EDUSAT PROGRAMME 15 Turbomachines (Unit 3) Axial … · EDUSAT PROGRAMME 15 Turbomachines (Unit 3)...

EDUSAT PROGRAMME 15

Turbomachines (Unit 3)

Axial flow compressors and pumps

Axial flow compressors and pumps are power absorbing turbomachines. These

machines absorb external power and thereby increase the enthalpy of the flowing

fluid. Axial flow turbomachines use large quantity of fluid compared to mixed and

centrifugal type of turbomachines. However, the pressure rise per stage is lower in

case of axial flow turbomachines than mixed and centrifugal flow turbomachines.

Axial flow compressors are used in aircraft engines, stand alone power generation

units, marine engines etc.

Design of axial flow compressors is more critical than the design of turbines

(power generating turbomachines). The reason is that, in compressors the flow

moves in the direction of increasing pressure (adverse pressure gradient). If the

flow is pressurized by supplying more power, the boundary layers attached to the

blades and casing get detached and reverse flow starts which leads to flow

instability leading to possible failure of the machine. However, in turbines the fluid

moves in the decreasing pressure (favourable pressure gradient). To transfer a

given amount of energy more number of stages are required in compressors than in

turbines. Generally, the fluid turning angles are limited to 20˚ in compressors and

is 150˚ to 165˚ in case of turbines. Thus the pressure rise per stage is limited in

case of compressors. Fluid flows in direction of increasing pressure, this increases

the density of the fluid, thus the height of the blade decreases from the entrance

to the exit. Axial flow compressors have inlet guide vanes at the entrance and

diffuser at the exit.

Figure 1 shows the flow through the compressor and fig. 2 shows the

corresponding inlet and exit velocity triangles. Generally for a compressor the

angles are defined with respect to axial direction (known as air angles). It can be

seen that the fluid turning angle is low in case of compressors. Figure 3 shows the

flow through the turbine blade and fig. 4 the corresponding inlet and exit velocity

triangles for a turbine.

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Fig. 1 Flow across a compressor blade

Fig. 2 Inlet and exit velocity triangles

Fig. 1 Flow across a compressor blade

Fig. 2 Inlet and exit velocity triangles

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Fig. 3 Flow across a turbine blade

Fig. 4 Inlet and exit velocity triangles for turbine

Fig. 3 Flow across a turbine blade



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Expression for degree of reaction for axial flow compressor

Degree of reaction is defined as a measure of static

rotor expressed as a percentage of the total static enthalpy

Assumptions:

1. The analysis is for 2-dimensional flow

2. The flow is assumed to take

peripheral velocities at inlet and outlet are same and there is no flow in the

radial direction

3. It is common to express blade angles w.r.t. axial direction in case of axial

flow compressor (air angles)

4. Axial velocity is assumed to remain constant

Figure 2 shows the inlet and exit velocity triangles for axial flow compressors

Static

StaticR =

o∆h =

V1

tanγa

VU +=

(tanVU a=

tantanγtanγ 21 =−

2(h

VR

02

2

r1 −=


s defined as a measure of static enthalpy rise that occurs in the

rotor expressed as a percentage of the total static enthalpy across the rotor.

dimensional flow

The flow is assumed to take place at a mean blade height where blade


It is common to express blade angles w.r.t. axial direction in case of axial

flow compressor (air angles)

y is assumed to remain constant

Figure 2 shows the inlet and exit velocity triangles for axial flow compressors

stageinriseenthalpyStatic

rotorinriseenthalpyStatic

1)u1

Vu2

U(V −−−−=

-2-)0

tanγ1

(tanγa

V0

tanγa

V +=

3)tanγ(tanγ 32 −−+

4tanγtanγ 03 −−−−

5h

V

01

2

r2 −−−−

−

)


enthalpy rise that occurs in the

across the rotor.

place at a mean blade height where blade


It is common to express blade angles w.r.t. axial direction in case of axial

Figure 2 shows the inlet and exit velocity triangles for axial flow compressors.

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Substituting eq. 6 and 7 in eq.5

6)γtanγ(tanVVV

γtanVVV

γtanVVV

22

122

a

2

r2

2

r1

222

a

2

a

2

r2

122

a

2

a

2

r1

−−−−=−

+=

+=

7)tanγ(tanγUV∆h

tanγVVandtanγVV

)VU(Vhh

03a0

0au13au2

U1u20102

−−−−=

==

−=−

)tanγ(tanγ2UV

)γtanγ(tanVR

03a

22

122

a

−

−=

4tanγtanγtanγtanγ 0321 −−−−=−

10)tanβtanβ

tanβtanβ(

2U

VR

)tanβ

1

tanβ

1(

2U

VR

9)cotβ(cotβ2U

VR

82U

)tanγ(tanγVR

)tanγ(tanγ2UV

)γtanγ(tanVR

21

21a

21

a

21a

21a

21a

22

122

a

−−−+

=

+=

−−−+=

−−−−+

=

−

−=

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( ) ((

−==

+−

−=

VVWDE

VVV

VVR

2

2

1

r2

2

2

2

1

2

r1

2

r2

)) ( )

−+

−

2

VV

V

2

r1

2

r2

2

2

r1

2

2

ADARSHA H G

ADARSHA H G

Problem 1

In a mixed flow pump absolute fluid velocity at the inlet is

radial velocity at the exit. Inlet hub

250 mm. Pump speed is 3000 rpm. Find the degree of reaction and

input to the fluid, if the relative velocity at the

speed.The fluid leaves the rotor in the radial direction

D1= 80 mm, D2= 250 mm, N=3000 rpm

The next step is to draw the velocity triangles

Problem 2

The total power input at a stage in an axial flow compressor with symmetric inlet

and outlet velocity triangles (R=0.5) is 27.85 kJ/kg of air flow. If the blade speed

is 180m/s throughout the rotor,

blade angles. Do you recommend the use of such compressor?. Assume the axial

velocity component to be 120m/s.

