JP2010051124A - Direction protective relay device - Google Patents

Abstract

<P>PROBLEM TO BE SOLVED: To provide a direction protective relay device for further reducing the number of current transformers and direction protective relays installed for protecting a three-phase AC circuit from a short circuit failure. <P>SOLUTION: The short circuit direction relay device for protecting a power transmission line from a short circuit failure includes: a cross-through current transformer 10 in which R- and S-phases of the power transmission line cross and penetrate reversely at an arbitrary angle through a circular core on which a secondary coil is wound; and a short circuit direction relay 4 for simultaneously interrupting breakers 2<SB>1</SB>-2<SB>3</SB>installed on each of phases of the power transmission line upon detecting a short circuit failure based on a short circuit current I<SB>Ry</SB>input from the cross-through current transformer 10 and interline voltages V<SB>RS</SB>, V<SB>ST</SB>, V<SB>TR</SB>of the power transmission line. <P>COPYRIGHT: (C)2010,JPO&INPIT

Description

  The present invention relates to a direction protection relay device, and more particularly to a direction protection relay device for protecting a three-phase AC circuit from a short circuit accident.

  Conventionally, in a three-phase AC circuit, in order to protect the three-phase AC circuit from a short-circuit accident, a short-circuit direction relay (DS) is installed for each phase (see Patent Document 1 below).

For example, as shown in FIG. 18, the first short directional relay 4 _1, enter the short-circuit current of the R-phase from the power distribution wires of the first current transformer installed in the R-phase (CT) 3 ₁ enter the phase voltage V _S of the phase voltage V _R and S phases of the R-phase from potential transformer (PT) 6 installed on the bus together, also in the second short directional relay 4 _2, transmission and distribution lines The S-phase short-circuit current is input from the _second current transformer 32 installed in the S-phase, and the S-phase phase voltage V _S and the T-phase phase voltage V _T are input from the instrument transformer 6. in addition, the third short-circuit direction relay 4 _3, from the third current transformer 3 ₃ installed in the T-phase of the transmission and distribution lines from the potential transformer 6 inputs the short-circuit current of the T-phase of the R-phase enter the phase voltage V _T of the phase voltage V _R and T phases, when the short-circuit fault occurs in the transmission and distribution lines, as shown below, the accident appearance Depending on transmission and distribution lines of R-phase, bulk blocked in the first to third circuit breaker 2 ₁ to 2 ₃ of the first to third short directional relay 41 _{to 3} of which are respectively installed on the S-phase and T-phase is doing.
(1) In case of short circuit between R phase and S phase Short circuit current in the internal direction (direction toward the end of the transmission and distribution line) flows in the R phase of the transmission and distribution line, and short circuit in the external direction to the S phase of the transmission and distribution line Current flows. Therefore, the first short-circuit direction relay 4 ₁ operates based on the R-phase short-circuit current and the R-phase to S-phase line voltage V _RS , thereby connecting the _first to _third circuit breakers 2 ₁ to 2 ₃ . Shut off all at once.
(2) In the case of a short-circuit accident between the S phase and the T phase An internal short circuit current flows in the S phase of the transmission and distribution line, and an external short circuit current flows in the T phase of the transmission and distribution line. Accordingly, the second short-circuit direction relay 4 ₂ operates based on the S-phase short-circuit current and the S-phase to T-phase line voltage V _ST , and the _first to _third circuit breakers 2 ₁ to 2 ₃ are connected. Shut off all at once.
(3) In the case of a short circuit accident between the T phase and the R phase An internal short circuit current flows in the T phase of the transmission and distribution line, and an external short circuit current flows in the R phase of the transmission and distribution line. Therefore, the third short-circuit direction relay 4 ₃ operates based on the T-phase short-circuit current and the T-phase to R-phase line voltage V _TR , and the _first to _third circuit breakers 2 ₁ to 2 ₃ are connected. Shut off all at once.
(4) In the case of a short circuit accident between the R phase, the S phase, and the T phase Short circuit currents in the internal direction flow in the R phase, the S phase, and the T phase of the transmission and distribution lines. Accordingly, the first to third short-circuit direction relays 4 ₁ to 4 ₃ are connected to the R-phase, S-phase, and T-phase short-circuit currents, the R-phase to S-phase line voltage V _RS , and the S-phase to T-phase line spacing. The _first to _third circuit breakers 2 ₁ to 2 ₃ are collectively cut off by operating on the basis of the voltage V _ST and the T-phase to R-phase line voltage V _TR .

As for the distance relay (DZ), as shown in FIG. 19, the first distance relay 20 _1, of the first current transformer 3 ₁ and the power distribution wires installed in the R-phase of the transmission and distribution lines the third current transformer 3 ₃ R-phase and T-phase of the short circuit current from voltage transformer 6 installed to the bus together with the respectively input R phase and T-phase phase voltage V _R of the installed in the T phase, V _T is input, and the second distance relay 20 ₂ is supplied with the R phase and S from the _first current transformer 3 ₁ and the second current transformer 3 ₂ installed in the S phase of the transmission and distribution line. The short-circuit current of the phase is input and the phase voltages V _R and V _S of the R phase and the S phase are input from the instrument transformer 6, and the second and third phase relays 20 ₃ are connected to the second and third phase relays 20 3. The S-phase and T-phase short-circuit currents are input from the current transformers 3 ₂ and 3 ₃ , respectively, and the S-phase phase voltage V _S is input from the instrument transformer 6. And enter the phase voltage V _T of the T-phase, when the short-circuit fault occurs in the transmission and distribution lines, as shown below, R-phase of the transmission and distribution lines according to the accident aspect, each S-phase and T-phase The installed _first to _third circuit breakers 2 ₁ to 2 ₃ are collectively disconnected by the _{first to third} distance relays 20 _{1 to} 20 ₃ .
(1) In the case of a short circuit accident between the R phase and the S phase A short circuit current in the internal direction flows in the R phase of the transmission and distribution line, and a short circuit current in the external direction flows in the S phase of the transmission and distribution line. Therefore, the second distance relay 20 _2, operates on the basis of the line voltage V _RS of the differential current and R-phase -S phase of the short circuit current of the short circuit current and the S-phase of the R-phase, the first to third All circuit breakers 2 _{1 to} 2 ₃ are collectively shut off.
(2) In the case of a short-circuit accident between the S phase and the T phase An internal short circuit current flows in the S phase of the transmission and distribution line, and an external short circuit current flows in the T phase of the transmission and distribution line. Accordingly, the third distance relay 20 ₃ operates based on the S-phase short-circuit current, the difference current between the T-phase short-circuit currents, and the S-phase to T-phase line voltage V _ST, and the first to third All circuit breakers 2 _{1 to} 2 ₃ are collectively shut off.
(3) In the case of a short circuit accident between the T phase and the R phase An internal short circuit current flows in the T phase of the transmission and distribution line, and an external short circuit current flows in the R phase of the transmission and distribution line. Therefore, first distance relay 20 ₁ operates on the basis of the line voltage V _TR of the differential current and the T phase -R phase of the short circuit current and short-circuit current of R-phase of T-phase, the first to third All circuit breakers 2 _{1 to} 2 ₃ are collectively shut off.
(4) In the case of a short circuit accident between the R phase, the S phase, and the T phase Short circuit currents in the internal direction flow in the R phase, the S phase, and the T phase of the transmission and distribution lines. Accordingly, the first to third distance relay 20 ₁ to 20 ₃ operate respectively, collectively blocking the first to third circuit breaker 2 ₁ to 2 _3.

For even line selection relay (SS), as shown in FIG. 20, in the first line selection relay 30 _1, the first and second transmission and distribution lines 1L, R-phase of 2L (equilibrium 2-line transmission and distribution lines) Is connected to the first and _fourth current transformers 3 ₁ and 3 ₄ respectively installed in the first and second current transmission and distribution lines 1L and 2L, and the R phase short-circuit current is input thereto, and the instrument transformer is installed on the bus. enter the phase voltage V _R of the R-phase from vessel 6, also in the second line selection relay 30 _2, second and disposed respectively first and second transmission and distribution lines 1L, the S phase of 2L The S-phase short-circuit current of the first and second power transmission lines 1L, 2L is input from the _fifth current transformers 3 ₂ , 3 ₅ and the S-phase phase voltage V _S is input from the instrument transformer 6. , further, the third line selection relay 30 _3, the first and second transmission and distribution lines 1L, respectively set to the T-phase of 2L Have been the third and sixth current transformer 3 _3, 3 to ₆ first and second transmission and distribution lines 1L, the phase voltage T-phase from the potential transformer 6 inputs the short-circuit current of the T-phase of 2L enter the V _T, the first or second transmission and distribution lines 1L, when the short-circuit failure occurs in 2L, as shown below, depending on the accident aspect, short-circuit fault in the first transmission and distribution lines 1L If the _first to _third circuit breakers 2 ₁ to 2 ₃ are determined to be collectively disconnected by the _first to _third line selection relays 30 ₁ to 30 ₃ , and short-circuited to the second transmission / distribution line 2L. If it is determined that an accident has occurred, the fourth to _sixth circuit breakers 2 ₁ to 2 ₆ are collectively disconnected by the _first to _third line selection relays 30 ₁ to 30 ₃ .
(1) In the case of a short circuit accident between the R phase and the S phase For example, when a short circuit accident occurs between the R phase and the S phase of the first transmission and distribution line 1L, the first and second transmission lines 1L and 2L While an internal short-circuit current flows in the R phase, an external short-circuit current flows in the S phase of the first and second transmission and distribution lines 1L and 2L. Thus, the first line selection relay 30 _1, first and second transmission line 1L, operates on the basis of the phase voltage V _R of the differential current and the R-phase of the short-circuit current of the R-phase of 2L, first To the third circuit breakers 2 ₁ to 2 ₃ are collectively cut off.
(2) In the case of a short circuit accident between the S phase and the T phase For example, when a short circuit accident occurs between the S phase and the T phase of the first transmission and distribution line 1L, the first and second transmission lines 1L and 2L A short-circuit current in the internal direction flows in the S phase, and a short-circuit current in the external direction flows in the T phase of the first and second power distribution lines 1L and 2. Accordingly, the second line selection relay 30 ₂ operates based on the difference current between the S-phase short-circuit currents of the first and second transmission lines 1L and 2L and the S-phase phase voltage V _S, and the first To the third circuit breakers 2 ₁ to 2 ₃ are collectively cut off.
(3) In the case of a short circuit accident between the T phase and the R phase For example, when a short circuit accident occurs between the T phase and the R phase of the first transmission and distribution line 1L, the first and second transmission lines 1L and 2 While a short-circuit current in the internal direction flows in the T phase, a short-circuit current in the external direction flows in the R phase of the first and second transmission and distribution lines 1L and 2. Thus, the third line selection relay 30 ₃ operates on the basis of the phase voltage V _R of the first and second transmission and distribution lines 1L, differential current 2 T-phase of the short-circuit current and T-phase, the 1 to the third circuit breakers 2 ₁ to 2 ₃ collectively blocking.
(4) In the case of a short circuit accident between the R phase, the S phase, and the T phase For example, when a short circuit accident occurs between the T phase and the R phase of the first transmission and distribution line 1L, the first and second transmission and distribution lines 1L , 2 short-circuit currents flow in the R phase, S phase, and T phase, respectively. Accordingly, the first to third line selection relays 30 ₁ to 30 ₃ operate to collectively shut off the _first to _third circuit breakers 2 ₁ to 2 ₃ .
JP 2001-16767 A

  However, since three sets of current transformers and direction protection relays (short-circuit direction relays, distance relays, and line selection relays) are installed for each transmission / distribution line, the installation cost is reduced by reducing the number of current transformers and direction protection relays installed. There is a request to reduce this.

  The objective of this invention is providing the direction protection relay apparatus which can reduce the installation number of the current transformer and direction protection relay for protecting a three-phase alternating current circuit from a short circuit accident.

The direction protection relay device of the present invention is a direction protection relay device for protecting a three-phase AC circuit from a short-circuit accident, and an arbitrary two of the three-phase AC circuit is mounted on an annular core wound with a secondary coil. A cross-through current transformer in which phases are crossed in reverse and at an arbitrary angle, and a short circuit based on a short-circuit current input from the cross-through current transformer and a line voltage of the three-phase AC circuit When an accident is detected, a directional protection relay is provided that collectively shuts off the circuit breakers installed in each phase of the three-phase AC circuit.
Here, the direction protection relay calculates a corrected short-circuit current by multiplying the short-circuit current input from the cross-through current transformer by a predetermined multiple, and the calculated corrected short-circuit current and the line of the three-phase AC circuit When a short-circuit accident is detected based on the voltage between the terminals, the circuit breakers installed in each phase of the three-phase AC circuit may be collectively disconnected.
The cross through current transformer (10) is installed in a power transmission / distribution line, and the first phase and the second phase of the transmission / distribution electric line are reversed in the annular iron core of the cross through current transformer and in any direction. The directional protection relay is crossed at an angle, and the direction protection relay is connected to the short circuit current (I _Ry ) input from the cross through current transformer and the line voltage (V _RS , V _ST , V _TR) ) And a short circuit direction relay (4) that collectively shuts off the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the transmission and distribution line. Also good.
The cross through current transformer (10) is installed in a power transmission / distribution line, and the first phase and the second phase of the transmission / distribution electric line are reversed in the annular iron core of the cross through current transformer and in any direction. The directional protection relay is crossed at an angle, and the direction protection relay is connected to the short circuit current (I _Ry ) input from the cross through current transformer and the line voltage (V _RS , V _ST , V _TR) ) And a distance relay (20) that collectively shuts off the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the transmission and distribution line. Good.
The cross-through current transformers are first and second cross-through current transformers (10 ₁ , 10 ₂ ) installed on the first and second transmission and distribution lines (1L, 2L), respectively. The first cross and the second phase of the first power transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of one cross-through current transformer, and the second cross The first phase and the second phase of the second transmission / distribution line are crossed in an opposite direction and crossed at an arbitrary angle through the annular iron core of the through-current transformer, and the direction protection relay is The short-circuit current (I _Ry ), which is the difference between the short-circuit current input from the first cross-through current transformer and the short-circuit current input from the second cross-through current transformer, and the line between the transmission and distribution lines Based on the voltage (V _RS , V _ST , V _TR ), a short circuit accident has occurred in the first transmission / distribution line. Is detected, the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the first transmission / distribution line are collectively disconnected, and the short-circuit current and the line between the transmission / distribution lines When it is detected that a short circuit accident has occurred in the second transmission / distribution line based on the voltage, _fourth to sixth circuit breakers (2 ₄ to 2) installed in each phase of the second transmission / distribution line ₆ ) A line selection relay (30) that cuts off all at once may be used.
One phase of the arbitrary two phases of the three-phase AC circuit and one other phase other than the arbitrary two phases are crossed in an opposite direction and at an arbitrary angle on an annular core around which a secondary coil is wound. Detecting a short-circuit fault based on another cross-through current transformer passed through, another short-circuit current input from the other cross-through current transformer, and a line voltage of the three-phase AC circuit, You may further comprise the other direction protection relay which interrupts | blocks the circuit breaker installed in each phase of a three-phase alternating current circuit collectively.
The other directional protection relay calculates another corrected short-circuit current by multiplying the other short-circuit current input from the other cross-through current transformer by a predetermined multiple, and the calculated other corrected short-circuit current and When a short circuit accident is detected based on the line voltage of the three-phase AC circuit, the circuit breakers installed in each phase of the three-phase AC circuit may be collectively shut off.
The cross-through current transformer is a first cross-through current transformer (10 ₁ ) installed on the transmission / distribution line, and the second cross-through current transformer is installed on the transmission / distribution line. A cross-through current transformer (10 ₂ ), wherein the first phase and the second phase of the transmission / distribution line are crossed in an opposite direction and at an arbitrary angle on the annular core of the first cross-through current transformer The second phase and the third phase of the transmission / distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the second cross-through current transformer. cage, said direction protective relay is the first of the first short-circuit current (I _Ry1) and the line voltage of the transmission and distribution lines to be input from the cross-through current transformer _{(V RS, V ST, V} TR) and Upon detection of a short circuit on the basis of the first to third circuit breakers installed in each phase of said transmission power distribution line (2 ₁ to 2 ₃₎ A first short directional relay for interrupting Batch (4 _1), wherein said another direction protective relays is a second short-circuit current which is input from the second cross through current transformer and (I _Ry2) transmission and distribution When a short circuit accident is detected based on the line voltage of the electric wire, the second short circuit direction relay (4 ₂ ) that collectively shuts off the first to third circuit breakers may be used.
The cross-through current transformer is a first cross-through current transformer (10 ₁ ) installed on the transmission / distribution line, and the second cross-through current transformer is installed on the transmission / distribution line. A cross-through current transformer (10 ₂ ), wherein the first phase and the second phase of the transmission / distribution line are crossed in an opposite direction and at an arbitrary angle on the annular core of the first cross-through current transformer The second phase and the third phase of the transmission / distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the second cross-through current transformer. And the direction protection relay includes a first short-circuit current (I _Ry1 ) input from the first cross-through current transformer and a line voltage (V _RS , V _ST , V _TR ) of the transmission / distribution line. Upon detection of a short circuit on the basis of the first to third circuit breakers installed in each phase of the transmission and distribution lines (2 ₁ to 2 ₃₎ A first distance relay for collectively blocking (20 _1), said other direction protective relays is, the second second short-circuit current (I _Ry2) and electric transmission input from the cross-through current transformer The second distance relay (20 ₂ ) that collectively shuts off the first to third circuit breakers when a short circuit accident is detected based on the line voltage.
The cross through current transformer is a first and second transmission and distribution lines (1L, 2L) to the first and second cross through current transformer installed respectively (10 _1, 10 _2),
The other cross through current transformers are third and fourth cross through current transformers (10 ₃ , 10 ₄ ) installed in the first and second transmission and distribution lines, respectively. A first phase and a second phase of the first transmission / distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the cross-through current transformer, The first cross-phase and the second phase of the second power transmission and distribution line are crossed and penetrated in an opposite direction and at an arbitrary angle through the annular iron core of the flow device. The second phase and the third phase of the first transmission and distribution line are crossed in an opposite direction and at an arbitrary angle, and penetrated through the annular iron core of the device, and the fourth cross-through current transformer The second phase and the third phase of the second transmission / distribution line are connected to the annular iron core in the opposite direction and at an arbitrary angle. The directional protection relay is a difference current between a short-circuit current input from the first cross-through current transformer and a short-circuit current input from the second cross-through current transformer. A short circuit accident occurred in the first transmission / distribution line based on the short circuit current (I _Ry1 ) of 1 and the line voltages (V _RS , V _ST , V _TR ) of the first and second transmission lines. When this is detected, the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in the respective phases of the first transmission and distribution line are collectively cut off, and the first short circuit current and the first and When it is detected that a short circuit accident has occurred in the second transmission / distribution line based on the line voltage of the second transmission / distribution line, the fourth to fourth installed in each phase of the second transmission / distribution line 6 is a first circuit selection relay (30 ₁ ) that collectively shuts off ₆ circuit breakers (2 ₄ to 2 ₆ ), and the other direction protection relay is , The second short-circuit current (I _Ry2 ) which is the difference between the short-circuit current input from the third cross-through current transformer and the short-circuit current input from the fourth cross-through current transformer, and the first And detecting the occurrence of a short-circuit accident in the first transmission / distribution line based on the line voltage of the second transmission / distribution line, the first to third circuit breakers are collectively cut off, and the second If it detects that the short circuit accident occurred in the 2nd transmission and distribution line based on the short circuit current and the line voltage of the 1st and 2nd transmission and distribution line, the 4th thru / or the 6th circuit breaker will be cut off collectively. The second line selection relay (30 ₂ ) may be used.

