CN102081132A - Two-end distance measuring method of transmission line fault under dynamic condition - Google Patents

Abstract

The invention relates to a two-end distance measuring method of transmission line fault under a dynamic condition, comprising the following steps of: acquiring two-end voltage and current signals through two-end phasor measurement units of the transmission line; then acquiring dynamic positive sequence impedance, admittance, wave impedance and propagation coefficients according to line parameter estimation and a dynamic line parameter method; and finally solving a nonlinear equation related to a fault distance according to the fact that positive sequence voltages derived from two ends to a fault point are equal by applying a Newton iteration method so as to obtain a dynamic distance measuring result of the transmission line fault. The two-end distance measuring method can more effectively solve the problem of line parameter and fault distance estimation under the dynamic condition and has accurate and reliable fault distance measuring result.

Description

Transmission line malfunction both-end distance measuring method under a kind of dynamic condition

Technical field

The present invention relates to the both-end distance measuring method of transmission line malfunction in the electric system.

Background technology

Along with the development of WAMS, increasing phasor measurement unit has been installed in the electrical network, for the Two-terminal Fault Location method based on synchronous phasor measurement provides the foundation.Both-end distance measuring method based on phasor measurement unit can utilize failure message fully, and online computational scheme parameter is avoided the influence of factors such as line parameter circuit value is worn out, environment temperature, humidity, can obtain localization of fault accurately.But in the application of reality, consider economy, electric system is in stable edge usually; Have disturbance inevitably on the power transmission line of long distance, the influence of uncertain factors such as fault causes electrical network to be in the current intelligence of vibration often.Difficult point based on the Two-terminal Fault Location of synchronous phasor measurement is when the amplitude frequency of voltage and current signal takes place along with system oscillation to change fast, the electric signal that original distance-finding method is described can't have the ability of spatial variations characteristic and time behavior, bigger error can appear in the fault localization result, be difficult to realize accurate localization of fault, influence the timely eliminating of fault.

Summary of the invention:

The purpose of this invention is to provide the transmission line malfunction both-end distance measuring method under a kind of dynamic condition, this method can be carried out more effectively line parameter circuit value and fault distance estimation under dynamic condition, and the fault localization result accurately, reliably; Can under various fault types, fault distance, fault angle situation, use.

The present invention is for solving its technical matters, and the technical scheme that is adopted is:

A, pre-service

Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after Filtering Processing, obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides _S1, I _R1With positive sequence failure voltage signal U _S1, U _R1, wherein S represents left end, R represents right-hand member;

B, line parameter circuit value are estimated

Homogeneous line Model Calculation according to distribution parameter obtains following positive order parameter:

B1, unit line length positive sequence wave impedance

Unit line length positive sequence propagation coefficient γ ₁=l ^-1Ln[(U _S1+ Z _C1I _S1)/(U _R1-Z _C1I _R1)], wherein l is a transmission line length;

B2, unit line length impedance Z ₁=Z _C1γ ₁, unit line length admittance Y ₁=γ ₁/ Z _C1

C, Two-terminal Fault Location

Under C1, the dynamic condition, the forward-order current amplitude of transmission line of electricity left end

The positive sequence voltage amplitude

The forward-order current amplitude of transmission line of electricity right-hand member

The positive sequence voltage amplitude

T is the time, ω ₀Angular frequency for system.

The dynamic positive sequence impedance of definition left end is

${\tilde{Z}}_{S1}=Z_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\∂\left[i_{S1}\left(t\right)\right]}{\∂t}=R_1+j{\mathrm{\ω}}_0{L}_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\∂\left[i_{S1}\left(t\right)\right]}{\∂t}$

The dynamic positive sequence admittance of definition left end is

${\tilde{Y}}_{S1}=Y_1+\frac{C_1}{U_{S1}\left(t\right)}\frac{\left[u_{S1}\left(t\right)\right]}{\∂t}=G_1+j{\mathrm{\ω}}_0{C}_1+\frac{C_1}{u_{S1}\left(t\right)}\frac{\∂\left[u_{S1}\left(t\right)\right]}{\∂t}$

The dynamic positive sequence impedance of definition right-hand member is

${\tilde{Z}}_{R1}=Z_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\∂\left[i_{R1}\left(t\right)\right]}{\∂t}=R_1+j{\mathrm{\ω}}_0{L}_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\∂\left[i_{R1}\left(t\right)\right]}{\∂t}$

