Gas Turbine Design Report

Concordia University Faculty of Engineering and Computer Science Department of Mechanical Engineering

Gas Turbine Design Project Report Winter 2009

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Summary 1. Part A: Engine Design 1.1 LPC exit (2) 1.2 HPC exit (3) 1.3 Combustor exit (4) 1.4 HPT exit (5) 1.5 Inter-turbine duct (6) 1.6 LPT exit (7) 1.7 Inter-turbine duct (8) 1.8 Power turbine (8) 1.9 Engine horsepower and SFC

2. Part B: Turbine Design 2.1 Mean Line design of HPT 2.1.1 Rotational speed and gas path 2.1.2 Velocity triangles and nozzle & rotor loss coefficients 2.2 Hub and tip velocity triangles 2.3 Vane and Blade parameters 2.4 Nozzles and rotor loss coefficients 2.5 Off-design performance 2.5.1 Reduction of the speed by 20% 2.5.2 Reduction of the pressure ratio by 20%

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Introduction Blah blah blah

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1. Part A: Engine Design In this part we have to determine the cycle points, temperatures, pressures, compressors and turbine work and the engine horsepower General assumptions at intake: To1 = 100oF= 310.78 oK

Altitude= 4000 ft =1219.2m 88100,15 Pa

P01 =

1.1 LPC exit (2) T T02 = T01+ 01 ( P.R η

γ −1 γ

Assumptions m = 12 lb/sec P.R=4.25 η=.86

− 1)

P02 = P.R * P01

Results

W2= Cp (T02 – T01) p0 2 T0 2 W 2

374425,6 3 Pa 497,02 K 186339,6 3 J.kg-1

1.2 HPC exit (3) m3= (1-10%) m2

T T03 = T02+ 02 ( P.R η

γ −1 γ

− 1)

Assumptions cooling air = 10% P.R=2.65 η=.84

P03 = P.R * P02

Results

W3= Cp (T02 – T01) m 3 p0 3 T0 3

5,44 Kg/s 992227,9 3 Pa 686,99 K

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W 3

190870,4 3 J.kg-1

m4= m3 (1+Fuel air ration) P04 = P03(1-ΔP/P) T04 =

(m

3 C p T03 +η * f .a.r * m 3 * HV )

1.3 Combustor exit (4)

C p ( m 3 + m fuel )

BTU/lb

Assumptions Fuel ration=0.02

η=.99

HV=16760

ΔP/P=0.02

Results m4 p04 T04

4,90 Kg/s 972383,3 7 Pa 1331,07 K

m5= m4+ (1+0.5*cooling air) γ

 T  γ −1 P05 = P04 1 + 1  05 −1   η  T04   T05 = T04 -

(m C i

p

1.4 HPT exit (5)

(T03 −T02 )

m5 C p

Assumptions η=.88

W5= Cp (T05 – T04)

Results m5 p05 T05 W5

5,17 505486,6 0 1154,45 -202945,7 0

Kg/s Pa K J.kg-1

m 6= m 5 

P06 = P05  1 +



∆p   p  

T06 = T05

1.5 Inter-turbine duct (6) Assumptions

ΔP/P=0.02

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Results m7= m6+ (1+0.5*cooling air)

p06 T06

γ

 T  γ −1 P07 = P06 1 + 1  07 −1    η  T06   T07 = T06 -

(m C i

p

(T02 − T01 )

495376, 87 Pa 1154,45 K

1.6 LPT exit (7)

m7 C p

W7= Cp (T07 – T06)

Assumptions η=.90

Results m7 p07 m 8= m 7 

P08 = P07  1 +



T07 W7

∆p   p  

5,44 244768,5 1 986,67 -192798,4 1

Kg/s Pa K J.kg-1

T08 = T07

1.7 Inter-turbine duct (8) Assumptions

ΔP/P=0.02

Results p08

m 9= m 2   P atm 09  ∆ P P =  1−   P

T08

     

   p 09 T09 = T08 - η * T08 1 −  p  08  

γ −1 γ

  

    

239873,1 4 Pa 986,67 K

1.8 Power turbine (8) Assumptions η=.93

W9= Cp (T09 – T08)

ΔP/P=0.02

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Results

HP= m9*W9

m9 p09 T09 W9

f .a.r SFC = m3* HP

5,44 89898,11 787,16 -229265,8 5

Kg/s Pa K J.kg-1

1.9 Engine horsepower and SFC

Engine Horspower

SFC

1247912,3 8 W kg/Kw/ 0,314046 h

1673,48 HP 516290, lb/hp/ 09 h

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2. Part B: Turbine Design 2.1 Mean Line design of HPT 2.1.1 Rotational speed and gas path Assumptions N=

