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Solucionario Cohen Turbinas a gas
Termodinamica (Universidad Nacional Mayor de San Marcos)
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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: 978-0-273-70934-3 978-0-273-70934-3
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
Supporting resources Visit www.pearsoned.co.uk/saravanamuttoo www.pearsoned.co.uk/saravanamuttooto to find other valuable online resources For instructor instructors s •
PowerPoint slides that can be downloaded and used for presentations presentations
For more information please contact your local Pearson Education sales representative or visit www.pearsoned.co.uk/saravanamuttoo www.pearsoned.co.uk/saravanamuttoo
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 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
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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 –1 T02 – T a = η c P a 1 288 = (11) 3.5 – 1 = 345.598 K 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 ( T0 3 − T04 )
1.005× 345.598 345.598 = 308.992 K 0.98×1.147
T 04 =11 =115 50 – 309 = 841K γ −1 γ 1 T03 − T04 = η t T 03 1 − P P 03 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 b ar P04 = 2.418 ba
, P05 = P a
1 1 4 = 148.254K T04 − T 05 = 0.89 × 841 1 − 2.418
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Specific power output: W N = 1.147 × 0.98 × 148.254 = 166.64 k kW Ws/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 Theoretical f = = 0.01415 (from Fig. 2.15) Actual f Actual f = 0.0 0.01415 0.99 = 0.01429 S.F.C =
3600 f W N
=
3600 × 0.01429 166.64
kg g/kW − hr = 0.308 k
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
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 ba 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 pg ( T03 − T04 ) −
mC pa (T02 − Ta )
η m(comp .rotor )
]
m × 1.005 × 157.3 200 = 0.98[( m − 1.5) × 1.147 × 256.87 − ] 0.99 200 = 288.73m − 433.1 − 156.5m (a) m = 4.788 kg/sec
With no bleed flow: Net work output = 0.98 × 4.788[1.147 × 256.87 −
1.005 × 157.3 0.99
]
= 633.11 kW (b) The power output output with no bleed = 633.11kW 633.11kW
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.4 A
B
C
n −1 n c
0.3287
0.3250
0.3213
n −1 n t
0.2225
0.2205
0.2205
T02 T 01
2.059
2.242
2.437
∆T 012 (K)
305
357.8
413.9
T 02 02 (K)
593
645.8
701.9
P 03 P 04 03/ P 04
8.55
11.40
15.2
1.612
1.708
1.820
T 03 03
1150
1400
1600
T 04 04
713.4
819.6
879.2
∆T 034
436.6
580.4
720.7
W c= (1.005 ×∆T 12 )/0.99
309.6
363.2
420.2
W t= 1.148(1 − B )∆T 034
501.2
649.6
786.0
W net Wc – net=W t W
191.6
286.4
365.8
Power
14,370
22,912
31,093
Base
+59.4%
+116%
557
754
898
0.0162
0.0219
0.0268
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 03 P 04
n −1
T 03 03/T 04 04=
n
∆T cc f /a
EGT (( C )
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Problem 2.5 T 03 03=1525 K P 03 03=29.69 bar P 04 04=13.00 bar
∴ P 03 P 04
= 2.284 ∴
T 03
= ( 2.284 )
T 04
0.2223
= 1.202
T 04 04=1268.7 K , ∆T hp = 256.3 P 05 P 06
=
13.00 × 0.96 1.02
= 12.24 ∴
T 05
= (12.24)
T 06
0.223
= 1.745
T 03 03 – T 04 04=256.3 T 05 05 T 06 06 T 05 T 06 05 – T 06 T 03 T 04 03 – T 04 ∆T total
∆T total ×1.148 × 0.99 – ∆T c × 1.004
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 f /a)2 ( f
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
0.02 0.0282 82 ×43 4310 100 0 37.7
0.024 0.0 247 7 ×43 431 100 38.5
0.022 0.0 227 7 ×43 4310 100 0 36.9
reduced. May be difficult to control low fuel: air ratio in 2 So η th remains high as power reduced. combustor. 9 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.6
288 (5.5)0.286 − 1 = 206.8 T2 = 494.8K ∆T 12 = 0.875 ∆T 34 =
300 0.286 ( 7.5) − 1 = 268.7 T 4 = 568.7K 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= ,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 bar P P 6 =
1
LPT , 182.9= 0.89 × 1300 1 −
R
0.25
39.19 2.248
= 17.43 bar
RLP =1.990 P RLP =1.990 P 7=8.76 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, 00000 ∴ m=211.3 kg/s f /a=
0.028 0.99
= 0.0283
Spe Specific output =1.148 ×0.99 × 416.3 = 473.1
∴ η th =
473.1 0.283 × 43100
= 38.8%
EGT =1117.7-416.3 =1117.7-416.3 =701.4 K =701.4 K =428.4 C 10 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 2.8 P a=1.013 bar , T a=15
∆T 12 =
C
288
8.50.286 − 1 = 279.5k 0.87
T =567.5 K 02
P 04 04=1.013 ×8. 5 × [1 − 0.015] × [1 − 0.042] = 8.125 bar
∆T 45 =
2 79.5 × 1.004 1.147 × 0.99
= 247.1 T 5 = 1037.8 K
247.1=0.87 ×1285 1 −
P P 5 = 2.991 bar ⇒ 4 = 2.716, P 0.25 ( P 4 / P 5) P 5 1
⇒
P 6=1.013 × 1.02 = 1.0333
∴ ∆T 56 = 0.88 × 1037.8 1 −
P 5 P 6
1 0.25
2.895
= 2.895
= 213.1 ∴ T 6=824.7 K
∴ Power = =1 112.0 ×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 in=799-273=526 C
( f f /a)id=0.0138
f /a=
0.0138 0.99
=0.0139
mf = 0.0 0.013 139 9 ×112 12..0 =1.56 kg/s Heat input = 1.56 × 43,100 43,100
kJ kg kg s
= 67,288 kW
∴ Efficiency =40.3 %
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
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 1.66 γ
n −1 n
− 1] = 310[20.4518 − 1] = 114K
T04 − T 03 = 300[20.4518 − 1] = 110.3K T04 = 410.3K T −T = 06
∴
05
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 b baar P 04 = 2 × 27.66 = 55.32 ba bar
P 06 = 55.32 − (0.27 + 1.03) = 54.02 ba 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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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Power output = mC p [∆T067 − ∆T034 − ∆T012 ] = 18 180 × 5, 5,19 × 22 229.0 = 213996 kW or 213.996 W W Thermal efficiency =
213.996 500
= 0.4279
or
42.8%
H.E.effectiveness = T05 − T 04 = 700 − 410.3 = 0.7798 T07 − T 04 781.8 − 410.3
or 78%
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 3.1
n −1
For compression:
For expansion:
n
n −1
1 0.87 × 3.5
0 .8 7
=
n
=
4
= 0.3284
= 0.2175
and
n n −1
= 4.5977
At 7000 m Pa = 0.411 bar; Ta = 242.7 K C a
2
2c p
=
2602 2 × 1.005 × 1000
= 33.63K
T01 = 242.7 + 33.63 = 276.33K P01 = P a [1 + η i
C a
2
2c pT a
P 01 = 0.441[1 +
]
γ γ −1
0.95 × 33.63
242.7 P 02 = 8 × 0.633 = 5.068bar
]3.5 = 0.633bar
T02 − T 01 = 276.33[8.00.3284 − 1] = 270.7K T 02 = 547.02K pa 02 01 T03 − T 04 = c (T − T ) = 1.005 × 270.7 = 239.6K 1.147 × 0.99 c pgη m
T04 = 1200 − 239.6
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
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/m3
1
1
C 5 = ( γ RT c ) 2 = (1.333 × 0.287 × 823.3 × 1000 ) 2 = 561.22 m/ m/sec A5=
m
15
=
ρ 5C 5
= 0.0708m2
0.377 × 561.22
F = m(C j − Ca ) + Aj ( Pj − P a ) F = 15(561.22 − 260) + 0.0708(0. (0.891 − 0.411)105 F = 4518.3 + 3398.4 = 7916.7 N
T 02 = 547.02K T03 − T 02 = 1200 − 547.02 = 652.8K Therefore theoretical f theoretical f = 0.01785 (Fig. 2.15) Actual f Actual f =
S .F .C =
0.01785 0.97
= 0.0184
0 .0 184 × 3600 × 15 7916.7
= 0.01255 kg/kN
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 3.2
From problem 3.1 P 3.1 P 04 04=1.710 bar P 05 = 1.17(1 − 0.03) = 1.658bar P 05 = 1.918 as be before P c P 05 P c
=
1.658 1.918
= 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/m3
= 0.1756
1
C 5 = (1.333 × 0.287 × 1714.5 × 1000) 2 = 809.88 15
m/s
2
A = 0.1756 × 809.88 = 0.1054
m
5
Percentage increase in area = 0.1054-0.0708=48.97% =15(809.88–260)+0.1054(0.8644–0.411) ×105 F =15(809.88–260)+0.1054(0.8644–0.411) F =8248.2+4778.83=13027.03 =8248.2+4778.83=13027.03 N Percentage increase in thrust =
13 1302 027. 7.03 03 − 791 916 6.7 7916.7
= 64.56%
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 3.3
Since P Since P 01 = P P a = 1 bar 01 = T 01 01 = T a = 288 K
T02
1 − T 01 = 288 9 3.5 − 1 = 306.77K 0.82
0.85mc pg (T03 − T04 ) = T03 − T 04 =
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 4 1 322.678 = 0.87 × 1275 1 − P03 / P 0 4
⇒
P 03 = 3.955 P 04
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
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2.161
= 1.126bar 1.918 T 04 = 1275 − 322.678 = 952.3K
P5 = P c =
2 2 × 952.322 = 816.4K × T 04 = 1 2 . 3 3 3 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/ 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) 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 =18.87 kN
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 3.4
Isentropic expansion in nozzle: P 04 P a
=
2.4 1.01
= 2.376
γ + 1 = P c 2
P 04
γ γ −1
4
2.333 = = 1.851 2
Choking ; So P5 = P c = T 04 T c
=
ρ 5 =
γ + 1 2
=
2 .333 2. 2
2.4 1.851 and
= 1.296
bar
T5 = Tc =
P c 100 × 1.296 = = 0 .5 2 6 RT c 0.287 × 857.26
1000 × 2 2 .3 3 3
= 857.26K
kg/m3
1
1
m//sec C 5= ( γ RT c ) 2 = (1.333 × 0.287 × 857.26 × 1000 ) 2 = 572.68 m A5=
m
ρ 5C 5
=
23 0.526 × 572.68
= 0.0763m2
F = 23 × 572.68 + 0.0763(1.296 − 1.01)105 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 = 1.768 bar
1 1 3.5 Ta P 07 = 288 (1.75 ) 3.5 − 1 = 56.74 K 1 T07 − T a = − 0.88 η c P a
mh C pg (T04 − T05 ) = mc C pa (T07 − Ta ) mc
But
= 2.0
mh
2 × 1.005 × 56.74
= 99.43K 1.147 T 05 = 1000 − 99.43 = 900.57 K T04 − T 05 =
1 4 P 1 99.43 = 0.90 × 1000 1 − ⇒ 04 = 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 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
2
2
C6 = 2C p (T05 − T06 ) = (2 × 1.147 × 85.21× 1000) C 6 = 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 :
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
= 1.753.5 = 1.1733
T8 =
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 × 1.005 × 50.94 × 1000 ) 2 = 320 m/s 2
Cold thrust Fc = 46 × 320 = 14720 N
To Tota tall th thru rust st:: =1472 =14720 0 + 101 016 68. 8.8 8 = 24 248 888N
or F F =24.89 =24.89 kN
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Problem 3.5 From solution of turbofan example in Chapter 3: T 03 03=800 K T 04 04-T 03 03=1550 – 800=750 K Actual f Actual f = =
0.0221 0.99
= 0.0223
m = 35.83 kg/s and F and F s=71,463 N S.F.C = =
0.0223 × 35.83 × 3600 71463
= 0.0403 kg/Nh
P 09 09= P 02 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 09 – T 8 = n jT 09 09 1 − P09 P a 1 3.5 = 0.95 × 1000 1 − =119.4 K 1.6 1
s C 8= ( 2 × 1.005 × 119.4 × 1000 ) 2 = 49 0 m/ s mc=179.2 kg/s New thrust: F thrust: F c=179.2 ×490 = 87, 8 80 08 N Total thrust = Unchanged F Unchanged F h+new +new F F c = 18931+87808=106,739 N T 02 and T 09 02=337.7 and T 09 – T 02 02=1000-337.7=662.3 K
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From Fig. 2.15, theoretical f theoretical f = 0.01715 Actual f Actual f =
S.F.C =
0.01715 0.97
TotalFuel
= 0.01768 for by-pass chamber
=
Totalthrust
( (0.01768 × 179.2 + 0.0223 × 35.83) × 3600
106,739
S.F.C =0.1338 =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 s
M =0.70 =0.70
1 0.4 n −1 0.333 n −1 = 0 .9 0 = 0.2248 = = 0.3210 n t 1.333 n c 0.89 1.4
T 01 01= Ta 1 +
γ − 1 2
27.61 61 C M 2 = 281.7(1 + 0.2 × 0.7 2 ) = 309.3K ∆T r = 27.
