CN109975653B  10kV distribution line fault location method  Google Patents
10kV distribution line fault location method Download PDFInfo
 Publication number
 CN109975653B CN109975653B CN201910139581.XA CN201910139581A CN109975653B CN 109975653 B CN109975653 B CN 109975653B CN 201910139581 A CN201910139581 A CN 201910139581A CN 109975653 B CN109975653 B CN 109975653B
 Authority
 CN
 China
 Prior art keywords
 fault
 current
 phase
 value
 time
 Prior art date
 Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
 Active
Links
 238000005070 sampling Methods 0.000 claims abstract description 124
 230000016507 interphase Effects 0.000 claims abstract description 55
 239000004020 conductor Substances 0.000 claims description 28
 238000004364 calculation method Methods 0.000 claims description 26
 230000000875 corresponding Effects 0.000 claims description 11
 238000011105 stabilization Methods 0.000 claims description 4
 238000004422 calculation algorithm Methods 0.000 claims description 3
 238000001514 detection method Methods 0.000 description 2
 238000010586 diagram Methods 0.000 description 2
 238000000034 method Methods 0.000 description 2
 230000001052 transient Effects 0.000 description 2
 150000001875 compounds Chemical class 0.000 description 1
 230000014509 gene expression Effects 0.000 description 1
 238000009434 installation Methods 0.000 description 1
 238000005259 measurement Methods 0.000 description 1
 230000001960 triggered Effects 0.000 description 1
Classifications

 G—PHYSICS
 G01—MEASURING; TESTING
 G01R—MEASURING ELECTRIC VARIABLES; MEASURING MAGNETIC VARIABLES
 G01R19/00—Arrangements for measuring currents or voltages or for indicating presence or sign thereof
 G01R19/02—Measuring effective values, i.e. rootmeansquare values

 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
Abstract
A10 kV distribution line fault location method comprises the following steps: (1) determining whether the fault type is a twophase short circuit or a threephase short circuit; (2) if the fault is a threephase shortcircuit fault, calculating the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state; (3) if the fault is an interphase shortcircuit fault, calculating a currentvoltage phase angle difference under the fault state; (4) respectively obtaining a threephase shortcircuit fault current effective value or a twophase shortcircuit fault current effective value according to the sampling information of the fault current; (5) and integrating the information of the distribution line, and calculating the distance between the threephase or twophase shortcircuit fault point and the line starting end. The method of the invention fully utilizes the fault information of the distribution line, can provide a relatively accurate fault location result under the condition of not increasing equipment, improves the operation and maintenance efficiency of the distribution line and reduces the operation and maintenance cost.
Description
Technical Field
The invention relates to a 10kV distribution line fault location method, and belongs to the technical field of power distribution networks.
Background
At present, the fault location technology of the power system is widely applied to the protection of lines with voltage levels of 110kV and above, for a 10kV distribution line, fault location and positioning need to be realized by additionally installing a switch, a fault indicator or other special detection devices and the like on the line, the complexity of the line is increased, and the fault location cost is increased. Meanwhile, the fault information of the distribution line is not paid enough attention in operation and maintenance practice, so that the fault characteristics contained in the fault information are not fully utilized.
Disclosure of Invention
The invention aims to increase the complexity of a line and improve the fault location cost by additionally installing a switch, a fault indicator or other special detection devices on the line according to the existing fault location and positioning of a 10kV distribution line, and the methods are still limited to determining a fault interval and cannot accurately give a fault distance.
The technical scheme of the invention is as follows: a10 kV distribution line fault location method comprises the following steps:
(1) determining whether the fault type is a twophase short circuit or a threephase short circuit;
(2) if the fault is a threephase shortcircuit fault, calculating the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state;
(3) if the fault is an interphase shortcircuit fault, calculating a currentvoltage phase angle difference under the fault state;
(4) respectively obtaining a threephase shortcircuit fault current effective value or a twophase shortcircuit fault current effective value according to the sampling information of the fault current;
(5) and integrating the information of the distribution line, and calculating the distance between the threephase shortcircuit fault point and the initial end of the line or the distance between the twophase shortcircuit fault point and the initial end of the line.
When the threephase short circuit fault occurs, the phase angle difference sigma of the voltage in the fault state lagging the voltage in the normal operation state is as follows:
wherein T is the system voltage period and the value is 0.02 s; δ t represents a time difference corresponding to the phase angle difference;
δt＝mod(Δt,T)
in the above formula, mod (a, b) is a remainder function, which means that parameter a is the remainder of parameter b; delta t is the time difference between the zero crossing point of the voltage waveform in the normal operation state and the zero crossing point of the voltage waveform in the fault stabilization state;
Δt＝t_{2}t_{1}
in the above formula, t_{1}The time value is from negative to positive zero crossing point under normal state; t is t_{2}The time value from negative to positive zero crossing point under the fault state;
t_{1}＝t_{k+1}arcsin(u(t_{k+1}))/100π
t_{2}＝t_{j+1}arcsin(u(t_{j+1}))/100π
wherein u (t)_{k+1}) T is in normal operation_{k+1}Voltage sampling value at time u (t)_{k+1}) And t_{k+1}Is sampled at the previous sampling instant t_{k}The voltage sampling value u (t) at a time_{k}) Satisfies the following formula:
u(t_{k}) Less than 0; and u (t)_{k+1})≥0；
u(t_{j+1}) T being a fault condition_{j+1}Voltage sampling value of a time, the sampling value and a previous sampling time t_{j}The voltage sampling value u (t) at a time_{j}) Satisfies the following formula:
u(t_{j}) Less than 0; and u (t)_{j+1})≥0。
The method for calculating the effective value of the threephase shortcircuit fault current comprises the following steps:
determining the moment of entering a fault state, carrying out effective value calculation on the sampled data in real time according to a halfperiod window, continuously comparing the sampled data with data before a half period, and if the ratio of the sampled data to the data before the half period exceeds a limit value, indicating that the current is in the fault state at the current moment; then, current sampling data after the time of entering the fault state is obtained, effective value calculation is continuously carried out in real time according to a halfperiod window, the data are continuously compared with the data before the half period, if the ratio of the two is within a limit value, the current is shown to be in the fault state at the moment, the current effective value at the moment is calculated, and the threephase fault current effective value is obtained;
and (3) solving an effective value of the Aphase fault current:
let the sampling value at time t be i_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in a system half cycle, namely 0.01s, and is determined by the sampling frequency f, and m is f/100; i.e. i_{A}(t) is a sampling value at the current moment t, and k is a counting parameter;
note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}By comparison, if: i is_{At}/I_{A(t0.01)}≥1.5，
Indicating that the t moment is in a fault state, and continuously calculating and analyzing subsequent sampling values of the current; if t is the time elapsed after t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following conditions:
the effective value I of the Aphase fault current is taken_{AK}Comprises the following steps:
in the same way, the solving method of the effective values of the fault currents of the B phase and the C phase is consistent with that of the A phase.
