CN103308823A - Method for locating single-phase disconnection non-ground fault of power distribution network - Google Patents

Abstract

The invention relates to a method for locating a single-phase disconnection non-ground fault of a power distribution network. After the single-phase disconnection non-ground fault occurs, the positive sequence voltage of the high-voltage side of each distribution transformer after a fault point is far lower than the positive sequence voltage of the high-voltage side of each distribution transformer before the fault point and the positive sequence voltage of the high-voltage side of each distribution transformer of a non-faulty line, and the positive sequence voltage of the low-voltage side of each distribution transformer after the fault point also correspondingly becomes about 50 percent of the positive sequence voltage before the fault point. For three-phase voltage information acquired by different types of measuring devices installed at load points, the positive-sequence component of the voltage of each load point is calculated in real time; and if the positive sequence voltage of a certain load point and the positive sequence voltages of all downstream load points of the load point are obviously reduced at one point, and the positive sequence voltages of upstream load points of the load point are constant, the single-phase disconnection non-ground fault can be determined in an upstream adjacent line connected with the load point. According to the method for locating the single-phase disconnection non-ground fault of the power distribution network disclosed by the invention, new equipment does not need to be equipped; the calculation is simple; and the fault can be located between two load points to realize the diagnosis and the location of the single-phase disconnection non-ground fault.

Description

A kind of power distribution network single-phase wire break phase to phase fault localization method

Technical field

The present invention relates to the Operation Technique of Electric Systems field, be specifically related to a kind of power distribution network single-phase wire break phase to phase fault localization method.

Background technology

The direct user oriented of power distribution network, broad covered area, running environment is different.Affected by operational outfit defective, human operational error, outside destroy and meteorological disaster etc., especially disconnection fault easily occurs in overhead transmission line.Wherein, the complex grounding fault with interruption can be differentiated according to earth fault transient state component and harmonic characteristic etc., and adopts impedance method or traveling wave method etc. to position.For phase to phase fault behind the broken string, there is not obvious fault signature, location difficulty.

Find through the retrieval to prior art; technology 1([1] You Yi; Liu Dong; Li Liang; Deng. based on the 10kV pole line single-phase wire break phase to phase fault regional determination [J] of load monitor. protecting electrical power system and control .2012; 40 (19): 144-149.[2] Liu Dong; You Yi; the topaz brightness; Deng. based on decision-making system and the method thereof of the single-phase wire break fault zone of load monitor: China; 201110457625.7[P] .2012-04-18.) the single-phase wire break phase to phase fault regional determination method based on load monitor has been proposed, the fault signature that the method adopts is that fault afterload point negative sequence voltage obviously becomes greatly.Technology 2(Cheng Hao is loyal, Wu Peng, Ma Zhoujun, Deng. the power distribution network single-phase wire break based on the load measuring-recording system is judged and addressing method: China, 201210106902.6[P] .2012-08-15.) power distribution network single-phase wire break fault verification and addressing method based on the load measuring-recording system proposed, the voltage that the method arrives according to the load measuring-recording system acquisition that is installed in load point, the voltage of current values and power distribution network single-phase wire break Mishap Database, current values compares, judge the disconnection fault type, adopt prospective method or rear pushing manipulation to carry out the single-phase wire break localization of fault, and process.Technology 3(Zhu tinkling of pieces of jades; Li Changkai; Zhang Huazhong; Deng. power distribution network single-phase wire break fault negative sequence current is analyzed and route selection [J]. protecting electrical power system and control; 2009; 37 (9): 35-38.) utilize negative-sequence current and fault phase voltage product and its method of carrying out the forward direction integration carried out failure line selection, the method can effectively pick out the broken string phase to phase fault, but can not carry out localization of fault.

Technology 1 and technology 2 can be judged that single-phase wire break is earth-free and simply locate, and the extra measuring equipment of needs outfit, have increased investment, installation and maintenance cost, and technology 3 can only pick out fault type, can not carry out localization of fault.

Along with the development of intelligent distribution network, power supply enterprise and power consumer are all improving the requirement of power supply quality.In case this just requires to break down, character, the scope that can occur according to accurately judgement accident of phenomenon of the failure, and can take rational method in time to dispose, realize safe and reliable, high-quality and efficient power supply.No matter be overhead transmission line, or cable line, disconnection fault happens occasionally.But for medium voltage distribution network, along the measuring equipment of feed configuration negligible amounts often, therefore, carry out fault diagnosis and location based on it, only can be positioned between certain two measuring equipment, bearing accuracy is lower.And limited measuring equipment only can be measured a certain or certain two alternate value usually, can't obtain each phase voltage current information, and imperfect information is carried out fault diagnosis accordingly, might cause the erroneous judgement or fail to judge.And in the power distribution network, the load density that distributes along feeder line is often greater than measuring equipment density, and because the generally application of the devices such as distribution transformer monitoring terminal (TTU)/load monitor, each phase voltage current value at load place all can obtain, and the accurate location that is disconnection fault provides sufficient data basis.

