Question

2.Двигун внутрішнього згоряння автомобіля має потужність 36,7 кВт, а його ККД становить 20%. Яку кількість бензину він споживає за 1 год роботи?

Answer 1

Ответ: 14,36кг

Объяснение:

Дано:

W=36,7кВт=36700Вт

η=20%=0,2

q=46000000Дж/кг

t=1ч=3600с

-----------------------

m-?

η=А/Q, где:

А-работа;

Q-теплота сгоревшего топлива.

А=W*t

Q=qm

Отсюда:

m=W*t/(qη)=36700*3600/(46000000*0,2)=14,36кг

Answer 2

#### Calculation of Fuel Consumption To calculate the fuel consumption of an internal combustion engine, we need to know the power output of the engine and its efficiency. In this case, the engine has a power output of 36.7 kW and an efficiency of 20%. The fuel consumption can be calculated using the formula: **Fuel Consumption = Power Output / (Efficiency * Energy Content of Fuel)** The energy content of gasoline is approximately 33.6 MJ/L However, we need to convert the power output from kilowatts to megajoules per hour to match the energy content unit. Converting 36.7 kW to megajoules per hour: **36.7 kW * 1 hour * 3.6 MJ/kWh = 132.12 MJ/h** Now we can calculate the fuel consumption: **Fuel Consumption = 132.12 MJ/h / (0.20 * 33.6 MJ/L)** Let's calculate the result: Fuel Consumption = 132.12 MJ/h / (0.20 * 33.6 MJ/L) = 132.12 MJ/h / 6.72 MJ/L = 19.64 L/h Therefore, the engine will consume approximately **19.64 liters of gasoline per hour** of operation. Please note that the above calculation assumes a constant power output and efficiency throughout the hour of operation. In real-world scenarios, the power output and efficiency may vary depending on various factors such as engine load, driving conditions, and maintenance.