Solucionario Cohen Turbinas a gas

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.

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The rights of HIH Saravanamuttoo, GFC Rogers, H Cohen and PV Straznicky to be identified 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, res old, hired out or otherwise disposed dis posed 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

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

3 © Pearson Education Limited 2009

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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


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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 = 345.598K 11 = ( )   0.82  

Compressor and turbine work required per unit mass flow is: Wtc =

C pa (T02 − T a )

T03 – T 04 =

η m

= C pg (T03 − T04 )

1.005× 345.598 0.98×1.147

= 308.992 K

T 04 =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 −     P P   03 04  

P 03 P 04

= 4.382

P 03 = 11.0 − 0.4 = 10.6 bar P04 = 2.418 bar

, P05 = P a

 1     4 1     = 148.254K T04 − T 05 = 0.89 × 841 1 −    2.418    


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

20 × 103 166.64

= 120.019kg/sec

T 02 = 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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Problem 2.3

T02 − T a =

288 0.85

1

[3.8

3.5

− 1] = 157.3K

P 03 = 3.8 − 0.12 = 3.68 bar P 03 P a

= 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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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

mf = ma × f × (1 − B )

4374

6150

7791

SFC (kg/kwhr)

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

EGT ( C ) 


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Problem 2.5 T 03=1525 K P 03=29.69 bar P 04=13.00 bar

∴ P 03 P 04

= 2.284

∴ T 03 = ( 2.284)0.2223 = 1.202 T 04

T 04=1268.7 K , ∆T hp = 256.3 P 05 P 06

=

13.00 × 0.96 1.02

= 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

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 M W = 240000 = 523.9 kg/s 458.1

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

η th

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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Problem 2.6

288

∆T 12 =

(5.5)0.286 − 1 = 206.8 T2 = 494.8K 0.875

∆T 34 =

300  0.286 − 1 = 268.7 T 4 = 568.7K 7.5) (   0.870

∆Tcc = 1550 − 569 = 981 C 

∆T 56 = ∆T 67 =

268.7 × 1.005 1.148 × 0.95 × 0.99 206.8 × 1.005 1.148 × 0.99

= 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 1   ∴ ∆T 78 = 0.89 × 1117.1 1 − = 416.3 K 0.25  8.76  

∴ 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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Problem 2.8 P a=1.013 bar , T a=15

∆T 12 =



C

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

 

∴ ∆T 56 = 0.88 ×1037.8 1 −

P 5 P 6

1 2.895

0.25

= 2.895

  = 213.1 ∴ T 6=824.7 K 

∴ Power =112.0 ×1.147 × 0.99 × 213.1 = 27,106 kW 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 × 43,100

kJ kg kg s

= 67,288 kW

∴ Efficiency =40.3 %


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Problem 2.9

n −1

For compression:

For expansion:

n

n −1 n

P T02 − T01 = T 01[( 02 ) P 01

=

1

η ∞c

= η ∞t (

(

0.66 γ − 1 = 0.4518 )= 0.88 × 1.66 γ

γ − 1 0.88 × 0.66 )= = 0.3498 γ 1.66

n −1 n

− 1] = 310[2 0.4518 − 1] = 114K

T04 − T 03 = 300[20.4518 − 1] = 110.3K T04 = 410.3K T06 − T05 =

∴

Q mC p

and

=

T05 = 700K

500 × 103 180 × 5.19

( given)

= 535.2 K

T 06 = 1235.2K

P 03 = 2 × 14.0 − 0.34 = 27.66 bar P 04 = 2 × 27.66 = 55.32 bar P 06 = 55.32 − (0.27 + 1.03) = 54.02 bar P 07 = 14.0 + 0.34 + 0.27 = 14.61bar

∴

P 06 P 07

= 3.697

T06 − T 07 = 1235.2[1 − (

1 3.697

) 0.3498 ] = 453.3K

T 07 = 781.82K


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Power output = mC p [∆T067 − ∆T034 − ∆T012 ] = 180 × 5,19 × 229.0 = 213996 kW or 213.996 W Thermal efficiency = H.E.effectiveness =

213.996 500

T05 − T 04 T07 − T 04

=

= 0.4279

700 − 410.3 781.8 − 410.3

or

42.8%

= 0.7798

or 78%


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Problem 3.1

n −1

For compression:

For expansion:

n

n −1

0.87 × 3.5

0.87

=

n

1

=

4

= 0.3284

= 0.2175

and

n n −1

= 4.5977

At 7000 m Pa = 0.411bar; Ta = 242.7K C a

2

2c p

=

2602 2 × 1.005 ×1000

= 33.63K

T01 = 242.7 + 33.63 = 276.33 K P01 = P a [1 + η i

C a

2

2c pT a

]

γ γ −1

0.95 × 33.63

P 01 = 0.441[1 +

242.7 P 02 = 8 × 0.633 = 5.068bar

]3.5 = 0.633bar

T02 − T 01 = 276.33[8.0 0.3284 − 1] = 270.7K T 02 = 547.02K T03 − T 04 =

c pa (T02 − T 01 ) c pgη m

T04 = 1200 − 239.6

1.005 × 270.7

=

1.147 × 0.99

T04 = 960.4K

n

P03  T 03  n −1  1200  =  =  P04  T 04   960.4   4.763  P 04 =   = 1.710bar  2.784  Nozzle pressure ratio

= 239.6K

P 03 = 5.068(1 − 0.06) = 4.763bar

4.5977

P 04 P a

=

= 2.784

1.710 0.411

= 4.16


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(Since the critical pressure ratio is of the order of 1.9 the nozzle is clearly choked) P 04 P c

=

1

 1  0.333   1 −  0.95  2.333     

P 5= P c=

1.710 1.918

ρ 5 =

= 0.891 bar

2

T5 = Tc =

γ + 1

P c RT c

T 04 =

2 × 960.4 2.333

100 × 0.891

=

=1.918

4

0.287 × 823.3

= 823.3K

= 0.377

kg/m 3

1 2

1 2

C 5 = ( γ RT c ) = (1.333 × 0.287 × 823.3 ×1000 ) = 561.22 m/sec A5=

m

ρ 5C 5

=

15 0.377 × 561.22

= 0.0708m 2

F = m(C j − Ca ) + A j ( Pj − P a ) F = 15(561.22 − 260) + 0.0708(0.891 − 0.411)10 5 F = 4518.3 + 3398.4 = 7916.7 N T 02 = 547.02K T03 − T 02 = 1200 − 547.02 = 652.8K Therefore theoretical f = 0.01785 (Fig. 2.15) Actual f =

S .F .C =

0.01785 0.97

= 0.0184

0.0184 × 3600 ×15 7916.7

= 0.01255 kg/kN


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Problem 3.2

From problem 3.1 P 04=1.710 bar P 05 = 1.17(1 − 0.03) = 1.658bar P 05 P c P 05 P c

= 1.918 =

1.658 1.918

as before

= 0.8644bar

Since T05 = 2000K,

ρ 5 =

T6 = T c =

100 × 0.8644 0.287 × 1714.5

2 2.333

× 2000 = 1714.5 K kg/m 3

= 0.1756

1

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

15 0.1756 × 809.88

m/s

m2

= 0.1054

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 7916.7

= 64.56%


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Problem 3.3

Since P 01 = P a = 1 bar T 01 = T a = 288 K T02 − T 01 =

288 

9 0.82 

1 3.5

0.85mc pg (T03 − T04 ) = T03 − T 04 =



− 1 = 306.77K



mC pa

η m

( T02 − T01 )

1.005 × 306.77 0.85 × 1.147 × 0.98

= 322.678K

γ −1   γ   1   T03 − T04 = η j T 03 1 −     P03 / P 04     1    1 4   322.678 = 0.87 ×1275 1 −    P03 / P 04    

