CN102116821A - Method for positioning transmission line fault based on power frequency phasor - Google Patents

Abstract

The invention relates to a method for positioning transmission line fault based on power frequency phasor, which comprises the following steps: confirming a characteristic impedance of propagation constant line according to a distribution parameter model of the transmission line; measuring the voltage Un and current In at tail end N of the transmission line by measuring the voltage Um and current Im at the starting end M of the transmission line, wherein DL is the actual length of transmission line; and calculating a distance DMf between the starting end M of the transmission line and the fault point according to the formula: DmF is equal to [th-1(BJ/AJ)]/gammaJ, thereby increasing the precision of positioning the transmission line fault.

Description

Localization method based on the failure point of power transmission line of power frequency phasor

One, technical field

The present invention relates to a kind of localization method of the failure point of power transmission line based on the power frequency phasor, especially be applicable to localization method the trouble spot of long distance transmission line.

Two, background technology

Guarantee the non-fault of conveying circuit, it is the important step that guarantees transmission of electric energy, therefore judge that the localization method of failure point of power transmission line is an important method, determine the position of failure point of power circuit quickly and accurately, can accelerate the reparation of permanent fault, in time remove a hidden danger to avoid the generation once more of a large amount of transient faults, safety and stability and the economical operation that guarantees electric system had crucial meaning.

General length surpasses the overhead power line of 300km and cable power circuit that length surpasses 100km is called long transmission line, existing impedance method distance-finding method adopts lumped parameter, can only be fit to find range than short-term road, and can not ignore for leakage conductance between the long transmission line lead and electric capacity, then inequality along lead electric current everywhere, the resistance of lead, inductance just can not be considered by lumped parameter, therefore everywhere voltage is also inequality between lead, electricity between line is led with electric capacity and can not be considered by lumped parameter, therefore the degree of accuracy of failure judgement point is not high, adopt the capable ripple principle of both-end can on-line automaticly provide position of failure point, but time dissemination system and given line length can influence its range finding result, to long also not high apart from the degree of accuracy of conveying circuit failure judgement point, and the instrument that uses involves great expense.

Three, summary of the invention

In order to overcome above-mentioned technical disadvantages, the present invention has found a kind of Fault Locating Method that is adapted to long distance transmission line, because this method has been considered parameter distributions, and is not subjected to the influence of time dissemination system and line length substantially, therefore improved the degree of accuracy of failure point of power transmission line location.

For achieving the above object, the technical scheme that the present invention takes is: first step: set up the circuit distributed parameter model, the circuit distributed parameter model includes top power supply E _m, terminal power supply E _n, circuit top M, line terminal N, circuit unit length impedance Z ₀With admittance Y ₀, top power supply E _mBe set to by one group of impedance Z ₀With terminal power supply E _nTandem connects, in each impedance Z ₀Two ends be provided with the admittance Y that is connected with ground ₀

According to the Transmission Line Distributed Parameter model, determine the circuit propagation constant

The characteristic impedance of circuit

D _LBe set to the circuit physical length, wherein, Z ₀With Y ₀Be known quantity.

Second step: the voltage U of measuring circuit top M _m, electric current I _m, the voltage U of measuring circuit terminal N _n, electric current I _n, according to D _MF=[th ^-1(B _J/ A _J)]/γ _JCalculate the distance D between trouble spot and the circuit top M _MF, wherein:

$A_J=Z_{\mathrm{cJ}}\mathrm{ch}\left({\mathrm{\γ}}_J{D}_L\right)I_{\mathrm{nJ}}-U_{\mathrm{nJ}}\mathrm{sh}\left({\mathrm{\γ}}_J{D}_L\right)+Z_{\mathrm{cJ}}{I}_{\mathrm{mJ}},$

$B_J=U_{\mathrm{mJ}}-U_{\mathrm{nJ}}\mathrm{ch}\left({\mathrm{\γ}}_J{D}_L\right)+Z_{\mathrm{CJ}}{I}_{\mathrm{nJ}}\mathrm{sh}\left({\mathrm{\γ}}_J{D}_L\right).$

Owing to used power frequency phasor and comparatively accurate Transmission Line Distributed Parameter Model Calculation trouble spot, overcome of the influence that be subjected to time dissemination system and given line length of traditional impedance method for the inaccurate and capable ripple principle of both-end of long distance transmission line range finding, not only can accurately calculate fault distance, and the economy advantage is obvious than the both-end traveling wave method.

