Inductive Load Over-Correction: Explained Mathematically

[QwertyXP]

I am looking for an explanation of why over correction of inductive loads can result in large circulating currents. Many sites give some form of vague verbal description but I have not been able to comprehend this phenomenon.

Can anyone explain it mathematically or provide a link that describes this concept in a mathematical manner?

 

[jim hardy]

Hmmm. Intuitively,
If you drive power factor to unity, have you not made X_C = X_L, which is a resonant circuit?

Here's as good an explanation as i could find with a cursory search...

http://www.lmphotonics.com/pwrfact.htm

As a large proportion of the inductive or lagging current on the supply is due to the magnetizing current of induction motors, it is easy to correct each individual motor by connecting the correction capacitors to the motor starters. With static correction, it is important that the capacitive current is less than the inductive magnetizing current of the induction motor. In many installations employing static power factor correction, the correction capacitors are connected directly in parallel with the motor windings. When the motor is Off Line, the capacitors are also Off Line. When the motor is connected to the supply, the capacitors are also connected providing correction at all times that the motor is connected to the supply. This removes the requirement for any expensive power factor monitoring and control equipment. In this situation, the capacitors remain connected to the motor terminals as the motor slows down. An induction motor, while connected to the supply, is driven by a rotating magnetic field in the stator which induces current into the rotor. When the motor is disconnected from the supply, there is for a period of time, a magnetic field associated with the rotor. As the motor decelerates, it generates voltage out its terminals at a frequency which is related to it's speed. The capacitors connected across the motor terminals, form a resonant circuit with the motor inductance. If the motor is critically corrected, (corrected to a power factor of 1.0) the inductive reactance equals the capacitive reactance at the line frequency and therefore the resonant frequency is equal to the line frequency. If the motor is over corrected, the resonant frequency will be below the line frequency. If the frequency of the voltage generated by the decelerating motor passes through the resonant frequency of the corrected motor, there will be high currents and voltages around the motor/capacitor circuit. This can result in severe damage to the capacitors and motor. It is imperative that motors are never over corrected or critically corrected when static correction is employed.
Static power factor correction should provide capacitive current equal to 80% of the magnetizing current, which is essentially the open shaft current of the motor.

The magnetizing current for induction motors can vary considerably. Typically, magnetizing currents for large two pole machines can be as low as 20% of the rated current of the motor while smaller low speed motors can have a magnetizing current as high as 60% of the rated full load current of the motor. It is not practical to use a "Standard table" for the correction of induction motors giving optimum correction on all motors. Tables result in under correction on most motors but can result in over correction in some cases. Where the open shaft current can not be measured, and the magnetizing current is not quoted, an approximate level for the maximum correction that can be applied can be calculated from the half load characteristics of the motor. It is dangerous to base correction on the full load characteristics of the motor as in some cases, motors can exhibit a high leakage reactance and correction to 0.95 at full load will result in over correction under no load, or disconnected conditions.

Any help ?

Sounds like "over-correction" means using a capacitor that draws close to a motor's magnetizing current, which is roughly the same as the motor's no load current . That's to stay away from resonance.

When disconnected from the source, the induction motor is an induction generator until its field dies away. If its inductance resonates with the capacitor recall that current and voltage depend on Q of the circuit, which is X/R and can be quite high.

Is that on the right track?

old jim

 

[QwertyXP]

I don't remember much about Q.

Yes, if power factor is unity and the load is purely inductive, we have a resonant circuit. I had the understanding that at resonance, capacitor provides only as much reactive power as is required by the inductor. But now, looking at it from a different perspective, net impedance of the resonant circuit is zero so there should be an infinite current circulating in the circuit (I=V/Z)! Did I go wrong somewhere?

If, with a purely inductive load, no problem occurs in a normal situation at pf=1, why should any problem occur when the motor has been switched off and resonates at a lower frequency because of overcorrection?

 

[Baluncore]

The difficulty here comes from the misleading term “circulating current”. Unfortunately, a mathematical explanation is more difficult than a literal description.

An induction motor can be operated as a generator by driving it faster than synchronous speed. When generating power to the grid, the grid will provide the synchronous excitation needed. Without a grid connection, capacitors can be used to self excite the motor as a generator. The frequency of the power generated when using capacitive “self excitation” will be determined by the resonant frequency of the motor inductance with the parallel capacitance. In order to change from motor to generator, the motor must be driven faster than the synchronous speed of the excitation LC resonance.

A reactive current has a phase of 90° to the supply voltage. The sign of that current is unimportant, it is still a reactive current. The inductive current “circulating” between the power station and the motor can be minimised by using parallel compensation capacitors to “neutralise” the inductive current. Zero circulating reactive current on the supply lines can be achieved when the inductive and capacitive currents are equal and opposite. That occurs when the LC circuit is resonant at the supply frequency.

When a compensated induction motor is connected to the supply, there are two separate reactive currents flowing. Those two currents are determined by the supply voltage divided by the inductive reactance of the motor, and the supply voltage divided by the capacitive reactance of the compensation capacitors. Those currents tend to cancel on the supply lines.

