CN103956741B - The straight algorithm of the many power supplys of three-phase symmetrical non-looped network electric power system of one word chain and attachment formula - Google Patents

Abstract

The invention discloses a kind of straight algorithm of the non-power flow of ring net of the many power supplys of three-phase symmetrical of a word chain and attachment formula, mainly solve the problems such as error calculated restrains greatly, or not arithmetic speed is slow of the iterative method of existing power flow algorithm application.The straight algorithm of the one many power supplys of word chain type three-phase symmetrical non-looped network electric power system, comprise the following steps: power system network is a word chain type, according in power system network element connect order successively by the matrix multiple of each element, obtain the global matrix on three rank: the origin or beginning voltage U calculating power system network _rise; According to element to be calculated present position in power system network, calculate the terminal voltage U of its previous element _{before 2}with end current I _{before 2}; Calculate the terminal voltage U of element to be calculated _{2 treat}, end current I _{2 treat}, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete calculating.

Description

The straight algorithm of the many power supplys of three-phase symmetrical non-looped network electric power system of one word chain and attachment formula

Technical field

The present invention relates to a kind of power flow algorithm of electric power system, specifically, relate to an a kind of word chain type for electric power system and a straight algorithm of the non-looped network of the many power supplys of chain type three-phase symmetrical.

Background technology

It is that the one studying power system mesomeric state ruuning situation substantially electrically calculates that electric power system tide calculates.Its task is the running status according to given service conditions and network configuration determination whole system, as the voltage (amplitude and phase angle) on each bus, the distribution of the power in network and power loss etc.The result that electric power system tide calculates is the basis of Model for Stability Calculation of Power System and accident analysis.In prior art, what the method that electric power system tide calculates adopted usually is iterative method, also claim to toss about in bed method, it is the process that a kind of old value recursion of continuous variable is newly worth, it utilizes the feature that the operational speed of a computer is fast, be applicable to doing repetitive operation, allows computer repeat one group of instruction (or certain step), when this group instruction (or these steps) of each execution, all release its a new value from the initial value of variable, modal iterative method is Newton method.The defect of iterative method is: need constantly newly to be worth by the old value recursion of variable, and its final error calculated is large, and accuracy is low; Secondly, the continuous recursion of iterative method in calculating process, when the value mistake adopted, then need again selected value to calculate, arithmetic speed is slow, and operation result is not restrained, and its principle of operation is as shown in Figure 1 again; In addition, the applicability of iterative method is not strong, be especially not suitable for macroreticular, and different power system networks needs to carry out iteration in different ways.

Summary of the invention

The object of the invention is to overcome above-mentioned defect, provide a kind of operation result accurately, the straight algorithm of the non-power flow of ring net of the many power supplys of three-phase symmetrical of fast operation.

To achieve these goals, based on same inventive concept, present invention employs two kinds of technical schemes:

One

The program is mainly for the power system network of a word chain type, and its particular content is as follows:

The straight algorithm of the one many power supplys of word chain type three-phase symmetrical non-looped network electric power system, comprises the following steps:

(I) power system network is a word chain type, according in power system network element connect order successively by the matrix multiple of each element, obtain the global matrix on three rank:

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ 0& 0& 1\end{array}\right]$

Element to comprise in generator, load, transformer, circuit any one or multiple, accordingly, the matrix of generator is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{r}& 1& -\frac{E}{r}\\ 0& 0& 1\end{array}\right],$ The matrix of load is $\left[\begin{array}{ccc}1& 0& 0\\ Y& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of circuit is $\left[\begin{array}{ccc}1& Z& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of transformer is $\left[\begin{array}{ccc}{n}_1/n_2& 0& 0\\ 0& n_2/n_1& 0\\ 0& 0& 1\end{array}\right];$ Wherein, E represents the desired voltage of generator, and r represents the internal impedance of generator, and Y represents the admittance of load, and Z represents the impedance of circuit, n ₁, n ₂the coil turn of the former secondary of indication transformer respectively;

1. for circuit (see Fig. 2):

According to I _{line 1}=I _{line 2}

U _{line 1}=U _{line 2}+ ZI _{line 2}

Then

2. for load (see Fig. 3):

According to U _{negative 1}=U _{negative 2}

I _{negative 1}=I _{negative 2}+ I

I=YU _{negative 2}

Then

I represents the operating current of load

3. for transformer (see Fig. 4):

Transformer is two-windingtransformer:

According to

Then

In the present invention, transformer is regarded ideal transformer.In actual applications, can using two-windingtransformer when as a load, a circuit and a desirable two-windingtransformer along company, its matrix can be drawn by the matrix multiple of load, circuit and ideal transformer accordingly.Three coil transformers equivalence can become corresponding three two coil transformers.

4. for generator (see Fig. 5):

According to U _{send out 1}=U _{send out 2}i _{send out 1}=I _{send out 2}+ I

I=(U _{send out 2}-E)/r

U _{send out 1}=Ir+E

I _{send out 1}=U _{send out 2}/ r+I _{send out 2}-E/r

Then

I represents the operating current of generator

When the matrix of calculating generator, the ideal source of to be a voltage by generator equivalent be E and impedance are the internal resistance series connection of r.

In actual applications, above-mentioned Y, E, r, Z, n ₁, n ₂be given value, corresponding, the matrix that each element is corresponding is also known.Traditional, the matrix notation for element is second order, and for the ease of application in the present invention, therefore the matrix of each element is expanded to three rank by second order, its extended mode is prior art, therefore does not repeat.

(II) global matrix is substituted into formula:

(III) the origin or beginning voltage U of power system network is calculated according to step (II) _rise;

(IV) according to element to be calculated present position in power system network, terminal voltage and the end current of its previous element is calculated.

(V) terminal voltage of the previous element of element to be calculated calculated in step (IV) and end current are as the origin or beginning voltage U of element to be calculated _{1 treats}with origin or beginning electric current I _{1 treats}, according to formula:

Calculate the terminal voltage U of element to be calculated _{2 treat}, end current I _{2 treat}, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete calculating;

Calculate the calculating of the next element adjacent with it after this element has calculated again, the rest may be inferred completes the Load flow calculation of electric power networks.