V12.57m/s;UV 21r2 ==

39.239.2V-UWDE

12.57m/s;VUV

u22

11rd2

×===

===

( )

0.5R

E2

V(VUUR

2

r1

2

2

2

1

=

×

−−−=

In a mixed flow pump absolute fluid velocity at the inlet is axial and equal to

radial velocity at the exit. Inlet hub diameter is 80 mm and impeller tip diameter is

speed is 3000 rpm. Find the degree of reaction and

input to the fluid, if the relative velocity at the exit equals the inlet tangential blade

the rotor in the radial direction Given: V1 = Va1= Vrd2 ;

mm, N=3000 rpm, R = ?, WD=?, Vr2=U1

The next step is to draw the velocity triangles



is 180m/s throughout the rotor, draw the velocity triangles and compute the rotor

Do you recommend the use of such compressor?. Assume the axial

velocity component to be 120m/s.

41.23m/sVU2

r2

2

2 =+=

1.543KJ/kg39.2

17.78m/sUVV2

1

2

1r1

=

=+=

)V2

r2

axial and equal to

diameter is 80 mm and impeller tip diameter is

speed is 3000 rpm. Find the degree of reaction and the energy

exit equals the inlet tangential blade

Given: V1 = Va1= Vrd2 ;



draw the velocity triangles and compute the rotor

Do you recommend the use of such compressor?. Assume the axial

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Given: Symmetric velocity triangles with R = 0.5; P= 27.85 kJ/kg of air flow;

U=180m/s; Va=120m/s; β1=? and

Draw the velocity triangles.

Since the R=0.5 and velocity triangles are symmetrical

α1=β2; α2=β1; V1=Vr2 and V2=Vr1

Problem 3

The axial component of air velocity at the exit of the nozzle of an axial flow

reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27

Find the rotor blades angles if the degree of reaction is 50% and the blade speed is

180m/s. Also, for the same blade speed, axial velocity and nozzle angle find degree

of reaction, if the absolute velocity at the rotor outlet should be axial and equal to

axial velocity at the inlet.

Given: 1) Va1=180m/s, α1=27

2) U=180m/s, Va1=180m/s,

Sol: 1) R = 50%, V1=Vr2, V2=Vr2,

Inlet velocity triangle is shown in the figure

154.72m/s∆V

27850kJ/Kg)VU(∆WD

u

u

=

==

12.64X

(UX;∆VX2U u

=

==×−

12.64180

120tanβ

83.9812.64

120tanβ

1

1

12

=

−=

=

=

−

−

o48.34ββ 12 =−


1=? and β2=?; Do you recommend such compressor?

Draw the velocity triangles.

Since the R=0.5 and velocity triangles are symmetrical

1; V1=Vr2 and V2=Vr1


reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27


. Also, for the same blade speed, axial velocity and nozzle angle find degree


1=27˚, R = 50%, U=180m/s, β1=?, β2=?

) U=180m/s, Va1=180m/s, α1=27˚, R=?, if V2=Va2=Va1

Sol: 1) R = 50%, V1=Vr2, V2=Vr2, α1=β2 and α2=β1

Inlet velocity triangle is shown in the figure

27850kJ/Kg

)/2∆Vu−

o

o

35.64

83.98

=


2=?; Do you recommend such compressor?


reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27˚.


. Also, for the same blade speed, axial velocity and nozzle angle find degree


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From the exit velocity triangles

Turbines

1.Utilization factor is

2.Adiabatic efficiency is the

3.Overall efficiency is product of adiabatic efficiency and

efficiency

4.Mechanical efficiency of majority of TM’s is nearly 100%

5.Therefore, overall efficiency is almost equal to

6. However, adiabatic efficiency is product of utilization factor

efficiency) and efficiency associated with various losses

7.Utilization factor deals with what is maximum

obtained from a turbine

8.Utilization factor is the ratio of ideal work

for conversion to work

9.Under ideal conditions it should be possible to utilize all the K.E. at inlet

and increase the K.E. due to reaction effect

o27ββα

173.26

180tanβ

173.26m/sUcosαV

V180;sinαV

221

11

11

111

==

==

=−

==

−

( ) (VVV

VVR

V396.48m/sV

180173.26V

254.55m/sV,45β

2

r2

2

212

2

r1

2

r2

21

2

r1

r22

+−

−=

==

+=

== o

From the exit velocity triangles

Turbines- Utilization factor

is defined only for PGTM- Turbines

Adiabatic efficiency is the quantity of interest in turbines

Overall efficiency is product of adiabatic efficiency and mechanical

Mechanical efficiency of majority of TM’s is nearly 100%

Therefore, overall efficiency is almost equal to adiabatic efficiency

However, adiabatic efficiency is product of utilization factor

efficiency) and efficiency associated with various losses.

Utilization factor deals with what is maximum energy that can be

obtained from a turbine without considering the losses in the turbine

Utilization factor is the ratio of ideal work output to the energy available

to work

Under ideal conditions it should be possible to utilize all the K.E. at inlet

ncrease the K.E. due to reaction effect

o46.09

173.26m/s

396.48m/s=

)0.0198

V

180m/s

249.83m/s180

254.55m/s

2

r1

2

2

=−

=

=

mechanical

efficiency

However, adiabatic efficiency is product of utilization factor (diagram

energy that can be

ut considering the losses in the turbine

output to the energy available

Under ideal conditions it should be possible to utilize all the K.E. at inlet

ADARSHA H G

10. The ideal energy available for conversion into work

11.The work output given by Euler’s Turbine Equation is

11. Utilization factor is given by

12. Utilization factor for modern TM’s is between 90% to 95%

Relation between utilization factor and degree of reaction

Utilization factor is given by

The degree of reaction is given by

Substituting the value of X in the expression for utilization factor and

simplifying

( ) ( )[ ]2

VVUUVw

2

r2

2

r1

2

2

2

1

2

1a

−−−+=

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]( ) ( )[ ]2

r2

2

r1

2

2

2

1

2

1

2

r2

2

r1

2

2

2

1

2

2

2

1

a

2

r2

2

r1

2

2

2

1

2

2

2

1

VVUUV

VVUUVV

w

wε

2

VVUUVVw

−−−+

−−−+−==

−−−+−=

( ) ( ) ( )[ ]( ) ( )[ ]2

r2

2

r1

2

2

2

1

2

1

2

r2

2

r1

2

2

2

1

2

2

2

1

a VVUUV

VVUUVV

w

wε

−−−+

−−−+−==

( ) ( )( ) ( ) ( )

