On the Calculation of Three-Phase Circuits with Local Asymmetry

V. N. SOSUNOVA

By local asymmetry is meant an asymmetry concentrated at a single point. Asymmetry may arise in short circuits, phase interruptions, and for other reasons. In the present work we shall confine ourselves only to considering local asymmetries due to short circuits and phase interruptions. For the sake of generality of the results, we shall assume that the circuit contains an arbitrary number \((n)\) of short circuits and phase interruptions.

General Provisions

When a circuit has one fault, then, as is known, the solution of the problem by the method of symmetrical components reduces to the calculation of three two-terminal networks: of positive, negative, and zero sequence. Each two-terminal network is characterized by one equation. Therefore, with one fault, the state of the circuit is determined by three equations connected with one another by three equations of boundary conditions. To solve the problem it is necessary to set up 6 equations.

With two faults, the network of each sequence will represent an ordinary (through) four-terminal network. Since such a four-terminal network is characterized by two equations, in the case of two faults the state of the circuit will be determined by six equations connected with one another by six equations of boundary conditions. To solve the problem there are 12 equations.

In the case of \(n\) faults, the network of each sequence can be represented in the form of a \(2n\)-terminal network with pairwise equal terminal currents. Such a \(2n\)-terminal network is characterized by \(n\) equations. Thus, the entire circuit will correspond to a system of \(3n\) equations, connected with one another by \(3n\) equations of boundary conditions. To solve the problem in this case there are \(6n\) equations.

Analyzing the relationships between the currents and voltages of the individual sequences (Table 1) for faults of the type under consideration (short circuits and interruptions), it can be established that each type of fault is characterized by one group of summing quantities and one common quantity. By summing quantities we mean the currents and voltages of the individual sequences which, for a given fault, are summed; and by common quantities—the currents and voltages which, for the multi-terminal networks of the individual sequences, are identical. For example, in a single-phase short circuit the summing quantities will be the voltages of the positive, negative, and zero sequences of the calculated phase, and the common quantity will be the current at the point of the short circuit.

Table 1

Boundary conditions for the currents and voltages of individual sequences in phase interruptions and short circuits

The table lists all types of short circuits and phase interruptions.

It is proposed that the equations of the \(2n\)-terminal networks of the individual sequences be composed so that the summing currents and voltages are expressed through the common ones, that is:

\[ \rho_c^{(\mu)}=\zeta^{(\mu)}\rho_0+\zeta_0^{(\mu)} \qquad (\mu=0,1,2), \tag{1} \]

where \(\mu\) is the sequence number,

\(\rho_c^{(\mu)}\) is a vector whose components are the summing currents and voltages of the \(\mu\)-th sequence,

\(\rho_0\) is a vector whose components are the common quantities,

\(\zeta^{(\mu)}\) and \(\zeta_0^{(\mu)}\) are a matrix and a vector depending on the circuit parameters for the \(\mu\)-sequence.

For faults of the type of short circuits and phase interruptions, the summing quantities, when added, prove to be equal to zero. Therefore, if the left- and right-hand sides of equations (1) are added, then in the case of one fault we obtain one equation with one unknown—the common quantity; in the case of two faults, two equations with two unknowns; and so on. With \(n\) faults, \(n\) equations with \(n\) unknowns are obtained. Thus, owing to a rational way of writing the equations, the problem is reduced to the smallest number of equations.

The final calculation equation will have the form:

\[ 0=\zeta_{\rho_0}+\zeta_0, \tag{2} \]

where \(\zeta\) is a certain matrix, which we shall hereafter call the matrix parameter.

\(\zeta_0\) is a vector whose value can be determined from the solution for the normal operation of the circuit. We shall agree to call \(\zeta_0\) the vector parameter.

The components of the matrix \(\zeta\) and of the vector \(\zeta_0\) have the form:

\[ \zeta_{ik}=\zeta^{(0)}_{ik}+\zeta^{(1)}_{ik}+\zeta^{(2)}_{ik}, \tag{3} \]

where \(i\) is the row number,
\(k\) is the column number, and the superscript is the sequence number.

In a circuit with symmetrically constructed energy sources (the usual cases)

\[ \zeta^{(0)}_{i0}=0,\qquad \zeta^{(2)}_{i0}=0. \]

The components \(\zeta_{ik}\) may be of type \(Z\), \(Y\), \(A\) (a chain matrix), or of other types. For simple \(2n\)-terminal networks these coefficients are calculated by the known methods of no-load and short-circuit operation, or by the superposition method, or are taken from tables [1].