The computing formula of each element

One, circuit

Operating voltage U=U ₁-U ₂operating current I=I ₂=I ₁

Power loss $S=U\·\stackrel{\‾}{I}$

Two, load

Operating voltage U=U ₁operating current I=I ₁-I ₂

Power loss $S=U\·\stackrel{\‾}{I}$

Three, power supply

Operating voltage U=U ₁operating current I=I ₁-I ₂

Generator voltage E=U ₂– I*r

Internal resistance power loss

Output of a generator (representing to network delivery electric power when the power output of generator is negative, for just namely representing consumption network electric power)

Four, transformer

U ₁, U ₂, I ₁, I ₂be the operating voltage of transformer, operating current

Power loss is zero

Wherein, U _risefor a terminal voltage of power system network, U _endfor the terminal voltage of power system network, I _risefor the origin or beginning electric current of power system network, its value equals 0, I _endfor the end current of power system network, its value equals 0, and [A] represents the matrix of element to be calculated, U _{1 treats}what represent element to be calculated plays terminal voltage, I _{1 treats}represent the origin or beginning electric current of element to be calculated, U _{2 treat}represent the terminal voltage of element to be calculated, I _{2 treat}represent the end current of element to be calculated, U _rise, I _risealso be terminal voltage, an electric current of network first element.

The above-mentioned formula enumerated and similar formula with it, as: etc. being known formula,

They are two years old

The program is mainly for the power system network of a chain type, and on the basis of a word chain, namely have additional the electric power networks of some side chains, its particular content is as follows:

Prop up the straight algorithm of the many power supplys of chain type three-phase symmetrical non-looped network electric power system, comprise the following steps:

(1) power system network is chain type, and being multiplied with the transition matrix of side chain by each element matrix successively according to its order of connection on main chain according to all elements on main chain and side chain obtains the global matrix of main chain.

The global matrix technical approach of main chain is as follows: the first step calculates the global matrix on side chain, is multiplied in order by the element matrix on side chain; Second step, calculates the transition matrix of side chain according to the global matrix of side chain; 3rd step calculates the global matrix of main chain, and being multiplied successively with the transition matrix of side chain by the element matrix on main chain obtains global matrix.If side chain divides secondary side chain again, then the global matrix account form of this side chain is identical with the account form of main chain, be this side chain in the presence of a main chain, secondary side chain carries out the calculating of matrix in the presence of side chain, if secondary side chain divides again level side chain again, then the account form of its matrix is identical with upper, and element matrix obtains the global matrix on three rank after being multiplied with the transition matrix of side chain:

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ 0& 0& 1\end{array}\right]$

Element to comprise in generator, load, transformer, circuit any one or multiple, accordingly, the matrix of generator is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{r}& 1& -\frac{E}{r}\\ 0& 0& 1\end{array}\right],$ The matrix of load is $\left[\begin{array}{ccc}1& 0& 0\\ Y& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of circuit is $\left[\begin{array}{ccc}1& Z& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of transformer is $\left[\begin{array}{ccc}{n}_1/n_2& 0& 0\\ 0& n_2/n_1& 0\\ 0& 0& 1\end{array}\right],$ The transition matrix of side chain is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{a_{21}^{\′}}{a_{11}^{\′}}& 1& a_{23}^{\′}-\frac{a_{21}^{\′}{a}_{13}^{\′}}{a_{11}^{\′}}\\ 0& 0& 1\end{array}\right],$ It is converted to for global matrix of side chain, and the transition matrix that the global matrix of side chain equals element matrixes all on side chain and branched chain is multiplied successively according to its order and draws: $\left[\begin{array}{ccc}{a}_{11}^{\′}& a_{12}^{\′}& a_{13}^{\′}\\ a_{21}^{\′}& a_{22}^{\′}& a_{23}^{\′}\\ 0& 0& 1\end{array}\right];$ Wherein, E represents the desired voltage of generator, and r represents the internal impedance of generator, and Y represents the admittance of load, and Z represents the impedance of circuit;

(2) global matrix is substituted into formula:

(3) the origin or beginning voltage U of power system network is calculated according to step (two) _rise;

(4) according to element to be calculated present position in power system network, the terminal voltage U of its previous element is calculated _{before 2}with end current I _{before 2}.

(5) with the terminal voltage U of the previous element of element to be calculated calculated in step (four) _{before 2}with end current I _{before 2}as the origin or beginning voltage U of element to be calculated _{1 treats}with origin or beginning electric current I _{1 treats}, according to formula:

Calculate the terminal voltage U of element to be calculated _{2 treat}, end current I _{2 treat}, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete the calculating of element to be calculated;

Wherein, U _risefor a terminal voltage of power system network, U _endfor the terminal voltage of power system network, I _risefor the origin or beginning electric current of power system network, its value equals 0, I _endfor the end current of power system network, its value equals 0, and [A] represents the matrix of element to be calculated, U _{1 treats}what represent element to be calculated plays terminal voltage, I _{1 treats}represent the origin or beginning electric current of element to be calculated, U _{2 treat}represent the terminal voltage of element to be calculated, I _{2 treat}represent the end current of element to be calculated, U _rise, I _risealso be terminal voltage, an electric current of network first element.

The transition matrix distortion derivation of side chain is as follows:

Due to

Namely

Due to I _{prop up 2}=0

Can be reduced to

Solve

According to above-mentioned formula, side chain can be equal to one generator

Can be obtained by the matrix of generator $\left(\begin{array}{ccc}1& 0& 0\\ \frac{1}{r}& 1& -\frac{E}{r}\\ 0& 0& 1\end{array}\right)=\left[\begin{array}{ccc}1& 0& 0\\ \frac{a_{21}^{\′}}{a_{11}^{\′}}& 1& a_{23}^{\′}-\frac{a_{21}^{\′}{a}_{13}^{\′}}{a_{11}^{\′}}\\ 0& 0& 1\end{array}\right],$

Relate to side chain, side chain equivalence can be become in a plant-grid connection main chain, namely the transition matrix of side chain be the equivalent matrix of a source element.So the operating current of the transition matrix institute equivalence element of side chain and terminal voltage are the origin or beginning electric current of side chain and play terminal voltage.

Calculate operating voltage, the electric current of next element in main chain according to the terminal voltage of side chain transition matrix equivalence element and end current, the like complete the Load flow calculation of main chain; Using the terminal voltage of side chain transition matrix equivalence element and the operating current origin or beginning electric current and voltage as side chain, then as the voltage and current calculating each element on side chain main chain successively, complete the Load flow calculation of side chain; Finally complete the Load flow calculation of whole electric power networks.

Wherein, U _{prop up 1}, I _{prop up 1}for terminal voltage, the electric current of side chain, U _{prop up 2}, I _{prop up 2}for terminal voltage, the electric current of side chain.

If side chain is also connected with secondary side chain, then calculated global matrix and the transition matrix of side chain at different levels successively by lowermost level side chain to highest side chain.

If one side chain separates side chain again, then the global matrix of this side chain transition matrix that equals the element matrix on side chain and secondary side chain is multiplied successively according to its order of connection on this side chain and draws.

