JPH0332285B2 - - Google Patents

Description

[Detailed description of the invention]

(a) Technical Field The present invention relates to improvements in ground fault line selection relays used in parallel multi-circuit transmission lines. (b) Prior art When a high-resistance grounded parallel two-circuit transmission line (hereinafter referred to as a protected transmission line) and another power transmission line (hereinafter referred to as an induced transmission line) are installed on the same tower,
Due to the induction caused by the current in the power transmission line, a circulating current flows between the two circuits of the protected power transmission line, making protection by a ground fault circuit selection relay difficult. Furthermore, if a large zero-sequence current flows due to a ground fault in a directly grounded transmission line, or if a large zero-sequence current flows due to a ground fault at a different point, in other cases, a large zero-sequence current flows in the protected transmission line due to induction. This generates voltage and current, causing the ground fault line selection relay to malfunction even if it would not malfunction under normal induction. This will be explained below using the drawings. FIG. 1 is a system diagram illustrating an example of a multi-circuit parallel power transmission line.
In the figure, 1L and 2L are induction transmission lines, 3L and 4L are protected transmission lines, T is a transformer, R is a resistor, A, B,
C and D are busbars, and the induced transmission line and protected transmission line are installed on the same tower between EF. The neutral point of the transformer T is directly grounded on the inductive power transmission line side, and is grounded via a resistor R on the protected power transmission line side. In the illustrated example, the bus D side of the lower system is not grounded. For simplicity, the protected power transmission line will be explained assuming that the load current between bus bars C and D is zero. Currents I and I' flow through the induction power transmission lines 1L and 2L, respectively. Due to this current and the mutual impedance of the parallel section,
Voltages υ _n and υ _n ' are induced in the power transmission lines 3L and 4L between EF, respectively. With this voltage, the power transmission line 3L
Currents i and i' flow through and 4L, respectively, and a voltage υ _o is generated across the bus line CD. FIG. 2 is a diagram illustrating an example of the arrangement of electric wires in the parallel section. A, B, C, A', B', C', a, b, c,
A', b', and c' are electric wires of phases a, b, and c of power transmission lines 1L, 2L, 3L, and 4L, respectively, and are arranged, for example, at the intervals shown (in meters).
In the induction side electric wires, the transmission lines 1L are arranged in reverse order, with the transmission lines 1L being A, B, C from the top, and 2L being C', B', A' from the top, and the protected transmission lines are arranged in the order of transmission lines 1L, 2.
The order of L is a, b, c and a', b', c' from the top. Due to this arrangement, even if the load current flows only through the protected transmission lines 3L and 4L, the current i
and i' are equal, and no current i-i' circulates through the power transmission lines 3L and 4L. Circulating currents are generated only by induced transmission line currents unless there is a fault on the protected transmission line. This will be explained below. The induced voltage generated in the protected power transmission line by the induced power transmission line current is expressed by the following equation. However, υ _an , υ _bn , υ _cn , υ _an ′, υ _bn ′ and υ _cn
′
are the values of the a, b, and c phases of the induced voltages υ _n and υ _n ′, respectively, and Z _Aa ...Z _Ccc ′ are the mutual impedances between the wires with the first letter and the wire with the second letter, for example, Z _Aa is the mutual impedance a, b, c, 'a, 'b, and 'c between the electric wires A and a, and the values of the current and each phase of the power transmission lines 1L and 2L, respectively. This induced voltage causes the protected power transmission lines 3L and 4 to
A current circulates through L, and the difference in current between power transmission lines 3L and 4L is expressed by equation (2). 1/2i _a −i′ _a i _b −i′ _b i _c −i′ _c = 1Z _pp υ _an −υ′ _an υ _bn −υ′ _bn υ _cn −υ′ _cn …(2) However, i _a , i _b , i _c , i′ _a , i′ _b and i′ _c are the values of each phase of currents i and i′ of transmission lines 3L and 4L, respectively, Z _pp
is the loop impedance between the power transmission lines 3L and 4L. Furthermore, the voltage generated between the bus bars CD due to the induced voltage is expressed by equation (3). υ _ao υ _bo υ _co =υ _an υ _bn υ _cn +Zoo/2i _a i _b i _c =υ _an ′ υ _bn ′ υ _cn ′+Zoo/2i _a ′ i _b ′ i _c ′ …(3) However, υ _ao , υ _bo , υ _co are each induced by bus line CD
These are the values of the a, b, and c phases of the voltage υ _o that occurs between the two. i _a = i _a ′, i _b = i _b ′, i _c = i _c ′, from equation (3), υ _ao υ _bo υ _co = 1/2υ _an +υ _an ′ υ _bn +υ _bn ′ υ _cn +υ _cn ′ …(4). The mutual impedance used in equation (1) is approximately expressed by the following equation. Z _Aa = μlog2H _e /D _Aa …(5) However, μ is the magnetic permeability in the air, is the parallel installation distance, H _e is the depth of the image to the ground (about 400 m), and D _Aa is the distance between electric wires A and a. It is. The values of the differential current due to the above circulating current and the zero-phase induced voltage at all times and at the time of a one-phase ground fault in the upper system are determined as follows for one system example. [Condition 1] 1L of induction power transmission line stopped, 2L in normal operation
The current in Equation (6) is _A = 2000A 0°, _B = 2000A 240°, _C = 2000A 120°...(6) However, □° indicates ε ^j 〓〓, and the same applies below. i _a −i _a ′＝98.8A 356゜i _b −i _b ′＝44.4A 348゜i _c −i _c ′＝26.4A 345゜i _p −i _p ′＝56.4A 352゜υ _po ＝2.84KV 67゜ ...(7) However, i _p = 1/3 (i _a + i _b + i _c ) i _p ′=1/3 (i _a ′+
i _b ′ + i _c ′) υ _po = 1/3 (υ _ao + υ _bo + υ _co ) [Condition 2] Both induction power transmission lines 1L and 2L are in normal operation, and the current is in the state of equation (8) _A = _A ′= 1000A 0゜_B ＝ _B ′＝1000A 240゜_C ＝＝ _C ′＝1000A 120゜ …(8) i _a −i _a ′＝94.4A 335゜i _b −i _b ′＝42.8A 333゜i _c −i _c ′＝25.6A 333゜i _p −i _p ′＝54.2A 334゜υ _po ＝0.52KV 129゜ …(9) [Condition 3] Induction transmission line 1L and 2L operation, 2L F
At point _A ', there is a 7000A A-phase 1-phase ground fault, and the A-phase voltage before the fault in the upper system is 0°. i _a −i _a ′＝474A 278゜i _b −i _b ′＝232A 277゜i _c −i _c ′＝150A 276゜i _p −i _p ′＝285A 277゜υ _po ＝31.7KV 169゜ …(10 ) Conditions 1 and 2 will be described below as the maximum load state, and Condition 3 as the maximum induction state due to fault current. Conventional ground fault line selection relays used to protect one-phase ground fault faults on high-resistance grounding power transmission lines are capable of detecting ground fault fault lines on the protected power transmission line due to the zero-sequence difference current i _p - i _p ' of parallel transmission lines. It is used to identify The terminals of the bus D will be explained as to what kinds of problems occur in typical cases. [Directional type] When installed on the generatrix D in Fig. 1,
If the effective component of the zero-sequence difference current i _p −i _p ' between the transmission lines 3L and 4L (component in the same phase as the zero-sequence voltage of the bus D) is greater than a certain positive value, the transmission line 3 is disconnected, and the negative If it is below a certain value, the power transmission line 4 is cut off. This cutting is generally performed on the condition that the zero-sequence voltage of the bus D is above a certain value (approximately 20 KV in this calculation example), and the same condition is applied to other line selection relays. In the constant operating state of conditions 1 and 2, the value of the zero-sequence difference current i _p -i _p ' is a little over 50 A. In this state, the zero-phase induced voltage υ _po is small and the zero-phase voltage υ _pD of the bus D is also small, so there is no malfunction. However, if a one-phase ground fault occurs in the system including the protected transmission line under this condition, zero-sequence voltage and zero-sequence current will be generated, and the resulting zero-sequence difference current will be expressed as Equations (7) and (9). It is superimposed on the zero-sequence difference current of the induced component shown. If the fault is outside (a part other than transmission lines 3L and 4L), the zero-sequence difference current for the fault is zero, and the zero-sequence difference current i _p -
i _p ′ becomes the value of equation (7) or equation (9). The ground fault line selection relay must not operate with this zero-sequence difference current, and if the above conditions 1 and 2 are set to the maximum load state, the operating value is slightly larger than the value of equation (7) or equation (9). value, for example 70A. In the case of condition 3, a one-phase ground fault in the induced power transmission line, (10)
A zero-sequence difference current i _p −i _p ′ and a zero-sequence dielectric voltage υ _po of the equation are generated. The zero-sequence voltage υ _pD of the bus D is expressed by the following equation. υ _pD ≒Z _pD /Z _pC +Z _cD υ _po (11) However, Z _pC and Z _pD are the zero-sequence impedances behind the bus lines C and D, respectively. As shown in FIG. 1, if there is no neutral grounding resistor on the bus D side and the capacitance is small, and Z _cD ≫ Z _pC , then υ _pD ≒ υ _po . In this case, in the calculation example for condition 3, the effective component of the zero-sequence difference current i _p -i _p ' is -88 A, which is smaller than the above-mentioned setting value example 70 A, and the relay operates and the power transmission line 4L is briefly disconnected. Further, when the capacitance on the side of the bus D is large and Z _pD is capacitive, Z _pC is resistive, so the zero-sequence voltage υ _pD of the bus D is delayed in phase from the zero-sequence induced voltage υ _po . FIG. 3 is a diagram showing an example of the relationship between the zero-sequence voltages υ _po and υ _pD and the zero-sequence difference currents i _p -i _p ' in such a case. If the voltage υ _pD lags the voltage υ _po by 30 degrees as shown in the figure, the effective portion of the zero-sequence difference current i _p −i _p ' in the above calculation example becomes -205 A, making the relay even more likely to malfunction. In order to avoid malfunctions in such cases, the operating value must be set to a value larger than the other 70A mentioned above. If such a large operating value cannot be achieved, a time-limited disconnection must be used so that a disconnection command is not issued until the induced power transmission line is temporarily disconnected and the ground fault line selection relay is restored. In this case, there is a disadvantage that even in the event of a ground fault in the protected power transmission line, the power outage will be interrupted for a similar period of time. [Variation type] In this type, the pre-failure value of the zero-sequence difference current i _p -i _p ' at the time of a one-phase ground fault responds to the effective portion of the variation with respect to the stored amount. In the case of a ground fault in a system that includes protected transmission lines (hereinafter referred to as protected system), the current is only the amount of change in the zero-sequence difference current i _p −i _p ′; in the case of an external fault, it becomes zero; 3L or 4L accident), the flow will prompt you to correctly select the accident line. Although the operating value can be reduced significantly in theory, it is actually limited by the amount of memory and errors in the calculation process. If this error is 10%, the lower limit of the operating value will be a little over 5A from equations (7) and (9), and the operating value will be, for example, 7A. Also, if condition 3 is considered, the zero-sequence difference current
