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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.

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th 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   – 1 T02 – Ta =   ηc  Pa     1  288      3.5  – 1 = 345.598K 11 = ( )   0.82  

Compressor and turbine work required per unit mass flow is:

Wtc =

C pa (T02 − Ta )

T03 – T04 =

ηm

= C pg (T03 − T04 )

1.005× 345.598 = 308.992 K 0.98 ×1.147

T04 = 1150 – 309 = 841K  γ −1      1  γ    T03 − T04 = ηt T03 1 −    P P   03 04   1     4   1   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    = 148.254 K 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: WN = 1.147 × 0.98 × 148.254 = 166.64 kWs/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

1

T02 − Ta =

288 [3.8 3.5 − 1] = 157.3K 0.85

P03 = 3.8 − 0.12 = 3.68bar P03 = 3.68 Pa T03 − T04 = 1050 × 0.88[1 − (

1 14 ) ] = 256.87K 3.68

Net work output

W = ηm ( load ) [( m − mc )C pg (T03 − T04 ) −

mC pa (T02 − Ta )

ηm ( 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 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

EGT ( D C )

8 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th 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 ) P06 T06 1.02 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

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

Problem 2.6

∆T12 =

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

∆T34 =

300  0.286 ( 7.5) − 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 8.760.25 

∴ 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

EGT =1117.7-416.3 =701.4 K=428.4 D 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

  P 1  ⇒ 4 = 2.716, P5 = 2.991 bar 247.1=0.87 ×1285 1 − 0.25 P5  ( P 4 / P5 )  ⇒

P6=1.013 × 1.02 = 1.0333

P5 = 2.895 P6

1   ∴ ∆T56 = 0.88 × 1037.8 1 − = 213.1 ∴ T6=824.7 K 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 = T01[(

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

n −1 γ − 1 0.88 × 0.66 )= = η ∞t ( = 0.3498 n 1.66 γ P02 nn−1 ) − 1] = 310[20.4518 − 1] = 114K P01

T04 − T03 = 300[20.4518 − 1] = 110.3K T04 = 410.3K T06 − T05 = ∴

T05 = 700K

and

( given)

Q 500 × 103 = = 535.2 K 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) = 54.02 bar 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 − ∆T012 ] = 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

n −1 1 = = 0.3284 n 0.87 × 3.5

For compression:

For expansion:

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 2602 = = 33.63K 2c p 2 × 1.005 × 1000 T01 = 242.7 + 33.63 = 276.33K 2

P01 = Pa [1 + ηi

γ

Ca γ −1 ] 2c pTa

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 pgηm

T04 = 1200 − 239.6 n

=

T04 = 960.4K

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

1.005 × 270.7 = 239.6K 1.147 × 0.99 P03 = 5.068(1 − 0.06) = 4.763bar

4.5977

= 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 = =1.918 4 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 T04 = = 823.3K 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 = 561.22 m/sec A5=

m 15 = = 0.0708m 2 ρ5C 5 0.377 × 561.22

F = m(C j − Ca ) + Aj ( Pj − Pa ) F = 15(561.22 − 260) + 0.0708(0.891 − 0.411)105 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 = = 0.8644 bar 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/m3 1

C5 = (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) ×105 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

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th 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    1 4   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 T5 = Tc =  = 816.4K  × T04 = 2.333  γ +1 P 100 × 1.126 ρ5 = c = = 0.480 kg/m3 RTc 0.287 × 816.4 1

1

C5 = ( γ RTc ) 2 = (1.333 × 0.287 × 816.4 × 1000 ) 2 = 555.86 m/s (m − mb ) = ρ5 A5C5 = 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

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th 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 Choking ; So P5 = Pc = = 1.296 bar 1.851 T04 γ + 1 2.333 1000 × 2 and T5 = Tc = = = = 857.26 K 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 A5 =

m 23 = = 0.0763m 2 ρ5C5 0.526 × 572.68

F = 23 × 572.68 + 0.0763(1.296 − 1.01)105 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   3.5 − 1 = 56.74 K −1 = T07 − Ta = 1.75 ( )       ηc  Pa  0.88     mh C pg (T04 − T05 ) = mc C pa (T07 − Ta )

But

mc = 2.0 mh

2 × 1.005 × 56.74 = 99.43K 1.147 T05 = 1000 − 99.43 = 900.57 K T04 − T05 =

1   4   1   ⇒ P04 = 1.597 99.43 = 0.90 × 1000 1 −    P04 P05   P05   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 4 =  = 1.488 = 1.1045 T6  P6  900.57 T6 = = 815.36 K 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 = 1.75 3.5 = 1.1733 T8

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

and

T07 = 344.74

1

1

C8 =  2C p (T07 − T8 )  2 = ( 2 × 1.005 × 50.94 × 1000 ) 2 = 320 m/s Cold thrust Fc = 46 × 320 = 14720 N Total thrust: =14720 + 10168.8 = 24888 N 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 From solution of turbofan example in Chapter 3: T03=800 K T04-T03=1550 – 800=750 K Actual f =

0.0221 = 0.0223 0.99

m = 35.83 kg/s and Fs=71,463 N S.F.C =

0.0223 × 35.83 × 3600 = 0.0403 kg/Nh 71463

P09=P02 – 0.05=1.65-0.05=1.60 bar  P  P So nozzle is unchoked and 09 = 1.60  < 05  Pa  Pc  3.5     T09 – T8 = njT09 1 −  1     P09 Pa  

  1 3.5  = 0.95 × 1000 1 −    =119.4 K   1.6   1

C8= ( 2 × 1.005 × 119.4 × 1000 ) 2 = 490 m/s mc=179.2 kg/s New thrust: Fc=179.2 ×490 = 87,808 N Total thrust = Unchanged Fh+new Fc = 18931+87808=106,739 N

T02=337.7 and T09 – T02=1000-337.7=662.3 K

22 © Pearson Education Limited 2009

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

From Fig. 2.15, theoretical f = 0.01715 Actual f =

S.F.C=

0.01715 = 0.01768 for by-pass chamber 0.97

TotalFuel ( (0.01768 × 179.2 + 0.0223 × 35.83) × 3600 = 106,739 Totalthrust

S.F.C=0.1338 kg/kN

23 © Pearson Education Limited 2009

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

Problem 3.6

Pa=0.899bar Ta=281.7K Ca=336.4 m/s

M=0.70

1  0.4   n −1  n −1  0.333    =   = 0.3210   = 0.90   = 0.2248  n c 0.89  1.4   n t  1.333   γ −1 2  T01= Ta 1 + M  = 281.7(1 + 0.2 × 0.7 2 ) = 309.3K ∆Tr = 27.61D C 2   1.4

P01  0.95 × 27.61  0.4 = 1 +  = 1.366 281.7  Pa 

∴ P01=1.228 bar

T02 0.3210 = ( 5.0 ) = 1.676 ∴ T02=472.2 T01

∆T12 = 162.9D C

∆T34 =

162.9 × 1.005 = 144.1 T04 = 1250 − 144.1 = 1105.9 K 1.148 × 0.99

T03  P03  =  T04  P04 

0.2248

1

P  1250  0.2248 ∴ 03 =  = 1.724  P04  1105.9 

P03=1.366 × 5.0 × 0.945 = 6.454 bar ∴ P04 = γ

 γ + 1  γ −1  1.333  Critical P.R =    =  2   2 

6.454 = 3.744 bar 1.724

4.003

=1.852

P04 3.744 = = 4.164 ∴ choked Pa 0.899

P4=

3.744 2 = 2.022 bar T4=1105.9 × =948.0 K 1.852 2.333

24 © Pearson Education Limited 2009

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

ρ=

100 × 2.022 = 0.7432 kg/m3 0.287 × 948.0

C = γ RT 4 = 1.333 × 0.287 × 1000 × 948.0 = 602.2 m/s

A=

π 4

( 0.15)

