On the Calculation of a Three-Phase Circuit with a Nonlinear Element

V. N. Sosunova

Let us consider the cases, shown in Fig. 1, of calculating three-phase circuits containing an inertial nonlinear element, whose presence in the circuit practically does not disturb the sinusoidal nature of the regime.

Such elements may approximately be taken to include shunt reactors when capacitances have little influence in the circuit. Fig. 1 corresponds to various cases of short circuits and phase interruptions at the point where the reactors are connected.

We shall show that all these cases of calculating a three-phase circuit can be reduced to the calculation of a single-phase circuit with one nonlinear element.

Fig. 1

We assume that the six-terminal network shown in Fig. 1 is a symmetrical three-phase circuit, and that the source of asymmetry, containing the nonlinear element, has been removed.

We consider the source of asymmetry as an unsymmetrical three-phase voltage system containing a nonlinear element; then the entire circuit may be regarded as a symmetrical three-phase circuit to which an unsymmetrical system of voltages is applied. For such a circuit the principle of independence of the action of symmetrical components is valid.

Any unbalanced three-phase system can be decomposed into 3 balanced three-phase systems. Applying the decomposition formulas, we find, for each case, the symmetrical components of the currents and voltages of the unbalanced source.

We obtain 3 nonlinear equations relating the currents and voltages of the separate sequences; in these equations the voltages are unknown functions of the currents.

Then, using the boundary conditions at the point of asymmetry, we reduce the problem to one nonlinear equation, in which the current and voltage are related by the given volt- and phase-ampere characteristics. Consequently, the problem is reduced to the calculation of a single-phase circuit with one nonlinear element. The solution of this problem is known \([1;2]\).

Let us consider the above on an example.

1. Circuit of Fig. 1a.

Boundary conditions:

\[ \dot U_A=\dot U,\quad \dot I_A=\dot I,\quad \dot I_B=\dot I_C=0. \tag{1} \]

We establish the relations between the symmetrical components of the unbalanced three-phase system:

\[ \dot I^{(0)}=\dot I^{(1)}=\dot I^{(2)}=\frac{\dot I}{3}, \tag{2} \]

\[ \dot U^{(0)}+\dot U^{(1)}+\dot U^{(2)}=\dot U. \]

The equivalent circuit of each sequence may be considered, from the terminals of the source of asymmetry of the given sequence, as a two-terminal network.

Let us write the equations relating the currents and voltage of the separate sequences:

\[ \begin{aligned} \dot U^{(0)}&=-z^{(0)}\dot I^{(0)},\\ \dot U^{(1)}&=-z^{(1)}\dot I^{(1)}+\dot{\mathcal E},\\ \dot U^{(2)}&=-z^{(2)}\dot I^{(2)}, \end{aligned} \tag{3} \]

where \(z^{(0)}, z^{(1)}, z^{(2)}\) are the resistances of the two-terminal networks of the equivalent circuits of the zero, positive, and negative sequences; \(\dot{\mathcal E}\) is the no-load voltage of the two-terminal network of the positive sequence.

Summing the left- and right-hand sides of equations (3) and taking the boundary conditions into account, we obtain one nonlinear equation

\[ \dot U=-\left(z^{(0)}+z^{(1)}+z^{(2)}\right)\frac{\dot I}{3}+\dot{\mathcal E}. \tag{4} \]

Denoting

\[ \frac{z^{(0)}+z^{(1)}+z^{(2)}}{3}=z, \]

we obtain

\[ \dot{\mathcal E}=\dot U+z\dot I. \tag{5} \]

Using equation (5) and the given characteristics of the nonlinear element (Figs. 2, 3), one can construct the curves \(\mathcal E=f_1(I)\) and \(\varphi_{\mathcal E}=f_2(I)\). To do this, we specify the current \(I\) and from the curve of Fig. 2 find the modulus of the voltage \(U\). From the curve of Fig. 3 we find the phase angle between the effective values of the current and voltage. Then, summing the quantities \(\dot U\) and \(z\dot I\),

we obtain the complex open-circuit voltage \(\dot{\mathcal E}\). By specifying a sufficient number of points, we obtain the dependences \(\mathcal E=f_1(I)\) and \(\varphi_{\mathcal E}=f_2(I)\), from which it is not difficult to find the value of the current corresponding to the calculated

Fig. 2

Fig. 3

preliminary open-circuit voltage \(\mathcal E\), and also \(\varphi_{\mathcal E}\)—the phase shift between the current \(\dot I\) and the voltage \(\dot{\mathcal E}\).