( ) ( )E2

VVUU

VVUUVV

VVUUR

2

r2

2

r1

2

2

2

12

r2

2

r1

2

2

2

1

2

2

2

1

2

r2

2

r1

2

2

2

1

×

−−−=

−−−+−

−−−=

( ) ( )

( )( )

R1

VVRX

XVV

XR

VVUUX

2

2

2

1

2

2

2

1

2

r2

2

r1

2

2

2

1

−

−=

+−=

−−−=

( )( )2

2

2

1

2

2

2

1

RVV

VVε

−

−=

ADARSHA H G

The above equation is valid for single rotor under the conditions where Euler’s

turbine equations are valid.

equation is valid in the following range of

Maximum Utilization factor


Utilization factor maximum if the exit absolute velocity is minimum. This is

possible when the exit absolute velocity is in axial direction

Maximum utilization factor is given by

(( 1V

Vε =

m

m

12

1ε

V

Vε

VV

=

=

=

m1

cosε

−=


The above equation is invalid when R = 1

equation is valid in the following range of R 0 ≤ R < 1

Maximum Utilization factor



possible when the exit absolute velocity is in axial direction

Maximum utilization factor is given by

))2

2

2

1

2

2

2

1

RV

VV

−

−

12

12

122

1

2

1

122

1

2

1

11

αRsin

αcos

αsinRVV

αsinVV

sinα

−

−

−

1

2

12

αRsin

αcos

−0αwhen1ε 1m ==


= 1. The above


ADARSHA H G

Maximum Utilization factor for impulse turbine

For an impulse turbine R = 0, thus

From the velocity triangles OAB and OBD are

similar. Thus AB = U,

α1 is made as small as possible(15

Maximum Utilization factor for 50% reaction turbines

50% reaction turbines have V1=Vr2, V2=Vr1,

maximum utilization V2 must be in axial direction. The corresponding

velocity triangles are

Speed2

cosα

V

U

UUcosαV

1

1

11

==

=+=

φcosαV

U

cosαVAlso,

1ε0.5,R

αRsin1

αcosε

1

1

11

m

1

2

1

2

m

===

=

==

−=


For an impulse turbine R = 0, thus

From the velocity triangles OAB and OBD are

similar. Thus AB = U,

is made as small as possible(15˚-20˚)


50% reaction turbines have V1=Vr2, V2=Vr1, α1=β2 and α2=


12

m αcosε =

φRatioSpeed

2U

=

=

ratiospeed

U

α0.5sin

αcos

12

1

2

1

=

=

−



2=β1 and for


ADARSHA H G

Comparison between Impulse and 50% reaction

A) When both have same blade speed

Let UI and UR be the blade speeds of impulse and 50% reaction

turbines. The velocity triangles for maximum utilization are

Comparing eq. 1 and 2, it is clear that impulse turb

of energy per unit mass than 50% reaction turbin

utilization is maximum.

• However, 50% reaction turbines are more efficient than impulse turbines

• But 50% reaction turbines transfer half the en

turbines.

• If only 50% reaction turbines are used more stages are required or if only

impulse turbines are used stages are less but efficiency is low

• In steam turbines where large pressure ratio is available it is common to use

one or two impulse stages followed by reaction stages

UE

VBut

UE

50%For

2UE

From

UE

R

R

I

1I

=

=

=

=

Comparison between Impulse and 50% reaction turbine at

maximum utilization

When both have same blade speed



Comparing eq. 1 and 2, it is clear that impulse turbine transfers twice the amount

of energy per unit mass than 50% reaction turbine for the same blade speed when

However, 50% reaction turbines are more efficient than impulse turbines

But 50% reaction turbines transfer half the energy compared to impulse

If only 50% reaction turbines are used more stages are required or if only

impulse turbines are used stages are less but efficiency is low.

In steam turbines where large pressure ratio is available it is common to use

ne or two impulse stages followed by reaction stages.

2U

UV

VU

turbinereaction50%

12U

2UVtrianglevelocity

V

2

R

Ru1

u1R

2

I

Iu1

u11

−−−

=

−−−

=

turbine at



ine transfers twice the amount

e for the same blade speed when

However, 50% reaction turbines are more efficient than impulse turbines.

ergy compared to impulse

If only 50% reaction turbines are used more stages are required or if only

In steam turbines where large pressure ratio is available it is common to use

ADARSHA H G


b) When both have same energy transfer

c) When V1 and α1 are same in both TM’s

Speed ratio for impulse and 50% reaction stage for maximum utilization

Problem 1

Find an expression for the utilization factor for an axial flow impulse turbine stage

which has equiangular rotor blades, in terms fixed inlet blade angle and speed ratio

ф.

Given: Equiangular rotor blades

turbine R = 0, Vr1=Vr2, find expression

for utilization factor in terms of

From the velocity triangles

1.414U2UU

2UU

EE

I

2

IR

2

I

2

R

IR

==

=

=

62UU

U;cosαφV

U

2U;2

cosαφ

V

U

IR

R1

I

R

I1

I

I

−−=

==

==

r222

r1

2

1

r222

r2

2

2

cos2UVUVV

cos2UVUVV

−+=

−+=


maximum utilization

b) When both have same energy transfer

1 are same in both TM’s




Given: Equiangular rotor blades β1=β2, axial flow turbine U1=U2=U, Impulse

turbine R = 0, Vr1=Vr2, find expression

for utilization factor in terms of α1 and ф.

3I −−−

5cosαV

4cosα

1I

1I

−−=

−−=

( )2

2

β180cos

cosβ

−





2, axial flow turbine U1=U2=U, Impulse

ADARSHA H G

Problem 2 For a 50% degree of reaction axial flow turbomachine, the inlet fluid velocity is 230m/s; outlet angle of inlet guide blade 3060˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial thrust and the power output/unit maGiven: R=50%; V1=230m/s;

ε = ?; Fax=?; P = ?

