ELECTRICAL CALCULATION OF ASYMMETRICAL POWER TRANSMISSION LINES

(WITH A GROUNDED PHASE)

SHCHERBAKOV V. K.

Professor, Doctor of Technical Sciences

I.

More than 1000 km of three-phase power-transmission lines operate in the USSR in an asymmetrical mode—with a grounded phase, thus using the earth as a current conductor. Since they do not provide protection against lightning strikes, these asymmetrical three-phase lines have no conductors in the grounded phase (the so-called “two wires—earth” system).

In the general case, high-voltage lines with a grounded phase must have conductors suitably arranged on the supports and in the grounded phase for protection of the line against lightning strikes (L1). These conductors may take the form of steel cables, usually suspended above the line solely for protection, or they may be made of nonferrous metal, like the conductors of the ungrounded phases, and in this case are used simultaneously for channeling part of the current of the grounded phase.

The methods known up to now for calculating lines with a grounded phase have in view only two-wire three-phase lines and short lines at nominal voltages up to 35 kV (L2); these methods of calculation do not take into account the capacitive currents of the lines and do not address the question of corona on lines with a grounded phase; nor is the question of determining negative-sequence currents in the generators and receivers of electrical systems having lines with a grounded phase considered.

The present work gives a method and the corresponding formulas for the electrical calculation of the general case of power transmission with a grounded phase. The proposed calculation method takes into account the capacitive currents of the line and makes it possible, comparatively simply, to determine the negative-sequence currents in any element of the power transmission. The work gives formulas for calculating the critical corona voltage and corona losses in lines with a grounded phase; an example is given of the calculation of an asymmetrical power transmission.

II.

The scheme of a three-phase line with a grounded phase is shown in Fig. 1a; a possible arrangement of the conductors on the support of a two-circuit line is shown in Fig. 1b.

Part of the current of the grounded phase flows in the earth, and therefore the resistance of this phase is different from the resistance of the two other phases. The electrical calculation of the line, because of this inequality of phase resistances, becomes more complicated, chiefly owing to the difficulties associated with determining the reactance encountered by the current of phase “a” in the earth.

Wishing to avoid difficulties in determining the resistance of the grounded phase, by the method indicated below the current in the conductor of the grounded phase is determined as part of the total current of this phase, and in what follows all calculations are connected only with the currents in the line conductors. The symmetrical components of the positive- and negative-sequence systems of currents in the conductors completely determine the required voltage drop in the line; determining the zero-sequence component in the phase voltages of the line is meaningless, since zero-sequence currents cannot penetrate into receivers of electric power through transformers with an isolated neutral.

Fig. 1

In the simplest case, when a single conductor–conductor loop has a parallel connection of the earth to one side of the loop (Fig. 2), the current distribution

Fig. 2

between the grounded side of the loop and the earth, under the assumption that the transient grounding resistance is equal to zero, is determined from the relation:

\[ (I_{\mathrm{p}} + I_{\mathrm{e}})\,R_{\mathrm{p}} + I_{\mathrm{p}}\,(R_{\mathrm{p}} + j\omega L_{\mathrm{pp}}) + I_{\mathrm{e}}\,j\omega M = (I_{\mathrm{p}} + I_{\mathrm{e}})\,R_{\mathrm{l}} + \]
\[ + I_{\mathrm{e}}(R_{\mathrm{e}} + j\omega L_{\mathrm{pe}}) + I_{\mathrm{p}}\,j\omega M, \tag{1} \]

where \(I_{\mathrm{p}}, I_{\mathrm{e}}\) and \(R_{\mathrm{p}}, R_{\mathrm{e}}\) are, respectively, the currents and active resistances of the conductor and of the earth;

\(L_{\mathrm{pp}}\) is the self-inductance coefficient of the conductor–conductor loop;

\(L_{\mathrm{pe}}\) is the self-inductance coefficient of the conductor–earth loop;

\(M\) is the mutual-inductance coefficient between the circuits: conductor–conductor and conductor–earth.

From relation (1)

\[ \frac{I_{\mathrm{p}}}{I_{\mathrm{e}}} = \frac{R_{\mathrm{e}} + j(\omega L_{\mathrm{pe}} - \omega M)} {R_{\mathrm{l}} + j(\omega L_{\mathrm{pp}} - \omega M)} = \frac{Z_{\mathrm{pe}} - 0.5 Z_{\mathrm{m}}}{0.5 Z_{\mathrm{nn}}}. \tag{2} \]

Passing to a three-phase line with a grounded phase and considering the current of the grounded phase (phase “\(a\)”) as a doubled zero-sequence current of an unsymmetrical system consisting of the currents of the two other phases:

\[ \left. \begin{array}{c} I_b\\ I_c \end{array} \right\}, \]

we may write, for the zero-sequence currents of each phase, relations similar to (1), whence, after some transformations, we obtain:

\[ \frac{I_{\mathrm{p}}^{(0)}}{I_{\mathrm{e}}^{(0)}} = \frac{Z_a^{(0)} - Z_a^{(1)}}{3Z_a^{(1)}}. \tag{3} \]

Here \(Z^{(0)}_a\) is the total impedance of the conductor of phase “a” to positive-sequence currents,

\(Z^{(0)}_a\) is the total impedance of phase “a” of the line to zero-sequence currents; \(Z^{(0)}_a\) is determined by Carson’s simplified formulas, proposed by Wagner and Evans, as

\[ Z^{(0)}_a = R_a + 0.002965 f + j\,0.008682 f \lg \frac{D_e}{\sqrt[3]{r_e \cdot D_{12} \cdot D_{13}}} \ \Omega/km, \tag{4} \]

or\(^1\)

\[ Z^{(0)}_a = R_a + (3 + 0.0493) + j \omega \left(4.6 \lg \frac{D_e^3}{r_a \cdot D_{12} \cdot D_{12}} + 0.5\right) \cdot 10^{-4}\ \Omega/km. \tag{5} \]

Here \(R_a\) is the active resistance of the conductor of the grounded phase (phase “a”);

\(r_a\) is the radius of the conductor of phase “a”;

\(D_e\) is the depth of the current axis in the earth; approximately, according to Carson,

\[ D_e = 664.1 \sqrt{\frac{\rho}{f}}, \]

where \(\rho\) is the specific resistivity of the earth, and \(f\) is the frequency of the alternating current;

\(D_{12}\) and \(D_{13}\) are the distances between the conductor of phase “a” and the conductors of the two other phases.

