Problem solving Heat Diffusion Equation

Click For Summary
SUMMARY

The discussion focuses on solving the Heat Diffusion Equation for a thick limestone layer subjected to sinusoidal temperature variations. The key equation used is the diffusion equation, represented as ∂T/∂t = -D∂²T/∂x². The solution involves separation of variables, leading to the expression T = Tcos(ωt - x√(ω/2D)) e^(-x√(ω/2D)), which describes the damped sinusoidal temperature oscillation within the layer. The decay length of the oscillation amplitude is determined by the term e^(-x√(ω/2D)).

PREREQUISITES
  • Understanding of the Heat Diffusion Equation
  • Familiarity with separation of variables method
  • Knowledge of sinusoidal functions and their properties
  • Basic concepts of boundary conditions in differential equations
NEXT STEPS
  • Study the derivation of the Heat Diffusion Equation in detail
  • Learn about boundary conditions and their impact on solutions
  • Explore the application of Fourier series in solving heat equations
  • Investigate the physical implications of decay length in thermal systems
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on thermal dynamics and heat transfer analysis.

XCBRA
Messages
18
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
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}}}
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
13
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 52 ·
2
Replies
52
Views
7K