CN102570419A - 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 the relay protection of power system technical field, be specifically related to a kind of electric transmission line longitudinal protection method based on the magnitude of current.

Background technology

Vertical join current differential protection have criterion simple, highly sensitive, have a natural advantage such as phase ability of selecting, extensively as the main protection of transmission line.Yet practical operating experiences shows this protection philosophy and has following defective: (1) this principle has been ignored the distributed capacitance that actual transmission line exists.Along with the raising of transmission line electric pressure and the growth of line length, distributed capacitance is with increasing, and this makes fault transient process more obvious, and transient current will influence the computational accuracy of traditional protection algorithm; (2) the stable state capacitance current can change the size and the phase place of both sides power current.In a word, the influence of ignoring distributed capacitance can cause protection sensitivity to reduce, maybe malfunction during external area error.

Processing method to the problems referred to above mainly is divided into two types at present: the first kind is the building-out capacitor electric current, for example shunt reactor penalty method, phasor backoff algorithm and time domain compensation algorithm.Preceding two kinds of methods can only the compensate for steady state capacitance current, can't compensate the transient state capacitance current, and shunt reactor and on-fixed insert, and compensation effect is limited.The time domain compensation algorithm can compensate whole capacitance currents, but the extra voltage of having introduced, yet that field operation experiences shows the progress of disease performance of voltage transformer and loop reliability is all relatively poor, so protective value can't be guaranteed.Second type is to study the protection philosophy that not influenced by capacitance current, for example the auspicious carina of shellfish road modelling, pattern recognition and comprehensive impedance method.These methods have been considered the influence of distributed capacitance on principle, thereby need not the building-out capacitor electric current.But owing to used both-end voltage data, require a terminal voltage amount to transmit, increased transinformation, communication port is had relatively high expectations to the opposite end.

Therefore, to line protection, the guard method of studying a kind of not working voltage amount and effectively solving the capacitance current problem is necessary.

Summary of the invention

In order to overcome the shortcoming that above-mentioned prior art exists, the object of the present invention is to provide a kind of electric transmission line longitudinal protection method based on the magnitude of current, this method has only utilizes the two ends current information to constitute the characteristics that not influenced by distributed capacitance.

For achieving the above object, the present invention adopts following technical scheme:

A kind of electric transmission line longitudinal protection method based on the magnitude of current may further comprise the steps:

Step 1, line protection are installed in the two ends m end and the n end of transmission line, and every cover protective device is gathered A, B, the C three-phase current of local terminal protection installation place current transformer;

Step 2, line protection carry out LPF, sampling maintenance and A/D conversion process to the three-phase current that collects, and obtain the sampled value of three-phase current;

Step 3, line protection transmit the three-phase current sampled value information of local terminal separately to the opposite end, receive the three-phase current sampled value information that transmits the opposite end simultaneously;

Step 4, line protection combine to select the phase result, and the sampled value of three-phase current is carried out phase-modular transformation, extract corresponding modulus information, and concrete steps are following:

A kind of formula of phase-modular transformation is following:

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

In the formula: subscript p represents m or n; The sequence number (1 modulus, 2 modulus, 3 modulus) of index number (1,2,3) expression modulus,

Three-phase fault has taken place, then extracted 1 modulus if select the phase result to show;

If AB phase short trouble or AB phase earth fault have taken place, then extract 1 modulus;

If BC phase short trouble or BC phase earth fault have taken place, then extract 2 modulus;

If CA phase short trouble or CA phase earth fault have taken place, then extract 3 modulus;

If A phase earth fault has taken place, then extract 1 modulus;

If B phase earth fault has taken place, then extract 2 modulus;

If C phase earth fault has taken place, then extract 3 modulus;

Step 5, line protection calculate two ends current failure component Δ i _m(t), Δ i _n(t), computational methods are following:

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

In the formula: subscript K represents the sequence number of the modulus that step 4 extracts, K=1,2,3; C gets positive integer, and numerical values recited is determined by protective device; T is the cycle of power current;

Step 6, line protection utilize the pencil of matrix algorithm to extract power frequency component and transient state component in the current failure component of two ends successively;

