JPH0627760B2 - Fault location method for power transmission system - Google Patents

Description

Detailed Description of the Invention A. TECHNICAL FIELD The present invention relates to a fault location system for a power transmission system.

B. SUMMARY OF THE INVENTION According to the present invention, in locating a fault point of a power transmission system according to a locating principle formula, a locator that compensates for a locating error due to a branch load from its own end to a fault point is used to locate the branch load. This is to eliminate the error.

C. BACKGROUND ART The applicant of the present application has already proposed a method for locating a fault point in a power transmission system, which is a method of locating a fault point by calculation using a voltage, a current, and a known line constant measured at an electric station at one end of a transmission line. (For example, Japanese Patent Application No. 59-143056,
Japanese Patent Application No. 59-143057).

An apparatus configuration based on the above method is as shown in FIG. Each phase voltage Va, Vb, Vc at each end and each phase current I
a, Ib, Ic are detected by the transformer PT and the current transformer CT,
These detection signals are taken into the first circuit D ₁ of the orienting device D ₁ as sampling data having a constant cycle, and the zero-phase voltage Vo and zero-phase current Io are also obtained in the _second circuit D ₂ by using these data, and the third circuit 3 Data of the self impedance Zs, the mutual impedance Zm, and the self admittance Ys per unit length are obtained from the circuit D _{3, and} from the data of the circuits D ₂ and D ₃ , the fourth circuit D ₄ is obtained from, for example, Calculate the fault point distance x at the time of phase-to-ground fault (track admittance Ys
Is ignored).

Real [A] / Imag [B] -Imag [A] / Real [B] = 0 ... (1) However, A = Va-{(Zs-Zm) / Ia + Zm / 3Io} x .... (2) B = 3Io …………………… (3) The above-mentioned orientation method is preferable when the power transmission system is a balanced line (a suspended system), and it is a general unbalanced line (a non-suspended system). ), The self-impedances of each phase Zaa, Zbb, Zcc and the mutual impedances Zab, Zbc, Zca are different, which causes localization error. So, for non-suspended system, 3 phase short circuit,
When there is a three-phase ground fault, a two-phase short circuit, or a two-phase ground fault, and Rf is the fault point resistance for the bc phase, for example, A = (Vb-Vc)-{(Zba-Zca) Ia + (Zbb-Zcb) Ib + (Zbc-Zcc) Ic} x …………………… (4) When B = Ib−Ic …………………… (5) and the transmission line has a one-phase ground fault, for example, Phase a As A = Va- {Zaa ・ Ia + Zab ・ Ib + Zac ・ Ic} x ………………………… (6) B = 3I ₀ …………………… (7) The applicant of the present application has proposed in the application on the same day to obtain the orientation value x.

D. Problems to be Solved by the Invention In the conventional fault point locating method, there is a problem that a locating error occurs when there is a load between the self-end and the fault point. Due to the occurrence of this error, it takes time and effort to find the failure point when observing the failure point.

E. Means and Actions for Solving Problems The present invention has been made in view of the above problems, and the orientation principle obtained from each phase voltage, current, zero-phase current, self impedance per unit length, and mutual impedance of a transmission line. The distance from the self-end to the failure point is obtained by an arithmetic expression in which the distance to the branch load, the load current and the load distribution ratio are corrected for each load.

By such an orientation method, the influence of the load current due to the presence of the branch load is removed from the detected current, and the orientation value in which the orientation error is reduced is obtained.

F. Embodiment FIG. 1 is a block diagram of an orientation apparatus showing an embodiment of the present invention. In the figure, the first circuit D ₁ of the orienting device D not only samples the phase voltages Va, Vb, Vc and the phase currents Ia, Ib, Ic, but also the phase currents Ia of the branch loads L _{1 to} Ln.
Each data of L ₁ , IbL ₁ , IcL _{1 to} IaLn, IbLn, IcLn is fetched from the transmission device TC. The third circuit D ₃ is for each phase per unit length of the transmission line, and line constants Raa, Xaa, Rbb, Xbb, Rc between the phases.
c, Xcc, Rab, Xab, Rbc, Xbc, Rca, Xca and load L ₁ ~
Data for distances _{1 to} n to Ln are generated. Here, Raa, Rbb, Rcc ... each phase resistance Xaa, Xbb, Xcc ... each phase reactance Rab, Rbc, Rca ... interphase resistance Xab, Xbc, Xca ... interphase reactance.

