CN108020753A - A kind of short-circuit current calculation method to break with cross-line simultaneous faults - Google Patents
A kind of short-circuit current calculation method to break with cross-line simultaneous faults Download PDFInfo
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Abstract
The invention discloses a kind of short-circuit current calculation method to break with cross-line simultaneous faults, comprise the following steps：The complex fault type of broken string and cross-line simultaneous faults is described with the method for fault bus, and the general fault topology of fault bus is accessed according to each trouble point, row write out the failure boundary conditional equation of broken string and cross-line simultaneous faults, the sequence net voltage equation for writing trouble point is arranged again, simultaneous solution finally is carried out to failure boundary conditional equation and each sequence network voltage equation, you can calculate the port Injection Current and port voltage of each trouble point on each faulty line.The method of the present invention, which compensate for conventional short-circuit current calculation method, can not handle broken string and the simultaneous defect of cross line fault.
Description
Technical field
The invention belongs to multiple-loop line power transmission failure Current calculation field, more particularly to a kind of broken string and cross-line simultaneous faults
Short-circuit current calculation method.
Background technology
Multiple-loop line power transmission mode has become the main trend of China's power grid, and crisscross circuit gradually increases.In recent years
Come, bad weather takes place frequently, and the probability of the complex fault type such as short increases after circuit generation broken string, broken string.It is various multiple
The possibility that miscellaneous failure occurs increasingly increases severely, and occurs being related to more back transmission line cross line faults, short trouble, short circuit grounding event
Barrier, disconnection fault one or more failure mode are in the simultaneous possibility of same abort situation.The serious journey of complex fault
Degree can exceed that system norm for civil defense, even if stability control device correctly acts, system, which may also lose, to be stablized and trigger large area
Have a power failure, it is necessary to give enough attention to the various complex faults being likely to occur.
Therefore, needed for the type selecting of electrical equipment, various relay protections and automatics parameter tuning, power grid operation etc.
The line fault type of consideration is compared in the past more harsh.But various fault type is limited by, traditional fault current
Computational methods are difficult to arrange the failure boundary condition for writing various fault types one by one, do not have versatility, it is impossible to meet that current electric grid is short
Road, which calculates, to be needed.
The content of the invention
It is an object of the invention to overcome the shortcomings of existing fault current computational methods, there is provided one kind is suitable for calculating disconnected
The short-circuit current calculation method of line and cross-line simultaneous faults, this method are also suitable for only existing cross line fault or only exist broken string event
Various fault types are had versatility by situation during barrier.
The purpose of the present invention is realized by following technical solution：
A kind of short-circuit current calculation method to break with cross-line simultaneous faults, comprises the following steps：
S1, the complex fault type to break with the method description of fault bus with cross-line simultaneous faults；
S2, the general fault topology according to each trouble point access fault bus, row write out broken string and cross-line simultaneous faults
Failure boundary conditional equation；
S3, arrange the sequence net voltage equation for writing trouble point again；
S4, carry out simultaneous solution to failure boundary conditional equation and each sequence network voltage equation, you can calculates each failure
The port Injection Current and port voltage of each trouble point on circuit.
Preferably, step S1 is specifically included：
S1.1 sets fault wire way as M, and the number of trouble point is N, and faulty line is numbered from 1 to M successively；Failure
The fault type of circuit is divided into broken string and does not break, and failure vertex type is divided into：Broken string front side fracture, broken string rear side fracture and constantly
Line trouble point；Trouble point is numbered from 1 to N successively according to faulty line order；If reference capacity is SB, line voltage benchmark
It is worth for UB, UBIt is defined as the voltage U of first trouble pointB1, line current a reference value isIf the end of each trouble point
Mouthful voltage perunit value isPort Injection Current perunit value is
S1.2 for disconnection fault type circuit, if i-th (i=1,2 ..., M) loop line road is disconnection fault type；k(k
=1,2,3 ..., N-1) and g (g=k+1) be respectively break front side fracture and broken string rear side fracture numbering；By faulty line i
On trouble point k and trouble point g access fault bus obtain disconnection fault type line general fault topology；
F1, F2 ..., Fn represent fault bus, the failure phase for connecting fault point；Switch SkgP(P=A, B, C) is disconnected