Taking into account the influence of the neighboring circuit, Fig. 3,

\[ Z^{(0)}_a = R_a + (6 \times 0.0493) + \]

\[ + j\omega \left(4.6 \lg \frac{D_e}{r_a \cdot D_{12} \cdot D_{13}} \cdot \frac{D_e^3}{D'_{11} \cdot D'_{12} \cdot D'_{13}} + 0.5\right) \cdot 10^{-4}\ \Omega/km. \tag{6} \]

For iron (steel) conductors it is also necessary to take additionally into account the internal inductive reactance from the corresponding tables.

In many cases, when there are no data concerning the conductivity of the earth, the external zero-sequence reactance may be taken approximately as:

a) for single-circuit lines

\[ x^{(0)} \approx 3.5\,x^{(1)}, \]

b) for double-circuit lines

\[ x^{(0)} \approx 5.5\,x^{(1)}, \]

where \(x^{(1)}\) is the reactance to positive-sequence currents.

Fig. 3

Using relation (3), we determine the current in the conductor (in the conductors) of the grounded phase as

\[ I_{a(n)} = I_a \frac{Z^{(0)}_a - Z^{(1)}_a} {3 Z^{(1)}_a + \left(Z^{(0)}_a - Z^{(1)}_a\right)} = m I_a. \tag{7} \]

III.

The currents in the load phases at the end of a power transmission line with a grounded phase constitute an unbalanced system of currents, as the sum of symmetrical systems of positive- and negative-sequence currents. The magnitude of the negative-sequence current, in accordance with the requirements of the “Rules for the Technical Operation of Electric Stations and Networks,” must not exceed, approximately, 5% of the magnitude of the positive-sequence current, and therefore in practical calculations it is entirely permissible to regard the three-phase system

\(^1\) At \(f = 50\) Hz.

the load currents symmetrical. In accordance with this, the following system of currents is represented in the line conductors:

\[ \begin{aligned} &I_a,\\ &I_b=\alpha I_a,\\ &I_c=\alpha^2 I_a, \end{aligned} \tag{8} \]

the symmetrical components of which are determined as

\[ \left. \begin{aligned} I^{(1)}_{a(n)}&=I_a\,\frac{m+2}{3},\\ I^{(2)}_{a(n)}&=I_a\,\frac{m-1}{3},\\ I^{(0)}_{a(n)}&=I_a\,\frac{m-1}{3}. \end{aligned} \right\} \tag{9} \]

The voltage drops in the phases of the line due to the load currents (without taking into account the line conductance currents) may be found as the sum of the corresponding voltage drops of the positive and negative sequences (the zero-sequence voltage drop is not taken into account), i.e.

\[ \left. \begin{aligned} \Delta U_a&=\Delta U^{(1)}_a+\Delta U^{(2)}_a,\\ \Delta U_b&=\alpha\,\Delta U^{(1)}_a+\alpha^2\,\Delta U^{(2)}_a,\\ \Delta U_c&=\alpha^2\,\Delta U^{(1)}_a+\alpha\,\Delta U^{(2)}_a. \end{aligned} \right\} \tag{10} \]

In turn, \(\Delta U^{(1)}_a\) and \(\Delta U^{(2)}_a\), according to the theory of symmetrical components, are determined from the relations:

\[ \left. \begin{aligned} \Delta U^{(1)}_a&=I^{(0)}_{a(n)}\cdot Z^{(1)}+I^{(1)}_{a(n)}\cdot Z^{(0)}+I^{(2)}_{a(n)}\cdot Z^{(2)},\\ \Delta U^{(2)}_a&=I^{(0)}_{a(n)}\cdot Z^{(2)}+I^{(1)}_{a(n)}\cdot Z^{(1)}+I^{(2)}_{a(n)}\cdot Z^{(0)}. \end{aligned} \right\} \tag{11} \]

Here

\[ \left. \begin{aligned} Z^{(0)}&=\frac{Z^{(1)}_a+Z^{(1)}_b+Z^{(1)}_c}{3},\\ Z^{(1)}&=\frac{Z^{(1)}_a+\alpha^2 Z^{(1)}_b+\alpha Z^{(1)}_c}{3},\\ Z^{(2)}&=\frac{Z^{(1)}_a+\alpha Z^{(1)}_b+\alpha^2 Z^{(1)}_c}{3}, \end{aligned} \right\} \tag{12} \]

where the mutual-induction impedances are taken to be the same for all conductors.

Substituting the values of the symmetrical components of the currents and impedances from (9) and (12) into expressions (11), and also taking into account that, upon transposition of the two lower conductors, \(Z^{(1)}_b=Z^{(1)}_c\), we obtain formulas for calculating the symmetrical components of the voltage drop in phase “\(a\)” in the following form:

\[ \left. \begin{aligned} \Delta U^{(1)}_a&=\frac{1}{3}I_a\left(mZ^{(1)}_a+2Z^{(1)}_b\right),\\ \Delta U^{(2)}_a&=\frac{1}{3}I_a\left(mZ^{(1)}_a-Z^{(1)}_b\right). \end{aligned} \right\} \tag{13} \]

From (13)—

\[ \Delta U_a=\Delta U^{(1)}_a+\Delta U^{(2)}_a=\frac{1}{3} I_a(2mZ_a+Z_b). \tag{14} \]

Here and below the signs at the resistances to positive-sequence currents are omitted.

