Motor Equivalent Circuit (Long & Detailed)

I've studied the white paper by MJ Melfi of Reliance Electric and am persuaded that the equivalent circuit he offers as Figure 3 is valid for a loaded induction motor.

It is clear from his Figure 3 that there is nothing in the equivalent circuit but inductance and resistance. There is no provision for any kind of internal generator in this circuit.

In equation #2, Melfi labels Eg as "internally generated voltage due to motor rotation". At the end of that same page and the beginning of the next page, he goes on to define Eg as Vt minus the voltage across the stator leakage resistance and the leakage reactance, none of which have anything to do with motor rotation. I believe that this is a labelling error of Eg and is probably due to Melfi being a former DC motor specialist (almost all Reliance motor engineers came up thru the DC ranks).

Further along in his paper under the heading "Field Oriented Control", Melfi defines motor lead current as the vector sum of Flux Producing Current and Torque Producing Current. This is a well-known definition and has been thoroughly tested in the field and found completely reliable. I have personally used this relationship numerous times to find flux amps. The procedure involves energizing the motor to rated volts and frequency and measuring current with no load on the shaft. Or, as I have often done, half the rated voltage and half the rated frequency can be applied, again with no load and the rotor turning at half the rated speed, with the same measured current. There is no provision in the definition and no findings in the measured current that would indicate an internally-generated voltage from rotation or any other source.

I conclude from the above that Melfi's equivalent circuit is valid and that there is no internal generator present in a loaded induction motor.

Having said all this, I realize that the above clearly does not explain, under overhauling conditions, how an induction motor regenerates power back to the supply, whether the AC line or an inverter.

I cannot claim to understand this mode of motor behavior and have never had it adequately explained either. As with all AC generation in a network, phasing of the AC wave is a critical indicator of energy flow. How this is brought to bear in an overhauled induction motor is not clear to me either.

I conclude that Melfi's equivalent circuit is valid only for a loaded motor and that there must be some internal generating element that appears only when the motor is overhauled or, expressed another way, when the motor is driven into positive slip.

I would like a more complete explanation with supporting documentation of the overhauling motor condition and accompanying regeneration.

Looking forward to further discussion on this, DickDV

Is there a link to the white paper?

Yes, but I don't know how to put the link in the posting. kamenges had the link in the "Inverter/Motor" thread of about a week ago.

I could stand to learn how to get these links in here.

Dick,

I see no mention of slip in your discussion.

At no load there is no slip, no induced current in the rotor windings and no torque produced.

Loading the motor slows the rotor down, induces slip which in turn induces current in the rotor windings. It is the interaction between the rotating stator field and the slightly slower rotor field that generates torque.

The same applies in overhauling conditions, but with the relative rotations reversed, and the energy flow reversed.

There is always slip in an induction motor, even at no load the motor will not run at synchronous speed.

Here is the link to MJ Melfi's white paper:

www.reliance.com/pdf_elements/b7113.pdf#search='ac%20motor%20equivalent%20circui t'

Philip, that there is slip in an induction motor is taken for a given in the discussion, in my view.

The concern is with the equivalent circuit, both under loaded and overhauled conditions.

DickDV - although the equivalent circuit contains inductance and resistance, the key point is that one of the resistances is actually a function of slip (which in turn will determine the back emf: e.g. if the motor is locked, the slip is unity and there is no back emf so the equiviant circuit shows the resistance of the rotor, if the motor is running with no slip, the back emf is equal to the applied volts and hence no current will flow in the equivalent resistance which is now open circuit as s=0
During overhaul, the slip will be negative so even though the applied volts are positive, the current flow will be in the opposite direction thus indicating that power is returning to the supply.

Good explanation Simon.

That the R2/s term is variable with respect to rotor slip is clear to me and is described in some detail in the paragraphs that follow Figure 3.

What is not described and is not at all clear to me is how a variable resistance is expected to generate a "back EMF". And, beyond that, how is that branch of the equivalent circuit going to produce voltage and phasing necessary to cause energy to flow out of the motor and into the supply circuit.

Understand now, I'm not saying that it doesn't happen. I'm simply looking for an appropriate modification in the equivalent circuit under overhauling conditions to explain the reverse flow of energy.