The calculation method of the distance between the threephase shortcircuit fault point and the line starting end is as follows:
when the threephase short circuit of the distribution line is failed, the relationship between the phase angle difference sigma of the voltage in the fault state lagging the voltage in the normal operation state and the related parameters meets the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a threephase fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}A transition resistance that is a point of failure; sigma is the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state;
is obtained by the formula:
obtaining the ratio eta of the reactance and the resistance of the threephase shortcircuit fault point as follows:
when threephase short circuit fault occurs, the distance l between the fault point and the initial end of the line_{K3}Comprises the following steps:
wherein, I_{AK}The effective value of the Aphase fault current is the effective value of the Aphase fault current in the case of threephase shortcircuit fault; u shape_{N}The rated voltage of the distribution line; x is l_{K3}A unit length reactance value of a position conductor model; b is determined by_{K3}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}；X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K3}And the line reactance value at the beginning of the conductor model.
When the interphase short circuit fault occurs, the current voltage phase angle difference under the fault stateComprises the following steps:
in the formula, T is the system voltage period and is 0.02 s; delta tau is a time difference corresponding to the phase angle difference;
δτ＝mod(Δτ,T)
in the formula, mod (a, b) is a remainder function, which represents that a parameter a is the remainder of a parameter b; delta tau is the time difference of the zero crossing point of the lagging voltage waveform of the current waveform in the phasetophase short circuit fault state;
Δτ＝τ_{2}τ_{1}
in the formula, τ_{1}The time value of the voltage from negative to positive zero crossing point under the phasetophase short circuit fault state; tau is_{2}The current is the time value from negative to positive zero crossing point under the condition of interphase short circuit fault;
τ_{1}＝τ_{j+1}arcsin(u(τ_{j+1}))/100π
τ_{2}＝τ_{k+1}arcsin(u(τ_{k+1}))/100π
in the formula, i (τ)_{k+1}) For tau in the course of interphase shortcircuit fault_{k+1}Sampled value of the current at a time, i (τ)_{k+1}) And τ_{k+1}Is the previous sampling instant t_{k}Current sample value i (tau) at a time_{k}) Satisfies the following formula:
i(τ_{k}) Less than 0; and i (τ)_{k+1})≥0；
u(τ_{j+1}) For tau in the course of interphase shortcircuit fault_{j+1}Voltage sample value at a time u (tau)_{j+1}) And τ_{j+1}Is the previous sampling instant t_{j}The voltage sample u (tau) at a time_{j}) Satisfies the following formula:
u(τ_{j}) Less than 0; and u (τ)_{j+1})≥0。
The calculation method of the interphase shortcircuit fault current effective value comprises the following steps:
determining the moment of entering a fault state, carrying out effective value calculation on the sampled data in real time according to a halfperiod window, continuously comparing the sampled data with data before a half period, and if the ratio of the sampled data to the data before the half period exceeds a limit value, indicating that the current is in the fault state at the current moment; then, current sampling data after the current enters the fault state to the moment are subjected to effective value calculation continuously in real time according to a halfperiod window and are continuously compared with data before the half period, if the ratio of the two is within a limit value, the current is indicated to be in the fault state at the moment, the current effective value at the moment is calculated, and the phasetophase fault current effective value is obtained;
the algorithm of the effective value of the Aphase fault current is as follows:
let the sampling value at time t be i_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in 0.01s which is the system half period, and is determined by the sampling frequency f, m is f/100, i_{A}(t) is a sampling value at the current moment t, and k is a counting parameter; note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}By comparison, if: i is_{At}/I_{A(t0.01)}≥1.5
Then it indicates that time t is atAnd in the fault state, continuously calculating and analyzing subsequent sampling values of the current, and if t is the elapsed time after t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following conditions:
the effective value I of the Aphase fault current is taken_{AK}Is composed of
In the same way, when an interphase fault occurs, the method for calculating the effective value of the fault current of the other phase is consistent with that of the phase A, and the effective values of the fault currents of the two phases of the interphase shortcircuit fault are equal under the general condition.
The calculation method of the distance between the interphase shortcircuit fault point and the line starting end is as follows:
phase angle difference of current waveform zero crossing point lagging voltage waveform zero crossing point in fault state during interphase short circuit fault of distribution lineAnd the related parameters satisfy the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}A transition resistance that is a point of failure;
is obtained by the formula:
obtaining the ratio lambda of the reactance and the resistance of the interphase shortcircuit fault point as follows:
when interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is l_{K2}Comprises the following steps:
wherein, I_{AK}The effective value of Aphase fault current is the effective value of the Aphase fault current in the interphase short circuit fault; u shape_{N}The rated voltage of the distribution line; b is determined by_{K2}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}；X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K2}The line reactance value at the beginning of the type of the conductor; x is the reactance value per unit length of the conductor type number.
The method for measuring the fault distance of the 10kV distribution line has the advantages that the fault type is judged according to the sampling information based on the self structure and parameter information of the distribution line, and the phase angle difference between fault voltage and normal operation voltage and the magnitude of the fault current are calculated for the threephase shortcircuit fault; for the interphase shortcircuit fault, calculating a phase angle difference between fault current and fault voltage and the magnitude of the fault current; and then, calculating the distance from the fault point to the line starting end by using a distribution line fault distance calculation algorithm, wherein the accurate position of the fault point can be determined by means of downstream power failure information due to the fact that the distribution line has branch lines.
The method provided by the invention fully utilizes the fault information of the distribution line, can provide a relatively accurate fault location result under the condition of not increasing equipment, improves the operation and maintenance efficiency of the distribution line and reduces the operation and maintenance cost.
Drawings
FIG. 1 is a schematic flow chart of a 10kV distribution line fault location method;
FIG. 2 is a schematic diagram showing a comparison of the fault current voltage waveforms of the distribution line in a normal operation state and a fault state;
in fig. 2, the upper part is a current waveform, the lower part is a voltage waveform, and Δ t is a time difference of voltage in a fault state lagging a zero crossing point by voltage in a normal operation state;
FIG. 3 is a schematic diagram of a distribution line fault current voltage waveform illustrating phase angle differences between fault state current voltage waveforms at a phasetophase fault;
in fig. 3, the upper part is a current waveform, the lower part is a voltage waveform, and Δ τ is a time difference between a current zerocrossing point and a lagging inphase voltage zerocrossing point.
Detailed Description
The specific implementation of the invention is shown in the flow chart of the fault location method of the 110kV distribution line.
The method for measuring the fault distance of the 10kV distribution line comprises the following steps:
(1) determining whether the fault type is a twophase short circuit or a threephase short circuit according to the sampling information of the fault current and voltage;
(2) if the fault is a threephase shortcircuit fault, calculating the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state;
(3) if the fault is an interphase shortcircuit fault, calculating a currentvoltage phase angle difference under the fault state;
(4) according to the sampling information of the fault current, solving a fault current effective value;
(5) and integrating the distribution line information, and calculating the distance between the shortcircuit fault point and the line starting end.
1. Determining fault type
As the 10kV distribution lines in China are all in a triangular connection mode, only two types of interphase short circuits and threephase short circuits are in fault, wherein the interphase short circuits comprise internal interphase short circuits and allopatric interphase short circuits, and the allopatric interphase short circuits refer to two lines which are in overline allopatric interphase short circuits.
The following installation method is carried out for judging the fault type:
(1) internal phase to phase short circuit fault
And if the effective value of the twophase voltage is less than 0.9 times of the rated value at the same time and the twophase current is greater than 800A, determining that the internal interphase short circuit fault occurs in the line.
The conditions are (taking AB phase twophase interphase short circuit as an example):
and is
In the formula of U_{A}、U_{B}Respectively obtaining A, B phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, u_{Ak}、u_{Bk}K is a counting parameter for the voltage instantaneous sample value.
I_{A}、I_{B}Respectively obtaining A, B phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, i_{Ak}、i_{Bk}K is a counting parameter for the current instantaneous sample value.
(2) Offsite twophase short circuit fault
And if the effective value of the twophase voltage is simultaneously less than 0.9 times of the rated value and only one phase current in the two phases is more than 800A, determining that the line has a twophase shortcircuit fault in different places.
The conditions are (taking A as an example of the interphase short circuit) as follows:
and is
In the formula of U_{A}、U_{B}Respectively obtaining A, B phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, u_{Ak}、u_{Bk}K is a counting parameter for the voltage instantaneous sample value.
I_{A}、I_{B}Respectively obtaining A, B phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, i_{Ak}、i_{Bk}K is a counting parameter for the current instantaneous sample value.
(3) Threephase shortcircuit fault
And if the effective value of the threephase voltage is simultaneously less than 0.9 time of the rated value and at least one phase current of the threephase current is more than 800A, determining that the threephase short circuit fault occurs in the line. The conditions are as follows:
and I_{A}Not less than 800A, or I_{B}Not less than 800A, or I_{C}≥800A
In the formula of U_{A}、U_{B}、U_{C}Respectively obtaining A, B, C phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, u_{Ak}、u_{Bk}、u_{Ck}K is a counting parameter for the voltage instantaneous sample value.
I_{A}、I_{B}、I_{C}Respectively obtaining A, B, C phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
wherein, N is the number of samples in 0.02s of the ac cycle, and is determined by the sampling frequency of the device, and if the sampling frequency is f, N is f/50, i_{Ak}、i_{Bk}、i_{Ck}K is a counting parameter for the current instantaneous sample value.
(4) Other faults
Faults which do not satisfy the above conditions are listed as other faults, and most other faults occur not due to short circuit on a 10kV line but due to faults on the lowvoltage side of a distribution transformer or overload of a user. And for other faults, fault distance measurement is not carried out.
The method of the present invention is limited to the three faults described in the cases (1) to (3).
2. Fault location method for threephase shortcircuit fault
I. Calculating the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state
When the distribution lines takes place looks or threephase short circuit fault, transient distortion can take place for the voltage and transition to stable fault state, because distribution lines short circuit capacity is very little for the major network capacity, therefore distribution lines trouble is minimum to the system influence, and when distribution lines fault state was stable, the voltage waveform appeared for the sine wave. The initial phase angle between the voltage waveforms is different between the normal operation state and the fault state, which is represented on the waveforms, i.e. the time difference between the zero crossing point of the voltage waveform in the normal operation state and the zero crossing point of the voltage waveform in the fault stabilization state is not equal to the integral multiple of the period of the alternating voltage, as shown in fig. 2. The phase angle corresponding to the time difference is the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state.
