CN103197204B - Mixed type method of multi-terminal circuit fault location - Google Patents

Abstract

Disclosed is a mixed type method of multi-terminal circuit fault location. The method comprises the following steps: step 1, symmetrical component voltage of a fault point is calculated according to symmetrical component voltage and current of a sending terminal; step 2, the symmetrical component voltage of the fault point is calculated according to symmetrical component voltage and current of a receiving terminal; step 3, delta is supposed to be a phase angle of the sending terminal lagging behind the receiving terminal, d is supposed to be the fault distance, and the phase angle displacement delta and the fault distance d are obtained through an iteration mode by using weighted least squares (WLS); step 4, sequence voltage and phasor voltage of a fault position can be measured and calculated from voltage and current at two terminals according to obtained fault points and phase angle difference; step 5, a threshold value Ith is set through the sum of the fault current and a pre-confirmed constant C, and the type of a fault is confirmed; and step 6, fault resistance is obtained by calculating the current and the voltage of the fault position.

Description

The mixed method of multi-point circuit fault location

Technical field

The present invention relates to the method for a kind of line fault location, particularly a kind of mixed method of multi-point circuit fault location.

Background technology

Utilities Electric Co. is necessary for users and provides reliable electrical power services, reduce operating cost, but, this target is threaten by the abnormal conditions constantly occurred, as electric network fault, along with the aging and ever-increasing electricity needs of electrical network, occur that the probability of serious electric network fault increases, in order to prevent the damage of power system device and reduce power off time, transmission line malfunction must be separated rapidly and excision, in fast recovery of power supply and prevention massive blackout, quick and accurate fault location is very important.

Fault location is studied to be counted for many years, and different FLTs is suggested, and in general, Fault Locating Method is divided into following two classes: row wave method and phase metering method.Row wave method carries out fault location by measuring fault wave head in the elapsed time of trouble point and the propagation of circuit two ends, because this method relies on high frequency fault wave head signal, so usually have special requirement to data collecting system, in addition, the signal that these methods can not rely on fault to generate completely, also needs extra equipment to generate impact signal and signal injection in power system.Based on the method for phasor, fault location is carried out by the impedance calculating trouble point and circuit two ends, the impedance calculated is directly proportional to the length of circuit one end to trouble point, phase metering method does not have special requirement to the data used, and the intelligent apparatus (IEDs) of present transformer station comprises digital relay and fault oscillograph (DFRs) can be good at meeting this point.Based in the method for phasor, there are single-ended algorithm and two side inputs, single-ended algorithm only uses a terminal voltage, electric current and do not need circuit two ends to carry out exchanges data, two side inputs uses voltage, the electric current at circuit two ends, so need circuit two end communication, in general, the accuracy of two side inputs is more much better than single-ended algorithm.

Great majority use simple circuit model based on the Fault Locating Method of phasor and only rely on the subset of an available voltage, current data.The precision that this FLT can reach inevitably is limited to the simplification that circuit model is unnecessary, ignores the performance that important information can reduce FLT further.

Summary of the invention

In order to solve the deficiencies in the prior art, the present invention specifically discloses a kind of mixed method of multi-point circuit fault location, this Fault Locating Method is a two side inputs, based on sampled data and phasor, and these data can very easily get from intelligent apparatus (IEDs), this new method is considered phasor with detailed long wire model together with, so be defined as mixed method.The advantage of mixed method is, cost is low and accuracy is high, and mixed method overcomes the drawback of the existing Fault Locating Method based on phasor, and the fault location result providing precision higher.

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

The mixed method of multi-point circuit fault location, comprises the following steps:

Step one, according to symmetrical components voltage, the electric current of transmitting terminal, calculates the symmetrical components voltage of trouble point;

Step 2, according to symmetrical components voltage, the electric current of receiving terminal, calculates the symmetrical components voltage of trouble point;

Step 3, because circuit two ends are asynchronous, so hypothesis δ is the phase angle that transmitting terminal lags behind receiving terminal, d fault distance, the symmetrical components voltage of the symmetrical components voltage of the trouble point utilizing transmitting terminal to calculate and the trouble point of receiving terminal obtains the not matching voltage of trouble point, utilizes weighted least-squares method WLS to obtain phase angle displacement δ and fault distance d by iterative manner the not matching voltage of trouble point and Jacobian matrix;

Step 4, according to getting fault distance, phase angle displacement, the sequence voltage of the position of fault and phasor voltage calculate from position of fault both end voltage and current measurement;

Step 5, sets a threshold value I by the summation of fault current and the functional relation of a predetermined constant C _th, determine fault type;

Step 6, calculates fault resstance by the electric current of abort situation and voltage.

