Question
Download Solution PDFThe Brake-specific fuel consumption (BSFC) of a diesel engine is 200 gm/kWh. If the calorific value of diesel is 40 MJ/kg, the brake thermal efficiency is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Brake specific fuel consumption (BSFC) =mf/BP
Where mf = mass flow rate of fuel, BP = Brake Power
\(Brake\;thermal\;efficiency\left( {{\eta _b}} \right) = \frac{{BP}}{{{m_f} \times CV}} = \frac{1}{{BSFC \times CV}}\)
CV = Calorific Value
Calculation:
Given:
CV = 40 MJ/kg = 40 × 106 J/kg.
\(BSFC = 200\;gm/kWh = \frac{{200 \times {{10}^{ - 3}}}}{{\left( {3600 \times {{10}^3}} \right)}}\;kg/J = \frac{1}{{18}} \times {10^{ - 6}}\;kg/J\)
\(\eta = \frac{1}{{\left( {\frac{1}{{18}}} \right) \times {{10}^{ - 6}} \times 40 \times {{10}^6}}} = \frac{{18}}{{40}} = 0.45 = 45\% \)
Last updated on May 28, 2025
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