CN109975653B - 10kV distribution line fault location method - Google Patents

10kV distribution line fault location method Download PDF

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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
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fault
phase
current
value
voltage
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CN109975653A (en
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郭亮
范瑞祥
李升健
王华云
郑蜀江
刘蓓
陈琛
戚沁雅
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Nanchang Kechen Electric Power Test Research Co ltd
State Grid Corp of China SGCC
Electric Power Research Institute of State Grid Jiangxi Electric Power Co Ltd
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Nanchang Kechen Electric Power Test Research Co ltd
State Grid Corp of China SGCC
Electric Power Research Institute of State Grid Jiangxi Electric Power Co Ltd
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    • 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. root-mean-square 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
    • 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
    • G01R31/50—Testing of electric apparatus, lines, cables or components for short-circuits, continuity, leakage current or incorrect line connections

Abstract

A10 kV distribution line fault location method comprises the following steps: (1) determining whether the fault type is a two-phase short circuit or a three-phase short circuit; (2) if the fault is a three-phase short-circuit 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 short-circuit fault, calculating a current-voltage phase angle difference under the fault state; (4) respectively obtaining a three-phase short-circuit fault current effective value or a two-phase short-circuit 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 three-phase or two-phase short-circuit 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

10kV distribution line fault location method
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 two-phase short circuit or a three-phase short circuit;
(2) if the fault is a three-phase short-circuit 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 short-circuit fault, calculating a current-voltage phase angle difference under the fault state;
(4) respectively obtaining a three-phase short-circuit fault current effective value or a two-phase short-circuit 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 three-phase short-circuit fault point and the initial end of the line or the distance between the two-phase short-circuit fault point and the initial end of the line.
When the three-phase 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:
Figure GDA0002804038460000021
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＝|t2-t1|
in the above formula, t1The time value is from negative to positive zero crossing point under normal state; t is t2The time value from negative to positive zero crossing point under the fault state;
t1＝tk+1-arcsin(u(tk+1))/100π
t2＝tj+1-arcsin(u(tj+1))/100π
wherein u (t)k+1) T is in normal operationk+1Voltage sampling value at time u (t)k+1) And tk+1Is sampled at the previous sampling instant tkThe voltage sampling value u (t) at a timek) Satisfies the following formula:
u(tk) Less than 0; and u (t)k+1)≥0；
u(tj+1) T being a fault conditionj+1Voltage sampling value of a time, the sampling value and a previous sampling time tjThe voltage sampling value u (t) at a timej) Satisfies the following formula:
u(tj) Less than 0; and u (t)j+1)≥0。
The method for calculating the effective value of the three-phase short-circuit 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 half-period 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 half-period 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 three-phase fault current effective value is obtained;
and (3) solving an effective value of the A-phase fault current:
let the sampling value at time t be iA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure GDA0002804038460000031
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. iA(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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtBy comparison, if: i isAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure GDA0002804038460000032
Satisfies the following conditions:
Figure GDA0002804038460000033
the effective value I of the A-phase fault current is takenAKComprises the following steps:
Figure GDA0002804038460000034
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 three-phase short-circuit fault point and the line starting end is as follows:
when the three-phase 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:
Figure GDA0002804038460000041
in the formula, R0、X0Resistance components and reactance components of system impedance when a three-phase fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGA 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:
Figure GDA0002804038460000042
obtaining the ratio eta of the reactance and the resistance of the three-phase short-circuit fault point as follows:
Figure GDA0002804038460000043
when three-phase short circuit fault occurs, the distance l between the fault point and the initial end of the lineK3Comprises the following steps:
Figure GDA0002804038460000044
wherein, IAKThe effective value of the A-phase fault current is the effective value of the A-phase fault current in the case of three-phase short-circuit fault; u shapeNThe rated voltage of the distribution line; x is lK3A unit length reactance value of a position conductor model; b is determined byK3Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT；X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K3And 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 state
Figure GDA0002804038460000045
Comprises the following steps:
Figure GDA0002804038460000046
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 phase-to-phase short circuit fault state;
Δτ＝|τ2-τ1|
in the formula, τ1The time value of the voltage from negative to positive zero crossing point under the phase-to-phase short circuit fault state; tau is2The current is the time value from negative to positive zero crossing point under the condition of interphase short circuit fault;
τ1＝τj+1-arcsin(u(τj+1))/100π
τ2＝τk+1-arcsin(u(τk+1))/100π
in the formula, i (τ)k+1) For tau in the course of interphase short-circuit faultk+1Sampled value of the current at a time, i (τ)k+1) And τk+1Is the previous sampling instant tkCurrent sample value i (tau) at a timek) Satisfies the following formula:
i(τk) Less than 0; and i (τ)k+1)≥0；
u(τj+1) For tau in the course of interphase short-circuit faultj+1Voltage sample value at a time u (tau)j+1) And τj+1Is the previous sampling instant tjThe voltage sample u (tau) at a timej) Satisfies the following formula:
u(τj) Less than 0; and u (τ)j+1)≥0。
The calculation method of the interphase short-circuit 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 half-period 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 half-period 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 phase-to-phase fault current effective value is obtained;
the algorithm of the effective value of the A-phase fault current is as follows:
let the sampling value at time t be iA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure GDA0002804038460000051
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, iA(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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtBy comparison, if: i isAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure GDA0002804038460000061
Satisfies the following conditions:
Figure GDA0002804038460000062
the effective value I of the A-phase fault current is takenAKIs composed of
Figure GDA0002804038460000063
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 short-circuit fault are equal under the general condition.
