Turbines and Compressors

8/11/2019 Turbines and Compressors

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Finite stage efficiency

A stage with finite pressure drop is a finite turbine stage.

In multi-stage turbines along with overall efficiency, theefficiencies of individual stages are also important

Different stages with same pressure ratio located in different

regions in the h-s plane will give different values of work

output. For a steady flow process

dw = -v dp

This implies that for the same pressure drop more work will be

done with higher values of v At each stage the work done is proportional to the initial

temperature of the gas


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Effect of reheat


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Total expansion process 1-2 is divided in to four stages of the

same efficiency (st) and pressure ratio

Consider the overall efficiency of the expansion is T

Actual work during the expansion process 1-2 is

wa= T ws

If the isentropic or ideal work in the stages are ws1, ws2 , ws3andws4


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The constant pressure lines in a h-s plane must diverge

towards right, therefore

This makes the overall efficiency of the turbine greater than

the individual stage efficiency.

T > st


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The quantity ws/ wsis known as the Reheat factor

This factor is always greater than unity

The effect depicted by T > st ,is due to a thermodynamic

effect called Reheat

This does not imply any heat transfer to the stages from

outside

It is the reappearance of stage losses as increased enthalpy

during the constant pressure heating (reheating) processes AX,

BY, CZ, D2


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Infinitesimal stage efficiency

T

-dts

T


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Expanding the binomial expression on RHS and ignoring theterms beyond the second


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This differential equation is valid along the actual expansion

process.

On integration eqn. 9 yields,

This relation defines the actual expansion line in a finite stageor a multistage machine between two given states

Here the value of infinitesimal or small stage efficiency (p) is

constant

(10)

(9)


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The value of p must be determined to use the above eqn. 10

for a given expansion between two states.

Integrating eqn.9 between the given two states 1 and 2,

(11)


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Irreversible adiabatic expansion process (the actual expansion

process) can be considered as equivalent to polytropic process

So eqn.11 can be written as

Equating the indices we have,

When p= 1, n = . The actual expansion line coincides withthe isentropic expansion and the above equations will be valid

for an isentropic process


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The efficiency of a finite stage can be expressed in terms of

the small stage efficiency.


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Variation of stage efficiencies with pressure

ratio at constantp


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Relationship between reheat factor, pressure ratio and

polytropic efficiency (n=1.3)


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Work and efficiencies in compressor stages

Concepts developed for diffusers can be employed in the case

of compressors also

But here we have consider the shaft work

Due to the energy transfer from the rotor to the gas, its

properties change from p1,p01,h1, etc. to p2,p02,h2etc.


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The actual work (wa) supplied during adiabatic compression is

given by the energy equations between the states O1and O2

For perfect gas


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Total to total efficiency

The stagnation pressure ratio is


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The efficiency is used in compressor stages where the gas

velocities at entry and exit are significant and the velocity

temperatues cant be ignored


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Static to static efficiency

The gas velocities at entry and exit are almost equal and

negligible . So the actual and ideal works are

Now efficiency of compressor is


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Effect of preheat


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The total compression pressure ratios between two states p1

and p2 are

Now the ideal work between in two states 1 and 2s is Ws andindividual stage works are

Overall compressor efficiency

1 2 3 4, , ,s s s sw w w w


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But we have

Thus for a compression process the isentropic efficiency of the

machine is less than the small stage efficiency, the difference

being dependent upon the divergence of the constant pressure

lines.

This is due to thermodynamic effect called preheating

This is the result of the reappearance of the effect of losses ofthe previous stage in the subsequent stages.


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Relationship between isentropic (overall) efficiency, pressure

ratio and small stage (polytropic) efficiency for a compressor

This figure shows that the

isentropic efficiency of a finite

compression process is less than

the efficiency of the small stages.

Comparison of the isentropicefficiency of two machines of

different pressure ratios is not a

valid procedure since, for equal

polytropic efficiency, the

compressor with the highestpressure ratio is penalised by the

thermodynamic effect, Preheat


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Infinitesimal stage efficiency


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Example 1

Gas enters the nozzles of a turbine stage at a stagnation

pressure and temperature of 4.0 bar and 1200K and leaveswith a velocity of 572 m/s and at a static pressure of 2.36 bar.

