CN102879668B - Asymmetric fault analysis method for power distribution network including inverted distribution type power supply - Google Patents
Asymmetric fault analysis method for power distribution network including inverted distribution type power supply Download PDFInfo
- Publication number
- CN102879668B CN102879668B CN201210344127.6A CN201210344127A CN102879668B CN 102879668 B CN102879668 B CN 102879668B CN 201210344127 A CN201210344127 A CN 201210344127A CN 102879668 B CN102879668 B CN 102879668B
- Authority
- CN
- China
- Prior art keywords
- mtd
- mtr
- msub
- mrow
- centerdot
- Prior art date
- Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
- Expired - Fee Related
Links
- 238000004458 analytical method Methods 0.000 title claims abstract description 19
- 238000000034 method Methods 0.000 claims description 9
- 239000002131 composite material Substances 0.000 claims description 7
- 239000011159 matrix material Substances 0.000 claims description 5
- 230000014509 gene expression Effects 0.000 claims description 4
- 230000009466 transformation Effects 0.000 claims description 3
- 238000002347 injection Methods 0.000 claims description 2
- 239000007924 injection Substances 0.000 claims description 2
- 238000011217 control strategy Methods 0.000 abstract description 5
- 230000008878 coupling Effects 0.000 abstract 1
- 238000010168 coupling process Methods 0.000 abstract 1
- 238000005859 coupling reaction Methods 0.000 abstract 1
- 230000001360 synchronised effect Effects 0.000 description 3
- 238000010586 diagram Methods 0.000 description 2
- 230000004048 modification Effects 0.000 description 2
- 238000012986 modification Methods 0.000 description 2
- 238000006467 substitution reaction Methods 0.000 description 2
- 230000009286 beneficial effect Effects 0.000 description 1
- 230000007547 defect Effects 0.000 description 1
- 230000006872 improvement Effects 0.000 description 1
- 230000008569 process Effects 0.000 description 1
- 230000001052 transient effect Effects 0.000 description 1
- 238000011144 upstream manufacturing Methods 0.000 description 1
Landscapes
- Supply And Distribution Of Alternating Current (AREA)
Abstract
The invention discloses an asymmetric fault analysis method for a power distribution network including an inverted distribution type power supply. The asymmetric fault analysis method comprises the steps as follows: firstly, calculating a voltage amplitude value UPCC of a point of common coupling PCC when the power distribution network runs normally in combination with a fault through control strategy and an output characteristic of the distribution type power supply; and secondly, building an equation set of a positive sequence voltage amplitude value U+<PCC.f> of the PCC when the power distribution network has a fault to calculate a positive sequence voltage of the PCC when the power distribution network has the fault so as to finish fault analysis of the power distribution network including the inverted distribution type power supply. Compared with the prior art, the accuracy of fault analysis is improved.
Description
Technical Field
The invention relates to a fault analysis method for a power system, in particular to an asymmetric fault analysis method for a power distribution network containing an inverter type distributed power supply.
Background
The symmetrical component method is the most important method for asymmetrical fault analysis, and for a traditional synchronous motor, the voltage source equivalent value can be adopted in a positive sequence network, and the impedance equivalent value can be adopted in a negative sequence network. However, the inverter type distributed power supply is completely different from the traditional synchronous motor type power supply, the output characteristic of the inverter type distributed power supply is completely determined by the control strategy of the inverter type distributed power supply under the asymmetric fault condition, and the traditional synchronous motor equivalent model cannot be adopted in a positive sequence network and a negative sequence network. Therefore, the conventional fault analysis method inevitably generates a large error. In order to realize accurate analysis of asymmetric faults of a power distribution network containing the inverter type distributed power supply, improvement must be carried out on an equivalent model of the inverter type power supply and a fault analysis model of the power distribution network.
Disclosure of Invention
In order to overcome the defects and shortcomings of the prior art, the invention provides a power distribution network asymmetric fault analysis method considering a fault ride-through control strategy and output characteristics aiming at an inverter type distributed power supply controlled by a positive sequence component, and realizes accurate analysis of asymmetric faults of the power distribution network containing the inverter type distributed power supply.