The dynamic positive sequence admittance of definition right-hand member is

${\tilde{Y}}_{R1}=Y_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\∂\left[u_{R1}\left(t\right)\right]}{\∂t}=G_1+j{\mathrm{\ω}}_0{C}_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\∂\left[u_{R1}\left(t\right)\right]}{\∂t}$

Wherein, R ₁, L ₁, G ₁And C ₁The positive sequence resistance, inductance, the electricity that are respectively unit line length are led and electric capacity;

C2, the dynamic positive sequence wave impedance of circuit right-hand member

Dynamic positive sequence propagation coefficient

Trouble spot positive sequence voltage U by the right-hand member derivation _RF1For:

$U_{\mathrm{RF}1}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{R1}x}$

Wherein x is the distance of trouble spot apart from the circuit right-hand member, i.e. fault distance;

The dynamic positive sequence wave impedance of circuit right-hand member

Dynamic positive sequence propagation coefficient Trouble spot positive sequence voltage U by the left end derivation _SF1For:

$U_{\mathrm{SF}1}=\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}$

C3, fault localization

Circuit is U under failure condition _RF1=U _SF1, the nonlinear equation of setting up about fault distance x is:

$\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{R1}x}$

Use the iteration function F (x) that Newton iteration method solves following formula:

$F\left(x\right)=x-{\widetilde{\mathrm{\γ}}}_{S1}\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}+{\widetilde{\mathrm{\γ}}}_{S1}\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{S1}(l-x)}$

$-({\widetilde{\mathrm{\γ}}}_{R1}\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\γ}}}_{R1}x}-{\widetilde{\mathrm{\γ}}}_{R1}\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\γ}}}_{R1}x})$

Following formula is carried out interative computation can obtain fault distance x accurately.

Compared with prior art, the invention has the beneficial effects as follows:

One, two ends of electric transmission line system voltage phasor and the electric current phasor that obtains by synchronized sampling, can calculate surge impedance of a line and propagation coefficient, thereby can estimate accurate line parameter circuit value, comprise that resistance, inductance, electricity lead and electric capacity, avoid the aging or different problems that cause variation of environment of line parameter circuit value, improved the reliability of fault localization.

Two, surge impedance of a line and propagation coefficient are subjected to frequency and signal amplitude variable effect, traditional time domain approach can not solve the problem of wave impedance and propagation coefficient variation, and the present invention is after sampling obtains positive sequential signal in transmission line of electricity two ends and frequency, can accurately calculate dynamic positive sequence wave impedance and dynamic positive sequence propagation coefficient, can obtain optimum fault distance by the introducing Newton iteration method at last separates, fast convergence rate and result are unique, improved system's counting accuracy to various failure conditions below dynamic condition greatly, its range finding result accurately, reliable.To in time searching and the processing circuitry fault, guarantee the safe operation of electrical network, improve stability of power system and reduce operating cost, have important society and economic worth.

Three, among the present invention, can think symmetry when electric system normally moves, promptly each element three-phase impedance phase with, three-phase voltage, size of current equate separately, have normal phase sequence.Under the uneven situation of calculating electric system, quoted symmetrical component method, any asymmetric three-phase phasor A, B, C can be decomposed into three groups of symmetrical components that phase sequence is different.It is promptly positive and negative, the zero phase-sequence of phasor composition that electric current, voltage or the impedance of any three-phase imbalance can be decomposed into three balances.Why hinder all existence under the type because of the As positive-sequence component is in office, so method of the present invention utilizes symmetrical component method to only adopting positive sequence voltage component U after the circuit decoupling zero ₁With forward-order current component I ₁Carry out fault localization, computing is simple, quick, and can carry out accurate fault localization under various short trouble type cases.

Four, time domain distance-finding method under the existing static condition, sample frequency was 2400Hz when the present invention extracted the transmission line malfunction signal, and sample devices is not had specific (special) requirements, convenient enforcement.