Overcompensation of an inductive current results in a capacitive current, which is not a big problem while the supply is connected, as the overcompensation of one machine will cancel part of the undercompensation on another. By overcompensating, the capacitance will be greater than required, and so the resonant frequency of the motor inductance with the compensation capacitance will therefore be lower than the supply frequency.

When the supply is disconnected however, there will be a short period of self generation while the motor decelerates from supply synchronous to below the LC synchronous speed. That deceleration will generate power at the resonant frequency of the motor inductance with the capacitive compensation. The self generating resonance can have a very high Q which may result in very high current in, and voltage across, the LC tuned circuit. That is not a reactive “circulating current” between the power station and the motor. It is a “circulating current” in the unloaded LC resonant circuit.

The mechanical energy available while the motor and mechanical load decelerate from mains synchronous to LC synchronous speed will be dumped into the unloaded LC circuit beginning at the moment of disconnection. The voltage and current can be very high which can damage both the motor insulation and the compensation capacitors.

That is the reason why overcompensation of an induction motor must be avoided.

 

[jim hardy]

I don't remember much about Q.

Yes, if power factor is unity and the load is purely inductive, we have a resonant circuit. I had the understanding that at resonance, capacitor provides only as much reactive power as is required by the inductor. But now, looking at it from a different perspective, net impedance of the resonant circuit is zero so there should be an infinite current circulating in the circuit (I=V/Z)! Did I go wrong somewhere?

If, with a purely inductive load, no problem occurs in a normal situation at pf=1, why should any problem occur when the motor has been switched off and resonates at a lower frequency because of overcorrection?

No, you're doing just fine.

Let's try a thought experiment to clarify the concepts, and see if somebody whose math is fresher than mine will elucidate further.

Initial condition:
Motor running with zero mechanical load so that its drawing basically just its magnetizing current,
meaning the motor looks like an inductor with not much resistance .
Capacitor in parallel with motor sized so pf= 1.0, ie X_c = X_L.
Motor's counter EMF is in parallel also, and very nearly equal to supply voltage.The supply lines set the voltage across all parallel elements.

--------------------------------------------------------------

Now remove the supply line. Voltage is no longer restrained, is it?
Energy is available from motor's rotating inertia, and as it's slowing down it acts as an induction generator.
And since the capacitor provides the magnetizing current, the voltage can rise to whatever is the motor's counter-emf capability.

I think that's why they want the capacitor sized slightly too small to provide full magnetizing current. That assures voltage decays rather than builds up.And i think "resonance" might be a red-herring here.---------------------------------------------------------------------------

In AC circuits we're taught that Q = X/R, also that it's the ratio of energy stored to energy dissipated per cycle.
It's tempting to think, as i did at first, that the circulating current goes high simply because of a high-Q resonant circuit. But upon working this in my alleged mind i changed my thoughts----
If the "induction generator" cannot find enough exciting current to maintain counter-emf, voltage will decay to zero from lack of excitation.
That's why the notes cited earlier have such strong precautions against overcorrecting either an induction generator or a motor that has an over-running load. If the machine is spun by its load, it becomes an induction generator, and if it can find enough excitation, you'll have a positive feedback mechanism: more voltage = more exciting current = more voltage yet... just like any other self-excited generator

So right now I'm thinking it's more a matter of feedback math than of resonance math, and the point at which you reach instability from feedback theory just happens to be the same point at which it's electrically resonant. [insert red herring icon]

I may experiment with a pair of washing machine motors and some run capacitors when get back home...

Meantime, perhaps a genuine 'power systems' PF member could correct me.

old jim

 

[jim hardy]

@ Baluncore:

That deceleration will generate power at the resonant frequency of the motor inductance with the capacitive compensation. The self generating resonance can have a very high Q which may result in very high current in, and voltage across, the LC tuned circuit. That is not a reactive “circulating current” between the power station and the motor. It is a “circulating current” in the unloaded LC resonant circuit.

I understand the statement, but I'm having a difficult time with the "very high voltage" as one would expect in a linear circuit.
Will the motor not saturate, taking us away from sinewaves ?

Is that the source of the high voltage and current, huge peaks as motor switches between peak values of flux as opposed to ordinary current and voltage gain of a linear resonant circuit ??

Or is it more of a feedback phenomenon as i proposed ?

Thanks for your insight -

old jim

 

[Baluncore]

I probably know less about this than you. I would like to do the disconnection experiment with an overcompensated three phase motor.

Now, thinking about your experiment...

Let's try a thought experiment to clarify the concepts, and see if somebody whose math is fresher than mine will elucidate further.
Initial condition:
Motor running with zero mechanical load so that its drawing basically just its magnetizing current,
meaning the motor looks like an inductor with not much resistance .
Capacitor in parallel with motor sized so pf= 1.0, ie Xc = XL.
Motor's counter EMF is in parallel also, and very nearly equal to supply voltage.
The supply lines set the voltage across all parallel elements.
--------------------------------------------------------------
Now remove the supply line. Voltage is no longer restrained, is it?
Energy is available from motor's rotating inertia, and as it's slowing down it acts as an induction generator.