Wherein, in step (five), if this computing element is positioned on side chain, what then first calculate this side chain plays terminal voltage and origin or beginning electric current, wherein, a terminal voltage of side chain is the terminal voltage of last element, and the origin or beginning electric current of side chain is the operating current of side chain transition matrix equivalence element; And then on the basis of this side chain, calculate terminal voltage and the end current of the last element of element to be calculated on this side chain, finally using the terminal voltage of the last element of element to be calculated and end current as terminal voltage and the origin or beginning electric current of element to be calculated, calculate the terminal voltage of element to be calculated, end current, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete calculating.

If element to be calculated is at the side chain of side chain, namely on secondary side chain, its account form is identical with upper, the account form of the element to be calculated again on level side chain is also the same, by that analogy, those skilled in the art, on the basis of technology contents disclosed in this invention, can unambiguously draw.

Compared with prior art, beneficial effect of the present invention is:

The present invention adopts straight calculation method, directly a terminal voltage of power system network and electric current can be calculated by this algorithm, and its value is exact value, and then according to electric power networks rise terminal voltage and electric current, what calculate each element successively plays terminal voltage, electric current, terminal voltage, electric current, the parameters such as power loss, the account form be newly worth by the old value recursion of variable that the iterative method that this method avoid existing application adopts, its result of calculation is accurate, error free, computational speed is faster, time is shorter, and be substantially applicable to multiple electric power networks, especially to macroreticular, complex network has very high practicality.

Accompanying drawing explanation

Fig. 1 is the schematic diagram of iterative method.

Fig. 2 is the schematic diagram of circuit in the present invention.

Fig. 3 is the schematic diagram of load in the present invention.

Fig. 4 is the schematic diagram of transformer in the present invention.

Fig. 5 is the schematic diagram of generator in the present invention.

Fig. 6 is the schematic diagram of the present invention-embodiment 1 one word chain type power system network.

Fig. 7 is the schematic diagram of the present invention-embodiment 2 chain type power system networks.

Embodiment

Below in conjunction with accompanying drawing, the invention will be further described.Embodiments of the present invention include but not limited to the following example.

Embodiment 1

First by reference to the accompanying drawings the element in electric power networks is described in the present embodiment:

As shown in Figure 2, for circuit:

According to I _{line 1}=I _{line 2}

U _{line 1}=U _{line 2}+ ZI _{line 2}

Then

As shown in Figure 3, for load:

According to U _{negative 1}=U _{negative 2}

I _{negative 1}=I _{negative 2}+ I

I=YU _{negative 2}

Then

I represents loaded work piece electric current

As shown in Figure 4, for transformer:

Transformer, transformer, for two-windingtransformer, when relating to three coil transformers, also can it can be used as three two-windingtransformers to calculate:

According to

Then

In actual applications, can using the two-windingtransformer of reality when as a load, a circuit and a desirable two-windingtransformer along company, its matrix can be drawn by the matrix multiple of load, circuit and ideal transformer accordingly.

As shown in Figure 5, for generator:

According to U _{send out 1}=U _{send out 2}i _{send out 1}=I _{send out 2}+ I

I=(U _{send out 2}-E)/r

U _{send out 1}=Ir+E

I _{send out 1}=U _{send out 2}/ r+I _{send out 2}-E/r

Then

When the matrix of calculating generator, the theoretical foundation that its matrix draws is: the series connection by generator equivalent being desired voltage E and internal resistance r.

Below for the NETWORK STRUCTURE PRESERVING POWER SYSTEM of a word chain, in conjunction with concrete example, the present invention is described:

As shown in Figure 6, the power system network of this word chain, from origin or beginning to end successively: load FZ ₁, circuit XL ₁, generator FDJ ₁, two-windingtransformer BYQ ₁, circuit XL ₂, load FZ ₂, circuit XL ₃, two-windingtransformer BYQ ₂, generator FDJ ₂, the relevant parameter because of each element is known, for the ease of calculating, existing by as follows for the setting parameter of each element:

Load FZ ₁impedance Y ₁=0.03-j0.018;

Circuit XL ₁: Z ₁=1.2+j2.4

Generator FDJ ₁internal resistance r=0.1+j1.5, desired voltage E=10500+j0;

Transformer BYQ ₁coil ratio n ₁: n ₂=10:35;

Circuit XL ₂: Z ₂=0.8+j1.2;

Load FZ ₂: Y ₂=2/350-j1/350;

Circuit XL ₃: Z ₃=0.6+j1.3;

Transformer BYQ ₂coil ratio n ₁: n ₂=35:6;

Generator FDJ ₂internal resistance r=0.2+j1.8, desired voltage E=6300+j0.

According to above-mentioned each element for impedance matrix, in Fig. 6 mono-word chain, the matrix of each element is as follows successively:

Load FZ ₁: $\left(\begin{array}{ccc}1& 0& 0\\ 0.03-j0.018& 1& 0\\ 0& 0& 1\end{array}\right);$

Circuit XL ₂: $\left(\begin{array}{ccc}1& 1.2+j2.4& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right);$

Generator FDJ ₁: $\left(\begin{array}{ccc}1& 0& 0\\ 1/(0.1+j1.5)& 1& -10500/(0.1+j1.5)\\ 0& 0& 1\end{array}\right);$

Transformer BYQ ₁: $\left(\begin{array}{ccc}10/35& 0& 0\\ 0& 35/10& 0\\ 0& 0& 1\end{array}\right);$

Circuit XL ₂ $\left(\begin{array}{ccc}1& 0.8+j1.2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right);$

Load FZ ₂: $\left(\begin{array}{ccc}1& 0& 0\\ 2/350-j1/350& 1& 0\\ 0& 0& 1\end{array}\right);$

Circuit XL ₃: $\left(\begin{array}{ccc}1& 0.6+j1.3& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right);$

Transformer BYQ ₂: $\left(\begin{array}{ccc}35/6& 0& 0\\ 0& 6/35& 0\\ 0& 0& 1\end{array}\right);$

Generator FDJ ₂: $\left(\begin{array}{ccc}1& 0& 0\\ 1/(0.2+j1.8)& 1& -6300/(0.2+j1.8)\\ 0& 0& 1\end{array}\right).$

