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I have a 3-phase system with an unbalanced delta load, as seen in the figure below.

I want to measure the total apparent and active power consumption and if possible, the respective power consumption for each of the phases, but I am a bit confused.

I am using a 3-phase equipment from National Instruments which shows that Vc = 380V, Vb = 380V and Vc = 0V. I assume that because of the absent neutral, it uses phase A as a reference, thus Vc = Vca, Vb = Vba.

I am also measuring all three line currents.

What is the total apparent power equation? I am aware of Blondel's Theorem, but it is valid only when measuring instant power right?

Also, is it possible to measure the power of each separate phase?

enter image description here

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    \$\begingroup\$ You cannot measure power without a wattmeter. End of story. \$\endgroup\$
    – Andy aka
    Commented Sep 22, 2016 at 22:25
  • \$\begingroup\$ You can do it without a wattmeter, it just takes more steps to do it (voltage, current, \$\cos\phi\$, harmonics, ...), that's why a wattmeter exists, to spare you the hassle. \$\endgroup\$
    – a concerned citizen
    Commented Apr 25, 2018 at 7:30

1 Answer 1

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  • The 3 Power meter method uses the meter impedance to create a virtual Y neutral node with 3 V readings
  • Thus 3 meter method can measure the P in each phase and total.
  • Using scalar VA products in each phase can get any result you need (VAR, PF)
  • the total apparent power (scalar product) can also be measured using true Vrms* true Irms
  • the reactive power can be computed by vector geometry.

enter image description here enter image description here

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  • \$\begingroup\$ I am aware of all these, but I have access only to the setup I described in the question. This means 1) I cannot connect any wattmeters as per figure 14 and 2) only RMS values can be measured. If I estimate though each line-to-neutral voltage as 220V, can I multiply this with the measured currents to get each line power? Would that be correct? \$\endgroup\$
    – DimP
    Commented Sep 22, 2016 at 21:35
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    \$\begingroup\$ you can only get apparent power without a wattmeter. \$\endgroup\$
    – Jasen Слава Україні
    Commented Dec 22, 2016 at 12:11

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