2

= 0.01767

m= ρ AC = 0.7432 × 0.01767 × 602.2 = 7.91 kg/s Ca= 0.70 × 336.4 = 235.5 m/s F= m ( C j − Ca ) + ( Pn − Pa ) A × 105 =7.91 ( 602.2 − 235.5 ) + ( 2.022 − 0.899 ) × 0.01767 × 105 =2900 N+1984 N=4884 N ∆Tcc = 1250 − 472 = 778, Tin=472 K → ( f / a )id = 0.0212 ≈ f/a = ∴mf= 7.91 ×0.02141 × 3600 = 609.8 kg/hr ∴SFC=

609.8 = 0.1249 kg/Nh 4884

25 © Pearson Education Limited 2009

0.0212 = 0.02141 0.99

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

Problem 3.7 M=0.75; BPR=3.8; FPR=1.7 Ca=0.75 ×295.1 = 221.3 m/s

 γ −1 2  T01= Ta 1 + M  = 216.7(1 + 0.2 × 0.752 ) = 241.08K 2   ∆Tr = 24.38°C P01  0.95 × 24.38  = 1 +  Pa  216.7  Fan n −1 T02 = (1.7 ) n = 1.1837 T01

3.5

= 1.427 ∴P01=0.2558 bar

n − 1 0.286 = = 0.3178 n 0.90

∴ T02=285.36 ∆T f = 44.28K P02=0.4400 HPC

 T03  0.3178 = 1.9287 T03=550.38 P03=3.4757   = ( 7.9 ) T  02  ∆T023 = 265.02 Combustor P04=3.4757(1 – 0.005)=3.3019 1.004 × 265.02 HP turbine ∆T45 = = 234.32 T05=985.7 K 1.147 × 0.99 LP turbine ∆T56 =

1.004 × 44.28(3.8 + 1) = 187.92 T06=797.8 K 1.147 × 0.99

n

P04  T04  n −1 =  P05  T05 

n −1 = 0.286 × 0.9 = 0.2232 n

n = 4.4803 n −1 P04  1220  =  P05  985.7 

4.4803

= 2.60

26 © Pearson Education Limited 2009

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

P05  985.7  =  P06  797.8 

∴ P06 =

4.4803

= 2.579

3.3019 = 0.4924 bar 2.60 × 2.579

Cold Stream

(both choked)

Hot Stream

P02 = 0.4400 T02=285.4

P06 = 0.4924bar T06=797.8K

P2 =

0.4400 = 0.2324 1.893

P6 =

T2 =

285.4 = 237.8 1.2

 2  T6 = 797.8   = 683.9  2.333 

ρ=

100 × 0.2324 = 0.3405 0.287 × 237.8

ρ=

mc =

9.6× 3.8 = 7.60 4.8

mh =

Cc = 1.4 × 287 × 237.8 = 309.1 m/s A=

7.6 = 0.0722 m2 0.3405 × 309.1

0.4924 = 0.2659 1.852

100 × 0.2659 = 0.1355 0.287 × 683.9 9.6 × 1 = 2.00 4.8

Ch = 1.4 × 287 × 683.9 = 511.5 m/s A=

2.0 = 0.02886 m2 0.1355 × 511.5

Fc = 7.6 ( 309.1 − 221.3) + 0.0722 ( 0.2324 − 0.1793) × 105 = 667.3 + 383.4=1050.7 Fh = 2.0 ( 511.5 − 221.3) + 0.02886 ( 0.2659 − 0.1793) × 105 = 580 + 250.0=830 F = 1880.7N T04 = 1220, T03=550 → f=0.0186 mf = 2.00 × 0.0186 × 3600 = 133.63 kg/hr SFC=

133.63 = 0.0711 kg/Nh 1880.7

27 © Pearson Education Limited 2009

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

Problem 3.8 M=0.85; BPR=5.7; FPR=1.77

 n − 1  0.286 = 0.3178   =  n c 0.90  n − 1  0.333 × 0.9 = 0.2248   = 1.333  n t  γ −1 2  T01= Ta 1 + M  = 216.8(1 + 0.2 × 0.852 ) = 248.13K 2   ∆Tr = 31.331C P01  0.95 × 31.33  = 1 +  Pa  216.8 

3.5

T02 0.31718 = (1.77 ) = 1.1989 T01

= 1.569 ∴P01=0.227 ×1.569 =0.3561 bar P02=0.6303 ∴∆T12 = 49.33K

∴T02=297.5

 P03  34 = 19.21  =  P02  1.77  T03  0.3178 = 2.558 , ∆T23 = 463.4   = (19.21) T  02 

T03=761K

∆T023 = 265.02 P04 = 0.3561 × 1.77 × 19.21 × 0.97 =11.74 bar ∆T45 =

1.004 × 463.4 = 409.4 T05=940.6 K 1.148 × 0.99 1

P04  1350  0.2248 = = 4.99  P05  940.6  ∆T56 =

1.004 × 49.33(5.7 + 1) = 292.0 T06=648.6 K 1.148 × 0.99

T6=556.0 K 28 © Pearson Education Limited 2009

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual 1

P05  940.6  0.2248 = = 5.22  P06  648.6  ∴ P06 =

P 11.74 = 0.4505 bar 06 = 1.984 ∴choked 4.99 × 5.22 Pa

Ca = 0.85 ×295.2 Cold Stream

(both choked)

Hot Stream

mc =

220 × 5.7 = 187.2 6.7

mh =

220 = 32.8 6.7

P2 =

P02 0.6303 = = 0.3330 Pcr 1.893

P6 =

0.4505 = 0.2433 1.852

T2 =

297.5 = 247.9 1.2

ρ=

100 × 0.3330 = 0.468 0.287 × 247.9

ρ=

100 × 0.2433 = 0.1525 0.287 × 556

Cc = 1.4 × 287 × 247.9 = 315.6 m/s A=

187.2 = 1.267 m2 0.468 × 315.6

Ch = 1.333 × 287 × 556 = 461.2 m/s A=

32.8 = 0.466 m2 0.1525 × 461.2

Fc = 187.2 ( 315.6 − 250.9 ) + 1.267 ( 0.333 − 0.227 ) × 105 =12.108+13430=25,538 Fh = 32.8 ( 461.2 − 250.9 ) + 0.466 ( 0.2433 − 0.227 ) × 105 = 6898+760=7658 F = 33,196N (Fs=150.9) f=

0.017 =0.0172 0.99

mf = 0.0172 × 32.8 × 3600 = 2028 kg/hr SFC =

2028 = 0.0611 kg/Nh 33196

29 © Pearson Education Limited 2009

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

Problem 3.9 M 1.4 Turbofan (BPR=2.0, FPR=2.2)

Assume full expansion, i.e. no pressure thrust 0.4  n −1 = 0.3175   =  n c 0.90 × 1.4

 n − 1  0.333 × 0.9 = 0.2248   = 1.333  n t  γ −1 2  T01= Ta 1 + M  = 216.7(1 + 0.2 × 1.42 ) = 301.65K 2   ∆Tr = 84.95K P01  0.93 × 84.95  = 1 +  Pa  216.7 

3.5

= 2.968 ∴P01=0.3594 bar

T02 0.3175 = ( 2.2 ) = 1.2845 T01

∴T02=387.46K ∴∆T12 = 85.8K

T03 0.3175 = ( 9.0 ) = 2.009 T02

∴T03=778.39K ∴∆T23 = 390.9 K

Assuming no cooling, 1.004 × 390.9 = 1.147 × 0.99 × ∆T45 ∆T45 = 345.6 T5=1154.3K 1

P04  1500  0.2248 = = 3.206  P05  1154.3  LP system (B+1) ×1.004 × 85.8 = 1.147 × 0.99 × ∆T 56 ∆T56 = 227.6 T6=926.7 K 1

P05  1154.3  0.2248 = = 2.656  P06  926.7  P04= 0.3594 × 2.2 × 9.0 × 0.93 = 6.618 bar P02=0.3594 ×2.2 = 0.7907 bar P06 =

P P 6.618 = 0.7772 bar 06 = 6.418 02 = 6.529 3.206 × 2.656 Pa Pa

30 © Pearson Education Limited 2009

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

With full expansion to Pa T 02 0.286 = ( 6.529 ) = 1.710 T2=226.6K ; Cc2 = 2 × 1.004 × 103 [387.4 − 226.6] = 568 m/s T2 T 06 0.25 = ( 6.418 ) = 1.582 T6=582.2K ; T6 Ch2 = 2 × 1.147 × 103 [926.7 − 582.2] = 888.9 m/s mc = 70 ×

