CN104218577B - A kind of three-phase load flow of active distribution network computational methods based on node voltage - Google Patents

Abstract

The present invention provides a kind of three-phase load flow of active distribution network computational methods based on node voltage, with the real part of node phase voltage with imaginary part as quantity of state, and to select the real part of node every phase injection current and imaginary part be known quantity, set up the relation of known quantity and unknown quantity by nodal voltage equation, directly calculated factor table by formula.The present invention can directly process many PV node, loop, and is not limited by loop number；Directly calculated the factor table of Jacobian matrix by formula, avoid factorisation in the calculation, thus greatly reduce amount of calculation；Jacobian matrix height is sparse, brings conveniently to storage and calculating；In each iteration, three-phase is full decoupled；Result of calculation ensure that zero injection-constraint strictly meets.

Description

A kind of three-phase load flow of active distribution network computational methods based on node voltage

Technical field

The invention belongs to Distribution Automation Technology field, be specifically related to a kind of three-phase load flow of active distribution network meter based on node voltage Calculation method.

Background technology

Distribution power system load flow calculation is calculation of distribution network and the basic tool in analysis, and it is according to network topology, root node voltage, feedback Specific electric load calculates each node voltage amplitude of network and phase angle, and provides full-mesh network loss.Before conventional Power Flow Calculation Methods For Distribution Network is Push back for method, Newton method etc..Owing to power distribution network closed loop design, open loop operation make the Load flow calculation of power distribution network have radial spy Point, forward-backward sweep method has that physical concept is distinct, simple and the easy feature such as realization, better astringency, is suitable for solving radial Distribution power flow.Newton method is better than forward-backward sweep method convergence, but owing to each iteration is required for re-forming Jacobian matrix, Computationally intensive, have impact on calculating speed.The many loops of active distribution network, multi-voltage grade, the feature of many PV node, to distribution The three-phase power flow of net brings difficulty.Forward-backward sweep method processes when processing loop with PV node the most difficult, it will big Big impact calculates speed and precision.Although Newton method can process many loops, many PV node, but amount of calculation is bigger.Therefore, Towards the rapidly and efficiently Three-phase Power Flow Calculation Method for Distribution System of the features such as multi-voltage grade, many PV node, many loops for active distribution The real time execution of net is the most necessary with control.

Summary of the invention

In order to overcome above-mentioned the deficiencies in the prior art, the present invention provides a kind of three-phase load flow of active distribution network meter based on node voltage Calculation method, it is possible to directly process many PV node and loop, and do not limited by loop number.

In order to realize foregoing invention purpose, the present invention adopts the following technical scheme that:

The present invention provides a kind of three-phase load flow of active distribution network computational methods based on node voltage, said method comprising the steps of:

Step 1: the three-phase voltage of each node in initialization system；

Step 2: according to the three-phase voltage of node each in system, determines that non-zero injects real part and the void of each phase injection current of node Portion；

Step 3: determine that non-zero injects each value of calculation injecting active power, reactive power and voltage magnitude mutually of node；

Step 4: determine that non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node；

Step 5: determine that non-zero injects the correction of a, b, c phase voltage of node；

Step 6: to non-zero inject node a, b, c phase magnitude of voltage be modified, obtain revised non-zero inject node a, The magnitude of voltage of b, c phase；

Step 7: determine the magnitude of voltage of zero injection node a, b, c phase；

Step 8: judge Δ W^a、ΔW^b、ΔW^cIn the absolute value of each element whether be respectively less than the threshold value set, if otherwise returning Step 2, if then terminating；

Wherein, Δ W^a、ΔW^b、ΔW^cIt is respectively non-zero and injects the amount of unbalance of node a, b, c phase.

In described step 2, set up system node voltage equation, have:

Y^abcU^abc=I^abc (1)

Wherein, Y^abcFor the three-phase bus admittance matrix under system rectangular coordinate, U^abcSend a telegram in reply for the node three-phase under rectangular coordinate Pressure column vector, I^abcSend a telegram in reply for the node three-phase under rectangular coordinate and flow column vector；

Obtain non-zero according to following formula and inject node three-phase telegram in reply stream column vector I under rectangular coordinate_N:

Y_NZU_Z+Y_NNU_N=I_N (2)

Wherein, Y_NZFor Y^abcIn inject node and zero corresponding to non-zero and inject the part of transadmittance, Y between node_NNFor Y^abc In corresponding to non-zero inject node part, U_ZIt is the three-phase complex voltage column vector of zero injection node, U_NNode is injected for non-zero Three-phase complex voltage column vector.

