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Capacitor Bank

09/14/2009 7:17 AM

we have consumers(induction motors-charger...) in oil & gas area with following spec.

p=430kw I=768A V=400v

we added 150 kvar capacitor bank for improve the PF but sometimes it reached to PF=0.77

Is it insufficient?how do we calculate the necessary kvar?

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#1

Re: Capacitor Bank

09/14/2009 10:39 AM

Due the figures you are showing include the capacitor? If not my calculations show that you should have a pf of about 0.91.

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#2
In reply to #1

Re: Capacitor Bank

09/14/2009 11:09 AM

The data you provided is

P=430 KW

I=768 A

V=400 V

From the data,

P.F = (430 x 1000) / (1.732 x 400 x 768) = 0.81

KVA = 430 / 0.81 = 531 KVA

Reactive power drawn by the motor = sqrt ( 5312 – 4302) = 311 KVAR (inductive)

To compensate this inductive KVAR (to make the P.F close to unity), you need almost 300 KVAR capacitor bank.

With your 150 KVAR capacitor bank, the effective reactive power is 311 – 150 = 161 KVAR

Effective KVA = sqrt (1612 + 4302) = 459 KVA

P.F = KW / KVA = 430 / 459 = 0.94.

You mentioned it is 0.77 and this value is strange to me. With 150 KVAR, the P.F should be close to 0.94.

-MS

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#3

Re: Capacitor Bank

09/14/2009 12:19 PM

Masamad

Thats the same figure I calculated. Thats why I asked if the data he posted was with the capacitors. The size he listed seems to be the correct size.

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#4
In reply to #3

Re: Capacitor Bank

09/14/2009 1:54 PM

Wareagle,

You and me are in same page answering this question. I provided the detail calculation to make the OP feel better to accept the answer (as he asked for the calculation). The P.F 0.94 is pretty good improvement, and improving more than this is not economical (as we can see, improve from 0.81 to 0.94 requires 150 KVAR capacitor bank, while to improve from 0.94 to unity requires another 150 KVAR). So, 150 KVAR bank is correct size in this consideration (if the target P.F 0.94 is ok). However, if some one is interested to improve it up to unity, they need to use 300 KVAR capacitor bank in total.

I am also curious about the PF 0.77 is with or without the capacitor bank.

- MS

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#6
In reply to #4

Re: Capacitor Bank

09/15/2009 3:17 PM

If they are using a fix capacitor bank and the load is reduced, the PF might be 0.77 leading...

This is a common problem when people just slap a capacitor bank on a feeder without having a controller to take it off as the load decreases.

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#5

Re: Capacitor Bank

09/15/2009 7:54 AM

"but sometimes it reached to PF=0.77" ?

Is this a static load or is the load varying (possibly a motor control center)? What is the range of PF's that you are logging? What is the range of power consumption in kW? As noted by the other gentlemen, based on the a PF of .77, assuming the same VA, would result in p=409kW. That's almost a 5% difference in kW.

What were your initial condtitions ___V ___A ___PF & what were your conditions after adding power factor correction ___V ___A ___PF ?

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