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
1 answer:
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 = 44 400 W.
C- Copyright.
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