資源簡介
matlab中5節點潮流計算,就是華中科技大學何仰贊版電力系統分析的例題
代碼片段和文件信息
X=[00011]‘;
F=zeros(51);
J=zeros(55);
Y=[1.0421?-?8.2429i??-0.5882?+?2.3529i????????0?+?3.6667i??-0.4539?+?1.8911i;
??-0.5882?+?2.3529i???1.0690?-?4.7274i????????0????????????-0.4808?+?2.4038i;
????????0?+?3.6667i????????0??????????????????0?-?3.3333i????????0??????????;
??-0.4539?+?1.8911i??-0.4808?+?2.4038i????????0?????????????0.9346?-?4.2616i;];
?
??for?k=1:10
????d1=X(1);d2=X(2);d3=X(3);
????V1=X(4);V2=X(5);
????
????F(1)=-0.30-V1*(V1*(real(Y(11))*cos(0)+imag(Y(11))*sin(0))+V2*(real(Y(12))*cos(d1-d2)+imag(Y(12))*sin(d1-d2))+1.1*real(Y(13))*cos(d1-d3)+1.1*imag(Y(13))*sin(d1-d3)+1.05*real(Y(14))*cos(d1)+1.05*imag(Y(14))*sin(d1));
????F(2)=-0.55-V2*(V1*(real(Y(12))*cos(d2-d1)+imag(Y(12))*sin(d2-d1))+V2*(real(Y(22))*cos(0)+imag(Y(22))*sin(0))+1.1*real(Y(23))*cos(d2-d3)+1.1*imag(Y(23))*sin(d2-d3)+1.05*real(Y(24))*cos(d2)+1.05*imag(Y(24))*sin(d2));
????F(3)=0.5-1.1*(V1*(real(Y(13))*cos(d3-d1)+imag(Y(13))*sin(d3-d1))+V2*(real(Y(23))*cos(d3-d2)+imag(Y(23))*sin(d3-d2))+1.1*real(Y(33))*cos(0)+1.1*imag(Y(33))*sin(0)+1.05*real(Y(34))*cos(d3)+1.05*imag(Y(34))*sin(d3));
????F(4)=-0.18-V1*(V1*(real(Y(11))*sin(0)-imag(Y(11))*cos(0))+V2*(real(Y(12))*sin(d1-d2)-imag(Y(12))*cos(d1-d2))+1.1*real(Y(13))*sin(d1-d3)-1.1*imag(Y(13))*cos(d1-d3)+1.05*real(Y(14))*sin(d1)-1.05*imag(Y(14))*cos(d1));
????F(5)=-0.13-V2*(V1*(real(Y(12))*sin(d2-d1)-imag(Y(12))*cos(d2-d1))+V2*(real(Y(22))*sin(0)-imag(Y(22))*cos(0))+1.1*real(Y(23))*sin(d2-d3)-1.1*imag(Y(23))*cos(d2-d3)+1.05*real(Y(24))*sin(d2)-1.05*imag(Y(24))*cos(d2));
?????
????J(11)=V1^2*imag(Y(11))+V1*(V1*(real(Y(11))*sin(0)-imag(Y(11))*cos(0))+V2*(real(Y(12))*sin(d1-d2)-imag(Y(12))*cos(d1-d2))+1.1*real(Y(13))*sin(d1-d3)-1.1*imag(Y(13))*cos(d1-d3)+1.05*real(Y(14))*sin(d1)-1.05*imag(Y(14))*cos(d1));
????J(12)=-V1*V2*(real(Y(12))*sin(d1-d2)-imag(Y(12))*cos(d1-d2));
????J(13)=-V1*1.1*(real(Y(13))*sin(d1-d3)-imag(Y(13))*cos(d1-d3));
????J(14)=-V1^2*real(Y(11))-V1*(V1*(real(Y(11))*cos(0)+imag(Y(11))*sin(0))+V2*(real(Y(12))*cos(d1-d2)+imag(Y(12))*sin(d1-d2))+1.1*real(Y(13))*cos(d1-d3)+1.1*imag(Y(13))*sin(d1-d3)+1.05*real(Y(14))*cos(d1)+1.05*imag(Y(14))*sin(d1));?
????J(15)=-V1*V2*(real(Y(12))*cos(d1-d2)+imag(Y(12))*sin(d1-d2));??
????
????J(21)=?-V1*V2*(real(Y(12))*sin(d2-d1)-imag(Y(12))*cos(d2-d1));
????J(22)=V2^2*imag(Y(22))+V2*(V1*(real(Y(12))*sin(d2-d1)-imag(Y(12))*cos(d2-d1))+V2*(real(Y(22))*sin(0)-imag(Y(22))*cos
- 上一篇:空時分組碼的仿真
- 下一篇:LDA人臉識別MATLAB含k近鄰算法
評論
共有 條評論