Thanks for clarifying that Dave.
Brad
-----Original Message-----
From: David Meeker
[mailto:dmeeker@xxxxxxxx]
Sent: Tuesday, May 06, 2003 9:20
AM
To: femm@xxxxxxxxxxxxxxx
Subject: Re: [femm] SRM Inductance
plot at constant current
Brad Frustaglio wrote:
> Hi All,
>
>
>
> I have been using FEMM for a few years now.
Many thanks to Dave,
> excellent program.
>
>
>
> I have question on evaluating the A-J
integral and then calculating
> the self inductance from the result.
>
>
>
> I am modeling a 4 phase SRM with 8 stator
poles and 6 rotor poles. I
> am attempting to extract the torque profile
and inductance profile
> with one phase excited at constant current as
the rotor moves through
> one stroke. The torque magnitude and shape
looks reasonable. However I
> am having trouble understanding how to
calculate the inductance
> correctly. The profile shape looks as
expected I am just concerned on
> the magnitude of inductance. My method is:
>
>
>
> 1. Highlight the blocks
with the current flowing ( In my case there
> are2
current blocks per pole (one on each side of the stator
> pole) for
a total of 4 current blocks per phase.
> 2. Evaluate the A-J
integral at each respective rotor position
> 3. The self inductance as
defined is int(A-J)dV /i^2) where i is
> thecoil
current
>
>
>
> I think I am running into trouble on the coil
current term. What is
> the correct coil current to use to evaluate
the inductance correctly?
>
>
>
> The value of one current block or the sum of
all current blocks. Also
> this is actually amp-turns. Not the actual
current flowing in the wire
> itself. Right?
>
>
>
> For instance: the current in one block in the
model is defined as
> 11.15 MA/m^2. The coil area is 2.6903 x10^-5
m^2. So the magnitude of
> current in the coil block is 300 A-T. Is this
the current I use for
> the evaluation of self inductance usingthe
A-J integral?
>
>
>
> Regards,
>
>
>
> Brad Frustaglio
>
The /i/ in the equation actually /is /the actual
current flowing in the
wire itself. Like in your example, saythat
your coil is made of 60
turns of wire. Since you have 300 Amp*Turns,
the current in the wire
would be 5 Amps. The 5 Amps is then the
value that you'd use as /i/ in
the equation. There's also an example that
might be relevant to you at
http://femm.berlios.de/induct1/induct1.htm
One last thought--switched reluctance motorsare
typically run in a
highly saturated condition. The quantity
that you are computing with
the A.J integral is (flux linkage/current), but it
doesn't have the same
relationship to stored energy as inductance in a linear
problem.
Furthermore, this "nonlinear inductance"
appears in a subtly different
way in any electric circuit equations that you
might write. For example,
the electric circuit equation that applies to your
case is:
D(flux linkage)/Dt + R i = v
where the Dt is meant to represent the /total/
derivative with respect
to time. If inductance, L, is not a
function of current, this
simplifies to the usual:
L di/dt + R i = v
However, if L is a function of i, we'd get:
(L(i) + i*dL/di)*di/dt + R i = v
Dave.
--
David Meeker
email: dmeeker@xxxxxxxx
www: http://femm.berlios.de/dmeeker
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