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

Answer:

$s =0.21\ kg/Kw.hr$

Explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

Where BP is the brake power

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

$0.379 =\dfrac{BP}{\dot{m_f}\times 44000}$

$\dfrac{BP}{\dot{m_f}}=16676$

Brake specific fuel consumption (s)

$s =\dfrac{\dot{m_f}}{BP}$

$s =\dfrac{3600\times \dot{m_f}}{BP}$

$s =\dfrac{3600}{16,676}\ kg/Kw.hr$

$s =0.21\ kg/Kw.hr$