PEMP RMD510. M.S. Ramaiah School of Advanced Studies, Bengaluru

Transcription

1 Design of Axial Flow Compressor- Session delivered by: Prof. Q. H. Nagpurwala

2 Session Objectives To learn design procedure for axial compressor stages To calculate flow properties across the blade rows and to determine velocity triangles To determine number of stages To determine annulus area from inlet to exit

3 Multistage Axial Flow Compressor 3

4 Velocity Triangles Two-Dimensional Approach U : Peripheral velocity at the mean blade height in a tangential plane C a : axial component C w : whirl or tangential component 4

5 Compression Process on T-S Diagram 5

6 Euler Turbine Equation Work done per unit mass flow rate or Specific Work W U C U w w For U = U, and C a = C a = C a, we can write Also W U C a U C a C U(C C ) ( w w UC a ( tan α tan α ) UC a ( tan β tan β ) tan tan β tan α tan β () () 6

7 Euler Turbine Equation The input energy is absorbed usefully in raising the pressure, temperature and velocity of the air and wastefully in overcoming various frictional losses. W C p ΔT ΔT os UC ( a UC c ( tan β tanβ ) tan β tanβ a os p And, if C 3 = C UCa ΔT os ΔT s ( tan β tan β ) Total pressure ratio, R s η s c p ΔT os γ γ- ) s = stage isentropic efficiency 7 T o T 0 = inlet stagnation temperature

8 Work Done Factor Radial distribution of axial velocity is not constant along the height h of the blade. The velocity profile settles down in the fourth stage. Axial velocity distributions from first to fourth stage are shown in the figure. 8

9 Work Done Factor Work kdone Factor is the ratio of the actual work absorbing capacity of the stage to its ideal value as calculated from the Euler turbine equation. W UCa tan β tan β U U C a tan α C tan β a β U U C a tan α tan β For a given rotor blade, α and β are almost constant Hence, less work is done at the region where C a is high and the actual temperature rise is given by λuca ΔTos ΔTs ( tan β tan β ) c p is the work done factor, which is less than unity. Its value may range from at the first stage to about 0.85 at the fourth and subsequent stages. 9

10 Degree of Reaction Degree of Reaction, R, is defined das the ratio of the static enthalpy rise in the rotor to the static enthalpy rise in the whole stage. Also, Since C p R c p W ΔT ΔT ΔT A A A ΔT B ΔT A ΔT B T 0 s c p Ts cpδta ΔTB cpδts UC a tan β tan β UC a ( tanα tanα ) A UC ( tanα It can be shown that Ca R ( tanα tanα U a tanα : Static temperature rise in the rotor : Static temperature rise in the stator )- C C 0 )

11 Symmetrical Blading By adding equations () and () U C a tan α tan β tan α tan C a U U R tan β tan β U Ca Ca C a tan β β tan U U If R then tan β tan β C a From equation () & () It is assumed that = α β β β α This results in symmetrical velocity triangles across the rotor β Since cannot be, the degree of reaction achieved will be slightly different from 0.5

12 Un-symmetrical Blading R > 50% ; β > α R < 50% ; β < α If R > 0.5, then β > α and the velocity diagram is skewed ed to the right. The static enthalpy rise in the rotor exceeds that in the stator (this is also true for the static ti pressure rise). If R < 0.5, then β < α and the velocity diagram is skewed to the left. The stator enthalpy (and pressure) rise exceeds that in the rotor.

13 Polytropic Efficiency Small Stage or Polytropic Efficiency of Compressor Polytropic efficiency is the efficiency of a compressor stage operating between infinitesimal pressure differential P. It is used in comparing the performance of two compressors having the same pressure ratio but operating at different temperature levels. In multistage compressors, the polytropic efficiency is used in defining the isentropic efficiency of individual stages. 3

14 Polytropic Efficiency Relation between Polytropic Efficiency and Isentropic Efficiency of a compressor 0.9 p = 0.9 p Isentropic 0.8 p = efficiency, c p p = 0.7 c p p p Pressure ratio, p /p Variation of small stage (polytropic) efficiency of compressor with pressure ratio 4

