turbine & boiler efficiency
TRANSCRIPT
POWER PLANT
PERFORMANCE
By
V . SRINIVASA RAO
A M
Efficiency of any plant or equipment is the ratio of
output to its input expressed as percentage . output
and input are expressed in same physical units.
The output is the electrical energy sent out to the
grid and input is the heat energy of the fuels fired in
boiler. This is normally termed as overall station
efficiency.
Thus Over all station efficiency
= (Output of ation/Input of station) X 100
= Energy sent out(KW) X 100
Fuel burnt (kg) X GCV fuel (Kcal / Kg)
( 1KW = 860 Kcal = 3600 KJ)
Energy sent out = 12,50,940 KWH
Fuel burnt = 1156 MT
calorific value of fuel = 3000 Kcal/Kg
Over all station efficiency
= Energy sent out (KW) X 100Fuel burnt (kg) X GCV of fuel (Kcal / Kg)
= 1250940 X 860 X 1001156 x 1000 x 3000
= 31.02 %.
Heat Rate:
Heat rate is more usual way of defining and
expressing overall turbo- alternator efficiency.
Heat Rate = Heat added to steam in boiler (Kcal)
Electrical energy sent out (kwhr)Unit of heat rate is thus Kcal/Kwhr
Heat rate is the heat input to turbine needed to
produce 1Kwh of electricity.
Efficiency = output
Input
= 1kwh
Input
kwh = 860 kcal
So efficiency = 860 x 100
Heat rate
Example for 50 MW turbine
Heat rate = 2330 kcal/kwh
Efficiency = 860 x 100
2330
= 36.91%
BOILER
F.P
FP
C
Turbine
1kg
hf2
A
B
C (1-m) kg
h3
condenser
(1-m)kg
hf3
m kg
h2
1kg
h1
Consider a regenerative cycle with single feed water heater.
The steam (at pressure p1) enters the turbine at point
A. Let a small amount of wet steam(say m kg) after
partial expansion (at pressure p2 ) be drained from the
turbine at point B and enters the feed water heater. The
remaining steam ( at pressure p2) is further expanded in
the turbine and leaves at point C .
This steam is then condensed in the condenser. The
condensate from the condenser, is pumped into the feed
water heater, where it mixes up with the steam extracted
from the turbine. The proportion of the steam extracted
is just sufficient to cause the steam leaving the feed
water heater to be saturated.
Now consider 1 kg of steam entering the turbine at point A.
Let h1 = Enthalpy or total heat of steam entering the
turbine at A,
h2 = Enthalpy or total heat of bled steam,
h3 = Enthalpy or total heat of steam leaving the
turbine at C
hf2 = Enthalpy or sensible heat of feed water
leaving the feed water heater
hf3 = Enthalpy or sensible heat of steam leaving
the condenser, and
m = Amount of bled steam per kg of steam
supplied
We know that heat lost by bled steam = Heat gained by feed
water
m(h2-hf2) = (1-m)(hf2-hf3)
mh2-mhf2 = hf2-hf3-mhf2+mhf3
Therefore m = (hf2-hf3)/ (h2-hf3)
We know that mass of steam in the turbine, per kg
of feed water, between A and B is 1 kg.
Therefore work done in the turbine per kg of feed
water between A and B = (h1-h2)
mass of steam in the turbine per kg of feed water
between B and C= (1-m) kg
Therefore,
Work done in the turbine between B and C
= (1-m) (h2-h3)
Total work done = (h1-h2)+(1-m)(h2-h3)
And total heat supplied per kg of feed water = h1-hf2
Therefore,
Efficiency of the cycle including the effect of bleeding,
= Total work done / Total heat supplied
= ((h1-h2) + (1-m)(h2-h3))/h1-hf2
NOTE:
If there had been no regenerative feed heating (or in other
words, m=1),then the efficiency of the cycle will be the
same, as that of rankine cycle. In this case, Rankine
efficiency,
Efficiency = (h1-h3)/(h1-hf3)
Problem : In a regenerative cycle, having one feed water,
the dry saturated steam is supplied from the boiler at a
pressure of 30 kg/cm2 and the condenser pressure is 1
kg/cm2. The steam is bled at a pressure of 5 5kg/cm2.
