Question

A 20 kw petrol engine consume 6 kg of petrol per hr find its efficiency ​

Answer 1

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%