Question

The high-performance version of the CV12 has a specific fuel consumption of 0.060 kg/MJ at maximum power, and a minimum specific fuel consumption of 0.052 kg/MJ. The calorific value of the fuel is 45MJ/kg.

1) calculate the fuel efficiency at maximum power.

2) calculate the fuel efficiency at minimum power.

Answer 1

Given data:-

• Specific fuel consumption at maximum power =
• Specific fuel consumption at minimum power =
• Calorific value of fuel = CV = 45 MJ/kg

Now, fuel efficiency is the one of the forms of thermal efficiency of engine.

Fuel efficiency can be given by,

$\eta_{fuel} = \frac{1}{SFC \times CV}$,

Where SFC = Specific fuel consumption ( kg/MJ )

CV = Calorific value of fuel

(1) Fuel efficiency at maximum power

$\eta_{fuel1} = \frac{1}{SFC_{1} \times CV}$

= 1 / (0.060 kg/MJ × 45 MJ/kg)

= 0.37037

so, $\eta_{fuel1} =$ 37.037 %

(2) Fuel efficiency at minimum power

  $\eta_{fuel2} = \frac{1}{SFC_{2} \times CV}$

= 1 / ( 0.052 kg/MJ × 45 MJ/Kg )

= 0.42735

so, $\eta_{fuel2} =$ 42.735 %