thermo 7e sm chap10-1
TRANSCRIPT
10-1
Solutions Manual for
Thermodynamics: An Engineering Approach Seventh Edition
Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011
Chapter 10 VAPOR AND COMBINED POWER CYCLES
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10-2
Carnot Vapor Cycle
10-1C 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.
10-2E 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
15.5%==−=−=
=°===°==
1553.0R 861R 2.727
11
R 2.727F2.267R 861F401
Cth,
psia40@sat
psia 025@sat
H
L
L
H
TT
TTTT
η
T
40 psia
250 psia
3
2
4
1qin
(b) Noting that s4 = s1 = sf @ 250 psia = 0.56784 Btu/lbm·R,
0.137=−
=−
=2845.1
3921.056784.044
fg
f
sss
x s
(c) The enthalpies before and after the heat addition process are
( )( ) Btu/lbm 3.116047.82595.009.376Btu/lbm 09.376
22
psia 025 @1
=+=+===
fgf
f
hxhhhh
Thus,
and ( )( ) Btu/lbm 122===
=−=−=
Btu/lbm 2.7841553.0
Btu/lbm 2.78409.3763.1160
inthnet
12in
qw
hhq
η
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10-3
10-3 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 TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermal efficiency becomes
36.3%==−=−= 3632.0K 523K 333.1
11Cth,H
L
TT
η T
20 kPa3
2
4
1qin
qout
(b) The heat supplied during this cycle is simply the enthalpy of vaporization,
250°C
Thus,
( ) kJ/kg 1092.3=⎟⎟⎠
⎞⎜⎜⎝
⎛===
==
kJ/kg 3.1715K 523K 333.1
kJ/kg 3.1715
inout
250 @in
qTT
hq
H
LL
Cfg o
s(c) The net work output of this cycle is
( )( ) kJ/kg623.0 kJ/kg 3.17153632.0inthnet === qw η
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 TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermal efficiency becomes
39.04%=−=−=K 523K 318.811C th,
H
L
TT
η T
10 kPa 3
2
4
1 qin
qout
(b) The heat supplied during this cycle is simply the enthalpy of vaporization,
Thus,
( ) kJ/kg 1045.6=⎟⎟⎠
⎞⎜⎜⎝
⎛===
== °
kJ/kg 3.1715K 523K 318.8
kJ/kg 3.1715
inout
C250 @in
qTT
hq
H
LL
fg
250°C
(c) The net work output of this cycle is s ( )( ) kJ/kg 669.7=== kJ/kg 3.17153904.0inthnet qw η
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10-4
10-5 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
η th, C60 273 K350 273 K
= − = −++
=1 1T
TL
H
46.5% T
3
2
4
1 (b) Note that 350°Cs2 = s3 = sf + x3sfg
= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K
Thus, 60°C
(Table A-6) MPa 1.40≅⎭⎬⎫
⋅=°=
22
2
KkJ/kg 1368.7C350
PsT s
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
Thus,
( )( )
( )( ) ( )( ) kJ/kg 1623=−−=−−==
⋅=+=+=
5390.11368.760350Area
KkJ/kg 5390.10769.71.08313.0
43net
44
ssTTw
sxss
LH
fgf
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10-5
The Simple Rankine Cycle
10-6C 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.
10-7C Heat rejected decreases; everything else increases.
10-8C Heat rejected decreases; everything else increases.
10-9C The pump work remains the same, the moisture content decreases, everything else increases.
10-10C 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.
10-11C 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.
10-12C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam will adversely affect the turbine work output.
10-13C 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.
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10-6
10-14 A simple ideal Rankine cycle with R-134a as the working fluid operates between the specified pressure limits. The mass flow rate of R-134a for a given power production 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 refrigerant tables (Tables A-11, A-12, and A-13),
kJ/kg 89.6495.094.63kJ/kg 95.0 mkPa 1
kJ 1 kPa)4001600)(/kgm 0007907.0()(
/kgm 0007907.0
kJ/kg 94.63
inp,12
33
121inp,
3MPa 0.4 @1
MPa 4.0 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
qin
qout
0.4 MPa
1
3
2
4
1.6 MPa
T
s
kJ/kg 21.273 MPa 4.0
KkJ/kg 9875.0kJ/kg 07.305
C80MPa 6.1
434
4
3
3
3
3
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
hss
P
sh
TP
Thus,
kJ/kg 91.3027.20918.240
kJ/kg 27.20994.6321.273kJ/kg 18.24089.6407.305
outinnet
14out
23in
=−=−==−=−==−=−=
qqwhhqhhq
The mass flow rate of the refrigerant and the thermal efficiency of the cycle are then
kg/s 24.26===kJ/kg 91.30kJ/s 750
net
net
wW
m&
&
0.129=−=−=18.24027.20911
in
outth q
qη
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10-7
10-15 A simple ideal Rankine cycle with R-134a as the working fluid is considered. The turbine inlet temperature, the cycle thermal efficiency, and the back-work ratio of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kJ/kg 21.6678.043.65kJ/kg 78.0 mkPa 1
kJ 1 kPa)89.4141400)(/kgm 0007930.0()(
/kgm 0007930.0
kJ/kg 43.65
kPa 89.414
inp,12
33
121inp,
3C10 @1
C10 @1
C10 @sat 1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
==
°
°
°
whh
PPw
hh
PP
f
f
v
vv
qin
qout
10°C
1
3
2
4
1.4 MPa
T
s
KkJ/kg 91295.0)67356.0)(98.0(25286.0kJ/kg 35.252)73.190)(98.0(43.65
98.0
C10
44
44
4
4
⋅=+=+==+=+=
⎭⎬⎫
=°=
fgf
fgf
sxsshxhh
xT
C53.0°==
⎭⎬⎫
⋅===
3
3
43
3 kJ/kg 91.276
KkJ/kg 91295.0kPa 1400
Th
ssP
Thus,
kJ/kg 92.18643.6535.252kJ/kg 70.21021.6691.276
14out
23in
=−=−==−=−=
hhqhhq
The thermal efficiency of the cycle is
0.113=−=−=70.21092.18611
in
outth q
qη
The back-work ratio is determined from
0.0318=−
=−
==kJ/kg )35.25291.276(
kJ/kg 78.0
43
inP,
outT,
inP,
hhw
Ww
rbw
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10-8
10-16 A simple ideal Rankine cycle with water as the working fluid is considered. The work output from the turbine, the heat addition 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 From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg 18.17665.853.167kJ/kg 65.8 mkPa 1
kJ 1 kPa)385.78588)(/kgm 001008.0()(
/kgm 001008.0
kJ/kg 53.167
kPa 8588
kPa 385.7
inp,12
33
121inp,
3C40 @1
C40 @1
C300 @sat 2
C04 @sat 1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
==
==
°
°
°
°
whh
PPw
hh
PP
PP
f
f
v
vv
qin
qout
40°C
1
3 2
4
300°C
s
T
kJ/kg 1.1775)0.2406)(6681.0(53.167
6681.06832.7
5724.07059.5
C40
KkJ/kg 7059.5kJ/kg 6.2749
1
C300
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
=°=
⋅==
⎭⎬⎫
=°=
fgf
fg
f
hxhhs
ssx
ssT
sh
xT
Thus,
kJ/kg 6.160753.1671.1775
18.1766.27491.17756.2749
14out
23in
43outT,
=−=−==−=−=
=−=−=
hhqhhqhhw
kJ/kg 2573.4kJ/kg 974.5
The thermal efficiency of the cycle is
0.375=−=−=4.25736.160711
in
outth q
qη
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10-9
10-17E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, 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),
Btu/lbm 81.11140.240.109Btu/lbm 40.2
ftpsia 5.404Btu 1 psia)3800)(/lbmft 01630.0(
)(
/lbmft 01630.0
Btu/lbm 40.109
inp,12
33
121inp,
3psia 3 @1
psia 3 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 24.975)8.1012)(8549.0(40.109
8549.06849.1
2009.06413.1
psia 3
RBtu/lbm 6413.1Btu/lbm 0.1456
F900psia 800
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
qin
qout
3 psia 1
3
2
4
800 psia
T
s
Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from
lbm/s 450.3kJ 1
Btu 0.94782)Btu/lbm24.975(1456.0
kJ/s 1750)(43
outT,43outT, =⎟
⎠⎞
⎜⎝⎛
−=
−=⎯→⎯−=
hhW
mhhmW&
&&&
The rates of heat addition and rejection are
Btu/s 2987Btu/s 4637
=−=−=
=−=−=
Btu/lbm)40.109.24lbm/s)(975 450.3()(
Btu/lbm)81.1110.6lbm/s)(145 450.3()(
14out
23in
hhmQ
hhmQ&&
&&
and the thermal efficiency of the cycle is
35.6%==−=−= 3559.04637298711
in
outth Q
Q&
&η
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10-10
10-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The turbine inlet temperature 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.
qin
qout
5 psia 1
3
2
4
2500 psia
T
s
Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),
Btu/lbm 76.13758.718.130Btu/lbm 58.7
ftpsia 5.404Btu 1 psia)52500)(/lbmft 01641.0(
)(
/lbmft 01641.0
Btu/lbm 18.130
inp,12
33
121inp,
3psia 5 @1
psia 5 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
RBtu/lbm 52203.1)60894.1)(80.0(23488.0Btu/lbm 58.930)5.1000)(80.0(18.130
80.0psia 5
44
44
4
4
⋅=+=+==+=+=
⎭⎬⎫
==
fgf
fgf
sxsshxhh
xP
F989.2°==
⎭⎬⎫
⋅===
3
3
43
3 Btu/lbm 8.1450
RBtu/lbm 52203.1psia 2500
Th
ssP
Thus,
Btu/lbm 4.80018.13058.930Btu/lbm 1313.076.1378.1450
14out
23in
=−=−==−=−=
hhqhhq
The thermal efficiency of the cycle is
0.390=−=−=0.13134.80011
in
outth q
qη
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10-11
10-19E 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),
Btu/lbm 18.7746.772.69Btu/lbm 46.7
ftpsia 5.404Btu 1 psia)12500)(/lbmft 01614.0(
)(
/lbmft 01614.0
Btu/lbm 72.69
inp,12
33
121inp,
3psia 6 @1
psia 1 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
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Btu/lbm 70.787)7.1035)(6932.0(72.69
6932.084495.1
13262.04116.1
psia 1
RBtu/lbm 4116.1Btu/lbm 0.1302
F800psia 2500
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fs
hxhhs
ssx
ssP
sh
TP
qin
qout
1 psia 1
3
2
4
2500 psia
4s
s
T
kJ/kg 13.839)70.7870.1302)(90.0(0.1302)( 4s3T3443
43T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
Thus,
Btu/lbm 39.45541.7698.1224
Btu/lbm 41.76972.6913.839Btu/lbm 8.122418.771302.0
outinnet
14out
23in
=−=−==−=−==−=−=
qqwhhqhhq
The mass flow rate of steam in the cycle is determined from
lbm/s 2.081=⎟⎠⎞
⎜⎝⎛==⎯→⎯=
kJ 1Btu 0.94782
Btu/lbm 455.39kJ/s 1000
net
netnetnet w
WmwmW
&&&&
The power output from the turbine and the rate of heat addition are
Btu/s 2549
kW 1016
===
=⎟⎠⎞
⎜⎝⎛−=−=
Btu/lbm) 4.8lbm/s)(122 081.2(
Btu 0.94782kJ 1Btu/lbm)13.8392.0lbm/s)(130 081.2()(
inin
43outT,
qmQ
hhmW
&&
&&
and the thermal efficiency of the cycle is
0.3718=⎟⎠⎞
⎜⎝⎛==
kJ 1Btu 0.94782
Btu/s 2549kJ/s 1000
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
10-12
10-20E 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),
Btu/lbm 18.7746.772.69Btu/lbm 46.7
ftpsia 5.404Btu 1 psia)12500)(/lbmft 01614.0(
)(
/lbmft 01614.0
Btu/lbm 72.69
inp,12
33
121inp,
3psia 6 @1
psia 1 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Btu/lbm 70.787)7.1035)(6932.0(72.69
6932.084495.1
13262.04116.1
psia 1
RBtu/lbm 4116.1Btu/lbm 0.1302
F800psia 2500
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fs
hxhhs
ssx
ssP
sh
TP
qin
qout
1 psia 1
3
2
4
2500 psia
4s
s
T
kJ/kg 13.839)70.7870.1302)(90.0(0.1302)( 4s3T3443
43T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
The mass flow rate of steam in the cycle is determined from
lbm/s 2.048kJ 1
Btu 0.94782Btu/lbm 839.13)(1302.0
kJ/s 1000)(43
net43net =⎟
⎠⎞
⎜⎝⎛
−=
−=⎯→⎯−=
hhW
mhhmW&
&&&
The rate of heat addition is
Btu/s 2508Btu 0.94782
kJ 1Btu/lbm)18.772.0lbm/s)(130 048.2()( 23in =⎟⎠⎞
⎜⎝⎛−=−= hhmQ &&
and the thermal efficiency of the cycle is
0.3779kJ 1
Btu 0.94782Btu/s 2508kJ/s 1000
in
netth =⎟
⎠⎞
⎜⎝⎛==
QW&
&η
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
1.64%=×−
= 1003718.0
3718.03779.0Error
preparation. If you are a student using this Manual, you are using it without permission.
10-13
10-21 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),
( )
( )( )
kJ/kg 198.8706.781.191
kJ/kg 7.06mkPa 1
kJ 1kPa 107,000/kgm 0.00101
/kgm 00101.0
kJ/kg .81191
in,12
33
121in,
3kPa 10 @1
kPa 10 @1
=+=+=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( ) kJ/kg 3.62151.23928201.081.191
8201.04996.7
6492.08000.6kPa 10
KkJ/kg 8000.6kJ/kg 411.43
C500MPa 7
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhh
sss
xss
P
sh
TP
qin
qout
10 kPa 1
3
2
4
7 MPa
T
s
Thus,
kJ/kg 7.12508.19615.3212kJ/kg 8.196181.1916.2153kJ/kg 5.321287.1984.3411
outinnet
14out
23in
=−=−=
=−=−=
=−=−=
qqwhhqhhq
and
38.9%===kJ/kg 3212.5kJ/kg 1250.7
in
netth q
wη
(b) skg 036 /.kJ/kg 1250.7kJ/s 45,000
net
net ===wWm&
&
(c) The rate of heat rejection to the cooling water and its temperature rise are
( )( )
( )( ) C8.4°=°⋅
==∆
===
CkJ/kg 4.18kg/s 2000kJ/s 70,586
)(
kJ/s ,58670kJ/kg 1961.8kg/s 35.98
watercooling
outwatercooling
outout
cmQ
T
qmQ
&
&
&&
preparation. If you are a student using this Manual, you are using it without permission.
10-14
10-22 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),
( )( )( ) ( )
kJ/kg 9.921911.881.191kJ/kg 8.11
0.87/mkPa 1
kJ 1kPa 107,000/kgm 00101.0
/
/kgm 00101.0
kJ/kg 91.811
in,12
33
121in,
3kPa 10 @1
kPa 10 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
pp
f
f
whh
PPw
hh
ηv
vv
T
( )( )
( )( )( ) kJ/kg 1.23176.21534.341187.04.3411
kJ/kg 6.21531.2392820.081.191
8201.04996.7
6492.08000.6kPa 10
KkJ/kg 8000.6kJ/kg .43411
C500MPa 7
433443
43
44
44
34
4
3
3
3
3
=−−=−−=⎯→⎯
−−
=
=+=+=
=−
=−
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
sTs
T
fgfs
fg
f
hhhhhhhh
hxhh
sss
xss
P
sh
TP
ηη
10 kPa 1
3
2
4
7 MPa
s
qout
qin2
4
Thus,
kJ/kg 2.10863.21255.3211kJ/kg 3.212581.1911.2317kJ/kg 5.321192.1994.3411
outinnet
14out
23in
=−=−==−=−==−=−=
qqwhhqhhq
and
33.8%===kJ/kg 3211.5kJ/kg 1086.2
in
netth q
wη
(b) skg 4341 /.kJ/kg 1086.2kJ/s 45,000
net
net ===wWm&
&
(c) The rate of heat rejection to the cooling water and its temperature rise are
( )( )
( )( ) C10.5°=°⋅
==∆
===
CkJ/kg 4.18kg/s 2000kJ/s 88,051
)(
kJ/s ,05188kJ/kg 2125.3kg/s 41.43
watercooling
outwatercooling
outout
cmQ
T
qmQ
&
&
&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-15
10-23 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The rate of heat addition in the boiler, the power input to the pumps, the net power, 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),
/kgm 001026.0kJ/kg 03.314
C753.63.813.6
kPa 503
C75 @1
C75 @ 1
kPa 50 @sat 1
1
==
=≅
⎭⎬⎫
°=−=−==
°
°
f
fhhTT
Pvv
kJ/kg 10.6 mkPa 1kJ 1 kPa)506000)(/kgm 001026.0(
)(
33
121inp,
=⎟⎠
⎞⎜⎝
⎛
⋅−=
−= PPw v
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kJ/kg 13.32010.603.314inp,12 =+=+= whh
kJ/kg 4.2336)7.2304)(8660.0(54.340
8660.05019.6
0912.17219.6
kPa 50
KkJ/kg 7219.6kJ/kg 9.3302
C450kPa 6000
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fs
hxhhs
ssx
ssP
sh
TP
qin
qout
50 kPa 1
3
2
4
6 MPa
4s
s
T
kJ/kg 4.2394)4.23369.3302)(94.0(9.3302)( 4s3T3443
43T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
Thus,
kW 18,050
kW 122
kW 59,660
=−=−=
===
=−=−=
=−=−=
122170,18
kJ/kg) kg/s)(6.10 20(
kW 18,170kJ/kg)4.2394.9kg/s)(3302 20()(
kJ/kg)13.320.9kg/s)(3302 20()(
inP,outT,net
inP,inP,
43outT,
23in
WWW
wmW
hhmW
hhmQ
&&&
&&
&&
&&
and
0.3025===660,59050,18
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
10-16
10-24 The change in the thermal efficiency of the cycle in Prob. 10-23 due to a pressure drop in the boiler is to be determined.
Analysis We use the following EES routine to obtain the solution.
"Given" P[2]=6000 [kPa] DELTAP=50 [kPa] P[3]=6000-DELTAP [kPa] T[3]=450 [C] P[4]=50 [kPa] Eta_T=0.94 DELTAT_subcool=6.3 [C] T[1]=temperature(Fluid$, P=P[1], x=x[1])-DELTAT_subcool m_dot=20 [kg/s] "Analysis" Fluid$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) v[1]=volume(Fluid$, P=P[1], T=T[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in Solution DELTAP=50 [kPa] DELTAT_subcool=6.3 [C] Eta_T=0.94 Eta_th=0.3022 Fluid$='steam_iapws' h[1]=314.11 [kJ/kg] h[2]=320.21 [kJ/kg] h[3]=3303.64 [kJ/kg] h[4]=2396.01 [kJ/kg] h_s[4]=2338.1 [kJ/kg] m_dot=20 [kg/s] P[1]=50 P[2]=6000 P[3]=5950 P[4]=50 q_in=2983.4 [kJ/kg] q_out=2081.9 [kJ/kg] s[3]=6.7265 [kJ/kg-K] s[4]=6.7265 [kJ/kg-K] T[1]=75.02 [C] T[3]=450 [C] v[1]=0.001026 [m^3/kg] w_net=901.5 [kJ/kg] w_p_in=6.104 [kJ/kg] x[1]=0
Discussion The thermal efficiency without a pressure drop was obtained to be 0.3025.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-17
10-25 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),
( )( )( )
kJ/kg 64.34510.554.340kJ/kg 10.5
mkPa 1kJ 1kPa 505000/kgm 001030.0
/kgm 001030.0
kJ/kg 54.340
in,12
33
121in,
3kPa 50 @1
kPa 05 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
T
1
2
3
4
Rankine cycle
s
( )(kJ/kg 2.2071
7.23047509.054.340
7509.05019.6
09120.19737.5kPa 50
KkJ/kg 9737.5kJ/kg 2.2794
1MPa 5
44
44
34
4
3
3
3
3
=
+=+=
=−
=−
=⎭⎬⎫
==
⋅==
⎭⎬⎫
==
fgf
fg
f
hxhh
sss
xss
P
sh
xP
)
kJ/kg 717.9=−=−=
=−=−==−=−=
7.17306.2448kJ/kg 7.173054.3402.2071kJ/kg 6.244864.3452.2794
outinnet
14out
23in
qqwhhqhhq
29.3%==−=−= 2932.06.24487.173011
in
outth q
qη
(b) Carnot Cycle analysis:
T
1
2 3
4
Carnot cycle
skJ/kg 05.989)7.2304)(2814.0(54.340
2814.05019.6
0912.19207.2kPa 50
KkJ/kg 9207.2kJ/kg 5.1154
0C9.263
C9.263kJ/kg 2.2794
1MPa 5
11
11
21
1
2
2
2
32
3
3
3
3
=+=
+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
=°==
°==
⎭⎬⎫
==
fgf
fg
f
hxhhs
ssx
ssP
sh
xTT
Th
xP
kJ/kg 557.5=−=−==−=−==−=−=
2.10827.1639kJ/kg 2.108254.3402.2071kJ/kg 7.16395.11542.2794
outinnet
14out
23in
qqwhhqhhq
34.0%==−=−= 3400.07.16392.108211
in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-18
10-26 A 120-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),
( )( )( )
kJ/kg .0523511.994.225kJ/kg .119
mkPa 1kJ 1kPa 519000/kgm 001014.0
/kgm 0010140.0
kJ/kg 94.225
in,12
33
121in,
3kPa 51 @1
kPa 15 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
T
( )( ) kJ/kg 8.22084.23728358.094.225
8358.02522.7
7549.08164.6kPa 15
KkJ/kg 8164.6kJ/kg 0.3512
C550MPa 9
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhh
sss
xss
P
sh
TP
Qin
Qout
15 kPa 1
3
2
4
9 MPa
·
·
s
The thermal efficiency is determined from
kJ/kg 9.198294.2258.2208kJ/kg 9.327605.2350.3512
14out
23in=−=−==−=−=
hhqhhq
and
Thus, ( )( )( ) 28.4%===××=
=−=−=
2843.096.075.03949.0
3949.09.32769.198211
gencombthoverall
in
outth
ηηηη
ηqq
(b) Then the required rate of coal supply becomes
and
tons/h 51.9====
===
kg/s 404.14kJ/kg 29,300
kJ/s 422,050
kJ/s 050,4222843.0
kJ/s 120,000
coal
incoal
overall
netin
CQ
m
WQ
&&
&&
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-19
10-27 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)
1661.0
210809.64014.990
kJ/kg 14.990kPa 500
kJ/kg 14.9900
C230
22
12
2
11
1
=−
=−
=⎭⎬⎫
===
=⎭⎬⎫
=°=
fg
f
hhh
xhh
P
hxT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
The mass flow rate of steam through the turbine is
kg/s 38.20=== kg/s) 230)(1661.0(123 mxm &&
(b) Turbine:
kJ/kg 7.2344)1.2392)(90.0(81.19190.0kPa 10
kJ/kg 3.2160kPa 10
KkJ/kg 8207.6kJ/kg 1.2748
1kPa 500
444
4
434
4
3
3
3
3
=+=+=⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
==
fgf
s
hxhhxP
hss
P
sh
xP
production well
2
reinjection well
separator
steam turbine
1
Flash chamber
65
4
3
condenser
0.686=−−
=−−
=3.21601.27487.23441.2748
43
43
sT hh
hhη
(c) The power output from the turbine is
kW 15,410=−=−= kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT, hhmW &&
(d) We use saturated liquid state at the standard temperature for dead state enthalpy
kJ/kg 83.1040
C250
0
0 =⎭⎬⎫
=°=
hxT
kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in =−=−= hhmE &&
7.6%==== 0.0757622,203
410,15
in
outT,th E
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
10-20
10-28 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)
1661.0kJ/kg 14.990
kPa 500
kJ/kg 14.9900
C230
212
2
11
1
=⎭⎬⎫
===
=⎭⎬⎫
=°=
xhh
P
hxT
kg/s 80.1911661.0230kg/s 38.20kg/s) 230)(1661.0(
316
123
=−=−====
mmmmxm
&&&
&&
productionwell
reinjectionwell
separator
steam turbine
1
3
condenser
4
5
6
Flash chamber
Flash chamber
9
separator
8
7kJ/kg 7.234490.0kPa 10
kJ/kg 1.27481
kPa 500
44
4
33
3
=⎭⎬⎫
==
=⎭⎬⎫
==
hxP
hxP
2
kJ/kg 1.26931
kPa 150
0777.0kPa 150
kJ/kg 09.6400
kPa 500
88
8
7
7
67
7
66
6
=⎭⎬⎫
==
=°=
⎭⎬⎫
==
=⎭⎬⎫
==
hxP
xT
hhP
hxP
C 111.35
(b) The mass flow rate at the lower stage of the turbine is
kg/s .9014kg/s) 80.191)(0777.0(678 === mxm &&
The power outputs from the high and low pressure stages of the turbine are
kW 15,410kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT1, =−=−= hhmW &&
kW 5191=−=−= kJ/kg)7.23443.1kJ/kg)(269 .9014()( 488outT2, hhmW &&
(c) We use saturated liquid state at the standard temperature for the dead state enthalpy
kJ/kg 83.1040
C250
0
0 =⎭⎬⎫
=°=
hxT
kW 621,203kJ/kg)83.10414kg/s)(990. 230()( 011in =−=−= hhmE &&
10.1%==+
== 0.101621,203
5193410,15
in
outT,th E
W&
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-21
10-29 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)
1661.0
kJ/kg 14.990kPa 500
kJ/kg 14.9900
C230
212
2
11
1
=⎭⎬⎫
===
=⎭⎬⎫
=°=
xhh
P
hxT
kg/s 80.19120.38230
kg/s 38.20kg/s) 230)(1661.0(
316
123
=−=−====
mmmmxm
&&&
&&
kJ/kg 7.234490.0kPa 10
kJ/kg 1.27481
kPa 500
44
4
33
3
=⎭⎬⎫
==
=⎭⎬⎫
==
hxP
hxP
steam turbine
isobutaneturbine
heat exchanger
pump
BINARY CYCLE
separator
air-cooled condenser
condenser
flash chamber
1
91
8
7
5
2
6
3
4
1reinjection well
production well
kJ/kg 04.377
0C90
kJ/kg 09.6400
kPa 500
77
7
66
6
=⎭⎬⎫
=°=
=⎭⎬⎫
==
hxT
hxP
The isobutane properties are obtained from EES:
/kgm 001839.0kJ/kg 83.270
0kPa 400
kJ/kg 01.691C80kPa 400
kJ/kg 05.755C145kPa 3250
310
10
10
10
99
9
88
8
==
⎭⎬⎫
==
=⎭⎬⎫
°==
=⎭⎬⎫
°==
v
hxP
hTP
hTP
( )( )( )
kJ/kg 65.27682.583.270kJ/kg. 82.5
90.0/mkPa 1
kJ 1kPa 4003250/kgm 001819.0
/
in,1011
33
101110in,
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
p
pp
whh
PPw ηv
An energy balance on the heat exchanger gives
kg/s 105.46=⎯→⎯=
−=−
isoiso
118iso766
kg276.65)kJ/-(755.05kg377.04)kJ/-09kg/s)(640. 81.191(
)()(
mm
hhmhhm
&&
&&
(b) The power outputs from the steam turbine and the binary cycle are
kW 15,410=−=−= kJ/kg)7.23448.1kJ/kg)(274 38.19()( 433steamT, hhmW &&
kW 6139=−=−=
=−=−=
)kJ/kg 82.5)(kg/s 46.105(6753
kW 6753kJ/kg)01.691.05kJ/kg)(755 .46105()(
,isoisoT,binarynet,
98isoT,
inp
iso
wmWW
hhmW
&&&
&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-22
(c) The thermal efficiencies of the binary cycle and the combined plant are
kW 454,50kJ/kg)65.276.05kJ/kg)(755 .46105()( 118isobinaryin, =−=−= hhmQ &&
12.2%==== 0.122454,50
6139
binaryin,
binarynet,binaryth, Q
W&
&η
kJ/kg 83.1040
C250
0
0 =⎭⎬⎫
=°=
hxT
kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in =−=−= hhmE &&
10.6%==+
=+
= 0.106622,203
6139410,15
in
binarynet,steamT,plantth, E
WW&
&&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-23
The Reheat Rankine Cycle
10-30C The pump work remains the same, the moisture content decreases, everything else increases.