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

At first, signal to be analyzed is generated sampling matrix Y according to certain rule ₁, Y ₂Secondly, compute matrix Y ₁ ⁺Y ₂Characteristic value, Y wherein ₁ ⁺Be Y ₁The Moore-Penrose pseudo inverse matrix, this characteristic value has comprised the frequency and the decay factor information of all subsignals in the signal to be analyzed; At last, through finding the solution amplitude and the initial phase information that least square problem obtains all subsignals;

After utilizing the pencil of matrix algorithm that the current failure component is carried out spectrum analysis, whether be that power frequency is classified as power frequency component or transient state component with subsignal according to the frequency of subsignal;

Relation below existing between current failure component, power frequency component and the 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 the formula: Δ i _Ms(t), Δ i _Mt(t) be the power frequency component and the transient state component of m end electric current respectively; Δ i _Ns(t), Δ i _Nt(t) be the power frequency component and the transient state component of n end electric current respectively;

Step 7, utilize the power frequency component and the transient state component of the two ends electric current that step 6 obtains, line protection calculates the power frequency component and the transient state component of protection installation place, two ends voltage, and concrete steps are following:

At first, explain that known two ends electric current calculates the conventional method of voltage:

Under complex frequency domain, adopt distributed parameter model, obey following relation between transmission line n terminal voltage electric current and the m terminal voltage electric current:

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

In the formula: d is a line length; Z _C0Be the line characteristics impedance, calculating formula does

r ₁, l ₁, g ₁, c ₁Positive sequence resistance, inductance, the electricity that is respectively the circuit unit length led, capacitance parameter; γ is the circuit propagation coefficient, and calculating formula does $\mathrm{\&gamma;}=\sqrt{({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">r</figure-callout>}}_1+{\mathrm{Sl}}_1)({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{Sc}}_1)};$

When the electric current of known two ends, can solve voltage by (4) formula, derive following computing formula:

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

When writing software program and realize, a kind of practical numerical computation method of (5) formula is following, and it has described terminal voltage at the variable quantity of 2 τ in the time interval:

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

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

Characteristic impedance when ignoring line loss, calculating formula does

R is the total resistance value of transmission line, and calculating formula is R=r ₁D;

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

The two ends electric current power frequency component Δ i that step 6 is obtained _Ms(t), Δ i _Ns(t) substitution (6) formula obtains the practical formula of 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), Δ i _Nt(t) substitution (6) formula obtains the practical formula of 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 utilizes electric current power frequency component that step 6 obtains and step 7 to obtain, line protection adopts two coefficient parameter recognition methods to calculate both sides system impedance parameter L successively _m, R _mAnd L _n, R _n, concrete steps are following:

At first, two coefficient parameter recognition methods calculating L is described _m, R _mMethod:

The current failure component of m side and voltage failure component are obeyed 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 are subtracted each other and are 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window after the taking-up fault, row are write equation successively, constitute the overdetermined equation group:

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_m\\ R_m\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ms}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_2-\mathrm{\&tau;})\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ms}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(11\right)$

In the following formula, the first derivative of electric current can replace calculating with diff,

(11) formula is abbreviated as:

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

Adopt the 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, two coefficient parameter recognition methods calculating L is described _n, R _nMethod:

The current failure component of n side and voltage failure component are obeyed 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 are subtracted each other and are 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window after the taking-up fault, row are write equation successively, constitute the overdetermined equation group:

$\left[\begin{array}{cc}\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})\\ M& M\\ \frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})}{\mathrm{dt}}-\frac{\mathrm{d\&Delta;}{i}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})}{\mathrm{dt}}& \mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})-\mathrm{\&Delta;}{i}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})\end{array}\right]\&CenterDot;\left[\begin{array}{c}{L}_n\\ R_n\end{array}\right]=\left[\begin{array}{c}\mathrm{\&Delta;}{u}_{\mathrm{ns}}({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_1-\mathrm{\&tau;})\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_2-\mathrm{\&tau;})\\ M\\ \mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_k+\mathrm{\&tau;})-\mathrm{\&Delta;}{u}_{\mathrm{ns}}(t_k-\mathrm{\&tau;})\end{array}\right]---\left(16\right)$

In the following formula, the first derivative of electric current can replace calculating with diff,

(16) formula is abbreviated as:

