CN103986132A - Electric transmission line traveling wave differential protection method - Google Patents

Abstract

The invention relates to an electric transmission line traveling wave differential protection method. It is supposed that the total length of a protected both-end electric transmission line is 1, setting valves of traveling wave differential protection are calculated according to the following steps: a zero mode unbalanced coefficient is calculated; the maximum zero mode forward-direction traveling wave current at the m end during an external earth fault in an electric transmission line area is determined, and the maximum zero mode backward-direction traveling wave current at the n end is determined; the setting valves of various phases of the forward-direction traveling wave differential protection and the backward-direction traveling wave differential protection are respectively calculated; the traveling wave differential protection is adjusted according to the setting valves of the various phases of the forward-direction traveling wave differential protection and the backward-direction traveling wave differential protection, failure recognition and judgment and protection movement on the traveling wave differential protection of the various phases are accordingly achieved. By means of the electric transmission line traveling wave differential protection method, protection error movement is not prone to being caused; the electric transmission line traveling wave differential protection method is easy to implement.

Description

A kind of transmission line travelling wave differential protecting method

Technical field

The relaying protection field that the present invention relates to power system transmission line, is specifically related to a kind of traveling-wave differential protection method.

Background technology

Transmission line differential current protection philosophy is simple, highly sensitive, is the main protection of spy/extra high voltage network.But; in ultra-high/extra-high voltage transmission line, capacitance current is large; selectivity and the reliability of this principle have greatly been affected; can improve to a certain extent protective value by condenser current compensation method; but because the frequency content of capacitance current in fault transient process is complicated, be difficult to effectively compensate ^[1-3].

Have scholar to propose the traveling-wave differential protection based on theory of travelling wave, this class protection is divided into again two kinds from the angle of utilizing of row ripple: 1. utilize the high fdrequency component fault transient.Amplitude, polarity and the time of advent of extracting the capable ripple of high frequency form to be protected ^[4,5]; 2. utilize power frequency (fundamental frequency) row ripple.Because the power current in system can be considered the stack of direct wave and returning wave, so utilize the power frequency row ripple at transmission line two ends to form differential protection, can identify internal fault external fault ^[6,7].In theory, above-mentioned two kinds of traveling-wave protections can both be eliminated the impact of capacitance current completely, and are not subject to the impact of load current.But in practical application, in the differential current of protecting when the normal operation of electric power system and external area error, can produce certain imbalance output unsymmetrical current, be also referred to as the material unaccounted for stream of differential protection.Because unsymmetrical current is directly connected to the selection principle of traveling-wave differential protection setting value, so should analyse in depth this electric current, but the current article of this being analyzed to and relate to setting principle is reported seldom.Document [8] has been analyzed several sources of transmission line unsymmetrical current, for uniline, has three sources, is respectively: circuit model error, interpolation truncated error and synchronously to time error.Traveling-wave differential protection principle is based on harmless circuit, and it is to damage circuit that actual transmission line contains resistance, and row ripple can be decayed because of resistance produce power, so circuit model error can cause material unaccounted for stream; The process need of asking for of row ripple differential current carries out interpolation to the sampled value of one end, and interpolation calculation error can produce material unaccounted for stream; Traveling-wave differential protection need to the data at circuit two ends carry out synchronous and to time, to time error, also can produce unsymmetrical current.But document [8] has only carried out qualitative analysis to this, do not carry out quantitative analysis.

List of references

[1] He Jiali, Song Congju. Power System Relay Protection. enlarged edition. Beijing, China Electric Power Publishing House: 2004

[2]YiningZ,JialeS.Phaselet-based current differential protection scheme based on transient capacitive current compensation.IET Proc Gener,Transm Distrib,2008,2(4):469–377

[3] Wu Tonghua, Zheng Yuping, Zhu Xiaotong. the line differential protection based on transient state condenser current compensation. Automation of Electric Systems, 2005,24:1-8

[4]Takagi T,Baba J,Uemura K,et al.Fault protection based on travelling wave theory,part I:theory.Ele Eng Japan,1978,98(1):79-86

[5]Takagi T,Baba J,Uemura K,et al.Fault protection based on travelling wave theory,partII:feasibility study.Ele Eng Japan,1978,98(4):711-718

[6] Su Bin, Dong Xinzhou, Sun Yuanzhang. the traveling-wave differential protection based on wavelet transformation [J]. Automation of Electric Systems, 2004,28 (18): 25-29.

[7] Su Bin, Dong Xinzhou, Sun Yuanzhang. be applicable to the differential protection distributed capacitive current algorithm [J] of extra high voltage line. Automation of Electric Systems, 2005,29 (8): 36-40.

[8] Zhang Wujun, what Pentium. the flow analysis of traveling-wave differential protection material unaccounted for and practical plan [J]. Automation of Electric Systems, 2007,31 (20): 49-55.

