The thermal efficiency of the ideal diesel cycle is given by equa
A. <span class="math-tex">\({\eta _{diesel}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\frac{{{(\rho ^\gamma } - 1)}}{{\gamma \left( {\rho - 1} \right)}}\)</span>
B. <span class="math-tex">\({\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)</span>
C. <span class="math-tex">\({\eta _{diesel}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\frac{{\gamma({\rho } - 1})}{{ \left( {\rho^\gamma - 1} \right)}}\)</span>
D. none of the above
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Right Answer is: A
SOLUTION
Concept:
Diesel cycle:
Processes involved in compression engine (diesel cycle):
Process 1-2: Reversible adiabatic compression
Process 2-3: Constant pressure heat addition
Process 3-4: Reversible adiabatic expansion
Process 4-1: Constant volume of heat rejection
Cut off ratio: \({\rho} = \frac{{{V_3}}}{{{V_2}}}\)
Compression ratio: \({r} = \frac{{{V_1}}}{{{V_2}}} \)
The efficiency of the diesel cycle is given by
\(\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}\left[ {\frac{{\rho^\gamma - 1}}{{\gamma \left( {{\rho} - 1} \right)}}} \right]\)
