Associate Professor R. A. Voronov.

On the Question of Measurements by Wattmeters Connected According to Aron’s Scheme

When two wattmeters are connected according to Aron’s scheme in a three-wire circuit of three-phase current, the algebraic sum of their readings gives the total power of all three phases. In this case it remains unknown whether the loads of these phases are equal and how they differ from one another. In practice, however, one sometimes has to face the necessity of solving such a question.

In the present article a method is presented which makes it possible, using only these two wattmeters without any changes in the circuit, to find the distribution, by phases, of the powers supplied from the network (the powers of the network phases), as well as the current strengths in the individual conductors and their phase-displacement angles with respect to the phase and line voltages of the network.

This method is very simple, but is valid only for symmetrical network voltages, i.e., for equal line voltages.

At the end of the article, as a consequence, a simple method is given for determining the correctness of the connection of the wattmeters and for determining the sign of the reading on the wattmeter with the smaller indication.

Taking, for equal line voltages, \(V_{13}\), \(V_{21}\), and \(V_{32}\), the same phase voltages of the network \(V_1\), \(V_2\), and \(V_3\), and arranging their vectors as this is done in Fig. 1, we obtain for the latter the following complex expressions through the phase voltage

Fig. 1.

\[ \dot V_1=V'_1+jV''_1=V \]

\[ \dot V_2=V'_2+jV''_2=\dot V_1\dot a^2=-\frac{1}{2}V-j\frac{\sqrt{3}}{2}V \tag{1} \]

\[ \dot V_3 = V'_3 + j V''_3 = \dot V_1 \dot a = -\frac{1}{2} V + j\frac{\sqrt{3}}{2} V \]

We shall also represent the currents of the individual phases in the form of complexes

\[ \begin{aligned} \dot I_1 &= I'_1 + j I''_1,\\ \dot I_2 &= I'_2 + j I''_2,\\ \dot I_3 &= I'_3 + j I''_3. \end{aligned} \tag{2} \]

The powers of the individual phases of the network are expressed through these complexes in the following form

\[ \begin{aligned} P_1 &= V_1 I_1 \operatorname{Cos}(\dot V_1;\dot I_1) = V'_1 I'_1 + V''_1 I''_1 = V I'_1,\\ P_2 &= V_2 I_2 \operatorname{Cos}(\dot V_2;\dot I_2) = V'_2 I'_2 + V''_2 I''_2 = -\frac{1}{2} V I'_2 - \frac{\sqrt{3}}{2} V I''_2,\\ P_3 &= V_3 I_3 \operatorname{Cos}(\dot V_3;\dot I_3) = V'_3 I'_3 + V''_3 I''_3 = -\frac{1}{2} V I'_3 + \frac{\sqrt{3}}{2} V I''_3. \end{aligned} \tag{3} \]

When two wattmeters are connected according to Aron’s circuit in the first and second phases (Fig. 2), their readings are expressed through the same complexes.

\[ \begin{aligned} P_{\mathrm{I}} &= V_{31} I_1 \operatorname{Cos}(\dot V_{31};\dot I_1) = V'_{31} I'_1 + V''_{31} I''_1,\\ P_{\mathrm{II}} &= V_{32} I_2 \operatorname{Cos}(\dot V_{32};\dot I_2) = V'_{32} I'_2 + V''_{32} I''_2. \end{aligned} \]

Since the voltages on the voltage coils will be equal to:

\[ \begin{aligned} \dot V_{31} &= -\dot V_{13} = \dot V_1 - \dot V_3 = \frac{3}{2} V - j\frac{\sqrt{3}}{2} V,\\ \dot V_{32} &= \dot V_2 - \dot V_3 = -j\sqrt{3}\,V, \end{aligned} \]

then the instrument readings are expressed in the form

\[ \begin{aligned} P_{\mathrm{I}} &= \frac{3}{2} V I'_1 - \frac{\sqrt{3}}{2} V I''_1 = \frac{3}{2} P_1 - \frac{\sqrt{3}}{2} V I''_1,\\ P_{\mathrm{II}} &= -\sqrt{3}\, V I''_2, \end{aligned} \tag{4} \]

where the power \(P_1\) has been introduced according to equation (3).

Fig. 2.

