Is the Heat Equation Model for an Insulated Rod Correct?

In summary: The equation\frac{ \partial{u}} { \partial{x} } = 0means nothing on its own; we need to specify a location:\left(\frac{ \partial{u}} { \partial{x} }\right)_{x=L} = 0
  • #1
dirk_mec1
761
13

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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  • #2
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.
 
  • #3
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.
 
  • #4
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!
 

Related to Is the Heat Equation Model for an Insulated Rod Correct?

1. What is the heat equation applied to a rod?

The heat equation applied to a rod is a mathematical model that describes the flow of heat through a rod over time. It takes into account the temperature at different points along the rod, as well as the thermal conductivity of the material.

2. How is the heat equation applied to a rod used in real-life applications?

The heat equation applied to a rod is used in various real-life applications, such as in the study of heat transfer in materials, thermal management in engineering, and temperature distribution in objects like pipes and rods.

3. What are the assumptions made in the heat equation applied to a rod?

The heat equation applied to a rod assumes that the rod is one-dimensional, has a uniform cross-section, and is made of a homogeneous material with constant thermal conductivity. It also assumes that the temperature variation along the length of the rod is small.

4. How is the heat equation applied to a rod solved?

The heat equation applied to a rod can be solved using various methods, such as the separation of variables method, finite difference method, or the Fourier transform method. The appropriate method depends on the specific boundary conditions and initial conditions of the problem.

5. What are the limitations of the heat equation applied to a rod?

The heat equation applied to a rod is a simplified model and may not accurately describe the behavior of materials with complex geometries or non-uniform thermal properties. It also assumes steady-state conditions and does not take into account any external heat sources or sinks. In some cases, it may be necessary to use more advanced models to accurately analyze heat transfer in real-life scenarios.

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