⇒

P 03 P 04

= 3.955

P 03 = 9.0 − 0.45 = 8.55bar P 04 = 2.161bar P04 Pa

. = 2.161

and

P 04 P a

=

1

 1  0.333   − 1  0.95  2.333     

4

So nozzle is choking even while landing


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P5 = P c =

2.161

= 1.126bar 1.918 T 04 = 1275 − 322.678 = 952.3K

 2  2 × 952.322 = 816.4K  × T 04 = 1 2.333 γ +  

T5 = Tc = 

ρ 5 =

P c RT c

=

100 × 1.126 0.287 × 816.4

= 0.480

kg/m 3

1

1

C5 = ( γ RT c ) 2 = (1.333 × 0.287 × 816.4 ×1000 ) 2 = 555.86 m/s (m − mb ) = ρ 5 A5C5 = 0.480 × 0.13 × 558.86 = 34.87 kg/sec mintake =

34.87 0.85

= 41.02

(N.B. – m b, 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) ×10 5 F = (19487.44-2256.1)+1638 F =18869.34 N

or

F =18.87 kN


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Problem 3.4

Isentropic expansion in nozzle: P 04 P a

=

2.4 1.01

= 2.376 γ

4

 γ + 1  γ −1  2.333  =  =  2  = 1.851 P c  2   

P 04

Choking ; So P5 = P c = T 04 T c

=

ρ 5 =

γ + 1 2 P c

RT c

=

=

2.333 2

2.4 1.851 and

100 × 1.296 0.287 × 857.26

= 1.296

bar

T5 = Tc =

= 0.526

1000 × 2 2.333

= 857.26K

kg/m 3

1

1

C 5= ( γ RT c ) 2 = (1.333 × 0.287 × 857.26 ×1000 ) 2 = 572.68 m/sec A5=

m

ρ 5C 5

=

23 0.526 × 572.68

= 0.0763m 2

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


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P 07 P a

= 1.75

and

P07 = 1.01 ×1.75 = 1.768bar

1   1 3.5   Ta  P 07 288    3.5 T07 − T a = 1.75 ) − 1 = 56.74 K −1 = (     0.88  η c  P a    

mh C pg (T04 − T05 ) = mc C pa (T07 − Ta ) mc

But

= 2.0

mh

T04 − T 05 =

2 × 1.005 × 56.74

= 99.43K 1.147 T 05 = 1000 − 99.43 = 900.57 K 1    1  4  P 04  99.43 = 0.90 × 1000 1 −  ⇒ = 1.597   P04 P05   P 05  

P 05 = P 05 P a

=

2.4

= 1.503bar

1.597 1.503

= 1.488

1.01

Hot nozzle is now unchoked, so P 6= P a=1.01 bar

 P  =  05  T6  P 6 

T05

T 6 =

γ −1 γ

1

= 1.488 4 = 1.1045

900.57

= 815.36K 1.1045 T05 − T 6 = 85.2K 1

1

C6 =  2C p (T05 − T06 )  = (2 × 1.147 × 85.21 ×1000) 2 C 6 = 442.12m/s 2

Hot nozzle thrust: Fh = 23 × 442.12 = 10,168.8 N The cold nozzle pressure ratio:=1.75, while the critical pressure ratio is :

 1.4 + 1  =  P c  2 

P 07

3.5

= 1.893


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So this nozzle is also unchoked. Since the expansion is isentropic: T 07 T 8

1

= 1.75

T8 =

3.5

= 1.1733

288 + 56.74

= 293.8 K 1.1733 T07 − T 8 = 50.94K

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 = 24888N or F =24.89 kN


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Problem 3.5 From solution of turbofan example in Chapter 3: T 03=800 K T 04-T 03=1550 – 800=750 K Actual f =

0.0221 0.99

= 0.0223

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

0.0223 × 35.83 × 3600 71463

= 0.0403 kg/Nh

P 09= P 02 – 0.05=1.65-0.05=1.60 bar So nozzle is unchoked and

P09 Pa

 P 05    P c 

= 1.60  <

  1 3.5  T 09 – T 8 = n jT 09 1 −      P09 P a     1 3.5  = 0.95 × 1000 1 −    =119.4 K   1.6   1

C 8= ( 2 × 1.005 ×119.4 ×1000 ) 2 = 490 m/ s mc=179.2 kg/s New thrust: F c=179.2 ×490 = 87, 808 N Total thrust = Unchanged F h+new F c = 18931+87808=106,739 N T 02=337.7 and T 09 – T 02=1000-337.7=662.3 K


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

S.F.C =

0.01715 0.97

TotalFuel Totalthrust

= 0.01768 for by-pass chamber

=

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

S.F.C =0.1338 kg/kN


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Problem 3.6

P a=0.899bar T a=281.7K C a=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 

 

T 01= Ta 1 + P 01 P a T 02 T 01

2

 

M 2  = 281.7(1 + 0.2 × 0.7 2 ) = 309.3 K ∆T r = 27.61 C 

1.4

 

= 1 +

0.95 × 27.61  0.4 281.7

= ( 5.0 )

∆T 34 =

γ − 1

0.3210

1.148 × 0.99 0.2248

= 1.366

∴ P 01=1.228 bar

= 1.676 ∴ T 02=472.2

162.9 × 1.005

T03  P 03  =  T04  P 04 

 

∆T 12 = 162.9 C 

= 144.1 T 04 = 1250 − 144.1 = 1105.9 K

P  1250  ∴ 03 =   P 04  1105.9 

1 0.2248

= 1.724

P 03=1.366 × 5.0 × 0.945 = 6.454 bar ∴ P 04 = γ

 γ + 1  γ −1  1.333  Critical P.R =   = 2  2     P04 Pa P 4=

=

3.744 0.899

3.744 1.852

6.454 1.724

= 3.744 bar

4.003

=1.852

= 4.164 ∴ choked

= 2.022 bar T 4=1105.9 ×

2 2.333

=948.0 K


th


ρ =

100 × 2.022 0.287 × 948.0

= 0.7432 kg/m3

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

A=

π 4

2

( 0.15) = 0.01767

m= ρ AC = 0.7432 × 0.01767 × 602.2 = 7.91 kg/s C a= 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 ×10 5 =2900 N+1984 N=4884 N

∆T cc = 1250 − 472 = 778, T in=472 K → ( f / a )id = 0.0212 ≈ f /a = ∴ mf = 7.91 ×0.02141 × 3600 = 609.8 kg/hr ∴ SFC =

609.8 4884

= 0.1249 kg/Nh


0.0212 0.99

= 0.02141

th


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

 

T 01= Ta 1 +

γ − 1 2

2

 

2

M  = 216.7(1 + 0.2 × 0.75 ) = 241.08 K

∆Tr = 24.38°C

 0.95 × 24.38  = 1 +  P a  216.7 

P 01

3.5

= 1.427 ∴ P 01=0.2558 bar

Fan T 02 T 01

= (1.7 )

n −1 n

n −1

= 1.1837

n

=

0.286 0.90

= 0.3178

∴ T 02=285.36 ∆T f = 44.28K P 02=0.4400 HPC

 T 03  0.3178 = 1.9287 T 03=550.38 P 03=3.4757   = ( 7.9 ) T  02  ∆T 023 = 265.02 Combustor P 04=3.4757(1 – 0.005)=3.3019 1.004 × 265.02

HP turbine ∆T 45 = LP turbine ∆T 56 = P04 P05

 T 04    T 05 

n −1

= 234.32 T 05=985.7 K

1.004 × 44.28(3.8 + 1)

n n −1

1.147 × 0.99 n −1

=

n

n

= 187.92 T 06=797.8 K

= 0.286 × 0.9 = 0.2232

= 4.4803

 1220  =  P 05  985.7  P 04

1.147 × 0.99

4.4803

= 2.60


th


 985.7  =  P 06  797.8  P 05

∴ P 06 =

4.4803

= 2.579

3.3019 2.60 × 2.579

Cold Stream

= 0.4924 bar

(both choked)