Since the value of i _p -i _p ' can be equivalent to the value of equation (10), malfunction is very likely to occur at the above operating value of 7A. [Compensation type] This type compensates the zero-sequence difference current due to induction with a healthy phase-difference current, and responds to the effective part of the compensated zero-sequence difference current. The compensated current i _pc is expressed, for example, by the following equation. (i) A-phase ground fault i _pc = i _p −i _p ′−R _c (i _c −i _c ′) …(12) (ii) B-phase ground fault i _pc = i _p −i _p ′−R _a (i _a −i _a ′) …(13) (iii) C-phase ground fault i _pc = i _p −i _p ′−R _b (i _b −i _b ′) …(14) However, R _a , R _b and R _c are complex constants. It is desirable that the constants R _a , R _{b ,} and R _c in equations (12) to (14) satisfy the following equation. R _a (i _a −i _a ′)=R _b (i _b −i _b ′) = R _c (i _c −i _c ′)=i _p −i _p ′ …(15) Under conditions 1 and 2, ( 15) Constants R _a , R _b , and R _c will be explained using the following values as values that approximately satisfy the equation. R _a = 0.57 3°, R _b = 1.26 3°, R _c = 2.12 4° …(16) However, □° indicates ε ^-j 〓〓〓. (Similarly below) In the case of a 1-phase ground fault in the protected system, the healthy phase difference current does not change, so the compensated current i _pc in the case of an external 1-phase ground fault can be calculated using equations (12) to (14) according to each condition. (7), (9)
Substituting the value of expression or (10), we get the following. (i) Condition 1 A-phase ground fault i _pc = 3.0A 74° B-phase ground fault i _pc = 0.99A 86° C-phase ground fault i _pc = 1.06 60° …(17) (ii) Condition 2 A-phase ground fault i _pc = 2.86A 128゜B phase ground fault i _pc = 1.94A 52゜C phase ground fault i _pc = 1.90A 106゜ ...(18) (vi) Condition 3 A phase ground fault i _pc = 37A 124゜B phase Ground fault i _pc = 20.4 126° C-phase ground fault i _pc = 16.9 A 162° (19) The operating value can be set to a value slightly larger than the maximum value of equations (17) and (18), for example, 5 A. However, if condition 3 is to be considered, the operating value must be determined based on the value of equation (19). As mentioned above, conventional ground fault line selection relay devices that use zero-sequence difference currents have sensitivity limitations due to the influence of zero-sequence difference currents caused by induction when used on multi-circuit parallel transmission lines. . (c) Purpose The present invention was made in view of the above, and it is an object of the present invention to provide a ground fault line selection relay with fewer restrictions on sensitivity due to induced current. (d) Outline of the Invention The present invention is provided with a line selection means for identifying a line in which a ground fault has occurred by using a detected current mainly consisting of a polarity quantity and a negative phase difference current of a parallel transmission line as calculation quantities, and a ground fault phase identification means is provided with one
The main point is that the phase-to-ground fault fault phase is identified, and the calculation amount selection means selects the amount of calculation to be used by the line selection means from among those set in advance according to the identification result of the earth-fault phase identification means. Since the influence of induction on negative phase difference currents is smaller than on zero-sequence difference currents, restrictions on sensitivity are relaxed accordingly, and fault lines can be identified by appropriate ground fault fault line identification means according to the single-phase ground fault fault phase. Can be identified with high sensitivity. The negative phase difference current is not only used alone, but also
It is used in the form of a current compensated for by the change in value relative to the pre-fault value or by other difference currents, allowing protection of sensitivity depending on each. (e) Embodiment (i) Configuration FIG. 4 is a diagram showing the configuration of an embodiment of the present invention. In the figure, 1 is a bus bar, and a, b, and c are phase symbols.
2 and 3 are power transmission lines, 4 and 5 are bridges and disconnectors, 6
and 7 is a current transformer, 8 is an instrument transformer, 9 is an input converter, 10 is a sample hold circuit, 11 is a multiplexer, 12 is an AD converter, and 13 is an arithmetic unit. A current corresponding to the current of the transmission lines 2 and 3 is obtained by the current transformers 6 and 7, and a,
B, c phase difference current i _a −i _a ′, i _b , i _o ′ and i _c −
A value 3 (i _p −i _p ′), which is three times the zero-sequence difference current as i _c ′, is obtained. In addition, the voltage υ _a , which corresponds to the terminal voltage of the power transmission lines 2 and 3 through the bus bar by the instrument transformer 8, is
υ _b , υ _c and zero-sequence voltage υ _p are obtained. These current voltages are applied to an input converter 9 and converted into voltages of appropriate values for the next sample and hold circuit 10. The input converter 9 is provided with a filter, and only the fundamental wave component of each voltage and current is obtained as the output of the sample and hold circuit 10. The sample and hold circuit samples and holds the input value at the same time and at a constant cycle (eg, 1/12 cycle of one cycle of voltage and current). This hold value is sequentially supplied to the AD converter 12 by the multiplexer 11 and converted into a digital value. Based on this digital value, the calculator 1
3 performs necessary calculations and instructs the disconnectors 4 and 5 to disconnect when predetermined conditions are met. (ii) Actions and Effects A flowchart of an embodiment of the present invention performed by the arithmetic unit 13 is shown in FIG. First, when a start command is given in step S1, one-phase ground fault detection is performed in step S2. No single-phase ground fault was detected here.
If a NO judgment is obtained, move on to other calculations. If a 1-phase ground fault is detected and a YES determination is obtained, the process advances to step S3 and a ground fault phase selection calculation is performed. In step S3, first, in step S3-1, an a-phase ground fault detection calculation is performed, and if an a-phase ground fault is detected, a determination 3Ya is obtained. If no a-phase ground fault is detected, step S3
Proceed to -2 to perform b-phase ground fault detection calculation. If a b-phase ground fault is detected, judgment 3Yb is obtained, and if not detected, the process proceeds to step S3-3. Perform phase-to-ground fault detection calculations. If a c-phase ground fault is detected, judgment 3Yc is obtained; if not detected, judgment 3N is obtained and the process moves on to other calculations. If any of the determinations 3Ya, 3Yb, or 3Yc is obtained in step S3, the amount of calculation is given in step S4. If judgment 3Ya is obtained, step S4-1
When the judgment 3Yb is obtained, the calculation amount b is given in step S4-2, and when the judgment 3Yc is obtained, the calculation amount c is given in step S4-3.
When the amount of calculation is given in step S4, step
Proceed to S5 to perform line selection calculation. In step S5, first, in step S5-1, a sample value of the calculation amount given in step S4 is calculated. Next, in step S5-2, the sample value obtained in S5-1 is used to perform a fault detection calculation on the power transmission line 2, and if a fault on the power transmission line 2 is detected, a judgment 5Y1 is obtained and the circuit breaker 4 is activated. I refuse. If an accident on the power transmission line 2 is not detected, the process proceeds to step S5-3, where an accident detection calculation on the power transmission line 3 is performed. If an accident with a calculation amount of 3 is detected, judgment 5Y2 is obtained and the breaker 5 is cut off.
If it is not detected, a judgment of 5N is obtained and the process moves on to other calculations. As described above, there are various means for the one-phase ground fault detection calculation in step S2. For example, when all of the following conditions (20) and (21) are satisfied, a one-phase ground fault is determined. ｜υ _p ｜K ₀ …(20) ｜υ _a −υ _b ｜＞K ₂ , ｜υ _b −υ _c ｜＞K ₂ , ｜υ _c −υ _a ｜＞K ₂ …(21) However, ｜υ _p ｜、｜υ _a −υ _b ｜、｜υ _b −υ _c ｜、｜υ _c
−υ _a
| is the zero-phase voltage υ _p and the inter-phase voltage of each of the three phases υ _a −υ _b ,
Indicates the magnitude of υ _b −υ _c , υ _c −υ _a (effective value, peak value, average value, etc.), and hereinafter the magnitude is indicated by the same symbol. Moreover, K ₁ and K ₂ are constants. Since the zero-sequence voltage υ _p occurs only when there is a one-phase ground fault or a two-phase ground fault, the occurrence of these faults is detected using equation (20), and in a high-resistance grounding system, the phase-to-phase voltage υ _a −υ _{b . _ _ _} separated. There are various means for the ground phase selection calculation in step S3, and one example is one that selects ground fault phases based on the following conditions. (i) Phase a ground fault υ _p・(υ _b −υ _c ) θ ₀ <O and υ _p・ (υ _b −υ _c ) θ ₂ <O …(22) (ii) Phase b ground fault υ _p・(υ _c −υ _a ) θ ₀ <O and υ _p・ (υ _c −υ _a ) θ ₂ >O …(23) (iii) C-phase ground fault υ _p・(υ _a −υ _b ) ＞O and υ _p・(υ _a −υ _b ) θ ₂ >O …(24) However, υ _p・(υ _b −υ _c ) θ ₀ is υ _p and (υ _b −υ _c )
ε ^-j 〓 ¹
Indicates the inner product of , and the same applies hereafter. Further, θ ₁ and θ ₂ are constant angles. In a single-phase ground fault in a high-resistance grounding system, the phase of the zero-phase voltage υ _p is approximately 90° with respect to the quadrature voltage (e.g., bc phase-to-phase voltage υ _b - υ _p ) with respect to the grounded phase (e.g., a phase). Since it is in a delayed phase, the ground fault phase can be identified from this phase relationship. Equation (22) detects an a-phase ground fault when the lagging phase of the voltage υ _p with respect to the bc phase voltage υ _b −υ _c is within the range of θ ₁ ±90° and also within the range of θ ₂ ±90°. So, if θ ₁ > θ ₂ , (θ ₁ −90°) ~ (θ ₂ +
90°), an A-phase fault is detected. The values of θ ₁ and θ ₂ are 120° to 150° and 60° to 30°, respectively, where θ ₁ = 120°, θ ₂ = 60
If υ _p is within the lagging phase range of 30° to 150° with respect to υ _b - υ _c , an a-phase fault will be detected. The above-described one-phase ground fault detection calculation in step S2 and the ground fault phase selection means in step S3 are both known means, so detailed explanations will be omitted for the sake of brevity. In addition, there are various other known means, and the purpose can be similarly achieved using these other means. In step S4, the detected current i _d and the polarity e _P are given as calculation amounts necessary for the line selection calculation in step 5. One example is shown in Table 1.

