kva formula with 150 fixed Cap bank

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raiderUM

Member
Location
Ohio
Hello everyone,

I am trying to figure out a way to see how much the capacitors I have installed are saving each month. The building I will be using as an example stands alone, so it makes it easy to see the saving. My utility does not charge for a leading PF and we only pay for a Energy Demand Charge (KVA) and KWH

This building is our central heat plant, so the KW is very consistent every month. From the history of the bills from the previous 4 years the KVA needed was 170, so this is were I got the number for the 150KVAR fixed bank.

Back in May before I installed 150kVAR fixed bank on building main. Copied directly off the Utility bill used for billing.

KVA = 280.35
KW = 231.3
PF = .8250

Last months Bill.

KVA = 250.5
KW = 250.5
PF = 1

To know exactly how much money the capacitors are saving seems very simple, but I feel like its to simple???

To find the money saved from the capacitors last month I would need to do the following:
(sqrt(250.5^2+150^2))=KVA 292
Then subtract last months KVA, which was KVA with fixed bank turned on 250.5KVA from new calculated KVA 292
292-250.5=41.5 KVA
Multiply this number by the amount the Utility charges per KVA in the Demand Charge.
41.5 x $9.7 = $402 dollars saved that month from Fixed cap bank.

Just wanting some feed back. This seems to easy. The reason I am doing this is to track the savings of the Fixed cap banks we currently have on campus and to show the admin that they are important and working. Thank you
 
ron

ron

Senior Member
Location
New York, 40.7514,-73.9925
Did the loads change? I would have expected the demand kW to have gone down. You may be saving kW demand charges, but it seems that your load went up, so it isn't a fair comaprison.
 
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raiderUM

Member
Location
Ohio
Phil Corso said:
RaiderUM...

Does your facility's load consist of primarily pumps, fans, and compessors?

Regards, Phil Corso
Click to expand...

Yes it does. It is a large Steam Plant with 3 - 100,000 pounds per hour boilers. The load is very consistent throughout the year as far as electrical usage goes. I have every bill for every month going back to 2006.
 
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raiderUM

Member
Location
Ohio
ron said:
Did the loads change? I would have expected the demand kW to have gone down. You may be saving kW demand charges, but it seems that your load went up, so it isn't a fair comaprison.
Click to expand...

Huh? The KW is only dependant on how much the buidling is consuming in electrical usage each month. The kw will only go down during the summer months when there is only one steam boiler operating, but even then the kw does not vary much. The Peak KW for 2012 was 356KW which was in Feb. The lowest KW was in July, which was 246KW and the PF has averaged in the low 80's each year.
 
R

raiderUM

Member
Location
Ohio
R

raiderUM

Member
Location
Ohio
I want to make sure I am doing the math correctly. I do not want to be overlooking something very simple when trying to calculate how much money a fixed 150kVAR bank is saving the university each month when the PF is at unity, and the KVA=250.5, and the KW is 250.5.

Obviously the right triangle is the easiest to understand, so my hypotenuse is 250.5 KVA, my "x" axis is 250.5 KW, and my "y" axis is 1.

To solve for capacitor savings I would say the "y"axis is 150kVAR? Calculate my new Hypotenuse using pythagorean theorem? Then subtract the old hypotenuse from the new. Giving me the difference in KVA that we would have had to pay that month if the Cap bank was not install?

This is all that I would like answered please. Thank you
 
mivey

mivey

Senior Member
raiderUM said:
I want to make sure I am doing the math correctly. I do not want to be overlooking something very simple when trying to calculate how much money a fixed 150kVAR bank is saving the university each month when the PF is at unity, and the KVA=250.5, and the KW is 250.5.
Click to expand...
Your math is correct.

The simple thing that you are missing is that since the POCO is not charging you for leading vars, it hides the fact that your 150 kvar bank value is being capped out before it realizes the full 150 kvar value.

It is very unusual for the kVA and kW to be the same and I suspect you actually ran leading for the corrected month peak kVA. That means some of the kvar was on the non-value side of the fence. If it was a perfect use of the 150 kvar, the un-corrected pf would have been 85.8% and this is different from the historic sample of 82.5%. That means your power factor does move around more than you might have been thinking. Perhaps it was because you looked at the average instead of the peak pf.

Whatever the reason, you only realize the full 150 kvar value some months. If the load is as steady as you are thinking, I would guess that you do get most of the $402. I would need more data to give you a better estimate. However, since you do have the math correct, you could probably do this yourself.
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
RaiderUM...

The increase in April-May kW increase could be explained if the installation of PFC (Power Factor Correction) equipment resulted in an increase in terminal voltage for Fan, Pump, and Compressor motors! Any increase in voltage will increase their rotational speed!

Power output increases as the cube of speed, that is, (Nn/No)^3, where Nn = new speed brought about by voltage increase, and No = original speed!

That increase follows what is called the Affinitry Law. For example, if speed inreased just 1%, then the corresponding power increase is 1.01^3 or 3%! For the May-April kW numbers you cited the increase is about 8%. Thus, the increase could be accounted for if voltage increased just 2.5%!

However, if flow-rate is controlled, then the Affinity Law is not applicable.

Regards, Phil Corso
 
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R

raiderUM

Member
Location
Ohio
mivey said:
Your math is correct.

The simple thing that you are missing is that since the POCO is not charging you for leading vars, it hides the fact that your 150 kvar bank value is being capped out before it realizes the full 150 kvar value.

It is very unusual for the kVA and kW to be the same and I suspect you actually ran leading for the corrected month peak kVA. That means some of the kvar was on the non-value side of the fence. If it was a perfect use of the 150 kvar, the un-corrected pf would have been 85.8% and this is different from the historic sample of 82.5%. That means your power factor does move around more than you might have been thinking. Perhaps it was because you looked at the average instead of the peak pf.

Whatever the reason, you only realize the full 150 kvar value some months. If the load is as steady as you are thinking, I would guess that you do get most of the $402. I would need more data to give you a better estimate. However, since you do have the math correct, you could probably do this yourself.
Click to expand...


Thank you mivey!

The PF does move around but its always in the low .80 range. For last year the peak pf was .8461 and the low was .7991. The average kVAR needed the last 4 years was 164 and the average this past year was 180kVAR from my calculations. Knowing we do not have to worry about leading pf except for higher voltages, which we monitor, I decided to install the 150kVAR bank. I have been working with the Utilities on this also, they actually did the harmonic testing at this building for me and found that it is below the 15%, so no filters were needed on the caps.
 
Besoeker

Besoeker

Senior Member
Location
UK
Phil Corso said:
RaiderUM...

The increase in April-May kW increase could be explained if the installation of PFC (Power Factor Correction) equipment resulted in an increase in terminal voltage for Fan, Pump, and Compressor motors! Any increase in voltage will increase their rotational speed!

Power output increases as the cube of speed, that is, (Nn/No)^3, where Nn = new speed brought about by voltage increase, and No = original speed!
Click to expand...
Most of the fans pumps, and compressors are likely to be driven by cage induction motors. Essentially these are fixed speed machines. A small change in terminal voltage, and it would be small if it is within supply voltage tolerances, is not likely to change the speed appreciably. Below are the curves for a motor that was/is used to drive a centrifugal compressor:

TorqueandCurrentvs.jpg


Various curves are shown including speed torque at rated voltage and 80% voltage which is much lower that you would ordinarily encounter.
 
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