CN105021955A - Method for calculating fault distance of series compensation circuit - Google Patents

Abstract

The invention discloses a method for calculating a fault distance of a series compensation circuit, and the method comprises the steps: 1, setting a voltage loop equation for a fault phase, which means that a fault phase end voltage is equal to the sum of the voltage drop at a fault resistor R, the voltage drop at a fault inductor L and the voltage drop at a fault capacitor during fault duration; 2, carrying out the integration of the voltage loop equation set at the step 1 in a time period of the fault duration; 3, respectively selecting at least four time periods in the fault duration after the moment t0, solving the integration equation, set at step 3, through the setting of an equation set, and obtaining a fault inductance L; 4, obtaining a fault reactance X through the fault inductance L and an angular frequency omega of a power transmission line; 5, calculating a fault distance D through the fault reactance X. The method gives full consideration to various conditions of actual series compensation circuit faults, gets rid of various types of unreasonable hypothesis in conventional fault distance measurement, is more accurate in fault distance measurement, and is not affected by the voltage and current reversion caused by a series compensation capacitor and the nonlinearity of an MOV.

Description

The computing method of Series compensation lines fault distance

Technical field:

The present invention relates to technical field of power systems, particularly a kind of computing method of Series compensation lines fault distance.

Background technology:

Series Compensated Transmission Lines is series in transmission line of alternation current by Capacitor banks, can improve transmission line of electricity power limit and stability of power system by the reactance compensating transmission line of alternation current.Serial compensation capacitance prevents superpotential by non-linear resistance (MOV) in parallel and portable protective gaps (GAP) electric discharge, and reverse and the non-linear of MOV of the voltage that serial compensation capacitance causes, electric current makes traditional fault distance-finding method no longer applicable.

Current Series compensation lines fault localization can adopt traveling wave method, estimate serial compensation capacitance voltage, both-end distance measuring etc., but traveling wave method too relies on the detection of wave head, and the wave head of reflection wave is difficult to detect, easily range finding by mistake, thus affects the reliability of fault localization; Estimate that serial compensation capacitance voltage independent is in the determination of fault initial time, because serial compensation capacitance may to be discharged short circuit by GAP because of superpotential when fault, so estimate the prerequisite of serial compensation capacitance voltage be serial compensation capacitance not by short circuit, therefore the range of application of this method has significant limitation; Both-end distance measuring adds the complexity of communication, and especially remote ultra-high-tension power transmission line two end data exists time delay, thus the effect of impact range finding.

Summary of the invention:

In view of this, the computing method that the reliable Series compensation lines fault distance of a kind of accurate stable is provided are necessary.

Computing method for Series compensation lines fault distance, comprising:

Step one: to fault phase column voltage loop equation, the pressure drop sum of voltage respectively in the pressure drop, fault capacitance C of the pressure drop of fault resstance R, fault inductance L duration that namely fault phase terminal voltage equaling fault; Wherein, the pressure drop on fault capacitance C is: wherein, i _bfor the electric current on fault capacitance C duration of fault, t0 be before fault or fault duration in any instant, u ₀for the voltage on t0 moment fault capacitance C;

It should be noted that u ₀there is material impact for result of calculation, establish serial compensation capacitance voltage initial time (i.e. t0 moment) voltage u here ₀for unknown number, thus do not need u ₀judge with additive method, then by subsequent step, the equation after integration is solved, solve four unknown numbers: R, L, C, u ₀.

And in the method that Series compensation lines fault localization adopts at present, usually ignore serial compensation capacitance voltage initial time voltage u ₀, or hypothesis u ₀be zero, this is all the computation model simplified, and does not conform to, the error of net result must be caused larger with actual conditions.

