mmmm feeling a bit thick today!! is the above formula gonna give me a wattage of
38277.2 or 38.3Kw? i used .85 as a pf but dont know if this is right, if someone could break down the above and give me amps per phase or total available amps and watts likewise that would be great!
well 45KVA is your apparant power, and is simply a product of supply volts by supply current
your watts (or KW) is your true power, and is a product of the supply current and v dropped accross the resistive element of the load
KVAR is your reactive power, and is current times v drop accroos reactive element of the load
of course, all this is theortical, using vectors etc etc
but what it means is your KW is smaller than your KVA
in fact true power / apparant power is p.f
so your p.f of o.85 x 45kVA = 38.25 KW
but of course, your loads may (or may not) be at 0.85 p.f
you should have your pf as high as possible, (close to unity), which means you are 'using' more of the KVA supplied
I have been in large installations running many inductive loads (motors etc) where they p.f correction capacitor banks fiited
anyway, lets say you have 38.25Kw
divide by three = 12.75Kw per phase divided by 230V (because its single phase) = 55.4 amps per phase
note this a very rough calc, and assumes perefctly balanced phases and loads, a fixed p.f of 0.85 etc etc
of course, your PF is not fixed
so using your 45KVA, divide by three = 15KVA per phase divide by 230V = 65A
this is 'available supply'
reality is, without a constantly chjanging p.f, so would your answer
oh, and i just made all that up because i am not an 'electrican' according to some on here
