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Lecturer’s Solutions Manual Gas Turbine Theory Sixth edition HIH Saravanamuttoo GFC Rogers H Cohen PV Straznicky For further instructor material please visit:

www.pearsoned.co.uk/saravanamuttoo ISBN: 978-0-273-70934-3

 Pearson Education Limited 2009 Lecturers adopting the main text are permitted to download and photocopy the manual as required.

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies around the world Visit us on the World Wide Web at: www.pearsoned.co.uk ---------------------------------This edition published 2009 The rights of HIH Saravanamuttoo, GFC Rogers, H Cohen and PV Straznicky to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. ISBN: 978-0-273-70934-3 All rights reserved. Permission is hereby given for the material in this publication to be reproduced for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without either the prior written permission of the Publishers or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6-10 Kirby Street, London EC1N 8TS. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the Publishers.

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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6 edition, Lecturer’s Solutions Manual

Supporting resources Visit www.pearsoned.co.uk/saravanamuttoo to find other valuable online resources For instructors •

PowerPoint slides that can be downloaded and used for presentations

For more information please contact your local Pearson Education sales representative or visit www.pearsoned.co.uk/saravanamuttoo

3 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Preface Since the introduction of the Second Edition in 1972 many requests for solutions have been received. The advent of modern word processing systems has now made it convenient to prepare these in an electronic format and I am glad to do so. All significant gas turbine calculations carried out in industry are universally done by digital computer and the purpose of these problems is to provide an understanding of the engineering principles involved. It is perhaps of historical significance that all of these calculations were originally done on a slide rule and many were previous examination questions. This manual will be available to instructors adopting the main text, who are then permitted to photocopy the material, but it is hoped that students will tackle the problems before looking at the manual. I will be very glad to hear of any corrections needed. My sincere thanks to Mr.Hariharan Hanumanthan, a doctoral student at Cranfield University, for his invaluable assistance in the preparation of this manual. H.I.H. Saravanamuttoo Ottawa, June 2008

4 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 2.2

 γ -1     Ta  P02  γ   T02 – Ta =  – 1  ηc  Pa   

=

 1  288  (11) 3.5 – 1 = 345.598K  0.82  

Compressor and turbine work required per unit mass flow is:

W tc =

C pa (T02 − T a )

T03 – T04 =

ηm

= C pg (T 03 −T 04 )

1.005× 345.598 = 308.992 K 0.98 ×1.147

T04 = 1150 – 309 = 841K γ −1      1  γ    T03 − T04 =η tT03 1 −    P P   03 04   1       1  4    308.992 = 0.87 × 1150 1 −    P P   03 04  

P03 = 4.382 P04 P03 = 11.0 − 0.4 = 10.6 bar P04 = 2.418 bar

, P05 = Pa

 1   1   4    = 148.254K T04 − T05 = 0.89 × 841 1 −    2.418    

5 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Specific power output: W N = 1.147 × 0.98 × 148.254 = 166.64kWs/kg Hence mass flow required =

20 × 103 = 120.019 kg/sec 166.64

T02 = 288 + 345.6 = 633.6K T03 − T02 = 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 3600 × 0.01429 = = 0.308kg/kW − hr WN 166.64

6 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 2.3

T02 − Ta =

1 288 [3.8 3.5 − 1] = 157.3K 0.85

P03 = 3.8 − 0.12 = 3.68bar P03 = 3.68 Pa 1 14 ) ] = 256.87K T03 − T04 = 1050 × 0.88[1 − ( 3.68 Net work output

W = η m( load) [(m − m c )C pg (T03 − T 04 ) −

mC pa (T02 − T a )

200 = 0.98[(m −1.5) ×1.147 × 256.87 −

ηm ( comp. rotor)

]

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

7 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th 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 T01

2.059

2.242

2.437

∆T012 (K)

305

357.8

413.9

T02 (K)

593

645.8

701.9

P03/P04

8.55

11.40

15.2

1.612

1.708

1.820

T03

1150

1400

1600

T04

713.4

819.6

879.2

∆T034

436.6

580.4

720.7

Wc= (1.005 ×∆T12 ) / 0.99

309.6

363.2

420.2

Wt= 1.148(1− B )∆T034

501.2

649.6

786.0

Wnet=Wt–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

mf = ma × f × (1 − B )

4374

6150

7791

SFC (kg/kwhr)

