CN103018635A - Fault distance detection method for transmission line containing series compensation element - Google Patents

Fault distance detection method for transmission line containing series compensation element Download PDF

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Publication number
CN103018635A
CN103018635A CN2012105381195A CN201210538119A CN103018635A CN 103018635 A CN103018635 A CN 103018635A CN 2012105381195 A CN2012105381195 A CN 2012105381195A CN 201210538119 A CN201210538119 A CN 201210538119A CN 103018635 A CN103018635 A CN 103018635A
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fault
electric current
spare
arc
voltage
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束洪春
蒋彪
董俊
田鑫萃
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Kunming University of Science and Technology
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Kunming University of Science and Technology
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Abstract

The invention relates to a fault distance detection method for a transmission line containing a series compensation element and belongs to the technical field of alternating current transmission line fault distance detection. Within a fault side, voltage and current on the right side of a fault point are calculated by deduction by means of electrical capacity at the measurement end and fault edge conditions, and current on the left side of the series compensation element is calculated by deduction; and simultaneously, within a non-fault side, current on the right side of the series compensation element is calculated by deduction by means of the electrical capacity at the measurement end. A fault location function is obtained according to the edge condition that currents on two sides of the series compensation element are the same, and accordingly the fault distance is worked out. According to the method, fault types do not need judging first for resistance faults and arc faults, a unified distance detection principle is adopted, the faults can be located accurately, and the method is good in reliability, enables calculation to be simple, and is high in practicality.

Description

Go here and there the fault positioning method for transmission line of complement spare a kind of containing
Technical field
The present invention relates to go here and there the fault positioning method for transmission line of complement spare a kind of containing, belong to transmission line of alternation current fault localization technical field.
Background technology
Because it is more and more extensive to contain the application of string complement spare transmission line of electricity, also become the object of recent domestic scholar extensive concern and research about the problem of its fault localization.If string complement spare is installed in the circuit two ends, the voltage at serial compensation capacitance two ends, electric current can directly measure, and then available general fault positioning method for transmission line carries out localization of fault.Otherwise when string complement spare was installed in the circuit middle part, the existence of string complement spare had destroyed the homogeneity of transmission line of electricity, and general distance-finding method is no longer valid to the transmission line of electricity that string complement spare has been installed.Existing many researchs about serial compensation capacitance line fault location are broadly divided into two classes according to its range measurement principle, i.e. traveling wave method and fault analytical method at present.Wherein fault analytical method is listed the range finding equation according to system's relevant parameters, the boundary condition of string benefit device and voltage, the electric current of measuring end, be generally voltage equation, current equation is also arranged, then it is carried out analytical calculation, obtain the trouble spot to distance between the measuring end.
But this string complement spare transmission line of electricity distance accuracy is low, and calculation of complex, practicality are not strong, therefore, necessary prior art is improved.
In order to overcome the low deficiency of tradition string complement spare transmission line of electricity distance accuracy, improve distance accuracy, the present invention is on the basis of Bei Jielong distributed parameter transmission line model, in conjunction with fault section measuring end electric parameters and failure boundary condition, derivation calculates string complement spare one side electric current in the fault section, derived by the measuring end electric parameters that perfects the section side simultaneously and calculate string complement spare opposite side electric current, equate this border according to string complement spare both sides electric current, can draw the localization of fault criterion.By this, propose to go here and there the fault positioning method for transmission line of complement spare a kind of containing.
Summary of the invention
In order to overcome the problem of the low deficiency of tradition string complement spare transmission line of electricity distance accuracy, the present invention proposes to go here and there the fault positioning method for transmission line of complement spare a kind of containing, on the basis of Bei Jielong distributed parameter transmission line model, in conjunction with fault section measuring end electric parameters and failure boundary condition, derivation calculates string complement spare one side electric current in the fault section, derived by the measuring end electric parameters that perfects the section side simultaneously and calculate string complement spare opposite side electric current, equate this border according to string complement spare both sides electric current, can draw the localization of fault criterion, can eliminate string complement spare to the impact of transmission line of electricity range finding, realize fast finding line fault, simultaneously, this method does not need first failure judgement type for Resistance Fault and arc fault, adopt unified range measurement principle, can be accurately to localization of fault, good reliability, calculate simple, practical.