Four, description of drawings

Fig. 1 is a uniform transmission line of the present invention road model

Fig. 2 is a simulation process process flow diagram of the present invention

Fig. 3 is a simulation process process flow diagram of the present invention

Power transmission line M end three-phase voltage oscillogram when Fig. 4 is A phase short circuit of the present invention

Power transmission line M end three-phase current oscillogram when Fig. 5 is A phase short circuit of the present invention

Power transmission line M end three-phase voltage oscillogram when Fig. 6 is BC phase short circuit of the present invention

Power transmission line N end three-phase current oscillogram when Fig. 7 is BC phase short circuit of the present invention

Five, embodiment

First embodiment of the present invention;

A kind of localization method of the failure point of power transmission line based on the power frequency phasor, first step: set up the circuit distributed parameter model, include top power supply E _m, terminal power supply E _n, circuit top M, line terminal N, circuit unit length impedance Z ₀With admittance Y ₀, top power supply E _mBe set to by one group of impedance Z ₀With terminal power supply E _nTandem connects, in each impedance Z ₀Two ends be provided with the admittance Y that is connected with ground ₀

According to the Transmission Line Distributed Parameter model, determine the circuit propagation constant

The characteristic impedance of circuit

D _LBe set to the circuit physical length, wherein, Z ₀With Y ₀Be known quantity.

Second step: the voltage U of measuring circuit top M _m, electric current I _m, the voltage U of measuring circuit terminal N _n, electric current I _n, according to D _MF=[th ^-1(B _J/ A _J)]/γ _JCalculate the distance D between trouble spot and the circuit top M _MF, wherein:

$A_J=Z_{\mathrm{cJ}}\mathrm{ch}\left({\mathrm{\γ}}_J{D}_L\right)I_{\mathrm{nJ}}-U_{\mathrm{nJ}}\mathrm{sh}\left({\mathrm{\γ}}_J{D}_L\right)+Z_{\mathrm{cJ}}{I}_{\mathrm{mJ}},$

$B_J=U_{\mathrm{mJ}}-U_{\mathrm{nJ}}\mathrm{ch}\left({\mathrm{\γ}}_J{D}_L\right)+Z_{\mathrm{CJ}}{I}_{\mathrm{nJ}}\mathrm{sh}\left({\mathrm{\γ}}_J{D}_L\right).$

In the present embodiment, total track length D _L=300km, Z ₀=0.1675+j17.0753, Y ₀=j0.0524, Z ₀, Y ₀Unit be an ohm Ω, voltage unit is a volt (V), current unit be ampere (A).

Work as voltage U _m=-49180-j378970, electric current I _m=-683.6-j1156.1, voltage U _n=-22780-j402200, electric current I _n=-132.59-j510.19 is by the calculating D of second step _MF=30.0544km, the distance between physical fault point and the circuit top M is 30km.

Work as voltage U _m=-40730-j382250, electric current I _m=-620.2-j958.7, voltage U _n=-23400-j399490, electric current I _n=-203.87-j525.63 is by the calculating D of second step _MF=60.0638km, the distance between physical fault point and the circuit top M is 60km.

Work as voltage U _m=-33250-j386640, electric current I _m=-524.29-j783.07, voltage U _n=-24840-j395880, electric current I _n=-291.96-j559.73 is by the calculating D of second step _MF=100.1112km, the distance between physical fault point and the circuit top M is 100km.