2. Circuit of Fig. 1b.

The boundary conditions at the point of asymmetry are:

\[ \dot U_A=\dot U,\qquad \dot I_A=\dot I,\qquad \dot U_B=\dot U_C=0. \]

We establish the relations between the symmetrical components of the unsymmetrical three-phase system

\[ \dot U^{(0)}=\dot U^{(1)}=\dot U^{(2)}=\frac{\dot U}{3}, \]

\[ \dot I^{(0)}+\dot I^{(1)}+\dot I^{(2)}=\dot I. \tag{6} \]

For this case of asymmetry it is convenient to write the equations of the separate sequences in the form (7):

\[ \dot I^{(0)}=-Y^{(0)}\dot U^{(0)}, \]

\[ \dot I^{(1)}=-Y^{(1)}\dot U^{(1)}+\dot I_k, \]

\[ \dot I^{(2)}=-Y^{(2)}\dot U^{(2)}, \tag{7} \]

where \(Y^{(0)}, Y^{(1)}, Y^{(2)}\) are the admittances of the two-terminal networks of the equivalent circuits of the zero, positive, and negative sequences; \(\dot I_k\) is the short-circuit current of the two-terminal network of the positive sequence.

Summing the left- and right-hand sides of equation (7) and taking into account the boundary conditions (6), we obtain:

\[ \dot I=-\left(Y^{(0)}+Y^{(1)}+Y^{(2)}\right)\frac{\dot U}{3}+\dot I_k. \tag{8} \]

Denoting

\[ \frac{Y^{(0)}+Y^{(1)}+Y^{(2)}}{3}=Y, \]

we obtain

\[ \dot I_k=\dot I+Y\dot U. \]

The solution of the obtained equation (8) corresponds to the calculation of a single-phase circuit with one nonlinear element. The calculation may be carried out analogously to that described for the circuit of Fig. 1a.

3. Circuit of Fig. 1c

It is evident that the circuit of Fig. 1c can be represented in the form of Fig. 4, and then in the form of Fig. 5, where \(\dot U_B=\dot U_C=0\).

The relations between the symmetrical components of the currents and voltages are: \(\dot U^{(1)}=\dot U^{(2)}\),

\[ \dot I^{(1)}=-\dot I^{(2)}=\frac{j\sqrt{3}}{3}\,\dot I . \tag{9} \]

Let us write the two-terminal equations of the equivalent circuits of positive and negative sequence:

\[ \dot U^{(1)}+\frac{z_H}{2}\dot I^{(1)}=-z^{(1)}\dot I^{(1)}+\dot{\mathcal E}, \tag{10} \]

\[ \dot U^{(2)}+\frac{z_H}{2}\dot I^{(2)}=-z^{(2)}\dot I^{(2)} . \tag{11} \]

Fig. 4

Fig. 5

Subtracting equation (11) from equation (10) and taking into account the relations between the symmetrical components (9), we obtain:

\[ z_H \dot I \,\frac{j}{\sqrt{3}} = -\left(z^{(1)}+z^{(2)}\right)\frac{\dot I\,j}{\sqrt{3}}+\dot{\mathcal E}. \tag{12} \]

Substituting \(z_H \dot I=\dot U\), we again obtain the equation corresponding to the calculation of a single-phase circuit with one nonlinear element:

\[ \dot{\mathcal E}=\frac{j}{\sqrt{3}}\,\dot U+\frac{j}{\sqrt{3}}\left(z^{(1)}+z^{(2)}\right)\dot I . \]

Calculations of a three-phase circuit with an inertial nonlinear element of type 1b and 1c were checked in the laboratory. The experimental results are satisfactory.

REFERENCES

1. Voronov R. A. and Ponomareva G. F. Circular diagrams in the study of nonlinear circuits. Elektrichestvo, No. 12, 1951.
2. Pukhov G. E. On the question of calculating an electric circuit with one nonlinear element in the steady-state sinusoidal regime. Izvestiya TPI, vol. 72.