Question

An automobile engine consumes fuel at a rate of 22 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 0.8 g/cm3, determine the efficiency of this engine.

Answer 1

Given Information:

Output Power = P = 55 kW

Fuel Consumption Rate = ν = 22 L/h

Fuel Heating Value q = 44,000 kJ/kg

Fuel Density = ρ = 0.8 g/cm³

Required Information:

Efficiency of the engine = η = ?

Answer:

Efficiency of the engine = 25.56%

Explanation:

The efficiency of the engine is given by

η = P/Q

Where P is the output power and Q is the input heat supply

The heat supply is given by

Q = ρνq

Where ν is fuel consumption rate and q is the heating value of the fuel and ρ is density

First convert density from g/cm³ to kg/L

0.8*1000L/1000kg = 0.8 kg/L

Q = 0.8*22*44,000

Q = 7.74x10⁵ kJ/h

Convert kJ/h to kW (kW and kJ/s are equivalent units)

Q = 7.74x10⁵/3600

Q = 215.11 kW

Finally, now we can find the efficiency

η = P/Q

η = 55/215.11

η = 0.2556

η = 25.56%

Therefore, the efficiency of this engine is 25.56%

Answer 2

= 25.6%

Step-by-step explanation:

Given that,

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P = 55 kW

heating value, H = 44,000 kJ/kg

Calculate energy intake

E = C * P * H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6%