The calculation method of the distance between the interphase short-circuit 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 line
Figure GDA0002804038460000064
And the related parameters satisfy the following formula:
Figure GDA0002804038460000065
in the formula, R0、X0Resistance components and reactance components of system impedance when a fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGA transition resistance that is a point of failure;
is obtained by the formula:
Figure GDA0002804038460000066
obtaining the ratio lambda of the reactance and the resistance of the interphase short-circuit fault point as follows:
Figure GDA0002804038460000067
when interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is lK2Comprises the following steps:
Figure GDA0002804038460000071
wherein, IAKThe effective value of A-phase fault current is the effective value of the A-phase fault current in the interphase short circuit fault; u shapeNThe rated voltage of the distribution line; b is determined byK2Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT；X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K2The 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 three-phase short-circuit fault; for the interphase short-circuit 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 phase-to-phase 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 zero-crossing point and a lagging in-phase voltage zero-crossing 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 two-phase short circuit or a three-phase short circuit according to the sampling information of the fault current and voltage;
(2) if the fault is a three-phase short-circuit 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 short-circuit fault, calculating a current-voltage 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 short-circuit 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 three-phase 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 two-phase voltage is less than 0.9 times of the rated value at the same time and the two-phase current is greater than 800A, determining that the internal inter-phase short circuit fault occurs in the line.
The conditions are (taking AB phase two-phase interphase short circuit as an example):
Figure GDA0002804038460000091
and is
Figure GDA0002804038460000092
In the formula of UA、UBRespectively obtaining A, B phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000093
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, uAk、uBkK is a counting parameter for the voltage instantaneous sample value.
IA、IBRespectively obtaining A, B phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000094
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, iAk、iBkK is a counting parameter for the current instantaneous sample value.
(2) Off-site two-phase short circuit fault
And if the effective value of the two-phase 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 two-phase short-circuit fault in different places.
The conditions are (taking A as an example of the interphase short circuit) as follows:
Figure GDA0002804038460000095
and is
Figure GDA0002804038460000096
In the formula of UA、UBRespectively obtaining A, B phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000097
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, uAk、uBkK is a counting parameter for the voltage instantaneous sample value.
IA、IBRespectively obtaining A, B phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000101
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, iAk、iBkK is a counting parameter for the current instantaneous sample value.
(3) Three-phase short-circuit fault
And if the effective value of the three-phase voltage is simultaneously less than 0.9 time of the rated value and at least one phase current of the three-phase current is more than 800A, determining that the three-phase short circuit fault occurs in the line. The conditions are as follows:
Figure GDA0002804038460000102
and IANot less than 800A, or IBNot less than 800A, or IC≥800A
In the formula of UA、UB、UCRespectively obtaining A, B, C phase voltage effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000103
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, uAk、uBk、uCkK is a counting parameter for the voltage instantaneous sample value.
