CN102749492B - Short-circuit current computing method for ring-shaped ship power grid - Google Patents

Abstract

According to the structure characteristic of an alternating current region power supply system of a ring-shaped bus, through improvement on several short circuit computing methods which are widely used in the prior art at home and abroad, the original short circuit current computing method for a tree-shaped power grid is deformed into a short-circuit current computing method suitable for a ring-shaped ship power grid. The method comprises the following steps of: firstly, computing the short-circuit current of a simple grid by a star-network transformation and local equal effect method; and secondly, providing a method in which a line impedance matrix is obtained by a graph theoretic approach and then the short circuit current is calculated. According to the method, the short-circuit current computing method for a ship ring-shaped region power distribution system can be explored, the thermal stability and power stability checkout when types of the equipment are selected can be realized, and the basis can be provided for fault protection when the equipment runs.

Description

The short-circuit current calculation method of annular network of ship

Technical field

The short-circuit current calculation method that the present invention relates to a kind of boats and ships region distribution system annular electric supply network, belongs to Ship Electrical Power System field.

Background technology

Along with the rising of Ship Electrical Power System capacity, network topology structure will become increasingly complex, and traditional boats and ships supply and distribution network, due to its structure, makes power supply reliability lower, if near short circuit bus, can make all loads afterwards of this point lose power supply; Due to the supply and distribution network topological structure of large ship electric system and the singularity of its mission, its network planning, load management, and the reconstructing method etc. of supply network all can not be indiscriminately imitated the method for land electrical network in the time breaking down, must preferentially ensure the power supply of important load, and also very strict to the requirement of its rapidity.

The form of traditional network of ship has become a great problem of restriction ship reliability, adopting block supply network is the good approach addressing this problem, block supply is on the basis based on loop network, according to load characteristic and load, whole Ship Electrical Power System is divided into some regions, between region, utilize connecting line or bus to form well-bedded multitiered network, each power station ensures the load supplying in one or several region.Loop network is between power station, to utilize longitudinal, horizontal connecting line or bus to form closed loop, and to the network of consumer power supply, its advantage is to form the path of more power station to load, has higher power supply reliability.

Block supply technology has become the Developing mainstream of following novel ship electric system, and it adopts unified supply network is the equipment power supply of full ship, has both reached the object that reduces electric system construction cost, can improve again the reliability of Ship Electrical Power System.

In various potential faults, the short circuit of electric system be each Ship Design and maintainer must pay close attention to a major issue.In the time of Ship Electrical Power System median generatrix or a certain main line short circuit, on bus or this main line, will occur than the short-circuit current of the large manyfold of normal value, the mechanical stress that excessive short-circuit current produces and thermal effect, can make the normal operation of generator and other electrical equipments be affected, even cause damage.Therefore in the time of design Ship Electrical Power System, must fully take into account the contingent short trouble of system, to reasonably select and verification relevant device and device, impact to ensure that marine electric equipment can bear short-circuit current on the one hand, guarantee on the other hand fast and effeciently cutting-off of short-circuit fault, make short-circuited region and isolation of system, short trouble is limited in minimum scope, and guarantee system is normally moved.

Calculation of short-circuit current is correctly to design shipboard power system, selects and check electric equipment, element, checks power stability and the thermal stability of bus-bar, checks the general power of parallel running genset and the basis of realizing Ship Electrical Power System reliably protecting.

Adopt the block supply destruction that Power System Shortcuts can be caused to be reduced to minimum zone, and each consumer can obtain power supply from other direction of circuit, so the fault of arbitrary circuit can not make circuit stop power supply, improve the reliability of power supply.But, the computing method of traditional AC electric power systems short-circuit current peculiar to vessel are to be based upon on the basis of tree-like electric network composition, introduce annular electric supply network peculiar to vessel and can produce larger impact to the short circuit calculation of electrical network, and the research of loop network short-circuit current is still in the junior stage at present.

At present, both at home and abroad the relatively more conventional method for boats and ships AC electric power systems calculation of short-circuit current mainly contains IEC(International Electrotechnical Commission), CNS, China national military standard, USN's standard etc.What this several method generally adopted is equivalence and approximation method, in engineering practice, there is certain practical significance, but wherein all to do feedback formula electrical network as hypothesis calculating object, clearly do not provide the computing method of the interchange region distribution network short-circuit current based on ring busbars, can not adapt to the development need of Ship Electrical Power System.

Summary of the invention

The object of the invention is to provide for the deficiencies in the prior art a kind of short-circuit current calculation method of annular network of ship.

The short-circuit current calculation method of a kind of annular network of ship of the present invention, determining after short dot, carry out Equivalent Calculation by generator end to short dot, after impedance network conversion and abbreviation equivalence, impedance network abbreviation becomes the form that several equivalent generators are directly connected with short dot through an impedance respectively the most at last, the peak value of short of asking each equivalent generator to be fed in short-circuit point, be then added is required maximal value.