In the method that current Series compensation lines fault localization adopts, in order to simplified model, also usually assumed fault electric capacity C is constant, or assumed fault electric capacity C is 0, and the error that the result of calculation of this hypothesis to fault localization causes is very large.Fault capacitance C has three duties: 1., the electric current of C below 2 times of ratings time, MOV and GAP can not affect C; 2., the electric current of C more than 2 times of rated current, MOV action, the voltage of restriction C is within 2.5 times of rated voltages; 3., the electric current of C too high time, the GAP electric discharge of C, C, by short circuit, does not at this moment have electric current to pass through C.In general, when line fault, cannot anticipation by the electric current of C, therefore, the duty of C is any in three kinds is also unknown; If suppose C not by short circuit, and in fact C is by short circuit, and the fault distance obtained is certainly not right; Otherwise suppose that C is by short circuit, in fact C is not by short circuit, solving at this moment is not right yet.Fault capacitance C is classified as unknown number by this method, is solved the equation after integration by subsequent step, has well met various objective circumstances, therefore result of calculation can be very accurate.

Step 2: on a period of time upper integral of the voltage circuit equation listed by step one in trouble duration with the impact of harmonic carcellation;

Step 3: be taken to few four periods in trouble duration and after the t0 moment respectively, solves the group that establishes an equation of the integral equation listed by step 3, obtains fault inductance L;

Step 4: try to achieve fault reactance X by fault inductance L and transmission line of electricity angular frequency, namely

X＝ωL (1)；

Wherein, ω=2 π f, f are the electrical network power frequency of transmission line of electricity;

Step 5: calculate fault distance D by fault reactance X:

$D=\frac{X}{x_1}---\left(2\right);$

Wherein, x ₁for the forward-sequence reactance of faulty line unit length.

Preferably, for single back line singlephase earth fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\φ}}=R(i_{\mathrm{\φ}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\φ}}+3k_x{i}_0)}{dt}+\frac{1}{C}{\∫}_{t0}^t{i}_{\mathrm{\φ}}dt+u_0---\left(3\right);$

Wherein, u _φ, i _φ, i ₀for the voltage current waveform sampling of line protective devices to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence; L is faulty line positive sequence inductance;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (3) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\∫}_{tm}^{tn}{u}_{\mathrm{\φ}}dt=R{\∫}_{tm}^{tn}(i_{\mathrm{\φ}}+3k_r{i}_0)dt+L(i_{\mathrm{\φ}}+3k_x{i}_0){|}_{tm}^{tn}+\frac{1}{C}{\∫}_{tm}^{tn}{\∫}_{t0}^t{i}_{\mathrm{\φ}}dt+u_0{\∫}_{tm}^{tn}dt---\left(4\right);$

Tm, tn are the moment in trouble duration, and t0<tm, $(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _φ[n] represents i _φin the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_1=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_1=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_1=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}}dt\end{array}\right]$

B ₁＝A ₁ ^-1C ₁(5)。

Preferably, for mutually indirectly or phase to phase fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\&phi;}\mathrm{\&psi;}}=R\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)+L\frac{d\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0---\left(6\right);$

Wherein, u _{φ ψ}, i _{φ ψ}for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ, ψ are two fault phases; L is faulty line positive sequence inductance;

u _φψ＝u _φ-u _ψ；i _φψ＝i _φ-i _ψ

Step 2: (6) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt=R{\&Integral;}_{tm}^{tn}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+{\mathrm{Li}}_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(7\right);$

Tm, tn are the moment in trouble duration, and t0<tm, wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _{φ ψ}[n] represents i _{φ ψ}in the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_2=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_2=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_2=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₂＝A ₂ ^-1C ₂(8)。

Preferably, for parallel double back serial supplementary line singlephase earth fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\&phi;}}=R(i_{\mathrm{\&phi;}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\&phi;}}+3k_x{i}_0)}{dt}+L_m\frac{{\mathrm{di}}_{30}}{dt}+\frac{1}{C}{\&Integral;}_{t0}^i{i}_{\mathrm{\&phi;}}dt+u_0---\left(9\right);$

Wherein, u _φ, i _φ, i ₀, i ₃₀for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence, and 30 represent an adjacent loop line road zero sequence, L _mfor double-circuit line protective device is to the zero-sequence mutual inductance of trouble spot;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (9) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}}dt=R{\&Integral;}_{tm}^{tn}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt+L(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}+L_m\left(i_{30}\right){|}_{tm}^{tn}\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(10\right);$

Tm, tn are the moment in trouble duration, and t0<tm;