The voltage drop in phase “b” is

\[ \Delta U_b=\alpha\cdot \Delta U^{(1)}_a+\alpha^2\cdot \Delta U^{(2)}_a, \]

or, superposing the system of current and voltage vectors of phase “b” and phase “a,” we obtain

\[ \alpha^2\cdot \Delta U_b=\Delta U^{(1)}_a+\alpha\cdot \Delta U^{(2)}_a, \tag{15} \]

or

\[ \alpha^2\cdot \Delta U_b=\left(\Delta U^{(1)}_a-\frac{1}{2}\Delta U^{(2)}_a\right)-j\frac{\sqrt{3}}{2}\Delta U^{(2)}_a. \tag{16} \]

The same holds for phase “c”:

\[ \Delta U=\alpha^2\cdot \Delta U^{(1)}_a+\alpha\cdot \Delta U^{(2)}_a; \tag{17} \]

or

\[ \alpha\cdot \Delta U_c=\left(\Delta U^{(1)}_a-\frac{1}{2}\Delta U^{(2)}_a\right)+j\frac{\sqrt{3}}{2}\Delta U^{(2)}_a. \tag{18} \]

IV.

In the particular case when there are no conductors in the grounded phase (Fig. 4), \(Z_a=\infty\), \(m=0\), and expressions (13) give an indeterminacy. To avoid this indeterminacy, let us write the expression for the product \(m\cdot Z_a\). According to (7),

\[ mZ_a=\frac{(Z^{(0)}_a-Z_a)\cdot Z_a}{3Z_a+(Z^{(0)}_a-Z_a)}; \]

in the limit, as \(Z_a\) tends to infinity,

\[ mZ_a=\frac{Z^{(0)}_a-Z_a}{3}=\frac{Z^{(0)}_b-Z_b}{3}, \tag{19} \]

since the difference \(Z^{(0)}_a-Z_a\) is a finite quantity and does not depend on the active resistance and radius of the conductor.

Fig. 4

After substituting \(mZ_a\) from (19), formulas (12) and (14) for calculating the voltage drops in a three-phase line with a grounded phase, when there are no conductors in the grounded phase, take the form:

\[ \left. \begin{aligned} \Delta U^{(1)}_a&=\frac{1}{9}I_a\left(Z^{(0)}_b+5Z_b\right),\\ \Delta U^{(2)}_a&=\frac{1}{9}I_a\cdot\left(Z^{(0)}_b-4Z_b\right); \end{aligned} \right\} \tag{20} \]

\[ \Delta U_a=\frac{1}{9}\left(2Z^{(0)}_b+Z_b\right). \tag{21} \]

Fig. 5

V.

For the purposes of calculating capacitive currents in a line with a grounded phase, it may be assumed that a symmetrical system of voltages is applied to the line. With reference to Fig. 5

the charges and potentials of the conductors of a three-phase single-circuit line are related through the partial capacitances \(C_{11}, C_{22}, C_{33}, C_{12}, C_{13}\), and \(C_{23}\) by the equations:

\[ \begin{aligned} q_1 &= C_{11} \cdot U_{10} + C_{12}\cdot (U_{10}-U_{20}) + C_{13}(U_{10}-U_{30}),\\ q_2 &= C_{22} \cdot U_{20} + C_{12}\cdot (U_{20}-U_{10}) + C_{23}(U_{20}-U_{30}),\\ q_3 &= C_{33} \cdot U_{30} + C_{23}\cdot (U_{30}-U_{20}) + C_{13}(U_{30}-U_{10}). \end{aligned} \tag{22} \]

Bearing in mind that in a line with a grounded phase

\[ U_{10}=U_a-U_a=0, \]

\[ U_{20}=U_b-U_a=(a-1)\cdot U_a, \]

\[ U_{30}=U_c-U_a=(a^2-1)U_a, \]

from (22) we find the expressions for the capacitive currents of the line phases:

\[ \begin{aligned} I_{a(c)} &= -(I_{b(c)}+I_{c(c)}) = j\,3(\omega C_{12}U_a+\omega C_{22}U_a),\\ I_{b(c)} &= \sqrt{3}\,[\omega C_{23}\cdot U_a + j\omega(C_{12}+C_{22})\cdot U_a \angle -150^\circ],\\ I_{c(c)} &= \sqrt{3}\,[-\omega C_{23}\cdot U_a + j\omega(C_{13}+C_{33})\cdot U_a \angle 150^\circ]. \end{aligned} \tag{23} \]

The vector diagram of the line capacitive currents is given in Fig. 6.

As with the load current, we separate the \(m\)-th part of the capacitive current of the grounded phase (phase “a”), flowing in the conductor of this phase, and then decompose the asymmetrical system of capacitive currents in the line conductors into symmetrical components:

Fig. 6

\[ \begin{aligned} I^{(1)}_{a(c)(n)} &= (m+1)(I_{c12}+I_{c22})+I_{c23},\\ I^{(2)}_{a(c)(n)} &= m(I_{c12}+I_{c22})-I_{c23},\\ I^{(0)}_{a(c)(n)} &= (m-1)(I_{c12}+I_{c22}). \end{aligned} \tag{24} \]

In expressions (24) it is assumed that:

\[ \begin{aligned} I_{c12} &= j\omega C_{12}\cdot U_a,\\ I_{c22} &= j\omega C_{22}\cdot U_a,\\ I_{c23} &= j\omega C_{23}\cdot U_a, \end{aligned} \tag{25} \]

and, bearing in mind the transposition of the conductors of the ungrounded phases, \(I_{c12}=I_{c13}\), \(I_{c22}=I_{c33}\).

