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Re: [femm] Re: Force calculation on an axisymmetric model




Paulo José Gameiro Pereirinha wrote:

> Thanks for all the suggestions.
>
> Indeed the main problem was that I hadn't set the relative permeability in
> the axial direction (the current density was correct). About the
> discrepancy between experimental and simulation results, I must say that I
> read the value on the "wrong column": it was 1.1 N instead of 1.35 N, but
> it still's a lot far away from the 0.5685 or 0.5456 N that Dave (David
> Meeker) calculated. Probably, as Israel Kehaty sugested, it is due to the
> side friction force (I'm not very shure of the accuracy of the mesurament
> device also).
>
> Two more questions if you don't mind:
>
> 1. If I compute the same device (as a time harmonic problem at) 50 Hz, I
> get the following results:
> -----------------
> DC Force:
> z-direction = -2.762e-001 N
> 100.000000 Hz Force:
> z-direction = -2.762e-001 + j 1.524e-002 N
> -----------------
>
> Shouldn't the DC force be equal to 0.5685 N as if it was calculated at 0
> Hz? And, if I'm performing a computation at 50 Hz, why does it give me the
> value at 100 Hz?

Although this behavior might seem strange, this is actually correct. The trick
here is that in a harmonic problem, the field varies sinusoidally, but the
force is related to the _square_ of the field, producing a dc component, and a
2x frequency component, but no force at the drive frequency.

To see this effect, consider for example a simple c-core electromagnet acting
on a slab of iron. The instantaneous force on one pole of the c-core can be
obtained via Maxwell's stress tensor:

F = B^2 * area /(2*mu_o)

Now, imagine that the problem is harmonic, and the gap flux density, B, is
defined as:

B=Bo*Cos[w t]

Substituting in the force gives:

F = Bo^2 * Cos[w t]^2 * area / (2*mu_o)

Now, using the trigonometric identity Cos[x]^2 = (1/2)*(1+Cos[2 t]) gives:

F= area*Bo^2/(4*mu_o) + area*Bo^2*Cos[2 w t]/(4*muo)

So there are 2 components of the force: a DC part of the force, and a part at
_twice_ the drive frequency. Now, the DC part is 1/2 of what it would be if
the problem were evaluated at 0 Hz, so that the _peak_ force is the same as in
the DC case (e.g when t=0, Pi/w, etc.) when the cosine part adds to the DC
component. Conversely, the minimum force is zero when the cosine part cancels
the DC part (at the instants where the current driving the fields passes
through zero).

However, for your problem, the amplitude of the 2 x part is not exactly the
same as the amplitude of the DC part, as one might expect from the above
electromagnet example. In your particular system, you have an additional
interesting effect, which is that a non-zero conductivity has been defined for
the iron bar. This means that eddy currents will be induced in the bar. The
field also acts on these induced currents, modifying the force. This is what
gives the phase and amplitude shift in the force. A very interesting
implication is that there is some instant during the cycle in which the force
is actually pushing the bar _away_ from the coil, due to the eddy currents.
Specifically, the maximum upward force from the 2x component is |-2.762e-001 +
j 1.524e-002| N minus the dc component of 2.762 pulling downwards yields a
whopping maximum upward force of 4.2e-4 N).

There are lots of devices out there that have sinusoidally varying currents and
produce a force at the same frequency. However, all of these machines have a
time-varying field interacting with a DC field. An example would be a voice
coil actuator (i.e. stereo speaker), in which force is created by a
sinusoidally varying coil current interacting with a DC field set up by a
permanent magnet. In your example, you could get forces at the drive frequency
by running a DC offset in your coil. Then, the coil produces forces at DC, at
the drive frequency, and at twice the drive frequency. (unfortunately, femm
can't analyze this situation all at the same time--oh, well...).

> 2. If I want the mesh to have smaller elements near the iron (or some other
> place), what is the best way of performing this? To define two or three
> blocks of air one inside the other's with different mesh size (and
> choosing in the "Segment property" to "Hide segment in postprocessor")?
> Perhaps it would be better with semi-circles than with rectangles (as I
> have done)...

This is generally the way that I do it--breaking the air into a few blocks with
different mesh sizes. Often, I do use semicircles, but this is highly
dependent on the problem you are analyzing.

An alternative is to impose a mesh density on particular lines in your geometry
(this is a definable property of each segment and arc segment). However, the
results aren't always that satisfactory, since the effects of defining a
maximum mesh size on a line or arc are usually confined pretty locally to the
line or arc.


> And finally a thing that is a little bit confusing to me: when I make some
> alteration to a model in Femm editor (without saving and reprocessing), and
> I clik on the "glasses symbol" it shows me the FE solution of a previous
> version of that model, without giving me any warning that the model as
> changed or that the solution presented is not for the model that its being
> changed. It can get confusing also because I can do the following: I have
> my model, I process, and see the solution. If I'm not satisfied, I can go
> to the Femm editor, change the model, save, reprocess and call the new
> solution. If I hadn't close the previous solution on the Femm View, I will
> have two solutions open, with the same name (or three as it just happened
> to me with "ECS10C_50Hz" model). I don't know if anyone else experiments
> this problem but for me it's confusing. Perhaps the program should not
> allow to open two solutions with the same name, or put some version number
> on the windows, maybe with the date and time.
> Well these are some sugestions for 3.0 Beta version.

There are still some rough edges to the program that need to be sanded away.
Thanks for the suggestion.

Dave.


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