2 = 46.67 ; mcCc=46.67 ×568 = 26508 N 2 +1

mh = 70 ×

1 = 23.33 ; mhCh=23.33 ×888.9 = 20738 N 2 +1 47246N (gross)

Ca=1.4 × 295.1 = 413.1 ram drag = 70 × 413.1 = 28920N

[ SpecificThrust

= 261.8 N / kg / hr ]

T04-T03=1500-778-722 ∴ fideal = 0.023 f =

18326 N (nett) 0.023 = 0.0232 0.99

∴ mf =23.33 × 0.0232 × 3600 = 1951 kg/hr SFC=

1951 = 0.1065 kg/Nh 18326

Ca=413 m/s = (413/1000) ×3600 km/h=1486 km/h ∴Flight time =

5600 = 3.766 hrs 1486.8

∴Fuel flow =3.766 ×2 × 1951 =14,697kg

31 © Pearson Education Limited 2009

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

Problem 3.10 Pa=1.013 bar ; Ta=303K

∆T12 =

303 150.286 − 1 = 437.5 0.81 

∴T2 = 740.5 K ∆Tcc = 1600 − 740.5 = 859.5D C ∆T34 =

437.5 × 1.005 1   = 398.8 ∴ 398.8 = 0.87 × 1600 1 − 0.25  1.148 × 0.99 × 0.97 R  

∴Rhp = 3.86 P3=1.013 ×15.0 × 0.95 = 14.43 bar P4=

14.43 = 3.74 bar 3.86

P4 3.74 = = 3.692 P5 1.013 T4=1600 – 398.8 =1201.2 K 1   ∴∆T45 = 1201.2 × 0.89 1 − = 297.8 K 0.25  3.692   ∴Specific output =1.148 ×297.8 × 0.99 × 0.985 = 33.4 ∆Tcc = 860, Tin=467 D C ∴ f/a=

kJ kg

( f / a )ideal = 0.0254

0.0254 = 0.0257 0.99

Airflow =11.0 kg/s ∴ Power =333.4 ×11.0 = 3667 kW mf= 0.0257 ×11.0 × 3600 = 1018.9 kg/hr ∴ SFC=0.278 kg/kWh 32 © Pearson Education Limited 2009

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

Problem 3.11

Load compressor ∆T =

308 3.650.286 − 1 = 156.9 K 0.88

Power = 3.5 ×1.004 × 156.9 = 551.2 kW For gas turbine ∆T12 =

308 120.286 − 1 = 379.6 0.84 

T02 = 688 K

P03 = (12 × 1.01 × 0.95 ) /1.04 = 11.07 P04 1   ∴ ∆T34 = 1390 × 0.87 1 − = 546 0.25   11.07  m × 1.148 × 546.0 × 0.99 = m × 1.004 × 379.6 + 200 + 551.2 m[620.5 − 381.1]=755.8 ∴m= f/a (690, ∆T = 700) =

755.8 = 3.16 kg/s 239.4

0.02 0.99

∴ mf =

0.02 × 3.16 × 3600 = 229.6 kg/hr 0.99

SFC =

229.6 = 0.304 kg/kwhr 755.8

33 © Pearson Education Limited 2009

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

Problem 4.1 Flow area = π (152 − 6.52 ) = 574 cm2= 0.0574 m2 C=

8.0 139.37 = m/s ρ × 0.0574 ρ 3.5

P T  C2 , 0 = 0  , Ts= 288 − 2010 Ps  Ts 

ρ=

θc =

C2 2c p

Ps 100 × Ps P = = 348.4 s RTs 0.287 × Ts Ts

We must find the axial velocity by iteration to satisfy continuity: Let Cmax=150 m/s C

θ

Ts

T 0 /T s

P 0 /P s

Ps

ρ

Ccalc

150

11.12

276.8

1.0405

1.149

0.870

1.095

127.3

127.3

8.05

279.94

1.0288

1.104

0.905

1.127

123.7

123.7

7.61

280.39

1.0271

1.098

0.910

1.131

123.1

123.1

7.54

280.46

1.0269

1.097

0.911

1.132

123.1

∴

Ca = 123.1 m/s,

Ts=280.46 K

At root, U= π × ( 0.065 × 2 ) × 270 = 110.3 m/s

123.1 = 1.116 110.3 α r = 48D10' tan α r =

34 © Pearson Education Limited 2009

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

At tip, U= π × ( 0.15 × 2 ) × 270 = 254.4 m/s

123.1 = 0.4839 254.4 α t = 25D 48' tan α t =

At tip V2=254.42+123.12=79720.42 ⇒ V=282.3 m/s

a= 1.4 × 0.287 × 280.46 × 103 = 335.7 m/s

∴ M=

282.3 = 0.841 335.7

35 © Pearson Education Limited 2009

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

Problem 4.2 T1=T01 –

C12 2302 =243.3 K ; T01 = 217 + 2Cp 2 × 1.005 × 103

 243.3  P01=0.23    217  A=

π

( 0.33 4

2

3.5

= 0.343 bar

− 0.182 ) = 0.060 m2

m= ρ ACa = ρ AC cos 25 ⇒ ρ C =

3.6 0.060 × 0.906

ρ C = 66.22 But based on stagnation conditions, for a first estimate

ρ=

100 P0 100 × 0.343 × = = 0.491 kg/m3 0.287 T0 0.287 × 243.3

∴

C=

66.22 = 134.87 m/s 0.491

First trial: Cmax=140 m/s C

θ

Ts

T0/Ts

P0/Ps

Ps

ρ

ρC

140

9.7

233.7

1.042

1.155

0.297

0.443

62

150

11.2

232.2

1.049

1.182

0.290

0.435

65.2

160

12.7

230.7

1.056

1.21

0.287

0.434

69.2

152.4

11.5

231.9

1.051

1.19

0.288

0.433

66.0

Take C=152.4 ×

66.22 = 152.90 m/s 66.0

C1w=152.9 × sin 25 = 152.9 × 0.4226 = 64.61 m/s 3.5

P02  ψηc  σ U 2 − Cw ( rω ) mean   = 1 + P01  C pT01 

 18 + 33  1 Ue = (r ω )mean=   × × 270 × 2π = 216.3 m/s  2 × 100  2

36 © Pearson Education Limited 2009

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

U = π × 270 × 0.54 = 458 m/s 3.5

P02  1.04 × 0.8  0.9 × 4582 − 64.61 × 216.3  = 1 + P01  1.005 × 243.3 × 103   P02 3.5 = (1.595 ) = 5.12 P01

∴ P02=5.12 × 0.343=1.756 bar

37 © Pearson Education Limited 2009

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

Problem 4.3 P01 = Pa = 0.99 bar and T01=Ta=288K γ −1   T01  P02  γ  T02-T01 = 429-288=   − 1 η  P01    1   288  2.97  3.5  1 − = 0.753 ∴ ηc =    141  0.99   

0.63π = 0.8835 17 16.5 U2 = π × × 765 = 397.1m / s 100

σ =1−

Hence: C2w = σ U 2 = 0.8835 × 397.1 = 351 m/s 2   Cr22 C2 3512 − =429 –  3  2C p  2 × 1.005 × 10  2C p

T2=T02-

Therefore T2=367.7 – θ r

ρ2 = A2 =

100 × 1.92 669 = T2 0.287 × T2

π × 16.5 × 1.0

Cr 2 =

104

= 51.83 × 10−4 m 2

m 0.60 1 115.76 = × = −4 ρ 2 A2 51.83 × 10 ρ2 ρ2

θ r is small – For the first trial let T2=366K ∴ ρ2 =

669 = 1.83 366

Hence Cr 2 = 63.26

m/ s

Final trial: Cr 2

θ

64.0

2.03

T2 366

ρ

Cr 2

1.827

63.5

∴ T2=366 K P2t  T2t  =  P2  T2 

3.5

 429  =   366 

3.5

= 1.743

38 © Pearson Education Limited 2009

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

P2t=1.92 × 1.743 = 3.347 bar P02  141  = 1 + ηi m p × P01  288 

3.5

3.347  141  = 1 + ηi m p × 0.99  288  ∴ ηimp = 0.850

3.5

Fraction of loss in impeller =

1 − 0.87 = 0.607 1 − 0.753

39 © Pearson Education Limited 2009

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

Problem 4.4 (a) U2= π × 0.50 × 270 =424.12 m/s 1.04 × 0.9 × 424.122 = 167.5K 1.005 × 103 T02 = 455.5K ∆T =

