# Work of Carnot Engine integral

1. May 20, 2013

### mathnerd15

I'm curious how do you choose the u and v transformations here to be equal to the constants and why is the f(u,v)=1 for the area- because you are summing infinitesimal x's and y's? I see that the area in xy is difficult to integrate because the sides are curved. is the transformation proven somewhere?
problem statement: the work done by an ideal Carnot engine is equal to the area enclosed by two isotherms and adiabatic curves. $$xy=a, xy=b, xy^{1.4}=c, xy^{1.4}=d$$$$\begin{bmatrix}\frac{\partial x}{\partial u}\ & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} &\frac{\partial y}{\partial v}\\ \end{bmatrix}=\frac{5}{2v}, \int_{c}^{d}\int_{a}^{b}\frac{5}{2v}dudv=\frac{5}{2}(b-a)ln\frac{d}{c}$$

by the way, how long does it take people to do these?

Last edited: May 20, 2013
2. May 20, 2013

### Staff: Mentor

You don't do it as a double integral, but rather as a difference between integrals. You have a total of four curves, which you can integrate to get the area under the curve (down to the x axis), then you take the sum of the upper isotherm and adiabat, and you substract the lower isotherm and adiabat, and what is left is the area enclosed by the curves.

The reason you can do that is that the extreme points along x are the same for the upper curves and the lower curves.

3. May 20, 2013

### mathnerd15

thanks very much! do you mean this derivation/equation isn't correct?

$$[\int_{0}^{\infty}\frac{a-b}{x}dx+\int_{0}^{\infty }\frac{c^{5/7}-d^{5/7}}{x^{5/7}}dx].... \int_{0}^{1000}(\frac{1}{x})^{\frac{5}{7}}dx\approx25.189$$
but then the integrals don't converge if you integrate to infinity though you can get numerical approximations...in some cases the adiabatic curves will be between the 2 hyperbolae

Last edited: May 20, 2013
4. May 20, 2013

### Staff: Mentor

I don't understand what $x$ represents.

5. May 20, 2013

### mathnerd15

as you know the isotherms seem to be represented by T(P,V)=xy=k graphs of constant temperature so I'd guess x is volume, y is pressure and on the P-V graph the adiabats asymptotically approach both the V and P axes so perhaps perturbation methods could be useful?
here it says for an ideal gas the work done is the integral of one adiabatic curve between 2 isotherms on a P-V graph

Last edited: May 20, 2013
6. May 20, 2013

### Staff: Mentor

Exactly. So why are you integrating from $0$ to $\infty$?

7. May 20, 2013

### mathnerd15

I was just curious about the integral calculation

8. Jun 1, 2013

### mathnerd15

can you construct a proof of the 3rd Law of Thermodynamics in this way since the integral doesn't converge at infinity it takes infinite work to transform a system into absolute zero state?

Last edited: Jun 1, 2013