1.4
0.95 × 27.61 0.4 = 1 + = 1.366 281.7 P a
P 01
T 02 T 01
= ( 5 .0 )
∆T 34 = T03
=
0.3210
1.148 × 0.99 0.2248
P 04
T04
= 1.676 ∴ T 02 02=472.2
1 62.9 × 1.005
P 03
∴ P 01 01=1.228 bar
∆T 12 = 162.9 C
= 144.1 T 04 = 1250 − 144.1 = 1105.9 K 1
P 1250 0.2248 ∴ 03 = = 1.724 P 04 1105.9
P 03 03= 1.366 × 5.0 × 0.945 = 6.454 bar ∴ P 04 =
γ + 1 Critical P.R Critical P.R = = 2
P04 Pa P 4=
=
3.744 0.899
3.744 1.852
γ γ −1
1.333 = 2
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
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ρ =
100 × 2.022 0.287 × 948.0
3 = 0.7432 kg/m
C = γ RT 4 = 1.333 × 0.287 × 1000 × 948.0 = 602 .2 m/ ss
π
2
A= 4 ( 0.15 ) = 0.01767 m= ρ AC = 0.7432 × 0.01767 × 602.2 = 7.91 kg/s C a= 0. 0.70 × 336.4 = 23 5.5 m/ ss 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
∆T cc = 1250 − 472 = 778, T iinn=472 K → ( f / a )id = 0.0212 ≈ f /a =
0.0212 0.99
∴ mf = 7.91 ×0.02141 × 3600 = 609.8 kg/hr ∴ SFC =
609.8 4884
= 0.1249 kg/Nh
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Problem 3.7 M =0.75; =0.75; BPR=3.8; FPR BPR=3.8; FPR=1.7 =1.7 C a=0.75 ×295.1 = 221.3 m/ s s
T 01 01= Ta 1 +
γ − 1 2
M 2 = 216.7(1 + 0.2 × 0.752 ) = 241.08K
∆Tr = 24.38°C
0.95 × 24.38 = 1 + P a 216.7
P 01
3.5
= 1.427 ∴ P 01 01=0.2558 bar
Fan T 02 T 01
= (1.7 )
n −1
n −1
= 1.1837
n
n
=
0.286 0.90
= 0.3178
P 02 ∴ T 02 02=0.4400 02=285.36 ∆T f = 44.28K P HPC
T 03 0.3178 =550.38 P P 03 = 1.9287 T 03 03=550.38 03=3.4757 = ( 7 .9 ) T 02 ∆T 023 = 265.02 Combustor P 04 Combustor P 04=3.4757(1 – 0.005)=3.3019 1 .004 × 265.02
HP turbine ∆T 45 = LP turbine ∆T 56 =
1.147 × 0.99
= 234.32 T 05 05=985.7 K
1 .004 × 44.28(3.8 + 1) 1.147 × 0.99
= 187.92 T 06 06=797.8 K
n
P04 T 04 n −1 = P05 T 05 n n −1
n
= 0.286 × 0.9 = 0.2232
= 4.4803
1220 = 05 P 985.7 P 04
n −1
4.4803
= 2.60
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985.7 = P 06 797.8 P 05
∴ P P 06 06 =
4.4803
= 2.579
3.3019 2.60 × 2.579
Cold Stream
= 0.4924 bar
(both choked)
Hot Stream
P 02 02 = 0.4400 T 02 02=285.4 P 2 =
T 2 =
ρ =
0.4400 1.893 285.4 1.2
P 06 06 = 0.4924bar T 0 06 6=797.8K
= 0.2324
P 6 =
0.287 × 237.8
1.852
= 0.2659
2 = 683.9 2.333
= 237.8
1 00 × 0.2324
0.4924
T 6 = 797.8
ρ =
= 0.3405
1 00 × 0.2659 0.287 × 683.9
= 0.1355
mc = 9.6×3.8 = 7.60 4.8
mh = 9.6 × 1 = 2.00 4.8
C c = 1.4 × 287 × 237.8 = 30 9.1 m/ s s
C h = 1.4 × 287 × 683.9 = 51 1.5 m/ s s
A =
7.6 0.3405 × 309.1
2 = 0.0722 m
A =
2.0 0.1355 × 511.5
2 = 0.02886 m
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 T 04 04 = 1220, T03=550 → f=0.0186 mf = 2.00 × 0.0186 × 3600 = 133.63 kg/hr SFC =
133.63 1880.7
= 0.0711 kg/Nh
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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 × 0.852 ) = 248.13K
∆T r = 31.331C
0.95 × 31.33 = 1 + 216.8 P a
P 01 T 02 T 01
= (1.77 )
0.31718
3.5
=0.227 27 ×1. 1.56 569 9 =0.3561 bar P bar P 0022=0.6303 = 1.569 ∴ P 01 01=0.2
= 1.1989
∴ ∆T 12 = 49.33K
∴ T02=297.5
P 03 34 = 19.21 = P 1.77 02 0.3178 T 03 = 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 =
1 .004 × 463.4 1.148 × 0.99
= 409.4 T 05 05=940.6 K
1
1350 0.2248 = = 4.99 P 05 940.6 P 04
∆T 56 =
1 .004 × 49.33(5.7 + 1) 1.148 × 0.99
= 292.0 T 06 06=648.6 K
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940.6 648.6
P 05
=
P 06
P 06 ∴ P 06 =
1 0.2248
= 5.22
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 =
220 6.7
= 32.8
0.4505 1.852
= 0.2433
= 247.9
1 00 × 0.3330 0.287 × 247.9
ρ =
= 0.468
C c = 1.4 × 287 × 247.9 = 31 5.6 m/ s s A =
Hot Stream
187.2 0.468 × 315.6
= 1.267 m2
1 00 × 0.2433 0.287 × 556
= 0.1525
C h = 1.333 × 287 × 556 = 46 1.2 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 ) × 105 =12.108+13430=25,538 F h = 32 .8 ( 461.2 − 250.9 ) + 0.466 ( 0.2433 − 0.227 ) × 105 = 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
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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 = 0.2248 1.333 n t
γ − 1
T 01 01= Ta 1 +
2
M 2 = 216.7(1 + 0.2 × 1.42 ) = 301.65K
∆Tr = 84.95K
0.93 × 84.95 = 1 + 216.7 P a
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 =926.7 K K 1
P 05 P 06
1154.3 0.2248
=
926.7
= 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 30
= 6.529
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With full expansion to P to P a T 02 T 2 T 06 T 6
= ( 6.529 ) = ( 6.418 )
0.286
0.25
s = 1.710 T 2=226.6 K ;; C c2 = 2 × 1 .004 × 103 [387.4 − 226.6] = 568 m/ s
= 1.582 T 6=582.2 K ;;
C h2 = 2 × 1.147 × 103 [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 = 265 08 N = 23.33 ; mhC h=23.33 ×888.9 = 207 38 N 47246N (gross)
C a= 1.4 × 295.1 = 413.1 r raam drag = 7 0 × 413.1 = 28920 N
[ Specific TThhrust
= 261.8 N / kg / hr ]
T 04 0.023 f = 04-T 03 03=1500-778-722 ∴ f ideal ideal = 0.023 f
18326 N (nett) 0.023 0.99
= 0.0232
∴ mf =23.33 × 0.0232 × 3600 = 1951 kg/hr SFC =
1951 18326
= 0.1065 kg/Nh
C a=413 m/ ss = (41 (413/ 3/10 1000 00)) ×36 3600 00 km/h=1486 km/h
∴ Flight time =
5600 1486.8
= 3.766 hrs
Fuel flow =3.766 ×2 × 1951 =14,697kg ∴ Fue
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Problem 3.10 P a=1.013 bar ; T a=303K
303 150.286 − 1 = 437.5 ∆T 12 = 0.81
∴ T 2 = 740.5 K 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 −
