CN103954885B - The single-ended alignment system of double line down and localization method based on distributed constant - Google Patents

Abstract

The invention discloses a kind of single-ended alignment system of the double line down based on distributed constant in line fault of electrical power system field of locating technology and localization method.System includes data acquisition module, bucking voltage solves module, fault place voltage solves module and fault location module；Method includes the voltage phasor and the electric current phasor that gather any one end of double back faulty line, solves the bucking voltage function of the function of zero-utility theory in the same direction of double back faulty line, reverse zero-utility theory function and double back faulty line；Solve fault place backward sequence electric current, further according to backward sequence Current calculation fault place of fault place voltage-phase, and solve fault place voltage magnitude, construct fault location function, by solving the SPA sudden phase anomalies point identification abort situation of fault location function.The present invention is applicable to multiple double line down, only need to gather single-ended power information, it is not necessary to the information of other transformer stations, and the impact by the factor such as abort situation, transition resistance is little, and positioning precision is high, it is easy to accomplish.

Description

Double-circuit line fault single-end positioning system and method based on distribution parameters

Technical Field

The invention belongs to the technical field of power system line fault positioning, and particularly relates to a double-circuit line fault single-terminal positioning system and a double-circuit line fault single-terminal positioning method based on distribution parameters.

Background

The high-voltage double-circuit line erected on the same tower has large transmission capacity, obvious economic benefit, and capability of improving the safety and stability of the operation of a power system, and is widely applied in recent years. However, in field operation, double circuit line faults are inevitable, and how to accurately position the faults has important significance for rapidly isolating the faults and recovering power supply in time.

At present, the fault location method for double circuit lines is mainly divided into a double-end method and a single-end method. The double-ended method can be divided into a double-ended algorithm based on synchronous sampling or synchronous processing at two ends and a double-ended algorithm without synchronous sampling or synchronous processing of data. The method adopts the electrical quantities at two ends of the line, has high precision in principle, is not influenced by the transition resistance, but has larger dependence degree on communication, and when the local end cannot acquire the data of the opposite end, the algorithm fails and the fault location cannot be realized. The single-ended method includes an algorithm using single-ended double loop information and an algorithm using single-ended single loop information. The single-end method does not need information of other stations, can perform fault location only by using the voltage and current information of the station, has low requirement on hardware, and is easy to realize. However, the conventional single-ended method usually adopts a centralized parameter model, neglects the influence of distributed capacitance, and cannot ensure the measurement accuracy with the increase of the fault distance.

In order to solve the problems, the invention provides a double-circuit line fault single-ended positioning system and a positioning method based on a distributed parameter based on a double-circuit line six-sequence network distributed parameter model and based on the electrical characteristics of double-circuit line overline faults. Firstly, considering mutual inductance influence in a double-circuit line six-sequence network distribution parameter model, and defining accurate compensation voltage by means of a same-direction zero-sequence compensation coefficient and a reverse zero-sequence compensation coefficient; on the basis, the voltage phase and the amplitude of the fault are calculated according to the relation between the reverse sequence current of the measuring end and the reverse sequence current phase of the fault; and finally, constructing a fault positioning function by using the measured voltage, the measured current and the calculated voltage at the fault position, and identifying the fault position by solving a phase catastrophe point of the fault positioning function. The PSCAD-based double-end system simulation model shows that the single-end positioning system and the method are suitable for various overline faults, do not need information of other transformer substations, are slightly influenced by factors such as fault positions and transition resistance, are high in positioning accuracy and are easy to realize.

Disclosure of Invention

The invention aims to provide a double-circuit fault single-end positioning system and a double-circuit fault single-end positioning method based on distribution parameters, which are used for solving the defects of the existing double-circuit fault positioning method.

In order to achieve the above object, the technical solution of the present invention is a double-circuit fault single-ended positioning system based on distributed parameters, which is characterized in that the system includes: the device comprises a data acquisition module, a compensation voltage solving module, a fault voltage solving module and a fault positioning module;

the data acquisition module is respectively connected with the compensation voltage solving module and the fault voltage solving module; the fault positioning module is respectively connected with the compensation voltage solving module and the fault position voltage solving module;

the data acquisition module is used for acquiring voltage phasor and current phasor at any end of the double-circuit fault line and respectively sending the acquired data to the compensation voltage solving module and the fault voltage solving module;

the compensation voltage solving module is used for solving a homodromous zero-sequence compensation coefficient function and a reverse zero-sequence compensation coefficient function of the double-circuit fault line according to a six-sequence network distribution parameter model of the double-circuit fault line, determining a compensation voltage function of the double-circuit fault line according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function, and sending the compensation voltage function of the double-circuit fault line to the fault positioning module;

the fault voltage solving module is used for solving the reverse sequence current of the fault, calculating the voltage phase of the fault according to the reverse sequence current of the fault, solving the voltage amplitude of the fault by using the acquired voltage phasor and current phasor to obtain the voltage of the fault, and then sending the voltage of the fault to the fault positioning module;

the fault positioning module is used for constructing a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifying the fault position by solving the phase catastrophe point of the fault positioning function.

A double-circuit fault single-end positioning method based on distribution parameters is characterized by comprising the following steps:

step 1: collecting voltage phasor and current phasor at any end of a double-circuit fault line;

step 2: solving a homodromous zero-sequence compensation coefficient function and a reverse zero-sequence compensation coefficient function of the double-circuit fault line according to the six-sequence network distribution parameter model of the double-circuit fault line, and determining a compensation voltage function of the double-circuit fault line according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function;

and step 3: solving reverse sequence current at the fault; calculating a voltage phase at the fault according to the reverse sequence current at the fault, and solving a voltage amplitude at the fault by using the acquired voltage phasor and current phasor to obtain a voltage at the fault;

and 4, step 4: and constructing a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifying the fault position by solving the phase catastrophe point of the fault positioning function.

The function for solving the equidirectional zero-sequence compensation coefficient of the double-circuit fault line adopts a formula:

$k_T\left(l_{\mathrm{mf}}\right)=\frac{z_{\mathrm{cT}0}\sinh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}+\frac{Z_{\mathrm{mT}0}\cosh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right)-Z_{\mathrm{mT}0}\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)};$

wherein l_mfThe distance from the acquisition end of each sequence network in the six sequence networks of the double-circuit fault line to the fault position;

z_cT0characteristic impedance of a same-direction zero sequence network in a six sequence network of a double-circuit fault line;

γ_T0the transmission coefficient of a homodromous zero sequence network in a six sequence network of a double-circuit fault line;

z_c1is the characteristic impedance of a forward sequence network in the same direction;

γ₁the forward sequence network propagation coefficient is the same direction;

Z_mT0is the same-direction zero sequence impedance;

sinh () is a hyperbolic sine function;

cosh () is a hyperbolic cosine function.

The reverse zero sequence compensation coefficient function for solving the double-circuit fault line adopts a formula:

$k_F\left(l_{\mathrm{mf}}\right)=\frac{z_{\mathrm{cF}0}\sinh \left({\mathrm{\γ}}_{F0}{l}_{\mathrm{mf}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)};$

wherein l_mfThe distance between the acquisition end of the double-circuit fault line and the fault position is obtained;

z_cF0is a reverse zero sequence net characteristic impedance;

γ_F0is a reverse zero-sequence net propagation coefficient;

z_c1is the characteristic impedance of a forward sequence network in the same direction;

γ₁the forward sequence network propagation coefficient is the same direction;

sinh () is a hyperbolic sine function.

