lecture 10 – axial compressors 2

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


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


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

11

21

21

0201

21

incidence

deflection

V

pploss


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

S 80.0*


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 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.


Annulus dimensionContinuity gives the compressor inlet area:

21

)1(2

)1(

211

101

01

21

11

1112

22

111

m 0.120

5000.02

11 0.45

2288

M

m/s 150 ,1

1

A

MMAP

RTm

cC

R

C

CIGVNoCCr

rrACm

T

p

aa

A

t

rta


Annulus dimensionA 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:

1696.0r

/ 2.3552 U 2261.05.0

1

m

t

2

2

smrratiotiprootAssume

rr

Ar t

t

r

t

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:

ratiotiprootr

r

r

rUardr

a

t

r

t

rt

bt

rr

bct

12

...2

22

max


Annulus dimension• A check on the tip Mach number gives: 165.1

21

2

,

RT

CUM at

tiprel 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:

2exit

)1(2

)1(

2

,0

,0

2

1

11

01

020102

044.0A 391.02

11 0.356

25.452

M

/ 150 ,5.452,

mMMAP

RTm

c

CR

C

smCCCP

PTT

exitexitexit

exitexit

T

p

exit

exitexit

exita

exit

c


Annulus dimension• The blade height at exit can now be calculated

assumed isdesign

diameter mean constant A

1489.0

1902.0

0413.0h 22

2

22

r

t

m

r

rt

h

rt

rtrtrtroottotannulus

r

r

mhrrr

rr

rrrrrrAAA

m


Estimation of the number of stages• Assumed polytropic efficiency gave: K 5.1642885.4520 T• 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:

m/s 6.2662 NrU m

p

wwa

p c

CCUUC

cT 12

210 tantan

• We have derived the stage temperature rise as:

• No IGV:

m/s 9.305cos

64.60)arctan(

11

1

a

a

CV

C

U


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:

m/s 22072.012 VV

01.47)arccos(2

2 V

Ca• Which gives a rotor blade outlet angle:

• Setting the work done factor λ = 1.0 yields:

K 28 tantan 0210 TUCc

T ap

stages 65.9/285.164Nstages

• 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 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6 Stage 7ΔT=20 ΔT=25 ΔT=25 ΔT=25 ΔT=25 ΔT=25 ΔT=20λ=0.98 λ=0.93 λ=0.88 λ=0.83 λ=0.83 λ=0.83 λ=0.83


Stage-by-stage design - Stage 1

14.27)arctan(

67.51)arctan( ,64.60)arctan(

IGV Nom/s 9.76

22

221

20

a

w

a

w

a

wp

w

C

C

C

CU

C

U

CU

TcC

• 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.

79.0cos

cos

cos

cos

2

1

1

2

1

2

a

a

C

C

V

V


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:

K

P

T

T

P

P StageStage

Stage

308T

bar 249.1

236.11

103

103

1

011 01

03

• 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 11.06 degrees. cos(11.06) =0.981)

856.02

1 1213

U

CCCC ww

• 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.


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:

21

210

tantan2

tantan

U

C

c

UCT

a

p

a

• Solving the two simultaneous equations for β 1 and β 2 gives:

19.42

70.57

2

1

• α 1 and α 2 are then obtained from (obtaining α 1 means that the design of the first stage is complete):

22

11

tantan

tantan

aa

aa

CCU

CCU

41.05

06.11

2

1


Stage-by-stage design - Stage 1• The design of the first stage is now complete:

β1 60.64β2 51.67α1 00.00α2 27.14α3 11.06

907.0

cos

cos

2

3

2

3

a

a

C

C

C

C

• The de Haller number in thestator is:


Stage-by-stage design – Stage 2• The stage pressure rise of stage 2 becomes:

K

P

T

T

P

P StageStage

333T

bar 599.1

280.11

203

203

1

01201

03

• 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.


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:

21

210

tantan2

tantan

U

C

c

UCT

a

p

a

• 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:

63.28

92.50

2

1

• 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:

β1 57.70β2 42.19α1 11.06α2 41.05α3 28.63

859.0

cos

cos

2

3

2

3

a

a

C

C

C

C



Stage-by-stage design – Stage 3• The stage pressure rise of stage 3 becomes:

K

P

T

T

P

P StageStage

357T

bar 992.1

246.11

303

303

1

01301

03

• 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.


Stage-by-stage design – Stage 4,5 and 6• We maintain Λ of 0.50 for stage 4, 5 and 6:

21

210

tantan2

tantan

U

C

c

UCT

a

p

a

• 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:

71.27

38.51

12

21


Stage-by-stage design - Stage 3• The design of the third stage is now complete:

β1 50.92β2 28.63α1 28.63α2 50.92α3 27.71

712.0

cos

cos

2

3

2

3

a

a

C

C

C

C



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.

Stage 4 5 6P03/P01 1.228 1.213 1.199P03 2.447 2.968 3.560T03 381 405 429


Stage-by-stage design - Stage 4,5,6• The design of stages 4,5 and 6 are now complete:

β1 51.38β2 27.71α1 27.71α2 51.38α3 27.71 (28.52

for stage 6)

705.0

cos

cos

2

3

2

3

a

a

C

C

C

C

• The de Haller number for thestators are:


Stage-by-stage design – stage 7• We maintain Λ of 0.50 for stage 7:

21

210

tantan2

tantan

U

C

c

UCT

a

p

a

• The pressure ratio of the seventh stage is set by the overall requirement of an r c = 4.15. The stage inlet pressure is 3.56 bar, which gives the required pressure ratio and temperature increase according to:

KT

T

T

P

P

StageStage

8.22

177.11

177.1560.3

192.4

0

1

01

701

03


Stage-by-stage design - Stage 7• The design of stage 7 is now complete:

β1 50.98β2 28.52α1 28.52α2 50.98α3 28.52

• 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

• 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: p

ww

c

CUCUT 1122

0


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

m/s 3.3552

m/s 6.2662

m/s 7.1772

NrU

NrU

NrU

tt

mm

rr

11.67)arctan(

64.60)arctan(

83.49)arctan(

1

1

1

a

tt

a

mm

a

rr

C

U

C

U

C

U


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:

2)1(2

)1(

2

0

0

2

03

3 1035.0A 4599.02

11 0.434

2

M,3

mMMAP

RTm

c

CTR

CexitT

p

exit

a

• Which gives the blade height, root and tip radii:

mr

mr

m

r

t

1210.0

2181.0

0971.0h


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:

mr

mr

t

r

2222.0

1171.0

m/s 0.3492

m/s 9.1832

NrU

NrU

tt

rr

• The free vortex condition gives:

m/s 7.58

m/s 4.111

,2,2

,2,2

t

mmwtw

r

mmwrw

r

rCC

r

rCC


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:

)arctan(

)arctan(

22

22

a

w

a

w

C

CU

C

C

69.62

67.51

80.25

37.21

14.27

60.36

,2

,2

,2

,2

,2

,2

t

m

r

t

m

r

• 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 ??


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:

A

dA

V

dVM )1( 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)


Supercritical bladingConcave portion => supercritical diffusionConcave section afterentrance region


Supercritical blading


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.

lecture 10 – axial compressors 2

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