Technical scheme of the present invention is: go here and there the fault positioning method for transmission line of complement spare a kind of containing, in the fault side, by measuring end electric parameters and failure boundary condition, deriving calculates right side, trouble spot voltage and current, and then derivation calculates string complement spare left side electric current; In non-fault side, calculate string complement spare right side electric current by the derivation of measuring end electric parameters simultaneously.Equate this border according to string complement spare both sides electric current, draw the localization of fault function, thereby calculate fault distance.
The concrete steps of fault positioning method for transmission line are as follows:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N , will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot
Figure 2012105381195100002DEST_PATH_IMAGE001
With left side, trouble spot electric current
Figure 567575DEST_PATH_IMAGE002
, calculate right side, trouble spot electric current according to the boundary condition of different faults respectively again
Figure 2012105381195100002DEST_PATH_IMAGE003
:
(1.1) for the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current
Figure 917785DEST_PATH_IMAGE004
For:
Figure 2012105381195100002DEST_PATH_IMAGE005
(1)
Wherein: Be right side, trouble spot electric current,
Figure 2012105381195100002DEST_PATH_IMAGE007
Be left side, trouble spot electric current,
Figure 753465DEST_PATH_IMAGE008
Be fault resistance.
(1.2) for the arcing ground fault, the boundary condition of arcing ground fault is:
Figure 2012105381195100002DEST_PATH_IMAGE009
； (2)
Figure 130089DEST_PATH_IMAGE010
(3)
Wherein,
Figure 2012105381195100002DEST_PATH_IMAGE011
Be the B phase current,
Figure 656010DEST_PATH_IMAGE012
Be the C phase current,
Figure 2012105381195100002DEST_PATH_IMAGE013
Be A phase fault point right side electric current,
Figure 159804DEST_PATH_IMAGE014
Be A phase fault point left side electric current,
Figure 2012105381195100002DEST_PATH_IMAGE015
Be fault point voltage,
Figure 586106DEST_PATH_IMAGE016
Be arc voltage;
This moment right side, trouble spot electric current
Figure 946680DEST_PATH_IMAGE004
For:
Figure 2012105381195100002DEST_PATH_IMAGE017
(4)
(2) again will
Figure 146848DEST_PATH_IMAGE004
,
Figure 454333DEST_PATH_IMAGE018
Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 2012105381195100002DEST_PATH_IMAGE019
; (5)
Wherein:
Figure 971027DEST_PATH_IMAGE020
; (6)
Figure 2012105381195100002DEST_PATH_IMAGE021
; (7)
Figure 955033DEST_PATH_IMAGE022
; (8)
Figure DEST_PATH_IMAGE023
(9)
Wherein, r,
Figure 501552DEST_PATH_IMAGE024
, vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length.
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Figure DEST_PATH_IMAGE025
Equate this border according to string complement spare both sides electric current, that is:
Figure 363460DEST_PATH_IMAGE026
(10)
So draw the localization of fault criterion be:
Figure DEST_PATH_IMAGE027
(11)
(3) calculate fault distance according to the localization of fault criterion;
(3.1) for Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f
(3.2) for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u Arc
For the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 498775DEST_PATH_IMAGE028
(12)
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l; (13)
Wherein: lBe total track length.
R f >0; (14)
u arc>0. (15)
(4) use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f
During described measurement transmission line of electricity both end sides electric current and voltage, the length of short data window is 3ms, and sample frequency is 10kHz.
Principle of the present invention is: break down when containing string complement spare transmission line of electricity left side section, in conjunction with fault section measuring end electric parameters and failure boundary condition, deriving calculates trouble spot string benefit component side electric current
Figure 404414DEST_PATH_IMAGE004
Then mend the derivation of component side electric current by fault point voltage and trouble spot string and calculate string complement spare left side P 1Place's electric current
Figure DEST_PATH_IMAGE029
, derived by the measuring end electric parameters that perfects the section side simultaneously and calculate string complement spare right side P 2Place's electric current
Figure 138363DEST_PATH_IMAGE025
, equate this border according to string complement spare both sides electric current, that is:
Figure 36918DEST_PATH_IMAGE030
, can draw the localization of fault criterion, by criterion function operation least square method and linear search are calculated optimum fault distance; Break down when containing string complement spare transmission line of electricity right side section, distance-finding method in like manner.
Beneficial effect of the present invention: the present invention is on the basis of Bei Jielong distributed parameter transmission line model, in conjunction with fault section measuring end electric parameters and failure boundary condition, derivation calculates string complement spare one side electric current in the fault section, derived by the measuring end electric parameters that perfects the section side simultaneously and calculate string complement spare opposite side electric current, equate this border according to string complement spare both sides electric current, can draw the localization of fault criterion, can eliminate string complement spare to the impact of transmission line of electricity range finding, realize fast finding line fault, simultaneously, this method does not need first failure judgement type for Resistance Fault and arc fault, adopt unified range measurement principle, can be accurately to localization of fault, good reliability, calculate simple, practical.
Description of drawings
Fig. 1 is transmission system structural representation of the present invention;
Among the figure, E M , E N Be the two ends power supply, u M , u N With i M , i N Be the transmission line of electricity two ends MThe point and NThe actual measurement voltage and current of some both sides, F 1Be circuit M-P 1 Single-phase electric arc earth fault occurs in the district, and the trouble spot is arrived MThe end distance is 20km; F 2Be circuit P 2 -NThe single-phase resistance earth fault of the outer generation in district, the trouble spot is arrived NThe end distance is 30km.
Embodiment
Below in conjunction with drawings and Examples this method is elaborated, understands to make things convenient for the technician.
As shown in Figure 1: go here and there the fault positioning method for transmission line of complement spare a kind of containing, and breaks down when containing string complement spare transmission line of electricity left side section, in accompanying drawing 1 F 1Point, in conjunction with fault section measuring end electric parameters and failure boundary condition, deriving calculates trouble spot string benefit component side electric current
Figure 839789DEST_PATH_IMAGE004
Then mend the derivation of component side electric current by fault point voltage and trouble spot string and calculate string complement spare left side P 1Place's electric current , derived by the measuring end electric parameters that perfects the section side simultaneously and calculate string complement spare right side P 2Place's electric current
Figure 250490DEST_PATH_IMAGE025
, equate this border according to string complement spare both sides electric current, that is:
Figure 437889DEST_PATH_IMAGE030
, can draw the localization of fault criterion.Break down when containing string complement spare transmission line of electricity right side section, in accompanying drawing F 2Point, distance-finding method are in like manner.
The concrete steps of fault positioning method for transmission line are as follows:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N , will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot
Figure 610113DEST_PATH_IMAGE001
With left side, trouble spot electric current
Figure 857555DEST_PATH_IMAGE002
, calculate right side, trouble spot electric current according to the boundary condition of different faults respectively again
Figure 616694DEST_PATH_IMAGE003
:
(1.1) for the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current For:
Figure 447564DEST_PATH_IMAGE005
(1)
Wherein:
Figure 928224DEST_PATH_IMAGE006
Be right side, trouble spot electric current,
Figure 407616DEST_PATH_IMAGE007
Be left side, trouble spot electric current,
Figure 202396DEST_PATH_IMAGE008
Be fault resistance.
