CN107064728A - The single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line - Google Patents

The single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line Download PDF

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CN107064728A
CN107064728A CN201610851981.XA CN201610851981A CN107064728A CN 107064728 A CN107064728 A CN 107064728A CN 201610851981 A CN201610851981 A CN 201610851981A CN 107064728 A CN107064728 A CN 107064728A
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mrow
msub
mfrac
omega
delta
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CN107064728B (en
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姚旭
康小宁
程蓉
智勇
崔力心
倪赛赛
陈新
赵杰
拜润卿
赵毅
王春光
梅姚
张坤贤
张大伟
郭文科
王永年
刘巍
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State Grid Corp of China SGCC
Xian Jiaotong University
State Grid Gansu Electric Power Co Ltd
Electric Power Research Institute of State Grid Gansu Electric Power Co Ltd
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State Grid Corp of China SGCC
Xian Jiaotong University
State Grid Gansu Electric Power Co Ltd
Electric Power Research Institute of State Grid Gansu Electric Power Co Ltd
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    • G—PHYSICS
    • G01—MEASURING; TESTING
    • G01R—MEASURING ELECTRIC VARIABLES; MEASURING MAGNETIC VARIABLES
    • G01R31/00—Arrangements for testing electric properties; Arrangements for locating electric faults; Arrangements for electrical testing characterised by what is being tested not provided for elsewhere
    • G01R31/08—Locating faults in cables, transmission lines, or networks
    • G01R31/081—Locating faults in cables, transmission lines, or networks according to type of conductors
    • G01R31/085—Locating faults in cables, transmission lines, or networks according to type of conductors in power transmission or distribution lines, e.g. overhead
    • G—PHYSICS
    • G01—MEASURING; TESTING
    • G01R—MEASURING ELECTRIC VARIABLES; MEASURING MAGNETIC VARIABLES
    • G01R31/00—Arrangements for testing electric properties; Arrangements for locating electric faults; Arrangements for electrical testing characterised by what is being tested not provided for elsewhere
    • G01R31/08—Locating faults in cables, transmission lines, or networks
    • G01R31/088—Aspects of digital computing
    • Y—GENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
    • Y04—INFORMATION OR COMMUNICATION TECHNOLOGIES HAVING AN IMPACT ON OTHER TECHNOLOGY AREAS
    • Y04S—SYSTEMS INTEGRATING TECHNOLOGIES RELATED TO POWER NETWORK OPERATION, COMMUNICATION OR INFORMATION TECHNOLOGIES FOR IMPROVING THE ELECTRICAL POWER GENERATION, TRANSMISSION, DISTRIBUTION, MANAGEMENT OR USAGE, i.e. SMART GRIDS
    • Y04S10/00—Systems supporting electrical power generation, transmission or distribution
    • Y04S10/50—Systems or methods supporting the power network operation or management, involving a certain degree of interaction with the load-side end user applications
    • Y04S10/52—Outage or fault management, e.g. fault detection or location

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  • Engineering & Computer Science (AREA)
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Abstract

The single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line, this method is from fault component network equation, if peer-to-peer system impedance, transition resistance and fault distance are undetermined parameter, the network equation of frequency domain under lumped parameter model is given, solution obtains systematic parameter and location information；The inventive method eliminates the Systematic Errors of one-terminal data ranging presence from principle, improves range accuracy.

Description

The single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line
Technical field
The present invention relates to power system transmission line technical field of relay protection, and in particular to ultra-high-tension power transmission line is single-ended complete Cease frequency domain Fault Locating Method.
Background technology
UHV transmission network delivery capacity is huge, and line fault reparation plays extremely great to the operational reliability of power network Effect.The quick reparation of line fault is decided by accurate fault location, and China's land resource is poor, and water power is sent outlet outside and walked Corridor is with a varied topography, while line walking equipment is simple and crude, more causes the degree of accuracy ever more important of fault location.
Summarize all kinds of algorithms of existing single-end electrical quantity, its common issue be due to peer-to-peer system impedance and transition resistance not Range equation caused by knowing is unsatisfactory for definite condition, it has to by assuming that supplementary condition realizes fault location, and various assumes The Protean running status of power network can not be applied to, so as to cause the generation of Systematic Errors.Here it is existing single-ended event Hinder distance-finding method distance measurement result not accurate enough, it is impossible to meet the basic reason of engineer applied requirement.
In recent years, intelligent substation construction is in full swing in China, and electronic mutual inductor is used widely.It has height The characteristics of acquisition rate, high linearity and high bandwidth, can real electric signal of the progress of disease.Due to electronic mutual inductor Excellent Transfer characteristic so that relay protection device is possible to apply real transient fault signal, and extracting more horn of plenty can The fault message leaned on.If transient information can provide more location conditions in itself, the deficiency of location condition can be supplemented, this Sample just need not be made any it is assumed that and directly giving accurate positioning result by single-end information to system.
