Answer:
Efficiency = 0.273 = 27.3%
Explanation:
The ideal energy that can be produced by certain mass of diesel fuel is given as follows:
[tex]E_{ideal} = (HV)(m)[/tex]
where,
HV = Heating Value of Diesel = 42 - 46 MJ = 44 MJ (Taking average value)
m = mass of diesel fuel = 0.3 kg
E(ideal) = Ideal energy produced by this mass of fuel = ?
Therefore,
[tex]E_{ideal} = (44\ MJ/kg)(0.3\ kg)\\E_{ideal} = 13.2\ MJ\\[/tex]
So, the efficiency can be given as:
[tex]Efficiency = \frac{E_{actual}}{E_{ideal}}\\\\Efficiency = \frac{3.6\ MJ}{13.2\ MJ}[/tex]
Efficiency = 0.273 = 27.3%
