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 .
Recently we had experienced following problem
All RCCB breakers (connected to the load as well as spares)
connected to various UPS systems at various floors at a
high rise building tripped during a power break down and
upon starting up of generator.This Generator is recently
inducted into the installation. Can any one explain the
reason behind this tripping.