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PROBLEM SOLVING
Carranza, Joem Q.
de la Fuente, Ceasar D.
Marcos, Ian Mark P.
PROBLEM 8.14 A combined gas-steam turbine powerplant is designed with four 50-
MW gas turbines and one 120-MW steam turbine. Each gas turbine
operates with compressor inlet temperature 505 R, turbine inlet
temperature 2450 R, pressure ratio for both compressor and turbine
5, and compressor and turbine polytropic efficiencies 0.87 and 0.96.
The gases leaving the turbine go to a heat-recovery boiler then to a
regenerator with an effectiveness of 0.87. The turbine gases
correspond to 200 percent of theoretical air. The steam cycle has a
turbine steam inlet at 1200 psia and 1460R, one open-type feedwater
heater (not optimally placed) with feedwater temperature to heat
recovery boiler at 920 R, condenser pressure at 1 psia, and polytropic
and mechanical efficiencies 0.87 and 0.96. all generator efficiencies
are 0.96. supplementary firing at full load raises the gas temperature
to 2000 R.
REQUIRED:
Draw the cycle flow ang T-s diagram and find (a) the required
steam mass flow rate in the steam turbine in pound mass per
hour, (b) the required air mass flow rate in each gas turbine in
pound mass per hour, (c) the heat added in the gas cycle and in
supplementary firing at full load, (d) the stack gas temperature in
degrees Fahrenheit, (e) the cycle efficiency at full load, and (f) the
efficiency at startup when only one gas turbine is used at its full
load with no supplementary firing or regeneration. (Ignore the
steam cycle pump work.)
AC
ST G
CFWH
Air in
1
2
8
9
10
12
13
14
11
SCHEMATIC
DIAGRAM 8.14
8
2 ‘
6 ‘
CC
GGT
SF
3
4
5
6
T-S DIAGRAM
T
s
1
2
2’
3
4
5
6
6’
6”
8
9
1011
12
13
14
B
CC
AC GGT
SFair
1
22 ‘ 3
4
5
6
6 ‘
6’
SCHEMATIC & T-s DIAGRAM (Brayton Cycle)
T
s
1
2
2’
3
4
5
6
6’
6”
ST G
CFWH
8
910
12
13
14
11
SCHEMATIC &
T-s DIAGRAM
(Rankine Cycle)
T
s
8
9
1011
12
13
14
B
9m
GIVEN: Gas Cycle
Working fluid: air
Four 50 MW gas turbine
Tcomp,inlet = 505 0R
Tturbine,inlet= 2450 0R
rPc=rPT= 5
ηc = 0.87
ηT =0.87
ηmech =0.96
εR = 0.87
Supplementary firing at 2000 0R
ηgen =0.96
The turbine gases correspond to
200 percent of theoretical air.
cp= cp,equiv =0.24 Btu/lbm 0F
Steam cycle
Working fluid: water
120 MW steam turbine
Pturbine,inlet = 1200 psia, Tturbine,inlet =1460 0R
One open-type feedwaterheater operating at 9200R
Pcond,inlet = 1 psia
ηT =0.87
ηmech =0.96
ηgen =0.96
Non ideal
Assume
Ignore pump work
Wn = 120MW
(A). FIND: REQUIRED STEAM MASS FLOW
RATE IN THE STEAM TURBINE (LBM/HR)
Assume
,
ns
total ST
Wm
w
, 8 9 9 9 10
9 9 13 9 13 12
( ) (1 )( )
( ) (1 )( )
total STw h h m h h
m h h m h h
8 1m
, Wn = 120MW
s
T
10
8
9
11
12
13
14
B
9m
ST G
8
9 10
Energy Balance on Feedwater Heater :
9m91 m
FWH
1213
9
9m
91 m
STEAM CYCLE
Finding the states starting from point 8.