Wiretap, SkgPP phases between=1 expression trouble point k and g do not break, SkgP=0 expression trouble point k breaks with the P phases of g；Switch
SkP、SgPSwitch is accessed for the fault bus of trouble point, with SkPExemplified by, SkP=x (x=1,2 ..., n) represents the P phases of trouble point k
Access fault bus Fx, SkPThe P phases not failure of=0 expression trouble point k；Switch SFx(x=1,2 ..., n) it is that corresponding failure is total
The earthing switch of line, SFx=1 represents fault bus Fx ground connection, SFx=0 represents that fault bus Fx is earth-free；KTk、KTgFor trouble point
The ideal transformer no-load voltage ratio of k and trouble point g, for handling the bridging failure between different voltages grade circuit, KTk=KTg=
UBi/UB；
Each variable subscript * represents to use perunit system；For electric current on front side of the fracture of fracture k-g and electricity on rear side of fracture
Stream, For trouble point k, the fault bus Injection Current of trouble point g,For the earth current of fault bus Fx；For
The voltage-to-ground of fault bus Fx；The alternate transition resistance of respectively trouble point k, trouble point g,It is total for failure
The ground connection transition resistance of line Fx；
S1.3 set jth (j=1,2 ..., M；J ≠ i) bar faulty line is do not break type line；H (h=1,2,3 ...,
N；H ≠ k, g) it is the numbering of trouble point of not breaking；It will be obtained constantly among trouble point h access fault bus on faulty line j
The general fault topology of line fault type circuit；Each switch, the definition of each variable and S1.2 are similar, wherein KTh=UBj/UB。
Preferably, step S2 is specifically included：
It is located at broken string and in cross-line simultaneous faults, the number of trouble point is N number of, fracture, broken string rear side on front side of its interrupt line
Fracture trouble point is p, and the trouble point that do not break is q, then N=2p+q, fault bus are n bars；
S2.1 row write the branch-current equation at transformer outlet；
Equation number is (6p+3q), wherein P=A, B, C；For broken string front side fracture, broken string rear side fracture trouble point, contain
Break branch current unknown quantity, then has
For the trouble point that do not break, without broken string branch current, then have
S2.2 row write broken string voltage, current equation；
Equation number is 6p, only broken string front side fracture, broken string rear side fracture trouble point k, g row is write, if SkgP=1, then have
If SkgP=0, then have
S2.3 row write voltage, current equation of the trouble point to fault bus；
Equation number is (6p+3q), travels through all trouble point l=1,2 ..., (2p+q), if SlP=0, then have
If SlP=x (x=1,2 ..., n), then have
S2.4 arranges the branch-current equation for writing fault bus；
Equation number is n, travels through x=1,2 ..., n, have
S2.5 row write fault bus voltage-to-ground, current equation；
Equation number is n, x=1,2 ..., n is traveled through, if SFx=0, then have
If SFx=1, then have
The row write circuit equation in the way of above-mentioned formula (1)~(12), equation sum is (18p+6q+2n), and above-mentioned formula
Unknown quantity in son (wherein l=1,2 ..., (2p+q)；X=1,2 ..., n；P=A, B, C) be respectively：
1. trouble point Injection Current(6p+3q) is a altogether；
2. fault point voltage(6p+3q) is a altogether；
3. branch current includesWith(12p+3q) is a altogether；
4. fault bus voltageCommon n；
5. fault bus are to earth-currentCommon n.
Further, in above-mentioned listed (18p+6p+2n) a equation, (24p+9q+2n) is a altogether for the number of unknown quantity,
The listed equation write is represented with the form of matrix, can be arranged and be write (18p+6p+2n) OK, the failure boundary condition of (24p+9q+2n) row
Equation initial matrixUnknown quantity is the column vector of (24p+9q+2n) dimension, is 0 on the right of matrix equation equation；
With the matrix elimination to initial matrixThe member processing that disappears is carried out, is eliminated except (6p+3q) a trouble point injection
Remaining all unknown quantity outside electric current and (6p+3q) a fault point voltage, the number for eliminating unknown quantity is (12p+3q+2n),
Then original matrix is changed into the failure boundary conditional equation matrix of (6p+3q) row (12p+6q) row after the member that disappearsWith trouble point
Number N (N=2p+q) is representedRow and column,The matrix arranged for 3N rows 6N,Write as
Matrix disappears after member processing, obtains the failure boundary conditional matrix equation of broken string and cross-line simultaneous faults, is represented by：
In formula (13)Represent the phase current phasor of injection trouble point,Represent the phase voltage phase of trouble point
Amount, putting in order for they should be consistent with the numbering of trouble point, O(3N×1)For complete zero column vector.
Preferably, further include, the failure boundary conditional equation of broken string and cross-line simultaneous faults is carried out symmetrical in step S2
Component converts.
Specifically, the step of failure boundary conditional equation of broken string and cross-line simultaneous faults is carried out symmetrical component transformation bag
Include：
Symmetrical component transformation is carried out to formula (13)：
In formula (14)
Form can be written as according to symmetrical component transformation principle formula (14)：
In formula (15)Represent the sequence electric current phasor of injection trouble point,Represent the sequence voltage phasor of trouble point, putting in order for they should be consistent with the numbering of trouble point.
