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Example

What would be the actual value of stator resistance and inductance of this motor in ohms, not in p.u?

Can someone give me a valid formula for calculating it? I get the "p.u" idea but I'm not sure what is "nominal value" of resistance for a motor and if the same value is used to calculate "p.u" for both resistance and reactance. Thanks in advance

Theresnoneed
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  • Perhaps, somewhere in the datasheet is also a note about p.u. meaning. – Marko Buršič Nov 26 '20 at 14:47
  • Attached motor params are not from the datasheet but from a scientific paper. Unfortunately they didn't mention how was the base value of resistance calculated so I assumed it must be common knowledge. – Theresnoneed Nov 26 '20 at 14:51
  • True, check this: https://link.springer.com/content/pdf/bbm%3A978-1-349-01615-0%2F1.pdf Difficult to find out, but at least you have some starting point. – Marko Buršič Nov 26 '20 at 15:03
  • Here another example: https://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html – Marko Buršič Nov 26 '20 at 15:07

3 Answers3

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The resistances and inductances of a 3-phase induction motor are normally the values of the per-phase equivalent circuit. The inductances used in the equivalent circuit are normally given is ohms as the reactances at the nominal frequency. The per phase equivalent circuit is the line-to neutral circuit for a wye connected motor regardless of the actual connection of the motor. If the motor is actually delta connected, the equivalent wye circuit is analyzed. A version of the standard equivalent circuit is shown in my answer to this question: Why does an induction motor draw more current when the load is increased?

The 1 per-unit inductance and reactance in ohms is the nominal line to neutral voltage divided by the nominal line current. For the data given in the question the base resistance or reactance in ohms is 231/350 = 0.66 ohms. That would make Rs = 0.66 * .11 = 0.0073 ohms. I believe that Rs is the equivalent circuit stator resistance. That would be the actual winding resistance for a wye connected motor. For a delta connected motor, the winding resistance would be 3 times that. The terminal resistance would be twice that in either case.

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$$S = \sqrt {3} \ V_{Line} \ I_{Line} = \sqrt {3} \times 400V \times 350A = 242.48kVA$$

$$Z_{Base} = \frac {V_{Line}^2}{S} = \frac {(400V)^2}{242.48kVA} = 660mΩ$$

Divide that by 3, for each of phases. \$Z_{Base} = 220mΩ\$. That should be pu quantity.

$$R_s = 0.011 pu = 0.011 \times 220mΩ = 2.42mΩ$$

And there are three of them.

StainlessSteelRat
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Source of example

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Base power = 200k/3 = 66.6k kVA
Base voltage = 400 / sqrt(3) = 230V
Base current = 66.6kVA / 230V = 290A
Base impedance = 230V / 290A = 0.793 ohm

Rs = 0.011 pu -> 0.011 * 0.793 = 0.0087 ohm

Edit:

Base voltage = 230 V
Base current = 350 A
Base impedance = 0.657 ohm

Rs = 0.011 pu -> 0.011 * 0.657 = 0.007227 ohm
Marko Buršič
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