Induction motor frequency of e.m.f. and current on the stator side 3

[jsmemphis]

Hi everyone,

I've been reading some texts about the induction motor. However I don't understand very well why the induction motor frequency of e.m.f. on the stator side remains the same as the grid (assuming that the induction motor is working as a motor connected to the grid).

Can anyone enlighten me about this particular issue.
Many thanks in advance.
jsm

 

[electricpete]

First lets look at magnetizing current. That magnetizing current is at grid frequency, so it creates a flux which induces a "magnetizing" emf in the stator winding at grid frequency. The magnitude of magnetizing current adjusts so that the associated induced voltage is roughly equal/opposite to applied voltage. (the "opposite" terminology arises from tracing a loop where voltages sum to zero...some would prefer to say it's the same voltage).

If the motor is at no-load, that's all you've got, simple story.

If you add a load to the motor, then the rotor slows down, which results in slip and current in the rotor. This rotor current and associated flux creates a deviation in the emf (which was previously balanced by magnetizing emf), such that a roughly equal/opposite stator load-component current is required to roughly cancel the emf from rotor current to restore total emf equal/opposite to of applied voltage.

Both the rotor and load-component stator currents (and associated emf's) mentioned above have their fundamental (working) component at grid frequency.

Why is the rotor component at grid frequency? Roughly speaking: the rotor speed is Fsync*(1-s), the frequency of current in the rotor (in rotor frame of reference) is given by s*Fysnc, and when we look at how that emf appears when seen in stator frame of reference, we have to add together rotor speed and rotor frequency to give Fsync*(1-s) + s*Fysnc, = Fsync. i.e. fundamental/working rotor current and emf as seen in the stator reference frame is at grid frequency.

Since the stator load component emf is generated in response to rotor emf, it also has the same frequency.

If you consider effect of slotting, there are other frequencies to consider, but above focused on the "fundamental/working" component.

Voltage associated with leakage reactances and stator winding resistance were omitted from above discussion for simplicity.

The term emf I think has been debated before in some circles of the forums. For my purposes it is the same as "induced voltage"


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(2B)+(2B)' ?

 

[jsmemphis]

Thank you very much electricpete!

If I understood correctly, the fundamental rotor current and emf frequency are independent of the slip, so no matter is the slip (s=1 or near 0) the frequency of these 2 rotor variables will be the same as the grid.

Regarding the rotating magnetic field created by the emf of the rotor, it will atenuate the one created in the stator/air gap by the grid voltage (and the current of the stator will compensate that).
What are the conditions for them to be partially canceled? Can I conclude that having the emf of the rotor the frequency of the grid (and the stator flux as well), his rotating flux has the same frequency and speed of the stator flux?

I'm not sure if I was clear enough. Anyway thank you once more.

Kind regards.
jsm

 

[electricpete]

If I understood correctly, the fundamental rotor current and emf frequency are independent of the slip, so no matter is the slip (s=1 or near 0) the frequency of these 2 rotor variables will be the same as the grid.

That is correct with the important qualifier that it applies when these rotor parameters are viewed in the stator reference frame.

Regarding the rotating magnetic field created by the emf of the rotor, it will atenuate the one created in the stator/air gap by the grid voltage (and the current of the stator will compensate that).
What are the conditions for them to be partially canceled? Can I conclude that having the emf of the rotor the frequency of the grid (and the stator flux as well), his rotating flux has the same frequency and speed of the stator flux?

I think you are asking about my discussion, which borrows from the simplified description of a transformer (with motor stator playing role of transformer primary and motor rotor playing role of transformer secondary). Namely: if we apply a voltage to primary with no load on secondary, then we have only a parimy magnetizing current which creates a primary emf that exactly balances the applied voltage. If we now connect some impedance to the secondary, it allows secondary load current to flow (I = V2/Zload). That secondary current would destroy the balance we had under load, so equal/opposite primary amp-turns (also called load component of primary current) flows to restore that balance. It's a simplified description that provides intuition imo. If it doesn't make sense, you might google for some transformer info... they may explain it better than me.

Conditions for partial cancellation? I'm not sure what you're asking. But your conclusion is correct that the *working* components of all the rotor and stator quantities when viewed in the stator reference frame have the same time-frequency and rotational speed (they rotate together synchronously).


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[electricpete]

A correction of an obvious (?) typographical error - I left out an important word which is added in bold here:
That secondary current would destroy the balance we had under no-[b/]load

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[jsmemphis]

Hi again,

Thanks for your reply.

Thanks for the comparison, In fact I understand how the transformers work, and, apart from the slip, they work in the same way!

What I've tried to say is that, like in the case of transformers the rotor flux weakens the stator flux.

...And weakens because, like in transformers, it opposes the stator rotating flux. How? Rotating in the opposite direction (having, like you wrote, same time-frequency and rotational speed). I think I got it now. Isn't that right?

Thanks again.
jsm

 

[cswilson]

At a very basic conceptual level, the "back emf" of a motor can be viewed as the "cost" to the electrical system of the transfer of power to the mechanical system of the motor. This is a consequence of conservation of energy, where the mechanical power generated by the motor (torque * speed) must be matched by the electrical power provided to the motor (back emf * current).

This is real power (not reactive) and so must be in phase with the stator electrical waveform. The equivalent circuit models of an induction motor represent the back emf as a speed-dependent variable resistance, a useful fiction for those used to thinking in circuit terms. A resistance represents real power (but dissipated, not transferred), in phase with the electrical waveform.

Curt Wilson
Delta Tau Data Systems

 

[electricpete]

What I've tried to say is that, like in the case of transformers the rotor flux weakens the stator flux.
...And weakens because, like in transformers, it opposes the stator rotating flux. How? Rotating in the opposite direction (having, like you wrote, same time-frequency and rotational speed). I think I got it now. Isn't that right?

I wouldn't say it quite that way. The working rotor and stator fields rotate in same direction. Any current that flows in the rotor has an associated flux component which (along with flux from stator current) determines the airgap flux. The total flux in the airgap is the sum of three components:
1 - Flux associated with magnetizing component of stator currrent.
2 - Flux associated with rotor currrent
3 - Flux associated with load component of stator current.

The airgap flux pattern 1 by itself induces an emf in the stator winding which equals (balances) the applied terminal voltage. Since the emf from the sum of the three must always balance applied terminal voltage, then we know that under load, #2 and #3 must be equal and opposite to each other.

(side note - Linearity is one assumption required to support the approach of assigning a flux to each individual current component)

Sorry I'm getting a little repetitive. Those are good comments by Curt. If Curt or anyone else wants to chime in further, please do.

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[jsmemphis]

Thank you both for your inputs. They were precious for my understanding!
I will dig a bit more on all this theory.

Cheers from Europe!
jsm