[Table] In Table 1, i _a2 −i _a2 ′, i _b2 −i _b2 ′ and i _c2 −i _c2 ′ are the negative phase difference currents υ _a1 , υ _b1 and υ _c1 with reference to phases a, b and c, respectively. is a positive-sequence voltage based on the a, b, and c phases, respectively, and is expressed by the following equation. i _a2 −i _a2 ′=1/3 [(i _a +a ² i _b +ai _c ) −(i _a ′+a ² i _b ′+ai _c ′)] …(25) i _b2 −i _b2 ′=1/3 [(i _b +a ² i _c +ai _a ) −(i _b ′+a ² i _c ′+ai _a ′)] …(26) i _c2 −i _c2 ′=1/3 [(i _c +a ² i _a +ai _b ) −(i _c ′+a ² i _a ′+ai _b ′)] …(27) υ _a1 = 1/3(υ _a +aυ _b +a ² υ _c ) …(28) υ _b1 = 1/3(υ _b +aυ _c + a ² υ _a ) ...(29) υ _c1 = 1/3 (υ _c + a υ _a + a ² υ _b ) ... (30) However, a = ε ^j120 °. In step S5, the sample value of the above calculation amount is calculated in step S5-1, and in steps S5-2 and S5-3, the calculations of equations (31) and (32) are performed, respectively, and the operation is performed when each equation holds true. and yield decisions 5Y1 and 5Y2, respectively. e _p・i _d ＞K ₃ ｜e _p ｜ …(31) e _p・i _d ＞−K ₃ ｜e _p ｜ …(32) In equations (31) and (32), e _p +i _d is ep and id (and so on) where K ₃ is a positive real constant.
Equations (31) and (32) are the polarity of the detection current i and _d , respectively.
When the effective component for e _p is greater than K ₃ and K ₃
Indicates that it works when it is smaller. Next, the operation of this embodiment will be explained. The a-phase reference negative phase difference current under conditions 1, 2, and 3 has the following values from equations (7), (9), (10), and (25), respectively. Condition 1: i _a2 −i _a2 ′=21.2A 13° …(33) Condition 2: i _a2 −i _a2 ′=20.7A 37° …(34) Condition 3: i _a2 −i _a2 ′=96.3A 95° ...(35) Furthermore, the negative sequence currents based on the b-phase reference and the c-phase reference are expressed by the following equations, and are equal in magnitude to the negative-sequence current based on the a-phase reference. i _b2 −i _b2 ′=a (i _a2 −i _a2 ′) i _c2 −i _c2 ′=a ² (i _a2 −i _a2 ′) …(36) System including protected transmission line (hereinafter referred to as protected system) ) The difference current that flows in the event of an external fault is only the grain conduction, and the value of the difference current in the maximum load state of conditions 1 and 2 is the maximum. As mentioned above, the negative phase difference current under conditions 1 and 2 is over 20A, and the operating value (value of K ₃ in equations (31) and (32)) to avoid malfunction in a protected system ground fault is, for example It can be 28A. This value is 40% of the allowable operating value for the conventional directional type responsive to the effective component of the zero-sequence difference current, and can provide extremely high sensitivity. Also, when selecting the operating value to avoid malfunction under condition 3, the value of the negative phase difference current (formula (35)) is the value of the zero sequence difference current ((10))
It is about 1/3 of that, and can be considered highly sensitive. We will explain the phenomenon when a single-phase ground fault occurs in the protected system. In this case, the fault point current and fault point voltage are expressed by the following equation using the symmetric coordinate method, as is well known. i _1F =i _2F =i _0F =e _1F /Z ₁ +Z ₂ +Z ₀ +3 _RF …(37) υ _1F =Z ₂ +Z ₀ +3 _RF /Z ₁ +Z ₂ +Z ₀ +3 _RF e _1F …(38) υ _2F = −Z ₂ /Z ₁ +Z ₂ +Z ₀ +3 _RF e _1F …(39) υ _0F =−Z ₀ /Z ₁ +Z ₂ +Z ₀ +3 _RF e _1F …(40) However, i _1F , i _2F and i _0F are the positive phases of the accident points, respectively,
Negative-sequence and zero-sequence current, υ _1F , υ _2F and υ _0F are positive-sequence, negative-sequence and zero-sequence voltage at the fault point, respectively, e _1F
is the positive phase voltage at the fault point before the fault, and both are based on the fault phase. Moreover, Z ₁ , Z ₂ and Z ₀ are the driving point impedances of the positive phase, negative phase and zero phase of the fault point, respectively, and _RF is the fault point resistance. In a resistive grounding system, the zero-sequence impedance Z ₀ is significantly larger than the positive-sequence and negative-sequence impedances Z ₁ and Z ₂ , so equations (37) to (40) approximately take the following form. i _1F ＝i _2F ＝i _0F ≒e _1F ／Z ₀ ＋3 _RF …(41) υ _1F ≒e _1F …(42) υ 2F ≒O …(43) _{υ 0F ≒} −Z ₀ ／Z ₀ ＋3 _RF e _1F ...(44) The phase relationship between the detected current i _d and the polarity e _p in Table 1 will be explained using equations (41) to (44) with reference to the drawings. The fault phase current i _F at the fault point is given by equation (45) from equation (41). i _F = i _1F + i _2F + i _0F ≒3e _1F /Z ₀ +3 _RF (45) In equation (45), Z ₀ is an impedance with a small reaction component and a large resistance component, so the current i _F is as shown in Figure 6. Fault point before fault phase reference positive sequence voltage
It has almost the same phase as e _1F . This current i _F flows from the fault section transmission line terminal. Currents i _a , i _b , i _c and i _a ′,
If the fault currents of the fault phases of i _b ′ and i _c ′ are respectively i and i′, the fault difference current i−i′ of the fault phase is the current in the fault of transmission line 2, as shown in Figure 6. Since it is in phase with i _F , it is almost in phase with voltage e _1F , and in the case of an accident on transmission line 3, it is in opposite phase with current i _F , so voltage
e It is almost in opposite phase to _1F . Such a fault phase fault difference current i-i' is superimposed on the induced difference current of the fault phase. Fault differential current does not flow in the healthy phase. Of the negative phase difference currents i _a2 −i _a2 ′, i _b2 −i _b2 ′, and i _c2 −i _c2 ′ based on each phase, let the fault negative phase difference current based on the fault phase be i ₂ −i ₂ ′, and The zero-sequence difference current is i ₀ −
When i ₀ ', each current is expressed by equation (46), and both of them in FIG. 6 are equal currents and have the same phase as current i-ii'. In both cases, in the case of an accident on transmission line 2, the phase is almost the same as that of voltage e _1F , and in the case of an accident on transmission line 3, it is almost in the opposite phase. i ₂ −i ₂ ′=i ₀ −i ₀ ′=1/3(i−i′)
...(46) On the other hand, in the case of a one-phase ground fault, the fault phase reference positive-sequence voltage υ _1F at the fault point is almost equal to the voltage e _1F and does not change from the value before the fault. Therefore, the fault phase reference positive-sequence voltage υ ₁ at the relay installation point also does not change due to the fault, and the phase difference with the voltage e _1F is small. Therefore, the fault current component i ₂ −i ₂ ′ of the negative phase difference current based on the fault phase is almost in phase with the positive sequence voltage υ ₁ based on the fault phase in the case of a fault on transmission line 2, and almost in phase in the case of a fault on transmission line 3. The phase is opposite. The calculation quantities a, b, and c in Table 1 are for a, b, and c phase 1-phase ground fault, respectively, the detected current i _d is the negative phase difference current based on the fault phase, and the polarity e _p is the positive phase difference current based on the fault phase. Since it is a voltage, from the above-mentioned relationship, the fault current component of the detected current i _d will be approximately in phase with respect to the polarity e _p in the case of a fault on the power transmission line 2, and will be in almost the opposite phase in the case of a fault on the power transmission line 3. Therefore, in the case of an accident on the transmission line 2 or 3, the effective component of the detected current i _d with respect to the polarity e _p becomes positive or negative, and the accident on the transmission line 2 or 3 is detected according to equation (31) or (32). can do. This detection sensitivity will be compared with the directional type of the conventional device.
As shown in equation (46), the fault current component i ₂ −i ₂ ' of the negative phase difference current used as the detection current i _d in this embodiment is the fault current component of the zero sequence difference current used in the conventional directional type.
Equal to i ₀ −i ₀ ′. On the other hand, the difference current due to induction has the values of equations (33) to (35) in this embodiment, whereas the conventional directional type has the values of i ₀ − of equations (7), (9), and (10).
i ₀ ', and the influence of induction is extremely small in this embodiment. In addition, as mentioned above, the operating value to avoid malfunction due to external accidents is 40% compared to the conventional directional type.
It can be taken as a small value. For these reasons, this embodiment can achieve protection with significantly higher sensitivity than the conventional directional type. (f) Other Examples (Part 1) The method of selecting the detection current i _d and the polarity e _p when using a negative phase difference current is not limited to those shown in Table 1, and various modifications may be made. A few examples of them
Shown in the table.