In accordance with the method already set forth above, for the Π-shaped equivalent circuit of the line, we obtain the voltage drop in phase “a” due to the capacitive current:

\[ \begin{aligned} \Delta U^{(1)}_{a(c)} &= mZ_a\cdot (I'_{c12}+I'_{c22})+Z_b\cdot I'_{c12}+I'_{c22}+I'_{c23});\\ \Delta U^{(2)}_{a(c)} &= mZ_a\,(I'_{c12}+I'_{c22})-Z_b I'_{c23};\\ \Delta U_{a(c)} &= \Delta U^{(1)}_{a(c)}+\Delta U^{(2)}_{a(c)} =(2mZ_a+Z_c)(I'+I'_{c22}); \end{aligned} \tag{26} \]

\[ I'_{c12}=\frac{I_{c12}}{2},\quad I'_{c22}=\frac{I_{c22}}{2}\ \text{and}\ I'_{c23}=\frac{I_{c23}}{2}. \]

The voltage drop in phases “b” and “c” due to the capacitive currents can be found through the symmetrical components of the voltage drop in phase “a”:

\[ \left. \begin{aligned} a^2\Delta U_{b(c)}&=\left(\Delta U^{(1)}_{a(c)}-\frac{1}{2}\Delta U^{(2)}_{a(c)}\right)-j\frac{\sqrt{3}}{2}\Delta U^{(2)}_{a(c)},\\ a\Delta U_{c(c)}&=\left(\Delta U^{(1)}_{a(c)}-\frac{1}{2}\Delta U^{(2)}_{a(c)}\right)+j\frac{\sqrt{3}}{2}\Delta U^{(2)}_{a(c)}. \end{aligned} \right\} \tag{27} \]

The total voltage drop in each phase (with the zero-sequence component excluded) will be obtained as the sum of the voltage drops due to the load current and the capacitive current.

For a two-wire three-phase line with a grounded phase, Fig. 4, \(I_{c12}=0\), and the capacitive currents in the phases are therefore, according to (23), calculated as

\[ \left. \begin{aligned} I_{b(c)}&=\sqrt{3}\left(\omega C_{23}U_a+\omega C_{22}U_a\angle -60^\circ\right),\\ I_{c(c)}&=\sqrt{3}\left(-\omega C_{23}U_a+\omega C_{22}U_a\angle -120^\circ\right), \end{aligned} \right\} \tag{28} \]

where

\[ C_{23}=\frac{-a_{23}}{a_{22}^{2}-a_{23}^{2}}\cdot\frac{1}{9\cdot10^{6}}\ \frac{\mathrm{F}}{\mathrm{km}}, \tag{29} \]

and

\[ C_{22}=\frac{1}{a_{22}+a_{23}}\cdot\frac{1}{9\cdot10^{6}}\ \frac{\mathrm{F}}{\mathrm{km}}, \tag{30} \]

where

\[ a_{23}=2\ln\frac{h_{23}}{D_{23}},\qquad a_{22}=\frac{h_{22}}{r}, \]

Fig. 7.

Fig. 7

The voltage drop in phase “a” due to the capacitive current, bearing in mind that

\[ mZ_a=\frac{Z_b^{(0)}-Z_b}{3}, \]

is obtained from (28) as

\[ \left. \begin{aligned} \Delta U^{(1)}_{a(c)}&=\frac{1}{3}\left[Z_b^{(0)}I'_{c22}+Z_b\left(2I'_{c22}+3I'_{c23}\right)\right],\\ \Delta U^{(2)}_{a(c)}&=\frac{1}{3}\left[Z_b^{(0)}I'_{c22}-Z_b\left(I'_{c22}+3I'_{c23}\right)\right],\\ \Delta U_{a(c)}&=\Delta U^{(1)}_{a(c)}+\Delta U^{(2)}_{a(c)} =\frac{1}{3}\left(2Z_b^{(0)}+Z_b\right)I'_{c22}. \end{aligned} \right\} \tag{31} \]

VI.

The “Rules for the Technical Operation of Electric Stations and Networks” permit a current asymmetry in generators of 10%, which is equivalent to the presence of a negative-sequence current component of slightly more than 5%. Therefore, the electrical calculation of an asymmetrical power transmission must provide for a mandatory check of the magnitude of current asymmetry in the elements of the electrical system supplying the asymmetrical power transmissions.

The negative-sequence current in a series electrical circuit can be determined, if the phase impedances are identical, as

\[ I^{(2)}=\frac{\Sigma U^{(2)}}{\Sigma Z^{(2)}}, \]

and, in turn, the negative-sequence voltage as

\[ \Sigma U^{(2)}=E^{(2)}-\Sigma \Delta U^{(2)}=0-\Sigma \Delta U^{(2)}=-\Sigma \Delta U^{(2)}. \]

The negative-sequence voltage drop in the line is determined by formulas (13), (20), (26), and (31), depending on whether the capacitive current is taken into account in the calculation of the line and whether there are conductors in the grounded phase. The negative-sequence voltage drop in transformers, generators, and symmetrical lines due to the unbalanced system of capacitive currents of the line with a grounded phase can be found as:

\[ \Delta U_{\mathrm{T}}^{(2)}=I_{\mathrm{T(c)}}^{(2)}\cdot Z_{\mathrm{T}}^{(2)},\quad \Delta U_{\mathrm{g}}^{(2)}=I_{\mathrm{g(c)}}^{(2)}\cdot Z_{\mathrm{g}}^{(2)} \quad \text{etc.}, \]

where \(I_{\mathrm{T(c)}}^{(2)}\), \(I_{\mathrm{g(c)}}^{(2)}\), etc. are the negative-sequence currents in transformers, generators, etc., due to the unbalanced system of capacitive currents of the line with a grounded phase; the negative-sequence component of the capacitive current in phase “a” of a line with a grounded phase (current in the conductor \(+\) current in the earth) is

\[ I_{\mathrm{a(c)}}^{(2)} = \frac{I_{\mathrm{a(c)}}+\alpha I_{\mathrm{b(c)}}+\alpha^{2}I_{\mathrm{c(c)}}}{3} = I_{\mathrm{c12}}+I_{\mathrm{c22}}-I_{\mathrm{c23}}. \tag{32} \]

The current \(I_{\mathrm{a(c)}}^{(2)}\) flows in the generators and transformers connected between the generators and the line with a grounded phase, and produces negative-sequence voltage drops.

Thus, when calculating the total negative-sequence current in any element of the system, it is necessary to take into account not only the current determined as

\[ I^{(2)}=\frac{\Sigma U^{(2)}}{\Sigma Z^{(2)}}, \]

but also the negative-sequence capacitive current, if it occurs in the given element of the system.