(1.2) for the arcing ground fault, the boundary condition of arcing ground fault is:
Figure 585098DEST_PATH_IMAGE009
； (2)
Figure 236660DEST_PATH_IMAGE010
(3)
Wherein,
Figure 219659DEST_PATH_IMAGE011
Be the B phase current,
Figure 801819DEST_PATH_IMAGE012
Be the C phase current, Be A phase fault point right side electric current,
Figure 376337DEST_PATH_IMAGE014
Be A phase fault point left side electric current, Be fault point voltage,
Figure 468369DEST_PATH_IMAGE016
Be arc voltage;
This moment right side, trouble spot electric current
Figure 58619DEST_PATH_IMAGE004
For:
Figure 317562DEST_PATH_IMAGE017
(4)
(2) again will
Figure 9575DEST_PATH_IMAGE004
,
Figure 435002DEST_PATH_IMAGE018
Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 896070DEST_PATH_IMAGE019
; (5)
Wherein:
Figure 512865DEST_PATH_IMAGE020
; (6)
Figure 426595DEST_PATH_IMAGE021
; (7)
Figure 655713DEST_PATH_IMAGE022
; (8)
Figure 236867DEST_PATH_IMAGE023
(9)
Wherein, r, , vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length.
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Equate this border according to string complement spare both sides electric current, that is:
Figure 566720DEST_PATH_IMAGE026
(10)
So draw the localization of fault criterion be:
Figure 2381DEST_PATH_IMAGE027
(11)
(3) calculate fault distance according to the localization of fault criterion;
(3.1) for Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f
(3.2) for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u Arc
For the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 508448DEST_PATH_IMAGE028
(12)
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l; (13)
Wherein: lBe total track length.
R f >0; (14)
u arc>0. (15)
(4) use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f
During described measurement transmission line of electricity both end sides electric current and voltage, the length of short data window is 3ms, and sample frequency is 10kHz.
Embodiment 1: analogue system as shown in Figure 1, among the figure, E M , E N Be the two ends power supply, u M , u N With i M , i N Be the transmission line of electricity two ends MThe point and NThe actual measurement voltage and current of some both sides, F 1Be circuit M-P 1 Single-phase electric arc earth fault occurs in the district, and the trouble spot is arrived MThe end distance is 20km; F 2Be circuit P 2 -NThe single-phase resistance earth fault of the outer generation in district, the trouble spot is arrived NThe end distance is 30km;
The 500kV AC power line M-NAdopt J.Marti according to frequently modified line road model, total track length 300km, string complement spare is installed in the circuit midpoint, and compensativity is 40%, and the serial compensation capacitance value is C=98uF. P 2- NIn the section, single-phase resistance earth fault occurs in distance NEnd 30km place is in Fig. 1 F 2, carrying out power system transient simulation with PSCAD, Matlab carries out algorithm simulating.When PSCAD emulation, the data sampling frequency is 10kHz, adopts the phase-model transformation matrix to extract modulus, because the uncertainty of transmission line of electricity zero mould parameter is got the α modulus, α mould wave impedance in the simulation calculation Z c =239.0203 Ω ,α mould resistance R=2.8143e-5 Ω/m, α mould velocity of wave v=2.96e8m/s.Avoid the transient state period of fault initial stage acute variation when taking data, computational data is got 10ms-20ms after the fault, and total effectively computable data window is long to be 3ms.This is a kind of to contain and goes here and there the step of fault positioning method for transmission line of complement spare and be:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N Will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot
Figure 413082DEST_PATH_IMAGE001
With left side, trouble spot electric current , below calculate right side, trouble spot electric current according to the boundary condition of resistance eutral grounding fault and arcing ground fault respectively again :
For the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current
Figure 652936DEST_PATH_IMAGE004
For:
Wherein:
Figure 188184DEST_PATH_IMAGE006
Be right side, trouble spot electric current,
Figure 598437DEST_PATH_IMAGE007
Be left side, trouble spot electric current,
Figure 446308DEST_PATH_IMAGE008
Be fault resistance.
