Question

An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumption with units of gkW/hr?

Answer 1

`s =0.21\ kg/Kw.hr`

Step-by-step explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

`\eta =\frac{BP}{\dot{m_f}* CV}`

Where BP is the brake power

`\eta =\frac{BP}{\dot{m_f}* CV}`

`0.379 =\frac{BP}{\dot{m_f}* 44000}`

`\frac{BP}{\dot{m_f}}=16676`

Brake specific fuel consumption (s)

`s =\frac{\dot{m_f}}{BP}`

`s =\frac{3600* \dot{m_f}}{BP}`

`s =(3600)/(16,676)\ kg/Kw.hr`

`s =0.21\ kg/Kw.hr`