The negative-sequence impedances of generators and transformers are usually known. The impedances of the phases of a line with a grounded phase are unequal and therefore, strictly speaking, it would be necessary to take into account the appearance also of positive-sequence currents when a negative-sequence voltage is applied to the line, but this would greatly complicate the solution of the problem. From a practical point of view, the accuracy of the solution will hardly suffer if, in the given case, it is assumed that the phase impedances of the line to positive- and negative-sequence currents are identical and equal to \(Z_b=Z_c\).

What has been said above regarding the calculation of the negative-sequence current in the elements of the power-transmission system (electrical system) is illustrated by the diagram of Fig. 8, in which

\[ I_{\mathrm{g}}^{(2)}=I_{\mathrm{T}}^{(2)}=I_{\mathrm{c}}^{(2)} + \frac{ U_{\mathrm{g}}^{(2)}+U_{\mathrm{T}}^{(2)}+U_{\mathrm{l}}^{(2)} }{ Z_{\mathrm{g}}^{(2)}+Z_{\mathrm{T}}^{(2)}+Z_{\mathrm{l}}^{(2)}+Z_{\mathrm{T}}^{(2)}+Z_{\mathrm{H}}^{(2)} }. \]

\(Z_{\mathrm{H}}^{(2)}\) is the load impedance to negative-sequence currents; if more exact data are unavailable, it may be taken as equal to \(\sim(15\text{–}20)+j(40\text{–}45)\%\). In the case of a complex closed system, the distribution of negative-sequence currents is conveniently found by the method of self and mutual impedances.

VII

In many cases it is sufficient to determine only the so-called longitudinal component of the voltage drop, which can then be equated to the voltage loss. Below are given the necessary formulas for determining

Fig. 8

the longitudinal component of the voltage drop in the line; in these formulas it is assumed that

\[ mZ_a = R'_a + jX'_a;\quad Z_b = R_b + jX_b;\quad \dot{\Psi}_a=\frac{P_2-jQ_2}{\sqrt{3}U_2}. \]

a) Without taking account of the capacitive currents of the line.
In phase “a” (grounded), from (13),

\[ \left. \begin{aligned} \sqrt{3}\,\Delta U^{(1)}_{a(\mathrm{long})} &=\frac{1}{3}\cdot \frac{P_2(R'_a+2R_b)+Q_2(X'_a+2X_b)}{U_2},\\[4pt] \sqrt{3}\,\Delta U^{(2)}_{a(\mathrm{long})} &=\frac{1}{3}\cdot \frac{P_2(R'_a-R_b)+Q_2(X'_a-X_b)}{U_2},\\[4pt] \sqrt{3}\,\Delta U_{a(\mathrm{long})} &=\frac{1}{3}\cdot \frac{P_2(2R'_a+R_b)+Q_2(2X'_a+X_b)}{U_2}. \end{aligned} \right\} \tag{33} \]

In phase “b” from (16),

\[ \sqrt{3}\,\Delta U_{b(\mathrm{long})} = \sqrt{3}\left(\Delta U^{(1)}_{a(\mathrm{long})} -\frac{1}{2}\Delta U^{(2)}_{a(\mathrm{long})}\right) + \frac{1}{2\sqrt{3}}\cdot \frac{P_2(X'_a-X_b)-Q_2(R'_a-R_b)}{U_2}. \tag{34} \]

In phase “c” from (18),

\[ \sqrt{3}\,\Delta U_{c(\mathrm{long})} = \sqrt{3}\left(\Delta U^{(1)}_{a(\mathrm{long})} -\frac{1}{2}\Delta U^{(2)}_{a(\mathrm{long})}\right) - \frac{1}{2\sqrt{3}}\cdot \frac{P_2(X'_a-X_b)-Q_2(R'_a-R_b)}{U_2}. \tag{35} \]

When there are no wires in the grounded phase (Fig. 4), in accordance with (20) and (21),

\[ \left. \begin{aligned} \sqrt{3}\,\Delta U^{(1)}_{a(\mathrm{long})} &=\frac{1}{9}\cdot \frac{P_2(R^{(0)}_b+5R_b)+Q_2(X^{(0)}_b+5X_b)}{U_2},\\[4pt] \sqrt{3}\,\Delta U^{(2)}_{a(\mathrm{long})} &=\frac{1}{9}\cdot \frac{P_2(R^{(0)}_b-4R_b)+Q_2(X^{(0)}_b-4X_b)}{U_2},\\[4pt] \sqrt{3}\,\Delta U_{a(\mathrm{long})} &=\frac{1}{9}\cdot \frac{P_2(2R^{(0)}_b+R_b)+Q_2(2X^{(0)}_b+X_b)}{U_2}. \end{aligned} \right\} \tag{36} \]

\[ \sqrt{3}\,\Delta U_{b(\text{long})} = \sqrt{3}\left(\Delta U^{(1)}_{a(\text{long})} -\frac{1}{2}\Delta U^{(2)}_{a(\text{long})}\right) + \frac{1}{6\sqrt{3}}\, \frac{P_2\left(X_b^{(0)}-4X_b\right)-Q_2\left(R_b^{(0)}-4R_b\right)}{U_2}; \tag{37} \]

\[ \sqrt{3}\,\Delta U_{c(\text{long})} = \sqrt{3}\left(\Delta U^{(1)}_{a(\text{long})} -\frac{1}{2}\Delta U^{(2)}_{a(\text{long})}\right) - \frac{1}{6\sqrt{3}}\, \frac{P_2\left(X_b^{(0)}-4X_b\right)-Q_2\left(R_b^{(0)}-4R_b\right)}{U_2}. \tag{38} \]

b) Taking into account capacitive currents, the longitudinal components of the voltage drops in the line:

In phase “a,” according to (26),

\[ \left. \begin{aligned} \Delta U^{(1)}_{a(c)(\text{long})} &= -\left[X'_a\left(I'_{c12}+I'_{c22}\right) +X_b\left(I'_{c12}+I'_{c22}+I'_{c23}\right)\right],\\ \Delta U^{(2)}_{a(c)(\text{long})} &= -\left[X'_a\left(I'_{c12}+I'_{c22}\right)-X_b I'_{c23}\right],\\ \Delta U_{a(c)(\text{long})} &= -\left[(2X'_a+X_b)\left(I'_{c12}+I'_{c22}\right)\right]. \end{aligned} \right\} \tag{39} \]