3.5

P02  0.90 × 167.5  3.5 = 1 + = (1.523) = 4.36  P01  288  P02 = 4.36 × 1.01 = 4.40bar

(b) Flow area at tip: =

π × 50 × 5 104

= 0.0785 m2

C2w=0.9 ×424.12 = 381.7 m/s 2

2

 381.7   96  If C2r=96 and θ c =   +  = 76.84  44.9   44.9  (noting

2 × 1.005 × 103 = 44.9 )

T2=455.5-76.84=378.66 K P02  455.5  =  P2  378.66 

ρ2 =

3.5

= 1.909

100 × 2.30 = 2.12 0.287 × 378.66

P2=

4.40 = 2.30 bar 1.909

ρ 2 A2 Cr 2 = 2.12 × 0.0785 × 96 = 15.97

The agreement is close since the required flow is 16.0 kg/sec. ∴ Cr 2 = 96 m/s a = 0.287 × 1.4 × 378.66 × 103 = 390 m/s C 22 = 381.7 2 + 962 ∴ C2 = 393.6 m / s C2 393.6 = = 1.01 390 a 96 = 0.251 tan α = 381.7 M=

∴

α = 14D5

40 © Pearson Education Limited 2009

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

(c) P3t=P2t=4.40 bar A=

π × 58 × 5 10

ρ3C3r =

4

= 0.0911m 2

16.0 = 175.6 0.0911

C3 w =

381.7 × 25 = 329 m/s 29

We must find Cr3 by trial and error; as a first guess assume: Cr3=96 ×

25 = 82.75m/s 29

T03=T02=455.5 K

∴

C3w

θw

C3r

θr

θ

T3

T0/Ts

P0/Ps

Ps

ρ

ρ Cr

329

53.6

82.75

3.4

57.0

398.4

1.143

1.596

2.755

2.41

199.5

73

2.64

56.2

399.2

1.140

1.581

2.78

2.43

177.5

72.4

2.60

56.2

399.1

1.140

1.581

2.78

2.43

176

C3r = 72.4 m/s, T3 = 399.1 K

C32 = 3292 + 72.42

⇒ C3 = 336.87 m/s

a= 1.4 × 0.287 × 399.1 × 103 = 400.44 m/s 336.87 = 0.841 M= 400.44 tan β =

72.4 = 0.22 β = 12D 24' 329

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Problem 4.5

∆T013 =

288 0.286 4 − 1 = 175.1 K 0.80 

∆T013 =

pσ U 22 = 175.1 c p × 103

U 22 =

1.005 × 175.1 × 103 = 1.876 × 105 1.04 × 0.90

∴U 2 = 433 m/s d=

433 = 0.689 m π × 200

T03 = T02 = 175.1+288=463.1 K At tip M=1.0 T2=T02 × 2 = 463.1 = 385.9 K γ + 1 1.2 ∴ ∴

a = 1.4 × 0.287 × 385.9 × 103 C2 = a = 393.7 m/s

C2w = 0.90 ×U 2 = 0.90 × 433 = 389.7 m/s C 22 = C 22r + C 22w ∴C 22r = 393.72 – 389.72=3133.6 ∴C 2r = 55.97 m/s With a 50% loss in impeller, η I = 0.90 ∴∆T '012 ∴

= 0.90 × 175.1 = 157.59 K

P02  157.59  = 1+ P01  288 

3.5

= 4.607bar

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∴

P02 = 1 × 4.607 = 4.607 bar γ

P02  γ + 1  γ −1 3.5 =  = 1.2 = 1.893 P2  2  4.607 = 2.434 bar P2 = 1.893

ρ2 =

100 × 2.434 = 2.198 kg/m3 0.287 × 385.9

m= ρ AC2 r ∴h =

∴A=

14.0 = 0.1138 2.198 × 55.97

m2

0.1138 × 100 = 5.257 cm π × 0.689

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Problem 5.1

At mean radius, m Ω∆Cw = mC p ∆T 1.005 × 20 × 103 = 108.1m/s 0.93 × 200

∴

∆Cw =

C0w=

200 − 108.1 = 45.95m/s 22

tan α 0 tan α1

45.95 = 0.307 α 0 = 17D 4' ( = α 2 ) 150 200 − 45.95 = = 1.027 α1 = 45D 44' ( = α 3 ) 150 =

With free vortex, Cwr=constant i.e. CwU= constant

Tip C0w ×250 = 46.0 × 200 ∴ C0 w

= 36.8 m/s

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

∆Cwu = constant ∴

∆Cw =

108.1 × 200 = 86.4 m/s 250

tan α 0 =

36.8 = 0.245, α 0 = 13D 46' 150

tan α1 =

250 − 36.8 = 1.421, α1 = 54D53' 150

tan α 2 =

36.8 + 86.4 = 0.845, α 2 = 40D14' 150

tan α 3 =

36.8 + 86.4 = 0.821, α 3 = 39D 26' 150

Root C0w ×150 = 46 × 200 ∆Cw =

108 × 200 = 144 150

∴

C0 w = 61.3 m / s m/ s

tan α 0 =

61.3 = 0.409, α 0 = 22D15' 150

tan α1 =

150 − 61.3 = 0.591, α1 = 30D36' 150

tan α 2 =

150 − (144 + 61.3) = −0.369, α 2 = −20D15' 150

tan α 3 =

144 + 61.3 = 1.369, α 3 = 53D51' 150

Λ=

static enthalpy rise in rotor Stagnation enthalpy rise in stage

Now (T0+

=

T1 − T0 20

 C2 C02 C2 C2 + 20) =T1+ 1 ∴ T1 − T0 = 20 +  0 − 1  2C 2C p 2C p  p 2C p 45 © Pearson Education Limited 2009

 1  3  10

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

Tip C0=

150 cos13D 46'

=

150 = 154.5 T1-T0=20+11.19-18.8=13.1 0.9712

C1=

150 cos39D 26'

=

150 = 194.1 0.7724

=

150 = 162.1 0.9255

∴Λ =

13.1 = 65.5% 20

Root C0=

150 cos 22D15'

T1-T0=20+

C1=

∴

1  162.12 254.22  −   103  2 × 1.005 2 × 1.005 

150 cos53D51'

=

T1 − T0 = 0.92,

150 = 254.2 0.5899 Λ=

T1-T0=20+13.07-32.15

0.92 = 4.6% 20

(N.B.: Λ is very sensitive to small errors in C0 and C1)

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Problem 5.2 Velocity diagram at mid span is as in previous question. ∆Cw : mean = 108, tip =86.3, root =144 m/s Equal work at all radius

Tip C0w=

250 − 86.3 = 81.85 m/s 2

tan α 0 =

81.85 = 0.546, α 0 = 28D36' ( = α 2 ) 150

tan α1 =

250 − 81.85 = 1.121, α1 = 48D16' ( = α 3 ) 150

47 © Pearson Education Limited 2009

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

Root C0w =

150 − 144 = 3 m/s 2

tan α 0 =

3 = 0.0020, α 0 = α 2 = 0°7 ' 150

tan α1 =

150 − 3 = 0.980, α1 = α 3 = 44°25' 150

48 © Pearson Education Limited 2009

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

Problem 5.3 (a) With no inlet guide vanes the inlet velocity is axial.