To sum up analyze, under the conditions of intelligent substation and electronic mutual inductor, application and trouble transient information, research has more High-performance, the more accurate novel power transmission line fault range measurement principle of ranging are significant.
The technical problem to be solved in the invention has：
1st, how fault transient information is utilized, more location conditions is supplemented, to avoid the systematicness of one-end fault ranging Error, reaches accurate ranging in principle.
2nd, the method for solving of the frequency domain range equation of thought is recognized based on parameter.
3rd, the computing technique of signal spectrum.
The content of the invention
In order to solve the above-mentioned technical problem, it is an object of the invention to provide a kind of single-ended holographic frequency domain of ultra-high-tension power transmission line Fault Locating Method, supplements more location conditions, to avoid the Systematic Errors of one-end fault ranging, standard is reached in principle True ranging.
In order to achieve the above object, the present invention is adopted the following technical scheme that：
The single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line, comprises the following steps：
A, B, C three-phase voltage u of step 1, collection protection installation place current transformer and voltage transformerA(k),uB(k), uC(k) with three-phase current iA(k),iB(k),iC(k)；
Step 2, A, B, C three-phase voltage and electric current to collecting, the event of three-phase voltage and electric current is calculated using following formula Hinder component Δ uA(k),ΔuB(k),ΔuC(k), Δ iA(k),ΔiB(k),ΔiC(k) with the sampled value 3i of zero-sequence component0(k)：
3i0(k)=Δ iA(k)+ΔiB(k)+ΔiC(k)
In above formula, k is sampling instant, and N is per cycle sampling number；A, B, C are calculated using all-round Fourier algorithm simultaneously The real and imaginary parts of three-phase voltage and electric current cosine phasor；
Step 3, definition constant：
z1=r1+jωl1
z0=r0+jωl0
In formula, ω0- power frequency angular frequency；- m is divider ratio, can be equal to 2,3 or 4 etc.；r1,l1,r0,l0- line Road unit length positive sequence, zero sequence resistance and inductance；Kr、Kl- zero-utility theory；
Step 4, calculate each mold component of this end system it is equivalent in inductance and interior resistance parameter RMi,LMi(i=0,1,2), takes two Not k three-phase current and three-phase voltage sampled value in the same time, the modulus of Current Voltage 0,1 modulus, 2 are obtained using Clarke transform above Modulus instantaneous value is brought into system of linear equations In, using linear least-squares Algorithm for Solving, this equation group obtains RMi,LMi(i=0,1,2)；
In formula, Δ iMiEach modulus fault component current instantaneous value of-M sides；ΔuMiMeasurement point M each modulus during-failure Fault component voltage；DT is sampling interval duration；
Step 5, determine fault type
(1) if singlephase earth fault, order,
K2=-j ω UMAHC-ω0UMAHS
K3=j ω (IMAHCr1+IMAHSω0l1)+ω0(IMAHSr1-IMAHCω0l1)
In formula, UMAHC, UMAHSThe phase of the real part of-measurement point M failure phase normal duty voltage cosine phasors with imaginary part in itself Anti- number；IMAHC, IMAHSThe opposite number of the real part of-measurement point M failure phase normal duty electric current cosine phasors with imaginary part in itself；
Calculate measurement point faulted phase current fault component Δ IMA(ω), failure phase busbar voltage fault component Δ UMA(ω) And zero mould electric current frequency ω spectrum component IM0(ω)：
In formula, Δ IMAC、ΔIMASThe opposite number of the real part of-measurement point M failure phase fault electric current phasors with imaginary part in itself； ΔiMA(k) sampled value of-measurement point M failure phase fault current components；ΔUMAC、ΔUMAS- measurement point M failure phase busbar voltages The opposite number of the real part of failure phasor with imaginary part in itself；ΔuMA(k) sampling of-measurement point M failure phase busbar voltage fault components Value；ΔIM0C、ΔIM0SThe opposite number of the real part of the mould electric currents of-measurement point M zero with imaginary part in itself；ΔiMA(k)-moulds of measurement point M zero The sampled value of electric current；The sampling number of N-mono- cycle；DT-sampling interval duration；K-it is sampling instant；M is divider ratio；
A0=K1ΔUMA(ω)z0D-K2z0D
A1=-K1ΔUMA(ω)z0-K1az0D+K2z0-K3z0D
A2=-3K1IM0(ω)(RM0+jωLM0+z0D)
A3=K1ΔUMA(ω)-K2
A4=j ω A3
A5=-K1a-K3
A6=j ω A5
A7=-3K1IM0(ω)
A8=j ω A7
A9=K1z0a+K3z0
A=r1(ΔIMA(ω)+KrIM0(ω))+jωl1(ΔIMA(ω)+KlIM0(ω))
