Design of a Multi-Stage Compressor
Motivation: Market research has shown the need for a low-cost turbojet
with a take-off thrust of 12,000N. Preliminary studies will show that a
single-spool all-axial flow machine is OK, using a low pressure ratio and
modest turbine inlet temperatures to keep cost down.
Problem: Design a suitable compressor operating at sea-level static
conditions with
• compressor pressure ratio = 4.15
• air mass flow
= 20 kg/s
• turbine inlet temperature = 1100K
Assume:
Pamb = 1.01 bar, Tamb = 288 K
Inlet rhub / rtip = 0.5
Mean radius is constant
Polytropic efficiency = 0.90
Constant axial velocity design
No swirl at exit of compressor
Utip = 350 m/s
Compressor has no inlet guide vanes
kJ
kg 0 K
kJ
c p  1.005 0
kg K
  1.4
R  0.287
1
Steps in the Meanline Design Process
Steps
1) Choice of rotational speed and annulus dimensions
2) Determine number of stages, using assumed efficiency
3) Calculate air angles for each stage at the mean radius meanline analysis
4) Determine variation of the air angles from root to tip - radial
equilibrium
5) Investigate compressibility effects
6) Select compressor blading, using experimentally obtained
cascade data or CFD
7) Check on efficiency previously assumed
8) Estimate off-design performance
2
Compressor Meanline Design Process
Given
Assume
PR
p0 , T0
m
p
T0in turbine
Cx  constant
1  0
U T  350 mps
Steps
1) Choose Cx1 and rH/rT to satisfy m and keep Mtip low and define
rT
2) Select N from rT and UT
3) Compute T0 across compressor and all exit flow conditions
[keep rm same through engine]
4) Estimate T0 for first stage from inlet condtions [Euler and de
Haller]
5) Select number of stages  T0comp / T0stage
6) …..
7) …..
3
Step 1- Choice of Rotational Speed & Annulus
Dimensions
• Construct table of inlet / exit properties and parametric study of
c1x vs. tip Mach number [next chart]
• Chose c1x from spread sheet to avoid high tip Mach numbers
and stresses
• Calculate 1 from inlet static pressure and temperature
• With mass flow = 20 kg/s and rhub/rtip = 0.5
  r 2 
m   ACx   rtip2 1   hub   Cx
  rtip  


rtip2 
m
  r 2 
 Cx 1   hub  
  rtip  


• and compute rotational speed and tip Mach number
N
2 rtip
U tip
M tip 
Wtip
 RT
2
where Wtip  U tip
 C x2
4
Calculate Tip Radius and Rotational Speed
Drive choice by compressor inlet conditions
Cx
T1 (degK) P1 (bar) P1 (kg/m2) rho1(kg/m3) rtip (m)
rhub/rtip
N rev/sec N (rpm)
W1tip
M1tip
100 283.0226 0.950214 9692.1879 1.17103756 0.254398876
0.4 218.9643 13137.86 364.0055 1.079401
0.261089548
0.45 213.3531 12801.19
0.269230464
0.5 206.9018 12414.11
0.279178986
0.55 199.5289 11971.73
0.291450527
0.6 191.1277 11467.66
150 276.8009 0.879091 8966.7278 1.10773699 0.213568228
0.4 260.8266 15649.6 380.7887 1.141789
0.219185056
0.45 254.1427 15248.56
0.226019368
0.5 246.458 14787.48
Pick this
0.234371166
0.55 237.6755 14260.53
0.244673143
0.6 227.6681 13660.09
200 268.0905 0.786018 8017.3822 1.0226367 0.192497422
0.4 289.3767 17362.6 403.1129 1.228207
0.19756009
0.45 281.9612 16917.67
0.203720123
0.5 273.4353 16406.12
0.211247926
0.55 263.6915 15821.49
0.220533502
0.6 252.5887 15155.32
250 256.8913 0.676972 6905.1159 0.91916112 0.181607916
0.4 306.7282 18403.69 430.1163 1.338742
0.186384191
0.45 298.868 17932.08
0.192195753
0.5 289.8309 17389.86
0.199297711
0.55 279.5028 16770.17
0.208058005
0.6 267.7344 16064.06
5
Compute Root (Hub) and Mean Radius
• Choose N = 250 rev/sec or 15,000 RPM and rhub/rtip = 0.5
• With hub/tip radius ratio and tip radius:
rhub 
rhub
rtip  0.5 0 .2262   0.1131 m
rtip
and
rmean  0.1697 m
U1 
2 r1 N
 266.6 m / s
60
6
Compressor Meanline Design
• Given: m, Utip, p01, T01, Pr, poly and c1x chosen to avoid high tip
Mach numbers and stresses
• Compressor inlet (1)

 T1   1
p1  p01  
 T01 
2
1x
C
T1  T01 
2c p
1 
p1
RT1
1/ 2






m
RT  

2



 RH  

c
1



 
 1 1x
R
  T   

 RH 
1
RH  RT 
R

 RH  RT 

m
R
2
 T 
Ctip
2

 C12x  U tip
1/ 2
M tip 
Select RH/RT and Utip (N)
for turbine issues
N
U tip
2 RT
Ctip
  RT1 
1/ 2
7
Compressor Meanline Design
•
Compressor exit (2)

T02  T01  Pr  poly  1
p02  p01 Pr
C x2
T2  T02 
2c p

 T2   1
p2
p2  p02  
2 
RT2
 T01 
m
A2 
 2C x
A2
blade height at exit  h  RT  RH 
2 Rm
8
Compute Compressor Exit Conditions
• Compute Compressor Exit Total Temperature
 P02 
T02  T01  
 P01 
(  1)
 p
• so that T02 = 288.0 (4.15)0.3175 = 452.5 0K,
T0 compressor= 452.5 - 288.0 = 164.5 0K and other conditions:
C2
1502
T2  T02 
 452.5 
 441.3
3
2c p
2 1.005 x10 
 T2  (
P2  P02  
 T02 

