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Thanks very much for the help/info!
What I actually did to calculate the currents used
in my model was:
I used a machine in the laboratory and conducted
some experiments to determine the equivalent circuit parameters (with help of
some experts to find the leakage reactances of the rotor and
stator)
Since the current through the magnetising reactance
in the equivalent circuit is responsible for the flux in the machine, I used the
reactive component of the no-load current for the lo-load
simulation.
For the full-load, I calculated the full-load
stator and rotor-bar current from the equivalent circuit for the measured input
voltage.
Ivan Slabbert
----- Original Message -----
Sent: Thursday, July 24, 2003 3:12
AM
Subject: Re: [femm] flux distribution in
an induction motor on full-load
Tielman Slabbert wrote:
Hallo
I wanted to see how the flux distribution
in an induction motor on full-load looks like. The results onno
load is below:
[...]
And on full-load:
[...]
The overall flux density in the motor decreased (I usedthe
same scaling for the flux density plot) and the field is distorted.
Is this how it should be???
Ivan Slabbert Yes, if you used the samestator
currents in each case.
If you neglected leakage and stator
resistance, you could think of the per-phase circuit model of an induction
motor as a voltage source driving two parallel impedances: j*w*M, representing
the mutual inductance between the rotor and stator, and a rotor impedance
w*Rr/ws, where w is the stator frequency, ws is the slip frequency, Rr isthe
rotor resistance, and j=sqrt(-1).
If you drive this trivial motor
model by voltage, it is easy to see that the current through the mutual
inductance path isn't affected by slip (or vice versa). The implication
is that the field would have the same orientation and magnitude on-load as
with no load--(similar idea to a transformer). The stator pulls a lot
more current on-load, however. When the motor is run on-load, ws increases,
making the rotor impedance smaller, drawing more current through the rotor
branch of the equivalent circuit. However, the field produced by the
extra stator current is cancelled out by the reaction field from the equal and
opposite currents induced in the rotor.
However, if you've made a FEMM
model, you've probably kept current constant. At any rate, you'd have to
adjust the currents carefully to get a full-load model that would requirethe
same voltage as the no-load model. Full-load won't be exactly the same
as the nameplate values, because FEMM neglects end effects (and some other
effects) and the values on the nameplate are just "nominal", so you'd have to
do a few iterations to find the "correct" full-load operating point in
FEMM). Ahyhow, if one assumes that the current is fixed, the current
gets divided between the rotor and stator branches. As slip (and load)
increase, the rotor draws current, from the mutual inductance branch,
reducing the magnitude and changing the phase of the flux.
Dave.
--
David Meeker
dmeeker@xxxxxxxx
http://femm.berlios.de/dmeeker
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