An automobile engine consumes fuel at a rate of 27.4 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.73 g/cm3, deter- mine the efficiency of this engine in percentage(

The efficiency of the engine is 22.5%.

Explanation:

Efficiency = power output ÷ power input

power output = 55 kW

power input = specific energy×volumetric flow rate×density

specific energy = 44,000 kJ/kg

volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s

density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3

power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW

Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%

answer:

yes that is right have a good day

answer:

yes

explanation: