Chapter 9

10-1
Chapter 10
VAPOR AND COMBINED POWER CYCLES
Carnot Vapor Cycle
10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture
content allowed is about 10%.
10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat
transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum
temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture
content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.
10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The
thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) We note that
and
η
T
T
H
L
th,C
= T
= T
sat@180 psia
sat@14.7psia
T
= 1−
T
L
H
= 373.1°
F = 833.1 R
= 212.0°
F = 672.0 R
672.0 R
= 1−
833.1 R
= 19.3%
T
1
180 psia
q in
(b) Noting that s 4 = s 1 = s f @ 180 psia = 0.53274 Btu/lbm·R,
14.7 psia
4 3
s4
− s f 0.53274 − 0.31215
x 4 = =
= 0.153
s
s
1.44441
fg
(c) The enthalpies before and after the heat addition process are
Thus,
and,
h = h
h
w
1
2
q
= h
in
net
f @ 180 psia
f
= h
+ x h
2
= η q
th
2
− h
in
1
=
fg
= 346.14 Btu/lbm
= 346.14 +
( 0.90)( 851.16) = 1112.2 Btu/lbm
= 1112.2 − 346.14 = 766.0 Btu/lbm
( 0.1934)( 766.0 Btu/lbm) = 148.1 Btu/lbm
2

10-2
10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The
thermal efficiency, the amount of heat rejected, and the net work output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Noting that T H = 250°C = 523 K and T L = T sat @ 20 kPa = 60.06°C = 333.1 K, the thermal
efficiency becomes
TL
333.1 K
η th,C = 1−
= 1−
= 0. 3632 = 36.3%
T
T 523 K
H
(b) The heat supplied during this cycle is simply the
enthalpy of vaporization ,
Thus,
q
q
in
out
= h
= q
fg@ 250 C
L
o
T
=
T
L
H
= 1715.3 kJ/kg
q
in
⎛ 333.1 K ⎞
=
⎜
⎟
⎝ 523 K ⎠
(c) The net work output of this cycle is
w η
( 1715.3 kJ/kg) = 1092.3 kJ/kg
( 0.3632)( 1715.3 kJ/kg) 623.0 kJ/kg
net = thqin
=
=
250°C
1
4
q in
20 kPa
q out
2
3
s
10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The
thermal efficiency, the amount of heat rejected, and the net work output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Noting that T H = 250°C = 523 K and T L = T sat @ 10 kPa = 45.81°C = 318.8 K, the thermal
efficiency becomes
TL
318.8 K
η th, C = 1 − = 1 − = 39.04%
T
T 523 K
H
(b) The heat supplied during this cycle is simply the
enthalpy of vaporization ,
Thus,
q
q
in = h fg@ 250°
C
out
= q
L
T
=
T
L
H
= 1715.3 kJ/kg
q
in
⎛ 318.8 K ⎞
=
⎜
⎟
⎝ 523 K ⎠
(c) The net work output of this cycle is
( 1715.3 kJ/kg) = 1045.6 kJ/kg
( 0.3904)( 1715.3 ) = 669.7 kJ/kg
w = η q =
kJ/kg
net
th
in
250°C
1
q in
10 kPa
4
q out
2
3
s

10-3
10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The
thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The thermal efficiency is determined from
(b) Note that
TL
60 + 273 K
η th, C = 1− = 1−
= 46.5%
T 350 + 273 K
H
s 2 = s 3 = s f + x 3 s fg
= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K
T
350°C
Thus , 60°C
T
s
2
2
= 350°
C ⎫
⎬ P
= 7.1368 kJ/kg ⋅ K ⎭
2
≅ 1.40 MPa (Table A-6)
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
Thus,
w
s
4
net
= s
f
+ x s
4
= Area =
fg
= 0.8313 +
( 0.1)( 7.0769)
= 1.5390 kJ/kg ⋅ K
( T − T )( s − s ) = ( 350 − 60)( 7.1368 −1.5390) = 1623 kJ/kg
H
L
3
4
1
4
2
3
s
The Simple Rankine Cycle
10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump,
(2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heat
rejection in a condenser.
10-8C Heat rejected decreases; everything else increases.
10-9C Heat rejected decreases; everything else increases.
10-10C The pump work remains the same, the moisture content decreases, everything else increases.
10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve
friction and pressure drops in various components and the piping, and heat loss to the surrounding medium
from these components and piping.
10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the
same as the boiler inlet pressure in ideal cycles.
10-13C We would reject this proposal because w turb = h 1 - h 2 - q out , and any heat loss from the steam will
adversely affect the turbine work output.
10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than
the temperature of the cooling water.

10-4
10-15 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits.
The thermal efficiency of the cycle and the net power output of the plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
Thus,
and
w
h = h
1
v = v
1
p,in
h
2
f @ 50 kPa
f @ 50 kPa
= 340.54 kJ/kg
= 0.001030 m
/kg
v 1( P2
− P1
)
3
( 0.001030 m /kg)( 3000 − 50)
=
⎛ 1 kJ
=
kPa⎜
1 kPa m
= 3.04 kJ/kg
⎝ ⋅
= h + w = 340.54 + 3.04 = 343.58 kJ/kg
1
p,in
P3
= 3 MPa ⎫ h3
⎬
T3
= 300 ° C ⎭ s3
P4
= 50 kPa ⎫
⎬ x
s4
= s3
⎭
h
q
w
η
q
out
net
th
&
in
= h
3
= h
4
= q
in
q
= 1−
q
&
− h
2
− h
1
− q
out
in
4
4
= 2994.3 kJ/kg
= 6.5412 kJ/kg ⋅ K
s4
− s f
=
s
= h
f
fg
+ x h
4
= 2272.3 kJ/kg
3
= 340.54 +
3
( 0.8382)(
⎞
⎟
⎠
6.5412 −1.0912
=
= 0.8382
6.5019
fg
= 2994.3 − 343.58 = 2650.7 kJ/kg
= 2272.3 − 340.54 = 1931.8 kJ/kg
out
= 2650.7 −1931.8
= 718.9 kJ/kg
1931.8
= 1−
= 27.1%
2650.7
(b) W = m = ( 35 kg/s)( 718.9 kJ/kg) = 25.2 MW
net w net
2304.7
)
T
2
3 MPa
q in
50 kPa
1
q out
3
4
s

10-5
10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the
steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
(b)
and
(c)
h = h
1
v = v
1
2
f @ 10 kPa
f @ 10 kPa
= 191.81 kJ/kg
= 0.00101 m /kg
v1( P2
− P1
)
3
( 0.00101 m /kg)( 10,000 −10
kPa)
wp,in
=
⎛ 1 kJ ⎞
=
⎜ ⎟
3
1 kPa m
= 10.09 kJ/kg
⎝ ⋅ ⎠
h = h + w = 191.81 + 10.09 = 201.90 kJ/kg
1
p,in
P3
= 10 MPa ⎫ h
⎬
T3
= 500 ° C ⎭ s
3
4
3
= 6.5995 kJ/kg ⋅ K
3
= 3375.1 kJ/kg
P4
= 10 kPa ⎫ s4
− s f 6.5995 − 0.6492
⎬ x4
= =
= 0.7934
s4
= s3
⎭ s fg 7.4996
h = h + x h = 191.81 +
q
w
q
in
out
net
= h
3
= h
4
= q
w
η th =
q
W&
m & = w
in
in
net
net
net
− h
2
− h
1
− q
f
4
fg
= 3375.1 − 201.90 = 3173.2 kJ/kg
= 2089.7 −191.81
= 1897.9 kJ/kg
out
= 3173.2 −1897.9
= 1275.4 kJ/kg
1275.4 kJ/kg
=
= 40.2%
3173.2 kJ/kg
210,000 kJ/s
=
= 164.7 kg/
s
1275.4 kJ/kg
( 0.7934)( 2392.1) = 2089.7 kJ/kg
T
2
10 MPa
q in
10 kPa
1
q out
3
4
s

