CN105510770A - Power transmission line one-terminal fault location method based on faulty traveling wave distribution characters along the line within two successive time windows - Google Patents

Abstract

The invention relates to a power transmission line one-terminal fault location method based on faulty traveling wave distribution characters along the line within two successive time windows and belongs to the technical field of the relay protection for an electrical power system. During the fault of a power transmission line, the fault traveling wave data is obtained at the measurement end of the faulted power transmission line, and then the line-mode traveling wave is obtained. The line-mode traveling wave is applied to calculate the distribution of the voltage traveling wave and the distribution of the current traveling wave along the line based on a Bergeron line transfer equation. The voltage traveling wave and the current traveling wave along the line are decomposed along the line direction, so that the directional traveling wave distributed along the line can be obtained. After that, the forward traveling wave is multiplied by the backward traveling wave and then a fault location function is constructed through the integral method within an observation time window. Therefore, the fault location function is realized. According to the technical scheme of the invention, the method realizes the fault location function for the power transmission line, and is simple in principle, wherein the wave head of the fault traveling wave is free from being calibrated. Meanwhile, the influence of the fault instantaneity, the fault transition resistance variation and other factors is avoided. The fault location result of the method is accurate and reliable.

Description

A kind of based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method

Technical field

The present invention relates to a kind of based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method, belong to Relay Protection Technology in Power System field.

Background technology

Fault distance-finding method is divided into impedance method, fault analytical method and traveling wave method by principle.Travelling wave ranging measures fault distance at the travel-time utilizing fault traveling wave between bus and trouble spot, and its distance accuracy is higher, and the scope of application is wider.Current travelling wave ranging method is based on fault traveling wave temporal signatures mostly and observes row ripple on time shaft, to portray and wave head is demarcated, and the calculating of fault distance.Wherein, time domain Single Terminal Traveling Wave Fault Location need to be expert at wave-wave leader fixed and the reliability of wave head identification, the robotization aspect of range finding analysis makes further research; Time domain both-end travelling wave ranging is owing to utilizing the initial row wave-wave arrival time difference of faulty line both sides, its initial row ripple demarcates reliability and accuracy is easy to get to ensure, and do not need to carry out identification to trouble spot reflection wave, but both-end travelling wave ranging synchronously requires higher to circuit two ends cycle accurate.Therefore, be badly in need of proposing a kind of new fault distance-finding method, be not subject to the effective identification of fault traveling wave wave head and range finding cycle accurate synchronously on the impact of localization of fault accuracy.

Summary of the invention

The object of the invention is to overcome the effective identification of conventional Time-domain travelling wave ranging requirement fault traveling wave and the synchronous limitation of range finding cycle accurate, propose a kind of based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method.

Technical scheme of the present invention is: a kind of based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method, when transmission line of electricity breaks down, line line ripple is asked for the fault traveling wave data that faulty line measuring end obtains, the line line ripple that application obtains also calculates voltage traveling wave along the line and current traveling wave distribution according to Bei Jielong circuit equation of transfer, voltage traveling wave along the line and current traveling wave along the line are carried out direction along the line row Wave Decomposition, obtain the direction row ripple of distribution along the line, recycle its direct wave be multiplied by backward-travelling wave and carry out integration to construct range function and to realize fault localization in window in time observing.

Concrete steps are:

(1) during transmission line of electricity generation singlephase earth fault, under sampling rate 1MHz, voltage, the electric current that measuring end obtains is sampled, obtains phase current sampling value sequence i _m,a(k), i _m,b(k), i _m,c(k), phase voltage sampled value sequence u _m,a(k), u _m,b(k), u _m,c(k), wherein k represents sampled point, k=1,2, M represents measuring end.

(2) the discrete series i of wire finishing die electric current and line mode voltage is asked respectively according to formula (1) and formula (2) _m,s(k) and u _m,s(k):

i _M,s(k)＝i _M,a(k)-i _M,b(k)(1)

u _M,s(k)＝u _M,a(k)-u _M,b(k)(2)

In formula, i _m,a(k), i _m,b(k), i _m,ck three-phase current sampled value sequence that () gets for measuring end, u _m,a(k), u _m,b(k), u _m,ck three-phase voltage sampled value sequence that () gets for measuring end; i _m,sk () is line mould electric current discrete series; u _m,s(k) line mode voltage discrete series.

