# Overcorrected Induction Motor And Resonance

### #1

Posted 15 September 2006 - 03:09 PM

### #2

Posted 15 September 2006 - 05:43 PM

Welcome to the forum.

I do not know af any model anywhere, but the mechanism you describe certainly is a major problem.

Essentially, when power is removed from a rotating induction motor, it acts as a generator due to the current flowing in the rotor circuit. The rotor has a relatively long time constant and with external influences ignored, the rotor field takes a reasonable period of time to decay. If you are reclosing onto a rotating machine, you need a delay of up to half a second to avoid any resynchronising transients, so that gives an indication of the typical rotor time constant.

If there is external capacitance connected across the stator, then the motor time constant can be extended as the capacitor can then provide the VARs to keep the rotor and stator excited for longer.

If the rotor continues to provide a rotating field in the stator as it pases through resonance, very high voltages will be developed which can cause insulation breakdown in the stator and the capacitor.

Essentially, you have a parallel tuned circuit with a generator in series, where the outptu of the generator has a falling frequency determined by the rate of deceleration of the rotor, and the generator voltage is exponentially decaying at a rate deteremined by the rotor time constant plus the effect of the VARs supplied by the capacitor.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

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### #3

Posted 15 September 2006 - 06:56 PM

I have searched the web as thoroughly as I know how, and I find numerous warnings and claims regarding the problem, but cannot find an analysis anywhere. I hope someone out there can put me on the right track.

### #4

Posted 17 September 2006 - 08:32 PM

Yes I follow your argument in regard to the Q of the windings due to the powerfactor, but this is only true when the motor is being driven by the supply. At that time, the secondary is sliping relative to the primary field and that is placing an impedance in the stator circuit that is dropping the Q.

Once the supply is removed,we no longer have the secondary loading the primary, and so the powerfactor is not relevant in determining the Q. The effective Q is very high as it is only the resistance of the primary that will have an effect. The voltage on the primary is in synch with the rotor as it is the rotor that is generating that voltage. - There is no slip.

The resistance of the stator is very low as is indicated by the full load efficiency. of the motor.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters

### #5

Posted 18 September 2006 - 01:33 AM

Hi tilg

Yes I follow your argument in regard to the Q of the windings due to the powerfactor, but this is only true when the motor is being driven by the supply. At that time, the secondary is sliping relative to the primary field and that is placing an impedance in the stator circuit that is dropping the Q.

Once the supply is removed,we no longer have the secondary loading the primary, and so the powerfactor is not relevant in determining the Q. The effective Q is very high as it is only the resistance of the primary that will have an effect. The voltage on the primary is in synch with the rotor as it is the rotor that is generating that voltage. - There is no slip.

The resistance of the stator is very low as is indicated by the full load efficiency. of the motor.

Best regards,

Thanks, Mark. Suppose the full-load equivalent resistance and reactance of a 0.7 pf motor are about 3ohms each. What might they be when the motor is taken off line and is coasting to a stop? By the way, I was able to model the circuit and get some overshoot even without reducing the equivalent resistance.

Til

### #6

Posted 20 September 2006 - 07:49 PM

Thanks, Mark. Suppose the full-load equivalent resistance and reactance of a 0.7 pf motor are about 3ohms each. What might they be when the motor is taken off line and is coasting to a stop? By the way, I was able to model the circuit and get some overshoot even without reducing the equivalent resistance.

Til

Mark, I am still engaged in this problem. I understand what you say about the effective resistance dropping when the motor is taken off line. The problem is, unless the motor is hugely overcorrected, the resonant frequency is then above the line frequency, because the resonant frequency of a parallel circuit depends upon the loop resistance. The series resonant frequency is higher still, so the frequency of the generated voltage never passes through a resonance. There is overshoot, nonetheless, but probably not as much as would be obtained near a resonance.

I can send you the results of some simulations, if you wish.

Til

### #7

Posted 20 September 2006 - 08:08 PM

I am sorry, I do not follow your argument.

If you correct a motor to unity power factor, the capcitive reqctance must equal the inductive reactance. At this point, the circuit is resonant at the supply frequency.

If you over correct, then the capacitive reactance is lower than the inductive reactance. The point where they are equal is then less than the supply frequency, therefore resonance is less than the supply frequency.

If you have a statically corrected motor that is corrected to unity power factor, then the moment that you disconnect the motor from the supply, the motor acts as a generator in the middle of a resonant circuit where the generator frequency is equal to the resonant frequency.

If the motor is over corrected, the motor will generate voltage as the motor slows and will pass through the resonant frequency.

