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10-77
Special Topic: Binary Vapor Cycles 10-82C 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. 10-83C 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
( ) ( 43123142
outin
(steady) 0systemoutin 0
hhmhhmorhmhmhmhm
hmhm
EE
EEE
BABABA
iiee
−=−+=+
=
=
=∆=−
∑∑&&&&&&
&&
&&
&&&
)Thus,
&
&
mm
h hh h
A
B=
−−
3 4
2 1
10-84C 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. 10-85C 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. 10-86C 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.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-78
Review Problems 10-87 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as where d are the thermal efficiencies of
the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. η η η η ηcc g s g s= + − ηg g i=W Q/ n an ηs s g,out=W Q/
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
= = −
= = −
= = −
WQ
WQ
WQ
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 can be expressed as η η η ηg s g+ − s
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
10-88 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 . It is to be shown that
the value of
η η η ηcc g s g s= + − η
ηcc is greater than either of . η ηg s or
Analysis By factoring out terms, the relation can be expressed as η η η ηcc g s g s= + − η
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.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-79
10-89 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
600°CTSINGLE
10 kPa
25 MPa 4
3
s 6
2
5
1
6
7DOUBL
10 kPa
25 MPa 4
3
8
2
5
1
600°C (b) Double Reheat :
C600 and
KkJ/kg 3637.6C600MPa 25
5
5
34
4
33
3
°==
==
⋅=
°==
TPP
ssPP
sTP
xx
s 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 MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-80
10-90E 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),
( )( )( )(
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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-81
10-91 A steam power plant operates on the simple ideal Rankine cycle. The turbine inlet temperature, the net power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg 2501.8kJ/kg 0.2574
kJ/kg 79.17404.675.168kJ/kg 6.04
mkPa 1kJ 1
kPa 7.56000/kgm 0.001008
C29.40/kgm 001008.0
kJ/kg 75.168
kPa 5.7 @ 4
kPa 5.7 @ 4
inp,12
33
121inp,
kPa 7.5 @ sat1
3kPa 5.7 @ 1
kPa 5.7 @ 1
====
=+=+==
⋅−=
−=
°======
g
g
f
f
sshh
whh
PPw
TT
hh
v
vv
T
6 MPa
4
2
3
17.5 kPa
qout
qin
s
C1089.2°==
==
3
3
43
3 kJ/kg 2.4852MPa 6Th
ssP
(b)
kJ/kg 1.22723.24054.4677kJ/kg 3.240575.1680.2574kJ/kg 4.467779.1742.4852
outinnet
14out
23in
=−=−==−=−==−=−=
qqwhhqhhq
and
48.6%===kJ/kg 4677.4kJ/kg 2272.1
in
netth q
wη
Thus,
( )( ) kJ/s 19,428=== kJ/s 40,0004857.0inthnet QW && η
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the condenser, which is 40.29°C,
( )( ) kg/s 194.6=°−°⋅
=∆
=
=−=−=
C1540.29CkJ/kg 4.18kJ/s 20,572
kJ/s 572,20428,19000,40
outcool
netinout
TcQm
WQQ
&&
&&&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-82
10-92 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 82.15207.1575.137kJ/kg 15.07
mkPa 1kJ 1kPa 515,000/kgm 0.001005
/kgm 001005.0kJ/kg 75.137
inp,12
33
121inp,
3kPa 5 @ 1
kPa 5 @ 1
=+=+==
⋅−=
−=
====
whh
PPw
hh
f
f
v
vv
T
( )( ) kJ/kg 9.23670.24239204.075.137
9204.09176.7
4762.07642.7kPa 5
