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

  • 2
    \$\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\$ Commented Apr 25, 2018 at 7:30

1 Answer 1

  • 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

  • \$\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
  • 1
    \$\begingroup\$ you can only get apparent power without a wattmeter. \$\endgroup\$ Commented Dec 22, 2016 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.