When the distribution line breaks down, the voltage sampling value and the corresponding sampling time before and after the fault are recorded, and the data in the voltage sampling value and the corresponding sampling time are processed.
And judging the data in the normal running state: if t_{k}The sampled value at the moment satisfies the following formula:
u(t_{k}) Less than 0; and u (t)_{k+1})≥0；
In the formula, u (t)_{k})、u(t_{k+1}) Are each t_{k}、t_{k+1}The voltage sample value at the time.
When the above conditions are satisfied, the zero crossing point indicating that the voltage changes from a negative value to a positive value is at t_{k}～t_{k+1}Between the moments, the value t of the moment from negative to positive zero crossing in the normal state can be determined_{1}Comprises the following steps:
t_{1}＝t_{k+1}arcsin(u(t_{k+1}))/100π
similarly, the data of the fault state is judged: if t_{j}The sampled value at the moment satisfies the following formula:
u(t_{j}) Less than 0; and u (t)_{j+1})≥0；
In the formula, u (t)_{j})、u(t_{j+1}) Are each t_{j}、t_{j+1}The voltage sample value at the time.
When the above conditions are satisfied, the zero crossing point indicating that the voltage changes from a negative value to a positive value is at t_{j}～t_{j+1}Between the moments, the value of the moment t from negative to positive zero crossing in the fault state can be determined_{2}Comprises the following steps:
t_{2}＝t_{j+1}arcsin(u(t_{j+1}))/100π
therefore, the time difference Δ t between the zero crossing point of the voltage waveform in the normal operation state and the zero crossing point of the voltage waveform in the fault stabilization state is: Δ t ═ t_{2}t_{1}
The time difference delta t is obtained by taking the remainder of the system voltage cycle, and the time difference delta t corresponding to the phase angle difference is obtained as follows:
δt＝mod(Δt,T)
in the formula, mod (a, b) is a remainder function, which means that parameter a is the remainder of parameter b, and T is the system voltage period, and has a value of 0.02 s.
Thus, the phase angle difference sigma of the voltage in the fault state lagging the voltage in the normal operation state is obtained as follows:
in the formula, T is the system voltage period, and its value is 0.02 s.
II. According to the sampling information of the fault current, the effective value of the fault current is obtained
The current effective value of the fault required by the calculation of the embodiment is not the current value when the protection device is triggered to trip, but the current effective value in a complete alternating current period in a fault stable state. In order to obtain the value, the sampling current of each phase needs to be calculated and analyzed, firstly, the time of entering the fault state is determined, effective value calculation is carried out on the sampling data according to a halfperiod window in real time, the sampling data is continuously compared with the data before the half period, and if the ratio of the sampling data to the sampling data exceeds the limit value, the current at the current time is in the fault state. And then, carrying out effective value calculation on current sampling data after the current sampling data enters the fault state to the moment, continuously carrying out effective value calculation according to a halfperiod window, continuously comparing the effective value with data before the half period, and if the ratio of the effective value to the data before the half period is within a limit value, indicating that the current is in the fault state at the moment, calculating the current effective value at the moment, and obtaining the threephase fault current effective value.
The fault state judging method comprises the following steps: realtime calculating current information in half period, and continuously calculating current in the last half period at a certain momentEffective value, for example, in phase A, i represents a sampling value at time t_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in a system half cycle, namely 0.01s, and is determined by the sampling frequency f, and m is f/100; i.e. i_{A}(t) is a sampling value at the current moment t; k is a counting parameter.
Note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}For comparison, if the following formula is satisfied:
I_{At}/I_{A(t0.01)}≥1.5
indicating that the time t is in a fault state, continuously calculating and analyzing subsequent sampling values of the current, and if t is the elapsed time after the time t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following formula:
the effective value I of the Aphase fault current is taken_{AK}Comprises the following steps:
in the same way, the solving method of the effective values of the fault currents of the B phase and the C phase is consistent with that of the A phase.
III, calculating the distance between the threephase shortcircuit fault point and the initial end of the line
After the phase angle difference of the voltage fundamental wave before and after the fault and the effective value of the fault current are obtained, the distance between the initial end of the line and the fault point, and the resistance component, the reactance component and the transition resistance of the wire can be obtained. The calculation principle and method are as follows.
The embodiment provides that, when a threephase short circuit fault occurs in a distribution line, the relationship between the phase angle difference sigma of the voltage in the fault state lagging the voltage in the normal operation state and the related parameters satisfies the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}Is the transition resistance of the fault point. σ is the phase angle difference of the fault state voltage lagging the normal operating state voltage.
From the formula (3)
Obtaining the ratio eta of the reactance and the resistance of the threephase shortcircuit fault point as follows:
meanwhile, taking phase A as an example, section 4.2 calculates the effective value of the Aphase fault current when the threephase shortcircuit fault occurs to be I_{AK}Thus, therefore, it is
In the formula of U_{N}The rated voltage of the distribution line, i.e. 10.5kV, can be obtained according to the equations (4) and (5)
In the formulae (6) and (7), there are three unknowns R_{l}、X_{l}、R_{G}；
Wherein R is_{l}、X_{l}Two unknowns, the distance between the threephase shortcircuit fault point of the power distribution line and the initial end of the lead is set to be l_{K3}When the conductor of the distribution line has impedance, the conductor resistance R between the fault point of the distribution line and the line starting end_{l}Comprises the following steps: r_{l}＝R_{0}+R_{T}+rl_{K3}
In the formula, R_{0}Is the resistance component of the system impedance when a fault occurs; r_{T}Is 1_{K3}The line resistance value at the initial end of the wire model; r is the resistance per unit length of the conductor pattern.
Wire reactance X between distribution line fault point and line starting end_{l}Comprises the following steps:
X_{l}＝X_{0}+X_{T}+xl_{K3}
in the formula, X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K3}The line reactance value at the beginning of the type of the conductor; x is the reactance value per unit length of the conductor type number.
