CN103560494A - Method for obtaining setting short-circuit currents protected by power distribution network - Google Patents

Abstract

The invention discloses a method for obtaining setting short-circuit currents protected by a power distribution network. The method includes the steps that 1, a three-phase network equivalent model of the power distribution network is established by calculating three-phase line parameters and equivalent transformation of a transformer T; 2, rated load currents of the power distribution network are calculated according to the three-phase network equivalent model; 3, fault currents of the power distribution network are calculated by means of equivalent impedance between fault points and root nodes in the three-phase network equivalent model; 4, rated load currents of circuit branches and fault current vector quantities are added to calculate short-circuit current values of the circuit branches. The method has the advantages that a calculation method is simple, practical and high in calculation speed, the calculation result is reliable, the method is suitable for open-loop and close-loop networks, special handling is needless, and imbalance of three-phase parameters and dissymmetry of three-phase loads are avoided.

Description

The short circuit current acquisition methods that distribution protection is adjusted

Technical field

The present invention relates to the short circuit current acquisition methods that a kind of distribution protection is adjusted.

Background technology

The accurate configuration of power distribution network relaying protection to power distribution network protective relaying device efficiently, played vital effect to fault section isolation quickly and accurately.And the configuration of short circuit current threshold value is an important component part of relaying protection configuration.Power distribution network has following characteristics: generally radially, R/X ratio is larger for network topology, and a way and nodes are more, three-phase imbalance etc.The adjust acquisition methods of short circuit current of distribution protection is varied, yet high pressure is asked for method for the larger distribution network line of R/X ratio, and its short circuit current is difficult to calculate accurately; Symmetrical component method needs the triphase parameter balance of three-phase system, and the load of power distribution network and line parameter circuit value are conventionally in unbalanced state; Based on Fault Compensation method and actual phase sequence method for expressing, decoupling zero component method etc., although these methods can be suitable for the short circuit current of power distribution network and obtain, method is more complicated, computational efficiency is lower.

Summary of the invention

The invention provides the short circuit current acquisition methods that a kind of distribution protection is adjusted; it has overcome and in background technology, has not been suitable for the protection large, triphase parameter unbalanced system of R/X ratio in the circuit short circuit current of adjusting and asks for, calculation of complex, the shortcoming that efficiency is lower.

The technical scheme adopting that the present invention solves its technical problem is:

The short circuit current acquisition methods that distribution protection is adjusted, it comprises step 1, obtains three-phase line parameter and transformer T shape equivalent transformation, and sets up according to this power distribution network three-phase network Equivalent Model; Step 2, asks for the rated load electric current of power distribution network according to three-phase network Equivalent Model; Step 3, utilizes fault point in three-phase network Equivalent Model to the mode of root node equiva lent impedance, to ask for the fault current of power distribution network; Step 4, by asking for each branch, short-circuit current value by the method for the rated load electric current of each branch road and fault current addition of vectors.

Among one embodiment, by step 1 obtain the unit resistance that following parameter: r is circuit, unit reactance, the l that x is circuit is branch road length, R _gfor ground resistance, X _sMfor two-phase mutual impedance;

The impedance matrix Z of any branch road of described three-phase line _lbe expressed as:

$Z_L=\left(\begin{array}{ccc}{Z}_{\mathrm{AA}}& Z_{\mathrm{AB}}& Z_{\mathrm{AC}}\\ Z_{\mathrm{BA}}& Z_{\mathrm{BB}}& Z_{\mathrm{BC}}\\ Z_{\mathrm{CA}}& Z_{\mathrm{CB}}& Z_{\mathrm{CC}}\end{array}\right)$

This Z _aA, Z _bB, Z _cCfor self-impedance, this circuit self-impedance is: Z _sS=rl+jxl;

This Z _aB, Z _bC, Z _cAfor mutual impedance, this circuit mutual impedance is: Z _sM=R _g+ jX _sm, R _g=π ²* 10 ^-4f,

