1
Solutions to Homework #10
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
th, C = 1
TL
60 + 273 K
= 1
= 46.5%
TH
350 + 273 K
(b) Note that
s2 = s3 = sf + x3sfg
350C
= 0.8313 + 0.891 7.0769 = 7.1368 kJ/kgK
Thus,
60C
T2 = 350C
P2 1.40 MPa (Table A-6)
s 2 = 7.1368 kJ/kg K
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
s 4 = s f + x 4 s fg = 0.8313 + (0.1)(7.0769 ) = 1.5390 kJ/kg K
Thus,
wnet = Area = (TH TL )(s 3 s 4 ) = (350 60)(7.1368 1.5390) = 1623 kJ/kg
10-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified
pressure limits. The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis For maximum thermal efficiency, the quality at state 4 would be at its minimum of 85% (most
closely approaches the Carnot cycle), and the properties at state 4 would be (Table A-5)
P4 = 30 kPa h4 = h f + x 4 h fg = 289.27 + (0.85)(2335.3) = 2274.3 kJ/kg
x 4 = 0.85 s 4 = s f + x 4 s fg = 0.9441 + (0.85)(6.8234) = 6.7440 kJ/kg K
Since the expansion in the turbine is isentropic,
P3 = 3000 kPa
h3 = 3115.5 kJ/kg
s 3 = s 4 = 6.7440 kJ/kg K
T
3
Other properties are obtained as follows (Tables A-4, A-5, and A-6),
h1 = h f @ 30 kPa = 289.27 kJ/kg
v 1 = v f @ 30 kPa = 0.001022 m 3 /kg
3 MPa
qin
30 kPa
wp,in = v 1 ( P2 P1 )
1 kJ
= (0.001022 m 3 /kg )(3000 30)kPa
1 kPa m 3
= 3.04 kJ/kg
h2 = h1 + wp,in = 289.27 + 3.04 = 292.31 kJ/kg
qout
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Thus,
q in = h3 h2 = 3115.5 292.31 = 2823.2 kJ/kg
q out = h4 h1 = 2274.3 289.27 = 1985.0 kJ/kg
and the thermal efficiency of the cycle is
th = 1
q out
1985.0
= 1
= 0.297
q in
2823.2
10-22E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate,
the power produced by the turbine, the rate of heat addition, 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-4E, A-5E, and A-6E),
h1 = h f @ 1 psia = 69.72 Btu/lbm
v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm
wp,in = v 1 ( P2 P1 )
1 Btu
= (0.01614 ft 3 /lbm)(2500 1)psia
5.404 psia ft 3
= 7.46 Btu/lbm
h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm
2500 psia
2
qin
1 psia
1
qout
P3 = 2500 psia h3 = 1302.0 Btu/lbm
T3 = 800F
s 3 = 1.4116 Btu/lbm R
s 4 s f 1.4116 0.13262
P4 = 1 psia x 4 s =
=
= 0.6932
s fg
1.84495
s 4 = s3
h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm
4s
f
4 s fg
T =
4s 4
s
h3 h4
h4 = h3 T (h3 h4s ) = 1302.0 (0.90)(1302.0 787.70) = 839.13 kJ/kg
h3 h4 s
The mass flow rate of steam in the cycle is determined from
W& net
1000 kJ/s
0.94782 Btu
W& net = m& (h3 h4 )
m& =
=
= 2.048 lbm/s
h3 h4 (1302.0 839.13) Btu/lbm
1 kJ
The rate of heat addition is
1 kJ
Q& in = m& (h3 h2 ) = (2.048 lbm/s)(1302.0 77.18)Btu/lbm
= 2508 Btu/s
0.94782 Btu
and the thermal efficiency of the cycle is
th =
W& net
1000 kJ/s 0.94782 Btu
=
= 0.3779
&
2508 Btu/s
1 kJ
Qin
The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermal
efficiency caused by neglecting the pump work is then
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Error =
0.3779 0.3718
100 = 1.64%
0.3718
10-25 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),
h1 = h f @ 10 kPa = 191.81 kJ/kg
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v 1 (P2 P1 )
1 kJ
= 0.00101 m 3 /kg (7,000 10 kPa )
1 kPa m 3
= 7.06 kJ/kg
h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg
3
7 MPa
qin
10 kPa
1
qout
P3 = 7 MPa h3 = 3411.4 kJ/kg
T3 = 500C s 3 = 6.8000 kJ/kg K
s 4 s f 6.8000 0.6492
P4 = 10 kPa
=
= 0.8201
x4 =
s 4 = s3
s fg
7.4996
h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg
Thus,
q in = h3 h2 = 3411.4 198.87 = 3212.5 kJ/kg
q out = h4 h1 = 2153.6 191.81 = 1961.8 kJ/kg
wnet = q in q out = 3212.5 1961.8 = 1250.7 kJ/kg
and
th =
(b)
m& =
wnet 1250.7 kJ/kg
=
= 38.9%
3212.5 kJ/kg
q in
W&net
45,000 kJ/s
=
= 36.0 kg/s
wnet 1250.7 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
Q& out = m& q out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s
Q& out
70,586 kJ/s
Tcooling water =
=
= 8.4C
&
(mc) cooling water (2000 kg/s )(4.18 kJ/kg C )
PROPRIETARY MATERIAL. 2008 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.
