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 PDF

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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
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CN102879668A (en
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王钢
吴争荣
李海锋
钟庆
梁远升
汪隆君
潘国清
高翔
王辉
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South China University of Technology SCUT
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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

Asymmetric fault analysis method for power distribution network containing inverter type distributed power supply
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:
<math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein, <math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>11</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>12</mn> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>1</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>21</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>22</mn> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>2</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mi>m</mi> <mn>1</mn> </mrow> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mi>m</mi> <mn>2</mn> </mrow> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</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;
<math> <mrow> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>1</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>2</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>-</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>m</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> Indicating the node voltage at fault, whereinPositive and negative sequence voltages of the PCC at fault, respectively;
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>1</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>2</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>m</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> indicating the injection current of the node at fault, whereinInjecting a current of a PCC for a Distributed Generation (DG) at fault;
expressed as: <math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>I</mi> <mrow> <mi>d</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>jI</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msub> <mi>I</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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:
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>jk</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>i</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>k</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mrow> <msub> <mi>Z</mi> <mi>jk</mi> </msub> </mfrac> </mrow> </math>
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:
<math> <mrow> <mo>[</mo> <mi>Y</mi> <mo>]</mo> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> </mrow> </math>
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>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</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>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <mi>Y</mi> <mrow> <mn>2</mn> <mi>n</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</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>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</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;
<math> <mrow> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mn>1</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mn>2</mn> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>PCC</mi> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>n</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> Which represents the node voltage during normal operation,is the voltage of the PCC during normal operation;
<math> <mrow> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mn>1</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mn>2</mn> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>DG</mi> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>n</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> representing the injected current of the node in normal operation,injecting a current of a PCC node into a DG in normal operation;
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:
<math> <mrow> <mo>[</mo> <mi>Y</mi> <mo>]</mo> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> </mrow> </math>
the node voltage equation in normal operation in this embodiment 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>L</mi> <mn>1</mn> </msub> </mfrac> </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> <msub> <mi>Z</mi> <mrow> <mi>L</mi> <mn>2</mn> </mrow> </msub> </mfrac> </mtd> </mtr> </mtable> </mfenced> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>M</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>PCC</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <mfrac> <msub> <mover> <mi>E</mi> <mo>&CenterDot;</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>&CenterDot;</mo> </mover> <mi>DG</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
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:
<math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein, <math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>11</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>12</mn> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>1</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>21</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>22</mn> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>2</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mi>m</mi> <mn>1</mn> </mrow> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mi>m</mi> <mn>2</mn> </mrow> </msub> </mtd> <mtd> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> <mo>&CenterDot;</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</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>&beta;</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>&beta;</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>&beta;</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>&beta;</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>&CenterDot;</mo> </mover> <mrow> <mi>M</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</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>&CenterDot;</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>&CenterDot;</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</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:
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>I</mi> <mrow> <mi>d</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>jI</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msub> <mi>I</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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>&CenterDot;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <msub> <mi>E</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>cos</mi> <mi>&alpha;</mi> <mo>+</mo> <mi>j</mi> <mi>sin</mi> <mi>&alpha;</mi> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math> To obtainSolving the system of equations:
