JPH0667105B2 - Zero-phase circulating current compensation method for ground relay - Google Patents

Description

TECHNICAL FIELD The present invention relates to a zero-phase circulating current compensation method for a ground fault relay,
More specifically, a ground fault that interrupts the transmission line that has a ground fault by detecting that a one-line ground fault has occurred in a high-resistance grounding parallel two-line transmission line that is parallel to at least one transmission line. The present invention relates to a zero-phase circulating current compensating method for removing a zero-phase circulating current that impairs the operation of the ground-fault relay so that the ground-fault relay can be operated only by a true fault current.

<Prior Art> In the ground fault protection of a high resistance grounding parallel two-line transmission line,
Concept of installing a ground fault circuit selection relay that selectively shuts down a failed circuit by using the differential current of both lines as a main protection relay, and a direction ground fault relay that is also applied to single-line transmission lines as a backup protection relay Is the mainstream. These ground fault relays are installed at both ends of the transmission line, and the current input to the relay is the zero phase difference current between both lines to the main protection ground fault circuit selection relay and the ground fault direction relay for backup protection. The zero-phase current of the line is input to.

On the other hand, due to the difficulty of securing a construction site for steel towers in recent years, one steel tower has different power transmission lines such as a direct grounding system and a high resistance grounding system, or transmission lines of the same system such as high resistance grounding systems. Are often put together. In such a parallel transmission line, there is a difference in the mutual inductance between the conductors due to the imbalance in the conductor spacing of the transmission lines of both systems, and the parallel 2
It is known that a so-called circulating current flows between lines in a line transmission line.

FIG. 3 is a diagram schematically showing the relationship between the parallel transmission lines,
A power transmission line (A) as an induction system and a high resistance grounding system parallel two-line power transmission line (B) as an induced system are provided together. In the figure, the induction system is shown as a parallel two-line transmission line.

In this high resistance grounding parallel two-line transmission line (B), the load current IL supplied to the load flows by IL / 2 by the lines B1 and B2, respectively, and the lines B1 and B2 are in the directions indicated by broken lines. Circulating current Ij flows. When a failure occurs, the failure current If flows at the failure point.
However, the load current, circulating current, and fault current are accurately expressed by the determinant corresponding to the number of phases in the line, but for simplicity, they are expressed as IL, If, Ij. There is. For example, And each element of each matrix is a complex number.

Therefore, the currents I1s and I2s of both lines B1 and B2 at the power source end of the high resistance grounding parallel two-line transmission line (B) and the currents I1r and I2r of both lines B1 and B2 at the load end are I1s = IL / 2-Ij, I2s = IL / 2 + Ij, I1r = -IL / 2 + Ij, I2r = -IL / 2-Ij, and at the time of failure, I1s = IL / 2 + (2l-x) If / 2I -Ij, I2s = IL / 2 + xIf / 2l + Ij, I1r = -IL / 2 + xIf / 2l + Ij, I2r = -IL / 2-xIf / 2l-Ij.

Here, 1 is the total length of the high resistance grounding parallel two-line transmission line (B), and x is the distance from the power source side to the failure point. The circulating current Ij is Ij = (Zs-Zm) ^-1 [(Z31-Z41) I1 + (Z32-Z42) I2] L / 21. Here, Zs is an impedance matrix per unit length of the lines B1 and B2 of the high resistance grounding parallel two-line transmission line (B), and Zm is the line B1 of the high resistance grounding parallel two-line transmission line (B). Z31 is a mutual inductance matrix per unit length between B2, and Z31 is a unit length between the line A1 of the parallel 2-line transmission line (A) and the line B1 of the high resistance grounding parallel 2-line transmission line (B). Is a mutual inductance matrix per hit, and Z41 is the line A1 of the parallel two-line power transmission line (A),
Z32 is a mutual inductance matrix per unit length between the high-resistance grounding system parallel 2-line transmission line (B) and the line B2. Z32 is parallel to the high-resistance grounding system A2 of the parallel 2-line transmission line (A). Z42 is a mutual inductance matrix per unit length between the two-line power transmission line (B) and the line B1. Z42 is a parallel two-line power transmission line (A) line A2 and a high-resistance grounding system parallel two.
A mutual inductance matrix per unit length between the line transmission line (B) and the line B2, and I1 and I2 are currents of the lines A1 and A2 of the parallel two-line transmission line (A) which is an induction system, L / l is the cross-section rate.