The direction protection relay device of the present invention has the following effects.
(1) By using a cross-through current transformer, the installation cost can be reduced by reducing the number of installed current transformers and direction protection relays to protect the three-phase AC circuit from a short circuit accident. .
(2) By using two sets of cross-through current transformers and direction protection relays, it is possible to reliably detect even a short-circuit accident spanning the own circuit and other circuits and prevent the expansion of the range of power failure.
(3) By using two sets of cross-through current transformers and direction protection relays, even if one direction protection relay becomes unusable due to failure or inspection, the short circuit accident of its own circuit will protect the direction of the other one It can be backed up by a relay to protect the three-phase AC circuit from short circuit accidents.

  For the above purpose, direction protection is achieved by using a cross-through current transformer in which any two phases of a three-phase AC circuit are crossed in opposite directions and at an arbitrary angle through an annular core wound with a secondary coil. When the relay detects a short-circuit accident based on the short-circuit current input from the cross-through current transformer and the line voltage of the three-phase AC circuit, the circuit breakers installed in each phase of the three-phase AC circuit are collectively shut off. It was realized.

Hereinafter, embodiments of the directional protection relay device of the present invention will be described with reference to the drawings.
As shown in FIG. 1, the direction protection relay device according to the first embodiment of the present invention includes a cross-through current transformer 10 that is penetrated so that the R phase and the S phase of the transmission / distribution line cross, and a bus Short-circuit current I _Ry input from the installed instrument transformer 6 and cross-through current transformer 10 and phase voltages V _R and V _{S of} R phase, S phase and T phase input from the instrument transformer 6 , V _T , R phase-S phase line voltage V _RS , S phase-T phase line voltage V _ST and T phase-R phase line voltage V _TR If an accident is detected, the short circuit direction relay 4 which cuts off the _1st thru | or 3rd circuit breakers 2 ₁ to 2 ₃ installed in the R phase, S phase and T phase of the transmission and distribution line, respectively, is provided.

Here, the cross-through current transformer 10 is a through-type current transformer in which an R-phase and an S-phase of a transmission / distribution wire are crossed in an opposite direction and at an arbitrary angle through an annular core around which a secondary coil is wound. It is.
That is, the R phase of the transmission / distribution line is penetrated in the polarity direction of the cross-through current transformer 10 (direction from the first opening surface of the annular core to the second opening surface of the annular core). The S phase of the cross through current transformer 10 is penetrated in the opposite polarity direction (direction from the second opening surface of the annular core to the first opening surface of the annular core).

Therefore, when the load currents flowing in the R-phase, S-phase, and T-phase of the transmission and distribution line when no short-circuit accident has occurred are represented by I _R , I _S , I _T , the R-phase load current I _R and the S-phase As shown in FIG. 2, the load current I _S flows through the annular core of the cross-through current transformer 10 in a reverse direction with a phase difference of 120 ° (that is, the R-phase load current I _R is cross-through variable). the toroids of Nagareki 10 flows through the polarity direction, the load current I _S of the S-phase flows through the annular core of the cross through the current transformer 10 in the opposite polarity direction). Therefore, the load current I input from the cross-through current transformer 10 to the short-circuit direction relay 4 is a vector difference between the R-phase load current I _R and the S-phase load current I _S, and the amplitude of the load current I is R-phase. The load current I _R (S-phase load current I _S ) is 3 ^1/2 times the amplitude.
I = I _R −I _S
| I | = | I _R −I _S | = 3 ^1/2 × | I _R | = 3 ^1/2 × | I _S |
Therefore, in order to make the short-circuit direction relay 4 the same as the amplitude of the load current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. The corrected load current I ′ is calculated by multiplying the load current I by 1/3 ^1/2 .
I ′ = I × 1/3 ^1/2
| I ′ | = | I | × 1/3 1/ ^2- = | I _R | = | I _S |

In addition, when a short circuit accident occurs in the transmission / distribution line, the short-circuit currents flowing in the R phase, S phase, and T phase of the transmission / distribution line are represented by I _FR , I _FS , I _FT (impedance angle is θ). The short-circuit direction relay 4 operates as follows according to the accident aspect.
(1) In the case of a short circuit accident between the R phase and the S phase When a short circuit accident between the R phase and the S phase occurs, the short circuit current I _FR of the R phase is generated in the R phase of the power transmission and distribution line as shown by the broken arrow in FIG. The S-phase short-circuit current I _FS flows in the S-phase of the transmission / distribution line, and the S-phase short-circuit current I _FS flows in the external direction. However, the T-phase short-circuit current I _FT does not flow in the T-phase of the transmission / distribution line.
Accordingly, the short-circuit current I _Ry input from the cross-through current transformer 10 to the short-circuit direction relay 4 is represented by the R-phase short-circuit current I _FR and the S-phase short-circuit current I _FS as indicated by the solid line thick arrow in FIG. Thus, the amplitude of the short-circuit current I _Ry is twice the amplitude of the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) (see FIG. 3A). Short-circuit currents I _FR , I _FS , and I _FT flowing in the internal direction of the transmission and distribution lines are indicated by solid arrows, and short-circuit currents I _FR , I _FS , and I _FT flowing in the external direction of the transmission and distribution lines are indicated by dashed-dotted arrows Yes.)
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, in order to make the short-circuit direction relay 4 the same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. The short-circuit current I _Ry is halved to calculate a corrected short-circuit current I _Ry '.
I _Ry '= I _Ry × 1/2
| I _Ry '| = | I _Ry | × 1 / 2- = | I _FR | = | I _FS |
The short-circuit direction relay 4 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the magnitude and direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the R phase and the S phase of the transmission and distribution line, as shown in FIG. 3A, the calculated corrected short circuit current I _Ry ′ and the line voltage V _RS between the R phase and the S phase Is used to determine the magnitude and direction of the short-circuit current I _Ry .
When it is determined that a short-circuit accident has occurred in the transmission / distribution line, the short-circuit direction relay 4 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ .
(2) In the case of a short-circuit accident between the S phase and the T phase When a short circuit accident occurs between the S phase and the T phase, as shown by the dashed arrows in FIG. 4, the S phase short circuit current I _FS There flow inside direction, but the short-circuit current I _FT T-phase to the T phase of the power distribution wires flows outward, the R-phase transmission and distribution lines does not flow a short-circuit current I _FR of R-phase.
Therefore, the short-circuit current I _Ry input from the cross-through current transformer 10 to the short-circuit direction relay 4 becomes an S-phase short-circuit current −I _FS having a negative polarity as shown by a solid line thick arrow in FIG. The amplitude of I _{Ry is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 3B).
I _Ry = −I _FS
| I _Ry | = | I _FS |
Therefore, in order to make the short-circuit direction relay 4 the same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. The corrected short-circuit current I _Ry 'is calculated by multiplying the short-circuit current I _Ry by 1.
I _Ry '= I _Ry × 1
| I _Ry '| = | I _Ry | × 1 = | I _FS |
The short-circuit direction relay 4 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the magnitude and direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident S phase -T phase of transmission and distribution lines, as shown in FIG. 3 (b), the calculated correction circuit current I _Ry 'and S phase -T phase of the line voltage V _ST ( The magnitude and direction of the short-circuit current I _Ry is determined based on the negative polarity.
When it is determined that a short-circuit accident has occurred in the transmission / distribution line, the short-circuit direction relay 4 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ .
(3) When T phase -R phase short fault when T-phase -R phase short-circuit accident occurs, as indicated by broken line arrow in FIG. 5, the short-circuit current of the T-phase to the T phase of the transmission and distribution lines I _FT There flow inside direction, but the short-circuit current I _FR of R-phase to R-phase of transmission and distribution lines to flow to the outside direction, the S-phase of the transmission and distribution lines does not flow a short-circuit current I _FS of S phase.
Therefore, the short-circuit current I _Ry input from the cross-through current transformer 10 to the short-circuit direction relay 4 becomes the R-phase short-circuit current I _{FR as} shown by the solid line thick arrow in FIG. 5, and the amplitude of the short-circuit current I _Ry is This is the amplitude of the R-phase short-circuit current I _FR (see FIG. 3C).
I _Ry = I _FR
| I _Ry | = | I _FR |
Therefore, in order to make the short-circuit direction relay 4 the same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. The corrected short-circuit current I _Ry 'is calculated by multiplying the short-circuit current I _Ry by 1.
I _Ry '= I _Ry × 1
| I _Ry '| = | I _Ry | × 1 = | I _FR |
The short-circuit direction relay 4 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the magnitude and direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the T phase and the R phase of the transmission and distribution line, as shown in FIG. 3C, the calculated corrected short circuit current I _Ry 'and the line voltage V _TR (T phase to R phase) The magnitude and direction of the short-circuit current I _Ry is determined based on the negative polarity.
When it is determined that a short-circuit accident has occurred in the transmission / distribution line, the short-circuit direction relay 4 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ .
(4) In the case of a short-circuit accident between R phase, S phase, and T phase When a short circuit accident between R phase, S phase, and T phase occurs, as shown by the dashed arrows in FIG. The R-phase short-circuit current I _FR , the S-phase short-circuit current I _FS, and the T-phase short-circuit current I _FT flow in the internal direction with a phase difference of 120 degrees.
Therefore, the short-circuit current I _Ry input from the cross-through current transformer 10 to the short-circuit direction relay 4 is represented by the R-phase short-circuit current I _FR and the S-phase short-circuit current I _{FS as} shown by the solid thick arrows in FIG. Therefore, the amplitude of the short-circuit current I _Ry is 3 ^1/2 times the amplitude of the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) (see FIG. 3D).
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, in order to make the short-circuit direction relay 4 the same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. The short-circuit current I _Ry is multiplied by 1/3 ^1/2 to calculate a corrected short-circuit current I _Ry '.
I _Ry '= I _Ry × 1/3 ^1/2
_{| I Ry '| = | I} Ry | × 1/3 1/2 - = | I FR | = | I FS |
The short-circuit direction relay 4 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the magnitude and direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the R phase, the S phase, and the T phase of the transmission / distribution line, as shown in FIG. _3D , the calculated corrected short circuit current I _Ry ′ and the line voltage between the R phase and the S phase The magnitude and direction of the short circuit current I _Ry is determined based on V _RS .
When it is determined that a short-circuit accident has occurred in the transmission / distribution line, the short-circuit direction relay 4 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ .

  In addition, although the R phase and S phase of the transmission / distribution line are crossed and penetrated through the cross-through current transformer 10, the S phase and T phase of the transmission / distribution line may be crossed and penetrated. The T phase and R phase of the electric wire may be crossed and penetrated.

However, the short-circuit direction relay 4 calculates the corrected short-circuit current I _Ry ′ at the magnification shown in Table 1 according to the accident aspect, depending on the combination of phases of the transmission and distribution lines to be crossed (hereinafter referred to as “CT connection”). At the same time, based on the calculated corrected short-circuit current I _Ry ′ and the line voltage shown in Table 2, it is determined whether or not a short-circuit accident has occurred in the transmission and distribution line.
In Tables 1 and 2, “+” in the CT connection indicates a phase that penetrates in the polarity direction of the cross-through current transformer, and “−” in the CT connection penetrates in the opposite polarity direction of the cross-through current transformer. The phase to be shown.

Next, a direction protection relay device according to a second embodiment of the present invention will be described with reference to FIG.
As shown in FIG. 7, the direction protection relay device according to the present embodiment is for a cross-through current transformer 10 that is penetrated so that the R-phase and S-phase of the transmission and distribution lines cross each other, and an instrument installed on the bus. Calculated from the short-circuit current I _Ry input from the transformer 6 and the cross-through current transformer 10 and the R-phase, S-phase, and T-phase phase voltages V _R , V _S , and V _T input from the instrument transformer 6. When a short circuit accident of the transmission / distribution line is detected based on the line voltage V _RS between the R phase and the S phase, the line voltage V _ST between the S phase and the T phase, and the line voltage V _TR between the T phase and the R phase, And a distance relay 20 that collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ installed in the R phase, S phase, and T phase of the transmission and distribution line, respectively.

Here, the cross-through current transformer 10 is a through-type current transformer in which an R-phase and an S-phase of a transmission / distribution wire are crossed in an opposite direction and at an arbitrary angle through an annular core around which a secondary coil is wound. It is.
That is, the R phase of the transmission and distribution line is penetrated in the polarity direction of the cross-through current transformer 10, while the S phase of the transmission and distribution line is penetrated in the opposite polarity direction of the cross-through current transformer 10.

Therefore, when the load currents flowing in the R phase, S phase, and T phase of the transmission and distribution line when no short-circuit accident has occurred are represented by I _R , I _S , and I _T , the distance relay 20 from the cross through current transformer 10 Is the vector difference between the R-phase load current I _R and the S-phase load current I _{S in the same} manner as in the case of the short-circuit direction relay 4 in the first embodiment described above. The amplitude of the current I is 3 ^1/2 times the amplitude of the R-phase load current I _R (S-phase load current I _S ).
I = I _R −I _S
| I | = | I _R −I _S | = 3 ^1/2 × | I _R | = 3 ^1/2 × | I _S |
Therefore, in order to make the distance relay 20 the same as the amplitude of the load current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. The corrected load current I ′ is calculated by multiplying I by 2/3 ^1/2 .
I ′ = I × 2/3 ^1/2
| I ′ | = | I | × 2/3 ^1/2 − = 2 × | I _R | = 2 × | I _S |

In addition, when a short circuit accident occurs in the transmission / distribution line, the short-circuit currents flowing in the R phase, S phase, and T phase of the transmission / distribution line are represented by I _FR , I _FS , I _FT (impedance angle is θ). The distance relay 20 operates as follows according to the accident aspect.
(1) When the short circuit in the case of a short-circuit accident R phase -S phase R phase -S phase occurs, as indicated by broken line arrow in FIG. 7, the short-circuit current of the R-phase to R-phase of transmission and distribution lines I _FR There flow inside direction, but the short-circuit current I _FS of S phase to the S phase of the transmission and distribution lines to flow to the outside direction, the T-phase of the transmission and distribution lines does not flow a short-circuit current I _FT T-phase.
Accordingly, the short-circuit current I _Ry input from the cross-through current transformer 10 to the distance relay 20 is the same as that of the short-circuit direction relay 4 in the first embodiment described above, and the R-phase short-circuit currents I _FR and S This is a vector difference from the phase short-circuit current I _FS, and the amplitude of the short-circuit current I _Ry is twice the amplitude of the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) (see FIG. 3A). .
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, distance relay 20, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, the short-circuit current as shown by the following formula The corrected short-circuit current I _Ry ′ is calculated by multiplying I _Ry by 1.
I _Ry '= I _Ry × 1
| I _Ry '| = | I _Ry | × 1− = 2 × | I _FR | = 2 × | I _FS |
The distance relay 20 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR Based on the above, the distance of the accident point and the direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the R phase and the S phase of the transmission and distribution line, the distance of the fault point and the short circuit current based on the calculated corrected short circuit current I _Ry ′ and the line voltage V _RS between the R phase and the S phase. The direction of I _Ry is determined (see FIG. 3A).
The distance relay 20 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(2) In the case of a short-circuit accident between the S phase and the T phase When a short circuit accident between the S phase and the T phase occurs, the S phase short circuit current I _FS flows in the S phase of the transmission and distribution line in the internal direction, and the T of the transmission and distribution line Although the T-phase short-circuit current I _FT flows outward in the phase, the R-phase short-circuit current I _FR does not flow in the R-phase of the transmission and distribution line (see FIG. 4).
Therefore, the short-circuit current I _Ry input from the cross-through current transformer 10 to the distance relay 20 is the same as the short-circuit direction relay 4 in the first embodiment described above, and the S-phase short-circuit current having a negative polarity. −I _{FS and} the amplitude of the short-circuit current I _Ry becomes the amplitude of the S-phase short-circuit current I _FS (see FIG. 3B).
I _Ry = −I _FS
| I _Ry | = | I _FS |
Therefore, distance relay 20, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, the short-circuit current as shown by the following formula by doubling the I _Ry to calculate a correction circuit current I _Ry '.
I _Ry '= I _Ry × 2
| I _Ry '| = | I _Ry | × 2 = 2 × | I _FS |
The distance relay 20 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR Based on the above, the distance of the accident point and the direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the S phase and the T phase of the transmission / distribution line, the accident point is based on the calculated corrected short circuit current I _Ry ′ and the line voltage V _ST (negative polarity) between the S phase and the T phase. And the direction of the short-circuit current I _Ry are determined (see FIG. 3B).
The distance relay 20 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(3) In the case of a short circuit accident between the T phase and the R phase When a short circuit accident occurs between the T phase and the R phase, a T phase short circuit current I _FT flows in the T phase of the transmission and distribution line, and the R of the transmission and distribution line. While the short-circuit current I _FR of R-phase to phase flows to the outside direction, the S-phase of the transmission and distribution lines does not flow a short-circuit current I _FS of S phase (see FIG. 5).
Accordingly, the short-circuit current I _Ry input from the cross-through current transformer 10 to the distance relay 20 becomes the R-phase short-circuit current I _{FR in the same} manner as the short-circuit direction relay 4 in the first embodiment described above. The amplitude of the short-circuit current I _{Ry is} the amplitude of the R-phase short-circuit current I _FR (see FIG. 3C).
I _Ry = I _FR
| I _Ry | = | I _FR |
Therefore, distance relay 20, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, the short-circuit current as shown by the following formula by doubling the I _Ry to calculate a correction circuit current I _Ry '.
I _Ry '= I _Ry × 2
| I _Ry '| = | I _Ry | × 2 = 2 × | I _FR |
The distance relay 20 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR Based on the above, the distance of the accident point and the direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the T phase and the R phase of the transmission and distribution line, the accident point is based on the calculated corrected short circuit current I _Ry ′ and the line voltage V _TR (negative polarity) between the T phase and the R phase. And the direction of the short-circuit current I _Ry are determined (see FIG. 3C).
The distance relay 20 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(4) In the case of a short circuit accident between R phase, S phase, and T phase When a short circuit accident between R phase, S phase, and T phase occurs, short circuit current I of R phase to R phase, S phase, and T phase of the transmission and distribution line _FR , S-phase short-circuit current I _FS and T-phase short-circuit current I _FT flow in the internal direction with a phase difference of 120 degrees (see FIG. 6).
Accordingly, the short-circuit current I _Ry input from the cross-through current transformer 10 to the distance relay 20 is the same as that of the short-circuit direction relay 4 in the first embodiment described above, and the R-phase short-circuit currents I _FR and S becomes a vector difference between the short-circuit current I _FS phase, short-circuit current amplitude I _Ry is three ^half the amplitude of the short-circuit current I _FR of R-phase (short-circuit current I _FS of S phase) (FIG. 3 (d )reference).
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, distance relay 20, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, the short-circuit current as shown by the following formula The corrected short-circuit current I _Ry 'is calculated by multiplying I _Ry by 2/3 ^1/2 .
I _Ry '= I _Ry × 2/3 ^1/2
| I _Ry '| = | I _Ry | × 2/3 ¹ /2-= 2 × | I _FR | = 2 × | I _FS |
The distance relay 20 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR Based on the above, the distance of the accident point and the direction of the short-circuit current I _Ry are determined. In the case of a short circuit accident between the R-phase, S-phase, and T-phase of the transmission / distribution line, the distance of the accident point is based on the calculated corrected short-circuit current I _Ry ′ and the R-S-phase line voltage V _RS. Then, the direction of the short-circuit current I _Ry is determined (see FIG. _3D ).
The distance relay 20 collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.

  In addition, although the R phase and S phase of the transmission / distribution line are crossed and penetrated through the cross-through current transformer 10, the S phase and T phase of the transmission / distribution line may be crossed and penetrated. The T phase and R phase of the electric wire may be crossed and penetrated.

However, according to the CT connection, the distance relay 20 calculates the corrected short-circuit current I _Ry ′ at the magnification shown in Table 3 according to the accident situation, and between the calculated corrected short-circuit current I _Ry ′ and the line shown in Table 2 Based on the voltage, the distance of the fault point and the direction of the short-circuit current I _Ry are discriminated to determine whether or not a short-circuit fault has occurred in the transmission and distribution line.

Next, a directional protection relay device according to a third embodiment of the present invention will be described with reference to FIG.
As shown in FIG. 8, the directional protection relay device according to the present embodiment has a first cross that is penetrated so that the R phase and the S phase of the first transmission / distribution line 1L of the balanced two-line transmission / distribution line cross each other. through a current transformer 10 _1, and the second cross through current transformer 10 ₂ that penetrate to R-phase and S-phase of the second transmission and distribution lines 2L equilibrium two-line transmission and distribution lines to cross, the bus The difference between the short-circuit current input from the installed instrument transformer 6 and the first cross-through current transformer 10 ₁ and the short-circuit current input from the second cross-through current transformer 10 ₂ (hereinafter, “ referred to as short-circuit current I _Ry ".) and the R-phase input from the potential transformer 6, S-phase and phase voltage V _R of the T-phase, V _S, the line voltage of the R phase -S phase determined from V _T V _RS, S phase -T phase line voltage V _ST and T-phase -R phase line voltage V _TR and the first or second transmission and distribution lines 1L based on the When a 2L short-circuit accident is detected, the R-phase, S-phase, and T-phase of the first and second transmission / distribution lines 1L, 2L in which the short-circuit accident has occurred (hereinafter referred to as “accident line”) are respectively detected. Installed circuit breakers ( _first to _third circuit breakers 2 ₁ to 2 ₃ installed in the R phase, S phase and T phase of the first transmission and distribution line 1L, respectively, or the second transmission and distribution line 2L And a line selection relay 30 that collectively shuts off the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ ) respectively installed in the R phase, the S phase, and the T phase.

Here, the first cross through the current transformer 10 ₁ is the toroids were wound secondary coil are crossed at any angle and the R-phase and S-phase of the first transmission and distribution lines 1L reversed This is a penetrating through-type current transformer.
Ie, R-phase of the first transmission and distribution lines 1L is through the first polarity direction of the cross through the current transformer 10 ₁ but, S phase of the first transmission and distribution lines 1L is first cross through varying It is penetrated in the opposite polarity direction of Nagareki 10 _1.
The second cross-through current transformer 10 ₂ is configured in the same manner as the first cross-through current transformer 10 ₁ , but is input to the line selection relay 30 from the second cross-through current transformer 10 _2. the polarity of the short-circuit current is connected to the line selection relay 30 so that the opposite polarity of the short-circuit current which is input from the first cross through current transformer 10 ₁ for line selection relay 30.

Therefore, the load current flowing in the R phase, S phase, and T phase of the first transmission / distribution line 1L when no short circuit accident has occurred is represented by I _R1 , I _S1 , I _T1 and the second transmission / distribution line 2L. When the load currents flowing in the R-phase, S-phase, and T-phase are represented by I _R2 , I _S2 , and I _T2 , the load current i ₁ input from the first cross-through current transformer 10 ₁ to the line selection relay 30 is As in the case of the short-circuit direction relay 4 in the first embodiment, the vector difference between the R-phase load current I _R1 and the S-phase load current I _S1 is obtained, and the amplitude of the load current i ₁ is R-phase. The load current I _R1 (S-phase load current I _S1 ) is 3 ^1/2 times the amplitude.
i ₁ = I _R1 −I _S1
| I ₁ | = | I _R1 −I _S1 | = 3 ^1/2 × | I _R1 | = 3 ^1/2 × | I _S1 |
Similarly, the load current i ₂ input from the second cross-through current transformer 10 ₂ to the line selection relay 30 inverts the polarity of the vector difference between the R-phase load current I _R2 and the S-phase load current I _S2. Thus, the amplitude of the load current i ₂ is 3 ^1/2 times the amplitude of the R-phase load current I _R2 (S-phase load current I _S2 ).
i ₂ = I _R2 −I _S2
| I ₂ | = | I _R2 −I _S2 | = 3 ^1/2 × | I _R2 | = 3 ^1/2 × | I _S2 |
As a result, the load current I input to the line selection relay 30 is a vector sum of the load current i ₁ and the load current i _2, and the amplitude of the load current I is “0” (I = i ₁ + i ₂ = 0). ).
In order to make the line selection relay 30 the same as the amplitude of the load current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. The corrected load current I ′ is calculated by multiplying the load current I by 1/3 ^1/2 .
I ′ = I × 1/3 ^1/2
| I '| = | I | × 1/3 1/ ^2-