Determine the amount of bled steam per kg of steam
supplied and the efficiency of the cycle. What would be the
efficiency without regenerative feed heating ? Also
determine the percentage increase in efficiency due to
regeneration.Given: p1 = 30 kg/cm2 , P2 = 5 kg/cm2 , P3 =1
kg/cm2
Enthalpy of steam at 30 kg/cm2, h1 = 670 kcal/kg
Enthalpy of steam at 5 kg/cm2, h2 = 588 kcal/kg
Enthalpy of steam at 1 kg/cm2, h3 = 531 kcal/kg
From steam tables, enthalpy or sensible heat of water at 5 kg/cm2 ,hf2 = 153 kcal/kg
And enthalpy or sensible heat of water at 1 bar,
hf3 = 100kcal/kg
Amount of bled steam per kg of steam supplied
We know that amount of bled steam per kg of steam
supplied
m = (hf2-hf3) / (h2-hf3)
= ( 153 - 100)/ (588-100)
= 0.11 kgs.
EFFICIENCY OF THE CYCLE
We know that efficiency of the cycle,
= (h1-h2)+(1-m)(h2-h3))/(h1-hf2) x100
=(670-588) + (1-0.11)(588-531)/(670-153) x100
=(82 + 0.89x57)/ 517 x100
= 25.67%.
EFFICIENCY OF THE CYCLE WITHOUT
REGENERATIVE FEED HEATING
We know that efficiency of the cycle,
= (h1-h3)/(h1-hf3) x100
= (670-531)/(670-100) x100
= 139/570 x100
= 0.243 OR 24.3 %
PERCENTAGE INCREASE IN EFFICIENCY DUE TO
REGENARATION
percentage increase in efficiency due to regeneration
= (0.257-0.243)/0.243
= 0.0535 or 5.35 %
BOILER
F.P
FP
C
Turbine
1kg
hf2
A
B
C (1-m1-m2) kg
h4
condenser
(1-m1-m2)kg
hf4
m1
h2
1kg
h1
Consider a regenerative cycle with two feed water heater.
B1
m2
h3
hf312
In this case, the steam is removed from the turbine at
two points B and B1.It is then fed into two open feed
water heaters 1 and 2
The steam (at pressure p1) enters the turbine at point
A. Let a small amount of steam (say m kg) after partial
expansion (at pressure p2 ) be drained from the
turbine at point B and enter the feed water heater
1.similarly,let another small amount of steam (say m2
kg ) after further expansion (at pressure p3 ) be
drained from the turbine at point B1 and enter the feed
water heater 2. The remaining steam equal to (1-m1-
m2) kg (at pressure p4) is further expanded in turbine,
and leaves it at point C .
The steam is then condensed in the condenser . The
condensate from the condenser is pumped into the feed
water heater , where it mixes up with the steam extracted
from the turbine. Now consider 1 kg of steam entering
into the turbine at A .
Let,
h1 = Enthalpy of steam entering the turbine at A
h2 = Enthalpy of steam bled at B
h3 = Enthalpy of steam bled at B1
h4 = Enthalpy of steam leaving the turbine at C
hf2 = Enthalpy of feed water leaving the feed water
heater 1
hf3 = Enthalpy of feed water leaving the feed water
heater 2
hf4 = Enthalpy of feed water leaving the condenser
m1 = Amount of steam bled at B per kg of steam
supplied, and
m2 = Amount of steam bled at B1 per kg of steam
supplied
As the that heat lost by bled steam at B
= Heat gained by feed water
m1 (h2-hf2) = (1-m1) (hf2-hf3)
m1h2-m1hf2 = hf2-hf3-m1hf2+m1hf3
m1 = (hf2-hf3) / (h2-hf3)
Similarly, heat lost by bled steam at B1
=Heat gained by feed water
m2 (h3-hf3) = (1-m1-m2) (hf3-hf4)
m2h3-m2hf3 = hf3-hf4-m1hf3+m1hf4-
m2hf3+m2hf4
m2 = ((1-m1) (hf3-hf4) / (h3-hf4)
We know that the mass of steam in the turbine per kg of feed
water between A and B is 1 kg.