The purpose of the invention is realized by the following technical scheme: the asymmetric fault analysis method of the power distribution network with the inverter-type distributed power supply comprises the following steps (all variables in the following steps represent per unit values):
s1, calculating the voltage amplitude U of the common junction Point (PCC) when the power distribution network is in normal operationPCC;
S2, establishing a composite sequence network when the power distribution network fails, wherein the distributed power source is only contained in the positive sequence network, and establishing the positive sequence voltage amplitude of the PCC when the power distribution network failsThe solving of the system of equations specifically comprises the following steps:
s21, obtaining a node voltage equation during the fault according to the composite sequence network during the fault of the power distribution network:
wherein, <math>
<mrow>
<mo>[</mo>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mo>]</mo>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>11</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>12</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>1</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>21</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>22</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>2</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>1</mn>
</mrow>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>2</mn>
</mrow>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mi>mm</mi>
</msub>
</mtd>
</mtr>
</mtable>
</mfenced>
</mrow>
</math> representing a node admittance matrix at the time of the fault; diagonal element YiiThe self-admittance is the self-admittance when the node i fails, and the value of the self-admittance is equal to the sum of the admittances of all branches connected to the node i when the node i fails; off diagonal element Y'ijFor mutual admittance in case of failure between nodes i, j, when a node is in operationi. Y 'when there is a branch between j'ijEqual to the negative of the branch admittance coupled directly between nodes i, j; y 'when no branch exists between nodes i and j'ij=0;
expressed as: <math>
<mrow>
<msub>
<mover>
<mi>I</mi>
<mo>·</mo>
</mover>
<mrow>
<mi>DG</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>I</mi>
<mrow>
<mi>d</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>-</mo>
<msub>
<mi>jI</mi>
<mrow>
<mi>q</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mi></mi>
<mrow>
<mo>(</mo>
<mn>2</mn>
<mo>)</mo>
</mrow>
</mrow>
</math>
wherein, <math>
<mfenced open='{' close=''>
<mtable>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>d</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>P</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>/</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</mrow>
</msub>
<mo>+</mo>
<msub>
<mi>k</mi>
<mi>q</mi>
</msub>
<mrow>
<mo>(</mo>
<msub>
<mi>U</mi>
<mi>PCC</mi>
</msub>
<mo>-</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
<mo>)</mo>
</mrow>
</mtd>
</mtr>
</mtable>
</mfenced>
</math>
in the formula Id·f、Iq·fRespectively representing the output active current and the output reactive current of the distributed power supply during fault; p(0)、Iq(0)Respectively representing active power and reactive current output by the distributed power supply in normal operation,Q(0)representing reactive power, k, of the distributed power supply output during normal operationqRepresenting the coefficients;
s22 linear transformation is carried out on the formula (1) to obtainSolving a system of equations;
s3 solving
S4 solving for Id·f、Iq·f;
S5 definition Iad=Id·f+Iq·fAnd judging IadWhether the rated current I of the inverter is exceededVSC·nIf not, proceedStep S6, otherwise, let Id·f=IVSC·n-Iq·fIs shown byd·f、Iq·fThe expression is substituted for the formula (1) and recalculatedThen step S6 is performed;
s6 calculation Using equation (1)Middle removingAn outer node voltage;
s7 calculates the branch current between node j and node k according to:
wherein,Zjkrespectively show the sectionsThe branch current between points j and k and the branch impedance,and respectively represent the voltages of the nodes j and k when the power distribution network fails.
Step S1, calculating the voltage amplitude U of the PCC during normal operation of the power distribution networkPCCThe method specifically comprises the following steps:
s11, obtaining a node voltage equation in normal operation according to the distribution network equivalent graph in normal operation:
wherein, <math>
<mrow>
<mo>[</mo>
<mi>Y</mi>
<mo>]</mo>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mn>11</mn>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mn>12</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mn>1</mn>
<mi>n</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mn>21</mn>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mn>22</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mn>2</mn>
<mi>n</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mi>n</mi>
<mn>1</mn>
</mrow>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mi>n</mi>
<mn>2</mn>
</mrow>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mi>nn</mi>
</msub>
</mtd>
</mtr>
</mtable>
</mfenced>
</mrow>
</math> representing a node admittance matrix at normal operation; diagonal element YijIs the self-admittance of the node i in normal operation, and the value is equal to all the nodes connected to the node i in normal operationSum of branch admittance; off diagonal element YijFor mutual admittance during normal operation between nodes i and j, when a branch exists between nodes i and j, YijEqual to the negative of the branch admittance coupled directly between nodes i, j; when no branch exists between nodes i and j, Yij=0;
s12 is obtained according to the node voltage equationThen, the absolute value is taken to obtain UPCC。
Compared with the prior art, the invention has the following advantages and beneficial effects: according to the method, a new distributed power supply transient equivalent model is established by considering the control strategy and the output characteristic of the inverter-type distributed power supply, so that the fault current characteristic of the distributed power supply can be reflected more truly; on the basis, a new power distribution network fault analysis model is established, and the accuracy of fault analysis is improved. The method provides scientific basis for the aspects of equipment model selection, protection setting and the like of the distribution network containing the inverter distributed power supply, and has strong practicability in engineering practice.
