A cyclic heat engine operates between a source temperature of 900°C and a sink temperature of 35°C. The least rate of heat rejection per kW net output of the engine will be:

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BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

The Carnot efficiency of a heat engine operating between two temperatures is:

\( \eta = 1 - \frac{T_2}{T_1} \)

Where, \( T_1 \) is source temperature and \( T_2 \) is sink temperature (in Kelvin).

Given:

Source temperature = 900°C = 1173 K

Sink temperature = 35°C = 308 K

Net output work = 1 kW

Calculation:

Efficiency, \( \eta = 1 - \frac{308}{1173} = 0.7375 \)

Heat supplied, \( Q_1 = \frac{1}{0.7375} = 1.356 \, \text{kW} \)

Heat rejected, \( Q_2 = Q_1 - W = 1.356 - 1 = 0.356 \, \text{kW} \)