Refresh My Memory

[wareagle]

I have 2 15 kva loads connected to a 3 phase Y generator 220
phase to phase voltage. Load connected A-B and A-C. The load current is 15 kva/(0.22 kva 1.73 = 39 amps. If the load was connected Phase to neutral the neutral current would be
39 x 1.73 = 64 amps. However in this case the load is connected phase to phase with phase A assuming the common position. I
want to know the current in Phase A. Please refresh my memory as to the proper calculation.(Should be simple I admit)
Thanks

 

[waross]

Try this for starters;
15 KVA @ 127 Volts = 118 Amps
15 KVA @ 220 Volts = 68 Amps
A phase 68 Amps x 1.73 = 118 Amps

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[wareagle]

Thanks Bill.

 

[7anoter4]

In principle I agree with waross. But the sum of currents depends a bit on the pf. If pf in both loads are equal then IA=SQRT(3)*68.18= 118.09.But if one pf=0.5 and the other will be 1 then IA=2*68.18=136.36 A.

 

[waross]

Good as far as it goes for line to line loads but line to neutral loads need a little more explanation:
When equal loads are connected from two phases to the neutral, the three currents (A phase, B phase and neutral) are equal.
Consider three equal loads each connected from a phase to neutral.
In the absence of harmonics the neutral current will be zero. If you remove one load, the remaining current will be equal but opposite in polarity.
For this reason the Canadian Code does not allow any reduction in the size of a neutral conductor serving two phases and requires that the neutral be counted as a current carrying conductor for the purpose of de-rating.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Thank you, waross. I don't know the Canadian Code, but your logic seems to me correct for two phases to neutral. I'm afraid I referred only to the first part of the wareagle o.p.