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10-30
10-46 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 fwh I Condenser
Boiler Turbine
6
5
4 3
2 1
10
9
8
7 T
10 MPa
1 - y
85
6
3y
4
0.2 MPa
5 kPa
0.6 MPa
91 - y - z
7
10
2
1
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 7.2618
6.22019602.071.504
9602.05968.5
5302.19045.6
MPa 2.0
kJ/kg 8.2821MPa 6.0
KkJ/kg 9045.6kJ/kg 8.3625
C600MPa 10
kJ/kg 73.68035.1038.670
99
99
79
9
878
8
7
7
7
7
in,56
=
+=+=
=−
=−
=
==
=
==
⋅==
°==
)
=+=+=
fgf
fg
f
pIII
hxhh
sss
x
ssP
hss
P
sh
TP
whh
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.
10-31
( )( ) 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 Q , 0∆pe∆ke ≅≅≅≅W&&
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) 47.1%=−=−=kJ/kg 2945.0kJ/kg 1556.8
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.
10-32
10-47 [Also solved by EES on enclosed CD] 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 250 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. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg 10.19229.081.191kJ/kg 29.0
mkPa 1kJ 1
kPa 10300/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 0.27555.20479935.087.720
9935.06160.4
0457.26317.6MPa 8.0
KkJ/kg 6317.6kJ/kg 5.3476
C550MPa 5.12
kJ/kg 727.83 MPa 12.5 ,
C4.170
/kgm 001115.0
kJ/kg 87.720
liquid sat.MPa 8.0
kJ/kg 52.57409.1343.561kJ/kg 13.09
mkPa 1kJ 1
kPa 30012,500/kgm 0.001073
/kgm 001073.0
kJ/kg 43.561
liquid sat.MPa 3.0
99
99
89
9
8
8
8
8
5556
MPa 0.8 @sat6
3MP 8.0 @6
MPa 8.0 @766
in,34
33
343in,
3MPa 3.0 @3
MPa 3.0 @33
=+=+=
=−
=−
=
==
⋅==
°==
=→==
°==
==
===
=
=+=+==
⋅−=
−===
==
=
fgf
fg
f
af
f
pII
pII
f
f
hxhh
sss
xss
P
sh
TP
hPTT
TT
hhhP
whh
PPw
hhP
vv
v
vv
1-y-zz
y
3 Closed
fwh P II
P I
Openfwh
Condenser
BoilerTurbine
5
6
4
7 2
1
11
10
9
8
T
7
95
6
3
40.3 MPa
z 10 kPa
0.8 MPa
12.5 MPa
y
10 1 - y - z
8
11
2
1s
( )( )
( )( ) kJ/kg 0.21001.23927977.081.191
7977.04996.7
6492.06317.kPa 10
kJ/kg 5.25785.21639323.043.561
9323.03200.5
6717.16317.6MPa 3.0
1111
1111
811
11
1010
1010
810
10
=+=+=
=−6
=−
=
==
=+=+=
=−
=−
=
==
fgf
fg
f
fgf
fg
f
hxhh
sss
x
ssP
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 Q , 0∆pe∆ke ≅≅≅≅W&&
( ) ( ) ( ) ( 4569455699
outin
(steady) 0systemoutin 0
hhhhyhhmhhmhmhm
EE
EEE
eeii −=−→−=−→=
=
=∆=−
∑∑ &&&&
&&
&&&
)
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.
10-33
where y is the fraction of steam extracted from the turbine ( 510 / mm &&= ). Solving for y,
0753.087.7200.275552.57483.727
69
45 =−−
=−−
=hhhhy
For the open FWH,
( ) ( 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,
( ) ( ) ( )( ) 1381.010.1925.2578
10.19287.7200753.010.19243.561
210
2723 =−
−−−=
−−−−
=hh
hhyhhz
Then,
( )( ) ( )( )kJ/kg 12491.15001.2749
kJ/kg 1.150081.1910.21001381.00753.011kJ/kg 1.274936.7275.3476
outinnet
111out
58in
=−=−==−−−=−−−=
=−=−=
qqwhhzyq
hhq
and
kg/s 200.2===kJ/kg 1249
kJ/s 250,000
net
net
wWm&
&
(b) 45.4%=−=−=kJ/kg 2749.1kJ/kg 1500.1
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.
10-34
10-48 EES Problem 10-47 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] = 12500 [kPa] T[8] = 550 [C] P[9] = 800 [kPa] "P_cfwh=300 [kPa]" P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=250 [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]"
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.