In formula, D-total track length；RM0、LM0- measurement point M side systems zero sequence resistance and inductance；
(2) if phase-to phase fault, order,
K2=-j ω UMBCHC-ω0UMBCHS
K3=j ω (IMBCHCr1+IMBCHSω0l1)+ω0(IMBCHSr1-IMBCHCω0l1)
In formula, UMBCHC, UMBCHSThe real part of the modulus normal duty network load voltage cosine phasors of-measurement point M 2 is in itself With the opposite number of imaginary part；IMBCHC, IMBCHSThe real part of the modulus normal duty electric current cosine phasors of-measurement point M 2 is in itself and imaginary part Opposite number；
Calculate the modulus fault current component Δ I of measurement point M 2MBCThe spectrum component of (ω), false voltage component in frequency ω ΔUMBC(ω)：
In formula, Δ IMBCC、ΔIMBCSThe opposite number of the real part of the modulus fault current phasors of-measurement point M 2 with imaginary part in itself； ΔiMBC(k) sampled value of-modulus fault current components of measurement point M 2；ΔUMBCC、ΔUMBCSThe modulus failures of-measurement point M 2 electricity Press the opposite number of the real part of phasor with imaginary part in itself；ΔuMBC(k) sampled value of-modulus false voltage components of measurement point M 2； The sampling number of N-mono- cycle；DT-sampling interval duration；
Order,
A0=K1ΔUMBC(ω)z1D-K2z1D
A1=-K1ΔUMBC(ω)z1-K1az1D+K2z1-K3z1D
A3=K1ΔUMBC(ω)-K2
A4=-j ω A3
A5=-K1a-K3
A6=j ω A5
A8=j ω A7
A9=K1z1a+K3z1
A=z1ΔIMBC(ω)
In formula, D-total track length；RM2、LM2- measurement point M side systems negative sequence resistance and inductance；
(3) if three-phase fault, order,
K2=-j ω UMAHC-ω0UMAHS
K3=j ω (IMAHCr1+IMAHSω0l1)+ω0(IMAHSr1-IMAHCω0l1)
In formula, UMAHC, UMAHSThe phase of the real part of-measurement point M failure phase normal duty voltage cosine phasors with imaginary part in itself Anti- number；IMAHC, IMAHSThe opposite number of the real part of-measurement point M failure phase normal duty electric current cosine phasors with imaginary part in itself；
Calculate measurement point faulted phase current fault component Δ IMA(ω), failure phase busbar voltage fault component are frequency ω's Spectrum component Δ UMA(ω)：
In formula, Δ IMAC、ΔIMASThe opposite number of the real part of-measurement point M failure phase fault electric current phasors with imaginary part in itself； ΔiMA(k) sampled value of-measurement point M failure phase fault current components；ΔUMAC、ΔUMAS- measurement point M failure phase busbar voltages The opposite number of the real part of failure phasor with imaginary part in itself；ΔuMA(k) sampling of-measurement point M failure phase busbar voltage fault components Value；The sampling number of N-mono- cycle；DT-sampling interval duration；
Order,
A0=K1ΔUMA(ω)z1D-K2z1D
A1=-K1ΔUMA(ω)z1-K1az1D+K2z1-K3z1D
A2=-K1ΔIMA(ω)(RM1+jωLM1+z1D)
A3=K1ΔUMA(ω)-K2
A4=-j ω A3
A5=-K1a-K3
A6=j ω A5
A7=-K1ΔIMA(ω)
A8=j ω A7
A9=K1z1a+K3z1
A=z1ΔIMA(ω)
In formula, D-total track length；RM1、LM1- measurement point M side systems positive sequence resistance and inductance
Step 6, range equation are：
A0+A1d+A2RF+A3RN+A4LN+A5RNd+A6LNd+A7RFRN+A8RFLN+A9d2=0
In formula, distance of the d-trouble point to measurement point M；RN,LNThe resistance and inductance parameters of-offside power supply；RF- failure Point transition resistance；A in above-mentioned equationiCalculated according to different faults type by the definition of step 5；
Under any frequency, fault current and the corresponding spectrum component of voltage strictly meet range equation, when ω is base Wave frequency rate ω0When, reality, 2 nonlinear equations of imaginary part are obtained according to range equation, when ω is other optional frequencies, according to survey Reality, 2 nonlinear equations of imaginary part can be also obtained away from equation, definite condition are met, using Newton iteration method or least square The estimation technique solves equation group, obtains the estimate of fault distance, transition resistance, opposite end power supply equivalent inductance and resistance parameter.
Compared to the prior art, the present invention possesses following advantage：
Prior art uses power frequency component, and unknown variable number is more than equation number, is unsatisfactory for definite condition, therefore generally root Do appropriate according to system features it is assumed that eliminating Partial Variable.So processing certainly will cause systematic error, because real system Vary, it is specific to assume that all scenario is adapted to.This method is also used in signal in addition to using power frequency component Other frequency components, therefore any hypothesis need not be done to system can solve fault distance, in the absence of the system of principle Error.
