Power Factor Correction

This was already asked by you and answered here

Except by the last line, which does not make any sense here.

What is that you want?

1. The APFC - if you go to any site of the supplier, give the data, they will select the correct system from their product catalogue. That is a far better option than telling you the value of C (or KVAr)

2. The calculations: refer any text book. Of course we are not going to do your home work even if the question is repeated again and again

3. If you have some clarification, you should have continued on same thread. There is no point in increasing number of new threads, result will be same, or may be worse. (When one opens a thread, before answering goes through previous posts, and then answers- in a new thread there may always be a wrong answer)

Mogahed,i dont really understand yr question,pls can u put it in a more understanding way(more specific),else i would just give u what i feel u may need.Also lets know yr voltage,kva or mva,kw,and yr present pf on load,is it O.7?

Awaiting yr response

Since no reply from u,kindly use the following to improve yr power factor .know yr kva,kw,frequency f,voltage E,and or yr present pf to be corrected,eg O.7.Use any of the parameters to solve yr problem,using the formulars below.

Kvar=√(kva² -kw²) using power triangle.

Q=E²/X; Kvar=E²/X; X=E²/Kvar =....Ω(capacitive reactance); Xc=1/2Πfc; C=1/2ΠfXc; C will be the value of yr capacitor to be installed in parallel to yr load.

However,yr total kvar now =inductive kvar-capacitive kvar.Your kvar must have been much more reduced.Also,yr kva will reduce to a value almost close to yr kw value,thus increasing yr pf to a higher value.This value may be up to O.95 compared to yr poor value of O.7.If u plot yr power triangle,kva and kw line will almost be close with the kvar which become very small.

If u take the steps properly,and do good calculation,u will get the right value or ratings of capacitor(in kvar and μf) to correct yr poor power factor.

Patrick Whowha