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authorprashantsinalkar2017-10-10 12:27:19 +0530
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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 : 3.10 :
+// Page number 135
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+f = 50.0 // Frequency(Hz)
+l = 125.0 // Line length(km)
+P_r = 40.0*10**6 // Load at receiving end(VA)
+V_r = 110.0*10**3 // Line voltage at receiving end(V)
+PF_r = 0.8 // Lagging load power factor
+R = 11.0 // Resistance(ohm/phase)
+X = 38.0 // Inductive reactance(ohm/phase)
+Y = 3.0*10**-4 // Capacitive susceptance(S)
+
+// Calculations
+// Case(i)
+E_r = V_r/3**0.5 // Receiving end phase voltage(V)
+Z = complex(R,X) // Total impedance(ohm/phase)
+I_c1 = E_r*(Y/2)*exp(%i*90.0*%pi/180) // Current through shunt admittance at receiving end(A)
+I_r = P_r/(3**0.5*V_r)*exp(%i*-acos(PF_r)) // Load current(A)
+I = I_r+I_c1 // Current through series impedance(A)
+E_s = I*Z+E_r // Voltage across shunt admittance at sending end(V)
+E_s_ll = 3**0.5*E_s/1000.0 // Line to line voltage at sending end(kV)
+I_c2 = E_s*(Y/2)*exp(%i*90.0*%pi/180) // Current through shunt admittance at sending end(A)
+// Case(ii)
+I_s = I_c2+I_r // Sending end current(A)
+angle_Er_Es = phasemag(E_s) // Angle between E_r and E_s(°)
+angle_Er_Is = phasemag(I_s) // Angle between E_r and I_s(°)
+angle_Es_Is = angle_Er_Es-angle_Er_Is // Angle between E_s and I_s(°)
+PF_s = cosd(angle_Es_Is) // Sending end power factor
+
+// Results
+disp("PART II - EXAMPLE : 3.10 : SOLUTION :-")
+printf("\nCase(i) : Line to line voltage at sending end, E_s = %.f kV", abs(E_s_ll))
+printf("\nCase(ii): Sending end power factor = %.3f \n", PF_s)
+printf("\nNOTE: Answers in the textbook are incomplete")