T = 288 − a=

140 × 140 1 × = 278.25 K 2 × 1.005 103

γ RT = 1.4 × 0.287 × 278.25 × 103 = 334.4 m/s

∴ V= 0.95 × 334.4 = 317.7 m/s V2=U2+C2 ∴ U2 = 317.7 2 − 1402 = 81333.3

∴ U=285.2 m/s

π × D × N = 285.2 ∴ D=

285.2

= 0.908 m

π × 100

∴Tip radius =0.454 m or 45.40 cm (b)

P0  T0  =  P T 

3.5

 288  =   278.25 

3.5

= 1.128

P=

1.01 = 0.895 bar 1.128

ρ=

0.895 × 100 = 1.121 kg/m3 0.287 × 278.25

Dhub = 0.60× 0.908 = 0.545 m A=

π 4

(0.9082 – 0.5452 ) = 0.4143m2

∴ m = 1.121 ×0.4143 × 140 =65.02 or m=65 kg/s 3.5

P  0.89 × 20  = 1.2335 (c) 02 = 1 + 288  P01 

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(d) At the root we have axial inlet velocity. Free vortex gives constant work at all radii and constant axial velocity.

Uh=0.60 ×285.2 = 171.1 m/s

1 = mC p ∆T 103 C p ∆T × 103 1.005 × 20 × 103 ∆Cw = = = 126.3m/s 171.1 × 0.93 uΩ

mU ∆Cw Ω

tan α1 =

171.1 = 1.2221, α1 = 50D 44' 140

tan α 2 =

171.1 − 126.3 = 0.320, α 2 = 17D 46' 140

At tip ∆Cw = 126.3 × 0.60 = 75.78m/s tan α1 =

285.2 = 2.0371, α1 = 63D50' 140

tan α 2 =

285.2 − 75.78 = 1.4958, α 2 = 56D14' 140

Power input =65 ×1.005 × 20 = 1306.5 kW

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

Problem 5.4 n − 1 0.286 = = 0.325 n 0.88 T02 0.325 = ( 4.0 ) = 1.569 T01 ∆T012 = 288[1.569 − 1] = 163.9K ∴ Number of stages = #

163.9 = 6.556 i.e. 7 stages 25

∆T 163.9 = = 23.4 K/stage 7 of stages

For first stage:

Tout 288 + 23.4 = = 1.0812 288 Tin

1.0812 = (R) 0.325 ∴ Rfirst=1.271 For last stage: Tout=288+163.9=451.9 K Tin = 451.9-23.4=428.5 K; ∴

Tout = 1.0546 Tin

1.0546= (R)0.325 ∴Rlast=1.178

mU ∆Cw Ω

1 = mC p ∆T ; (note that Ca=165 cos20) 103

∆Cw = U − 2C sin 20 = U − 330sin 20 ΩU (U − 112.9)

1 = C p ∆T 103

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U2 – 112.9U=

1.005 × 23.4 × 103 0.83

U2 – 112.9U-28.3 ×103 = 0 ∴U =

112.9 ± 112.92 + 4 × 28.3 × 103 112.9 ± 354.9 = 2 2

∴ U = 233.9 m/s π DN 60

= 233.9 ∴ N=

60 × 233.9 = 24817.5 rpm π × 0.18

At the entry to last rotor: T0=428.5; P0=

4 × 1.01 = 3.43 bar 1.178

C=165 m/s ∴ T=428.5 P0  T0  =  P T  P=

3.5

 428.5  =   414.96 

1652 = 414.96 K 2.01 × 103 3.5

= 1.1189

3.43 = 3.07 bar 1.1189

∴ ρ=

100 × 3.07 = 2.58 kg/m3 0.287 × 414.96

(π Dh ) ρ × Ca = m π × 0.18 × h × 2.58 × 165cos 20 = 3.0 ∴h =

3.0

π × 0.18 × 2.58 × 165 × 0.9397

= 0.01326 m

h = 1.326 cm

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Problem 5.5 Axial:

n − 1 0.286 n − 1 0.286 = = 0.311 Centrifugal: = = 0.3446 n 0.92 n 0.83

∴ T2=288+120=408 K

For axial ∆T = 4 × 30 = 120 K

P  T02 408 = = 1.417 =  02  T01 288  P01 

∴

0.311

∴

P02 = 3.07 P01

P03 10.0 = = 3.257 P02 3.07

Axial : mU ∆Cw Ω

1 = mC p ∆T 103

∆Cw = U − 2Ca tan 20 = U-300 ×0.364 = U − 109.2 103 × 1.005 × 30 =35.05 ×103 0.86

U(U-109.2)=

U2-109.2U-35.05 ×103 =0 109.22 + 4 × 35.05 × 103 109.2 ± 390 = 2 2

∴ U= 109.2 ± U=249.6 m/s 249.6=

π DN 60

∴ N=

60 × 249.6 = 19068 rpm π × 0.25

Centrifugal: T03  P03  =  T02  P02 

0.3446

= ( 3.257 )

0.3446

= 1.502

∆T023 = 0.502 × 408 = 204.82 K ∆T =

σψ U 22 Cp

=

0.90 × 1.04 × U 22 1.005 × 103

∴

U 22 =

1.005 × 103 × 204.82 0.90 × 1.04

U2=468.95 m/s

∴ N(Centrifugal) =

60 × 468.95 = 27140 rpm π × 0.33

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Problem 6.1 (a) From Fig. 2.15, for T1=298 K and ∆T = 583 K, theoretical f=0.0147

ηb =

0.0147 = 0.980 0.0150

(b) From Fig. 2.15, for f=0.0150 and T1=298 K, theoretical ∆T = 593 K

ηb =

583 = 0.983 593

(c) mass of CO per kg of fuel: m=

28 × 0.04 × 0.8608 = 0.0803 kg 12

Actual released energy = 43100–(0.0803 ×10110) Efficiency based on released energy: 43100 − (0.0803 × 10110) = 1 − 0.0188 43100 η = 0.9812

η=

54 © Pearson Education Limited 2009

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Problem 6.2 The combustion equation: C10H12 +65O2+244.5N2 → 10CO2+6H2O+52O2+244.5N2 132kg 2080

6846

440

108

1664

Fuel/ air ratio =

132 = 0.0148 ( 2080 + 6846 )

6846

Since the H2O will be in vapour phase, we require ∆H vap , given by: ∆H vap = −42500 +

108 × 2442 = −40500 KJ/kg of C10H12 132

Energy equation is:

(

)

(Hp2-Hp0)+ ∆H 0 + H R0 − H R1 =0 Where the first stage is 325 K for C10H12 and 450 K for air Mean temperature of reactants O2 and N2 is

450 + 298 = 374 K 2

From fig. 2.15 for f=0.0148 and for an initial temperature of 450 K For air, temperature rise ∆T = 565 K. Hence approximate final temperature T2=565+450 =1015 K First guess at mean temperature of products = (1015+298)/2=656.5 K. Mean values of Cp from tables are : R- reactants, P- products.

O2 N2

R’s 0.934 1.042

P’s CO2 H2O O2 N2

1st Guess 1.105 2.051 1.019 1.088

2nd Guess 1.047 2.041 1.014 1.084

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mCp 460.7 220.4 1687.2 7421.1 9789.4

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

(T2 –298) {( 440 × 1.105 ) + (108 × 2.051) + (1664 × 1.019 ) + (6846 × 1.088)} – (132 × 40500 ) − ( 450 − 298 ) ( 2080 × 0.934 ) + ( 6846 × 1.042 )  – ( 325 − 198 ) [132 × 1.945] = (T2 − 298 ) × 985.5 − 5,35,000 − 152 × 9073 − 27 × 257 = 0

∴ T2=298+

6736000 = 982 K 9855

Second guess at mean temperature of products=1280/2=640 K

∴ T2=298+

6736000 = 986.1 K 9789.4

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Problem 6.3 Under design conditions: T01=T1+

475=

475=

C12 P m2 = 1 + 2 2 2C p R ρ 1 2 ρ1 A1 C p

4.47 × 100 9.02 + 2 0.287 ρ 1 2 ρ1 × 0.03892 × 1.005 × 103 1557

ρ1

+

26.70

ρ12

475 ρ12 − 1557 ρ1 − 26.70 = 0 From which : ρ1 =

1557 ± 1557 2 − 4 × 475 × (−26.76) 2 × 475

ρ1 = 3.294 kg/m3 Hence:

0.27 × 105  1023  =19.0+ K2  − 1 2 2 9.0 / 2 × 3.294 × 0.0975  475  20.88=19.0+1.153 K2

∴ K2=1.630 Part-load conditions: 439=

3.52 × 100 7.42 1226.5 18.004 + 2 = + 2 3 ρ1 ρ12 0.287 ρ1 2 ρ1 × 0.0389 × 1.005 × 10

