CN106202915A - Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule - Google Patents

Abstract

The invention belongs to stable state grid power technical field, particularly relate to a kind of Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule.Including setting up basic link power transmission equation, the power transmission equation of basic link is utilized to build the equation group describing whole grid power transmission relation；State equation group with matrix form, this matrix inversion is obtained the matrix description form of stable state grid power transmission；Build the matrix description form that power supply injecting power is independent variable, branch road loss and system loading is dependent variable, based on Cramer's rule, obtain each power supply the relation between supply and demand in the whole network and quantity delivered, and analytical Calculation goes out each composition share of Power System Steady-state power.The present invention solves steady state power transmission problem in circuit and power system theory, steady state power theory has been carried out necessity and has supplemented；In application aspect, work for Operation of Electric Systems and correlational study and provide new theories integration, and provide the mathematical measure of a large-scale calculations.

Description

Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule

Technical field

The invention belongs to stable state grid power technical field, particularly relate to a kind of Power System Steady-state merit based on Cramer's rule Rate constituent relation Analytic Calculation Method.

Background technology

For deepening power system reform further, the conspicuous contradiction and the profound level that solve restriction power industry scientific development are asked Topic, pushing structure transition and industrial upgrading, State Council of the Central Committee of the Communist Party of China issued " about in-depth further on March 15th, 2015 Some suggestions of power system reform ".Power industry market-oriented reform is in the right season, and studies economic problems in the urgent need to accurately Physical basis, thus electricity market is the same with power system, basis is all physical problem.

For Circuit theory and power system theory, though the conveying relation of stable state grid power amount be one basic Problem, but its inherent law not to be the most people definitely known；And the analytical Calculation of the corresponding quantity delivered of each power supply in steady state power Method is not the most suggested.Stable state grid power amount is one of basis supporting power generation, operation and related science research, If problem above is addressed, all working necessarily related to above-mentioned field produces great pushing effect.

Summary of the invention

In order to solve the problems referred to above, the present invention proposes Power System Steady-state power constituent relation based on Cramer's rule and resolves Computational methods, it is characterised in that described method includes

Step 1, definition node, branch road are power delivery unit minimum in electrical network, referred to as the basic link of power transmission, Use circuit analytic method, set up the power transmission equation of basic link, describe the injecting power of basic link and send power Ratio relation；

Step 2, the power of sending taking each basic link are independent variable, and the injecting power of power supply is dependent variable, utilizes base The power transmission equation of this link builds the equation describing the whole electrical network power transmission relation relevant with the injecting power of power supply Group；

Step 3, describing the equation group in step 2 with matrix form, this matrix is Invertible Square Matrix；

Step 4, to the matrix inversion in step 3, obtain with power supply injecting power as independent variable, the sending of each basic link Power is the matrix description form of the stable state grid power transmission of dependent variable；

Step 5, foundation branch road loss power, load power and basic link send the compound proportion relation of power, in step 4 On the basis of, build the matrix description form that power supply injecting power is independent variable, branch road loss and system loading is dependent variable；

Step 6, based on Cramer's rule, obtain the confession in the whole network of each power supply by the power transmission equation of matrix form To relation and quantity delivered, and analytical Calculation goes out each composition share of Power System Steady-state power.

Described basic link power transmission equation is divided into the power transmission equation of branch road and the power transmission equation of node；

The power transmission equation of branch road is

${\stackrel{\·}{S}}_{ij}^{in}=(1+{\stackrel{\·}{H}}_{ij}){\stackrel{\·}{S}}_{ij}^{out}$

Wherein,For the injecting power of branch road i-j,For the power of sending of branch road i-j, compound proportion coefficient The equiva lent impedance that power is corresponding, z is sent for branch road i-j_ijFor branch impedance；

The power transmission equation of node is divided into the power transmission equation of convergent type node and the power transmission of conveying type node Equation；The power transmission equation of convergent type node is

${\stackrel{\·}{S}}_j={\stackrel{\·}{S}}_{1j}^{out}+{\stackrel{\·}{S}}_{2j}^{out}+...+{\stackrel{\·}{S}}_{nj}^{out}$

For at node j to ground leg power consumption；The injecting power provided for nth bar Zhi Luxiang node j.