∴ Rhp = 3.86 P 3=1.013 ×15.0 × 0.95 = 14.43 bar bar P 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 = 297.8 K 3.6920.25
kJ ∴ Specific output =1.148 ×297.8 × 0.99 × 0.985 = 33.4 kg
∆T cc = 860, T in = 0.0254 in=467 C ( f / a )ideal
f /a= ∴ f
0.0254 0.99
= 0.0257
Airflow =11.0 kg/s ∴ P Po ower =333.4 ×11.0 = 3667 kW mf = 0.0257 ×11.0 × 3600 = 1018.9 kg/hr
∴ SFC =0.278 =0.278 kg/kWh 32
1
R 0.25
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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 = 308 120.286 − 1 = 379.6 T 02 02 = 688 K 0.84 P 03 P 04
= (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=
(690, ∆T = 700) = f /a (6
∴ mf =
SFC =
0.02 0.99
0.02 0.99
755.8 239.4
= 3.16 kg/s
× 3.16 × 3600 = 229.6 kg/hr
229.6 755.8
= 0.304 kg/kwhr
33
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Problem 4.1 Flow area = π (1 52 − 6.52 ) = 57 4 cm2= 0.0574 m2 8.0
139.37
C = ρ × 0.0574 =
m/ s s
ρ
3.5
2 T 0 C T s= 288 − , = , θ c = 2010 P s Ts 2c p 2
P0
C
ρ =
P s 100 × Ps P = = 348.4 s RT s 0.287 × Ts T s
We must find the axial velocity by iteration to satisfy continuity: Let C max s max=150 m/ s C
θ
T s
T 0/T s
P 0/ P P s
P s
ρ
C calc 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, s,
T s=280.46 K
At root, U = π × ( 0.065 × 2 ) × 270 = 110 .3 m/ s s
tan α r =
123.1 110.3
α r = 4810'
= 1.116
34
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At tip, U = π × ( 0.15 × 2 ) × 270 = 254.4 m/ s s tan α t =
123.1 254.4
= 0.4839
α t = 25 48'
At tip V =254.4 +123.1 =79720.42 ⇒ V =282.3 =282.3 m/ s s 2
2
2
a= 1.4 × 0.287 × 280.46 × 103 = 33 5.7 m/ s s 282.3
∴ M M =
335.7
= 0.841
35
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Problem 4.2 T 1=T 01 01 –
C 12 2Cp
; T 01 01 = 217 +
243.3 P 01 01=0.23 217 A A= =
π
( 0.33 4
2
2302 2 × 1.005 × 103
=243.3 K
3.5
= 0.343 bar
− 0.182 ) = 0.060 m
2
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 100
P
100 × 0.343
ρ = 0.287 × T 0 = 0.287 × 243.3 = 0.491 kg/m3 0 ∴
C =
66.22 0.491
= 134.87 m/ s s
First trial: C m max ax=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 =152.4 ×
66.22 66.0
= 152.90 m/ s s
C 1w 25 5 = 152.9 × 0.4226 = 64 .61 m/ s s 1w=152.9 × sin 2 3.5
ψη c σ U 2 − Cw (r ω ) mean = 1 + P01 C pT 01
P 02
18 + 33 1 s × × 270 × 2π = 216.3 m/ s 2 × 100 2
U e = (r ω )mean=
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U = π × s × 270 × 0.54 = 458 m/ s 3.5
1.04 × 0.8 0.9 × 4582 − 64.61 × 21 216.3 = 1 + 3 P 01 1.005 × 243.3 × 10
P 02
P 02
P 01
3.5
= (1.595)
= 5.12
P 02 ∴ P 02=5.12 × 0.343=1.756 bar
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Problem 4.3 P 01 01 = P a = 0.99 bar and T 01 01=T a=288K γ −1 T01 P 02 γ T 02-T 01 = 429-288= η P 01 − 1
∴ 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 m/s 2 = 0.8835 × 397.1 = 351 m/s 2w = σ U
C r 22 3512 − =429 – T 2=T 02 023 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 3.5
T 429 = 2t = 2 2 P T 366
P2 t
3.5
= 1.743
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P 2t 2t=1.92 × 1.743 = 3.347 bar 141 = 1 + η i m p × 288 P 01
P 02
3.347
= 1 + η i
×
3.5
141
3.5
0.99 288 ∴ η imp = 0.850 m p
Fraction of loss in impeller =
1 − 0.87 1 − 0.753
= 0.607
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Problem 4.4 (a) U 2= π × × 0.50 × 270 =424.12 m/ s s
∆T =
1.04 × 0.9 × 424.122
1.005 × 10 T 02 = 455.5K
= 167.5K
3
0.90 × 167.5 = 1 + P 01 288
P 02
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 s 2w=0.9 ×424.12 = 381.7 m/ s 2
2
381.7 96 If C 22r r =96 and θ c = + = 76.84 44.9 44.9 (noting
2 × 1.005 × 103 = 44.9 )
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. s ∴ C = 96 m/ s r 2
a = 0.287 × 1.4 × 378.66 × 103 = 390 m/ s s C 22 = 381.7 2 + 962 ∴ C2 = 393.6 M = tan α
C 2 a
=
=
393.6 390 96
381.7
m/s
= 1.01
= 0.251
∴
α = 14 5
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(c) P 3t 3t= P 2t 2t=4.40 bar A A= =
π × × 58 × 5 10
ρ 3C3r =
4
= 0.0911m 2
16.0
= 175.6
C 3 w =
381.7 × 2 25 5
0.0911
= 329 m/s
29
We must find C r3 r3 by trial and error; as a first guess assume: C r3 r3=96 ×
25 29
= 82.75m/s
T 03 =455.5 K K 03=T 02 02=455.5 C 3w 3w
θ w
C 3r 3r
θ r
θ
T 3
T 0/T s
P 0/ P 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 = 3292 + 72.4 2
⇒ C 3 = 336.87 m/s
a= 1.4 × 0.287 × 399.1 × 103 = 400.44 m/s M =
336.87 400.44
tan β =
72.4 329
= 0.841
= 0.22
β = 12 24 '
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Problem 4.5
288 0.286 4 ∆T 013 = − 1 = 175.1 K 0.80
∆T013 =
pσ U 22 c p × 10
3
= 175.1
2 2
U =
1.005 × 175.1× 103 1.04 × 0.90
∴U 2 = 433 m/s d = =
433
= 0.689m
π × 200
T 03 03 = T 02 02 = 175.1+288=463.1 K At tip M tip M =1.0 =1.0 T 2=T 02 02 ×
2
γ + 1
=
463.1 1 .2
= 385.9K
∴
a = 1.4 × 0.287 × 385.9 × 103
∴
C2 = a = 393.7m/s
C 2w .7 m/ s s 2w = 0.90 ×U 2 = 0.90 × 433 = 389 C 22 = C 22 r + C 22 w
∴ C 22 r = 393.72 – 389.72=3133.6 ∴ C 2 r = 55.97 m/s With a 50% loss in impeller, η I = 0.90
∴ ∆T '012 P 02
= 0.90 × 175.1 = 157.59 K 157.59
∴ P 01 = 1 + 288
3.5
= 4.607bar
= 1.876 × 105
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P 02 = 1× 4.607 = 4.607 bar
∴
γ
γ + 1 γ −1 = = 1.23.5 = 1.893 P 2 2
P 02
4.607 P = 1.893 = 2.434bar 2
ρ 2 =
100 × 2.434 0.287 × 385.9
m= ρ AC2 r
∴h =
∴A=
0.1138
π × 0.689
= 2.198 kg/m3 14.0
2.198 × 55.97
× 100
= 5.257 cm
= 0.1138
m2