$X_{\mathrm{SM}}=0.1445\lg \frac{D_g}{D_{\mathrm{SM}}}.$

Among one embodiment: also obtain following parameter by step 1: Δ P _krated load loss for transformer; U _nrated voltage for transformer; S _nrated capacity for transformer; U _k% is the percentage of transformer impedance drop; Δ P ₀specified no-load loss for transformer; I ₀% is the percentage of the specified no-load current of transformer;

Described transformer each phase parameter after T-shaped equivalent transformation can be expressed as:

$R_T=\frac{\mathrm{\&Delta;}{P}_K{U}_N^2\&times;{10}^3}{S_N^2},X_T=\frac{U_k\%U_N^2\&times;10}{S_N},G_T=\frac{\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="0"\; label="P"\; filenames="HDA0000407738620000051.png,HDA0000407738620000052.png"\; state="\{\{state\}\}">P</figure-callout>}}_0\&times;{10}^{-3}}{U_N^2},B_T=\frac{{\mathrm{<figure-callout\; id="0"\; label="I"\; filenames="HDA0000407738620000051.png,HDA0000407738620000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_0\%S_N}{U_N^2}\&times;{10}^{-5},$ Wherein: R _tall-in resistance for transformer high-low-voltage winding; X _ttotal impedance for transformer high-low-voltage winding; G _tfor the electricity of transformer is led; B _tsusceptance for transformer.

Among one embodiment: in described step 2, the rated load electric current step of utilizing three-phase network Equivalent Model to ask for each phase of power distribution network comprises:

Step 21, from root node, carries out node serial number according to BFS method to network node, simultaneously from top to bottom by branch road layering;

Step 22, arranges root node three-phase voltage initial value;

Step 23, calculates three phase of impedance under each branch road rated condition and transformer reduction on high-tension side equiva lent impedance and admittance according to distribution network line and transformer equivalent model;

Step 24, utilize distribution trend before push back the rated load electric current of asking for each branch road for method.

Among one embodiment: in described step 3, the step that in three-phase network Equivalent Model, distribution network failure electric current is asked for to the mode of root node equiva lent impedance in fault point comprises:

Step 31, calculates three phase of impedance of each branch road under the fault condition that is short-circuited according to distribution network line and transformer equivalent model;

Step 32, calculates be short-circuited three-phase voltage after fault calculate fault point to the fault current of each branch road of root node of place, fault point.

Among one embodiment: described each branch, short-circuit electric current is the stack of each branch road rated load electric current and fault current vector.

The technical program is compared with background technology, and its tool has the following advantages:

1, by first obtaining three-phase line parameter and transformer T shape equivalent transformation, and set up according to this power distribution network three-phase network Equivalent Model; Then according to three-phase network Equivalent Model, ask for the rated load electric current of power distribution network; In recycling three-phase network Equivalent Model, the fault current of power distribution network is asked in fault point to the mode of root node equiva lent impedance; Finally; by the method for the rated load electric current of each branch road and fault current addition of vectors being asked for to the distribution protection of each branch, short-circuit current value short circuit current acquisition methods of adjusting; go for the distribution network systems that line parameter circuit value is asymmetric, circuit R/X ratio is larger; the distribution network protection that is simultaneously applicable to open loop and weak ring distribution network the obtaining of short circuit current of adjusting; it is few that the method is asked for step, and method is simple, applicability is wide.

2, the stack that described each branch, short-circuit electric current is each branch road rated load electric current and fault current vector; be rated load electric current and the difference of the fault current after the fault independent process before short trouble; can adapt to engineering calculation needs, the short circuit current that is applicable to extensive power distribution network relaying protection obtains.

Accompanying drawing explanation

Below in conjunction with drawings and Examples, the invention will be further described.

Fig. 1 has illustrated the distribution protection short circuit current of adjusting and has obtained flow chart.

Fig. 2 has illustrated feed connection node annexation and the hierarchical relational figure after layering numbering.

Fig. 3 has illustrated power distribution network rated load electric current and has obtained flow chart.

The simulation model figure of this feeder line is built in the PSB tool box that Fig. 4 has illustrated MATLAB.

