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...

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.