# Lecture 10 – Axial compressors 2

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Lecture 10 – Axial compressors 2
Blade design Preliminary design of a seven stage compressor choice of rotational speed and annulus dimension estimation of the number of stages and stage by stage design variation of air angles from root to tip Compressibility effects in axial compressors Problem 5.1

Blade design For a test camber angle θ, chord c and pitch s will be fixed and the stagger angle ζ is changed by the turn table. Pressures and velocities are measured downstream and upstream by traversing instruments

Blade design Measurements are recorded as pressure losses and deflection Incidence is varied by turning the turn-table Mean deflection and loss are computed

Blade design Selecting more than 80% of stall deflection means risk of stalling at part load Select Stall reached when loss is twice of minimum loss ε* is dependent mainly on s/c and α2 for a given cascade. Thus, data can be reduced into one diagram

Blade design Air angles decided from design Pitch/Chord from Fig. 5-26
Blade heigth from area requirement. Assume h/c (methods for selecting h/c are discussed in section 5.9 [which is less relevant to course]) s, c and h/c will then be known The blade outlet angle is determined from air outlet angle and empirical rule for deviation. Assume camber-line shape (for instance circular arc)

Rotational speed and annulus dimensions
Compressor for low cost, turbojet Design point specification rc = 4.15, m = 20 kg/s, T3 = 1100 K No inlet guide vanes (α1=0.0) Annulus dimension? Assume values on: blade tip speed: range m/s. Values close to 350 m/s will limit stress problems axial velocity: range m/s. Try 150 m/s to reduce difficulties associated with shock losses root-tip ratio: range You should know that these ranges represent typical values that you can assume on exam. No inlet guide vanes will keep noise and weight down.

Annulus dimension Continuity gives the compressor inlet area:
Some matlab code: T1 = ((C1^2)/(2*1005)) astar = sqrt(gam*R*T1) M1=C1/astar= sqrt(gam)*M1*(1+((gam-1)/2)*M1^2)^(-((gam+1)/(2*(gam-1))))= A = (m*sqrt(R*T0))/(P0*X) = Root tip ratio can be further increased without producing a too high Ut.

Annulus dimension A single stage turbine can designed to drive this compressor if a rotational speed of 250 rev/s is chosen (Chapter 7). N = 250 rev/s gives: Considering the relatively low Ut centrifugal stresses in the root will not be critical and the choice of a root-tip ratio of 0.5 will be considered a good starting point for the design. Recall the approximative formula:

Annulus dimension A check on the tip Mach number gives:
This is a suitable level of Mach number. Relative Mach numbers over about would require supercritical (controlled diffusion blading). Too low values would result in a low stage temperature rise. The compressor exit temperature is estimated assuming a polytropic efficiency, η c,polytropic, of 90%, which gives the exit area:

Annulus dimension The blade height at exit can now be calculated

Estimation of the number of stages
Assumed polytropic efficiency gave: Reasonable stage temperature rises are K. Up to 45 is possible for high-performance transonic stages. Let us estimate what we could get in our case: We have derived the stage temperature rise as: No IGV:

Estimation of the number of stages
An overestimation of the temperature rise is obtained for a de Haller number equal to the minimum allowable limit = 0.72: Which gives a rotor blade outlet angle: Setting the work done factor λ = 1.0 yields: We could not expect to achieve the design target unless we use: Since 6 is close to achievable aerodynamic limits, seven is a reasonable assumption!

Design goal Stage 1 and stage 7 are somewhat less loaded to allow for:
Stage 1: highest Mach numbers occur in first stage rotor tip => difficult aerodynamic design. Inlet distortion of flow may be substantial. Less aerodynamic loading may alleviate these difficulties Stage 7: it is desirable to have an axial flow exiting the stator of the last stage => a higher deflection is necessary in this stage which may be easier to design for if a reduction in goal temperature rise is allowed for This gives the following design criteria (assuming a typical work done distribution)

Stage-by-stage design - Stage 1
The change in whirl velocity for the first stage is: A check on the de Haller number gives: which is satisfactory. The diffusion factors will be checked at a later stage.

Stage-by-stage design – Stage 1
For small pressure ratios isentropic and polytropic efficiencies are close to equal. Approximate the isentropic stage efficiency to be This gives the stage pressure rise as: We have finally to choose α 3. Since α 1 in stage 2 will equal α 3 in stage 1, this will be done as part of the design process for the second stage.

