The ship is equipped with a 80 kW diesel engine with an efficiency of 30%.

The ship is equipped with a 80 kW diesel engine with an efficiency of 30%. How many kilometers of the way will it have enough diesel fuel weighing 1 ton at a speed of 20 km / h.

Np = 80 kW = 80,000 W.

m = 1 t = 1000 kg.

V = 20 km / h = 5.6 m / s.

q = 42.7 * 106 J / kg.

Efficiency = 30%.

S -?

During the combustion of diesel fuel, the amount of heat Q is released in the motor ship engine. Only efficiency = 30% of this amount of heat is spent on the movement of the motor ship, that is, on performing useful work Ap.

Efficiency = Ap * 100% / Q.

Q = q * m.

When the motor ship is moving, we will express its useful work by the formula: Ap = Np * t, where Np is the useful power of the engine, t is the time of movement of the motor ship.

With uniform rectilinear movement of the motor ship, the time of its movement t will be the ratio of the distance traveled S to the speed of its movement V: t = S / V.

Ap = Np * S / V.

Efficiency = Np * S * 100% / V * q * m.

S = efficiency * V * q * m / Np * 100%.

S = 30% * 5.6 m / s * 42.7 * 106 J / kg * 1000 kg / 80,000 W * 100% = 896,700 m = 896.7 km.

Answer: the diesel fuel will be enough for the motor ship to cover the distance S = 896.7 km.