In the above-mentioned A step pre-service, carry out decoupling change with symmetrical component method and get transmission line of electricity positive sequence fault current signal I in return ₁With positive sequence failure voltage signal U ₁The decoupling zero transformation for mula be:

$\left[\begin{array}{c}{U}_1\\ U_2\\ {\mathrm{<figure-callout\; id="0"\; label="U"\; filenames="HDA0000036122520000011.png"\; state="\{\{state\}\}">U</figure-callout>}}_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]$

$\left[\begin{array}{c}{I}_1\\ I_2\\ {\mathrm{<figure-callout\; id="0"\; label="I"\; filenames="HDA0000036122520000011.png"\; state="\{\{state\}\}">I</figure-callout>}}_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

In the formula, U _a, U _b, U _cAnd I _a, I _b, I _cBe phase voltage and the phase current that collects, U ₁, U ₂, U ₀And I ₁, I ₂, I ₀Be respectively sequence voltage and preface electric current, and

The present invention is described in further detail below in conjunction with the drawings and specific embodiments.

Description of drawings

Fig. 1 be dynamic fault distance-finding method of the present invention with existing static failure distance-finding method to the relative error curve of a concrete circuit in different moment range finding results, wherein horizontal ordinate is the time, ordinate is range finding result's a relative error.

Fig. 2 is dynamic fault distance-finding method of the present invention and the existing static failure distance-finding method relative error curve to the range finding result of concrete circuit under different faults position and fault type, wherein horizontal ordinate is the fault distance, and ordinate is the average relative error of (t=50ms-1000ms) range finding result in a period of time.

Embodiment

Embodiment

Transmission line malfunction both-end distance measuring method under a kind of dynamic condition of the present invention the steps include:

A, pre-service

Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after Filtering Processing, obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides _S1, I _R1With positive sequence failure voltage signal U _S1, U _R1, wherein S represents left end, R represents right-hand member;

Carry out decoupling change with symmetrical component method and get transmission line of electricity positive sequence fault current signal I in return ₁With positive sequence failure voltage signal U ₁The decoupling zero transformation for mula be:

$\left[\begin{array}{c}{U}_1\\ U_2\\ {\mathrm{<figure-callout\; id="0"\; label="U"\; filenames="HDA0000036122520000011.png"\; state="\{\{state\}\}">U</figure-callout>}}_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]$

$\left[\begin{array}{c}{I}_1\\ I_2\\ {\mathrm{<figure-callout\; id="0"\; label="I"\; filenames="HDA0000036122520000011.png"\; state="\{\{state\}\}">I</figure-callout>}}_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

In the formula, U _a, U _b, U _cAnd I _a, I _b, I _cBe phase voltage and the phase current that collects, U ₁, U ₂, U ₀And I ₁, I ₂, I ₀Be respectively sequence voltage and preface electric current, and

B, line parameter circuit value are estimated

Homogeneous line Model Calculation according to distribution parameter obtains following positive order parameter:

B1, unit line length positive sequence wave impedance

Unit line length positive sequence propagation coefficient γ ₁=l ^-1Ln[(U _S1+ Z _C1I _S1)/(U _R1-Z _C1I _R1)], wherein l is a transmission line length;

B2, unit line length impedance Z ₁=Z _C1γ ₁, unit line length admittance Y ₁=γ ₁/ Z _C1

C, Two-terminal Fault Location

Under C1, the dynamic condition, the forward-order current amplitude of transmission line of electricity left end

The positive sequence voltage amplitude

The forward-order current amplitude of transmission line of electricity right-hand member

The positive sequence voltage amplitude

T is the time, ω ₀Angular frequency for system.

Can obtain on the circuit positive sequence voltage U of 1 m (to left end, the span of m corresponds to 0 to l from the right-hand member of circuit) under dynamic condition according to the homogeneous line model of distribution parameter ₁Differential equation:

$\frac{\&PartialD;U_1}{\&PartialD;m}=I_1{R}_1+L_1\frac{\&PartialD;I_1}{\&PartialD;t}$

$=I_1{R}_1+L_1\frac{j{\mathrm{\&omega;}}_0{i}_1\left(t\right)e^{j{\mathrm{\&omega;}}_{0}t}\&PartialD;t+e^{j{\mathrm{\&omega;}}_{0}t}\&PartialD;\left[i_1\left(t\right)\right]}{\&PartialD;t}$

$=I_1{R}_1+j{\mathrm{\&omega;}}_0{L}_1{I}_1+L_1\frac{\&PartialD;\left[i_1\left(t\right)\right]}{i_1\left(t\right)\&PartialD;t}{i}_1\left(t\right)e^{j{\mathrm{\&omega;}}_{0}t}$