 1)
 441.3 
 4.19 
 452.5 
0
K
3.5
 3.838 bar
kg 

3.838bar 10200 2

P2
m
bar


2 

 3.03 kg / m3
kJ 
kg m 
RT2
0
0.287
102
441.3
K


kg 0 K 
kJ 
9
Compute Compressor Exit Conditions
• Exit area, hub and tip radius:
20
m
 0.044 m 2

A2 
 2Cx 3.031150 
r  rtip  rhub 
0.044
A2
 0.0413 m

2 0.1697
2 rmean
0.0413
r
 0.1903 m
 0.1697 
rtip  rmean 
2
2
0.0413
r
 0.1491 m
 0.1697 
rhub  rmean 
2
2
10
Step 2 - Estimate the Number of Stages
• From Euler’s Turbine Equation:
T0stage 
U (CU 2  CU 1 )
UCx (tan  2  tan 1 )

cP
cp
and (tan  2  tan 1 )  (tan 1  tan  2 )
WU  Cx tan( )
CU  Cx tan( )
U1 
2 r1 N
 266.6 m / s
60
• With no inlet guide vane (Cu1=0, 1 = 0, and Wu1= -U), the
relative flow angle is:
tan(  1 ) 
U 266.6

  1  60.640
Cx
150
• And the relative inlet velocity to the 1st rotor is:
W1 
Cx
150

 305.9 m / s
cos(  1 ) cos(60.64)
11
Maximum Diffusion Across Compressor Blade-Row
• There are various max. diffusion criteria. Every engine company has
it’s own rules. Lieblein’s rule is one example. Another such rule is
the de Haller criterion that states:
W2
0.72 
 1.0
W1
Note that de Haller’s criterion is simpler than
Lieblein’s rule since it does not involve relative
circumferential velocities or solidity. To first order,
this is same as a 0<Dfactor<~0.4. Could use
Lieblein’s rule but would have to iterate.
• This criteria can also take the form of max. pressure ratio with
correlations for relative total pressure loss across the blade row as a
function of Mach number, incidence, thickness/chord, etc. Taking the
maximum diffusion (de Haller), leads to:
W2  0.72 W1  0.72 305.9  220 m / s
cos(  2 ) 
T0stage 
U (CU 2  CU 1 )
cP
C x 150

  2   47.010
W2 220
266.6 150  (tan(60.64)  tan(47.01))
0


28
K
3
12
1.005 x10
Choose Number of Stages
• Given poly and T0out/T0in  T0 = T0out -T0in, so the number of
stages is T0 compressor / T0stage = 164.5/28 = 5.9
• Typically (T0)stage  40K (subsonic) - 100K (transonic)
• Therefore we choose to use six or seven stages. To be
conservative (account for losses, ie. a<1),
– Choose 7 stages
• Recalculate the T0stage = 164.5/7 = 23.5
• So 1st stage temperature ratio is T0 ratio = 288 + 23.5/288 =1.0816
• The stage pressure ratio is then P0 ratio = (T0 ratio ) p /   1 = 1.2803
13
Compressor Meanline Design
Develop estimate of the number of stages
• Assuming Cx = constant
 Cx
 T0 stage

–
–
–
–
–
cp
 tan 1  tan 2 
for axial inflow tan1 = Um/Cx
V1 = Cx / cos 1
de Haller criterion (like Dfactor) V2/V1  0.72
cos 2 = Cx/V2
neglect work done factor (=1)  (T0)stage = ….
– (T0)stage Nstages  T0out -T0in
– Select Nstages and select nearly constant set of (T0)stage
Develop Stage by Stage Design
• Assume that continual blockage buildup due to boundary layers reduces
work done, therefore
stage
1
2
3
4
5
6
7

0.98
0.93
0.88
0.83
0.83
0.83
0.83
14
Compressor Meanline Design
Develop Stage by Stage Design
• C = absolute velocity,
CU = absolute velocity in U direction
CU 1  0 axial inf low
tan 1 
tan  2 
Um
cx
U m  CU

cx
W2
C2
c p T0
U m
cx
Um 
C1
Constant Cx
W1
U
15
Step 3 - Calculate Velocity Triangles of 1st Stage
at Mean Radius
• So from Euler Turbine Equation:
T0stage 
U (CU 2  CU 1 ) UCx (tan  2  tan 1 )

 23.5
cP
cp
• We can re-calculate the relative angles for the 1st stage:
1  0
tan  2 
c p T0 stage
UCx
tan(  1 ) 
1.005 x103 (23.5)
 2 
 30.57 0
266.6 150 
U 266.6

  1  60.640
Cx
150
tan( 1 )  tan( 2 )  tan( 2 )  tan(1 )  2  49.890
which leads to :
W2
 0.76 (acceptable according to de Haller criterion)
W1
16
Velocity Components and Reaction of 1st Stage
• The velocity components for the 1st stage (rotor) are therefore:
C x 2  150 m / s
C x  150 m / s
CU 2  C x 2 tan  2  88.6 m / s
CU 1  0.0
C1  150 m / s
C 2  C x2  CU2 2  174.21 m / s
U 1  266.6 m / s
U 2  266.6 m / s
WU 1   U 1   266.6 m / s
WU 2  CU 2  U 2   178.0 m / s
W1  C x2  WU21  305.9 m / s
W2  C x2  WU22  232.77 m / s
• The Reaction of the 1st stage is given by:
Wu 2  Wu1 
 266.6 190.31
 0.836
2U
2 (266.6)
(which is high for typica l designs)
R