Due to the above relationships, particularly in the case of a one-line ground fault, the fault current If is zero in the sound two-phase components other than the fault phase. If the above relationship is expressed in terms of each phase and zero phase, I1sa = ILa / 2 + (2l-x) Ifa / 2l-Ija I1sb = ILb in each line B1 and B2 at the power source end. / 2--Ijb I1sc = ILc / 2-Ijc 3I1s0 = I1sa + I1sb + I1sc = (2l-x) Ifa / 2l-3Ij0 I2sa = ILa / 2 + xIfa / 2l + Ija I2sb = Ila + I2s + Ic2Is + Ic2Is + Ic2Is xIfa / 2l + 3Ij0 In each line B1 and B2 at the load end, I1ra = -ILa / 2 + xIfa / 2l + Ija I1rb = -ILb / 2 + Ijb I1rc = -ILc / 2 + Ijc3I1r0 = I1ra + I1rb + I1rc2ra = Ijla / 2l + iJla / 2I + 3Ij2 / 2 -xIfa / 2l-Ija I2rb = -ILb / 2-Ijb I2rc = -ILc / 2-Ijc3I2r0 = I2ra + I2rb + I2rc = -xIfa / 2l-3Ij0. Here, 3Ij0 = Ija + Ijb + Ijc, which is a so-called zero-phase circulating current. In addition, since the load current does not include the zero phase component, ILa + ILb
+ Ilc = 0. Further, regarding the difference current between both lines at the power source end and the load end, and the sum current, at the power source end, Isda = I1sa-I2sa = (l-x) Ifa / l-2Ija Isdb = I1sb-I2sb = -2Ijb Isdc = I1sc-I2sc = -2Ijc 3Isd0 = 3I1s0-3I2s0 = (l-x) Ifa / l-2 * 3Ij0 Issa = I1sa + I2sa = ILa + Ifa Issb = I1sb + I2s0 = I2s0 = I1s0 = I1s0 = I1s0 = I1s0 = I1s0 = I1s0 I1ra-I2ra = xIfa / l + 2Ija Irdb = I1rb-I2rb = + 2Ijb Irdc = I1rc-I2rc = + 2Ijc3Ird0 = 3I1r0-3I2r0 = xIfa / Irb + Irb = Irb + Irb + Ira + I2ra1 + Ibra + I2b1Irb = I2b = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb + Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb = Irb + Irb 3Irs0 = 3I1r0 + 3I2r0 = 0.

As explained above, the zero-phase difference current between both lines, which is the current input to the ground fault line selection relay, or the zero-phase current of the line which is the current input to the direction ground fault relay, is ,
Since the zero-phase circulating current is included, and the fault current at the time of a one-line ground fault in a high-resistance grounding parallel two-line transmission line is quite small, it may depend on the magnitude and phase of the zero-phase circulating current. This may cause inconveniences such as a decrease in the detection sensitivity of the fault relay and a wrong direction (erroneously disconnecting a healthy line other than the line in which the ground fault has occurred).

The problems of the conventional relay will be described in detail below first with respect to the ground fault line selection relay which is the main protection relay.

Focusing on the fact that the zero-phase sum current at the power supply end does not include the zero-phase circulating current, but only the fault current, the direction ground fault relay that inputs the zero-phase sum current is used in combination with the ground fault line selection relay. Things are also being done. However, in this case, since the zero-phase sum current at the load end becomes 0, it can be used only at the power supply end, and at the load end, the current sensitivity is lowered, or a time limit is set to reduce the current sensitivity. However, there is no choice but to use the ground-fault line selection relay at the expense of its original performance such as high sensitivity and high-speed operation, such as a time-off after the circulating current has disappeared due to the preceding interruption.