In described step 6, the power transmission equation of matrix form is

$\stackrel{\·}{A}\·\stackrel{\·}{S}={\stackrel{\·}{S}}_s$

Wherein,It is the relational matrix of power transmission, is reversible square formation, if it is n rank；It it is Power System Steady-state power square Battle array, including each branch road loss power and each load bus power consumption over the ground, i.e. For the loss power of branch road i-j,Being pth load bus power consumption over the ground, they together constituteMatrix i.e. electricity Net steady state power matrix, has a m bar branch road, p load, and m+p=n, thenThere is n element；It it is the injecting power square of power supply Battle array, i.e.There is q power supply；

For equationI.e.According to the impedance in electrical network, admittance Set up transmission equation, thus relational matrixWith power supply injecting power matrixIt is known that seek Power System Steady-state power matrixEach unit Element, by Cramer's rule, has

$\mathrm{\Δ}=\left|\stackrel{\·}{A}\right|;$

${\mathrm{\Δ}}_k=\left|\stackrel{\·}{A_k}\right|;$

Wherein, Δ is relational matrixDeterminant, Δ_kIt it is matrixDeterminant；MatrixIt is WithSubstituteMiddle kth column element A_1k,A_2k,……,A_nkThe new matrix obtained；ThenMiddle kth element i.e. kth amount to be asked

Relational matrixReversible, trying to achieve its inverse matrix isAlso it is n rank square formation, has

Wherein,It it is the quantity delivered corresponding for power supply q constituting Power System Steady-state power；Power System Steady-state power matrixIn ElementConstituent relation and each power supply of Power System Steady-state power are reflected Quantity delivered, obtains each power supply the relation between supply and demand in the whole network and quantity delivered further to the analysis and solution of the whole network steady state power.

Beneficial effect

The present invention, on the premise of known electrical network steady-state load flow, uses circuit analytic method, establishes node, branch road two kinds The power conveying relation of the ratio type relation sending power and injecting power of power conveying basic link, referred to as basic link, The electric energy transport behavior at this basic link is described whereby.Owing to this power conveying relation embodies the impedance of energy transport Matching properties, has uniqueness, and then can carry relation by setting up the power of each basic link, and use matrix form The power conveying relation of whole electrical network is described.Sending power in view of branch road loss power and this branch road is compound proportion relation, joint Point power consumption over the ground sends power also in compound proportion relation with this node, each node, branch road can be sent power and is converted to Branch road loss power or over the ground power consumption, hereby it is possible to set up with branch road loss power, node power consumption over the ground for output Amount, the matrix description with power supply injecting power as input quantity.Based on Cramer's rule, by the power transmission equation of matrix form Each power supply the relation between supply and demand in the whole network and quantity delivered can be obtained, and analytical Calculation goes out each composition part of Power System Steady-state power Volume.Applying above-mentioned conclusion, quantitative relationship and each power supply that can quickly calculate in the matrix form in electrical network between each steady state power exist The relation between supply and demand of the whole network, is especially suitable for the power calculation of large-scale power grid, carries for the work of Operation of Electric Systems, operation and correlational study Supply new theories integration, and provide the mathematical measure of a large-scale calculations.

Accompanying drawing explanation

Fig. 1 is the flow chart of the inventive method；

Fig. 2 is branch power mode；

Fig. 3 is convergent type nodal analysis method；

Fig. 4 is one to enter line one load three outlet nodal analysis method；

Fig. 5 be one enter line one load three outlet node power transmission Integrated Models；

Fig. 6 is two-shipper 6 node steady-state load flow figure.

Detailed description of the invention

The present invention proposes a kind of Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule, method Flow chart is as shown in Figure 1.

Step 1, definition node, branch road are power delivery unit minimum in electrical network, referred to as the basic link of power transmission, Use circuit analytic method, set up the power transmission equation of basic link, describe the injecting power of basic link and send power Ratio relation；

Step 2, the power of sending taking each basic link are independent variable, and the injecting power of power supply is dependent variable, utilizes base The power transmission equation of this link builds the equation describing the whole electrical network power transmission relation relevant with the injecting power of power supply Group；

Step 3, describing the equation group in step 2 with matrix form, this matrix is Invertible Square Matrix；

Step 4, to the matrix inversion in step 3, obtain with power supply injecting power as independent variable, the sending of each basic link Power is the matrix description form of the stable state grid power transmission of dependent variable；

Step 5, foundation branch road loss power, load power and basic link send the compound proportion relation of power, in step 4 On the basis of, build the matrix description form that power supply injecting power is independent variable, branch road loss and system loading is dependent variable；

Step 6, based on Cramer's rule, obtain the confession in the whole network of each power supply by the power transmission equation of matrix form To relation and quantity delivered, and analytical Calculation goes out each composition share of Power System Steady-state power.

The power transmission equation of 1 branch road

Definition node, branch road are the basic link of stable state grid power transmission, and concrete conveying relation lies in basic ring Joint injecting power and send between power.

In Fig. 2,Injecting power for branch road i-j；The power i.e. injecting power of node j is sent for branch road i-j； z_ijFor branch impedance.

If: injecting power and the loss power that difference is branch road i-j sending powerThe electric current of branch road i-j isSend The equiva lent impedance that power is corresponding is

This purpose derived is exactly the parsing relation set up branch road injecting power with send power.