Determine the nozzle efficiency assuming the gas has the

average properties over the temperature range of the expansion

of Cp

= 1.160 kJ/kg K and = 1.33.


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Solution


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Example 2

An axial flow air compressor is designed to provide an overalltotal-to-total pressure ratio of 8 to 1. At inlet and outlet the

stagnation temperatures are 300K and 586.4 K, respectively.

Determine the overall total-to-total efficiency and the

polytropic efficiency for the compressor. Assume that for air

is 1.4.


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Solution


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Example 3

Air at 2.05 bar and 417 K is expanded through a row of blades

to a pressure of 1.925 bar. The stagnation pressure loss across

the nozzle is measured at 10 mm Hg. Determine the efficiency

of this nozzle and the velocity of air at the exit.

Take cp= 1005 J/kg K and = 1.4. State assumptions used.


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Solution

Assuming the flow through the nozzle as incompressible

For small pressure ratios

N= 1- (p0)/(p1-p2).

p1= 2.05 bar;

T1= 417 K

p2= 1.925 bar;

C2= ?

N= ?

q = 0

N= 89.5%


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Assuming adiabatic expansion, and negligible inlet velocity

(V1= 0)

h1+ C12/2 = h2+ C2

2/2 C22/2 = (h1h2).

C2= [2(h1h2)]1/2

Also, N= (h1h2)/(h1h2s)

h1-h2= 0.895 cp[T1T2s]

h1-h2=0.895 x 1005 x T1x [1-(p2/p1)(-1)/]

h1-h2=0.895 x 1005 x 417 x [1-(1.925/2.05)0.4/1.4]

h1-h2=6688.61 J/kg K

So the exit velocity is C2= 115.66 m/s


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Example 4

An air stream (= 1.25 kg/m3) is decelerated from 100 m/s to

75 m/s in a diffuser giving a pressure rise of 250 mm (W.G).Calculate the diffuser efficiency.

Solution

V1=100 m/s

1=1.25 kg/m3

V2= 75 m/s

' 2 1

2 2

1 2

3'

2 2

2( )

( )

2(0.25 9.81 10 )89.7%

1.25 (100 75 )

D

D

p p

V V


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Example 5 A diffuser at the exit of a gas turbine has an area ratio of 2.0 If

the static pressure at the diffuser exit is 1.013 bar and the velocityof gas 30 m/s, calculate the static pressure of the gas at the

turbine exit. Take diffuser efficiency equal to 77% and the

density of gas as 1.25 kg/m3 (constant). And what will be the

value in manometer gauge ?(132.5 mm WG)

1

2

p3= 1.013 barV3= 30 m/s

3

G.T

D= 77%

= 1.25 kg/m3

A3/A2= 2.0

Diffuser


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Solution

' 3 2

2 22 3

2 2 3 3

3 32

2

3 2

2 2

3 2

2

2( )

( )

30 2 60 /

2( )0.77

1.25 (60 30 )

1299.38

1.0

D

p p

V V

Continuity

V A V A

V AV m sA

p p

p p Pa

p bar


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Example 6

A nozzle expands air from P1=8.0 bar,T1=540K to a pressureof 5.8 bar with efficiency of 95%. The air is then passed

through diffuser of area ratio 4.0. The total pressure loss

across the diffuser is 367 mm of Hg.

Determine the efficiency of diffuser and velocities of air at its

entry and exit . What is the static pressure at exit?


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Example 7


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Aerofoil section and cascading of blades


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Aerofoil blades

An aerofoil is a streamlined body having a thick, rounded

leading edge and a thin trailing edge.

Its max. thickness occurs somewhere near the mid point of the

chord

An array of blades representing the blade ring of an actual

turbo machinery is called the cascade

The back bone lying midway between the upper and lower

surfaces is known as the camber line.

It is possible to achieve very high lift to drag ratio when such a

blade is suitably shaped and properly oriented in the flow

Aerofoil shapes are used for aircraft wing sections and blades

of various turbo-machines.


