JPH0481145B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a method for locating fault points in power transmission lines.

Currently, the fault point locating method that is widely used in practice involves applying and transmitting a pulse from one end of the line when a fault occurs on the line, measuring the time until the reflected wave is received from the fault point, and measuring the time from which the fault point is located. There is a pulse transmission method for determining distance. However, this method suffers from large waveform distortion and attenuation during the line propagation process.
Furthermore, the waveforms of surges or reflected pulses differ depending on the failure mode, making it difficult to locate the failure point with high precision.

Another method is the distance relay method, which has been put into practical use as a line protection device, but this method has large errors due to the influence of fault point resistance, and even if it is put into practical use as a protection method, it is not practical as a location method. There is no way to change it.

An object of the present invention is to provide a locating method that can locate a fault point with high accuracy without being affected by the fault point.

The present invention is characterized in that a failure point is located by calculation using voltage and current values measured at an electric station at one end of the line and known line constants.

The method of the present invention will be explained in detail below.

Figure 1 shows a circuit that treats a single-phase line as a distributed constant. Point S is the substation, point F is the fault point, R _F
indicates the fault point resistance, which is the resistance per unit length of the line.
r', reactance L', susceptance _C1 ', distance x from point S to fault point F, and total line length l, then in terms of lumped constants, it is represented by the π-type equivalent circuit shown in the figure, and as seen from point s. Admittance Y and impedance Z are as follows.

Y=jxC ₁ 'Z=xr'+jxL' The three-phase power transmission line is also represented by the equivalent circuit in Figure 2, which is treated as a similar distributed constant, and the point S
Self-impedance of each phase of a, b, and c as seen from
Z _s , mutual impedance Z _n , and self-admittance Y _s1 are as follows.

Z _s = r _s + jωl _s = Z′ _s・x = (r′ _s + jωl _s ′)・x Z _n = r _n + jωl _n = Z′ _n・x = (r′ _n + jωl _n ′)・x Y _s1 ＝jωC _s1 ＝Y _s ′ ₁・x＝jωC′ _s1・x However, Z _s ′: Self-impedance per unit length Z _n ′: Mutual impedance per unit length Y _s ′: Self-admittance per unit length Y _s ' ₁ : Self-admittance per unit length from point S to failure point F Y _s ' ₂ : Self-admittance per unit length from failure point F to the load end r _s : Resistance from point S to failure point F r _s ′: Resistance per unit length from point S to failure point F _{l s} : Self-inductance from point S to failure point F l _s ′: Self-inductance per unit length from point S to failure point F r _n : Point Mutual resistance from S to failure point F r _n ′: Mutual resistance per unit length from point S to failure point F l _n : Mutual inductance from point S to failure point F l _n ′: From point S to failure point F Mutual inductance per unit length C _s1 : Capacitance per unit length from point S to fault point F C' _s1 : Capacitance per unit length from point S to fault point ω : Power source angular frequency This three-phase circuit For example, when one wire is grounded in
As shown in Figure A, when phase a has a fault point resistance R _F and is grounded, the equivalent circuit according to the object coordinate method is the third
As shown in Figure B, zero-sequence current _{I p} flows through the series circuit of zero-sequence circuit O, positive-phase circuit P, negative-phase circuit N, and _3RF , and the zero-sequence current I _p as the fault point target, The positive sequence current I ₁ and the negative sequence current I ₂ are as follows.