(3) calculating of distribution along the line: utilize the voltage's distribiuting along the line of formula (3) and formula (4) difference computing electric power line and distribution of current along the line.

$\begin{array}{c}{u}_{M,x,s}(x,k)=\frac{1}{2}{\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k+\frac{x}{v_s})-i_{M,s}(k+\frac{x}{v_s})(Z_{c,s}+\frac{r_{s}x}{4})\]\\ +\frac{1}{2}{\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k-\frac{x}{v_s})+i_{M,s}(k-\frac{x}{v_s})(Z_{c,s}-r_{s}x)\]\\ -{\left(\frac{r_{s}x/4}{Z_{c,s}}\right)}^2{u}_{M,s}\left(k\right)-\frac{r_{s}x}{4}\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)i_{M,s}\left(k\right)\end{array}---\left(3\right)$

$\begin{array}{c}{i}_{M,x,s}(x,k)=\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k+x/v_s)-i_{M,s}(k+x/v_s)\·(Z_{c,s}+r_{s}x/4)\]\\ -\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k-x/v_s)+i_{M,s}(k-x/v_s)\·(Z_{c,s}-r_{s}x/4)\]\\ -\frac{1}{2Z_{c,s}}\·\frac{r_{s}x}{2Z_{c,s}}\[u_{M,s}\left(k\right)-i_{M,s}\left(k\right)(r_{s}x/4)\]\end{array}---\left(4\right)$

In formula, s is Aerial mode component; X is the distance of any point along the line to measuring end; V is the wave velocity of circuit; K represents current time; Z _c,sfor the characteristic impedance of circuit; r _sfor circuit resistance per unit length; u _m,sk line mode voltage row ripple that () is measuring end; i _m,sk line mould current traveling wave that () is measuring end; u _{m, x, s}(x, k) is for the k moment is apart from the voltage at measuring end x place; i _{m, x, s}(x, k) is for the k moment is apart from the electric current at measuring end x place.

(4) direct wave and the backward-travelling wave of distribution along the line is calculated: according to formula (5) and formula (6) the capable ripple of forward voltage, the capable ripple of reverse voltage that distributes that distribute of computing electric power line respectively, namely along the line along the line

u ⁺ _M,x,s＝(u _M,x,s+Z _c,si _M,x,s)/2(5)

u ^- _M,x,s＝(u _M,x,s-Z _c,si _M,x,s)/2(6)

In formula, u ⁺ _{m, x, s}for the capable ripple of forward voltage that circuit distributes along the line; u ^- _{m, x, s}for the capable ripple of reverse voltage that circuit distributes along the line.

(5) the direct wave gradient of distribution along the line and the calculating of backward-travelling wave gradient: the forward voltage gradient utilizing the difference structure distribution along the line of adjacent two sampled values of the capable ripple of forward voltage of distribution along the line, namely

c ⁺ _M,dif—u(k)＝u ⁺ _k,x,s(k)-u ⁺ _k,x,s(k-1)(7)

Utilize the reverse voltage gradient of the difference structure distribution along the line of adjacent two sampled values of the capable ripple of reverse voltage of distribution along the line, namely

c ^- _M,dif—u(k)＝u ^- _k,x,s(k)-u ^- _k,x,s(k-1)(8)

In formula, c ⁺ _{m, dif-u}t forward voltage gradient that () distributes along the line for circuit; c ^- _{m, dif-u}t reverse voltage gradient that () distributes along the line for circuit.

(6) direct wave of calculating distribution along the line suddenlys change and backward-travelling wave sudden change: the capable ripple sudden change of forward voltage distributed along the line according to formula (9) extraction faulty line, namely

${S^{+}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{+}}_{M,dif\_u}\left(k\right)\]}^3---\left(9\right)$

According to the capable ripple sudden change of reverse voltage that formula (10) extraction faulty line distributes along the line, namely

${S^{-}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{-}}_{M,dif\_u}\left(k\right)\]}^3---\left(\mathrm{10}\right)$

In formula, R is taken as 3; S ⁺ _{k, 2u}the sudden change of the capable ripple of forward voltage that (x, t) distributes along the line for faulty line; S ^- _{k, 2u}the sudden change of the capable ripple of reverse voltage that (x, t) distributes along the line for faulty line.

