JPH04319674A - Method for locating earth fault point - Google Patents

Abstract

PURPOSE:To perform work searching a fault point by reduced labor by specifying the fault point by judging which of branches the fault point on and after a branch point is present in by the use of the data of a power supply terminal and two power receiving terminals. CONSTITUTION:The distance from the power terminal A of a resistance- grounded type three-terminal single circuit transmission line to a fault point is calculated and, when the calculated distance exceeds the distance from the power supply terminal A to a rotary branch point T, the voltages VB, VC and currents IB, IC of a fault phase of other terminals B, C are measured. The positive phase impedance ZB of the circuit in the section TB between the branch point T and a terminal B and the positive phase impedance ZC in the section TC between the branch point T and a terminal C are used to respectively calculate the values of formulae VTB=VB-ZBIB and VTC=VC-ZCIC. When the value VTB is smaller than the value VTC, the fault point is judged to be present in the section TB and, when the value VTC is smaller than the value VTB, the fault point is judged to be present in the section TC.

Description

[Detailed description of the invention]

0001

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a method for locating a one-wire ground fault point in a resistance-grounded three-terminal single-circuit power transmission line. Here, the "single-circuit power transmission line" may be one that was originally installed as a single-circuit line, or it may be one that is operated as a single line due to a failure in one of the parallel two-circuit power transmission lines.

0002

[Prior Art] Compared to substations that are maintained and managed inside buildings, power transmission lines between substations are more susceptible to failures caused by external sources (insulation breakdown due to lightning strikes, contact with birds or trees, etc.). It is inevitable. When a failure occurs, it is necessary to search for the failure point, but searching for the failure point can be extremely difficult, especially in mountainous areas.

[0003] Therefore, if the position and range of the failure point are specified (orientated) by calculation in advance, the failure point can be searched for within that range, leading to improved work efficiency. Conventionally, there has been a method of calculating the component in the right angle direction as a method for locating the ground fault point in resistance-grounded power transmission lines (see Japanese Patent Publication No. 58-36743).
has been adopted. This method consists of terminal voltage, terminal current,
This method identifies the failure point based on the impedance of the line. Using this method as an example of a two-terminal single-circuit transmission line,
A case where an a-phase ground fault occurs will be explained. As shown in FIG. 4, a power transmission line L is connected between a power transmission end A and a power reception end B, a power supply TR is connected to the power transmission end A, and a load L is connected to the power reception end B.
B is connected. d: Distance from transmitting end A to receiving end B, xd: Transmitting end A
Distance from to the fault point, Va: A-phase voltage at the sending end A, Ia: A-phase current at the sending end A, Vaf: A-phase voltage at the fault point, Iaf: A-phase current at the fault point, Zf: A-phase current at the fault point Ground fault impedance. x
, Vaf, Iaf, and Zf are unknown values. Note that in this specification, notations V and I respectively represent vectors.

0004

[Outside 1]

0005. The circuit in FIG. 4 is equivalently transformed as shown in FIG. 5 by using the object coordinate method. Here, V0, V1, V2: Zero-sequence, positive-sequence, and negative-sequence voltages at transmission end A, I0, I1, I2; Zero-phase, positive-sequence, and negative-sequence currents of power transmission line L at transmission end A, Vof, V1f , V2f; zero phase, positive phase at the fault point,
Negative sequence voltage, I0f, I1f, I2f: Zero-sequence, positive-sequence, and negative-sequence currents of the power transmission line L at the failure point.

The zero-sequence current flowing through the fault point flows through an impedance element 3Zf having a value equivalently three times the ground fault impedance Zf. In the above equivalent circuit diagram (FIG. 5), Vof=V0 -xZ0 I0 V1f=V1 -xZ1 I1 V2f=V2 -xZ1 I2 3Zf I0f=Vof+V1f+V2f. Here, Z1 is the positive sequence impedance of the power transmission line L, and Z
0 is the zero-sequence impedance of the power transmission line L. In addition, in this specification, the notation Z is a vector

00
07]

[Outside 2]

[0008] Using the relational expression Va =V0 +V1 +V2, Ia1=I0 +I1 +I2, 3Zf I0f=V0 +V1 +V2
-xZ0 I0 -xZ1 (I1 +I2)
=Va −x[(Z0 −
Z1 )I0 +Z1 Ia1] is derived. Rewriting this formula, Va = 3Zf I0f+x[Z1 Ia
+(Z0 −Z1)I0] (1).

