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Irreversible Adiabatic Expansion And Compression

Consider the ...

Question

Consider the following reversible cycle shown below.

It operates a diatomic gas. It's efficiency is given as If its now replaced by a monatomic gas then the change in percentage efficiency of the cycle is $η = \frac{γ - 1}{\frac{γ P_{2}}{P_{2} - P_{1}} + \frac{V}{V_{2} - V_{1}}}$

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

A

4.77

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B

3.87

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C

2.35

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D

4.07

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Solution

The correct option is A 4.77
$Efficiency with diatomic gas (γ = 1.4)$

$η = \frac{1.4 - 1}{\frac{1.4 \times 2}{(2 - 1)} + (\frac{1}{2 - 1})} = 0.105 = 10.5 %$

$Efficiency with monatomic gas (γ = 1.06)$

$η^{^{'}} = \frac{1.66 - 1}{\frac{1.66 \times 2}{(2 - 1)} + \frac{1}{(2 - 1)}} = 0.1527 = 15.27 %$

$∴ η^{^{'}} - η = 15.27 - 10.5 = 4.77 %$

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