If the motor is unterminated when disconnected from the supply, the voltage generated will decay faster than if the motor has capacitors connected to it's terminals.

When the motor is connected to the supply, the resonant circuit behaves like a parrallel resonat circuit with a very low shunt resistance due to the shaft load on the motor and the impedance of the supply. In most cases, the effect of critical correction whil under load and connected to the supply, is minimal due to the very low Q.

When the motor is disconnected from the supply, the load is no longer reflected back into the primary circuit and the circuit behaves as a series resonant circuit with a very low series resistance due to the resistance of the stator windings. Because this resistance is very low, the Q is very high.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters

### #8

Posted 20 September 2006 - 08:18 PM

Hi Til

I am sorry, I do not follow your argument.

If you correct a motor to unity power factor, the capcitive reqctance must equal the inductive reactance. At this point, the circuit is resonant at the supply frequency.

If you over correct, then the capacitive reactance is lower than the inductive reactance. The point where they are equal is then less than the supply frequency, therefore resonance is less than the supply frequency.

If you have a statically corrected motor that is corrected to unity power factor, then the moment that you disconnect the motor from the supply, the motor acts as a generator in the middle of a resonant circuit where the generator frequency is equal to the resonant frequency.

If the motor is over corrected, the motor will generate voltage as the motor slows and will pass through the resonant frequency.

If the motor is unterminated when disconnected from the supply, the voltage generated will decay faster than if the motor has capacitors connected to it's terminals.

When the motor is connected to the supply, the resonant circuit behaves like a parrallel resonat circuit with a very low shunt resistance due to the shaft load on the motor and the impedance of the supply. In most cases, the effect of critical correction whil under load and connected to the supply, is minimal due to the very low Q.

When the motor is disconnected from the supply, the load is no longer reflected back into the primary circuit and the circuit behaves as a series resonant circuit with a very low series resistance due to the resistance of the stator windings. Because this resistance is very low, the Q is very high.

Best regards,

### #9

Posted 20 September 2006 - 08:30 PM

It is the VARs that must be equal, not the reactances. Equating the reactances leads to loop resonance (RLC loop) at 60 HZ, but not to unity pf.

The correcting capacitor is connected in parallel with an RL series circuit, and the impedance seen at the terminals of the capacitor must be real (for unity pf) at 60 Hz. That is different from requiring loop resonance, which means the loop impedance must be real. For example, for a 50 hp motor having loaded pf = 0.7, the inductance is approx 8.2 mH. Requiring loop resonance at 60 Hz yields -- from f = 1/ [2pi*sqrt(LC)] -- C = 859 uF. But requiring unity pf leads to C = 438 uF. 859 uF would be a huge overcorrection.

Overcorrecting to say, 518 uF, gives a loop resonance at about 77 Hz and parallel resonance at about 49 Hz.

Til

### #10

Posted 20 September 2006 - 09:07 PM

If the reactances are equal, wont the VARs be equal also??

If the voltage is the same, then the current is the voltage over the reactance and thus provided the reactances are equal, the V X A will be equal.

Best regards,

Skype Contact = markempson | phone +64 274 363 067

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### #11

Posted 20 September 2006 - 09:23 PM

In normal pf correction, the C is in

__parallel__with RL, and the voltage across C is the same as the voltage across RL, but the currents through L and C are different. In that case, you must compute the VARs differently, where V^2*wC = (V/ZRL)^2*wL (for unity pf). This calculation does not lead to series resonance at the line frequency, and usually (again, unless an overcorrection is huge) does not yield series (loop) resonance below the line frequency.

I wish I knew how to do math symbols and present graphics in this space.

Til

### #12

Posted 20 September 2006 - 09:54 PM

Til

#### Attached Files

### #13

Posted 20 September 2006 - 10:29 PM

My understanding of unity power factor is that the displacement power factor is at unity when the capacitive current is equal in magnitude to the inductive current. i.e. the resultant current vector is in phase with the supply voltage.

For this to be the case, the voltage across the magnetising inductance equals the voltage across the capacitor and so both must have equal reactances. - this is a simplified diagram in that it does not include the leakage reactance and winding resistance, but these are very small relative to the load and magnetising impedances.

So we have equal VARs due to equal reactances.

When we include the series copper loss and leakage reactance, there will be a small shift, but very small.

Best regards,

Skype Contact = markempson | phone +64 274 363 067

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### #14

Posted 20 September 2006 - 10:43 PM

I do not follow your calculation of the equivilent inductance.

I calculate the inductance as V x V / (2 x PI x F x Q) and get an inductance of 16mH

This then works!

What have I done wrong??

Best regards,

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters

### #15

Posted 21 September 2006 - 04:19 PM

Hi tilg

I do not follow your calculation of the equivilent inductance.