According to branch power balance principle, have

${\stackrel{\·}{S}}_{ij}^{in}={\stackrel{\·}{S}}_{ij}+{\stackrel{\·}{S}}_{ij}^{out}---\left(1\right)$

From formula (1) it can be seen that work as branch road loss powerBy sending powerDuring expression, then target is reached.First build Vertical branch road loss power and the analytic expression sending power, have

${\stackrel{\·}{S}}_{ij}=z_{ij}{\stackrel{\·}{I}}_{ij}{\stackrel{\·}{I}}_{ij}^{*}---\left(2\right)$

${\stackrel{\·}{S}}_{ij}^{out}={\stackrel{\·}{U}}_j{\stackrel{\·}{I}}_{ij}^{*}=z_{ij}^{out}{\stackrel{\·}{I}}_{ij}{\stackrel{\·}{I}}_{ij}^{*}---\left(3\right)$

Take the ratio of two formulas againTherefore branch road loss power with the relation sending power is(1) formula of substitution, can obtain branch power transmission equation, as shown in (4) formula.

${\stackrel{\·}{S}}_{ij}^{in}=(1+{\stackrel{\·}{H}}_{ij}){\stackrel{\·}{S}}_{ij}^{out}---\left(4\right)$

Due to branch impedance each in circuit it is known that thereforeBeing a compound proportion coefficient determined, now branch road injects merit Rate and send between power linear, describes equation (4) formula of this power transmission relation, is a linear algebraic equation.Right In all of branch road, the form describing power transmission relation equation is consistent, but the value of compound proportion coefficient is different.Due to one Power transmission relation exists only in a specific branch road, and therefore (4) formula has stand-alone nature, it may have local features.By InWhereinSend, for branch road i-j, the equiva lent impedance that power is corresponding, be unique in whole electrical network, therefore The branch power conveying relation that equation (4) describes also is unique.Certainly, above-mentioned derivation is also applied for direct current network.

The power transmission equation of 2 nodes

Owing to the connection type complexity of node is various, it is impossible to enumerate, the effect played in therefore carrying by power, will Node is divided into two big classes: convergent type node and conveying type node.The former is the terminal of power conveying, and the latter is holding that power carries Load person.

The power transmission equation of 2.1 convergent type nodes

In Fig. 3,For at node j to ground leg power consumption；It is the q article Zhi Luxiang The injecting power that node j provides.

Due to the terminal that convergent type node is power conveying, the most there is not transmission relation, only exist power-balance relation, For:

${\stackrel{\·}{S}}_j={\stackrel{\·}{S}}_{1j}^{out}+{\stackrel{\·}{S}}_{2j}^{out}+...+{\stackrel{\·}{S}}_{nj}^{out}---\left(5\right)$

The power transmission equation of 2.2 conveying type nodes

Conveying type node may have a plurality of enter line, it is also possible to only one enters line.Owing to the most only describing the power of node Conveying relation, the most unrelated with entering line, therefore, by a plurality of enter after linear heat generation rate summation, then ask for node power transmission relation.In order to Consider generality, choose 1 enter line 1 load 3 outlet node as a example by, its power transmission equation of deriving.

In Fig. 4, i, j, k, l, m are node；Load for node j；The power sent to branch road j-k for node j；The power sent to branch road j-l for node j；The power sent to branch road j-m for node j.

Make y_jFor load powerCorresponding admittance；ForCorresponding admittance；ForCorresponding admittance；ForCorresponding admittance.

According to node power balance principle, have

${\stackrel{\·}{S}}_{ij}^{out}={\stackrel{\·}{S}}_j+{\stackrel{\·}{S}}_{jk}^{in}+{\stackrel{\·}{S}}_{jl}^{in}+{\stackrel{\·}{S}}_{jm}^{in}---\left(6\right)$

The another kind of form of equation (6) is

${\stackrel{\·}{S}}_{ij}^{out}={\stackrel{\·}{U}}_j{\stackrel{\·}{U}}_j^{*}{(y_j+y_{jk}^{in}+y_{jl}^{in}+y_{jm}^{in})}^{*}---\left(7\right)$

OrderThen (6) formula becomes

${\stackrel{\·}{S}}_{ij}^{out}={\stackrel{\·}{S}}_j+{\stackrel{\·}{S}}_x---\left(8\right)$

TakeThen have(8) formula of substitution, can obtain load merit Parsing relation between rate and injecting power, such as (9) formula.

${\stackrel{\·}{S}}_{ij}^{out}=\[1+{\left(\frac{y_{jk}^{in}+y_{jl}^{in}+y_{jm}^{in}}{y_j}\right)}^{*}\]{\stackrel{\·}{S}}_j---\left(9\right)$

In like manner can obtain three transmission relation sending between power and injecting power, for

${\stackrel{\·}{S}}_{ij}^{out}=\[1+{\left(\frac{y_j+y_{jl}^{in}+y_{jm}^{in}}{y_{jk}^{in}}\right)}^{*}\]{\stackrel{\·}{S}}_{jk}^{in}---\left(10\right)$