$=I_1(R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{i_1\left(t\right)}\frac{\&PartialD;\left[i_1\left(t\right)\right]}{\&PartialD;t})$

${=I}_1{\tilde{Z}}_1$

$\frac{\&PartialD;I_1}{\&PartialD;m}=U_1{G}_1+C_1\frac{\&PartialD;U_1}{\&PartialD;t}$

$=U_1{G}_1+C_1\frac{j{\mathrm{\&omega;}}_0{u}_1\left(t\right)e^{j{\mathrm{\&omega;}}_{0}t}\&PartialD;t+e^{j{\mathrm{\&omega;}}_{0}t}\&PartialD;\left[u_1\left(t\right)\right]}{\&PartialD;t}$

$=U_1{G}_1+j{\mathrm{\&omega;}}_0{C}_1{U}_1+C_1\frac{\&PartialD;\left[u_1\left(t\right)\right]}{u_1\left(t\right)\&PartialD;t}{u}_1\left(t\right)e^{j{\mathrm{\&omega;}}_{0}t}$

$=U_1(G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{u_1}\frac{\&PartialD;\left[u_1\left(t\right)\right]}{\&PartialD;t})$

$=U_1{\tilde{Y}}_1$

According to above derivation, definable goes out following parameter:

The dynamic positive sequence impedance of definition left end is

${\tilde{Z}}_{S1}=Z_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\&PartialD;\left[i_{S1}\left(t\right)\right]}{\&PartialD;t}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\&PartialD;\left[i_{S1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence admittance of definition left end is

${\tilde{Y}}_{S1}=Y_1+\frac{C_1}{U_{S1}\left(t\right)}\frac{\&PartialD;\left[u_{S1}\left(t\right)\right]}{\&PartialD;t}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{u_{S1}\left(t\right)}\frac{\&PartialD;\left[u_{S1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence impedance of definition right-hand member is

${\tilde{Z}}_{R1}=Z_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\&PartialD;\left[i_{R1}\left(t\right)\right]}{\&PartialD;t}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\&PartialD;\left[i_{R1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence admittance of definition right-hand member is

${\tilde{Y}}_{R1}=Y_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\&PartialD;\left[u_{R1}\left(t\right)\right]}{\&PartialD;t}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\&PartialD;\left[u_{R1}\left(t\right)\right]}{\&PartialD;t}$

Wherein, R ₁, L ₁, G ₁And C ₁The positive sequence resistance, inductance, the electricity that are respectively unit line length are led and electric capacity.

As the t=0 dynamic positive sequence impedance of left end constantly

The dynamic positive sequence admittance of left end

The dynamic positive sequence impedance of right-hand member With the dynamic positive sequence admittance of right-hand member

Specifically be calculated as:

Report the positive sequence voltage signal U of (τ reporting period) constantly according to continuous 3 phasor measurement unit that circuit right-hand member (m=0) is located to gather _R1(t) and forward-order current signal I _R1(t), obtain t=0, positive sequence voltage amplitude change rate during m=0 With the forward-order current amplitude change rate

For:

$\frac{\&PartialD;\left[u_{R1}\left(0\right)\right]}{\&PartialD;t}=\frac{U_{R1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-U_{R1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&sigma;}}}{2\mathrm{\&tau;}}$

$\frac{\&PartialD;\left[i_{R1}\left(0\right)\right]}{\&PartialD;t}=\frac{I_{R1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-I_{R1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&sigma;}}}{2\mathrm{\&tau;}}$

So, the dynamic positive sequence impedance of circuit right-hand member (m=0)

With dynamic positive sequence admittance

When t=0, be respectively:

${\tilde{Z}}_{R1}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{I_{R1}\left(0\right)}\frac{I_{R1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-I_{R1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}}{2\mathrm{\&tau;}}$

${\tilde{Y}}_{R1}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{U_{R1}\left(0\right)}\frac{U_{R1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-U_{R1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}}{2\mathrm{\&tau;}}$

In like manner, can obtain the dynamic positive sequence impedance of circuit left end (m=l)

With dynamic positive sequence admittance

When t=0, be respectively:

${\tilde{Z}}_{S1}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{I_{S1}\left(0\right)}\frac{I_{S1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-I_{S1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}}{2\mathrm{\&tau;}}$

