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.

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%