Question

An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine the power output and the rate of heat rejection.

Answer 1

Efficiency = Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

= 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts

Power input = 0.061667 MW = 61 667 W

From Efficiency = Power Output / Power Input

28% = Power Output / 61667

Power Output = 0.28 * 61667

Power Output = 17266.76 W

Power Output ≈ 17 267 W

Rate of heat rejection = Power Input - Power Output

= 61667 - 17267 = 44400 W

Rate of heat rejection = 44 400W.