In equation (1), since the value of ground fault impedance Zf is a real number (pure resistance), I on both sides
By using the relationship that the components in the direction perpendicular to 0f are equal, the distance x to the failure point can be found regardless of the value of Zf. To do this, it is necessary to know the a-phase voltage Va, the a-phase current Ia, and the zero-sequence current I0, as well as the positive-sequence impedance Z1 and zero-sequence impedance Z0, but these values may be constants or , which can be measured at the power transmission end A.

(1) To find the orthogonal component of the equation,
Assuming that the phases of the zero-sequence fault current I0f and the zero-sequence current I0 at the transmission end A are equal, the complex conjugate I0 of the zero-sequence current I0
* Multiply to find the imaginary component Im. As a result, x=Im Va I0 * /Im
{Z1 Ia + (Z0 - Z1 )I0 }I0
*

(2) The following equation is obtained, and the distance xd to the failure point can be determined.

[0011]

[Problems to be Solved by the Invention] It is assumed that the above method is applied to a three-terminal single-circuit power transmission line. FIG. 6 is a system diagram of a typical three-terminal single-line power transmission line, with each end designated as A, B, and C, and a branching point as T. A power source whose neutral point is grounded is connected to the A terminal, and a load or a power source whose neutral point is not grounded is connected to the B and C ends, although not shown.

[0012] It is assumed that the failure point is located between TBs (the handling is the same even if it is between TCs), but the transmission line manager does not know where the failure point is. The administrator is responsible for the above (
2) I try to search for the fault point using the formula, but the distance of the fault point is larger than the distance d1 between TAs, so I know that it is farther than the branch point T, but the fault point is I don't know whether it is between TBs or between TCs.

[0013] Therefore, it is necessary to search both between TBs and between TCs, and the effort required to search for a failure point is doubled. The present invention has been made in view of the above-mentioned problem, and is a resistance-grounded type having one power supply end A that is resistance-grounded, and the other ends B and C are connected to a power supply without load or neutral point grounding. The purpose of this invention is to provide a method that can accurately locate a fault point even if the fault point is far from a branch point when locating a single line ground fault fault point in a three-terminal single circuit power transmission line. do.

[0014]

[Means for Solving the Problems] The method of the present invention calculates the distance from the power supply terminal A to the failure point, and the calculated distance is
When exceeding the distance from the power source end A to the line branch point T, measure the fault phase voltages VB, VC and fault phase currents IB, IC at the other ends B and C, and check the distance between the branch point T and the line branch point B. Positive sequence impedance ZB of the line in section TB, positive sequence impedance Z of the line in section TC between branch point T and end C
Using C, the formula VTB=VB −ZB IB,

(3) VTC=VC-ZC IC

(4) Calculate each value and obtain the value V
When TB is smaller than the value VTC, it is determined that the fault point is in the section TB, and when the value VTC is smaller than the value VTB,
This method determines that the failure point is in the section TC, and locates the failure point in the determined section as described above.

[0015]

[Operation] This will be explained with reference to FIGS. 1 and 2. FIG. 2 shows a resistance-grounded three-terminal single-line power transmission line circuit to which the invention is applied. A T-type 3-phase line is installed between power supply end A, power receiving end B, and power receiving end C. The distance between power end A and branch point T is d1, and the distance between power receiving end B and branch point T is d1. The distance between them is d2, and the distance between the receiving end C and the branch point T is d3.
shall be. The positive phase impedance of the line is ZA between power supply end A and branch point T, and ZB between power receiving end B and branch point T.
, the distance between the power receiving end C and the branch point T is ZC.