Furthermore, the circulating current is determined only based on the current of the steel tower structure and the parallel two-line transmission line (A), and the high resistance grounding system parallel two
Paying attention to the fact that it is irrelevant to the presence or absence of a failure in the line transmission line (B), under the assumption that the zero-phase circulating current before and after the failure of the high resistance grounding parallel two-line transmission line (B) is constant. It is conceivable to extract the fault current component from the amount of change in the zero-phase difference current before and after the fault. For example, in a parallel 2-line transmission line (B) with a high resistance grounding system or in a multi-terminal system, one terminal is Since it is possible that the zero-phase circulating current changes when a failure occurs, such as when the circuit is cut off in advance, there is the inconvenience that a reliable one-line ground fault cannot be detected.

On the other hand, the directional ground fault relay, which is a backup protection relay, uses the zero-phase current of the line, and this zero-phase current also contains the zero-phase circulating current, as described above. Therefore, it can be said that the directional ground fault relay has the above-mentioned inconveniences.

That is, using the ground fault relay, the fault current can be reliably detected,
It was extremely difficult to reliably and quickly shut off the ground fault circuit.

<Purpose> The present invention has been made in view of the above problems,
An object of the present invention is to provide a zero-phase circulating current compensating method for a ground fault relay, which can remove a zero-phase circulating current and apply only a fault current to the ground fault relay.

<Structure> The zero-phase circulating current compensating method of the first aspect of the invention for achieving the above-mentioned object is a method of changing the line-to-line differential current of the leading phase to the line-to-line of the delayed phase among the line-to-line differential currents of sound two phases. The difference current advanced by 120 degrees is subtracted, and the zero-phase circulation current compensation value is calculated by multiplying by a constant determined by a fixed factor such as the tower structure and conductor arrangement, and the zero-phase circulation current is calculated from the line-to-line zero-phase difference current. The fault current is obtained by subtracting the current compensation value. As the constant, one determined by a fixed factor such as the steel tower structure and conductor arrangement is used.

That is, as described above, the difference current and zero phase difference current of each phase between lines at the time of a one-line ground fault are generally Ida = Ifa + Ija Idb = Ijb Idc = Ijc 3Id0 = without distinction between the power source end and the load end. It is understood that it is expressed in the form Ifa + 3Ij0. Because (l-
This is because xfa Ifa / l can be considered as new Ifa and -2Ija as new Ija and the like. As can be seen from this equation, each phase difference current does not include the load current. However, when there is an unbalanced branch in the parallel two-line power transmission line or when one end of the multi-terminal system is cut off in advance, each phase difference current includes a load current. Therefore, the above equation can be more generally expressed as Ida = ILa + Ifa + Ija Idb = ILb + Ijb Idc = ILc + Ijc 3Id0 = Ida + Idb + Idc = Ifa + 3Ij0. Since the load current does not include the zero phase component, it is clear that ILa + ILb + ILc = 0.

This expression is for each phase current at the time of the one-line ground fault shown above,
It will also be appreciated that it is consistent with the expression for zero-phase current. Therefore, the zero-phase circulating current compensation method described below can be applied not only to a ground fault circuit selection relay that inputs a zero-phase difference current between lines, but also to a direction ground fault relay that inputs a zero-phase current of a line. Needless to say.

In addition, the above equation shows the case of a phase 1 line ground fault,
A failure current component appears in the b-phase difference current in the case of the b-phase and in the c-phase difference current in the case of the c-phase, but the case of the a-phase single-line failure will be described below. This does not lose any generality.

An object of the present invention is to compensate a zero-phase circulating current and obtain a fault current. However, the present invention focuses on circulating currents included in two healthy phases other than the fault phase, and based on these, , The zero-phase circulating current included in the zero-phase difference current is compensated to extract the fault current.

Considering that most of the load currents ILa, ILb, and ILc are positive phase components, α = −1 / 2 + j√3 / 2, and ILa = IL, I
It can be expressed as Lb = α ² IL and ILc = α IL. Therefore, it can be expressed as Ida = IL + Ifa + Ija Idb = α ² IL + Ijb Idc = αIL + Ijc 3Id0 = Ifa + 3Ij0. Here, 3Ij0 is a so-called zero-phase circulating current, and 3Ij0 = Ija + Ijb + Ijc.

