Hi All.
I need to work the load flow in the MATPOWER to the bus isolation condition
(due to a fault) for the restoration process. I did this by using the
extract_islands function. The procedure I did Was as follows: Taking as an
example the case16 below:
I considered a fault in bar 8, with all the connections this bar were turned
off, ie:(2-8), (8-9) and (8-10) were withdrawn to be isolated. To do this, I
used The following routine:
mpc_array = extract_islands(case16)
groups = find_islands(case16)
mpc1 = extract_islands(case16, groups, 1)
mpc2 = extract_islands(case16, groups, 2)
runpf(mpc1)
runpf(mpc2)
It turns out that my work is an optimization process. I need something somewhat
automatic.
Does anyone know a more concise way to conduct such a study for any isolated
bar?
mpc.bus = [
1 3 0.0000 0.0000 0 0 1 1.00 0
Vb 1 1.05 0.950;
2 3 0.0000 0.0000 0 0 1 1.00 0
Vb 1 1.05 0.950;
3 3 0.0000 0.0000 0 0 1 1.00 0
Vb 1 1.05 0.950;
4 1 2.0000 1.6000 0 0 1 1.00 0
Vb 1 1.05 0.950;
5 1 3.0000 1.5000 0 1.1 1 1.00 0 Vb
1 1.05 0.950;
6 1 2.0000 0.8000 0 1.2 1 1.00 0 Vb
1 1.05 0.950;
7 1 1.5000 0.2000 0 0 1 1.00 0
Vb 1 1.05 0.950;
8 1 4.0000 2.7000 0 0 1 1.00 0
Vb 1 1.05 0.950;
9 1 5.0000 3.0000 0 1.2 1 1.00 0 Vb
1 1.05 0.950;
10 1 1.0000 0.9000 0 0 1 1.00 0
Vb 1 1.05 0.950;
11 1 0.6000 0.1000 0 0.6 1 1.00 0 Vb
1 1.05 0.950;
12 1 4.5000 2.0000 0 3.7 1 1.00 0 Vb
1 1.05 0.950;
13 1 1.0000 0.9000 0 0 1 1.00 0
Vb 1 1.05 0.950;
14 1 1.0000 0.7000 0 1.8 1 1.00 0 Vb
1 1.05 0.950;
15 1 1.0000 0.9000 0 0 1 1.00 0
Vb 1 1.05 0.950;
16 1 2.1000 1.0000 0 1.8 1 1.00 0 Vb
1 1.05 0.950;
];
.
.
.
%% MATRIZ ORIGINAL valors das impedâncias estao em ohms
% fbus tbus r x b rateA rateB rateC ratio angle
status angmin angmax
mpc.branch = [
1 4 0.075 0.10 0 0 0 0 0 0 0 -360 360;
4 5 0.080 0.11 0 0 0 0 0 0 0 -360 360;
4 6 0.090 0.18 0 0 0 0 0 0 0 -360 360;
6 7 0.040 0.04 0 0 0 0 0 0 1 -360 360;
2 8 0.110 0.11 0 0 0 0 0 0 1 -360 360;
8 9 0.080 0.11 0 0 0 0 0 0 1 -360 360;
8 10 0.110 0.11 0 0 0 0 0 0 0 -360 360;
9 11 0.110 0.11 0 0 0 0 0 0 1 -360 360;
9 12 0.080 0.11 0 0 0 0 0 0 1 -360 360;
3 13 0.110 0.11 0 0 0 0 0 0 1 -360 360;
13 14 0.090 0.12 0 0 0 0 0 0 1 -360 360;
13 15 0.080 0.11 0 0 0 0 0 0 1 -360 360;
15 16 0.040 0.04 0 0 0 0 0 0 1 -360 360;
5 11 0.040 0.04 0 0 0 0 0 0 1 -360 360;
10 14 0.040 0.04 0 0 0 0 0 0 1 -360 360;
7 16 0.090 0.12 0 0 0 0 0 0 1 -360 360;
];
mpc.branch(:,3:4) = mpc.branch(:,3:4)*((mpc.baseMVA)/(Vb.^2)); % Deixa todas as
impedâncias na base de
% de potência 'mpc.baseMVA' dada em MVA e tensão 'Vb' dada em kV.