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Problem 5.1
At mean radius, m Ω∆Cw = mC p ∆ T
∴
∆C w =
1 .005 × 20 × 103
= 108.1m/s
0.93 × 200 200 − 108.1 C 0w = 45.95m/s 0w= 22 tan α 0
=
tan α1
=
45.95
= 0.307
150 200 − 45.95 150
α 0 = 17 4 ' ( = α 2 )
= 1 .0 2 7
α1 = 45 44 ' ( = α 3 )
With free vortex, C vortex, C wr=constant i.e. C i.e. C wU = constant
Tip C 0w 0w ×250 = 46.0 × 200
∴ C 0 w
= 36.8 m/s m/s
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∆C wu = constant ∴ tan α 0 =
36.8 150
∆C w =
1 08.1 × 200 250
s = 86.4 m/ s
= 0.245, α 0 = 13 46'
50 − 36.8 = 1.421, α 1 = 54 53' tan α 1 = 2 150
tan α 2 = tan α 3 =
3 6.8 + 86.4 150 3 6.8 + 86.4 150
= 0.845, α 2 = 40 14'
= 0.821, α 3 = 39 26 '
Root C 0w 0w ×150 = 46 × 200
∆Cw =
108 × 200 150
= 144
C0 w = 61.3 m / s
∴
m/ s
61.3
15 ' tan α = 150 = 0.409, α 0 = 22 15 0
tan α 1 = tan α 2 = tan α 3 =
Λ=
1 50 − 61.3 150
= 0.591, α 1 = 30 36'
150 − (144 + 61.3) 150 1 44 + 61.3 150
= −0.369, α 2 = −2015'
51'' = 1.369, α 3 = 53 51
static enthalpy rise in rotor Stagnation enthalpy rise in stage C 02
C 12
=
T1 − T 0 20
C 02 C 12 1
Now (T0+ 2C p + 20) =T1+ 2C p ∴ T1 − T 0 = 20 + 2C p − 2C p 103
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Tip C 0=
150
=
cos13 46 46 '
150 C 1=
150 0.9712
= 154.5 T 1-T 0=20+11.19-18.8=13.1
150
=
cos 39 26 '
0.7724
13.1
= 194.1 ∴ Λ =
20
= 65.5%
Root C 0=
150 cos 22 15 15 '
T 1-T 0=20+
C 1=
∴
=
150 0.9255
1 162.12
103 2 × 1.005
150 cos 53 51 51'
=
T1 − T 0 = 0.92,
−
150 0.5899
Λ=
= 162.1
254.22
2 × 1.005
= 254.2 T 1-T 0=20+13.07-32.15
0.92 20
= 4.6%
(N.B.: Λ is very sensitive to small errors in C 0 and and C C 1)
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Problem 5.2 Velocity diagram at mid span is as in previous question. s ∆C w : mean = 108, tip =86.3, root =144 m/ s Equal work at all radius
Tip C 0w 0w=
250 − 86.3
tan α 0 =
2 81.85 150
= 81.85 m/ s s
= 0.546, α 0 = 28 36' ( = α 2 )
2 50 − 81.85 tan α 1 =
150
= 1.121, α1 = 4816' ( = α 3 )
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Root C 0w 0w =
150 − 144
tan α 0 = tan α 1 =
2 3
s = 3 m/ s
= 0.0020, α 0 = α 2 = 0°7'
150 1 50 − 3 150
= 0.980, α1 = α 3 = 44°25'
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Problem 5.3 (a) With no inlet guide vanes the inlet velocity is axial .
T = 288 −
a=
140 × 140 2 × 1.005
×
1
K = 278.25 K
103
= 1.4 × 0.287 × 278.25 × 103 = 334.4 m/ s s γ RT
V= = 0. 0.95 × 334.4 = 317.7 m/ s s ∴ V 2
2
2
2
V =U +C ∴ U = 317.7 2 − 1402 = 81333.3 =285.2 m/s ∴ U =285.2 D= = π × × D × N = 285.2 ∴ D
285.2
π × 100
= 0.908 m
∴ Tip radius =0.454 m or 45.40 cm (b)
P0
3.5
T 288 = 0 =
P P =
ρ =
T 1.01 1.128
3.5
= 1.128
278.25 = 0.895 bar
0 .895 × 100 0.287 × 278.25
3 = 1.121 kg/m
Dhub = 0.60× 0.908 = 0.545 m A A = =
π 4
(0.9082 – 0.5 452 ) = 0.4143m2
∴ m = 1. 1.121 ×0.4143 × 140 =65.02 or m=65 kg/s
(c)
P 02 P 01
0.89 × 20
288
= 1+
3.5
= 1.2335
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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.
U h=0.60 ×285.2 = 171.1 m/ s 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 × 103 = = 126.3m/s 171.1 × 0.93
= 1.2221, α 1 = 50 44 '
1 71.1 − 126.3 140
= 0.320, α 2 = 17 46 '
At tip
∆C w = 126.3 × 0.60 = 75.78 m m//s 285.2 = 2.0371, α 1 = 63 50' tan α 1 = 140
tan α 2 =
2 85.2 − 75.78 140
= 1.4958, α 2 = 56 14'
Power iin nput = =6 65 ×1.005 × 20 = 1306.5 kW
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Problem 5.4 n −1 n
=
0.286 0.88
= 0.325
T 02 = ( 4.0 )0.325 = 1.569 T 01
∆T 012 = 288[1.569 − 1] = 163.9K 163.9
∴ Number of stages = ∆T #
of
st stag agees
For first stage:
=
T out T in
25
163.9
=
7
= 6.556 i.e. 7 stages
= 23.4 K/stage
288 + 23.4 288
= 1.0812
1.0812 = ( R R)) 0.325 ∴ R Rfirst=1.271 For last stage: T out =288+163.9=451.9 K K out=288+163.9=451.9 T in 451.9-23.4=428.5 K ; ∴ in = 451.9-23.4=428.5 K 1.0546= ( R) R)
0.325
mU ∆Cw Ω
T out T in
= 1.0546
∴ Rlast=1.178
1 103
= mC p ∆T ; (note that C a=165 cos20)
∆Cw = U − 2C sin 20 = U − 330 sin 20 ΩU (U − 112.9)
1 103
= C p ∆T
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2
U – 112.9U 112.9U =
1.005 × 23.4 × 103 0.83
U 2 – 112.9U-28.3 ×103 = 0 2
3
∴ U = 112.9 ± 112.9 + 4 × 28.3 × 10 = 112.9 ± 354.9 2 2
∴ U = 233.9 m/ s s π DN 60
= 233.9 ∴ N N =
60 × 233.9
π × 0.18
= 24817.5 rpm
At the entry to last rotor: T 0=428.5; =428.5; P P 0=
4 × 1 .0 1 1.178
= 3.43 bar
C =165 =165 m/ s s ∴ T =428.5 =428.5 3.5
T 0 428.5 P = T = 414.96
P0
P =
3.43 1.1189
∴ ρ =
1652 2 .0 1 × 1 0 3
= 414.96 K
3.5
= 1.1189
= 3.07 bar
1 0 0 × 3 .0 7 0.287 × 414.96
= 2.58 kg/m3
(π Dh ) ρ × Ca = m π × 0.18 × h × 2.58 ×165 cos 20 = 3.0 ∴h =
3.0
π × 0.18 × 2.58 × 165 × 0.9397
h = 1.326 cm
= 0.01326 m
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Problem 5.5 n −1
Axial:
n
0.286
=
0.92
P = = 1.417 = 02 T01 288 P 01
∴
P 03 P 02
408
=
10.0 3.07
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
Axial : mU ∆Cw Ω
1
= mC p ∆T
103
∆Cw = U − 2C a tan 20 = U-300 ×0.364 = U − 109.2 U (U -109.2)= -109.2)=
103 × 1.005 × 30 0.86
=35.05 ×103
U 2-109.2U-35.05 ×103 =0
∴ U =
109.2 ± 109.22 + 4 × 35.05 × 103 2
=
109.2 ± 390 2
U =249.6 =249.6 m/ s s 249.6=
π DN 60
∴ N 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 0.90 × 1.04 × U 22 C p
=
1.005 × 103
∴ U = 2 2
