In a combined cycle for power generation, if the topping cycle has an efficiency of 45% and for bottoming cycle 30%, then the combined efficiency of the cycle is
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For combined cycles, always remember the formula $\eta_{cc} = \eta_1 + \eta_2 - \eta_1\eta_2$. Alternatively, you can think in terms of waste: if the first cycle rejects $(1 - 0.45) = 55\%$ of heat, the second cycle converts $30\%$ of that waste into useful work: $0.30 \times 0.55 = 0.165$. Adding this to the original efficiency gives $0.45 + 0.165 = 0.615$ or $61.5\%$.
Step 1: Track an actual quantity of heat through both cycles.
Suppose the topping cycle receives \(100\) units of heat. With \(45\%\) efficiency, its work output is:
\[
W_1 = 0.45 \times 100 = 45 \text{ units}
\]
The heat it rejects, which becomes the heat input to the bottoming cycle, is:
\[
Q_{rejected} = 100 - 45 = 55 \text{ units}
\]
Step 2: Find the work from the bottoming cycle.
With \(30\%\) efficiency acting on this 55 units of input:
\[
W_2 = 0.30 \times 55 = 16.5 \text{ units}
\]
Step 3: Add up total work and divide by the original heat input.
\[
W_{total} = 45 + 16.5 = 61.5 \text{ units}
\]
\[
\eta_{cc} = \frac{W_{total}}{Q_{in}} = \frac{61.5}{100} = 0.615
\]
\[
\boxed{\eta_{cc} = 61.5\%}
\]
This confirms option 4.