A heat engine working on carnot cycle receives heat at the rate of 40 kW from a source at 1200 K rejects to a sink at 300 K. The heat rejected is _______.

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JKSSB JE Mechanical 02 Nov 2021 Shift 1 Official Paper

Answer 1

Concept:

Clausius inequality for a reversible engine

\(\oint \frac{{dQ}}{T} = 0\)

For irreversible engine

\(\oint \frac{{dQ}}{T} < 0\)

The carnot cycle is a reversible cycle

Calculation:

Given:

Q₁ = 40 kW, T₁ = 1200 K, T₂ = 300 K

For reversible cycle

\(\begin{array}{l} \frac{{{Q_1}}}{{{T_1}}} - \frac{{{Q_2}}}{{{T_2}}} = 0 \end{array}\)

\( \frac{{40}}{{1200}} - \frac{{{Q_2}}}{{300}} = 0\)

\(\\ {Q_2} = \frac{{40 \times 300}}{{1200}} = 10\;kW\)