For an ideal Otto cycle, T₁ and T₃ are the lower and upper limits of absolute temperature respectively. What will be compression ratio(R) for maximum work output of the cycle? [γ = Ratio of specific heat \(\left.=\frac{C_p}{C_v}\right]\)

Concept:

In an ideal Otto cycle, the net work output is the difference between heat added and heat rejected. To maximize this work output for a given temperature range, we find the optimum compression ratio \(R\).

The net work is expressed in terms of temperatures as:

\( W = C_v \left[ T_3 - T_4 - (T_2 - T_1) \right] \)

Since processes 1-2 and 3-4 are isentropic, we use the relations:

\( T_2 = T_1 R^{\gamma - 1} \) and \( T_4 = \frac{T_3}{R^{\gamma - 1}} \)

Substitute these into the work expression:

\( W = C_v \left[ T_3 - \frac{T_3}{R^{\gamma - 1}} - (T_1 R^{\gamma - 1} - T_1) \right] \)

Simplifying,

\( W = C_v \left[ T_3 \left(1 - \frac{1}{R^{\gamma - 1}} \right) - T_1 \left( R^{\gamma - 1} - 1 \right) \right] \)

Calculation:

To find the compression ratio for maximum work, we differentiate W with respect to R and set it to zero:

\( \frac{dW}{dR} = 0 \)

This gives the condition for maximum work output:

\( R^{2(\gamma - 1)} = \frac{T_3}{T_1} \)

Taking power on both sides:

\( R = \left( \frac{T_3}{T_1} \right)^{\frac{1}{2(\gamma - 1)}} \)