CN102570419B - Power transmission line pilot protection method based on magnitude of current - Google Patents

Abstract

The invention relates to a power transmission line pilot protection method based on magnitude of current, which comprises the following steps: firstly, collecting A, B and C three-phase current of a current transformer by use of protective devices installed at two ends of a power transmission line, then treating the collected three-phase current to obtain sampling values; after that, sending the sampling values at the corresponding end of the power transmission line to the opposite end by each protective device which receives the sampling value sent from the opposite end; after that, performing phrase-to-modulus conversion on the sampling values of the three-phase current to obtain corresponding modulus information; computing current fault component at the two ends of the power transmission line as delta im(t) and delta in(t); after that, collecting power frequency component and transient component of the current fault component at the two ends of the power transmission line in sequence; computing power frequency component and transient component of voltage at installing parts of the protective devices at the two ends of the power transmission line; computing system impedance parameters at the two ends of the power transmission line as Lm, Rm and Ln, Rn; computing transient current energy; computing model error En and Em to judge the fault position if the transient current energy is relative big; and judging the fault position according to the changing amplitude of the system impedance parameters and the voltage and power frequency component. The power transmission line pilot protection method provided by the invention has the characteristics of only using information composition of two-end current and not being influenced by distributed capacitance.

Description

A kind of electric transmission line longitudinal protection method based on the magnitude of current

Technical field

The present invention relates to Relay Protection Technology in Power System field, and in particular to a kind of electric transmission line longitudinal protection method based on the magnitude of current.

Background technology

Current comparison pilot protection has the advantages that criterion is simple, sensitivity is high, with natural phase selection ability, is widely used as the main protection of transmission line of electricity.But practical operating experiences show that the protection philosophy has following defect：(1) principle have ignored the distribution capacity that actual transmission line of electricity is present.With the raising and the growth of line length of transmission line of electricity voltage class, distribution capacity will be increasing, and this causes fault transient process more obvious, and transient current will influence the computational accuracy of traditional protection algorithm；(2) stable state capacitance current can change the size and phase of both sides power current.In a word, ignoring the influence of distribution capacity can cause protection sensitivity to reduce, may malfunction during external area error.

Processing method currently for above mentioned problem is broadly divided into two classes：The first kind is compensating electric capacity electric current, such as shunt reactor penalty method, phasor backoff algorithm and time domain compensation algorithm.First two method can only compensate stable state capacitance current, it is impossible to compensate transient state capacitance current, and shunt reactor and on-fixed access, and compensation effect is limited.Time domain compensation algorithm can compensate whole capacitance currents, but additionally introduce voltage, but field operation experiences show that the progress of disease performance and loop reliability of voltage transformer are all poor, therefore protective value can not be guaranteed.Equations of The Second Kind is to study the protection philosophy not influenceed by capacitance current, the auspicious carina road modelling of such as shellfish, pattern recognition and comprehensive impedance method.These methods consider the influence of distribution capacity in principle, thus without compensating electric capacity electric current.But due to having used both-end voltage data, it is desirable to which one end voltage is transmitted to opposite end, adds transinformation, higher is required to communication port.

Therefore, for line protection, studying a kind of can solve the problems, such as without using voltage and effectively that the guard method of capacitance current is necessary.

The content of the invention

In order to overcome the shortcoming that above-mentioned prior art is present, it is an object of the invention to provide a kind of electric transmission line longitudinal protection method based on the magnitude of current, this method has to be constituted merely with two ends current information, the characteristics of not influenceed by distribution capacity.

To reach above-mentioned purpose, the present invention uses following technical scheme：

A kind of electric transmission line longitudinal protection method based on the magnitude of current, comprises the following steps：

Step 1: line protection is arranged on the two ends m ends and n ends of transmission line of electricity, A, B, C three-phase current that protection device collection local terminal protects installation place current transformer are often covered；

Kept and A/D conversion process Step 2: line protection carries out LPF, sampling to the three-phase current collected, obtain the sampled value of three-phase current；

Step 3: line protection each transmits the three-phase current sampling value information of local terminal to opposite end, while receiving the three-phase current sampling value information transmitted opposite end；

Step 4: line protection combination phase selection result, the sampled value to three-phase current carries out phase-modal transformation, corresponding modulus information is extracted, is comprised the following steps that：

A kind of formula of phase-modal transformation is as follows：

$\left[\begin{array}{c}{i}_{p1}\\ i_{p2}\\ i_{p3}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right]\left[\begin{array}{c}{i}_{\mathrm{pA}}\\ i_{\mathrm{pB}}\\ i_{\mathrm{pC}}\end{array}\right]---\left(1\right)$

In formula：Subscript p represents m or n；Index number (1,2,3) represents the sequence number (1 modulus, 2 modulus, 3 modulus) of modulus,

If phase selection result shows to there occurs three-phase fault, 1 modulus is extracted；

If there occurs AB phases short trouble or AB phase earth faults, 1 modulus is extracted；

If there occurs BC phases short trouble or BC phase earth faults, 2 modulus are extracted；

If there occurs CA phases short trouble or CA phase earth faults, 3 modulus are extracted；

If there occurs A phase earth faults, 1 modulus is extracted；

If there occurs B phase earth faults, 2 modulus are extracted；

If there occurs C phase earth faults, 3 modulus are extracted；

Step 5: line protection calculates two ends current failure component Δ i_m(t), Δ i_n(t), computational methods are as follows：

$\left\{\begin{array}{c}\mathrm{\Δ}{i}_m\left(t\right)=i_{\mathrm{mK}}\left(t\right)-i_{\mathrm{mK}}(t-\mathrm{cT})\\ \mathrm{\Δ}{i}_n\left(t\right)=i_{\mathrm{nK}}\left(t\right)-i_{\mathrm{nK}}(t-\mathrm{cT})\end{array}\right.---\left(2\right)$

In formula：Subscript K represents the sequence number for the modulus that step 4 is extracted, K=1,2,3；C takes positive integer, and numerical values recited is determined by protection device；T is the cycle of power current；

Step 6: line protection extracts power frequency component and transient state component in the current failure component of two ends using matrix pencil algorithm successively；

The step of matrix pencil algorithm carries out spectrum analysis to signal is summarized as follows：

First, it is analysed to signal and generates sampling matrix Y according to certain rule₁, Y₂；Secondly, calculating matrix Y₁ ⁺Y₂Characteristic value, wherein Y₁ ⁺It is Y₁Moore-Penrose pseudo inverse matrixs, this feature value contains the frequency of all subsignals and decay factor information in signal to be analyzed；Finally, the amplitude and initial phase information of all subsignals are obtained by solving least square problem；

Current failure component is carried out after spectrum analysis using matrix pencil algorithm, whether the frequency according to subsignal is that subsignal is classified as power frequency component or transient state component by power frequency；

There is following relation between current failure component, power frequency component and transient state component three：

$\left\{\begin{array}{c}\mathrm{\Δ}{i}_m\left(t\right)=\mathrm{\Δ}{i}_{\mathrm{ms}}\left(t\right)+\mathrm{\Δ}{i}_{\mathrm{mt}}\left(t\right)\\ \mathrm{\Δ}{i}_n\left(t\right)=\mathrm{\Δ}{i}_{\mathrm{ns}}\left(t\right)+\mathrm{\Δ}{i}_{\mathrm{nt}}\left(t\right)\end{array}\right.---\left(3\right)$

In formula：Δi_ms(t), Δ i_mt(t) be respectively m ends electric current power frequency component and transient state component；Δi_ns(t), Δ i_nt(t) be respectively n ends electric current power frequency component and transient state component；