Said method is a kind of method based on detailed circuit model, in circuit modeling, utilize distributed line parameter circuit value and do not make any hypothesis, being used properly in order to ensure all available information, method applies weighted least-squares method (WLS).Based on weighted least-squares method, fault location can be counted as a minimized problem, minimize because its target is the unmatched weighted sum of symmetrical components voltage making to calculate in trouble point, different weight coefficients is selected for different symmetrical components, so in solution fault-location problem, for parameter more accurate sequence network, select larger weight coefficient, thus it is larger that corresponding order components is play a part.

Synchronized sampling is not needed, because locking phase angular difference and fault location can solve simultaneously from different line scan pickup coil side data.After calculating position of failure point and synchronous phase angle, can synchronously be calculated at the symmetrical components voltage and current of trouble point, symmetrical components can be converted into phasor and carry out failure mode analysis, after confirming fault type, can calculate the fault impedance of trouble point according to phase voltage and phase current.

Beneficial effect of the present invention: the advantage of mixed method is, cost efficiency is high and accuracy is also higher, by using suitable system model and intelligently must utilizing existing information, mixed method overcomes the drawback of the existing Fault Locating Method based on phasor, and the fault location result providing precision higher.

Accompanying drawing explanation

Fig. 1 three phase line wiring diagram;

Fig. 2 fault resstance Rag calculates schematic diagram;

Fig. 3 fault resstance Rbg calculates schematic diagram;

Fig. 4 fault resstance Rcg calculates schematic diagram;

Fig. 5 fault resstance Rab calculates schematic diagram;

Fig. 6 fault resstance Rbc calculates schematic diagram;

Fig. 7 fault resstance Rac calculates schematic diagram;

Fig. 8 fault resstance Ra, Rb and Rabg calculate schematic diagram;

Fig. 9 fault resstance Rb, Rc and Rbcg calculate schematic diagram;

Figure 10 fault resstance Ra, Rc, Racg calculate schematic diagram;

Figure 11 fault resstance Ra, Rb, Rc, Rabc calculate schematic diagram.

Detailed description of the invention

With reference to the accompanying drawings adopted technical scheme is further elaborated below.

As shown in Figure 1, the length of circuit is represented by l, and the length of trouble point distance transmitting terminal s is d, and the length of distance receiving terminal r is l-d.

According to the symmetrical components voltage of transmitting terminal, electric current, characteristic impedance, the symmetrical components voltage of trouble point can calculate like this:

V _fs,1＝V _s,1cosh(dγ ₁)-I _s,1Z _c1sinh(dγ ₁)

V _fs,2＝V _s,2cosh(dγ ₂)-I _s,2Z _c2sinh(dγ ₂)

V _fs,0＝V _s,0cosh(dγ ₀)-I _s,0Z _c0sinh(dγ ₀)

Wherein, sinh () and cosh () is hyperbolic sine and hyperbolic cosine function;

V _{s, 1}, V _{s, 2}and V _{s, 0}represent positive sequence, negative phase-sequence and residual voltage that transmitting terminal is measured;

I _{s, 1}, I _{s, 2}and I _{s, 0}represent positive sequence, negative phase-sequence and zero-sequence current that transmitting terminal is measured;

V _{fs, 1}, V _{fs, 2}and V _{fs, 0}represent the positive sequence of the trouble point calculated according to transmitting terminal measured value, negative phase-sequence and residual voltage;

γ ₁, γ ₂and γ ₀represent positive sequence, negative phase-sequence and zero sequence propagation constant respectively;

Z _c1, Z _c2and Z _c0represent the positive sequence of circuit, negative phase-sequence and zero sequence characteristic impedance respectively.

According to the symmetrical components voltage of receiving terminal, electric current, characteristic impedance, can calculate like this at the symmetrical components voltage of trouble point:

V _fr,1＝V _r,1cosh((l-d)γ ₁)-I _r,1Z _c1sinh((l-d)γ ₁)

V _fr,2＝V _r,2cosh((l-d)γ ₂)-I _r,2Z _c2sinh((l-d)γ ₂)

V _fr,0＝V _r,0cosh((l-d)γ ₀)-I _r,0Z _c0sinh((l-d)γ ₀)

Wherein, V _{r, 1}, V _{r, 2}and V _{r, 0}represent positive sequence, negative phase-sequence and residual voltage that receiving terminal is measured;

U _{r, 1}, I _{r, 2}and I _{r, 0}represent positive sequence, negative phase-sequence and zero-sequence current that receiving terminal is measured;

V _{fr, 1}, V _{fr, 2}and V _{fr, 0}represent the positive sequence of the trouble point calculated according to receiving terminal measured value, negative phase-sequence and residual voltage.

