An ideal engine operates on the Carnot cycle using a perfects gas as the working fluid. The ratio of the greatest to the least volume is fixed and is x : 1, the lower temperature of the cycle is also fixed, but the volume compression ratio ‘r’ of the reversible adiabatic compression is variable. The ratio of the specific heats is γ.
Show that if the work done in the cycle is a maximum then,

Question

[latex] (γ - 1) \log _{e} \frac{x}{r} + \frac{1}{r^{γ - 1}} - 1 [/latex] = 0.

Accepted Answer

Refer Fig. 1.

[latex] \frac{V_{3}}{V_{1}} = x ; \frac{V_{4}}{V_{1}} = r[/latex]

During isotherms, since compression ratio = expansion ratio

∴ [latex] \frac{V_{3}}{V_{4}} = \frac{V_{2}}{V_{1}} [/latex]

Also [latex] \frac{V_{3}}{V_{4}} = \frac{V_{3}}{V_{1}} × \frac{V_{1}} {V_{4}} = x × \frac{1}{r} = \frac{x}{r}[/latex]

Work done per kg of the gas

= Heat supplied – Heat rejected = [latex]R T_{1} \log _{e} \frac{x}{r} - R T_{2} \log _{e} \frac{x}{r} [/latex]

= [latex]R (T_{1} - T_{2} ) \log _{e} \frac{x}{r} = R T_{2} (\frac{T_{1} }{T_{2}} - 1) \log _{e} \frac{x}{r} [/latex]

But [latex] \frac{T_{1} }{T_{2}} = ( \frac{V_{4}}{V_{1}})^{γ - 1} = (r)^{γ - 1} [/latex]

∴ Work done per kg of the gas,

[latex]W = R T_{2} ( r^{γ - 1} - 1) \log _{e} \frac{x}{r} [/latex]

Differentiating W w.r.t. ‘r’ and equating to zero

[latex] \frac{dW}{dr} = R T_{2} [ ( r^{γ - 1} - 1) \left\{ \frac{r}{x} × (-x r^{-2}) \right\} + \log _{e} \frac{x}{r} \left\{ (γ - 1) r^{γ - 2} \right\}] [/latex] = 0

or [latex] ( r^{γ - 1} - 1) (- \frac{1}{r} ) + (γ - 1) × r^{γ - 2} \log _{e} \frac{x}{r} [/latex] = 0

or [latex]- r^{γ - 2} + \frac{1}{r} + r^{γ - 2} (γ - 1) \log _{e} \frac{x}{r} [/latex] = 0

or [latex]r^{γ - 2} \left\{ - 1 + \frac{1}{r . r^{γ - 2}} + (γ - 1) \log _{e} \frac{x}{r} \right\} [/latex] = 0

or [latex] - 1 + \frac{1}{r . r^{γ - 2}} + (γ - 1) \log _{e} \frac{x}{r} [/latex] = 0

[latex] (γ - 1) \log _{e} \frac{x}{r} + \frac{1}{r^{γ - 1}}[/latex] - 1 = 0. ........proved.

Question 8.6: An ideal engine operates on the Carnot cycle using a perfect...

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