10-31C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the low-pressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%.
0 20 40 60 80 100 120 140 160 180200
300
400
500
600
700
800
900
s [kJ/kmol-K]
T [K
]
6000 kPa 4000 kPa
500 kPa
20 kPa 0.2 0.4 0.6 0.8
SteamIAPWS
1
2
3
4
5
6
7
10-32C 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.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-24
10-33 An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, 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),
kJ/kg 88.19907.881.191kJ/kg 07.8 mkPa 1
kJ 1 kPa)108000)(/kgm 001010.0()(
/kgm 001010.0
kJ/kg 81.191
inp,12
33
121inp,
3kPa 10 @1
kPa 10 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
T
kJ/kg 4.2636)0.2108)(9470.0(09.640
9470.09603.4
8604.15579.6
kPa 500
KkJ/kg 5579.6kJ/kg 3.3273
C045kPa 8000
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
1
5
2
6
3
4
500 kPa
10 kPa
8 MPa
s
kJ/kg 9.2564)1.2392)(9921.0(81.191
9921.04996.7
6492.00893.8
kPa 10
KkJ/kg 0893.8kJ/kg 5.3484
C500kPa 500
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Thus,
kJ/kg 5.15481.23735.3921
kJ/kg 1.237381.1919.2564kJ/kg 5.39214.26365.348488.1993.3273)()(
outinnet
16out
4523in
=−=−==−=−=
=−+−=−+−=
qqwhhq
hhhhq
The mass flow rate of steam in the cycle is determined from
kg/s 3.229===⎯→⎯−=kJ/kg 1548.5kJ/s 5000)(
net
net43net w
WmhhmW
&&&&
and the thermal efficiency of the cycle is
0.395=−=−=5.39211.237311
in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-25
10-34 An ideal reheat steam Rankine cycle produces 2000 kW power. The mass flow rate of the steam, the rate of heat transfer in the reheater, the power used by the pumps, 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 or EES),
kJ/kg 05.43354.1551.417kJ/kg 54.15 mkPa 1
kJ 1 kPa)10015000)(/kgm 001043.0()(
/kgm 001043.0
kJ/kg 51.417
inp,12
33
121inp,
3kPa 100 @1
kPa 100 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
T
kJ/kg 3.2703)8.1889)(9497.0(47.908
9497.08923.3
4467.21434.6
kPa 2000
KkJ/kg 1434.6kJ/kg 9.3157
C045
kPa 000,15
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
1
5
2
6
4
2 MPa
100 kPa
15 MPa 3
s
kJ/kg 0.2648)5.22257)(9880.0(51.417
9880.00562.6
3028.12866.7
kPa 100
KkJ/kg 2866.7kJ/kg 2.3358
C450kPa 2000
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Thus,
kJ/kg 2.11495.22308.379
kJ/kg 5.223051.4170.2648kJ/kg 8.33793.27032.335805.4339.3157)()(
outinnet
16out
4523in
=−=−==−=−=
=−+−=−+−=
qqwhhq
hhhhq
The power produced by the cycle is
kW 2000=== )kJ/kg 1149.2)(kg/s 1.74(netnet wmW &&
The rate of heat transfer in the rehetaer is
kW 27
kW 1140===
=−=−=
kJ/kg) 4kg/s)(15.5 740.1(
kJ/kg )3.2703.2kg/s)(3358 740.1()(
inP,inP,
45reheater
wmW
hhmQ&&
&&
and the thermal efficiency of the cycle is
0.340=−=−=8.33795.223011
in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-26
10-35 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),
( )( )( )
kJ/kg .5025708.642.251
kJ/kg .086mkPa 1
kJ 1kPa 206000/kgm 001017.0
/kgm 700101.0
kJ/kg 42.251
in,12
33
121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
T
( )( ) kJ/kg 7.23495.23578900.042.251
8900.00752.7
8320.01292.7kPa 20
KkJ/kg 1292.7kJ/kg 4.3248
C400MPa 2
kJ/kg 0.2901MPa 2
KkJ/kg 5432.6kJ/kg 3.3178
C400MPa 6
66
66
56
6
5
5
5
5
434
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=⎭⎬⎫
==
⋅=
=
⎭⎬⎫
°=
=
fgf
fg
f
hxhh
sss
x
ssP
sh
TP
hss
P
sh
TP
1
5
2
6
3
4
20 kPa
6 MPa
s
The turbine work output and the thermal efficiency are determined from
and ( ) ( )
( ) ( ) kJ/kg 32680.29014.324850.2573.3178
7.23494.32480.29013.3178
4523in
6543outT,
=−+−=−+−=
=−+−=−+−=
hhhhq
hhhhw kJ/kg 1176
Thus,
35.8%====
=−=−=
358.0kJ/kg 3268kJ/kg 1170
kJ/kg 117008.61176
in
netth
in,,net
qw
www poutT
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-27
10-36 Problem 10-35 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] = 6000 [kPa] T[3] = 400 [C] P[4] = 2000 [kPa] T[5] = 400 [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$='(superheated)' if (x6<0) then x6$='(subcooled)' end Fluid$='Steam_IAPWS' P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,P=P[2],h=h[2]) s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) v[3]=volume(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,T=T[4],P=P[4]) v[4]=volume(Fluid$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" P[5]=P[4] s[5]=entropy(Fluid$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6]) vs[6]=volume(Fluid$,s=s_s[6],P=P[6]) Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"
preparation. If you are a student using this Manual, you are using it without permission.
10-28
x[6]=QUALITY(Fluid$,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
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
6000 kPa
2000 kPa
20 kPa
SteamIAPWS
1,2
3
4
5
6
Ideal Rankine cycle with reheat
SOLUTION Eff=0.358 Eta_p=1 Eta_t=1 Fluid$='Steam_IAPWS' Q_in=3268 [kJ/kg] Q_out=2098 [kJ/kg] W_net=1170 [kJ/kg] W_p=6.083 [kJ/kg] W_p_s=6.083 [kJ/kg] W_t_hp=277.2 [kJ/kg] W_t_lp=898.7 [kJ/kg] x6s$=''
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-29
10-37E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, 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 or EES),
Btu/lbm 06.16381.125.161Btu/lbm 81.1
ftpsia 5.404Btu 1 psia)10600)(/lbmft 01659.0(
)(
/lbmft 01659.0
Btu/lbm 25.161
inp,12
33
121inp,
3psia 10 @1
psia 10 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
T
Btu/lbm 5.1187)33.843)(9865.0(46.355
9865.000219.1
54379.05325.1
psia 200
RBtu/lbm 5325.1Btu/lbm 9.1289
F600psia 600
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
1
5
2
6
4
200 psia
10 psia
600 psia 3
s
Btu/lbm 0.1071)82.981)(9266.0(25.161
9266.050391.1
28362.06771.1
psia 10
RBtu/lbm 6771.1Btu/lbm 3.1322
F600psia 200
66
46
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Thus,
Btu/lbm 0.3528.9097.1261
Btu/lbm 7.90925.1610.1071Btu/lbm 7.12615.11873.132206.1631289.9)()(
outinnet
16out
4523in
=−=−==−=−=
=−+−=−+−=
qqwhhq
hhhhq
The mass flow rate of steam in the cycle is determined from
lbm/s 47.13kJ 1
Btu 0.94782Btu/lbm 352.0
kJ/s 5000
net
netnetnet =⎟
⎠⎞
⎜⎝⎛==⎯→⎯=
wW
mwmW&
&&&
The rates of heat addition and rejection are
Btu/s 12,250Btu/s 16,995
===
===
Btu/lbm) .7lbm/s)(909 47.13(
Btu/lbm) 1.7lbm/s)(126 47.13(
outout
inin
qmQ
qmQ&&
&&
and the thermal efficiency of the cycle is
0.2790=⎟⎠⎞
⎜⎝⎛==
kJ 1Btu 0.94782
Btu/s 16,990kJ/s 5000
in
netth Q
W&
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-30
10-38E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined for a reheat pressure of 100 psia.
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 or EES),
Btu/lbm 06.16381.125.161Btu/lbm 81.1
ftpsia 5.404Btu 1 psia)10600)(/lbmft 01659.0(
)(
/lbmft 01659.0
Btu/lbm 25.161
inp,12
33
121inp,
3psia 6 @1
psia 10 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
T
Btu/lbm 9.1131)99.888)(9374.0(51.298
9374.012888.1
47427.05325.1
psia 100
RBtu/lbm 5325.1Btu/lbm 9.1289
F600psia 600
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
1
5
2
6
4
100 psia
10 psia
600 psia 3
s
Btu/lbm 2.1124)82.981)(9808.0(25.161
9808.050391.1
28362.07586.1
psia 10
RBtu/lbm 7586.1Btu/lbm 4.1329
F600
psia 100
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Thus,
Btu/lbm 5.3619.9624.1324
Btu/lbm 9.96225.1612.1124Btu/lbm 4.13249.11314.132907.1631289.9)()(
outinnet
16out
4523in
=−=−==−=−=
=−+−=−+−=
qqwhhq
hhhhq
The mass flow rate of steam in the cycle is determined from
lbm/s 11.13kJ 1
Btu 0.94782Btu/lbm 361.5
kJ/s 5000
net
netnetnet =⎟
⎠⎞
⎜⎝⎛==⎯→⎯=
wW
mwmW&
&&&
The rates of heat addition and rejection are
Btu/s 12,620Btu/s 17,360
===
===
Btu/lbm) .9lbm/s)(962 11.13(
Btu/lbm) 4.4lbm/s)(132 11.13(
outout
inin
qmQ
qmQ&&
&&
and the thermal efficiency of the cycle is
0.2729=⎟⎠⎞
⎜⎝⎛==
kJ 1Btu 0.94782
Btu/s 17,360kJ/s 5000
in
netth Q
W&
&η
Discussion The thermal efficiency for 200 psia reheat pressure was determined in the previous problem to be 0.2790. Thus, operating the reheater at 100 psia causes a slight decrease in the thermal efficiency.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-31
10-39 An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inlet of both turbines, 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.
T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg 87.19806.781.191kJ/kg 06.7 mkPa 1
kJ 1 kPa)107000)(/kgm 001010.0()(
/kgm 001010.0
kJ/kg 81.191
inp,12
33
121inp,
3kPa 10 @1
kPa 10 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛⋅
−=
−=
==
==
whh
PPw
hh
f
f
v
vv
C373.3°==
⎭⎬⎫
==
⋅=+=+==+=+=
⎭⎬⎫
==
3
3
43
3
44
44
4
4
kJ/kg 5.3085
kPa 7000
KkJ/kg 3385.6)6160.4)(93.0(0457.2kJ/kg 0.2625)5.2047)(93.0(87.720
93.0
kPa 800
Th
ssP
sxsshxhh
xP
fgf
fgf
1
5
2
6
3
4
800 kPa
10 kPa
7 MPa
s
C416.2°==
⎭⎬⎫
==
⋅=+=+==+=+=
⎭⎬⎫
==
5
5
65
5
66
66
6
6
kJ/kg 0.3302
kPa 800
KkJ/kg 6239.7)4996.7)(93.0(6492.0kJ/kg 4.2416)1.2392)(93.0(81.191
90.0kPa 10
Th
ssP
sxsshxhh
xP
fgf
fgf
Thus,
kJ/kg 6.222481.1914.2416
kJ/kg 6.35630.26250.330287.1985.3085)()(
16out
4523in
=−=−==−+−=−+−=
hhqhhhhq
and
37.6%==−=−= 3757.06.35636.222411
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
10-32
10-40 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),
T
( )( )( )
kJ/kg 95.20614.1581.191
kJ/kg .1415mkPa 1
kJ 1kPa 10000,15/kgm 00101.0
/kgm 00101.0
kJ/kg 81.191
in,12
33
121in,
3kPa 10 @sat1
kPa 10 @sat1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
whh
PPw
hh
v
vv
( )( )( )( )
( )
kJ/kg 2.2817MPa 15.2
kJ/kg 61.3466pressurereheat theC500
KkJ/kg 3988.74996.790.06492.0
kJ/kg 7.23441.239290.081.191kPa 10
KkJ/kg 3480.6kJ/kg .83310
C500MPa 15
434
4
5
5
65
5
66
66
56
6
3
3
3
3
=⎭⎬⎫
==
==
⎭⎬⎫
=°=
⋅=+=+=
=+=+=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
hss
P
hP
ssT
sxss
hxhh
ssP
sh
TP
fgf
fgf
kPa 2150
1
5
2
6
3
4
10 kPa
15
s
(b) The rate of heat supply is
( ) ( )[ ]
( )( )kW 45,039=
−+−=−+−=
kJ/kg2.281761.346695.2068.3310kg/s 124523in hhhhmQ &&
(c) The thermal efficiency is determined from
Thus, ( ) ( )( )
42.6%=−=−=
=−=−=
kJ/s 45,039kJ/s 25,834
11
kJ/s ,83525kJ/kg81.1917.2344kJ/s 12
in
outth
16out
hhmQ
&
&
&&
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-33
10-41 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),
( )( )( )
( ) ( )( ) 3) (Eq. 2.335885.02.3358
2) (Eq. ?
1) (Eq. 95.0
?
KkJ/kg 2815.7kJ/kg 2.3358
C450 MPa2
kJ/kg 3.30271.29485.347685.05.3476
kJ/kg 1.2948 MPa2
KkJ/kg 6317.6kJ/kg 5.3476
C550 MPa5.12
6655665
65
656
6
66
6
5
5
5
5
4334
43
43
434
4
3
3
3
3
ssTs
T
s
sT
sT
ss
hhhhhhhhh
hss
P
hxP
sh
TP
hhhh
hhhh
hss
P
sh
TP
−−=−−=⎯→⎯−−
=
=⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=−−=
−−=→
−−
=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
ηη
η
η
1
5
2s
6s
3
4s 12.5 MPa
P = ? 6
4
2
T
3
1 2
Turbine Boiler
CondenserPump
5
4
6
s
The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above three equations:
P6 = 9.73 kPa, h6 = 2463.3 kJ/kg,
(b) Then,
( )( )( ) ( )
kJ/kg 59.20302.1457.189
kJ/kg 14.020.90/
mkPa 1kJ 1kPa 73.912,500/kgm 0.00101
/
/kgm 001010.0
kJ/kg 57.189
in,12
33
121in,
3kPa 10 @1
kPa 73.9 @1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
pp
f
f
whh
PPw
hh
ηv
vv
Cycle analysis:
( ) ( )
kW 10,242==−=
=−=−=
=−+−=−+−=
kg2273.7)kJ/-.8kg/s)(3603 7.7()(
kJ/kg 7.227357.1893.2463
kJ/kg 8.36033.24632.335859.2035.3476
outinnet
16out
4523in
qqmW
hhq
hhhhq
&&
(c) The thermal efficiency is
36.9%==−=−= 369.0kJ/kg 3603.8kJ/kg 2273.7
11in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-34
Regenerative Rankine Cycle
10-42C Moisture content remains the same, everything else decreases.
10-43C 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.
10-44C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing.
10-45C Both cycles would have the same efficiency.
10-46C 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.
Boiler exit
s
TBoiler inlet
qin
qout
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-35
10-47E Feedwater is heated by steam in a feedwater heater of a regenerative Rankine cycle. The ratio of the bleed steam mass flow rate to the inlet feedwater mass flow rate 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4E through A-6E or EES),
h1 ≅ hf @ 110°F = 78.02 Btu/lbm
Btu/lbm 2.1167 F250
psia 202
2
2 =⎭⎬⎫
°==
hTP
h3 ≅ hf @ 225°F = 193.32 Btu/lbm
Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as
Mass balance:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
321
outin
(steady) 0systemoutin
0
mmmmm
mmm
&&&
&&
&&&
=+=
=∆=−
Energy balance:
)(0)peke (since
0
3212211
332211
outin
energies etc. potential, kinetic, internal,in change of Rate
(steady) 0system
mass and work,heat,by nsferenergy tranet of Rate
outin
hmmhmhmWQhmhmhm
EE
EEE
&&&&
&&&&&
&&
444 3444 21&
43421&&
+=+≅∆≅∆===+
=
=∆=−
Water 110°F 20 psia
3 2
1
Steam 20 psia 250°F
Water 20 psia 225°F
Solving for the bleed steam mass flow rate to the inlet feedwater mass flow rate, and substituting gives
0.118=−−
=−−
=kJ/kg )2.116732.193(kJ/kg )32.19302.78(
23
31
1
2
hhhh
mm&
&
preparation. If you are a student using this Manual, you are using it without permission.
10-36
10-48 In a regenerative Rankine cycle, the closed feedwater heater with a pump as shown in the figure is arranged so that the water at state 5 is mixed with the water at state 2 to form a feedwater which is a saturated liquid. The amount of bleed steam required to heat 1 kg of feedwater 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through A-6),
1
7000 kPasat. liq.
Feedwater 7000 kPa 260°C
6
5
3
Steam 6000 kPa 325°C
4
2
kJ/kg 5.1267 0
kPa 7000
kJ/kg 5.2969 C325kPa 6000
kJ/kg 8.1134 C260kPa 7000
kPa 7000 @ 66
6
33
3
C260 @ 11
1
==⎭⎬⎫
==
=⎭⎬⎫
°==
=≅⎭⎬⎫
°==
°
f
f
hhxP
hTP
hhTP
Analysis We take the entire unit 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
631inP,33311
66inP,33311
outin
energies etc. potential, kinetic, internal,in change of Rate
(steady) 0system
mass and work,heat,by nsferenergy tranet of Rate
outin
)(
0
hmmwmhmhmhmwmhmhm
EE
EEE
&&&&&
&&&&
&&
444 3444 21&
43421&&
+=++
=++=
=∆=−
Solving this for , 3m&
kg/s 0.0779=+−
−=
+−−
=319.15.12675.2969
8.11345.1267kg/s) 1()( inP,63
1613 whh
hhmm &&
where
kJ/kg 319.1
mkPa 1kPa 1kPa )60007000)(/kgm 001319.0(
)()(
33
45kPa 6000 @ 454inP,
=⎟⎠
⎞⎜⎝
⎛⋅
−=
−=−= PPPPw fvv
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-37
10-49E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat rejected in the condenser 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),
Btu/lbm 29.13011.018.130Btu/lbm 11.0
ftpsia 5.404Btu 1 psia)540)(/lbmft 01641.0(
)(
/lbmft 01641.0
Btu/lbm 18.130
inpI,12
33
121inPI,
3psia 5 @1
psia 5 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
1
qin
2
7
1-y 6
qout
500 psia
5 psia
y 40 psia
4
3
5
s
T
Btu/lbm 60.23746.114.236Btu/lbm 46.1
ftpsia 5.404Btu 1 psia)40500)(/lbmft 01715.0(
)(
/lbmft 01715.0
Btu/lbm 14.236
inpII,34
33
343inPII,
3psia 40 @3
psia 40 @3
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 63.953)5.1000)(8230.0(18.130
8230.060894.1
23488.05590.1
psia 5
Btu/lbm 4.1084)69.933)(9085.0(14.236
9085.028448.1
39213.05590.1
psia 40
RBtu/lbm 5590.1Btu/lbm 6.1298
F600psia 500
77
47
57
7
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that , 0≅∆≅∆≅≅ pekeWQ &&
326332266
(steady) 0
1)1(
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
outin
systemoutin
=−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
where y is the fraction of steam extracted from the turbine ( 36 / mm &&= ). Solving for y,
1109.029.1304.108429.13014.236
26
23 =−−
=−−
=hhhh
y
Then,
Btu/lbm 732.1Btu/lbm 1.57
Btu/lbm 330.5
=−−=−−=
=+=+=
=−−+−=−−+−=
)18.13063.953)(1109.01())(1(
46.111.0)63.9534.1084)(1109.01(4.10846.1298))(1(
17out
inPII,inPI,inP,
7665outT,
hhyq
wwwhhyhhw
Also, Btu/lbm 106160.2376.129845in =−=−= hhq
3100.01061
1.73211in
outth =−=−=
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-38
10-50E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The change in thermal efficiency when the steam supplied to the open feedwater heater is at 60 psia rather than 40 psia is 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),
Btu/lbm 35.13017.018.130Btu/lbm 17.0
ftpsia 5.404Btu 1 psia)560)(/lbmft 01641.0(
)(
/lbmft 01641.0
Btu/lbm 18.130
inp,12
33
121inPI,
3psia 5 @1
psia 5 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
1
qin
2
7
1-y 6
qout
500 psia
5 psia
y 60 psia
4
3
5
s
T
Btu/lbm 62.26342.120.262Btu/lbm 42.1
ftpsia 5.404Btu 1 psia)60500)(/lbmft 01738.0(
)(
/lbmft 01738.0
Btu/lbm 20.262
inp,34
33
343inPII,
3psia 60 @3
psia 60 @3
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 63.953)5.1000)(8230.0(18.130
8230.060894.1
23488.05590.1
psia 5
Btu/lbm 7.1113)61.915)(9300.0(20.262
9300.021697.1
42728.05590.1
psia 60
RBtu/lbm 5590.1Btu/lbm 6.1298
F600psia 500
77
47
57
7
66
46
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that , 0≅∆≅∆≅≅ pekeWQ &&
326332266
(steady) 0
1)1(
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
outin
systemoutin
=−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
where y is the fraction of steam extracted from the turbine ( 36 / mm &&= ). Solving for y,
1341.035.1307.111335.13020.262
26
23 =−−
=−−
=hhhh
y
Then,
Btu/lbm 713.0)18.13063.953)(1341.01())(1(Btu/lbm 103562.2636.1298
17out
45in
=−−=−−==−=−=
hhyqhhq
and 0.3111=−=−=1035
0.71311in
outth q
qη
When the reheater pressure is increased from 40 psia to 60 psia, the thermal efficiency increases from 0.3100 to 0.3111, which is an increase of 0.35%.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-39
10-51E The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES.