10-35
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]" 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
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.
10-36
ηturb ηturb ηth m [kg/s] 0.7 0.7 0.3916 231.6
0.75 0.75 0.4045 224.3 0.8 0.8 0.4161 218
0.85 0.85 0.4267 212.6 0.9 0.9 0.4363 207.9
0.95 0.95 0.4452 203.8 1 1 0.4535 200.1
0.7 0.75 0.8 0.85 0.9 0.95 1
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46
200
205
210
215
220
225
230
235
ηturb
ηth
m [kg/s]
ηturb =ηpump
0 2 4 6 8 10 120
100
200
300
400
500
600
s [kJ/kg-K]
T[C] 12500 kPa
800 kPa
300 kPa
10 kPa
Steam
1,2
3,4
5,6
7
8
910
11
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.
10-37
10-49 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater heater. The mass flow rate of steam through 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),
( ) ( )( )
kJ/kg 61.19280.081.191kJ/kg 0.80
mkPa 1kJ 1
kPa 10800/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
T
( ) ( )( )
( )( ) kJ/kg 7.24941.23929627.081.191
9627.04996.7
6492.08692.7kPa 10
KkJ/kg 8692.7kJ/kg 3.3481
C500MPa 8.0
kJ/kg 1.2812MPa 8.0
KkJ/kg 7585.6kJ/kg 0.3502
C550MPa 10
kJ/kg 12.73126.1087.720kJ/kg 10.26
mkPa 1kJ 1
kPa 80010,000/kgm 0.001115
/kgm 001115.0
kJ/kg 87.720
liquidsat.MPa 8.0
88
88
78
8
7
7
7
7
656
6
5
5
5
5
in,34
33
343in,
3MPa 8.0 @3
MPa 8.0 @33
=+=+=
=−
=−
=
==
⋅==
°==
=
==
⋅==
°==
=+=+==
⋅−=−=
==
==
=
fgf
fg
f
pII
pII
f
f
hxhh
sss
xss
P
sh
TP
hss
P
sh
TP
whh
PPw
hhP
v
vv
10 MPa
1 - y
63
4
y
10 kPa
0.8 MPa 7
5
8
2
1s
6
1-y 7
6
P II P I
Openfwh
Condenser
BoilerTurbine
4
3 2
1
8
5
y
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&&
( ) ( 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,
2017.061.1921.281261.19287.720
26
23 =−−
=−−
=hhhh
y
Then, ( ) ( )( ) ( ) ( )( )( )( ) ( )( )
kJ/kg 6.14665.18381.3305kJ/kg 5.183881.1917.24942017.011
kJ/kg 1.33051.28123.34812017.0112.7310.35021
outinnet
18out
6745in
=−=−==−−=−−=
=−−+−=−−+−=
qqwhhyq
hhyhhq
and kg/s 54.5===kJ/kg 1466.1kJ/s 80,000
net
net
wW
m&
&
(b) 44.4%===kJ/kg 3305.1kJ/kg 1466.1
in
netth q
wη
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.
10-38
10-50 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater. The mass flow rate of steam through 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
9
10
Mixing chamber
1-y
7 y
P II P I
Closed fwh
Condenser
Boiler Turbine
4
3 2
1
8
5
6
T
10 MPa
1 - y
6
3
410
9y
10 kPa
0.8 MPa 7
5
8
2
1
s (a) From the steam tables (Tables A-4, A-5, and A-6),
( ) ( )( )
( ) ( )( )
kJ/kg 13.73126.1087.720kJ/kg 10.26
mkPa 1kJ 1kPa 80010,000/kgm 0.001115
/kgm 001115.0
kJ/kg 87.720
liquidsat.MPa 8.0
kJ/kg 90.20109.1081.191kJ/kg 10.09
mkPa 1kJ 1kPa 1010,000/kgm 0.00101
/kgm 00101.0
kJ/kg 81.191
in,34
33
343in,
3MP 8.0 @3
MPa 8.0 @33
in,12
33
121in,
3kPa 10 @1
kPa 10 @1
=+=+==
⋅−=−=
==
==
=
=+=+==
⋅−=−=
==
==
pII
pII
af
f
pI
pI
f
f
whh
PPw
hhP
whh
PPw
hh
v
vv
v
vv
Also, h4 = h9 = h10 = 731.12 kJ/kg since the two fluid streams that are being mixed have the same enthalpy.
