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16-02-2017, 08:07 PM
Post: #1
Request for codes

Hello,I'm Anahita and I want coding for backward and forward sweep method and for bibc and bcbv method for radial distribution system.Kindly send the program to my email id:1995singhanahita[at]gmail.com
17-02-2017, 09:28 AM
Post: #2
RE: Request for codes

clear all
%branch no sending reciving R(PU) X(pu)
LD=[ 1 1 2 0.04997 0.06644;
2 2 3 0.02332 0.03310 ;
3 1 4 0.04664 0.06201;
4 1 5 0.02082 0.02768;
5 5 6 0.02500 0.03322;
6 1 7 0.02665 0.03543;
7 7 8 0.02748 0.03654;
8 1 9 0.03331 0.04430;
9 1 10 0.02082 0.02768;
10 2 11 0.02082 0.02768];

% bus no activepower reactivepower
BD=[ 1 0 0
2 1.22 0.916
3 0.032 0.024
4 0.778 0.584
5 0.673 0.595
6 1.22 0.916
7 0.0488 0.0366
8 0.956 0.717
9 0.698 0.523
10 1.265 0.949
11 0.265 0.0949];

F=LD(:,2:3);
M=max(LD(:,2:3));
N=max(M);
f=[1:N]';
for i=1:N
g=find(F(:,Smile==i);
h(i)=length(g);
end
k(:,1)=f;
k(:,2)=h';
cent=input('central bus ');
% this section of the code is to adjust line data to the standard
NLD=zeros(N,size(LD,2));
c=find(LD(:,2:3)==cent);
NLD=LD(c,Smile;
LD(c,Smile=[];
t=find(k(:,1)==cent);
k(t,2)=k(t,2)-size(c,1);
j=size(c,1);
i=1;
while sum(k(:,2))>0
c=[];
b=[];
t=[];
[c e]=find(LD(:,2:3)==NLD(i,3));
if size(c,2)~=0
b=LD(c,Smile;
LD(c,Smile=[];
t=find(k(:,1)==NLD(i,3));
k(t,2)=k(t,2)-(size(c,1)+1);
d=find(b(:,3)==NLD(i,3));
b(d,2:3)=[b(d,3),b(d,2)];
NLD(j+1:j+size(c,1),Smile=b;
j=j+size(c,1);
end
i=i+1;
end
LD=NLD;
%LD=sortrows(NLD,3);

% end the data is represented in standard format

%code for bus-injection to branch-current matrix and bcbv
bibc=zeros(size(LD,1),size(LD,1));
for i=1:size(LD,1)

if LD(i,2)==1
bibc(LD(i,3)-1,LD(i,3)-1)=1;
BCBV(LD(i,3)-1,LD(i,3)-1)=complex(LD(i,4),LD(i,5));
else
bibc(:,LD(i,3)-1)=bibc(:,LD(i,2)-1);
bibc(LD(i,3)-1,LD(i,3)-1)=1;
BCBV(LD(i,3)-1,Smile=BCBV(LD(i,2)-1,Smile;
BCBV(LD(i,3)-1,LD(i,3)-1)=complex(LD(i,4),LD(i,5));
end
end
S=complex(BD(:,2),BD(:,3));% complex power
Vo=ones(size(LD,1),1);% initial bus votage% 10 change to specific data value
S(1)=[];
VB=Vo;
iteration=input('number of iteration : ');

for i=1:iteration
%backward sweep
I=conj(S./VB);% injected current
%Z=complex(LD(:,4),LD(:,5));%branch impedance
%ZD=diag(Z);%makeing it diagonal
IB=bibc*I; %branch current

%forward sweep
TRX=BCBV*bibc;
VB=Vo-TRX*I;
end

Vbus=[1;VB];
display(abs(Vbus));
display(abs(IB));
% reference
%1)An Effective Approach for Distribution System Power Flow Solution A. Alsaadi, and B. Gholam
%2)The Standard Backward/Forward Sweep Power Flow Paulo M. De Oliveira-De Jesus,

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