For complex \(2n\)-terminal networks, the theory of composite electric circuits should be used.

In the case of faults in different phases, the components \(\zeta^{(1)}_{ik}\) and \(\zeta^{(2)}_{ik}\) must be preceded by the factor

\[ a^{\gamma^{(\mu)}_{ik}}\left(a=e^{j120^\circ}\right). \]

Thus, the formula for finding the components of the matrix \(\zeta\) and the vector \(\zeta_0\) is expressed in general form by equation (4):

\[ \zeta_{ik}=\zeta_{ik}(0)+a^{\gamma^{(1)}_{ik}}\zeta^{(1)}_{ik}+a^{\gamma^{(2)}_{ik}}\zeta^{(2)}_{ik}. \tag{4} \]

The exponent \(\gamma^{(\mu)}_{ik}\) can take the values \(\gamma^{(\mu)}_{ik}=0,1,2\) and is calculated by formula (5)

\[ \gamma^{(\mu)}_{ik}=\gamma^{(\mu)}_i-\gamma^{(\mu)}_k, \tag{5} \]

where \(\gamma^{(\mu)}_i\) is the exponent of the operator of the summated quantity of the \(\mu\)-sequence, \(\gamma^{(\mu)}_k\) is the exponent of the operator of the corresponding common quantity; \(\gamma^{(\mu)}_i\) and \(\gamma^{(\mu)}_k\) are taken from the boundary conditions (Table 1). We shall show how \(\gamma^{(\mu)}_{ik}\) is determined by an example.

Example 1.
Phase \(A\) is open, and phase \(B\) is shorted to ground (Fig. 1). From the table of boundary conditions we write down the summated quantities

\[ \dot I^{(0)}_p+\dot I^{(1)}_p+\dot I^{(2)}_p=0, \]

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

and the common quantities

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

\[ \dot I^{(0)}_k=a^2\dot I^{(1)}_k=a\dot I^{(2)}_k=\frac{\dot I_k}{3}. \]

The equations of the equivalent circuits will be written in the form of equation (5′)

\[ \begin{pmatrix} i_p\\ U'_k \end{pmatrix} = [\zeta] \begin{pmatrix} U'_p\\ i_k \end{pmatrix} +\zeta_0 \tag{5′} \]

Fig. 1

\[ \begin{aligned} \zeta_{11} &= \zeta_{11}^{(0)}+\zeta_{11}^{(1)}+\zeta_{11}^{(2)},& \zeta_{12} &= \zeta_{12}^{(0)}+a\zeta_{12}^{(1)}+a^2\zeta_{12}^{(2)},\\ \zeta_{21} &= \zeta_{21}^{(0)}+a^2\zeta_{21}^{(1)}+a\zeta_{21}^{(2)},& \zeta_{22} &= \zeta_{22}^{(0)}+\zeta_{22}^{(1)}+\zeta_{22}^{(2)},\\ \zeta_{10} &= \zeta_{10}^{(1)},& \zeta_{20} &= \zeta_{20}^{(1)}. \end{aligned} \]

Calculation procedure

The following technique may be recommended for calculating circuits with the types of faults under consideration.

1. The asymmetrical circuit is replaced by a calculated symmetrical one by introducing fictitious emf sources and current sources at the fault locations.

Such a replacement is necessary in order subsequently to be able to use the principle of independence of action of the symmetrical components of currents and voltages, which is known to be valid only for symmetrically constructed circuits.

1. The relations between the symmetrical components of currents and voltages at the fault locations are established, and these quantities are subdivided into summing and common quantities.

2. Equivalent circuits of the individual sequences are drawn up in the usual manner.

3. The equations of the \(2n\)-terminal networks of the individual sequences are drawn up in the form of equation (1), i.e., in such a way that the summing currents and voltages are expressed through the common ones. Then, after the corresponding summation of the coefficients (formula 3 or 4), a system of equations with \(n\) unknowns is obtained, these unknowns being the common currents and voltages.

4. The resulting system is solved for the unknown quantities.

Example 2.

Let us examine in detail the proposed method for setting up the calculation equations on the simplest example of a short circuit of phase \(A\) (Fig. 2).