The fourth circuit D ₄ receives each phase current I from the _second circuit D _2.
a, Ib, Ic, each phase voltage Va, Vb, Vc, zero phase current
And 3Io (= Ia + Ib + Ic ) and the phase current IaL ₁ ~IcLn of each load _L 1 Ln, each phase of the self-impedance Zaa obtained from line constants from the third circuit _{D 3,} Zbb, Zcc, mutual impedance Zab, Zbc , Zca and loads L _{1 to} Ln
Calculation of the following formula according to the distance _{1 to} n The distance x to the fault point for the a-phase ground fault is obtained by The same formula is used for the b and c phases.

In addition, the fourth circuit D ₄ calculates the following equation for a short-circuit fault. Therefore, the distance x to the failure point for the bc phase short circuit failure is
Ask for. The ab phase and the ca phase are also obtained from the same formula.

With such orientation, the orientation value x with reduced orientation error is obtained even when the branch load is between the fault point.

The reason why the orientation error is reduced by the above equations (8) and (9) is explained in detail below.

As shown in FIG. ₂ , for a single-line power transmission line that has a terminal load L ₂ with the other end as a non-power supply end and is connected to a three-phase balanced power supply via a rear impedance, a branch load is created between the power supply line and the fault point. With L ₁ and ignoring the ground capacitance of the line, each phase voltage Va, Vb, Vc, current Ia, Ib,
Ic and distance x to accident point, distance K to branch load L ₁
x, the fault point The following basic circuit equations hold between the phase voltages Vfa, Vfb, and Vfc.

However The accident point impedance Zf in FIG. 2 is a pure resistance that does not change with time and has the form shown in FIG. 3 according to the type of accident. Based on the relational expressions established for each accident type and assumptions, the arithmetic logical expressions for one-phase ground fault and other failures are as follows.

(3.1) About the three-phase ground fault, the three-phase short circuit, the two-phase ground fault, and the two-phase short circuit, when using the b and c phase data, the relational expression that holds at the accident point is However By assumption, Rfb ＝ Rfc ＝ Rf ………………… (15) From circuit equations (11), (12) and (13), (14), Vb-Vc ＝ {(Zba-Zca) ・ Ia + (Zbb- Zcb) ・ Ib + (Zbc-Zcc) ・ Ic} ・ x-{(Zba-Zca) ・ I _aLl + (Zbb-Zcb) ・ I _bL1 + (Zbc-Zcc) ・ I _cL1 } ・ (1-K)・ X + Rf ・ {(Ib-Ic)-(I _bL -I _cL )} …………… (16) Comparing Eq. (16) with theoretical formula (4),-{in Eq. (16) (Zba-Zca) ・ I _aL1 + (Zbb-Zcb) ・ I _bL1 + (Zbc-Zcc) ・ I _cL1 } ・ (1-K) ・ x and − (I _bL −I _cL ). .

However, the fault point resistance Rfb, when Rfc is small, (4) Rf · (Ib-Ic) and (16) Rf · of _{{(Ib-Ic) - (} I bL -I cL)} is Has no meaning.

(3.2) Regarding 1-phase ground fault, in the case of a-phase ground fault, the relational expression established at the accident point is Vfa = Rf / 3Io ………… (17) Therefore, Va = {Zaa ・ Ia + Zab・ Ib + Zac ・ Ic} ・ x = {Zaa ・ I _aL1 + Zab ・ I _bL1 + Zac ・ I _cL1 } ・ (1-K) ・ x + Rf ・ 3Io
………… (18) Comparing Eq. (18) with theoretical formula (6),-{Zaa ・ I _aL1 + Zab ・ I _bL1 + Zac ・ I _cL1 } ・ (1-K) in Eq. (18)・ The term of x becomes an error.

As is apparent from the above description, when the orientation calculation including the error due to the branch load has n branch loads from the above equations (8) and (9), the error due to each branch load is calculated. Can be compensated. In this compensation procedure, when a failure occurs, the orientation value x is first obtained by the conventional equations (1), (4), and (6), and the orientation value x is compared with the self-end by comparing the distance between this x and the branch load point. It is determined whether or not there is a branch load between them, and when there is at least one branch load, compensation calculation is performed for these branch loads.

In the embodiment, the case where each load current is taken into the locating device is shown. However, if it is difficult to install the transmission device TC from the economical point of view, the current information at one end of the line and the distribution ratio to each load will The load current can be estimated according to the following equation (in the case of a-phase ground fault). In this case, the load distribution ratio can be determined by the equipment capacity ratio or the past load record, and is given from the third circuit D ₃ .

