Heat rate is the amount of energy used by an electrical generator/power plant to generate one kilo Watthour (kWh) of electricity
Heat rate (HR) = Heat input /
Power generation =Kcal / Kwh
Total heat input:
The
chemical energy available in the fuel (coal, biomass, oil, gas etc) is
converted into heat energy in Boilers, this process is called as oxidation. The
heat available in the fuel is measured in terms of Kcal/kg, KJ/kg or BTU units.
The part of this fuel is used as useful heat and rest is lost as dry flue gas
loss, moisture loss, un burnt loss, radiation/convection losses
etc.Based on Boiler efficiency this heat energy from the fuel is utilised,
generally fuel heat utilisation is in the range of 60 to 90%.
This
heat generated in the boilers due to oxidation of fuel is used to generate high
pressure & temperature steam. Thus generated steam is fed into the steam
Turbine, where this heat energy also called thermal energy gets converted into
Kinetic energy then into Mechanical energy in steam turbine, finally mechanical
energy into electrical energy in Generator.
So
total heat input to power plant =Chemical energy + Thermal energy + Kinetic
energy+Mechanical energy
Output
=Electrical power in Kwh
Heat
rate =Heat input / Power generation
Efficiency:
Efficiency
is nothing but the ratio of the useful work done to the heat provided. This
means the friction and other losses are subtracted from the work done by
thermodynamic cycles.
In
Boilers efficiency = Heat out from the Boiler/Heat input to the Boiler
Heat
output is Thermal energy in steam and heat in put is calorific value present in
the fuel
In case
of Turbine, Efficiency =860 X 100/Turbine heat rate
Case1:
Gross heat rate of Thermal power plant
In
thermal power plants all the thermal energy generated from the steam
generators/Boilers is used for only power generation.
Example: A
100 MW thermal power plant is running on 100% PLF, which consumes around 55 MT
of coal having GCV 4500 kcal/kg per hour, then calculate the Gross station heat
rate of the plant
We
have,
Gross
station heat rate =Heat input to the plant / Power generation
=Fuel consumed (MT)X GCV
(kcal/kg)of fuel/Power generation/MWH
= (55 X 4500)/100
= 2475 kcal/kwh
Above
problem can be solved by converting fuel consumed in kg/hr & power
generation in
Kwh,
then heat rate can be calculated as,
=55
X 1000 X 4500/(100 X 1000) =2475 kcal/kwh
CaseII
Station
heat rate of Cogeneration plant
In
cogeneration plant thermal energy is used for process requirement and power
generation. In cogeneration plant there are various sources of heat input &
output to the station & from the station, where as in Thermal power plants
heat input & output sources are only one.
Heat
input to the station is in the form of heat energy present in fuel, make
up water and return condensate from the process.
Heat
output from the station is in the form of heat energy in process steam
& power generation
Cogeneration
heat rate = (Fuel consumed (MT) X GCV of fuel (kcal/kg + Quantity of
return condensate from process (MT) X its enthalpy (kcal/kg) + Quantity of
makeup water (MT) x its enthalpy kcal/kg)(Process steam quantity (MT) X its
enthalpy in kcal/kg) /Power generation in MW
Example:
A process based cogeneration plant has following data on full day case study.
Calculate the station heat rate
Sl No.