[Table] Each voltage used as the polarity e _p in Table 2 will be explained using the drawings. FIG. 7 is a vector diagram showing the voltage when an A-phase 1-phase ground fault occurs at a point close to the relay installation point. Since this is a close point accident, each symmetrical voltage component υ _a1 , υ _a2 , υ _p is υ _1F in equations (42) to (44),
It is similar to υ _2F and υ _pF . (Since the values of zero-sequence voltage υ _p and zero-sequence current do not change no matter which phase is used as the reference, the reference phase is omitted.) In equation (44), Z ₀ is resistance,
Since R _F is a resistance, the zero-sequence voltage υ _pF at the fault point is almost in opposite phase to the pre-fault positive-sequence voltage e _1F .
Therefore, in Figure 7, the zero-sequence voltage υ _p is the positive-sequence voltage before the accident.
e has almost the opposite phase to _a1 . As shown in equation (43), the negative sequence voltage can be ignored, and the positive sequence voltage of the fault phase reference does not change from before the fault. Therefore, the a-phase reference positive-sequence voltage υ _a1 is equal to the voltage e _a1 . Therefore, the phase voltages υ _a , υ _b and υ _c are respectively the voltages υ _a1 , a ² υ _a1 and aυ _a1
and the zero-sequence voltage υ _p , resulting in a vector as shown. The phase-to-phase voltages υ _b −υ _c , υ _c −υ _a and υ _a −υ _b do not change from before the accident. Even if the accident point is far away,
The phase change in the voltage that does not change from the value before the accident is slight, and can be considered to be an approximation to the state shown in FIG. In combinations 1 to 3 in Table 2, the detected current i _d is the same as in Table 1, and only the polarity amount e _P is different. The difference υ _a −υ _p between the fault phase voltage and the zero-sequence voltage, which is used as the polarity e _P of the calculation amount a in combination 1, is expressed by a vector from point n to point a in Figure 7, and the voltage υ _a1
be equivalent to. Voltage υ _b used for the a-phase quantity of combination 2
−υ _c lags the voltage υ _a1 by 90° as shown, and the quadrature voltage for the fault phase (υ _b −υ _c )
90° is in phase with the voltage υ _a1 . Polar amount of combination 3
As shown in the figure, the voltage υ _p used as e _P is in the opposite phase to the voltage υ _a1 in the case of an a-phase fault, and the voltage -υ _p is in the same phase as υ _a1 . As mentioned above, the polarity of the a-phase quantity e _P in the a-phase 1-phase ground fault
All combinations 1 to 3 are in phase with those in Table 1. Similarly, in a b-phase or c-phase one-phase ground fault, the polarity e _P of the b-phase quantity or c-phase quantity is all in phase with the combinations 1 to 3 of Table 1. From this relationship, even if the polarity e _P of combinations 1 to 3 is used instead of Table 1, there is no change in the calculation results in equations (31) and (32).
Can be used in the same way as a table. One of the detection current i _d or the polarity e _P is the calculation amount a,
b and c can be the same. Combination 4 has the same detection current i _d , and combination 5 has the same polarity amount e _P . In combination 4, There is a relationship, and in combination 5 In both cases, the relative relationship between the polarity amount e _P and the detected current i _d is exactly the same as in Table 1, and they respond in exactly the same way. The calculation amounts in Tables 1 and 2 above are for the case where the inner product calculations of equations (31) and (32) are performed in steps S5-2 and S5-3.
Steps S5-2 and S5-3 are for detecting what kind of relationship the detected current i _d has with the polarity amount e _P , and not only calculates the inner product but also calculates other Various calculations can be performed. An example of this is to calculate the cross product of the detected current i _d and the polarity e _P , and in this case it is assumed that one of the calculation amounts in Tables 1 and 2 is shifted in phase by 90°.
For example, in combination 2 of Table 2, when using the above-mentioned cross product operation, the polarity e _P is set to υ _a -υ _b , υ _b -υ _c and υ _c -υ _a with the calculation quantities a, b and c, respectively, and 90° The number of steps needed to advance the phase can be saved. Each of the above-mentioned embodiments uses the negative phase difference current of the parallel power transmission lines as the detection current, and can achieve significantly higher sensitivity than the conventional directional type described above. Although they do not have the same high sensitivity as the conventional variation type and compensation type, they have the disadvantage of a poor response when the mating terminal is disconnected. Or,
Or, it is necessary to detect and lock a disconnection of the other end. Each of the above-mentioned embodiments is free from such drawbacks and can achieve the sensitivity described above. This problem will be explained below using FIG. FIG. 8 shows only the protected power transmission line portion of FIG. 1, and the same parts are indicated by the same symbols. Further, F is the accident point. Assume that an accident occurs at the closest point F to bus line D, as shown in Figure a. In this case, the power transmission line 3L on the bus C side
The terminal currents of terminals 4L and 4L have the same amount of fault current, and neither the variation type nor the compensation type operates, and only the terminal of power transmission line 3L on the bus D side is disconnected. Incidentally, a circulating current i _i due to induction flows until the shear breaks, and this appears as a differential current on the bus D side. When the terminal on the bus D side of the power transmission line 3L is disconnected, the state shown in FIG. b occurs. The current i in the power transmission line 3L on the bus C side is only for the fault current, and the terminal current of 4L is for the load current and the fault current flowing from the bus D toward C. The induced current is dissipated. The zero-sequence difference current used in the variation type includes only the fault current component and the induced component, and when moving from figure a to b, the induced component is lost and the fault current component is generated. At this time, the fault current changes to detect a fault in the power transmission line 3L. However, if the change in the induced component is in the direction of detecting a fault in the power transmission line 4L and is larger than the change in the fault current, the variation type will falsely detect a fault in the power transmission line 4L. In the compensation type, the zero-sequence difference current is compensated with a healthy phase difference current. In the state shown in FIG. b, this compensation is performed using the load current, so there is a risk that an accident in the power transmission line 4 will be detected by mistake. (Part 2)

