MEE-304MEE-304Turbomachines Turbomachines

Unit 3Unit 3STEAM AND GAS TURBINES

Lecture 23Lecture 23

Topics to be discussed

Axial turbine stages - Stage velocity triangle, work, single stage impulse turbine, speed ratio maximum utilization factor, multistage velocity compounded impulse, multi stage pressure compounded impulse, reaction stages, degree of reaction, zero reaction stages, fifty percent reaction stages, hundred percent reaction, negative reaction, free and forced vortex flow

Inward flow radial turbine stages, IFR Turbine, T-s diagram, degree of reaction - Steam turbine governing – Features of steam turbine and gas turbine

Axial turbine stages

Stage velocity triangles

From the velocity triangle,

2 2 2 2 2

2 2 2 2 2 2

2 2

2 2 2 2

2 2

2 2

3 3 3 3 3

y3 3 3 3 3 3

2 3 2 2 3 3

2 3 2 2 3 3

2

cos w cos

sin w sin

sin( ) sin(90 ) cos

sin( )

cos

cos cos

c c sin w sin

sin sin

(w sin ) (w sin )

x

y y

x

y y

y y

y

c c

c c w u u

c cu

u

c

c w c

w u u

c c c c

c c u u

c

3 2 2 3 3

2 3 2 3

w sin w siny

y y y y

c

c c w w

x3x22 3

2 3

32y2 y3 x

2 3

2 3 x 2 3

x3x 22 3

2 3

32x x 2 3

2 3

2 3 2 3

ccBut w ; and v

cos cos

sinsin c c c ( )

cos cos

c (tan tan )

cSimilarly, c ; c

cos cos

sinsinc ( ) c (tan tan )

cos cos

tan tan tan tan

y yc c

c

Stage work

.

2 3 2 3

.2 32

2

y3

.22 2 2 2

.2 2

w ( ( )) ( )

w ( )

Defining the blade-gas speed ratio :

u c

For axial discharge at exit (c 0)

sinw ( ) ( )

sinw ( )

st y y y y

y yst

yst

st

u c c u c c

c cu

u u

c cu u

u u

u

Blade Loading and Flow Coefficients

.

2

.2

2 3 2 3 2 3

.

2 3 2 3 22

( ) (tan tan ) ( )(tan tan )

( )(tan tan ) (tan tan ) (tan ta

st

x

xst y y x

st x

w

uc

uc

w u c c uc uu

cw

u u

3n )

Stage reaction

The stage reaction is defined as the ratio of the static enthalpy drop in the rotor to the static enthalpy drop in the stage

Blade and Stage Efficiencies

b

2 2 3 3

.2 2 2 2 2 22 3 2 3 3 2

Blade Efficiency or Utilization factor ( )

rotor blade work = =

energy supplied to the rotor blades

1 1 1( ) ( ) (w w )

2 2 2

Energy supplied to the rotor

e

st y y

st

rotorbladework

w u c u c

w u u c c

i

2 2 2 2 2i 2 2 3 3 2

absolute K.E supplied + change in K.E in rotor blades

1 1 1e ( ) + (w w )

2 2 2c u u

Stage efficiency

2 2 2 2 2 22 3 2 3 3 2

2 2 2 2 22 2 3 3 2

2 3

2 2 2 22 3 3 2

b 2 2 22 3 2

1 1 1( ) ( ) (w w )

2 2 2 1 1 1

( ) + (w w )2 2 2

,

( ) (w w ) =

c (w w )

u u c c

c u u

for axial machines u u u

c c

Blade and stage efficiencies are different on account of the variable energy inputs and losses.

Further, stage efficiency accounts for the stage losses. While, blade efficiency or blade utilization factor doesn’t consider this.

Single impulse stage

w3 = w2

w2

c2cx2

cy2

u

c3cx3

cy3

2β2

u

3β3

There is no pressure drop in the in the rotor blades, so the relative velocities at their entry and exit are the same (w2=w3) for frictionless flow

So the utilization factor is given by

We have

Also w2= w3 and β2 = β3

2 3 2 2 3 3w sin w siny yc c

2 2 2 22

2 22

22 222

2

(2sin ) 4 sin12

4 (c sin )4( sin )

uw uw

cc

u u

c

Thus the utilization factor is a function of the blade-to-gas speed ratio and the nozzle angle

ε=f(σ,α2)

Maximum Blade Utilization Factor

22

2

2

2

2 2 2

0

(4 sin 4 ) 0

4sin 8 0

sin

2

sin 2

opt

y

d

dd

d

u

c

c c u

So the maximum utilization factor requires the exit from the stage in the axial direction

C3=cx3

y2 2 2 2 2

2 2

2 3 2 3

2 2 3 3

y3 3 3

Also c sin w sin 2

w sin

Since w w ; and [no pressure drop in rotor]

v sin v sin

c w sin 0

c u u

u

u

u u u

w w

Stage Velocity Triangles for εmax

w3 = w2

w2

c2cx2

cy2=2u

u

cx3=c3

2β2

u

β3

2=0

So the maximum utilization factor

2max 2

2max 2 2 2

2max 2

4( sin )

1 14( sin .sin sin )

2 4

sin

opt opt

Multi Stage Velocity Compounding in impulse turbines(Curtis stages or velocity stages)

N R1 F R2

V

p

Velocity triangles for a two-stage velocity compounded impulse turbine with εmax

Assumptions for Curtis Stages1.Equiangular flow through rotor and guide blade

(β2 = β3, 3 = ´2, β´

2 = β´3)

2.Frictionless flow over the bladesw2 = w3, c3 = c´

2, w´2 = w

´3

• Such stages are known as Curtis stages or velocity stages Assumptions for Curtis Stages

Equiangular flow through rotor and guide blade

(β2 = β3, 3 = ´2, β´

2 = β´3)

Frictionless flow over the blades

w2 = w3, c3 = c´2, w´

2 = w´3

For maximum utilization factor :

c´y3 = 0; u = w´

3sinβ´3 = w´

2 sinβ´2

c´y2 = 2u = c´

2 sin´2 = c3 sin3

cy3 = 3u = w3 sinβ3 = w2 sinβ2

cy2 = c2sin2 = w2sinβ2+u

cy2 = 4u

Similarly for a three stage velocity compounded turbine

opt= (1/6) sin2

For n velocity stages.

opt = u/c2 = ¼ sin2

opt =(1/2n) sin2

Work Output

Stage I : wI = u(Cy2 + Cy3)

= u(w2sinβ2+w3sinβ3)

= 2uw2sinβ2 = 6u2

Stage II

wII = 2uw�2sin�2

wII = 2u2

Total work = 6u2 + 2u2 = 8u2.

For an n-stage velocity compounded turbine

Total turbine work

Maximum utilization factor

n2 2

T ii=1

w w 2n u

2 22 2 2 2

max 22 22 2

2max 2

2 14 4 ( sin )

1 1 2( )2 2

sin

Tw n un n

nc c

Wi = 2[2(n-i)+1]u2

Velocity compounded impulse stages have the following disadvantages : Nozzles have to be of the converging-diverging type to

generate high velocities. Hence, blade design is more difficult and expensive.

High velocity at nozzle exit leads to higher cascade losses due to the formation of shock waves.

Pressure compounding eliminates these problems by : Dividing the pressure drop between many stages, and thus

reducing leakage losses. Lower stage velocities at subsonic speeds also reduce the

stage losses.

Multistage Pressure Compounded Impulse (Rateau Stages)

N1 R1N2 R2

V

p

Reaction Stages

NR-1

FR-2

V

p

Enthalpy-entropy diagram

p1

p22

3

3s3ss

p01 p02

O1 O2

h01 = h02

Entropy

En

thal

py

1

½ c12

p03

p3

wa

ws

p03ssO3ss

½ c22

p03relh02rel = h03rel

½ w32

p02rel

2s

½ c32

Isentropic stage work :

ws = h01 – h03ss = cp (T01 – T03ss)

Nozzle :

h01 = h02 = h2 + ½ c22 [No work transfer]

Actual work

wa = h02 – h03 = h01 – h03 = cp (T01 – T03)

Losses due to irreversibility is given by the enthalpy loss coefficient

Stagnation Pressure Loss coefficient

2 22 22

2 22

2-( - )

12

psN s

ch hT T

cc

3 33 32

2 33

2-( - )

1 ww2

psR s

ch hT T

01 02 0

2 22 2

( )1 12 2

NN

p p pY

c c

02 03 0

2 23 3

( )1 1

w w2 2

rel rel RR

p p pY

R R

For subsonic applications

Y

Degree of reaction

2 3

01 03

1 3

2 3

1 3

2 2 2 22 3 3 2 2 3

01 03 st 2 2

static enthalpy change in rotorR =

stagnation enthalpy change in stage

h h R =

h h

if

h h R =

h h

1 1Since h h (w w ) (u u )