Summary of the invention

For the deficiencies in the prior art, the purpose of this invention is to provide a kind of power distribution network single-phase wire break phase to phase fault localization method, the present invention need not be equipped with new equipment, and calculates simple, can with between localization of fault to two load point, realize diagnosis and the location of single-phase wire break phase to phase fault.

The objective of the invention is to adopt following technical proposals to realize:

The invention provides a kind of power distribution network single-phase wire break phase to phase fault localization method, its improvements are, described method comprises the steps:

(1) the distribution network topology is numbered;

(2) determine the positive sequence voltage of fault afterload point;

(3) judge that positive sequence voltage is whether less than setting valve (setting valve can be set as load point positive sequence voltage under the non-malfunction 110%): if not, then the single-phase wire break phase to phase fault does not occur in load point and supply path thereof; Otherwise, enter step (4);

Whether the positive sequence voltage of the adjacent load in upstream (electric current flows to the B load from the A load, and then A is called the upstream load of B load) of (4) judging load point is less than setting valve: if push away the positive sequence voltage of determining the adjacent load in upstream before the continuation, repeating step (4); Otherwise, enter step (5);

(5) determine to occur single-phase wire break phase to phase fault zone, output result of calculation.

Preferably, in the described step (2), after the single-phase wire break phase to phase fault occurs, determine respectively the positive sequence voltage of load point high-pressure side and low-pressure side, comprising:

1) determine the on high-tension side positive sequence voltage of load point, comprising:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}=(\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}}{{\mathrm{jX}}_{G\left(1\right)}}-\stackrel{\&CenterDot;}{V_{f\left(1\right)}}).---<1>;$

$\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}$

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{f\left(1\right)}}+\stackrel{\&CenterDot;}{V_{3\left(1\right)}}$

$=\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(\stackrel{\&CenterDot;}{E}\&CenterDot;{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+\stackrel{\&CenterDot;}{V_{f\left(1\right)}})}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}---<2>;$

Wherein:

---supply voltage;

JX _{G (1)}---the power supply positive sequence impedance;

JX _{L01 (1)}, jX _{L12 (1)}, jX _{L14 (1)}---branch road 01,12,14 positive sequence impedance;

JX _{2f (1)}, jX _{F3 (1)}---be respectively the positive sequence impedance of node 2 to the circuit positive sequence impedance of trouble spot, trouble spot to node 3;

---the trouble spot positive sequence voltage;

JX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD3 (1)}, jX _{LD4 (1)}---the positive sequence impedance of node 1,2,3,4 connected loads;

---the positive sequence voltage of node i, i=0,1,2,3,4;

JX _{G (1)}//jX _{LD1 (1)}//jX _{LD2 (1)}//jX _{LD4 (1)}---impedance jX _{G (1)}, jX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD4 (1)}Between in parallel;

Boundary condition by the single-phase wire break fault is known:

$\stackrel{\&CenterDot;}{V_{f\left(1\right)}}=\stackrel{\&CenterDot;}{V_{f\left(2\right)}}=\stackrel{\&CenterDot;}{V_{f\left(0\right)}=\frac{Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}}\&CenterDot;\stackrel{\&CenterDot;}{E}---<3>;$

In the formula:

Z _{Σ 1}Positive sequence impedance for from breakpoint (breakpoint refers to the position of circuit generation disconnection fault) equivalence represents with following expression formula:

Z _Σ1＝jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)+jX _LD3(1) <4>；

Z _{Σ 2}Be the negative sequence impedance from the breakpoint equivalence, represent with following expression formula:

Z _Σ2＝jX _G(2)//jX _LD1(2)//jX _LD2(2)//jX _LD4(2)+jX _LD3(2) <5>；

Z _{Σ 0}Be the negative sequence impedance from the breakpoint equivalence,

Then:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}}}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})-\frac{Z_{\mathrm{\&Sigma;}2}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}}{\mathrm{jX}}_{G\left(1\right)})$ $=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})-Z_{\mathrm{\&Sigma;}2}j{X}_{G\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}j{X}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ <6>；

Because:

jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)＜jX _G(1) <7>；

So:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}---<8>;$

The voltage of other each node is;

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}$ $\&ap;\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E\&CenterDot;}({\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}{Z}_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ $=\frac{X_{\mathrm{LD}3\left(1\right)}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LdD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}+---<9>;$ $\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}(Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}---<9>$

By formula＜8〉and＜9:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}---<10>;$

2) determine the positive sequence voltage of load point low-pressure side, comprise and determine Yyn0 type substation transformer and Dyn11 type substation transformer low-pressure side positive sequence voltage.

