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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