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# Re: [femm] skin effect in the rotor bars of a squirrel cage induction motor

- To: femm@yahoogroups.com
- Subject: Re: [femm] skin effect in the rotor bars of a squirrel cage induction motor
- From: David Meeker <dmeeker@ieee.org>
- Date: Tue, 08 Apr 2003 22:27:47 -0400
- In-reply-to: <000e01c2fde7$2bdef560$0100a8c0@mshome.net>
- References: <000e01c2fde7$2bdef560$0100a8c0@mshome.net>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.3; MultiZilla v1.1.34 (d)) Gecko/20030312

Tielman Slabbert wrote:

> I want to determine the resistance of a rotor bar in a squirrel-cage induction motor, including the skin effect. I tried using FEMM but the results are not looking right.
> I drawed the rotor bar in FEMM, created a boundary (Prescribed A with all the parameters zero) so that no flux should cross to the outside of the bar, as was done in the example on skin effect on the website. (Is this the right method for this case, or must I draw the whole induction motor?)

I think that you'd want to draw the bar in the slot so that you get losses due to the slot leakage flux passing through the conductor.

> I took a bar length of 1 metre for simplicity sake (Problem -- depth = 1000), and used a circuit property to have a current of 50 Ampere flow through the bar. (Bar material = copper)
> To calculate the DC resistance, I used the Block integral in FEMMview to find the area, and then calculated the resistance to be 359 microOHM
> When using the block integral for resistive losses (problem frequency = 0Hz) I get 0,898 Watt which corresponds to my calculation using P = I^2*R
> However, when I change the frequency to 50 Hz, the resistive losses are 0,449 Watt, which is less than the loss at 0 Hz! (I wanted to use this loss to determine the effective rotor-bar resistance at 50 Hz)
> Am I doing anything wrong? or understanding anything wrong?

The convention that FEMM uses is that all current, voltages, current densities, flux densities, potentials, etc. are represented as amplitude rather than RMS quantities. If currrent "I" is in terms of amplitude the time average power loss is P = (1/2)*I^2*R for frequencies not equal to zero. That means your 50 Hz calculation is off by a factor of 2.
Dave.
--

    David Meeker
    dmeeker@ieee.org
    http://femm.berlios.de/dmeeker

- References:
  - skin effect in the rotor bars of a squirrel cage induction motor
    - From: "Tielman Slabbert" <tielman01@telkomsa.net>

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