he 6-bus power system is given with one-line diagram in Fig. 1 and line/transfor
ID: 2085855 • Letter: H
Question
he 6-bus power system is given with one-line diagram in Fig. 1 and line/transformer parameters in Table 2 and solved load-flow data in Table 3. Using MATLAB function bldybus.m calculate Ybus matrix. Include the load as a constant impedance in Ybus matrix. 18 / 230 CO bus 18 k 30/13.8 G2 230 k 230 k 230 k 5 230/69 Figure 1: Problem 3: system one-line diagram Table 2: Problem 3: Line and Transformer Data for the system Series Z [pu] Shunt Y [pul Bus to bus R Transf. Tap Line 3-4 0.015 0.111 Line 3-5 (10.02 0.148 Line 3-5 (2 0.02 0.148 Line 4-5 0.015 0.111 0.13 0.10 0.08 0.223 0.298 0.298 0.223 Trans. 1-3 Trans. 2-4 Trans. 5-6Explanation / Answer
% Program to form Admittance Bus Formation with Transformer Tap setting..
num = 6; % IEEE-6, IEEE-14, IEEE-30, IEEE-57 bus systems..
linedata = linedatas(num); % Calling "linedata6.m" for Line Data...
fb = linedata(:,1); % From bus number...
tb = linedata(:,2); % To bus number...
r = linedata(:,3); % Resistance, R...
x = linedata(:,4); % Reactance, X...
b = linedata(:,5); % Ground Admittance, B/2...
a = linedata(:,6); % Tap setting value..
z = r + i*x; % Z matrix...
y = 1./z; % To get inverse of each element...
b = i*b; % Make B imaginary...
nbus = max(max(fb),max(tb)); % no. of buses...
nbranch = length(fb); % no. of branches...
Y = zeros(nbus,nbus); % Initialise YBus...
% Formation of the Off Diagonal Elements...
for k=1:nbranch
Y(fb(k),tb(k)) = Y(fb(k),tb(k))-y(k)/a(k);
Y(tb(k),fb(k)) = Y(fb(k),tb(k));
end
% Formation of Diagonal Elements....
for m =1:nbus
for n =1:nbranch
if fb(n) == m
Y(m,m) = Y(m,m) + y(n)/(a(n)^2) + b(n);
elseif tb(n) == m
Y(m,m) = Y(m,m) + y(n) + b(n);
end
end
end
Y % Bus Admittance Matrix..
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