AN

2

AN2

π( rt 2 −rh 2 )

rt = rh

π* AN 2 30 2 * U h

rh =U h

60 2πN

2

Uh +1

M1 M3 α1 α3

4E+1 0 rpm.in² 1100 ft/sec 0,1 0,4 10 deg 10 deg

Results Hub to tip ratio

rh/r t

Hub radius

rh

Tip radius

rt

Rotational speed

N

0.71 0.10 3.79

m in

0.14 5.34 30019.0 51

m in rp m

The radius rh is calculated from the blade hub speed. The radius rt is calculated from the area A1 which is calculated from the formula A1 =

in the report.

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m1RT1 calculated later P1V1

2.1.2 Velocity triangles and nozzle & rotor loss coefficients T1 =

P1 =

T01 γ −1 2 1+ M1 2

T3 =

P01  γ −1 2  1 + M 1   2  

γ γ −1

V1 = M 1 . γ .R.T1

P3 =

T03 γ −1 2 1+ M3 2

2

T2 = T3 +

P03  γ −1 2  1 + M 3   2  

γ γ −1

V3 = M 3 . γ .R.T3

Va1= V1 cos α1

P2 =

Vr 3 − Vr 2 2Cp

P01

2

γ

 T01  γ −1    T '2 

T’2=T2-λN(T02-T2)

Va3= V3 cos α3

Va2

Vu3= V3 sin α3

Vu2=-

=Vu2/tanα2 Vu1= V1 sin α1 WHPT/Um –Vu3 A1 =

 U + Vu 3   α r 3 = tan −1  m  Va 3 

m1RT1 P1V1

Vr2=Vu2-Um

Vr3=Va3/cos αr3 2

V r 2 = V ru + V a 2

2

A3 =

m3 RT3 P3V3

2

V 2 = Vu + V a 2

ρ2 =

2

P2 R.T2

Results

at 1 V1 Vu 1 Va 1

at 2 71.3 1 12.3 8 70.2 3

m/ s m/ s m/ s

Vu2 Va2 V2 V ru2 Vr2

at 3 511. 36 186. 12 544. 18 146. 95 237. 14

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m/ s m/ s m/ s m/ s m/ s

V3 Vu 3 Va 3 Vr3

262. 39 45.5 6 258. 40 484. 61

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YN =

P0 r 2

P01 - P02 P02 - P2

T = P2  0 r 2  T2

YR =

γ

γ

 γ −1  

V 2  T0 r 2 = T2 +  r 2   2C p   

P0r2 - P0r3 P0r3 - P3

P0 r 3

 T  γ −1 = P3  0 r 3   T3 

V 2  T0 r 3 = T3 +  r 3   2C p   

The nozzle and rotor loss coefficients are: YN YR

0,111 0,166

Efficie ncy ξN ξR ηtt

0,0768 0,1204 0,87

These losses were verified by the efficiency calculated from the losses, which is 87% knowing that the efficiency should be 88%.

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2.2 Hub and tip velocity triangles Free vortex : Vu1= cste1/r Vu2= cste2/r Vu3= cste3/r U=N.r 2

V1 = Vu1 + V a1

2

2

V 2 = Vu 2 + V a 2

2

2

V3 = Vu 3 + Va 3

2

Vru 2 = Vu 2 − U 2

Vr 2 = Vru 2 + Va 2 V  α 1 = tan −1  u1   Va1 

V α 2 = cos −1  a 2  V2

   

V 2 T2 = T02 −  2  2C  p

  

   

V 2 T0 r 2 = T2 +  r 2  2C  p

   

R=

 U + Vu 3   α r 3 = tan −1   Va 3   (V 2 − Vr 2 2 T3 = T2 −  r 3  2C p  V 2 T0 r 3 = T3 +  r 3  2C  p P3 = ρ 3 .R.T3

P2 = ρ 2 .R.T2 T P02 = P2  02  T2

Vr 3 = Va 3 / cos α r 3 V  α 3 = tan −1  u 3   Va 3 

  

V α r 2 = tan −1  ru 2  Va 2 V2 T1 = T01 −  1  2C  p

2

γ

T P03 = P3  03  T3

 γ −1  

γ

 γ −1  

T2 − T3 T1 − T3

M = V / γ .R.T The Results are:

r

0.10

0.11

0.12

0.13

0.14

T2

1151.00

1180.15

1202.21

1219.31

1232.82

T3 p2

1124.09 597057.7

1124.32 612182.9

1124.50 623626.3

1124.63 632492.5

1124.74 639501.3

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   

   

