Previous in Forum: New Motor Technology need a Drive   Next in Forum: What is the heat output from HP laser diodes?
Close
Close
Close
8 comments
Associate

Join Date: Jan 2007
Location: South Africa
Posts: 43

Power Factor Correction

02/14/2007 1:01 PM

Hi I'm still confusing because I'd like to calculate the saving money:in fact,my company has a 50KVA power supply,a power factor of 0.86 and after one year,my company needs to use 80KVA.my question is how much should I save if I want to improve my new power factor to 0.86.(NB:the cost per 1KVA is 30$).

kind regards

__________________
Impossible n'est pas Camerounais
Register to Reply
Pathfinder Tags: Power Factor power supply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Anonymous Poster
#1

Re: Power Factor Correction

02/15/2007 12:11 AM

I have also same question to ask. Kindly reply if any answer

Thank you

Kemsindia@gmail.com

Register to Reply
Anonymous Poster
#3
In reply to #1

Re: Power Factor Correction

02/15/2007 2:10 AM

Dear Sir,

We have high speed real time automatic power factor correction system

if your are interested pl. send your requirement on automac.paras@gmail.com

Paras Patel

Register to Reply
Anonymous Poster
#2

Re: Power Factor Correction

02/15/2007 2:07 AM

dear sir,

we have high speed real time automatic power factor correction system,

this system is very smooth & transient free

if you are interested pl. contact in india on 09377542899 or mail your requirement on automac.paras@gmail .com

Paras Patel

Register to Reply
Member

Join Date: Feb 2007
Location: India
Posts: 6
#4

Re: Power Factor Correction

02/15/2007 2:29 AM

POWER FACTOR IMPROVEMENT 80kVA

The present average power factor = 0.86

The present Recorded Maximum Demand = 80KVA

Therefore the average max. load = 80x0.86 =69KW

Recommended PF = 1.00

RMD at improved PF = 69/1.00 =69KVA

Reduction in KVA = 80-69 =11.2KVA

Annual Saving due to reduction in RMD = 11x$30 x 12 =$3,000

Additional KVAr requirement = KW x (Tan f1-Tan f2)

= 69(0.593-0.000)

= 41KVAr

POWER FACTOR IMPROVEMENT 50kVA

The present average power factor = 0.86

The present Recorded Maximum Demand = 50KVA

Therefore the average max. load = 50x0.86 = 43KW

Recommended PF = 1.00

RMD at improved PF = 43/1.00 = 43KVA

Reduction in KVA = 50-43 = 7.0KVA

Annual Saving due to reduction in RMD = 7x$30 x 12 = $2,000

Additional KVAr requirement = KW x (Tan f1-Tan f2)

= 43x(0.593-0.000)

= 26KVAr

Register to Reply
Associate

Join Date: Jan 2007
Location: South Africa
Posts: 43
#5
In reply to #4

Re: Power Factor Correction

02/15/2007 6:37 AM

Happy!! thank you verry much

__________________
Impossible n'est pas Camerounais
Register to Reply
Anonymous Poster
#6
In reply to #4

Re: Power Factor Correction

02/15/2007 7:13 AM

Where do you come up with the angles f1 and f2.

I know they are the angle of the power factor but how does that relate to the power factor of .86?

Register to Reply
Participant

Join Date: Feb 2007
Posts: 1
#7
In reply to #6

Re: Power Factor Correction

02/15/2007 7:17 AM

I assume a power factor of 1.00 has an angle of zero?

a Power facotor of .50 would be ????

Register to Reply
Anonymous Poster
#8
In reply to #7

Re: Power Factor Correction

02/28/2007 8:20 AM

Use your calculator or look up tables to find the Cosine of 0.5!

Register to Reply
Register to Reply 8 comments
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Copy to Clipboard

Users who posted comments:

Anonymous Poster (5); electrical@wonderla.com (1); fchild (1); nganpet (1)

Previous in Forum: New Motor Technology need a Drive   Next in Forum: What is the heat output from HP laser diodes?

Advertisement