More preferably, determine the positive sequence voltage of YYn0 type substation transformer:

Suppose that A breaks mutually, then YYn0 type substation transformer A phase current

If

Then

Use symmetrical component method to calculate the three-phase symmetrical component of A, B, C three-phase current:

The positive sequence of A phase current, negative phase-sequence and zero-sequence component are as follows:

In the formula:

The positive sequence, negative phase-sequence and the zero-sequence component that represent respectively the A phase current;

Represent respectively A, B, C phase current; I represents electric current.

The positive sequence of B phase current, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase current, negative phase-sequence and zero-sequence component are as follows:

Suppose that every phase of impedance is jX, then the positive sequence of A phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of B phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase voltage, negative phase-sequence and zero-sequence component are as follows:

Described YYn0 type substation transformer low-pressure side positive sequence voltage is identical with the high-pressure side phase place with negative sequence voltage, and each order component of low-pressure side a phase voltage is:

In the formula: n---substation transformer high-pressure side, low-pressure side no-load voltage ratio;

Yyn0 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage becomes original

Doubly.

More preferably, determine Dyn11 type substation transformer low-pressure side positive sequence voltage, comprising:

Described each order component of Dyn11 type substation transformer low-pressure side a phase voltage is:

Dyn11 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage is original

Doubly.

More preferably, described load point is divided into load point, the load point of low-pressure side outfit load monitor and load point three classes that low-pressure side is equipped with the distribution transformer monitoring terminal that the high-pressure side is equipped with intelligent monitoring terminal, when if A breaks mutually, determine respectively the positive sequence voltage of above-mentioned 3 type load points, comprising:

I, determine that the high-pressure side is equipped with the load point positive sequence voltage of intelligent monitoring terminal:

The outfit intelligent monitoring terminal collects on high-tension side three-phase voltage and is respectively

Then substation transformer high-pressure side positive sequence voltage mould value is:

$\left|U_{A\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_A+\&PartialD;{\stackrel{\&CenterDot;}{U}}_B+{\&PartialD;}^2\stackrel{\&CenterDot;}{U_C})\right|---<19>;$

II, determine that low-pressure side is equipped with the load point positive sequence voltage of load monitor:

Load monitor is installed in the substation transformer low-pressure side, gather three-phase voltage, the three-phase current of distribution low-voltage side by the voltage and current mutual inductor, quantize through A/D converter, microprocessor carries out control, data storage, the realization of communication protocol and the output of data of sequential; Load monitor can collect the three-phase voltage phasor of substation transformer low-pressure side, and low-pressure side positive sequence voltage mould value is:

$\left|U_{a\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_a+\&PartialD;{\stackrel{\&CenterDot;}{U}}_b+{\&PartialD;}^2{\stackrel{\&CenterDot;}{U}}_c)\right|---<20>;$

III, determine that low-pressure side is equipped with the load point positive sequence voltage of distribution transformer monitoring terminal:

Distribution transformer monitoring terminal carries out AC sampling by electric current and voltage transformer (VT) to three-phase voltage and three-phase current, and low-pressure side phase voltage mean value is:

$U_{\mathrm{ave}}=\frac{U_a+U_b+U_c}{3}---<21>;$

The positive-sequence component mould value of a phase voltage is as follows:

|U _a(1)|≈|U _ave| <22>；

In the formula:

Preferably, in the described step (5), determine that power distribution network single-phase wire break phase to phase fault zone comprises: according to the voltage positive-sequence component of each load point that calculates, if the positive sequence voltage of this load point and downstream load point thereof becomes original about 50%, and this load point upstream load point positive sequence voltage remains unchanged, and the single-phase wire break phase to phase fault occurs in the adjacent lines highway section, upstream that then is connected with this load point.

Compared with the prior art, the beneficial effect that reaches of the present invention is:

(1) the present invention can only realize distribution line single-phase wire break phase to phase fault location according to existing substation transformer measuring equipment under the condition that does not increase measuring equipment, has higher economy and practicality.

