CN103345551A - Method for calculating tower electric potential in back flashover based on vector matching method - Google Patents

Abstract

The invention relates to a method for calculating tower electric potential in back flashover based on a vector matching method and belongs to the technical field of power system electromagnetic transient calculation. Firstly, tower impact self-impedance response and mutual impedance response of any two points in complex frequency domains on a typical tower are analyzed through electromagnetic simulation, and on the basis, impedance response of each branch of a tower T-type circuit is obtained; then, an impedance function of each branch of the tower is fitted through the vector matching method, complex number pole pairs, real number pole pairs and constant terms of the impedance response function of each branch are expressed through circuits respectively, and the circuits are in series connection respectively so that a branch equivalent circuit corresponding to each branch can be obtained. According to the T-type circuit, with the combination of each branch equivalent circuit, a tower two-port equivalent circuit is established. Therefore, the electric potential of any point on the tower is obtained through the tower two-port equivalent circuit. According to the method for calculating tower electric potential in back flashover based on the vector matching method, simulation calculation accuracy in key links of a power system in the lightning stroke process can be improved and lightning protection performance calculation accuracy and assessment accuracy of an electric transmission line in the prior art can be effectively improved.

Description

A kind of based on shaft tower voltage calculating method in the back flashover of vector matching method

Technical field

The present invention relates to a kind of based on shaft tower voltage calculating method in the back flashover of vector matching method, relate in particular in a kind of emulation back flashover based on the electric power line pole tower Potential distribution two port equivalent electrical circuit of vector matching method, belong to electromagnetic transient in power system computing technique field.

Background technology

After thunder and lightning hits transmission line of electricity, shaft tower will become one of main route of transmission of lightning wave, and the formation of lightning surge has been played key effect.In addition, the statistics to the back flashover accident shows that most of thunderbolt occurs in the transmission pole top of tower.Therefore set up rational electric power line pole tower current potential computing method and in circuit is strikeed back the research of anti-thunder performance, occupy critical role.The current potential computing method of existing electric power line pole tower lack research targetedly relatively for the widely used wineglass tower of China, cat head tower and a large amount of compact shaft towers that use etc.The potential change that existing computing method generally only can analog insulation place can't be analyzed the clearance potential change that other may discharge simultaneously, and then also just can't analyze in the back flashover the especially potential change situation of clearance of the many discharge channels of shaft tower.And in back flashover calculates, be very necessary for above-mentioned tower and the potentiometric analysis of many discharge channels.

Summary of the invention

The objective of the invention is to propose a kind of based on shaft tower voltage calculating method in the back flashover of vector matching method, according to electromagnetic field numerical simulation result, in complex frequency domain, set up two port equivalent electrical circuit based on the vector matching method any on the electric power line pole tower at 2, in the time of can calculating thunderbolt shaft tower optional position, the potential change of arbitrfary point on the shaft tower, to simulate possible discharge channel over-voltage condition, and then judgement thunderbolt back shaft tower discharge scenario, improve the simulation calculation precision of key link in the electric system lightning stroke process, and effectively promote present simulation accuracy to the transmission line of electricity lightning protection properties.

The present invention propose based on shaft tower voltage calculating method in the back flashover of vector matching method, may further comprise the steps:

(1) for any 2 point on the electric power line pole tower, be designated as and second point respectively at first, apply excitation to any first, adopt the electromagnetic field numerical algorithm, obtain this any complex frequency domain voltage V (s) on the electric power line pole tower at first, and any second complex frequency domain electric current I (s) on the electric power line pole tower, wherein, V (s) is the magnitude of voltage of complex frequency domain that first time domain voltage v (t) on the shaft tower is obtained after Fourier transform, I (s) is the current value of complex frequency domain that second time domain current i (t) on the shaft tower is obtained after Fourier transform, s is the complex frequency domain value corresponding with time domain frequency values f, and s=j2 π f, j are imaginary unit;

Obtain the complex frequency domain shaft tower impact impedance of electric power line pole tower If first and second coincidence, then Z (s) is called this complex frequency domain shaft tower self-surge impedance of first; If first and second do not overlap, then Z (s) is called the mutual impact impedance of complex frequency domain shaft tower of and point to point at this first;

(2) repeating step (1) obtains the complex frequency domain shaft tower self-surge impedance Z corresponding with and respectively at first at second ₁₁And Z ₂₂, applying excitation to this first on the electric power line pole tower, repeating step (1) obtains this first mutual impact impedance Z of the complex frequency domain shaft tower with second on the electric power line pole tower ₁₂

(3) form a T type circuit, with above-mentioned first as a point in T type circuit first port, the earth is as another point in first port, with above-mentioned second as a point in T type circuit second port, the earth is as another point in second port, and the branch impedance in the T type circuit is respectively z ¹¹, z ¹²And z ²²

(4) be respectively z according to the branch impedance in the T type circuit ¹¹, z ¹²And z ²², obtain complex frequency domain shaft tower self-surge impedance Z ₁₁And Z ₂₂With the mutual impact impedance Z of complex frequency domain shaft tower ₁₂Simultaneous equations:

$\left\{\begin{array}{c}{Z}_{11}=z^{11}+z^{12}\\ {\mathrm{<figure-callout\; id="22"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{22}=z^{22}+z^{12}\\ {\mathrm{<figure-callout\; id="12"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{12}=z^{12}\end{array}\right.$

Find the solution above-mentioned simultaneous equations, obtain:

$\left\{\begin{array}{c}{z}^{11}=Z_{11}-Z_{12}\\ z^{22}=Z_{22}-Z_{12}\\ z^{12}={\mathrm{<figure-callout\; id="12"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{12}\end{array}\right.;$

(5) adopt the vector matching method, respectively in complex frequency domain to branch impedance z ¹¹, z ¹²And z ²²Carry out match with following rational function form:

$z^m\left(s\right)=\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}(m=\mathrm{11,12,22})$

Wherein, a _nBe limit, c _nBe the residual corresponding with this limit, N is the limit number, and d and h are respectively the real constant item, and limit and residual are respectively real number or conjugate complex number is right,

To complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) it is identical to carry out the process of match, and z in the following fit procedure (s) represents complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s), triplicate obtains whole complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s), fit procedure may further comprise the steps:

(5-1) during initialization, establishing initial pole value is a ' ₁, a ' ₂... a ' _N, N is the limit number,

Setting one has the unknown function σ (s) of same form with above-mentioned complex frequency domain branch impedance function z (s), and the limit of this unknown function σ (s) is a ' _n(n=1,2 ... N), multiply each other with unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s), and make the relation of unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s) satisfy following system of equations:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$

D+sh item among the unknown function σ (s) is forced to 1, again the row of second in the following formula is multiply by z (s), obtain:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)z\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ (\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)\end{array}\right]$

In the following formula, equal sign right-hand member two row up and down is equal, that is:

$(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh})=(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)$

(5-2) to above-mentioned equation

Find the solution, all complex frequency domain value s in the traversal complex frequency domain obtain one group with c _n, d, h and c ' _n(n=1,2 ... N) be the equation of variable, find the solution this equation, obtain unknown quantity c _n, d, h and c ' _n(n=1,2 ... N), c wherein _nFor with limit a _nCorresponding residual, c ' _nFor with initial limit a ' _nCorresponding residual, d and h are respectively the real constant item;

(5-3) will $\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$ In σ (s) z (s) function and σ (s) the function fraction that turns to following form:

$\mathrm{\&sigma;}\left(s\right)z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

$\mathrm{\&sigma;}\left(s\right)=\frac{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

Wherein, z _nFor unknown function σ (s) z (s) zero of a function (n=1,2 ... N), z ' _nFor unknown function σ (s) zero of a function (n=1,2 ... N), a _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), a ' _nFor initial limit (n=1,2 ... N),

Two formulas are divided by obtain:

$z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}$

Wherein, z ' _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), it is identical with the zero point of σ (s) to draw the limit of z (s) thus;

The z ' at zero point of unknown function σ (s) _nBe exactly matrix [A-bc ' ^T] eigenwert, wherein A is with initial limit a ' _n(n=1,2 ... N) be the diagonal matrix of diagonal entry, b is that element is 1 column vector, c ' ^TFor with unknown function σ (s) residual c ' _n(n=1,2 ... N) be the row vector of element.By matrix [A-bc ' ^T] eigenwert just can obtain the z ' at zero point of unknown function σ (s) _n, i.e. the new limit a ' ' of complex frequency domain branch impedance function z (s) _n

Adopt new limit a ' ' _n(n=1,2 ... N), repeating step (5-1) and step (5-2), when unknown function σ (s) becomes 1, i.e. all residual c ' of unknown function σ (s) _n(n=1,2 ... N) become at 0 o'clock, obtain the limit a of complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h, carry out step (5-4);

(5-4) according to the limit a of above-mentioned complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h,, find the solution above-mentioned formula

In residual c _n(n=1,2 ... N),, obtain complex frequency domain branch impedance function z (s);

(5-5) repeating step (5-1) is to step (5-4), and match obtains complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s);

(6) represent and above-mentioned complex frequency domain branch impedance function z with equivalent electrical circuit respectively ¹¹(s), z ¹²(s) and z ²²(s) And (d+sh), with the equivalent electrical circuit series connection, obtain respectively and complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) corresponding equivalent electrical circuit;

Represent and above-mentioned complex frequency domain branch impedance function with equivalent electrical circuit respectively

In

(n=1,2 ... N) and (d+sh), with the equivalent electrical circuit series connection, obtain the equivalent electrical circuit corresponding with complex frequency domain branch impedance function z (s), establish N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, wherein limit is right two of conjugate complex number

The sum equivalence is the z in the circuit _1n(s) (n=1 ..., K), limit is real number

The item equivalence is the z in the circuit _2n(s) (n=1 ..., N-2K), (d+sh) equivalence is the z in the circuit ₃(s);

If N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, and establish the conjugate complex number limit of K and be: $\left\{\begin{array}{c}{a}_{2n-1}=-p_{\mathrm{rn}}+{\mathrm{jp}}_{\mathrm{in}}\\ a_{2n}=-p_{\mathrm{rn}}-{\mathrm{jp}}_{\mathrm{in}}\end{array}\right.n=\mathrm{1,2},...,K,$ Wherein, p _RnAnd p _InBe arithmetic number, with the corresponding residual of conjugate complex number limit be

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="2n"\; label="c"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">c</figure-callout>}}_{2n-1}=c_{\mathrm{rn}}+{\mathrm{jc}}_{\mathrm{in}}\\ a_{2n}=c_{\mathrm{rn}}-{\mathrm{jc}}_{\mathrm{in}}\end{array}\right.n=\mathrm{1,2},...,K,$ Wherein, c _RnAnd c _InBe arithmetic number,

Establishing N-2K real pole again is: a _n＜0n=2K+1 ..., N, corresponding residual is: c _nN=2K+1 ..., N, then complex frequency domain branch impedance function z (s) can turn to

$z\left(s\right)=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+\underset{n=2K+1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}$

$=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}{z}_{1n}\left(s\right)+\underset{n=2K+1}{\overset{N}{\mathrm{\&Sigma;}}}{z}_{2n}\left(s\right)+z_3\left(s\right)$

In the following formula:

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

${\mathrm{<figure-callout\; id="2n"\; label="z"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">z</figure-callout>}}_{2n}\left(s\right)=\frac{c_n}{s-a_n}$

z ₃(s)＝d+sh

z _1n(s), z _2n(s) and z ₃(s) equivalence is first circuit, second circuit and tertiary circuit respectively;

Z in the following formula _1n(s), n=1,2 ..., K,

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

$=\frac{s\frac{{2c}_{\mathrm{<figure-callout\; id="2"\; label="rn\; p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">rn</figure-callout>}}}{p_{\mathrm{rn}}^{}}+\frac{{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}}{\frac{s^2}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+\frac{{2p}_{\mathrm{rn}}s}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+1}$

Get

$R_{1n}=\frac{2(c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}})}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

${\mathrm{<figure-callout\; id="2n"\; label="G"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">G</figure-callout>}}_{2n}=\frac{2p_{\mathrm{rn}}^2}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

$L_{1n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}{({\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

${\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}{({\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

$C_{1n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}$

${\mathrm{<figure-callout\; id="2n"\; label="C"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">C</figure-callout>}}_{2n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

Z then _1n(s) be designated as:

$z_{1n}\left(s\right)=\frac{s(L_{1n}+{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n})+R_{1n}}{s^2{L}_{1n}{C}_{1n}+s{R}_{1n}{C}_{1n}+1}$

$=\frac{s{L}_{1n}+R_{1n}}{s^2{L}_{1n}{C}_{1n}+s{R}_{1n}{C}_{1n}+1}+\frac{s{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n}}{s^2{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2\mathrm{<figure-callout\; id="2n"\; label="n\; C"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">n</figure-callout>}}{C}_{}{}_{2n}+s{\mathrm{<figure-callout\; id="2n"\; label="G"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">G</figure-callout>}}_{2\mathrm{<figure-callout\; id="2n"\; label="n\; L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">n</figure-callout>}}{L}_{}{}_{2n}+1}$

z _1n(s) be first circuit;

Z in the following formula _2n(s), n=2K+1 ..., N gets

$C_n=\frac{1}{c_n}$

$G_n=-\frac{a_n}{c_n}$

Z then _2n(s) be designated as:

${\mathrm{<figure-callout\; id="2n"\; label="z"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">z</figure-callout>}}_{2n}\left(s\right)=\frac{c_n}{s-a_n}=\frac{1}{s{C}_n+G_n}$

z _2n(s) equivalence is second circuit;

Z in the following formula ₃(s), get

R ₃＝d

L ₃＝h

Z then ₃(s) be designated as:

z ₃(s)＝d+sh＝R ₃+sL ₃

z ₃(s) equivalence is tertiary circuit;

As the individual and z K _1n(s) Dui Ying equivalent electrical circuit, (N-2K) individual and z _2n(s) Dui Ying equivalent electrical circuit and and z ₃(s) Dui Ying equivalent electrical circuit is together in series, and obtains the corresponding equivalent electrical circuit of complex frequency domain branch impedance function z (s);

(7) apply the lightning impulse current of random waveform at 2 at this any T type equivalent electrical circuit first port of electric power line pole tower, try to achieve the potential change of second port, the current potential of second port is current potential any on the shaft tower at second.

The present invention propose based on shaft tower voltage calculating method in the back flashover of vector matching method, compare with original computing method, have the following advantages:

1, shaft tower current potential computing method of the present invention, the possibility discharging gap is set up two port equivalent electrical circuit of shaft tower respectively on transmission line of electricity, can be in the hope of also comparing the Potential distribution of each discharge channel, and then the flashover situation of a plurality of insulators gap and clearance when can be used for analyzing back flashover, be conducive to analyze and definite power transmission line lightning shielding counterattack culprit.

2, shaft tower current potential computing method of the present invention for the electric power line pole tower of various electric pressures and tower, can both be set up equivalent electrical circuit analysis, have applicability preferably.

3, shaft tower current potential computing method of the present invention can be analyzed the shaft tower Potential distribution situation under the excitation of random waveform lightning current, have applicability preferably.

Description of drawings

Fig. 1 is in the inventive method, the T type equivalent electrical circuit corresponding with the shaft tower impact impedance.

Fig. 2 is in the inventive method, the equivalent electrical circuit corresponding with complex frequency domain branch impedance function.

Fig. 3 is in the inventive method, with conjugate complex number limit in the complex frequency domain branch impedance function to corresponding equivalent electrical circuit.

Fig. 4 is in the inventive method, the equivalent electrical circuit corresponding with real pole in the complex frequency domain branch impedance function.

Fig. 5 is in the inventive method, the equivalent electrical circuit corresponding with constant term in the complex frequency domain branch impedance function.

Embodiment

The present invention propose based on shaft tower voltage calculating method in the back flashover of vector matching method, may further comprise the steps:

(1) for any 2 point on the electric power line pole tower, be designated as and second point respectively at first, apply excitation to any first, adopt the electromagnetic field numerical algorithm, obtain this any complex frequency domain voltage V (s) on the electric power line pole tower at first, and any second complex frequency domain electric current I (s) on the electric power line pole tower, wherein, V (s) is the magnitude of voltage of complex frequency domain that first time domain voltage v (t) on the shaft tower is obtained after Fourier transform, I (s) is the current value of complex frequency domain that second time domain current i (t) on the shaft tower is obtained after Fourier transform, s is the complex frequency domain value corresponding with time domain frequency values f, and s=j2 π f, j are imaginary unit.

Obtain the complex frequency domain shaft tower impact impedance of electric power line pole tower

If first and second coincidence, then Z (s) is called this complex frequency domain shaft tower self-surge impedance of first; If first and second do not overlap, then Z (s) is called the mutual impact impedance of complex frequency domain shaft tower of and point to point at this first;

Time domain impact impedance z (t) can be defined as

$z\left(t\right)=\frac{v\left(t\right)}{i\left(t\right)}$

V in the formula (t) is function of voltage any first on the shaft tower, and i (t) is dash current function any second on the shaft tower.If first and second coincidence, then z (t) is called this time domain shaft tower self-surge impedance of first; If first and second does not overlap, then z (t) is called the mutual impact impedance of time domain shaft tower of and point to point at this first.Utilize Fourier transform that the time domain impact impedance is transformed to complex frequency domain, can obtain the shaft tower impact impedance Z (s) in the complex frequency domain

$Z\left(s\right)=\frac{V\left(s\right)}{I\left(s\right)}$

If first and second coincidence, then Z (s) is called this complex frequency domain shaft tower self-surge impedance of first; If first and second does not overlap, then Z (s) is called the mutual impact impedance of complex frequency domain shaft tower of and point to point at this first.

(2) repeating step (1) obtains the complex frequency domain shaft tower self-surge impedance Z corresponding with and respectively at first at second ₁₁And Z ₂₂, applying excitation to this first on the electric power line pole tower, repeating step (1) obtains this first mutual impact impedance Z of the complex frequency domain shaft tower with second on the electric power line pole tower ₁₂

(3) form a T type circuit, as shown in Figure 1, with above-mentioned first as a point in T type circuit first port, the earth is as another point in first port, with above-mentioned second as a point in T type circuit second port, the earth is as another point in second port, and the branch impedance in the T type circuit is respectively z ¹¹, z ¹²And z ²²

(4) be respectively z according to the branch impedance in the T type circuit ¹¹, z ¹²And z ²², obtain complex frequency domain shaft tower self-surge impedance Z ₁₁And Z ₂₂With the mutual impact impedance Z of complex frequency domain shaft tower ₁₂Simultaneous equations:

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="11"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{11}={\mathrm{<figure-callout\; id="11"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{11}+z^{12}\\ {\mathrm{<figure-callout\; id="22"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{22}={\mathrm{<figure-callout\; id="22"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{22}+z^{12}\\ {\mathrm{<figure-callout\; id="12"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{12}={\mathrm{<figure-callout\; id="12"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{12}\end{array}\right.$

Find the solution above-mentioned simultaneous equations, obtain:

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="11"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{11}=Z_{11}-Z_{12}\\ {\mathrm{<figure-callout\; id="22"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{22}=Z_{22}-Z_{12}\\ {\mathrm{<figure-callout\; id="12"\; label="z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">z</figure-callout>}}^{12}={\mathrm{<figure-callout\; id="12"\; label="Z"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">Z</figure-callout>}}_{12}\end{array}\right.;$

(5) adopt the vector matching method, respectively in complex frequency domain to branch impedance z ¹¹, z ¹²And z ²²Carry out match with following rational function form:

$z^m\left(s\right)=\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}(m=\mathrm{11,12,22})$

Wherein, a _nBe limit, c _nBe the residual corresponding with this limit, N is the limit number, and d and h are respectively the real constant item, and limit and residual are respectively real number or conjugate complex number is right,

To complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) it is identical to carry out the process of match, and z in the following fit procedure (s) represents complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s), triplicate obtains whole complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s), fit procedure may further comprise the steps:

(5-1) during initialization, establishing initial pole value is a ' ₁, a ' ₂... a ' _N, N is the limit number,

Setting one has the unknown function σ (s) of same form with above-mentioned complex frequency domain branch impedance function z (s), and the limit of this unknown function σ (s) is a ' _n(n=1,2 ... N), multiply each other with unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s), and make the relation of unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s) satisfy following system of equations:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$

D+sh item among the unknown function σ (s) is forced to 1, again the row of second in the following formula is multiply by z (s), obtain:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)z\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ (\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)\end{array}\right]$

In the following formula, equal sign right-hand member two row up and down is equal, that is:

$(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh})=(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)$

(5-2) to above-mentioned equation

Find the solution, all complex frequency domain value s in the traversal complex frequency domain obtain one group with c _n, d, h and c ' _n(n=1,2 ... N) be the equation of variable, find the solution this equation, obtain unknown quantity c _n, d, h and c ' _n(n=1,2 ... N), c wherein _nFor with limit a _nCorresponding residual, c ' _nFor with initial limit a ' _nCorresponding residual, d and h are respectively the real constant item.