KkJ/kg 7642.7kJ/kg 1.3479
C500MPa 1
kJ/kg 3.2971MPa 1
KkJ/kg 9781.6kJ/kg 7.3434
C500MPa 5
kJ/kg 4.3007MPa 5
KkJ/kg 3480.6kJ/kg 8.3310
C500MPa 15
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 MPa 5 MPa
6
7
5 kPa
15 MPa 4
3
8
2
5
1s
Then,
( ) ( ) ( )
kJ/kg 9.18622.22301.4093kJ/kg 2.223075.1379.2367
kJ/kg 1.40933.29711.34794.30077.343482.1528.3310
outinnet
18out
674523in
=−=−==−=−=
=−+−+−=−+−+−=
qqwhhq
hhhhhhq
Thus,
45.5%===kJ/kg 4093.1kJ/kg 1862.9
in
netth q
wη
(b) kg/s 64.4===kJ/kg 1862.9
kJ/s 120,000
net
net
wWm&
&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-83
10-93 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
1-y 6
P II P I
Open fwh
Condenser
Boiler Turbine
4
3 2
1
7
5
y
T
6
7 qout
3 y
4
1-y 0.5 MPa
10 kPa
10 MPa
6s
5
7s
2
qin
1
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 MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-84
( )(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
qsmsmTsTiL
iiee
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-85
10-94 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
1-y 6
P II P I
Open fwh
Condenser
Boiler Turbine
4
3 2
1
7
5
y
T
qout
3 y
4
1-y
0.5 MPa
10 kPa
10 MPa
6
5
7
2
qin
1
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 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
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-86
The fraction of steam extracted is determined from the steady-flow energy equation applied to the feedwater heaters. Noting that Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( 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,
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
qsmsmTsTiL
iiee
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-87
10-95 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
7
6
1-y 8
P II P I
Openfwh
Condens.
BoilerTurbine
4
3 2
1
9
5
y
3
48
1 MPa
0.6 MPa
15 kPa
10 MPa6
5
9
2
7
1
T
s
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( 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 7.196294.2258.25181440.011kJ/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 MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-88
10-96 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
7
6
1-y 8
P II P I
Open fwh
Condenser
Boiler Turbine
4
3 2
1
9
5
y
T
1-y3
4 8 6 6s 8s y
5
9 9s
2
7
1
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
)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-89
( ) ( )(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
shP
)
T
( )( )
( ) ( )(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 Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( 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 MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-90
10-97 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for the open feedwater heater, the thermal efficiency of the cycle, and the net power output 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),
High-P Turbine
11 1-y-z
13
12 z
10
4
Closed fwh
5 y
P II
P I
Open fwh I
Condenser
Boiler Low-P Turbine
6 3
2 1
7
8
9
T 1 MPa
10
11 15 MPa
z
8
56
3
y
4
1 - y - z 5 kPa 7
0.2 MPa
0.6 MPa
12
9
13
2
1s
( )
( )( )
kJ/kg 95.13720.075.137kJ/kg 0.20
mkPa 1kJ 1
kPa 5200/kgm 0.001005
/km 001005.0kJ/kg 75.137
inpI,12
33
121inpI,
3kPa 5 @ 1
kPa 5 @ 1
=+=+==
⋅−=
−=