IA、IB、ICRespectively obtaining A, B, C phase current effective values obtained by sampling, wherein the calculation formulas are respectively as follows:
Figure GDA0002804038460000104
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, iAk、iBk、iCkK 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 low-voltage 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 three-phase short-circuit 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 three-phase 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 tkThe sampled value at the moment satisfies the following formula:
u(tk) Less than 0; and u (t)k+1)≥0；
In the formula, u (t)k)、u(tk+1) Are each tk、tk+1The 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 tk～tk+1Between the moments, the value t of the moment from negative to positive zero crossing in the normal state can be determined1Comprises the following steps:
t1＝tk+1-arcsin(u(tk+1))/100π
similarly, the data of the fault state is judged: if tjThe sampled value at the moment satisfies the following formula:
u(tj) Less than 0; and u (t)j+1)≥0；
In the formula, u (t)j)、u(tj+1) Are each tj、tj+1The 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 tj～tj+1Between the moments, the value of the moment t from negative to positive zero crossing in the fault state can be determined2Comprises the following steps:
t2＝tj+1-arcsin(u(tj+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 ═ t2-t1|
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:
Figure GDA0002804038460000121
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 half-period 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 half-period 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 three-phase fault current effective value.
The fault state judging method comprises the following steps: real-time 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 tA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure GDA0002804038460000131
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. iA(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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtFor comparison, if the following formula is satisfied:
IAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure GDA0002804038460000132
Satisfies the following formula:
Figure GDA0002804038460000133
the effective value I of the A-phase fault current is takenAKComprises the following steps:
Figure GDA0002804038460000134
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 three-phase short-circuit 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 three-phase 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:
Figure GDA0002804038460000141
in the formula, R0、X0Resistance components and reactance components of system impedance when a fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGIs 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)
Figure GDA0002804038460000142
Obtaining the ratio eta of the reactance and the resistance of the three-phase short-circuit fault point as follows:
Figure GDA0002804038460000143
meanwhile, taking phase A as an example, section 4.2 calculates the effective value of the A-phase fault current when the three-phase short-circuit fault occurs to be IAKThus, therefore, it is
Figure GDA0002804038460000144
In the formula of UNThe rated voltage of the distribution line, i.e. 10.5kV, can be obtained according to the equations (4) and (5)
Figure GDA0002804038460000145
Figure GDA0002804038460000146
In the formulae (6) and (7), there are three unknowns Rl、Xl、RG；
Wherein R isl、XlTwo unknowns, the distance between the three-phase short-circuit fault point of the power distribution line and the initial end of the lead is set to be lK3When the conductor of the distribution line has impedance, the conductor resistance R between the fault point of the distribution line and the line starting endlComprises the following steps: rl＝R0+RT+rlK3
In the formula, R0Is the resistance component of the system impedance when a fault occurs; rTIs 1K3The 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 endlComprises the following steps:
Xl＝X0+XT+xlK3
in the formula, X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K3The 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 lK3Can be respectively expressed as:
Rl＝a+rlK3 (8)
Xl＝b+xlK3 (9)
wherein a is determined byK3A constant of line resistance at the beginning of the conductor type, a ═ R0+RT(ii) a b is determined byK3Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT(ii) a r and x are each lK3The 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 beginninglReactance XlRespectively as follows:
Figure GDA0002804038460000151
Figure GDA0002804038460000152
transition resistance R of fault pointGComprises the following steps:
Figure GDA0002804038460000153
finally, when three-phase short circuit fault occurs, the distance l between the fault point and the initial end of the line is obtainedK3Comprises the following steps:
Figure GDA0002804038460000154
3. fault location method for interphase short-circuit 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 phase-to-phase 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 phase-to-phase fault state.
Distribution line generationAnd when the fault occurs, recording a current and voltage sampling value and corresponding sampling time in the interphase short-circuit fault process, and processing data in the interphase short-circuit fault process. Judging voltage sampling data when an interphase short-circuit fault occurs: if taujThe sampled value at the moment satisfies the following formula:
u(τj) Less than 0; and u (τ)j+1)≥0；
In the formula, u (τ)j)、u(τj+1) Are each tauj、τj+1The 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 τj～τj+1Between the moments, the moment value tau of the voltage from negative to positive zero crossing point under the interphase short-circuit fault state can be obtained1Comprises the following steps:
τ1＝τj+1-arcsin(u(τj+1))/100π
judging current sampling data when an interphase short-circuit fault occurs: if taukThe sampling value at the moment satisfies:
i(τk) Less than 0; and i (τ)k+1)≥0；
In the formula, i (τ)k)、i(τk+1) Are each tauk、τk+1Current 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 τk～τk+1Between the moments, the moment value tau of the current from negative to positive zero crossing point under the interphase short-circuit fault state can be obtained2Comprises the following steps:
τ2＝τk+1-arcsin(u(τk+1))/100π
therefore, the time difference delta tau of the zero-crossing point of the current waveform lagging voltage waveform in the phase-to-phase short-circuit fault state is as follows: Δ τ ═ τ2-τ1|
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 phase-to-phase short circuit fault state
Figure GDA0002804038460000171
Comprises the following steps:
Figure GDA0002804038460000172
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 phase-to-phase fault current is obtained
The method for solving the phase-to-phase short-circuit fault current effective value is the same as the method for solving the three-phase 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 half-period 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 half-period 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 phase-to-phase 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 asA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure GDA0002804038460000181
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, iAAnd (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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtFor comparison, if the following formula is satisfied:
IAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure GDA0002804038460000182
Satisfies the following formula:
Figure GDA0002804038460000183
the effective value I of the A-phase fault current is takenAKComprises the following steps:
Figure GDA0002804038460000184
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 short-circuit fault are equal under the general condition.