${\tilde{Y}}_{S1}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{U_{S1}\left(0\right)}\frac{U_{S1}\left(\mathrm{\&tau;}\right)e^{-j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}-U_{S1}(-\mathrm{\&tau;})e^{j{\mathrm{\&omega;}}_0\&CenterDot;\mathrm{\&tau;}}}{2\mathrm{\&tau;}}$

C2, the dynamic positive sequence wave impedance of circuit right-hand member

Dynamic positive sequence propagation coefficient

Trouble spot positive sequence voltage U by the right-hand member derivation _RF1For:

$U_{\mathrm{RF}1}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x}$

Wherein x is the distance of trouble spot apart from the circuit right-hand member, i.e. fault distance;

The dynamic positive sequence wave impedance of circuit left end

Dynamic positive sequence propagation coefficient

Trouble spot positive sequence voltage U by the left end derivation _SF1For:

$U_{\mathrm{SF}1}=\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}$

C3, fault localization

Circuit is U under failure condition _RF1=U _SF1, the nonlinear equation of setting up about fault distance x is:

$\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x}$

Can obtain function f (x) by following formula is:

$f\left(x\right)=\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}-\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}-\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x}$

According to the iteration function F (x) in the Newton iteration method=x-f (x)/f ' (x), set up iteration function F (x):

$F\left(x\right)=x-{\widetilde{\mathrm{\&gamma;}}}_{S1}\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+{\widetilde{\mathrm{\&gamma;}}}_{S1}\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}$

$-({\widetilde{\mathrm{\&gamma;}}}_{R1}\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}-{\widetilde{\mathrm{\&gamma;}}}_{R1}\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x})$

Following formula is carried out interative computation can obtain fault distance x accurately.

Adopt the method for present embodiment, the result of the fault localization that carries out for a concrete transmission line of electricity system is as follows:

This transmission line of electricity system is the dual power supply double-circuit system, in order to force system to be in current intelligence a ground short circuit fault F need be set on circuit 1 ₁, power transmission line total length 200km, the circuit pattern is selected TOWER 3H5 shaft tower for use, power supply 1: voltage 230 ∠ 0kV, frequency 50Hz, power supply 2: voltage 230 ∠ 10kV, frequency 50Hz, the positive sequence resistance R of circuit unit ₁=0.03468 Ω, unit positive sequence inductance L ₁=1.3476mH, unit positive sequence electricity are led G ₁=1.0 * 10 ^-7S, unit positive sequence capacitor C ₁=8.6771 * 10 ^-9F, the zero sequence resistance R of unit ₀=0.30000 Ω, unit zero sequence inductance L ₀=3.6371mH, unit zero sequence electricity are led G ₀=1.0 * 10 ^-7S and unit zero sequence capacitor C ₀=6.1610 * 10 ^-9F, soil resistivity R are 100 Ω * m, the power supply 1 equiva lent impedance Z of place ₁=0, distance measuring equipment is installed in respectively on the bus of power supply 1 and power supply 2 and carries out both-end distance measuring.

Fault localization result's relative error E _EstBe calculated as follows:

$E_{\mathrm{est}}=\frac{x-x_0}{l}\&times;100\%$

Wherein, x ₀Fault distance for reality.

For verifying the transmission line malfunction both-end distance measuring method accuracy under a kind of dynamic condition of the present invention, carry out the test under 4 kinds of situations, wherein adopted average relative error and the maximum relative error that each result that constantly finds range in the t=50ms-1000ms of back takes place fault to estimate.

Test 1: the test of different faults position and transition resistance;

A phase earth fault takes place at t=0ms in circuit constantly, and the fault angle is 0 °, and fault resstance is respectively Rf=0 Ω, 5 Ω, 20 Ω, 50 Ω, physical fault apart from the circuit right-hand member apart from x ₀Be 0km～200km, the result is as shown in table 1.

Table 1: the average relative error that back 50-1000ms carries out fault localization takes place in the fault at diverse location and transition resistance

As can be seen from Table 1, its range error of dynamic fault distance-finding method of the present invention is between 0.0002% to 0.1765%, and its range error scope of existing static failure distance-finding method is 0.0001% to 0.5375%, and method of the present invention error on the whole is littler, and it is more accurate to find range.

