Transcript No Slide Title

Chalmers University of Technology
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
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Blade design
• We want:
– blade must achieve required turning at
maximum efficiency over a range of
rotational speeds
• Correlated experiments are very
valuable to estimate performance
– tests on single blades (effect of
adjacent blades must be estimated)
– tests on rows of blades - so called
cascades
• Linear cascades (rectilinear cascade).
Mechanically simpler than annular to
build. Flow patterns are simpler to
interpret. More frequent.
• Annular. Many root tip ratios would
be required. Does not satisfactorily
reproduce flow in actual compressor
Annular cascade
Linear cascade
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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
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Blade design
• Measurements are recorded
as pressure losses and
deflection
• Incidence is varied by
turning the turn-table
• Mean deflection and loss
are computed
p01  p02
1
V12
2
deflection   1   2
incidence 1  1
loss 
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Blade design
• Selecting more than 80%
of stall deflection means
risk of stalling at part load
*
• Select   0.80  S
• 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
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• 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)
Blade design
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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 350-450 m/s.
Values close to 350 m/s will limit
stress problems
– axial velocity: range 150-200 m/s.
Try 150 m/s to reduce difficulties
associated with shock losses
– root-tip ratio: range 0.4-0.6
• You should know that these ranges
represent typical values that you can
assume on exam.
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Annulus dimension
Continuity gives the compressor inlet area:
  rr2 
m  1 ACa1  1 rt 1   2 Ca1 ,
rt 




2
C1  No IGV   Ca1  150 m/s
A
M1 
C1
T1

2 

C
R 288 1 
2c p 

A1  0.120m 2
 0.45 
m RT01
  1 2 
  M 1 1 
M1 
P01 A1
2



( 1)
2 ( 1)
 0.5000 
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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:
rt 
A
  rr2 
 1   2 
  rt 
 Assum eroot  tip ratio  0.5  0.2261 
rm  0.1696
Ut  2  rt  355.2 m / s
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:
2



b
b 2
rr 
 ct max 
ardr  ...  U t 1    

r
ar
2
  rt  
rr
 root  tip ratio
rt
2
t
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Annulus dimension
• A check on the tip Mach number gives: M rel ,tip 
U t2  Ca21
RT
 1.165
This is a suitable level of Mach number. Relative Mach numbers over
about 1.2-1.3 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:
P 
T02  T01  02 
 P01 
M exit 
1  1
c , 
 452.5,
Cexit
Texit



2

C 
R 452.5  exit 
2c p 

C1  Ca  Cexit  150 m / s
 0.356 
mexit RT0,exit
P0,exit A
  1 2 
  M exit 1 
M exit 
2



( 1)
2 ( 1)
 0.391  A exit  0.044 m 2
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Annulus dimension
• The blade height at exit can now be calculated
Aannulus  Atot  Aroot  rt 2  rr2   rt  rr rt  rr  

rt  rr 
2 rt  rr 
 2hrm

h
2

rm
rt  0.1902
rr  0.1489
A constantmean diamet er
design is assumed
 h  0.0413m
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Estimation of the number of stages
• Assumed polytropic efficiency gave: T0  452.5  288  164.5 K
• Reasonable stage temperature rises are 10-30 K. Up to 45 is
possible for high-performance transonic stages. Let us estimate
what we could get in our case:
U  2  rm  N  266.6 m/s
• We have derived the stage temperature rise as:
U C w2  C w1 

T0  UCa tan 1  tan  2  
cp
• No IGV:
cp
U
Ca
1  arct an( )  60.64
V1 
Ca
 305.9 m/s
cos 1
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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:
V2  V1  0.72  220 m/s
• Which gives a rotor blade outlet angle:
Ca
 2  arccos( )  47.01
V2
• Setting the work done factor λ = 1.0 yields:

T0  UCa tan 1  tan  2   T0  28 K
cp
• We could not expect to achieve the design target unless we use:
Nstages  164.5/28  5.9  6 stages
• Since 6 is close to achievable aerodynamic limits, seven is a
reasonable assumption!
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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 1
ΔT=20
λ=0.98
Stage 2
ΔT=25
λ=0.93
Stage 3
ΔT=25
λ=0.88
Stage 4
ΔT=25
λ=0.83
Stage 5
ΔT=25
λ=0.83
Stage 6
ΔT=25
λ=0.83
Stage 7
ΔT=20
λ=0.83
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Stage-by-stage design - Stage 1
• The change in whirl velocity for the first stage is:
c p T0
C w 
 76.9 m/s  No IGV  Cw 2
U
U  Cw 2
U
1  arctan( )  60.64,  2  arctan(
)  51.67
Ca
Ca
Cw 2
 2  arctan( )  27.14
Ca
• A check on the de Haller number gives:
which is satisfactory. The diffusion
factors will be checked at a later stage.
Ca
V2 cos  2 cos 1


 0.79
C
V1
cos  2
a
cos 1
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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 0.90. This gives the stage pressure rise as:

 P03 
  Stage  TStage   1


  1.236
 1 
T01
 P01  Stage 1 

P03 1  1.249 bar
T03 1  308 K
• 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.
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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 11.06
degrees. cos(11.06) =0.981)
Cw 2  Cw1
  C3  C1   1 
 0.856
2U
• 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.
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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:
T0 

UCa
cp
t an 1  t an  2 
Ca
t an 1  t an  2 
2U
• Solving the two simultaneous equations for
β 1 and β 2 gives:   57.70
1
 2  42.19
• α 1 and α 2 are then obtained from (obtaining α 1 means that the
design of the first stage is complete):
U  Ca tan1  Ca tan 1
1  11.06
U  Ca tan 2  Ca tan  2