[Table] A modified embodiment of the present invention uses the amount of change in the negative phase current as the detection current, and Table 3 shows the amount of calculation. In the table, (i _a2 −i _a2 ′) _n , (i _b2 −i _b2 ′) _n and (i _c
2 −
i _c2 ′) _n is the negative phase difference current (i _a2 − i _a2 ′) based on each phase,
(i _b2 −i _b2 ′) and (i _c2 −i _c2 ′) These are the values of _n stored before the accident. These memorized values are stored under conditions where there are no short circuits or ground faults in the protected system, and
It is determined using constantly updated sample values. When an accident is detected in the protected system, updating of this sample value is stopped and retained as a stored value.
When a 1-phase ground fault is detected, using this stored sample value and the sample value after the ground fault occurred,
Line selection calculations are performed with the amount of calculations depending on the ground fault phase. The detected current i _d of calculation quantities a, b, and c in Table 3 is the negative phase difference current based on the fault phase until some kind of interruption occurs after the occurrence of the fault when a one-phase ground fault occurs in phases a, b, or c, respectively. It should be equal to the fault current of As mentioned above, this fault current component is in phase with each polarity e _P in the case of a fault in transmission line 2, and in opposite phase in a fault in transmission line 3, and the fault is correctly determined by calculating equations (31) and (32). Identify the line. In this embodiment, there is no change in the detected current i _d in the event of an external fault in the protected system, so it is possible to achieve extremely high sensitivity. The operating value is limited by the amount of memory and other errors in the calculation process, and if this error is set to 10%, which is the same as the negative phase difference current due to induction as in the conventional variable amount system, the lower limit of the operating value is (33) and It is 10% of the value of formula (34), or a little over 2A, and the operating value is set with a margin.
2.8A, making it significantly more sensitive than conventional variable amount types. Furthermore, if condition 3 is considered, the amount of change in the negative phase difference current when condition 3 occurs reaches 96.3A, which is the value of equation (35), and it is difficult to avoid this operation with the sensitivity described above. be. However, it can be applied to a protected system in which the zero-phase induced voltage υ _po is slightly smaller than the value of equation (10) and the one-phase ground fault detection in step 2 of FIG. 5 is not performed. Similar to the case of Table 1, the calculation amounts in Table 3 are only one example of many examples in which the detected current using the amount of change is used. i _a2 −i _a2 ′, i _b2 of the detected current i _d in Table 2 above
−i _b2 ′ and i _c2 −i _c2 ′ respectively (i _a2 −i _a2 ′)−(i _a2
−
i _a2 ′) _n , (i _b2 −i _b2 ′)−(i _b2 −i _b2 ′) _n and (i
_c2 −
If you rewrite it as i _c2 ′) − (i _c2 − i _c2 ′) _n , you can apply it while keeping everything else the same. In each of the above embodiments, the amount of change in value at the time of a one-phase ground fault with respect to the value before the occurrence of an accident in the negative phase difference current of the parallel power transmission line is used as the detection current, and it is different from the conventional zero-sequence difference current. It can be made extremely sensitive to variations. (Part 3)

