0367_GTurbineT_6ebySaravanamuttoo.pdf

Lecturer’s SolutionsManual Gas Turbine Theory Sixth edition HIH Saravanamuttoo GFC Rogers H Cohen PV Straznicky For further instructor material please visit: www.pearsoned.co.uk/saravanamuttoo ISBN: 978-0-273-70934-3  Pearson Education Limited 2009 Lecturers adopting the main text are permitted to download and photocopy the manual as required. Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies around the world Visit us on the World Wide Web at: www.pearsoned.co.uk ---------------------------------This edition published 2009 The rights of HIH Saravanamuttoo, GFC Rogers, H Cohen and PV Straznicky to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. ISBN: 978-0-273-70934-3 All rights reserved. Permission is hereby given for the material in this publication to be reproduced for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without either the prior written permission of the Publishers or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6-10 Kirby Street, London EC1N 8TS. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the Publishers. HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual Supporting resources Visit www.pearsoned.co.uk/saravanamuttoo to find other valuable online resources For instructors • PowerPoint slides that can be downloaded and used for presentations For more information please contact your local Pearson Education sales representative or visit www.pearsoned.co.uk/saravanamuttoo 3 © Pearson Education Limited 2009 HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual Preface Since the introduction of the Second Edition in 1972 many requests for solutions have been received. The advent of modern word processing systems has now made it convenient to prepare these in an electronic format and I am glad to do so. All significant gas turbine calculations carried out in industry are universally done by digital computer and the purpose of these problems is to provide an understanding of the engineering principles involved. It is perhaps of historical significance that all of these calculations were originally done on a slide rule and many were previous examination questions. This manual will be available to instructors adopting the main text, who are then permitted to photocopy the material, but it is hoped that students will tackle the problems before looking at the manual. I will be very glad to hear of any corrections needed. My sincere thanks to Mr.Hariharan Hanumanthan, a doctoral student at Cranfield University, for his invaluable assistance in the preparation of this manual. H.I.H. Saravanamuttoo Ottawa, June 2008 4 © Pearson Education Limited 2009 98 ×1.HIH Saravanamuttoo.82   Compressor and turbine work required per unit mass flow is: Wtc = C pa (T02 − Ta ) T03 – T04 = ηm = C pg (T03 − T04 ) 1.147 T04 = 1150 – 309 = 841K  γ −1      1  γ    T03 − T04 = ηt T03 1 −    P P   03 04   1     4   1   308.0 − 0. Gas Turbine Theory.382 P04 P03 = 11. Lecturer’s Solutions Manual Problem 2. 6th edition.4 = 10. H Cohen.992 = 0.254 K T04 − T05 = 0.87 × 1150 1 −     P P   03 04   P03 = 4.5  – 1 = 345. PV Straznicky.992 K 0.2  γ -1      Ta  P02  γ   – 1 T02 – Ta =   ηc  Pa     1  288      3.89 × 841 1 −    2.598K 11 = ( )   0. P05 = Pa 1     1  4    = 148.418     5 © Pearson Education Limited 2009 .6 bar P04 = 2.005× 345.598 = 308. GFC Rogers.418 bar . C = 3600 f 3600 × 0.01415 (from Fig. H Cohen.01429 S. 2.254 = 166.64 T02 = 288 + 345. 6th edition. Gas Turbine Theory.01429 = = 0.64 kWs/kg Hence mass flow required = 20 × 103 = 120. GFC Rogers.64 6 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Specific power output: WN = 1. PV Straznicky.98 × 148.4K Theoretical f = 0.6K T03 − T02 = 1150 − 633.6 = 633.15) Actual f = 0.019 kg/sec 166.HIH Saravanamuttoo.01415 0.99 = 0.147 × 0.F.6 = 516.308kg/kW − hr WN 166. 8 − 0.rotor ) 200 = 0.99 = 633.1 − 156.73m − 433.88[1 − ( 1 14 ) ] = 256.147 × 256. PV Straznicky.788 kg/sec With no bleed flow: Net work output = 0.87 − 1.8 3.5) × 1.11kW 7 © Pearson Education Limited 2009 .3 1 T02 − Ta = 288 [3. GFC Rogers.68 Net work output W = ηm ( load ) [( m − mc )C pg (T03 − T04 ) − mC pa (T02 − Ta ) ηm ( comp.68 Pa T03 − T04 = 1050 × 0. 6th edition.HIH Saravanamuttoo.98 × 4.12 = 3.5 − 1] = 157. Gas Turbine Theory.3K 0.788[1.147 × 256.68bar P03 = 3.005 × 157.87K 3.98[( m − 1.3 ] 0. H Cohen.005 × 157.85 P03 = 3.87 − ] m × 1.11 kW (b) The power output with no bleed = 633.99 200 = 288. Lecturer’s Solutions Manual Problem 2.5m (a) m = 4.3 ] 0. 2205 T02 T01 2.9 T02 (K) 593 645.8 701.9 P03/P04 8.0219 0.3287 0.2 Wt= 1.6 786.HIH Saravanamuttoo.8% 0.7 Wc= (1.0162 0.0268 mf = ma × f × (1 − B ) 4374 6150 7791 SFC (kg/kwhr) 0.99 309.4% +116% 557 754 898 0. GFC Rogers.2 ∆T034 436. 6th edition.6 879.4 A B C  n −1    n c 0.2205 0.6 580.4 365.242 2.3250 0.912 31.2 420.820 T03 1150 1400 1600 T04 713. Lecturer’s Solutions Manual Problem 2.304 base 0.268 11.2 1.005 ×∆T12 ) / 0.40 15.4 819.6 286.4 720.0 Wnet=Wt–Wc 191.059 2. Gas Turbine Theory. H Cohen.3213  n −1    n t 0.55 11.2 649.251 17.437 ∆T012 (K) 305 357.708 1.2225 0.6 363.8 Power 14.370 22.093 Base +59.148(1 − B) ∆T034 501. PV Straznicky.4% 440 547 606 P  T03/T04=  03   P04  n −1 n ∆Tcc f/a EGT ( D C ) 8 © Pearson Education Limited 2009 .612 1.8 413. 4 256.9 So ηth remains high as power reduced.0085 0.2 573.1 214.6 608. H Cohen.2223 = 1.3 907.0 1325 759. May be difficult to control low fuel: air ratio in 2nd combustor.284 )0.284 P04 ∴ T03 = ( 2.1 MW (f/a)1 (f/a)2 ηth 240.0282 458.223 = 1.7 0.0197 0.0197 0.9 kg/s 458.2 0.69 bar P04=13. GFC Rogers.7 189.7 256.3 T05 T06 T05–T06 T03–T04 ∆T total ∆T total ×1.0 934.99 – ∆Tc × 1.24 ) P06 T06 1.0050 0.HIH Saravanamuttoo.0 1425 816. 6th edition.0 0.148 × 0.0 458.2 ∴ m for 240 MW = 240000 = 523.7 573.00 bar ∴ P03 = 2.3 1031.3 822.3 565.5 0.0247 ×43100 38.3 864. PV Straznicky.3 P05 13. Gas Turbine Theory.0227 ×43100 36.00 × 0.2 0.6 0.0 256.24 ∴ 05 = (12. 9 © Pearson Education Limited 2009 .0030 0.0197 0.0282 ×43100 37. Lecturer’s Solutions Manual Problem 2.7 K.004 1525 874 651.96 T 0.1 409.0227 361.5 T03=1525 K P03=29.7 361.1 573.7 982. ∆Thp = 256.0247 409.202 T04 T04=1268.745 = = 12.02 T03 – T04=256. H Cohen.9= 0.1 = 38.99 ∆T67 = 206.89 × 1300 1 − 0.8% 0.00 ×5.3 = 100. Lecturer’s Solutions Manual Problem 2.3 =701.89 × 1117. Gas Turbine Theory.25  ∴ m ×1.148 × 0.248  R  CDP=1.7 T4 = 568. PV Straznicky.25  Rhp=2.95 = 39.HIH Saravanamuttoo.95 × 0.286 − 1 = 206.250.19 bar P6 = 39.248 1   LPT. 6th edition.0283 0.000 ∴ m=211.028 = 0.99 × 416.8K 0. GFC Rogers.43 bar 2.1K 1   HPT.3 kg/s f/a = 0.148 ×0.286 ( 7.7K  0.1 1 − 1  = 416.99 Specific output =1.5) − 1 = 268.5 × 0.005 = 182.148 × 0.148 × 0. 182.1 T6=1550 – 250 = 1300K 1.9 = 1117.3 K 8.8 × 1.6 ∆T12 = 288 (5.1= 0.8 T2 = 494.283 × 43100 EGT =1117.19 = 17.00   ∴ ∆T78 = 0.870 ∆Tcc = 1550 − 569 = 981D C ∆T56 = 268.7-416.99 T7=1300 – 182.875  ∆T34 = 300  0.7 × 1.3 = 473.990 P7=8.1 ∴ ηth = 473.4 D C 10 © Pearson Education Limited 2009 .5 × 7.25  RLP=1.005 = 250.88 × 1550 1 − 0.4 K=428.5)0.76 bar R   P8=P1=1.99 × 416.760.9 1. 