(7) structure of range function: adopt formula (11) and formula (12), window [k when direct wave step (6) obtained sudden change is multiplied with backward-travelling wave sudden change and observes respectively at row ripple ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in carry out integration, obtain range function f _uI(x) and f _uIIthe row ripple sudden change along the line of (x).

$\begin{array}{cc}{f}_{uI}\left(x\right)={\∫}_{k_0}^{k_0+l/2v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[0,l/2\]\end{array}---\left(11\right)$

$\begin{array}{cc}{f}_{uII}\left(x\right)={\∫}_{k_0+l/2v}^{k_0+l/v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[l/2,l\]\end{array}---\left(12\right)$

In formula, k ₀represent the fault initial row ripple due in that measuring end M detects; L is faulty line length; f _uI(x) and f _uIIwindow [k when () is two observation x ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in range function.

(8) structure of localization of fault criterion:

[k is calculated according to step (7) ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] two in succession time window in, range function f _uI(x) and f _uIIx the distribution catastrophe point along the line of (), its respective distances is designated as [x respectively _i1, x _i2... ] and [x _iI1, x _iI2... ].

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be negative, x ^* _iIcatastrophe point polarity be just, then fault is positioned within half line length, and the distance of trouble spot distance measuring end is x ^* _i;

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be just, x ^* _iIcatastrophe point polarity be negative, then fault is positioned at outside half line length, and the distance of trouble spot distance measuring end is x ^* _iI.

x ^* _I+x ^* _II＝l(13)。

The invention has the beneficial effects as follows:

This method carries out localization of fault for transmission line of electricity, and its principle is simple, does not need to demarcate fault traveling wave wave-wave head, and not by the impact of the factor such as fault instantaneity, fault resistance change, range measurement accurately and reliably.

Accompanying drawing explanation

Fig. 1 is the transmission line structure figure of embodiment 1, embodiment 2;

Fig. 2 is the capable ripple of false voltage under fault within embodiment 1 half line length;

Fig. 3 is the capable ripple of fault current under fault within embodiment 1 half line length;

Fig. 4 is within half line length under fault, [k ₀, k ₀+ l/ (2v)] time window in range function f _uthe sudden change distribution results of (x);

Fig. 5 is within half line length under fault, [k ₀+ l/ (2v), k ₀+ l/v] time window in range function f _uthe sudden change distribution results of (x);

Fig. 6 is the capable ripple of false voltage under fault outside embodiment 2 half line length;

Fig. 7 is the capable ripple of fault current under fault outside embodiment 2 half line length;

Fig. 8 is outside half line length under fault, [k ₀, k ₀+ l/ (2v)] time window in range function f _uthe sudden change distribution results of (x);

Fig. 9 is outside half line length under fault, [k ₀+ l/ (2v), k ₀+ l/v] time window in range function f _uthe sudden change distribution results of (x).

Embodiment

Below in conjunction with the drawings and specific embodiments, the invention will be further described.

A kind of based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method, when transmission line of electricity breaks down, line line ripple is asked for the fault traveling wave data that faulty line measuring end obtains, the line line ripple that application obtains also calculates voltage traveling wave along the line and current traveling wave distribution according to Bei Jielong circuit equation of transfer, voltage traveling wave along the line and current traveling wave along the line are carried out direction along the line row Wave Decomposition, obtain the direction row ripple of distribution along the line, recycle its direct wave be multiplied by backward-travelling wave and carry out integration to construct range function and to realize fault localization in window in time observing.

Concrete steps are:

(1) during transmission line of electricity generation singlephase earth fault, under sampling rate 1MHz, voltage, the electric current that measuring end obtains is sampled, obtains phase current sampling value sequence i _m,a(k), i _m,b(k), i _m,c(k), phase voltage sampled value sequence u _m,a(k), u _m,b(k), u _m,c(k), wherein k represents sampled point, k=1,2, M represents measuring end.