Brief description of the drawings
Fig. 1 (a)-Fig. 1 (c) is transmission line fault network, and wherein Fig. 1 (a) is failure whole network, and Fig. 1 (b) is normal negative Lotus network, Fig. 1 (c) is fault component network.
Embodiment
The present invention is described in further detail below in conjunction with the accompanying drawings.
The present invention is that a kind of ultra-high-tension power transmission line fault localization based on parameter recognition principle is new under the conditions of intelligent substation Method, its main purpose is to solve the problem of traditional one-terminal data fault distance-finding method has Systematic Errors, in intelligent substation Under the conditions of the abundant transient information that is provided using electronic mutual inductor, the principle recognized with parameter carries out transmission line malfunction survey Away from.
The basic ideas of parameter identification distance-finding method proposed by the invention are to obtain fault network equation, trouble point side On the basis of boundary's conditional equation, the current in the fault point and opposite end electric current, voltage in equation are eliminated, obtains including home terminal current, electricity The network equation of pressure, offside system impedance, trouble point transition resistance and fault distance, then using local terminal sample rate current, voltage, The estimation of unknown parameter in equation is realized, so as to realize fault location.
The widely used R-L models of single-end electrical quantity Fault Location Algorithm, ignore the influence of distribution capacity.Under normal circumstances, For low pressure overhead transmission line or high pressure short-term road, carry out approximately, the requirement of engineer applied being reached using R-L models. General principle is introduced by two-shipper list loop line system model first, shown in model such as Fig. 1 (a)-Fig. 1 (c), it is assumed that system model is Single-phase, circuit uses R-L models.
For Linear Network, failure whole network can be decomposed into normal duty network and fault component using principle of stacking Network.Obtain measurement point M electric currents, voltage failure component and trouble point superimposed current, the network equation of voltage satisfaction
Calculated according to circuit both sides and arrive the equal condition of fault point voltage, obtained circuit two ends electric current, voltage and meet as off line Network equation
There is following failure boundary conditional equation for trouble point
ΔiM+ΔiN=iFFormula (3)
In formula：ΔuM- M sides voltage failure component instantaneous value；ΔiM,ΔiN- M sides and N side lines road current failure component wink Duration；uFThe instantaneous value in-fault component network superimposed voltage source；iF- current in the fault point instantaneous value；D-trouble point is to measurement point M distance；D-total track length；The resistance and inductance parameters of r, l-circuit unit length；RM,LM,RN,LN- both sides power supply Resistance and inductance parameters；RF- trouble point transition resistance.
Fourier transform is carried out to formula (1) to formula (3) both sides, corresponding frequency domain equation is obtained as follows
ΔUM(ω)=rd Δs IM(ω)+jωldΔIM(ω)+RFIF(ω)+UF(ω) formula (4)
(RM+dr+jω(LM+dl))ΔIM(ω)=(RN+(D-d)r+jωLN+jω(D-d)l)ΔIN(ω) formula (5)
ΔIM(ω)+ΔIN(ω)=IF(ω) formula (6)
Simultaneous obtains eliminating the network equation of circuit opposite end electric current, voltage and current in the fault point, with following general type
f(ΔUM(ω),ΔIM(ω),d,RM,LM,RN,LN,RF,UF(ω))=0 formula (7)
In formula：ω-investigate the corresponding angular frequencies of frequency f；ΔUM(ω), Δ IM(ω)-M sides voltage, current failure point Measure the spectrum component in frequency ω；ΔUN(ω), Δ INThe frequency spectrum point of (ω)-N sides voltage, current failure component in frequency ω Amount；UFSpectrum component of the trouble point superimposed voltage source in frequency ω in (ω)-fault component network；IF(ω)-current in the fault point In frequency ω spectrum component.
Equation (7) is nonlinear equation.Wherein Δ UM(ω),ΔIM(ω) can calculate frequency spectrum by this side measurement data and obtain Arrive；RM,LMFault component network can be solved by local terminal sampled data, be obtained by parameter identification method, is considered as known Amount.
According to principle of stacking, uFTime-domain expression it is as follows
uF(t)=- uH(t) u (t)=- (UHCcosω0t+UHSsinω0T) u (t) formula (8)
Wherein, uH(t) it is the voltage of trouble point under normal operating conditions, u (t) represents unit-step function.UF(ω's) Analytical expression is
With voltage phasor before failure at measurement point MAnd electric current phasorRelation be
So as to obtain
As can be seen here, each variable in formula (7), except RN,LN,RF, beyond d, remaining variables can be before measurement point M failure Measurement data afterwards is calculated and obtained, or is expressed as expression formula (such as U of local measurement data and fault distanceF(ω)), therefore can Using abbreviation as the range equation containing only following known variables
f(d,RN,LN,RF)|ω=0 formula (12)
After failure, obtained Current Voltage frequency spectrum can be calculated with fault transient process sampled data and is set up in different frequent points 2 or more than 2 range equations corresponding with formula (12).Because formula (12) is complex number equation, thus 4 or 4 can be obtained The equation group of real number equation composition above, solves this Nonlinear System of Equations by least square optimization, can obtain Fault distance d, while unknown parameter R in R-L models can also be tried to achieveN,LN,RFValue.