Work as voltage U _m=-27570-j391580, electric current I _m=-408.51-j648.59, voltage U _n=-28120-j391250, electric current I _n=-400.95-j635.34 is by the calculating D of second step _MF=149.9917km, the distance between physical fault point and the circuit top M is 150km.

Work as voltage U _m=-24310-j396030, electric current I _m=-299.86-j570.41, voltage U _n=-33820-j386130, electric current I _n=-517.00-j764.92 is by the calculating D of second step _MF=199.8569km, the distance between physical fault point and the circuit top M is 200km.

Work as voltage U _m=-22660-j400330, electric current I _m=-189.65-j529.67, voltage U _n=-43770-j380360, electric current I _n=-637.6-j989.3 is by the calculating D of second step _MF=249.8966km, the distance between physical fault point and the circuit top M is 250km.

The computing formula of second step of the present invention is to obtain by Fourier filtering calculating and phase-model transformation, and is specific as follows:

The calculating of power frequency phasor, all-wave Fourier difference algorithm.During the system failure, often produce bigger attenuating dc component, used Fourier's difference algorithm.

Voltage and current signal when setting up departments the system fault is:

Wherein:

Be signal attenuation DC component, f _m(k),

Amplitude and initial phase for the k subharmonic.In the formula:

$a_n=\frac{2}{N}\·\frac{1}{2\sin (\mathrm{\π}/N)}\left|\underset{k=1}{\overset{N-1}{\mathrm{\Σ}}}(x_{k+1}-x_k)\sin (2\mathrm{kn\π}/N)\right|$

$b_n=\frac{2}{N}\·\frac{1}{2\cos (\mathrm{\π}/N)}\left|\underset{k=1}{\overset{N-1}{\mathrm{\Σ}}}(x_{k+1}-x_k)\cos (2\mathrm{kn\π}/N)\right|$

2X ²＝a ²+b ²

$\tan \mathrm{\α}=\frac{b}{a}$

Suppose At sampling interval T _sVariation during this time is little, but the therefore influence of filtering attenuating dc component.

Transmission line of electricity is any voltage and electric current phasor calculating arbitrarily

According to the basic thought of phasor method, will wait that the amount of asking comes out with its phasor representation.Present amount to be asked is

U (x, t) and i (x, t), its phasor representation is as follows

$u(x,t)=\mathrm{Im}\left[\sqrt{2}\stackrel{\·}{U}\left(x\right)e^{\mathrm{j\ωt}}\right]$

(4)

$i(x,t)=\mathrm{Im}\left[\sqrt{2}\stackrel{\·}{I}\left(x\right)e^{\mathrm{j\ωt}}\right]$

$-\frac{\∂u}{\∂x}=R_{0}i+L_0\frac{\∂i}{\∂t}$

(5)

$-\frac{\∂i}{\∂x}=G_{0}i+C_0\frac{\∂u}{\∂t}$

Wherein

Be respectively u (x, t), (x, phasor t) they are the functions of x to i.For convenient, will

Be abbreviated as

With formula (4) substitution uniform transmission line equation, promptly in the formula (5), can contain and wait to ask the equation of phasor to be

$-\frac{d\stackrel{\·}{U}}{\mathrm{dx}}=(R_0+\mathrm{j\ω}{L}_0)\stackrel{\·}{I}=Z_0\stackrel{\·}{I}$

(6)

$-\frac{d\stackrel{\·}{I}}{\mathrm{dx}}=(G_0+\mathrm{j\ω}{C}_0)\stackrel{\·}{I}=Y_0\stackrel{\·}{U}$

Z wherein ₀=R ₀+ j ω L ₀Be the impedance of unit length, Y ₀=G ₀+ j ω C ₀Admittance for unit length.