(2) the discrete series i of wire finishing die electric current and line mode voltage is asked respectively according to formula (1) and formula (2) _m,s(k) and u _m,s(k):

i _M,s(k)＝i _M,a(k)-i _M,b(k)(1)

u _M,s(k)＝u _M,a(k)-u _M,b(k)(2)

In formula, i _m,a(k), i _m,b(k), i _m,ck three-phase current sampled value sequence that () gets for measuring end, u _m,a(k), u _m,b(k), u _m,ck three-phase voltage sampled value sequence that () gets for measuring end; i _m,sk () is line mould electric current discrete series; u _m,s(k) line mode voltage discrete series.

(3) calculating of distribution along the line: utilize the voltage's distribiuting along the line of formula (3) and formula (4) difference computing electric power line and distribution of current along the line.

$\begin{array}{c}{u}_{M,x,s}(x,k)=\frac{1}{2}{\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k+\frac{x}{v_s})-i_{M,s}(k+\frac{x}{v_s})(Z_{c,s}+\frac{r_{s}x}{4})\]\\ +\frac{1}{2}{\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k-\frac{x}{v_s})+i_{M,s}(k-\frac{x}{v_s})(Z_{c,s}-r_{s}x)\]\\ -{\left(\frac{r_{s}x/4}{Z_{c,s}}\right)}^2{u}_{M,s}\left(k\right)-\frac{r_{s}x}{4}\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)i_{M,s}\left(k\right)\end{array}---\left(3\right)$

$\begin{array}{c}{i}_{M,x,s}(x,k)=\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k+x/v_s)-i_{M,s}(k+x/v_s)\·(Z_{c,s}+r_{s}x/4)\]\\ -\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k-x/v_s)+i_{M,s}(k-x/v_s)\·(Z_{c,s}-r_{s}x/4)\]\\ -\frac{1}{2Z_{c,s}}\·\frac{r_{s}x}{2Z_{c,s}}\[u_{M,s}\left(k\right)-i_{M,s}\left(k\right)(r_{s}x/4)\]\end{array}---\left(4\right)$

In formula, s is Aerial mode component; X is the distance of any point along the line to measuring end; V is the wave velocity of circuit; K represents current time; Z _c,sfor the characteristic impedance of circuit; r _sfor circuit resistance per unit length; u _m,sk line mode voltage row ripple that () is measuring end; i _m,sk line mould current traveling wave that () is measuring end; u _{m, x, s}(x, k) is for the k moment is apart from the voltage at measuring end x place; i _{m, x, s}(x, k) is for the k moment is apart from the electric current at measuring end x place.

(4) direct wave and the backward-travelling wave of distribution along the line is calculated: according to formula (5) and formula (6) the capable ripple of forward voltage, the capable ripple of reverse voltage that distributes that distribute of computing electric power line respectively, namely along the line along the line

u ⁺ _M,x,s＝(u _M,x,s+Z _c,si _M,x,s)/2(5)

u ^- _M,x,s＝(u _M,x,s-Z _c,si _M,x,s)/2(6)

In formula, u ⁺ _{m, x, s}for the capable ripple of forward voltage that circuit distributes along the line; u ^- _{m, x, s}for the capable ripple of reverse voltage that circuit distributes along the line.

(5) the direct wave gradient of distribution along the line and the calculating of backward-travelling wave gradient: the forward voltage gradient utilizing the difference structure distribution along the line of adjacent two sampled values of the capable ripple of forward voltage of distribution along the line, namely

c ⁺ _M,dif—u(k)＝u ⁺ _k,x,s(k)-u ⁺ _k,x,s(k-1)(7)

Utilize the reverse voltage gradient of the difference structure distribution along the line of adjacent two sampled values of the capable ripple of reverse voltage of distribution along the line, namely

c ^- _M,dif—u(k)＝u ^- _k,x,s(k)-u ^- _k,x,s(k-1)(8)

In formula, c ⁺ _{m, dif-u}t forward voltage gradient that () distributes along the line for circuit; c ^- _{m, dif-u}t reverse voltage gradient that () distributes along the line for circuit.

(6) direct wave of calculating distribution along the line suddenlys change and backward-travelling wave sudden change: the capable ripple sudden change of forward voltage distributed along the line according to formula (9) extraction faulty line, namely

${S^{+}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{+}}_{M,dif\_u}\left(k\right)\]}^3---\left(9\right)$

According to the capable ripple sudden change of reverse voltage that formula (10) extraction faulty line distributes along the line, namely

${S^{-}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{-}}_{M,dif\_u}\left(k\right)\]}^3---\left(10\right)$

In formula, R is taken as 3; S ⁺ _{k, 2u}the sudden change of the capable ripple of forward voltage that (x, t) distributes along the line for faulty line; S ^- _{k, 2u}the sudden change of the capable ripple of reverse voltage that (x, t) distributes along the line for faulty line.