Formula (2) is obtained by following formula:

$\left[\begin{array}{cc}{Y}_{\mathrm{ZZ}}& Y_{\mathrm{ZN}}\\ Y_{\mathrm{NZ}}& Y_{\mathrm{NN}}\end{array}\right]\left[\begin{array}{c}{U}_Z\\ U_N\end{array}\right]=\left[\begin{array}{c}0\\ I_N\end{array}\right]---\left(3\right)$

The most available by formula (3):

Y_ZZU_Z+Y_ZNU_N=0 (4)

Wherein, Y_ZZFor Y^abcIn corresponding to zero injection node part, Y_ZNFor Y^abcIn note corresponding to zero injection node and non-zero The part of transadmittance between ingress.

In described step 3, non-zero injects each of node and injects the value of calculation of active power, reactive power and voltage magnitude mutually respectively WithWithRepresent, and have:

$P_i^k=U_{i,r}^k{I}_{i,r}^k-U_{i,x}^k{I}_{i,x}^k---\left(5\right)$

$Q_i^k=U_{i,r}^k{I}_{i,x}^k+U_{i,x}^k{I}_{i,r}^k---\left(6\right)$

$U_i^k=\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}---\left(7\right)$

Wherein,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase injection current of node i is injected for non-zero,The imaginary part of the k phase injection current of node i is injected for non-zero, and K=a, b, c, i=1,2 ..., n-1, during wherein n is system, non-zero injects node total number.

In described step 4, non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node and divides Do not useWithRepresent, specifically have:

$\mathrm{\Δ}{P}_i^k=P_i^k-P_{i,s}^k---\left(8\right)$

$\mathrm{\Δ}{Q}_i^k=Q_i^k-Q_{i,s}^k---\left(9\right)$

$\mathrm{\Δ}{U}_i^k=U_i^k-U_{i,s}^k---\left(10\right)$

Wherein,WithThe k phase being respectively non-zero injection node i injects the set-point of active power,Joint is injected for non-zero The k phase of some i injects the set-point of reactive power,The k phase voltage amplitude set-point of node i is injected for non-zero.

In described step 5, non-zero injects the update equation of node k phase amount of unbalance and is defined as:

ΔW^k=-H^kΔV^k (11)

Wherein, $\mathrm{\Δ}{W}^k={\left[\begin{array}{ccccccc}\mathrm{\Δ}{P}_1^k& \mathrm{\Δ}{Q}_1^k& \mathrm{\Δ}{P}_2^k& \mathrm{\Δ}{U}_2^k& ...& \mathrm{\Δ}{P}_{n-1}^k& \mathrm{\Δ}{Q}_{n-1}^k\end{array}\right]}^T,$ ΔW^kThe k of node is injected for non-zero Phase amount of unbalance；ΔV^kThe correction of the k phase voltage of node is injected for non-zero；H_kThe k phase Jacobi square of node is injected for non-zero Battle array；

Can obtain according to formula (5) and (6):

$\left\{\begin{array}{c}\frac{\∂P_i^k}{\∂U_{i,r}^k}=I_{i,r}^k\\ \frac{\∂P_i^k}{\∂U_{i,x}^k}=-I_{i,x}^k\end{array}\right.---\left(12\right)$

$\left\{\begin{array}{c}\frac{\∂Q_i^k}{\∂U_{i,r}^k}=I_{i,x}^k\\ \frac{\∂Q_i^k}{\∂U_{i,x}^k}=-I_{i,r}^k\end{array}\right.---\left(13\right)$

Can obtain according to formula (7):

$\left\{\begin{array}{c}\frac{\∂U_i^k}{\∂U_{i,r}^k}=\frac{U_{i,r}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\\ \frac{\∂U_i^k}{\∂U_{i,x}^x}=\frac{U_{i,x}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\end{array}\right.---\left(14\right)$

Can obtain according to formula (12), (13) and (14):

$H^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& {-I}_{1,x}^k& 0& 0& ...& 0& 0\\ I_{1,x}^k& I_{1,r}^k& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& -I_{2,x}^k& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& I_{n-1,r}^k& {-I}_{n-1,x}^k\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k\end{array}\right]---\left(15\right)$

To H_kCarrying out LU decomposition, formatting by professional etiquette and eliminating computing obtains matrix

$H_1^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& \frac{{-I}_{1,x}^k}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& \frac{-I_{2,x}^k}{I_{2,x}^k}& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_{2,x}^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& I_{n-1,r}^k& \frac{-I_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right.2---\left(16\right)$

