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

4.1

005.1

287.0

0

0

Kkg

kJc

Kkg

kJR

p

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

1) Choose Cx1 and rH/rT to satisfy m and keep Mtip low and define

rT

2) Select N from rT and UT

3) Compute T0 across compressor and all exit flow conditions

[keep rm same through engine]

4) Estimate T0 for first stage from inlet condtions [Euler and de

Haller]

5) Select number of stages T0comp / T0stage

6) ..

7) ..

0 0

0

1

,

0 350

p in turbine x

T

Given PR p T m

T C constant

Assume U mps

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

2

2 1 hubx tip xtip

rm AC r C

r

2

2

1

tip

hubx

tip

mr

rC

r

2 tip

tip

rN

U

2 2tip

tip tip tip x

WM where W U C

RT

5

Calculate Tip Radius and Rotational Speed

Cx T1 (degK) P1 (bar) P1 (kg/m2) rho1(kg/m3) rtip (m) rhub/rtip N rev/sec N (rpm) W1tip M1tip

100 283.0226 0.950214 9692.1879 1.17103756 0.254398876 0.4 218.9643 13137.86 364.0055 1.079401

0.261089548 0.45 213.3531 12801.19

0.269230464 0.5 206.9018 12414.11

0.279178986 0.55 199.5289 11971.73

0.291450527 0.6 191.1277 11467.66

150 276.8009 0.879091 8966.7278 1.10773699 0.213568228 0.4 260.8266 15649.6 380.7887 1.141789

0.219185056 0.45 254.1427 15248.56

0.226019368 0.5 246.458 14787.48 Pick this

0.234371166 0.55 237.6755 14260.53

0.244673143 0.6 227.6681 13660.09

200 268.0905 0.786018 8017.3822 1.0226367 0.192497422 0.4 289.3767 17362.6 403.1129 1.228207

0.19756009 0.45 281.9612 16917.67

0.203720123 0.5 273.4353 16406.12

0.211247926 0.55 263.6915 15821.49

0.220533502 0.6 252.5887 15155.32

250 256.8913 0.676972 6905.1159 0.91916112 0.181607916 0.4 306.7282 18403.69 430.1163 1.338742

0.186384191 0.45 298.868 17932.08

0.192195753 0.5 289.8309 17389.86

0.199297711 0.55 279.5028 16770.17

0.208058005 0.6 267.7344 16064.06

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:

mr

and

mrr

rr

mean

tip

tip

hubhub

1697.0

1131.02262.05.0

11

2266.6 /

60

r NU m s

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)

2 11 1 1

1 01 1 01 1

01 1

1/ 2

2

1 1

1/ 22 2

1 1/ 2

1

2

1

1

2 2

x

p

T

Hx

T

tipHH T m H T

T T

tip

tip x tip tip

C T pT T p p

c T RT

mR

Rc

R

URR R R R R N

R R

CC C U M

RT

Select RH/RT and Utip (N)

for turbine issues

8

Compressor Meanline Design

Compressor exit (2)

2

102 01 02 01 2 02

12 2

2 02 2 2

01 2 2

2

Pr Pr2

2

polyx

p

x

T H

m

Cp p T T T T

c

T p mp p A

T RT C

Ablade height at exit h R R

R

9

Compute Compressor Exit Conditions

Compute Compressor Exit Total Temperature

so that T02 = 288.0 (4.15)0.3175 = 452.5 0K,

T0 compressor= 452.5 - 288.0 = 164.5 0K and other conditions:

( 1)

0202 01

01

pPT T

P

2 20

2 02 3

3.5( 1)

22 02

02

232

202

0

150452.5 441.3

2 2 1.005 10

441.34.19 3.838

452.5

3.838 10200

3.03 /

0.287 102 441.3

p

CT T K

c x

TP P bar

T

kgbar

P m barkg m

kJ kg mRTK

kg K kJ

10

Compute Compressor Exit Conditions

Exit area, hub and tip radius:

2

2

2

2

200.044

3.031 150

0.0440.0413

2 2 0.1697

0.04130.1697 0.1903

2 2

0.04130.1697 0.1491

2 2

x

tip hub

mean

tip mean

hub mean

mA m

C

Ar r r m

r

rr r m

rr r m

11

Step 2 - Estimate the Number of Stages

From Eulers Turbine Equation:

With no inlet guide vane (Cu1=0, 1 = 0, and Wu1= -U), the relative flow angle is:

And the relative inlet velocity to the 1st rotor is:

2 1 2 10

2 1 1 2

( ) (tan tan )

(tan tan ) (tan tan )

stage

U U x

P p

U C C UCT

c c

and

tan( )U xC C

0

1 1

266.6tan( ) 60.64

150x

U

C

1

1

150305.9 /

cos( ) cos( 60.64)

xCW m s

tan( )U xW C 11

2266.6 /

60

r NU m s

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:

0.172.01

2 W

W

0

2

2

2 01.47220

150)cos(

W

Cx

smWW /2209.30572.072.0 12

2 1 00 3

( ) 266.6 150 (tan(60.64) tan(47.01))28

1.005 10stageU U

P

U C CT K

c x

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 0

13

Choose Number of Stages

Given poly and T0out/T0in T0 = T0out -T0in, so the number of stages is T0 compressor / T0stage = 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. a

14

Compressor Meanline Design Develop estimate of the number of stages

Assuming Cx = constant

for axial inflow tan1 = 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

0 1 2tan tanxstagep

CT

c

stage 1 2 3 4 5 6 7

0.98 0.93 0.88 0.83 0.83 0.83 0.83

15

Compressor Meanline Design Develop Stage by Stage Design

C = absolute velocity, CU = absolute velocity in U direction

1

1

0

2

0 inf

tan

tan

U

m

x

p

m

m U m

x x

C axial low

U

c

c TU

U C U

c c

Constant Cx

U

C1 W2

W1

C2

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:

0

1 1

266.6tan( ) 60.64

150x

U

C

01

0

1 2 2 1 2tan( ) tan( ) tan( ) tan( ) 49.89

30 0

2 2

1.005 10 (23.5)tan 30.57

266.6 150

p stage

x

c T x

UC

2 1 2 10

( ) (tan tan )23.5

stage

U U x

P p

U C C UCT

c c

criterion)Haller de toaccording acceptable(76.0: toleadswhich 1

2 W

W

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:

smWCW

smUW

smU

smC

C

smC

Ux

U

U

x

/9.305

/6.266

/6.266

/150

0.0

/150

2

1

2

1

11

1

1

1

designs) lfor typicahigh is(which

836.0)6.266(2

31.1906.266

2

12

U

WWR uu

smWCW

smUCW

smU

smCCC

smCC

smC

Ux

UU

Ux

xU

x

/77.232

/0.178

/6.266

/21.174

/6.88tan

/150

2

2

2

2

222

2

2

2

2

2

222

2

18

Velocity Components for Stator of 1st Stage

Now consider the stator of the 1st stage. The h0 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:

1 3

3 2

3

0

0 for stator, so 0

U U

x x

U

C C

C C

U W

rotorstatorstator

rotorstatorstator

122

211

19

2=49.89

Velocity Triangles of 1st Stage Using Repeating Stage Assumption

U=- WU1 =266.6

1=0

W1=305.9

Cx1=150

W2=232.77

Cx2=150

U=266.6

2=30.57

WU2=178.0

CU2=88.6

1=60.64

U=266.6

3=0

W3=305.9

Cx3=150

3=60.64

W = C - U

ROTOR

STATOR

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.

C2=174.21

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 the stage temperature rise for

the first and last stages of the compressor and increase it for the

middle stages. This strategy is typically used to:

reduce the loading of the first stage to allow for a wide variation in angle of attack due to various aircraft flight conditions

reduce the turning required in the last stage to provide for zero swirl flow going into the combustor

With this in mind, lets change the work distribution in the compressor to:

0 23.5stageT

1 2,3, 4, 5, 6 70 0 020.0 25.0 20.0stage stage stageT T T

22

1st Stage Design for Non-Repeating Stages

We can re-calculate the relative angles for the 1st stage:

0

1 1

266.6tan( ) 60.64 (same as before)

150x

U

C

01

0

1 2 2 1 2tan( ) tan( ) tan( ) tan( ) 51.89

30 0

2 2

1.005 10 (20)tan 26.68

266.6 150

p stage

x

c T x

UC

t)requiremen work reduced the todue turningreduced thenote (also

0.72) i.e. ,acceptable still(79.0: toleadswhich 1

2 W

W

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:

smWCW

smUW

smU

smC

C

smC

Ux

U

U

x

/9.305

/6.266

/6.266

/150

0.0

/150

2

1

2

1

11

1

1

1

increased)reaction that the(note.C that assumption with the

859.0)6.266(2

31.1906.266

2

13

12

C

U

WWR uu

smWCW

smUCW

smU

smCCC

smCC

smC

Ux

UU

Ux

xU

x

/03.243

/22.191

/6.266

/87.167

/38.75tan

/150

2

2

2

2

222

2

2

2

2

2

222

2

24

Design of 1st Stage Stator

The pressure ratio for this design with a temperature change, T0 = 20 is:

So P03= P02 = 1.01 (1.236) = 1.248 bar and T03= 288+20=308 0K

Now we must choose a value of 3 leaving the stator.