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Total upward force acting on the aerofoil is equal to the

projected area times the pressure difference on the two sides

Due to the large area aircraft wing will provide high lift even

for very small pressure difference over its aerofoil section Since the projected areas of the turbo-machine blades are

much smaller, considerable difference of static pressure across

the section is required in turbo-machines

This can be only achieved by providing highly cambered bladesections

Blade camber in the compressor blades is between those of

aircraft wings and turbine blades

Lift forces exerted by fluid on the aircraft wings and theturbine rotor blades

In power absorbing turbo-machines the lift force exerted by

their rotor blades on the fluid


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Energy transfer in turbomachines

r2

r1

Entry

Exit

Rotor

Control Surface

u1

c1 w1

c2

w2

12

u2

C1

Cr1

Cr2

C2Control

volume


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These equations 12 & 13 are known asEulers pump and

turbine equationsrespectively

(12)

(13)


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The Euler relation state mathematically an important fact

If the turbomachine has to act as a head or pressure producing

machine, its flow passages should be designed to obtain an

increase in the quantity ucwhereas if it is to act as a power

producing machine, there must be a decrease in the quantity

ucbetween its entry and exit


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Forces on the rotor bladesTangential Thrust

Tangential thrust is the force generated by the peripheral

change in momentum of the fluid.

Tangential thrust produces the net work done by a turbine or

results in head generation in a compressor.

Hence, tangential thrust has to be maximized in a

turbomachine for given flow conditions

Axial Thrust

Axial thrust is the force due to both the axial change in fluid

momentum, and also static pressure variation in a

turbomachine.

Axial thrust doesnt contribute to the rotors motion. It is

borne by the rotor bearings and has to be minimized.


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Radial Thrust

Radial thrust is generated by static pressure and fluid

momentum variation in the radial direction.

This thrust doesnt contribute to rotary motion and has to be

absorbed by the rotor shaft bearings.

Hence, radial thrust must also be minimized in a

turbomachine.

This thrust appears as the load on the rotor shaft bearings


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Components of Energy Transfer

Cr1

w1C1

u1

C1

2 2 2 2 2

1 1 1 1 1 1

2 2 2 2 2

1 1 1 1 1 1 1

2 2 2

1 1 1 1 1

2 2 2

1 1 1 1 1

- - ( - )

- - - 2

2 -1

( - )2

rc c c w u c

c c w u c u c

u c c u w

u c c u w


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Similarly

Thus the total energy transfer in turbomachines is made up ofthree componentschange in KE in absolute frame of

coordinates, change in centrifugal energy, change in KE in the

relative frame of coordinates

1

2 2 2

2 2 2 2 2

2 2 2 2 2 2

1 2 2 1 1 1 2 2 2

2 2 2 2 2 2

T 1 2 1 2 1 2

2 2 2

1( )

2

1

[( ) ( - )]2

1w [( ) ( ) ( )]

2

Euler's Work in differential form

1 1 1 ( ) ( ) ( ) ( )

2 2 2T

u c c u w

u c u c c u w c u w

c c u u w w

dw d uc d c d u d w


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Flow Through Cascades

In the development of the highly efficient modern axial flow

compressor or turbine, the study of the two-dimensional flowthrough a cascade of airfoils has played an important part

An array of blades representing the blade ring of an actual

turbo machinery is called the cascade

If the blades are arranged in a straight line, the cascade isknown as a rectilinear cascade.

If the blades are arranged in an annulus, it is known as anannular cascade.

Rectilinear and annular cascades are deployed for axial flowmachines.

If the flow is completely radial, the cascade is known as aradial cascade.


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Cascade is constructed by assembling a number of blades of a

given shape at the required pitch and stagger angle

The assembly is then fixed on the test section of a wind tunnel Air at near ambient condition is blown over the cascade of

blades to simulate the flow over an actual blade row in a

turbomachine.

Information through cascade tests is useful in predicting theperformance of blade rows in actual machine

These tests can also be employed in determining optimum

design of a blade row for prescribed conditions

Figure shows a compressor blade cascade tunnel As the air


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Figure shows a compressor blade cascade tunnel. As the air

stream is passed through the cascade, the direction of air is

turned.

Pressure and velocity measurements are made at up anddownstream of cascade as shown.