Step 7: the power frequency component and transient state component of the two ends electric current obtained using step 6, line protection calculate the power frequency component and transient state component that installation place voltage is protected at two ends, comprise the following steps that：

First, the conventional method of known two ends Current calculation both end voltage is illustrated：

Under complex frequency domain, using distributed parameter model, following relation is obeyed between transmission line of electricity n terminal voltages electric current and m terminal voltage electric currents：

$\left[\begin{array}{c}{U}_n\left(s\right)\\ I_n\left(s\right)\end{array}\right]=\left[\begin{array}{cc}\cosh \left(\mathrm{\γd}\right)& Z_{C0}\·\sinh \left(\mathrm{\γd}\right)\\ \sinh \left(\mathrm{\γd}\right)/Z_{C0}& \cosh \left(\mathrm{\γd}\right)\end{array}\right]\left[\begin{array}{c}{U}_m\left(s\right)\\ -I_m\left(s\right)\end{array}\right]---\left(4\right)$

In formula：D is line length；Z_C0For line characteristic impedance, calculating formula isr₁, l₁, g₁, c₁The respectively positive sequence resistance of circuit unit length, inductance, conductance, capacitance parameter；γ is line propagation coefficient, and calculating formula is $\mathrm{\&gamma;}=\sqrt{({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">r</figure-callout>}}_1+{\mathrm{sl}}_1)({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{sc}}_1)};$

When known two ends electric current, by (4), formula can solve both end voltage, derive and formula is calculated as below：

$\left[\begin{array}{c}{U}_m\left(s\right)\\ U_n\left(s\right)\end{array}\right]=\left[\begin{array}{cc}{Z}_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)\\ Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)\end{array}\right]\left[\begin{array}{c}{I}_m\left(s\right)\\ I_n\left(s\right)\end{array}\right]---\left(5\right)$

When writing software program and realizing, a kind of practical numerical computation method of (5) formula is as follows, and it describes variable quantity of the terminal voltage in 2 τ time intervals：

$\left\{\begin{array}{c}{u}_m(t+\mathrm{\&tau;})-u_m(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_m(t+\mathrm{\&tau;})+i_m(t-\mathrm{\&tau;})+{2i}_n\left(t\right)]+R/2\&CenterDot;[i_m(t+\mathrm{\&tau;})-i_m(t-\mathrm{\&tau;})]\\ u_n(t+\mathrm{\&tau;})-u_n(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_n(t+\mathrm{\&tau;})+i_n(t-\mathrm{\&tau;})+{2i}_m\left(t\right)]+R/2\&CenterDot;[i_n(t+\mathrm{\&tau;})-i_n(t-\mathrm{\&tau;})]\end{array}\right.---\left(6\right)$

In formula：τ is time-consuming for propagation of the electromagnetic wave on power transmission line, and calculating formula is

To ignore characteristic impedance during line loss, calculating formula is

R is the total resistance value of transmission line of electricity, and calculating formula is R=r₁d；

Secondly, the practical calculation method of both end voltage power frequency component and transient state component is illustrated：

The two ends electric current power frequency component Δ i that step 6 is obtained_ms(t), Δ i_ns(t) (6) formula is substituted into, the practical formula of both end voltage power frequency component variable quantity is obtained：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{ns}}\left(t\right)]+R/2\&CenterDot;[\mathrm{\&Delta;}{i}_{\mathrm{ms}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t-\mathrm{\&tau;})]\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{ns}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[\mathrm{\&Delta;}{i}_{\mathrm{ns}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{ms}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t-\mathrm{\&tau;})]\end{array}\right.---\left(7\right)$

The two ends current temporary state component Δ i that step 6 is obtained_mt(t), Δ i_nt(t) (6) formula is substituted into, the practical formula of both end voltage transient state component variable quantity is obtained：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{mt}}(t+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{mt}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{nt}}\left(t\right)]+R/2\&CenterDot;[\mathrm{\&Delta;}{i}_{\mathrm{mt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t-\mathrm{\&tau;})]\\ \mathrm{\&Delta;}{u}_{\mathrm{nt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{nt}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[\mathrm{\&Delta;}{i}_{\mathrm{nt}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{mt}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t-\mathrm{\&tau;})]\end{array}\right.---\left(8\right)$

Step 8: the voltage power frequency component variable quantity that the electric current power frequency component and step 7 that are obtained using step 6 are obtained, line protection calculates both sides system impedance parameter L successively using two figure parameters method of identifications_m, R_mAnd L_n, R_n, comprise the following steps that：

First, illustrate that two figure parameters method of identifications calculate L_m, R_mMethod：

The current failure component and voltage failure component of m sides obey following relation：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{u}_m(t+\mathrm{\&tau;})=-L_m\frac{\mathrm{d\&Delta;}{i}_m(t+\mathrm{\&tau;})}{\mathrm{dt}}-R_m\mathrm{\&Delta;}{i}_m(t+\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_m(t-\mathrm{\&tau;})=-L_m\frac{\mathrm{d\&Delta;}{i}_m(t-\mathrm{\&tau;})}{\mathrm{dt}}-R_m\mathrm{\&Delta;}{i}_m(t-\mathrm{\&tau;})\end{array}\right.---\left(9\right)$

Two formulas, which are subtracted each other, to be obtained：

$[\frac{\mathrm{d\&Delta;}{i}_m(t-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_m(t+\mathrm{\&tau;})}{\mathrm{dt}}]L_m+[\mathrm{\&Delta;}{i}_m(t-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_m(t+\mathrm{\&tau;})]R_m=\mathrm{\&Delta;}{u}_m(t+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_m(t-\mathrm{\&tau;})---\left(10\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, row write equation successively, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(11\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

(11) formula is abbreviated as：

$\left[i^{\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[u^{\&prime;}\right]---\left(12\right)$

Using least-squares calculation parameter L_m, R_m：

$\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]={({\left[i^{\&prime;}\right]}^T\&CenterDot;[i^{\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;}\right]---\left(13\right)$

Secondly, illustrate that two figure parameters method of identifications calculate L_n, R_nMethod：

The current failure component and voltage failure component of n sides obey following relation：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{u}_n(t+\mathrm{\&tau;})=-L_n\frac{\mathrm{d\&Delta;}{i}_n(t+\mathrm{\&tau;})}{\mathrm{dt}}-R_n\mathrm{\&Delta;}{i}_n(t+\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_n(t-\mathrm{\&tau;})=-L_n\frac{\mathrm{d\&Delta;}{i}_n(t-\mathrm{\&tau;})}{\mathrm{dt}}-R_n\mathrm{\&Delta;}{i}_n(t-\mathrm{\&tau;})\end{array}\right.---\left(14\right)$

Two formulas, which are subtracted each other, to be obtained：

$[\frac{\mathrm{d\&Delta;}{i}_n(t-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_n(t+\mathrm{\&tau;})}{\mathrm{dt}}]L_n+[\mathrm{\&Delta;}{i}_n(t-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_n(t+\mathrm{\&tau;})]R_n=\mathrm{\&Delta;}{u}_n(t+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_n(t-\mathrm{\&tau;})---\left(15\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, row write equation successively, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(16\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

(16) formula is abbreviated as：

$\left[i^{\&prime;\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[u^{\&prime;\&prime;}\right]---\left(17\right)$

Using least-squares calculation parameter L_n, R_n：

$\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]={({\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;[i^{\&prime;\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;\&prime;}\right]---\left(18\right)$

Step 9: the two ends current temporary state component that step 6 is obtained is substituted into following formula by line protection, transient current energy is calculated,