10-6
10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality of
the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
(b)
and
(c)
w
h = h
1
v = v
1
p,in
h
2
=
f @ 10 kPa
f @ 10 kPa
= 191.81 kJ/kg
= 0.00101 m
/kg
v 1( P2
− P1
)/
η p
⎛ 1 kJ ⎞
3
( 0.00101 m /kg)( 10,000 −10
kPa) ⎜ ⎟/
( 0.85)
=
⎜1 kPa m
= 11.87 kJ/kg
⎝ ⋅
= h + w = 191.81+
11.87 = 203.68 kJ/kg
P
s
P
h
4
4
1
4s
4s
p,in
P3
= 10 MPa ⎫ h3
⎬
T3
= 500 ° C ⎭ s3
= 10 kPa ⎫
⎬ x
= s3
⎭
h
h3
− h4
ηT
=
h − h
q
q
w
η
out
net
th
3
4s
4s
⎯⎯→
= h
= 10 kPa ⎫
⎬ x
= 2282.5 kJ/kg⎭
in
= h
= h
= q
3
4
w
=
q
W&
m&
=
w
in
in
net
net
3
= 3375.1 kJ/kg
= 6.5995 kJ/kg ⋅ K
4s
s4s
− s f
=
s
4
h
f
4
fg
+ x h
4
= h3
−ηT
= 3375.1−
h4
− h f
=
h
fg
3
⎟
⎠
6.5995 − 0.6492
=
= 0.7934
7.4996
fg
= 191.81+
( 0.7934)( 2392.1)
( h3
− h4s
)
( 0.85)( 3375.1−
2089.7)
2282.
5 −19181
.
=
= 0.874
23921 .
− h2
= 3375.1−
203.68 = 3171.4 kJ/kg
− h1
= 2282.5 −191.81
= 2090.7 kJ/kg
− q = 3171.4 − 2090.7 = 1080.7 kJ/kg
net
out
1080.7 kJ/kg
=
= 34.1%
3171.5 kJ/kg
210,000 kJ/s
=
1080.7 kJ/kg
= 194.3 kg/s
T
2
2
1
= 2089.7 kJ/kg
= 2282.5 kJ/kg
10 MPa
q in
10 kPa
q out
3
4
4
s

10-7
10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressure
limits is considered. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the
thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist.
2 Kinetic and potential energy changes are
negligible.
Analysis (a) From the steam tables (Tables A-4E,
A-5E, and A-6E),
h = h
1
v = v
1
2
f @ 2 psia
f @ 2 psia
= 94.02 Btu/lbm
= 0.01623 ft /lbm
v1( P2
− P1
)
3
( 0.01623 ft /lbm)( 1250 − 2 psia)
wp,in
=
⎛ 1 Btu
=
⎜
3
5.4039 psia ft
= 3.75 Btu/lbm
⎝
⋅
h = h + w = 94.02 + 3.75 = 97.77 Btu/lbm
h
s
4
4
= h
= s
f
f
1
+ x h
4
+ x s
4
p,in
fg
fg
= 94.02 +
= 0.17499
3
( 0.9)( 1021.7)
=
+ ( 0.9)( 1.74444)
P3
= 1250 psia ⎫ h3
= 1693.4 Btu/lbm
⎬
s3
= s4
⎭T3
= 1337°F
&
⎞
⎟
⎠
1013.6 Btu/lbm
T
= 1.7450 Btu/lbm ⋅ R
(b) Q = m( h − h ) = ( 75 lbm/s)( 1693.4 − 97.77) = 119,672 Btu/s
in
&
3
2
(c) Q&
= m&
( h − h ) = ( 75 lbm/s)( 1013.6 − 94.02)
out
4
Q&
ηth
= 1 −
Q&
out
in
1
68,967 Btu/s
= 1 −
= 42.4%
119,672 Btu/s
= 68,967 Btu/s
2
1
1250 psia
·
Q in
2 psia
·
Q out
3
4
x 4 = 0.9
s

10-8
10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure
limits. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the thermal
efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic
and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4E, A-5E,
and A-6E),
h = h
1
v = v
1
2
f @ 2 psia
f @ 2 psia
= 94.02 Btu/lbm
= 0.01623 ft
/lbm
v 1( P2
− P1
)/
η P
3
( 0.01623 ft /lbm)( 1250 − 2 psia)
w p,in
=
⎛ 1 Btu
=
⎜
5.4039 psia ft
= 4.41 Btu/lbm
⎝
⋅
h = h + w = 94.02 + 4.41 = 98.43 Btu/lbm
1
h
s
4
4
= s
p,in
= h
f
f
+ x h
4
+ x s
4
fg
fg
3
= 94.02 +
= 0.17499
3
T
⎞
⎟ / 0.85
⎠
2
2s
( 0.9)( 1021.7)
= 1013.6 Btu/lbm
+ ( 0.9)( 1.74444) = 1.7450 Btu/lbm ⋅ R
The turbine inlet temperature is determined by trial and error ,
Try 1:
Try 2:
x
h
4s
4s
P3
= 1250 psia ⎫ h
⎬
T3
= 900°
F ⎭ s
s4s
− s f
=
s
= h
f
3
fg
+ x
4s
h3
− h4
ηT
=
h − h
x
h
4s
4s
4s
3
3
s3
− s f
=
s
h
fg
fg
= 94.02 +
P3
= 1250 psia ⎫ h
⎬
T3
= 1000°
F ⎭ s
s4s
− s f
=
s
= h
f
3
fg
+ x
4s
h3
− h4
ηT
=
h − h
4s
= 1439.0 Btu/lbm
= 1.5826 Btu/lbm.R
1.5826 − 0.17499
=
= 0.8069
1.74444
( 0.8069)( 1021.7)
1439.0 −1013.6
=
= 0.8171
1439.0 − 918.4
3
3
s3
− s f
=
s
h
fg
fg
= 94.02 +
= 1498.6 Btu/lbm
= 1.6249 Btu/lbm.R
( 0.8312)( 1021.7)
1498.6 −1013.6
=
= 0.8734
1498.6 − 943.3
= 918.4 Btu/lbm
1.6249 − 0.17499
=
= 0.8312
1.74444
= 943.3
Btu/lbm
1
1250 psia
·
Q in
2 psia
·
Q out
3
4s 4
x 4 = 0.9
By linear interpolation, at η T = 0.85 we obtain T 3 = 958.4°F. This is approximate. We can determine state 3
exactly using EES software with these results: T 3 = 955.7°F, h 3 = 1472.5 Btu/lbm.
&
(b) Q = m( h − h ) = ( 75 lbm/s)( 1472.5 − 98.43) = 103,055 Btu/s
in
&
3
2
(c) Q&
= m&
( h − h ) = ( 75 lbm/s)( 1013.6 − 94.02)
out
4
Q&
η th = 1−
Q&
out
in
1
68,969 Btu/s
= 1−
103,055 Btu/s
= 33.1%
= 68,969 Btu/s
s

