Automatic Cap Banks

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raiderUM

Member
Location
Ohio
RaiderUM...

I should have asked for Hp or voltage!

Phil

Voltage is 460V
Full load Amps is 465
PF is around .8
I dont know what %eff is?? I'm guessing 92.4?

So HP is around 367?

Three phase HP = I x E x 1.73 x %eff x pf/746 (%eff = efficiency of the motor), (pf = power factor)
 

Phil Corso

Senior Member
RaiderUM...

Using a time honored R-O-T (Rule of Thumb) PF correction should offset about 80% of the no-load kvA of a 2-pole, TEFC, induction motor.

Presuming full-load kVA is about equal to rated Hp, and no-load kVA is, say, 15% of full-load,then the "safest" capacitive reactance, kVAr(c), should not exceed some 60 kVAr!

Thus, the value you detrmined is much too high!

Regards, Phil

PS: what are the approximate bus-voltages used in your facility?
 

raiderUM

Member
Location
Ohio
RaiderUM...

Using a time honored R-O-T (Rule of Thumb) PF correction should offset about 80% of the no-load kvA of a 2-pole, TEFC, induction motor.

Presuming full-load kVA is about equal to rated Hp, and no-load kVA is, say, 15% of full-load,then the "safest" capacitive reactance, kVAr(c), should not exceed some 60 kVAr!

Thus, the value you detrmined is much too high!

Regards, Phil

PS: what are the approximate bus-voltages used in your facility?

Thank you for the correction.

Bus-voltages vary depending on building. We have some 12470V, which is 480V and a lot of 4160V which is 460V. This particular location we are talking about with the chillers is 12470 to 480V

Our electric system is crazy. We have a loop system which is 12470V. It feeds all the newer buildings and the 4160V distribution centers.
 

Phil Corso

Senior Member
RaiderUM...

If you intend to "switch" capacitors at the bus or panel level, then following are additional quantitative detail regarding switch-device and cable ratings:

o Because of the possibility of a "restrike" phenomenon automatically-operated breakers, contactors, and motor-operated switches should have an Ampere-rating appreciably higher than the capacitor's rating!

o Cables must be sized to accomodate the fact that the load-factor of a capacitor is 100%!

Regards, Phil
 

Phil Corso

Senior Member
Boeseker...

I never said, in my example, that I used a "typical" value. In fact, the no-load current is a funtcion of construction, speed (# poles), and voltage.

I do stress that for proper application the "preferred" method is to check with the the motor mfgr!

Regards, Phil
 

bob

Senior Member
Location
Alabama
As I said in my note, 12 kv primary capacitors are easy to install. You can get then in 1000 kvar and smaller if needed. You can leave these on line until you feel it is time to take them off line. You use load-break fused disconnects and a hot stick. This would be much simpler that what you had proposed. Pay back maybe 1 year.
 

raiderUM

Member
Location
Ohio
As I said in my note, 12 kv primary capacitors are easy to install. You can get then in 1000 kvar and smaller if needed. You can leave these on line until you feel it is time to take them off line. You use load-break fused disconnects and a hot stick. This would be much simpler that what you had proposed. Pay back maybe 1 year.

We already have primary caps at the 4160V level though, and at two different locations. That a total of 1200 KVAR on primary cable. Wouldnt it be easiest to figure out why these two banks are currently turned off verse installing new 12.47KV primary caps?
 

bob

Senior Member
Location
Alabama
According to the demand and PF you posted you need a total of 5000 kvar to get the PF to 0.96. If these caps
were off line during billing, then they should be online and that would leave you 3800 kvar additional to add.
 

topgone

Senior Member
I took time back-reading and really believe your problem is about finding the best setup that allows your BAS to respond to whatever load PF you will be experiencing for a wide range seasons! Without knowing how your building automation system is capable of (does it initiate 'raise' or 'lower' signal to your PFC controls, are these power factor devices operate on its own, etc.), it would be best to tabulate each of your building loads and it's PFs and start your analysis from there.

My own experience with a 900 kVAr, 12-step, automatic power factor control on a multi-function rooms convention center gave me headaches as under and over-correction always occurred when the building was first used. I managed to solve the frequent capacitor contactors hunting when I tweaked the amount of kVAr that can be switched-in at different PFs caused by HVAC cycling. It was a 75 kVAr per step but that setup led to over/under correction. We have to lump more caps on one bank than the others while some were made to cut-in less than 75 kVArs.

Hope that helps you.
 

Besoeker

Senior Member
Location
UK
Voltage is 460V
Full load Amps is 465
PF is around .8
I dont know what %eff is?? I'm guessing 92.4?
Since you're correcting the supply power factor, motor efficiency doesn't have a direct bearing on the correction required.
From current and voltage, you get 370 kVA. (sqrt3)*VL*IL/1000)
With the pf of 0.8, you then get 296kW from the supply. (370*0.8)
This gives an uncorrected reactive component in the supply of 222 kVAr (sqrt (370^2-296^2)). Graphically, you might think of it as two sides of a right triangle and solve for the third

If you want a new cosphi 0f 0.95 you need a new kVA of 311kVA. (296/0.95)
From the above method, you get a required kVar of 95kVAr. (sqrt (311^2-296^2))
Required correction is thus (222-92)kVar or 127kVar.

The numbers have been rounded and the actual calculation I do is on a spreadsheet like this:

PFC02j.jpg


But it uses the same basic mathematics.
 

Besoeker

Senior Member
Location
UK
Boeseker...

I never said, in my example, that I used a "typical" value. In fact, the no-load current is a funtcion of construction, speed (# poles), and voltage.
Of course it varies.
I was simply pointing out that your suggested15% was rather atypical of no load kVA.
 

Besoeker

Senior Member
Location
UK
Besoeker,
Your saying that my Kvar should be 116 Kvar then?

140*1.732*480=116390
See post #54.
I don't know if I explained it very well.
As it happens, I was a question I was asked on site today.
I've spent the last three days on site with my guys getting a couple of 3,000HP, 11,000V variable speed fan drives back up and running after a major supply problem had taken out much of the electronics.
Centrifugal fans have a cube law curve for power. Double the speed, eight times the power. (2^3)
So, at half rated speed, the motor ought to be at one eighth of rated power.
In this case, the speed range is from half speed to full speed. Ought to be under 400kW at minimum speed.
One guy, a sharp cookie from the customer's side asked why the stator current was still relatively high at such a relatively light load.
The answer is simply that the power factor is so much worse on lighter loads.
 

Phil Corso

Senior Member
Besoeker...

Tit-For-Tat, huh :happyyes:

How about an 11,400 kW (15,300 Hp) ethelene compressor-drive, having a no-load current of 135A and a full-load current of 750A. NL/FL ratio is 18%. Closer to 15% than 30%.

Regards, Phil
 

Besoeker

Senior Member
Location
UK
Besoeker...

Tit-For-Tat, huh :happyyes:

How about an 11,400 kW (15,300 Hp) ethelene compressor-drive, having a no-load current of 135A and a full-load current of 750A. NL/FL ratio is 18%. Closer to 15% than 30%.

Regards, Phil
Care to supply actual drive motor name plate data?
 
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