CN103986132B - A kind of transmission line travelling wave differential protecting method - Google Patents

Abstract

The present invention relates to a kind of transmission line travelling wave differential protecting method, if the total length of protected both-end transmission line of electricity is l, calculate the setting valve of traveling-wave differential protection according to following steps: calculate 0 mould unbalance factor；The maximum 0 mould direct wave electric current of m end when determining transmission line of electricity district external ground fault, and determine the maximum 0 mould backward-travelling wave electric current of n end；Calculate direct wave differential protection and each phase setting valve of backward-travelling wave differential protection respectively；Traveling-wave differential protection is adjusted by direct wave differential protection and each phase setting valve of backward-travelling wave differential protection according to gained, thus realizes judging each phase traveling-wave differential protection Fault Identification and protection act.The present invention has and is not susceptible to protection misoperation, it is easy to accomplish advantage.

Description

A kind of transmission line travelling wave differential protecting method

Technical field

The present invention relates to the relay protection field of power system transmission line, be specifically related to a kind of traveling-wave differential protection side Method.

Background technology

Transmission line of electricity differential current protection philosophy is simple, highly sensitive, is the main guarantor of spy/extra high voltage network Protect.But, in ultra-high/extra-high voltage transmission line of electricity, capacitance current is big, significantly impacts selectivity and the reliability of this principle, can To improve protective value to a certain extent by condenser current compensation method, but due to capacitance current in fault transient process Frequency content is complicated, it is difficult to effectively compensate^[1-3]。

Having scholar to propose traveling-wave differential protection based on theory of travelling wave, this kind of protection is divided into again from the angle that utilizes of row ripple Two kinds: 1. utilize the high fdrequency component in fault transient.Extract and the amplitude of high frequency row ripple, polarity and the time of advent constitute protection^[4,5]； 2. power frequency (fundamental frequency) row ripple is utilized.Because the power current in system can be considered the superposition of direct wave and backward-travelling wave, so The power frequency row ripple utilizing transmission line of electricity two ends constitutes differential protection and can recognize that internal fault external fault^[6,7].In theory, above two row ripple Protection can be completely eliminated the impact of capacitance current, and is not affected by load current.But in actual application, power train Unite properly functioning and external area error time protection difference current in can produce certain uneven output i.e. out-of-balance current, also by It is referred to as the material unaccounted for stream of differential protection.Due to out-of-balance current, to be directly connected to the selection of traveling-wave differential protection setting valve former Then, therefore this electric current should be analysed in depth, but this is analyzed and relates to the article report of setting principle at present Seldom.Document [8] analyzes several sources of transmission line of electricity out-of-balance current, has three sources for uniline, is respectively as follows: Circuit model error, interpolated truncation error and synchronization are to time error.Traveling-wave differential protection principle is based on lossless circuit, reality transmission of electricity It is to damage circuit that circuit contains resistance, and row ripple can produce energy attenuation because of resistance, and therefore circuit model error can cause imbalance Difference stream；The process of asking for of row ripple difference current needs the sampled value to one end to carry out interpolation, and interpolation calculation error can produce injustice Weighing apparatus difference stream；Traveling-wave differential protection need the data at circuit two ends are synchronized and pair time, time error also can be produced imbalance Electric current.But, document [8] has only carried out qualitative analysis to this, does 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. line differential protection based on transient state condenser current compensation. power system is certainly Dynamicization, 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. traveling-wave differential protection [J] based on wavelet transformation. Automation of Electric Systems, 2004,28(18):25-29.

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

[8] Zhang Wujun, what Pentium. the analysis of traveling-wave differential protection material unaccounted for flow point and practical plan [J]. power system is automatic Change, 2007,31 (20): 49-55.

Summary of the invention

It is an object of the invention to overcome the above-mentioned deficiency of prior art, the out-of-balance current of traveling-wave differential protection is being carried out On the basis of quantitative analysis, it is provided that a kind of to realize simple, to be not susceptible to protection misoperation transmission line travelling wave differential protection whole Determine method, then realize each phase traveling-wave differential protection Fault Identification is judged and protection act.

Technical scheme is as follows:

A kind of transmission line travelling wave differential protecting method, if the total length of protected both-end transmission line of electricity is l, according to following The setting valve of step calculating traveling-wave differential protection:

(1). by transmission line parameter, calculate 0 mould amplitude attenuation factor alpha according to the following formula₀；

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

In formula, ω is power frequency angular velocity, L₀It is 0 mould unit length inductance, C₀It is 0 mould unit length ground capacity, R₀It is 0 Mould resistance per unit length；

(2). by 0 mould amplitude attenuation factor alpha₀Calculate 0 mould unbalance factor

(3). according to transmission system parameter, the maximum 0 mould direct wave electricity of m end when determining transmission line of electricity district external ground fault StreamAnd determine the maximum 0 mould backward-travelling wave electric current of n end

(4). by calculated for above step data substitute into following formula, following formula calculate respectively direct wave differential protection and Each phase setting valve of backward-travelling wave differential protection:

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

In formula, I_op1.a、I_op1.b、I_op1.cIt is respectively direct wave differential protection each phase setting valve, I_op2.a、I_op2.b、I_op2.c It is respectively backward-travelling wave differential protection each phase setting valve.

(5) differential to row ripple according to the direct wave differential protection of gained and each phase setting valve of backward-travelling wave differential protection Protection is adjusted, thus realizes judging each phase traveling-wave differential protection Fault Identification and protection act.

Beneficial effects 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 is given is abundant, adjusts simple easy OK；

2, the setting method of the present invention can guarantee that power system is properly functioning, external area error and during line no-load switching, OK Ripple differential protection reliable operation without misoperation.

Accompanying drawing explanation

The single-phase lossless circuit row ripple schematic diagram of Fig. 1.

Detailed description of the invention

The present invention will be described with embodiment below in conjunction with the accompanying drawings.

Mn in Fig. 1 is single-phase lossless circuit, a length of l, velocity of wave be v, τ=l/v be row ripple biography on total track length Broadcasting time delay, natural impedance is Z_c, circuit two ends electric current positive direction is for be flowed to circuit by bus, and this current visible is by forward current row Ripple and reverse current row ripple superposition are constituted.Forward current row ripple is defined as left end and flows to right-hand member, otherwise be reverse current row ripple, Then 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,It is respectively m end and the forward of n end, reverse current row ripple；i_m(t)、i_n (t)、u_m(t)、u_nT () is m end and the electric current of n end, voltage.Definition forward and reverse row ripple difference 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 lossless circuit, the direct wave at m point is propagated through τ time delay and is arrived n point, becomes the direct wave of n point；N point The backward-travelling wave at place travels to m point through duration τ, becomes the backward-travelling wave of m point, so i during properly functioning and external fault_D1(t) And i_D2T () is zero.During line-internal fault, due to the existence of trouble point, circuit is no longer uniform transmission line, i_D1(t) and i_D2 T () is not zero, its amplitude is equal to current in the fault point amplitude, and traveling-wave differential protection action is tripped.

Describing the partial differential equation of voltage x current variation characteristic on lossy uniform transmission lines road is:

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

In above formula, R, L, G, C are the parameter of circuit unit length, and (x, t), (x t) is the electricity of x point t on circuit to i to u Pressure and electric current.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, WithDetermined by boundary condition.Because circuit top is installed for protection Place, its voltage x current is known asThen:

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

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

In formula,It is direct wave component, can makeOwing to row ripple amplitude during advancing occurs Decay, phase place produce displacement, therefore, after traveling unit length byBecomeHave:

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

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

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

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, for the amplitude attenuation coefficient of unit length upstream ripple, β is the phase coefficient of unit length upstream ripple.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 characterizes the relation between row wave amplitude and the Changing Pattern of phase place and line parameter circuit value.R is the biggest, and row ripple is at circuit The energy of upper loss is the biggest, and amplitude attenuation coefficient is the biggest；L, C are the biggest, and the alternately conversion of electric field and magnetic field is the most violent, phase coefficient β is the biggest.