After described network transformation, star network is transformed into mesh network, in equivalent net form circuit, between node i and node j, the impedance computation method of branch road is:

star network is made up of node n and other m node, and the each node in a node n and m node has a branch road to join; StarNet's conversion is transformed into node n cancellation with 1 exactly, 2. ... m is the mesh network on summit.

Short-circuit current calculation method carries out according to following step:

The first step, determines the typical node that need to calculate short-circuit current;

Second step, the short-circuit conditions of analysis generator end;

The 3rd step, analyzes the short-circuit conditions on port and starboard bus: the both sides of first analyzing isolating switch are short-circuit current when short circuit respectively, and it is compared, and chooses larger one and carries out calculation of short-circuit current; Now, each power station, front and back equivalence is become to a generator, calculate respectively this two short-circuit currents that generator is fed to isolating switch, then ask its algebraic sum;

The 4th step, the outlet short circuit of calculating two ends, left and right main distribution board load;

The 5th step, calculates the short-circuit conditions in a certain interval: first draw the impedance network of power station to short dot, then carry out network transformation, become power end and be directly connected to the form of short dot through impedance, then calculate.

Adopt the short-circuit current calculation method based on graph theoretic approach as follows: to have the impedance matrix of the passive network of m isolated node to be:

$\stackrel{\‾}{Z}=\left[\begin{array}{cccc}{\stackrel{\‾}{Z}}_{11}& {\stackrel{\‾}{Z}}_{12}& ...& {\stackrel{\‾}{Z}}_{1m}\\ .& .& .& .\\ .& .& .& .\\ .& .& .& .\\ {\stackrel{\‾}{Z}}_{m1}& {\stackrel{\‾}{Z}}_{m2}& ...& {\stackrel{\‾}{Z}}_{\mathrm{mm}}\end{array}\right]$

On the node p of this passive network, add branch pk, the dimension of former impedance matrix will add 1, become the matrix shown in following formula:

Wherein original impedance matrix, the mutual resistance matrix between original matrix and new branch pk, it is the matrix of new branch pk;

Order

, in the time there is transimpedance between pk and already present branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}},j=\mathrm{1,2},...m,j\≠k$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xk}}-{\stackrel{\‾}{Z}}_{\mathrm{yk}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}}$

If there is not transimpedance between pk and already present arbitrary branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+{\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

If new branch is to be added between p and reference mode 0,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

On link pk, establish a new node e, E _erepresent the voltage of node e to node k,

The impedance that contains subscript e in formula has following definition:

Z _1ewhen=unitary current injects k node, the voltage of circuit 1 to reference zero,

Z _e1when=unitary current injects circuit 1, the voltage of circuit e to node k,

Z _ee=unitary current in the time that k flows into circuit e, the voltage of circuit e to node k,

Above formula is transformed to:

$\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{xy}}\\ {\stackrel{\‾}{I}}_{\mathrm{pe}}\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{pe}}\\ {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{xy}}\\ {\stackrel{\‾}{V}}_{\mathrm{pe}}\end{array}\right]$

? ${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}},j=\mathrm{1,2},...,\mathrm{m\; j}\≠e$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pe}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}+\frac{1+{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xe}}-{\stackrel{\‾}{Z}}_{\mathrm{ye}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}}$

If nothing coupling between pe and other branch or link, and p is as a reference point, has:

${\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}$

The voltage that then can make node e is 0, the artificial node e increasing of cancellation;

$\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{bus}}\\ 0\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Z}}_{\mathrm{bus}}& {\stackrel{\‾}{Z}}_{\mathrm{je}}\\ {\stackrel{\‾}{Z}}_{\mathrm{ej}}& {\stackrel{\‾}{Z}}_{\mathrm{ee}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{bus}}\\ {\stackrel{\‾}{I}}_e\end{array}\right]$

${\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{mod}\mathrm{ified}}={\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{primitive}}-\frac{{\stackrel{\‾}{Z}}_{\mathrm{je}}{\stackrel{\‾}{Z}}_{\mathrm{je}}^{\′}}{{\stackrel{\‾}{Z}}_{\mathrm{ee}}};$

The self-impedance of trouble spot D ${\stackrel{\‾}{Z}}_{\mathrm{dd}}=(1-l){\stackrel{\‾}{Z}}_{\mathrm{di}}+l{\stackrel{\‾}{Z}}_{\mathrm{dj}}+l(1-l)z_{\mathrm{ij}}$

