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Power Factor Optimization


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#1 brucecranene

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Posted 09 November 2003 - 04:40 AM

Power Factor Optimization can be done on line, at the inductive load side contactor. Using a extech power factor meter, one can measure the amps on the supply side, introduce capacitance on the load side. This continues until enough capacitance is added to start raising the amps after "bottoming out". The previous capacitance is then used to permanently optimize the power factor. an example initial 3ph,206.6v,45.9a,85pf,8.62kvar,16.36kva,14.01kw.
final 39.6a,99.2pf,1.61kvar,14.16kva,14.08kw.

#2 marke

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Posted 09 November 2003 - 06:39 AM

Hello brucecranene

Yes, this is known as static correction.
I would be very wary of doing this in the case of a motor load as you could be very close to critical correction. It is very important with static correction on motors, where the motor and capacitors are controlled by the same contactor, that you only correct to 80% of the magnetising current. You could use your method described to find 100% of the correction required, and then apply 80% in the final installation. If you are making these tests with an induction motor as the inductive load, I would strongly recommend using a separate contactor to control the capacitors until the 80% value has been determined.

Best regards,

#3 brucecranene

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Posted 25 November 2003 - 08:35 AM

You can go to my new web site (wyesenergy.peachhost.com) I'm just getting it up. Reference: compensators equals capacitance. Why do you say to load at 80%. When the contactor shuts the power off the capacitor discharges, the motor is constantly slowing down so current is less and less. I have not seen any adverse effects of many field installations. What type of problems are you referring to that is present in the field?:D;p;

#4 marke

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Posted 25 November 2003 - 09:23 AM

Hello Brucecranene


QUOTE

Why do you say to load at 80%.

That is a good question and confuses many people.
This only applies to static correction of induction motors.
When the motor becomes disconnected from the supply, it acts as a generator, generating voltage while there is a field in the rotor. Imediately after switch OFF, this is significant and the generated voltage will be close to line voltage. The rotor current falls,and the generated voltage also falls. The rate of decay is dependant on a number of factors, but can be measured over several seconds in some cases. As the motor slow, the frequency of the voltage generated by the motor also falls. (frequency is directly proportional to the rotational speed of the rotor.)

If you apply 100% correction, that means that the inductive reactive current and the capacitive reactive current are equal. This means that the magnitude of the capacitive reactance is equal to the magnitude of the inductive reactance. At this point, we have a resonant circuit equal to the line frequency. If we now disconnect the motor from the supply (and leave the capacitors connected to the motor) we will have a generator operating into a tuned circuit at resonance. This will amplify the voltage and result in a very high voltage being developed across the motor terminals. This can cause an insulation breakdown in the motor, and in the capacitors. - it is a very dangerous situation.
If the motor is over corrected, then as the motor slows down, it will pass through resonance and damage will occur.
While the motor is connected to the supply, 100% correction is not an issue as the supply dampens and detunes the resonance.
In order to avoid resonance when using static correction, only apply cpacitive current equal to or less than 80% of the inductive current. (the resonant frequency is above line frequency and the motor will not go there)

Does this answer your question??

Best regards,




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