Actual circuit is three-phase system, and the location algorithm of the present invention is introduced exemplified by occurring singlephase earth fault.Still with Fig. 1 (a)-Fig. 1 (c) carries out algorithmic derivation, but the figure should regard three-phase system as.Assuming that singlephase earth fault occurs for trouble point, And assume that fault type is A phase earth faults.Then for fault component network, electric current, the electricity of measurement point M failure phases can be obtained Press the network equation of fault component and trouble point superimposed current, voltage satisfaction as follows
In formula：rs,rmResistance and the mutual resistance certainly of-circuit unit length；ls,lmThe self-induction of-circuit unit length and mutual inductance； ΔuMA(t)-measurement point M fault component instantaneous voltages；ΔiMA(t),ΔiMB(t),ΔiMC(t)-measurement point M three-phase currents Fault component instantaneous value；uFA(t)-fault component network trouble point superimposed voltage instantaneous value；iFA(t)-trouble point fault current Instantaneous value.
Notice r1=rs-rm, l1=ls-lmWith Δ iMA+ΔiMB+ΔiMC=3 Δ iM0=3iM0And define zero sequence compensation system NumberFormula (13) is rewritten as
For zero lay wire network, the equal condition of voltage for protecting installation place to calculate to trouble point by circuit both sides is obtained
For A phase earth faults, its current in the fault point meets such as downstream condition
iFA(t)=3 (iM0(t)+iN0(t)) formula (16)
Carry out Fourier transform, you can obtain corresponding frequency domain equation as follows
ΔUMA(ω)=UFA(ω)+IFA(ω)RF+r1d(ΔIMA(ω)+krIM0(ω))+jωl1d(ΔIMA(ω)+ klIM0(ω)) formula (17)
(RM0+r0d+jω(LM0+l0d))IM0(ω)=(RN0+r0D-r0d+jω(LN0+l0D-l0d))IN0(ω) formula (18)
IFA(the I of (ω)=3M0(ω)+IN0(ω)) formula (19)
In formula：The π f of the ω=2-corresponding angular frequencies of frequency f；ΔUMA(ω),ΔIMA(ω),IM0(ω)-measurement point bus The spectrum component of voltage failure component, failure phase A phase currents fault component and zero mould electric current in frequency ω；IFA(ω),IN0 The spectrum component of (ω)-trouble point fault current and N sides zero-sequence current in frequency ω；r1,l1,r0,l0- circuit unit length Positive sequence, zero sequence resistance and inductance；RN0=RNs+2RNm,LN0=LNs+2LNm- N side systems zero sequence resistance and inductance；RNs,RNm,LNs, LNm- N side systems are each mutually from resistance, self-induction and alternate mutual resistance, mutual inductance.
The load voltage electric current phasor that the entire spectrum and measurement point M of fault component network trouble point superimposed voltage are normally run Between relation be
It is above-mentioned
UFAC=-UMAHC+IMAHCr1d+IMAHSω0l1d
UFAS=-UMAHS+IMAHSr1d-IMAHCω0l1D formula (21)
In formula：UFAC, UFASThe phase of the real part of-fault component network trouble point superimposed voltage cosine phasor with imaginary part in itself Anti- number；UMAHC, UMAHSThe opposite number of the real part of-measurement point M failure phase normal duty voltage cosine phasors with imaginary part in itself； IMAHC, IMAHSThe opposite number of the real part of-measurement point M failure phase normal duty electric current cosine phasors with imaginary part in itself.
Define constant
K2=-j ω UMAHC-ω0UMAHS
K3=j ω (IMAHCr1+IMAHSω0l1)+ω0(IMAHSr1-IMAHCω0l1)
It is collated to obtain following nonlinear equation
A0+A1d+A2RF+A3RN0+A4LN0+A5RN0d+A6LN0d+A7RFRN0+A8RFLN0+A9d2=0 formula (22)
It is above-mentioned
A0=K1ΔUMA(ω)z0D-K2z0D
A1=-K1ΔUMA(ω)z0-K1az0D+K2z0-K3z0D
A2=-3K1IM0(ω)(RM0+jωLM0+z0D)
A3=K1ΔUMA(ω)-K2, A4=j ω A3
A5=-K1a-K3, A6=j ω A5
A7=-3K1IM0(ω), A8=j ω A7
A9=K1z0a+K3z0, z0=r0+jωl0
A=r1(ΔIMA(ω)+KrIM0(ω))+jωl1(ΔIMA(ω)+KlIM0(ω))
Under any frequency, fault current and the corresponding spectrum component of voltage strictly meet formula (22), certainly for base Ripple, each harmonic and Fractional Frequency are equally set up.When ω is fundamental frequency ω0When, reality, imaginary part 2 can be obtained by stable state network non- Linear equation.When ω is other optional frequencies, such as ω=ω0/ 2 or ω0/ 4, from transient network equation also can obtain it is real, 2 nonlinear equations of imaginary part, so as to meet definite condition, can solve the party using Newton iteration method or least squares estimate Journey group, obtains the estimate of the parameters such as fault distance.For two-phase short-circuit fault and three phase short circuit fault, its location algorithm is pushed away Lead process similar.