In addition, when a short circuit accident occurs in the first or second transmission / distribution lines 1L, 2L, the short-circuit current flowing in the R phase, S phase, and T phase of the first and second transmission / distribution lines 1L, 2L is expressed as I _FR , I _FS , I _FT (impedance angle is θ), the line selection relay 30 operates as follows according to the aspect of the accident.
(1) In the case of a short circuit accident between the R phase and the S phase When a short circuit accident between the R phase and the S phase occurs, the R of the first and second transmission / distribution lines 1L and 2L, as shown by the dashed arrows in FIG. An R-phase short-circuit current _IFR flows in the internal direction, and an S-phase short-circuit current _IFS flows in the S-phase of the first and second transmission and distribution lines 1L and 2L. The T-phase short circuit current I _FT does not flow in the T-phase of the transmission and distribution lines 1L and 2L.
Therefore, the short-circuit current I _Ry input to the line selection relay 30 is the same as that of the short-circuit direction relay 4 in the first embodiment described above, and the R-phase short-circuit current I _FR (the first transmission / distribution line 1L). The short-circuit current flowing in the R phase of the second and the short-circuit current flowing in the R-phase of the second transmission and distribution line 2L) and the S-phase short-circuit current I _FS (short-circuit current flowing in the S-phase of the first transmission and distribution line 1L) And the amplitude of the short-circuit current I _Ry is the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ). It becomes twice the amplitude (see FIG. 3A).
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, in order to make the line selection relay 30 the same as the amplitude of the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. The short-circuit current I _Ry is halved to calculate a corrected short-circuit current I _Ry '.
I _Ry '= I _Ry × 1/2
| I _Ry '| = | I _Ry | × 1 / 2- = | I _FR | = | I _FS |
The line selection relay 30 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the accident line is determined. In the case of a short-circuit accident between the R-phase and S-phase of the first or second transmission and distribution lines 1L, 2L, the calculated corrected short-circuit current I _Ry ′ and the R-phase to S-phase line voltage V _RS Based on this, an accident line is determined (see FIG. 3A).
When the line selection relay 30 determines that a short-circuit accident has occurred in the first transmission / distribution line 1L, the line selection relay 30 disconnects the _first to _third circuit breakers 2 ₁ to 2 _{3 all} together, and the second transmission / distribution line 2L If it is determined that a short circuit accident has occurred, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(2) In the case of a short circuit accident between the S phase and the T phase When a short circuit accident between the S phase and the T phase occurs, the S phase short circuit current I _FS is internally contained in the S phase of the first and second power transmission lines 1L and 2L. The T-phase short circuit current I _FT flows outward in the T phase of the first and second transmission / distribution lines 1L, 2L, but the R phase of the first and second transmission / distribution lines 1L, 2L. No R-phase short-circuit current I _FR flows (see FIG. 4).
Accordingly, the short-circuit current I _Ry input to the line selection relay 30 is the same as the short-circuit direction relay 4 in the first embodiment described above, and the S-phase short-circuit current −I _FS (first The polarity of the difference current between the short-circuit current flowing in the S-phase of the transmission / distribution line 1L and the short-circuit current flowing in the S-phase of the second transmission / distribution line 2L is inverted), and the amplitude of the short-circuit current I _Ry is The amplitude of the short-circuit current I _FS (see FIG. 3B).
I _Ry = −I _FS
| I _Ry | = | I _FS |
Therefore, in order to make the line selection relay 30 the same as the amplitude of the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. The corrected short-circuit current I _Ry 'is calculated by multiplying the short-circuit current I _Ry by 1.
I _Ry '= I _Ry × 1
| I _Ry '| = | I _Ry | × 1 = | I _FS |
The line selection relay 30 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the accident line is determined. In the case of a short circuit accident between the S phase and the T phase of the first or second transmission / distribution lines 1L, 2L, the calculated corrected short circuit current I _Ry ′ and the line voltage V _ST (polarity between the S phase and the T phase) Is negative), the accident line is determined (see FIG. 3B).
When the line selection relay 30 determines that a short-circuit accident has occurred in the first transmission / distribution line 1L, the line selection relay 30 disconnects the _first to _third circuit breakers 2 ₁ to 2 _{3 all} together, and the second transmission / distribution line 2L If it is determined that a short circuit accident has occurred, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(3) In the case of a short circuit accident between the T phase and the R phase When a short circuit accident occurs between the T phase and the R phase, the T phase short circuit current I _FT is internally generated in the T phase of the first and second power transmission lines 1L and 2L. The R-phase short-circuit current _IFR flows to the R phase of the first and second transmission and distribution lines 1L and 2L, but flows to the S phase of the first and second transmission and distribution lines 1L and 2L. No S-phase short-circuit current I _FS flows (see FIG. 5).
Therefore, the short-circuit current I _Ry input to the line selection relay 30 is the same as that of the short-circuit direction relay 4 in the first embodiment described above, and the R-phase short-circuit current I _FR (the first transmission / distribution line 1L). 3) and the amplitude of the short-circuit current I _{Ry is} the amplitude of the R-phase short-circuit current I _FR (FIG. 3). (See (c)).
I _Ry = I _FR
| I _Ry | = | I _FR |
Therefore, in order to make the line selection relay 30 the same as the amplitude of the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. The corrected short-circuit current I _Ry 'is calculated by multiplying the short-circuit current I _Ry by 1.
I _Ry '= I _Ry × 1
| I _Ry '| = | I _Ry | × 1 = | I _FR |
The line selection relay 30 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the accident line is determined. In the case of a short-circuit accident between the T-phase and R-phase of the first or second transmission / distribution lines 1L, 2L, the calculated corrected short-circuit current I _Ry ′ and the T-phase-R phase line voltage V _TR (polarity Is negative), the accident line is determined (see FIG. 3C).
When the line selection relay 30 determines that a short-circuit accident has occurred in the first transmission / distribution line 1L, the line selection relay 30 disconnects the _first to _third circuit breakers 2 ₁ to 2 _{3 all} together, and the second transmission / distribution line 2L If it is determined that a short circuit accident has occurred, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(4) In the case of a short circuit accident between the R phase, the S phase, and the T phase When a short circuit accident between the R phase, the S phase, and the T phase occurs, the R phase, the S phase, and the first and second power transmission lines 1L and 2L An R-phase short-circuit current I _FR , an S-phase short-circuit current I _FS and a T-phase short-circuit current I _FT flow in the T phase in the internal direction with a phase difference of 120 ° (see FIG. 6).
Therefore, the short-circuit current I _Ry input to the line selection relay 30 is the same as that of the short-circuit direction relay 4 in the first embodiment described above, and the R-phase short-circuit current I _FR (the first transmission / distribution line 1L). The short-circuit current flowing in the R phase of the second and the short-circuit current flowing in the R-phase of the second transmission and distribution line 2L) and the S-phase short-circuit current I _FS (the short-circuit current flowing in the S-phase of the first transmission and distribution line 1L) And the amplitude of the short-circuit current I _Ry is the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ). 3 ^1/2 times the amplitude (see FIG. 3D).
I _Ry = I _FR −I _FS
| I _Ry | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, in order to make the line selection relay 30 the same as the amplitude of the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. The short-circuit current I _Ry is multiplied by 1/3 ^1/2 to calculate a corrected short-circuit current I _Ry '.
I _Ry '= I _Ry × 1/3 ^1/2
_{| I Ry '| = | I} Ry | × 1/3 1/2 - = | I FR | = | I FS |
The line selection relay 30 includes the calculated corrected short-circuit current I _Ry ′, the R-phase / S-phase line voltage V _RS , the S-phase / T-phase line voltage V _ST, and the T-phase / R-phase line voltage V _TR. Based on the above, the accident line is determined. In the case of a short-circuit accident between the R-phase, S-phase, and T-phase of the first or second transmission / distribution lines 1L, 2L, the calculated corrected short-circuit current I _Ry 'and the R-phase-S-phase line voltage V _An accident line is determined based on the _RS (see FIG. 3D).
When the line selection relay 30 determines that a short-circuit accident has occurred in the first transmission / distribution line 1L, the line selection relay 30 disconnects the _first to _third circuit breakers 2 ₁ to 2 _{3 all} together, and the second transmission / distribution line 2L If it is determined that a short circuit accident has occurred, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.

Note that the first cross through the current transformer 10 ₁ has been passed through by cross the R-phase and S-phase of the first transmission and distribution lines 1L, the S phase and T-phase of the first transmission and distribution lines 1L It may be crossed and penetrated, or the T phase and R phase of the first power transmission and distribution line 1L may be crossed and penetrated.
The same applies to the second cross-through current transformer 10 ₂ .

However, according to the CT connection, the line selection relay 30 calculates the corrected short-circuit current I _Ry ′ at the magnification shown in Table 1 according to the accident aspect, and the calculated corrected short-circuit current I _Ry ′ and the table shown in Table 2. The fault line is determined based on the line voltage, and it is determined whether or not a short-circuit fault has occurred in the first or second transmission / distribution lines 1L, 2L.

Next, a directional protection relay device according to a fourth embodiment of the present invention will be described with reference to FIGS.
Direction protective relay apparatus according to this embodiment, as shown in FIG. 9, the first cross through current transformer 10 ₁ R-phase and S-phase of the transmission and distribution lines are through to cross, transmission and distribution lines the second cross through current transformer 10 ₂ that penetrate to S-phase and T-phase is cross, the instrument transformer 6 installed to the bus, the first cross through input from current transformer 10 ₁ R-S phase line voltage obtained from the first short-circuit current I _Ry1 and the R-phase, S-phase, and T-phase voltages V _R , V _S , V _T input from the instrument transformer 6 When a short circuit fault is detected on the transmission / distribution line based on V _RS , S-phase / T-phase line voltage V _ST and T-phase / R-phase line voltage V _TR , the R-phase, S-phase and a first short directional relay 4 ₁ collectively blocking the first to third circuit breaker 2 ₁ to 2 ₃ of which are respectively installed on the T-phase, a second cross through current transformer 1 The second short-circuit current I _Ry2 and R-phase inputted from the potential transformer 6 which is input from the _2, S-phase and T-phase phase voltage V _R of, V _S, the R-phase -S phase determined from V _T If a short circuit accident of the transmission / distribution line is detected based on the line voltage V _RS , the S-phase / T-phase line voltage V _ST and the T-phase / R-phase line voltage V _TR , the first to third interruptions are detected. vessels to 2 ₁ to 2 ₃ comprises a second and short directional relay 4 ₂ for collectively blocking.

Here, the first cross through the current transformer 10 _1, through which is passed through by a cross at any angle and R-phase and S-phase of the transmission and distribution lines to the annular core formed by winding a secondary coil in the opposite direction It forms a current transformer, a second cross through current transformer 10 _2, by the cross at any angle and S-phase and T-phase of the transmission and distribution lines to the annular core formed by winding a secondary coil in the opposite direction This is a penetrating through-type current transformer.
Ie, R-phase of the electric transmission has been through the first polarity direction of the cross through the current transformer 10 _1, S-phase first opposite polarity direction of the cross through the current transformer 10 _first transmission and distribution lines It is penetrated by. Similarly, S-phase of the electric transmission has been through the second polarity direction of the cross through the current transformer 10 _2, T-phase of the electric transmission a second opposite polarity cross the through current transformer 10 ₂ Penetrated in the direction.

Therefore, when the load currents flowing in the R phase, S phase, and T phase of the transmission / distribution line when no short circuit accident has occurred are represented by I _R , I _S , and I _T , as shown in FIG. The current I _R and the S-phase load current I _S flow through the annular core of the first cross-through current transformer 101 _{1 in} a reverse direction with a phase difference of 120 ° (that is, the R-phase load current I _R Flows through the annular iron core of the first cross-through current transformer 10 _{1 in} the polarity direction, and the S-phase load current I _S penetrates the annular iron core of the first cross-through current transformer 10 _{1 in} the opposite polarity direction. to flow), the first load current I ₁ is the load of the R-phase current I _R and S phases of the load current inputted from the first cross through current transformer 10 ₁ in the first short directional relay 4 ₁ It becomes a vector difference between the I _S, a first amplitude of the load current I ₁ is a 3 ^1/2 times the amplitude of the load R-phase current I _R (load current I _S of the S-phase) Become.
I ₁ = I _R −I _S
| I ₁ | = | I _R −I _S | = 3 ^1/2 × | I _R | = 3 ^1/2 × | I _S |
Therefore, in order first short directional relay 4 _1, be the same as the amplitude of the load current inputted to the first to third short directional relay 41 _{to 3} of prior art shown in FIG. 18, the following equation As shown in FIG. 1, the first corrected load current I ₁ ′ is calculated by multiplying the first load current I ₁ by 1/3 ^1/2 .
I ₁ '= I ₁ × 1/3 ^1/2
| I ₁ '| = | I ₁ | × 1/3 1/ ^2- = | I _R | = | I _S |
Similarly, through the load current I _S and T phases of the load current I _T and the second cross through current transformer 10 _{and second} annular core with a phase difference of 120 ° in the S-phase, as shown in FIG. 10 in the opposite direction (That is, the S-phase load current I _S flows through the annular core of the second cross-through current transformer 102 _{2 in} the polarity direction, and the T-phase load current I _T is the second cross-through current change). Nagareki 10 ₂ toroid flows through the opposite polarity direction), the second load current I ₂ which is input from the second cross through current transformer 10 ₂ in the second short directional relay 4 ₂ It becomes a vector difference between the S-phase load current I _S and the T-phase load current I _T, and the amplitude of the _second load current I _{2 is} the amplitude of the S-phase load current I _S (T-phase load current I _T ). 3 ^1/2 times.
I ₂ = I _S −I _{T
| I 2 | = | I S} -I T | = 3 1/2 × | I S | = 3 1/2 × | I T |
Therefore, in order to make the second short-circuit direction relay 4 ₂ have the same amplitude as the load current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown, the second load current I ₂ ′ is calculated by multiplying the second load current I ₂ by 1/3 ^1/2 .
I ₂ '= I ₂ × 1/3 ^1/2
_{| I 2 '| = | I} 2 | × 1/3 1/2 - = | I S | = | I T |

In addition, when a short circuit accident occurs in the transmission / distribution line, the short-circuit currents flowing in the R phase, S phase, and T phase of the transmission / distribution line are expressed as I _FR , I _FS , I _FT (impedance angle is θ) The first and second short-circuit direction relays 4 ₁ and 4 ₂ operate as follows according to the accident aspect.
(1) When the short circuit in the case of a short-circuit accident R phase -S phase R phase -S phase occurs, as indicated by broken line arrow in FIG. 9, the short-circuit current of the R-phase to R-phase of transmission and distribution lines I _FR There flow inside direction, but the short-circuit current I _FS of S phase to the S phase of the transmission and distribution lines to flow to the outside direction, the T-phase of the transmission and distribution lines does not flow a short-circuit current I _FT T-phase.
Accordingly, the first short-circuit current I _Ry1 inputted from the first cross through current transformer 10 ₁ in the first short directional relay 4 _1, short-circuit current of the R-phase as shown by the solid line in bold arrow in FIG. 9 The vector difference between I _FR and the S-phase short-circuit current I _FS, and the amplitude of the first short-circuit current I _Ry1 is twice the amplitude of the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) ( 11 (a-1) In addition, in Fig. 11 and Fig. 12, the short-circuit currents I _FR , I _FS , I _FT flowing in the inner direction of the transmission / distribution line are solid arrows, The flowing short-circuit currents I _FR , I _FS , and I _FT are indicated by dashed-dotted arrows)).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, in order to make the first short-circuit direction relay 4 _{1 the} same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown in FIG. 1, the first corrected short-circuit current I _Ry1 ′ is calculated by multiplying the first short-circuit current I _Ry1 by ½.
I _Ry1 '= I _Ry1 × 1/2
| I _Ry1 '| = | I _Ry1 | × 1 / 2- = | I _FR | = | I _FS |
The first short directional relay 4 _1, the first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The magnitude and direction of the first short-circuit current I _Ry1 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the R phase and the S phase of the transmission and distribution line, as shown in FIG. 11 (a-1), the calculated first corrected short circuit current I _Ry1 ′ and the R phase to S phase line The magnitude and direction of the first short-circuit current I _Ry1 is determined based on the inter-voltage V _RS .
The first short-circuit direction relay 4 ₁ collectively cuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short-circuit accident has occurred in the transmission and distribution line.
Further, the second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second short-circuit direction relay 4 ₂ is an S-phase short-circuit current as indicated by a broken thick arrow in FIG. I _FS , and the amplitude of the second short-circuit current I _{Ry2 is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 11A-2).
I _Ry2 = I _FS
| I _Ry2 | = | I _FS |
Therefore, in order to make the second short-circuit direction relay 4 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by one.
I _Ry2 '= I _Ry2 × 1
_{| I Ry2 '| = | I} Ry2 | × 1- = | I FS |
Second short directional relay 4 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the R phase and the S phase of the transmission and distribution line, as shown in FIG. 11 (a-2), the calculated second corrected short circuit current I _Ry2 ′ and the R phase to S phase line The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the inter-voltage V _RS (polarity is negative).
The second short circuit direction relay 4 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(2) S-phase -T For short circuit between the phases when the short circuit of the S-phase -T phase occurs, as indicated by broken line arrow in FIG. 13, the short-circuit current of the S-phase to the S phase of the transmission and distribution lines I _FS There flow inside direction, but the short-circuit current I _FT T-phase to the T phase of the transmission and distribution lines to flow to the outside direction, the R-phase of the transmission and distribution lines does not flow a short-circuit current I _FR of R-phase.
Accordingly, the first short-circuit current I _Ry1 inputted from the first cross through current transformer 10 ₁ in the first short directional relay 4 _1, polar, as indicated by the solid line in bold arrow in FIG. 13 is negative S The short-circuit current −I _{FS of the} phase is obtained, and the amplitude of the first short-circuit current I _{Ry1 is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 11B-1).
I _Ry1 = -I _FS
| I _Ry1 | = | I _FS |
Therefore, in order to make the first short-circuit direction relay 4 _{1 the} same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown, the first short-circuit current I _Ry1 is multiplied by 1 to calculate the first corrected short-circuit current I _Ry1 ′.
I _Ry1 '= I _Ry1 × 1
| I _Ry1 '| = | I _Ry1 | × 1 = | I _FS |
The first short directional relay 4 _1, the first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The magnitude and direction of the first short-circuit current I _Ry1 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the S phase and the T phase of the power transmission and distribution line, as shown in FIG. 11 (b-1), the calculated first corrected short circuit current I _Ry1 ′ and the S phase-T phase line The magnitude and direction of the first short-circuit current I _Ry1 is determined based on the inter-voltage V _ST (polarity is negative).
The first short-circuit direction relay 4 ₁ collectively cuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short-circuit accident has occurred in the transmission and distribution line.
Further, the second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second short-circuit direction relay 4 ₂ is an S-phase short-circuit current as indicated by a broken thick arrow in FIG. The vector difference between I _FS and the T-phase short-circuit current I _FT, and the amplitude of the second short-circuit current I _Ry2 is twice the amplitude of the S-phase short-circuit current I _FS (T-phase short-circuit current I _FT ) ( (Refer FIG.11 (b-2)).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 2 × | I _FS | = 2 × | I _FT |
Therefore, in order to make the second short-circuit direction relay 4 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown in FIG. 2, the second short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by ½.
I _Ry2 '= I _Ry2 × 1/2
| I _Ry2 '| = | I _Ry2 | × 1 / 2- = | I _FS | = | I _FT |
Second short directional relay 4 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the S phase and the T phase of the transmission and distribution line, as shown in FIG. 11 (b-2), the calculated second corrected short circuit current I _Ry2 ′ and the line between the S phase and the T phase The magnitude and direction of the second short-circuit current I _Ry2 is determined based on the inter-voltage V _ST .
The second short circuit direction relay 4 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(3) When T phase -R phase short fault when T-phase -R phase short-circuit accident occurs, as indicated by broken line arrow in FIG. 14, the short-circuit current of the T-phase to the T phase of the transmission and distribution lines I _FT There flow inside direction, but the short-circuit current I _FR of R-phase to R-phase of transmission and distribution lines to flow to the outside direction, the S-phase of the transmission and distribution lines does not flow a short-circuit current I _FS of S phase.
Accordingly, the first short-circuit current I _Ry1 inputted from the first cross through current transformer 10 ₁ in the first short directional relay 4 _1, short-circuit current of the R-phase as shown by the solid line in bold arrow in FIG. 14 I _FR , and the amplitude of the first short-circuit current I _{Ry1 is} the amplitude of the R-phase short-circuit current I _FR (see FIG. 12A-1).
I _Ry1 = I _FR
｜ I _Ry1 ｜ ＝ ｜ I _FR ｜
Therefore, in order to make the first short-circuit direction relay 4 _{1 the} same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown, the first short-circuit current I _Ry1 is multiplied by 1 to calculate the first corrected short-circuit current I _Ry1 ′.
I _Ry1 '= I _Ry1 × 1
| I _Ry1 '| = | I _Ry1 | × 1 = | I _FR |
The first short directional relay 4 _1, the first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The magnitude and direction of the first short-circuit current I _Ry1 are determined based on the phase line voltage V _TR . In addition, in the case of a short circuit accident between the T phase and the R phase of the transmission and distribution line, as shown in FIG. 12 (a-1), the calculated first corrected short circuit current I _Ry1 ′ and the T phase-R phase line The magnitude and direction of the first short-circuit current I _Ry1 is determined based on the inter-voltage V _TR (polarity is negative).
The first short-circuit direction relay 4 ₁ collectively cuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short-circuit accident has occurred in the transmission and distribution line.
Further, the second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second short-circuit direction relay 4 ₂ has a negative polarity T as shown by the broken thick arrow in FIG. The short-circuit current −I _{FT of the} phase is obtained, and the amplitude of the second short-circuit current I _{Ry2 is} the amplitude of the short-circuit current I _FT of the T-phase (see FIG. 12A-2).
I _Ry2 = −I _FT
| I _Ry2 | = | I _FT |
Therefore, in order to make the second short-circuit direction relay 4 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by one.
I _Ry2 '= I _Ry2 × 1
| I _Ry2 '| = | I _Ry2 | × 1 = | I _FT |
Second short directional relay 4 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the T phase and the R phase of the transmission and distribution line, as shown in FIG. 12 (a-2), the calculated second corrected short circuit current I _Ry2 ′ and the T phase to R phase line The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the inter-voltage V _TR (polarity is negative).
The second short circuit direction relay 4 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(4) In the case of a short circuit accident between R phase, S phase, and T phase When a short circuit accident occurs between R phase, S phase, and T phase, as indicated by the dashed arrows in FIG. The R-phase short-circuit current I _FR , the S-phase short-circuit current I _FS, and the T-phase short-circuit current I _FT flow in the internal direction with a phase difference of 120 ° in the phase and the T phase, respectively.
Accordingly, a first short-circuit current I _Ry1 inputted from the first cross through current transformer 10 ₁ in the first short directional relay 4 _1, short-circuit current of the R-phase as shown by the solid line in bold arrow in FIG. 15 The vector difference between I _FR and S-phase short-circuit current I _FS, and the amplitude of first short-circuit current I _Ry1 is 3 ^1/2 times the amplitude of R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) (See FIG. 12B-1).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, in order to make the first short-circuit direction relay 4 _{1 the} same as the amplitude of the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown in FIG. 1, the first corrected short-circuit current I _Ry1 ′ is calculated by multiplying the first short-circuit current I _Ry1 by 1/3 ^1/2 .
I _Ry1 '= I _Ry1 × 1/3 ^1/2
_{| I Ry1 '| = | I} Ry1 | × 1/3 1/2 - = | I FR | = | I FS |
The first short directional relay 4 _1, the first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The magnitude and direction of the first short-circuit current I _Ry1 are determined based on the phase line voltage V _TR . In the case of a short circuit accident R phase -S phase -T phase of electric transmission, FIG. 12 (b-1) as shown in, the first correction circuit current I _Ry1 'and R phase -S calculated The magnitude and direction of the first short-circuit current I _Ry1 are determined based on the phase line voltage V _RS .
The first short-circuit direction relay 4 ₁ collectively cuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short-circuit accident has occurred in the transmission and distribution line.
Further, the second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second short-circuit direction relay 4 ₂ is the S-phase short-circuit current as indicated by the broken thick arrow in FIG. The vector difference between I _FS and T-phase short-circuit current I _FT, and the amplitude of second short-circuit current I _Ry2 is 3 ^1/2 times the amplitude of S-phase short-circuit current I _FS (T-phase short-circuit current I _FT ) (See FIG. 12B-2).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 3 ^1/2 × | I _FS | = 3 ^1/2 × | I _FT |
Therefore, in order to make the second short-circuit direction relay 4 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third short-circuit direction relays 4 ₁ to 4 ₃ shown in FIG. As shown in FIG. 2, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by 1/3 ^1/2 .
I _Ry2 '= I _Ry2 × 1/3 ^1/2
_{| I Ry2 '| = | I} Ry2 | × 1/3 1/2 - = | I FS | = | I FT |
Second short directional relay 4 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the phase line voltage V _TR . In the case of a short circuit accident between the R phase, the S phase, and the T phase of the transmission and distribution line, as shown in FIG. 12B-2, the calculated second corrected short circuit current I _Ry2 ′ and the S phase-T The magnitude and direction of the second short-circuit current I _Ry2 are determined based on the phase line voltage V _ST .
The second short circuit direction relay 4 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.