$\mathrm{\&Delta;}{I}_{\mathrm{mt}}=\sqrt{\frac{1}{t_2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1}{\&Integral;}_{t_1}^{t_2}\mathrm{\&Delta;}{i}_{\mathrm{mt}}^2\left(t\right)\mathrm{dt}}---\left(19\right)$

$\mathrm{\&Delta;}{I}_{\mathrm{nt}}=\sqrt{\frac{1}{t_2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">t</figure-callout>}}_1}{\&Integral;}_{t_1}^{t_2}\mathrm{\&Delta;}{i}_{\mathrm{nt}}^2\left(t\right)\mathrm{dt}}---\left(20\right)$

In formula：t₁, t₂It is the initial time and end time of 10ms data window length respectively；

If Δ I_mt＞ I_setAnd Δ I_nt＞ I_set, then transient current energy is larger, goes to step ten, starts the main criterion based on model error；If Δ I_mt≤I_setOr Δ I_nt≤I_set, then transient current energy is not enough, goes to step 11, starts in the assistant criteria based on system impedance and voltage power frequency component, formula, I_setFor current calibration threshold, 0.1 times of circuit rated current is set to；

Step 10: the system impedance parameter that voltage transient component variation amount and step 8 that the current temporary state component obtained using step 6, step 7 are obtained are obtained, line protection computation model error E_n, E_m, and failure judgement position, comprise the following steps that：

First, model error E is illustrated_nComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_m, R_mFollowing formula is substituted into, a is calculated respectively_nj, b_nj：

$\left\{\begin{array}{c}{a}_{\mathrm{nj}}=(\frac{\mathrm{d\&Delta;}{i}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{mt}}(t_j+\mathrm{\&tau;})}{\mathrm{dt}})L_m+({\mathrm{\&Delta;i}}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{mt}}(t_j+\mathrm{\&tau;}))R_m\\ b_{\mathrm{nj}}=\mathrm{\&Delta;}{u}_{\mathrm{mt}}(t_j+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})\end{array}\right.j=\mathrm{1,2},L,k---\left(21\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_nj, b_njFollowing formula is substituted into, calculating obtains E_n：

$E_n=\frac{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}|a_{\mathrm{nj}}-b_{\mathrm{nj}}|}{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{nj}}\right|+\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{nj}}\right|}---\left(22\right)$

Secondly, model error E is illustrated_mComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_n, R_nFollowing formula is substituted into, a is calculated respectively_mj, b_mj：

$\left\{\begin{array}{c}{a}_{\mathrm{mj}}=(\frac{\mathrm{d\&Delta;}{i}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{nt}}(t_j+\mathrm{\&tau;})}{\mathrm{dt}})L_n+({\mathrm{\&Delta;i}}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{nt}}(t_j+\mathrm{\&tau;}))R_n\\ b_{\mathrm{mj}}=\mathrm{\&Delta;}{u}_{\mathrm{nt}}(t_j+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})\end{array}\right.j=\mathrm{1,2},L,k---\left(23\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_mj, b_mjFollowing formula is substituted into, calculating obtains E_m：

$E_m=\frac{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}|a_{\mathrm{mj}}-b_{\mathrm{mj}}|}{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{mj}}\right|+\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{mj}}\right|}---\left(24\right)$

Finally, the result of calculation of model error is substituted into Protection criteria, failure judgement position：

If E_n＜ ξ or E_m＜ ξ, are judged to external area error, and protection is failure to actuate；

If E_n>=ξ and E_m>=ξ, is judged to troubles inside the sample space, protection act；

In criterion, ξ is action threshold, is adjusted according to the maximum amount of unbalance of model error when occurring external area error；

Step 11: the voltage power frequency component variable quantity that system impedance parameter and step 7 that line protection is obtained using step 8 are obtained, failure judgement position is comprised the following steps that：

If L_m＜ 0 or L_n＜ 0, is judged to external area error, and protection is failure to actuate；

If L_m>=0 and L_n>=0, it is necessary to could failure judgement position with reference to voltage power frequency component variable quantity；

Extract the amplitude Δ U of both end voltage power frequency component variable quantity respectively using matrix pencil algorithm_ms, Δ U_ns,

If Δ U_ms＜ U_setAnd Δ U_ns＜ U_set, external area error is judged to, protection is failure to actuate；

If Δ U_ms≥U_setOr Δ U_ns≥U_set, it is judged to troubles inside the sample space, protection act；

In criterion, U_setThreshold is adjusted for voltage, numerical value is

Wherein K_relFor safety factor, 1.1-1.2, U are taken as_NFor the voltage class of transmission line of electricity.

The present invention mainly proposes four kinds of new methods：(1) method that both end voltage is calculated using two ends current information; this method is based on distributed parameter model; voltage result of calculation has higher computational accuracy; so that protection is not necessarily based on voltage transformer information; the requirement to protecting hardware configuration is reduced, while improving the reliability of protection philosophy；(2) based on parameter recognize system impedance computational methods, using least square method computing system impedance parameter, result of calculation have it is precise and stable, not by Effect of Transient Component the characteristics of；(3) the line protection criterion based on model error, the criterion make use of the transient information in fault current, it is sensitive it is reliable, quick acting can be realized, as main criterion；(4) assistant criteria based on system impedance and voltage power frequency component, when transient information is not enough in fault current, the main criterion based on model error is exited, assistant criteria input.

Compared with prior art, the present invention mainly has following five advantages：

1st, this method employs Transmission Line Distributed Parameter model, is not influenceed by distribution capacity, without compensating electric capacity electric current, solve current comparison pilot protection due to the influence by line distribution capacitance and fault transient thus performance reduce the problem of.

2nd, this method is based only upon the information of current transformer, it is not necessary to use voltage transformer information, compared with existing guard method, the requirement to protecting hardware configuration is reduced, while also improving the reliability of protection philosophy.

3rd, this method is applied to the transmission line of electricity of various voltage class and various length, it is adaptable to the various system operation modes such as single-ended big power-supply system, the big power-supply system of both-end, weak feedback system, single supply on-load circuit, and tolerance transition resistance ability is stronger.

4th, this method make use of the information in broad frequency band in fault current, therefore require relatively low to the performance indications of low pass filter, and it is higher that cut-off frequency can be set, and only need to meet sampling thheorem.

5th, this method only need using 10ms data windows it is long, protection can realize quick acting.

Brief description of the drawings

Fig. 1 is the hardware block diagram of transmission line of electricity microcomputer pilot protection.

Fig. 2 is flow chart of data processing figure.

Fig. 3 is distributed constant model of power transmission system figure.

Fig. 4 is troubles inside the sample space illustraton of model, and wherein Fig. 4 a are failure general quantity model figure, and Fig. 4 b are fault component illustraton of model.

Fig. 5 is external area error illustraton of model, and wherein Fig. 5 a are failure general quantity model figure, and Fig. 5 b are fault component illustraton of model.

The calculating waveform of model error when Fig. 6 is three-phase fault outside m lateral areas.

In Tu7Wei areas during the three-phase fault of midpoint model error calculating waveform.

The calculating waveform of system inductance when Fig. 8 is A phase earth faults outside m lateral areas.

In Tu9Wei areas during midpoint A phase earth faults voltage power frequency component variable quantity amplitude calculating waveform.

Embodiment

The present invention is described in more detail with reference to the accompanying drawings and detailed description.