(5-3) will $\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$ In σ (s) z (s) function and σ (s) the function fraction that turns to following form:

$\mathrm{\&sigma;}\left(s\right)z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

$\mathrm{\&sigma;}\left(s\right)=\frac{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

Wherein, z _nFor unknown function σ (s) z (s) zero of a function (n=1,2 ... N), z ' _nFor unknown function σ (s) zero of a function (n=1,2 ... N), a _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), a ' _nFor initial limit (n=1,2 ... N),

Two formulas are divided by obtain:

$z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}$

Wherein, z ' _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), it is identical with the zero point of σ (s) to draw the limit of z (s) thus;

The z ' at zero point of unknown function σ (s) _nBe exactly matrix [A-bc ' ^T] eigenwert, wherein A is with initial limit a ' _n(n=1,2 ... N) be the diagonal matrix of diagonal entry, b is that element is 1 column vector, c ' ^TFor with unknown function σ (s) residual c ' _n(n=1,2 ... N) be the row vector of element.By matrix [A-bc ' ^T] eigenwert just can obtain the z ' at zero point of unknown function σ (s) _n, they also are the new limit a ' ' of complex frequency domain branch impedance function z (s) simultaneously _n

Adopt new limit a ' ' _n(n=1,2 ... N), repeating step (5-1) and step (5-2), when unknown function σ (s) becomes 1, i.e. all residual c ' of unknown function σ (s) _n(n=1,2 ... N) become at 0 o'clock, obtain the limit a of complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h, carry out step (5-4);

(5-4) according to the limit a of above-mentioned complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h,, find the solution above-mentioned formula In residual c _n(n=1,2 ... N),, obtain complex frequency domain branch impedance function z (s);

(5-5) repeating step (5-1) is to step (5-4), and match obtains complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s).

(6) represent and above-mentioned complex frequency domain branch impedance function z with equivalent electrical circuit respectively ¹¹(s), z ¹²(s) and z ²²(s)

And (d+sh), with the equivalent electrical circuit series connection, obtain respectively and complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) corresponding equivalent electrical circuit, as shown in Figure 2;

Because of respectively to complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) it is identical to carry out the process of circuit equivalent, and z in the following equivalent process (s) represents complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s), triplicate can access whole complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) equivalent electrical circuit.

Represent and above-mentioned complex frequency domain branch impedance function with equivalent electrical circuit respectively

In

(n=1,2 ... N) and (d+sh), with the equivalent electrical circuit series connection, obtain the equivalent electrical circuit corresponding with complex frequency domain branch impedance function z (s), establish N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, wherein limit is right two of conjugate complex number

The sum equivalence is the z in the circuit _1n(s) (n=1 ..., K), limit is real number

The item equivalence is the z in the circuit _2n(s) (n=1 ..., N-2K), (d+sh) equivalence is the z in the circuit ₃(s);

If N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, and establish the conjugate complex number limit of K and be: $\left\{\begin{array}{c}{a}_{2n-1}=-p_{\mathrm{rn}}+{\mathrm{jp}}_{\mathrm{in}}\\ a_{2n}=-p_{\mathrm{rn}}-{\mathrm{jp}}_{\mathrm{in}}\end{array}\right.n=\mathrm{1,2},...,K,$ Wherein, p _RnAnd p _InBe arithmetic number, with the corresponding residual of conjugate complex number limit be

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="2n"\; label="c"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">c</figure-callout>}}_{2n-1}=c_{\mathrm{rn}}+{\mathrm{jc}}_{\mathrm{in}}\\ a_{2n}=c_{\mathrm{rn}}-{\mathrm{jc}}_{\mathrm{in}}\end{array}\right.n=\mathrm{1,2},...,K,$ Wherein, c _RnAnd c _InBe arithmetic number,

Establishing N-2K real pole again is: a _n＜0n=2K+1 ..., N, corresponding residual is: c _nN=2K+1 ..., N, then complex frequency domain branch impedance function z (s) can turn to

$z\left(s\right)=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+\underset{n=2K+1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}$

$=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}{z}_{1n}\left(s\right)+\underset{n=2K+1}{\overset{\mathrm{<figure-callout\; id="2n"\; label="N\; z"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">N</figure-callout>}}{\mathrm{\&Sigma;}}}{z}_{}{}_{2n}\left(s\right)+z_3\left(s\right)$

In the following formula:

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

${\mathrm{<figure-callout\; id="2n"\; label="z"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">z</figure-callout>}}_{2n}\left(s\right)=\frac{c_n}{s-a_n}$

z ₃(s)＝d+sh

z _1n(s), z _2n(s) and z ₃(s) equivalence is as Fig. 3 respectively, Fig. 4 and first circuit, second circuit and tertiary circuit shown in Figure 5;

Z in the following formula _1n(s), n=1,2 ..., K,

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+2c_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+2p_{\mathrm{rns}}+{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

$=\frac{s\frac{{2c}_{\mathrm{<figure-callout\; id="2"\; label="rn\; p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">rn</figure-callout>}}}{p_{\mathrm{rn}}^{}}+\frac{{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-2p_{\mathrm{in}}{c}_{\mathrm{in}}}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}}{\frac{s^2}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+\frac{{2p}_{\mathrm{rn}}s}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}+1}$

Get

$R_{1n}=\frac{2(c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}})}{{\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2}$

${\mathrm{<figure-callout\; id="2n"\; label="G"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">G</figure-callout>}}_{2n}=\frac{2p_{\mathrm{rn}}^2}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

$L_{1n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}{({\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

${\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}{({\mathrm{<figure-callout\; id="2"\; label="p\; rn"\; filenames="HDA00003434287200011.png"\; state="\{\{state\}\}">p</figure-callout>}}_{\mathrm{rn}}^{}+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