(2) the present invention can accurately be positioned to the part of path between the two station power distribution transformers, reduces track walker's labour intensity, in the Reduction of failure processing time, improves power supply reliability.

(3) the present invention calculates simply, is applicable to most of distribution lines, has wide application space.

Description of drawings

Fig. 1 is distribution network line schematic diagram provided by the invention;

Fig. 2 is power distribution network positive sequence network figure provided by the invention;

Fig. 3 is the power distribution network positive sequence network figure behind the abbreviation provided by the invention;

Fig. 4 is Yyn0 type substation transformer winding connection mode figure provided by the invention;

Fig. 5 is Yyn0 type substation transformer high side voltage phasor graph provided by the invention

Fig. 6 is Yyn0 type substation transformer low-pressure side voltage phasor-diagram provided by the invention;

Fig. 7 is Dyn11 type substation transformer winding connection mode figure provided by the invention;

Fig. 8 is Dyn11 type substation transformer high side voltage phasor graph provided by the invention;

Fig. 9 is Dyn11 type substation transformer low-pressure side voltage phasor-diagram provided by the invention;

Figure 10 is power distribution network single-phase wire break phase to phase fault localization method overview flow chart provided by the invention.

Embodiment

Below in conjunction with accompanying drawing the specific embodiment of the present invention is described in further detail.

Power distribution network single-phase wire break phase to phase fault localization method provided by the invention, after the single-phase wire break phase to phase fault occurs, each substation transformer high-pressure side positive sequence voltage is far smaller than each substation transformer and each substation transformer high-pressure side positive sequence voltage of non-fault line before the trouble spot behind the trouble spot, and substation transformer low-pressure side positive sequence voltage also can correspondingly become about 50% of the front positive sequence voltage of fault after the fault.The characteristics that positive sequence voltage changed after this patent occured according to the single-phase wire break phase to phase fault, the three-phase voltage information that collects for the different measuring equipments that are installed in load point high-pressure side or low-pressure side, calculate in real time the positive-sequence component of each load point voltage, if all load point positive sequence voltages of a certain load point of a certain moment and downstream thereof obviously diminish, and the positive sequence voltage of its upstream load point is constant, can determine that then the single-phase wire break phase to phase fault appears in the adjacent lines highway section, upstream that is connected with this load point.

Power distribution network single-phase wire break phase to phase fault localization method overall procedure provided by the invention comprises the steps: as shown in figure 10

(1) the distribution network topology is numbered;

(2) determine the positive sequence voltage of fault afterload point:

In the step (2), after the single-phase wire break phase to phase fault occurs, determine respectively the positive sequence voltage of load point high-pressure side and low-pressure side, process comprises:

1) determine the on high-tension side positive sequence voltage of load point, comprising:

For the distribution line shown in Fig. 1, set up its positive sequence network figure, as shown in Figure 2, among Fig. 2:

JX _{G (1)}---the power supply positive sequence impedance;

JX _{L01 (1)}, jX _{L12 (1)}, jX _{L14 (1)}---branch road 01,12,14 positive sequence impedance;

JX _{2f (1)}, jX _{F3 (1)}---be respectively the positive sequence impedance of node 2 to the circuit positive sequence impedance of trouble spot, trouble spot to node 3;

---the trouble spot positive sequence voltage;

JX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD3 (1)}, jX _{LD4 (1)}---the positive sequence impedance of node 1,2,3,4 connected loads;

---the positive sequence voltage of node i, i=0,1,2,3,4;

JX _{G (1)}//jX _{LD1 (1)}//jX _{LD2 (1)}//jX _{LD4 (1)}---impedance jX _{G (1)}, jX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD4 (1)}Between in parallel;

Because each branch road is shorter in the power distribution network, and the positive sequence impedance of respectively loading is larger, can ignore each branch road positive sequence impedance, the positive sequence network figure behind the abbreviation as shown in Figure 3.Can be calculated by Fig. 3:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}=(\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}}{{\mathrm{jX}}_{G\left(1\right)}}-\stackrel{\&CenterDot;}{V_{f\left(1\right)}}).---<1>;$

$\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}$

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{f\left(1\right)}}+\stackrel{\&CenterDot;}{V_{3\left(1\right)}}$

$=\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(\stackrel{\&CenterDot;}{E}\&CenterDot;{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+\stackrel{\&CenterDot;}{V_{f\left(1\right)}})}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}---<2>;$

