Proving the circle shape of I.M. circle diagram

[electricpete]

An induction motor circle diagram is also known as Heyland circle or Osanna circle / diagram.

It is a relatively straightforward phasor diagram representing currents in the equivalent circuit, and quantities derived from those.

The semi-circle shape arises from a plot of the following vector quantity (*):
I2 = V / (j* X2 + R2/s)

Specifically, if we plot I2 in the complex plane while we vary s, the tip of I2 traces a semicircle. The range of s to trace out a full semi-circle would be 0 to infinity (only 0 to 1 generally relevant for our purposes).

I can verify numerically that the plot of I2 = V / (j* X2 + R2/s) in the complex plane is a semi-circle for any arbitrary values of R2 and X2 that I choose. But I can’t come up with a proof why that is so.

QUESTION: Can you prove mathematically why a plot of I2 = V / (j* X2 + R2/s) in the complex plane traces out a semi-circle as we vary s from 0 to infinity

(* aside: I am neglecting the magnetizing branch component and focusing only on rotor banch current to simplify the question for my purposes... obviously we have to add in the magnetizing branch component to get total stator current).

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[cswilson]

pete:

I think I can get you started, although I don't have time now to go through it completely.

Complex numbers in the denominator are impossible to work with, so the first step is to multiply numerator and denominator by the "complex conjugate" of (R2/s - j*X2). This gives you a real denominator of (R2^2/s^2 + X2^2).

At s = 0, both real and imaginary terms are 0.

At s = inf, the real term is 0, and the imaginary term is -j*V/X2.

So if the curve is a semi-circle, the center would have to be at [0, -j*V/(2*X2)].

Now use the Pythagorean theorem to solve for the distance from this supposed center point. If the curve is truly a semi-circle, all of the terms with "s" should cancel out, leaving a constant value.

Good luck!

Curt Wilson
Delta Tau Data Systems

P.S. For those not particularly interested in the algebra, this curve still has interesting implications for how the power factor of an induction motor varies with slip. It is not at all intuitive (to me at least).

 

[cswilson]

Yeah, it works. Just a bunch of algebra. The radius of the semi-circle does solve as V/(2*X2).

Curt Wilson
Delta Tau Data Systems

 

[electricpete]

Thanks Curt. What you wrote in your 1st post is straightforward. Perhaps you can describe the strategy by which the expression for I2 is transformed into the equation for a circle.

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[electricpete]

Aha. I found it here on page 143:

http://books.google.com/books?id=fd...resnum=10&ved=0CD8Q6AEwCQ#v=onepage&q&f=false

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[electricpete]

Although it has not much practical importance, there was another case where plot of a computed circuit variable (voltage in the linked case, current in the Osanna circle case) turns out to be a circle in complex plane as we vary a parameter related to a circuit impedance:
thread238-240139
(see post 27 Mar 09 16:51)
I suspect the circle for the linked case can be proved in similar manner although I haven't done the math excercize.

In both cases it is a little mysterious and non-intuitive why it would do that. Although I can follow the math I certainly don't expect to see perfect circle in complex plane jump up out of the blue from circuit analysis.


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(2B)+(2B)' ?

 

[electricpete]

I guess there are other examples. I vaguely remember impedance circle appearing in diagrams used for setting distance relays. Is it similar phenomenon?

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(2B)+(2B)' ?