100 kJ kg = 67.99   P 1  ⇒ 4 = 2.106 kW T3 – T2=0.013 ×8.0139 ×112.5 × 1.1 ∴ T6=824.013 bar .56 kg/s Heat input = 1. PV Straznicky.99 × 213.0139 0.HIH Saravanamuttoo.02 = 1.5 × [1 − 0.87  T02=567.125 bar ∆T45 = 279. Gas Turbine Theory.7 K 0.8 1 − = 213.50.1 = 27.0138 =0.5 K P04=1. P5 = 2.895   ∴ Power =112.042] = 8.5)=231.5=799K ∆Tcc = 1285 − 799 = 486D C Tin=799-273=526 D C (f/a)id=0.5+567.8 Pa=1.87 ×1285 1 − 0.0 =1.716.147 × 0.3 % 11 © Pearson Education Limited 2009 .147 × 0.0 ×1.991 bar 247.1 T5 = 1037.90(824.25 P5  ( P 4 / P5 )  ⇒ P6=1.015] × [1 − 0. Ta=15 ∆T12 = D C 288 8.286 − 1 = 279.99 mf= 0.5 T3=231.013 × 1. H Cohen.288 kW kg s ∴ Efficiency =40. GFC Rogers. 6th edition.25  2.88 × 1037.895 P6 1   ∴ ∆T56 = 0.7-567.8 K 1. Lecturer’s Solutions Manual Problem 2.0333 P5 = 2.56 × 43.1=0.5k 0.004 = 247.0138 f/a= 0. 0 − 0.66 γ P02 nn−1 ) − 1] = 310[20.32 bar P06 = 55.88 η ∞c γ × 1.3498 ) ] = 453.34 = 27.34 + 0.3K T06 − T05 = ∴ T05 = 700K and ( given) Q 500 × 103 = = 535.4518 − 1] = 110.19 T06 = 1235.82K 12 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 2.66 = 55.3K T04 = 410.697 P07 T06 − T07 = 1235.66 ( )= = = 0.2K P03 = 2 × 14.02 bar P07 = 14.4518 n 0.27 + 1.61bar ∴ P06 = 3.32 − (0. Gas Turbine Theory.2[1 − ( 1 0.27 = 14.697 T07 = 781. 6th edition.66 n −1 γ − 1 0.66 bar P04 = 2 × 27. H Cohen.3K 3.0 + 0. PV Straznicky.88 × 0.4518 − 1] = 114K P01 T04 − T03 = 300[20.9 For compression: For expansion: T02 − T01 = T01[( n −1 1 γ −1 0.66 )= = η ∞t ( = 0.3498 n 1.03) = 54.2 K mC p 180 × 5.HIH Saravanamuttoo. GFC Rogers. Gas Turbine Theory.7798 T07 − T04 781.3 or 78% 13 © Pearson Education Limited 2009 .E. PV Straznicky.19 × 229. GFC Rogers. H Cohen.996MW Thermal efficiency = H.8 − 410.4279 500 or 42. Lecturer’s Solutions Manual Power output = mC p [∆T067 − ∆T034 − ∆T012 ] = 180 × 5.996 = 0.0 = 213996 kW or 213.effectiveness = 213.3 = = 0.8% T05 − T04 700 − 410. 6th edition.HIH Saravanamuttoo. 63K 2c p 2 × 1.4   4.068(1 − 0. PV Straznicky. 6th edition.87 = = 0.147 × 0.633bar 242.411 14 © Pearson Education Limited 2009 .5 ] = 0.3284 − 1] = 270.3284 n 0.7 P02 = 8 × 0.16 Pa 0.068bar P01 = 0.763bar 4. H Cohen.99 P03 = 5.33K 2 P01 = Pa [1 + ηi γ Ca γ −1 ] 2c pTa 0.4K P03  T03  n −1  1200  =  =  P04  T04   960.87 × 3.00.HIH Saravanamuttoo.633 = 5.02K T03 − T04 = c pa (T02 − T01 ) c pgηm T04 = 1200 − 239.33[8.7K 2 Ca 2602 = = 33.005 × 270.7 + 33.763  P04 =   = 1. Ta = 242.411bar.784  Nozzle pressure ratio 1. GFC Rogers.7 = 239.06) = 4.95 × 33.710 bar  2.5977 = 2. Lecturer’s Solutions Manual Problem 3.63 3.710 = = 4.7K T02 = 547.5 For compression: For expansion: n − 1 0.005 × 1000 T01 = 242.6K 1. Gas Turbine Theory.1 n −1 1 = = 0.2175 n 4 and n = 4.6 n = T04 = 960.784 P04 1.5977 n −1 At 7000 m Pa = 0.63 = 276.441[1 + T02 − T01 = 276. PV Straznicky.95  2.0708m 2 ρ5C 5 0.377 × 561. 6th edition.02 = 652.333 Pc 100 × 0.3 × 1000 ) 2 = 561.333      P5=Pc= 1.3 kg/m3 1 1 C5 = ( γ RTc ) 2 = (1. GFC Rogers.891 − 0.891 = = 0. Lecturer’s Solutions Manual (Since the critical pressure ratio is of the order of 1.7 15 © Pearson Education Limited 2009 .01785 = 0.0708(0.4 = 7916.0184 0.411)105 F = 4518.377 RTc 0.97 0.333   1 − 0.01255 kg/kN 7916.287 × 823.0184 × 3600 × 15 = 0.01785 (Fig. Gas Turbine Theory.22 F = m(C j − Ca ) + Aj ( Pj − Pa ) F = 15(561.710 = 0.3K γ +1 2.891 bar 1.15) Actual f = S .02K T03 − T02 = 1200 − 547.333 × 0. H Cohen.HIH Saravanamuttoo.F .7N T02 = 547.918 4 Pc  1  0. 2.918 T5 = Tc = ρ5 = 2 2 × 960.4 T04 = = 823.C = 0.9 the nozzle is clearly choked) P04 1 = =1.22 − 260) + 0.3 + 3398.287 × 823.22 m/sec A5= m 15 = = 0.8K Therefore theoretical f = 0. 7 = 64.658 = = 0.2 From problem 3.1756 × 809.83=13027. 6th edition.918 Since T05 = 2000K.710 bar P05 = 1.03 N Percentage increase in thrust = 13027. Gas Turbine Theory.88–260)+0.8644 = 0. H Cohen. Lecturer’s Solutions Manual Problem 3.03 − 7916.333 × 0.411) ×105 F=8248.97% F=15(809.1 P04=1.17(1 − 0. ρ5 = T6 = Tc = 2 × 2000 = 1714.1054 0. GFC Rogers.1054-0.8644 bar Pc 1.1054(0.8644–0.HIH Saravanamuttoo.88 A5 = 15 = 0.5K 2.56% 7916.658bar P05 = 1. PV Straznicky.5 kg/m3 1 C5 = (1.287 × 1714.03) = 1.88 m/s m2 Percentage increase in area = 0.287 × 1714.1756 0.7 16 © Pearson Education Limited 2009 .333 100 × 0.0708=48.5 × 1000) 2 = 809.918 Pc as before P05 1.2+4778. 955 P04 P03 = 9. H Cohen.5  − 1 = 306.678K 0.333   1 −  0.55bar P04 = 2. PV Straznicky.87 × 1275 1 −    P03 / P04     ⇒ P03 = 3.85mc pg (T03 − T04 ) = T03 − T04 = mC pa ηm (T02 − T01 ) 1.95  2.HIH Saravanamuttoo.82   0.678 = 0.77 = 322.005 × 306.98 γ −1   γ   1   T03 − T04 = η jT03 1 −    P03 / P04     1    1 4   322. Lecturer’s Solutions Manual Problem 3.0 − 0. 6th edition.147 × 0.77K 9  0.333      So nozzle is choking even while landing 17 © Pearson Education Limited 2009 .3 Since P01 = Pa = 1 bar T01 = Ta = 288 K T02 − T01 = 1 288  3.85 × 1.161bar P04 . GFC Rogers.161 Pa and P04 1 = 4 Pa  1  0. Gas Turbine Theory.45 = 8. = 2. 13 × 558.87 ×558. PV Straznicky.02 0. Gas Turbine Theory.HIH Saravanamuttoo. Lecturer’s Solutions Manual 2.13(1.480 kg/m3 RTc 0.333 × 0. GFC Rogers.480 × 0.34 N or F=18.4K  × T04 = 2. bleed at 90º produces no thrust.678 = 952.85 (N.918 T04 = 1275 − 322.86 = 34. – mb. H Cohen.B.4 1 1 C5 = ( γ RTc ) 2 = (1.87 kN 18 © Pearson Education Limited 2009 .1)+1638 F =18869.4 × 1000 ) 2 = 555.333  γ +1 P 100 × 1.126 − 1) × 105 F = (19487.287 × 816. Also.87 kg/sec mintake = 34.02 × 55) + 0.322 T5 = Tc =  = 816.86 m/s (m − mb ) = ρ5 A5C5 = 0.3K P5 = Pc =  2  2 × 952.87 = 41. ram drag based on intake flow) F = (34.126 ρ5 = c = = 0.287 × 816.44-2256.126bar 1. 6th edition.86 − 41.161 = 1. H Cohen.01)105 F = 13171.851 T04 γ + 1 2.296 bar 1.851 Pc  2   2  2.4 Choking . PV Straznicky. Gas Turbine Theory.376 Pa 1. Lecturer’s Solutions Manual Problem 3.333 ρ5 = Pc 100 × 1.287 × 857.526 RTc 0.296 − 1.526 × 572.35kN 19 © Pearson Education Limited 2009 .68m/sec A5 = m 23 = = 0.82N or F = 15.68 + 0.4 Isentropic expansion in nozzle: P04 2.26 × 1000 ) 2 = 572.68 F = 23 × 572.0763m 2 ρ5C5 0. 6th edition.287 × 857.26 K Tc 2 2 2.296 = = 0.333 × 0.4 = = 2. So P5 = Pc = = 1.333 1000 × 2 and T5 = Tc = = = = 857.01 γ 4 P04  γ + 1  γ −1  2.64 + 2182.HIH Saravanamuttoo.0763(1. GFC Rogers.26 kg/m3 1 1 C5= ( γ RTc ) 2 = (1.333  =  =  = 1.18 = 15353. 4 = 1.21 × 1000) 2 C6 = 442.005 × 56. GFC Rogers.168.43 = 900.12 = 10.43K 1. Gas Turbine Theory.147 T05 = 1000 − 99.5  288   3.893 20 © Pearson Education Limited 2009 .2K 1 1 C6 =  2C p (T05 − T06 )  2 = (2 × 1.01 × 1.4 + 1  =  Pc  2  3.75 Pa and P07 = 1.43 = 0.HIH Saravanamuttoo.88     mh C pg (T04 − T05 ) = mc C pa (T07 − Ta ) But mc = 2.1045 T05 − T6 = 85.597 99.8 N The cold nozzle pressure ratio:=1.74 K −1 = T07 − Ta = 1.5 − 1 = 56.5 = 1.75 ( )       ηc  Pa  0.57 T6 = = 815.90 × 1000 1 −    P04 P05   P05   2. so P6=Pa=1. Lecturer’s Solutions Manual P07 = 1. H Cohen.147 × 85.503 = = 1.12m/s Hot nozzle thrust: Fh = 23 × 442.1045 T6  P6  900.597 P05 1.488 = 1. PV Straznicky.36 K 1.75 = 1. while the critical pressure ratio is : P07  1.503bar P05 = 1. 6th edition.0 mh 2 × 1.74 = 99.01 Hot nozzle is now unchoked.768bar 1   1 Ta  P07  3.75.01 bar γ −1 1 T05  P05  γ 4 =  = 1.488 Pa 1.57 K T04 − T05 = 1   4   1   ⇒ P04 = 1. Lecturer’s Solutions Manual So this nozzle is also unchoked.74 = 293.89 kN 21 © Pearson Education Limited 2009 . PV Straznicky.94 K T8 = and T07 = 344.74 1 1 C8 =  2C p (T07 − T8 )  2 = ( 2 × 1.1733 T8 288 + 56.005 × 50. Since the expansion is isentropic: 1 T07 = 1.HIH Saravanamuttoo. Gas Turbine Theory. 