PDF Solution Manual For Gas Turbine Theory 6th Edition Saravanamuttoo Rogers Compress [PDF]

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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 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  = 0.01415 (from Fig. 2.15) Actual f = 0.01415 0.99 = 0.01429 S.F.C  =

3600 f  W  N 

=

3600 × 0.01429 166.64

= 0.308 kg/kW − hr  

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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

= 3.8 − 0.12 = 3.68 bar  

 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 ( 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 with no bleed = 633.11kW

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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 (K)

593

645.8

701.9

 P 03/ P 04

8.55

11.40

15.2

1.612

1.708

1.820

T 03

1150

1400

1600

T 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=W  –  t Wc 

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   T 03/T 04=  03   P 04 

n −1 n

∆T cc

 f /a mf = ma × f

× (1 − B )

SFC (kg/kwhr)

 EGT  ( C ) 

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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=1525 K  P 03=29.69 bar  P 04=13.00 bar

∴

 P 03  P 04

=

2.284 ∴

T 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 – T 04=256.3 T 05 T 06 T 05 – T 06 T 03 – T 04 ∆T  total ∆T  total ×1.148 × 0.99

 – ∆T c × 1.004

∴ m for

 MW ( f /a)1 ( f /a)2 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

0.0282×43100   37.7

0.0247 × 43100   38.5

0.0227 ×43100 36.9

So η th remains high as power reduced. May be difficult to control low fuel: air ratio in 2nd combustor. 9 © 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 2.6

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= 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 6 =

 

 LPT , 182.9= 0.89 × 1300 1 −

1 

 R 0.25 

39.19 2.248

= 17.43 bar

 RLP =1.990 P 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, 000 ∴ m=211.3 kg/s

 f /a=

0.028 0.99

= 0.0283

Specific output =1.148 ×0.99 × 416.3 = 473.1 ∴ η th =

473.1 0.283 × 43100

= 38.8%

 EGT  =1117.7-416.3 =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=567.5 K  P 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

=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=799-273=526 C 

( f /a)id=0.0138

 f /a=

0.0138 0.99

=0.0139

mf = 0.0139 ×112.0  =1.56 kg/s Heat input = 1.56 ∴  Efficiency

× 43,100

kJ kg kg s

= 67,288 kW

=40.3 %

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