If we now disconnect the voltmeter windings of the instruments from the third phase, leaving them connected to each other, opening the switch \(m\) inserted in the circuit (Fig. 2), or simply disconnecting the conductor, then the wattmeter readings will change because of the change in voltage at their terminals. Since usually the resistances of the windings of both instruments are chosen equal to each other, the voltage on each of them will be equal to half the line voltage \(V_{21}\). In this case the new readings will be expressed through complexes in the form

\[ P'_{\mathrm{I}} = \left(\frac{1}{2}V_{21}\right)\cdot I_1\cdot \operatorname{Cos}(V_{21}; I_1) = \frac{1}{2}\left(V'_{21}\cdot I'_1+V''_{21} I''_1\right) \]

\[ P'_{\mathrm{II}} = \left(-\frac{1}{2}V_{21}\right)\cdot I_2\cdot \operatorname{Cos}(\dot V_{12}; \dot I_2) = -\frac{1}{2}\left(V'_{21} I'_2+V''_{21} I''_2\right) \]

or, since according to equation (1)

\[ \dot V_{21}=\dot V_1-\dot V_2=\frac{3}{2}V+j\frac{\sqrt{3}}{2} \]

they will acquire the expressions

\[ 2P'_{\mathrm{I}} = \frac{3}{2}VI'_1+\frac{\sqrt{3}}{2}VI''_1 = \frac{3}{2}P_1+\frac{\sqrt{3}}{2}VI''_1 \]

\[ 2P'_{\mathrm{II}} = -\frac{3}{2}VI'_2-\frac{\sqrt{3}}{2}VI''_2 = 3P_2+\sqrt{3}VI''_2 \tag{5} \]

Adding equations (4) and (5) pairwise, we find

\[ P_{\mathrm{I}}+2P'_{\mathrm{I}}=3P_1 \]

\[ P_{\mathrm{II}}+2P'_{\mathrm{II}}=3P_2 \]

whence we obtain expressions for determining the powers of the first two phases of the network

\[ P_1=\frac{1}{3}\left(P_{\mathrm{I}}+2P'_{\mathrm{I}}\right) \]

\[ P_2=\frac{1}{3}\left(P_{\mathrm{II}}+2P'_{\mathrm{II}}\right) \tag{6-а} \]

for the third phase we obtain an equation, proceeding from the total power

\[ P_3=(P_{\mathrm{I}}+P_{\mathrm{II}})-(P_1+P_2) = \frac{2}{3}\left[P_{\mathrm{I}}+P_{\mathrm{II}}-P'_{\mathrm{I}}-P'_{\mathrm{II}}\right] \tag{6-б} \]

Having the wattmeter readings with the switch \(m\) closed (normal circuit) and after opening it, it is easy, using these equations, to calculate the powers of each of the phases of the network separately.

The values \(P_1,\ P_2\), and \(P_3\), equal to the powers supplied from the individual phases of the network, will of course not be equal to the powers consumed by each of the load phases, since, owing to the inequality of the voltage drop, the neutral point of the load (or of the star equivalent to it) will be displaced relative to the neutral point of the network.

The voltages on the load phases will turn out to be unequal to one another and asymmetrically situated, and it is not possible to take into account the power corresponding to them without any additional measurements pertaining to the neutral point of the load.

Nevertheless, the inequality of the powers obtained immediately indicates the nonuniformity of the phase loading and which of them is loaded more or less.

Using equations (4) and (5), we find expressions for the components of the currents of the first and second phases

\[ I'_1=\frac{P_1+2P'_1}{3V};\qquad I''_1=\frac{2P'_1-P_1}{\sqrt{3}\,V} \]

\[ I'_2=\frac{P_{\mathrm{II}}-4P'_{\mathrm{II}}}{3V}\qquad I''_2=-\frac{P_{\mathrm{II}}}{\sqrt{3}\,V} \tag{7-a} \]

For the third phase we find the components from the equations

\[ I'_1+I'_2+I'_3=0 \qquad I''_1+I''_2+I''_3=0 \]

whence

\[ I'_3=-\frac{P_1+P_{\mathrm{II}}+2P'_1-4P'_{\mathrm{II}}}{3V}; \qquad I''_3=\frac{P_1+P_{\mathrm{II}}-2P_1}{\sqrt{3}\,V} \tag{7-b} \]