6 454826.2 9 1068336. 66 506035.9 1

7 454919.9 5 991016.3 7 505723.1 0

0 454990.8 2 937404.7 0 505486.6 0

9 455045.7 2 898495.0 0 505303.4 6

1 455089.1 3 869247.2 0 505158.7 6

73.18

71.58

70.00

68.45

66.94

11.99

10.91

10.00

9.23

8.57

V2

643.30

588.92

544.18

506.80

475.18

V3

264.16

263.15

262.39

261.79

261.32

Vu2

615.79

558.74

511.36

471.39

437.21

Vu3

54.87

49.78

45.56

42.00

38.96

Va2

186.12

186.12

186.12

186.12

186.12

Va3

258.40

258.40

258.40

258.40

258.40

Vr2

364.31

292.18

237.14

201.07

186.44

Vru2

313.18

225.23

146.95

76.08

11.00

Vr3

441.09

462.26

484.61

507.95

532.12

0.13

0.27

0.37

0.45

0.52

p3 p02 p03 α2 deg α3 deg

R

This table shows:  A radial increase in temperature and pressure for the leading edge, this is due to the change in the radius so a change in the speed. However for the trailing edge, the radial conditions are almost the same, since we are at the exit of the HPT,  A temperature and a pressure drop between leading and trailing edge, which is normal since we transfer energy from the flow to the blades in a turbine,  The turning is lower as the radius increases, in order to reduce the losses on the tip (tip leakage, secondary…)  The reaction increases with the radius since the temperature follows the same pattern.

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2.3 Vane and Blade parameters Stagger angle is calculated from Fig5

(rt - rh ) h/c

Chord

c=

Axial chord

ca=c*cos(γ)

Pitch of vanes

s = ca

ψ 2.(tanα 1 + tan α 2 ) cos 2 α 2 N=

Number of vanes/blades

2π .rm s

Results To be validated Vane h/c Ψ TAT C Ca γ s Nv

Blade 0,7 0,75 0,04 0,056 0,036 50 0,040 18

in m m deg m vane

h/c Ψ TAT C Ca γ s Nb

So the final design will be 18 vanes and 44 blades.

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1,3 0,9 0,02 0,028 0,022 38 0,018 44

in m m deg m blade

2.4 Nozzles and rotor loss coefficients  YN = YP * fRE + YS + YTET +YTC 2  YP , AMDC K P + Yshock  3  

YP = 0.914 

YP, AMDC =

   t /c  β 1  β1  YP( β 1=0 ) +    YP( β 1=α 2 ) - YP( β1 =0 )   max  α 2  α 2     0.2

[

]

   

β1 α2

With Kp = 1 – K2 (1- K1), K1 = 1-1.25(M2-0.2), K2 = (M1/M2)2 YS = 1.2 YS, AMDCKS

YS, AMDC = 0.0334 f(AR)

With

 cosα 2   cosβ1

CL s/c = 2(tanα

  C L  2 cos 2α 2    3   s/c  cos α m −1

1

+ tanα2) cosαm

αm = tan

1  (tan α1 − tan α2 )  2 

f(AR) =

1 − 0.25 2 − h c h c

Using AMDC loss system, the loss coefficients for the stator are: secondary losses

profile losses

Trailing

edge

losses book page 330

tmax/c s/c Yp (β1=0)

0,2

assumption

from graph

Yp (β1=α2)

Yshock

0 0,10

M2

0,80

K1

0,25

0,99

Re

1,38E+06

Yp

M1 is too small

0,015

Kp

f(Re)

from graph

0,0387

M1

K2

-0,91

rad

-52,12

deg

t/s correction factor

CL/(s/c) 0,14

Yp AMDC

1,02

αm

0,705 0,04

fAR

3,59 Ys AMDC

0,08

K3

0,84

Ks

0,99

Ys

0,093

reynolds > 10^6

0,938 0,022

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YTET

0,02 6

≠ 0,02 from 1,05 graph 0 ,00 6

Total Pressure loss

Total AMDC loss YN

0,12

compared to

0,11

Assuming an unshrouded blade with a radial tip clearance of 2% of blade height, the loss coefficients for the rotor are: profile losses tmax/c

0,200

s/c

0,525

Yp (β2=0)

0,040

Yp (β2=αr3)

0,100

Yp AMDC

secondary losses

Tip clearance losses

fAR

book page 329 assumption of unshrouded blades with radial tip clearance of 2% of blade height

0,6083

αm from graph from graph

CL/(s/c)

