Question

Coal consumption of the thermal power plant is 100 tonnes of anthracite coal per day. CV of
caol is 21000KJ/kg. If the power generation in one day is 500KW. Find the overall
efficiency of the thermal power plant

Answer 1

The power engineer must do some fundamental calculations involving the power plant's essential characteristics. The amount and cost of gasoline required are the most essential factors.

The power plant's primary job is to transform coal energy into electricity. As a result, the first thing we need to know is how much energy coal contains.

Given:

Quantity of coal used in a day = m = 100000 kg

Calorific Value of coal used = CV = 21000 kJ/kg

Power generation = $Power_{out}$ = 500 kW

∴ Total energy produced by coal = $Power_{in}$ = m × CV

∴ $Power_{in}$ = 2,10,00,00,000 kJ for 24 hours

1 kW = 3600 kJ/hr

∴ $Power_{in}$ = 24,305 kW

Efficiency of the thermal power plant = $\frac{Power_{out} }{Power_{in} }$

∴ Efficiency of the thermal power plant = $\frac{500}{24305 }$

∴ Efficiency of the thermal power plant = 0.0206

∴ Efficiency of the thermal power plant = 2.06%