2 2and h h w y

c c

u c

3 3

2 2 2 23 2 2 3

2 2 3 3

2 3 y3

2 23 2

2 3

1 1(w w ) ( )

2 2 R =

For axial turbine with u , and negative swirl (-c )

(w w ) R =

2 ( )

y

y y

y y

u c

u u

u c u c

u

u c c

2 2 2 2 2 22 2 2

2 2 2 23 3

2 2 23 2 3 2 3 2

y2 3 2 3 2 3

2 23 2 2 3 3 2

From velocity triangles

w ( ) tan

Similarly

w tan

w w (tan tan )(tan tan )

But c ( tan ) ( tan ) (tan tan )

w w ( )(tan tan

x y x x

x x

x

y x x x

x y y

c c u c c

c c

c

c c u c u c

c c c

2 3 3 23 2

2 3

3 2

m 3 2

)

( )(tan tan ) 1(tan tan )

2 ( ) 2

1 (tan tan ) tan

21

where tan (tan tan )2

x y y x

y y

m

c c c cR

u c c u

R

2 2

2 2

3 2

3 2

3 2

tan tan

tan tan

1(tan tan )

21

= (tan tan )2

1 1 R= (tan tan )

2 2

x x

x

x

x

x

x

c c u

u

c

cR

uc u

u c

c

u

Zero degree reaction stages

Fifty percent reaction stages Velocity triangles will be symmetric

β2=α3 & β3=α2

w2=c3 & w3=c2

Utilization factor

Optimum blade to gas speed

σopt = sin α2

Maximum utilization factor

Hundred percent reaction

Negative reaction

MEE-304 TURBOMACHINESMEE-304 TURBOMACHINES

AXIAL TURBINESAXIAL TURBINES

NUMERICAL PROBLEMS – Set 7NUMERICAL PROBLEMS – Set 7

Example -1 In an axial turbine stage, the absolute velocity entering and

leaving the stage are in the axial direction. The degree of reaction is 0.55 and nozzle angle is 22º. Calculate the flow coefficient, loading coefficient and relative flow angles.

Example - 2 Following data refer to a two-stage velocity compounded impulse

turbine operating on hot air : Flow rate = 1.0 kg/s Mean blade diameter = 75 cm Rotational speed = 3600 rpm Nozzle blade angle = 80° from axial direction Deviation = 5°

Assuming optimum utilization factor and constant axial velocity, calculate

1. blade to gas speed ratio

2. utilization factor

3. rotor blade air angles at entry and exit in the two stages

4. flow coefficient

5. the loading coefficients in the two stages

6. power developed separately in the two stages

Given :

m = 1 kg/s; dm = 75 cm; N = 3600 rpm;

2´ = 80°; δ = 5°

'2 2

'2 2

2

2 2max 2

5

5 80 5 75

1) blade to gas speed ratio

1 1 sin (sin 75) 0.241

4 42) Utilization factor

sin sin 75 0.933 (Ans.)

opt

Solution

22

m

2

2

x 2 2

2 2

33

3

3) Stage-1

3tan

d 0.75 3600; u = =

60 60

141.4 m/s

141.4 c 586.6 m/s

0.241 0.241c cos 586.6 cos75 151.82 m/s

3 141.4 tan 70.31

151.82

3Similarly tan

y

x x

opt

m

y

x

c u u

c c

Nu

c

u

u

c

c u

c

3

3 141.4

151.82

70.31x

u

c

'2'

2 '2

'2 '

2

'2

'' 33 '

3

'3

Similarly tan

141.4tan

151.82

42.96

141.4and tan

151.82

42.96

y

x

x

x

c u

c

u

c

u

c

x

stage-I 2 31 2 2 2

stage-22 2 2

. .2

stage-I 2 3

stage-I

4) Flow coefficient

c 151.82 = = 1.07 (Ans.)

u 141.45) Blade loading coefficient

w ( ) (4 2 ) = 6

uw (2 0)

and 2u

6) Power ( ) = m(6 )

P 1

y y

y y

u c c u u u

u u

u u

u

mu c c u

2

.2 2

stage-II

.0 6.0 141.4 119.96 kW

P (2 ) = 1.0 2 141.4 39.99 kWm u

Example-3 A low pressure turbine within a turbojet engine consists of five

repeating stages. The turbine inlet stagnation temperature is 1200 K and the inlet stagnation pressure is 213 kPa. It operates with a mass flow of 15 kg/s and generates 6.64 MW of mechanical power. The stator in each turbine stage turns the flow 15º at stator inlet to 70º at stator outlet. The turbine mean radius is 0.46 m and the rotational shaft speed is 5600 rpm.