Fig. 5 has illustrated the value of asking for of each branch road three-phase shortcircuit electric current and the error curve diagram of standard value.

Fig. 6 has illustrated each branch road AB line to line fault, the error curve of the A phase short circuit current value of asking for and standard value.

Embodiment

Please refer to Fig. 1 to Fig. 6, the short circuit current acquisition methods that distribution protection is adjusted, it comprises step 1, obtains three-phase line parameter and transformer T shape equivalent transformation, and sets up according to this power distribution network three-phase network Equivalent Model; Step 2, asks for the rated load electric current of power distribution network according to three-phase network Equivalent Model; Step 3, utilizes fault point in three-phase network Equivalent Model to the mode of root node equiva lent impedance, to ask for the fault current of power distribution network; Step 4, by asking for each branch, short-circuit current value by the method for the rated load electric current of each branch road and fault current addition of vectors.

(1) described in step 1, set up power distribution network three-phase network Equivalent Model, first will obtain the unit resistance that following parameter: r is circuit, unit reactance, the l that x is circuit is branch road length, R _gfor ground resistance, X _sMfor two-phase mutual impedance;

The impedance matrix Z of any branch road of described three-phase line _lbe expressed as:

$Z_L=\left(\begin{array}{ccc}{Z}_{\mathrm{AA}}& Z_{\mathrm{AB}}& Z_{\mathrm{AC}}\\ Z_{\mathrm{BA}}& Z_{\mathrm{BB}}& Z_{\mathrm{BC}}\\ Z_{\mathrm{CA}}& Z_{\mathrm{CB}}& Z_{\mathrm{CC}}\end{array}\right)$

This Z _aA, Z _bB, Z _cCfor self-impedance, this circuit self-impedance is: Z _sS=rl+jxl;

This Z _aB, Z _bC, Z _cAfor mutual impedance, this circuit mutual impedance is: Z _sM=R _g+ jX _sm, R _g=π ²* 10 ^-4f,

$X_{\mathrm{SM}}=0.1445\lg \frac{D_g}{D_{\mathrm{SM}}}.$

Moreover, obtain following parameter: Δ P _krated load loss for transformer; U _nrated voltage for transformer; S _nrated capacity for transformer; U _k% is the percentage of transformer impedance drop; Δ P ₀specified no-load loss for transformer; I ₀% is the percentage of the specified no-load current of transformer;

Transformer in power distribution network each phase parameter after T-shaped equivalent transformation can be expressed as:

$R_T=\frac{\mathrm{\&Delta;}{P}_K{U}_N^2\&times;{10}^3}{S_N^2},X_T=\frac{U_k\%U_N^2\&times;10}{S_N},G_T=\frac{\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="0"\; label="P"\; filenames="HDA0000407738620000051.png,HDA0000407738620000052.png"\; state="\{\{state\}\}">P</figure-callout>}}_0\&times;{10}^{-3}}{U_N^2},B_T=\frac{{\mathrm{<figure-callout\; id="0"\; label="I"\; filenames="HDA0000407738620000051.png,HDA0000407738620000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_0\%S_N}{U_N^2}\&times;{10}^{-5},$ Wherein: R _tall-in resistance for transformer high-low-voltage winding; X _ttotal impedance for transformer high-low-voltage winding; G _tfor the electricity of transformer is led; B _tsusceptance for transformer.

(2), as described in step 2, the rated load electric current step of utilizing three-phase network Equivalent Model to ask for each phase of power distribution network comprises:

Step 21, from root node, carries out node serial number according to BFS method to network node, simultaneously from top to bottom by branch road layering;

Step 22, makes root node three-phase voltage initial value be respectively:

wherein U is the measurement effective value of root node line voltage.