$={(1-l)}^2{\stackrel{\‾}{Z}}_{\mathrm{ii}}+l^2{\stackrel{\‾}{Z}}_{\mathrm{jj}}+2l(1-l){\stackrel{\‾}{Z}}_{\mathrm{ij}}+l(1-l)z_{\mathrm{ij}};$

When the voltage of known fault point before short circuit occurs according to the self-impedance of trouble spot D just can obtain the short-circuit current of trouble spot:

${\stackrel{\‾}{I}}_d=-\frac{{\stackrel{\‾}{V}}_d^{\left(0\right)}}{\stackrel{\‾}{Z}}.$

The present invention adopts equivalent generator method and cy-pres doctrine, for several short circuit calculation methods that generally use at present both at home and abroad, improved, adopt star-net transformation and impedance matrix acquiring method, the short-circuit current calculation method of original tree-like electrical network is transform as to the short-circuit current calculation method that is applicable to annular electrical network, short dot judgement and system of selection are proposed simultaneously, and according to be short-circuited not in the same time, calculate the short-circuit current value of different short dots of this moment, explored the short-circuit current calculation method of annular electric system.

In Ship Electrical Power System, short circuit is one of modal electric fault in the operational process of Ship Power Station.In this case; because electric system peculiar to vessel does not generally expect to have longer dynamic process; but taking protective device successfully cutting-off of short-circuit electric current as object; so can the short-circuit analysis master in the present invention be transitioned into another stable operating point if it were not for analyzing system after short circuit; but the transient state process that whole annular electric system is carried out after short circuit is calculated; thermally-stabilised, power while realizing lectotype selection are stablized verification, and provide foundation for run time fault protection.

Brief description of the drawings

Fig. 1 is star network figure;

Fig. 2 is mesh network figure;

Fig. 3 is annular electrical network figure;

Tu4Shi StarNet conversion 1;

Tu5Shi StarNet conversion 2;

Fig. 6 StarNet conversion 3;

Fig. 7 adds branch schematic diagram;

Fig. 8 adds link schematic diagram.

Embodiment

The equivalence principle of the region distribution calculation of short-circuit current based on ring busbars is:

1) the interchange region supply network based on ring busbars, is connected the distribution center in Yu Ge region, each power station (block supply plate) with ring busbars by isolating switch.In computation process, the equivalence of generator and motor can not be indiscriminately imitated the equivalent method in existing standard, and need be changed according to actual conditions.

2) for generator, each power station is relatively independent running unit, therefore can carry out equivalence at bus-bar place, each power station to all generators in this power station respectively, the fault if certain power station is short-circuited (short dot is between generator and main busbar, on distribution center at different levels and feeder line), the each generator in this power station is calculated respectively, and be that unit carries out respectively equivalence by the generator in all the other power stations by power station; If short circuit occurs in ring busbars or load area, the generator in each power station is equivalent to respectively a generator, has several power stations just to have the generator of several equivalences.

3) for motor, if short circuit occurs in the port of certain motor, other motor in this region are equivalent to a motor, and the motor in other regions carries out equivalence by region; When short circuit occurs on ring busbars, in power station or on block supply bus time, all motor carry out equivalence by region respectively.

Region distribution calculation of short-circuit current based on ring busbars adopts star-net transformation method:

Determining after short dot, carry out Equivalent Calculation by generator end to short dot, after impedance network conversion and abbreviation equivalence layer by layer, impedance network abbreviation becomes the form that several equivalent generators are directly connected with short dot through an impedance respectively the most at last, the peak value of short that can ask each equivalent generator to be fed in short-circuit point, be then added is required maximal value.

For the regional distribution network network based on ring busbars, line impedance cannot directly calculate as specifying in existing standard.In the time of short circuit, because short-circuit current can be fed to short dot through different circuits from different power ends, be also that short-circuit current may flow to short dot from two or more directions, its electrical distance difference of different circuits, therefore line impedance is also different.Due to the regional distribution network network structure relative complex based on ring busbars, the line impedance from each power end to short dot cannot directly calculate, and for addressing this problem, can carry out network abbreviation by Adoption Network converter technique, then tries to achieve short-circuit impedance.

Network transformation is transformed into simple network according to circuit theory by complex network exactly, is exactly complex electric network is transformed into the simple network that each power end is directly connected through a short-circuit impedance with short dot while specifically calculating.

Can reduce on the one hand the exponent number of electric power networks modal equation by network transformation, thereby reduce the operand that solves network node equation; In some calculates, sometimes need to obtain the direct relation between some node on the other hand, this just need to be by node cancellation irrelevant in network.Final object is all that former complex network is transformed into the simple network of being convenient to calculating.

The equivalent transformation of network is a method the most basic of simplified network, and its requirement is: the state (referring to voltage and electric current) that network is not transformed part distributes and should remain unchanged.