Preferably, step S3 is specifically included：
Row write each sequence network voltage equation of trouble point, have
Subscript (1) in above-mentioned formula, (2), (0) represent positive and negative, zero sequence respectively；Represent trouble point l
(l=1,2 ..., N) and k (k=1,2 ..., N) from, mutual impedance；Represent trouble point
Positive and negative, the residual voltage phasor of l；Represent positive and negative, the zero-sequence current phase of trouble point k
Amount；Represent voltage before the failure of trouble point l, meter and during trend,The fault point voltage for taking Load flow calculation to export, takes no account of
During trend,Take the reference voltage of faulty line；
The form of matrix is write as in wushu (16)~(18), the order of trouble point sequence electric current and sequence voltage unknown quantity also according to
The numbering of trouble point sorts successively, is consistent, can obtain with putting in order for unknown quantity in formula (15)：
In formula (19)
(E(3N×3N)For the unit diagonal matrix of 3N rows 3N row)
Preferably, step S4 is specifically included：
Wushu (15) can be obtained with formula (19) simultaneous：
By (20) as it can be seen that for broken string and cross-line simultaneous faults, when the number of trouble point is N, it can arrange and write out 6N bar sides
Journey, each sequence voltage and each sequence electric current of the unknown quantity for each trouble point, unknown quantity number is also 6N, therefore equation can solve
The voltage of each each sequence in trouble point and the electric current of each sequence, and then the phase electricity of all trouble points can be tried to achieve by symmetrical components inverse transformation
Pressure and phase current.
Compared with prior art, the present invention having the following advantages that and beneficial effect：
1. the fault current computational methods of broken string proposed by the invention and cross-line simultaneous faults, compensate for conventional short-circuit electricity
Flow calculation methodologies can not handle broken string and the simultaneous defect of cross line fault.
2. the method proposed by the invention with fault bus addition trouble point, it is same with cross-line to can be used for description broken string
When failure complex fault type, can be used for description only exist disconnection fault or only exist the situation of cross line fault, solve
This inenarrable problem of complex fault type, while the failure boundary condition for also solving complex fault type is difficult to arrange
The problem write.
Brief description of the drawings
Fig. 1 is disconnection fault type line schematic diagram.
Fig. 2 is the general fault topology schematic diagram of disconnection fault type line.
Fig. 3 is the fault type conspectus that do not break.
Fig. 4 is the general fault topology schematic diagram of fault type circuit of not breaking.
Fig. 5 is example model topology figure.
Embodiment
With reference to embodiment and attached drawing, the present invention is described in further detail, but embodiments of the present invention are unlimited
In this.
Embodiment 1
As shown in Figure 1, a kind of short-circuit current calculation method for being suitable for calculating broken string and cross-line simultaneous faults, this method purport
The fault current of broken string and cross-line simultaneous faults is being calculated, is mainly passing through each sequence of simultaneous solution broken string and cross-line simultaneous faults
Network voltage equation and failure boundary conditional equation, obtain the voltage and current of each trouble point.Wherein failure boundary is write in order to arrange
Conditional equation, proposition are suitable for the general fault topology schematic diagram of broken string and cross-line simultaneous faults, and event is utilized according to fault message
Hinder bus and access each trouble point to describe corresponding complex fault type, formula is then pressed by the on off state in failure topology
(1)~(12) row write failure boundary conditional equation.
Specifically include following steps：
(1) disconnection fault and cross line fault type are described respectively with fault bus method
1.1 set fault wire way as M, and the number of trouble point is N.Faulty line is numbered from 1 to M successively.Failure
The fault type of circuit is divided into broken string and does not break.Failure vertex type is divided into：Broken string front side fracture, broken string rear side fracture and constantly
Line trouble point.Trouble point is numbered from 1 to N successively according to faulty line order.If reference capacity is SB, line voltage benchmark
It is worth for UB(UBIt is defined as the voltage U of first trouble pointB1), line current a reference value isIf each trouble point
Port voltage perunit value isPort Injection Current perunit value is
1.2 for disconnection fault type circuit, if i-th (i=1,2 ..., M) loop line road is disconnection fault type, it shows
It is intended to as shown in Figure 1.
K in Fig. 1 (k=1,2,3 ..., N-1) and g (g=k+1) is respectively break front side fracture and broken string rear side fracture
Numbering.Trouble point k on faulty line i and trouble point g access fault bus are obtained into the general event of disconnection fault type line
Barrier topology, its schematic diagram are as shown in Figure 2.
F1 in Fig. 2, F2 ..., Fn represent fault bus, the failure phase for connecting fault point.Switch SkgP(P=A, B, C)
For wire-breaking switch, SkgP=1 (closure) represents that the P phases between trouble point k and g do not break, SkgP=0 (opening) represents trouble point k
Break with the P phases of g.Switch SkP、SgPSwitch is accessed for the fault bus of trouble point, with SkPExemplified by, SkP=x (x=1,2 ..., n)
Represent that the P phases of trouble point k access fault bus Fx, SkPThe P phases not failure of=0 expression trouble point k.Switch SFx(x=1,
2 ..., n) be corresponding failure bus earthing switch, SFx=1 represents fault bus Fx ground connection, SFx=0 represents fault bus Fx
It is earth-free.KTk、KTgFor the ideal transformer no-load voltage ratio of trouble point k and trouble point g, for handling between different voltages grade circuit
Bridging failure, KTk=KTg=UBi/UB。
Each variable subscript * represents to use perunit system in Fig. 2.For electric current and fracture on front side of the fracture of fracture k-g
Rear side electric current,For trouble point k, the fault bus Injection Current of trouble point g,Enter ground electricity for fault bus Fx
Stream；For the voltage-to-ground of fault bus Fx；The alternate transition resistance of respectively trouble point k, trouble point g,
For the ground connection transition resistance of fault bus Fx.
