Understanding 3-phase measurements

I would like to ask you to check if my conclusions are correct. Thank you.

In every case, 3 phase 4 line AC system is measured using 3 CTs (measuring each individual phase) connected to Emontx v3 with:
a) 1 phase sketch, Emontx v3 powered by AA batteries
The result: the apparent power of each individual phase showed separately.
The sum of all 3 phases (summed in emonCMS) is apparent energy consumption within a building.
b) 1 phase sketch, Emontx v3 powered by AC adapter
The result: The real power of the phase, which AC adapter is fed from. The power measurements of other two phases are totally incorrect, because current (data from CT) and voltage (acquired by AC adapter) are shifted by ±120deg. (the calculated power has values as if there is a big reactive load, which actually is not there) .
c) 3 phase sketch, Emontx v3 powered by AA batteries
The result: is the same as in a)
d) 3 phase sketch, Emontx v3 powered by AC adapter
The result: The real power of the phase, which AC adapter is fed from. The apparent power of the other two phases.

Two are correct, two are wrong.
(c) is quite wrong because the 3-phase sketch needs the a.c. adapter. So you cannot do that. So you must use (a) instead.
(a) is totally correct.
(b) is also correct as far as it goes, but the apparent power is available inside the sketch, it is not transmitted by default. You can edit the sketch to replace the real power for the other two phases with the apparent power. So you would then have one real power and two apparent powers.
(d) is wrong too. You will get a good value of real power for the phase that has it’s voltage measured, you will get a good approximation for the real power for the other two phases, depending on how accurately the three voltages balance.


Thank you so much :slight_smile:

a), b), c) are clear to me.
Also d) I hope: So in case d) the other 2 phases are actually real powers only if there are no reactive loads on those 2 phases (so when power factor=1, which means real power = apparent power). Am I right?

In case b), the apparent power values on the other phases 2&3 would be very different from real power values on those phases, due to ±120deg. shift between U from phase 1 (from AC adapter) and I from CT on phase 2&3.
I quite do not understand, why anything in the sketch has to be modified, since Real power and Apparent power are from the same calculation = Irms x Urms.

No, it will read the real power whatever the power factor. The three-phase sketch gets its voltage reference for the second and third phases by delaying the voltage of the first phase, but it has to hope that the amplitude is correct. Thus it can calculate real power for all three phases.

Again, no - if you mean what I understand you to mean. You cannot measure the real power of phases 2 & 3, but it exists nevertheless. The only quantity you can measure for those phases (or more correctly, estimate in the hope that the amplitude of the voltage is the same for all three phases) is apparent power. This is because the single phase sketch does not know about the voltages of phases 2 & 3.

The real power and apparent power for phases 2 & 3 will be different, according to the power factor of the load and any difference between the amplitudes of the voltages.

No they are not. Real power is average (over time) of Isample × Vsample. Apparent power is, as you say, Irms × Vrms
Where the averaging is done is the difference. For real power, you calculate the instantaneous power for each pair of samples, then find the average. For apparent power, you average separately the voltage samples and the current samples and then multiply to get the power.

And before you ask, in general
    power factor = V.I.cos(φ)
is NOT true. It is only true when both the voltage and current waveforms are perfect sinusoids.
The definition of power factor, which is how we measure it, is

    Power factor = (real power) / (apparent power)