Setting up Fick's Law with Internal heating

[jonthebaptist]

I am solving for the flow of heat on a resistor, say it has a cross-section in the x-z plane and is extended along the y-axis. Fick's Law with internal heating is simply
$\frac{\partial T}{\partial t}=A\nabla^{2}T+BQ$
My PDE text gives Q as power delivered per unit volume. So substituting in the electrical power delivered divided by the differential volume element, I get
$\frac{\partial T}{\partial t}=A\nabla^{2}T+B\frac{I^{2}\rho}{dx^{2}dz^{2}}$
What I did makes sense to me physically, but I am not sure if it makes sense mathematically as a PDE with a term having differential elements instead of derivatives.

 

[hunt_mat]

My advice would be to choose a control volume, V and look for the heat entering and leaving this control volume, use some vector analysis and it will give you a sensible differential equations. The PDE you write down is wrong mathematically.

 

[timthereaper]

Can you really apply Fick's laws to heat? From what I understand, the laws concern concentrations of actual objects. I would think you could measure energy density that way, considering it it also dependent somewhat on position and time.

I would personally use Fourier's Law and other traditional heat transfer equations to measure heat flow.

 

[jonthebaptist]

@hunt_mat Thanks for confirming my suspicion about my PDE

@timthereaper I believe Fick's laws were discovered empirically from diffusion experiements, but Asmar's PDE text gives a derivation for a PDE describing heat flow on a rod based on Fourier's law and arrives at Fick's law.

Either way, I notice that COMSOL has a module that models the physics of heat flow on parts with Joules internal heating, so someone has figured this out. I'll think about it a little more and see if I come up with anything...

 

[timthereaper]

Hmm...I'll have to look at that text. You've piqued my interest.

 

[osturk]

john, the last term in heat conduction equation should be J^2*rho. This is the general form for the local heat production (and you don't need the derivatives). J is the current density (charge per time per area), and rho is the resitivity (resistance*length). However, if the cross section and rho are constant through the wire, then J is uniform everywhere and you can substitute I=J*A. The term becomes I^2*rho/A - no derivatives.

 

[jonthebaptist]

My first take was to use $J^{2}\rho$, but that only works if you are modeling heat flow in the one dimension longitudinal along the resistor, so all transverse flow effects are ignored. I need to account for those effects for what I'm doing, which is why I am taking my infinitesimal volume element to have infinitesimal lengths in all three dimensions.