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// Book - Power System: Analysis & Design 5th Edition
// Authors - J. Duncan Glover, Mulukutla S. Sharma, Thomas J. Overbye
// Chapter - 5 : Example 5.8
// Scilab Version 6.0.0 : OS - Windows
clc;
clear;
N = 6; // Number of transmission lines
Vs = 765; // Transmission voltage in kV
l = 167; // Intermediate substations distance in km
Pl=9000; //Load power , value taken from Example 5.7
lambda = 5000; // Wavelength in km
Beta = (2*%pi)/lambda; // Taken from Eq 5.4.15
L = 500; // Eqivalent pi circuit lenght in km
Zc = 266; // Characteristic impedance of the line in Ohm
X = Zc*sin(Beta*L); // Series reactance in Ohm ; taken from Eq 5.4.10
Xeq = ((1/5)*((2/3)*X))+((1/4)*(X/3)); // Equivalent reactance of five lines with one line section out of service in Ohm
Vr = 0.95*765; // Receiving end voltage in kV
delta = 35; // Phase angle in degree
P = ((Vs*Vr)/Xeq)*sind(delta); // Real power delivered in MW ; taken from Eq 5.4.26
printf('Series reactance is (X) = %0.2f Ohm', X);
printf('\nEquivalent reactance of five lines with one line section out of service is (Xeq) = %0.2f Ohm', Xeq);
printf('\nReal power delivered is (P) = %0.0f MW', P);
if 0.97*P>Pl //Assuming 3% as losses
printf('\n\nThe five instead of six 765-kV lines can transmit the required power in Example 5.7')
end
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