The amount of back EMF is based on the amount of current flowing thru the R2/s. More slip more back emf, less slip less back emf. This would be at any steady state of applied volts and frequency . As you increase the load, the slip increases as does the current. As current rises, so does developed torque.

Im should remain constant and It varies with load requirements.

BTW DickDV, nice explaination. Have you noticed if you read all of the information available for motors they sometimes contradict each other?

This is a link to a Marathon Electric motor:

http://www.marathonelectric.com/motors/datasheet.asp?REV=R1&WIND=K1844217&MOD=184THTS8028&VOLTAGE=460

Notice the values of inductive reactance for the stator coils and the rotor as well as the mutual inductive reactance.

This is a link to one of many web pages that deals with mutual inductance:

http://www.uoguelph.ca/~antoon/gadgets/coils/coils.html

About 3/4 of the way down the page they show a formula that will allow one to calculate the mutual inductance given the two inductance values and the coefficient of coupling. I have seen this same equation in other locations so I'm pretty confident it is correct. The equations is:

M = k*SQRT(L₁*L₂)

It also states that k can't be greater than 1.

So this leaves us a little bit of a quandry. If we take the model as a free-standing entity we need to assume that the frequency seen by any if the coils is the same. If we assume this then we can say that the ratio of the inductive reactance values in the circuit are the same as the ratio of the inductance values in the circuit. There are no individual values of inductive reactance in the motor parameters listed that is above 5 Ohms. Yet the mutual inductive reactance is is 77 Ohms, which is way too high given the individual reactance values.

I still believe that the motor model is just that; a model. It contains all the elements that can effectively model the voltage, current and flux vectors in the motor. The mutual inductance shown in the motor model is an effective modelling tool and does correctly model the actual values in the motor. But at the end of the day mutual inductance is really nothing more than an explanation of how one coil producing a varying magnetic field can induce a current in another coil and how that induced current can affect the exciting coil. There are no inherent constraints on how that effect on the exciting coil is achieved. We tend to think of mutual inductance in terms of transformers because that is always the example given. That doesn't mean that is the only physical model that can exhibit mutual inductance.

Keith

Just to throw a little additional confusion into this discussion, take a look at this web page:

http://www.ene.ttu.ee/elektriajamid/teadus/artiklid/Ungar965/

Relatively early in the page reference is made to five different AC motor models. Check out 'e'. I have to admit that this is the first time I've ever seen this representation myself. But someone must see things that way or that particular model wouldn't exist.

As Dick said in another thread, I think magnetising current is as low as the current level in an AC induction motor can go. I'll have to take a peak at the vector transforms but I think the voltage to current phase angle smoothly changes as the motor passes from driving to braking. If you look at it purely in terms of the I_d/I_q relationship, I_q simply gets shorter and shorter as motoring torque decreases until it hits zero length and then starts growing in the other direction as braking torque increases. All this time I_d maintains the same vector length. This would equate to a voltage/current phase angle that grows to 90^o at no load and then increases further as braking torque is generated. The total motor current would correspondingly decrease to magentizing current at no load and then increase again as braking increases. The thing that I hate about this is that most of the AC motor speed/torque curves I have seen lie as they show the motor current going to zero along with the torque at synchronous speed.

Keith

Yes, Keith, your description of the torque vector going to zero and then swinging back in the opposite direction is how I visualize it too.

Exactly how this results in energy flowing backward to the source is a bit of a puzzle.

Thanks for the links. I've been traveling and not very available to read and ponder the links. But I surely will.

I find this exchange very beneficial so, thanks. I hate to just argue and, in fact, won't. But this has been a thoughtful discussion and my hope is that we all can benefit from it.

I have some connections with motor people in the ABB and Marathon organizations and might be able to gather something useful through those sources. I'll advise when I have anything.

Thanks again.

I came across this website as I was poking around. It's a brief write-up about steering power assist motors in cars and why brushless DC motors are the way to go. However, about half way down the author gets into a brief description about the basic concept of field oriented control, complete with diagrams and transforms.

http://www.automotivedesignline.com/howto/165600237

Based on this I put together a little spreadsheet and started playing around. It appears to me that as I_q goes positione the quadrature current magnitudes increase and the stator current vector to rotor flux vector angle increases. The same happens as I_q goes negative except the vector angle decreases.

I attached the spreadsheet I put together so anyone who is interested can play around.

Keith