From the above two formulas, the resistance and reactance of the wire between the fault point of the distribution line and the initial end of the line are respectively l_{K3}Can be respectively expressed as:
R_{l}＝a+rl_{K3} (8)
X_{l}＝b+xl_{K3} (9)
wherein a is determined by_{K3}A constant of line resistance at the beginning of the conductor type, a ═ R_{0}+R_{T}(ii) a b is determined by_{K3}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}(ii) a r and x are each l_{K3}The resistance value and the reactance value of the unit length of the conductor are determined.
Based on the formulae (6) to (9), the compounds can be obtainedConductor resistance R between fault point and line beginning_{l}Reactance X_{l}Respectively as follows:
transition resistance R of fault point_{G}Comprises the following steps:
finally, when threephase short circuit fault occurs, the distance l between the fault point and the initial end of the line is obtained_{K3}Comprises the following steps:
3. fault location method for interphase shortcircuit fault
I. Calculating the current voltage phase angle difference under the fault state
When the distribution lines takes place interphase short circuit fault, transient distortion can take place for the voltage and transition to stable fault state, because distribution lines short circuit capacity is very little for the major network capacity, therefore the distribution lines trouble is minimum to the system influence, and when distribution lines fault state was stable, voltage and current waveform all appeared as the sine wave. When the interphase short circuit occurs, the phase angle difference between the voltage waveforms of the fault current reflects the magnitude relation between the reactance and the resistance of the fault point; the phase angle difference between the fault state current and voltage waveforms during the phasetophase fault is represented on the waveforms, i.e. the time difference between the zero crossing point of the fault current waveform and the zero crossing point of the voltage waveform is not equal to the integral multiple of the period of the alternating voltage, as shown in fig. 3. The phase angle corresponding to the time difference is the phase angle difference of the current and voltage waveforms in the phasetophase fault state.
Distribution line generationAnd when the fault occurs, recording a current and voltage sampling value and corresponding sampling time in the interphase shortcircuit fault process, and processing data in the interphase shortcircuit fault process. Judging voltage sampling data when an interphase shortcircuit fault occurs: if tau_{j}The sampled value at the moment satisfies the following formula:
u(τ_{j}) Less than 0; and u (τ)_{j+1})≥0；
In the formula, u (τ)_{j})、u(τ_{j+1}) Are each tau_{j}、τ_{j+1}The voltage sample value at the time.
When the above conditions are satisfied, the zero crossing point indicating that the voltage changes from a negative value to a positive value is at τ_{j}～τ_{j+1}Between the moments, the moment value tau of the voltage from negative to positive zero crossing point under the interphase shortcircuit fault state can be obtained_{1}Comprises the following steps:
τ_{1}＝τ_{j+1}arcsin(u(τ_{j+1}))/100π
judging current sampling data when an interphase shortcircuit fault occurs: if tau_{k}The sampling value at the moment satisfies:
i(τ_{k}) Less than 0; and i (τ)_{k+1})≥0；
In the formula, i (τ)_{k})、i(τ_{k+1}) Are each tau_{k}、τ_{k+1}Current sample values at time instants.
When the above conditions are satisfied, the zero crossing point indicating that the voltage changes from a negative value to a positive value is at τ_{k}～τ_{k+1}Between the moments, the moment value tau of the current from negative to positive zero crossing point under the interphase shortcircuit fault state can be obtained_{2}Comprises the following steps:
τ_{2}＝τ_{k+1}arcsin(u(τ_{k+1}))/100π
therefore, the time difference delta tau of the zerocrossing point of the current waveform lagging voltage waveform in the phasetophase shortcircuit fault state is as follows: Δ τ ═ τ_{2}τ_{1}
The time difference delta tau is obtained by taking the remainder of the system voltage cycle, and the time difference delta tau corresponding to the phase angle difference is obtained as:
δτ＝mod(Δτ,T)
in the formula, mod (a, b) is a remainder function, which means that parameter a is the remainder of parameter b, and T is the system voltage period, and has a value of 0.02 s.
Thereby obtaining the phase angle difference of the zero crossing point of the current waveform and the lagging voltage waveform in the phasetophase short circuit fault stateComprises the following steps:
in the formula, T is the system voltage period, and its value is 0.02 s.
II. According to the sampling information of the fault current, the effective value of the phasetophase fault current is obtained
The method for solving the phasetophase shortcircuit fault current effective value is the same as the method for solving the threephase fault current effective value, firstly, the moment of entering the fault state is determined, the effective value of the sampled data is calculated in real time according to a halfperiod window and is continuously compared with the data before the half period, and if the ratio of the sampled data to the data before the half period exceeds the limit value, the current at the moment is in the fault state. And then, continuously carrying out effective value calculation on current sampling data after the current sampling data enters the fault state to the moment according to a halfperiod window, continuously comparing the current sampling data with the data before the half period, and if the ratio of the current sampling data to the data before the half period is within a limit value, indicating that the current is in the fault state at the moment, calculating the current effective value at the moment, and obtaining the phasetophase fault current effective value.
The fault state judging method comprises the following steps: calculating current information in a half period in real time, continuously calculating an effective value of a certain moment in the latest half period, taking phase A as an example, and setting a sampling value i at moment t as_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in the system half period, i.e. 0.01s, and is determined by the samplingFrequency f, m is f/100, i_{A}And (t) is a sampling value at the current moment t, and k is a counting parameter.
Note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}For comparison, if the following formula is satisfied:
I_{At}/I_{A(t0.01)}≥1.5
indicating that the time t is in a fault state, continuously calculating and analyzing subsequent sampling values of the current, and if t is the elapsed time after the time t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following formula:
the effective value I of the Aphase fault current is taken_{AK}Comprises the following steps:
in the same way, when an interphase fault occurs, the method for calculating the effective value of the fault current of the other phase is consistent with that of the phase A, and the effective values of the fault currents of the two phases of the interphase shortcircuit fault are equal under the general condition.