Summary of the invention

The object of the invention is to overcome the above-mentioned deficiency of prior art; the unsymmetrical current of traveling-wave differential protection is being carried out on the basis of quantitative analysis; a kind of simple, to be difficult for occurring protection misoperation transmission line travelling wave differential protection setting method of realizing is provided, then realizes each phase traveling-wave differential protection Fault Identification judgement and protection action.

Technical scheme of the present invention is as follows:

A transmission line travelling wave differential protecting method, the total length of establishing protected both-end transmission line is l, calculates the setting value of traveling-wave differential protection according to following steps:

(1). by transmission line parameter, calculate according to the following formula 0 mould amplitude attenuation factor alpha ₀;

${\mathrm{\α}}_0=\sqrt{-\frac{1}{2}{\mathrm{\ω}}^2{L}_0{C}_0+\frac{1}{2}\sqrt{{\mathrm{\ω}}^2{C}_0^2(R_0^2+{\mathrm{\ω}}^2{C}_0^2)}}$

In formula, ω is power frequency angular speed, L ₀be 0 mould unit length inductance, C ₀be 0 mould unit length ground capacity, R ₀it is 0 mould resistance per unit length;

(2). by 0 mould amplitude attenuation factor alpha ₀calculate 0 mould unbalanced coefficient

(3). according to transmission system parameter, the maximum 0 mould direct wave electric current of m end while determining transmission line district external ground fault and the maximum 0 mould returning wave electric current of definite n end

(4). the data substitution following formula that above step is calculated, is calculated respectively each phase setting value of direct wave differential protection and returning wave differential protection by following formula:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{\mathrm{op}1.a}={\mathrm{\μ}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.b}=\frac{4}{3}{\mathrm{\μ}}_0{I}_{m-0-\mathrm{m\; ax}}^{+}\\ I_{\mathrm{op}1.c}=\frac{4}{3}{\mathrm{\μ}}_0{I}_{m-0-\max }^{+}\end{array}\right.\left\{\begin{array}{c}{I}_{\mathrm{op}2.a}={\mathrm{\μ}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.b}=\frac{4}{3}{\mathrm{\μ}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.c}=\frac{4}{3}{\mathrm{\μ}}_0{I}_{n-0-\max }^{-}\end{array}\right.& \end{array}$

In formula, I _op1.a, I _op1.b, I _op1.cbe respectively each phase setting value of direct wave differential protection, I _op2.a, I _op2.b, I _op2.cbe respectively each phase setting value of returning wave differential protection.

(5) according to each phase setting value of the direct wave differential protection of gained and returning wave differential protection, traveling-wave differential protection is adjusted, thereby realized each phase traveling-wave differential protection Fault Identification judgement and protection action.

Beneficial effect of the present invention is as follows:

The setting method theoretical foundation of the transmission line travelling wave differential protection that 1, the present invention provides is abundant, adjusts simple;

When 2, setting method of the present invention can guarantee normally operation of electric power system, external area error and line no-load switching, traveling-wave differential protection reliable operation without misoperation.

Accompanying drawing explanation

The capable ripple schematic diagram of the single-phase harmless circuit of Fig. 1.

Embodiment

Below in conjunction with drawings and Examples, the present invention will be described.

Mn in Fig. 1 is single-phase harmless circuit, and length is l, and velocity of wave is v, and τ=l/v is the propagation delay of row ripple on total track length, and wave impedance is Z _c, circuit two ends electric current positive direction is for to flow to circuit by bus, and this current visible is for consisting of the capable ripple of forward current and the capable ripple stack of reverse current.The capable ripple of forward current is defined as to left end and flows to right-hand member, otherwise be the capable ripple of reverse current, two ends (2 times) current traveling wave is:

$\left\{\begin{array}{c}{i}_m^{+}\left(t\right)=u_m\left(t\right)/Z_c+i_m\left(t\right)\\ i_m^{-}\left(t\right)=u_m\left(t\right)/Z_c-i_m\left(t\right)\\ i_n^{+}\left(t\right)=u_n\left(t\right)/Z_c-i_n\left(t\right)\\ i_n^{-}\left(t\right)=u_n\left(t\right)/Z_c+i_n\left(t\right)\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(1\right)$

In above formula, be respectively the capable ripple of forward, reverse current of m end and n end; i _m(t), i _n(t), u _m(t), u _n(t) for m holds and electric current, the voltage of n end.Definition forward and reverse row ripple differential current are ^[6,7]:

$\left\{\begin{array}{c}{i}_{D1}\left(t\right)=i_m^{+}(t-\mathrm{\τ})-i_n^{+}\left(t\right)\\ i_{D2}\left(t\right)=i_n^{-}(t-\mathrm{\τ})-i_m^{-}\left(t\right)\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(2\right)$

For harmless circuit, the direct wave at m point place is propagated and is arrived n point through τ time delay, becomes the direct wave that n is ordered; The returning wave at n point place propagates into m point through duration τ, becomes the returning wave that m is ordered, so normally operation and i during external fault _d1and i (t) _d2(t) be zero.During line-internal fault, due to the existence of fault point, circuit is no longer uniform transmission line, i _d1and i (t) _d2(t) non-vanishing, its amplitude equals current in the fault point amplitude, traveling-wave differential protection action tripping operation.