17
Velocity Components for Stator of 1st Stage
• Now consider the stator of the 1st stage. The h0 of the stator is
zero so from Euler’s eqn.:
• If design uses assumption of “repeating stage”, then inlet angle
to stator is absolute air angle coming out of rotor and, exit
absolute angle of stator is inlet absolute angle of rotor:
CU 1  CU 3  0
Cx 3  Cx 2
U  0 for stator, so WU 3  0
1 stator  1 stator   2 rotor
 2 stator   2 stator  1 rotor
18
Velocity Triangles of 1st Stage Using “Repeating Stage”
Assumption
1=0
Cx1=150
W=C-U
1=60.64
W1=305.9
U=- WU1 =266.6
Notice that the velocity triangles
are not “symmetric” between the
rotor and stator due to the high
reaction design of the rotor. The
rotor is doing most of the static
pressure (temperature) rise.
STATOR
3=0
Cx3=150
3=60.64
C2=174.21
2=30.57
ROTOR
CU2=88.6
Cx2=150
2=49.89
W3=305.9
U=266.6
U=266.6
WU2=178.0
W2=232.77
19
Stage Design Repeats for Stages 2-7
• Then the mean radius velocity triangles “essentially”
stay the same for stages 2-7, provided:
– mean radius stays constant
– hub/tip radius ratio and annulus area at the exit of each
stage varies to account for compressibility (density variation)
– stage temperature rise stays constant
– reaction stays constant
• If, however, we deviate from the “repeating stage”
assumption, we have more flexibility in controlling
each stage reaction and temperature rise.
20
Non- “Repeating Stage” Design Strategy
• Instead of taking a constant temperature rise
T0stage  23.5
across each stage, we could reduce the stage temperature rise for
the first and last stages of the compressor and increase it for the
middle stages. This strategy is typically used to:
– reduce the loading of the first stage to allow for a wide variation
in angle of attack due to various aircraft flight conditions
– reduce the turning required in the last stage to provide for zero
swirl flow going into the combustor
• With this in mind, lets change the work distribution in the
compressor to:
T0stage  20.0
1
T0stage
 25.0
2, 3, 4, 5, 6
T0stage  20.0
7
21
1st Stage Design for “Non-Repeating Stages”
• We can re-calculate the relative angles for the 1st stage:
1  0
tan  2 
c p T0 stage
UCx
1.005 x103 (20)
 2 
 26.68 0
266.6 150 
U 266.6
tan(  1 ) 

  1  60.640 (same as before)
Cx
150
tan( 1 )  tan( 2 )  tan( 2 )  tan(1 )  2  51.890
which leads to :
W2
 0.79 (still acceptable , i.e.  0.72)
W1
(also note the reduced turning due to the reduced work requiremen t)
22
Velocity Components and Reaction of 1st
Stage with Non-Repeating Stages
• The new velocity components for the 1st stage (rotor) are
therefore:
C x 2  150 m / s
C x  150 m / s
CU 2  C x 2 tan  2  75.38 m / s
CU 1  0.0
C 2  C x2  CU2 2  167.87 m / s
C1  150 m / s
U 2  266.6 m / s
U 1  266.6 m / s
WU 1   U 1   266.6 m / s
WU 2  CU 2  U 2   191.22 m / s
W1  C  W
W2  C x2  WU22  243.03 m / s
2
x
2
U1
 305.9 m / s
• The Reaction of the 1st stage is given by:
R
Wu 2  Wu1 
2U
 266.6 190.31

 0.859
2 (266.6)
with the assumption that C 3  C1 . (note that the reaction increased)
23
Design of 1st Stage Stator
• The pressure ratio for this design with a temperature
change, T0 = 20 is:
 T02 
P02
  
P01
 T01 
 P
(  1)
 288  20 


 288 
0.9 (1.4 )
.4
 1.236
• So P03= P02 = 1.01 (1.236) = 1.248 bar and T03=
288+20=308 0K
• Now we must choose a value of 3 leaving the stator.
– When we designed with repeating stages, 3= 1.
– But now we have the flexibility to change 3.
24
Design of the 1st Stage Stator & the 2nd Stage
• Change 3 so that there is swirl going into the second stage and
thereby reduce the reaction of our second stage design.
• Design the second stage to have a reaction of 0.7, then from the
equation for reaction:
Rstage 2 
Cx
(tan 1  tan  2 )
2U
• And if we design the second stage to a temperature rise of 25 0,
the Euler’s equation:
T0 stage 2
UCx

(tan 1  tan 2 )
cp
• Which can be solved simultaneously for 1and 2
 1 stage2   57.310
 2 stage2   42.92 0
25
Design of 1st Stage Stator & 2nd Stage Rotor
• Note that this is the same as specifying E, n, and R as
in one of your homeworks and computing the angles.
• And the absolute flow angles of the second stage can
be found from
U
 tan 1  tan 1  tan  2  tan  2
Cx
• So
 1 stage2  12.36 0   3 stage1
 2 stage2  40.27 0
• Therefore, we have determined the velocity triangles of
the 1st stage stator and the second stage rotor
26
Velocity Triangles of 1st Rotor Using “NonRepeating Stage” Assumption
1=0
Cx1=150
W=C-U
1=60.64
W1=305.9
Notice that the velocity triangles are
not “symmetric” due to the high
reaction design of the rotor. Also,
there is swirl now leaving the stator.
U=- WU1 =266.6
C3=153.56
STATOR
3=12.36
Cx3=150
3=57.31
C2=167.87
2=26.68
U=266.6
CU2=75.38
Cx2=150
ROTOR
2=51.89
W2=242.03
CU3=32.87
W3=277.73
U=266.6
WU2=191.2
2
WU3=233.7
7
27
Design of 2nd Stage Stator & 3rd Stage Rotor
• Design of the 2nd stage stator and 3rd stage rotor can
be done in the same manner as the 1st stage stator
and 2nd stage rotor.
• A choice of 50% reaction and a temperature rise of 25
degrees for the 3rd stage will lead to increased work by
the stage but a more evenly balanced rotor/stator
design. The velocity triangle of the stator will be a
mirror of the rotor.
• This stage design will then be repeated for stages 4 - 6.
28
Class 12 - The 7-Stage Compressor Design So
Far Has Lead to 1st and 2nd Stages:
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
1
2
3
1
2
3
Cx
Cu1
C1
W1
W u1
0
12.36
26.68
40.27
12.36
60.64
57.31
-51.89
-42.92
-57.31
150
150
0
32.87
150
153.56
305.9
277.73
-266.6
-233.77
M2
Mr2
Cu2
C2
W2
W u2
75.38
127.07
167.87
196.59
243.03
204.86
-191.22
-139.53
M3
Mr3
P01
P02
0.445072 0.804962
1.01
1.248
1.248
1.596
0.488438 0.707125
0.553668 0.576959
P03
1.248
1.596
T01
288
308
M1
Mr1
0.449778 0.917248
0.445072 0.804962
Cu3
C3
W3
W u3
T03/T01
153.56
277.73
-233.77
R
0.874
0.7
P03/P01
32.87
1.236
1.279
1.069
1.081
T02
T03
P1
P2
P3
T1
T2
308
333
T3
308 0.879088 1.060143 1.08932261 276.806 293.9799 296.2683
333 1.089323 1.295942
296.2683 313.7723
29
Design of 2nd Stage Stator
• The pressure ratio for the 2nd stage design with a temperature
change, T0 = 25 is:
 T02 
P02
  