$i_{30}{|}_{tm}^{tn}=i_{30}\&lsqb;n\&rsqb;-i_{30}\&lsqb;m\&rsqb;$

$(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively ₃₀[n] represents i ₃₀in the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_3=\left[\begin{array}{ccccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t1}^{t2}& \left(i_{30}\right){|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t3}^{t4}& \left(i_{30}\right){|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t5}^{t6}& \left(i_{30}\right){|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t7}^{t8}& \left(i_{30}\right){|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\\ {\&Integral;}_{t9}^{t10}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t9}^{t10}& \left(i_{30}\right){|}_{t9}^{t10}& {\&Integral;}_{t9}^{t10}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t9}^{t10}dt\end{array}\right]$

$B_3=\left[\begin{array}{c}R\\ L\\ L_m\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_{\text{3}}=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t9}^{t10}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₃＝A ₃ ^-1C ₃(11)。

Method of the present invention adopts single end distance measurement; and consider the various situations of actual Series compensation lines fault comprehensively; abandon all unreasonable hypothesis of conventional failure range finding; the fault distance that this method is calculated is more accurate; the voltage do not caused by series compensation capacitance, electric current oppositely and the nonlinear of MOV affect, thus provide strong basis for the correctness of distance protection action.

Accompanying drawing illustrates:

Accompanying drawing 1 is the circuit diagram of single time Series compensation lines fault.

Accompanying drawing 2 is circuit diagrams of double back Series compensation lines fault.

Accompanying drawing 3 is the local circuit schematic diagram of fault distance D in accompanying drawing 1,2.

In figure: M, N are the two ends of circuit and M/N end is equipped with line protective devices respectively; S1, S2 are voltage source; k is trouble spot; D represents the fault distance from line protective devices to trouble spot; TA, TV are mutual inductor; C is series compensation capacitance, and MOV is the non-linear resistance of protection series compensation capacitance, and GAP is the discharging gap for the protection of series compensation capacitance.

Embodiment:

The implication of term once as used in this specification is:

The distance of fault distance D: trouble spot k to M end or N end.

Fault resstance R: be fault distance D internal fault line equivalent resistance.

Fault inductance L: be the inductance comprised in fault distance D.

Fault capacitance C: be the series compensation capacitance comprised in fault distance D.

Embodiment 1

As shown in Figure 1, a kind of computing method of Series compensation lines fault distance, for single back line singlephase earth fault, comprising:

Step one: to fault phase column voltage loop equation:

$u_{\mathrm{\&phi;}}=R(i_{\mathrm{\&phi;}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\&phi;}}+3k_x{i}_0)}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0\text{---}\left(3\right);$

Wherein, u _φ, i _φ, i ₀for the voltage current waveform sampling of line protective devices to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence; L is faulty line positive sequence inductance;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (3) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}}dt=R{\&Integral;}_{tm}^{tn}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt+L(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(4\right);$

Tm, tn are the moment in trouble duration, and t0<tm, $(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _φ[n] represents i _φin the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_1=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_1=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_{\text{1}}=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}}dt\end{array}\right]$

B ₁＝A ₁ ^-1C ₁(5)；

Step 4: try to achieve fault reactance X by fault inductance L and transmission line of electricity angular frequency, namely

X＝ωL (1)；

Wherein, ω=2 π f, f are the electrical network power frequency of transmission line of electricity;

Step 5: calculate fault distance D by fault reactance X:

$D=\frac{X}{x_1}\text{---}\left(2\right);$

Wherein, x ₁for the forward-sequence reactance of faulty line unit length.