If the voltage and current at circuit two ends is synchronous, so should be just the same from the symmetrical components voltage x current of the trouble point that the measurement data at circuit two ends calculates, this is true to consider the impossible Complete Synchronization in circuit two ends, we suppose that the phase angle displacement of non-synchronous sampling is δ, we suppose that δ is the phase angle that transmitting terminal lags behind receiving terminal further, then weighted least-squares method (WLS) is used, fault-location problem can be summed up as solution δ and d, so method can minimize below:

w ₁|F ₁(δ,d)| ²+w ₂|F ₂(δ,d)| ²+w ₀|F ₀(δ,d)| ²

Wherein, F ₁(δ, d), F ₂(δ, d) and F ₀what (δ, d) represented that trouble point calculates does not mate positive sequence, negative phase-sequence and residual voltage;

W ₁, w ₂and w ₀represent the weight coefficient not mating positive sequence, negative phase-sequence and residual voltage respectively;

F ₁(δ, d), F ₂(δ, d) and F ₀(δ, d) is calculated as follows:

F ₁(δ,d)＝V _fs,1e ^jδ-V _fr,1

F ₂(δ,d)＝V _fs,2e ^jδ-V _fr,2

F ₀(δ,d)＝V _fs,0e ^jδ-V _fr,0

Wherein, V _{fr, 1}, V _{fr, 2}and V _{fr, 0}represent the positive sequence of the trouble point calculated according to receiving terminal measured value, negative phase-sequence and residual voltage;

V _{fs, 1}, V _{fs, 2}and V _{fs, 0}represent the positive sequence of the trouble point calculated according to transmitting terminal measured value, negative phase-sequence and residual voltage;

E is the truth of a matter of natural logrithm, the square root of j representative-1.

Based on weighted least-squares method (WLS), phase angle displacement δ and fault distance d can be solved by such iteration:

$\left[\begin{array}{c}{\mathrm{\δ}}_{k+1}\\ d_{k+1}\end{array}\right]=\left[\begin{array}{c}{\mathrm{\δ}}_k\\ d_k\end{array}\right]-{\left[J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WJ}({\mathrm{\δ}}_k,d_k)\right]}^{-1}J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WF}({\mathrm{\δ}}_k,d_k)$

Wherein, k represents iterations, and F δ, d represent not matching voltage matrix, does not mate positive sequence, negative phase-sequence and residual voltage F1 δ respectively by true and hypothesis, and d, F2 δ, d and F0 δ, s form:

$F(\mathrm{\δ},d)=\left[\begin{array}{c}{F}_{1,\mathrm{real}}(\mathrm{\δ},d)\\ F_{1,\mathrm{imag}}(\mathrm{\δ},d)\\ F_{2,\mathrm{real}}(\mathrm{\δ},d)\\ F_{2,\mathrm{imag}}(\mathrm{\δ},d)\\ F_{0,\mathrm{real}}(\mathrm{\δ},d)\\ F_{0,\mathrm{imag}}(\mathrm{\δ},d)\end{array}\right]$

J (δ, d) represents Jacobian matrix, is made up of the partial derivative not mating symmetrical components voltage that is true and hypothesis about phase angle displacement δ and fault distance d:

$J(\mathrm{\δ},d)=\left[\begin{array}{cc}{\left(\frac{{\∂F}_1}{\∂\mathrm{\δ}}\right)}_{\mathrm{real}}& {\left(\frac{{\∂F}_1}{\∂d}\right)}_{\mathrm{real}}\\ {\left(\frac{{\∂F}_1}{\∂\mathrm{\δ}}\right)}_{\mathrm{imag}}& {\left(\frac{{\∂F}_1}{\∂d}\right)}_{\mathrm{imag}}\\ {\left(\frac{{\∂F}_2}{\∂\mathrm{\δ}}\right)}_{\mathrm{real}}& {\left(\frac{{\∂F}_2}{\∂d}\right)}_{\mathrm{real}}\\ {\left(\frac{{\∂F}_2}{\∂\mathrm{\δ}}\right)}_{\mathrm{imag}}& {\left(\frac{{\∂F}_2}{\∂d}\right)}_{\mathrm{imag}}\\ {\left(\frac{{\∂F}_0}{\∂\mathrm{\δ}}\right)}_{\mathrm{real}}& {\left(\frac{{\∂F}_0}{\∂d}\right)}_{\mathrm{real}}\\ {\left(\frac{{\∂F}_0}{\∂\mathrm{\δ}}\right)}_{\mathrm{imag}}& {\left(\frac{{\∂F}_0}{\∂d}\right)}_{\mathrm{imag}}\end{array}\right]$

