Question

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

Answer 1

The formula for efficiency is this one:
Efficiency =  Power Output / Power Input

Power Input is just equal to Rate of Energy input
= 44.4 MJ/kg * 5 kg/h
= 222 MJ/h

Take note that 1 hour = 3600seconds
= 222 MJ/h
= 222 MJ/3600s
=  0.061667 MW            

J / S = Watts

Power input = 0.061667 MW = 61 667 W

Going back to formula:
Efficiency =  Power Output / Power Input
28% =  Power Output / 61667
Power Output = 0.28 * 61667
Power Output = 17266.76 W

Power Output ≈ 17 267 W 

The power output is approximately 17, 267 W.

Rate of heat rejection = Power Input - Power Output
Rate of heat rejection =   61667 - 17267
Rate of heat rejection =  44400 W.

So the heat of rejection is 44,400 W.