Analysis The EES program used to solve this problem as well as the solutions are given below.
P[5]=500 [psia] T[5]=600 [F] P[6]=60 [psia] P[7]=5 [psia] x[3]=0 "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia-ft^3, Btu) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out Eta_th=1-q_out/q_in
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-40
P 6
[kPa] ηth
10 0.3011 20 0.3065 30 0.3088 40 0.3100 50 0.3107 60 0.3111 70 0.31126 80 0.31129 90 0.3112 100 0.3111 110 0.3109 120 0.3107 130 0.3104 140 0.3101 150 0.3098 160 0.3095 170 0.3091 180 0.3087 190 0.3084 200 0.3080
0 40 80 120 160 2000.3
0.302
0.304
0.306
0.308
0.31
0.312
Bleed pressure [kPa]
ηth
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-41
10-52 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net power output of the power plant 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
P III P II P I
fwh II fwh I Condenser
Boiler Turbine
6
5
4 3
2 1
10
9
7
8
T
1
2
10
1 - y
8
91 - y - z
0.6 MPa
5 kPa
0.2 MPa
4y
3
10 MPa
5
6
7
s
(a) From the steam tables (Tables A-4, A-5, and A-6),
( ) ( )( )kJ/kg 95.13720.075.137
kJ/kg 02.0mkPa 1
kJ 1kPa 5200/kgm 0.001005
/kgm 001005.0
kJ/kg 75.137
in,12
33
121in,
3kPa 5 @1
kPa 5 @1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
==
==
pI
pI
f
f
whh
PPw
hh
v
vv
( ) ( )( )
( ) ( )( )kJ/kg 10.35
mkPa 1kJ 1
kPa 60010,000/kgm 0.001101
/kgm 001101.0
kJ/kg 38.670
liquidsat.MPa 6.0
kJ/kg 13.50542.071.504kJ/kg 0.42
mkPa 1kJ 1
kPa 200600/kgm 0.001061
/kgm 001061.0
kJ/kg 71.504
liquidsat.MPa 2.0
33
565in,
3MPa 6.0 @5
MPa 6.0 @55
in,34
33
343in,
3MPa 2.0 @3
MPa 2.0 @33
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
==
==
⎭⎬⎫=
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
==
==
⎭⎬⎫=
PPw
hhP
whh
PPw
hhP
pIII
f
f
pII
pII
f
f
v
vv
v
vv
kJ/kg 8.2821MPa 6.0
KkJ/kg 9045.6kJ/kg 8.3625
C600MPa 10
kJ/kg 73.68035.1038.670
878
8
7
7
7
7
in,56
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=+=+=
hss
P
sh
TP
whh pIII
( )( ) kJ/kg 7.26186.22019602.071.504
9602.05968.5
5302.19045.6MPa 2.0
99
99
79
9
=+=+=
=−
=−
=
⎭⎬⎫
==
fgf
fg
f
hxhh
sss
xss
P
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-42
( )( ) kJ/kg 0.21050.24238119.075.137
8119.09176.7
4762.09045.6kPa 5
1010
1010
710
10
=+=+=
=−
=−
=
⎭⎬⎫
==
fgf
fg
f
hxhh
sss
x
ssP
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that , 0∆pe∆ke ≅≅≅≅WQ &&
FWH-2:
( ) ( 548554488
outin
(steady) 0outin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii
system
=−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
)
where y is the fraction of steam extracted from the turbine ( = & / &m m8 5 ). Solving for y,
07133.013.5058.282113.50538.670
48
45 =−−
=−−
=hhhhy
FWH-1:
( ) ( ) 329332299 11 hyhzyzhhmhmhmhmhm eeii −=−−+⎯→⎯=+⎯→⎯=∑∑ &&&&&
where z is the fraction of steam extracted from the turbine ( = & / &m m9 5 ) at the second stage. Solving for z,
( ) ( ) 1373.007136.0195.1377.261895.13771.5041
29
23 =−−−
=−−−
= yhhhh
z
Then,
( )( ) ( )( )kJ/kg 2.13888.15560.2945
kJ/kg 1556.8137.752105.01373.007133.011kJ/kg 0.294573.6808.3625
outinnet
110out
67in
=−=−==−−−=−−−=
=−=−=
qqwhhzyq
hhq
and
( )( ) MW 30.5≅=== kW 540,30kJ/kg 1388.2kg/s 22netnet wmW &&
(b) The thermal efficiency is
47.1%=−=−=kJ/kg 2945.0kJ/kg 1556.8
11in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-43
10-53 An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat added in the boiler 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),
kJ/kg 45.25403.342.251kJ/kg 03.3 mkPa 1
kJ 1 kPa)203000)(/kgm 001017.0()(
/kgm 001017.0
kJ/kg 42.251
inp,12
33
121inp,
3kPa 20 @1
kPa 20 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
kJ/kg 7.2221)5.2357)(8357.0(42.251
8357.00752.7
8320.07450.6
kPa 20
kJ/kg 9.2851 kPa 1000
KkJ/kg 7450.6kJ/kg 1.3116
C350kPa 3000
66
66
46
6
545
5
4
4
4
4
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
hss
P
sh
TP
5
Condenser
12 3
4 Turbine
Boiler
Closed fwh
8 7
Pump
6
For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.
1
qin
2
6
1-y 5
3 MPa
20 kPa
qout
y 3 1 MPa 7
8
4
s
T
kJ/kg 53.763 C9.209
kPa 3000kJ/kg 51.762
C9.179kJ/kg 51.762
0
kPa 1000
373
3
78
7
7
7
7
=⎭⎬⎫
°===
==
°==
⎭⎬⎫
==
hTT
Phh
Th
xP
An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( ) for closed feedwater heater:
45 / mm &&=
7325
77332255
11 yhhhyhhmhmhmhm
hmhm eeii
+=++=+
=∑∑&&&&
&&
Rearranging,
2437.051.7629.285145.25453.763
75
23 =−−
=−−
=hhhh
y
Then,
kJ/kg 2353
kJ/kg 3.03kJ/kg 740.9
=−=−=
=
=−−+−=−−+−=
53.7631.3116
)7.22219.2851)(2437.01(9.28511.3116))(1(
34in
inP,
6554outT,
hhqw
hhyhhw
Also, kJ/kg 8.73703.39.740inP,outT,net =−=−= www
3136.02353
8.737
in
netth ===
qw
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-44
10-54 Problem 10-53 is reconsidered. The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES.
Analysis The EES program used to solve this problem as well as the solutions are given below.
"Given" P[4]=3000 [kPa] T[4]=350 [C] P[5]=600 [kPa] P[6]=20 [kPa] P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-45
P 6 [kPa]
ηth
100 0.32380 110 0.32424 120 0.32460 130 0.32490 140 0.32514 150 0.32534 160 0.32550 170 0.32563 180 0.32573 190 0.32580 200 0.32585 210 0.32588 220 0.32590 230 0.32589 240 0.32588 250 0.32585 260 0.32581 270 0.32576 280 0.32570 290 0.32563
100 140 180 220 260 3000.3238
0.324
0.3243
0.3245
0.3248
0.325
0.3253
0.3255
0.3258
0.326
Bleed pressure [kPa]
ηth
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-46
10-55 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is 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 or EES),
kJ/kg 45.25403.342.251kJ/kg 03.3 mkPa 1
kJ 1 kPa)203000)(/kgm 001017.0()(
/kgm 001017.0
kJ/kg 42.251
inp,12
33
121inp,
3kPa 20 @1
kPa 20 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛⋅
−=
−=
==
==
whh
PPw
hh
f
f
v
vv
kJ/kg 7.2221)5.2357)(8357.0(42.251
8357.00752.7
8320.07450.6
kPa 20
kJ/kg 9.2851 kPa 1000
KkJ/kg 7450.6kJ/kg 1.3116
C350kPa 3000
66
66
46
6
545
5
4
4
4
4
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fss
s
ss
hxhhs
ssx
ssP
hss
P
sh
TP
5 6
12 3
4 Turbine
Boiler
Condenser
Closed fwh
7
Pump
kJ/kg 3.2878)9.28511.3116)(90.0(1.3116)( 5s4T4554
54T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
kJ/kg 1.2311)7.22211.3116)(90.0(1.3116)( 6s4T4664
64T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.
1
qin
2
6
4
5s
3 MPa
20 kPa
3 1 MPa 7
1-y
y
qout
5
6s
s
T
kJ/kg 53.763 C9.209
kPa 3000
C9.179kJ/kg 51.762
0
kPa 1000
373
3
7
7
7
7
=⎭⎬⎫
°===
°==
⎭⎬⎫
==
hTT
P
Th
xP
An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( ) for closed feedwater heater:
45 / mm &&=
7325
77332255
11 yhhhyhhmhmhmhm
hmhm eeii
+=++=+
=∑∑&&&&
&&
Rearranging,
2406.051.7623.287845.25453.763
75
23 =−−
=−−
=hhhh
y
Then,
kJ/kg 235353.7631.3116kJ/kg 3.03
kJ/kg 668.5)1.23113.2878)(2406.01(3.28781.3116))(1(
34in
inP,
6554outT,
=−=−=
=
=−−+−=−−+−=
hhqw
hhyhhw
Also, kJ/kg 5.66503.35.668inP,outT,net =−=−= www
28.3%==== 0.28292353
5.665
in
netth q
wη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-47
10-56 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is 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 or EES),
5
Condenser
12 3
4 Turbine
Boiler
Closed fwh
7
Pump
6
1
qin
2
6
4
5s
3 MPa
20 kPa
1 MPa
1-y 7
y 3
qout
5
6s
s
T
When the liquid enters the pump 10°C cooler than a saturated liquid at the condenser pressure, the enthalpies become
/kgm 001012.0
kJ/kg 34.209
C501006.6010kPa 20
3C50 @ 1
C50 @ 1
kPa 20 @sat 1
1
=≅
=≅
⎭⎬⎫
°≅−=−==
°
°
f
fhhTT
Pvv
kJ/kg 02.3 mkPa 1
kJ 1 kPa)203000)(/kgm 001012.0()(
33
121inp,
=⎟⎠
⎞⎜⎝
⎛
⋅−=
−= PPw v
kJ/kg 36.21202.334.209inp,12 =+=+= whh
kJ/kg 7.2221)5.2357)(8357.0(42.251
8357.00752.7
8320.07450.6
kPa 20
kJ/kg 9.2851 kPa 1000
KkJ/kg 7450.6kJ/kg 1.3116
C350kPa 3000
66
66
46
6
545
5
4
4
4
4
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fss
s
ss
hxhhs
ssx
ssP
hss
P
sh
TP
kJ/kg 3.2878)9.28511.3116)(90.0(1.3116)( 5s4T4554
54T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
kJ/kg 1.2311)7.22211.3116)(90.0(1.3116)( 6s4T4664
64T =−−=−−=⎯→⎯
−−
= hhhhhhhh
sηη
For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.
kJ/kg 53.763 C9.209
kPa 3000
C9.179kJ/kg 51.762
0
kPa 1000
373
3
7
7
7
7
=⎭⎬⎫
°===
°==
⎭⎬⎫
==
hTT
P
Th
xP
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-48
An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( ) for closed feedwater heater:
45 / mm &&=
7325
77332255
11 yhhhyhhmhmhmhm
hmhm eeii
+=++=+
=∑∑&&&&
&&
Rearranging,
2605.051.7623.287836.21253.763
75
23 =−−
=−−
=hhhh
y
Then,
kJ/kg 235353.7631.3116
kJ/kg 3.03kJ/kg 657.2)1.23113.2878)(2605.01(3.28781.3116))(1(
34in
inP,
6554outT,
=−=−=
=
=−−+−=−−+−=
hhqw
hhyhhw
Also,
kJ/kg 2.65403.32.657inP,outT,net =−=−= www
0.2781===2353
2.654
in
netth q
wη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-49
10-57 The effect of pressure drop and non-isentropic turbine on the rate of heat input is to be determined for a given power plant.
Analysis The EES program used to solve this problem as well as the solutions are given below.
"Given" P[3]=3000 [kPa] DELTAP_boiler=10 [kPa] P[4]=P[3]-DELTAP_boiler T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] eta_T=0.90 P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-50
Solution with 10 kPa pressure drop in the boiler: DELTAP_boiler=10 [kPa] eta_T=0.9 Eta_th=0.2827 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=2990 [kPa] q_in=2352.8 [kJ/kg] w_net=665.1 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] w_T_out=668.1 [kJ/kg] y=0.2405 Solution without any pressure drop in the boiler: DELTAP_boiler=0 [kPa] eta_T=1 Eta_th=0.3136 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=3000 [kPa] q_in=2352.5 [kJ/kg] w_net=737.8 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] w_T_out=740.9 [kJ/kg] y=0.2437
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-51
10-58 A steam power plant operates on an ideal regenerative Rankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam through the boiler for a net power output of 400 MW 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.
1-y-z
Condenser
Turbine
1
z 11
8
9
10
2 7
4
6
5
Open fwh
P I
P II
3
y
Closedfwh
BoilerAnalysis (a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg 40.19260.081.191kJ/kg 60.0
mkPa 1kJ 1
kPa 10600/kgm 0.00101
/kgm 00101.0
kJ/kg 81.191
in,12
33
121in,
3kPa 10 @1
kPa 10 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
pI
pI
f
f
whh
PPw
hh
v
vv
( )
( )( )
kJ/kg 73.68035.1038.670kJ/kg 10.35
mkPa 1kJ 1
kPa 00610000/kgm 0.001101
/kgm 001101.0
kJ/kg 38.670
liquid sat.MPa 6.0
in,34
33
343in,
3MPa 3.0 @3
MPa 3.0 @33
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−===
==
⎭⎬⎫=
pII
pII
f
f
whh
PPw
hhP
v
vv
T
1
2
11
9
1 - y - z 10
10 MPa
1.2 MPa
10 kPa
0.6 MPa
4 y
37 z
5 6
8
s
kJ/kg 5.2974MPa 2.1
KkJ/kg 9045.6kJ/kg 8.3625
C600MPa 10
kJ/kg 798.33 MPa 10 ,
C0.188
kJ/kg 33.798
liquid sat.MPa 2.1
989
9
8
8
8
8
5556
MPa .21 @sat6
MPa 2.1 @766
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=→==
°==
===
⎭⎬⎫=
hss
P
sh
TP
hPTT
TT
hhhP f
( )( ) kJ/kg 0.21871.23928341.081.191
8341.04996.7
6492.09045.kPa 10
kJ/kg 9.2820MPa 6.0
1111
1111
811
11
10810
10
=+=+=
=−6
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
fgf
fg
f
hxhh
sss
x
ssP
hss
P
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that , 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( ) ( ) ( 4569455699
outin
(steady) 0systemoutin 0
hhhhyhhmhhmhmhm
EE
EEE
eeii −=−⎯→⎯−=−⎯→⎯=
=
=∆=−
∑∑ &&&&
&&
&&&
)
where y is the fraction of steam extracted from the turbine ( 510 / mm &&= ). Solving for y,
05404.033.7985.297473.68033.798
69
45 =−−
=−−
=hhhh
y
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-52
For the open FWH,
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
)
( ) ( 310273310102277
outin
(steady) 0systemoutin
11
0
hzhhzyyhhmhmhmhmhmhm
EE
EEE
eeii =+−−+⎯→⎯=++⎯→⎯=
=
=∆=−
∑∑ &&&&&&
&&
&&&
where z is the fraction of steam extracted from the turbine ( = & / &m m9 5 ) at the second stage. Solving for z,
( ) ( ) ( )( ) 1694.0
40.1929.282040.19233.79805404.040.19238.670
210
2723 =−
−−−=
−−−−
=hh
hhyhhz
Then,
( )( ) ( )( )kJ/kg 127815492827
kJ/kg 154981.1910.21871694.005404.011kJ/kg 282733.7988.3625
outinnet
111out
58in
=−=−==−−−=−−−=
=−=−=
qqwhhzyq
hhq
and
kg/s 313.0===kJ/kg 1278
kJ/s 400,000
net
net
wW
m&
&
(b) 45.2%==−=−= 452.0kJ/kg 2827kJ/kg 154911
in
outth q
qη
preparation. If you are a student using this Manual, you are using it without permission.
10-53
10-59 Problem 10-58 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" P[8] = 10000 [kPa] T[8] = 600 [C] P[9] = 1200 [kPa] P_cfwh=600 [kPa] P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=400 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages" Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages" Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages" Eta_pump = 100/100 "Pump isentropic efficiency" "Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[11] P[2]=P[10] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[5]=P[8] P[4] = P[5] P[3]=P[10] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Closed Feedwater Heater analysis" P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy" h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]" T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]" s[5]=entropy(Fluid$,P=P[6],h=h[5]) h[6]=enthalpy(Fluid$,P=P[6],x=0) T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]" PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-54
s[6]=entropy(Fluid$,P=P[6],x=0) "Trap analysis" P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "Boiler analysis" q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler" h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) "Turbine analysis" ss[9]=s[8] hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9]) Ts[9]=temperature(Fluid$,s=ss[9],P=P[9]) h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9])"Definition of turbine efficiency for high pressure stages" T[9]=temperature(Fluid$,P=P[9],h=h[9]) s[9]=entropy(Fluid$,P=P[9],h=h[9]) ss[10]=s[8] hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10]) Ts[10]=temperature(Fluid$,s=ss[10],P=P[10]) h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10])"Definition of turbine efficiency for Intermediate pressure stages" T[10]=temperature(Fluid$,P=P[10],h=h[10]) s[10]=entropy(Fluid$,P=P[10],h=h[10]) ss[11]=s[8] hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11]) Ts[11]=temperature(Fluid$,s=ss[11],P=P[11]) h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11])"Definition of turbine efficiency for low pressure stages" T[11]=temperature(Fluid$,P=P[11],h=h[11]) s[11]=entropy(Fluid$,P=P[11],h=h[11]) h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine" "Condenser analysis" (1-y-z)*h[11]=q_out+(1-y-z)*h[1]"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net
ηturb ηth m [kg/s] 0.7
0.75 0.8
0.85 0.9
0.95 1
0.3834 0.397
0.4096 0.4212 0.4321 0.4423 0.452
369 356.3 345.4 335.8 327.4 319.8 313
ηpump ηth m [kg/s] 0.7
0.75 0.8
0.85 0.9
0.95 1
0.4509 0.4511 0.4513 0.4515 0.4517 0.4519 0.452
313.8 313.6 313.4 313.3 313.2 313.1 313
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-55
Pcfwh [kPa] y z
100 200 300 400 500 600 700 800 900
1000
0.1702 0.1301 0.1041
0.08421 0.06794 0.05404 0.04182 0.03088 0.02094 0.01179
0.05289 0.09634 0.1226 0.1418 0.1569 0.1694 0.1801 0.1895 0.1979 0.2054
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°C
]
10000 kPa
1200 kPa 600 kPa
10 kPa
SteamIAPWS
1,2
3,45,6
7
8
9
10
11
0.7 0.75 0.8 0.85 0.9 0.95 10.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46
ηturb
ηth
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10-56
0.7 0.75 0.8 0.85 0.9 0.95 1310
320
330
340
350
360
370
η turb
m [
kg/s
]
0.7 0.75 0.8 0.85 0.9 0.95 10.4508
0.451
0.4512
0.4514
0.4516
0.4518
0.452
0.4522
312.9
313
313.1
313.2
313.3
313.4
313.5
313.6
313.7
313.8
ηpump
ηth
m [
kg/s
]
100 200 300 400 500 600 700 800 900 10000
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0.22
Pcfwh [kPa]
y
z
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-57
10-60 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, 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),
( )
kJ/kg 65.260
223.942.251
kJ/kg .22988.01)kPa 200080)(/kgm 0.001017(
/
/kgm 001017.0
kJ/kg 42.251
in,12
3
121in,
3kPa 20 @1
kPa 20 @1
=
+=
+=
=
−=
−=
==
==
pI
ppI
f
f
whh
PPw
hh
ηv
vv
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
/kgm 001127.0
kJ/kg 51.762
liquid sat.MPa 1
3MPa 1 @3
MPa 1 @33
==
==
⎭⎬⎫=
f
fhhPvv
( )
kJ/kg 48.77197.851.762kJ/kg 97.8
88.0/)kPa 00108000)(/kgm 001127.0(
/
in,311
3
3113in,
=+=+==
−=
−=
pII
ppII
whh
PPw ηv
6
5
8
1 2
3
High-P turbine
Boiler
Cond. Closed
fwh
PI PII
10
11
MixingCham.
y
7 1-y9
Low-P turbine
4
Also, h4 = h10 = h11 = 771.48 kJ/kg since the two fluid streams which are being mixed have the same enthalpy.
( )( )( ) kJ/kg 1.31407.31045.339988.05.3399
kJ/kg 7.3104MPa 3
KkJ/kg 7266.6kJ/kg 5.3399
C500MPa 8
655665
65
656
6
5
5
5
5
=−−=−−=⎯→⎯
−−
=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
sTs
T
s
hhhhhhhh
hss
P
sh
TP
ηη
( )( )( )
C349.9°=⎭⎬⎫
==
=−−=−−=⎯→⎯
−−
=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
88
8
877887
87
878
8
7
7
7
7
kJ/kg 9.3157MPa 1
kJ/kg 9.31571.31172.345788.02.3457
kJ/kg 1.3117MPa 1
KkJ/kg 2359.7kJ/kg 2.3457
C500MPa 3
ThP
hhhhhhhh
hss
P
sh
TP
sTs
T
s
ηη
preparation. If you are a student using this Manual, you are using it without permission.
10-58
( )( )( ) kJ/kg 9.25132.23852.345788.02.3457
kJ/kg 2.2385kPa 20
977997
97
979
9
=−−=
−−=⎯→⎯−−
=
=⎭⎬⎫
==
sTs
T
s
hhhhhhhh
hss
P
ηη
The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that , & &Q W ke pe≅ ≅ ≅ ≅∆ ∆ 0
( )( ) ( )1758.0)51.7629.3157()65.26048.771)(1(
1 38210
=⎯→⎯−=−−
−=−−
yyy
hhyhhy
The corresponding mass flow rate is
kg/s 2.637=== kg/s) 15)(1758.0(58 mym &&
(c) Then,
( )( ) ( )( ) kJ/kg 8.186442.2519.25131758.011kJ/kg 2.29451.31402.345748.7715.3399
19out
6745in
=−−=−−==−+−=−+−=
hhyqhhhhq
and
kW 16,206=−=−= kJ/kg)8.18648.2945)(kg/s 15()( outinnet qqmW &&
(b) The thermal efficiency is determined from
36.7%==−=−= 3668.0kJ/kg 2945.8kJ/kg 1864.811
in
outth q
qη
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10-59
10-61 A Rankine steam cycle modified with two closed feedwater heaters is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass extracted for the closed feedwater heater z and the cooling water flow rate are to be determined. Also, the net power output 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 (b) Using the data from the problem statement, the enthalpies at various states are
( )( )( )
kJ/kg 1.2561.5251kJ/kg 1.5
mkPa 1kJ 1
kPa 205000/kgm 0.00102
/kgm 00102.0kJ/kg 251
inpI,12
33
121inpI,
3kPa 20 @ 1
kPa 20 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
1
2
9
s
5
6 245 kPa
20 kPa
1.4 MPa
8 4
31-y-z
1-y
y
5 MPa
11
121-y-z
y
10 z
y+z
8
7
T
Also,
operation) valve(throttle
kJ/kg 533
1112
kPa 245 @ 113
hhhhh f
====
operation) valve(throttle
kJ/kg 830
910
kPa 1400 @ 94
hhhhh f
====
An energy balance on the closed feedwater heater gives 1131072 )(11 hzyhyhzhh ++=++
where z is the fraction of steam extracted from the low-pressure turbine. Solving for z,
0.1017=−
−+−=
−−(+−(
=5332918
)830533)(1153.0()1.256533())
117
101123
hhhhyhh
z
(c) An energy balance on the condenser gives
1112128822
2111212188
)( hmhmhmhhmhmhmhmhmhm
www
wwww
&&&&
&&&&&
−+=−+=++
Solving for the mass flow rate of cooling water, and substituting with correct units,
[ ]
[ ]
kg/s 2158=
−++−−=
∆−++−−
=
)10)(18.4()251(1)533)(1017.01153.0()2477)(1017.01153.01()50(
1)()1( 11285
wpww Tc
hhzyhzymm
&&
(d) The work output from the turbines is
kJ/kg 1271
)2477)(1017.01153.01()2918)(1017.0()3406)(1153.0(3900)1( 8765outT,
=−−−−−=
−−−−−= hzyzhyhhw
The net work output from the cycle is
kJ/kg 9.12651.51271
inP,outT,net
=−=
−= www
The net power output is
MW 63.3==== kW 300,63kJ/kg) .9kg/s)(1265 50(netnet wmW &&
The rate of heat input in the boiler is
kW 500,153kJ/kg )8303900)(kg/s 50()( 45in =−=−= hhmQ &&
The thermal efficiency is then
41.2%==== 412.0kW 500,153kW 300,63
in
netth Q
W&
&η
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10-60
Second-Law Analysis of Vapor Power Cycles
10-62C In the simple ideal Rankine cycle, irreversibilities occur during heat addition and heat rejection processes in the boiler and the condenser, respectively, and both are due to temperature difference. Therefore, the irreversibilities can be decreased and thus the 2nd law efficiency can be increased by minimizing the temperature differences during heat transfer in the boiler and the condenser. One way of doing that is regeneration.