The compensation voltage function of the double-circuit fault line is determined according to the same-direction zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function by adopting a formula:

${\stackrel{\·}{U}}_{\mathrm{op}}\left(l_{\mathrm{mx}}\right)=\stackrel{\·}{U}\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right){\stackrel{\·}{I}}^{\′};$

wherein,is a compensation voltage function of the double-circuit fault line and is used for calculating the distance between the acquisition end l and the double-circuit fault line_mxA compensation voltage of;

z_c1is the characteristic impedance of a forward sequence network in the same direction;

γ₁the forward sequence network propagation coefficient is the same direction;

sinh () is a hyperbolic sine function;

single phase voltage as a return lineInterphase voltage of same loopOr voltage between phases of overline

When in useSingle phase voltage as a return lineWhen the temperature of the water is higher than the set temperature, ${\stackrel{\·}{I}}^{\′}={\stackrel{\·}{I}}_m+k_T\left(l_{\mathrm{mx}}\right){\stackrel{\·}{I}}_{\mathrm{mT}0}+k_F\left(l_{\mathrm{mx}}\right){\stackrel{\·}{I}}_{\mathrm{mF}0};$

when in useInter-phase voltage of the same loopWhen the temperature of the water is higher than the set temperature,

when in useAs a voltage between phases of the overlineWhen the temperature of the water is higher than the set temperature,

collecting the phase current;

the same loop line interphase current is used as the acquisition end;

cross-line interphase current is collected at the end;

sequence current of a homodromous zero sequence network acquisition end;

the sequence current of the reverse zero sequence network acquisition end.

The formula for solving the reverse current of each sequence at the fault position is as follows:

${\stackrel{\·}{I}}_{\mathrm{fFi}}=\cosh \left({\mathrm{\γ}}_{\mathrm{Fi}}{l}_{\mathrm{mf}}\right)\·[1+\frac{\tanh \left({\mathrm{\γ}}_{\mathrm{Fi}}{l}_{\mathrm{mf}}\right)}{\tanh \left({\mathrm{\γ}}_{\mathrm{Fi}}(l_{\mathrm{mn}}-l_{\mathrm{mf}})\right)}]\·{\stackrel{\·}{I}}_{\mathrm{mFi}};$

wherein, i is 0,1,2, respectively representing zero sequence, positive sequence and negative sequence;

is a reverse sequence current at fault;

γ_Fithe propagation coefficients of all reverse order nets;

l_mnthe distance between two ends of the double-circuit fault line;

l_mfthe distance between the acquisition end of the double-circuit fault line and the fault position is obtained;

cosh () is a hyperbolic cosine function;

tanh () is a hyperbolic tangent function;

the sequence current of the acquisition end in each sequence network is reversed.

The step of calculating the voltage phase at the fault according to the reverse sequence current at the fault specifically comprises the following steps:

substep A1: determining the fault boundary current condition of the double-circuit fault line according to the fault type of the double-circuit fault line;

substep A2: and calculating the voltage phase at the fault according to the fault boundary current condition of the double-circuit fault line and the reverse sequence current at the fault.

The voltage amplitude at the fault position is solved by using the collected voltage phasor and current phasor by adopting a formula:

wherein, U_fIs the fault location voltage amplitude;

single phase voltage as a return lineInterphase voltage of same loopOr voltage between phases of overline

The impedance angle of the double-circuit fault line;

theta isAndthe included angle of (A);

gamma isAndthe included angle of (A);

when in useSingle phase voltage as a return lineWhen the temperature of the water is higher than the set temperature, ${\stackrel{\·}{I}}^{\′}={\stackrel{\·}{I}}_m+k_T\left(l_{\mathrm{mx}}\right){\stackrel{\·}{I}}_{\mathrm{mT}0}+k_F\left(l_{\mathrm{mx}}\right){\stackrel{\·}{I}}_{\mathrm{mF}0};$

when in useInter-phase voltage of the same loopWhen the temperature of the water is higher than the set temperature,

when in useAs a voltage between phases of the overlineWhen the temperature of the water is higher than the set temperature,

collecting the phase current;

the same loop line interphase current is used as the acquisition end;

cross-line interphase current is collected at the end;

sequence current of a homodromous zero sequence network acquisition end;

the sequence current of the reverse zero sequence network acquisition end.

The fault location function is: $f\left(l_{\mathrm{mx}}\right)=\frac{{\stackrel{\·}{U}}_{\mathrm{op}}\left(l_{\mathrm{mx}}\right)-{\stackrel{\·}{U}}_f\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right)}{{\stackrel{\·}{I}}^{\′}};$

wherein,a compensation voltage function for a double-circuit fault line;

is the fault location voltage;

l_mxis the distance from the acquisition end;

γ₁the forward sequence network propagation coefficient is the same direction;

cosh () is a hyperbolic cosine function.

The step of identifying the fault location by solving the phase discontinuity point of the fault location function specifically includes:

substep B1: dividing the double-circuit fault line n into equal parts, wherein n is a set value;

substep B2: calculating the phases of the fault locating functions of the end points at the two sides of each equally divided interval according to the fault locating functions, and if the phases of the fault locating functions of the end points at the two sides of the equally divided intervals are one larger than zero and one smaller than zero, judging that phase mutation points are in the equally divided intervals;

substep B3: starting from an endpoint on the left side of an equal division interval where a phase catastrophe point is located, selecting a point every set step length delta l and calculating the phase of a fault positioning function of the selected point;

substep B4: taking a point with the phase of the first fault positioning function being less than 0 as a reference point which is far away from the acquisition end l_f+If the fault position is far from the acquisition end, it is l_f＝l_f+-0.5·Δl。

According to the method, a fault positioning function based on a double-circuit line distribution parameter model is established based on the electrical characteristics of double-circuit line faults, and the fault position is judged by utilizing the phase mutation of the fault positioning function. Simulation results show that the method is applicable to various double circuit line faults, only single-end power information needs to be acquired, information of other transformer substations is not needed, influence of factors such as fault positions and transition resistance is small, positioning accuracy is high, and the method is easy to achieve.

Drawings

FIG. 1 is a diagram of a distributed parameter based double-circuit fault single-ended positioning system;

FIG. 2 is a flow chart of a double-circuit fault single-end positioning method based on distribution parameters;

FIG. 3 is a schematic diagram of the electrical quantities of the off-line networks in case of failure;

FIG. 4 is a schematic view of aPhase estimationA phase error graph;

FIG. 5 is a schematic view of aPhase estimationA phase error graph;

FIG. 6 is a voltage phasor diagram at fault;

FIG. 7 is a graph of the phase characteristics of the localization function at 100km fault;

FIG. 8 is a diagram of a simulation model of a two-wire system;

FIG. 9 is a schematic diagram illustrating the effect of fault location and fault type on cross-line ungrounded fault location;

FIG. 10 is a schematic diagram of the effect of fault location on over-the-wire two-wire ground fault location;

FIG. 11 is a schematic diagram illustrating the effect of fault location and fault type on over-the-wire three-wire ground fault location;

FIG. 12 is a schematic diagram of the effect of fault location and fault type on single-phase ground localization of a three-wire fault;

FIG. 13 is a schematic illustration of the effect of transition resistance on over-the-wire two-wire ground fault localization;

FIG. 14 is a schematic illustration of the effects of transition resistance and fault type on over-the-wire three-wire ground fault localization;

fig. 15 is a schematic diagram of the effect of transition resistance and fault type on single-phase ground localization of a three-wire fault.