The first cross through current transformer 10 ₁ crosses and penetrates the R phase and S phase of the power transmission and distribution line, and the second cross through current transformer 10 ₂ has the S phase and T of the power transmission and distribution line. While phases were allowed to penetrate by the cross, two phases of transmission and distribution lines to penetrate by cross the first and second cross through current transformer 10 _1, 10 ₂ may be other combinations.

However, due to the CT connection, the first and second short-circuit direction relays 4 ₁ and 4 ₂ have the first and second corrected short-circuit currents I _Ry1 ′ and I _Ry2 at the magnification shown in Table 1 according to the accident _situation. 'calculates the first and second correction circuit current I _Ry1 calculated', first and second short-circuit current I _Ry1 on the basis of the line voltage shown as in Table 2 I _Ry2 ', I _Ry2 It is determined whether or not a short circuit accident has occurred in the transmission and distribution line.

Next, a directional protection relay device according to a fifth embodiment of the present invention will be described with reference to FIG.
Direction protective relay apparatus according to this embodiment, as shown in FIG. 16, a first cross through current transformer 10 ₁ R-phase and S-phase of the transmission and distribution lines are through to cross, transmission and distribution lines The second cross-through current transformer 10 ₂ penetrated so that the S phase and the T phase cross each other, the instrument transformer 6 installed on the bus, and the first cross-through current transformer 10 _1. the first short-circuit current I _Ry1 and R-phase input from the potential transformer 6, S-phase and phase voltage V _R of the T-phase, V _S, the line voltage of the R phase -S phase determined from V _T to be When a short circuit fault is detected on the transmission / distribution line based on V _RS , S-phase / T-phase line voltage V _ST and T-phase / R-phase line voltage V _TR , the R-phase, S-phase and first distance relay 20 ₁ for collectively blocking the first to third circuit breaker 2 ₁ to 2 ₃ of which are respectively installed on the T-phase, a second cross through current transformer 1 The second short-circuit current I _Ry2 and R-phase inputted from the potential transformer 6 which is input from the _2, S-phase and T-phase phase voltage V _R of, V _S, the R-phase -S phase determined from V _T If a short circuit accident of the transmission / distribution line is detected based on the line voltage V _RS , the S-phase / T-phase line voltage V _ST and the T-phase / R-phase line voltage V _TR , the first to third interruptions are detected. And a second distance relay 20 ₂ that collectively shuts off the devices 2 _{1 to} 2 ₃ .

Here, the first cross through the current transformer 10 _1, through which is passed through by a cross at any angle and R-phase and S-phase of the transmission and distribution lines to the annular core formed by winding a secondary coil in the opposite direction It forms a current transformer, a second cross through current transformer 10 _2, by the cross at any angle and S-phase and T-phase of the transmission and distribution lines to the annular core formed by winding a secondary coil in the opposite direction This is a penetrating through-type current transformer.
Ie, R-phase of the electric transmission has been through the first polarity direction of the cross through the current transformer 10 _1, S-phase first opposite polarity direction of the cross through the current transformer 10 _first transmission and distribution lines It is penetrated by. Similarly, S-phase of the electric transmission has been through the second polarity direction of the cross through the current transformer 10 _2, T-phase of the electric transmission a second opposite polarity cross the through current transformer 10 ₂ Penetrated in the direction.

Therefore, when the load currents flowing in the R-phase, S-phase, and T-phase of the transmission / distribution line when no short circuit accident has occurred are represented by I _R , I _S , I _T , the first cross-through current transformer 10 ₁ from the first load current I ₁ which is input to the first distance relay 20 _1, as in the case of the fourth embodiment the first short directional relay in example 4 ₁ described above, the R-phase load current I The vector difference between the _R and S phase load currents I _S, and the amplitude of the _first load current I ₁ is 3 ^1/2 times the amplitude of the R phase load current I _R (S phase load current I _S ). Become.
I ₁ = I _R −I _S
| I ₁ | = | I _R −I _S | = 3 ^1/2 × | I _R | = 3 ^1/2 × | I _S |
Therefore, first distance relay 20 _1, to the same as the amplitude of the load current inputted to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, indicated by the following formula Thus, the first corrected load current I ₁ ′ is calculated by multiplying the first load current I ₁ by 2/3 ^1/2 .
I ₁ '= I ₁ × 2/3 ^1/2
| I ₁ ′ | = | I ₁ | × 2/3 ^1/2 − = 2 × | I _R | = 2 × | I _S |
Similarly, the second load current I ₂ input from the second cross-through current transformer 10 ₂ to the second distance relay 20 ₂ is the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. In the same manner as described above, the vector difference between the S-phase load current I _S and the T-phase load current I _T results in the amplitude of the _second load current I _{2 being} the S-phase load current I _S (the T-phase load 3 ^1/2 times the amplitude of the current I _T ).
I ₂ = I _S −I _{T
| I 2 | = | I S} -I T | = 3 1/2 × | I S | = 3 1/2 × | I T |
Therefore, the second distance relay 20 ₂ is expressed by the following equation in order to make it the same as the amplitude of the load current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. Thus, the second corrected load current I ₂ ′ is calculated by multiplying the second load current I ₂ by 2/3 ^1/2 .
I ₂ '= I ₂ × 2/3 ^1/2
| I ₂ '| = | I ₂ | × 2/3 ¹ /2-= 2 × | I _S | = 2 × | I _T |

