1

Design of a Multi-Stage CompressorMotivation: 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 compressor4.1

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

• Choose N = 250 rev/sec or 15,000 RPM and rhub/rtip = 0.5, calculated rhub, rtip, rmean

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 M1tip100 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 12801.190.269230464 0.5 206.9018 12414.110.279178986 0.55 199.5289 11971.730.291450527 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 1.1417890.219185056 0.45 254.1427 15248.560.226019368 0.5 246.458 14787.48 Pick this0.234371166 0.55 237.6755 14260.530.244673143 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 1.2282070.19756009 0.45 281.9612 16917.67

0.203720123 0.5 273.4353 16406.120.211247926 0.55 263.6915 15821.490.220533502 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 1.3387420.186384191 0.45 298.868 17932.080.192195753 0.5 289.8309 17389.860.199297711 0.55 279.5028 16770.170.208058005 0.6 267.7344 16064.06

Drive choice by compressor inlet conditions

6

Compressor Meanline Design

• c1x chosen to avoid high tip Mach numbers and stresses

• Calculate compressor inlet details (1)

• Given: m, Utip, p01, T01, Pr, poly calculate compressor exit details including Aexit, h exit (2)

• Calculate 1 from U, Cx

• Calculate W2=0.72W1, 2 from Cx, W2

• From Euler, 1 and 2, calculate T0-12

7

Step 3-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

• To be conservative (account for losses, i.e. a<1), – Choose 7 stages

• Recalculate the T0stage = 164.5/7 = 23.5

• Calculate 1st stage temperature ratio is T0 ratio and pressure ratio

Step 3 - Calculate Velocity Triangles of 1st Stage at Mean Radius [Rm]

• Use Euler Turbine Equation to recalculate 2 and check de Haller

• Calculate 1st stage rotor exit conditions [2]

• Calculate Reaction [R1=0.84, high]

• Consider the stator of the 1st stage, assume “repeating stage”, then inlet angle to stator [α1=0] is absolute air angle coming out of rotor [α3] and, exit absolute angle of stator is inlet absolute angle of rotor

Constant Cx

U

C1W2

W1C2

9

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=150U=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

10

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.

11

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 23.5stage

T

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

T T T

12

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=150U=266.6

2=26.68

WU2=191.22

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 arenot “symmetric” due to the highreaction design of the rotor. Also, there is swirl now leaving the stator.

CU3=32.87

WU3=233.77

C3=153.56

C2=167.87

13

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=150U=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

14

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.

15

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

16

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 Euler’s equation:

• Which can be solved simultaneously for 1and

3 1 2(tan tan )2x

stage

CR

U

0 3 1 2(tan tan )xstage

p

UCT

c

032

031

88.29

26.50

stage

stage

17

2=42.92

Velocity Triangles of 2nd Stage

W2=204.86

Cx2=150U=266.6

2=40.27

WU2=139.53

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=57.31CU3=32.87

WU3=233.77

C3=153.56

U=266.6

3=29.88

W3=234.63

Cx3=150

3=50.26

CU3=86.18

WU3=180.42

C3=172.99

18

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)

19

2=29.88

Velocity Triangles of Stages 3 - 6

W2=172.99

Cx2=150U=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=180.42

C3=172.99

U=266.6

3=29.88

W3=234.63

Cx3=150

3=50.26

CU3=86.18

WU3=180.42

C3=172.99

20

Stage 7 Design

• For stage 7, we have P01= 3.65 and T01 = 433. For this 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

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

• or:

5.0)tan(tan2 217 U

CR xstage

5.19)tan(tan 2170 p

xstage c

CUT

01 7

02 7

48.59

32.77

stage

stage

21

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.

072

730

71

59.48

77.32

stage

stagestage

2211 tantantantan xC

U

22

Summary of Compressor DesignSTAGE

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 0.9172482 12.36 40.27 29.88 -57.31 -42.92 -50.26 150 32.87 153.56 277.73 -233.77 0.445072 0.8049623 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.483867 0.6562794 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.465906 0.6319185 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.449807 0.6100836 29.88 50.26 29.88 -50.26 -29.88 -50.26 150 86.18 172.99 234.63 -180.42 0.435269 0.5903657 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 1.0692 127.07 196.59 204.86 -139.53 0.553668 0.576959 86.18 172.99 234.63 -180.42 0.7 1.279 1.0813 180.42 234.63 172.99 -86.18 0.643754 0.474632 86.18 172.99 234.63 -180.42 0.5 1.256 1.0754 180.42 234.63 172.99 -86.18 0.620713 0.457644 86.18 172.99 234.63 -180.42 0.5 1.237 1.075 180.42 234.63 172.99 -86.18 0.599981 0.442359 86.18 172.99 234.63 -180.42 0.5 1.22 1.0656 180.42 234.63 172.99 -86.18 0.581197 0.42851 86.18 172.99 234.63 -180.42 0.5 1.206 1.0617 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 296.26832 0.483867 0.656279 1.24836 1.596652 1.596652 308 333 333 1.089637 1.296472 1.36034905 296.2683 313.7723 318.11173 0.465906 0.631918 1.596652 2.005395 2.005395 333 358 358 1.360349 1.517622 1.72823259 318.1117 330.6113 343.11174 0.449807 0.610083 2.005395 2.480674 2.480674 358 383 383 1.728233 1.913086 2.15910202 343.1117 355.6113 368.11175 0.435269 0.590365 2.480674 3.026423 3.026423 383 408 408 2.159102 2.372762 2.65703431 368.1117 380.6113 393.11176 0.422056 0.572443 3.026423 3.649866 3.649866 408 433 433 2.657034 2.903398 3.22898653 393.1117 405.6113 418.11177 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

23

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

atio

n P

res

su

re

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 15Blade Row #

Sta

gn

atio

n T

em

pe

ratu

re

T0

24

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