State 8
P8 = 1200 psia, T8 = 1460 0R = 1000 0F,
(superheated)
h8=1499.4 Btu/lbm
s8=1.6298
State 9s (ideal)
T9s = 920 OR = 4600F, s8=s9=1.6298
(superheated)
P9s=162.5676 psia
h9s=1272.5916 Btu/lbm
s
T
10
8
9
11
12
13
14
B
Double interpolation, Table A-3:
STEAM CYCLE
State 9
P9=P9s=162.5676 psia
8 9
8 9
9
9
1499.40.87
1499.4 1272.5196
1302.014
T
s
h h
h h
h
h
s
T
10
8
9
11
12
13
14
B
STEAM CYCLE
State 10s
P10s=1 psia
s10s=s9s=s8=1.6298, mixed phase
T10s=Tsat=101.740F
10 ,10 10 ,10
10 ,10
10
,10
10 ,10 10 ,10
10
10
( )
1.6298 0.13260.81127
1.8455
( )
69.73 (0.81127)(1036.1)
910.28753 /
s f s s fg s
s f s
s
fg s
s f s s fg s
s
s m
s s x s
s sx
s
h h x h
h
h Btu lb
s
T
10
8
9
11
12
13
14
B
STEAM CYCLE
State 10
P10=P10s=1 psia
9 10
9 10
10
10
1302.0140.87
1302.014 910.28573
961.22
T
s
h h
h h
h
h
T
s
8
9
11
12
13
14
B
STEAM CYCLE
State 11
P11=P10=P10s=1 psia, saturated liquid
T11=101.740F
v11= 0.01614 ft3/lbm
h11=hf =69.73 Btu/lbm
s
T
10
8
9
11
12
13
14
B
STEAM CYCLE
State 12
P12=P9=162.5676 psia
11 12 1112 11
12
12
( )
0.01614(162.5676 1)*14469.73
788.16
70.2123
v P Ph h
J
h
h
s
T
10
8
9
11
12
13
14
B
STEAM CYCLE
State 13
P13=P12=P9=162.5676psia
T13=364.8 0F
v13=vf=0.01816 ft3/lbm
h13=hf=337.409 Btu/lbm
s
T
10
8
9
11
12
13
14
B
Interpolation, Table A-1:
STEAM CYCLE
State 14
P14= P8 = 1200 psia
13 13 1414 13
14
14
( )
0.01816(1200 162.5676)*144337.409
788.16
340.851 / m
v P Ph h
J
h
h Btu lb
s
T
10
8
9
11
12
13
14
B
(a) Steam mass flow rate
9 9 13 9 13 12
9 9
9
( ) (1 )( )
(1302.01 337.409) (1 )(337.409 70.2123)
0.2169056
m h h m h h
m m
m
8 9 9 9 10
6
( ) (1 )( )
(1499.4 1302.01) (1 0.2169056)(1302.01 961.22)
464.26 /
( )( )
464.26(0.96)(0.96)
427.863 /
120(3.412)(10 )
427.863
9.5694(
T
T
T m
total T M G
total
total m
nw
Total
w
w h h m h h
w
w Btu lb
w w
w
w Btu lb
Wm
w
m
510 ) /mlb h
s
T
10
8
9
11
12
13
14
B
FWH
1213
9Energy Balance on
Feedwater Heater :
Steam cycle efficiency
, 8 14
427.863 /
(1499.4 340.851) 1158.549 /
427.8630.3693
1158.549
total m
a hrb m
totalTH
a
w Btu lb
q h h Btu lb
w
q
s
T
10
8
9
11
12
13
14
B
< Ignore steam cycle pump work
REQUIRED AIR MASS FLOW RATE IN EACH
GAS TURBINE(LBM/HR)
Note:
Wn = 50MW for each gas turbine
1MW = 3.412 (106) Btu/lbm
3 4 3 4
2 1 2 1
( )
( )( )
( ) ( )
( ) ( )
na
tot
tot GT AC gen M
GT P
AC P
Wm
w
w w w
w h h c T T
w h h c T T
s
T
1
2
2’
3
4
5
6
6’
6”
GAS CYCLE
State 1
T1 = 505 0R, dry air
Pr,1 = 1.09695
h1 = 120. 675 Btu/lbm.mol
State 2s
rpc = 5
Pr,2s=rpc(Pr,1)=5(1.09695)=5.48475
T2s = 798.2883817 0R
h2s= 191. 3940768 Btu/lbm.mol
s
T
1
2
2’
3
4
5
6
6’
6”
Interpolation, Table I-1:
Interpolation, Table I-1:
GAS CYCLE
State 2
2 1
2 1 2
2
191.3940768 120.6750.87
120.675
201.9612952 / .
sc
m
h h
h h h
h Btu lb mol
2 1
2 1
2
0
2
798.2883817 5050.87
505
842.113082
sc
T T
T T
T
T R
s
T
1
2
2’
3
4
5
6
6’
6”
GAS CYCLE
State 3
T3 = 2450 0R, 200 percent theoretical air
Pr3 = 511.9
h3= 19080.7 Btu/lbm.mol
State 4s
200 percent theoretical air
rPT = 5
,3
,4
4
4
511.9102.38
5
1683.083004
12539.51146 / .
