In the boiler house of the power plant for 10 hours of operation, 100 tons of coal with a heat of combustion

In the boiler house of the power plant for 10 hours of operation, 100 tons of coal with a heat of combustion Qрн = 29300 kJ / kg were burned. Determine the amount of generated electricity in the average capacity of the station, if the efficiency is the process of converting thermal energy into electrical energy is 20%.

Let’s translate all the data into SI units:

10 h = 10 h * 3600 s = 36,000 s,

100 t = 100,000 kg,

29,300 KJ = 29,300,000 J.

Let’s calculate how much energy is generated as a result of the combustion of all coal:

Q = Qpn * m = 29300000 * 100000 = 2,930,000,000 J.

Because Efficiency = Apol / Atot = 0.2 (20%), then Apol = 0.2 * Q = 586,000,000 J.

The electric power of the station, taking into account the efficiency, will be equal to:

N = Apol / t = 586,000,000 / 36,000 = 16,277.8 W = 16.3 kW.

The amount of electricity generated is calculated taking into account the fact that 1 kWh = 1000 W * 3600 s = 3 600 000 J, therefore:

A = 586,000,000 / 3,600,000 = 162.8 kWh.

Answer: N = 16.3 KW, A = 162.8 KWh.