10-42 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 h3 = 3476.5 kJ/kg
T3 = 550C
3
s 3 = 6.6317 kJ/kg K
Turbine
Boiler
P4 = 2 MPa
4
h4 s = 2948.1 kJ/kg
s 4s = s3
6
h3 h4
T =
5
Condenser
h3 h4 s
Pump
2
h4 = h3 T (h3 h4 s )
1
= 3476.5 (0.85)(3476.5 2948.1)
= 3027.3 kJ/kg
T
P5 = 2 MPa h5 = 3358.2 kJ/kg
3 5
T5 = 450C s 5 = 7.2815 kJ/kg K
12.5 MPa
P6 = ?
h
=
4s 4
6
x 6 = 0.95
2
2s
P6 = ?
h6 s =
s6 = s5
P=?
1
6s 6
h h
s
T = 5 6 h6 = h5 T (h5 h6 s )
h5 h6 s
= 3358.2 (0.85)(3358.2 2948.1)
= 3027.3 kJ/kg
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:
P6 = 9.73 kPa, h6 = 2463.3 kJ/kg,
(b) Then,
h1 = h f @ 9.73 kPa = 189.57 kJ/kg
v1 = v f @ 10 kPa = 0.001010 m3 /kg
w p ,in = v1 (P2 P1 ) / p
1 kJ
/ (0.90)
= 0.00101 m3 /kg (12,500 9.73 kPa )
3
1 kPa m
= 14.02 kJ/kg
h2 = h1 + w p ,in = 189.57 + 14.02 = 203.59 kJ/kg
Cycle analysis:
q in = (h3 h2 ) + (h5 h4 ) = 3476.5 3027.3 + 3358.2 2463.3 = 3603.8 kJ/kg
q out = h6 h1 = 3027.3 189.57 = 2273.7 kJ/kg
W& net = m& (q in q out ) = (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW
(c) The thermal efficiency is
q
2273.7 kJ/kg
th = 1 out = 1
= 0.369 = 36.9%
3603.8 kJ/kg
q in
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.
The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net
work input are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78F = 532.8 R and TL = Tsat @ 30 psia = 15.37F = 475.4 R.
COPR,C =
1
1
=
= 8.28
TH / TL 1 (532.8 R )/ (475.4 R ) 1
(b) Process 4-1 is isentropic, and thus
s1 = s 4 = s f + x 4 s fg
) @ 90 psia = 0.07481 + (0.05)(0.14525)
= 0.08207 Btu/lbm R
s1 s f
x1 =
s fg
QH
0.08207 0.03793
=
= 0.2374
0.18589
@ 30 psia
QL
(c) Remembering that on a T-s diagram the area enclosed
represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbmR,
wnet,in = (T H T L )(s 3 s 4 ) = (72.78 15.37)(0.22006 0.08207 ) Btu/lbm R = 7.92 Btu/lbm
11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is
considered. The COP and the power requirement are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the
refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser
as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13),
T1 = 4C h1 = h g @ 4C = 252.77 kJ/kg
sat. vapor s1 = s g @ 4C = 0.92927 kJ/kg K
P2 = 1 MPa
h2 = 275.29 kJ/kg
s 2 = s1
P3 = 1 MPa
h = hf
sat. liquid 3
@ 1 MPa
QH
3 1 MPa
Win
= 107.32 kJ/kg
h4 h3 = 107.32 kJ/kg ( throttling)
4C
4s
QL
The mass flow rate of the refrigerant is
Q& L = m& (h1 h4 )
m& =
Q& L
400 kJ/s
=
= 2.750 kg/s
h1 h4 (252.77 107.32) kJ/kg
The power requirement is
W& in = m& (h2 h1 ) = (2.750 kg/s)(275.29 252.77) kJ/kg = 61.93 kW
The COP of the refrigerator is determined from its definition,
COPR =
Q& L
400 kW
=
= 6.46
&
Win 61.93 kW
PROPRIETARY MATERIAL. 2008 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.
11-17 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is
considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the
rate of heat rejection to the environment, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the
refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser
as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),
P1 = 120 kPa h1 = h g @ 120 kPa = 236.97 kJ/kg
s = s
sat. vapor
g @ 120 kPa = 0.94779 kJ/kg K
1
P2 = 0.9 MPa
s 2 = s1
h2 = 278.93 kJ/kg (T2 = 44.45C )
P3 = 0.9 MPa
h3 = h f
sat. liquid
@ 0.9 MPa
QH
3 0.9 MPa
Win
= 101.61 kJ/kg
h4 h3 = 101.61 kJ/kg (throttling )
Then the rate of heat removal from the
refrigerated space and the power input to the
compressor are determined from
0.12 MPa
4s
QL
1
s
Q& L = m& (h1 h4 ) = (0.05 kg/s )(236.97 101.61) kJ/kg = 6.77 kW
and
W& in = m& (h2 h1 ) = (0.05 kg/s )(278.93 236.97 ) kJ/kg = 2.10 kW
(b) The rate of heat rejection to the environment is determined from
Q& H = Q& L + W& in = 6.77 + 2.10 = 8.87 kW
(c) The COP of the refrigerator is determined from its definition,
COPR =
Q& L
6.77 kW
=
= 3.23
W& in 2.10 kW
PROPRIETARY MATERIAL. 2008 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.