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

Adopt the least-squares calculation parameter L _n, R _n:

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

The two ends current temporary state component substitution following formula that step 9, line protection obtain step 6 calculates the transient current energy,

$\mathrm{\&Delta;}{I}_{\mathrm{mt}}=\sqrt{\frac{1}{t_2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1}{\&Integral;}_{t_1}^{t_2}\mathrm{\&Delta;}{i}_{\mathrm{mt}}^2\left(t\right)\mathrm{dt}}---\left(19\right)$

$\mathrm{\&Delta;}{I}_{\mathrm{nt}}=\sqrt{\frac{1}{t_2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">t</figure-callout>}}_1}{\&Integral;}_{t_1}^{t_2}\mathrm{\&Delta;}{i}_{\mathrm{nt}}^2\left(t\right)\mathrm{dt}}---\left(20\right)$

In the formula: t ₁, t ₂It is respectively the long initial moment of 10ms data window and the termination moment;

If Δ I _Mt＞I _SetAnd Δ I _Nt＞I _Set, then the transient current energy is bigger, changes step 10, starts the main criterion based on model error; If Δ I _Mt≤I _SetOr Δ I _Nt≤I _Set, then the transient current energy shortage changes step 11, starts the assistant criteria based on system impedance and voltage power frequency component, in the formula, and I _SetBe the current calibration threshold, be set to 0.1 times of circuit rated current;

Step 10, the system impedance parameter of utilizing current temporary state component that step 6 obtains, voltage transient component variation amount that step 7 obtains and step 8 to obtain, line protection computation model error E _n, E _m, and the failure judgement position, concrete steps are following:

At first, model error E is described _nComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _m, R _mThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Nj, b _NjThe substitution following formula calculates 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 described _mComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _n, R _nThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Mj, b _MjThe substitution following formula calculates E _m:

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

At last, with the result of calculation substitution protection criterion of model error, the failure judgement position:

If E _n＜ξ or E _m＜ξ is 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, the protection action;

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

The voltage power frequency component variable quantity that step 11, line protection utilize system impedance parameter that step 8 obtains and step 7 to obtain, the failure judgement position, concrete steps are following:

If L _m＜0 or L _n＜0, be judged to external area error, protection is failure to actuate;

If L _m>=0 and L _n>=0, need to combine voltage power frequency component variable quantity ability failure judgement position;

Utilize the pencil of matrix algorithm to extract the amplitude Δ U of voltage power frequency component variable quantity respectively _Ms, Δ U _Ns,

If Δ U _Ms＜U _SetAnd Δ U _Ns＜U _Set, being judged to external area error, protection is failure to actuate;

If Δ U _Ms>=U _SetOr Δ U _Ns>=U _Set, be judged to troubles inside the sample space, the protection action;

In the criterion, U _SetBe the voltage threshold of adjusting, numerical value does

K wherein _RelBe safety factor, be taken as 1.1-1.2, U _NElectric pressure for transmission line.

The present invention has mainly proposed four kinds of new methods: (1) utilizes the two ends current information to calculate the method for voltage; This method is based on distributed parameter model; Voltage result of calculation has higher computational accuracy; Feasible protection need not based on voltage transformer information, has reduced the requirement to the protection Hardware configuration, has improved the reliability of protection philosophy simultaneously; (2), utilize least square method computing system impedance parameter, the characteristics that result of calculation has is precise and stable, not influenced by transient process based on the system impedance computational methods of parameter recognition; (3) based on the line protection criterion of model error, this criterion has been utilized the transient state amount information in the fault current, and is sensitive reliable, can realize quick acting, with it as main criterion; (4), when transient state amount information in the fault current is not enough, withdraw from the assistant criteria input based on the main criterion of model error based on the assistant criteria of system impedance and voltage power frequency component.

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

1, this method has adopted the Transmission Line Distributed Parameter model, not influenced by distributed capacitance, need not the building-out capacitor electric current, has solved vertical current differential protection owing to receive the influence of line distribution capacitance and fault transient thereby the problem that performance reduces.

2, this method does not need working voltage instrument transformer information only based on the information of current transformer, compares with existing guard method, has reduced the requirement to the protection Hardware configuration, has improved the reliability of protection philosophy simultaneously yet.