It is assumed that the positive-sequence voltage at the power source end A is VA, the positive-sequence voltage at the power-receiving end B is VB, the positive-sequence voltage at the power-receiving end C is VC, and the positive-sequence voltage at the branch point is VT. Positive sequence current I from power supply terminal A
It is assumed that B+IC begins to flow, a positive sequence current IB flows to the power receiving end B, and a positive sequence current IC flows to the power receiving end C. The voltage VB at the power receiving end B is calculated using the voltage VA at the power supply end A, the positive sequence impedances ZA, ZB ZC, the current IB, and the current IC, as follows: VB = VA - (IB + I
C) ZA +IB ZB
(5) It is expressed as In addition, the voltage VC at the receiving end C is VC = VA - (IB + IC
)ZA +IC ZC
(6) It is expressed as

[0017] Also, when viewed from the power receiving end B, the voltage VT at the branch point is as follows: VTB=VB -IB ZB

(7) Viewed from the receiving end C, the voltage VT at the branch point is: VTC=VC -IC ZC

(8) It is expressed as When there is no failure, both (7) and (8) are naturally equal.

However, as shown in FIG. 1, when a ground fault occurs in the section TB and the ground fault current If begins to flow, the voltage VB at the receiving end B becomes VB = VA - (IB + IC + If) Z
A + (IB - If) x ZB
+IB (1-x)ZB
(9) The voltage VC at the receiving end C is VC = VA - (IB +IC +If)Z
A +IC ZC (10)
becomes.

If we calculate the voltage VTB at the branch point based on equation (7) without knowing that a ground fault has occurred, we get:
VTB=VB-IB ZB
=VA −(IB +IC +If)ZA +(IB
-If)xZB
+IB (1-x)ZB -IB ZB
=VA −(IB +IC +If)Z
A - IfxZB (11), and when the voltage VTC at the branch point is calculated based on formula (8),
VTC=VC-IC ZC=VA
−(IB +IC +If)ZA +IC ZC −
IC ZC =VA −(IB +I
C +If)ZA
(12).

Difference between equation (11) and equation (12) VTC-V
Taking TB leaves the term IfxZB. This term is based on the fault current, and since one resistance-grounded power supply terminal A is installed, If will flow out (If>0
), VTC>VTB. If a ground fault (not shown) occurs in the section TC and a ground fault current If begins to flow, if a similar formula is derived, VTC<VTB.

Therefore, by calculating the equations (7) and (8) and taking the difference between them, it is possible to determine whether the point where the failure occurs is in the section TB or the section TC. After it is determined that it is in section TB,
Using the orthogonal component method, the formula (xd2 −d1 )Im {Z1BIa +
(Z0B-Z1B)I0 }I0 * = Im
VaI0*-d1
Im {ZA Ia + (Z0A-Z1A) I0
}I0 * The failure point can be calculated by (13).

After it is determined that it is in the interval TC, the formula (yd3 - d1 )Im {Z1CIa +
(Z0C-Z1C)I0 }I0 * = Im
VaI0*-d1
Im {ZA Ia + (Z0A-Z1A) I0
}I0 * The failure point can be calculated by (14). Here, yd3 is the distance from branch point C in section TC to the fault point, Z1B is the positive sequence impedance in section TB, Z0B is the zero sequence impedance in section TB, Z1A is the positive sequence impedance in section TA, and Z0A is the section Zero-sequence impedance in TA, Z1C is positive-sequence impedance in section TC, Z0
C is the zero-sequence impedance in section TC.

[0023]

DESCRIPTION OF THE PREFERRED EMBODIMENTS The failure point locating method of the present invention will be explained in detail below with reference to the accompanying drawings. Note that the same reference numerals are used for the same parts as in FIGS. 1 and 2 described above. FIG. 3 is a diagram showing a general three-terminal single-circuit power transmission line and a fault point calculation device applied to the fault point locating method according to the present invention.
A terminal single-line power transmission line (hereinafter referred to as a three-terminal system) L has a power source TR grounded by a high resistance R on the power source end A side, a load LB on the power receiving end B side, and an ungrounded power source on the power receiving end C side. Power supply LC
are placed. The failure point calculation device 1 is placed on the power supply end A side. Note that the placement of the load LB and the placement of the power source whose neutral point is not grounded may be reversed, and the receiving end B
, a load may be placed on both the power receiving end B and the power receiving end C, and a power source whose neutral point is not grounded may be placed on both the power receiving end B and the power receiving end C.

