电力网节点导纳矩阵计算例题与程序

1、电力网节点导纳矩阵计算例题与程序余名寰 编与用计算机解算电力网潮流电压和短路电流问题首先需确定电力网的节点导纳矩阵或节点阻抗矩阵。本文通过例题介绍用网络拓扑法计算节点导纳矩阵的方法和程序，程序考虑了线路并联电容和变压器支路标么变比不为1时的影响。程序用 MATLAB语言编写，线路参数均采用标么值。本文稿用office word 2007版编写，可供电气专业人员计算相关问题时参考。1 .用网络拓扑计算节点导纳矩阵1.1 网络拓扑矩阵：【例1.1】例图1-1是有5个节点和5条支路的网络，节点5作为基准参考点，1 ,2, 3, 4 为独立节点，支路编号和方向图中已标识。Z21=0.04+J0.25Z

2、23=0.08+J0.30Z13=0.1+J0.35Z42=J0.015Z53=J0.03对于具有n个节点b条支路的有向图，它的关联矩阵为一个NX B的矩阵A:A=a ij 若支路j与节点i相关，且箭头背离节点i,则aj =1,若箭头指向节点则aj =-1 ,若 支路j与节点i无关，则aij =0,图1-1所示的有向图的关联矩阵为支路编号-?-?Aij= ? -? -? -?.? ?行编号从上到下为1 2 3 4 5 节点编号(5 为参考节点)去掉第 5 行即为独立节点的关联矩阵。以下介绍生成网络关联矩阵的M 函数文件ffm.m ：% M FUNCTION ffm.m% Np is numbe

3、r of node point,Nb is number of braches% nstart-the start point of branches ,nend - the end point,% A - network incidence matrixfunction A=ffm(nstart,nend)global Np Nbn=length(nstart);A=zeros(Np,Nb);for i=1:nA(nstart(i),i)=1;A(nend(i),i)=-1;end以例图 1-1 网络为例调用ffm.m 文件求其关联矩阵运算以上程序可得关联矩阵mmij 如下：mm =-101

4、001 10-100-1-10-10001000001Mmij 明显与Aij 是相同的。1.2 生成节点导纳矩阵程序： 由网络原始矩阵计算节点导纳矩阵公式Y=AYs0At(1-1)Y 节点导纳矩阵A网络关联矩阵AtA 的转置矩阵YS0 网络原始导纳矩阵若网络各支总阻抗为Zb=Zb1 ,Zb2, ,Zbnzb100000zb2000则ZS0= 000000.000000zbn Ys0=Zs0-1(1-2)Y=A Zs0-1At(1-3) 节点导纳计算程序以例 1-1 网络为例，在不计对地电容和变压器变比假定为1 条件下， 节点导纳矩阵计算程序如下：clearglobal Np Nb% Np is

5、 number of node point,Nb is number of braches, Np=5;Nb=5;% nstart-the start point of branches ,nend - the end point,% mm - network incidence matrixnstart=2,2,1,4,5;nend=1,3,3,2,3;mm=ffm(nstart,nend);% zb1,the series impedances of transmission line% yb1,the series admittances of transmission line zb1

6、r=0.04,0.08,0.1,0.0,0.0;zb1i=0.25,0.30,0.35,0.015,0.03;zb1=zb1r+zb1i*j;yb1=zb1.A(-1);yb=diag(conj(yb1');y=mm*yb*(mm)'运算程序后可得节点导纳矩阵y=1.3787 - 6.5417i-0.6240 + 3.9002i-0.7547 + 2.6415i00-0.6240 + 3.9002i1.4539 -73.6789i-0.8299 + 3.1120i0 +66.6667i0-0.7547 +2.6415i-0.8299+ 3.1120i1.5846 -39.086

7、9i00 +33.3333i00 +66.6667i00 -66.6667i0000 +33.3333i00 -33.3333i 线路对地并联电容的处理例图1-1网络中，线路，两端对地电容导纳均为j0.25,节点1,2,3自导纳相应增加j0.25,j0.5,j0.25 。程序处理如下：% yb0 ,the shunt admittances of transmission lineyb0i=0.25,0.25,0.0,0.0,0.0;yb0=0+yb0i*j;y0=diag(conj(yb0');yg=mm*y0*(mm)'yn=diag(diag(yg);yy=yn-yg;y

8、=y+yn;程序中 yb0 为各线路对地电容导纳，y0 为原始导纳矩阵，yn 对角矩阵对应各节点自导纳增加值，yy 相应各支路两端对地并联导纳，用于计算支路电流。y 为计及对地电容后的节点导纳矩阵。程序运算结果yn =0 + 0.2500i000000 + 0.5000i000000 + 0.2500i000000000000yy =00 + 0.2500i0000 + 0.2500i00 + 0.2500i0000 + 0.2500i0000000000000y =1.3787 - 6.2917i-0.6240 + 3.9002i-0.7547 + 2.6415i-0.6240 + 3.90