$C_{1n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}$

${\mathrm{<figure-callout\; id="2n"\; label="C"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">C</figure-callout>}}_{2n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

Z then _1n(s) be designated as:

$z_{1n}\left(s\right)=\frac{s(L_{1n}+{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n})+R_{1n}}{s^2{L}_{1n}{C}_{1n}+s{R}_{1n}{C}_{1n}+1}$

$=\frac{s{L}_{1n}+R_{1n}}{s^2{L}_{1n}{C}_{1n}+s{R}_{1n}{C}_{1n}+1}+\frac{s{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2n}}{s^2{\mathrm{<figure-callout\; id="2n"\; label="L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">L</figure-callout>}}_{2\mathrm{<figure-callout\; id="2n"\; label="n\; C"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">n</figure-callout>}}{C}_{}{}_{2n}+s{\mathrm{<figure-callout\; id="2n"\; label="G"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">G</figure-callout>}}_{2\mathrm{<figure-callout\; id="2n"\; label="n\; L"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">n</figure-callout>}}{L}_{}{}_{2n}+1}$

z _1n(s) be as shown in Figure 3 first circuit;

Z in the following formula _2n(s), n=2K+1 ..., N gets

$C_n=\frac{1}{c_n}$

$G_n=-\frac{a_n}{c_n}$

Z then _2n(s) be designated as:

${\mathrm{<figure-callout\; id="2n"\; label="z"\; filenames="HDA00003434287200013.png"\; state="\{\{state\}\}">z</figure-callout>}}_{2n}\left(s\right)=\frac{c_n}{s-a_n}=\frac{1}{s{C}_n+G_n}$

z _2n(s) equivalence is second circuit as shown in Figure 4;

Z in the following formula ₃(s), get

R ₃＝d

L ₃＝h

Z then ₃(s) be designated as:

z ₃(s)＝d+sh＝R ₃+sL ₃

z ₃(s) equivalence is tertiary circuit as shown in Figure 5;

As the individual and z K _1n(s) Dui Ying equivalent electrical circuit, (N-2K) individual and z _2n(s) Dui Ying equivalent electrical circuit and and z ₃(s) Dui Ying equivalent electrical circuit is together in series, and obtains the corresponding equivalent electrical circuit of complex frequency domain branch impedance function z (s);

Repeat said process, obtain complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s) Dui Ying equivalent electrical circuit according to Fig. 1 combination, can obtain this any T type equivalent electrical circuit on the electric power line pole tower at 2.

(7) apply lightning impulse current at this any T type equivalent electrical circuit first port of electric power line pole tower at 2, try to achieve the potential change of second port, the current potential of second port is current potential any on the shaft tower at second.Because said process can set up any 2 T type equivalent electrical circuit relative to the earth on the shaft tower, in the time of then can trying to achieve that on the electric power line pole tower be struck by lightning in the arbitrfary point according to above step, the potential change situation of other arbitrfary points on the shaft tower.

Claims (1)

1. one kind based on shaft tower voltage calculating method in the back flashover of vector matching method, it is characterized in that this method may further comprise the steps:

(1) for any 2 point on the electric power line pole tower, be designated as and second point respectively at first, apply excitation to any first, adopt the electromagnetic field numerical algorithm, obtain this any complex frequency domain voltage V (s) on the electric power line pole tower at first, and any second complex frequency domain electric current I (s) on the electric power line pole tower, wherein, V (s) is the magnitude of voltage of complex frequency domain that first time domain voltage v (t) on the shaft tower is obtained after Fourier transform, I (s) is the current value of complex frequency domain that second time domain current i (t) on the shaft tower is obtained after Fourier transform, s is the complex frequency domain value corresponding with time domain frequency values f, and s=j2 π f, j are imaginary unit;

Obtain the complex frequency domain shaft tower impact impedance of electric power line pole tower

If first and second coincidence, then Z (s) is called this complex frequency domain shaft tower self-surge impedance of first; If first and second do not overlap, then Z (s) is called the mutual impact impedance of complex frequency domain shaft tower of and point to point at this first;

(2) repeating step (1) obtains the complex frequency domain shaft tower self-surge impedance Z corresponding with and respectively at first at second ₁₁And Z ₂₂, applying excitation to this first on the electric power line pole tower, repeating step (1) obtains this first mutual impact impedance Z of the complex frequency domain shaft tower with second on the electric power line pole tower ₁₂

(3) form a T type circuit, with above-mentioned first as a point in T type circuit first port, the earth is as another point in first port, with above-mentioned second as a point in T type circuit second port, the earth is as another point in second port, and the branch impedance in the T type circuit is respectively z ¹¹, z ¹²And z ²²

(4) be respectively z according to the branch impedance in the T type circuit ¹¹, z ¹²And z ²², obtain complex frequency domain shaft tower self-surge impedance Z ₁₁And Z ₂₂With the mutual impact impedance Z of complex frequency domain shaft tower ₁₂Simultaneous equations:

$\left\{\begin{array}{c}{Z}_{11}=z^{11}+z^{12}\\ Z_{22}=z^{22}+z^{12}\\ Z_{12}=z^{12}\end{array}\right.$

Find the solution above-mentioned simultaneous equations, obtain:

$\left\{\begin{array}{c}{z}^{11}=Z_{11}-Z_{12}\\ z^{22}=Z_{22}-Z_{12}\\ z^{12}=Z_{12}\end{array}\right.;$