Because

Be the function of x only, so partial derivative becomes total derivative.For conveniently finding the solution, the derivative of formula (6) two ends being got again an x gets

$-\frac{d^2\stackrel{\&CenterDot;}{U}}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HSA00000397887900021.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=Z_0\frac{d\stackrel{\&CenterDot;}{I}}{\mathrm{dx}}$

(7)

$-\frac{d^2\stackrel{\&CenterDot;}{I}}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HSA00000397887900021.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=Y_0\frac{d\stackrel{\&CenterDot;}{U}}{\mathrm{dx}}$

With in the formula (6)

With The right-hand member of expression formula substitution following formula, can get

$-\frac{d^2\stackrel{\&CenterDot;}{U}}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HSA00000397887900021.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=Z_0{Y}_0\stackrel{\&CenterDot;}{U}={\mathrm{\&gamma;}}^2\stackrel{\&CenterDot;}{U}$

$-\frac{d^2\stackrel{\&CenterDot;}{I}}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HSA00000397887900021.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=Z_0{Y}_0\stackrel{\&CenterDot;}{I}={\mathrm{\&gamma;}}^2\stackrel{\&CenterDot;}{I}---\left(8\right)$

Wherein,

Be a plural number that does not have unit, be called the propagation constant of uniform transmission line.Following formula is the differential equation of two linear constant coefficients, so its form of separating should be

$\stackrel{\&CenterDot;}{U}=A_1{e}^{-\mathrm{\&gamma;x}}+A_2{e}^{\mathrm{\&gamma;x}}$

(9)

$\stackrel{\&CenterDot;}{I}=B_1{e}^{-\mathrm{\&gamma;x}}+B_2{e}^{\mathrm{\&gamma;x}}$

A wherein ₁, A ₂, B ₁, B ₂Be coefficient undetermined, have according to formula (6)

$\stackrel{\&CenterDot;}{I}=-\frac{1}{Z_0}\frac{d\stackrel{\&CenterDot;}{U}}{\mathrm{dx}}=-\frac{1}{Z_0}(-{\mathrm{<figure-callout\; id="1"\; label="A"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">A</figure-callout>}}_1{\mathrm{\&gamma;e}}^{-\mathrm{\&gamma;x}}+A_2{\mathrm{\&gamma;e}}^{\mathrm{\&gamma;x}})$

$=\frac{A_1}{\sqrt{\frac{Z_0}{Y_0}}}{e}^{-\mathrm{\&gamma;x}}-\frac{A_2}{\sqrt{\frac{Z_0}{Y_0}}}{e}^{\mathrm{\&gamma;x}}=B_1{e}^{-\mathrm{\&gamma;x}}+B_2{e}^{\mathrm{\&gamma;x}}---\left(1\right)$

Order

As seen A ₁And B ₁, A ₂And B ₂The pass be

So electric current

Can be written as again

$\stackrel{\&CenterDot;}{I}=\frac{A_1}{Z_c}{e}^{-\mathrm{\&gamma;x}}-\frac{A_2}{Z_c}{e}^{\mathrm{\&gamma;x}}---\left(11\right)$

Z in the formula _cCharacteristic impedance for transmission line.In sum, find the solution u (x, t) and i (x t), can solve its phasor earlier

With

, its expression formula is

$\stackrel{\&CenterDot;}{U}=A_1{e}^{-\mathrm{\&gamma;x}}+A_2{e}^{\mathrm{\&gamma;x}}$

$\stackrel{\&CenterDot;}{I}=\frac{A_1}{Z_c}{e}^{-\mathrm{\&gamma;x}}-\frac{A_2}{Z_c}{e}^{\mathrm{\&gamma;x}}---\left(12\right)$

γ in the formula, Z _cBe referred to as the second parameter of uniform transmission line, behind the raw parameter of known uniform transmission line, can draw, so as long as obtain undetermined coefficient A ₁, A ₂The back just can be write out

Expression formula.A ₁, A ₂Can try to achieve by two kinds of methods.