(7) structure of range function: adopt formula (11) and formula (12), window [k when direct wave step (6) obtained sudden change is multiplied with backward-travelling wave sudden change and observes respectively at row ripple ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in carry out integration, obtain range function f _uI(x) and f _uIIthe row ripple sudden change along the line of (x).

$\begin{array}{cc}{f}_{uI}\left(x\right)={\∫}_{k_0}^{k_0+l/2v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[0,l/2\]\end{array}---\left(11\right)$

$\begin{array}{cc}{f}_{uII}\left(x\right)={\∫}_{k_0+l/2v}^{k_0+l/v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[l/2,l\]\end{array}---\left(12\right)$

In formula, k ₀represent the fault initial row ripple due in that measuring end M detects; L is faulty line length; f _uI(x) and f _uIIwindow [k when () is two observation x ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in range function.

(8) structure of localization of fault criterion:

[k is calculated according to step (7) ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] two in succession time window in, range function f _uI(x) and f _uIIx the distribution catastrophe point along the line of (), its respective distances is designated as [x respectively _i1, x _i2... ] and [x _iI1, x _iI2... ].

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be negative, x ^* _iIcatastrophe point polarity be just, then fault is positioned within half line length, and the distance of trouble spot distance measuring end is x ^* _i;

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be just, x ^* _iIcatastrophe point polarity be negative, then fault is positioned at outside half line length, and the distance of trouble spot distance measuring end is x ^* _iI.

x ^* _I+x ^* _II＝l(13)。

Embodiment 1:

Adopt transmission line of electricity as shown in Figure 1, M end is " two enter one goes out " bus connection type, perfects circuit l _mk1=50km, perfects the line length l=150km that line end is III class bus connection type, faulty line MN.Now suppose that within circuit MN half line length, distance M holds 70km place that A phase earth fault occurs, the voltage traveling wave that measuring end obtains and current traveling wave are as shown in Figures 2 and 3.Material calculation along the line is 0.1km, respectively at two in succession time window [k ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] calculate the range function f of measuring end M _uI(x) and f _uIIx the distribution results in the long scope in all fronts, () edge as shown in Figure 4 and Figure 5.As shown in Figure 4, [k ₀, k ₀+ l/ (2v)] time window in, f _uIcatastrophe point A (the x)=69.8km of (x), and polarity is negative; As shown in Figure 5, [k ₀+ l/ (2v), k ₀+ l/v] time window in, f _uIIcatastrophe point B (the x)=79.6km of (x), and polarity is just.Because A (x)+B (x)=69.8+79.6=149.4km ≈ 150km, meets the line length constraint condition shown in formula (13), fault is positioned within half line length, and the distance of trouble spot distance measuring end M is 69.8km.

Embodiment 2:

Adopt transmission line of electricity as shown in Figure 1, M end is " two enter one goes out " bus connection type, perfects circuit l _mk1=50km, perfects the line length l=150km that line end is III class bus connection type, faulty line MN.Now suppose that outside circuit MN half line length, distance M holds 80km place that A phase earth fault occurs, the voltage traveling wave that measuring end obtains and current traveling wave are as shown in Figure 6 and Figure 7.Material calculation along the line is 0.1km, respectively at two in succession time window [k ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] calculate the range function f of measuring end M _uI(x) and f _uIIx the distribution results in the long scope in all fronts, () edge as shown in Figure 8 and Figure 9.As shown in Figure 8, [k ₀, k ₀+ l/ (2v)] time window in, f _uIcatastrophe point B (the x)=69.7km of (x), and polarity is just; As shown in Figure 9, [k ₀+ l/ (2v), k ₀+ l/v] time window in, f _uIIcatastrophe point A (the x)=79.8km of (x), and polarity is negative.Because A (x)+B (x)=69.7+79.8=149.5km ≈ 150km, meets the line length constraint condition shown in formula (13), fault is positioned at outside half line length, and the distance of trouble spot distance measuring end M is 79.8km.