I _p = V _Fa /Z _p +Z ₁ +Z ₂ +3R _F I ₁ = I _p I ₂ = I _p However, Z _p : Zero-sequence impedance to the commercial frequency viewed from the failure point Z ₁ : Positive-sequence impedance Z ₂ : Anti-phase impedance V _Fa : A-phase voltage before failure at the failure point From this equivalent circuit and the equivalent circuit of FIG. 2, the following equations (1) to (7) hold true when an a-phase ground fault occurs. Furthermore, Z _p = Z _s − Z _n = r _p + jωl _s = (r _p ′+jωl _p ′) x Z ₁ = Z _s +2Z _n = r ₁ + jωl ₁ = (r ₁ ′+ jωl ₁ ′) x However, r _p : Zero-sequence resistance r _p ′: Zero-sequence resistance per unit length l _p ′: Zero-sequence inductance per unit length r ₁ : Positive sequence resistance r ₁ ′: Positive sequence resistance per unit length l ₁ : Positive sequence inductance l ₁ ′: Positive sequence inductance per unit length V _a = Z _s I _a1 + Z _n (I _b1 + I _c1 ) + V _Fa = Z ₁ I _a1 +3I _p1 Z _p −Z ₁ /3 + V _Fa …(1) V _Fa = I _F・R _F = R _F (3I _p2 −△3I _pc ) = R _F {I _a2 + I _b2 + I _c2 −Y _s2 (V _Fa +V _Fb +V _Fc )} = R _F (3I _p2 −3V _pF・Y _s2 ) …(2) 3I _p1 =3I _p −Y _s1 /2(V _a +V _b +V _c ) =3I _p −Y _s1 /2・3V _p 3I _p2 =3I _p1 −Y _s1 /2(V _Fa +V _Fb +V _Fc )=3I _p1 −Y _s1 /
23V _pF =3I _p −Y _s1 /2 (3V _p +3V _pF )…(3) V _Fa =V _a −Z ₁・I _a1 −3I _p1・Z _p −Z ₁ /3 3V _pF =3V _p −Z・3I _p1 −3I _p1 (Z _p − Z ₁ ) = 3V _p −Z _p・3I _p1 = 3V _p −Z _p (3I _p −Y _s1 /2・3V _p ) = (1+Z _p Y _s1 /2) 3V _p −Z _p・3I _p …(4) From formulas (2) and (3), V _Fa = R _F {3I _p −Y _s1 /2 (3V _p +3V _pF )−3V _pF Y _s 2} = R _F
{3I _p −Y _s1 /23V _p −3V _pF (Y _s1 /2+Y _s2 }}…(5) From equations (4) and (5), V _Fa = R _F [3I _p −Y _s1 /2・3V _p −(Y _s1 /2+Y _s2 ) {(1
+Z _p Y _s1 /2)・3V _p −Z _p 3I _p }] = R _F [3I _p {1+Z _p (Y _s1 /2+Y _s2 )}−3V _p {Y _s1 /
2+(Y _s1 /2+Y _s2 )・(1+Z _p Y _s1 /2)}]...(6) From equations (1) and (6), V _a =Z ₁ (I _a −V _a Y _s1 /2)+ (3I _p −Y _s1 /2・3V _p )Z _p −
Z ₁ /3+R _F [3I _p {1+Z _p (Y _s1 /2+Y _s2 )} −3V _p {Y _s1 /2+ (Y _s1 /2+Y _s2 ) (1+Z _p Y _s1 /2
)}] ∴(1+Z ₁ Y _s1 /2)V _a =Z ₁ I _a +3I _p [Z _p −Z ₁ /3+{1
+Z _p (Y _s1 /2+Y _s2 )｝R _F ] −3V _p [Y _s1 /2・Z _p −Z ₁ /3+{Y _s1 /2+(Y _s1 /2
+Y _s2 )・(1+Z _p Y _s1 /2)}R _F ]...(7) In equations (1) to (7), V _Fa , V _Fb , and V _Fc are the accident point F
I _F is the zero-sequence current flowing from the fault point to the ground, and I _a1 , I _b1 , I _c1 are the voltages of the a, b, and c phases flowing at point A1 in Figure 2. The current, I _a2 ,
I _b2 and I _c2 are the phase currents flowing through point A2, 3I _p1 and 3I _p2 are the zero-sequence currents flowing through points A1 and A2, respectively, V _pF is the zero-sequence voltage at the fault point, and △3I _pC is from the fault point to the load side 3I _p1 = I _{a1 +} I _b1 + I _c1 3I p2 = I _a2 + I _b2 + I _c2 3V _pF = V _Fa + V _Fb + V _Fc △3I _pc = Y _s2 (V _Fa + V _Fb + V _Fc ) There is a relationship between

