前推回代法潮流计算IEEE33节点算例-凡亿课堂

1 前推回代法原理对于如图所示的辐射状配电馈线,k,i分别为父、子节点, i,j分别为父,子节点,Ci为由节点i的子节点构成的节点集.

配网潮流前推回代潮流算法第n 1步的迭代公式如下:

节点i的前推计算公式为

2 算例

IEEE33节点结构

3 程序运行结果

4 matlab程序

clcclearclose all%% 前推回代法

%% 系统负荷参数节点有功无功Bus=[1,0,0 ;2,100, 60;3,90,40;4,120,80;5,60,30;6,60,20;7, 200, 100 ;8,200, 100 ;9, 60, 20;10,60, 20;11,45,30;12,60, 35;13,60, 35;14,120,80;15,60, 10;16,60, 20;17,60, 20;18,90, 40;19,90, 40;20,90, 40;21,90, 40;22,90, 40; 23,90, 40; 24,420, 200;25,420, 200 ;26,60, 25;27,60 ,25;28,60, 20;29,120, 70;30,200,600 ;31,150, 70;32,210, 100 ;33,-1060,40;];%% 系统支路参数节点节点电阻电抗Branch=[1 ,1 ,2 ,0.0922,0.0407;2 ,2 ,3 ,0.4930,0.2511;3 ,3 ,4 ,0.3660,0.1864;4 ,4 ,5 ,0.3811,0.1941;5 ,5 ,6 ,0.8190,0.7070;6 ,6 ,7 ,0.1872,0.6188;7 ,7 ,8 ,0.7144,0.2351;8 ,8 ,9 ,1.0300,0.7400;9 ,9 ,10,1.0440,0.7400;10,10,11,0.1966,0.065;11,11,12,0.3744,0.1238;12,12,13,1.4680,1.1550;13,13,14,0.5416,0.7129;14,14,15,0.5910,0.5260;15,15,16,0.7463,0.5450;16,16,17,1.2890,1.7210;17,17,18,0.7320,0.5740;18,2, 19,0.1640,0.1565;19,19,20,1.5042,1.3554;20,20,21,0.4095,0.4784;21,21,22,0.7089,0.9373;22,3, 23,0.4512,0.3083;23,23,24,0.8980,0.7091;24,24,25,0.8960,0.7011;25,6, 26,0.2030,0.1034;26,26,27,0.2842,0.1447;27,27,28,1.0590,0.9337;28,28,29,0.8042,0.7006;29,29,30,0.5075,0.2585;30,30,31,0.9744,0.9630;31,31,32,0.3105,0.3619;32,32,33,0.3410,0.5302;];%% 数据初始化[busnum,row]=size(Bus);[branchnum,row]=size(Branch);soubus=Branch(:,2);mobus=Branch(:,3);Vbus=ones(busnum,1);Vbus(:,1)=12.66;Vbus1=Vbus;Ploss=zeros(busnum,1);Qloss=zeros(busnum,1);e=1;k=0;Branch1=Branch;n=1;%% 前推回代循环while ~isempty(Branch1)m=1;[s,row]=size(Branch1);while s>0t=find(Branch1(:,2)== Branch1(s,3));if isempty(t)T1(n,:)= Branch1(s,:);n=n 1;elseT2(m,:)= Branch1(s,:);m=m 1;end;s=s-1;end;Branch1=T2;T2=[];end;while e>1.0e-05P=zeros(busnum,1);Q=zeros(busnum,1);for s=1:branchnumi=T1(s,2);j=T1(s,3);R=T1(s,4);X=T1(s,5);Pload=Bus(j,2);Qload=Bus(j,3);II=((Pload P(j))^2 (Qload Q(j))^2)/(Vbus(j)^2*1000);Ploss(i,j)=II*R;Qloss(i,j)=II*X;P(i,j)=Pload Ploss(i,j) P(j);Q(i,j)=Qload Ploss(i,j) Q(j);P(i)=P(i) P(i,j);Q(i)=Q(i) Q(i,j);end;for s=branchnum:-1:1i=T1(s,2);j=T1(s,3);R=T1(s,4);X=T1(s,5);Vbus(j)=(Vbus(i)-(P(i,j)*R Q(i,j)*X)/(Vbus(i)*1000))^2 ((P(i,j)*X-Q(i,j)*R)/(Vbus(i)*1000))^2;Vbus(j)=sqrt(Vbus(j));end;e=max(abs(Vbus1-Vbus));Vbus1=Vbus;k=k 1;end;%% 输出结果numc=[1:1:33];figure;plot(numc,Vbus1);xlabel('节点')ylabel('电压幅值/kv')

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