Motor Equivalent Circuit (Long & Detailed)

[rsdoran]

Sometimes I believe y'all overthink an issue, my problem is not knowing how to explain what I understand.

First thing I understand is that ANY MOTOR can also be a generator...ie if electric power is applied a motor will convert it to rotational energy...a generator uses rotational energy and converts it to electrical energy. I understand it gets a little more involved but basically this is a fact.

MY SIMPLE EXPLANATION:
A motor has 2 rotational aspects, the current
applied with AC creates a rotating magnetic field in the stator....for a 4 pole motor this "field" rotation equates to 1800 rpm....ie the poles will change "polarity" from North to South at this rate.

This magnetic field "induces" current into the rotor which in turn creates a magnetic field that is converted into rotational energy. This is where "slip" is involved. You only create electrical current when you pass a conductor through magnetic field lines of force. The stator has its magnetic field rotating within the motor housing at 1800 rpm, if the rotor were also rotating at 1800 rpm, the rotor conductors would not be passing through any magnetic field lines but rather moving with them. So, in order to generate a magnetic field in the rotor, the rotor must be moving slower than the stator magnetic field for the motor to produce mechanical power. Basically the rotor is being pushed/pulled around by the stators rotating field.

With that stated it should be understood that if a motor were running at synchronous speed there would be no magnetic lines being crossed so technically you would not produce any mechanical energy...this does not mean that there can not be rotation though but torque etc does have a proportional linear dropoff.

If you have "rotation" in a motor but the "rotor" starts rotating faster, over-haul or over-running condition, than the "stator" then the magnetic fields will be getting crossed in the opposite direction....ie now the rotor is inducing current into the stator. The rotor is no longer being pushed/pulled by the stator, the rotational energy, of the rotor, is now being converted into electrical energy that is greater that the electrical energy in the stator. Personally I do not believe this would be CEMF nor is it technically "internally generated".

I know this is all very simplistic but it helps(me)to understand. This is the basic principle behind regenerative drives.

I was going to use an analogy but they always create other questions.

 

[TimothyMoulder]

Thanks Ron, for a clear and simple explanation that reaffirmed my grasp of the subject.

There are times that listening to guys like DickDV is like listening to Yoko Kanno - lovely to hear, no clue what it means

I'm not being critical of anyone, just the opposite. I hope to reach this level on motors some day.

TM

 

[rsdoran]

Thanks Tim, I appreciate knowing that I may have offered an understandable explanation that may be basically accurate. I appreciate DickDv, Keith etc being able to apply/show how the formulas work but sometimes the equations get to in-depth which makes it difficult for some of us to follow.

EMF is electromotive force and is measured in volts.
Inductance is a measure of the amount of magnetic flux produced for a given electric current.
Properties of inductance: For a steady applied voltage v, the current changes in a linear manner, at a rate proportional to the applied voltage, but inversely proportionally to the inductance. Conversely, if the current through the inductor is changing at a constant rate, the induced voltage is constant.

When AC current flows through an inductor, a sinusoidal alternating voltage emf is induced. The amplitude of the emf is equal to the amplitude of the current and to the AC frequency.

THIS induced EMF is called back-emf or CEMF because it "opposes" the applied voltage. The problem is that we do not think of the applied voltage being the applied EMF (but it is) so the induced EMF (voltage) is a voltage/emf that is counter (opposing) to the applied...ie CEMF

A motor, AC or DC, is basically nothing but a "group" of inductors designed to convert electrical energy to rotational energy.

There are "types" of inductance, the 3 main ones are coupled, mutual, and self. With a motor you are basically using coupled/mutual induction to covert electrical energy to rotational energy.
BUT
There is rotational energy that can "also" self-induce an EMF...I believe this is what they mean by "internally-generated". In general though these "induced emf's" are designed/created to develop the impedance that develops the HP/torque for the motor when running at nameplate speed under nameplate conditions.
ALSO
When a motor "over-runs", runs faster than rated,..ie a 1750 rpm motor is turning 1825 rpm then it becomes a generator. A voltage is already present so it has an "excitation" from the magnetic fields developed therefore the rotational energy is converted to electrical that can now "induce" a current into the stator. As I mentioned above once the motor reaches synchronous speed no field lines are being cut therefore no energy is being created so when it exceeds synchronous speed the energy will flow in the opposite direction, from the rotor to the stator.

I know this is a very simple answer but I am trying to answer it without using formulas/equations. One of these days though....