Vector expansion of equation (7) above gives V _a {(1−ω ² l ₁ C _s1 /2) + jωC _s1 /2r ₁ }=I _a (r ₁ +
jωl ₁ ) +3I _p [{r _p −r ₁ 3+(1−ω ₂ l _p (C _s1 /2+C _s2 ))
R _F }+jω(l _p −l ₁ /3+r _p (C _s1 /2+C _s2 )R _F }] −3V _p [−ωC _s1 {ωl _p −l ₁ /3+r _p ω(C _s1 /2+C _s2
) R _F } +jω{C _s1 /2・r _p −r ₁ 3+(C _s1 /2+C(C _S1 /2 _s
2・(1−ω ² l _p C _s1 /2)R _F )}]…(8) Here, if we generally rearrange equation (8) from the following relationship, 1−ω ² l ₁ C _s1 /2≒ 1 1−ω ² l _p C _s1 /2≒1 V _a =I _a (r ₁ +jωl ₁ )+3I _p [r _p −r ₁ /3+R _F +jω{l _p
−l ₁ /3+r _p (C _s1 /2+C _s2 )R _F }] −3V _p [−ωC _s1 /2{ωl _p −l ₁ /3+r _p ω(C _s1 /2
+C _s2 ) R _F }+jω{C _s1 /2・r _p −r ₁ /3+(C _s1 +C _s2
)R _F }]...(9) Also, if the fault point resistance R _F in equation (9) is large, then in the term 3I _p , r _p −r ₁ /3 + R _F ≫ω{l _p −l ₁ /3 + r _p (C _s1 +C _s2 )R _F } 3V In _{the p} term, ω{C _s1 /2・r _p −r ₁ /3 + (C _s1 +C _s2 )R _F }≫ωC _s1 /
2{ωl _p −l ₁ /3+r _p ω(C _s1 /2+C _s2 )R _F }, and if R _F is small, in the term of 3I _p , r _p −r ₁ /3+R _F ≒ωl _p −l ₁ / 3 In the term of 3V _p , ωC _s1 /2・ωl _p −l ₁ /3≒ωC _s1 /2・r _p −r ₁ /3, so equation (9) can be expressed as the following equation (10). .

V _a = I _a (r ₁ + jωl ₁ ) + 3I _p {(r _p − _{r 1} /3 + R _F ) + jωl
_p −l ₁ /3} −3V _p [−ωC _s1 /2・ω・l _p −l ₁ /3+jω{C _s1 /2
・r _p −r ₁ /3+(C _s1 +C _s2 )・R _F }]...(10) If we rearrange this equation (10) into a general form, V _a =Z ₁ I _a +Z _n・3I _p +3I _p R _F −1/2Y _s Z _n・3V _p −Y _s ′lR _F
・3V _p =Z ₁ I _a +Z _n (3I _p −1/2Y _s・3V _p )+R _F (3I _p −Y _s ′
l・3V _p )...(11) From this equation (11), in order to locate the fault point that is not affected by the fault point resistance R _F , the following (12-1) and (12 -2) It can be found from equation (12) by eliminating R _F from equation. Note that Z ₁ = Z ₁ ′x, Z _n =
The transformation is Z _n ′x, Y _s = Y _s ′x.

Real〔V _a −{Z ₁ ′xI _a +Z _n ′x(3I _p −1/2Y _s ′x・3V
_p )}]=R _F・Real(3I _p −Y _s ′l・3V _p )…(12−1) Imag[V _a −{Z ₁ ′xI _a +Z _n ′x(3I _p −1/2Y _s 'x・3V
_p )}] = R _F・Imag(3I _p −Y _s ′l・3V _p )…(12-2) Real [A]・Imag [B] − Imag [A]・Real [B] =
0 …(12) However, A and B are A=V _a −{Z ₁ ′xIa+Z _n ′x(3I _p −1/2Y _s ′x・3V _p
)} B=(3I _p −Y _s ′l・3V _p ) As mentioned above, the fault point location according to the present invention is performed by calculating the zero-sequence current from each phase current and each phase voltage of the line flowing to its own end (point S). I _p , zero-sequence voltage V _p and fault phase current
I _a is calculated using equation (12) using the known self-impedance Z _s ′, mutual impedance Z _n ′, self-admittance Y _s ′ per unit length of the line, and total line length l, and the fault point distance x seek.

