1 Turbomachinery Class 11. 2 Axial Flow Compressors: Efficiency Loss: Centrifugal Compressors...

1

Turbomachinery

Class 11

2

• Axial Flow Compressors: • Efficiency Loss:

• Centrifugal Compressors • Efficiency Loss:

• Axial Flow turbines: • Efficiency Loss:

1.4h

4b

1.75h

3.63[ .294] 1 0.586

[ .360] 1 10 ...cos

Tw tip

m

tip

Baskharone

rTurbine p K K Z

h r

ECompressor p

h E h

3

Configuration Selection & Multidisciplinary Decisions

• Turbomachinery Design Requires Balance Between:

Performance

Weight

Cost

Optimization Approach

• A Strategy:– Find feasible solution(s) within each discipline– Use each as starting points for multi-disciplined optimization

• Single vs. Multi-Disciplinary Optimization– A discipline’s potential vs. a balanced design– Trading away potential in one discipline to improve another (often

to find feasible design space)

• Pointers– Design variable count: less is more– Initially utilize large scale perturbations to identify gradients– Variable side constraints: consult with other disciplines for input

5

Turbomachinery Design

• Consider Turbine Efficiency & Stress

• Performance - Smith Correlation for simplicity– "A Simple Correlation of Turbine Efficiency" S. F. Smith, Journal

of Royal Aeronautical Society, Vol 69, July 1965– Correlation of Rolls Royce data for 70 Turbines– Shows shape of velocity diagram is important for turbine efficiency– Correlation conditions

- Cx approximately constant

- Mach number - low enough

- Reaction - high enough

- Zero swirl at nozzle inlet

- "Good" airfoil shapes

- Corrected to zero clearance

7

Smith Turbine Efficiency Correlation

94% 92% 90% 88%

0.8

1.2

1.6

2.0

2.4

2.8

0.4 0.6 0.8 1.0 1.2 1.4

Cx/u

E

Increasing

Note: The sign of E should be negative

8

Turbomachinery Design

• Efficiency Variation on Smith Curve

– Increasing E from 1.33 to 2.4 [more negative] (at Cx/U=0.6):

• Higher turning increasing profile loss faster than work.

– Raising Cx/U from 0.76 to 1.13 (at E=1.2):• Higher velocity causes higher profile loss with no

additional work

– Remember - Mach number will also matter!

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Smith Turbine Efficiency Correlation

94% 92% 90% 88%

0.8

1.2

1.6

2.0

2.4

2.8

0.4 0.6 0.8 1.0 1.2 1.4

Cx/u

E

Increasing

Note: The sign of E should be negative

Typical Optimization Formulations

Aero Structures

Efficiency Weight, Pull

Design Variables Objective Function(s) Thickness distributionChord distributionCG offsets (stacking)

Design Constraints

Design point flow & pressure profile Off-design lapse StabilityCasing clearance

Material propertiesStressTuningFlutter

Airfoil Structural Overview

• Tools• Hand calculations, finite element analysis

• Design responses: stress, deflection, frequencies, mode shapes

• Design constraints• Strength, life• Tuning• Aero-elastic stability (flutter)• HCF [High Cycle Fatigue] margin

Low Cycle Fatigue [LCF] Considerations

• Life Limited Parts Vs Limited Useful Life– Disks & high pressure cases – removed at end of certified life– Blades – removed for cause / wear out modes, such as airfoil

erosion• Assessment

– Attachment fillet Kt’s available via Peterson’s or FEA– Nominal stress– S-N curve

Blade Vibration

• Cantilevered structures attain various modes: bending, torsion, coupled bending / torsion

