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I have 3 x AT 50 B10 current transformers (CT), one on each phase (240V RMS), Vp, going into a CNC machine.

The output of each is 0-10V where each volt corresponds to 5A RMS passing through the CT.

I want to calculate power over a given time period, t0 to tn, for each phase (P1, P2, P3) & for all phases added together (PT).

My thoughts on how I do this,

Covert my acquired voltage to current, Ia = Va x 5

Sum Vp x Ia over t0 to tn for each phase

Then PT = P1 + P2 + P3

This is RMS power?

Is it that simple? Am I missing something here? Have I actually calculated electrical energy use here?

DrBwts
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    Rms power does not exist per se. If you multiply the two rms values of current and voltage, you obtain *apparent* power, expressed in VA. If you want to compute power in watts - the power that performs the *work* in the physical sense - you first plot the instantaneous power \$p(t)=i(t)v(t)\$ then *average* it across a line cycle: you obtain watts. – Verbal Kint Aug 03 '23 at 11:36
  • @VerbalKint " average it across a line cycle" what do you mean by line cycle? – DrBwts Aug 03 '23 at 12:41
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    A line cycle is one complete sine wave period. – Aaron Aug 03 '23 at 14:25
  • Oui, a complete mains cycle period, merci *Aaron* : ) – Verbal Kint Aug 03 '23 at 14:54

2 Answers2

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In general yes, but its not that simple. There is "Active" and "Reactive" power involved. Good explanation to your question is given here https://wiraelectrical.com/balanced-three-phase-power/

Kiper
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Is it that simple? Am I missing something here? Have I actually calculated electrical energy use here?

No, Yes and no.

An instantaneous power waveform is produced by the multiplication of the current and voltage waveforms. A few examples of what I mean from here: -

enter image description here

If you then average the power waveform you get the average power consumed per phase.

The three average power values (from each of the three phases) can be summed to give you total average power into your load. Average power is what you are generally billed on.

Andy aka
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