6th edition.8 K 1.1733 T07 − T8 = 50. GFC Rogers.8 = 24888 N or F=24. H Cohen.94 × 1000 ) 2 = 320 m/s Cold thrust Fc = 46 × 320 = 14720 N Total thrust: =14720 + 10168.75 3.5 = 1. 463 N S.808 N Total thrust = Unchanged Fh+new Fc = 18931+87808=106.5     T09 – T8 = njT09 1 −  1     P09 Pa     1 3.05=1.739 N T02=337.C = 0.5 From solution of turbofan example in Chapter 3: T03=800 K T04-T03=1550 – 800=750 K Actual f = 0.0221 = 0. Lecturer’s Solutions Manual Problem 3.99 m = 35.7 and T09 – T02=1000-337. Gas Turbine Theory.0403 kg/Nh 71463 P09=P02 – 0.F.83 kg/s and Fs=71.60 bar  P  P So nozzle is unchoked and 09 = 1. H Cohen.3 K 22 © Pearson Education Limited 2009 .005 × 119. 6th edition.6   1 C8= ( 2 × 1.0223 0.60  < 05  Pa  Pc  3. PV Straznicky.0223 × 35.65-0.95 × 1000 1 −    =119. GFC Rogers.2 ×490 = 87.2 kg/s New thrust: Fc=179.7=662.5  = 0.4 × 1000 ) 2 = 490 m/s mc=179.4 K   1.05=1.83 × 3600 = 0.HIH Saravanamuttoo. 01715 Actual f = S.1338 kg/kN 23 © Pearson Education Limited 2009 . 6th edition.01768 × 179. GFC Rogers.83) × 3600 = 106. Gas Turbine Theory. PV Straznicky.HIH Saravanamuttoo. 2.97 TotalFuel ( (0.2 + 0.01715 = 0.01768 for by-pass chamber 0.15. H Cohen. Lecturer’s Solutions Manual From Fig.C=0.739 Totalthrust S.F.0223 × 35.C= 0. theoretical f = 0.F. 852 2.333    =   = 0.022 bar T4=1105.164 ∴ choked Pa 0.852 P04 3.724 4.7 2 ) = 309.70 1  0.744 bar 1.4   n −1  n −1  0. GFC Rogers.899bar Ta=281. 6th edition.366 × 5.454 bar ∴ P04 = γ  γ + 1  γ −1  1.3210 = ( 5.61  0.1 = 1105.744 = = 4.2 × 0.2248 ∴ 03 =  = 1.4 P01  0.724  P04  1105.7(1 + 0.945 = 6. Gas Turbine Theory.4 m/s M=0.9 K 1.6 Pa=0.9  P03=1.9 × =948.3210   = 0.148 × 0. Lecturer’s Solutions Manual Problem 3.228 bar T02 0.899 P4= 3.676 ∴ T02=472.0 × 0.454 = 3.2248 1 P  1250  0.4 = 1 +  = 1. H Cohen.2248  n c 0.0 K 1.89  1.7  Pa  ∴ P01=1.9 × 1.R =    =  2   2  6.3K ∆Tr = 27.7K Ca=336.4   n t  1.1 T04 = 1250 − 144.005 = 144.90   = 0.744 2 = 2.333   γ −1 2  T01= Ta 1 + M  = 281.0 ) = 1.95 × 27.2 T01 ∆T12 = 162.HIH Saravanamuttoo.366 281.9D C ∆T34 = 162.003 =1. PV Straznicky.61D C 2   1.99 T03  P03  =  T04  P04  0.333 24 © Pearson Education Limited 2009 .333  Critical P. 02141 0.02141 × 3600 = 609.0212 = 0.0 = 602.70 × 336.01767 × 105 =2900 N+1984 N=4884 N ∆Tcc = 1250 − 472 = 778.5 m/s F= m ( C j − Ca ) + ( Pn − Pa ) A × 105 =7. H Cohen.15) 2 = 0. PV Straznicky. Tin=472 K → ( f / a )id = 0.01767 × 602.2 m/s A= π 4 ( 0.2 = 7.7432 kg/m3 0.287 × 1000 × 948.5 ) + ( 2.0212 ≈ f/a = ∴mf= 7.022 − 0.91 ( 602.99 .HIH Saravanamuttoo.022 = 0.8 = 0.91 kg/s Ca= 0.8 kg/hr ∴SFC= 609.7432 × 0. GFC Rogers.287 × 948.0 C = γ RT 4 = 1.333 × 0.899 ) × 0.01767 m= ρ AC = 0.2 − 235.91 ×0. Lecturer’s Solutions Manual ρ= 100 × 2.4 = 235. Gas Turbine Theory. 6th edition.1249 kg/Nh 4884 25 © Pearson Education Limited 2009 0. 004 × 265.4757   = ( 7.4803 = 2.7 ) n = 1.7 Ca=0.8 K 1.7 K 1.8 + 1) = 187.286 = = 0. FPR=1.92 T06=797.99 n P04  T04  n −1 =  P05  T05  n −1 = 0.75 ×295. BPR=3.08K 2   ∆Tr = 24. Lecturer’s Solutions Manual Problem 3.3178 n 0.1837 T01 3.5 = 1. Gas Turbine Theory.752 ) = 241.95 × 24.3019 1.427 ∴P01=0.38°C P01  0.90 ∴ T02=285.HIH Saravanamuttoo.28(3. PV Straznicky.147 × 0.2558 bar n − 1 0. H Cohen.9287 T03=550.2 × 0. GFC Rogers.4400 HPC  T03  0.286 × 0.004 × 44.8.3 m/s  γ −1 2  T01= Ta 1 + M  = 216.75.7(1 + 0.38 P03=3.7 M=0.3178 = 1.9 ) T  02  ∆T023 = 265.1 = 221.2232 n n = 4.147 × 0.7  4.02 HP turbine ∆T45 = = 234.7  Fan n −1 T02 = (1.9 = 0.005)=3.60 26 © Pearson Education Limited 2009 .36 ∆T f = 44.02 Combustor P04=3.38  = 1 +  Pa  216.99 LP turbine ∆T56 = 1.32 T05=985.4803 n −1 P04  1220  =  P05  985.4757(1 – 0. 6th edition.28K P02=0. 287 × 237.579 Cold Stream (both choked) Hot Stream P02 = 0. H Cohen.00 4. Lecturer’s Solutions Manual P05  985.7  =  P06  797.8K P2 = 0.3 + 383.0 ( 511.3405 × 309.0 = 0.8  ∴ P06 = 4.00 × 0.8 = 7.1355 × 511. PV Straznicky.0186 × 3600 = 133.5 m/s A= 2.4=1050.7 27 © Pearson Education Limited 2009 .893 P6 = T2 = 285.02886 ( 0.333  ρ= 100 × 0.60 × 2.7N T04 = 1220.9 = 511.4803 = 2.6 = 0.HIH Saravanamuttoo.2  2  T6 = 797.287 × 683.8 1.6 ( 309.4400 T02=285.4 P06 = 0.3) + 0.2324 1.4924bar T06=797.63 kg/hr SFC= 133.4400 = 0.8   = 683.5 Fc = 7.8 mh = Cc = 1.1355 0.6 × 1 = 2.4 × 287 × 237.2324 = 0.4 = 237.1793) × 105 = 667.60 4. Gas Turbine Theory.9  2.0711 kg/Nh 1880.63 = 0.852 100 × 0.4 × 287 × 683.7 Fh = 2.1 − 221.2659 − 0.579 3.1 0.5 − 221.2659 = 0.8 = 309.3405 0. T03=550 → f=0.4924 bar 2.2324 − 0.1 m/s A= 7.02886 m2 0.6× 3.2659 1.8 Ch = 1.3) + 0. GFC Rogers.0=830 F = 1880.0722 m2 0. 6th edition.1793) × 105 = 580 + 250.0186 mf = 2.8 ρ= mc = 9.9 9.0722 ( 0.4924 = 0.3019 = 0. HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual Problem 3.8 M=0.85; BPR=5.7; FPR=1.77  n − 1  0.286 = 0.3178   =  n c 0.90  n − 1  0.333 × 0.9 = 0.2248   = 1.333  n t  γ −1 2  T01= Ta 1 + M  = 216.8(1 + 0.2 × 0.852 ) = 248.13K 2   ∆Tr = 31.331C P01  0.95 × 31.33  = 1 +  Pa  216.8  3.5 T02 0.31718 = (1.77 ) = 1.1989 T01 = 1.569 ∴P01=0.227 ×1.569 =0.3561 bar P02=0.6303 ∴∆T12 = 49.33K ∴T02=297.5  P03  34 = 19.21  =  P02  1.77  T03  0.3178 = 2.558 , ∆T23 = 463.4   = (19.21) T  02  T03=761K ∆T023 = 265.02 P04 = 0.3561 × 1.77 × 19.21 × 0.97 =11.74 bar ∆T45 = 1.004 × 463.4 = 409.4 T05=940.6 K 1.148 × 0.99 1 P04  1350  0.2248 = = 4.99  P05  940.6  ∆T56 = 1.004 × 49.33(5.7 + 1) = 292.0 T06=648.6 K 1.148 × 0.99 T6=556.0 K 28 © Pearson Education Limited 2009 HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual 1 P05  940.6  0.2248 = = 5.22  P06  648.6  ∴ P06 = P 11.74 = 0.4505 bar 06 = 1.984 ∴choked 4.99 × 5.22 Pa Ca = 0.85 ×295.2 Cold Stream (both choked) Hot Stream mc = 220 × 5.7 = 187.2 6.7 mh = 220 = 32.8 6.7 P2 = P02 0.6303 = = 0.3330 Pcr 1.893 P6 = 0.4505 = 0.2433 1.852 T2 = 297.5 = 247.9 1.2 ρ= 100 × 0.3330 = 0.468 0.287 × 247.9 ρ= 100 × 0.2433 = 0.1525 0.287 × 556 Cc = 1.4 × 287 × 247.9 = 315.6 m/s A= 187.2 = 1.267 m2 0.468 × 315.6 Ch = 1.333 × 287 × 556 = 461.2 m/s A= 32.8 = 0.466 m2 0.1525 × 461.2 Fc = 187.2 ( 315.6 − 250.9 ) + 1.267 ( 0.333 − 0.227 ) × 105 =12.108+13430=25,538 Fh = 32.8 ( 461.2 − 250.9 ) + 0.466 ( 0.2433 − 0.227 ) × 105 = 6898+760=7658 F = 33,196N (Fs=150.9) f= 0.017 =0.0172 0.99 mf = 0.0172 × 32.8 × 3600 = 2028 kg/hr SFC = 2028 = 0.0611 kg/Nh 33196 29 © Pearson Education Limited 2009 HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual Problem 3.9 M 1.4 Turbofan (BPR=2.0, FPR=2.2) Assume full expansion, i.e. no pressure thrust 0.4  n −1 = 0.3175   =  n c 0.90 × 1.4  n − 1  0.333 × 0.9 = 0.2248   = 1.333  n t  γ −1 2  T01= Ta 1 + M  = 216.7(1 + 0.2 × 1.42 ) = 301.65K 2   ∆Tr = 84.95K P01  0.93 × 84.95  = 1 +  Pa  216.7  3.5 = 2.968 ∴P01=0.3594 bar T02 0.3175 = ( 2.2 ) = 1.2845 T01 ∴T02=387.46K ∴∆T12 = 85.8K T03 0.3175 = ( 9.0 ) = 2.009 T02 ∴T03=778.39K ∴∆T23 = 390.9 K Assuming no cooling, 1.004 × 390.9 = 1.147 × 0.99 × ∆T45 ∆T45 = 345.6 T5=1154.3K 1 P04  1500  0.2248 = = 3.206  P05  1154.3  LP system (B+1) ×1.004 × 85.8 = 1.147 × 0.99 × ∆T 56 ∆T56 = 227.6 T6=926.7 K 1 P05  1154.3  0.2248 = = 2.656  P06  926.7  