Measuring 3-Wire AC Power

Thank you for your prompt reply.

1. As mentioned, it is the total power that is required, so it stands to reason that the inductive (lagging PF) power or capacitive (leading PF) power is unknown, even in terms of cos φ.
2. The term "phase ange difference between volts and amps" is meaningless as the current flows through a single feed line but the voltage is measured between feed lines.
3. As mentioned, a power meter is unavailable. We accept the fact that without a power meter (or a 4-channel oscilloscope with two current probes for current measurement) we will be measuring apparent power. A concise measuring method using a scope is found in http://www.lecroy.com/tm/library/Labs/lab741/default.asp, but a scope, too is unavailable.
4. The background theory is found in section 12.10
5. Back to real life. Knowing only the scalar values (as measured with a clip-around current meter) we measure V1-2, V2-3, V3-1 and I1, I2 and I3. If all 3 voltages are identical, we can choose any 2 of the 3 current readings, multiply the currents by the voltage and come up with an "apparent power." But if we choose the two highest current readings, the power calculation will be high; if we choose the two lowest current readings, the result will be low.
   ==> Will the "average" of the currents yield the correct power result? Or perhaps we should choose the highest and lowest currents, excluding the mid-range one. Neither approach seems true to the degenerative case where when all voltages are equal, at 120°120° from one another, and the currents are identical and with no reactive part and therefore P=√3VI.

Your considered reply would be appreciated. (What seemed to be trivial at first has me stumped.)

Sincerely,

dnardi

The phase difference between voltage and current (φ) is different from the difference of nominally 120° between the line phases. If you measure V₁₂ and I₁₂ with an oscilloscope, the two waves may differ in phase.

In the unbalanced case, I think that to compute power you need to do:

(V₁₂ I₁₂ cos φ₁₂ + V₂₃ I₂₃ cos φ₂₃ + V₃₁ I₃₁ cos φ₃₁)/√3.

The three φ's are not necessarily equal, hence the subscripts on them as well. I am not sure that the √3 is exactly equal among all three phases, either.

In each phase, φ or cos φ must be measured or computed from the portion of inductive/resistive/capacitive loads in the circuit.

Thank you.

See my reply (#2 thread) in reply to #1's for a more comprehensive discussion.

In brief, if only a voltmeter and ampmeter are used, and we assume that the load is purely resistive (though the feed lines are unbalanced), how should the "apparent" power be calculated?

Sincerely,

dnardi

If the load is entirely resistive, φ = 0, cos φ = power factor = 1; and apparent and actual power are identical. The other calculations enter only if reactive (inductive or capacitive) loads are present.

Thank you.

Quite. But how should the calculation be done when the currents are unequal (assuming that the load is non-reactive)?

Sincerely,

dnardi

Except for the lingering question of whether √3 is the correct divisor in the unbalanced case, I think post #4 is the proper calculation.

Sometimes a graphic solution helps. If you draw a wye or delta diagram with line segment lengths proportional to the measured voltages, even if the supposed center is displaced off-center owing to ground faults or unbalanced loads, the angles between the line segments will conform to the phase angles of the matching circuits. It all then turns into vector algebra. The corresponding wye or delta of current will nearly overlie the voltage diagram, but will be rotated by φ according the presence of reactive loads.

It has been quite some time since I have worked with this, and thus my description may lack precision. Maybe we'll get some up-to-date EE input.

I think for a purely resistive load (PF=1), draw a phase diagram with the three currents as vectors 120 degrees apart from the origin. The phase-to-phase current would be the vector difference between the currents (distance between the heads of the vectors). The power would be the sum of the products of the phase-to-phase voltage times the phase-to-phase currents, i.e. P=V12(I2-I1)+V23(I3-I2)+V31(I1-I3). I don't see how you have enough information to measure power factor.

You are correct. What we want is to calculate the net power consumption of the device when only the phase-to-phase volatages and the three phase currents are known (scaler, not vector values).

Be glad to hear what you have to say.

dnardi

As it stands this question is not answerable, Power Factor and kWatts require the relationship between Votage Frequency and Current Frequency to be known.

If you have a Wattmeter + a Voltmeter + an Ohmmeter you could calculate the PF.

kW/kVA=PF

Or as others have posted, you will need an oscilliscope to see the angle.

With Only Voltage and Current information, it is Not possible to get the information you see.

Regards,
Sapper

Thank you.

If the voltages are balanced, the formula would be

P(apparent)=V(i-j)*[I(1)+I(2)+I(3)]/sqrt(3). It is obvious that the PF is unobtainable without phase shift information as you state. And in today's world with non-sinusoidal current waveforms in the vogue, this is an even more difficult problem requiring an oscilloscope or equivilant digital manipulation to do the point-by-point calculation.

But how does one approach the case of unbalanced feed voltages?

dnardi

If your Suppliers incoming feeder Voltages are not the same you have a more immediate problem and I would be talking to the supply authority about getting that fixed.

If your wave form is not sinusoidal, again I would be looking to your supplier or generator manufacturer.

The only place you are likely to find non-sinusoidal forms is in Cheap portable Generators or Inverters, out side of a lab that is.
They are definately not "in vogue" as you put it.

Most, if not all, Power authorities use rotating machines which by thier nature produce sinusoidal wave forms, I've yet to meet a square turbine or alternator.

Regards,
Sapper

Thank you Sapper.

If we limit our discussion to direct feed from a utility, the voltage is usually rather clean (when measured with a harmonic analyzer such as a Fluke 42B), but the current waveforms are frequently non-sinsusoidal in switching power supply loads. Things frequently get so bad that 3-phase 4-wire feed systems today usually use cables whose neutral line have the same cross-section as the phases, even though the phases are balanced. (Sinusoidal currents "cancel" each other out so that there is no neutral current and the neutral line cross section can be about one half that of the hot lines in traditional systems. High neutral currents on "balanced" feeds in modern system loads changed that to require the neutral line be equal in cross section to the hot lines.)

As a practical matter, the feed line voltages may be unequal if one is far down on the power distribution totem-pole. Another application would be power from an electronic source. Anyway, the question remains open -- how does one calculate the apparent power using only a voltmeter and ammeter on the unbalanced 3-wire feed.

dnardi,

With the equipment you have....You can Not get the information you want.

Regards,
Sapper

http://en.wikipedia.org/wiki/Power_factor