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

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

6. 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:

7. Compressor Meanline Design
Given: m, Utip, p01, T01, Pr, poly and c1x chosen to avoid high tip Mach numbers and stresses
Compressor inlet (1)
Select RH/RT and Utip (N)
for turbine issues

8. Compressor Meanline Design
Compressor exit (2)

9. Compute Compressor Exit Conditions
Compute Compressor Exit Total Temperature
so that T02 = (4.15) = K,
DT0 compressor= = K and other conditions:

10. Compute Compressor Exit Conditions
Exit area, hub and tip radius:

11. Step 2 - Estimate the Number of Stages
From Euler’s Turbine Equation:
With no inlet guide vane (Cu1=0, a1 = 0, and Wu1= -U), the relative flow angle is:
And the relative inlet velocity to the 1st rotor is:

12. 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:
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:
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.

13. 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
Typically (T0)stage  40K (subsonic) K (transonic)
Therefore we choose to use six or seven stages. To be conservative (account for losses, ie. ha<1),
Choose 7 stages
Recalculate the DT0stage = 164.5/7 = 23.5
So 1st stage temperature ratio is T0 ratio = /288 =1.0816
The stage pressure ratio is then P0 ratio = (T0 ratio ) hpg / (g - 1) =

14. Compressor Meanline Design
Develop estimate of the number of stages
Assuming Cx = constant
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

15. Compressor Meanline Design
Develop Stage by Stage Design
C = absolute velocity, CU = absolute velocity in U direction W2 C2 C1
Constant Cx W1 U

16. Step 3 - Calculate Velocity Triangles of 1st Stage at Mean Radius
So from Euler Turbine Equation:
We can re-calculate the relative angles for the 1st stage:

17. Velocity Components and Reaction of 1st Stage
The velocity components for the 1st stage (rotor) are therefore:
The Reaction of the 1st stage is given by:

18. Velocity Components for Stator of 1st Stage
Now consider the stator of the 1st stage. The Dh0 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:

19. 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

20. 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.

21. Non- “Repeating Stage” Design Strategy
Instead of taking a constant temperature rise
across each stage, we could reduce stage temperature rise for first and last stages of the compressor and increase it for the middle stages. This strategy is typically used to:
reduce loading of first stage to allow for a wide variation in angle of attack due to various aircraft flight conditions
reduce turning required in last stage to provide for zero swirl flow going into combustor
With this in mind, let’s change the work distribution in the compressor to:

22. 1st Stage Design for “Non-Repeating Stages”
We can re-calculate the relative angles for the 1st stage:

23. Velocity Components and Reaction of 1st Stage with Non-Repeating Stages
The new velocity components for the 1st stage (rotor) are therefore:
The Reaction of the 1st stage is given by:

24. Design of 1st Stage Stator
The pressure ratio for this design with a temperature change, DT0 = 20 is:
So P03= P02 = 1.01 (1.236) = bar and T03= =308 0K
Now we must choose a value of a3 leaving the stator.
When we designed with repeating stages, a3= a1.
But now we have the flexibility to change a3.

25. Design of the 1st Stage Stator & the 2nd Stage
Change a3 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:
And if we design the second stage to a temperature rise of 25 0, the Euler’s equation:
Which can be solved simultaneously for b1and b2

26. Design of 1st Stage Stator & 2nd Stage Rotor
Note that this is same as specifying E, n, and R as in one of your homeworks and computing the angles.
And absolute flow angles of second stage can be found from So
Therefore, we have determined the velocity triangles of the 1st stage stator and the second stage rotor

27. 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

28. Design of 2nd Stage Stator & 3rd Stage Rotor
Design of 2nd stage stator and 3rd stage rotor can be done in same manner as 1st stage stator and 2nd stage rotor.
A choice of 50% reaction and a temperature rise of 25 degrees for 3rd stage will lead to increased work by 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

29. Class 12 - The 7-Stage Compressor Design So Far Has Lead to 1st and 2nd Stages:

30. 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 2nd stage stator that provides for the desired Reaction and Work in 3rd stage using a similar technique as previously used.

31. Design of 2nd Stage Stator & 3rd Stage
We can change a3 so that there is swirl going into third stage and thereby reduce reaction of second stage design. If we design third stage to have a reaction of 0.5, then from equation for reaction:
And if we design third stage to a temperature rise of 25 0, Euler’s equation:
Which can be solved simultaneously for b1and b2

32. Design of 2nd Stage Stator & 3rd Stage Rotor
And the absolute flow angles of the second stage can be found from So
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:

33. 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).

34. Summary of Conditions for Stages 1 - 3

35. 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)

36. 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.