The conversion of so-called StarNet is transformed into mesh network by star network exactly.If certain part of network is made up of node n and other m node, the each node in a node n and m node has a branch road to join, and as shown in Figure 1, this network is star network.StarNet's conversion is transformed into node n cancellation with 1 exactly, 2. ... m is the network on summit, and as shown in Figure 2, this network is mesh network.

After StarNet's conversion, in equivalent net form circuit, between node i and node j, the impedance computation formula of branch road is:

$Z_{\mathrm{ij}}^{\′}=Z_{\mathrm{in}}{Z}_{\mathrm{jn}}\underset{k=1}{\overset{m}{\mathrm{\Σ}}}(1/Z_{\mathrm{kn}})$

In the time of m=3, above formula just turns to common Y-△ transformation for mula.

A kind of equivalent network figure of simple ring busbars distribution network as shown in Figure 3, has four main electrical power plants in this distribution network, in figure, EA, EB, EC, ED are the electrical source voltage of each power station equivalent generator group; Z " _a, Z " _b, Z " _c, Z " _dfor the super transient impedance of each power station equivalent generator group.Z ' _a, Z ' _b, Z ' _c, Z ' _dfor the transient impedance (not marking in Fig. 3) of each power station equivalent generator group.In super transient impedance and transient impedance, all comprise the cable resistance of genset ring busbars; A, b, c, d is the interface of each power station and ring busbars; Z _ab, Z _ac, Z _bc, Z _bd, Z _cdfor the impedance between each power station.

Be located in the B of power station the contiguous interface b place fault that is short-circuited, short dot is K.Want the short-circuit current of calculating K point should first calculate respectively the short-circuit current that four power stations are fed to K point, first this just need to obtain the line impedance between each power station and K point.As seen from Figure 3, flow to the short-circuit current that K orders can pass through different circuits from each power station, the short-circuit current that flows to K as C power station has three-line: from C through Z _ac, Z _abto K, from C through Z _cd, Z _bdto K, from C through Z _bcto K, therefore the line impedance from bus-bar to short dot cannot directly calculate.This loop power network is carried out abbreviation by Adoption Network converter technique for this reason, makes that it is transformed into taking four equivalent sources as summit, star network centered by short dot K tries to achieve short-circuit impedance.

Below for ask for the step of short-circuit impedance with the equivalent converter technique of network:

The first step: by Z _bd, Z _cdand Z " _dand Z _ab, Z _ac, Z " _atwo star circuits that form respectively equivalence are transformed to delta circuit, and the impedance on its three limit is respectively Z _dc, Z _db, Z _bc2and Z _ac, Z _ab, Z _bc1.So node a, d are eliminated, as shown in Figure 4.Then by Z _bc1, Z _bc, Z _bc2article three, parallel branch merges, and obtaining equivalent impedance is Z ' _bc.

Second step: will be by Z " _c, Z _dc, Z _ac, Z _bc1article four, the star circuit equivalence that branch road forms is transformed into E _a, K, E _c, E _dfor the net form circuit on summit, the impedance of its six branch roads is Z ' _ab, Z _ac, Z _aD, Z ' _db, Z _dc, Z _bc, node C is eliminated, and whole network reduction is to only have the impedance between power end and short dot K and each power end and short dot, each power end E _a, E _c, E _dbetween impedance, as shown in Figure 5.

Finally draw the impedance being directly connected between each power end and short dot K, Z as shown in Figure 6 _ak, Z _bk, Z _ck, Z _dk, these four impedances are asked for line impedance.Wherein:

$Z_{\mathrm{Ak}}=Z_{\mathrm{Ab}}^{/}*Z_{\mathrm{Ab}}/Z_{\mathrm{Db}}^{/}+Z_{\mathrm{Ab}}$

$Z_{\mathrm{Bk}}=Z_B^{//}$

Z _Ck＝Z _bc

$Z_{\mathrm{Dk}}=Z_{\mathrm{Db}}^{/}*Z_{\mathrm{Db}}/Z_{\mathrm{Db}}^{/}+Z_{\mathrm{Db}}$

In addition the Z shown in Fig. 5 _aC, Z _aD, Z _dCfor the branch impedance between each power supply, they are nonsensical to short circuit calculation, therefore their cancellations are disregarded in Fig. 6.

Just can try to achieve short-circuit current according to the circuit of abbreviation gained and correlation parameter.

Concrete calculation of short-circuit current is carried out according to following step:

The first step, determines the typical node that need to calculate short-circuit current.In electrical network structure, symmetrical node is only got one and is calculated.

Second step, the short-circuit conditions of analysis generator end.