Wherein, line fault occur in apart from the circuit right-hand member apart from x ₀Be the 40km place, fault type is an A phase earth fault, and the fault angle is 0 °, and transition resistance is 20 Ω, utilizes existing static failure distance-finding method and dynamic fault distance-finding method of the present invention at the different relative error curves of testing the result that obtains finding range constantly, sees Fig. 1.Among Fig. 1, dotted line is the relative error curve of existing static failure distance-finding method, and solid line is the relative error curve of dynamic fault distance-finding method of the present invention.

As can be seen from Figure 1, the range error of static failure distance-finding method is vibrated along with the dynamic oscillation of system, and variation range is big, is difficult to obtain accurate fault localization result; And method range error of the present invention is stable, and the dynamic oscillation with system does not change, and error is little, the distance accuracy height.

Test 2: the test of different faults type and transition resistance;

The fault distance circuit right-hand member that circuit takes place constantly at t=0 apart from x ₀Be 80km, the fault angle is 0 °, and fault resstance is respectively Rf=0 Ω, 5 Ω, 20 Ω, 50 Ω, fault type has A phase earth fault (Ag), AB phase-to phase fault (AB), AB double earthfault (ABg), ABC three-phase earth fault (ABCg), and the result is as shown in table 2.

Table 2: in the fault of different faults type and transition resistance back 50ms-1000ms takes place and carry out fault localization

Average relative error and maximum relative error

As can be seen from Table 2, the average relative error scope that existing static failure distance-finding method obtains is 0.0661% to 1.0812%, and the average relative error scope that dynamic fault distance-finding method of the present invention obtains only is 0.0058% to 0.0525%, and its relative error significantly reduces.

Test 3: the test under the different noises;

A phase earth fault takes place at t=0 in circuit constantly, and the fault angle is 0 °, and fault resstance is respectively Rf=50 Ω, physical fault apart from the circuit right-hand member apart from x ₀Be 0km～200km, noise is 40dB, 50dB, and 60dB, the result is as shown in table 3.

Table 3: the average relative error that back 50ms-1000ms carries out fault localization takes place in the fault at different faults position and noise

As can be seen from Table 3, the average relative error scope that existing static failure distance-finding method obtains is 0.15990% to 0.56697%, and the average relative error scope that dynamic fault distance-finding method of the present invention obtains only is 0.02366% to 0.21288%.

Test 4: the test of different faults position and fault type.

Circuit breaks down constantly at t=0ms, the fault angle is 0 °, and fault resstance is respectively Rf=50 Ω, and physical fault is 0km～200km apart from the bus of power supply R, fault type has A phase earth fault, AB phase-to phase fault, AB double earthfault, ABC three-phase earth fault, the results are shown in Figure 2.Horizontal ordinate is a fault distance among Fig. 2, and ordinate is range finding result's average relative error; The top solid line is the average relative error curve of existing fault distance-finding method when A phase earth fault, dotted line is the average relative error curve of existing fault distance-finding method when the AB double earthfault, and dot-and-dash line is the average relative error curve of existing fault distance-finding method when ABC three-phase earth fault; The lower part solid line is the average relative error curve of fault distance-finding method of the present invention when A phase earth fault, dotted line is the average relative error curve of fault distance-finding method of the present invention when the AB double earthfault, and dot-and-dash line is the average relative error curve of fault distance-finding method of the present invention when ABC three-phase earth fault.

As can be seen from Figure 2, under all fault distances of 3 kinds of fault types, the average relative error of dynamic fault distance-finding method of the present invention all is starkly lower than existing static failure distance-finding method.

In a word, more than four evidences, under the same line system condition, method of the present invention is than existing method, it is subjected to fault distance, fault type, fault angle and The noise little, the distance accuracy height.

Claims (2)

1. the transmission line malfunction both-end distance measuring method under the dynamic condition the steps include:

A, pre-service

Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after Filtering Processing, obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides _S1, I _R1With positive sequence failure voltage signal U _S1, U _R1, wherein S represents left end, R represents right-hand member;

B, line parameter circuit value are estimated

Homogeneous line Model Calculation according to distribution parameter obtains following positive order parameter:

B1, unit line length positive sequence wave impedance

Unit line length positive sequence propagation coefficient γ ₁=l ^-1Ln[(U _S1+ Z _C1I _S1)/(U _R1-Z _C1I _R1)], wherein l is a transmission line length;

B2, unit line length impedance Z ₁=Z _C1γ ₁, unit line length admittance Y ₁=γ ₁/ Z _C1