FromIn obtain inferior triangular flap L_kWith unit upper triangular matrix U_k, wherein L_kByUnit below middle diagonal and diagonal Element composition, U_kDiagonal element be all 1, on it triangle element byMore than middle diagonal elementary composition, is expressed as:

$L^k\left[\begin{array}{ccccccc}{I}_{1,r}^k& 0& 0& 0& ...& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& 0& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{{\∂U}_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_2^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ 0& 0& 0& 0& ...& ...& ...\\ ...& ...& 0& 0& ...& I_{n-1,r}^k& 0\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right]---\left(17\right)$

$U^k=\left[\begin{array}{ccccccc}1& \frac{-I_{1,x}^k}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 1& 0& 0& ...& 0& 0\\ 0& 0& 1& \frac{{-I}_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ 0& 0& 0& 1& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& 1& \frac{{-I}_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& ...& 0& 1\end{array}\right]---\left(18\right)$

Decompose according to LU, H_kIt is expressed as:

H^k=L^kU^k (19)

Then, formula (11) is expressed as:

L^kU^kΔV^k=-Δ W^k (20)

By formula (20), utilize the method for solving of the former generation back substitution of system of linear equations, obtain

$\mathrm{\Δ}{V}^k={\left[\begin{array}{ccccccc}\mathrm{\Δ}{U}_{1,r}^k& \mathrm{\Δ}{U}_{1,x}^k& \mathrm{\Δ}{U}_{2,r}^k& \mathrm{\Δ}{U}_{2,x}^k& ...& \mathrm{\Δ}{U}_{n-1,r}^k& \mathrm{\Δ}{U}_{n-1,x}^k\end{array}\right]}^T,k=a,b,c.$

In described step 6, by following formula, the magnitude of voltage of non-zero power node k phase is modified, obtains revised non-zero power The magnitude of voltage V of rate node k phase^k:

$V^k=V_{\mathrm{old}}^k+\mathrm{\Δ}{V}^k---\left(21\right)$

Wherein,The magnitude of voltage of node k phase, k=a, b, c is injected for non-zero before revising.

In described step 7, if U_N=U_N,R+jU_N,X, U_Z=U_Z,R+jU_Z,X, Y_ZZ=G_ZZ+jB_ZZ, Y_ZN=G_ZN+jB_ZN；Wherein U_Z,RAnd U_Z,XFor U_ZReal part and imaginary part；(4) are converted to the reality under following rectangular coordinate Number equation:

$\left[\begin{array}{cc}{B}_{\mathrm{ZZ}}& G_{\mathrm{ZZ}}\\ G_{\mathrm{ZZ}}& -B_{\mathrm{ZZ}}\end{array}\right]\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]+\left[\begin{array}{cc}{B}_{\mathrm{ZN}}& G_{\mathrm{ZN}}\\ G_{\mathrm{ZN}}& {-B}_{\mathrm{ZN}}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]=0---\left(22\right)$

Then zero injects node voltage value V_ZIt is expressed as:

$V_Z=\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]=-{\left[\begin{array}{cc}{B}_{\mathrm{ZZ}}& G_{\mathrm{ZZ}}\\ G_{\mathrm{ZZ}}& -B_{\mathrm{ZZ}}\end{array}\right]}^{-1}\left[\begin{array}{cc}{B}_{\mathrm{ZN}}& G_{\mathrm{ZN}}\\ G_{\mathrm{ZN}}& {-B}_{\mathrm{ZN}}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]---\left(23\right)$

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

The present invention with the real part of node phase voltage with imaginary part as quantity of state, and select the real part of node every phase injection current with imaginary part to be Known quantity, sets up the relation of known quantity and unknown quantity by nodal voltage equation, is directly calculated factor table by formula.Its advantage For:

1), many PV node can directly be processed；

2), can directly process loop, not limited by loop number；

3), directly calculated the factor table of Jacobian matrix by formula, it is to avoid factorisation, reduce amount of calculation, improve Calculating speed；

4), Jacobian matrix height sparse, bring conveniently to storage and calculating；

5), in the iteration often walked, three-phase is full decoupled；

6), result of calculation ensure that zero injection-constraint can strictly meet.

Accompanying drawing explanation

Fig. 1 is three-phase load flow of active distribution network computational methods flow chart based on node voltage.

Detailed description of the invention

Below in conjunction with the accompanying drawings the present invention is described in further detail.

