Indirect measure of reactive power in three phase line

There is a method to detect the reactive power in three phase line, that use only two wattmeter (used to measure real part).

I have problem to understand this method and resultant expression of the reactive power.
I have attached a representation of this method: connection of two instruments and the formula valid for reactive power. In the figure “Rv” is the resistance added to the third line, his value equals the internal resistance of wattmeter.

I desired to ask you if you can describe me the derivation of the Q expression, involving phasors present in the line.

Reactive power.png

Q is simply derived from the word "Quantity". The Q-hour meter was intended to facilitate VA measurement by using electromechanical technology.

A VAR meter has maximum response when the voltage and current are 90 degrees out of phase. A Q meter has maximum response when the voltage and current are 60 degrees out of phase.

A VAR meter's range of +'ve response (2-quadrants) 0 to 180 degrees. A Q meter's range of +'ve response (2-quadrants) 30 degrees (leading) to 150 degrees (lagging).

A Wattmeter connected with voltage shifted by 60 degrees lagging becomes a Q-meter. This is easy to do in 3-phase systems.

Personally I preferred Q measurement because it can measure most all practical loads - lighting (with leading PF) and lightly loaded motors (highly lagging PF).

Electronic metering has basically made Q measurement redundant.

Can you tell us the source of the drawing, and what it's intended to represent? Your post suggests that it may be a scheme for measuring reactive power. The two-meter/three-line scheme evokes Blondel's Theorem. The fact that the voltage between lines 2 and 3 appear to be reacted with the current through line 1 suggests that the scheme is intended to phase shift the voltages, presumably to measure VAR flow as opposed to power flow. That suggests that there are some assumptions involved - particularly, that the voltages are precisely equal, and that the phase angles are precisely 2*pi/3, or 120 degrees.

I think that, for you, Q is the symbol for reactive power, and it means something different to you than it does to dlloyd. Maybe you can tell us what Q means to you in this context?

Also interested in the OP's comments.

The point labelled O is at the midpoint of the phases (since each meter's V-sense terminals and the Rv have the same resistance), so each meter measures one phase's current against the voltage of another (120 degrees different)

The power meters measure Vrms x Irms x cos(a) where a is the phase difference between V and I, but in this configuration the meters measure

V2.I1.cos (r+120) and -V1.I2.cos (r-120)

So if all the voltages are the same for each phase and all the currents are the same, you get to measure cos(r+120)-cos(r-120) and a bit of trigonometry will show this is equal to -sin(r), and thus the reactive power directly. I think the sqrt(3) is bogus, it should be 3, since we need to multiply by the number of phases.

Of course if the currents or voltages are imbalanced or the phase angle isn't 120 it shows erroneous readings. It takes 11 values to characterize 3-phase power (excluding harmonics)!

I think that, for you, Q is the symbol for reactive power, and it means something different to you than it does to dlloyd. Maybe you can tell us what Q means to you in this context?

Yes, sorry, what I was referring to has mostly gone the way of the dinosaur … fooled by the eerily similar connections and using the same terminology.

From here, “for the balanced condition, the magnitude three phase reactive power can be found using . . .”

Q = √3 (W1 - W2)

Another resource is here

Q = √3 (WC - WA)

For the connections shown in the OP’s diagram, maybe the second wattmeter is connected for negative readings (its not very clear), then the formula shown where the readings are added seems valid.

Ah, I got the cos(120) and sin(120) wrong, thought they were -1/2 and +1/2, whereas they are root-3 times larger in magnitude, so all is well.

Ok thank you, I was imprecise. I didn't specify the voltage condition, they are balanced (0°, -120°, +120° i.e. direct sequence). The current is unbalanced, but there is "Rv" that is equal to the internal impedance of wattmeter. So the connection appears like balanced (in current and voltage).

I try to develop in phasor diagram as MarkT said. If I consider three phase current and voltage equal in amplitude and voltage, it seems verify.