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1、6电力网潮流电压计算例题与MATLAB程序 编写 佘名寰 本文介绍了两个电力网潮流计算例题。一例为5个节点5条支路；另一例有6个节点7条支路，有PQ负荷节点也有PV发电机节点，变压器支路标么变比不为1。本文给出了完整的计算程序和计算结果。程序包括计算网络节点导纳矩阵和潮流电压两部分，程序中引用的M函数见参考文献1。例题选自研究生教材，比较典型，计算结果与教材核对基本一致。本文可供电力专业师生和基层技术人员潮流计算时参考。【例1.3】如图 1-4所示，五个节点三条母线简单系统，系统参数均用标么值（pu.）表示。节点4、5通过变压器与节点2、3相连，变比为1.05:1，节点1、2、3为PQ负荷节点
2、，节点4为PV发电机节点，节点5为参考节点，节点4、5电压均为1.05，图中箭头表示各支路的正方向，各个节点功率如表2.1所示,计算各节点电压、发电机无功功率和系统潮流分布：4Z23Z13Z21YC2YC3YC1Z42Z531:1.051.05:142315S3S2S1Z21=0.04+J0.25Z23=0.08+J0.30Z13=0.1+J0.35Z42=J0.015Z53=J0.03YC1=J0.25YC2=J0.50YC3=J0.25S1=1.6+J0.8S2=2+J1S3=3.7+J1.3P4=5.0V4=1.05V5=1.05P4=5.0V4=1.05vV5=1.05 图1-4 例1
3、.3网络接线图表 1-1母线有功负载无功负载发电机有功发电机无功1PD1=1.6QD1=0.82PD2=2.0QD2=1.03PD3=3.7QD3=1.345PG4=5.0PG5=？QG4=未定QG5=？【例1.3】 程序说明： 本例题与参考文献2的【例2.3.1】不同之处在于有两个变比不为1的变压器支路，并且引入了PV发电机节点。由于网络结构与例图1-1完全相同，网络节点导纳矩阵已在文献1之1.3节求出不再赘述。潮流计算程序参考【例2.3.1】节点电压符号矩阵：一共五个节点，独立节点数N1=4，PV节点数N2=1 节点电压幅值符号矩阵为 u=sym(u1,u2,u3,u4,u5); 节点电压
4、相角符号矩阵为 delt=sym(d1,d2,d3,d4,d5);节点功率给定值为：三个PQ节点，一个PV 节点p=-1.6,-2,-3.7,5;q=-0.8,-1.0,-1.3;【例1.3】 源程序 NU114N.m%* NU114N.m , example 2-3-2, fig 2-2 * % %The following Program for load flow calculation is based on MATLAB2007clear% bus 1,2,3 is PQ bus,bus 4 is PV bus,bus 5 is slack busglobal Np Nb bt p
5、m k% Np is number of node point,Nb is number of braches,% p the transformer turns ratio,with off-nominal tap-setting,fig 2-4 ,PSCLF p6, % bt is two-dimensional array ,line one is point 1,line two is point 2,% m is number of node point, k is number of transformers Np=5;Nb=5; bt=2,3;4,5;p=1/1.05,1/1.0
6、5;m=5;k=2;% nstart-the start point of branches ,nend - the end point,% mm - network incidence matrix nstart=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 zb1r=0.04,0.08,0.1,0.0,0.0; zb1i=0.25,0.30,0.3
7、5,0.015,0.03; zb1=zb1r+zb1i*j; yb1=zb1.(-1); yb=diag(conj(yb1); y=mm*yb*(mm); % yb0 ,the shunt admittances of transmission line yb0i=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=y+yn; % yn0,the node-admittance of shunt capacitor %yn0i=; %yn0=0+
8、yn0i*j; %ync=diag(conj(yn0); %y=y+ync; Y,YY=fdt1(y,yy); G=real(Y);B=imag(Y);% YY is the shund admittances of the line and transformers at each end,% Y is node-admittance matrix for network%*u=sym(u1,u2,u3,u4,u5);delt=sym(d1,d2,d3,d4,d5);% u - node voltage magnitude,delt - anglep=-1.6,-2,-3.7,5;q=-0.
9、8,-1.0,-1.3;k=0;precision=1;N1=4;%the N1 is the amount of the PQ and PV busN2=1;%the N2 is the amount of the PV bus for m=1:N1 for n=1:N1+1 pt(n)=u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n)+B(m,n)*sin(delt(m)-delt(n); end pp(m)=p(m)-sum(pt); end for m=1:N1-N2 for n=1:N1+1 qt(n)=u(m)*u(n)*(G(m,n)*sin(delt(
10、m)-delt(n)-B(m,n)*cos(delt(m)-delt(n); end qq(m)=q(m)-sum(qt); end J1=jacobian(pp,d1,d2,d3,d4,u1,u2,u3); J2=jacobian(qq,d1,d2,d3,d4,u1,u2,u3); J=vertcat(J1,J2); uu=1.0,1.0,1.0,1.05,1.05;dd=0,0,0,0,0; while precision0.00001 u1=uu(1);u2=uu(2);u3=uu(3);u4=uu(4);u5=uu(5); d1=dd(1);d2=dd(2);d3=dd(3);d4=d
11、d(4);d5=dd(5); for m=1:N1 PP(m)=eval(pp(m); end for m=1:N1-N2 PP(N1+m)=eval(qq(m); end JJ=eval(J); du=-inv(JJ)*PP;precision=max(abs(du); for n=1:N1 dd(n)=dd(n)+du(n); endfor n=1:N1-N2 uu(n)=uu(n)+du(N1+n); end k=k+1;endk-1,dd*180/pi,uu%*% the following program is used to calculate the Sm and Smnfor
12、n=1:N1+1 U(n)=uu(n)*(cos(dd(n)+j*sin(dd(n);end Um=conj(U); I=Y*Um;Sm=diag(Um)*conj(I)for m=1:N1+1 for n=1:N1+1 Smn(m,n)=U(m)*(conj(U(m)-conj(U(n)*conj(-Y(m,n)+U(m)*conj(U(m)*YY(m,n); endendSmn【例1.3】 潮流计算结果：迭代次数ans = 4节点1至5电压相角（度）ans = -4.7785 17.8535 -4.2819 21.8433 0节点1至5电压幅值（pu）uu = 0.8622 1.0779
13、1.0364 1.0500 1.0500节点1至5功率（pu）Sm = -1.6000 - 0.8000i -2.0000 - 1.0000i -3.7000 - 1.3000i 5.0000 + 1.8131i 2.5794 + 2.2994i支路功率（行列编号均为节点1至5）Smn = 0 -1.4662 - 0.4091i -0.1338 - 0.3909i 0 0 1.5845 + 0.6726i 0 1.4155 - 0.2443i -5.0000 - 1.4282i 0 0.1568 + 0.4713i -1.2774 + 0.2032i 0 0 -2.5794 - 1.9745i 0 5.0000 + 1.8131i 0 0 0 0 0