It should be noted that: in the method that Series compensation lines fault localization adopts at present, in order to simplified model, also usually assumed fault electric capacity C is constant, or assumed fault electric capacity C is 0, and the error that the result of calculation of this hypothesis to fault localization causes is very large.Fault capacitance C has three duties: 1., the electric current of C below 2 times of ratings time, MOV and GAP can not affect C; 2., the electric current of C more than 2 times of rated current, MOV action, the voltage of restriction C is within 2.5 times of rated voltages; 3., the electric current of C too high time, the GAP electric discharge of C, C, by short circuit, does not at this moment have electric current to pass through C.In general, when line fault, cannot anticipation by the electric current of C, therefore, the duty of C is any in three kinds is also unknown; If suppose C not by short circuit, and in fact C is by short circuit, and the fault distance obtained is certainly not right; Otherwise suppose that C is by short circuit, in fact C is not by short circuit, solving at this moment is not right yet.Fault capacitance C is classified as unknown number by this method, is solved the equation after integration by subsequent step, has well met various objective circumstances, therefore result of calculation can be very accurate.

Embodiment 2

Computing method for Series compensation lines fault distance, for mutually indirectly or phase to phase fault:

Step one: to fault phase column voltage loop equation:

$u_{\mathrm{\&phi;}\mathrm{\&psi;}}=R\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)+L\frac{d\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0---\left(6\right);$

Wherein, u _{φ ψ}, i _{φ ψ}for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ, ψ are two fault phases, and L is faulty line positive sequence inductance;

u _φψ＝u _φ-u _ψ；i _φψ＝i _φ-i _ψ

Step 2: (6) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt=R{\&Integral;}_{tm}^{tn}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+{\mathrm{Li}}_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(7\right);$

Tm, tn are the moment in trouble duration, and t0<tm, wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _{φ ψ}[n] represents i _{φ ψ}in the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_2=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& i_{\mathrm{\&phi;}\mathrm{\&psi;}}{|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_2=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_2=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₂＝A ₂ ^-1C ₂(8)；

Step 4: try to achieve fault reactance X by fault inductance L and transmission line of electricity angular frequency, namely

X＝ωL (1)；

Wherein, ω=2 π f, f are the electrical network power frequency of transmission line of electricity;

Step 5: calculate fault distance D by fault reactance X:

$D=\frac{X}{x_1}---\left(2\right);$

Wherein, x ₁for the forward-sequence reactance of faulty line unit length.

Embodiment 3

As shown in Figure 2, a kind of computing method of Series compensation lines fault distance, for parallel double back serial supplementary line singlephase earth fault:

Step one: to fault phase column voltage loop equation:

$u_{\mathrm{\&phi;}}=R(i_{\mathrm{\&phi;}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\&phi;}}+3k_x{i}_0)}{dt}+L_m\frac{{\mathrm{di}}_{30}}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0---\left(9\right);$

Wherein, u _φ, i _φ, i ₀, i ₃₀for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence, and 30 represent an adjacent loop line road zero sequence; L _mfor double-circuit line protective device is to the zero-sequence mutual inductance of trouble spot;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (9) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}}dt=R{\&Integral;}_{tm}^{tn}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt+L(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}+L_m\left(i_{30}\right){|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(10\right);$

Tm, tn are the moment in trouble duration, and t0<tm;

$i_{30}{|}_{tm}^{tn}=i_{30}\&lsqb;n\&rsqb;-i_{30}\&lsqb;m\&rsqb;$

$(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively ₃₀[n] represents i ₃₀in the sampled value in tn moment, the rest may be inferred;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_3=\left[\begin{array}{ccccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t1}^{t2}& \left(i_{30}\right){|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t3}^{t4}& \left(i_{30}\right){|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t5}^{t6}& \left(i_{30}\right){|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t7}^{t8}& \left(i_{30}\right){|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\\ {\&Integral;}_{t9}^{t10}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& (i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{t9}^{t10}& \left(i_{30}\right){|}_{t9}^{t10}& {\&Integral;}_{t7}^{t10}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t9}^{t10}dt\end{array}\right]$

$B_3=\left[\begin{array}{c}R\\ L\\ L_m\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_{\text{3}}=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t9}^{t10}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₃=A ₃ ^-1c ₃(11); Step 4: try to achieve fault reactance X by fault inductance L and transmission line of electricity angular frequency, namely

X=ω L (1); Wherein, ω=2 π f, f are the electrical network power frequency of transmission line of electricity;

Step 5: calculate fault distance D by fault reactance X:

$D=\frac{X}{x_1}\text{---}\left(2\right);$

Wherein, x ₁for the forward-sequence reactance of faulty line unit length.