Work done in the turbine per kg of feed water between A and B
= h1-h2
And mass of steam in the turbine per kg of feed water between B
and B1
= (1-m1) kg
Work done in the turbine between B and B1
= (1-m1) (h2-h3)
Similarly, mass of steam in the turbine per kg of feed water
between B1 and C
= (1-m1-m2) kg
Work done in the turbine between B1 and C
= (1-m1-m2) (h3-h4)
Thus total work done per kg of feed water
= (h1-h2) + (1-m1) (h2-h3) + (1-m1-m2) (h3-h4)
And total heat supplied per kg of feed water
= h1-hf2
Efficiency of the plant including the effect of bleeding,
Efficiency = Total work done/Total heat supplied
= (h1-h2) + (1-m1) (h2-h3) + (1-m1-m2) (h3-h4)
h1-hf2
NOTE: When the bleeding takes place at more than two points,
the efficiency of the plant may be obtained by proceeding in the
same way as explained above.
In a steam turbine plant the steam is generated and
supplied to the turbine at 50 kg/cm2 and 370deg C, the
condenser pressure is 0.1kg/cm2 . Two feed heaters are
used, the steam in the heaters bled at 5kg/cm2 and
0.5kg/cm2. In Each heater the feed water is heated to
saturation temperature of the bled steam. The
condensate is also pumped at this temperature into the
feed line immediately after the heater. Find the masses
of the steam bled in the turbine per 1kg of steam
entering the turbine. calculate the thermal efficiency of
the cycle
SOLUTION:
Given: P1 = 50 kg/cm2 , P2 = 5 kg/cm2 , P3 = 0.5
kg/cm2, P4= 0.1 kg/cm2, T1= 370C
Enthalpy of steam at 50 kg/cm2, h1 = 742.77kcal/kg
Enthalpy of steam at 5 kg/cm2, h2 = 663.95 kcal/kg
Enthalpy of steam at 0.5 kg/cm2, h3 = 599.47kcal/kg
Enthalpy of steam at 0.1 kg/cm2, h4 = 561.26 kcal/kg
hf2 = 152.87 kcal/kg(at 5kg/cm2)
hf3 = 81.34 kcal/kg(at 0.5 kg/cm2)
hf4 = 45.80 kcal/kg(at 0.1 kg/cm2)
Mass of steam bled in turbine:
We know that mass of steam bled at B,
m1 = (hf2-hf3) / (h2-hf3)
= (152.87 – 81.34) / (561.26-81.34)
= 0.123 kgs.
And mass of steam bled at B1,
m2 = (1-m1)(hf3-hf4) / h3-hf4
= (1-0.123)(81.34-45.8) / (59.47-45.8)
= 0.056 kg
Thermal Efficiency of the Cycle:
We know the work done from A to B per kg of feed water
= h1-h2
= 742.77-663.95 = 78.81 kcal/kg
Similarly work done from B to B1 per kg of feed water
= (1-m1)(h2-h3)
= (1-0.123)(663.95-599.47)
= 56.55 kcal /kg
And work done from B1 to C per kg of feed water
= (1-m1-m2)(h3-h4)
= (1-0.123-0.056)(599.47-561.26)
= 31.38 kcal/kg
Therefore total work done = A to B + B to B1 + B1 to C
= 78.81+ 56.55+ 31.38
= 166.74 kcal/kg
Heat supplied = h1-hf2
= 742.77- 152.87
= 589.9 kcal/kg
Therefore,
work done
thermal efficiency of the cycle = -------------- x100
Heat Supplied
166.74
= ------------ x 100
589.9
= 28 %
50 MW Turbine efficiency calculation
MS flow = 1kg
MS enthalpy = 823.7 kcal
HP:6 Extraction flow = 0.02kg
HP:6 Extraction enthalpy = 767.0 kcal
HP:5 Extraction flow = 0.07kg
HP:5 Extraction enthalpy = 737.8 kcal
Deaerator Extraction flow = 0.02 kg
Deaerator Extraction enthalpy = 718.3 kcal