[Table] In a further modified embodiment of the present invention, a current compensated by a negative phase difference current is used as a detection current, and Table 4 shows an example of the calculation amount. In the table, K ₄ , K _5a , K _5b and K _5c are complex constants. Now, the constant K ₄ is set to the following value, and this value will be used in the following explanation. K ₄ =0.38 8゜ (49) At this time, the values of the detected current i _d of the calculation amount a in Table 4 due to induction under conditions 1, 2, and 3 are (7), (9), ( 10), (33), (34), and (35), resulting in the following value. [Condition 1] i _da = (21.2 13°−0.38×56.4 16°)
A=1.15A 87゜...(50) [Condition 2] i _da = (20.7 37゜−0.38×54.2 34゜)
A=1.08A 122゜...(51) [Condition 3] i _da = (96.3 95゜−0.38×285 91゜)
A=14.0A 118° (52) where i _da is the detected current i _d of the calculation amount a. In addition, the detected current i _d for the calculation amounts b and c in Table 4
From the relationship in equation (36), there is the following relationship. i _db = ai _da (53) i _dc = a ² i _da (54) where i _db and i _dc are the detection currents i _d of the calculation quantities b and c, respectively. As shown in equations (53) and (54), the detection currents i _db and i _dc are each equal in magnitude to i _da and lead in phase by 120° or 240°. Next, the phenomenon that occurs when a one-phase ground fault occurs on a protected power transmission line will be explained. At the time of a one-phase ground fault, there is a relationship expressed by equation (46) between the fault phase difference current and the fault phase difference current based on the fault phase and the fault fault phase difference current. From this relationship, the constant K ₄ is set as (49) for the fault portion of each detected current.
Assuming the value of the equation, the following relational expression holds true between the fault phase difference current i _f −i _f ′. (i) Detected current i _da i _da = (i _a2 −i _a2 ′)−K ₄ (i _p −i _p ′) =0.63 5°×1/3(i _f −i _f ′) at the time of a-phase ground fault ) …(53) (ii) Detection current i _db at b-phase ground fault i _db = (i _b2 −i _b2 ′) − aK ₄ (i _p −i _p ′) = 1.19 17°×1/3(i _f −i _f ′) …(54) (iii) Detection current i _dc at c-phase ground fault i _dc = (i _c2 −i _c2 ′) − a ² K ₄ (i _p −i _p ′) = 1.27 14゜×1/3 (i _f −i _f ′) ...(55) In Table 4, the constants K _5a , K _5b and K _5c of the polarity e _P are given the following values. K _5a = 1 5°, K _5b = 1 17°, K _5c = 1 14°
…(56) In a one-phase ground fault in a, b or c phase, the fault phase difference current i _f −i _f ′ is almost in phase with the voltage e _a1 , e _b1 or e _c1 in the case of a fault in transmission line 2. , in the fault on transmission line 3, the phases are almost opposite, so the constants K _5a , K _5b and K _5c
If is the value of equation (56), then from the relationship of equations (53) to (55), the fault portion of the detected current i _d of the calculation quantities a, b and _c will be a, b or In the case of an accident on transmission line 2 due to a phase c phase 1 ground fault, almost the same phase occurs on transmission line 3.
In the case of an accident, the phase is almost opposite. From this relationship, it is possible to detect a faulty line based on the faulty current using the calculation amount shown in Table 4. Next, the sensitivity relationship will be compared with that of the conventional compensation type device. Regarding the differential current due to induction, the maximum value of the compensated current of the compensation type under conditions 1 and 2 is 3.0A from equations (17) and (18). On the other hand, the maximum value of the detected current i _d in this example is calculated from equations (50) and (51).
The current is 1.15A, which is 0.38 times that of the compensated type. On the other hand, the fault current is (if − i _{f ′} )/3 for the compensated type due to the relationship of equation (46), whereas the detected current in this example differs depending on the ground fault phase. , the minimum value is Equation (53) for the a-phase ground fault, which is 0.63 times that of the compensated type. As mentioned above, the detection current of this implementation has 0.38 times the influence of induction compared to the conventional compensation type, and the fault current is minimal.
Since it is 0.63 times, the influence of induction on fault current is improved to 0.38/0.63=0.6 times. If this effect is small, the operating value to avoid malfunction due to an external fault can be made small, and when the induced current is in the opposite phase to the fault current due to an internal fault, the fault current can be canceled and the fault line can be detected. Since the interfering effect is small, highly sensitive protection is possible. Also, when considering condition 3, the detected current due to induction is 14.0A as shown in equation (52), which is 0.38 times the maximum value of 37A for the compensated type in equation (19). As in cases 1 and 2, high sensitivity can be achieved. In this embodiment, there is also less possibility of a defective response in the state shown in FIG. 8b. In this case, there is no zero-sequence component in the load current component of the current i' of the power transmission line 4L, and there is almost no negative-sequence current. Therefore, the detected current i _d in Table 4 is almost only the fault current, and the fault line can be correctly selected even in this state. As in the case of Table 1, the calculation amount in Table 4 is only one example of many examples in which the detected current is a current obtained by compensating the negative phase difference current with the zero sequence difference current. i _a2 −i _a2 ′, i _b2 −i _b2 ′ and i _c2 −i _c2 ′ of the detection current i _d in Table 2 above are respectively (i _a2 −i _a2 ′)−K ₄ (i _p −i _p ′), (i _b2 −i _b2 ′)
−aK ₄
(i _p −i _p ′) and (i _c2 −i _c2 ′)−a ² K ₄ (i _p −i _p ′)
, and multiplying the polarity e _P of the calculation quantities a, b, and c by the constants K _5a , K _5b , and K _5c , respectively, is the detected current i _d and the polarity of each calculation quantity a, b, and c. The relationship for the quantity e _P is the same as that in Table 4 and can be applied in the same way. In each of the above embodiments, the detection current is a current obtained by compensating the negative phase difference current of the parallel power transmission line with the zero-sequence difference current, and it is possible to provide highly sensitive protection compared to the conventional compensation type using the zero-sequence difference current. , and even after the opposite terminal is accidentally disconnected as shown in FIG. 8b, the faulty line can be accurately identified. (Part 4) In a further modified embodiment of the present invention, the current obtained by compensating the negative phase difference current with the difference current of one of the healthy phases is used as the detection current, and Table 5 shows an example of the calculation amount. . In the table, K _6a , K _6b and K _6c are complex constants.

[Table] The constants K _6a , K _6b and K _6c are as follows, and will be explained below using these values. K _6a = 0.48 5°, K _6b = 0.82 116°, K _6c = 0.22 130°
...(57) At this time, the values of the detected current i _d of each calculation amount due to induction under conditions 1, 2, and 3 are (7), (9), (10),
By substituting the values of equations (33) and (35) into the equations in Table 5, the following values are obtained. [Condition 1] i _da = 1.5A 81゜i _db = 2.3A 153゜i _dc = 0.7A 10゜ ...(58) [Condition 2] i _da = 1.8A 122゜i _db = 2.2A 15゜i _dc = 0.7A 104゜ …(59) [Condition 3] i _da = 20.8A 127゜i _db = 30A 235゜i _dc = 9.3A 0゜ …(60) When a one-phase ground fault occurs on the protected transmission line , since the difference current corresponding to the fault does not occur in the healthy phase, the difference current corresponding to the fault with the calculation amount i _d in Table 5 is only for the negative sequence current, which is the same as the example using the calculation amount in Table 1. , select the fault line correctly based on the fault current. Furthermore, from equation (46), this fault difference current is equal to the fault zero-sequence difference current, which is the detection current of the conventional compensation type. The difference current due to induction in this example is (58) under conditions 1 and 2.
From equation (59), the maximum is 2.3A, which is the compensated type.
This is 0.77 times the 3.0A (17) (18) formula, which allows for highly sensitive protection. Also, when considering condition 3, the difference current due to induction is the maximum from equation (60).
At 30A, it is smaller than the compensated type 37A (formula (19)) and can achieve higher sensitivity. The calculation amount in Table 5 is only an example of a case where the detected current is a current compensated by the difference current of one of the healthy phases of the negative phase difference current. Combinations 1~ in Table 2
Even if the polarity quantity e _P of 3 is used in place of the polarity quantity e _P of Table 5, the response will be exactly the same. Furthermore, it is also possible to use any phase current of the healthy phase to compensate for each detected current, and for example, the following equation can be used. i _da = (i _a2 −i _a2 ′)−K _7a (i _c −i _c ′) i _db = (i _b2 −i _b2 ′)−K _7b (i _a −i _a ′) i _dc = (i _c2 −i _c2 ′) −K _7c (i _b −i _b ′) …(61) However, K _7a , K _7b and K _7c are complex constants. (Part 5)

