The 19 kW heat engine consumed 12 kg in 3 hours of operation. kerosene. Determine the efficiency?

To find out the efficiency of the presented heat engine, consider the formula: ηx = (N * t) / (qk * mk), where N is the power of the presented heat engine (N = 19 kW = 19 * 10 ^ 3 W); t is the duration of the engine operation (t = 3 h = 10.8 * 10 ^ 3 s); qk – beats. combustion heat of kerosene (qk = 43.7 * 10 ^ 6 J / kg); mk is the mass of kerosene consumed by the heat engine (mk = 12 kg).

Let’s perform the calculation: ηx = (N * t) / (qk * mk) = (19 * 10 ^ 3 * 10.8 * 10 ^ 3) / (43.7 * 10 ^ 6 * 12) ≈ 0.391 = 39.1 %.

Answer: The efficiency of the presented heat engine is 39.1%.