Drawings
Fig. 1 is a single line diagram of a power distribution network according to an embodiment of the present invention.
Fig. 2 is a flowchart of an asymmetric fault analysis method for a power distribution network including an inverter-type distributed power supply according to an embodiment of the present invention.
Fig. 3 is a composite grid sequence diagram of the power distribution network shown in fig. 1 in case of a fault according to an embodiment of the present invention.
Detailed Description
The present invention will be described in further detail with reference to examples and drawings, but the present invention is not limited to these examples.
Examples
In this embodiment, taking the power distribution network shown in fig. 1 as an example, the method for analyzing the asymmetric fault of the power distribution network including the inverter-type distributed power supply according to the present invention includes, as shown in fig. 2, the following steps:
s1 calculating the voltage amplitude U of PCC when the power distribution network operates normallyPCCThe method specifically comprises the following steps:
s11, obtaining a node voltage equation in normal operation according to the distribution network equivalent graph in normal operation:
the node voltage equation in normal operation in this embodiment is:
whereinRespectively the self-admittance of the normal runtime node M, PCC,is the transadmittance between nodes M, PCC; zs、ZL1、ZL2Respectively representing the equivalent impedance of the system, the impedance of a PCC upstream line L1 and the impedance of a PCC downstream line L2;represents the equivalent potential of the system and is,taking the value as the rated current of the distributed power supply;is the voltage of node M in normal operation;is the voltage of the PCC during normal operation;is the current injected into point M.
S12 is obtained according to the node voltage equationThen, the absolute value is taken to obtain UPCC;
S2, establishing a composite sequence network (as shown in figure 3) when the power distribution network fails, wherein the distributed power source is only contained in the positive sequence network, and establishing PCC of the PCC when the power distribution network failsPositive sequence voltageThe equation system is solved by the following specific steps:
s21, obtaining a node voltage equation during the fault according to the composite sequence network during the fault of the power distribution network:
wherein, <math>
<mrow>
<mo>[</mo>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mo>]</mo>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>11</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>12</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>1</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>21</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>22</mn>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>2</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>1</mn>
</mrow>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>2</mn>
</mrow>
</msub>
</mtd>
<mtd>
<mo>·</mo>
<mo>·</mo>
<mo>·</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mi>mm</mi>
</msub>
</mtd>
</mtr>
</mtable>
</mfenced>
</mrow>
</math> an admittance matrix representing the node at the time of the fault; diagonal element YiiThe self-admittance is the self-admittance when the node i fails, and the value of the self-admittance is equal to the sum of the admittances of all branches connected to the node i when the node i fails; off diagonal element Y'ijIs mutual admittance in case of fault between nodes i and j, and Y 'when a branch circuit exists between nodes i and j'ijEqual to the negative of the admittance of the branch directly coupled between nodes i, j, Y 'when no branch is present between nodes i, j'ij=0;
Namely, it is <math>
<mrow>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mi>s</mi>
</msub>
</mfrac>
<mo>+</mo>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>1</mn>
</mrow>
</msub>
</mfrac>
</mtd>
<mtd>
<mo>-</mo>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>1</mn>
</mrow>
</msub>
</mfrac>
</mtd>
<mtd>
<mn>0</mn>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>-</mo>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>1</mn>
</mrow>
</msub>
</mfrac>
</mtd>
<mtd>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>1</mn>
</mrow>
</msub>
</mfrac>
<mo>+</mo>
<mfrac>
<mn>1</mn>
<mrow>
<mn>2</mn>
<mi>β</mi>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>2</mn>
</mrow>
</msub>
</mrow>
</mfrac>
</mtd>
<mtd>
<mo>-</mo>
<mfrac>