Such as Fig. 1, sky of the present invention provides a kind of three-phase load flow of active distribution network computational methods based on node voltage, described method bag Include following steps:

Step 1: the three-phase voltage of each node in initialization system；

Step 2: according to the three-phase voltage of node each in system, determines that non-zero injects real part and the void of each phase injection current of node Portion；

Step 3: determine that non-zero injects each value of calculation injecting active power, reactive power and voltage magnitude mutually of node；

Step 4: determine that non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node；

Step 5: determine that non-zero injects the correction of a, b, c phase voltage of node；

Step 6: to non-zero inject node a, b, c phase magnitude of voltage be modified, obtain revised non-zero inject node a, The magnitude of voltage of b, c phase；

Step 7: determine the magnitude of voltage of zero injection node a, b, c phase；

Step 8: judge Δ W^a、ΔW^b、ΔW^cIn the absolute value of each element whether be respectively less than the threshold value set, if otherwise returning Step 2, if then terminating；

Wherein, Δ W^a、ΔW^b、ΔW^cIt is respectively non-zero and injects the amount of unbalance of node a, b, c phase.

In described step 2, set up system node voltage equation, have:

Y^abcU^abc=I^abc (1)

Wherein, Y^abcFor the three-phase bus admittance matrix under system rectangular coordinate, U^abcSend a telegram in reply for the node three-phase under rectangular coordinate Pressure column vector, I^abcSend a telegram in reply for the node three-phase under rectangular coordinate and flow column vector；

Obtain non-zero according to following formula and inject node three-phase telegram in reply stream column vector I under rectangular coordinate_N:

Y_NZU_Z+Y_NNU_N=I_N (2)

Wherein, Y_NZFor Y^abcIn inject node and zero corresponding to non-zero and inject the part of transadmittance, Y between node_NNFor Y^abcIn The part of node, U is injected corresponding to non-zero_ZIt is the three-phase complex voltage column vector of zero injection node, U_NNode is injected for non-zero Three-phase complex voltage column vector.

Formula (2) is obtained by following formula:

$\left[\begin{array}{cc}{Y}_{\mathrm{ZZ}}& Y_{\mathrm{ZN}}\\ Y_{\mathrm{NZ}}& Y_{\mathrm{NN}}\end{array}\right]\left[\begin{array}{c}{U}_Z\\ U_N\end{array}\right]=\left[\begin{array}{c}0\\ I_N\end{array}\right]---\left(3\right)$

The most available by formula (3):

Y_ZZU_Z+Y_ZNU_N=0 (4)

Wherein, Y_ZZFor Y^abcIn corresponding to zero injection node part, Y_ZNFor Y^abcIn note corresponding to zero injection node and non-zero The part of transadmittance between ingress.

In described step 3, non-zero injects each of node and injects the value of calculation of active power, reactive power and voltage magnitude mutually respectively WithWithRepresent, and have:

$P_i^k=U_{i,r}^k{I}_{i,r}^k-U_{i,x}^k{I}_{i,x}^k---\left(5\right)$

$Q_i^k=U_{i,r}^k{I}_{i,x}^k+U_{i,x}^k{I}_{i,r}^k---\left(6\right)$

$U_i^k=\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}---\left(7\right)$

Wherein,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase injection current of node i is injected for non-zero,The imaginary part of the k phase injection current of node i is injected for non-zero, and K=a, b, c, i=1,2 ..., n-1, during wherein n is system, non-zero injects node total number.

In described step 4, non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node and divides Do not useWithRepresent, specifically have:

$\mathrm{\Δ}{P}_i^k=P_i^k-P_{i,s}^k---\left(8\right)$

$\mathrm{\Δ}{Q}_i^k=Q_i^k-Q_{i,s}^k---\left(9\right)$

$\mathrm{\Δ}{U}_i^k=U_i^k-U_{i,s}^k---\left(10\right)$

Wherein,WithThe k phase being respectively non-zero injection node i injects the set-point of active power,Joint is injected for non-zero The k phase of some i injects the set-point of reactive power,The k phase voltage amplitude set-point of node i is injected for non-zero.