P01
 T01 
 P
(  1)
 308  25 


 308 
0.9 (1.4 )
.4
 1.279
• So P03= P02= 1.248 (1.279) = 1.596 bar and T03= 308+25=333 0K
• Now we must choose a value of 3 leaving the 2nd stage stator
that provides for the desired Reaction and Work in the 3rd stage
using a similar technique as previously used.
30
Design of 2nd Stage Stator & 3rd Stage
• We can change 3 so that there is swirl going into the third
stage and thereby reduce the reaction of our second stage
design. If we design the third stage to have a reaction of 0.5,
then from the equation for reaction:
Rstage 3 
Cx
(tan 1  tan  2 )
2U
• And if we design the third stage to a temperature rise of 25 0, the
Euler’s equation:
T0 stage 3
UCx

(tan 1  tan 2 )
cp
• Which can be solved simultaneously for 1and 2
 1 stage3   50.26 0
 2 stage3   29.88 0
31
Design of 2nd Stage Stator & 3rd Stage Rotor
• And the absolute flow angles of the second stage can be found
from
U
Cx
• So
 tan 1  tan 1  tan  2  tan  2
 1 stage3  29.88 0   3 stage2
 1 stage3   50.26 0
 2 stage3  50.26 0
 2 stage3   29.88 0
• Note the symmetry in angles for 3rd stage due to the 50%
reaction !
• Therefore, we have determined the velocity triangles of the 2nd
stage stator and the third stage rotor. Check the de Haller
number for the 3rd stage rotor:
W2
cos 1