10-63E The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-17E are to be determined for the specified source and sink temperatures.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Problem 10-17E,
qin
qout
3 psia 1
3
2
4
800 psia
T
s
Btu/lbm 8.86540.10924.975Btu/lbm 2.134481.1110.1456
RBtu/lbm 6413.1
RBtu/lbm 2009.0
14out
23in
43
psia 3 @ 21
=−=−==−=−=
⋅==
⋅===
hhqhhq
ss
sss f
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
Btu/lbm 146
Btu/lbm 377
=⎟⎠⎞
⎜⎝⎛ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
=⎟⎠⎞
⎜⎝⎛ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
R 500Btu/lbm 8.8656413.12009.0)R 500(
R 1960Btu/lbm 2.13442009.06413.1)R 500(
sink
out41041 destroyed,
source
in23023 destroyed,
Tq
ssTx
Tq
ssTx
Processes 1-2 and 3-4 are isentropic, and thus
00
=
=
34 destroyed,
12 destroyed,
x
x
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10-61
10-64 The exergy destruction associated with the heat rejection process in Prob. 10-21 is to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Problem 10-21,
kJ/kg 8.1961kJ/kg 4.3411
KkJ/kg 8000.6
KkJ/kg 6492.0
out
3
43
kPa10@21
==
⋅==
⋅===
qh
ss
sss f
The exergy destruction associated with the heat rejection process is
( ) kJ/kg 178.0=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−=
K 290kJ/kg 1961.8
8000.66492.0K 29041,41041destroyed,
R
R
Tq
ssTx
The exergy of the steam at the boiler exit is simply the flow exergy,
( ) ( )( ) ( )03003
03
023
030033 2ssThhqzssThh
−−−=++−−−=
Vψ
where
( )( ) KkJ/kg 2533.0
kJ/kg 95.71
K 290 @ kPa 100,K 290@0
K 290 @ kPa 100 ,K 290@0
⋅=≅==≅=
f
f
ssshhh
Thus,
( ) ( )( ) kJ/kg 1440.9=⋅−−−=ψ KkJ/kg 2532.0800.6K 290kJ/kg 95.714.34113
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10-62
10-65 The component of the ideal reheat Rankine cycle described in Prob. 10-33 with the largest exergy destruction is to be identified.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Prob. 10-33,
kJ/kg 1.2373kJ/kg 1.8484.26364.3485kJ/kg 4.307388.1993.3273
KkJ/kg 0893.8KkJ/kg 5579.6
KkJ/kg 6492.0
out
455-4 in,
233-2 in,
65
43
kPa 10 @ 21
=
=−=−=
=−=−=⋅==⋅==
⋅===
qhhqhhq
ssss
sss f T
1
5
2
6
3
4
500 kPa
10 kPa
8 MPa
s
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
kJ/kg 267.6
kJ/kg 161.6
kJ/kg 687.1
=⎟⎠
⎞⎜⎝
⎛ +−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
=⎟⎠
⎞⎜⎝
⎛ −−=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
=⎟⎠
⎞⎜⎝
⎛ −−=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
K 283kJ/kg 1.23730893.86492.0)K 283(
K 883kJ/kg 1.8485579.60893.8)K 283(
K 883kJ/kg 4.30736492.05579.6)K 283(
sink
out61061 destroyed,
source
5-4 in,45045 destroyed,
source
3-2 in,23023 destroyed,
Tq
ssTx
Tq
ssTx
Tq
ssTx
Processes 1-2, 3-4, and 5-6 are isentropic, and thus,
000
=
=
=
56 destroyed,
34 destroyed,
12 destroyed,
x
x
x
The greatest exergy destruction occurs during heat addition process 2-3.
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10-63
10-66 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-35 are to be determined for the specified source and sink temperatures.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Problem 10-35,
kJ/kg 3.209842.2517.2349
kJ/kg 3.3470.29014.3248
kJ/kg 8.292050.2573.3178KkJ/kg 1292.7KkJ/kg 5432.6
KkJ/kg 8320.0
16out
in,45
in,23
65
43
kPa20@21
=−=−=
=−=
=−=⋅==⋅==
⋅===
hhq
q
qssss
sss f
Processes 1-2, 3-4, and 5-6 are isentropic. Thus, i12 = i34 = i56 = 0. Also,
( )
( )
( ) kJ/kg 240.6
kJ/kg 104.6
kJ/kg 1110
=⎟⎠
⎞⎜⎝
⎛ +−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −+−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−=
=⎟⎠
⎞⎜⎝
⎛ −+−=⎟
⎟⎠
⎞⎜⎜⎝
⎛+−=
K 295kJ/kg 2098.31292.78320.0K 295
K 1500kJ/kg 347.35432.61292.7K 295
K 1500kJ/kg .829208320.05432.6K 295
61,61061destroyed,
45,45045destroyed,
23,23023destroyed,
R
R
R
R
R
R
Tq
ssTx
Tq
ssTx
Tq
ssTx
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10-64
10-67 Problem 10-66 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. Also, the T-s diagram is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)' end "Input Data - from diagram window" {P[6] = 20 [kPa] P[3] = 6000 [kPa] T[3] = 400 [C] P[4] = 2000 [kPa] T[5] = 400 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Data for the irreversibility calculations:" T_o = 295 [K] T_R_L = 295 [K] T_R_H = 1500 [K] "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,P=P[2],h=h[2]) s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) v[3]=volume(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,T=T[4],P=P[4]) v[4]=volume(Fluid$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" P[5]=P[4] s[5]=entropy(Fluid$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6]) vs[6]=volume(Fluid$,s=s_s[6],P=P[6]) Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"
preparation. If you are a student using this Manual, you are using it without permission.
10-65
x[6]=QUALITY(Fluid$,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 "The irreversibilities (or exergy destruction) for each of the processes are:" q_R_23 = - (h[3] - h[2]) "Heat transfer for the high temperature reservoir to process 2-3" i_23 = T_o*(s[3] -s[2] + q_R_23/T_R_H) q_R_45 = - (h[5] - h[4]) "Heat transfer for the high temperature reservoir to process 4-5" i_45 = T_o*(s[5] -s[4] + q_R_45/T_R_H) q_R_61 = (h[6] - h[1]) "Heat transfer to the low temperature reservoir in process 6-1" i_61 = T_o*(s[1] -s[6] + q_R_61/T_R_L) i_34 = T_o*(s[4] -s[3]) i_56 = T_o*(s[6] -s[5]) i_12 = T_o*(s[2] -s[1])
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
6000 kPa
2000 kPa
20 kPa
SteamIAPWS
1,2
3
4
5
6
Ideal Rankine cycle with reheat
SOLUTION Eff=0.358 Eta_p=1 Eta_t=1 Fluid$='Steam_IAPWS' i_12=0.007 [kJ/kg] i_23=1110.378 [kJ/kg] i_34=-0.000 [kJ/kg] i_45=104.554 [kJ/kg] i_56=0.000 [kJ/kg] i_61=240.601 [kJ/kg] Q_in=3268 [kJ/kg] Q_out=2098 [kJ/kg] q_R_23=-2921 [kJ/kg] q_R_45=-347.3 [kJ/kg] q_R_61=2098 [kJ/kg] T_o=295 [K] T_R_H=1500 [K] T_R_L=295 [K] W_net=1170 [kJ/kg] W_p=6.083 [kJ/kg] W_p_s=6.083 [kJ/kg] W_t_hp=277.2 [kJ/kg] W_t_lp=898.7 [kJ/kg]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-66
10-68E The component of the ideal regenerative Rankine cycle described in Prob. 10-49E with the largest exergy destruction is to be identified.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Prob. 10-49E,
0.1109Btu/lbm 1.732
kJ/kg 10616.2376.1298RBtu/lbm 5590.1
RBtu/lbm 39213.0
RBtu/lbm 23488.0
out
45in
765
psia 40 @ 43
psia 5 @ 21
==
=−=−=⋅===
⋅===
⋅===
yq
hhqsss
sss
sss
f
f
1
qin
2
7
1-y 6
qout
500 psia
5 psia
y 40 psia
4
3
5
s
T
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
Btu/lbm 43.6
Btu/lbm 168.9
=⎟⎠⎞
⎜⎝⎛ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
=⎟⎠⎞
⎜⎝⎛ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
R 520Btu/lbm 1.7325590.123488.0)R 520(
R 1260Btu/lbm 106139213.05590.1)R 520(
sink
out71071 destroyed,
source
in45045 destroyed,
Tq
ssTx
Tq
ssTx
For open feedwater heater, we have
[ ][ ]
Btu/lbm 5.4=−−−=
−−−=
)23488.0)(1109.01()5590.1)(1109.0(39213.0)R 520(
)1( 2630FWH destroyed, syyssTx
Processes 1-2, 3-4, 5-6, and 6-7 are isentropic, and thus
00
=
=
34 destroyed,
12 destroyed,
x
x
00
=
=
67 destroyed,
56 destroyed,
x
x
The greatest exergy destruction occurs during heat addition process 4-5.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-67
10-69 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The power output from the turbine, the thermal efficiency of the plant, the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions and exergy efficiencies for the flash chamber, the turbine, and the entire 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, A-5, and A-6)
kJ/kg.K 6841.21661.0
kJ/kg 14.990kPa 500
kJ/kg.K 6100.2kJ/kg 14.990
0C230
2
2
12
2
1
1
1
1
==
⎭⎬⎫
===
==
⎭⎬⎫
=°=
sx
hhP
sh
xT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
kg/s 38.19kg/s) 230)(1661.0(
123
=== mxm &&
KkJ/kg 7739.7kJ/kg 3.2464
95.0kPa 10
KkJ/kg 8207.6kJ/kg 1.2748
1kPa 500
4
4
4
4
3
3
3
3
⋅==
⎭⎬⎫
==
⋅==
⎭⎬⎫
==
sh
xP
sh
xP
KkJ/kg 8604.1kJ/kg 09.640
0kPa 500
6
6
6
6
⋅==
⎭⎬⎫
==
sh
xP
2
productionwell
reinjectionwell
separator
steam turbine
1
Flash chamber
65
4
3
condenser
kg/s 81.19119.38230316 =−=−= mmm &&&
The power output from the turbine is
kW 10,842=−=−= kJ/kg)3.24648.1kJ/kg)(274 38.19()( 433T hhmW &&
We use saturated liquid state at the standard temperature for dead state properties
kJ/kg 3672.0kJ/kg 83.104
0C25
0
0
0
0
==
⎭⎬⎫
=°=
sh
xT
kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in =−=−= hhmE &&
5.3%==== 0.0532622,203
842,10
in
outT,th E
W&
&η
(b) The specific exergies at various states are
kJ/kg 53.216kJ/kg.K)3672.0K)(2.6100 (298kJ/kg)83.104(990.14)( 010011 =−−−=−−−= ssThhψ
kJ/kg 44.194kJ/kg.K)3672.0K)(2.6841 (298kJ/kg)83.104(990.14)( 020022 =−−−=−−−= ssThhψ
kJ/kg 10.719kJ/kg.K)3672.0K)(6.8207 (298kJ/kg)83.104(2748.1)( 030033 =−−−=−−−= ssThhψ
kJ/kg 05.151kJ/kg.K)3672.0K)(7.7739 (298kJ/kg)83.104(2464.3)( 040044 =−−−=−−−= ssThhψ
kJ/kg 97.89kJ/kg.K)3672.0K)(1.8604 (298kJ/kg)83.104(640.09)( 060066 =−−−=−−−= ssThhψ
The exergy of geothermal water at state 6 is
kW 17,257=== kJ/kg) 7kg/s)(89.9 .81191(666 ψmX &&
preparation. If you are a student using this Manual, you are using it without permission.
10-68
(c) Flash chamber:
kW 5080=−=−= kJ/kg)44.19453kg/s)(216. 230()( 211FC dest, ψψmX &&
89.8%==== 0.89853.21644.194
1
2FCII, ψ
ψη
(d) Turbine:
kW 10,854=−=−−= kW 10,842-kJ/kg)05.15110kg/s)(719. 19.38()( T433Tdest, WmX &&& ψψ
50.0%==−
=−
= 0.500)kJ/kg05.15110kg/s)(719. 19.38(
kW 842,10)( 433
TTII, ψψ
ηm
W&
&
(e) Plant:
kW 802,49kJ/kg) 53kg/s)(216. 230(11Plantin, === ψmX &&
kW 38,960=−=−= 842,10802,49TPlantin,Plantdest, WXX &&&
21.8%==== 0.2177kW 802,49kW 842,10
Plantin,
TPlantII, X
W&
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-69
Cogeneration
10-70C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purpose to the total energy supplied. It could be unity for a plant that does not produce any power.
10-71C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity.
10-72C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference.
10-73C Cogeneration is the production of more than one useful form of energy from the same energy source. Regeneration is the transfer of heat from the working fluid at some stage to the working fluid at some other stage.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-70
10-74 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. The net power produced and the utilization factor of the plant 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),
( )( )( )
kJ/kg 38.670
kJ/kg 41.19260.081.191kJ/kg 0.60
mkPa 1kJ 1
kPa 10600/kgm 0.00101
/kgm 00101.0kJ/kg 81.191
MPa 0.6 @ 3
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
f
f
f
hh
whh
PPw
hh
v
vv
P II P I
Condenser
Boiler
6
7 Turbine
Process heater
3
2 4
5
1
8
Mixing chamber:
332244
outin(steady) 0
systemoutin 0
hmhmhmhmhm
EEEEE
eeii &&&&&
&&&&&
+=⎯→⎯=
=⎯→⎯=∆=−
∑∑
or, ( )( ) ( )( )
( )( )( )
kJ/kg 47.31857.690.311kJ/kg 6.57
mkPa 1kJ 1kPa 6007000/kgm 0.001026
/kgm 001026.0
kJ/kg 90.31130
38.67050.741.19250.22
inII,45
33
454inII,
3kJ/kg 90.311 @ 4
4
33224
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
=≅
=+
=+
=
=
p
p
hf
whh
PPw
mhmhmh
f
v
vv&
&&
T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
KkJ/kg 8000.6
kJ/kg 4.3411C500
MPa 76
6
6
6⋅=
=
⎭⎬⎫
°==
sh
TP
( )( ) kJ/kg 6.21531.23928201.081.191
8201.04996.7
6492.08000.6kPa 10
kJ/kg 6.2774MPa 6.0
88
88
68
8
767
7
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
fgf
fg
f
hxhhs
ssx
ssP
hss
P
1
2
8
7
7 MPa
10 kPa
0.6 MPa 4
5
3
Qin·
Qout·
Qproces·
6
s
Then,
( ) ( )( )( ) ( )( )
( )( ) ( )( )
kW 32,866=−=−=
=+=+=
=−+−=−+−=
6.210077,33
kW 210.6kJ/kg 6.57kg/s 30kJ/kg 0.60kg/s 22.5
kW 077,33kJ/kg6.21536.2774kg/s 22.5kJ/kg6.27744.3411kg/s 30
inp,outT,net
inpII,4inpI,1inp,
878766outT,
WWW
wmwmW
hhmhhmW
&&&
&&&
&&&
Also, ( ) ( )( ) kW 782,15kJ/kg 38.6706.2774kg/s 7.5377process =−=−= hhmQ &&
( ) ( )( ) kW 788,9247.3184.3411kg/s 30565in =−=−= hhmQ &&
and 52.4%=+
=+
=788,92
782,15866,32
in
processnet
Q
QWu &
&&ε
preparation. If you are a student using this Manual, you are using it without permission.
10-71
10-75E A large food-processing plant requires steam at a relatively high pressure, which is extracted from the turbine of a cogeneration plant. The rate of heat transfer to the boiler and the power output of the cogeneration 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-4E, A-5E, and A-6E),
( )
Btu/lbm 92.324
Btu/lbm 50.9448.002.94Btu/lbm 0.48
ftpsia 5.4039Btu 1
psia2)014)(/lbmft 0.01623(86.01
/
/lbmft 01623.0Btu/lbm 02.94
psia 140 @ 3
inpI,12
3
3
121inpI,
3psia 2 @ 1
psia 2 @ 1
==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅×
−=
−=
==
==
f
p
f
f
hh
whh
PPw
hh
ηv
vv
P II P I
Condenser
Boiler
6
7 Turbine
Process heater
3
2 4
5
1
8
Mixing chamber: T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
& & &
& &
& & & & &
E E E
E E
m h m h m h m h m hi i e e
in out system (steady)
in out
− = =
=
= ⎯→⎯ = +∑ ∑
∆ 0
4 4 2 2 3 3
0
or,
( )( ) ( )( )
( )( )( ) ( )
Btu/lbm 39.13133.207.129Btu/lbm .332
86.0/ftpsia 5.4039
Btu 1psia 014800/lbmft 0.01640
/
/lbmft 01640.0
Btu/lbm 07.12910
92.3245.150.945.8
in,45
33
454inpII,
3Btu/lbm 07.129 @ 4
4
33224
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
=≅
=+
=+
=
=
pII
p
hf
whh
PPw
mhmhm
h
f
ηv
vv
&
&&
1
2
8
6
7s
800psia
2 psia
140 psia 4
5
3
Qin·
Qprocess·
8s
7
s
RBtu/lbm 6812.1
Btu/lbm 2.1512F1000
psia 8006
6
6
6⋅=
=
⎭⎬⎫
°==
sh
TP
( )( ) Btu/lbm 21.9767.10218634.002.94
8634.074444.1
17499.06812.1psia 2
Btu/lbm 5.1287psia 140
88
88
68
8
767
7
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
fgsfs
fg
fss
s
s
ss
s
hxhhs
ssx
ssP
hss
P
Then, ( ) ( )( ) Btu/s 13,810=−=−= Btu/lbm39.1312.1512lbm/s 10565in hhmQ &&
(b) ( ) ( )[ ]( ) ( )( ) ( )( )[ ]
kW 4440==−+−=
−+−==
Btu/s 4208Btu/lbm 976.211287.5lbm/s 1.5Btu/lbm 1287.51512.2lbm/s 0186.0
878766,outT, sssTsTT hhmhhmWW &&&& ηη
preparation. If you are a student using this Manual, you are using it without permission.
10-72
10-76 A cogeneration plant has two modes of operation. In the first mode, all the steam leaving the turbine at a relatively high pressure is routed to the process heater. In the second mode, 60 percent of the steam is routed to the process heater and remaining is expanded to the condenser pressure. The power produced and the rate at which process heat is supplied in the first mode, and the power produced and the rate of process heat supplied in the second mode 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),
( )( )( )
( )( )( )
kJ/kg 47.65038.1009.640kJ/kg 10.38
mkPa 1kJ 1
kPa 50010,000/kgm 0.001093
/kgm 001093.0kJ/kg 09.640
kJ/kg 57.26115.1042.251kJ/kg 10.15
mkPa 1kJ 1
kPa 0210,000/kgm 0.001017
/kgm 001017.0kJ/kg 42.251
inpII,34
33
343inpII,
3MPa 5.0 @ 3
MPa 5.0 @ 3
inpI,12
33
121inpI,
3kPa 20 @ 1
kPa 20 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
whh
PPw
hh
f
f
f
f
v
vv
v
vv
2 PI
Condens.
6 Turbine
5
Boiler
Process heater
PII
4
3
7
1
8
T
1
2
8
7
5
4 3
6
Mixing chamber:
& & & & &
& & & & &
E E E E E
m h m h m h m h m hi i e e
in out system (steady)
in out − = = → =
= ⎯→⎯ = +∑ ∑∆ 0
5 5 2 2 4 4
0
s
or, ( )( ) ( )( ) kJ/kg 91.4945
47.650357.2612
5
44225 =
+=
+=
mhmhm
h&
&&
KkJ/kg 4219.6kJ/kg 4.3242
C450MPa 10
6
6
6
6⋅=
=
⎭⎬⎫
°==
sh
TP
( )( )
( )( ) kJ/kg 0.21145.23577901.042.251
7901.00752.7
8320.04219.6kPa 20
kJ/kg 6.25780.21089196.009.640
9196.09603.4
8604.14219.6MPa 5.0
88
88
68
8
77
77
67
7
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
When the entire steam is routed to the process heater,
( ) ( )( )( ) ( )( ) kW 9693
kW 3319
=−=−=
=−=−=
kJ/kg09.6406.2578kg/s 5
kJ/kg6.25784.3242kg/s 5
377process
766outT,
hhmQ
hhmW
&&
&&
(b) When only 60% of the steam is routed to the process heater,
( ) ( )
( )( ) ( )( )( ) ( )( ) kW 5816
kW 4248
=−=−=
=−+−=−+−=
kJ/kg 09.6406.2578kg/s 3
kJ/kg 0.21146.2578kg/s 2kJ/kg 6.25784.3242kg/s 5
377process
878766outT,
hhmQ
hhmhhmW
&&
&&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-73
10-77 A cogeneration plant modified with regeneration is to generate power and process heat. The mass flow rate of steam through the boiler for a net power output of 25 MW is 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),
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )( )
( )( )( )
kJ/kg 02.86757.844.858kJ/kg 57.8
mkPa 1kJ 1kPa 4009000/kgm 0.001159
/kgm 001159.0
kJ/kg 44.858
kJ/kg 41.19361.181.191kJ/kg .611
mkPa 1kJ 1kPa 101600/kgm 0.00101
/kgm 00101.0kJ/kg 81.191
inpII,45
33
454inpII,
3MPa 6.1 @ 4
MPa 6.1 @ 943
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
====
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hhhh
whh
PPw
hh
f
f
f
f
v
vv
v
vv
( )( )
( )( ) kJ/kg 2.19901.23927518.081.191
7518.04996.7
6492.02876.6kPa 10
kJ/kg 0.27304.19349675.044.858
9675.00765.4
3435.22876.6MPa 6.1
KkJ/kg 2876.6kJ/kg 8.3118
C400MPa 9
88
88
68
8
77
77
67
7
6
6
6
6
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
1
2
8
T
6
7
10 kPa
9 MPa
1.6 MPa
5
3,4,9
Boiler
6
7 Turbine
8
1
5 Condenser
Process heater
PI PII
9
2
3
fwh
4
s
Then, per kg of steam flowing through the boiler, we have
( ) ( )( ) ( )(
kJ/kg 7.869kJ/kg 2.19900.273035.01kJ/kg 0.27308.3118
)1( 8776outT,
=−−+−=
−−+−= hhyhhw)
( )( ) ( )
kJ/kg 1.86062.97.869
kJ/kg .629kJ/kg 8.57kJ/kg 1.6135.01
)1(
inp,outT,net
inpII,inpI,inp,
=−=−=
=+−=
+−=
www
wwyw
Thus,
kg/s 29.1===kJ/kg 860.1
kJ/s 25,000
net
net
wW
m&
&
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10-74
10-78 Problem 10-77 is reconsidered. The effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" y = 0.35 "fraction of steam extracted from turbine for feedwater heater and process heater" P[6] = 9000 [kPa] T[6] = 400 [C] P_extract=1600 [kPa] P[7] = P_extract P_cond=10 [kPa] P[8] = P_cond W_dot_net=25 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_pump = 100/100 "Pump isentropic efficiency" P[1] = P[8] P[2]=P[7] P[3]=P[7] P[4] = P[7] P[5]=P[6] P[9] = P[7] "Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis:" z*h[7] + (1- y)*h[2] = (1- y + z)*h[3] "Steady-flow conservation of energy" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[3]=entropy(Fluid$,P=P[3],x=0) "Process heater analysis:" (y - z)*h[7] = q_process + (y - z)*h[9] "Steady-flow conservation of energy" Q_dot_process = m_dot*(y - z)*q_process"[kW]" h[9]=enthalpy(Fluid$,P=P[9],x=0) T[9]=temperature(Fluid$,P=P[9],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[9]=entropy(Fluid$,P=P[9],x=0) "Mixing chamber at 3, 4, and 9:" (y-z)*h[9] + (1-y+z)*h[3] = 1*h[4] "Steady-flow conservation of energy" T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Condensate leaves heater as sat. liquid at P[3]" s[4]=entropy(Fluid$,P=P[4],h=h[4]) "Boiler condensate pump or Pump 2 analysis" v4=volume(Fluid$,P=P[4],x=0) w_pump2_s=v4*(P[5]-P[4])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[4]+w_pump2= h[5] "Steady-flow conservation of energy"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-75
s[5]=entropy(Fluid$,P=P[5],h=h[5]) T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Boiler analysis" q_in + h[5]=h[6]"SSSF conservation of energy for the Boiler" h[6]=enthalpy(Fluid$, T=T[6], P=P[6]) s[6]=entropy(Fluid$, T=T[6], P=P[6]) "Turbine analysis" ss[7]=s[6] hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7]) Ts[7]=temperature(Fluid$,s=ss[7],P=P[7]) h[7]=h[6]-Eta_turb*(h[6]-hs[7])"Definition of turbine efficiency for high pressure stages" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) ss[8]=s[7] hs[8]=enthalpy(Fluid$,s=ss[8],P=P[8]) Ts[8]=temperature(Fluid$,s=ss[8],P=P[8]) h[8]=h[7]-Eta_turb*(h[7]-hs[8])"Definition of turbine efficiency for low pressure stages" T[8]=temperature(Fluid$,P=P[8],h=h[8]) s[8]=entropy(Fluid$,P=P[8],h=h[8]) h[6] =y*h[7] + (1- y)*h[8] + w_turb "SSSF conservation of energy for turbine" "Condenser analysis" (1- y)*h[8]=q_out+(1- y)*h[1]"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1- y)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
9000 kPa
1600 kPa
10 kPa
SteamIAPWS
1,2
3,4,5,9
6
7
8
Pextract [kPa]
ηth m [kg/s]
Qprocess
[kW] 200 400 600 800
1000 1200 1400 1600 1800 2000
0.3778 0.3776 0.3781 0.3787 0.3794 0.3802 0.3811 0.3819 0.3828 0.3837
25.4 26.43 27.11 27.63 28.07 28.44 28.77 29.07 29.34 29.59
2770 2137 1745 1459 1235 1053 900.7 770.9 659
561.8
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. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-76
200 400 600 800 1000 1200 1400 1600 1800 20000.377
0.378
0.379
0.38
0.381
0.382
0.383
0.384
Pextract [kPa]
ηth
200 400 600 800 1000 1200 1400 1600 1800 200025
26
27
28
29
30
500
1000
1500
2000
2500
3000
Pextract [kPa]
m [
kg/s
]
Qpr
oces
s [k
W]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-77
10-79E A cogeneration plant is to generate power while meeting the process steam requirements for a certain industrial application. The net power produced, the rate of process heat supply, and the utilization factor of this 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-4E, A-5E, and A-6E),
P
Boiler
2
6
1
Turbine
Process heater
3 4 5
7
Btu/lbm 5.1229psia 120
RBtu/lbm 6348.1Btu/lbm 0.1408
F800psia 600
Btu/lbm 49.208
737
7
6543
753
3
3
3
12
F240 @ 1
=⎭⎬⎫
==
===
⋅====
⎭⎬⎫
°==
≅=≅ °
hss
P
hhhh
sssh
TP
hhhh f
( )( )( )
kW 2260==−=
−=
Btu/s 2142Btu/lbm 5.12290.1408lbm/s 12
755net hhmW &&
T
1 6
7
600 psia
120 psia
2
3,4,5 (b)
( )( ) ( )( ) ( )( )Btu/s 19,450=
−+=−−+=
−= ∑∑
49.208185.1229120.14086117766
process
hmhmhm
hmhmQ eeii
&&&
&&&
(c) εu = 1 since all the energy is utilized.
