Re: [femm] Shielding effectiveness

["Emiliano" <emiliano.menghi@xxxxxxxxx>]

But if i use a second conductor (at distance of 5 meters) for the return,
the problem is solved?

----- Original Message -----
From: "David Meeker" <dmeeker@xxxxxxxxxxxxxxxxx>
To: <femm@xxxxxxxxxxxxxxx>
Sent: Wednesday, April 25, 2001 9:34 PM
Subject: Re: [femm] Shielding effectiveness

Hi Dave, i know the problem of non-zero total current and for this, i used
this configuration:

A wire with R= 7.1 mm
A dielectric (ext rad= 16.5 mm)
A tubolar shield (ext rad= 19.5 mm)
A second wire with R= 7.1 mm at 5 mt from the first
A circuit property with I=324 A for wire 1
A circuit property with I=-324 A for the return (wire 2).
A=0 at R=25 mt
Wires and shield are in alluminium



And the question about the lenght of the wire (and the shield)?

>
> 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
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
>
> 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
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>