0.304 base

0.268 11.8%

0.251 17.4%

440

547

606

P  T03/T04=  03   P04 

n−1 n

∆Tcc f/a

D

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 T03=1525 K P03=29.69 bar P04=13.00 bar

∴

P03 = 2.284 P04

∴ T03 = (2.284 )0.2223 = 1.202 T04

T04=1268.7 K, ∆Thp = 256.3 P05 13.00 × 0.96 T 0.223 =1.745 = = 12.24 ∴ 05 = (12.24 ) T06 1.02 P06 T03 – T04=256.3 T05 T06 T05–T06 T03–T04 ∆T total ∆T total ×1.148 × 0.99 – ∆Tc × 1.004

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

∴ m for 240 MW = 240000 = 523.9 kg/s 458.1

MW (f/a)1 (f/a)2

ηth

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 ηth 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

∆T12 =

288  (5.5) 0.286 − 1 = 206.8 T2 = 494.8K  0.875 

∆T34 =

300  ( 7.5)0.286 − 1 = 268.7 T4 = 568.7K 0.870 

∆Tcc = 1550 − 569 = 981D C ∆T56 =

268.7 ×1.005 = 250.1 T6=1550 – 250 = 1300K 1.148× 0.95× 0.99

∆T67 =

206.8 ×1.005 =182.9 1.148 × 0.99

T7=1300 – 182.9 = 1117.1K

 1  HPT,250.1= 0.88× 1550 1− 0.25  Rhp=2.248  R  CDP=1.00 ×5.5 × 7.5 × 0.95 = 39.19 bar P6 =

39.19 = 17.43 bar 2.248

 1  LPT, 182.9= 0.89 ×1300 1 − 0.25  RLP=1.990 P7=8.76 bar  R  P8=P1=1.00

 

∴ ∆T78 = 0.89 × 1117.1 1 −

1  = 416.3 K 8.76 0.25 

∴ m ×1.148 × 0.99 × 416.3 = 100, 000 ∴ m=211.3 kg/s f/a=

0.028 = 0.0283 0.99

Specific output =1.148 ×0.99 × 416.3 = 473.1

∴ ηth =

473.1 = 38.8% 0.283 × 43100 D

EGT =1117.7-416.3 =701.4 K=428.4 C 10 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 2.8 Pa=1.013 bar , Ta=15

∆T12 =

D

C

288  8.50.286 − 1 = 279.5k 0.87 

T02=567.5 K P04=1.013 ×8.5 ×[ 1 − 0.015] ×[ 1 − 0.042] = 8.125 bar ∆T45 =

279.5 × 1.004 = 247.1 T5 = 1037.8 K 1.147× 0.99

 1 247.1=0.87 × 1285 1− 0.25  ( P 4 / P5)

 P  ⇒ 4 = 2.716, P5 = 2.991 bar P5 

⇒

P6=1.013 × 1.02 = 1.0333

P5 = 2.895 P6

1   = 213.1 ∴ T6=824.7 K ∴ ∆T56 = 0.88 × 1037.8 1− 0.25  2.895  

∴ Power =112.0 ×1.147 × 0.99 × 213.1 = 27,106 kW T3 – T2=0.90(824.7-567.5)=231.5 T3=231.5+567.5=799K

∆Tcc = 1285 − 799 = 486D C Tin=799-273=526 D C (f/a)id=0.0138

f/a=

0.0138 =0.0139 0.99

mf= 0.0139 ×112.0 =1.56 kg/s Heat input = 1.56 × 43,100

kJ kg = 67,288 kW kg s

∴ Efficiency =40.3 % 11 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 2.9

For compression:

For expansion:

T02 −T01 =T 01[(

0.66 n −1 1 γ −1 ( )= = = 0.4518 n 0.88 × 1.66 η∞c γ

γ −1 0.88 × 0.66 n −1 )= = η∞t ( = 0.3498 n 1.66 γ P02 ) P01

n −1 n

−1] = 310[2 0.4518 −1] =114K

T04 −T03 = 300[20.4518 −1] = 110.3K T04 = 410.3K and T 05 = 700K T06 −T05 =

( given )