1.3 set jth (j=1,2 ..., M；J ≠ i) bar faulty line is does not break type line, its schematic diagram such as Fig. 3 institutes
Show.
H in Fig. 3 (h=1,2,3 ..., N；H ≠ k, g) it is the numbering of trouble point of not breaking.By the failure on faulty line j
Obtain not breaking the general fault topology of fault type circuit among point h access fault bus, its schematic diagram is as shown in Figure 4.
Respectively switch, the definition of each variable and Fig. 2 are similar in Fig. 4, wherein KTh=UBj/UB。
(2) row write the failure boundary conditional equation of failure
The various fault types of broken string and cross-line simultaneous faults can be described using the general fault topology of step (1), below
According to the state respectively switched in general fault topology, row write the failure boundary conditional equation of corresponding failure.It is located at broken string and cross-line
In simultaneous faults, the number of trouble point is N number of, and fracture, broken string rear side fracture trouble point are p on front side of its interrupt line, constantly
Line trouble point is q, then N=2p+q, fault bus are n bars.
1. row write the branch-current equation at transformer outlet
Equation number is (6p+3q), wherein P=A, B, C.For broken string front side fracture, broken string rear side fracture trouble point (Fig. 2
K, g), containing broken string branch current unknown quantity, then have
For the trouble point that do not break (h of Fig. 4), without broken string branch current, then have
2. row write broken string voltage, current equation
Equation number is 6p, only broken string front side fracture, broken string rear side fracture trouble point k, g row is write, if SkgP=1, then have
If SkgP=0, then have
3. row write voltage, current equation of the trouble point to fault bus
Equation number is (6p+3q), travels through all trouble point l=1,2 ..., (2p+q), if SlP=0, then have
If SlP=x (x=1,2 ..., n), then have
4. row write the branch-current equation of fault bus
Equation number is n, travels through x=1,2 ..., n, have
5. row write fault bus voltage-to-ground, current equation
Equation number is n, x=1,2 ..., n is traveled through, if SFx=0, then have
If SFx=1, then have
The row write circuit equation in the way of above-mentioned formula (1)~(12), equation sum is (18p+6q+2n), and above-mentioned formula
Unknown quantity in son (wherein l=1,2 ..., (2p+q)；X=1,2 ..., n；P=A, B, C) be respectively：
1. trouble point Injection Current(6p+3q) is a altogether；
2. fault point voltage(6p+3q) is a altogether；
3. branch current includesWith(12p+3q) is a altogether；
4. fault bus voltageCommon n；
5. fault bus are to earth-currentCommon n.
It can be obtained by above-mentioned analysis, in listed (18p+6p+2n) a equation, (24p+9q+2n) is a altogether for the number of unknown quantity,
The listed equation write is represented with the form of matrix, can arrange and write (18p+6p+2n) OK, the failure boundary bar of (24p+9q+2n) row
Part equation initial matrixUnknown quantity is the column vector of (24p+9q+2n) dimension, is 0 on the right of matrix equation equation.
With the matrix elimination to initial matrixThe member processing that disappears is carried out, is eliminated except (6p+3q) a trouble point injection
Remaining all unknown quantity outside electric current and (6p+3q) a fault point voltage, the number for eliminating unknown quantity is (12p+3q+2n),
Then original matrix is changed into the failure boundary conditional equation matrix of (6p+3q) row (12p+6q) row after the member that disappearsWith trouble point
Number N (N=2p+q) is representedRow and column,The matrix arranged for 3N rows 6N,Write as
Matrix disappears after member processing, obtains the failure boundary conditional matrix equation of broken string and cross-line simultaneous faults, is represented by：
In formula (13)Represent the phase current phasor of injection trouble point,Represent the phase voltage phase of trouble point
Amount, putting in order for they should be consistent with the numbering of trouble point.O(3N×1)For complete zero column vector.
(3) the failure boundary conditional equation of broken string and cross-line simultaneous faults is carried out symmetrical component transformation
To solve trouble point Injection Current, the common 6N unknown quantity of fault point voltage, 3N bar failure boundaries condition side has been arranged to obtain
Journey, also lacks the sequence net voltage equation of 3N bars trouble point.And sequence net voltage equation is made not with the sequence electric current and sequence voltage of trouble point
The amount of knowing row are write, therefore need first to make symmetrical component transformation to 3N bar failure boundary conditional equations, and the unknown quantity under phase component is changed
For the unknown quantity under order components.
Symmetrical component transformation is carried out to formula (13)：
In formula (14)
Form can be written as according to symmetrical component transformation principle formula (14)：
In formula (15)Represent the sequence electric current phasor of injection trouble point,Represent the sequence voltage phasor of trouble point, putting in order for they should be consistent with the numbering of trouble point.