The cascade is mounted on a turn-table so that its angular

direction relative to the direction of inflow can be changed,

which enables tests to be made for a range of incidence angle. As the flow passes through the cascade, it is deflected and

there will be a circulation and thus the lift generated

Cascade variables


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Cascade variables

Reynolds number

Mach number

Pitch-Chord ratio : s/L Aspect Ratio : h/L

Blade geometry and profile : , etc.

Boundary layer and turbulence

Angle of Incidence , CD, CL, , Y = f (Cascade Variables)

In developing the design of a blade row some of the

parameters are fixed by the given conditions and couple of

significant variables can be identified Cascade testing facilities have become almost inseparable

sector of all big companies which design and manufacture

turbomachines

A i l bi d Bl d l


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Axial turbine cascadeBlade angles

T.E

1

1

2

2

L.E

1

1

22


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: Blade Camber Angle;

1 & 2: Inlet and exit camber angles;

: Blade stagger angle;

1& 2: Absolute air angles;

1& 2: Relative air angles or blade angles.

= 1+ 2 = 1+ 2

1= 1-

2= 2+


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Angle of incidence i = 1- 1

angle of deviation = 2- 2

The total angle of deflection then is

= 1+ 2

= (i + 1) + (2- )

= [(1- ) + (2+ )] + i -

= + i -

V l i i A i l M hi


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Velocity components in an Axial Machine

V l i i l A i l bi d


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Velocity trianglesAxial turbine cascade

p01

p02

b

c1

cy1

cx1 1

Fy

L

D

FxFr

p01p2

Fy,max

cx2

cy2

c22 s

Pressure

Side

Suction

Side

mcxm

cym

cm


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Mean Flow Parameters

cxm

= (cx1

+ cx2

)

cym= (cy2cy1)

tan m= cym/cxm

for constant axial velocity,

tan m= (tan 2tan 1)

Bl d F


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Blade Forces

Continuity equation :

Considering unit height of the blade and assuming,

T ti l F


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Tangential Force

Rewriting the above equation

1 2

1 1 2 2

2

1 2

[ ]

tan , tan

[tan tan ]

y xm y y

y xm y xm

y xm

F sc c c

c c c c

F sc

2

1 2

2

1 2

1 22

2( )( )[tan tan ]

2

1 2( )[tan tan ]2

2( )[tan tan ]1

2y

y xm

y xm

y

F

xm

s lF c

l

sF lc l

F sC

llc

p01

p02

c1

cy1

cx1 1

Fy

L

D

FxFr

cx2

cy2

c22

mcxm

cym

cm


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It can also be written as

2 2

2 2 1 2

' 2

2 1 2

22

1( ) 2 ( ) cos (tan tan )2

2 ( ) cos (tan tan )

12

y

y

y

F

sF lc

L

F sC

llc

A ial force


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Axial force

Axial force is due to the static pressure drop across the cascade

and due to the change of momentum in the axial directionFx= (p1p2) (s1) +cxm(s 1) (cx1cx2)

.

.

Fx= (scxm2) tanm(tan2+tan1) + sp0

Pressure loss coeff icient

0

2

2

1

2

pY

lc


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Axial thrust coeff icient

2 2 2 02 2 1

2

2

cos (tan tan )1

2

xF

s psC

llc

Lift f r


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Lift force

L = Fy

cosm

+ Fx

sinm

Substituting the expressions for Fxand Fy,

L = scxm2(tan1+ tan2)cosm

+ scxm2tanm(tan1+ tan2)sinm

+ sp0sinm

L = scm2cosm(tan1+ tan2) + sp0sinm

Fy

LFx

Fr

m

cxm

cym

cmm

Drag force


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Drag force

D = Fxcosm+ Fysinm

D = scxm2 tanm(tan1+ tan2) cosm

+ sp0cosm scxm2(tan1+ tan2) sinm

D = sp0cosm

Drag coefficient is


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Since cmcosm= c2cos2= Vxm

cm

Fy

LFx

Fr

m

cxm

cym

m


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For reversible flow through cascade p0=0, CD= 0


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Circulation

Circulation around the blade contained in the control surface is

the line integral of the velocity around the closed circuit.

= cy1s + cy2s

The line integral along the curved branches of the circuit

cancel each other

So the relation between the circulation and the lift is

L =cm

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Turbines and Compressors