change.
7-10C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel work
while losing an equivalent amount of heat.
7-11C The value of this integral is always larger for reversible processes.
7-12C No. Because the entropy of the surrounding air increases even more during that process, making
the total entropy change positive.
7-13C It is possible to create entropy, but it is not possible to destroy it.
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7-2
7-14C Sometimes.
7-15C Never.
7-16C Always.
7-17C Increase.
7-18C Increases.
7-19C Decreases.
7-20C Sometimes.
7-21C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of this
heat loss is equal to the increase in entropy as a result of irreversibilities.
7-22C They are heat transfer, irreversibilities, and entropy transport with mass.
7-23C Greater than.
7-24 A rigid tank contains an ideal gas that is being stirred by a paddle wheel. The temperature of the gas
remains constant as a result of heat transfer out. The entropy change of the gas is to be determined.
Assumptions The gas in the tank is given to be an ideal gas.
Analysis The temperature and the specific volume of the gas remain
constant during this process. Therefore, the initial and the final states
of the gas are the same. Then s2 = s1 since entropy is a property.
200 kJ
Therefore,
IDEAL GAS
40°C
Heat
30°C
∆S sys = 0
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7-3
7-25 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.
Properties Noting that h = h(T) for ideal gases, we have h1 = h2 since T1 = T2 = 25°C.
Analysis We take the compressor as the system. Noting that the enthalpy of air remains constant, the
energy balance for this steady-flow system can be expressed in the rate form as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W&in = Q& out
P2
·
Q
AIR
T = const.
12 kW
Therefore,
Q& out = W&in = 12 kW
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
Q&
12 kW
∆S&air = − out,air = −
= −0.0403 kW/K
298 K
Tsys
P1
7-26 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
∆S =
Q
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
∆Sfluid =
(b)
Qfluid Qin,fluid 900 kJ
=
=
= 1.337 kJ/K
Tfluid
Tfluid
673 K
∆Ssource =
Q
Qsource
900 kJ
= − out, source = −
= −1.337 kJ/K
Tsource
Tsource
673 K
(c) Thus the total entropy change of the process is
Source
400°C
900 kJ
Sgen = ∆S total = ∆Sfluid + ∆Ssource = 1.337 − 1.337 = 0
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7-4
7-27 EES Problem 7-26 is reconsidered. The effects of the varying the heat transferred to the working fluid
and the source temperature on the entropy change of the working fluid, the entropy change of the source,
and the total entropy change for the process as the source temperature varies from 100°C to 1000°C are to
be investigated. The entropy changes of the source and of the working fluid are to be plotted against the
source temperature for heat transfer amounts of 500 kJ, 900 kJ, and1300 kJ.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
{T_H = 400 [C]}
Q_H = 1300 [kJ]
T_Sys = T_H
"Analysis:
(a) & (b) This is a reversible isothermal process, and the entropy change during such a process
is given by
DELTAS = Q/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 "
DELTAS_source = -Q_H/(T_H+273)
DELTAS_fluid = +Q_H/(T_Sys+273)
"(c) entropy generation for the process:"
S_gen = DELTAS_source + DELTAS_fluid
∆Sfluid
[kJ/K]
3.485
2.748
2.269
1.932
1.682
1.489
1.336
1.212
1.108
1.021
∆Ssource
[kJ/K]
-3.485
-2.748
-2.269
-1.932
-1.682
-1.489
-1.336
-1.212
-1.108
-1.021
Sgen
[kJ/K]
0
0
0
0
0
0
0
0
0
0
TH
[C]
100
200
300
400
500
600
700
800
900
1000
4
3.5
]
K
/
J
K
[
di
ul
f
S
∆
∆Ssource = -∆Sfluid
3
2.5
QH = 1300 kJ
2
1.5
1
0.5
QH = 900 kJ
QH = 500 kJ
0
100 200 300 400 500 600 700 800 900 1000
TH [C]
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7-5
7-28E 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.
Analysis (a) This is a reversible isothermal process, and the
Heat
entropy change during such a process is given by
∆S =
Q
T
SINK
95°F
Noting that heat transferred from the working fluid is equal to
the heat transferred to the sink, the heat transfer become
95°F
Carnot heat engine
Qfluid = Tfluid ∆Sfluid = (555 R )(−0.7 Btu/R ) = −388.5 Btu → Qfluid,out = 388.5 Btu
(b) The entropy change of the sink is determined from
∆Ssink =
Qsink,in
Tsink
=
388.5 Btu
= 0.7 Btu/R
555 R
(c) Thus the total entropy change of the process is
Sgen = ∆S total = ∆Sfluid + ∆Ssink = −0.7 + 0.7 = 0
This is expected since all processes of the Carnot cycle are reversible processes, and no entropy is
generated during a reversible process.
7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred to
the refrigerant from the cooled space, and the liquid is vaporized. The entropy change of the refrigerant, the
entropy change of the cooled space, and the total entropy change for this process are to be determined.
Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such as
friction. 2 Any temperature change occurs within the wall of the tube, and thus both the refrigerant and the
cooled space remain isothermal during this process. Thus it is an isothermal, internally reversible process.
Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, the
entropy change for them can be determined from
∆S =
Q
T
(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerant
also remains constant at the saturation value,
T = Tsat @160 kPa = −15.6°C = 257.4 K
(Table A-12)
Then,
∆S refrigerant =
Qrefrigerant,in
Trefrigerant
=
180 kJ
= 0.699 kJ/K
257.4 K
180 kJ
-5°C
(b) Similarly,
∆S space = −
R-134a
160 kPa
Qspace,out
Tspace
=−
180 kJ
= −0.672 kJ/K
268 K
(c) The total entropy change of the process is
Sgen = S total = ∆S refrigerant + ∆Sspace = 0.699 − 0.672 = 0.027 kJ/K
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7-6
Entropy Changes of Pure Substances
7-30C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.
7-31 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),
P1 = 200 kPa  v 1 = 0.95986 m 3 /kg

T1 = 150°C  s1 = 7.2810 kJ/kg ⋅ K
v 2 − v f 0.95986 − 0.001008
T2 = 40°C 
= 0.04914
=
 x2 =
v 2 = v1
v fg
19.515 − 0.001008

H2O
200 kPa
150°C
s 2 = s f + x 2 s fg = 0.5724 + (0.04914 )(7.6832 ) = 0.9499 kJ/kg ⋅ K
The mass of the steam is
m=
0.020 m 3
V
=
= 0.02084 kg
v 1 0.95986 m 3 /kg
Then the entropy change of the steam during this process becomes
∆S = m(s 2 − s1 ) = (0.02084 kg )(0.9499 − 7.2810 ) kJ/kg ⋅ K = −0.132 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
Q
7-7
7-32 A rigid tank is initially filled with a saturated mixture of R-134a. Heat is transferred to the tank from a
source until the pressure inside rises to a specified value. The entropy change of the refrigerant, entropy
change of the source, and the total entropy change for this process are to be determined. √
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There
are no work interactions.
Analysis (a) From the refrigerant tables (Tables A-11 through A-13),
P1 = 200 kPa 

x1 = 0.4

u1 = u f + x1u fg = 38.28 + (0.4 )(186.21) = 112.76 kJ/kg
s1 = s f + x1 s fg = 0.15457 + (0.4 )(0.78316 ) = 0.4678 kJ/kg ⋅ K
v 1 = v f + x1v fg = 0.0007533 + (0.4 )(0.099867 − 0.0007533) = 0.04040 m 3 /kg
P2 = 400 kPa 

v 2 = v1

x2 =
v 2 −v f
v fg
u 2 = u f + x 2 u fg
0.04040 − 0.0007907
= 0.7857
0.051201 − 0.0007907
= 63.62 + (0.7857 )(171.45) = 198.34 kJ/kg
=
s 2 = s f + x 2 s fg = 0.24761 + (0.7857 )(0.67929 ) = 0.7813 kJ/kg ⋅ K
The mass of the refrigerant is
m=
0.5 m 3
V
=
= 12.38 kg
v 1 0.04040 m 3 /kg
Then the entropy change of the refrigerant becomes
∆Ssystem = m(s2 − s1 ) = (12.38 kg )(0.7813 − 0.4678) kJ/kg ⋅ K = 3.880 kJ/K
(b) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the
volume of the system is constant and thus there is no boundary work, the energy balance for this stationary
closed system can be expressed as
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
R-134a
200 kPa
Qin = ∆U = m(u2 − u1 )
Q
Source
35°C
Substituting,
Qin = m(u2 − u1 ) = (12.38 kg )(198.34 − 112.76 ) = 1059 kJ
The heat transfer for the source is equal in magnitude but opposite in direction. Therefore,
Qsource, out = - Qtank, in = - 1059 kJ
and
∆Ssource = −
Qsource,out
Tsource
=−
1059 kJ
= −3.439 kJ/K
308 K
(c) The total entropy change for this process is
∆S total = ∆S system + ∆S source = 3.880 + (− 3.439) = 0.442 kJ/K
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7-8
7-33 EES Problem 7-32 is reconsidered. The effects of the source temperature and final pressure on the
total entropy change for the process as the source temperature varies from 30°C to 210°C, and the final
pressure varies from 250 kPa to 500 kPa are to be investigated. The total entropy change for the process is
to be plotted as a function of the source temperature for final pressures of 250 kPa, 400 kPa, and 500 kPa.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
P_1 = 200 [kPa]
x_1 = 0.4
V_sys = 0.5 [m^3]
P_2 = 400 [kPa]
{T_source = 35 [C]}
"Analysis: "
" Treat the rigid tank as a closed system, with no work in, neglect changes in KE and PE of the
R134a."
E_in - E_out = DELTAE_sys
E_out = 0 [kJ]
E_in = Q
DELTAE_sys = m_sys*(u_2 - u_1)
u_1 = INTENERGY(R134a,P=P_1,x=x_1)
v_1 = volume(R134a,P=P_1,x=x_1)
V_sys = m_sys*v_1
"Rigid Tank: The process is constant volume. Then P_2 and v_2 specify state 2."
v_2 = v_1
u_2 = INTENERGY(R134a,P=P_2,v=v_2)
"Entropy calculations:"
s_1 = entropy(R134a,P=P_1,x=x_1)
s_2 = entropY(R134a,P=P_2,v=v_2)
DELTAS_sys = m_sys*(s_2 - s_1)
"Heat is leaving the source, thus:"
DELTAS_source = -Q/(T_source + 273)
"Total Entropy Change:"
DELTAS_total = DELTAS_source + DELTAS_sys
∆Stotal
[kJ/K]
0.3848
0.6997
0.9626
1.185
1.376
1.542
1.687
Tsource
[C]
30
60
90
120
150
180
210
3
2.5
]
K
/
J
k[
l
at
ot
S
∆
2
P2 = 250 kPa
= 400 kPa
= 500 kPa
1.5
1
0.5
0
25
65
105
145
185
Tsource [C]
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225
7-9
7-34 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An
electric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of the
water during this process is to be determined.
Analysis From the steam tables (Tables A-4 through A-6)
P1 = 100 kPa  v1 = v f + x1v fg = 0.001 + (0.25)(1.6941 − 0.001) = 0.4243 m3/kg

x1 = 0.25
 s1 = s f + x1s fg = 1.3028 + (0.25)(6.0562) = 2.8168 kJ/kg ⋅ K
v 2 = v1

 s2 = 6.8649 kJ/kg ⋅ K
sat. vapor 
H 2O
2 kg
100 kPa
We
Then the entropy change of the steam becomes
∆S = m(s 2 − s1 ) = (2 kg )(6.8649 − 2.8168) kJ/kg ⋅ K = 8.10 kJ/K
7-35 [Also solved by EES on enclosed CD] A rigid tank is divided into two equal parts by a partition. One
part is filled with compressed liquid water while the other side is evacuated. The partition is removed and
water expands into the entire tank. The entropy change of the water during this process is to be determined.
Analysis The properties of the water are (Table A-4)
P1 = 300 kPa  v1 ≅ v f @ 60°C = 0.001017 m3/kg

T1 = 60°C
 s1 = s f @ 60°C = 0.8313 kJ/kg ⋅ K
Noting that v 2 = 2v1 = (2 )(0.001017 ) = 0.002034 m3/kg
1.5 kg
compressed
liquid
Vacuum
300 kPa
60°C
v 2 − v f 0.002034 − 0.001014

x2 =
=
= 0.0001018
10.02 − 0.001014
v fg

3
v 2 = 0.002034 m /kg  s = s + x s = 0.7549 + (0.0001018)(7.2522) = 0.7556 kJ/kg ⋅ K
2
f
2 fg
P2 = 15 kPa
Then the entropy change of the water becomes
∆S = m(s2 − s1 ) = (1.5 kg )(0.7556 − 0.8313) kJ/kg ⋅ K = −0.114 kJ/K
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7-10
7-36 EES Problem 7-35 is reconsidered. The entropy generated is to be evaluated and plotted as a function
of surroundings temperature, and the values of the surroundings temperatures that are valid for this
problem are to be determined. The surrounding temperature is to vary from 0°C to 100°C.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Input Data"
P[1]=300 [kPa]
T[1]=60 [C]
m=1.5 [kg]
P[2]=15 [kPa]
Fluid$='Steam_IAPWS'
V[1]=m*spv[1]
spv[1]=volume(Fluid$,T=T[1], P=P[1]) "specific volume of steam at state 1, m^3/kg"
s[1]=entropy(Fluid$,T=T[1],P=P[1]) "entropy of steam at state 1, kJ/kgK"
V[2]=2*V[1] "Steam expands to fill entire volume at state 2"
"State 2 is identified by P[2] and spv[2]"
spv[2]=V[2]/m "specific volume of steam at state 2, m^3/kg"
s[2]=entropy(Fluid$,P=P[2],v=spv[2]) "entropy of steam at state 2, kJ/kgK"
T[2]=temperature(Fluid$,P=P[2],v=spv[2])
DELTAS_sys=m*(s[2]-s[1]) "Total entopy change of steam, kJ/K"
"What does the first law tell us about this problem?"
"Conservation of Energy for the entire, closed system"
E_in - E_out = DELTAE_sys
"neglecting changes in KE and PE for the system:"
DELTAE_sys=m*(intenergy(Fluid$, P=P[2], v=spv[2]) - intenergy(Fluid$,T=T[1],P=P[1]))
E_in = 0
"How do you interpert the energy leaving the system, E_out? Recall this is a constant volume
system."
Q_out = E_out
"What is the maximum value of the Surroundings temperature?"
"The maximum possible value for the surroundings temperature occurs when we set
S_gen = 0=Delta S_sys+sum(DeltaS_surr)"
Q_net_surr=Q_out
S_gen = 0
S_gen = DELTAS_sys+Q_net_surr/Tsurr
"Establish a parametric table for the variables S_gen, Q_net_surr, T_surr, and DELTAS_sys. In
the Parametric Table window select T_surr and insert a range of values. Then place '{' and '}'
about the S_gen = 0 line; press F3 to solve the table. The results are shown in Plot Window 1.
What values of T_surr are valid for this problem?"
Sgen
[kJ/K]
0.02533
0.01146
0.0001205
-0.009333
-0.01733
Qnet,surr
[kJ]
37.44
37.44
37.44
37.44
37.44
Tsurr
[K]
270
300
330
360
390
∆Ssys
[kJ/K]
-0.1133
-0.1133
-0.1133
-0.1133
-0.1133
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-11
0.030
S gen [kJ/K]
0.020
0.010
-0.000
-0.010
-0.020
260
280
300
320
340
360
380
400
Tsurr [K]
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-12
7-37E A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed
at constant pressure. The entropy change of refrigerant during this process is to be determined
Analysis From the refrigerant tables (Tables A-11E through A-13E),
P1 = 120 psia 
s1 = 0.22361 Btu/lbm ⋅ R

T1 = 100°F
T2 = 50°F

 s 2 ≅ s f @90o F = 0.06039 Btu/lbm ⋅ R
P2 = 120 psia 
R-134a
120 psia
100°F
Then the entropy change of the refrigerant becomes
Q
∆S = m(s 2 − s1 ) = (2 lbm )(0.06039 − 0.22361)Btu/lbm ⋅ R = −0.3264 Btu/R
7-38 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water is
heated electrically at constant pressure. The entropy change of the water during this process is to be
determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated
and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis From the steam tables (Tables A-4 through A-6),
v 1 = v f @150 kPa = 0.001053 m 3 /kg
P1 = 150 kPa 
 h1 = h f @150 kPa = 467.13 kJ/kg
sat.liquid
 s =s
1
f @150 kPa = 1.4337 kJ/kg ⋅ K
Also,
m=
H2O
2200 kJ
150 kPa
Sat. liquid
0.005 m 3
V
=
= 4.75 kg
v 1 0.001053 m 3 /kg
We take the contents of the cylinder as the system. This is a closed system since no mass enters or
leaves. The energy balance for this stationary closed system can be expressed as
E − Eout
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
We,in − Wb,out = ∆U
We,in = m(h2 − h1 )
since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process. Solving for h2,
h2 = h1 +
We,in
m
= 467.13 +
2200 kJ
= 930.33 kJ/kg
4.75 kg
Thus,
h2 − h f 930.33 − 467.13
=
= 0.2081
x2 =

2226.0
h fg

h2 = 930.33 kJ/kg 
s 2 = s f + x 2 s fg = 1.4337 + (0.2081)(5.7894) = 2.6384 kJ/kg ⋅ K
P2 = 150 kPa
Then the entropy change of the water becomes
∆S = m(s 2 − s1 ) = (4.75 kg )(2.6384 − 1.4337 )kJ/kg ⋅ K = 5.72 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-13
7-39 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The
refrigerant expands in a reversible manner until the pressure drops to a specified value. The final
temperature in the cylinder and the work done by the refrigerant are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated
and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
process is stated to be reversible.
Analysis (a) This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant
tables (Tables A-11 through A-13),
v1 = v g @ 0.8 MPa = 0.025621 m3/kg
P1 = 0.8 MPa 
 u1 = u g @ 0.8 MPa = 246.79 kJ/kg
sat. vapor
s = s
1
g @ 0.8 MPa = 0.91835 kJ/kg ⋅ K
Also,
m=
0.05 m 3
V
=
= 1.952 kg
v 1 0.025621 m 3 /kg
and
R-134a
0.8 MPa
0.05 m3
s 2 − s f 0.91835 − 0.24761
P2 = 0.4 MPa 
=
= 0.9874
x2 =
0.67929
s fg

s 2 = s1
 u = u + x u = 63.62 + (0.9874)(171.45) = 232.91 kJ/kg
2
f
2 fg
T2 = Tsat @ 0.4 MPa = 8.91°C
(b) We take the contents of the cylinder as the system. This is a closed system since no mass enters or
leaves. The energy balance for this adiabatic closed system can be expressed as
E − Eout
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Wb,out = ∆U
Wb,out = m(u1 − u2 )
Substituting, the work done during this isentropic process is determined to be
Wb, out = m(u1 − u2 ) = (1.952 kg )(246.79 − 232.91) kJ/kg = 27.09 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-14
7-40 EES Problem 7-39 is reconsidered. The work done by the refrigerant is to be calculated and plotted as
a function of final pressure as the pressure varies from 0.8 MPa to 0.4 MPa. The work done for this process
is to be compared to one for which the temperature is constant over the same pressure range.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Procedure
IsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isother
m)
T_isotherm=Temperature(R134a,P=P_1,x=x_1)
T=T_isotherm
u_1 = INTENERGY(R134a,P=P_1,x=x_1)
v_1 = volume(R134a,P=P_1,x=x_1)
s_1 = entropy(R134a,P=P_1,x=x_1)
u_2 = INTENERGY(R134a,P=P_2,T=T)
s_2 = entropy(R134a,P=P_2,T=T)
"The process is reversible and Isothermal thus the heat transfer is determined by:"
Q_isotherm = (T+273)*m_sys*(s_2 - s_1)
DELTAE_isotherm = m_sys*(u_2 - u_1)
E_in = Q_isotherm
E_out = DELTAE_isotherm+E_in
Work_out_isotherm=E_out
END
"Knowns:"
P_1 = 800 [kPa]
x_1 = 1.0
V_sys = 0.05[m^3]
"P_2 = 400 [kPa]"
"Analysis: "
" Treat the rigid tank as a closed system, with no heat transfer in, neglect
changes in KE and PE of the R134a."
"The isentropic work is determined from:"
E_in - E_out = DELTAE_sys
E_out = Work_out_isen
E_in = 0
DELTAE_sys = m_sys*(u_2 - u_1)
u_1 = INTENERGY(R134a,P=P_1,x=x_1)
v_1 = volume(R134a,P=P_1,x=x_1)
s_1 = entropy(R134a,P=P_1,x=x_1)
V_sys = m_sys*v_1
"Rigid Tank: The process is reversible and adiabatic or isentropic.
Then P_2 and s_2 specify state 2."
s_2 = s_1
u_2 = INTENERGY(R134a,P=P_2,s=s_2)
T_2_isen = temperature(R134a,P=P_2,s=s_2)
Call
IsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_is
otherm)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-15
P2
[kPa]
400
500
600
700
800
Workout,isen
[kJ]
27.09
18.55
11.44
5.347
0
Workout,isotherm
[kJ]
60.02
43.33
28.2
13.93
0
Qisotherm
[kJ]
47.08
33.29
21.25
10.3
0
60
50
]
J
k[
40
kr
o
W
20
t
u
o
Isentropic
30
10
0
400
Isothermal
450
500
550
600
650
700
750
800
P2 [kPa]
50
Qisentropic = 0 kJ
40
]
J
k[
m
r
e
ht
o
si
Q
30
20
10
0
400
450
500
550
600
650
700
750
800
P2 [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-16
7-41 Saturated Refrigerant-134a vapor at 160 kPa is compressed steadily by an adiabatic compressor. The
minimum power input to the compressor 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 device is adiabatic and thus heat transfer is negligible.
Analysis The power input to an adiabatic compressor will be a minimum when the compression process is
reversible. For the reversible adiabatic process we have s2 = s1. From the refrigerant tables (Tables A-11
through A-13),
v1 = v g @160 kPa = 0.12348 m3/kg
P1 = 160 kPa 
 h1 = hg @160 kPa = 241.11 kJ/kg
sat. vapor
s = s
1
g @160 kPa = 0.9419 kJ/kg ⋅ K
2
P2 = 900 kPa 
 h2 = 277.06 kJ/kg

R-134a
s2 = s1
Also,
m& =
V&1
2 m 3 /min
=
= 16.20 kg/min = 0.27 kg/s
v 1 0.12348 m 3 /kg
1
&1 = m
&2 = m
& . We take the compressor as the system,
There is only one inlet and one exit, and thus m
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
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& 1 = mh
& 2 (since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W& in + mh
W& in = m& (h2 − h1 )
Substituting, the minimum power supplied to the compressor is determined to be
W&in = (0.27 kg/s )(277.06 − 241.11) kJ/kg = 9.71 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-17
7-42E Steam expands in an adiabatic turbine. The maximum amount of work that can be done by the
turbine is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential
energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.
Analysis The work output of an adiabatic turbine is maximum when the expansion process is reversible.
For the reversible adiabatic process we have s2 = s1. From the steam tables (Tables A-4E through A-6E),
P1 = 800 psia  h1 = 1456.0 Btu/lbm

T1 = 900°F  s1 = 1.6413 Btu/lbm ⋅ R
s2 − s f
1.6413 − 0.39213
P2 = 40 psia 
=
= 0.9725
x2 =
1.28448
s fg

s 2 = s1
 h = h + x h = 236.14 + (0.9725)(933.69) = 1144.2 Btu/lbm
2
f
2 fg
&1 = m
&2 = m
& . We take the
There is only one inlet and one exit, and thus m
turbine 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
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
1
H2O
2
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& 1 = W& out + mh
& 2
mh
W& out = m& (h1 − h2 )
Dividing by mass flow rate and substituting,
wout = h1 − h2 = 1456.0 − 1144.2 = 311.8 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-18
7-43E EES Problem 7-42E is reconsidered. The work done by the steam is to be calculated and plotted as a
function of final pressure as the pressure varies from 800 psia to 40 psia. Also the effect of varying the
turbine inlet temperature from the saturation temperature at 800 psia to 900°F on the turbine work is to be
investigated.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
P_1 = 800 [psia]
T_1 = 900 [F]
P_2 = 40 [psia]
T_sat_P_1= temperature(Fluid$,P=P_1,x=1.0)
Fluid$='Steam_IAPWS'
"Analysis: "
" Treat theturbine as a steady-flow control volume, with no heat transfer in, neglect
changes in KE and PE of the Steam."
"The isentropic work is determined from the steady-flow energy equation written per unit mass:"
e_in - e_out = DELTAe_sys
350
E_out = Work_out+h_2 "[Btu/lbm]"
e_in = h_1 "[Btu/lbm]"
Isentropic Process
300
DELTAe_sys = 0 "[Btu/lbm]"
]
P1 = 800 psia
h_1 = enthalpy(Fluid$,P=P_1,T=T_1) m 250
bl
s_1 = entropy(Fluid$,P=P_1,T=T_1)
T1 = 900 F
/
200
ut
B
[
"The process is reversible
150
t
and adiabatic or isentropic.
u
o 100
Then P_2 and s_2 specify state 2."
kr
s_2 = s_1 "[Btu/lbm-R]"
o
50
h_2 = enthalpy(Fluid$,P=P_2,s=s_2) W
T_2_isen=temperature(Fluid$,P=P_2,s=s_2) 0
0
100 200 300 400 500 600 700 800
P2 [psia]
Workout
T1
[Btu/lbm]
[F]
320
520
219.3
Isentropic Process
560
229.6
298
P1 = 800 psia
600
239.1
]
650
690
730
770
820
860
900
250.7
260
269.4
279
291.3
301.5
311.9
m
bl
/
ut
B
[
276
kr
o
W
232
t
u
o
P2 = 40 psia
254
210
500
550
600
650
700
750
800
850
900
T1 [F]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-19
7-44 An insulated cylinder is initially filled with superheated steam at a specified state. The steam is
compressed in a reversible manner until the pressure drops to a specified value. The work input during this
process is to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated
and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
process is stated to be reversible.
Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the steam tables
(Tables A-4 through A-6),
v = 0.63402 m 3 /kg
P1 = 300 kPa  1
 u1 = 2571.0 kJ/kg
T1 = 150°C 
s1 = 7.0792 kJ/kg ⋅ K
P2 = 1 MPa 
 u 2 = 2773.8 kJ/kg
s 2 = s1

Also,
m=
0.05 m 3
V
=
= 0.0789 kg
v 1 0.63402 m 3 /kg
H2O
300 kPa
150°C
We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves.
The energy balance for this adiabatic closed system can be expressed as
E − Eout
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Wb,in = ∆U = m(u2 − u1 )
Substituting, the work input during this adiabatic process is determined to be
Wb,in = m(u2 − u1 ) = (0.0789 kg )(2773.8 − 2571.0 ) kJ/kg = 16.0 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-20
7-45 EES Problem 7-44 is reconsidered. The work done on the steam is to be determined and plotted as a
function of final pressure as the pressure varies from 300 kPa to 1 MPa.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
P_1 = 300 [kPa]
T_1 = 150 [C]
V_sys = 0.05 [m^3]
"P_2 = 1000 [kPa]"
"Analysis: "
Fluid$='Steam_IAPWS'
" Treat the piston-cylinder as a closed system, with no heat transfer in, neglect
changes in KE and PE of the Steam. The process is reversible and adiabatic thus isentropic."
"The isentropic work is determined from:"
E_in - E_out = DELTAE_sys
E_out = 0 [kJ]
E_in = Work_in
DELTAE_sys = m_sys*(u_2 - u_1)
u_1 = INTENERGY(Fluid$,P=P_1,T=T_1)
v_1 = volume(Fluid$,P=P_1,T=T_1)
s_1 = entropy(Fluid$,P=P_1,T=T_1)
V_sys = m_sys*v_1
" The process is reversible and adiabatic or isentropic.
Then P_2 and s_2 specify state 2."
s_2 = s_1
u_2 = INTENERGY(Fluid$,P=P_2,s=s_2)
T_2_isen = temperature(Fluid$,P=P_2,s=s_2)
Workin
[kJ]
0
3.411
6.224
8.638
10.76
12.67
14.4
16
16
14
12
W ork in [kJ]
P2
[kPa]
300
400
500
600
700
800
900
1000
10
W ork on Steam
P = 300 kPa
1
T = 150 C
1
8
6
4
2
0
300
400
500
600
P
2
700
800
900
1000
[kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-21
7-46 A cylinder is initially filled with saturated water vapor at a specified temperature. Heat is transferred
to the steam, and it expands in a reversible and isothermal manner until the pressure drops to a specified
value. The heat transfer and the work output for this process are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated
and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
process is stated to be reversible and isothermal.
Analysis From the steam tables (Tables A-4 through A-6),
T1 = 200°C u1 = u g @ 200°C = 2594.2 kJ/kg

sat.vapor  s1 = s g @ 200°C = 6.4302 kJ/kg ⋅ K
P2 = 800 kPa  u 2 = 2631.1 kJ/kg

 s 2 = 6.8177 kJ/kg ⋅ K
T2 = T1
H2O
200°C
sat. vapor
T = const
The heat transfer for this reversible isothermal process can be determined from
Q
Q = T∆S = Tm(s 2 − s1 ) = (473 K )(1.2 kg )(6.8177 − 6.4302)kJ/kg ⋅ K = 219.9 kJ
We take the contents of the cylinder 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
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin − Wb,out = ∆U = m(u2 − u1 )
Wb,out = Qin − m(u2 − u1 )
Substituting, the work done during this process is determined to be
Wb, out = 219.9 kJ − (1.2 kg )(2631.1 − 2594.2) kJ/kg = 175.6 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-22
7-47 EES Problem 7-46 is reconsidered. The heat transferred to the steam and the work done are to be
determined and plotted as a function of final pressure as the pressure varies from the initial value to the
final value of 800 kPa.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
T_1 = 200 [C]
x_1 = 1.0
m_sys = 1.2 [kg]
{P_2 = 800"[kPa]"}
"Analysis: "
Fluid$='Steam_IAPWS'
" Treat the piston-cylinder as a closed system, neglect changes in KE and PE of the Steam. The
process is reversible and isothermal ."
T_2 = T_1
200
E_in - E_out = DELTAE_sys
E_in = Q_in
E_out = Work_out
160
DELTAE_sys = m_sys*(u_2 - u_1)
]
P_1 = pressure(Fluid$,T=T_1,x=1.0)
J 120
u_1 = INTENERGY(Fluid$,T=T_1,x=1.0) K
[
v_1 = volume(Fluid$,T=T_1,x=1.0)
t
u
s_1 = entropy(Fluid$,T=T_1,x=1.0)
o
80
kr
V_sys = m_sys*v_1
o
W
40
" The process is reversible and isothermal.
Then P_2 and T_2 specify state 2."
u_2 = INTENERGY(Fluid$,P=P_2,T=T_2)
s_2 = entropy(Fluid$,P=P_2,T=T_2)
Q_in= (T_1+273)*m_sys*(s_2-s_1)
P2
[kPa]
800
900
1000
1100
1200
1300
1400
1500
1553
Qin
[kJ]
219.9
183.7
150.6
120
91.23
64.08
38.2
13.32
219.9
Workout
[kJ]
175.7
144.7
117
91.84
68.85
47.65
27.98
9.605
175.7
0
800
1000
1200
1400
1600
1400
1600
P2 [kPa]
200
160
]
J
k[
ni
120
80
Q
40
0
800
1000
1200
P2 [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-23
7-48 A cylinder is initially filled with saturated water vapor mixture at a specified temperature. Steam
undergoes a reversible heat addition and an isentropic process. The processes are to be sketched and heat
transfer for the first process and work done during the second process are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The thermal energy stored in the
cylinder itself is negligible. 3 Both processes are reversible.
Analysis (b) From the steam tables (Tables A-4 through A-6),
T1 = 100°C
 h1 = h f + xh fg = 419.17 + (0.5)(2256.4) = 1547.4 kJ/kg
x = 0.5

h2 = h g = 2675.6 kJ/kg
H2O
100°C
x=0.5
T2 = 100°C
 u 2 = u g = 2506.0 kJ/kg
x2 = 1
 s = 7.3542 kJ/kg ⋅ K
2
Q
P3 = 15 kPa 
u 3 = 2247.9 kJ/kg
s3 = s 2

We take the contents of the cylinder 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
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin − Wb,out = ∆U = m(u2 − u1 )
For process 1-2, it reduces to
Q12,in = m(h2 − h1 ) = (5 kg)(2675.6 - 1547.4)kJ/kg = 5641 kJ
(c) For process 2-3, it reduces to
W23, b,out = m(u2 − u3 ) = (5 kg)(2506.0 - 2247.9)kJ/kg = 1291 kJ
SteamIAPWS
700
600
500
]
C
°[
T
400
101.42 kPa
300
200
100
0
0.0
15 kPa
2
1
3
1.1
2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9
11.0
s [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-24
7-49 A rigid tank contains saturated water vapor at a specified temperature. Steam is cooled to ambient
temperature. The process is to be sketched and entropy changes for the steam and for the process are to be
determined.
Assumptions 1 The kinetic and potential energy changes are negligible.
Analysis (b) From the steam tables (Tables A-4 through A-6),
v = v g = 1.6720 kJ/kg
T1 = 100°C 1
 u1 = u g = 2506.0 kJ/kg
x =1
 s = 7.3542 kJ/kg ⋅ K
1
H2O
100°C
x=1
x = 0.0386
T2 = 25°C 2
 u = 193.78 kJ/kg
v 2 = v1  2
s = 1.0715 kJ/kg ⋅ K
Q
2
The entropy change of steam is determined from
∆S w = m( s 2 − s1 ) = (5 kg)(1.0715 - 7.3542)kJ/kg ⋅ K = -31.41 kJ/K
(c) We take the contents of the tank 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
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = m(u2 − u1 )
That is,
Qout = m(u1 − u2 ) = (5 kg)(2506.0 - 193.78)kJ/kg = 11,511 kJ
The total entropy change for the process is
Sgen = ∆S w +
11,511 kJ
Qout
= −31.41 kJ/K +
= 7.39 kJ/K
298 K
Tsurr
SteamIAPWS
700
600
500
]
C
°[
T
400
300
200
100
0
10-3
1
101.4 kPa
2
3.17 kPa
10-2
10-1
100
101
102
103
3
v [m /kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-25
7-50 Steam expands in an adiabatic turbine. Steam leaves the turbine at two different pressures. The
process is to be sketched on a T-s diagram and the work done by the steam per unit mass of the steam at the
inlet are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible.
P1 = 6 MPa
T1 = 500°C
Analysis (b) From the steam tables (Tables A-4 through A-6),
P2 = 1 MPa 
T1 = 500°C  h1 = 3423.1 kJ/kg
h2 s = 2921.3 kJ/kg

