When burning gasoline in a car engine for 2 seconds, energy losses amounted to 300 kJ, while the engine
| August 25, 2021 | education
When burning gasoline in a car engine for 2 seconds, energy losses amounted to 300 kJ, while the engine developed a power of 50 kW. Determine the efficiency of the motor.
t = 2 s.
Wpot = 300 kJ = 3 * 10 ^ 5 J.
Np = 50 kW = 5 * 10 ^ 4 W.
Efficiency -?
The formula for determining the efficiency coefficient of the engine efficiency will be as follows: efficiency = Ap * 100% / Az.
We will express the useful work Ap by the formula: Ap = Az – Wpot.
Az = An + Wpot.
The expended work Az is expressed through the power: An = Nz * t.
Ap = 5 * 10 ^ 4 W * 2 s = 10 * 10 ^ 4 J = 1 * 10 ^ 5 J.
Az = 1 * 10 ^ 5 J +
Az = 1 * 10 ^ 5 J + 3 * 10 ^ 5 J = 4 * 10 ^ 5 J /
Efficiency = 1 * 10 ^ 5 J * 100% / 4 * 10 ^ 5 J = 25%.
Answer: a gasoline engine has an efficiency of 25%.
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