Detailed Description

The preferred embodiments will be described in detail below with reference to the accompanying drawings. It should be emphasized that the following description is merely exemplary in nature and is not intended to limit the scope of the invention or its application.

Fig. 1 is a structural diagram of a distributed parameter-based double-circuit fault single-ended positioning system, and as shown in fig. 1, the distributed parameter-based double-circuit fault single-ended positioning system provided by the present invention includes: the device comprises a data acquisition module, a compensation voltage solving module, a fault voltage solving module and a fault positioning module. The data acquisition module is respectively connected with the compensation voltage solving module and the fault voltage solving module, and the fault positioning module is respectively connected with the compensation voltage solving module and the fault voltage solving module.

The data acquisition module is used for acquiring voltage phasor and current phasor at any end of the double-circuit fault line and respectively sending the acquired data to the compensation voltage solving module and the fault voltage solving module.

The compensation voltage solving module is used for solving the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function of the double-circuit fault line according to the six-sequence network distribution parameter model of the double-circuit fault line, determining the compensation voltage function of the double-circuit fault line according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function, and sending the compensation voltage function of the double-circuit fault line to the fault positioning module.

The fault voltage solving module is used for solving the reverse sequence current of the fault, calculating the voltage phase of the fault according to the reverse sequence current of the fault, solving the voltage amplitude of the fault by using the collected voltage phasor and current phasor to obtain the voltage of the fault, and sending the voltage of the fault to the fault positioning module.

And the fault positioning module is used for constructing a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifying the fault position by solving the phase catastrophe point of the fault positioning function.

The invention also provides a double-circuit line fault single-terminal positioning method based on distribution parameters, as shown in fig. 2, the method comprises the following steps:

step 1: and collecting the voltage phasor and the current phasor at any end of the double-circuit fault line.

The data acquisition module acquires voltage phasor and current phasor at any end of the double-circuit fault line and sends the voltage phasor and the current phasor to the compensation voltage solving module and the fault position voltage solving module.

Step 2: in the compensation voltage solving module, according to a six-sequence network distribution parameter model of the double-circuit fault line, a homodromous zero-sequence compensation coefficient function and a reverse zero-sequence compensation coefficient function of the double-circuit fault line are solved, and then according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function, a relation function of the compensation voltage of the double-circuit fault line and the distance from a collection end to a fault is determined, namely a compensation voltage function of the double-circuit fault line.

For voltage and current phasors of the symmetrically coupled double-circuit line, decoupling is realized through a matrix T to obtain a six-sequence network distribution parameter model, as shown in a formula (1).

$T=\frac{1}{6}\left[\begin{array}{cccccc}1& \mathrm{\α}& {\mathrm{\α}}^2& 1& \mathrm{\α}& {\mathrm{\α}}^2\\ 1& {\mathrm{\α}}^2& \mathrm{\α}& 1& {\mathrm{\α}}^2& \mathrm{\α}\\ 1& 1& 1& 1& 1& 1\\ 1& \mathrm{\α}& {\mathrm{\α}}^2& -1& -\mathrm{\α}& {-\mathrm{\α}}^2\\ 1& {\mathrm{\α}}^2& \mathrm{\α}& -1& {-\mathrm{\α}}^2& -\mathrm{\α}\\ 1& 1& 1& -1& -1& -1\end{array}\right]---\left(1\right)$

Wherein, α ═ e^j120°And j is an imaginary unit. After the double circuit line fails, the electrical quantity of each grid is as shown in fig. 3, and in fig. 3, the voltage and the current of each grid satisfy the relation:

$\left[\begin{array}{c}{\stackrel{\·}{U}}_{\mathrm{mi}}\\ {\stackrel{\·}{I}}_{\mathrm{mi}}\end{array}\right]=\left[\begin{array}{cc}\cosh \left({\mathrm{\γ}}_i{l}_{\mathrm{mf}}\right)& z_{\mathrm{ci}}\sinh \left({\mathrm{\γ}}_i{l}_{\mathrm{mf}}\right)\\ \sinh \left({\mathrm{\γ}}_i{l}_{\mathrm{mf}}\right)/z_{\mathrm{ci}}& \cosh \left({\mathrm{\γ}}_i{I}_{\mathrm{mf}}\right)\end{array}\right]\left[\begin{array}{c}{\stackrel{\·}{U}}_{\mathrm{mfi}}\\ {\stackrel{\·}{I}}_{\mathrm{mfi}}\end{array}\right]---\left(2\right)$

in formula (2), i ═ T1, T2, T0, F1, F2, and F0 denote a net of the same-direction positive order, a net of the same-direction negative order, a net of the same-direction zero order, a net of the reverse positive order, a net of the reverse negative order, and a net of the reverse zero order, respectively.Sequence voltage and sequence current collected by the M end in each sequence network respectively;for the fault-handling voltage in each of the sequencing nets,the sequence short-circuit current provided for the M end at the fault position in each sequence network,for fault-handling short-circuit currents in the respective grid,/_mfAnd the distance from the M end to the fault position in each sequence network.

Six-sequence voltage collected by M end in each sequence network of formula (2)The addition can result in:

$\begin{array}{c}{\stackrel{\·}{U}}_m=\frac{{\stackrel{.}{U}}_{\mathrm{mfT}1}}{\cosh \left({\mathrm{\γ}}_{T1}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cT}1}\tanh \left({\mathrm{\γ}}_{T1}{l}_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mT}1}+\frac{{\stackrel{\·}{U}}_{\mathrm{mfT}2}}{\cosh \left({\mathrm{\γ}}_{T2}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cT}2}\tanh \left({\mathrm{\γ}}_{T2}{l}_{\mathrm{mf}}\right){\stackrel{.}{I}}_{\mathrm{mT}2}+\\ \frac{{\stackrel{.}{U}}_{\mathrm{mfT}0}}{\cosh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cT}0}\tanh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mT}0}+\frac{{\stackrel{\·}{U}}_{\mathrm{mfF}1}}{\cosh \left({\mathrm{\γ}}_{F1}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cF}1}\tanh \left({\mathrm{\γ}}_{F1}{l}_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}1}+\\ \frac{{\stackrel{\·}{U}}_{\mathrm{mfF}2}}{\cosh \left({\mathrm{\γ}}_{F2}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cF}2}\tanh \left({\mathrm{\γ}}_{F2}{l}_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}2}+\frac{{\stackrel{\·}{U}}_{\mathrm{mfF}0}}{\cosh \left({\mathrm{\γ}}_{F0}{l}_{\mathrm{mf}}\right)}+z_{\mathrm{cF}0}\tanh \left({\mathrm{\γ}}_{F0}{l}_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}0}\\ =\frac{{\stackrel{\·}{U}}_{\mathrm{mf}}}{\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}+z_{c1}\tanh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)[{\stackrel{\·}{I}}_m+k_T\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mT}0}+k_F\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}0}]\end{array}---\left(3\right)$

wherein,respectively phase voltage and phase current of the acquisition end M,to the phase voltage at fault, z_c1、γ₁Characteristic impedance and propagation coefficient, k, of the same-direction positive sequence network_T(l_mf)、k_F(l_mf) The zero sequence compensation coefficient function is represented by a same-direction zero sequence compensation coefficient function and a reverse zero sequence compensation coefficient function, and the two functions are represented by:

$k_T=\left(l_{\mathrm{mf}}\right)=\frac{z_{\mathrm{cT}0}\sinh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}+\frac{Z_{\mathrm{mT}0}\cosh \left({\mathrm{\γ}}_{T0}{l}_{\mathrm{mf}}\right)-Z_{\mathrm{mT}0}\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}---\left(4\right)$

$k_F\left(l_{\mathrm{mf}}\right)=\frac{z_{\mathrm{cF}0}\sinh \left({\mathrm{\γ}}_{F0}{l}_{\mathrm{mf}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}{z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mf}}\right)}---\left(5\right)$

in formulae (4) and (5), z_cT0Characteristic impedance of homodromous zero-sequence network in six-sequence network of double-circuit fault line_T0Homodromous zero sequence net propagation coefficient, z, in six sequence net of double-circuit fault line_c1Characteristic impedance of homodromous positive sequence network, gamma₁Is the co-directional forward sequence net propagation coefficient, z_cF0For reverse zero sequence net characteristic impedance, gamma_F0For reverse zero-order net propagation coefficient, Z_mT0For the homodromous zero sequence impedance, sinh () is a hyperbolic sine function, and cosh () is a hyperbolic cosine function.

As shown in the formula (3), if the two phase voltages on a return line are subtracted from each other, the zero sequence current in the same direction and the zero sequence current in the opposite direction are both cancelled out, so that the interphase voltage of the return lineVoltage between phases and faultAnd inter-phase currentThe relationship between them is:

similarly, if the two phase voltages of the overline are subtracted and only the zero sequence current in the same direction is offset, the voltage between the overline phasesVoltage between the fault point and the overline phaseAnd a cross-line interphase currentThe relationship between them is:

define distance M to end l_mxFunction of compensation voltage ofComprises the following steps:

${\stackrel{\·}{U}}_{\mathrm{op}}\left(l_{\mathrm{mx}}\right)=\stackrel{\·}{U}\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right)-z_{c1}\sinh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right){\stackrel{\·}{I}}^{\′}---\left(8\right)$

wherein,value and voltage ofIs related to the type of (c), as shown in formula (9):

if it isSingle phase voltage as a return lineThenIs a single phase current of the return lineThe sum of the zero sequence current in the same direction and the zero sequence current in the opposite direction; if it isInter-phase voltage of the same loopThenAs phase currentIf it isAs a voltage between phases of the overlineThenFor current flowing between phases of overlinesAnd the sum of the reverse zero sequence currents.

And step 3: in a fault voltage solving module, solving reverse sequence current at the fault; and calculating the voltage phase of the fault according to the reverse sequence current of the fault, and solving the voltage amplitude of the fault by using the acquired voltage phasor and current phasor to obtain the voltage of the fault.

(i) First, the reverse sequence current at the fault is solved.

According to the reverse network, the reverse positive/negative/zero sequence current of the M end (acquisition end) can be obtainedPositive/negative/zero sequence current reverse to faultThe relationship of (1):

$\frac{{\stackrel{\·}{I}}_{\mathrm{fF}012}}{{\stackrel{\·}{I}}_{\mathrm{mF}012}}=\cosh \left({\mathrm{\γ}}_{F012}{l}_{\mathrm{mf}}\right)\·\{1+\frac{\tanh \left({\mathrm{\γ}}_{F012}{l}_{\mathrm{mf}}\right)}{\tanh \left[{\mathrm{\γ}}_{F012}(l_{\mathrm{mn}}-l_{\mathrm{mf}})\right]}\}---\left(10\right)$

from equation (10), the reverse sequence current at fault is ${\stackrel{\·}{I}}_{\mathrm{fFi}}=\cosh \left({\mathrm{\γ}}_{\mathrm{Fi}}{l}_{\mathrm{mf}}\right)\·[1+\frac{\tanh \left({\mathrm{\γ}}_{\mathrm{Fi}}{l}_{\mathrm{mf}}\right)}{\tanh \left[{\mathrm{\γ}}_{\mathrm{Fi}}(l_{\mathrm{mn}}-l_{\mathrm{mf}})\right]}]\·{\stackrel{\·}{I}}_{\mathrm{mFi}}.$

Wherein, i is 0,1,2, respectively represent a zero sequence, a positive sequence and a negative sequence,for reverse sequence current at fault, gamma_FiFor reversing the propagation coefficients of the respective sequence nets,/_mnIs the distance between two ends of the double-circuit fault line, l_mfThe distance from the acquisition end of the double-circuit fault line to the fault position is shown, cosh () is a hyperbolic cosine function, tanh () is a hyperbolic tangent function,the sequence current of the acquisition end in each sequence network is reversed. To be provided withPhase estimationThe phase error curve of (2) is shown in FIG. 4, considering that the negative sequence net and the positive sequence net have the same parameters, soPhase estimationThe phase error curve is the same as the positive sequence net. In the same way, withPhase estimationThe phase error curve is shown in fig. 5, and it can be seen that the maximum error does not exceed 0.2 °.

Location of failurePhase separate utilizationPhase estimation, noted as:

${\stackrel{\·}{I}}_{\mathrm{fF}1}\≈k_1\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}1},{\stackrel{\·}{I}}_{\mathrm{fF}2}\≈k_2\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}2}---\left(11\right)$

wherein,from equation (11), the phase of the reverse positive-negative sequence current at the fault can be estimated from the phase of the reverse positive-negative sequence current at the acquisition end M:

$\begin{array}{c}\arg (a_1{\stackrel{\·}{I}}_{\mathrm{fF}1}+a_2{\stackrel{\·}{I}}_{\mathrm{fF}2})\≈\arg [a_1{k}_1\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}1}+a_2{k}_2\left(l_{\mathrm{mf}}\right){\stackrel{\·}{I}}_{\mathrm{mF}2}]\\ =\arg \left[k_1\left(l_{\mathrm{mf}}\right)(a_1{\stackrel{\·}{I}}_{\mathrm{mF}1}+a_2{\stackrel{\·}{I}}_{\mathrm{mF}2})\right]\\ =\arg \left[(a_1{\stackrel{\·}{I}}_{\mathrm{mF}1}+a_2{\stackrel{\·}{I}}_{\mathrm{mF}2})\right]\end{array}---\left(12\right)$

wherein, a₁、a₂Is any real number.

(ii) And secondly, calculating the voltage phase at the fault position by utilizing the reverse sequence current at the fault position according to the boundary condition of the double-circuit line crossing fault and the six-sequence component of the double-circuit line current.