( )( ) kJ/kg 7.24941.23929627.081.191
9627.04996.7
6492.08692.7kPa 10
KkJ/kg 8692.7kJ/kg 3.3481
C500MPa 8.0
kJ/kg 7.2812MPa 8.0
KkJ/kg 7585.6kJ/kg 0.3502
C550MPa 10
88
88
78
8
7
7
7
7
656
6
5
5
5
5
=+=+=
=−
=−
=
==
⋅==
°==
=
==
⋅==
°==
fgf
fg
f
hxhh
sss
x
ssP
sh
TP
hss
P
sh
TP
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.
10-39
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q , 0∆peke ≅≅∆≅≅W&&
( ) ( ) ( )( ) ( 3629363292
outin
(steady) 0systemoutin
1
0
hhyhhyhhmhhmhmhm
EE
EEE
eeii −=−−→−=−→=
=
=∆=−
∑∑ &&&&
&&
&&&
)
where y is the fraction of steam extracted from the turbine ( = & / &m m3 4 ). Solving for y,
( ) ( ) 2019.090.20113.73187.7207.2812
90.20113.731
2936
29 =−+−
−=
−+−−
=hhhh
hhy
Then,
( ) ( )( ) ( ) ( )( )( )( ) ( )( )
kJ/kg 6.14668.18375.3304kJ/kg 9.183781.1917.24942019.011
kJ/kg 5.33047.28123.34812019.0113.7310.35021
outinnet
18out
6745in
=−=−==−−=−−=
=−−+−=−−+−=
qqwhhyq
hhyhhq
and
kg/s 54.5===kJ/kg 1467.1kJ/s 80,000
net
net
wW
m&
&
(b) 44.4%=−=−=kJ/kg 3304.5kJ/kg 1837.8
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.
10-40
10-51E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of the 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
High-P Turbine
5 P III
1-y-z 1211
z
10
4
Open fwh II
1-y
y
P II P I
Open fwh I
Condenser
Boiler Low-P Turbine
6
3 2
1
8
7
9
T
9
10z
1500 8 1 - y 5
6
3
y 4 250 psia
1 psia
40 psia 140 psia
1 - y - z
11
7
12
2
1
s (a) From the steam tables (Tables A-4E, A-5E, and A-6E),
( )
( )( )
Btu/lbm 84.6912.072.69Btu/lbm 0.12
ftpsia 5.4039Btu 1
psia 140/lbmft 0.01614
/lbmft 01614.0
Btu/lbm 72.69
in,12
33
121in,
3psia 1 @1
psia 1 @1
=+=+==
⋅−=
−=
==
==
pI
pI
f
f
whh
PPw
hh
v
vv
( )( )( )
( )( )( )
Btu/lbm 41.38031.409.376Btu/lbm 4.31
ftpsia 5.4039Btu 1
psia 2501500/lbmft 0.01865
/lbmft 01865.0Btu/lbm 09.376
liquid sat.psia 250
Btu/lbm 81.23667.014.236
Btu/lbm 0.67ftpsia 5.4039
Btu 1psia 40250/lbmft 0.01715
/lbmft 01715.0Btu/lbm 14.236
liquid sat.psia 40
in,56
33
565in,
3psia 250 @5
psia 250 @55
i,34
33
343in,
3psia 40 @3
psia 40 @33
=+=+==
⋅−=
−=
==
==
=
=+=+=
=
⋅−=
−=
==
==
=
pIII
pIII
f
f
npII
pII
f
f
whh
PPw
hhP
whh
PPw
hhP
v
vv
v
vv
Btu/lbm 5.1308psia 250
RBtu/lbm 6402.1Btu/lbm 5.1550
F1100psia 1500
878
8
7
7
7
7
=
==
⋅==
°==
hss
P
sh
TP
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.