(1) undetermined according to the top condition.If the voltage and current at known top is respectively

With

Reference direction as shown in Figure 2.Because top is the position of x=0 on the transmission line, so will have in the x=0 substitution formula (12)

$\stackrel{\&CenterDot;}{U}=A_1+A_2={\stackrel{\&CenterDot;}{U}}_1$

$\stackrel{\&CenterDot;}{I}=\frac{A_1}{Z_c}-\frac{A_2}{Z_c}={\stackrel{\&CenterDot;}{I}}_1---\left(13\right)$

From top equation, can solve undetermined coefficient A ₁, A ₂For

$A_1=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_1+Z_c{\stackrel{\&CenterDot;}{I}}_1)---\left(14\right)$

$A_2=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_1-Z_c{\stackrel{\&CenterDot;}{I}}_1)$

The substitution formula has in (12)

$\stackrel{\&CenterDot;}{U}=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_1+Z_c{\stackrel{\&CenterDot;}{I}}_1)e^{-\mathrm{\&gamma;x}}+\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_1-Z_c{\stackrel{\&CenterDot;}{I}}_1)e^{\mathrm{\&gamma;x}}$

(15)

$\stackrel{\&CenterDot;}{I}=\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_1}{Z_c}+{\stackrel{\&CenterDot;}{I}}_1)e^{-\mathrm{\&gamma;x}}-\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_1}{Z_c}-{\stackrel{\&CenterDot;}{I}}_1)e^{\mathrm{\&gamma;x}}$

Formula (15) is exactly the sinusoidal steady state solution of uniform transmission line equation.Utilize hyperbolic function

$\cosh \left(\mathrm{\&gamma;x}\right)=\frac{1}{2}(e^{\mathrm{\&gamma;x}}+e^{-\mathrm{\&gamma;x}})$

(16)

$\sinh \left(\mathrm{\&gamma;x}\right)=\frac{1}{2}(e^{\mathrm{\&gamma;x}}-e^{-\mathrm{\&gamma;x}})$

Formula (15) can be written as again

$\stackrel{\&CenterDot;}{U}={\stackrel{\&CenterDot;}{U}}_1\cosh \left(\mathrm{\&gamma;x}\right)-Z_c{\stackrel{\&CenterDot;}{I}}_1\sinh \left(\mathrm{\&gamma;x}\right)$

$\stackrel{\&CenterDot;}{I}={\stackrel{\&CenterDot;}{I}}_1\cosh \left(\mathrm{\&gamma;x}\right)-\frac{{\stackrel{\&CenterDot;}{U}}_1}{Z_c}\sinh \left(\mathrm{\&gamma;x}\right)---\left(17\right)$

(2) undetermined according to terminal condition.If the voltage and current at known terminal place is respectively

Reference direction as shown in Figure 2.Because terminal is the position of x=l on the transmission line (l is a line length),, have in the x=l substitution formula (12)

$\stackrel{\&CenterDot;}{U}=A_1{e}^{-\mathrm{\&gamma;l}}+A_2{e}^{\mathrm{\&gamma;l}}={\stackrel{\&CenterDot;}{U}}_2$

$\stackrel{\&CenterDot;}{I}=\frac{A_1}{Z_c}{e}^{-\mathrm{\&gamma;l}}-\frac{A_2}{Z_c}{e}^{\mathrm{\&gamma;l}}={\stackrel{\&CenterDot;}{I}}_2---\left(18\right)$

From following formula, can solve

$A_1=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2+Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{\mathrm{\&gamma;l}}$

(19)

$A_2=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2-Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{-\mathrm{\&gamma;l}}$

The another kind of expression formula that can get the sinusoidal steady state solution of uniform transmission line equation in the substitution formula (12) is

$\stackrel{\&CenterDot;}{U}=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2+Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{\mathrm{\&gamma;}(l-x)}+\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2-Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{-\mathrm{\&gamma;}(l-x)}$

(20)

$\stackrel{\&CenterDot;}{I}=\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_2}{Z_c}+{\stackrel{\&CenterDot;}{I}}_2)e^{\mathrm{\&gamma;}(l-x)}-\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_2}{Z_c}-{\stackrel{\&CenterDot;}{I}}_2)e^{-\mathrm{\&gamma;}(l-x)}$