Although R = 50%, α1≠

Inlet and exit velocity triangles are

drawn

cosβ4UVVV 1r1

2

2

2

1 =−

(

1

1

1

2

1

2

2

2

1

2

2

2

1

2

2

2

1

V

Uφ where,

φ)(cosα4φε

cosV4U

V

VVε

RVV

VVε

=

−=

=−

=

−

−=

132.8m/scosβVcosαVU

132.8m/ssinβ

VV

115m/ssinαVV

1r111

1

a1r1

11a1

=−=

==

==

( ) (

233.6m/sV

26450120.36VV

17635.8240.72V2V

cos2UVUVV

132.8VVVV

VVVV

VVR

r2

r2

2

r2

r2

2

r2

r2

22

r2

2

2

2

2

2

r2

2

1

2

r1

r1

2

r2

2

2

2

1

2

r1

2

r2

=

−−

+−

−+=

=+=+

−+−

−=

For a 50% degree of reaction axial flow turbomachine, the inlet fluid velocity is 230m/s; outlet angle of inlet guide blade 30˚; inlet rotor angle

˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial thrust and the power output/unit mass flow. Given: R=50%; V1=230m/s; α1=30˚; β1=60˚; β2=25˚

≠ β2; as Va1≠Va2

Inlet and exit velocity triangles are

( )UcosαV4U 11 −=

)2

1

1

V

Ucosα −

132.8m/s

)

026450

70535.817635.8

cosβ

70535.8230132.8

0.5

2

22

2

r1

=

=

=+

=

For a 50% degree of reaction axial flow turbomachine, the inlet fluid ˚; inlet rotor angle

˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial

ADARSHA H G

0.822126.360.5230

126.36230

RVV

VVε

126.36m/sV

70535.8VV

22

22

2

2

2

1

2

2

2

1

2

2

2

2

r2

=×−

−=

−

−=

=

=+

( )

( )

s

kg

N16.28

m

F

sinβVSinαVVVm

F

VVmThrustAxial

ax

2r211a2a1ax

a2a1

=

−=−=

−=

&

&

&

( )

( ) 37KJ/kg/s78.91199.98132.8m

P

78.91m/sUcosβVV

199.98m/scosαVV

VVUmP

2r2u2

11u1

u2u1

=+=

=−=

==

−=

&

&

ADARSHA H G

UNIT 4

Thermodynamics of fluid

Sonic velocity and Mach number

1.Sonic velocity is the speed of propagation of pressure wave in a medium.

2.The speed of sound in a fluid at a local temperature for isentropic flow is given

by where γ is the ratio of specific heats = 1.4, R is characteristics gas constant =

287 J/kg K and T is the local temperature in kelvins. At 15˚ the speed of sound is

340 m/s.

3.As altitude increases temperature decreases and speed of sound decreases.

4.Mach Number is defined as the ratio of local velocity of fluid to the sonic

velocity of sound in that fluid

5. Many turbines and compressors experience high Mach numbers

6.High Mach numbers give rise to some special problems such as shock waves

which leads to irreversibility and cause loss in stagnation pressure and increase in

entropy

7.Using continuity equation, Euler’s equation and isentropic equation following

two equations are derived

8.The above equations decide the variations in velocity, pressure and area for

different Mach numbers

9.The three basic classifications are a) Subsonic flow M<1

b) Sonic flow M=1 c) Supersonic flow M>1

γRT

V

c

VM ==

γRTc =

2γM

M1

p

dp

A

dA

1Mγp

dp

V

dV

2

2

2

−−−

−=

−−−−=

ADARSHA H G

Classification fluid flow based on Mach number

a) Subsonic flow (M<1) :

Nozzle

Diffuser

anddecreasesarea

negativeisp

dpwhen

andincreasesarea

increasesisp

dpwhen


Subsonic flow (M<1) :

increasesvelocityand

negativeisA

dAnegative

γ

1

p

dp

A

dA

γp

dp

V

dV

=

−=

decreasesvelocityand

positiveisA

dAincreases


2γM

M1

1M

dp

2

2

2

−−−

−

−−−

ADARSHA H G

b) Supersonic flow (M>1) :

Divergent nozzle

Convergent Diffuser

velocityandincreasesarea

negativeisp

dpwhen

velocityanddecreasearea

positiveisp

dpwhen

Supersonic flow (M>1) :

γM

1

p

dp

A

dA

Mγp

dp

V

dV

−=

−=increasesvelocity

positiveisA

dA

decreasesvelocity

negativeisA

dApositive

2γM

M

1M

dp

2

2

2

−−−

−

−−−

ADARSHA H G

C) Sonic flow (M=1) :

At the throat portion of the convergent divergent nozzle the velocity is sonic.

Static and Stagnation states

Turbomachines involve the use of compressible and incompressible fluids. In compressible TM’s fluid move with velocities more than Mach one at many locations. In incompressible TM’s the fluid velocities are generally low, however, K.E. and P.E. of the moving fluid are very large and cannot be neglected To formulate equations based on actual state of the fluid based on laws of thermodynamics two states are used.

The states are static state and stagnation states.

Static State If the measuring instrument is static with respect to the fluid, the measured quantity is known as static property. The measured static property could be pressure, velocity temperature, enthalpy etc.,. The state of the particle fixed by a set of static properties is called static state.

Stagnation State

It is defined as the terminal state of a fictitious, isentropic work-free and steady flow process during which the final macroscopic P.E. and K.E. of the fluid particle are reduced to zero. Real process does not lead to stagnation state because no real process is isentropic. Stagnation property changes provide ideal value against which the real machine performance can be compared. It is possible to obtain stagnation properties in terms of static properties by using the definition of stagnation state. Consider the steady flow process given by the first law of thermodynamics.

sonicisvelocityandconstantisarea

zeroisA

dA

2γM

M1

p

dp

A

dA

1Mγp

dp

V

dV

2

2

2

−−−

−=

−−−−=

ADARSHA H G

Static state is the initial state in a fictitious isentropic work free , steady flow process and the stagnation state is the terminal state in which the ke and pe are reduced to zero, one can define a stagnation state at the initial static state.