Next, when formula (12) is expanded specifically, V _a cosθ ₁ = I _a (r ₁ ′cosθ ₂ −ωl ₁ ′SINθ ₂ )x +3V _p (ωC′ _s1 /2・ω・l _p ′− l ₁ ′ ^／ _{3cosθ 3 + ωC ′ s1} /2_{・ r p} ′−r ₁ ′／ _{3SINθ 3 ) 3} ) …(13-1) V _a SINθ ₁ = I _a (r ₁ ′SINθ ₂ +ωl ₁ ′cosθ ₂ )x +3I _p ωl _p ′−l ₁ ′/3x +3V _p {(ωC′ _s1 /2・ω・l _p ′−l ₁ ′／3・SINθ ₃ −ωC′ _s1 /2・r _p ′−r ₁ ′／3cosθ ₃ )x ² +R _F ωC _s
′l・cosθ ₃ } …(13-2) However, θ ₁ is the phase difference between V _a and 3I _p , θ ₂ is I _a and 3I _p , and θ ₃ is
Phase difference between 3V _p and 3I _p From these two equations, the following quadratic equation for the fault point distance x is given:
It can be obtained using equation (13).

A(1)x ² +A(2)x+A(3)=0 ...(13) However, A(1), A(2), A(3) are the following (14-1), (14
-2), (14-3), where 3V _p , 3I _p , I _a , V _a
are their respective absolute values.

A(1)=ωC _s ′/2 [3V _p・3I _p・SINθ ₃・ωl _p ′−l ₁ ′
/3 −3V _p・3I _p・cosθ ₃・r _p ′−r ₁ ′/3+3V _p ² ωC _s ′l
・ωl _p ′−l ₁ ′/3〕…(14−1) A(2)=3I _p・I _a・SINθ ₂・r ₁ ′＋3I _p・I _a cosθ ₂ ωl ₁ ′
+3I _p ² ωl _p ′−l ₁ ′/3 +ωC′ _s・l [3V _p・I _a cos(θ ₃ −θ ₂ ) r ₁ ′+3V _p I _a S
IN(θ ₃ −θ ₂ )ωl′ ₁ +3V _p・3I _p SINθ ₃・ωl _p ′−l ₁ ′/3+3V _p・3I _p cos
θ ₃ r _p ′−r ₁ ′/3]…(14-2) A(3)=−[3I _p・V _a・SINθ ₁ +3V _p・V _a・cos(θ ₃ −θ ₁
) ωC _s ′・l]…(14-3) Next, in equation (13), if A(1)≪A(2), then A
The fault point distance x can be found from the following equation (15) where (1)=0.

_{A ( 2 ) _}・3I _p +R _F・3I _p , and in the same way as the calculation of equation (12), Imag[V _a − (Z ₁ ′・I _a +Z _n ′・3I _p )x]=0 …From the calculation of (16) It can be obtained as a linear expression of the fault point distance x.

Furthermore, the failure point distance x can be obtained from B(1)x+B(2)=0 (17) by expanding equation (16) in the same way as equation (13). B in the formula
(1) and B(2) are as follows.

B(1)=3I _p・I _a・SINθ ₂・r ₁ ′＋3I _p・I _a・cosθ ₂・ωl
₁ ′+3I _p ²・ω・l _p ′−l ₁ ′/3…(17−1) B(2)=−3I _p・V _a・SINθ ₁ …(17−2) Figure 4 shows the method of the present invention The configuration of the orientation device based on the following is shown. Each phase voltage V _a , V _b , V _c and each phase current at own end
I _a , I _b , and I _c are detected by transformer PT and current transformer CT.
This detection signal is taken into the first circuit D ₁ of the location device D as sampling data at a constant period, and using this data, the second circuit D ₂ outputs the zero-phase voltage V _p ,
From the zero-sequence current V _p, 3I _p = I _a + I _b + I _c 3V _p = V _a + V _b + I _c , and from the third circuit D ₅ , the line constants per unit length Z _s ′, Z _n ′, Y _s ′ or Z _p ′, Z ₁ ′, Y _s ′ and data of total length l are generated, and the above-mentioned (12) to (17) are generated from the data from circuits D ₂ and _{D 3 by} the fourth circuit D 4. ) Calculate the fault point distance x according to one of the following equations.
In equations (13) to (17), the vector inner product and outer product are VI cosθ=V(1)・I(1)+V(3)・I(3)+V(5)・I(5) VI SINθ=V (2) {I(1)-I(3)} +V(4){I(3)-I(5)}+V(6){I(5)-I(7)} However, V(1), V(2),...V(6) and I(1), I(2),...I(7) represent sampling data sequences at 30° intervals as shown in Figure 5. .

