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I have to calculate the work done in this thermodynamic cycle: LINK.

enter image description here

But the bottom bit of the cycle is something I've never seen before. I guess the curve in $pV$ space is given by

$$\left( V- \frac{V_2 -V_1}{2} \right) ^2 + \left( p-p_1 \right) ^2 = \left( \frac{V_2 -V_1}{2} \right) ^2$$

Which is physically meaningless because of units...

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    $\begingroup$ Seems to me like an exercise on integration. This is no thermodynamic proces for as far as I know. $\endgroup$
    – Nick
    Jun 23 '13 at 21:53
  • $\begingroup$ Suitable constants can probably be introduced to the equation to make it dimensionally self-consistent. $\endgroup$
    – leongz
    Jun 23 '13 at 22:10
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    $\begingroup$ Your equation isn't completely correct, since the curve is elliptic, not circular. That explains the dimensional problem. $\endgroup$
    – Wouter
    Jun 23 '13 at 22:11
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    $\begingroup$ The correct equation (shifted to center on the origin) would be $V^2/V_0^2 + p^2/p_0^2 = 1$ where $V_0 = (V_2-V_1)/2$ and $p_0 = p_2-p_1$. But I think @Nick is right, this is most likely intended as an exercise mainly on integration. To give an example for which it is a little bit harder to find the work. $\endgroup$
    – Wouter
    Jun 23 '13 at 22:22
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The work done in a reversible cyclic process equals $ -\int PdV$ which is also equal to the negative area under the graph of that cyclic process.

From the looks of it, process $D \to A$ appears to be a semi-ellipse, whose area equals $1/2\pi ab$ , where a and b are lengths of semi-axes ($p_2-p_1$ and $(v_2-v_1)/2$).

So using that, you wont need to integrate the function.

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