Question
Download Solution PDFA diesel engine has a brake thermal efficiency of 30%. If the calorific value of fuel is 40 MJ/kg, then what will be brake specific fuel consumption?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Brake thermal efficiency is given as:
\(η_b = \frac{B.P}{\dot{m}\;×\;C.V}\) ...(i)
Where, ṁ = mass flow rate of fuel, B.P = Brake Power, C.V = Calorific value
Brake specific fuel consumption is given as:
\(bsfc = \frac{\dot{m}}{B.P}\) ...(ii)
Combining (i) and (ii) we have
\(bsfc = \frac{1}{η_b\;×\;C.V}\) ...(iii)
Calculation:
Given:
ηb = 30% = 0.3, C.V = 40 MJ/kg = 40000 kJ/kg
Using equation (iii)
\(bsfc = \frac{1}{η_b\;×\;C.V}\)
\(\Rightarrow bsfc = \frac{1}{0.3\;×\;40000} =8.33\times10^-5 \ \rm{kg/kWsec } \)
\(\therefore bsfc=8.33\times10^-5\times3600=0.3\ \rm{kg/kWh} \)
Last updated on Sep 23, 2024
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