Answer:
[tex]s =0.21\ kg/Kw.hr[/tex]
Explanation:
Given that
Calorific value (CV) = 44 MJ/Kg
CV= 44,000 KJ/kg
Brake thermal efficiency(η) = 37.9 %
We know that
[tex]\eta =\dfrac{BP}{\dot{m_f}\times CV}[/tex]
Where BP is the brake power
[tex]\eta =\dfrac{BP}{\dot{m_f}\times CV}[/tex]
[tex]0.379 =\dfrac{BP}{\dot{m_f}\times 44000}[/tex]
[tex]\dfrac{BP}{\dot{m_f}}=16676[/tex]
Brake specific fuel consumption (s)
[tex]s =\dfrac{\dot{m_f}}{BP}[/tex]
[tex]s =\dfrac{3600\times \dot{m_f}}{BP}[/tex]
[tex]s =\dfrac{3600}{16,676}\ kg/Kw.hr[/tex]
[tex]s =0.21\ kg/Kw.hr[/tex]