Also, the circulating currents Ija, Ijb, Ijc have already been described, but in general You can think of it as Because (L / 2l) I1, (L / 2
l) I2 should be considered as new I1 and I2.
Furthermore, the above formula is as shown in FIG. In FIG. 4B, In FIG. 4C, Is easily proved to be Can be expressed as

In addition, as a result of diligent efforts by the present inventor, the zero-phase circulating current 3I
j0 = Ija + Ijb + Ijc is obtained by subtracting 120 degrees of the circulating current component of the lagging phase difference current from the circulating current component of the leading phase difference current of the two healthy phases (Ijb- Based on αIjc), we have found that it can be approximated with extremely high accuracy.

Then, (Ijb-αIjc) is Idb-αIdc = (α ² I
L + Ijb) −α (αIL + Ijc) = Ijb−αIjc As shown by the relationship, since it can be calculated from the difference currents Idb and Idc of the healthy two phases, it is possible to measure the difference current Idb of the healthy two phases. , Idc, it is possible to obtain a compensation current value that is very accurately approximated to the zero-phase circulating current.

That is, if ΔIj = 3Ij0-k (Ijb-αIjc),
By expanding this equation, ΔIj = Ija + (1-k) Ijb + (1 + αk) Ijc is obtained.

Here, the circulating currents Ija, Ijb, and Ijc greatly change in magnitude and phase under the influence of the current of the induction system which changes significantly during normal operation of the induction system, failure, or phase loss. However, if the constant k that satisfies ΔIj = 0 is obtained without being affected by the magnitude and phase of the circulating currents Ija, Ijb, and Ijc, the zero-phase circulating current 3I _j0 can be accurately calculated. it can.

On the other hand, the circulating current Ij is, as described above, Is expressed as Becomes However, ka = − (k21−αk31), ka ′ = (k11
+ K21 + k31), kb =-(k22-αk32), kb '= (k
12 + k22 + k32), kc =-(k23-αk33), kc '=
(K13 + k23 + k33).

Therefore, ΔIj = (kaK + ka ') I12a + (kbK + k
b ') I12b + (kcK + kc') I12c.

Here, in order to set ΔIj = 0 without being influenced by the magnitude and phase of the currents I12a, I12b, I12c of the induction system, a constant k that sets all the coefficients kaK + ka ', kbK + kb', kcK + kc 'to 0. However, it cannot be expected that there is a constant k that sets all the coefficients to 0. Therefore, for example, according to the principle of least squares, f (k) = lkaK + ka′l ² + lkbK + kb′l ² + lkcK
By selecting k that minimizes + kc′l ² , the compensation value k (I _jb −αI _jc ) that approximates the zero-phase circulating current can be calculated.

Here, the constant k selected according to the principle of the least squares method is k =-(ka '+ kb' + kc ') / (lkal ² +1
It is represented by kbl ² + lkcl ² ). However,
Are the conjugate complex numbers of ka, kb, and kc, respectively. Further, each element kij of the matrix (Zs-Zm) ^-1 (Z31-Z41) can be considered as a real number, as will be described later in an embodiment, and thus the constant k is k = Σ (k2i + k3i / 2) ・ (k1i ＋ k2i ＋ k3i) / Σ
(K2i ² + k2i k3i + k3i ² ) + jΣ (√3 / 2) k3i (k1
i + k2i + k3i) / Σ ( will be represented by ^{k2i 2 + k2ik3i + k3i 2)} . However, Σ indicates addition up to i = 1 to 3.

In the above, the case of the a-phase one-line ground fault has been described, but in the case of the b-phase one-line ground fault, Id = IL + Ija Idb = α ² IL + Ifd + Ijb Idc = αIL + Ijc 3Id0 = Ifb + 3Ij0 3Ij0 = Ija + Ijb = Ijc = Ijc = Ijb = Ijc = Ijc ) (K2i + k3i / 2) · (k1i + k2i + k3i) / Σ (k2i
² + k2i k3i + k3i ² ) + jΣ (√3 / 2) k3i (k1i + k2
i + k3i) / Σ (k2i ² + k2i k3i + k3i ² ) where k1i, k2i, k3i are replaced by k2i, k3, respectively.
It can be calculated by cyclically replacing i and k1i.