10-41
( )(Btu/lbm 4.1052
7.10359488.072.69
9488.084495.1
13262.08832.1
psia 1
Btu/lbm 0.1356psia 40
RBtu/lbm 8832.1Btu/lbm 3.1531
F1000psia 140
Btu/lbm 8.1248psia 140
1212
1212
1012
12
111011
11
10
10
10
10
979
9
=
+=+=
=−
=−
=
==
=
==
⋅==
°==
=
==
fgf
fg
f
hxhh
sss
x
ssP
hss
P
sh
TP
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&&
FWH-2:
( ) ( 548554488
outin
(steady) 0systemoutin
11
0
hhyyhhmhmhmhmhm
EE
EEE
eeii =−+→=+→=
=
=∆=−
∑∑ &&&&&
&&
&&&
where y is the fraction of steam extracted from the turbine ( = & / &m m8 5 ). Solving for y,
1300.081.2365.130881.23609.376
48
45 =−−
=−−
=hhhh
y
FWH-1
( ) ( 321133221111
outin
(steady) 0systemoutin
11
0
hyhzyzhhmhmhmhmhm
EE
EEE
eeii −=−−+→=+→=
=
=∆=−
∑∑ &&&&&
&&
&&&
)
where z is the fraction of steam extracted from the turbine ( = & / &m m9 5 ) at the second stage. Solving for z,
( ) ( ) 1125.01300.0184.690.135684.6914.2361
211
23 =−−−
=−−−
= yhhhh
z
Then,
( )( ) ( )( )
( )( ) ( )( )Btu/lbm 4.6714.7448.1415
Btu/lbm 744.469.721052.41125.01300.011Btu/lbm 8.14158.12483.15311300.0141.3805.15501
outinnet
112out
91067in
=−=−==−−−=−−−=
=−−+−=−−+−=
qqwhhzyq
hhyhhq
and
lbm/s 282.5=×
==Btu/lbm 1415.8
Btu/s 104 5
in
in
m&
&
(b) ( )( ) MW 200.1=
==
Btu 1kJ 1.055
Btu/lbm 671.4lbm/s 282.5netnet wmW &&
(c) 47.4%=−=−=Btu/lbm 1415.8Btu/lbm 744.4
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.
10-42
10-52 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 85.26543.1442.251
kJ/kg .431488.0/)kPa 2012,500)(/kgm 0.001017(
/
/kgm 001017.0
kJ/kg 42.251
in,12
3121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=−=
−=
==
==
pI
ppI
f
f
whh
PPw
hh
ηv
vvLow-P turbine
910
11
MixingCham.
1-y7
y
PII PI
Closed fwh Cond.
Boiler
High-P turbine
4
3 2
1
8
5
6
/kgm 001127.0
kJ/kg 51.762
liquid sat.MPa 1
3MPa 1 @3
MPa 1 @33
==
==
=
f
fhhPvv
( )
kJ/kg 25.77773.1451.762kJ/kg 73.14
88.0/)kPa 001012,500)(/kgm 001127.0(
/
in,311
3
3113in,
=+=+==
−=
−=
pII
ppII
whh
PPw ηv
Also, h4 = h10 = h11 = 777.25 kJ/kg since the two fluid streams which are being mixed have the same enthalpy.
( )( )( ) kJ/kg 5.32206.31855.347688.05.3476
kJ/kg 6.3185MPa 5
KkJ/kg 6317.6kJ/kg 5.3476
C550MPa 5.12
655665
65
656
6
5
5
5
5
=−−=−−=→
−−
=
=
==
⋅==
°==
sTs
T
s
hhhhhhhh
hss
P
sh
TP
ηη
( )( )( )
C328°=
==
=−−=−−=→
−−
=
=
==
⋅==
°==
88
8
877887
87
878
8
7
7
7
7
kJ/kg 1.3111MPa 1
kJ/kg 1.31111.30519.355088.09.3550
kJ/kg 1.3051MPa 1
KkJ/kg 1238.7kJ/kg 9.3550
C550MPa 5
ThP
hhhhhhhh
hss
P
sh
TP
sTs
T
s
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.
10-43
( )( )( ) kJ/kg 2.24929.23479.355088.09.3550
kJ/kg 9.2347kPa 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
( )( ) ( )1788.0)51.7621.3111()85.26525.777)(1(
1 38210
=→−=−−
−=−−
yyy
hhyhhy
The corresponding mass flow rate is kg/s 4.29=== kg/s) 24)(1788.0(58 mym &&
(c) Then,
( )( ) ( )( ) kJ/kg 1.184042.2512.24921788.011kJ/kg 7.30295.32209.355025.7775.3476
19out
6745in
=−−=−−==−+−=−+−=
hhyqhhhhq
and
kW 28,550=−=−= kJ/kg)1.18407.3029)(kg/s 24()( outinnet qqmW &&
(b) The thermal efficiency is determined from
39.3%==−=−= 393.0kJ/kg 3029.7kJ/kg 1840.1
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.
