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  • 大小: 5KB
    文件類型: .m
    金幣: 1
    下載: 0 次
    發布日期: 2021-05-29
  • 語言: Matlab
  • 標簽: matlab??IEEE??

資源簡介

matlab代碼編寫的可以求電力系統中標準IEEE6節點潮流,可在matlab里直接運行。

資源截圖

代碼片段和文件信息

%%%IEEE6算例初始潮流計算
n=6;
nl=7;
zn=[0?0?0?0?0?0];
ph=[3?4?2?3?1?2?4];??
qh=[5?6?5?4?3?6?1];
zz=[0.133i?0.3i?0.723+1.05i?0.097+0.407i?0.08+0.37i?0.282+0.64i?0.123+0.518i];
zx=[0?0?0?0?0?0?0];
gd=[1?1?0?0?0?0?0];
sg=[0?0.501?0?0?0?0];
sl=[0?0?0?0.5+0.05i?0.55+0.13i?0.3+0.18i];
jl=[1?3?2?2?2?2];
pr=0.00001;
ijl=[1?3?2?2?2?2];
uo=[1.05?1.1?1?1?1?1];%1.05?1.1????1.1?1.025????
kk=[1.1??1.025?1?1?1?1?1];?????????????????????????????????
qb=[0?0?0?0?0?0];????????????????????????
y=zeros(n);
?for?i=1:n
????if?zn(i)~=0;
????????y(ii)=1./zn(i);
????end
?end
?for?ji=1:nl
????if?gd(ji)==0???
????????p=ph(ji);q=qh(ji);%某支路首端號p某支路末端號q???
????else?p=qh(ji);q=ph(ji);%反之
????end
????y(pq)=?y(pq)-1./(zz(ji)*kk(ji));
????y(qp)=?y(pq);
????y(qq)=?y(qq)+1./(zz(ji)*kk(ji)^2)+zx(ji);
????y(pp)=?y(pp)+1./zz(ji)+zx(ji);???
?end
b=imag(y);
gg=real(y);bb=imag(y);
bj=zeros(n-1);
for?i=1:n??
????if?jl(i)==1??????
?????????b(i:)=[];??
?????????b(:i)=[];?????????????????
????end????
end
bj=b;
jl(jl==1)=[];
ind=find(jl==2);
?w=length(ind);
for?i=1:n-1?????
????if?jl(i)==3;?????
????????b(i:)=[];
????????b(:i)=[];
????end
end
bjj=zeros(w);
bjj=b;
%disp(bjj)
for?i=1:n
???s(i)=sg(i)-sl(i)+qb(i);
end
p=real(s);q=imag(s);
v=zeros(1n);o=zeros(1n);
for?i=1:n?
????e(i)=real(uo(i));
????f(i)=imag(uo(i));
????v(i)=uo(i);?
????o(i)=atan(f(i)/e(i));
end????
????
?jd1=1;jd2=1;dt=0;kp=1;kq=1;
????while??jd1~=0|jd2~=0;
???????????jd1=0;jd2=0;?
???????????dp=zeros(size(ijl));dpl=zeros(size(ijl));
???????for?i=1:n
???????????if??ijl(i)~=1
??????????????c(i)=0;
???????????????for?k=1:n
????????????????c(i)=c(i)+v(k)*(gg(ik)*cos((o(i)-o(k))*pi./180)+bb(ik)*sin((o(i)-o(k))*pi./180));?????????????
???????????????end??????????
??????????????dpl(i)=p(i)-v(i)*c(i);
??????????????dp(i)=dpl(i)./?v(i);
???????????end
???????end?
????????????dp(ijl==1)=[];
????????????dpl(ijl==1)=[];???????????????
????????????for?i=1:n-1
????????????detp(i)=abs(dpl(i));?
????????????end
??????????if??detp>=pr
???????????????jd1=jd1+1;???
??????????end??????
????if??jd1~=0??????????
??????????kq=1;???????
?????????dto=-dp*inv(bj)*180./pi;
?????????po=zeros(size(ijl));
?????????po(ijl~=1)=dto;??????????
??????for?i=1:n
????????deto(i)=po(i)./v(i);
????????o(i)=o(i)+deto(i);?
??????end
??????dq=zeros(size(ijl));dql=zeros(size(ijl));
??????for?i=1:n
???????if?ijl(i)==2
????????????d(i)=0;
?????????????for?k=1:n
???????????????????d(i)=d(i)+v(k)*(gg(ik)*sin((o(i

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