Hot Stream

P 02 = 0.4400 T 02=285.4 0.4400

P 2 =

1.893 285.4

T 2 =

ρ =

1.2

= 0.2324

P 6 =

0.287 × 237.8 9.6×3.8 4.8

ρ =

= 0.3405

= 7.60

7.6 0.3405 × 309.1

1.852

= 0.2659

T 6 = 797.8 

mh =

C c = 1.4 × 287 × 237.8 = 309.1 m/ s A =

0.4924

 2   = 683.9  2.333 

= 237.8

100 × 0.2324

mc =

P 06 = 0.4924bar T 06=797.8K

= 0.0722 m2

100 × 0.2659 0.287 × 683.9 9.6 × 1 4.8

= 0.1355

= 2.00

C h = 1.4 × 287 × 683.9 = 511.5 m/ s A =

2.0 0.1355 × 511.5

= 0.02886 m2

Fc = 7.6 ( 309.1 − 221.3 ) + 0.0722 (0.2324 − 0.1793 ) ×10 5 = 667.3 + 383.4=1050.7 Fh = 2.0 ( 511.5 − 221.3 ) + 0.02886 (0.2659 − 0.1793 ) ×10 5 = 580 + 250.0=830 F = 1880.7N T 04 = 1220, T 03=550 → f=0.0186 mf = 2.00 × 0.0186 × 3600 = 133.63 kg/hr SFC =

133.63 1880.7

= 0.0711 kg/Nh


th


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

 n − 1  0.286  n  = 0.90 = 0.3178  c  n − 1  0.333 × 0.9  n  = 1.333 = 0.2248  t  

T 01 01= Ta  1 +

γ − 1 2

 

M 2  = 216.8(1 + 0.2 × 0.85 2 ) = 248.13K

∆T r = 31.331C

 0.95 × 31.33  = 1 +  P a  216.8 

P 01 T 02 T 01

= (1.77 )

0.31718

3.5

=0.227 27 ×1.56 1.569 9 =0.3561 bar P bar P 02 = 1.569 ∴ P 01 01=0.2 02=0.6303

= 1.1989

∴ ∆T 12 = 49.33K

∴ T02=297.5

 P 03  34 = 19.21  = P 1.77  02   T 03  0.3178 = 2.558 , ∆T 23 = 463.4   = (19.21) T  02 

T03=761K

∆T 023 = 265.02 P04 = 0.3561 × 1.77 ×19.21 × 0.97 =11.74 bar

∆T 45 = P 04 P 05

1.004 × 463.4 1.148 × 0.99

 1350    940.6 

=

∆T 56 =

= 409.4 T 05 05=940.6 K

1 0.2248

= 4.99

1.004 × 49.33(5.7 + 1) 1.148 × 0.99

= 292.0 T 06 06=648.6 K

T 6=556.0 K


th

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

 940.6  0.2248

P 05

= 5.22

=

  648.6 

P 06

∴ P 06 06 =

11.74 4.99 × 5.22

= 0.4505 bar

P 06 P a

= 1.984 ∴ choked

C a = 0.85 ×295.2 Cold Stream mc = P 2=

220 × 5.7 6.7

P 02 Pcr

T 2 =

ρ =

=

297.5 1.2

(both choked)

= 187.2

0.6303 1.893

mh =

= 0.3330

P 6 =

= 0.468

ρ =

220 6.7

= 32.8

0.4505 1.852

= 0.2433

= 247.9

100 × 0.3330 0.287 × 247.9

C c = 1.4 × 287 × 247.9 = 315.6 m/ s s A =

Hot Stream

187.2 0.468 × 315.6

= 1.267 m2

100 × 0.2433 0.287 × 556

= 0.1525

C h = 1.333 × 287 × 556 = 461.2 m/ s s A =

32.8 0.1525 × 461.2

= 0.466 m2

F c = 187.2 ( 315.6 − 250.9 ) + 1.267 (0.333 − 0.227 ) ×10 5 =12.108+13430=25,538 F h = 32.8 ( 461.2 − 250.9 ) + 0.466 (0.2433 − 0.227 ) ×10 5 = 6898+760=7658 F = 33,196N ( F s=150.9) f =

0.017 0.99

=0.0172

mf = 0. 0.0172 × 32.8 × 3600 = 2028 kg/hr SFC =

2028 33196

= 0.0611 kg/Nh


th


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

Assume full expansion, i.e. no pressure thrust 0.4  n −1   n  = 0.90 ×1.4 = 0.3175  c

 n − 1  0.333 × 0.9  n  = 1.333 = 0.2248  t  

γ − 1

T 01 01= Ta  1 +

2

 

M 2  = 216.7(1 + 0.2 ×1.4 2 ) = 301.65 K

∆Tr = 84.95 K

 0.93 × 84.95  = 1 +  P a  216.7 

P 01 T 02 T 01 T 03 T 02

= ( 2.2 ) = ( 9.0 )

0.3175

0.3175

3.5

= 2.968 ∴ P 01 01=0.3594 bar

= 1.2845

∴ T 02 02=387.46 K ∴ ∆T12 = 85.8 K

= 2.009

∴ T 03 03=778.39 K ∴ ∆T23 = 390.9 K

Assuming no cooling, 1.004 × 390.9 = 1.147 × 0.99 × ∆T 45 ∆T 45 = 345.6 T 5=1154.3 K 1

 1500  0.2248 = = 3.206  P 05  1154.3  P 04

LP sy system (B+1) ×1.004 × 85.8 = 1.147 × 0.99 × ∆T 56 ∆T 56 = 227.6 T 6=926.7 K =926.7 K P 05 P 06

 1154.3    926.7 

=

1 0.2248

= 2.656

P 04 0.3594 × 2.2 × 9.0 × 0.93 = 6.618 bar 04= 0. P 02 02=0.3594 ×2.2 = 0.7907 bar P 06 =

6.618 3.206 × 2.656

= 0.7772 bar

P 06 P a

= 6.418

P 02 P a

= 6.529


th


With full expansion to P to P a T 02 T 2 T 06 T 6

= ( 6.529 ) = ( 6.418 )

0.286

0.25

; C c2 = 2 × 1.004 ×10 3 [387.4 − 226.6 ] = 568 m/ s = 1.710 T 2=226.6 K ; s

= 1.582 T 6=582.2 K ; ;

C h2 = 2 × 1.147 × 10 3 [926.7 − 582.2 ] = 888.9 m/ s s mc = 70 × mh = 70 ×

2 2 +1 1 2 +1

= 46.67 ; m ; mcC c=46.67 ×568 = 26508 N = 23.33 ; mhC h=23.33 ×888.9 = 20738 N 47246N (gross)

C a= 1.4 × 295.1 = 413.1 ra ram drag = 7 0 × 413.1 = 28920 N Thrust [ Specific Th

= 261.8 N / kg / hr ]

T 04 04-T 03 03=1500-778-722

18326 N (nett)

0.023 ∴ f ideal = 0.0232 0.023 f = ideal = 0.023 f 0.99

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

1951 18326

= 0.1065 kg/Nh

C a=413 m/ s s = (41 (413/ 3/10 1000 00)) ×3600 3600 km/h=1486 km/h

∴Flight time =

5600 1486.8

= 3.766 hrs

Fuel flow =3.766 ×2 × 1951 =14,697kg ∴ Fue


th


Problem 3.10 P a=1.013 bar ; T a=303K

∆T 12 =

303

150.286 − 1 = 437.5 0.81

∴ T 2 = 740.5 K ∆T cc = 1600 − 740.5 = 859.5 C 

∆T 34 =

437.5 × 1.005 1.148 × 0.99 × 0.97

 

= 398.8 ∴ 398.8 = 0.87 ×1600 1 −

1 

R 0.25 

∴ Rhp = 3.86 P 3=1.013 ×15.0 × 0.95 = 14.43 bar P 4= P 4 P 5

=

3.74 1.013

14.43 3.86

= 3.74 bar

= 3.692

T 4=1600 – 398.8 =1201.2 K

 

∴ ∆T 45 = 1201.2 × 0.89 1 −

1 3.692

0.25

  = 297.8 K

∴ Specific output =1.148 ×297.8 × 0.99 × 0.985 = 33.4 ∆T cc = 860, T in=467 C 

∴ f /a=

0.0254 0.99

kJ kg

( f / a )ideal = 0.0254

= 0.0257

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


th


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 ∆T 12 = P 03 P 04

308

120.286 − 1 = 379.6 0.84

T 02 = 688 K

= (12 × 1.01 × 0.95 ) /1.04 = 11.07

 

∴ ∆T 34 = 1390 × 0.87 1 −

1 11.07

0.25

  = 546

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) =

∴ mf =

SFC =

0.02 0.99

755.8 239.4

= 3.16 kg/s

0.02 0.99

× 3.16 × 3600 = 229.6 kg/hr

229.6 755.8

= 0.304 kg/kwhr


th


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

C=

8.0

=

ρ × 0.0574

139.37

m/ s

ρ

3.5

 T  = 0  , T s= 288 − , 2010 P s  Ts  C 2

ρ =

P s RT s

=

P0

100 × Ps 0.287 × Ts

= 348.4

θ c =

C 2 2c p

P s T s

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

θ

T s

T 0/T s

P 0/ P s

P s

ρ

C calc

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

∴

C a = 123.1 m/ s,

T s=280.46 K

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

tan α r =

123.1 110.3

= 1.116

α r = 4810' 


th


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

123.1 254.4

= 0.4839

α t = 25 48' 

At tip V 2=254.42+123.12=79720.42 ⇒ V =282.3 m/ s

a= 1.4 × 0.287 × 280.46 ×10 3 = 335.7 m/ s

∴ M =

282.3 335.7

= 0.841


th


Problem 4.2 T 1=T 01 –

C 12 2Cp

; T 01 = 217 +

 243.3  P 01=0.23    217  A=

π

( 0.33 4

2

2302 2 × 1.005 × 10 3

=243.3 K

3.5

= 0.343 bar

− 0.182 ) = 0.060 m2

m= ρ ACa = ρ AC cos25 ⇒ ρ C =

3.6 0.060 × 0.906

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

ρ =

100 0.287

∴

C =

×

P 0

=

T 0

66.22 0.491

100 × 0.343 0.287 × 243.3

= 0.491 kg/m3

= 134.87 m/ s

First trial: C max=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 66.0

= 152.90 m/ s

C 1w=152.9 × sin 25 = 152.9 × 0.4226 = 64.61 m/ s 3.5

 ψη c  σ U 2 − Cw (r ω ) mean   = 1 + P01    C pT 01

P 02

 18 + 33  1  × × 270 × 2π = 216.3 m/ s  2 × 100  2

U e = (r ω )mean= 


th


U = π × 270 × 0.54 = 458 m/ s 3.5

1.04 × 0.8   2   0.9 458 64.61 216.3 = 1 + × − ×   P 01  1.005 × 243.3 ×10 3 

P 02

P 02 P 01

= (1.595 )

3.5

= 5.12

∴ P 02=5.12 × 0.343=1.756 bar


th


Problem 4.3 P 01 = P a = 0.99 bar and T 01=T a=288K γ −1     T01  P 02 γ T 02-T 01 = 429-288= − 1    η  P 01   



∴ η c =

1

288  2.97  3.5





141  0.99 



0.63π

σ = 1 −

U 2 = π ×

17 16.5 100

 − 1 = 0.753  

= 0.8835 × 765 = 397.1 m / s

Hence: C 2w = σ U 2 = 0.8835 × 397.1 = 351 m/s

  C r 22 3512 − T 2=T 02 =429 –  3  2C p 2 1.005 10 × ×   2C p C 2

2

Therefore T 2=367.7 – θ r

ρ 2 = A2 =

100 × 1.92 0.287 × T2

669

=

T 2

π × 16.5 × 1.0

C r 2 =

10 m

ρ 2 A2

4

=

= 51.83 ×10 −4 m 2

0.60 51.83 × 10 −4

×

1

ρ2

=

115.76

ρ 2

θ r is small – For the first trial let T 2=366K ∴ ρ 2 =

669 366

= 1.83

Hence Cr 2 = 63.26

m/ s

Final trial: Cr 2

θ

64.0

2.03

T2 366

ρ

C r 2

1.827

63.5

∴ T 2=366 K

 T  =  2t  P2  T 2 

P2 t

3.5

 429  =   366 

3.5

= 1.743


th


P 2t=1.92 × 1.743 = 3.347 bar 141   = 1 + η i m p × P 01  288 

P 02

3.5

141   = 1 + η i m p × 0.99  288  ∴ η imp = 0.850 3.347

3.5

Fraction of loss in impeller =

1 − 0.87 1 − 0.753

= 0.607


th


Problem 4.4 (a) U 2= π × 0.50 × 270 =424.12 m/ s

∆T =

1.04 × 0.9 × 424.12 2

1.005 ×103 T 02 = 455.5K

 0.90 × 167.5  = 1 +  P 01  288 

P 02

= 167.5K

3.5

= (1.523 )

3.5

= 4.36

P 02 = 4.36 × 1.01 = 4.40bar

(b) Flow area at tip: =

π × 50 × 5 10

4

= 0.0785 m2

C 2w=0.9 ×424.12 = 381.7 m/ s 2

2

 381.7   96  If C 2r =96 and θ c =   +  = 76.84  44.9   44.9  2 × 1.005 × 103 = 44.9 )

(noting

T 2=455.5-76.84=378.66 K

 455.5  =  P 2  378.66 

P 02

ρ 2 =

3.5

= 1.909

100 × 2.30

P 2=

1.909

= 2.30 bar

ρ 2 A2 C r 2 = 2.12 × 0.0785 × 96 = 15.97

= 2.12

0.287 × 378.66

4.40

The agreement is close since the required flow is 16.0 kg/sec.

∴ C r 2 = 96 m/ s a = 0.287 × 1.4 × 378.66 ×10 3 = 390 m/ s C 22 = 381.7 2 + 96 2 ∴ C2 = 393.6 M = tan α

C 2 a

=

=

393.6 390 96

381.7

m/ s

= 1.01

= 0.251

∴

α = 14 5 


th


(c) P 3t= P 2t=4.40 bar A=

π × 58 × 5 10

ρ 3C3r =

4

= 0.0911 m 2

16.0 0.0911

= 175.6

C 3 w =

381.7 × 25 29

= 329 m/s

We must find C r3 by trial and error; as a first guess assume: C r3=96 ×

25 29

= 82.75m/s

T 03=T 02=455.5 K C 3w

θ w

C 3r

θ r

θ

T 3

T 0/T s

P 0/ P s

P s

ρ

ρ C r

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

∴

C3 r = 72.4 m/s, T3 = 399.1 K

C32 = 329 2 + 72.42

⇒ C 3 = 336.87 m/s

a= 1.4 × 0.287 × 399.1 ×10 3 = 400.44 m/s M =

336.87 400.44

tan β =

72.4 329

= 0.841

= 0.22

β = 12 24' 


th


Problem 4.5

∆T 013 = ∆T013 =

288

 40.286 − 1 = 175.1 K 0.80 pσ U 22 c p × 10

3

U 22 =

= 175.1

1.005 ×175.1 ×10 3 1.04 × 0.90

= 1.876 ×10 5

∴U 2 = 433 m/s d =

433

π × 200

= 0.689m

T 03 = T 02 = 175.1+288=463.1 K At tip M =1.0 T 2=T 02 ×

2

γ + 1

=

463.1 1.2

= 385.9K

∴

a = 1.4 × 0.287 × 385.9 ×10 3

∴

C2 = a = 393.7m/s

C 2w = 0.90 ×U 2 = 0.90 × 433 = 389.7 m/ s C 22 = C 22 r + C 22 w

∴ C 22 r = 393.7 2 – 389.72=3133.6 ∴ C 2 r = 55.97 m/s With a 50% loss in impeller, η I = 0.90

∴∆T '012

= 0.90 ×175.1 = 157.59 K

P  157.59  ∴ 02 = 1 + P 01  288 

3.5

= 4.607bar


th


P 02 = 1 × 4.607 = 4.607 bar

∴

γ

 γ + 1  γ −1 = = 1.23.5 = 1.893  P 2  2 

P 02

P 2 =

ρ 2 =

4.607 1.893

100 × 2.434 0.287 × 385.9

m= ρ AC2 r

∴h =

= 2.434bar

∴A=

0.1138

π × 0.689

= 2.198 kg/m3 14.0

2.198 × 55.97

× 100

= 0.1138

m2

= 5.257 cm


th


Problem 5.1

At mean radius, m Ω∆Cw = mC p ∆ T

∆C w =

∴ C 0w=

1.005 × 20 × 10 3 0.93 × 200

200 − 108.1 22

tan α 0

=

tan α1

=

45.95

= 108.1m/s

= 45.95m/s = 0.307

150 200 − 45.95 150

α 0 = 17 4 ' ( = α 2 ) 

= 1.027

α1 = 45 44' ( = α 3 ) 

With free vortex, C wr=constant i.e. C wU = constant

Tip C 0w ×250 = 46.0 × 200

∴ C 0 w

= 36.8 m/s


th


∆C wu = constant ∴ tan α 0 = tan α 1 = tan α 2 = tan α 3 =

36.8 150

∆C w =

108.1 × 200 250

= 86.4 m/ s

= 0.245, α 0 = 13 46' 

250 − 36.8

= 1.421, α 1 = 54 53' 

150

36.8 + 86.4 150 36.8 + 86.4 150

= 0.845, α 2 = 40 14' 

= 0.821, α 3 = 39 26 ' 

Root C 0w ×150 = 46 × 200

∆Cw =

108 × 200 150

tan α 0 = tan α 1 = tan α 2 = tan α 3 =

Λ=

61.3 150

C0 w = 61.3 m / s

∴

= 144

m/ s

= 0.409, α 0 = 22 15' 

150 − 61.3 150

= 0.591, α 1 = 30 36' 

150 − (144 + 61.3) 150 144 + 61.3 150

= −0.369, α 2 = −2015' 

= 1.369, α 3 = 53 51' 

static enthalpy rise in rotor

=

Stagnation enthalpy rise in stage

Now (T0+

C 02 2C p

+ 20) =T1+

C 12 2C p

T1 − T 0 20

 C 02 C 2 − 1  2C p 2C p 

∴ T1 − T 0 = 20 + 


 1  3  10

th


Tip C 0=

C 1=

150

=

cos13 46' 

150

=

cos 39 26 ' 

150 0.9712 150 0.7724

= 154.5 T 1-T 0=20+11.19-18.8=13.1

= 194.1

∴Λ =

13.1 20

= 65.5%

Root C 0=

150 cos 22 15' 

T 1-T 0=20+

C 1=

∴

=

150 0.9255

1  162.12



103  2 × 1.005

150 cos 53 51' 

=

T1 − T 0 = 0.92,

−

150 0.5899

Λ=

= 162.1 254.2 2 



2 ×1.005 

= 254.2

0.92 20

T 1-T 0=20+13.07-32.15

= 4.6%

(N.B.: Λ is very sensitive to small errors in C 0 and C 1)


th


Problem 5.2 Velocity diagram at mid span is as in previous question.

∆C w : mean = 108, tip =86.3, root =144 m/ s Equal work at all radius

Tip C 0w=

250 − 86.3

tan α 0 = tan α 1 =

2 81.85 150

= 81.85 m/ s

= 0.546, α 0 = 28 36' ( = α 2 )

250 − 81.85 150



= 1.121, α1 = 4816' ( = α 3 ) 


th


Root C 0w =

150 − 144

tan α 0 = tan α 1 =

2 3 150

= 3 m/ s

= 0.0020, α 0 = α 2 = 0°7'

150 − 3 150

= 0.980, α1 = α 3 = 44°25'


th


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

T = 288 −

a=

140 ×140 2 × 1.005

×

1

= 278.25 K

10 3

γ RT = 1.4 × 0.287 × 278.25 ×10 3 = 334.4 m/ s

∴ V= 0.95 × 334.4 = 317.7 m/ s V 2=U 2+C 2 ∴ U 2 = 317.7 2 − 140 2 = 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

 T  (b) = 0  P  T  P0

P =

ρ =

1.01 1.128

3.5

 288  =   278.25 

3.5

= 1.128

= 0.895 bar

0.895 ×100 0.287 × 278.25

= 1.121 kg/m3

Dhub = 0.60× 0.908 = 0.545 m A =

π 4

(0.9082 – 0.545 2 ) = 0.4143 m2

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

 0.89 × 20  = 1 + (c) = 1.2335 P 01  288  P 02


th


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

U h=0.60 ×285.2 = 171.1 m/ s mU ∆Cw Ω

∆C w =

1

= mC p ∆T

103

C p ∆T × 103

tan α 1 =

tan α 2 =

uΩ 171.1 140

=

1.005 × 20 × 10 3 171.1 × 0.93

= 126.3m/s

= 1.2221, α 1 = 50 44 ' 

171.1 − 126.3 140

= 0.320, α 2 = 17 46 ' 

At tip

∆C w = 126.3 × 0.60 = 75.78 m/s tan α 1 = tan α 2 =

285.2 140

= 2.0371, α 1 = 63 50'

285.2 − 75.78 140



= 1.4958, α 2 = 56 14' 

Power input =65 ×1.005 × 20 = 1306.5 kW


th


Problem 5.4 n −1 n T 02 T 01

=

0.286 0.88

= ( 4.0 )

= 0.325

0.325

= 1.569

∆T 012 = 288 [1.569 − 1] = 163.9K 163.9

∴ Number of stages = ∆T #

of

stages

For first stage:

=

T out T in

163.9

=

= 6.556 i.e. 7 stages

25 7

= 23.4 K/stage

288 + 23.4

= 1.0812

288

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

T out T in

= 1.0546

1.0546= ( R)0.325 ∴ Rlast=1.178

mU ∆Cw Ω

1 103

= mC p ∆T ; (note that C a=165 cos20)

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

1 103

= C p ∆T


th


2

U – 112.9U =

1.005 × 23.4 ×10 3 0.83

U 2 – 112.9U-28.3 ×103 = 0 112.9 ± 112.9 2 + 4 × 28.3 ×10 3

∴U =

2

=

112.9 ± 354.9 2

∴ U = 233.9 m/ s π DN 60

= 233.9 ∴ N =

60 × 233.9

π × 0.18

= 24817.5 rpm

At the entry to last rotor: T 0=428.5; P 0=

4 × 1.01 1.178

= 3.43 bar

C =165 m/ s ∴ T =428.5 -

 T  = 0  P  T 

P0

P =

3.43 1.1189

∴ ρ =

3.5

 428.5  =   414.96 

1652 2.01 × 103

= 414.96 K

3.5

= 1.1189

= 3.07 bar

100 × 3.07 0.287 × 414.96

= 2.58 kg/m3

(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


th


Problem 5.5 n −1

Axial:

n

0.286

=

0.92

 P  = = 1.417 =  02  T01 288  P 01 

∴

408

P 03 P 02

=

10.0

n

=

0.286 0.83

= 0.3446

∴ T 2=288+120=408 K

For axial ∆T = 4 × 30 = 120 K T02

n −1

= 0.311 Centrifugal:

0.311

∴

P 02 P 01

= 3.07

= 3.257

3.07

Axial : mU ∆Cw Ω

1

= mC p ∆T

103

∆Cw = U − 2C a tan 20 = U-300 ×0.364 = U − 109.2 103 × 1.005 × 30

U (U -109.2)=

0.86

=35.05 ×103

U 2-109.2U-35.05 ×103 =0

∴

U =

109.2 ± 109.2 2 + 4 × 35.05 ×10 3 2

=

109.2 ± 390 2

U =249.6 m/ s 249.6=

π DN 60

∴ N =

60 × 249.6

π × 0.25

= 19068 rpm

Centrifugal:

 P  =  03  T02  P 02  T03

0.3446

= ( 3.257 )

0.3446

= 1.502

∆T 023 = 0.502 × 408 = 204.82 K ∆T =

σψ U 22 C p

=

0.90 × 1.04 × U 22 1.005 × 103

∴

2 U 2 =

1.005 × 103 × 204.82 0.90 × 1.04

U 2=468.95 m/ s

∴ N (Centrifugal) =

60 × 468.95

π × 0.33

= 27140 rpm


th


Problem 6.1 (a) From Fig. 2.15, for T 1=298 K and ∆T = 583 K , theoretical f=0.0147

η b =

0.0147 0.0150

= 0.980

(b) From Fig. 2.15, for f=0.0150 and T 1=298 K , theoretical ∆T = 593 K

η b =

583 593

= 0.983

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

28 12

× 0.04 × 0.8608 = 0.0803 kg

Actual released energy = 43100–(0.0803 ×10110) Efficiency based on released energy:

η =

43100 − (0.0803 ×10110) 43100

= 1 − 0.0188

η = 0.9812


th


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

6846

440 132

( 2080 + 6846 )

108

1664

6846

= 0.0148

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

∆ H vap = −42500 +

108 132

× 2442 = −40500 KJ/kg of C10H12

Energy equation is: ( H p2- H p0)+ ∆ H 0 + ( H R0 − H R1 ) =0 Where the first stage is 325 K for C 10H12 and 450 K for air Mean temperature of reactants O 2 and N2 is

450 + 298 2

= 374 K

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 T 2=565+450 =1015 K First guess at mean temperature of products = (1015+298)/2=656.5 K . Mean values of C p from tables are : R- reactants, P- products. R’s

P ’s

1st Guess

2nd Guess

mC p

O2

0.934

CO2

1.105

1.047

460.7

N2

1.042

H 2O

2.051

2.041

220.4

O2

1.019

1.014

1687.2

N2

1.088

1.084

7421.1 9789.4


th


(T 2 –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 ] = (T 2 − 298 ) × 985.5 − 5,35, 000 −152 × 9073 − 27 × 257 = 0

∴ T 2=298+

6736000 9855

= 982 K

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

∴ T 2=298+

6736000 9789.4

= 986.1 K


th


Problem 6.3 Under design conditions: T 01=T 1+

475=

475=

C 12 2C p

=

P1 R ρ 1

4.47 ×100

+

0.287 ρ 1 1557

ρ1

+

+

m2 2

2 ρ 12 A1 C p 9.0 2

2 ρ 12 × 0.0389 2 ×1.005 ×10 3

26.70

ρ 12

475 ρ12 − 1557 ρ 1 − 26.70 = 0

From which : ρ 1 =

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

ρ 1 = 3.294 kg/m3 Hence:

0.27 × 10

 1023  − 1  475 

5

2

9.0 / 2 × 3.294 × 0.0975

2

=19.0+ K 2 

20.88=19.0+1.153 K 2

∴ K 2=1.630 Part-load conditions: 439=

3.52 × 100 0.287 ρ1

+

7.4 2 2 1

2

2 ρ × 0.0389 ×1.005 ×10

3

=

1226.5

ρ1

+

18.004

ρ 12

439 ρ12 − 1226.5 ρ 1 − 18 = 0

∴ ρ 1 = 2.808 kg/m3 105 ∆ P 0 =

  900   19.0 1.630 + −1    2 × 2.808 × 0.0975 2   439   7.4

2

∆ P 0 = 0.213 bar At design : C 1=

m

ρ 1 A1

At part load: C 1=

=

9.0 3.294 × 0.0389

=70.24 m/ s

7.4

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


th


 T  P 01= P 1  01   T 1  ∆ P 0 P 01

=

0.27 4.543

γ / γ −1

 475  = 4.47    472.8 

3.52 × 100 0.287 × 2.808

 439  P 01=3.52    436.78 

P 01

=

0.213 3.583

= 4.543 bar

=0.0594

At part load: T 1=

∆ P 0

3.5

= 436.78 K

3.5

= 3.583bar

= 0.0595


th


Problem 7.1

C 2=

C a cos 65

260

=

= 615.2 m/ s

0.4226

C w2= 615.2 2 − 260 2 = 557.5 m/ s 557.5 − 360

Tan β 2 =

260

= 0.7596 ∴ β 2 = 37 13' 

C w3=260 tan 10=260 ×0.1763 = 45.84 m/ s 360 + 45.84

Tan β 3 =

260

=1.5609 ∴ β 3 = 57 22 ' 

Since C1=C3 and C a is constant:

Λ=

C a 2U

260

( tan β3 − tan β 2 ) =

2 × 360

(1.5609 − 0.7596 )

Λ = 0.289 Temperature drop coefficient:

ψ =

2C p ∆T 0 s 2

U

=

2U ∆C w 2

U

=

2 ( 557.5 + 45.84 ) 360

=3.35

Power output =mU ∆C w = 20 × 360 × (557.5 + 45.84 ) P =4343760W or 4344 KW T 2=T 01-

C 22 2C p

=1000 –

615.2 2 2294

= 835 K


th

HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky,Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual ' 2

T 2 – T = λ N

C 22 2C p

= 0.05 ×

615.22 2294

= 8.25 K

T '2 = 835 − 8.25 = 826.75 K γ / γ −1

 T  =  01'  P2  T 2 

P01

4

P 2 =

∴

2.140

4

 1000  =  = 2.14 826.75   = 1.869bar

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

∴

P 01 P 2

> 1.853. ∴ The nozzle is choking

The throat conditions are: P c=

4.0 1.853

ρ c =

P c RT c

 2  2 × T 01 = × 1000 = 857 K  2.333  γ + 1 

= 2.158 bar, T c= 

=

2.158 × 100 0.287 × 857

= 0.877 kg/m3

C c= 1.333 × 0.287 × 857 ×10 3 = 572.6 m/ s Athroat=

m

ρ c C c

=

20 0.877 × 572.6

= 0.0398 m2


th


Problem 7.2

∆T013

γ −1   γ   1  = η t T 01 1 −    P01 P 03     1   4 1    =0.88 ×1050 1 −    = 147 K  2   

T 03 = 1050 – 147=903 K With zero outlet swirl: W s=UC w2=C p ∆T 013 And with free vortex, C wr = constant Hence at root (C w2)r =

C p ∆T 013 U

=

1.147 × 147 ×10 3 300

Now at root, Λ = 0 ∴

∴ T 02-

Hence

C 22r 2C p C 22r 2C p

=T 03 –

=

T2 − T 3 T1 − T 3

C 32 2C p

2752 2294

= 562 m/ s

= 0 hence T 2=T 3

; T 02=T 01 and C 3=C a3=275 m/ s

+ 147 = 180 K

∴ C 2r =642.6 m/ s


th


α 2 r = sin −1

562

= sin −1 0.8746 = 61



642.6

(C a2)r = 642.62 − 562 2 = 311.6 m/ s tan β 2 r =

( Cw2 )r − U r 562 − 300 = = 0.8406 311.6 ( C a 2 )r

β 2r = 40 4 ' 

r t

(C a2)t=(C a2)r =311.6 m/ s and

(C w2)t=

562

r r

= 1.4

=401.4 m/ s

1.4

 401.4  α 2t = tan −1  = tan −1 (1.2882 )   311.6  ∴ α 2t = 52°10' C 2t= 401.4 2 + 311.6 2 = 508.15 m/ s With no exit swirl T 3t=T 3r =903 –

2752 2294

= 870 K

508.152 = 937.5 K T 2t=1050 – 2294 (T 1-T 3)= ∆T 013 because C 1=C 3 with no inlet or exit swirl

Λ tip =

T2t − T 3t T1 − T 3

=

937.5 − 870 147

At the root T 2r -T ’2r = λ n

C 22r 2Cp

=0.46

=0.05 ×

642.62 2294

T 2r – T ’2r = 9 K T ’2r =(1050 – 180) – 9.0 =861 K γ / γ −1

 T  =  01  P2  T '2 

P01

γ / γ −1

 T  =  03  P3  T '3 

P03

4

3.8  1050  = = 2.21 ∴ P 2 = = 1.719 bar  2.21  861  4

3.8/2  903  = = 1.16 ∴ P 3 = = 1.638 bar  1.16  870 


th


Problem 7.3 C 2 4002 T 3=T 03 – 2 = (1200 − 150) − = 980.2K 2C p 2294 U m =2 π rm N or r m=