As shown in figure 1, the invention belongs to circuit Microcomputer Protection, circuit both sides are respectively arranged with the protection device using the principle of the invention, the protection device of both sides exchanges data by communication port, and Fig. 1 depicts the annexation between hardware, and Fig. 2 describes flow chart of data processing.By taking the protection device of m sides as an example; side current transformer TA1 collects the analog quantity of this side three-phase current; digital quantity is obtained after LPF, sampling are kept and A/D is changed; input DSP1; DSP1 is connected equipped with communication interface and with communication port simultaneously; on the one hand the current digital amount locally collected is sent to n sides, on the other hand receives the current digital amount that n sides are transmitted.The core content of the present invention is achieved by programming in microsystem DSP.DSP is once it is determined that failure occurs in area (shown in Fig. 4) or outside area (shown in Fig. 5), to send corresponding control signal and send relay to by Phototube Coupling, relay, which is received, makes corresponding action after control signal.

A kind of electric transmission line longitudinal protection method based on the magnitude of current, when being applied to but being not limited to the long 750kV transmission lines of electricity of 500km, specifically includes following steps：

First, the protection device installed in transmission line of electricity two ends (being designated as m ends and n ends) gathers A, B, C three-phase current i of this side current transformer respectively_mA(t), i_mB(t), i_mCAnd i (t)_nA(t), i_nB(t), i_nC(t)；

2nd, line protection carries out LPF, sampling holding and A/D conversion process to the three-phase current collected, obtains the sampled value i of three-phase current_mA(k), i_mB(k), i_mCAnd i (k)_nA(k), i_nB(k), i_nC(k).The cut-off frequency of low pass filter is set to 600Hz, primarily to eliminating the influence of noise, makes algorithm more stable, sample frequency is set to 10kHz, now per cycle sampling number N=200, sampling time interval T_S=0.1ms；

3rd, after failure occurs, starting element starts, and is put into using the pilot protection element of this method, and line protection each transmits the three-phase current sampling value information of local terminal to opposite end, while receiving the three-phase current sampling value information transmitted opposite end；

4th, line protection combination phase selection result, the sampled value to three-phase current carries out phase-modal transformation, extracts corresponding modulus information, comprises the following steps that：

A kind of formula of phase-modal transformation is as follows：

$\left[\begin{array}{c}{i}_{p1}\\ i_{p2}\\ i_{p3}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right]\left[\begin{array}{c}{i}_{\mathrm{pA}}\\ i_{\mathrm{pB}}\\ i_{\mathrm{pC}}\end{array}\right]---\left(1\right)$

In formula：Subscript p represents m or n；Index number (1,2,3) represents the sequence number (1 modulus, 2 modulus, 3 modulus) of modulus；

If phase selection result shows to there occurs three-phase fault, 1 modulus is extracted；

If there occurs AB phases short trouble or AB phase earth faults, 1 modulus is extracted；

If there occurs BC phases short trouble or BC phase earth faults, 2 modulus are extracted；

If there occurs CA phases short trouble or CA phase earth faults, 3 modulus are extracted；

If there occurs A phase earth faults, 1 modulus is extracted；

If there occurs B phase earth faults, 2 modulus are extracted；

If there occurs C phase earth faults, 3 modulus are extracted；

5th, line protection calculates two ends failure of the current component Δ i_m(k), Δ i_n(k), computational methods are as follows：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{i}_m\left(k\right)=i_{\mathrm{mK}}\left(k\right)-i_{\mathrm{mK}}(k-\mathrm{cT})\\ \mathrm{\&Delta;}{i}_n\left(k\right)=i_{\mathrm{nK}}\left(k\right)-i_{\mathrm{nK}}(k-\mathrm{cT})\end{array}\right.---\left(2\right)$

In formula：Subscript K represents the sequence number for the modulus that the 4th step is extracted, K=1,2,3；C numerical values recited is determined by protection device, and 2 are taken as here；N is per cycle sampling number, N=200；

6th, line protection extracts the power frequency component and transient state component in the current failure component of two ends using matrix pencil algorithm successively；

The step of matrix pencil algorithm carries out spectrum analysis to signal is summarized as follows：

First, it is analysed to signal and generates sampling matrix Y according to certain rule₁, Y₂；Secondly, calculating matrix Y₁ ⁺Y₂Characteristic value, wherein Y₁ ⁺It is Y₁Moore-Penrose pseudo inverse matrixs, this feature value contains the frequency of all subsignals and decay factor information in signal to be analyzed；Finally, the amplitude and initial phase information of all subsignals are obtained by solving least square problem；

Current failure component is carried out after spectrum analysis using matrix pencil algorithm, whether the frequency according to subsignal is that subsignal is classified as power frequency component or transient state component by power frequency；

There is following relation between current failure component, power frequency component and transient state component three：

$\left\{\begin{array}{c}\mathrm{\&Delta;}{i}_m\left(k\right)=\mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(k\right)+\mathrm{\&Delta;}{i}_{\mathrm{mt}}\left(k\right)\\ \mathrm{\&Delta;}{i}_n\left(k\right)=\mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(k\right)+\mathrm{\&Delta;}{i}_{\mathrm{nt}}\left(k\right)\end{array}\right.---\left(3\right)$

In formula：Δi_ms(k), Δ i_mt(k) be respectively m ends electric current power frequency component and transient state component；Δi_ns(k), Δ i_nt(k) be respectively n ends electric current power frequency component and transient state component；

7th, the power frequency component and transient state component of the two ends electric current obtained using the 6th step, line protection calculates the power frequency component and transient state component that installation place voltage is protected at two ends, comprises the following steps that：

First, the conventional method of known two ends Current calculation both end voltage is illustrated：

Under complex frequency domain, using distributed parameter model, following relation is obeyed between transmission line of electricity n terminal voltages electric current and m terminal voltage electric currents：

$\left[\begin{array}{c}{U}_n\left(s\right)\\ I_n\left(s\right)\end{array}\right]=\left[\begin{array}{cc}\cosh \left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\sinh \left(\mathrm{\&gamma;d}\right)\\ \sinh \left(\mathrm{\&gamma;d}\right)/Z_{C0}& \cosh \left(\mathrm{\&gamma;d}\right)\end{array}\right]\left[\begin{array}{c}{U}_m\left(s\right)\\ -I_m\left(s\right)\end{array}\right]---\left(4\right)$

In formula：D is line length；Z_C0For line characteristic impedance, calculating formula is

r₁, l₁, g₁, c₁The respectively positive sequence resistance of circuit unit length, inductance, conductance, capacitance parameter；γ is line propagation coefficient, and calculating formula is $\mathrm{\&gamma;}=\sqrt{({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">r</figure-callout>}}_1+{\mathrm{sl}}_1)({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="HDA0000126081780000033.png,HDA0000126081780000042.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{sc}}_1)};$

When known two ends electric current, by (4), formula can solve both end voltage, derive and formula is calculated as below：

$\left[\begin{array}{c}{U}_m\left(s\right)\\ U_n\left(s\right)\end{array}\right]=\left[\begin{array}{cc}{Z}_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)\\ Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)\end{array}\right]\left[\begin{array}{c}{I}_m\left(s\right)\\ I_n\left(s\right)\end{array}\right]---\left(5\right)$

When writing software program and realizing, a kind of practical numerical computation method of (5) formula is as follows, and it describes variable quantity of the terminal voltage in 2 τ time intervals：

$\left\{\begin{array}{c}{u}_m(t+\mathrm{\&tau;})-u_m(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_m(t+\mathrm{\&tau;})+i_m(t-\mathrm{\&tau;})+{2i}_n\left(t\right)]+R/2\&CenterDot;[i_m(t+\mathrm{\&tau;})-i_m(t-\mathrm{\&tau;})]\\ u_n(t+\mathrm{\&tau;})-u_n(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_n(t+\mathrm{\&tau;})+i_n(t-\mathrm{\&tau;})+{2i}_m\left(t\right)]+R/2\&CenterDot;[i_n(t+\mathrm{\&tau;})-i_n(t-\mathrm{\&tau;})]\end{array}\right.---\left(6\right)$

