Thermal Power Plant Performance
The performance of any thermal power plant depends on so many factors like coal quality, heat rate, cycle efficiency, design etc. The efficiency of any plant is not constant always but reduces years of years. Generally a coal fired thermal power plant efficiently performs for 25 years if all the required maintainance is done time to time.
Heat rate - “Heat rate is the heat required to generate one unit of electrical energy.” Heat rate in a thermal power plant is measure constitute for its thermal efficiency. The plant output should be higher with minimum heat requirement so we can say that the heat rate must be as low as possible.
The property of heat generation by fuel/coal depends on the calorific value. Also the CV(Calorific Value) depends on the elemental and non-elemental composition of coal. The analysis of coal composition are divided into two categories:
- Proximate analysis – In this analysis non-elemental composition are calculated. The non-elemental composition are fixed carbon, volatile matter, moisture and ash contents.
- Ultimate analysis – In this analysis elemental composition are calculated. The elemental composition are carbon(C), hydrogen(H), sulphur(S), nitrogen(N), oxygen(O), phosphorus(P), chlorine(Cl) etc.
Grading of coal is the measurement of heat content of the coal. It is measured in calorific value for each kilogram of coal weight.
Calorific Value (kCal/kg)
5601 – 6200
4941 – 5600
4201 – 4940
3361 – 4200
2401 – 3360
1301 – 2400
Below ‘G’ grade coal are generally not used in any power plant to maintain the efficiency. High grade coals are generally mixed and blended with low grade coal. Before using the coal, it is passed from coal washery system where lot of waste substances of coal are removed by washing.
In a thermal power plant, unit's heat rate is required to power the turbine but practically it is not possible because heat loss occured in most of the parts like turbine extraction, vents, drains, tubes and a major part of heat is consumed in condenser side. Hence unit heat rate can never be equalize turbine heat rate.
Let's find out which formula that can be used to calculate the terms like heat rate, calorific values, efficiency etc. Then we'ill do some calculation on all these.
We know that Efficiency η = output/input
Efficiency η = THR/UHR where THR = turbine heat rate and UHR = unit heat rate
UHR = (Q x CV)/MW where Q = coal flow rate, CV = calorific value and MW = plant output in MW
GCV is calculated by Dulong’s formula, GCV = (33820 x C) + 144000( H – O/8) + (9.204 x S)
Qu. A steam power plant has 20% overall efficiency. To generate each unit of electricity 900gm of coal is required. Calculate the calorific value of coal.
Sol. CV(Calorific Value) = ?
900gm = 0.9 kg
Heat content by 0.9kg of coal = 0.9 x CV
Efficiency η = output/input= 860kCal/Heat input
0.20 = 860/(0.9 x CV)
CV= 860/(0.9 x 0.2)
CV = 4777.78 kCal/kg
Qu. A 250MWsteam power plant consumes 160T/hr(tonns per hour) with calorific value of coal is 3300kCal/kg. Calculate the overall efficiency.
Sol. Efficiency η = ?
CV(Calorific value) = 3300kCal/kg
Coal flow Q = 160T/hr
Plant capacity MW = 250
Heat content by 0.9kg of coal = 0.9 x CV
UHR(unit heat rate) = (Q x CV)/MW
= (160 x 3300)/250
= 2112 kCal/kWh
Efficiency η = output/input = 860kCal/unit heat rate
η = 860/2112 = 0.40719
η = 40.719%
Qu. Calculate the amount of coal required in TPH(Tonns per hour) for a 600MW, 37.5% efficient thermal power plant if the calorific value of coal is 3150kCal/kg.
Sol. CV(Calorific value) = 3150kCal/kg
Coal flow Q = ?
Efficiency η = 37.5% = 0.375
Plant capacity = 600 MW
UHR(unit heat rate)= 860/η
UHR = 860/0.375 =2293.33 kCal/kWh
UHR = (Q x CV)/MW
Q = (MW x UHR)/CV
Q = (600 x 2293.33)/3150 = 436.8 T/hr
Qu. Calculate the GCV(gross calorific value) of coal for which the following are the percentage obtained by ultimate analysis.
Sol. GCV is calculated by Dulong’s formula :
GCV = (33820 x C) + 144000( H – O/8) + (9.204 x S)
In the above formula, we can see that we need only percentage of C, H, O and S to calculate the GCV.
Now put the values in the Dulong’s formula,
GCV = (33820 x 59) + 144000 x (3.7 – (8.4/8)) + (9.204 x 1.7)
= 2377 kJ/kg