Question

1. A n+350-KVA 2300+n - 230+n; volt, 60-hz transformer has percent core loss , resistance and percent leakage reactance of m+0.56 ,m+ 33 and m+5.57 respectively. For 10 increments of power-factor angle, tabulate and plot percent regulation, and efficiency. Power factor for operation between 0.5 lagging and 0.5 leading. Use Matlab

n=20

m=.2

Answer 1

clc;
close all;
clear all;
pfangle=acosd(0.5);
theta=pfangle:-10:-pfangle;
ironloss=0.56;
r=1.33;
x=5.57;
Vreg=zeros(1,length(theta));
effi=zeros(1,length(theta));
for n=1:length(theta)
Vreg(n)=(r*cosd(theta(n)))+(x*sind(theta(n)));
effi(n)=(cosd(theta(n)))/(cosd(theta(n))+1.33+0.56);
end
subplot(2,1,1);
title('Voltage Regulation');
plot(theta,Vreg,'r');
subplot(2,1,2);
title('Efficiency Vs Powerfactor angle');
plot(theta,effi,'g');

clc;
close all;
clear all;
pfangle=acosd(0.5);
theta=pfangle:-10:-pfangle;
ironloss=0.56;
r=1.33;
x=5.57;
Vreg=zeros(1,length(theta));
effi=zeros(1,length(theta));
for n=1:length(theta)
Vreg(n)=(r*cosd(theta(n)))+(x*sind(theta(n)));
effi(n)=(cosd(theta(n)))/(cosd(theta(n))+1.33+0.56);
end
subplot(2,1,1);
title('Voltage Regulation');
plot(theta,Vreg,'r');
subplot(2,1,2);
title('Efficiency Vs Powerfactor angle');
plot(theta,effi,'g');clc;
close all;
clear all;
pfangle=acosd(0.5);
theta=pfangle:-10:-pfangle;
ironloss=0.56;
r=1.33;
x=5.57;
Vreg=zeros(1,length(theta));
effi=zeros(1,length(theta));
for n=1:length(theta)
Vreg(n)=(r*cosd(theta(n)))+(x*sind(theta(n)));
effi(n)=(cosd(theta(n)))/(cosd(theta(n))+1.33+0.56);
end

table=[cosd(theta); Vreg; effi]';

subplot(2,1,1);
title('Voltage Regulation');
plot(theta,Vreg,'r');
subplot(2,1,2);
title('Efficiency Vs Powerfactor angle');

plot(theta,effi,'g');

Table =

0.5000 5.4888 0.2092
0.6428 5.1218 0.2538
0.7660 4.5992 0.2884
0.8660 3.9368 0.3142
0.9397 3.1548 0.3321
0.9848 2.2770 0.3426
1.0000 1.3300 0.3460
0.9848 0.3426 0.3426
0.9397 -0.6553 0.3321
0.8660 -1.6332 0.3142
0.7660 -2.5615 0.2884
0.6428 -3.4120 0.2538
0.5000 -4.1588 0.2092