Previous in Forum: Residual Current   Next in Forum: Differential Relay
Close
Close
Close
4 comments
Rate Comments: Nested
Associate

Join Date: Feb 2009
Posts: 33

Power Factor Panel Issue

02/01/2011 4:56 AM

Dear Engineer; Actually i was working on installing a power factor correcting on a factory and i heard an info i just want to confirm it: We all know that power factor correcting panels reduces the Kvar. and you get some discounts on the bill What I heard is that with each 1 Kvar u SAVE, 10 Watta increases in ur consumption. Which means reducing Kvar will increase ur Watta which results a higher bill..but with the discounts u gain ...the bill will be approximately the same... What do u think?

Register to Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.

Good Answers:

These comments received enough positive votes to make them "good answers".
Guru
United Kingdom - Member - New Member

Join Date: Aug 2010
Location: Resting under the Major Oak
Posts: 4347
Good Answers: 181
#1

Re: Power Factor Panel issue

02/01/2011 5:11 AM

What do u think?

Your talking out of your hat

__________________
The reports of my death are greatly exaggerated.
Register to Reply
2
Associate

Join Date: May 2010
Posts: 31
Good Answers: 3
#2

Re: Power Factor Panel issue

02/01/2011 6:52 AM

Yes Capacitor Bank reduces the " KVar " .

(From The Thread http://cr4.globalspec.com/thread/30702/power-factor)

Lets take an example of to understand somewhat the concept of power factor correction.

We have a 500 VA UPS. It runs 2 computers each having Load of 0.9 A and P.F 0.6 at 220 VAC system. When we add a third computer, UPS fails to provide power to these three computers.

UPS Fails Why?

For a Single Computer at P.F = 0.6

Real Power = 220 x 0.9 x 0.6 = 119 W

Apparent Power = 220 x 0.9 = 198 VA

When both computers were on we have

Real Power = 238 W

Apparent Power = 396 VA

When the Third Computers was added then

Real Power = 357 W

Apparent Power = 594 VA

Since Apparent Power of UPS is 500 VA and at this P.F i-e 0.6, 594 VA are required so UPS fails to fulfill the requirement.

Now a Capacitor is added to improve the Power Factor from 0.6 to 0.95 then Current Reduces to 0.5693 A.

Now for a single system

Real Power = 220 x 0.5693 x 0.95 = 119 W

Apparent Power = 220 x 0.5693 = 125.246 VA

Now If three Computers are added to the system then

Real Power = 220 x 0.5693 x 0.95 = 357 W

Apparent Power = 220 x 0.5693 = 375.738 VA

System will run quite smoothly as Requirement reduces from 594 VA to 376 VA.

By improving the P.F

1.We reduce the size of Transformers, cables etc.

2.We are very efficiently utilizing the System Power.

Also see the video

http://www.youtube.com/watch?v=AI_e3dF14eg

Register to Reply Good Answer (Score 2)
Associate

Join Date: Feb 2009
Posts: 33
#3
In reply to #2

Re: Power Factor Panel issue

02/01/2011 7:34 AM

thank you very much for ur reply...actually u made things clearer...but till now i couldn't get an answer. Does the Capacitors we may say..consumes 10 watt for each 10 KVAR they save...which means for the current we save in VA..do we consumes current i n Kw to make these capacitors work maybe?

Register to Reply
Guru

Join Date: Oct 2009
Posts: 2061
Good Answers: 169
#4
In reply to #3

Re: Power Factor Panel issue

02/01/2011 8:24 AM

There is something called "dielectrc watt loss" in each capacitor. It could vary from 0.2W/kVAr to even 2 to 3W/kVAr, depending upon the technology. You have to select a capacitor with a low watt loss. But, that could cost more. Check with capacitor manufacturers' catalogues.

Register to Reply
Register to Reply 4 comments

Good Answers:

These comments received enough positive votes to make them "good answers".
Copy to Clipboard

Users who posted comments:

electricalexpert65 (1); Salar (1); therealabdo (1); TonyS (1)

Previous in Forum: Residual Current   Next in Forum: Differential Relay

Advertisement