P1 = 6 MPa  s1 = 6.8826 kJ/kg ⋅ K s2 = s1

Turbine
P3 = 10 kPa h3s = 2179.6 kJ/kg

x3s = 0.831
s3 = s1

A mass balance on the control volume gives
m& 2 = 0.1m& 1
m& 1 = m& 2 + m& 3 where
m& 3 = 0.9m& 1
We take the turbine as the system, which is a
control volume. The energy balance for this
steady-flow system can be expressed in the
rate form as
E& = E&
in
out
m& 1h1 = W& s ,out + m& 2 h2 + m& 3h3
m& 1h1 = W& s ,out + 0.1m& 1h2 + 0.9m& 1h3
P2 = 1 MPa
P3 = 10 kPa
SteamIAPWS
700
600
1
500
]
C
°[
T
or
400
300
6000 kPa
200
100
h1 = ws ,out + 0.1h2 + 0.9h3
ws ,out = h1 − 0.1h2 − 0.9h3
3
10 kPa
0
0.0
2
1000 kPa
1.1
2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9
11.0
s [kJ/kg-K]
= 3423.1 − (0.1)(2921.3) − (0.9)(2179.6) = 1169.3 kJ/kg
The actual work output per unit mass of steam at the inlet is
wout = ηT ws ,out = (0.85)(1169.3 kJ/kg ) = 993.9 kJ/kg
7-51E 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
(Tables A-11E through A-13E)
R-134
P1 = 140 psia 
140 psia
 s1 ≅ s f @70° F = 0.07306 Btu/lbm ⋅ R
T1 = 70°F
70°F

Leak
s2 − s f
0.07306 − 0.02605
x2 =
=
= 0.2355
P2 = 20 psia 
s fg
0.19962

s 2 = s1
 v = v + x v = 0.01182 + (0.2355)(2.2772 − 0.01182) = 0.5453 ft 3 /lbm
2
f
2
fg
Thus the final mass of the refrigerant in the can is
m=
1.2 ft 3
V
=
= 2.201 lbm
v 2 0.5453 ft 3 /lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-26
Entropy Change of Incompressible Substances
7-52C No, because entropy is not a conserved property.
7-53 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of
the tank and the total entropy change are to be determined.
Assumptions 1 Both the water and the copper block are incompressible substances with constant specific
heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are
negligible. 3 The tank is well-insulated and thus there is no heat transfer.
Properties The density and specific heat of water at 25°C are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°C. The
specific heat of copper at 27°C is cp = 0.386 kJ/kg.°C (Table A-3).
Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closed
system since no mass crosses the system boundary during the process. The energy balance for this system
can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
WATER
Copper
50 kg
0 = ∆U
or,
∆U Cu + ∆U water = 0
120 L
[mc(T2 − T1 )]Cu + [mc(T2 − T1 )]water = 0
where
mwater = ρV = (997 kg/m3 )(0.120 m3 ) = 119.6 kg
Using specific heat values for copper and liquid water at room temperature and substituting,
(50 kg)(0.386 kJ/kg ⋅ °C)(T2 − 80)°C + (119.6 kg)(4.18 kJ/kg ⋅ °C)(T2 − 25)°C = 0
T2 = 27.0°C
The entropy generated during this process is determined from
T 
 300.0 K 
 = −3.140 kJ/K
∆Scopper = mcavg ln 2  = (50 kg )(0.386 kJ/kg ⋅ K ) ln
 353 K 
 T1 
T 
 300.0 K 
 = 3.344 kJ/K
∆S water = mcavg ln 2  = (119.6 kg )(4.18 kJ/kg ⋅ K ) ln
T
 298 K 
 1
Thus,
∆S total = ∆Scopper + ∆S water = −3.140 + 3.344 = 0.204 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-27
7-54 A hot iron block is dropped into water in an insulated tank. The total entropy change during this
process is to be determined.
Assumptions 1 Both the water and the iron block are incompressible substances with constant specific
heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are
negligible. 3 The tank is well-insulated and thus there is no heat transfer. 4 The water that evaporates,
condenses back.
Properties The specific heat of water at 25°C is cp = 4.18 kJ/kg.°C. The specific heat of iron at room
temperature is cp = 0.45 kJ/kg.°C (Table A-3).
Analysis We take the entire contents of the tank, water + iron block, as the system. This is a closed system
since no mass crosses the system boundary during the process. The energy balance for this system can be
expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U
WATER
18°C
Iron
350°C
or,
∆U iron + ∆U water = 0
[mc(T2 − T1 )]iron + [mc(T2 − T1 )]water = 0
Substituting,
(25 kg )(0.45 kJ/kg ⋅ K )(T2 − 350o C) + (100 kg )(4.18 kJ/kg ⋅ K )(T2 − 18o C) = 0
T2 = 26.7°C
The entropy generated during this process is determined from
T 
 299.7 K 
 = −8.232 kJ/K
∆Siron = mcavg ln 2  = (25 kg )(0.45 kJ/kg ⋅ K ) ln
 623 K 
 T1 
T 
 299.7 K 
 = 12.314 kJ/K
∆S water = mcavg ln 2  = (100 kg )(4.18 kJ/kg ⋅ K ) ln
 291 K 
 T1 
Thus,
Sgen = ∆S total = ∆Siron + ∆S water = −8.232 + 12.314 = 4.08 kJ/K
Discussion The results can be improved somewhat by using specific heats at average temperature.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-28
7-55 An aluminum block is brought into contact with an iron block in an insulated enclosure. The final
equilibrium temperature and the total entropy change for this process are to be determined.
Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specific
heats. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The system is
well-insulated and thus there is no heat transfer.
Properties The specific heat of aluminum at the anticipated average temperature of 450 K is cp = 0.973
kJ/kg.°C. The specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45
kJ/kg.°C (Table A-3).
Analysis We take the iron+aluminum blocks as the system, which is a closed system. The energy balance
for this system can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U
or,
Iron
20 kg
100°C
Aluminum
20 kg
200°C
∆U alum + ∆U iron = 0
[mc(T2 − T1 )]alum + [mc(T2 − T1 )]iron = 0
Substituting,
(20 kg )(0.45 kJ/kg ⋅ K )(T2 − 100o C) + (20 kg)(0.973 kJ/kg ⋅ K )(T2 − 200o C) = 0
T2 = 168.4o C = 441.4 K
The total entropy change for this process is determined from
T 
 441.4 K 
 = 1.515 kJ/K
∆Siron = mcavg ln 2  = (20 kg )(0.45 kJ/kg ⋅ K ) ln
 373 K 
 T1 
T 
 441.4 K 
 = −1.346 kJ/K
∆Salum = mcavg ln 2  = (20 kg )(0.973 kJ/kg ⋅ K ) ln
 473 K 
 T1 
Thus,
∆S total = ∆Siron + ∆Salum = 1.515 − 1.346 = 0.169 kJ/K
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7-29
7-56 EES Problem 7-55 is reconsidered. The effect of the mass of the iron block on the final equilibrium
temperature and the total entropy change for the process is to be studied. The mass of the iron is to vary
from 1 to 10 kg. The equilibrium temperature and the total entropy change are to be plotted as a function
of iron mass.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
T_1_iron = 100 [C]
{m_iron = 20 [kg]}
T_1_al = 200 [C]
m_al = 20 [kg]
C_al = 0.973 [kJ/kg-K] "FromTable A-3
at the anticipated average temperature of
450 K."
C_iron= 0.45 [kJ/kg-K] "FromTable A-3
at room temperature, the only value
available."
198
196
194
192
T2
190
188
"Analysis: "
186
" Treat the iron plus aluminum as a
184
closed system, with no heat transfer in,
182
no work out, neglect changes in KE and
PE of the system. "
180
1
2
"The final temperature is found from the
energy balance."
E_in - E_out = DELTAE_sys
E_out = 0
E_in = 0
DELTAE_sys = m_iron*DELTAu_iron + m_al*DELTAu_al
DELTAu_iron = C_iron*(T_2_iron - T_1_iron)
DELTAu_al = C_al*(T_2_al - T_1_al)
3
4
5
6
7
8
9
8
9
10
m iron [kg]
"the iron and aluminum reach thermal equilibrium:"
T_2_iron = T_2
T_2_al = T_2
DELTAS_iron = m_iron*C_iron*ln((T_2_iron+273) / (T_1_iron+273))
DELTAS_al = m_al*C_al*ln((T_2_al+273) / (T_1_al+273))
DELTAS_total = DELTAS_iron + DELTAS_al
miron
[kg]
1
2
3
4
5
6
7
8
9
10
T2
[C]
197.7
195.6
193.5
191.5
189.6
187.8
186.1
184.4
182.8
181.2
0.1
0.09
0.08
0.07
∆ S total [kJ/kg]
∆Stotal
[kJ/kg]
0.01152
0.0226
0.03326
0.04353
0.05344
0.06299
0.07221
0.08112
0.08973
0.09805
0.06
0.05
0.04
0.03
0.02
0.01
1
2
3
4
5
m
iron
6
7
[kg]
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10
7-30
7-57 An iron block and a copper block are dropped into a large lake. The total amount of entropy change
when both blocks cool to the lake temperature is to be determined.
Assumptions 1 Both the water and the iron block are incompressible substances with constant specific
heats at room temperature. 2 Kinetic and potential energies are negligible.
Properties The specific heats of iron and copper at room temperature are ciron = 0.45 kJ/kg.°C and ccopper =
0.386 kJ/kg.°C (Table A-3).
Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron
and the copper blocks will drop to the lake temperature (15°C) when the thermal equilibrium is established.
Then the entropy changes of the blocks become
T 
 288 K 
 = −4.579 kJ/K
∆Siron = mcavg ln 2  = (50 kg )(0.45 kJ/kg ⋅ K )ln
 353 K 
 T1 
T 
 288 K 
 = −1.571 kJ/K
∆Scopper = mcavg ln 2  = (20 kg )(0.386 kJ/kg ⋅ K )ln
 353 K 
 T1 
We take both the iron and the copper blocks, as the
system. This is a closed system since no mass crosses
the system boundary during the process. The energy
balance for this system can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
Iron
50 kg
80°C
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = ∆U iron + ∆U copper
or,
Lake
15°C
Copper
20 kg
80°C
Qout = [mc(T1 − T2 )]iron + [mc(T1 − T2 )]copper
Substituting,
Qout = (50 kg )(0.45 kJ/kg ⋅ K )(353 − 288)K + (20 kg )(0.386 kJ/kg ⋅ K )(353 − 288)K
= 1964 kJ
Thus,
∆Slake =
Qlake,in
Tlake
=
1964 kJ
= 6.820 kJ/K
288 K
Then the total entropy change for this process is
∆S total = ∆Siron + ∆Scopper + ∆Slake = −4.579 − 1.571 + 6.820 = 0.670 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-31
7-58 An adiabatic pump is used to compress saturated liquid water in a reversible manner. The work input
is to be determined by different approaches.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3
Heat transfer to or from the fluid is negligible.
Analysis The properties of water at the inlet and exit of the pump are (Tables A-4 through A-6)
h1 = 191.81 kJ/kg
P1 = 10 kPa 
 s1 = 0.6492 kJ/kg
x1 = 0
 v = 0.001010 m 3 /kg
1
15 MPa
P2 = 15 MPa  h2 = 206.90 kJ/kg

3
v 2 = 0.001004 m /kg
s 2 = s1
(a) Using the entropy data from the compressed liquid water table
10 kPa
pump
wP = h2 − h1 = 206.90 − 191.81 = 15.10 kJ/kg
(b) Using inlet specific volume and pressure values
wP = v 1 ( P2 − P1 ) = (0.001010 m 3 /kg)(15,000 − 10)kPa = 15.14 kJ/kg
Error = 0.3%
(b) Using average specific volume and pressure values
[
]
wP = v avg ( P2 − P1 ) = 1 / 2(0.001010 + 0.001004) m3/kg (15,000 − 10)kPa = 15.10 kJ/kg
Error = 0%
Discussion The results show that any of the method may be used to calculate reversible pump work.
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7-32
Entropy Changes of Ideal Gases
7-59C For ideal gases, cp = cv + R and
P2V 2 P1V1
TP
V
=
→ 2 = 2 1
T2
T1
V1 T1P2
Thus,
V 
T 
s2 − s1 = cv ln 2  + R ln 2 
 V1 
 T1 
T P 
T 
= cv ln 2  + R ln 2 1 
 T1P2 
 T1 
P 
T 
T 
= cv ln 2  + R ln 2  − R ln 2 
T
T
 P1 
 1
 1
P 
T 
= c p ln 2  − R ln 2 
 P1 
 T1 
7-60C For an ideal gas, dh = cp dT and v = RT/P. From the second Tds relation,
dh v dP c p dP RT dP
dT
dP
ds =
−
=
−
= cp
−R
T
T
T
P T
T
P
Integrating,
P
T 
s 2 − s1 = c p ln 2  − R ln 2
 P1
 T1 
Since cp is assumed to be constant.



7-61C No. The entropy of an ideal gas depends on the pressure as well as the temperature.
7-62C Setting ∆s = 0 gives
P
T 
c p ln 2  − R ln 2
T
 P1
 1

T
 = 0 
→ ln 2

 T1
P 
 R  P2 
T
 =
ln  
→ 2 =  2 
c
P
T
1
p
 P1 

 1
But
c p − cv
1 k −1
R
=
= 1− =
cp
cp
k
k
since
k = c p / cv . Thus,
T2  P2 
= 
T1  P1 
R Cp
(k −1) k
7-63C The Pr and vr are called relative pressure and relative specific volume, respectively. They are
derived for isentropic processes of ideal gases, and thus their use is limited to isentropic processes only.
7-64C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends
on the pressure as well as the temperature.
7-65C The entropy change relations of an ideal gas simplify to
∆s = cp ln(T2/T1) for a constant pressure process
and
∆s = cv ln(T2/T1) for a constant volume process.
Noting that cp > cv, the entropy change will be larger for a constant pressure process.
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7-33
7-66 Oxygen gas is compressed from a specified initial state to a specified final state. The entropy change
of oxygen during this process is to be determined for the case of constant specific heats.
Assumptions At specified conditions, oxygen can be treated as an ideal gas.
Properties The gas constant and molar mass of oxygen are R = 0.2598 kJ/kg.K and M = 32 kg/kmol (Table
A-1).
Analysis The constant volume specific heat of oxygen at the average temperature is (Table A-2)
Tavg =
298 + 560
= 429 K 
→ cv ,avg = 0.690 kJ/kg ⋅ K
2
Thus,
V
T2
+ R ln 2
V1
T1
0.1 m3/kg
560 K
= (0.690 kJ/kg ⋅ K ) ln
+ (0.2598 kJ/kg ⋅ K ) ln
298 K
0.8 m3/kg
= −0.105 kJ/kg ⋅ K
s2 − s1 = cv ,avg ln
O2
0.8 m3/kg
25°C
7-67 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank
stirs the gas, and the pressure and temperature of CO2 rises. The entropy change of CO2 during this process
is to be determined using constant specific heats.
Assumptions At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at
room temperature.
Properties The specific heat of CO2 is cv = 0.657 kJ/kg.K (Table A-2).
Analysis Using the ideal gas relation, the entropy change is determined to be
P2V
PV
T
P
150 kPa
= 1 
→ 2 = 2 =
= 1.5
T2
T1
T1
P1 100 kPa
CO2
1.5 m3
100 kPa
2.7 kg
Thus,

T
V ©0 
T
∆S = m(s2 − s1 ) = m cv ,avg ln 2 + R ln 2  = mcv ,avg ln 2


T
V
T1
1
1


= (2.7 kg )(0.657 kJ/kg ⋅ K ) ln (1.5)
= 0.719 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-34
7-68 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder
is turned on, and air is heated for 15 min at constant pressure. The entropy change of air during this process
is to be determined for the cases of constant and variable specific heats.
Assumptions At specified conditions, air can be treated as an ideal gas.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis The mass of the air and the electrical work done during this process are
(
)
(120 kPa ) 0.3 m3
P1V1
=
= 0.4325 kg
RT1
0.287 kPa ⋅ m3/kg ⋅ K (290 K )
We,in = W&e,in ∆t = (0.2 kJ/s )(15 × 60 s ) = 180 kJ
m=
(
)
The energy balance for this stationary closed system can be expressed as
E − Eout
=
∆Esystem
1in
424
3
1
424
3
Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
AIR
0.3 m3
120 kPa
17°C
We
We,in − Wb,out = ∆U 
→We,in = m(h2 − h1 ) ≅ c p (T2 − T1 )
since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process.
(a) Using a constant cp value at the anticipated average temperature of 450 K, the final temperature
becomes
W
180 kJ
Thus, T2 = T1 + e,in = 290 K +
= 698 K
mc p
(0.4325 kg )(1.02 kJ/kg ⋅ K )
Then the entropy change becomes

T
P ©0 
T
∆Ssys = m(s2 − s1 ) = m c p ,avg ln 2 − R ln 2  = mc p ,avg ln 2


T
P
T1
1
1


 698 K 
 = 0.387 kJ/K
= (0.4325 kg )(1.020 kJ/kg ⋅ K ) ln
 290 K 
(b) Assuming variable specific heats,
We,in = m(h2 − h1 ) 
→ h2 = h1 +
We,in
m
= 290.16 kJ/kg +
180 kJ
= 706.34 kJ/kg
0.4325 kg
From the air table (Table A-17, we read s2o = 2.5628 kJ/kg·K corresponding to this h2 value. Then,
(
)

P ©0 
∆Ssys = m s2o − s1o + R ln 2  = m s2o − s1o = (0.4325 kg )(2.5628 − 1.66802 )kJ/kg ⋅ K = 0.387 kJ/K

P1 

7-69 A cylinder contains N2 gas at a specified pressure and temperature. It is compressed polytropically
until the volume is reduced by half. The entropy change of nitrogen during this process is to be determined.
Assumptions 1 At specified conditions, N2 can be treated as an ideal gas. 2 Nitrogen has constant specific
heats at room temperature.
Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The constant volume specific
heat of nitrogen at room temperature is cv = 0.743 kJ/kg.K (Table A-2).
Analysis From the polytropic relation,
n −1
n −1
v 
T2  v1 

→ T2 = T1 1  = (300 K )(2 )1.3−1 = 369.3 K
=  
T1  v 2 
v2 
Then the entropy change of nitrogen becomes

V 
T
∆S N 2 = m cv ,avg ln 2 + R ln 2 
T1
V1 

N2
PV1.3 = C


369.3 K
= (1.2 kg ) (0.743 kJ/kg ⋅ K ) ln
+ (0.297 kJ/kg ⋅ K ) ln (0.5) = −0.0617 kJ/K
300
K


PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-35
7-70 EES Problem 7-69 is reconsidered. The effect of varying the polytropic exponent from 1 to 1.4 on the
entropy change of the nitrogen is to be investigated, and the processes are to be shown on a common P-v
diagram.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Function BoundWork(P[1],V[1],P[2],V[2],n)
"This function returns the Boundary Work for the polytropic process. This function is required
since the expression for boundary work depens on whether n=1 or n<>1"
If n<>1 then
BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1"
else
BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1"
endif
end
n=1
P[1] = 120 [kPa]
T[1] = 27 [C]
m = 1.2 [kg]
V[2]=V[1]/2
Gas$='N2'
MM=molarmass(Gas$)
R=R_u/MM
R_u=8.314 [kJ/kmol-K]
"System: The gas enclosed in the piston-cylinder device."
"Process: Polytropic expansion or compression, P*V^n = C"
P[1]*V[1]=m*R*(T[1]+273)
P[2]*V[2]^n=P[1]*V[1]^n
W_b = BoundWork(P[1],V[1],P[2],V[2],n)
"Find the temperature at state 2 from the pressure and specific volume."
T[2]=temperature(gas$,P=P[2],v=V[2]/m)
"The entropy at states 1 and 2 is:"
s[1]=entropy(gas$,P=P[1],v=V[1]/m)
s[2]=entropy(gas$,P=P[2],v=V[2]/m)
DELTAS=m*(s[2] - s[1])
"Remove the {} to generate the P-v plot data"
{Nsteps = 10
VP[1]=V[1]
PP[1]=P[1]
Duplicate i=2,Nsteps
VP[i]=V[1]-i*(V[1]-V[2])/Nsteps
PP[i]=P[1]*(V[1]/VP[i])^n
END }
∆S [kJ/kg]
-0.2469
-0.2159
-0.1849
-0.1539
-0.1229
-0.09191
-0.06095
-0.02999
0.0009849
n
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
Wb [kJ]
-74.06
-75.36
-76.69
-78.05
-79.44
-80.86
-82.32
-83.82
-85.34
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-36
350
n=1
= 1.4
300
PP[i]
250
200
150
100
0.4
0.5
0.6
0.7
0.8
0.9
1
VP[i]
0.05
∆ S = 0 kJ/k
0
∆ S [kJ/K]
-0.05
-0.1
-0.15
-0.2
-0.25
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
n
-72.5
-75.5
W b [kJ]
-78.5
-81.5
-84.5
-87.5
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
n
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-37
7-71E A fixed mass of helium undergoes a process from one specified state to another specified state. The
entropy change of helium is to be determined for the cases of reversible and irreversible processes.
Assumptions 1 At specified conditions, helium can be treated as an ideal gas. 2 Helium has constant
specific heats at room temperature.
Properties The gas constant of helium is R = 0.4961 Btu/lbm.R (Table A-1E). The constant volume
specific heat of helium is cv = 0.753 Btu/lbm.R (Table A-2E).
Analysis From the ideal-gas entropy change relation,

v 
T
∆S He = m cv ,ave ln 2 + R ln 2 
v1 
T1


 10 ft 3/lbm  
660 R

= (15 lbm) (0.753 Btu/lbm ⋅ R) ln
+ (0.4961 Btu/lbm ⋅ R ) ln
 50 ft 3/lbm  

540 R



= −9.71 Btu/R
He
T1 = 540 R
T2 = 660 R
The entropy change will be the same for both cases.
7-72 Air is compressed in a piston-cylinder device in a reversible and isothermal manner. The entropy
change of air and the work done are to be determined.
Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is specified to be
reversible.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis (a) Noting that the temperature remains constant, the entropy change of air is determined from
∆Sair = c p ,avg ln
T2 ©0
P
P
− R ln 2 = − R ln 2
T1
P1
P1
 400 kPa 
 = −0.428 kJ/kg ⋅ K
= −(0.287 kJ/kg ⋅ K )ln
 90 kPa 
Also, for a reversible isothermal process,
AIR
T = const
q = T∆s = (293 K )(− 0.428 kJ/kg ⋅ K ) = −125.4 kJ/kg 
→ qout = 125.4 kJ/kg
(b) The work done during this process is determined from the closed system energy balance,
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Win − Qout = ∆U = mcv (T2 − T1 ) = 0
win = qout = 125.4 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
Q
7-38
7-73 Air is compressed steadily by a 5-kW compressor from one specified state to another specified state.
The rate of entropy change of air is to be determined.
Assumptions At specified conditions, air can be treated as an ideal gas. 2 Air has variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
P2 = 600 kPa
T2 = 440 K
Analysis From the air table (Table A-17),
T1 = 290 K
 o
s1 = 1.66802 kJ/kg ⋅ K
P1 = 100 kPa 
T2 = 440 K
 o
s 2 = 2.0887 kJ/kg ⋅ K
P2 = 600 kPa 
AIR
COMPRESSOR
5 kW
Then the rate of entropy change of air becomes

P 
∆S&sys = m&  s2o − s1o − R ln 2 
P1 


 600 kPa  
 
= (1.6/60 kg/s ) 2.0887 − 1.66802 − (0.287 kJ/kg ⋅ K ) ln


 100 kPa  

= −0.00250 kW/K
P1 = 100 kPa
T1 = 290 K
7-74 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature and
pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. The
total entropy change during this process is to be determined.
Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply.
Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = m(u2 − u1 )
u2 = u1
T2 = T1
IDEAL
GAS
5 kmol
40°C
since u = u(T) for an ideal gas. Then the entropy change of the gas becomes

V 
V
T ©0
∆S = N  cv ,avg ln 2 + Ru ln 2  = NRu ln 2


V1 
V1
T1

= (5 kmol)(8.314 kJ/kmol ⋅ K ) ln (2)
= 28.81 kJ/K
This also represents the total entropy change since the tank does not contain anything else, and there are
no interactions with the surroundings.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-39
7-75 Air is compressed in a piston-cylinder device in a reversible and adiabatic manner. The final
temperature and the work are to be determined for the cases of constant and variable specific heats.
Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is given to be
reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air at low
to moderately high temperatures is k = 1.4 (Table A-2).
Analysis (a) Assuming constant specific heats, the ideal gas isentropic relations give
P 
T2 = T1 2 
 P1 
(k −1) k
 800 kPa 

= (290 K )
 100 kPa 
0.4 1.4
= 525.3 K
Then,
Tavg = (290 + 525.3)/2 = 407.7 K 
→ cv ,avg = 0.727 kJ/kg ⋅ K
We take the air in the cylinder as the system. The energy balance for
this stationary closed system can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
AIR
Reversible
Change in internal, kinetic,
potential, etc. energies
Win = ∆U = m(u2 − u1 ) ≅ mcv (T2 − T1 )
Thus,
win = cv ,avg (T2 − T1 ) = (0.727 kJ/kg ⋅ K )(525.3 − 290 ) K = 171.1 kJ/kg
(b) Assuming variable specific heats, the final temperature can be determined using the relative pressure
data (Table A-17),
T1 = 290 K 
→
and
Pr2 =
Pr1 = 1.2311
u1 = 206.91 kJ/kg
T = 522.4 K
P2
800 kPa
(1.2311) = 9.849 → 2
Pr =
u2 = 376.16 kJ/kg
P1 1 100 kPa
Then the work input becomes
win = u2 − u1 = (376.16 − 206.91) kJ/kg = 169.25 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-40
7-76 EES Problem 7-75 is reconsidered. The work done and final temperature during the compression
process are to be calculated and plotted as functions of the final pressure for the two cases as the final
pressure varies from 100 kPa to 800 kPa.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Procedure ConstPropSol(P_1,T_1,P_2,Gas$:Work_in_ConstProp,T2_ConstProp)
C_P=SPECHEAT(Gas$,T=27)
MM=MOLARMASS(Gas$)
R_u=8.314 [kJ/kmol-K]
R=R_u/MM
C_V = C_P - R
k = C_P/C_V
T2= (T_1+273)*(P_2/P_1)^((k-1)/k)
T2_ConstProp=T2-273 "[C]"
DELTAu = C_v*(T2-(T_1+273))
Work_in_ConstProp = DELTAu
End
"Knowns:"
P_1 = 100 [kPa]
T_1 = 17 [C]
P_2 = 800 [kPa]
"Analysis: "
" Treat the piston-cylinder as a closed system, with no heat transfer in, neglect
changes in KE and PE of the air. The process is reversible and adiabatic thus isentropic."
"The isentropic work is determined from:"
e_in - e_out = DELTAe_sys
e_out = 0 [kJ/kg]
e_in = Work_in
DELTAE_sys = (u_2 - u_1)
u_1 = INTENERGY(air,T=T_1)
v_1 = volume(air,P=P_1,T=T_1)
s_1 = entropy(air,P=P_1,T=T_1)
" The process is reversible and
adiabatic or isentropic.
Then P_2 and s_2 specify state 2."
s_2 = s_1
u_2 = INTENERGY(air,P=P_2,s=s_2)
T_2_isen=temperature(air,P=P_2,s=s_2) 180
Gas$ = 'air'
160
Call ConstPropSol(P_1,T_1,P_2,Gas$:
Work_in_ConstProp,T2_ConstProp)
140
]
g 120
P2
Workin
Workin,ConstProp k/
100
J
[kPa]
[kJ/kg]
[kJ/kg]
k[
80
100
0
0
ni
200
45.63
45.6
60
kr
300
76.84
76.77
o
40
400
101.3
101.2
W
20
500
121.7
121.5
600
139.4
139.1
0
100 200 300 400 500 600 700
700
155.2
154.8
800
169.3
168.9
P2 [kPa]
800
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-41
7-77 Helium gas is compressed in a piston-cylinder device in a reversible and adiabatic manner. The final
temperature and the work are to be determined for the cases of the process taking place in a piston-cylinder
device and a steady-flow compressor.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is given to be reversible
and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.
Properties The specific heats and the specific heat ratio of helium are cv = 3.1156 kJ/kg.K, cp = 5.1926
kJ/kg.K, and k = 1.667 (Table A-2).
Analysis (a) From the ideal gas isentropic relations,
P 
T2 = T1 2 
 P1 
(k −1) k
 450 kPa 

= (303 K )
 90 kPa 
2
0.667 1.667
= 576.9 K
(a) We take the air in the cylinder as the system. The
energy balance for this stationary closed system can be
expressed as
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
He
Rev.
He
Rev.
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
1
Win = ∆U = m(u2 − u1 ) ≅ mcv (T2 − T1 )
Thus,
win = cv (T2 − T1 ) = (3.1156 kJ/kg ⋅ K )(576.9 − 303)K = 853.4 kJ/kg
(b) If the process takes place in a steady-flow device, the final temperature will remain the same but the
work done should be determined from an energy balance on this steady-flow device,
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W&in + m& h1 = m& h2
W&in = m& (h2 − h1 ) ≅ m& c p (T2 − T1 )
Thus,
win = c p (T2 − T1 ) = (5.1926 kJ/kg ⋅ K )(576.9 − 303)K = 1422.3 kJ/kg
7-78 An insulated rigid tank contains argon gas at a specified pressure and temperature. A valve is opened,
and argon escapes until the pressure drops to a specified value. The final mass in the tank is to be
determined.
Assumptions 1 At specified conditions, argon can be treated as an ideal gas. 2 The process is given to be
reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.
Properties The specific heat ratio of argon is k = 1.667 (Table A-2).
Analysis From the ideal gas isentropic relations,
P
T2 = T1  2
 P1



(k −1) k
 200 kPa 

= (303 K )
 450 kPa 
0.667 1.667
= 219.0 K
The final mass in the tank is determined from the ideal gas relation,
ARGON
4 kg
450 kPa
30°C
P1V
m RT
PT
(200 kPa )(303 K ) (4 kg ) = 2.46 kg
= 1 1 
→ m 2 = 2 1 m1 =
(450 kPa )(219 K )
P2V m 2 RT2
P1T2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-42
7-79 EES Problem 7-78 is reconsidered. The effect of the final pressure on the final mass in the tank is to
be investigated as the pressure varies from 450 kPa to 150 kPa, and the results are to be plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"UNIFORM_FLOW SOLUTION:"
"Knowns:"
C_P = 0.5203"[kJ/kg-K ]"
C_V = 0.3122 "[kJ/kg-K ]"
R=0.2081 "[kPa-m^3/kg-K]"
P_1= 450"[kPa]"
T_1 = 30"[C]"
m_1 = 4"[kg]"
P_2= 150"[kPa]"
"Analysis:
We assume the mass that stays in the tank undergoes an isentropic expansion
process. This allows us to determine the final temperature of that gas at the final
pressure in the tank by using the isentropic relation:"
k = C_P/C_V
T_2 = ((T_1+273)*(P_2/P_1)^((k-1)/k)-273)"[C]"
V_2 = V_1
P_1*V_1=m_1*R*(T_1+273)
P_2*V_2=m_2*R*(T_2+273)
P2
[kPa]
150
200
250
300
350
400
450
m2
[kg]
2.069
2.459
2.811
3.136
3.44
3.727
4
4
3.6
m 2 [kg]
3.2
2.8
2.4
2
150
200
250
300
P
2
350
400
450
[kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-43
7-80E Air is accelerated in an adiabatic nozzle. Disregarding irreversibilities, the exit velocity of air is to
be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is given to be reversible and
adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 2 The nozzle operates
steadily.
Analysis Assuming variable specific heats, the inlet and exit properties are determined to be
T1 = 1000 R 
→
and
Pr2 =
Pr1 = 12.30
h1 = 240.98 Btu/lbm
T2 = 635.9 R
P2
12 psia
(12.30) = 2.46 →
Pr1 =
h2 = 152.11 Btu/lbm
P1
60 psia
1
AIR
2
We take the nozzle as the system, which is a control volume. The energy
balance for this steady-flow system can be expressed in the rate form as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12 / 2) = m& (h2 + V22 /2)
h2 − h1 +
V22 − V12
=0
2
Therefore,
 25,037 ft 2 /s 2
V 2 = 2(h1 − h2 ) + V12 = 2(240.98 − 152.11)Btu/lbm
 1 Btu/lbm

= 2119 ft/s

 + (200 ft/s )2


PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-44
7-81 Air is accelerated in an nozzle, and some heat is lost in the process. The exit temperature of air and
the total entropy change during the process are to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily.
Analysis (a) Assuming variable specific heats, the inlet properties are determined to be,
T1 = 350 K

→
s1o
h1 = 350.49 kJ / kg
= 1.85708 / kJ / kg ⋅ K
(Table A-17)
We take the nozzle as the system, which is a control volume. The energy
balance for this steady-flow system can be expressed in the rate form as
E& − E& out
1in
424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
3.2 kJ/s
1
AIR
2
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12 / 2) = m& (h2 + V22 /2) + Q& out
0 = qout + h2 − h1 +
V22 − V12
2
Therefore,
h2 = h1 − qout −
V22 − V12
(320 m/s)2 − (50 m/s)2  1 kJ/kg 
= 350.49 − 3.2 −
 1000 m 2 /s 2 
2
2


