求帮忙分析电力系统潮流计算的MATLAB程序

求帮忙分析电力系统潮流计算的MATLAB程序

% 配电网前推回代潮流计算程序 % 使用IEEE 33节点配电系统作为算例，可实现弱环网情况下的潮流计算countnum=0;BranchData = [ 1 2 0.0922 0.0470; 2 3 0.4930 0.2511; 3 4 0.3660 0.1864; 4 5 0.3811 0.1941; 5 6 0.8190 0.7070; 6 7 0.1872 0.6188; 7 8 0.7114 0.2351; 8 9 1.0300 0.7400; 9 10 1.0440 0.7400; 10 11 0.1966 0.0650; 11 12 0.3744 0.1238; 12 13 1.4680 1.1550; 13 14 0.5416 0.7129; 14 15 0.5910 0.5260; 15 16 0.7463 0.5450; 16 17 1.2890 1.7210; 17 18 0.7320 0.5740; 2 19 0.1640 0.1565; 19 20 1.5042 1.3554; 20 21 0.4095 0.4784; 21 22 0.7089 0.9373; 3 23 0.4512 0.3083; 23 24 0.8980 0.7091; 24 25 0.8960 0.7011; 6 26 0.2030 0.1034; 26 27 0.2842 0.1447; 27 28 1.0590 0.9337; 28 29 0.8042 0.7006; 29 30 0.5075 0.2585; 30 31 0.9744 0.9630; 31 32 0.3105 0.3619; 32 33 0.3410 0.5302;]; % 支路，阻抗NodeData = [ 2 100.00 60.00; 3 90.00 40.00; 4 120.00 80.00; 5 60.00 30.00; 6 60.00 20.00; 7 200.00 100.00; 8 200.00 100.00; 9 60.00 20.00; 10 60.00 20.00; 11 45.00 30.00; 12 60.00 35.00; 13 60.00 35.00; 14 120.00 80.00; 15 60.00 10.00; 16 60.00 20.00; 17 60.00 20.00; 18 90.00 40.00; 19 90.00 40.00; 20 90.00 40.00; 21 90.00 40.00; 22 90.00 40.00; 23 90.00 50.00; 24 420.00 200.00; 25 420.00 200.00; 26 60.00 25.00; 27 60.00 25.00; 28 60.00 20.00; 29 120.00 70.00; 30 200.00 600.00; 31 150.00 70.00; 32 210.00 100.00; 33 60.00 40.00;]; % 节点，负荷UB = 12.66; % 电压基准 kVSB = 10; % 功率基准 MVAZB = UB^2/SB; % 阻抗基准 ohmBranchData(:,[3,4]) = BranchData(:,[3,4]) / ZB; % 阻抗标幺化NodeData(:,[2,3]) = NodeData(:,[2,3]) / SB / 1000;% 功率标幺化NN = 33; % 节点数A0 = zeros(NN);for n = 1:NN-1 A0(BranchData(n,1),BranchData(n,2)) = 1;end % 形成 A0AssociatedMatrix=0;for n=2:NN-1 AssociatedMatrix(n,n)=1; temp=BranchData(n-1,1); AssociatedMatrix(n,1:n-1)=AssociatedMatrix(temp,1:n-1);endA0T = A0'; % 形成 A0 的转置S = [0;-NodeData(:,2) - i*NodeData(:,3)]; % 形成 SZL = [0;BranchData(:,3) + i*BranchData(:,4)]; % 形成 ZLV = ones(NN,1);V(1) = 1; % 各个节点电压赋初值IL(NN,1) = -conj(S(NN) / V(NN)); % 最末支路电流赋初值Delta = 1; % 收敛判据赋初值TempV = V; % 赋初值，用于记忆上次迭代结果while Delta > 1e-8 countnum=countnum+1; IN = conj(S ./ V); % 节点注入电流 for n = 1:NN-1 IL(NN-n) = A0(NN-n,NN-n+1:end) * IL(NN-n+1:end) - IN(NN-n); end % 电流回代过程 for n = 2:NN V(n) = A0T(n,1:n-1) * V(1:n-1) - ZL(n) * IL(n); end % 电压前推过程 Delta = max(abs(V-TempV)); % 更新收敛判据 TempV = V; % 记忆迭代结果endVangle(:,1)=abs(V);Vangle(:,2)=angle(V)/3.1415*180;for i=1:NN-1 st=BranchData(i,1); en=BranchData(i,2); Sij(i,1)=V(st)*conj((V(st)-V(en))/ZL(i+1)); Sji(i,1)=V(en)*conj((V(en)-V(st))/ZL(i+1));end 希望你能用到

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