10-9
10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the
specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
w
h = h
1
v = v
1
p,in
h
2
f @ 25 kPa
f @ 25 kPa
= 271.96 kJ/kg
= 0.001020 m
/kg
v 1( P2
− P1
)
3
( 0.00102 m /kg)( 5000 − 25 kPa)
=
⎛ 1 kJ
=
⎜
1 kPa m
= 5.07 kJ/kg
⎝ ⋅
= h + w = 271.96 + 5.07 = 277.03 kJ/kg
1
p,in
P3
= 5 MPa ⎫ h3
⎬
T3
= 450°
C ⎭ s3
= 3317.2 kJ/kg
= 6.8210 kJ/kg ⋅ K
P4
= 25 kPa ⎫ s4
− s f 6.8210 − 0.8932
⎬ x4
= =
= 0.8545
s4
= s3
⎭ s fg 6.9370
h = h + x h = 271.96 +
The thermal efficiency is determined from
and
Thus,
q
q
η
in
out
η
= h
= h
th
overall
3
4
− h
− h
= η
2
1
q
= 1−
q
th
4
f
4
fg
= 3317.2 − 277.03 = 3040.2 kJ/kg
= 2276.2 − 271.96 = 2004.2 kJ/kg
out
in
× η
= 1−
comb
× η
2004.2
3040.2
gen
=
3
= 0.3407
(b) Then the required rate of coal supply becomes
and
m&
Q&
in
coal
W&
=
η
Q&
=
C
net
overall
in
coal
3
⎞
⎟
⎠
( 0.8545)( 2345.5) = 2276.2 kJ/kg
( 0.3407)( 0.75)( 0.96) = 24.5%
300,000 kJ/s
=
= 1,222,992 kJ/s
0.2453
1,222,992 kJ/s ⎛ 1 ton ⎞
=
⎜
⎟ = 0.04174
29,300 kJ/kg ⎝1000
kg ⎠
T
2
1
5 MPa
·
Q in
25 kPa
Q·
out
tons/s = 150.3 tons/h
3
4
s

10-10
10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the
working fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13),
w
h = h
1
v = v
1
p,in
h
2
= v
=
f @ 0.7 MPa
f @ 0.7 MPa
1
( P − P )
3
( 0.0008331 m /kg)( 1400 − 700 kPa)
= 0.58 kJ/kg
= h
1
2
+ w
1
p,in
P3
= 1.4 MPa ⎫ h3
= h
⎬
sat.vapor ⎭ s3
= s
4
= 88.82
kJ/kg
= 0.0008331 m
/kg
= 88.82 + 0.58 = 89.40 kJ/kg
g@ 1.4 MPa
g@ 1.4 MPa
f
4
3
= 276.12 kJ/kg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
= 0.9105 kJ/kg ⋅ K
P4
= 0.7 MPa ⎫ s4
− s f 0.9105 − 0.33230
⎬ x4
= =
= 0.9839
s4
= s3
⎭ s fg 0.58763
h = h + x h = 88.82 +
Thus ,
and
q
w
η
q
in
out
net
th
&
= h
3
= h
4
= q
w
=
q
in
net
&
in
− h
2
− h
1
− q
fg
= 276.12 − 89.40 = 186.72 kJ/kg
= 262.20 − 88.82 = 173.38 kJ/kg
out
= 186.72 −173.38
= 13.34 kJ/kg
13.34 kJ/kg
=
= 7.1%
186.72 kJ/kg
(b) W = m = ( 3 kg/s)( 13.34 kJ/kg) = 40.02 kW
net w net
3
⎞
⎟
⎠
( 0.9839)( 176.21) = 262.20 kJ/kg
T
2
1
1.4 MPa
q in
R-134a
0.7 MPa
q out
3
4
s

10-11
10-22 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits.
The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling
water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
w
Thus,
h = h
1
v = v
1
p,in
h
2
= v
=
f @ 10 kPa
f @ 10 kPa
1
( P − P )
3
( 0.00101 m /kg)( 7,000 −10
kPa)
= 7.06 kJ/kg
= h
1
2
+ w
1
p,in
P3
= 7 MPa ⎫ h3
⎬
T3
= 500°
C ⎭ s3
= 191.81 kJ/kg
= 0.00101 m
/kg
= 191.81+
7.06 = 198.87
4
= 6.8000 kJ/kg ⋅ K
3
= 3411.4 kJ/kg
f
4
fg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
kJ/kg
P4
= 10 kPa ⎫ s4
− s f 6.8000 − 0.6492
⎬ x4
= =
= 0.8201
s4
= s3
⎭ s fg 7.4996
h = h + x h = 191.81 +
q
w
q
in
out
net
= h
3
= h
4
= q
in
− h
2
− h
1
− q
= 2153.6 −191.81
= 1961.8 kJ/kg
3
⎞
⎟
⎠
= 3411.4 −198.87
= 3212.5 kJ/kg
out
= 3212.5 −1961.8
= 1250.7 kJ/kg
wnet
1250.7 kJ/kg
and η th = =
= 38.9%
q 3212.5 kJ/kg
(b)
W&
m & = w
net
in
45,000 kJ/s
=
1250.7 kJ/kg
net
=
36 . 0 kg/
s
( 0.8201)( 2392.1) = 2153.6 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
∆T
Q&
out
coolingwater
= mq &
out =
Q&
=
( mc & )
( 35.98 kg/s)( 1961.8 kJ/kg)
out
coolingwater
=
70,586 kJ/s
( 2000 kg/s)( 4.18 kJ/kg ⋅° C)
T
= 70,586 kJ/s
2
= 8.4°C
7 MPa
q in
10 kPa
1
q out
3
4
s

10-12
10-23 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure
limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the
cooling water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
w
Thus,
h = h
1
v = v
1
p,in
h
2
=
=
f @ 10 kPa
f @ 10 kPa
v 1( P2
− P1
)/
η p
⎛ 1 kJ ⎞
3
( 0.00101 m /kg)( 7,000 −10
kPa) ⎜ ⎟/
( 0.87)
= 8.11 kJ/kg
= h
1
+ w
3
p,in
P3
= 7 MPa ⎫ h3
⎬
T3
= 500°
C ⎭ s3
= 191.81 kJ/kg
= 0.00101 m
3
/kg
⎜
⎝1 kPa ⋅ m
= 191.81+
8.11 = 199.92 kJ/kg
4s
= 3411.4 kJ/kg
= 6.8000 kJ/kg ⋅ K
P4
= 10 kPa ⎫ s4
− s f 6.8000 − 0.6492
⎬ x4
= =
= 0.8201
s4
= s3
⎭ s fg 7.4996
h = h + x h = 191.81 +
h3
− h4
ηT
=
h − h
q
w
q
in
out
net
= h
3
= h
4
= q
in
− h
2
− h
1
− q
4s
⎯⎯→
f
h
4
4
fg
= h3
−ηT
= 3411.4
= 2317.1−191.81
= 2125.3 kJ/kg
3
⎟
⎠
= 3411.4 −199.92
= 3211.5 kJ/kg
out
( 0.820)( 2392.1)
( h3
− h4s
)
− ( 0.87)( 3411.4 − 2153.6) = 2317.1 kJ/kg
= 3211.5 − 2125.3 = 1086.2 kJ/kg
wnet
1086.2 kJ/kg
and η th = =
= 33.8%
q 3211.5 kJ/kg
(b)
W&
m & = w
net
in
45,000 kJ/s
=
1086.2 kJ/kg
net
=
41 . 43 kg/
s
T
2
2
1
= 2153.6 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
∆T
Q&
out
coolingwater
= mq &
out
=
Q&
=
( mc & )
( 41.43 kg/s)( 2125.3 kJ/kg)
out
coolingwater
=
88,051 kJ/s
= 88,051 kJ/s
( 2000 kg/s)( 4.18 kJ/kg ⋅° C)
= 10.5°C
7 MPa
q in
10 kPa
q out
3
4
4
s