 

[AutomaticLeigh]

This is a link to a Marathon Electric motor:

http://www.marathonelectric.com/mot...028&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.

This worried me for a bit, but I don't think there is anything inconsistent here. The values X1 and X2 are leakage reactances, not the stator and rotor reactances. Imagine the rotor bars are disconnected. We now have the equivalent of a transformer with open circuit secondary. The equivalent circuit is a 'T' (or WYE), with the arms of the T as X1 and X2, and the leg Xm. Looking in the primary side you see (X1 + Xm), which is the reactance of the stator (no current flows in the rotor bars, so the 'secondary' winding has no effect).

 

[kamenges]

OK, you have me there. Back to the drawing board, or more correctly, the message board.

Thanks for the clarification. Like Dick said in an earlier post, this isn't about being personally right. It's about finding the right answer. Granted, as Ron said, at the end of the day as long as it works, it works. But at this point I'm in this deep enough that I just want to know.

I found an interesting PowerPoint on a website that discussed AC induction motor operation in the overdriven mode using the typical equivalent circuit. All they did was let the slip go negative. This forced the variable rotor resistance value to go negative leading to circuit current in the opposite direction. Negative resistance is just a mathmatical artifact of either the voltage or current changing sign.

This may be the standard way to model this. But, as I said in a previous post, this just reinforces to me that the equivalent circuit is a model that allows one to calculate actual values from a set of conditions. I'm still not fully convinced that it is a complete representation of the physical phenomena occuring inside the motor.

 

[Leadfoot]

We all have our comfort zone of knowledge on how motors function and their inner workings. Then a thread like this comes along and stimulates the thought processes. I just may have to dig my books out of storage.

I do not feel like I have learned anything new from this discussion so much as it has been an EXCELLENT review. It is pressing me to do better. I like that.

Ron, you have the gift of the K.I.S.S. principle when explaining things. Do not appologize for how you write your posts. They are good, basic and easy to follow. You may have a future in training or teaching.

 

[katratzi]

I agree with Leadfoot, Ron, you expalin it well without
all the equations, etc. But like kamenges, I'd still like
to know. Speaking of knowing, can anyone tell me why, if
you connect a switch between the positive of one battery,
and the negative of another, then close the switch you
get no current flow?

 

[AutomaticLeigh]

Originally posted by kamenges:

This may be the standard way to model this. But, as I said in a previous post, this just reinforces to me that the equivalent circuit is a model that allows one to calculate actual values from a set of conditions. I'm still not fully convinced that it is a complete representation of the physical phenomena occuring inside the motor.

Unless we get a motor manufacturer in here, I don't think we're going to get a good answer. My interest is a little bit more than academic, as I was once involved in a case where a motor pushing a centrifugal pump wouldn't accelerate past a certain speed. I wasn't resposible for sizing the motor, but as the electrical contractor, suddenly everyone is looking at me to get it going.

What happens is, the load torque gets too close to the pull-up torque, and there's not enough left to accelerate the machine. If you could get it to full speed it would be quite happy, but there's no way to get it there. Big panic, get the motor data sheet, calculate equivalent circuit, calculate torque curve, all the numbers look good. Then the motor manufacturer produces a graph of the motor torque which looks absolutely like nothing I'd seen before! How can you say he's wrong? he made the damn thing! Talking to a motor guy later, he said "yeah, it's always about 70% speed when they stick", so I guess I'm not the first.

It's funny how, when everything goes right, you don't learn much. When it goes wrong, you find out an awful lot in a big hurry.

 

[rsdoran]

I do not understand what is "needed to know".

Nslip = Nsync - N
T = k x Nslip = k x (Nsync - N)
An example: You have an induction motor rated at 2 hp at 1750 rpm (Nslip = 50 rpm).
T = 63000 x 2 / 1750 = 72 in-lb.
Now we find that the motor is actually running at 1775 rpm (Nslip = 25 rpm). With Nslip cut in half, torque is cut in half to 36 in-lb. HP = 36 x 1775 / 63000 = 1.01 HP

Using standard equations will show a motor loses torque/HP as slip is removed. When a motor reaches synchronous speed it basically "balances" out so there can be no torque/HP or energy put into the rotor.

If the rotor exceeds the 1800 rpm (4 pole motor) then the stator does not provide the energy anymore, the rotor does via rotational energy.