<mn>1</mn>
<mrow>
<mn>2</mn>
<mi>β</mi>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>2</mn>
</mrow>
</msub>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr>
<mtd>
<mn>0</mn>
</mtd>
<mtd>
<mo>-</mo>
<mfrac>
<mn>1</mn>
<mrow>
<mn>2</mn>
<mi>β</mi>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>2</mn>
</mrow>
</msub>
</mrow>
</mfrac>
</mtd>
<mtd>
<mfrac>
<mn>1</mn>
<mrow>
<mn>2</mn>
<mi>β</mi>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>2</mn>
</mrow>
</msub>
</mrow>
</mfrac>
<mo>+</mo>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mi>s</mi>
</msub>
</mfrac>
<mo>+</mo>
<mfrac>
<mn>1</mn>
<msub>
<mi>Z</mi>
<mrow>
<mi>L</mi>
<mn>1</mn>
</mrow>
</msub>
</mfrac>
</mtd>
</mtr>
</mtable>
</mfenced>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<mover>
<mi>U</mi>
<mo>·</mo>
</mover>
<mrow>
<mi>M</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>U</mi>
<mo>·</mo>
</mover>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msubsup>
<mover>
<mi>U</mi>
<mo>·</mo>
</mover>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>-</mo>
</msubsup>
</mtd>
</mtr>
</mtable>
</mfenced>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<mfrac>
<msub>
<mover>
<mi>E</mi>
<mo>·</mo>
</mover>
<mi>s</mi>
</msub>
<msub>
<mi>Z</mi>
<mi>s</mi>
</msub>
</mfrac>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mover>
<mi>I</mi>
<mo>·</mo>
</mover>
<mrow>
<mi>DG</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mn>0</mn>
</mtd>
</mtr>
</mtable>
</mfenced>
</mrow>
</math>
WhereinThe self-admittance of the positive sequence voltage node and the PCC negative sequence voltage node of the node M, PCC in the fault state respectively;mutual admittance between positive sequence voltage nodes of node M, PCC at fault; ,the mutual admittance between the PCC positive sequence voltage node and the PCC negative sequence voltage node is represented, beta is represented by the fault position in a line L2, and the value is 0-100%;positive and negative sequence voltages of the PCC at fault, respectively;injecting a current of PCC for DG in fault; in accordance with the distributed power control strategy,expressed as:
wherein, <math>
<mfenced open='{' close=''>
<mtable>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>d</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>P</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>/</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</mrow>
</msub>
<mo>+</mo>
<msub>
<mi>k</mi>
<mi>q</mi>
</msub>
<mrow>
<mo>(</mo>
<msub>
<mi>U</mi>
<mi>PCC</mi>
</msub>
<mo>-</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
<mo>)</mo>
</mrow>
</mtd>
</mtr>
</mtable>
</mfenced>
</math>
in the formula,Id·f、Iq·frespectively representing fault current, active current and reactive current output by the distributed power supply during fault; p(0)、Iq(0)Respectively representing active power and reactive current output by the distributed power supply in normal operation,Q(0)representing reactive power, k, of the distributed power supply output during normal operationqRepresenting the coefficients;
s22 linear transformation is carried out on the formula (1) to obtainThe solving equation set of (1) is specifically as follows:
to be provided withThe voltage phase is taken as a reference, then <math>
<mrow>
<msub>
<mover>
<mi>E</mi>
<mo>·</mo>
</mover>
<mi>s</mi>
</msub>
<mo>=</mo>
<msub>
<mi>E</mi>
<mi>s</mi>
</msub>
<mrow>
<mo>(</mo>
<mi>cos</mi>
<mi>α</mi>
<mo>+</mo>
<mi>j</mi>
<mi>sin</mi>
<mi>α</mi>
<mo>)</mo>
</mrow>
<mo>,</mo>
</mrow>
</math> To obtainSolving the system of equations:
wherein a + jb = [ 1/(2. beta. Z)L2+Zs+ZL1)+1/(Zs+L1)-jkq/Un](Zs+ZL1),c+jd=UPCCId(0)(Zs+ZL1),e+jf=j(Iq(0)+kqUPCC/Un)(Zs+ZL1);
S3 solving
S4 solving for Id·f、Iq·f;
S5 definition Iad=Id·f+Iq·fAnd judging IadWhether the rated current I of the inverter is exceededVAC·nIf not, go to step S6, otherwise, letId·f=IVSC·n-Iq·fIs shown byd·f、Iq·fThe expressions are substituted for equations (1) and (2), and recalculatedThen step S6 is performed;
wherein recalculatingThe process of (2) is as follows:
will Id·f、Iq·fThe expressions are substituted for the formula (1) and the formula (2) to obtainSolving the system of equations:
wherein a + jb = 1/(2. beta. Z)L2+Zs+ZL1)+1/(Zs+ZL1),c+jd=1/(Zs+ZL1) (ii) a S6 calculation Using equation (1)Middle removingAn outer node voltage;
s7 calculates the branch current between nodes j and k according to:
wherein,Zjkrepresenting the branch current and branch impedance between nodes j and k,and respectively represent the voltages of the nodes j and k when the power distribution network fails.