9、02i1.4539 -73.1789i-0.8299 + 3.1120i-0.7547 + 2.6415i-0.8299 + 3.1120i1.5846 -38.8369i00 +66.6667i0000 +33.3333i00 +66.6667i00 -66.6667i0000 +33.3333i00 -33.3333i 非标准变压器支路的处理1-2所示。图中当支路有变压器并且标么变比不为1时，可将变压器化作n形等值电路，如图Zt与节点j相连。理想变压器变比为 k:1,接入节点i,高压侧经漏抗图1-2变压器等值电路等值电路图中 Yt=1/Zt。i侧自导变为 yii=kyt+k(k-1)yt=

10、k2yt,改变量为A yii=(k2-i)yt；I j间互导变为yj=-kyt, A yj =-ky t+yt=-(k-1)y t。j 侧自导变为 yjj=kyt+(1-k)yt=yt,与 K=1 时相同； 变压器变比不为1时修正导纳矩阵的M函数文件fdtl.m% M FUNCTION fdt1.m% the transformer turns ratio,with off-nominal tap-setting,fig 1-2% bt口 two-dimensional array ,line one is point i,line two is point j% p is the trans

11、former turns ratio,k is number of transformersfunction A,B=fdt1(a,b)global bt p m k for i=1:m for j=1:m A(i,j)=a(i,j);B(i,j)=b(i,j); end end for i1=1:k k1=bt(1,i1);k2=bt(2,i1);t=-a(k1,k2);p1=p(i1);for i=1:mfor j=1:mif i=k1if j=k1A(i,j)=A(i,j)+(p1*p1-1)*t;elseif j=k2A(i,j)=A(i,j)-(p1-1)*t;B(i,j)=p1*(

12、p1-1)*t;endelseif i=k2if j=k1A(i,j)=A(i,j)-(p1-1)*t;B(i,j)=(1-p1)*t;endendendendend程序中bt是2彳f K列数组，第一行为变压器接入点i的节点编号，第二行为经漏抗后所接j节点编号 ,k 为变压器支路数。P 为一维k 列数组，输入各个变压器的变比；m 为矩阵 a,b,A,B 的阶数a, A是进行变比修正前后的节点导纳矩阵，b， B 是变比修正前后支路对地并联导纳矩阵。1.3 例图 1-1 节点导纳和支路对地并联导纳计算程序例图 1-1 共有 5 个节点，5 条支路，所以Np=5， Nb=5支路，为变压器支路，其 i

13、节点编号为2, j节点编号为4,变比为1: 1/1.05;支路也为变压器支路，其i节点编号为3, j节点编号为5,变比为1: 1/1.05;所以数组bt=2,3;4,5 ， p=1/1.05,1/1.05 ， k=2;5 条支路按箭头方向首末端节点分别为2,2,1,4,5 和 1,3,3,2,3，所以 nstart=2,2,1,4,5;*nend=1,3,3,2,3; 输入以上数据可得以下程序：%* NU11.m , example 1-1, fig 1-1 % The following Program for node impedances calculation is based on

14、MATLAB2007 clear % bus 1,2,3 is PQ bus,bus 4 is PV bus,bus 5 is slack busglobal Np Nb bt p m k% Np is number of node point,Nb is number of braches,% p the transformer turns ratio,with off-nominal tap-setting,fig 1-1% bt is two-dimensional array ,line one is point i,line two is point j,% m is number

15、of node point, k is number of transformersNp=5;Nb=5;bt=2,3;4,5;p=1/1.05,1/1.05;m=5;k=2;% nstart-the start point of branches ,nend - the end point,% mm - network incidence matrixnstart=2,2,1,4,5;nend=1,3,3,2,3;mm=ffm(nstart,nend);% zb1,the series impedances of transmission line% yb1,the series admitt

16、ances of transmission linezb1r=0.04,0.08,0.1,0.0,0.0;zb1i=0.25,0.30,0.35,0.015,0.03;zb1=zb1r+zb1i*j;yb1=zb1.A(-1);yb=diag(conj(yb1');y=mm*yb*(mm)'% yb0 ,the shunt admittances of transmission lineyb0i=0.25,0.25,0.0,0.0,0.0;yb0=0+yb0i*j;y0=diag(conj(yb0');yg=mm*y0*(mm)'yn=diag(diag(yg)

17、;yy=yn-yg;y=y+yn;Y,YY=fdt1(y,yy);% YY is the shund admittances of the line and transformers at each end,% Y is node-admittance matrix for network运行结果为：Y =1.3787 - 6.2917i-0.6240 + 3.9002i-0.7547 + 2.6415i00-0.6240 + 3.9002i1.4539 -66.9808i-0.8299 + 3.1120i0 +63.4921i0-0.7547 +2.6415i-0.8299 + 3.1120i1.5846 -35.7379i00 +31.7460i00 +63.4921i00 -66.6667i0