How much coal is required for a steam locomotive developing a power of 1470 kW

| February 13, 2021 | education

How much coal is required for a steam locomotive developing a power of 1470 kW at efficiency? 7.5% to cover 1000 km at an average speed of 54 km / h?

N = 1470 kW = 1.47 * 10 ^ 6 W.

S = 1000 km = 10 ^ 6 m.

V = 54 km / h = 15 m / s.

Efficiency = 7.5%.

λ = 2.7 * 10 ^ 7 J / kg.

m -?

When coal is burned, the amount of heat Q is released from it, which we express by the formula: Q = λ * m.

Only 7.5% of this amount of heat is converted into useful work by Apol.

Q * efficiency / 100% = Apol.

Find useful work Apol by the formula: Apol = F * S, where F is the traction force of the train, S is the movement of the train.

We express the power N by the formula: N = F * V.

F = N / V.

Apol = N * S / V.

λ * m * efficiency / 100% = N * S / V.

m = N * S * 100% / V * λ * efficiency.

m = 1.47 * 10 ^ 6 W * 10 ^ 6 m * 100% / 15 m / s * 2.7 * 10 ^ 7 J / kg * 7.5% = 48390 kg.

Answer: for the train to move, it is necessary to burn coal m = 48390 kg.



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