Why is one current reading ~20% higher than the other two in a balanced high-leg delta load?

Question

We are using eGuage watthourmeters to monitor the outdoor unit on five VRF systems powered by two different 3P4W 120/208V high-leg delta panels. The neutral is not connected to the outdoor units.

We have connected the meters according to the manufacturer's instructions as shown in the diagram below:

We are satisfied that these are installed correctly and accurately metering the devices, however I'm struggling to understand how to interpret the current readings. Here's a sample of some of the readings we are seeing (one for each of the five units), including the angle between the current and voltage phasors (note that the eGuage only reports power factor, not leading or lagging -- I've converted it to degrees here):

Quantity [units]	Reading 1	Reading 2	Reading 3	Reading 4	Reading 5
L1 [V]	120.8	120.8	120.8	120.8	120.8
L2 [V]	211.9	211.9	211.9	211.9	212.1
L3 [V]	120.7	120.7	120.4	120.4	120.4
S1 [A]	10.658	15.574	23.364	12.067	12.871
S2 [A]	15.337	19.767	26.674	15.757	17.057
S3 [A]	11.866	15.711	22.992	12.661	12.805
|L1-S1| [deg]	50.2	43.9	35.9	46.4	46.8
|L2-S2| [deg]	21.6	19.9	16.3	22.9	21.7
|L3-S3| [deg]	62.6	53.8	44.8	58.7	58.9

I found a Power Measurements Handbook which includes this phasor diagram for the distribution side of a high leg delta, to which I've mapped the labels of what we're reading:

The voltage is as I would expect. Assuming a purely resistive load, the angle between S2 and L2 should be zero. Then, the S3 and S1 currents lag their respective voltages by 30 degrees (just like in a delta). On the phasor diagram S1 appears to lead L1, because the reference direction (L1-N) is the reverse of the "normal" delta phase voltage.

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What I can't explain is why the S2 current reading is higher than the other two. In my head it makes sense -- each leg is using an equivalent amount of current (call it x), but the L1-L3 leg is split in half, so these two legs each use x + x/2, while the high leg uses 2x. This roughly matches what I'm seeing in terms of readings.

But is this correct? Mathematically, how can I determine what the three current measurements should be for a balanced high-leg delta load to verify what I'm measuring?

Answer 1

Assuming an A-B-C rotation system, is my assumed connection diagram below correct (including assumed currents being measured)?

Assuming that your terminology "L1-S1 [deg]" refers to the angle by which the current S1 lags the voltage measured from L1 to N (where N is the center tap) I get the following phasor diagram:

It is clear from this diagram that the three currents do not add up to zero - which they must do if no loads are connected between the phases (L1, L2, L3) and the center tap T (I think you call this neutral). Adding these up you get \$3I_0=12.66\angle-19.8° A\$.

Can you correct my assumed connections?

UPDATE 1: Answering question about how derived.

1. I assumed the system has A-B-C rotation so I can draw the 3 phase-neutral phasors, and from them derive the 3 phase-to-phase phasors. e.g. To find \$V_{ab}\$ just take the \$V_a\$ phasor and subtract the \$V_b\$ phasor from it. To do that negate the \$V_b\$ phasor (flip it 180°) and add it head-to-tail with \$V_a\$).

2. I wrote the voltage phasors on the connection diagram so it is easier to write KVL (head of the arrow is the +). For example, following the blue path in the figure below - starting at \$V_a\$ at top-right and following a closed path back:

$$+V_{aT} - \frac{1}{2}V_{bc} -V_{ab}=0$$

So,

$$V_{aT} = \frac{1}{2}V_{bc} + V_{ab}=0$$

And letting ph-ph voltage equal 1.0 per unit (e.g. 240V) then we have,

$$V_{aT} = \frac{1}{2}\angle-90° + 1.0\angle30° =\frac{\sqrt3}{2}\angle0° pu =208 V$$

$$ $$

Similarly we find,

$$V_{bT} = \frac{1}{2}V_{bc} =\frac{1}{2}\angle-90° pu = 120V$$

$$V_{bT} = -\frac{1}{2}V_{bc} =\frac{1}{2}\angle90° pu = 120V$$

Plotting these 3 guys on the original phasor diagram gives us the following: