Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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...

share|improve this question
1  
Seems to me like an exercise on integration. This is no thermodynamic proces for as far as I know. –  Nick Jun 23 '13 at 21:53
    
Suitable constants can probably be introduced to the equation to make it dimensionally self-consistent. –  leongz Jun 23 '13 at 22:10
2  
Your equation isn't completely correct, since the curve is elliptic, not circular. That explains the dimensional problem. –  Wouter Jun 23 '13 at 22:11
2  
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. –  Wouter Jun 23 '13 at 22:22
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Thanks to everyone for the input! –  DepeHb Jun 25 '13 at 19:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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