[Table] The amount of computation of further modified embodiments of the present invention is shown in the sixth table.
Shown in the table. In the table, K _8a , K _8b and K _8c are complex constants, and these constants will be explained as the following values. K _8a = 0.48 5°, K _8b = 0.17 110°, K _8c = 0.48 125°
...(61) At this time, the values of each calculation amount i _d by induction under conditions 1, 2, and 3 are as follows. [Condition 1] i _da = 1.5A 81゜i _db = 1.1A 161゜i _dc = 1.5A 39゜ ...(62) [Condition 2] i _da = 1.8A 122゜i _db = 0.75A 16 i _dc = 1.8 A 118゜ ...(63) [Condition 3] i _da = 20.8A 127゜i _db = 9.9A 145゜i _dc = 20.8A 7 ...(64) The maximum detected current due to induction is under conditions 1 and 2.
1.8A and 20.8A under condition 3, which is further improved than the example shown in Table 5. Further, as in Table 5, compensation is performed using a healthy phase difference current, and the fault current is the same as in the embodiments in Table 5, so this embodiment can protect with higher sensitivity than the embodiments in Table 5. The amount of calculation in Table 6 is to convert the negative phase difference current into the b phase difference current in the case of an a-phase or c-phase fault, and the a and c phase difference current in the case of a b-phase fault.
This is only one embodiment in which the current compensated by the sum of the phase difference currents is used as the detection current, and like the other embodiments, the amount of polarity can be modified in various ways. (Part 6)

[Table] Table 7 shows the amount of calculations for further different embodiments of the present invention, and each formula in the table is shown below. i _da = (i _a2 −i _a2 ′)−K _9a [(i _b −i _b ′) −a(i _c −i _c ′)] …(65) i _db = (i _b2 −i _b2 ′)− K _9b [(i _c −i _c ′) −a(i _a −i _a ′)] …(66) i _dc =(i _c2 −i _c2 ′)−K _9c [(i _a −i _a ′) − a( _ib −i _b ′)] …(67) However, K _9a , K _9b and K _9c are complex constants. Let each constant be K _9a = 0.34 17°, K _9b = 0.19 157°, K _9c = 0.17
111°...(68) Then, the values of each calculation amount i _d due to induction under conditions 1, 2, and 3 are as follows. [Condition 1] i _da = 2.0A 65゜i _db = 1.4A 175 i _dc = 0.9A 98゜ …(69) [Condition 2] i _da = 1.8A 38゜i _db = 1.5A 89゜i _dc = 1.1 A 129゜ …(70) [Condition 3] i _da = 18.2A 128゜i _db = 12.2A 135゜i _dc = 10.2A 3゜ …(71) The maximum detected current due to induction is under conditions 1 and 2.
2.0A, and 18.4A under condition 3, which is approximately the same level as the example shown in Table 6. Further, since the compensation is based on the healthy phase difference current and not the fault current, protection can be achieved with almost the same high sensitivity as the embodiments shown in Table 6. The detection currents in Table 7 can also be used in combination with various amounts of polarity as in other embodiments. In all of the examples shown in Tables 5 to 7 above, the detection current is a current obtained by compensating the negative phase difference current of the parallel transmission line with a healthy phase difference current, and the conventional zero-sequence current is compensated with a healthy phase difference current. Highly sensitive protection is possible for compensated types. (Part 7) FIG. 9 shows a further different embodiment of the present invention, and the same parts as in FIG. 5 are indicated by the same symbols. After the calculation amount is given in step S4 in the same way as in the embodiment of FIG. 5, the suppression amount _er is given in step S6. When the calculation amount a, b or c is given, the suppression amount a, b or c is given respectively. step
After S6, a sample value of the suppression amount is calculated in step S7. Thereafter, as in the embodiment shown in FIG. 5, sample values of calculation amounts are calculated in step S5, and fault detection calculations for power transmission lines 2 and 3 are performed. Examples of calculations performed in steps S5-2 and S5-3 of this embodiment are shown in equations (72) and (73), and when each equation holds true, judgments 5Y1 and 5Y2 are obtained. e _P・i _d ＞｜e _p | (K ₃ | e _r | and the maximum value of K ₁₀ ) …(72) e _P・i _d ＜−｜e _p | (K ₃ | e _r | and K ₁₀ (maximum value) ...(73) where e _r is the amount of suppression, K ₁₀ is a positive constant Equations (72) and (73) are the effective portion of the detected current i _d with respect to the polarity e _P , respectively, or their signs are changed. is larger than the maximum value of K ₃ | _er | and K ₁₀ (determination 5Y1 or 5Y2 is obtained). FIG _. 10 is a diagram showing the operating characteristics in this case. When the _magnitude of the suppression amount _| e _r
When | is large, it will not work unless the absolute value of the effective part of i _d is large. As the suppression amount e _r , various differential currents or a composite value of differential currents are used, examples of which are shown in Table 8. The various quantities listed in the table are all in a constant induction state, and their magnitudes are approximately proportional to the magnitude of the detected current i _d due to induction, and as shown in Condition 3, the magnitude of the detected current i _d due to induction is When the value is large, the suppression amount e _r is large, and no malfunction will occur even if the constant K ₁₀ is small or the constant K ₃ is small.

[Table] In addition, when the magnitude of the detected current i _d due to induction is small as in conditions 1 and 2, the magnitude of the suppression amount e _r is also small, and in this state, a one-phase ground fault can occur on the protected transmission line. When the following conditions are met, there is an advantage that protection can be achieved with high sensitivity. This condition will be explained. Size of detected current of induced component | i _d | / Size of suppression amount |
e _r ｜=R _i …(74) Magnitude of detected current for fault caused by one-phase ground fault on protected transmission line｜i _d ｜/Size of suppression amount｜e _r ｜=
When R _f ...(75), if R _i is sufficiently smaller than R _f, protection can be achieved with high sensitivity. That is, if the constant K ₃ is set to a value slightly larger than the ratio R _i under conditions 1 to 3, equations (72) and (73) will not hold, and there is no risk of malfunction. On the other hand, in the case of a protected transmission line fault, the magnitude of the detected current |i _d | is the magnitude of the suppression amount |i _r |
is multiplied by R _f , which is a sufficiently large value for the suppressing force (=K ₃ | e _r |). Furthermore, since the detection current i _d is substantially in phase with the polarity amount e _P , it is possible to operate with high sensitivity under the above conditions. For example, when R _i <0.38R _f , if K ₃ =0.4R _f , equations (72) and (73) will not hold and there is no risk of malfunction. In addition, for the protected transmission line accident, the suppressing force (=K ₃ | e _r |) is 0.4R _f | e _r
|, whereas the magnitude of the detected current |i _d | is
Since R _f |e _r |, it can work satisfactorily. The example in Table 8 illustrates meeting these conditions. First, Table 9 shows the magnitude of the detected current | i _d | and the magnitude of the negative phase difference current | i _a2 −i _a2 ′| = | i _b2 | −i _b2 ′｜=｜
The ratio to i _c2 −i _c2 '| is shown for the induced component (maximum value of conditions 1 to 3) and accident component.

[Table] Next, Table 10 shows the magnitude of the suppression amount |e _r | and the magnitude of the negative sequence current |i _a2 −i _a2 ′|= |i _b2 |−i _b2 ′|= |i _c2 −
The ratio to i _c2 '| is shown for the induced component (minimum value of conditions 1 to 3) and the accident component. However, e _ra , e _rb
and e _rc indicate the suppression amounts a, b, and c, respectively.

[Table] When considering a certain combination of calculation amount and suppression amount, the ratio of the induced value in Table 9 to the induced value in Table 10 is the ratio R _i of equation (74), which is A similar ratio becomes the ratio R _f in equation (75). For example, when the calculation amount is shown in Table 4 and the suppression amount is Example 3, for the a phase, the ratio R _i is 0.073 = 0.145/2, and the ratio R _f is 0.46 =
0.63/1.38, which is suitable for the conditions because R _f is sufficiently larger than R _i . Table 11 shows the ratios R _i and R _f for each suppression amount when the calculation amount is shown in Table 1 or Table 2. In this case, since the values for each phase are almost equal in Tables 9 and 10, the values of R _i and R _f are also almost equal for each phase, and the tables show approximate values.