Above-mentioned impedance matrix is multiplied successively,

$\begin{array}{c}\left(\begin{array}{ccc}1& 0& 0\\ 0.03-j0.018& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 1.2+j2.4& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 1/\left(0.1+j1.5\right)& 1& -10500/(0.1+j1.5)\\ 0& 0& 1\end{array}\right)*\\ \left(\begin{array}{ccc}10/35& 0& 0\\ 0& 35/10& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0.8+j1.2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 2/350-j1/350& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0.6+j1.3& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\\ \left(\begin{array}{ccc}35/6& 0& 0\\ 0& 6/35& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 1/(0.2+j1.8)& 1& -6300/(0.2+j1.8)\\ 0& 0& 1\end{array}\right)\\ =\\ \left(\begin{array}{ccc}1& 0& 0\\ 0.03-j0.018& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 1.2+j2.4& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ (0.0442477876106195-j0.663716814159292)& 1& -464.601769911504+j6969.02654867257\\ 0& 0& 1\end{array}\right)*\end{array}$

$\begin{array}{c}\left(\begin{array}{ccc}0.285714285714286& 0& 0\\ 0& 3.5& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0.8+j1.2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 0.00571428571428571-j0.00285714285714286& 1& 0\\ 0& 0& 1\end{array}\right)*\\ \left(\begin{array}{ccc}1& 0.6+j1.3& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}5.83333333333333& 0& 0\\ 0& 0.171428571428571& 0\\ 0& 0& 1\end{array}\right)*\\ \left(\begin{array}{ccc}1& 0& 0\\ 0.0609756097560976-j0.548780487804878& 1& -384.146341463415+j3457.31707317073\\ 0& 0& 1\end{array}\right)\\ =\\ \left(\begin{array}{ccc}5.74145175408486-j1.36337336264092& 0.98317403441781+j1.73281164116721& -23651.7478206478+j9981.27873331072\\ 0.365690642741442-j1.69763805397826& 0.749701904259654-j0.00231248925926884& -1132.6374032535+j10090.403049798\\ 0& 0& 1\end{array}\right)\end{array}$

In conjunction with the ownership matrix that above-mentioned multiplied result draws, then according to formula:

and I _rise=I _end=0

Draw

Namely

U _rise=8950.58467473813-j1314.63420344468V.

What namely calculated this word chain power system network plays terminal voltage, draws the origin or beginning voltage magnitude=9046.6142345368V of power system network according to above formula.

According to a terminal voltage of the power system network calculated, then carry out lower column count:

Load FZ ₁:

Left end

U _{negative 1}=U _rise=8950.58467473813-j1314.63420344468 amplitude=9046.6142345368

I _{negative 1}=I _rise=0+j0

I=Y*U _{negative 2}=Y*U _{negative 1}

＝(0.03-j0.018)*(8950.58467473813-j1314.63420344468)

=244.85412458014-j200.549550248627 amplitude=316.502234476845

Right-hand member

U _{negative 2}=U _{negative 1}=8950.58467473813-j1314.63420344468 amplitude=9046.6142345368

I _{negative 2}=I _{negative 1}– I

＝(0+j0)–(244.85412458014-j200.549550248627)

=-244.85412458014+j200.549550248627 amplitude=316.502234476845

Bearing power

Circuit XL ₁:

Left end, this end is also circuit XL ₁origin or beginning

U _{line 1}=load FZ ₁right-hand member voltage U _{negative 2}=8950.58467473813-j1314.63420344468

Amplitude=9046.61423453679

I _{line 1}=load FZ ₁right-hand member electric current I _{negative 2}=-244.85412458014+j200.549550248627

Amplitude=316.502234476845

Right-hand member, this end is also circuit XL ₁end

The implication of other element left ends following, right-hand member also with circuit XL ₁unanimously, thus following in repeat no more.

U _{line 2}=U _{line 1}– I _{line 1}* Z ₁

＝(8950.58467473813-j1314.63420344468)-(-244.85412458014+j200.549550248627)*(1.2+j2.4)

＝(8950.58467473813-j1314.63420344468)-(-775.143870092881-j346.990438693988)

=9725.72854483101-j967.643764750692 amplitude=9773.74698788554

I _{line 2}=I _{line 1}=-244.85412458014+j200.549550248627 amplitude=316.502234476845

Line loss

Generator FDJ ₁:

Left end

U _{send out 1}=circuit 1 right-hand member voltage U _{line 2}=9725.72854483101-j967.643764750692

Amplitude=9773.74698788554

I _{send out 1}=circuit 1 right-hand member electric current I _{line 2}=-244.85412458014+j200.549550248627

Amplitude=316.502234476845

I=(U _{send out 1}– E)/r

＝((9725.72854483101-j967.643764750692)-(10500+j0))/(0.1+j1.5)

＝(-774.27145516899-j967.643764750692)/(0.1+j1.5)

=-676.501235682715+j471.080887733812 amplitude=824.361040241663

Right-hand member

U _{send out 2}=U _{send out 1}=9725.72854483101-j967.643764750692 amplitude=9773.74698788554

I _{send out 1}=I _{send out 1}– I

＝(-244.85412458014+j200.549550248627)-(-676.501235682715+j471.080887733812)

=431.647111102575-j270.531337485185 amplitude=509.417739271731

Generator internal resistance loss

$\begin{array}{c}S=I*\stackrel{\‾}{I}*r=\\ =\left(-676.501235682715+j471.080887733812\right)*\left(\stackrel{\‾}{-676.501235682715+j471.080887733812}\right)*\left(0.1+j1.5\right)\\ 67957.1124668317+j1019356.68700248\end{array}$

Output of a generator

Transformer BYQ ₁:

Left end

U _{become 1}=generator right-hand member voltage U _{send out 2}=9725.72854483101-j967.643764750692

Amplitude=9773.74698788554

I _{become 1}=generator right-hand member voltage I _{send out 2}=431.647111102575-j270.531337485185

Amplitude=509.417739271731

Right-hand member

U _{become 2}=(n ₂/ n ₁) * U ₁

＝(35/10)*(9725.72854483101-j967.643764750692)

=34040.0499069085-j3386.75317662742 amplitude=34208.1144575993

I _{become 2}=(n ₁/ n ₂) * I ₁

＝(10/35)*(431.647111102575-j270.531337485185)

=123.327746029307-j77.29466785291 amplitude=145.547925506209

Circuit XL ₂

Left end

U _{line 1}the right-hand member voltage U of=transformer 1 _{become 2}=34040.0499069085-j3386.75317662742

Amplitude=34208.1144575993

I _{line 1}the right-hand member electric current I of=transformer 1 _{become 2}=123.327746029307-j77.29466785291

Amplitude=145.547925506209

Right-hand member

U _{line 2}=U _{line 1}-I _{line 1}* Z ₂=U _{line 1}– I _{line 2}* Z ₂

＝(34040.0499069085-j3386.75317662742)-(123.327746029307-j77.29466785291)*(0.8+j1.2)