 .74 which is OK
W1 cos  2
32
Velocity Triangles of 2nd Stage
C3=153.56
3=12.36
W=C-U
Cx3=150
3=57.31
CU3=32.87
C3=172.99
U=266.6
3=29.88
W3=277.73
STATOR
WU3=233.7
7
3=50.26
U=266.6
C2=196.59
2=40.27
ROTOR
Cx3=150 CU3=86.18
W3=234.63
CU2=127.07
Cx2=150
2=42.92
W2=204.86
U=266.6
WU3=180.4
2
WU2=139.5
3
Notice that the velocity triangles are
not “symmetric” for the second
33
stage due to 70%reaction design
but
will be for 3rd stage (50% reaction).
Summary of Conditions for Stages 1 - 3
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
1
2
3
1
2
3
Cx
Cu1
C1
W1
W u1
0
12.36
29.88
26.68
40.27
50.26
12.36
29.88
29.88
-60.64
-57.31
-50.26
-51.89
-42.92
-29.88
-57.31
-50.26
-50.26
150
150
150
0
32.87
86.18
150
153.56
172.99
305.9
277.73
234.63
-266.6
-233.77
180.42
Cu2
C2
W2
W u2
M2
Mr2
Cu3
C3
W3
W u3
P03/P01
T03/T01
75.38
127.07
180.42
167.87
196.59
234.63
243.03
204.86
172.99
-191.22
-139.53
-86.18
32.87
86.18
86.18
153.56
172.99
172.99
277.73
234.63
234.63
-233.77
-180.42
-180.42
R
0.874
0.7
0.5
1.236
1.279
1.256
1.069
1.081
1.075
M3
Mr3
P01
P02
T02
T03
P1
P2
P3
T1
T2
0.445072 0.804962
0.483867 0.656279
0.465906 0.631918
1.01
1.248
1.596
1.248
1.596
2.005
0.488438 0.707125
0.553668 0.576959
0.643754 0.474632
P03
1.248
1.596
2.005
T01
288
308
333
308
333
358
M1
Mr1
0.449778 0.917248
0.445072 0.804962
0.483867 0.656279
T3
308 0.879088 1.060143 1.08932261 276.806 293.9799 296.2683
333 1.089323 1.295942 1.35979317 296.2683 313.7723 318.1117
358 1.359793 1.517323 1.72789178 318.1117 330.6113 343.1117
34
Design of Stages 4-6
• The velocity triangles of stages 4 through 6 will
essentially be repeats of stage 3 since all have a 50%
reaction and a temperature rise of 25 degrees.
• Stagnation and static pressure as well as stagnation
and static temperature of these stages will increase as
you go back through the machine.
• As a result, density will also change and will have to be
compensated for by changing the spanwise radius
difference (area) between the hub and tip (i.e. hub/tip
radius ratio)
35
Velocity Triangles of Stages 3 - 6
C3=172.99
3=29.88
W=C-U
Cx3=150 CU3=86.18
C3=172.99
3=50.26
3=29.88
U=266.6
Cx3=150 CU3=86.18
STATOR
3=50.26
W3=234.63
U=266.6
WU3=180.4
2
ROTOR
W3=234.63
C2=234.63
2=50.26
Cx2=150
2=29.88
W2=172.99
CU2=180.42
WU3=180.4
2
U=266.6
WU2=86.18
Notice that the velocity triangles
are “symmetric” due to the
36
50%reaction design.
Summary of Conditions for Stages 1 - 6
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
STAGE
1
2
3
4
5
6
7
1
2
3
1
2
3
Cx
Cu1
C1
W1
W u1
M1
Mr1
0
12.36
29.88
29.88
29.88
29.88
26.68
40.27
50.26
50.26
50.26
50.26
12.36
29.88
29.88
29.88
29.88
29.88
-60.64
-57.31
-50.26
-50.26
-50.26
-50.26
-51.89
-42.92
-29.88
-29.88
-29.88
-29.88
-57.31
-50.26
-50.26
-50.26
-50.26
-50.26
150
150
150
150
150
150
0
32.87
86.18
86.18
86.18
86.18
150
153.56
172.99
172.99
172.99
172.99
305.9
277.73
234.63
234.63
234.63
234.63
-266.6
-233.77
180.42
180.42
180.42
180.42
0.449778
0.445072
0.483867
0.465906
0.449807
0.435269
0.917248
0.804962
0.656279
0.631918
0.610083
0.590365
Cu2
C2
W2
W u2
M2
Mr2
Cu3
C3
W3
W u3
P03/P01
T03/T01
75.38
127.07
180.42
180.42
180.42
180.42
167.87
196.59
234.63
234.63
234.63
234.63
243.03
204.86
172.99
172.99
172.99
172.99
-191.22
-139.53
-86.18
-86.18
-86.18
-86.18
0.488438
0.553668
0.643754
0.620713
0.599981
0.581197
0.707125
0.576959
0.474632
0.457644
0.442359
0.42851
32.87
86.18
86.18
86.18
86.18
86.18
153.56
172.99
172.99
172.99
172.99
172.99
277.73
234.63
234.63
234.63
234.63
234.63
-233.77
-180.42
-180.42
-180.42
-180.42
-180.42
R
0.874
0.7
0.5
0.5
0.5
0.5
1.236
1.279
1.256
1.237
1.22
1.206
1.069
1.081
1.075
1.07
1.065
1.061
M3
Mr3
P01
P02
P03
T01
T02
T03
P1
P2
P3
T1
T2
T3
0.445072
0.483867
0.465906
0.449807
0.435269
0.422056
0.804962
0.656279
0.631918
0.610083
0.590365
0.572443
1.01
1.24836
1.596652
2.005395
2.480674
3.026423
1.24836
1.596652
2.005395
2.480674
3.026423
3.649866
1.24836
1.596652
2.005395
2.480674
3.026423
3.649866
0.879088
1.089637
1.360349
1.728233
2.159102
2.657034
1.060449
1.296472
1.517622
1.913086
2.372762
2.903398
1.08963684
1.36034905
1.72823259
2.15910202
2.65703431
3.22898653
276.806
296.2683
318.1117
343.1117
368.1117
393.1117
293.9799
313.7723
330.6113
355.6113
380.6113
405.6113
296.2683
318.1117
343.1117
368.1117
393.1117
418.1117
288
308
333
358
383
408
308
333
358
383
408
433
308
333
358
383
408
433
37
Stage 7 Design
• So going into stage 7, we have P01= 3.65 and T01 = 433. The
requirements for our 7-stage compressor design we have
– P0 exit = 4.15 * 1.01 = 4.19 bar
– T0 exit = 288.0 (4.15)0.3175 = 452.5 0K
• This makes the requirements for stage 7:
T02
452.5

 1.045 or T0  452.5  433  19.5
T01
433
P02
 1.045
P01
 0.9  1.4
0.4
 1.149
38
Stage 7 Design
• If we assume a Reaction = 0.5 for the 7th stage:
R stage7 
T0 stage7
Cx
(tan 1  tan  2 )  0.5
2U
U Cx

