Answer / joseph

It is better to express the `Load' in kW. In your case it
is assumed that the average load at a P.F. of 0.66 is about
132 kW. Please note that excessive capacitors in circuit
can act as `Load'. Hence, firstly ascertain whether the
Power Factor is 0.66 lag (lesser capacitors than required)
or 0.66 lead (more capacitors than required). The best way
is to switch off all the capacitors for some time and take
the P.F., kW, KVA readings from your meter. Based on this
suitable solution can be provided.

Answer / ashish

Consider your load in kw i.e 200*p.f.=200*.66=132kw
Generaly capacitor bank used for improving power factor
then by using this formula kw & kvar relation formula. it
will help u to decide exact kavar for ur load.

Answer / khalidshaikh

A CONSUMER HAVING TOTAL INSTALLED CAPACITY OR DEMAND UP TO
200 KVA. THE CONSUMER ARRANGED 250 KVAR CAPACTOR BANK, I
THINK CONSUMER IS MISS GUIDED OR HIS CONNECTION POINT IS AT
FOR DISTANCE FROM THE SUPPLY SOURCE, WHERE HE RECEIVE
DROPED VOLTAGE AND HIS LOAD IS TOTALLY INDUCTIVE WHICH
DECREASES POWER FACTOR. OR LEADING POWER FACTOR WHICH
CAUSES ALSO LINE LOSSES.
IT IS SUGGESTED THAT CONSUMER GOT INSTALL KVARH METER AND
KWH METER WITHOUT CAPACITOR BANK IN CIRCUIT THAN AFTER A
CERTAIN OBSERVATION TIME GOT CALCULAT POWER FACOR AND
ARRANGE FOR THE REQUIRED VALUE CAPACITOR BANK.
THANKS

Answer / zahid noor

Connecting a capacitor bank to laod is not sufficient to
control power factor, it is controlled by increasing &
decreasing value of capacitor on increase & decrease of
your load, which needs to be calculated & automatic control
can be given by APFC panels. Please contact a suitable
party .

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