Calculate the turbine stage loading coefficient and flow coefficient. Hence show that the reaction is 0.5 and sketch the velocity triangles for one complete stage.

(take γ=1.333, R=287.2 J/kgK and cp=1150 J/kgK)

Example-4 The high pressure stage of an axial turbine has the following

data : Degree of reaction = 50% Exit air angle of the fixed blade ring = 70° Mean diameter of the stage = 1 m Rotational speed = 3000 rpm Power developed = 5 MW

Assuming maximum utilization factor determine :A. Blade to gas speed ratio

B. Utilization factor

C. Flow coefficient

D. Inlet and exit air angles for the rotor

E. Mass flow rate of gas

solution Given :

R = 50%; 2 = 70°; dm = 1 m;

N = 3000 rpm; Power = 5 MW

1. Blade to gas speed ratio

opt = sin2 = sin70 = 0.9397

2. Utilization factor

2 22

max 2 22

2sin 2 sin 700.937

1 sin 1 sin 70

x

m

opt 22

2

x 2 2

c3) Flow coefficient = .

ud 1 3000

= 60 60157.1 m/s

157.1167.2 m/s

0.9397c cos (167.2 cos 70) 57.19 m/s

57.190.364

157.1

opt

x

Nu

u

u uc

c

c

c

c

u

22

2

3 2

3

3

.

2 3

.

2

tan 57.19 tan 70 157.14) tan

57.19

0 (Ans.)

Degree of reaction

1 1R = (tan tan )

2 2 257.19

0 (tan tan 70)2 157.170 (Ans.)

5) Power = 5MW = m ( )

157.1 [( tan ( tan

x

x

x

y y

x x

c u

c

c

u

u c c

m c c

63

.6

.

)] 5 10

157.1 [(57.19 tan 70) (57.19 tan 70 157.1)] 5 10

202.52 kg/s

u

m

m

u

C2=w3

Cx2 w2

u

2

β2

Cx3

w3

C3

3

β3

u

Radial Turbines

Inward flow radial (IFR) turbine Applications: automotive turbochargers, aircraft auxiliary

power units, expansion units in gas liquefaction and cryogenic systems

over a limited range of specific speed, IFR turbines provide an efficiency about equal to that of the best axial-flow turbines.

The significant advantages offered by the IFR turbine compared with the axial-flow turbine is

the greater amount of work that can be obtained per stage

the ease of manufacture and its superior ruggedness.

In the centripetal turbine energy is transferred from the fluid to the rotor in passing from a large radius to a small radius.

For the production of positive work the product of ucθ at entry

to the rotor must be greater than ucθ at rotor exit

This is usually arranged by imparting a large component of

tangential velocity at rotor entry, using single or multiple

nozzles, and allowing little or no swirl in the exit absolute

flow.

If the KE at rotor exit is high, a part of it can be recovered by

passing the gas through an exhaust – diffuser whose action is

like that of a draught tube in a hydroturbine.

Types of inward flow radial turbineCantilever turbine

r2 2 2 2

r2 2 r2 2 2

22 2

2

r22

2 2

2 2 2 22

2 2 2 2 2

3 3 33

3 3 3 3 3

From inlet velocity triangle

c (cot cot )

c cot c cot

cot cot

cAlso, tan

c

sin sintan

cos

Similarly

sintan

cos

r

r

u

u

u

c

u

c c

c u c u

c c

c u c u

Stage work2 2 3 3

3

2 2 2 2 2

2 2 2 2 22 22 2 2

22

2 22 2 2

2 2

w = u

For zero exit swirl, c 0

w = u cos

Pr coefficient

cos cos

cossin

cot cot

r

r

c u c

c u c

essure

u c cw

u u u

cc

u u

2 2 2 2

2 22

2 2

2

2

2 2

Also,

c cot

1 cot

cFor a rotor with zero exit swirl,

u

1 cot

r

r

u c

c c

u u

2 2 2 3 3 3

.