= 297.34 kJ/kg
At this h2 value we read, from Table A-17,
T2 = 297.2 K,
s2o = 1.6924 kJ / kg ⋅ K
(b) The total entropy change is the sum of the entropy changes of the air and of the surroundings, and is
determined from
∆stotal = ∆sair + ∆ssurr
where
∆sair = s2o − s1o − R ln
P2
85 kPa
= 1.6924 − 1.85708 − (0.287 kJ/kg ⋅ K ) ln
= 0.1775 kJ/kg ⋅ K
P1
280 kPa
and
∆ssurr =
Thus,
qsurr,in
Tsurr
=
3.2 kJ/kg
= 0.0109 kJ/kg ⋅ K
293 K
∆stotal = 0.1775 + 0.0109 = 0.1884 kJ/kg ⋅ K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-45
7-82 EES Problem 7-76 is reconsidered. The effect of varying the surrounding medium temperature from
10°C to 40°C on the exit temperature and the total entropy change for this process is to be studied, and the
results are to be plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Function HCal(WorkFluid$, Tx, Px)
"Function to calculate the enthalpy of an ideal gas or real gas"
If 'Air' = WorkFluid$ then
HCal:=ENTHALPY('Air',T=Tx) "Ideal gas equ."
else
HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ."
endif
end HCal
"System: control volume for the nozzle"
"Property relation: Air is an ideal gas"
"Process: Steady state, steady flow, adiabatic, no work"
"Knowns - obtain from the input diagram"
WorkFluid$ = 'Air'
T[1] = 77 [C]
P[1] = 280 [kPa]
Vel[1] = 50 [m/s]
P[2] = 85 [kPa]
Vel[2] = 320 [m/s]
q_out = 3.2 [kJ/kg]
"T_surr = 20 [C]"
"Property Data - since the Enthalpy function has different parameters
for ideal gas and real fluids, a function was used to determine h."
h[1]=HCal(WorkFluid$,T[1],P[1])
h[2]=HCal(WorkFluid$,T[2],P[2])
"The Volume function has the same form for an ideal gas as for a real fluid."
v[1]=volume(workFluid$,T=T[1],p=P[1])
v[2]=volume(WorkFluid$,T=T[2],p=P[2])
"If we knew the inlet or exit area, we could calculate the mass flow rate. Since we don't know
these areas, we write the conservation of energy per unit mass."
"Conservation of mass: m_dot[1]= m_dot[2]"
"Conservation of Energy - SSSF energy balance for neglecting the change in potential energy, no
work, but heat transfer out is:"
h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg) = h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)+q_out
s[1]=entropy(workFluid$,T=T[1],p=P[1])
s[2]=entropy(WorkFluid$,T=T[2],p=P[2])
"Entropy change of the air and the surroundings are:"
DELTAs_air = s[2] - s[1]
q_in_surr = q_out
DELTAs_surr = q_in_surr/(T_surr+273)
DELTAs_total = DELTAs_air + DELTAs_surr
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-46
∆stotal
[kJ/kg-K]
0.189
0.1888
0.1886
0.1884
0.1882
0.188
0.1879
Tsurr
[C]
10
15
20
25
30
35
40
T2
[C]
24.22
24.22
24.22
24.22
24.22
24.22
24.22
0.189
0.1888
∆ s total [kJ/kg-K]
0.1886
0.1884
0.1882
0.188
0.1878
10
15
20
25
T
surr
30
35
40
[C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-47
7-83 A container is filled with liquid water is placed in a room and heat transfer takes place between the
container and the air in the room until the thermal equilibrium is established. The final temperature, the
amount of heat transfer between the water and the air, and the entropy generation are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constant
specific heats. 3 The room is well-sealed and there is no heat transfer from the room to the surroundings. 4
Sea level atmospheric pressure is assumed. P = 101.3 kPa.
Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv =
0.718 kJ/kg.K. The specific heat of water at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3).
Analysis (a) The mass of the air in the room is
ma =
(101.3 kPa)(90 m 3 )
PV
=
= 111.5 kg
RTa1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(12 + 273 K)
Room
90 m3
12°C
An energy balance on the system that consists of the water in the
container and the air in the room gives the final equilibrium
temperature
Water
45 kg
95°C
0 = m w c w (T2 − Tw1 ) + m a cv (T2 − Ta1 )
0 = (45 kg)(4.18 kJ/kg.K)(T2 − 95) + (111.5 kg)(0.718 kJ/kg.K)(T2 − 12) 
→ T2 = 70.2°C
(b) The heat transfer to the air is
Q = m a cv (T2 − Ta1 ) = (111.5 kg)(0.718 kJ/kg.K)(70.2 − 12) = 4660 kJ
(c) The entropy generation associated with this heat transfer process may be obtained by calculating total
entropy change, which is the sum of the entropy changes of water and the air.
∆S w = mwcw ln
P2 =
ma RT2
V
=
(70.2 + 273) K
T2
= (45 kg)(4.18 kJ/kg.K)ln
= −13.11 kJ/K
Tw1
(95 + 273) K
(111.5 kg)(0.287 kPa ⋅ m3/kg ⋅ K)(70.2 + 273 K)
(90 m3 )
= 122 kPa

P 
T
∆S a = m a  c p ln 2 − R ln 2 
P1 
T
a1

S gen

(70.2 + 273) K
122 kPa 
= (111.5 kg) (1.005 kJ/kg.K)ln
− (0.287 kJ/kg.K)ln
 = 14.88 kJ/K
(12 + 273) K
101.3 kPa 

= ∆S total = ∆S w + ∆S a = −13.11 + 14.88 = 1.77 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-48
7-84 Air is accelerated in an isentropic nozzle. The maximum velocity at the exit is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 The nozzle operates steadily.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, k = 1.4 (Table A-2a).
Analysis The exit temperature is determined from ideal gas isentropic relation to be,
P
T2 = T1  2
 P1



( k −1) / k
 100 kPa 

= (400 + 273 K )
 800 kPa 
0.4/1.4
= 371.5 K
We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow
system can be expressed in the rate form as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12
/ 2) =
0=
=0
1
AIR
2
m& (h2 + V22 /2)
V2 −0
h2 − h1 + 2
2
0 = c p (T2 − T1 ) +
V22
2
Therefore,
V 2 = 2c p (T2 − T1 ) = 2(1.005 kJ/kg.K)(673 - 371.5)K = 778.5 m/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-49
7-85 An ideal gas is compressed in an isentropic compressor. 10% of gas is compressed to 400 kPa and
90% is compressed to 600 kPa. The compression process is to be sketched, and the exit temperatures at the
two exits, and the mass flow rate into the compressor are to be determined.
Assumptions 1 The compressor operates steadily. 2 The process is reversible-adiabatic (isentropic)
Properties The properties of ideal gas are given to be cp = 1.1 kJ/kg.K and cv = 0.8 kJ/kg.K.
Analysis (b) The specific heat ratio of the gas is
k=
cp
cv
=
1.1
= 1.375
0.8
P3 = 600 kPa
The exit temperatures are determined from ideal gas isentropic
relations to be,
P
T2 = T1  2
 P1



( k −1) / k
P
T3 = T1  3
 P1



( k −1) / k
 400 kPa 

= (27 + 273 K )
 100 kPa 
 600 kPa 

= (27 + 273 K )
 100 kPa 
COMPRESSOR
0.375/1.375
= 437.8 K
32 kW
= 489.0 K
P2 = 400 kPa
0.375/1.375
P1 = 100 kPa
T1 = 300 K
(c) A mass balance on the control volume gives
m& 1 = m& 2 + m& 3
where
m& 2 = 0.1m& 1
m& 3 = 0.9m& 1
T
P2
We take the compressor as the system, which is a control
volume. The energy balance for this steady-flow system can
be expressed in the rate form as
E& − E& out
1in
424
3
Rate of net energy transfer
by heat, work, and mass
P3
=
∆E& system©0 (steady)
1442443
P1
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1h1 + W&in = m& 2 h2 + m& 3h3
m& 1c pT1 + W&in = 0.1m& 1c pT2 + 0.9m& 1c pT3
Solving for the inlet mass flow rate, we obtain
W&in
m& 1 =
c p [0.1(T2 − T1 ) + 0.9(T3 − T1 )]
32 kW
(1.1 kJ/kg ⋅ K)[0.1(437.8 - 300) + 0.9(489.0 - 300)]
= 0.158 kg/s
=
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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s
7-50
7-86 Air contained in a constant-volume tank s cooled to ambient temperature. The entropy changes of the
air and the universe due to this process are to be determined and the process is to be sketched on a T-s
diagram.
Assumptions 1 Air is an ideal gas with constant specific heats.
Properties The specific heat of air at room temperature is
cv = 0.718 kJ/kg.K (Table A-2a).
Air
5 kg
327°C
100 kPa
Analysis (a) The entropy change of air is determined from
∆Sair = mcv ln
T2
T1
= (5 kg)(0.718 kJ/kg.K)ln
(27 + 273) K
(327 + 273) K
= −2.488 kJ/K
(b) An energy balance on the system gives
T
327ºC
Qout = mcv (T2 − T1 )
= (5 kg)(0.718 kJ/kg.K)(327 − 27)
= 1077 kJ
air
2
27ºC
The entropy change of the surroundings is
∆ssurr =
1
1
surr
2
Qout 1077 kJ
=
= 3.59 kJ/K
Tsurr
300 K
The entropy change of universe due to this process is
Sgen = ∆S total = ∆Sair + ∆Ssurr = −2.488 + 3.59 = 1.10 kJ/K
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s
7-51
Reversible Steady-Flow Work
7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas.
Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the
compressor.
7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the work
output of the turbine. Therefore, this is not a good proposal.
7-89C We would not support this proposal since the steady-flow work input to the pump is proportional to
the specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.
7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to the
pump is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is reversible.
Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg
(Table A-5).
Analysis The power input to the pump can be determined directly from the
2
steady-flow work relation for a liquid,

W&in = m& 

∫
2
1
©0
vdP + ∆ke
©0 
+ ∆pe
 = m& v1 (P2 − P1 )

45 kg/s
H2O
Substituting,
 1 kJ 
 = 274 kW
W&in = (45 kg/s)(0.001017 m3/kg )(6000 − 20) kPa 
3
 1 kPa ⋅ m 
1
7-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the water
can be pumped to is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is assumed to be reversible since we will determine the limiting case.
Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.
Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversible
steady-flow work relation for a liquid,
Thus,

W&in = m& 

∫
2
1

 1 kJ 

25 kJ/s = (5 kg/s)(0.001 m3/kg )( P2 − 100) k Pa 
3
⋅
1
kPa
m


It yields
P2

vdP + ∆ke©0 + ∆pe©0  = m& v1 (P2 − P1 )
25 kW
PUMP
P2 = 5100 kPa
100 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-52
7-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The power
input to the compressor is to be determined, and it is also to be compared to the work input for the liquid
case.
Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changes
are negligible. 3 The process is reversible. 4 The compressor is adiabatic.
Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then the
properties of the refrigerant are (Tables A-11E through A-13E)
P1 = 15 psia  h1 = 100.99 Btu/lbm

sat. vapor  s1 = 0.22715 Btu/lbm ⋅ R
P1 = 80 psia 
 h2 = 115.80 Btu/lbm
s2 = s1

R-134a
The work input to this isentropic compressor is
determined from the steady-flow energy balance to be
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
2
2
R-134a
=0
Rate of change in internal, kinetic,
potential, etc. energies
1
1
E& in = E& out
W&in + m& h1 = m& h2
W&in = m& (h2 − h1 )
Thus,
win = h2 − h1 = 115.80 − 100.99 = 14.8 Btu/lbm
If the refrigerant were first condensed at constant pressure before it was compressed, we would use a pump
to compress the liquid. In this case, the pump work input could be determined from the steady-flow work
relation to be
win =
∫
2
1
v dP + ∆ke©0 + ∆pe©0 = v1 (P2 − P1 )
where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting,


1 Btu
 = 0.14 Btu/lbm
win = (0.01165 ft 3/lbm)(80 − 15) psia 
3
 5.4039 psia ⋅ ft 
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-53
7-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of the
turbine work to the pump work is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic.
Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg
(Table A-5).
Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,
s1 = s2 and s3 = s4. Then the properties of the steam are
2
P4 = 20 kPa  h4 = h g @ 20 kPa = 2608.9 kJ/kg

sat.vapor  s 4 = s g @ 20 kPa = 7.9073 kJ/kg ⋅ K
P3 = 10 MPa 
 h3 = 4707.2 kJ/kg
s3 = s 4

H2O
3
H2O
Also, v1 = vf @ 20 kPa = 0.001017 m3/kg.
The work output to this isentropic turbine is determined
from the steady-flow energy balance to be
E& − E& out
1in
424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
1
4
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h3 = m& h4 + W&out
W&out = m& (h3 − h4 )
Substituting,
wturb,out = h3 − h4 = 4707.2 − 2608.9 = 2098.3 kJ/kg
The pump work input is determined from the steady-flow work relation to be
wpump,in =
2
∫ v dP + ∆ke
1
©0
+ ∆pe©0 = v1 (P2 − P1 )
 1 kJ 
= (0.001017 m3/kg )(10,000 − 20) kPa 

3
 1 kPa ⋅ m 
= 10.15 kJ/kg
Thus,
wturb,out
wpump,in
=
2098.3
= 206.7
10.15
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-54
7-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the net
work output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to be
plotted as a function of this quality.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"System: control volume for the pump and turbine"
"Property relation: Steam functions"
"Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible or
isentropic"
"Since we don't know the mass, we write the conservation of energy per unit mass."
"Conservation of mass: m_dot[1]= m_dot[2]"
"Knowns:"
WorkFluid$ = 'Steam_IAPWS'
P[1] = 20 [kPa]
x[1] = 0
P[2] = 10000 [kPa]
x[4] = 1.0
"Pump Analysis:"
T[1]=temperature(WorkFluid$,P=P[1],x=0)
v[1]=volume(workFluid$,P=P[1],x=0)
h[1]=enthalpy(WorkFluid$,P=P[1],x=0)
s[1]=entropy(WorkFluid$,P=P[1],x=0)
s[2] = s[1]
h[2]=enthalpy(WorkFluid$,P=P[2],s=s[2])
T[2]=temperature(WorkFluid$,P=P[2],s=s[2])
"The Volume function has the same form for an ideal gas as for a real fluid."
v[2]=volume(WorkFluid$,T=T[2],p=P[2])
"Conservation of Energy - SSSF energy balance for pump"
"
-- neglect the change in potential energy, no heat transfer:"
h[1]+W_pump = h[2]
"Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"
W_pump_incomp = v[1]*(P[2] - P[1])
"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potential
energy, no heat transfer:"
P[4] = P[1]
P[3] = P[2]
h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])
s[4]=entropy(WorkFluid$,P=P[4],x=x[4])
T[4]=temperature(WorkFluid$,P=P[4],x=x[4])
s[3] = s[4]
h[3]=enthalpy(WorkFluid$,P=P[3],s=s[3])
T[3]=temperature(WorkFluid$,P=P[3],s=s[3])
h[3] = h[4] + W_turb
W_net_out = W_turb - W_pump
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-55
Wnet,out
[kJ/kg]
557.1
734.7
913.6
1146
1516
2088
Wpump
[kJ/kg]
10.13
10.13
10.13
10.13
10.13
10.13
Wpump,incomp
[kJ/kg]
10.15
10.15
10.15
10.15
10.15
10.15
Wturb
[kJ/kg]
567.3
744.8
923.7
1156
1527
2098
x4
0.5
0.6
0.7
0.8
0.9
1
Steam
1100
3
1000
900
T [°C]
800
700
x = 1.0
600
= 0.5
4
500
10000 kPa
400
300
200
1, 2
100
0
0.0
0.2 20 kPa
1.1
2.2
3.3
0.6
4.4
5.5
4
0.8
6.6
7.7
8.8
9.9
11.0
s [kJ/kg-K]
2250
W net,out [kJ/kg]
1900
1550
1200
850
500
0.5
0.6
0.7
0.8
0.9
1
x[4]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-56
7-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highest
possible mass flow rate of water is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic
energy changes are negligible, but potential energy changes may be
significant. 3 The process is assumed to be reversible since we will
determine the limiting case.
Properties The specific volume of liquid water is given to be v1 = 0.001
m3/kg.
P2 = 5 MPa
PUMP
Analysis The highest mass flow rate will be realized when the entire
process is reversible. Thus it is determined from the reversible steadyflow work relation for a liquid,
Thus,

W&in = m& 

∫
2
1

v dP + ∆ke©0 + ∆pe  = m& {v (P2 − P1 ) + g (z2 − z1 )}

Water
P1 = 120 kPa

 1 kJ 
 1 kJ/kg 
 + (9.8 m/s 2 )(10 m)

7 kJ/s = m& (0.001 m3/kg )(5000 − 120)kPa 
3
 1000 m 2 /s 2 

 1 kPa ⋅ m 


It yields
m& = 1.41 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-57
7-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. The
power input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, and
two-stage compression.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kinetic
and potential energy changes are negligible.
Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R =
0.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667
(Table A-2E).
Analysis The mass flow rate of helium is
m& =
(
·
)
)
P1V&1
(14 psia ) 5 ft 3 /s
=
= 0.0493 lbm/s
RT1
2.6805 psia ⋅ ft 3 /lbm ⋅ R (530 R )
(
(a) Isentropic compression with k = 1.667:
kRT1  P2 
W&comp,in = m&
 
k − 1  P1 

2
He
W
5 ft3/s
(k −1) / k
1


− 1

0.667/1.667

(
1.667 )(0.4961 Btu/lbm ⋅ R )(530 R )  120 psia 


1
−
= (0.0493 lbm/s)



1.667 − 1

 14 psia 
= 44.11 Btu/s
= 62.4 hp
since 1 hp = 0.7068 Btu/s
(b) Polytropic compression with n = 1.2:
(n −1) / n 
nRT1  P2 



W&comp,in = m&
− 1

n − 1  P1 


0.2/1.2

(
1.2)(0.4961 Btu/lbm ⋅ R )(530 R )  120 psia 

− 1
= (0.0493 lbm/s)

1.2 − 1
14 psia 


= 33.47 Btu/s
= 47.3 hp
since 1 hp = 0.7068 Btu/s
(c) Isothermal compression:
P
120 psia
W&comp,in = m& RT ln 2 = (0.0493 lbm/s)(0.4961 Btu/lbm ⋅ R )(530 R )ln
= 27.83 Btu/s = 39.4 hp
P1
14 psia
(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each
stage is the same, and its value is determined from
Px = P1P2 =
(14 psia )(120 psia ) = 41.0 psia
The compressor work across each stage is also the same, thus total compressor work is twice the
compression work for a single stage:
(n −1) / n 
nRT1  Px 



− 1
W&comp,in = 2m& wcomp,I = 2m&

n − 1  P1 


0.2/1.2

(
1.2)(0.4961 Btu/lbm ⋅ R )(530 R )  41 psia 

= 2(0.0493 lbm/s)
− 1

1.2 − 1
14 psia 


= 30.52 Btu/s
= 43.2 hp
since 1 hp = 0.7068 Btu/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-58
7-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium is
to be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Procedure FuncPoly(m_dot,k, R,
T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out
_2stagePoly)
If n =1 then
T2=T1
W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"
W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"
Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"
Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"
Else
C_P = k*R/(k-1) "[Btu/lbm-R]"
T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"
W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"
Px=(P1*P2)^0.5
T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"
W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1T2x)"[Btu/s]"
endif
END
R=0.4961[Btu/lbm-R]
k=1.667
n=1.2
P1=14 [psia]
T1=70 [F]
P2=120 [psia]
V_dot = 5 [ft^3/s]
P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)
W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_isentropic = 0"[Btu/s]"
Call FuncPoly(m_dot,k, R,
T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out
_2stagePoly)
W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-59
n
Wcomp2StagePoly
[hp]
39.37
41.36
43.12
44.68
46.09
47.35
49.19
1
1.1
1.2
1.3
1.4
1.5
1.667
Wcompisentropic
[hp]
62.4
62.4
62.4
62.4
62.4
62.4
62.4
Wcompisothermal
[hp]
39.37
39.37
39.37
39.37
39.37
39.37
39.37
Wcomppolytropic
[hp]
39.37
43.48
47.35
50.97
54.36
57.54
62.4
65
W comp,polytropic [hp]
60
55
50
Polytropic
Isothermal
Isentropic
2StagePoly
45
40
35
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
n
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-60
7-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. The
mass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,
polytropic, isothermal, and two-stage compression.
Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kinetic
and potential energy changes are negligible.
Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio of
nitrogen is k = 1.4 (Table A-2).
2
Analysis (a) Isentropic compression:
{
or,
}
kRT1
(P2 P1 )(k −1) / k − 1
W&comp,in = m&
k −1
10 kJ/s = m&
(1.4)(0.297 kJ/kg ⋅ K )(300 K ) {(480 kPa
N2
·
m
}
10 kW
80 kPa )0.4/1.4 − 1
1.4 − 1
It yields
1
m& = 0.048 kg / s
(b) Polytropic compression with n = 1.3:
{
or,
}
nRT1
(P2 P1 )(n −1) / n − 1
W&comp,in = m&
n −1
10 kJ/s = m&
(1.3)(0.297 kJ/kg ⋅ K )(300 K ) {(480 kPa
1.3 − 1
}
80 kPa )0.3/1.3 − 1
It yields
m& = 0.051 kg / s
(c) Isothermal compression:
 480 kPa 
P

W&comp,in = m& RT ln 1 
→ 10 kJ/s = m& (0.297 kJ/kg ⋅ K )(300 K ) ln
P2
 80 kPa 
It yields
m& = 0.063 kg / s
(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each
stage is the same, and its value is determined to be
Px =
P1P2 =
(80 kPa )(480 kPa ) = 196 kPa
The compressor work across each stage is also the same, thus total compressor work is twice the
compression work for a single stage:
{
or,
}
nRT1
(Px P1 )(n −1) / n − 1
W&comp,in = 2m& wcomp,I = 2m&
n −1
10 kJ/s = 2m&
(1.3)(0.297 kJ/kg ⋅ K )(300 K ) {(196 kPa
1.3 − 1
}
80 kPa )0.3/1.3 − 1
It yields
m& = 0.056 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-61
7-99 Water mist is to be sprayed into the air stream in the compressor to cool the air as the water
evaporates and to reduce the compression power. The reduction in the exit temperature of the compressed
air and the compressor power saved are to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is reversible. 3 Kinetic and
potential energy changes are negligible. 3 Air is compressed isentropically. 4 Water vaporizes completely
before leaving the compressor. 4 Air properties can be used for the air-vapor mixture.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air is k =
1.4. The inlet enthalpies of water and air are (Tables A-4 and A-17)
hw1 = hf@20°C = 83.29 kJ/kg , hfg@20°C = 2453.9 kJ/kg and ha1 = h@300 K =300.19 kJ/kg
Analysis In the case of isentropic operation (thus no cooling or water spray), the exit temperature and the
power input to the compressor are
T2  P2
=
T1  P1



( k −1) / k
 1200 kPa 
→ T2 = (300 K)

 100 kPa 
{
(1.4 −1) / 1.4
= 610.2 K
}
kRT1
(P2 P1 )(k −1) / k − 1
W&comp,in = m&
k −1
(1.4)(0.287 kJ/kg ⋅ K )(300 K ) (1200 kPa/100 kPa )0.4/1.4 − 1 = 654.3 kW
= (2.1 kg/s)
1.4 − 1
{
When water is sprayed, we first need to check the accuracy of the
assumption that the water vaporizes completely in the compressor. In the
limiting case, the compression will be isothermal at the compressor inlet
temperature, and the water will be a saturated vapor. To avoid the
complexity of dealing with two fluid streams and a gas mixture, we
disregard water in the air stream (other than the mass flow rate), and
assume air is cooled by an amount equal to the enthalpy change of water.
The rate of heat absorption of water as it evaporates at the inlet
temperature completely is
}
1200 kPa
2
·
He
Water
20°C
Q& cooling,max = m& w h fg @ 20°C = (0.2 kg/s)(2453.9 kJ/kg) = 490.8 kW
1
W
100 kPa
300 K
The minimum power input to the compressor is
 1200 kPa 
P
 = 449.3 kW
W&comp,in, min = m& RT ln 2 = (2.1 kg/s)(0.287 kJ/kg ⋅ K)(300 K) ln
P1
 100 kPa 
This corresponds to maximum cooling from the air since, at constant temperature, ∆h = 0 and thus
Q& out = W&in = 449.3 kW , which is close to 490.8 kW. Therefore, the assumption that all the water vaporizes
is approximately valid. Then the reduction in required power input due to water spray becomes
∆W&comp,in = W&comp, isentropic − W&comp, isothermal = 654.3 − 449.3 = 205 kW
Discussion (can be ignored): At constant temperature, ∆h = 0 and thus Q& out = W& in = 449.3 kW corresponds
to maximum cooling from the air, which is less than 490.8 kW. Therefore, the assumption that all the water
vaporizes is only roughly valid. As an alternative, we can assume the compression process to be polytropic
and the water to be a saturated vapor at the compressor exit temperature, and disregard the remaining
liquid. But in this case there is not a unique solution, and we will have to select either the amount of water
or the exit temperature or the polytropic exponent to obtain a solution. Of course we can also tabulate the
results for different cases, and then make a selection.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-62
Sample Analysis: We take the compressor exit temperature to be T2 = 200°C = 473 K. Then,
hw2 = hg@200°C = 2792.0 kJ/kg and ha2 = h@473 K = 475.3 kJ/kg
Then,
T2  P2 
= 
T1  P1 
( n −1) / n
→
{
473 K  1200 kPa 
=

300 K  100 kPa 
( n −1) / n
→ n = 1.224
}
nRT1
(P2 P1 )(n−1) / n − 1 = m& nR (T2 − T1 )
W& comp,in = m&
n −1
n −1
1.224 )(0.287 kJ/kg ⋅ K )
(
= (2.1 kg/s)
(473 − 300)K = 570 kW
1.224 − 1
Energy balance:
W&comp,in − Q& out = m& (h2 − h1 ) → Q& out = W&comp,in − m& (h2 − h1 )
= 569.7 kW − (2.1 kg/s)(475.3 − 300.19) = 202.0 kW
Noting that this heat is absorbed by water, the rate at which water evaporates in the compressor becomes
Q&
202.0 kJ/s
→ m& w = in, water =
= 0.0746 kg/s
Q& out,air = Q& in, water = m& w (hw2 − hw1 ) 
hw2 − hw1 (2792.0 − 83.29) kJ/kg
Then the reductions in the exit temperature and compressor power input become
∆T2 = T2,isentropic − T2,water cooled = 610.2 − 473 = 137.2° C
∆W&comp,in = W&comp,isentropic − W&comp,water cooled = 654.3 − 570 = 84.3 kW
Note that selecting a different compressor exit temperature T2 will result in different values.
7-100 A water-injected compressor is used in a gas turbine power plant. It is claimed that the power output
of a gas turbine will increase when water is injected into the compressor because of the increase in the
mass flow rate of the gas (air + water vapor) through the turbine. This, however, is not necessarily right
since the compressed air in this case enters the combustor at a low temperature, and thus it absorbs much
more heat. In fact, the cooling effect will most likely dominate and cause the cyclic efficiency to drop.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-63
Isentropic Efficiencies of Steady-Flow Devices
7-101C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. The
adiabatic efficiencies of these devices are defined as
actual work output
insentropic work input
actual exit kineticenergy
,η =
, and η N =
ηT =
insentropic work output C
actual work input
insentropic exit kinetic energy
7-102C No, because the isentropic process is not the model or ideal process for compressors that are
cooled intentionally.
7-103C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a result
of irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exit
state
7-104 Steam enters an adiabatic turbine with an isentropic efficiency of 0.90 at a specified state with a
specified mass flow rate, and leaves at a specified pressure. The turbine exit temperature and power output
of the turbine are to be determined.
Assumptions 1 This is a steady-flow process since there is no change
P1 = 8 MPa
with time. 2 Kinetic and potential energy changes are negligible. 3
T
The device is adiabatic and thus heat transfer is negligible.
1 = 500°C
Analysis (a) From the steam tables (Tables A-4 through A-6),
P1 = 8 MPa  h1 = 3399.5 kJ/kg

T1 = 500°C  s1 = 6.7266 kJ/kg ⋅ K
P2 s = 30 kPa 

s 2 s = s1
h
s2s − s f
STEAM
TURBINE
ηT = 90%
6.7266 − 0.9441
= 0.8475
6.8234
s fg
= h f + x 2 s h fg = 289.27 + (0.8475)(2335.3) = 2268.3 kJ/kg
x2s =
2s
=
P2 = 30 kPa
From the isentropic efficiency relation,
h −h
η T = 1 2a 
→ h2 a = h1 − η T (h1 − h2 s ) = 3399.5 − (0.9)(3399.5 − 2268.3) = 2381.4 kJ/kg
h1 − h2 s
Thus,
P2 a = 30 kPa

 T2 a = Tsat @ 30 kPa = 69.09°C
h2 a = 2381.4 kJ/kg 
&1 = m
&2 = m
& . We take the actual turbine as the system,
(b) There is only one inlet and one exit, and thus m
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& system©0 (steady)
=0
E& − E& out
1in
424
3
1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h1 = W&a,out + m& h2
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&a,out = m& (h1 − h2 )
Substituting,
W&
a, out
= (3kg/s )(3399.5 − 2381.4) kJ/kg = 3054 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-64
7-105 EES Problem 7-104 is reconsidered. The effect of varying the turbine isentropic efficiency from
0.75 to 1.0 on both the work done and the exit temperature of the steam are to be investigated, and the
results are to be plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"System: control volume for turbine"
"Property relation: Steam functions"
"Process: Turbine: Steady state, steady flow, adiabatic, reversible or isentropic"
"Since we don't know the mass, we write the conservation of energy per unit mass."
"Conservation of mass: m_dot[1]= m_dot[2]=m_dot"
600
T [°C]
400
300
8000 kPa
0.4
0.6
0.2
30 kPa
1.0
2.0
3.0
4.0
5.0
2
0.8
2
s
6.0
7.0
8.0
9.0
10.0
s [kJ/kg-K]
3400
3300
3200
3100
W turb [kW ]
Wturb [kW]
2545
2715
2885
3054
3224
3394
1
500
"Conservation of Energy - SSSF energy
200
balance for turbine -- neglecting the
change in potential energy, no heat
100
transfer:"
h[1]=enthalpy(WorkFluid$,P=P[1],T=T[1])
0
0.0
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])
T_s[1] = T[1]
s[2] =s[1]
s_s[2] = s[1]
h_s[2]=enthalpy(WorkFluid$,P=P[2],s=s_s[2])
T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])
eta_turb = w_turb/w_turb_s
h[1] = h[2] + w_turb
h[1] = h_s[2] + w_turb_s
T[2]=temperature(WorkFluid$,P=P[2],h=h[2])
W_dot_turb = m_dot*w_turb
ηturb
0.75
0.8
0.85
0.9
0.95
1
Steam
700
"Knowns:"
WorkFluid$ = 'Steam_iapws'
m_dot = 3 [kg/s]
P[1] = 8000 [kPa]
T[1] = 500 [C]
P[2] = 30 [kPa]
"eta_turb = 0.9"
3000
2900
2800
2700
2600
2500
0.75
0.8
0.85
η
0.9
0.95
turb
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
1
7-65
7-106 Steam enters an adiabatic turbine at a specified state, and leaves at a specified state. The mass flow
rate of the steam and the isentropic efficiency are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible.
Analysis (a) From the steam tables (Tables A-4 and A-6),
P1 = 7 MPa  h1 = 3650.6 kJ/kg

T1 = 600°C  s1 = 7.0910 kJ/kg ⋅ K
P2 = 50 kPa 
 h2 a = 2780.2 kJ/kg
T2 = 150°C 
&1 = m
&2 = m
& . We take the actual turbine as the system, which
There is only one inlet and one exit, and thus m
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
1in
424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
=0
1
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
6 MW
H2O
m& (h1 + V12 / 2) = W&a,out + m& (h2 + V12 /2) (since Q& ≅ ∆pe ≅ 0)

V 2 − V12 
W&a,out = − m&  h2 − h1 + 2


2


2
Substituting, the mass flow rate of the steam is determined to be

(140 m/s) 2 − (80 m/s) 2  1 kJ/kg
6000 kJ/s = −m&  2780.2 − 3650.6 +


2
 1000 m 2 /s 2

m& = 6.95 kg/s
 


(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are
P2 s = 50 kPa 

s 2 s = s1
h
and
(
2s
s 2s − s f
7.0910 − 1.0912
=
= 0.9228
6.5019
s fg
= h f + x 2 s h fg = 340.54 + (0.9228)(2304.7 ) = 2467.3 kJ/kg
x 2s =
{(
) })
W&s, out = − m& h2 s − h1 + V22 − V12 / 2

(140 m/s) 2 − (80 m/s) 2  1 kJ/kg  
W&s, out = −(6.95 kg/s ) 2467.3 − 3650.6 +


2 2

2
 1000 m /s  

= 8174 kW
Then the isentropic efficiency of the turbine becomes
W&
6000 kW
ηT = a =
= 0.734 = 73.4%
&
W s 8174 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-66
7-107 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a
specified pressure. The isentropic efficiency of 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 device is adiabatic and thus heat transfer is negligible. 4 Argon is an
ideal gas with constant specific heats.
Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp =
0.5203 kJ/kg.K (Table A-2).
&1 = m
&2 = m
& . We take the isentropic turbine as the
Analysis There is only one inlet and one exit, and thus m
system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
1
E& in = E& out
m& h1 = W& s ,out + m& h2 s
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&s, out = m& (h1 − h2 s )
ηT
From the isentropic relations,
(k −1) / k
0.667/1.667
P 
 200 kPa 

= 479 K
T2 s = T1  2 s 
= (1073 K )
P
 1500 kPa 
 1 
Then the power output of the isentropic turbine becomes
W&
= m& c (T − T ) = (80/60 kg/min )(0.5203 kJ/kg ⋅ K )(1073 − 479) = 412.1 kW
s, out
p
1
370 kW
Ar
2
2s
Then the isentropic efficiency of the turbine is determined from
370 kW
W&
η T = &a =
= 0.898 = 89.8%
412.1 kW
W
s
7-108E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specified
state, and leave at a specified pressure. The work output of 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 device is adiabatic and thus heat transfer is negligible. 4 Combustion
gases can be treated as air that is an ideal gas with variable specific heats.
Analysis From the air table and isentropic relations,
1
h1 = 504.71 Btu / lbm
T1 = 2000 R

→
Pr1 = 174.0
P 
 60 psia 
(174.0) = 87.0 
→ h2s = 417.3 Btu/lbm
Pr2 =  2  Pr1 = 
P
 120 psia 
 1
&1 = m
&2 = m
& . We take the
There is only one inlet and one exit, and thus m
actual turbine as the system, which is a control volume since mass crosses the
boundary. The energy balance for this steady-flow system can be expressed
as
E& = E&
in
AIR
ηT = 82%
2
out
m& h1 = W&a,out + m& h2
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&a,out = m& (h1 − h2 )
Noting that wa = ηTws, the work output of the turbine per unit mass is determined from
wa = (0.82 )(504.71 − 417.3)Btu/lbm = 71.7 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-67
7-109 [Also solved by EES on enclosed CD] Refrigerant-134a enters an adiabatic compressor with an
isentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specified
pressure. The compressor exit temperature and power input to the compressor 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.
2
Analysis (a) From the refrigerant tables (Tables A-11E through A-13E),
h1 = hg @120 kPa = 236.97 kJ/kg
P1 = 120 kPa 
 s1 = s g @120 kPa = 0.94779 kJ/kg ⋅ K
sat. vapor 
v =v
= 0.16212 m3/kg
g @120 kPa
1
R-134a
ηC = 80%
P2 = 1 MPa 
 h2 s = 281.21 kJ/kg

s2 s = s1
0.3 m3/min
From the isentropic efficiency relation,
ηC =
1
h2 s − h1

→ h2 a = h1 + (h2 s − h1 ) /η C = 236.97 + (281.21 − 236.97 )/0.80 = 292.26 kJ/kg
h2 a − h1
Thus,
P2 a = 1 MPa
h2 a