• Each mode has its own natural frequency• Effect of rotation [shaft] is to stiffen structure and raise natural

frequency• Structural design should be resonance free operating condition

at: design speed, idle speed and other key operating points• Campbell diagram shows possible matches [Excitation] between

vibrational mode frequencies and multiples of shaft rotation [N]• Multiples of N caused by stators, blades, struts in neighboring

rows

• Examples: – Forced spring – mass damping– Chinook helicopter

13

Motion of a damped spring-mass system

14

/ / /0

220

220

220

0

: 1 02

2 02

3 02

kmy cy ky where natural frequency of system

m

cCases signal decays overdamped

m

ccritically damped

m

cunderdamped

m

Forced motion: Damped spring-mass system

15

/ / / cos[ ]my cy ky A t

CHINOOK HELICOPTER

16

Airfoil Tuning Represented on Campbell Diagram

• Airfoil frequency vs. rpm• Excitation orders

– Static flow disturbance relative to the rotating frame

– Source = inlet distortions– Freq = EO*RPM/60

• Project Requirements– 1st bending @ RL > 20%– 2nd & 3rd modes @ RL > 5%

HCF Strength Assessed with Goodman Diagram

Steady Stress (ksi)Steady Stress (ksi)

Vib

rato

ry S

tres

s (k

si,

0-p

eak)

Vib

rato

ry S

tres

s (k

si,

0-p

eak)

UltimateUltimateStrengthStrength

AlternatingAlternatingStrengthStrength

AMS4928 R=-1 Goodman DiagramAMS4928 R=-1 Goodman DiagramSmooth, Minimum PropertiesSmooth, Minimum Properties

Vibratory LimitVibratory Limit

Steady LimitSteady Limit

Stresses

21

22

Secondary Air Systems

23

S S RR

24

Turbomachinery Design Structural Considerations

Centrifugal stresses in rotating components• Rotor airfoil stresses

– Centrifugal due to blade rotation [cent]• Rim web thickness

– Rotating airfoil inserted into solid annulus (disk rim). – Airfoil hub tensile stress smeared out over rim

• Disk stress [disk]– Torsional: Tangential disk stress required to transfer

shaft horsepower to the airfoils– Thermal: Stresses arising from radial thermal

gradients• Cyclic effect called low-cycle fatigue (LCF)

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Turbomachinery DesignStructural Considerations

• Blade pitch [s] at Rmean chosen for performance s/b, h/b values• Need to check if [s] too small for disc rim attachment

• number of blades have an upper limit• Fir tree holds blade from radial movement, cover plates for axial

• slight movement allowed to damp unwanted vibrations• manufacturing tolerances critical in fir tree region

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Structural Design Considerations• Airfoil Centrifugal Stress

Blade of constant cross section has mass:

2BMPull r

g

2RT DD

h

4

T Rm

D Dr

2 2

2.

222.

2 2 2.

[ ]

sec

12

0.5 2 2

T

H

centrifugal m

centrifugalcent

m m

R

cent T H

m TR

cent m T H

dF Rdm R AdR

dFdRdR

A

for constant blade cross tional area

U RRdR

R

N r r

An

27


Centrifugal stress is limited by blade material properties

2

2

3

[ ][ ]

[ ]

2[ / ]

2 2 12 2 2 12 60 30

0.28 / [ ]

[ ]2

ccs

B

T H T Hmean

metal metal cs

T Hblade

ccs

Blade Pull P lbfStress psi

Blade cross section area A in

MPull r

g

D D D D N NR rad s

M mass L A lbm in for steel

D DL in

PStress

A

2

2

2 2 12 900 2 790,000metal anT H T H A ND D D D

Ng

Aan

29


Centrifugal stress is limited by blade material properties

Gas bending

Cent. bending

L

From Rear

Mechanical Design – Minimizing Root Moments

Blade is balanced about rim to minimizeBlade is balanced about rim to minimize

Bearing stress maldistributionBearing stress maldistribution

Bending stress on disk webBending stress on disk web

Disk rim rollingDisk rim rolling

Blade airfoil is tilted to offset root bending stressesBlade airfoil is tilted to offset root bending stresses