Because the x representative is that any is to the distance at top on the line, l represents line length, so the l-x representative is exactly the distance that a bit arrives end on the line.Make x '=l-x, then following formula can be written as again

$\stackrel{\&CenterDot;}{U}=\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2+Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{{\mathrm{\&gamma;x}}^{\&prime;}}+\frac{1}{2}({\stackrel{\&CenterDot;}{U}}_2-Z_c{\stackrel{\&CenterDot;}{I}}_2)e^{-{\mathrm{\&gamma;x}}^{\&prime;}}$

(21)

$\stackrel{\&CenterDot;}{I}=\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_2}{Z_c}+{\stackrel{\&CenterDot;}{I}}_2)e^{{\mathrm{\&gamma;x}}^{\&prime;}}-\frac{1}{2}(\frac{{\stackrel{\&CenterDot;}{U}}_2}{Z_c}-{\stackrel{\&CenterDot;}{I}}_2)e^{-\mathrm{\&gamma;}{x}^{\&prime;}}$

X ' in the formula refers to the distance of terminal.Utilize hyperbolic function, formula (21) can be expressed as again

$\stackrel{\&CenterDot;}{U}={\stackrel{\&CenterDot;}{U}}_2\cosh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)+Z_c{\stackrel{\&CenterDot;}{I}}_2\sinh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)$

$\stackrel{\&CenterDot;}{I}={\stackrel{\&CenterDot;}{I}}_2\cosh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)+\frac{{\stackrel{\&CenterDot;}{U}}_2}{Z_c}\sinh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)---\left(22\right)$

As seen, as long as the condition of top or terminal is known one arbitrarily, just can obtain undetermined coefficient A ₁, A ₂Thereby, try to achieve the voltage phasor of transmission line

And electric current phasor

Data sync

When calculating fault distance with formula (26), require M, the voltage at N two ends, current data synchronous, at this moment, the voltage and current before available arbitrary end fault for example can write out modulus 1 before fault as the reference phasor synchronous with the opposite end measurement data:

${\stackrel{\&CenterDot;}{U}}_{\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">m</figure-callout>}1}={\stackrel{\&CenterDot;}{U}}_{\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">n</figure-callout>}1}\mathrm{ch}\left({\mathrm{\&gamma;}}_1{D}_L\right)-Z_{c1}{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">n</figure-callout>}1}\mathrm{sh}\left({\mathrm{\&gamma;}}_1{D}_L\right)$

${\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">m</figure-callout>}1}=\frac{{\stackrel{\&CenterDot;}{U}}_{R1}}{{\mathrm{<figure-callout\; id="1"\; label="Z\; c"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">Z</figure-callout>}}_c}\mathrm{sh}\left({\mathrm{\&gamma;}}_1{D}_L\right)-{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="HSA00000397887900031.png,HSA00000397887900032.png"\; state="\{\{state\}\}">n</figure-callout>}1}\mathrm{ch}\left({\mathrm{\&gamma;}}_1{D}_L\right)$

Following formula has provided between the circuit both end voltage and the phase relation between the electric current, and this concerns proof.Under the existing electrical quantities measurement condition of electric system, the electric current at two ends, voltage can be used as the normative reference of two ends time synchronized, stopped clock to the time influence.

Power frequency phasor method range finding ultimate principle

Suppose to break down at M end x place, so according to above analysis, can draw fault point voltage can be expressed as with the two ends electric parameters:

$\stackrel{\&CenterDot;}{U}x={\stackrel{\&CenterDot;}{U}}_m\cosh \left(\mathrm{\&gamma;x}\right)-Z_c{\stackrel{\&CenterDot;}{I}}_m\sinh \left(\mathrm{\&gamma;x}\right)---\left(23\right)$

$\stackrel{\&CenterDot;}{U}x={\stackrel{\&CenterDot;}{U}}_n\cosh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)+Z_c{\stackrel{\&CenterDot;}{I}}_n\sinh \left({\mathrm{\&gamma;x}}^{\&prime;}\right)$