(4) row write each sequence network voltage equation of broken string and cross-line simultaneous faults
In order to be solved to breaking with the fault current of cross-line simultaneous faults, it is necessary to arrange each sequence network electricity for writing trouble point
Equation is pressed, is had
Subscript (1) in above-mentioned formula, (2), (0) represent positive and negative, zero sequence respectively；Represent trouble point l
(l=1,2 ..., N) and k (k=1,2 ..., N) from, mutual impedance；Represent trouble point
Positive and negative, the residual voltage phasor of l.Represent positive and negative, the zero-sequence current phase of trouble point k
Amount；Represent voltage before the failure of trouble point l, meter and during trend,The fault point voltage for taking Load flow calculation to export, takes no account of
During trend,Take the reference voltage of faulty line.
The form of matrix is write as in wushu (16)~(18), the order of trouble point sequence electric current and sequence voltage unknown quantity also according to
The numbering of trouble point sorts successively, is consistent, can obtain with putting in order for unknown quantity in formula (15)：
In formula (19)
(E(3N×3N)For the unit diagonal matrix of 3N rows 3N row)
(5) failure boundary condition and each sequence network voltage equation of the simultaneous broken string with cross-line simultaneous faults, solves trouble point
Voltage and current.
Wushu (15) can be obtained with formula (19) simultaneous：
By (20) as it can be seen that for broken string and cross-line simultaneous faults, when the number of trouble point is N, it can arrange and write out 6N bar sides
Journey, each sequence voltage and each sequence electric current of the unknown quantity for each trouble point, unknown quantity number is also 6N, therefore equation can solve
The voltage of each each sequence in trouble point and the electric current of each sequence, and then the phase electricity of all trouble points can be tried to achieve by symmetrical components inverse transformation
Pressure and phase current.
This method is described in further detail with reference to the example model shown in Fig. 5, but the embodiment party of the present invention
Formula not limited to this.
Example model shown in Fig. 5 adds on the basis of tri- machines of IEEE9, nine nodal analysis method between busbar 1 and busbar A
Common-tower double-return coupling power transmission line (coupling zero-sequence mutual inductance Z1m0=0.03+j0.1), respectively with circuit 1-A1 and circuit 1-A2 areas
Point, remaining parameter is consistent with IEEE9 classical models.Network parameter represents that a reference value takes S using perunit valueB=100MVA, UB=
230kV, system frequency 50Hz.
If disconnection fault occurs for the A phases of circuit 3-C, the distance of abort situation to busbar 3 accounts for the 70% of total track length, fracture
The A phases of front side fracture drop to be occurred to bridge failure, while the A phases of fracture and circuit 1-A2 on rear side of fracture with the B phases of circuit 1-A1
C phases cross-line earth fault occurs.The distance of abort situation to busbar 1 on circuit 1-A1 and circuit 1-A2 accounts for total track length
40%.
S1. failure Turbo codes corresponding failure type is utilized
Above-mentioned fault type totally 3 faulty lines.Circuit 3-C for broken string circuit, numbering 1；Circuit 1-A1 is not break
Circuit, numbering 2；Circuit 1-A2 is does not break circuit, numbering 3.The number of trouble point is 4, before including broken string on circuit 1
Side fracture, broken string rear side fracture, trouble point numbering is respectively 1,2；It is the trouble point that do not break on circuit 2, trouble point numbering is 3；
It is the trouble point that do not break on circuit 3, trouble point numbering is 4.
Fault bus are accessed in trouble point 1, trouble point 2 in the way of Fig. 2, trouble point 3 and trouble point 4 are according to Fig. 4's
Mode accesses fault bus.Alternate transition resistance, ground resistance are zero, simultaneously because faulty line voltage class is consistent, reason
It is 1 to think transformer voltage ratio.After trouble point access fault bus, the closure situation respectively switched is：S12B=S12C=1, S1A=S3B
=1, S2A=S4C=2；SF2=1；Rest switch S=0.
S2. row write the failure boundary conditional equation of corresponding failure type
Failure boundary conditional equation is write according to the principle of formula (1)~(12) row, to above-mentioned specific failure p=1, q=2, n=
2, equation number is 34, and the quantity of unknown quantity is 46.Therefore obtained failure boundary conditional equation initial matrix
For the matrix of 34 rows 46 row.All unknown quantitys in addition to the phase current of trouble point and phase voltage are eliminated using the matrix elimination,
22 unknown quantitys are eliminated altogether, are obtained after the member that disappearsIt is as follows for the matrix of 12 rows 24 row, its expression formula：
To the specific failure set in example, failure boundary conditional equation can be arranged and is written as：
Trouble point phase current in formulaPhase voltageSort according to the order of trouble point numbering.
S3. to breaking symmetrical component transformation is carried out with the failure boundary conditional equation of cross-line simultaneous faults
Symmetrical component transformation is carried out to formula (21), is obtained
In formula (22)
To the specific failure in exampleReal part expression formula it is as follows：
Imaginary part expression formula it is as follows：
S4. row write each sequence network voltage equation of broken string and cross-line simultaneous faults
The sequence net voltage equation that each trouble point in example is write according to formula (16)~(18) row is as follows：
Wherein
And for the specific failure in exampleReal part expression formula be：
Imaginary part expression formula be：
S5. simultaneous failure boundary conditional equation and each sequence network voltage equation solve fault point voltage electric current
Formula (22) and formula (23) totally 24 equations, simultaneous solution totally 24 unknown quantitys, can solve voltage, the electricity of each trouble point
Stream.