Particular

Value

1

Power
generation

977 MW

2

Total
coal consumption Q1

875 MT

3

Gross
calorific value of coal G

5100
kcal/kg

4

Steam
given to process1 plant at 2 kg/cm2g and 135 0C Q2

3720
MT

5

Steam
given to process2 plant at 7 kg/cm2g and 175 0C Q3

192 MT

6

Return
condensate from process1 plant at temperature 120 0C Q4

3350
MT

7

Return
condensate from process2 plant at temperature 85 0C Q5

135 MT

8

DM
water make up to boiler at temperature 25 0C Q6

490 MT

From
above given data we have,
Enthalpy
of steam given to process plant1 h2= 666.71 kcal/kg……..Refer steam table
Enthalpy
of steam given to process plant2 h3= 651.68kcal/kg
Enthalpy
of process1 return condensate h4 = 120.3 kcal/kg
Enthalpy
of process2 return condensate h5 = 85 kcal/kg
Enthalpy
of makeup water h6 = 25 kcal/kg
We have
station heat rate = ((Fuel consumption X GCV +Heat content in return condensate
+ Heat content in makeup waterSum of heat content in process steam))/Power generation.
= ((
Q1X G + Q4 X h4+Q5X h5 + Q6X h6)(Q2 X h2+Q3 X h3))/Power generation
= ((875
X 5100+3350 x 120.3+135 x 85 +490 x 25)(3720 x 666.71+192 x 651.68))/977
= 2337.71
kcal/kg
Turbine heat rate and efficiency:
CaseI:
Turbine heat rate of a thermal power plant during performance guarantee (PG)
test
Turbine
Heat Rate (THR) = Steam flow X (Enthalpy of steamEnthalpy of feed water)/Power
generation
CaseII:
Turbine heat rate of a thermal power plant during normal O&M condition
Turbine
Heat Rate (THR) = (Steam flow X Enthalpy of steamFeed water flow X Enthalpy of
feed water)/Power generation
Turbine
efficiency is given by
Turbine
efficiency =860 X 100/Turbine heat rate
Example:
A 22 MW Turbine has inlet steam flow 100 TPH at pressure & temperature 110
kg/cm2 & 535 Deg C respectively, then calculate the Turbine heat rate in
both PG test case & O&M condition, also calculate Turbine efficiency in
both the cases. Consider feed water temperature at economiser inlet is 195 deg
c & flow is 102 TPH.
Solution:
Turbine
inlet steam enthalpy at operating pressure & temperature H1 =824 kcal/kg
Feed
water enthalpy =H2=198.15 kcal/kg
Steam
flow =100TPH
Power
generation =22 MW
Turbine
heat rate of a thermal power plant during performance guarantee (PG) test
Turbine
Heat Rate (THR) =(100 X (824198.15)/22) =2844.77 kcal/kwh
Turbine
efficiency =(860 X 100)/2844.77 =30.23%
Turbine
heat rate of a thermal power plant during normal O&M condition
Turbine
Heat Rate (THR) =(100 X 824102 X 198.15)/22 =2826.25 kcal/kg
Turbine
efficiency =(860 X 100)/2826.25 =30.42%
CaseIII:
Cogeneration Turbine heat rate
In case
of Cogen, Turbine heat rate is calculated by considering extractions and
return condensate received.
Formula1
CogenTHR
=((Turbine inlet steam flow X its Enthalpy)(Process steam flow X Enthalpy
Exhaust steam flow X Enthalpy))Power generation
Formula2
CogenTHR
=((Turbine inlet steam flow X its Enthalpy + Process return condensate flow X
its Enthalpy + Make up water flow X Its enthalpy)(Process steam flow X
Enthalpy + Feed water flow X Enthalpy)) / Power generation
Example:21
MW condensing cum extraction turbine has inlet steam flow 120 TPH at 88 kg/cm2g
pressure and 520 0C temperature, it has two extraction first, at 16 kg/cm2g
pressure and temperature 280 0C at flow 25 TPH and second at 2.5 kg/cm2g
pressure and temperature 150 0C at flow 75 TPH.Remaining steam goes to
condenser at exhaust pressure 0.09 kg/cm2a.Calculate the turbine heat rate and
thermal efficiency by using both formulae. Consider steam given to process is
10 TPH less than each extraction, the return condensate from the process is 70
TPH at temperature 90 deg c, feed water flow 122 TPH at temperature 195 deg c
& make up water flow 13 TPH at temp 28 deg c.
Given
that,
Power
generating capacity of turbine = 21 MWH
Q1 =
120 TPH
Enthalpy
h1 at 88 kg/cm2g and 5200C = 820.66 kcal/kg
Q2 = 25
TPH
h2 at
16 kg/cm2g and 2800C = 715.88 kcal/kg
Q3 = 75
TPH
h3 at
2.5 kg/cm2g and 1500C = 658.40 kcal/kg
Condenser
steam flow Q4 = Q1Q2Q3 = 1202575 = 20 TPH
h4 at
exhaust pressure = 44.06 kcal/kg
Formula1
CogenTurbine
heat rate (THR) = (Input heat to turbine Sum of extraction and exhaust
heat)/Power generation
= ((Q1 X h1)(Q2 X h2+Q3 X h3
+Q4 X h4))/Power generation
= ((120 X 820.66)(25 X 715.88 +75 X
658.40 + 20 X 44.06))/21
= 1443.85 kcal/kwh
Turbine
thermal efficiency = (860 X 100)/Turbine heat rate
= (860
x100)/1443.85
= 59.56%
Formula2
CogenTHR
=((Turbine inlet steam flow X its Enthalpy + Process return condensate flow X
its Enthalpy + Make up water flow X Its enthalpy)(Process steam flow X
Enthalpy + Feed water flow X Enthalpy))Power generation
THR
=((120 X 820.66 +90 X 90 +13 x 28)(15 X 715.88 +65 X 658.40 + 120 X 198.15))/21
THR
=1495.73 kcal/kwh
Turbine
thermal efficiency = (860 X 100)/Turbine heat rate
= (860 x100)/1495.73
= 57.49%
In the example ,heat rate during PG test is higher than normal O&M condition with increased feed water flow of 102 t. how it can be possible? In my calculation the heat rate will become 2882.52 kcal/kwh from 2844kcal/kwh and turbine efficiency will also decreased from 30.23% to 29.83%.
ReplyDeleteFeed water flow will be always more than steam generation..Considered 2 TPH i.e 2% as blow down loss
DeleteSir according to my experience mean of feed flow is less than the mean of steam flow
DeleteIn such case, attemperator water flow might have calculated separately.
DeleteIn all cases, steam generated = Feed water flow + Attemperator water flow
By calculating plant heat rate indirect method, I am getting 2691kcal/kwh. Where as direct method I am getting 2760 kcal/kwh. Why I am getting this difference and what are the reasons.
ReplyDeleteVery very useful
ReplyDeleteNice to be visiting your blog again, it has been months for me. Well this article that i've been waited for so long. I need this article to complete my assignment in the college, and it has same topic with your article. Thanks, great share. carbon monoxide
ReplyDeleteHave anyone solve this math problem?
ReplyDeleteIn a coal fired power plant the boiler used 3000t/h steam generator at certain perameter & boiler efficiency .9 with electrical efficiency 45% how much water consumption on cooling tower?
in coal fired power plant the boiler used 3000t/h steam generator & boiler efficiency .9 with electrical efficiency 45% how much water consumption on cooling tower?
ReplyDelete