10-13
10-24 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine
cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),
h = h
1
v = v
1
2
f @ 20 kPa
f @ 20 kPa
= 251.42 kJ/kg
= 0.001017 m /kg
v1( P2
− P1
)
3
( 0.001017 m /kg)( 10,000 − 20)
wp,in
=
⎛ 1 kJ ⎞
=
kPa⎜
⎟
3
1 kPa m
= 10.15 kJ/kg
⎝ ⋅ ⎠
h = h + w = 251.42 + 10.15 = 261.57 kJ/kg
1
p,in
P3
= 10 MPa ⎫ h
⎬
x3
= 1 ⎭ s
P4
= 20 kPa ⎫
⎬ x
s4
= s3
⎭
h
q
w
η
q
in
out
net
th
= h
3
= h
= q
3
4
4
4
3
in
q
= 1−
q
= 5.6159 kJ/kg ⋅ K
= 1845.3 kJ/kg
3
= 2725.5 kJ/kg
s4
− s f
=
s
= h
− h
2
− h
1
− q
out
in
f
fg
+ x h
4
5.6159 − 0.8320
=
= 0.6761
7.0752
fg
= 251.42 +
( 0.6761)(
2357.5
= 2725.5 − 261.57 = 2463.9 kJ/kg
= 1845.3 − 251.42 = 1594.0 kJ/kg
out
(b) Carnot Cycle analysis:
x1
P1
= 20 kPa ⎫
⎬
s1
= s2
⎭ h1
= 2463.9 −1594.0
= 869.9 kJ/kg
1594.0
= 1−
= 0.353
2463.9
P3
= 10 MPa ⎫ h3
= 2725.5 kJ/kg
⎬
x3
= 1 ⎭T3
= 311.0 ° C
T2
= T3
= 311.0 ° C ⎫ h2
= 1407.8 kJ/kg
⎬
x2
= 0
⎭ s2
= 3.3603 kJ/kg ⋅ K
q
w
q
in
out
net
= h
3
= h
4
= q
in
− h
− h
q
η th = 1−
q
1
− q
out
in
s1
− s f 3.3603 − 0.8320
= =
= 0.3574
s fg 7.0752
= h + x h
f
1
fg
= 251.42 + (0.3574)(2357.5) = 1093.9 kJ/kg
2
= 2725.5 −1407.8
= 1317.7 kJ/kg
= 1845.3 −1093.9
= 751.4 kJ/kg
out
= 1317.7 − 752.3 = 565.4 kJ/kg
751.4
= 1−
1317.7
= 0.430
)
T
2
Rankine
cycle
T
1
Carnot
cycle
2 3
1
4
3
4
s
s

10-14
10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the working
fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle
are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Properties The specific heat of geothermal water is taken to be 4.18 kJ/kg.ºC.
Analysis (a) We need properties of isobutane,
which are not available in the book. However,
we can obtain the properties from EES.
Turbine:
P3
= 3250 kPa ⎫h
⎬
T3
= 147°
C ⎭ s
P4
= 410 kPa ⎫
⎬h
s4
= s3
⎭
P4
= 410 kPa ⎫
⎬h4
T4
= 179.5°
C ⎭
h3
− h4
η T =
h − h
3
(b) Pump:
1
1
2
4s
h = h
v = v
3
3
4s
= 761.54 kJ/kg
= 2.5457 kJ/kg ⋅ K
= 670.40 kJ/kg
= 689.74 kJ/kg
761.54 − 689.74
=
= 0.788
761.54 − 670.40
f @ 410 kPa
f @ 410 kPa
= 273.01kJ/kg
= 0.001842 m /kg
v1( P2
− P1
)/
ηP
3
( 0.001842 m /kg)( 3250 − 410)
1
p,in
3
turbine
4
3 2
Geothermal
water in
wp,in
=
⎛ 1 kJ
=
kPa⎜
3
1 kPa m
= 5.81 kJ/kg
⎝ ⋅
h = h + w = 273.01+
5.81 = 278.82 kJ/kg
⎞
⎟
/ 0.90
⎠
T
air-cooled
condenser
heat exchanger
2
2s
W &
T, out = m&
( h3
− h4 ) = (305.6 kJ/kg)(761.54 − 689.74)kJ/kg = 21,941kW
W &
P, in = m&
( h2
− h1
) = m&
wp,
in = (305.6 kJ/kg)(5.81kJ/kg) = 1777 kW
W &
net
Heat Exchanger:
= W&
−W&
= 21,941−1777
= 20,165 kW
T,out
P,in
1
3.25 MPa
q in
410 kPa
Q &
in = m&
geocgeo( Tin
− Tout
) = (555.9 kJ/kg)(4.18 kJ/kg. ° C)(160 − 90) ° C = 162,656 kW
W&
net 20,165
(c) η th = = = 0.124 = 12.4%
Q&
162,656
in
q out
1
Geothermal
water out
3
4s 4
pump
s

10-15
10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The
mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from
the turbine, and the thermal efficiency of the plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) We use properties of water for
geothermal water (Tables A-4 through A-6)
T1
= 230°
C⎫
⎬ h1
= 990.14 kJ/kg
x1
= 0 ⎭
P2
= 500 kPa ⎫ h2
− h f
⎬x2
=
h2
= h1
= 990.14 kJ/kg⎭
h fg
The mass flow rate of steam through the turbine is
m &
= x & = (0.1661)(230 kg/s) = 38.20 kg/s
3 2m1
(b) Turbine:
P3
= 500 kPa ⎫ h
⎬
x3
= 1 ⎭ s
P4
= 10 kPa ⎫
⎬h
s4
= s3
⎭
3
3
4s
P4
= 10 kPa ⎫
⎬h4
x4
= 0.90 ⎭
3
= 2748.1 kJ/kg
= 6.8207 kJ/kg ⋅ K
= 2160.3 kJ/kg
= h
h3
− h4
η T =
h − h
4s
f
+ x h
4
fg
990.14 − 640.09
=
= 0.1661
2108
production
well
Flash
chamber
= 191.81+
(0.90)(2392.1) = 2344.7 kJ/kg
2748.1−
2344.7
=
= 0.686
2748.1−
2160.3
(c) The power output from the turbine is
W &
2
1
6
separator
= m&
( h − 4 ) = (38.20 kJ/kg)(2748.1−
2344.7) kJ/kg = 15,410 kW
T, out 3 3 h
(d) We use saturated liquid state at the standard temperature for dead state enthalpy
T0
= 25°
C⎫
⎬ h0
x0
= 0 ⎭
= 104.83 kJ/kg
E &
in = m&
1( h1
− h0 ) = (230 kJ/kg)(990.14 −104.83)kJ/kg
= 203,622 kW
W&
T,out 15,410
η th = = = 0.0757 = 7.6%
E&
203,622
in
3
steam
turbine
condenser
4
5
reinjection
well

