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10 November, 07:10

An automobile engine consumes fuel at a rate of 22 L/hr and delivers 50kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and adensity of 2 g/cm3, determine the efficiency of this engine.

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  1. 10 November, 07:37
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    The efficiency of the engine is 9.3%

    Explanation:

    Efficiency = power output/power input * 100

    Power output = 50 kW

    Power input = heating value * density * volumetric flow rate

    Heating value = 44,000 kJ/kg

    Density = 2 g/cm^3 = 2/1000 = 0.002 kg/cm^3

    Volumetric flow rate = 22 L/hr = 22 L/hr * 1000 cm^3/1 L * 1 hr/3600 s = 6.11 cm^3/s

    Power input = 44,000 kJ/kg * 0.002 kg/cm^3 * 6.11 cm^3/s = 537.68 kJ/s = 537.68 kW

    Efficiency = 50 kW/537.68 kW * 100 = 9.3%
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