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12 March, 17:21

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?

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  1. 12 March, 17:39
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    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.
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