Concordia University

Faculty of Engineering and Computer Science

Department of Mechanical Engineering

Gas Turbine DesignProject Report

Winter 2009

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Summary1. 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 DesignIn this part we have to determine the cycle points, temperatures, pressures, compressors and turbine work and the engine horsepower

General assumptions at intake:

Altitude= 4000 ft =1219.2m To1 = 100oF= 310.78 oK P01 = 88100,15 Pa

1.1 LPC exit (2)

Assumptionsm = 12 lb/sec P.R=4.25 η=.86

Results

p02 374425,63 PaT02 497,02 KW2 186339,63 J.kg-1

1.2 HPC exit (3)

Assumptionscooling air = 10% P.R=2.65 η=.84

Results

m3 5,44 Kg/sp03 992227,93 PaT03 686,99 K

W3 190870,43 J.kg-1

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m3= (1-10%) m2

T03 = T02+

P03 = P.R * P02

W3= Cp (T02 – T01)

T02 = T01+

P02 = P.R * P01

W2= Cp (T02 – T01)

1.3 Combustor exit (4)

AssumptionsFuel ration=0.02 HV=16760

BTU/lb η=.99 ΔP/P=0.02

Results

m4 4,90 Kg/sp04 972383,37 Pa

T04 1331,07 K

1.4 HPT exit (5)

Assumptionsη=.88

Results

m5 5,17 Kg/sp05 505486,60 PaT05 1154,45 K

W5 -202945,70 J.kg-1

1.5 Inter-turbine duct (6)

AssumptionsΔP/P=0.02

Results

p06 495376,87 PaT06 1154,45 K

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m5= m4+ (1+0.5*cooling air)

P05 = P04

T05 = T04 -

W5= Cp (T05 – T04)

m6= m5

P06 = P05

T06 = T05

m4= m3 (1+Fuel air ration)

P04 = P03(1-ΔP/P)

T04 =

1.6 LPT exit (7)

Assumptionsη=.90

Results

m7 5,44 Kg/sp07 244768,51 PaT07 986,67 K

W7 -192798,41 J.kg-1

1.7 Inter-turbine duct (8)

AssumptionsΔP/P=0.02

Results

p08 239873,14 Pa

T08 986,67 K

1.8 Power turbine (8)

Assumptions η=.93

ΔP/P=0.02

Results

m9 5,44 Kg/sp09 89898,11 PaT09 787,16 K

W9 -229265,85 J.kg-1

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

P07 = P06

T07 = T06 -

W7= Cp (T07 – T06)

m8= m7

P08 = P07

T08 = T07

m9= m2

P

09

=

T09 = T08 -

W9= Cp (T09 – T08)

1.9 Engine horsepower and SFC

Engine Horspower 1247912,38 W 1673,48 HP

SFC 0,314046 kg/Kw/h 516290,09lb/hp/h

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HP= m9*W9

SFC = m3*

2. Part B: Turbine Design

2.1 Mean Line design of HPT

2.1.1 Rotational speed and gas path

Assumptions

AN2 4E+10 rpm.in²

Uh 1100 ft/secM1 0,1 M3 0,4 α1 10 deg

α3 10 deg

Results

Hub to tip ratio rh/rt 0.71

Hub radius rh 0.10 m3.79 in

Tip radius rt 0.14 m5.34 in

Rotational speed N 30019.051 rpm

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 calculated later in the report.

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N =

2.1.2 Velocity triangles and nozzle & rotor loss coefficients

Results

at 1 at 2 at 3

V1 71.31 m/s Vu2 511.36 m/s V3 262.39Vu1 12.38 m/s Va2 186.12 m/s Vu3 45.56

Va1 70.23 m/s V2 544.18 m/s Va3 258.40

Vru2 146.95 m/s Vr3 484.61

Vr2 237.14 m/s

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T’2=T2-λN(T02-T2)

Va1= V1 cos 1 Va3= V3 cos 3 Va2 =Vu2/tanα2

Vu1= V1 sin 1 Vu3= V3 sin 3 Vu2=-WHPT/Um –Vu3

Vr2=Vu2-Um

Vr3=Va3/cos r3

A2=A3

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The nozzle and rotor loss coefficients are:

YN 0,111YR 0,166

Efficiency

ξN 0,0768 ξR 0,1204 ηtt 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

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 1124.09 1124.32 1124.50 1124.63 1124.74

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Free vortex : Vu1= cste1/r Vu2= cste2/r Vu3= cste3/r U=N.r

p2 597057.76 612182.97 623626.30 632492.59 639501.31

p3 454826.29 454919.95 454990.82 455045.72 455089.13

p02 1068336.66 991016.37 937404.70 898495.00 869247.20

p03 506035.91 505723.10 505486.60 505303.46 505158.76

α2 deg 73.18 71.58 70.00 68.45 66.94

α3 deg 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

R 0.13 0.27 0.37 0.45 0.52

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.

