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Measuring Riddle No.8 - Power factor correction
let suppose we have one Main distribution board with incoming breaker TP 40A , having two motors 10 kw Each as a load...
according to motors we have two TP 20A breakers on outgoing of the Distribution board and one breaker of 20A TP for capacitor bank of (10KVAR) so we have total three outgoing breakers..
let suppose, i switch off the outgoing breakers for the motors... now only main breaker and one outgoing breaker for capacitor bank is switch on..
please tell me what will be the power factor on the power factor regulator at no load????
Author : Asad Ali - From: Pakistan
 
#1
Tue, June 19th, 2012 - 19:03
Power factor regulator should be fed from incoming current transformer generally. Therefore the regulator will detect the load with lead power factor in mentioned case and will send an open command to capacitor bank circuit breaker related to regulator setting. 
Author : Hamid - From: Iran
 
#2
Fri, June 29th, 2012 - 14:26
can you please tell me how it will go towards leading side if we have no loading condition means

our load current is zero so KW=0
power factor=kw/KVA
 
#3
Sat, June 30th, 2012 - 08:42
For inductive or capacitive loading we can write:

KW =0

Cosφ=0

φ=90° leading in capaciotive loading condition
φ=90° lagging in inductive loading condition

Generally there are some limitation for capacitive loading of a power distribution panel. For more information you can refer to General Electrical Riddle No.31 
Author : Hamid - From: Iran
 
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