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Power Factor Correction


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#1 Mathusalum

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Posted 11 September 2009 - 04:02 AM

Hello;
I'm new to this topic and have a few general questions.
(1) At our plant we have a wide variety of motors ranging from pumps to conveyors and vibrator units. The horsepower range is from 1 to 100 HP. Within this range is about 20 motors. What can be done to have a better power factor?

(2) At the moment, only a few of the motors utilize VFD's, mostly the smaller motors, Does the VFD's provide a better Power factor, or do they benefit in any way our power bill? Its my understanding that by slowly applying speed to the motor the inrush of high starting current is eliminated, is this true?

(3) Is it a favorable option to utilize Soft Starters on high HP motors to also eliminate high starting currents, thus lowering the power bill.

(4) Does Power Factor Capacitors really lower the power bill? Right now we are paying over $15,000 per month to the power company and we are attempting to lower this bill somewhat. Any Suggestions?

Thanks! blink.gif

#2 marke

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Posted 11 September 2009 - 07:25 AM

Hello Mathusalum

Welcome to the forum.

1) If the displacement power factor is less than 0.95, you can improve the power factor by the addition of power factor correction capacitors.
For more information see http://www.LMPhotonics.com/pwrfact.htm

2) VFDs have a displacement power factor of better than 0.95, but a distortion power factor that could be less than 0.8
The operation of a VFD on some machines will enable the speed of the machine to be optimised for best operating efficiency and save energy. The reduction in start current does not actually save money off the power bill.

3) Soft starters will reduce the start current but lengthen the start time. This results in a lower current for a longer time and the same KW input. The soft starter will not reduce your power bil unless there is a penalty for the starting current. - Maximum demand metering integration time is usually much longer than the starttime and the start current will have no influence on the metered maximum demand.

4) Power factor correction, correctly used, will improve the power factor and if there is a penalty for poor power factor, this penalty will be reduced by the additon of power factor correction. If the bill is based on KWHrs only, then there will be no cost reduction.
Power factor correction will reduce the current but not the KW.

Best regards,
Mark.

#3 Mathusalum

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Posted 12 September 2009 - 01:36 AM

Thanks Mark!

I really appreciate your reply.

According to my Boss, the Power Company does charge a penalty for low Power Factor. I'm not sure how to really read the Power Bill, but they do not charge by the straight KW, they reserve that for residential use. Evidently, they have some kind of structured rate for commercial users. Monday I will attempt to get hold of someone at the Power company so they can enlighten me.

It appears that a good power factor not only can save on the power bill, but also increase the efficiancy of the motors. Is this correct?

I guess what I'm trying to do here is justify the cost of the power factor equipment.

Thanks!

#4 marke

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Posted 12 September 2009 - 06:40 AM

There are many different ways that the tariffs are structured so you do need to look at your bill to work out the potential savings.

Adding power factor correction will affect the current drawn from the supply only. The current into the motors will not change, neither will the motor efficiency.
The reduction in current will be between the point where the capacitors are connected and the supply. down stream of the capacitor connection, the current will not change.

Best regards,
Mark.

#5 yuri

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Posted 12 September 2009 - 09:08 AM

With great respect to Mark I would like to offer my view of P.f correction devices and (some) power economy.

1. When the devices are positioned as close as possible to the power consuming units the reduced current will lead to reduced losses in the form of heat between the units and the power meter;

2. The reduced current will reduce voltage level drop in the line so the units will work with a "benevolantly" higher voltage (speaking about fully loaded units).

These, of course, refer only to induction loads of stable character, and, I agree, the power economy would be (very) small.

Best regards.




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