When we designed with repeating stages, 3= 1.

But now we have the flexibility to change 3.

236.1288

20288 4.)4.1(9.0

)1(

01

02

01

02

P

T

T

P

P

25

Design of the 1st Stage Stator & the 2nd Stage

Change 3 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 1and 2

2 1 2(tan tan )2

xstage

CR

U

0 2 1 2(tan tan )x

stage

p

UCT

c

0

22

0

21

92.42

31.57

stage

stage

26

Design of 1st Stage Stator & 2nd Stage Rotor

Note that this is the same as specifying E, n, and R as in one of your homeworks and computing the angles.

And the absolute flow angles of the second stage can be found from

So

Therefore, we have determined the velocity triangles of the 1st stage stator and the second stage rotor

0

22

13

0

21

27.40

36.12

stage

stagestage

1 1 2 2tan tan tan tanx

U

C

27

2=51.89

Velocity Triangles of 1st Rotor Using Non-Repeating Stage Assumption

U=- WU1 =266.6

1=0

W1=305.9

Cx1=150

W2=242.03

Cx2=150

U=266.6

2=26.68

WU2=191.2

2

CU2=75.38

1=60.64

U=266.6

3=12.36

W3=277.73

Cx3=150

3=57.31

W = C - U

ROTOR

STATOR

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.

CU3=32.87

WU3=233.7

7

C3=153.56

C2=167.87

28

Design of 2nd Stage Stator & 3rd Stage Rotor

Design of the 2nd stage stator and 3rd stage rotor can be done in the same manner as the 1st stage stator and 2nd stage rotor.

A choice of 50% reaction and a temperature rise of 25 degrees for the 3rd stage will lead to increased work by the 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:

STAGE 1 2 3 1 2 3 Cx Cu1 C1 W1 Wu1 M1 Mr1

1 0 26.68 12.36 60.64 -51.89 -57.31 150 0 150 305.9 -266.6 0.449778 0.917248

2 12.36 40.27 57.31 -42.92 150 32.87 153.56 277.73 -233.77 0.445072 0.804962

3

4

5

6

7

STAGE Cu2 C2 W2 Wu2 M2 Mr2 Cu3 C3 W3 Wu3 R P03/P01 T03/T01

1 75.38 167.87 243.03 -191.22 0.488438 0.707125 32.87 153.56 277.73 -233.77 0.874 1.236 1.069

2 127.07 196.59 204.86 -139.53 0.553668 0.576959 0.7 1.279 1.081

3

4

5

6

7

STAGE M3 Mr3 P01 P02 P03 T01 T02 T03 P1 P2 P3 T1 T2 T3

1 0.445072 0.804962 1.01 1.248 1.248 288 308 308 0.879088 1.060143 1.08932261 276.806 293.9799 296.2683

2 1.248 1.596 1.596 308 333 333 1.089323 1.295942 296.2683 313.7723

3

4

5

6

7

30

Design of 2nd Stage Stator

The pressure ratio for the 2nd stage design with a temperature change, T0 = 25 is:

So P03= P02= 1.248 (1.279) = 1.596 bar and T03= 308+25=333 0K

Now we must choose a value of 3 leaving the 2nd stage stator that provides for the desired Reaction and Work in the 3rd stage

using a similar technique as previously used.