Step 23, calculates three phase of impedance under each branch road rated condition and transformer reduction on high-tension side equiva lent impedance and admittance according to distribution network line and transformer equivalent model, and specifically each calculation of parameter is as follows;

1) transformer resistance loss (kW) is:

wherein: φ represents each phase of three-phase, U _φfor the magnitude of voltage of transformer T shape equivalent electric circuit load side, its initial value is identical with root node voltage, P _φfor the active power of transformer T shape equivalent electric circuit load side, Q _φreactive power for transformer;

2) transformer reactance loss (kvar) is: $\mathrm{\&Delta;}{Q}_{\mathrm{zT\&phi;}}=\frac{P_{\mathrm{\&phi;}}^2+Q_{\mathrm{\&phi;}}^2}{U_{\mathrm{\&phi;}}^2}{X}_T\&times;{10}^{-3};$

3) the vertical component of transformer voltage landing is: $\mathrm{\&Delta;}{U}_{\mathrm{T\&phi;}}=\frac{P_{\mathrm{\&phi;}}{R}_T+Q_{\mathrm{\&phi;}}{X}_T}{U_{\mathrm{\&phi;}}}\&times;{10}^{-3};$

The horizontal component of voltage-drop is: $\mathrm{\&delta;\&Delta;}{U}_{\mathrm{T\&phi;}}=\frac{P_{\mathrm{\&phi;}}{X}_T-Q_{\mathrm{\&phi;}}{R}_T}{U_{\mathrm{\&phi;}}}\&times;{10}^{-3};$

The voltage magnitude of transformer T shape equivalent electric circuit power end is:

Phase place between transformer T shape equivalent electric circuit power end and load side voltage is:

${\mathrm{\&delta;}}_{\mathrm{T\&phi;}}=\arctan \frac{\mathrm{\&delta;\&Delta;}{U}_{\mathrm{T\&phi;}}}{U_{\mathrm{\&phi;}}+\mathrm{\&Delta;}{U}_{\mathrm{T\&phi;}}};$

4) transformer electricity is led loss (kW) and is:

5) transformer susceptance loss (kW) is:

6) the total injecting power (kVA) of transformer is:

$S_{\mathrm{\&phi;}}^{\&prime;}=S_{\mathrm{\&phi;}}+\mathrm{\&Delta;}{S}_{\mathrm{T\&phi;}}=(P_{\mathrm{\&phi;}}+\mathrm{\&Delta;}{P}_{\mathrm{zT\&phi;}}+\mathrm{\&Delta;}{P}_{\mathrm{yT\&phi;}})+j(Q_{\mathrm{\&phi;}}+\mathrm{\&Delta;}{Q}_{\mathrm{zT\&phi;}}+\mathrm{\&Delta;}{Q}_{\mathrm{yT\&phi;}});$

7) electric current of calculating transformer T shape equivalent electric circuit power end, in an embodiment, distribution transformer is that Dyn connects, by each phase power end voltage magnitude U _φ' and voltage phase angle δ _{t φ}the line voltage (kV) that obtains high-pressure side D end is: $\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_{1\mathrm{AB}}\\ {\stackrel{\&CenterDot;}{U}}_{1\mathrm{BC}}\\ {\stackrel{\&CenterDot;}{U}}_{1\mathrm{CA}}\end{array}\right)=\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_A^{\&prime;}-{\stackrel{\&CenterDot;}{U}}_B^{\&prime;}\\ {\stackrel{\&CenterDot;}{U}}_B^{\&prime;}-{\stackrel{\&CenterDot;}{U}}_C^{\&prime;}\\ {\stackrel{\&CenterDot;}{U}}_C^{\&prime;}-{\stackrel{\&CenterDot;}{U}}_A^{\&prime;}\end{array}\right);$

According to the conservation of energy, by each phase injecting power of transformer S _φ', the line injecting power (kVA) that can obtain high-pressure side D end is: $\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">S</figure-callout>}}_{1\mathrm{<figure-callout\; id="1"\; label="AB\; S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">AB</figure-callout>}}& \\ S_{}{}_{1\mathrm{<figure-callout\; id="1"\; label="BC\; S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">BC</figure-callout>}}& \\ S_{}{}_{1\mathrm{CA}}\end{array}\right)=\left(\begin{array}{c}{S}_A^{\&prime;}\\ S_B^{\&prime;}\\ S_C^{\&prime;}\end{array}\right);$