(tan 1  tan  2 )  19.5
cp
• Then, solving equations:
1stage 7   48.590
 2 stage 7   32.770
39
Stage 7 Design
• And from:
U
 tan  1  tan  1  tan  2  tan  2
Cx
or from symmetry of the velocity triangles for 50% reaction:
 1 stage7  32.77 0   3 stage7
 2 stage7  48.59 0
• Note that the absolute angles going into stage 7 have changed
from those computed for stages 3 - 6 and that the exit absolute
air angle leaving the compressor is 32.770. This means that a
combustor pre-diffuser is required to take all of the swirl out of
the flow prior to entering the combustor.
40
Summary of Compressor Design
STAGE
1
2
3
4
5
6
7
1
2
3
1
2
3
Cx
Cu1
C1
W1
W u1
M1
Mr1
0
12.36
29.88
29.88
29.88
29.88
32.77
26.68
40.27
50.26
50.26
50.26
50.26
48.59
12.36
29.88
29.88
29.88
29.88
29.88
32.77
-60.64
-57.31
-50.26
-50.26
-50.26
-50.26
-48.59
-51.89
-42.92
-29.88
-29.88
-29.88
-29.88
-32.77
-57.31
-50.26
-50.26
-50.26
-50.26
-50.26
-48.59
150
150
150
150
150
150
150
0
32.87
86.18
86.18
86.18
86.18
96.56
150
153.56
172.99
172.99
172.99
172.99
178.39
305.9
277.73
234.63
234.63
234.63
234.63
226.78
-266.6
-233.77
-180.42
-180.42
-180.42
-180.42
-170.08
0.449778
0.445072
0.483867
0.465906
0.449807
0.435269
0.435723
0.917248
0.804962
0.656279
0.631918
0.610083
0.590365
0.553917
STAGE
1
2
3
4
5
6
7
Cu2
C2
W2
W u2
M2
Mr2
Cu3
C3
W3
W u3
T03/T01
167.87
196.59
234.63
234.63
234.63
234.63
226.78
243.03
204.86
172.99
172.99
172.99
172.99
178.39
-191.22
-139.53
-86.18
-86.18
-86.18
-86.18
-96.56
0.488438
0.553668
0.643754
0.620713
0.599981
0.581197
0.547558
0.707125
0.576959
0.474632
0.457644
0.442359
0.42851
0.430721
32.87
86.18
86.18
86.18
86.18
86.18
96.56
153.56
172.99
172.99
172.99
172.99
172.99
178.39
277.73
234.63
234.63
234.63
234.63
234.63
226.78
-233.77
-180.42
-180.42
-180.42
-180.42
-180.42
-170.08
R
0.874
0.7
0.5
0.5
0.5
0.5
0.5
P03/P01
75.38
127.07
180.42
180.42
180.42
180.42
170.08
1.236
1.279
1.256
1.237
1.22
1.206
1.149
1.069
1.081
1.075
1.07
1.065
1.061
1.045
STAGE
1
2
3
4
5
6
7
M3
Mr3
P01
P02
P03
T01
T02
T03
0.445072
0.483867
0.465906
0.449807
0.435269
0.422056
0.425883
0.804962
0.656279
0.631918
0.610083
0.590365
0.572443
0.541407
1.01
1.24836
1.596652
2.005395
2.480674
3.026423
3.649866
1.24836
1.596652
2.005395
2.480674
3.026423
3.649866
4.193696
1.24836
1.596652
2.005395
2.480674
3.026423
3.649866
4.193696
288
308
333
358
383
408
433
308
333
358
383
408
433
452.5
308
333
358
383
408
433
452.5
P1
P2
P3
T1
T2
T3
0.879088
1.089637
1.360349
1.728233
2.159102
2.657034
3.20353
1.060449
1.296472
1.517622
1.913086
2.372762
2.903398
3.42041
1.08963684
1.36034905
1.72823259
2.15910202
2.65703431
3.22898653
3.70197943
276.806
296.2683
318.1117
343.1117
368.1117
393.1117
417.1677
293.9799
313.7723
330.6113
355.6113
380.6113
405.6113
426.9133
296.2683
318.1117
343.1117
368.1117
393.1117
418.1117
436.6677
41
Hub and Tip Radii for Each Blade Row
• From the pressure and temperature, we can compute the
density from the equation of state:
STAGE
1
2
3
4
5
6
7
P1
P2
P3
T1
T2
T3
0.879088
1.089637
1.360349
1.728233
2.159102
2.657034
3.20353
1.060449
1.296472
1.517622
1.913086
2.372762
2.903398
3.42041
1.089637
1.360349
1.728233
2.159102
2.657034
3.228987
3.701979
276.806
296.2683
318.1117
343.1117
368.1117
393.1117
417.1677
293.9799
313.7723
330.6113
355.6113
380.6113
405.6113
426.9133
296.2683
318.1117
343.1117
368.1117
393.1117
418.1117
436.6677
1
1.10656
1.281488
1.490009
1.755031
2.043674
2.355046
2.675693
2
3
1.25687
1.439682
1.599425
1.874463
2.172154
2.494105
2.791621
1.281488
1.490009
1.755031
2.043674
2.355046
2.690866
2.953936
42
Hub and Tip Radii for Each Blade Row
• From Continuity:
rtip2 
m
  r 2 
 Cx 1   hub  
  rtip  


• and our design value of rmean = 0.1697,
rmean  0.5(rtip  rhub )  0.1697  rhub  0.3394  rtip
• we can calculate the hub and tip radii (i.e. area) at the entrance
and exit of each blade row:
rtip2 
m
  .3394  r
tip
 Cx 1  
rtip
 




2




 rtip 
m
 0.1697
 Cx (.6788)
43
Hub & Tip Radii for All Stages of Compressor
• So we get:
Station
rotor
stator
rotor
stator
rotor
stator
rotor
stator
rotor
stator
rotor
stator
rotor
stator
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
P
P0abs
T
0.879088
1.060449
1.089637
1.296472
1.360349
1.517622
1.728233
1.913086
2.159102
2.372762
2.657034
2.903398
3.20353
3.42041
3.701979
1.01
1.24836
1.24836
1.596652
1.596652
2.005395
2.005395
2.480674
2.480674
3.026423
3.026423
3.649866
3.649866
4.193696
4.193696
276.806
293.9799
296.2683
313.7723
318.1117
330.6113
343.1117
355.6113
368.1117
380.6113
393.1117
405.6113
417.1677
426.9133
436.6677
T0abs
288
308
308
333
333
358
358
383
383
408
408
433
433
452.5
452.5
rtip
rhub
0.226203
0.219446
0.21849
0.213129
0.211662
0.208792
0.205326
0.203056
0.200294
0.198484
0.196249
0.194769
0.193067
0.192097
0.190866
0.113197
0.119954
0.12091
0.126271
0.127738
0.130608
0.134074
0.136344
0.139106
0.140916
0.143151
0.144631
0.146333
0.147303
0.148534
44
Conditions in Compressor
Temperature vs. Blade Row #
500
450
400
T0
300
250
350
T
300
250
Blade Row #
15
13
11
9
Blade Row #
Stagnation Pressure vs. Blade Row #
Pressure vs. Blade Row #
4
Blade Row #
15
13
11
9
7
15
13
11
9
0
7
0
5
1
3
1
Blade Row #
P
2
5
P0
2
3
3
3
Pressure
4
1
5
1
Stagnation Pressure
7
1
5
200
15
13
11
9
7
5
3
200
400
3
350
Tem perature
450
1
Stagnation Tem perature
Stagnation Temperature vs. Blade Row #
45
Hub & Tip Radii Distribution - Flow Path Area
Hub and Tip Radii vs. Blade Row #
0.25
0.15
rtip
rhub
0.1
0.05
15
13
11
9
7
5
3
0
1
Radius
0.2
Blade Row #
46
Spanwise Variations
• Blade wheel speeds vary with radius leading to a change in
velocity triangles with span for each blade row. For instance, the
first blade row has
rhub = .1131,
rmean = .1697, rtip = .2262 m and
Uhub = 177.7 , Umean = 266.6, Utip = 355.3 m/s
leading to relative flow angles:
1 hub
 49.83
1 mean  60.64
1 tip
 67.11
1 hub
 1 mean  1 tip  0
• Next, we must choose the type of radial design strategy from:
•
– free vortex where CU r = constant (dh0/dr = 0)
– constant reaction where U CU = constant
– exponential where CU1 = a - (b/R) and CU2 = a + (b/R)
The exit radial pressure gradient will be different for each of the designs47
Nonisentropic Designs
• Include losses (nonisentropic effects)
– see Mattingly
– Common Meanline (CML)
– bleeds, Dfactor, shocks, compressibility, ...
48
Real World Effects
• 3-Dimensional effects
– radial equilibrium
– free vortex designs
– secondary flows
• Tip speed limitations  maximum blade stresses (later)
• Axial velocity  compressibility, shocks, losses
• High flow deflection  Dfactor, de Haller, Carter’s rule
• Blockage (Kbar) due to boundary layers  work done factor 
49
Axisymmetric Flow Analyses