∆T013

320 2π × 250

= 0.2037 m

1   4   1   = η t T 01 1 −     P01 / P 03     1

 1 4 ∆T 013 150 =1− = 0.8611   =1− η t T 01 0.90 ×1200  P01 / P03  P 01 P 03

= 1.818 ∴ P 03= γ / γ −1

 T  =  03  P3  T 3 

P03

ρ 3 = A3= h= r t r r

P 3 RT 3 m

ρ 3C 3 A

2π r m

=

=

= 4.4bar

1.818

4

4.4  1050  = = 1.316 ∴ P 3= = 3.343 bar  1.316 980.25  

3.343 × 100

=

=

8

0.287 × 980.25 36 1.19 × 400

=1.19 kg/m3

=0.0756 m2 (N.B. C3=Ca3)

0.0756 2π × 0.2037

=0.0591 m

0.2037 + (0.0591/2) 0.2037 − (0.0591/2)

=

0.233 0.1741

= 1.34

A2 = A3 C a2 must satisfy C a2=

m

ρ 2 A2

and C a2= C 2 − C w2 2

W s=C w2U =C p ∆T 013 (because α 3 = 0 ) C w2=

1.147 × 150 ×10 3 320

= 537.6 m/ s

If C a2=346 m/ s C 22 = 346 2 + 537.6 2 T 2=T 02-

C 22

T 2’=T 2 – λ N P 2=

=T 01-

2C p

C 22 2C p

P 01

(T01 T 2' )

4

=

C 22 2C p

=1200-

⇒ C 2=639 m/ s

6392 2294

= 1022 K

=1022-0.07 ×178 = 1009.5 K 8.0

(1200/1009.5 )

4

=

8.0 1.995

= 4.008 bar


th


ρ 2 =

4.008 × 100 0.287 × 1022 m

C a2=

= 1.366 kg/m3

C2 A3

36

=

1.366 × 0.0756

= 348 m/ s (agrees with given 346 m/ s)

At the root, for free vortex design, we have: r m r t

=

0.2037 0.1741

= 1.17

(Cw2 )r =(Cw2 )m ×

r m r r

= 537.6 × 1.17 =629 m/ s

C a2 is constant at 346 m/ s. Hence: C 2r = 6292 + 346 2 = 717.9 m/ s T 2r =1200 –

U R=

320 1.17

7182 2294

=975.3 K

= 273.5 m/ s

V 2r = 346 2 + (629 − 273.5) 2 = 496.1 m/ s a2r = γ RT 2 r = 1.333 × 0.287 × 975.3 ×10 3 =610.86 m/ s ( M v2)r =

496.1 610.86

=0.812


th


Problem 7.4

C r2=constant

(1)

C w2r=constant

(2)

From (2), since C a2=C w2 cot α 2

 r m    r  2

C a2r=constant and C a2=(C a2)m 

 r    rm 2

Also U =U m 

Now,

U C a 2

= tan α 2 − tan β 2

∴ tan β 2 =tan α 2 -

Um  r 



2



( Ca 2 )m  r m 

tan α 2 =(tan α 2 ) m = tan 58 23' =1.624 

U m

( C a 2 )m

= tan 58 23’-tan 20 29' =1.624-0.374=1.25 



(N.B. Flow coefficient (C a/U )m=0.8 for this mean diameter design) 2

 r  Hence, tan β 2 = 1.624 − 1.25    r m 2


th


2

 r m   r   2

 r m   r   2

tan β 2

β 2

Root

1.164

1.357

0.709

35 20’

Tip

0.877

0.769

0

0

We therefore have untwisted nozzles. C w2 r x=constant where x= sin 2 α 2 = sin 2 58 23' =0.726 With 

α 2 = constant, we also have C a2 r x = constant. x

 r  Hence, C a2= (C a2)m  m   r 2 This with

 r   yields r  m 2

U =U m 

x +1

 r   M  tan β 2 = tan α 2 -      r m 2  C a 2  m

1.726

 r m   r   

tan β 2

β 2

Root

1.164

0.662

35 20’

Tip

0.797

0.056

3 12'








th


Problem 7.5 m= ρ 3 AC3 =

=

P 3 RT 3

A

P 3 RT 3

A 2C p × T03 − T 3

2C p × T01 − ∆Tw − T3

Assuming isentropic expansion, ( γ / γ −1)

 T  P 3= P 01  3   T 01  m=

m=

A 2C p R A 2C p R

×

×

2

P 01

γ −1

T3

T 01(γ / γ −1)

(T01 − ∆Tw − T3 )

γ +1 γ −1

2

P 01

γ −1

T3

T 01(γ / γ −1)

(T01 − ∆Tw ) − T3

For the given inlet conditions m is a maximum when

2

γ −1

dm dT 3

= 0 i.e when:

2

(3−γ ) /(γ −1) 3

(T01 − ∆Tw ) T

γ + 1 γ −1 − T3 = 0 γ − 1

∴ T 3= 2 (T01 − ∆T w ) γ + 1

Hence, writing K =

A 2C p

mmax= K (T01 − ∆T w )

mmax= K (T01 − ∆T w )

mmax= K (T01 − ∆T w )

R γ +1 γ −1

γ +1 γ −1

γ +1 γ −1

×

P 01 T 01(γ / γ −1)

γ +1 2   γ γ − −1 1     2 2   −    γ + 1   γ + 1     2 2  γ γ 1 − −1  2  −  2   2       γ + 1  γ + 1   γ + 1   

   

2

 2  γ −1  λ − 1       γ + 1   γ + 1 


th


But

C p R

mmax =

mmax

mmax

=

γ γ − 1

AP 01 (γ / γ −1) 01

T

T 01 AP 01

T 01 AP 01

=

=

(T01 − ∆T w )

γ  2    R  γ + 1 

γ  2    R  γ + 1 

γ +1 γ −1

γ +1 γ −1

γ +1 γ −1

2 γ     R γ − 1 

 ∆Tw  1 −  T 01  

 ∆T 1−  T 

w

01

γ +1 γ −1

2

γ −1

2

γ +1 γ −1

(γ − 1)

(γ + 1)

γ +1 γ −1

γ +1 γ −1

T 01

γ +1 γ −1

T 01

γ +1

 γ −1  

Hence maximum mass flow will vary with N / T 01 because ∆T w varies with P 01/ P 03.


th


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

σ r = =

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 ≈ 300 Pa -high -Allowable approximately 1 ≅ 100 MPa 3


th


Problem 8.2 N =250 rev/s R0=0.113m; R i=0.105m Width=0.025m Blade m ≅ 0.020 kg r cg ≅ 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 =

F rim 2π Aring

0.113 + 0.105 2

= 0.109m

Aring =0.025(0.113-0.105) =2 ×10 −4 m2 F rim =43 ( mr ω 2 ) = 43 ( 0.02 × 0.163 ×1570.5 2 ) ≅ 345,750N 2

2

σ t = 4500 (1570.5 ) (0.109 ) +

345,750 2π ( 2 × 10 −4 )

=131.87 ×106 + 275 ×10 6 ≅ 407.06 ×10 6 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.


th


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

t 1%

2.029 × 10-4

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


th


Problem 8.4 Haynes 188 at 1033K

∆ε = 0.01 N 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

∆ε 1 = 0.009 → Nf1 = 2 ×103 cycles ∆ε 2 = 0.007 → N f2 = 1 ×104 cycles Using Miner’s rule (linear damage accumulation) n N f 1

1=

+

n N f 2

=1=

n 2 × 103

1 × 10 4 n + 2 × 103 n 2 × 107 n

+

n 1 × 10 4

 12 ×10 3  7   2 ×10 

= n

∴ n ≅ 1,670 cycles of each loading.