I calculate the inductance as V x V / (2 x PI x F x Q) and get an inductance of 16mH

This then works!

What have I done wrong??

Best regards,

Hi, Mark. Normally, an induction motor would be modeled as a series RL circuit, not parallel RL as you have done. For the series model, the impedance of the motor at 60 Hz is (very approximately, I forget the exact values) 3+j3 ohms. The 3 ohm reactance of the inductor translates to

L = 3 ohms / 377 = about 8 mH.

I am putting together a pdf file of a simulation, which I will post soon, if it will fit.

Cheers, Til

### #16

Posted 21 September 2006 - 06:04 PM

I pulled the change from 3 ohms to 0.1 ohms out of thin air. Also, the simulation is somewhat inexact because the motor remains excited for a time after being disconnected from the line. Thus the effective resistance doesn't drop immediately to the smaller value.

When I get time, I will do a more elaborate model and simulation.

It's been fun.

Til

#### Attached Files

### #17

Posted 21 September 2006 - 06:43 PM

You can model the induction motor as either series or parallel, most of the equivilent circuits that I have used have been a combination of both.

If you look at the equivilent circuit, the major inductive current is the magnetising current and it is constant and independant of load. - Hence the shunt component. There is some additional inductive current resulting from the load current through leakage reactance and this varies with load.

The problem with using the series circuit is that both the inductive element and the resistive element are not constant. They both must have a slip component in their value. Hence in trying to model the behaviour as a two component network, you must take the slip or load into account. The inductive element will be different open shaft than when the motor is loaded.

In the shunt model, only the resistive element changes with slip.

I believe that the difficulty that you are haveing modelling the resonance is because you are applying the values calculated from a partially loaded motor to an unloaded motor without slip compensation.

If you look at the current components of the motor under varying load conditions, you find that the resistive component has a small fixed value due to the iron loss and friction, but the major part of the resistive current is load dependant and is due to the shaft power and to a much smaller amount, the copper loss.

The inductive components are made up of a large fixed value due to the magnetising current (essentially load independant) and a small variable amount due to the leakage reactance.

Hence, for the pruposes of modelling the motor, it is common to model it as a shunt inductance of fixed value with a shunt resistance that is load dependent.

Bottom line, put a motor on the bench and look at the inductive current with changing load. You will find that the inductive current is almost constant and load independant. You series model will not give that result if the inductive element is constant, the shunt model will.

Best regards,

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters

### #18

Posted 21 September 2006 - 08:12 PM

As I said, I'll do a more complete simulation when I get a chance.

All in all, this is a complicated problem, isn't it?

Thanks, Mark for your patient responses. You have helped me greatly as I have attempted to understand this issue.

Til

#### Attached Files

### #19

Posted 21 September 2006 - 08:44 PM

Yes, I agree that you can look at the circuit as either series or parallel.

If you use the series circuit then you have to consider both the inductance and the resistance as variables dependant on the slip. The reactance is also a function of frequency.

In the case of the paralel circuit, the inductance is constant and the resistance is slip dependant, so the model becomes simpler.

In terms of impedance, you must apply changes in frequency only ot the inductive reactance.

Both ways are correct, with the shunt circuit you have a shunt to series conversion to make it easy to express the result in R + jWL, but in the series circuit you have a formular to express L rather than a value.

I believe that once you either apply the "formula" to L in the series model, or use constant L with the shunt model, you will find that the motor will actually go through resonance when slowing if overcorrected.

Best regards,

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters

### #20

Posted 22 September 2006 - 02:27 PM

Hello tilg

Yes, I agree that you can look at the circuit as either series or parallel.

If you use the series circuit then you have to consider both the inductance and the resistance as variables dependant on the slip. The reactance is also a function of frequency.

In the case of the paralel circuit, the inductance is constant and the resistance is slip dependant, so the model becomes simpler.

In terms of impedance, you must apply changes in frequency only ot the inductive reactance.

Both ways are correct, with the shunt circuit you have a shunt to series conversion to make it easy to express the result in R + jWL, but in the series circuit you have a formular to express L rather than a value.

I believe that once you either apply the "formula" to L in the series model, or use constant L with the shunt model, you will find that the motor will actually go through resonance when slowing if overcorrected.

Best regards,

Mark, I am told that the (transformer) model I am using is valid for only steady-state analyses. If that is true, do you know of a model that can be used for transient analysis?

Thanks

Til

Mark, I am told that the (transformer) model I am using is valid for only steady-state analyses. If that is true, do you know of a model that can be used for transient analysis?

Thanks

Til

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