Basically there are two methods of measuring three-phase powers: one is to use three wattmeters; the other using two wattmeters.
Let's study each method separately. At this site we will use the symbol W to represent the power measured by a wattmeter ,
and the symbol P to refer to the power active or real.
We can measure the power supplied to a load, balanced or not, connected to Y with four wires using the
three wattmeters method. This is possible using suitable connections as shown in the figure below.
It is noticed that each wattmeter is measuring the power consumed by each phase. Note that the wattmeter voltage coils
are connected in parallel with the load, and the current coils in series with the load.
Figure 85-01
Therefore, the total average power of the system can be calculated by adding the readings of the three wattmeters.
For any circuit we can calculate the total power supplied since we know n - 1 currents and voltages, because according to the laws of Kirchhoff, there are only n - 1 currents and independent voltages. So, in the case of a three-conductor circuit, balanced or not, we need only two wattmeters to measure the total power. In the case of a circuit
to four conductors, we need three wattmeters if the circuit is unbalanced. In the case of the balanced circuit, we need only two wattmeters, since, the neutral current will be null. In this case, the total power supplied to the load is the algebraic sum of the readings of the two wattmeters.
Let's look at the case of a balanced three-wire circuit. In this case, we can represent the load as an impedance of the type Z∠φ. Of course this is a general case, since any delta circuit can be replaced by an equivalent star circuit. It should be noted that in the case of balanced circuits, the impedance angle φ, is not altered by the delta-star transformation. In the circuit shown in the figure below, the load is connected in the configuration
star and the voltage of point B is the reference voltage for the two wattmeters.
Figure 85-02
Based on the circuit shown in the Figure 85-02, we can write the power that the wattmeter W1 will measure.
Let us assume that the phase difference between the measured voltage and the current is represented by the angle
θ. Like this:
W1 = |VAB| |IA| cos θ1
Where θ1 is the angle between the line voltage VL = VAB
and the line current IL = IA. So we can simplify the equation and write:
W1 = VL IL cos θ1
eq. 85-02
The same we can do for the second wattmeter and write:
W2 = |VCB| |IC| cos θ2
In this case, θ2 is the angle of lag between the line voltage
VL = VCB and the line current IL = IC. Then:
W2 = VL IL cos θ2
eq. 85-03
For the measures of W1 and W2, let's express θ1 and θ2 as a function of the angle of the impedance φ, which is equal to the angle between the phase voltage and the phase current. In the case of a direct or
positive phase sequence, remember that the line voltage is 30° ahead of the phase voltage. Therefore, we must add this angle to the impedance phase. Then we have:
θ1 = φ + 30°
In an analogous way, we can show that the angle θ2 can be expressed as:
θ2 = φ - 30°
This is due to the fact that we are taking as reference the phase voltage B. For the calculation of W2, note that we use the line voltage VCB = VBC∠180°. Then VCB is advanced of VAB of an angle equal to 60°. This explains θ2 = φ - 30°.
Therefore, substituting these last two equations into equations eq. 85-02 e eq. 85-03 we have:
W1 = VL IL cos (φ + 30°)
eq. 85-04
W2 = VL IL cos (φ - 30°)
eq. 85-05
Now that we know the angles, voltages and currents, we can sum ALGEBRAICALLY the values read by
W1 and W2 to find the total power consumed by the load.
PT = W1 + W2 = VL IL [cos (φ + 30°) + cos (φ - 30°)]
In the above equation two trigonometric identities appear, already studied in
Important Trigonometric Relations. Applying the correct relations
we find the equation that will allow the calculation of the total power, or:
PT = √3 VL IL cos φ
eq. 85-06
Note that the eq. 85-05 is exactly equal to eq. 82-03, this equation that we developed in the chapter
Three-Phase Power to find the average power in a balanced three-phase load.
Attention
Observing more closely the equations 85-03 and 85-04 and remembering that the power factor
of a circuit is given by cos (φ), we can draw the following conclusions regarding the readings
of the two wattmeters:
1 - If the power factor is greater than 0.5, the reading of the two wattmeters are
positive.
2 - If the power factor is exactly equal to 0.5, the reading of one of the wattmeters is
zero, since arccos (0.5) = 60° and then cos (60° + 30°) = cos (90°) = 0.
3 - If the power factor is less than 0.5, the reading of one of the wattmeters is
negative, since in this case we have the total angle greater than 90°. Thus, the value of
cos is negative.
4 - If the phase sequence is inverted, then the readings of the
two wattmeters will also be inverted.
Just as we calculate the real power by adding algebraically the readings of the two wattmeters,
we can calculate the reactive power of the circuit by performing the subtraction
of the readings of the two wattmeters. Soon we have:
W1 - W2 = VL IL [cos (φ + 30°) - cos (φ - 30°)]
And again, using the trigonometric relations and developing the equation above we arrive at:
W1 - W2 = VL IL sin φ
Then comparing the equations 82-02 and 82-05 with the latter, we conclude that the wattmeter
reading difference is proportional to the total reactive power. Soon:
Analyzing the equations presented above we can draw the following conclusions:
1 - If W2 = W1, the load is resistive .
2 - If W2 > W1, the load is inductive.
3 - If W2 < W1, the load is capacitive.
Although this data is derived from a connected load in star, they are
equally valid for a load connected in delta or triangle balanced. However, the two wattmeters method is not applicable in the power measurement of an unbalanced quadrifilar circuit. In this case we have to use the method of
three wattmeters, as showed item 2. (above).