P04= 0.3594 × 2.2 × 9.0 × 0.93 = 6.618 bar P02=0.3594 ×2.2 = 0.7907 bar P06 = P P 6.618 = 0.7772 bar 06 = 6.418 02 = 6.529 3.206 × 2.656 Pa Pa 30 © Pearson Education Limited 2009 HIH Saravanamuttoo, GFC Rogers, H Cohen, PV Straznicky, Gas Turbine Theory, 6th edition, Lecturer’s Solutions Manual With full expansion to Pa T 02 0.286 = ( 6.529 ) = 1.710 T2=226.6K ; Cc2 = 2 × 1.004 × 103 [387.4 − 226.6] = 568 m/s T2 T 06 0.25 = ( 6.418 ) = 1.582 T6=582.2K ; T6 Ch2 = 2 × 1.147 × 103 [926.7 − 582.2] = 888.9 m/s mc = 70 × 2 = 46.67 ; mcCc=46.67 ×568 = 26508 N 2 +1 mh = 70 × 1 = 23.33 ; mhCh=23.33 ×888.9 = 20738 N 2 +1 47246N (gross) Ca=1.4 × 295.1 = 413.1 ram drag = 70 × 413.1 = 28920N [ SpecificThrust = 261.8 N / kg / hr ] T04-T03=1500-778-722 ∴ fideal = 0.023 f = 18326 N (nett) 0.023 = 0.0232 0.99 ∴ mf =23.33 × 0.0232 × 3600 = 1951 kg/hr SFC= 1951 = 0.1065 kg/Nh 18326 Ca=413 m/s = (413/1000) ×3600 km/h=1486 km/h ∴Flight time = 5600 = 3.766 hrs 1486.8 ∴Fuel flow =3.766 ×2 × 1951 =14,697kg 31 © Pearson Education Limited 2009 HIH Saravanamuttoo.43 bar P4= 14.8 K 0.74 = = 3.97 R   ∴Rhp = 3.95 = 14.99 Airflow =11. Ta=303K ∆T12 = 303 150.0 kg/s ∴ Power =333.0254 0.985 = 33.692 P5 1.0257 0. H Cohen.2 × 0.5 = 859. Tin=467 D C ∴ f/a= kJ kg ( f / a )ideal = 0.8 × 0.148 × 0. Gas Turbine Theory.25  3.148 ×297.8 =1201.013 ×15.2 K 1   ∴∆T45 = 1201.81  ∴T2 = 740.89 1 − = 297.5D C ∆T34 = 437. PV Straznicky.286 − 1 = 437.10 Pa=1.86 P4 3.87 × 1600 1 − 0.74 bar 3. 6th edition.278 kg/kWh 32 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 3. GFC Rogers.013 bar .5 K ∆Tcc = 1600 − 740.0 × 0.4 ×11.43 = 3.25  1.692   ∴Specific output =1.0 = 3667 kW mf= 0.9 kg/hr ∴ SFC=0.0254 = 0.8 = 0.86 P3=1.99 × 0.005 1   = 398.5 0.0257 ×11.5 × 1.99 × 0.8 ∴ 398.4 ∆Tcc = 860.0 × 3600 = 1018.013 T4=1600 – 398. 8 = 3.1]=755.02 0.286 − 1 = 379. H Cohen. Gas Turbine Theory.01 × 0.6 kg/hr 0.004 × 379.2 m[620.99 = m × 1.5 ×1.0 × 0.HIH Saravanamuttoo.16 kg/s 239.4 0. PV Straznicky.99 ∴ mf = 0.99 SFC = 229.07  m × 1.84  T02 = 688 K P03 = (12 × 1.004 × 156.87 1 − = 546 0.304 kg/kwhr 755.88 Power = 3.02 × 3.16 × 3600 = 229.2 kW For gas turbine ∆T12 = 308 120.5 − 381.8 ∴m= f/a (690.8 33 © Pearson Education Limited 2009 .11 Load compressor ∆T = 308 3.07 P04 1   ∴ ∆T34 = 1390 × 0.95 ) /1.6 0.6 + 200 + 551.9 = 551.286 − 1 = 156.650. ∆T = 700) = 755. GFC Rogers. Lecturer’s Solutions Manual Problem 3.04 = 11.25   11. 6th edition.6 = 0.9 K 0.148 × 546. 7 123.46 K At root.1 = 1. 6th edition.4 s RTs 0.3 m/s 123.104 0.37 = m/s ρ × 0.0271 1.0405 1.54 280. PV Straznicky.116 110.1 Flow area = π (152 − 6.5 P T  C2 .94 1.3 α r = 48D10' tan α r = 34 © Pearson Education Limited 2009 .0288 1.097 0.3 127.61 280.46 1.132 123.910 1.095 127.287 × Ts Ts We must find the axial velocity by iteration to satisfy continuity: Let Cmax=150 m/s C θ Ts T 0 /T s P 0 /P s Ps ρ Ccalc 150 11. Ts=280.0574 ρ 3. Ts= 288 − 2010 Ps  Ts  ρ= θc = C2 2c p Ps 100 × Ps P = = 348.0269 1.098 0.39 1.3 8.1 ∴ Ca = 123. H Cohen.149 0.05 279. 0 = 0  .0 139.8 1.7 7.911 1.1 123. GFC Rogers.131 123.1 7.0574 m2 C= 8. U= π × ( 0. Gas Turbine Theory.12 276.1 m/s.127 123.52 ) = 574 cm2= 0.HIH Saravanamuttoo.065 × 2 ) × 270 = 110.870 1.905 1. Lecturer’s Solutions Manual Problem 4. 287 × 280.42 ⇒ V=282.4839 254.46 × 103 = 335.HIH Saravanamuttoo.7 m/s ∴ M= 282.4 × 0.7 35 © Pearson Education Limited 2009 . H Cohen.12=79720.4 α t = 25D 48' tan α t = At tip V2=254.15 × 2 ) × 270 = 254.3 m/s a= 1. U= π × ( 0.1 = 0.841 335.42+123. GFC Rogers.4 m/s 123. 6th edition. PV Straznicky. Lecturer’s Solutions Manual At tip. Gas Turbine Theory.3 = 0. 056 1.3 K .049 1.9 1. for a first estimate ρ= 100 P0 100 × 0.060 m2 m= ρ ACa = ρ AC cos 25 ⇒ ρ C = 3.22 But based on stagnation conditions.5 = 0.2 232.434 69.491 First trial: Cmax=140 m/s C θ Ts T0/Ts P0/Ps Ps ρ ρC 140 9.4226 = 64.9 × 0.5 231.6 0.042 1.297 0.287 T0 0.HIH Saravanamuttoo.343 × = = 0. Gas Turbine Theory.90 m/s 66. PV Straznicky.443 62 150 11.288 0.33 4 2 3.61 m/s 3.22 = 152.7 1.005 × 103  243.906 ρ C = 66. H Cohen.060 × 0.87 m/s 0.435 65.23    217  A= π ( 0.182 ) = 0.3 m/s  2 × 100  2 36 © Pearson Education Limited 2009 .290 0.5 P02  ψηc  σ U 2 − Cw ( rω ) mean   = 1 + P01  C pT01   18 + 33  1 Ue = (r ω )mean=   × × 270 × 2π = 216. 6th edition.9 × sin 25 = 152.3  P01=0.7 233. T01 = 217 + 2Cp 2 × 1.2 152.22 = 134. Lecturer’s Solutions Manual Problem 4.4 11.2 1.0 C1w=152.051 1.21 0.491 kg/m3 0.3 ∴ C= 66.287 × 243.433 66.2 T1=T01 – C12 2302 =243.287 0.19 0.182 0.2 160 12.155 0.343 bar − 0.0 Take C=152.4 × 66.7 230.7 1. GFC Rogers. 9 × 4582 − 64.3 × 103   P02 3.005 × 243.61 × 216.04 × 0. Lecturer’s Solutions Manual U = π × 270 × 0.595 ) = 5.5 P02  1.HIH Saravanamuttoo.12 × 0.3  = 1 + P01  1.12 P01 ∴ P02=5. Gas Turbine Theory. PV Straznicky.343=1.756 bar 37 © Pearson Education Limited 2009 .54 = 458 m/s 3. H Cohen. 6th edition. GFC Rogers.5 = (1.8  0. 5 = 1.63π = 0.1 = 351 m/s 2   Cr22 C2 3512 − =429 –  3  2C p  2 × 1.83 366 Hence Cr 2 = 63.005 × 10  2C p T2=T02- Therefore T2=367.83 × 10 ρ2 ρ2 θ r is small – For the first trial let T2=366K ∴ ρ2 = 669 = 1.827 63.743 38 © Pearson Education Limited 2009 .0 Cr 2 = 104 = 51.0 2.26 m/ s Final trial: Cr 2 θ 64.8835 17 16.83 × 10−4 m 2 m 0.76 = × = −4 ρ 2 A2 51.92 669 = T2 0. 6th edition.7 – θ r ρ2 = A2 = 100 × 1.5 × 1.8835 × 397.99    0. H Cohen.5 ∴ T2=366 K P2t  T2t  =  P2  T2  3. GFC Rogers. PV Straznicky.287 × T2 π × 16.5  429  =   366  3.97  3. Lecturer’s Solutions Manual Problem 4. Gas Turbine Theory.03 T2 366 ρ Cr 2 1.5 U2 = π × × 765 = 397.5  1 − = 0.3 P01 = Pa = 0.1m / s 100 σ =1− Hence: C2w = σ U 2 = 0.HIH Saravanamuttoo.99 bar and T01=Ta=288K γ −1   T01  P02  γ  T02-T01 = 429-288=   − 1 η  P01    1   288  2.753 ∴ ηc =    141  0.60 1 115. PV Straznicky.HIH Saravanamuttoo. Gas Turbine Theory.753 39 © Pearson Education Limited 2009 .99  288  ∴ ηimp = 0.743 = 3. H Cohen.87 = 0. Lecturer’s Solutions Manual P2t=1. GFC Rogers. 6th edition.347  141  = 1 + ηi m p × 0.850 3.607 1 − 0.347 bar P02  141  = 1 + ηi m p × P01  288  3.5 3.5 Fraction of loss in impeller = 1 − 0.92 × 1. 12 0.84=378.5-76.40bar (b) Flow area at tip: = π × 50 × 5 104 = 0.0785 × 96 = 15.6 = = 1.12 m/s 1. PV Straznicky. H Cohen.287 × 1.909 100 × 2.0785 m2 C2w=0.9 ) T2=455.5 P02  0.HIH Saravanamuttoo. 6th edition.7 2 + 962 ∴ C2 = 393.9  (noting 2 × 1.30 bar 1.251 tan α = 381.005 × 103 = 44.36 × 1.7 m/s 2 2  381.0 kg/sec.01 = 4.5 = 1 + = (1.66  ρ2 = 3.01 390 a 96 = 0.30 = 2. ∴ Cr 2 = 96 m/s a = 0.12 = 381. Lecturer’s Solutions Manual Problem 4.9 ×424.4 × 378.5  =  P2  378.7   96  If C2r=96 and θ c =   +  = 76.005 × 103 T02 = 455.909 ρ 2 A2 Cr 2 = 2.5  3.50 × 270 =424.84  44.66 × 103 = 390 m/s C 22 = 381.66 P2= 4.04 × 0.12 × 0.122 = 167.5 = 1.9 × 424.90 × 167.40 = 2.5K 1.6 m / s C2 393.523) = 4.66 K P02  455. GFC Rogers.9   44.36  P01  288  P02 = 4.5K ∆T = 3.97 The agreement is close since the required flow is 16.4 (a) U2= π × 0.7 M= ∴ α = 14D5 40 © Pearson Education Limited 2009 . Gas Turbine Theory.287 × 378. 