GATE AG 2024 • Q30 NAT • 1 Mark

Section 3: Farm Power

Farm Power

Subtopic: Thermodynamic principles of I.C. engines

Question Statement:

The air standard cycle efficiency of the Otto cycle is 56% and heat supplied is 12.5 kJ, the heat rejected during the cycle, in kJ, is __________. (Rounded off to 2 decimal places)

Official Answer Key:

5.40 to 5.60

Step-by-Step Solution & Explanation

For any cycle, thermal efficiency is: η = Work done/Heat supplied = (Q_supplied − Q_rejected)/Q_supplied = 1 − Q_rejected/Q_supplied Rearranging: Q_rejected = Q_supplied (1 − η) Substituting: Q_rejected = 12.5 × (1 − 0.56) = 12.5 × 0.44 = 5.50 kJ (Check: work done = 12.5 − 5.5 = 7.0 kJ; 7.0/12.5 = 0.56 ✔)

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