The 3rd step, analyzes the short-circuit conditions on port and starboard bus.Because bus is longer, when diverse location short circuit, caused short-circuit current value is slightly different, and on bus, the breaking capacity of isolating switch also has difference a little.In the time selecting isolating switch, can only consider according to the maximum short circuit current that may pass through.Therefore the both sides of first analyzing isolating switch are short-circuit current when short circuit respectively, and it is compared, choose larger one and carry out calculation of short-circuit current.Now, each power station, front and back equivalence is become to a generator, calculate respectively this two short-circuit currents that generator is fed to isolating switch, then ask its algebraic sum.

The 4th step, the outlet short circuit of calculating two ends, left and right main distribution board load.

The 5th step, calculates the short-circuit conditions in a certain interval.In the time being short-circuited in a certain interval, situation is comparatively complicated, and generator is fed to short-circuit current through an impedance network to short dot, can not directly calculate, and need to carry out impedance network conversion.First draw the impedance network of power station to short dot, then carry out network transformation, become power end and be directly connected to the form of short dot through impedance, then calculate.

Based on the calculation of short-circuit current of graph theory

The Impedance Moment tactical deployment of troops

Boats and ships regional distribution network network is enclosed electrical network, and it is a kind of multiple feed mode, i.e. load can obtain electric power from multiple directions.If now still adopt star-net transformation method will make computation process comparatively complicated.The present invention proposes the method that application graph theoretic approach obtains calculating after line impedance short-circuit current again.

Graph theoretic approach is asked line impedance matrix

Impedance matrix can not directly form according to network, and in the time that network node quantity is very large, calculated amount is very huge.For the more complex network of feeder number, the acquiring method of the line impedance matrix based on graph theory is realized by progressively increasing node (branch) and link, from arbitrary ground connection branch road, under the prerequisite of network-in-dialing, append one by one according to first appending the principle of appending branch after link.

Suppose that one has the impedance matrix of the passive network of m isolated node to be

$\stackrel{\‾}{Z}=\left[\begin{array}{cccc}{\stackrel{\‾}{Z}}_{11}& {\stackrel{\‾}{Z}}_{12}& ...& {\stackrel{\‾}{Z}}_{1m}\\ .& .& .& .\\ .& .& .& .\\ .& .& .& .\\ {\stackrel{\‾}{Z}}_{m1}& {\stackrel{\‾}{Z}}_{m2}& ...& {\stackrel{\‾}{Z}}_{\mathrm{mm}}\end{array}\right]$

On the node p of this passive network, add branch pk, as shown in Figure 7, the dimension of former impedance matrix will add 1, become the matrix shown in following formula.

Wherein original impedance matrix, the mutual resistance matrix between original matrix and new branch pk, it is the matrix of new branch pk.

Order

, in the time there is transimpedance between pk and already present branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}},j=\mathrm{1,2},...m,j\≠k$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xk}}-{\stackrel{\‾}{Z}}_{\mathrm{yk}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}}$

If there is not transimpedance between pk and already present arbitrary branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+{\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

If new branch is to be added between p and reference mode 0,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

If add a link in former network structure, as shown in Fig. 8 (a).Because k is not the new node of system, the dimension of line impedance matrix is constant, but the element of line impedance will change.In order to retain former impedance matrix (original matrix element does not change), on link pk, establish a new node e, as shown in Fig. 8 (b), E _erepresent the voltage of node e to node k.?

In this case, can regard pe as the branch newly increasing process as front.The impedance that contains subscript e in formula has following definition:

Z _1ewhen=unitary current injects k node, the voltage of circuit 1 to reference zero

Z _e1when=unitary current injects circuit 1, the voltage of circuit e to node k

Z _ee=unitary current in the time that k flows into circuit e, the voltage of circuit e to node k

Above formula is transformed to

$\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{xy}}\\ {\stackrel{\‾}{I}}_{\mathrm{pe}}\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{pe}}\\ {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{xy}}\\ {\stackrel{\‾}{V}}_{\mathrm{pe}}\end{array}\right]$

? ${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}},j=\mathrm{1,2},...,\mathrm{m\; j}\≠e$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pe}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}+\frac{1+{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xe}}-{\stackrel{\‾}{Z}}_{\mathrm{ye}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}}$

If nothing coupling between pe and other branch or link, and p is as a reference point, has:

${\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}$

The voltage that then can make node e is 0, the artificial node e increasing of cancellation.