The example model such as Fig. 5 is built in PSCAD/EMTDC, sets identical fault type to be emulated, emulation
As a result compared with this method calculated results, the results are shown in Table 1.
1 simulation result of table and the fault point voltage of result of calculation, electric current deck watch
By table 1 as it can be seen that the voltage of each trouble point, electric current maximum relative error are no more than 1.75%, phase angle error does not surpass
0.56 ° is crossed, there is higher precision, demonstrates the correctness of proposed method.
Above-described embodiment is the preferable embodiment of the present invention, but embodiments of the present invention and from above-described embodiment
Limitation, other any Spirit Essences without departing from the present invention with made under principle change, modification, replacement, combine, simplification,
Equivalent substitute mode is should be, is included within protection scope of the present invention.
Claims (8)
1. a kind of short-circuit current calculation method to break with cross-line simultaneous faults, it is characterised in that comprise the following steps：
S1, the complex fault type to break with the method description of fault bus with cross-line simultaneous faults；
S2, the general fault topology according to each trouble point access fault bus, row write out the failure of broken string and cross-line simultaneous faults
Boundary condition equation；
S3, arrange the sequence net voltage equation for writing trouble point again；
S4, carry out simultaneous solution to failure boundary conditional equation and each sequence network voltage equation, you can calculates each faulty line
On each trouble point port Injection Current and port voltage.
2. short-circuit current calculation method according to claim 1, it is characterised in that step S1 is specifically included：
S1.1 sets fault wire way as M, and the number of trouble point is N, and faulty line is numbered from 1 to M successively；Faulty line
Fault type be divided into broken string and do not break, failure vertex type is divided into：Broken string front side fracture, broken string rear side fracture and the event that do not break
Barrier point；Trouble point is numbered from 1 to N successively according to faulty line order；If reference capacity is SB, line voltage a reference value is
UB, UBIt is defined as the voltage U of first trouble pointB1, line current a reference value isIf the port electricity of each trouble point
Pressure perunit value bePort Injection Current perunit value is
S1.2 for disconnection fault type circuit, if i-th (i=1,2 ..., M) loop line road is disconnection fault type；K (k=1,
2,3 ..., N-1) and g (g=k+1) be respectively break front side fracture and broken string rear side fracture numbering；By on faulty line i
Trouble point k and trouble point g access fault bus obtain the general fault topology of disconnection fault type line；
F1, F2 ..., Fn represent fault bus, the failure phase for connecting fault point；Switch SkgP(P=A, B, C) opens for broken string
Close, SkgPP phases between=1 expression trouble point k and g do not break, SkgP=0 expression trouble point k breaks with the P phases of g；Switch SkP、
SgPSwitch is accessed for the fault bus of trouble point, with SkPExemplified by, SkP=x (x=1,2 ..., n) represent that the P phases of trouble point k access
Fault bus Fx, SkPThe P phases not failure of=0 expression trouble point k；Switch SFx(x=1,2 ..., n) it is corresponding failure bus
Earthing switch, SFx=1 represents fault bus Fx ground connection, SFx=0 represents that fault bus Fx is earth-free；KTk、KTgFor trouble point k and
The ideal transformer no-load voltage ratio of trouble point g, for handling the bridging failure between different voltages grade circuit, KTk=KTg=UBi/
UB；
Each variable subscript * represents to use perunit system；Fracture front side electric current for fracture k-g and electric current on rear side of fracture, For trouble point k, the fault bus Injection Current of trouble point g,For the earth current of fault bus Fx；For failure
The voltage-to-ground of bus Fx；The alternate transition resistance of respectively trouble point k, trouble point g,For fault bus Fx
Ground connection transition resistance；
S1.3 set jth (j=1,2 ..., M；J ≠ i) bar faulty line is do not break type line；H (h=1,2,3 ..., N；h≠
K, g) it is the numbering of trouble point of not breaking；The failure that do not break will be obtained among trouble point h access fault bus on faulty line j
The general fault topology of type line；Each switch, the definition of each variable and S1.2 are similar, wherein KTh=UBj/UB。
3. short-circuit current calculation method according to claim 2, it is characterised in that step S2 is specifically included：
It is located at broken string and in cross-line simultaneous faults, the number of trouble point is N number of, fracture, broken string rear side fracture on front side of its interrupt line
Trouble point is p, and the trouble point that do not break is q, then N=2p+q, fault bus are n bars；
S2.1 row write the branch-current equation at transformer outlet；
Equation number is (6p+3q), wherein P=A, B, C；For broken string front side fracture, broken string rear side fracture trouble point, contain broken string
Branch current unknown quantity, then have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>/</mo>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>k</mi>
</mrow>
</msub>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>g</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>/</mo>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>g</mi>
</mrow>
</msub>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>k</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>2</mn>
<mo>)</mo>
</mrow>
</mrow>
For the trouble point that do not break, without broken string branch current, then have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>h</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>/</mo>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>h</mi>
