Diesel engine 35 kW with 40% efficiency. What amount of fuel (Kerosene) does it consume per hour?

Diesel engine 35 kW with 40% efficiency. What amount of fuel (Kerosene) does it consume per hour? How much heat is dissipated during its operation into the environment during this time?

Data: N (diesel engine power) = 35 kW (35 * 10 ^ 3 W); η (efficiency) = 40% (0.4); t (duration of operation) = 1 hour (3600 s).

Constants: q (specific heat of combustion of kerosene) = 46 MJ / kg (46 * 10 ^ 6) J / kg.

1) Kerosene consumption: η = N / (Q / t) = N * t / (q * m), whence m = N / (q * η) = N * t / (q * η) = 35 * 10 ^ 3 * 3600 / (46 * 10 ^ 6 * 0.4) = 6.85 kg.

2) Heat losses: η = Qp / (Qp + Q) and Q = Qp / η – Qp = N * t / η – N * t = N * t * (1 / η – 1) = 35 * 10 ^ 3 * 3600 * (1 / 0.4 – 1) = 189 * 10 ^ 6 J.