The winding current of high voltage side of transformer D end is:

$\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{1\mathrm{AB}}\\ {\stackrel{\&CenterDot;}{I}}_{1\mathrm{BC}}\\ {\stackrel{\&CenterDot;}{I}}_{1\mathrm{CA}}\end{array}\right)=\begin{array}{}\end{array}\left(\begin{array}{c}{\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">S</figure-callout>}}_{1\mathrm{AB}}}{{\stackrel{\&CenterDot;}{U}}_{1\mathrm{BC}}}\right)}^{*}\\ {\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">S</figure-callout>}}_{1\mathrm{BC}}}{{\stackrel{\&CenterDot;}{U}}_{1\mathrm{BC}}}\right)}^{*}\\ {\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="S"\; filenames="HDA0000407738620000031.png"\; state="\{\{state\}\}">S</figure-callout>}}_{1\mathrm{CA}}}{{\stackrel{\&CenterDot;}{U}}_{1\mathrm{CA}}}\right)}^{*}\end{array}\right);$

The load phase current of high voltage side of transformer D end is:

$\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{1A}\\ {\stackrel{\&CenterDot;}{I}}_{1B}\\ {\stackrel{\&CenterDot;}{I}}_{1C}\end{array}\right)=\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{1\mathrm{AB}}-{\stackrel{\&CenterDot;}{I}}_{1\mathrm{CA}}\\ {\stackrel{\&CenterDot;}{I}}_{1\mathrm{BC}}-{\stackrel{\&CenterDot;}{I}}_{1\mathrm{AB}}\\ {\stackrel{\&CenterDot;}{I}}_{1\mathrm{CA}}-{\stackrel{\&CenterDot;}{I}}_{1\mathrm{BC}}\end{array}\right)$

Step 24, utilize distribution trend before push back the rated load electric current of asking for each branch road for method, concrete steps comprise;

1) the front journey that pushes through

If the end-node that the tail node j of this branch road (j) is circuit, the electric current that flows through this branch road equals the load current of this node, that is:

$\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)A}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)B}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)C}\end{array}\right)=\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\mathrm{jA}}\\ {\stackrel{\&CenterDot;}{I}}_{\mathrm{jB}}\\ {\stackrel{\&CenterDot;}{I}}_{\mathrm{jC}}\end{array}\right)$

If the tail node j of branch road (j) is not the end-node of circuit, known according to Kirchhoff's current law (KCL), branch current

should be this branch road load current

with along direction of tide, it connects the electric current sum of sub-branch road after all, that is:

$\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)A}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)B}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)C}\end{array}\right)=\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\mathrm{jA}}\\ {\stackrel{\&CenterDot;}{I}}_{\mathrm{jB}}\\ {\stackrel{\&CenterDot;}{I}}_{\mathrm{jC}}\end{array}\right)+\underset{h\&Element;j}{\mathrm{\&Sigma;}}\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(h\right)A}\\ {\stackrel{\&CenterDot;}{I}}_{\left(h\right)B}\\ {\stackrel{\&CenterDot;}{I}}_{\left(h\right)C}\end{array}\right)$

In formula, h is the set that node j connects sub-branch road after all.

2) backward steps

According to Kirchhoff's second law, the voltage relationship of the head and the tail node of branch road (j) is:

$\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_{\mathrm{jA}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jB}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jC}}\end{array}\right)=\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_{\mathrm{iA}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{iB}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{iC}}\end{array}\right)-\left(\begin{array}{ccc}{Z}_{\left(j\right)\mathrm{AA}}& Z_{\left(j\right)\mathrm{AB}}& Z_{\left(j\right)\mathrm{AC}}\\ Z_{\left(j\right)\mathrm{BA}}& Z_{\left(j\right)\mathrm{BB}}& Z_{\left(j\right)\mathrm{BC}}\\ Z_{\left(j\right)\mathrm{CA}}& Z_{\left(j\right)\mathrm{CB}}& Z_{\left(j\right)\mathrm{CC}}\end{array}\right)\left(\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)A}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)B}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)C}\end{array}\right)\&times;{10}^{-3}$ Wherein:

voltage for the first node of branch road (j).

