power factor correction capacitors - Resonant condition ? 3

[edison123]

Has anyone encountered resonant condition where capacitive reactance of the power factor correction capacitor is equal to inductive reactance of the motor ? If so, what were your observations during such resonance ? (excessive current, current oscillation etc.)

 

[Shortstub]

To avoid resonance, the PF-correction capacitive reactance should be sized to offset 80-85% of the inductive reactance of an unloaded motor.

 

[Marke]

Hello edison123

If you apply enough capacitance to fully correct a motor, or to over correct a motor, then that will be a major problem if the capacitors are permanently connected to the motor terminals. If the capacitors are disconnected from the motor when the supply is disconnected from the motor (i.e. vai a second contactor) this should not pose a problem. When the supply is connected to the motor, the impedance of the supply will swamp the resonant circuit formed by the motor and capacitors and drop the Q very low as well as shifting the resonant frequency up. If the motor is operating at full speed, and then disconnected from the supply, it behaves as a generator, generating voltage at the speed that it is rotating. If the capacitors are still connected across the motor terminals, then resonance will occur and the voltage generated will be very high. The only thing limiting the voltage, is the insulation breakdown. As the insulation breaks down, there will be high current surges and resultant torques. The net result of this situation is insulation damage to the motor windings and terminations, damage to the capaciotrs, and often, damage to the mechanical components in the system due to the transients. This damage can be shattered couplings, broken shafts etc. Definitley undesirable!!

For static correction, only apply 80% of the correction.
Best regards,

Mark Empson

http://www.lmphotonics.com

 

[busbar]

The resonant harmonic order of a capacitor set is approximately sqrt(MVA_SC/MVAR).

 

[edison123]

Thanks shortstub, mark and busbar.

In a parallel LC resonant circuit, since both the impedances (capacitive and inductive) are same, then the total impedance is infnity. That means, the current flowing thru' this circuit is zero (ignoring the resistive components of both the motor and the capacitor). So, theoretically, a resonant condition is ideal from the standpoint of improving pf. So, why we should restrict the capacitive compensation to only 80% ?

 

[Marke]

Hello edison123

When the motor is connecte to the supply, there is no problem at all. The problem occurs if the capacitors are permanently connected to the motor terminals and the motor is then disconnected to the supply. At that point, you have a parallel resonant circuit with a generator driving energy into it at the resonant frequency.
Resonant circuits act as amplifiers to any information at their resonant frequency. In this situation, the voltage gnerated by the motor (could be line voltage) will be amplified considerably. The resulting voltage will exceed insulation ratings etc.
Fit a second contactor, one to control the motor, and one to control the capacitors, and the problem goes away.
The 80% rule only applies to situations where the motor is likely to become disconnected from the supply with the capacitors still connected across its terminals.
Best regards,

Mark Empson

http://www.lmphotonics.com

 

[edison123]

Thanks Mark, that cleared some cobweb.

 

[jbartos]

Suggestion: A damping resistor may be added in capacitor circuit to the existing resistances of motor winding and conductors to suppress or control the oscillation waves magnitudes.

 

[edison123]

Another question, after compnesating for 100% pf correction, will the motor pf change from no-load to full load ? Theoritically, it should not since we have fully compensated for the no load lagging current of the motor fully and the balance current is doing only active work. Am I right ?

 

[electricpete]

Motor requires slightly more vars in the loaded than unloaded condition.... due to the current passing through series leakage reactances.

If you completely compensated the vars in the unloaded condition, you would go slightly lagging in the loaded condition.

 

[Shortstub]

Edison123,
Yes, the theoretical-case parameters you have chosen, i.e., L & C but no R, do result in an infinite impedance. But, only at the resonance frequency. No so at line frequency. The two frequencies are only coincident if all R is ignored.

Thus, line current will be zero at resonance, but the L and C branch currents are equal in magnitude, but opposite in phase.

 

[edison123]

thanks shortstub.

While the resonance in a parallel LC circuit is created when both the reactances are the same, the frequency at which the resonance occurs need not be the power line frequency. thanks again for making me use my brains.

 

[electricpete]

I have one more comment on the subject.

If you have corrected to pf=1.0, that means the resonant frequency IS power line frequency.

Now as Mark said if caps are still connected when you cut of the power, the motor continues for a short time to act like a generator initially generating power line frequency. Draw out the circuit and you see that you now have a SERIES lc loop/circuit being excited by that motor voltage. Series impedance approaches zero. That voltage drives the current very high. Voltage accross the terminals can get very high.

 

[edison123]

pete,

in parallel resonant condition, the resonant frequency is 1/[2*PI*sqrt(LC)]. how does this create a resonant frequency of power line frequency ? with 100 mH and 10 microfarad, the resonance will occur at about 159.155 HZ and not at powerline ferquency.

 

[electricpete]

where did 100 mH and 10 microfarad come from? It does not correspond to full correction to 1.0.

Let me use a simplified model ignoring series leakage reactance. Model the motor running at low load as just a parallel combination of:
capacitance (correction cap)
inductance (magnetizing branch)
resistance (rotor/torque branch).

If you correct it to pf=1.0 by adjusting the capacitance, that means you have selected the capacitor which exactly "cancels" the inductance ans satisfies your relation: power frequency is 1/[2*PI*sqrt(LC)]

 

[electricpete]

One more thing, going back to your previous statement:
"While the resonance in a parallel LC circuit is created when both the reactances are the same, the frequency at which the resonance occurs need not be the power line frequency"

Reactance is a function of frequency. If you have a parallel LC circuit where reactances are the same at power frequency, then power frequency is the resonant frequency of that circuit.

 

[Shortstub]

Electricpete,
PF is not 1.0 for the parameters chosen by Edison123.

 

[electricpete]

Shortstub - No kidding. My point exactly. My comments:

"where did 100 mH and 10 microfarad come from? It does not correspond to full correction to 1.0"

What's your point?

 

[Shortstub]

Electricpete,
Are you addressing me? Or Edison123?

 

[edison123]

pete and shortstub,

You are right. I stand corrected.