W is diagonal matrix, represents the weight coefficient of not matching voltage:

W＝diagonal[w ₁w ₁w ₂w ₂w ₀w ₀]

In order to find abort situation d and phase angle difference δ, equation

$\left[\begin{array}{c}{\mathrm{\δ}}_{k+1}\\ d_{k+1}\end{array}\right]=\left[\begin{array}{c}{\mathrm{\δ}}_k\\ d_k\end{array}\right]-{\left[J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WJ}({\mathrm{\δ}}_k,d_k)\right]}^{-1}J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WF}({\mathrm{\δ}}_k,d_k)$ Iterative, until some d _k+1with d _kbetween difference be less than predetermined value, as 0.1 mile or kilometer or iteration sum reach its predetermined maximum restriction, as 8.

Once get abort situation and phase angle difference, the sequence voltage of abort situation and phasor voltage can calculate from both end voltage and current measurement, and we utilize the measured value of transmitting terminal here, therefore,

V _f,1＝V _s,1e ^jδcosh(dγ ₁)-I _s,1e ^jδZ _c1sinh(dγ ₁)

V _f,2＝V _s,2e ^jδcosh(dγ ₂)-I _s,2e ^jδZ _c2sinh(dγ ₂)

V _f,0＝V _s,0e ^jδcpsh(dγ ₀)-I _s,0e ^jδZ _c0sinh(dγ ₀)

V _fa＝V _f,1+V _f,2+V _f,0

V _fb＝a ²V _f,1+aV _f,2+V _f,0

V _fc＝aV _f,1+a ²V _f,2+V _f,0

Wherein, a represents 120 degree of phase shifts, V _{f, 1}, V _{f, 2}and V _{f, 0}represent the positive sequence, negative phase-sequence and the residual voltage that calculate in fault location; V _fa, V _fband V _fcrepresentative calculates the false voltage of phasor A, phasor B and phasor C in trouble point.

In order to obtain fault type result, introduce a threshold value I _th.This threshold value is defined by the summation of fault current and a predetermined constant C:

$I_{\mathrm{th}}=\frac{\left|I_{\mathrm{fa}}\right|+\left|I_{\mathrm{fb}}\right|+\left|I_{\mathrm{fc}}\right|}{C}$

Wherein, I _faa phase fault electric current, I _fbit is B phase fault electric current I _fcc phase fault electric current, I _f0it is fault current at zero point.

Through test, 15, as the value of constant C, achieve satisfied result, as long as this just means the electric current not matching degree fault current that is less than 6.6% perfecting phase, so fault type just can be properly determined.Fault type recognition standard: if any one phase current or earth current are greater than threshold value I _th, so just think this phase or ground relevant to this fault; Otherwise fault type is defined as unclear.If namely | I _fa| >I _th, be defined as the fault that A phase participates in; If | I _fb| >I _th, be defined as the fault that B phase participates in; If | I _fc| >I _th, be defined as the fault that C phase participates in; If | I _f0| >I _th, be defined as the fault that Ground phase participates in; If below all incorrect, then fault type cannot judge, is designated as not clear.

In order to more fully understand to take preventive measures a kind of character of specific fault type, know that fault resstance and fault type are very useful, fault resstance can be calculated by the electric current of abort situation and voltage.

As shown in Figure 2, fault resstance Rag can by obtaining in the ratio calculation of the voltage of abort situation phasor A and the fault current of phasor A.

$R_{\mathrm{ag}}=\left|\frac{V_{\mathrm{fa}}}{I_{\mathrm{fa}}}\right|$

As shown in Figure 3, fault resstance Rbg can by obtaining in the ratio calculation of the voltage of abort situation phasor B and the fault current of phasor B.

$R_{\mathrm{bg}}=\left|\frac{V_{\mathrm{fb}}}{I_{\mathrm{fb}}}\right|$

As shown in Figure 4, fault resstance Rcg can by obtaining in the ratio calculation of the voltage of trouble point phasor C and the fault current of phasor C.