${\stackrel{\·}{S}}_{ij}^{out}=\[1+{\left(\frac{y_j+y_{jk}^{in}+y_{jm}^{in}}{y_{jl}^{in}}\right)}^{*}\]{\stackrel{\·}{S}}_{jl}^{in}---\left(11\right)$

${\stackrel{\·}{S}}_{ij}^{out}=\[1+{\left(\frac{y_j+y_{jk}^{in}+y_{jl}^{in}}{y_{jm}^{in}}\right)}^{*}\]{\stackrel{\·}{S}}_{jm}^{in}---\left(12\right)$

Formula (10), (11), (12) are the power transmission equation of node j, equivalent admittance hereWithHave Uniqueness so that (10), (11), the power conveying relation that (12) three formulas describe is unique.Owing to derivation not drawn Entering public parameter, therefore three power transmission equations are independent of one another.The node power in the case of three outlets although only having derived Transmission equation, but during for outgoing lines, still can process the most equally.

In Fig. 5, have 3 power carrying paths: Article 1 is branch road i-j to node j to branch road j-k；Article 2 is for propping up Road i-j to node j to branch road j-l；Article 3 is branch road i-j to node j to branch road j-m.With Article 1 power carrying path it is Example, resolves and describes power conveying relation.

According to (4) formula, there is the power transmission relation equation of branch road i-jNode j is had to propping up according to (10) formula The power transmission equation of road j-kAgain according to (4) formula, the power of branch road j-k is had to transmit Relation equationIf setting up injecting powerPower is sent with lastBetween parsing relation, Have only to three power transmission equation simultaneous, and eliminate the quantity of power of centre it is achieved that be

${\stackrel{\·}{S}}_{ij}^{in}=(1+{\stackrel{\·}{H}}_{ij})\[1+{\left(\frac{y_j+y_{jl}^{in}+y_{jm}^{in}}{y_{jk}^{in}}\right)}^{*}\](1+{\stackrel{\·}{H}}_{jk}){\stackrel{\·}{S}}_{jk}^{out}---\left(13\right)$

If regarding the power of sending of three equations as independent variable, then the equation group that these three equations independent of each other are constituted must Surely there is one group of solution.Meanwhile, equation group can be described by matrix form, as shown in (14) formula.

$\left[\begin{array}{c}{\stackrel{\·}{S}}_{ij}^{in}\\ 0\\ 0\end{array}\right]=\left[\begin{array}{ccc}1+{\stackrel{\·}{H}}_{ij}& 0& 0\\ -1& 1+{\left(\frac{y_j+y_{jl}^{in}+y_{jm}^{in}}{y_{jk}^{in}}\right)}^{*}& 0\\ 0& -1& 1+{\stackrel{\·}{H}}_{jk}\end{array}\right]\left[\begin{array}{c}{\stackrel{\·}{S}}_{ij}^{out}\\ {\stackrel{\·}{S}}_{jk}^{in}\\ {\stackrel{\·}{S}}_{jk}^{out}\end{array}\right]---\left(14\right)$

The power transmission relation that this explanation stable state electrical network is total, can be by setting up each node, the respective merit of branch road respectively Rate transmission equation, then the mode carrying out simultaneous solves.Owing to the power transmission equation of basic link is independent of one another, and equation Quantity is consistent with independent variable number, and the Algebraic Equation set now describing grid power transmission relation certainly exists solution.Such equation Group can be described with matrix form and require supplementation with loss and load.

The matrix description of 3 grid power transmission relation

In Fig. 6,I=1,3,4,5 is that node is to ground leg power consumption；I=2,6 is power supply injecting power； WithIt is respectively the injecting power of branch road i-j and sends power；Arrow instruction direction of tide.This is the figure of a Load flow calculation Shape result.

Initially set up the power transmission equation of each node, branch road.

3.1 branch power transmission equations

Branch road i-j loss powerPower is sent with branch roadRatio be compound proportion coefficientComputing formula isThe power transmission equation of branch road i-j isSpecifically Formula and result of calculation are as follows:

${\stackrel{\·}{H}}_{21}={\stackrel{\·}{U}}_{21}/{\stackrel{\·}{U}}_1=0.0172+j0.0232,{\stackrel{\·}{S}}_{21}^{in}=(1+{\stackrel{\·}{H}}_{21}){\stackrel{\·}{S}}_{21}^{out}=(1.0172+j0.0232){\stackrel{\·}{S}}_{21}^{out};$

${\stackrel{\·}{H}}_{23}={\stackrel{\·}{U}}_{23}/{\stackrel{\·}{U}}_3=0.0174+j0.0059,{\stackrel{\·}{S}}_{23}^{in}=(1+{\stackrel{\·}{H}}_{23}){\stackrel{\·}{S}}_{23}^{out}=(1.0174+j0.0059){\stackrel{\·}{S}}_{23}^{out};$

