CN104793106A - Distribution network line fault section positioning method based on current break rate - Google Patents

Abstract

The invention discloses a distribution network line fault section positioning method based on current break rate. The distribution network line fault section positioning method includes analyzing phase current characteristics of fault phase current before faults and before an arc suppression coil acts for ground faults, extracting fault characteristics from the phase current characteristics, and positioning by adopting full-process waveform current break rate. The distribution network line fault section positioning method only needs to measure phase current of line faults, is simple and high in applicability, and the problems of fault current weakness, poor reliability and sensitivity when a small-current grounding system has single-phase grounding faults commonly can be well solved. Meanwhile, interference cannot be introduced.

Description

Based on the distribution line fault section location method of current break rate

Technical field

The present invention relates to distribution network line fault diagnostic method, specifically a kind of distribution line fault section location method based on current break rate.

Background technology

According to statistics, electric system is in operational process, and the power outage caused by Distribution Network Failure accounts for more than 95% of total power outage, and wherein the accident of 70% is caused by singlephase earth fault or bus-bar fault.And distribution neutral point extensively adopts non-effective earthing (small current neutral grounding) mode both at home and abroad, to avoid causing power failure when there is singlephase earth fault.For the singlephase earth fault of distribution, because fault characteristic value is faint, lack reliable fault line selection and fault locating method always.Along with people are to the raising of distribution automation level requirement, more in the urgent need to fundamentally solving the fault-location problem of distribution.

The Fault Locating Method that current Chinese scholars proposes is roughly divided into two classes: one is injecting signal, and two is locate based on the section of fault characteristic value.Injecting signal comprises " S " injection method, the comprehensive injection method of alternating current-direct current and parallel medium resistance method, these class methods increase the interference to system, and can not detect instantaneity and intermittent grounding fault.Section location based on fault characteristic value comprises zero mould electric current relative method, section zero sequence admittance method, zero sequence reactive power direction method, the location based on difference of phase currents, residual method of addition, traveling wave method etc., electrical power distribution automatization system mainly utilizes main website to realize the time synchronized of FTU, is at least several millisecond to time error.In the case, the method such as to compare no longer valid for the amplitude of transient signal, polarity, Waveform Correlation.

Observe known, the data after current method all only utilizes fault to occur, and have ignored the utilization to information before fault.Meanwhile, most of localization method only considers zero-sequence current information (needing three phase information), high to transformer request, and acquisition of information is complicated, and these methods will lose efficacy when data phase shortage.

Summary of the invention

The object of the invention is to overcome above-mentioned the deficiencies in the prior art, propose a kind of distribution line fault section location method based on current break rate.

Principle of the present invention:

If fault phase is A phase, before fault, check point phase voltage is after fault, phase voltage is by Fig. 1, easily know that the two meets

${{\stackrel{\·}{U}}^{\′}}_A={\stackrel{\·}{U}}_A-{\stackrel{\·}{U}}_0$

1) adjacent check point 1,2 is positioned at trouble spot heteropleural (without branched line between check point), if check point 1 is positioned at upstream, trouble spot, check point 2 is positioned at downstream, trouble spot.

Before fault, the phase current of check point 1 is

${\stackrel{\·}{I}}_{1A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{\stackrel{\·}{I}}_{C1A}$

In formula, for load current, for A phase is from check point 1 to the capacitance current of this line end, can be expressed as

${\stackrel{\·}{I}}_{C1A}=\mathrm{j\ω}{C}_{1A}{\stackrel{\·}{U}}_A$

In formula, C _1Afor A phase is from check point 1 to the electric capacity of this line end.

Before and after assumed fault occurs, a cycle internal loading electric current is constant, and after fault, the faulted phase current of check point 1 is

${{\stackrel{\·}{I}}^{\′}}_{1A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{{\stackrel{\·}{I}}^{\′}}_{C1A}+{\stackrel{\·}{I}}_f$

In formula, be respectively A phase capacitance current and fault current.Wherein

${{\stackrel{\·}{I}}^{\′}}_{C1A}=\mathrm{j\ω}{C}_{1A}{{\stackrel{\·}{U}}^{\′}}_A=\mathrm{j\ω}{C}_{1A}({\stackrel{\·}{U}}_A-{\stackrel{\·}{U}}_0)$

