A heat engine operating on a reversible Carnot cycle receives heat from a thermal reservoir at $T_H = 768\text{ K}$ and rejects heat to a sink at $T_L = 300\text{ K}$. If the rate of heat supplied is $Q_{in} = 68\text{ kW}$, what is the maximum power output $W_{net}$ developed by the engine (in kW)?

Answer Key:

41.21,41.61

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Governing Concept & Domain Context:** * This question evaluates **Brayton Cycle with Real Component Efficiencies** under *Thermodynamic Cycles of Gas Turbines* in GATE AE. 2. **Stage 2: Mathematical Derivation:** * Thermal efficiency of Carnot cycle: $\eta_{th} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{768} = 0.609$. Power developed $W_{net} = \eta_{th} \times Q_{in} = 0.609 \times 68 = 41.41\text{ kW}$. 3. **Stage 3: Distractor Elimination Forensics:** * The evaluated correct answer is derived using standard physical equations. * Distractor values correspond to typical unit conversion omissions or inverted formulas. 4. **Stage 4: Common Student Trap:** * *Trap:* Using temperatures in Celsius instead of Kelvin in thermodynamic cycle efficiency. *Reference: Philip G. Hill, Carl R. Peterson, Mechanics and Thermodynamics of Propulsion (Pearson).*

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