15 Polytropic Index Pl Polytropic index id n is defined dsuch that n n or p n n p From consideration o of small stage efficiency e cy For ideal compression process Stage polytropic p efficiency can now be written as p T p p T T T ' n ln p p n ln T T p p p 5

16 Blade Loading Criteria De Haller Number V 0.7 for rotor V V C C for stator Lieblein s Diffusion Factor V D V V V w s c cos cos s tan tan cos c D for incompressible flow D > (at rotor tip) ; > 0.6 (at rotor hub) 6

17 Design Procedure. Make appropriate assumptions about the efficiency, tip speed, axial velocity and related parameters. Size the annulus at inlet and outlet of the compressor 3. Calculate the air angles required for each stage at the mean diameter 4. Choose a suitable vortex theory and calculate the air angles at various radii from root to tip 5. Check for blade stresses, rate of diffusion and Mach number levels 6. Choose incidence angles, solidity, blade thickness, etc. from cascade data. Estimate deviation angles. Calculate blade metal angles 7. Choose suitable aerofoil shape. Generate and stack blade sections from hub to tip 8. Evaluate the compressor performance through empirical relations or CFD analysis or experiments 7

18 Design of Axial Flow Compressor Task: To design an Axial Flow Compressor assuming that the compressor has no inlet guide vanes Design Data Sea level static conditions : p a =.0 bar, T a = 88 K Compressor pressure ratio : 4.5 Air mass flow rate : 0 kg/s 8

19 Design Process The complete design process for the compressor will encompass the following steps: Choice of rotational speed and annulus dimensions Determination of number of stages, using an assumed efficiency Calculation of the air angles for each stage at the mean radius Determination of the variation of the air angles from root to tip Investigation of compressibility effects Selection of compressor blading, using experimentally obtained cascade data Check on efficiency previously assumed, using the cascade data Estimation of off-design performance Rig testing 9

20 Design Process ( contd.) In practice, the process of design will be one of continued refinement, coupled with feedback from other groups such as: Designers of the combustion system Designers of turbine module Material specialists Mechanical designers Stress analysts Experts in rotor dynamics Experts in bearings and lubrication system Personnel responsible for fabrication and assembly 0

21 Rotational Speed and Annulus Dimensions Assume from previous experience Tip speed: U t = 350 m/s Axial velocity C a = 50 m/s and no IGV at inlet so that there will be no whirl component of velocity at inlet (see figure in next slide to understand the effect of IGV) Hub/tip diameter ratio ~ 0.4 to 0.6 at the entrance For a specified annulus area the tip radius will be a function of the hub-tip ratio For a fixed blade speed, the rotational speed will also be a function of hub-tip ratio Thus the designer has a wide range of solutions and must use engineering judgment to select the most promising one

22 Inlet Relative Mach Number

23 Mass Flow Continuity To satisfy continuity: r t a t r t a C r r r AC m r r t m r t r a t r r C πρ r t : Tip radius ; r r : Root M.S. Ramaiah School of Advanced Studies, Bengaluru 05 3

24 Compression Process on T-S Diagram 4

25 Rotational Speed At sea-level static conditions, T 0 =T= a = 88 K. Assuming no loss in intake, p 0 = p a =.0 bar. C T C a 50m / s ( C w 0 ) *.005 * K 3. 5 T 76.8 p p bar T * kg / m 0.87* rt.06*50 r r r r * r t r t The tip speed, U t, is related to r t by U t = r t N,, and hence if U t is chosen to be 350 m/s, then 350 N 5 r t

26 Variation of N with Radius Ratio r t and N are evaluated over a range of hub-tip ratios Results are given in the following table : r rt t r t m N rev s

27 Turbine Data We need to consider the turbine to which this compressor could be coupled. Let us say a turbine designed for this purpose has a speed of 50 rev/s and the outer radius of the turbine inlet is 0.39m Referring to the table above, a hub-tip ratio of 0.50 would give a compatible compressor tip radius of 0.6 m although the rotational speed is 46.3 rev/s. There was nothing sacrosanct about the choice of 350 m/s for the tip speed, and the design could be adjusted for a rotational speed of 50 rev/s. With the speed slightly altered, then U t = *0.6*5050 = m/s 7

