用matlab进行牛顿拉夫逊法的潮流计算的编程,

问题描述:

用matlab进行牛顿拉夫逊法的潮流计算的编程,
1个回答 分类:综合 2014-12-12

问题解答:

我来补答
B2=input('请输入各节点参数形成的矩阵:B2=');
%本程序的功能是用牛顿-拉夫逊法进行潮流计算
n=input('请输入节点数:n=');
nl=input('请输入支路数:nl=');
isb=input('请输入平衡母线节电号:isb=');
pr=input('请输入误差精度:pr=');
B1=input('请输入由支路参数形成的矩阵:B1=');%变压器侧为1,否则为0
B2=input('请输入各节点参数形成的矩阵:B2=');
X=input('请输入由节点号及其对地阻抗形成的矩阵:X=');
X=input('请输入由节点号及其对地阻抗形成的矩阵:X=');
Y=zeros(n);U=zeros(1,n);cta=zeros(1,n);V=zeros(1,n);O=zeros(1,n);S1=zeros(nl);
for i=1:n
if X(i,2)~=0;
p=X(i,1);
Y(p,p)=X(i,2);
end
end
for i=1:nl
if B1(i,6)==0
p=B1(i,1);q=B1(i,2);
else p=B1(i,2);q=B1(i,1);
end
Y(p,q)=Y(p,q)-1./(B1(i,3)*B1(i,5));
Y(q,p)=Y(p,q);
Y(q,q)=Y(q,q)+1./(B1(i,3)*B1(i,5)^2)+B1(i,4)./2;
Y(p,p)=Y(p,p)+1./B1(i,3)+B1(i,4)./2;
end %求导纳矩阵
G=real(Y);B=imag(Y);
for i=1:n
cta(i)=angle(B2(i,3));
U(i)=abs(B2(i,3));
%V(i)=B2(i,4);
end
for i=1:n
S(i)=B2(i,1)-B2(i,2);
B(i,i)=B(i,i)+B2(i,5);
end
P=real(S);Q=imag(S);
ICT1=0;IT2=1;
while IT2~=0
IT2=0;t1=1;t2=1;
for i=1:n
if i~=isb
C(i)=0;
D(i)=0;
for j1=1:n
C(i)=C(i)+U(i)*U(j1)*(G(i,j1)*cos(cta(i)-cta(j1))+B(i,j1)*sin(cta(i)-cta(j1)));
D(i)=D(i)+U(i)*U(j1)*(G(i,j1)*sin(cta(i)-cta(j1))-B(i,j1)*cos(cta(i)-cta(j1)));
end
DP(t1)=P(i)-C(i);
t1=t1+1;
if B2(i,6)==2
DQ(t2)=Q(i)-D(i);
t2=t2+1;
end
end
end
t1=t1-1;t2=t2-1;
DPQ=[DP';DQ']; %求DP,DQ
for i=1:t1+t2
if abs(DPQ(i))>pr
IT2=IT2+1;
end
end
H=zeros(t1,t1);N=zeros(t1,t2);K=zeros(t2,t1);L=zeros(t2,t2);
for i=1:t1
for j1=1:t1
if j1~=isb&j1~=i
H(i,j1)=0-U(i)*U(j1)*(G(i,j1)*sin(cta(i)-cta(j1))-B(i,j1)*cos(cta(i)-cta(j1)));
elseif j1~=isb&j1==i
H(i,j1)=U(i)^2*B(i,j1)+D(i);
end
end
end
for i=1:t1
for j1=1:t2
if j1~=isb&j1~=i
N(i,j1)=0-U(i)*U(j1)*(G(i,j1)*cos(cta(i)-cta(j1))+B(i,j1)*sin(cta(i)-cta(j1)));
elseif j1~=isb&j1==i
N(i,j1)=0-U(i)^2*G(i,j1)-C(i);
end
end
end
for i=1:t2
for j1=1:t1
if j1~=isb&j1~=i
K(i,j1)= U(i)*U(j1)*(G(i,j1)*cos(cta(i)-cta(j1))+B(i,j1)*sin(cta(i)-cta(j1)));
elseif j1~=isb&j1==i
K(i,j1)=U(i)^2*G(i,j1)-C(i);
end
end
end
for i=1:t2
for j1=1:t2
if j1~=isb&j1~=i
L(i,j1)=0-U(i)*U(j1)*(G(i,j1)*sin(cta(i)-cta(j1))-B(i,j1)*cos(cta(i)-cta(j1)));
elseif j1~=isb&j1==i
L(i,j1)=U(i)^2*B(i,j1)-D(i);
end
end
end
J=[H,N;K,L];%求雅可比矩阵
modify=-J\DPQ;
Dcta=modify([1:t1],:);
t3=U(:,[1:t2]);
DU=diag(t3,0)*modify([t1+1:t1+t2],:);
t4=1;
for i=1:t1
if B2(i,6)~=1
cta(1,i)=cta(1,i)+Dcta(t4,1);
t4=t4+1;
end
end
t5=1;
for i=1:t2
if B2(i,6)==2
U(1,i)=U(1,i)+DU(t5,1);
t5=t5+1;
end
end
ICT1=ICT1+1;
end %修正原值
for i=1:n
UU(i)=U(i)*cos(cta(i))+1i*U(i)*sin(cta(i));
end
for p=1:n
c(p)=0;
for q=1:n
c(p)=c(p)+conj(Y(p,q))*conj(UU(q));
end
s(p)=UU(p)*c(p);
end
disp('--------------------------------------------------------------------------------');
disp('各节点电压U为(节点从小到大排列):');
disp(UU);
disp('--------------------------------------------------------------------------------');
disp('各节点电压相角为(节点从小到大排列):');
disp(180*angle(UU)/pi);
disp('--------------------------------------------------------------------------------');
disp('按公式计算全部线路功率,结果如下:');
for i=1:nl
if B1(i,6)==0
p=B1(i,1);q=B1(i,2);
else p=B1(i,2);q=B1(i,1);
end
Si(p,q)=UU(p)*(conj(UU(p))*conj(B1(i,4)./2)+(conj(UU(p)*B1(i,5))-conj(UU(q)))*conj(1./(B1(i,3)*B1(i,5))));%各条支路首端功率Si
f=[p,q,Si(p,q)];
disp(f);
end
for i=1:nl
if B1(i,6)==0
p=B1(i,1);q=B1(i,2);
else p=B1(i,2);q=B1(i,1);
end
Sj(q,p)=UU(q)*(conj(UU(q))*conj(B1(i,4)./2)+(conj(UU(q)./B1(i,5))-conj(UU(p)))*conj(1./(B1(i,3)*B1(i,5))));%各条支路末端功率Sj
f=[q,p,Sj(q,p)];
disp(f);
end
disp('--------------------------------------------------------------------------------');
disp('各条支路的功率损耗DS为(顺序同您输入B1时一样):');
for i=1:nl
if B1(i,6)==0
p=B1(i,1);q=B1(i,2);
else p=B1(i,2);q=B1(i,1);
end
DS(i)=Si(p,q)+Sj(q,p);%各条支路功率损耗DS
disp(DS(i));
end
Sp=0;
for i=1:n
Sp=Sp+UU(isb)*conj(Y(isb,i))*conj(UU(i));
end
disp('平衡节点的功率:');
disp(Sp);
------------
在网上找到的一个,希望能有帮助.
 
 
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