In fig. 1 of the present embodiment, the equivalent impedance Z of the power distribution grid systems1.3j (Ω), equivalent impedance Z of line L1L1L2 equivalent impedance ZL21.18+3.56j (omega) and 0.59+1.78j (omega), respectively, and the rated capacity of the distributed power supply and the rated capacity of the inverter interface are respectively 4MW and 8MVA, kqIs 2. From the above conditions I can be obtainedVSC·n0.46 kA. The active output of the distributed power supply is a rated value before the fault, and the reactive output is zero.
Two different short circuit conditions are listed below for illustration:
case 1:
the two-phase short-circuit fault occurs at the tail end of the L2 line, namely beta =100%, the steps S1-S4 are carried out, and the calculation result is obtainedHas a value of 7.26kV, UPCCIs 10.1 kV. By UPCC、Obtainable of Id·f、Iq·f、IDG·fRespectively at 0.32kA, 0.13kA and 0.35 kA. Due to Id·f+Iq·f<IVSC·nIs shown byd·f、Iq·fSubstituting into the equation of node voltage when the power distribution network is in faultThe voltage of the power supply is 4.10kV,respectively 0.365kA and 0.48kA, are all 0.48 kA.
Case 2:
the two-phase short-circuit fault occurs at 70% of the L2 line, namely beta =70%, the steps S1-S4 are carried out, and the two-phase short-circuit fault is obtainedSolving the system of equations, calculatingHas a value of 6.92kV, UPCCIs 10.1 kV. By UPCC、Obtainable of Id·f、Iq·f0.33kA and 0.14kA, respectively. Due to Id·f+Iq·f>IVSC·nRecalculating Id·f、IDG·fThe concentration was 0.32kA and 0.35 kA. Finally, find outThe voltage of the power supply is 4.10kV,are respectively 0.39kA and 0.52kA,are all 0.52 Ka.
The above embodiments are preferred embodiments of the present invention, but the present invention is not limited to the above embodiments, and any other changes, modifications, substitutions, combinations, and simplifications which do not depart from the spirit and principle of the present invention should be construed as equivalents thereof, and all such changes, modifications, substitutions, combinations, and simplifications are intended to be included in the scope of the present invention.
