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clc
// Fundamental of Electric Circuit
// Charles K. Alexander and Matthew N.O Sadiku
// Mc Graw Hill of New York
// 5th Edition
// Part 2 : AC Circuits
// Chapter 11 : AC power Analysis
// Example 11 - 15
clear; clc; close;
//
// Given data
pf_old = 0.8000;
pf_new = 0.9500;
f = 60.0000;
Vrms_mag = 120.0000;
Vrms_angle = 0.0000;
P_load = 4.0000;
Vrms = complex(Vrms_mag*cosd(Vrms_angle),Vrms_mag*sind(Vrms_angle))
// Calculations S1 dan Q1
S1 = (P_load*1000)/pf_old;
Q1 = S1 * sind(acosd(pf_old));
// Calculations S2 dan Q2
S2 = (P_load*1000)/pf_new;
Q2 = S2 * sind(acosd(pf_new));
// Calculations Reactive Power of Capacitors and Capacitance of Capacitors
Qc = Q1 - Q2;
C = Qc/(2*%pi*f*(Vrms_mag)^2);
//
disp("Example 11-15 Solution : ");
printf(" \n a. Qc = Reactive Power of Capacitors = %.3f VAR",Qc)
printf(" \n a. C = Capacitance of Capacitors = %.7f MikroFarad",C*1000000)
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