LP:4 Extraction flow = 0.04 kg
LP:4 Extraction enthalpy = 676.4 kcal
LP:3 Extraction flow = 0.04 kg
LP:3 Extraction enthalpy = 649.7 kcal
LP:2 Extraction flow = 0.02 kg
LP:2 Extraction enthalpy = 617.7 kcal
LP:1 Extraction flow = 0.02 kg
LP:1 Extraction enthalpy = 49.6 kcal
Work done between turbine inlet and HP-6 Extrn
= 1(Enthalpy of steam at turbine inlet - Enthalpy of steam at HP-6
Extrn)
= 1( 823.7-767.0)
= 56.7 kcal/kg ---------------------------(1)
Work done between HP-6 Extrn point and HP-5 Extrn point
= (1- HP-6 Extrn flow)(Enthalpy of steam at HP-6 Extrn - Enthalpy of
steam at HP-5 Extrn)
=(1-0.02)( 767.0 – 737.8)
= 0.98 x 29.2
= 28.62 kcal/kg ---------------------------(2)
Work done between HP-5 Extrn point and D’tr Extrn point
= (1- HP-6 Extrn flow- HP-5 Extrn flow )(Enthalpy of steam at HP-5 Extrn
- Enthalpy of steam at D’tr Extrn)
=(1-0.02-0.07)(737.8 – 718.3)
= 0.91 x 19.5
= 17.75 kcal/kg --------------------------(3)
Work done between D’tr Extrn point and LP-4 Extrn point
= (1- HP-6 Extrn flow- HP-5 Extrn flow -D’tr Extrn flow )(Enthalpy of
steam at D’tr Extrn - Enthalpy of steam at LP-4 Extrn)
=(1-0.02-0.07-0.02)( 718.3 – 676.4)
= 0.89 x 41.9
= 37.29 kcal/kg ---------------------------- (4)
Work done between LP-4 Extrn point and LP-3 Extrn point
= (1- HP-6 Extrn flow- HP-5 Extrn flow -D’tr Extrn flow – LP-4 Extrn
flow)(Enthalpy of steam at LP-4 Extrn - Enthalpy of steam at LP- 3
Extrn)
=(1-0.02-0.07-0.02- 0.04)( 676.4 – 649.7)
= 0.85 x 26.7
= 22.70 kcal/kg-----------------------------( 5)
Work done between LP-3 Extrn point and LP-2 Extrn point
= (1- HP-6 Extrn flow- HP-5 Extrn flow -D’tr Extrn flow – LP-4 Extrn
flow- LP-3 Extraction flow)(Enthalpy of steam at LP-3 Extrn - Enthalpy
of steam at LP- 2 Extrn)
=(1-0.02-0.07-0.02- 0.04 – 0.04)( 649.7 – 617.7)
= 0.81 x 32
=25.92 kcal/kg ------------------------------ (6)
Work done between LP-2 Extrn point and LP-1 Extrn point
= (1- HP-6 Extrn flow- HP-5 Extrn flow -D’tr Extrn flow – LP-4 Extrn
flow- LP-3 Extraction flow – LP-2 Extraction flow )(Enthalpy of steam at
LP- 2 Extrn - Enthalpy of steam at LP- 1 Extrn)
=(1-0.02-0.07-0.02- 0.04 – 0.04 – 0.02)( 617.7 – 598.4)
= 0.79 x 19.3
=15.25 kcal/kg ------------------------------- (7)
Work done between LP-1 Extrn point and last stage of turbine
= (1- HP-6 Extrn flow- HP-5 Extrn flow -D’tr Extrn flow – LP-4 Extrn
flow- LP-3 Extraction flow – LP-2 Extraction flow – LP-1 Extraction
flow)(Enthalpy of steam at LP-1 Extrn - Enthalpy of steam at last
stage of turbine)
=(1-0.02-0.07-0.02- 0.04 – 0.04 – 0.02-0.02)( 598.4 – 556.4)
= 0.77 x 42
= 32.34 kcal/kg ------------------------------ (8)
Total work done = (1) + (2) + (3) + (4) +(5) +(6) +(7)+(8)
= 56.7 + 28.62 + 17.75 + 37.29 + 22.70 + 25.92 + 15.25 + 32.34
= 236.57 kcal/kg
Heat supplied = Enthalpy of main steam – enthalpy of feed water after
HP-5
= (823.7- 214.74) kcal/kg
= 608.96 kcal/kg
Efficiency = Total work done x 100
Heat input
Efficiency = 236.57 x 100
608.96
38.84 %