In the above eq. subscript o represents stagnation state and subscript i represents initial static state If subscript i is removed from the initial static state Thus stagnation state has been expressed as the sum of three static properties. Since the process is isentropic, the final entropy is same as initial enthalpy. Final entropy is stagnation and initial entropy is static. Any two independent properties at a specific state is sufficient to fix the state of simple compressible substance by using thermodynamic relations.

a) Incompressible Fluid: (Density is constant)

Final state is stagnation with subscript o and initial state is Static without subscript

∆pe∆ke∆hwq

gZ2

VhwgZ

2

Vhq o

2

ooi

2

ii

++=−

+++=

+++

∆pe∆ke∆hwq ++=−

iooo hh∆h0;pe0;ke0;w0;q −=====

( ) 1−−−++= iiio pekehh

( ) 2pekehho −−−++=

3---sso =

∫∫ =

=

==

−=⋅

oo

o

dpρ

1dh

vdpdh

ssbecause0dsbut

vdpdhdsT ADARSHA H G

Thus stagnation pressure of an incompressible fluid is expressed in terms of static

pressure, velocity and height above a datum line.

Thus using thermodynamic relations stagnation pressure, Stagnation pressure,

stagnation temperature and stagnation internal energy are found.

Problem 1

Liquid water at standard density flows at a temperature of 20˚, a static pressure of 10 bar and a velocity of 20m/s. Find the total pressure and total temperature of the water. Given: T=20˚, p=10 bar and V=20m/s, To=? and Po=? , Water is incompressible with ρ = 1000 kg/m3

( )

( )( )

pgZ)2

Vρ(p

phpekehρp

phhρp

ppρ

1hh

2

o

o

oo

oo

++=

+−++=

+−=

−=−

TT0;dT

dTCduAlso,

uu0;du

constantρ0dv

pdvdu

ssas0ds

pdvdudsT

o

v

o

o

==∴

=

==∴

==

−=

==

+=⋅

o20TT

12barp

10102

201000p

pgZ)2

Vρ(p

o

o

52

o

2

o

==

=

×+

=

++=ADARSHA H G

Problem 2

A turbomachine handling liquid water is located 8m above the sump level and

delivers the liquid to a tank located 15m above the pump. The water velocities in

the inlet and outlet pipes are 2m/s and 4m/s respectively. Find the power required

to drive the pump if it delivers 100 kg/min of water.

Given: z1=8m; z2=15m; V1=2m/s; V2=4m/s; mass flow rate=100 kg/min

Find the Power P=?

b) Perfect gas

( )

( )

W386wmPPower

J/kg231.6pumponwork

J/kg231.68159.812

240w

zzg2

VV

ρ

ppw

ρ

∆p∆hqw

22

12

2

1

2

212

oo

===

=

−=

++

−+−=

−+

−+

−−=

−=−=

&

1γ

Rγc

cgeliminatin

Rccandγc

c

p

v

vp

v

p

−=

=−=

p

2

o2c

VTT +=

ADARSHA H G

Substituting the value of cp in the above equation and simplifying

Efficiencies of Turbomachines

Efficiency of a turbomachine is given by

( )

−+=

2

M1γ1TT

2

o

( )

( ) 1γ

γ2

o

γ

1

o

o

2

oo

ooo

o

oo

2

M1γ1

p

p

p

p

v

v

2

M1γ1pvvp

pv

vp

T

T

T

pv

T

vp

−

−+=

=

−+=

=

=

a

m

ma

ma

ηη

100%almostareη

efficiencymechanicalη;efficiencyadiabaticηwhere,

ηηη

≈∴

==

= ADARSHA H G

The adiabatic efficiency of a TM can be calculated from the h

the expansion and compression process. The ideal

using either static or stagnation states.

a) Power Generating Turbomachines (PGTM)

Actual work output for PGTM =

The proper equation is determined by the conditions of Turbomachine in question.

For example in a turbine if the inlet

production of mechanical energy somewhere else, then static to total definition is

used. If the exit ke is wasted then static to static definition is used.

48;h50;hLet 101 ==

35

30

hh

hhη

02'01

0201tt ==

−

−=−

33

30

hh

hhη

02'1

0201ts =

−

−=−

The adiabatic efficiency of a TM can be calculated from the h-s diagram for both

the expansion and compression process. The ideal work input or output can be

r static or stagnation states.

Power Generating Turbomachines (PGTM)

Actual work output for PGTM =


For example in a turbine if the inlet ke is negligible and exit ke is used for


is wasted then static to static definition is used.

0201 hh −

1

01s-s2'1ss

1

01ts02'1ts

01

01st2'01st

01

01tt02'01tt

h

hηhhw

h

hηhhw

h

hηhhw

h

hηhhw

=−=

=−=

=−=

=−=

−

−−

−−

−−

h10;h15;h20;h48; 202'02 ===

85.71% 66.67%45

30

hh

hhη

2'01

0201st ==

−

−=−

90.9%= 69.76%43

30

hh

hhη

2'1

0201s-s ==

−

−=

diagram for both

work input or output can be


is used for


2'1

0201

02'1

0201

2'01

0201

02'01

0201

h

h

h

h

h

h

h

h

−

−

−

−

−

−

−

−

5h2' =

66.67%

69.76%

ADARSHA H G

b) Power Absorbing Turbomachines (PATM)

Actual work input for PATM =

Problem 1

Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 atm., and a static temperature of temperature of air at this point inRepeat the problem with a flow velocity of 500 m/s.Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K Find (a) To = ? and po=? (b) To = ? and po=? Solution: Equations used are

p

2

o2c

VTT += o

p

p

bar2.07p

2871.4

60

γRT

VM

301.79K10052

60300T

0

2

0

=

×==

=×

+=

Power Absorbing Turbomachines (PATM)

Actual work input for PATM =

Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 atm., and a static temperature of 300 K. (a) Find total pressure and total temperature of air at this point in the duct. Assume ratio of specific heats as 1.4. (b) Repeat the problem with a flow velocity of 500 m/s. Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K Find (a) To = ? and po=? (b) To = ? and po=? Solution: Equations used are

0102 hh −

s-s12'ss

ts102'ts

st012'st

tt0102'tt

h

hηhhw

h

hηhhw

h

hηhhw

h

hηhhw

=−=

=−=

=−=

=−=

−

−−

−−

−−

( ) 1γ

γ2

2

M1γ1

−

−+=

0.1728300

301.79K

=×

Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 300 K. (a) Find total pressure and total the duct. Assume ratio of specific heats as 1.4. (b)

Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K

0102

12'

0102

102'

0102

012'

0102

0102'

hh

hh

hh

hh

hh

hh

hh

hh

−

−

−

−

−

−

−

−

ADARSHA H G

Problem 2 Air enters a compressor at a static pressure of 15 bar, static temperature of 15and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature of 100˚C and flow velocity of 10isentropic change in total enthalpy and b) Actual change in total enthalpy.Given: p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100z1=1m. Find a) Isentropic change in total enthalpySolution: Plot the h-s or T-s diagram as shown in fig.p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100

bar6.9p

2871.4

500

γRT

VM

424.38K10052

500300T

0

2

0

=

×==

=×

+=

(

++=

−=−

p

2

p

2

22'p

0102'p0102'

c

gz

2c

VTc

TTchh

inchangeIsentropic

T;p

p

T

T2'

γ

1γ

1

2

1

2'

=

−

(Tchh 02'p0102' −=−

Air enters a compressor at a static pressure of 15 bar, static temperature of 15and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature

˚C and flow velocity of 100 m/s. The outlet is 1 m above the inlet. Find a) isentropic change in total enthalpy and b) Actual change in total enthalpy.Given: p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100˚C; V2=100m/s; z2

Find a) Isentropic change in total enthalpy, b) Actual change in total enthalpys diagram as shown in fig.

p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100˚C; V2=100m/s; z2

1.44300287

500

424.38K

=×

)

++−

−=

p

1

p

2

11

01

0102'

c

gz

2c

VT

hhenthalpytotalin

351.07K=

) KJ/kg67084T01 =−

Air enters a compressor at a static pressure of 15 bar, static temperature of 15˚C and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature

0 m/s. The outlet is 1 m above the inlet. Find a) isentropic change in total enthalpy and b) Actual change in total enthalpy.

˚C; V2=100m/s; z2-

tual change in total enthalpy

˚C; V2=100m/s; z2-z1=1m

ADARSHA H G

Finite stage efficiency

1. A stage with a finite pressure drop is a finite stage 2. In a multi-stage turbine along with the overall isentropic efficiency the

efficiencies of individual stages are important 3. On account of large pressure drop and associated thermodynamic effect the

overall isentropic efficiency is not a true index of aerodynamic or hydraulic performance of machine

4. Different stages with the same pressure drop located in different regions of h-s plane will give different values of work output

Effect of Reheat (Turbines)

Consider four number of stages between two states as shown in fig. It is assumed that the pressure ratio and stage efficiency are same for all the four stages.

The actual work during expansion from state 1 to state 2 is

( )

++−

++=

−=−

−=

p

1

p

2

11

p

2

p

2

22p

0102p0102

0102

c

gz

2c

VT

c

gz

2c

VTc

TTchh

hhenthalpytotalinchangeActual

( ) KJ/kg89099.8TTchh 0102p0102 =−=−

s

aT

2

z

z

y

y

x

x

1

W

WηefficiencyOverall

1---Constantp

p

p

p

p

p

p

p

==

====

2WηW sTa −−−=

ADARSHA H G

The values of ideal or isentropic work in the stages are

The total value of actual work done in these stages are

Equating eq. 2 and eq. 3

The slope of constant pressure lines on h

s2s1 ∆,∆W,∆W

η∆WW staa ==∑2WηW sTa −−−=

W

∆W

ηη

ηWηW

s

4

1istT

stsTa

=

==

∑=

Ts

h

p

−=

∂

∂

The values of ideal or isentropic work in the stages are

The total value of actual work done in these stages are

Equating eq. 2 and eq. 3

The slope of constant pressure lines on h-s plane is given by

s4s3 ∆W,∆W

( )∆W∆W∆W∆W s4s3s2s1st −−−+++

2

4

W

∆W

si

4

1i

sist

−−−

∑=

5−−−

3−

ADARSHA H G

The above equation shows that constant pressure lines must diverge towards the

right

This makes the overall efficiency of the turbine greater than the individual stage

efficiencies

The quantity is known as reheat factor and is always greater than

one.

Reheat is due to the reappearance of stage losses as increased enthalpy during

constant pressure heating process.

Infinitesimal stage efficiency or Polytropic efficiency

(Turbines)

• To obtain the true aerodynamic performance of a stage the concept of small or infinitesimal stage is used

• This is an imaginary stage with infinitesimal pressure drop and is therefore independent of reheat effect

• Fig. shows a small stage between pressures p and p-dp The efficiency of this stage is

6---1W

∆W

s

4

1i

si

>∴∑

=

stT ηη >

s

4

1i

si

W

∆W∑=

1>=stη

ηRF

s

pdT

dT

dropetemperaturisentropic

dropetemperaturactualη ==

ADARSHA H G

For infinitesimal isentropic expansion

Expanding the terms on r.h.s. using bionomial expression and neglecting terms

beyond second

γ

1γ

s

γ

γ

s

p

dp1

T

dT1

p

dpp

T

dTT

−

−

−=−

−=

−

−−=−

γ

1γ1

T

dT1 s

ηp

dp

γ

1γ

T

dT

dp1γ

γ

T

dTη

1dT

dTηBut

p

dp

γ

1γ

T

dT

p

s

p

s

−=

−=∴

−−−=

−=

For infinitesimal isentropic expansion

the terms on r.h.s. using bionomial expression and neglecting terms

1

γ

1−

p

dp1

2η

dp

p

1

p −−−

the terms on r.h.s. using bionomial expression and neglecting terms

ADARSHA H G

Integrating the eq.2 between limits 1 and 2

The irreversible adiabatic (actual) expansion process can be considered as equivalent to a polytropic process with index n.

