Is the Heat Equation Model for an Insulated Rod Correct?

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Homework Help Overview

The discussion revolves around the heat equation model for an insulated rod, focusing on the implications of boundary conditions and the interpretation of heat supply terms in the context of the equation provided by the original poster.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive a solution for the heat equation and questions the correctness of their derivative at a specific point. Other participants seek to clarify the meaning of "total heat supply" and differentiate between total heat and heat per length in the context of the equation.

Discussion Status

Participants are engaging in a detailed examination of the mathematical expressions and their implications. Some guidance has been offered regarding the need for specificity in boundary conditions and the distinction between different heat terms, indicating a productive exploration of the topic.

Contextual Notes

There is a noted confusion regarding the notation used in the equations and the definitions of heat supply terms, which are under discussion but not resolved.

dirk_mec1
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Homework Statement



http://img444.imageshack.us/img444/7641/20240456gw8.png

Homework Equations


http://img14.imageshack.us/img14/5879/63445047rj2.png

Note that the rightside of the rod is insulated.

The Attempt at a Solution


I get this model:

[tex]\frac{ \partial{u} }{ \partial{t} } = \kappa \frac{ \partial{ ^2 u} }{ \partial{x^2} } +s[/tex]

[tex]u(0,t)=u_0[/tex]
[tex]\frac{ \partial{u}} { \partial{x} } = 0[/tex]In steady state this gives: [tex]u(x) = \frac{- s}{ \kappa} \frac{1}{2}x^2 + \frac{s}{ \kappa } L x + u_0[/tex]

But if I calcute than the asked u' at x=0:

I get:

[tex]\frac{du}{dx} = \frac{s}{ \kappa} L[/tex]

Is this correct?
 
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What I don't understand is what do they mean by "total heat supply"? I presume they mean s (=source). But I get a different answer out of my equation.
 
Your answer looks fine. Note, though, that the equation

[tex] \frac{ \partial{u}} { \partial{x} } = 0[/tex]

means nothing on its own; we need to specify a location:

[tex] \left(\frac{ \partial{u}} { \partial{x} }\right)_{x=L} = 0[/tex]

For the heat supply question: we need to distinguish the total heat S from the heat per length [itex]s=S/L[/itex] that goes into the differential equation. By applying Fourier's conduction law, your answer indicates a total heat flow of [itex]sL=S[/itex], which is correct. The units will always confirm whether S or s is being used appropriately.
 
Mapes said:
Your answer looks fine. Note, though, that the equation

[tex] \frac{ \partial{u}} { \partial{x} } = 0[/tex]

means nothing on its own; we need to specify a location:

[tex] \left(\frac{ \partial{u}} { \partial{x} }\right)_{x=L} = 0[/tex]
You're right but I couldn't get this in latex. Note that the notation you are using isn't the right one either there should be a large bar at the right hand side something like this: [tex]|_{x=L}[/tex]

For the heat supply question: we need to distinguish the total heat S from the heat per length [itex]s=S/L[/itex] that goes into the differential equation. By applying Fourier's conduction law, your answer indicates a total heat flow of [itex]sL=S[/itex], which is correct. The units will always confirm whether S or s is being used appropriately.
Of ocurse, how could I overlooked that!
 

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