Energy Saving with PF Improvement Capacitor

.98 is a very high PF to achieve, not impossible but expensive. Aim for .95.

If the plant loading is constant go for static PF correction, otherwise variable PF correction is available.

A form of variable PF correction would be to fit static capacitors to selected large drives, as the drive starts additional correction is introduced. It's an old way of doing it but tried and tested.

Unfortunately it all comes down to money, you've got to spend to save.

Two types of PF correction are adopted

1) Static correction.

2) Bulk correction.

In first case, PF corrects at the load by connecting capacitor at motor terminal. In this way, we provide reactive power of load at the spot and the reactive current components don't pass in the motor input cable and thus we save I²R losses from motor to transformer.

In second case, PF corrects at main distribution board by connecting a PF controller and capacitors. In this way, reactive current components of load passes through main supply cable from distribution board to load which increases I²R losses as compare with 1^st case.

But if don't correct PF, these I²R losses increases as transformer/generator has to provide both active and reactive current components to load. These reactive current components starts from source to load increase our electricity bill but not 6%. It is a little amount.

May some power expert correct me and you.

As usual we get half a story.

From the fact you had a PF of .92 some correction had to be in place. You made no mention of I²R losses. If this is the loss your concerned with then say so.

Increase the static correction in the distribution network.

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There's times when I wonder why I bother

I am sorry and I agree with Signode.

The losses saving is depending on the current.

With other things remaining constant

I₁PF₁ = I₂PF₂

By improving the pf from 0.92 to 0.98 you are decreasing the current by about 6% (0.92/0.98=0.939)

But then the loss is depending upon the I² and not I and a 6% reduction in I will have about 11% in I².

But then this is applicable for the resistive portion of the circuit between the compensator and the source

Let us assume your source is the utility transformer. at a distance of say 150mt from it you have the compensator put and at 10mt from it is the load.

In this case you must neglect the 10mt and the load since those are least bothered about the compensator.

It is the 150mt that will bear any improvement and that will be negligible since it is only the cabling and will have (or if properly designed should have) very low resistance and will save you a few watts only.

Unless you are working at very high power and currents, I don't really find any significant improvement in 0.92 to 0.98.

Of course for low pfs say 0.8 or even 0.85 it might be worth a try.

This has been discussed many times in this forum. However, again JRaef explained it nicely. GA.

- MS

Quote Aikhh "I was told if we improve PF will also energy saving, example if exisiting PF is 0.92 and we increase the capacitor to 0.98. We are saving of 6% of total Kwh monthly."

No + JR's comments

Dear aikhh ,

My 2 cents (or rupees):

Since pf ≠ kWh, then you can be certain that what you heard is a LIE.

Your equipment will consume EXACTLY the SAME energy at ANY power factor, and the kWh meter will not see a difference either.

The only claimed savings that has any basis in reality is related to heat in the conductors by the formula I²R, where I is the current and R is the resistance. What that means is, that even if you did see a magical 6% reduction in your current, the resistance value planned into most circuits with proper wiring is almost ZERO.

The math then; {(your original amps) - (your adjusted amps)}² X zero = zero (rounded to the nearest 1,000th).

Go ahead, touch a nearby cord or two and see if you can detect any heating. Most cords will in fact feel quite cold when properly sized. If you are not challenging a conductor's capacity, neither you or your meter will notice anything. Only under overloaded conditions for your conductor might you begin to notice any heating.

In many places, ice forming on power lines in the winter pulls them down causing outages etc. There isn't enough I²R heating in properly sized conductors that feed entire neighborhoods to pay for whatever device someone is trying to claim will benefit you from the claimed "6%" of your tiny (by comparison) load.

Again, 6% pf improvement does not equal 6% kWh improvement. Any claim that it does is a flagrant lie.

Regards, CJM.

If you use bulk(automatic) compensation connected to the Switchboard there will be a small saving which depends on the resistance of the main cable and the magnitude of the current consumed from the utility. If you use static capacitors to motors above ,say,5hp the current consumed by each motor connected to a capacitor will be reduced and in some cases smaller cables could be used.The savings will be more but selection of capacitor should be done by a prefessional.