-0,379

rad

-21,73

deg

4.4146 0,098

Ys AMDC 0,069

Δp/q1 hub

0,027

K3

0,5397

Δp/q1 shock

0,019

Ks

0,92

Yshock

0,005

Mr2

0,350

Mr3

0,739

K1

0,327

K2

0,224

Kp

0,849

Re

5,16E+05

f(Re)

Yp

1

1

0,04

0,108

Ys

Trailing edge losses

(cosβ3/cosβ2) ^2 /(1+rm/rt)

0,263

λ

0,008

B(k/h)

0,010

βm

0,354

Ys+Yk

0,121

YTC

rad

0,01 3

book page 330 t/s correctio n factor

0,032 1,05

YTET

Total AMDC loss YR 0,17

≠ 0,02 from graph

0,009

Total Pressure loss compared to

0,17

We notice that the loss coefficients calculated with the AMDC method are very close to those calculated with the pressure losses. We also notice that the most predominant losses are the secondary.

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Blade metal area ration

τ K1 K2 K3 K4 ρ

300 15 55,6 -5,2 0,6 0,315

h

lb/in3 -

Lm2

45,2

KSI

Lm1

45,2

KSI

maximum life Actual life

σc

247,3

23,48

KSI

K5 AH/AT

175,7

16,68

≈1

2.3

Since the Lm1 equation is a second order equation, we have 2 solutions for σc and so for K5 , but only one of them seems to be reasonable. We chose 2.3 as AH/AT.

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2.5 Off-design performance 2.5.1 Reduction of the speed by 20%  ioff = α r 2 off - β2 with β2 is metal angle (unchanged) Incidence on the rotor Umean off β2 αr2 off

291.5 2 metal  γ −1 angle 2iscst 1 40.29  1− M 

  Y’=0,88

50.39 i off

10,10

−γ

γ−1 −1 −1  

r3  2 2 φ rad −γ 1 γ −1  2 γ− 1 −1 + M r3  2  deg

incidence on the rotor

x’=

−1.6

d    s 

−2

 cos β2     cos β3 

* 

* (α r 2 off − α r 2 )

Incident Loss on rotor 2

and

φ 2 = 1 − ∆φ p

∆φ p =0.778*10 x’ + 0.56*10 x’ + 0.4*10 x’ + 2.054*10 incident profile incident secondary losses losses assumed 2 χ'' x ”) d/c 0,044 0,18 Y’’= Ys,des(exp(0.9 x ”) + 13 x ” + 400 -5

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s/c

0.52

d/s

0.08

β2

40.29

deg

0,70

rad

β3

57.78

deg

1,01

rad

αr2

38.29

deg

Mr 3

0.75

χ' ∆φ²P YiP

312.23 0,0093 0 0,0117 3

x” =

αoff r 2 − β2 180 − (β2 + β3 )

0 < χ' < 800

2

(Y/Ydes−)1S.5

 cos β2 *  cos iS 3  Yβ

   

-10

2,08−0.3

d    * C 0,225 

3

0 < χ'' < ,3

Assume same TET and Tip clearance losses

Total loss YR

20

0,31 7

-19

x’6

2

Efficiency YN drop ξ N = 1 + 0.5 * γ * M ξN ξR ηtt



2 2

0,0690 YR 2 0,2298 1 + 0. 5 * γ * M r 3 Knowing that we originally had to 0,8256

ξR =

have 0,88

dηtt



η tt off

1 2 ξ .V + ξ R (Vr3 ) 2 1+ N 2 2.(h 01 − h03 )

6,19 % =

With (h01-h03)=U2Vu2-U3Vu3=Um(Vu2-Vu3)

Reducing the speed by 20% will lead to:  an increase of αr2 and then generate a positive incidence of 10 degrees  an increase of losses on the rotor  a reduction of the efficiency by 6.19%

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2.5.2 Reduction of the pressure ratio by 20%

Design conditions

80% Pressure ratio

22

% decrease 1.67

Design 71.31

80% PR 70.12

Vu1

12.38

12.18

1.67

Va1

70.23

69.05

1.67

Vu2

471.32

369.81

21.53

Va2

171.55

134.60

21.54

V2

501.56

393.54

21.54

78.68

5.40

93.13

Vr2

188.73

134.71

28.62

V3

262.39

262.47

-0.03

Vu3

45.56

45.58

-0.03

Va3

258.40

258.48

-0.03

V1

Vru2

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Vr3

508.71

484.66

4.73

The effect of reducing the pressure ratio by 20% on the velocity triangle is:  a reduction of the incidence (so a reduction of losses),  a reduction of speeds at the stator and the rotor,  a reduction of the component Vu, which reduces the energy transfer.

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Conclusion

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