439 ρ12 − 1226.5ρ1 − 18 = 0

∴ ρ1 = 2.808 kg/m3 105 ∆P0 =

7.42 2 × 2.808 × 0.09752

  900   19.0 + 1.630  439 − 1    

∆P0 = 0.213 bar At design : C1=

m 9.0 = =70.24 m/s ρ1 A1 3.294 × 0.0389

7.4 =67.75 m/s 2.808 × 0.0389 P 4.47 × 100 At design: T1= 1 = = 472.8 K R ρ1 0.287 × 3.294

At part load: C1=

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T  P01=P1  01   T1 

γ / γ −1

 475  = 4.47    472.8 

3.5

= 4.543 bar

∆P0 0.27 = =0.0594 P01 4.543 At part load: T1=

3.52 × 100 = 436.78 K 0.287 × 2.808

 439  P01=3.52    436.78 

3.5

= 3.583bar

∆P0 0.213 = = 0.0595 P01 3.583

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Problem 7.1

C2=

Ca 260 = = 615.2 m/s cos 65 0.4226

Cw2= 615.22 − 2602 = 557.5 m/s Tan β 2 =

557.5 − 360 = 0.7596 ∴ β 2 = 37D13' 260

Cw3=260 tan 10=260 ×0.1763 = 45.84 m/s Tan β 3 =

360 + 45.84 =1.5609 ∴ β 3 = 57D 22' 260

Since C1=C3 and Ca is constant: Λ=

Ca 260 ( tan β3 − tan β 2 ) = (1.5609 − 0.7596 ) 2U 2 × 360

Λ = 0.289 Temperature drop coefficient:

ψ=

2C p ∆T0 s U

2

=

2U ∆Cw 2 ( 557.5 + 45.84 ) = =3.35 U2 360

Power output =mU ∆Cw = 20 × 360 × (557.5 + 45.84 ) P=4343760W or 4344 KW T2=T01-

C22 615.22 = 835 K =1000 – 2294 2C p

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T2 – T2' = λN

C22 615.22 = 0.05 × = 8.25 K 2C p 2294

T '2 = 835 − 8.25 = 826.75 K

P01  T01  =  P2  T2'  ∴

P2 =

γ / γ −1

4

 1000  =  = 2.14  826.75 

4 = 1.869 bar 2.140

For isentropic flow, critical pressure ratio Pc/Pa=1.853.

∴

P01 > 1.853. ∴ The nozzle is choking P2

The throat conditions are: Pc=

 2  2 4.0 = 2.158 bar, Tc=  × 1000 = 857 K  × T01 = 1.853 2.333  γ +1

ρc =

Pc 2.158 × 100 = = 0.877 kg/m3 RTc 0.287 × 857

Cc= 1.333 × 0.287 × 857 × 103 = 572.6 m/s Athroat=

m

ρc Cc

=

20 = 0.0398 m2 0.877 × 572.6

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Problem 7.2 ∆T013

γ −1   γ   1   = ηt T01 1 −    P01 P03     1   4 1    =0.88 ×1050 1 −    = 147 K  2   

T03 = 1050 – 147=903 K With zero outlet swirl: Ws=UCw2=Cp ∆T013 And with free vortex, Cwr = constant Hence at root (Cw2)r =

C p ∆T013 U

=

1.147 × 147 × 103 = 562 m/s 300

Now at root, Λ = 0 ∴

T2 − T3 = 0 hence T2=T3 T1 − T3

∴ T02-

C2 C22r =T03 – 3 ; T02=T01 and C3=Ca3=275 m/s 2C p 2C p

Hence

C22r 2752 + 147 = 180 K = 2C p 2294

∴ C2r=642.6 m/s

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α 2 r = sin −1

562 = sin −1 0.8746 = 61D 642.6

(Ca2)r = 642.62 − 5622 = 311.6 m/s tan β 2 r =

( Cw 2 ) r − U r ( Ca 2 ) r

=

562 − 300 = 0.8406 311.6

β 2 r = 40D 4' (Ca2)t=(Ca2)r=311.6 m/s and (Cw2)t=

rt = 1.4 rr

562 =401.4 m/s 1.4  401.4  −1  = tan (1.2882 ) 311.6  

α 2t = tan −1 

∴ α 2t = 52°10' C2t= 401.42 + 311.62 = 508.15 m/s With no exit swirl T3t=T3r=903 –

T2t=1050 –

2752 = 870 K 2294

508.152 = 937.5 K 2294

(T1-T3)= ∆T013 because C1=C3 with no inlet or exit swirl Λ tip =

T2t − T3t 937.5 − 870 = =0.46 147 T1 − T3

At the root T2r-T’2r= λn

C22r 642.62 =0.05 × 2294 2Cp

T2r – T’2r = 9 K T’2r=(1050 – 180) – 9.0 =861 K P01  T01  =  P2  T'2 

γ / γ −1

P03  T03  =  P3  T'3 

γ / γ −1

4

3.8  1050  = 1.719 bar =  = 2.21 ∴ P2 = 2.21 861   4

3.8 / 2  903  = 1.638 bar =  = 1.16 ∴ P3 = 1.16  870 

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Problem 7.3 T3=T03 –

C22 4002 = (1200 − 150) − = 980.2K 2C p 2294

Um =2 π rm N or rm= ∆T013

320 = 0.2037 m 2π × 250

1   4   1  = ηt T01 1 −    P01 / P03     1

 1 4 ∆T013 150 =1− = 0.8611   =1− 0.90 × 1200 ηt T01  P01 / P03  P01 8 = 4.4bar = 1.818 ∴ P03= 1.818 P03 P03  T03  =  P3  T3 

ρ3 = A3= h=

γ / γ −1

4

4.4  1050  = 3.343 bar =  = 1.316 ∴ P3= 1.316  980.25 

P3 3.343 × 100 =1.19 kg/m3 = RT3 0.287 × 980.25

m 36 = =0.0756 m2 (N.B. C3=Ca3) ρ3C3 1.19 × 400

A 0.0756 = =0.0591 m 2π rm 2π × 0.2037

rt 0.2037 + (0.0591/ 2) 0.233 = = = 1.34 rr 0.2037 − (0.0591/ 2) 0.1741 A2 = A3 Ca2 must satisfy Ca2=

m and Ca2= C 2 − Cw2 2 ρ 2 A2

Ws=Cw2U=Cp ∆T013 (because α 3 = 0 ) Cw2=

1.147 × 150 × 103 = 537.6 m/s 320

If Ca2=346 m/s C22 = 3462 + 537.62 ⇒ C2=639 m/s C22 C2 6392 = 1022 K =T01- 2 =12002294 2C p 2C p

T2=T02-

T2’=T2 – λN P2=

C22 =1022-0.07 ×178 = 1009.5 K 2C p

P01

(T01

T2 ' )

4

=

8.0

(1200 /1009.5 )

4

=

8.0 = 4.008 bar 1.995

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ρ2 =

4.008 × 100 = 1.366 kg/m3 0.287 × 1022

Ca2=

m 36 = = 348 m/s (agrees with given 346 m/s) C2 A3 1.366 × 0.0756

At the root, for free vortex design, we have: rm 0.2037 = = 1.17 rt 0.1741

(C w2 )r =(C w2 )m ×

rm = 537.6 × 1.17 =629 m/s rr

Ca2 is constant at 346 m/s. Hence: C2r= 6292 + 3462 = 717.9 m/s T2r=1200 –

UR=

7182 =975.3 K 2294

320 = 273.5 m/s 1.17

V2r= 3462 + (629 − 273.5) 2 = 496.1 m/s a2r= γ RT2 r = 1.333 × 0.287 × 975.3 × 103 =610.86 m/s (Mv2)r=

496.1 =0.812 610.86

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Problem 7.4

Cr2=constant

(1)

Cw2r=constant

(2)