C, Two-terminal Fault Location

Under the C1 dynamic condition, the forward-order current amplitude of transmission line of electricity left end

The positive sequence voltage amplitude

The forward-order current amplitude of transmission line of electricity right-hand member

The positive sequence voltage amplitude

T is the time, ω ₀Angular frequency for system:

The dynamic positive sequence impedance of definition left end is

${\tilde{Z}}_{S1}=Z_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\&PartialD;\left[i_{S1}\left(t\right)\right]}{\&PartialD;t}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{i_{S1}\left(t\right)}\frac{\&PartialD;\left[i_{S1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence admittance of definition left end is

${\tilde{Y}}_{S1}=Y_1+\frac{C_1}{U_{S1}\left(t\right)}\frac{\&PartialD;\left[u_{S1}\left(t\right)\right]}{\&PartialD;t}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{u_{S1}\left(t\right)}\frac{\&PartialD;\left[u_{S1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence impedance of definition right-hand member is

${\tilde{Z}}_{R1}=Z_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\&PartialD;\left[i_{R1}\left(t\right)\right]}{\&PartialD;t}=R_1+j{\mathrm{\&omega;}}_0{L}_1+\frac{L_1}{i_{R1}\left(t\right)}\frac{\&PartialD;\left[i_{R1}\left(t\right)\right]}{\&PartialD;t}$

The dynamic positive sequence admittance of definition right-hand member is

${\tilde{Y}}_{R1}=Y_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\&PartialD;\left[u_{R1}\left(t\right)\right]}{\&PartialD;t}=G_1+j{\mathrm{\&omega;}}_0{C}_1+\frac{C_1}{u_{R1}\left(t\right)}\frac{\&PartialD;\left[u_{R1}\left(t\right)\right]}{\&PartialD;t}$

Wherein, R ₁, L ₁, G ₁And C ₁The positive sequence resistance, inductance, the electricity that are respectively unit line length are led and electric capacity;

C2, the dynamic positive sequence wave impedance of circuit right-hand member Dynamic positive sequence propagation coefficient

Trouble spot positive sequence voltage U by the right-hand member derivation _RF1For:

$U_{\mathrm{RF}1}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x}$

Wherein x is the distance of trouble spot apart from the circuit right-hand member, i.e. fault distance;

The dynamic positive sequence wave impedance of circuit right-hand member

Dynamic positive sequence propagation coefficient Trouble spot positive sequence voltage U by the left end derivation _SF1For:

$U_{\mathrm{SF}1}=\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}$

C3, fault localization

Circuit is U under failure condition _RF1=U _SF1, the nonlinear equation of setting up about fault distance x is:

$\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}=\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}+\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x}$

Use the iteration function F (x) that Newton iteration method solves following formula:

$F\left(x\right)=x-{\widetilde{\mathrm{\&gamma;}}}_{S1}\frac{U_{S1}-I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}+{\widetilde{\mathrm{\&gamma;}}}_{S1}\frac{U_{S1}+I_{S1}{\tilde{Z}}_{\mathrm{SC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{S1}(l-x)}$

$-({\widetilde{\mathrm{\&gamma;}}}_{R1}\frac{U_{R1}-I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{{\widetilde{\mathrm{\&gamma;}}}_{R1}x}-{\widetilde{\mathrm{\&gamma;}}}_{R1}\frac{U_{R1}+I_{R1}{\tilde{Z}}_{\mathrm{RC}1}}{2}{e}^{-{\widetilde{\mathrm{\&gamma;}}}_{R1}x})$

Following formula is carried out interative computation can obtain fault distance x accurately.

2. the transmission line malfunction both-end distance measuring method under a kind of dynamic condition as claimed in claim 1 is characterized in that: in the described A step pre-service, carry out decoupling change with symmetrical component method and get transmission line of electricity positive sequence fault current signal I in return ₁With positive sequence failure voltage signal U ₁The decoupling zero transformation for mula be:

$\left[\begin{array}{c}{U}_1\\ U_2\\ U_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]$

$\left[\begin{array}{c}{I}_1\\ I_2\\ I_0\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& a& a^2\\ 1& a^2& a\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

In the formula, U _a, U _b, U _cAnd I _a, I _b, I _cBe phase voltage and the phase current that collects, U ₁, U ₂, U ₀And I ₁, I ₂, I ₀Be respectively sequence voltage and preface electric current, and

Images