3) convergence decision condition

The voltage that before and after each node, adjacent twice iterative computation goes out on circuit subtracts each other when difference is by the time less than given error ε restrains, that is:

$|{\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_{\mathrm{jA}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jB}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jC}}\end{array}\right)}^{\left(k\right)}-{\left(\begin{array}{c}{\stackrel{\&CenterDot;}{U}}_{\mathrm{jA}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jB}}\\ {\stackrel{\&CenterDot;}{U}}_{\mathrm{jC}}\end{array}\right)}^{(k-1)}|\&le;\left(\begin{array}{c}\mathrm{\&epsiv;}\\ \mathrm{\&epsiv;}\\ \mathrm{\&epsiv;}\end{array}\right)$

In formula: k is iterations.

When iteration convergence, each branch road load current of the k time iteration gained be the rated load electric current of each branch road.

(3) in described step 3, the step that in three-phase network Equivalent Model, distribution network failure electric current is asked for to the mode of root node equiva lent impedance in fault point comprises:

Step 31, according to distribution network line and transformer equivalent model, calculate three phase of impedance of each branch road under the fault condition that is short-circuited:

$\left(\begin{array}{ccc}{Z}_{\mathrm{aa}}& Z_{\mathrm{ab}}& Z_{\mathrm{ac}}\\ Z_{\mathrm{ba}}& Z_{\mathrm{bb}}& Z_{\mathrm{bc}}\\ Z_{\mathrm{ca}}& Z_{\mathrm{cb}}& Z_{\mathrm{cc}}\end{array}\right)=\underset{j\&Element;L}{\mathrm{\&Sigma;}}\left(\begin{array}{ccc}{Z}_{\left(j\right)\mathrm{AA}}& Z_{\left(j\right)\mathrm{AB}}& Z_{\left(j\right)\mathrm{AC}}\\ Z_{\left(j\right)\mathrm{BA}}& Z_{\left(j\right)\mathrm{BB}}& Z_{\left(j\right)\mathrm{BC}}\\ Z_{\left(j\right)\mathrm{CA}}& Z_{\left(j\right)\mathrm{CB}}& Z_{\left(j\right)\mathrm{CC}}\end{array}\right),$ Wherein: L is that fault point is to the set of all branch roads of root node, Z _aa, Z _bb, Z _ccfor the self-impedance sum of all branch roads from fault point to root node, Z _ab, Z _cb, Z _ac, Z _bc, Z _ba, Z _camutual impedance sum for all branch roads from fault point to root node;

Step 32, calculate be short-circuited three-phase voltage after fault calculate fault point to the fault current of each branch road of root node of place, fault point:

$\left[\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fA}}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fB}}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fC}}\end{array}\right]={\left(\begin{array}{ccc}{Z}_{\mathrm{aa}}& Z_{\mathrm{ab}}& Z_{\mathrm{ac}}\\ Z_{\mathrm{ba}}& Z_{\mathrm{bb}}& Z_{\mathrm{bc}}\\ Z_{\mathrm{ca}}& Z_{\mathrm{cb}}& Z_{\mathrm{cc}}\end{array}\right)}^{-1}\left[\begin{array}{c}-{\stackrel{\&CenterDot;}{U}}_{A\left|0\right|}\\ -{\stackrel{\&CenterDot;}{U}}_{B\left|0\right|}\\ -{\stackrel{\&CenterDot;}{U}}_{C\left|0\right|}\end{array}\right],$ Wherein:

with

for fault point is in the three-phase voltage before breaking down,

for the three-phase fault current vector of fault point to all branch roads of root node.