In phases “b” and “c,” according to 27,

\[ \Delta U_{b(c)(\text{long})} = \Delta U^{(1)}_{a(c)(\text{long})} -\frac{1}{2}\Delta U^{(2)}_{a(c)(\text{long})} + \frac{\sqrt{3}}{2}\left[R'_a\left(I'_{c12}+I'_{c22}\right)-R_b I'_{c23}\right]; \tag{40} \]

\[ \Delta U_{c(c)(\text{long})} = \Delta U^{(1)}_{a(c)(\text{long})} -\frac{1}{2}\Delta U^{(2)}_{a(c)(\text{long})} - \frac{\sqrt{3}}{2}\left[R'_a\left(I'_{c12}+I'_{c22}\right)-R_b I'_{c23}\right]. \tag{41} \]

In cases where there are no conductors in the grounded phase, Fig. 4, according to (31),—

\[ \left. \begin{aligned} \Delta U^{(1)}_{a(c)(\text{long})} &= \frac{1}{3}\left[\left(X_b^{(0)}+2X_b\right)I'_{c22} +3X_b I'_{c23}\right],\\ \Delta U^{(2)}_{a(c)(\text{long})} &= \frac{1}{3}\left[\left(X_b^{(0)}-X_b\right)I'_{c22} -3X_b I'_{c23}\right],\\ \Delta U_{a(c)(\text{long})} &= \frac{1}{3}\left(2X_b^{(0)}+X_b\right)I'_{c22}. \end{aligned} \right\} \tag{42} \]

VIII.

Corona on lines with a grounded phase, from the quantitative standpoint, manifests itself differently than on symmetrical lines, and therefore other relations are required for taking corona into account in the electrical calculation of lines with a grounded phase.

In practical calculations it is customary to assume that the electric field around the line conductors is uniform. On this basis, the relation between the electric-field intensity and the charge on the conductor may be written in the form

\[ E=\frac{q}{2\pi r}, \]

where \(r\) is the radius of the conductor. Since, on the other hand,

\[ q=C_{\text{phase}}U_{\text{ph}}, \]

\[ \dot U_{\phi}=2\pi\cdot \frac{r}{C_{\text{op}}}\cdot \dot E . \]

The critical voltage will correspond to the critical disruptive field strength

\[ U_{\text{cr}(\phi)}=2\pi\cdot \frac{r}{C_{\text{op}}}\cdot E_{\text{cr}} . \]

And if, according to Peek, \(E_{\text{cr}}=21.1\delta\) (\(\delta\) is the air density), \(r\) is substituted in centimeters, \(C_{\text{op}}\) in farads per kilometer, and coefficients are introduced that take into account the weather condition, \(m_n\), the conductor surface, \(m_0\), and the conversion to practical units of measurement, \(8.84\cdot 10^{-6}\), then we obtain the calculation formula for the critical corona voltage in the following form:

\[ U_{\text{cr}(\phi)} = m_0\cdot m_n\cdot \delta\cdot 8.84\cdot 10^{-6}\cdot 2\pi\cdot \frac{r}{C_{\text{op}}}\cdot 10^{-3} = \]

\[ = m_0\cdot m_n\cdot \delta\cdot 1.17\,\frac{r}{C_{\text{op}}}\cdot 10^{-6}\ \text{kV}. \tag{43} \]

Formula (43) is equally suitable both for a symmetrical line and for a line with a grounded phase. The operating capacitance of the ungrounded conductors of a line with a grounded phase may be calculated by the following formula, obtained from relation (22):

\[ C_{\text{op}}= \frac{1}{2} \sqrt{ 9(C_{12}+C_{22})^2+3(C_{12}+C_{22}+C_{23})^2 }. \tag{44} \]

Calculations of operating capacitances for specific examples of high-voltage power-transmission lines show that, all other conditions being equal, the operating capacitance of the ungrounded conductors of a line with a grounded phase is greater than the operating capacitance of the conductors of a symmetrical line by 24–32% in the case of single-circuit lines, and by 15–25% in the case of double-circuit lines. Consequently, according to (43), in the same ratio the critical corona voltage of the conductors of a line with a grounded phase is lower than in the case of symmetrical lines, which creates difficulties in the use of AC-95 conductors in lines with a grounded phase in the case of single-circuit lines, and AC-70 conductors in the case of double-circuit lines.

Power losses due to corona in a line are usually calculated by Peek’s formula:

\[ P_{\text{cor}}= \frac{241}{\delta}(f+25)\sqrt{\frac{r}{\delta}}\,(U-U_{\text{cr}})^2\cdot 10^{-5} \ \text{kW/km}. \]

This formula, being experimental, does not directly indicate the mechanism of formation of corona losses. However, from a whole series of considerations, on which there is no possibility of dwelling in this brief article, and also using Holm’s reasoning on this question, one may come to the conclusion that corona losses in a line are proportional to the charges of the line conductors, all other conditions remaining unchanged. This proposition was partially verified by the author by experiment on one laboratory line; the experimental results closely coincided with those calculated by (13). Therefore, for calculating corona losses in the conductors of lines with a grounded phase, Peek’s formula may be recommended, supplemented by a factor in the form of the ratio

\[ \frac{C_{\text{op}(\text{asym})}}{C_{\text{op}(\text{sym})}} = \frac{C'_p}{C_p}, \]

where \(C_r\) and \(C_p\) are the operating capacitances, respectively, of the asymmetrical and symmetrical lines, for identical line dimensions.

The operating capacitance of a symmetrical three-phase line is

\[ C_p = C_{22} + 3C_{12}. \]

IX.