In addition, the same can be calculated for a c-phase one-line ground fault.

As is clear from the above description, the constant k for calculating the compensation value k (I _jb −αI _jc ) that approximates the zero-phase circulating current is not affected by the current in the induction system at all, and the tower structure,
The above matrix (Z
s-Zm) ^-1 (Z31-Z41) can be determined only based on each element kij, and as will be apparent from the examples described later, zero-phase circulating current is compensated with high accuracy. It is possible to obtain a true fault current with high accuracy, and if the ground fault line selection relay is driven by the value after compensation, it is possible to selectively and quickly cut off only the line that has a ground fault fault. You can

Furthermore, this method can be applied to a power source end and a load end, or can be applied to a multi-terminal system, and even if there is a change in circulating current when a failure occurs, it is not affected at all.

The zero-phase circulating current compensating method of the second aspect of the invention for achieving the above object is to subtract, from the currents of the normal two phases, the currents of the lagging phase by 120 degrees from the currents of the leading phase. Then
Calculate the zero-phase circulating current compensation value by multiplying the constant determined by fixed factors such as the steel tower structure and conductor layout, and subtract the above zero-phase circulating current compensation value from the zero-phase current to obtain the fault current and obtain the direction It drives a relay relay, and as the above constant, one determined by a fixed factor such as a steel tower structure and conductor arrangement is used.

In this case as well, a value that can be approximated to the true fault current can be obtained in the same manner as above, and the direction ground fault relay can be driven by the obtained value.

<Example> Hereinafter, detailed description will be given with reference to the accompanying drawings illustrating an example.

FIG. 3 is a diagram showing a state in which a parallel 2-circuit power transmission line (A) of a different system is laid on the parallel 2-circuit power transmission line (B) of a high-resistance ground system. The power supply side of (B) is connected to a power supply (not shown) through a main transformer (T1) of Y-Y connection, and the load side is connected to a main transformer (T2) of Y-Δ connection. Is connected to a load (not shown). Then, the neutral point on the secondary side of the main transformer (T1) is grounded via a resistor (R).

FIG. 5 shows a general circuit configuration for obtaining each phase difference current and zero phase difference current on the power source side of the high resistance grounding parallel two-line transmission line (B). The current is supplied to each phase of the first line B1. While installing the transformer (CT1) (CT2) (CT3),
Current transformer (CT4) (CT5) (C for each phase of the second line B2
T6) is installed. Then, the current transformers for each phase of both lines B1 and B2 are connected to each other in a difference circuit, and one terminal of the current transformers (CT4) (CT5) (CT6) is connected to an auxiliary current transformer for detecting a zero phase difference current. (CT7) is connected to one end, and one terminal of the current transformer (CT1) (CT2) (CT3) is connected to each auxiliary current transformer for phase difference current detection.
(CT8) (CT9) (CT10) connected to each one end, auxiliary current transformer (CT8) (CT9) (CT10) each end connected to the other end of the current transformer (CT7) There is.

Fig. 2 shows parallel 2-circuit transmission line (A) and high-resistance grounding system parallel 2
It is a figure which shows the state which put together the line transmission line (B) and the same steel tower (P), the parallel 2-line transmission line (A) is installed in the upper part of a steel tower (P), and a high resistance earthing | grounding system is provided in the lower part. A parallel two-line power transmission line (B) is installed. And in the figure, A, B, C, D
Shows an example of transmission line towers with different tower heights and conductor intervals.

FIG. 1 is a block diagram for taking measures against circulating current,
Apply the output signal of the current transformer (CT8) (CT9) (CT10) to the zero-phase circulating current compensation value calculation circuit (1) (2) (3), and the zero-phase circulating current compensation value calculation circuit (1) (2) ) (3) output signal selection circuit (4)
To the selection circuit (4) by applying the output signal from the ground fault phase discrimination circuit (5) to the selection circuit (4), Only the output signal of is selectively output. Then, the output signal of the auxiliary current transformer (CT7) and the output signal from the selection circuit (4) are applied to the zero-phase circulating current compensation circuit (6),
By subtracting the zero-phase circulating current compensation value from the zero-phase difference current, a value that can be approximated to the true fault current with high accuracy can be obtained. However, the zero-phase circulating current compensation value calculation circuits (1) (2) (3) are based on the differential current between the lines of each two phases, and k1 (Ida-αIdb), k2 (Idb-αIdc), k3 ( Idc-αIda) is calculated.