 T2 a = 58.9°C
= 292.26 kJ/kg 
(b) The mass flow rate of the refrigerant is determined from
m& =
V&1
0.3/60 m 3 /s
=
= 0.0308 kg/s
v 1 0.16212 m 3 /kg
&1 = m
&2 = m
& . We take the actual compressor as the system,
There is only one inlet and one exit, and thus m
which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system
can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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 )
Substituting, the power input to the compressor becomes,
W&a,in = (0.0308 kg/s )(292.26 − 236.97 )kJ/kg = 1.70 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-68
7-110 EES Problem 7-109 is reconsidered. The problem is to be solved by considering the kinetic energy
and by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe inside
diameter is 2 cm.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data from diagram window"
{P[1] = 120 "kPa"
P[2] = 1000 "kPa"
Vol_dot_1 = 0.3 "m^3/min"
Eta_c = 0.80 "Compressor adiabatic efficiency"
A_ratio = 1.5
d_2 = 2/100 "m"}
"System: Control volume containing the compressor, see the diagram window.
Property Relation: Use the real fluid properties for R134a.
Process: Steady-state, steady-flow, adiabatic process."
Fluid$='R134a'
"Property Data for state 1"
T[1]=temperature(Fluid$,P=P[1],x=1)"Real fluid equ. at the sat. vapor state"
h[1]=enthalpy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
s[1]=entropy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
v[1]=volume(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
"Property Data for state 2"
s_s[1]=s[1]; T_s[1]=T[1] "needed for plot"
s_s[2]=s[1] "for the ideal, isentropic process across the compressor"
h_s[2]=ENTHALPY(Fluid$, P=P[2], s=s_s[2])"Enthalpy 2 at the isentropic state 2s and
pressure P[2]"
T_s[2]=Temperature(Fluid$, P=P[2], s=s_s[2])"Temperature of ideal state - needed only for
plot."
"Steady-state, steady-flow conservation of mass"
m_dot_1 = m_dot_2
m_dot_1 = Vol_dot_1/(v[1]*60)
Vol_dot_1/v[1]=Vol_dot_2/v[2]
Vel[2]=Vol_dot_2/(A[2]*60)
A[2] = pi*(d_2)^2/4
A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]"
A_ratio=A[1]/A[2]
"Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagram
window"
m_dot_1*(h[1]+(Vel[1])^2/(2*1000)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/(2*1000))
"Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act"
Eta_c = (h_s[2]-h[1])/(h[2]-h[1])
"Knowing h[2], the other properties at state 2 can be found."
v[2]=volume(Fluid$, P=P[2], h=h[2])"v[2] is found at the actual state 2, knowing P and h."
T[2]=temperature(Fluid$, P=P[2],h=h[2])"Real fluid equ. for T at the known outlet h and P."
s[2]=entropy(Fluid$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P."
T_exit=T[2]
"Neglecting the kinetic energies, the work is:"
m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-69
SOLUTION
s_s[1]=0.9478 [kJ/kg-K]
s_s[2]=0.9478 [kJ/kg-K]
T[1]=-22.32 [C]
T[2]=58.94 [C]
T_exit=58.94 [C]
T_s[1]=-22.32 [C]
T_s[2]=48.58 [C]
Vol_dot_1=0.3 [m^3 /min]
Vol_dot_2=0.04244 [m^3 /min]
v[1]=0.1621 [m^3/kg]
v[2]=0.02294 [m^3/kg]
Vel[1]=10.61 [m/s]
Vel[2]=2.252 [m/s]
W_dot_c=1.704 [kW]
W_dot_c_noke=1.706 [kW]
A[1]=0.0004712 [m^2]
A[2]=0.0003142 [m^2]
A_ratio=1.5
d_2=0.02 [m]
Eta_c=0.8
Fluid$='R134a'
h[1]=237 [kJ/kg]
h[2]=292.3 [kJ/kg]
h_s[2]=281.2 [kJ/kg]
m_dot_1=0.03084 [kg/s]
m_dot_2=0.03084 [kg/s]
P[1]=120.0 [kPa]
P[2]=1000.0 [kPa]
s[1]=0.9478 [kJ/kg-K]
s[2]=0.9816 [kJ/kg-K]
R134a
125
T-s diagram for real and ideal com pressor
100
Tem perature [C]
75
Ideal Compressor
Real Com pressor
50
1000 kPa
25
0
-25
-50
0.0
120 kPa
0.2
0.4
0.6
0.8
1.0
1.2
Entropy [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-70
7-111 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, and
leaves at a specified temperature. The exit pressure of air and the power input to the compressor 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. 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)
2
Analysis (a) From the air table (Table A-17),
T1 = 290 K 
→ h1 = 290.16 kJ/kg, Pr1 = 1.2311
T2 = 530 K 
→ h2 a = 533.98 kJ/kg
From the isentropic efficiency relation η C =
AIR
ηC = 84%
h2 s − h1
,
h2 a − h1
h2 s = h1 + η C (h2 a − h1 )
2.4 m3/s
1
= 290.16 + (0.84 )(533.98 − 290.16 ) = 495.0 kJ/kg 
→ Pr2 = 7.951
Then from the isentropic relation ,
 Pr 
P2 Pr2
 7.951 
=

→ P2 =  2  P1 = 
(100 kPa ) = 646 kPa


P1 Pr1
 1.2311 
 Pr1 
&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 since mass crosses the boundary. The energy balance for this steadyflow system can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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 )
where
m& =
P1V&1
(100 kPa )(2.4 m 3 /s)
=
= 2.884 kg/s
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(290 K )
Then the power input to the compressor is determined to be
W&a,in = (2.884 kg/s)(533.98 − 290.16) kJ/kg = 703 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-71
7-112 Air is compressed by an adiabatic compressor from a specified state to another specified state. The
isentropic efficiency of the compressor and the exit temperature of air for the isentropic case 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. 4 Air is an
ideal gas with variable specific heats.
Analysis (a) From the air table (Table A-17),
T1 = 300 K

→
h1 = 300.19 kJ / kg,
T2 = 550 K

→
h2 a = 554.74 kJ / kg
2
Pr1 = 1.386
From the isentropic relation,
AIR
P 
 600 kPa 
(1.386) = 8.754 
→ h2 s = 508.72 kJ/kg
Pr2 =  2  Pr1 = 
P
 95 kPa 
 1
Then the isentropic efficiency becomes
ηC =
h2 s − h1 508.72 − 30019
.
=
= 0.819 = 81.9%
h2a − h1 554.74 − 30019
.
1
(b) If the process were isentropic, the exit temperature would be
h2 s = 508.72 kJ / kg

→
T2 s = 505.5 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-72
7-113E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, and
leaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with
constant specific heats.
Properties The specific heat ratio of argon is k = 1.667. The constant
pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E).
2
Analysis (a) The isentropic exit temperature T2s is determined from
P 
T2 s = T1 2 s 
 P1 
(k −1) / k
 200 psia 

= (550 R )
 20 psia 
Ar
0.667/1.667
= 1381.9 R
ηC = 80%
The actual kinetic energy change during this process is
∆ke a =
V 22 − V12 (240 ft/s )2 − (60 ft/s )2
=
2
2
 1 Btu/lbm

 25,037 ft 2 /s 2


 = 1.08 Btu/lbm


1
The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kinetic
energy changes for the actual and isentropic cases to be same in efficiency calculations. From the
isentropic efficiency relation, including the effect of kinetic energy,
ηC =
It yields
w s (h2 s − h1 ) + ∆ke c p (T2 s − T1 ) + ∆ke s
0.1253(1381.9 − 550 ) + 1.08
=
=

→ 0.8 =
wa (h2 a − h1 ) + ∆ke c p (T2 a − T1 ) + ∆ke a
0.1253(T2 a − 550 ) + 1.08
T2a = 1592 R
&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 since mass crosses the boundary. The energy balance for this steadyflow system can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W&a,in + m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ ∆pe ≅ 0)

V 2 − V12 
W&a,in = m&  h2 − h1 + 2

→ wa,in = h2 − h1 + ∆ke


2


Substituting, the work input to the compressor is determined to be
wa,in = (0.1253 Btu/lbm ⋅ R )(1592 − 550)R + 1.08 Btu/lbm = 131.6 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-73
7-114 CO2 gas is compressed by an adiabatic compressor from a specified state to another specified state.
The isentropic efficiency of the compressor 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 device is adiabatic and thus heat transfer is negligible. 4 CO2 is an
ideal gas with constant specific heats.
Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specific
2
heat and the specific heat ratio of CO2 are k = 1.260 and cp = 0.917 kJ/kg.K (Table A-2).
Analysis The isentropic exit temperature T2s is
(k −1) / k
0.260/1.260
P 
 600 kPa 

= (300 K )
= 434.2 K
T2 s = T1 2 s 
 100 kPa 
 P1 
From the isentropic efficiency relation,
c p (T2 s − T1 ) T2 s − T1 434.2 − 300
w
h −h
η C = s = 2s 1 =
= 0.895 = 89.5%
=
=
wa h2 a − h1 c p (T2 a − T1 ) T2 a − T1
450 − 300
CO2
1.8 kg/s
1
7-115E Air is accelerated in a 90% efficient adiabatic nozzle from low velocity to a specified velocity. The
exit temperature and pressure of the air are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with
variable specific heats.
Analysis From the air table (Table A-17E),
T1 = 1480 R

→
h1 = 363.89 Btu / lbm,
Pr1 = 53.04
&1 = m
&2 = m
& . We take the nozzle as the system, which is a
There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be
expressed as
E& − E& out
=
∆E& system©0 (steady)
=0
1in
424
3
1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
1
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since W& = Q& ≅ ∆pe ≅ 0)
AIR
ηN = 90%
2
©0
V22 − V12
2
Substituting, the exit temperature of air is determined to be
h2 = h1 −
(800 ft/s )2 − 0 
1 Btu/lbm 
 = 351.11 Btu/lbm
2 2

2
 25,037 ft /s 
From the air table we read
T2a = 1431.3 R
h −h
From the isentropic efficiency relation η N = 2 a 1 ,
h2 s − h1
h2 = 363.89 kJ/kg −
h2 s = h1 + (h2 a − h1 ) /η N = 363.89 + (351.11 − 363.89 )/ (0.90 ) = 349.69 Btu/lbm 
→ Pr2 = 46.04
Then the exit pressure is determined from the isentropic relation to be
 Pr
P2 Pr2
=

→ P2 =  2
 Pr
P1 Pr1
 1

 P =  46.04 (60 psia ) = 52.1 psia
 1  53.04 

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-74
7-116E EES Problem 7-115E is reconsidered. The effect of varying the nozzle isentropic efficiency from
0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to be
plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
WorkFluid$ = 'Air'
P[1] = 60 [psia]
T[1] = 1020 [F]
Vel[2] = 800 [ft/s]
Vel[1] = 0 [ft/s]
eta_nozzle = 0.9
"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potential
energy, no heat transfer:"
h[1]=enthalpy(WorkFluid$,T=T[1])
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])
T_s[1] = T[1]
s[2] =s[1]
s_s[2] = s[1]
h_s[2]=enthalpy(WorkFluid$,T=T_s[2])
T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])
eta_nozzle = ke[2]/ke_s[2]
ke[1] = Vel[1]^2/2
ke[2]=Vel[2]^2/2
h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm)
h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm)
T[2]=temperature(WorkFluid$,h=h[2])
P_2_answer = P[2]
T_2_answer = T[2]
0.8
0.85
0.9
0.95
1
P2
[psia
51.09
51.58
52.03
52.42
52.79
T2
[F
971.4
971.4
971.4
971.4
971.4
Ts,2
[F]
959.2
962.8
966
968.8
971.4
980
970
T s[2]
ηnozzle
T
960
T
950
0.8
0.84
s [2 ]
0.88
η
2
0.92
0.96
nozzle
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
1
7-75
7-117 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a
specified velocity. The exit velocity and the exit temperature are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustion gases can be
treated as air that is an ideal gas with variable specific heats.
Analysis From the air table (Table A-17),
P1 = 260 kPa
T1 = 747°C
V1 = 80 m/s
T1 = 1020 K 
→ h1 = 1068.89 kJ/kg, Pr1 = 123.4
From the isentropic relation ,
AIR
ηN = 92%
P2 = 85 kPa
P 
 85 kPa 
(123.4 ) = 40.34 
Pr2 =  2  Pr1 = 
→ h2 s = 783.92 kJ/kg
P
 260 kPa 
 1
&1 = m
&2 = m
& . We take the nozzle as the system, which is a
There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system for the
isentropic process can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12 / 2) = m& (h2 s + V 22s /2) (since W& = Q& ≅ ∆pe ≅ 0)
h2 s = h1 −
V 22s − V12
2
Then the isentropic exit velocity becomes
V 2 s = V12 + 2(h1 − h2 s ) =

/s 2
1 kJ/kg
(80 m/s )2 + 2(1068.89 − 783.92)kJ/kg 1000 m

2

 = 759.2 m/s


Therefore,
V2 a = η N V2 s = 0.92 (759.2 m/s ) = 728.2 m/s
The exit temperature of air is determined from the steady-flow energy equation,
h2 a = 1068.89 kJ/kg −
From the air table we read
(728.2 m/s)2 − (80 m/s)2 
2


 1000 m /s  = 806.95 kJ/kg


1 kJ/kg
2
2
T2a = 786.3 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-51
Reversible Steady-Flow Work
7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas.
Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the
compressor.
7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the work
output of the turbine. Therefore, this is not a good proposal.
7-89C We would not support this proposal since the steady-flow work input to the pump is proportional to
the specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.
7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to the
pump is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is reversible.
Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg
(Table A-5).
Analysis The power input to the pump can be determined directly from the
2
steady-flow work relation for a liquid,

W&in = m& 

∫
2
1
©0
vdP + ∆ke
©0 
+ ∆pe
 = m& v1 (P2 − P1 )

45 kg/s
H2O
Substituting,
 1 kJ 
 = 274 kW
W&in = (45 kg/s)(0.001017 m3/kg )(6000 − 20) kPa 
3
 1 kPa ⋅ m 
1
7-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the water
can be pumped to is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is assumed to be reversible since we will determine the limiting case.
Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.
Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversible
steady-flow work relation for a liquid,
Thus,

W&in = m& 

∫
2
1

 1 kJ 

25 kJ/s = (5 kg/s)(0.001 m3/kg )( P2 − 100) k Pa 
3
⋅
1
kPa
m


It yields
P2

vdP + ∆ke©0 + ∆pe©0  = m& v1 (P2 − P1 )
25 kW
PUMP
P2 = 5100 kPa
100 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-52
7-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The power
input to the compressor is to be determined, and it is also to be compared to the work input for the liquid
case.
Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changes
are negligible. 3 The process is reversible. 4 The compressor is adiabatic.
Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then the
properties of the refrigerant are (Tables A-11E through A-13E)
P1 = 15 psia  h1 = 100.99 Btu/lbm

sat. vapor  s1 = 0.22715 Btu/lbm ⋅ R
P1 = 80 psia 
 h2 = 115.80 Btu/lbm
s2 = s1

R-134a
The work input to this isentropic compressor is
determined from the steady-flow energy balance to be
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
2
2
R-134a
=0
Rate of change in internal, kinetic,
potential, etc. energies
1
1
E& in = E& out
W&in + m& h1 = m& h2
W&in = m& (h2 − h1 )
Thus,
win = h2 − h1 = 115.80 − 100.99 = 14.8 Btu/lbm
If the refrigerant were first condensed at constant pressure before it was compressed, we would use a pump
to compress the liquid. In this case, the pump work input could be determined from the steady-flow work
relation to be
win =
∫
2
1
v dP + ∆ke©0 + ∆pe©0 = v1 (P2 − P1 )
where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting,


1 Btu
 = 0.14 Btu/lbm
win = (0.01165 ft 3/lbm)(80 − 15) psia 
3
 5.4039 psia ⋅ ft 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-53
7-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of the
turbine work to the pump work is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are
negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic.
Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg
(Table A-5).
Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,
s1 = s2 and s3 = s4. Then the properties of the steam are
2
P4 = 20 kPa  h4 = h g @ 20 kPa = 2608.9 kJ/kg

sat.vapor  s 4 = s g @ 20 kPa = 7.9073 kJ/kg ⋅ K
P3 = 10 MPa 
 h3 = 4707.2 kJ/kg
s3 = s 4

H2O
3
H2O
Also, v1 = vf @ 20 kPa = 0.001017 m3/kg.
The work output to this isentropic turbine is determined
from the steady-flow energy balance to be
E& − E& out
1in
424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
1
4
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h3 = m& h4 + W&out
W&out = m& (h3 − h4 )
Substituting,
wturb,out = h3 − h4 = 4707.2 − 2608.9 = 2098.3 kJ/kg
The pump work input is determined from the steady-flow work relation to be
wpump,in =
2
∫ v dP + ∆ke
1
©0
+ ∆pe©0 = v1 (P2 − P1 )
 1 kJ 
= (0.001017 m3/kg )(10,000 − 20) kPa 

3
 1 kPa ⋅ m 
= 10.15 kJ/kg
Thus,
wturb,out
wpump,in
=
2098.3
= 206.7
10.15
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-54
7-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the net
work output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to be
plotted as a function of this quality.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"System: control volume for the pump and turbine"
"Property relation: Steam functions"
"Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible or
isentropic"
"Since we don't know the mass, we write the conservation of energy per unit mass."
"Conservation of mass: m_dot[1]= m_dot[2]"
"Knowns:"
WorkFluid$ = 'Steam_IAPWS'
P[1] = 20 [kPa]
x[1] = 0
P[2] = 10000 [kPa]
x[4] = 1.0
"Pump Analysis:"
T[1]=temperature(WorkFluid$,P=P[1],x=0)
v[1]=volume(workFluid$,P=P[1],x=0)
h[1]=enthalpy(WorkFluid$,P=P[1],x=0)
s[1]=entropy(WorkFluid$,P=P[1],x=0)
s[2] = s[1]
h[2]=enthalpy(WorkFluid$,P=P[2],s=s[2])
T[2]=temperature(WorkFluid$,P=P[2],s=s[2])
"The Volume function has the same form for an ideal gas as for a real fluid."
v[2]=volume(WorkFluid$,T=T[2],p=P[2])
"Conservation of Energy - SSSF energy balance for pump"
"
-- neglect the change in potential energy, no heat transfer:"
h[1]+W_pump = h[2]
"Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"
W_pump_incomp = v[1]*(P[2] - P[1])
"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potential
energy, no heat transfer:"
P[4] = P[1]
P[3] = P[2]
h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])
s[4]=entropy(WorkFluid$,P=P[4],x=x[4])
T[4]=temperature(WorkFluid$,P=P[4],x=x[4])
s[3] = s[4]
h[3]=enthalpy(WorkFluid$,P=P[3],s=s[3])
T[3]=temperature(WorkFluid$,P=P[3],s=s[3])
h[3] = h[4] + W_turb
W_net_out = W_turb - W_pump
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-55
Wnet,out
[kJ/kg]
557.1
734.7
913.6
1146
1516
2088
Wpump
[kJ/kg]
10.13
10.13
10.13
10.13
10.13
10.13
Wpump,incomp
[kJ/kg]
10.15
10.15
10.15
10.15
10.15
10.15
Wturb
[kJ/kg]
567.3
744.8
923.7
1156
1527
2098
x4
0.5
0.6
0.7
0.8
0.9
1
Steam
1100
3
1000
900
T [°C]
800
700
x = 1.0
600
= 0.5
4
500
10000 kPa
400
300
200
1, 2
100
0
0.0
0.2 20 kPa
1.1
2.2
3.3
0.6
4.4
5.5
4
0.8
6.6
7.7
8.8
9.9
11.0
s [kJ/kg-K]
2250
W net,out [kJ/kg]
1900
1550
1200
850
500
0.5
0.6
0.7
0.8
0.9
1
x[4]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-56
7-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highest
possible mass flow rate of water is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic
energy changes are negligible, but potential energy changes may be
significant. 3 The process is assumed to be reversible since we will
determine the limiting case.
Properties The specific volume of liquid water is given to be v1 = 0.001
m3/kg.
P2 = 5 MPa
PUMP
Analysis The highest mass flow rate will be realized when the entire
process is reversible. Thus it is determined from the reversible steadyflow work relation for a liquid,
Thus,

W&in = m& 

∫
2
1

v dP + ∆ke©0 + ∆pe  = m& {v (P2 − P1 ) + g (z2 − z1 )}

Water
P1 = 120 kPa

 1 kJ 
 1 kJ/kg 
 + (9.8 m/s 2 )(10 m)

7 kJ/s = m& (0.001 m3/kg )(5000 − 120)kPa 
3
 1000 m 2 /s 2 

 1 kPa ⋅ m 


It yields
m& = 1.41 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-57
7-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. The
power input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, and
two-stage compression.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kinetic
and potential energy changes are negligible.
Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R =
0.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667
(Table A-2E).
Analysis The mass flow rate of helium is
m& =
(
·
)
)
P1V&1
(14 psia ) 5 ft 3 /s
=
= 0.0493 lbm/s
RT1
2.6805 psia ⋅ ft 3 /lbm ⋅ R (530 R )
(
(a) Isentropic compression with k = 1.667:
kRT1  P2 
W&comp,in = m&
 
k − 1  P1 

2
He
W
5 ft3/s
(k −1) / k
1


− 1

0.667/1.667

(
1.667 )(0.4961 Btu/lbm ⋅ R )(530 R )  120 psia 


1
−
= (0.0493 lbm/s)



1.667 − 1

 14 psia 
= 44.11 Btu/s
= 62.4 hp
since 1 hp = 0.7068 Btu/s
(b) Polytropic compression with n = 1.2:
(n −1) / n 
nRT1  P2 



W&comp,in = m&
− 1

n − 1  P1 


0.2/1.2

(
1.2)(0.4961 Btu/lbm ⋅ R )(530 R )  120 psia 

− 1
= (0.0493 lbm/s)

1.2 − 1
14 psia 


= 33.47 Btu/s
= 47.3 hp
since 1 hp = 0.7068 Btu/s
(c) Isothermal compression:
P
120 psia
W&comp,in = m& RT ln 2 = (0.0493 lbm/s)(0.4961 Btu/lbm ⋅ R )(530 R )ln
= 27.83 Btu/s = 39.4 hp
P1
14 psia
(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each
stage is the same, and its value is determined from
Px = P1P2 =
(14 psia )(120 psia ) = 41.0 psia
The compressor work across each stage is also the same, thus total compressor work is twice the
compression work for a single stage:
(n −1) / n 
nRT1  Px 



− 1
W&comp,in = 2m& wcomp,I = 2m&

n − 1  P1 


0.2/1.2

(
1.2)(0.4961 Btu/lbm ⋅ R )(530 R )  41 psia 

= 2(0.0493 lbm/s)
− 1

1.2 − 1
14 psia 


= 30.52 Btu/s
= 43.2 hp
since 1 hp = 0.7068 Btu/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-58
7-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium is
to be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Procedure FuncPoly(m_dot,k, R,
T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out
_2stagePoly)
If n =1 then
T2=T1
W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"
W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"
Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"
Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"
Else
C_P = k*R/(k-1) "[Btu/lbm-R]"
T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"
W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"
Px=(P1*P2)^0.5
T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"
W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1T2x)"[Btu/s]"
endif
END
R=0.4961[Btu/lbm-R]
k=1.667
n=1.2
P1=14 [psia]
T1=70 [F]
P2=120 [psia]
V_dot = 5 [ft^3/s]
P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)
W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) 1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_isentropic = 0"[Btu/s]"
Call FuncPoly(m_dot,k, R,
T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out
_2stagePoly)
W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"
Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-59
n
Wcomp2StagePoly
[hp]
39.37
41.36
43.12
44.68
46.09
47.35
49.19
1
1.1
1.2
1.3
1.4
1.5
1.667
Wcompisentropic
[hp]
62.4
62.4
62.4
62.4
62.4
62.4
62.4
Wcompisothermal
[hp]
39.37
39.37
39.37
39.37
39.37
39.37
39.37
Wcomppolytropic
[hp]
39.37
43.48
47.35
50.97
54.36
57.54
62.4
65
W comp,polytropic [hp]
60
55
50
Polytropic
Isothermal
Isentropic
2StagePoly
45
40
35
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
n
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-60
7-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. The
mass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,
polytropic, isothermal, and two-stage compression.
Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kinetic
and potential energy changes are negligible.
Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio of
nitrogen is k = 1.4 (Table A-2).
2
Analysis (a) Isentropic compression:
{
or,
}
kRT1
(P2 P1 )(k −1) / k − 1
W&comp,in = m&
k −1
10 kJ/s = m&
(1.4)(0.297 kJ/kg ⋅ K )(300 K ) {(480 kPa
N2
·
m
}
10 kW
80 kPa )0.4/1.4 − 1
1.4 − 1
It yields
1
m& = 0.048 kg / s
(b) Polytropic compression with n = 1.3:
{
or,
}
nRT1
(P2 P1 )(n −1) / n − 1
W&comp,in = m&
n −1
10 kJ/s = m&
(1.3)(0.297 kJ/kg ⋅ K )(300 K ) {(480 kPa
1.3 − 1
}
80 kPa )0.3/1.3 − 1
It yields
m& = 0.051 kg / s
(c) Isothermal compression:
 480 kPa 
P

W&comp,in = m& RT ln 1 
→ 10 kJ/s = m& (0.297 kJ/kg ⋅ K )(300 K ) ln
P2
 80 kPa 
It yields
m& = 0.063 kg / s
(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each
stage is the same, and its value is determined to be
Px =
P1P2 =
(80 kPa )(480 kPa ) = 196 kPa
The compressor work across each stage is also the same, thus total compressor work is twice the
compression work for a single stage:
{
or,
}
nRT1
(Px P1 )(n −1) / n − 1
W&comp,in = 2m& wcomp,I = 2m&
n −1
10 kJ/s = 2m&
(1.3)(0.297 kJ/kg ⋅ K )(300 K ) {(196 kPa
1.3 − 1
}
80 kPa )0.3/1.3 − 1
It yields
m& = 0.056 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-61
7-99 Water mist is to be sprayed into the air stream in the compressor to cool the air as the water
evaporates and to reduce the compression power. The reduction in the exit temperature of the compressed
air and the compressor power saved are to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is reversible. 3 Kinetic and
potential energy changes are negligible. 3 Air is compressed isentropically. 4 Water vaporizes completely
before leaving the compressor. 4 Air properties can be used for the air-vapor mixture.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air is k =
1.4. The inlet enthalpies of water and air are (Tables A-4 and A-17)
hw1 = hf@20°C = 83.29 kJ/kg , hfg@20°C = 2453.9 kJ/kg and ha1 = h@300 K =300.19 kJ/kg
Analysis In the case of isentropic operation (thus no cooling or water spray), the exit temperature and the
power input to the compressor are
T2  P2
=
T1  P1



( k −1) / k
 1200 kPa 
→ T2 = (300 K)

 100 kPa 
{
(1.4 −1) / 1.4
= 610.2 K
}
kRT1
(P2 P1 )(k −1) / k − 1
W&comp,in = m&
k −1
(1.4)(0.287 kJ/kg ⋅ K )(300 K ) (1200 kPa/100 kPa )0.4/1.4 − 1 = 654.3 kW
= (2.1 kg/s)
1.4 − 1
{
When water is sprayed, we first need to check the accuracy of the
assumption that the water vaporizes completely in the compressor. In the
limiting case, the compression will be isothermal at the compressor inlet
temperature, and the water will be a saturated vapor. To avoid the
complexity of dealing with two fluid streams and a gas mixture, we
disregard water in the air stream (other than the mass flow rate), and
assume air is cooled by an amount equal to the enthalpy change of water.
The rate of heat absorption of water as it evaporates at the inlet
temperature completely is
}
1200 kPa
2
·
He
Water
20°C
Q& cooling,max = m& w h fg @ 20°C = (0.2 kg/s)(2453.9 kJ/kg) = 490.8 kW
1
W
100 kPa
300 K
The minimum power input to the compressor is
 1200 kPa 
P
 = 449.3 kW
W&comp,in, min = m& RT ln 2 = (2.1 kg/s)(0.287 kJ/kg ⋅ K)(300 K) ln
P1
 100 kPa 
This corresponds to maximum cooling from the air since, at constant temperature, ∆h = 0 and thus
Q& out = W&in = 449.3 kW , which is close to 490.8 kW. Therefore, the assumption that all the water vaporizes
is approximately valid. Then the reduction in required power input due to water spray becomes
∆W&comp,in = W&comp, isentropic − W&comp, isothermal = 654.3 − 449.3 = 205 kW
Discussion (can be ignored): At constant temperature, ∆h = 0 and thus Q& out = W& in = 449.3 kW corresponds
to maximum cooling from the air, which is less than 490.8 kW. Therefore, the assumption that all the water
vaporizes is only roughly valid. As an alternative, we can assume the compression process to be polytropic
and the water to be a saturated vapor at the compressor exit temperature, and disregard the remaining
liquid. But in this case there is not a unique solution, and we will have to select either the amount of water
or the exit temperature or the polytropic exponent to obtain a solution. Of course we can also tabulate the
results for different cases, and then make a selection.
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7-62
Sample Analysis: We take the compressor exit temperature to be T2 = 200°C = 473 K. Then,
hw2 = hg@200°C = 2792.0 kJ/kg and ha2 = h@473 K = 475.3 kJ/kg
Then,
T2  P2 
= 
T1  P1 
( n −1) / n
→
{
473 K  1200 kPa 
=

300 K  100 kPa 
( n −1) / n
→ n = 1.224
}
nRT1
(P2 P1 )(n−1) / n − 1 = m& nR (T2 − T1 )
W& comp,in = m&
n −1
n −1
1.224 )(0.287 kJ/kg ⋅ K )
(
= (2.1 kg/s)
(473 − 300)K = 570 kW
1.224 − 1
Energy balance:
W&comp,in − Q& out = m& (h2 − h1 ) → Q& out = W&comp,in − m& (h2 − h1 )
= 569.7 kW − (2.1 kg/s)(475.3 − 300.19) = 202.0 kW
Noting that this heat is absorbed by water, the rate at which water evaporates in the compressor becomes
Q&
202.0 kJ/s
→ m& w = in, water =
= 0.0746 kg/s
Q& out,air = Q& in, water = m& w (hw2 − hw1 ) 
hw2 − hw1 (2792.0 − 83.29) kJ/kg
Then the reductions in the exit temperature and compressor power input become
∆T2 = T2,isentropic − T2,water cooled = 610.2 − 473 = 137.2° C
∆W&comp,in = W&comp,isentropic − W&comp,water cooled = 654.3 − 570 = 84.3 kW
Note that selecting a different compressor exit temperature T2 will result in different values.
7-100 A water-injected compressor is used in a gas turbine power plant. It is claimed that the power output
of a gas turbine will increase when water is injected into the compressor because of the increase in the
mass flow rate of the gas (air + water vapor) through the turbine. This, however, is not necessarily right
since the compressed air in this case enters the combustor at a low temperature, and thus it absorbs much
more heat. In fact, the cooling effect will most likely dominate and cause the cyclic efficiency to drop.
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7-63
Isentropic Efficiencies of Steady-Flow Devices
7-101C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. The
adiabatic efficiencies of these devices are defined as
actual work output
insentropic work input
actual exit kineticenergy
,η =
, and η N =
ηT =
insentropic work output C
actual work input
insentropic exit kinetic energy
7-102C No, because the isentropic process is not the model or ideal process for compressors that are
cooled intentionally.
7-103C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a result
of irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exit
state
7-104 Steam enters an adiabatic turbine with an isentropic efficiency of 0.90 at a specified state with a
specified mass flow rate, and leaves at a specified pressure. The turbine exit temperature and power output
of the turbine are to be determined.
Assumptions 1 This is a steady-flow process since there is no change
P1 = 8 MPa
with time. 2 Kinetic and potential energy changes are negligible. 3
T
The device is adiabatic and thus heat transfer is negligible.
1 = 500°C
Analysis (a) From the steam tables (Tables A-4 through A-6),
P1 = 8 MPa  h1 = 3399.5 kJ/kg

T1 = 500°C  s1 = 6.7266 kJ/kg ⋅ K
P2 s = 30 kPa 

s 2 s = s1
h
s2s − s f
STEAM
TURBINE
ηT = 90%
6.7266 − 0.9441
= 0.8475
6.8234
s fg
= h f + x 2 s h fg = 289.27 + (0.8475)(2335.3) = 2268.3 kJ/kg
x2s =
2s
=
P2 = 30 kPa
From the isentropic efficiency relation,
h −h
η T = 1 2a 
→ h2 a = h1 − η T (h1 − h2 s ) = 3399.5 − (0.9)(3399.5 − 2268.3) = 2381.4 kJ/kg
h1 − h2 s
Thus,
P2 a = 30 kPa