＝(34040.0499069085-j3386.75317662742)-(191.415798246938+j86.1575609528404)

=33848.6341086616-j3472.91073758026 amplitude=34026.3300991049

I _{line 2}=I _{line 1}=123.327746029307-j77.29466785291 amplitude=145.547925506209

Line loss

Load FZ ₂:

Left end

U _{negative 1}=circuit 2 right-hand member voltage U _{line 2}=33848.6341086616-j3472.91073758026

Amplitude=34026.3300991049

I _{negative 1}=circuit 2 right-hand member electric current I _{line 2}=123.327746029307-j77.29466785291

Amplitude=145.547925506209

I=Z*U _{negative 2}=Z*U _{negative 1}

＝(2/350–j1/350)*(33848.6341086616-j3472.91073758026)

=183.498164227837-j116.555587382349 amplitude=217.386248932702

Right-hand member

U _{negative 2}=U _{negative 1}=33848.6341086616-j3472.91073758026 amplitude=34026.3300991049

I _{negative 2}=I _{negative 1}– I

＝(123.327746029307-j77.29466785291)-(183.498164227837-j116.555587382349)

=-60.17041819853+j39.260919529439 amplitude=71.8463570995988

Bearing power

Circuit XL ₃:

Left end

U _{line 1}=load 2 right-hand member voltage U _{negative 2}=33848.6341086616-j3472.91073758026

Amplitude=34026.3300991049

I _{line 1}=load 2 right-hand member electric current I _{negative 2}=-60.17041819853+j39.260919529439

Amplitude=71.8463570995988

Right-hand member

U _{line 2}=U _{line 1}-I _{line 1}* Z ₃

＝(33848.6341086616-j3472.91073758026)-(-60.17041819853+j39.260919529439)*(0.6+j1.3)

＝(33848.6341086616-j3472.91073758026)-(-87.1414463073887-j54.6649919404256)

=33935.775554969-j3418.24574563983 amplitude=34107.4957523242

I _{line 2}=I _{line 1}=-60.17041819853+j39.260919529439 amplitude=71.8463570995988

Line loss

Transformer BYQ ₂:

Left end

U _{become 1}=circuit 3 right-hand member voltage U _{line 3}=33935.775554969-j3418.24574563983

Amplitude=34107.4957523242

I _{become 1}=circuit 3 right-hand member electric current I _{line 3}=-60.17041819853+j39.260919529439

Amplitude=71.8463570995988

Right-hand member

U _{become 2}=(n ₂/ n ₁) * U _{become 1}

＝(6/35)*(33935.775554969-j3418.24574563983)

=5817.56152370897-j585.984984966828 amplitude=5846.99927182701

I _{become 2}=(n ₁/ n ₂) * I _{become 1}

＝(35/6)*(-60.17041819853+j39.260919529439)

=-350.994106158092+j229.022030588394 amplitude=419.10374974766

Generator FDJ ₂

Left end

U _{send out 1}=transformer 2 right-hand member voltage U _{become 2}=5817.56152370897-j585.984984966828

Amplitude=5846.99927182701

I _{send out 1}=transformer 2 right-hand member electric current I _{become 2}=-350.994106158092+j229.022030588394

Amplitude=419.10374974766

I＝(U ₁–E)/r

＝((5817.56152370897-j585.984984966828)-(6300+j0))/(0.2+j1.8)

＝(-482.43847629103-j585.984984966828)/(0.2+j1.8)

=-350.994106158078+j229.022030588564 amplitude=419.103749747741

Right-hand member

U _{send out 2}=U _{send out 1}=5817.56152370897-j585.984984966828 amplitude=5846.99927182701

I _{send out 2}=I _{send out 1}– I

＝(-350.994106158092+j229.022030588394)–(-350.994106158078+j229.022030588564)

＝-0.000000000014–j0.00000000017

Internal resistance loss

$\begin{array}{c}S=I*\stackrel{\‾}{I}*r\\ =(-350.994106158078+j229.022030588564)*\left(\stackrel{\‾}{-350.994106158078+j229.022030588564}\right)*(0.2+j1.8)\\ =35129.5906105098+j316166.315494588\end{array}$

Output of a generator

Σ power=0.0000000223517-j0.00000108592

Embodiment 2

As shown in Figure 7, the present embodiment, based on the power system network of a chain type, that includes main chain A, side chain B and secondary side chain C, wherein, side chain C is lowermost level side chain, no longer separates side chain thereon, and side chain B is highest side chain, namely side chain C separates from it, for convenience of description, be numbered respective element according to element orders, situation is as follows:

1. generator 1, uses alphabetical FDJ ₁refer to, internal resistance r=0.08+j1.1, E=10500+j0; 2. transformer 1, uses letter b YQ ₁refer to, coil ratio is n ₁: n ₂=10/35; 3. circuit 1, uses alphabetical XL ₁refer to, Z ₁=0.8+j1.2; 4. load 1, uses alphabetical FZ ₁refer to, Y ₁=0.005-j0.002; 5. circuit 2, uses alphabetical XL ₂refer to, Z ₂=0.7+j1.1; 6. transformer 2, uses letter b YQ ₂refer to, coil ratio is n ₁: n ₂=35/6; 7. generator 2, uses alphabetical FDJ ₂refer to, internal resistance r=0.1+j0.8, desired voltage E=6300+j0; 8. circuit 3, uses alphabetical XL ₃refer to, Z ₃=0.2+j0.8; 9. load 2, uses alphabetical FZ ₂refer to, Y ₂=0.004-j0.002.

Accordingly, the impedance matrix of each element is as follows:

FDJ ₁fabric error! Do not find Reference source.；

A fabric error of transformer 1! Do not find Reference source.；

An impedance mistake of circuit 1! Do not find Reference source.；

A fabric error of load 1! Do not find Reference source.；

A fabric error of circuit 2! Do not find Reference source.

A fabric error of transformer 2! Do not find Reference source.；

A fabric error of generator 2! Do not find Reference source.；

A fabric error of circuit 3! Do not find Reference source.；

A fabric error of load 2! Do not find Reference source.；

Side chain point to have side chain, first need the global matrix of the side chain of most final stage to obtain, in the present embodiment, the global matrix ZC by C chain obtains, and it is long-pending with matrix multiple of load 2 that it equals circuit 3:

$ZC=\left(\begin{array}{ccc}1& 0.2+j0.8& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 0.004-j0.002& 1& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1.0024+j0.0028& 0.2+j0.8& 0\\ 0.004-j0.002& 1& 0\\ 0& 0& 1\end{array}\right)$

Again above-mentioned matrix ZC distortion is converted to the transition matrix of C chain

$YC=\left(\begin{array}{ccc}1& 0& 0\\ 0.00398481867681727-j0.00200634227084506& 1& 0\\ 0& 0& 1\end{array}\right),$ This distortion conversion has explanation above-mentioned, and therefore not to repeat here.