<math> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mi>a</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>e</mi> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mo>-</mo> <mi>c</mi> <mo>=</mo> <msub> <mi>E</mi> <mi>s</mi> </msub> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mi>cos</mi> <mi>&alpha;</mi> </mtd> </mtr> <mtr> <mtd> <mi>b</mi> <msup> <mrow> <mo>(</mo> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mi>f</mi> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mo>-</mo> <mi>d</mi> <mo>=</mo> <msub> <mi>E</mi> <mi>s</mi> </msub> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mi>sin</mi> <mi>&alpha;</mi> </mtd> </mtr> </mtable> </mfenced> </math>
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:
<math> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mrow> <mo>(</mo> <mi>a</mi> <mo>-</mo> <msub> <mi>K</mi> <mi>q</mi> </msub> <mo>/</mo> <msub> <mi>U</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mi>c</mi> <msub> <mi>E</mi> <mi>s</mi> </msub> <mi>cos</mi> <mi>&alpha;</mi> <mo>+</mo> <mi>d</mi> <msub> <mi>E</mi> <mi>s</mi> </msub> <mi>sin</mi> <mi>&alpha;</mi> <mo>=</mo> <msub> <mi>I</mi> <mi>max</mi> </msub> <mo>-</mo> <mrow> <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> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mo>/</mo> <msub> <mi>U</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>U</mi> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <msub> <mi>K</mi> <mi>q</mi> </msub> <mo>/</mo> <msub> <mi>U</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>d</mi> <msub> <mi>E</mi> <mi>s</mi> </msub> <mi>cos</mi> <mi>&alpha;</mi> <mo>-</mo> <mi>c</mi> <msub> <mi>E</mi> <mi>s</mi> </msub> <mi>sin</mi> <mi>&alpha;</mi> <mo>=</mo> <mo>-</mo> <mrow> <mo>(</mo> <msub> <mi>I</mi> <mrow> <mi>q</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <msub> <mi>K</mi> <mi>q</mi> </msub> <msub> <mi>U</mi> <mi>PCC</mi> </msub> <mo>/</mo> <msub> <mi>U</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mtd> </mtr> </mtable> </mfenced> </math>
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:
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>jk</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>i</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>k</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mrow> <msub> <mi>Z</mi> <mi>jk</mi> </msub> </mfrac> </mrow> </math>
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 UPCCObtainable 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 UPCCObtainable 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:
[ Y ] U . = I .
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>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</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;
<math> <mrow> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mn>1</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mn>2</mn> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>PCC</mi> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>n</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> Which represents the node voltage during normal operation,a voltage of a common junction point PCC for normal operation;
<math> <mrow> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mn>1</mn> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mn>2</mn> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>DG</mi> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>n</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> representing the injected current of the node in normal operation,injecting current of a PCC node at a common junction point for the distributed power supply during normal operation;
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:
<math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein, <math> <mrow> <mo>[</mo> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mo>]</mo> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>11</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>12</mn> </msub> </mtd> <mtd> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>1</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>21</mn> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mn>22</mn> </msub> </mtd> <mtd> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mn>2</mn> <mi>m</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> <mtd> </mtd> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</mo> </msup> <mrow> <mi>m</mi> <mn>1</mn> </mrow> </msub> </mtd> <mtd> <msub> <msup> <mi>Y</mi> <mo>&prime;</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>&prime;</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;
<math> <mrow> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>1</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>2</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msubsup> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>PCC</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>-</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>m</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> Indicating the node voltage at fault, whereinPositive and negative sequence voltages of a common junction point PCC at fault are respectively;
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mi>f</mi> </msub> <mo>=</mo> <mfenced open='[' close=']'> <mtable> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>1</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mn>2</mn> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> </mtd> </mtr> <mtr> <mtd> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>m</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> indicating the injection current of the node at fault, whereinInjecting current of a common junction point PCC for the distributed power supply in fault;
expressed as: <math> <mrow> <msub> <mover> <mi>I</mi> <mo>.</mo> </mover> <mrow> <mi>DG</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>I</mi> <mrow> <mi>d</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>jI</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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>&CenterDot;</mo> <mi>f</mi> </mrow> <mo>+</mo> </msubsup> </mtd> </mtr> <mtr> <mtd> <msub> <mi>I</mi> <mrow> <mi>q</mi> <mo>&CenterDot;</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>&CenterDot;</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:
<math> <mrow> <msub> <mover> <mi>I</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>jk</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>j</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mover> <mi>U</mi> <mo>&CenterDot;</mo> </mover> <mrow> <mi>k</mi> <mo>&CenterDot;</mo> <mi>f</mi> </mrow> </msub> </mrow> <msub> <mi>Z</mi> <mi>jk</mi> </msub> </mfrac> </mrow> </math>
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.
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