Description damages the power on partial differential equation of current voltage variation characteristic of uniform transmission circuit:

$\left\{\begin{array}{c}-\frac{\∂u(x,t)}{\∂x}=\mathrm{Ri}(x,t)+L\frac{\∂i(x,t)}{\∂t}\\ -\frac{\∂i(x,t)}{\∂x}=\mathrm{Gi}(x,t)+C\frac{\∂u(x,t)}{\∂t}\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(3\right)$

In above formula, R, L, G, C are the parameter of circuit unit length, and u (x, t), i (x, t) are x point t voltage and current constantly on circuit.For power frequency component, solution above formula can obtain electric current steady state solution and be:

$\stackrel{\·}{I}\left(x\right)={\stackrel{\·}{A}}_1{e}^{-\mathrm{\γx}}-{\stackrel{\·}{A}}_2{e}^{\mathrm{\γx}},\backslash \mathrm{MERGEFORMAT}---(4)$

In formula, with by boundary condition, determined.Because circuit top is protection installation place, its electric current and voltage is known as :

$\left\{\begin{array}{c}{\stackrel{\·}{A}}_1=({\stackrel{\·}{U}}_m+Z_c{\stackrel{\·}{I}}_m)/\left(2Z_c\right)\\ {\stackrel{\·}{A}}_1=({\stackrel{\·}{U}}_m-Z_c{\stackrel{\·}{I}}_m)/\left(2Z_c\right)\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(5\right)$

$Z_c=\sqrt{(R+\mathrm{j\ωL})/(G+\mathrm{j\ωC})}$ For surge impedance of a line.

In formula, be direct wave component, can make because row ripple amplitude in traveling process decays, phase place produces displacement, therefore, after the unit length of advancing by become have:

$\frac{{\stackrel{\·}{I}}^{+}\left(x\right)}{{\stackrel{\·}{I}}^{+}(x+1)}=\frac{{\stackrel{\·}{A}}_1{e}^{-\mathrm{\γx}}}{{\stackrel{\·}{A}}_1{e}^{-\mathrm{\γ}(x+1)}}=\frac{{\stackrel{\·}{A}}_1{e}^{-\mathrm{\γx}}}{{\stackrel{\·}{A}}_1{e}^{-\mathrm{\γx}}{e}^{-\mathrm{\γ}}}=\frac{1}{e^{-\mathrm{\γ}}}=e^{\mathrm{\γ}},\backslash *\mathrm{MERGEFORMAT}---\left(6\right)$

Order ${\stackrel{\·}{I}}^{+}\left(x\right)=I^{+}\left(x\right)\∠{\mathrm{\θ}}_x^{+},{\stackrel{\·}{I}}^{+}(x+1)=I^{+}(x+1)\∠{\mathrm{\θ}}_{x+1}^{+},\mathrm{\γ}=\mathrm{\α}+\mathrm{j\β},$ :

$e^{\mathrm{\γ}}=\frac{{\stackrel{\·}{I}}^{+}\left(x\right)}{{\stackrel{\·}{I}}^{+}(x+1)}=\frac{I^{+}\left(x\right)\∠{\mathrm{\θ}}_x^{+}}{I^{+}(x+1)\∠{\mathrm{\θ}}_{x+1}^{+}}=\frac{I^{+}\left(x\right)}{I^{+}(x+1)}{e}^{j({\mathrm{\θ}}_x^{+}-{\mathrm{\θ}}_{x+1}^{+})},\backslash *\mathrm{MERGEFORMAT}---\left(7\right)$

Abbreviation can obtain:

$\mathrm{\γ}=\ln \frac{I^{+}\left(x\right)}{I^{+}(x+1)}+j({\mathrm{\θ}}_x^{+}-{\mathrm{\θ}}_{x+1}^{+}),\backslash *\mathrm{MERGEFORMAT}---\left(8\right)$

Therefore,

$\left\{\begin{array}{c}\mathrm{\α}=\ln \frac{I^{+}\left(x\right)}{I^{+}(x+1)}\\ \mathrm{\β}={\mathrm{\θ}}_x^{+}-{\mathrm{\θ}}_{x+1}^{+}\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(9\right)$

In above formula, be the amplitude attenuation coefficient of unit length upgoing wave, β is the phase coefficient of unit length upgoing wave.By $\mathrm{\γ}=\mathrm{\α}+\mathrm{j\β}=\sqrt{(R+\mathrm{j\ωL})(G+\mathrm{j\ωC})}$ Solve:

$\left\{\begin{array}{c}\mathrm{\α}=\sqrt{\frac{1}{2}(\mathrm{RG}-{\mathrm{\ω}}^2\mathrm{LC})+\frac{1}{2}\sqrt{(R^2+{\mathrm{\ω}}^2{L}^2)(G^2+{\mathrm{\ω}}^2{C}^2)}}\\ \mathrm{\β}=\sqrt{\frac{1}{2}({\mathrm{\ω}}^2\mathrm{LC}-\mathrm{RG})+\frac{1}{2}\sqrt{(R^2+{\mathrm{\ω}}^2{L}^2)(G^2+{\mathrm{\ω}}^2{C}^2)}}\end{array}\right.,\backslash *\mathrm{MERGEFORMAT}---\left(10\right)$

Formula has characterized the Changing Pattern of row wave amplitude and phase place and the relation between line parameter circuit value.R is larger, and the energy that row ripple loses is on the line larger, and amplitude attenuation coefficient is larger; L, C are larger, and Electric and magnetic fields alternately conversion is strong with regard to Shaoxing opera, and phase coefficient β is larger.

So, at mn for damaging in transmission line situation, two ends direct wave with amplitude between meet:

$\begin{array}{c}\ln \frac{I_m^{+}}{I_n^{+}}=\ln \frac{I_m^{+}}{I_{m+1}^{+}}\frac{I_{m+1}^{+}}{I_{m+2}^{+}}...\frac{I_{m+h}^{+}}{I_{m+h+1}^{+}}...\frac{I_{m+L-2}^{+}}{I_{m+L-1}^{+}}\frac{I_{n+L-1}^{+}}{I_n^{+}}\\ =\ln \frac{I_m^{+}}{I_{m+1}^{+}}+...+\ln \frac{I_{m+h}^{+}}{I_{m+h+1}^{+}}+...+\ln \frac{I_{n+L-1}^{+}}{I_n^{+}}\\ =\mathrm{\α}+...+\mathrm{\α}+...+\mathrm{\α}=\mathrm{l\α}\end{array},\backslash *\mathrm{MERGEFORMAT}---\left(11\right)$

In above formula, expression is apart from the direct wave amplitude of m end h km.Abbreviation formula obtains:

$I_n^{+}=e^{-\mathrm{l\α}}{I}_m^{+},\backslash *\mathrm{MERGEFORMAT}---\left(12\right)$

with phase difference be:

$\begin{array}{c}{\mathrm{\θ}}_m^{+}-{\mathrm{\θ}}_n^{+}={\mathrm{\θ}}_m^{+}-{\mathrm{\θ}}_{m+1}^{+}+{\mathrm{\θ}}_{m+1}^{+}-...-{\mathrm{\θ}}_{m+L-1}^{+}+{\mathrm{\θ}}_{m+L-1}^{+}-{\mathrm{\θ}}_n^{+}\\ =\mathrm{\β}+...+\mathrm{\β}+...+\mathrm{\β}=\mathrm{l\β}\end{array},\backslash *\mathrm{MERGEFORMAT}---\left(13\right)$

Abbreviation formula obtains:

${\mathrm{\θ}}_n^{+}={\mathrm{\θ}}_m^{+}-\mathrm{l\β},\backslash *\mathrm{MERGEFORMAT}---\left(14\right)$

Composite type and formula, can obtain and between the direct wave of circuit two ends, have following relation:

${\stackrel{\·}{I}}_n^{+}=I_n^{+}\∠{\mathrm{\θ}}_n^{+}=e^{-\mathrm{l\α}}{I}_m^{+}\∠({\mathrm{\θ}}_m^{+}-\mathrm{l\β}){=e}^{-\mathrm{l\α}}{\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{jl\β}}\backslash *\mathrm{MERGEFORMAT}---\left(15\right)$

Therefore, normally operation or during external area error the poor stream of direct wave be:

${\stackrel{\·}{I}}_{D1}={\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{j\ω\τ}}-{\stackrel{\·}{I}}_n^{+}={\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{j\ω\τ}}-e^{-\mathrm{L\α}}{\stackrel{\·}{I}}_n^{+}{e}^{-\mathrm{jL\β}}={\stackrel{\·}{I}}_{\mathrm{ub}}^{+},\backslash *\mathrm{MERGEFORMAT}---\left(16\right)$

be forward unsymmetrical current.Because ω τ=ω l/v and v=ω/β, ω τ=l β, time shift compensated the phase shift producing in row ripple traveling process.So formula becomes:

${\stackrel{\·}{I}}_{\mathrm{ub}}^{+}={\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{j\ω\τ}}-e^{-\mathrm{l\α}}{\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{jl\β}}=(1-e^{-\mathrm{l\α}}){\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{jl\β}},\backslash *\mathrm{MERGEFORMAT}---\left(17\right)$

Definition unbalanced coefficient μ=1-e ^{-l α}:

${\stackrel{\·}{I}}_{\mathrm{ub}}^{+}=\mathrm{\μ}{\stackrel{\·}{I}}_m^{+}{e}^{-\mathrm{jl\β}},\backslash *\mathrm{MERGEFORMAT}---\left(18\right)$

For returning wave, through similar derivation, can obtain and between the returning wave of circuit two ends, have following relation:

${\stackrel{\·}{I}}_m^{-}=e^{-\mathrm{l\α}}{\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{jl\β}},\backslash *\mathrm{MERGEFORMAT}---\left(19\right)$

When normal operation or external area error, the poor stream of returning wave is material unaccounted for stream for:

$\begin{array}{c}{\stackrel{\·}{I}}_{D2}={\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{j\ω\τ}}-{\stackrel{\·}{I}}_m^{-}={\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{j\ω\τ}}-e^{-\mathrm{l\α}}{\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{jl\β}}\\ =(1-e^{-\mathrm{l\α}}){\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{jl\β}}={\stackrel{\·}{I}}_{\mathrm{ub}}^{-}\end{array},\backslash *\mathrm{MERGEFORMAT}---\left(20\right)$

Abbreviation obtains:

${\stackrel{\·}{I}}_{\mathrm{ub}}=\mathrm{\μ}{\stackrel{\·}{I}}_n^{-}{e}^{-\mathrm{jl\β}},\backslash *\mathrm{MERGEFORMAT}---\left(21\right)$

Unbalanced coefficient μ is relevant with the amplitude attenuation factor alpha of line length and row ripple.Circuit is longer, the resistance of unit length is larger, and μ is also larger, with also larger.Material unaccounted for stream is also relevant with the top amplitude of forward and reverse row ripple at circuit two ends, the direct wave that m end flows through it is larger, larger; The returning wave that n end flows through it is larger, larger.

Upper surface analysis be single-phase transmission line, for three phase line, to block human relations Bauer transfer pair three-phase, realize the decoupling zero of phase mould, transformation matrix is:

$S^{-1}=\left[\begin{array}{ccc}1& 1& 1\\ 1& -1& 0\\ 1& 0& -1\end{array}\right],\backslash *\mathrm{MERGEFORMAT}---\left(22\right)$

0 mould, 1 mould and 2 mould electric currents that conversion obtains are respectively

$\left\{\begin{array}{c}{i}^0\left(t\right)=i^a\left(t\right)+i^b\left(t\right)+i^c\left(t\right)\\ i^1\left(t\right)=i^a\left(t\right)-i^b\left(t\right)\\ i^2\left(t\right)=i^a\left(t\right)-i^c\left(t\right)\end{array}\right.,/*\mathrm{MERGEFORMAT}---\left(23\right)$

The direct wave differential current of 0 mould, 1 mould and 2 moulds is:

$\left\{\begin{array}{c}{i}_{D1-0}=i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{n-0}^{+}\left(t\right)\\ i_{D1-1}=i_{m-1}^{+}(t-{\mathrm{\τ}}_1)-i_{n-1}^{+}\left(t\right)\\ i_{D1-2}=i_{m-2}^{+}(t-{\mathrm{\τ}}_1)-i_{n-2}^{+}\left(t\right)\end{array}\right.---\left(24\right)$

The unsymmetrical current that can be obtained the poor stream of each modulus forward by the derivation of unsymmetrical current is above:

$\left\{\begin{array}{c}{\stackrel{\·}{I}}_{\mathrm{ub}-0}^{+}={\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_0}-{\stackrel{\·}{I}}_{n-0}^{+}={\mathrm{\μ}}_0{\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_0}\\ {\stackrel{\·}{I}}_{\mathrm{ub}-1}^{+}={\stackrel{\·}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}-{\stackrel{\·}{I}}_{n-1}^{+}={\mathrm{\μ}}_1{\stackrel{\·}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}\\ {\stackrel{\·}{I}}_{\mathrm{ub}-2}^{+}={\stackrel{\·}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}-{\stackrel{\·}{I}}_{n-2}^{+}={\mathrm{\μ}}_1{\stackrel{\·}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}\end{array}\right.---\left(25\right)$

In formula, μ ₀, μ ₁, β ₀, β ₁be respectively unbalanced coefficient and the phase coefficient of 0 mould and 1 mould.From transformation matrix formula (22), 2 modular transformations are the different of separate combination from 1 modular transformation, and the modulus parameter after conversion is identical, so the unbalanced coefficient of 2 moulds is identical with 1 mould with phase coefficient.In actual track, unit length line conductance G=0, can be obtained by 0 mould and 1 mould parameter of formula and circuit:

$\left\{\begin{array}{c}{\mathrm{\α}}_0=\sqrt{-\frac{1}{2}{\mathrm{\ω}}^2{L}_0{C}_0+\frac{1}{2}\sqrt{{\mathrm{\ω}}^2{C}_0^2(R_0^2+{\mathrm{\ω}}^2{L}_0^2)}}\\ {\mathrm{\α}}_1=\sqrt{-\frac{1}{2}{\mathrm{\ω}}^2{L}_1{C}_1+\frac{1}{2}\sqrt{{\mathrm{\ω}}^2{C}_1^2(R_1^2+{\mathrm{\ω}}^2{L}_1^2)}}\\ {\mathrm{\β}}_0=\sqrt{\frac{1}{2}{\mathrm{\ω}}^2{L}_0{C}_0+\frac{1}{2}\sqrt{{\mathrm{\ω}}^2{C}_0^2(R_0^2+{\mathrm{\ω}}^2{L}_0^2)}}\\ \mathrm{\β}=\sqrt{\frac{1}{2}{\mathrm{\ω}}^2{L}_1{C}_1+\frac{1}{2}\sqrt{{\mathrm{\ω}}^2{C}_1^2(R_1^2+{\mathrm{\ω}}^2{L}_1^2)}}\end{array}\right.---\left(26\right)$

The unbalanced coefficient of 0 mould and 1 mould is:

$\left\{\begin{array}{c}{\mathrm{\μ}}_0=1-e^{-l{\mathrm{\α}}_0}\\ {\mathrm{\μ}}_1=1-e^{-l{\mathrm{\α}}_1}\end{array}\right.---\left(27\right)$

Line parameter circuit value from ultra-high/extra-high voltage transmission line 0 mould and 1 mould: μ ₀> > μ ₁, so when the normal operation of system and external area error, in the maximum imbalance current of three kinds of modulus, 0 mould unsymmetrical current is maximum.So, when generating region external ground fault, obtain the peaked while, also obtain the maximum imbalance current in modulus.In like manner, for reverse poor stream, when generating region external ground fault obtain the peaked while, correspondingly also obtain the maximum imbalance current in modulus.

For phase-splitting traveling-wave differential protection, during due to fault, between each phase, there is the effect of intercoupling, so the direct wave differential current of each phase can be by not existing the differential current of 0 mould, 1 mould and 2 moulds of coupling to express through inverse transformation.Therefore, by phase mould inverse transformation, obtained the direct wave differential current i of a phase _d1.afor:

$\begin{array}{c}{i}_{D1.a}=\frac{1}{3}{i}_{D1-0}+\frac{1}{3}{i}_{D1-1}+\frac{1}{3}{i}_{D1-2}\\ =\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{n-0}^{+}\left(t\right)]+\frac{1}{3}[i_{m-1}^{+}(t-{\mathrm{\τ}}_1)-i_{n-1}^{+}\left(t\right)]+\frac{1}{3}[i_{m-2}^{+}(t-{\mathrm{\τ}}_1)-i_{n-2}^{+}\left(t\right)]\\ =i_{m-a}^{+}(t-{\mathrm{\τ}}_1)-i_{n-a}^{+}\left(t\right)+\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{m-0}^{+}(t-{\mathrm{\τ}}_1)]\end{array}---\left(28\right)$

The direct wave differential current of b phase and c phase also obtains as follows by phase mould inverse transformation:

$\left\{\begin{array}{c}{i}_{D1.b}=\frac{1}{3}{i}_{D1-0}-\frac{2}{3}{i}_{D1-1}+\frac{1}{3}{i}_{D1-2}\\ =\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{n-0}^{+}\left(t\right)]-\frac{2}{3}[i_{m-1}^{+}(t-{\mathrm{\τ}}_1)-i_{n-1}^{+}\left(t\right)]+\frac{1}{3}[i_{m-2}^{+}(t-{\mathrm{\τ}}_1)-i_{n-2}^{+}\left(t\right)]\\ =i_{m-b}^{+}(t-{\mathrm{\τ}}_1)-i_{n-b}^{+}\left(t\right)+\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{m-0}^{+}(t-{\mathrm{\τ}}_1)]\\ i_{D1.c}=\frac{1}{3}{i}_{D1-0}+\frac{1}{3}{i}_{D1-1}-\frac{2}{3}{i}_{D1-2}\\ =\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{n-0}\left(t\right)]+\frac{1}{3}[i_{m-1}^{+}(t-{\mathrm{\τ}}_1)-i_{n-1}^{+}\left(t\right)]-\frac{2}{3}[i_{m-2}^{+}(t-{\mathrm{\τ}}_1)-i_{n-2}^{+}\left(t\right)]\\ =i_{m-c}^{+}(t-{\mathrm{\τ}}_1)-i_{n-c}^{+}\left(t\right)+\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\τ}}_0)-i_{m-0}^{+}(t-{\mathrm{\τ}}_1)]\end{array}---\left(29\right)\right.$

Actual track is for damaging circuit, and amplitude attenuation and phase deviation occur row ripple in transmitting procedure, and the row ripple of each phase is differential has unsymmetrical current output under the normal operation of system and external fault condition.Phase-splitting is below analyzed.