$\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{bus}}\\ 0\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Z}}_{\mathrm{bus}}& {\stackrel{\‾}{Z}}_{\mathrm{je}}\\ {\stackrel{\‾}{Z}}_{\mathrm{ej}}& {\stackrel{\‾}{Z}}_{\mathrm{ee}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{bus}}\\ {\stackrel{\‾}{I}}_e\end{array}\right]$

${\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{mod}\mathrm{ified}}={\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{primitive}}-\frac{{\stackrel{\‾}{Z}}_{\mathrm{je}}{\stackrel{\‾}{Z}}_{\mathrm{je}}^{\′}}{{\stackrel{\‾}{Z}}_{\mathrm{ee}}};$

Based on the short-circuit current calculation method of graph theory

Generally, load current is much smaller compared with short-circuit current, in order to simplify calculating, can ignore the impact of load current in calculation of short-circuit current, at this moment the load impedance of computing node not.Another kind method is the network state after fault to be regarded as to the stack of two kinds of situations, and a kind of is state before fault, i.e. the result of calculation of normal operating mode; Another kind is that each generator electromotive force is equal to zero, and only adds an electromotive force at trouble spot place, and this potential value just equals the magnitude of voltage of trouble spot in the first situation, but polarity is contrary.Calculate the electric current obtaining by the second situation, i.e. fault component in short-circuit current, with the result of calculation stack in the first situation, just can wait until total short-circuit current.

In the time of the direct short circuit of certain malfunctioning node D three-phase, be equivalent to connect at this node the ground connection link that an impedance is null value.If the impedance matrix elements of having obtained before fault occurs is by formula impedance matrix elements after node D access null value ground connection link should be modified to:

${\stackrel{\‾}{Z}}_{\mathrm{pq}}^{-}={\stackrel{\‾}{Z}}_{\mathrm{pq}}-\frac{{\stackrel{\‾}{Z}}_{\mathrm{dp}}{\stackrel{\‾}{Z}}_{\mathrm{dq}}}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}$

When short dot is not to occur on node, but certain 1 D in circuit ij is when upper, needs to increase a new node D, and corresponding impedance matrix increases single order.Before null value impedance ground does not access, this newly-increased node does not affect the element of original matrix, but demand goes out to increase newly node to other internodal transimpedance and self-impedance thereof.Suppose that newly-increased node D is long l × 100% in road completely apart from the distance of one end node i of faulty line, (1-l) × 100% that is full line length apart from the distance of the other end node i of circuit, the at this moment transimpedance of node D and arbitrary node k magnitude of voltage when value equals to inject a unitary current on node k on node D:

${\stackrel{\‾}{Z}}_{\mathrm{dk}}={\stackrel{\‾}{V}}_d={\stackrel{\‾}{V}}_i-{\stackrel{\‾}{I}}_{\mathrm{ij}}{\mathrm{lz}}_{\mathrm{ij}}={\stackrel{\‾}{Z}}_{\mathrm{ik}}-\frac{{\stackrel{\‾}{Z}}_{\mathrm{ik}}-{\stackrel{\‾}{Z}}_{\mathrm{jk}}}{z_{i}j}{\mathrm{lz}}_{\mathrm{ij}}=(1-l){\stackrel{\‾}{Z}}_{\mathrm{ik}}+l{\stackrel{\‾}{Z}}_{\mathrm{jk}}$

Z in formula _ijit is the impedance of circuit between node i and j.

The self-impedance value of node D equals the magnitude of voltage of node D in the time that node D injects a unitary current.

$\frac{{\stackrel{\‾}{V}}_d-{\stackrel{\‾}{V}}_i}{{\mathrm{lz}}_{\mathrm{ij}}}+\frac{{\stackrel{\‾}{V}}_d-{\stackrel{\‾}{V}}_j}{(1-i)z_{\mathrm{ij}}}=1$ , $\frac{{\stackrel{\‾}{Z}}_{\mathrm{dd}}-{\stackrel{\‾}{Z}}_{\mathrm{di}}}{{\mathrm{lz}}_{\mathrm{ij}}}+\frac{{\stackrel{\‾}{Z}}_{\mathrm{dd}}-{\stackrel{\‾}{Z}}_{\mathrm{dj}}}{(1-l)z_{\mathrm{ij}}}=1$

? ${\stackrel{\‾}{Z}}_{\mathrm{dd}}=(1-l){\stackrel{\‾}{Z}}_{\mathrm{di}}+l{\stackrel{\‾}{Z}}_{\mathrm{dj}}+l(1-l)z_{\mathrm{ij}}$

$={(1-l)}^2{\stackrel{\‾}{Z}}_{\mathrm{ii}}+l^2{\stackrel{\‾}{Z}}_{\mathrm{jj}}+2l(1-l){\stackrel{\‾}{Z}}_{\mathrm{ij}}+l(1-l)z_{\mathrm{ij}};$

So, on computational scheme when difference generation three phase short circuit fault, only need know branch road two end nodes from, transimpedance, just can obtain malfunctioning node and other internodal transimpedance and self-impedance value thereof.

