This document provides instructions and data for solving the power flow of a 26-bus power system using MATLAB. It includes: 1) The bus data including voltage magnitudes, generation schedules, and reactive power limits. 2) The line data including resistance, reactance, and line charging values. 3) Instructions to use MATLAB routines to obtain the power flow solution using three different methods: Gauss-Seidel, Newton-Raphson, and Fast Decoupled. 4) Commands to output the power flow solution, calculate line losses, economic dispatch, and print final results.

1. Please use the MATLAB routines provided in the text (Saadat), and use the Matlab Report
format for your report. Problem Description The - 26-bus power system of an electric utility is
shown in Figure 5 below. Obtain a power flow solution by using the following methods Method
1: Gauss-Seidel power low method; Method 2: Newton-Raphson power flow method; and
Method 3: Fast decoupled power flow method The load data is given in Figure 1. The Voltage
magnitude, genera- tion schedule, and the reactive power limits for the regulated buses are
tabulated in Figure 2. Bus 1, whose voltage is specified as Vi = 1.032
Solution
clear
basemva = 100; accuracy = 0.00001; maxiter = 50;
% no code Mag Deg MW MVAR MW MVAR Qmin Qmax Mvar
busdata=[
1 1 1.025 0.0 51.0 41.0 0 0 0 0 4.0
2 2 1.020 0.0 22.0 15.0 79.0 0 40.0 250.0 0
3 3 1.025 0.0 64.0 50.0 20.0 0 40.0 150.0 0
4 4 1.050 0.0 25.0 10.0 100.0 0 40.0 80.0 2.0
5 5 1.045 0.0 50.0 30.0 300.0 0 40.0 160.0 5.0
6 0 1.000 0.0 76.0 29.0 0 0 0 0 2.0
7 0 1.000 0.0 0.0 0.0 0 0 0 0 0
8 0 1.000 0.0 0.0 0.0 0 0 0 0 0
9 0 1.000 0.0 89.0 50.0 0 0 0 0 0
10 0 1.000 0.0 0.0 0.0 0 0 0 0 0
11 0 1.000 0.0 25.0 15.0 0 0 0 0 1.5
12 0 1.000 0.0 89.0 48.0 0 0 0 0 2.0
13 0 1.000 0.0 31.0 15.0 0 0 0 0 0
14 2 1.000 0.0 24.0 12.0 0 0 0 0 0
15 0 1.000 0.0 70.0 31.0 0 0 0 0 0.5
16 2 1.000 0.0 55.0 27.0 0 0 0 0 0
17 3 1.000 0.0 78.0 38.0 0 0 0 0 0
18 4 1.000 0.0 153.0 67.0 0 0 0 0 0
19 5 1.000 0.0 75.0 15.0 0 0 0 0 5.0
20 0 1.000 0.0 48.0 27.0 0 0 0 0 0
21 2 1.000 0.0 46.0 23.0 0 0 0 0 0
22 0 1.000 0.0 45.0 22.0 0 0 0 0 0

2. 23 0 1.000 0.0 25.0 12.0 0 0 0 0 0
24 0 1.000 0.0 54.0 27.0 0 0 0 0 0
25 0 1.000 0.0 28.0 13.0 0 0 0 0 0
26 6 1.015 0.0 40.0 20.0 60.0 0 15.0 50.0 0];
% Line code
% Bus bus R X 1/2 B 1 for line code or
% nl nr p.u. p.u. p.u. tap setting value
linedata=[
1 2 0.0005 0.0048 0.0300 1
1 18 0.0013 0.0110 0.0600 1
2 3 0.0014 0.0513 0.0500 1
2 7 0.0103 0.0586 0.0180 1
2 8 0.0074 0.0321 0.0390 1
2 13 0.0007 0.0054 0.0005 1
2 26 0.0323 0.1967 0.0000 1
3 13 0.0007 0.0054 0.0005 1
4 8 0.0008 0.0240 0.0001 1
4 12 0.0016 0.0207 0.0150 1
5 6 0.0069 0.0300 0.0990 1
6 7 0.0053 0.0306 0.0010 1
6 11 0.0097 0.0570 0.0001 1
6 18 0.0037 0.0222 0.0012 1
6 19 0.0035 0.0660 0.0450 1
6 21 0.0050 0.0900 0.0226 1
7 8 0.0012 0.0069 0.0001 1
7 9 0.0009 0.0429 0.0250 1
8 12 0.0020 0.0180 0.0200 1
9 10 0.0010 0.0493 0.0010 1
10 12 0.0024 0.0132 0.0100 1
10 19 0.0547 0.2360 0.0000 1
10 20 0.0066 0.0160 0.0010 1
10 22 0.0069 0.0298 0.005 1
11 25 0.0960 0.2700 0.010 1
11 26 0.0165 0.0970 0.004 1
12 14 0.0327 0.0802 0.000 1
12 15 0.0180 0.0598 0.000 1

3. 13 14 0.0046 0.0271 0.001 1
13 15 0.0116 0.0610 0.000 1
13 16 0.0179 0.0888 0.001 1
14 15 0.0069 0.0382 0.000 1
15 16 0.0209 0.0512 0.000 1
16 17 0.0990 0.0600 0.000 1
16 20 0.0239 0.0585 0.000 1
17 18 0.0032 0.0600 0.038 1
17 21 0.2290 0.4450 0.000 1
19 23 0.0300 0.1310 0.000 1
19 24 0.0300 0.1250 0.002 1
19 25 0.1190 0.2249 0.004 1
20 21 0.0657 0.1570 0.000 1
20 22 0.0150 0.0366 0.000 1
21 24 0.0476 0.1510 0.000 1
22 23 0.0290 0.0990 0.000 1
22 24 0.0310 0.0880 0.000 1
23 25 0.0987 0.1168 0.000 1];
Lfybus
% form the bus admittance matrix
Lfnewton
% Power flow solution by Newton-Raphson method
Busout
% Prints the power flow solution on the screen
%lineflow
bloss
% Obtains the loss formula coefficients
gencost
dispatch
while dpslack > 0.001
lfnewton
bloss
dispatch
end
busout
gencost

4. lineflow
The coomand used at the last are from the text