Question

An automobile engine consumes fuel at a rate of 22 L/h and delivers 85 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine.

Answer 1

Answer:

Efficiency of the engine is 39.51 %

Explanation:

Mass of fuel consumed per hour = 22 L/h * 0.08 Kg/L = 17.6 Kg/h

Total Energy Consumed per hour = Mass of fuel consumed per hour * Heating  value of fuel

Total Energy Consumed per hour = 17.6 Kg/h * 44,000 KJ/Kg = 774,400 KJ/h

since, 1 KW = 3600 KJ/h

774,400KJ/h = 774,400/3600 kW = 215.1 kW

$Efficiency of the engine = \frac{Power to the wheels}{Total Power}$

Efficiency = 85/215.1 = 39.51 %

Answer 2

Answer:

The efficiency of the engine is 39.53%

Explanation:

data given by the exercise

HV = heating value = 44000 kJ/kg

Vfuel = volume flow rate of fuel = 22 L/h = 0.0061 L/s

W = power output = 55 kW

p = density = 0.8 g/cm³

the mass flow rate of fuel is equal to

$m_{fuel} =V*p=0.0061*0.8=4.89x10^{-3} kg/s$

the rate of heat supplied is

$Q_{in} =m_{fuel} *HV=4.89x10^{-3} *44000=215kW$

the efficiency is equal to

$n=\frac{W}{Q_{in} } *100=\frac{85}{215} *100=39.53$%