2 2 2 2 2 3 3 3 3 3

2

2

Continuity equation :

m = =

m ( ) ( )

r r

r r

r

c A c A

V d nt b V d nt b

Flow coefficient

c

u

The 90 degree IFR turbine

Axial discharge, c3= 0

w = u2c2 = u2c2cos2 = u22

Radial (shock less) entry, c2 = u2,

Ψ = w/u22 = 1

.2 2

2 2 2 2 2 3 3 3 3

3

m ( ) [ ( ) ]4

Blade exit height

1b ( )

2

r x t h

t h

c d nt b c d d nt b

d d

01 02

2 21 2

1 2

01 02

2a 02 03 2 3

Since there is no work done in the nozzle:

h h

2 2Also, assuming no pressure drop in the nozzle

p p

Actual work transfer = Actual change in stagnation enthalpy

1w h h (

2

c ch h

c c

2 2 2 2 22 3 3 2

2 2 2 2 2 202 2 2 2 03 3 3 3

2 202,rel 2 03,rel 3

1 1) ( ) (w -w )

2 21 1 1 1 1 1

(h c ) w u (h c ) w u2 2 2 2 2 2

1 1 h u h u

2 2

u u

Stage Efficiency

3

a 01 03 02 03 2 2 3 3 2 2

a 2 2 2

2 2a 01 03 02 03 2 2

2

a 02

Total-to-Total Efficiency

(Assuming zero swirl at the exit, i.e., c 0)

w

w (1 cot )

Assuming a perfect gas

w ( ) ( ) (1 cot )

w (

r

rp p

p

h h h h u c u c u c

u c

cc T T c T T u

u

c T

2 203 2 2 2 2) (1 cot )T u u

03s 01 03 01

01

1

03s 01

01

2 2a 2 2

1 1s 03

01 01 001

Isentropic work :

w [1 ( )]

w [1 ( ) ]

w

w[1 ( ) ] [1 ]

ssss p

p

tt

p p r

Th h c T

T

pc T

p

u u

pc T c T p

p

Total-Static Efficiency

3

a 01 3 02 3 2 2 3 3 2 2

a 2 2 2

2 2a 01 3 02 3 2 2

2

2 2a 02 3 2 2 2 2

(Assuming zero swirl at the exit, i.e., c 0)

w

w (1 cot )

Assuming a perfect gas

w ( ) ( ) (1 cot )

w ( ) (1 cot )

r

rp p

p

h h h h u c u c u c

u c

cc T T c T T u

u

c T T u u

3s 01 3 01

01

1

3s 01

01

2a 2

1s 01

013

Isentropic work :

w [1 ( )]

w [1 ( ) ]

w

w[1 ( ) ]

ssss p

p

ts

p

Th h c T

T

pc T

p

u

pc T

p

Spouting Velocity It is defined as that velocity which has an associated kinetic

energy equal to the isentropic enthalpy drop from turbine inlet stagnation pressure p01 to the final exhaust pressure.

The exhaust pressure here can have several interpretations depending upon whether total or static conditions are used in the related efficiency

Total

Static

20 01 03

2 030 01 03 01

01

12 030 01

01

1

030 01

01

1

2Assuming a perfect gas

1( ) (1 )

2

1[1 ( ) ]

2

c 2 [1 ( )

ss

ssp ss p

p

p

c h h

Tc c T T c T

T

pc c T

p

pc T

p

Total

3

20 01 3

1

30 01

01

With exit pressure equal to p

1

2

2 [1 ( )

ss

p

c h h

pc c T

p

Static

Degree of Reaction

2 2 2 22 3 3 2

R

0 2 2 3 3

2 2 2 2 2 22 2 2 2 2 2 2 2 2

2 2 23 3 3 3

1 1( ) (w w )( h) 2 2R =

( h )

From inlet triangle

w ( ) 2

Similarly from the exit triangle

w ( )

stage

r r

r

u u

u c u c

c c u c c u c u

c u c

2 2 2 2 2 2 23 2 3 3 3 2 2 2 2 2

2 2 2 2 2 2 2 23 2 3 3 3 3 3 2 2 2 2 2

3

2 2 2 2 23 2 3 2 2 2 2

2 2 2 2 22 3 3 2 2 2 2

2 2

w w ( ) [ 2 ]

w w 2 2

Assuming c 0,

w w (2 )