28 Hub -Tip Radius Ratio For a simple engine of the type under consideration, there would be no merit in using a low hub-tip ratio; this would merely increase the mismatch between the compressor and turbine diameters, and also complicate both the mechanical and aerodynamic design of the first stage. On the other hand, using a high hub-tip ratio would unnecessarily increase the compressor diameter and weight. But it should be realized that the choice of 0.50 for hub-tip ratio is arbitrary, and merely provides a sensible starting point. Later considerations following detailed analysis could cause an adjustment, and a considerable amount of design optimization is called for. 8

29 Inlet Mach Number At this stage it is appropriate to check the Mach number relative to the rotor tip at inlet to the compressor. Assuming the axial velocity to be constant across the annulus, which will be the case where there are no inlet guide vanes, the relative velocity and V t U t C a V t = m/s a M t RT V a t.4*0.87 *000 * , 33.0m / s Thus, the relative Mach number at rotor tip is.65 and the first stage is transonic; this level of Mach number should not present any serious problem. 9

30 Annulus Dimensions at Exit Hub-tip ratio = 0.50 Tip radius = 0.6m Root radius = 0.3m Mean radius = 0.697m To estimate the annulus dimensions at exit from the compressor, it will be assumed that the mean radius is kept constant for all stages. Thecompressor delivery pressure, p 0 = 4.5*.0= 4.9 bar To estimate the compressor delivery temperature, it will be assumed that the polytropic efficiency of the compressor is Then T 0 n n p 0 T ( n ) * p0 n so that T 0 = 88.0 (4.5) = 45.5K 30

31 Annulus Dimensions at Exit ( contd.) Assuming that the air leaving the stator of the last stage has an axial velocity of 50 m/s and no swirl, the static temperature, pressure and density at exit can readily be calculated as follows: T K * 005. * γ γ T 443. p p0 49. T * ρ kg/m 0. 87* bar Please note: Suffix stands for inlet of the compressor and suffix stand for outlet of the compressor. Do not mix with the accompanying figure which is shown for a stage. 3

32 Annulus Dimensions at Exit ( contd.) The exit annulus area is thus given by A = 0/3.03*50 = m With r m = m, the blade height at exit, h, is given by h r r m * m 0.043m 3

33 Annulus Dimensions The radii at exit from the last stator are then r t = (0.043/) = m r r = (0.043/) = 0.49 m At this point we have established the rotational speed and the annulus dimensions at inlet and outlet, on the basis of a constant mean diameter. To summarise N = 50 rev/s r t = 0.6 m inlet U t = m/s r r = 0.3 m C a =50 m/s r t = m r m = m (constant) r r = 0.49 m outlet 33

34 Annulus Dimensions Casing Many stages Inlet 0.6 m Exit 0.697m 0.3 m Mean radius Hub Compressor Axis m 0.49m 34

35 Estimation of Number of Stages Assumed polytropic efficiency is The overall stagnation ti temperature t rise through h the compressor is = 64.5 K. The stage temperature rise T os can vary widely in different compressor designs, depending on the application and the importance or otherwise of low weight: values may vary from 0 to 30 K for subsonic stages and may be 45 K or high performance transonic stages. Rather choosing a value at random, T os can be estimated based on the mean blade speed. 35

36 Estimation of Number of Stages U =* * * 50 = 66.6 m/s We will adopt the simple design condition C a = C a = C a throughout the compressor, so the temperature rise is given by UC tan t U C C T a w w os cp cp tan With a purely axial velocity at entry to the first stage, in the absence of IGVs, U 66.6 tan C a C V a m / s cos cos U

37 Estimation of Number of Stages In order to estimate the maximum possible deflection in the rotor, we will apply the de Haller criterion V / V > 0.7. On this basis the minimum allowable value of V = 305.9*0.7 = 0 m/s, and the corresponding rotor blade outlet angle is given by C 50 cos a, V