0911m 2 16.60 56.1 × 103 = 400.4 = 0.1 K C32 = 3292 + 72.841 M= 400. H Cohen. GFC Rogers.4 2.44 tan β = 72.22 β = 12D 24' 329 41 © Pearson Education Limited 2009 .78 2.287 × 399.581 2.44 m/s 336.43 177.755 2.2 1.140 1. as a first guess assume: Cr3=96 × 25 = 82. T3 = 399.4 57.87 = 0.75m/s 29 T03=T02=455.5 K ∴ C3w θw C3r θr θ T3 T0/Ts P0/Ps Ps ρ ρ Cr 329 53.4 m/s.5 73 2.78 2. Gas Turbine Theory.0 398.1 1.143 1.0 = 175.596 2.140 1.87 m/s a= 1. PV Straznicky.581 2.5 72.7 × 25 = 329 m/s 29 We must find Cr3 by trial and error.43 176 C3r = 72.6 82.HIH Saravanamuttoo.4 1. Lecturer’s Solutions Manual (c) P3t=P2t=4.75 3.41 199. 6th edition.4 × 0.0911 C3 w = 381.42 ⇒ C3 = 336.40 bar A= π × 58 × 5 10 ρ3C3r = 4 = 0.64 56.2 399.6 0.2 399. 1 = 157.286 4 − 1 = 175.97 m/s With a 50% loss in impeller.59 K P02  157.90 ×U 2 = 0.7 m/s C 22 = C 22r + C 22w ∴C 22r = 393.HIH Saravanamuttoo. H Cohen. η I = 0.72 – 389.9 K γ + 1 1.005 × 175.59  = 1+ P01  288  3.9 × 103 C2 = a = 393. Gas Turbine Theory.5 ∆T013 = 288 0.72=3133.1 c p × 103 U 22 = 1. GFC Rogers. PV Straznicky.90 ∴∆T '012 ∴ = 0.1 K 0.689 m π × 200 T03 = T02 = 175.6 ∴C 2r = 55.04 × 0.4 × 0.80  ∆T013 = pσ U 22 = 175.876 × 105 1. Lecturer’s Solutions Manual Problem 4.607bar 42 © Pearson Education Limited 2009 .1 × 103 = 1.5 = 4.90 × 175.0 T2=T02 × 2 = 463.287 × 385.1+288=463.90 ∴U 2 = 433 m/s d= 433 = 0.7 m/s C2w = 0.1 = 385.1 K At tip M=1.90 × 433 = 389.2 ∴ ∴ a = 1. 6th edition. 287 × 385.257 cm π × 0.0 = 0.97 m2 0.607 bar γ P02  γ + 1  γ −1 3.HIH Saravanamuttoo.198 × 55. H Cohen.434 bar P2 = 1.5 =  = 1. Lecturer’s Solutions Manual ∴ P02 = 1 × 4.198 kg/m3 0.607 = 4.434 = 2. PV Straznicky.689 43 © Pearson Education Limited 2009 .2 = 1.9 m= ρ AC2 r ∴h = ∴A= 14. GFC Rogers.1138 2.607 = 2.893 P2  2  4. Gas Turbine Theory.893 ρ2 = 100 × 2.1138 × 100 = 5. 6th edition. PV Straznicky.1 = 45.e.95 = 0.027 α1 = 45D 44' ( = α 3 ) 150 = With free vortex. m Ω∆Cw = mC p ∆T 1.95 = = 1.8 m/s 44 © Pearson Education Limited 2009 . H Cohen. 6th edition.95m/s 22 tan α 0 tan α1 45.005 × 20 × 103 = 108.1 At mean radius.0 × 200 ∴ C0 w = 36. Gas Turbine Theory.1m/s 0. Cwr=constant i.93 × 200 ∴ ∆Cw = C0w= 200 − 108. GFC Rogers. CwU= constant Tip C0w ×250 = 46.307 α 0 = 17D 4' ( = α 2 ) 150 200 − 45. Lecturer’s Solutions Manual Problem 5.HIH Saravanamuttoo. 4 = 0.409.3 = 0.8 = 1.4 = 0.1 × 200 = 86. α 2 = −20D15' 150 tan α 3 = 144 + 61. 6th edition.369.3 = 0. α 3 = 53D51' 150 Λ= static enthalpy rise in rotor Stagnation enthalpy rise in stage Now (T0+ = T1 − T0 20  C2 C02 C2 C2 + 20) =T1+ 1 ∴ T1 − T0 = 20 +  0 − 1  2C 2C p 2C p  p 2C p 45 © Pearson Education Limited 2009  1  3  10 .8 = 0.3) = −0. α 2 = 40D14' 150 tan α 3 = 36. GFC Rogers.3 = 1. α 0 = 22D15' 150 tan α1 = 150 − 61. α1 = 54D53' 150 tan α 2 = 36.8 + 86. α 3 = 39D 26' 150 Root C0w ×150 = 46 × 200 ∆Cw = 108 × 200 = 144 150 ∴ C0 w = 61.591.821. H Cohen.369.421. Gas Turbine Theory. α1 = 30D36' 150 tan α 2 = 150 − (144 + 61.3 m / s m/ s tan α 0 = 61.245.4 m/s 250 tan α 0 = 36.845.HIH Saravanamuttoo. α 0 = 13D 46' 150 tan α1 = 250 − 36.8 + 86. PV Straznicky. Lecturer’s Solutions Manual ∆Cwu = constant ∴ ∆Cw = 108. 5 T1-T0=20+11.005 2 × 1.07-32.1 0.2 0.B.5899 Λ= T1-T0=20+13.7724 = 150 = 162.12 254.1 0.92 = 4. Gas Turbine Theory. 6th edition.9712 C1= 150 cos39D 26' = 150 = 194.19-18. H Cohen.HIH Saravanamuttoo. Lecturer’s Solutions Manual Tip C0= 150 cos13D 46' = 150 = 154.1 0.9255 ∴Λ = 13.22  −   103  2 × 1. 150 = 254.: Λ is very sensitive to small errors in C0 and C1) 46 © Pearson Education Limited 2009 .6% 20 (N.1 = 65.92.005  150 cos53D51' = T1 − T0 = 0. PV Straznicky. GFC Rogers.15 0.8=13.5% 20 Root C0= 150 cos 22D15' T1-T0=20+ C1= ∴ 1  162. 546. PV Straznicky.85 = 1. α 0 = 28D36' ( = α 2 ) 150 tan α1 = 250 − 81. ∆Cw : mean = 108.HIH Saravanamuttoo. α1 = 48D16' ( = α 3 ) 150 47 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 5.121.2 Velocity diagram at mid span is as in previous question.3. root =144 m/s Equal work at all radius Tip C0w= 250 − 86.3 = 81. H Cohen. 6th edition.85 = 0.85 m/s 2 tan α 0 = 81. Gas Turbine Theory. GFC Rogers. tip =86. 6th edition. α1 = α 3 = 44°25' 150 48 © Pearson Education Limited 2009 .HIH Saravanamuttoo. Gas Turbine Theory. PV Straznicky. Lecturer’s Solutions Manual Root C0w = 150 − 144 = 3 m/s 2 tan α 0 = 3 = 0. H Cohen. GFC Rogers.980. α 0 = α 2 = 0°7 ' 150 tan α1 = 150 − 3 = 0.0020. 5 P  0.4 = 317. T = 288 − a= 140 × 140 1 × = 278.908 m π × 100 ∴Tip radius =0.2 ∴ D= 285.895 × 100 = 1.908 = 0.3 (a) With no inlet guide vanes the inlet velocity is axial. GFC Rogers.5452 ) = 0. PV Straznicky.4143m2 ∴ m = 1.5 = 1.287 × 278.60× 0. Gas Turbine Theory.2335 (c) 02 = 1 + 288  P01  49 © Pearson Education Limited 2009 .2 m/s π × D × N = 285.454 m or 45.25 Dhub = 0. 6th edition.40 cm (b) P0  T0  =  P T  3.89 × 20  = 1.895 bar 1.25  3.7 2 − 1402 = 81333.121 kg/m3 0.25 K 2 × 1.4 m/s ∴ V= 0.4143 × 140 =65.01 = 0.25 × 103 = 334.95 × 334.02 or m=65 kg/s 3. Lecturer’s Solutions Manual Problem 5.2 = 0.287 × 278.9082 – 0. H Cohen.HIH Saravanamuttoo.128 ρ= 0.4 × 0.3 ∴ U=285.121 ×0.005 103 γ RT = 1.545 m A= π 4 (0.128 P= 1.5  288  =   278.7 m/s V2=U2+C2 ∴ U2 = 317. 2 = 171. α1 = 63D50' 140 tan α 2 = 285.320.HIH Saravanamuttoo.1 = 1.2 = 2. α 2 = 17D 46' 140 At tip ∆Cw = 126.78m/s tan α1 = 285.3 = 0. GFC Rogers.93 uΩ mU ∆Cw Ω tan α1 = 171.005 × 20 × 103 ∆Cw = = = 126. α 2 = 56D14' 140 Power input =65 ×1.3m/s 171.5 kW 50 © Pearson Education Limited 2009 .3 × 0. Lecturer’s Solutions Manual (d) At the root we have axial inlet velocity.1 m/s 1 = mC p ∆T 103 C p ∆T × 103 1.1 × 0. Uh=0.2 − 75.4958.0371. Gas Turbine Theory.78 = 1. Free vortex gives constant work at all radii and constant axial velocity. PV Straznicky.60 ×285.1 − 126. H Cohen. α1 = 50D 44' 140 tan α 2 = 171. 6th edition.005 × 20 = 1306.60 = 75.2221. 286 = = 0.9K ∴ Number of stages = # 163. GFC Rogers.4 K/stage 7 of stages For first stage: Tout 288 + 23.e.325 ∴ Rfirst=1.4=428.9 = 6.556 i.569 − 1] = 163.88 T02 0.0546 Tin 1.0546= (R)0.325 ∴Rlast=1. Lecturer’s Solutions Manual Problem 5.569 T01 ∆T012 = 288[1.HIH Saravanamuttoo.0 ) = 1.0812 288 Tin 1.9) 1 = C p ∆T 103 51 © Pearson Education Limited 2009 .9=451. ∴ Tout = 1.0812 = (R) 0. Gas Turbine Theory.4 n − 1 0.5 K. (note that Ca=165 cos20) 103 ∆Cw = U − 2C sin 20 = U − 330sin 20 ΩU (U − 112. 7 stages 25 ∆T 163.9-23.9 K Tin = 451.271 For last stage: Tout=288+163.325 = ( 4.178 mU ∆Cw Ω 1 = mC p ∆T . H Cohen.325 n 0. 6th edition.4 = = 1. PV Straznicky.9 = = 23. 9 ± 354. PV Straznicky.96 K 2.96  1652 = 414.5 P0  T0  =  P T  P= 3.5  428.18 At the entry to last rotor: T0=428.58 × 165 × 0.18 × h × 2. Lecturer’s Solutions Manual U2 – 112.9 ± 112.0 π × 0.9397 = 0.01 × 103 3.43 = 3. H Cohen.1189 3.01 = 3.18 × 2.9 m/s π DN 60 = 233.9U-28.4 × 103 0.9 ∴ N= 60 × 233.92 + 4 × 28.287 × 414.9 = 2 2 ∴ U = 233. P0= 4 × 1.3 × 103 112.07 bar 1.83 U2 – 112.01326 m h = 1.5.07 = 2.HIH Saravanamuttoo.0 ∴h = 3.5 rpm π × 0.96 (π Dh ) ρ × Ca = m π × 0.3 ×103 = 0 ∴U = 112.43 bar 1.5 = 1.9U= 1. GFC Rogers.178 C=165 m/s ∴ T=428.58 kg/m3 0.58 × 165cos 20 = 3.5  =   414.9 = 24817.005 × 23.1189 ∴ ρ= 100 × 3. 