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• 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
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Turbomachinery
Class 11
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Configuration Selection & Multidisciplinary Decisions
• Turbomachinery Design Requires Balance Between:
Performance
Weight
Cost
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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
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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
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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
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Dixon
Thi
s is
E
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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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Secondary Air Systems
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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
• 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
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Turbomachinery DesignStructural Considerations
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
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Turbomachinery DesignStructural Considerations
• 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
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Typical Centrifugal Stress Values
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Typical Centrifugal Stress Values
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
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Typical Centrifugal Stress Values
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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Turbomachinery DesignStructural Considerations
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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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S S RR
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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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Turbomachinery DesignStructural Considerations
• 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
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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
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• 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
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Turbomachinery DesignStructural Considerations
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
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Turbomachinery DesignStructural Considerations
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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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Turbomachinery DesignStructural Considerations
• 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
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38Plot shows effect of +20% change in N, D & stress on Cx/U, E, and Efficiency. Stress changes allowable blade height or annulus area.
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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
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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)
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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
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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
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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
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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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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
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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