Problem solving Heat Diffusion Equation

[XCBRA]

Homework Statement

One face of a thick uniform layer is subject to a sinusoidal temperature variation of angular frequency ω. SHow that the damped sinusoidal temperature oscillation propagate into eh layer and give an expression for the decay length of the oscillation amplitude.

A cellar is built underground covered by a ceiling which 3m thick made of limestone. The Outside temperature is subject to daily fluctuations of amplitude 10 C and annual fluctuations of 20 C. Estimate the magnitude of the daily and annual temperature variation within the cellar.

Homework Equations

The Attempt at a Solution

I am unable to solve the first part of this question.

Take the diffusion equation

$$\frac{\partial T}{\partial t} = -D\frac{\partial^2T}{\partial x^{2}}$$

Using separation of variable method:

Let $$T=X(x)F(t)$$

$$X \frac{dF}{dt} = -DF\frac{d^2X}{DX^2}$$

$$-D\frac{dF}{dt}=\frac{1}{X}\frac{d^2X}{DX^2}= k$$

where k is the separation constant.

These separate into two equation which I solve to give

$$X=Ae^{\sqrt{k}x}+Be^{-\sqrt{k}x}$$

$$F=Ce^{-Dkt}$$

By superpositon principle

$$T=\sum (A_ke^{\sqrt{k}x}+B_ke^{-\sqrt{k}x})e^{-Dkt} + A_0 + B_0x$$

where C has been absorbed into A and B.

Then taking the boundary conditions:

At x →∞ T→0, which shows A_k → 0

$$T=\sum B_ke^{-\sqrt{k}x}e^{-Dkt} + B_0x$$
Then apply conditon that at x=0 T $\propto$ sinωt.
$$e^{-Dkt} = sinwt$$

However here is where I am stuck, I do not see how proceed further. Is my solution so far correct, it does not seem so as I seem to have the wrong form. Or have I chosen the wrong form for my separation coefficient?

Any help would be greatly appreciated.

 

[LawrenceC]

The transient heat equation is generally written without the negative sign as you have written it. When you use separation of variables, use -k as your separation constant. That causes the X(x) portion to be sines and cosines.

 

[XCBRA]

Ahh that makes better sense. Thanks for the help, I was able to solve to give:

$$Tcos(wt-x\sqrt{\frac{w}{2D}}) e^{-x\sqrt{\frac{w}{2D}}}$$