In formula：τ is time-consuming for propagation of the electromagnetic wave on power transmission line, and calculating formula is

To ignore characteristic impedance during line loss, calculating formula is

R is the total resistance value of transmission line of electricity, and calculating formula is R=r₁d；

Secondly, the practical calculation method of both end voltage power frequency component is illustrated：

Write (6) formula as discrete-time version, and substitute into the two ends electric current power frequency component Δ i that the 6th step is obtained_ms(k), Δ i_ns(k) practical formula of both end voltage power frequency component variable quantity, is obtained：

For the long 750kV transmission lines of electricity of 500km, τ ≈ 1.7ms,R≈6.4Ω.Therefore (7) formula is changed into：

Finally, the practical calculation method of both end voltage transient state component is illustrated：

Write (6) formula as discrete-time version, and substitute into the two ends current temporary state component Δ i that the 6th step is obtained_mt(k), Δ i_nt(k) practical formula of both end voltage transient state component variable quantity, is obtained：

Substitute into τ,

R design parameter, (9) formula is changed into：

8th, the voltage power frequency component variable quantity that the electric current power frequency component and the 7th step obtained using the 6th step is obtained, line protection calculates both sides system impedance parameter L successively using two figure parameters method of identifications_m, R_mAnd L_n, R_n.Comprise the following steps that：

First, illustrate that two figure parameters method of identifications calculate L_m, R_mMethod：

The current failure component and voltage failure component of m sides obey following relation：

$[\frac{\mathrm{d\&Delta;}{i}_m(k-\mathrm{\&tau;}/T_S)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_m(k+\mathrm{\&tau;}/T_S)}{\mathrm{dt}}]L_m+[\mathrm{\&Delta;}{i}_m(k-\mathrm{\&tau;}/T_S)-\mathrm{\&Delta;}{i}_m(k+\mathrm{\&tau;}/T_S)]R_m=\mathrm{\&Delta;}{u}_m(k+\mathrm{\&tau;}/T_S)-\mathrm{\&Delta;}{u}_m(k-\mathrm{\&tau;}/T_S)---\left(11\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, can arrange and write out 64 equations, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(1\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(35\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(1\right)-\mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(35\right)\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(2\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(36\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(2\right)-\mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(36\right)\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(64\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(98\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(64\right)-\mathrm{\&Delta;}{i}_{\mathrm{ms}}\left(98\right)\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(35\right)-\mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(1\right)\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(36\right)-\mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(2\right)\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(98\right)-\mathrm{\&Delta;}{u}_{\mathrm{ms}}\left(64\right)\end{array}\right]---\left(12\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff；

(12) formula is abbreviated as：

$\left[i^{\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[u^{\&prime;}\right]---\left(13\right)$

Using least-squares calculation parameter L_m, R_m：

$\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]={({\left[i^{\&prime;}\right]}^T\&CenterDot;[i^{\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;}\right]---\left(14\right)$

Secondly, illustrate that two figure parameters method of identifications calculate L_n, R_nMethod：

The current failure component and voltage failure component of n sides obey following relation：

$[\frac{\mathrm{d\&Delta;}{i}_n(k-\mathrm{\&tau;}/T_S)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_n(k+\mathrm{\&tau;}/T_S)}{\mathrm{dt}}]L_n+[\mathrm{\&Delta;}{i}_n(k-\mathrm{\&tau;}/T_S)-\mathrm{\&Delta;}{i}_n(k+\mathrm{\&tau;}/T_S)]R_n=\mathrm{\&Delta;}{u}_n(k+\mathrm{\&tau;}/T_S)-\mathrm{\&Delta;}{u}_n(k-\mathrm{\&tau;}/T_S)---\left(15\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, can arrange and write out 64 equations, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}\left(1\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}\left(35\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(1\right)-\mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(35\right)\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}\left(2\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}\left(36\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(2\right)-\mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(36\right)\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}\left(64\right)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}\left(98\right)}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(64\right)-\mathrm{\&Delta;}{i}_{\mathrm{ns}}\left(98\right)\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(35\right)-\mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(1\right)\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(36\right)-\mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(2\right)\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(98\right)-\mathrm{\&Delta;}{u}_{\mathrm{ns}}\left(64\right)\end{array}\right]---\left(16\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff；

(16) formula is abbreviated as：

$\left[i^{\&prime;\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[u^{\&prime;\&prime;}\right]---\left(17\right)$

Using least-squares calculation parameter L_n, R_n：

$\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]={({\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;[i^{\&prime;\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;\&prime;}\right]---\left(18\right)$

9th, the two ends current temporary state component that the 6th step is obtained is substituted into following formula by line protection, calculates transient current energy,

$\mathrm{\&Delta;}{I}_{\mathrm{mt}}=\sqrt{\frac{1}{100}\underset{k=0}{\overset{99}{\mathrm{\&Sigma;}}}\mathrm{\&Delta;}{i}_{\mathrm{mt}}^2\left(k\right)}---\left(19\right)$

$\mathrm{\&Delta;}{I}_{\mathrm{nt}}=\sqrt{\frac{1}{100}\underset{k=0}{\overset{99}{\mathrm{\&Sigma;}}}\mathrm{\&Delta;}{i}_{\mathrm{nt}}^2\left(k\right)}---\left(20\right)$

If Δ I_mt＞ I_setAnd Δ I_nt＞ I_set, then transient current energy is larger, goes to step ten, starts the main criterion based on model error；If Δ I_mt≤I_setOr Δ I_nt≤I_set, then transient current energy is not enough, goes to step 11, starts the assistant criteria based on system impedance and voltage power frequency component；In formula, I_setFor current calibration threshold, 0.1 times of circuit rated current is set to；

Tenth, the system impedance parameter that the voltage transient component variation amount and the 8th step that the current temporary state component that is obtained using the 6th step, the 7th step are obtained are obtained, line protection computation model error E_n, E_m, failure judgement position is comprised the following steps that：

First, model error E is illustrated_nComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_m, R_mFollowing formula is substituted into, a is calculated respectively_nk, b_nk：

$\left\{\begin{array}{c}{a}_{\mathrm{nk}}=(\frac{\mathrm{d\&Delta;}{i}_{\mathrm{mt}}(k-17)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{mt}}(k+17)}{\mathrm{dt}})L_m+({\mathrm{\&Delta;i}}_{\mathrm{mt}}(k-17)-\mathrm{\&Delta;}{i}_{\mathrm{mt}}(k+17))R_m\\ b_{\mathrm{nk}}=\mathrm{\&Delta;}{u}_{\mathrm{mt}}(k+17)-\mathrm{\&Delta;}{u}_{\mathrm{mt}}(k-17)\end{array}\right.kk=\mathrm{18,19},L,81---\left(21\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_nk, b_nkFollowing formula is substituted into, calculating obtains E_n：

$E_n=\frac{\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}|a_{\mathrm{nk}}-b_{\mathrm{nk}}|}{\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{nk}}\right|+\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{nk}}\right|}---\left(22\right)$

Secondly, model error E is illustrated_mComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_n, R_nFollowing formula is substituted into, a is calculated respectively_mk, b_mk：