10-16
10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The
temperature of the steam at the exit of the second flash chamber, the power produced from the second
turbine, and the thermal efficiency of the plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)
T1
= 230°
C⎫
⎬ h1
= 990.14 kJ/kg
x1
= 0 ⎭
P2
= 500 kPa ⎫
⎬x2
= 0.1661
h2
= h1
= 990.14 kJ/kg⎭
m&
3
m&
6
= x m&
2
= m&
1
1
− m&
P3
= 500 kPa ⎫
⎬ h
x3
= 1 ⎭
= (0.1661)(230 kg/s) = 38.20 kg/s
3
P4
= 10 kPa ⎫
⎬h4
x4
= 0.90 ⎭
P6
= 500 kPa ⎫
⎬ h
x6
= 0 ⎭
P7
= 150 kPa ⎫ T7
⎬
h7
= h6
⎭ x7
P8
= 150 kPa ⎫
⎬h
x8
= 1 ⎭
= 230 − 0.1661 = 191.80 kg/s
8
3
= 2748.1 kJ/kg
= 2344.7 kJ/kg
6
= 640.09 kJ/kg
= 111.35 ° C
= 0.0777
= 2693.1 kJ/kg
2
production
well
6
Flash
chamber
(b) The mass flow rate at the lower stage of the turbine is
m &
8 = x7m&
6 = (0.0777)(191.80 kg/s) = 14.90 kg/s
1
separator
7
Flash
chamber
The power outputs from the high and low pressure stages of the turbine are
W &
T1, out = m&
3( h3
− h4 ) = (38.20 kJ/kg)(2748.1−
2344.7)kJ/kg = 15,410 kW
W &
= m&
( h − 4 ) = (14.90 kJ/kg)(2693.1−
2344.7) kJ/kg = 5191 kW
T2, out 8 8 h
8
3
separator
(c) We use saturated liquid state at the standard temperature for the dead state enthalpy
T0
= 25°
C⎫
⎬ h0
x0
= 0 ⎭
= 104.83 kJ/kg
E &
in = m&
1( h1
− h0)
= (230 kg/s)(990.14 −104.83)kJ/kg
= 203,621kW
W&
T,out 15,410 + 5193
η th = =
= 0.101 = 10.1%
E&
203,621
in
9
4
condenser
5
reinjection
well
steam
turbine

10-17
10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat
source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine
and the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)
T1
= 230°
C⎫
⎬ h1
= 990.14 kJ/kg
x1
= 0 ⎭
P2
= 500 kPa ⎫
⎬x2
= 0.1661
h2
= h1
= 990.14 kJ/kg⎭
m&
3
m&
6
= x m&
2
= m&
1
1
− m&
P3
= 500 kPa ⎫
⎬ h
x3
= 1 ⎭
= (0.1661)(230 kg/s) = 38.20 kg/s
3
P4
= 10 kPa ⎫
⎬h4
x4
= 0.90 ⎭
= 230 − 38.20 = 191.80 kg/s
3
= 2748.1 kJ/kg
= 2344.7 kJ/kg
P6
= 500 kPa ⎫
⎬ h6
= 640.09 kJ/kg
x6
= 0 ⎭
T7
= 90°
C ⎫
⎬ h7
= 377.04 kJ/kg
x7
= 0 ⎭
The isobutene properties
are obtained from EES:
P8
= 3250 kPa ⎫
⎬ h
T8
= 145°
C ⎭
P9
= 400 kPa ⎫
⎬ h9
T9
= 80°
C ⎭
P
x
w
10
10
= 400 kPa ⎫ h
⎬
= 0 ⎭v
p,in
h
11
8
= 755.05 kJ/kg
= 691.01 kJ/kg
10
10
= 270.83 kJ/kg
= 0.001839 m
3
/kg
v10( P11
− P10
)/
η p
3
( 0.001819 m /kg)( 3250 − 400)
2
flash
chamber
1
separator
production
well
=
⎛ 1 kJ
=
kPa⎜
1 kPa m
= 5.82 kJ/kg.
⎝ ⋅
= h + w = 270.83 + 5.82 = 276.65 kJ/kg
10
p,in
An energy balance on the heat exchanger gives
m&
( h
− h
) = m&
(191.81kg/s)(640.09 - 377.04)kJ/kg = m&
6
6
7
iso
iso
3
( h
6
3
⎞
⎟
/ 0.90
⎠
8
− h
11
)
steam
turbine
8
7
4
9
isobutane
turbine
heat exchanger
(755.05- 276.65)kJ/kg
(b) The power outputs from the steam turbine and the binary cycle are
W &
⎯⎯→
= m&
( h − 4 ) = (38.19 kJ/kg)(2748.1−
2344.7) kJ/kg = 15,410 kW
T, steam 3 3 h
air-cooled
condenser
1
m&
BINARY
CYCLE
iso
condenser
pump
1
= 105.46 kg/s
5
reinjection
well

10-18
W&
W&
T,iso
= m&
net,binary
iso
( h
= W&
8
T,iso
− h
9
− m&
) = (105.46 kJ/kg)(755.05 − 691.01)kJ/kg = 6753 kW
iso
w
p,
in
= 6753 − (105.46 kg/s)(5.82 kJ/kg) = 6139 kW
(c) The thermal efficiencies of the binary cycle and the combined plant are
Q &
in, binary = m&
iso ( h8
− h11 ) = (105.46 kJ/kg)(755.05 − 276.65)kJ/kg = 50,454 kW
η
th,binary
T0
= 25°
C⎫
⎬ h0
x0
= 0 ⎭
W&
net,binary 6139
= = = 0.122 = 12.2%
Q&
50,454
in,binary
= 104.83 kJ/kg
E &
in = m&
1( h1
− h0 ) = (230 kJ/kg)(990.14 −104.83)kJ/kg
= 203,622 kW
η
th,plant
W&
T,steam + W&
net,binary 15,410 + 6139
= =
= 0.106 = 10.6%
E&
203,622
in
The Reheat Rankine Cycle
10-29C The pump work remains the same, the moisture content decreases, everything else increases.
10-30C The T-s diagram of the ideal Rankine
cycle with 3 stages of reheat is shown on the
side. The cycle efficiency will increase as the
number of reheating stages increases.
T
3 5 7 9
I II III
4 6 8
2
1
10
s
10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average
temperature at which heat is added will be higher in this case.

10-19
10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine
cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
w
h = h
1
v = v
1
p,in
2
=
=
f @ 20 kPa
f @ 20 kPa
v1( P2
− P1
)
3
( 0.001017 m /kg)( 8000 − 20 kPa)
= 8.12 kJ/kg
h = h
1
+ w
p,in
P3
= 8 MPa ⎫ h
⎬
T3
= 500°
C ⎭ s
P5
= 3 MPa ⎫ h
⎬
T5
= 500°
C ⎭ s
P6
= 20 kPa ⎫ x6
⎬
s6
= s5
⎭
h
3
3
P4
= 3 MPa ⎫
⎬ h4
s4
= s3
⎭
5
5
6
= 251.42 kJ/kg
= 0.001017 m /kg
3
⎛ 1 kJ ⎞
⎜ ⎟
3
1 kPa m
⎝ ⋅ ⎠
= 251.42 + 8.12 = 259.54 kJ/kg
= 3399.5 kJ/kg
= 6.7266 kJ/kg ⋅ K
= 3105.1 kJ/kg
= 3457.2 kJ/kg
= 7.2359 kJ/kg ⋅ K
s6
− s f
=
s
= h
f
fg
+ x h
6
7.2359 − 0.8320
=
= 0.9051
7.0752
fg
= 251.42 +
( 0.9051)( 2357.5) = 2385.2 kJ/kg
The turbine work output and the thermal efficiency are determined from
and
Thus,
w
w
T,out
q
net
in
=
=
= w
w
η th =
q
( h − h ) + ( h − h )
3
T
2
1
8 MPa
20 kPa
( h − h ) + ( h − h ) = 3399.5 − 259.54 + 3457.2 − 3105.1 = 3492.0 kJ/kg
3
T , out
net
in
4
2
− w
p,in
5
5
1358.3 kJ/kg
=
= 38.9%
3492.5 kJ/kg
6
4
= 3399.5 − 3105.1+
3457.2 − 2385.2 = 1366.4 kJ/kg
= 1366.4 − 8.12 = 1358.3 kJ/kg
3
4
5
6
s

