2011/05/08 at 5:57 pm #10478ikram shehzadParticipant
on 1000 kw load our power factor is 0.85. how much kvar power is required to improve power factor up to 0.98. please tell me the formula to improve power factor by capacitors. thanks.2011/05/09 at 2:21 pm #12064shahidhamidParticipant
p.f = real power / apparant power
real power = load as u connect it..
p.f = cos @2011/05/09 at 2:24 pm #12065adminKeymaster
You draw it's power triangle, then combine the power triangle at the new desired power factor so that from that power triangle you may found the kvar. the Difference between two triangles is the your required kvar for desired power factor.
At 0.85 power factor kvar=619.74
At 0.98 power factor kvar=203.05
so you required 416.68 kvar to improve your power factor to 0.98 from 0.852011/05/10 at 6:08 pm #12078ikram shehzadParticipant
thank u gys.2011/06/02 at 3:56 pm #12129Spir Georges GHALIParticipant
ikram shehzad said:
on 1000 kw load our power factor is 0.85. how much kvar power is required to improve power factor up to 0.98. please tell me the formula to improve power factor by capacitors. thanks.
Dear Ikram ;
We have a very simple formula to calculate the Q ( kVAR ) that's :
Q = P ( tg φ1 – tg φ2 ) where :
– Q : is the power of Capacitors in ” kVAR “
– P : is the real power load in ” kW “
– tg φ1 : tangent φ1 ( before compensation ) ( calculated from Cos φ1 )
– tg φ2 : tangent φ2 ( after compensation ) ( calculated from Cos φ2 )
So, in your case :
– Cos φ1 = 0.85 → tg φ1 = 0.619
– Cos φ2 = 0.98 → tg φ2 = 0.203
Q = 1000 ( 0.619 – 0.203 ) = 416 kVAR
Noting that you should add at least 10-15% from the calculated Q, because if you have a drop voltage, then the real reactive power of capacitors is less than mentioned on.
Regards2011/06/04 at 6:35 am #12135adminKeymaster
In our plant we have load of 750kw we are using 400kvar capacitor bank ….we are getting 0.91-0.97 pf sum tymes we are getting less pf like o.87….why getting like this>is capacitor bank has to increase ?2011/06/04 at 8:45 am #12138HamdullahParticipant
Disply your circuit diagram of capacitor bank…. then it will be easier to solve and discuse .2011/06/09 at 12:39 pm #12168Spir Georges GHALIParticipant
In our plant we have load of 750kw we are using 400kvar capacitor bank ….we are getting 0.91-0.97 pf sum tymes we are getting less pf like o.87….why getting like this>is capacitor bank has to increase ?
Dear Mr. Rafi ;
Firts, you should do the calculation, as I mentioned before for the same topic, to know if ” 400kVAR ” is enough or not for your installation.
In all case, there some reasons for your case that are :
– Normally, and depending to the connection's type of the controller, the Controller of capacitors is connected to ” one phase or 2 phases ” accordingly the the connection's type, but allways with ” one Current Transformer ” that's installed on the same phase ( for the 1st connection type ) or on the third phase ( for the 2nd connection type ), that means the measuring of the angle ” φ ” is doing between the current of one phase only and the voltage.
– In some time we have on the same controller the value of PF ” 0.91 or 0.96 or 0.87 ” but we adjusted the controller to achieve ” 0.95 ” : as the methode of working of Capacitors bank is doing by a controller where it needs a time ( at least 50sec ) to Switch-On or Switch-Off ” the steps accordingly to the difference between the measuring value and the reference value, in these times it's possible to see on the controller a smaller or bigger value than the reference one.
– If you compare the PF value of the controller with the PF value mentioned on the Bill, there's certainly some difference, because the ” kVARh meter ” works on 3 phase for currents & Voltage.
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