Design of a Multi-Stage Compressor

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 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
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
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 288.0 (4.15)0.3175 452.5 0K,
• DT0 compressor 452.5 - 288.0 164.5 0K and
  other conditions

10
Compute Compressor Exit Conditions

• Exit area, hub and tip radius

11
Step 2 - Estimate the Number of Stages

• From Eulers Turbine Equation
• With no inlet guide vane (Cu10, 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 its own rules. Liebleins
  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 Hallers criterion is simpler than
Liebleins rule since it does not involve
relative circumferential velocities or solidity.
To first order, this is same as a 0ltDfactorlt0.4.
Could use Liebleins 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) - 100K
  (transonic)
• Therefore we choose to use six or seven stages.
  To be conservative (account for losses, ie.
  halt1),
• Choose 7 stages
• Recalculate the DT0stage 164.5/7 23.5
• So 1st stage temperature ratio is T0 ratio 288
  23.5/288 1.0816
• The stage pressure ratio is then P0 ratio (T0
  ratio ) hpg / (g - 1) 1.2803

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
Constant Cx
C1
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 Eulers 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.
Cx1150
a10
W C - U
b160.64
U- WU1 266.6
W1305.9
Cx3150
STATOR
a30
b360.64
C2174.21
a230.57
U266.6
CU288.6
W3305.9
Cx2150
ROTOR
U266.6
b249.89
WU2178.0
W2232.77
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, lets 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) 1.248 bar and T03
  28820308 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 Eulers 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
Cx1150
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.
a10
W C - U
b160.64
U- WU1 266.6
C3153.56
W1305.9
STATOR
a312.36
Cx3150
CU332.87
b357.31
C2167.87
U266.6
a226.68
CU275.38
W3277.73
Cx2150
U266.6
ROTOR
b251.89
WU3233.77
WU2191.22
W2242.03
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 4 - 6.

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 1.248 (1.279) 1.596 bar and T03
  30825333 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, Eulers 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
C3153.56
W C - U
a312.36
Cx3150
CU332.87
b357.31
C3172.99
U266.6
a329.88
Cx3150
CU386.18
W3277.73
STATOR
b350.26
U266.6
WU3233.77
C2196.59
W3234.63
a240.27
CU2127.07
Cx2150
ROTOR
WU3180.42
U266.6
b242.92
WU2139.53
W2204.86
Notice that the velocity triangles are not
symmetric for the second stage due to
70reaction 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
C3172.99
W C - U
a329.88
C3172.99
Cx3150
CU386.18
b350.26
a329.88
Cx3150
CU386.18
STATOR
U266.6
b350.26
U266.6
W3234.63
W3234.63
WU3180.42
C2234.63
CU2180.42
WU3180.42
a250.26
Cx2150
ROTOR
U266.6
b229.88
WU286.18
W2172.99
Notice that the velocity triangles are
symmetric due to the 50reaction 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
  433. The requirements for our 7-stage
  compressor design we have
• P0 exit 4.15 1.01 4.19 bar
• T0 exit 288.0 (4.15)0.3175 452.5 0K
• 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 3 - 6
  and that the exit absolute air angle leaving the
  compressor is 32.770. 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 0.1697,
• 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 rotor stator rotor stator rotor stato
r rotor stator rotor stator 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
  .2262 m and
• Uhub 177.7 , Umean 266.6, Utip 355.3
  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,
  Carters 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
If Cr0 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
  Cr0)

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 CXconstant 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 doesnt
  have to be !)

56
1st Stage Blade Spanwise Variations (Free Vortex
Design)
W C - U
Cx1150
a10
W1h232.5
b1h49.83
Uh- WU1h 177.7
C2h184.02
Um- WU1m 266.6
W1m305.9
b1m60.64
C2m167.87
Ut- WU1t 355.3
W1t385.7
b1t67.11
C2t160.93
a2h35.4
a2m26.68
CU2h106.6
a2t21.24
CU2m75.38
CU2t58.3
Uh177.7
Cx2150
ROTOR
Um266.6
WU2h71.1
b2h25.36
W2m166.0
Ut355.3
b2m51.89
W2m242.03
WU2m191.22
b2t63.2
W2m332.73
WU2t297.0
57
Spanwise Variation in Reaction with Free Vortex
Design

• Remember that Reaction is given by
• If R50 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 0.918 (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 0.918 (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
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/ctip max3 for structure, 10 for hub,
  linear variation between, t/cm6.5

62

• Pick cchord for turbulent flow, Rec gt 300,000 at
  all hs.
• 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 Carters rule
• Define stagger angle from
• Set number of blades