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Steady-State Temperature Distribution of a circular disk

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A circular disc of radius a is heated in such a way that its perimeter [itex]r=a[/itex] has a steady temperature distribution [itex]A+B \cos ^2 \phi [/itex] where r and [itex]\phi[/itex] are plane polar coordiantes and A and B are constants. Find the temperature [itex]T(\rho, \phi)[/itex] everywhere in the region [itex]\rho < a[/itex]

    2. The attempt at a solution
    I have been able to come to a few conclusions. First i assumed we should use the diffusion equation
    [itex]\nabla ^2 u = \frac{1}{\alpha ^2} \frac{\partial u}{\partial t}[/itex]

    Since we are talking about a steady state problem
    [itex] \frac{\partial u}{\partial t} = 0[/itex]
    and thus we get the Laplace
    [itex]\nabla^2 u = 0 [/itex]
    Laplaces equation in 2d polar coordinates is
    [itex]\nabla ^2 u = \frac{1}{r}\frac{\partial}{\partial r}\left ( r \frac{\partial u}{\partial r} \right ) + \frac{1}{r^2}\frac{\partial ^2 u}{\partial \phi ^2} [/itex]
    and we assume a solution of the form
    [itex]u = R(r) \Phi(\phi) [/itex]
    and thus
    [itex] \frac{1}{R}\frac{1}{r}\frac{\partial}{\partial r}\left ( r \frac{\partial R}{\partial r} \right ) + \frac{1}{\Phi}\frac{1}{r^2}\frac{\partial ^2 \Phi}{\partial \phi ^2} = 0 [/itex]
    [itex] \frac{r^2}{R}\frac{1}{r}\frac{\partial}{\partial r}\left ( r \frac{\partial R}{\partial r} \right ) + \frac{1}{\Phi}\frac{\partial ^2 \Phi}{\partial \phi ^2} = 0 [/itex]
    and thus
    [itex] \frac{\partial^2 \Phi}{\partial \phi^2} + n^2 \Phi = 0 \rightarrow \Phi = c_1 cos(n\phi) + c_2 sin(n\phi) [/itex]
    and
    [itex]r^2\frac{\partial^2 R}{\partial r^2} - n^2 R = 0 \rightarrow R = e^n + e^{-n} [/itex]
    and the solutions for u become
    [itex]u = \{ (e^n + e^{-n})cos(n\phi) \}, \{ (e^n + e^{-n})sin(n\phi) \} [/itex]

    This is how far i have gotten. I am not sure if it is completley correct. If it is any hints on where to go from here would be great. Thanks a bunch!
     
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 14, 2012 #2
    I am currently working on the same problem exactly and I came up with the same general solution. I tried to apply the boundary conditions when r=a, but that got me nowhere useful and I can't think of any other useful boundary conditions to apply.

    The only place where I was able to eliminate one part of the solution is when you look at the exponential solutions when n goes to infinity, one of the exponentials gives an unphysical result.
     
  4. Apr 14, 2012 #3
    How i looked at it was the temperature at the center of the disk could no approach infinity so i eliminated the [itex]e^n[/itex] term. Since the Distribution is based on a [itex]cos^2[/itex] term maby we can eliminate the [itex]sin[/itex] term as well. Giving us [itex] u = c_1e^{-n}cos(n \phi) [/itex] where [itex]c_1[/itex] is some constant. Not sure where to go from there then though.
     
  5. Apr 14, 2012 #4
    I saw somewhere online that the fourier series expansion can be used to find the coefficients. If you notice, the solution with only the e-n term but including both the sine and cosine terms resembles the fourier series expansion. Then the coefficients can be found using the formula to find the coefficients for the fourier series. Unfortunately when I tried that I had both an and bn terms come out to zero, which I don't think is correct.
     
  6. Apr 14, 2012 #5
    ok, so I reread the online source, first the exponential should be a power function in the form [itex]r^{n}[/itex] and [itex]r^{-n}[/itex] and then the [itex]r^{-n}[/itex] term should be removed because this blows up when r=0.
     
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