(4) by obtaining each branch road rated load electric current and short-circuit current, obtain each branch, short-circuit electric current, described each branch, short-circuit electric current is the stack of each branch road rated load electric current and fault current vector, and fault point to the short circuit current of all branch roads of root node is:

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dA}}={\stackrel{\&CenterDot;}{I}}_{\left(j\right)A}+{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fA}}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dB}}\text{=}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)B}+{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fB}}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dC}}={\stackrel{\&CenterDot;}{I}}_{\left(j\right)C}+{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{fC}}\end{array}\right.,$ Wherein, j ∈ L, L is that fault point is to the set of all branch roads of root node.

The short circuit current of other non-fault branches of system is:

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dA}}={\stackrel{\&CenterDot;}{I}}_{\left(j\right)A}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dB}}\text{=}{\stackrel{\&CenterDot;}{I}}_{\left(j\right)B}\\ {\stackrel{\&CenterDot;}{I}}_{\left(j\right)\mathrm{dC}}={\stackrel{\&CenterDot;}{I}}_{\left(j\right)C}\end{array}\right.$

Wherein

l is that fault point is to the set of all branch roads of root node.

In one specific embodiment, a feeder system of certain transformer station, application BFS layering number algorithm is numbered feed connection node, as shown in Figure 2, wherein node 8,11,13,15,16,17,18,19,20,21,25,26 and 27 is transformer node for the node annexation after numbering and hierarchical relational.In Fig. 2, the line length of each branch road, circuit unit resistance and the reactance of circuit unit are as shown in table 1; The numbering of each transformer node in Fig. 2, transformer parameter are as shown in table 2.

The three-phase voltage initial value of supposing root node 0 is respectively: by flow chart shown in Fig. 3, can obtain the rated load electric current of system, the rated load electric current of each branch road is as shown in table 3.And then by fault point to each branch impedance of root node and ask for the fault current of fault point, the electric current of each branch road after finally asking for system and break down by the rated load electric current of each branch road and the method for fault current addition of vectors, in each branch road generation three-phase shortcircuit situation, the short circuit current of each branch road is as shown in table 4; In each step down side generation three-phase shortcircuit situation, flow through on high-tension side short circuit current as shown in table 5.When circuit AB mutually phase fault occurs, the short circuit current of each branch road is as shown in table 6; When AB phase phase fault occurs each step down side, the short circuit current that flows through fault point is as shown in table 7.

For verifying the validity of this short circuit current acquisition methods, utilize the PSB tool box of MATLAB to build the simulation model of this feeder line, as shown in Figure 4.Utilize simulation model emulation to obtain, in each branch road generation three-phase shortcircuit situation, the three-phase shortcircuit electric current of each branch road is as shown in table 8; In each step down side generation three-phase shortcircuit situation, flow through on high-tension side short circuit current as shown in table 9.When each branch road AB phase phase fault, the short circuit current that flows through fault point is as shown in table 10; When AB phase phase fault occurs step down side, the short circuit current that flows through fault point is as shown in table 11.

Each branch road simulation model gained short circuit current can be used as the standard value of this feeder line short circuit current, the correctness of the short circuit current that method is asked for being added for application rated load electric current and fault current, Fig. 5 visual representation the value of obtaining of each branch road three-phase shortcircuit electric current and the error of standard value; Fig. 6 visual representation each branch road AB phase phase fault, the value of asking for of A phase short circuit current and the error of standard value.From Fig. 5 and Fig. 6; the method of short circuit current that method is asked for that application rated load electric current and fault current are added and the error of standard value, in 5% error range, can be applied to during distribution protection adjusts, and this distribution short circuit current setting method is simply effective; reliably, and applicability wide.

Table 1

Table 2

Table 3

Table 4

Table 5

Table 6

Table 7

Table 8

Table 9

Table 10

Table 11

The above, only for preferred embodiment of the present invention, therefore can not limit according to this scope of the invention process, the equivalence of doing according to the scope of the claims of the present invention and description changes and modifies, and all should still belong in the scope that the present invention contains.