Think about it this way, the stator is physically stationary BUT it rotates...ie the magnetic field rotates at a rate of 1800rpm (4 pole motor), so IMAGINE it as moving like the rotor. If/when the rotor rotates faster then the stator then the "roles" are reversed...ie the rotor now becomes the power source.

Dang Nabbit y'all done made me use formuals.

 

[DickDV]

After reviewing the linked material and doing a Google search for "induction motor equivalent circuit", it is clear to me that Melfi's circuit is a bit simplified. In fact, it has to be since the motors we are considering are three-phase and his circuit is single-phase.

Despite that, it now seems to me that it is a useful circuit for understanding motor operation including when a motor is overhauled and regenerates power. The key for me is that resistive element in the rotor circuit R2/s and understanding that it is resistance divided by some function of slip (s). Even the circuit shown in www.ene.ttu.ee/electriajamid..... offered by kamenges, as Figure 2e, which shows a voltage generator in the rotor branch can be accommodated by Melfi's circuit by allowing s to swing both + and -.

While I am not comfortable in the world of vectors and phasing, I feel like I gain some clarity when the slip (s) is allowed to swing positive and negative. This was mentioned earlier in this thread by Simon Goldsworthy and I must thank him for his insight. It started my thoughts down a trail that is, at least for me, very helpful.

If we look at Melfi's equivalent circuit again, Eg is the voltage across the stator windings and also across the rotor branch. Eg is nearly equivalent to the applied voltage (Vt) except for minor losses thru the R1 and X1. If we define slip to be positive when the rotor is turning slower than the spinning field (motor under load) and negative when the rotor speed is higher than the spinning field (motor overhauled) then the current thru R2/s is in phase with Vt since X2 is always very small and the stator current thru Xm is 90 degrees lagging. This gives us the familiar current vectors that produce total motor lead current. Since the rotor branch circuit current is approximately equal to
Eg/(R2/s) which simplifies to Eg x s/R2, it is clear that, as the load varies, slip varies in direct proportion causing the rotor branch current to also vary in direct proportion. It follows then that, as load approaches zero, the rotor branch circuit current also approaches zero just as expected from the familiar motor lead vector diagram. As the rotor speed matches the spinning field speed, s becomes zero and so does rotor branch circuit current. This leaves only the magnetizing amps thru the stator flowing in the motor leads which matches field experience.

As the motor becomes overhauled, s now becomes negative with its absolute value increasing as the overhauling torque increases. From the rotor circuit current equation above, it can be seen that current also becomes negative and its magnetude increases as the rotor is driven further into negative slip.

With this current being 180 degrees out of phase with the applied voltage Vt, it follows that the flow is backward to the source also. Regeneration is the result of this backward flow.

A couple of curious things happen when a motor regenerates like this. First, since the regen current is about 90 degrees out of phase from the stator magnetizing current Im, the stator is not properly magnetized by the regen current. This would explain why a properly-phased external AC voltage must be applied to an overhauled motor to get it to regenerate. It will not "self-excite".

Second, I have heard of one special condition where an overhauled induction motor can self-excite which involves the installation of capacitors on the motor leads. This is often done for power-factor correction or as a sine filter on a VFD output. While highly undesireable, the right combination of capacitance and inductance can result in precisely the phase shift needed to magnetize the motor stator and regeneration can occur without an externally applied voltage. Since this is basically an uncontrolled and unexpected condition, it is to be avoided.

I'm pleased with this new insight into motor operation and am indebted to those who posted on this subject. It is kind of fun to reach for something and "strain my brain" once in a while and that is what this thread was for me. Thanks again.

Finally, I noted some comment earlier in the thread about how un-understandable I can be. I make it my goal to be as simple and clear as possible when answering questions in this forum or in the many training classes I conduct on drives and motors. Most of the students in these classes are maintenance techs and communicating effectively with them is personally important to me. (It is also what they came to the class for!) However, this particular thread was not of that type. Rather, it was an attempt for me to gain an understanding in an area that wasn't clear to me. For that reason, I put in the original thread heading "(long and detailed)" as a warning that the contents of the thread might get a little messy. For other situations, please stop me or wack me in some way if I am not clear. I very much want to be helpful. We've all sat in a class where the instructor is trying to explain something he doesn't himself understand and it sure is uncomfortable and a waste of time. I don't want to be one of those.