1 .005 × 103 × 204.82
U 2=468.95 m/ s s (Centrifugal) = ∴ N (Centrifugal)
60 × 468.95
π × 0.33
= 27140 rpm
0.90 × 1.04
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Problem 6.1 (a) From Fig. 2.15, 2.15, for T 1=298 =298 K K and ∆T = 583 K K , theoretical f=0.0147
η b =
0.0147
= 0.980
0.0150 (b) From Fig. 2.15, for f=0.0150 and T 1=298 =298 K K , theoretical ∆T = 593 K
η b =
583 593
= 0.983
(c) mass of CO per kg of of fuel: m=
28 12
× 0.04 × 0.8608 = 0.0803 kg
Actual released released energy energy = 43100–(0 43100–(0.0803 .0803 ×10110) 10110) Efficiency based on released energy:
η =
43100 − (0.0803 × 10110)
η = 0.9812
43100
= 1 − 0.0188
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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 H p2- H 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 O2 and N2 is
450 + 298 2
= 374 K
From fig. 2.15 for f for f =0.0148 =0.0148 and for an initial temperature of 450 K 450 K For air, temperature rise ∆T = 565 K. Hence approximate final temperature T 2=565+450 =1015 K =1015 K First guess at mean temperature of products = (1015+298)/2=656.5 K (1015+298)/2=656.5 K . Mean values of C p from tables are : R- reactants, P- products. R’s R’s
’s P ’s
1st Guess
2nd Guess
mC p
O2
0.934
CO2
1.105
1.047
460.7
N2
1.042
H2O
2.051
2.041
220.4
O2
1.019
1.014
1687.2
N2
1.088
1.084
7421.1 9789.4
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(T 2 –298) { ( 440 × 1.105 ) + (108 × 2.051) + (1664 ×1.019 ) + (6846 × 1.088)} – (132 × 40500 ) − ( 450 − 29 8 ) ( 20 80 × 0.934 ) + ( 6846 × 1.0 42 )
– ( 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 K
Second guess at mean temperature of products=1280/2=640 K
∴ T 2=298+
6736000 9789.4
= 986.1 K K
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Problem 6.3 Under design conditions: C 12
m2
P1
T 01=T 1+ 2C p = R ρ + 2 ρ 12 A12C p 1
475=
475=
4.47 ×10 100
+
0.287 ρ 1 1557
ρ1
+
9 .0 2 2 ρ 12 × 0.03892 × 1.005 × 103
26.70
ρ 12
475 ρ12 − 1557 ρ 1 − 26.70 = 0
From which : ρ 1 =
1 557 ± 1557 2 − 4 × 475 × (−26.76) 2 × 475
ρ 1 = 3.294 kg/m3 Hence:
0 .2 7 × 1 0
5
2
9.0 / 2 × 3.294 × 0.0975
2
1023 − 1 475
=19.0+ K =19.0+ K 2
20.88=19.0+1.153 K 20.88=19.0+1.153 K 2
∴ K K 2=1.630 Part-load conditions: 439=
3.52 × 100 0.287 ρ1
+
7.42 2 1
2
2 ρ × 0.0389 × 1.005 × 10
3
=
1226.5
ρ1
439 ρ12 − 1226.5ρ 1 − 18 = 0
∴ ρ 1 = 2.808 kg/m3 105 ∆ P 0 =
900 1 9 . 0 1 . 6 3 0 + 439 − 1 2 × 2.808 × 0.09752 7.42
∆ P 0 = 0.213 bar At design : C 1=
m
ρ 1 A1
=
9.0 3.294 × 0.0389 7.4
=70.24 m/ s s
=67.75 m/ s s 2.808 × 0.0389 4.47 × 100 P At design: T 1= 1 = = 472.8 K At part load: C 1=
R ρ 1
0.287 × 3.294
+
18.004
ρ 12
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T 01 P 01 01= P 1 T 1 ∆ P 0
=
P 01
0.27
γ / γ −1
475 = 4.47 472.8
4.543 3.52 × 100 0.287 × 2.808
439 P 01 01=3.52 436.78
P 01
= 4.543 bar
=0.0594
At part load: T 1=
∆ P 0
3.5
=
0.213 3.583
= 436.78 K
3.5
= 3.583bar
= 0.0595
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Problem 7.1
C 2=
C a cos 6 65 5
260
=
0.4226
= 615.2 m/ s s
2 2 C w s w2 2= 615.2 − 260 = 557.5 m/ s
Tan β 2 =
5 57.5 − 360 260
= 0.7596 ∴ β 2 = 37 13'
C w tan 10 10=260 ×0.1763 = 45.84 m/ s s w3 3=260 ta Tan β 3 =
360 + 45.84 260
=1.5609 ∴ β 3 = 57 22 '
Since C1=C3 and Ca is constant: C a
260
Λ = 2U ( tan β − tan β ) = 2 × 360 (1.5609 − 0.7596 ) 3
2
Λ = 0.289 Temperature drop coefficient:
ψ =
2C p ∆T 0 s U 2
=
2U ∆C w U 2
=
2 ( 557.5 + 45.84 ) 360
=3.35
Power output =mU =mU ∆C w = 20 × 360 × (557.5 + 45.84 ) P =4343760W =4343760W or 4344 KW C 2 T 2=T 01-
2
615.22 =1000 – = 835 K
2294
2C p
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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 =
∴
4
1000 = 826.75 = 2.14
2.140
= 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:
2 2 01 = × × 1000 = 857 K T 2.333 γ + 1
4.0 P c=
1.853
ρ c =
= 2.158 bar, T c=
P c 2.158 × 100 3 = = 0.877 kg/m RT c 0.287 × 857
C c= 1.333 × 0.287 × 857 × 103 = 572 .6 m/ s s Athroat=
m
ρ c C c
=
20 0.877 × 572.6
2 = 0.0398 m
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Problem 7.2 γ −1 1 γ ∆T013 = η t T 01 1 − P01 P 03
1 4 1 =0.88 ×1050 1 − = 147 K 2
T 03 03 = 1050 – 147=903 K With zero outlet swirl: W s=UC w w2 2=C p ∆T 013 And with free vortex, C wr = constant Hence at root (C w w2 2)r =
C p ∆T 013 U
=
1.147 × 147 × 103 300
Now at root, Λ = 0 ∴
∴ T 02 02-
Hence
C 22r 2C p C 22r 2C p
=T 03 03 –
=
T1 − T 3
C 32 2C p
2752 2294
∴ C 2r s 2r =642.6 m/ s
T2 − T 3
= 562 m/ s s
= 0 hence T 2=T 3
s ; T 02 02=T 01 01 and C 3=C aa3 3=275 m/ s
+ 147 = 180 K
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α 2 r = sin −1
562
= sin −1 0.8746 = 61
642.6
(C aa22)r = 642.62 − 5622 = 311 .6 m/ s s w 2 r
) − U r = 562 − 300 = 0.8406 311.6 ( C a 2 )r
tan β 2 r = ( C
β 2 r = 40 4 '
r t
(C aa22)t=(C =(C aa22)r =311.6 m/ s s and
(C w w2 2)t=
562
r r
= 1.4
=401.4 m/ s s
1.4
401.4 α 2t = tan −1 = tan −1 (1.2882 ) 311.6 ∴ α 2 t = 52°10' 2 2 C 2t s 2t= 401.4 + 311.6 = 508.15 m/ s
2752
With no exit swirl T 33tt=T 3r = 870 K 3r =903 – 2294 508.152 T 2t = 937.5 K 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
937.5 − 870
= T1 − T 3 =
147
At the root T 22r r -T ’’2r 2r = λ n
=0.46
C 22r 2Cp
=0.05 ×
642.6
2
2294
T 22rr – T ’2r ’2r = 9 K T ’’2r 2r =(1050 – 180) – 9.0 =861 K γ / γ −1
