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 .
what is the difference between earthed system and unearthed
system.. where do we use unearthed system because in
electrical catalogues always cable for earthed and unearthed
system is given..pls elaborate
If one d.g. is unable to take full load of system, then
second d.g. will take care of partly Load.for this
condition RPM of both DG set should be equal.By using
Synchronization panel RPM of dg sets are match, is it true?