Q13Basic Mechanical Engineering

Question

Q.3. Derive an expression for the air standard efficiency of Otto cycle. Draw neat P-V and T-S diagrams.

Answer 1

The Air Standard Otto Cycle: The Otto cycle consists of four internally reversible thermodynamic processes operating on an ideal gas (air):

Process 1-2: Reversible Adiabatic (Isentropic) Compression. The piston moves from BDC to TDC. Volume decreases from $V_{1}$ to $V_{2}$ .
Process 2-3: Constant Volume Heat Addition. Spark ignition occurs instantly at TDC. Heat ( $Q_{in}$ ) is supplied. $V_{2} = V_{3}$ .
Process 3-4: Reversible Adiabatic (Isentropic) Expansion. The high-pressure gas pushes the piston down to BDC (Power stroke).
Process 4-1: Constant Volume Heat Rejection. Exhaust valve opens instantly at BDC. Heat ( $Q_{o u t}$ ) is rejected. $V_{4} = V_{1}$ .

Derivation of Efficiency: The thermal efficiency is defined as $η = \frac{Net Work}{Heat Supplied} = \frac{Q _{in} - Q _{o u t}}{Q _{in}} = 1 - \frac{Q _{o u t}}{Q _{in}}$ Using the specific heat at constant volume ( $C_{v}$ ): $Q_{in} = m C_{v} (T_{3} - T_{2})$ $Q_{o u t} = m C_{v} (T_{4} - T_{1})$ Therefore, $η = 1 - \frac{m C _{v} ( T _{4} - T _{1} )}{m C _{v} ( T _{3} - T _{2} )} = 1 - \frac{T _{4} - T _{1}}{T _{3} - T _{2}}$ Pulling out $T_{1}$ and $T_{2}$ : $η = 1 - \frac{T _{1} ( T _{4} / T _{1} - 1 )}{T _{2} ( T _{3} / T _{2} - 1 )}$

For the isentropic processes 1-2 and 3-4, utilizing the compression ratio $r = \frac{V _{1}}{V _{2}} = \frac{V _{4}}{V _{3}}$ : $\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{γ - 1} = r^{γ - 1}$ and $\frac{T _{3}}{T _{4}} = (\frac{V _{4}}{V _{3}})^{γ - 1} = r^{γ - 1}$ Thus, $\frac{T _{2}}{T _{1}} = \frac{T _{3}}{T _{4}}$ , which rearranges to $\frac{T _{4}}{T _{1}} = \frac{T _{3}}{T _{2}}$ . Substituting this back into the efficiency equation cancels out the bracketed terms, leaving: $η_{O tt o} = 1 - \frac{T _{1}}{T _{2}} = 1 - \frac{1}{r ^{γ - 1}}$

Q13Basic Mechanical Engineering

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