279.1308

25308 4.)4.1(9.0

)1(

01

02

01

02

P

T

T

P

P

31

Design of 2nd Stage Stator & 3rd Stage

We can change 3 so that there is swirl going into the third stage and thereby reduce the reaction of our second stage

design. If we design the third stage to have a reaction of 0.5,

then from the equation for reaction:

And if we design the third stage to a temperature rise of 25 0, the Eulers equation:

Which can be solved simultaneously for 1and 2

3 1 2(tan tan )2

xstage

CR

U

0 3 1 2(tan tan )x

stage

p

UCT

c

0

32

0

31

88.29

26.50

stage

stage

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:

0

32

23

0

31

26.50

88.29

stage

stagestage

1 1 2 2tan tan tan tanx

U

C

0

32

0

31

88.29

26.50

stage

stage

OK iswhich 74.cos

cos

2

1

1

2

W

W

33

2=42.92

Velocity Triangles of 2nd Stage

W2=204.86

Cx2=150

U=266.6

2=40.27

WU2=139.5

3

CU2=127.07

W = C - U

ROTOR

STATOR

Notice that the velocity triangles are

not symmetric for the second stage due to 70%reaction design but

will be for 3rd stage (50% reaction).

C2=196.59

U=266.6

3=12.36

W3=277.73

Cx3=150

3=57.31 CU3=32.87

WU3=233.7

7

C3=153.56

U=266.6

3=29.88

W3=234.63

Cx3=150

3=50.26

CU3=86.18

WU3=180.4

2

C3=172.99

34

Summary of Conditions for Stages 1 - 3

STAGE 1 2 3 1 2 3 Cx Cu1 C1 W1 Wu1 M1 Mr1

1 0 26.68 12.36 -60.64 -51.89 -57.31 150 0 150 305.9 -266.6 0.449778 0.917248

2 12.36 40.27 29.88 -57.31 -42.92 -50.26 150 32.87 153.56 277.73 -233.77 0.445072 0.804962

3 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 180.42 0.483867 0.656279

4

5

6

7

STAGE Cu2 C2 W2 Wu2 M2 Mr2 Cu3 C3 W3 Wu3 R P03/P01 T03/T01

1 75.38 167.87 243.03 -191.22 0.488438 0.707125 32.87 153.56 277.73 -233.77 0.874 1.236 1.069

2 127.07 196.59 204.86 -139.53 0.553668 0.576959 86.18 172.99 234.63 -180.42 0.7 1.279 1.081

3 180.42 234.63 172.99 -86.18 0.643754 0.474632 86.18 172.99 234.63 -180.42 0.5 1.256 1.075

4

5

6

7

STAGE M3 Mr3 P01 P02 P03 T01 T02 T03 P1 P2 P3 T1 T2 T3

1 0.445072 0.804962 1.01 1.248 1.248 288 308 308 0.879088 1.060143 1.08932261 276.806 293.9799 296.2683

2 0.483867 0.656279 1.248 1.596 1.596 308 333 333 1.089323 1.295942 1.35979317 296.2683 313.7723 318.1117

3 0.465906 0.631918 1.596 2.005 2.005 333 358 358 1.359793 1.517323 1.72789178 318.1117 330.6113 343.1117

4

5

6

7

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

2=29.88

Velocity Triangles of Stages 3 - 6

W2=172.99

Cx2=150 U=266.6

2=50.26

WU2=86.18

CU2=180.42

W = C - U

ROTOR

STATOR

Notice that the velocity triangles

are symmetric due to the 50%reaction design.

C2=234.63

U=266.6

3=29.88

W3=234.63

Cx3=150

3=50.26

CU3=86.18

WU3=180.4

2

C3=172.99

U=266.6

3=29.88

W3=234.63

Cx3=150

3=50.26

CU3=86.18

WU3=180.4

2

C3=172.99

37

Summary of Conditions for Stages 1 - 6

STAGE 1 2 3 1 2 3 Cx Cu1 C1 W1 Wu1 M1 Mr1

1 0 26.68 12.36 -60.64 -51.89 -57.31 150 0 150 305.9 -266.6 0.449778 0.917248

2 12.36 40.27 29.88 -57.31 -42.92 -50.26 150 32.87 153.56 277.73 -233.77 0.445072 0.804962

3 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 180.42 0.483867 0.656279

4 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 180.42 0.465906 0.631918

5 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 180.42 0.449807 0.610083

6 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 180.42 0.435269 0.590365

7

STAGE Cu2 C2 W2 Wu2 M2 Mr2 Cu3 C3 W3 Wu3 R P03/P01 T03/T01

1 75.38 167.87 243.03 -191.22 0.488438 0.707125 32.87 153.56 277.73 -233.77 0.874 1.236 1.069

2 127.07 196.59 204.86 -139.53 0.553668 0.576959 86.18 172.99 234.63 -180.42 0.7 1.279 1.081

3 180.42 234.63 172.99 -86.18 0.643754 0.474632 86.18 172.99 234.63 -180.42 0.5 1.256 1.075

