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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.2
γ -1 Ta P 02 γ T02 – T a = –1 η c P a 1 288 3.5 – 1 = 345.598 K 1 1 = ( ) 0.82
Compressor and turbine work required per unit mass flow is: Wtc
=
C pa (T02
− T a
η m
T03 – T 04 =
)
(T0 3 − T04 )
= C pg
1.005× 345.598 345.598 0.98×1.147
= 308.992 K
T 04 =115 =1150 – 309 = 841K γ −1 1 γ T03 − T04 = η t T 03 1 − P03 P 04 1 4 1 308.992 = 0.87 ×1150 1 − P03 P 04
P 03 P 04
=
4.382
P 03
= 11.0 − 0.4 = 10.6
P04
=
2.418 ba b ar
bar
, P05
=
P a
1 4 1 = 148.254K T04 − T 05 = 0.89 × 841 1 − 2.418
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Specific power output: W N = 1.147 × 0.98 × 148.254 = 166.64 kW kWs/kg Hence mass flow required = T 02
=
20 × 103 166.64
= 120.019kg/sec
288 + 345.6 = 633.6K
T03 − T 02
= 1150 − 633.6 = 516.4K
Theoretical f Theoretical f = = 0.01415 (from Fig. 2.15) Actual f Actual f = = 0.0 0.01415 0.99 = 0.01429 S.F.C =
3600 f W N
=
3600 × 0.01429 166.64
kg/kW − hr = 0.308 kg
6 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.3
288
1
T02
− T a =
P 03
bar = 3.8 − 0.12 = 3.68 ba
P 03 P a
0.85
[3.8
3.5
− 1] = 157.3K
= 3.68
T03 − T 04
= 1050 × 0.88[1 − (
1 3.68
1
) 4 ] = 256.87K
Net work output W
= η m (load ) [( m − mc )C pg pg ( T03 − T04 ) −
mC pa (T02
− Ta )
]
η m (comp .rotor )
200 = 0.98[( m − 1.5) ×1.147 × 256.87 −
m × 1.005 ×157.3 0.99
]
200 = 288.73m − 433.1 − 156.5m (a) m = 4.788 kg/sec
With no bleed flow: Net work output = 0.98 × 4.788[1.147 × 256.87 −
1.005 × 157.3 0.99
]
= 633.11 kW (b) The power output output with no bleed = 633.11kW 633.11kW
7 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.4 A
B
C
n −1 n c
0.3287
0.3250
0.3213
n −1 n t
0.2225
0.2205
0.2205
T02 T 01
2.059
2.242
2.437
∆T 012 (K)
305
357.8
413.9
T 02 02 (K)
593
645.8
701.9
P 03 P 04 03/ P 04
8.55
11.40
15.2
1.612
1.708
1.820
T 03 03
1150
1400
1600
T 04 04
713.4
819.6
879.2
∆T 034
436.6
580.4
720.7
W c= (1.005 ×∆T 12 )/0.99
309.6
363.2
420.2
W t= 1.148(1 − B )∆T 034
501.2
649.6
786.0
W net Wc – net=W t W
191.6
286.4
365.8
Power
14,370
22,912
31,093
Base
+59.4%
+116%
557
754
898
0.0162
0.0219
0.0268
4374
6150
7791
0.304
0.268
0.251
base
11.8%
17.4%
440
547
606
P 03 T 03 03/T 04 04= P 04
n −1 n
∆T cc
f /a mf = ma × f
× (1 − B )
SFC (kg/kwhr)
EGT ( ( C )
8 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.5 T 03 03=1525 K P 03 03=29.69 bar P 04 04=13.00 bar
∴
P 03 P 04
=
2.284 ∴
T 04 04=1268.7 K , P 05 P 06