As described above, according to the present invention, the fault point resistance _R
The failure point can be located with high precision without being affected by the admittance _Ys located further away (on the load side) than the failure point.

[Brief explanation of drawings]

Figure 1 is an equivalent circuit diagram of a single-phase line, Figure 2 is an equivalent circuit of a three-phase line, Figure 3A is a fault circuit diagram of a single-line ground fault, and Figure 3B is the symmetrical coordinate method in Figure 3A. FIG. 4 is an equivalent circuit diagram based on the method of the present invention, and FIG. 5 is a waveform diagram for explaining sampling data of voltage V and current I in FIG. 4. D...Location equipment, PT...Transformer, CT...Current transformer.

Claims (1)

[Claims] 1. Each phase current _I a of the own ends a, b, c of the power transmission line,
Zero-sequence current calculated from I _b , I _c and each phase voltage V _a , V _b , V _c
I _p and zero-sequence voltage V _p , self impedance Z′ _s per unit length of the transmission line, mutual impedance
Z' _n , self-admittance Y' _s and total line length l, the following formula Real[A]・Imag[B] −Imag[A]・Real[B]=0 However, A=V _a −{(Z′ _s −Z _{′ n )} X・_{I a} +Z′ _n X(3I _p −1/2Y′ _s X・3V _p )} A method for locating a fault point on a power transmission line, characterized by determining the distance x from the end to the fault point. 2 In the first claim, the following formula A(1)x ² +A(2)x+A(3)=0 However, A(1)=ωC _s ′/2 [3V _p・3I _p・SINθ ₃ l′ _p −l′
₁ ／3 −3V _p・3I _p・cosθ ₃・r _p ′−r′ ₁ ／3＋3V _p ² ωC
′ _sl・ωl _p ′−l ₁ ′/3] A(2)=3I _p・I _a SINθ ₂・r′ ₁ +3I _p・I _a cosθ ₂ ωl′ ₁ +3
I ² _p・ωl _p ′−l ₁ ′/3 +ωC _s ′・l [3V _p・I _a COS (θ ₃ −θ ₂ ) r ₁ ′+3V _p I _a S
IN(θ ₃ −θ ₂ )ωl ₁ ′ +3V _p・3I _p SINθ ₃・ωl _p ′−l′/3+3V _p・3I _p cosθ
₃ r _p ′−r ₁ /3] A(3)=−[3I _p・V _a・SINθ ₁ +3V _p・V _a・cos (θ ₃ −θ ₁
)ωC _s ′・l] where V _p , I _p , I _a , and V _a are each absolute value, ω is the power source angular frequency, C _s ′: capacitance per unit length l _p ′: unit Sum of self inductance per unit length l _s ′ and mutual inductance l _n ′ × 2 l ₁ ′: l _s ′−l _n ′ r _p ′: Resistance per unit length r _s ′ and mutual resistance r _n ′ × 2 Sum r ₁ ′: r _s ′−r _n ′ θ ₁ : Phase difference between V _a and 3I _p θ ₂ : Phase difference between I _a and 3I _p θ ₃ : Failure point according to the phase difference between 3V _p and 3I _p A method for locating a fault point on a power transmission line, characterized by determining a distance x. 3 In claim 1 or 2,
A fault point locating method for a power transmission line, characterized in that distance x is determined according to the following formula: A(2)x+A(3)=0. 4 In claim 1, the distance x of the line with small admittance Y _s ' is calculated by the following formula Imag[V _a - (Z ₁ '・x・I _a +Z _n '・x・3I _p )]=0 A fault point locating method for a power transmission line, characterized in that the fault point is determined according to the following. 5 In claim 4, the following formula B(1)x+B(2)=0 However, B(1)=3I _p・I _a・SINθ ₂・r ₁ ′ +3I _p・I _a・cosθ ₂・ωl ₁ ′ +3I _p ²・ω・l _p ′−l ₁ ′/3 B(2)=−3I _p・V _a・SINθ ₁ Fault point of a power transmission line characterized by finding distance x according to 1 Orientation method.