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Heat equation applied to a rod

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img444.imageshack.us/img444/7641/20240456gw8.png [Broken]

    2. Relevant equations
    http://img14.imageshack.us/img14/5879/63445047rj2.png [Broken]

    Note that the rightside of the rod is insulated.

    3. The attempt at a solution
    I get this model:

    [tex] \frac{ \partial{u} }{ \partial{t} } = \kappa \frac{ \partial{ ^2 u} }{ \partial{x^2} } +s [/tex]

    [tex]\frac{ \partial{u}} { \partial{x} } = 0[/tex]

    In steady state this gives: [tex]u(x) = \frac{- s}{ \kappa} \frac{1}{2}x^2 + \frac{s}{ \kappa } L x + u_0 [/tex]

    But if I calcute than the asked u' at x=0:

    I get:

    [tex]\frac{du}{dx} = \frac{s}{ \kappa} L [/tex]

    Is this correct?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 12, 2009 #2
    What I don't understand is what do they mean by "total heat supply"? I presume they mean s (=source). But I get a different answer out of my equation.
  4. Feb 12, 2009 #3


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    Your answer looks fine. Note, though, that the equation

    \frac{ \partial{u}} { \partial{x} } = 0

    means nothing on its own; we need to specify a location:

    \left(\frac{ \partial{u}} { \partial{x} }\right)_{x=L} = 0

    For the heat supply question: we need to distinguish the total heat S from the heat per length [itex]s=S/L[/itex] that goes into the differential equation. By applying Fourier's conduction law, your answer indicates a total heat flow of [itex]sL=S[/itex], which is correct. The units will always confirm whether S or s is being used appropriately.
  5. Feb 12, 2009 #4
    You're right but I couldn't get this in latex. Note that the notation you are using isn't the right one either there should be a large bar at the right hand side something like this: [tex] |_{x=L} [/tex]

    Of ocurse, how could I overlooked that!
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