 T2 a = Tsat @ 30 kPa = 69.09°C
h2 a = 2381.4 kJ/kg 
&1 = m
&2 = m
& . We take the actual turbine as the system,
(b) There is only one inlet and one exit, and thus m
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& system©0 (steady)
=0
E& − E& out
1in
424
3
1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h1 = W&a,out + m& h2
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&a,out = m& (h1 − h2 )
Substituting,
W&
a, out
= (3kg/s )(3399.5 − 2381.4) kJ/kg = 3054 kW
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-64
7-105 EES Problem 7-104 is reconsidered. The effect of varying the turbine isentropic efficiency from
0.75 to 1.0 on both the work done and the exit temperature of the steam are to be investigated, and the
results are to be plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"System: control volume for turbine"
"Property relation: Steam functions"
"Process: Turbine: Steady state, steady flow, adiabatic, reversible or isentropic"
"Since we don't know the mass, we write the conservation of energy per unit mass."
"Conservation of mass: m_dot[1]= m_dot[2]=m_dot"
600
T [°C]
400
300
8000 kPa
0.4
0.6
0.2
30 kPa
1.0
2.0
3.0
4.0
5.0
2
0.8
2
s
6.0
7.0
8.0
9.0
10.0
s [kJ/kg-K]
3400
3300
3200
3100
W turb [kW ]
Wturb [kW]
2545
2715
2885
3054
3224
3394
1
500
"Conservation of Energy - SSSF energy
200
balance for turbine -- neglecting the
change in potential energy, no heat
100
transfer:"
h[1]=enthalpy(WorkFluid$,P=P[1],T=T[1])
0
0.0
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])
T_s[1] = T[1]
s[2] =s[1]
s_s[2] = s[1]
h_s[2]=enthalpy(WorkFluid$,P=P[2],s=s_s[2])
T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])
eta_turb = w_turb/w_turb_s
h[1] = h[2] + w_turb
h[1] = h_s[2] + w_turb_s
T[2]=temperature(WorkFluid$,P=P[2],h=h[2])
W_dot_turb = m_dot*w_turb
ηturb
0.75
0.8
0.85
0.9
0.95
1
Steam
700
"Knowns:"
WorkFluid$ = 'Steam_iapws'
m_dot = 3 [kg/s]
P[1] = 8000 [kPa]
T[1] = 500 [C]
P[2] = 30 [kPa]
"eta_turb = 0.9"
3000
2900
2800
2700
2600
2500
0.75
0.8
0.85
η
0.9
0.95
turb
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1
7-65
7-106 Steam enters an adiabatic turbine at a specified state, and leaves at a specified state. The mass flow
rate of the steam and the isentropic efficiency are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible.
Analysis (a) From the steam tables (Tables A-4 and A-6),
P1 = 7 MPa  h1 = 3650.6 kJ/kg

T1 = 600°C  s1 = 7.0910 kJ/kg ⋅ K
P2 = 50 kPa 
 h2 a = 2780.2 kJ/kg
T2 = 150°C 
&1 = m
&2 = m
& . We take the actual turbine as the system, which
There is only one inlet and one exit, and thus m
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
1in
424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
=0
1
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
6 MW
H2O
m& (h1 + V12 / 2) = W&a,out + m& (h2 + V12 /2) (since Q& ≅ ∆pe ≅ 0)

V 2 − V12 
W&a,out = − m&  h2 − h1 + 2


2


2
Substituting, the mass flow rate of the steam is determined to be

(140 m/s) 2 − (80 m/s) 2  1 kJ/kg
6000 kJ/s = −m&  2780.2 − 3650.6 +


2
 1000 m 2 /s 2

m& = 6.95 kg/s
 


(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are
P2 s = 50 kPa 

s 2 s = s1
h
and
(
2s
s 2s − s f
7.0910 − 1.0912
=
= 0.9228
6.5019
s fg
= h f + x 2 s h fg = 340.54 + (0.9228)(2304.7 ) = 2467.3 kJ/kg
x 2s =
{(
) })
W&s, out = − m& h2 s − h1 + V22 − V12 / 2

(140 m/s) 2 − (80 m/s) 2  1 kJ/kg  
W&s, out = −(6.95 kg/s ) 2467.3 − 3650.6 +


2 2

2
 1000 m /s  

= 8174 kW
Then the isentropic efficiency of the turbine becomes
W&
6000 kW
ηT = a =
= 0.734 = 73.4%
&
W s 8174 kW
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-66
7-107 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a
specified pressure. The isentropic efficiency of 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 device is adiabatic and thus heat transfer is negligible. 4 Argon is an
ideal gas with constant specific heats.
Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp =
0.5203 kJ/kg.K (Table A-2).
&1 = m
&2 = m
& . We take the isentropic turbine as the
Analysis There is only one inlet and one exit, and thus m
system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
1
E& in = E& out
m& h1 = W& s ,out + m& h2 s
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&s, out = m& (h1 − h2 s )
ηT
From the isentropic relations,
(k −1) / k
0.667/1.667
P 
 200 kPa 

= 479 K
T2 s = T1  2 s 
= (1073 K )
P
 1500 kPa 
 1 
Then the power output of the isentropic turbine becomes
W&
= m& c (T − T ) = (80/60 kg/min )(0.5203 kJ/kg ⋅ K )(1073 − 479) = 412.1 kW
s, out
p
1
370 kW
Ar
2
2s
Then the isentropic efficiency of the turbine is determined from
370 kW
W&
η T = &a =
= 0.898 = 89.8%
412.1 kW
W
s
7-108E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specified
state, and leave at a specified pressure. The work output of 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 device is adiabatic and thus heat transfer is negligible. 4 Combustion
gases can be treated as air that is an ideal gas with variable specific heats.
Analysis From the air table and isentropic relations,
1
h1 = 504.71 Btu / lbm
T1 = 2000 R

→
Pr1 = 174.0
P 
 60 psia 
(174.0) = 87.0 
→ h2s = 417.3 Btu/lbm
Pr2 =  2  Pr1 = 
P
 120 psia 
 1
&1 = m
&2 = m
& . We take the
There is only one inlet and one exit, and thus m
actual turbine as the system, which is a control volume since mass crosses the
boundary. The energy balance for this steady-flow system can be expressed
as
E& = E&
in
AIR
ηT = 82%
2
out
m& h1 = W&a,out + m& h2
(since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
W&a,out = m& (h1 − h2 )
Noting that wa = ηTws, the work output of the turbine per unit mass is determined from
wa = (0.82 )(504.71 − 417.3)Btu/lbm = 71.7 Btu/lbm
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-67
7-109 [Also solved by EES on enclosed CD] Refrigerant-134a enters an adiabatic compressor with an
isentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specified
pressure. The compressor exit temperature and power input to the compressor 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.
2
Analysis (a) From the refrigerant tables (Tables A-11E through A-13E),
h1 = hg @120 kPa = 236.97 kJ/kg
P1 = 120 kPa 
 s1 = s g @120 kPa = 0.94779 kJ/kg ⋅ K
sat. vapor 
v =v
= 0.16212 m3/kg
g @120 kPa
1
R-134a
ηC = 80%
P2 = 1 MPa 
 h2 s = 281.21 kJ/kg

s2 s = s1
0.3 m3/min
From the isentropic efficiency relation,
ηC =
1
h2 s − h1

→ h2 a = h1 + (h2 s − h1 ) /η C = 236.97 + (281.21 − 236.97 )/0.80 = 292.26 kJ/kg
h2 a − h1
Thus,
P2 a = 1 MPa
h2 a

 T2 a = 58.9°C
= 292.26 kJ/kg 
(b) The mass flow rate of the refrigerant is determined from
m& =
V&1
0.3/60 m 3 /s
=
= 0.0308 kg/s
v 1 0.16212 m 3 /kg
&1 = m
&2 = m
& . We take the actual compressor as the system,
There is only one inlet and one exit, and thus m
which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system
can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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 )
Substituting, the power input to the compressor becomes,
W&a,in = (0.0308 kg/s )(292.26 − 236.97 )kJ/kg = 1.70 kW
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7-68
7-110 EES Problem 7-109 is reconsidered. The problem is to be solved by considering the kinetic energy
and by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe inside
diameter is 2 cm.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data from diagram window"
{P[1] = 120 "kPa"
P[2] = 1000 "kPa"
Vol_dot_1 = 0.3 "m^3/min"
Eta_c = 0.80 "Compressor adiabatic efficiency"
A_ratio = 1.5
d_2 = 2/100 "m"}
"System: Control volume containing the compressor, see the diagram window.
Property Relation: Use the real fluid properties for R134a.
Process: Steady-state, steady-flow, adiabatic process."
Fluid$='R134a'
"Property Data for state 1"
T[1]=temperature(Fluid$,P=P[1],x=1)"Real fluid equ. at the sat. vapor state"
h[1]=enthalpy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
s[1]=entropy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
v[1]=volume(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"
"Property Data for state 2"
s_s[1]=s[1]; T_s[1]=T[1] "needed for plot"
s_s[2]=s[1] "for the ideal, isentropic process across the compressor"
h_s[2]=ENTHALPY(Fluid$, P=P[2], s=s_s[2])"Enthalpy 2 at the isentropic state 2s and
pressure P[2]"
T_s[2]=Temperature(Fluid$, P=P[2], s=s_s[2])"Temperature of ideal state - needed only for
plot."
"Steady-state, steady-flow conservation of mass"
m_dot_1 = m_dot_2
m_dot_1 = Vol_dot_1/(v[1]*60)
Vol_dot_1/v[1]=Vol_dot_2/v[2]
Vel[2]=Vol_dot_2/(A[2]*60)
A[2] = pi*(d_2)^2/4
A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]"
A_ratio=A[1]/A[2]
"Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagram
window"
m_dot_1*(h[1]+(Vel[1])^2/(2*1000)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/(2*1000))
"Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act"
Eta_c = (h_s[2]-h[1])/(h[2]-h[1])
"Knowing h[2], the other properties at state 2 can be found."
v[2]=volume(Fluid$, P=P[2], h=h[2])"v[2] is found at the actual state 2, knowing P and h."
T[2]=temperature(Fluid$, P=P[2],h=h[2])"Real fluid equ. for T at the known outlet h and P."
s[2]=entropy(Fluid$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P."
T_exit=T[2]
"Neglecting the kinetic energies, the work is:"
m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2]
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-69
SOLUTION
s_s[1]=0.9478 [kJ/kg-K]
s_s[2]=0.9478 [kJ/kg-K]
T[1]=-22.32 [C]
T[2]=58.94 [C]
T_exit=58.94 [C]
T_s[1]=-22.32 [C]
T_s[2]=48.58 [C]
Vol_dot_1=0.3 [m^3 /min]
Vol_dot_2=0.04244 [m^3 /min]
v[1]=0.1621 [m^3/kg]
v[2]=0.02294 [m^3/kg]
Vel[1]=10.61 [m/s]
Vel[2]=2.252 [m/s]
W_dot_c=1.704 [kW]
W_dot_c_noke=1.706 [kW]
A[1]=0.0004712 [m^2]
A[2]=0.0003142 [m^2]
A_ratio=1.5
d_2=0.02 [m]
Eta_c=0.8
Fluid$='R134a'
h[1]=237 [kJ/kg]
h[2]=292.3 [kJ/kg]
h_s[2]=281.2 [kJ/kg]
m_dot_1=0.03084 [kg/s]
m_dot_2=0.03084 [kg/s]
P[1]=120.0 [kPa]
P[2]=1000.0 [kPa]
s[1]=0.9478 [kJ/kg-K]
s[2]=0.9816 [kJ/kg-K]
R134a
125
T-s diagram for real and ideal com pressor
100
Tem perature [C]
75
Ideal Compressor
Real Com pressor
50
1000 kPa
25
0
-25
-50
0.0
120 kPa
0.2
0.4
0.6
0.8
1.0
1.2
Entropy [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-70
7-111 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, and
leaves at a specified temperature. The exit pressure of air and the power input to the compressor 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. 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)
2
Analysis (a) From the air table (Table A-17),
T1 = 290 K 
→ h1 = 290.16 kJ/kg, Pr1 = 1.2311
T2 = 530 K 
→ h2 a = 533.98 kJ/kg
From the isentropic efficiency relation η C =
AIR
ηC = 84%
h2 s − h1
,
h2 a − h1
h2 s = h1 + η C (h2 a − h1 )
2.4 m3/s
1
= 290.16 + (0.84 )(533.98 − 290.16 ) = 495.0 kJ/kg 
→ Pr2 = 7.951
Then from the isentropic relation ,
 Pr 
P2 Pr2
 7.951 
=

→ P2 =  2  P1 = 
(100 kPa ) = 646 kPa


P1 Pr1
 1.2311 
 Pr1 
&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 since mass crosses the boundary. The energy balance for this steadyflow system can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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 )
where
m& =
P1V&1
(100 kPa )(2.4 m 3 /s)
=
= 2.884 kg/s
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(290 K )
Then the power input to the compressor is determined to be
W&a,in = (2.884 kg/s)(533.98 − 290.16) kJ/kg = 703 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-71
7-112 Air is compressed by an adiabatic compressor from a specified state to another specified state. The
isentropic efficiency of the compressor and the exit temperature of air for the isentropic case 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. 4 Air is an
ideal gas with variable specific heats.
Analysis (a) From the air table (Table A-17),
T1 = 300 K

→
h1 = 300.19 kJ / kg,
T2 = 550 K

→
h2 a = 554.74 kJ / kg
2
Pr1 = 1.386
From the isentropic relation,
AIR
P 
 600 kPa 
(1.386) = 8.754 
→ h2 s = 508.72 kJ/kg
Pr2 =  2  Pr1 = 
P
 95 kPa 
 1
Then the isentropic efficiency becomes
ηC =
h2 s − h1 508.72 − 30019
.
=
= 0.819 = 81.9%
h2a − h1 554.74 − 30019
.
1
(b) If the process were isentropic, the exit temperature would be
h2 s = 508.72 kJ / kg

→
T2 s = 505.5 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-72
7-113E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, and
leaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with
constant specific heats.
Properties The specific heat ratio of argon is k = 1.667. The constant
pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E).
2
Analysis (a) The isentropic exit temperature T2s is determined from
P 
T2 s = T1 2 s 
 P1 
(k −1) / k
 200 psia 

= (550 R )
 20 psia 
Ar
0.667/1.667
= 1381.9 R
ηC = 80%
The actual kinetic energy change during this process is
∆ke a =
V 22 − V12 (240 ft/s )2 − (60 ft/s )2
=
2
2
 1 Btu/lbm

 25,037 ft 2 /s 2


 = 1.08 Btu/lbm


1
The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kinetic
energy changes for the actual and isentropic cases to be same in efficiency calculations. From the
isentropic efficiency relation, including the effect of kinetic energy,
ηC =
It yields
w s (h2 s − h1 ) + ∆ke c p (T2 s − T1 ) + ∆ke s
0.1253(1381.9 − 550 ) + 1.08
=
=

→ 0.8 =
wa (h2 a − h1 ) + ∆ke c p (T2 a − T1 ) + ∆ke a
0.1253(T2 a − 550 ) + 1.08
T2a = 1592 R
&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 since mass crosses the boundary. The energy balance for this steadyflow system can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W&a,in + m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ ∆pe ≅ 0)

V 2 − V12 
W&a,in = m&  h2 − h1 + 2

→ wa,in = h2 − h1 + ∆ke


2


Substituting, the work input to the compressor is determined to be
wa,in = (0.1253 Btu/lbm ⋅ R )(1592 − 550)R + 1.08 Btu/lbm = 131.6 Btu/lbm
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-73
7-114 CO2 gas is compressed by an adiabatic compressor from a specified state to another specified state.
The isentropic efficiency of the compressor 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 device is adiabatic and thus heat transfer is negligible. 4 CO2 is an
ideal gas with constant specific heats.
Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specific
2
heat and the specific heat ratio of CO2 are k = 1.260 and cp = 0.917 kJ/kg.K (Table A-2).
Analysis The isentropic exit temperature T2s is
(k −1) / k
0.260/1.260
P 
 600 kPa 

= (300 K )
= 434.2 K
T2 s = T1 2 s 
 100 kPa 
 P1 
From the isentropic efficiency relation,
c p (T2 s − T1 ) T2 s − T1 434.2 − 300
w
h −h
η C = s = 2s 1 =
= 0.895 = 89.5%
=
=
wa h2 a − h1 c p (T2 a − T1 ) T2 a − T1
450 − 300
CO2
1.8 kg/s
1
7-115E Air is accelerated in a 90% efficient adiabatic nozzle from low velocity to a specified velocity. The
exit temperature and pressure of the air are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with
variable specific heats.
Analysis From the air table (Table A-17E),
T1 = 1480 R

→
h1 = 363.89 Btu / lbm,
Pr1 = 53.04
&1 = m
&2 = m
& . We take the nozzle as the system, which is a
There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be
expressed as
E& − E& out
=
∆E& system©0 (steady)
=0
1in
424
3
1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
1
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since W& = Q& ≅ ∆pe ≅ 0)
AIR
ηN = 90%
2
©0
V22 − V12
2
Substituting, the exit temperature of air is determined to be
h2 = h1 −
(800 ft/s )2 − 0 
1 Btu/lbm 
 = 351.11 Btu/lbm
2 2

2
 25,037 ft /s 
From the air table we read
T2a = 1431.3 R
h −h
From the isentropic efficiency relation η N = 2 a 1 ,
h2 s − h1
h2 = 363.89 kJ/kg −
h2 s = h1 + (h2 a − h1 ) /η N = 363.89 + (351.11 − 363.89 )/ (0.90 ) = 349.69 Btu/lbm 
→ Pr2 = 46.04
Then the exit pressure is determined from the isentropic relation to be
 Pr
P2 Pr2
=

→ P2 =  2
 Pr
P1 Pr1
 1

 P =  46.04 (60 psia ) = 52.1 psia
 1  53.04 

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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-74
7-116E EES Problem 7-115E is reconsidered. The effect of varying the nozzle isentropic efficiency from
0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to be
plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
WorkFluid$ = 'Air'
P[1] = 60 [psia]
T[1] = 1020 [F]
Vel[2] = 800 [ft/s]
Vel[1] = 0 [ft/s]
eta_nozzle = 0.9
"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potential
energy, no heat transfer:"
h[1]=enthalpy(WorkFluid$,T=T[1])
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])
T_s[1] = T[1]
s[2] =s[1]
s_s[2] = s[1]
h_s[2]=enthalpy(WorkFluid$,T=T_s[2])
T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])
eta_nozzle = ke[2]/ke_s[2]
ke[1] = Vel[1]^2/2
ke[2]=Vel[2]^2/2
h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm)
h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm)
T[2]=temperature(WorkFluid$,h=h[2])
P_2_answer = P[2]
T_2_answer = T[2]
0.8
0.85
0.9
0.95
1
P2
[psia
51.09
51.58
52.03
52.42
52.79
T2
[F
971.4
971.4
971.4
971.4
971.4
Ts,2
[F]
959.2
962.8
966
968.8
971.4
980
970
T s[2]
ηnozzle
T
960
T
950
0.8
0.84
s [2 ]
0.88
η
2
0.92
0.96
nozzle
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
1
7-75
7-117 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a
specified velocity. The exit velocity and the exit temperature are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustion gases can be
treated as air that is an ideal gas with variable specific heats.
Analysis From the air table (Table A-17),
P1 = 260 kPa
T1 = 747°C
V1 = 80 m/s
T1 = 1020 K 
→ h1 = 1068.89 kJ/kg, Pr1 = 123.4
From the isentropic relation ,
AIR
ηN = 92%
P2 = 85 kPa
P 
 85 kPa 
(123.4 ) = 40.34 
Pr2 =  2  Pr1 = 
→ h2 s = 783.92 kJ/kg
P
 260 kPa 
 1
&1 = m
&2 = m
& . We take the nozzle as the system, which is a
There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system for the
isentropic process can be expressed as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12 / 2) = m& (h2 s + V 22s /2) (since W& = Q& ≅ ∆pe ≅ 0)
h2 s = h1 −
V 22s − V12
2
Then the isentropic exit velocity becomes
V 2 s = V12 + 2(h1 − h2 s ) =

/s 2
1 kJ/kg
(80 m/s )2 + 2(1068.89 − 783.92)kJ/kg 1000 m

2

 = 759.2 m/s


Therefore,
V2 a = η N V2 s = 0.92 (759.2 m/s ) = 728.2 m/s
The exit temperature of air is determined from the steady-flow energy equation,
h2 a = 1068.89 kJ/kg −
From the air table we read
(728.2 m/s)2 − (80 m/s)2 
2