==
==
whh
PPw
ghh
f
f
v
vv
( )( )( )
/kgm 100110.0kJ/kg 38.670
liquidsat.MPa 6.0
kJ/kg 41.52070.1571.504kJ/kg 15.70
mkPa 1kJ 1kPa 20015,000/kgm 0.001061
/kgm 001061.0kJ/kg 71.504
liquidsat.MPa 2.0
3MPa 6.0 @ 6
MPa 6.0 @ 766
inpII,34
33
343inpII,
3MPa 2.0 @ 3
MPa 2.0 @ 33
==
===
=
=+=+==
⋅−=
−=
==
==
=
f
f
f
f
hhhP
whh
PPw
hhP
vv
v
vv
( )( )( )
KkJ/kg 7642.7kJ/kg 1.3479
C500MPa 0.1
kJ/kg 8.2820MPa 0.1
KkJ/kg 6796.6kJ/kg 1.3583
C600MPa 15
kJ/kg 686.23 mkPa 1kJ 1kPa 60015,000/kgm 0.00110138.670
10
10
10
10
989
9
8
8
88
33
6566556
⋅==
°==
=
==
⋅==
°==
=
⋅−+=
−+=→=
sh
TP
hss
P
sh
TP
PPvhhTT
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-91
( )( )kJ/kg 1.2368
0.24239205.075.137
9205.0
9176.74762.07642.7
kPa 5
kJ/kg 9.3000MPa 2.0
kJ/kg 2.3310MPa 6.0
1313
1313
1013
13
121012
12
111011
11
=+=
+==
−=
−=
==
=
==
=
==
fgf
fg
f
hxhh
sss
x
ssP
hss
P
hss
P
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( ) ( ) ( 4561145561111
outin
(steady) 0systemoutin 0
hhhhyhhmhhmhmhm
EE
EEE
eeii −=−→−=−→=
=
=∆=−
∑∑ &&&&
&&
&&&
)
where y is the fraction of steam extracted from the turbine ( = & / &m m11 5 ). Solving for y,
06287.038.6702.331041.52023.686
611
45 =−−
=−−
=hhhhy
For the open FWH,
( ) ( ) 31227
3312122277
outin
(steady) 0systemoutin
11
0
hzhhzyyhhmhmhmhm
hmhm
EE
EEE
eeii
=+−−+=++
=
=
=∆=−
∑∑&&&&
&&
&&
&&&
where z is the fraction of steam extracted from the turbine ( = & / &m m12 5 ) at the second stage. Solving for z,
( ) ( ) ( )( ) 0.1165=
−−−−
=−
−−−=
95.1370.300095.13738.67006287.095.13771.504
212
2723
hhhhyhh
z
(b) ( ) ( )( ) ( )( )( ) ( )( )
kJ/kg 9.17244.18303.3555kJ/kg 4.183075.1370.23681165.006287.011
kJ/kg 3.35558.28201.347923.6861.3583
outinnet
113out
91058in
=−=−==−−−=−−−=
=−+−=−+−=
qqwhhzyq
hhhhq
and
48.5%=−=−=kJ/kg 3555.3kJ/kg 1830.4
11in
outth q
qη
(c) ( )( ) kW 72,447=== kJ/kg 1724.9kg/s 42netnet wmW &&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-92
10-98 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies 7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating 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 8.2518
3.23729665.094.225
9665.02522.7
7549.07642.7
kPa 15
KkJ/kg 7642.7kJ/kg 1.3479
C500MPa 1
kJ/kg 7.2843MPa 1
KkJ/kg 7266.6kJ/kg 5.3399
C500MPa 8
kJ/kg 81.503
kJ/kg 94.225
99
99
89
9
8
8
8
8
767
7
6
6
6
6
C 120 @ 3
12
kPa 15 @ 1
=+=+=
=−
=−
=
==
⋅==
°==
=
==
⋅==
°==
==≅
==
°
fgf
fg
f
f
f
hxhhs
ssx
ssP
shP
hssP
sh
TP
hhhhhh
T 4
7
8
P II P I
Process heater
Condenser
BoilerTurbine
5
3
2
1
9
6
)
T
6 8 8 MPa
31 MPa5 7
42The mass flow rate through the process heater is
( ) kg/s 929.2kJ/kg 81.5037.2843
kJ/s 7,000
37
process3 =
−=
−=
hhQ
m&
& 15 kPa9 1
s Also,
or, ( ) ( ) ( ) ( )( )
( ) ( )( )8.25181.3479992.27.28435.3399kJ/s 3000
993.2
66
986766989766
−−+−=
−−+−=−+−=
mm
hhmhhmhhmhhmWT
&&
&&&&&
It yields kg/s 873.36 =m&
and kg/s 881.0992.2873.3369 =−=−= mmm &&& 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
4 4 2 2 3 3
0
&
or, ( )( ) ( )( ) kJ/kg 60.440873.3
81.503992.294.225881.0
4
332254 =
+=
+=≅
mhmhmhh
&
&&
Then,
( ) ( )( )( ) ( )(
kW 12,020=−+−=
−+−=kJ/kg 2843.73479.1kg/s 0.881kJ/kg 440.603399.5kg/s 3.873
788566in hhmhhmQ &&&
)
(b) The fraction of steam extracted for process heating is
77.3%===kg/s 3.873kg/s 2.992
total
3
mm
y&
&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.