Axial & tangential tiltsAxial & tangential tilts

CGCG

Air pressureAir pressure

PullPull

CG OffsetCG Offset


• Bending stress on a cantilevered bead under aerodynamic loading [Kerrebrock]

• Centrifugal stress is typically larger than bending stress

31

2

max

12

x Tbending s

T

C rspU c t

c/s=

32

Typical Centrifugal Stress Values

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0 0

20 3 2 3 3 2 2

3 2 3 2

3 2

: 1200 4.0

0.75 0.51 10,500 / min

50% 0.7 2.5 [ ]

1 2 3

/ /

/ tan tan

tan tan /

T H

mean

u u u u u u

u u

First stage turbine T K p bar

r m r m N rev

R E

stator inlet stator exit rotor exit

E h U C C U C W U C W U

E W W U

R

3 2

2 3 2 2

2 2

2 68.2 46.98

50%

/ 2 0.315 2 346.4

242.45 / cos 652.86m T H m m

x m x

For R

at r r r U Nr mps

C U mps C C mps

34


22 02 2

/ 1

2 22

01 01

2 2 2

3

2 2

96%

/ 2 1016.3

11 1.986

39.1 /

8,000 /

412.32 0.518,000 1 2.437

3 2 0.75

stator

p

x

m

c

Given

T T C c K

p Tp bar

p T

m A C kg s

For tapered blade of material kg m

MPa

Need to determine if blade with this stress level will last 1000hr to rupture

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• Airfoils inserted into slots of otherwise solid annulus [rim]

• Airfoil tensile stress is treated as ‘smeared out’ over rim

• Disk supports rim and connects to shaft

Turbo Design - Structural Considerations

2 [ ]c blades hub

disk bladesrim

n A

r x

36


• The average tangential stress due to inertia then is:

• The contribution of the external force to the average tangential stress is

• so that the total average tangential stress becomes:

2

2V

t

F I

A A

A

Frim2

A

F

A

I rimt

2

2

37


• For the same speed and pull, the average tangential stress can be reduced by:

– increasing disk cross sectional area

– decreasing disk polar moment of inertia - moving mass to ID of disk

A

F

A

I rimt

2

2

38


• Stress and major flow design parameters (, E) relate directly to achievable

• Recalling from Dimensional Analysis:

• Higher stress () at constant N and Dmean occurs on longer blades and lower flow coefficient ()

2

1

1

x

x

C m m N

U AU AN D

C m N

U D

m N

D

39


• Also :

• Flow, Density & Work are set by cycle requirements

• Stress (P/A) capability is set by material, temperature, & blade configuration

• Parametric effects– increased N increased (to first order), decreased E (to 2nd

order)– increased D decreased (to first order), decreased E (to 2nd

order)

02 2

1xh C m NE

N D U D

40Plot shows effect of +20% change in N, D & stress on Cx/U, E, and Efficiency. Stress changes allowable blade height or annulus area.

41

Turbomachinery Gaspath Design Problem• Objective: to illustrate interaction of several design parameters

, stress level (cent), x, cost, weight flowpath dimensions

• Design a baseline turbine and 3 alternative configurations

– Dmean or weight and cost on

– Aan or Cx or weight on

– Stress level on • All turbine designs have the following conditions

1 2

01 01

1 2

0

2

1 1 1

50 /

200 28,800 2200

50%

1.0

2 cossin 1.0

cos /

x x

mean mean

x x

x x b xw x

x mean b mean

m lbm s C C

p psia psf T R

D D R

span LAR h same

b b

b b n bZ where

s D n D

42

Turbomachinery Gaspath Design Problem• Design: fill in the missing blanks in the table below

• Account for tip clearance losses as a 2% debit in efficiency

• Remember cent AanN2 and cost blade count (nb)

43

Turbomachinery Gaspath Design Problem• Base Case: Assume only for this case M1=0.8 is given.