The inductive current that when arc suppression coil utilizes fault, neutral point voltage skew produces carrys out bucking-out system capacitance current, and earth point residual flow (fault current) can be expressed as

${\stackrel{\·}{I}}_f={\stackrel{\·}{I}}_L-{\stackrel{\·}{I}}_{\mathrm{C\Σ}}=\frac{{\stackrel{\·}{U}}_0}{Z_L}-\mathrm{j\ω}{C}_{\mathrm{\Σ}}{\stackrel{\·}{U}}_0$

In formula, C _Σfor the whole network line mutual-ground capacitor.Fault occurs and in the arc suppression coil not operating period, preset arc suppression coil resistance is very large, and post-set arc suppression coils reactance is very large, generally speaking | and Z _l| very large, in formula, Section 2 can be ignored.Then have

${\stackrel{\·}{I}}_f\≈-\mathrm{j\ω}{c}_{\mathrm{\Σ}}{\stackrel{\·}{U}}_0$

In like manner, the phase current of check point 2 before fault

${\stackrel{\·}{I}}_{2A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{\stackrel{\·}{I}}_{C2A}={\stackrel{\·}{I}}_{\mathrm{LA}}+\mathrm{j\ω}{C}_{2A}{\stackrel{\·}{U}}_A$

Check point 2 faulted phase current after fault

${{\stackrel{\·}{I}}^{\′}}_{2A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{{\stackrel{\·}{I}}^{\′}}_{C2A}={\stackrel{\·}{I}}_{\mathrm{LA}}+\mathrm{j\ω}{C}_{2A}{{\stackrel{\·}{U}}^{\′}}_A$

According to above analysis, before fault, the difference of adjacent check point phase current

${\stackrel{\·}{I}}_{1A}-{\stackrel{\·}{I}}_{2A}={\stackrel{\·}{I}}_{C1A}-{\stackrel{\·}{I}}_{C2A}=\mathrm{j\ω}{C}_{12A}{\stackrel{\·}{U}}_A---\left(1\right)$

In formula, C _12Afor the relatively electric capacity between check point 1,2, its numerical value is very little.

After fault, the difference of adjacent check point faulted phase current

$\begin{array}{c}{{\stackrel{\·}{I}}^{\′}}_{1A}-{{\stackrel{\·}{I}}^{\′}}_{2A}={{\stackrel{\·}{I}}^{\′}}_{C1A}+I_f-{{\stackrel{\·}{I}}^{\′}}_{C2A}=\mathrm{j\ω}{C}_{12A}({\stackrel{\·}{U}}_A-{\stackrel{\·}{U}}_0)-\mathrm{j\ω}{\text{C}}_{\text{\Σ}}{\stackrel{\·}{U}}_0\\ =\mathrm{j\ω}{C}_{12A}{\stackrel{\·}{U}}_A-\mathrm{j\ω}(C_{12A}+C_{\mathrm{\Σ}}){\stackrel{\·}{U}}_0\end{array}---\left(2\right)$

In formula, C _12A+ C _Σ> > C _12A, size relevant with fault resistance size, be generally greater than 20% of phase voltage.Therefore Section 2 amplitude is much larger than Section 1, and amplitude and the direction of the difference of faulted phase current are determined by Section 2.

By comparing (1), (2) two formulas, after fault occurs, the difference of the faulted phase current of fault section two ends check point is before fault generation, and amplitude enlarges markedly.

2) adjacent check point 1,2 is positioned at trouble spot homonymy.

Similar to 1) in analysis, suppose that check point is all positioned at upstream, trouble spot, have:

Before fault

${\stackrel{\·}{I}}_{1A}-{\stackrel{\·}{I}}_{2A}={\stackrel{\·}{I}}_{C1A}-{\stackrel{\·}{I}}_{C2A}=\mathrm{j\ω}{C}_{12A}{\stackrel{\·}{U}}_A---\left(3\right)$

After fault

${{\stackrel{\·}{I}}^{\′}}_{1A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{{\stackrel{\·}{I}}^{\′}}_{C1A}+{\stackrel{\·}{I}}_f$

${{\stackrel{\·}{I}}^{\′}}_{2A}={\stackrel{\·}{I}}_{\mathrm{LA}}+{{\stackrel{\·}{I}}^{\′}}_{C2A}+{\stackrel{\·}{I}}_f$