When ηp=1, n=γ the expansion line coincides with the isentropic expansion. The efficiency of a finite stage can now be expressed in terms of small stage efficiency. Taking static values of Tand p and assuming perfect gas

4p

p

T

T

3

p

plog

T

Tlog

1γ

γ

η

p

plogη

γ

1γ

T

Tlog

p

dp

γ

1γη

T

dT

p

2

1

2

1

ηγ

1γ

1

2

1

2

1

2e

1

2e

p

1

2ep

1

2e

p

p

p

T

T

−−−

=

−−−

−=

−=

−=

−

∫∫

( )6

η1γγ

γn

isprocessactualinexpansionofindexThe

51γ

γ

n

1nη

n

1nη

γ

1γ

indicesEquating

p

p

p

p

T

T

p

p

p

n

1n

1

2

ηγ

1γ

1

2

1

2

p

−−−−−

=

−−−

−

−=

−=

−

=

=

−

−

ADARSHA H G

Equations 7 and 8 can give the efficiencies of various finite expansion processes

with different values of pressure ratio and small stage efficiency

Problem 1

The overall pressure ratio through a 3 stage gas turbThe temperature at inlet is 1400 K. If the temperature rise in each stage is same, determine for each stage (a) pressure ratio (b) stage efficiency.Given: T1=1400 K; p1/p2=10; To find: pressure ratio for each stage and Draw the T-s diagram and assume the gas to be perfect gas

−=−

−

−=

121

2s1

21st

1TTT

TT

TTη

γ

γ

γ

1γ

st

12s1

pr1

pr1η

1TTT

−

−=

−=−

−

−−

ratiopressureif


with different values of pressure ratio and small stage efficiency

The overall pressure ratio through a 3 stage gas turbine is 10 and efficiency is 86%. The temperature at inlet is 1400 K. If the temperature rise in each stage is same, determine for each stage (a) pressure ratio (b) stage efficiency. Given: T1=1400 K; p1/p2=10; η=86%; ∆Tst= constant To find: pressure ratio for each stage and ηst

s diagram and assume the gas to be perfect gas

( )

−=

−

−pη

1γ

γ1

1

2

Pr

11T

T

T

( )

2

1

γ

1

η1

γ

1γ1

1

2s

p

pprwhere-7-;-

pr

11T

T

T

p

=

−=

−

−

−

8

pr1

pr1η;

p

pprratio

γ

1γ

ηγ

1γ

st

1

2

p

−−

−

−==

−

−


ine is 10 and efficiency is 86%. The temperature at inlet is 1400 K. If the temperature rise in each stage is same,

8

ADARSHA H G

First stage efficiency is given by

1013.1KTand1206.55KT

193.45K3

580.37∆T

sameisstageeachinriseeTemperatur

580.37KTT

819.6KT

725.1KT;10

1

p

p

T

T

0.86TT

TTη

yx

st

21

2

2'

1.4

0.4γ

1γ

1

2

1

2'

2'1

21

==

==

=−

=∴

=

=

=

=−

−=

−

( )

1.895p

p

0.5271400

1206.55

T

T

p

p

p

p

T

T

0.814

10

1log

1400

819.6log

0.4

1.4

p

plog

T

Tlog

1γ

γ

η

x

1

.40.814

1.4η1γ

γ

1

x

1

x

ηγ

1γ

1

x

1

x

e

e

1

2e

1

2e

p

p

p

=

=

=

=

=

=

×=

−=

×−

−

0.8η

p

pprwhere;

pr1

pr1η

st

x

1

γ

1γ

ηγ

1γ

st

p

=

=

−

−=

−−

−−ADARSHA H

G

Pressure ratio and stage efficiency of 2 stage is given by

Pressure ratio and stage efficiency of 3 stage is given by

Problem 2 In a 3 stage turbine the pressure ratio of each stage is 2 and stage efficiency is 75%. Calculate the overall efficiency and power developed if the air is initially at a temperature of 600˚C and flows through it at the rate of 25kg/s. Also Find the reheat factor. Given: ηst=75%; 3 stage turbine; pressure ratio across each stage=2; T1=600˚C; mass flow rate = 25 kg/s To find: η=?; P=?; RF=? Draw the T-s diagram

1013.1KT1206.55K;T yx ==

( )

0.4711206.55

1013.1

T

T

p

p .40.814

1.4η1γ

γ

x

y

x

y p

=

=

=

×−

2.12p

p0.471;

p

p

y

x

x

y==

y

x

γ

1γ

ηγ

1γ

stp

pprwhere;

pr1

pr1η

p

=

−

−=

−−

−−

0.797ηst =

0.7938η

2.487p

p0.402

p

p

819.6KT1013.1K;T

st

2

y

y

2

2y

=

==

==

ADARSHA H G

Finite stage efficiency (Compressor)

• A compressor with a finite pressure rise is known as a finite stage • Stage work is a function of initial temperature and pressure ratio • For the same pressure ratio, a stage requires a higher value of work with

higher temperature • Thus compressor stages in the higher temperature region suffer on account

of this The above factors have a cumulative effect on the efficiency of multistage compressor

( )( )73.07%η

0.750.51

0.5-1

0.75

pr1

pr1η

p

0.2857

η.2857

γ

1γ

ηγ

1γ

st

p

p

=

=−

=

−

−=

×

−

−

( )

1.0480.75

0.7862

η

ηRF

7724.25KW8

118731.00525TTcmP

78.62%η

p

p1

p

p1

p

p

p

p

p

p1

p

p

p

p

p

p1

TT

TTη

st

.2857.7307

21p

3γ

1γ

1

x

3ηγ

1γ

1

x

γ

1γ

y

2

x

y

1

x

ηγ

1γ

y

2

x

y

1

x

2'1

21

pp

===

=

−×××=−=

=

−

−

=

××−

××−

=−

−=

×

×−

×−

−

−

&

ADARSHA H G

Effect of preheat (Compressor)Consider a compressor with four stages as shown. It is assumed that all the stages have the same efficiencies and pressure ratios.

The total isentropic work from state 1 to 2s is WsThe isentropic work in the individual stages are The overall efficiency of the compressor is

This makes the overall efficiency of the compressor smaller than stage efficiency

This is due to the thermodynamic effect called ‘Preintentionally heated( preheated) at the end of eachin small constant pressure processes is only an internal phenomena and the compression process still remains an adiabatic process.