=
T 03 T 04
∆T hp =
13.00 × 0.96 1.02
=
( 2.284 )
0.2223
= 1.202
256.3
= 12.24 ∴
T 05 T 06
=
(12.24)
0.223
= 1.745
T 03 03 – T 04 04=256.3 T 05 05 T 06 06 T 05 T 06 05 – T 06 T 03 T 04 03 – T 04 ∆T total ∆T total ×1.148 × 0.99
– ∆T c × 1.004
∴ m for
MW f /a)1 ( f f /a)2 ( f η th
240 M W =
240000 458.1
1525 874 651.0 256.3 907.3 1031.1 573.0
1425 816.6 608.4 256.3 864.7 982.7 573.0
1325 759.3 565.7 256.3 822.0 934.2 573.0
458.1
409.7
361.2
= 523.9 kg/s
240.0 0.0197 0.0085 0.0282 458.1
214.6 0.0197 0.0050 0.0247 409.7
189.2 0.0197 0.0030 0.0227 361.2
0.02 0.0282 82×4310 43100 0 37.7
0.0 0.024 247 7 ×431 43100 38.5
0.0 0.022 227 7 ×4310 43100 0 36.9
So η th remains high as power reduced. reduced. May be difficult to control low fuel: air ratio in 2nd combustor. 9 © Pearson Education Limited 2009
th
HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.6
288
∆T 12 =
(5.5)0.286 − 1 = 206.8 T2 0.875
∆T 34 =
300 0.286 7 .5 ) ( 0.870
∆Tcc = 1550 − 569 = ∆T 56 =
∆T 67
=
− 1 =
268.7 T 4 = 568.7K
1.148 × 0.95 × 0.99 1.148 × 0.99
494.8K
981 C
268.7 × 1.005 206.8 × 1.005
=
=
250.1 T 6=1550 – 250 = 1300K
= 182.9
HPT ,250.1= ,250.1= 0.88 ×1550 1 −
T 7=1300 – 182.9 = 1117.1K 1
R 0.25
Rhp=2.248
CDP =1.00 ×5.5 × 7.5 × 0.95 = 39.19 bar P bar P 6 =
LPT , 182.9= 0.89 × 1300 1 −
1
R 0.25
39.19 2.248
= 17.43 bar
RLP =1.990 P =1.990 P 7=8.76 7=8.76 bar
P 8= P 1=1.00
∴ ∆T 78 = 0.89 × 1117.1 1 − ∴
= 416.3 K 8.760.25 1
m ×1.148 × 0.99 × 416.3 = 100, 00 000 ∴ m=211.3 kg/s
f /a=
0.028 0.99
= 0.0283
Spe Specific output =1.148 ×0.99 × 416.3 = 473.1 ∴ η th =
473.1 0.283 × 43100
= 38.8%
EGT =1117.7-416.3 =1117.7-416.3 =701.4 K =701.4 K =428.4 C
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.8 P a=1.013 bar , T a=15 C
∆T 12 =
288
8.50.286 − 1 = 279.5k 0.87
T 02 02=567.5 K P 04 04=1.013 ×8. 5 × [1 − 0.015 ] × [1 − 0.042 ] = 8.125 bar ∆T 45 =
279.5 × 1.004 1.147 × 0.99
=
247.1 T 5 = 1037.8 K
1 P 4 ⇒ = 2.716, P 5 = 2.991 bar 247.1=0.87 ×1285 1 − 0.25 P 5 ( P 4 / P 5 ) ⇒
P 6=1.013 × 1.02 = 1.0333
P 5 P 6
1
=
2.895
= 213.1
∴ T 6=824.7
∴ ∆T 56 =
0.88 × 1037.8 1 −
∴ Power
=1 =112.0 ×1.147 × 0.99 × 213.1 = 27,106 kW
2.895
0.25
K
T 3 – T 2=0.90(824.7-567.5)=231.5 T 3=231.5+567.5=799K ∆T cc = 1285 − 799 =
486 C
T in in=799-273=526 C
( f f /a)id=0.0138
f /a=
0.0138 0.99
=0.0139
mf = 0.0 0.013 139 9 ×112. 12.0 =1.56 kg/s Heat input = 1.56 ∴ Efficiency
43,100 × 43,100
kJ kg kg s
= 67,288 kW
=40.3 %
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