After the global matrix of C chain calculates, then calculate the global matrix of B chain, in the present embodiment, the global matrix ZB of B chain equal circuit 2, the transition matrix of C chain, transformer 2, generator 2 matrix be multiplied successively and draw:

$\begin{array}{c}ZB=\left(\begin{array}{ccc}1& 0.7+j1.1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 0.00398481867681727-j0.00200634227084506& 1& 0\\ 0& 0& 1\end{array}\right)*\\ \left(\begin{array}{ccc}\frac{35}{6}& 0& 0\\ 0& \frac{6}{35}& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ \frac{1}{0.1+j0.8}& 1& \frac{6300}{0.1+j0.8}\\ 0& 0& 1\end{array}\right)\\ =\left(\begin{array}{ccc}6.11302815638438-j0.1013046297777692& 0.12+j0.188571428571429& -1578.46153846154+j474.692307692308\\ 0.0496184019883938-j0.222692674235607& 0.171428571428571& -166.153846153846+j1329.23076923077\\ 0& 0& 1\end{array}\right)\end{array}$

Again above-mentioned matrix distortion is converted to the transition matrix of B chain

$YB=\left(\begin{array}{ccc}1& 0& 0\\ 0.00871813613185083-j0.0362847153666& 1& -179.522406149059+j1265.43825826737\\ 0& 0& 1\end{array}\right)$

After the transition matrix of B chain calculates, finally calculate the global matrix ZA of main chain A, its equal generator 1, transformer 1, B chain transition matrix, circuit 1, load 1 impedance matrix be multiplied successively and draw:

$\begin{array}{c}ZA=\left(\begin{array}{ccc}1& 0& 0\\ \frac{1}{0.08+j1.1}& 1& \frac{10500}{0.08+j1.1}\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}\frac{10}{35}& 0& 0\\ 0& \frac{35}{10}& 0\\ 0& 0& 1\end{array}\right)*YB*\\ \left(\begin{array}{ccc}1& 0.8+j1.2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)*\left(\begin{array}{ccc}1& 0& 0\\ 0.005-j0.002& 1& 0\\ 0& 0& 1\end{array}\right)\\ =\left(\begin{array}{ccc}0.287542857142857+j0.00125714285714286& 0.228571428571429+j0.342857142857143& 0\\ 0.0688154638962749-j0.394619589217566& 4.0018876210826-j0.249130981946827& -1318.89073654966+j13924.2657355701\\ 0& 0& 1\end{array}\right)\end{array}$

After the global matrix of trying to achieve A chain, can according to the relevant parameter of each element of following calculating:

For convenience of description, in the present embodiment by each element rise the equal called after U of terminal voltage ₁, origin or beginning electric current is I ₁, terminal voltage is U ₂, end current is I ₂, then by described separately for each element, therefore can not misleading be caused.

One, generator 1

$\begin{array}{c}{U}_1=a{13}_{}-\frac{a_{11}*a_{23}}{a_{21}}\\ =0-\frac{\left(0.287542857142857+j0.001257142857142856\right)\left(-1318.890736549966+j13924.2657355701\right)}{\left(0.0688154638962749-j0.394619589217566\right)}\end{array}$

A mistake! Do not find Reference source.Volt

I ₁=0+j0 amplitude=0 is pacified

Output of a generator

$\begin{array}{c}S=E*\stackrel{\‾}{I}=\left(10500+j0\right)\left(-701.84704801192-j391.984675432677\right)\\ =-7369394.00412516-j4115839.09204311\end{array}$

P=-7369394.00412516 watt of Q=-4115839.09204311 is weary

Internal resistance loss

$\begin{array}{c}S=\left(U_1-E\right)*\stackrel{\‾}{I}=\left(\left(10012.6690931831-j740.672978778498\right)-\left(10500+j0\right)\right)*\left(-701.84704801192-j391.984675432677\right)\\ =51699.3011661686+j710865.391034818\end{array}$

P＝51699.3011661686Q＝710865.391034818

A mistake! Do not find Reference source.Volt

Two, transformer 1

U ₁=10012.6690931831-j740.672978778497 amplitude=10040.0267341811

I ₁＝(701.84704801192-j391.984675431677)

Due to the transition matrix of B chain

A YB=mistake! Do not find Reference source.

U ₁=35044.3418261409-j2592.35542572474 amplitude=35140.0939196339 volt

I ₁=200.527728003406-j111.995621552193 amplitude=229.68323609935 are pacified

U ₂=U ₁=35044.3418261409-j2592.35542572474 amplitude=35140.0939196339 volt

I＝a ₂₁*U ₂+a ₂₃

＝(0.00871813613185083-j0.0362847153666)(35044.3418261409-j2592.35542572474)+(-179.522406149059+j1265.43825826737)

=31.9360577908673-j28.7362176075982 amplitude=42.9614011597952 are pacified

I ₂＝I ₁-I＝

(200.527728003406-j111.995621552193)-(31.9360577908673-j28.7362176075982)

=168.591670212538-j83.2594039445952 amplitude=188.029996570394 are pacified

YBU=U ₂=35044.3418261409-j2592.35542572474 amplitude=35140.0939196339 volt

YBI=I=31.9360577908673-j28.7362176075982 amplitude=42.9614011597952 are pacified

In formula, YBU refers to a terminal voltage of B chain, and YBI refers to the origin or beginning electric current of B chain

Three, circuit 1

U ₁=35044.3418261409-j2592.35542572474 amplitude=35140.0939196339 volt

I ₁=168.591670212538-j83.2594039445952 amplitude=188.029996570394 are pacified

$\begin{array}{c}{U}_2=U_1-I_1*Z\\ =(35044.3418261409-2592.35542572474)-\\ (168.591670212538-j83.2594039445952)*(0.8+j1.2)\end{array}$ A mistake! Do not find Reference source.Volt

I ₂=I ₁=168.591670212538-j83.25594039445952 amplitude=188.029996570394 are pacified

Line loss

$S=(U_1-U_2)*\stackrel{\‾}{I_1}$

$\begin{array}{c}=\left(\left(65044.3418261409-j2592.35542572474\right)-\left(34809.5572052373-j2728.05790682411\right)\right)*\\ \stackrel{\‾}{\left(168.591670212538-j83.2594039445952\right)}\\ =28284.22368821+j424246.3355323151\end{array}$