 1000 m /s  = 806.95 kJ/kg


1 kJ/kg
2
2
T2a = 786.3 K
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7-107
Special Topic: Reducing the Cost of Compressed Air
7-150 The total installed power of compressed air systems in the US is estimated to be about 20 million
horsepower. The amount of energy and money that will be saved per year if the energy consumed by
compressors is reduced by 5 percent is to be determined.
Assumptions 1 The compressors operate at full load during one-third of the time on average, and are shut
down the rest of the time. 2 The average motor efficiency is 85 percent.
Analysis The electrical energy consumed by compressors per year is
Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency
= (20×106 hp)(0.746 kW/hp)(1/3)(365×24 hours/year)/0.85
= 5.125×1010 kWh/year
2
Then the energy and cost savings corresponding to a 5% reduction in
energy use for compressed air become
Energy Savings = (Energy consumed)(Fraction saved)
= (5.125×1010 kWh)(0.05)
Air
Compressor
W=20×106 hp
= 2.563×109 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy)
= (2.563×109 kWh/year)($0.07/kWh)
= $0.179×109 /year
1
Therefore, reducing the energy usage of compressors by 5% will save $179 million a year.
7-151 The total energy used to compress air in the US is estimated to be 0.5×1015 kJ per year. About 20%
of the compressed air is estimated to be lost by air leaks. The amount and cost of electricity wasted per
year due to air leaks is to be determined.
2
Assumptions About 20% of the compressed air is lost by air leaks.
Analysis The electrical energy and money wasted by air leaks are
Energy wasted = (Energy consumed)(Fraction wasted)
= (0.5×1015 kJ)(1 kWh/3600 kJ)(0.20)
Air
Compressor
W=0.5×1015 kJ
= 27.78×109 kWh/year
Money wasted = (Energy wasted)(Unit cost of energy)
= (27.78×109 kWh/year)($0.07/kWh)
1
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-108
= $1.945×109 /year
Therefore, air leaks are costing almost $2 billion a year in electricity costs. The environment also suffers
from this because of the pollution associated with the generation of this much electricity.
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7-109
7-152 The compressed air requirements of a plant is being met by a 125 hp compressor that compresses air
from 101.3 kPa to 900 kPa. The amount of energy and money saved by reducing the pressure setting of
compressed air to 750 kPa is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are
negligible. 3 The load factor of the compressor is given to be 0.75. 4 The pressures given are absolute
pressure rather than gage pressure.
Properties The specific heat ratio of air is k = 1.4 (Table A-2).
Analysis The electrical energy consumed by this compressor per year is
Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency
= (125 hp)(0.746 kW/hp)(0.75)(3500 hours/year)/0.88
900 kPa
= 278,160 kWh/year
The fraction of energy saved as a result of reducing the
pressure setting of the compressor is
Power Reduction Factor = 1 −
Air
Compressor
( P2, reduced / P1 )( k −1) / k − 1
2
W=125 hp
( P2 / P1 )( k −1) / k − 1
(750 / 101.3)(1.4 −1) / 1, 4 − 1
(900 / 101.3)(1.4 −1) / 1,4 − 1
= 0.1093
= 1−
1
101 kPa
15°C
That is, reducing the pressure setting will result in about 11 percent savings from the energy consumed by
the compressor and the associated cost. Therefore, the energy and cost savings in this case become
Energy Savings = (Energy consumed)(Power reduction factor)
= (278,160 kWh/year)(0.1093)
= 30,410 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy)
= (30,410 kWh/year)($0.085/kWh)
= $2585/year
Therefore, reducing the pressure setting by 150 kPa will result in annual savings of 30.410 kWh that is
worth $2585 in this case.
Discussion Some applications require very low pressure compressed air. In such cases the need can be met
by a blower instead of a compressor. Considerable energy can be saved in this manner, since a blower
requires a small fraction of the power needed by a compressor for a specified mass flow rate.
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7-110
7-153 A 150 hp compressor in an industrial facility is housed inside the production area where the average
temperature during operating hours is 25°C. The amounts of energy and money saved as a result of
drawing cooler outside air to the compressor instead of using the inside air are to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are
negligible.
Analysis The electrical energy consumed by this compressor per year is
2
Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency
= (150 hp)(0.746 kW/hp)(0.85)(4500 hours/year)/0.9
= 475,384 kWh/year
T0 = 10°C
Also,
Air
Compressor
W=150 hp
Cost of Energy = (Energy consumed)(Unit cost of energy)
= (475,384 kWh/year)($0.07/kWh)
= $33,277/year
The fraction of energy saved as a result of drawing in cooler outside air is
Power Reduction Factor = 1 −
1
101 kPa
25°C
Toutside
10 + 273
= 1−
= 0.0503
Tinside
25 + 273
That is, drawing in air which is 15°C cooler will result in 5.03 percent savings from the energy consumed
by the compressor and the associated cost. Therefore, the energy and cost savings in this case become
Energy Savings = (Energy consumed)(Power reduction factor)
= (475,384 kWh/year)(0.0503)
= 23,929 kWh/year
Cost Savings
= (Energy savings)(Unit cost of energy)
= (23,929 kWh/year)($0.07/kWh)
= $1675/year
Therefore, drawing air in from the outside will result in annual savings of 23,929 kWh, which is worth
$1675 in this case.
Discussion The price of a typical 150 hp compressor is much lower than $50,000. Therefore, it is
interesting to note that the cost of energy a compressor uses a year may be more than the cost of the
compressor itself.
The implementation of this measure requires the installation of an ordinary sheet metal or PVC
duct from the compressor intake to the outside. The installation cost associated with this measure is
relatively low, and the pressure drop in the duct in most cases is negligible. About half of the
manufacturing facilities we have visited, especially the newer ones, have the duct from the compressor
intake to the outside in place, and they are already taking advantage of the savings associated with this
measure.
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7-111
7-154 The compressed air requirements of the facility during 60 percent of the time can be met by a 25 hp
reciprocating compressor instead of the existing 100 hp compressor. The amounts of energy and money
saved as a result of switching to the 25 hp compressor during 60 percent of the time are to be determined.
Analysis Noting that 1 hp = 0.746 kW, the electrical energy consumed by each compressor per year is
determined from
(Energy consumed)Large = (Power)(Hours)[(LFxTF/ηmotor)Unloaded + (LFxTF/ηmotor)Loaded]
= (100 hp)(0.746 kW/hp)(3800 hours/year)[0.35×0.6/0.82+0.90×0.4/0.9]
= 185,990 kWh/year
(Energy consumed)Small =(Power)(Hours)[(LFxTF/ηmotor)Unloaded + (LFxTF/ηmotor)Loaded]
= (25 hp)(0.746 kW/hp)(3800 hours/year)[0.0×0.15+0.95×0.85]/0.88
2
= 65,031 kWh/year
Therefore, the energy and cost savings in this case become
Energy Savings = (Energy consumed)Large- (Energy consumed)Small
= 185,990 - 65,031 kWh/year
W=100 hp
Air
Compressor
= 120,959 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy)
= (120,959 kWh/year)($0.075/kWh)
1
= $9,072/year
Discussion Note that utilizing a small compressor during the times of reduced compressed air requirements
and shutting down the large compressor will result in annual savings of 120,959 kWh, which is worth
$9,072 in this case.
7-155 A facility stops production for one hour every day, including weekends, for lunch break, but the 125
hp compressor is kept operating. If the compressor consumes 35 percent of the rated power when idling,
the amounts of energy and money saved per year as a result of turning the compressor off during lunch
break are to be determined.
Analysis It seems like the compressor in this facility is kept on unnecessarily for one hour a day and thus
365 hours a year, and the idle factor is 0.35. Then the energy and cost savings associated with turning the
compressor off during lunch break are determined to be
Energy Savings = (Power Rating)(Turned Off Hours)(Idle Factor)/ηmotor
2
= (125 hp)(0.746 kW/hp)(365 hours/year)(0.35)/0.84
= 14,182 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy)
= (14,182 kWh/year)($0.09/kWh)
Air
Compressor
W=125 hp
= $1,276/year
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
1
7-112
Discussion Note that the simple practice of turning the compressor off
during lunch break will save this facility $1,276 a year in energy costs.
There are also side benefits such as extending the life of the motor and
the compressor, and reducing the maintenance costs.
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7-156 It is determined that 40 percent of the energy input to the compressor is removed from the
compressed air as heat in the aftercooler with a refrigeration unit whose COP is 3.5. The amounts of the
energy and money saved per year as a result of cooling the compressed air before it enters the refrigerated
dryer are to be determined.
Assumptions The compressor operates at full load when operating.
Analysis Noting that 40 percent of the energy input to the compressor is removed by the aftercooler, the
rate of heat removal from the compressed air in the aftercooler under full load conditions is
Q& aftercooling = (Rated Power of Compressor)(Load Factor)(Aftercooling Fraction)
= (150 hp)(0.746 kW/hp)(1.0)(0.4) = 44.76 kW
The compressor is said to operate at full load for 1600 hours a
year, and the COP of the refrigeration unit is 3.5. Then the energy
and cost savings associated with this measure become
Energy Savings = ( Q& aftercooling )(Annual Operating Hours)/COP
= (44.76 kW)(1600 hours/year)/3.5
Qaftercooling
Aftercooler
Air
Compressor
W=150 hp
= 20,462 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy saved)
= (20,462 kWh/year)($0.06/kWh)
= $1227/year
Discussion Note that the aftercooler will save this facility 20,462 kWh of electrical energy worth $1227
per year. The actual savings will be less than indicated above since we have not considered the power
consumed by the fans and/or pumps of the aftercooler. However, if the heat removed by the aftercooler is
utilized for some useful purpose such as space heating or process heating, then the actual savings will be
much more.
7-157 The motor of a 150 hp compressor is burned out and is to be replaced by either a 93% efficient
standard motor or a 96.2% efficient high efficiency motor. It is to be determined if the savings from the
high efficiency motor justify the price differential.
Assumptions 1 The compressor operates at full load when operating. 2 The life of the motors is 10 years. 3
There are no rebates involved. 4 The price of electricity remains constant.
Analysis The energy and cost savings associated with the installation of the high efficiency motor in this
case are determined to be
Energy Savings = (Power Rating)(Operating Hours)(Load Factor)(1/ηstandard - 1/ηefficient)
= (150 hp)(0.746 kW/hp)(4,368 hours/year)(1.0)(1/0.930 - 1/0.962)
= 17,483 kWh/year
Cost Savings
= (Energy savings)(Unit cost of energy)
Air
Compressor
= (17,483 kWh/year)($0.075/kWh)
= $1311/year
The additional cost of the energy efficient motor is
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150 hp
7-114
Cost Differential = $10,942 - $9,031 = $1,911
Discussion The money saved by the high efficiency motor will pay for this cost difference in
$1,911/$1311 = 1.5 years, and will continue saving the facility money for the rest of the 10 years of its
lifetime. Therefore, the use of the high efficiency motor is recommended in this case even in the absence of
any incentives from the local utility company.
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7-158 The compressor of a facility is being cooled by air in a heat-exchanger. This air is to be used to heat
the facility in winter. The amount of money that will be saved by diverting the compressor waste heat into
the facility during the heating season is to be determined.
Assumptions The compressor operates at full load when operating.
Analysis Assuming operation at sea level and taking the density of air to be 1.2 kg/m3, the mass flow rate
of air through the liquid-to-air heat exchanger is determined to be
Mass flow rate of air = (Density of air)(Average velocity)(Flow area)
= (1.2 kg/m3)(3 m/s)(1.0 m2)
= 3.6 kg/s = 12,960 kg/h
Noting that the temperature rise of air is 32°C, the rate at which heat can be
recovered (or the rate at which heat is transferred to air) is
Rate of Heat Recovery
= (Mass flow rate of air)(Specific heat of air)(Temperature rise)
= (12,960 kg/h)(1.0 kJ/kg.°C)(32°C)
= 414,720 kJ/h
The number of operating hours of this compressor during the
heating season is
Air
20°C
3 m/s
Hot
Compressed
air
52°C
Operating hours = (20 hours/day)(5 days/week)(26 weeks/year)
= 2600 hours/year
Then the annual energy and cost savings become
Energy Savings = (Rate of Heat Recovery)(Annual Operating Hours)/Efficiency
= (414,720 kJ/h)(2600 hours/year)/0.8
= 1,347,840,000 kJ/year
= 12,776 therms/year
Cost Savings = (Energy savings)(Unit cost of energy saved)
= (12,776 therms/year)($1.0/therm)
= $12,776/year
Therefore, utilizing the waste heat from the compressor will save $12,776 per year from the heating costs.
Discussion The implementation of this measure requires the installation of an ordinary sheet metal duct
from the outlet of the heat exchanger into the building. The installation cost associated with this measure is
relatively low. A few of the manufacturing facilities we have visited already have this conservation system
in place. A damper is used to direct the air into the building in winter and to the ambient in summer.
Combined compressor/heat-recovery systems are available in the market for both air-cooled
(greater than 50 hp) and water cooled (greater than 125 hp) systems.
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7-116
7-159 The compressed air lines in a facility are maintained at a gage pressure of 850 kPa at a location
where the atmospheric pressure is 85.6 kPa. There is a 5-mm diameter hole on the compressed air line.
The energy and money saved per year by sealing the hole on the compressed air line.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are
negligible.
Properties The gas constant of air is R = 0.287 kJ/kg.K. The specific heat ratio of air is k = 1.4 (Table A2).
Analysis Disregarding any pressure losses and noting that the absolute pressure is the sum of the gage
pressure and the atmospheric pressure, the work needed to compress a unit mass of air at 15°C from the
atmospheric pressure of 85.6 kPa to 850+85.6 = 935.6 kPa is determined to be
wcomp, in
=
⎡⎛ P
kRT1
⎢⎜ 2
η comp (k − 1) ⎢⎜⎝ P1
⎣
=
(1.4)(0.287 kJ/kg.K)(288 K) ⎡⎛ 935.6 kPa ⎞
⎢⎜
⎟
(0.8)(1.4 − 1)
⎣⎢⎝ 85.6 kPa ⎠
⎞
⎟⎟
⎠
( k −1) / k
⎤
− 1⎥
⎥
⎦
(1.4 −1) / 1.4
⎤
− 1⎥
⎦⎥
= 354.5 kJ/kg
Patm = 85.6 kPa, 15°C
The cross-sectional area of the 5-mm diameter hole is
Air leak
A = πD 2 / 4 = π (5 × 10 −3 m) 2 / 4 = 19.63 × 10 −6 m 2
Noting that the line conditions are T0 = 298 K
and P0 = 935.6 kPa, the mass flow rate of the air
leaking through the hole is determined to be
⎛ 2 ⎞
m& air = C loss ⎜
⎟
⎝ k +1 ⎠
1 /( k −1)
⎛ 2 ⎞
= (0.65)⎜
⎟
⎝ 1.4 + 1 ⎠
Compressed air line
850 kPa, 25°C
P0
⎛ 2 ⎞
A kR⎜
⎟T0
RT0
⎝ k +1⎠
1 /(1.4 −1)
935.6 kPa
3
(19.63 × 10 − 6 m 2 )
(0.287 kPa.m / kg.K)(298 K)
⎛ 1000 m 2 / s 2
× (1.4)(0.287 kJ/kg.K)⎜
⎜ 1 kJ/kg
⎝
= 0.02795 kg/s
⎞⎛ 2 ⎞
⎟⎜
(298 K)
⎟⎝ 1.4 + 1 ⎟⎠
⎠
Then the power wasted by the leaking compressed air becomes
Power wasted = m& air wcomp,in = (0.02795 kg / s)(354.5 kJ / kg) = 9.91 kW
Noting that the compressor operates 4200 hours a year and the motor efficiency is 0.93, the annual energy
and cost savings resulting from repairing this leak are determined to be
Energy Savings = (Power wasted)(Annual operating hours)/Motor efficiency
= (9.91 kW)(4200 hours/year)/0.93
= 44,755 kWh/year
Cost Savings = (Energy savings)(Unit cost of energy)
= (44,755 kWh/year)($0.07/kWh)
= $3133/year
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7-117
Therefore, the facility will save 44,755 kWh of electricity that is worth $3133 a year when this air leak is
sealed.
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Review Problems
7-160 A piston-cylinder device contains steam that undergoes a reversible thermodynamic cycle composed
of three processes. The work and heat transfer for each process and for the entire cycle are to be
determined.
Assumptions 1 All processes are reversible. 2 Kinetic and potential energy changes are negligible.
Analysis The properties of the steam at various states
are (Tables A-4 through A-6)
u = 2884.5 kJ/kg
P1 = 400 kPa ⎫ 1
3
⎬ v 1 = 0.71396 m /kg
T1 = 350°C ⎭
s = 7.7399 kJ/kg.K
P = const.
3
1
s = const.
1
P2 = 150 kPa ⎫ u 2 = 2888.0 kJ/kg
⎬
T2 = 350°C ⎭ s 2 = 8.1983 kJ/kg.K
P3 = 400 kPa
⎫ u 3 = 3132.9 kJ/kg
⎬
s 3 = s 2 = 8.1983 kJ/kg.K ⎭ v 3 = 0.89148 m 3 /kg
T = const.
2
The mass of the steam in the cylinder and the volume at state 3 are
m=
V1
0.3 m 3
=
= 0.4202 kg
v 1 0.71396 m 3 /kg
V 3 = mV 3 = (0.4202 kg)(0.89148 m 3 /kg) = 0.3746 m 3
Process 1-2: Isothermal expansion (T2 = T1)
∆S1− 2 = m( s2 − s1 ) = (0.4202 kg)(8.1983 − 7.7399)kJ/kg.K = 0.1926 kJ/kg.K
Qin,1− 2 = T1 ∆S1− 2 = (350 + 273 K)(0.1926 kJ/K) = 120 kJ
Wout,1− 2 = Qin,1− 2 − m(u 2 − u1 ) = 120 kJ − (0.4202 kg)(2888.0 − 2884.5)kJ/kg = 118.5 kJ
Process 2-3: Isentropic (reversible-adiabatic) compression (s3 = s2)
Win,2 − 3 = m(u3 − u2 ) = (0.4202 kg)(3132.9 - 2888.0)kJ/kg = 102.9 kJ
Q2-3 = 0 kJ
Process 3-1: Constant pressure compression process (P1 = P3)
Win,3−1 = P3 (V3 − V1 ) = (400 kPa)(0.3746 - 0.3) = 29.8 kJ
Qout,3−1 = Win,3−1 − m(u1 − u 3 ) = 29.8 kJ - (0.4202 kg)(2884.5 - 3132.9) = 134.2 kJ
The net work and net heat transfer are
Wnet,in = Win,3−1 + Win,2 −3 − Wout,1− 2 = 29.8 + 102.9 − 118.5 = 14.2 kJ
Qnet,in = Qin,1− 2 − Qout,3−1 = 120 − 134.2 = −14.2 kJ ⎯
⎯→ Qnet,out = 14.2 kJ
Discussion The results are not surprising since for a cycle, the net work and heat transfers must be equal to
each other.
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7-119
7-161 The work input and the entropy generation are to be determined for the compression of saturated
liquid water in a pump and that of saturated vapor in a compressor.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3
Heat transfer to or from the fluid is zero.
Analysis Pump Analysis: (Properties are obtained from EES)
P1 = 100 kPa ⎫ h1 = 417.51 kJ/kg
⎬
x1 = 0 (sat. liq.)⎭ s1 = 1.3028 kJ/kg.K
h2 = h1 +
h2 s − h1
ηP
= 417.51 +
P2 = 1 MPa ⎫
⎬h2 s = 418.45 kJ/kg
s2 = s1
⎭
418.45 − 417.51
= 418.61 kJ/kg
0.85
1 MPa
pump
100 kPa
P2 = 1 MPa
⎫
⎬s 2 = 1.3032 kJ/kg.K
h2 = 418.61 kJ/kg ⎭
wP = h2 − h1 = 418.61 − 417.51 = 1.10 kJ/kg
s gen,P = s 2 − s1 = 1.3032 − 1.3028 = 0.0004 kJ/kg.K
Compressor Analysis:
P1 = 100 kPa
⎫ h1 = 2675.0 kJ/kg
⎬
x1 = 1 (sat. vap.)⎭ s1 = 7.3589 kJ/kg.K
h2 = h1 +
h2 s − h1
ηC
= 2675.0 +
P2 = 1 MPa ⎫
⎬h2 s = 3193.6 kJ/kg
s2 = s1
⎭
1 MPa
3193.6 − 2675.0
= 3285.1 kJ/kg
0.85
Compresso
P2 = 1 MPa
⎫
⎬s 2 = 7.4974 kJ/kg.K
=
3285
.
1
kJ/kg
h2
⎭
100 kPa
wC = h2 − h1 = 3285.1 − 2675.0 = 610.1 kJ/kg
s gen,C = s 2 − s1 = 7.4974 − 7.3589 = 0.1384 kJ/kg.K
7-162 A paddle wheel does work on the water contained in a rigid tank. For a zero entropy change of
water, the final pressure in the tank, the amount of heat transfer between the tank and the surroundings, and
the entropy generation during the process are to be determined.
Assumptions The tank is stationary and the kinetic and potential energy changes are negligible.
Analysis (a) Using saturated liquid properties for the compressed liquid at the initial state (Table A-4)
T1 = 120°C
⎫ u1 = 503.60 kJ/kg
⎬
x1 = 0 (sat. liq.)⎭ s1 = 1.5279 kJ/kg.K
The entropy change of water is zero, and thus at the final state we have
T2 = 95°C
⎫ P2 = 84.6 kPa
⎬
s 2 = s1 = 1.5279 kJ/kg.K ⎭ u 2 = 492.63 kJ/kg
Water
120°C
500 kPa
Wpw
(b) The heat transfer can be determined from an energy balance on the tank
Qout = W Pw,in − m(u 2 − u1 ) = 22 kJ − (1.5 kg)(492.63 − 503.60)kJ/kg = 38.5 kJ
(c) Since the entropy change of water is zero, the entropy generation is only due to the entropy increase of
the surroundings, which is determined from
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S gen = ∆S surr =
Qout
38.5 kJ
=
= 0.134 kJ/K
Tsurr (15 + 273) K
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7-163 A horizontal cylinder is separated into two compartments by a piston, one side containing nitrogen
and the other side containing helium. Heat is added to the nitrogen side. The final temperature of the
helium, the final volume of the nitrogen, the heat transferred to the nitrogen, and the entropy generation
during this process are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Nitrogen and helium are ideal gases
with constant specific heats at room temperature. 3 The piston is adiabatic and frictionless.
Properties The properties of nitrogen at room temperature are R = 0.2968 kPa.m3/kg.K, cp = 1.039
kJ/kg.K, cv = 0.743 kJ/kg.K, k = 1.4. The properties for helium are R = 2.0769 kPa.m3/kg.K, cp = 5.1926
kJ/kg.K, cv = 3.1156 kJ/kg.K, k = 1.667 (Table A-2).
Analysis (a) Helium undergoes an isentropic compression
process, and thus the final helium temperature is
determined from
( k −1) / k
⎛P ⎞
THe,2 = T1 ⎜⎜ 2 ⎟⎟
⎝ P1 ⎠
= 321.7 K
⎛ 120 kPa ⎞
= (20 + 273)K⎜
⎟
⎝ 95 kPa ⎠
(1.667 −1) / 1.667
N2
0.2 m3
He
0.1 kg
(b) The initial and final volumes of the helium are
V He,1 =
mRT1 (0.1 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K)
=
= 0.6406 m 3
95 kPa
P1
V He,2 =
mRT2 (0.1 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(321.7 K)
=
= 0.5568 m 3
120 kPa
P2
Then, the final volume of nitrogen becomes
V N2,2 = V N2,1 + V He,1 −V He,2 = 0.2 + 0.6406 − 0.5568 = 0.2838 m 3
(c) The mass and final temperature of nitrogen are
m N2 =
P1V 1
(95 kPa)(0.2 m 3 )
=
= 0.2185 kg
RT1 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K)
T N2,2 =
P2V 2
(120 kPa)(0.2838 m 3 )
=
= 525.1 K
mR
(0.2185 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)
The heat transferred to the nitrogen is determined from an energy balance
Qin = ∆U N2 + ∆U He
= [mcv (T2 − T1 )]N2 + [mcv (T2 − T1 )]He
= (0.2185 kg)(0.743 kJ/kg.K)(525.1 − 293) + (0.1 kg)(3.1156 kJ/kg.K)(321.7 − 293)
= 46.6 kJ
(d) Noting that helium undergoes an isentropic process, the entropy generation is determined to be
⎛
T
P ⎞ − Qin
S gen = ∆S N2 + ∆S surr = m N2 ⎜⎜ c p ln 2 − R ln 2 ⎟⎟ +
T1
P1 ⎠ TR
⎝
525.1 K
120 kPa ⎤
− 46.6 kJ
⎡
= (0.2185 kg) ⎢(1.039 kJ/kg.K)ln
− (0.2968 kJ/kg.K)ln
+
⎥
293 K
95 kPa ⎦ (500 + 273) K
⎣
= 0.057 kJ/K
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7-164 An electric resistance heater is doing work on carbon dioxide contained an a rigid tank. The final
temperature in the tank, the amount of heat transfer, and the entropy generation are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Carbon dioxide is ideal gas with
constant specific heats at room temperature.
Properties The properties of CO2 at an anticipated average temperature of 350 K are R = 0.1889
kPa.m3/kg.K, cp = 0.895 kJ/kg.K, cv = 0.706 kJ/kg.K (Table A-2b).
Analysis (a) The mass and the final temperature of CO2
may be determined from ideal gas equation
CO2
250 K
100 kPa
PV
(100 kPa)(0.8 m 3 )
m= 1 =
= 1.694 kg
RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(250 K)
T2 =
We
P2V
(175 kPa)(0.8 m 3 )
=
= 437.5 K
mR (1.694 kg)(0.1889 kPa ⋅ m 3 /kg ⋅ K)
(b) The amount of heat transfer may be determined from an energy balance on the system
Q = E& ∆t − mc (T − T )
out
e, in
v
2
1
= (0.5 kW)(40 × 60 s) - (1.694 kg)(0.706 kJ/kg.K)(437.5 - 250)K = 975.8 kJ
(c) The entropy generation associated with this process may be obtained by calculating total entropy
change, which is the sum of the entropy changes of CO2 and the surroundings
⎛
T
P ⎞ Q
S gen = ∆S CO2 + ∆S surr = m⎜⎜ c p ln 2 − R ln 2 ⎟⎟ + out
T1
P1 ⎠ Tsurr
⎝
437.5 K
175 kPa ⎤ 975.8 kJ
⎡
= (1.694 kg) ⎢(0.895 kJ/kg.K)ln
− (0.1889 kJ/kg.K)ln
+
250 K
100 kPa ⎥⎦ 300 K
⎣
= 3.92 kJ/K
7-165 Heat is lost from the helium as it is throttled in a throttling valve. The exit pressure and temperature
of helium and the entropy generation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3
Helium is an ideal gas with constant specific heats.
Properties The properties of helium are R = 2.0769
kPa.m3/kg.K, cp = 5.1926 kJ/kg.K (Table A-2a).
Analysis (a) The final temperature of helium may be
determined from an energy balance on the control volume
q out = c p (T1 − T2 ) ⎯
⎯→ T2 = T1 −
q
Helium
500 kPa
70°C
q out
2.5 kJ/kg
= 70°C −
= 342.5 K = 69.5°C
cp
5.1926 kJ/kg.°C
The final pressure may be determined from the relation for the entropy change of helium
∆sHe = c p ln
T2
P
− R ln 2
T1
P1
0.25 kJ/kg.K = (5.1926 kJ/kg.K)ln
342.5 K
P2
− (2.0769 kJ/kg.K)ln
343 K
500 kPa
P2 = 441.7 kPa
(b) The entropy generation associated with this process may be obtained by adding the entropy change of
helium as it flows in the valve and the entropy change of the surroundings
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7-123
sgen = ∆sHe + ∆ssurr = ∆sHe +
qout
2.5 kJ/kg
= 0.25 kJ/kg.K +
= 0.258 kJ/kg.K
Tsurr
(25 + 273) K
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7-166 Refrigerant-134a is compressed in a compressor. The rate of heat loss from the compressor, the exit
temperature of R-134a, and the rate of entropy generation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties of R-134a at the inlet of the compressor are (Table A-12)
3
P1 = 200 kPa ⎫ v 1 = 0.09987 m /kg
⎬ h1 = 244.46 kJ/kg
x1 = 1
⎭ s = 0.93773 kJ/kg.K
1
The mass flow rate of the refrigerant is
m& =
V&
1
v1
=
3
0.03 m /s
0.09987 m 3 /kg
700 kPa
Compressor
= 0.3004 kg/s
Given the entropy increase of the surroundings, the heat
lost from the compressor is
∆S& surr =
Q
R-134a
200 kPa
sat. vap.
Q& out
⎯
⎯→ Q& out = Tsurr ∆S& surr = (20 + 273 K)(0.008 kW/K) = 2.344 kW
Tsurr
(b) An energy balance on the compressor gives
W& in − Q& out = m& (h2 − h1 )
10 kW - 2.344 kW = (0.3004 kg/s)(h2 - 244.46) kJ/kg ⎯
⎯→ h2 = 269.94 kJ/kg
The exit state is now fixed. Then,
P2 = 700 kPa
⎫ T2 = 31.5°C
⎬
h2 = 269.94 kJ/kg ⎭ s 2 = 0.93620 kJ/kg.K
(c) The entropy generation associated with this process may be obtained by adding the entropy change of
R-134a as it flows in the compressor and the entropy change of the surroundings
S& gen = ∆S& R + ∆S& surr = m& ( s 2 − s1 ) + ∆S& surr
= (0.3004 kg/s)(0.93620 - 0.93773) kJ/kg.K + 0.008 kW/K
= 0.00754 kJ/K
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7-167 Air flows in an adiabatic nozzle. The isentropic efficiency, the exit velocity, and the entropy
generation are to be determined.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible.
Analysis (a) (b) Using variable specific heats, the properties can be determined from air table as follows
h1 = 400.98 kJ/kg
⎯→ s10 = 1.99194 kJ/kg.K
T1 = 400 K ⎯
Pr1 = 3.806
⎯→
T2 = 350 K ⎯
Pr 2 =
Air
500 kPa
400 K
30 m/s
h2 = 350.49 kJ/kg
s20 = 1.85708 kJ/kg.K
300 kPa
350 K
P2
300 kPa
(3.806) = 2.2836 ⎯⎯→ h2 s = 346.31 kJ/kg
Pr1 =
P1
500 kPa
Energy balances on the control volume for the actual and isentropic processes give
h1 +
400.98 kJ/kg +
V12
V2
= h2 + 2
2
2
(30 m/s)2 ⎛ 1 kJ/kg ⎞
V22 ⎛ 1 kJ/kg ⎞
350
.
49
kJ/kg
=
+
⎟
⎜
⎜
⎟
2 2
2
2 ⎝ 1000 m 2 /s 2 ⎠
⎝ 1000 m /s ⎠
V2 = 319.1 m/s
h1 +
400.98 kJ/kg +
V12
V2
= h2 s + 2s
2
2
(30 m/s)2 ⎛ 1 kJ/kg ⎞
V 2 ⎛ 1 kJ/kg ⎞
⎟
⎜
⎟ = 346.31 kJ/kg + 2s ⎜
2 2
2
2 ⎝ 1000 m 2 /s 2 ⎠
⎝ 1000 m /s ⎠
V2s = 331.8 m/s
The isentropic efficiency is determined from its definition,
ηN =
V 22
V 2s2
=
(319.1 m/s) 2
(331.8 m/s) 2
= 0.925
(c) Since the nozzle is adiabatic, the entropy generation is equal to the entropy increase of the air as it
flows in the nozzle
sgen = ∆sair = s20 − s10 − R ln
P2
P1
= (1.85708 − 1.99194)kJ/kg.K − (0.287 kJ/kg.K)ln
300 kPa
= 0.0118 kJ/kg.K
500 kPa
7-168 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 ) = P dv + v dP = δ wb − δ w flow = δ wb − w flow
)
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7-126
Therefore, the difference between the reversible steady-flow work and the reversible boundary work is the
flow energy.
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7-127
7-169 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 s2 = s1. From the
steam tables (Tables A-4 through A-6),
v 1 = v g @500 kPa = 0.37483 m 3 /kg
P1 = 500 kPa ⎫
⎬ u1 = u g @500 kPa = 2560.7 kJ/kg
sat.vapor
⎭s = s
g @ 500 kPa = 6.8207 kJ/kg ⋅ K
1
T2, A = Tsat @150 kPa = 111.35 °C
s 2, A − s f
6.8207 − 1.4337
=
=
= 0.9305
s fg
5.7894
P2 = 150 kPa ⎫
x 2, A
⎪
s 2 = s1
⎬
(sat.mixture ) ⎪⎭ v 2, A = v f + x 2, Av fg = 0.001053 + (0.9305)(1.1594 − 0.001053) = 1.0789 m 3 /kg
u 2, A = u f + x 2, A u fg = 466.97 + (0.9305)(2052.3 kJ/kg ) = 2376.6 kJ/kg
The initial and the final masses in tank A are
m1, A =
Thus,
0.4 m3
VA
=
= 1.067 kg
v1, A 0.37483 m3/kg
and m2, A =
0.4 m3
VA
=
= 0.371 kg
v 2, A 1.0789 m3/kg
m2, B = m1, A − m2, A = 1.067 − 0.371 = 0.696 kg
(b) The boundary work done during this process is
Wb,out =
∫
(
2
1
)
P dV = PB V 2, B − 0 = PB m 2, Bv 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 − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Sat.
vapor
500 kPa
0 4 m3
150 kPa
Change in internal, kinetic,
potential, etc. energies
− Wb, out = ∆U = (∆U )A + (∆U )B
Wb, out + (∆U )A + (∆U )B = 0
or,
PB m2, Bv 2, B + (m2u2 − m1u1 )A + (m2u2 )B = 0
m2, B h2, B + (m2u2 − m1u1 )A = 0
Thus,
h 2, B =
(m1u1 − m 2 u 2 ) A
m 2, B
=
(1.067)(2560.7 ) − (0.371)(2376.6) = 2658.8 kJ/kg
0.696
At 150 kPa, hf = 467.13 and hg = 2693.1 kJ/kg. Thus at the final state, the cylinder will contain a saturated
liquid-vapor mixture since hf < h2 < hg. Therefore,
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7-128
T2, B = Tsat @150 kPa = 111.35°C
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7-129
7-170 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 (a) The volume and the mass of the air in the room are
V = 4x5x7 = 140 m³
mair =
(
)
(100 kPa ) 140 m3
P1V1
=
= 165.4 kg
RT1
0.2870 kPa ⋅ m3/kg ⋅ K (295 K )
(
)
4m×5m×7m
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 − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
or
∆Esystem
1
424
3
→ 0 = ∆U = (∆U )water + (∆U )air
Heat
Water
80°C
Change in internal, kinetic,
potential, etc. energies
[mc(T2 − T1 )]water + [mcv (T2 − T1 )]air
=0
Substituting,
(1000 kg )(4.18 kJ/kg⋅o C)(T f
)
(
)(
)
− 80 o C + (165.4 kg ) 0.718 kJ/kg⋅o C T f − 22 o C = 0
It gives the final equilibrium temperature in the room to be
Tf = 78.4°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
∆Sair = mcv ln
∆S water = mc ln
351.4 K
T2
V ©0
+ mR ln 2
= (165.4 kg )(0.718 kJ/kg ⋅ K ) ln
= 20.78 kJ/K
295 K
T1
V1
351.4 K
T2
= (1000 kg )(4.18 kJ/kg ⋅ K ) ln
= −18.99 kJ/K
353 K
T1
Substituting, the total entropy change is determined to be
∆Stotal = 20.78 - 18.99 = 1.79 kJ/K
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7-130
7-171E 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. The specific heats
of helium are cv = 0.753 and cp = 1.25 Btu/lbm.R (Table A-2E).
Analysis (a) The mass of helium is
m=
(
)
)
(25 psia ) 15 ft 3
P1V1
=
= 0.264 lbm
RT1
2.6805 psia ⋅ ft 3/lbm ⋅ R (530 R )
(
HELIUM
15 ft3
n
PV = const
Then the entropy change of helium becomes
⎛
T
P ⎞
∆Ssys = ∆S helium = m⎜⎜ c p ,avg ln 2 − R ln 2 ⎟⎟
P1 ⎠
T1
⎝
Q
⎡
760 R
70 psia ⎤
= (0.264 lbm) ⎢(1.25 Btu/lbm ⋅ R ) ln
− (0.4961 Btu/lbm ⋅ R ) ln
⎥ = −0.016 Btu/R
530 R
25 psia ⎦
⎣
(b) The exponent n and the boundary work for this polytropic process are determined to be
P1V1 P2V 2
T P
(760 R )(25 psia )
=
⎯
⎯→V 2 = 2 1 V1 =
15 ft 3 = 7.682 ft 3
(530 R )(70 psia )
T1
T2
T1 P2
(
⎛P
P2V 2n = P1V1n ⎯
⎯→ ⎜⎜ 2
⎝ P1
⎞ ⎛ V1
⎟⎟ = ⎜⎜
⎠ ⎝V 2
n
⎞
⎛ 70 ⎞ ⎛ 15 ⎞
⎟⎟ ⎯
⎯→ ⎜ ⎟ = ⎜
⎟
⎝ 25 ⎠ ⎝ 7.682 ⎠
⎠
)
n
⎯
⎯→ n = 1.539
Then the boundary work for this polytropic process can be determined from
∫
2
Wb,in = − P dV = −
1
=−
P2V 2 − P1V1
mR(T2 − T1 )
=−
1− n
1− n
(0.264 lbm)(0.4961 Btu/lbm ⋅ R )(760 − 530)R
1 − 1.539
= 55.9 Btu
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 − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout + Wb,in = ∆U = m(u2 − u1 )
− Qout = m(u2 − u1 ) − Wb,in
Qout = Wb,in − mcv (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,
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7-131
∆S total = ∆Ssys + ∆Ssurr = −0.016 + 0.019 = 0.003 Btu/R > 0
Therefore, the process is irreversible.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-132
7-172 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
and
Pr2 =
P2
700 kPa
Pr1 =
(1.2311) = 8.6177
P1
100 kPa
→
T2 = 503.3 K
h2 = 506.45 kJ/kg
The energy balance for the compressor, which is a steady-flow
system, can be expressed in the rate form as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
2
=0
2
AIR
Rev.
Rate of change in internal, kinetic,
potential, etc. energies
T = const
E& in = E& out
W& in + m& h1 = m& h2 → W& in = m& (h2 − h1 )
1
1
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
E& in = E& out → W&in + m& h1 − Q& out = m& h2 → W&in = Q& out + m& (h2 − h1 )©0 = Q& out
since h = h(T) for ideal gases, and thus the enthalpy change in this case is zero. Also, for a reversible
isothermal process,
Q& out = m& T (s1 − s2 ) = −m& T (s2 − s1 )
where
(
s2 − s1 = s2o − s1o
)
©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
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7-133
7-173 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
Px =
The compressor work across each stage is the
same, thus total compressor work is twice the
compression work for a single stage:
(
100 kPa
27°C
(100 kPa )(900 kPa ) = 300 kPa
P1P2 =
)
(
W
Stage I
Stage II
)
kRT1
(Px P1 )(k −1) / k − 1
k −1
(1.4)(0.287 kJ/kg ⋅ K )(300 K ) ⎛⎜ ⎛ 300 kPa ⎞0.4/1.4 − 1⎞⎟
=2
⎜
⎟
⎜ ⎝ 100 kPa ⎠
⎟
1.4 − 1
⎝
⎠
= 222.2 kJ/kg
wcomp,in = (2 ) wcomp,in, I = 2
900 kPa
27°C
Heat
and
W&in = m& wcomp,in = (0.02 kg/s )(222.2 kJ/kg ) = 4.44 kW
The work input to a single-stage compressor operating between the same pressure limits would be
wcomp,in =
⎛⎛
⎞
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
⎟
⎠
and
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.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-134
7-174 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:
n −1
−1
n
∂w
n − 1 ⎛ 1 ⎞⎛ Px
⎜ ⎟⎜
=0⎯
⎯→ −
∂ Px
n ⎜⎝ P1 ⎟⎠⎜⎝ P1
⎞
⎟⎟
⎠
⎞⎛ Py
⎟⎜
⎟⎜ P
⎠⎝ x
⎞
⎟
⎟
⎠
∂w
n −1⎛ 1
⎜
=0⎯
⎯→ −
∂ Py
n ⎜⎝ Px
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 ⎞
= n ⎜⎜ ⎟⎟
n ⎜P ⎟
Px ⎝ y ⎠ P2 ⎝ P2 ⎠
1− 2 n
(
⎯
⎯→ Px2(1− n ) = P1 Py
)1−n
⎯
⎯→ Py2(1− n ) = (Px P2 )1− n
which yield
( )
= (P P )
Px2 = P1 Px P2 ⎯⎯→ Px = P12 P2
Py2 = P2 P1 Py ⎯⎯→ Py
13
2 13
1 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-135
7-175 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 = 6 MPa ⎫ h1 = 3423.1 kJ/kg
⎬
T1 = 500°C ⎭ s1 = 6.8826 kJ/kg ⋅ K
P2 = 1.2 MPa ⎫
⎬ h2 = 2962.8 kJ / kg
s2 = s1
⎭
s3s − s f
6.8826 − 0.8320
=
= 0.8552
P3 = 20 kPa ⎫ x3s =
s fg
7.0752
⎬
s3 = s1
⎭
h3s = h f + x3s h fg = 251.42 + (0.8552 )(2357.5) = 2267.5 kJ/kg
Analysis (a) The mass flow rate through the second stage is
m& 3 = 0.9m& 1 = (0.9)(15 kg/s) = 13.5 kg/s
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
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
6 MPa
500°C
I
II
1.2 MPa
20 kPa
Rate of change in internal, kinetic,
potential, etc. energies
10%
E& in = E& out
STEAM
13.5 kg/s
STEAM
15 kg/s
90%
m& 1h1 = (m& 1 − m& 3 )h2 + W&out + m& 3h3
W& = m& h − (m& − m& )h − m& h
out
1 1
1
3
2
3 3
= m& 1 (h1 − h2 ) + m& 3 (h2 − h3 )
Substituting, the power output of the turbine is
W&out = (15 kg/s )(3423.1 − 2962.8)kJ/kg + (13.5 kg )(2962.8 − 2267.5)kJ/kg = 16,291 kW
(b) If the turbine has an adiabatic efficiency of 88%, then the power output becomes
W&a = ηTW&s = (0.88)(16,291 kW ) = 14,336 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-136
7-176 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 = 3521.8 kJ/kg
⎬
T1 = 550°C ⎭ s1 = 6.8800 kJ/kg ⋅ K
2 MPa
P2 s = 2 MPa ⎫
⎬ h2 s = 3089.7 kJ/kg
s 2 s = s1
⎭
Stage I
2 MPa
550°C
Stage II
80 MW
P3 = 2 MPa ⎫ h3 = 3579.0 kJ/kg
⎬
T3 = 550°C ⎭ s 3 = 7.5725 kJ/kg ⋅ K
P4 s = 200 kPa ⎫
⎬ h4 s = 2901.7 kJ/kg
s 4s = s3
⎭
8 MPa
550°C
200 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 / ηT = (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
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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& [(3521.8 − 3089.7) kJ/kg + (3579.0 − 2901.7) kJ/kg ]
which gives
m& = 85.8 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-137
7-177 Refrigerant-134a is compressed by a 0.7-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 1 = 0.14605 m 3 /kg
P1 = 140 kPa ⎫
⎬ h1 = 246.36 kJ/kg
T1 = −10°C ⎭
s1 = 0.9724 kJ/kg ⋅ K
2
0.7 kW
R-134a
P2 = 700 kPa ⎫
⎬ h2 = 288.53 kJ/kg
⎭
T2 = 50°C
P2 = 700 kPa ⎫
⎬ h2 s = 281.16 kJ/kg
s 2 s = s1
⎭
·
V1
1
Analysis (a) The isentropic efficiency is determined from its definition,
ηC =
h2 s − h1 281.16 − 246.36
=
= 0.825 = 82.5%
h2 a − h1 288.53 − 246.36
&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
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=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
=
0.7 kJ/s
= 0.0166 kg/s
(288.53 − 246.36)kJ/kg
(
)
V&1 = m& v1 = (0.0166 kg/s ) 0.14605 m3/kg = 0.00242 m3/s = 145 L/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 =
W&s,in
=
0.7 kJ/s
= 0.0201 kg/s
(281.16 − 246.36)kJ/kg
V&1, max = m& maxv1 = (0.0201 kg/s )(0.14605 m3/kg ) = 0.00294 m3/s = 176
h2 s − h1
L/min
Discussion Note that the raising the isentropic efficiency of the compressor to 100% would increase the
volumetric flow rate by more than 20%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-138
7-178E 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
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
1
HELIUM
ηN = 94%
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ W& ≅ ∆pe ≅ 0)
0 = h2 − h1 +
V22 − V12
V 2 − V12
⎯
⎯→ 0 = c p , avg (T2 − T1 ) + 2
2
2
Solving for T1 and substituting,
T1 = T2 a
V 2 − V12
+ 2s
2C p
©0
= 180°F +
(1000 ft/s )2
⎛ 1 Btu/lbm
⎜
2(1.25 Btu/lbm ⋅ R ) ⎜⎝ 25,037 ft 2 /s 2
⎞
⎟ = 196.0°F = 656 R
⎟
⎠
From the isentropic efficiency relation,
ηN =
or,
h2 a − h1 cP (T2 a − T1 )
=
h2 s − h1 cP (T2 s − T1 )
T2 s = T1 + (T2 a − T1 ) / η N = 656 + (640 − 656 ) / (0.94 ) = 639 R
From the isentropic relation,
⎛ T
P1 = P2 ⎜⎜ 1
⎝ T2 s
⎞
⎟
⎟
⎠
k / (k −1)
T2 s ⎛ P2 ⎞
=⎜ ⎟
T1 ⎜⎝ P1 ⎟⎠
(k −1) / k
⎛ 656 R ⎞
⎟⎟
= (14 psia )⎜⎜
⎝ 639 R ⎠
1.667/0.667
= 14.9 psia
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2
7-139
7-179 [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 The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). From the steam tables (Tables A-4
through 6) and air table (Table A-17),
⎯→ h1 = 295.17 kJ/kg, s1o = 1.68515 kJ/kg ⋅ K
T1 = 295 K ⎯
⎯→ h1 = 628.07 kJ/kg, s2o = 2.44356 kJ/kg ⋅ K
T2 = 620 K ⎯
P3 = 12.5 MPa ⎫ h3 = 3343.6 kJ/kg
⎬
⎭ s3 = 6.4651 kJ/kg ⋅ K
T3 = 500°C
P4 = 10 kPa ⎫ h4 = h f + x4 h fg = 191.81 + (0.92 )(2392.1) = 2392.5 kJ/kg
⎬
x4 = 0.92
⎭ s4 = s f + x4 s fg = 0.6492 + (0.92 )(7.4996 ) = 7.5489 kJ/kg ⋅ K
Analysis There is only one inlet and one exit for either device, and thus m& in = m& out = m& . We take either
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
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system©0 (steady)
1442443
= 0 → E& in = E& out
12.5 MPa
500°C
1 MPa
620 K
Rate of change in internal, kinetic,
potential, etc. energies
For the turbine and the compressor it becomes
Compressor: W&comp, in + m& air h1 = m& air h2 → W&comp, in = m& air (h2 − h1 )
Air
comp
Steam
turbine
Turbine: m& steam h3 = W& turb, out + m& steam h4 → W& turb, out = m& steam (h3 − h4 )
Substituting,
W&comp,in = (10 kg/s )(628.07 − 295.17 )kJ/kg = 3329 kW
98 kPa
295 K
10 kPa
W& turb,out = (25 kg/s )(3343.6 − 2392.5)kJ/kg = 23,777 kW
Therefore,
W& net,out = W& turb,out − W& comp,in = 23,777 − 3329 = 20,448 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&gen = m& air ( s2 − s1 ) + m& steam ( s4 − s3 )
where
⎛
P ⎞
m& air (s2 − s1 ) = m& ⎜⎜ s2o − s1o − R ln 2 ⎟⎟
P1 ⎠
⎝
⎛
1000 kPa ⎞
⎟kJ/kg ⋅ K = 0.92 kW/K
= (10 kg/s )⎜⎜ 2.44356 − 1.68515 − 0.287 ln
98 kPa ⎟⎠
⎝
m& steam (s4 − s3 ) = (25 kg/s )(7.5489 − 6.4651)kJ/kg ⋅ K = 27.1 kW/K
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7-140
Substituting, the total rate of entropy generation is determined to be
S&gen, total = S&gen, comp + S&gen, turb = 0.92 + 27.1 = 28.02 kW/K
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-141
7-180 EES Problem 7-179 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.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-142
"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 becomes:"
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 Parmetric 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
Parmetric Table, then press F3 to solve the table. EES then solves for the h_st[2] for each
Eta_turb."
Wnet
[kW]
20124
21745
23365
24985
26606
Sgentotal
[kW/K]
27.59
22.51
17.44
12.36
7.281
ηturb
ηcomp
0.75
0.8
0.85
0.9
0.95
0.6665
0.6665
0.6665
0.6665
0.6665
Wnet
[kW]
19105
19462
19768
20033
20265
Sgentotal
[kW/K]
30
29.51
29.07
28.67
28.32
ηturb
ηcomp
0.7327
0.7327
0.7327
0.7327
0.7327
0.6
0.65
0.7
0.75
0.8
Effect of Compressor Efficiency on Net Work and Entropy Generated
20400
30.0
ηturb =0.7333
]
W
k[
t
e
n
29.6
]
K/
29.2 W
k[
20000
19600
28.8
W
28.4 S
19200
0.60
l
at
ot,
n
e
g
0.65
0.70
0.75
0.80
ηcompb
Effect of Turbine Efficiency on Net Work and Entropy Generated
30
ηcomp = 0.6665
26000
]
W
k[
t
e
n
W
25
]
K/
20 W
k[
24000
l
15 at
ot,
n
e
10 g
22000
S
20000
0.75
5
0.78
0.81
0.84
0.87
0.90
0.93
ηturbb
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-143
7-181 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
m, c
Sgen = (∆S )source + (∆S )engine©0 + (∆S )sink = 0
T1
mcln
ln
Tf
T1
+ ln
Tf
T2
Tf
T1
+ 0 + mcln
=0 ⎯
⎯→ ln
Tf Tf
T1 T2
Tf
T2
=0
QH
W
HE
=0 ⎯
⎯→
T f2
= T1T2
and thus
T f = T1T2
QL
m, c
T2
for maximum power production.
7-182 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 @ 2 MPa = 0.001177 m3/kg
P1 = 2 MPa ⎫
⎬ u1 = u f @ 2 MPa = 906.12 kJ/kg
⎭
s1 = s f @ 2 MPa = 2.4467 kJ/kg ⋅ K
sat. liquid
P2 = 100 kPa ⎫ u f = 417.40, u fg = 2088.2 kJ/kg
⎬
s2 = s1
⎭ s f = 1.3028, s fg = 6.0562 kJ/kg ⋅ K
x2 =
s2 − s f
s fg
=
Water
Tank
2 MPa
2.4467 − 1.3028
= 0.1889
6.0562
u2 = u f + x2u fg = 417.40 + (0.1889)(2088.2 ) = 811.83 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 − Eout
=
∆Esystem
1in
424
3
1
424
3
Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
− Wb,out = ∆U = m(u2 − u1 )
Eexp = Wb,out = m(u1 − u2 )
V
0.080 m3
=
= 67.99 kg
v1 0.001177 m3/kg
where
m=
Substituting,
E exp = (67.99 kg )(906.12 − 811.83)kJ/kg = 6410 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-144
which is equivalent to
mTNT =
6410 kJ
= 1.972 kg TNT
3250 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-145
7-183 A 0.35-L canned drink explodes at a pressure of 1.2 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 @1.2 MPa = 0.001138 m3/kg
P1 = 1.2 MPa ⎫
⎬ u1 = u f @1.2 MPa = 796.96 kJ/kg
Comp. liquid ⎭
s1 = s f @1.2 MPa = 2.2159 kJ/kg ⋅ K
P2 = 100 kPa ⎫ u f = 417.40, u fg = 2088.2 kJ/kg
⎬
s2 = s1
⎭ s f = 1.3028, s fg = 6.0562 kJ/kg ⋅ K
x2 =
s2 − s f
s fg
=
2.2159 − 1.3028
= 0.1508
6.0562
COLA
1.2 MPa
u2 = u f + x2u fg = 417.40 + (0.1508)(2088.2) = 732.26 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 − Eout
1in
424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Wb, out = ∆U = m(u2 − u1 )
Eexp = Wb,out = m(u1 − u2 )
where
m=
0.00035 m 3
V
=
= 0.3074 kg
v 1 0.001138 m 3 /kg
Substituting,
E exp = (0.3074 kg )(796.96 − 732.26)kJ/kg = 19.9 kJ
which is equivalent to
mTNT =
19.9 kJ
= 0.00612 kg TNT
3250 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-135
7-184 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 δQ/T for the reversible and irreversible heat engine cycles become
δ QH
δ QL
1
δQ 
−
=
 =
T
T
T
 rev
H
L
H
∫  T
∫
∫
∫δQ
H
−
1
TL
∫δQ
L
=
QH QL
−
=0
TH
TL
since (QH/TH) = (QL/TL) for reversible cycles. Also,
Q L ,irrev Q H Q L Qdiff
Q
Q
δQ 
= H −
=
−
−
= − diff < 0

T
T
T
T
T
TL
 irrev
H
L
H
L
L
∫  T
since Qdiff is a positive quantity. Thus,
δQ 
≤ 0.