$\left\{\begin{array}{c}{a}_{\mathrm{mk}}=(\frac{\mathrm{d\&Delta;}{i}_{\mathrm{nt}}(k-17)}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{nt}}(k+17)}{\mathrm{dt}})L_n+({\mathrm{\&Delta;i}}_{\mathrm{nt}}(k-17)-\mathrm{\&Delta;}{i}_{\mathrm{nt}}(k+17))R_n\\ b_{\mathrm{mk}}=\mathrm{\&Delta;}{u}_{\mathrm{nt}}(k+17)-\mathrm{\&Delta;}{u}_{\mathrm{nt}}(k-17)\end{array}\right.kk=\mathrm{18,19},L,81---\left(23\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_mk, b_mkFollowing formula is substituted into, calculating obtains E_m：

$E_m=\frac{\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}|a_{\mathrm{mk}}-b_{\mathrm{mk}}|}{\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{mk}}\right|+\underset{k=18}{\overset{81}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{mk}}\right|}---\left(24\right)$

Finally, the result of calculation of model error is substituted into Protection criteria, failure judgement position：

If E_n＜ ξ or E_m＜ ξ, are judged to external area error, and protection is failure to actuate；

If E_n>=ξ and E_m>=ξ, is judged to troubles inside the sample space, protection act；

In criterion, ξ is action threshold, is adjusted according to the maximum amount of unbalance of model error when occurring external area error, for the long 750kV transmission lines of electricity of 500km, ξ is adjusted as 0.2；

11, the voltage power frequency component variable quantity that the system impedance parameter and the 7th step that line protection is obtained using the 8th step are obtained, failure judgement position is comprised the following steps that：

If L_m＜ 0 or L_n＜ 0, is judged to external area error, and protection is failure to actuate；

If L_m>=0 and L_n>=0, it is necessary to could failure judgement position with reference to voltage power frequency component variable quantity；

Extract the amplitude Δ U of both end voltage power frequency component variable quantity respectively using matrix pencil algorithm_ms, Δ U_ns,

If Δ U_ms＜ U_setAnd Δ U_ns＜ U_set, external area error is judged to, protection is failure to actuate；

If Δ U_ms≥U_setOr Δ U_ns≥U_set, it is judged to troubles inside the sample space, protection act；

In criterion, U_setThreshold is adjusted for voltage, numerical value is

Wherein K_relFor safety factor, 1.1-1.2, U are taken as_NFor the voltage class of transmission line of electricity.For the long 750kV transmission lines of electricity of 500km, U_set=685kV.

It is the implementing procedure of the electric transmission line longitudinal protection method proposed by the present invention based on the magnitude of current above.

The long 750kV transmission lines of electricity of 500km are built using electromagnetic transient simulation software (EMTP), various fault types are set in different faults position, fault current data are generated, the performance of the present invention is tested.Fig. 6 to Fig. 9 is given in partial test result, test process, and data window length is taken as 10ms, and slides 40ms after failure generation.

Fig. 6 is that there occurs three-phase fault outside m lateral areas, and transient current energy is larger, when main criterion starts, the calculating waveform of model error.Due to E_m＜ ξ, main criterion is reliably judged to external area error.

Midpoint there occurs three-phase fault in Tu7Wei areas, and transient current energy is larger, when main criterion starts, the calculating waveform of model error.Due to E_m＞ ξ and E_n＞ ξ, main criterion is reliably judged to troubles inside the sample space, outlet tripping operation.

Fig. 8 is that there occurs A phase earth faults outside m lateral areas, and transient current energy is smaller, when assistant criteria starts, the calculating waveform of system inductance.Due to L_m＜ 0, assistant criteria is reliably judged to external area error.

Midpoint there occurs A phase earth faults in Tu9Wei areas, and transient current energy is smaller, when assistant criteria starts, the calculating waveform of voltage power frequency component variable quantity amplitude.Due to Δ U_ms＞ U_setAnd Δ U_ns＞ U_set, assistant criteria is reliably judged to troubles inside the sample space, outlet tripping operation.

From test result, the present invention is not influenceed by distribution capacity completely merely with current transformer information structuring Protection criteria, can be with sensitive reliable rapidly differentiation troubles inside the sample space and external area error, and protective value is superior.

Claims (1)

1. a kind of electric transmission line longitudinal protection method based on the magnitude of current, comprises the following steps：

Step 1: the protection device installed in transmission line of electricity two ends m ends and n ends gathers A, B, C three-phase current of this side current transformer respectively；

Kept and A/D conversion process Step 2: line protection carries out LPF, sampling to the three-phase current collected, obtain the sampled value of three-phase current；

Step 3: line protection each transmits the three-phase current sampling value information of local terminal to opposite end, while receiving the three-phase current sampling value information transmitted opposite end；

Step 4: line protection combination phase selection result, the sampled value to three-phase current carries out phase-modal transformation, corresponding modulus information is extracted, is comprised the following steps that：

A kind of formula of phase-modal transformation is as follows：

$\left[\begin{array}{c}{i}_{p1}\\ i_{p2}\\ i_{P3}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right]\left[\begin{array}{c}{i}_{\mathrm{pA}}\\ i_{\mathrm{pB}}\\ i_{\mathrm{pC}}\end{array}\right]---\left(1\right)$

In formula：Subscript p represents m or n；Index number（1,2,3）Represent the sequence number of modulus（1 modulus, 2 modulus, 3 modulus）,

If phase selection result shows to there occurs three-phase fault, 1 modulus is extracted；

If there occurs AB phases short trouble or AB phase earth faults, 1 modulus is extracted；

If there occurs BC phases short trouble or BC phase earth faults, 2 modulus are extracted；

If there occurs CA phases short trouble or CA phase earth faults, 3 modulus are extracted；

If there occurs A phase earth faults, 1 modulus is extracted；

If there occurs B phase earth faults, 2 modulus are extracted；

If there occurs C phase earth faults, 3 modulus are extracted；

Step 5: line protection calculates two ends current failure component Δ i_m(t),Δi_n(t), computational methods are as follows：

$\left\{\begin{array}{c}{\mathrm{\&Delta;i}}_m\left(t\right)=i_{\mathrm{mK}}\left(t\right)-i_{\mathrm{mK}}(t-\mathrm{cT})\\ {\mathrm{\&Delta;i}}_n\left(t\right)=i_{\mathrm{nK}}\left(t\right)-i_{\mathrm{nK}}(t-\mathrm{cT})\end{array}---\left(2\right)\right.$

In formula：Subscript K represents the sequence number for the modulus that step 4 is extracted, K=1,2,3；C takes positive integer, and numerical values recited is determined by protection device；T is the cycle of power current；

Step 6: line protection extracts power frequency component and transient state component in the current failure component of two ends using matrix pencil algorithm successively；

The step of matrix pencil algorithm carries out spectrum analysis to signal is summarized as follows：

First, it is analysed to signal and generates sampling matrix Y according to certain rule₁,Y₂；Secondly, calculating matrix Y₁ ⁺Y₂Characteristic value, wherein Y₁ ⁺It is Y₁Moore-Penrose pseudo inverse matrixs, this feature value contains the frequency of all subsignals and decay factor information in signal to be analyzed；Finally, the amplitude and initial phase information of all subsignals are obtained by solving least square problem；

Current failure component is carried out after spectrum analysis using matrix pencil algorithm, whether the frequency according to subsignal is that subsignal is classified as power frequency component or transient state component by power frequency；

There is following relation between current failure component, power frequency component and transient state component three：