38 Estimation of Number of Stages Using this deflection and neglecting the work-done factor for this crude estimate 66.6 *50(tan tan 47.0) T os 8K 3.005*0 A temperature rise of 8 K per stage implies 64.5/8 = 5.9 stages. It is likely that the compressor will require six or seven stages; and in view of fthe influence of fthe work-done kd factor, seven is more likely. l An attempt will, therefore, be made to design a seven-stage compressor. 38

39 Stage Temperature Rise With seven stages and an overall temperature rise of 64.5 K the average temperature rise is 3.5 K per stage. It is normal to design for a somewhat lower temperature rise in the first and last stages. A good starting point would be to assume T 0 0K for the first and last stages, leaving a requirement for remaining stages as T0 5K Having determined the rotational speed and annulus dimensions, and estimated the number of stages required, the next step is to evaluate the air angles for each stage at the mean radius. It will then be possible to check that the estimated number of stages is likely to result in an acceptable design. 39

40 Air Angles at Mean Radius Stage-by-stage design: From the velocity triangles, we get C tan w C a Cw For the first stage = 0 because there are no inlet guide vanes. C w C The stator outlet angle for each stage, 3, will be the inlet angle for the following rotor. Calculations of stage temperature rise are based on rotor considerations only, but care must be taken to ensure that the diffusion in the stator is kept to a reasonable level. The work-done factors will vary through the compressor and reasonable values for the seven stages would be 0.98 for the first stage, 0.93 for the second, 0.88 for the third and 0.83 for the remaining four stages. 40 w

41 Design of Stages & First stage nd and 3 rd stage with symmetric blading 7-stage axial compressor 4

42 Design of Stage Recalling the equation for the stage temperature rise in terms of change in whirl velocity ΔC w = C w -C w, we have 3 cpt0.005*0 * 0 C w 76.9 m s U 0.98* 66.6 / β Ca V Since C w = 0, C w = 76.9 m/s and hence tan β = U/C a = 66.6/50 =.7773 U β =

43 Design of Stage ( contd.) tan U C C a w , C 76.9 tan w 0.53, 7. 4 C 50 a The velocity diagram for the first stage therefore appears as 0 43

44 Design of Stage ( contd.) The deflection in the rotor blades is β - β = , which is modest. The diffusion can readily be checked using the de Haller number as follows: V V C / cos / cos cos cos a C a This value of de Haller number indicates a relatively light aerodynamic loading, i.e, a low rate of diffusion. It is not necessary to calculate the diffusion factor at this stage, because the de Haller number gives an adequate preliminary check. After the pitch chord ratio (s/c) is determined from cascade data, the diffusion factor can be calculated readily from the known velocities. 44

45 Design of Stage ( contd.) At this point, it is convenient to calculate the pressure ratio of the stage (P 03 /P 0 ), the suffix outside the parentheses denoting the number of the stage; and then the pressure and temperature at exit which will also be the values at inlet tto the second stage. The isentropic efficiency of the stage is approximately equal to the polytropic efficiency of the compressor, which has been assumed to be 0.90, so we have: P P P * * bar 03 T K 03 45

46 Design of Stage ( contd.) We have finally to choose a value for the air angle at outlet from the stator row, 3,which will also be the direction of flow,, into the second stage. Here it is useful to consider the degree of reaction. For this first stage, with the prescribed axial inlet velocity, C 3 will not equal C (unless 3 is made zero) whereas our equations for R were derived on the assumption of this equality of inlet and outlet velocities. Nevertheless, C 3 will not differ markedly from C, and we can arrive at an approximate value of R by using equation Cw Cw 76.9 R U * 66.6 The degree of reaction is high, h but this is necessary with low hub-tip ratios to avoid a negative value at the root radius. We shall hope to be able to use 50 percent reaction stages from the third or fourth stage onwards, and an appropriate value of frr for the second stage may be about

47 Design of Stage For the second stage, T os = 5 K and = 0.93 and we can determine and using equations * 66. * 50 5 ( tan β 3 tan 005. * 0 tan tan β ) and from Ca R tan β tan UU β (tan * 66.6 tan tan.4883 tan ) 0 47