∫  T
TH
·
QH
Rev
HE
QH
·
Wnet, rev
Irrev
HE
·
QL
·
Wnet, irrev
·
QL, irrev
TL
7-185 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
Glass
Q = Q& cond ∆t = (3.2 kJ/s)(5 × 3600 s) = 57,600 kJ
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
1424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
= ∆S& system ©0 = 0
14243
Rate of change
of entropy
10°C
3°C
Q&
Q& in
− 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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-136
7-186 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

v 1, A = v f + x1v fg = 0.001084 + (0.8)(0.46242 − 0.001084 ) = 0.37015 m 3 /kg
u1, A = u f + x1u fg = 604.22 + (0.8)(1948.9 ) = 2163.3 kJ/kg
s1, A = s f + x1 s fg = 1.7765 + (0.8)(5.1191) = 5.8717 kJ/kg ⋅ K
T2, A = Tsat @300 kPa = 133.52 °C
s 2, A − s f
5.8717 − 1.6717
=
=
= 0.7895
5.3200
s fg
P1 = 300 kPa 
x 2, A

s 2 = s1

(sat. mixture)  v 2, A = v f + x 2, Av fg = 0.001073 + (0.7895)(0.60582 − 0.001073) = 0.47850 m 3 /kg
u 2, A = u f + x 2, A u fg = 561.11 + (0.7895)(1982.1 kJ/kg ) = 2125.9 kJ/kg
Tank B :
600 kJ
v = 1.1989 m 3 /kg
P1 = 200 kPa  1, B
 u1, B = 2731.4 kJ/kg
T1 = 250°C 
s1, B = 7.7100 kJ/kg ⋅ K
A
V = 0.2 m3
steam
P = 400 kPa
x = 0.8
The initial and the final masses in tank A are
m1, A =
VA
0.2 m 3
=
= 0.5403 kg
v 1, A 0.37015 m 3 /kg
m 2, A =
VA
0.2 m 3
=
= 0.4180 kg
v 2, A 0.47850 m 3 /kg
and
×
B
m = 3 kg
steam
T = 250°C
P = 200 kPa
Thus, 0.5403 - 0.4180 = 0.1223 kg of mass flows into tank B. Then,
m 2, B = m1, B + 0.1223 = 3 + 0.1223 = 3.1223 kg
The final specific volume of steam in tank B is determined from
v 2, B =
VB
m 2, B
=
(m1v 1 )B
m 2, B
=
(3 kg )(1.1989 m 3 /kg ) = 1.1519 m 3 /kg
3.1223 kg
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
1in424out
3
Net energy transfer
by heat, work, and mass
=
∆E system
1
424
3
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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-137
Substituting,
{
}
− 600 = {(0.418)(2125.9 ) − (0.5403)(2163.3)} + (3.1223)u 2, B − (3)(2731.4 )
u 2, B = 2522.0 kJ/kg
Thus,
T2, B = 113.2°C
v 2, B = 1.1519 m 3 /kg 

u 2, B = 2522.0 kJ/kg  s 2, B = 7.2274 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 − S out
1in424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
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
S gen = ∆S A + ∆S B +
Qout
Q
= (m 2 s 2 − m1 s1 ) A + (m 2 s 2 − m1 s1 ) B + out
Tb,surr
Tb,surr
= {(0.418)(5.8717 ) − (0.5403)(5.8717 )} + {(3.1223)(7.2274 ) − (3)(7.7100)} +
600 kJ
273 K
= 0.916 kJ/K
7-187 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& − S& out
1in424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
104°C
= ∆S& system ©0 = 0
14243
Rate of change
of entropy
Q&
Q& in
− out + S& gen,system = 0
Tb,in Tb,out
500 W
105°C
500 W 500 W &
−
+ S gen,system = 0 → S& gen,system = 0.00351 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.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-138
7-188 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 −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
We,in = (∆U ) water
Water
40 kg
W& e,in ∆t = mc(T2 − T1 ) water
Heater
Substituting,
(1200 J/s)∆t = (40 kg)(4180 J/kg·°C)(50 - 20)°C
Solving for ∆t gives
∆t = 4180 s = 69.7 min = 1.16 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
1424
3
Net entropy transfer
by heat and mass
+ S gen = ∆S system
{
1
424
3
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
323 K
= (40 kg )(4.18 kJ/kg ⋅ K ) ln
= 16.3 kJ/K
293 K
T1
7-189 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
1424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
= ∆S& system ©0 = 0
14243
80°C
Rate of change
of entropy
Q&
Q& in
− out + S& gen,system = 0
Tb,in Tb,out
2200 W 2200 W &
−
+ S gen,system = 0 → S& gen,system = 1.68 W/K
353 K
278 K
Q
Air, 5°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-139
7-190 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)
P1 = 1 MPa  h1 = 2828.3 kJ/kg

T1 = 200°C  s1 = 6.6956 kJ/kg ⋅ K
1
Steam
from
turbine
h2 = h f @1 MPa = 762.51 kJ/kg
P2 = 1 MPa 
s
 2 = s f @1 MPa = 2.1381 kJ/kg ⋅ K
sat. liquid 
T2 = 179.88°C
P3 = 2.5 MPa  h3 ≅ h f @50o C = 209.34 kJ/kg
s ≅ s
= 0.7038 kJ/kg ⋅ K
T3 = 50°C
f @ 50o C
 3
1 MPa
200°C
Feedwater
3
2.5
MPa
4
 h4 ≅ h f @170o C = 719.08 kJ/kg

T4 = T2 − 10 C ≅ 170 C  s 4 ≅ s f @170o C = 2.0417 kJ/kg ⋅ K
P4 = 2.5 MPa
o
o
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:
2
sat. liquid
Mass balance (for each fluid stream):
m& in − m& out = ∆m& system Ê0 (steady) = 0 → m& in = m& out → m& 1 = m& 2 = m& s and m& 3 = m& 4 = m& fw
Energy balance (for the heat exchanger):
E& − E& out
1in
424
3
∆E&systemÊ0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1h1 + m& 3h3 = m& 2h2 + m& 4h4 (since Q& = W& = ∆ke ≅ ∆pe ≅ 0)
Combining the two,
m& s (h2 − h1 ) = m& fw (h3 − h4 )
Dividing by m& fw and substituting,
m& s
h − h3 (719.08 − 209.34) kJ/kg
= 4
=
= 0.247
m& fw h1 − h2 (2828.3 − 762.51) 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
1424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
= ∆S& system ©0 = 0
14243
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 − s 3 ) = (0.247 )(2.1381 − 6.6956) + (2.0417 − 0.7038)
&
m fw
= 0.213 kJ/K per kg of feedwater
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-140
7-191 EES Problem 7-190 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.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Knowns:"
WorkFluid$ = 'Steam_iapws'
"P[3] = 1000 [kPa]" "place {} around P[3] and T[3] eqations to solve the table"
T[3] = 200 [C]
P[4] = P[3]
x[4]=0
T[4]=temperature(WorkFluid$,P=P[4],x=x[4])
P[1] = 2500 [kPa]
T[1] = 50 [C]
P[2] = 2500 [kPa]
T[2] = T[4] - 10"[C]"
"Since we don't know the mass flow rates and we want to determine the ratio of mass flow rate of
the extracted steam and the feedwater, we can assume the mass flow rate of the feedwater is 1
kg/s without loss of generality. We write the conservation of energy."
"Conservation of mass for the steam extracted from the turbine: "
m_dot_steam[3]= m_dot_steam[4]
"Conservation of mass for the condensate flowing through the feedwater heater:"
m_dot_fw[1] = 1
m_dot_fw[2]= m_dot_fw[1]
"Conservation of Energy - SSSF energy balance for the feedwater heater -- neglecting the
change in potential energy, no heat transfer, no work:"
h[3]=enthalpy(WorkFluid$,P=P[3],T=T[3])
"To solve the table, place {} around s[3] and remove them from the 2nd and 3rd equations"
s[3]=entropy(WorkFluid$,P=P[3],T=T[3])
{s[3] =6.693 [kJ/kg-K] "This s[3] is for the initial T[3], P[3]"
T[3]=temperature(WorkFluid$,P=P[3],s=s[3]) "Use this equation for T[3] only when s[3] is given."}
h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])
s[4]=entropy(WorkFluid$,P=P[4],x=x[4])
h[1]=enthalpy(WorkFluid$,P=P[1],T=T[1])
s[1]=entropy(WorkFluid$,P=P[1],T=T[1])
h[2]=enthalpy(WorkFluid$,P=P[2],T=T[2])
s[2]=entropy(WorkFluid$,P=P[2],T=T[2])
"For the feedwater heater:"
E_dot_in = E_dot_out
E_dot_in = m_dot_steam[3]*h[3] +m_dot_fw[1]*h[1]
E_dot_out= m_dot_steam[4]*h[4] + m_dot_fw[2]*h[2]
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
DELTAS_dot_sys = 0 "[KW/K]" "steady-flow result"
S_dot_in = m_dot_steam[3]*s[3] +m_dot_fw[1]*s[1]
S_dot_out= m_dot_steam[4]*s[4] + m_dot_fw[2]*s[2]
S_gen_PerUnitMassFWH = S_dot_gen/m_dot_fw[1]"[kJ/kg_fw-K]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-141
mratio
0.2109
0.2148
0.219
0.223
0.227
0.2309
0.2347
0.2385
0.2422
0.2459
]
K
w
f
0.22
H
W
F
s
s
a
M
ti
n
U
r
e
P,
n
e
g
0.2
g
k/
J
k[
S
Sgen,PerUnitMass
[kJ/kg-K]
0.1811
0.185
0.189
0.1929
0.1968
0.2005
0.2042
0.2078
0.2114
0.2149
P3
[kPa]
732
760
790
820
850
880
910
940
970
1000
P3 = 1000 kPa
0.21
For P3 < 732 kPa
0.19
0.18
0.21
Sgen < 0
P3 = 732 kPa
0.22
0.23
0.24
0.25
mratio
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-142
7-192E 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 110°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)
v 1 = v g @100 psia = 0.47760 ft 3 /lbm
P1 = 100 psia 
140 psia
R-134a
 u1 = u g @100 psia = 104.99 Btu/lbm
sat. vapor
s = s
80°F
g @100 psia = 0.2198 Btu/lbm ⋅ R
1
P2 = 120 psia 

sat. liquid

v 2 = v f @120 psia = 0.01360 ft 3 /lbm
u 2 = u f @120 psia = 41.49 Btu/lbm
s 2 = s f @120 psia = 0.08589 Btu/lbm ⋅ R
R-134a
110°F
3 ft3
Q
Pi = 140 psia  hi ≅ h f @ 80°F = 38.17 Btu/lbm

Ti = 80°F
 s i ≅ s f @ 80° F = 0.07934 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 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 − m out = ∆msystem → mi = m 2 − m1
Energy balance:
E −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin + mi hi = m 2 u 2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)
The initial and the final masses in the tank are
V
V
3 ft 3
3 ft 3
m1 =
=
=
6
.
28
lbm
m
=
=
= 220.55 lbm
2
v1 0.4776 ft 3/lbm
v 2 0.01360 ft 3/lbm
Then from the mass balance,
mi = m 2 − m1 = 220.55 − 6.28 = 214.3 lbm
(b) The heat transfer during this process is determined from the energy balance to be
Qin = −mi hi + m 2 u 2 − m1u1
= −(214.3 lbm )(38.17 Btu/lbm) + (220.55 lbm )(41.49 Btu/lbm) − (6.28 lbm )(104.99 Btu/lbm) = 312 Btu
(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
Q
S in − S out + S gen = ∆S system 
→ in + mi s i + S gen = ∆S tank = m 2 s 2 − m1 s1
1424
3
{
1
424
3
Tb,in
Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
Therefore, the total entropy generated during this process is
Q
S gen = −mi s i + (m 2 s 2 − m1 s1 ) − in
Tb,in
= −(214.3)(0.07934 ) + (220.55)(0.08589 ) − (6.28)(0.2198) −
312 Btu
= 0.0169 Btu/R
570 R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-143
7-193 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 T1. 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
→ mc =
Q = mc(T2 − T1 ) 
and
∆S = mc ln
Q
T2 − T1
T2
ln (T2 / T1 )
=Q
T1
T2 − T1
Q
m, c, T
Taking the limit as T2 → T1 by applying the L'Hospital's rule,
∆S = Q
Thermal energy
reservoir
1 / T1 Q
=
1
T1
which is the desired result.
7-194 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
Q&
110 W
= 0.378 W/K
S& left = left =
291 K
Tleft
Q& right 110 W
=
= 0.394 W/K
S& right =
Tright
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
1424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
Air
Q·
= ∆S& system ©0 = 0
14243
Rate of change
of entropy
Q&
Q& in
− 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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
6°C
7-144
7-195 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 −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = m(u 2 − u1 )
(since W = KE = PE = 0)
Qout = m(u1 − u 2 )
Using data from the steam tables (Tables A-4 through A-6), some
properties are determined to be
10°C
4m×4m×5m
Steam
radiator
3
P1 = 200 kPa  v 1 = 1.0805 m /kg
 u1 = 2654.6 kJ/kg
T1 = 200°C 
s1 = 7.5081 kJ/kg.K
v = 0.001043, v g = 1.6941 m 3 /kg
P2 = 100 kPa  f
u = 417.40, u fg = 2088.2 kJ/kg
(v 2 = v 1 )  f
s f = 1.3028,
s fg = 6.0562 kJ/kg.K
x2 =
v 2 −v f
v fg
=
1.0805 − 0.001043
= 0.6376
1.6941 − 0.001043
u 2 = u f + x 2 u fg = 417.40 + 0.6376 × 2088.2 = 1748.7 kJ/kg
s 2 = s f + x 2 s fg = 1.3028 + 0.6376 × 6.0562 = 5.1642 kJ/kg.K
m=
V1
0.015 m 3
=
= 0.01388 kg
v 1 1.0805 m 3 /kg
Substituting,
Qout = (0.01388 kg)( 2654.6 - 1748.7)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 =
P1V 1
(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 = (0.120 kJ/s)(30 × 60 s) = 216kJ
We now take the air in the room as the system. The energy balance for this closed system is expressed as
E in − E out = ∆E system
Qin + Wfan,in − W b,out = ∆U
Qin + Wfan,in = ∆H ≅ mc p (T2 − T1 )
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-145
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 min.
(b) The entropy change of the steam is
∆S steam = m(s 2 − s1 ) = (0.01388 kg )(5.1642 − 7.5081)kJ/kg ⋅ K = −0.0325 kJ/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
= (98.5 kg )(1.005 kJ/kg ⋅ K ) ln
285.3 K
= 0.8013 kJ/K
283 K
(d) 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
1424
3
Net entropy transfer
by heat and mass
+ S gen = ∆S system
{
1
424
3
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 = −0.0325 + 0.8013 = 0.7688 kJ/K
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7-146
7-196 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 ρ = 1 kg/L and c = 4.18
kJ/kg·°C (Table A-3).
Analysis The total mass of water is
m w = ρV = (1 kg/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 −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
50,000 kJ/h
Change in internal, kinetic,
potential, etc. energies
We,in − Qout = ∆U = (∆U ) water + (∆U ) air
= (∆U ) water
22°C
= mc(T2 − T1 ) water
or,
W& e,in ∆t − Qout = [mc(T2 − T1 )] water
water
80°C
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 − S out
1in424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
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
Qout 
Q
T 
+ out
=  mc ln 2 
Tb,out 
T1  water Tsurr
295 K 500,000 kJ
= (1000 kg )(4.18 kJ/kg ⋅ K )ln
+
353 K
276 K
S gen = ∆S water +
= −750 + 1811 = 1061 kJ/K
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7-147
7-197E 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. The density of water at room temperature is 62.1 lbm/ft³ (Table A-3E).
and
Analysis The mass of the water is
m water = ρVV = (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 −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = ∆U container + ∆U water
Steel
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⋅ o 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 − S out
1in424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
Qout
+ S gen = ∆S container + ∆S water
Tb,out
where
∆S container = mc avg ln
∆S water = mc avg 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
580 R
T1
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
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7-148
7-198 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 + x1 h fg = 22.49 + 0.3 × 214.48 = 86.83 kJ/kg

x1 = 0.3
 s1 = s f + x1 s fg = 0.09275 + 0.3(0.85503) = 0.3493 kJ/kg ⋅ K
T2 = 120 kPa  h2 = hg @ 120

sat. vapor
 s2 = hg @ 120
kPa
= 236.97 kJ/kg
kPa
= 0.9478 kJ/kg ⋅ K
Analysis (a) The mass flow rate of air is
m& air =
P V&
3 3
=
RT3
(100 kPa )(6 m /min )
(0.287 kPa ⋅ m3/kg ⋅ K )(300 K ) = 6.97 kg/min
3
6 m3/min
R-134a
AIR
3
1
2 kg/min
We take the entire heat exchanger as the system, which is a
control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as
2
4
sat. vapor
Mass balance ( for each fluid stream):
m& 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
Energy balance (for the entire heat exchanger):
E& − E& out
1in424
3
∆E& system©0 (steady)
1442443
=
Rate of net energy transfer
by heat, work, and mass
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1h1 + m& 3h3 = m& 2 h2 + m& 4 h4 (since Q& = W& = ∆ke ≅ ∆pe ≅ 0)
Combining the two,
m& R (h2 − h1 ) = m& air (h3 − h4 ) = m& air c p (T3 − T4 )
m& R (h2 − h1 )
m& air c p
Solving for T4,
T4 = T3 −
Substituting,
T4 = 27°C −
(2 kg/min)(236.97 − 86.83) kJ/kg
= −15.9°C = 257.1 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& − S&out
1in
424
3
Rate of net entropy transfer
by heat and mass
+
S&gen
{
Rate of entropy
generation
= ∆S&system©0 (steady)
1442443
Rate of change
of entropy
m& 1s1 + m& 3 s3 − m& 2 s2 − m& 4 s4 + S&gen = 0 (since Q = 0)
m& R s1 + m& air s3 − m& R s2 − m& air s4 + S&gen = 0
or,
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7-149
S&gen = m& R (s2 − s1 ) + m& air (s4 − s3 )
where
0
s 4 − s 3 = c p ln
T4
P©
T
257.1 K
− R ln 4 = c p ln 4 = (1.005 kJ/kg ⋅ K) ln
= −0.1551 kJ/kg ⋅ K
T3
P3
T3
300 K
Substituting,
S&gen = (2 kg/min )(0.9478 - 0.3493 kJ/kg ⋅ K ) + (6.97 kg/min)(−0.1551 kJ/kg ⋅ K)
= 0.116 kJ/min ⋅ K
= 0.00193 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 )
Q& in − m& R (h2 − h1 )
m& air c p
Solving for T4,
T4 = T3 +
Substituting,
T4 = 27°C +
(30 kJ/min) − (2 kg/min)(236.97 − 86.83) kJ/kg
= −11.6°C = 261.4 K
(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
S&in − S&out
1424
3
+
Rate of net entropy transfer
by heat and mass
S&gen
{
Rate of entropy
generation
= ∆S&system©0 (steady)
1442443
Rate of change
of entropy
Qin
+ m& 1s1 + m& 3 s3 − m& 2 s2 − m& 4 s4 + S&gen = 0
Tb,out
Qin
+ m& R s1 + m& air s3 − m& R s2 − m& air s4 + S&gen = 0
Tsurr
or
Q&
S&gen = m& R (s2 − s1 ) + m& air (s4 − s3 ) − in
T0
where
s 4 − s 3 = c p ln
0
T4
PÊ
261.4 K
− R ln 4 = (1.005 kJ/kg ⋅ K) ln
= −0.1384 kJ/kg ⋅ K
T3
P3
300 K
Substituting,
30 kJ/min
S&gen = (2 kg/min )(0.9478 − 0.3493) kJ/kg ⋅ K + (6.97 kg/min )(− 0.1384 kJ/kg ⋅ K ) −
305 K
= 0.1340 kJ/min ⋅ K
= 0.00223 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.
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7-150
7-199 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 − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
→ − Qout = ∆U = (∆U )water + (∆U )air©0
Change in internal, kinetic,
potential, etc. energies
10,000 kJ/h
or
-Qout = [mc(T2 - T1)]water
20°C
Substituting,
-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
Sin − Sout
1424
3
Net entropy transfer
by heat and mass
−
+ Sgen = ∆Ssystem
{
1
424
3
Entropy
generation
Change
in entropy
Qout
+ Sgen = ∆S water + ∆Sair©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
Qout 
T 
Q
=  mc ln 2 
+ out
Tb,out 
T1  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
Sgen = ∆S water +
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7-151
7-200 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 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
( )
(
)
(500 kPa )(1 m )
=
(2.0769 kPa ⋅ m /kg ⋅ K )(298 K ) = 0.808 kg
 PV 
(500 kPa ) 1 m3
mN 2 =  1 1  =
= 4.77 kg
0.2968 kPa ⋅ m3/kg ⋅ K (353 K )
 RT1  N 2
 PV 
mHe =  1 1 
 RT1  He
N2
1 m3
500 kPa
80°C
3
He
1 m3
500 kPa
25°C
3
Taking the entire contents of the cylinder as our system, the 1st law relation can be written as
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = (∆U )N 2 + (∆U )He 
→ 0 = [mcv (T2 − T1 )]N 2 + [mcv (T2 − T1 )]He
Substituting,
(4.77 kg ) 0.743 kJ/kg⋅o C T f − 80 °C + (0.808 kg )(3.1156 kJ/kg ⋅ °C) T f − 25 °C = 0
(
)(
)
(
)
It gives 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
Sin − Sout
1424
3
Net entropy transfer
by heat and mass
+ Sgen = ∆Ssystem
{
1
424
3
Entropy
generation
Change
in entropy
0 + Sgen = ∆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 4 kg/kmol
P2 =
N total RuT
V total
=
(0.372 kmol)(8.314 kPa ⋅ m3/kmol ⋅ K )(330.2 K ) = 510.6 kPa
2 m3
Then,

T
P 
∆S N 2 = m c p ln 2 − R ln 2 
T1
P1  N

2

330.2 K
510.6 kPa 
= (4.77 kg )(1.039 kJ/kg ⋅ K )ln
− (0.2968 kJ/kg ⋅ K )ln

353 K
500 kPa 

= −0.361 kJ/K
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7-152

P 
T
∆S He = m c p ln 2 − R ln 2 
P1  He
T1


330.2 K
510.6 kPa 
= (0.808 kg )(5.1926 kJ/kg ⋅ K )ln
− (2.0769 kJ/kg ⋅ K )ln

298 K
500 kPa 

= 0.395 kJ/K
Sgen = ∆S N 2 + ∆S He = −0.361 + 0.395 = 0.034 kJ/K
If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas would
remain constant in this case:

330.2 K
T
V ©0 
= −0.237 kJ/K
∆S N 2 = m cv ln 2 − R ln 2  = (4.77 kg )(0.743 kJ/kg ⋅ K )ln


353 K
T1
V1 

N2

330.2 K
V ©0 
T
= 0.258 kJ/K
∆S He = m cv ln 2 − R ln 2  = (0.808 kg )(3.1156 kJ/kg ⋅ K )ln


298 K
V1 
T1

He
Sgen = ∆S N 2 + ∆S He = −0.237 + 0.258 = 0.021 kJ/K
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7-153
7-201 EES Problem 7-200 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.
Analysis The problem is solved using EES, and the results are given below.
"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
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
"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])
"Total volume of gases:"
V_total=V_N2[1]+V_He[1]
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)
S_gen_PistonMoving = DELTAS_He_PM+DELTAS_N2_PM
DELTAS_He_PM=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1]))
DELTAS_N2_PM=m_N2*(Cp_N2*ln((T_2+273)/(T_N2[1]+273))-R_N2*ln(P_2/P_N2[1]))
"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 are given by:"
S_gen_PistNotMoving = DELTAS_He_PNM+DELTAS_N2_PNM
DELTAS_He_PNM=m_He*(Cv_He*ln((T_2+273)/(T_He[1]+273)))
DELTAS_N2_PNM=m_N2*(Cv_N2*ln((T_2+273)/(T_N2[1]+273)))
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-154
"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])
"Final Pressure for moving piston:"
P_2_VP*V_total=N_total*R_u*(T_2_VP+273)
S_gen_PistMoving_VP = DELTAS_He_PM_VP+DELTAS_N2_PM_VP
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]))
DELTAS_He_PM_VP=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1]))
"Fianl 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
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]))
DELTAS_He_PNM_VP=m_He*(Cv_He*ln((T_2_VP+273)/(T_He[1]+273)))
SOLUTION
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]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-155
7-202 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 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
( )
(
)
(500 kPa )(1m )
=
(2.0769 kPa ⋅ m /kg ⋅ K )(298 K ) = 0.808 kg
 PV 
(500 kPa ) 1 m3
mN 2 =  1 1  =
= 4.77 kg
0.2968 kPa ⋅ m3/kg ⋅ K (353 K )
 RT1  N 2
 PV 
mHe =  1 1 
 RT1  He
3
N2
1 m3
500 kPa
80°C
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 − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = (∆U )N 2 + (∆U )He + (∆U )Cu
0 = [mcv (T2 − T1 )]N 2 + [mcv (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
)
(
)
+ (5.0 kg )(0.386 kJ/kg ⋅ °C )(T f − 52.5)°C = 0
− 80 °C + (0.808 kg )(3.1156 kJ/kg ⋅ °C ) T f − 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
Sin − Sout
1424
3
Net entropy transfer
by heat and mass
+ Sgen = ∆Ssystem
{
1
424
3
Entropy
generation
Change
in entropy
0 + Sgen = ∆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  N 2  M  He 28 kg/kmol 4 kg/kmol
P2 =
N total RuT
V total
=
(0.372 kmol)(8.314 kPa ⋅ m3/kmol ⋅ K )(329 K ) = 508.8 kPa
2 m3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-156
Then,

T
P 
∆S N 2 = m c p ln 2 − R ln 2 
P1  N
T1

2
∆S He
∆S piston

329 K
508.8 kPa 
= (4.77 kg )(1.039 kJ/kg ⋅ K )ln
− (0.2968 kJ/kg ⋅ K )ln

353
K
500 kPa 

= −0.374 kJ/K

T
P 
= 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

298 K
500 kPa 

= 0.386 kJ/K

329 K
T 
=  mc ln 2 
= (5 kg )(0.386 kJ/kg ⋅ K )ln
= 0.021 kJ/K
325.5
K
T
1  piston

Sgen = ∆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:

329 K
V ©0 
T
= −0.250 kJ/K
∆S N 2 = m cv ln 2 − R ln 2  = (4.77 kg )(0.743 kJ/kg ⋅ K ) ln


353 K
V
T
1
1
N2


329 K
V ©0 
T
= 0.249 kJ/K
∆S He = m cv ln 2 − R ln 2  = (0.808 kg )(3.1156 kJ/kg ⋅ K ) ln


298 K
V1 
T1

He
Sgen = ∆S N 2 + ∆S He + ∆S piston = −0.250 + 0.249 + 0.021 = 0.020 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-157
7-203 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 is R = 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17)
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
Mass balance:
min − mout = ∆msystem → me = m1 − m2
E − Eout
1in
424
3
Energy balance:
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
We,in − me he = m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0)
The initial and the final masses of air in the tank are
(500 kPa ) 5 m3
PV
= 26.40 kg
m1 = 1 =
RT1
0.287 kPa ⋅ m3/kg ⋅ K (330 K )
(
PV
m2 = 2
RT2
)
(
)
(200 kPa )(5 m )
=
(0.287 kPa ⋅ m /kg ⋅ K )(330 K ) = 10.56 kg
3
3
Then from the mass and energy balances,
me = m1 − m2 = 26.40 − 10.56 = 15.84 kg
We,in = me he + m2u2 − m1u1
AIR
5 m3
500 kPa
57°C
We
= (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
Sin − Sout + Sgen = ∆Ssystem
1424
3
{
1
424
3
Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
− me se + Sgen = ∆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
T ©0
P
P
200 kPa
s 2 − s e = c p ln 2
− R ln 2 = − R ln 2 = −(0.287 kJ/kg ⋅ K )ln
= 0.1606 kJ/kg ⋅ K
Te
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
Sgen = (10.56 kg )(0.1606 kJ/kg ⋅ K ) − (26.40 kg )(− 0.1024 kJ/kg ⋅ K ) = 4.40 kJ/K
s1 − s e = c p ln
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-158
7-204 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
1in424
3
=
Net energy transfer
by heat, work, and mass
ice
-5°C
80 kg
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U
0 = ∆U ice + ∆U water
WATER
1 ton
[mc(0°C − T1 ) solid + mhif + mc(T2 − 0°C) liquid ] ice + [mc(T2 − T1 )] water = 0
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 T2 = 12.42°C
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
Sin − Sout
1424
3
Net entropy transfer
by heat and mass
+ Sgen = ∆Ssystem
{
1
424
3
Entropy
generation
Change
in entropy
0 + Sgen = ∆Sice + ∆S water
where

285.42 K
T 
= (1000 kg )(4.18 kJ/kg ⋅ K ) ln
= −109.6 kJ/K
∆S water =  mc ln 2 
293 K
T1  water

∆Sice = ∆Ssolid + ∆S melting + ∆Sliquid ice
(
)


Tmelting 
mhif

T 


 mc ln 2 
+
+
=   mc ln





T
T
T
1
melting
1  liquid 
solid


ice

273 K 333.7 kJ/kg
285.42 K 

= (80 kg ) (2.11 kJ/kg ⋅ K )ln
+
+ (4.18 kJ/kg ⋅ K ) ln
268 K
273 K
273 K 

= 115.8 kJ/K
Then,
Sgen = ∆S water + ∆Sice = −109.6 + 115.8 = 6.2 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-159
7-205 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
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3

