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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+// 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 : 3.2 :
+// Page number 128-129
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 10.0                 // Length(km)
+V_s = 11.0*10**3         // Sending end voltage(V)
+P = 1000.0*10**3         // Load delivered at receiving end(W)
+PF_r = 0.8               // Receiving end lagging power factor
+r = 0.5                  // Resistance of each conductor(ohm/km)
+x = 0.56                 // Reactance of each conductor(ohm/km)
+
+// Calculations
+// Case(a)
+R = r*l                             // Resistance per phase(ohm)
+X = x*l                             // Reactance per phase(ohm)
+E_s = V_s/3**0.5                    // Phase voltage(V)
+I = P/(3**0.5*V_s*PF_r)             // Line current(A)
+// Case(b)
+sin_phi_r = (1-PF_r**2)**0.5        // Sinφ_R
+E_r = E_s-I*R*PF_r-I*X*sin_phi_r    // Receiving end voltage(V)
+E_r_ll = 3**0.5*E_r/1000            // Receiving end line to line voltage(kV)
+// Case(c)
+loss = 3*I**2*R                     // Loss in the transmission line(W)
+P_s = P+loss                        // Sending end power(W)
+n = P/P_s*100                       // Transmission efficiency(%)
+// Alternate method
+Z = R**2+X**2
+P_A = 1.0/3*P                       // Load delivered(W/phase)
+Q = 1.0*P*sin_phi_r/(3*PF_r)        // Reactive load delivered(VAR/phase)
+A = (V_s**2/3.0)-2*(P_A*R+Q*X)      // Constant
+B = (1/9.0)*P**2*Z/PF_r**2          // Constant
+const = (A**2-4*B)**0.5             // sqrt(A^2-4B)
+E_r_A = ((A+const)/2)**0.5/1000.0   // Receiving end voltage(kV/phase)
+E_r_A_ll = 3**0.5*E_r_A             // Receiving end line-line voltage(kV)
+I_A = P/(3**0.5*E_r_A_ll*1000*PF_r) // Line current(A)
+loss_A = 3*I_A**2*R                 // Loss in the transmission line(W)
+P_s_A = P+loss_A                    // Sending end power(W)
+n_A = P/P_s_A*100                   // Transmission efficiency(%)
+
+// Results
+disp("PART II - EXAMPLE : 3.2 : SOLUTION :-")
+printf("\nCase(a): Line current, |I| = %.1f A", I)
+printf("\nCase(b): Receiving end voltage, E_r = %.f V (line-to-neutral) = %.2f kV (line-to-line)", E_r,E_r_ll)
+printf("\nCase(c): Efficiency of transmission = %.2f percent \n", n)
+printf("\nAlternative solution by mixed condition:")
+printf("\nCase(a): Line current, |I| = %.1f A", I_A)
+printf("\nCase(b): Receiving end voltage, E_r = %.3f kV/phase = %.2f kV (line-line)", E_r_A,E_r_A_ll)
+printf("\nCase(c): Efficiency of transmission = %.2f percent", n_A)
-- 
cgit