III, calculating the distance between the interphase shortcircuit fault point and the initial end of the line
After the phase angle difference of the zero crossing point lagging voltage waveform zero crossing point and the phase shortcircuit fault current effective value are obtained in the phase shortcircuit fault state, the resistance component, the reactance component and the transition resistance between the initial end of the line and the fault point can be obtained.
The calculation principle and method are as follows:
the invention provides a phase angle difference of current waveform zero crossing points lagging voltage waveform zero crossing points in a fault state when a distribution line interphase short circuit fault occursAnd the related parameters satisfy the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}Is the transition resistance of the fault point.
From formula (12):
obtaining the ratio lambda of the reactance and the resistance of the interphase shortcircuit fault point as follows:
meanwhile, assuming that an AB or AC phase interphase short circuit occurs, taking the A phase as an example, the effective value of the Aphase fault current is I when the interphase short circuit fault is calculated in section 5.2_{AK}And therefore, the first and second electrodes are,
in the formula of U_{N}The rated voltage of the distribution line, i.e., 10.5kV, can be obtained according to the equations (13) and (14)
In the formulae (15) and (16), there are three unknowns R_{l}、X_{l}、R_{G}；
Wherein R is_{l}、X_{l}Two unknowns, the distance between the interphase shortcircuit fault point of the power distribution line and the initial end of the lead is set to be l_{K2}When the conductor of the distribution line has impedance, the conductor resistance R between the fault point of the distribution line and the line starting end_{l}Comprises the following steps: r_{l}＝R_{0}+R_{T}+rl_{K2}
In the formula, R_{0}Is the resistance component of the system impedance when a fault occurs; r_{T}Is 1_{K2}The line resistance value at the initial end of the wire model; r is the resistance per unit length of the conductor pattern.
Wire reactance X between distribution line fault point and line starting end_{l}Comprises the following steps:
X_{l}＝X_{0}+X_{T}+xl_{K2}
in the formula, X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K2}The line reactance value at the beginning of the type of the conductor; x is the reactance value per unit length of the conductor type number.
From the above two formulas, the resistance and reactance of the wire between the fault point of the distribution line and the initial end of the line are respectively l_{K2}Can be respectively expressed as:
R_{l}＝a+rl_{K2} (17)
X_{l}＝b+xl_{K2} (18)
wherein a is determined by_{K2}A constant of line resistance at the beginning of the conductor type, a ═ R_{0}+R_{T}(ii) a b is determined by_{K2}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}(ii) a r and x are each l_{K2}The resistance value and the reactance value of the unit length of the conductor are determined.
Based on the expressions (15) to (18), the wire resistance R between the fault point and the line start end can be obtained_{l}Reactance X_{l}Respectively as follows:
transition resistance R of fault point_{G}Comprises the following steps:
finally, when the interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is l_{K2}Comprises the following steps:
Claims (5)
1. a10 kV distribution line fault location method is characterized by comprising the following steps:
(1) determining whether the fault type is a twophase short circuit or a threephase short circuit;
(2) if the fault is a threephase shortcircuit fault, calculating the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state;
(3) if the fault is an interphase shortcircuit fault, calculating a currentvoltage phase angle difference under the fault state;
(4) respectively obtaining a threephase shortcircuit fault current effective value or a twophase shortcircuit fault current effective value according to the sampling information of the fault current;
(5) integrating the information of the distribution line, and calculating the distance between the threephase or twophase shortcircuit fault point and the initial end of the line;
the calculation method of the distance between the threephase shortcircuit fault point and the line starting end is as follows:
when the threephase short circuit of the distribution line is failed, the relationship between the phase angle difference sigma of the voltage in the fault state lagging the voltage in the normal operation state and the related parameters meets the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a threephase fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}A transition resistance that is a point of failure; sigma is the phase angle difference of the voltage in the fault state lagging the voltage in the normal operation state;
is obtained by the formula:
obtaining the ratio eta of the reactance and the resistance of the threephase shortcircuit fault point as follows:
when threephase short circuit fault occurs, the distance l between the fault point and the initial end of the line_{K3}Comprises the following steps:
wherein, I_{AK}The effective value of the Aphase fault current is the effective value of the Aphase fault current in the case of threephase shortcircuit fault; u shape_{N}The rated voltage of the distribution line; x is l_{K3}A unit length reactance value of a position conductor model; b is determined by_{K3}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}；X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K3}The line reactance value at the beginning of the type of the conductor;
the calculation method of the distance between the interphase shortcircuit fault point and the line starting end is as follows:
when the interphase short circuit fault of the distribution line occurs, the current waveform zero crossing point lags the voltage waveform in the fault statePhase angle difference of zero crossing pointAnd the related parameters satisfy the following formula:
in the formula, R_{0}、X_{0}Resistance components and reactance components of system impedance when a fault occurs; r_{l}、X_{l}The resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; r_{G}A transition resistance that is a point of failure;
is obtained by the formula:
obtaining the ratio lambda of the reactance and the resistance of the interphase shortcircuit fault point as follows:
when interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is l_{K2}Comprises the following steps:
wherein, I_{AK}The effective value of Aphase fault current is the effective value of the Aphase fault current in the interphase short circuit fault; u shape_{N}The rated voltage of the distribution line; b is determined by_{K2}Constant of line reactance value at the beginning of the conductor type, b ═ X_{0}+X_{T}；X_{0}Is the reactive component of the system impedance at the time of the fault; x_{T}Is 1_{K2}The line reactance value at the beginning of the type of the conductor; x is the reactance value per unit length of the conductor type number.