For the arcing ground fault, the boundary condition of arcing ground fault is:
Figure 558489DEST_PATH_IMAGE009
；
Figure 782797DEST_PATH_IMAGE010
Wherein,
Figure 47556DEST_PATH_IMAGE011
Be the B phase current,
Figure 331907DEST_PATH_IMAGE012
Be the C phase current,
Figure 432849DEST_PATH_IMAGE013
Be A phase fault point right side electric current,
Figure 132952DEST_PATH_IMAGE014
Be A phase fault point left side electric current, Be fault point voltage,
Figure 956737DEST_PATH_IMAGE016
Be arc voltage.
This moment right side, trouble spot electric current
Figure 856560DEST_PATH_IMAGE004
For:
Figure 360354DEST_PATH_IMAGE017
(2) again will
Figure 396443DEST_PATH_IMAGE004
,
Figure 711012DEST_PATH_IMAGE018
Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 98131DEST_PATH_IMAGE019
;
Wherein:
Figure 326987DEST_PATH_IMAGE020
;
;
Figure 906315DEST_PATH_IMAGE022
;
Figure 515151DEST_PATH_IMAGE023
Wherein, r,
Figure 626327DEST_PATH_IMAGE024
, vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length.
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Equate this border according to string complement spare both sides electric current, that is:
Figure 791915DEST_PATH_IMAGE026
So draw the localization of fault criterion be:
Figure 825730DEST_PATH_IMAGE027
(3) calculate fault distance according to the localization of fault criterion.For Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f And for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u ArcFor the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 271755DEST_PATH_IMAGE028
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l;
Wherein: lBe total track length.
R f >0;
u arc>0.
Use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f =29.87km;
Embodiment 2: analogue system as shown in Figure 1, the 500kV AC power line M-NAdopt J.Marti according to frequently modified line road model, total track length 300km, systematic parameter such as embodiment 1, P 2- NIn the section, single-phase electric arc earth fault occurs in distance NEnd 50km place.This is a kind of to contain and goes here and there the step of fault positioning method for transmission line of complement spare and be:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N Will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot With left side, trouble spot electric current
Figure 229794DEST_PATH_IMAGE002
, below calculate right side, trouble spot electric current according to the boundary condition of resistance eutral grounding fault and arcing ground fault respectively again
Figure 750906DEST_PATH_IMAGE003
:
For the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current
Figure 187572DEST_PATH_IMAGE004
For:
Figure 110529DEST_PATH_IMAGE005
Wherein:
Figure 108703DEST_PATH_IMAGE006
Be right side, trouble spot electric current,
Figure 179427DEST_PATH_IMAGE007
Be left side, trouble spot electric current, Be fault resistance.
For the arcing ground fault, the boundary condition of arcing ground fault is:
；
Figure 677907DEST_PATH_IMAGE010
Wherein,
Figure 970348DEST_PATH_IMAGE011
Be the B phase current, Be the C phase current,
Figure 459416DEST_PATH_IMAGE013
Be A phase fault point right side electric current,
Figure 799392DEST_PATH_IMAGE014
Be A phase fault point left side electric current,
Figure 782392DEST_PATH_IMAGE015
Be fault point voltage,
Figure 177601DEST_PATH_IMAGE016
Be arc voltage.
This moment right side, trouble spot electric current
Figure 913345DEST_PATH_IMAGE004
For:
Figure 627485DEST_PATH_IMAGE017
(2) again will
Figure 894518DEST_PATH_IMAGE004
,
Figure 765522DEST_PATH_IMAGE018
Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 621352DEST_PATH_IMAGE019
;
Wherein:
Figure 880295DEST_PATH_IMAGE020
;
Figure 306728DEST_PATH_IMAGE021
;
Figure 309319DEST_PATH_IMAGE022
;
Figure 532839DEST_PATH_IMAGE023
Wherein, r,
Figure 634787DEST_PATH_IMAGE024
, vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length.