</mrow>
</msub>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>h</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mo>)</mo>
</mrow>
</mrow>
S2.2 row write broken string voltage, current equation；
Equation number is 6p, only broken string front side fracture, broken string rear side fracture trouble point k, g row is write, if SkgP=1, then have
<mrow>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>k</mi>
</mrow>
</msub>
<mo>-</mo>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>g</mi>
</mrow>
</msub>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>4</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>g</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>+</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>k</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>5</mn>
<mo>)</mo>
</mrow>
</mrow>
If SkgP=0, then have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>k</mi>
<mi>g</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>6</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>g</mi>
<mi>k</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>7</mn>
<mo>)</mo>
</mrow>
</mrow>
S2.3 row write voltage, current equation of the trouble point to fault bus；
Equation number is (6p+3q), travels through all trouble point l=1,2 ..., (2p+q), if SlP=0, then have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>l</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>8</mn>
<mo>)</mo>
</mrow>
</mrow>
If SlP=x (x=1,2 ..., n), then have
<mrow>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>l</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<msub>
<mi>K</mi>
<mrow>
<mi>T</mi>
<mi>l</mi>
</mrow>
</msub>
<mo>-</mo>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mi>x</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>-</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>l</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<msubsup>
<mi>Z</mi>
<mrow>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>9</mn>
<mo>)</mo>
</mrow>
</mrow>
S2.4 arranges the branch-current equation for writing fault bus；
Equation number is n, travels through x=1,2 ..., n, have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mi>x</mi>
<mi>G</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mover>
<mi>&Sigma;</mi>
<mrow>
<msub>
<mi>S</mi>
<mrow>
<mi>l</mi>
<mi>P</mi>
</mrow>
</msub>
<mo>=</mo>
<mi>x</mi>
</mrow>
</mover>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>l</mi>
<mi>F</mi>
<mi>P</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>10</mn>
<mo>)</mo>
</mrow>
</mrow>
S2.5 row write fault bus voltage-to-ground, current equation；
Equation number is n, x=1,2 ..., n is traveled through, if SFx=0, then have
<mrow>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mi>x</mi>
<mi>G</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>11</mn>
<mo>)</mo>
</mrow>
</mrow>
If SFx=1, then have
<mrow>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mi>x</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>-</mo>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mi>x</mi>
<mi>G</mi>
</mrow>
<mo>*</mo>
</msubsup>
<msubsup>
<mi>Z</mi>
<mrow>
<mi>F</mi>
<mi>x</mi>
<mi>G</mi>
</mrow>
<mo>*</mo>
</msubsup>
<mo>=</mo>
<mn>0</mn>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>12</mn>
<mo>)</mo>
</mrow>
</mrow>
The row write circuit equation in the way of above-mentioned formula (1)~(12), equation sum is (18p+6q+2n), and in above-mentioned formula
Unknown quantity (wherein l=1,2 ..., (2p+q)；X=1,2 ..., n；P=A, B, C) be respectively：
1. trouble point Injection Current(6p+3q) is a altogether；
2. fault point voltage(6p+3q) is a altogether；
3. branch current includes(6p+3q)、(3p) and(3p), (12p+3q) is a altogether；
4. fault bus voltageCommon n；
5. fault bus are to earth-currentCommon n.
4. short-circuit current calculation method according to claim 3, it is characterised in that listed (18p+6p+2n) a equation
In, (24p+9q+2n) is a altogether for the number of unknown quantity, and the listed equation write is represented with the form of matrix, can arrange and write (18p+6p+
2n) row, the failure boundary conditional equation initial matrix of (24p+9q+2n) rowUnknown quantity is (24p+9q+2n) dimension
Column vector, matrix equation equation the right are 0；
With the matrix elimination to initial matrixThe member processing that disappears is carried out, cancellation removes (6p+3q) a trouble point Injection Current
Remaining all unknown quantity outside (6p+3q) a fault point voltage, the number for eliminating unknown quantity are (12p+3q+2n), then former
Matrix is changed into the failure boundary conditional equation matrix of (6p+3q) row (12p+6q) row after the member that disappearsWith the number of trouble point
N (N=2p+q) is representedRow and column,The matrix arranged for 3N rows 6N,Write asMatrix disappears
After member processing, the failure boundary conditional matrix equation of broken string and cross-line simultaneous faults is obtained, is represented by：
<mrow>
<msubsup>
<mi>J</mi>
<mrow>
<mi>F</mi>
<mi>a</mi>
<mi>u</mi>
<mi>l</mi>
<mi>t</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>6</mn>
<mi>N</mi>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</msubsup>
<mo>&times;</mo>
<mfenced open = "[" close = "]">
<mtable>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>*</mo>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</mrow>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>*</mo>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</mrow>
</msubsup>
</mtd>
</mtr>
</mtable>
</mfenced>
<mo>=</mo>
<msub>
<mi>O</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>13</mn>
<mo>)</mo>
</mrow>
</mrow>
In formula (13)Represent the phase current phasor of injection trouble point,Represent the phase voltage phasor of trouble point, they
Put in order should be consistent with the numbering of trouble point, O(3N×1)For complete zero column vector.