th


Problem 8.5 Rigid bearings : K r = rotor stiffness 1

ω =

1

 N  2  kgm  2 1 =  =  s 2 mkg  = s m  mkg   

K r

m=20 kg ; ω =

π N 30

= 5,235 rad/s

1

 K  2 5,235=  r  ⇒ K r =548 × 106 N/m  20  Soft bearings: total stiffness 1 K t

=

K r + K s K r K s

=

548 × 10 6 + 1 ×10 7 548 × 10 6 × 1 × 10 7

= 1.018 × 10 –7 K t = 9.821 × 10 6 N/m First critical speed

ω =

9.821×106 20

= 700.7 rad/s = 6,693 ≅ 6,700 rpm

Plot the critical speed as a function of support stiffness!


th


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 T 1 P1

=

m T 3 P3

P 3

×

P1

×

T 1 T 3

And P3 P1

∴

P 2 P 1

=

P2 P1

−

P2 − P 3 P1

=

P2 P1

− 0.05

P2 P1

= 0.95

P 2 P 1

m T 1

P 288 P =14.2 ×0.95 2 × = 6.903 2 P P1 1100 P 1 1

6.903

P 2

m T 1

P 1

P 1

5.0

34.515

32.9

4.7

32.444

33.8

4.5

31.064

34.3

From curves, equilibrium point is at: P 2 P 1

=4.835,

m T 1 P 1

=33.4,

η c = 0.795


th


P 3 P 1

=0.95 ×4.835 =4.59

And from turbine characteristic graph: η t = 0.8497 m=

33.4 × 1.01 288

= 1.988 kg/s

Net power output = mC pg ∆T34 −

1

η m

mC pa ∆T12

1    1 4   P output = 1.988 × 1.147 × 0.8497 ×1100 1 −    4.59    

1 288   3.5 −1.988 × 1.005 × ( 4.835 ) −1  0.795  

P output = 675.189 – 411.620 = 263.989KW Net power output = 264 KW


th


Problem 9.2

m T 4 P4 T 4 T3

m T 3

=

=1−

P3

×

P3

×

P4

T 4 T 3

∆T 34 T 3

  1   = η t 1 − 1 ;   T 3  r 4 

∆T 34

r=

P 3 P 4

; and η t = 0.85

1

∆T 34

T 4

T 4

m T 3

m T 4

P 4

 P 3  4    P 4 

T 3

T 3

T 3

P 3

P 4

2.00

1.189

0.135

0.865

0.930

88.2

164.05

2.25

1.2247

0.156

0.844

0.918

90.2

186.31

2.50

1.257

0.174

0.826

0.909

90.2

205.0

P 3

∴ ∴

m T 4 P 4 m T 3 P 3

=188

when

P 3 P 4

=2.270 (graphical solution)

=90.2

1   4 1     = 0.85 1 −    = 0.1575  T 3  2.27   

∆T 34


th


From compatibility of flow: m T 1

=

P1 T 3

∴

T 1

T 3

P3

=

(m

×

P3

T 1

×

P1

T 3

2

1

(

and P 3= P 2

) ( P / P ) ( m T / P ) T3 / P3

90.2 ( P2 / P 1)

=

T 1

m T 3

1

1

)

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

m T1 / P 1

From compatibility of work:

∆T T1

=

∆T12 T1

∆T T3 C pgη m

=

T3 T1 C pa

=

0.286 1  P 2 

  η c  P 1  

0.1575 ×1.147 × 0.98 T3 1.005

T1

= 0.1762

T 3 T1

-------(B)

 − 1 

P 2

m T 1

T 3

T 3

P 1

P 1

T 1

T 1

η c

 P 2     P 1 

0.286

∆T 12 T 1

(A)

T 3   1   ∆T   =    T 1   0.1762   T 1   (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

∴

P 2 P 1

= 5.105 and

T 3 T 1

= 4.06, from graphical solution

Hence T 3=1170 K


th


Problem 9.3 At outlet of gas generator turbine m T 4 P4

=

m T 3 P3

×

P 3 P4

×

T 4 T 3

1   4   T 4 1  ; = 1 − η t 1 −  and    P / P   T3 3 4  

1   4 m T 4 m T 3 P 3   1  = × × 1 − η t 1 −  ∴  ;  P4 P3 P4 P / P   3 4  

η t = 0.85

1

1

1

  2 4   1       1 − η t 1 −  P 3 / P 4         

P 4

 P 3  4    P 4 

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

P 3

 1  4  1 −  P P  3 4 

1

The value of P 4/ P a in the table is found from P 4 P4 P 3 P 2 P4 P = = × 0.96 × 2.60 = 2.496 4 P a P3 P2 P a P3 P 3 Hence curves – from which power turbine pressure ratio is r = 1.55


m T 3

m T 4

P 4

P 3

P 4

P a

104.90

1.387

th


And gas generator turbine pressure ratio r=1.61 Use work compatibility equation to find T 3

∆T 12 T1

=

∆T34 T3

×

T 3  C pgη m 





T1  C pa 

1 1     3.5  1 4  T1  P 2    ∴ C pa    − 1 = η mC pg T 3 1 −  P3 / P 4   η c  P1      

T1 must be given. Compressor efficiency from compressor characteristic once operating point found from

m T 1 P1

=

m T 3 P3

×

P 3 P1

×

T 1 T 3

where T3 unknown, so trial and error method required.


th


Problem 9.4 For gas generator turbine m T 4 P4

=

m T 3 P3

×

P3 P4

×

T 4 T 3

γ −1   γ   ∆T 34 T 4 1  where = 1− = 1 − η t 1 −   ;  T3 T3 P /P   3 4  

and η t is constant.

Thus:

m T 4 P3

 P 3   from gas generator turbine characteristics. P  4 

= f 

Since power turbine is choking at all conditions considered, the gas generator turbine is ∆T 34 operating at a fixed pressure ratio and hence fixed value of . T 3 (a) At 95% of speed, work compatibility yields γ −1     T1  P 2 γ η m C pg ∆T34 = C pa − 1    η c  P 1    1   3.5 ∆T34 = C pa ( 4.0 ) − 1  η m C pg 0.863  

T 1

=

C paT 1 0.486

η m C pg 0.863

At 100% speed we have similarly: 1   C paT 1 0.5465 3.5 4.6 1 ∆T34 = C pa − ( )  = η C 0.859 η m C pg 0.859   m pg

T 1

 ∆T   ∆T34  =  34    1075 95% speed  T 3 100% speed

But: 

Therefore; T 3= 1075 ×

0.5465 0.859

×

0.863 0.486

= 1214.5 K

(b) 95% mechanical speed at 273 K.


th


N T 1

(% design) = 95

288 273

From the operating line

∆T =

m=

=97.57 %

m T 1 P 1

= 439,

p2 P 1

= 4.29 and η c = 0.862

273

(4.29) 0.286 − 1 = 163.6 K 0.862

439 × 0.76 273

= 20.19 kg/s

Power=20.19 × 1.005 × 163.5 = 3318 kw.


th


Problem 9.5 Work compatibility yields: γ −1   γ   1   = m T1 mη t T3 1 −   c   P3 / P4   η c  

P3

Let:

P4

=

P 2 P 1

1   3.5    P 2 − 1    P 1    

= r

At design point: 1  3.51   3.5  − T1  r − 1 4 1   mc − mb 1 1 1     = = × × 1 1 mc 0.8 0.85 3.3 ηtη c  ×    3.5 3.5 1 1        T 3 1 −   1−      r    4     

m mc mb mc

=

mc − mb mc

= 0.6622

= 1 − 0.6622 = 0.3378 =

mb

T1 / P 1

mc

T1 / P 1

Therefore at design point: mb T 1 P 1

= 0.3378 × 22.8 = 7.70

   (m − m ) T  ( m − mb ) T3 b 3  c  = c 1  1  1 1 − − P P 3 3 2 2   r r   d And

T 3 is constant.

mc − mb

( mc − mb )d

=

P 3

( P3 ) d

1− 1−

1 2

r 1 −

2

r d 1 −

r = 1 r d

mc T1 / P1 − mb T1 / P1 r 1 − 1 r 2 = = 22.8 − 7.7 4 1 − 1/16

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

r d

r 2 − 1 15

mc T1 / P1 − mb T1 / P1 = 3.899 r 2 − 1 --------------------(1)


Solucionario Cohen Turbinas a gas

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