2. The method of claim 1, wherein the phase angle difference σ between the voltage in the fault state and the voltage in the normal operation state at the time of the threephase short circuit fault is:
wherein T is the system voltage period and the value is 0.02 s; δ t represents a time difference corresponding to the phase angle difference;
δt＝mod(Δt,T)
in the above formula, mod (a, b) is a remainder function, which means that parameter a is the remainder of parameter b; delta t is the time difference between the zero crossing point of the voltage waveform in the normal operation state and the zero crossing point of the voltage waveform in the fault stabilization state;
Δt＝t_{2}t_{1}
in the above formula, t_{1}The time value is from negative to positive zero crossing point under normal state; t is t_{2}The time value from negative to positive zero crossing point under the fault state;
t_{1}＝t_{k+1}arcsin(u(t_{k+1}))/100π
t_{2}＝t_{j+1}arcsin(u(t_{j+1}))/100π
wherein u (t)_{k+1}) T is in normal operation_{k+1}Voltage sampling value at time u (t)_{k+1}) And t_{k+1}Is sampled at the previous sampling instant t_{k}The voltage sampling value u (t) at a time_{k}) Satisfies the following formula:
u(t_{k}) Less than 0; and u (t)_{k+1})≥0；
u(t_{j+1}) T being a fault condition_{j+1}Voltage sampling value of a time, the sampling value and a previous sampling time t_{j}The voltage sampling value u (t) at a time_{j}) Satisfies the following formula:
u(t_{j}) Less than 0; and u (t)_{j+1})≥0。
3. The method of claim 1, wherein the effective value of the threephase shortcircuit fault current is calculated as follows:
determining the moment of entering a fault state, carrying out effective value calculation on the sampled data in real time according to a halfperiod window, continuously comparing the sampled data with data before a half period, and if the ratio of the sampled data to the data before the half period exceeds a limit value, indicating that the current is in the fault state at the current moment; then, current sampling data after the time of entering the fault state is obtained, effective value calculation is continuously carried out in real time according to a halfperiod window, the data are continuously compared with the data before the half period, if the ratio of the two is within a limit value, the current is shown to be in the fault state at the moment, the current effective value at the moment is calculated, and the threephase fault current effective value is obtained;
and (3) solving an effective value of the Aphase fault current:
let the sampling value at time t be i_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in a system half cycle, namely 0.01s, and is determined by the sampling frequency f, and m is f/100; i.e. i_{A}(t) is a sampling value at the current moment t, and k is a counting parameter;
note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}By comparison, if: i is_{At}/I_{A(t0.01)}≥1.5，
Indicating that the t moment is in a fault state, and continuously calculating and analyzing subsequent sampling values of the current; if t is the time elapsed after t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following conditions:
the effective value I of the Aphase fault current is taken_{AK}Comprises the following steps:
in the same way, the solving method of the effective values of the fault currents of the B phase and the C phase is consistent with that of the A phase.
4. The method of claim 1, wherein the phasetophase short circuit fault is a fault condition in which a current voltage phase angle difference existsComprises the following steps:
in the formula, T is the system voltage period and is 0.02 s; delta tau is a time difference corresponding to the phase angle difference;
δτ＝mod(Δτ,T)
in the formula, mod (a, b) is a remainder function, which represents that a parameter a is the remainder of a parameter b; delta tau is the time difference of the zero crossing point of the lagging voltage waveform of the current waveform in the phasetophase short circuit fault state;
Δτ＝τ_{2}τ_{1}
in the formula, τ_{1}The time value of the voltage from negative to positive zero crossing point under the phasetophase short circuit fault state; tau is_{2}The current is the time value from negative to positive zero crossing point under the condition of interphase short circuit fault;
τ_{1}＝τ_{j+1}arcsin(u(τ_{j+1}))/100π
τ_{2}＝τ_{k+1}arcsin(u(τ_{k+1}))/100π
in the formula, i (τ)_{k+1}) For tau in the course of interphase shortcircuit fault_{k+1}Sampled value of the current at a time, i (τ)_{k+1}) And τ_{k+1}Is the previous sampling instant t_{k}Current sample value i (tau) at a time_{k}) Satisfies the following formula:
i(τ_{k}) Less than 0; and i (τ)_{k+1})≥0；
u(τ_{j+1}) For tau in the course of interphase shortcircuit fault_{j+1}Voltage sample value at a time u (tau)_{j+1}) And τ_{j+1}Is the previous sampling instant t_{j}The voltage sample u (tau) at a time_{j}) Satisfies the following formula:
u(τ_{j}) Less than 0; and u (τ)_{j+1})≥0。
5. The method of claim 1, wherein the effective interphase shortcircuit fault current value is calculated as follows:
determining the moment of entering a fault state, carrying out effective value calculation on the sampled data in real time according to a halfperiod window, continuously comparing the sampled data with data before a half period, and if the ratio of the sampled data to the data before the half period exceeds a limit value, indicating that the current is in the fault state at the current moment; then, current sampling data after the current enters the fault state to the moment are subjected to effective value calculation continuously in real time according to a halfperiod window and are continuously compared with data before the half period, if the ratio of the two is within a limit value, the current is indicated to be in the fault state at the moment, the current effective value at the moment is calculated, and the phasetophase fault current effective value is obtained;
the algorithm of the effective value of the Aphase fault current is as follows:
let the sampling value at time t be i_{A}(t), effective value I in the most recent half cycle of the current_{At}Comprises the following steps:
wherein f is the current sampling frequency, m is the sampling number in 0.01s which is the system half period, and is determined by the sampling frequency f, m is f/100, i_{A}(t) is a sampling value at the current moment t, and k is a counting parameter; note that the effective value in the most recent half cycle of time (t0.01s) is I_{A(t0.01)}Effective value I in the last half cycle of time t_{At}By comparison, if: i is_{At}/I_{A(t0.01)}≥1.5
Indicating that the time t is in a fault state, continuously calculating and analyzing subsequent sampling values of the current, and if t is the elapsed time after the time t_{0}At time t + t_{0}Valid value in the last half cycleSatisfies the following conditions:
the effective value I of the Aphase fault current is taken_{AK}Is composed of
In the same way, when an interphase fault occurs, the method for calculating the effective value of the fault current of the other phase is consistent with that of the phase A, and the effective values of the fault currents of the two phases of the interphase shortcircuit fault are equal under the general condition.