r
r s
pt
o
s
s m
PP
r
T R
h Btu lb mol
s
T
1
2
2’
3
4
5
6
6’
6”
Interpolation, Table I-2: 4s
GAS CYCLE
State 4
200 percent theoretical air
3 4 4
3 4
4
19080.70.87
19080.7 12539.51146
13389.86597 / .
c
s
m
h h h
h h
h Btu lb mol
3 4
3 4
4
0
4
24500.87
2450 1683.083004
1782.782213
T
s
T T
T T
T
T R
s
T
1
2
2’
3
4
5
6
6’
6”
GAS CYCLE
State 5
5
5
5
2000
Pr 211.6
15189.3 / .
o
m
T R
h Btu lb mol
s
T
1
2
2’
3
4
5
6
6’
6”
200 percent theoretical air, Table I-2
(B) AIR MASS FLOW RATE
2 1
3 4
( )
(0.24)(842.113082 505) 80.907
( )
(0.24)(2450 1683.083004) 184.06
( )( )
(184.06 80.907)(0.96)(0.96) 95.0657
95.0657 95.0657
50(3.412)(10
AC P
GT P
total GT AC gen M
n na
W mc T T
m m
W mc T T
m m
W W W
m m
W Wm
66)
1.7945(10 ) /95.0657
mlb h
s
T
1
2
2’
3
4
5
6
6’
6”
GAS CYCLE
State 6
, , 8 14 5 6
5 6
6
0
6
( ) ( )
(1158.549)(9.5694)(10 ) (4)(1.7945)(10 )(0.24)(2000 )
1356.462
a hrb a hrb w w a PQ q m h h m m c T T
T
T R
s
T
1
2
2’
3
4
5
6
6’
6”
(D) STACK GAS TEMPERATURE
s
T
1
2
2’
3
4
5
6
6’6”
2”
2 ‘
6 ‘
6
2
0 0
6" 2
2' 2
6 2
2'
0 0
2'
842.113082 382.44 °
842.1130820.87
1362.587515 842.113082
1294.9258 835.26 °
R
Tstack T T R F
T T
T T
T
T R F
State T2’
(C) HEAT ADDED BY THE CC
, 3 2'
6
,
8
,
( )
(1.7945(10 ))(0.24)(2450 1294.9258)
4.923(10 ) /
a cc a P
a cc
a cc
Q m c T T
Q
Q Btu h
5 4
6
7
( )
(1.7945(10 ))(0.24)(2000 1782.782213)
9.355(10 ) /
SF a P
SF
SF
Q m c T T
Q
Q Btu h
(C) HEAT ADDED IN SUPPLEMENTARY
FIRING
s
T
1
2
2’
3
4
5
6
6’
6”
CC
GGT
SF
3
4
5
2’
(E) GAS CYCLE EFFICIENCY
6 6
8
,
7
6
8 7
,
50(3.412)(10 ) 170.6(10 )
4.923(10 ) /
9.355(10 ) /
170.6(10 )0.2912
4.923(10 ) 9.355(10 )
N
a cc
SF
totalTH
a cc SF
W
Q Btu h
Q Btu h
W
Q Q
OVERALL PLANT CYCLE EFFICIENCY
6 9
8 8
,
7 8
10
8 8
,
[(4*50) (120)](3.412)(10 )(0.96)(0.96) 1.006(10 ) /
4*4.923(10 ) 19.69(10 ) /
4*9.355(10 ) 3.74(10 ) /
1.006(10 ). 0.429
19.69(10 ) 3.74(10 )
total
a cc
SF
total
a cc SF
W Btu h
Q Btu h
Q Btu h
Wplant eff
Q Q
CC
AC GGT
SG G
CFWH
air
1
23
4
6
8
910
1112
13
14
SCHEMATIC DIAGRAM
(no SF and regeneration)
6
, 3 2
6 8
,
6
8
,
170.6(10 )
( )
1.7945(10 )(0.24)(2450 842.113082) 6.75(10 ) /
170.6(10 )0.253
6.75(10 )
total
a G a P
a G
netTH
a G
W
Q m c T T
Q Btu h
W
Q
(E) GAS CYCLE EFFICIENCY WHEN ONLY 1 GAS TURBINE
IS USED WITH NO SF AND REGENERATION
s
T
1
2
3
46
