电力系统稳态分析课程设计

第五章 牛顿—拉夫逊直角坐标潮流计算

Matlab程序及运行结果

第一节 程序框图

- 23 -

第一节 Matlab程序

0 . 45 + j 0 . 15 0 . 4 + j 0 . 05 G 5 6 0 . 0 j + 2 0 . 0 1 0 . 08 + j 0 . 24 8 1 . 0 j + .0 6 0 2 0 . 01 + j 0 . 03 1 8 0 . + j 0 6 0 . 3 4 2 . 0 j + 8 0 . 0 4 0 . 04 + j 0 . 12 G - ( 0 . 2 + j 0 . 2 ) 0 . 6 + j 0 . 1

clear;clc

%重新编号，把原题中的节点1,2,3,4,5重新依次编号为5,1,2,3,4，其中1-4号为PQ节点，5号为平衡节点

y=0;

%输入原始数据，求节点导纳矩阵

y (1,2)=1/(0.06+0.18i); y (1,3)=1/(0.06+0.18i); y (1,4)=1/(0.04+0.12i); y(1,5)=1/(0.02+0.06i);

y(2,3)=1/(0.01+0.03i);y(2,5)=1/(0.08+0.24i); y(3,4)=1/(0.08+0.24i); y(4,5)=0; for i=1:5 for j=i:5 y(j,i)=y(i,j); end end Y=0; %求互导纳 for i=1:5 for j=1:5 if i~=j Y(i,j)=-y(i,j);

- 24 -

end end end %求自导纳 for i=1:5

Y(i,i)=sum(y(i,:)); end

Y %Y 为导纳矩阵 G=real(Y); B=imag(Y); %原始节点功率 S(1)=0.2+0.2i; S(2)=-0.45-0.15i; S(3)=-0.4-0.05i; S(4)=-0.6-0.1i; S(5)=0; P=real(S); Q=imag(S);

%赋初值

U=ones(1,5);U(5)=1.06; e=zeros(1,5);

ox=ones(8,1);fx=ones(8,1); count=0 %计算迭代次数 while max(fx)>1e-5 for i=1:4

for j=1:4

H(i,j)=0;N(i,j)=0;M(i,j)=0;L(i,j)=0;oP(i)=0;oQ(i)=0; end

end for i=1:4 for j=1:5

oP(i)=oP(i)-U(i)*U(j)*(G(i,j)*cos(e(i)-e(j))+B(i,j)*sin(e(i)-e(j))); oQ(i)=oQ(i)-U(i)*U(j)*(G(i,j)*sin(e(i)-e(j))-B(i,j)*cos(e(i)-e(j))); end

oP(i)=oP(i)+P(i); oQ(i)=oQ(i)+Q(i); end

fx=[oP,oQ]'; %求雅克比矩阵

%当i~=j时候求H,N,M,L 如下： for i=1:4 for j=1:4 if i~=j

H(i,j)=-U(i)*U(j)*(G(i,j)*sin(e(i)-e(j))-B(i,j)*cos(e(i)-e(j))); N(i,j)=-U(i)*U(j)*(G(i,j)*cos(e(i)-e(j))+B(i,j)*sin(e(i)-e(j))); L(i,j)=H(i,j);

- 25 -

M(i,j)=-N(i,j); end end end H,N,M,L

%当i=j 时H,N,M,L如下： for i=1:4 for j=1:5 if i~=j

H(i,i)=H(i,i)+U(i)*U(j)*(G(i,j)*sin(e(i)-e(j))-B(i, j)*cos (e(i)-e(j))); N(i,i)=N(i,i)-U(i)*U(j)*(G(i,

j)*cos(e(i)-e(j))+B(i,j)*sin(e(i)-e(j)));

M(i,i)=M(i,i)-U(i)*U(j)*(G(i,j)*cos(e(i)-e(j))+B(i,j)*sin(e(i)-e(j))); L(i,i)=L(i,i)-U(i)*U(j)*(G(i,j)*sin(e(i)-e(j))-B(i,j)*cos(e(i)-e(j))); end end

N(i,i)=N(i,i)-2*(U(i))^2*G(i,i); L(i,i)=L(i,i)+2*(U(i))^2*B(i,i); end

J=[H,N;M,L] %J 为雅克比矩阵 ox=-((inv(J))*fx); for i=1:4

oe(i)=ox(i); oU(i)=ox(i+4)*U(i); end for i=1:4

e(i)=e(i)+oe(i); U(i)=U(i)+oU(i); end

count=count+1; end

ox,U,e,count %求节点注入的净功率 i=5; for j=1:5

P(i)=U(i)*U(j)*(G(i,j)*cos(e(i)-e(j))+B(i,j)*sin(e(i)-e(j)))+P(i); Q(i)=U(i)*U(j)*(G(i,j)*sin(e(i)-e(j))-B(i,j)*cos(e(i)-e(j)))+Q(i); end

S(5)=P(5)+Q(5)*sqrt(-1); S

%求节点注入电流

I=Y*U'

- 26 -