10-20
10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entry
feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality
at the low-pressure turbine exit Also, the T-s diagram is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data - from diagram window"
{P[6] = 20 [kPa]
P[3] = 8000 [kPa]
T[3] = 500 [C]
P[4] = 3000 [kPa]
T[5] = 500 [C]
Eta_t = 100/100 "Turbine isentropic efficiency"
Eta_p = 100/100 "Pump isentropic efficiency"}
"Pump analysis"
function x6$(x6) "this function returns a string to indicate the state of steam at point 6"
x6$=''
if (x6>1) then x6$='(superheated)'
if (x6

10-21
vs[6]=volume(Fluid$,s=s_s[6],P=P[6])
Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"
h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"
x[6]=QUALITY(Fluid$,h=h[6],P=P[6])
"Boiler analysis"
Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler"
"Condenser analysis"
h[6]=Q_out+h[1]"SSSF First Law for the Condenser"
T[6]=temperature(Fluid$,h=h[6],P=P[6])
s[6]=entropy(Fluid$,h=h[6],P=P[6])
x6s$=x6$(x[6])
"Cycle Statistics"
W_net=W_t_hp+W_t_lp-W_p
Eff=W_net/Q_in
700
600
Ideal Rankine cycle with reheat
500
3
5
T [C]
400
300
200
8000 kPa
3000 kPa
4
100
1,2
20 kPa
6
0
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0
s [kJ/kg-K]
SOLUTION
Eff=0.389 Eta_p=1 Eta_t=1
Fluid$='Steam_IAPWS' h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg]
h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg]
h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg]
P[1]=20 [kPa] P[2]=8000 [kPa] P[3]=8000 [kPa]
P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa]
Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K]
s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K]
s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K]
s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C]
T[3]=500 [C] T[4]=345.2 [C] T[5]=500 [C]
T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C]
v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg]
v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg]
W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg]
W_t_lp=1072 [kJ/kg] x6s$='' x[1]=0
x[6]=0.9051

10-22
10-34 A steam power plant that operates on a reheat Rankine cycle is considered. The quality (or
temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the
mass flow rate of the steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
w
h = h
1
v = v
1
p,in
h
2
=
=
f @ 10 kPa
f @ 10 kPa
v 1( P2
− P1
)/
η p
⎛ 1 kJ ⎞
3
( 0.00101 m /kg)( 10,000 −10
kPa) ⎜ ⎟/
( 0.95)
= 10.62
= h
1
+ w
kJ/kg
p,in
P3
= 10 MPa ⎫ h
⎬
T3
= 500°
C ⎭ s
P
s
4s
4s
η
T
= 1 MPa ⎫
⎬ h
= s3
⎭
h3
− h
=
h − h
3
= 191.81 kJ/kg
= 0.001010
m
/kg
= 191.81+
10.62 = 202.43
4
4s
P5
= 1 MPa ⎫ h5
⎬
T5
= 500°
C ⎭ s5
P
s
6s
6s
= 10 kPa ⎫ x
⎬
= s5
⎭
h
3
3
4s
⎯→
= 6.5995 kJ/kg ⋅ K
= 7.7642 kJ/kg ⋅ K
3
= 3375.1 kJ/kg
= 2783.8 kJ/kg
h = h − η
4
3
= 3375.1
= 3479.1 kJ/kg
6s
6s
s
=
= h
6s
f
s
− s
fg
+ x
f
6s
T
⎜
⎝1
kPa ⋅ m
kJ/kg
3
⎟
⎠
( h3
− h4
s )
− ( 0.80)( 3375.1−
2783.7)
7.7642 − 0.6492
=
= 0.9487
7.4996
h
fg
= 191.81 +
( 0.9487)( 2392.1)
T
2
2
1
= 2902.0 kJ/kg
10 MPa
( at turbine exit)
= 2461.2 kJ/kg
10 kPa
3
4s 4
6
5
6
s
η
T
h
=
h
5
5
− h
− h
6
6s
⎯⎯→
h = h −η
6
5
T
= 3479.1 −
= 2664.8 kJ/kg
( h5
− h6s
)
( 0.80)( 3479.1 − 2461.2)
> h ( superheated vapor)
g
From steam tables at 10 kPa we read T 6 = 88.1°C.
(b) w = ( h − h ) + ( h − h )
T,out
w
q
in
net
=
3
( h − h ) + ( h − h )
= w
3
T,out
4
2
− w
p,in
Thus the thermal efficiency is
η
th
w
=
q
net
in
5
5
6
4
= 3375.1−
2902.0 + 3479.1−
2664.8 = 1287.4 kJ/kg
= 3375.1−
202.43 + 3479.1−
2902.0 = 3749.8 kJ/kg
= 1287.4 −10.62
= 1276.8 kJ/kg
1276.8 kJ/kg
=
= 34.1%
3749.8 kJ/kg
(c) The mass flow rate of the steam is
W
m & =
&
w
net
net
=
80,000 kJ/s
1276.9 kJ/kg
= 62.7 kg/s

10-23
10-35 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or
temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the
mass flow rate of the steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
w
h = h
1
v = v
1
p,in
h
2
=
=
f @ 10 kPa
f @ 10 kPa
v 1( P2
− P1
)
3
( 0.00101 m /kg)( 10,000 −10
kPa)
= 10.09
= h
1
+ w
kJ/kg
p,in
P3
= 10 MPa ⎫ h
⎬
T3
= 500°
C ⎭ s
P4
= 1 MPa ⎫
⎬ h4
s4
= s3
⎭
P5
= 1 MPa ⎫ h5
⎬
T5
= 500°
C ⎭ s5
P6
= 10 kPa ⎫ x6
⎬
s6
= s5
⎭
h
6
3
3
= 191.81 kJ/kg
= 0.00101 m
/kg
= 191.81+
10.09 = 201.90
= 6.5995 kJ/kg ⋅ K
= 2783.8 kJ/kg
= 3479.1 kJ/kg
= 7.7642 kJ/kg ⋅ K
3
= 3375.1 kJ/kg
s6
− s f
=
s
= h
f
fg
+ x h
(b) w = ( h − h ) + ( h − h )
T,out
w
q
in
net
=
3
( h − h ) + ( h − h )
= w
3
T,out
4
2
−
5
5
w p ,in
Thus the thermal efficiency is
η
th
w
=
q
net
in
6
6
4
= 191.81+
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
kJ/kg
3
⎞
⎟
⎠
7.7642 − 0.6492
=
= 0.9487
7.4996
fg
= 1609.4 −10.09
= 1599.3 kJ/kg
1599.3 kJ/kg
=
= 41.3%
3868.5 kJ/kg
(c) The mass flow rate of the steam is
W&
m & = w
net
net
=
80,000 kJ/s
1599.3 kJ/kg
T
( at turbine exit)
( 0.9487)( 2392.1) = 2461.2 kJ/kg
2
1
10 MPa
10 kPa
= 3375.1−
2783.7 + 3479.1−
2461.2 = 1609.3 kJ/kg
= 3375.1−
201.90 + 3479.1−
2783.7 = 3868.5 kJ/kg
= 50 . 0 kg/
s
3
4
5
6
s