The three-terminal system L includes a current transformer CT1 connected to the a phase, b phase, and c phase of the line L on the power supply terminal A side.
a, 1b, 1c, and a transformer 2 that is connected to the bus bar on the power supply end A side and detects line voltage.

The failure point calculation device 1 converts each phase voltage/current, zero-sequence voltage/current, positive-sequence voltage/current, and negative-sequence voltage/current read through the phase advancer 3 into a voltage signal of a predetermined level. An auxiliary transformer 4, a sample hold circuit 5 that samples the voltage signal converted by the auxiliary transformer 4 at every predetermined electrical angle (for example, 30 degrees), an A/D converter 6, and a power receiving end B.
, C, a receiver 12 that receives data of measured values at wireless, optical, etc., a digital value converted by the A/D converter 6, and a power receiving end B that reads the data through the receiver 12.
, a data memory 7 that stores digital values of measured values at C, a ground fault detection section 8 (consisting of, for example, 64 relays) that detects ground faults based on the digital values, and a three-terminal system section AT. Positive sequence impedance Z1A
value, value of zero-sequence impedance Z0A, value of positive-sequence impedance Z1B of section BT, zero-sequence impedance Z0B
, the value of the positive sequence impedance Z1C of the section CT, and the value of the zero sequence impedance Z0C as constants. 11
each impedance stored in the data memory 7, each phase voltage/current, zero-sequence voltage/current, positive-sequence voltage/current, negative-sequence voltage/current, and the power receiving end B read through the receiver 12. Calculate the distance from the power supply terminal A to the fault point by calculating the equation (2) above using each phase voltage and current, zero-sequence voltage and current, positive-sequence voltage and current, and negative-sequence voltage and current at C and C as elements. If the calculated distance from the power supply end A to the fault point is greater than the distance d1 from the power supply end A to the branch point T, calculate the above equations (3) and (4) to find the fault point. Determine whether it is in section TB or section TC,
If it is in the section TB, then the formula (13) is used, and if it is in the section TC, it is the formula (14).
) A calculation unit 9 that calculates the failure point after the branch point T based on the formula, and a display unit 10 that displays information such as the failure point calculated by the calculation unit 9 are provided.

[0026] The power receiving end B also has a current transformer CT1 connected to the a phase, b phase, and c phase of the line L on the power receiving end B side.
4a, 14b, 14c, connected to the bus bar on the power supply end A side,
A transformer 15 for detecting line voltage, and the current transformer CT14
a, 14b, 14c, and each phase voltage/current measured by the transformer 15, zero-sequence voltage/current, positive-sequence voltage/current,
A transmitter 13 is provided that detects current, negative phase voltage/current, and transmits the detected data via radio, light, or the like.

The transmitter 13 includes a phase advancer for detecting zero-sequence voltage/current, positive-sequence voltage/current, and negative-sequence voltage/current, a sample-and-hold circuit for digitally converting data, and an A/
Built-in D converter. Sampling synchronization is provided between this sample and hold circuit and the sample and hold circuit 5 of the failure point calculation device 1, as will be described later, so as to prevent calculation errors from occurring.

In addition, the power receiving end C has a current transformer CT1 connected to the a phase, b phase, and c phase of the line L on the power receiving end C side.
6a, 16b, 16c, connected to the bus bar on the power supply end A side,
A transformer 17 for detecting line voltage and the current transformer CT16
Based on the voltage and current of each phase measured by a, 16b, 16c and transformer 17, zero-sequence voltage and current, positive-sequence voltage and
A transmitter 18 is provided that detects current, negative phase voltage/current, and transmits the detected data via radio, light, or the like.

The transmitter 18 includes a phase advancer for detecting zero-sequence voltage/current, positive-sequence voltage/current, and negative-sequence voltage/current, a sample-and-hold circuit for digitally converting data, and an A/
Built-in D converter. Sampling synchronization is provided between this sample and hold circuit and the sample and hold circuit 5 of the failure point calculation device 1, as will be described later, so as to prevent calculation errors from occurring.

