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1 dimensional heat transfer question

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    zNE6DL6.png

    2. Relevant equations
    d/dx(kdT/dx) + S = 0 (for steady state)
    q_cond = -kAdT/dx
    q_gen = S*Vol

    3. The attempt at a solution
    ksn9AyQ.jpg

    So I tried to equate q_cond to q_gen but it seemed to give me the wrong answer. Does anyone know why? I believe to get the right answer I have to use the first equation I wrote in the relevant equations section.

    Does my method not work because S is not uniform? But then it tells me that at x = 0, S = So so it shouldn't matter since we have a boundary condition?
     
  2. jcsd
  3. Jan 29, 2014 #2

    BvU

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    The quick reply would be that there is no reason to equate q_cond to q_gen: there are two mechanisms that cause temperature change, namely conduction and heat generation. The sum of the contributions to dT/dt has to cause a steady state, hence the = 0.
     
    Last edited: Jan 29, 2014
  4. Jan 29, 2014 #3

    BvU

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    Could you also check q_gen = S * Vol ? This looks too much as if S is constant all over the place.
     
  5. Jan 29, 2014 #4
    Hmm, I am not too sure about the equations you use for S, but I think you need to do something about the constants, C2 and C3 as well.

    I would suggest looking at the boundary conditions. You have already been given an expression for the temperature, which you solved correctly. When T=T2 you get 1 (as you also did).

    But the deriviative of the equation also has to be continuous over the boundary. (Note: can you explain why?)

    You then have 2 equations with 2 unknowns. Remember that at this boundary x=L! (which you also wrote down).

    So you could then express the constants C2 and C3 in terms of L. I think that is what you should start by doing.



    EDIT: I solved it now (phew), but the equation you need would be:

    $$ \frac{d^2T}{dx^2}+\frac{S}{k}=0 $$

    Are you familiar with this expression, or?
     
    Last edited: Jan 29, 2014
  6. Jan 29, 2014 #5
    This is exactly the same equation you wrote down (correctly) in your original post. So you shouldn't be too surprised that it gives you the right answer.
     
  7. Jan 29, 2014 #6
    But not the one used in the attached paper :-) but yeah, then you must be familiar with it!
     
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