Q13Basic Mechanical Engineering

Question

Q.3 Derive efficiency of Otto cycle.

Answer 1

The Otto cycle constitutes the absolute fundamental ideal thermodynamic model strictly governing all modern spark-ignition internal combustion engines (petrol engines). The cycle rigorously consists of four totally distinct, theoretically reversible thermodynamic processes traversing a fixed mass of ideal air continuously operating within a sealed cylinder.

The Four Thermodynamic Processes

Process 1-2 (Isentropic Compression): The piston systematically compresses the air adiabatically and reversibly from the maximum volume ( $V_{1}$ ) to the minimum clearance volume ( $V_{2}$ ). The absolute compression ratio is rigidly defined as $r = \frac{V _{1}}{V _{2}}$ .
Process 2-3 (Constant-Volume Heat Addition): External thermal heat ( $Q_{s}$ ) is instantaneously supplied strictly at constant geometric volume ( $V_{2} = V_{3}$ ).
Process 3-4 (Isentropic Expansion): The extremely high-pressure, hot air expands adiabatically and reversibly, forcefully driving the piston back completely to its initial physical volume ( $V_{4} = V_{1}$ ), actively producing all the useful mechanical work.
Process 4-1 (Constant-Volume Heat Rejection): Residual waste thermal heat ( $Q_{r}$ ) is instantaneously rejected strictly at constant volume, returning the exact fluid absolutely back to state 1.

Rigorous Mathematical Derivation

The fundamental universal thermal efficiency ( $η$ ) of absolutely any heat engine is strictly the mathematical ratio of the net mechanical work produced directly to the total heat supplied:

η = \frac{W _{net}}{Q _{s}} = \frac{Q _{s} - Q _{r}}{Q _{s}} = 1 - \frac{Q _{r}}{Q _{s}}

Because both the thermal heat addition and the thermal heat rejection physically occur strictly at absolutely constant geometric volumes, we can meticulously express these exact quantities explicitly using the specific heat at constant volume ( $C_{v}$ ):

Q_{s} = m C_{v} (T_{3} - T_{2}) and Q_{r} = m C_{v} (T_{4} - T_{1})

We seamlessly substitute these rigid expressions directly back into the primary efficiency equation:

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}}

We carefully algebraically factor out $T_{1}$ explicitly from the entire numerator and exactly $T_{2}$ from the denominator:

η = 1 - \frac{T _{1} ( \frac{T _{4}}{T _{1}} - 1 )}{T _{2} ( \frac{T _{3}}{T _{2}} - 1 )}

For the two mathematically perfectly isentropic processes (1-2 and 3-4), the rigorous fundamental temperature-volume relationship universally holds:

\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{γ - 1} = r^{γ - 1}

\frac{T _{3}}{T _{4}} = (\frac{V _{4}}{V _{3}})^{γ - 1} = (\frac{V _{1}}{V _{2}})^{γ - 1} = r^{γ - 1}

Because both exact isentropic ratios inherently equal strictly $r^{γ - 1}$ , it becomes mathematically undeniably proven that $\frac{T _{2}}{T _{1}} = \frac{T _{3}}{T _{4}}$ , which algebraically systematically rearranges exactly to $\frac{T _{4}}{T _{1}} = \frac{T _{3}}{T _{2}}$ .

Since exactly $\frac{T _{4}}{T _{1}} = \frac{T _{3}}{T _{2}}$ , the complex bracketed terms in our previous factored efficiency equation rigorously, mathematically cancel out absolutely perfectly:

η = 1 - \frac{T _{1}}{T _{2}}

Finally, we strictly substitute the inverse of our previously derived isentropic compression ratio relation $(\frac{T _{1}}{T _{2}} = \frac{1}{r ^{γ - 1}})$ strictly back into the equation, yielding the final, mathematically unassailable formula:

η = 1 - \frac{1}{r ^{γ - 1}}

Q13Basic Mechanical Engineering

Question

Answer

The Four Thermodynamic Processes

Rigorous Mathematical Derivation