wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
That's a lengthy way of saying that when vector X and Y are 60 degrees apart, and |Y| << |X|, then |X+Y| = |X| + 0.5 |Y|, to good approximation. Since |X+Y| = sqrt( (|X|+0.5|Y|)2 + 0.75|Y|2 ), and |Y|2 << |X|2, so the last term is negligible.I like your idea of putting the MSB single phase loads across just A-C and B-C. These load currents IA-C ≈ IB-C would each be 60° from the 904A on A-C. Because each of these load currents will be substantially less than IA-B = 904A, their component in quadrature (at 90°) to IA-B will end up phase shifting the total line currents on A and B a small amount, but add little to its magnitude. The components of the load currents IA-C and IB-C that are in-phase with IA-B will each be down by a factor of sin(30°) = 1/2. And if these two in-phase current components are equal, they will flow through one load across to the other without impacting the current on C. And so their current component in-phase with IA-B = 904A will be the same as if these two loads were in connected in series across the 240V on A-B, and not connected to high leg phase C. So, for example, if IA-C = IB-C = 100A, then they would contribute a total of 50A to the IA-B current.
If C is the high leg, and the supply is open delta with, say, A-C open, what is best as far as the relative loading of A-C vs B-C? 1:1, or more heavily load B-C, which has the transformer? Assuming all single phase loads that may not be concurrent. [Versus a 3 phase load that will load A-C and B-C equally.]
Cheers, Wayne