In this example, IA (phase A of I loop) is selected as special phase, and fault boundary six-phase current I_fAT,FDecoupled into six-sequence component I by T-matrix_fAI,II：

I_fAT,F＝TI_fAI,II(13)

Wherein, $I_{\mathrm{fAT},F}={\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_{\mathrm{fAT}1}& {\stackrel{\·}{I}}_{\mathrm{fAT}2}& {\stackrel{\·}{I}}_{\mathrm{fAT}0}& {\stackrel{\·}{I}}_{\mathrm{fAT}1}& {\stackrel{\·}{I}}_{\mathrm{fAF}2}& {\stackrel{\·}{I}}_{\mathrm{fAF}0}\end{array}\right]}^T;$ $I_{\mathrm{fAI},\mathrm{II}}={\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_{\mathrm{fAI}}& {\stackrel{\·}{I}}_{\mathrm{fBI}}& {\stackrel{\·}{I}}_{\mathrm{fCI}}& {\stackrel{\·}{I}}_{\mathrm{fAII}}& {\stackrel{\·}{I}}_{\mathrm{fBII}}& {\stackrel{\·}{I}}_{\mathrm{fCII}}\end{array}\right]}^T.$

the relationship between the phase of the voltage at the fault and the phase of the reverse-sequence current can be derived from the boundary conditions in the case of a two-wire two-phase fault, a three-wire two-phase fault and a three-wire three-phase fault and equation (13).

(A) For a two-wire two-phase overwire interphase fault:

when IBIIC (B phase of I loop and C phase of II loop) overline ungrounded fault occurs, the current conditions of the fault boundary are as follows:

$I_{\mathrm{fAI},\mathrm{II}}{\left[\begin{array}{cccccc}0& {\stackrel{\·}{I}}_{\mathrm{fBI}}& 0& 0& 0& {\stackrel{\·}{I}}_{\mathrm{fBI}}\end{array}\right]}^T---\left(14\right)$

by substituting equation (14) for equation (13), the relationship between the reverse positive sequence current and the phase current at the fault can be obtained:

${\stackrel{\·}{I}}_{\mathrm{fAF}1}=\frac{1}{6}(\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fBI}}-{\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fCII}})=\frac{1}{6}(\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fBI}}+{\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fBI}})=-\frac{1}{6}{\stackrel{\·}{I}}_{\mathrm{fBI}}---\left(15\right)$

according to the fault type, the interphase voltage of two phases (I loop B phase and II loop C phase) with fault can be obtainedPhase and reverse positive sequence current ofRelationship between phases:

(B) for a two-wire two-phase overwire ground fault:

when an IBIIC (B phase of an I loop and C phase of a II loop) overline ground fault occurs, the current conditions of the fault boundary are as follows:

$I_{\mathrm{fAI},\mathrm{II}}{\left[\begin{array}{cccccc}0& {\stackrel{\·}{I}}_{\mathrm{fBI}}& 0& 0& 0& {\stackrel{\·}{I}}_{\mathrm{fCII}}\end{array}\right]}^T---\left(17\right)$

by substituting the formula (17) into the formula (13), the relation between the reverse zero-sequence current and the phase current at the fault can be obtained:

${\stackrel{\·}{I}}_{\mathrm{fAF}0}=\frac{1}{6}({\stackrel{\·}{I}}_{\mathrm{fBI}}-{\stackrel{\·}{I}}_{\mathrm{fCII}})---\left(18\right)$

according to the fault type, the fault two-phase interphase voltage can be obtainedPhase and reverse zero sequence currentThe phase relationship is:

$\arg \left({\stackrel{\·}{U}}_{\mathrm{fBICII}}\right)=\arg ({\stackrel{\·}{I}}_{\mathrm{fBI}}-{\stackrel{\·}{I}}_{\mathrm{fCII}})=\arg \left({\stackrel{\·}{I}}_{\mathrm{fAF}0}\right)---\left(19\right)$

(C) for three-wire two-phase overline interphase fault:

when IBCIIC (BC phase of loop I and C phase of loop II) overline ungrounded fault occurs, the current conditions of the fault boundary are as follows:

$\left\{\begin{array}{c}{I}_{\mathrm{fAI},\mathrm{II}}={\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_{\mathrm{fBI}}& 0& {\stackrel{\·}{I}}_{\mathrm{fCI}}& 0& 0& {\stackrel{\·}{I}}_{\mathrm{fCII}}\end{array}\right]}^T\\ {\stackrel{\·}{I}}_{\mathrm{fBI}}+{\stackrel{\·}{I}}_{\mathrm{fCI}}+{\stackrel{\·}{I}}_{\mathrm{fCII}}=0\end{array}\right.---\left(20\right)$

by substituting formula (20) for formula (13), the reverse positive sequence current at fault can be obtainedAnd reverse negative sequence currentExpression (c):

$\left\{\begin{array}{c}{\stackrel{\·}{I}}_{\mathrm{fAF}1}=\frac{1}{6}(\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fBI}}+{\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fCI}}-{\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fAII}})\\ {\stackrel{\·}{I}}_{\mathrm{fAF}2}=\frac{1}{6}({\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fBI}}+\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fCI}}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAII}})\end{array}\right.---\left(21\right)$

elimination of formula (21)The relationship between the reverse positive and negative sequence current and the phase current at the fault can be obtained:

${\stackrel{\·}{I}}_{\mathrm{fAF}1}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAF}2}=\frac{1}{6}(\mathrm{\α}-1){\stackrel{\·}{I}}_{\mathrm{fBI}}---\left(22\right)$

according to the fault type, the inter-phase voltage of the I loop at the fault position can be obtainedPhase of (d) and reverse positive-negative sequence current phase relationship:

$\arg \left({\stackrel{\·}{U}}_{\mathrm{fBCI}}\right)=\arg \left({\stackrel{\·}{I}}_{\mathrm{fBI}}\right)=\arg \left(\frac{{\stackrel{\·}{I}}_{\mathrm{fAF}1}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAF}2}}{\mathrm{\α}-1}\right)---\left(23\right)$

(D) for a three-wire two-phase overline ground fault:

when IBCIIC (BC phase of I loop and C phase of II loop) overline ground fault occurs, the relation between B phase fault current and reverse positive-negative sequence current represented by formula (22) is still satisfied, and at the moment, the B phase voltage of I loop at fault position is in voltageCan be controlled byPhase representation:

$\arg \left({\stackrel{\·}{U}}_{\mathrm{fBI}}\right)=\arg \left({\stackrel{\·}{I}}_{\mathrm{fBI}}\right)=\arg \left(\frac{{\stackrel{\·}{I}}_{\mathrm{fAF}1}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAF}2}}{\mathrm{\α}-1}\right)---\left(24\right)$

(E) for a three-wire two-phase non-overwire single-phase earth fault:

when the BC phase-to-phase of the I loop has an ungrounded fault, the C phase of the II loop has an earthed fault (IBC-IICG), the relation shown in the formula (22) is still satisfied, and at the moment, the voltage between the BC phases of the I loop at the fault position isCan be controlled byPhase representation:

$\arg \left({\stackrel{\·}{U}}_{\mathrm{fBCI}}\right)=\arg \left({\stackrel{\·}{I}}_{\mathrm{fBI}}\right)=\arg \left(\frac{{\stackrel{\·}{I}}_{\mathrm{fAF}1}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAF}2}}{\mathrm{\α}-1}\right)---\left(25\right)$

(F) for three-wire two-phase overline single-phase earth fault:

when IBIIC overline ungrounded fault occurs and the C phase of the I return line has grounded fault (IBIIC-ICG), the relation shown in the formula (22) is still satisfied, and the phase-to-phase voltage at the fault is still satisfied at the momentThe phases of (phase B of loop I and phase C of loop II) can be controlled byThe phase is represented as:

$\arg \left({\stackrel{\·}{U}}_{\mathrm{fBICII}}\right)=\arg \left({\stackrel{\·}{I}}_{\mathrm{fBI}}\right)=\arg \left(\frac{{\stackrel{\·}{I}}_{\mathrm{fAF}1}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAF}2}}{\mathrm{\α}-1}\right)---\left(26\right)$

(G) for a three-wire three-phase overline interphase fault:

when an IBCIIA overline ungrounded fault occurs, the current conditions of the fault boundary are as follows:

$I_{\mathrm{fAI},\mathrm{II}}={\left[\begin{array}{cccccc}0& {\stackrel{\·}{I}}_{\mathrm{fBI}}& {\stackrel{\·}{I}}_{\mathrm{fCI}}& {\stackrel{\·}{I}}_{\mathrm{fAII}}& 0& 0\end{array}\right]}^T---\left(27\right)$

by substituting equation (27) for equation (13), expressions of reverse positive-sequence current and reverse negative-sequence current at fault can be obtained:

$\left\{\begin{array}{c}{\stackrel{\·}{I}}_{\mathrm{fAF}1}=\frac{1}{6}(\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fBI}}+{\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fCI}}-{\stackrel{\·}{I}}_{\mathrm{fAII}})\\ {\stackrel{\·}{I}}_{\mathrm{fAF}2}=\frac{1}{6}({\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fBI}}+\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fCI}}-{\stackrel{\·}{I}}_{\mathrm{fAII}})\end{array}\right.---\left(28\right)$

by elimination in formula (28)The relationship between the reverse positive and negative sequence current and the phase current at the fault can be obtained:

${\stackrel{\·}{I}}_{\mathrm{fAF}1}-{\stackrel{\·}{I}}_{\mathrm{fAF}2}=\frac{1}{6}(\mathrm{\α}-{\mathrm{\α}}^2)({\stackrel{\·}{I}}_{\mathrm{fBI}}-{\stackrel{\·}{I}}_{\mathrm{fCI}})=\frac{1}{6}j\sqrt{3}({\stackrel{\·}{I}}_{\mathrm{fBI}}-{\stackrel{\·}{I}}_{\mathrm{fCI}})---\left(29\right)$

according to the fault type, the inter-phase voltage of the I loop at the fault position can be obtainedPhase of (d) and reverse positive-negative sequence current phase relationship:

(H) for a three-wire three-phase overline ground fault:

when an IBCIIA overline ground fault occurs, the phase relation (30) derived in the IBCIIA three-wire three-phase overline interphase fault is still true.

(I) For a three-wire three-phase non-overwire single-phase earth fault:

when the phase-to-phase ungrounded fault occurs in the BC phase of the loop I and the phase A of the loop II has a grounded fault (IBC-IIAG), the following equations (27) are satisfied at the same time:

${\stackrel{\·}{I}}_{\mathrm{fBI}}+{\stackrel{\·}{I}}_{\mathrm{fCI}}=0---\left(31\right)$

substituting equation (31) into (12) can obtain the relation between the reverse positive-negative sequence current and the phase current at the fault:

${\stackrel{\·}{I}}_{\mathrm{fAF}1}-{\stackrel{\·}{I}}_{\mathrm{fAF}2}=\frac{1}{3}(\mathrm{\α}-{\mathrm{\α}}^2){\stackrel{\·}{I}}_{\mathrm{fBI}}=\frac{1}{3}j\sqrt{3}{\stackrel{\·}{I}}_{\mathrm{fBI}}---\left(32\right)$

according to the fault type, the inter-phase voltage of the I loop at the fault position can be obtainedPhase of (d) and reverse positive-negative sequence current phase relationship:

it should be noted that the above-mentioned (a) - (I) give specific examples of calculating the voltage phase at the fault according to different types of double-circuit fault lines with IA as a special phase. For the same fault type, the fault line phase is different, the selected special phase is different, and the current condition of the fault boundary is different. For example, when the two-wire two-phase overline interphase fault in (a) occurs between the a phase of the I loop and the C phase of the II loop, that is, when the IAIIC overline non-ground fault occurs, IB is selected as the special phase, and the current condition of the fault boundary condition is $I_{\mathrm{fI},\mathrm{II}}={\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_{\mathrm{fAI}}& 0& 0& 0& 0& -{\stackrel{\·}{I}}_{\mathrm{fAI}}\end{array}\right]}^T$ The relation between reverse sequence current and phase current at fault can be obtained by substituting formula (13) ${\stackrel{\·}{I}}_{\mathrm{fBF}1}=\frac{1}{6}({\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fAI}}-\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fCII}})=\frac{1}{6}({\mathrm{\α}}^2{\stackrel{\·}{I}}_{\mathrm{fAI}}+\mathrm{\α}{\stackrel{\·}{I}}_{\mathrm{fAI}})=-\frac{1}{6}{\stackrel{\·}{I}}_{\mathrm{fAI}},$ According to the fault type, the interphase voltage of two phases (I loop A phase and II loop C phase) with faults can be obtainedPhase and reverse positive sequence current ofRelationship between phases:for another example, in the three-wire two-phase overline interphase fault in (C), when the IACIIC overline ungrounded fault occurs, IB is used as the special phase, and the current condition at the fault boundary is set as $\left\{\begin{array}{c}{I}_{\mathrm{fI},\mathrm{II}}={\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_{\mathrm{fAI}}& 0& {\stackrel{\·}{I}}_{\mathrm{fCI}}& 0& 0& {\stackrel{\·}{I}}_{\mathrm{fCII}}\end{array}\right]}^T\\ {\stackrel{\·}{I}}_{\mathrm{fAI}}+{\stackrel{\·}{I}}_{\mathrm{fCI}}+{\stackrel{\·}{I}}_{\mathrm{fCII}}=0\end{array}\right.,$ The relation between the reverse sequence current and the phase current at the fault can be obtained by substituting formula (13)Available fault phase-to-phase voltageRelation to reverse sequence current phaseAfter the fault type is obtained, determining the current condition of the fault boundary according to the fault type, selecting a corresponding special phase, and deducing by utilizing six-sequence components of the current of the double-circuit line to obtain the fault voltage phaseThe relationship between the reverse sequence current and the fault voltage is a common technique for those skilled in the art, and for reasons of space, the relationship between the fault voltage phase and the reverse sequence current is derived only for the IA as the special phase in various fault types, and the derivation process in other cases is not described in detail.

(iii) And finally, calculating the voltage amplitude of the fault position according to the M terminal voltage and the current. According to the phasor relationship shown in fig. 6, the voltage amplitude expression at the fault can be obtained:

wherein, U_fFor single-phase voltage at faultOr voltage between phasesOr voltage between phases of overline Is the line impedance angle, theta isAndis gamma ofAndthe included angle of (a).

And 4, step 4: and the fault positioning module constructs a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifies the fault position by solving the phase catastrophe point of the fault positioning function.