Example of an electrical calculation of a power transmission line with a grounded phase (Fig. 9).
Variant 1: the grounded phase has conductors—\(2 \times \mathrm{AC}\text{-}95\). The resistances of the line phases:

Fig. 9.

\[ Z_a = R_a + jX_a = \frac{1}{2}\cdot 100(0{,}310 + j0{,}450)=15{,}5+j22{,}5\ \Omega; \]

\[ Z_b = R_b + jX_b = \frac{1}{2}\cdot 100(0{,}165 + j0{,}42)=8{,}25+j21{,}0\ \Omega. \]

For \(\rho_z = 100\ \Omega/\mathrm{m}^3\), \(D_e \approx 1000\ \mathrm{m}\), and by formula (5),

\[ Z_a^{(0)}=\frac{1}{2}\cdot 100(0{,}310+6\cdot 0{,}0493+j2{,}28)=30{,}3+j114\ \Omega. \]

The fraction of current in the conductors of the grounded phase (phase “\(a\)”) according to (7) is

\[ m=\frac{Z_a^{(0)}-Z_a}{3Z_a+\left(Z_a^{(0)}-Z_a\right)} = \frac{(30{,}3+j114{,}0)-(15{,}5+j22{,}5)} {3(15{,}5+j22{,}5)+14{,}8+j91{,}5} = \]

\[ =0{,}553\angle 12^\circ 15'. \]

\[ mZ_a = R'_a+jX'_a = 0{,}553\angle 12^\circ 15'\,(15{,}5+j22{,}5) = 15{,}1\angle 67^\circ 15' = 5{,}82+j13{,}9\ \Omega. \]

From Maxwell’s potential equations (see any course in the theoretical foundations of electrical engineering) we find the determinants and the corresponding minors, from which the partial capacitances of the line are then easily determined:

\[ C_{12}=C_{13}=0{,}00154\cdot 10^{-6}\ \mathrm{F/km}, \]

assuming that the ungrounded conductors are transposed,

\[ C_{22}=C_{33}=0{,}00343\cdot 10^{-6}\ \mathrm{F/km}, \]

\[ C_{23}=0{,}00170\cdot 10^{-6}\ \mathrm{F/km}; \]

and the partial capacitive currents:

\[ I_{c12}=j2\cdot100\cdot314\cdot0,00154\cdot10^{-6}\cdot\frac{110}{3}=j0,0062\ \text{kA}, \]

\[ I_{c22}=j0,0137\ \text{kA}, \]

\[ I_{c23}=0,0068\ \text{kA}. \]

The symmetrical components of the system of capacitive currents are obtained, according to (24):

\[ I_{a(c)}^{(1)}=2(I_{c12}+I_{c22})+I_{c23}=j0,0268\ \text{kA}, \]

\[ I_{a(c)}^{(2)}=(I_{c12}+I_{c22}-I_{c23})=j0,0131\ \text{kA}. \]

Let us find the voltage drops in the line phases. In accordance with (13) and (26), for a Π-shaped equivalent circuit of the line,—

\[ \Delta U_a^{(1)}= \frac{1}{3}I_a(mZ_a+2Z_b)+ \left[mZ_a(I'_{c12}+I'_{c22})+Z_b(I'_{c12}+I'_{c22}+I'_{c23})\right]= \]

\[ =\frac{1}{3}\cdot \frac{40(0,85-j0,52)}{0,85\sqrt{3}\cdot105} (5,82+j13,9+2\cdot8,25+j2\cdot21)+ \]

\[ +\left[(5,82+j13,9)\cdot\frac{1}{2}(0,0062+0,0137) +(8,25+j21,0)\cdot\frac{1}{2}(0,0062+\right. \]

\[ \left.+0,0137+0,0068)\right] =(4,15+j3,33)+(-0,42+j0,17)=3,73+j3,50\ \text{kV}; \]

\[ \Delta U_a^{(2)}= \frac{1}{3}I_a(mZ_a-Z_b)+ \left[mZ_a(I'_{c12}+I'_{c23})-Z_bI'_{c23}\right]= \]

\[ =(-0,50-j0,41)+(-0,03+j0,02)= \]

\[ =-0,53-j0,39\ \text{kV}; \]

voltage drops in the grounded phase—

\[ \Delta U_a=\Delta U_a^{(1)}+\Delta U_a^{(2)}=3,20+j3,11\ \text{kV}; \]

voltage drops in the ungrounded phases:

\[ a^2\Delta U_b=\Delta U_a^{(1)}+\frac{1}{2}\Delta U_a^{(2)} -j\frac{\sqrt{3}}{2}\Delta U_a^{(2)} =(3,73+j3,50)- \]

\[ -\frac{1}{2}(-0,53-j0,39) -j\frac{\sqrt{3}}{2}(-0,53-j0,39)= \]

\[ =3,66+j4,16\ \text{kV}; \]

\[ a\Delta U_c=\Delta U_a^{(1)}-\frac{1}{2}\Delta U_a^{(2)} +j\frac{\sqrt{3}}{2}\Delta U_a^{(2)} =4,34+j3,24\ \text{kV}. \]

The voltage drop in transformer \(T_1\) and in the generator due to the negative-sequence system of capacitive currents:

\[ \Delta U_{aT}^{(1)}=Z_T\cdot I_{a(c)}^{(1)} =(1,57+j27)\cdot j0,0268\approx-0,724\ \text{kV}; \]

\[ \Delta U_{aT}^{(2)}=Z_T\cdot I_{a(c)}^{(2)} =(1,57+j27)\cdot j0,0131\approx-0,354\ \text{kV}; \]

\[ \Delta U_{aT}=(-0,724)+(-0,354)=-1,078\ \text{kV}; \]

\[ a^2\Delta U_{bT}=\Delta U_{aT}^{(1)}-\frac{1}{2}\Delta U_{aT}^{(2)}-j\frac{\sqrt{3}}{2}\Delta U_{aT}^{(2)}=-0.547+j0.306\ \text{kV}; \]

\[ a\Delta U_{cT}=\Delta U_{aT}^{(1)}-\frac{1}{2}\Delta U_{aT}^{(2)}+j\frac{\sqrt{3}}{2}\Delta U_{aT}^{(2)}=-0.547-j0.306\ \text{kV}; \]

\[ \Delta U_{a(T)}^{(1)}=Z_{\Gamma}\cdot I_{a(c)}^{(1)}\simeq j34.7\cdot j0.0268=-0.93\ \text{kV}; \]

\[ \Delta U_{aT}^{(2)}Z_{\Gamma}\cdot I_{a(c)}^{(2)}\simeq j34.7\cdot j0.0131=-0.455\ \text{kV}. \]

Bearing in mind that in generators \(Z_{\Gamma}^{(2)}\) is, approximately, one tenth of \(Z_{\Gamma}^{(1)}\) (synchronous reactance), while \(I_{\Gamma}^{(1)}\) is tens of times greater than \(I_{\Gamma}^{(2)}\), there is no point in practical calculations in taking into account the negative-sequence voltage drop in the generators; it may be assumed that at the generator terminals there is a symmetrical system of phase voltages.

