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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:

eGuage connection wiring for high leg delta

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:

High leg delta meter phasors

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.


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?

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  • \$\begingroup\$ I have opened a ticket with eGuage to get a better understanding of how to interpret the phase data that the meter is reporting. I'll edit or answer once I reach a conclusion. \$\endgroup\$
    – LShaver
    May 10, 2022 at 15:09

1 Answer 1

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

enter image description here

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:

enter image description here

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\$).

enter image description here

  1. 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,

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

enter image description here $$ $$

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:

enter image description here

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  • \$\begingroup\$ Because the VcT reference is reverse of a "normal" delta, I believe the angle for Is1 should be leading instead of lagging -- see the updated phasor diagram I added. Otherwise this looks correct. How did you develop these diagrams and do the phasor addition? \$\endgroup\$
    – LShaver
    May 7, 2022 at 22:02
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    \$\begingroup\$ Hi @LShaver, i added an update to my answer to show you how i found the voltages & phasor diagram. I'm sorry, i don't understand what you are saying about VcT. Maybe you can take my phasor diagram and edit it to show me - and add to bottom of your question. Also, i'll be traveling all day tomorrow so may be a while for i get back to this...but i will. \$\endgroup\$ May 8, 2022 at 1:40
  • \$\begingroup\$ Looking at your last phasor diagram; in my thinking, when converting a delta to a high leg delta, VaT "maps" to Va, but both VbT and VcT map to Vb, except VcT is mirrored. So assuming the currents lag the voltage throughout, when we mirror Ic it the reference switches from lagging to leading. If I'm correct, it would look like this. \$\endgroup\$
    – LShaver
    May 8, 2022 at 2:30
  • \$\begingroup\$ What matters is the polarity on the CTs. Looking at your top figure I would assume that the polarity marks/dots are facing the same direction for all CTs. I would also assume they are facing toward the transformer. Can you verify? \$\endgroup\$ May 8, 2022 at 12:29
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    \$\begingroup\$ The manual you linked in your post shows a function that should show the phase angle with polarity (-/+). "FFTarg2(f, c, r) Returns phase-angle of the spectral component at frequency f of the signal at input channel c relative to the reference channel r". Pretty odd to have a meter that measures phase angle but doesn't show you +/- w.r.t the reference. \$\endgroup\$ May 16, 2022 at 15:48

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