6th edition.326 cm 52 © Pearson Education Limited 2009 . Gas Turbine Theory. Lecturer’s Solutions Manual Problem 5.3446 n 0.05 ×103 0. H Cohen.86 U(U-109.92 n 0.95 m/s ∴ N(Centrifugal) = 60 × 468.311 ∴ P02 = 3. Gas Turbine Theory.6= π DN 60 ∴ N= 60 × 249.82 K ∆T = σψ U 22 Cp = 0.502 × 408 = 204.04 U2=468.22 + 4 × 35.6 m/s 249.82 0.364 = U − 109.04 × U 22 1.5 Axial: n − 1 0.502 ∆T023 = 0.311 Centrifugal: = = 0.005 × 30 =35.286 n − 1 0.83 ∴ T2=288+120=408 K For axial ∆T = 4 × 30 = 120 K P  T02 408 = = 1.95 = 27140 rpm π × 0.3446 = ( 3.005 × 103 ∴ U 22 = 1.05 ×103 =0 109.2 103 × 1. GFC Rogers.90 × 1.33 53 © Pearson Education Limited 2009 .07 P01 P03 10. 6th edition.005 × 103 × 204. PV Straznicky.HIH Saravanamuttoo.3446 = 1.257 P02 3.417 =  02  T01 288  P01  ∴ 0.2U-35.05 × 103 109.90 × 1.0 = = 3.2 ± U=249.2)= U2-109.25 Centrifugal: T03  P03  =  T02  P02  0.257 ) 0.286 = = 0.6 = 19068 rpm π × 0.07 Axial : mU ∆Cw Ω 1 = mC p ∆T 103 ∆Cw = U − 2Ca tan 20 = U-300 ×0.2 ± 390 = 2 2 ∴ U= 109. 9812 η= 54 © Pearson Education Limited 2009 .1 (a) From Fig. for f=0. 2. Gas Turbine Theory. for T1=298 K and ∆T = 583 K.0150 (b) From Fig. PV Straznicky. theoretical f=0.0147 = 0. Lecturer’s Solutions Manual Problem 6.HIH Saravanamuttoo.0188 43100 η = 0.15.8608 = 0. 2.0147 ηb = 0.0803 ×10110) Efficiency based on released energy: 43100 − (0.983 593 (c) mass of CO per kg of fuel: m= 28 × 0.0803 kg 12 Actual released energy = 43100–(0. H Cohen.0150 and T1=298 K.04 × 0. 6th edition.15. GFC Rogers. theoretical ∆T = 593 K ηb = 583 = 0.0803 × 10110) = 1 − 0.980 0. 051 1. O2 N2 R’s 0. Mean values of Cp from tables are : R.934 1. 2.0148 and for an initial temperature of 450 K For air.4 1687.7 220.5N2 → 10CO2+6H2O+52O2+244.1 9789. Hence approximate final temperature T2=565+450 =1015 K First guess at mean temperature of products = (1015+298)/2=656. GFC Rogers.084 55 © Pearson Education Limited 2009 mCp 460.042 P’s CO2 H2O O2 N2 1st Guess 1. P.041 1.15 for f=0.5 K.4 . given by: ∆H vap = −42500 + 108 × 2442 = −40500 KJ/kg of C10H12 132 Energy equation is: ( ) (Hp2-Hp0)+ ∆H 0 + H R0 − H R1 =0 Where the first stage is 325 K for C10H12 and 450 K for air Mean temperature of reactants O2 and N2 is 450 + 298 = 374 K 2 From fig. PV Straznicky. Gas Turbine Theory.HIH Saravanamuttoo. 6th edition.105 2.088 2nd Guess 1.reactants.047 2.products. Lecturer’s Solutions Manual Problem 6. we require ∆H vap .0148 ( 2080 + 6846 ) 6846 Since the H2O will be in vapour phase.2 The combustion equation: C10H12 +65O2+244.5N2 132kg 2080 6846 440 108 1664 Fuel/ air ratio = 132 = 0.2 7421.014 1. temperature rise ∆T = 565 K.019 1. H Cohen. 5 − 5.4 56 © Pearson Education Limited 2009 .000 − 152 × 9073 − 27 × 257 = 0 ∴ T2=298+ 6736000 = 982 K 9855 Second guess at mean temperature of products=1280/2=640 K ∴ T2=298+ 6736000 = 986. H Cohen. PV Straznicky.1 K 9789. 6th edition.051) + (1664 × 1.019 ) + (6846 × 1.105 ) + (108 × 2.HIH Saravanamuttoo.945] = (T2 − 298 ) × 985. Lecturer’s Solutions Manual (T2 –298) {( 440 × 1.042 )  – ( 325 − 198 ) [132 × 1.35.934 ) + ( 6846 × 1. Gas Turbine Theory. GFC Rogers.088)} – (132 × 40500 ) − ( 450 − 298 ) ( 2080 × 0. 76) 2 × 475 ρ1 = 3.42 1226.287 × 3.153 K2 ∴ K2=1.808 × 0.0389 × 1.3 Under design conditions: T01=T1+ 475= 475= C12 P m2 = 1 + 2 2 2C p R ρ 1 2 ρ1 A1 C p 4.0389 P 4.70 = 0 From which : ρ1 = 1557 ± 1557 2 − 4 × 475 × (−26.0+ K2  − 1 2 2 9.213 bar At design : C1= m 9.88=19.5ρ1 − 18 = 0 ∴ ρ1 = 2.42 2 × 2.287 ρ 1 2 ρ1 × 0.294 × 0.294 kg/m3 Hence: 0.70 ρ12 475 ρ12 − 1557 ρ1 − 26.5 18.0 = =70.8 K R ρ1 0. PV Straznicky.0 / 2 × 3.27 × 105  1023  =19.47 × 100 At design: T1= 1 = = 472.HIH Saravanamuttoo.24 m/s ρ1 A1 3.4 =67.0975  475  20.294 At part load: C1= 57 © Pearson Education Limited 2009 .287 ρ1 2 ρ1 × 0.0+1.0389 7.808 kg/m3 105 ∆P0 = 7.52 × 100 7.808 × 0. Lecturer’s Solutions Manual Problem 6. GFC Rogers.75 m/s 2. Gas Turbine Theory.0 + 1. H Cohen.294 × 0. 6th edition.47 × 100 9.03892 × 1.02 + 2 0.005 × 103 1557 ρ1 + 26.630  439 − 1     ∆P0 = 0.09752   900   19.005 × 10 439 ρ12 − 1226.004 + 2 = + 2 3 ρ1 ρ12 0.630 Part-load conditions: 439= 3. 8  3.583 58 © Pearson Education Limited 2009 .5 = 4.47    472.0594 P01 4.287 × 2.543 At part load: T1= 3.583bar ∆P0 0.52    436.543 bar ∆P0 0. Gas Turbine Theory. Lecturer’s Solutions Manual T  P01=P1  01   T1  γ / γ −1  475  = 4. PV Straznicky.52 × 100 = 436.HIH Saravanamuttoo. 6th edition. GFC Rogers. H Cohen.808  439  P01=3.78 K 0.27 = =0.78  3.5 = 3.0595 P01 3.213 = = 0. 5 m/s Tan β 2 = 557.5 + 45.84 =1.22 − 2602 = 557.1 C2= Ca 260 = = 615.5609 − 0.5 + 45.84 m/s Tan β 3 = 360 + 45. GFC Rogers.7596 ∴ β 2 = 37D13' 260 Cw3=260 tan 10=260 ×0.35 U2 360 Power output =mU ∆Cw = 20 × 360 × (557. Lecturer’s Solutions Manual Problem 7. Gas Turbine Theory. 6th edition. PV Straznicky.2 m/s cos 65 0.84 ) = =3.HIH Saravanamuttoo.289 Temperature drop coefficient: ψ= 2C p ∆T0 s U 2 = 2U ∆Cw 2 ( 557.84 ) P=4343760W or 4344 KW T2=T01- C22 615.7596 ) 2U 2 × 360 Λ = 0.4226 Cw2= 615.5 − 360 = 0.22 = 835 K =1000 – 2294 2C p 59 © Pearson Education Limited 2009 . H Cohen.5609 ∴ β 3 = 57D 22' 260 Since C1=C3 and Ca is constant: Λ= Ca 260 ( tan β3 − tan β 2 ) = (1.1763 = 45. ∴ P01 > 1.877 kg/m3 RTc 0.869 bar 2. ∴ The nozzle is choking P2 The throat conditions are: Pc=  2  2 4.853 2.877 × 572. 6th edition. H Cohen. Gas Turbine Theory.0398 m2 0.287 × 857 × 103 = 572. critical pressure ratio Pc/Pa=1.158 bar.25 K 2C p 2294 T '2 = 835 − 8.853. GFC Rogers.75 K P01  T01  =  P2  T2'  ∴ P2 = γ / γ −1 4  1000  =  = 2.HIH Saravanamuttoo.333  γ +1 ρc = Pc 2. PV Straznicky. Lecturer’s Solutions Manual T2 – T2' = λN C22 615.14  826.75  4 = 1.6 60 © Pearson Education Limited 2009 . Tc=  × 1000 = 857 K  × T01 = 1.0 = 2.333 × 0.22 = 0.05 × = 8.158 × 100 = = 0.287 × 857 Cc= 1.853.6 m/s Athroat= m ρc Cc = 20 = 0.140 For isentropic flow.25 = 826. Gas Turbine Theory.HIH Saravanamuttoo.2 ∆T013 γ −1   γ   1   = ηt T01 1 −    P01 P03     1   4 1    =0. Lecturer’s Solutions Manual Problem 7.147 × 147 × 103 = 562 m/s 300 Now at root.88 ×1050 1 −    = 147 K  2    T03 = 1050 – 147=903 K With zero outlet swirl: Ws=UCw2=Cp ∆T013 And with free vortex. H Cohen. T02=T01 and C3=Ca3=275 m/s 2C p 2C p Hence C22r 2752 + 147 = 180 K = 2C p 2294 ∴ C2r=642. Cwr = constant Hence at root (Cw2)r = C p ∆T013 U = 1.6 m/s 61 © Pearson Education Limited 2009 . 6th edition. PV Straznicky. GFC Rogers. Λ = 0 ∴ T2 − T3 = 0 hence T2=T3 T1 − T3 ∴ T02- C2 C22r =T03 – 3 . 4 m/s 1. Gas Turbine Theory.05 × 2294 2Cp T2r – T’2r = 9 K T’2r=(1050 – 180) – 9. Lecturer’s Solutions Manual α 2 r = sin −1 562 = sin −1 0.6 (Ca2)r = 642.6 β 2 r = 40D 4' (Ca2)t=(Ca2)r=311.16  870  62 © Pearson Education Limited 2009 .15 m/s With no exit swirl T3t=T3r=903 – T2t=1050 – 2752 = 870 K 2294 508.21 861   4 3.62 − 5622 = 311.4  401. 6th edition.4 rr 562 =401.HIH Saravanamuttoo.152 = 937.2882 ) 311.6   α 2t = tan −1  ∴ α 2t = 52°10' C2t= 401.6 m/s tan β 2 r = ( Cw 2 ) r − U r ( Ca 2 ) r = 562 − 300 = 0.8746 = 61D 642.0 =861 K P01  T01  =  P2  T'2  γ / γ −1 P03  T03  =  P3  T'3  γ / γ −1 4 3.4  −1  = tan (1. H Cohen.638 bar =  = 1.5 K 2294 (T1-T3)= ∆T013 because C1=C3 with no inlet or exit swirl Λ tip = T2t − T3t 937.8  1050  = 1.719 bar =  = 2.46 147 T1 − T3 At the root T2r-T’2r= λn C22r 642.6 m/s and (Cw2)t= rt = 1.42 + 311.62 =0.8 / 2  903  = 1.8406 311.16 ∴ P3 = 1.21 ∴ P2 = 2. GFC Rogers.5 − 870 = =0.62 = 508. PV Straznicky. 343 × 100 =1.19 × 400 A 0.0 = 4.6 m/s 320 If Ca2=346 m/s C22 = 3462 + 537.5 ) 4 = 8.0591/ 2) 0.316  980.2 =12002294 2C p 2C p T2=T02- T2’=T2 – λN P2= C22 =1022-0.0756 = =0.2037 m 2π × 250 1   4   1  = ηt T01 1 −    P01 / P03     1  1 4 ∆T013 150 =1− = 0.1741 A2 = A3 Ca2 must satisfy Ca2= m and Ca2= C 2 − Cw2 2 ρ 2 A2 Ws=Cw2U=Cp ∆T013 (because α 3 = 0 ) Cw2= 1.B. Gas Turbine Theory.316 ∴ P3= 1.2K 2C p 2294 Um =2 π rm N or rm= ∆T013 320 = 0.008 bar 1.2037 + (0.8611   =1− 0.0 (1200 /1009. PV Straznicky.818 P03 P03  T03  =  P3  T3  ρ3 = A3= h= γ / γ −1 4 4.343 bar =  = 1. GFC Rogers.34 rr 0. C3=Ca3) ρ3C3 1.287 × 980.4  1050  = 3. H Cohen.19 kg/m3 = RT3 0.147 × 150 × 103 = 537.3 T3=T03 – C22 4002 = (1200 − 150) − = 980.0591/ 2) 0.62 ⇒ C2=639 m/s C22 C2 6392 = 1022 K =T01.07 ×178 = 1009.25  P3 3.0756 m2 (N.0591 m 2π rm 2π × 0.HIH Saravanamuttoo.2037 − (0.5 K 2C p P01 (T01 T2 ' ) 4 = 8.4bar = 1. 6th edition.233 = = = 1.2037 rt 0.995 63 © Pearson Education Limited 2009 .818 ∴ P03= 1. Lecturer’s Solutions Manual Problem 7.90 × 1200 ηt T01  P01 / P03  P01 8 = 4.25 m 36 = =0. 333 × 0. Hence: C2r= 6292 + 3462 = 717. Lecturer’s Solutions Manual ρ2 = 4.366 kg/m3 0.86 64 © Pearson Education Limited 2009 .5 m/s 1.1 m/s a2r= γ RT2 r = 1.1741 (C w2 )r =(C w2 )m × rm = 537. H Cohen.287 × 975. GFC Rogers. Gas Turbine Theory.0756 At the root.1 =0.812 610. we have: rm 0. for free vortex design. 6th edition.366 × 0.17 V2r= 3462 + (629 − 273.9 m/s T2r=1200 – UR= 7182 =975.287 × 1022 Ca2= m 36 = = 348 m/s (agrees with given 346 m/s) C2 A3 1.2037 = = 1.17 rt 0.86 m/s (Mv2)r= 496.6 × 1.17 =629 m/s rr Ca2 is constant at 346 m/s.3 × 103 =610. PV Straznicky.3 K 2294 320 = 273.008 × 100 = 1.HIH Saravanamuttoo.5) 2 = 496. 624 − 1. U = tan α 2 − tan β 2 Ca 2 Um  r  ∴tan β 2 =tan α 2 ( Ca 2 )m  rm  2 tan α 2 =(tan α 2 ) m = tan 58D 23' =1. since Ca2=Cw2 cot α 2  rm    r 2 Ca2r=constant and Ca2=(Ca2)m   r  Also U=Um    rm  2 Now.4 Cr2=constant (1) Cw2r=constant (2) From (2).25 ( Ca 2 ) m (N. GFC Rogers. Gas Turbine Theory. Flow coefficient (Ca/U)m=0. Lecturer’s Solutions Manual Problem 7. H Cohen. tan β 2 = 1.624-0.HIH Saravanamuttoo.374=1.25    rm  2 65 © Pearson Education Limited 2009 .8 for this mean diameter design) 2  r  Hence. 6th edition.624 Um = tan 58 D 23’-tan 20 D 29' =1.B. PV Straznicky. 726  rm     r  tan β 2 β2 Root 1.357 0. Gas Turbine Theory. GFC Rogers.877 0.769 0 We therefore have untwisted nozzles. PV Straznicky.164 1.056 3 D 12' 66 © Pearson Education Limited 2009 β2 35 D 20’ 0 . we also have Ca2 rx = constant. H Cohen. 6th edition. Lecturer’s Solutions Manual 2  rm   r   2  rm   r   2 tan β 2 Root 1. x r  Hence.164 0.     rm  2  Ca 2  m 1.709 Tip 0. Ca2= (Ca2)m  m   r 2 This with   U=Um  r  yields  rm  2 x +1  r   M  tan β 2 = tan α 2 . Cw2 rx=constant where x= sin 2 α 2 = sin 2 58D 23' =0.662 35 D 20’ Tip 0.HIH Saravanamuttoo.797 0.726 With α 2 = constant. T  P3=P01  3   T01  m= m= A 2C p R A 2C p R ( γ / γ −1) × × 2 P01 γ −1 T3 T01(γ / γ −1) (T01 − ∆Tw − T3 ) γ +1 γ −1 2 P01 γ −1 T3 T01(γ / γ −1) (T01 − ∆Tw ) − T3 For the given inlet conditions m is a maximum when dm = 0 i. Lecturer’s Solutions Manual Problem 7.e when: dT3 2 2 γ + 1 γ −1 T3 = 0 (T01 − ∆Tw ) T3(3−γ ) /(γ −1) − γ −1 γ −1 ∴ T3= 2 (T01 − ∆Tw ) γ +1 Hence. GFC Rogers. PV Straznicky. Gas Turbine Theory. writing K= A 2C p mmax= K (T01 − ∆Tw ) mmax= K (T01 − ∆Tw ) mmax= K (T01 − ∆Tw ) R γ +1 γ −1 γ +1 γ −1 γ +1 γ −1 × P 01 ( γ / γ −1) T01 2 γ +1   γ −1 γ −1  2  −  2      γ + 1   γ + 1     2 2  γ −1 γ −1  2  −  2   2       γ + 1   γ +1  γ +1      2  2  γ −1  λ − 1       γ +1  γ +1 67 © Pearson Education Limited 2009 . H Cohen.HIH Saravanamuttoo.5 m= ρ3 AC3 = = P3 A RT3 P3 A 2C p × T03 − T3 RT3 2C p × T01 − ∆Tw − T3 Assuming isentropic expansion. 6th edition. Lecturer’s Solutions Manual But Cp R mmax = mmax mmax = γ γ −1 AP01 T01(γ / γ −1) T01 AP01 T01 AP01 (T01 − ∆Tw ) γ +1 γ −1 γ +1 γ +1 = γ +1 2  2 γ  γ −1 2 γ −1 (γ − 1)   γ +1  R γ −1 ( γ + 1) γ −1 γ +1 γ  2  γ −1  ∆Tw  γ −1 T01γ −1    1 − γ +1 R  γ +1  T01  γ −1 T01 γ +1 = γ  2  γ −1  ∆T   1 − R  γ + 1   T w 01 γ +1  γ −1   Hence maximum mass flow will vary with N/ T01 because ∆Tw varies with P01/P03.HIH Saravanamuttoo. 68 © Pearson Education Limited 2009 . PV Straznicky. 6th edition. Gas Turbine Theory. GFC Rogers. H Cohen. 11312 ) ( 0.HIH Saravanamuttoo. PV Straznicky.1131m Rt=0. Gas Turbine Theory. GFC Rogers.2262 m ρ = 4500 kg/m3 (Ti-6Al-4V) ref table page 407 Centrifugal stress at the root σr = = ρω 2 2 2 2 ( Rt − Rr ) K 4500 (1570. H Cohen.1 N=250 rev/s ⇒ Angular velocity ω = 2π N = 1570. Lecturer’s Solutions Manual Problem 8.8 MPa Material UTS=880 MPa Endurance limit =350 MPa Assess the vibratory stress at failure – Goodman diagram -vibratory stress at failure ≈ 300MPa -high -Allowable approximately 1 ≅ 100 MPa 3 69 © Pearson Education Limited 2009 .5 ) 2 2 ( 0.6 ) ≅ 127.5 rad/s Rr=0.2262 2 − 0. 6th edition. 6th edition.109m 2 Aring =0.2 N=250 rev/s R0=0. PV Straznicky. 70 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 8. Gas Turbine Theory. GFC Rogers.025m Blade m ≅ 0. but acceptable for a conceptual design. H Cohen.025(0.750 2π ( 2 × 10−4 ) =131.750N σ t = 4500 (1570.52 ) ≅ 345. Material UTS = 880 MPa For a thin ring rotor (eq 8.113-0.02 × 0.105 = 0. for UTS = 880MPa and burst margin of 1.06 × 106 Pa σ t = 407 MPa Considering burst.2 ) 2 = 611 MPa >> 407MPa The design is conservative.163 × 1570.113m.5 ) ( 0.2.87 ×106 + 275 × 106 ≅ 407.105) =2 ×10−4 m2 Frim =43 ( mrω 2 ) = 43 ( 0.105m Width=0.HIH Saravanamuttoo. Ri=0.109 ) + 2 2 345.113 + 0.29) 2 σ t = ρω 2 Rring + Rring = Frim 2π Aring 0.163m ρ = 4500 kg/m3. the maximum allowable σ t (σ t )allowable = 880 (1.020 kg rcg ≅ 0. 6th edition. for CMSX-10 71 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 8.7.929 flights PLM is obtained from figure 8. Gas Turbine Theory. PV Straznicky. H Cohen. GFC Rogers.5 1050 124.000 5 × 10-5 Descent+L 0.520 4 × 10-6 t1% 2.3 t t[hrs] σ [MPa] PLM T[K] t1%[hrs] TO+Climb 0.489 × 10-4 Cruise 5 220 28 1000 100.HIH Saravanamuttoo.5 150 29.8 1150 2.015 1.029 × 10-4 ∴ 4.33 300 26. org ∆ε1 = 0. www.15(c). All rights reserved. Gas Turbine Theory.670 cycles of each loading. Fatigue and durability of structural materials (ASM International.72 Source: Manson. Lecturer’s Solutions Manual Problem 8. 72 © Pearson Education Limited 2009 .009 → Nf1 = 2 ×103 cycles ∆ε 2 = 0.007 → Nf2 = 1 ×104 cycles Using Miner’s rule (linear damage accumulation) n n n n + =1= + 3 2 × 10 1 × 104 N f1 N f 2 1=  12 × 103  1 × 104 n + 2 × 103 n n =  7  2 × 107 n  2 × 10  ∴ n ≅ 1.01N −f 0..7 N −f 0.06 + 0. Reprinted with permission of ASM International ®.asminternational. S. S. G.4 Haynes 188 at 1033K ∆ε = 0. GFC Rogers. where it is Figure 6. 2006). R. H Cohen. PV Straznicky. and Halford. 6th edition.HIH Saravanamuttoo. 5 Rigid bearings : Kr= rotor stiffness 1 ω= 1 K r  N  2  kgm  2 1 = = 2  =s m  mkg   s mkg  m=20 kg . PV Straznicky.235=  r  ⇒ Kr=548 × 106 N/m  20  Soft bearings: total stiffness 1 Kr + Ks 548 × 106 + 1 × 107 = = Kt K r Ks 548 × 106 × 1 × 107 = 1.HIH Saravanamuttoo. ω = πN 30 = 5.821×106 = 700.700 rpm 20 Plot the critical speed as a function of support stiffness! 73 © Pearson Education Limited 2009 . Lecturer’s Solutions Manual Problem 8. GFC Rogers.821 × 106 N/m First critical speed ω= 9.018 × 10–7 Kt = 9. 