By following formula, cancellation

After can try to achieve fault distance

x＝Dmf＝th ^-1(B/A)/γ (24)

In the formula, $A=Z_c\mathrm{ch}\left({\mathrm{\&gamma;D}}_L\right)=\stackrel{\&CenterDot;}{I_n}-\stackrel{\&CenterDot;}{U_n}\mathrm{sh}\left(\mathrm{\&gamma;}{D}_L\right)+Z_c\stackrel{\&CenterDot;}{I_m}$

$B=\stackrel{\&CenterDot;}{U_m}-\stackrel{\&CenterDot;}{U_n}\mathrm{ch}\left({\mathrm{\&gamma;D}}_L\right)+Z_C\stackrel{\&CenterDot;}{I_n}\mathrm{sh}\left({\mathrm{\&gamma;D}}_L\right)$

Because phase component is difficult for finding the solution, and utilizes coordinate transform that phase space is become other coordinate spaces usually.This new space is called the modular space, and voltage wherein, electric current etc. are called mode voltage, mould electric current.Because each mold component is separate, therefore all available formula of each mold component (23) is expressed, just wherein propagation constant and wave impedance Zc all corresponding to the value of this mold component.

Mold component transformation matrix commonly used has the symmetrical components transformation matrix, Clarke matrix, triumphant human relations boolean conversion etc.Three phase line is carried out can obtaining three kinds of mold components 0,1,2 after the modular transformation.These three kinds of mold components have three groups of formula corresponding with formula (23), can unify to be expressed as

$\stackrel{\&CenterDot;}{U_{\mathrm{FJ}}}=A_{\mathrm{mJ}}\stackrel{\&CenterDot;}{U_{\mathrm{mJ}}}-B_{\mathrm{mJ}}\stackrel{\&CenterDot;}{I_{\mathrm{mJ}}}---\left(25\right)$

$\stackrel{\&CenterDot;}{U_{\mathrm{FJ}}}=A_{\mathrm{nJ}}\stackrel{\&CenterDot;}{U_{\mathrm{nJ}}}-B_{\mathrm{nJ}}\stackrel{\&CenterDot;}{I_{\mathrm{nJ}}}$

J in the formula---pattern number, J=0,1,2; A _MJ, A _NJ, B _MJ, B _NJ---the coefficient of corresponding mold component J;

I _NJ---be respectively m, the voltage of n end mold component J, electric current are by formula (25) cancellation

Can get fault distance is

D _mF＝[th ^-1(B _J/A _J)]/γ _J (26)

In the formula,

$A_J=Z_{\mathrm{cJ}}\mathrm{ch}\left({\mathrm{\&gamma;}}_J{D}_L\right)\stackrel{\&CenterDot;}{I_{\mathrm{nJ}}}-\stackrel{\&CenterDot;}{U_{\mathrm{nJ}}}\mathrm{sh}\left({\mathrm{\&gamma;}}_J{D}_L\right)+Z_{\mathrm{cJ}}\stackrel{\&CenterDot;}{I_{\mathrm{mJ}}}$

$B_J=\stackrel{\&CenterDot;}{U_{\mathrm{mJ}}}-\stackrel{\&CenterDot;}{U_{\mathrm{nJ}}}\mathrm{ch}\left({\mathrm{\&gamma;}}_J{D}_L\right)+Z_{\mathrm{CJ}}\stackrel{\&CenterDot;}{I_{\mathrm{nJ}}}\mathrm{sh}\left({\mathrm{\&gamma;}}_J{D}_L\right)$

γ _J---the propagation coefficient of mould J,

Z _CJ---the mould J wave impedance of power transmission line,

This shows, all can obtain fault distance with arbitrary modulus.Since to used for transmission line be distribution parameter, therefore circuit model is accurate, long and short line all is suitable for.