We find the negative-sequence current in the generator due to the asymmetry of the transmission phase voltages from the equivalent circuit of the transmission line (Fig. 10). In this circuit the negative-sequence voltages of the transmission-line elements, calculated above, and the resistances to negative-sequence currents are indicated.

Fig. 10

From the circuit we find the current in the generator, equal to

\[ 3.15-j5.58\ \text{A}, \]

where the negative-sequence resistance of the loads is taken as

\[ Z_{\mathrm{H}}=(17+j45)\%. \]

The total negative-sequence current in the generator, taking into account the capacitive current of the line, is

\[ I_{\Gamma}^{(2)}=3.15-j5.58+j\,0.0131\cdot 10^3=3.15+j7.52=8.17\angle\ \text{A}. \]

Current asymmetry in the generator—

\[ \frac{I_{\Gamma}^{(2)}}{I_{\Gamma}^{(1)}}\cdot 100 \simeq \frac{I_{\Gamma}^{(2)}}{I_{(n)}}\cdot 100 = \frac{8.17\cdot \sqrt{3}\cdot 121\cdot 100}{58.9\cdot 10^3} =2.91\%. \]

Variant 2: the grounded phase has no conductors.

In this case

\[ mZ_a=\frac{1}{3}\left(Z_b^{(0)}-Z_b\right); \]

\[ Z_b^{(0)}=\frac{1}{2}\left(8.25+100\cdot 6\cdot 0.0493+j2.25\right)=23+j112.5\ \Omega. \]

Partial capacitances—

\[ C_{22}= \frac{1}{(a_{22}+a_{22}')+(a_{23}+a_{23}')} \cdot \frac{1}{9\cdot 10^6} = 0.00428\cdot 10^{-6}\ \text{F/km}; \]

\[ C_{23}=-\frac{-(a_{23}+a'_{23})}{(a_{22}+a'_{22})^2-(a_{23}+a'_{23})^2}\cdot \frac{1}{9\cdot 10^6} =0.00263\cdot 10^{-6}\ \mathrm{F/km}. \]

Partial capacitive currents—

\[ I_{c23}=j\,0.0102\ \mathrm{kA},\qquad I_{c22}=j\,0.0171\ \mathrm{kA}. \]

Symmetrical components of the system of capacitive currents

\[ I^{(1)}_{a(c)}=j\,0.0444\ \mathrm{kA},\qquad I^{(2)}_{a(c)}=j\,0.0069\ \mathrm{kA}. \]

Voltage drops in the phases of the line in accordance with (20) and (31), for the $\Pi$-shaped equivalent circuit of the line—

\[ \begin{aligned} \Delta U^{(1)}_a &=\frac{1}{9} I_a\left(Z_b^{(0)}+5Z_b\right) +\left[\frac{1}{3}\left(Z_b^{(0)}-Z_b\right) I_{c22} +Z_b\left(I_{c23}+I_{c22}\right)\right] \\ &=\frac{1}{9}(0.220-j0.135)(23+j112.5+5\cdot 8.25+j5\cdot 21.0)+ \\ &\quad+\left[(23.0+j112.5-8.25-j21.0)\,j0.0171 +(8.25+j21.0)\,j(0.0171+0.0102)\right] \\ &=3.78+j4.66\ \mathrm{kV}; \end{aligned} \]

\[ \Delta U^{(2)}_a =\frac{1}{9} I_a\left(Z_b^{(0)}-4Z_b\right) +\left[\frac{1}{3}\left(Z_b^{(0)}-Z_b\right) I_{c22} -Z_b I_{c23}\right] =-0.124+j0.845\ \mathrm{kV}; \]

\[ \Delta U_a=3.66+j5.59\ \mathrm{kV}; \]

\[ a^2\Delta U_b=4.57+j4.34\ \mathrm{kV}; \]

\[ a\Delta U_c=3.01+j4.13\ \mathrm{kV}. \]

Voltage drop in the transformer due to capacitive currents—

\[ U\Delta^{(1)}_{a(c)}=Z_T I^{(1)}_{a(c)}\simeq j27\cdot j0.044=-1.24\ \mathrm{kV}; \]

\[ U\Delta^{(2)}_{a(c)}=Z_T I^{(2)}_{a(c)}\simeq j27\cdot j0.0069=-0.186\ \mathrm{kV}. \]

Negative-sequence voltage drop in the generator—

\[ \Delta U^{(2)}_{a(c)}=Z_p I^{(2)}_{a(c)}\simeq j34.7\cdot j0.0069=-0.24\ \mathrm{kV}. \]

Analogously to the first variant, we find the negative-sequence current in the generator—

\[ I^{(2)}_g=-3.1+j3.7\ \mathrm{A}; \]

\[ \frac{I^{(2)}_g}{I^{(1)}_g}\,100=1.7\%. \]

LITERATURE

1. V. K. Shcherbakov.—Lines using the conductor of one of the phases also as protection against lightning strokes. Proceedings of the Tomsk Industrial Institute, vol. 59, 1941.

2. Eng. L. E. Ebin and Eng. Sh. M. Aluker.—A guide to the electrical calculation of three-phase networks using the earth as one of the conductors. Publishing House of the NKZem of the USSR, 1940.

3. V. K. Shcherbakov.—Corona on a line with a grounded phase. Proceedings of the Tomsk Polytechnic Institute, vol. 63.