Claims (4)

1. computing method for Series compensation lines fault distance, comprising:

Step one: to fault phase column voltage loop equation, the pressure drop sum of voltage respectively in the pressure drop, fault capacitance C of the pressure drop of fault resstance R, fault inductance L duration that namely fault phase terminal voltage equaling fault; Wherein, the pressure drop on fault capacitance C is: wherein, i _bfor the electric current on fault capacitance C duration of fault, t0 be before fault or fault duration in any instant, u ₀for the voltage on t0 moment fault capacitance C;

Step 2: to a period of time upper integral of the voltage circuit equation listed by step one in trouble duration;

Step 3: be taken to few four periods in trouble duration and after the t0 moment respectively, solves the group that establishes an equation of the integral equation listed by step 3, obtains fault inductance L;

Step 4: try to achieve fault reactance X by fault inductance L and transmission line of electricity angular frequency, namely

X＝ωL (1)；

Wherein, ω=2 π f, f are the electrical network power frequency of transmission line of electricity;

Step 5: calculate fault distance D by fault reactance X:

$D=\frac{X}{x_1}---\left(2\right);$

Wherein, x ₁for the forward-sequence reactance of faulty line unit length.

2. the computing method of Series compensation lines fault distance as claimed in claim 1, is characterized in that: for single back line singlephase earth fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\&phi;}}=R(i_{\mathrm{\&phi;}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\&phi;}}+3k_x{i}_0)}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0---\left(3\right);$

Wherein, u _φ, i _φ, i ₀for the voltage current waveform sampling of line protective devices to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence; L is faulty line positive sequence inductance;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (3) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}}dt=R{\&Integral;}_{tm}^{tn}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt+L{(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(4\right);$

Tm, tn are the moment in trouble duration, and t0<tm,

${(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _φ[n] represents i _φin the sampled value in tn moment;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_1=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_1=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_1=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}}dt\end{array}\right]$

B ₁＝A ₁ ^-1C ₁(5)。

3. the computing method of Series compensation lines fault distance as claimed in claim 1, is characterized in that: for mutually indirectly or phase to phase fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\&phi;}\mathrm{\&psi;}}=R\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)+L\frac{d\left(i_{\mathrm{\&phi;}\mathrm{\&psi;}}\right)}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0---\left(6\right);$

Wherein, u _{φ ψ}, i _{φ ψ}for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ, ψ are two fault phases, and L is faulty line positive sequence inductance;

u _φψ＝u _φ-u _ψ；i _φψ＝i _φ-i _ψ

Step 2: (6) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt=R{\&Integral;}_{tm}^{tn}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt{+{\mathrm{Li}}_{\mathrm{\&phi;}\mathrm{\&psi;}}|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(7\right);$

Tm, tn are the moment in trouble duration, and t0<tm, wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively _{φ ψ}[n] represents i _{φ ψ}in the sampled value in tn moment;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_2=\left[\begin{array}{cccc}{\&Integral;}_{t1}^{t2}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {i_{\mathrm{\&phi;}\mathrm{\&psi;}}|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {i_{\mathrm{\&phi;}\mathrm{\&psi;}}|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {i_{\mathrm{\&phi;}\mathrm{\&psi;}}|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {i_{\mathrm{\&phi;}\mathrm{\&psi;}}|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt& {\&Integral;}_{t7}^{t8}dt\end{array}\right]$

$B_2=\left[\begin{array}{c}R\\ L\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_2=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₂＝A ₂ ^-1C ₂(8)。

4. the computing method of Series compensation lines fault distance as claimed in claim 1, is characterized in that: for parallel double back serial supplementary line singlephase earth fault:

Step one: voltage circuit equation is:

$u_{\mathrm{\&phi;}}=R(i_{\mathrm{\&phi;}}+3k_r{i}_0)+L\frac{d(i_{\mathrm{\&phi;}}+3k_x{i}_0)}{dt}+L_m\frac{{\mathrm{di}}_{30}}{dt}+\frac{1}{C}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0---\left(9\right);$