F   Vdv   V V  dA
t


• Consider axisymmetric flows [  /   0 ] and steady flow [ / t  0 ]
• In the radial direction,
 CU2 CS2

dCs

cos  s 
sin  s   rdrd  Fr

rs
dt
 r

1  centripetal force due to circumferential flow
2  force due to streamline curvature
3  force for linear acceleration along streamline
where Fr 
 pdA
r
50
Radial Equilibrium
P+dp
Mass = dm =  r d dr
dr
CU
P+1/2 dp
P+1/2 dp
P
r
If Cr=0 and Pressure Balances
Centrifugal Forces, and streamline
curvature effects neglected then:
d
1
d dmCU2
 p  dp  r  dr  d  prd  2( p  dp)dr  51
2
2
r
Simple Radial Equilibrium
• So from the radial momentum equation:
dmCU2
1
 p  dp  r  dr  d  prd  ( p  dp)drd 
2
r
• reduces to:
1 dp CU2

 dr
r
Radial equilibrium equation
1 dp CU2 dh
ds


T
 dr
r
dr
dr
• Tds/dr represents the spanwise variation in relative loss. If this
is assumed to be zero, then (Simple Radial Equilibrium Equation
with Cr=0):
1 dp C 2 dh
 dr

U
r

dr
52
Simple Radial Equilibrium
• Now:

C x2  CU2
C2
h0  h 
h
2
2

• So that we get (vortex energy equation):
dh0 CU2
dCx
dCU

 Cx
 CU
dr
r
dr
dr
• If dh0/dr = 0 (work is constant with r) and we assume that
CX=constant as a function of span, then:
CU2
dCU CU d (rCU )
 CU

0
r
dr
r
dr
53
Simple Radial Equilibrium
• One important variations / solutions to this equation
CU d (rCU )
0
r
dr
– Free vortex flow
CU dCU

r
dr
•
dCU dr
or

 CU r  constant
CU
r
Later we will see more general form of this equaion will lead to
another solution for forced vortex flow
CU
 constant
r
54
Consider Free Vortex Design for 1st Stage Rotor
• CU r = constant so that:
 rmean 
CUtip  CUmean 

 rtip 
 rmean 
CUhub  CUmean 

r
 hub 
• So that at the exit of the first stage rotor we have:
 .1697 
CU 2 tip  75.38 
 58.3

 .2194 
 .1697 
CU 2 hub  75.38 
 106.6

 .12 
1  58.3 
0

tan

21.24
tip


 150 
 106.6 
0
 2 hub  tan 1 
  35.4
 150 
2
55
Free Vortex Design for 1st Stage Rotor
• The axial velocity Cx is assumed to be constant as a function of
radius (Note that it doesn’t have to be !)
WU  CU U
 58.3  355.3 
0
 2 tip  tan 


63.2

150


 106.6  177.7 
0
 2 hub  tan 1 


25.36

150


1
56
1st Stage Blade Spanwise Variations
(Free Vortex Design)
W=C-U
1=0
W1h=232.5
W1m=305.
9
W1t=385.7
Cx1=150
1h=49.83
Uh=- WU1h =177.7
1m=60.64
Um=- WU1m =266.6
Ut=- WU1t =355.3
1t=67.11
2h=35.4
C2h=184.02
C2m=167.87
C2t=160.93
2m=26.68
CU2h=106.6
2t=21.24
CU2m=75.38
CU2t=58.3
ROTOR
Cx2=150
Uh=177.7
Um=266.6
W2m=166.0
2h=25.36
W2m=242.03
2m=51.89
W2m=332.73
2t=63.2
WU2h=71.1
Ut=355.3
WU2m=191.22
57
WU2t=297.0
Spanwise Variation in Reaction with Free
Vortex Design
• Remember that Reaction is given by:
Cx
Cx
tan