→Wb,in = ∆U → ∆H = 0 
→ ∆H ice + ∆H water = 0
Change in internal, kinetic,
potential, etc. energies
[mc(0°C − T1 ) solid + mhif + mc(T2 − 0°C) liquid ] ice + [m(h2 − h1 )] water = 0
or
The properties of water at 100°C are (Table A-4)
v f = 0.001043, v g = 1.6720 m 3 /kg
h f = 419.17, h fg = 2256.4 kJ.kg
s f = 1.3072
ice
-18°C
s fg = 6.0490 kJ/kg.K
v1 = v f + x1v fg = 0.001043 + (0.1)(1.6720 − 0.001043) = 0.16814 m3/kg
WATER
0.02 m3
100°C
h1 = h f + x1h fg = 419.17 + (0.1)(2256.4) = 644.81 kJ/kg
s1 = s f + x1s fg = 1.3072 + (0.1)(6.0470 ) = 1.9119 kJ/kg ⋅ K
h2 = h f @100o C = 419.17 kJ/kg
s2 = s f @100o C = 1.3072 kJ/kg ⋅ K
msteam =
V1
0.02 m3
=
= 0.119 kg
v1 0.16814 m3/kg
Noting that T1, ice = -18°C and T2 = 100°C and substituting gives
m{(2.11 kJ/kg.K)[0-(-18)] + 333.7 kJ/kg + (4.18 kJ/kg·°C)(100-0)°C}
+(0.119 kg)(419.17 – 644.81) kJ/kg = 0
m = 0.034 kg = 34.0 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
Sin − Sout + Sgen = ∆Ssystem
1424
3
{
1
424
3
Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
0 + Sgen = ∆Sice + ∆Ssteam
∆Ssteam = m(s2 − s1 ) = (0.119 kg )(1.3072 − 1.9119 )kJ/kg ⋅ K = −0.0719 kJ/K
(
∆Sice = ∆Ssolid + ∆S melting + ∆Sliquid



)ice =   mc ln Tmelting
T 

1
solid
+
mhif
Tmelting


T 

+  mc ln 2 

T
1  liquid 

ice

273.15 K 333.7 kJ/kg
373.15 K 
 = 0.0907 kJ/K
= (0.034 kg ) (2.11 kJ/kg ⋅ K )ln
+
+ (4.18 kJ/kg ⋅ K )ln
255.15 K
273.15 K
273.15 K 

Then,
Sgen = ∆Ssteam + ∆Sice = −0.0719 + 0.0907 = 0.0188 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-160
7-206 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
E − Eout
1in424
3
Energy balance:
=
Net energy transfer
by heat, work, and mass
(since mout = minitial = 0)
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin + mi hi = m2u2 (since W ≅ Eout = Einitial = ke ≅ pe ≅ 0)
Combining the two balances:
Qin = m2 (u2 − hi )
10 kPa
17°C
where
m2 =
(
)
P2V
(100 kPa ) 0.005 m 3
=
= 0.0060 kg
RT2
0.287 kPa ⋅ m 3 /kg ⋅ K (290 K )
(
-17
Ti = T2 = 290 K Table
 A
→
)
5L
Evacuated
hi = 290.16 kJ/kg
u 2 = 206.91 kJ/kg
Substituting,
Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ
→
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
1424
3
Net entropy transfer
by heat and mass
mi si −
+ Sgen = ∆Ssystem
{
1
424
3
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
Sgen = −mi si + m2 s2 +
0.5 kJ
Qout
Q
Q
= m2 (s2 − si )©0 + out = out =
= 0.0017 kJ/K
Tb,out
Tb, out Tsurr 290 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-161
7-207 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 ∆E CV = 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 ρ = 1 kg/L. The specific heat of water at room temperature is
c = 4.18 kJ/kg·°C (Table A-3).
Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the system
boundary during the process. We observe that there is only one inlet and one exit and thus m& 1 = m& 2 = m& .
Then the energy balance for this steady-flow system can be expressed in the rate form as
E& − E& out
=
∆E& systemÊ0 (steady)
= 0 → E& in = E& out
1in
424
3
1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
W&e,in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0)
W&e,in = m& (h2 − h1 ) = m& c(T2 − T1 )
where
Substituting,
WATER
16°C
43°C
m& = ρV& = (1 kg/L)(10 L/min) = 10 kg/min
W&
= (10/60 kg/s )(4.18 kJ/kg ⋅ °C )(43 − 16 )°C = 18.8 kW
e,in
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& − S&out
+
S&gen
= ∆S&system©0 = 0
1in
424
3
{
14243
Rate of net entropy transfer
by heat and mass
Rate of entropy
generation
Rate of change
of entropy
m& s1 − m& s2 + S&gen = 0 
→ S&gen = m& ( s2 − s1 )
Noting that water is an incompressible substance and substituting,
316 K
T
= 0.0622 kJ/K
S&gen = m& c ln 2 = (10/60 kg/s )(4.18 kJ/kg ⋅ K )ln
289 K
T1
(b) The energy recovered by the heat exchanger is
Q& saved = εQ& max = εm& 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 from
Q& = m& c(Tc, out − Tc,in )
Substituting,
8 kJ/s = (10/60 kg/s )( 4.18 kJ/kg ⋅o C)(Tc, out − 16o C)
It yields
Tc, out = 27.5o C = 300.5 K
The rate of entropy generation in the heating section in this case is determined similarly to be
T
316 K
= 0.0350 kJ/K
S&gen = m& cln 2 = (10/60 kg/s )(4.18 kJ/kg ⋅ K ) ln
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
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7-162
7-208 EES 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.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
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
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)
Nstage
1
2
3
4
5
6
7
8
9
10
Wcomp [kJ/kg]
229.4
198.7
189.6
185.3
182.8
181.1
179.9
179
178.4
177.8
230
220
W comp [kJ/kg]
210
200
190
180
170
1
2
3
4
5
6
7
8
9
10
Nstage
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7-163
7-209 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed of
three processes. The work and heat transfer for each process are to be determined.
Assumptions 1 All processes are reversible. 2 Kinetic
and potential energy changes are negligible. 3 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).
P = const.
1
3
s = const.
Analysis Using variable specific heats, the properties
can be determined using the air table as follows
u1 = u2 = 214.07 kJ/kg
0
0
s
T1 = T2 = 300 K 
→ 1 = s2 = 1.70203 kJ/kg.K
Pr1 = Pr 2 = 1.3860
Pr 3 =
T = const.
2
u = 283.71 kJ/kg
P3
400 kPa
(1.3860) = 3.696 → 3
Pr 2 =
T3 = 396.6 K
P2
150 kPa
The mass of the air and the volumes at the various states are
m=
P1V1
(400 kPa)(0.3 m 3 )
=
= 1.394 kg
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(300 K)
V2 =
mRT2 (1.394 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(300 K)
=
= 0.8 m 3
P2
150 kPa
V3 =
mRT3 (1.394 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(396.6 K)
=
= 0.3967 m 3
P3
400 kPa
Process 1-2: Isothermal expansion (T2 = T1)
∆S1− 2 = −mR ln
P2
150 kPa
= (1.394 kg)(0.287 kJ/kg.K)ln
= 0.3924 kJ/kg.K
400 kPa
P1
Qin,1− 2 = T1 ∆S1− 2 = (300 K)(0.3924 kJ/K) = 117.7 kJ
Wout,1− 2 = Qin,1− 2 = 117.7 kJ
Process 2-3: Isentropic (reversible-adiabatic) compression (s2 = s1)
Win,2 −3 = m(u3 − u2 ) = (1.394 kg)(283.71 - 214.07) kJ/kg = 97.1 kJ
Q2-3 = 0 kJ
Process 3-1: Constant pressure compression process (P1 = P3)
Win,3−1 = P3 (V3 − V1 ) = (400 kg)(0.3924 - 0.3) kJ/kg = 37.0 kJ
Qout,3−1 = Win,3−1 − m(u1 − u3 ) = 37.0 kJ - (1.394 kg)(214.07 - 283.71) kJ/kg = 135.8 kJ
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7-164
7-210 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust
gases and a compressor driven by the turbine is considered. The air temperature at the compressor exit and
the isentropic efficiency of the compressor are to be determined.
Assumptions 1 Steady operating conditions exist. 2
Kinetic and potential energy changes are negligible. 3
Exhaust gases have air properties and air is an ideal gas
with constant specific heats.
Properties The specific heat of exhaust gases at the
average temperature of 425ºC is cp = 1.075 kJ/kg.K and
properties of air at an anticipated average temperature of
100ºC are cp = 1.011 kJ/kg.K and k =1.397 (Table A-2).
Analysis (a) The turbine power output is determined from
W& = m& c (T − T )
T
exh
p
1
2
400°C
Turbine
Exh. gas
450°C
0.02 kg/s
Air, 70°C
95 kPa
0.018 kg/s
Compressor
135 kPa
= (0.02 kg/s)(1.075 kJ/kg.°C)(450 - 400)°C = 1.075 kW
For a mechanical efficiency of 95% between the turbine and the compressor,
W& C = η mW& T = (0.95)(1.075 kW) = 1.021 kW
Then, the air temperature at the compressor exit becomes
W& = m& c (T − T )
C
air p
2
1
1.021 kW = (0.018 kg/s)(1.011 kJ/kg.°C)(T2 - 70)°C
T2 = 126.1°C
(b) The air temperature at the compressor exit for the case of isentropic process is
T2 s
P
= T1  2
 P1



( k −1) / k
 135 kPa 
= (70 + 273 K)

 95 kPa 
(1.397 -1)/1.397
= 379 K = 106°C
The isentropic efficiency of the compressor is determined to be
ηC =
T2 s − T1
106 − 70
= 0.642
=
T2 − T1 126.1 − 70
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7-165
7-211 Air is compressed in a compressor that is intentionally cooled. The work input, the isothermal
efficiency, and the entropy generation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and
potential energy changes are negligible. 3 Air is an ideal gas with
constant specific heats.
Q
Properties The gas constant of air is R = 0.287 kJ/kg.K and the
specific heat of air at an average temperature of (20+300)/2 = 160ºC
= 433 K is cp = 1.018 kJ/kg.K (Table A-2).
Compressor
Analysis (a) The power input is determined from an energy balance
on the control volume
W& = m& c (T − T ) + Q&
C
p
2
1
out
300°C
1.2 MPa
Air
20°C, 100 kPa
= (0.4 kg/s)(1.018 kJ/kg.°C)(300 − 20)°C + 15 kW
= 129.0 kW
(b) The power input for a reversible-isothermal process is given by
P
 1200 kPa 
W& T =const. = m& RT1 ln 2 = (0.4 kg/s)(0.287 kJ/kg.K)(20 + 273 K)ln
 = 83.6 kW
P1
 100 kPa 
Then, the isothermal efficiency of the compressor becomes
W&
83.6 kW
η T = T =const. =
= 0.648
&
129.0 kW
WC
(c) The rate of entropy generation associated with this process may be obtained by adding the rate of
entropy change of air as it flows in the compressor and the rate of entropy change of the surroundings
T
P Q&
S& gen = ∆S& air + ∆S& surr = c p ln 2 − R ln 2 + out
T1
P1 Tsurr
= (1.018 kJ/kg.K)ln
300 + 273 K
1200 kPa
15 kW
− (0.287 kJ/kg.K)ln
+
20 + 273 K
100 kPa (20 + 273) K
= 0.0390 kW/K
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7-166
7-212 Air is allowed to enter an insulated piston-cylinder device until the volume of the air increases by
50%. The final temperature in the cylinder, the amount of mass that has entered, the work done, and the
entropy generation are to be determined.
Assumptions 1 Kinetic and potential energy changes
are negligible. 2 Air is an ideal gas with constant
specific heats.
Air
0.25 m3
0.7 kg
20°C
Properties The gas constant of air is R = 0.287
kJ/kg.K and the specific heats of air at room
temperature are cp = 1.005 kJ/kg.K, cv = 0.718
kJ/kg.K (Table A-2).
Analysis The initial pressure in the cylinder is
P1 =
m2 =
m1 RT1
V1
=
(0.7 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K)
0.25 m 3
Air
500 kPa
70°C
= 235.5 kPa
P2V 2 (235.5 kPa)(1.5 × 0.25 m 3 ) 307.71
=
=
RT2
T2
(0.287 kPa ⋅ m 3 /kg ⋅ K)T2
A mass balance on the system gives the expression for the mass entering the cylinder
mi = m 2 − m1 =
307.71
− 0.7
T2
(c) Noting that the pressure remains constant, the boundary work is determined to be
W b,out = P1 (V 2 −V1 ) = (235.5 kPa)(1.5 × 0.25 - 0.5)m 3 = 29.43 kJ
(a) An energy balance on the system may be used to determine the final temperature
mi hi − Wb, out = m2u2 − m1u1
mi c pTi − Wb, out = m2cv T2 − m1cv T1
 307.71 

 307.71
(0.718)T2 − (0.7)(0.718)(20 + 273)

− 0.7 (1.005)(70 + 273) − 29.43 = 

 T
2
 T2 


There is only one unknown, which is the final temperature. By a trial-error approach or using EES, we find
T2 = 308.0 K
(b) The final mass and the amount of mass that has entered are
m2 =
307.71
= 0.999 kg
308.0
mi = m 2 − m1 = 0.999 − 0.7 = 0.299 kg
(d) The rate of entropy generation is determined from
S gen = m 2 s 2 − m1 s1 − mi s i = m 2 s 2 − m1 s1 − (m 2 − m1 ) s i = m 2 ( s 2 − s i ) − m1 ( s1 − s i )


T
P 
T
P 
= m 2  c p ln 2 − R ln 2  − m1  c p ln 1 − R ln 1 
Ti
Pi 
Ti
Pi 



 308 K 
 235.5 kPa 
= (0.999 kg) (1.005 kJ/kg.K)ln
 − (0.287 kJ/kg.K)ln

 343 K 
 500 kPa 


 293 K 
 235.5 kPa 
− (0.7 kg) (1.005 kJ/kg.K)ln
 − (0.287 kJ/kg.K)ln

 343 K 
 500 kPa 

= 0.0673 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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7-167
7-213 A cryogenic turbine in a natural gas liquefaction plant produces 350 kW of power. The efficiency of
the turbine is to be determined.
Assumptions 1 The turbine operates steadily. 2 The properties of methane is used for natural gas.
Properties The density of natural gas is given to be 423.8 kg/m3.
Analysis The maximum possible power that can be obtained from this
turbine for the given inlet and exit pressures can be determined from
m&
W&max = ( Pin − Pout ) =
ρ
(55 kg/s)
423.8 kg/m3
(4000 − 300)kPa = 480.2 kW
Given the actual power, the efficiency of this cryogenic turbine becomes
350 kW
W&
η= &
=
= 0.729 = 72.9%
480.2 kW
W
max
3 bar
Cryogenic
turbine
LNG, 40 bar
-160°C, 55 kg/s
This efficiency is also known as hydraulic efficiency since the cryogenic
turbine handles natural gas in liquid state as the hydraulic turbine handles
liquid water.
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7-168
Fundamentals of Engineering (FE) Exam Problems
7-214 Steam is condensed at a constant temperature of 30°C as it flows through the condenser of a power
plant by rejecting heat at a rate of 55 MW. The rate of entropy change of steam as it flows through the
condenser is
(a) –1.83 MW/K
(b) –0.18 MW/K
(c) 0 MW/K
(d) 0.56 MW/K
(e) 1.22 MW/K
Answer (b) –0.18 MW/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
T1=30 "C"
Q_out=55 "MW"
S_change=-Q_out/(T1+273) "MW/K"
"Some Wrong Solutions with Common Mistakes:"
W1_S_change=0 "Assuming no change"
W2_S_change=Q_out/T1 "Using temperature in C"
W3_S_change=Q_out/(T1+273) "Wrong sign"
W4_S_change=-s_fg "Taking entropy of vaporization"
s_fg=(ENTROPY(Steam_IAPWS,T=T1,x=1)-ENTROPY(Steam_IAPWS,T=T1,x=0))
7-215 Steam is compressed from 6 MPa and 300°C to 10 MPa isentropically. The final temperature of the
steam is
(a) 290°C
(b) 300°C
(c) 311°C
(d) 371°C
(e) 422°C
Answer (d) 371°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=6000 "kPa"
T1=300 "C"
P2=10000 "kPa"
s2=s1
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
T2=TEMPERATURE(Steam_IAPWS,s=s2,P=P2)
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P2"
W3_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P1"
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7-169
7-216 An apple with an average mass of 0.15 kg and average specific heat of 3.65 kJ/kg.°C is cooled from
20°C to 5°C. The entropy change of the apple is
(a) –0.0288 kJ/K
(b) –0.192 kJ/K
(c) -0.526 kJ/K
(d) 0 kJ/K
(e) 0.657 kJ/K
Answer (a) –0.0288 kJ/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
C=3.65 "kJ/kg.K"
m=0.15 "kg"
T1=20 "C"
T2=5 "C"
S_change=m*C*ln((T2+273)/(T1+273))
"Some Wrong Solutions with Common Mistakes:"
W1_S_change=C*ln((T2+273)/(T1+273)) "Not using mass"
W2_S_change=m*C*ln(T2/T1) "Using C"
W3_S_change=m*C*(T2-T1) "Using Wrong relation"
7-217 A piston-cylinder device contains 5 kg of saturated water vapor at 3 MPa. Now heat is rejected from
the cylinder at constant pressure until the water vapor completely condenses so that the cylinder contains
saturated liquid at 3 MPa at the end of the process. The entropy change of the system during this process is
(a) 0 kJ/K
(b) -3.5 kJ/K
(c) -12.5 kJ/K
(d) -17.7 kJ/K
(e) -19.5 kJ/K
Answer (d) -17.7 kJ/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=3000 "kPa"
m=5 "kg"
s_fg=(ENTROPY(Steam_IAPWS,P=P1,x=1)-ENTROPY(Steam_IAPWS,P=P1,x=0))
S_change=-m*s_fg "kJ/K"
7-218 Helium gas is compressed from 1 atm and 25°C to a pressure of 10 atm adiabatically. The lowest
temperature of helium after compression is
(a) 25°C
(b) 63°C
(c) 250°C
(d) 384°C
(e) 476°C
Answer (e) 476°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
k=1.667
P1=101.325 "kPa"
T1=25 "C"
P2=10*101.325 "kPa"
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7-170
"s2=s1"
"The exit temperature will be lowest for isentropic compression,"
T2=(T1+273)*(P2/P1)^((k-1)/k) "K"
T2_C= T2-273 "C"
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=T1*(P2/P1)^((k-1)/k) "Using C instead of K"
W3_T2=(T1+273)*(P2/P1)-273 "Assuming T is proportional to P"
W4_T2=T1*(P2/P1) "Assuming T is proportional to P, using C"
7-219 Steam expands in an adiabatic turbine from 8 MPa and 500°C to 0.1 MPa at a rate of 3 kg/s. If steam
leaves the turbine as saturated vapor, the power output of the turbine is
(a) 2174 kW
(b) 698 kW
(c) 2881 kW
(d) 1674 kW
(e) 3240 kW
Answer (a) 2174 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=8000 "kPa"
T1=500 "C"
P2=100 "kPa"
x2=1
m=3 "kg/s"
h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)
h2=ENTHALPY(Steam_IAPWS,x=x2,P=P2)
W_out=m*(h1-h2)
"Some Wrong Solutions with Common Mistakes:"
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
h2s=ENTHALPY(Steam_IAPWS, s=s1,P=P2)
W1_Wout=m*(h1-h2s) "Assuming isentropic expansion"
7-220 Argon gas expands in an adiabatic turbine from 3 MPa and 750°C to 0.2 MPa at a rate of 5 kg/s. The
maximum power output of the turbine is
(a) 1.06 MW
(b) 1.29 MW
(c) 1.43 MW
(d) 1.76 MW
(e) 2.08 MW
Answer (d) 1.76 MW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=0.5203
k=1.667
P1=3000 "kPa"
T1=750 "C"
m=5 "kg/s"
P2=200 "kPa"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-171
"s2=s1"
T2=(T1+750)*(P2/P1)^((k-1)/k)
W_max=m*Cp*(T1-T2)
"Some Wrong Solutions with Common Mistakes:"
Cv=0.2081"kJ/kg.K"
W1_Wmax=m*Cv*(T1-T2) "Using Cv"
T22=T1*(P2/P1)^((k-1)/k) "Using C instead of K"
W2_Wmax=m*Cp*(T1-T22)
W3_Wmax=Cp*(T1-T2) "Not using mass flow rate"
T24=T1*(P2/P1) "Assuming T is proportional to P, using C"
W4_Wmax=m*Cp*(T1-T24)
7-221 A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining
heat from the surroundings at temperature T in the amount of q. If the entropy of the substance is s1 at state
1, and s2 at state 2, the entropy change of the substance ∆s during this process is
(b) ∆s > s2 – s1
(c) ∆s = s2 – s1
(d) ∆s = s2 – s1 + q/T
(a) ∆s < s2 – s1
(e) ∆s > s2 – s1 + q/T
Answer (c) ∆s = s2 – s1
7-222 A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure
P1 to pressure P2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas
constant of the gas is R, the entropy change of the gas ∆s during this process is
(b) ∆s = R ln(P2/P1)- q/T
(c) ∆s =R ln(P1/P2)
(d) ∆s =R ln(P1/P2)-q/T
(a) ∆s =R ln(P2/P1)
(e) ∆s= 0
Answer (c) ∆s =R ln(P1/P2)
7-223 Air is compressed from room conditions to a specified pressure in a reversible manner by two
compressors: one isothermal and the other adiabatic. If the entropy change of air is ∆sisot during the
reversible isothermal compression, and ∆sadia during the reversible adiabatic compression, the correct
statement regarding entropy change of air per unit mass is
(b) ∆sisot= ∆sadia>0
(c) ∆sadia> 0
(d) ∆sisot < 0
(e) ∆sisot= 0
(a) ∆sisot= ∆sadia=0
Answer (d) ∆sisot < 0
7-224 Helium gas is compressed from 15°C and 5.4 m3/kg to 0.775 m3/kg in a reversible adiabatic manner.
The temperature of helium after compression is
(a) 105°C
(b) 55°C
(c) 1734°C
(d) 1051°C
(e) 778°C
Answer (e) 778°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
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7-172
k=1.667
v1=5.4 "m^3/kg"
T1=15 "C"
v2=0.775 "m^3/kg"
"s2=s1"
"The exit temperature is determined from isentropic compression relation,"
T2=(T1+273)*(v1/v2)^(k-1) "K"
T2_C= T2-273 "C"
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
W2_T2=T1*(v1/v2)^(k-1) "Using C instead of K"
W3_T2=(T1+273)*(v1/v2)-273 "Assuming T is proportional to v"
W4_T2=T1*(v1/v2) "Assuming T is proportional to v, using C"
7-225 Heat is lost through a plane wall steadily at a rate of 600 W. If the inner and outer surface
temperatures of the wall are 20°C and 5°C, respectively, the rate of entropy generation within the wall is
(a) 0.11 W/K
(b) 4.21 W/K
(c) 2.10 W/K
(d) 42.1 W/K
(e) 90.0 W/K
Answer (a) 0.11 W/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Q=600 "W"
T1=20+273 "K"
T2=5+273 "K"
"Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives"
Q/T1-Q/T2+S_gen=0 "W/K"
"Some Wrong Solutions with Common Mistakes:"
Q/(T1+273)-Q/(T2+273)+W1_Sgen=0 "Using C instead of K"
W2_Sgen=Q/((T1+T2)/2) "Using avegage temperature in K"
W3_Sgen=Q/((T1+T2)/2-273) "Using avegage temperature in C"
W4_Sgen=Q/(T1-T2+273) "Using temperature difference in K"
7-226 Air is compressed steadily and adiabatically from 17°C and 90 kPa to 200°C and 400 kPa. Assuming
constant specific heats for air at room temperature, the isentropic efficiency of the compressor is
(a) 0.76
(b) 0.94
(c) 0.86
(d) 0.84
(e) 1.00
Answer (d) 0.84
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=1.005 "kJ/kg.K"
k=1.4
P1=90 "kPa"
T1=17 "C"
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-173
P2=400 "kPa"
T2=200 "C"
T2s=(T1+273)*(P2/P1)^((k-1)/k)-273
Eta_comp=(Cp*(T2s-T1))/(Cp*(T2-T1))
"Some Wrong Solutions with Common Mistakes:"
T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K in finding T2s"
W1_Eta_comp=(Cp*(T2sW1-T1))/(Cp*(T2-T1))
W2_Eta_comp=T2s/T2 "Using wrong definition for isentropic efficiency, and using C"
W3_Eta_comp=(T2s+273)/(T2+273) "Using wrong definition for isentropic efficiency, with K"
7-227 Argon gas expands in an adiabatic turbine steadily from 500°C and 800 kPa to 80 kPa at a rate of 2.5
kg/s. For an isentropic efficiency of 80%, the power produced by the turbine is
(a) 194 kW
(b) 291 kW
(c) 484 kW
(d) 363 kW
(e) 605 kW
Answer (c) 484 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=0.5203 "kJ/kg-K"
k=1.667
m=2.5 "kg/s"
T1=500 "C"
P1=800 "kPa"
P2=80 "kPa"
T2s=(T1+273)*(P2/P1)^((k-1)/k)-273
Eta_turb=0.8
Eta_turb=(Cp*(T2-T1))/(Cp*(T2s-T1))
W_out=m*Cp*(T1-T2)
"Some Wrong Solutions with Common Mistakes:"
T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K to find T2s"
Eta_turb=(Cp*(T2W1-T1))/(Cp*(T2sW1-T1))
W1_Wout=m*Cp*(T1-T2W1)
Eta_turb=(Cp*(T2s-T1))/(Cp*(T2W2-T1)) "Using wrong definition for isentropic efficiency, and
using C"
W2_Wout=m*Cp*(T1-T2W2)
W3_Wout=Cp*(T1-T2) "Not using mass flow rate"
Cv=0.3122 "kJ/kg.K"
W4_Wout=m*Cv*(T1-T2) "Using Cv instead of Cp"
7-228 Water enters a pump steadily at 100 kPa at a rate of 35 L/s and leaves at 800 kPa. The flow
velocities at the inlet and the exit are the same, but the pump exit where the discharge pressure is measured
is 6.1 m above the inlet section. The minimum power input to the pump is
(a) 34 kW
(b) 22 kW
(c) 27 kW
(d) 52 kW
(e) 44 kW
Answer (c) 27 kW
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7-174
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
V=0.035 "m^3/s"
g=9.81 "m/s^2"
h=6.1 "m"
P1=100 "kPa"
T1=20 "C"
P2=800 "kPa"
"Pump power input is minimum when compression is reversible and thus w=v(P2-P1)+Dpe"
v1=VOLUME(Steam_IAPWS,T=T1,P=P1)
m=V/v1
W_min=m*v1*(P2-P1)+m*g*h/1000 "kPa.m^3/s=kW"
"(The effect of 6.1 m elevation difference turns out to be small)"
"Some Wrong Solutions with Common Mistakes:"
W1_Win=m*v1*(P2-P1) "Disregarding potential energy"
W2_Win=m*v1*(P2-P1)-m*g*h/1000 "Subtracting potential energy instead of adding"
W3_Win=m*v1*(P2-P1)+m*g*h "Not using the conversion factor 1000 in PE term"
W4_Win=m*v1*(P2+P1)+m*g*h/1000 "Adding pressures instead of subtracting"
7-229 Air at 15°C is compressed steadily and isothermally from 100 kPa to 700 kPa at a rate of 0.12 kg/s.
The minimum power input to the compressor is
(a) 1.0 kW
(b) 11.2 kW
(c) 25.8 kW
(d) 19.3 kW
(e) 161 kW
Answer (d) 19.3 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=1.005 "kJ/kg.K"
R=0.287 "kJ/kg.K"
Cv=0.718 "kJ/kg.K"
k=1.4
P1=100 "kPa"
T=15 "C"
m=0.12 "kg/s"
P2=700 "kPa"
Win=m*R*(T+273)*ln(P2/P1)
"Some Wrong Solutions with Common Mistakes:"
W1_Win=m*R*T*ln(P2/P1) "Using C instead of K"
W2_Win=m*T*(P2-P1) "Using wrong relation"
W3_Win=R*(T+273)*ln(P2/P1) "Not using mass flow rate"
7-230 Air is to be compressed steadily and isentropically from 1 atm to 25 atm by a two-stage compressor.
To minimize the total compression work, the intermediate pressure between the two stages must be
(a) 3 atm
(b) 5 atm
(c) 8 atm
(d) 10 atm
(e) 13 atm
Answer (b) 5 atm
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
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Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=1 "atm"
P2=25 "atm"
P_mid=SQRT(P1*P2)
"Some Wrong Solutions with Common Mistakes:"
W1_P=(P1+P2)/2 "Using average pressure"
W2_P=P1*P2/2 "Half of product"
7-231 Helium gas enters an adiabatic nozzle steadily at 500°C and 600 kPa with a low velocity, and exits at
a pressure of 90 kPa. The highest possible velocity of helium gas at the nozzle exit is
(a) 1475 m/s
(b) 1662 m/s
(c) 1839 m/s
(d) 2066 m/s
(e) 3040 m/s
Answer (d) 2066 m/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
k=1.667
Cp=5.1926 "kJ/kg.K"
Cv=3.1156 "kJ/kg.K"
T1=500 "C"
P1=600 "kPa"
Vel1=0
P2=90 "kPa"
"s2=s1 for maximum exit velocity"
"The exit velocity will be highest for isentropic expansion,"
T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "C"
"Energy balance for this case is h+ke=constant for the fluid stream (Q=W=pe=0)"
(0.5*Vel1^2)/1000+Cp*T1=(0.5*Vel2^2)/1000+Cp*T2
"Some Wrong Solutions with Common Mistakes:"
T2a=T1*(P2/P1)^((k-1)/k) "Using C for temperature"
(0.5*Vel1^2)/1000+Cp*T1=(0.5*W1_Vel2^2)/1000+Cp*T2a
T2b=T1*(P2/P1)^((k-1)/k) "Using Cv"
(0.5*Vel1^2)/1000+Cv*T1=(0.5*W2_Vel2^2)/1000+Cv*T2b
T2c=T1*(P2/P1)^k "Using wrong relation"
(0.5*Vel1^2)/1000+Cp*T1=(0.5*W3_Vel2^2)/1000+Cp*T2c
7-232 Combustion gases with a specific heat ratio of 1.3 enter an adiabatic nozzle steadily at 800°C and
800 kPa with a low velocity, and exit at a pressure of 85 kPa. The lowest possible temperature of
combustion gases at the nozzle exit is
(a) 43°C
(b) 237°C
(c) 367°C
(d) 477°C
(e) 640°C
Answer (c) 367°C
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7-176
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
k=1.3
T1=800 "C"
P1=800 "kPa"
P2=85 "kPa"
"Nozzle exit temperature will be lowest for isentropic operation"
T2=(T1+273)*(P2/P1)^((k-1)/k)-273
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1*(P2/P1)^((k-1)/k) "Using C for temperature"
W2_T2=(T1+273)*(P2/P1)^((k-1)/k) "Not converting the answer to C"
W3_T2=T1*(P2/P1)^k "Using wrong relation"
7-233 Steam enters an adiabatic turbine steadily at 400°C and 3 MPa, and leaves at 50 kPa. The highest
possible percentage of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is
(a) 5%
(b) 10%
(c) 15%
(d) 20%
(e) 0%
Answer (b) 10%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=3000 "kPa"
T1=400 "C"
P2=50 "kPa"
s2=s1
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
x2=QUALITY(Steam_IAPWS,s=s2,P=P2)
misture=1-x2
"Checking x2 using data from table"
x2_table=(6.9212-1.091)/6.5029
7-234 Liquid water enters an adiabatic piping system at 15°C at a rate of 8 kg/s. If the water temperature
rises by 0.2°C during flow due to friction, the rate of entropy generation in the pipe is
(a) 23 W/K
(b) 55 W/K
(c) 68 W/K
(d) 220 W/K
(e) 443 W/K
Answer (a) 23 W/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=4180 "J/kg.K"
m=8 "kg/s"
T1=15 "C"
T2=15.2 "C"
S_gen=m*Cp*ln((T2+273)/(T1+273)) "W/K"
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7-177
"Some Wrong Solutions with Common Mistakes:"
W1_Sgen=m*Cp*ln(T2/T1) "Using deg. C"
W2_Sgen=Cp*ln(T2/T1) "Not using mass flow rate with deg. C"
W3_Sgen=Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with deg. C"
7-235 Liquid water is to be compressed by a pump whose isentropic efficiency is 75 percent from 0.2 MPa
to 5 MPa at a rate of 0.15 m3/min. The required power input to this pump is
(a) 4.8 kW
(b) 6.4 kW
(c) 9.0 kW
(d) 16.0 kW
(e) 12.0 kW
Answer (d) 16.0 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
V=0.15/60 "m^3/s"
rho=1000 "kg/m^3"
v1=1/rho
m=rho*V "kg/s"
P1=200 "kPa"
Eta_pump=0.75
P2=5000 "kPa"
"Reversible pump power input is w =mv(P2-P1) = V(P2-P1)"
W_rev=m*v1*(P2-P1) "kPa.m^3/s=kW"
W_pump=W_rev/Eta_pump
"Some Wrong Solutions with Common Mistakes:"
W1_Wpump=W_rev*Eta_pump "Multiplying by efficiency"
W2_Wpump=W_rev "Disregarding efficiency"
W3_Wpump=m*v1*(P2+P1)/Eta_pump "Adding pressures instead of subtracting"
7-236 Steam enters an adiabatic turbine at 8 MPa and 500°C at a rate of 18 kg/s, and exits at 0.2 MPa and
300°C. The rate of entropy generation in the turbine is
(a) 0 kW/K
(b) 7.2 kW/K
(c) 21 kW/K
(d) 15 kW/K
(e) 17 kW/K
Answer (c) 21 kW/K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
P1=8000 "kPa"
T1=500 "C"
m=18 "kg/s"
P2=200 "kPa"
T2=300 "C"
s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)
s2=ENTROPY(Steam_IAPWS,T=T2,P=P2)
S_gen=m*(s2-s1) "kW/K"
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
7-178
"Some Wrong Solutions with Common Mistakes:"
W1_Sgen=0 "Assuming isentropic expansion"
7-237 Helium gas is compressed steadily from 90 kPa and 25°C to 600 kPa at a rate of 2 kg/min by an
adiabatic compressor. If the compressor consumes 70 kW of power while operating, the isentropic
efficiency of this compressor is
(a) 56.7%
(b) 83.7%
(c) 75.4%
(d) 92.1%
(e) 100.0%
Answer (b) 83.7%
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on
a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical
values).
Cp=5.1926 "kJ/kg-K"
Cv=3.1156 "kJ/kg.K"
k=1.667
m=2/60 "kg/s"
T1=25 "C"
P1=90 "kPa"
P2=600 "kPa"
W_comp=70 "kW"
T2s=(T1+273)*(P2/P1)^((k-1)/k)-273
W_s=m*Cp*(T2s-T1)
Eta_comp=W_s/W_comp
"Some Wrong Solutions with Common Mistakes:"
T2sA=T1*(P2/P1)^((k-1)/k) "Using C instead of K"
W1_Eta_comp=m*Cp*(T2sA-T1)/W_comp
W2_Eta_comp=m*Cv*(T2s-T1)/W_comp "Using Cv instead of Cp"
7-238 … 7-241 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.