$R_{\mathrm{cg}}=\left|\frac{V_{\mathrm{fc}}}{I_{\mathrm{fc}}}\right|$

As shown in Figure 5, fault resstance Rab is calculated as follows:

$R_{\mathrm{ab}}=\left|\frac{V_{\mathrm{fa}}}{I_{\mathrm{fa}}}\right|$

As shown in Figure 6, fault resstance Rbc is calculated as follows:

$R_{\mathrm{bc}}=\left|\frac{V_{\mathrm{fbc}}}{I_{\mathrm{fb}}}\right|$

As shown in Figure 7, fault resstance Rac is calculated as follows:

$R_{\mathrm{ac}}=\left|\frac{V_{\mathrm{fac}}}{I_{\mathrm{fa}}}\right|$

As shown in Figure 8, fault resstance R _a, R _band R _abgsolve as follows:

I _fg＝I _fa+I _fb

Wherein I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fa}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fb}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_b\\ R_{\mathrm{abg}}\end{array}\right]=\begin{array}{}\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fb}}\end{array}\right]\end{array}$

So

$\left[\begin{array}{c}{R}_a\\ R_b\\ R_{\mathrm{abg}}\end{array}\right]={\left({M1}^{T}M1\right)}^{-1}{M1}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\end{array}\right]$

M1 is that 4 × 3 matrixes are made up of the part of fault current that is real and that suppose:

$M1=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

As shown in Figure 9, fault resstance Rb, Rc, Rbcg and solve as follows:

I _fg＝I _fb+I _fc

Wherein I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fb}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_b\\ R_c\\ R_{\mathrm{bcg}}\end{array}\right]=\begin{array}{c}\left[\begin{array}{c}{V}_{\mathrm{fb}}\\ V_{\mathrm{fc}}\end{array}\right]\end{array}$

So

$\left[\begin{array}{c}{R}_b\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]={\left({M2}^{T}M2\right)}^{-1}{M2}^T\left[\begin{array}{c}{V}_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{real}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M2 is that 4 × 3 matrixes are made up of the part of fault current that is real and that suppose:

$M2=\left[\begin{array}{ccc}{I}_{\mathrm{fb},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fb},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

As shown in Figure 10, fault resstance Ra, Rc, Racg is calculated as follows:

I _fg＝I _fa+I _fc

I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fa}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]=\begin{array}{c}\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fc}}\end{array}\right]\end{array}$

So

$\left[\begin{array}{c}{R}_a\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]={\left({M3}^{T}M3\right)}^{-1}{M3}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{real}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M3 is that 4 × 3 matrixes are made up of the part of fault current that is real and that suppose:

$M3=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

As shown in figure 11, fault resstance Ra, Rb, Rc, Rabc is calculated as follows:

I _fg＝I _fa+I _fb+I _fc

I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{cccc}{I}_{\mathrm{fa}}& 0& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fb}}& 0& I_{\mathrm{fg}}\\ 0& 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_b\\ R_c\\ R_{\mathrm{abc}}\end{array}\right]=\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fb}}\\ V_{\mathrm{fc}}\end{array}\right]$

So

$\left[\begin{array}{c}{R}_a\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]={\left({M4}^{T}M4\right)}^{-1}{M4}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\\ V_{\mathrm{fb},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M4 is that 4 × 3 matrixes are made up of the part of fault current that is real and that suppose:

$M4=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{real}}& I_{\mathrm{fg},\mathrm{real}}\\ 0& I_{\mathrm{fb},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& 0& I_{\mathrm{fg},\mathrm{real}}\\ 0& 0& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right].$

Claims (6)

1. a mixed method for multi-point circuit fault location, is characterized in that, comprises the following steps:

Step one, according to symmetrical components voltage, the electric current of transmitting terminal, calculates the symmetrical components voltage of trouble point;

Step 2, according to symmetrical components voltage, the electric current of receiving terminal, calculates the symmetrical components voltage of trouble point;

Step 3, because circuit two ends are asynchronous, so hypothesis δ is the phase angle that transmitting terminal lags behind receiving terminal, d fault distance, the symmetrical components voltage of the symmetrical components voltage of the trouble point utilizing transmitting terminal to calculate and the trouble point of receiving terminal obtains the not matching voltage of trouble point, utilizes weighted least-squares method WLS to obtain phase angle displacement δ and fault distance d by iterative manner the not matching voltage of trouble point and Jacobian matrix;

Step 4, according to getting fault distance, phase angle displacement, the sequence voltage of the position of fault and phasor voltage calculate from position of fault both end voltage and current measurement;

Step 5, sets a threshold value I by the summation of fault current and the functional relation of a predetermined constant C _th, determine fault type;

Step 6, calculates fault resstance by the electric current of abort situation and voltage;

Described step 3 to obtain the process of phase angle displacement δ and fault distance d by iterative manner as follows:

$\left[\begin{array}{c}{\mathrm{\δ}}_{k+1}\\ d_{k+1}\end{array}\right]=\left[\begin{array}{c}{\mathrm{\δ}}_k\\ d_k\end{array}\right]-{\left[J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WJ}({\mathrm{\δ}}_k,d_k)\right]}^{-1}J{({\mathrm{\δ}}_k,d_k)}^T\mathrm{WF}({\mathrm{\δ}}_k,d_k)$ Iterative

F ₁(δ, d), F ₂(δ, d) and F ₀(δ, d) is calculated as follows:

F ₁(δ,d)＝V _fs,1e ^jδ-V _fr,1

F ₂(δ,d)＝V _fs,2e ^jδ-V _fr,2

F ₀(δ,d)＝V _fs,0e ^jδ-V _fr,0

F (δ, d) represents not matching voltage matrix, does not mate positive sequence, negative phase-sequence and residual voltage F respectively by the real part of corresponding vector parameters and imaginary part ₁(δ, d), F ₂(δ, d) and F ₀(δ, d) forms;

$F(\mathrm{\δ},d)=\left[\begin{array}{c}{F}_{1,\mathrm{real}}(\mathrm{\δ},d)\\ F_{1,\mathrm{imag}}(\mathrm{\δ},d)\\ F_{2,\mathrm{real}}(\mathrm{\δ},d)\\ F_{2,\mathrm{imag}}(\mathrm{\δ},d)\\ F_{0,\mathrm{real}}(\mathrm{\δ},d)\\ F_{0,\mathrm{imag}}(\mathrm{\δ},d)\end{array}\right]$

J (δ, d) represents Jacobian matrix, is made up of the real part of the corresponding vector parameters about phase angle displacement δ and fault distance d and the partial derivative not mating symmetrical components voltage of imaginary part:

$J(\mathrm{\δ},d)=\left[\begin{array}{cc}\left(\frac{{\∂F}_1}{\∂\mathrm{\δ}}\right)\mathrm{real}& \left(\frac{{\∂F}_1}{\∂d}\right)\mathrm{real}\\ \left(\frac{{\∂F}_1}{\∂\mathrm{\δ}}\right)\mathrm{imag}& \left(\frac{{\∂F}_1}{\∂d}\right)\mathrm{imag}\\ \left(\frac{{\∂F}_2}{\∂\mathrm{\δ}}\right)\mathrm{real}& \left(\frac{{\∂F}_2}{\∂d}\right)\mathrm{real}\\ \left(\frac{{\∂F}_2}{\∂\mathrm{\δ}}\right)\mathrm{imag}& \left(\frac{{\∂F}_2}{\∂d}\right)\mathrm{imag}\\ \left(\frac{{\∂F}_0}{\∂\mathrm{\δ}}\right)\mathrm{real}& \left(\frac{{\∂F}_0}{\∂d}\right)\mathrm{real}\\ \left(\frac{{\∂F}_0}{\∂\mathrm{\δ}}\right)\mathrm{imag}& \left(\frac{{\∂F}_0}{\∂d}\right)\mathrm{imag}\end{array}\right]$

W is diagonal matrix, represents the weight coefficient of not matching voltage:

W＝diagonal[w ₁w ₁w ₂w ₂w ₀w ₀]

K represents iterations, and e is the truth of a matter of natural logrithm, the square root of j representative-1, w ₁, w ₂and w ₀represent the weight coefficient not mating positive sequence, negative phase-sequence and residual voltage respectively;

V _{fs, 1}, V _{fs, 2}and V _{fs, 0}represent the positive sequence of the trouble point calculated according to transmitting terminal measured value, negative phase-sequence and residual voltage;

V _{fr, 1}, V _{fr, 2}and V _{fr, 0}represent the positive sequence of the trouble point calculated according to receiving terminal measured value, negative phase-sequence and residual voltage.

2. the mixed method of a kind of multi-point circuit fault location as claimed in claim 1, is characterized in that, the symmetrical components voltage computing formula of the trouble point of described step one:

V _fs,1＝V _s,1cosh(dγ ₁)-I _s,1Z _c1sinh(dγ ₁)

V _fs,2＝V _s,2cosh(dγ ₂)-I _s,2Z _c2sinh(dγ ₂)

V _fs,0＝V _s,0cosh(dγ ₀)-I _s,0Z _c0sinh(dγ ₀)

Wherein, sinh () and cosh () is hyperbolic sine and hyperbolic cosine function;

V _{s, 1}, V _{s, 2}and V _{s, 0}represent positive sequence, negative phase-sequence and residual voltage that transmitting terminal is measured;

I _{s, 1}, I _{s, 2}and I _{s, 0}represent positive sequence, negative phase-sequence and zero-sequence current that transmitting terminal is measured;

γ ₁, γ ₂and γ ₀represent positive sequence, negative phase-sequence and zero sequence propagation constant respectively;

The length of trouble point distance transmitting terminal s is d;

Z _c1, Z _c2and Z _c0represent the positive sequence of circuit, negative phase-sequence and zero sequence characteristic impedance respectively.

