潮流計(jì)算matlab程序

1、精品好資料學(xué)習(xí)推薦clear;%各節(jié)點(diǎn)參數(shù)：節(jié)點(diǎn)編號(hào)，類型，電壓幅值，電壓相位，注入有功，注入無功%類型：1PQ節(jié)點(diǎn)，2PV節(jié)點(diǎn)，3平衡節(jié)點(diǎn)%本程序中將最后一個(gè)節(jié)點(diǎn)設(shè)為平衡節(jié)點(diǎn)R_1=1 1 1.0 0 0.20.2j;2 1 1.0 0 -0.45 -0.15j;3 1 1.0 0 -0.45 -0.05j;4 1 1.0 0 -0.6 -0.1j;5 3 1.0 0 0 0;%支路號(hào)首端節(jié)點(diǎn)末端節(jié)點(diǎn)支路導(dǎo)納R_2=1 5 2 1.25-3.75j;2 2 3 10.00-30.00j;3 3 4 1.25-3.75j;4 1 4 2.50-7.50j;5 1 5 5.00-15.00j;6

2、 1 2 1.667-5.00j;n=5;L=6;%需要改變的到此為止i=0;j=0;a=0;precision=1;k=0;Y=zeros(n,n);u=zeros(1,n);delt=zeros(1,n);P=zeros(1,n);Q=zeros(1,n);G=;B=;PP=;uu=;U=;dp=;dq=;for a=1:Li=R_2(a,2); j=R_2(a,3);Y(i,j)=-R_2(a,4);Y(j,i)=Y(i,j);endfor a=1:nfor b=1:nif a=bY(a,a)=Y(a,a)+Y(a,b);endendendfori=1:nfor j=1:nifi=jY(

3、i,j)=-Y(i,j);endendendY %形成導(dǎo)納矩陣fori=1:nfor j=1:nG(i,j)=real(Y(i,j);B(i,j)=imag(Y(i,j);endendfor a=1:nu(a)=R_1(a,3);P(a)=R_1(a,5);Q(a)=R_1(a,6);delt(a)=R_1(a,4);endwhile precision0.0001 %判斷是否滿足精度要求fora=1:n-1forb=1:n pt(b)=u(a)*u(b)*(G(a,b)*cos(delt(a)-delt(b)+B(a,b)*sin(delt(a)-delt(b); qt(b)=u(a)*u(

4、b)*(G(a,b)*sin(delt(a)-delt(b)-B(a,b)*cos(delt(a)-delt(b);endpt,qtpi(a)=sum(pt);qi(a)=sum(qt); %計(jì)算PQ節(jié)點(diǎn)的注入功率dp(a)=P(a)-pi(a);dq(a)=Q(a)-qi(a); %計(jì)算PQ節(jié)點(diǎn)的功率不平衡量endfora=1:n-1for b=1:n-1ifa=b H(a,a)=-qi(a)-u(a)2*B(a,a); N(a,a)=pi(a)+u(a)2*G(a,a);J(a,a)=pi(a)-u(a)2*G(a,a); L(a,a)=qi(a)-u(a)2*B(a,a); JJ(2*a

5、-1,2*a-1)=H(a,a); JJ(2*a-1,2*a)=N(a,a); JJ(2*a,2*a-1)=J(a,a); JJ(2*a,2*a)=L(a,a);else H(a,b)=u(a)*u(b)*(G(a,b)*sin(delt(a)-delt(b)-B(a,b)*cos(delt(a)-delt(b); J(a,b)=-u(a)*u(b)*(G(a,b)*cos(delt(a)-delt(b)+B(a,b)*sin(delt(a)-delt(b); N(a,b)=-J(a,b);L(a,b)=H(a,b); JJ(2*a-1,2*b-1)=H(a,b);JJ(2*a-1,2*b)=

6、N(a,b); JJ(2*a,2*b-1)=J(a,b); JJ(2*a,2*b)=L(a,b);endendend %計(jì)算jocbi各項(xiàng),并放入統(tǒng)一矩陣JJ中，對JJ下標(biāo)統(tǒng)一編號(hào)JJfora=1:n-1PP(2*a-1)=dp(a);PP(2*a)=dq(a); end %按統(tǒng)一矩陣形成功率不平衡uu=inv(JJ)*PP;precision=max(abs(uu); %判斷是否收斂forb=1:n-1delt(b)=delt(b)+uu(2*b-1); u(b)=u(b)+uu(2*b)*u(b); %將結(jié)果分解為電壓幅值和角度end %求解修正方程，得電壓幅值變化量（標(biāo)幺值）和角度變化量k=k+1;endfora=1:nU(a)=u(a)*(cos(delt(a)+j*sin(delt(a);endforb=1:n I(b)=Y(n,b)*U(b);%求平衡節(jié)點(diǎn)的注入電流endS5=U(n)*sum(conj(I)%求平衡節(jié)點(diǎn)的注入功率fora=1:n