Problem 2: Study MatPower case9.m

1. a. provide a one-line diagram to describe the case9 system
2. b. Power Flow Study: Use Newton's method to solve AC power flow
3. i) Include a table to report input data of the case9.m system
4. ii) Include a table to report power flow results

Below is the MatPower Code, and the the results of the code.

function mpc = case9 %CASE9 Power flow data for 9 bus, 3 generator case. % Please see CASEFORMAT for details on the case file format. % % Based on data from p. 70 of: % % Chow, J. H., editor. Time-Scale Modeling of Dynamic Networks with % Applications to Power Systems. Springer-Verlag, 1982. % Part of the Lecture Notes in Control and Information Sciences book % series (LNCIS, volume 46) % % which in turn appears to come from: % % R.P. Schulz, A.E. Turner and D.N. Ewart, "Long Term Power System % Dynamics," EPRI Report 90-7-0, Palo Alto, California, 1974.

% MATPOWER

%% MATPOWER Case Format : Version 2 mpc.version = '2';

%%----- Power Flow Data -----%% %% system MVA base mpc.baseMVA = 100;

%% bus data % bus_i type Pd Qd Gs Bs area Vm Va baseKV zone Vmax Vmin mpc.bus = [ 1 3 0 0 0 0 1 1 0 345 1 1.1 0.9; 2 2 0 0 0 0 1 1 0 345 1 1.1 0.9; 3 2 0 0 0 0 1 1 0 345 1 1.1 0.9; 4 1 0 0 0 0 1 1 0 345 1 1.1 0.9; 5 1 90 30 0 0 1 1 0 345 1 1.1 0.9; 6 1 0 0 0 0 1 1 0 345 1 1.1 0.9; 7 1 100 35 0 0 1 1 0 345 1 1.1 0.9; 8 1 0 0 0 0 1 1 0 345 1 1.1 0.9; 9 1 125 50 0 0 1 1 0 345 1 1.1 0.9; ];

%% generator data % bus Pg Qg Qmax Qmin Vg mBase status Pmax Pmin Pc1 Pc2 Qc1min Qc1max Qc2min Qc2max ramp_agc ramp_10 ramp_30 ramp_q apf mpc.gen = [ 1 72.3 27.03 300 -300 1.04 100 1 250 10 0 0 0 0 0 0 0 0 0 0 0; 2 163 6.54 300 -300 1.025 100 1 300 10 0 0 0 0 0 0 0 0 0 0 0; 3 85 -10.95 300 -300 1.025 100 1 270 10 0 0 0 0 0 0 0 0 0 0 0; ];

%% branch data % fbus tbus r x b rateA rateB rateC ratio angle status angmin angmax mpc.branch = [ 1 4 0 0.0576 0 250 250 250 0 0 1 -360 360; 4 5 0.017 0.092 0.158 250 250 250 0 0 1 -360 360; 5 6 0.039 0.17 0.358 150 150 150 0 0 1 -360 360; 3 6 0 0.0586 0 300 300 300 0 0 1 -360 360; 6 7 0.0119 0.1008 0.209 150 150 150 0 0 1 -360 360; 7 8 0.0085 0.072 0.149 250 250 250 0 0 1 -360 360; 8 2 0 0.0625 0 250 250 250 0 0 1 -360 360; 8 9 0.032 0.161 0.306 250 250 250 0 0 1 -360 360; 9 4 0.01 0.085 0.176 250 250 250 0 0 1 -360 360; ];

%%----- OPF Data -----%% %% generator cost data % 1 startup shutdown n x1 y1 ... xn yn % 2 startup shutdown n c(n-1) ... c0 mpc.gencost = [ 2 1500 0 3 0.11 5 150; 2 2000 0 3 0.085 1.2 600; 2 3000 0 3 0.1225 1 335; ];

case9

ans =

struct with fields:

version: '2' baseMVA: 100 bus: [913 double] gen: [321 double] branch: [913 double] gencost: [37 double]