We shall adhere to the calculation procedure indicated above.

We replace the asymmetrical circuit by a calculated symmetrical one by introducing fictitious current sources (Fig. 3).

We establish (or take from Table 1) the relations between the symmetrical components of the currents and voltages at the point of the short circuit, and immediately subdivide these currents and voltages into summable and total ones.

Group of summable quantities

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

Total quantity

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

Fig. 2

Then we construct the equivalent circuits of the positive-, negative-, and zero-sequence networks. Since there is one fault in the circuit, the equivalent circuits are represented in the form of two-terminal networks (Fig. 4). We write the equations characterizing the two-terminal networks:

\[ \begin{pmatrix} \dot U^{(0)}\\ \dot U^{(1)}\\ \dot U^{(2)} \end{pmatrix} = \begin{pmatrix} \zeta_{11}^{(0)} & 0 & 0\\ 0 & \zeta_{11}^{(1)} & 0\\ 0 & 0 & \zeta_{11}^{(2)} \end{pmatrix} \begin{pmatrix} \dot I^{(0)}\\ \dot I^{(1)}\\ \dot I^{(2)} \end{pmatrix} + \begin{pmatrix} \zeta_{10}^{(1)}\\ 0\\ 0 \end{pmatrix}. \tag{6} \]

Adding the left- and right-hand sides of equation (6) and taking into account the conditions characterizing the fault, we obtain equation (7), and from it the well-known solution for the short-circuit current \(I\).

Fig. 3

Fig. 4

Equation (7) could also have been obtained simply by using formula (3):

\[ 0=\zeta_{11}^{(0)}+\zeta_{11}^{(1)}+\zeta_{11}^{(2)}\frac{\dot I}{3}+\zeta_{10} \tag{7} \]

\[ I=\frac{3\zeta_{10}}{\zeta_{11}^{(0)}+\zeta_{11}^{(1)}+\zeta_{11}^{(2)}}. \tag{8} \]

Let us consider a sample calculation of a three-phase circuit when it contains three faults.

Example 3.

Let us consider the double-ended disconnection of one phase of a line, with a short circuit of the disconnected phase to ground\(^1\), Fig. 5.

The numerical values of the parameters of the original circuit, in relative units, are:

\[ \dot E_M = j2,\qquad \dot E_N = j1,\qquad \dot Z_M^{(1)} = j2,\qquad \dot Z_N^{(1)} = \dot Z_M^{(2)} = \dot Z_N^{(2)} = \dot Z_M^{(0)} = \dot Z_N^{(0)} = j1, \]

\[ \dot Z_T = j0.4, \]

where the line quantities enter into the resistances of the corresponding parts of the system.

Fig. 5

The unsymmetrical calculation circuit is replaced by a calculated symmetrical one by introducing fictitious e.m.f. sources at the points of the break and introducing a fictitious current source at the point of the short circuit.

For the resulting symmetrical circuit, equivalent circuits are drawn up.

The configuration of the equivalent circuits of positive-, negative-, and zero-sequence components is the same for the given problem (Fig. 6).

Fig. 6

From Table 1, the dependences between the symmetrical components of currents and voltages at the fault locations are taken.

Quantities to be summed:

\[ \dot I_p^{(0)} + \dot I_p^{(1)} + \dot I_p^{(2)} = 0, \]

\[ \dot I_0^{(0)} + \dot I_0^{(1)} + \dot I_0^{(2)} = 0, \]

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

Common quantities:

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

\[ \dot U_0^{(0)} = \dot U_0^{(1)} = \dot U_0^{(2)} = \frac{\dot U_0}{3}, \]

\[ \dot I_k^{(0)} = \dot I_k^{(1)} = \dot I_k^{(2)} = \frac{\dot I_k}{3}. \]

\(^1\) The data of the example are borrowed from [2].

