奥托循环热效率

The derivation of the Otto cycle thermal efficiency formula begins with the general definition of thermal efficiency for any heat engine: [latex]\eta_{th} = \frac{W_{net}}{Q_{in}} = 1 – \frac{Q_{out}}{Q_{in}}[/latex], where [latex]W_{net}[/latex] is the net work output, [latex]Q_{in}[/latex] is the heat added, and [latex]Q_{out}[/latex] is the heat rejected. For the Otto cycle, heat is added at constant volume (process 2-3) and rejected at constant volume (process 4-1). Therefore, [latex]Q_{in} = m c_v (T_3 – T_2)[/latex] and [latex]Q_{out} = m c_v (T_4 – T_1)[/latex], where [latex]m[/latex] is the mass of the working fluid, [latex]c_v[/latex] is the specific heat at constant volume, and [latex]T[/latex] represents the temperature at the numbered states of the cycle.

Substituting these into the efficiency equation gives [latex]\eta_{th} = 1 – \frac{T_4 – T_1}{T_3 – T_2}[/latex]. To simplify this in terms of volumes, we use the relationships for the isentropic processes (1-2 and 3-4). For an isentropic process, [latex]TV^{\gamma-1} = \text{constant}[/latex]. Thus, [latex]\frac{T_2}{T_1} = (\frac{V_1}{V_2})^{\gamma-1} = r^{\gamma-1}[/latex] and [latex]\frac{T_3}{T_4} = (\frac{V_4}{V_3})^{\gamma-1} = r^{\gamma-1}[/latex]. This implies [latex]\frac{T_2}{T_1} = \frac{T_3}{T_4}[/latex] or [latex]\frac{T_4}{T_1} = \frac{T_3}{T_2}[/latex]. Rearranging the efficiency equation to [latex]\eta_{th} = 1 – \frac{T_1(T_4/T_1 – 1)}{T_2(T_3/T_2 – 1)}[/latex] and substituting the temperature ratio equality, the terms in the parentheses cancel out. This leaves [latex]\eta_{th} = 1 – \frac{T_1}{T_2}[/latex]. Finally, using the isentropic relation [latex]\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma-1} = \frac{1}{r^{\gamma-1}}[/latex], we arrive at the final formula: [latex]\eta_{th} = 1 – \frac{1}{r^{\gamma-1}}[/latex].