According to a function of the compensation voltageAnd estimated fault location voltageFrom equations (3) and (6), the following relationships can be obtained:

$\frac{{\stackrel{\·}{U}}_{\mathrm{op}}\left(l_{\mathrm{mx}}\right)-{\stackrel{\·}{U}}_f\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right)}{{\stackrel{\·}{I}}^{\′}}=z_{c1}\sinh \left[{\mathrm{\γ}}_1(l_{\mathrm{mf}}-l_{\mathrm{mx}})\right]---\left(35\right)$

constructing a fault location function f (l)_mx)：

$f\left(l_{\mathrm{mx}}\right)=\frac{{\stackrel{\·}{U}}_{\mathrm{op}}\left(l_{\mathrm{mx}}\right)-{\stackrel{\·}{U}}_f\cosh \left({\mathrm{\γ}}_1{l}_{\mathrm{mx}}\right)}{{\stackrel{\·}{I}}^{\′}}---\left(36\right)$

Set fault distance l_mfThe phase characteristics of the fault localization function are shown in fig. 7 for 100 km.

As can be seen from FIG. 7, when l_mx<l_mfWhen the phase of the fault localization function is around 90 DEG, when_mx>l_mfWhen the phase of the fault localization function is around-90 DEG, when_mx＝l_mfThe phase of the fault locating function is 0. Thus, the fault localization function phase is at the fault distance l_mfThe method has the following specific sub-steps that (1) sudden change occurs, and only one sudden change point exists, so that the fault position can be identified by solving the phase sudden change point of the fault positioning function:

substep B1: and equally dividing the double-circuit fault line n into equal parts, wherein n is a set value.

Substep B2: and calculating the phase of the fault locating function of the end points at the two sides of each equally divided interval according to the fault locating function. And if the phases of the fault positioning functions at the end points on the two sides of the equal partition interval are one larger than zero and one smaller than zero, judging that the phase mutation point is in the equal partition interval.

Substep B3: and starting from the end point on the left side of the equal division where the phase catastrophe point is located, selecting one point every set step length delta l and calculating the phase of the fault positioning function of the selected point.

Substep B4: taking a point with the phase of the first fault positioning function being less than 0 as a reference point which is far away from the acquisition end l_f+If the fault position is far from the acquisition end, it is l_f＝l_f+-0.5·Δl。

The number n of equal partitions in the substep B1 is reasonably selected, and only a certain reduced equal partition is needed to search for a phase discontinuity. Therefore, on the premise of not influencing the fault location precision, the operation time required by the fault location algorithm is shortened to 1/n of the original operation time, and the solving speed is high.

Example 1

Fig. 8 shows a simulation model of a two-terminal system with a voltage level of 330kV and a length of 300km, which was constructed using PSCAD software.

The phase angle difference of the equivalent power supplies at the two ends of the circuit MN is 1.05 and 1pu respectively. Two-sided system parameters of：Z_M1＝1.0515+j43.1749Ω，Z_M0＝0.6+j29.0911Ω，Z_N1＝26+j44.9185ΩΩ，Z_N0＝20+j37.4697Ω。

The single loop positive (negative) sequence parameters are: r₁＝0.05468Ω/km，L₁＝1.0264mH/km，C₁＝0.01095μF/km。

The zero sequence parameter of the single circuit is as follows: r₀＝0.2931Ω/km，L₀＝3.9398mH/km，C₀＝00.05473μF/km。

The zero sequence mutual impedance parameters of the double circuit lines are as follows: r_m0＝0.2385Ω/km，L_m0＝2.6274mH/km，C_m0＝0.00026μF/km。

Defining fault location relative error:

under the conditions of the two-wire two-phase overline ungrounded fault, the three-wire two-phase overline ungrounded fault and the three-wire three-phase overline ungrounded fault, the influence degree of different fault positions on the fault positioning precision is shown in fig. 9, and it can be seen that the relative distance measurement error reaches the maximum at 290km, and the maximum relative errors are respectively 0.055%, 0.0517% and 0.055%, and are not more than 0.1%.

Fig. 10 shows the degree of influence of different fault positions on the fault location accuracy in the case of a two-phase overline ground fault. Fig. 11 shows the degree of influence of different fault positions on the fault location accuracy in the case of a three-wire two-phase overline ground fault and a three-wire three-phase overline ground fault. As can be seen from fig. 10 and 11, the relative distance measurement error increases with the increase of the fault distance in each fault condition, and the absolute values of the maximum relative distance measurement error are 0.2283%, 0.0783% and 0.125%, respectively, and do not exceed 0.3%.

The influence degree of the fault position on the fault positioning accuracy under the conditions of the three-wire two-phase non-overline single-phase ground fault, the three-wire two-phase overline single-phase ground fault and the three-wire three-phase non-overline single-phase ground fault is as shown in fig. 12, and as can be seen from fig. 12, the maximum relative distance measurement errors are respectively 0.0783%, 0.0783% and 0.055%, and are not more than 0.1%.

At 290km from the measurement (acquisition) position, two-wire two-phase overline ground fault occurs, the influence degree of the transition resistance on the fault positioning precision is shown in fig. 13, and it can be seen that when the transition resistance is 300 Ω, the relative distance measurement error is the largest, and the absolute value is 2.7717%.

Fig. 14 shows the influence degree of the transition resistance on the positioning accuracy when a three-wire two-phase overline ground fault and a three-wire three-phase overline ground fault occur at a distance measurement position 290 km. Similarly, when the transition resistance is 300 Ω, the absolute value of the relative range error is also the largest, 1.2683% and 1.6183%, respectively.

When a three-wire two-phase non-overline single-phase ground fault, a three-wire two-phase overline single-phase ground fault and a three-wire three-phase non-overline single-phase ground fault occur at 290km from the protection measurement position, the influence degree of the transition resistance on the fault positioning accuracy is as shown in fig. 15, and it can be known from fig. 15 that the maximum relative distance measurement errors are 0.095%, 0.215% and 0.055%, respectively. Therefore, among various faults connected to the ground via the transition resistor, the absolute value of the maximum relative error is only 2.7717% and does not exceed 3%.

The above description is only for the preferred embodiment of the present invention, but the scope of the present invention is not limited thereto, and any changes or substitutions that can be easily conceived by those skilled in the art within the technical scope of the present invention are included in the scope of the present invention. Therefore, the protection scope of the present invention shall be subject to the protection scope of the claims.

Claims (10)

1. A double-circuit line fault single-end positioning system based on distributed parameters is characterized by comprising: the device comprises a data acquisition module, a compensation voltage solving module, a fault voltage solving module and a fault positioning module;

the data acquisition module is respectively connected with the compensation voltage solving module and the fault voltage solving module; the fault positioning module is respectively connected with the compensation voltage solving module and the fault position voltage solving module;

the data acquisition module is used for acquiring voltage phasor and current phasor at any end of the double-circuit fault line and respectively sending the acquired data to the compensation voltage solving module and the fault voltage solving module;

the compensation voltage solving module is used for solving a homodromous zero-sequence compensation coefficient function and a reverse zero-sequence compensation coefficient function of the double-circuit fault line according to a six-sequence network distribution parameter model of the double-circuit fault line, determining a compensation voltage function of the double-circuit fault line according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function, and sending the compensation voltage function of the double-circuit fault line to the fault positioning module;

the fault voltage solving module is used for solving the reverse sequence current of the fault, calculating the voltage phase of the fault according to the reverse sequence current of the fault, solving the voltage amplitude of the fault by using the acquired voltage phasor and current phasor to obtain the voltage of the fault, and then sending the voltage of the fault to the fault positioning module;

the fault positioning module is used for constructing a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifying the fault position by solving the phase catastrophe point of the fault positioning function.