${\stackrel{\·}{H}}_{63}={\stackrel{\·}{U}}_{63}/{\stackrel{\·}{U}}_3=0.0077+j0.0128,{\stackrel{\·}{S}}_{63}^{in}=(1+{\stackrel{\·}{H}}_{63}){\stackrel{\·}{S}}_{63}^{out}=(1.0077+j0.0128){\stackrel{\·}{S}}_{63}^{out};$

${\stackrel{\·}{H}}_{14}={\stackrel{\·}{U}}_{14}/{\stackrel{\·}{U}}_4=0.0058+j0.0100,{\stackrel{\·}{S}}_{14}^{in}=(1+{\stackrel{\·}{H}}_{14}){\stackrel{\·}{S}}_{14}^{out}=(1.0058+j0.0100){\stackrel{\·}{S}}_{14}^{out};$

${\stackrel{\·}{H}}_{45}={\stackrel{\·}{U}}_{45}/{\stackrel{\·}{U}}_5=0.0237+j0.0248,{\stackrel{\·}{S}}_{45}^{in}=(1+{\stackrel{\·}{H}}_{45}){\stackrel{\·}{S}}_{45}^{out}=(1.0237+j0.0248){\stackrel{\·}{S}}_{45}^{out};$

${\stackrel{\·}{H}}_{65}={\stackrel{\·}{U}}_{65}/{\stackrel{\·}{U}}_5=0.0358+j0.0662,{\stackrel{\·}{S}}_{65}^{in}=(1+{\stackrel{\·}{H}}_{65}){\stackrel{\·}{S}}_{21}^{out}=(1.0358+j0.0662){\stackrel{\·}{S}}_{65}^{out}.$

3.2 node power transmission equations

Node 1,4 only one enters line, a load and an outlet.When node j have an injection branch i-j, one Send branch road j-k, one to ground leg time, the electric current sending branch road j-k isInjecting power isTo ground leg Electric current bePower consumption isTo ground leg power consumptionWith send branch road j-k injecting powerRatio be Compound proportion coefficientComputing formula isNode j is to branch road j-k Power transmission equation beConcrete formula and result of calculation are as follows:

${\stackrel{\·}{H}}_{Jb14}={({\stackrel{\·}{I}}_1/{\stackrel{\·}{I}}_{14})}^{*}=1.9460+j0.3884,S_{21}^{out}=(1+{\stackrel{\·}{H}}_{Jb14}){\stackrel{\·}{S}}_{14}^{in}=(2.9460+j0.388){\stackrel{\·}{S}}_{14}^{in};$

${\stackrel{\·}{H}}_{Jb45}={({\stackrel{\·}{I}}_4/{\stackrel{\·}{I}}_{45})}^{*}=-0.2283-j0.2682,{\stackrel{\·}{S}}_{14}^{out}=(1+{\stackrel{\·}{H}}_{Jb45}){\stackrel{\·}{S}}_{45}^{in}=(0.7717-j0.2682){\stackrel{\·}{S}}_{45}^{in}.$

Node 2,6 is slightly different with node 1,4, has two outlets, and not to ground leg, is therefore calculating a branch road Time, another branch road is regarded as to ground leg, re-use the computing formula of node 1,4.Concrete power transmission equation such as 2.2 joint Shown in.Then the power transmission compound proportion coefficient calculations of node 2,6 is as follows:

${\stackrel{\·}{H}}_{Jb23}={({\stackrel{\·}{I}}_{21}/{\stackrel{\·}{I}}_{23})}^{*}=2.7043-j2.0434,{\stackrel{\·}{S}}_{s2}=(1+{\stackrel{\·}{H}}_{Jb23}){\stackrel{\·}{S}}_{23}^{in}=(3.7043-j2.0434){\stackrel{\·}{S}}_{23}^{in};$

${\stackrel{\·}{H}}_{Jb21}={({\stackrel{\·}{I}}_{23}/{\stackrel{\·}{I}}_{21})}^{*}=0.2354+j0.1779,{\stackrel{\·}{S}}_{s2}=(1+{\stackrel{\·}{H}}_{Jb21}){\stackrel{\·}{S}}_{21}^{in}=(1.2354+j0.1779){\stackrel{\·}{S}}_{21}^{in};$

Take two formula power ratio, have after arrangement

${\stackrel{\·}{H}}_{Jb63}={({\stackrel{\·}{I}}_{65}/{\stackrel{\·}{I}}_{63})}^{*}=1.1801+j0.1687,{\stackrel{\·}{S}}_{s6}=(1+{\stackrel{\·}{H}}_{Jb63}){\stackrel{\·}{S}}_{63}^{in}=(2.1801+j0.1687){\stackrel{\·}{S}}_{63}^{in};$

${\stackrel{\·}{H}}_{Jb65}={({\stackrel{\·}{I}}_{63}/{\stackrel{\·}{I}}_{65})}^{*}=0.8304-j0.1187,{\stackrel{\·}{S}}_{s6}=(1+{\stackrel{\·}{H}}_{Jb65}){\stackrel{\·}{S}}_{65}^{in}=(1.8304-j0.1187){\stackrel{\·}{S}}_{65}^{in};$

