Math Modeling Problem

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Homework Statement


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

Homework Equations





The Attempt at a Solution



Any hints on how to even start?
 

Answers and Replies

  • #2
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anyone????
 
  • #3
lanedance
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for 1) how about trying to perform the given integral to get the average conductivity
 
  • #4
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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:
  • #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?
 
  • #6
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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?
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] ?
 
  • #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
 
  • #8
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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.
 
  • #9
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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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