Question

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

• A. \(45 \% \)
• B. \(30 \% \)
• C. \(75 \% \)
• D. \(61.5 \% \)

Hint:

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 $61.5\%$.

Answer 1

The Correct Option is D

Concept: A combined cycle power plant links two thermodynamic cycles together to achieve greater overall thermal efficiency than either cycle operating independently. The high-temperature exhaust gas from the first cycle (topping cycle) serves as the input heat source for the second cycle (bottoming cycle). Let the thermal efficiency of the topping cycle be $\eta_1$ and the thermal efficiency of the bottoming cycle be $\eta_2$. The overall or combined thermal efficiency ($\eta_{cc}$) of the system can be derived from basic energy balances and is expressed mathematically as: \[ \eta_{cc} = \eta_1 + \eta_2 - \eta_1 \cdot \eta_2 \] This equation shows that the net efficiency is the sum of both individual efficiencies reduced by their product, reflecting that the bottoming cycle only processes the heat rejected by the topping cycle.

Step 1: Identify the given individual efficiencies.
From the problem description, we have the following percentage values for each constituent thermodynamic cycle:
• Efficiency of the topping cycle, \(\eta_1 = 45\% = 0.45\)
• Efficiency of the bottoming cycle, \(\eta_2 = 30\% = 0.30\)

Step 2: Substitute values into the combined efficiency formula.
Let's substitute the decimal representations into our standard algebraic expression: \[ \eta_{cc} = 0.45 + 0.30 - (0.45 \times 0.30) \] Now we execute the arithmetic operations systematically step-by-step:
• Sum of the individual efficiencies: \[ 0.45 + 0.30 = 0.75 \]
• Product of the individual efficiencies: \[ 0.45 \times 0.30 = 0.135 \] Subtracting the product from the sum yields: \[ \eta_{cc} = 0.75 - 0.135 = 0.615 \]

Step 3: Convert the decimal form back to percentage.
To state the efficiency as a percentage value, we multiply by 100: \[ \eta_{cc} = 0.615 \times 100\% = 61.5\% \] Thus, the combined thermal efficiency of the power plant is precisely 61.5%, matches Option (D).