1

∆W

Wbut

∆Wη

1W

W

∆WηHowever

4

1i

si

s

4

1i

si

st

a

1x

s1st

<

=

==

∑

∑

=

=

η

Effect of preheat (Compressor) Consider a compressor with four stages as shown. It is assumed that all the stages have the same efficiencies and pressure ratios.

otal isentropic work from state 1 to 2s is Ws. The isentropic work in the individual stages are ∆Ws1, ∆Ws2, ∆Ws3 and The overall efficiency of the compressor is η=Ws/Wa


This is due to the thermodynamic effect called ‘Pre-reheating ; the gas is not intentionally heated( preheated) at the end of each compression stage. The preheat in small constant pressure processes is only an internal phenomena and the compression process still remains an adiabatic process.

W

∆W

W

∆W

W

∆W

z2

s4

yz

s3

xy

s2 == ;

( s2s1

st

a ∆W∆W∆Wη

1W ++==

stηη <

Consider a compressor with four stages as shown. It is assumed that all the stages

Ws3 and ∆Ws4


reheating ; the gas is not compression stage. The preheat

in small constant pressure processes is only an internal phenomena and the

)s4s3 ∆W∆W +

ADARSHA H G

Infinitesimal or Polytropic Efficiency (Compressor)• A finite compressor stage can be made up of infinite number of small stages • Each of these infinitesimal stages have an efficiency

efficiency or infinitesimal stage efficiency• It is independent of thermodynamic effect and is therefor

the aerodynamic performance of the compressor• Consider a stage in which air is compressed from state 1 to state 2. It also

shows an infinite stage operating between pressures

T

dT

T

dT1

using

T

dT1

T

dTT

dT

dTη

s

s

s

s

p

=

+

+

+

=

Infinitesimal or Polytropic Efficiency (Compressor)A finite compressor stage can be made up of infinite number of small stages Each of these infinitesimal stages have an efficiency ηp called polytropic efficiency or infinitesimal stage efficiency It is independent of thermodynamic effect and is therefore a true measure of the aerodynamic performance of the compressor Consider a stage in which air is compressed from state 1 to state 2. It also shows an infinite stage operating between pressures p and p+dp

p

dp

γ

1γ

p

dp

γ

1γ1

expansionbionomial

p

dp1

p

dpp

1dT

dT

γ

1γ

γ

1γ

s

−

−+=

+=

+=

−−

−

−

Infinitesimal or Polytropic Efficiency (Compressor) A finite compressor stage can be made up of infinite number of small stages

called polytropic

e a true measure of

Consider a stage in which air is compressed from state 1 to state 2. It also p+dp

ADARSHA H G

Substituting the value of dTs into equation 1

Integrating eq.2 between state 1 to 2

Assuming the irreversible adiabatic compression as equivalent to a polytropic

process with index n, equation 3 can be written as

The efficiency of finite compressor stage can be related to small stage efficiency

The actual temperature rise is given by

2η

1

p

dp

γ

1γ

T

dT

p

dp

γ

1γ

T

dTη

p

p

−−×

−=

−=

3

T

Tlog

p

plog

γ

1γ

ηorp

plog

η

1

γ

1γ

T

Tlog

1

2e

1

2e

p

1

2e

p1

2e −−

−

=

−=

( )4

η1γ1

γη

1n

n

γ

1γη

n

1n

γη

1γ

p

p

p

p

p

p

p

p

n

1n

2

1γη

1γ

2

1p

−−−−−

=−

×−

=

−=

−

=

−−

5

1pr

1pr

1T

T

1T

T

TT

TTη

p

pprwhere1prT1

T

TTTT

p

p

γη

1γ

γ

1γ

1

2

1

2s

12

12sst

1

2γη

1γ

1

1

2112

−−−

−

−=

−

−

=−

−=

=

−=

−=−

−

−

−ADARSHA H G

Problem1

A 16 stage axial flow compressor is to have a pressure ratio of 6.3, with a stage efficiency of 89.5%. Intake conditions are 288K and 1 bar. Find (a) Overall efficiency (b) Polytropic efficiency ( c) Preheat factor. Assume pressure ratio per stage is same. Given: 16 stage axial flow compressor, pressure ratio = 6.3, ηst=89.5%, T1=288K and p1=1bar Solution:

Problem 2 An air compressor has 8 stages of equal pressure ratios of 1.3. The flow rate through the compressor and its overall efficiency are 45 kg/s and 80% respectively. If the conditions of air at entry are 1 bar and 35˚C determine (a) State of compressed air at exit (b) polytropic efficiency (c) Stage efficiency Given: 8 stages of equal pressure ratio of 1.3, mass flow rate of 45 kg/s, η = 80%, p1= 1bar, T1= 35˚C To find: (a) p2=?, T2=? (b) ηp=? (c) ηst=?

1.1219stageperratiopressurex

x6.3p

p

xconstantp

p

p

p...

p

p

p

p

16

1

17

1

2

2

3

15

16

16

17

==

==

======

( )

( )

( )( )

( )

( )

( )( )

0.970.9095

0.8775

η

ηfactorPreheat

87.75%16.3

16.3

1pr

1prη

0.9045η

11.1219

11.12190.895

1pr

1prη

st

0.2857/η

0.2857

γη

1γ

o

γ

1γ

o

p

0.2857/η

0.2857

γη

1γ

γ

1γ

st

p

p

p

p

===

=−

−=

−

−=

=

−

−==

−

−=

−

−

−

−

ADARSHA H G

Stage pressure ratio = 1.3; Overall pressure ratio = (1.3) = 8.157

8.157barp624.26KT

308T

3085610.8

TT

TTη

561KT

1

8.157

p

p

T

T

22

212

12'

2'

0.2857γ

1γ

1

2

1

2'

==

−

−==

−

−=

=

=

=

−

84.88%

308

624.26log

1

8.157log

0.2857

T

Tlog

p

plog

γ

1γ

η

e

e

1

2e

1

2e

p =

=

−

=

( )

( )84.3%

11.3

11.3

1pr

1prη

0.8488

0.2857

0.2857

γη

1γ

γ

1γ

st

p

=

−

−=

−

−=

−

−

ADARSHA H G

EDUSAT PROGRAMME 15 Turbomachines (Unit 3) Axial … · EDUSAT PROGRAMME 15 Turbomachines (Unit 3)...

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Transcript of EDUSAT PROGRAMME 15 Turbomachines (Unit 3) Axial … · EDUSAT PROGRAMME 15 Turbomachines (Unit 3)...