235 rad/s 1  K 2 5.7 rad/s = 6. Gas Turbine Theory. 6th edition. H Cohen.693 ≅ 6. 4. 6th edition.9 4.515 32.835.444 33. Lecturer’s Solutions Manual Problem 9.3 From curves. P1 ηc = 0.0 34. GFC Rogers. Flow compatibility is expressed by: m T1 P1 = m T3 P3 × P3 T × 1 P1 T3 And P3 P2 P2 − P3 P2 P P = − = − 0.5 31.HIH Saravanamuttoo.8 4.7 32.1 Note: In the problems for chapter 9. equilibrium point is at: P2 =4.064 34. H Cohen. P1 m T1 =33.95 2 P1 P1 P1 P1 P1 P1 ∴ m T1 P 288 P =14. PV Straznicky.95 2 × = 6.903 P2 P1 5.2 ×0. we are dealing with stagnation conditions at all times and the suffix ‘0’ has dropped throughout.05 2 = 0.903 2 P1 P1 P1 1100 m T1 P1 P2 P1 6. Gas Turbine Theory.795 74 © Pearson Education Limited 2009 . 988 × 1. PV Straznicky.5 − 1 4. Lecturer’s Solutions Manual P3 =0.835 =4.988 × 1. H Cohen.189 – 411.989KW Net power output = 264 KW 75 © Pearson Education Limited 2009 .8497 m= 33.59     −1.4 × 1. GFC Rogers.147 × 0.HIH Saravanamuttoo.795   Poutput = 675.005 × 1 288   3.59 P1 And from turbine characteristic graph: ηt = 0.620 = 263.835 ( )   0.01 = 1. 6th edition.988 kg/s 288 Net power output = mCpg ∆T34 − 1 mC pa ∆T12 ηm 1    1 4   Poutput = 1.8497 × 1100 1 −    4. Gas Turbine Theory.95 ×4. 135 0.156 2.0 T4 T3 P3 =2.HIH Saravanamuttoo.174 ∴ ∴ m T4 =188 P4 m T3 P3 when m T3 P3 m T4 P4 0.2 186.2 164.31 0.27   T3   76 © Pearson Education Limited 2009 .826 0.909 90.865 2. Gas Turbine Theory.930 88. PV Straznicky.50 1. GFC Rogers.270 (graphical solution) P4 =90.257 0.189 0.2 1   ∆T34  1 4   = 0.2247 0. 6th edition.1575   2.25 1.00 r= P3 . T3    r4  P3 P4  P3   P4 2. Lecturer’s Solutions Manual Problem 9.85 1 −  = 0.844 0.2 m T4 P4 = m T3 P3 × P3 T × 4 P4 T3 ∆T T4 = 1 − 34 T3 T3  ∆T34 1 = ηt 1 − 1  . and ηt = 0.2 205.85 P4 1 4   ∆T34 T3 T4 T3 1.918 90. H Cohen.05 0. 82 1.147 × 0.584 0. from graphical solution P1 T1 Hence T3=1170 K 77 © Pearson Education Limited 2009     .99 4. Gas Turbine Theory. 6th edition.77 3.105 and 3 = 4.602 0.2 ( P2 / P1 ) = T1 m T1 / P1 ( -----------------------------(A) ) From compatibility of work: T ∆T ∆T T3 C pgη m 0. Lecturer’s Solutions Manual From compatibility of flow: m T1 P1 ∴ = m T3 P3 ( × P3 T × 1 and P3=P2 P1 T3 ) m T3 / P3 ( P2 / P1 ) T3 = T1 m T1 / P1 ( ) T3 90. H Cohen. GFC Rogers.92 ∴ T P2 = 5.005 0.15 0.1575 × 1.566 0.704 3.1762   T1 (B) 5.98 T3 = = = 0.91 3.1762 3 -------(B) T1 T3 T1 C pa T1 T1 1.17 5.06.2 220 2.690 3.8 244 1. PV Straznicky.82 1.734 4.54 0.13 4.83 1.0 236 1.65 0.286  ∆T12 1  P2  =   − 1 T1 ηc  P1    P2 P1 m T1 P1 T3 T1 T3 T1 ηc  P2   P1    0.HIH Saravanamuttoo.286 ∆T12 T1 (A) T3 T1   1   ∆T  =     0. 55 78 © Pearson Education Limited 2009 m T3 P3 m T4 P4 P4 Pa .5 1.3 At outlet of gas generator turbine m T4 P4 = m T3 P3 × P3 T × 4 P4 T3 1   4   T4 1  .0678 0.0 63. 6th edition. Gas Turbine Theory.097 0. GFC Rogers.958 44.972 20.1067 0. × × 1 − ηt 1 −    P3 / P4   P4   ηt = 0.1583 0. H Cohen.664 1.85 1 1 4  1   1 −  P P  3 4 1 4 1  2  4   1       1 − ηt 1 −   P /P    3 4     P3 P4  P3     P4  1.3 1.496 4 Pa P3 P2 Pa P3 P3 Hence curves – from which power turbine pressure ratio is r = 1. PV Straznicky. = 1 − ηt 1 −  and   P3 / P4   T3   ∴ m T4 P4 = m T3 P3 1   4   P3 1  . Lecturer’s Solutions Manual Problem 9.940 62.96 × 2.0 104.92 1.60 = 2.387 The value of P4/Pa in the table is found from P4 P4 P3 P2 P4 P = = × 0.137 0.90 1.27 1.8 1.0 25.22 1.HIH Saravanamuttoo.064 0. 6th edition. PV Straznicky.5  T 1   ∴ Cpa    − 1 = ηm C pg T3 1 −  ηc  P1  P3 / P4        T1 must be given. Compressor efficiency from compressor characteristic once operating point m T1 m T3 P3 T = × × 1 where T3 unknown. Gas Turbine Theory. GFC Rogers. Lecturer’s Solutions Manual And gas generator turbine pressure ratio r=1.HIH Saravanamuttoo.61 Use work compatibility equation to find T3 ∆T12 ∆T34 T3  C pgη m  = ×   T1 T3 T1  C pa  1 1      1 4   P2  3. found from P1 P3 P1 T3 79 © Pearson Education Limited 2009 . H Cohen. so trial and error method required. 486 (b) 95% mechanical speed at 273 K. where =1− = 1 − ηt 1 −    P3 / P4   T3 T3   and ηt is constant.486 ηm C pg 0.5 K 0.5465 0. GFC Rogers.859   m pg  ∆T   ∆T  =  34  But:  34   1075 95% speed  T3 100% speed Therefore. Thus: m T4 P3  P3  = f  from gas generator turbine characteristics. T3= 1075 × 0.0 ( )   η m C pg 0.5 − 1 = 4.863   C paT1 0.5465 3. work compatibility yields γ −1   T1  P2  γ  ηm C pg ∆T34 = C pa   − 1 ηc  P1    ∆T34 = C pa = 1 T1   3.HIH Saravanamuttoo. Lecturer’s Solutions Manual Problem 9. 6th edition.5 − 1 4.  P4  Since power turbine is choking at all conditions considered. 80 © Pearson Education Limited 2009 . PV Straznicky. H Cohen.859 0. T3 (a) At 95% of speed.859 ηm C pg 0.863 At 100% speed we have similarly: ∆T34 = C pa 1 T1   C paT1 0. the gas generator turbine is ∆T34 operating at a fixed pressure ratio and hence fixed value of .4 For gas generator turbine m T4 P4 = m T3 P3 × P3 T × 4 P4 T3 γ −1   γ   ∆T34 T4 1  .6 ( )   η C 0. Gas Turbine Theory.863 × = 1214. 57 % 273 From the operating line ∆T = m= m T1 p = 439. GFC Rogers.862 439 × 0.5 = 3318 kw. Lecturer’s Solutions Manual N T1 (%design) = 95 288 =97.286 − 1 = 163.862 P1 P1 273  (4.76 273 = 20.005 × 163. 6th edition. 81 © Pearson Education Limited 2009 .6K 0.29)0.HIH Saravanamuttoo. 2 = 4. H Cohen. PV Straznicky.19 kg/s Power=20. Gas Turbine Theory.29 and ηc = 0.19 × 1. 5  T1  r − 1  4 − 1 mc − mb 1 1 1   =  = × ×  1 1 mc ηtηc  0.899 r 2 − 1 --------------------(1) 82 © Pearson Education Limited 2009 .70 P1    (m − m ) T  (m − mb ) T3 b 3  c  = c 1  1  P3 1 − 2  P3 1 − r 2  r  d And T3 is constant.3 ×    3. Gas Turbine Theory.3378 × 22. PV Straznicky.8 − 7.3378 = b mc mc T1 / P1 T1 / P1 Therefore at design point: mb T1 = 0.5 1 −  1   T3 1 −     r   4      m mc − mb = = 0. Lecturer’s Solutions Manual Problem 9. GFC Rogers.6622 = 0.5   3.HIH Saravanamuttoo. H Cohen.5   P  2 − 1  P1     P3 P2 = =r P4 P1 At design point: 1 1  3.5 Work compatibility yields: γ −1   γ   1   = m T1 mηt T3 1 −   c   P3 / P4   ηc   Let: 1   3.6622 mc mc mb m = 1 − 0.7 4 1 − 1/16 15 mc T1 / P1 − mb T1 / P1 = 3.85 3.5  1  3.8 = 7. 6th edition. mc − mb P = 3 ( mc − mb )d ( P3 )d 1 1 r 1− 2 2 r = r Thus : P =P 3 2 1 1 1 − 2 rd 1 − 2 rd rd 1− mc T1 / P1 − mb T1 / P1 r 1 − 1 r 2 r2 −1 = = 22.8 0. 85 0.299 0.638 3.261 0.5 − 1 1    1 − 1 3.865 13. 6th edition.5 2.116 0.05 16.775 4 From (1): mc T1 / P1 = 5.899 2.0 2.5 12.5 1.5   r  r r 3.4457(r1 3.0 9.192 0.5 1 −  1    r    3 Now: mb T1 / P1 = × 7.899 r 2 − 1 0.230 0.207 0.775+ 3.775 5.94 14.83 11.430 0.4457( r1 3.63 0.164 0.270 0.25 1.886 83 © Pearson Education Limited 2009 right-hand-side .694 2.369 0. r = 2.0 1.015 7. Gas Turbine Theory.899 r 2 − 1 r r2 r2 −1 3.06 2.4457  r1 3.578 2.775 + 3. mc T1 / P1 = 5. solve by trial and error 1   3.5 − 1) ------------------(2) 1   3. PV Straznicky.5 − 1) .08 18.71 0.35 3.301 0.25 2. Lecturer’s Solutions Manual Also from work compatibility: mc T1 / P1 − mb T1 / P1 mc T1 / P1 = 0.899 r 2 − 1 left-hand-side 13.657 6.775 + 3.HIH Saravanamuttoo.576 1 3.5 1.560 From curves.7 = 5.607 2.5 1 −  1    r    5.08 Hence.5 3.082 − 1 = 12.775 + 3.889 r 2 − 1 (2) becomes: 1− 5.25 3. GFC Rogers.133 0.82 0.607 5.25 = 0. H Cohen.29 8.