Question 2.48: An engine working on Carnot cycle absorbs Q1 units of heat f...
An engine working on Carnot cycle absorbs Q_1 units of heat from a source at T_1 and rejects Q_2 units of heat to a sink at T_2 . The temperature of the working fluid is \theta_1 and \theta_2 , where \theta_1<T_1 and \theta_2>T_2.
If \theta_1=T_1-k Q_1 \text { and } \theta_2=T_2+k Q_2
where k is constant, then show that efficiency of the engine is given by :
\eta=1-\frac{T_2}{T_1-2 k Q_1}
\eta=1-\frac{Q_2}{Q_1}=1-\frac{\theta_2}{\theta_1}
or \quad \frac{Q_2}{Q_1}=\frac{\theta_2}{\theta_1}
Also \left\lgroup \frac{\theta_2-T_2}{T_1-\theta_1}\right\rgroup=\frac{Q_2}{Q_1} (given)
\begin{aligned}\left\lgroup \frac{\theta_2-T_2}{T_1-\theta_1}\right\rgroup &=\frac{\theta_2}{\theta_1} \\ \\\theta_1 \theta_2-\theta_1 T_2 &=\theta_2 T_1-\theta_1 \theta_2 \\ \\\theta_2\left(2 \theta_1-T_1\right) &=\theta_1 T_2\end{aligned}
\begin{aligned}\theta_2 &=\left\lgroup \frac{\theta_1 T_2}{2 \theta_1-T_1}\right\rgroup=\frac{\theta_1 T_2}{2\left(T_1-k Q_1\right)-T_1} \\ \\&=\frac{\theta_1 T_2}{T_1-2 k Q_1}\end{aligned}
\therefore \quad \eta=1-\frac{T_2}{T_1-2 k Q_1}
Hence Proved.