A Carnot engine C_1 operates between temperat | vedclass

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Question

(A) Let $Q_1$ be the heat absorbed by engine $C_1$ from the source at $T_1$, and $Q_2$ be the heat rejected by $C_1$ at $T_2$.The efficiency of engine

Vedclass · Accepted Answer

$1 - \frac{T_3}{T_1}$

Vedclass · Answer

(A) Let $Q_1$ be the heat absorbed by engine $C_1$ from the source at $T_1$, and $Q_2$ be the heat rejected by $C_1$ at $T_2$.The efficiency of engine $C_1$ is $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{Q_1 - Q_2}{Q_1}$.Engine $C_2$ absorbs the heat $Q_2$ rejected by $C_1$ and rejects heat $Q_3$ at $T_3$.The efficiency of engine $C_2$ is $\eta_2 = 1 - \frac{T_3}{T_2} = \frac{Q_2 - Q_3}{Q_2}$.The total work done by the combined system is $W = W_1 + W_2 = (Q_1 - Q_2) + (Q_2 - Q_3) = Q_1 - Q_3$.The total heat absorbed by the combined system is $Q_1$.The efficiency of the combined engine is $\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_3}{Q_1} = 1 - \frac{Q_3}{Q_1}$.From the efficiency formulas: $Q_2 = Q_1(1 - \eta_1) = Q_1 \frac{T_2}{T_1}$ and $Q_3 = Q_2(1 - \eta_2) = Q_2 \frac{T_3}{T_2}$.Substituting $Q_2$ into the expression for $Q_3$: $Q_3 = \left(Q_1 \frac{T_2}{T_1}\right) \frac{T_3}{T_2} = Q_1 \frac{T_3}{T_1}$.Therefore, the efficiency of the combined engine is $\eta = 1 - \frac{Q_3}{Q_1} = 1 - \frac{T_3}{T_1}$.

Similar Questions

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