s
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-78
10-80 A Rankine steam cycle modified for a closed feedwater heater and a process heater is considered. The T-s diagram for the ideal cycle is to be sketched; the mass flow rate of the cooling water; and the utilization efficiency of the plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (b) Using the data from the problem statement, the enthalpies at various states are
( )( )( )
kJ/kg 9.2011.1081.191kJ/kg 1.10
mkPa 1kJ 1
kPa 1010000/kgm 0.00101
/kgm 00101.0kJ/kg 81.191
inpI,12
33
121inpI,
3kPa 20 @ 1
kPa 10 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
1
2
8
s
4
5 700 kPa
10 kPa
2 MPa
3
1-y-z
1-y
y
10 MPa
10
111-y-z
9 z
y
7
6
T
kJ/kg 00.697kJ/kg 47.908
kPa 700 @ 1110
kPa 2000 @ 983
===
====
f
f
hhhhhhh
An energy balance on the closed feedwater heater gives
3495.0
47.90829309.20147.908
)(
85
23
2385
=−−
=−−
=
−=−
hhhh
y
hhhhy
The process heat is expressed as
kg/s 60.3=°°⋅
−=
∆−
=
∆=−=
C)40(C)kJ/kg 18.4(kJ/kg )00.6972714(kg/s) 100(05.0)(
)(
106
106process
wpw
wpw
Tchhmz
m
TcmhhmzQ&
&
&&&
(c) The net power output is determined from
[ ]
kW 970,94kJ/kg) 1.10(kJ/kg )20893374)(05.03495.01(kJ/kg )27143374(05.0kJ/kg )29303374(3495.0
)kg/s 100(
))(1()()( 746454
PTnet
=
⎥⎦
⎤⎢⎣
⎡−−−−+−+−
=
−−−−+−+−=−=
PwhhzyhhzhhymWWW
&
&&&
The rate of heat input in the boiler is
kW 296,550kJ/kg )47.9083874)(kg/s 100()( 34in =−=−= hhmQ &&
The rate of process heat is
kW 10,085kJ/kg )00.6972714(kg/s) 100(05.0)(05.0 106process =−=−= hhmQ &&
The utilization efficiency of this cogeneration plant is
35.4%==+
=+
= 354.0kW 550,296
kW )085,10970,94(
in
processnet
Q
QWu &
&&ε
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-79
Combined Gas-Vapor Power Cycles
10-81C The energy source of the steam is the waste energy of the exhausted combustion gases.
10-82C Because the combined gas-steam cycle takes advantage of the desirable characteristics of the gas cycle at high temperature, and those of steam cycle at low temperature, and combines them. The result is a cycle that is more efficient than either cycle executed operated alone.
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10-80
10-83 A 450-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is an ideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle 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 variable specific heats.
Analysis (a) The analysis of gas cycle yields (Table A-17)
( )( )
( )
kJ/kg 02462 K460
kJ/kg 873518325450141
5450kJ/kg 421515K 1400
kJ/kg 56354019386114
3861kJ/kg 19300K 300
1212
1110
11
1010
98
9
88
1011
10
89
8
.hT
.h..PPPP
.P.hT
.h..PPP
P
.P.hT
rr
r
rr
r
=⎯→⎯=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯=
=⎯→⎯===
==⎯→⎯=
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
From the steam tables (Tables A-4, A-5, A-6),
( )
( )( )
kJ/kg 01.25259.042.251kJ/kg 0.59
mkPa 1kJ 1
kPa 20600/kgm 0.001017
/kgm 001017.0kJ/kg 42.251
inpI,12
33
121inpI,
3kPa 20 @ 1
kPa 20 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
1
2
7
1400 K
300 K 6
8 MPa
20 kPa 0.6 MPa
4
9
3
Qout·
Qin·
460 K STEAM CYCLE
GAS CYCLE
10
11
12
8
5 400°C
T
s
( )
( )( )
kJ/kg 53.67815.838.670kJ/kg 8.15
mkPa 1kJ 1kPa 6008,000/kgm 0.001101
/kgm 001101.0kJ/kg 38.670
inpI,34
33
343inpII,
3MPa 6.0 @ 3
MPa 6.0 @ 3
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
( )( )
( )( ) kJ/kg 2.20955.23577821.042.251
7821.00752.7
8320.03658.6kPa 20
kJ/kg 1.25868.20859185.038.670
9185.08285.4
9308.13658.6MPa 6.0
KkJ/kg 3658.6kJ/kg 4.3139
C400MPa 8
77
77
57
7
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
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10-81
( ) ( )
steam kg air / kg8.99 02.46280.73553.6784.3139
0
1211
45air
1211air45
outin
(steady) 0systemoutin
=−−
=−−
=
−=−⎯→⎯=
=
=∆=−
∑∑
hhhh
mm
hhmhhmhmhm
EE
EEE
s
seeii
&
&
&&&&
&&
&&&
(b) Noting that for the open FWH, the steady-flow energy balance equation yields & &Q W ke pe≅ ≅ ≅ ≅∆ ∆ 0
( ) ( ) 326336622
outin
(steady) 0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii =−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
Thus,
( )
( )( )( )( ) kJ/kg 23.9562.20951.25861792.011.25864.3139
1
extracted steam offraction the 1792.001.2521.258601.25238.670
7665
26
23
=−−+−=−−+−=
=−−
=−−
=
hhyhhw
hhhhy
T
( )( )( )
( ) ( )( ) kJ/kg 3.44419.3005.6358.73542.1515
kJ/kg 56.94815.859.01792.0123.9561
891110in,gasnet,
,,in,steamnet,
=−−−=−−−=−=
=−−−=−−−=−=
hhhhwww
wwywwww
CT
IIpIpTpT
The net work output per unit mass of gas is
( ) kJ/kg 8.54956.9483.444 99.81
steamnet,99.81
gasnet,net =+=+= www
and
( ) ( )( ) kW 720,215=−=−=
===
kJ/kg 5.63542.1515kg/s 818.5
kg/s 7.188kJ/kg 549.7
kJ/s 450,000
910airin
net
netair
hhmQ
wWm
&&
&&
(c) 62.5%===kW 720,215kW 450,000
in
net
QW
ηth &
&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-82
10-84 Problem 10-83 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input data" T[8] = 300 [K] "Gas compressor inlet" P[8] = 14.7 [kPa] "Assumed air inlet pressure" "Pratio = 14" "Pressure ratio for gas compressor" T[10] = 1400 [K] "Gas turbine inlet" T[12] = 460 [K] "Gas exit temperature from Gas-to-steam heat exchanger " P[12] = P[8] "Assumed air exit pressure" W_dot_net=450 [MW] Eta_comp = 1.0 Eta_gas_turb = 1.0 Eta_pump = 1.0 Eta_steam_turb = 1.0 P[5] = 8000 [kPa] "Steam turbine inlet" T[5] =(400+273) "[K]" "Steam turbine inlet" P[6] = 600 [kPa] "Extraction pressure for steam open feedwater heater" P[7] = 20 [kPa] "Steam condenser pressure" "GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11])
PROPRIETARY MATERIAL
preparation. If you are a student using this Manual, you are using it without permission. . © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-83
"Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"
preparation. If you are a student using this Manual, you are using it without permission.
10-84
Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam
Pratio MassRatio gastosteam
Wnetgas[kW]
Wnetsteam[kW]
ηth[%]
NetWorkRatio gastosteam
10 7.108 342944 107056 59.92 3.203 11 7.574 349014 100986 60.65 3.456 12 8.043 354353 95647 61.29 3.705 13 8.519 359110 90890 61.86 3.951 14 9.001 363394 86606 62.37 4.196 15 9.492 367285 82715 62.83 4.44 16 9.993 370849 79151 63.24 4.685 17 10.51 374135 75865 63.62 4.932 18 11.03 377182 72818 63.97 5.18 19 11.57 380024 69976 64.28 5.431 20 12.12 382687 67313 64.57 5.685
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
s [kJ/kg-K]
T [K
]
8000 kPa
600 kPa
20 kPa
Com bined Gas and Steam Pow er Cycle
8
9
10
11
12
1,23,4
5
6
7
Steam CycleSteam Cycle
Gas CycleGas Cycle
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-85
5 9 13 17 21 2555
57 .2
59 .4
61 .6
63 .8
66
P ra tio
ηth
[%
]
5 9 14 18 232.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Pratio
Net
Wor
kRat
ioga
stos
team
W dot,gas / W dot,steam vs Gas Pressure Ratio
5 9 14 18 235.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
Pratio
Mas
sRat
ioga
sto
stea
m
Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-86
10-85 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis Working around the topping cycle gives the following results:
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
K 8.530K)(8) 293( 0.4/1.4/)1(
5
656 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
K 8.57285.0
2938.530293
)()(
5656
56
56
56
56
=−
+=
−+=⎯→⎯
−
−=
−−
=
C
s
p
spsC
TTTT
TTcTTc
hhhh
η
η
K 0.75881K) 1373(
0.4/1.4/)1(
7
878 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
K 5.819)0.7581373)(90.0(1373
)()()(
877887
87
87
87
=−−=
−−=⎯→⎯−
−=
−−
= sTsp
p
sT TTTT
TTcTTc
hhhh
ηη
1
2
4s 4
3
6 MPa
20 kPa
6s
1373 K
293 K
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
320°C6
8
T
s
K 548.6 C6.275kPa 6000 @sat 9 =°== TT
Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6):
kJ/kg 5.25708.642.251kJ/kg 08.6 mkPa 1
kJ 1 kPa)206000)(/kgm 001017.0()(
/kgm 001017.0
kJ/kg 42.251
inp,12
33
121inp,
3kPa 20 @1
kPa 20 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
kJ/kg 8.2035 kPa 20
KkJ/kg 1871.6kJ/kg 6.2953
C320kPa 6000
434
4
3
3
3
3
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
shss
P
sh
TP
kJ/kg 6.2127)8.20356.2953)(90.0(6.2953
)( 433443
43
=−−=
−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη
preparation. If you are a student using this Manual, you are using it without permission.
10-87
The net work outputs from each cycle are
kJ/kg 2.275K)2937.5725.8191373)(KkJ/kg 1.005(
)()( 5687
inC,outT,cycle gas net,
=+−−⋅=
−−−=
−=
TTcTTc
www
pp
kJ/kg 9.81908.6)6.21276.2953(
)( inP,43
inP,outT,cycle steam net,
=−−=
−−=
−=
whh
www
An energy balance on the heat exchanger gives
aap
wwpa mm-hh
TTcm-hhmTTcm &&&&& 1010.0
5.2576.2953)6.5485.819)(005.1()(
)()(23
982398 =
−−
=−
=⎯→⎯=−
That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is
kg/s 279.3=×+×
==air kJ/kg )9.8191010.02.2751(
kJ/s 000,100
net
net
wW
ma
&&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-88
10-86 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
1
2
4s 4
3
6 MPa
20 kPa
6s
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
1373 K
320°C
6a6
8
293 K
T
Analysis With an ideal regenerator, the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit. Representing this state by “6a”
K 5.81986 == TT a
The rate of heat addition in the cycle is
kW 370,155K )5.8191373(C)kJ/kg 005.1(kg/s) 3.279(
)( 67in
=−°⋅=
−= apa TTcmQ &&
The thermal efficiency of the cycle is then
0.6436===kW 370,155kW 000,100
in
netth Q
W&
&η
s
Without the regenerator, the rate of heat addition and the thermal efficiency are
kW 640,224K )7.5721373(C)kJ/kg 005.1(kg/s) 3.279()( 67in =−°⋅=−= TTcmQ pa&&
0.4452===kW 640,224kW 000,100
in
netth Q
W&
&η
The change in the thermal efficiency due to using the ideal regenerator is
0.1984=−=∆ 4452.06436.0thη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-89
10-87 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob. 10-86 is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis From Problem 10-86,
kg/s 21.28)3.279(1010.01010.0kJ/kg 2.1876kJ/kg 1.2696
kJ/kg 3.804)(KkJ/kg 4627.6KkJ/kg 1871.6
KkJ/kg 8320.0K 293
K 5.819
K 1373
14out
23in,23
67in,67
4
3
kPa 20 @ 21
sink
8cycle steam source,
cycle gas source,
====−=
=−=
=−=⋅=⋅=
⋅====
==
=
aw
p
f
mmhhqhhq
TTcqss
sssT
TT
T
&&
( ) kW 2278)1871.64627.6)(K 293(kg/s) 21.28(
process) c(isentropi 0
34034 destroyed,
12 destroyed,
=−=−=
=
ssTmX
X
w&&
&
1
2
4s 4
3
6 MPa
20 kPa
6s
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
320°C
86
1373 K
293 K
s
kW 8665K 293kJ/kg 2.1876
1871.68320.0)K 293(kg/s) 21.28(
sink
out41041 destroyed,
=⎟⎠
⎞⎜⎝
⎛ +−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
Tq
ssTmX w&&
kW 11260
)8320.01871.6)(293)(21.28(5.8196.548ln)005.1()293)(3.279(
)(ln 2308
90230890exchangerheat destroyed,
=
−+⎥⎦⎤
⎢⎣⎡=
−+⎟⎟⎠
⎞⎜⎜⎝
⎛=∆+∆= ssTm
TT
cTmsTmsTmX apawa &&&&&
kW 6280)8ln()287.0(293
572.7(1.005)ln)293)(3.279(lnln5
6
5
6056 destroyed, =⎥⎦
⎤⎢⎣⎡ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
PP
RTT
cTmX pa&&
kW 23,970=⎥⎦⎤
⎢⎣⎡ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
13733.804
572.71373(1.005)ln)293)(3.279(ln
source
in
6
7067 destroyed, T
qTT
cTmX pa&&
kW 639681ln)287.0(
1373819.5(1.005)ln)293)(3.279(lnln
7
8
7
8078 destroyed, =⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
PP
RTT
cTmX pa&&
The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle.
preparation. If you are a student using this Manual, you are using it without permission.
10-90
10-88 A 280-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle 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 variable specific heats.
Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields T
( )( )
( )( ) ( )
1
2
7s
5
6s
4
9s
3
Qout·
Qin·
8
12
11s
10
GAS CYCLE
STEAM CYCLE 6
7
11
9
( )
( )( )(
kJ/kg 26.421K 420
kJ/kg .4067418.59507.116186.007.1161
kJ/kg 18.59519.151.167111
1.167kJ/kg 07.1161K 1100
kJ/kg .7466082.0/19.30084.59519.300
/
kJ/kg 84.59525.15386.111
386.1kJ/kg 19.300K 300
1212
111010111110
1110
1110
11
1010
898989
89
98
9
88
1011
10
89
8
=⎯→⎯=
=−−=
−−=⎯→⎯−−
=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯=
=−+=
−+=⎯→⎯−−
=
=⎯→⎯===
==⎯→⎯=
hT
hhhhhhhh
hPPP
P
PhT
hhhhhhhh
hPPP
P
PhT
sTs
T
srr
r
Css
C
srr
r
ηη
ηη
s )
From the steam tables (Tables A-4, A-5, and A-6),
( )
( )( )
kJ/kg 60.19280.081.191kJ/kg 0.80
mkPa 1kJ 1
kPa 01800/kgm 0.00101
/kgm 00101.0kJ/kg 81.191
inpI,12
33
121inpI,
3kPa 01 @ 1
kPa 10 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
( )
( )( )
kJ/kg 55.72568.487.720kJ/kg .684
mkPa 1kJ 1kPa 0085000/kgm 0.001115
/kgm 001115.0kJ/kg 87.720
inpI,34
33
343inpII,
3MPa 8.0 @ 3
MPa 8.0 @ 3
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
KkJ/kg 4516.6kJ/kg 3.3069
C350MPa 5
5
5
5
5⋅=
=
⎭⎬⎫
°==
sh
TP
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-91
( )( )
( ) ( )( )
( )( )
( ) ( )( ) kJ/kg 3.21865.20423.306986.03.3069
kJ/kg 5.20421.23927737.081.191
7737.04996.7
6492.04516.6kPa 10
kJ/kg 3.27301.26753.306986.03.3069
kJ/kg 1.26758.20859545.087.720
9545.06160.4
0457.24516.6MPa 8.0
755775
75
77
77
57
7
655665
65
66
66
56
6
=−−=−−=⎯→⎯−−
=
=+=+=
=−
=−
=
⎭⎬⎫
==
=−−=−−=⎯→⎯−−
=
=+=+=
=−
=−
=
⎭⎬⎫
==
sTs
T
fgfs
fg
fs
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
hhhhhhhh
hxhhs
ssx
ssP
ηη
ηη
Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( )
steam kg air / kg 9.259=−−
=−−
=
−=−⎯→⎯=
=
=∆=−
∑∑
26.42140.67455.7253.3069
0
1211
45air
1211air45
outin
(steady) 0systemoutin
hhhh
mm
hhmhhmhmhm
EE
EEE
s
seeii
&
&
&&&&
&&
&&&
(b) Noting that for the open FWH, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( ) 326336622
outin(steady) 0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii =−+⎯→⎯=+⎯→⎯=
=→=∆=−
∑∑ &&&&&
&&&&&
Thus,
( )
( )( )[ ]( ) ( )( )[ ] kJ/kg 8.7693.21863.27302082.013.27303.306986.0
1
extracted steam offraction the 2082.060.1923.273060.19287.720
7665
26
23
=−−+−=−−+−=
=−−
=−−
=
hhyhhw
hhhh
y
TT η
( )( )( )
( ) ( )( ) kJ/kg 12.12619.30074.66040.67407.1161
kJ/kg 5.76468.480.02082.018.7691
891110in,gasnet,
IIp,Ip,inp,steamnet,
=−−−=−−−=−=
=−−−=−−−=−=
hhhhwww
wwywwww
CT
TT
The net work output per unit mass of gas is
( ) kJ/kg 69.2085.764259.9112.126
425.61
steamnet,gasnet,net =+=+= www
kg/s 1341.7kJ/kg 208.69
kJ/s 280,000
net
netair ===
wW
m&
&
and ( ) ( )( ) kW 671,300=−=−= kJ/kg 74.66007.1161kg/s 1341.7910airin hhmQ &&
(c) 41.7%==== 4171.0kW 671,300kW 280,000
in
net
QW
th &
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-92
10-89 Problem 10-88 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input data" T[8] = 300 [K] "Gas compressor inlet" P[8] = 100 [kPa] "Assumed air inlet pressure" "Pratio = 11" "Pressure ratio for gas compressor" T[10] = 1100 [K] "Gas turbine inlet" T[12] = 420 [K] "Gas exit temperature from Gas-to-steam heat exchanger " P[12] = P[8] "Assumed air exit pressure" W_dot_net=280 [MW] Eta_comp = 0.82 Eta_gas_turb = 0.86 Eta_pump = 1.0 Eta_steam_turb = 0.86 P[5] = 5000 [kPa] "Steam turbine inlet" T[5] =(350+273.15) "K" "Steam turbine inlet" P[6] = 800 [kPa] "Extraction pressure for steam open feedwater heater" P[7] = 10 [kPa] "Steam condenser pressure" "GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11])
PROPRIETARY MATERIAL
preparation. If you are a student using this Manual, you are using it without permission. . © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-93
"Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"
preparation. If you are a student using this Manual, you are using it without permission.
10-94
Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam
0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
s [kJ/kg-K]
T [K
]
5000 kPa
800 kPa
10 kPa
Combined Gas and Steam Power CyclePratio MassRatiogastosteam ηth[%]
10 11 12 13 14 15 16 17 18 19 20
8.775 9.262 9.743 10.22 10.7
11.17 11.64 12.12 12.59 13.07 13.55
42.03 41.67 41.22 40.68 40.08 39.4 38.66 37.86 36.99 36.07 35.08
8
9
10
11
121,2
3,4
5
6
7
Steam CycleSteam Cycle
Gas CycleGas Cycle
10 12 14 16 18 208
9
10
11
12
13
14
Pratio
Mas
sRat
ioga
stos
team
10 12 14 16 18 2035
36
37
38
39
40
41
42
43
Pratio
ηth
[%
]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-95
10-90 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressure turbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of the combined cycle 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 variable specific heats.
Analysis (a) We obtain the air properties from EES. The analysis of gas cycle is as follows
( )( ) ( )
( )( )(
kJ/kg 62.475C200
kJ/kg 98.87179.7638.130480.08.1304
kJ/kg 79.763kPa 100
kJ/kg 6456.6kPa 700C950
kJ/kg 8.1304C950
kJ/kg 21.55780.0/16.29047.50316.290
/
kJ/kg 47.503kPa 700
kJ/kg 6648.5kPa 100
C15kJ/kg 50.288C15
1111
109910109
109
10910
10
99
9
99
787878
78
878
8
77
7
77
=⎯→⎯°=
=−−=
−−=⎯→⎯−−
=
=⎭⎬⎫
==
=⎭⎬⎫
=°=
=⎯→⎯°=
=−+=
−+=⎯→⎯−−
=
=⎭⎬⎫
==
=⎭⎬⎫
=°=
=⎯→⎯°=
hT
hhhhhhhh
hss
P
sPT
hT
hhhhhhhh
hss
P
sPT
hT
sTs
T
s
Css
C
s
ηη
ηη
)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
From the steam tables (Tables A-4, A-5, and A-6 or from EES),
( )( )( )
kJ/kg 37.19965.781.191kJ/kg .567
80.0/mkPa 1
kJ 1kPa 106000/kgm 0.00101
/
/kgm 00101.0kJ/kg 81.191
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
p
f
f
ηv
vv
( )( ) kJ/kg 4.23661.23929091.081.191
9091.04996.7
6492.04670.7kPa 10
KkJ/kg 4670.7kJ/kg 5.3264
C400MPa 1
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fss
s hxhhs
ssx
ssP
sh
TP
1
2
6ss
T
3
6 MPa
10 kPa
8s
Qout·
Qin·
7
11
10s
9
GAS CYCLE
STEAM CYCLE
950°C
15°C
5
4
1 MPa
10
8
4s
6
3
1 2
Steam turbine
Gas turbine
Condenserpump
5
4
Combustion chamber
11
Compressor
Heat exchanger
10
9 8
7
6
preparation. If you are a student using this Manual, you are using it without permission.