So, at mn for damaging under power line condition, two ends direct waveWithAmplitude 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,Represent the direct wave amplitude of distance m end h km.Abbreviation formula obtains:

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

WithPhase contrast 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 the following relation that exists between the direct wave of circuit two ends:

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

Therefore, during properly functioning or external area error, the poor stream of direct wave is:

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

It is forward out-of-balance current.Because ω is τ=ω l/v and v=ω/β, so ω τ=l β, i.e.Time Move the phase shift that compensate for 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 unbalance factor μ=1-e^-lα, then:

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

For backward-travelling wave, can obtain, through similar derivation, the following relation that exists between the backward-travelling wave of circuit two ends:

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

During properly functioning or external area error, the poor stream of backward-travelling wave is material unaccounted for streamFor:

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

Abbreviation obtains:

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

Unbalance factor μ is relevant with the amplitude attenuation factor alpha of line length and row ripple.Circuit is the longest, the electricity of unit length Hindering the biggest, μ is the biggest,WithThe biggest.Material unaccounted for stream also top amplitude with forward and reverse row ripple at circuit two ends has Close, the direct wave that m end flows throughIt is the biggest,The biggest；The backward-travelling wave that n end flows throughIt is the biggest,The biggest.

Be analyzed above is single-phase power transmission line, for three phase line, realizes three-phase with the conversion of card human relations Bauer Phase mould decouples, and 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 the 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 difference 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 out-of-balance current that can be obtained each modulus forward difference stream by the derivation of out-of-balance current above is:

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

In formula, μ₀、μ₁、β₀、β₁It is respectively 0 mould and the unbalance factor of 1 mould and phase coefficient.From transformation matrix formula (22) Understanding, 2 modular transformations are the different of separate combination from 1 modular transformation, and the modulus parameter after conversion is identical, so the imbalance of 2 moulds Coefficient is identical with phase coefficient and 1 mould.In actual track, unit length line conductance G=0, by formula and 0 mould of circuit and 1 mould Parameter can obtain:

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

The unbalance factor of 0 mould and 1 mould is:

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

From ultra-high/extra-high voltage transmission line of electricity 0 mould and the line parameter circuit value of 1 mould: μ₀＞ ＞ μ₁, therefore system properly functioning and During external area error, in the maximum imbalance current of three kinds of modulus, 0 mould out-of-balance current is maximum.So, when generating region external ground event Barrier,While obtaining maximum,Also the maximum imbalance current in modulus is obtained.In like manner, for reverse poor stream, when Generating region external ground faultWhile obtaining maximum, correspondinglyAlso the maximum imbalance current in modulus is obtained.

For split-phase traveling-wave differential protection, owing to there is, between phase each during fault, the effect of intercoupling, so each phase Direct wave difference current can be expressed through inverse transformation by the difference current that there are not 0 mould of coupling, 1 mould and 2 moulds.Therefore, Direct wave difference current i of a phase is obtained by phase mould inverse transformation_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 difference current of b phase and c phase is also obtained 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 row ripple occurs amplitude attenuation and phase offset, the row of each phase in transmitting procedure Ripple is differential properly functioning in system and has out-of-balance current to export under external fault condition.Split-phase is analyzed below.

For a phase direct wave difference 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,WithIt is respectively the out-of-balance current of each mould,'s Maximum is more thanWithOrderMaximum beThen 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 a phase direct wave protection properly functioning and outside break down in the case of ensure not malfunction Make, its setting valve I_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 difference 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 b phase direct wave protection properly functioning and outside break down in the case of ensure not malfunction Make, its setting valve I_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 difference 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 c phase direct wave protection properly functioning and outside break down in the case of ensure not malfunction Make, its setting valve I_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, the backward-travelling wave differential protection for each phase carries out similarity analysis can to obtain setting valve 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 backward-travelling wave electric current of n end during transmission line of electricity district external ground fault.

In sum, split-phase forward, backward-travelling wave differential protection for three-phase ultra-high/extra-high voltage transmission line of electricity should be according to such as Lower principle is adjusted:

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

For known both-end transmission line of electricity, the setting valve of traveling-wave differential protection can be calculated according to following steps.

