From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

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 3472/CH10/EX10.14/Example10_14.sce | 48 ++++++++++++++++++++++++++++++++++++++
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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 : 3.14 :
+// Page number 144
+clear ; clc ; close ; // Clear the work space and console
+funcprot(0)
+
+// Given data
+f = 50.0                      // Frequency(Hz)
+r = 0.1                       // Resistance(ohm/km)
+l = 1.4*10**-3                // Inductance(H/km)
+c = 8.0*10**-9                // Capacitance(F/km)
+g = 4.0*10**-8                // conductance(mho/km)
+V_r = 400.0                   // Receiving end voltage(kV)
+x = 200.0                     // Length of line(km)
+
+// Calculations
+V_2 = V_r/3**0.5                         // Receiving end phase voltage(kV)
+z = r+%i*2*%pi*f*l                       // Total impedance(ohm/km)
+y = g+%i*2*%pi*f*c                       // Total susceptance(mho/km)
+Z_c = (z/y)**0.5                         // Surge impedance(ohm)
+gamma = (z*y)**0.5                       // γ
+// Case(i)
+V_0_plus = V_2/2                         // Incident voltage to neutral at receiving end(kV)
+// Case(ii)
+V_0_minus = V_2/2                        // Reflected voltage to neutral at receiving end(kV)
+// Case(iii)
+gamma_l = gamma*x                        // γl
+V_1_plus = (V_2/2)*exp(gamma_l)          // Incident voltage to neutral at 200 km from receiving end(kV)
+V_1_minus = (V_2/2)*exp(-gamma_l)        // Reflected voltage to neutral at 200 km from receiving end(kV)
+// Case(iv)
+V_1 = V_1_plus+V_1_minus                 // Resultant voltage to neutral(kV)
+V_L = abs(V_1)                           // Resultant voltage to neutral(kV)
+V_L_ll = 3**0.5*V_L                      // Line to line voltage at 200 km from receiving end(kV)
+
+// Results
+disp("PART II - EXAMPLE : 3.14 : SOLUTION :-")
+printf("\nCase(i)  : Incident voltage to neutral at receiving end, V_0_plus = %.1f∠%.f° kV", abs(V_0_plus),phasemag(V_0_plus))
+printf("\nCase(ii) : Reflected voltage to neutral at receiving end, V_0_minus = %.1f∠%.f° kV", abs(V_0_minus),phasemag(V_0_minus))
+printf("\nCase(iii): Incident voltage to neutral at 200 km from receiving end, V_1_plus = (%.3f+%.2fj) kV", real(V_1_plus),imag(V_1_plus))
+printf("\nCase(iv) : Resultant voltage to neutral at 200 km from receiving end, V_L = %.2f kV", V_L)
+printf("\n           Line to line voltage at 200 km from receiving end = %.2f kV", V_L_ll)
-- 
cgit