III, calculating the distance between the interphase short-circuit 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 short-circuit fault current effective value are obtained in the phase short-circuit 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 occurs
Figure GDA0002804038460000185
And the related parameters satisfy the following formula:
Figure GDA0002804038460000191
in the formula, R0、X0Resistance components and reactance components of system impedance when a fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGIs the transition resistance of the fault point.
From formula (12):
Figure GDA0002804038460000192
obtaining the ratio lambda of the reactance and the resistance of the interphase short-circuit fault point as follows:
Figure GDA0002804038460000193
meanwhile, assuming that an AB or AC phase interphase short circuit occurs, taking the A phase as an example, the effective value of the A-phase fault current is I when the interphase short circuit fault is calculated in section 5.2AKAnd therefore, the first and second electrodes are,
Figure GDA0002804038460000194
in the formula of UNThe rated voltage of the distribution line, i.e., 10.5kV, can be obtained according to the equations (13) and (14)
Figure GDA0002804038460000195
Figure GDA0002804038460000196
In the formulae (15) and (16), there are three unknowns Rl、Xl、RG；
Wherein R isl、XlTwo unknowns, the distance between the inter-phase short-circuit fault point of the power distribution line and the initial end of the lead is set to be lK2When the conductor of the distribution line has impedance, the conductor resistance R between the fault point of the distribution line and the line starting endlComprises the following steps: rl＝R0+RT+rlK2
In the formula, R0Is the resistance component of the system impedance when a fault occurs; rTIs 1K2The 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 endlComprises the following steps:
Xl＝X0+XT+xlK2
in the formula, X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K2The 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 lK2Can be respectively expressed as:
Rl＝a+rlK2 (17)
Xl＝b+xlK2 (18)
wherein a is determined byK2A constant of line resistance at the beginning of the conductor type, a ═ R0+RT(ii) a b is determined byK2Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT(ii) a r and x are each lK2The 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 obtainedlReactance XlRespectively as follows:
Figure GDA0002804038460000201
Figure GDA0002804038460000202
transition resistance R of fault pointGComprises the following steps:
Figure GDA0002804038460000203
finally, when the interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is lK2Comprises the following steps:
Figure GDA0002804038460000204

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 two-phase short circuit or a three-phase short circuit;
(2) if the fault is a three-phase short-circuit 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 short-circuit fault, calculating a current-voltage phase angle difference under the fault state;
(4) respectively obtaining a three-phase short-circuit fault current effective value or a two-phase short-circuit 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 three-phase or two-phase short-circuit fault point and the initial end of the line;
the calculation method of the distance between the three-phase short-circuit fault point and the line starting end is as follows:
when the three-phase 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:
Figure FDA0002804038450000011
in the formula, R0、X0Resistance components and reactance components of system impedance when a three-phase fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGA 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:
Figure FDA0002804038450000012
obtaining the ratio eta of the reactance and the resistance of the three-phase short-circuit fault point as follows:
Figure FDA0002804038450000013
when three-phase short circuit fault occurs, the distance l between the fault point and the initial end of the lineK3Comprises the following steps:
Figure FDA0002804038450000021
wherein, IAKThe effective value of the A-phase fault current is the effective value of the A-phase fault current in the case of three-phase short-circuit fault; u shapeNThe rated voltage of the distribution line; x is lK3A unit length reactance value of a position conductor model; b is determined byK3Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT；X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K3The line reactance value at the beginning of the type of the conductor;
the calculation method of the distance between the interphase short-circuit 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 point
Figure FDA0002804038450000022
And the related parameters satisfy the following formula:
Figure FDA0002804038450000023
in the formula, R0、X0Resistance components and reactance components of system impedance when a fault occurs; rl、XlThe resistance and reactance of a wire between a fault point of the distribution line and the initial end of the line; rGA transition resistance that is a point of failure;
is obtained by the formula:
Figure FDA0002804038450000024
obtaining the ratio lambda of the reactance and the resistance of the interphase short-circuit fault point as follows:
Figure FDA0002804038450000025
when interphase short circuit fault occurs, the distance between the fault point and the initial end of the lead is lK2Comprises the following steps:
Figure FDA0002804038450000026
wherein, IAKThe effective value of A-phase fault current is the effective value of the A-phase fault current in the interphase short circuit fault; u shapeNThe rated voltage of the distribution line; b is determined byK2Constant of line reactance value at the beginning of the conductor type, b ═ X0+XT；X0Is the reactive component of the system impedance at the time of the fault; xTIs 1K2The 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 three-phase short circuit fault is:
Figure FDA0002804038450000031
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＝|t2-t1|
in the above formula, t1The time value is from negative to positive zero crossing point under normal state; t is t2The time value from negative to positive zero crossing point under the fault state;
t1＝tk+1-arcsin(u(tk+1))/100π
t2＝tj+1-arcsin(u(tj+1))/100π
wherein u (t)k+1) T is in normal operationk+1Voltage sampling value at time u (t)k+1) And tk+1Is sampled at the previous sampling instant tkThe voltage sampling value u (t) at a timek) Satisfies the following formula:
u(tk) Less than 0; and u (t)k+1)≥0；
u(tj+1) T being a fault conditionj+1Voltage sampling value of a time, the sampling value and a previous sampling time tjThe voltage sampling value u (t) at a timej) Satisfies the following formula:
u(tj) Less than 0; and u (t)j+1)≥0。
3. The method of claim 1, wherein the effective value of the three-phase short-circuit 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 half-period 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 half-period 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 three-phase fault current effective value is obtained;
and (3) solving an effective value of the A-phase fault current:
let the sampling value at time t be iA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure FDA0002804038450000041
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. iA(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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtBy comparison, if: i isAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure FDA0002804038450000042
Satisfies the following conditions:
Figure FDA0002804038450000043
the effective value I of the A-phase fault current is takenAKComprises the following steps:
Figure FDA0002804038450000044
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 phase-to-phase short circuit fault is a fault condition in which a current voltage phase angle difference exists
Figure FDA0002804038450000045
Comprises the following steps:
Figure FDA0002804038450000046
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 phase-to-phase short circuit fault state;
Δτ＝|τ2-τ1|
in the formula, τ1The time value of the voltage from negative to positive zero crossing point under the phase-to-phase short circuit fault state; tau is2The current is the time value from negative to positive zero crossing point under the condition of interphase short circuit fault;
τ1＝τj+1-arcsin(u(τj+1))/100π
τ2＝τk+1-arcsin(u(τk+1))/100π
in the formula, i (τ)k+1) For tau in the course of interphase short-circuit faultk+1Sampled value of the current at a time, i (τ)k+1) And τk+1Is the previous sampling instant tkCurrent sample value i (tau) at a timek) Satisfies the following formula:
i(τk) Less than 0; and i (τ)k+1)≥0；
u(τj+1) For tau in the course of interphase short-circuit faultj+1Voltage sample value at a time u (tau)j+1) And τj+1Is the previous sampling instant tjThe voltage sample u (tau) at a timej) Satisfies the following formula:
u(τj) Less than 0; and u (τ)j+1)≥0。
5. The method of claim 1, wherein the effective interphase short-circuit 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 half-period 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 half-period 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 phase-to-phase fault current effective value is obtained;
the algorithm of the effective value of the A-phase fault current is as follows:
let the sampling value at time t be iA(t), effective value I in the most recent half cycle of the currentAtComprises the following steps:
Figure FDA0002804038450000061
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, iA(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 (t-0.01s) is IA(t-0.01)Effective value I in the last half cycle of time tAtBy comparison, if: i isAt/IA(t-0.01)≥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 t0At time t + t0Valid value in the last half cycle
Figure FDA0002804038450000062
Satisfies the following conditions:
Figure FDA0002804038450000063
the effective value I of the A-phase fault current is takenAKIs composed of
Figure FDA0002804038450000064
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 short-circuit fault are equal under the general condition.
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