Is Replacing a 25 kVAr Unit with Two 12.5 kVAr Units More Efficient?

AI Thread Summary
Replacing a 25 kVAr unit with two 12.5 kVAr units can enhance efficiency based on site demand data, which indicates that peak requirements rarely exceed 30 kVAr and often only need 12.5 kVAr. The recommendation stems from the analysis of interval data, suggesting that smaller units provide better power factor correction for varying loads. Economic justification for this change largely depends on how the energy supplier bills for power factor; utilities may charge for capacitive power factor, influencing the decision. If the utility does not impose such charges, the cost-effectiveness of the new arrangement may not be justified. Ultimately, the decision should be based on a thorough analysis of energy costs and power factor billing practices.
AbdullahS
Messages
6
Reaction score
1
TL;DR Summary
Advice needed on Power Factor Correction Unit upgrade
Hi all,

The power factor correction unit for one of the buildings that I am working on needs maintenance. The service report recommends replacing the existing 25 kVAr unit with 2 x 12.5 kVAr units as "the demand on your site according to the data supplied shows us that at most there is only 30kVAr required at peak demand and quite often only requires 12.5kVAr".

This has been recommended after the analysis of interval data. The following graph shows the KVAr at the site for a period of 3 months.

1581896853254.png


Could you please advice if replacing the larger bank with two 2 smaller capacitor banks justified? If so, what is the reason? Shouldn't the old one do the job even when the kVAr requirement is low?

A bit of background about myself:
I am a mechanical engineer who is working in sustainability industry. I do have basic knowledge of reactive power but not detailed enough to answer questions such as these.

Thanks in advance.

Regards,
Abdullah
 
Engineering news on Phys.org
Whether the change is a good economic decision (or not) depends primarily on how your energy supplier bills for Power Factor. In the simplest case, Power Factor correction systems add capacitance to 'offset' the inductive character of most loads. A 'too large' correction will leave your power factor 'too capacitive.' 2 smaller systems would allow closer to ideal (net zero) correction for a range of loads. As I understand it, some utilities charge for capacitive PF, and some don't - if yours does bill for capacitive PF, you'll need to compare the 'do nothing' cost to the cost of the new arrangement. If they don't bill for capacitive PF, you probably can't (economically) justify the change.
 
  • Like
Likes berkeman
Thanks @Dullard!
 
Last edited by a moderator:
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I have recently moved into a new (rather ancient) house and had a few trips of my Residual Current breaker. I dug out my old Socket tester which tell me the three pins are correct. But then the Red warning light tells me my socket(s) fail the loop test. I never had this before but my last house had an overhead supply with no Earth from the company. The tester said "get this checked" and the man said the (high but not ridiculous) earth resistance was acceptable. I stuck a new copper earth...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top