Then the equations are set up so that the summed quantities are expressed through the common ones. Since in the problem under consideration there are three faults, the equivalent circuit of each sequence must be regarded as a six-terminal network. The equations characterizing this six-terminal network are easily composed by the superposition method.

\[ \begin{aligned} \dot I_p={}& -\frac{1}{\dot Z_N+\dfrac{\dot Z_M\dot Z_T}{\dot Z_M+\dot Z_T}}\,\dot U_p +\frac{\dfrac{\dot Z_T}{\dot Z_N+\dot Z_T}} {\dot Z_M+\dfrac{\dot Z_N\dot Z_T}{\dot Z_N+\dot Z_T}}\,\dot U_0 +\frac{\dfrac{\dot Z_T\dot Z_M}{\dot Z_T+\dot Z_M}} {\dot Z_M+\dfrac{\dot Z_N\dot Z_T}{\dot Z_T+\dot Z_M}}\,\dot I_k \\ &+\frac{\dot E_N} {\dot Z_N+\dfrac{\dot Z_M\dot Z_T}{\dot Z_M+\dot Z_T}} -\frac{\dot E_M} {\dot Z_M+\dfrac{\dot Z_N\dot Z_T}{\dot Z_N+\dot Z_T}} \cdot \frac{\dot Z_T}{\dot Z_N+\dot Z_T}. \tag{9} \end{aligned} \]

\[ \begin{aligned} \dot I_0={}& \frac{\dfrac{\dot Z_T}{\dot Z_M+\dot Z_T}} {\dot Z_N+\dfrac{\dot Z_T\dot Z_M}{\dot Z_M+\dot Z_T}}\,\dot U_p -\frac{1} {\dot Z_M+\dfrac{\dot Z_N\dot Z_T}{\dot Z_N+\dot Z_T}}\,\dot U_0 +\frac{\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_N}} {\dot Z_M+\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_N}}\,\dot I_k \\ &-\frac{\dot E_N} {\dot Z_N+\dfrac{\dot Z_M\dot Z_T}{\dot Z_M+\dot Z_T}} \cdot \frac{\dot Z_T}{\dot Z_M+\dot Z_T} +\frac{\dot E_M} {\dot Z_M+\dfrac{\dot Z_N\dot Z_T}{\dot Z_N+\dot Z_T}} . \tag{10} \end{aligned} \]

\[ \begin{aligned} \dot U_k={}& \frac{\dfrac{\dot Z_T\dot Z_M}{\dot Z_T+\dot Z_M}} {\dot Z_N+\dfrac{\dot Z_T\dot Z_M}{\dot Z_T+\dot Z_M}}\,\dot U_p -\frac{\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_{MN}}} {\dot Z_M+\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_N}}\,\dot U_0 \\ &+\frac{\dot Z_N\dot Z_T\dot Z_M} {\dot Z_N\dot Z_T+\dot Z_M\dot Z_T+\dot Z_N\dot Z_M}\,\dot I_k +\frac{\dfrac{\dot Z_T\dot Z_M}{\dot Z_T+\dot Z_M}} {\dot Z_N+\dfrac{\dot Z_T\dot Z_M}{\dot Z_T+\dot Z_M}}\,\dot E_N \\ &+\frac{\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_N}} {\dot Z_M+\dfrac{\dot Z_T\dot Z_N}{\dot Z_T+\dot Z_N}}\,\dot E_M . \tag{11} \end{aligned} \]

Substituting the numerical values of the parameters into equations 9, 10, 11, we calculate the coefficients of the six-terminal networks of the positive-, negative-, and zero-sequence networks. Then, by formula (3), we calculate the coefficients of the computational six-terminal network. The results of the calculation are collected in Table 2.

The calculation equations have the form:

\[ j\frac{83}{36.3}\dot U_p -j\frac{41}{72.3}\dot U_0 +\frac{25}{36.3}\dot I_k +\frac{1}{2}=0, \]

\[ -j\frac{41}{72.3}\dot U_p +j\frac{7.41}{144.3}\dot U_0 +\frac{41}{72.3}\dot I_k +\frac{3}{4}=0, \]

\[ -\frac{25}{36.3}\dot U_p -\frac{41}{72.3}\dot U_0 +j\frac{25}{36.3}\dot I_k +j\frac{1}{2}=0. \]

Table 2

The obtained equations are solved for the unknowns—the total quantities

\[ I_k=0;\qquad \dot U_p=j;\qquad \dot U_0=j\dfrac{58}{41}. \]

We note that the results obtained do not coincide with the results of the example in paper [2]. Apparently, an error was made in that paper, since the results of example [2] do not satisfy the voltage equilibrium equation.

References

1. Zelyakh E. V. Fundamentals of the general theory of linear electrical circuits, 1951.
2. Sokolov N. I. Construction and application of complex substitution circuits for complicated asymmetrical circuits. Electricity, No. 8, 1949.