More specifically, FIG. 2A shows that the lines A1 and A2 of the parallel two-line power transmission line (A) are installed in opposite directions to each other.
The conductors adjacent to each other in the height direction have an interval of 8.5 m, the conductors adjacent to each other in the horizontal direction have an interval of 11.0 m, and the lines B1 and B2 of the high resistance grounding parallel two-line power transmission line (B) are installed in the same direction. The distance between adjacent conductors in the height direction is 3.5 m, and the distance between adjacent conductors in the horizontal direction is 10.0 m, 10.
5 m, 11.0 m, and the distance between the parallel 2-line transmission line (A) and the high resistance grounding parallel 2-line transmission line (B) in the height direction is 8.
This is the case when each is set to 0 m.

The matrices (Z31-Z41) and (Zs-Zm) ^-1 (Z31-Z41) in this case are shown below, and it is known that each element of the matrix is given according to the following calculation formula.

Diagonal elements Zaa and Zb of the four matrices of
b and Zcc represent the self-impedance of the conductor, and the resistance component can be almost ignored, and is given by Zmm = jω × [0.10 + 0.4605 log ₁₀ (2He / r)].

Elements Zmn other than Zaa, Zbb, and Zcc are conductors m and n.
The mutual inductance between them is given by Zmn = jω × [0.05 + 0.4605 log ₁₀ (2He / Dmn)].

In these formulas, ω = 2π ,: system frequency (Hz) r: conductor radius (m) Dmn = distance between conductors m and n (m) He: equivalent ground depth (m), The unit of Zmm and Zmn is mΩ / km.

According to these formulas, Is calculated.

Also, (However, the line type of the high resistance grounding parallel two-line transmission line (B) is ACSR610 _sq , the conductor radius is 17.1 mm, 2He =
It is 800m. ) Is calculated, Becomes

Based on the above calculation results, the constant k used in the zero-phase circulating current compensation calculation at the time of the b-phase one-line ground fault is calculated.

The coefficient ka = − (0.617-1.047α) × 10 ⁻² , ka ′ = 2.3
76 × 10 ^-2 , kb =-(1.045-2.237α) × 10 ^-2 , kb '
= 4.622 × 10 ^-2 , kc =-(2.052-6.806α) × 10 ^-2 ,
Since kc ′ = 12.050 × 10 ⁻² , k = 1.513e ^j46.3 ° can be calculated based on these coefficients.

Zero-phase circulating current compensation value k (Ijc-
αIja) and true zero-phase circulating current 3Ij0 = Ija + Ijb + I
Comparing and jc, a 3Ij0 = Ija + Ijb + Ijc = (k11 + k21 + k31) I12a + (k12 + k22 + k32) I12b + (k13 + k23 + k33) I12c = (2.376I12a + 4.622I12b + 12.050I12c) × 10 -2, k (Ijc-αIja ) = K (k31−αk11) I12a + k (k32−αk12) I12b + k (k33−αk13) I12c = (2.205e ^j7.8 ° × I12a + 4.395e ^j4.5 ° × I12b + 12.153e ^j0.9 ° × Since I12c × 10 ^-2 , each current I of the induction system is
The coefficients of 12a, I12b, and I12c match with high precision. Actually, if the load current of the induction system is 2000 A, for example, I12a = 2000, I12b = 2000α ² , I12c = 2
Since it is 000α, 3Ij0 = 175.4e ^j132.8 ° A, k (Ijc-αIja) = 178.8e ^j128.9 ° A, which compensates the zero-phase circulating current with extremely high accuracy and is extremely close to the true fault current. It is possible to operate the ground fault relay without a fear of malfunction by obtaining a fault current that causes

The steel tower structure of FIGS. 2B to 2D is not particularly illustrated, but the constant k can be calculated in the same manner as in the case of the steel tower structure of FIG. 2A, and the zero-phase circulation is based on the constant k. By calculating the current compensation value and subtracting it from the zero phase difference current, it is possible to obtain a fault current and operate the ground fault relay without fear of malfunction.