 2  41.05
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Stage-by-stage design - Stage 1
• The design of the first stage is now complete:
β1
β2
α1
α2
α3
60.64
51.67
00.00
27.14
11.06
• The de Haller
number in the
stator is:
Ca
C3 cos 3

 0.907
Ca
C2
cos 2
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Stage-by-stage design – Stage 2
• The stage pressure rise of stage 2 becomes:

 P03    Stage  TStage   1

  1 
  1.280
T01
 P01  2 

P03 2  1.599 bar
T03 2  333 K
• 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.
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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:
UCa
t an 1  t an  2 
T0 
cp

Ca
t an 1  t an  2 
2U
• Solving the two simultaneous equations for β 1 and β 2 gives
β 1=51.24 and β 2=28.00. 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:
1  50.92
 2  28.63
• which is produces an ok de Haller number. α 1 and α 2 are
obtained from symmetry which is obtained when Λ = 0.50.
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Stage-by-stage design - Stage 2
• The design of the second stage is now complete:
β1
β2
α1
α2
α3
57.70
42.19
11.06
41.05
28.63
• The de Haller
number in the
stator is:
Ca
C3 cos 3

 0.859
Ca
C2
cos 2
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Stage-by-stage design – Stage 3
• The stage pressure rise of stage 3 becomes:

 P03    Stage  TStage   1

  1 
  1.246
T01
 P01 3 

P03 3  1.992 bar
T03 3  357 K
• 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.
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Stage-by-stage design – Stage 4,5 and 6
• We maintain Λ of 0.50 for stage 4, 5 and 6:
T0 

UCa
cp
t an 1  t an  2 
Ca
t an 1  t an  2 
2U
• 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:
1   2  51.38
 2  1  27.71
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Stage-by-stage design - Stage 3
• The design of the third stage is now complete:
β1
β2
α1
α2
α3
50.92
28.63
28.63
50.92
27.71
• The de Haller
number in the
stator is:
Ca
C3 cos 3

 0.712
Ca
C2
cos 2
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Stage-by-stage design – stage 4,5,6
• The stage pressure rise and exit temperature and pressure of
stages 4,5,6 become:
Stage
P03/P01
P03
T03
4
1.228
2.447
381
5
1.213
2.968
405
6
1.199
3.560
429
• 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.
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Stage-by-stage design - Stage 4,5,6
• The design of stages 4,5 and 6 are now complete:
β1
β2
α1
α2
α3
51.38
27.71
27.71
51.38
27.71 (28.52
for stage 6)
• The de Haller
number for the
stators are:
Ca
C3 cos 3

 0.705
Ca
C2
cos 2
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Stage-by-stage design – stage 7
• We maintain Λ of 0.50 for stage 7:
T0 

UCa
cp
t an 1  t an  2 
Ca
t an 1  t an  2 
2U
• The pressure ratio of the seventh stage is set by the overall
requirement of an rc = 4.15. The stage inlet pressure is 3.56 bar,
which gives the required pressure ratio and temperature increase
according to:
 P03 
4.192

 
 1.177
 P01  7 3.560
  Stage  TStage 
1 

T01


T0  22.8 K

 1

 1.177

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Stage-by-stage design - Stage 7
• The design of stage 7 is now complete:
β1
β2
α1
α2
α3
50.98
28.52
28.52
50.98
28.52
• Exit guide
vanes can be
incorporated to
straighten flow
before it enters
the burner
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Annulus shape
• The main types of annulus
designs exists
– Constant mean diameter
– Constant outer diameter
– Constant inner diameter
• Constant outer 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:
T0  
U 2Cw2 U 1Cw1
cp
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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
U r  2  rr  N  177.7 m/s
U m  2  rm  N  266.6 m/s
U t  2  rt  N  355.3 m/s

Ur
1r  arctan( )  49.83
Ca
1m
Um
 arctan( )  60.64
Ca
U
Ca
1t  arctan( t )  67.11
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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:
m RT0
Ca
  1 2 
M3 
 0.434 
  M 1 
M 
T3 ,exit
P
A
2


0

2 

C
exit

R T03 
2c p 


( 1)
2 ( 1)
 0.4599  A  0.1035m 2
• Which gives the blade height, root and tip radii:
h  0.0971m
rt  0.2181m
rr  0.1210m
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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:
rr  0.1171m
rt  0.2222m

U r  2  rr  N  183.9 m/s
U t  2  rt  N  349.0 m/s
• The free vortex condition gives:
Cw 2,r  Cw 2, m
rm
 111.4 m/s
rr
C w 2 ,t  C w 2 , m
rm
 58.7 m/s
rt
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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:
C
 2  arctan( w 2 )
Ca
U  Cw 2
 2  arctan(
)
Ca
 2 ,r  36.60

 2 ,m  27.14
 2 ,t  21.37
 2 , r  25.80
 2 , m  51.67
 2 ,t  62.69
• Note the necessary radial blade twist
for air and blade angles to agreee
• The reaction at the root is 0.697.
The high value at the mean radius
ensured a high enough value at the
root
• Where do the highest stator
Mach numbers occur ??
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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!!!
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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
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Supercritical design - Mrel > 1.3
• Recall lecture 5:
dV dA
( M  1)

V
A
2
• 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)
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Concave portion =>
supercritical diffusion
Concave section after
entrance region
Supercritical blading
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Supercritical blading
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Learning goals
• Know how to determine a multistage
compressor design for a certain
specification.
– This includes making assumptions design
parameters
• Have an understanding of how
compressor design must be adjusted for
high Mach number effects.