axial, radial turbines
DESCRIPTION
Axial Turbine infoTRANSCRIPT
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MEE-304MEE-304Turbomachines Turbomachines
Unit 3Unit 3STEAM AND GAS TURBINES
Lecture 23Lecture 23
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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
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Axial turbine stages
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Stage velocity triangles
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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
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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
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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
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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 )
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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
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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
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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
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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.
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Single impulse stage
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w3 = w2
w2
c2cx2
cy2
u
c3cx3
cy3
2β2
u
3β3
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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)
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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
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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
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Stage Velocity Triangles for εmax
w3 = w2
w2
c2cx2
cy2=2u
u
cx3=c3
2β2
u
β3
2=0
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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
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Multi Stage Velocity Compounding in impulse turbines(Curtis stages or velocity stages)
N R1 F R2
V
p
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Velocity triangles for a two-stage velocity compounded impulse turbine with εmax
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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
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• 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
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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
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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.
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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
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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)
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N1 R1N2 R2
V
p
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Reaction Stages
NR-1
FR-2
V
p
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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
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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)
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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
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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
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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
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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
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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
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Hundred percent reaction
Negative reaction
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MEE-304 TURBOMACHINESMEE-304 TURBOMACHINES
AXIAL TURBINESAXIAL TURBINES
NUMERICAL PROBLEMS – Set 7NUMERICAL PROBLEMS – Set 7
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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.
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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
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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
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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
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'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
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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
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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)
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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
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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
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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
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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
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Radial Turbines
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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.
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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.
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Types of inward flow radial turbineCantilever turbine
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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
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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
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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
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The 90 degree IFR turbine
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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MEE-304 TURBOMACHINESMEE-304 TURBOMACHINES
RADIAL TURBINESRADIAL TURBINES
NUMERICAL PROBLEMS – Set 8NUMERICAL PROBLEMS – Set 8
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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
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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.
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Soln. Inlet Velocity Triangle
Exit velocity triangle
c2w2
cr2
2β2
c2
u2
cr3 = c3
w3
u3
β3
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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
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2. Reaction
2 2
1R = (1 cot )
21
(1 0.639cot 60)2
0.3155 31.55%
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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%
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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
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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
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.
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
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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.
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