(5) adopt the vector matching method, respectively in complex frequency domain to branch impedance z ¹¹, z ¹²And z ²²Carry out match with following rational function form:

$z^m\left(s\right)=\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}$ (m=11,12,22)

Wherein, a _nBe limit, c _nBe the residual corresponding with this limit, N is the limit number, and d and h are respectively the real constant item, and limit and residual are respectively real number or conjugate complex number is right,

To complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) it is identical to carry out the process of match, and z in the following fit procedure (s) represents complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s), triplicate obtains whole complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s), fit procedure may further comprise the steps:

(5-1) during initialization, establishing initial pole value is a' ₁, a' ₂... a' _N, N is the limit number,

Setting one has the unknown function σ (s) of same form with above-mentioned complex frequency domain branch impedance function z (s), and the limit of this unknown function σ (s) is a' _n(n=1,2 ... N), multiply each other with unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s), and make the relation of unknown function σ (s) and above-mentioned complex frequency domain branch impedance function z (s) satisfy following system of equations:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$

D+sh item among the unknown function σ (s) is forced to 1, again the row of second in the following formula is multiply by z (s), obtain:

$\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)z\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ (\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)\end{array}\right]$

In the following formula, equal sign right-hand member two row up and down is equal, that is:

$(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh})=(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)$

(5-2) to above-mentioned equation $(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh})=(\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1)z\left(s\right)$ Find the solution, all complex frequency domain value s in the traversal complex frequency domain obtain one group with c _n, d, h and c' _n(n=1,2 ... N) be the equation of variable, find the solution this equation, obtain unknown quantity c _n, d, h and c' _n(n=1,2 ... N), c wherein _nFor with limit a _nCorresponding residual, c' _nFor with initial limit a ' _nCorresponding residual, d and h are respectively the real constant item;

(5-3) will $\left[\begin{array}{c}\mathrm{\&sigma;}\left(s\right)z\left(s\right)\\ \mathrm{\&sigma;}\left(s\right)\end{array}\right]\&ap;\left[\begin{array}{c}\underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-{a^{\&prime;}}_n}+d+\mathrm{sh}\\ \underset{n=1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{{c^{\&prime;}}_n}{s-{a^{\&prime;}}_n}+1\end{array}\right]$ In σ (s) z (s) function and σ (s) the function fraction that turns to following form:

$\mathrm{\&sigma;}\left(s\right)z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

$\mathrm{\&sigma;}\left(s\right)=\frac{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{a^{\&prime;}}_n)}$

Wherein, z _nFor unknown function σ (s) z (s) zero of a function (n=1,2 ... N), z' _nFor unknown function σ (s) zero of a function (n=1,2 ... N), a _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), a ' _nFor initial limit (n=1,2 ... N),

Two formulas are divided by obtain:

$z\left(s\right)=h\frac{\underset{n=1}{\overset{N+1}{\mathrm{\&Pi;}}}(s-z_n)}{\underset{n=1}{\overset{N}{\mathrm{\&Pi;}}}(s-{z^{\&prime;}}_n)}$

Wherein, z' _nFor the limit of complex frequency domain branch impedance function z (s) (n=1,2 ... N), the limit that draws z (s) thus is identical with the zero point of σ (s);

The z' at zero point of unknown function σ (s) _nBe exactly matrix [A-bc' ^T] eigenwert, wherein A is with initial limit a' _n(n=1,2 ... N) be the diagonal matrix of diagonal entry, b is that element is 1 column vector, c' ^TFor with unknown function σ (s) residual c' _n(n=1,2 ... N) be the row vector of element.By matrix [A-bc' ^T] eigenwert just can obtain the z' at zero point of unknown function σ (s) _n, i.e. the new limit a ' ' of complex frequency domain branch impedance function z (s) _n

Adopt new limit a ' ' _n(n=1,2 ... N), repeating step (5-1) and step (5-2), when unknown function σ (s) becomes 1, i.e. all residual c' of unknown function σ (s) _n(n=1,2 ... N) become at 0 o'clock, obtain the limit a of complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h, carry out step (5-4);

(5-4) according to the limit a of above-mentioned complex frequency domain branch impedance function z (s) _n(n=1,2 ... N) and constant term d and h,, find the solution above-mentioned formula

In residual c _n(n=1,2 ... N),, obtain complex frequency domain branch impedance function z (s);

(5-5) repeating step (5-1) is to step (5-4), and match obtains complex frequency domain branch impedance function z respectively ¹¹(s), z ¹²(s) and z ²²(s);

(6) represent and above-mentioned complex frequency domain branch impedance function z with equivalent electrical circuit respectively ¹¹(s), z ¹²(s) and z ²²(s)

And (d+sh), with the equivalent electrical circuit series connection, obtain respectively and complex frequency domain branch impedance function z ¹¹(s), z ¹²(s) and z ²²(s) corresponding equivalent electrical circuit;

Represent and above-mentioned complex frequency domain branch impedance function with equivalent electrical circuit respectively

In

(n=1,2 ... N) and (d+sh), with the equivalent electrical circuit series connection, obtain the equivalent electrical circuit corresponding with complex frequency domain branch impedance function z (s), establish N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, wherein limit is right two of conjugate complex number

The sum equivalence is the z in the circuit _1n(s) (n=1 ..., K), limit is real number The item equivalence is the z in the circuit _2n(s) (n=1 ..., N-2K), (d+sh) equivalence is the z in the circuit ₃(s);

