Delta & Wye imbalanced loads

[ptonsparky]

Thanks to "Water Heater Circuit" I now want to know:
What is the current on B given that A-B=5kw, B-C=5kw, C-A=2kw (Unbalanced Wye @208 for this one)
..........................".......................................(Unbalanced Delta 240V)

I know this has been touched on before, but now I have time and am curious. References to the threads would help.

 

[kwired]

from post 12 in that other thread:

The currents on the L1-L2 and L2-L3 loads are 60° apart. If one is Ia and the other is Ib, then the resultant current on L2 would be
√ [ Ia2 + Ia x Ib + Ib2 ] . Note that if Ia = Ib, then the current will be the familiar √ 3 x Ia .
And so for your example above, the current on L2 would be √ [ 242 + 24 x 9.6 + 9.62 ] = 29.98A, or 30A to 2 significant figures.

I think you just need to extend the formula to include the c line values to get the results when there is current on all three supply lines. Maybe a little more to it than that? Pretty sure that same formula extended with the c line values would give you the neutral current if this were all line to neutral loads, but not quite certain how to figure unbalanced line to line loads other than possibly you figure out what balanced loading value is and then run the calculation I quoted and and add it's result to the unbalanced line?

 

[ptonsparky]

Extend the formula...I'll try it.
Wye neutral loads, I'm good.

 

[synchro]

Thanks to "Water Heater Circuit" I now want to know:
What is the current on B given that A-B=5kw, B-C=5kw, C-A=2kw (Unbalanced Wye @208 for this one)

Clue: Because the 2kW load across C-A is not connected to B, it does not affect the current on B. This is assuming that the wye source has a zero impedance, and so the voltages on C and A are unchanged by having a C-A load.

 

[ptonsparky]

Extend the formula most certainly did not work out.

 

[kwired]

Clue: Because the 2kW load across C-A is not connected to B, it does not affect the current on B. This is assuming that the wye source has a zero impedance, and so the voltages on C and A are unchanged by having a C-A load.

I guess that was my thinking in what I posted at end of post #2, but this is kind of a simpler way of saying it.