Firstly, nit picking, Xc = XL is not possible. I believe you mean Xc + XL = 0
Secondly, with a pf = 1, Xc = – XL , if the supply is removed there will be no generation. When an induction generator is at or below synchronous speed, generation is not possible. It effectively short circuits it's self excitation.

Induction generator current does not begin to rise until a speed greater than synchronous is reached. The generator speed is then limited by the available mechanical power input. However, when the supply is removed from an overcompensated induction motor, it immediately switches to become an induction generator with a very significant over-synchronous speed. I think the problem arises from that sudden transition to an extreme generation mode that is not normally attainable with an induction generator.

Conceptually, the available rotational energy is dumped into the resistive component of the LCR circuit. There is also circulating energy in the LC component. I suspect that the waveform will be distorted and that saturation will result in a loss of voltage control and so high voltages.

Without a better understanding of the behaviour of over-driven synchronous generators I find it difficult to predict exactly what will happen.

 

[jim hardy]

Well you're not nit-picking , just pointing out my sloppy wording and inattention to details like signs. Thank you.
And i think you're ahead of me on this subject.

However, when the supply is removed from an overcompensated induction motor, it immediately switches to become an induction generator with a very significant over-synchronous speed.

That's logical, for the oversize capacitor would seem to place resonant frequency below line frequency.

I suspect that the waveform will be distorted and that saturation will result in a loss of voltage control and so high voltages.

We're in agreement there .

This is an interesting phenomenon and i haven't yet wrapped my pea-brain around it.
From http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/content/ct_168468.pdf

The most significant issue with motor applied capacitors is
“self-excitation.” This occurs when a motor has enough inertia
to keep spinning once disconnected from the power system
and the capacitor is large enough to supply the reactive power
needs of the motor. Self-excitation can cause damaging
overvoltages. It can also cause damage to the motor shaft if
the motor is reconnected to the system while still rotating due
to self-excitation. This issue is well documented in many
technical papers. Due to self-excitation concerns, all motor
applied capacitors are smaller than the total required kvar of
the motor.

Clearly i have some reading to do on "self excited induction generator"

Without a better understanding of the behaviour of over-driven synchronous generators I find it difficult to predict exactly what will happen.

Did you meany asynchronous generators?

A synchronous generator that's heavily overexcited offline makes a highly distorted waveform and the induction heating melts holes in the stator iron.
Picture courtesy these guys: http://www.exponent.com/Generator-Core-Failure/

We wrecked a big one ca 1981.

I don't know yet whether a self excited induction generator can do the same.

Cheers, and thanks !

old jim

 

[jim hardy]

Did quite a bit of searching on this subject, and information is sparse.

In principle one could make an induction machine be self exciting by attaching capacitors that would "overcompensate" a motor. I actually found a paper where students did exactly that with some small machines built for early wind turbines. The machine walks up its saturation curve just like any other self excited machine, and saturation is what stops the increase making the machine stable..

If i can find that link again, will post it.

A manufacturer of induction generators has this on his site:
http://www.usmotors.com/TechDocs/ProFacts/Induction-Generator.aspx

There are some definite differences in Induction Generator use that should be accounted for in the application:

The indiscriminate use of Induction Motors as generators should be avoided. It is possible that a particular motor would not work well as a generator due to internal magnetic saturation. The internal voltage as a generator can be higher than it would be as a motor with the same terminal voltage. The magnetic densities in the machine are determined by the voltage at the equivalent circuit air gap. For a motor, the air gap voltage is typically 85-95 percent of the terminal voltage. For a generator, the air gap voltage is typically 100-105 percent of the terminal voltage. This higher air gap voltage may make the machine over-saturate magnetically, and have high core losses and draw high magnetizing currents. It is conceivable that the machine could overheat at very low load output. If an Induction Motor is to be used as a generator then this information should be known by the designer so he can make appropriate allowances in the magnetic densities.
Induction Motors are usually rated 460 volts for use on a 480 volt system. Induction Generators should be rated for the nominal system voltage, or slightly higher rather than lower, because the generator is now the power source rather than being a load on the power system.
Power factor correction capacitors can be used to correct the generator power factor in the same manner as for an Induction Motor. However, if there is any chance that the generator may over-speed, whether connected to the power system or not, the capacitors should be connected to the system through a separate breaker so that when the generator breaker is opened, the capacitors will not be connected to the generator. Under overspeed conditions, the capacitors can overexcite the generator and cause uncontrolled high voltages to occur. These voltages can destroy generator insulation systems and also can be hazardous to other equipment and to personnel.

So what I'm taking away is that one can design a stand alone induction machine that's self excited through external capacitors
but garden variety motors are not built for that service, they need somewhat different magnetics.

old jim