10-24
10-36E A steam power plant that operates on the ideal reheat Rankine cycle is considered. The pressure at
which reheating takes place, the net power output, the thermal efficiency, and the minimum mass flow rate
of the cooling water required are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E),
w
h = h
1
v = v
1
T = T
1
p,in
h
2
=
=
sat@ 1 psia
sat@ 1 psia
sat@ 1 psia
v 1( P2
− P1
)
3
( 0.01614 ft /lbm)( 800 −1 psia)
= 2.39 Btu/lbm
= h
1
+ w
p,in
= 69.72 Btu/lbm
= 0.01614
= 101.69°
F
ft
3
/lbm
⎛ 1 Btu ⎞
⎜
⎟
3
5.4039 psia ft
⎝
⋅ ⎠
= 69.72 + 2.39 = 72.11 Btu/lbm
T
2
1
800 psia
1 psia
3
4
5
6
s
P
T
s
3
3
4
= 800 psia
= 900 ° F
= s
3
( sat. vapor )
⎫ h
⎬
⎭ P
⎫ h
⎬
⎭ s
4
4
3
3
= h
= P
= 1456 .0 Btu/lbm
= 1.6413 Btu/lbm ⋅ R
g @ s g = s4
sat @ s g = s4
= 1178 .5 Btu/lbm
= 62.23 psia (the
reheat
pressure)
P
T
P
s
5
5
6
6
= 62 .23 psia
= 800 ° F
= 1 psia
= s
5
⎫ x
⎬
⎭
h
6
6
⎫ h
⎬
⎭ s
=
5
5
s
= h
6
f
= 1431 .4 Btu/lbm
= 1.8985 Btu/lbm ⋅ R
− s
s
fg
f
+ x h
(b) q = ( h − h ) + ( h − h )
Thus,
q
η
in
out
th
= h
6
3
− h
q
= 1−
q
out
in
1
2
5
4
6
1.8985 − 0.13262
=
1.84495
fg
= 69 .72 +
= 1061.0 − 69.72 = 991.3 Btu/lbm
= 0.9572
( 0.9572 )( 1035 .7) = 1061 .0 Btu/lbm
= 1456.0 − 72.11 + 1431.4 −1178.5
= 1636.8 Btu/lbm
991.3 Btu/lbm
= 1−
= 39.4%
1636.8 Btu/lbm
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the
steam in the condenser, which is 101.7°F,
Q&
m&
out
cool
= Q&
in −W&
Q&
out
= =
c∆T
net
=
4
( 1−η
) Q&
= ( 1−
0.3943)( 6×
10 Btu/s)
th
in
3.634×
10
Btu/s
( 1.0 Btu/lbm⋅°
F)( 101.69 − 45)
4
= 641.0 lbm/s
° F
= 3.634×
10
4
Btu/s

10-25
10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure
limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler,
and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h = h
1
sat@ 10 kPa
= 191.81 kJ/kg
T
w
v = v
1
p,in
h
2
=
=
sat@ 10 kPa
v 1( P2
− P1
)
3
( 0.00101 m /kg)( 15,000 −10
kPa)
= 15.14 kJ/kg
= h
1
+ w
p,in
P3
= 15 MPa ⎫ h
⎬
T3
= 500°
C ⎭ s
3
3
= 0.00101 m
3
/kg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
= 191.81+
15.14 = 206.95 kJ/kg
10 kPa
1
= 3310.8 kJ/kg
= 6.3480 kJ/kg ⋅ K
P6
= 10 kPa ⎫ h6
= hf
+ x6h
⎬
s6
= s5
⎭ s6
= s f + x6s
T5
= 500°
C ⎫ P5
= 2161kPa
⎬
s5
= s6
⎭ h5
= 3466.53 kJ/kg
fg
fg
= 191.81+
= 0.6492 +
3
( 0.90)( 2392.1)
( 0.90)( 7.4996)
( the reheat pressure)
15 MPa
⎞
⎟
⎠
2
= 2344.7 kJ/kg
= 7.3988 kJ/kg ⋅ K
3
4
5
6
s
P4
= 2.161 MPa ⎫
⎬ h4
s4
= s3
⎭
= 2817.2 kJ/kg
(b) The rate of heat supply is
Q &
in
=
=
m&
[( h3
− h2
) + ( h5
− h4
)]
( 12 kg/s)( 3310.8 − 206.95 + 3466.53 − 2817.2) kJ/kg = 45,038 kW
(c) The thermal efficiency is determined from
Thus,
Q&
out
= m&
h
( − h ) = ( 12 kJ/s)( 2344.7 − )
6
Q&
η th = 1−
Q&
out
in
1
25,834 kJ/s
= 1−
= 42.6%
45,039 kJ/s
191.81 kJ/kg = 25,835 kJ/s

10-26
10-38 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure,
the net power output, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
P3
= 12.5 MPa ⎫ h
⎬
T3
= 550°
C ⎭ s
P4
= 2 MPa ⎫
⎬ h
s4s
= s3
⎭
η
T
→h
h3
− h
=
h − h
4
3
= h
= 3476.5
= 3027.3 kJ/kg
P5
= 2 MPa ⎫ h
⎬
T5
= 450°
C ⎭ s
T
5
3
P6
= ⎫
⎬ h
s6
= s5⎭
6s
4
4s
− η
h5
− h
=
h − h
6
T
P6
= ⎫
⎬ h6
=
x6
= 0.95⎭
η
=
6s
4s
= 3476.5
= 6.6317 kJ/kg ⋅ K
= 2948.1 kJ/kg
( h3
− h4
s )
− ( 0.85)( 3476.5 − 2948.1)
5
5
3
3
= 3358.2 kJ/kg
kJ/kg
= 7.2815 kJ/kg ⋅ K
⎯→
h = h − η
6
5
T
= 3358.2
Boiler
2
2s
12.5 MPa
( h5
− h6
s )
− ( 0.85)( 3358.2 − 2948.1) = 3027.3 kJ/kg
2
T
1
4
5
Pump
P =
3
Condenser
3
4
4s
5
6s 6
Turbine
The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES
from the above equations:
P 6 = 9.73 kPa, h 6 = 2463.3 kJ/kg,
(b) Then,
w
h = h
1
v = v
1
p,in
h
2
=
=
Cycle analysis:
q =
q
W&
in
out
net
f @ 9.73 kPa
f @ 10 kPa
v1( P2
− P1
)/
η p
3
⎛ 1 kJ ⎞
( 0.00101 m /kg)( 12,500 − 9.73 kPa) ⎜ ⎟/
( 0.90)
= 14.02
= h
1
+ w
kJ/kg
p,in
= 189.57 kJ/kg
( h − h ) + ( h − h )
= h
6
3
= m&
( q
− h
in
1
2
− q
= 0.001010 m /kg
= 189.57 + 14.02 = 203.59
3
⎜
3
1 kPa m ⎟
⎝ ⋅ ⎠
kJ/kg
= 3476.5 − 3027.3 + 3358.2 − 2463.3 = 3603.8 kJ/kg
= 3027.3 −189.57
= 2273.7 kJ/kg
out
5
4
) = (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW
(c) The thermal efficiency is
qout
2273.7 kJ/kg
η th = 1−
= 1−
= 0. 369 = 36.9%
q 3603.8 kJ/kg
in
1
s
6