Q 500 × 10 3 = = 535.2K mC p 180 × 5.19

∴ T06 = 1235.2K P03 = 2 ×14.0 − 0.34 = 27.66 bar P04 = 2 × 27.66 = 55.32 bar

P06 = 55.32 − (0.27 +1.03) = 54.02bar P07 = 14.0 + 0.34 + 0.27 = 14.61bar ∴

P06 = 3.697 P07

T06 −T07 = 1235.2[1 − (

1 0.3498 ) ] = 453.3K 3.697

T07 = 781.82K 12 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Power output = mC p[∆ T067 − ∆T034 − ∆T 012 ] = 180 × 5,19 × 229.0 = 213996 kW or 213.996MW Thermal efficiency = H.E.effectiveness =

213.996 = 0.4279 500

or

42.8%

T05 − T04 700 − 410.3 = = 0.7798 T07 − T04 781.8 − 410.3

or 78%

13 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 3.1

For compression:

For expansion:

1 n −1 = = 0.3284 n 0.87 ×3.5

n −1 0.87 = = 0.2175 n 4

and

n = 4.5977 n −1

At 7000 m Pa = 0.411bar; Ta = 242.7K 2

Ca 260 2 = = 33.63K 2c p 2 × 1.005 × 1000 T01 = 242.7 + 33.63 = 276.33K γ

P01 = Pa [1 +ηi

C a2 γ −1 ] 2 cp Ta

0.95 × 33.63 3.5 ] = 0.633bar 242.7 P02 = 8 × 0.633 = 5.068bar P01 = 0.441[1 +

T02 −T01 = 276.33[8.00.3284 −1] = 270.7K T02 = 547.02K T03 − T04 =

c pa (T02 − T01 ) c pgη m

T04 = 1200− 239.6

1.005 × 270.7 = 239.6K 1.147 × 0.99

=

T04 = 960.4K

P03 = 5.068(1− 0.06)= 4.763bar

n 4.5977

P03  T03 n −1  1200  =  =  P04  T04   960.4   4.763  P04 =   = 1.710 bar  2.784  Nozzle pressure ratio

= 2.784

P04 1.710 = = 4.16 Pa 0.411

14 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

(Since the critical pressure ratio is of the order of 1.9 the nozzle is clearly choked)

P04 1 = 4 =1.918 Pc  1  0.333     1 −  0.95  2.333   P5=Pc=

1.710 = 0.891 bar 1.918

T5 = Tc =

ρ5 =

2 2 × 960.4 = 823.3K T04 = + γ 1 2.333

Pc 100 × 0.891 = = 0.377 RTc 0.287 × 823.3

kg/m3

1

1

C5 = ( γ RTc ) 2 = (1.333× 0.287 × 823.3× 1000 ) 2 = 561.22m/sec A 5=

m

ρ 5C

= 5

15 = 0.0708m2 0.377 × 561.22

F = m( Cj − Ca ) + Aj ( Pj − Pa ) F =15(561.22 − 260) + 0.0708(0.891 − 0.411)10 5 F = 4518.3 + 3398.4 = 7916.7N T02 = 547.02K T03 − T02 = 1200 − 547.02 = 652.8K Therefore theoretical f = 0.01785 (Fig. 2.15) Actual f =

S .F .C =

0.01785 = 0.0184 0.97

0.0184× 3600× 15 = 0.01255 kg/kN 7916.7

15 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 3.2

From problem 3.1 P04=1.710 bar

P05 = 1.17(1− 0.03) = 1.658bar P05 = 1.918 Pc

as before

P05 1.658 = = 0.8644bar Pc 1.918 Since T05 = 2000K,

ρ5 =

T6 = Tc =

2 × 2000 = 1714.5K 2.333

100 × 0.8644 = 0.1756 0.287 × 1714.5

kg/m 3 1

C 5 = (1.333× 0.287× 1714.5× 1000)2 = 809.88 A5 =

15 = 0.1054 0.1756× 809.88

m/s

m2

Percentage increase in area = 0.1054-0.0708=48.97% F=15(809.88–260)+0.1054(0.8644–0.411)×10 5 F=8248.2+4778.83=13027.03 N Percentage increase in thrust =

13027.03 − 7916.7 = 64.56% 7916.7

16 © 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 3.3

Since P01 = Pa = 1 bar T01 = Ta = 288 K T02 − T01 =

1  288  3.5 − 1 = 306.77K 9  0.82  

0.85mc pg (T03 − T04 ) = T03 − T04 =

mC pa

ηm

(T02 −T01 )