However, Pi: i-th load distribution ratio Ia '=-(Ib + Ic): total of fault phase load currents In the above equation (19), the total Ia' of fault phase load currents in the case of one-wire ground fault is given. It is limited to 1-wire ground fault because it can be obtained from sound phase b and c currents, and is applicable to failures involving short-circuit between phases (2-phase ground fault, 2-phase short circuit, 3-phase ground fault, 3-phase short circuit) Can not. However, since the magnitude of the branch load current is sufficiently smaller than the fault phase current in the case of a fault involving a short circuit, the error due to the branch load does not become as large as the one-phase ground fault.

In the above embodiments, the zero-phase current 3Io is the current transformer CT.
Detected from the residual circuit or the tertiary circuit of the second circuit D
_The calculation by ₂ can be omitted.

G. EFFECTS OF THE INVENTION As described above, according to the present invention, in order to obtain the orientation value from the orientation principle formula according to each phase current, voltage, zero-phase current and line constant of the power transmission system, from the self-end to the failure point. Since the orientation is determined based on the calculation that removes the influence of the branch load on the detected current, the orientation error due to the branch load can be reduced and accurate orientation can be performed. Further, along with this, there is an effect that the fault point search work by orientation is facilitated.

[Brief description of drawings]

FIG. 1 is an apparatus configuration diagram showing an embodiment of the present invention, and FIGS. 2 and 3 are basic circuit diagrams of a power transmission system (FIG. 2) for explaining the principle of the present invention and the fault types. An equivalent circuit diagram (FIG. 3) of the fault impedance Zf and FIG. 4 are conventional device configuration diagrams. D ...... locating system, PT ...... instrument transformer, CT ...... current transformer, TC ...... transmission _{device, L 1,} Ln ...... branch load.

 ─────────────────────────────────────────────────── ─── Continuation of the front page (72) Inventor Toshihisa Funabashi 2-1-1-17 Osaki, Shinagawa-ku, Tokyo Stock Company Inside the company Meidensha (72) Norio Suda 2-1-1-17 Osaki, Shinagawa-ku, Tokyo Stock Association Shameidensha (72) Inventor Tetsuya Mizutori 2-17 Osaki, Shinagawa-ku, Tokyo Stock company Shameidensha (56) References JP-A-60-164264 (JP, A) JP-A-60-204220 (JP) , A) JP 61-22264 (JP, A)

Claims (2)

[Claims] 1. A phase current Ia of each end a, b, c of a transmission line,
Ib, Ic, phase voltages Va, Vb, Vc, zero-phase current I
o, self-impedance Zaa, Zab, Zac, mutual impedance Zab, Zbc, Zca of each phase per unit length of the transmission line, distances ₁ to _n to a branch load connected to the transmission line, and load current _{IaL1 〜I aLn} , I _{bL1 〜I bLn} , I _{cL1 〜I cLn} From the following formula Real [A] ・ Imag [B] −Imag [A] ・ Real [B] = 0 However, the transmission line has a one-line ground fault. Then, for the a phase, A = Va− {Zaa · Ia + Zab · Ib + Zac · Ic} x B = 3I _{0 When the} transmission line has a short circuit fault, for the bc phase, A = (Vb-Vc)-{(Zba -Zca) Ia + (Zbb-Zcb) Ib + (Zbc-Zcc)} x According to B = Ib-Ic, the distance x from the self-end to the failure point is calculated, and the distance x and the distance ₁ are compared, and x> ₁ Then the following formula About bc phase when the transmission line has a short circuit fault A fault location method for a power transmission system, characterized in that the distance x from the self-end to the fault point is obtained in accordance with. 2. A phase current Ia of each end a, b, c of the transmission line,
Ib, Ic, phase voltages Va, Vb, Vc, zero-phase current I ₀ , self-impedance Zaa, Zab, Zac, mutual impedance Zab, Zbc, Zca for each phase per unit length of the transmission line, and the transmission line Based on the distances ₁ to _n to the branch load connected to and the distribution ratios P _{1 to} P _{n of the} respective branch loads, the following expression Real [A] · Imag [B] −Imag [A] · Real [B] = 0 , When the transmission line has a one-line ground fault, the distance x to the fault point is calculated according to A = Va− {Zaa · Ia + Zab · Ib + Zac · Ic} x B = 3I _{0 for the} phase a and the distance x distance
₁ is compared, and when x> _{1, the} following equation A fault location method for a power transmission system, characterized in that the distance x from the self-end to the fault point is obtained in accordance with.