In addition, when a short circuit accident occurs in the transmission / distribution line, the short-circuit currents flowing in the R phase, S phase, and T phase of the transmission / distribution line are expressed as I _FR , I _FS , I _FT (impedance angle is θ). The first and second distance relays 20 ₁ and 20 ₂ operate as follows according to the accident aspect.
(1) If a short circuit accident R phase -S phase R phase -S phase short accident occurs, as indicated by broken line arrow in FIG. 16, the short-circuit current of the R-phase to R-phase of transmission and distribution lines I _FR There flow inside direction, but the short-circuit current I _FS of S phase to the S phase of the transmission and distribution lines to flow to the outside direction, the T-phase of the transmission and distribution lines does not flow a short-circuit current I _FT T-phase.
Thus, the first cross through current transformer 10 ₁ first short-circuit current I _Ry1 which is input to the first distance relay 20 _1, the first in the fourth embodiment described above short directional relay 4 ₁ As in the case, the vector difference between the R-phase short-circuit current I _FR and the S-phase short-circuit current I _FS is obtained, and the amplitude of the first short-circuit current I _Ry1 is the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) is twice the amplitude (see FIG. 11 (a-1)).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, first distance relay 20 _1, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, indicated by the following formula Thus, the first corrected short circuit current I _Ry1 ′ is calculated by multiplying the first short circuit current I _Ry1 by one.
I _Ry1 '= I _Ry1 × 1
| I _Ry1 '| = | I _Ry1 |-× 1 = 2 × | I _FR | = 2 × | I _FS |
First distance relay 20 _1, a first correction circuit current I _Ry1 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated Based on the line voltage V _TR , the distance of the _fault point and the direction of the first short-circuit current I _Ry1 are determined. In the case of a short circuit accident between the R phase and S phase of the transmission / distribution line, the distance of the fault point is based on the calculated first corrected short circuit current I _Ry1 ′ and the line voltage V _RS between the R phase and the S phase. And the direction of 1st short circuit current _IRy1 is discriminate _| determined (refer FIG. 11 (a-1)).
First distance relay 20 _1, short circuit to the electric transmission is collectively cut off to the first to third circuit breaker 2 ₁ to 2 ₃ determined to have occurred.
The second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second distance relay 20 ₂ is the same as that of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. Similarly to the case, the S-phase short-circuit current I _{FS is obtained} , and the amplitude of the second short-circuit current I _{Ry2 is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 11A-2).
I _Ry2 = I _FS
| I _Ry2 | = | I _FS |
Therefore, the second distance relay 20 ₂ is expressed by the following equation in order to make it the same as the amplitude of the short-circuit current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. a second short-circuit current I _Ry2 calculating a second correction circuit current I _Ry2 'is doubled as.
I _Ry2 '= I _Ry2 × 2
| I _Ry2 '| = | I _Ry2 | × 2− = 2 × | I _FS |
Second distance relay 20 _2, the second correction circuit current I _Ry2 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated determine the direction of the distance of the fault point and the second short-circuit current I _Ry2 based inter line to the voltage V _TR. In the case of a short-circuit accident between the R phase and S phase of the transmission and distribution line, based on the calculated second corrected short circuit current I _Ry2 ′ and the R-S phase line voltage V _RS (negative polarity). Thus, the distance of the accident point and the direction of the second short-circuit current I _Ry2 are determined (see FIG. 11 (a-2)).
The second distance relay 20 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(2) In the case of a short-circuit accident between the S phase and the T phase When a short circuit accident between the S phase and the T phase occurs, the S phase short circuit current I _FS flows in the S phase of the transmission and distribution line in the internal direction, and the T of the transmission and distribution line Although the T-phase short-circuit current I _FT flows in the external direction in the phase, the R-phase short-circuit current I _FR does not flow in the R-phase of the transmission and distribution line (see FIG. 13).
Thus, the first cross through current transformer 10 ₁ first short-circuit current I _Ry1 which is input to the first distance relay 20 _1, the first in the fourth embodiment described above short directional relay 4 ₁ Similarly to the case, the negative polarity S-phase short-circuit current −I _{FS is obtained} , and the amplitude of the first short-circuit current I _{Ry1 is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 11B-1). ).
I _Ry1 = -I _FS
| I _Ry1 | = | I _FS |
Therefore, first distance relay 20 _1, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, indicated by the following formula the first short-circuit current I _Ry1 to calculate a first correction circuit current I _Ry1 'is doubled as.
I _Ry1 '= I _Ry1 × 2
| I _Ry1 '| = | I _Ry1 | × 2 = 2 × | I _FS |
First distance relay 20 _1, a first correction circuit current I _Ry1 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated Based on the line voltage V _TR , the distance of the _fault point and the direction of the first short-circuit current I _Ry1 are determined. In the case of a short circuit accident between the S phase and the T phase of the transmission and distribution line, based on the calculated first corrected short circuit current I _Ry1 ′ and the line voltage V _ST (negative polarity) between the S phase and the T phase. Thus, the distance of the accident point and the direction of the first short-circuit current I _Ry1 are determined (see FIG. 11B-1).
First distance relay 20 _1, short circuit to the electric transmission is collectively cut off to the first to third circuit breaker 2 ₁ to 2 ₃ determined to have occurred.
The second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second distance relay 20 ₂ is the same as that of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. As in the case, the vector difference between the S-phase short-circuit current I _FS and the T-phase short-circuit current I _FT is obtained, and the amplitude of the second short-circuit current I _Ry2 is the S-phase short-circuit current I _FS (T-phase short-circuit current I _FT ) is twice the amplitude (see FIG. 11B-2).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 2 × | I _FS | = 2 × | I _FT |
Therefore, the second distance relay 20 ₂ is expressed by the following equation in order to make it the same as the amplitude of the short-circuit current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. Thus, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by 1.
I _Ry2 '＝ I _Ry2 × 1-
| I _Ry2 '| = | I _Ry2 | × 1− = 2 × | I _FS | = 2 × | I _FT |
Second distance relay 20 _2, the second correction circuit current I _Ry2 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated determine the direction of the distance of the fault point and the second short-circuit current I _Ry2 based inter line to the voltage V _TR. In the case of a short circuit accident between the S phase and the T phase of the transmission / distribution line, the distance of the accident point based on the calculated second corrected short circuit current I _Ry2 ′ and the line voltage V _ST between the S phase and the T phase. And the direction of 2nd short circuit current _IRy2 is discriminate _| determined (refer _FIG.11 (b-2)).
The second distance relay 20 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(3) In the case of a short circuit accident between the T phase and the R phase When a short circuit accident occurs between the T phase and the R phase, a T phase short circuit current I _FT flows in the T phase of the transmission and distribution line, and the R of the transmission and distribution line. While the short-circuit current I _FR of R-phase to phase flows to the outside direction, the S-phase of the transmission and distribution lines does not flow a short-circuit current I _FS of S-phase (see Figure 14).
Thus, the first cross through current transformer 10 ₁ first short-circuit current I _Ry1 which is input to the first distance relay 20 _1, the first in the fourth embodiment described above short directional relay 4 ₁ Similarly to the case, the R-phase short-circuit current I _{FR is obtained} , and the amplitude of the first short-circuit current I _{Ry1 is} the amplitude of the R-phase short-circuit current I _FR (see FIG. 12A-1).
I _Ry1 = I _FR
｜ I _Ry1 ｜ ＝ ｜ I _FR ｜
Therefore, first distance relay 20 _1, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, indicated by the following formula the first short-circuit current I _Ry1 to calculate a first correction circuit current I _Ry1 'is doubled as.
I _Ry1 '= I _Ry1 × 2
| I _Ry1 '| = | I _Ry1 | × 2 = 2 × | I _FR |
First distance relay 20 _1, a first correction circuit current I _Ry1 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated Based on the line voltage V _TR , the distance of the _fault point and the direction of the first short-circuit current I _Ry1 are determined. In the case of a short circuit accident between the T phase and the R phase of the transmission and distribution line, based on the calculated first corrected short circuit current I _Ry1 ′ and the T-R phase line voltage V _TR (polarity is negative). Thus, the distance of the accident point and the direction of the first short-circuit current I _Ry1 are determined (see FIG. 12 (a-1)).
First distance relay 20 _1, short circuit to the electric transmission is collectively cut off to the first to third circuit breaker 2 ₁ to 2 ₃ determined to have occurred.
The second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second distance relay 20 ₂ is the same as that of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. Similarly to the case, the T-phase short-circuit current −I _FT having a negative polarity is obtained, and the amplitude of the second short-circuit current I _{Ry2 is} the amplitude of the T-phase short-circuit current I _FT (see FIG. 12A-2). ).
I _Ry2 = −I _FT
| I _Ry2 | = | I _FT |
Therefore, the second distance relay 20 ₂ is expressed by the following equation in order to make it the same as the amplitude of the short-circuit current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. a second short-circuit current I _Ry2 calculating a second correction circuit current I _Ry2 'is doubled as.
I _Ry2 '= I _Ry2 × 2
| I _Ry2 '| = | I _Ry2 | × 2 = 2 × | I _FT |
Second distance relay 20 _2, the second correction circuit current I _Ry2 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated determine the direction of the distance of the fault point and the second short-circuit current I _Ry2 based inter line to the voltage V _TR. In the case of a short-circuit accident between the T-phase and R-phase of the power transmission and distribution line, based on the calculated second corrected short-circuit current I _Ry2 ′ and the T-phase-R line voltage V _TR (negative polarity). Thus, the distance of the accident point and the direction of the second short-circuit current I _Ry2 are determined (see FIG. 12A-2).
The second distance relay 20 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.
(4) In the case of a short circuit accident between R phase, S phase, and T phase When a short circuit accident between R phase, S phase, and T phase occurs, short circuit current I of R phase to R phase, S phase, and T phase of the transmission and distribution line _FR , S-phase short-circuit current I _FS and T-phase short-circuit current I _FT flow in the internal direction with a phase difference of 120 ° (see FIG. 15).
Thus, the first cross through current transformer 10 ₁ first short-circuit current I _Ry1 which is input to the first distance relay 20 _1, the first in the fourth embodiment described above short directional relay 4 ₁ As in the case, the vector difference between the R-phase short-circuit current I _FR and the S-phase short-circuit current I _FS is obtained, and the amplitude of the first short-circuit current I _Ry1 is the R-phase short-circuit current I _FR (S-phase short-circuit current I _FS ) is 3 ^1/2 times the amplitude (see FIG. 12B-1).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, first distance relay 20 _1, to the same as the amplitude of the short-circuit current which is input to the first to third distance relay 20 ₁ to 20 ₃ of the prior art shown in FIG. 19, indicated by the following formula Thus, the first corrected short-circuit current I _Ry1 ′ is calculated by multiplying the first short-circuit current I _Ry1 by 2/3 ^1/2 .
I _Ry1 '= I _Ry1 × 2/3 ^1/2
| I _Ry1 '| = | I _Ry1 | × 2/3 ¹ /2-= 2 × | I _FR | = 2 × | I _FS |
First distance relay 20 _1, a first correction circuit current I _Ry1 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated Based on the line voltage V _TR , the distance of the _fault point and the direction of the first short-circuit current I _Ry1 are determined. In the case of a short circuit accident between the R phase, the S phase, and the T phase of the transmission and distribution line, the accident is based on the calculated first corrected short circuit current I _Ry1 ′ and the line voltage V _RS between the R phase and the S phase. The distance between the points and the direction of the first short-circuit current I _Ry1 are determined (see FIG. 12A-1).
First distance relay 20 _1, short circuit to the electric transmission is collectively cut off to the first to third circuit breaker 2 ₁ to 2 ₃ determined to have occurred.
The second short-circuit current I _Ry2 input from the second cross-through current transformer 10 ₂ to the second distance relay 20 ₂ is the same as that of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. As in the case, the vector difference between the S-phase short-circuit current I _FS and the T-phase short-circuit current I _FT is obtained, and the amplitude of the second short-circuit current I _Ry2 is the S-phase short-circuit current I _FS (T-phase short-circuit current I _FT ) is 3 ^1/2 times the amplitude (see FIG. 12B-2).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 3 ^1/2 × | I _FS | = 3 ^1/2 × | I _FT |
Therefore, the second distance relay 20 ₂ is expressed by the following equation in order to make it the same as the amplitude of the short-circuit current input to the conventional first to third distance relays 20 _{1 to} 20 ₃ shown in FIG. Thus, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by 2/3 ^1/2 .
I _Ry2 '= I _Ry2 × 2/3 ^1/2
| I _Ry2 '| = | I _Ry2 | × 2/3 ¹ /2-= 2 × | I _FS | = 2 × | I _FT |
Second distance relay 20 _2, the second correction circuit current I _Ry2 'and line voltage V _RS of the R-phase -S phase, S phase -T phase line voltage V _ST and T-phase -R phase calculated determine the direction of the distance of the fault point and the second short-circuit current I _Ry2 based inter line to the voltage V _TR. In the case of a short circuit accident between the R phase, the S phase, and the T phase of the transmission and distribution line, the accident is based on the calculated second corrected short circuit current I _Ry2 ′ and the line voltage V _ST between the S phase and the T phase. The distance between the points and the direction of the second short-circuit current I _Ry2 are determined (see FIG. 12B-2).
The second distance relay 20 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 ₃ when it is determined that a short circuit accident has occurred in the transmission and distribution line.

The first cross through current transformer 10 ₁ crosses and penetrates the R phase and S phase of the power transmission and distribution line, and the second cross through current transformer 10 ₂ has the S phase and T of the power transmission and distribution line. While phases were allowed to penetrate by the cross, two phases of transmission and distribution lines to penetrate by cross the first and second cross through current transformer 10 _1, 10 ₂ may be other combinations.

However, the CT connection, first and second distance relay 20 _1, 20 _2, depending on the accident aspect, the magnification at the first and second correction circuit current I _Ry1 shown in Table 3 ', I _Ry2' calculates the first and second correction circuit current I calculated _Ry1 ', I _Ry2' the distance of the accident point on the basis of the line voltage as shown in tables 2 and first and second short-circuit current I _Ry1, to determine the direction of the I _Ry2, determines whether the short-circuit fault occurs in the transmission and distribution lines.

Next, a directional protection relay device according to a sixth embodiment of the present invention will be described with reference to FIG.
As shown in FIG. 17, the directional protection relay device according to the present embodiment has a first cross that is penetrated so that the R phase and the S phase of the first transmission / distribution line 1L of the balanced two-line transmission / distribution line cross each other. through a current transformer 10 _1, and the second cross through current transformer 10 ₂ that penetrate to R-phase and S-phase of the second transmission and distribution lines 2L equilibrium two-line transmission and distribution lines to cross, the first A third cross-penetrating current transformer 10 ₃ penetrated so that the S phase and the T phase of the transmission / distribution line 1L cross, and the S phase of the transmission / distribution line of the second transmission / distribution line 2L and a fourth cross through current transformer 10 ₄ T-phase is through to cross, the instrument transformer 6 placed bus, short-circuit current which is input from the first cross through current transformer 10 ₁ and a second difference between the short-circuit current input from the cross through current transformer 10 ₂ current (hereinafter, "first short-circuit current I _Ry1" . Referred to) and R-phase input from the potential transformer 6, S-phase and T-phase phase voltage V _R of, V _S, the line voltage than obtained R-phase -S phase V _T V _RS, S phase - the first or second transmission and distribution lines 1L based on the line voltage V _TR of the T-phase line voltage V _ST and T-phase -R phase, upon detecting a short circuit of 2L, feeding of the first and second Circuit breakers (R phase of the first transmission line 1L) installed in the R phase, S phase, and T phase of the distribution line 1L, 2L where the short circuit accident occurred (hereinafter referred to as "accident line") , _First to _third circuit breakers 2 ₁ to 2 ₃ installed in S phase and T phase, respectively, or fourth to fourth circuit breakers installed in R phase, S phase and T phase of second transmission line 2L, respectively. first the line selection relay 30 ₁ for collectively blocking 6 breaker 2 _{4-2 6)} of the short-circuit current which is input from the third cross through current transformer 10 ₃ Preliminary fourth difference of the short circuit current that is input from the cross-through current transformer 10 ₄ current (hereinafter, referred to as a "second short-circuit current I _Ry2".) And the R-phase input from the potential transformer 6, S R-phase to S-phase line voltage V _RS , S-phase to T-phase line voltage V _ST and T-phase to R-phase line voltage obtained from phase voltages T _R , V _S , and V _T the first or second transmission and distribution lines 1L on the basis of the voltage V _TR, upon detecting a short circuit of 2L, R-phase of the accident line breaker installed respectively in S-phase and T-phase (first to third And a second circuit selection relay 30 ₂ that collectively cuts off the circuit breakers 2 _{1 to} 2 ₃ or the fourth to _sixth circuit breakers 2 ₄ to 2 ₆ ).

Here, the first cross through the current transformer 10 ₁ is the toroids were wound secondary coil are crossed at any angle and the R-phase and S-phase of the first transmission and distribution lines 1L reversed a through-type current transformer which is passed through the second cross through current transformer 10 _2, the annular core formed by winding a secondary coil of S-phase and T-phase of the first transmission and distribution lines 1L reversed The third cross-through current transformer 10 ₃ is a through-type current transformer crossed at an arbitrary angle, and the third cross-through current transformer 10 ₃ has an R of the second power transmission and distribution line 2L around an annular core around which a secondary coil is wound. a phase and a through type current transformer which is penetrated by a cross at any angle and the S-phase in the opposite direction, a fourth cross through current transformer ₁₀₄ of the first to toroids were wound secondary coil 2 is a through-type current transformer in which the S phase and the T phase of the second transmission / distribution line 2L are crossed in an opposite direction and crossed at an arbitrary angle.
Ie, R-phase of the first transmission and distribution lines 1L is through the first polarity direction of the cross through the current transformer 10 ₁ but, S phase of the first transmission and distribution lines 1L is first cross through varying It is penetrated in the opposite polarity direction of Nagareki 10 _1.
The second cross through current transformer 10 ₂ is configured in the same manner as the first cross through current transformer 10 ₁ , but from the second cross through current transformer 10 ₂ to the first line selection relay 30 _1. The first line selection relay 30 ₁ is such that the polarity of the short-circuit current input to the first line-through current transformer 10 ₁ is opposite to the polarity of the short-circuit current input to the first line selection relay 30 _1. Connected with.
In addition, the S phase of the first transmission / distribution line 1L is penetrated in the polarity direction of the third cross-through current transformer 10 ₃ , but the T-phase of the first transmission / distribution line 1L is the third cross-penetration change. It penetrates in the opposite direction of the flow device 10 ₃ .
The fourth cross through current transformer 10 ₄ is configured in the same manner as the third cross through current transformer 10 ₃ , but the fourth cross through current transformer 10 _{4 is connected} to the second line selection relay 30 _2. the polarity of the short-circuit current to be input first to the third cross through current transformer ₁₀₃ from the second line selection relay 30 so that the opposite polarity of the short circuit current that is input to the ₂ second line selection relay 30 ₂ Connected with.