10-44
Second-Law Analysis of Vapor Power Cycles 10-53C 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. 10-54 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-15 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 10-15,
kJ/kg 8.1931kJ/kg 72.2650
KkJ/kg 5412.6
KkJ/kg 0912.1
out
43
kPa 50 @21
==
⋅==
⋅===
ss
sss
in
f
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,
( )
( ) kJ/kg 351.3
kJ/kg 1068
=
+−=
+−=
=
−+−=
+−=
K 290kJ/kg 1931.85412.60912.1K 290
K 1500kJ/kg 2650.80912.15412.6K 290
41,41041destroyed,
23,23023destroyed,
R
R
R
R
Tq
ssTx
Tq
ssTx
10-55 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-16 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 10-16,
kJ/kg 9.1897kJ/kg 2.3173
KkJ/kg 5995.6
KkJ/kg 6492.0
out
in
43
kPa10@21
==
⋅==
⋅===
ss
sss f
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,
( )
( ) kJ/kg 172.3
kJ/kg 1112.1
=
+−=
+−=
=
−+−=
+−=
K 290kJ/kg 1897.9
5995.66492.0K 290
K 1500kJ/kg 3173.2
6492.05995.6K 290
41,41041destroyed,
23,23023destroyed,
R
R
R
R
Tq
ssTx
Tq
ssTx
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.
10-45
10-56 The exergy destruction associated with the heat rejection process in Prob. 10-22 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 10-22,
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
10-57 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-32 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 10-32,
kJ/kg 8.213342.2512.2385
kJ/kg 1.3521.31052.3457
kJ/kg 0.314054.2595.3399KkJ/kg 2359.7KkJ/kg 7266.6
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 212.6
kJ/kg 94.1
kJ/kg 1245.0
=
+−=
+−=
=
−+−=
+−=
=
−
+−=
+−=
K 300kJ/kg 2133.82359.78320.0K 300
K 1800kJ/kg 352.57266.62359.7K 300
K 1800kJ/kg 3140.08320.07266.6K 300
61,61061destroyed,
45,45045destroyed,
23,23023destroyed,
R
R
R
R
R
R
Tq
ssTx
Tq
ssTx
Tq
ssTx
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.
10-46
10-58 EES Problem 10-57 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$='(subcooled)' end "Input Data - from diagram window" {P[6] = 20 [kPa] P[3] = 8000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Data for the irreversibility calculations:" T_o = 300 [K] T_R_L = 300 [K] T_R_H = 1800 [K] "Pump analysis" Fluid$='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" h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid$,h=h[6],P=P[6])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-47
"Boiler analysis" Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler" "Condenser analysis" h[6]=Q_out+h[1]"SSSF First Law for the Condenser" T[6]=temperature(Fluid$,h=h[6],P=P[6]) s[6]=entropy(Fluid$,h=h[6],P=P[6]) x6s$=x6$(x[6]) "Cycle Statistics" W_net=W_t_hp+W_t_lp-W_p Eff=W_net/Q_in "The irreversibilities (or exergy destruction) for each of the processes are:" q_R_23 = - (h[3] - h[2]) "Heat transfer for the high temperature reservoir to process 2-3" i_23 = T_o*(s[3] -s[2] + q_R_23/T_R_H) q_R_45 = - (h[5] - h[4]) "Heat transfer for the high temperature reservoir to process 4-5" i_45 = T_o*(s[5] -s[4] + q_R_45/T_R_H) q_R_61 = (h[6] - h[1]) "Heat transfer to the low temperature reservoir in process 6-1" i_61 = T_o*(s[1] -s[6] + q_R_61/T_R_L) i_34 = T_o*(s[4] -s[3]) i_56 = T_o*(s[6] -s[5]) i_12 = T_o*(s[2] -s[1])
0 .0 1 .1 2 .2 3 .3 4 .4 5 .5 6 .6 7 .7 8.8 9 .9 11 .00
100
200
300
400
500
600
700
s [kJ /kg -K ]
T [C
]
8000 kP a
300 0 kP a
20 kP a
3
4
5
6
Id ea l R an k in e cyc le w ith reh eat
1 ,2
SOLUTION Eff=0.389 Eta_p=1 Eta_t=1 Fluid$='Steam_IAPWS' h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg] h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg] h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg] i_12=0.012 [kJ/kg] i_23=1245.038 [kJ/kg] i_34=-0.000 [kJ/kg] i_45=94.028 [kJ/kg] i_56=0.000 [kJ/kg] i_61=212.659 [kJ/kg] P[1]=20 [kPa] P[2]=8000 [kPa]
P[3]=8000 [kPa] P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa] Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] q_R_23=-3140 [kJ/kg] q_R_45=-352.5 [kJ/kg] q_R_61=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K] s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K] s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K] s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C] T[3]=500 [C]
T[4]=345.2 [C] T[5]=500 [C] T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C] T_o=300 [K] T_R_H=1800 [K] T_R_L=300 [K] v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg] v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg] W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg] W_t_lp=1072 [kJ/kg] x6s$='' x[1]=0
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.