Boundary condition by the single-phase wire break fault is known:

$\stackrel{\&CenterDot;}{V_{f\left(1\right)}}=\stackrel{\&CenterDot;}{V_{f\left(2\right)}}=\stackrel{\&CenterDot;}{V_{f\left(0\right)}=\frac{Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}}\&CenterDot;\stackrel{\&CenterDot;}{E}---<3>;$

In the formula:

Z _{Σ 1}Positive sequence impedance for from breakpoint (breakpoint refers to the position of circuit generation disconnection fault) equivalence represents with following expression formula:

Z _Σ1＝jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)+jX _LD3(1) <4>；

Z _{Σ 2}Be the negative sequence impedance from the breakpoint equivalence, represent with following expression formula:

Z _Σ2＝jX _G(2)//jX _LD1(2)//jX _LD2(2)//jX _LD4(2)+jX _LD3(2) <5>；

Be the negative sequence impedance from the breakpoint equivalence, because China's power distribution network mostly is small current neutral grounding system, therefore

It is very large, Then:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}}}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})-\frac{Z_{\mathrm{\&Sigma;}2}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}}{\mathrm{jX}}_{G\left(1\right)})$ $=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})-Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{G\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ <6>；

Because:

jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)＜jX _G(1) <7>；

So:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}---<8>;$

The voltage of other each node is;

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}$ $\&ap;\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E\&CenterDot;}({\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}{Z}_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ $=\frac{X_{\mathrm{LD}3\left(1\right)}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LdD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}+---<9>;$ $\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}(Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$

By formula＜8〉and＜9:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}---<10>;$

That is to say, after the single-phase wire break phase to phase fault occurs, behind the trouble spot each point positive sequence voltage less than the trouble spot before the positive sequence voltage of each load point and each load point of non-fault line.

2) determine the positive sequence voltage of load point low-pressure side, in China, substation transformer connects group take Yyn0 as main, secondly is Dyn11, and the present invention changes Yyn0 type substation transformer and Dyn11 type substation transformer low-pressure side positive sequence voltage respectively and discusses.

1. determine the positive sequence voltage of YYn0 type substation transformer:

Yyn0 type substation transformer winding connection mode as shown in Figure 4.Suppose that A breaks mutually, then YYn0 type substation transformer A phase current

If

Then

Use symmetrical component method to calculate the three-phase symmetrical component of A, B, C three-phase current:

The substation transformer high-pressure side can be regarded pure inductance as, supposes that every phase of impedance is jX, and the positive sequence of A phase current, negative phase-sequence and zero-sequence component are as follows:

In the formula:

The positive sequence, negative phase-sequence and the zero-sequence component that represent respectively the A phase current;

Represent respectively A, B, C phase current; I represents electric current

The positive sequence of B phase current, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase current, negative phase-sequence and zero-sequence component are as follows:

Suppose that every phase of impedance is jX, then the positive sequence of A phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of B phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase voltage, negative phase-sequence and zero-sequence component are as follows:

Each order component of A, B, C three-phase voltage as shown in Figure 5 and Figure 6.

Because YYn0 type substation transformer low-pressure side positive sequence voltage is identical with the high-pressure side phase place with negative sequence voltage, so each order component of low-pressure side a phase voltage is:

In the formula: n---substation transformer high-pressure side, low-pressure side no-load voltage ratio;

Therefore, by following formula as can be known, Yyn0 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage becomes original Doubly.

2. determine Dyn11 type substation transformer low-pressure side positive sequence voltage, comprising:

Dyn11 substation transformer winding connection mode as shown in Figure 7.Mutually in the broken string situation, substation transformer high voltage side current, voltage computing method connect the substation transformer computing method of group with Yyn0 at A.

Because it is Dyn11 that substation transformer connects group, residual voltage does not induce residual voltage in low-pressure side; Under the positive sequence voltage effect, 330 ° of low-pressure side phase voltage hysteresis high-pressure side phase voltages; Under the negative sequence voltage effect, 30 ° of low-pressure side phase voltage hysteresis high-pressure side phase voltages.The voltage vector diagram of high-pressure side, low-pressure side such as Fig. 8 and shown in Figure 9.As seen from the figure, described each order component of Dyn11 type substation transformer low-pressure side a phase voltage is:

Dyn11 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage is original Doubly.

