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// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART II : TRANSMISSION AND DISTRIBUTION
// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES
// EXAMPLE : 3.7 :
// Page number 132-133
clear ; clc ; close ; // Clear the work space and console
// Given data
V_s = 66.0 // Voltage(kV)
f = 50.0 // Frequency(Hz)
l = 150.0 // Line length(km)
r = 0.25 // Resistance of each conductor(ohm/km)
x = 0.5 // Inductive reactance of each conductor(ohm/km)
y = 0.04*10**-4 // Capacitive admittance(s/km)
// Calculations
// Case(a)
R = r*l // Total resistance(ohm)
X = x*l // Inductive reactance(ohm)
Y = y*l // Capacitive resistance(s)
Y_2 = Y/2 // 1/2 of Capacitive resistance(s)
// Case(b)
Z = complex(R,X) // Total impedance(ohm)
A = 1+(Y*exp(%i*90.0*%pi/180)*Z/2) // Line constant
V_R_noload = V_s/abs(A) // Receiving end voltage at no-load(kV)
// Results
disp("PART II - EXAMPLE : 3.7 : SOLUTION :-")
printf("\nCase(a): Total resistance, R = %.1f ohm", R)
printf("\n Inductive reactance, X = %.1f ohm", X)
printf("\n Capacitive resistance, Y = %.1e s", Y)
printf("\n Capacitive resistance, Y/2 = %.1e s", Y_2)
printf("\nCase(b): Receiving end voltage at no-load, V_R = %.2f kV", V_R_noload)
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