For the correctness and validity of verification method, the emulation of fault localization is carried out to Beijing-Tianjin-Tangshan 500kV transmission lines of electricity, is Unite wiring such as Fig. 1 (a)-Fig. 1 (c), and sample frequency 500kHz, simulation system parameters are as follows.
Line length：D=300km
Line parameter circuit value：r1=0.02083 Ω/km；l1=0.8984mH/km；
r0=0.1148 Ω/km；l0=2.2886mH/km；
M side system parameters：RM1=1.0515 Ω；LM1=0.13743H；
RM0=0.6 Ω；LM0=0.0926H；
N side system parameters；RN1=26 Ω；LN1=0.14298H；
RN0=20 Ω；LN0=0.11927H；The advanced N side systems of M side systems
Range error：
Table 1 simulates circuit different location and occurs the single-phase distance measurement result through during 200 Ω transition resistance failures.Can from table 1 To find out, when under the practical operation situation deviation algorithm of the system assumed condition basically identical to both sides system impedance angle, this hair Bright range accuracy under various fault ' conditions is all higher, can meet requirement of engineering.
Distance measurement result (R of the different faults of table 1 underF=200 Ω)
Table 2 is simulated in the line away from generation singlephase earth fault, sheet when transition resistance takes different value at measurement point 200km The distance measurement result of invention.From table 2 it can be seen that using modal identification algorithm carry out ranging, distance measurement result almost with transition resistance It is unrelated, and range accuracy is also to meet engine request.
The asynchronous emulation distance measurement result (d=200km) of the trouble point transition resistance of table 2
Table 3 is given when different type failure occurs for circuit diverse location, and the emulation of location algorithm is recognized using parameter As a result, wherein transition resistance is used uniformly 50 Ω.From table 3 it can be seen that parameter recognizes location algorithm in different faults type feelings Accurate distance measurement result can be provided under condition.
Emulation distance measurement result (RF=50 Ω) during 3 different faults type of table
To sum up, traditional power frequency single end distance measurement algorithm, can only list two equations, and range equation is to owe fixed, all directly or Indirectly to variable RN,LN,RFCarried out simplify it is assumed that thus there are Systematic Errors.The present invention is by RN,LN,RFAlso serve as treating Identification parameter, this is different from traditional power frequency single end distance measurement algorithm, fundamentally avoids the systematic error of single end distance measurement. Simulation result also demonstrates the present invention can overcome systematic error from principle, and distance measurement result is not by peer-to-peer system impedance and transition The influence of resistance, adapts to the requirement applied in intelligent substation.

Claims (1)

1. the single-ended holographic frequency domain Fault Locating Method of ultra-high-tension power transmission line, it is characterised in that：Comprise the following steps：
A, B, C three-phase voltage u of step 1, collection protection installation place current transformer and voltage transformerA(k),uB(k),uC(k) With three-phase current iA(k),iB(k),iC(k)；
Step 2, A, B, C three-phase voltage and electric current to collecting, three-phase voltage and failure of the current point are calculated using following formula Measure Δ uA(k),ΔuB(k),ΔuC(k), Δ iA(k),ΔiB(k),ΔiC(k) with the sampled value 3i of zero-sequence component0(k)：
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>u</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>u</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>u</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>u</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>u</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>u</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> </mtable> </mfenced>
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>i</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>i</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>i</mi> <mi>A</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>i</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>i</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>i</mi> <mi>B</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mi>&amp;Delta;</mi> <msub> <mi>i</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>=</mo> <msub> <mi>i</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>)</mo> <mo>-</mo> <msub> <mi>i</mi> <mi>C</mi> </msub> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mi>N</mi> <mo>)</mo> </mtd> </mtr> </mtable> </mfenced>
3i0(k)=Δ iA(k)+ΔiB(k)+ΔiC(k)