Accordingly, the load current flowing in the R phase, S phase, and T phase of the first transmission / distribution line 1L when no short circuit accident has occurred is represented by I _R1 , I _S1 , I _T1 and the second transmission / distribution line 2L. When the load currents flowing in the R phase, S phase, and T phase are represented by I _R2 , I _S2 , and I _T2 , the load that is input from the first cross-through current transformer 10 ₁ to the first line selection relay 30 ₁ current i _11, as in the case of the first short directional relay 4 ₁ of the fourth embodiment described above, it is the vector difference between the load current I _S1 of the R-phase load current I _R1 and S-phase, load The amplitude of the current i ₁₁ is 3 ^1/2 times the amplitude of the R-phase load current I _R1 (S-phase load current I _S1 ).
i ₁₁ = I _R1 −I _{S1
| I 11 | = | I R1} -I S1 | = 3 1/2 × | I R1 | = 3 1/2 × | I S1 |
Similarly, the load current i ₁₂ input from the _second cross-through current transformer 10 ₂ to the first line selection relay 30 ₁ is the vector difference between the R-phase load current I _R2 and the S-phase load current I _S2. The amplitude of the load current i ₁₂ is 3 ^1/2 times the amplitude of the R-phase load current I _R2 (S-phase load current I _S2 ).
i ₁₂ = I _R2 −I _S2
| I ₁₂ | = | I _R2 −I _S2 | = 3 ^1/2 × | I _R2 | = 3 ^1/2 × | I _S2 |
As a result, the first load current I ₁ which is input from the first and second cross through current transformer 10 _1, 10 ₂ to the first line selection relay 30 ₁ of the load current i ₁₁ and the load current i ₁₂ The amplitude of the first load current I ₁ is “0” (I ₁ = i ₁₁ + i ₁₂ = 0).
In order to make the first line selection relay 30 _{1 the} same as the amplitude of the load current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown in FIG. 1, the first corrected load current I ₁ ′ is calculated by multiplying the first load current I ₁ by 1/3 ^1/2 .
I ₁ '= I ₁ × 1/3 ^1/2
| I ₁ '| = | I ₁ | × 1/3 1/ ^2-
In addition, the load current i ₂₁ input from the _third cross-through current transformer 10 ₃ to the second line selection relay 30 ₂ is the same as that of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. Similarly, the vector difference between the S-phase load current I _S1 and the T-phase load current I _T1 is obtained, and the amplitude of the load current i _{21 is} the amplitude of the S-phase load current I _S1 (T-phase load current I _T1 ). 3 ^1/2 times as much.
i ₂₁ = I _S1 −I _T1
| I ₂₁ | = | I _S1 −I _T1 | = 3 ^1/2 × | I _S1 | = 3 ^1/2 × | I _T1 |
Similarly, the load current i ₂₂ input from the _fourth cross-through current transformer 10 ₄ to the second line selection relay 30 ₂ is a vector difference between the S-phase load current I _S2 and the T-phase load current I _T2. The amplitude of the load current i ₂₂ is 3 ^1/2 times the amplitude of the S-phase load current I _S2 (T-phase load current I _T2 ).
i ₂₂ = I _S2 −I _T2
| I ₂₂ | = | I _S2 −I _T2 | = 3 ^1/2 × | I _S2 | = 3 ^1/2 × | I _T2 |
As a result, the second load current I ₂ that is input from the third and fourth cross through current transformer _{103, 104} to the second line selection relay 30 ₂ of the load current i ₂₁ and the load current i ₂₂ The amplitude of the second load current I ₂ is “0” (I ₂ = i ₂₁ + i ₂₂ = 0).
In order to make the second line selection relay 30 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown, the second load current I ₂ ′ is calculated by multiplying the second load current I ₂ by 1/3 ^1/2 .
I ₂ '= I ₂ × 1/3 ^1/2
| I ₂ '| = | I ₂ | × 1/3 ^1/2

Further, when a short circuit accident occurs in the first or second transmission / distribution lines 1L, 2L, the short-circuit current flowing in the R phase, S phase, and T phase of the first and second transmission / distribution lines 1L, 2L is expressed as I _FR , I _FS , I _FT (impedance angle is θ), the first and second line selection relays 30 ₁ , 30 ₂ operate as follows according to the accident aspect.
In the case of a short circuit accident between the R phase and the S phase When a short circuit accident between the R phase and the S phase occurs, the R phase of the first and second transmission / distribution lines 1L and 2L is changed as shown by the broken arrows in FIG. The R-phase short-circuit current I _FR flows in the internal direction, and the S-phase short-circuit current I _FS flows in the S-phase of the first and second transmission and distribution lines 1L and 2L. The T-phase short circuit current I _FT does not flow in the T-phase of the distribution lines 1L and 2L.
Accordingly, the first short-circuit current I _Ry1 input to the first line selection relay 30 ₁ is short-circuited in the R phase in the same manner as the _first short-circuit direction relay ₄₁ in the fourth embodiment described above. The current I _FR (the difference between the short-circuit current flowing in the R phase of the first transmission / distribution line 1L and the short-circuit current flowing in the R phase of the second transmission / distribution line 2L) and the S-phase short-circuit current I _FS (first Vector difference between the short-circuit current flowing in the S phase of the transmission / distribution line 1L and the short-circuit current flowing in the S-phase of the second transmission / distribution line 2L), and the amplitude of the first short-circuit current I _Ry1 is R-phase This is twice the amplitude of the short-circuit current I _FR (S-phase short-circuit current I _FS ) (see FIG. 11 (a-1)).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 2 × | I _FR | = 2 × | I _FS |
Therefore, in order to make the first line selection relay 30 ₁ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown in FIG. 1, the first corrected short-circuit current I _Ry1 ′ is calculated by multiplying the first short-circuit current I _Ry1 by ½.
I _Ry1 '= I _Ry1 × 1/2
| I _Ry1 '| = | I _Ry1 |-× 1/2 = | I _FR | = | I _FS |
The first line selecting relay 30 _1, a first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the R phase and the S phase of the first or second power transmission line 1L, 2L, the calculated first corrected short circuit current I _Ry1 ′ and the R phase to S phase line voltage V _RS. Based on the above, the accident line is determined (see FIG. 11 (a-1)).
If the first line selection relay 30 determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the first circuit breaker 2 ₁ -2 ₃ are collectively disconnected, and the second transmission / distribution When it is determined that a short circuit accident has occurred in the electric wire 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
The second short-circuit current I _Ry2 input to the second line selection relay 30 ₂ is short-circuited in the S phase in the same manner as the _second short-circuit direction relay 4 ₂ in the fourth embodiment described above. The current I _FS (the difference between the short circuit current flowing in the S phase of the first transmission / distribution line 1L and the short circuit current flowing in the S phase of the second transmission / distribution line 2L), and the amplitude of the second short circuit current I _Ry2 is The amplitude of the S-phase short-circuit current I _FS (see FIG. 11A-2).
I _Ry2 = I _FS
| I _Ry2 | = | I _FS |
Therefore, in order to make the second line selection relay 30 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by one.
I _Ry2 '= I _Ry2 × 1
_{| I Ry2 '| = | I} Ry2 | × 1- = | I FS |
The second line selection relay 30 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the R phase and the S phase of the first or second transmission / distribution lines 1L and 2L, the calculated second corrected short circuit current I _Ry2 ′ and the line voltage V between the R phase and the S phase are calculated. _{The fault} line is determined based on _RS (negative polarity) (see FIG. 11 (a-2)).
When the second line selection relay 30 ₂ determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the _second circuit selection relay 30 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 _3, and When it is determined that a short circuit accident has occurred in the distribution line 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(2) In the case of a short circuit accident between the S phase and the T phase When a short circuit accident occurs between the S phase and the T phase, the S phase short circuit current I _FS is internally generated in the S phase of the first and second power transmission lines 1L and 2L. The T-phase short circuit current I _FT flows outward in the T phase of the first and second transmission / distribution lines 1L, 2L, but the R phase of the first and second transmission / distribution lines 1L, 2L. No R-phase short circuit current I _FR flows (see FIG. 13).
Therefore, the first short-circuit current I _Ry1 input to the first line selection relay 30 ₁ is negative in polarity as in the case of the first short-circuit direction relay 4 ₁ in the fourth embodiment described above. S-phase short-circuit current -I _FS (the polarity of the difference current between the short-circuit current flowing in the S-phase of the first transmission / distribution line 1L and the short-circuit current flowing in the S-phase of the second transmission / distribution line 2L) The amplitude of the first short-circuit current I _{Ry1 is} the amplitude of the S-phase short-circuit current I _FS (see FIG. 11B-1).
I _Ry1 = -I _FS
| I _Ry1 | = | I _FS |
Therefore, in order to make the first line selection relay 30 ₁ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown, the first short-circuit current I _Ry1 is multiplied by 1 to calculate the first corrected short-circuit current I _Ry1 ′.
I _Ry1 '= I _Ry1 × 1
| I _Ry1 '| = | I _Ry1 | × 1 = | I _FS |
The first line selecting relay 30 _1, a first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the S phase and the T phase of the first or second power transmission line 1L, 2L, the calculated first corrected short circuit current I _Ry1 ′ and the line voltage V _ST between the S phase and the T phase. The accident line is determined based on (the polarity is negative) (see FIG. 11B-1).
If the first line selection relay 30 determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the first circuit breaker 2 ₁ -2 ₃ are collectively disconnected, and the second transmission / distribution When it is determined that a short circuit accident has occurred in the electric wire 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
Further, the second short-circuit current I _Ry2 input to the second line selection relay 30 ₂ is short-circuited in the S phase in the same manner as the _second short-circuit direction relay 4 ₂ in the fourth embodiment described above. The current I _FS (the difference between the short-circuit current flowing in the S phase of the first transmission / distribution line 1L and the short-circuit current flowing in the S phase of the second transmission / distribution line 2L) and the T-phase short-circuit current I _FT (first Vector difference between the short-circuit current flowing in the T phase of the transmission and distribution line 1L and the short-circuit current flowing in the T phase of the second transmission and distribution line 2L), and the amplitude of the second short-circuit current I _Ry2 is This is twice the amplitude of the short-circuit current I _FS (T-phase short-circuit current I _FT ) (see FIG. 11B-2).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 2 × | I _FS | = 2 × | I _FT |
Therefore, in order to make the second line selection relay 30 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown in FIG. 2, the second short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by ½.
I _Ry2 '= I _Ry2 × 1 / 2-
| I _Ry2 '| = | I _Ry2 | × 1 / 2- = | I _FS | = | I _FT |
The second line selection relay 30 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the S phase and the T phase of the first or second transmission / distribution lines 1L, 2L, the calculated second corrected short circuit current I _Ry2 ′ and the line voltage V between the S phase and the T phase _The accident line is determined based on the _ST (see FIG. 11B-2).
When the second line selection relay 30 ₂ determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the _second circuit selection relay 30 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 _3, and When it is determined that a short circuit accident has occurred in the distribution line 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(3) In the case of a short circuit accident between the T phase and the R phase When a short circuit accident occurs between the T phase and the R phase, the T phase short circuit current I _FT is internally generated in the T phase of the first and second power transmission lines 1L and 2L. The R-phase short-circuit current _IFR flows to the R phase of the first and second transmission and distribution lines 1L and 2L, but flows to the S phase of the first and second transmission and distribution lines 1L and 2L. No S-phase short-circuit current _IFS flows (see FIG. 14).
Accordingly, the first short-circuit current I _Ry1 input to the first line selection relay 30 ₁ is short-circuited in the R phase in the same manner as the _first short-circuit direction relay ₄₁ in the fourth embodiment described above. The current I _FR (the difference between the short-circuit current flowing in the R phase of the first transmission / distribution line 1L and the short-circuit current flowing in the R-phase of the second transmission / distribution line 2L), and the amplitude of the first short-circuit current I _Ry1 is This is the amplitude of the R-phase short-circuit current I _FR (see FIG. 12A-1).
I _Ry1 = I _FR
｜ I _Ry1 ｜ ＝ ｜ I _FR ｜
Therefore, in order to make the first line selection relay 30 ₁ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown, the first short-circuit current I _Ry1 is multiplied by 1 to calculate the first corrected short-circuit current I _Ry1 ′.
I _Ry1 '= I _Ry1 × 1
| I _Ry1 '| = | I _Ry1 | × 1 = | I _FR |
The first line selecting relay 30 _1, a first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the T phase and the R phase of the first or second power transmission line 1L, 2L, the calculated first corrected short circuit current I _Ry1 ′ and the line voltage V _TR between the T phase and the R phase. The accident line is determined based on (the polarity is negative) (see FIG. 12A-1).
If the first line selection relay 30 determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the first circuit breaker 2 ₁ -2 ₃ are collectively disconnected, and the second transmission / distribution When it is determined that a short circuit accident has occurred in the electric wire 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
The second short-circuit current I _Ry2 input to the second line selection relay 30 ₂ is negative in polarity as in the case of the second short-circuit direction relay 4 ₂ in the fourth embodiment described above. T-phase short-circuit current −I _FT (the polarity of the difference current between the short-circuit current flowing in the T-phase of the first transmission / distribution line 1L and the short-circuit current flowing in the T-phase of the second transmission / distribution line 2L) The amplitude of the second short-circuit current I _{Ry2 is} the amplitude of the T-phase short-circuit current I _FT .
I _Ry2 = −I _FT
| I _Ry2 | = | I _FT |
Therefore, in order to make the second line selection relay 30 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. a second short-circuit current I _Ry2 1 times as shown by calculating a second correction circuit current I _Ry2 'with (see FIG. 12 (a-2)).
I _Ry2 '= I _Ry2 × 1
| I _Ry2 '| = | I _Ry2 | × 1 = | I _FT |
The second line selection relay 30 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the T phase and the R phase of the first or second transmission and distribution lines 1L and 2L, the calculated second corrected short circuit current I _Ry2 ′ and the line voltage V between the T phase and the R phase are calculated. _{The fault} line is determined based on _TR (negative polarity) (see FIG. 12 (a-2)).
When the second line selection relay 30 ₂ determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the _second circuit selection relay 30 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 _3, and When it is determined that a short circuit accident has occurred in the distribution line 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
(4) In the case of a short circuit accident between the R phase, the S phase, and the T phase When a short circuit accident between the R phase, the S phase, and the T phase occurs, the R phase, the S phase, and the first and second power transmission lines 1L and 2L An R-phase short-circuit current I _FR , an S-phase short-circuit current I _FS, and a T-phase short-circuit current I _FT flow in the T phase in the internal direction with a phase difference of 120 ° (see FIG. 15).
Accordingly, the first short-circuit current I _Ry1 input to the first line selection relay 30 ₁ is short-circuited in the R phase in the same manner as the _first short-circuit direction relay ₄₁ in the fourth embodiment described above. The current I _FR (the difference between the short-circuit current flowing in the R phase of the first transmission / distribution line 1L and the short-circuit current flowing in the R phase of the second transmission / distribution line 2L) and the S-phase short-circuit current I _FS (first Vector difference between the short-circuit current flowing in the S phase of the transmission / distribution line 1L and the short-circuit current flowing in the S-phase of the second transmission / distribution line 2L), and the amplitude of the first short-circuit current I _Ry1 is R-phase This is 3 ^1/2 times the amplitude of the short-circuit current I _FR (S-phase short-circuit current I _FS ) (see FIG. 12B-1).
I _Ry1 = I _FR -I _FS
| I _Ry1 | = | I _FR −I _FS | = 3 ^1/2 × | I _FR | = 3 ^1/2 × | I _FS |
Therefore, in order to make the first line selection relay 30 ₁ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown in FIG. 1, the first corrected short-circuit current I _Ry1 ′ is calculated by multiplying the first short-circuit current I _Ry1 by 1/3 ^1/2 .
I _Ry1 '= I _Ry1 × 1/3 ^1/2
_{| I Ry1 '| = | I} Ry1 | × 1/3 1/2 - = | I FR | = | I FS |
The first line selecting relay 30 _1, a first correction circuit current I _Ry1 'and R phase -S phase line voltage V _RS of, S interphase -T phase of the line voltage V _ST and T phases calculated -R The _fault line is determined based on the phase line voltage _VTR . In the case of a short circuit accident between the R phase, the S phase, and the T phase of the first or second transmission lines 1L, 2L, the calculated first corrected short circuit current I _Ry1 ′ and the line between the R phase and the S phase line An accident line is determined based on the voltage V _RS (see FIG. 12B-1).
If the first line selection relay 30 determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the first circuit breaker 2 ₁ -2 ₃ are collectively disconnected, and the second transmission / distribution When it is determined that a short circuit accident has occurred in the electric wire 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.
Further, the second short-circuit current I _Ry2 input to the second line selection relay 30 ₂ is short-circuited in the S phase in the same manner as the _second short-circuit direction relay 4 ₂ in the fourth embodiment described above. The current I _FS (the difference between the short-circuit current flowing in the S phase of the first transmission / distribution line 1L and the short-circuit current flowing in the S phase of the second transmission / distribution line 2L) and the T-phase short-circuit current I _FT (first Vector difference between the short-circuit current flowing in the T phase of the transmission and distribution line 1L and the short-circuit current flowing in the T phase of the second transmission and distribution line 2L), and the amplitude of the second short-circuit current I _Ry2 is This is 3 ^1/2 times the amplitude of the short-circuit current I _FS (T-phase short-circuit current I _FT ) (see FIG. 12B-2).
I _Ry2 = I _FS -I _FT
| I _Ry2 | = | I _FS −I _FT | = 3 ^1/2 × | I _FS | = 3 ^1/2 × | I _FT |
Therefore, in order to make the second line selection relay 30 ₂ have the same amplitude as the short-circuit current input to the conventional _first to _third line selection relays 30 ₁ to 30 ₃ shown in FIG. As shown in FIG. 2, the second corrected short-circuit current I _Ry2 ′ is calculated by multiplying the second short-circuit current I _Ry2 by 1/3 ^1/2 .
I _Ry2 '= I _Ry2 × 1/3 ^1/2
_{| I Ry2 '| = | I} Ry2 | × 1/3 1/2 - = | I FS | = | I FT |
The second line selection relay 30 _2, the second correction circuit current I _Ry2 'and R phase -S phase line voltage V _RS of the line voltage of the S-phase -T phase V _ST and T phases -R calculated The _fault line is determined based on the phase line voltage _VTR . In the case of a short-circuit accident between the R phase, the S phase, and the T phase of the first or second transmission and distribution lines 1L, 2L, the calculated second corrected short circuit current I _Ry2 ′ and the S phase-T phase line The _fault line is determined based on the inter-voltage V _ST (see FIG. 12B-2).
When the second line selection relay 30 ₂ determines that a short circuit accident has occurred in the first transmission / distribution line 1L, the _second circuit selection relay 30 ₂ collectively shuts off the _first to _third circuit breakers 2 ₁ to 2 _3, and When it is determined that a short circuit accident has occurred in the distribution line 2L, the _fourth to _sixth circuit breakers 2 ₄ to 2 ₆ are collectively disconnected.

