Re: [femm] flux distribution in an induction motor on full-load

[David Meeker <dmeeker@xxxxxxxx>]

Tielman Slabbert wrote:

Hallo

I wanted to see how the flux distribution in an induction motor on full-load looks like. The results on no load is below:

[...]

And on full-load:

[...]

The overall flux density in the motor decreased (I used the 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 same stator 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 is the
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 require the 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