Q. 4.18

Question

Derive equation $4.10$ for the efficiency of the Otto cycle.

Step-by-Step Solution

Verified

Answer

The efficiency of the Otto cycle is $η = 1 - {(\frac{V_{2}}{V_{1}})}^{γ - 1}$

1Step 1 : Introduction

Efficiency $(η) = \frac{Net work done (W_{net})}{Heat supplied (Q_{1})}$

Heat supplied:

$Q_{1} = m C_{v} (T_{3} - T_{2}) - - - (1)$

Hear rejected:

$\begin{aligned} Q_{2} = m C_{v} (T_{4} - T_{1}) - - - (2) \\ η = \frac{Q_{1} - Q_{2}}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}} - - (3) \\ From (1) and (2) \\ η = 1 - \frac{m C_{v} (T_{4} - T_{1})}{m C_{4} (T_{3} - T_{2})} \\ η = 1 - \frac{(T_{4} - T_{1})}{(T_{3} - T_{2})} - - (4) \end{aligned}$

2Step 2 : Calculation

As it is an adiabatic process:

$\begin{aligned} T_{2} V_{2}^{γ - 1} = T_{1} V_{1}^{γ - 1} \\ \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \end{aligned}$

Similarly,

$\begin{aligned} T_{2} V_{2}^{γ - 1} = T_{1} V_{1}^{γ - 1} \\ \frac{T_{3}}{T_{4}} = {(\frac{V_{4}}{V_{3}})}^{γ - 1} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \\ ∴ \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{4}} \end{aligned}$

$\frac{T_{3}}{T_{2}} = \frac{T_{4}}{T_{1}} - - - - (5)$

Subtract the above equation by 1 both sides:

$\begin{aligned} \frac{T_{3}}{T_{2}} - 1 = \frac{T_{4}}{T_{1}} - 1 \\ \frac{T_{3} - T_{2}}{T_{2}} = \frac{T_{4} - T_{1}}{T_{1}} \\ \frac{T_{1}}{T_{2}} = \frac{T_{4} - T_{1}}{T_{3} - T_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ - 1} \end{aligned}$