From (2), since Ca2=Cw2 cot α 2

 rm    r 2

Ca2r=constant and Ca2=(Ca2)m   r  Also U=Um    rm  2 Now,

U = tan α 2 − tan β 2 Ca 2

Um  r  ∴tan β 2 =tan α 2 ( Ca 2 )m  rm 

2

tan α 2 =(tan α 2 ) m = tan 58D 23' =1.624 Um = tan 58 D 23’-tan 20 D 29' =1.624-0.374=1.25 ( Ca 2 ) m (N.B. Flow coefficient (Ca/U)m=0.8 for this mean diameter design) 2

 r  Hence, tan β 2 = 1.624 − 1.25    rm  2

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2

 rm   r   2

 rm   r   2

tan β 2

Root

1.164

1.357

0.709

Tip

0.877

0.769

0

We therefore have untwisted nozzles. Cw2 rx=constant where x= sin 2 α 2 = sin 2 58D 23' =0.726 With

α 2 = constant, we also have Ca2 rx = constant. x

r  Hence, Ca2= (Ca2)m  m   r 2 This with   U=Um  r  yields  rm  2 x +1

 r   M  tan β 2 = tan α 2 -      rm  2  Ca 2  m 1.726

 rm     r 

tan β 2

β2

Root

1.164

0.662

35 D 20’

Tip

0.797

0.056

3 D 12'

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β2

35 D 20’ 0

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

Problem 7.5 m= ρ3 AC3 =

=

P3 A RT3

P3 A 2C p × T03 − T3 RT3

2C p × T01 − ∆Tw − T3

Assuming isentropic expansion, T  P3=P01  3   T01  m=

m=

A 2C p R A 2C p R

( γ / γ −1)

×

×

2

P01

γ −1

T3

T01(γ / γ −1)

(T01 − ∆Tw − T3 ) γ +1 γ −1

2

P01

γ −1

T3

T01(γ / γ −1)

(T01 − ∆Tw ) − T3

For the given inlet conditions m is a maximum when

dm = 0 i.e when: dT3

2

2 γ + 1 γ −1 T3 = 0 (T01 − ∆Tw ) T3(3−γ ) /(γ −1) − γ −1 γ −1

∴ T3= 2 (T01 − ∆Tw ) γ +1 Hence, writing K=

A 2C p

mmax= K (T01 − ∆Tw )

mmax= K (T01 − ∆Tw )

mmax= K (T01 − ∆Tw )

R γ +1 γ −1

γ +1 γ −1

γ +1 γ −1

×

P

01 ( γ / γ −1)

T01

2 γ +1   γ −1 γ −1  2  −  2      γ + 1   γ + 1     2 2  γ −1 γ −1  2  −  2   2       γ + 1   γ +1  γ +1 

   

2

 2  γ −1  λ − 1       γ +1  γ +1

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But

Cp R

mmax =

mmax

mmax

=

γ γ −1

AP01 T01(γ / γ −1)

T01 AP01

T01 AP01

(T01 − ∆Tw )

γ +1 γ −1

γ +1

γ +1

=

γ +1

2

 2 γ  γ −1 2 γ −1 (γ − 1)   γ +1  R γ −1 ( γ + 1) γ −1 γ +1

γ  2  γ −1  ∆Tw  γ −1 T01γ −1    1 − γ +1 R  γ +1  T01  γ −1 T01

γ +1

=

γ  2  γ −1  ∆T   1 − R  γ + 1   T

w

01

γ +1

 γ −1  

Hence maximum mass flow will vary with N/ T01 because ∆Tw varies with P01/P03.

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Problem 8.1 N=250 rev/s ⇒ Angular velocity ω = 2π N = 1570.5 rad/s Rr=0.1131m Rt=0.2262 m

ρ = 4500 kg/m3 (Ti-6Al-4V) ref table page 407 Centrifugal stress at the root

σr = =

ρω 2 2

2

2

( Rt − Rr ) K

4500 (1570.5 ) 2

2

( 0.2262

2

− 0.11312 ) ( 0.6 )

≅ 127.8 MPa Material UTS=880 MPa Endurance limit =350 MPa Assess the vibratory stress at failure – Goodman diagram -vibratory stress at failure ≈ 300MPa -high -Allowable approximately 1 ≅ 100 MPa 3

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Problem 8.2 N=250 rev/s R0=0.113m; Ri=0.105m

Width=0.025m Blade m ≅ 0.020 kg rcg ≅ 0.163m

ρ = 4500 kg/m3,

Material

UTS = 880 MPa

For a thin ring rotor (eq 8.29) 2 σ t = ρω 2 Rring +

Rring =

Frim 2π Aring

0.113 + 0.105 = 0.109m 2

Aring =0.025(0.113-0.105)

=2 ×10−4 m2 Frim =43 ( mrω 2 )

= 43 ( 0.02 × 0.163 × 1570.52 ) ≅ 345,750N

σ t = 4500 (1570.5 ) ( 0.109 ) + 2

2

345,750 2π ( 2 × 10−4 )

=131.87 ×106 + 275 × 106 ≅ 407.06 × 106 Pa

σ t = 407 MPa Considering burst, for UTS = 880MPa and burst margin of 1.2, the maximum allowable σ t

(σ t )allowable =

880

(1.2 )

2

= 611 MPa >> 407MPa

The design is conservative; but acceptable for a conceptual design.

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Problem 8.3

t

t[hrs]

σ [MPa]

PLM

T[K]

t1%[hrs]

TO+Climb

0.33

300

26.8

1150

2,015

1.489 × 10-4

Cruise

5

220

28

1000

100,000

5 × 10-5

Descent+L

0.5

150

29.5

1050

124,520

4 × 10-6

t1%

2.029 × 10-4

∴ 4,929 flights PLM is obtained from figure 8.7, for CMSX-10

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Problem 8.4 Haynes 188 at 1033K ∆ε = 0.01N −f 0.06 + 0.7 N −f 0.72

Source: Manson, S. S. and Halford, G. R., Fatigue and durability of structural materials (ASM International, 2006), where it is Figure 6.15(c). Reprinted with permission of ASM International ®. All rights reserved. www.asminternational.org

∆ε1 = 0.009 → Nf1 = 2 ×103 cycles ∆ε 2 = 0.007 → Nf2 = 1 ×104 cycles Using Miner’s rule (linear damage accumulation) n n n n + =1= + 3 2 × 10 1 × 104 N f1 N f 2

1=

 12 × 103  1 × 104 n + 2 × 103 n n =  7  2 × 107 n  2 × 10 

∴ n ≅ 1,670 cycles of each loading.

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Problem 8.5 Rigid bearings : Kr= rotor stiffness 1

ω=

1

K r  N  2  kgm  2 1 = = 2  =s m  mkg   s mkg 

m=20 kg ; ω =

πN 30

= 5, 235 rad/s

1

 K 2 5,235=  r  ⇒ Kr=548 × 106 N/m  20 

Soft bearings: total stiffness 1 Kr + Ks 548 × 106 + 1 × 107 = = Kt K r Ks 548 × 106 × 1 × 107 = 1.018 × 10–7 Kt = 9.821 × 106 N/m

First critical speed

ω=

9.821×106 = 700.7 rad/s = 6,693 ≅ 6,700 rpm 20

Plot the critical speed as a function of support stiffness!

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Problem 9.1 Note: In the problems for chapter 9, we are dealing with stagnation conditions at all times and the suffix ‘0’ has dropped throughout.