3, this method is applicable to the transmission line of various electric pressures and all lengths, is applicable to various system operation modes such as single-ended large power supply system, both-end large power supply system, weak feedback system, single supply on-load circuit, and tolerance transition resistance ability is stronger.

4, this method has been utilized the information in the broad frequency band in the fault current, and therefore lower to the performance index requirement of low pass filter, it is higher that cut-off frequency can be provided with, and only need satisfy sampling thheorem and get final product.

5, it is long that this method only need be used the 10ms data window, and protection can realize quick acting.

Description of drawings

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

Fig. 2 is flow chart of data processing figure.

Fig. 3 is a distributed constant transmission line illustraton of model.

Fig. 4 is the troubles inside the sample space illustraton of model, and wherein Fig. 4 a is a fault full dose illustraton of model, and Fig. 4 b is the fault component illustraton of model.

Fig. 5 is the external area error illustraton of model, and wherein Fig. 5 a is a fault full dose illustraton of model, and Fig. 5 b is the fault component illustraton of model.

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

Fig. 7 is the calculating waveform of model error during the mid point three-phase fault in the district.

Fig. 8 is the calculating waveform of system's inductance during A phase earth fault outside the m lateral areas.

Fig. 9 is the calculating waveform of voltage power frequency component variable quantity amplitude during mid point A phase earth fault in the district.

Embodiment

Below in conjunction with accompanying drawing and embodiment the present invention is described in more detail.

As shown in Figure 1, the invention belongs to the circuit Microcomputer Protection, the protective device that uses the principle of the invention all is equipped with in the circuit both sides, and the protective device of both sides is through the communication port swap data, and Fig. 1 has described the annexation between the hardware, and Fig. 2 has described flow chart of data processing.Protective device with the m side is an example; This side current transformer TA1 collects the analog quantity of this side three-phase current; Through obtaining digital quantity after LPF, sampling maintenance and the A/D conversion, import DSP1, DSP1 is equipped with communication interface and links to each other with communication port simultaneously; The current digital amount that on the one hand this locality is collected is sent to the n side, receives the current digital amount that the n side transmits on the other hand.Core content of the present invention is achieved in microsystem DSP through programming.DSP occurs in district's interior (shown in Figure 4) still district outer (shown in Figure 5) in case confirm fault, and isolation sends relay to through photoelectricity just can to send control signal corresponding, and relay is made corresponding action after receiving control signal.

A kind of electric transmission line longitudinal protection method based on the magnitude of current is applied to but when being not limited to the long 750kV transmission line of 500km, specifically may further comprise the steps:

One, the protective device that is installed in transmission line two ends (being designated as m end holds with n) is gathered A, B, the C three-phase current i of this side current transformer respectively _MA(t), i _MB(t), i _MC(t) and i _NA(t), i _NB(t), i _NC(t);

Two, line protection carries out LPF, sampling maintenance and A/D conversion process to the three-phase current that collects, and obtains the sampled value i of three-phase current _MA(k), i _MB(k), i _MC(k) and i _NA(k), i _NB(k), i _NC(k).The cut-off frequency of low pass filter is made as 600Hz, mainly is in order to eliminate The noise, to make algorithm more stable, and sample frequency is made as 10kHz, and this moment is ripple sampling number N=200 weekly, sampling time interval T _S=0.1ms;

Three, after fault took place, starting component started, and adopted the pilot protection element of this method to drop into, and line protection transmits the three-phase current sampled value information of local terminal separately to the opposite end, receive the three-phase current sampled value information that transmits the opposite end simultaneously;

Four, line protection combines to select the phase result, and the sampled value of three-phase current is carried out phase-modular transformation, extracts corresponding modulus information, and concrete steps are following:

A kind of formula of phase-modular transformation is following:

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

In the formula: subscript p represents m or n; The sequence number (1 modulus, 2 modulus, 3 modulus) of index number (1,2,3) expression modulus;

Three-phase fault has taken place, then extracted 1 modulus if select the phase result to show;

If AB phase short trouble or AB phase earth fault have taken place, then extract 1 modulus;

If BC phase short trouble or BC phase earth fault have taken place, then extract 2 modulus;