In above formula, k is sampling instant, and N is per cycle sampling number；A, B, C three-phase are calculated using all-round Fourier algorithm simultaneously The real and imaginary parts of voltage and current cosine phasor；
Step 3, definition constant：
<mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> <mo>=</mo> <msubsup> <mi>&amp;omega;</mi> <mn>0</mn> <mn>2</mn> </msubsup> <mo>-</mo> <msup> <mi>&amp;omega;</mi> <mn>2</mn> </msup> </mrow>
z1=r1+jωl1
z0=r0+jωl0
<mrow> <msub> <mi>K</mi> <mi>r</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>r</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>r</mi> <mn>1</mn> </msub> </mrow> <msub> <mi>r</mi> <mn>1</mn> </msub> </mfrac> </mrow>
<mrow> <msub> <mi>K</mi> <mi>l</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>l</mi> <mn>1</mn> </msub> </mrow> <msub> <mi>l</mi> <mn>1</mn> </msub> </mfrac> </mrow>
In formula, ω0- power frequency angular frequency；- m is divider ratio, can be equal to 2,3 or 4 etc.；r1,l1,r0,l0- circuit list Bit length positive sequence, zero sequence resistance and inductance；Kr、Kl- zero-utility theory；
Step 4, calculate each mold component of this end system it is equivalent in inductance and interior resistance parameter RMi,LMi(i=0,1,2), take two with On not k three-phase current and three-phase voltage sampled value in the same time, obtain the modulus of Current Voltage 0,1 modulus, 2 using Clarke transform Modulus instantaneous value is brought into system of linear equations
In, using linear least-squares algorithm Solve this equation group and obtain RMi,LMi(i=0,1,2)；
In formula, Δ iMiEach modulus fault component current instantaneous value of-M sides；ΔuMiMeasurement point M each modulus failure during-failure Component voltage；DT is sampling interval duration；
Step 5, determine fault type
(1) if singlephase earth fault, order,
K2=-j ω UMAHC-ω0UMAHS
K3=j ω (IMAHCr1+IMAHSω0l1)+ω0(IMAHSr1-IMAHCω0l1)
In formula, UMAHC, UMAHSThe opposite number of the real part of-measurement point M failure phase normal duty voltage cosine phasors with imaginary part in itself； IMAHC, IMAHSThe opposite number of the real part of-measurement point M failure phase normal duty electric current cosine phasors with imaginary part in itself；
Calculate measurement point faulted phase current fault component Δ IMA(ω), failure phase busbar voltage fault component Δ UMA(ω) and zero mould Spectrum component I of the electric current in frequency ωM0(ω)：
<mrow> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;i</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
<mrow> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;u</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
<mrow> <msub> <mi>I</mi> <mrow> <mi>M</mi> <mn>0</mn> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;I</mi> <mrow> <mi>M</mi> <mn>0</mn> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>I</mi> <mrow> <mi>M</mi> <mn>0</mn> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>i</mi> <mrow> <mi>M</mi> <mn>0</mn> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
In formula, Δ IMAC、ΔIMASThe opposite number of the real part of-measurement point M failure phase fault electric current phasors with imaginary part in itself；ΔiMA (k) sampled value of-measurement point M failure phase fault current components；ΔUMAC、ΔUMAS- measurement point M failure phase busbar voltage failures The opposite number of the real part of phasor with imaginary part in itself；ΔuMA(k) sampled value of-measurement point M failure phase busbar voltage fault components； ΔIM0C、ΔIM0SThe opposite number of the real part of the mould electric currents of-measurement point M zero with imaginary part in itself；ΔiMA(k)-mould electric currents of measurement point M zero Sampled value；The sampling number of N-mono- cycle；DT-sampling interval duration；K-it is sampling instant；M is divider ratio；
A0=K1ΔUMA(ω)z0D-K2z0D
A1=-K1ΔUMA(ω)z0-K1az0D+K2z0-K3z0D
A2=-3K1IM0(ω)(RM0+jωLM0+z0D)
A3=K1ΔUMA(ω)-K2
A4=j ω A3
A5=-K1a-K3
A6=j ω A5
A7=-3K1IM0(ω)
A8=j ω A7
A9=K1z0a+K3z0
A=r1(ΔIMA(ω)+KrIM0(ω))+jωl1(ΔIMA(ω)+KlIM0(ω))
In formula, D-total track length；RM0、LM0- measurement point M side systems zero sequence resistance and inductance；
(2) if phase-to phase fault, order,
K2=-j ω UMBCHC-ω0UMBCHS
K3=j ω (IMBCHCr1+IMBCHSω0l1)+ω0(IMBCHSr1-IMBCHCω0l1)
In formula, UMBCHC, UMBCHSThe real part of-measurement point M2 modulus normal duty network load voltage cosine phasors is in itself and imaginary part Opposite number；IMBCHC, IMBCHSThe real part of-measurement point M2 modulus normal duty electric current cosine phasors is opposite with imaginary part in itself Number；
Calculate measurement point M2 modulus fault current component Δs IMBCThe spectrum component Δ U of (ω), false voltage component in frequency ωMBC (ω)：
<mrow> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;i</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
<mrow> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;u</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
