1. 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: P amb = 1.01 bar, T amb = 288 KU tip = 350 m/s Inlet r hub / r tip = 0.5 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. 2 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

3. 3 Compressor Meanline Design Process Steps 1)Choose Cx1 and r H /r T to satisfy m and keep Mtip low and define r T 2)Select N from r T and U T 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)…..

4. 4 Step 1- Choice of Rotational Speed & Annulus Dimensions Construct table of inlet / exit properties and parametric study of c 1x vs. tip Mach number [next chart] Chose c 1x 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 r hub /r tip = 0.5 and compute rotational speed and tip Mach number

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

6. 6 Compute Root (Hub) and Mean Radius Choose N = 250 rev/sec or 15,000 RPM and r hub /r tip = 0.5 With hub/tip radius ratio and tip radius:

7. 7 Compressor Meanline Design Given: m, U tip, p 01, T 01, Pr,  poly and c 1x chosen to avoid high tip Mach numbers and stresses Compressor inlet (1) Select R H /R T and U tip (N) for turbine issues

8. 8 Compressor Meanline Design Compressor exit (2)

9. 9 Compute Compressor Exit Conditions Compute Compressor Exit Total Temperature so that T 02 = 288.0 (4.15) 0.3175 = 452.5 0 K,  T 0 compressor = 452.5 - 288.0 = 164.5 0 K and other conditions:

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

11. 11 Step 2 - Estimate the Number of Stages From Euler’s Turbine Equation: With no inlet guide vane (C u1 =0,   = 0, and W u1 = -U), the relative flow angle is: And the relative inlet velocity to the 1st rotor is:

12. 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. 13 Choose Number of Stages Given  poly and T 0out /T 0in   T 0 = T 0out -T 0in, so the number of stages is  T 0 compressor /  T 0stage = 164.5/28 = 5.9 Typically (  T 0 ) 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  T 0stage = 164.5/7 = 23.5 So 1st stage temperature ratio is T 0 ratio = 288 + 23.5/288 =1.0816 The stage pressure ratio is then P 0 ratio = (T 0 ratio )  p  = 1.2803

14. 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)  (  T 0 ) stage = …. –(  T 0 ) stage N stages  T 0out -T 0in –Select N stages and select nearly constant set of (  T 0 ) stage Develop Stage by Stage Design Assume that continual blockage buildup due to boundary layers reduces work done, therefore

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

16. 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. 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. 18 Velocity Components for Stator of 1st Stage Now consider the stator of the 1st stage. The  h 0 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. 19  2 =49.89 Velocity Triangles of 1st Stage Using “Repeating Stage” Assumption U=- W U1 =266.6  1 =0 W 1 =305.9 C x1 =150 W 2 =232.77 C x2 =150 U=266.6  2 =30.57 W U2 =178.0 C U2 =88.6  1 =60.64 U=266.6  3 =0 W 3 =305.9 C x3 =150  3 =60.64 W = C - U ROTOR STATOR 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. C 2 =174.21

20. 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. 21 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:

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

23. 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. 24 Design of 1st Stage Stator The pressure ratio for this design with a temperature change,  T 0 = 20 is: So P 03 = P 02 = 1.01 (1.236) = 1.248 bar and T 03 = 288+20=308 0 K 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.

25. 25 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: And if we design the second stage to a temperature rise of 25 0, the Euler’s equation: Which can be solved simultaneously for  1 and  

26. 26 Design of 1st Stage Stator & 2nd Stage Rotor Note that this is the same as specifying E,, 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 So Therefore, we have determined the velocity triangles of the 1st stage stator and the second stage rotor

27. 27  2 =51.89 Velocity Triangles of 1st Rotor Using “Non- Repeating Stage” Assumption U=- W U1 =266.6  1 =0 W 1 =305.9 C x1 =150 W 2 =242.03 C x2 =150 U=266.6  2 =26.68 W U2 =191.2 2 C U2 =75.38  1 =60.64 U=266.6  3 =12.36 W 3 =277.73 C x3 =150  3 =57.31 W = C - U ROTOR STATOR 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. C U3 =32.87 W U3 =233.7 7 C 3 =153.56 C 2 =167.87

28. 28 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.

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

30. 30 Design of 2nd Stage Stator The pressure ratio for the 2nd stage design with a temperature change,  T 0 = 25 is: So P 03 = P 02 = 1.248 (1.279) = 1.596 bar and T 03 = 308+25=333 0 K 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.

31. 31 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: And if we design the third stage to a temperature rise of 25 0, the Euler’s equation: Which can be solved simultaneously for  1 and  

32. 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. 33  2 =42.92 Velocity Triangles of 2nd Stage W 2 =204.86 C x2 =150 U=266.6  2 =40.27 W U2 =139.5 3 C U2 =127.07 W = C - U ROTOR STATOR 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). C 2 =196.59 U=266.6  3 =12.36 W 3 =277.73 C x3 =150  3 =57.31 C U3 =32.87 W U3 =233.7 7 C 3 =153.56 U=266.6  3 =29.88 W 3 =234.63 C x3 =150  3 =50.26 C U3 =86.18 W U3 =180.4 2 C 3 =172.99

34. 34 Summary of Conditions for Stages 1 - 3

35. 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. 36  2 =29.88 Velocity Triangles of Stages 3 - 6 W 2 =172.99 C x2 =150 U=266.6  2 =50.26 W U2 =86.18 C U2 =180.42 W = C - U ROTOR STATOR Notice that the velocity triangles are “symmetric” due to the 50%reaction design. C 2 =234.63 U=266.6  3 =29.88 W 3 =234.63 C x3 =150  3 =50.26 C U3 =86.18 W U3 =180.4 2 C 3 =172.99 U=266.6  3 =29.88 W 3 =234.63 C x3 =150  3 =50.26 C U3 =86.18 W U3 =180.4 2 C 3 =172.99

37. 37 Summary of Conditions for Stages 1 - 6

38. 38 Stage 7 Design So going into stage 7, we have P 01 = 3.65 and T 01 = 433. The requirements for our 7-stage compressor design we have –P 0 exit = 4.15 * 1.01 = 4.19 bar –T 0 exit = 288.0 (4.15) 0.3175 = 452.5 0 K This makes the requirements for stage 7:

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

40. 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 3 - 6 and that the exit absolute air angle leaving the compressor is 32.77 0. 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. 41 Summary of Compressor Design

42. 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. 43 Hub and Tip Radii for Each Blade Row From Continuity: and our design value of r mean = 0.1697, we can calculate the hub and tip radii (i.e. area) at the entrance and exit of each blade row:

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

45. 45 Conditions in Compressor

46. 46 Hub & Tip Radii Distribution - Flow Path Area