In described step 5, non-zero injects the update equation of node k phase amount of unbalance and is defined as:

ΔW^k=-H^kΔV^k (11)

Wherein, $\mathrm{\Δ}{W}^k={\left[\begin{array}{ccccccc}\mathrm{\Δ}{P}_1^k& \mathrm{\Δ}{Q}_1^k& \mathrm{\Δ}{P}_2^k& \mathrm{\Δ}{U}_2^k& ...& \mathrm{\Δ}{P}_{n-1}^k& \mathrm{\Δ}{Q}_{n-1}^k\end{array}\right]}^T,$ ΔW^kThe k of node is injected for non-zero Phase amount of unbalance；ΔV^kThe correction of the k phase voltage of node is injected for non-zero；H_kThe k phase Jacobi square of node is injected for non-zero Battle array；

Can obtain according to formula (5) and (6):

$\left\{\begin{array}{c}\frac{\∂P_i^k}{\∂U_{i,r}^k}=I_{i,r}^k\\ \frac{\∂P_i^k}{\∂U_{i,x}^k}=-I_{i,x}^k\end{array}\right.---\left(12\right)$

$\left\{\begin{array}{c}\frac{\∂Q_i^k}{\∂U_{i,r}^k}=I_{i,x}^k\\ \frac{\∂Q_i^k}{\∂U_{i,x}^k}=-I_{i,r}^k\end{array}\right.---\left(13\right)$

Can obtain according to formula (7):

$\left\{\begin{array}{c}\frac{\∂U_i^k}{\∂U_{i,r}^k}=\frac{U_{i,r}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\\ \frac{\∂U_i^k}{\∂U_{i,x}^x}=\frac{U_{i,x}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\end{array}\right.---\left(14\right)$

Can obtain according to formula (12), (13) and (14):

$H^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& {-I}_{1,x}^k& 0& 0& ...& 0& 0\\ I_{1,x}^k& I_{1,r}^k& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& -I_{2,x}^k& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& I_{n-1,r}^k& {-I}_{n-1,x}^k\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k\end{array}\right]---\left(15\right)$

To H_kCarrying out LU decomposition, formatting by professional etiquette and eliminating computing obtains matrix

$H_1^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& \frac{{-I}_{1,x}^k}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& \frac{-I_{2,x}^k}{I_{2,x}^k}& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_{2,x}^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& I_{n-1,r}^k& \frac{-I_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right.2---\left(16\right)$

FromIn obtain inferior triangular flap L_kWith unit upper triangular matrix U_k, wherein L_kByUnit below middle diagonal and diagonal Element composition, U_kDiagonal element be all 1, on it triangle element byMore than middle diagonal elementary composition, is expressed as:

$L^k\left[\begin{array}{ccccccc}{I}_{1,r}^k& 0& 0& 0& ...& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 0& I_{2,r}^k& 0& ...& 0& 0\\ 0& 0& \frac{\∂U_2^k}{{\∂U}_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_2^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ 0& 0& 0& 0& ...& ...& ...\\ ...& ...& 0& 0& ...& I_{n-1,r}^k& 0\\ 0& 0& 0& 0& ...& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right]---\left(17\right)$

$U^k=\left[\begin{array}{ccccccc}1& \frac{-I_{1,x}^k}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 1& 0& 0& ...& 0& 0\\ 0& 0& 1& \frac{{-I}_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ 0& 0& 0& 1& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& 1& \frac{{-I}_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& ...& 0& 1\end{array}\right]---\left(18\right)$

Decompose according to LU, H_kIt is expressed as:

H^k=L^kU^k (19)

Then, formula (11) is expressed as:

L^kU^kΔV^k=-Δ W^k (20)

By formula (20), utilize the method for solving of the former generation back substitution of system of linear equations, obtain

$\mathrm{\Δ}{V}^k={\left[\begin{array}{ccccccc}\mathrm{\Δ}{U}_{1,r}^k& \mathrm{\Δ}{U}_{1,x}^k& \mathrm{\Δ}{U}_{2,r}^k& \mathrm{\Δ}{U}_{2,x}^k& ...& \mathrm{\Δ}{U}_{n-1,r}^k& \mathrm{\Δ}{U}_{n-1,x}^k\end{array}\right]}^T,k=a,b,c.$

In described step 6, by following formula, the magnitude of voltage of non-zero power node k phase is modified, obtains revised non-zero power The magnitude of voltage V of rate node k phase^k:

$V^k=V_{\mathrm{old}}^k+\mathrm{\Δ}{V}^k---\left(21\right)$

Wherein,The magnitude of voltage of node k phase, k=a, b, c is injected for non-zero before revising.