Claims (2)

1. one kind based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method, it is characterized in that: when transmission line of electricity breaks down, line line ripple is asked for the fault traveling wave data that faulty line measuring end obtains, the line line ripple that application obtains also calculates voltage traveling wave along the line and current traveling wave distribution according to Bei Jielong circuit equation of transfer, voltage traveling wave along the line and current traveling wave along the line are carried out direction along the line row Wave Decomposition, obtain the direction row ripple of distribution along the line, recycle its direct wave be multiplied by backward-travelling wave and carry out integration to construct range function and to realize fault localization in window in time observing.

2. according to according to claim 1 based on two in succession time the capable ripple of window internal fault distribution character along the line transmission line of electricity one-end fault ranging method, it is characterized in that concrete steps are as follows:

(1) during transmission line of electricity generation singlephase earth fault, under sampling rate 1MHz, voltage, the electric current that measuring end obtains is sampled, obtains phase current sampling value sequence i _m,a(k), i _m,b(k), i _m,c(k), phase voltage sampled value sequence u _m,a(k), u _m,b(k), u _m,c(k), wherein k represents sampled point, k=1,2, M represents measuring end;

(2) the discrete series i of wire finishing die electric current and line mode voltage is asked respectively according to formula (1) and formula (2) _m,s(k) and u _m,s(k):

i _M,s(k)＝i _M,a(k)-i _M,b(k)(1)

u _M,s(k)＝u _M,a(k)-u _M,b(k)(2)

In formula, i _m,a(k), i _m,b(k), i _m,ck three-phase current sampled value sequence that () gets for measuring end, u _m,a(k), u _m,b(k), u _m,ck three-phase voltage sampled value sequence that () gets for measuring end; i _m,sk () is line mould electric current discrete series; u _m,s(k) line mode voltage discrete series;

(3) calculating of distribution along the line: utilize the voltage's distribiuting along the line of formula (3) and formula (4) difference computing electric power line and distribution of current along the line:

$\begin{array}{c}{u}_{M,x,s}(x,k)=\frac{1}{2}{\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k+\frac{x}{v_s})-i_{M,s}(k+\frac{x}{v_s})(Z_{c,s}+\frac{r_{s}x}{4}\mathrm{)\]}\\ +\frac{1}{2}{\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)}^2\[u_{M,s}(k-\frac{x}{v_s})+i_{M,s}(k-\frac{x}{v_s})(Z_{c,s}-r_{s}x\mathrm{)\]}\\ -{\left(\frac{r_{s}x/4}{Z_{c,s}}\right)}^2{u}_{M,s}\left(k\right)-\frac{r_{s}x}{4}\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)i_{M,s}\left(k\right)\end{array}---\left(3\right)$

$\begin{array}{c}{i}_{M,x,s}(x,k)=\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}+r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k+x/v_s)-i_{M,s}(k+x/v_s)\·(Z_{c,s}+r_{s}x/4\mathrm{)\]}\\ -\frac{1}{2Z_{c,s}}\left(\frac{Z_{c,s}-r_{s}x/4}{Z_{c,s}}\right)\[u_{M,s}(k-x/v_s)+i_{M,s}(k-x/v_s)\·(Z_{c,s}-r_{s}x/4\mathrm{)\]}\end{array}$

$-\frac{1}{2Z_{c,s}}\·\frac{r_{s}x}{2Z_{c,s}}\[u_{M,s}\left(k\right)-i_{M,s}\left(k\right)(r_{s}x/4)\]---\left(4\right)$

In formula, s is Aerial mode component; X is the distance of any point along the line to measuring end; V is the wave velocity of circuit; K represents current time; Z _c,sfor the characteristic impedance of circuit; r _sfor circuit resistance per unit length; u _m,sk line mode voltage row ripple that () is measuring end; i _m,sk line mould current traveling wave that () is measuring end; u _{m, x, s}(x, k) is for the k moment is apart from the voltage at measuring end x place; i _{m, x, s}(x, k) is for the k moment is apart from the electric current at measuring end x place;