$\left\{\begin{array}{c}{\mathrm{\&Delta;i}}_m\left(t\right)={\mathrm{\&Delta;i}}_{\mathrm{ms}}\left(t\right)+{\mathrm{\&Delta;i}}_{\mathrm{mt}}\left(t\right)\\ {\mathrm{\&Delta;i}}_n\left(t\right)={\mathrm{\&Delta;i}}_{\mathrm{ns}}\left(t\right)+{\mathrm{\&Delta;i}}_{\mathrm{nt}}\left(t\right)\end{array}\right.---\left(3\right)$

In formula：Δi_ms(t),Δi_mt(t) be respectively m ends electric current power frequency component and transient state component；Δi_ns(t),Δi_nt(t) be respectively n ends electric current power frequency component and transient state component；

Step 7: the power frequency component and transient state component of the two ends electric current obtained using step 6, line protection calculate the power frequency component and transient state component that installation place voltage is protected at two ends, comprise the following steps that：

First, the conventional method of known two ends Current calculation both end voltage is illustrated：

Under complex frequency domain, using distributed parameter model, following relation is obeyed between transmission line of electricity n terminal voltages electric current and m terminal voltage electric currents：

$\left[\begin{array}{c}{U}_n\left(s\right)\\ I_n\left(s\right)\end{array}\right]\left[\begin{array}{cc}\cosh \left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\sinh \left(\mathrm{\&gamma;d}\right)\\ \sinh \left(\mathrm{\&gamma;d}\right)/Z_{C0}& \cosh \left(\mathrm{\&gamma;d}\right)\end{array}\right]\left[\begin{array}{c}{U}_m\left(s\right)\\ -I_m\left(s\right)\end{array}\right]---\left(4\right)$

In formula：D is line length；Z_C0For line characteristic impedance, calculating formula is

r₁,l₁,g₁,c₁The respectively positive sequence resistance of circuit unit length, inductance, conductance, capacitance parameter；γ is line propagation coefficient, and calculating formula is $\mathrm{\&gamma;}=\sqrt{(r_1+{\mathrm{sl}}_1)(g_1+{\mathrm{sc}}_1)};$

When known two ends electric current, by（4）Formula can solve both end voltage, derive and formula is calculated as below：

$\left[\begin{array}{c}{U}_m\left(s\right)\\ U_n\left(s\right)\end{array}\right]=\left[\begin{array}{cc}{Z}_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)\\ Z_{C0}\&CenterDot;\mathrm{csch}\left(\mathrm{\&gamma;d}\right)& Z_{C0}\&CenterDot;\coth \left(\mathrm{\&gamma;d}\right)\end{array}\right.-\left[\begin{array}{c}{I}_m\left(s\right)\\ I_n\left(s\right)\end{array}\right]---\left(5\right)$

When writing software program and realizing,（5）A kind of practical numerical computation method of formula is as follows, and it describes variable quantity of the terminal voltage in 2 τ time intervals：

$\left\{\begin{array}{c}{u}_m(t-\mathrm{\&tau;})-u_m(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_m(t+\mathrm{\&tau;})+i_m(t-\mathrm{\&tau;})+{2i}_n\left(t\right)]+R/2\&CenterDot;[i_m(t+\mathrm{\&tau;})-i_m(t+\mathrm{\&tau;})]\\ u_n(t+\mathrm{\&tau;})-u_n(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[i_n(t+\mathrm{\&tau;})+i_n(t-\mathrm{\&tau;})+{2i}_m\left(t\right)]+R/2\&CenterDot;[i_n(t+\mathrm{\&tau;})-i_n(t-\mathrm{\&tau;})]\end{array}\right.---\left(6\right)$

In formula：τ is time-consuming for propagation of the electromagnetic wave on power transmission line, and calculating formula is

Z′_C0To ignore characteristic impedance during line loss, calculating formula isR is the total resistance value of transmission line of electricity, and calculating formula is R=r₁d；

Secondly, the practical calculation method of both end voltage power frequency component and transient state component is illustrated：

The two ends electric current power frequency component Δ i that step 6 is obtained_ms(t),Δi_ns(t) substitute into（6）Formula, obtains the practical formula of both end voltage power frequency component variable quantity：

$\left\{\begin{array}{c}{\mathrm{\&Delta;u}}_{\mathrm{ms}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{ms}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{ns}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t-\mathrm{\&tau;})]\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{ns}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{ms}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t-\mathrm{\&tau;})]\end{array}\right.---\left(7\right)$

The two ends current temporary state component Δ i that step 6 is obtained_mt(t),Δi_nt(t) substitute into（6）Formula, obtains the practical formula of both end voltage transient state component variable quantity：

$\left\{\begin{array}{c}{\mathrm{\&Delta;u}}_{\mathrm{mt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{mt}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{nt}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t-\mathrm{\&tau;})]\\ {\mathrm{\&Delta;u}}_{\mathrm{nt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{nt}}(t-\mathrm{\&tau;})=Z_{C0}^{\&prime;}[{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t+\mathrm{\&tau;})+{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t-\mathrm{\&tau;})+{2\mathrm{\&Delta;i}}_{\mathrm{mt}}\left(t\right)]+R/2\&CenterDot;[{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t+\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t-\mathrm{\&tau;})]\end{array}\right.---\left(8\right)$

Step 8: the voltage power frequency component variable quantity that the electric current power frequency component and step 7 that are obtained using step 6 are obtained, line protection calculates both sides system impedance parameter L successively using two figure parameters method of identifications_m,R_mAnd L_n,R_n, comprise the following steps that：

First, illustrate that two figure parameters method of identifications calculate L_m,R_mMethod：

The current failure component and voltage failure component of m sides obey following relation：

$\left\{\begin{array}{c}{\mathrm{\&Delta;u}}_m(t+\mathrm{\&tau;})={-L}_m\frac{\mathrm{d\&Delta;}{i}_m(t+\mathrm{\&tau;})}{\mathrm{dt}}-R_m{\mathrm{\&Delta;i}}_m(t+\mathrm{\&tau;})\\ {\mathrm{\&Delta;u}}_m(t-\mathrm{\&tau;})={-L}_m\frac{{\mathrm{d\&Delta;i}}_m(t-\mathrm{\&tau;})}{\mathrm{dt}}-R_m\mathrm{\&Delta;}{i}_m(t-\mathrm{\&tau;})\end{array}\right.---\left(9\right)$

Two formulas, which are subtracted each other, to be obtained：

$[\frac{{\mathrm{d\&Delta;i}}_m(t-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_m(t+\mathrm{\&tau;})}{\mathrm{dt}}]L_m+[{\mathrm{\&Delta;i}}_m(t-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_m(t+\mathrm{\&tau;})]R_m={\mathrm{\&Delta;u}}_m(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_m(t-\mathrm{\&tau;})---\left(10\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, row write equation successively, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_1+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_1+\mathrm{\&tau;})\\ \frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})\\ \&CenterDot;& \&CenterDot;\\ \&CenterDot;& \&CenterDot;\\ \&CenterDot;& \&CenterDot;\\ \frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]\text{=}\left[\begin{array}{c}{\mathrm{\&Delta;i}}_{\mathrm{ms}}(t_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})\\ \&CenterDot;\\ \&CenterDot;\\ \&CenterDot;\\ {\mathrm{\&Delta;u}}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(11\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Will（11）Formula is abbreviated as：

$\left[i^{\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[u^{\&prime;}\right]---\left(12\right)$

Using least-squares calculation parameter L_m,R_m：

$\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]={({\left[i^{\&prime;}\right]}^T\&CenterDot;[i^{\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;}\right]---\left(13\right)$

Secondly, illustrate that two figure parameters method of identifications calculate L_n,R_nMethod：

The current failure component and voltage failure component of n sides obey following relation：

