Now just connect a 240V water heater across B-C, and model the open wye primary side.
So I think it is interesting to determine the coil currents (in magnitude and phase) from the various ways of linearly loading a high leg open delta. I'm going to express the phasors (sinewaves of a fixed period) in polar form X < Y, where X is the magnitude and Y is the phase angle in degrees (i.e. the sinewave f(x) = X * sin(2*pi*(t/T + Y/360)), where T is the fixed period, t is time).
For the high leg open delta as pictured in post #40 (with the undepicted neutral point N as the midpoint of A and C), and V
XY meaning the voltage from X to Y, i.e the voltage at Y minus the voltage at X, we have:
V
AB = 240V < 60
V
BC = 240V < -60 = 240V < 300
V
CN = V
NA = 120V < 180
V
CA = 240V < 180
V
NB = 208V < 90
If we apply an impedance Z across a voltage V
XY, we get a current through the impedance of I
yx = V
XY / Z. Here if Z is just a resistance, its magnitude divides the magnitude of V, and the angle of I matches the angle of V; if Z has a non-zero phase angle associated with it (it has a reactive component), then the phase angle of I is the phase angle of V minus the phase angle of Z. (This is just complex multiplication/division in polar form.)
Now if the voltage V
XY is supplied by a coil directly, the coil current I
XY = the load current I
yx. Otherwise the current I
yx travels through multiple coils to get from X to Y, and we add I
yx to each of those coil currents.
Then with those conventions, and for the case that the impedances are all resistances, we get (if I tracked all the angles correctly; the most likely error would be a sign error, equivalently having one of the angles 180 degrees off):
I
AB = |I
ba| < 60 + |I
cb| < 120 + |I
bn| < 90
I
CN = |I
nc| < 180 + |I
cb| < 120
I
NA = |I
an| < 180 + |I
cb| < 120 + |I
bn| < 90
Here |X| = magnitude of X, and so |X| < Y is just a way to explicitly specify the phase angle of X. The above just says that each coil sees the current from loads connected directly across it, plus the current from B-C connected loads, plus the current from N-B connected loads (except the C-N coil doesn't see N-B connected loads). Also, if the impedances aren't purely resistive, their phase angles would subtract from the associated fixed phase angle specified above, but that would make the formulas a lot messier.
In applying the above to a balanced 3-phase load, we need to consider it as 3 identical impedances delta connected A-B, B-C, and C-A. Likewise, the load current from a C-A connected load shows up in both I
nc and I
an.
Cheers, Wayne