1.005 ×306.77 = 322.678K 0.85 ×1.147 × 0.98

γ −1   γ   1  T03 − T04 =η jT03 1 −    P03 / P04     1   4   1   322.678 = 0.87 ×1275 1 −    P03 / P04    

⇒

P03 = 3.955 P04

P03 = 9.0 − 0.45 = 8.55bar P04 = 2.161bar P04 . = 2.161 Pa

and

P04 1 = 4 Pa  1  0.333  − 1  0.95  2.333    

So nozzle is choking even while landing 17 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

2.161 = 1.126bar 1.918 T04 = 1275− 322.678= 952.3K P5 = Pc =

 2  2 × 952.322 = 816.4K T5 = Tc =  × T04 = 2.333  γ +1  P 100 ×1.126 ρ5 = c = = 0.480 kg/m 3 × RTc 0.287 816.4 1

1

C 5 = (γ RT c )2 = (1.333× 0.287× 816.4× 1000 )2 = 555.86m/s ( m − mb ) = ρ 5A 5C 5 = 0.480 × 0.13 × 558.86 = 34.87 kg/sec mintake = 34.87 = 41.02 0.85 (N.B. – mb, bleed at 90º produces no thrust. Also, ram drag based on intake flow) F = (34.87 ×558.86 − 41.02 × 55) + 0.13(1.126 − 1) × 105 F = (19487.44-2256.1)+1638 F =18869.34 N

or

F=18.87 kN

18 © 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 3.4

Isentropic expansion in nozzle: P04 2.4 = = 2.376 Pa 1.01 γ

4

P04  γ + 1  γ − 1  2.333  = =  = 1.851 Pc  2   2  2.4 = 1.296 bar Choking ; So P5 = Pc = 1.851 T04 γ + 1 2.333 1000 × 2 = = = 857.26 K and T5 = Tc = Tc 2 2 2.333

ρ5 =

Pc 100 × 1.296 = = 0.526 RTc 0.287× 857.26

kg/m3

1

1

C5= ( γ RTc ) 2 = ( 1.333 × 0.287 × 857.26 ×1000) 2 = 572.68m/sec A 5=

m 23 = = 0.0763m 2 C 0.526 572.68 × ρ5 5

F = 23 ×572.68 + 0.0763(1.296 −1.01)10 5 F =13171.64 + 2182.18 =15353.82N or F =15.35kN

19 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

P07 = 1.75 Pa

and

P07 = 1.01 × 1.75 = 1.768bar

1   1 Ta  P07 3.5  = 288 ( 1.75) 3.5 − 1 = 56.74K − T07 − Ta = 1      0.88  ηc  Pa     mhC pg (T04 − T05 ) = m cC pa (T07 − T a )

mc = 2.0 mh 2 ×1.005 × 56.74 = 99.43K T04 − T05 = 1.147 T05 =1000 − 99.43 = 900.57 K But

1   4   1   P04 = 1.597 99.43 = 0.90 × 1000 1 −   ⇒  P P P05   04 05   2.4 = 1.503bar P05 = 1.597 P05 1.503 = = 1.488 Pa 1.01

Hot nozzle is now unchoked, so P6=Pa=1.01 bar γ −1

1 T05  P05  γ =  = 1.4884 = 1.1045 T6  P6  900.57 = 815.36K T6 = 1.1045 T05 − T6 = 85.2K 1

1

C6 = 2C p (T05 − T06 )  2 = (2 × 1.147 × 85.21× 1000) 2 C6 = 442.12m/s Hot nozzle thrust: Fh = 23 ×442.12 =10,168.8 N The cold nozzle pressure ratio:=1.75, while the critical pressure ratio is : P07  1.4 + 1  =  Pc  2 

3.5

= 1.893

20 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

So this nozzle is also unchoked. Since the expansion is isentropic: 1 T07 = 1.753.5 = 1.1733 T8

288 + 56.74 = 293.8 K 1.1733 T07 − T8 = 50.94K T8 =

and

T07 = 344.74

1

1

C8 = 2C p (T07 − T8 ) 2 = ( 2 ×1.005 × 50.94 × 1000) 2 = 320m/s Cold thrust Fc = 46 × 320 = 14720 N Total thrust: =14720 + 10168.8 = 24888N or F=24.89 kN

21 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual

Problem 3.5 Fr...