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Figure 876412DEST_PATH_IMAGE025
Equate this border according to string complement spare both sides electric current, that is:
So draw the localization of fault criterion be:
Figure 981958DEST_PATH_IMAGE027
(3) calculate fault distance according to the localization of fault criterion.For Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f And for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u ArcFor the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 520386DEST_PATH_IMAGE028
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l;
Wherein: lBe total track length.
R f >0;
u arc>0.
Use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f =50.31km;
Embodiment 3: analogue system as shown in Figure 1, the 500kV AC power line M-NAdopt J.Marti according to frequently modified line road model, total track length 300km, systematic parameter such as embodiment 1. M- P 1In the section, distance MSingle-phase electric arc earth fault occurs in end 20km place, in Fig. 1 F 1This is a kind of to contain and goes here and there the step of fault positioning method for transmission line of complement spare and be:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N Will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot
Figure 983729DEST_PATH_IMAGE001
With left side, trouble spot electric current , below calculate right side, trouble spot electric current according to the boundary condition of resistance eutral grounding fault and arcing ground fault respectively again
Figure 452198DEST_PATH_IMAGE003
:
For the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current
Figure 145217DEST_PATH_IMAGE004
For:
Wherein:
Figure 870039DEST_PATH_IMAGE006
Be right side, trouble spot electric current,
Figure 160206DEST_PATH_IMAGE007
Be left side, trouble spot electric current,
Figure 837175DEST_PATH_IMAGE008
Be fault resistance.
For the arcing ground fault, the boundary condition of arcing ground fault is:
Figure 462060DEST_PATH_IMAGE009
；
Wherein, Be the B phase current,
Figure 94795DEST_PATH_IMAGE012
Be the C phase current,
Figure 20025DEST_PATH_IMAGE013
Be A phase fault point right side electric current,
Figure 182016DEST_PATH_IMAGE014
Be A phase fault point left side electric current,
Figure 696043DEST_PATH_IMAGE015
Be fault point voltage, Be arc voltage.
This moment right side, trouble spot electric current
Figure 878074DEST_PATH_IMAGE004
For:
Figure 843756DEST_PATH_IMAGE017
(2) again will ,
Figure 169006DEST_PATH_IMAGE018
Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 68829DEST_PATH_IMAGE019
;
Wherein:
Figure 572623DEST_PATH_IMAGE020
;
Figure 608712DEST_PATH_IMAGE021
; ;
Figure 543356DEST_PATH_IMAGE023
Wherein, r,
Figure 850840DEST_PATH_IMAGE024
, vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length.
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Figure 7015DEST_PATH_IMAGE025
Equate this border according to string complement spare both sides electric current, that is:
Figure 492485DEST_PATH_IMAGE026
So draw the localization of fault criterion be:
Figure 101321DEST_PATH_IMAGE027
(3) calculate fault distance according to the localization of fault criterion.For Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f And for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u ArcFor the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 946918DEST_PATH_IMAGE028
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l;
Wherein: lBe total track length.
R f >0;
u arc>0.
Use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f =19.71km;
The present invention describes by specific implementation process, without departing from the present invention, can also carry out various conversion and be equal to replacement patent of the present invention, therefore, patent of the present invention is not limited to disclosed specific implementation process, and should comprise the whole embodiments that fall in the Patent right requirement scope of the present invention.

