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Homework Help: Math Modeling Problem

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A slab is in a steady state with temperature T0 at x = 0, and T1 at x = 1. The thermal conductivity is given by K(x) = K0e[tex]\epsilon[/tex]x where |[tex]\epsilon[/tex]| << 1. The governing
    equation is given by, [tex]\frac{d}{dx}[/tex](K0e[tex]\epsilon[/tex]x [tex]\frac{dT}{dx}[/tex]) = 0

    (1). Obtain an approximation solution to the temperature distribution by replacing K(x)
    with its average value [tex]\bar{K}[/tex] = [tex]\frac{\int K(x) dx}{\int dx}[/tex] over the slab (integrals from 0 to 1)

    (2). Otain an exact solution to the temperature distribution.

    (3). Rewrite K(x) = K(x) − [tex]\bar{K}[/tex] + [tex]\bar{K}[/tex] then term K(x) − [tex]\bar{K}[/tex] is neglected while replacing K(x) with [tex]\bar{K}[/tex]. Check consistency, i.e, prove that |[tex]\frac{K(x) - \bar{K}}{\bar{K}}[/tex]| << 1

    2. Relevant equations



    3. The attempt at a solution

    Any hints on how to even start?
     
  2. jcsd
  3. Apr 14, 2010 #2
    anyone????
     
  4. Apr 14, 2010 #3

    lanedance

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    for 1) how about trying to perform the given integral to get the average conductivity
     
  5. Apr 14, 2010 #4
    ok, I got the average conductivity to be the constant: (e[tex]\epsilon[/tex] - 1) ([tex]\frac{K0}{\epsilon}[/tex]) for [tex]\overline{K}[/tex]

    So when I sub this back in I get: [tex]\frac{d}{dx}[/tex] ( [tex]\overline{K}[/tex] [tex]\frac{dT}{dx}[/tex] ) = 0

    or simply [tex]\overline{K}[/tex] [tex]\frac{d2T}{d2x}[/tex] = 0

    Now the solution of this equation should be T(x) = c1x
     
    Last edited: Apr 14, 2010
  6. Apr 14, 2010 #5

    lanedance

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    ok, your tex is a little difficult to read (click on equation below to see code), but i agree if this is what you meant, which i suppose makes it a little redundant ;)
    [tex] \overline{K} = \int_0^1 dx K(x) = \int_0^1 dx K_0 e^{\epsilon x} = \frac{K}{\epsilon}(e^{\epsilon}-1)[/tex]

    whats next?
     
  7. Apr 14, 2010 #6
    Yeah, thats what I get.

    So when I sub this back in I get: [tex]\frac{d}{dx}[/tex] ( [tex]\overline{K}[/tex] [tex]\frac{dT}{dx}[/tex] ) = 0

    or simply [tex]\overline{K}[/tex] [tex]\frac{d2T}{d2x}[/tex] = 0

    Now the solution of this equation should be T(x) = c1x with c1 being [tex]\overline{K}[/tex] ?
     
  8. Apr 14, 2010 #7

    lanedance

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    first you can write a whole equation between the tex tags

    now second what are you trying to do? i would think a constant conductivity would lead to a constant temperature gradient, whilst the conductivity dependent on x will lead to slightly more complex temperture distribution. What is the exact solution of? I would take it to be the full temp gradient, but haven't tried to solve it
     
  9. Apr 14, 2010 #8
    Well for one, I am supposed to just get an approximate solution for the Temp distribution.

    For part II, Im supposed to get the exact solution. Let me work on that for a bit.
     
  10. Apr 14, 2010 #9
    So for part 2, I "expanded" the D.E. and rewrote it like:

    K0 [tex]\epsilon[/tex] e[tex]\epsilon[/tex]x [tex]\frac{dT}{dx}[/tex] + K0e[tex]\epsilon[/tex]x[tex]\frac{d2T}{dx2}[/tex] = 0 or
    T'' + [tex]\epsilon[/tex] T' = 0 ...correct?
     
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