(4) direct wave and the backward-travelling wave of distribution along the line is calculated: according to formula (5) and formula (6) the capable ripple of forward voltage, the capable ripple of reverse voltage that distributes that distribute of computing electric power line respectively, that is: along the line along the line

u ⁺ _M,x,s＝(u _M,x,s+Z _c,si _M,x,s)/2(5)

u ^- _M,x,s＝(u _M,x,s-Z _c,si _M,x,s)/2(6)

In formula, u ⁺ _{m, x, s}for the capable ripple of forward voltage that circuit distributes along the line; u ^- _{m, x, s}for the capable ripple of reverse voltage that circuit distributes along the line;

(5) the direct wave gradient of distribution along the line and the calculating of backward-travelling wave gradient: the forward voltage gradient utilizing the difference structure distribution along the line of adjacent two sampled values of the capable ripple of forward voltage of distribution along the line, that is:

c ⁺ _{M,dif_u}(k)＝u ⁺ _k,x,s(k)-u ⁺ _k,x,s(k-1)(7)

Utilize the reverse voltage gradient of the difference structure distribution along the line of adjacent two sampled values of the capable ripple of reverse voltage of distribution along the line, that is:

c ^- _{M,dif_u}(k)＝u ^- _k,x,s(k)-u ^- _k,x,s(k-1)(8)

In formula, c ⁺ _{m, dif_u}t forward voltage gradient that () distributes along the line for circuit; c ^- _{m, dif_u}t reverse voltage gradient that () distributes along the line for circuit;

(6) direct wave of calculating distribution along the line suddenlys change and backward-travelling wave sudden change: the capable ripple sudden change of forward voltage distributed along the line according to formula (9) extraction faulty line, that is:

${S^{+}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{+}}_{M,dif\_u}\left(k\right)\]}^3---\left(9\right)$

According to the capable ripple sudden change of reverse voltage that formula (10) extraction faulty line distributes along the line, that is:

${S^{-}}_{M,2u}(x,k)=\underset{n=k-R+1}{\overset{k}{\Σ}}{\[{c^{-}}_{M,dif\_u}\left(k\right)\]}^3---\left(10\right)$

In formula, R is taken as 3; S ⁺ _{k, 2u}the sudden change of the capable ripple of forward voltage that (x, t) distributes along the line for faulty line; S ^- _{k, 2u}the sudden change of the capable ripple of reverse voltage that (x, t) distributes along the line for faulty line;

(7) structure of range function: adopt formula (11) and formula (12), window [k when direct wave step (6) obtained sudden change is multiplied with backward-travelling wave sudden change and observes respectively at row ripple ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in carry out integration, obtain range function f _uI(x) and f _uIIthe row ripple sudden change along the line of (x);

$\begin{array}{cc}{f}_{uI}\left(x\right)={\∫}_{k_0}^{k_0+l/2v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[0,l/2\]\end{array}---\left(11\right)$

$\begin{array}{cc}{f}_{uII}\left(x\right)={\∫}_{k_0+l/2v}^{k_0+l/v}{S}_{M,2u}^{+}(x,k)\×S_{M,2u}^{-}(x,k)dk& x\∈\[l/2,l\]\end{array}---\left(12\right)$

In formula, k ₀represent the fault initial row ripple due in that measuring end M detects; L is faulty line length; f _uI(x) and f _uIIwindow [k when () is two observation x ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] in range function;

(8) structure of localization of fault criterion:

[k is calculated according to step (7) ₀, k ₀+ l/ (2v)] and [k ₀+ l/ (2v), k ₀+ l/v] two in succession time window in, range function f _uI(x) and f _uIIx the distribution catastrophe point along the line of (), its respective distances is designated as [x respectively _i1, x _i2... ] and [x _iI1, x _iI2... ];

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be negative, x ^* _iIcatastrophe point polarity be just, then fault is positioned within half line length, and the distance of trouble spot distance measuring end is x ^* _i;

If [x _i1, x _i2... ] in mutation distance x ^* _i[x _iI1, x _iI2... ] in mutation distance x ^* _iImeet the line length constraint condition shown in formula (13), and x ^* _icatastrophe point polarity be just, x ^* _iIcatastrophe point polarity be negative, then fault is positioned at outside half line length, and the distance of trouble spot distance measuring end is x ^* _iI;

x ^* _I+x ^* _II＝l(13)。