P11. Write down the binary connectivity matrix (BCM) for PMU placement for IEEE

Question

P11. Write down the binary connectivity matrix (BCM) for PMU placement for IEEE 30 bus system, and formulate the LP constraints. Run the code in AMPL and obtain the optimal solution of PMUs for the following system. Assume the cost of a PMU is $3000 per unit Your overall goal is to minimize the total cost of PMU while finding the optimal placement of PMUs in appropriate bus locations. Note: The buses 6, 9,11, 25, and 28 in the following figure are zero injection buses. 1. Find the optimal placement of PMUs with no zero injection constraints considered. Show binary connectivity matrix, and LP formulations. Also, compute SORI index (See class notes on SORI). 2. Find the optimal placement with zero injection constraints considered. Show binary connectivity matrix, and LP formulations. Also compute SORI index. 3. Show the AMPL run with model and data files for question 2. 29 27 28 26-25- 30 23 24 15 18 19 20 21 7 416 10 1312 19 IEEE 30 bus system

Explanation / Answer

function [baseMVA, bus, gen, branch, areas, gencost] = c14
%CASE14    Power flow data for IEEE 14 bus test case.
%   Please see 'help caseformat' for details on the case file format.
%   This data was converted from IEEE Common Data Format
%   (ieee14cdf.txt) on 20-Sep-2004 by cdf2matp, rev. 1.11
%   See end of file for warnings generated during conversion.
%
%   Converted from IEEE CDF file from:
%       http://www.ee.washington.edu/research/pstca/
%
% 08/19/93 UW ARCHIVE           100.0 1962 W IEEE 14 Bus Test Case

%   MATPOWER
%   $Id: case14.m,v 1.5 2004/09/21 01:46:23 ray Exp $

%% system MVA base
baseMVA = 100;

%% bus data
% bus_i type Pd Qd Gs Bs area Vm Va baseKV zone Vmax Vmin
bus = [
1 3 0 0 0 0 1 1.06 0 0 1 1.06 0.94;
2 2 21.7 12.7 0 0 1 1.045 -4.98 0 1 1.06 0.94;
3 2 94.2 19 0 0 1 1.01 -12.72 0 1 1.06 0.94;
4 1 47.8 -3.9 0 0 1 1.019 -10.33 0 1 1.06 0.94;
5 1 7.6 1.6 0 0 1 1.02 -8.78 0 1 1.06 0.94;
6 2 11.2 7.5 0 0 1 1.07 -14.22 0 1 1.06 0.94;
7 1 0 0 0 0 1 1.062 -13.37 0 1 1.06 0.94;
8 2 0 0 0 0 1 1.09 -13.36 0 1 1.06 0.94;
9 1 29.5 16.6 0 19 1 1.056 -14.94 0 1 1.06 0.94;
10 1 9 5.8 0 0 1 1.051 -15.1 0 1 1.06 0.94;
11 1 3.5 1.8 0 0 1 1.057 -14.79 0 1 1.06 0.94;
12 1 6.1 1.6 0 0 1 1.055 -15.07 0 1 1.06 0.94;
13 1 13.5 5.8 0 0 1 1.05 -15.16 0 1 1.06 0.94;
14 1 14.9 5 0 0 1 1.036 -16.04 0 1 1.06 0.94;
];

%% generator data
% bus Pg Qg Qmax Qmin Vg mBase status Pmax Pmin
gen = [
1 232.4 -16.9 10 0 1.06 100 1 332.4 0;
2 40 42.4 50 -40 1.045 100 1 140 0;
3 0 23.4 40 0 1.01 100 1 100 0;
6 0 12.2 24 -6 1.07 100 1 100 0;
8 0 17.4 24 -6 1.09 100 1 100 0;
];

%% branch data
% fbus tbus r x b rateA rateB rateC ratio angle status
branch = [
1 2 0.01938 0.05917 0.0528 9900 1 1 1 0 1;
1 5 0.05403 0.22304 0.0492 9900 1 1 1 0 1;
2 3 0.04699 0.19797 0.0438 9900 1 1 1 0 1;
2 4 0.05811 0.17632 0.034 9900 1 1 1 0 1;
2 5 0.05695 0.17388 0.0346 9900 1 1 1 0 1;
3 4 0.06701 0.17103 0.0128 9900 1 1 1 0 1;
4 5 0.01335 0.04211 0 9900 0 1 1 1 1;
4 7 0 0.20912 0 9900 0 0 0.978 0 1;
4 9 0 0.55618 0 9900 0 0 0.969 0 1;
5 6 0 0.25202 0 9900 0 0 0.932 0 1;
6 11 0.09498 0.1989 0 9900 0 1 1 0 1;
6 12 0.12291 0.25581 0 9900 0 1 1 0 1;
6 13 0.06615 0.13027 0 9900 0 1 1 0 1;
7 8 0 0.17615 0 9900 0 0 1 1 1;
7 9 0 0.11001 0 9900 0 0 1 1 1;
9 10 0.03181 0.0845 0 9900 0 1 1 0 1;
9 14 0.12711 0.27038 0 9900 0 0 1 0 1;
10 11 0.08205 0.19207 0 9900 0 0 1 0 1;
12 13 0.22092 0.19988 0 9900 0 0 1 0 1;
13 14 0.17093 0.34802 0 9900 0 0 1 0 1;
];

%% area data
areas = [
1 1;
];

%% generator cost data
% 1 startup shutdown n x0 y0 ... xn yn
% 2 startup shutdown n c(n-1) ... c0
gencost = [
2 0 0 3 0.0430293 20 0;
2 0 0 3 0.25 20 0;
2 0 0 3 0.01 40 0;
2 0 0 3 0.01 40 0;
2 0 0 3 0.01 40 0;
];

return;

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