Problem solving Heat Diffusion Equation

In summary, the problem involves a thick uniform layer with a sinusoidal temperature variation on one face. The damped sinusoidal temperature oscillation propagates into the layer and the decay length of the oscillation amplitude is given by the diffusion equation. In the second part, the cellar is subject to daily and annual temperature variations and the magnitude of these variations are estimated using the diffusion equation. The solution involves using separation of variables and the boundary conditions of the problem. The final solution involves a combination of sines, cosines, and exponential functions.
  • #1
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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

[tex] \frac{\partial T}{\partial t} = -D\frac{\partial^2T}{\partial x^{2}} [/tex]

Using separation of variable method:

Let [tex] T=X(x)F(t) [/tex]

[tex]X \frac{dF}{dt} = -DF\frac{d^2X}{DX^2} [/tex]

[tex]-D\frac{dF}{dt}=\frac{1}{X}\frac{d^2X}{DX^2}= k [/tex]

where k is the separation constant.

These separate into two equation which I solve to give

[tex] X=Ae^{\sqrt{k}x}+Be^{-\sqrt{k}x} [/tex]

[tex]F=Ce^{-Dkt} [/tex]

By superpositon principle

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

where C has been absorbed into A and B.

Then taking the boundary conditions:

At x →∞ T→0, which shows A_k → 0

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

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.
 
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  • #2
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.
 
  • #3
Ahh that makes better sense. Thanks for the help, I was able to solve to give:

[tex]Tcos(wt-x\sqrt{\frac{w}{2D}}) e^{-x\sqrt{\frac{w}{2D}}}[/tex]
 

What is the Heat Diffusion Equation?

The Heat Diffusion Equation is a mathematical model used to describe how heat is transferred through a medium over time. It takes into account factors such as the temperature gradient, thermal conductivity, and specific heat capacity of the medium.

Why is the Heat Diffusion Equation important?

The Heat Diffusion Equation has many practical applications in fields such as physics, engineering, and materials science. It allows us to predict and understand how heat will be distributed in various systems, which is crucial for designing and optimizing processes, materials, and technologies.

What are the key assumptions of the Heat Diffusion Equation?

The Heat Diffusion Equation assumes that the medium is homogeneous, isotropic, and has a constant thermal conductivity. It also assumes that there are no heat sources or sinks within the medium, and that there is no internal heat generation.

How is the Heat Diffusion Equation solved?

The Heat Diffusion Equation can be solved using various numerical and analytical methods. Some common techniques include finite difference methods, separation of variables, and Fourier series. The specific method used will depend on the complexity of the problem and the desired level of accuracy.

What are some real-world examples of the Heat Diffusion Equation in action?

The Heat Diffusion Equation can be seen in action in a wide range of applications, such as predicting the temperature distribution in a heated building, analyzing the thermal behavior of electronic devices, and understanding the heat transfer in chemical reactions. It is also used in medical imaging techniques like magnetic resonance imaging (MRI) and positron emission tomography (PET) to reconstruct images based on the diffusion of heat through tissues.

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