In formula, Δ IMBCC、ΔIMBCSThe opposite number of the real part of-measurement point M2 modulus fault current phasors with imaginary part in itself；ΔiMBC (k) sampled value of-measurement point M2 modulus fault current components；ΔUMBCC、ΔUMBCS- measurement point M2 modulus false voltage phasors Real part opposite number with imaginary part in itself；ΔuMBC(k) sampled value of-measurement point M2 modulus false voltage components；N-mono- week The sampling number of ripple；DT-sampling interval duration；
Order,
A0=K1ΔUMBC(ω)z1D-K2z1D
A1=-K1ΔUMBC(ω)z1-K1az1D+K2z1-K3z1D
<mrow> <msub> <mi>A</mi> <mn>2</mn> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>K</mi> <mn>1</mn> </msub> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>R</mi> <mrow> <mi>M</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>j&amp;omega;L</mi> <mrow> <mi>M</mi> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>z</mi> <mn>1</mn> </msub> <mi>D</mi> <mo>)</mo> </mrow> </mrow>
A3=K1ΔUMBC(ω)-K2
A4=-j ω A3
A5=-K1a-K3
A6=j ω A5
<mrow> <msub> <mi>A</mi> <mn>7</mn> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>K</mi> <mn>1</mn> </msub> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>B</mi> <mi>C</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> </mrow>
A8=j ω A7
A9=K1z1a+K3z1
A=z1ΔIMBC(ω)
In formula, D-total track length；RM2、LM2- measurement point M side systems negative sequence resistance and inductance；
(3) if three-phase fault, order,
K2=-j ω UMAHC-ω0UMAHS
K3=j ω (IMAHCr1+IMAHSω0l1)+ω0(IMAHSr1-IMAHCω0l1)
In formula, UMAHC, UMAHSThe opposite number of the real part of-measurement point M failure phase normal duty voltage cosine phasors with imaginary part in itself； IMAHC, IMAHSThe opposite number of the real part of-measurement point M failure phase normal duty electric current cosine phasors with imaginary part in itself；
Calculate measurement point faulted phase current fault component Δ IMAThe frequency spectrum of (ω), failure phase busbar voltage fault component in frequency ω Component Δ UMA(ω)：
<mrow> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;I</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;i</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
<mrow> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>j&amp;omega;&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>C</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mrow> <msub> <mi>&amp;omega;</mi> <mn>0</mn> </msub> <msub> <mi>&amp;Delta;U</mi> <mrow> <mi>M</mi> <mi>A</mi> <mi>S</mi> </mrow> </msub> </mrow> <msub> <mi>K</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>m</mi> <mi>N</mi> </mrow> </munderover> <msub> <mi>&amp;Delta;u</mi> <mrow> <mi>M</mi> <mi>A</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <msup> <mi>e</mi> <mfrac> <mrow> <mo>-</mo> <msub> <mi>jk&amp;omega;</mi> <mn>0</mn> </msub> <mi>D</mi> <mi>T</mi> </mrow> <mi>m</mi> </mfrac> </msup> <mi>D</mi> <mi>T</mi> </mrow>
In formula, Δ IMAC、ΔIMASThe opposite number of the real part of-measurement point M failure phase fault electric current phasors with imaginary part in itself；ΔiMA (k) sampled value of-measurement point M failure phase fault current components；ΔUMAC、ΔUMAS- measurement point M failure phase busbar voltage failures The opposite number of the real part of phasor with imaginary part in itself；ΔuMA(k) sampled value of-measurement point M failure phase busbar voltage fault components； The sampling number of N-mono- cycle；DT-sampling interval duration；
Order,
A0=K1ΔUMA(ω)z1D-K2z1D
A1=-K1ΔUMA(ω)z1-K1az1D+K2z1-K3z1D
A2=-K1ΔIMA(ω)(RM1+jωLM1+z1D)
A3=K1ΔUMA(ω)-K2
A4=-j ω A3
A5=-K1a-K3
A6=j ω A5
A7=-K1ΔIMA(ω)
A8=j ω A7
A9=K1z1a+K3z1
A=z1ΔIMA(ω)
In formula, D-total track length；RM1、LM1- measurement point M side systems positive sequence resistance and inductance；
Step 6, range equation are：
A0+A1d+A2RF+A3RN+A4LN+A5RNd+A6LNd+A7RFRN+A8RFLN+A9d2=0
In formula, distance of the d-trouble point to measurement point M；RN,LNThe resistance and inductance parameters of-offside power supply；RF- trouble point mistake Cross resistance；A in above-mentioned range equationiCalculated according to different faults type by the definition of step 5；
Under any frequency, fault current and the corresponding spectrum component of voltage strictly meet range equation, when ω is fundamental wave frequency Rate ω0When, reality, 2 nonlinear equations of imaginary part are obtained according to range equation, when ω is other optional frequencies, according to ranging side Journey can also obtain reality, 2 nonlinear equations of imaginary part, definite condition be met, using Newton iteration method or least-squares estimation Method solves equation group, obtains the estimate of fault distance, transition resistance, opposite end power supply equivalent inductance and resistance parameter.
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