The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

\frac{1}{4}

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\frac{1}{2}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
$Q_{1} = T_{0} S_{0} + \frac{1}{2} T_{0} S_{0} = \frac{3}{2} T_{0} S_{0}$
$Q_{2} = T_{0} (2 S_{0} - S_{0}) = T_{0} S_{0}$ and $Q_{3} = 0$
$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$= 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{0} S_{0}}{\frac{3}{2} T_{0} S_{0}} = \frac{1}{3}$

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Similar questions

100

- 273^{o} C

η = \frac{T_{2} - T_{1}}{T_{2}}

η = \frac{γ - 1}{\frac{γ P_{2}}{P_{2} - P_{1}} + \frac{V}{V_{2} - V_{1}}}

Take P_{1} = 1 bar, P_{2} = 2 bar, v_{1} = 1 m^{3}, v_{2} = 2 m^{3}

\frac{1}{6}

62^{o} C

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