The engine consumed 6 kg of kerosene in 2 hours. What is the power of this engine if its efficiency is 25%?

univerkov | February 20, 2021 | education

We calculate the engine efficiency using the formula:

η = (A / Q) * 100%.

η = 25%.

Useful engine work, A = N * t, where N is engine power, t is movement time (t = 2 h = 2 * 3600 = 7200 s).

Energy expended, Q = q * m, q is the specific heat of combustion of kerosene (q = 46 MJ = 46 * 10 ^ 6 J / kg), m is the mass of consumed kerosene (m = 6 kg).

η = N * t / (q * m) * 100%.

N = η * q * m / t = 0.25 * 46 * 10 ^ 6 * 6/7200 = 9583.3 W ≈ 9.6 kW.

Answer: The motor power is 9.6 kW.

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