<Effect> As described above, according to the present invention, there is no distinction between the power source end and the load end.
Even in a multi-terminal system, a current that is close to the true fault current can be obtained with high accuracy by performing simple four arithmetic operations based on the sound currents in two phases, and the ground fault relay is activated only by this current. The unique effect that the ground fault occurrence line can be cut off accurately and promptly is achieved.

[Brief description of drawings]

Fig. 1 is a block diagram for taking measures against circulating current, Fig. 2 is a diagram showing a tower structure and conductor arrangement, and Fig. 3 is a parallel two-line transmission line (A) and a high resistance grounding system parallel 2
FIG. 4 is a diagram showing the relationship with the line transmission line (B), and FIG. 4 is a parallel 2 line transmission line (A) and a high resistance grounding system parallel 2
FIG. 5 is a diagram showing a state in which a line power transmission line (B) is placed side by side, and FIG. 5 is a diagram showing electrical connection for obtaining each phase difference current. (1) (2) (3) ... Zero-phase circulating current compensation value calculation circuit, (4) ... Selecting circuit, (6) ... Zero-phase circulating current compensating circuit, (B) ... High resistance grounding parallel 2-line transmission line

Claims (4)

[Claims] 1. A ground fault line selection for selectively disconnecting a ground fault failure line by detecting that a one-line ground fault has occurred in a high resistance grounding parallel two-line power transmission line that is installed together with at least one transmission line. In relays, constants (k _a , k _b , k _c ) determined by fixed factors such as steel tower structure and conductor arrangement are multiplied by a constant (k), and the result of multiplication is determined by fixed factors such as steel tower structure and conductor arrangement. The constant (k) is found by a predetermined method, paying attention to the fact that there is a constant (k) that minimizes the sum of the constants (k _a ′, k _b ′, k _c ′) of , 1 line (for example, a phase) When a ground fault occurs, a sound 2 phase (b, c
Line between difference current _(I db _phase) of the _{I dc),} the process proceeds phase line between differential current _{(I db),} the line between the difference current delay phase that is 120 DoSusumusho the (.alpha. I _dc) The subtracted current is multiplied by the constant (k) to calculate the zero phase circulating current compensation value (3I _j0 ), and the line zero phase difference current (3I _j
A zero-phase circulating current compensating method for a ground fault line selection relay, wherein a true fault current (I _fa ) is obtained by subtracting the zero-phase circulating current compensation value (3I _j0 ) from _d0 ). 2. The zero-phase circulating current compensating method for a ground fault line selection relay according to claim 1, wherein the predetermined method for obtaining the constant (k) is a least square method. 3. A directional ground fault relay that cuts off a transmission line by detecting the occurrence of a single-line ground fault in a high resistance grounding parallel two-line transmission line that is laid together with at least one transmission line. , A constant (k _a , k _b , k _c ) determined by a fixed factor such as conductor arrangement is multiplied by a constant (k), and the result of multiplication is another constant (k _a determined by a fixed factor such as steel tower structure and conductor arrangement). Paying attention to the fact that there is a constant (k) that minimizes the sum of the values of ′, k _b ′ and k _c ′), the constant (k) is obtained by a predetermined method, and a straight line (for example, a Phase) Sound 2 phase (b, c
Line between difference current _(I db _phase) of the _{I dc),} the process proceeds phase line between differential current _{(I db),} the line between the difference current delay phase that is 120 DoSusumusho the (.alpha. I _dc) The subtracted current is multiplied by the constant (k) to calculate the zero phase circulating current compensation value (3I _j0 ), and the line zero phase difference current (3I _j
A zero-phase circulating current compensation method for a directional ground fault relay, characterized in that a true fault current (I _fa ) is obtained by subtracting the zero-phase circulating current compensation value (3I _j0 ) from _d0 ). 4. The zero-phase circulating current compensating method for a directional ground fault relay according to claim 3, wherein the predetermined method for obtaining the constant (k) is a least square method.