10-96
( )( )(
1.6%==−=−=
=⎭⎬⎫
==
=−−=
−−=⎯→⎯−−
=
0158.09842.011Percentage Moisture
9842.0kJ/kg 5.2546kPa 10
kJ/kg 0.25464.23665.326480.05.3264
6
66
6
655665
65
x
xhP
hhhhhhhh
sTs
T ηη)
(b) Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( ) ( )
[ ] kJ/kg 0.2965)62.47598.871)(10()5.3264()37.1995.3346()15.1( 44
1110air4523
outin
=⎯→⎯−=−+−
−=−+−
=
=
∑∑
hh
hhmhhmhhm
hmhm
EE
ss
eeii
&&&
&&
&&
Also,
( )sTs
T
ss
hhhhhhhh
hssP
sh
TP
433443
43
434
4
3
3
3
3 MPa 1 ?MPa 6
−−=⎯→⎯−−
=
=⎭⎬⎫
==
==
⎭⎬⎫
==
ηη
The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or using EES from the above relations. The answer is T3 = 468.0ºC. Then, the enthalpy at state 3 becomes: h3 = 3346.5 kJ/kg
(c) ( ) ( )( ) kW 4328kJ/kg 98.8718.1304kg/s 10109airgasT, =−=−= hhmW &&
( ) ( )( ) kW 2687kJ/kg 50.28821.557kg/s 1078airgasC, =−=−= hhmW &&
kW 164126874328gasC,gasT,gasnet, =−=−= WWW &&&
( ) ( )( ) kW 1265kJ/kg 0.25465.32640.29655.3346kg/s 1.156543ssteamT, =−+−=−+−= hhhhmW &&
( )( ) kW 7.8kJ/kg 564.7kg/s 1.15ssteamP, === pumpwmW &&
kW 12567.81265steamP,steamT,steamnet, =−=−= WWW &&&
kW 2897=+=+= 12561641steamnet,gasnet,plantnet, WWW &&&
(d) ( ) ( )( ) kW 7476kJ/kg 21.5578.1304kg/s 1089airin =−=−= hhmQ &&
38.8%==== 0.388kW 7476kW 2897
in
plantnet,th Q
W&
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-97
Special Topic: Binary Vapor Cycles
10-91C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.
10-92C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency.
10-93C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low.
10-94C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization.
10-95 Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottoming cycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potential energies, and assuming the heat exchanger is well-insulated, the steady-flow energy balance relation yields
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
)
( ) ( 43123142
outin
(steady) 0systemoutin 0
hhmhhmorhmhmhmhm
hmhm
EE
EEE
BABABA
iiee
−=−+=+
=
=
=∆=−
∑∑&&&&&&
&&
&&
&&&
Thus,
&
&
mm
h hh h
A
B=
−−
3 4
2 1
preparation. If you are a student using this Manual, you are using it without permission.
10-98
Review Problems
10-96 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The thermal efficiency of the cycle is to be compared when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters as a subcooled liquid.
determined power produced by the turbine and consumed by the pump 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),
kJ/kg 67.34613.654.340kJ/kg 13.6 mkPa 1
kJ 1 kPa)506000)(/kgm 001030.0()(
/kgm 001030.0
kJ/kg 54.340
inp,12
33
121inp,
3kPa 20 @1
kPa 50 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
kJ/kg 0.2495)7.2304)(9348.0(54.340
9348.05019.6
0912.11693.7
kPa 50
KkJ/kg 1693.7kJ/kg 8.3658
C600kPa 6000
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
qin
qout
50 kPa 1
3
2
4
6 MPa
T
s
Thus,
kJ/kg 5.215454.3400.2495kJ/kg 1.331267.3468.3658
14out
23in
=−=−==−=−=
hhqhhq
and the thermal efficiency of the cycle is
0.3495=−=−=1.33125.215411
in
outth q
qη
When the liquid enters the pump 11.3°C cooler than a saturated liquid at the condenser pressure, the enthalpies become
/kgm 001023.0
kJ/kg 07.293
C703.113.813.11kPa 50
3C70 @ 1
C70 @ 1
kPa 50 @sat 1
1
=≅
=≅
⎭⎬⎫
°=−=−==
°
°
f
fhhTT
Pvv
kJ/kg 09.6 mkPa 1
kJ 1 kPa)506000)(/kgm 001023.0()(
33
121inp,
=⎟⎠
⎞⎜⎝
⎛
⋅−=
−= PPw v
kJ/kg 16.29909.607.293inp,12 =+=+= whh
Then,
kJ/kg 9.220109.2930.2495kJ/kg 6.335916.2998.3658
14out
23in
=−=−==−=−=
hhqhhq
0.3446=−=−=6.33599.220111
in
outth q
qη
The thermal efficiency slightly decreases as a result of subcooling at the pump inlet.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-99
10-97E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as the working fluid is considered. The exit temperature of the geothermal water from the vaporizer, the rate of heat rejection from the working fluid in the condenser, the mass flow rate of geothermal water at the preheater, and the thermal efficiency of the Level I cycle of this plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine),
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
)
)
)
( )( )( )(
F267.4°=°−°⋅=−
−=
2
2
12brinebrine
F325FBtu/lbm 1.03lbm/h 384,286Btu/h 000,790,22T
T
TTcmQ p&&
(b) The rate of heat rejection from the working fluid to the air in the condenser is determined from the steady-flow energy balance on air,
( )( )( )(
MBtu/h 29.7=°−°⋅=
−=
F555.84FBtu/lbm 0.24lbm/h 4,195,10089airair TTcmQ p&&
(c) The mass flow rate of geothermal water at the preheater is determined from the steady-flow energy balance on the geothermal water,
( )( )(
lbm/h 187,120=
°−°⋅=−
−=
geo
geo
inoutgeogeo
F8.2110.154FBtu/lbm 1.03Btu/h 000,140,11
m
m
TTcmQ p
&
&
&&
(d) The rate of heat input is
and
& & & , , , ,
, ,
&
Q Q Q
W
in vaporizer reheater
net
Btu / h
kW
= + = +
=
= − =
22 790 000 11140 000
33 930 000
1271 200 1071
Then,
10.8%=⎟⎟⎠
⎞⎜⎜⎝
⎛==
kWh 1Btu 3412.14
Btu/h 33,930,000kW 1071
in
netth Q
W&
&η
preparation. If you are a student using this Manual, you are using it without permission.
10-100
10-98 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. 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),
( )( )( )
kJ/kg 37.29919.1018.289kJ/kg .1910
mkPa 1kJ 1
kPa 3010,000/kgm 0.001022
/kgm 001022.0kJ/kg 18.289
inp,12
33
121inp,
3kPa 30 @ 1
kPa 30 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
T
( )( ) kJ/kg 1.25573.23359711.027.289
9711.08234.6
9441.05706.7kPa 30
KkJ/kg 5706.7kJ/kg 4.3578
C550MPa 2
kJ/kg 1.3321MPa 2
KkJ/kg 2335.7kJ/kg 7.3559
C550MPa 4
kJ/kg 9.3204MPa 4
KkJ/kg 7561.6kJ/kg 9.3500
C550MPa 10
88
88
78
8
7
7
7
7
656
6
5
5
5
5
434
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
hssP
sh
TP
hssP
sh
TP
1
5
2
8
3
4
7
6
2 MPa 4 MPa
30 kPa
10 MPa
s
Then,
( ) ( ) ( )
kJ/kg 8.15459.22677.3813kJ/kg 9.226718.2891.2557
kJ/kg 7.38131.33214.35789.32047.355937.2999.3500
outinnet
18out
674523in
=−=−==−=−=
=−+−+−=−+−+−=
qqwhhq
hhhhhhq
Thus,
40.5%==== 4053.0kJ/kg 3813.7kJ/kg 1545.8
in
netth q
wη
(b) The mass flow rate of the steam is then
kg/s 48.5===kJ/kg 1545.8kJ/s 75,000
net
net
wW
m&
&
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-101
10-99 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )( )
kPa 2780=⎭⎬⎫
=°=
⋅=+=+==+=+=
⎭⎬⎫
==
565
5
66
66
6
6
C600
KkJ/kg 5488.74996.792.06492.0kJ/kg 5.23921.239292.081.191
92.0kPa 10
Pss
T
sxsshxhh
xP
fgf
fgf
T
1
5
2
6
s
3
425
MPa
10 kPa
SINGLE600°C
1
5
2
8
3
4
7
6
DOUBLE
10 kPa
25 MPa
s
T600°C
(b) Double Reheat:
C600 and
KkJ/kg 3637.6C600MPa 25
5
5
34
4
33
3
°==
==
⋅=⎭⎬⎫
°==
TPP
ssPP
sTP
xx
Any pressure Px selected between the limits of 25 MPa and 2.78 MPa will satisfy the requirements, and can be used for the double reheat pressure.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-102
10-100 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis
T
y
P II P I
Open fwh
Condenser
Boiler
4
3 2
1
6
Turbine
7
1-y
5
1
qin
2
7s 7
5
6s
10 MPa
10 kPa
0.5 MPa
1-y
4
y 3
qout
6
s
(a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( ) ( )
/kgm 001093.0kJ/kg 09.640
liquidsat.MPa 5.0
kJ/kg 33.19252.081.191kJ/kg 0.52
95.0/mkPa 1
kJ 1kPa 10500/kgm 0.00101
/
/kgm 00101.0kJ/kg 81.191
3MPa 5.0 @ 3
MPa 5.0 @ 33
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
====
⎭⎬⎫=
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
η−=
==
==
f
f
p
f
f
hhP
whh
PPw
hh
vv
v
vv
( )( )( ) ( )
kJ/kg 02.65193.1009.640kJ/kg 10.93
95.0/mkPa 1
kJ 1kPa 50010,000/kgm 0.001093
/
inpII,34
33
343inpII,
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
η−=
whh
PPw pv
( )(kJ/kg 1.2654
0.21089554.009.640
9554.09603.4
8604.15995.6
MPa 5.0
KkJ/kg 5995.6kJ/kg 1.3375
C500MPa 10
66
66
56
6
5
5
5
5
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgsfs
fg
fss
s
s hxhhs
ssx
ssP
sh
TP
)
( )( )(
kJ/kg 3.27981.26541.337580.01.3375
655665
65
=−−=
−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη)
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-103
( )(kJ/kg 7.2089
1.23927934.081.191
7934.04996.7
6492.05995.6
kPa 1077
77
57
7
=+=+=
=−
=−
=
⎭⎬⎫
==
fgsfs
fg
fss
s
s hxhhs
ssx
ssP )
( )( )(
kJ/kg 8.23467.20891.337580.01.3375
755775
75
=−−=
−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη)
)
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that , & &Q W ke pe≅ ≅ ≅ ≅∆ ∆ 0
( ) ( 326332266
outin
(steady) 0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii =−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
where y is the fraction of steam extracted from the turbine ( = & / &m m6 3 ). Solving for y,
1718.033.1923.279833.19209.640
26
23 =−−
=−−
=hhhh
y
Then, ( )( ) ( )( )
kJ/kg 4.9397.17841.2724kJ/kg 7.178481.1918.23461718.011
kJ/kg 1.272402.6511.3375
outinnet
17out
45in
=−=−==−−=−−=
=−=−=
qqwhhyq
hhq
and
skg 7159 /.kJ/kg 939.4
kJ/s 150,000
net
net ===wW
m&
&
(b) The thermal efficiency is determined from
34.5%=−=−=kJ/kg 2724.1kJ/kg 1784.7
11in
outth q
qη
Also,
KkJ/kg 6492.0
KkJ/kg 8604.1
KkJ/kg 9453.6kJ/kg 3.2798
MPa 5.0
kPa 10 @ 12
MPa 5.0 @ 3
66
6
⋅===⋅==
⋅=⎭⎬⎫
==
f
f
sssss
shP
Then the irreversibility (or exergy destruction) associated with this regeneration process is
( )[ ]
( ) ( )( ) ( )( )[ ]kJ/kg 39.25=
−−−=
−−−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−== ∑∑
6492.01718.019453.61718.08604.1K 303
1 2630
0surr
0gen0regen syyssTT
qsmsmTsTi
Liiee
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-104
10-101 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis
T
1
qin
2
7
1-y
5
6
qout
10 MPa
10 kPa
0.5 MPa
4
y 3
s
y
P II P I
Open fwh
Condenser
Boiler
4
3 2
1
6
Turbine
7 1-y
5
(a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
/kgm 001093.0kJ/kg 09.640
liquidsat.MPa 5.0
kJ/kg 30.19250.081.191
kJ/kg 0.50mkPa 1
kJ 1kPa 10500/kgm 0.00101
/kgm 00101.0kJ/k 81.191
3MPa 5.0 @ 3
MPa 5.0 @ 33
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
====
⎭⎬⎫=
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
f
f
f
f
hhP
whh
PPw
ghh
vv
v
vv
( )( )( )
( )( )
( )( ) kJ/kg 7.20891.23927934.081.191
7934.04996.7
6492.05995.6kPa 10
kJ/kg 1.26540.21089554.009.640
9554.09603.4
8604.15995.6MPa 5.0
KkJ/kg 5995.6kJ/kg 1.3375
C500MPa 10
kJ/kg 47.65038.1009.640
kJ/kg .3810mkPa 1
kJ 1kPa 50010,000/kgm 0.001093
77
77
57
7
66
66
56
6
5
5
5
5
inpII,34
33
343inpII,
=+=+=
=−
=−
=
⎭⎬⎫
==
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
fgf
fg
f
fgf
fg
f
hxhhs
ssx
ssP
hxhhs
ssx
ssP
sh
TP
whh
PPw v
The fraction of steam extracted is determined from the steady-flow energy equation applied to the feedwater heaters. Noting that , 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( 326332266
outin(steady) 0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii =−+⎯→⎯=+⎯→⎯=
=→=∆=−
∑∑ &&&&&
&&&&&
)
where y is the fraction of steam extracted from the turbine ( = & / &m m6 3 ). Solving for y,
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-105
1819.031.1921.265431.19209.640
26
23 =−−
=−−
=hhhh
y
Then, ( )( ) ( )( )
kJ/kg 0.11727.15526.2724kJ/kg 7.155281.1917.20891819.011
kJ/kg 6.272447.6501.3375
outinnet
17out
45in
=−=−==−−=−−=
=−=−=
qqwhhyq
hhq
and skg 0128 /.kJ/kg 1171.9
kJ/s 150,000
net
net ===wWm&
&
(b) The thermal efficiency is determined from
%0.43=−=−=kJ/kg 2724.7kJ/kg 1552.7
11in
outth q
qη
Also,
KkJ/kg 6492.0
KkJ/kg 8604.1KkJ/kg 5995.6
kPa 10 @ 12
MPa 5.0 @ 3
56
⋅===
⋅==⋅==
f
f
sss
ssss
Then the irreversibility (or exergy destruction) associated with this regeneration process is
( )[ ]
( ) ( )( ) ( )( )[ ]kJ/kg 39.0=
−−−=
−−−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−== ∑∑
6492.01819.015995.61819.08604.1K 303
1 2630
0surr
0gen0regen syyssTT
qsmsmTsTi
Liiee
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-106
10-102 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration 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),
( )( )( )
/kgm 001101.0kJ/k 38.670
liquid sat.MPa 6.0
kJ/kg 53.22659.094.225kJ/kg 0.59
mkPa 1kJ 1
kPa 15600/kgm 0.001014
/kgm 001014.0kJ/kg 94.225
3MPa 6.0 @ 3
MPa 6.0 @ 33
inpI,12
33
121inpI,
3kPa 15 @ 1
kPa 15 @ 1
====
⎭⎬⎫=
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
f
f
f
f
ghhP
whh
PPw
hh
vv
v
vv
( )( )( )
( )( ) kJ/kg 8.25183.23729665.094.225
9665.02522.7
7549.07642.7kPa 15
kJ/kg 2.3310MPa 6.0
KkJ/kg 7642.7kJ/kg 1.3479
C500MPa 0.1
kJ/kg 8.2783MPa 0.1
KkJ/kg 5995.6kJ/kg 1.3375
C500MPa 10
kJ/kg 73.68035.1038.670kJ/kg 10.35
mkPa 1kJ 1kPa 60010,000/kgm 0.001101
99
99
79
9
878
8
7
7
7
7
656
6
5
5
5
5
inpII,34
33
343inpII,
=+=+=
=−
=−
=
⎭⎬⎫
==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
fgf
fg
f
hxhhs
ssx
ssP
hssP
sh
TP
hssP
sh
TP
whh
PPw v
y
Condens.
5
1 2
3
4
Turbine Boiler
Open fwh
P I P II
7
6
8 9
1-y
1
7
2
9
5
6 8
1 MPa
15 kPa
10 MPa
0.6 MPa
4
3
T
s
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that , 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( 328332288
outin(steady) 0
systemoutin
11
0
hhyyhhmhmhmhmhm
EEEEE
eeii =−+⎯→⎯=+⎯→⎯=
=→=∆=−
∑∑ &&&&&
&&&&&
)
where y is the fraction of steam extracted from the turbine ( = & / &m m8 3 ). Solving for y,
0.144=−−
=−−
=53.2262.331053.22638.670
28
23
hhhhy
(b) The thermal efficiency is determined from
( ) ( ) ( ) ( )( )( ) ( )( ) kJ/kg 7.196294.2258.25181440.011
kJ/kg 7.33898.27831.347973.6801.3375
19out
6745in
=−−=−−==−+−=−+−=
hhyqhhhhq
and 42.1%=−=−=kJ/kg 3389.7kJ/kg 1962.7
11in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-107
10-103 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration 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
T
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
y
P II P I
Open fwh
Condenser4
3 2
1
5
8
Boiler
7
6 Turbine
9 1-y
1
7
2
9
5
6 8 4
1-y
9s
6s 8sy
3
s
(a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
/kgm 001101.0kJ/kg 38.670
liquid sat.MPa 6.0
kJ/kg 54.22659.094.225kJ/kg 0.59
mkPa 1kJ 1
kPa 15600/kgm 0.001014
/kgm 001014.0kJ/kg 94.225
3MPa 6.0 @3
MPa 6.0 @33
in,12
33
121in,
3kPa 15 @1
kPa 15 @1
====
⎭⎬⎫=
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
f
f
pI
pI
f
f
hhP
whh
PPw
hh
vv
v
vv
( )( )( )
kJ/kg 8.2783MPa 0.1
KkJ/kg 5995.6kJ/kg 1.3375
C500MPa 10
kJ/kg 73.68035.1038.670kJ/kg 10.35
mkPa 1kJ 1kPa 60010,000/kgm 0.001101
656
6
5
5
5
5
inpII,34
33
343inpII,
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
ss
s hssP
sh
TP
whh
PPw v
( )( )(
kJ/kg 4.28788.27831.337584.01.3375
655665
65
=−−=
−η−=⎯→⎯−−
=η sTs
T hhhhhhhh
)
( ) ( )(kJ/kg 2.3337
2.33101.347984.01.3479
kJ/kg 2.3310MPa 6.0
KkJ/kg 7642.7kJ/kg 1.3479
C500MPa 0.1
877887
87
878
8
7
7
7
7
=−−=−η−=⎯→⎯
−−
=η
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
sTs
T
ss
s
hhhhhhhh
hssP
sh
TP
)
preparation. If you are a student using this Manual, you are using it without permission.
10-108
( )( )
( ) ( )(kJ/kg 5.2672
8.25181.347984.01.3479
kJ/kg 8.25183.23729665.094.225
9665.02522.7
7549.07642.7kPa 15
977997
97
99
99
79
9
=−−=−−=⎯→⎯
−−
=
=+=+=
=−
=−
=
⎭⎬⎫
==
sTs
T
fgsfs
fg
fss
s
s
hhhhhhhh
hxhhs
ssx
ssP
ηη )
)
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that , 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( 328332288
outin
(steady) 0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii =−+⎯→⎯=+⎯→⎯=
=
=∆=−
∑∑ &&&&&
&&
&&&
where y is the fraction of steam extracted from the turbine ( = & / &m m8 3 ). Solving for y,
0.1427=−−
=−−
=53.2263.333553.22638.670
28
23
hhhhy
(b) The thermal efficiency is determined from
( ) ( )( ) ( )( )( ) ( )( ) kJ/kg 2.209794.2255.26721427.011
kJ/kg 1.32954.28781.347973.6801.3375
19out
6745in
=−−=−−==−+−=
−+−=
hhyq
hhhhq
and
36.4%=−=−=kJ/kg 3295.1kJ/kg 2097.2
11in
outth q
qη
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-109
10-104 A steam power plant operating on the ideal reheat-regenerative Rankine cycle with three feedwater heaters is considered. Various items for this system per unit of mass flow rate through the boiler are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
16
n
13
1
Low-P Turbine
Boiler
Condenser
High-P Turbine
14
2
19
15m
z
P IV
P I
Closed FWH I
12
11 10 9
8
6 34 5
7
17
ClosedFWH II
OpenFWH P II
yx
P III
18
Analysis The compression processes in the pumps and the expansion processes in the turbines are isentropic. Also, the state of water at the inlet of pumps is saturated liquid. Then, from the steam tables (Tables A-4, A-5, and A-6),
kJ/kg 8.1011kJ/kg 3.1008kJ/kg 86.885kJ/kg 46.884kJ/kg 28.419kJ/kg 51.417kJ/kg 84.168kJ/kg 75.168
10
9
6
5
4
3
2
1
========
hhhhhhhh
kJ/kg 7.2454kJ/kg 5.2891kJ/kg 3.3481kJ/kg 0.2871kJ/kg 6.3063kJ/kg 5.3204kJ/kg 1.3423
19
18
17
16
15
14
13
=======
hhhhhhh
For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. Then,
kJ/kg 8.1008 C9.233
kPa 3000
kJ/kg 91.884 C1.207
kPa 1800
11911
11
757
7
=⎭⎬⎫
°===
=⎭⎬⎫
°===
hTT
P
hTT
P
Enthalpies at other states and the fractions of steam extracted from the turbines can be determined from mass and energy balances on cycle components as follows:
Mass Balances:
znm
zyx=+=++ 1
Open feedwater heater:
3218 zhnhmh =+
Closed feedwater heater-II:
57154 yhzhyhzh +=+
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. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-110
Closed feedwater heater-I:
911148 )()( xhhzyxhhzy ++=++
Mixing chamber after closed feedwater heater II:
867 )( hzyyhzh +=+
Mixing chamber after closed feedwater heater I:
121110 1)( hhzyxh =++
Substituting the values and solving the above equations simultaneously using EES, we obtain
0.708820.07124
0.16670.05334
=======
nmzyx
hh
78000.0
kJ/kg 0.1009kJ/kg 08.885
12
8
Note that these values may also be obtained by a hand solution by using the equations above with some rearrangements and substitutions. Other results of the cycle are
43.9%
kJ/kg 1620kJ/kg 2890
kJ/kg 769.6kJ/kg 502.3
==−=−=
=−==−+−=
=−+−=
=−+−+−=
4394.02890162011
)()(
)()()()()(
in
outth
119out
16171213in
19171817LPout,T,
161315131413HPout,T,
hhnqhhzhhq
hhnhhmwhhzhhyhhxw
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-111
10-105 The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined using EES.
Analysis The EES program used to solve this problem as well as the solutions are given below.
