Question

Двигатель мощностью 18кВт за 3 часа потратил 12 кг киросина. Какой КПД этого двигателя

Answer 1

1 кг керосина даёт при сгорании 40.8*10^3 кДж тепла, стало быть 12 кг горючего - 4.989*10^5 кДж. По закону сохранения энергии, это максимально возможная работа двигателя (А_max). Реальная работа двигателя (A_real) равна произведению мощности на время: 18*3*60*60 = 1.944*10^5 кДж. Тогда КПД двигателя = (A_real /A_max)*100% = 39.

Answer 2

Calculating the Efficiency of the Engine

To calculate the efficiency of the engine, we can use the formula:

Efficiency (η) = (Useful energy output / Energy input) x 100%

In this case, the useful energy output is the work done by the engine, and the energy input is the energy provided by the fuel.

Given Information

- Power of the engine (P) = 18 kW - Time taken (t) = 3 hours - Fuel consumption = 12 kg

Calculating Energy Input

The energy input can be calculated using the fuel consumption and its energy content. The energy content of kerosene is approximately 43 MJ/kg.

Energy Input = Fuel consumption x Energy content of kerosene

Energy Input = 12 kg x 43 MJ/kg

Energy Input = 516 MJ

Calculating Useful Energy Output

The useful energy output can be calculated using the power of the engine and the time taken.

Useful Energy Output = Power x Time

Useful Energy Output = 18 kW x 3 hours

Useful Energy Output = 54 kWh

Calculating Efficiency

Now, we can calculate the efficiency using the formula mentioned earlier.

Efficiency (η) = (Useful energy output / Energy input) x 100%

Efficiency (η) = (54 kWh / 516 MJ) x 100%

Efficiency (η) ≈ 0.1047%

So, the efficiency of this engine is approximately 0.1047%.

The search results did not provide specific information related to the efficiency of an engine with the given parameters. Therefore, the efficiency calculation is based on standard formulas and assumptions about the energy content of kerosene and the power and time parameters provided.