In described step 7, if U_N=U_N,R+jU_N,X, U_Z=U_Z,R+jU_Z,X, Y_ZZ=G_ZZ+jB_ZZ, Y_ZN=G_ZN+jB_ZN；Wherein U_Z,RAnd U_Z,XFor U_ZReal part and imaginary part；(4) are converted to the reality under following rectangular coordinate Number equation:

$\left[\begin{array}{cc}{B}_{\mathrm{ZZ}}& G_{\mathrm{ZZ}}\\ G_{\mathrm{ZZ}}& -B_{\mathrm{ZZ}}\end{array}\right]\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]+\left[\begin{array}{cc}{B}_{\mathrm{ZN}}& G_{\mathrm{ZN}}\\ G_{\mathrm{ZN}}& {-B}_{\mathrm{ZN}}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]=0---\left(22\right)$

Then zero injects node voltage value V_ZIt is expressed as:

$V_Z=\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]=-{\left[\begin{array}{cc}{B}_{\mathrm{ZZ}}& G_{\mathrm{ZZ}}\\ G_{\mathrm{ZZ}}& -B_{\mathrm{ZZ}}\end{array}\right]}^{-1}\left[\begin{array}{cc}{B}_{\mathrm{ZN}}& G_{\mathrm{ZN}}\\ G_{\mathrm{ZN}}& {-B}_{\mathrm{ZN}}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]---\left(23\right).$

Finally should be noted that: above example only in order to illustrate that technical scheme is not intended to limit, art Those of ordinary skill still the detailed description of the invention of the present invention can be modified or equivalent with reference to above-described embodiment, These are without departing from any amendment of spirit and scope of the invention or equivalent, the claim of the present invention all awaited the reply in application Within protection domain.

Claims (7)

1. three-phase load flow of active distribution network computational methods based on node voltage, it is characterised in that: described method includes following Step:

Step 1: the three-phase voltage of each node in initialization system；

Step 2: according to the three-phase voltage of node each in system, determines that non-zero injects real part and the void of each phase injection current of node Portion；

Step 3: determine that non-zero injects each value of calculation injecting active power, reactive power and voltage magnitude mutually of node；

Step 4: determine that non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node；

Step 5: determine that non-zero injects the correction of a, b, c phase voltage of node；

Step 6: to non-zero inject node a, b, c phase magnitude of voltage be modified, obtain revised non-zero inject node a, The magnitude of voltage of b, c phase；

Step 7: determine the magnitude of voltage of zero injection node a, b, c phase；

Step 8: judge Δ W^a、ΔW^b、ΔW^cIn the absolute value of each element whether be respectively less than the threshold value set, if otherwise returning Step 2, if then terminating；

Wherein, Δ W^a、ΔW^b、ΔW^cIt is respectively non-zero and injects the amount of unbalance of node a, b, c phase；

In described step 2, set up system node voltage equation, have:

Y^abcU^abc=I^abc (1)

Wherein, Y^abcFor the three-phase bus admittance matrix under system rectangular coordinate, U^abcSend a telegram in reply for the node three-phase under rectangular coordinate Pressure column vector, I^abcSend a telegram in reply for the node three-phase under rectangular coordinate and flow column vector；

Obtain non-zero according to following formula and inject node three-phase telegram in reply stream column vector I under rectangular coordinate_N:

Y_NZU_Z+Y_NNU_N=I_N (2)

Wherein, Y_NZFor Y^abcIn inject node and zero corresponding to non-zero and inject the part of transadmittance, Y between node_NNFor Y^abcIn The part of node, U is injected corresponding to non-zero_ZIt is the three-phase complex voltage column vector of zero injection node, U_NNode is injected for non-zero Three-phase complex voltage column vector.

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 1, it is characterised in that: Formula (2) is obtained by following formula:

$\left[\begin{array}{cc}{Y}_{ZZ}& Y_{ZN}\\ Y_{NZ}& Y_{NN}\end{array}\right]\left[\begin{array}{c}{U}_Z\\ U_N\end{array}\right]=\left[\begin{array}{c}0\\ I_N\end{array}\right]---\left(3\right)$

The most available by formula (3):

Y_ZZU_Z+Y_ZNU_N=0 (4)

Wherein, Y_ZZFor Y^abcIn corresponding to zero injection node part, Y_ZNFor Y^abcIn note corresponding to zero injection node and non-zero The part of transadmittance between ingress.

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 1, it is characterised in that: In described step 3, non-zero injects each value of calculation injecting active power, reactive power and voltage magnitude mutually of node and uses respectively WithRepresent, and have:

$P_i^k=U_{i,r}^k{I}_{i,r}^k-U_{i,x}^k{I}_{i,x}^k---\left(5\right)$

$Q_i^k=U_{i,r}^k{I}_{i,x}^k+U_{i,x}^k{I}_{i,r}^k---\left(6\right)$

$U_i^k=\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}---\left(7\right)$

Wherein,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase voltage of node i is injected for non-zero,The real part of the k phase injection current of node i is injected for non-zero,The imaginary part of the k phase injection current of node i is injected for non-zero, and K=a, b, c, i=1,2 ..., n-1, during wherein n is system, non-zero injects node total number.