For a phase direct wave differential current i _d1.a, the amplitude of its uneven output is:

$\begin{array}{c}{I}_{D1.a}=|{\stackrel{\·}{I}}_{m-a}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}-{\stackrel{\·}{I}}_{n-a}^{+}+\frac{1}{3}[{\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_0}-{\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}\left]\right|\\ =\frac{1}{3}|({\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_0}-{\stackrel{\·}{I}}_{n-0}^{+})+({\stackrel{\·}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}-{\stackrel{\·}{I}}_{n-1}^{+})+({\stackrel{\·}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_2}-{\stackrel{\·}{I}}_{n-2}^{+})|\\ =\frac{1}{3}|{\mathrm{\μ}}_0{\stackrel{\·}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_0}+{\mathrm{\μ}}_1{\stackrel{\·}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}+{\mathrm{\μ}}_1{\stackrel{\·}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\β}}_1}|\end{array}---\left(29\right)$

In above formula, with be respectively the unsymmetrical current of each mould, maximum be greater than with order maximum be have:

$\begin{array}{c}{I}_{D1.a}=\frac{1}{3}|({\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}-{\stackrel{\&CenterDot;}{I}}_{n-0}^{+})+({\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{n-1}^{+})+({\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_2}-{\stackrel{\&CenterDot;}{I}}_{n-2}^{+})|\\ =\frac{1}{3}|{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}|\\ <\frac{1}{3}({\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}+{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1})<{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}\end{array}---\left(30\right)$

Therefore, in order to allow the direct wave protection of a phase guarantee not misoperation, its setting value I under breaking down situation in normal operation and outside _op1.ashould be taken as:

$I_{\mathrm{op}1.a}={\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}---\left(31\right)$

For b phase direct wave differential current i _d1.b, the amplitude of its uneven output is:

$\begin{array}{c}{I}_{D1.b}=|i_{m-b}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{n-b}^{+}+\frac{1}{3}[{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}\left]\right|\\ =\frac{1}{3}|({\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}-{\stackrel{\&CenterDot;}{I}}_{n-0}^{+})-2({\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{n-1}^{+})+({\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_2}-{\stackrel{\&CenterDot;}{I}}_{n-2}^{+})|\\ =\frac{1}{3}|{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}-2{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}+{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}|\\ <\frac{1}{3}({\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+2{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0})=\frac{4}{3}{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}\end{array}---\left(32\right)$

Therefore, in order to allow the direct wave protection of b phase guarantee not misoperation, its setting value I under breaking down situation in normal operation and outside _op1.bshould be taken as:

$I_{\mathrm{op}1.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}---\left(33\right)$

For c phase direct wave differential current i _d1.c, the amplitude of its uneven output is:

$\begin{array}{c}{I}_{D1.c}=|i_{m-c}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{n-c}^{+}+\frac{1}{3}[{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}\left]\right|\\ =\frac{1}{3}|({\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}-{\stackrel{\&CenterDot;}{I}}_{n-0}^{+})+({\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-{\stackrel{\&CenterDot;}{I}}_{n-1}^{+})-2({\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_2}-{\stackrel{\&CenterDot;}{I}}_{n-2}^{+})|\\ =\frac{1}{3}|{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-1}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}-2{\mathrm{\&mu;}}_1{\stackrel{\&CenterDot;}{I}}_{m-2}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_1}|\\ <\frac{1}{3}({\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}+2{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0})=\frac{4}{3}{\mathrm{\&mu;}}_0{\stackrel{\&CenterDot;}{I}}_{m-0}^{+}{e}^{-\mathrm{jl}{\mathrm{\&beta;}}_0}\end{array}---\left(34\right)$

Therefore, in order to allow the direct wave protection of c phase guarantee not misoperation, its setting value I under breaking down situation in normal operation and outside _op1.cshould be taken as:

$I_{\mathrm{op}1.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}---\left(35\right)$

In like manner, for the returning wave differential protection of each phase, carrying out similarity analysis, can to obtain setting value as follows:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{\mathrm{op}2.a}={\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\end{array}\right.& \end{array}---\left(36\right)$

In above formula, the maximum 0 mould returning wave electric current of n end during transmission line district external ground fault.

In sum, for phase-splitting forward, the returning wave differential protection of three-phase ultra-high/extra-high voltage transmission line, should adjust according to following principle:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{\mathrm{op}1.a}={\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\end{array}\right.\left\{\begin{array}{c}{I}_{\mathrm{op}2.a}={\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\end{array}\right.& \end{array}---\left(37\right)$

For known both-end transmission line, can calculate according to following steps the setting value of traveling-wave differential protection.