If N the limit a of complex frequency domain branch impedance function z (s) _nIn, the conjugate complex number limit of K and N-2K real pole are arranged, and establish the conjugate complex number limit of K and be: $\left\{\begin{array}{c}{a}_{2n-1}={-p}_{\mathrm{rn}}+{\mathrm{jp}}_{\mathrm{in}}\\ a_{2n}={-p}_{\mathrm{rn}}-{\mathrm{jp}}_{\mathrm{in}}\end{array}\right.$ N=1,2 ..., K, wherein, p _RnAnd p _InBe arithmetic number, with the corresponding residual of conjugate complex number limit be

$\left\{\begin{array}{c}{c}_{2n-1}=c_{\mathrm{rn}}+{\mathrm{jc}}_{\mathrm{in}}\\ c_{2n}=c_{\mathrm{rn}}-{\mathrm{jc}}_{\mathrm{in}}\end{array}\right.$ N=1,2 ..., K, wherein, c _RnAnd c _InBe arithmetic number,

Establishing N-2K real pole again is: a _n＜0n=2K+1 ..., N, corresponding residual is: c _nN=2K+1 ..., N,

Then complex frequency domain branch impedance function z (s) can turn to

$z\left(s\right)=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}\frac{{2c}_{\mathrm{rn}}s+{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-{2p}_{\mathrm{in}}{c}_{\mathrm{in}}}{s^2+{2p}_{\mathrm{rns}}+p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}+\underset{n=2K+1}{\overset{N}{\mathrm{\&Sigma;}}}\frac{c_n}{s-a_n}+d+\mathrm{sh}$

$=\underset{n=1}{\overset{K}{\mathrm{\&Sigma;}}}{z}_{1n}\left(s\right)+\underset{n=2K+1}{\overset{N}{\mathrm{\&Sigma;}}}{z}_{2n}\left(s\right)+z_3\left(s\right)$

In the following formula:

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-{2p}_{\mathrm{in}}{c}_{\mathrm{in}}}{s^2+{2p}_{\mathrm{rns}}+p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}$

$z_{2n}\left(s\right)=\frac{c_n}{s-a_n}$

z ₃(s)＝d+sh

z _1n(s), z _2n(s) and z ₃(s) equivalence is first circuit, second circuit and tertiary circuit respectively;

Z in the following formula _1n(s), n=1,2 ..., K,

$z_{1n}\left(s\right)=\frac{{2c}_{\mathrm{rn}}s+{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-{2p}_{\mathrm{in}}{c}_{\mathrm{in}}}{s^2+{2p}_{\mathrm{rns}}+p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}$

$\text{=}\frac{s\frac{{2c}_{\mathrm{rn}}}{p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}+\frac{{2c}_{\mathrm{rn}}{p}_{\mathrm{rn}}-{2p}_{\mathrm{in}}{c}_{\mathrm{in}}}{p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}}{\frac{s^2}{p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}+\frac{{2p}_{\mathrm{rn}}s}{p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}+1}$

Get

$R_{1n}=\frac{2(c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}})}{p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2}$

$G_{2n}=\frac{{2p}_{\mathrm{rn}}^2}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

$L_{1n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}{(p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

$L_{2n}=\frac{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}{(p_{\mathrm{rn}}^2+p_{\mathrm{in}}^2)p_{\mathrm{rn}}}$

$C_{1n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}-p_{\mathrm{in}}{c}_{\mathrm{in}}}$

$C_{2n}=\frac{p_{\mathrm{rn}}}{c_{\mathrm{rn}}{p}_{\mathrm{rn}}+p_{\mathrm{in}}{c}_{\mathrm{in}}}$

Z then _1n(s) be designated as:

$z_{1n}\left(s\right)=\frac{s(L_{1n}+L_{2n})+R_{1n}}{s^2{L}_{1n}{C}_{1n}+{\mathrm{sR}}_{1n}{C}_{1n}+1}$

$=\frac{{\mathrm{sL}}_{1n}+R_{1n}}{s^2{L}_{1n}{C}_{1n}+{\mathrm{sR}}_{1n}{C}_{1n}+1}+\frac{{\mathrm{sL}}_{2n}}{s^2{L}_{2n}{C}_{2n}+{\mathrm{sG}}_{2n}{L}_{2n}+1}$

z _1n(s) be first circuit;

Z in the following formula _2n(s), n=2K+1 ..., N gets

$C_n=\frac{1}{c_n}$

$G_n=-\frac{a_n}{c_n}$

Z then _2n(s) be designated as:

$z_{2n}\left(s\right)=\frac{c_n}{s-a_n}=\frac{1}{{\mathrm{sC}}_n+G_n}$

z _2n(s) equivalence is second circuit;

Z in the following formula ₃(s), get

R ₃＝d

L ₃＝h

Z then ₃(s) be designated as:

z ₃(s)＝d+sh＝R ₃+sL ₃

z ₃(s) equivalence is tertiary circuit;

As the individual and z K _1n(s) Dui Ying equivalent electrical circuit, (N-2K) individual and z _2n(s) Dui Ying equivalent electrical circuit and and z ₃(s) Dui Ying equivalent electrical circuit is together in series, and obtains the corresponding equivalent electrical circuit of complex frequency domain branch impedance function z (s);

(7) apply the lightning impulse current of random waveform at 2 at this any T type equivalent electrical circuit first port of electric power line pole tower, try to achieve the potential change of second port, the current potential of second port is current potential any on the shaft tower at second.

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