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 3, it is characterised in that: In described step 4, non-zero injects each amount of unbalance injecting active power, reactive power and voltage magnitude mutually of node and uses respectivelyWithRepresent, specifically have:

${\mathrm{\ΔP}}_i^k=P_i^k-P_{i,s}^k---\left(8\right)$

${\mathrm{\ΔQ}}_i^k=Q_i^k-Q_{i,s}^k---\left(9\right)$

${\mathrm{\ΔU}}_i^k=U_i^k-U_{i,s}^k---\left(10\right)$

Wherein,WithThe k phase being respectively non-zero injection node i injects the set-point of active power,Joint is injected for non-zero The k phase of some i injects the set-point of reactive power,The k phase voltage amplitude set-point of node i is injected for non-zero.

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 4, it is characterised in that: In described step 5, non-zero injects the update equation of node k phase amount of unbalance and is defined as:

ΔW^k=-H^kΔV^k (11)

Wherein, ${\mathrm{\ΔW}}^k={\left[\begin{array}{ccccccc}{\mathrm{\ΔP}}_1^k& {\mathrm{\ΔQ}}_1^k& {\mathrm{\ΔP}}_2^k& {\mathrm{\ΔU}}_2^k& \mathrm{...}& {\mathrm{\ΔP}}_{n-1}^k& {\mathrm{\ΔQ}}_{n-1}^k\end{array}\right]}^T,$ ΔW^kThe k of node is injected for non-zero Phase amount of unbalance；ΔV^kThe correction of the k phase voltage of node is injected for non-zero；H_kThe k phase Jacobi square of node is injected for non-zero Battle array；

Can obtain according to formula (5) and (6):

$\left\{\begin{array}{c}\frac{\∂P_i^k}{\∂U_{i,r}^k}=I_{i,r}^k\\ \frac{\∂P_i^k}{\∂U_{i,x}^k}=-I_{i,x}^k\end{array}\right.---\left(12\right)$

$\left\{\begin{array}{c}\frac{\∂Q_i^k}{\∂U_{i,r}^k}=I_{i,x}^k\\ \frac{\∂Q_i^k}{\∂U_{i,x}^k}=I_{i,r}^k\end{array}\right.---\left(13\right)$

Can obtain according to formula (7):

$\left\{\begin{array}{c}\frac{\∂U_i^k}{\∂U_{i,r}^k}=\frac{U_{i,r}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\\ \frac{\∂U_i^k}{\∂U_{i,x}^k}=\frac{U_{i,x}^k}{\sqrt{{\left(U_{i,r}^k\right)}^2+{\left(U_{i,x}^k\right)}^2}}\end{array}\right.---\left(14\right)$

Can obtain according to formula (12), (13) and (14):

$H^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& -I_{1,x}^k& 0& 0& \mathrm{...}& 0& 0\\ I_{1,x}^k& I_{1,r}^k& 0& 0& \mathrm{...}& 0& 0\\ 0& 0& I_{2,r}^k& -I_{2,x}^k& \mathrm{...}& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}& \mathrm{...}& 0& 0\\ \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}\\ 0& 0& 0& 0& \mathrm{...}& I_{n-1,r}^k& -I_{n-1,x}^k\\ 0& 0& 0& 0& \mathrm{...}& I_{n-1,x}^k& I_{n-1,r}^k\end{array}\right]---\left(15\right)$

To H_kCarrying out LU decomposition, formatting by professional etiquette and eliminating computing obtains matrix

$H_1^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& \frac{-I_{1,x}^k}{I_{1,r}^k}& 0& 0& \mathrm{...}& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& \mathrm{...}& 0& 0\\ 0& 0& I_{2,r}^k& \frac{-I_{2,x}^k}{I_{2,r}^k}& \mathrm{...}& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_2^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& \mathrm{...}& 0& 0\\ \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}& \mathrm{...}\\ 0& 0& 0& 0& \mathrm{...}& I_{n-1,r}^k& \frac{-I_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& \mathrm{...}& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right]---\left(16\right)$

FromIn obtain inferior triangular flap L_kWith unit upper triangular matrix U_k, wherein L_kByUnit below middle diagonal and diagonal Element composition, U_kDiagonal element be all 1, on it triangle element byMore than middle diagonal elementary composition, is expressed as:

$L^k=\left[\begin{array}{ccccccc}{I}_{1,r}^k& 0& 0& 0& \mathrm{...}& 0& 0\\ I_{i,x}^k& I_{i,r}^k+\frac{{\left(I_{1,x}^k\right)}^2}{I_{1,r}^k}& 0& 0& \mathrm{...}& 0& 0\\ 0& 0& I_{2,r}^k& 0& \mathrm{...}& 0& 0\\ 0& 0& \frac{\∂U_2^k}{\∂U_{2,r}^k}& \frac{\∂U_2^k}{\∂U_{2,x}^k}+\frac{\∂U_2^k}{\∂U_{2,r}^k}\frac{I_{2,x}^k}{I_{2,r}^k}& \mathrm{...}& 0& 0\\ 0& 0& 0& 0& \mathrm{...}& \mathrm{...}& \mathrm{...}\\ \mathrm{...}& \mathrm{...}& 0& 0& \mathrm{...}& I_{n-1,r}^k& 0\\ 0& 0& 0& 0& \mathrm{...}& I_{n-1,x}^k& I_{n-1,r}^k+\frac{{\left(I_{n-1,x}^k\right)}^2}{I_{n-1,r}^k}\end{array}\right]---\left(17\right)$

$U^k=\left[\begin{array}{ccccccc}1& \frac{-I_{1,x}^k}{I_{1,r}^k}& 0& 0& ...& 0& 0\\ 0& 1& 0& 0& ...& 0& 0\\ 0& 0& 1& \frac{-I_{2,x}^k}{I_{2,r}^k}& ...& 0& 0\\ 0& 0& 0& 1& ...& 0& 0\\ ...& ...& ...& ...& ...& ...& ...\\ 0& 0& 0& 0& ...& 1& \frac{-I_{n-1,x}^k}{I_{n-1,r}^k}\\ 0& 0& 0& 0& ...& 0& 1\end{array}\right]---\left(18\right)$

Decompose according to LU, H_kIt is expressed as:

H^k=L^kU^k (19)

Then, formula (11) is expressed as:

L^kU^kΔV^k=-Δ W^k (20)

By formula (20), utilize the method for solving of the former generation back substitution of system of linear equations, obtain

${\mathrm{\ΔV}}^k={\left[\begin{array}{ccccccc}{\mathrm{\ΔU}}_{1,r}^k& {\mathrm{\ΔU}}_{1,x}^k& {\mathrm{\ΔU}}_{2,r}^k& {\mathrm{\ΔU}}_{2,x}^k& ...& {\mathrm{\ΔU}}_{n-1,r}^k& {\mathrm{\ΔU}}_{n-1,x}^k\end{array}\right]}^T,k=a,b,c.$

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 5, it is characterised in that: In described step 6, by following formula, the magnitude of voltage of non-zero power node k phase is modified, obtains revised non-zero power joint The magnitude of voltage V of some k phase^k:

$V^k=V_{old}^k+{\mathrm{\ΔV}}^k---\left(21\right)$

Wherein,The magnitude of voltage of node k phase, k=a, b, c is injected for non-zero before revising.

Three-phase load flow of active distribution network computational methods based on node voltage the most according to claim 6, it is characterised in that: In described step 7, if U_N=U_N,R+jU_N,X, U_Z=U_Z,R+jU_Z,X, Y_ZZ=G_ZZ+jB_ZZ, Y_ZN=G_ZN+jB_ZN； Wherein U_Z,RAnd U_Z,XFor U_ZReal part and imaginary part；(4) are converted to the real number equation under following rectangular coordinate:

$\left[\begin{array}{cc}{B}_{ZZ}& G_{ZZ}\\ G_{ZZ}& -B_{ZZ}\end{array}\right]\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]+\left[\begin{array}{cc}{B}_{ZN}& G_{ZN}\\ G_{ZN}& -B_{ZN}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]=0---\left(22\right)$

The then magnitude of voltage V of zero injection node_ZIt is expressed as:

$V_Z=\left[\begin{array}{c}{U}_{Z,R}\\ U_{Z,X}\end{array}\right]=-{\left[\begin{array}{cc}{B}_{ZZ}& G_{ZZ}\\ G_{ZZ}& -B_{ZZ}\end{array}\right]}^{-1}\left[\begin{array}{cc}{B}_{ZN}& G_{ZN}\\ G_{ZN}& -B_{ZN}\end{array}\right]\left[\begin{array}{c}{U}_{N,R}\\ U_{N,X}\end{array}\right]---\left(23\right).$