3. the mixed method of a kind of multi-point circuit fault location as claimed in claim 1, is characterized in that,

The symmetrical components voltage computing formula of the trouble point of described step 2:

V _fr,1＝V _r,1cosh((l-d)γ ₁)-I _r,1Z _c1sinh((l-d)γ ₁)

V _fr,2＝V _r,2cosh((l-d)γ ₂)-I _r,2Z _c2sinh((l-d)γ ₂)

V _fr,0＝V _r,0cosh((l-d)γ ₀)-I _r,0Z _c0sinh((l-d)γ ₀)

Wherein, V _{r, 1}, V _{r, 2}and V _{r, 0}represent positive sequence, negative phase-sequence and residual voltage that receiving terminal is measured;

I _{r, 1}, I _{r, 2}and I _{r, 0}represent positive sequence, negative phase-sequence and zero-sequence current that receiving terminal is measured;

L-d is the length of trouble point distance receiving terminal r;

Z _c1, Z _c2and Z _c0represent the positive sequence of circuit, negative phase-sequence and zero sequence characteristic impedance respectively;

γ ₁, γ ₂and γ ₀represent positive sequence, negative phase-sequence and zero sequence propagation constant respectively.

4. the mixed method of a kind of multi-point circuit fault location as claimed in claim 1, is characterized in that, sequence voltage and the phasor voltage computing formula of the position of fault of described step 4 are as follows:

V _f,1＝V _s,1e ^jδcosh(dγ ₁)-I _s,1e ^jδZ _c1sinh(dγ ₁)

V _f,2＝V _s,2e ^jδcosh(dγ ₂)-I _s,2e ^jδZ _c2sinh(dγ ₂)

V _f,0＝V _s,0e ^jδcosh(dγ ₀)-I _s,0e ^jδZ _c0sinh(dγ ₀)

V _fa＝V _f,1+V _f,2+V _f,0

V _fb＝a ²V _f,1+aV _f,2+V _f,0

V _fc＝aV _f,1+a ²V _f,2+V _f,0

Wherein, a represents 120 degree of phase shifts, V _{f, 1}, V _{f, 2}and V _{f, 0}represent the positive sequence, negative phase-sequence and the residual voltage that calculate in fault location; V _fa, V _fband V _fcrepresentative calculates the false voltage of phasor A, phasor B and phasor C in trouble point;

V _{s, 1}, V _{s, 2}and V _{s, 0}represent positive sequence, negative phase-sequence and residual voltage that transmitting terminal is measured;

I _{s, 1}, I _{s, 2}and I _{s, 0}represent positive sequence, negative phase-sequence and zero-sequence current that transmitting terminal is measured; Z _c1, Z _c2and Z _c0represent the positive sequence of circuit, negative phase-sequence and zero sequence characteristic impedance respectively; γ ₁, γ ₂and γ ₀represent positive sequence, negative phase-sequence and zero sequence propagation constant respectively.

5. the mixed method of a kind of multi-point circuit fault location as claimed in claim 1, is characterized in that, the threshold value I of described step 5 _thand fault type is defined as follows:

$I_{\mathrm{th}}=\frac{\left|I_{\mathrm{fa}}\right|+\left|I_{\mathrm{fb}}\right|+\left|I_{\mathrm{fc}}\right|}{C}$

Wherein, I _faa phase fault electric current, I _fbb phase fault electric current, I _fcbe C phase fault electric current, C is constant 15;

According to | I _fa|, | I _fb|, | I _fc|, | I _f0| with I _thmagnitude relationship determination fault type, I _f0it is fault current at zero point;

If | I _fa| >I _th, be defined as the fault that A phase participates in; If | I _fb| >I _th, be defined as the fault that B phase participates in; If | I _fc| >I _th, be defined as the fault that C phase participates in; If | I _f0| >I _th, be defined as the fault that Ground phase participates in; If below all incorrect, then fault type cannot judge, is designated as not clear.