[Table] Example 1 of the amount of suppression is that R _i and R _f cannot be used equally.
In Example 3, R _i is smaller than R _f , but the difference is not so large that it can be considered unusable. In Example 6, R _i is 1/2 of R _f , and although it is not impossible to use, it is not very suitable. In Example 1, it can be seen that R _i can be used at 0.38 times R _f . Examples 5, 8 and 9 have R _f of infinity and are the most suitable. Example 2 of the suppression amount is the most difficult to use because the values in Table 10 are equal for the induced portion and the accident portion. However, it can be used satisfactorily for the calculation amounts shown in Tables 4 to 7. In this case, the values of the induction component and accident component in Table 9 become the ratios R _i and R _f as they are,
The ratio R _i is sufficiently small compared to R _f . For suppression amounts other than Example 2, the values for accidents in Table 10 are smaller than the induced values, so
The ratio of R _i to R _f is smaller than in Example 2, and can be used satisfactorily in combination with the calculation amounts shown in Tables 4 to 7. As mentioned above, the embodiment shown in FIG. 9 uses the difference current or composite current of the same phase or the same symmetry of parallel transmission lines as the suppression amount, as shown in the example shown in Table 8, to reduce the difference current due to induction. This has the advantage of being able to protect protected power transmission line accidents with high sensitivity, without malfunctioning even when the amount is extremely large, compared to the case where these are not used for the suppression amount. (Part 8) The calculation using the suppression amount performed in each of steps S5-2 and S5-3 in Fig. 9 is not limited to equations (72) and (73), and examples thereof are shown below. . e _P・i _d ＞｜e _P ｜(K ₃ ｜ _er ｜＋K ₁₀ ) …(76) −e _P・i _d ＜｜−e _P ｜(K ₃ ｜ _er ｜＋K ₁₀ ) …(77) Equation (76) is performed at step S5-2, and equation (77) is performed at step S5-3. The operating characteristics in this case are shown in FIG. In addition, the following equation (78) and the above equation (31) are calculated in step S5-2, and the equations (78) and (32) are calculated in step S5-3, and both equations are satisfied. It is also possible to obtain the time determinations 5Y1 and 5Y2. |i _d | > K ₃ | e _r | and the maximum value of K ₁₀ (or K ₃ | e _r | + K ₁₀ )...(78) In either case, the magnitude of the detected current | i _d | is K ₃ |
If it is smaller than e _r ｜, it will not operate and the restraining force K ₃
When |e _r | is small, highly sensitive protection is possible. (Part 9) Compensation constants for the detection current i _d for the calculation amount in Tables 4 to 7 K ₄ , K _5a , K _5b , K _5c , K _6a , K _6b , K _6c , K _8a ,
The explanation was given by assuming that K _8b , K _8c , K _9a , K _9b , and K _9c are all constant values obtained from the calculated values of the difference currents. In this method, the value of the detected current due to induction can be reduced by setting each constant as the average value of values that slightly fluctuate under various operating conditions, but the calculation depends on the wire arrangement of parallel transmission lines. It is necessary to If it is desired to omit such calculations, these constants can be determined by calculations using differential currents due to constant induction. In other words, in the case of the calculation amount in Table 4, there is no short circuit or ground fault in the protected system, and the difference current i _a2 −
The constant K ₄ is calculated as follows when at least one of i _a2 ′ or i _p −i _p ′ is greater than a certain value. K ₄ = i _a2 −i _a2 ′/i _p −i _p ′ during constant operation…(76) For example, in the case of condition 1, the value of K ₄ is
From the values of equations (7) and (33), K ₄ =21.2A 13°/56.4A 352°=0.376 5° (77) is calculated, and the detected current i _d is calculated using this constant. This calculated value is held and used to calculate the detected current i _d when the value of the differential current becomes small or when a short circuit or ground fault occurs in the protected system.
The calculated value of equation (77) has a small difference from the value K ₄ = 0.38 8° of equation (49) given earlier as a fixed constant, and the difference current due to induction during a single-phase ground fault in the protected system The effect of reducing the influence of is almost the same as in the case of a fixed constant. Other compensation constants K _5a to _{K 9c} similar to the above example are also provided on the condition that there is no accident in the protected system and at least one of the difference currents used for the detection current i _d is equal to or higher than the predetermined value. Even if the value of the compensation constant is calculated from the value of the difference current during operation, this calculated value is held when the above conditions are not met, and the detected current i _d is calculated using this calculated value as the value of the compensation constant. , which shows almost the same effect as when using a fixed constant. (g) Overall effect As described above, according to the present invention, it is possible to provide a ground fault line selection relay that reduces the influence of circulating current due to induction.

[Brief explanation of drawings]

Figure 1 is a system diagram illustrating an example of a multi-circuit parallel transmission line, Figure 2 is a diagram showing an example of wire arrangement in a parallel section,
FIG. 3 is a vector diagram showing an example of the zero-sequence voltage and zero-sequence difference current of the protected power transmission line when a one-phase ground fault occurs in the induced power transmission line, and FIG. 4 is a circuit diagram showing the configuration of an embodiment of the present invention. , FIG. 5 is a flowchart showing the flow of one embodiment of the present invention, FIG. 6 is a vector diagram showing the fault differential current, FIG. 7 is a vector diagram showing the voltage at the time of a phase 1 ground fault, and FIG. Fig. 8 is a system diagram for explaining the response when the other party is disconnected, Fig. 9 is a flow chart showing another embodiment of the present invention, and Figs. 10 and 11 are characteristics of the embodiment of Fig. 9. FIG. 4 and 5 are disconnectors, 9 is an input converter, 10
1 is a sample hold circuit, 11 is a multiplexer, 12 is an AD converter, and 13 is an arithmetic unit.

Claims (1)

[Scope of Claims] 1. Ground fault phase identification means that performs ground fault phase selection calculations according to changes in the zero-phase voltage and phase-to-phase voltage of a power transmission line and identifies a single-phase ground fault phase; and this ground fault phase identification means. calculation amount selection means for selecting, as calculation amounts, a polarity amount synthesized by the positive sequence voltage and zero-sequence voltage and a detection current synthesized by the negative phase difference current of the parallel power transmission line, which are preset according to the identification result of; A ground fault line selection relay comprising a line selection means for calculating a sample value of the calculation amount selected by the calculation amount selection means and selecting a ground fault fault line based on the polarity amount and the value of the detected current. 2. The ground fault line selection relay according to claim 1, wherein the detected current is the amount of change in the value of the negative phase difference current of the parallel power transmission line at the time of the accident with respect to the stored value before the occurrence of the accident. 3. The ground fault line selection relay according to claim 1, wherein the detection current is a current obtained by compensating the negative phase difference current of the parallel power transmission line with the zero sequence difference current. 4. The ground fault line selection relay according to claim 1, wherein the detection current is a current obtained by compensating the negative phase difference current of the parallel power transmission line with a healthy phase difference current at the time of a ground fault accident. 5 Ground fault phase identification means that performs calculations for ground fault phase sorting according to changes in the zero-sequence voltage and phase-to-phase voltage of the transmission line and identifies a single-phase ground fault fault phase, and according to the identification result of this ground fault phase identification means. a calculation amount selection means for selecting a polarity amount synthesized by a preset positive-sequence voltage and a zero-sequence voltage, and a detection current synthesized by a negative-phase difference current of a parallel power transmission line as a calculation amount; Suppression amount selection means for selecting, as a suppression amount, a composite current composed of the same phase or the same symmetric differential current of parallel power transmission lines set in advance according to the identification result; the calculation amount selection means and the suppression amount selection means. and a line selection means for calculating sample values of the selected calculation amount and suppression amount, and selecting a ground fault fault line from the values of the polarity amount, the detected current, and the composite current. relay.