Power Factor

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rmonroe

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This question is in regards the Power Factor Correction. I received from our power company a spreadsheet, upon my request, a yearly study of our power usage. This was broken down into monthly usage for 2003. On this spreadsheet they show kWh, Actual demand kW, Actual kVA, and Power Factor. After reviewing this information I seen that the majority of the time we had a low power factor (86 to 89). On the same spreadsheet they also show in Kvar the amount needed to correct the power factor to 90 percent for that peak demand during that billing cycle.

I know were our problem area was in our plant in regard to power factor correction. We have two 250 hp compressor that had no power capacitor on them, which base on our P.F. and to achieve a P.F. of 90 percent plus we installed on each compressor 75 Kvar. We have a PowerLogic meter and how we have a P.F. of 91 to 93 percent.

I would like to know what formula did the power company use to figure out that monthly Kvar correction needed to obtain the 90 percent P.F. I?m trying to do a cost comparison between before capacitor were installed and after.

Thanks

rmonroe
 
Re: Power Factor

You can start with the Pythagorean Theorem. The hypotenuse of a right triangle is the KVA, and one leg is the KW. The power company gave you those two numbers. The other leg, KVAR, can be found from
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(KVA)*2 = (KW)*2 + (KVAR)*2</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">So (KVAR) = the Square Root of [ (KVA)*2 - (KW)*2 ]</font>
<font size="2" face="Verdana, Helvetica, sans-serif">That tells you the total KVAR that you have at present. It also tells you the negative KVAR (i.e., capacitive) needed to bring the power factor to 1.0. Let?s call this number ?the existing KVAR.?

But you want to bring it to 0.9, not to 1.0. The thing that will remain constant is the KW, and the thing you will be changing is the KVA. Let us suppose that you have added the right amount of capacitors, and that the power factor is now 0.9. Now use the following formula (it is one way to define power factor),
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Power factor is equal to KW divided by KVA.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">So the ?new value of KVA? is equal to KW divided by 0.9 (i.e., the ?new power factor?).</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Now go back to the second equation shown above, and calculate what we will call ?the new KVAR.?

The amount of negative (capacitive) KVAR you need to add to get a power factor of 0.9 will be ?the existing KVAR? minus ?the new KVAR.?

One final note: The results of this calculation process do not necessarily give you the appropriate value of capacitors to install. Nor do they mean that installing that amount is the most economical choice. The economics of the situation is a far more complicated calculation. But this is an answer to your specific question.
 
Re: Power Factor

How about giving us your KW, KVA and the utilities demand charge. Most utilities charge around $5.00 per kva demand. It is very likely that you will benefit for adding additional capacitor to bring the PF up to .98 or .99. You can do a cost study to get your ROI. It is cheaper to install a large bank at the main rather that installing them at each motor. The disadvantage is that you do not reduce the amp load on the feeders when the caps are on the main. Put you number up and lets see what the results are.
Bob

[ May 27, 2004, 02:10 PM: Message edited by: bob ]
 
Re: Power Factor

Bob,

Here is some numbers that our power company supplied me for 2003.

June:
428,826kwh
Actual-Demand 915kw
Actual-kva 1,043
P.F. 0.877

July:
516,017kwh
Actual-Demand 991kw
Actual-kva 1,136
P.F. 0.872

Delivery Charges = $3.1956 per/Kw
Generation Charges = $0.04814500
Generation-Demand Charges = 1,000 kw @ $7.6574
Generation-Demand Charges over 1,000 = @$6.0574

I really appreciate your time and the information that both you and Charlie have supplied me. Hope this will help with the comparison.

Thanks,


rmonroe
 
Re: Power Factor

July:
Actual-Demand 991kw
Actual-kva 1136 kva
P.F. 0.87
You have listed a demand charge of $7.65 for
1000 kw. I think that should be $7.65 per kva if you are being bill on a KVA demand. If you are being billed on a kw demand the added caps would not reduce your billing.
Assume you are being billed on a KVA demand at
$7.65 per KVA. Using you July demand of 991 kw and
1136 KVA, improving you PF to 1.0 would reduce your demand to 991 kva or a reduction of 145 kva.
145 kva x $7.65 = $1100 per month estimated reduction in you bill. This is only an estimate.
You should check with your utility to get the actual savings. It appears to be worth the investment.
If you have variable speed drives or other harmonic producing equipment get some assistance before installing the caps. They can cause you problems if this type equipment is present.

[ May 28, 2004, 02:59 PM: Message edited by: bob ]
 
Re: Power Factor

Ditto on that bob, capacitor switching can lead to other problems of a transient nature.
 
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