The operation of the failure point calculation device 1 is as follows. When the ground fault detection section 8 detects a fault, it causes the calculation section 9 to start a fault point locating operation. The calculation unit 9 retrieves current and voltage data corresponding to the failure detection phase stored in the data memory 7. The calculation unit 9 takes in each of the above data and calculates the faulty phase current IA and faulty phase voltage VA of the power supply terminal A.
, the faulty phase current IB of the power receiving end B, the faulty phase voltage VB, the faulty phase current IB of the power receiving end B, and the faulty phase voltage VB are detected. Then, the equations (3) and (4) shown above are given absolute values: VTB=|VB −IB ZB |

(15) VTC=|VC
-IC ZC |
Substituting into (16),
The distance x from the power supply end A to the fault point is calculated numerically, and if the calculated distance x from the power supply end A to the fault point is greater than the distance d1 from the power supply end A to the branch point T, the above (3)
) (4) to determine whether the failure point is in the interval TB or the interval TC, and if it is in the interval TB, (1
If the equation (3) is in the section TC, the failure point after the branch point T can be calculated based on the equation (14).

[0031] Note that the transmitters 13, 18 and the receiver 12
High speed and high reliability are required for data transmission between the two. Therefore, as a data transmission method, for example, P
It is preferable to use a CM transmission method and to use a communication channel with a large capacity. In particular, if the sampling synchronization between the data at power receiving end A and the data at power end B is not accurately achieved, errors will occur in the calculation results, so so-called SP synchronization control is used to accurately measure and correct the sampling time difference that occurs during data transmission. Technology (Technology to send a signal back and forth between the transmitter 13 and receiver 12, measure the time taken for the round trip, and find the sampling time difference. Mitsubishi Electric Technical Report Vol. 63, No. 8
, 1989, p. p. 27-31) is preferable.

It should be noted that the present invention is not limited to the above-mentioned embodiments; for example, transmitters may be installed at the power source end A and power receiving ends B and C to transmit data, and power can also be received from the power source end A. It is also possible to install the failure point calculation device 1 including the receiver 12 at a location away from the ends B and C. Various other changes may be made without departing from the gist of the present invention.

[0033]

As described above, according to the present invention, by using information on the power source end and the two power receiving ends, it is possible to determine in which branch the failure point after the branch point is located. It becomes possible to identify the failure point, and search for the failure point can be performed with less effort.

[Brief explanation of drawings]

FIG. 1 is a fault phase equivalent circuit diagram of a three-terminal single-line power transmission line for explaining the principle of the present invention.

FIG. 2 is a fault phase equivalent circuit diagram of a three-terminal single-line power transmission line for explaining the principle of the present invention.

FIG. 3 is a diagram showing a fault point calculation device applied to a fault point locating method in a three-terminal single-circuit power transmission line.

FIG. 4 is a circuit diagram of a general two-terminal single-circuit power transmission line.

FIG. 5 is an equivalent circuit diagram for calculating a failure point.

FIG. 6 is a circuit diagram of a general three-terminal single-circuit power transmission line.

[Explanation of symbols]

1 Failure point calculation device 9 Arithmetic section L　Resistance grounding type 3 terminal single circuit power transmission line A　Power supply end B Power receiving end C Power receiving end

Claims (1)

[Claims] Claim 1: Has one power supply terminal A that is resistively grounded, and the other terminal B
and C is a method for locating a single-line ground fault fault point in a resistance-grounded three-terminal single-circuit transmission line connected to a power supply without load or neutral point grounding, and the distance from power supply end A to the fault point. Calculate the fault phase voltages VB and VC at the other ends B and C.
and currents IB and IC of the faulty phase are measured, and if the calculated distance exceeds the distance from the power supply end A to the line branch point T, the positive phase of the line in the section TB between the branch point T and end B is determined. Impedance ZB, section T between branch point T and end C
Using the positive sequence impedance ZC of the line at C, the formula V
Calculate the values of TB = VB - ZB IB, VTC = VC - ZC IC, and when the value VTB is smaller than the value VTC, it is determined that the failure point is in the section TB, and when the value VTC is smaller than the value VTB. A ground fault fault point locating method characterized by determining that the fault point is in the section TC, and locating the fault point in the section determined as described above.