Note that the first cross through the current transformer 10 ₁ in the first transmission and distribution lines 1L of R-phase and S-phase third cross through current transformer ₁₀₃ causes penetrate by cross the feed first The S phase and the T phase of the distribution line 1L are crossed and penetrated, but the two phases of the transmission and distribution lines crossed and penetrated by the first and third cross through current transformers 10 ₁ and 10 ₃ are in other combinations. But you can.
The same applies to the second and fourth cross-through current transformers 10 ₂ and 10 ₄ .

However, due to the CT connection, the first and second line selection relays 30 ₁ and 30 ₂ have the first and second corrected short-circuit currents I _Ry1 ′ and I _Ry2 at the magnification shown in Table 1 according to the accident _situation. 'calculates the, the first and second calculated correction circuit current I _Ry1', to determine the accident line based on the line voltage, indicated as I _Ry2 'in Table 2, the first or second It is determined whether or not a short circuit accident has occurred in the transmission and distribution lines 1L and 2L.

  As described above, in the fourth to sixth embodiments, by using two sets of cross-through current transformers and directional protection relays for transmission and distribution lines, it is possible to ensure even a short-circuit accident extending over the own circuit and other circuits. Even if one directional protection relay can be detected and becomes unusable due to a failure or inspection, the short circuit accident of its own circuit can be backed up by another directional short circuit protection relay.

In the above, the cross-through current transformer has been described in combination with the direction protection relay used in the transmission and distribution lines. However, the cross-through current transformer is used for the direction protection used in the three-phase AC circuit other than the transmission and distribution lines. Even when combined with an apparatus, the same effect can be obtained.
In addition, any two phases of the three-phase AC circuit are crossed once in the opposite direction through the annular core of the cross-through current transformer, but any two phases of the three-phase AC circuit cross two or more times. Thus, any two phases of the three-phase AC circuit may be wound around the annular core of the cross-through current transformer the same number or different times so as to penetrate the cross-through current transformer.

The accident aspect can be determined based on, for example, three line voltages as shown in Table 4. In Table 4, a circle indicates a line voltage at which a voltage drop is detected based on voltage information from an undervoltage relay installed on the bus, and a cross indicates a voltage from the undervoltage relay. The line voltage in which no voltage drop was detected based on the information is shown (the voltage drop detection sensitivity is about 75 to 80% of the rated voltage).

It is a figure for demonstrating the direction protection relay apparatus by 1st Example of this invention. It is a figure for demonstrating the load current input into the short circuit direction relay 4 from the cross penetration current transformer 10 shown in FIG. 1 when the short circuit accident has not occurred. It is a figure for demonstrating short circuit current _IRy inputted into short circuit direction relay 4 from cross penetration current transformer 10 shown in Drawing 1 when a short circuit accident occurs. It is a figure for demonstrating the short circuit current _IRy input into the short circuit direction relay 4 from the cross penetration current transformer 10 when the short circuit accident generate | occur | produces between the S phase-T phases of the power transmission and distribution line shown in FIG. It is a figure for demonstrating the short circuit current _IRy input into the short circuit direction relay 4 from the cross penetration current transformer 10 when the short circuit accident generate | occur | produces between the T phase-R phases of the power transmission and distribution line shown in FIG. The figure for demonstrating the short circuit current _IRy input into the short circuit direction relay 4 from the cross penetration current transformer 10 when the short circuit accident generate | occur | produces between the R phase-S phase-T phase of the power transmission and distribution line shown in FIG. It is. It is a figure for demonstrating the direction protection relay apparatus by 2nd Example of this invention. It is a figure for demonstrating the direction protection relay apparatus by the 3rd Example of this invention. It is a figure for demonstrating the direction protection relay apparatus by the 4th Example of this invention. The first and second short-circuiting current transformers 10 ₁ and 10 ₂ shown in FIG. 9 and the first and second short-circuit direction relays 4 ₁ and 4 ₂ are respectively input to the first and second short-circuit direction relays 4 ₁ and 4 ₂ when a short-circuit accident occurs. and is a diagram for explaining the second load current I _1, I ₂ for. The first and second short-circuiting current transformers 10 ₁ and 10 ₂ shown in FIG. 9 and the first and second short-circuit direction relays 4 ₁ and 4 ₂ are respectively input to the first and second short-circuit direction relays 4 ₁ and 4 ₂ when a short circuit accident occurs. and second short-circuit current I _Ry1, is a diagram for explaining I _Ry2. The first and second short-circuiting current transformers 10 ₁ and 10 ₂ shown in FIG. 9 and the first and second short-circuit direction relays 4 ₁ and 4 ₂ are respectively input to the first and second short-circuit direction relays 4 ₁ and 4 ₂ when a short circuit accident occurs. and second short-circuit current I _Ry1, is a diagram for explaining I _Ry2. When a short circuit accident occurs between the S phase and the T phase of the transmission and distribution line shown in FIG. 9, the first and second cross through current transformers 10 ₁ and 10 ₂ to the first and second short circuit direction relays 4 _1. is a diagram for explaining a first and second short-circuit current I _Ry1, I _Ry2 which are input to the 4 _2. When a short circuit accident occurs between the T phase and the R phase of the power transmission and distribution line shown in FIG. 9, the first and second cross through current transformers 10 ₁ and 10 ₂ to the first and second short circuit direction relays 4 _1. is a diagram for explaining a first and second short-circuit current I _Ry1, I _Ry2 which are input to the 4 _2. First and second short-circuit directions from the first and _second cross-through current transformers 10 ₁ and 10 ₂ when a short-circuit accident occurs between the R-phase, S-phase, and T-phase of the power transmission and distribution line shown in FIG. relay 4 ₁ is a diagram for explaining a 4 first and are input to the ₂ second short-circuit current I _Ry1, I _Ry2. It is a figure for demonstrating the direction protection relay apparatus by the 5th Example of this invention. It is a figure for demonstrating the direction protection relay apparatus by the 6th Example of this invention. It is a figure for demonstrating the conventional short circuit direction relay. It is a figure for demonstrating the conventional distance relay. It is a figure for demonstrating the conventional line selection relay.

Explanation of symbols

1 Power supply 2 _{1 to} 2 _{6 1st} to _6th circuit breakers 3 _{1 to} 3 _{6 1st} to _6th current transformers 4 Short circuit direction relays 4 ₁ to 4 _{3 1st} to _3rd short circuit direction relays 6 Transformer 10 Cross-through current transformers 10 ₁ to 10 ₄ First to _fourth cross-through current transformers 20 Distance relays 20 _{1 to} 20 ₃ First to third distance relays 30 Line selection relays 30 ₁ to 30 ₃ 1 to 3 line selection relays 1L, 2L The first and second transmission and distribution lines I, I _R , I _S , I _T , I _R1 , I _S1 , I _T1 , I _R2 , I _S2 , I _T2 , i ₁ , i ₂ , i ₁₁ , i ₁₂ , i ₂₁ , i ₂₂ Load current I ′ correction Load current I ₁ , I ₂ First and second load currents I ₁ ′, I ₂ ′ First and second correction load current _{I Ry, I FR, I FS} , I FT short-circuit current I _Ry 'correction circuit current I _Ry1, I _Ry2 first and second short-circuit current I _Ry1', I _Ry2 'first and second correction short conductive V _R, V _S, V _T-phase voltage V _RS, V _ST, V _TR line voltage θ impedance angle

Claims (10)

A directional protection relay device for protecting a three-phase AC circuit from a short circuit accident,
A cross through current transformer in which an arbitrary two phases of the three-phase AC circuit are crossed in an opposite direction and at an arbitrary angle through an annular iron core wound with a secondary coil;
When a short circuit fault is detected based on the short circuit current input from the cross-through current transformer and the line voltage of the three-phase AC circuit, the circuit breakers installed in each phase of the three-phase AC circuit are collectively shut off. A directional protection relay,
A directional protection relay device comprising:   The direction protection relay calculates a corrected short-circuit current by multiplying the short-circuit current input from the cross-through current transformer by a predetermined multiple, and calculates the corrected short-circuit current and the line voltage of the three-phase AC circuit. 2. The direction protection relay device according to claim 1, wherein when a short circuit accident is detected based on the circuit breaker, the circuit breakers installed in each phase of the three-phase AC circuit are collectively disconnected. The cross-through current transformer (10) is installed in the transmission and distribution line,
The first phase and the second phase of the transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the cross-through current transformer,
When the direction protection relay detects a short-circuit accident based on a short-circuit current (I _Ry ) input from the cross-through current transformer and a line voltage (V _RS , V _ST , V _TR ) of the transmission and distribution line , A short-circuit direction relay (4) that collectively shuts off the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the transmission and distribution line.
The direction protection relay device according to claim 1 or 2, characterized in that. The cross-through current transformer (10) is installed in the transmission and distribution line,
The first phase and the second phase of the transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the cross-through current transformer,
The directional protection relay is the cross through the short-circuit current which is input from the current transformer (I _Ry) and the line voltage of the transmission and distribution lines _{(V RS, V ST, V} TR) upon detecting a short circuit based on the , A distance relay (20) that collectively shuts off the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the transmission and distribution line.
The direction protection relay device according to claim 1 or 2, characterized in that. The cross through current transformer is a first and second transmission and distribution lines (1L, 2L) to the first and second cross through current transformer installed respectively (10 _1, 10 _2),
The first phase and the second phase of the first transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the first cross-through current transformer;
The first phase and the second phase of the second transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the second cross-penetrating current transformer,
The direction protection relay includes a short-circuit current (I _Ry ) that is a difference between a short-circuit current input from the first cross-through current transformer and a short-circuit current input from the second cross-through current transformer; When it is detected that a short circuit accident has occurred in the first transmission / distribution line based on the line voltages (V _RS , V _ST , V _TR ) of the transmission / distribution line, each phase of the first transmission / distribution line is detected. The installed _first to _third circuit breakers (2 ₁ to 2 ₃ ) are collectively cut off, and a short circuit accident occurs in the second transmission and distribution line based on the short circuit current and the line voltage of the transmission and distribution line. A line selection relay (30) that collectively shuts off the _fourth to _sixth circuit breakers (2 ₄ to 2 ₆ ) installed in each phase of the second transmission / distribution line when it is detected that they have occurred.
The direction protection relay device according to claim 1 or 2, characterized in that. One annular phase of the three-phase AC circuit and one other phase other than the arbitrary two phases are crossed in an opposite direction and at an arbitrary angle on an annular core around which a secondary coil is wound. With other cross-penetrating current transformers,
When a short-circuit fault is detected based on another short-circuit current input from the other cross-through current transformer and a line voltage of the three-phase AC circuit, a circuit breaker installed in each phase of the three-phase AC circuit With other directional protection relay that cuts off at once,
The directional protection relay device according to claim 1, further comprising:   The other directional protection relay calculates another corrected short-circuit current by multiplying the other short-circuit current input from the other cross-through current transformer by a predetermined multiple, and the calculated other corrected short-circuit current and 7. The direction protection according to claim 6, wherein when a short-circuit accident is detected based on a line voltage of the three-phase AC circuit, circuit breakers installed in each phase of the three-phase AC circuit are collectively shut off. Relay device. The cross-through current transformer is a first cross-through current transformer (10 ₁ ) installed on a transmission and distribution line,
The other cross-through current transformer is a second cross-through current transformer (10 ₂ ) installed in the transmission and distribution line;
The first phase and the second phase of the transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the first cross-through current transformer;
The second phase and the third phase of the power transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the second cross-through current transformer;
The direction protection relay is based on a first short-circuit current (I _Ry1 ) input from the first cross-through current transformer and a line voltage (V _RS , V _ST , V _TR ) of the transmission and distribution line. When a short circuit accident is detected, the first short circuit direction relay (4 ₁ ) that collectively shuts off the _first to _third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of the transmission and distribution line,
When the other directional protection relay detects a short-circuit accident based on the second short-circuit current (I _Ry2 ) input from the second cross-through current transformer and the line voltage of the transmission and distribution line, A second short-circuit direction relay (4 ₂ ) that collectively shuts off the first to third circuit breakers;
The direction protection relay device according to claim 6 or 7, characterized by the above. The cross-through current transformer is a first cross-through current transformer (10 ₁ ) installed on a transmission and distribution line,
The other cross-through current transformer is a second cross-through current transformer (10 ₂ ) installed in the transmission and distribution line;
The first phase and the second phase of the transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the first cross-through current transformer;
The second phase and the third phase of the power transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the second cross-through current transformer;
The direction protection relay is based on a first short-circuit current (I _Ry1 ) input from the first cross-through current transformer and a line voltage (V _RS , V _ST , V _TR ) of the transmission and distribution line. Upon detection of a short circuit Te, a first distance relay for collectively blocking the first to third circuit breakers installed in each phase of the transmission and distribution lines (2 ₁ to 2 ₃₎ (20 _1),
When the other directional protection relay detects a short-circuit accident based on the second short-circuit current (I _Ry2 ) input from the second cross-through current transformer and the line voltage of the transmission and distribution line, A second distance relay (20 ₂ ) that collectively shuts off the first to third circuit breakers;
The direction protection relay device according to claim 6 or 7, characterized by the above. The cross through current transformer is a first and second transmission and distribution lines (1L, 2L) to the first and second cross through current transformer installed respectively (10 _1, 10 _2),
The other cross through current transformers are third and fourth cross through current transformers (10 ₃ , 10 ₄ ) installed in the first and second transmission and distribution lines, respectively.
The first phase and the second phase of the first transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the first cross-through current transformer;
The first phase and the second phase of the second transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the second cross-penetrating current transformer,
The second phase and the third phase of the first power transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular core of the third cross-through current transformer,
The second phase and the third phase of the second power transmission and distribution line are crossed in an opposite direction and at an arbitrary angle through the annular iron core of the fourth cross-through current transformer,
The directional protection relay has a first short-circuit current (I _Ry1) that is a difference between a short-circuit current input from the first cross-through current transformer and a short-circuit current input from the second cross-through current transformer. ) And the line voltages (V _RS , V _ST , V _TR ) of the first and second transmission / reception lines, it is detected that a short-circuit accident has occurred in the first transmission / distribution line. First to third circuit breakers (2 ₁ to 2 ₃ ) installed in each phase of one transmission / distribution line are collectively cut off, and the first short circuit current and the first and second transmission / distribution lines When it is detected that a short circuit accident has occurred in the second transmission / distribution line based on the line voltage, the fourth to sixth circuit breakers (2 ₄ ) installed in each phase of the second transmission / distribution line. to 2 ₆₎ is a first line selection relay for collectively blocking the (30 _1),
The other short-circuit protection relay is a second short-circuit current (a difference current between a short-circuit current input from the third cross-through current transformer and a short-circuit current input from the fourth cross-through current transformer). I _Ry2 ) and the line voltage of the first and second transmission / distribution lines, when it is detected that a short circuit accident has occurred in the first transmission / distribution line, the first to third circuit breakers are collectively When the second and short-circuit currents are detected and a short-circuit accident has occurred in the second transmission and distribution line based on the second short-circuit current and the line voltage of the first and second transmission and distribution lines, the fourth to fourth The second circuit selection relay (30 ₂ ) that collectively shuts off the 6 circuit breakers.
The direction protection relay device according to claim 6 or 7, characterized by the above.