Claims (6)

1. the short circuit current acquisition methods that distribution protection is adjusted, is characterized in that: comprise

Step 1, obtains three-phase line parameter and transformer T shape equivalent transformation, and sets up according to this power distribution network three-phase network Equivalent Model;

Step 2, asks for the rated load electric current of power distribution network according to three-phase network Equivalent Model;

Step 3, utilizes fault point in three-phase network Equivalent Model to the mode of root node equiva lent impedance, to ask for the fault current of power distribution network;

Step 4, by asking for each branch, short-circuit current value by the method for the rated load electric current of each branch road and fault current addition of vectors.

2. the short circuit current acquisition methods that distribution protection according to claim 1 is adjusted, is characterized in that: in described step 1, obtain the unit resistance that following parameter: r is circuit, unit reactance, the l that x is circuit is branch road length, R _gfor ground resistance, X _sMfor two-phase mutual impedance;

The impedance matrix Z of any branch road of described three-phase line _lbe expressed as:

$Z_L=\left(\begin{array}{ccc}{Z}_{\mathrm{AA}}& Z_{\mathrm{AB}}& Z_{\mathrm{AC}}\\ Z_{\mathrm{BA}}& Z_{\mathrm{BB}}& Z_{\mathrm{BC}}\\ Z_{\mathrm{CA}}& Z_{\mathrm{CB}}& Z_{\mathrm{CC}}\end{array}\right)$

This Z _aA, Z _bB, Z _cCfor self-impedance, this circuit self-impedance is: Z _sS=rl+jxl;

This Z _aB, Z _bC, Z _cAfor mutual impedance, this circuit mutual impedance is: Z _sM=R _g+ jX _sm, R _g=π ²* 10 ^-4f,

$X_{\mathrm{SM}}=0.1445\lg \frac{D_g}{D_{\mathrm{SM}}}.$

3. the short circuit current acquisition methods that distribution protection according to claim 1 is adjusted, is characterized in that: in described step 1, obtain following parameter: Δ P _krated load loss for transformer; U _nrated voltage for transformer; S _nrated capacity for transformer; U _k% is the percentage of transformer impedance drop; Δ P ₀specified no-load loss for transformer; I ₀% is the percentage of the specified no-load current of transformer;

Described transformer each phase parameter after T-shaped equivalent transformation can be expressed as:

$R_T=\frac{\mathrm{\&Delta;}{P}_K{U}_N^2\&times;{10}^3}{S_N^2},X_T=\frac{U_k\%U_N^2\&times;10}{S_N},G_T=\frac{\mathrm{\&Delta;}{P}_0\&times;{10}^{-3}}{U_N^2},B_T=\frac{I_0\%S_N}{U_N^2}\&times;{10}^{-5},$ Wherein: R _tall-in resistance for transformer high-low-voltage winding; X _ttotal impedance for transformer high-low-voltage winding; G _tfor the electricity of transformer is led; B _tsusceptance for transformer.

4. the short circuit current acquisition methods that distribution protection according to claim 1 is adjusted, is characterized in that: described step 2, and the rated load electric current step of utilizing three-phase network Equivalent Model to ask for each phase of power distribution network comprises:

Step 21, from root node, carries out node serial number according to BFS method to network node, simultaneously from top to bottom by branch road layering;

Step 22, arranges root node three-phase voltage initial value;

Step 23, calculates three phase of impedance under each branch road rated condition and transformer reduction on high-tension side equiva lent impedance and admittance according to distribution network line and transformer equivalent model;

Step 24, utilize distribution trend before push back the rated load electric current of asking for each branch road for method.

5. the short circuit current acquisition methods that distribution protection according to claim 1 is adjusted, is characterized in that: described step 3, and the step that in three-phase network Equivalent Model, distribution network failure electric current is asked for to the mode of root node equiva lent impedance in fault point comprises:

Step 31, calculates three phase of impedance of each branch road under the fault condition that is short-circuited according to distribution network line and transformer equivalent model;

Step 32, calculates be short-circuited three-phase voltage after fault calculate fault point to the fault current of each branch road of root node of place, fault point.

6. the short circuit current acquisition methods that distribution protection according to claim 1 is adjusted, is characterized in that: described each branch, short-circuit electric current is the stack of each branch road rated load electric current and fault current vector.

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