tan


1


 tan  2  tan 1 
2
1
2U
2U
CU 1  CU 2
CU 1r  CU 2 r
R  1
 1
2U
2U m r 2 / rm2
R
where U  U m
R  1
r
rm
and CU r  constant
constant
r2
• If R=50% at rm, radial variation may make root too small and tip too
large for good efficiency
• The free vortex design, Rh =0.7, Rm = 0.859, and Rt = 0.918 (high at
the tip !!). This is why designers sometimes move away from free
vortex design in favor of a different strategy, like constant spanwise
58
Reaction distribution or “forced vortex” design.
Consideration When Diverting from Free
Vortex Design
• The free vortex design,
Rh =0.7,
Rm = 0.859, and
Rt = 0.918 (high at the tip !!).
This is why designers sometimes move away from free vortex
design in favor of a different strategy, like constant spanwise
Reaction distribution or “forced vortex” design.
• When using a radial design strategy other than free vortex:
– radial equilibrium is not satisfied, some error in velocities arise
– total work of stage may not deliver design intent exactly
– axial velocity may not be constant across span
• mass flow may not reach design intent exactly
• hub and tip radii would need to be re-computed
59
Non- Free Vortex Designs
• Consider rotor inlet / exit swirl velocity distributions of the form
CU 1  aR n 
b
R
and CU 2  aR n 
b
R
where a, b, n are constants and R  r / rm
1) if n  1, reduces to free vortex design approach
2) if n  1, called first  power or forced  vortex design approach
Equation of motion must be rederived with C x  constant
C x21   C x21   2  a 2 R 2  2ab ln R  a 2  and
m
2a ln R a
R  1

Um
Um
60
Airfoil Design
• Once the velocity triangles for that blade-row are established
from the meanline analysis, then the job remaining is to design
the airfoil that will deliver the required exit velocity triangle given
the inlet velocity triangle
Correlations For Deviation and Loss are
Derived from Cascade Data
Deviation
Loading Coefficient Gives Solidity

2Cos 2  2  C X 1

z 

Tan1  Tan 2 
x
 CX 2

Solidity and Velocity Gives Dfactor
D factor  1 
Loss
V2 Vu1  Vu 2

V1
2V1
Dfactor Gives /c, Loss, and Efficiency

c
 0.006  0.0002  e
 
2  
c
   
cos  2
7.5 D f
 C x 2 cos  1 


C
cos

2 
 x1
2
2

 R 61
1  Cx    S
  1

   2

2 E  U   cos  2 cos 2  1 
Streamline Curvature Method
• For Simple Radial Equilibrium (Cr=0), the solution is possible at
any given axial plane with no information from up- or downstream!
• Streamline Curvature Methods - Cr  0 (Axisymmetric Analysis
with Models for Loss and Deviation)
• 3D Velocity Diagram:
62
2D Throughflow Analyses
Wu’s S1 & S2 surface method
63
Streamline Curvature Method
• Inviscid Momentum Equation:
P

Substantial Derivative:
• Using:

DC
0
Dt
DC  C  C  C  C




Dt
t  x  y  z
x  r cos 
y  r sin 
zz
64
Streamline Curvature Method
• Radial, tangential & axial components of acceleration in
cylindrical coordinates.
2
dC r CU
ar 

dt
r
dCU CU C r
a 

dt
r
dC z
az 
dt
• Using the chain rule in cylindrical coordinates:
d

 dr 



dt  t  r dt 
C
d



 Cr
 U
dt  t
r
r
d
 dz

dt  z dt


 Cz

z
65
Streamline Curvature Method
• The radial component of the momentum equation becomes:
C r CU C r
C r CU
1 P C r

 Cr

 Cz

 r
t
r
r 
z
r
2
0
• The radial pressure gradient for steady, axisymmetric flow
becomes:
1 P CU

 r
r
2
 Cr
C r
C r
 Cz
r
z
• Modification to Simple Radial Equilibrium depends entirely on
radial velocity!
66
Streamline Curvature Method
• Axial (z) gradient - can't solve flow with information at only one
axial location. Multiple stations are used in streamline curvature
codes.
• Derivative in meridional direction, along streamline in the r-z
plane:

r 
z 
Cm
 Cm
 Cm
m
mr
mz



Cm
 Cm sin 
 Cm cos 
m
r
z



Cm
 Cr
 Cz
m
r
z
C r
1 P Cu

 Cm
 r
r
m
2
67
Streamline Curvature Method
• Further
C r  C m sin  

m
m
C m
C r

 sin 
 C m cos 
m
m
m
Cm
Cr

2
Cm
 Cm sin 
 Cm cos 
m
m
m
C m C m 2 cos 
C r
Cm
 Cr

m
m
rm
• Meridional Momentum Equation:
C m cos 
C m
1 P C u


 Cr
 r
r
rm
m
2
2
68
Streamline Curvature Method
• Above form shows effects of streamline curvature & meridional
acceleration
• Method used with cascade data for design & off design analysis
• Method used to convert measured P & T into velocities, angles
etc.
• Extension of theory accounts for radial blade force terms at
intra-blade stations
• Same approach used for centrifugal designs, but in streamwise
& normal coordinate system
69
Streamline Curvature Method
• "Streamline" codes are only as good as the loss & blockage
models in the code
• Inherently weak prediction of endwall flows - secondary flow,
seal & tip leakage effects
• Streamline solutions quickly generate velocity diagrams for
initial design
• Reduce measured P & T into velocity diagrams
70
Potential Homeworks
Problem 1: Re-design the first 3 stages of our 7-stage compressor:
– holding the tip radius constant (half the class)
– holding the hub radius constant (half the class)
– keep the existing assumptions (constant Cx with r, constant work with r
(free vortex), stage T (20,25,25), N, etc.
– sketch the velocity triangles and airfoil shapes
• Compare your results to Mattingly’s COMPR code
• Compare your results with the constant mean radius and other design
for this homework
– how do airfoil shapes change?
– which design is the best?
Problem 2: Use Euler’s turbine equation to verify the hub and tip velocity
triangles for the 1st stage rotor obtained in class from radial equilibrium
– work is assumed to be constant with r (free vortex)
– note that there is radius change at the hub and tip
71