1 1( ) ( 2 )

2 2 R =

r r

r r

c u c c c u c u

c u c u c c c u c u

u u u c c

u u u u u c c

u c

22 2 2

2 2

2

2

2 2

1u

2R =

1R=1- ( )

2

12

1Also R = (1- cot )

2

c c

u c

c

u

R

MEE-304 TURBOMACHINESMEE-304 TURBOMACHINES

RADIAL TURBINESRADIAL TURBINES

NUMERICAL PROBLEMS – Set 8NUMERICAL PROBLEMS – Set 8

Example-1 A cantilever blade type IFR receives air at p01 = 3 bar, T01 = 373

K. Other data for this turbine are : Rotor tip diameter 50 cm Rotor exit diameter 30 cm Speed 7200 rpm Rotor blade width at entry 3 cm Air angle at rotor entry 60° Air angle at nozzle exit 25° Nozzle efficiency 97% Stage pressure ratio (p01/p3) 2.0

The radial velocity is constant and the swirl at the rotor exit is zero. Determine : 1)the flow and loading coefficients (2)the degree of reaction and stage efficiency (ηts) (3)the air angle and width at the rotor exit (4)the mass flow rate and power developed

Given :

p01 = 3 bar; T01 = 373 K;

dt = 0.5 m; N = 7200 rpm;

d3 = 0.3 m; b2 = 3 cm;

2 = 25°; β2 = 60°;

ηN = 97%,;

p01/p3 = 2.0.

Soln. Inlet Velocity Triangle

Exit velocity triangle

c2w2

cr2

2β2

c2

u2

cr3 = c3

w3

u3

β3

22

r2

r2

r2

22

2

2 2

1) flow and loading coefficient

0.5 7200u 188.5 m/s

60 60 60c (cot 25 cot60) = 188.5

(2.145 0.58) 188.5

120.44 m/s

120.440.639

188.5

1c cot 25 = 120.44 258.3 m/s

tan25

Si

t

r

r

d Nd N

c

c

c

u

c

2 2 3 3 22 32

2 2

u c u 258.3milarly 1.37 [ c 0]

u 188.5

c c

u

2. Reaction

2 2

1R = (1 cot )

21

(1 0.639cot 60)2

0.3155 31.55%

Total to static efficiency

02 03 01 03

02 3 01 3

10.28701 01

33 3 3

2 2 01 03

2 203 01

ts

373( ) (2.0) 305.7 K

u ( )

188.5 258.3373 324.55

1005

373-324.6 48.471

373 305.7 67.3

tss s

ss s

p

p

h h T T

h h T T

p TT

p T T

c c T T

u cT T K

c

.92%

33

3

3

2 2 2 2 3 3 3 3

rotor exit

120.44tan

113.1

46.8

( ) ( )

O

r

r r

air angle at

c

u

Width at rotor exit

c d b c d b

2222 2

2p

2

2

2

532

22

3 3

533

33

( )c sin

T 373 3732c 2

120.44( )

sin 25T 3732 1005

T 332.59 K

1.98 1.01325 102.1 kg/m

287 332.59

3Similarly p 1.5 bar; T 311.02 K

2

1.5 1.01325 101.7 kg/m

287 311.02

2.1

r

p

c

c

p

RT

p

RT

3

3

0.5 0.03 1.7 0.3

6.18 cm

b

b

.

3 3 3 3

.

.

02 03

( )=1.7 120.4 ( 0.3 0.0618)

m 11.92 kg/s

Power

P = m ( ) 11.92 1005 (373 324.6)

Power = 579.81 kW

r

p

mass flow rate

m c d b

c T T

Example-2 A single stage ninety degree IFR turbine fitted with an exhaust

diffuser has the following data : Overall stage pressure ratio 4.0 Temperature at entry 557 K Diffuser exit pressure 1 bar Mass flow rate of air 6.5 kg/s Flow coefficient 0.3 Rotor tip diameter 42 cm Mean diameter at Rotor exit 21 cm Speed 18000 rpm Enthalpy losses in the nozzles and the rest of the stage are equal.

Assuming negligible velocities at the nozzle entry and diffuser exit, determine : (a) the nozzle exit air angle (b) the rotor width at the entry (c ) the power developed (d) the stage efficiency (e) the rotor blade height at the exit (f) Mach numbers at nozzle and rotor (relative) exits and (g) the nozzle and rotor loss coefficients.