"Given" P_boiler=6000 [kPa] P_cfwh1=3000 [kPa] P_cfwh2=1800 [kPa] P_reheat=800 [kPa] "P_ofwh=100 [kPa]" P_condenser=7.5 [kPa] T_turbine=500 [C] "Analysis" Fluid$='steam_iapws' "turbines" h[13]=enthalpy(Fluid$, P=P_boiler, T=T_turbine) s[13]=entropy(Fluid$, P=P_boiler, T=T_turbine) h[14]=enthalpy(Fluid$, P=P_cfwh1, s=s[13]) h[15]=enthalpy(Fluid$, P=P_cfwh2, s=s[13]) h[16]=enthalpy(Fluid$, P=P_reheat, s=s[13]) h[17]=enthalpy(Fluid$, P=P_reheat, T=T_turbine) s[17]=entropy(Fluid$, P=P_reheat, T=T_turbine) h[18]=enthalpy(Fluid$, P=P_ofwh, s=s[17]) h[19]=enthalpy(Fluid$, P=P_condenser, s=s[17]) "pump I" h[1]=enthalpy(Fluid$, P=P_condenser, x=0) v[1]=volume(Fluid$, P=P_condenser, x=0) w_pI_in=v[1]*(P_ofwh-P_condenser) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P_ofwh, x=0) v[3]=volume(Fluid$, P=P_ofwh, x=0) w_pII_in=v[3]*(P_cfwh2-P_ofwh) h[4]=h[3]+w_pII_in "pump III" h[5]=enthalpy(Fluid$, P=P_cfwh2, x=0) T[5]=temperature(Fluid$, P=P_cfwh2, x=0) v[5]=volume(Fluid$, P=P_cfwh2, x=0) w_pIII_in=v[5]*(P_cfwh1-P_cfwh2) h[6]=h[5]+w_pIII_in "pump IV" h[9]=enthalpy(Fluid$, P=P_cfwh1, x=0) T[9]=temperature(Fluid$, P=P_cfwh1, x=0) v[9]=volume(Fluid$, P=P_cfwh1, x=0) w_p4_in=v[5]*(P_boiler-P_cfwh1) h[10]=h[9]+w_p4_in "Mass balances" x+y+z=1 m+n=z
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10-112
"Open feedwater heater" m*h[18]+n*h[2]=z*h[3] "closed feedwater heater 2" T[7]=T[5] h[7]=enthalpy(Fluid$, P=P_cfwh1, T=T[7]) z*h[4]+y*h[15]=z*h[7]+y*h[5] "closed feedwater heater 1" T[11]=T[9] h[11]=enthalpy(Fluid$, P=P_boiler, T=T[11]) (y+z)*h[8]+x*h[14]=(y+z)*h[11]+x*h[9] "Mixing chamber after closed feedwater heater 2" z*h[7]+y*h[6]=(y+z)*h[8] "Mixing chamber after closed feedwater heater 1" x*h[10]+(y+z)*h[11]=1*h[12] "cycle" w_T_out_high=x*(h[13]-h[14])+y*(h[13]-h[15])+z*(h[13]-h[16]) w_T_out_low=m*(h[17]-h[18])+n*(h[17]-h[19]) q_in=h[13]-h[12]+z*(h[17]-h[16]) q_out=n*(h[19]-h[1]) Eta_th=1-q_out/q_in
P open fwh
[kPa] ηth
10 20 30 40 50 60 70 80 90
100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300
0.429828 0.433780 0.435764 0.436978 0.437790 0.438359 0.438768 0.439065 0.439280 0.439432 0.439536 0.439602 0.439638 0.439647 0.439636 0.439608 0.439565 0.439509 0.439442 0.439367 0.439283 0.439192 0.439095 0.438993 0.438887 0.438776 0.438662 0.438544 0.438424 0.438301
0 50 100 150 200 250 3000.428
0.43
0.432
0.434
0.436
0.438
0.44
Pofwh [kPa]
ηth
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10-113
10-106 A cogeneration plant is to produce power and process heat. There are two turbines in the cycle: a high-pressure turbine and a low-pressure turbine. The temperature, pressure, and mass flow rate of steam at the inlet of high-pressure turbine 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),
( )( )
( )( )(
kJ/kg 1.23446.20479.278860.09.2788
kJ/kg 6.20471.23927758.081.191
7758.04996.7
6492.04675.6
kPa 10
KkJ/kg 4675.6kJ/kg 9.2788
vapor sat.MPa 4.1
544554
54
55
45
45
5
MPa 4.1 @ 4
MPa 4.1 @ 44
=−−=
−−=⎯→⎯−−
=
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅====
⎭⎬⎫=
sTs
T
fgsfs
fg
fss
s
g
g
hhhhhhhh
hxhhs
ssx
ssP
sshhP
ηη)
1
2
5s
T
5
44
3
and
kg/min 9.107kg/s 799.1
kJ/kg 444.8kJ/s 800
kJ/kg 8.4441.23449.2788
lowturb,
IIturb,turblow
54lowturb,
====
=−=−=
wW
m
hhw&
&
Therefore ,
kJ/kg 0.28439.278815.54
kJ/kg 54.15kg/s 18.47kJ/s 1000
=kg/min 11081081000
4highturb,3
43turbhigh,
,turbhighturb,
total
=+=+=
−====
=+=
hwh
hhmW
w
m
I
&
&
& kg/s 18.47
( )( ) ( )
( )( ) KkJ/kg 4289.61840.49908.02835.2
9908.09.1958
96.8298.2770MPa 4.1
kJ/kg 8.277075.0/9.27880.28430.2843
/
44
44
34
4
433443
43
⋅=+=+=
=−
=−
=
⎭⎬⎫
==
=−−=
−−=⎯→⎯−−
=
fgsfs
fg
fss
s
s
Tss
T
sxssh
hhx
ssP
hhhhhhhh
ηη
Then from the tables or the software, the turbine inlet temperature and pressure becomes
C227.5MPa 2
°==
⎭⎬⎫
⋅==
3
3
3
3KkJ/kg 4289.6
kJ/kg 0.2843TP
sh
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10-114
10-107 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. The rate of process heat, the net power produced, and the utilization factor of the plant are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
1
2
8s 8
T
6
7
8 MPa
20 kPa
2 MPa 4
5
3
Qout·
Qin·
Qprocess·
Conden.
6 Turbine
Processhe
ater
I 3
7
4 2 P I P I
Boiler
5
1
8
s
Analysis From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg 47.908
kJ/kg 71.25329.242.251kJ/kg .292
88.0/mkPa 1
kJ 1kPa 022000/kgm 0.001017
/
/kgm 001017.0kJ/kg 42.251
MPa 2 @ 3
inpI,12
33
121inpI,
3kPa 02 @ 1
kPa 20 @ 1
==
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
f
p
f
f
hh
whh
PPw
hh
ηv
vv
Mixing chamber:
kJ/kg 81.491kg/s) 11()kJ/kg) 71kg/s)(253. 411(kJ/kg) 47kg/s)(908. 4( 44
442233
=⎯→⎯=−+
=+
hh
hmhmhm &&&
( )( )( )
kJ/kg 02.49921.781.491kJ/kg .217
88.0/mkPa 1
kJ 1kPa 00208000/kgm 0.001058
/
/kgm 001058.0
inII,45
33
454inII,
3kJ/kg 81.491 @ 4
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
=≅ =
p
pp
hf
whh
PPw
f
ηv
vv
KkJ/kg 7266.6kJ/kg 5.3399
C500MPa 8
6
6
6
6⋅=
=
⎭⎬⎫
°==
sh
TP
kJ/kg 4.3000MPa 2
767
7 =⎭⎬⎫
==
shss
P
( ) ( )( ) kJ/kg 3.30484.30005.339988.05.3399766776
76 =−−=−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη
kJ/kg 5.2215kPa 20
868
8 =⎭⎬⎫
==
shss
P
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-115
( ) ( )( ) kJ/kg 6.23575.22155.339988.05.3399866886
86 =−−=−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη
Then,
( ) ( )( ) kW 8559=−=−= kJ/kg 47.9083.3048kg/s 4377process hhmQ &&
(b) Cycle analysis:
( ) ( )( )( ) ( )( )
( )( ) ( )( )
kW 8603=−=−=
=+=+=
=−+−=
−+−=
958698
kW 95kJ/kg 7.21kg/s 11kJ/kg 2.29kg/s 7
kW 8698kJ/kg6.23575.3399kg/s 7kJ/kg3.30485.3399kg/s 4
inp,outT,net
inpII,4inpI,1inp,
868767outT,
WWW
wmwmW
hhmhhmW
&&&
&&&
&&&
(c) Then,
and ( ) ( )( )
53.8%==+
=+
=
=−=−=
538.0905,31
85598603
kW 905,3102.4995.3399kg/s 11
in
processnet
565in
Q
QW
hhmQ
u &
&&
&&
ε
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-116
10-108E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable for Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
Analysis Working around the topping cycle gives the following results:
R 1043R)(10) 540( 0.4/1.4/)1(
5
656 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
R 109990.0
5401043540
)()(
5656
56
56
56
56
=−
+=
−+=⎯→⎯
−
−=
−−
=
C
s
p
spsC
TTTT
TTcTTc
hhhh
η
η
R 1326101R) 2560(
0.4/1.4/)1(
7
878 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
R 1449)13262560)(90.0(2560
)()()(
877887
87
87
87
=−−=
−−=⎯→⎯−
−=
−−
= sTsp
p
sT TTTT
TTcTTc
hhhh
ηη 1
2
4s 4
3
800 psia
5 psia
6s
2560 R
540 R
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
600°F 6
8
T
s
R 102850R 3.97850psia 800 @sat 9 =+=+= TT
Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E):
Btu/lbm 59.13241.218.130Btu/lbm 41.2
ftpsia 5.404Btu 1 psia)5800)(/lbmft 01641.0(
)(
/lbmft 01641.0
Btu/lbm 18.130
inp,12
33
121inp,
3psia 5 @1
psia 5 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 6.908 psia 5
RBtu/lbm 4866.1Btu/lbm 9.1270
F600
psia 800
434
4
3
3
3
3
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
shss
P
sh
TP
Btu/lbm 7.926)6.9089.1270)(95.0(9.1270
)( 433443
43
=−−=
−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη
The net work outputs from each cycle are
Btu/lbm 5.132R)540109914492560)(RBtu/lbm 240.0(
)()( 5687
inC,outT,cycle gas net,
=+−−⋅=
−−−=
−=
TTcTTc
www
pp
preparation. If you are a student using this Manual, you are using it without permission.
10-117
Btu/lbm 8.34141.2)7.9269.1270(
)( inP,43
inP,outT,cycle steam net,
=−−=
−−=
−=
whh
www
An energy balance on the heat exchanger gives
aap
wwpa mm-hh
TTcm-hhmTTcm &&&&& 08876.0
59.1329.1270)10281449)(240.0()(
)()(23
982398 =
−−
=−
=⎯→⎯=−
That is, 1 lbm of exhaust gases can heat only 0.08876 lbm of water. Then the heat input, the heat output and the thermal efficiency are
Btu/lbm 8.187Btu/lbm )18.1307.926(08876.0R)5401028)(RBtu/lbm 240.0(1
)()(
Btu/lbm 6.350R)10992560)(RBtu/lbm 240.0()(
1459out
67in
=−×+−⋅×=
−+−=
=−⋅=−=
hhmm
TTcmm
q
TTcmm
q
a
wp
a
a
pa
a
&
&
&
&
&
&
0.4643=−=−=6.3508.18711
in
outth q
qη
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10-118
10-109E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
1
2
4s 4
3
800 psia
10 psia
6s
2560 R
540 R
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
600°F 6
8
TAnalysis Working around the topping cycle gives the following results:
R 1043R)(10) 540( 0.4/1.4/)1(
5
656 ==⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT
R 109990.0
5401043540
)()(
5656
56
56
56
56
=−
+=
−+=⎯→⎯
−
−=
−−
=
C
s
p
spsC
TTTT
TTcTTc
hhhh
η
η
R 1326101R) 2560(
0.4/1.4/)1(
7
878 =⎟
⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
s PP
TT s
R 1449)13262560)(90.0(2560
)()()(
877887
87
87
87
=−−=
−−=⎯→⎯−
−=
−−
= sTsp
p
sT TTTT
TTcTTc
hhhh
ηη
R 102850R 3.97850psia 800 @sat 9 =+=+= TT
Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E):
Btu/lbm 7.16343.225.161Btu/lbm 43.2
ftpsia 5.404Btu 1 psia)10800)(/lbmft 01659.0(
)(
/lbmft 01659.0
Btu/lbm 25.161
inp,12
33
121inp,
3psia 10 @1
psia 10 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 6.946 psia 10
RBtu/lbm 4866.1Btu/lbm 9.1270
F600
psia 800
434
4
3
3
3
3
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
shss
P
sh
TP
Btu/lbm 8.962)6.9469.1270)(95.0(9.1270
)( 433443
43
=−−=
−−=⎯→⎯−−
= sTs
T hhhhhhhh
ηη
The net work outputs from each cycle are
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10-119
Btu/lbm 5.132R)540109914492560)(RBtu/lbm 240.0(
)()( 5687
inC,outT,cycle gas net,
=+−−⋅=
−−−=
−=
TTcTTc
www
pp
Btu/lbm 7.30543.2)8.9629.1270(
)( inP,43
inP,outT,cycle steam net,
=−−=
−−=
−=
whh
www
An energy balance on the heat exchanger gives
aap
wwpa mmhh
TTcmhhmTTcm &&&&& 09126.0
7.1639.1270)10281449)(240.0()(
)()(23
982398 =
−−
=−
−=⎯→⎯−=−
That is, 1 lbm of exhaust gases can heat only 0.09126 lbm of water. Then the heat input, the heat output and the thermal efficiency are
Btu/lbm 3.190Btu/lbm )25.1618.962(09126.0R)5401028)(RBtu/lbm 240.0(1
)()(
Btu/lbm 6.350R)10992560)(RBtu/lbm 240.0()(
1459out
67in
=−×+−⋅×=
−+−=
=−⋅=−=
hhmm
TTcmm
q
TTcmm
q
a
wp
a
a
pa
a
&
&
&
&
&
&
0.4573=−=−=6.3503.19011
in
outth q
qη
When the condenser pressure is increased from 5 psia to 10 psia, the thermal efficiency is decreased from 0.4643 to 0.4573.
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10-120
10-110E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The cycle supplies a specified rate of heat to the buildings during winter. The mass flow rate of air and the net power output from the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable to Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
Analysis The mass flow rate of water is
lbm/h 2495Btu/lbm 161.25)-(962.8
Btu/h 102 6
14
buildings =×
=−
=hh
Qmw
&&
The mass flow rate of air is then
lbm/h 27,340===0.09126
249509126.0
wa
mm
&&
The power outputs from each cycle are
kW 1062Btu/h 3412.14
kW 1R)540109914492560)(RBtu/lbm 240.0(lbm/h) 340,27(
)()(
)(
5687
inC,outT,cycle gas net,
=
⎟⎠⎞
⎜⎝⎛+−−⋅=
−−−=
−=
TTcmTTcm
wwmW
papa
a
&&
&&
1
2
4s 4
3
800 psia
10 psia
6s
2560 R
540 R
Qout·
Qin·
5
9
8s
7
GAS CYCLE
STEAM CYCLE
600°F6
8
T
s
kW 224Btu/h 3412.14
kW 1)43.28.9629.1270(lbm/h) 2495(
)(
)(
inP,43
inP,outT,cycle steam net,
=
⎟⎠⎞
⎜⎝⎛−−=
−−=
−=
whhm
wwmW
a
a
&
&&
The net electricity production by this cycle is then
kW 1286=+= 2241062netW&
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10-121
10-111 A combined gas-steam power plant is considered. The topping cycle is an ideal gas-turbine cycle and the bottoming cycle is an ideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle 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 variable specific heats. Analysis (a) The analysis of gas cycle yields
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
( )( )
( )
kJ/kg 63.523K 520
kJ/kg 38.76854.375.450121
5.450kJ/kg 42.1515K 1400
kJ/kg 18.63066.185546.112
5546.1kJ/kg 24.310K 310
1111
109
10
99
87
8
77
910
9
78
7
=⎯→⎯=
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯=
=⎯→⎯===
==⎯→⎯=
hT
hPPP
P
PhT
hPPP
P
PhT
rr
r
rr
r
From the steam tables (Tables A-4, A-5, and A-6),
( ) ( )( )
gwhh
PPw
hh
f
f
kJ/k 42.20462.1281.191
kJ/kg .6212mkPa 1
kJ 1kPa 1012,500/kgm 0.00101
/kgm 00101.0
kJ/kg 81.191
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=−=
==
==
v
vv
1
26
312.5 MPa
10 kPa
8
Qout·
Qin·
7
11
10
9
GAS CYCLE
STEAM CYCLE
1400 K
310 K
5
4
2.5 MPa
T
s
( )( ) kJ/kg 8.23651.23929089.081.191
9089.04996.7
6492.04653.7kPa 10
KkJ/kg 4653.7kJ/kg 4.3574
C550MPa 5.2
kJ/kg 6.2909MPa 5.2
KkJ/kg 4651.6kJ/kg 6.3343
C500MPa 5.12
66
66
56
6
5
5
5
5
434
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
hssP
sh
TP
Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( )
( ) kg/s 153.9=−−
=−−
=
−=−⎯→⎯=⎯→⎯= ∑∑kg/s 12
63.52338.76842.2046.3343
1110
23air
1110air23outin
s
seeii
mhhhh
m
hhmhhmhmhmEE
&&
&&&&&&
(b) The rate of total heat input is ( ) ( )
( )( ) ( )( )
kW 101.44 5×≅=
−+−=−+−=+=
kW 200,144kJ/kg6.29094.3574kg/s 12kJ/kg 18.63042.1515kg/s 153.9
45reheat89airreheatairin hhmhhmQQQ &&&&&
(c) The rate of heat rejection and the thermal efficiency are then ( ) ( )
( )( ) ( )( )
59.1%
,
==−=−=
=−+−=
−+−=+=
5913.0kW 144,200kW 58,930
11
kW 93058kJ/kg 81.1918.2365kg/s 12kJ/kg 24.31063.523kg/s 153.9
in
outth
16711airsteamout,airout,out
hhmhhmQQQ s
&
&
&&&&&
η
preparation. If you are a student using this Manual, you are using it without permission.
10-122
10-112 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle 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 variable specific heats.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Analysis (a) The analysis of gas cycle yields (Table A-17)
( )( )
( )( ) ( )
( )
( )( )( )
kJ/kg 63.523K 520
kJ/kg 4.95835.86042.151585.042.1515
kJ/kg 35.8603.565.45081
5.450kJ/kg 42.1515K 1400
kJ/kg 1.58580.0/16.29012.52616.290
/
kJ/kg 12.526849.92311.18
2311.1kJ/kg 16.290K 290
1111
109910109
109
109
10
99
787878
78
87
8
77
910
9
78
7
=⎯→⎯=
=−−=
−η−=⎯→⎯−−
=η
=⎯→⎯=⎟⎠⎞
⎜⎝⎛==
==⎯→⎯=
=−+=
η−+=⎯→⎯−−
=η
=⎯→⎯===
==⎯→⎯=
hT
hhhhhhhh
hPP
PP
PhT
hhhhhhhh
hPPPP
PhT
sTs
T
srs
r
r
Css
C
srs
r
r
s
s
1
26s
3
15 MPa
10 kPa
8
Qout·
Qin·
7
11
10s
9
GAS CYCLE
STEAM CYCLE
1400 K
290 K
5
4
3 MPa
6
4
81
T
s
From the steam tables (Tables A-4, A-5, and A-6),
( )
( )( )
kJ/kg 95.20614.1581.191kJ/kg 15.14
mkPa 1kJ 1kPa 1015,000/kgm 0.00101
/kgm 00101.0kJ/kg 81.191
inpI,12
33
121inpI,
3kPa 10 @ 1
kPa 10 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
( )( )
( )( )( )
kJ/kg 1.28387.27819.315785.09.3157
kJ/kg 7.27819.17949879.03.1008
9880.05402.3
6454.21434.6MPa 3
KkJ/kg 1428.6kJ/kg 9.3157
C450MPa 15
433443
43
44
44
34
4
3
3
3
3
=−−=
−η−=⎯→⎯−−
=η
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
sh
TP
preparation. If you are a student using this Manual, you are using it without permission.