Take two formula power ratio, have after arrangement

The matrix description of 3.3 grid power transmission relation

3.3.1 basic link sends power transmission relation between power and power supply injecting power

Above-mentioned 12 the power transmission equations that there are, described in it, the equation of branch power transmission has 6, describes node The equation of power transmission has 6.Having 4 power supply links from power supply to load, they are node 2 → branch road 2-3 → nodes 3； Node 6 → branch road 6-3 → node 3；Node 6 → branch road 6-5 → node 5；Node 2 → branch road 1-2 → node 1 → branch road 1-4 → Node 4 → branch road 4-5 → node 5.The power (i.e. equation right side power amount) of sending taking each equation is independent variable, and takes two electricity Source injecting power is dependent variable, and by power link and the node of each of the links, the branch road order of connection of 1 to 4, sets up power and pass Defeated equation group.Owing to 12 equations are independent of one another, equation number is consistent with independent variable number again, therefore can be described this net by matrix form The power transmission relation of network is

$\left[\begin{array}{c}{\stackrel{\·}{S}}_{s2}\\ 0\\ {\stackrel{\·}{S}}_{s6}\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]=\left[\begin{array}{cccccccccccc}1+{\stackrel{\·}{H}}_{Jb23}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -1& 1+{\stackrel{\·}{H}}_{23}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1+{\stackrel{\·}{H}}_{Jb63}& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& -1& 1+{\stackrel{\·}{H}}_{63}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1+{\stackrel{\·}{H}}_{Jb63}& 0& -1-{\stackrel{\·}{H}}_{Jb65}& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& -1& 1+{\stackrel{\·}{H}}_{65}& 0& 0& 0& 0& 0& 0\\ 1+{\stackrel{\·}{H}}_{Jb23}& 0& 0& 0& 0& 0& -1-{\stackrel{\·}{H}}_{Jb21}& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& -1& 1+{\stackrel{\·}{H}}_{21}& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& -1& 1+{\stackrel{\·}{H}}_{Jb14}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& -1& 1+{\stackrel{\·}{H}}_{14}& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& -1& 1+{\stackrel{\·}{H}}_{Jb45}& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1+{\stackrel{\·}{H}}_{45}\end{array}\right]\left[\begin{array}{c}{\stackrel{\·}{S}}_{23}^{in}\\ {\stackrel{\·}{S}}_{23}^{out}\\ {\stackrel{\·}{S}}_{23}^{in}\\ {\stackrel{\·}{S}}_{23}^{out}\\ {\stackrel{\·}{S}}_{65}^{in}\\ {\stackrel{\·}{S}}_{65}^{out}\\ {\stackrel{\·}{S}}_{21}^{in}\\ {\stackrel{\·}{S}}_{21}^{out}\\ {\stackrel{\·}{S}}_{14}^{in}\\ {\stackrel{\·}{S}}_{14}^{out}\\ {\stackrel{\·}{S}}_{45}^{in}\\ {\stackrel{\·}{S}}_{45}^{out}\end{array}\right]$

Finding out from the arrangement feature of matrix element, relational matrix is an inferior triangular flap, has inverse matrix.To above-mentioned relation square Battle array is inverted, and can be able to node, the power of sending of branch road is output, the matrix description shape with power supply injecting power as input quantity Formula.

3.3.2 branch road loss, over the ground power transmission relation between power consumption and power supply injecting power

In branch power transmission relation derivation, the compound proportion relation establishing loss power with sending power, altogether may be used Obtain the equation of 6 description branch road loss powers and branch road injecting power relation；In node power transmission relation derivation, build Found power consumption and the compound proportion relation of injecting power over the ground, 4 can have been set up altogether and describe branch power and node injection over the ground The equation of power relation, is write as matrix form as follows:

$\left[\begin{array}{c}{\stackrel{\·}{S}}_{s2}\\ {\stackrel{\·}{S}}_{s6}\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]=\left[\begin{array}{cccccccccc}(1+{\stackrel{\·}{H}}_{Jb23})(1+\frac{1}{{\stackrel{\·}{H}}_{23}})& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& (1+{\stackrel{\·}{H}}_{Jb63})(1+\frac{1}{{\stackrel{\·}{H}}_{63}})& 0& 0& 0& 0& 0& 0& 0& 0\\ -\frac{1}{{\stackrel{\·}{H}}_{23}}& -\frac{1}{{\stackrel{\·}{H}}_{63}}& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& (1+{\stackrel{\·}{H}}_{Jb63})(1+\frac{1}{{\stackrel{\·}{H}}_{63}})& 0& -(1+{\stackrel{\·}{H}}_{Jb65})(1+\frac{1}{{\stackrel{\·}{H}}_{65}})& 0& 0& 0& 0& 0& 0\\ (1+{\stackrel{\·}{H}}_{Jb23})(1+\frac{1}{{\stackrel{\·}{H}}_{23}})& 0& 0& 0& -(1+{\stackrel{\·}{H}}_{Jb21})(1+\frac{1}{{\stackrel{\·}{H}}_{21}})& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{21}}& 1+\frac{1}{{\stackrel{\·}{H}}_{Jb14}}& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{Jb14}}& (1+{\stackrel{\·}{H}}_{Jb14})(1+\frac{1}{{\stackrel{\·}{H}}_{14}})& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{14}}& (1+{\stackrel{\·}{H}}_{14})(1+\frac{1}{{\stackrel{\·}{H}}_{Jb45}})& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{Jb45}}& (1+{\stackrel{\·}{H}}_{Jb45})(1+\frac{1}{{\stackrel{\·}{H}}_{45}})& 0\\ 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{65}}& 0& 0& 0& 0& -\frac{1}{{\stackrel{\·}{H}}_{45}}& 1\end{array}\right]\left[\begin{array}{c}{\stackrel{\·}{S}}_{23}\\ {\stackrel{\·}{S}}_{63}\\ {\stackrel{\·}{S}}_3\\ {\stackrel{\·}{S}}_{65}\\ {\stackrel{\·}{S}}_{21}\\ {\stackrel{\·}{S}}_1\\ {\stackrel{\·}{S}}_{14}\\ {\stackrel{\·}{S}}_4\\ {\stackrel{\·}{S}}_{45}\\ {\stackrel{\·}{S}}_5\end{array}\right]$

Relational matrix above is inverted, such that it is able to it is defeated for obtaining with branch road loss power, node power consumption over the ground Output, the matrix description form with power supply injecting power as input quantity.

The Analytic Calculation Method of 4 steady state power constituent relations

According to said method, the power transmission equation having matrix form to describe is as follows:

$\stackrel{\·}{A}\·\stackrel{\·}{S}={\stackrel{\·}{S}}_s$

Wherein,It is relational matrix, is reversible square formation；It is Power System Steady-state power matrix, including each branch road loss power With each load bus power consumption over the ground, i.e.There are m bar branch road, p load, and m+p= N, thenThere is n element (amount to be asked)；It is power supply injecting power matrix, i.e. There is q power supply.

For equationAssume that relational matrix is n rank, i.e.Close It it is matrixIt is known that power supply injecting power matrixIt is known that want to claim to obtain Power System Steady-state power matrixEach element, by Gramer Rule, has:

$\mathrm{\Δ}=\left|\stackrel{\·}{A}\right|$

${\mathrm{\Δ}}_i=\left|{\stackrel{\·}{A}}_i\right|$

Wherein, matrixIt is to useSubstituteIn the new matrix that obtains of the i-th column element, thenMiddle i-th element (amount to be asked)For:

${\stackrel{\·}{S}}_i=\frac{\mathrm{\Δ}}{{\mathrm{\Δ}}_i}=\frac{\left|\stackrel{\·}{A}\right|}{\left|{\stackrel{\·}{A}}_i\right|}$

Note, hereIt isMiddle i-th element (amount to be asked), is not load bus power consumption over the ground.

Simultaneously as relational matrixReversible, trying to achieve inverse matrix isAlso it is n rank square formation, has:

$\stackrel{\·}{S}=\stackrel{\·}{B}\·{\stackrel{\·}{S}}_s$

Now,Middle i-th element (amount to be asked)It is represented by:

${\stackrel{\·}{S}}_i={\stackrel{\·}{B}}_{i1}\·{\stackrel{\·}{S}}_{s1}+{\stackrel{\·}{B}}_{i2}\·{\stackrel{\·}{S}}_{s2}+...+{\stackrel{\·}{B}}_{iq}\·{\stackrel{\·}{S}}_{sq}$

Wherein,It it is the quantity delivered corresponding for power supply q constituting Power System Steady-state power.Reflect the constituent relation of Power System Steady-state power and the quantity delivered of each power supply, And the analysis and solution of the whole network steady state power can be obtained each power supply the relation between supply and demand in the whole network and quantity delivered further.

Thus, this patent analytical Calculation has gone out the constituent relation of Power System Steady-state power and each composition share, has obtained each electricity Source the relation between supply and demand in the whole network and quantity delivered.Wherein, the injecting power of each branch road i-j, send power, can be according to each branch road Loss power and each load bus power consumption over the ground are tried to achieve.

According to theory analysis above, the application Power System Steady-state based on Cramer's rule power can be proposed and constitute Relation decomposing computational methods.This patent have found the power conveying pass in stable state electrical network between all quantity of power and power supply injecting power System, i.e. cutting link really between each quantity of power of power system, and can be described by matrix form；Based on Gramer's method Then, by power transmission equation can analytical Calculation go out the injection of all branch roads of system send power, branch road loss power and Each load bus power consumption over the ground；This patent energy analytical Calculation goes out the constituent relation of Power System Steady-state power and each composition share, Obtain each power supply the relation between supply and demand in the whole network and quantity delivered.