1., by transmission line parameter, calculate 0 mould amplitude attenuation factor alpha according to the following formula₀；

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

In above formula, ω is power frequency angular velocity, L₀It is 0 mould unit length inductance, C₀It is 0 mould unit length ground capacity, R₀For 0 mould resistance per unit length.

2. by 0 mould amplitude attenuation factor alpha₀Calculate 0 mould unbalance factor

3. according to transmission system parameter, the maximum 0 mould direct wave electric current of m end when determining transmission line of electricity district external ground faultAnd determine the maximum 0 mould backward-travelling wave electric current of n end

4. calculated for above step data are substituted into following formula, following formula calculate calculating direct wave differential protection respectively Each phase setting valve with backward-travelling wave differential protection:

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

In above formula, I_op1.a、I_op1.b、I_op1.cIt is respectively direct wave differential protection each phase setting valve, I_op2.a、I_op2.b、 I_op2.cIt is respectively backward-travelling wave differential protection each phase setting valve；μ₀It it is 0 mould unbalance factor；For m during district's external ground fault The maximum 0 mould direct wave electric current of end,For the maximum 0 mould backward-travelling wave electric current of n end during district's external ground fault.

Below, as a example by direct wave differential protection a phase is differential, the mistake that traveling-wave differential protection Fault Identification judges is described Journey.Direct wave differential protection b, c be biphase and each phase of backward-travelling wave differential protection to realize process the most similar, repeat no more.

1. by circuit m end measurement to a phase voltage u_m-a、i_m-aIt is calculated m end a phase forward current row rippleBy circuit A phase voltage u that n end measurement is arrived_n-a、i_n-aIt is calculated n end a phase forward current row ripple

2., by line parameter circuit value, the propagation delay τ of circuit 0 mould and 1 mould can be obtained₀、τ₁；

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

4. obtained circuit m end 0 mould direct wave by the voltage of circuit m end 0 mould, Current calculation

5. calculated for above step data are substituted into following formula, can extreme a phase row ripple difference current:

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

6. compare a phase row ripple difference current and a fix threshold:

I_D1.a＞ I_op1.a (41)

If above formula meets, then it is assumed that a phase fault；If above formula is unsatisfactory for, it is believed that a phase does not has fault.

Claims (1)

1. a transmission line travelling wave differential protecting method, if the total length of protected m, n both-end transmission line of electricity is l, according to The setting valve of lower step calculating traveling-wave differential protection:

(1). by transmission line parameter, calculate 0 mould amplitude attenuation factor alpha according to the following formula₀；

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

In formula, ω is power frequency angular velocity, L₀It is 0 mould unit length inductance, C₀It is 0 mould unit length ground capacity, R₀It it is 0 mould list Bit length resistance；

(2). by 0 mould amplitude attenuation factor alpha₀Calculate 0 mould unbalance factor

(3). according to transmission system parameter, the maximum 0 mould direct wave electric current of m end when determining transmission line of electricity district external ground faultAnd determine the maximum 0 mould backward-travelling wave electric current of n end

(4). calculated for above step data are substituted into following formula, following formula calculates direct wave differential protection respectively with reverse Each phase setting valve of traveling-wave differential protection:

$\begin{array}{cc}\left\{\begin{array}{c}{I}_{op1.a}={\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{op1.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\\ I_{op1.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{m-0-\max }^{+}\end{array}\right.& \left\{\begin{array}{c}{I}_{op2.a}={\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{op2.b}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\\ I_{op2.c}=\frac{4}{3}{\mathrm{\&mu;}}_0{I}_{n-0-\max }^{-}\end{array}\right.\end{array}$

In formula, I_op1.a、I_op1.b、I_op1.cIt is respectively direct wave differential protection each phase setting valve, I_op2.a、I_op2.b、I_op2.cRespectively For backward-travelling wave differential protection each phase setting valve；

(5) according to the direct wave differential protection of gained and each phase setting valve of backward-travelling wave differential protection to traveling-wave differential protection Adjust, thus realize each phase traveling-wave differential protection Fault Identification is judged and protection act.