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
R  0.287
kJ
kg 0 K
c p  1.005
  1.4
kJ
kg 0 K
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 and
compute rotational speed and tip Mach number
• Choose N = 250 rev/sec or 15,000 RPM and rhub/rtip
= 0.5, calculated rhub, rtip, rmean
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
Compressor Meanline Design
• c1x chosen to avoid high tip Mach numbers and stresses
• Calculate compressor inlet details (1)
• Given: m, Utip, p01, T01, Pr, poly calculate compressor exit
details including Aexit, h exit (2)
• Calculate 1 from U, Cx
• Calculate W2=0.72W1, 2 from Cx, W2
• From Euler, 1 and 2, calculate T0-12
6
Step 3-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
• To be conservative (account for losses, i.e. a<1),
– Choose 7 stages
• Recalculate the T0stage = 164.5/7 = 23.5
• Calculate 1st stage temperature ratio is T0 ratio and
pressure ratio
7
Step 3 - Calculate Velocity Triangles of 1st Stage
at Mean Radius [Rm]
• Use Euler Turbine Equation to recalculate 2 and check
de Haller
• Calculate 1st stage rotor exit conditions [2]
• Calculate Reaction [R1=0.84, high]
• Consider the stator of the 1st stage, assume “repeating
stage”, then inlet angle to stator [α1=0] is absolute air
angle coming out of rotor [α3] and, exit absolute angle of
stator is inlet absolute angle of rotor
W2
C2
W1
U
C1
Constant Cx
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
9
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.
10
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
11
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
12
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
13
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.
14
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.
15
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:
UCx
T0 stage 3 
(tan 1  tan 2 )
cp
• Which can be solved simultaneously for 1and 2
 1 stage3   50.26 0
 2 stage3   29.88 0
16
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
17
stage due to 70%reaction design
but
will be for 3rd stage (50% reaction).
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)
18
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
19
50%reaction design.
Stage 7 Design
• For stage 7, we have P01= 3.65 and T01 = 433. For
this 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
• If we assume a Reaction = 0.5 for the 7th stage:
Cx
R stage7 
(tan 1  tan  2 )  0.5
2U
T0 stage7
• or:
U Cx

(tan 1  tan  2 )  19.5
cp
1stage 7   48.590
 2 stage 7   32.770
20
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.
21
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
22
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 #
23
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 #
24