4 180.42 234.63 172.99 -86.18 0.620713 0.457644 86.18 172.99 234.63 -180.42 0.5 1.237 1.07

5 180.42 234.63 172.99 -86.18 0.599981 0.442359 86.18 172.99 234.63 -180.42 0.5 1.22 1.065

6 180.42 234.63 172.99 -86.18 0.581197 0.42851 86.18 172.99 234.63 -180.42 0.5 1.206 1.061

7

STAGE M3 Mr3 P01 P02 P03 T01 T02 T03 P1 P2 P3 T1 T2 T3

1 0.445072 0.804962 1.01 1.24836 1.24836 288 308 308 0.879088 1.060449 1.08963684 276.806 293.9799 296.2683

2 0.483867 0.656279 1.24836 1.596652 1.596652 308 333 333 1.089637 1.296472 1.36034905 296.2683 313.7723 318.1117

3 0.465906 0.631918 1.596652 2.005395 2.005395 333 358 358 1.360349 1.517622 1.72823259 318.1117 330.6113 343.1117

4 0.449807 0.610083 2.005395 2.480674 2.480674 358 383 383 1.728233 1.913086 2.15910202 343.1117 355.6113 368.1117

5 0.435269 0.590365 2.480674 3.026423 3.026423 383 408 408 2.159102 2.372762 2.65703431 368.1117 380.6113 393.1117

6 0.422056 0.572443 3.026423 3.649866 3.649866 408 433 433 2.657034 2.903398 3.22898653 393.1117 405.6113 418.1117

7

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:

020

01

0.9 1.4

02 0.4

01

452.51.045 or 452.5 433 19.5

433

1.045 1.149

TT

T

P

P

39

Stage 7 Design

If we assume a Reaction = 0.5 for the 7th stage:

Then, solving equations:

5.0)tan(tan2

217 U

CR xstage

5.19)tan(tan 2170 p

xstage

c

CUT

0

1 7

0

2 7

48.59

32.77

stage

stage

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.

0

72

73

0

71

59.48

77.32

stage

stagestage

2211 tantantantan xC

U

41

Summary of Compressor Design STAGE 1 2 3 1 2 3 Cx Cu1 C1 W1 Wu1 M1 Mr1

1 0 26.68 12.36 -60.64 -51.89 -57.31 150 0 150 305.9 -266.6 0.449778 0.917248

2 12.36 40.27 29.88 -57.31 -42.92 -50.26 150 32.87 153.56 277.73 -233.77 0.445072 0.804962

3 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.483867 0.656279

4 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.465906 0.631918

5 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.449807 0.610083

6 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.435269 0.590365

7 32.77 48.59 32.77 -48.59 -32.77 -48.59 150 96.56 178.39 226.78 -170.08 0.435723 0.553917

STAGE Cu2 C2 W2 Wu2 M2 Mr2 Cu3 C3 W3 Wu3 R P03/P01 T03/T01

1 75.38 167.87 243.03 -191.22 0.488438 0.707125 32.87 153.56 277.73 -233.77 0.874 1.236 1.069

2 127.07 196.59 204.86 -139.53 0.553668 0.576959 86.18 172.99 234.63 -180.42 0.7 1.279 1.081

3 180.42 234.63 172.99 -86.18 0.643754 0.474632 86.18 172.99 234.63 -180.42 0.5 1.256 1.075

4 180.42 234.63 172.99 -86.18 0.620713 0.457644 86.18 172.99 234.63 -180.42 0.5 1.237 1.07

5 180.42 234.63 172.99 -86.18 0.599981 0.442359 86.18 172.99 234.63 -180.42 0.5 1.22 1.065

6 180.42 234.63 172.99 -86.18 0.581197 0.42851 86.18 172.99 234.63 -180.42 0.5 1.206 1.061

7 170.08 226.78 178.39 -96.56 0.547558 0.430721 96.56 178.39 226.78 -170.08 0.5 1.149 1.045

STAGE M3 Mr3 P01 P02 P03 T01 T02 T03 P1 P2 P3 T1 T2 T3

1 0.445072 0.804962 1.01 1.24836 1.24836 288 308 308 0.879088 1.060449 1.08963684 276.806 293.9799 296.2683

2 0.483867 0.656279 1.24836 1.596652 1.596652 308 333 333 1.089637 1.296472 1.36034905 296.2683 313.7723 318.1117

