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Re: [femm] Shielding effectiveness




emiliano.menghi@xxxxxxxxx wrote:

> Hi. I use femm to calculate shielding effectiveness of various
> material (such as alluminium, copper, iron, mumetal etc,) at 50 Hz,
> but i have some problem.
> My configuration is simple:
> A alluminium wire with 1000 A 50 Hz (r=7.1 mm)
> A dielectric around the wire (external radius r=16.5 mm)
> A tubolar shield around (external radius r= 19.5 mm)
>
> I use A=0 as boundary condition
>
> If i don't simulate the shield all is OK, but when i use for example
> an alluminum shield, i have these problems:
>
> 1- Shielding effectiveness depends on the position of the boundary
> condition
> 2- I compared femm results with real measures: in the measure setup
> the wire and the shield lenghts were 10 mt. The wire was grounded
> (for femm is grounded at infinity, i think)
> Results ( distance from wire: 20 cm)
>
> without shield
> B femm= 323 microT B measured=321
>
> with alluminium shield
> B femm= 101 microT B measured= 275 microT
> I in the shield femm= 200A measured=35 A
> with shield floating B measured= 303 microT
> Why?
> Is The problem, perhaps, the lenght of the wire (and the shield)
> infinite for femm and finite (10 mt) for the measure?
> Can i simulate with femm terminations of shield (floating or
> grounded)?
> Thanks and exuse me for my poor english
>
> Menghi Emiliano
>
>
>
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You have to be extremely careful in interpreting 2-D planar problems in which
the current is not conserved (all current doesn't add up to zero). This is
totally independent the way femm solves problems, but is instead a
fundamental attribute of the problems themselves.

Basically, the trick is that A doesn't go to zero for an exact solution of
your problem as the distance from the currents goes to zero (i.e. on an
unbounded domain). Instead, A goes to + or - infinity. Consider the exact
solution for the A resulting from a point current, which is proportional to
log(r))--that's what's happening here. For the magnetostatic problem, things
work OK because there are no dynamics and flux distribution just has to
satisfy Ampere's loop law--specific values of A aren't important. For the
eddy current problem where your shield is "grounded", however, the inductance
of the wire and of the shield are dependent on the radius at which you
declare A=0. The bigger the the radius, the bigger the shield's inductance.
If the impedance that the shield sees is different, the induced currents in
the shield will be different as well. It is interesting to note that as your
radius goes to infinity, so does the inductance. (The case in which the total
shield current is constrained to be zero ("floating shield") is different--it
doesn't "talk" to the outside region, so the A=0 location isn't important as
long as the wire current that drives the shield is constrained to be the
"right" value).

An interesting way of looking at is that by defining A=0 at some radius R,
it's like having a very conductive shell of radius R that carries all of the
return currents. The inductance that you are driving is associated with the
flux in the air between your wire, shields, and the shell of radius R.
Outside the radius R shell there is no field, since the enclosed currents for
any loop that lies beyond radius R are zero.

One way or another, you have to model how the eddy currents complete their
electric circuit in the finite element geometry. (or the program ends up
doing it for you by virtue of the equations that it is trying to solve,
perhaps returning the currents in a way that you did not intend). You can
use the "circuit" properties to define constraints on the net current carried
in a section or group of sections. For example, for your shield, you'd
define a circuit property where the total current is constrained to add up to
zero. Then, you'd apply this property to both the shield and its return path
to get the "grounded" case. To get the floating case, you'd apply the zero
total current constraint to just the shield itself. You could also define a
second circuit property that contains the total current in the wire to make
sure that the total current is what you want it to be in the presence of eddy
currents in the wire.

Dave Meeker


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