5. short-circuit current calculation method according to claim 1, it is characterised in that further included in step S2, broken string with
The failure boundary conditional equation of cross-line simultaneous faults carries out symmetrical component transformation.
6. short-circuit current calculation method according to claim 4 or 5, it is characterised in that broken string and cross-line simultaneous faults
Failure boundary conditional equation carry out symmetrical component transformation the step of include：
Symmetrical component transformation is carried out to formula (13)：
<mrow>
<msubsup>
<mi>J</mi>
<mrow>
<mi>F</mi>
<mi>a</mi>
<mi>u</mi>
<mi>l</mi>
<mi>t</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>6</mn>
<mi>N</mi>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</msubsup>
<msubsup>
<mi>Q</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>3</mn>
<mi>N</mi>
<mo>)</mo>
</mrow>
<mrow>
<mo>-</mo>
<mn>1</mn>
</mrow>
</msubsup>
<mo>&times;</mo>
<msub>
<mi>Q</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>3</mn>
<mi>N</mi>
<mo>)</mo>
</mrow>
</msub>
<mfenced open = "[" close = "]">
<mtable>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mi>A</mi>
<mi>B</mi>
<mi>C</mi>
<mo>)</mo>
</mrow>
</msubsup>
</mtd>
</mtr>
</mtable>
</mfenced>
<mo>=</mo>
<msub>
<mi>O</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>14</mn>
<mo>)</mo>
</mrow>
</mrow>
In formula (14)
Form can be written as according to symmetrical component transformation principle formula (14)：
<mrow>
<msubsup>
<mi>J</mi>
<mrow>
<mi>F</mi>
<mi>a</mi>
<mi>u</mi>
<mi>l</mi>
<mi>t</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>6</mn>
<mi>N</mi>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mn>120</mn>
<mo>)</mo>
</mrow>
</msubsup>
<mo>&times;</mo>
<mfenced open = "[" close = "]">
<mtable>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>I</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mn>120</mn>
<mo>)</mo>
</mrow>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>U</mi>
<mo>&CenterDot;</mo>
</mover>
<mrow>
<mi>F</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</mrow>
<mrow>
<mo>(</mo>
<mn>120</mn>
<mo>)</mo>
</mrow>
</msubsup>
</mtd>
</mtr>
</mtable>
</mfenced>
<mo>=</mo>
<msub>
<mi>O</mi>
<mrow>
<mo>(</mo>
<mn>3</mn>
<mi>N</mi>
<mo>&times;</mo>
<mn>1</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>15</mn>
<mo>)</mo>
</mrow>
</mrow>
In formula (15) Represent the sequence electric current phasor of injection trouble point,
Represent the sequence voltage phasor of trouble point, putting in order for they should be consistent with the numbering of trouble point.
7. short-circuit current calculation method according to claim 6, it is characterised in that step S3 is specifically included：
Row write each sequence network voltage equation of trouble point, have
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Subscript (1) in above-mentioned formula, (2), (0) represent positive and negative, zero sequence respectively；Expression trouble point l (l=1,
2 ..., N) with k (k=1,2 ..., N) from, mutual impedance；(l=1,2 ..., N) represent trouble point l's
Positive and negative, residual voltage phasor；(k=1,2 ..., N) represents positive and negative, the zero-sequence current phasor of trouble point k；Represent voltage before the failure of trouble point l, meter and during trend,The fault point voltage for taking Load flow calculation to export, takes no account of tide
During stream,Take the reference voltage of faulty line；
The form of matrix is write as in wushu (16)~(18), and the order of trouble point sequence electric current and sequence voltage unknown quantity is also according to failure
The numbering of point sorts successively, is consistent, can obtain with putting in order for unknown quantity in formula (15)：
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In formula (19)
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(E(3N×3N)For the unit diagonal matrix of 3N rows 3N row)
8. short-circuit current calculation method according to claim 7, it is characterised in that step S4 is specifically included：
Wushu (15) can be obtained with formula (19) simultaneous：
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By (20) as it can be seen that for broken string and cross-line simultaneous faults, when the number of trouble point is N, it can arrange and write out 6N bar equations, not
Each sequence voltage and each sequence electric current of the amount of knowing for each trouble point, unknown quantity number is also 6N, therefore equation can solve each event
The voltage of each sequence of barrier point and the electric current of each sequence, so can be tried to achieve by symmetrical components inverse transformation all trouble points phase voltage and
Phase current.
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| CN112630587A (en) * | 2020-11-11 | 2021-04-09 | 大唐水电科学技术研究院有限公司 | Single-circuit three-phase disconnection fault analysis method for four-circuit line erected on same pole |
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Application publication date: 20180511 |