資源簡介
一個簡單的最優潮流實現程序。
代碼片段和文件信息
function?[?]?=?optflow(??)
%UNtitleD?Summary?of?this?function?goes?here
%???Detailed?explanation?goes?here
y=[2.1569-8.6274i??????????0?????????-0.5882+2.3529i????-0.3922+1.5686i???-1.1765+4.7059i;
?????????0??????????2.3530-9.4118i???-1.1765+4.7059i???????????0??????????-1.1765+4.7059i;
??-0.5882+2.3529i???-1.1765+4.7059i???2.3529-9.4117i?????-0.5882+2.3529i?????????0;
??-0.3922+1.5686i??????????0?????????-0.5882+2.3529i?????0.9804-3.9215i??????????0;
??-1.1765+4.7059i???-1.1765+4.7059i?????????0??????????????????0???????????2.3530-9.4118i]??;%導納矩陣
g=real(y);
b=imag(y);
u(:1)=[1;1;1;1;1];????????%各節點初始電壓值
ct(:1)=[0;0;0;0;0];???????%初始角度
l=ones(114);
uu=ones(114);
z=ones(114);
w=-0.5*ones(114);
y=ones(110);
y(1:2:9)=(1e-10)*y(1:2:9);
y(2:2:10)=(-1e-10)*y(2:2:10);
p0=[1;0;-0.6;-0.4;-0.6];???%各節點的輸入有功功率
q0=[0.5;0;-0.3;-0.1;-0.2];?%各節點的輸入無功功率
hPG=zeros(210);
hPG(14*2-1)=-1;
hPG(25*2-1)=-1;%發電機1在4節點上,發電機2在5節點上
hQR(14*2)=-1;
hQR(25*2)=-1;
hx=zeros(1010);
hxx=zeros(1410);
gxx=zeros(1414);
k=1;
?for?i=1:5
????for?j=1:5
????????if?i~=j
????????????hh(ij)=-u(ik)*u(jk)*(g(ij)*sin(ct(ik)-ct(jk))-b(ij)*cos(ct(ik)-ct(jk)));
????????????jj(ij)=-u(ik)*u(jk)*(g(ij)*cos(ct(ik)-ct(jk))+b(ij)*sin(ct(ik)-ct(jk)));
????????????nn(ij)=-u(ik)*(g(ij)*cos(ct(ik)-ct(jk))+b(ij)*sin(ct(ik)-ct(jk)));
??
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