1/ 220 01 1 1 1 1

1 1 11 0 1 0 1 0 0

11

0

01 1 2

1 1 1

10.8 ( ) 1

2

0.7532 1731.9

2 2 2 ( 2) 2 0.5tan 1.666 59

2 2 0.9

cos 1731.9 cos(59) 891.0x

a TC C C C CM f M M

a a a a T a a

CC fps

a

E R

C C fps

44

Turbomachinery Gaspath Design Problem• Base Case: Assume only for this case M1=0.8 is given.

01 2 21

01 1 1

/ 891.0 / 0.9 990

1202 2 1.2605 15.126

2 / 60 2 15,000

0.3087 44.45cos ( )

/( ) 44.45 /( 15.126) 0.93

/ 0.93

x

mean mean

an

an mean

x

U C fps

U UD R ft in

N

m TA ft in

p MFP M

L A D in

b L AR L in

45

Turbomachinery Gaspath Design Problem• Base Case:

2 2 10 2 2

01 1

44.45 15,000 1 10 [ / min ]

2 2 0.5 2tan 0.5555 29.0 [ ]

2 2 0.9

29 ( 59) 88

2 cos59sin88 1.177

cos 29

60.14 60

an

xw

x meanb

x

A N x in

R Eby convention

Z

Dn Number of airfoils

b

46

Turbomachinery Gaspath Design Problem• Base Case:

0

2

0 78.28 /

2.0, 0.9 90.7 2.0( ) 88.7

Find h

EUh Btu lbm

gJ

Find from Smith turbine correlation

E tip clearance

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Turbomachinery Gaspath Design Problem

• Baseline Design:

• Account for tip clearance losses as a 2% debit in efficiency

• Remember cent AanN2 and cost blade count (nb)

2

[ ]790,000

anc

A NStress psi

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Turbomachinery Gaspath Design Problem• Alternate Design 1: Given N, Aan1N2, Dmean1

2

102 2

2

15% 1.15 1.15 990 1139.0

15% 1.15 15.126 1.15 17.39

1 10/( ) 0.813

17.39 15,000

base

mean mean base

an

an mean

an mean

U increased by U U fps

D increased by D D in

A N constant, therefore compute new span L

xL A N D N in

A D L

2

02 2

1

17.39 0.813 44.42

/ 0.813

78.28 32.174 7781.511

/( ) 1139

2 2 2 ( 1.511) 2(0.5) 1.255tan

2 2

x

in

b L AR in

hE

U gJ

E R

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Turbomachinery Gaspath Design Problem• Alternate Design 1:

1

010 1

01 1 1 1 1

1/ 221

1 1 1 10

011 1 11 1 1

0 0 0

11

1.0883 1.0883 50 2200 0.2873

cos 200 17.14 0.825cos cos

1( ) 1

2

49.02 2200cos cos 2.018 cos

1139.0

tan

an

x

Guess

m TFP Get M

p A

Cf M M M Get C

a

RTC C C CGet

U a U a a

11 1

0

01 2

(1.255 / )

: , , , 4

: 58.8 / 0.7527x

CUnknowns M with equations set up iteration

a

Solution C U

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Turbomachinery Gaspath Design Problem• Alternate Design 1:

01 2

1 1

0

58.8 / 0.7527

2 1.511 2 0.5tan 18.75

2 2 0.7527

18.75 ( 58.8) 77.55

2 cos( 58.8)sin(77.55) 1.068

cos(18.75)

x

w

xw

x meanb

C U

E R

Determine solidity from Z

Z

Determine the number of airfoils

Dn

b

1.068 17.39

71.76 720.8

[ ] 93.3% 2%[ ] 91.3%

bx

n

Determine turbine efficiency

from Smith chart tip clarance

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Turbomachinery Gaspath Design Problem• Summary

1 Turbomachinery Class 11. 2 Axial Flow Compressors: Efficiency Loss: Centrifugal Compressors...

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