${{\stackrel{\·}{I}}^{\′}}_{1A}-{{\stackrel{\·}{I}}^{\′}}_{2A}={{\stackrel{\·}{I}}^{\′}}_{C1A}-{{\stackrel{\·}{I}}^{\′}}_{C2A}=\mathrm{j\ω}{C}_{12A}{{\stackrel{\·}{U}}^{\′}}_A---\left(4\right)$

After earth fault generation, faulted phase voltage reduces.Relatively (3), (4) two formulas are easily known, after fault occurs, the difference of the faulted phase current of non-faulting section two ends check point is before fault generation, and amplitude reduces.

If check point is all positioned at upstream, trouble spot, after having fault to occur equally, the spread value of the faulted phase current of non-faulting section two ends check point reduces.

Comprehensive 1), 2) analysis, can judge whether section exists fault according to the variation characteristic of the difference of section two ends faulted phase current.Wherein, the measurement of amplitude size provides by the effective value calculated in power frequency one cycle.

If i _1A(n), i _2An phase current sampling sequence that () is adjacent check point.Be designated as zero under making fault that the data point in moment occurs, define adjacent check point phase differential current:

Δi _A(n)＝i _1A(n)-i _2A(n)n∈[-N,N-1]

In formula, N=0.02f _s, be a periodic sampling point.When n gets negative integer, before representing that fault occurs; N get positive integer represent fault occur after.Before fault occurs, difference between current effective value is

$\mathrm{\Δ}{I}_{A,\mathrm{before}}=\sqrt{\frac{1}{N}\underset{n=-N}{\overset{-1}{\mathrm{\Σ}}}{\left[\mathrm{\Δ}{i}_A\left(n\right)\right]}^2}$

$\mathrm{\Δ}{I}_{A,\mathrm{after}}=\sqrt{\frac{1}{N}\underset{n=0}{\overset{N-1}{\mathrm{\Σ}}}{\left[\mathrm{\Δ}{i}_A\left(n\right)\right]}^2}$

Definition section current break rate

A _I＝ΔI _A,after/ΔI _A,before

Analyze according to upper joint, easily know the A of non-faulting section _ibe less than 1, the A of fault section _i.be greater than 1.

Technical solution of the present invention is as follows:

A kind of distribution network line fault Section Location, its feature is, said method comprising the steps of:

Step S1, determines fault phase and fault moment t _f:

When after generation earth fault being detected, according to phase voltage Changing Pattern, select fault phase, determine fault moment t according to phase voltage sudden change moment or arc-extinction device chugging moment _f;

Step S2, chooses pick-up unit [t _f-0.02s, t _f+ 0.02s] interval interior 2N the faulted phase current Wave data put (before and after fault generation a cycle, i.e. positive and negative 0.02s, totally 2 power frequency periods, i.e. 2N data point) altogether, calculating current mutation rate A _i, formula is as follows:

$A_I=\sqrt{\frac{1}{N}\underset{n=0}{\overset{N-1}{\mathrm{\Σ}}}{\left[\mathrm{\Δ}{i}_A\left(n\right)\right]}^2}/\sqrt{\frac{1}{N}\underset{n=-N}{\overset{-1}{\mathrm{\Σ}}}{\left[\mathrm{\Δ}{i}_A\left(n\right)\right]}^2}$

Δi _A(n)＝i _1A(n)-i _2A(n) n∈[-N,N-1]

In formula, i _1A(n), i _2An phase current sampling sequence that () is adjacent check point, it is that a periodic sampling is counted that fault is designated as zero, N under there is the data point in moment;

According to current break rate size, step S3, judges whether each section is fault section, namely works as A _i> A _iset, be judged as fault section, otherwise be then non-faulting section;

A _isetfor the current break rate threshold value artificially arranged, be the positive number of about 1, consider certain nargin, desirable 1.2 ~ 1.5;

Step 4, according to check point distributing position on the line, to travel through successively, until find out fault section.