If CA phase short trouble or CA phase earth fault have taken place, then extract 3 modulus;

If A phase earth fault has taken place, then extract 1 modulus;

If B phase earth fault has taken place, then extract 2 modulus;

If C phase earth fault has taken place, then extract 3 modulus;

Five, line protection calculates two ends failure of the current component Δ i _m(k), Δ i _n(k), computational methods are following:

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

In the formula: subscript K represents the sequence number of the 4th modulus that extract of step, K=1,2,3; The numerical values recited of c is determined by protective device, is taken as 2 here; N is ripple sampling number weekly, N=200;

Six, line protection utilizes the pencil of matrix algorithm to extract power frequency component and transient state component in the current failure component of two ends successively;

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

At first, signal to be analyzed is generated sampling matrix Y according to certain rule ₁, Y ₂Secondly, compute matrix Y ₁ ⁺Y ₂Characteristic value, Y wherein ₁ ⁺Be Y ₁The Moore-Penrose pseudo inverse matrix, this characteristic value has comprised the frequency and the decay factor information of all subsignals in the signal to be analyzed; At last, through finding the solution amplitude and the initial phase information that least square problem obtains all subsignals;

After utilizing the pencil of matrix algorithm that the current failure component is carried out spectrum analysis, whether be that power frequency is classified as power frequency component or transient state component with subsignal according to the frequency of subsignal;

Relation below existing between current failure component, power frequency component and the 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 the formula: Δ i _Ms(k), Δ i _Mt(k) be the power frequency component and the transient state component of m end electric current respectively; Δ i _Ns(k), Δ i _Nt(k) be the power frequency component and the transient state component of n end electric current respectively;

Seven, utilize the power frequency component and the transient state component of the 6th two ends electric current that obtain of step, line protection calculates the power frequency component and the transient state component of protection installation place, two ends voltage, and concrete steps are following:

At first, explain that known two ends electric current calculates the conventional method of voltage:

Under complex frequency domain, adopt distributed parameter model, obey following relation between transmission line n terminal voltage electric current and the m terminal voltage electric current:

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

In the formula: d is a line length; Z _C0Be the line characteristics impedance, calculating formula does

r ₁, l ₁, g ₁, c ₁Positive sequence resistance, inductance, the electricity that is respectively the circuit unit length led, capacitance parameter; γ is the circuit propagation coefficient, and calculating formula does $\mathrm{\&gamma;}=\sqrt{({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">r</figure-callout>}}_1+{\mathrm{Sl}}_1)({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="HDA0000126081780000043.png,HDA0000126081780000052.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{Sc}}_1)};$

When the electric current of known two ends, can solve voltage by (4) formula, derive following computing formula:

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

When writing software program and realize, a kind of practical numerical computation method of (5) formula is following, and it has described terminal voltage at the variable quantity of 2 τ in the time interval:

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

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

Characteristic impedance when ignoring line loss, calculating formula does R is the total resistance value of transmission line, and calculating formula is R=r ₁D;

Secondly, the practical calculation method of voltage power frequency component is described:

Write (6) formula as the discrete time form, and substitution the 6th goes on foot the two ends electric current power frequency component Δ i that obtains _Ms(k), Δ i _Ns(k), obtain the practical formula of voltage power frequency component variable quantity:

For the long 750kV transmission line of 500km; τ ≈ 1.7ms,

R ≈ 6.4 Ω.Therefore (7) formula becomes:

The practical calculation method of voltage transient state component is described at last:

Write (6) formula as the discrete time form, and substitution the 6th goes on foot the two ends current temporary state component Δ i that obtains _Mt(k), Δ i _Nt(k), obtain the practical formula of voltage transient state component variable quantity:

Substitution τ; The concrete parameter of

R, (9) formula becomes:

Eight, the electric current power frequency component and the 7th that utilized for the 6th step obtained goes on foot the voltage power frequency component variable quantity that obtains, and line protection adopts two coefficient parameter recognition methods to calculate both sides system impedance parameter L successively _m, R _mAnd L _n, R _nConcrete steps are following:

At first, two coefficient parameter recognition methods calculating L is described _m, R _mMethod:

The current failure component of m side and voltage failure component are obeyed 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window can be listed as and write out 64 equations after the taking-up fault, constitutes the overdetermined equation group:

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

In the following formula, the first derivative of electric current can replace calculating with diff;

(12) formula is abbreviated as:

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

Adopt the 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, two coefficient parameter recognition methods calculating L is described _n, R _nMethod:

The current failure component of n side and voltage failure component are obeyed 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window can be listed as and write out 64 equations after the taking-up fault, constitutes the overdetermined equation group:

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

In the following formula, the first derivative of electric current can replace calculating with diff;

(16) formula is abbreviated as:

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

Adopt the least-squares calculation parameter L _n, R _n:

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

Nine, line protection calculates the transient current energy with the two ends current temporary state component substitution following formula that the 6th step obtained,

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

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

If Δ I _Mt＞I _SetAnd Δ I _Nt＞I _Set, then the transient current energy is bigger, changes step 10, starts the main criterion based on model error; If Δ I _Mt≤I _SetOr Δ I _Nt≤I _Set, then the transient current energy shortage changes step 11, starts the assistant criteria based on system impedance and voltage power frequency component; In the formula, I _SetBe the current calibration threshold, be set to 0.1 times of circuit rated current;

Ten, the current temporary state component, the 7th that utilized for the 6th step obtained goes on foot the voltage transient component variation amount and the 8th that obtains and goes on foot the system impedance parameter that obtains, line protection computation model error E _n, E _m, the failure judgement position, concrete steps are following:

At first, model error E is described _nComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _m, R _mThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Nk, b _NkThe substitution following formula calculates 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 described _mComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _n, R _nThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Mk, b _MkThe substitution following formula calculates E _m:

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

At last, with the result of calculation substitution protection criterion of model error, the failure judgement position:

If E _n＜ξ or E _m＜ξ is 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, the protection action;

In the criterion, ξ is the action threshold, and the maximum amount of unbalance of model error is adjusted during according to the generation external area error, and for the long 750kV transmission line of 500km, it is 0.2 that ξ adjusts;

11, line protection utilizes the voltage power frequency component variable quantity that the 8th system impedance parameter that obtain of step and the 7th step obtain, the failure judgement position, and concrete steps are following:

If L _m＜0 or L _n＜0, be judged to external area error, protection is failure to actuate;

If L _m>=0 and L _n>=0, need to combine voltage power frequency component variable quantity ability failure judgement position;

Utilize the pencil of matrix algorithm to extract the amplitude Δ U of voltage power frequency component variable quantity respectively _Ms, Δ U _Ns,

If Δ U _Ms＜U _SetAnd Δ U _Ns＜U _Set, being judged to external area error, protection is failure to actuate;

If Δ U _Ms>=U _SetOr Δ U _Ns>=U _Set, be judged to troubles inside the sample space, the protection action;

In the criterion, U _SetBe the voltage threshold of adjusting, numerical value does

K wherein _RelBe safety factor, be taken as 1.1-1.2, U _NElectric pressure for transmission line.For the long 750kV transmission line of 500km, U _Set=685kV.

The implementing procedure that is more than that the present invention proposes based on the electric transmission line longitudinal protection method of the magnitude of current.

Utilize electromagnetic transient simulation software (EMTP) to build the long 750kV transmission line of 500km, various fault types are set, generate the fault current data, test performance of the present invention in the different faults position.Fig. 6 to Fig. 9 has provided the partial test result, and in the test process, data window length is taken as 10ms, and in fault slip 40ms in back takes place.

Three-phase fault has taken place for outside the m lateral areas in Fig. 6, and the transient current energy is bigger, when main criterion starts, and the calculating waveform of model error.Because E _m＜ξ, main criterion reliably is judged to external area error.

Three-phase fault has taken place for mid point in the district in Fig. 7, and the transient current energy is bigger, when main criterion starts, and the calculating waveform of model error.Because E _m＞ξ and E _n＞ξ, main criterion reliably is judged to troubles inside the sample space, the outlet tripping operation.

A phase earth fault has taken place for outside the m lateral areas in Fig. 8, and the transient current energy is less, when assistant criteria starts, and the calculating waveform of system's inductance.Because L _m＜0, assistant criteria reliably is judged to external area error.