Active loss P=28284.22368821 watt of reactive loss Q=42426.3355323151 is weary

Four, load 1

U ₁=34809.5572052373-j2728.05790682411 amplitude=34916.2938005693 volt

I ₁=168.591670312538-j83.2594039445942 amplitude=158.029996570394 are pacified

A mistake! Do not find Reference source.Volt

I＝Y*U ₂＝(0.005-j0.002)*(34809.5572052373-j2728.05790682411)

=168.591670212538-j83.2594039445942 amplitude=188.029996570394

I ₂＝I ₁-I＝(168.591670212538-j83.2594039445952)-(168.591670212538-j83.2594039445952)

=5.55111512312578E-17-j2.77555756156289E-16 amplitude=2.83052443340184E-16

Load power consumption

$\begin{array}{c}S=U_2*\stackrel{\‾}{I}=(34809.5572052373-j2728.25790682411)*\left(168.591670212538+j83.2594039445952\right)\\ =6095737.86383837+j2438295.14553535\end{array}$

The meritorious idle Q=2438295.14553535 of P=6095737.86383837

Five, circuit 2

U ₁=YBU=35044.3418161409-j2592.35542572474 amplitude=35140.0939196339 volt

I ₁=YBI=31.9360577908673-j28.7362176075982 amplitude=42.9614011597951 are pacified

U ₂＝U ₁-t ₁*Z＝(35044.3418261409-j2592.35542572474)-(31.9360577908673-j28.7362176075982)*(0.7+j1.1)

=34990.3767463189-j2607.36973696938 amplitude=35087.3886431378

I ₂=I ₁=31.9360577908673-j28.7362176075982 amplitude=42.9614011597952 are pacified

In above-mentioned formula, Z represents line impedance.

Line loss

Active loss P=1291.977392729 watt of reactive loss Q=2030.25018857414 is weary

Due to the transition matrix of C chain, YC:

A YC=mistake! Do not find Reference source.

U ₁=34990.3767462189-j2607.36973696938 amplitude=35087.3886431378

I ₁=31.9360577908673-j28.7362176075982 amplitude=42.9614011597952 are pacified

U ₂=U ₁=34990.3767463189-j2607.36973696938 amplitude=35087.3886431378

I＝a ₂₁U ₂+a ₂₃

＝(0.00398481867681727-j0.00200634227084506)*(34990.3767463189-j2607.36973696938)+0

=134.1990306486-j80.5925675641773 amplitude=156.539265915289

I ₂＝I ₁-I＝

(31.9360577908673-j28.7362176075982)-(134.1990306486-j80.5925675641773)

=-102.262972857733+j51.8563499565792 amplitude=114.659481284893 are pacified

YCU=U ₂=34990.3767462189-j2607.36973696938 amplitude=35087.3886431378 volt

YCI=I=134.1990306486-j80.5925675641773 amplitude=156.539265916289 are pacified

Six, transformer 2

U ₁=34990.3767463189-j2607.36973696938 amplitude=35087.3886431378

I ₁=-102.262972857733+j51.8563499565792 amplitude=114.659481284893

Seven, generator 2

U ₁=5998.35029936895-j446.97766919475 amplitude=6014.9809102522 volt

I ₁=-596.534008336777+j302.495374746712 amplitude=668.846974161875 are pacified

Output of a generator

$\begin{array}{c}S=E*\stackrel{\‾}{I}=(6300+j0)(-596.534008336777-j302.495374746712)\\ =-3758164.25252169-j1905720.86090428\end{array}$

P=-3758164.25252169 watt of Q=-1905720.86090428 is weary

Internal resistance loss

$\begin{array}{c}S=(U_1-E)*\stackrel{\‾}{I}=\left(\left(5998.35029936895-j446.97766919475\right)-\left(6300+j0\right)\right)*\left(-596.534008336777-j302.495374746712\right)\\ =44735.6274845495+j357885.019876396\end{array}$

P＝44735.6274845495Q＝357885.019876396

U ₂=U ₁=5998.35029936795-j446.97766919475 amplitude=6014.9809102522 volt

I ₂＝I ₁＝I＝(-596.534008336777+j302.495374746712)-(-596.534008336777+j302.495374746712)

=1.11022302462516E-16-j1.11022302462516E-15 amplitude=1.11576033091875E-15 peace

Eight, circuit 3

U ₁=YCU=34990.3767463189-j2607.36973696938 amplitude=35087.3886431378 volt

I ₁=YCI=134.1990306486-j80.5925675641773 amplitude=156.539265916289 are pacified

U ₂＝u ₁-I ₁·Z＝(34990.3767463189-j2607.36973696938)-(134.1990306486-j80.5925675641773)*(0.2+j0.8)

=34899.0628861378-j2698.61044797542 amplitude=35003.2439736737

I ₂=I ₁=134.1990306486-j80.5925675641773 amplitude=156.539265916289 are pacified

Line loss

$\begin{array}{c}S=(U_1-U_2*\stackrel{\‾}{I_1})\\ =\left(\left(34990.3767463189-j2607.36973696938\right)-\left(34899.0628861378-j2698.61044797542\right)\right)*\left(134.1990306486+j80.5925675641773\right)\\ =4900.9083547221+j19603.6334188884\end{array}$

Active loss P=4900.9083547221 watt of reactive loss Q=19603.6334188884 is weary

Nine, load 2

U ₁=34899.0628861378-j2698.61044797542 amplitude=35003.2439736737 volt

I ₁=134.1990306486-j80.5925675641773 amplitude=156.539265916289 are pacified

A mistake! Do not find Reference source.Volt

I＝Y*U ₂＝(0.004-j0.002)*(34899.0628861378-j2698.61044797542)

=134.1990306486-j80.5925675641773 amplitude=156.539265916289

I ₂＝I ₁-I＝(134.1990306486-j80.5925675641773)-(134.1990306486-j80.5925675641773)

=0-j1.38777878078145E-17 amplitude=1.38777878078145E-17

|I ₂|＝1.38*10 ^-17＜1.5*10 ^-17

Power consumption

$\begin{array}{c}S=U_2*\stackrel{\‾}{I}=(34899.0628861378-j2698.61044797542)*(134.1990306486+j80.5925675641773)\\ =4900908.3547221+j2450454.17736105\end{array}$

The meritorious idle Q=2450454.17736105 of P=4900908.3547221

According to above-described embodiment, just the present invention can be realized well.What deserves to be explained is; under prerequisite based on above-mentioned design principle; for solving same technical problem; even if some making on architecture basics disclosed in this invention are without substantial change or polishing; the essence of the technical scheme adopted is still the same with the present invention, therefore it also should in protection scope of the present invention.