10-31
x[6]=0.9051
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.
10-48
10-59 The exergy destruction associated with the heat addition process and the expansion process in Prob. 10-34 are 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 10-34,
( )
( )
kJ/kg 8.3749kJ/kg 1.3375
kJ/kg 8.2664 ,kPa 10 KkJ/kg 3870.8KkJ/kg 7642.7
kJ/kg 0.2902 ,MPa 1 KkJ/kg 8464.6KkJ/kg 5995.6
KkJ/kg 6492.0
in
3
666
5
444
3
kPa 10 @ 21
==
==⋅=⋅=
==⋅=⋅=
⋅===
qh
hPss
hPss
sss f
The exergy destruction associated with the combined pumping and the heat addition processes is
( ) kJ/kg 5.1289K 1600kJ/kg 3749.8
8464.67642.76492.05995.6K 285
15,45130destroyed
=
−+−+−=
+−+−=
R
R
Tq
ssssTx
The exergy destruction associated with the pumping process is
Thus, kJ/kg 1289=−=−=
=−=∆−=−≅
5.05.1289
kJ/kg53.009.1062.10
12destroyed,destroyedheating destroyed,
,,,12destroyed,
xxx
Pvwwwx apspap
The exergy destruction associated with the expansion process is
( ) ( )
( )( ) kJ/kg 247.9=⋅−+−=
+−+−=
KkJ/kg7642.73870.85995.68464.6K 285
036,
5634034destroyed,R
R
Tq
ssssTx
The exergy of the steam at the boiler exit is determined from
( ) ( )( ) ( )03003
03
023
030033 2ssThhqz
VssThh
−−−=++−−−=ψ
where
( )( ) KkJ/kg 1806.0
kJ/kg 51.50
K 285 @ kPa 100 K, 285 @ 0
K 285 @ kPa 100 K, 285 @ 0
⋅=≅==≅=
f
f
ssshhh
Thus, ( ) ( )( ) kJ/kg 1495=⋅−−−= KkJ/kg 1806.05995.6K 285kJ/kg 51.501.33753ψ
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.
10-49
10-60 The exergy destruction associated with the regenerative cycle described in Prob. 10-44 is 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 10-44, qin = 2692.2 kJ/kg and qout = 1675.7 kJ/kg. Then the exergy destruction associated with this regenerative cycle is
( ) kJ/kg 1155=
−=
−=
K 1500kJ/kg 2692.2
K 290kJ/kg 1675.7
K 290inout0destroyed,
HLcycle T
qTq
Tx
10-61 The exergy destruction associated with the reheating and regeneration processes described in Prob. 10-49 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 10-49 and the steam tables,
kJ/kg 6.6687.28123.3481
KkJ/kg 6492.0KkJ/kg 8692.7
KkJ/kg 7585.6
KkJ/kg 0457.22016.0
67reheat
kPa10@21
7
65
MPa8.0@3
=−=−=
⋅===⋅=
⋅==
⋅===
hhq
ssss
ss
ssy
f
f
Then the exergy destruction associated with reheat and regeneration processes are
( )
( )( )
( ) ( )( ) ( )( )[ ] kJ/kg 47.8
kJ/kg 214.3
=−−−=
−−−=
+−==
=
−+−=
+−=
∑∑6492.02016.017585.62016.00457.2K 290
1
K 1800kJ/kg 668.6
7585.68692.7K 290
2630
0
0
surr0gen0regendestroyed,
67,670reheatdestroyed,
syyssTT
qsmsmTsTx
Tq
ssTx
iiee
R
R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-50
10-62 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
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 8604.1
kJ/kg 09.6400
kPa 500
6
6
6
6
⋅==
==
sh
xP
kg/s 81.19119.38230316 =−=−= mmm &&&
2
production well
reinjection well
separator
steam turbine
1
Flash chamber
65
4
3
condenser
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 &&
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10-51
(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
Cogeneration 10-63C 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. 10-64C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity. 10-65C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference. 10-66C 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 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.