 

[Leadfoot]

The typical "motor equivalent circuit" is always shown as a single phase. I have seen the 3 phase one and it is basically 3 single phase circuits connected, much like taking 3 single phase transformers and connecting them for 3 phase operation. I saw this circuit while observing an engineer do the calculations for preseting the parameters in a flux vector drive. I worked the formulas and got values. We set them and then ran the auto tune. Drive's today usually do not need values calculated and entered prior to auto tuning, for this I thank the software guys.

The advances made in controls has made it so much easier to do things today than 10 to 15 years ago.

I find it amazing just how much some engineers get stuck on buzz words and feel they need to have things based on their date with Miss Information.

 

[RMA]

I'd just like to say thank you to DickDV

for starting this Thread. I've never really been heavily involved in motors (or transformers for that matter) and what little knowledge I do have goes back 40+ years to school and University.

This is all knowledge which can be very useful when troubleshooting in the field (regardless of whether the problem is "mine" or not).

It never hurts to increase your knowledge of related (and even unrelated) subjects - and gee, it's fun too!

 

[Leadfoot]

for starting this Thread. I've never really been heavily involved in motors (or transformers for that matter) and what little knowledge I do have goes back 40+ years to school and University.

This is all knowledge which can be very useful when troubleshooting in the field (regardless of whether the problem is "mine" or not).

It never hurts to increase your knowledge of related (and even unrelated) subjects - and gee, it's fun too!

Well said. BTW, did you ever have much experience with mag amps? You may be old enough to remember them.

 

[RMA]

BTW, did you ever have much experience with mag amps?

KNow them by name, but never worked with them, sorry!

 

[kamenges]

Originally posted by rsdoran:

I do not understand what is "needed to know".

What you say in this post is all true, Ron. I think everyone will agree with that. The question is why is it true. The common concensus in this thread and the one that spawned it has been that there is no such thing as back EMF in an AC induction motor. If that is true then what causes the energy to flow back to the drive (or the line) when an AC motor is overhaled? Does this really matter in our everyday lives? No. I'm just curious.

Originally posted by DickCV:

Second, I have heard of one special condition where an overhauled induction motor can self-excite which involves the installation of capacitors on the motor leads.

That would truly suck. I'm pretty sure I wouldn't want to see that. It could lead to some nasty surprises.

Dick, I think you may be winning me over. I still have a couple of things to personally resolve but that's OK. For example, it is 'apparent' rotor resistance that changes with slip. We all know that a squirrel cage motor's true resistance can't change since it is nothing but a set of aluminum bars shorted on both ends. The case of the rotor spinning at exactly synchronous speed pretty much throws back-EMF as a current limiting factor right out the window. I guess the apparent resistance would be infinite if there was no induced voltage in the rotor since there is no relative motion between the stator field and the rotor. As slip increases you would get higher flux change rates, giving you higher rotor voltages and higher rotor currents. Thus the lower apparent resistance.

The only thing I'm still lost on is the rotor to stator induced voltage thing. I think we can all agree that the rotor has a magnetic field that rotates at roughly the same speed relative to the stator that the rotor does. Lenz's Law says that this magnetic field should be inducing a voltage in the stator. Also according to Lenz's Law this voltage should be in opposition to the voltage that created the magnetic field. Now we could discuss which voltage created the magnetic field, which changes which voltage the induced voltage is in opposition to. But at this point we can say this this voltage will be somewhere between 0 and 180 degrees out of phase. If it is 0 degrees out of phase it will add to the fundamental voltage, if it is 180 out it will fully subtract. Also, this voltage component should increase with mechanical load since the higher slip will increase the strength of the magnetic field on the rotor. First of all, does this induced voltage exist? If not, why not? This would seem to violate Lenz's Law. If it does exist, what is its phase and how does it affect the fundamental voltage?

Originally posted by Leadfoot:

I find it amazing just how much some engineers get stuck on buzz words and feel they need to have things based on their date with Miss Information.

Hey, I resemble that remark.
With most of what we have been saying so far, the basic justification has been 'Well, the mtor model says...'. Well the motor model also says that the stator resistance goes negative. I don't know about you but my meter doesn't go that way when I switch it to Ohms. So, while the motor model accurately predits what will happen in all operational cases, I think it's a bit more involved that a couple of inductors and a negative resistance. ANY model is nothing more than a mathematical analogy of a physical system. It is not necessarily an accurate facsimile of that same system. I'm just trying to figure out what that 'little bit more involved' is all about.

Keith