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.