According to load point measuring equipment configuring condition calculated load point positive sequence voltage, because China's each department Distribution Network Equipment level is uneven, load point place measure configuration complete degree also is not quite similar.The present invention is according to the actual conditions of China's load point, load point is divided into load point, the load point of low-pressure side outfit load monitor and load point three classes that low-pressure side is equipped with the distribution transformer monitoring terminal that the high-pressure side is equipped with intelligent monitoring terminal, when if A breaks mutually, determine respectively the positive sequence voltage of above-mentioned 3 type load points, comprising:

I, determine that the high-pressure side is equipped with the load point positive sequence voltage of intelligent monitoring terminal:

The substation transformer high-pressure side is equipped with intelligent monitoring terminal, can collect on high-tension side three-phase voltage and be respectively

Then substation transformer high-pressure side positive sequence voltage mould value is:

$\left|U_{A\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_A+\&PartialD;{\stackrel{\&CenterDot;}{U}}_B+{\&PartialD;}^2\stackrel{\&CenterDot;}{U_C})\right|---<19>;$

II, determine that low-pressure side is equipped with the load point positive sequence voltage of load monitor:

Load monitor is installed in the substation transformer low-pressure side, gather three-phase voltage, the three-phase current of distribution low-voltage side by the voltage and current mutual inductor, quantize through A/D converter, microprocessor carries out control, data storage, the realization of communication protocol and the output of data of sequential; Load monitor can collect the three-phase voltage phasor of substation transformer low-pressure side, and low-pressure side positive sequence voltage mould value is:

$\left|U_{a\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_a+\&PartialD;{\stackrel{\&CenterDot;}{U}}_b+{\&PartialD;}^2{\stackrel{\&CenterDot;}{U}}_c)\right|---<20>;$

III, determine that low-pressure side is equipped with the load point positive sequence voltage of distribution transformer monitoring terminal:

Distribution transformer monitoring terminal (TTU) carries out AC sampling by electric current and voltage transformer (VT) to three-phase voltage and three-phase current, but only collect the three-phase electricity pressing mold value of substation transformer low-pressure side, can not collect phase angle, substation transformer low-pressure side positive sequence voltage adopts approximate treatment, and low-pressure side phase voltage mean value is:

$U_{\mathrm{ave}}=\frac{U_a+U_b+U_c}{3}---<21>;$

The positive-sequence component mould value of a phase voltage is as follows:

|U _a(1)|≈|U _ave| <22>；

In the formula:

(3) judge that positive sequence voltage is whether less than setting valve (setting valve can be set as load point positive sequence voltage under the non-malfunction 110%): if not, then the single-phase wire break phase to phase fault does not occur in load point and supply path thereof; Otherwise, enter step (4);

Whether the positive sequence voltage of the adjacent load in upstream of (4) judging load point is less than setting valve: if push away the positive sequence voltage of determining the adjacent load in upstream before the continuation, repeating step (4); Otherwise, enter step (5);

(5) determine to occur single-phase wire break phase to phase fault zone, output result of calculation:

According to each load point voltage positive-sequence component that calculates, if the positive sequence voltage of this load point and downstream load point thereof obviously diminishes (become original about 50%), and this load point upstream load point positive sequence voltage remains unchanged, and the single-phase wire break phase to phase fault occurs in the adjacent lines highway section, upstream that then is connected with this load point.

The invention provides a kind of feature of utilizing broken string afterload point positive sequence voltage to reduce and realize the method for single-phase wire break phase to phase fault location.The present invention need not be equipped with new equipment, and calculates simply, can with between localization of fault to two load point, realize diagnosis and the location of single-phase wire break phase to phase fault.

Should be noted that at last: above embodiment is only in order to illustrate that technical scheme of the present invention is not intended to limit, although with reference to above-described embodiment the present invention is had been described in detail, those of ordinary skill in the field are to be understood that: still can make amendment or be equal to replacement the specific embodiment of the present invention, and do not break away from any modification of spirit and scope of the invention or be equal to replacement, it all should be encompassed in the middle of the claim scope of the present invention.

Claims (6)

1. a power distribution network single-phase wire break phase to phase fault localization method is characterized in that, described method comprises the steps:

(1) the distribution network topology is numbered;

(2) determine the positive sequence voltage of fault afterload point;

(3) judge that positive sequence voltage is whether less than setting valve: if not, then the single-phase wire break phase to phase fault does not occur in load point and supply path thereof; Otherwise, enter step (4);

Whether the positive sequence voltage of the adjacent load in upstream of (4) judging load point is less than setting valve: if push away the positive sequence voltage of determining the adjacent load in upstream before the continuation, repeating step (4); Otherwise, enter step (5);

(5) determine to occur single-phase wire break phase to phase fault zone, output result of calculation.