Claims (1)
1. The asymmetric fault analysis method of the power distribution network containing the inverter type distributed power supply is characterized by comprising the following steps of:
s1, calculating the voltage amplitude U of the PCC at the common junction point when the power distribution network operates normallyPCCThe method specifically comprises the following steps:
s11, obtaining a node voltage equation in normal operation according to the distribution network equivalent graph in normal operation:
wherein, <math>
<mrow>
<mo>[</mo>
<mi>Y</mi>
<mo>]</mo>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mn>11</mn>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mn>12</mn>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mn>1</mn>
<mi>n</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mn>21</mn>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mn>22</mn>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mn>2</mn>
<mi>n</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mi>n</mi>
<mn>1</mn>
</mrow>
</msub>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mrow>
<mi>n</mi>
<mn>2</mn>
</mrow>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<mi>Y</mi>
<mi>nn</mi>
</msub>
</mtd>
</mtr>
</mtable>
</mfenced>
<mo>,</mo>
</mrow>
</math> representing a node admittance matrix at normal operation; diagonal element YiiIs the self-admittance of the node i in normal operation, and the value of the self-admittance is equal to the sum of all branch admittance connected with the node i in normal operation; off diagonal element YijFor mutual admittance during normal operation between nodes i and j, when a branch exists between nodes i and j, YijEqual to the negative of the branch admittance coupled directly between nodes i, j; when no branch exists between nodes i and j, Yij=0;
s12 is obtained according to the node voltage equationThen, the absolute value is taken to obtain UPCC;
S2, a composite sequence network is established when the power distribution network fails, the distributed power sources are only contained in the positive sequence network, and the positive sequence voltage amplitude value of the PCC of the common junction point is established when the power distribution network failsThe solving of the system of equations specifically comprises the following steps:
s21, obtaining a node voltage equation during the fault according to the composite sequence network during the fault of the power distribution network:
wherein, <math>
<mrow>
<mo>[</mo>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mo>]</mo>
<mo>=</mo>
<mfenced open='[' close=']'>
<mtable>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>11</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>12</mn>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>1</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>21</mn>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mn>22</mn>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mn>2</mn>
<mi>m</mi>
</mrow>
</msub>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
<mtd>
</mtd>
<mtd>
<mo>·</mo>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>1</mn>
</mrow>
</msub>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mrow>
<mi>m</mi>
<mn>2</mn>
</mrow>
</msub>
</mtd>
<mtd>
<mo>.</mo>
<mo>.</mo>
<mo>.</mo>
</mtd>
<mtd>
<msub>
<msup>
<mi>Y</mi>
<mo>′</mo>
</msup>
<mi>mm</mi>
</msub>
</mtd>
</mtr>
</mtable>
</mfenced>
</mrow>
</math> representing a node admittance matrix at the time of the fault; diagonal element Y'iiThe self-admittance is the self-admittance when the node i fails, and the value of the self-admittance is equal to the sum of the admittances of all branches connected to the node i when the node i fails; off diagonal element Y'ijIs mutual admittance in case of fault between nodes i and j, and Y 'when a branch circuit exists between nodes i and j'ijEqual to the negative of the admittance of the branch directly coupled between nodes i, j, Y 'when no branch is present between nodes i, j'ij=0;
expressed as: <math>
<mrow>
<msub>
<mover>
<mi>I</mi>
<mo>.</mo>
</mover>
<mrow>
<mi>DG</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>I</mi>
<mrow>
<mi>d</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>-</mo>
<msub>
<mi>jI</mi>
<mrow>
<mi>q</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>-</mo>
<mo>-</mo>
<mo>-</mo>
<mrow>
<mo>(</mo>
<mn>2</mn>
<mo>)</mo>
</mrow>
</mrow>
</math>
wherein, <math>
<mfenced open='{' close=''>
<mtable>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>d</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>P</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</msub>
<mo>/</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
</mtd>
</mtr>
<mtr>
<mtd>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>I</mi>
<mrow>
<mi>q</mi>
<mrow>
<mo>(</mo>
<mn>0</mn>
<mo>)</mo>
</mrow>
</mrow>
</msub>
<mo>+</mo>
<msub>
<mi>k</mi>
<mi>q</mi>
</msub>
<mrow>
<mo>(</mo>
<msub>
<mi>U</mi>
<mi>PCC</mi>
</msub>
<mo>-</mo>
<msubsup>
<mi>U</mi>
<mrow>
<mi>PCC</mi>
<mo>·</mo>
<mi>f</mi>
</mrow>
<mo>+</mo>
</msubsup>
<mo>)</mo>
</mrow>
</mtd>
</mtr>
</mtable>
</mfenced>
</math>
in the formula Id·f、Iq·fRespectively representing active current and reactive current output by the distributed power supply during fault; p(0)、Iq(0)Respectively representing active power and reactive current output by the distributed power supply in normal operation,Q(0)representing reactive power, k, of the distributed power supply output during normal operationqRepresenting the coefficients;
s22 linear transformation is carried out on the formula (1) to obtainSolving a system of equations;
s3 solving
S4 solving for Id·f、Iq·f;
S5 definition Iad=Id·f+Iq·fAnd judging IadWhether the rated current I of the inverter is exceededVSC·nIf not, go to step S6, otherwise, let Id·f=IVSC·n-Iq·fIs shown byd·f、Iq·fThe expressions are substituted for equations (1) and (2), and recalculatedThen step S6 is performed;
s6 calculation Using equation (1)Middle removingAn outer node voltage;
s7 calculates the branch current between node j and node k according to:
wherein,representing the branch current and branch impedance between nodes j and k,and respectively represent the voltages of the nodes j and k when the power distribution network fails.