In the time adopting the computing method of superposition principle, when the voltage of known fault point before short circuit occurs according to the self-impedance of trouble spot D just can obtain the short-circuit current of trouble spot:

${\stackrel{\‾}{I}}_d=-\frac{{\stackrel{\‾}{V}}_d^{\left(0\right)}}{\stackrel{\‾}{Z}}.$

Again by formula the trouble spot short-circuit current of trying to achieve is as unique non-zero Injection Current substitution:

$\stackrel{\‾}{Z}\stackrel{\‾}{I}=\stackrel{\‾}{V}$

Just obtain the component of voltage being produced at each node by trouble spot short-circuit current:

${\stackrel{\‾}{V}}_{i\left(d\right)}=-\frac{{\stackrel{\‾}{V}}_d^{\left(0\right)}}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}{\stackrel{\‾}{Z}}_{\mathrm{di}},i=\mathrm{1,2},...,m$

In formula for the transimpedance between short dot D and node i.By the component of voltage of this node before this component of voltage and fault be added, obtain the node voltage after short trouble:

${\stackrel{\‾}{V}}_i={\stackrel{\‾}{V}}_i^{\left(0\right)}+{\stackrel{\‾}{V}}_{i\left(d\right)}$

Then obtain and in the time of short trouble, pass through the electric current of each branch road:

${\stackrel{\‾}{I}}_{\mathrm{ij}}=\frac{{\stackrel{\‾}{V}}_i-{\stackrel{\‾}{V}}_j}{z_{\mathrm{ij}}}$

In the simplification calculation of short-circuit current of disregarding load, suppose that approx fault front nodal point voltage perunit value equals 1, and ignore normal by the electric current of branch road.The calculating of short-circuit current is just equivalent to make generator electromotive force to equal zero, and trouble spot calculation of short-circuit current when add a perunit value in trouble spot be 1 electromotive force:

${\stackrel{\‾}{I}}_d=-\frac{1}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}$

The corresponding each node voltage component being produced by trouble spot short-circuit current:

${\stackrel{\‾}{V}}_{i\left(d\right)}=-\frac{{\stackrel{\‾}{Z}}_{\mathrm{di}}}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}$

The voltage of each node should be fault occur before component of voltage and the stack of fault component.Because the perunit value of generator electromotive force is equal to 1, also equal 1 so the perunit value of front each node voltage occurs fault, so:

${\stackrel{\‾}{V}}_i=1+{\stackrel{\‾}{V}}_{i\left(d\right)}=1-\frac{{\stackrel{\‾}{Z}}_{\mathrm{di}}}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}$

According to formula can obtain branch current:

Claims (2)

1. the short-circuit current calculation method of an annular network of ship, determining after short dot, carry out Equivalent Calculation by generator end to short dot, after impedance network conversion and abbreviation equivalence, impedance network abbreviation becomes the form that several equivalent generators are directly connected with short dot through an impedance respectively the most at last, the peak value of short of asking each equivalent generator to be fed in short-circuit point, be then added is required maximal value; After described network transformation, star network is transformed into mesh network, in equivalent net form circuit, between node i and node j, the impedance computation method of branch road is:

star network is made up of node n and other m node, and the each node in a node n and m node has a branch road to join; StarNet's conversion is transformed into node n cancellation with 1 exactly, 2. ... m is the mesh network on summit; It is characterized in that short-circuit current calculation method carries out according to following step:

The first step, determines the typical node that need to calculate short-circuit current;

Second step, the short-circuit conditions of analysis generator end;

The 3rd step, analyzes the short-circuit conditions on port and starboard bus: the both sides of first analyzing isolating switch are short-circuit current when short circuit respectively, and it is compared, and chooses larger one and carries out calculation of short-circuit current; Now, each power station, front and back equivalence is become to a generator, calculate respectively this two short-circuit currents that generator is fed to isolating switch, then ask its algebraic sum;

The 4th step, the outlet short circuit of calculating two ends, left and right main distribution board load;

The 5th step, calculates the short-circuit conditions in a certain interval: first draw the impedance network of power station to short dot, then carry out network transformation, become power end and be directly connected to the form of short dot through impedance, then calculate.

