Approximate and Exact Solutions for a Thermal Conductivity Problem

[hitmeoff]

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

Homework Equations

The Attempt at a Solution

Any hints on how to even start?

 

[hitmeoff]

anyone?

 

[lanedance]

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

 

[hitmeoff]

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

 

[lanedance]

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?

 

[hitmeoff]

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, that's 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]

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

 

[hitmeoff]

Well for one, I am supposed to just get an approximate solution for the Temp distribution.

For part II, I am supposed to get the exact solution. Let me work on that for a bit.

 

[hitmeoff]

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?