Consider two Carnot engines of efficiencies $\eta_1$ and $\eta_2$. The first engine absorbs heat $Q_1$ from a heat reservoir A and releases heat $Q_2$ to a heat reservoir B. The second engine takes heat $Q_2$ from B and releases heat $Q_3$ to a heat reservoir C. If $Q_1 > Q_2 > Q_3$, what is the net efficiency of this combination of the two Carnot engines?
Show Hint
For any number of engines operating in series, the overall fraction of heat remaining is the product of the individual remaining fractions:
\[ (1 - \eta_{net}) = (1 - \eta_1)(1 - \eta_2)\dots(1 - \eta_n) \]
Expanding this product for two stages immediately yields the correct net efficiency.
Step 1: Understanding the Question:
We are given two Carnot engines operating in series.
The first engine takes heat $Q_1$ from reservoir A and rejects $Q_2$ to reservoir B.
The second engine takes this rejected heat $Q_2$ from reservoir B and rejects $Q_3$ to reservoir C.
We need to find the overall efficiency of this combined multi-stage system. Step 2: Key Formula or Approach:
The efficiency of a heat engine is defined as:
\[ \eta = 1 - \frac{Q_{rejected}}{Q_{absorbed}} \]
The individual efficiencies of the two engines are:
\[ \eta_1 = 1 - \frac{Q_2}{Q_1} \implies \frac{Q_2}{Q_1} = 1 - \eta_1 \]
\[ \eta_2 = 1 - \frac{Q_3}{Q_2} \implies \frac{Q_3}{Q_2} = 1 - \eta_2 \]
Step 3: Detailed Explanation:
• The net efficiency of the combined system is the ratio of total work done to the heat absorbed from the primary source (reservoir A).
• The net heat absorbed from reservoir A is $Q_1$, and the final heat rejected to reservoir C is $Q_3$.
• Thus, the net efficiency of the combined system is:
\[ \eta_{net} = 1 - \frac{Q_3}{Q_1} \]
• We can express the ratio $\frac{Q_3}{Q_1}$ as the product of the individual ratios:
\[ \frac{Q_3}{Q_1} = \frac{Q_3}{Q_2} \cdot \frac{Q_2}{Q_1} \]
• Substituting the expressions in terms of efficiency:
\[ \frac{Q_3}{Q_1} = (1 - \eta_2)(1 - \eta_1) \]
\[ \frac{Q_3}{Q_1} = 1 - \eta_1 - \eta_2 + \eta_1\eta_2 \]
• Now, we substitute this back into the formula for net efficiency:
\[ \eta_{net} = 1 - (1 - \eta_1 - \eta_2 + \eta_1\eta_2) \]
\[ \eta_{net} = \eta_1 + \eta_2 - \eta_1\eta_2 \]
Step 4: Final Answer:
The net efficiency of the combination is $\eta_1 + \eta_2 - \eta_1\eta_2$.