2.3 Vane and Blade parameters

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Stagger angle is calculated from Fig5

Chord

Axial chord ca=c*cos(γ)

Pitch of vanes

Number of vanes/blades N

Results

To be validated

Vane Blade

h/c 0,7 - h/c 1,3 -Ψ 0,75 - Ψ 0,9 -TAT 0,04 in TAT 0,02 inC 0,056 m C 0,028 mCa 0,036 m Ca 0,022 mγ 50 deg γ 38 degs 0,040 m s 0,018 mNv 18 vane Nb 44 blade

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

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2.4 Nozzles and rotor loss coefficients

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

prof ile losses secondary losses T railing edge losses

book page 330

tmax/c 0,2 assumption fAR 1,02 t/s 0,026 ≠ 0,02 s/c 0,705 αm -0,91 rad correction factor 1,05 from graph

Yp (β1=0) 0,04 from graph -52,12 deg Yp (β1=α2) 0,14 from graph CL/(s/c) 3,59 YTET 0,006

Yp AMDC 0,0387 Ys AMDC 0,08

Yshock 0 M1 is too small K3 0,84

M1 0,10 Ks 0,99

M2 0,80

K1 0,25

K2 0,015 Ys 0,093

Kp 0,99

Re 1,38E+06 reynolds > 10^6

f(Re) 0,938 Yp 0,022

Total AMDC loss Total Pressure loss

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YN = YP * fRE + YS + YTET +YTC

YP = 0.914

YP, AMDC =

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 = 2(tan1 + tan2) cosm m = tan f(AR) =

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 secondary losses Tip clearance losses

tmax/c 0,200 fAR 0,6083 book page 329

s/c 0,525 αm -0,379 rad assumption of unshrouded blades

Yp (β2=0) 0,040 from graph -21,73 degwith radial tip clearance of 2% of blade height

Yp (β2=αr3) 0,100 from graph CL/(s/c) 4.4146

Yp AMDC 0,069 Ys AMDC 0,098 (cosβ3/cosβ2)^2/(1+rm/rt) 0,263

Δp/q1 hub 0,027 K3 0,5397 λ 0,008

Δp/q1 shock 0,019 Ks 0,92 B(k/h) 0,010

Yshock 0,005 βm 0,354 rad

Mr2 0,350 Ys 0,108 Ys+Yk 0,121

Mr3 0,739

K1 0,327 Trailing edge losses YTC 0,013

K2 0,224

Kp 0,849 book page 330

Re 5,16E+05 t/s 0,032 ≠ 0,02

f(Re) 1correction factor 1,05 from graph

Yp 0,041 YTET 0,009

Total AMDC loss Total Pressure lossYR 0,17 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 300 h K1 15 K2 55,6 K3 -5,2

K4 0,6 lb/in3 ρ 0,315 - Lm2 45,2 KSI maximum life Lm1 45,2 KSI Actual life

σc 247,3 23,48 KSI K5 175,7 16,68 AH/AT ≈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%

Incidence on the rotor

Umean off 291.52

β2 40.29 metal angle is cst

αr2 off 0,88 rad

50.39 deg

i off 10,10 incidence on the rotor

Incident Loss on rotor

incident profile losses incident secondary losses

d/c 0,044 assumed '' 0,18

s/c 0.52 (Y/Ydes)S 2,08 0 < '' < ,3

d/s 0.08 YiS 0,225

β2 40.29 deg 0,70 rad

β3 57.78 deg 1,01 rad

αr2 38.29 deg Assume same TET and Tip clearance losses

Mr3 0.75

' 312.23

²P 0,00930 0 < ' < 800 Total loss

YiP 0,01173 YR 0,317

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= - β2 with β2 is metal angle (unchanged)

Y’=

x’= * *

and

=0.778*10-5 x’ + 0.56*10-7 x’2 + 0.4*10-10 x’3 + 2.054*10-19 x’6

Y’’= Ys,des(exp(0.9 x ”) + 13 x ” + 400 x ”)

x” = * *

Efficiency drop

ξN 0,0690 ξR 0,2298 ηtt 0,8256 Knowing that we originally had to have 0,88

dηtt 6,19 %

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

2.5.2 Reduction of the pressure ratio by 20%

Design conditions 80% Pressure ratio

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Design 80% PR % decreaseV1 71.31 70.12 1.67

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

Vru2 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

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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