37. Summary of Conditions for Stages 1 - 6

38. Stage 7 Design
So going into stage 7, we have P01= 3.65 and T01 = The requirements for our 7-stage compressor design we have
P0 exit = 4.15 * 1.01 = 4.19 bar
T0 exit = (4.15) = K
This makes the requirements for stage 7:

39. Stage 7 Design If we assume a Reaction = 0.5 for the 7th stage:
Then, solving equations:

40. 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.

41. Summary of Compressor Design

42. Hub and Tip Radii for Each Blade Row
From the pressure and temperature, we can compute the density from the equation of state:

43. Hub and Tip Radii for Each Blade Row
From Continuity:
and our design value of rmean = ,
we can calculate the hub and tip radii (i.e. area) at the entrance and exit of each blade row:

44. Hub & Tip Radii for All Stages of Compressor
So we get:
rotor
stator

45. Conditions in Compressor

46. Hub & Tip Radii Distribution - Flow Path Area

47. 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 = m and
Uhub = , Umean = 266.6, Utip = m/s
leading to relative flow angles:
Next, we must choose the type of radial design strategy from:
free vortex where CU r = constant (dh0/dr = 0)
constant reaction where U DCU = constant
exponential where CU1 = a - (b/R) and CU2 = a + (b/R)
The exit radial pressure gradient will be different for each of the designs

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
Consider axisymmetric flows [ ] and steady flow [ ]
In the radial direction,

50. Radial Equilibrium
P+1/2 dp P
P+dp r dr
Mass = dm = r r dq dr CU dq
If Cr=0 and Pressure Balances Centrifugal Forces, and streamline curvature effects neglected then:

51. Simple Radial Equilibrium
So from the radial momentum equation:
reduces to:
Tds/dr represents the spanwise variation in relative loss. If this is assumed to be zero, then (Simple Radial Equilibrium Equation with Cr=0):

52. Simple Radial Equilibrium
Now:
So that we get (vortex energy equation):
If dh0/dr = 0 (work is constant with r) and we assume that CX=constant as a function of span, then:

53. Simple Radial Equilibrium
One important variations / solutions to this equation
Free vortex flow
Later we will see more general form of this equaion will lead to another solution for forced vortex flow

54. Consider Free Vortex Design for 1st Stage Rotor
CU r = constant so that:
So that at the exit of the first stage rotor we have:

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 !)

56. 1st Stage Blade Spanwise Variations (Free Vortex Design)
W = C - U
Cx1=150
a1=0
W1h=232.5
b1h=49.83
Uh=- WU1h =177.7
C2h=184.02
Um=- WU1m =266.6
W1m=305.9
b1m=60.64
C2m=167.87
Ut=- WU1t =355.3
W1t=385.7
b1t=67.11
C2t=160.93
a2h=35.4
a2m=26.68
CU2h=106.6
a2t=21.24
CU2m=75.38
CU2t=58.3
Uh=177.7
Cx2=150
ROTOR
Um=266.6
WU2h=71.1
W2m=166.0
b2h=25.36
Ut=355.3
W2m=242.03
b2m=51.89
WU2m=191.22
W2m=332.73
b2t=63.2
WU2t=297.0

57. Spanwise Variation in Reaction with Free Vortex Design
Remember that Reaction is given by:
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 = (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.

58. Consideration When Diverting from Free Vortex Design
The free vortex design,
Rh =0.7,
Rm = 0.859, and
Rt = (high at the tip !!).
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

60. Correlations For Deviation and Loss are Derived from Cascade Data
Airfoil Design
Once velocity triangles for that blade-row are established from meanline analysis, then job remaining is to design the airfoil that will deliver required exit velocity triangle given the inlet velocity triangle
Loading Coefficient Gives Solidity
Correlations For Deviation and Loss are Derived from Cascade Data
Solidity and Velocity Gives Dfactor
Deviation
Dfactor Gives q/c, Loss, and Efficiency
Loss

61. Determined kinematics and thermodynamics for compressor
Determined radii, i.e. rh, rm, rT
Now need to define actual airfoils, c, =stagger, shape, Nb based on Class 6 and 8 notes.
Pick =solidity from choice of Df and velocities
Choose shape from family, e.g. DCA [double circular arc] for supersonic tip
[t/c]tip max=3% for structure, 10% for hub, linear variation between, [t/c]m=6.5%

62. Pick c=chord for turbulent flow, Rec > 300,000 at all h’s.
Make bigger for structure and vibration issues
Re at 50 kft about 1/5 SLTO
Define incidence such that flow is aligned with upper surface to produce no shock in supersonic flow

63. Estimate deviation from Carter’s rule
Define stagger angle from
Set number of blades