The Volga car engine develops a power of 55 kW and consumes 0.31 kg

| April 9, 2021 | education

The Volga car engine develops a power of 55 kW and consumes 0.31 kg of gasoline per 1 kWh of energy. Determine the efficiency of the engine.

N = 55 kW = 55000 W.

N1 = 1 kW = 1000 W.

m1 = 0.31 kg.

t1 = 1 h = 3600 s.

q = 4.6 * 10 ^ 7 J / kg.

Efficiency -?

Let us write down the definition for the efficiency of a car engine: efficiency = Npol * 100% / Nzat, where Npol is the useful engine power, Nzat is the power that is obtained from the thermal energy of fuel combustion.

Npol = N.

Nzat = Q1 / t1.

Let’s find the amount of heat energy Q1, which is obtained during the combustion of gasoline: Q1 = q * m1.

Q1 = 4.6 * 10 ^ 7 J / kg * 0.31 kg = 1.426 * 10 ^ 7 J.

Nzat = 1.426 * 10 ^ 7 J / 3600 s = 3961.1 W.

Efficiency = 55000 W * 100% / 3961.1 W = 1388.5%.

The efficiency cannot be more than 100%; an error was made in the problem statement.

Answer: efficiency = 1388.5%.



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