Priority Applications (1)
Application Number | Priority Date | Filing Date | Title |
---|---|---|---|
CN201210344127.6A CN102879668B (en) | 2012-09-17 | 2012-09-17 | Asymmetric fault analysis method for power distribution network including inverted distribution type power supply |
Applications Claiming Priority (1)
Application Number | Priority Date | Filing Date | Title |
---|---|---|---|
CN201210344127.6A CN102879668B (en) | 2012-09-17 | 2012-09-17 | Asymmetric fault analysis method for power distribution network including inverted distribution type power supply |
Publications (2)
Publication Number | Publication Date |
---|---|
CN102879668A CN102879668A (en) | 2013-01-16 |
CN102879668B true CN102879668B (en) | 2014-12-03 |
Family
ID=47481064
Family Applications (1)
Application Number | Title | Priority Date | Filing Date |
---|---|---|---|
CN201210344127.6A Expired - Fee Related CN102879668B (en) | 2012-09-17 | 2012-09-17 | Asymmetric fault analysis method for power distribution network including inverted distribution type power supply |
Country Status (1)
Country | Link |
---|---|
CN (1) | CN102879668B (en) |
Families Citing this family (13)
Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
CN103487698B (en) * | 2013-09-29 | 2016-01-20 | 国家电网公司 | A kind of failure analysis methods of distributed power source access power distribution network |
CN104022500B (en) * | 2014-05-14 | 2016-02-03 | 华南理工大学 | The micro-capacitance sensor failure analysis methods of inverse distributed power is controlled containing V/f |
CN104331619B (en) * | 2014-10-30 | 2017-07-14 | 国网上海市电力公司 | A kind of Analysis of Short-Circuit Current method of the power system of interface power containing inverter |
CN104698396A (en) * | 2015-03-17 | 2015-06-10 | 天津大学 | Fault characteristic analyzing method for grid-connected inverter type distributed power supply |
CN104698346A (en) * | 2015-03-31 | 2015-06-10 | 国网内蒙古东部电力有限公司通辽供电公司 | Method and device for analyzing faults of source-containing power distribution network |
CN105305392B (en) * | 2015-10-21 | 2018-12-21 | 国家电网公司 | Short circuit calculation symmetrical component method suitable for the type IIDG power distribution network of control containing voltage |
CN106353564B (en) * | 2016-09-30 | 2019-01-15 | 广东电网有限责任公司电力调度控制中心 | The Power System Shortcuts electric current acquisition methods of meter and V/X Connection Traction Transformer |
CN106841850B (en) * | 2016-12-24 | 2019-04-05 | 国网吉林省电力有限公司培训中心 | A kind of distribution network failure analysis method containing inverse distributed power |
CN107064736B (en) * | 2017-03-22 | 2019-04-09 | 华南理工大学 | A kind of Fault Locating Method connecing inverse distributed power power distribution network containing more T |
CN107576886A (en) * | 2017-09-13 | 2018-01-12 | 华南理工大学 | The single-phase-to-earth fault analysis method of the small resistance grounding system containing inverse distributed power |
CN108387818A (en) * | 2018-01-23 | 2018-08-10 | 中国石油大学(华东) | A kind of fault distance-finding method suitable for the tree-shaped catalogue containing distributed generation resource |
CN110018411B (en) * | 2019-05-14 | 2021-06-15 | 集美大学 | Inverter circuit fault diagnosis method based on symmetric component analysis |
CN110880764B (en) * | 2019-10-30 | 2023-05-23 | 华南理工大学 | Fault processing method for unbalanced distribution network containing inversion type distributed power supply |
Citations (3)
Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
EP0276181A1 (en) * | 1987-01-22 | 1988-07-27 | Schlumberger Industries | Method and device for checking faulty conductors of an electrical line |
CN101562332A (en) * | 2009-05-27 | 2009-10-21 | 天津大学 | Self-adaptive current fast tripping protection method for distribution network comprising inverse distributed power |
CN101813739A (en) * | 2010-04-14 | 2010-08-25 | 天津大学 | Adaptive three-phase symmetric fault phase selecting method for ultra high voltage transmission line |
-
2012
- 2012-09-17 CN CN201210344127.6A patent/CN102879668B/en not_active Expired - Fee Related
Patent Citations (3)
Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
EP0276181A1 (en) * | 1987-01-22 | 1988-07-27 | Schlumberger Industries | Method and device for checking faulty conductors of an electrical line |
