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Copy pathGauss_Seidel_no_shunt.m
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Gauss_Seidel_no_shunt.m
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% BEng Electrical and Electronic Engineering
% Module 6EJ528
% Assignment – ACADEMIC YEAR …
% MODELLING LOAD FLOW STUDIES USING MATLAB
% USING THE GAUSS SEIDAL METHOD - NO SHUNT
% 100448597
% NETWORK (see Fig 1)
% Input (1)
% Insert the Network Admittance Matrix
% (see page 2 of supplied paperwork ignoring shunt capacitive admittance)
format short
Y(1,1) = 10.958905 - 26.027398i;
Y(1,2) = -3.424658 + 7.534247i;
Y(1,3) = -3.424658 + 7.534247i;
Y(1,4) = 0.00;
Y(1,5) = -4.109589 + 10.958904i;
Y(2,1) = -3.424658 + 7.534247i;
Y(2,2) = 11.672080 - 26.090949i;
Y(2,3) = -4.123711 + 9.278351i;
Y(2,4) = 0.00;
Y(2,5) = -4.123711 + 9.278351i;
Y(3,1) = -3.424658 + 7.534247i;
Y(3,2) = -4.123711 + 9.278351i;
Y(3,3) = 10.475198 - 23.154061i;
Y(3,4) = -2.926829 + 6.341463i;
Y(3,5) = 0.00;
Y(4,1) = 0.00;
Y(4,2) = 0.00;
Y(4,3) = -2.926829 + 6.341463i;
Y(4,4) = 7.050541 - 15.619814i;
Y(4,5) = -4.123711 + 9.278351i;
Y(5,1) = -4.109589 + 10.958904i;
Y(5,2) = -4.123711 + 9.278351i;
Y(5,3) = 0.00;
Y(5,4) = -4.123711 + 9.278351i;
Y(5,5) = 12.357012 - 29.515606i;
% Input (2)
% Input the given node loadings .....see diagram on page 00 ....
% .... of the supplied paperwork
% A positive (+ sign) is for generated power,
% The negative (- sign) indicates load taken from the network.
% Per unit loading is used with a base loading taken at 100MVA
P2 = - 0.4; % p.u. active power loading
P3 = - 0.25; % p.u. active power loading
P4 = - 0.4; % p.u. active power loading
P5 = - 0.5; % p.u. active power loading
Q2 = - 0.2i; % p.u. reactive power loading
Q3 = - 0.15i; % p.u. reactive power loading
Q4 = - 0.2i; % p.u. reactive power loading
Q5 = - 0.2i; % p.u. reactive power loading
% The voltage at the generation busbar, node (1) is 1.0 p.u.
% This voltage is fixed and therefore determines the reference ....
% ... or SLACK BUSBAR must be Node(1)
% JACOBI SOLUTION
% The number of iteration will need to be set
% Set the value of 'l' the chosen number of iterations
% Typically l=40 should be sufficient for this problem.
l = 40; % sets the number of iterations
m = l + 1; % allocates the space used to store each iteration
vector = [1:m]; % assigns a row with the appropriate spaces
row = ones(size(vector));
Vnode1 = row; % each node voltage has the required storage spaces
Vnode2 = row;
Vnode3 = row;
Vnode4 = row;
Vnode5 = row;
for n = 1:m
Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltages
Vnode2(n) = 1.0 + 0.00i;
Vnode3(n) = 1.0 + 0.00i;
Vnode4(n) = 1.0 + 0.00i;
Vnode5(n) = 1.0 + 0.00i;
end
S2star = P2 - Q2; % congugates of the specified node loadings
S3star = P3 - Q3;
S4star = P4 - Q4;
S5star = P5 - Q5;
% Now Proceed with the JACOBI SOLUTION
% Following the same method as in the worked example
% Page 5 of the assignment notes
for n = 1:l
V2star(n) = conj(Vnode2(n)); % conjugate of V2
V3star(n) = conj(Vnode3(n)); % conjugate of V3
V4star(n) = conj(Vnode4(n)); % conjugate of V4
V5star(n) = conj(Vnode5(n)); % conjugate of V5
% Solving each of the values and replacing the value as soon as it is
% calculated
I2(n) = S2star/V2star(n);
Sum2(n) = (Y(2,1) * Vnode1(n)) + (Y(2,3) * Vnode3(n)) + (Y(2,5) * Vnode5(n));
V2(n) = ( I2(n) - Sum2(n) ) / Y(2,2)
Vnode2(n) = V2(n);
I3(n) = S3star/V3star(n);
Sum3(n) = (Y(3,1) * Vnode1(n)) + (Y(3,2) * Vnode2(n)) + (Y(3,4) * Vnode4(n));
V3(n) = ( I3(n) - Sum3(n) ) / Y(3,3);
Vnode3(n) = V3(n);
I4(n) = S4star/V4star(n);
Sum4(n) = (Y(4,3) * Vnode3(n)) + (Y(4,5) * Vnode5(n));
V4(n) = ( I4(n) - Sum4(n) ) / Y(4,4);
Vnode4(n) = V4(n);
I5(n) = S5star/V5star(n);
Sum5(n) = (Y(5,1) * Vnode1(n)) + (Y(5,2) * Vnode2(n)) + (Y(5,4) * Vnode4(n));
V5(n) = ( I5(n) - Sum5(n) ) / Y(5,5);
Vnode5(n) = V5(n);
u = 1 + n;
Vnode2(u) = V2(n);
Vnode3(u) = V3(n);
Vnode4(u) = V4(n);
Vnode5(u) = V5(n);
end
disp ('GAUSS SIEDAL - NO SHUNT')
disp (' Vnode2 Vnode3 Vnode4 Vnode5 ')
disp(':')
d = 0;
for c = 1:l
fprintf(' %1.4f %1.4fi\t, %1.4f %1.4fi\t, %1.4f %1.4fi\t, %1.4f %1.4fi\n',...
real(Vnode2(1+d)), imag(Vnode2(1+d)), real(Vnode3(1+d)),...
imag(Vnode3(1+d)), real(Vnode4(1+d)), imag(Vnode4(1+d)), real(Vnode5(1+d)), imag(Vnode5(1+d)));
d=d+1;
end
figure(1);
clf;
subplot(2,1,1)
XAxis=0:1:l;
plot(XAxis,Vnode2,'black',XAxis,Vnode3,'blue',XAxis,Vnode4,'green',XAxis,Vnode5,'red');
grid on;
xlabel('Iterations');
ylabel('Node Voltage (p.u.)');
legend('Node 2','Node 3','Node 4','Node 5');
title('Convergence of node voltages (per unit) - GAUSS SEIDAL (NO SHUNT CAPACITANCE)');
subplot(2,1,2)
XAxis=1:1:l;
plot(XAxis,I2,'black',XAxis,I3,'blue',XAxis,I4,'green',XAxis,I5,'red');
grid on;
xlabel('Iterations');
ylabel('Line current (p.u.)');
legend('Line 2','Line 3','Line 4','Line 5');
title('Convergence of node voltages (per unit) - GAUSS SEIDAL (NO SHUNT CAPACITANCE)');