The mini-power plant generates electricity to power the 800 W. If, in one hour of operation, diesel fuel weighing 0.4 kg
| February 8, 2021 | education
The mini-power plant generates electricity to power the 800 W. If, in one hour of operation, diesel fuel weighing 0.4 kg with a calorific value of 30 MJ / kg is consumed, then the efficiency of the power plant is?
Initial data: N (power required to power the installation) = 800 W; t (unit operation time) = 1 h (3600 s); m (mass of consumed diesel fuel) = 0.4 kg.
Reference data: q (specific heat of combustion of diesel fuel) = 30 MJ / kg = 30 * 10 ^ 6 J / kg.
The efficiency of the power plant is determined by the formula: η = Ap / Az = N * t / (q * m) = 800 * 3600 / (30 * 10 ^ 6 * 0.4) = 0.24 (24%).
Answer: The efficiency of the mini-power plant is 24%.
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