Wherein, u _φ, i _φ, i ₀, i ₃₀for the voltage current waveform sampling of the line protective devices nearest apart from trouble spot to mutual inductor obtains voltage/current instantaneous value, u is voltage, and i is electric current, and φ is fault phase, and 0 represents zero sequence, and 30 represent an adjacent loop line road zero sequence, L _mfor double-circuit line protective device is to the zero-sequence mutual inductance of trouble spot;

k _r＝(r ₀-r ₁)/3r ₁，k _x＝(x ₀-x ₁)/3x ₁

R ₀for faulty line unit length zero sequence resistance; r ₁for faulty line unit length positive sequence resistance; x ₀for faulty line unit length zero-sequence reactance;

Step 2: (9) formula a period of time in trouble duration [tm, tn] upper integral, obtains:

${\&Integral;}_{tm}^{tn}{u}_{\mathrm{\&phi;}}dt=R{\&Integral;}_{tm}^{tn}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt+L(i_{\mathrm{\&phi;}}+3k_x{i}_0){|}_{tm}^{tn}+L_m\left(i_{30}\right){|}_{tm}^{tn}+\frac{1}{C}{\&Integral;}_{tm}^{tn}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt+u_0{\&Integral;}_{tm}^{tn}dt---\left(10\right);$

Tm, tn are the moment in trouble duration, and t0<tm;

${i_{30}|}_{tm}^{tn}=i_{30}\&lsqb;n\&rsqb;-i_{30}\&lsqb;m\&rsqb;$

${(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{tm}^{tn}=(i_{\mathrm{\&phi;}}\&lsqb;n\&rsqb;+3k_x{i}_0\&lsqb;n\&rsqb;)-(i_{\mathrm{\&phi;}}\&lsqb;m\&rsqb;+3k_x{i}_0\&lsqb;m\&rsqb;),$ Wherein, [n], [m] represent that the sampling instant of electric current is tn, tm, i.e. i respectively ₃₀[n] represents i ₃₀in the sampled value in tn moment;

Step 3: make [tm, tn] to be taken to few 4 concrete time periods [t1, t2], [t3, t4], [t5, t6], [t7, t8] respectively, the group that establishes an equation solves L by matrix inversion: wherein,

$A_3=\left[\begin{array}{ccccc}{\&Integral;}_{t1}^{t2}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t1}^{t2}& {\left(i_{30}\right)|}_{t1}^{t2}& {\&Integral;}_{t1}^{t2}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t1}^{t2}dt\\ {\&Integral;}_{t3}^{t4}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t3}^{t4}& {\left(i_{30}\right)|}_{t3}^{t4}& {\&Integral;}_{t3}^{t4}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t3}^{t4}dt\\ {\&Integral;}_{t5}^{t6}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t5}^{t6}& {\left(i_{30}\right)|}_{t5}^{t6}& {\&Integral;}_{t5}^{t6}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t5}^{t6}dt\\ {\&Integral;}_{t7}^{t8}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t7}^{t8}& {\left(i_{30}\right)|}_{t7}^{t8}& {\&Integral;}_{t7}^{t8}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t7}^{t8}dt\\ {\&Integral;}_{t9}^{t10}(i_{\mathrm{\&phi;}}+3k_r{i}_0)dt& {(i_{\mathrm{\&phi;}}+3k_x{i}_0)|}_{t9}^{t10}& {\left(i_{30}\right)|}_{t9}^{t10}& {\&Integral;}_{t9}^{t10}{\&Integral;}_{t0}^t{i}_{\mathrm{\&phi;}}dt& {\&Integral;}_{t9}^{t10}dt\end{array}\right]$

$B_3=\left[\begin{array}{c}R\\ L\\ L_m\\ \frac{1}{C}\\ u_0\end{array}\right]$

$C_3=\left[\begin{array}{c}{\&Integral;}_{t1}^{t2}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t3}^{t4}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t5}^{t6}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t7}^{t8}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\\ {\&Integral;}_{t9}^{t10}{u}_{\mathrm{\&phi;}\mathrm{\&psi;}}dt\end{array}\right]$

B ₃＝A ₃ ^-1C ₃(11)。