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Question

(C) Let the heat input to the first engine be $Q_1$ and the work done by it be $W_1$. The heat rejected by the first engine is $Q_2 = Q_1(1 - \eta_1)$

Vedclass · Accepted Answer

$\eta_{net} = \eta_1 + (1 - \eta_1)\eta_2$

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(C) Let the heat input to the first engine be $Q_1$ and the work done by it be $W_1$. The heat rejected by the first engine is $Q_2 = Q_1(1 - \eta_1)$.This $Q_2$ acts as the input for the second engine. The work done by the second engine is $W_2 = Q_2 \eta_2 = Q_1(1 - \eta_1)\eta_2$.The total work done by the combination is $W_{net} = W_1 + W_2 = Q_1 \eta_1 + Q_1(1 - \eta_1)\eta_2$.The net efficiency is $\eta_{net} = \frac{W_{net}}{Q_1} = \frac{Q_1 \eta_1 + Q_1(1 - \eta_1)\eta_2}{Q_1}$.Therefore,$\eta_{net} = \eta_1 + (1 - \eta_1)\eta_2$.

Suppose that two heat engines are connected in series,such that the heat exhaust of the first engine is used as the heat input of the second engine as shown in the figure. The efficiencies of the engines are $\eta_1$ and $\eta_2$,respectively. The net efficiency of the combination is given by

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