CN101562332A (en) * | 2009-05-27 | 2009-10-21 | 天津大学 | Self-adaptive current fast tripping protection method for distribution network comprising inverse distributed power |
CN101813739A (en) * | 2010-04-14 | 2010-08-25 | 天津大学 | Adaptive three-phase symmetric fault phase selecting method for ultra high voltage transmission line |
Non-Patent Citations (2)
Title |
---|
张健.逆变型分布式电源故障特性分析及配电网保护策略研究.《中国优秀硕士学位论文全文数据库工程科技Ⅱ辑 》.2012,(第7期),C042-284. * |
逆变型分布式电源故障特性分析及配电网保护策略研究;张健;《中国优秀硕士学位论文全文数据库工程科技Ⅱ辑 》;20120715(第7期);C042-284 * |
Also Published As
Publication number | Publication date |
---|---|
CN102879668A (en) | 2013-01-16 |
Similar Documents
Publication | Publication Date | Title |
---|---|---|
CN102879668B (en) | Asymmetric fault analysis method for power distribution network including inverted distribution type power supply | |
CN102866315B (en) | Symmetric fault analyzing method containing inversion type distributed power distribution network | |
CN104485684B (en) | A kind of power system fault current computational methods containing double-fed fan motor unit | |
Haileselassie et al. | Power flow analysis of multi-terminal HVDC networks | |
CN104022500B (en) | The micro-capacitance sensor failure analysis methods of inverse distributed power is controlled containing V/f | |
CN101635456B (en) | Method for joint state estimation of transmission network and distribution network | |
CN104333244B (en) | Positive sequence component-based three-phase inverter control method and device | |
CN107272645B (en) | The photovoltaic electric station grid connection fault model and analysis method of Neutral Grounding through Resistance in Electrical | |
CN103345162A (en) | Power level digital-analog hybrid simulation system | |
CN105162099A (en) | Operational curved surface method for determining asymmetric short-circuit current of distributed generation access power grid | |
CN107576886A (en) | The single-phase-to-earth fault analysis method of the small resistance grounding system containing inverse distributed power | |
CN104698346A (en) | Method and device for analyzing faults of source-containing power distribution network | |
CN110880764B (en) | Fault processing method for unbalanced distribution network containing inversion type distributed power supply | |
CN103956735A (en) | Harmonic power flow analysis method of distributed power generation system | |
CN106291046A (en) | Mixed pressure common-tower double-return line single-phase across single-phase across voltage failure current calculation method | |
CN105305392B (en) | Short circuit calculation symmetrical component method suitable for the type IIDG power distribution network of control containing voltage | |
CN105977962B (en) | Transmission network-distribution network joint fault analysis method based on improved node method | |
CN107167707A (en) | A kind of double circuit lines distance-finding method and device based on unknown parameters | |
CN104917197A (en) | Method for parallel computation of unbalanced three-phase power flow of active power distribution network | |
CN110208634B (en) | Method for acquiring asymmetric short-circuit current direct-current component of complex power system | |
CN104092213B (en) | A kind of uncertain trend branch power analytical method based on optimization method | |
CN113030643B (en) | Fault analysis method and system for distribution network voltage source type distributed power supply | |
CN105098776A (en) | Calculation method for three-phase power flow of active power distribution network | |
CN113447803A (en) | Short-circuit current calculation voltage coefficient value taking method for checking on-off capacity of circuit breaker | |
CN104167732A (en) | Grid equivalent method based on phase angle difference of call lines |
Legal Events
Date | Code | Title | Description |
---|---|---|---|
C06 | Publication | ||
PB01 | Publication | ||
C10 | Entry into substantive examination | ||
SE01 | Entry into force of request for substantive examination | ||
C14 | Grant of patent or utility model | ||
GR01 | Patent grant | ||
CF01 | Termination of patent right due to non-payment of annual fee |
Granted publication date: 20141203 Termination date: 20200917 |
|
CF01 | Termination of patent right due to non-payment of annual fee |