Stage-by-stage design – Stage 2
Since we do not know α1 for the second stage we need further design requirements. We will use the degree of reaction. The degree of reaction for the first stage is (α3 in first stage is degrees. cos(11.06) =0.981) Since the root-tip ratio of the first stage is the lowest, the greatest difficulties with low degrees of reaction will be experienced in the first stage rotor. Thus, a good margin to 0.50 has to be accepted. Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design
Due to the increase in root-tip ratio for the second stage we hope to be able to use a Λ of 0.70: Solving the two simultaneous equations for β 1 and β 2 gives: α 1 and α 2 are then obtained from (obtaining α 1 means that the design of the first stage is complete):

Stage-by-stage design - Stage 1
The design of the first stage is now complete: The de Haller number in the stator is:

Stage-by-stage design – Stage 2
The stage pressure rise of stage 2 becomes: We have finally to choose α 3. Since α 1 in stage 3 will equal α 3 in stage 2, this will be done as part of the design process for the third stage. Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design – Stage 3
Due to the further decrease in root-tip ratio to the third stage we hope to be able to use a Λ of 0.50: Solving the two simultaneous equations for β 1 and β 2 gives β 1=51.24 and β 2= This gives a to low de Haller number which can be dealt with by reducing the temperature increase over the stage to 24K. Repeating the calculation gives: Cos11.06=0.981=Ca/C3=C1/C3 which is produces an ok de Haller number. α 1 and α 2 are obtained from symmetry which is obtained when Λ = 0.50.

Stage-by-stage design - Stage 2
The design of the second stage is now complete: The de Haller number in the stator is:

Stage-by-stage design – Stage 3
The stage pressure rise of stage 3 becomes: We have finally to choose α 3. Since α 1 in stage 4 will equal α 3 in stage 3, this will be done as part of the design process for the fourth. Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design – Stage 4,5 and 6
We maintain Λ of 0.50 for stage 4, 5 and 6: Since λ is 0.83 for the remaining stages, the stages 4, 5 and 6 are all designed with the same angles. Again the stage temperature rise is reduced to 24 K to maintain the de Haller number at an high enough number. Solving the two equations give: Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design - Stage 3
The design of the third stage is now complete: The de Haller number in the stator is:

Stage-by-stage design – stage 4,5,6
The stage pressure rise and exit temperature and pressure of stages 4,5,6 become: Stage 4, 5 and 6 are repeating stages, except for the stator outlet angle of stage 6 which is governed by the design of stage 7. Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design - Stage 4,5,6
The design of stages 4,5 and 6 are now complete: The de Haller number for the stators are:

Stage-by-stage design – stage 7
We maintain Λ of 0.50 for stage 7: The pressure ratio of the seventh stage is set by the overall requirement of an rc = The stage inlet pressure is 3.56 bar, which gives the required pressure ratio and temperature increase according to: Cos11.06=0.981=Ca/C3=C1/C3

Stage-by-stage design - Stage 7
The design of stage 7 is now complete: Exit guide vanes can be incorporated to straighten flow before it enters the burner

Annulus shape The main types of annulus designs exists
Constant mean diameter Constant outer diameter Constant inner diameter Mean blade speed increases with stage number Less deflection required => de Haller number will be greater U1 and U2 will not be the same!!! It will be necessary to use:

Variations from root to tip
Calculate angles at root, mid and tip using the free vortex design principle, i.e.: Cwr = constant The requirement is satisfied at inlet since Cw= 0 (no IGV) Blade speed at root, mean and tip are

Variations from root to tip
Cw velocities are computed using the whirl velocity at the mid, Cw2,m = 76.9 m/s together with the free vortex condition. : Cwr = constant The root and tip radii will have changed due to the increase in density of the gas. Compute the exit area according to: Which gives the blade height, root and tip radii:

Variations from root to tip
The rotor exit tip and root radii are assumed to be the average of the stage exit and inlet radii: The free vortex condition gives:

Variations from root to tip
The stator inlet angles α 2,r , α 2,m , α 2,t and rotor exit angles β 2,r , β 2,m , β 2,t are obtained from: Note the necessary radial blade twist for air and blade angles to agreee The reaction at the root is The high value at the mean radius ensured a high enough value at the root Where do the highest stator Mach numbers occur ??

Compressibility effects
Fan tip Mach numbers of more than 1.5 are today frequent in high bypass ratio turbofans At some free-stream Mach number a local Mach number exceeding 1.0 will occur over the blade This Mach number is called the critical Mach number Mcr. A turbulent boundary layer will separate if the pressure rise across the shock exceeds that for a normal shock with an upstream Mach number of 1.3 (Schlichting 1979). Keep this in mind!!!

Effect of Mach number on losses
As the Mach number passes the critical Mach number: minimum loss increases and the range of incidence for which losses are acceptable is drastically reduced Simplified shock model: The turning determines the Mach number at station B (Prandtl-Meyer relations - Appendix A.8 - you may skip that section). The more turning the higher the Mach number Shock loss = Normal shock loss taken at averageMach number at A and B. Why less loaded first stage in example Less loaded first stage => less turning => reduced shock losses

Supercritical design - Mrel > 1.3
Recall lecture 5: How would you design a blade operating at Mach number 1.6 remembering that: A turbulent boundary layer will separate if the pressure rise across the shock exceeds that for a normal shock with an upstream Mach number of 1.3 (Schlichting 1979)