1. 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 Utip = 350 m/s
Inlet rhub / rtip = Compressor has no inlet guide vanes
Mean radius is constant
Polytropic efficiency = 0.90
Constant axial velocity design
No swirl at exit of compressor

2. Steps in the Meanline Design Process
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

3. Compressor Meanline Design Process
Steps
Choose Cx1 and rH/rT to satisfy m and keep Mtip low and define rT
Select N from rT and UT
Compute T0 across compressor and all exit flow conditions [keep rm same through engine]
Estimate T0 for first stage from inlet condtions [Euler and de Haller]
Select number of stages  T0comp / T0stage
…..

4. 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

5. Calculate Tip Radius and Rotational Speed
Drive choice by compressor inlet conditions

6. 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

7. Step 3-Choose Number of Stages
Given poly and T0out/T0in  T0 = T0out -T0in, so the number of stages is
DT0 compressor / DT0stage = 164.5/28 = 5.9
To be conservative (account for losses, i.e. ha<1),
Choose 7 stages
Recalculate the DT0stage = 164.5/7 = 23.5
Calculate 1st stage temperature ratio is T0 ratio and pressure ratio

8. 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
Constant Cx U C1 W2 W1 C2

9. Velocity Triangles of 1st Stage Using “Repeating Stage” Assumption
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.
Cx1=150
a1=0
W = C - U
b1=60.64
U=- WU1 =266.6
W1=305.9
STATOR
Cx3=150
a3=0
b3=60.64
C2=174.21
a2=30.57
U=266.6
CU2=88.6
W3=305.9
Cx2=150
ROTOR
U=266.6
b2=49.89
WU2=178.0
W2=232.77

10. 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.

11. Non- “Repeating Stage” Design Strategy
Instead of taking a constant temperature rise
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:

12. Velocity Triangles of 1st Rotor Using “Non-Repeating Stage” Assumption
Cx1=150
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.
a1=0
W = C - U
b1=60.64
U=- WU1 =266.6
C3=153.56
W1=305.9
STATOR
a3=12.36
Cx3=150
CU3=32.87
b3=57.31
C2=167.87
U=266.6
a2=26.68
CU2=75.38
W3=277.73
Cx2=150
ROTOR
U=266.6
b2=51.89
WU3=233.77
WU2=191.22
W2=242.03

13. Velocity Triangles of 1st Stage Using “Repeating Stage” Assumption
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.
Cx1=150
a1=0
W = C - U
b1=60.64
U=- WU1 =266.6
W1=305.9
STATOR
Cx3=150
a3=0
b3=60.64
C2=174.21
a2=30.57
U=266.6
CU2=88.6
W3=305.9
Cx2=150
ROTOR
U=266.6
b2=49.89
WU2=178.0
W2=232.77 13 13

14. 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

15. Design of 2nd Stage Stator
The pressure ratio for the 2nd stage design with a temperature change, DT0 = 25 is:
So P03= P02= (1.279) = bar and T03= =333 0K
Now we must choose a value of a3 leaving the 2nd stage stator that provides for the desired Reaction and Work in the 3rd stage using a similar technique as previously used.

16. Design of 2nd Stage Stator & 3rd Stage
We can change a3 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:
And if we design the third stage to a temperature rise of 25 0, the Euler’s equation:
Which can be solved simultaneously for b1and b2

17. Velocity Triangles of 2nd Stage
W = C - U
Cx3=150
CU3=32.87
b3=57.31
C3=172.99
U=266.6
a3=29.88
Cx3=150
CU3=86.18
W3=277.73
STATOR
b3=50.26
U=266.6
WU3=233.77
C2=196.59
W3=234.63
a2=40.27
CU2=127.07
Cx2=150
ROTOR
WU3=180.42
U=266.6
b2=42.92
WU2=139.53
W2=204.86
Notice that the velocity triangles are not “symmetric” for the second stage due to 70%reaction design but will be for 3rd stage (50% reaction).

18. 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)

19. Velocity Triangles of Stages 3 - 6
W = C - U
Cx3=150
CU3=86.18
C3=172.99
b3=50.26
a3=29.88
Cx3=150
CU3=86.18
U=266.6
STATOR
b3=50.26
W3=234.63
U=266.6
W3=234.63
WU3=180.42
C2=234.63
CU2=180.42
WU3=180.42
a2=50.26
Cx2=150
ROTOR
U=266.6
b2=29.88
WU2=86.18
W2=172.99
Notice that the velocity triangles are “symmetric” due to the 50%reaction design.

20. Stage 7 Design
For stage 7, we have P01= 3.65 and T01 = For this our 7-stage compressor design we have
P0 exit = 4.15 * 1.01 = 4.19 bar
T0 exit = (4.15) = K
If we assume a Reaction = 0.5 for the 7th stage:
or:

21. Stage 7 Design
And from:
or from symmetry of the velocity triangles for 50% reaction:
Note that the absolute angles going into stage 7 have changed from those computed for stages and that the exit absolute air angle leaving the compressor is This means that a combustor pre-diffuser is required to take all of the swirl out of the flow prior to entering the combustor.

22. Summary of Compressor Design

23. Conditions in Compressor

24. Hub & Tip Radii Distribution - Flow Path Area