2. A double-circuit fault single-end positioning method based on distribution parameters is characterized by comprising the following steps:

step 1: collecting voltage phasor and current phasor at any end of a double-circuit fault line;

step 2: solving a homodromous zero-sequence compensation coefficient function and a reverse zero-sequence compensation coefficient function of the double-circuit fault line according to the six-sequence network distribution parameter model of the double-circuit fault line, and determining a compensation voltage function of the double-circuit fault line according to the homodromous zero-sequence compensation coefficient function and the reverse zero-sequence compensation coefficient function;

and step 3: solving reverse sequence current at the fault; calculating a voltage phase at the fault according to the reverse sequence current at the fault, and solving a voltage amplitude at the fault by using the acquired voltage phasor and current phasor to obtain a voltage at the fault;

and 4, step 4: and constructing a fault positioning function according to the compensation voltage function of the double-circuit fault line and the voltage at the fault, and identifying the fault position by solving the phase catastrophe point of the fault positioning function.

3. The method as claimed in claim 2, wherein the function for solving the co-directional zero-sequence compensation coefficient of the double-circuit fault line adopts the formula:

$k_T\left(l_{mf}\right)=\frac{z_{cT0}\sinh \left({\mathrm{\&gamma;}}_{T0}{l}_{mf}\right)-z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)}{z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)}+\frac{Z_{mT0}\cosh \left({\mathrm{\&gamma;}}_{T0}{l}_{mf}\right)-Z_{mT0}\cosh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)}{z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)};$

wherein l_mfThe distance from the acquisition end of each sequence network in the six sequence networks of the double-circuit fault line to the fault position;

z_cT0characteristic impedance of a same-direction zero sequence network in a six sequence network of a double-circuit fault line;

γ_T0the transmission coefficient of a homodromous zero sequence network in a six sequence network of a double-circuit fault line;

z_c1is the characteristic impedance of a forward sequence network in the same direction;

γ₁the forward sequence network propagation coefficient is the same direction;

Z_mT0is the same-direction zero sequence impedance;

sinh () is a hyperbolic sine function;

cosh () is a hyperbolic cosine function.

4. The method as claimed in claim 3, wherein the inverse zero sequence compensation coefficient function for solving the double-circuit fault line adopts the formula:

$k_F\left(l_{mf}\right)=\frac{z_{cF0}\sinh \left({\mathrm{\&gamma;}}_{F0}{l}_{mf}\right)-z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)}{z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mf}\right)};$

wherein z is_cF0Is a reverse zero sequence net characteristic impedance;

γ_F0is a reverse zero-order net propagation coefficient.

5. The method as claimed in claim 4, wherein the determining the compensation voltage function of the double-circuit fault line according to the same-direction zero-sequence compensation coefficient function and the reverse-direction zero-sequence compensation coefficient function adopts the formula:

${\stackrel{\&CenterDot;}{U}}_{op}\left(l_{mx}\right)=\stackrel{\&CenterDot;}{U}\cosh \left({\mathrm{\&gamma;}}_1{l}_{mx}\right)-z_{c1}\sinh \left({\mathrm{\&gamma;}}_1{l}_{mx}\right){\stackrel{\&CenterDot;}{I}}^{\&prime;};$

wherein,is a compensation voltage function of the double-circuit fault line and is used for calculating the distance between the acquisition end l and the double-circuit fault line_mxA compensation voltage of;

single phase voltage as a return lineInterphase voltage of same loopOr voltage between phases of overline

When in useSingle phase voltage as a return lineWhen the temperature of the water is higher than the set temperature, ${\stackrel{\&CenterDot;}{I}}^{\&prime;}={\stackrel{\&CenterDot;}{I}}_m+k_T\left(l_{mx}\right){\stackrel{\&CenterDot;}{I}}_{mT0}+k_F\left(l_{mx}\right){\stackrel{\&CenterDot;}{I}}_{mF0};$

when in useInter-phase voltage of the same loopWhen the temperature of the water is higher than the set temperature,

when in useAs a voltage between phases of the overlineWhen the temperature of the water is higher than the set temperature,

collecting the phase current;

the same loop line interphase current is used as the acquisition end;

cross-line interphase current is collected at the end;

sequence current of a homodromous zero sequence network acquisition end;

the sequence current of the reverse zero sequence network acquisition end.

6. The method of claim 5, wherein said solving for reverse sequence currents at fault uses the formula:

${\stackrel{\&CenterDot;}{I}}_{fFi}=\cosh \left({\mathrm{\&gamma;}}_{Fi}{l}_{mf}\right)\&CenterDot;\&lsqb;1+\frac{\tanh \left({\mathrm{\&gamma;}}_{Fi}{l}_{mf}\right)}{\tanh \&lsqb;{\mathrm{\&gamma;}}_{Fi}(l_{mn}-l_{mf})\&rsqb;}\&rsqb;\&CenterDot;{\stackrel{\&CenterDot;}{I}}_{mFi};$

wherein, i is 0,1,2, respectively representing zero sequence, positive sequence and negative sequence;

is a reverse sequence current at fault;

γ_Fithe propagation coefficients of all reverse order nets;

l_mnthe distance between two ends of the double-circuit fault line;

tanh () is a hyperbolic tangent function;

the sequence current of the acquisition end in each sequence network is reversed.

7. The method of claim 6, wherein said calculating the voltage phase at the fault from the reverse sequence current at the fault is by:

substep A1: determining the fault boundary current condition of the double-circuit fault line according to the fault type of the double-circuit fault line;

substep A2: and calculating the voltage phase at the fault according to the fault boundary current condition of the double-circuit fault line and the reverse sequence current at the fault.

8. The method of claim 7, wherein the using the collected voltage phasor and current phasor to solve for the voltage magnitude at the fault is by the formula:

wherein, U_fIs the fault location voltage amplitude;

single phase voltage as a return lineInterphase voltage of same loopOr voltage between phases of overline

The impedance angle of the double-circuit fault line;

theta isAndthe included angle of (A);

gamma isAndthe angle of,is the fault location voltage.

9. The method of claim 8, wherein the fault localization function is:

$f\left(l_{mx}\right)=\frac{{\stackrel{\&CenterDot;}{U}}_{op}\left(l_{mx}\right)-{\stackrel{\&CenterDot;}{U}}_f\cosh \left({\mathrm{\&gamma;}}_1{l}_{mx}\right)}{{\stackrel{\&CenterDot;}{I}}^{\&prime;}}.$

10. the method of claim 9, wherein said identifying a fault location by solving for a phase discontinuity of a fault localization function specifically comprises:

substep B1: dividing the double-circuit fault line n into equal parts, wherein n is a set value;

substep B2: calculating the phases of the fault locating functions of the end points at the two sides of each equally divided interval according to the fault locating functions, and if the phases of the fault locating functions of the end points at the two sides of the equally divided intervals are one larger than zero and one smaller than zero, judging that phase mutation points are in the equally divided intervals;

substep B3: starting from an endpoint on the left side of an equal division interval where a phase catastrophe point is located, selecting a point every set step length delta l and calculating the phase of a fault positioning function of the selected point;

substep B4: taking a point with the phase of the first fault positioning function being less than zero as a reference point, wherein the reference point is far away from the acquisition end l_f+If the fault position is far from the acquisition end, it is l_f＝l_f+-0.5·Δl。