Although power conveying relation objective reality, but before this patent, people is not had clearly to propose and conclude；And pass through This patent, not only can carry out the relation of each quantity of power resolving describing, and in known injecting power and electricity grid network parameter In the case of, it is possible to calculate Power System Steady-state power rapidly.

The invention provides the Analytic Calculation Method of a kind of steady state power constituent relation, divide for electricity market and power system Analysis research lays the first stone, for such as electrical network profit distributed problem solving etc. deeper into physics and economic analysis provide and provide powerful support for, right Accelerate power market reform, promote the side such as the operation for electricity consumption cause and maintenance, the quality of raising electric service and cost benefit Face is of great advantage.

Claims (3)

1. Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule, it is characterised in that described method bag Include

Step 1, definition node, branch road are power delivery unit minimum in electrical network, and referred to as the basic link of power transmission, uses Circuit analytic method, sets up the power transmission equation of basic link, describes the injecting power of basic link and the ratio sending power Value relation；

Step 2, the power of sending taking each basic link are independent variable, and the injecting power of power supply is dependent variable, utilizes basic ring The power transmission equation of joint builds the equation group describing the whole electrical network power transmission relation relevant with the injecting power of power supply；

Step 3, describing the equation group in step 2 with matrix form, this matrix is Invertible Square Matrix；

Step 4, to the matrix inversion in step 3, obtain with power supply injecting power as independent variable, each basic link send power The matrix description form transmitted for the stable state grid power of dependent variable；

Step 5, foundation branch road loss power, load power and basic link send the compound proportion relation of power, in the base of step 4 On plinth, build the matrix description form that power supply injecting power is independent variable, branch road loss and system loading is dependent variable；

Step 6, based on Cramer's rule, obtain the supply in the whole network of each power supply by the power transmission equation of matrix form and close It is and quantity delivered, and analytical Calculation goes out each composition share of Power System Steady-state power.

Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule the most according to claim 1, its Being characterised by, described basic link power transmission equation is divided into the power transmission equation of branch road and the power transmission equation of node；

The power transmission equation of branch road is

${\stackrel{\·}{S}}_{ij}^{in}=(1+{\stackrel{\·}{H}}_{ij}){\stackrel{\·}{S}}_{ij}^{out}$

Wherein,For the injecting power of branch road i-j,For the power of sending of branch road i-j, compound proportion coefficientThe equiva lent impedance that power is corresponding, z is sent for branch road i-j_ijFor branch impedance；

The power transmission equation of node is divided into power transmission equation and the power transmission equation of conveying type node of convergent type node； The power transmission equation of convergent type node is

${\stackrel{\·}{S}}_j={\stackrel{\·}{S}}_{1j}^{out}+{\stackrel{\·}{S}}_{2j}^{out}+...+{\stackrel{\·}{S}}_{nj}^{out}$

For at node j to ground leg power consumption；The injecting power provided for nth bar Zhi Luxiang node j.

Power System Steady-state power constituent relation Analytic Calculation Method based on Cramer's rule the most according to claim 1, its Being characterised by, in described step 6, the power transmission equation of matrix form is

$\stackrel{\·}{A}\·\stackrel{\·}{S}={\stackrel{\·}{S}}_s$

Wherein,It is the relational matrix of power transmission, is reversible square formation, if it is n rank；It is Power System Steady-state power matrix, bag Include each branch road loss power and each load bus power consumption over the ground, i.e.For propping up The loss power of road i-j,Being pth load bus power consumption over the ground, they together constituteMatrix i.e. Power System Steady-state Power matrix, has a m bar branch road, p load, and m+p=n, thenThere is n element；It is the injecting power matrix of power supply, i.e.There is q power supply；

For equationI.e.Set up according to the impedance in electrical network, admittance Transmission equation, thus relational matrixWith power supply injecting power matrixIt is known that seek Power System Steady-state power matrixEach element, by Cramer's rule, has

$\mathrm{\Δ}=\left|\stackrel{\·}{A}\right|;$

$\mathrm{\Δ}k=\left|{\stackrel{\·}{A}}_k\right|;$

Wherein, Δ is relational matrixDeterminant, Δ_kIt it is matrixDeterminant；MatrixIt is to useSubstituteMiddle kth column element A_1k,A_2k,……,A_nkThe new matrix obtained；ThenIn Kth element i.e. kth amount to be askedRelational matrixReversible, trying to achieve its inverse matrix isAlso it is n rank side Battle array, has

Wherein,It it is the quantity delivered corresponding for power supply q constituting Power System Steady-state power；Power System Steady-state power matrixIn unit ElementThe constituent relation of Power System Steady-state power and the supply of each power supply are reflected Amount, obtains each power supply the relation between supply and demand in the whole network and quantity delivered further to the analysis and solution of the whole network steady state power.