# Math Modeling Problem

## 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$$\epsilon$$x where |$$\epsilon$$| << 1. The governing
equation is given by, $$\frac{d}{dx}$$(K0e$$\epsilon$$x $$\frac{dT}{dx}$$) = 0

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

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

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

## The Attempt at a Solution

Any hints on how to even start?

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anyone????

lanedance
Homework Helper
for 1) how about trying to perform the given integral to get the average conductivity

ok, I got the average conductivity to be the constant: (e$$\epsilon$$ - 1) ($$\frac{K0}{\epsilon}$$) for $$\overline{K}$$

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

or simply $$\overline{K}$$ $$\frac{d2T}{d2x}$$ = 0

Now the solution of this equation should be T(x) = c1x

Last edited:
lanedance
Homework Helper
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 ;)
$$\overline{K} = \int_0^1 dx K(x) = \int_0^1 dx K_0 e^{\epsilon x} = \frac{K}{\epsilon}(e^{\epsilon}-1)$$

whats next?

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 ;)
$$\overline{K} = \int_0^1 dx K(x) = \int_0^1 dx K_0 e^{\epsilon x} = \frac{K}{\epsilon}(e^{\epsilon}-1)$$

whats next?
Yeah, thats what I get.

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

or simply $$\overline{K}$$ $$\frac{d2T}{d2x}$$ = 0

Now the solution of this equation should be T(x) = c1x with c1 being $$\overline{K}$$ ?

lanedance
Homework Helper
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

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.

So for part 2, I "expanded" the D.E. and rewrote it like:

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