6. the mixed method of a kind of multi-point circuit fault location as claimed in claim 1, is characterized in that, institute

State the R of the fault resstance of step 6 _a, R _band R _abgsolve as follows:

I _fg＝I _fa+I _fb

Wherein, I _fgrepresent earth current,

Relation in fault location between electric current and voltage is described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fa}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fb}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_b\\ R_{\mathrm{abg}}\end{array}\right]=\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fb}}\end{array}\right]$

So

$\left[\begin{array}{c}{R}_a\\ R_b\\ R_{\mathrm{abg}}\end{array}\right]={\left({M1}^{T}M1\right)}^{-1}{M1}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\end{array}\right]$

M1 is that 4 × 3 matrixes are made up of the part of the real part of corresponding vector parameters and the fault current of imaginary part:

$M1=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

The account form of three resistance Rb, Rc, Rbcg and three resistance Ra, Rc, Racg and four resistance Ra, Rb, Rc, Rabc and R _a, R _band R _abgcalculating consistent;

Fault resstance Rb, Rc, Rbcg and solve as follows:

I _fg＝I _fb+I _fc

Wherein I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fb}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_b\\ R_c\\ R_{\mathrm{bcg}}\end{array}\right]=\left[\begin{array}{c}{V}_{\mathrm{fb}}\\ V_{\mathrm{fc}}\end{array}\right]$

So

$\left[\begin{array}{c}{R}_b\\ R_c\\ R_{\mathrm{bcg}}\end{array}\right]={\left({M2}^{T}M2\right)}^{-1}{M2}^T\left[\begin{array}{c}{V}_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{real}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M2 is that 4 × 3 matrixes are made up of the part of the real part of corresponding vector parameters and the fault current of imaginary part:

$M2=\left[\begin{array}{ccc}{I}_{\mathrm{fb},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fb},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

Fault resstance Ra, Rc, Racg are calculated as follows:

I _fg＝I _fa+I _fc

I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{ccc}{I}_{\mathrm{fa}}& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]=\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fc}}\end{array}\right]$

So

$\left[\begin{array}{c}{R}_a\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]={\left({M3}^{T}M3\right)}^{-1}{M3}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{real}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M3 is that 4 × 3 matrixes are made up of the part of the real part of corresponding vector parameters and the fault current of imaginary part:

$M3=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{real}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fc},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right]$

Fault resstance Ra, Rb, Rc, Rabc are calculated as follows:

I _fg＝I _fa+I _fb+I _fc

I _fgrepresent earth current,

Relation in fault location between electric current and voltage can be described to:

$\left[\begin{array}{cccc}{I}_{\mathrm{fa}}& 0& 0& I_{\mathrm{fg}}\\ 0& I_{\mathrm{fb}}& 0& I_{\mathrm{fg}}\\ 0& 0& I_{\mathrm{fc}}& I_{\mathrm{fg}}\end{array}\right]\left[\begin{array}{c}{R}_a\\ R_b\\ R_c\\ R_{\mathrm{abc}}\end{array}\right]=\left[\begin{array}{c}{V}_{\mathrm{fa}}\\ V_{\mathrm{fb}}\\ V_{\mathrm{fc}}\end{array}\right]$

So

$\left[\begin{array}{c}{R}_a\\ R_b\\ R_c\\ R_{\mathrm{acg}}\end{array}\right]={\left({M4}^{T}M4\right)}^{-1}{M4}^T\left[\begin{array}{c}{V}_{\mathrm{fa},\mathrm{real}}\\ V_{\mathrm{fa},\mathrm{imag}}\\ V_{\mathrm{fb},\mathrm{real}}\\ V_{\mathrm{fb},\mathrm{imag}}\\ V_{\mathrm{fc},\mathrm{real}}\\ V_{\mathrm{fc},\mathrm{imag}}\end{array}\right]$

M4 is that 4 × 3 matrixes are made up of the part of the real part of corresponding vector parameters and the fault current of imaginary part:

$M4=\left[\begin{array}{ccc}{I}_{\mathrm{fa},\mathrm{real}}& 0& I_{\mathrm{fg},\mathrm{real}}\\ I_{\mathrm{fa},\mathrm{imag}}& 0& I_{\mathrm{fg},\mathrm{imag}}\\ 0& I_{\mathrm{fb},\mathrm{real}}& I_{\mathrm{fg},\mathrm{real}}\\ 0& I_{\mathrm{fb},\mathrm{imag}}& I_{\mathrm{fg},\mathrm{imag}}\\ 0& 0& I_{\mathrm{fg},\mathrm{real}}\\ 0& 0& I_{\mathrm{fg},\mathrm{imag}}\end{array}\right];$

I _faa phase fault electric current, I _fbb phase fault electric current, I _fcc phase fault electric current, V _fa, V _fband V _fcrepresentative calculates the false voltage of phasor A, phasor B and phasor C in trouble point.