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

- To: <femm@yahoogroups.com>
- Subject: Re: [femm] flux distribution in an induction motor on full-load
- From: "Tielman Slabbert" <tielman01@telkomsa.net>
- Date: Thu, 24 Jul 2003 21:56:11 +0200
- References: <000401c350ec$38d39480$6b74ef9b@mshome.net> <3F1F3266.30703@ieee.org>

> 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 ----- From: David Meeker To: femm@yahoogroups.com 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@ieee.org
>     http://femm.berlios.de/dmeeker

- References:
  - flux distribution in an induction motor on full-load
    - From: "Tielman Slabbert" <tielman01@telkomsa.net>
  - Re: [femm] flux distribution in an induction motor on full-load
    - From: David Meeker <dmeeker@ieee.org>

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