2. power distribution network single-phase wire break phase to phase fault localization method as claimed in claim 1 is characterized in that, in the described step (2), after the single-phase wire break phase to phase fault occurs, determines respectively the positive sequence voltage of load point high-pressure side and low-pressure side, comprising:

1) determine the on high-tension side positive sequence voltage of load point, comprising:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}=(\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}}{{\mathrm{jX}}_{G\left(1\right)}}-\stackrel{\&CenterDot;}{V_{f\left(1\right)}}).---<1>;$

$\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}$

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{f\left(1\right)}}+\stackrel{\&CenterDot;}{V_{3\left(1\right)}}$

$=\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(\stackrel{\&CenterDot;}{E}\&CenterDot;{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+\stackrel{\&CenterDot;}{V_{f\left(1\right)}})}{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})+{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}}---<2>;$

Wherein:

---supply voltage;

JX _{G (1)}---the power supply positive sequence impedance;

JX _{L01 (1)}, jX _{L12 (1)}, jX _{L14 (1)}---branch road 01,12,14 positive sequence impedance;

JX _{2f (1)}, jX _{F3 (1)}---be respectively the positive sequence impedance of node 2 to the circuit positive sequence impedance of trouble spot, trouble spot to node 3;

---the trouble spot positive sequence voltage;

JX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD3 (1)}, jX _{LD4 (1)}---the positive sequence impedance of node 1,2,3,4 connected loads;

---the positive sequence voltage of node i, i=0,1,2,3,4;

JX _{G (1)}//jX _{LD1 (1)}//jX _{LD2 (1)}//jX _{LD4 (1)}---impedance jX _{G (1)}, jX _{LD1 (1)}, jX _{LD2 (1)}, jX _{LD4 (1)}Between in parallel;

Boundary condition by the single-phase wire break fault is known:

$\stackrel{\&CenterDot;}{V_{f\left(1\right)}}=\stackrel{\&CenterDot;}{V_{f\left(2\right)}}=\stackrel{\&CenterDot;}{V_{f\left(0\right)}=\frac{Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}//Z_{\mathrm{\&Sigma;}0}}}\&CenterDot;\stackrel{\&CenterDot;}{E}---<3>;$

In the formula:

Z _{Σ 1}Be the positive sequence impedance from the breakpoint equivalence, represent with following expression formula:

Z _Σ1＝jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)+jX _LD3(1) <4>；

Z _{Σ 2}Be the negative sequence impedance from the breakpoint equivalence, represent with following expression formula:

Z _Σ2＝jX _G(2)//jX _LD1(2)//jX _LD2(2)//jX _LD4(2)+jX _LD3(2) <5>；

Z _{Σ 0}Be the negative sequence impedance from the breakpoint equivalence,

Then:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}}}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})-\frac{Z_{\mathrm{\&Sigma;}2}}{Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}}{\mathrm{jX}}_{G\left(1\right)})$

$=\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}(({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})-Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{G\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ <6>；

Because:

jX _G(1)//jX _LD1(1)//jX _LD2(1)//jX _LD4(1)＜jX _G(1) <7>；

So:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\frac{{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}{\mathrm{jX}}_{G\left(1\right)}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}---<8>;$

The voltage of other each node is;

$\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}$ $\&ap;\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E\&CenterDot;}({\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}{Z}_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$ $=\frac{X_{\mathrm{LD}3\left(1\right)}\&CenterDot;Z_{\mathrm{\&Sigma;}1}\&CenterDot;\stackrel{\&CenterDot;}{E}\&CenterDot;({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LdD}4\left(1\right)})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}+---<9>;$ $\frac{({\mathrm{jX}}_{G\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}1\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}2\left(1\right)}//{\mathrm{jX}}_{\mathrm{LD}4\left(1\right)})\&CenterDot;\stackrel{\&CenterDot;}{E}(Z_{\mathrm{\&Sigma;}2}{\mathrm{jX}}_{\mathrm{LD}3\left(1\right)}+Z_{\mathrm{\&Sigma;}2})}{Z_{\mathrm{\&Sigma;}1}(Z_{\mathrm{\&Sigma;}1}+Z_{\mathrm{\&Sigma;}2})}$

By formula＜8〉and＜9:

$\stackrel{\&CenterDot;}{V_{3\left(1\right)}}<\stackrel{\&CenterDot;}{V_{0\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{1\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{2\left(1\right)}}\&ap;\stackrel{\&CenterDot;}{V_{4\left(1\right)}}---<10>;$

2) determine the positive sequence voltage of load point low-pressure side, comprise and determine Yyn0 type substation transformer and Dyn11 type substation transformer low-pressure side positive sequence voltage.

