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Double Integrals, Jacobians, Thermodynamics

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    An important problem in thermodynamics is to find the work done by an Ideal Carnot engine. A cycle consists of alternating expansion and compression of gas in a piston. The work done by the engine is equal to the area of the region R enclosed by two isothermal curves [itex] xy = a[/itex], [itex] xy = b [/itex] and two adiabatic curves [itex] xy^{1.4} = c [/itex], [itex] xy^{1.4} = d [/itex], where [itex] 0 < a < b [/itex] and [itex] 0 < c < d [/itex]. Compute the work done by determining the area of R.

    2. Relevant equations
    [itex] \int \int f(x,y),dA = \int \int f(x(u,v),y(u,v))\frac{\partial (x,y)}{\partial (u,v)} du dv [/itex]
    [itex] \frac{\partial (x,y)}{\partial (u,v)} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} [/itex]


    3. The attempt at a solution
    First, transforming to u and v:
    [itex] u = xy [/itex]
    [itex] v = xy^{1.4} [/itex]
    Now, finding the inverse transformation by solving the above equations:
    [itex] x = \frac{u}{y} [/itex]
    [itex] v = \frac{u}{y}y^{1.4} [/itex]
    [itex] v = uy^{\frac{2}{5}} [/itex]
    [itex] y^{\frac{2}{5}} = \frac{v}{u} [/itex]
    [itex] y = (\frac{v}{u})^{\frac{5}{2}} [/itex]
    I plugged this value into my equation for x, I won't show the work because it is hard for me to write in latex :D. This is my result:
    [itex] x = \frac{u^{\frac{7}{2}}}{v^{\frac{5}{2}}} [/itex]
    [itex] x(u,v) = \frac{u^{\frac{7}{2}}}{v^{\frac{5}{2}}} [/itex]
    [itex] y(u,v) = \frac{v^{\frac{5}{2}}}{u^{\frac{5}{2}}} [/itex]
    Next, I took the partial derivatives to find the Jacobian:
    [itex] \frac{\partial x}{\partial u} = \frac{7u^{\frac{5}{2}}}{2v^{\frac{5}{2}}} [/itex]
    [itex] \frac{\partial y}{\partial u} = -\frac{5v^{\frac{5}{2}}}{2u^{\frac{7}{2}}} [/itex]
    [itex] \frac{\partial x}{\partial v} = -\frac{5u^{\frac{7}{2}}}{2v^{\frac{7}{2}}} [/itex]
    [itex] \frac{\partial y}{\partial v} = \frac{5v^{\frac{3}{2}}}{2u^{\frac{5}{2}}} [/itex]
    Then, using the formula to find the Jacobian (which I also won't include here because it would be hard to write in Latex) I found:
    [itex] \frac{35}{4v} - \frac{25}{4v} [/itex]
    [itex] = \frac{5}{2v} [/itex]
    I know that since the problem is asking for area, the value of the function being integrated is equal to 1. So I obtained the following double integral:
    [itex] \int_{c}^{d} \int_{a}^{b} \frac{5}{2v} du \hspace{1 mm} dv [/itex]
    [itex] \int_{c}^{d} \frac{5b-5a}{2v}\,dv [/itex]
    [itex] \frac{5b-5a}{2} \int_{c}^{d} \frac{1}{v}\,dv [/itex]
    So my final answer is:
    [itex] \frac{5b-5a}{2}(ln(d)-ln(c)) [/itex]
    I was hoping someone here would be kind enough to check my work. I asked a few people in the chat if this was bad form (asking to have your work checked) and I was told that it was not, However, if it is frowned upon please let me know. Thanks :)
     
  2. jcsd
  3. Nov 14, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 14, 2014 #3
    Thanks for the reply, this was actually an extra credit assignment that was handed in Tuesday...I'm pretty sure it's right but I suppose I'll find out soon enough :D
     
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