We have worked out the computer program process flow diagram according to the present invention, have worked out computer program, and have carried out test:

(1) single-phase earthing fault emulation

By the closure time of realistic model switch is set, the fault type of circuit can be set, circuit might as well be set A phase short trouble has taken place.Set and just can utilize after the parameter Matlab to obtain three-phase voltage, the current waveform of transmission route survey end, as Fig. 4, shown in Figure 5.

(2) heterogeneous short trouble emulation

Might as well establish circuit the BC phase fault takes place, obtain the measuring junction three-phase voltage, three-phase current waveform such as Fig. 6 are shown in 7.Three-phase voltage after the emulation, current data are carried out Fourier transform, by the phasor location algorithm, can calculate fault distance then, result of calculation is as shown in table 1.Wherein, fault distance unit is km.Maximum error is 0.8126/300*100%=0.2709%.

By analyzing as can be known, utilize arbitrary preface amount can carry out fault localization.Yet in positive and negative, 03 prefaces, have only positive sequence can adapt to all types of faults.Under the condition of utilizing the positive sequence amount to find range, for unsymmetrical short-circuit and ground short circuit, also can be simultaneously with the calculating of finding range of negative phase-sequence amount or zero sequence amount, and compare, so that the range finding result is more reliable with its result and range finding result with the positive sequence amount.Find that through test result of calculation conforms to actual.

The characteristics that the present invention has are: adopted comparatively accurate Transmission Line Distributed Parameter model, be adapted to long transmission line, improved the degree of accuracy of failure point of power transmission line location.

In localization method technical field based on the failure point of power transmission line of power frequency phasor; Every including, determine the circuit propagation constant according to the Transmission Line Distributed Parameter model

The characteristic impedance of circuit Voltage U by measuring circuit top M _m, electric current I _m, the voltage U of measuring circuit terminal N _n, electric current I _n, D wherein _LFor the circuit physical length, according to D _MF=[th ^-1(B _J/ A _J)]/γ _JCalculate the distance D between trouble spot and the circuit top M _MFTechnology contents all in protection scope of the present invention.

Claims (1)

1. localization method based on the failure point of power transmission line of power frequency phasor, it is characterized in that: first step: set up the circuit distributed parameter model, the circuit distributed parameter model includes top power supply E _m, terminal power supply E _n, circuit top M, line terminal N, circuit unit length impedance Z ₀With admittance Y ₀, top power supply E _mBe set to by one group of impedance Z ₀With terminal power supply E _nTandem connects, in each impedance Z ₀Two ends be provided with the admittance Y that is connected with ground ₀

According to the Transmission Line Distributed Parameter model, determine the circuit propagation constant

The characteristic impedance of circuit

D _LBe set to the circuit physical length, wherein, Z ₀With Y ₀Be known quantity.

Second step: the voltage U of measuring circuit top M _m, electric current I _m, the voltage U of measuring circuit terminal N _n, electric current I _n, according to D _MF=[th ^-1(B _J/ A _J)]/γ _JCalculate the distance D between trouble spot and the circuit top M _MF, wherein:

$A_J=Z_{\mathrm{cJ}}\mathrm{ch}\left({\mathrm{\&gamma;}}_J{D}_L\right)\stackrel{\&CenterDot;}{I_{\mathrm{nJ}}}-\stackrel{\&CenterDot;}{U_{\mathrm{nJ}}}\mathrm{sh}\left({\mathrm{\&gamma;}}_J{D}_L\right)+Z_{\mathrm{cJ}}\stackrel{\&CenterDot;}{I_{\mathrm{mJ}}},$

$B_J=U_{\mathrm{mJ}}-U_{\mathrm{nJ}}\mathrm{ch}\left({\mathrm{\&gamma;}}_J{D}_L\right)+Z_{\mathrm{CJ}}\stackrel{\&CenterDot;}{I_{\mathrm{nJ}}}\mathrm{sh}\left({\mathrm{\&gamma;}}_J{D}_L\right).$

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