3. power distribution network single-phase wire break phase to phase fault localization method as claimed in claim 2 is characterized in that, determines the positive sequence voltage of YYn0 type substation transformer:

Suppose that A breaks mutually, then YYn0 type substation transformer A phase current

If

Then Use symmetrical component method to calculate the three-phase symmetrical component of A, B, C three-phase current:

The positive sequence of A phase current, negative phase-sequence and zero-sequence component are as follows:

In the formula:

The positive sequence, negative phase-sequence and the zero-sequence component that represent respectively the A phase current;

Represent respectively A, B, C phase current; I represents electric current.

The positive sequence of B phase current, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase current, negative phase-sequence and zero-sequence component are as follows:

Suppose that every phase of impedance is jX, then the positive sequence of A phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of B phase voltage, negative phase-sequence and zero-sequence component are as follows:

The positive sequence of C phase voltage, negative phase-sequence and zero-sequence component are as follows:

Described YYn0 type substation transformer low-pressure side positive sequence voltage is identical with the high-pressure side phase place with negative sequence voltage, and each order component of low-pressure side a phase voltage is:

In the formula: n---substation transformer high-pressure side, low-pressure side no-load voltage ratio;

Yyn0 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage becomes original

Doubly.

4. power distribution network single-phase wire break phase to phase fault localization method as claimed in claim 2 is characterized in that, determines Dyn11 type substation transformer low-pressure side positive sequence voltage, comprising:

Described each order component of Dyn11 type substation transformer low-pressure side a phase voltage is:

Dyn11 connects group type substation transformer at A mutually in the broken string situation, and substation transformer low-pressure side a phase positive sequence voltage is original

Doubly.

5. power distribution network single-phase wire break phase to phase fault localization method as claimed in claim 2, it is characterized in that, described load point is divided into load point, the load point of low-pressure side outfit load monitor and load point three classes that low-pressure side is equipped with the distribution transformer monitoring terminal that the high-pressure side is equipped with intelligent monitoring terminal, when if A breaks mutually, determine respectively the positive sequence voltage of above-mentioned 3 type load points, comprising:

I, determine that the high-pressure side is equipped with the load point positive sequence voltage of intelligent monitoring terminal:

The outfit intelligent monitoring terminal collects on high-tension side three-phase voltage and is respectively

Then substation transformer high-pressure side positive sequence voltage mould value is:

$\left|U_{A\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_A+\&PartialD;{\stackrel{\&CenterDot;}{U}}_B+{\&PartialD;}^2\stackrel{\&CenterDot;}{U_C})\right|---<19>;$

II, determine that low-pressure side is equipped with the load point positive sequence voltage of load monitor:

Load monitor is installed in the substation transformer low-pressure side, gather three-phase voltage, the three-phase current of distribution low-voltage side by the voltage and current mutual inductor, quantize through A/D converter, microprocessor carries out control, data storage, the realization of communication protocol and the output of data of sequential; Load monitor can collect the three-phase voltage phasor of substation transformer low-pressure side, and low-pressure side positive sequence voltage mould value is:

$\left|U_{a\left(1\right)}\right|=\left|\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_a+\&PartialD;{\stackrel{\&CenterDot;}{U}}_b+{\&PartialD;}^2{\stackrel{\&CenterDot;}{U}}_c)\right|---<20>;$

III, determine that low-pressure side is equipped with the load point positive sequence voltage of distribution transformer monitoring terminal:

Distribution transformer monitoring terminal carries out AC sampling by electric current and voltage transformer (VT) to three-phase voltage and three-phase current, and low-pressure side phase voltage mean value is:

$U_{\mathrm{ave}}=\frac{U_a+U_b+U_c}{3}---<21>;$

The positive-sequence component mould value of a phase voltage is as follows:

|U _a(1)≈|U _ave| <22>；

6. power distribution network single-phase wire break phase to phase fault localization method as claimed in claim 1, it is characterized in that, in the described step (5), determine that power distribution network single-phase wire break phase to phase fault zone comprises: according to the voltage positive-sequence component of each load point that calculates, if the positive sequence voltage of this load point and downstream load point thereof becomes original about 50%, and this load point upstream load point positive sequence voltage remains unchanged, and the single-phase wire break phase to phase fault occurs in the adjacent lines highway section, upstream that then is connected with this load point.

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