A 20 kw petrol engine consume 6 kg of petrol per hr find its efficiency
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Answer:
About 31%
Power In:
Pin = 11*10^3 kcal/kg * 5 kg/hr
Pin = 55*10^3 kcal/hr
unit conversion:
1 kcal/hr = 0.001163 kW
Pin = 55*10^3 kcal/hr * 0.001163 kW /kcal/hr
Pin = 63.965 kW
Power Out:
Pout = 20 kW
Efficiency:
Eff = Pout / Pin
Eff = 20 kW / 63.965 kW
Eff = 0.3127 or 31.27%
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