Find the efficiency of a tractor engine that develops a power of 110 kW and consumes 28 kg of diesel fuel per hour.

Initial data: N (the power that the tractor engine develops) = 110 kW (110 * 10 ^ 3 W); t (duration of the tractor engine) = 1 hour (3600 s); m (mass of consumed diesel fuel) = 28 kg.

Reference data: q (specific heat of combustion of diesel fuel) = 42.7 * 10 ^ 6 J / kg.

Let us derive a formula for calculating the efficiency of a tractor engine: η = Ap / Az = N * t / (q * m).

Calculation: η = 110 * 10 ^ 3 * 3600 / (42.7 * 10 ^ 6 * 28) = 0.331 or 33.1%.

Answer: The efficiency of the tractor engine is 33.1%.