Claims (3)

1. one kind contains and goes here and there the fault positioning method for transmission line of complement spare, and it is characterized in that: in the fault side, by measuring end electric parameters and failure boundary condition, deriving calculates right side, trouble spot voltage and current, and then derives and calculate string complement spare left side electric current; In non-fault side, calculate string complement spare right side electric current by the derivation of measuring end electric parameters simultaneously;
Equate this border according to string complement spare both sides electric current, draw the localization of fault function, thereby calculate fault distance
2, go here and there the fault positioning method for transmission line of complement spare a kind of containing according to claim 1, it is characterized in that: the concrete steps of fault positioning method for transmission line are as follows:
(1) after containing string complement spare transmission line of electricity lateral areas section and breaking down, measures protection installation place, transmission line of electricity two ends MThe point, NPoint voltage u M , u N Electric current i M , i N , will MTerminal voltage u M , electric current i M Substitution voltage along the line, distribution of current expression formula calculate the voltage of trouble spot
Figure 375988DEST_PATH_IMAGE001
With left side, trouble spot electric current
Figure 878776DEST_PATH_IMAGE002
, calculate right side, trouble spot electric current according to the boundary condition of different faults respectively again
Figure 701239DEST_PATH_IMAGE003
:
(1.1) for the resistance eutral grounding fault, can derive calculates right side, trouble spot electric current
Figure 905955DEST_PATH_IMAGE004
For:
Figure 104855DEST_PATH_IMAGE005
(1)
Wherein:
Figure 960684DEST_PATH_IMAGE006
Be right side, trouble spot electric current,
Figure 954048DEST_PATH_IMAGE007
Be left side, trouble spot electric current,
Figure 646061DEST_PATH_IMAGE008
Be fault resistance;
(1.2) for the arcing ground fault, the boundary condition of arcing ground fault is:
； (2)
Figure 594873DEST_PATH_IMAGE010
(3)
Wherein,
Figure 24718DEST_PATH_IMAGE011
Be the B phase current,
Figure 204026DEST_PATH_IMAGE012
Be the C phase current,
Figure 666101DEST_PATH_IMAGE013
Be A phase fault point right side electric current,
Figure 247255DEST_PATH_IMAGE014
Be A phase fault point left side electric current, Be fault point voltage,
Figure 999758DEST_PATH_IMAGE016
Be arc voltage;
This moment right side, trouble spot electric current
Figure 78573DEST_PATH_IMAGE004
For:
Figure 514233DEST_PATH_IMAGE017
(4)
(2) again will
Figure 285880DEST_PATH_IMAGE004
, Substitution current formula along the line calculates string complement spare left side P 1The electric current at place:
Figure 305974DEST_PATH_IMAGE019
; (5)
Wherein:
Figure 596142DEST_PATH_IMAGE020
; (6)
Figure 227105DEST_PATH_IMAGE021
; (7)
Figure 602723DEST_PATH_IMAGE022
; (8)
Figure 23340DEST_PATH_IMAGE023
(9)
Wherein, r,
Figure 495910DEST_PATH_IMAGE024
, vRespectively resistance per unit length, characteristic impedance, the wave velocity of circuit, l 1Be total track length;
In like manner will NTerminal voltage u N , electric current i N Substitution current formula along the line calculates string complement spare right side P 2The electric current at place
Figure 796310DEST_PATH_IMAGE025
Equate this border according to string complement spare both sides electric current, that is:
Figure 721541DEST_PATH_IMAGE026
(10)
So draw the localization of fault criterion be:
Figure 883532DEST_PATH_IMAGE027
(11)
(3) calculate fault distance according to the localization of fault criterion;
(3.1) for Resistance Fault, there are two unknown numbers, i.e. fault distances x f And transition resistance R f
(3.2) for arc fault, there are three unknown numbers, i.e. fault distance x f And transition resistance R f And arc voltage u Arc
For the ease of calculating Resistance Fault is regarded as u Arc=0 arc fault, the structure mapping function is as follows when therefore calculating:
Figure 633444DEST_PATH_IMAGE028
(12)
In this optimization problem, containing three variablees is fault distance x f , transition resistance R f , arc voltage u Arc, its constraint condition is:
0< x f < l; (13)
Wherein: lBe total track length.
2. 7. R f >0; (14)
u arc>0. (15)
(4) use least square method to calculate transition resistance R f And arc voltage u Arc, use again the linear search method to calculate fault distance x f
3. go here and there the fault positioning method for transmission line of complement spare a kind of containing according to claim 1 and 2, it is characterized in that: when measuring transmission line of electricity both end sides electric current and voltage, the length of short data window is 3ms, and sample frequency is 10kHz.
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