$\left\{\begin{array}{c}{\mathrm{\&Delta;u}}_n(t+\mathrm{\&tau;})=-L_n\frac{{\mathrm{d\&Delta;i}}_n(t+\mathrm{\&tau;})}{\mathrm{dt}}-R_n{\mathrm{\&Delta;i}}_n(t+\mathrm{\&tau;})\\ {\mathrm{\&Delta;u}}_n(t-\mathrm{\&tau;})={-L}_n\frac{{\mathrm{d\&Delta;i}}_n(t-\mathrm{\&tau;})}{\mathrm{dt}}-R_n{\mathrm{\&Delta;i}}_n(t-\mathrm{\&tau;})\end{array}\right.---\left(14\right)$

Two formulas, which are subtracted each other, to be obtained：

$[\frac{{\mathrm{d\&Delta;i}}_n(t-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{d{\mathrm{\&Delta;i}}_n(t+\mathrm{\&tau;})}{\mathrm{dt}}]L_n+[{\mathrm{\&Delta;i}}_n(t-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_n(t+\mathrm{\&tau;})]R_n={\mathrm{\&Delta;u}}_n(t+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_n(t-\mathrm{\&tau;})---\left(15\right)$

During specific calculating, the electric current power frequency component and voltage power frequency component variable quantity of 10ms data window length after failure are taken out, row write equation successively, constitute over-determined systems：

$\left[\begin{array}{cc}\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_1+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_1+\mathrm{\&tau;})\\ \frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})\\ \&CenterDot;& \&CenterDot;\\ \&CenterDot;& \&CenterDot;\\ \&CenterDot;& \&CenterDot;\\ \frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& {\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]\text{=}\left[\begin{array}{c}{\mathrm{\&Delta;i}}_{\mathrm{ns}}(t_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})\\ \&CenterDot;\\ \&CenterDot;\\ \&CenterDot;\\ {\mathrm{\&Delta;u}}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(16\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Will（16）Formula is abbreviated as：

$\left[i^{\&prime;\&prime;}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[u^{\&prime;\&prime;}\right]---\left(17\right)$

Using least-squares calculation parameter L_n,R_n：

$\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]={({\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;[i^{\&prime;\&prime;}\left]\right)}^{-1}\&CenterDot;{\left[i^{\&prime;\&prime;}\right]}^T\&CenterDot;\left[u^{\&prime;\&prime;}\right]---\left(18\right)$

Step 9: the two ends current temporary state component that step 6 is obtained is substituted into following formula by line protection, transient current energy is calculated,

${\mathrm{\&Delta;I}}_{\mathrm{mt}}=\sqrt{\frac{1}{t_2-t_1}{\&Integral;}_{t_1}^{t_2}{\mathrm{\&Delta;i}}_{\mathrm{mt}}^2\left(t\right)\mathrm{dt}}---\left(19\right)$

${\mathrm{\&Delta;I}}_{\mathrm{nt}}=\sqrt{\frac{1}{t_2-t_1}{\&Integral;}_{t_1}^{t_2}{\mathrm{\&Delta;i}}_{\mathrm{nt}}^2\left(t\right)\mathrm{dt}}---\left(20\right)$

In formula：t₁,t₂It is the initial time and end time of 10ms data window length respectively；

If Δ I_mt＞ I_setAnd Δ I_nt＞ I_set, then transient current energy is larger, goes to step ten, starts the main criterion based on model error；If Δ I_mt≤I_setOr Δ I_nt≤I_set, then transient current energy is not enough, goes to step 11, starts the assistant criteria based on system impedance and voltage power frequency component；In formula, I_setFor current calibration threshold, 0.1 times of circuit rated current is set to；

Step 10: the system impedance parameter that voltage transient component variation amount and step 8 that the current temporary state component obtained using step 6, step 7 are obtained are obtained, line protection computation model error E_n,E_m, and failure judgement position, comprise the following steps that：

First, model error E is illustrated_nComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_m,R_mFollowing formula is substituted into, a is calculated respectively_nj,b_nj：

$\left\{\begin{array}{c}{a}_{\mathrm{nj}}=(\frac{{\mathrm{d\&Delta;i}}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{mt}}(t_j+\mathrm{\&tau;})}{\mathrm{dt}})L_m({\mathrm{\&Delta;i}}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{mt}}(t_j+\mathrm{\&tau;}))R_m\\ b_{\mathrm{nj}}={\mathrm{\&Delta;u}}_{\mathrm{mt}}(t_j+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{mt}}(t_j-\mathrm{\&tau;})\end{array}\right.j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,k---\left(21\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_nj,b_njFollowing formula is substituted into, calculating obtains E_n：

$E_n=\frac{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}|a_{\mathrm{nj}}-b_{\mathrm{nj}}|}{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{nj}}\right|+\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{nj}}\right|}---\left(22\right)$

Secondly, model error E is illustrated_mComputational methods：

By the current temporary state component of 10ms data windows length, voltage transient component variation amount and L after failure_n,R_nFollowing formula is substituted into, a is calculated respectively_mj,b_mj：

$\left\{\begin{array}{c}{a}_{\mathrm{mj}}=(\frac{{\mathrm{d\&Delta;i}}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{{\mathrm{d\&Delta;i}}_{\mathrm{nt}}(t_j+\mathrm{\&tau;})}{\mathrm{dt}})L_n+({\mathrm{\&Delta;i}}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})-{\mathrm{\&Delta;i}}_{\mathrm{nt}}(t_j+\mathrm{\&tau;}))R_n\\ b_{\mathrm{mj}}={\mathrm{\&Delta;u}}_{\mathrm{nt}}(t_j+\mathrm{\&tau;})-{\mathrm{\&Delta;u}}_{\mathrm{nt}}(t_j-\mathrm{\&tau;})\end{array}\right.j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,k---\left(23\right)$

In above formula, the first derivative of electric current can be replaced calculating with diff,

Again by a_mj,b_mjFollowing formula is substituted into, calculating obtains Em：

$E_m=\frac{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}|a_{\mathrm{mj}}-b_{\mathrm{mj}}|}{\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|a_{\mathrm{mj}}\right|+\underset{j=1}{\overset{k}{\mathrm{\&Sigma;}}}\left|b_{\mathrm{mj}}\right|}---\left(24\right)$

Finally, the result of calculation of model error is substituted into Protection criteria, failure judgement position：

If E_n＜ ξ or E_m＜ ξ, are judged to external area error, and protection is failure to actuate；

If E_n>=ξ and E_m>=ξ, is judged to troubles inside the sample space, protection act；

In criterion, ξ is action threshold, is adjusted according to the maximum amount of unbalance of model error when occurring external area error；

Step 11: the voltage power frequency component variable quantity that system impedance parameter and step 7 that line protection is obtained using step 8 are obtained, failure judgement position is comprised the following steps that：

If L_m＜ 0 or L_n＜ 0, is judged to external area error, and protection is failure to actuate；

If L_m>=0 and L_n>=0, it is necessary to could failure judgement position with reference to voltage power frequency component variable quantity；

Extract the amplitude Δ U of both end voltage power frequency component variable quantity respectively using matrix pencil algorithm_ms,ΔU_ns,

If Δ U_ms＜ U_setAnd Δ U_ns＜ U_set, external area error is judged to, protection is failure to actuate；

If Δ U_ms≥U_setOr Δ U_ns≥U_set, it is judged to troubles inside the sample space, protection act；

In criterion, U_setThreshold is adjusted for voltage, numerical value isWherein K_relFor safety factor, 1.1-1.2, U are taken as_NFor the voltage class of transmission line of electricity.

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