A phase earth fault has taken place for mid point in the district in Fig. 9, and the transient current energy is less, when assistant criteria starts, and the calculating waveform of voltage power frequency component variable quantity amplitude.Because Δ U _Ms＞U _SetAnd Δ U _Ns＞U _Set, assistant criteria reliably is judged to troubles inside the sample space, the outlet tripping operation.

Can be known that by test result the present invention only utilizes current transformer information structuring protection criterion, not influenced by distributed capacitance, can sensitivity reliably differentiate troubles inside the sample space and external area error apace, protective value is superior.

Claims (1)

1. electric transmission line longitudinal protection method based on the magnitude of current may further comprise the steps:

Step 1, the protective device that is installed in transmission line two ends m end and n end are gathered A, B, the C three-phase current of this side current transformer respectively;

Step 2, line protection carry out LPF, sampling maintenance and A/D conversion process to the three-phase current that collects, and obtain the sampled value of three-phase current;

Step 3, line protection transmit the three-phase current sampled value information of local terminal separately to the opposite end, receive the three-phase current sampled value information that transmits the opposite end simultaneously;

Step 4, line protection combine to select the phase result, and the sampled value of three-phase current is carried out phase-modular transformation, extract corresponding modulus information, and concrete steps are following:

A kind of formula of phase-modular transformation is following:

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

In the formula: subscript p represents m or n; The sequence number (1 modulus, 2 modulus, 3 modulus) of index number (1,2,3) expression modulus,

Three-phase fault has taken place, then extracted 1 modulus if select the phase result to show;

If AB phase short trouble or AB phase earth fault have taken place, then extract 1 modulus;

If BC phase short trouble or BC phase earth fault have taken place, then extract 2 modulus;

If CA phase short trouble or CA phase earth fault have taken place, then extract 3 modulus;

If A phase earth fault has taken place, then extract 1 modulus;

If B phase earth fault has taken place, then extract 2 modulus;

If C phase earth fault has taken place, then extract 3 modulus;

Step 5, line protection calculate two ends current failure component Δ i _m(t), Δ i _n(t), computational methods are following:

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

In the formula: subscript K represents the sequence number of the modulus that step 4 extracts, K=1,2,3; C gets positive integer, and numerical values recited is determined by protective device; T is the cycle of power current;

Step 6, line protection utilize the pencil of matrix algorithm to extract power frequency component and transient state component in the current failure component of two ends successively;

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

At first, signal to be analyzed is generated sampling matrix Y according to certain rule ₁, Y ₂Secondly, compute matrix Y ₁ ⁺Y ₂Characteristic value, Y wherein ₁ ⁺Be Y ₁The Moore-Penrose pseudo inverse matrix, this characteristic value has comprised the frequency and the decay factor information of all subsignals in the signal to be analyzed; At last, through finding the solution amplitude and the initial phase information that least square problem obtains all subsignals;

After utilizing the pencil of matrix algorithm that the current failure component is carried out spectrum analysis, whether be that power frequency is classified as power frequency component or transient state component with subsignal according to the frequency of subsignal;

Relation below existing between current failure component, power frequency component and the 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 the formula: Δ i _Ms(t), Δ i _Mt(t) be the power frequency component and the transient state component of m end electric current respectively; Δ i _Ns(t), Δ i _Nt(t) be the power frequency component and the transient state component of n end electric current respectively;

Step 7, utilize the power frequency component and the transient state component of the two ends electric current that step 6 obtains, line protection calculates the power frequency component and the transient state component of protection installation place, two ends voltage, and concrete steps are following:

At first, explain that known two ends electric current calculates the conventional method of voltage:

Under complex frequency domain, adopt distributed parameter model, obey following relation between transmission line n terminal voltage electric current and the m terminal voltage electric current:

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

In the formula: d is a line length; Z _C0Be the line characteristics impedance, calculating formula does

r ₁, l ₁, g ₁, c ₁Positive sequence resistance, inductance, the electricity that is respectively the circuit unit length led, capacitance parameter; γ is the circuit propagation coefficient, and calculating formula does $\mathrm{\&gamma;}=\sqrt{(r_1+{\mathrm{Sl}}_1)(g_1+{\mathrm{Sc}}_1)};$