Priority Applications (1)
Application Number  Priority Date  Filing Date  Title 

CN201910139581.XA CN109975653B (en)  20190226  20190226  10kV distribution line fault location method 
Applications Claiming Priority (1)
Application Number  Priority Date  Filing Date  Title 

CN201910139581.XA CN109975653B (en)  20190226  20190226  10kV distribution line fault location method 
Publications (2)
Publication Number  Publication Date 

CN109975653A CN109975653A (en)  20190705 
CN109975653B true CN109975653B (en)  20210302 
Family
ID=67077305
Family Applications (1)
Application Number  Title  Priority Date  Filing Date 

CN201910139581.XA Active CN109975653B (en)  20190226  20190226  10kV distribution line fault location method 
Country Status (1)
Country  Link 

CN (1)  CN109975653B (en) 
Families Citing this family (2)
Publication number  Priority date  Publication date  Assignee  Title 

CN110456219A (en) *  20190812  20191115  国网江西省电力有限公司电力科学研究院  A kind of appraisal procedure of distribution line short voltage dip degree 
CN112946416A (en) *  20210121  20210611  国网山东省电力公司沂南县供电公司  Distribution network line fault range distinguishing method and system 
Family Cites Families (10)
Publication number  Priority date  Publication date  Assignee  Title 

CH613570A5 (en) *  19760325  19790928  Bbc Brown Boveri & Cie  
JPS5829471B2 (en) *  19781030  19830622  Tokyo Denryoku Kk  
JPH065255B2 (en) *  19851025  19940119  東京電力株式会社  Fault location method for power transmission system 
CN1100997C (en) *  20000915  20030205  清华大学  Method and equipment for positioning failure point on electric power transmission line 
CN1142443C (en) *  20011228  20040317  清华大学  Highaccuracy failure waverecording device and its transmission line combined failure distancemeasuring method 
CN102570428B (en) *  20120229  20140716  山东电力集团公司莱芜供电公司  Fault location and distance protection method based on differential output of electronic mutual inductor 
CN103149502B (en) *  20130220  20150812  保定浪拜迪电气股份有限公司  Based on the fault positioning method for transmission line of synchronized sampling unit 
CN103226175A (en) *  20130321  20130731  江苏省电力公司泰州供电公司  Method for achieving doubleended ranging by virtue of resistance characteristics of ground resistor 
CN105445622B (en) *  20151228  20190625  武汉精伦电气有限公司  A kind of resistance algorithm method based on prediction phaseshifting technique improvement 
CN109270402B (en) *  20180919  20220304  中国电力科学研究院有限公司  Timelimited distance protection auxiliary ranging method and system for series compensation circuit 

2019
 20190226 CN CN201910139581.XA patent/CN109975653B/en active Active
Also Published As
Publication number  Publication date 

CN109975653A (en)  20190705 
Similar Documents
Publication  Publication Date  Title 

CN103018627B (en)  Adaptive fault type fault line detection method for noneffectively earthed system  
CN108957244B (en)  Singlephase earth fault line selection positioning method for distribution network main station  
CN109975653B (en)  10kV distribution line fault location method  
CN103226176B (en)  A kind of wire selection method for power distribution network single phase earthing failure  
CN109283430B (en)  Power distribution network fault location method based on voltage distribution principle  
CN104777397A (en)  Distribution line singlephase break line judgment and positioning method based on line voltage vector criterion  
Styvaktakis  Automating power quality analysis  
CN107102236A (en)  A kind of fault line selection method for singlephasetoground fault based on waveform correlation analysis after failure  
CN108254657A (en)  Power distribution network section with lowcurrent ground faults localization method based on Study of Transient Energy  
CN111077471B (en)  NPC threelevel inverter opencircuit fault diagnosis method based on instantaneous frequency  
CN107045093B (en)  Lowcurrent singlephase earth fault line selection method based on quick Stransformation  
CN105067948A (en)  Smallcurrent grounding line selection device and singlephase grounding detection method  
CN107526010A (en)  A kind of distributed small current earthing wireselecting method based on double CT samplings  
CN105974254B (en)  The temporary stable state selection method of comprehensive weight is calculated based on voltage  
CN105203911A (en)  Open conductor fault detection method and device for threephase power source and automatic transfer switch  
CN108490310A (en)  Method for small electric current grounding system of distribution network earth fault line selection  
CN107765076B (en)  Magnetizing inrush current identification method and device  
CN108152680B (en)  Method for detecting commutation failure of directcurrent transmission  
CN107179465B (en)  Extrahigh voltage direct current phase selection closing device performance and secondary circuit field test method  
CN109964136B (en)  Method and control system for fault direction detection  
CN110412400B (en)  PT broken line and PT threephase reverse sequence fault judgment method and fault protection device  
CN112505490A (en)  Power distribution network singlephase earth fault line selection method based on mutual difference absolute value sum  
CN112034307A (en)  Cable early fault detection method based on stationary wavelet transform and symmetric component method  
CN112649694A (en)  Method for judging singlephase earth fault of smallcurrent grounding system  
CN111157827A (en)  Method for detecting state of direct current converter valve based on port current time sequence characteristics 
Legal Events
Date  Code  Title  Description 

PB01  Publication  
PB01  Publication  
SE01  Entry into force of request for substantive examination  
SE01  Entry into force of request for substantive examination  
GR01  Patent grant  
GR01  Patent grant 