Page 1
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
Page 2
2
Turbomachinery
Class 11
Page 3
3
Configuration Selection & Multidisciplinary Decisions
• Turbomachinery Design Requires Balance Between:
Performance
Weight
Cost
Page 4
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
Page 5
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
Page 6
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
Page 7
7
Dixon
Thi
s is
E
Page 8
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!
Page 9
9
Secondary Air Systems
Page 10
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)
Page 11
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
Page 12
12
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
Page 13
13
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
Page 14
14
Typical Centrifugal Stress Values
Page 15
15
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
Page 16
16
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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17
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
Page 19
19
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
Page 23
23
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
Page 24
24
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
Page 25
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
Page 26
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
Page 27
27
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
Page 28
28
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
Page 29
29
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
Page 30
30
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
Page 31
31
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
Page 32
32
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
Page 33
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
Page 34
34
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
Page 35
35
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
Page 36
36
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
Page 37
37
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
Page 38
38Plot shows effect of +20% change in N, D & stress on Cx/U, E, and Efficiency. Stress changes allowable blade height or annulus area.
Page 39
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
Page 40
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)
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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
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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
Page 43
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
Page 44
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
Page 45
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
Page 46
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
Page 47
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
Page 48
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
Page 49
49
Turbomachinery Gaspath Design Problem• Summary