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//Book - Power System: Analysis & Design 5th Edition
//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye
//Chapter - 10 ; Example 10.2
//Scilab Version - 6.0.0 ; OS - Windows
clc;
clear;
Irelay=200 //Current through the relay in Amperes
CTratio=100/5; //CT ratio
Zs=0.082; //Secondary resistance of a 100:5 CT in Ohm
IZB=[8 0.8; 8 3]; //Secondary output current in Amperes and burden resistance in Ohm
E=(Zs+IZB(1,2))*IZB(1,1); //Secondary Excitation voltage in Volts
Ie=0.40 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
I=CTratio*(IZB(1,1)+Ie); //Primary current of the CT in Amperes
printf('\nCase: a');
if (Irelay>I) then
printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(1,2),I)
else
printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(1,2),I);
end
E=(Zs+IZB(2,2))*IZB(2,1); //Secondary Excitation voltage in Volts
Ie=30 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
I=CTratio*(IZB(2,1)+Ie); //Primary current of the CT in Amperes
printf('\n\nCase: b');
if (Irelay>I) then
printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(2,2),I)
else
printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(2,2),I);
end
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