T = 01 P2 T '2
P01
P 03
γ / γ −1
T 03
4
3.8 1050 P = = 1.719 bar ∴ 2.21 = = 2 861 2.21
903
4
3.8/2
=
T '3
P3
= 870 = 1.16 ∴ P 3 = 1.16 = 1.638 bar
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Problem 7.3 C 22
4002
T 3=T 03 = (1200 − 150) − = 980.2K 03 – 2C p 2294 U m =2 π rm N or r m=
320 = 0.2037 m 2π × 250
1 1 4 ∆T013 = η t T 01 1 − P01 / P 03 1
1 4 ∆T 013 150 =1− = 0.8611 =1− / 0 . 9 0 1 2 0 0 × P P η T t 01 01 03 P 01 P 03
P 03 = 1.818 ∴ P 03=
ρ 3 = A3=
r t r r
= 4.4bar
1.818
γ / γ −1
T = 03 P3 T 3
P03
h=
8
P 3 RT 3 m
ρ 3C 3 A
=
2π r m
=
3.343 × 100
=
=
4
4.4 1050 = 3.343 bar = = 1.316 ∴ P 3= 1.316 980.25
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. (0.0591/2) 0.2037 − (0.0591/2)
=
0.233 0.1741
= 1.34
A2 = A3 m C a2 must
satisfy
C a2=
2
C a2=
ρ 2 A2 and
2 w2
C − C
W s=C w (because α 3 = 0 ) w2 2U =C p ∆T 013 C w w2 2=
1.147 × 150 × 103 320
s = 537.6 m/ s
If C a2 s C 22 = 3462 + 537 .62 a2=346 m/ s T 2=T 02 02-
C 22 2C p
T 22’’=T 2 – λ N P 2=
=T 01 01C 22 2C p
P 01 4
=
C 22 2C p
=1200-
s ⇒ C 2=639 m/ s
6392 2294
= 1022 K
=1022-0.07 ×178 = 1009.5 K 8.0 4
=
8 .0
= 4.008 bar
(T01 T 2 ' )
(1200/1009.5 )
1.995
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ρ 2 =
4 .008 × 100 0.287 × 1022 m
C aa22=
C2 A3
3
= 1.366 kg/m 36
=
1.366 × 0.0756
s (agrees with given 346 m/ s) s) = 348 m/ s (agrees
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 s
C aa22 is constant at 346 m/ s. s. Hence: 2 2 C 2r s 2r = 629 + 346 = 717.9 m/ s
T 22r r =1200 –
U R R=
320 1.17
7182 =975.3 K 2294
= 273.5 m/ s s
V 22r r = 3462 + (629 − 273.5) 2 = 496.1 m/ s s a2r = γ RT 2 r = 1.333 × 0.287 × 975.3 × 103 =610.86 m/ s s ( M M v2 v2)r =
496.1 610.86
=0.812
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Problem 7.4
C r2 r2=constant
(1)
C w w2 2r=constant
(2)
From (2), since C aa22=C w w2 2 cot α 2
r m r 2
C aa22r=constant and C aa22=(C =(C aa22)m
r rm 2
Also U =U m
Now,
U C a 2
= tan α 2 − tan β 2
Um r
2
∴ tan β 2 =tan α 2 - ( Ca 2 ) r m m tan α 2 =(tan α 2 ) m = ta tan n 58 58 23' =1.624
U m
( C a 2 )m
= tan 58 23’23’-ta tan n 20 29' =1.624-0.374=1.25
(N.B. Flow coefficient (C (C a/U )m=0.8 for this mean diameter design)
2
r Hence, tan β 2 = 1.624 − 1.2 5 r m 2
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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. x 2 2 C w w2 2 r =constant where x= sin α 2 = sin 58 23' =0.726 With
x α 2 = constant, we also have C a2 a2 r = constant. x
r m Hence, C aa22= (C (C a2 a2)m r 2 This with
r yields r m 2
U =U m
x +1
r M tan β 2 = tan α 2 - C r m 2 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'
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 7.5 m= ρ 3 AC 3=
=
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 3 P 3= P 01 01 T 01 m=
A 2C p R
×
P 01 T 01(γ / γ −1)
2
γ −1
T3 (T01 − ∆Tw − T3 )
γ +1
2 01 (γ / γ −1) 01
m= A 2C × P T R p
γ −1
T3 (T01 − ∆Tw ) − T3γ − 1
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 − T3γ −1 = 0 γ − 1
∴ T 3= 2 (T01 − ∆T w ) γ + 1 A 2C p Hence, writing K writing K =
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)
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 γ +1
2
γ −1
λ − 1 γ + 1
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But
C p R
mmax =
=
γ γ − 1
AP 01 T (γ / γ −1)
(T01 − ∆T w )
01
mmax
mmax
T 01
=
AP 01
2 γ γ −1 2 (γ − 1) γ +1 R γ − 1 −1 ( γ + 1) γ
γ 2 ∆Tw 1 − R γ + 1 T 01
γ +1 γ −1
=
γ +1 γ −1
2
γ +1 γ −1
AP 01
T 01
γ +1 γ −1
γ 2 ∆T 1 − R γ + 1 T w
γ +1 γ −1
γ +1 γ −1
T 01
γ +1 γ −1
T 01
γ +1 γ −1
01
Hence maximum mass flow will vary with N with N / T 01 because ∆T w varies with P with P 0011/ P P 0033.
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Problem 8.1 N =250 =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
45 4500 1570 70.5 .5) 00(15 2
2
( 0.2262
2
− 0.11312 ) ( 0.6 )
≅ 127.8 MPa Material UTS =880 =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
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Problem 8.2 N =250 =250 rev/s R =0.113m; R =0.105m 0
i
Width=0.025m Blade m ≅ 0.020 kg r cg cg ≅ 0.163m
= 4500 kg/m3, ρ =
Material
UTS = 880 MPa
For a thin ring rotor (eq 8.29)
σ t = ρω 2 Rr2ing +
Rring =
F rim 2π Aring
0.113 + 0.105 2
= 0.109m
Aring =0.025(0.113-0.105) 2
=2 ×10−4 m F rim ω 2 ) rim =43 ( mr ω = 4 3 ( 0.02 × 0.163 × 1570.52 ) ≅ 345, 750 N 2
2
σ t = 4500 (1570.5 ) ( 0.109 ) +
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
Cruise
5
220
28
1000
100,000
5 × 10
Descent+L
0.5
150
29.5
1050
124,520
4 × 10
t 1%
-4
-5
-6
-4
2.029 × 10
∴ 4,929 flights PLM is obtained from figure 8.7, for CMSX-10
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Problem 8.4 Haynes 188 at 1033K −0.06
∆ε = 0.01 N
f
−0.72
+ 0.7 N f
Source:: Manson, S. S. and Halford, G. R., Fatigue and durability of structural materials (ASM International, 2006), Source where it is Figure 6.15(c). Reprinted with permission of ASM International ®. All rights reserved. www.asminternational.org
→ Nf1 = 2 ×103 cycles ∆ε 1 = 0.009
N ff22 = 1 ×104 cycles ∆ε 2 = 0.007 → N Using Miner’s rule (linear damage accumulation) n N f 1
1=
+
n N f 2
=1=
n 2 × 10
1 × 10 4 n + 2 × 103 n 2 × 10 7 n
3
+
n 1 × 10
4
12 × 103 = n 7 2 × 10
∴ n ≅ 1,670 cycles of each loading.