10-123
( )( )
( )( )( )
kJ/kg 5.24678.22922.345785.02.3457
kJ/kg 8.22921.23928782.081.191
8783.04996.7
6492.02359.7kPa 10
KkJ/kg 2359.7kJ/kg 2.3457
C500MPa 3
655665
65
66
66
56
6
5
5
5
5
=−−=
−η−=⎯→⎯−−
=η
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
sTs
T
fgsfs
fg
fss
s
hhhhhhhh
hxhhs
ssx
ssP
sh
TP
Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&
( ) ( )
( ) kg/s 203.6=−−
=−−
=
−=−⎯→⎯=
=
=∆=−
∑∑kg/s 30
63.5234.95895.2069.3157
0
1110
23air
1110air23
outin
(steady) 0systemoutin
s
seeii
mhhhhm
hhmhhmhmhm
EE
EEE
&&
&&&&
&&
&&&
(b) ( ) ( )( )( ) ( )( )
kW 207,986=−+−=
−+−=+=kJ/kg 1.28382.3457kg/s 30kJ/kg 1.58542.1515kg/s 203.6
45reheat89airreheatairin hhmhhmQQQ &&&&&
(c) ( ) ( )( )( ) ( )( )
44.3%=−=−=
=−+−=
−+−=+=
kW 207,986kW 115,805
11
kW 115,805kJ/kg 81.1915.2467kg/s 30kJ/kg 16.29063.523kg/s 203.6
in
outth
16711airsteamout,airout,out
hhmhhmQQQ s
&
&
&&&&&
η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-124
10-113 A Rankine steam cycle modified with two closed feedwater heaters and one open feed water heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass extracted for the open feedwater heater y and the cooling water flow temperature rise are to be determined. Also, the rate of heat rejected in the condenser 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 (b) Using the data from the problem statement, the enthalpies at various states are
1
2
12
s
7
8 140 kPa
20 kPa
1910 kPa
4
3
56
5 MPa
9
11
14
15
1-y-z-w 13
z
w
w
y
10
w+z
kJ/kg 898
kJ/kg 676
kJ/kg 458
kJ/kg 251
kPa 1910@ 126
kPa 062@ 4
kPa 140 @ 14315
kPa 02@ 1
===
==
====
==
f
f
f
f
hhh
hh
hhhh
hh
T
620 kPa
An energy balance on the open feedwater heater gives
439 1)1( hhyyh =−+
where z is the fraction of steam extracted from the low-pressure turbine. Solving for z,
0.08086=−−
=−−
=4583154458676
39
34
hhhh
y
(c) An energy balance on the condenser gives
[ ] wpwwwww Tcmhhmhyhzwhzywm ∆=−=−−++−−− &&& )()1()()1( 22115117
Solving for the temperature rise of cooling water, and substituting with correct units,
[ ]
[ ]
C9.95°=
−−++−−−=
−−++−−−=∆
)18.4)(4200()251)(08086.01()458)(0655.00830.0()2478)(0655.008086.00830.01()100(
)1()()1( 115117
pwww cm
hyhzwhzywmT
&
&
(d) The rate of heat rejected in the condenser is
kW 174,700=°°⋅=∆= C)95.9)(CkJ/kg kg/s)(4.18 4200(out wpww TcmQ &&
The rate of heat input in the boiler is
kW 200,300kJ/kg )8983900)(kg/s 100()( 67in =−=−= hhmQ &&
The thermal efficiency is then
41.8%==−=−= 418.0kW 200,300kW 700,17411
in
outth Q
Q&
&η
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-125
10-114 A Rankine steam cycle modified for reheat, two closed feedwater heaters and a process heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass, w, that is extracted for the closed feedwater heater is to be determined. Also, the mass flow rate through the boiler, the rate of process heat supplied, and the utilization efficiency of this cogeneration plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (b) Using the data from the problem statement, the enthalpies at various states are
( )( )( )
kJ/kg 5.2561.54.251kJ/kg 1.5
mkPa 1kJ 1
kPa 205000/kgm 0.00102
/kgm 00102.0kJ/kg 4.251
inpI,12
33
121inpI,
3kPa 20 @ 1
kPa 20 @ 1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
12
14
s 17 11
5
6
3
245 kPa
20 kPa
1.4 MPa
7 4
5 MPa
9 12
1315
z
1-y-z-w
y 8
w 16 10
1.2 MPa
150 kPa
T
kJ/kg 467
kJ/kg 830kJ/kg 533
kPa 150 @ 163
kPa 1400 @ 14154
kPa 245 @ 1213
=======
===
f
f
f
hhhhhhh
hhh
An energy balance on the closed feedwater heater gives 1631513102 )(11 hwzyhyhzhwhh +++=+++
where w is the fraction of steam extracted from the low-pressure turbine. Solving for z,
0.0620=−
−−++−=
−−−++−(
=
4673023)830)(1160.0()533)(15.0()467)(15.01160.0()5.256467(
)()
1610
15131623
hhyhzhhzyhh
w
(c) The work output from the turbines is
kJ/kg 6.1381)2620)(0620.015.01160.01()3023)(0620.0()3154)(15.0()3692)(1160.01(
)3349)(1160.01()3400)(1160.0(3894)1()1()1( 111098765outT,
=−−−−−−−+
−−−=
−−−−−−−+−−−= hwzywhzhhyhyyhhw
The net work output from the cycle is kJ/kg 5.13761.56.1381inP,outT,net =−=−= www
The mass flow rate through the boiler is
kg/s 217.9===kJ/kg 1376.5
kW 000,300
net
net
wW
m&
&
The rate of heat input in the boiler is
kW 700,733
kJ/kg )33493692)(kg/s 9.217)(1160.01(kJ/kg )8303894)(kg/s 9.217()()1()( 7845in
=−−+−=
−−+−= hhmyhhmQ &&&
The rate of process heat and the utilization efficiency of this cogeneration plant are
kW 85,670=−=−= kJ/kg )5333154)(kg/s 9.217)(15.0()( 129process hhmzQ &&
52.6%==+
=+
= 526.0kW 700,733
kW )670,85000,300(
in
processnet
Q
QWu &
&&ε
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-126
10-115 The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$='' if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)' end P[3] = 10000 [kPa] T[3] = 550 [C] "P[4] = 5 [kPa]" Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[4] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) x[4]=quality(Fluid$,h=h[4],P=P[4]) h[3] =W_t+h[4]"SSSF First Law for the turbine" x4s$=x4$(x[4]) "Boiler analysis" Q_in + h[2]=h[3]"SSSF First Law for the Boiler" "Condenser analysis" h[4]=Q_out+h[1]"SSSF First Law for the Condenser" "Cycle Statistics" W_net=W_t-W_p Eta_th=W_net/Q_in
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-127
P4
[kPa] ηth Wnet
[kJ/kg] 5
15 25 35 45 55 65 75 85
100
0.4268 0.3987 0.3841 0.3739 0.3659 0.3594 0.3537 0.3488 0.3443 0.3385
1432 1302 1237 1192 1157 1129 1105 1084 1065 1040
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]
T [°
C]
10 MPa
5 kPa
SteamIAPWS
1,2
3
4
0 20 40 60 80 1000.32
0.34
0.36
0.38
0.4
0.42
0.44
P[4] [kPa]
ηth
0 20 40 60 80 1001000
1050
1100
1150
1200
1250
1300
1350
1400
1450
P[4] [kPa]
Wne
t [k
J/kg
]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-128
10-116 The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$='' if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)' end P[3] = 3000 [kPa] {T[3] = 600 [C]} P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[4] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) x[4]=quality(Fluid$,h=h[4],P=P[4]) h[3] =W_t+h[4]"SSSF First Law for the turbine" x4s$=x4$(x[4]) "Boiler analysis" Q_in + h[2]=h[3]"SSSF First Law for the Boiler" "Condenser analysis" h[4]=Q_out+h[1]"SSSF First Law for the Condenser" "Cycle Statistics" W_net=W_t-W_p Eta_th=W_net/Q_in
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-129
0 2 4 6 8 10 120
100
200
300
400
500
600
700
s [kJ/kg-K]T
[°C
]
3000 kPa
10 kPa
Steam
1
2
3
4
T3
[C] ηth Wnet
[kJ/kg] x4
250 0.3241 862.8 0.752 344.4 0.3338 970.6 0.81 438.9 0.3466 1083 0.8536 533.3 0.3614 1206 0.8909 627.8 0.3774 1340 0.9244 722.2 0.3939 1485 0.955 816.7 0.4106 1639 0.9835 911.1 0.4272 1803 100 1006 0.4424 1970 100 1100 0.456 2139 100
200 300 400 500 600 700 800 900 1000 11000.32
0.34
0.36
0.38
0.4
0.42
0.44
0.46
T[3] [C]
ηth
200 300 400 500 600 700 800 900 1000 1100750
1050
1350
1650
1950
2250
T[3] [C]
Wne
t [k
J/kg
]
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
10-130
10-117 The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)' end Procedure Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6) P3=P[3] T5=T[5] h4=h[4] Q_in_reheat =0 W_t_lp = 0 R_P=(1/Pratio)^(1/(NoRHStages+1)) imax:=NoRHStages - 1 i:=0 REPEAT i:=i+1 P4 = P3*R_P P5=P4 P6=P5*R_P Fluid$='Steam_IAPWS' s5=entropy(Fluid$,T=T5,P=P5) h5=enthalpy(Fluid$,T=T5,P=P5) s_s6=s5 hs6=enthalpy(Fluid$,s=s_s6,P=P6) Ts6=temperature(Fluid$,s=s_s6,P=P6) vs6=volume(Fluid$,s=s_s6,P=P6) "Eta_t=(h5-h6)/(h5-hs6)""Definition of turbine efficiency" h6=h5-Eta_t*(h5-hs6) W_t_lp=W_t_lp+h5-h6"SSSF First Law for the low pressure turbine" x6=QUALITY(Fluid$,h=h6,P=P6) Q_in_reheat =Q_in_reheat + (h5 - h4) P3=P4 UNTIL (i>imax) END "NoRHStages = 2" P[6] = 10"kPa" P[3] = 15000"kPa" P_extract = P[6] "Select a lower limit on the reheat pressure" T[3] = 500"C" T[5] = 500"C" Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" Pratio = P[3]/P_extract P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa" Fluid$='Steam_IAPWS'
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-131
"Pump analysis" P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,P=P[2],h=h[2]) s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) v[3]=volume(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) v[4]=volume(Fluid$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" Call Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6) h[6]=h6 {P[5]=P[4] s[5]=entropy(Fluid$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6]) 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]) W_t_lp_total = NoRHStages*W_t_lp Q_in_reheat = NoRHStages*(h[5] - h[4])} "Boiler analysis" Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler" Q_in = Q_in_boiler+Q_in_reheat "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]) x[6]=QUALITY(Fluid$,h=h[6],P=P[6]) x6s$=x6$(x[6]) "Cycle Statistics" W_net=W_t_hp+W_t_lp - W_p Eta_th=W_net/Q_in
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10-132
ηth NoRH
Stages Qin
[kJ/kg] Wnet
[kJ/kg] 0.4097 1 4085 1674 0.4122 2 4628 1908 0.4085 3 5020 2051 0.4018 4 5333 2143 0.3941 5 5600 2207 0.386 6 5838 2253
0.3779 7 6058 2289 0.3699 8 6264 2317 0.3621 9 6461 2340 0.3546 10 6651 2358
1 2 3 4 5 6 7 8 9 101600
1700
1800
1900
2000
2100
2200
2300
2400
NoRHStages
Wne
t [k
J/kg
]
1 2 3 4 5 6 7 8 9 100.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
NoRHStages
ηth
1 2 3 4 5 6 7 8 9 104000
4500
5000
5500
6000
6500
7000
NoRHStages
Qin
[kJ
/kg]
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10-118 The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
Procedure Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Fluid$='Steam_IAPWS' Tcond = temperature(Fluid$,P=P_cond,x=0) Tboiler = temperature(Fluid$,P=P[5],x=0) P[7] = P_cond s[5]=entropy(Fluid$, T=T[5], P=P[5]) h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) h[1]=enthalpy(Fluid$, P=P[7],x=0) P4[1] = P[5] "NOTICE THIS IS P4[i] WITH i = 1" DELTAT_cond_boiler = Tboiler - Tcond If NoFWH = 0 Then "the following are h7, h2, w_net, and q_in for zero feedwater heaters, NoFWH = 0" h7=enthalpy(Fluid$, s=s[5],P=P[7]) h2=h[1]+volume(Fluid$, P=P[7],x=0)*(P[5] - P[7])/Eta_pump w_net = Eta_turb*(h[5]-h7)-(h2-h[1]) q_in = h[5] - h2 else i=0 REPEAT i=i+1 "The following maintains the same temperature difference between any two regeneration stages." T_FWH[i] = (NoFWH +1 - i)*DELTAT_cond_boiler/(NoFWH + 1)+Tcond"[C]" P_extract[i] = pressure(Fluid$,T=T_FWH[i],x=0)"[kPa]" P3[i]=P_extract[i] P6[i]=P_extract[i] If i > 1 then P4[i] = P6[i - 1] UNTIL i=NoFWH P4[NoFWH+1]=P6[NoFWH] h4[NoFWH+1]=h[1]+volume(Fluid$, P=P[7],x=0)*(P4[NoFWH+1] - P[7])/Eta_pump i=0 REPEAT i=i+1 "Boiler condensate pump or the Pumps 2 between feedwater heaters analysis" h3[i]=enthalpy(Fluid$,P=P3[i],x=0) v3[i]=volume(Fluid$,P=P3[i],x=0) w_pump2_s=v3[i]*(P4[i]-P3[i])"SSSF isentropic pump work assuming constant specific volume" w_pump2[i]=w_pump2_s/Eta_pump "Definition of pump efficiency" h4[i]= w_pump2[i] +h3[i] "Steady-flow conservation of energy" s4[i]=entropy(Fluid$,P=P4[i],h=h4[i]) T4[i]=temperature(Fluid$,P=P4[i],h=h4[i]) Until i = NoFWH i=0 REPEAT
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i=i+1 "Open Feedwater Heater analysis:" {h2[i] = h6[i]} s5[i] = s[5] ss6[i]=s5[i] hs6[i]=enthalpy(Fluid$,s=ss6[i],P=P6[i]) Ts6[i]=temperature(Fluid$,s=ss6[i],P=P6[i]) h6[i]=h[5]-Eta_turb*(h[5]-hs6[i])"Definition of turbine efficiency for high pressure stages" If i=1 then y[1]=(h3[1] - h4[2])/(h6[1] - h4[2]) "Steady-flow conservation of energy for the FWH" If i > 1 then js = i -1 j = 0 sumyj = 0 REPEAT j = j+1 sumyj = sumyj + y[ j ] UNTIL j = js y[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1]) ENDIF T3[i]=temperature(Fluid$,P=P3[i],x=0) "Condensate leaves heater as sat. liquid at P[3]" s3[i]=entropy(Fluid$,P=P3[i],x=0) "Turbine analysis" T6[i]=temperature(Fluid$,P=P6[i],h=h6[i]) s6[i]=entropy(Fluid$,P=P6[i],h=h6[i]) yh6[i] = y[i]*h6[i] UNTIL i=NoFWH ss[7]=s6[i] hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7]) Ts[7]=temperature(Fluid$,s=ss[7],P=P[7]) h[7]=h6[i]-Eta_turb*(h6[i]-hs[7])"Definition of turbine efficiency for low pressure stages" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) sumyi = 0 sumyh6i = 0 wp2i = W_pump2[1] i=0 REPEAT i=i+1 sumyi = sumyi + y[i] sumyh6i = sumyh6i + yh6[i] If NoFWH > 1 then wp2i = wp2i + (1- sumyi)*W_pump2[i] UNTIL i = NoFWH "Condenser Pump---Pump_1 Analysis:" P[2] = P6 [ NoFWH] P[1] = P_cond h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[2]=w_pump1+ h[1] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Boiler analysis" q_in = h[5] - h4[1]"SSSF conservation of energy for the Boiler" w_turb = h[5] - sumyh6i - (1- sumyi)*h[7] "SSSF conservation of energy for turbine" PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-135
"Condenser analysis" q_out=(1- sumyi)*(h[7] - h[1])"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1- sumyi)*w_pump1+ wp2i) endif END "Input Data" NoFWH = 2 P[5] = 10000 [kPa] T[5] = 500 [C] P_cond=10 [kPa] Eta_turb= 1.0 "Turbine isentropic efficiency" Eta_pump = 1.0 "Pump isentropic efficiency" P[1] = P_cond P[4] = P[5] "Condenser exit pump or Pump 1 analysis" Call Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Eta_th=w_net/q_in
No FWH
ηth wnet [kJ/kg]
qin [kJ/kg]
0 1 2 3 4 5 6 7 8 9
10
0.4019 0.4311 0.4401 0.4469 0.4513 0.4544 0.4567 0.4585 0.4599 0.4611 0.462
1275 1125 1061 1031 1013 1000 990.5 983.3 977.7 973.1 969.4
3173 2609 2411 2307 2243 2200 2169 2145 2126 2111 2098
0 1 2 3 4 5 6 7 8 9 100.4
0.41
0.42
0.43
0.44
0.45
0.46
0.47
NoFwh
ηth
0 1 2 3 4 5 6 7 8 9 10950
1000
1050
1100
1150
1200
1250
1300
NoFwh
wne
t [k
J/kg
]
0 1 2 3 4 5 6 7 8 9 10
2000
2200
2400
2600
2800
3000
3200
NoFwh
q in
[kJ/
kg]
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10-136
10-119 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as η η η η ηcc g s g s= + − where ηg g i=W Q/ n and ηs s g,out=W Q/ are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained.
Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as
η
η
η
cctotal
in
out
in
gg
in
g,out
in
ss
g,out
out
g,out
= = −
= = −
= = −
W
Q
Q
Q
W
Q
Q
Q
W
Q
Q
Q
1
1
1
where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle.
Using the relations above, the expression η η η ηg s g s+ − can be expressed as
cc
η
ηηηη
=
−=
−++−−+−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛−−⎟
⎟⎠
⎞⎜⎜⎝
⎛−+⎟
⎟⎠
⎞⎜⎜⎝
⎛−=−+
in
out
in
out
outg,
out
in
outg,
outg,
out
in
outg,
outg,
out
in
outg,
outg,
out
in
outg,sgsg
1
111
1111
Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combined cycle is determined to be
η η η η ηcc g s g s= + − = + − × =0 4 0 30 0 40 0 30. . . . 0.58
10-120 The thermal efficiency of a combined gas-steam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as η η η η ηcc g s g s= + − . It is to be shown that the value of ηcc is greater
than either of η ηg s or .
Analysis By factoring out terms, the relation η η η η ηcc g s g s= + − can be expressed as
η η η η η η η η η
η
cc g s g s g s g
Positive since <1
g
g
= + − = + − >( )11 24 34
or η η η η η η η η η
η
cc g s g s s g s
Positive since <1
s
s
= + − = + − >( )11 24 34
Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbine cycles alone.
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10-137
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b10-121 It is to e shown that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as ( )thinqx ηη −= Carnotth,destroyed , where ηth is efficiency of the Rankine cycle and ηth, Carnot is the efficiency of the Carnot
cycle operating between the same temperature limits.
Analysis The exergy destruction associated with a cycle is given on a unit mass basis as
∑=R
R
Tq
Tx 0destroyed
where the direction of qin is determined with respect to the reservoir (positive if to the reservoir and negative if from the reservoir). For a cycle that involves heat transfer only with a source at TH and a sink at T0, the irreversibility becomes
( ) ( )[ ] ( )thCthCthth
HHH
qqTT
qqqq
TTq
Tq
TqTx
ηηηη −=−−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
,in,in
0
in
outinin
0out
in
0
out0destroyed
11
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10-138
Fundamentals of Engineering (FE) Exam Problems
10-122 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet state the same, (select the correct statement)
(a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the pump work input will decrease.
Answer (b) the amount of heat rejected will decrease.
10-123 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (select the correct statement)
(a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease.
Answer (d) the moisture content at turbine exit will decrease.
10-124 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with reheating, (select the correct statement) (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease. Answer (d) the moisture content at turbine exit will decrease.
10-125 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with regeneration that involves one open feed water heater, (select the correct statement per unit mass of steam flowing through the boiler) (a) the turbine work output will decrease. (b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease. (d) the quality of steam at turbine exit will decrease. (e) the amount of heat input will increase. Answer (a) the turbine work output will decrease.
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10-139
10-126 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 3 MPa and 10 kPa. Water changes from saturated liquid to saturated vapor during the heat addition process. The net work output of this cycle is
(a) 666 kJ/kg (b) 888 kJ/kg (c) 1040 kJ/kg (d) 1130 kJ/kg (e) 1440 kJ/kg
Answer (a) 666 kJ/kg
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=33000 "kPa" P2=10 "kPa" h_fg=ENTHALPY(Steam_IAPWS,x=1,P=P1)-ENTHALPY(Steam_IAPWS,x=0,P=P1) T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1)+273 T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273 q_in=h_fg Eta_Carnot=1-T2/T1 w_net=Eta_Carnot*q_in "Some Wrong Solutions with Common Mistakes:" W1_work = Eta1*q_in; Eta1=T2/T1 "Taking Carnot efficiency to be T2/T1" W2_work = Eta2*q_in; Eta2=1-(T2-273)/(T1-273) "Using C instead of K" W3_work = Eta_Carnot*ENTHALPY(Steam_IAPWS,x=1,P=P1) "Using h_g instead of h_fg" W4_work = Eta_Carnot*q2; q2=ENTHALPY(Steam_IAPWS,x=1,P=P2)-ENTHALPY(Steam_IAPWS,x=0,P=P2) "Using h_fg at P2"
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10-140
10-127 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa, with a turbine inlet temperature of 600°C. Disregarding the pump work, the cycle efficiency is
(a) 24% (b) 37% (c) 52% (d) 63% (e) 71%
Answer (b) 37%
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=10 "kPa" P2=3000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) "kJ/kg" h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 q_out=h4-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-(h44-h1)/(h3-h2); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Using h_g for h4" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = (h3-h4)/q_in "Disregarding pump work"
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10-141
10-128 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600°C. The mass fraction of steam that condenses at the turbine exit is
(a) 6% (b) 9% (c) 12% (d) 15% (e) 18%
Answer (c) 12%
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=10 "kPa" P2=5000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) x4=QUALITY(Steam_IAPWS,s=s4,P=P4) moisture=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_moisture = x4 "Taking quality as moisture" W2_moisture = 0 "Assuming superheated vapor"
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10-142
10-129 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 5 kPa and 10 MPa, with a turbine inlet temperature of 600°C. The rate of heat transfer in the boiler is 300 kJ/s. Disregarding the pump work, the power output of this plant is
(a) 93 kW (b) 118 kW (c) 190 kW (d) 216 kW (e) 300 kW
Answer (b) 118 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=10 "kPa" P2=5000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 Q_rate=300 "kJ/s" m=Q_rate/q_in h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 "pump work is neglected" "v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 W_turb=m*(h3-h4) "Some Wrong Solutions with Common Mistakes:" W1_power = Q_rate "Assuming all heat is converted to power" W3_power = Q_rate*Carnot; Carnot = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_power = m*(h3-h44); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Taking h4=h_g"
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10-143
10-130 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Water enters the heat exchanger at 200°C and 8 MPa and leaves at 350°C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heat exchanger becomes
(a) 11 kg/s (b) 24 kg/s (c) 46 kg/s (d) 53 kg/s (e) 60 kg/s
Answer (a) 11 kg/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).
m_gas=60 "kg/s" Cp=1.005 "kJ/kg.K" T3=800 "K" T4=400 "K" Q_gas=m_gas*Cp*(T3-T4) P1=8000 "kPa" T1=200 "C" P2=8000 "kPa" T2=350 "C" h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) Q_steam=m_steam*(h2-h1) Q_gas=Q_steam "Some Wrong Solutions with Common Mistakes:" m_gas*Cp*(T3 -T4)=W1_msteam*4.18*(T2-T1) "Assuming no evaporation of liquid water" m_gas*Cv*(T3 -T4)=W2_msteam*(h2-h1); Cv=0.718 "Using Cv for air instead of Cp" W3_msteam = m_gas "Taking the mass flow rates of two fluids to be equal" m_gas*Cp*(T3 -T4)=W4_msteam*(h2-h11); h11=ENTHALPY(Steam_IAPWS,x=0,P=P1) "Taking h1=hf@P1"
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10-144
10-131 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500°C, and the enthalpy of steam is 3185 kJ/kg at the exit of the high-pressure turbine, and 2247 kJ/kg at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is
(a) 29% (b) 32% (c) 36% (d) 41% (e) 49%
Answer (d) 41%
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=10 "kPa" P2=8000 "kPa" P3=P2 P4=4000 "kPa" P5=P4 P6=P1 T3=500 "C" T5=500 "C" s4=s3 s6=s5 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 h44=3185 "kJ/kg - for checking given data" h66=2247 "kJ/kg - for checking given data" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) h5=ENTHALPY(Steam_IAPWS,T=T5,P=P5) s5=ENTROPY(Steam_IAPWS,T=T5,P=P5) h6=ENTHALPY(Steam_IAPWS,s=s6,P=P6) q_in=(h3-h2)+(h5-h4) q_out=h6-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-q_out/(h3-h2) "Disregarding heat input during reheat" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = 1-q_out/(h5-h2) "Using wrong relation for q_in"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-145
10-132 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 2 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ/kg and the enthalpy of extracted steam is 2810 kJ/kg, the mass fraction of steam extracted from the turbine is
(a) 10% (b) 14% (c) 26% (d) 36% (e) 50%
Answer (c) 26%
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).
h_feed=252 "kJ/kg" h_extracted=2810 "kJ/kg" P3=2000 "kPa" h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) "Energy balance on the FWH" h3=x_ext*h_extracted+(1-x_ext)*h_feed "Some Wrong Solutions with Common Mistakes:" W1_ext = h_feed/h_extracted "Using wrong relation" W2_ext = h3/(h_extracted-h_feed) "Using wrong relation" W3_ext = h_feed/(h_extracted-h_feed) "Using wrong relation"
10-133 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater. The enthalpy of the steam is 3374 kJ/kg at the turbine inlet, 2797 kJ/kg at the location of bleeding, and 2346 kJ/kg at the turbine exit. The net power output of the plant is 120 MW, and the fraction of steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rate of steam at the turbine inlet is
(a) 117 kg/s (b) 126 kg/s (c) 219 kg/s (d) 288 kg/s (e) 679 kg/s
Answer (b) 126 kg/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).
h_in=3374 "kJ/kg" h_out=2346 "kJ/kg" h_extracted=2797 "kJ/kg" Wnet_out=120000 "kW" x_bleed=0.172 w_turb=(h_in-h_extracted)+(1-x_bleed)*(h_extracted-h_out) m=Wnet_out/w_turb "Some Wrong Solutions with Common Mistakes:" W1_mass = Wnet_out/(h_in-h_out) "Disregarding extraction of steam" W2_mass = Wnet_out/(x_bleed*(h_in-h_out)) "Assuming steam is extracted at trubine inlet" W3_mass = Wnet_out/(h_in-h_out-x_bleed*h_extracted) "Using wrong relation"
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-146
10-134 Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450°C at a rate of 20 kg/s and expands to a pressure of 0.4 MPa. At this pressure, 60% of the steam is extracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steam is used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in the condenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabatic condenser at a rate of 463 kg/s.
Condenser
Boiler
6
7 10
8 Turbine
5 Process heater
P IP II
9
2
3 fwh
4 1
11
h1 = 191.81 h2 = 192.20 h3 = h4 = h9 = 604.66 h5 = 610.73 h6 = 3302.9 h7 = h8 = h10 = 2665.6 h11 = 2128.8
1. The total power output of the turbine is
(a) 17.0 MW (b) 8.4 MW (c) 12.2 MW (d) 20.0 MW (e) 3.4 MW
Answer (a) 17.0 MW
2. The temperature rise of the cooling water from the river in the condenser is
(a) 8.0°C (b) 5.2°C (c) 9.6°C (d) 12.9°C (e) 16.2°C
Answer (a) 8.0°C
3. The mass flow rate of steam through the process heater is
(a) 1.6 kg/s (b) 3.8 kg/s (c) 5.2 kg/s (d) 7.6 kg/s (e) 10.4 kg/s
Answer (e) 10.4 kg/s
4. The rate of heat supply from the process heater per unit mass of steam passing through it is
(a) 246 kJ/kg (b) 893 kJ/kg (c) 1344 kJ/kg (d) 1891 kJ/kg (e) 2060 kJ/kg
Answer (e) 2060 kJ/kg
5. The rate of heat transfer to the steam in the boiler is
(a) 26.0 MJ/s (b) 53.8 MJ/s (c) 39.5 MJ/s (d) 62.8 MJ/s (e) 125.4 MJ/s
Answer (b) 53.8 MJ/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).
Note: The solution given below also evaluates all enthalpies given on the figure.
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
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10-147
P1=10 "kPa" P11=P1 P2=400 "kPa" P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P5=6000 "kPa" P6=P5 T6=450 "C" m_total=20 "kg/s" m7=0.6*m_total m_cond=0.4*m_total C=4.18 "kJ/kg.K" m_cooling=463 "kg/s" s7=s6 s11=s6 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) h4=h3; h9=h3 v4=VOLUME(Steam_IAPWS,x=0,P=P4) w_pump2=v4*(P5-P4) h5=h4+w_pump2 h6=ENTHALPY(Steam_IAPWS,T=T6,P=P6) s6=ENTROPY(Steam_IAPWS,T=T6,P=P6) h7=ENTHALPY(Steam_IAPWS,s=s7,P=P7) h8=h7; h10=h7 h11=ENTHALPY(Steam_IAPWS,s=s11,P=P11) W_turb=m_total*(h6-h7)+m_cond*(h7-h11) m_cooling*C*T_rise=m_cond*(h11-h1) m_cond*h2+m_feed*h10=(m_cond+m_feed)*h3 m_process=m7-m_feed q_process=h8-h9 Q_in=m_total*(h6-h5)
10-135 ··· 10-142 Design and Essay Problems
PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.
. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course