1.Existing P.F of Load (tanQ1)(0.92)
2.Required P.F (Improvement in P.F)-(tanQ2)(0.99)
3.Supply Voltage - for reference(Voltage variation if any)
4.Load in KW (exam- availabl with you i.e 500KW)
KVAR= 500(tan0.92- tan0.99)
KVAR = 500(0.2890)
KVAR = 141.75 KVAR (@ 142KVAR)
if you want to calculate by amp need following data
efficiency of motor, required P.F, KW,maganatising current
of motor,formula find in Electrical Machine Engineering
Reactive power (KVAR) is increase due to run inductive load.as a result power factor is lagging. so have to use capacitive power to reduce this inductive or reactive load. generally we have to use 63% capacitive load to remove the reactive load.i.e.
so 60% of this will be=(60*500/100) = 300 KVAR
so we have to use 300KVAr capacitive capacitor bank to remove this reactive power .
I BOUGHT A HAVELLS ACCL WITH GENERATOR STOP FASCILITY(i.e-
BASICALLY IT IS A AUTO CHANGE OVER SWITCH) FOR MY HONDA EXK
2000S HOME GENERATOR (WHICH START IN PETROL & RUN IN
KEROSENE). ACCL HAS 8 POINT. 2 FOR MAINS(PH & NUE), 2 FOR
LOAD, 2 FOR GEN AND 2 FOR IGNITION POINT(X1 & X2). THE
IGNITION POINT IS SPECIALLY FOR AUTO ENGINE STOP WHEN MAINS
POWER RESTORE. I CONNECT THIS X1 & X2 POINT TO THE
GENERATOR STATER MOTOR & EARTHING POINT. BUT WHEN MAINS
FAILS THE GENERATOR COULD NOT START, BUT WHEN DISCONNECT
THOSE X1 OR X2 POINT, THEN THE GENERATOR START. BUT WHEN
MAINS ON THEN THE GENERATOR SHOULD START, AND WHEN I OFF
THE MAINS THE GENERATOR AUTOMATICALLY STOP.
PLESE HELP ME HOW I STOP THE GENERATOR AUTOMATICALLY WHEN
THE MAINS RESTORE.