When the electric current of known two ends, can solve voltage by (4) formula, derive following computing formula:

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

When writing software program and realize, a kind of practical numerical computation method of (5) formula is following, and it has described terminal voltage at the variable quantity of 2 τ in the time interval:

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

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

Characteristic impedance when ignoring line loss, calculating formula does

R is the total resistance value of transmission line, and calculating formula is R=r ₁D;

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

The two ends electric current power frequency component Δ i that step 6 is obtained _Ms(t), Δ i _Ns(t) substitution (6) formula obtains the practical formula of 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), Δ i _Nt(t) substitution (6) formula obtains the practical formula of 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 utilizes electric current power frequency component that step 6 obtains and step 7 to obtain, line protection adopts two coefficient parameter recognition methods to calculate both sides system impedance parameter L successively _m, R _mAnd L _n, R _n, concrete steps are following:

At first, two coefficient parameter recognition methods calculating L is described _m, R _mMethod:

The current failure component of m side and voltage failure component are obeyed 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 are subtracted each other and are 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window after the taking-up fault, row are write equation successively, constitute the overdetermined equation group:

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

In the following formula, the first derivative of electric current can replace calculating with diff,

(11) formula is abbreviated as:

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

Adopt the 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, two coefficient parameter recognition methods calculating L is described _n, R _nMethod:

The current failure component of n side and voltage failure component are obeyed 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 are subtracted each other and are 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 concrete calculating, long electric current power frequency component and the voltage power frequency component variable quantity of 10ms data window after the taking-up fault, row are write equation successively, constitute the overdetermined equation group:

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

In the following formula, the first derivative of electric current can replace calculating with diff,

(16) formula is abbreviated as:

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

Adopt the least-squares calculation parameter L _n, R _n:

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

The two ends current temporary state component substitution following formula that step 9, line protection obtain step 6 calculates the transient current energy,

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

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

In the formula: t ₁, t ₂It is respectively the long initial moment of 10ms data window and the termination moment;

If Δ I _Mt＞I _SetAnd Δ I _Nt＞I _Set, then the transient current energy is bigger, changes step 10, starts the main criterion based on model error; If Δ I _Mt≤I _SetOr Δ I _Nt≤I _Set, then the transient current energy shortage changes step 11, starts the assistant criteria based on system impedance and voltage power frequency component; In the formula, I _SetBe the current calibration threshold, be set to 0.1 times of circuit rated current;

Step 10, the system impedance parameter of utilizing current temporary state component that step 6 obtains, voltage transient component variation amount that step 7 obtains and step 8 to obtain, line protection computation model error E _n, E _m, and the failure judgement position, concrete steps are following:

At first, model error E is described _nComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _m, R _mThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Nj, b _NjThe substitution following formula calculates 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 described _mComputational methods:

Current temporary state component, voltage transient component variation amount and L that 10ms data window after the fault is long _n, R _nThe substitution following formula calculates a 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 the following formula, the first derivative of electric current can replace calculating with diff,

Again with a _Mj, b _MjThe substitution following formula calculates E _m:

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

At last, with the result of calculation substitution protection criterion of model error, the failure judgement position:

If E _n＜ξ or E _m＜ξ is 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, the protection action;

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

The voltage power frequency component variable quantity that step 11, line protection utilize system impedance parameter that step 8 obtains and step 7 to obtain, the failure judgement position, concrete steps are following:

If L _m＜0 or L _n＜0, be judged to external area error, protection is failure to actuate;

If L _m>=0 and L _n>=0, need to combine voltage power frequency component variable quantity ability failure judgement position;

Utilize the pencil of matrix algorithm to extract the amplitude Δ U of voltage power frequency component variable quantity respectively _Ms, Δ U _Ns,

If Δ U _Ms＜U _SetAnd Δ U _Ns＜U _Set, being judged to external area error, protection is failure to actuate;

If Δ U _Ms>=U _SetOr Δ U _Ns>=U _Set, be judged to troubles inside the sample space, the protection action;

In the criterion, U _SetBe the voltage threshold of adjusting, numerical value does

K wherein _RelBe safety factor, be taken as 1.1-1.2, U _NElectric pressure for transmission line.

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