Understanding Power Factor
As with any equipment, an electrical system handles its job to some degree of efficiency ranging from poor to excellent. The measure of electrical efficiency is known as Power Factor. The motors and other inductive equipment in a plant require two kinds of electric power.
One type is working power, measured by the kilowatt (kW). This is what actually powers the equipment and performs useful work. Secondly, inductive equipment needs magnetizing power to produce the flux necessary for the operation of inductive devices. The unit of measurement of magnetizing or reactive power is the kilovar (kVAR).
Most AC power systems require both kW (kilowatts) and kVAR (kilovars). Capacitors installed near the loads in a plant are the most economical and efficient way of supplying these kilovars. Low voltage capacitors are traditionally a high reliability maintenance-free device.
On the spot delivery of magnetizing current provided by capacitors means that kilovars do not have to be sent all the way from the utility generator to you. This relieves both you and your utility of the cost of carrying this extra kilovar load. The utility charges you for this reactive power in the form of a direct, or indirect power factor penalty charge.
In addition, you’ll gain system capacity, improve voltage and reduce your power losses.
|Title:||How Power Factor Corection Works – GE|
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I have a motor 3hp. My average power factor is.63 to .65, but suddenly its shows .04 is it possible or not.
Or my meter is correct or not
Thank you and the EE team, I have gotten really useful guidance and information.
It’s fixed, now you can download.
Could you please fix the link to this book?
The link for download was removed in the box site. Hope to get the update about this issue. Thank you.
The download link does not work. The file needs to be reuploaded or link needs to be updated
Excuse me ! Is there any diference between the theorical calculation and mesured result? About power factor and reactive energie?
sir, i have a doubt,is anycalculated time to reach the desired power factor for example a load of 630 kw and capacitor bank is 575 kvar present p.f is 0.13 required p.f is 0.98 at what time the p.f will reach.
So Sorry I forgot to thank you for the interesting information
I think you can provide us with new books of electronics and electricity
very good info…