10-27
Regenerative Rankine Cycle
10-39C Moisture content remains the same, everything else decreases.
10-40C This is a smart idea because we waste little work potential but we save a lot from the heat input.
The extracted steam has little work potential left, and most of its energy would be part of the heat rejected
anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work
output.
10-41C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no
mixing.
10-42C Both cycles would have the same efficiency.
10-43C To have the same thermal efficiency as the Carnot
cycle, the cycle must receive and reject heat isothermally.
Thus the liquid should be brought to the saturated liquid
state at the boiler pressure isothermally, and the steam must
be a saturated vapor at the turbine inlet. This will require
an infinite number of heat exchangers (feedwater heaters),
as shown on the T-s diagram.
T
Boiler
inlet
q in
Boiler
exit
q out
s

10-28
10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwater
heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h = h = 251.42 kJ/kg
T
1
f @ 20 kPa
w
w
pII ,in
v = v
1
pI ,in
h
4
h
2
=
=
f @ 20 kPa
v1( P2
− P1
)
3
( 0.001017 m /kg)( 400 − 20 kPa)
= 0.39 kJ/kg
= h
1
+ w
pI ,in
= 0.001017
P3
= 0.4 MPa ⎫ h3
= h f
⎬
sat.liquid ⎭v
= v
= v
= h
3
3
m
3
/kg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
= 251.42 + 0.39 = 251.81 kJ/kg
3
( P − P ) = ( 0.001084 m /kg)( 6000 − 400 kPa)
4
+ w
3
pII ,in
3
@ 0.4 MPa
f @ 0.4 MPa
= 604.66 + 6.07 = 610.73 kJ/kg
3
= 604.66 kJ/kg
= 0.001084 m
⎞
⎟
⎠
3
/kg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
3
4
3 0.4 MPa y
2
1
⎞
⎟ = 6.07
⎠
q in
20 kPa
kJ/kg
6
MPa
q out
5
6
1-y
7
s
P5
= 6 MPa ⎫ h5
⎬
T5
= 450°
C ⎭ s5
P6
= 0.4 MPa ⎫ x6
⎬
s6
= s5
⎭
h
P7
= 20 kPa ⎫ x7
⎬
s7
= s5
⎭
h
7
= 3302.9 kJ/kg
= 6.7219 kJ/kg ⋅ K
6
s6
− s f
=
s
= h
s7
− s f
=
s
= h
f
f
fg
fg
+ x h
7
6
+ x h
6.7219 −1.7765
=
= 0.9661
5.1191
= 604.66 +
= 251.42 +
( 0.9661)( 2133.4)
6.7219 − 0.8320
=
= 0.8325
7.0752
fg
fg
= 2665.7 kJ/kg
( 0.8325)( 2357.5) = 2214.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the
feedwater heater. Noting that Q & ≅W & ≅ ∆ ke ≅ ∆ pe ≅ 0 ,
E&
in
− E&
∑
out
E&
i
in
i
= ∆E&
= E&
m&
h =
0 (steady)
system
out
∑
m&
h
e
e
⎯→
= 0
m&
h
6
6
+ m&
h
2
2
= m&
h
3
3
⎯→
yh
6
+
( 1−
y) h 2 = 1( h 3 )
where y is the fraction of steam extracted from the turbine ( = m&
6 / m&
3). Solving for y,
h3
− h2
604.66 − 251.81
y = =
= 0.1462
h6
− h2
2665.7 − 251.81
Then,
q = h − h = 3302.9 − 610.73 = 2692.2 kJ/kg
q
in
out
=
5
4
( 1 − y)( h − h ) = ( 1 − 0.1462)( 2214.0 − 251.42) = 1675.7 kJ/kg
7
1
And w net = qin
− qout
= 2692.2 −1675.
7 = 1016.5 kJ/kg
(b) The thermal efficiency is determined from
qout
1675.7 kJ/kg
η th = 1−
= 1−
= 37.8%
q 2692.2 kJ/kg
in

10-29
10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater
heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h = h = 251.42 kJ/kg
T
w
1
v = v
1
pI ,in
h
2
=
=
f @ 20 kPa
f @ 20 kPa
v 1( P2
− P1
)
3
( 0.001017 m /kg)( 6000 − 20 kPa)
= 6.08 kJ/kg
= h
1
+ w
pI ,in
P3
= 0.4 MPa ⎫ h3
= h f
⎬
sat. liquid ⎭v
= v
w
pII ,in
h
9
= v
3
= h
3
= 0.001017
3
m
3
/kg
⎛
⎜
1 kJ
⎝1 kPa ⋅ m
= 251.42 + 6.08 = 257.50 kJ/kg
@ 0.4 MPa
f @ 0.4 MPa
= 604.66 kJ/kg
= 0.001084 m
3
/kg
3
( P − P ) = ( 0.001084 m /kg)( 6000 − 400 kPa)
9
+ w
3
pII ,in
( P − P ) = h = 610.73 kJ/kg
3
⎞
⎟
⎠
= 604.66 + 6.07 = 610.73 kJ/kg
q in
6
4
9
MPa
2 8 0.4 MPa y
3
6
1-y
20 kPa
1
7
q ou
⎛ 1 kJ ⎞
⎜ ⎟
= 6.07 kJ/kg
3
1 kPa m
⎝ ⋅ ⎠
h8
= h3
+ v3
8 3 9
Also, h 4 = h 9 = h 8¸ = 610.73 kJ/kg since the two fluid streams which are being mixed have the same
enthalpy.
P5
= 6 MPa ⎫ h5
= 3302.9 kJ/kg
⎬
T5
= 450°
C ⎭ s5
= 6.7219 kJ/kg ⋅ K
s6
− s f 6.7219 −1.7765
P 0.4 MPa 6
0.9661
6 = ⎫ x = =
=
⎬ s fg 5.1191
s6
= s5
⎭
h = h + x h = 604.66 +
7
6
f
f
7
6
fg
fg
( 0.9661)( 2133.4)
s7
− s f 6.7219 − 0.8320
P 20 kPa 7
0.8325
7 = ⎫ x = =
=
⎬ s fg 7.0752
s7
= s5
⎭
h = h + x h = 251.42 +
= 2665.7 kJ/kg
( 0.8325)( 2357.5) = 2214.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the
feedwater heater. Noting that Q ≅ W & ≅ ∆ke ≅ ∆pe ≅ 0 ,
E&
in
∑
− E&
out
E&
i
in
i
= ∆E&
= E&
m&
h =
out
∑
system
m&
e
h
& 0 (steady)
= 0
⎯⎯→
m&
( ) ( ) ( )( ) ( 2 h8
− h2
= m&
6 h6
− h3
⎯⎯→
1−
y h8
− h2
= y h6
− h ) 3
e
where y is the fraction of steam extracted from the turbine ( = m&
6 / m&
5). Solving for y,
h8
− h2
610.73 − 257.50
y =
=
= 0.1463
h − h + h − h 2665.7 − 604.66 + 610.73 − 257.50
Then,
q
q
in
out
( ) ( )
=
6
= h
5
3
− h
4
8
2
= 3302.9 − 610.73 = 2692.2 kJ/kg
( 1 − y)( h − h ) = ( 1 − 0.1463)( 2214.0 − 251.42) = 1675.4 kJ/kg
7
1
And w net = qin
− qout
= 2692.2 −1675.
4 = 1016.8 kJ/kg
(b) The thermal efficiency is determined from
qout
1675.4 kJ/kg
η th = 1 − = 1 −
= 37.8%
q 2692.2 kJ/kg
in
5
s