- #1

j_phillips

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I'm a little rusty with this solution. Any help?

Thanks.

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- Thread starter j_phillips
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In summary, the conversation discusses modeling a system where the material varies along its length and its conductivity coefficient is a function of both temperature and position. The heat diffusion equation is considered, and it is noted that the correct equation should include the heat capacity, density, and conductivity of the material. The equation for steady-state is also mentioned, and an example with a specific function for k(x,T) is discussed. The conversation ends with a clarification on the reference point for x.

- #1

j_phillips

- 6

- 0

I'm a little rusty with this solution. Any help?

Thanks.

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- #2

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Hi j_phillips. Welcome to physics forums!j_phillips said:

I'm a little rusty with this solution. Any help?

Thanks.

In your equation, what does the lower case t represent? Is this supposed to be a steady state problem, or an unsteady state problem?

Chet

- #3

dauto

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j_phillips said:

I'm a little rusty with this solution. Any help?

Thanks.

That's not the correct equation. It should be

[tex]C \rho\frac{\partial}{\partial t}T =\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T,[/tex]

where [itex]C, \rho,[/itex] and [itex] k [/itex] are the heat capacity, density, and conductivity of the material

- #4

j_phillips

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This would be steady-state so the right side of the equation would go to zero. If k=x, then d/dx(k) would equal a constant K, but then there is t d/dx*T term which would be unaccounted for?

Instead of a constant (K), let's say k(x,T)=100+cos(2x) for 0<x<(pi/2) (I'm just trying to liven up the problem.) then d'K=-2*sin(2x) then 0=2*sin(2x)*d/dx(T)?

- #5

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[tex]\frac{d}{dx} \left(k(x)\frac{dT}{d x}\right)=0[/tex]

Integrating this once with respect to x gives:

[tex]k(x)\frac{dT}{d x}=C[/tex]

where C is minus the (constant) heat flux.

Integrating again gives:

[tex]T=C\int_0^x{\frac{dζ}{k(ζ)}}+T(0)[/tex]

where ζ is a dummy variable of integration. If the temperature is specified at x = 0 and x = L, that is enough information to get the heat flux C.

Chet

- #6

dauto

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- 201

j_phillips said:

This would be steady-state so the right side of the equation would go to zero. If k=x, then d/dx(k) would equal a constant K, but then there is t d/dx*T term which would be unaccounted for?

Instead of a constant (K), let's say k(x,T)=100+cos(2x) for 0<x<(pi/2) (I'm just trying to liven up the problem.) then d'K=-2*sin(2x) then 0=2*sin(2x)*d/dx(T)?

For a steady state the equation becomes

[tex]\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T = 0[/tex]

[tex]k(x,T)\frac{\partial}{\partial x} T = A[/tex]

Where A is some constant.

If k = x then

[tex]x\frac{\partial}{\partial x} T = A[/tex]

[tex]\frac{\partial}{\partial x} T = \frac{A}{x}[/tex]

[tex]T = A ln(x) + B[/tex]

- #7

dauto

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- 201

- #8

j_phillips

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Thanks! Much appreciated!

The thermal conductivity of a material typically decreases with increasing length. This is due to the fact that longer lengths of a material allow for more opportunities for heat to dissipate, resulting in a lower overall thermal conductivity.

The relationship between thermal conductivity and length is inverse. As the length of a material increases, the thermal conductivity decreases.

Yes, thermal conductivity can be manipulated by changing the length of a material. Shorter lengths will result in higher thermal conductivity, while longer lengths will result in lower thermal conductivity.

Varying thermal conductivity with length can significantly impact heat transfer. Materials with longer lengths and lower thermal conductivity will have a slower rate of heat transfer, while materials with shorter lengths and higher thermal conductivity will have a faster rate of heat transfer.

The relationship between thermal conductivity and length can be affected by factors such as the type of material, its composition, and its physical properties. Additionally, external factors such as temperature and pressure can also impact this relationship.

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