Flow compatibility is expressed by: m T1 P1

=

m T3 P3

×

P3 T × 1 P1 T3

And P3 P2 P2 − P3 P2 P P = − = − 0.05 2 = 0.95 2 P1 P1 P1 P1 P1 P1

∴

m T1 P 288 P =14.2 ×0.95 2 × = 6.903 2 P1 P1 P1 1100

m T1 P1

P2 P1

6.903

P2 P1

5.0

34.515

32.9

4.7

32.444

33.8

4.5

31.064

34.3

From curves, equilibrium point is at: P2 =4.835, P1

m T1 =33.4, P1

ηc = 0.795

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P3 =0.95 ×4.835 =4.59 P1

And from turbine characteristic graph: ηt = 0.8497 m=

33.4 × 1.01 = 1.988 kg/s 288

Net power output = mCpg ∆T34 − 1 mC pa ∆T12

ηm

1    1 4   Poutput = 1.988 × 1.147 × 0.8497 × 1100 1 −    4.59    

−1.988 × 1.005 ×

1 288   3.5 − 1 4.835 ( )   0.795  

Poutput = 675.189 – 411.620 = 263.989KW

Net power output = 264 KW

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Problem 9.2

m T4 P4

=

m T3 P3

×

P3 T × 4 P4 T3

∆T T4 = 1 − 34 T3 T3  ∆T34 1 = ηt 1 − 1  ; T3    r4 

P3 P4

 P3   P4

2.00

r=

P3 ; and ηt = 0.85 P4

1

4  

∆T34 T3

T4 T3

1.189

0.135

0.865

2.25

1.2247

0.156

2.50

1.257

0.174

∴ ∴

m T4 =188 P4 m T3 P3

when

m T3 P3

m T4 P4

0.930

88.2

164.05

0.844

0.918

90.2

186.31

0.826

0.909

90.2

205.0

T4 T3

P3 =2.270 (graphical solution) P4

=90.2

1   ∆T34  1 4   = 0.85 1 −  = 0.1575   2.27   T3  

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From compatibility of flow: m T1 P1

∴

=

m T3 P3

(

×

P3 T × 1 and P3=P2 P1 T3

)

m T3 / P3 ( P2 / P1 ) T3 = T1 m T1 / P1

(

)

T3 90.2 ( P2 / P1 ) = T1 m T1 / P1

(

-----------------------------(A)

)

From compatibility of work: T ∆T ∆T T3 C pgη m 0.1575 × 1.147 × 0.98 T3 = = = 0.1762 3 -------(B) T1 T3 T1 C pa T1 T1 1.005 0.286  ∆T12 1  P2  =   − 1 T1 ηc  P1   

P2 P1

m T1 P1

T3 T1

T3 T1

ηc

 P2   P1

  

0.286

∆T12 T1

(A)

T3 T1

  1   ∆T  =     0.1762   T1 (B)

5.2

220

2.13

4.54

0.82

1.602

0.734

4.17

5.0

236

1.91

3.65

0.83

1.584

0.704

3.99

4.8

244

1.77

3.15

0.82

1.566

0.690

3.92

∴

T P2 = 5.105 and 3 = 4.06, from graphical solution P1 T1

Hence T3=1170 K

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   

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

Problem 9.3 At outlet of gas generator turbine m T4 P4

=

m T3 P3

×

P3 T × 4 P4 T3

1   4   T4 1  ; = 1 − ηt 1 −  and   P3 / P4   T3  

∴

m T4 P4

=

m T3 P3

1   4   P3 1  ; × × 1 − ηt 1 −    P3 / P4   P4  

ηt = 0.85 1

1 4

 1   1 −  P P  3 4

1 4

1  2  4   1       1 − ηt 1 −   P /P    3 4    

P3 P4

 P3     P4 

1.3

1.0678

0.064

0.972

20.0

25.27 1.92

1.5

1.1067

0.097

0.958

44.0

63.22 1.664

1.8

1.1583

0.137

0.940

62.0

104.90 1.387

The value of P4/Pa in the table is found from P4 P4 P3 P2 P4 P = = × 0.96 × 2.60 = 2.496 4 Pa P3 P2 Pa P3 P3 Hence curves – from which power turbine pressure ratio is r = 1.55

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m T3 P3

m T4 P4

P4 Pa

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

And gas generator turbine pressure ratio r=1.61 Use work compatibility equation to find T3 ∆T12 ∆T34 T3  C pgη m  = ×   T1 T3 T1  C pa  1 1      1 4   P2  3.5  T 1   ∴ Cpa    − 1 = ηm C pg T3 1 −  ηc  P1  P3 / P4       

T1 must be given. Compressor efficiency from compressor characteristic once operating point m T1 m T3 P3 T = × × 1 where T3 unknown, so trial and error method required. found from P1 P3 P1 T3

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Problem 9.4 For gas generator turbine m T4 P4

=

m T3 P3

×

P3 T × 4 P4 T3

γ −1   γ   ∆T34 T4 1  ; where =1− = 1 − ηt 1 −    P3 / P4   T3 T3  

and ηt is constant. Thus:

m T4 P3

 P3  = f  from gas generator turbine characteristics.  P4 

Since power turbine is choking at all conditions considered, the gas generator turbine is ∆T34 operating at a fixed pressure ratio and hence fixed value of . T3 (a) At 95% of speed, work compatibility yields γ −1   T1  P2  γ  ηm C pg ∆T34 = C pa   − 1 ηc  P1   

∆T34 = C pa

=

1 T1   3.5 − 1 4.0 ( )   η m C pg 0.863  

C paT1 0.486

ηm C pg 0.863

At 100% speed we have similarly: ∆T34 = C pa

1 T1   C paT1 0.5465 3.5 − 1 = 4.6 ( )   η C 0.859 ηm C pg 0.859   m pg

 ∆T   ∆T  =  34  But:  34   1075 95% speed  T3 100% speed Therefore; T3= 1075 ×

0.5465 0.863 × = 1214.5 K 0.859 0.486

(b) 95% mechanical speed at 273 K.

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N T1

(%design) = 95

288 =97.57 % 273

From the operating line

∆T = m=

m T1 p = 439, 2 = 4.29 and ηc = 0.862 P1 P1

273  (4.29)0.286 − 1 = 163.6K 0.862

439 × 0.76 273

= 20.19 kg/s

Power=20.19 × 1.005 × 163.5 = 3318 kw.

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Problem 9.5 Work compatibility yields: γ −1   γ   1   = m T1 mηt T3 1 −   c   P3 / P4   ηc  

Let:

1   3.5   P  2 − 1  P1    

P3 P2 = =r P4 P1

At design point: 1 1  3.5   3.5  T1  r − 1  4 − 1 mc − mb 1 1 1   =  = × ×  1 1 mc ηtηc  0.8 0.85 3.3 ×    3.5  1  3.5 1 −  1   T3 1 −     r   4     

m mc − mb = = 0.6622 mc mc mb m = 1 − 0.6622 = 0.3378 = b mc mc

T1 / P1 T1 / P1

Therefore at design point: mb T1 = 0.3378 × 22.8 = 7.70 P1

   (m − m ) T  (m − mb ) T3 b 3  c  = c 1  1  P3 1 − 2  P3 1 − r 2  r  d And

T3 is constant.

mc − mb P = 3 ( mc − mb )d ( P3 )d

1 1 r 1− 2 2 r = r Thus : P =P 3 2 1 1 1 − 2 rd 1 − 2 rd rd 1−

mc T1 / P1 − mb T1 / P1 r 1 − 1 r 2 r2 −1 = = 22.8 − 7.7 4 1 − 1/16 15 mc T1 / P1 − mb T1 / P1 = 3.899 r 2 − 1 --------------------(1)

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Also from work compatibility: mc T1 / P1 − mb T1 / P1 mc T1 / P1

=

0.4457( r1 3.5 − 1) ------------------(2) 1   3.5 1 −  1    r   

3 Now: mb T1 / P1 = × 7.7 = 5.775 4 From (1): mc T1 / P1 = 5.775 + 3.889 r 2 − 1 (2) becomes: 1−

5.775 5.775 + 3.899 r 2 − 1

r

r2

r2 −1

3.5

12.25

3.35

3.0

9.0

2.5 2.25

=

0.4457(r1 3.5 − 1) , solve by trial and error 1   3.5 1 −  1    r    5.775 + 3.899 r 2 − 1

left-hand-side

13.08

18.85

0.694

2.83

11.05

16.82

0.657

6.25

2.29

8.94

14.71

0.607

5.06

2.015

7.865

13.63

0.576

1 3.5

3.899 r 2 − 1

0.4457  r1 3.5 − 1

1    1 − 1 3.5   r 

r

r

3.5

1.430

0.192

0.301

0.638

3.0

1.369

0.164

0.270

0.607

2.5

1.299

0.133

0.230

0.578

2.25

1.261

0.116

0.207

0.560

From curves, r = 2.08 Hence; mc T1 / P1 = 5.775+ 3.899 2.082 − 1 = 12.886

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right-hand-side