Hello Eric & PIC.ers,

Eric said:

>Date:    Fri, 20 Apr 2001 16:55:59 -0700
>From:    Eric Smith <eric@BROUHAHA.COM>
>Subject: [EE] three-phase power measurement, and rectifier output
>
>I posed the following two questions on sci.electronics, but got no replies.
>I'm posting them here because I know that there are people here who have
>used PICs for power measurement.
>
>
>
>1:
>
>I've been trying to figure out how to measure the power consumption on
>a three-phase wye circuit, and I'm stumped.
>
>Things I can measure:
>
>Vx, Vy, Vz:   voltage on each phase relative to neutral
>Ix, Iy, Iz:   current on each phase
>In:           current on neutral
>
>Wanted:  equation giving power consumption as a function of these variables
>
>For instance, with a normal single-phase device I measure the power
>consumed by integrating Vx*Ix.  This is simpler because there's only a
>single path for the current.  In the three-phase case, there may be varying
>phase-to-phase loads as well as phase-to-neutral loads.
>
>My first thought was to integrate Vx*Ix + Vy*Iy + Vz*Iz.  But this doesn't
>take into account any neutral current, and I'm not convinced that it would
>be right even for a three-phase delta circuit (no neutral).

This might be a pointer rather than a  complete answer, but traditionally
it has always been possible to accumulate the energy consumption
of a 3-phase, 3-wire load (whether balanced or not) by the
`two-wattmeter' method.

Essentially this picks one of the incoming lines (i.e. *not* the neutral)
as a voltage reference node and integrates the instantaneous V x I as
seen by each wattmeter. Summing the consumption of the _two_
meters then gives you the total consumption by all _three_ phases.
This is a cheaper way to measure energy use than by having
a wattmeter on each phase (needs three). Note that only two currents are
monitored, not all three - and not the current through the voltage reference
node for that matter. It is just the resultant of the other two so carries
no independant information.

Now...  if these results are valid, then is it not equally valid for you
to do your Vx*Ix + Vy*Iy + Vz*Iz using the star point (neutral) as the
voltage reference, irrespective of the (resultant) neutral line current?


    best regards,   John

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