48 Design of Stage ( contd.) Solving these simultaneous equations we get 57.7 and Finally, using equations U C a 4.9 U tan α tan β tan α tan β C a.06 and 4.05 The whirl velocities at inlet and outlet are readily found from the velocity diagram, Cw Ca tan α 50 tan m/s Cw Ca tan α 50 tan m/s 48

49 Design of Stage ( contd.) The required change in whirl velocity is 0.3 m/s, compared with 76.9 m/s for the first stage; this is due to the higher stage temperature rise and the lower work-done factor. The fluid deflection in the rotor blades has increased to It appears that 3 for the first stage should be.06. This design gives a de Haller number for the second-stage rotor blades of cos57.70 /cos4.9 = 0.7, which is satisfactory. With the stator outlet angle for the first-stage stator now known, the de Hll Haller number for the first-stage t stator t would ldbe C cos C cos 3 cos 7.5 cos implying a small amount of diffusion. This is a consequence of the high degree of reaction in the first stage. 49

50 Design of Stage ( contd.) The velocity diagram for the second stage appears as shown in the Figure and the outlet pressure and temperature become P P P * * bar 03. T K 03 50

51 Design of Stage ( contd.) At this point we do not know 3 for the second stage, but it will be determined from the fact that it is equal to for the third stage. It is useful to point out that the degree of reaction is directly related to the shape of the velocity diagram. It is known that for 50 percent reaction the velocity diagram is symmetrical. Writing C wm = (C w +C w )/, degree of reaction can be rewritten in the form R = -(C wm /U). When C wm /U is small, and the corresponding reaction is high, the velocity diagram is highly skewed; the high degree of reaction in the first stage is a direct consequence of the decision to dispense with inlet guide vanes and use a purely axial inlet velocity. The degree of reaction is reduced in the second stage, and we would eventually like to achieve 50 percent reaction in the later stages where the hub-tip ratios are higher. 5

52 Design of Stage 3 Using a stage temperature rise of 5K and a work-done factor of 0.88, an attempt will be made to use a 50 percent reaction design for the third stage. Proceeding as before 3 ΔTosC p 5* 005. * 0 tan β tan β λuc 0. 88* 66. 6* 50 U 0. 5* * tan β tan β R C 50 Yielding 5.4 and 8.0. The corresponding value of the de Haller number is given by cos5.4/cos8.0 = a a This is rather low, but could be deemed satisfactory for a preliminary design. It is instructive, however, to investigate the possibilities available to the designer for reducing the diffusion. 5

53 Design of Stage 3 ( contd.) One possibility is to consider changing the degree of reaction, but it is found that the de Haller number is not strongly influenced by the degree of reaction chosen; as R had a value of = 0.70 for the second stage it might appear that a suitable value for the third stage might be between 0.70 and Repeating the above calculations for a range of R, however, shows that R = 0.55 results in a further decrease of the de Haller number to 0.706; For a specified axial ilvelocity, the required ddiffusion i increases with ihreaction. Ade Hll Haller number of 0.75 can be achieved for R = 0.40, but it is undesirable to use such a low degree of reaction. A more useful approach might be to accept a slightly lower temperature rise in the stage, and reducing ΔT os from 5 K to 4 K while keeping R = 0.50 gives tan β -tan β = Yielding β = 50.9, β = 8.63 and a de Haller number of 0.78, which is satisfactory for this preliminary design. Other methods of reducing the aerodynamic loading include increases in blade speed or axial velocity, which could readily be accommodated. 53

54 Design of Stage 3 ( contd.) With a stage temperature rise of 4 K, the performance of the third stage is then given by p p * p * T K bar From the symmetry of the velocity diagram α = β = 8.63 and α = β = The whirl velocities are given by C w =50 tan 8.63 = 8.9 m/s C w =50 tan 50.6 =84.7 m/s 54