Equations (7-a) and (7-b), together with equations (2), make it possible to find the currents in the form of complexes, from which it is not difficult to find the current values themselves. Dividing these complexes by the complexes of the corresponding phase voltages of the network, we obtain

\[ \frac{\dot I_1}{\dot V_1} = \frac{P_1+2P'_1}{3V^2} + j\,\frac{2P'_1-P_1}{\sqrt{3}\,V^2} \]

\[ \frac{\dot I_2}{\dot V_2} = \frac{P_{\mathrm{II}}+2P'_{\mathrm{II}}}{3V^2} + j\,\frac{P_{\mathrm{II}}-2P'_{\mathrm{II}}}{\sqrt{3}\,V^2}; \qquad \frac{\dot I_3}{\dot V_2} = \frac{2(P_1+P_{\mathrm{II}}-P'_1-P'_{\mathrm{II}})}{3V^2} + \]

\[ + j\,\frac{2(P'_{\mathrm{II}}-P'_1)}{\sqrt{3}\,V^2} \]

From these equations, dividing the second terms by the first, we obtain expressions for the tangents of the phase-shift angles between the currents and the network voltages.

\[ \operatorname{tg}\varphi_1=\sqrt{3}\,\frac{2P'_1-P_1}{2P'_1+P_1};\qquad \operatorname{tg}\varphi_2=\sqrt{3}\,\frac{P_{II}-2P'_{II}}{P_{II}+2P'_{II}}; \]

\[ \operatorname{tg}\varphi_3=\sqrt{3}\,\frac{P'_{II}-P'_1}{P_1+P_{II}-P'_1-P'_{II}} \tag{8} \]

from which it is already easy to find the angles themselves and the corresponding power factors, i.e. their cosines. These phase-shift angles and power factors, of course, likewise refer to the phases of the current source, and not to the load.

Having found the latter, one can obtain the values of the currents from the usual equations

\[ I_n=\frac{P_n}{V\cdot \operatorname{Cos}\varphi_n} \]

In the case of inequality of the resistances of the wattmeter windings, a correction must be introduced, taking in all equations, instead of the readings \(P'_I\) and \(P'_{II}\), their corrected values \(\dfrac{r_1+r_2}{2r_1}P'_I\) and \(\dfrac{r_1+r_2}{2r_2}P'_{II}\),

where \(r_1\) and \(r_2\) are the resistances of the windings.

Let us consider also the case of symmetry of the load, in which all currents, all powers, and phase-shift angles in all phases are the same. In this case the current of the second phase may be expressed in terms of the current of the first phase

\[ i_2=I'_2+jI''_2=i_1\dot a^2 =\left(-\frac{1}{2}I'_1+\frac{\sqrt{3}}{2}I''_1\right) +j\left(-\frac{1}{2}I''_1+\frac{\sqrt{3}}{2}I'_1\right) \]

which gives for the powers the values

\[ P_I=\frac{3}{2}VI'_1-\frac{\sqrt{3}}{2}VI''_1 \qquad P_{II}=\frac{3}{2}VI'_1+\frac{\sqrt{3}}{2}VI''_1 \]

\[ 2P'_I=\frac{3}{2}VI'_1+\frac{\sqrt{3}}{2}VI''_1 \qquad 2P'_{II}=\frac{3}{2}VI'_1-\frac{\sqrt{3}}{2}VI''_1 \]

Comparing them with one another, we find

\[ P_I=2P'_{II};\qquad P_{II}=2P'_I \tag{9} \]

i.e., with symmetry of the loads, the wattmeter readings upon opening switch \(m\) interchange places and are reduced by half in comparison with those obtained in the normal circuit. For the tangents of the phase-shift angles in all phases, one obtains one and the same expression

\[ \operatorname{tg}\varphi=\sqrt{3}\,\frac{P_{II}-P_I}{P_{II}+P_I} \]

which is the generally known equation for Aron’s circuit.

Relations (9) are well suited for checking the correctness of the connection of instruments. Since under ordinary conditions the loads of the individual phases differ little from one another, when the circuit is opened at point \(m\) the instrument readings should, after decreasing by approximately half, interchange places. If this does not occur, then the circuit has been connected incorrectly and must be corrected. These same relations also apply to checking the connection of two single-phase meters (with respect to the speed and direction of their rotation), as well as to three-phase wattmeters and meters (a decrease of the readings by approximately half without a change in direction).

Proceeding from these relations, it is also easy to obtain a rule for determining the sign of the reading on the instrument with the smaller indication. It is not difficult to see that if, when the circuit is opened at point \(m\), its indication remains positive, then its reading should be taken with a plus sign; if, however, the indication changes to the opposite one, then with a minus sign.