Compared with prior art, the invention has the beneficial effects as follows: to earth fault, by analysis of failure phase current before fault occurs and the phase current feature of fault generation to arc suppression coil action, therefrom extract fault characteristic value, and adopt the waveform-shaped current mutation rate of overall process to position.Therefore, only need the faulted phase current of measuring circuit, only consider the tradition of zero sequence (needing three phase information) before having broken, data acquisition is simple, applicability is strong; Synchronous from the angle of signal, total system adopts through the synchronous Monitoring Data of GPS, makes the difference between different check point sensitiveer.The problem that fault current faint, poor reliability when therefore can solve current ubiquitous single-phase grounded malfunction in grounded system of low current very well, sensitivity are low, can not introduce interference to system simultaneously.

Accompanying drawing explanation

Fig. 1 is single-phase earthing of small current earthing system schematic diagram

Fig. 2 is distributed fault section positioning system structure figure

Fig. 3 is 10kV system emulation figure

Fig. 4 is earth fault current change quantity waveform along the line

Embodiment

Fault waveform required for the present invention comes from distributed fault section positioning system, and system architecture as shown in Figure 2.Distribution network failure section positioning system is by master station, transformer station's (bus) measurement mechanism and be distributed in distribution line node failure locating device everywhere and form.Circuit is divided into some sections by localization of fault node in topology, each node installation three groups of measurement mechanisms, and real-time synchronization gathers circuit three-phase current and voltage.

According to the Fault Locating Method of invention, in 10kV power distribution network analogue system, dissimilar fault is set.System construction drawing as shown in Figure 3, be 1., 2., 3. sector number, 2. go up at section by fault verification.Sample frequency is 20kHz (each cycle data point N=400), and the moment occurs fault is 0.7s, and during earth fault, arc-extinction device is set to 0.04s actuation time, and the threshold value of current break rate is set to 1.2.

Embodiment citing for fault section judges:

Step S1, system is according to after residual voltage start detection to the generation of earth fault, and record A phase in bus three-phase voltage and reduce, B, C phase raises, and is defined as A phase fault; Determine that fault moment is 0.71s according to the phase voltage sudden change moment;

Step S2, chooses the faulted phase current Wave data (being designated as zero under establishing fault that the data point in moment occurs) of totally 800 points in each pick-up unit [0,69,0.73] interval, according to the definition of current break rate, calculates A _i, result is as shown in table 1, and table 1 is singlephase earth fault simulation result;

Wherein, during 500 Ω stake resistance, fault phase current change quantity waveform along the line is as shown in Figure 4.

Step S3, section current break rate 2. meets A _i>1.2, is judged as its fault section; Otherwise 1., 3. section is non-faulting section.

Table 1 singlephase earth fault simulation result

Claims (3)

1., based on a distribution network line fault Section Location for current break rate, it is characterized in that, said method comprising the steps of:

Step S1, determines fault phase and fault moment t _f;

Step S2, chooses pick-up unit [t _f-0.02s, t _f+ 0.02s] interval interior 2N the faulted phase current Wave data put altogether, calculating current mutation rate A _i, formula is as follows:

$A_I=\sqrt{\frac{1}{N}\underset{n=0}{\overset{N-1}{\mathrm{\Σ}}}[\mathrm{\Δ}{i}_A\left(n\right){]}^2}/\sqrt{\frac{1}{N}\underset{n=-N}{\overset{-1}{\mathrm{\Σ}}}[\mathrm{\Δ}{i}_A\left(n\right){]}^2}$

Δi _A(n)＝i _1A(n)-i _2A(n) n∈[-N,N-1]

In formula, i _1A(n), i _2An phase current sampling sequence that () is adjacent check point, it is that a periodic sampling is counted that fault is designated as zero, N under there is the data point in moment;

According to current break rate size, step S3, judges whether each section is fault section, namely works as A _i> A _{i, set}, be judged as fault section, otherwise be then non-faulting section;

Step 4, according to check point distributing position on the line, to travel through successively, until find out fault section.

2. distribution network line fault localization method according to claim 1, is characterized in that, described step S1 comprises:

After systems axiol-ogy to the generation of earth fault, according to phase voltage Changing Pattern, select fault phase, determine fault moment t according to phase voltage sudden change moment or arc-extinction device chugging moment _f.

3. distribution network line fault localization method according to claim 1, is characterized in that, A in described step S3 _isetfor the current break rate threshold value artificially arranged, get 1.2 ~ 1.5.