3 0.465906 0.631918 1.596652 2.005395 2.005395 333 358 358 1.360349 1.517622 1.72823259 318.1117 330.6113 343.1117

4 0.449807 0.610083 2.005395 2.480674 2.480674 358 383 383 1.728233 1.913086 2.15910202 343.1117 355.6113 368.1117

5 0.435269 0.590365 2.480674 3.026423 3.026423 383 408 408 2.159102 2.372762 2.65703431 368.1117 380.6113 393.1117

6 0.422056 0.572443 3.026423 3.649866 3.649866 408 433 433 2.657034 2.903398 3.22898653 393.1117 405.6113 418.1117

7 0.425883 0.541407 3.649866 4.193696 4.193696 433 452.5 452.5 3.20353 3.42041 3.70197943 417.1677 426.9133 436.6677

42

Hub and Tip Radii for Each Blade Row

From the pressure and temperature, we can compute the density from the equation of state:

STAGE P1 P2 P3 T1 T2 T3 1 2 3

1 0.879088 1.060449 1.089637 276.806 293.9799 296.2683 1.10656 1.25687 1.281488

2 1.089637 1.296472 1.360349 296.2683 313.7723 318.1117 1.281488 1.439682 1.490009

3 1.360349 1.517622 1.728233 318.1117 330.6113 343.1117 1.490009 1.599425 1.755031

4 1.728233 1.913086 2.159102 343.1117 355.6113 368.1117 1.755031 1.874463 2.043674

5 2.159102 2.372762 2.657034 368.1117 380.6113 393.1117 2.043674 2.172154 2.355046

6 2.657034 2.903398 3.228987 393.1117 405.6113 418.1117 2.355046 2.494105 2.690866

7 3.20353 3.42041 3.701979 417.1677 426.9133 436.6677 2.675693 2.791621 2.953936

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:

2

2

1

tip

hubx

tip

mr

rC

r

0.5( ) 0.1697 0.3394mean tip hub hub tipr r r r r

2

20.1697

(.6788).33941

tip tip

xtip

x

tip

m mr r

CrC

r

44

Hub & Tip Radii for All Stages of Compressor

So we get:

Station P P0abs T T0abs rtip rhub

1 0.879088 1.01 276.806 288 0.226203 0.113197

2 1.060449 1.24836 293.9799 308 0.219446 0.119954

3 1.089637 1.24836 296.2683 308 0.21849 0.12091

4 1.296472 1.596652 313.7723 333 0.213129 0.126271

5 1.360349 1.596652 318.1117 333 0.211662 0.127738

6 1.517622 2.005395 330.6113 358 0.208792 0.130608

7 1.728233 2.005395 343.1117 358 0.205326 0.134074

8 1.913086 2.480674 355.6113 383 0.203056 0.136344

9 2.159102 2.480674 368.1117 383 0.200294 0.139106

10 2.372762 3.026423 380.6113 408 0.198484 0.140916

11 2.657034 3.026423 393.1117 408 0.196249 0.143151

12 2.903398 3.649866 405.6113 433 0.194769 0.144631

13 3.20353 3.649866 417.1677 433 0.193067 0.146333

14 3.42041 4.193696 426.9133 452.5 0.192097 0.147303

15 3.701979 4.193696 436.6677 452.5 0.190866 0.148534

rotor

stator

rotor

stator

rotor

stator

rotor

stator

rotor

stator

rotor

stator

rotor

stator

45

Conditions in Compressor

Pressure vs. Blade Row #

0

1

2

3

4

1 3 5 7 9

11

13

15

Blade Row #

Pre

ss

ure

P

Stagnation Pressure vs. Blade Row #

0

1

2

3

4

5

1 3 5 7 9

11

13

15

Blade Row #

Sta

gn

ati

on

Pre

ss

ure

P0

Temperature vs. Blade Row #

200

250

300

350

400

450

1 3 5 7 9

11

13

15

Blade Row #

Te

mp

era

ture

T

Stagnation Temperature vs. Blade Row #

200

250

300

350

400

450

500

1 3 5 7 9

11

13

15

Blade Row #

Sta

gn

ati

on

Te

mp

era

ture

T0

46

Hub & Tip Radii Distribution - Flow Path Area

Hub and Tip Radii vs. Blade Row #

0

0.05

0.1

0.15

0.2

0.251 3 5 7 9

11

13

15

Blade Row #

Rad

ius rtip

rhub