2. the short-circuit current calculation method of annular network of ship according to claim 1, is characterized in that adopting the short-circuit current calculation method based on graph theoretic approach as follows: to have the impedance matrix of the passive network of m isolated node to be:

$\stackrel{\‾}{Z}=\left[\begin{array}{cccc}{\stackrel{\‾}{Z}}_{11}& {\stackrel{\‾}{Z}}_{12}& \·\·\·& {\stackrel{\‾}{Z}}_{1m}\\ \·& \·& \·& \·\\ \·& \·& \·& \·\\ \·& \·& \·& \·\\ {\stackrel{\‾}{Z}}_{m1}& {\stackrel{\‾}{Z}}_{m2}& \·\·\·& {\stackrel{\‾}{Z}}_{\mathrm{mm}}\end{array}\right]$

On the node p of this passive network, add branch pk, the dimension of former impedance matrix will add 1, become the matrix shown in following formula:

Wherein original impedance matrix, the mutual resistance matrix between original matrix and new branch pk, it is the matrix of new branch pk;

Order

, in the time there is transimpedance between pk and already present branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}},j=\mathrm{1,2},...m,j\≠k$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xk}}-{\stackrel{\‾}{Z}}_{\mathrm{yk}})}{{\stackrel{\‾}{Y}}_{\mathrm{pk},\mathrm{pk}}}$

If there is not transimpedance between pk and already present arbitrary branch or link,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk}}+{\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

If new branch is to be added between p and reference mode 0,

${\stackrel{\‾}{Z}}_{\mathrm{kj}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{kk}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}$

On link pk, establish a new node e, E _erepresent the voltage of node e to node k,

The impedance that contains subscript e in formula has following definition:

Z _1ewhen=unitary current injects k node, the voltage of circuit 1 to reference zero,

Z _e1when=unitary current injects circuit 1, the voltage of circuit e to node k,

Z _ee=unitary current in the time that k flows into circuit e, the voltage of circuit e to node k,

Above formula is transformed to:

$\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{xy}}\\ {\stackrel{\‾}{I}}_{\mathrm{pe}}\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{xy},\mathrm{pe}}\\ {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}& {\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{xy}}\\ {\stackrel{\‾}{V}}_{\mathrm{pe}}\end{array}\right]$

? ${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}+\frac{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xj}}-{\stackrel{\‾}{Z}}_{\mathrm{yj}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}},j=\mathrm{1,2},...,mj\≠e$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pe}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}+\frac{1+{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{xy}}({\stackrel{\‾}{Z}}_{\mathrm{xe}}-{\stackrel{\‾}{Z}}_{\mathrm{ye}})}{{\stackrel{\‾}{Y}}_{\mathrm{pe},\mathrm{pe}}}$

If nothing coupling between pe and other branch or link, and p is as a reference point, has:

${\stackrel{\‾}{Z}}_{\mathrm{pj}}=0$

${\stackrel{\‾}{Z}}_{\mathrm{ej}}={\stackrel{\‾}{Z}}_{\mathrm{pj}}-{\stackrel{\‾}{Z}}_{\mathrm{kj}}$

${\stackrel{\‾}{Z}}_{\mathrm{ee}}={\stackrel{\‾}{Z}}_{\mathrm{pk},\mathrm{pk}}-{\stackrel{\‾}{Z}}_{\mathrm{ke}}$

The voltage that then can make node e is 0, the artificial node e increasing of cancellation;

$\left[\begin{array}{c}{\stackrel{\‾}{V}}_{\mathrm{bus}}\\ 0\end{array}\right]=\left[\begin{array}{cc}{\stackrel{\‾}{Z}}_{\mathrm{bus}}& {\stackrel{\‾}{Z}}_{\mathrm{je}}\\ {\stackrel{\‾}{Z}}_{\mathrm{ej}}& {\stackrel{\‾}{Z}}_{\mathrm{ee}}\end{array}\right]\left[\begin{array}{c}{\stackrel{\‾}{I}}_{\mathrm{bus}}\\ {\stackrel{\‾}{I}}_e\end{array}\right]$

${\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{mod}\mathrm{ified}}={\stackrel{\‾}{Z}}_{\mathrm{bus},\mathrm{primitive}}-\frac{{\stackrel{\‾}{Z}}_{\mathrm{je}}{\stackrel{\‾}{Z}}_{\mathrm{je}}^{\′}}{{\stackrel{\‾}{Z}}_{\mathrm{ee}}};$

The self-impedance of trouble spot D ${\stackrel{\‾}{Z}}_{\mathrm{dd}}=(1-l){\stackrel{\‾}{Z}}_{\mathrm{di}}+l{\stackrel{\‾}{Z}}_{\mathrm{dj}}+l(1-l)z_{\mathrm{ij}}$

$={(1-l)}^2{\stackrel{\‾}{Z}}_{\mathrm{ii}}+l^2{\stackrel{\‾}{Z}}_{\mathrm{jj}}+2l(1-l){\stackrel{\‾}{Z}}_{\mathrm{ij}}+l(1-l)Z_{\mathrm{ij}};$

When the voltage of known fault point before short circuit occurs according to the self-impedance of trouble spot D just can obtain the short-circuit current of trouble spot:

${\stackrel{\‾}{I}}_d=-\frac{{{\stackrel{\‾}{V}}_d}^{\left(0\right)}}{{\stackrel{\‾}{Z}}_{\mathrm{dd}}}.$