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tjatherton
Junior Member
Username: tjatherton

Post Number: 3
Registered: 06-2005
Posted on Monday, August 23, 2010 - 08:58 pm:   

We're trying to compute the power dissipated by eddy currents in a thin conducting nonmagnetic metal plates subject to an applied driving field. In the Coulomb gauge, and within the harmonic approximation the equation to be solved is:

DEL2(A)=I * omega * sigma * mu * A

where I is Sqrt(-1), omega is the driving frequency of the applied field and sigma is the conductivity of the material. A represents the components of the complex vector potential (i.e. there is an identical equation for Ax, Ay, Az).

I've attached a problem descriptor file, with a rectangular plate (the real shape is much more complicated and not relevant).

The A and derived B fields look reasonable, but the eddy current density does not, apparently because the divergence of A is not constrained to be zero.

What I'd like to know is how to constrain A so that DIV(A)=0 locally---is this possible in FlexPDE?
application/octet-streamEddy Current Script
EddyCurrentCalculation.pde (2.0 k)
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rgnelson
Moderator
Username: rgnelson

Post Number: 1397
Registered: 06-2003
Posted on Wednesday, August 25, 2010 - 02:41 pm:   

If you plot div(Ar) on z=zmid, you will see that the greatest deviation from zero is at the corners of the domain. This would imply that the cause is in the applied boundary conditions. Are you sure these are right? Try computing div(Ar) for the imposed conditions at the corners and see if they are consistent.

There is also a deviation introduced at the surface of the sheet, where sigma is discontinuous. Are you sure that the implied internal condition Normal(grad(A))=0 is correct on this surface?
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rgnelson
Moderator
Username: rgnelson

Post Number: 1399
Registered: 06-2003
Posted on Thursday, August 26, 2010 - 12:51 pm:   

Correction:
Normal(grad(A)) is CONTINUOUS on internal boundaries, not zero. (ie, the DIFFERENCE between the values on two sides is zero).
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amc1
Member
Username: amc1

Post Number: 10
Registered: 03-2008
Posted on Friday, September 17, 2010 - 04:17 pm:   

The eddy current problem has some extremely sophisticated experts with their own code, but they have gone way beyond those of us who want to use FlexPDE and I cannot connect with them.
I have a number of problems with this as my material is very non-linear but I may be able to help here.
Your problem is not making div A zero, this only changes the gauge and has no significance. If I read your results right your problem is that you have current crossing the boundary, so that div J is not zero which is unphysical.
This is because you need an electrostatic charge with a potential V which turns the current round the sharp corners of the conductor.
In general E=-dt(A)-grad(V) and in this geometry you need the V.
This requires an extra equation div(sE)=0 with s the conductivity.
It might be thought you also need to include the charge density and and Poisson's equation, div J =- dq/dt and d2V=-q/eo. (q charge density). However I have tried this and get the same answers with a longer time. I think this is because eo is so small we can assume div(J)=0, although we cannot ignore the electrostatic fields from the surface charges.
I hope this is helpful, and would appreciate any feed-back as I have some similar problems. It is not a FlexPde problem, but a lack of useful information in the literature.




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