Claims (4)

1. the straight algorithm of the many power supplys of a word chain type three-phase symmetrical non-looped network electric power system, is characterized in that, comprise the following steps:

(I) power system network is a word chain type, according in power system network element connect order successively by the matrix multiple of each element, obtain the global matrix on three rank:

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ 0& 0& 1\end{array}\right]$

Element to comprise in generator, load, transformer, circuit any one or multiple, accordingly, the matrix of generator is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{r}& 1& -\frac{E}{r}\\ 0& 0& 1\end{array}\right],$ The matrix of load is $\left[\begin{array}{ccc}1& 0& 0\\ Y& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of circuit is $\left[\begin{array}{ccc}1& Z& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of transformer is $\left[\begin{array}{ccc}{n}_1/n_2& 0& 0\\ 0& n_2/n_1& 0\\ 0& 0& 1\end{array}\right];$ Wherein, E represents the desired voltage of generator, and r represents the internal impedance of generator, and Y represents the admittance of load, and Z represents the impedance of circuit, n ₁, n ₂the coil turn of the former secondary of indication transformer respectively;

(II) global matrix is substituted into formula:

(III) the origin or beginning voltage U of power system network is calculated according to step (II) _rise;

(IV) according to element to be calculated present position in power system network, the terminal voltage U of its previous element is calculated _{before 2}with end current I _{before 2};

(V) with the terminal voltage U of the previous element of element to be calculated calculated in step (IV) _{before 2}with end current I _{before 2}as the origin or beginning voltage U of element to be calculated _{1 treats}with origin or beginning electric current I _{1 treats}, according to formula:

Calculate the terminal voltage U of element to be calculated _{2 treat}, end current I _{2 treat}, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete the calculating of element to be calculated;

Wherein, U _risefor a terminal voltage of power system network, U _endfor the terminal voltage of power system network, I _risefor the origin or beginning electric current of power system network, its value equals 0, I _endfor the end current of power system network, its value equals 0; [A] represents the matrix of element to be calculated, U _{1 treats}what represent element to be calculated plays terminal voltage, I _{1 treats}represent the origin or beginning electric current of element to be calculated, U _{2 treat}represent the terminal voltage of element to be calculated, I _{2 treat}represent the end current of element to be calculated.

2. prop up the straight algorithm of the many power supplys of chain type three-phase symmetrical non-looped network electric power system, it is characterized in that, comprise the following steps:

(1) power system network is chain type, is multiplied by the matrix of each element successively, obtains the global matrix on three rank according to all elements on main chain and side chain according to its order of connection on main chain with the transition matrix of side chain:

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ 0& 0& 1\end{array}\right]$

Element to comprise in generator, load, transformer, circuit any one or multiple, accordingly, the matrix of generator is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{r}& 1& -\frac{E}{r}\\ 0& 0& 1\end{array}\right],$ The matrix of load is $\left[\begin{array}{ccc}1& 0& 0\\ Y& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of circuit is $\left[\begin{array}{ccc}1& Z& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],$ The matrix of transformer is $\left[\begin{array}{ccc}{n}_1/n_2& 0& 0\\ 0& n_2/n_1& 0\\ 0& 0& 1\end{array}\right],$ The transition matrix of side chain is $\left[\begin{array}{ccc}1& 0& 0\\ \frac{a_{21}^{\′}}{a_{11}^{\′}}& 1& a_{23}^{\′}-\frac{a_{21}^{\′}{a}_{13}^{\′}}{a_{11}^{\′}}\\ 0& 0& 1\end{array}\right],$ It is converted to for global matrix of side chain, and the global matrix of side chain equals element matrixes all on side chain and is multiplied successively according to its order and draws: $\left[\begin{array}{ccc}{a}_{11}^{\′}& a_{12}^{\′}& a_{13}^{\′}\\ a_{21}^{\′}& a_{22}^{\′}& a_{23}^{\′}\\ 0& 0& 1\end{array}\right];$ Wherein, E represents the desired voltage of generator, and r represents the internal impedance of generator, and Y represents the admittance of load, and Z represents the anti-of circuit, n ₁, n ₂the coil turn of the former secondary of indication transformer respectively;

(2) global matrix is substituted into formula:

(3) the origin or beginning voltage U of power system network is calculated according to step (two) _rise;

(4) according to element to be calculated present position in power system network, the terminal voltage U of its previous element is calculated _{before 2}with end current I _{before 2};

(5) terminal voltage of the previous element of element to be calculated calculated in step (four) and end current are as the origin or beginning voltage U of element to be calculated _{1 treats}with origin or beginning electric current I _{1 treats}, according to formula:

Calculate the terminal voltage U of element to be calculated _{2 treat}, end current I _{2 treat}, under terminal voltage, electric current and the terminal voltage, all fixed prerequisite of electric current of element to be calculated, the operating voltage of element to be calculated, operating current and power loss can be calculated, complete the calculating of element to be calculated;

Wherein, U _risefor a terminal voltage of power system network, U _endfor the terminal voltage of power system network, I _risefor the origin or beginning electric current of power system network, its value equals 0, I _endfor the end current of power system network, its value equals 0, and [A] represents the matrix of element to be calculated, U _{1 treats}what represent element to be calculated plays terminal voltage, I _{1 treats}represent the origin or beginning electric current of element to be calculated, U _{2 treat}represent the terminal voltage of element to be calculated, I _{2 treat}represent the end current of element to be calculated.

3. according to claim 2 the straight algorithm of the many power supplys of chain type three-phase symmetrical non-looped network electric power system, it is characterized in that, if a branched chain again on side chain, then the global matrix of this side chain equals element on side chain and the transition matrix separating side chain and is multiplied according to the order of connection and obtains; Wherein, then the transition matrix of branched chain be converted to by its global matrix.

4. according to claim 3 the straight algorithm of the many power supplys of chain type three-phase symmetrical non-looped network electric power system, it is characterized in that, in described step (five), if this element to be calculated is positioned on side chain, what then first calculate this side chain plays terminal voltage and origin or beginning electric current, and then on the basis of this side chain, calculate terminal voltage and the end current of the last element of element to be calculated on this side chain, finally using the terminal voltage of the last element of element to be calculated and end current as terminal voltage and the origin or beginning electric current of element to be calculated, calculate the terminal voltage of element to be calculated, end current, in a terminal voltage of element to be calculated, electric current and terminal voltage, under all fixed prerequisite of electric current, the operating voltage of element to be calculated can be calculated, operating current and power loss, complete the Load flow calculation of side chain.