Question

As shown in the figure, five Carnot engines, each with efficiency \(\eta\) and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider \(Q_0\) to be the amount of heat absorbed per cycle by the first engine and \(W\) as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be \[ \eta_{\mathrm{net}} = \frac{W}{Q_0} = \frac{211}{243} \] The value of \(\eta\) is _______.

Hint:

For cascaded engines: \[ Q_n=Q_0(1-\eta)^n \] and: \[ \eta_{\mathrm{net}} = 1-(1-\eta)^n \] for \(n\) identical stages.

Answer 1

Step 1: Write efficiency relation for each engine.
For each Carnot engine: \[ \eta=\frac{W_i}{Q_{i-1}} \] Hence: \[ W_i=\eta Q_{i-1} \] Heat rejected by first engine: \[ Q_1=Q_0-W_1 \] \[ Q_1=Q_0(1-\eta) \] Similarly: \[ Q_2=Q_1(1-\eta) \] Thus: \[ Q_n=Q_0(1-\eta)^n \]

Step 2: Find total work done.
Work done by: \[ i^{\text{th}} \] engine: \[ W_i=\eta Q_{i-1} \] Thus: \[ W = \eta Q_0 \left[ 1+(1-\eta)+(1-\eta)^2+(1-\eta)^3+(1-\eta)^4 \right] \] Using geometric series: \[ W = \eta Q_0 \cdot \frac{1-(1-\eta)^5}{1-(1-\eta)} \] \[ W = Q_0\left[1-(1-\eta)^5\right] \] Hence: \[ \eta_{\mathrm{net}} = \frac{W}{Q_0} = 1-(1-\eta)^5 \]

Step 3: Use given net efficiency.
Given: \[ 1-(1-\eta)^5 = \frac{211}{243} \] Thus: \[ (1-\eta)^5 = 1-\frac{211}{243} \] \[ = \frac{32}{243} \] \[ = \left(\frac23\right)^5 \] Therefore: \[ 1-\eta=\frac23 \] \[ \eta=\frac13 \]

Step 4: Identify the final answer.
Therefore: \[ \boxed{\frac13} \]