55 Design of Stage 4, 5 and 6 A work-done factor of 0.83 is appropriate for all stages from the fourth onwards, and 50 percent reaction can be used. The design can be simplified by using the same mean diameter velocity diagrams for stages 4 to 6, although each blade will have a different length due to the continuous increase in density. The seventh and final stage can then be designed to give the required overall pressure ratio. It is not necessary to repeat all the calculations for stages 4-6, but it should be noted that the reduction in work-done factor to 0.83, combined with the desired stage temperature rise of 5K, results in an unacceptably low de Haller number of Reducing the stage temperature rise to 4 K increases the de Haller number to 0.705, which is considered to be just acceptable for the preliminary design. 55

56 Design of Stage 4, 5 and 6 ( contd.) Proceeding as bf before, tan 3 tan tan tan 4*.005* * 66.6* * * Yielding β = (= α )andβ = 7.7 (= α ). The performance of the three stages can be summarized below: stage p 0 (bar) T 0 (K) (p 03 /p 0 ) p 03 (bar) T 03 (K) p 03 -p 0 (bar)

57 Design of Stage 4, 5 and 6 ( contd.) It should be noted that although each stage is designed for the same temperature rise, the pressure ratio decreases with stage number; this is a direct consequence of the increasing inlet temperature as flow progresses through the compressor. The pressure rise, however, increases steadily. 57

58 Design of Stage 7 At entry to the final stage the pressure and temperature are bar and 49 K. The required compressor delivery pressure is 4.5*.0 = 4.9 bar. The pressure ratio of the seventh stage is thus given by p p The temperature rise required to give this pressure ratio can be determined from T os giving.8k T os The corresponding air angles, assuming 50 percent reaction, are then β = (= α ), β = 8.5 (= α ) with a satisfactory de Haller number of

59 Design of Stage 7 With a 50 percent reaction design used for the final stage, the fluid will leave the last stator with an angle α 3 = α = 8.5, whereas ideally the flow should be axial at entry to the combustion chamber. The flow can be straightened by incorporating vanes after the final compressor stage and these can form part of the necessary diffuser at entry to the combustion chamber. 59

60 Comments All the preliminary calculations have been carried out on the basis of a constant mean diameter. Another problem now arises: a sketch, approximately to scale, of the compressor and turbine annuli shows that the combustor will have an awkward shape, the required changes in flow direction causing additional pressure losses. 60

61 Comments A more satisfactory solution might be to design the compressor for a constant outer diameter. The use of a constant outer diameter results in the mean blade speed increasing with stage number, and this in turn implies that for a given temperature rise, ΔC w is reduced. The fluid deflection is correspondingly reduced with a beneficial increase in de Haller number. Alternatively, because of the higher blade speed, a higher temperature rise could be achieved in the later stages ; this might permit the required pressure ratio to be obtained in six stages rather than seven. Note that the simple equations derived on the basis of U = constant are then not valid, and it would be necessary to use the appropriate values of U and U ; the stage temperature rise would then be given by λ(u C w - U C w )/C p. Compressors which use constant inner diameter, constant mean diameter or constant outer diameter will all be found in service. 6

62 Comments The use of a constant inner diameter is often found in industrial i units, permitting the use of rotor discs of the same diameter, which lowers the cost. Constant outer diameter compressors are used where the minimum number of stages is required, and these are commonly found in aircraft engines. The compressor annulus of the Olympus 593 engine used in Concorde employs a combination of these approaches; the LP compressor annulus has a virtually constant inner diameter, while the HP compressor has a constant outer diameter. The accessories are packed around the HP compressor annulus and the engine when fully equipped is almost cylindrical in shape, with the compressor inlet and turbine exit diameters almost equal. In this application, frontal area is of critical importance because of the high h supersonic speed. 6

63 Olympus 593 Mk 60 Engine Concorde aircraft Compressor: Axial 7 high pr. stages; 7 low pr. stages Turbine: low pr. stage; high pr. stage Weight: 380 kg Length: 7.m Diameter:.m Thrust: 70kN 63

64 Session Summary The design procedure for multistage compressor is explained. The calculation of annulus area and importance of hub and tip flares are explained. Detailed procedures for estimation of number of stages and blade velocity triangles are presented. 64