Download - 1 Axial Flow Compressors: Efficiency Loss: Centrifugal Compressors Efficiency Loss: Axial Flow turbines: Efficiency Loss:

1

• 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

2

Turbomachinery

Class 11

3

Configuration Selection & Multidisciplinary Decisions

• Turbomachinery Design Requires Balance Between:

Performance

Weight

Cost

4

Turbomachinery Design

• Several Aspects to "Cost" as seen by customer

First Cost - Price

Operating Cost - Fuel & Maintenance

Efficiency

Weight

No. of Parts

Complexity

Manufacturing

Materials

Life; Stress & Temperature

5


• 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

6

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

7

Dixon

Thi

s is

E

8


• 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!

9

Secondary Air Systems

10

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)

11

Turbomachinery DesignStructural Considerations

• Airfoil Centrifugal Stress

Blade of constant cross section has mass:

2BMPull r

g

2RT DD

h

4

RT DDr

2 2

2.

222.

[ ]

sec

12

T

H

centrifugal m

centrifugalcent

m m

R

cent T H

m TR

dF Rdm R AdR

dFdRdR

A

for constant blade cross tional area

U RRdR

R

12


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

13


• For centrifugal stress of 40,000 lbf/in2,

– AanN2 = 790,000 x 40,000=3.16 x 1010

– Design practice for AN2 is from (2.5-3.5) x 1010

• Since N is fixed, this places upper limit on annulus area

• In another, more basic form:

Where: Ut Blade Tip Speed,ft/sec

m Metal Density, lbm/in3

cent Centrifugal Stress, lbf/in2

hub/tip radius ratio

-16 2

2

g

UTmcent From chart 11

14

Typical Centrifugal Stress Values

15


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 / 2

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

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

16


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

17


Centrifugal stresses due to torsional disk stresses• The force from the change in angular momentum of gas in the

tangential direction which produces useful torque.

• Mw = bending moment about axial direction

• Ma=gas bending moment about tangential direction [If Cx constant, pressure force produced in axial direction]

• Mw is largest bending moment

• Approximate form for bending stress

• Design blade with centroids of cross section slightly off-center– gas bending moment is of opposite sign to centrifugal bending moment

2 3( ) 1

2 ( )blade U U

bsblades xx

M C C h

n f I

18


• Disk & Blade Stress considerations influence selection of work and flow coefficients – from above

• Selection of work and flow coefficients greatly effects blade cross sections

• Following chart from former Pratt&Whitney turbine designers illustrate blade shape variation

• Their meanline doesn’t exactly match Smith data

19


• Allowable stress levels are set by material properties, material temperature, time of operation and cycles of strain

• Stress level measures– Ultimate stress: part fails if this level is reached– 1000 hrs rupture life: part fails after 1000 hrs at a

given temperature– 1000 hrs creep life: part will stretch a certain

percentage (0.1 - 0.2%) at a given temperature

21

S S RR

23


• 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

24


• External load due to:

– airfoil, attachment & platform pull– disk lug– side plates, seals, etc.

• Inertial loads due to:

– centrifugal force from bore to live rim

25

• 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

26

• Average Tangential Stress

Consider radial inertia load on disk element:

• Noting that , an element of area in the disk cross section:

Turbo Design - Structural Considerations

o

i

o

i

r

rr

r

rr

r

r

drxrF

drdxrF

drdxrdF

drxrddm

rdmdF

22

2

0

22

22

2

2

dAxdr

o

i

r

rr dArF 222

FR

r

Disk Front View

FR

r

Disk Front View

FR

r

Disk Front View

X

dr

Tangential disk stresses: forces on itself due to rotation + external (blade pull ) forces

27


is the polar moment of inertia of disk cross section about the center line.

• The total radial force becomes:

• Design disk for constant stress… as r decreases, increase thickness x

• Force normal to any given diameter is needed for average tangential force:

IdAro

i

r

r 2

IFr22

28


FR

Fv

Fv

Fv/2 IF

drxrF

drdrxF

dFdF

V

r

rV

r

rV

rV

o

i

o

i

2

2

0

2

2

0

2

2

cos

sin

:half topover the gIntegratin

sin

r

V

FF • note that

29


• 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

30


• 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

31


• Rim Stress - Consider a thin ring.

Neglecting the external force, the rim inertial tangential

force is:

X

r

r

A

It

2

22

rrx

rxr

A

I

2Ut

32


• Important Thoughts About Tangential Stress in a Ring

– Wheel Speed Drives Stress, not RPM !

– Hoop Stress Low at Low Wheel Speed

– Ring Cannot Support Itself at High Speeds (needs a bore!)

– Hoop Stress Equation Has form of Dynamic Head, a Pressure Term

33

High Disk Stress in Advanced HPTs

1000

1200

1400

1600

1800

100 200 300 400 500 600 700

A*N2 X 10-8

RimSpeedft/sec

34


• Average Tangential Stress in HPT disks is Increasing

En g i n e

E x t e r n a l ks i

I n e r t i a l T o t a l

Tot

Ext

1982 32 68 100 32

1980 43 70 113 38

2000 52 62 114 46

2010 46 64 110 42

2015 54 71 125 43

35


• Conclusions:

– Disk Stress Driven by Wheel Speed & Radius Ratio.

– Mass at Bore Strengthens Disks

– Mass at Rim Difficult to Carry

– At Some Thickness, Bore is Impractical

– Direct Relation Between Flow & Work Coefficients & Disk Stress

36


• 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

37


• 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

38Plot shows effect of +20% change in N, D & stress on Cx/U, E, and Efficiency. Stress changes allowable blade height or annulus area.

39

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

40

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)

41

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

42

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

43

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

44

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

45

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

46

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

47

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

48

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

49

Turbomachinery Gaspath Design Problem• Summary