1. by transmission line parameter, calculate according to the following formula 0 mould amplitude attenuation factor alpha ₀;

${\mathrm{\&alpha;}}_0=\sqrt{-\frac{1}{2}{\mathrm{\&omega;}}^2{L}_0{C}_0+\frac{1}{2}\sqrt{{\mathrm{\&omega;}}^2{C}_0^2(R_0^2+{\mathrm{\&omega;}}^2{C}_0^2)}}---\left(38\right)$

In above formula, ω is power frequency angular speed, L ₀be 0 mould unit length inductance, C ₀be 0 mould unit length ground capacity, R ₀it is 0 mould resistance per unit length.

2. by 0 mould amplitude attenuation factor alpha ₀calculate 0 mould unbalanced coefficient

3. according to transmission system parameter, the maximum 0 mould direct wave electric current of m end while determining transmission line district external ground fault and the maximum 0 mould returning wave electric current of definite n end

4. the data substitution following formula above step being calculated, is calculated each phase setting value that calculates respectively direct wave differential protection and returning wave differential protection by following formula:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{\mathrm{op}1.a}={\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\end{array}\right.\left\{\begin{array}{c}{I}_{\mathrm{op}2.a}={\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\end{array}\right.& \end{array}---\left(39\right)$

In above formula, I _op1.a, I _op1.b, I _op1.cbe respectively each phase setting value of direct wave differential protection, I _op2.a, I _op2.b, I _op2.cbe respectively each phase setting value of returning wave differential protection; μ ₀it is 0 mould unbalanced coefficient; the maximum 0 mould direct wave electric current of m end during for district's external ground fault, the maximum 0 mould returning wave electric current of n end during for district's external ground fault.

Below, the direct wave differential protection a of take is mutually differential is example, and the process of traveling-wave differential protection Fault Identification judgement is described.The implementation procedure of direct wave differential protection b, c two-phase and each phase of returning wave differential protection substantially roughly the same, repeats no more.

1. a phase voltage u being measured by circuit m end _m-a, i _m-acalculate the capable ripple of m end a phase forward current the a phase voltage u being measured by circuit n end _n-a, i _n-acalculate the capable ripple of n end a phase forward current

2. by line parameter circuit value, can obtain the propagation delay τ of circuit 0 mould and 1 mould ₀, τ ₁;

3. by circuit m terminal voltage electric current, through phase-model transformation, obtain voltage, the electric current of 0 mould.

4. by circuit m, hold voltage, the Current calculation of 0 mould to obtain circuit m and hold 0 mould direct wave

5. the data substitution following formula above step being calculated, extremely a row ripple differential current mutually:

$i_{D1.a}=i_{m-a}^{+}(t-{\mathrm{\&tau;}}_1)-i_{n-a}^{+}\left(t\right)+\frac{1}{3}[i_{m-0}^{+}(t-{\mathrm{\&tau;}}_0)-i_{m-0}^{+}(t-{\mathrm{\&tau;}}_1)]$

Relatively a mutually row ripple differential current and a fix threshold:

I _D1.a＞I _op1.a (41)

If above formula meets, think a phase fault; If above formula does not meet, think that a does not have fault mutually.

Claims (1)

1. a transmission line travelling wave differential protecting method, establishes protected m, and the total length of n both-end transmission line is l, calculates the setting value of traveling-wave differential protection according to following steps:

(1). by transmission line parameter, calculate according to the following formula 0 mould amplitude attenuation factor alpha ₀;

${\mathrm{\&alpha;}}_0=\sqrt{-\frac{1}{2}{\mathrm{\&omega;}}^2{L}_0{C}_0+\frac{1}{2}\sqrt{{\mathrm{\&omega;}}^2{C}_0^2(R_0^2+{\mathrm{\&omega;}}^2{C}_0^2)}}$

In formula, ω is power frequency angular speed, L ₀be 0 mould unit length inductance, C ₀be 0 mould unit length ground capacity, R ₀it is 0 mould resistance per unit length;

(2). by 0 mould amplitude attenuation factor alpha ₀calculate 0 mould unbalanced coefficient

(3). according to transmission system parameter, the maximum 0 mould direct wave electric current of m end while determining transmission line district external ground fault and the maximum 0 mould returning wave electric current of definite n end

(4). the data substitution following formula that above step is calculated, is calculated respectively each phase setting value of direct wave differential protection and returning wave differential protection by following formula:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{\mathrm{op}1.a}={\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{\mathrm{op}1.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\mathrm{m\; ax}}^{+}\\ I_{\mathrm{op}1.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\end{array}\right.\left\{\begin{array}{c}{I}_{\mathrm{op}2.a}={\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{\mathrm{op}2.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\end{array}\right.& \end{array}$

In formula, I _op1.a, I _op1.b, I _op1.cbe respectively each phase setting value of direct wave differential protection, I _op2.a, I _op2.b, I _op2.cbe respectively each phase setting value of returning wave differential protection.

(5) according to each phase setting value of the direct wave differential protection of gained and returning wave differential protection, traveling-wave differential protection is adjusted, thereby realized each phase traveling-wave differential protection Fault Identification judgement and protection action.