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Problem 8.5 Rigid bearings : K : K r r = rotor stiffness 1
ω =
1
2 K r = N = kgm = 1 2 m mkg s s mkg 2
m=20 kg ; ω = =
π N 30
= 5,235 rad/s
1
K r 2
⇒ K r r =548 × 10 N/m 20 6
5,235=
Soft bearings: total stiffness K r + K s 548 × 106 + 1 × 107 = = K K K 548 × 106 × 1 × 107 1
t
r
s
= 1.018 × 10 –7 K t = 9.821 × 106 N/m First critical speed
ω = =
9.821×106 20
= 700.7 rad/s = 6,693 ≅ 6, 6,700 700 rpm
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 deali ng with wit h stagnation s tagnation conditions at all al l times ti mes and the suffix ‘0’ has dropped throughout.
Flow compatibility is expressed by: m T 1 P1
=
m T 3 P3
P 3 T 1 × P1 T 3
×
And P3 P1
∴
P 2 P 1
=
P2 P1
−
P2 − P 3 P1
=
P2 P1
− 0 .0 5
P2 P1
= 0 .9 5
P 2 P 1
m T 1
288 P P =14.2 ×0.95 2 × = 6.903 2 P1 1100 P P 1 1
6.903
P 2 P 1
m T 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 =4.835, P 1
m T 1 =33.4, P 1
η c = 0.795
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P 3 P 1
=0.95 ×4.835 =4.59
And from turbine characteristic graph: η t = 0.8497 m=
3 3.4 × 1.01 288
= 1.988 kg/s
Net power output = mC pg ∆T34 −
1
η m
mC pa ∆T12
1 4 1 P output 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 output = 675.189 – 411.620 = 263.989KW
Net power output = 264 KW
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Problem 9.2
m T 4
P4
=
m T 3 × P3 × T 4 P3 P4 T 3
T 4
∆T = 1 − 34 T3 T 3
∆T 34
= η t 1 −
T 3
; 1 r 4 1
r=
P 3 P 4
; and η t = 0.85
1
P 3 4 P 4
∆T 34
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 P 4
∴ ∴
m T 4 P 4 m T 3 P 3
=188
=90.2
T 3
when
P 3 P 4
T 4 T 3
T 4 T 3
=2.270 (graphical solution)
m T 3 P 3
m T 4 P 4
1 1 4 34 T 3 = 0.85 1 − 2.27 = 0.1575
∆T
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From compatibility of flow: m T 1
=
P1 T 3
∴
T 1
m T 3 P3
=
(m
×
P 3 T 1 × and P 3= P 2 and P P1 T 3
) ( P / P ) / m T P ( ) T3 / P3
2
1
1
1
90.2 ( P2 / P 1) = T 1 m T1 / P 1
T 3
(
-----------------------------(A)
)
From compatibility of work:
∆T T1
=
∆T12 T1
P 2 P 1
∆T T3 C pgη m
=
T3 T1 C pa
=
1 P 2
P 1
T1
1.005
= 0.1762
T 3 T1
-------(B)
0.286
η c P 1
m T 1
0.1575 × 1.147 × 0.98 T3
− 1
T 3 T 1
T 3 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
Hence T 3=1170 K
T 3 T 1
= 4.06, from graphical solution
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Problem 9.3 At outlet of gas generator turbine m T 4 = m T 3 × P 3 × T 4 P4 P3 P4 T 3 1 4 T 4 1 ; = 1 − η t 1 − and P3 / P 4 T3
∴
m T 4 P4
=
m T 3 P3
1 4 P 3 1 × × 1 − η t 1 − ; P4 P / P 3 4
η t = 0.85
1
1
P 3 P 4
P 3 P 4 4
1
1 1 − P P 3 4 4
2 4 1 m T 3 m T 4 P 4 1 − η t 1 − P a P P P 3 P 4 / 3 4 1
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
The value of P of P 4/ P P a in the table is found from P P 4 P4 P 3 P 2 P4 = = × 0.96 × 2.60 = 2.496 4 P 3 P a P3 P2 P a P3 Hence curves – from which power turbine pressure ratio is r is r = 1.55
104.90
1.387
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And gas generator turbine pressure ratio r=1.61 Use work compatibility equation to find T 3
∆T 12 T1
=
∆T34 T3
T × 3 T1
C pgη m C pa
1 1 4 3.5 1 T1 P 2 ∴ C pa − 1 = η mC pg T 3 1 − P / P η c P1 3 4
T1 must be given. Compressor efficiency from compressor characteristic once operating point found from
m T 1 P1
=
m T 3 P3
×
P 3 T 1 where T3 unknown, so trial and error method required. × P1 T 3
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 9.4 For gas generator turbine m T 4
P4
m T = 3 × P3 × T 4 P3 P4 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 = f from gas generator turbine characteristics. P 4
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 T 95% speed 3 100% speed
But:
Therefore; T 3= 1075 ×
0.5465
×
0.863
= 1214.5 K
0.859
0.486
(b) 95% mechanical speed at 273 K.
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
N T 1
(% design) = 95 95
288 273
From the operating line
=97.57 %
m T 1
= 439,
1 P
∆T =
m=
p2
= 4.29 and η c = 0.862
1 P
273
(4.29)0.286 − 1 = 163.6K 0.862
4 39 × 0.76 273
= 20.19 kg/s
Power=20.19 × 1.005 × 163.5 = 3318 kw.
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Problem 9.5 Work compatibility yields: 1 −
1 γ γ mη t T3 1 − P3 / P4 P3
Let:
P4
=
P 2 P 1
= m T1 c η c
3.5 P 2 − 1 P 1 1
= r
At design point:
31.5 31.5 T1 r − 1 4 − 1 mc − mb 1 1 1 = = × × 1 1 mc ηtη c 0.8 × 0.85 3.3 1 3.5 1 3.5 1− T 3 1 − r 4 m mc mb mc
=
mc − mb mc
= 0.6622 mb
= 1 − 0.6622 = 0.3378 =
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 (mc − mb ) T3 3 c b 1 = 1 P 3 1 − 2 P3 1 − r 2 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 = =
1 r 2 Thus : P : P 3= P 2 1 2
r d
r 2 − 1
22.8 7.7
4 1 − 1 / 16
15
mc T1 / P1 − mb T1 / P1 = 3.899 r 2 − 1 --------------------(1) --------------------(1)
82 © Pearson Education Limited 2009
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HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory , 6 edition, Lecturer’s Solutions Manual
Also from work compatibility: 3.5 5 mc T1 / P1 − mb T1 / P1 0.4 457(r 1 3. − 1) = ------------------(2) ------------------(2) 1 mc T1 / P 1 3.5 1 1 − r
Now: mb T1 / P = P 1=
3 4
× 7.7 = 5.775
From (1): mc T1 / P1 = 5.775 + 3.889 r 2 − 1 (2) becomes: 5.775
1−
2
5.775 + 3.899 r − 1
r
r 2
r 2 − 1
3.5
12.25
3.35
3.0
9.0
2.5 2.25
=
3.5 5 0.4457(r 1 3. − 1) 1 3.5 1 1 − r
, solve by trial and error
3.899 r 2 − 1
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
3.5 5 0.4457 r 1 3. − 1
1 1 − 1 3.5 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
right-hand-side
