From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

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 1820/CH5/EX5.3/Example5_3.sce | 31 +++++++++++++++++++++++++++++++
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+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+D = 1.235 ; // Inside diameter of sheath in inch
+d = 0.575 ; // Conductor diameter in inch
+kv = 115 ; // Voltage in kV
+l = 6000 ; // Length of cable in feet
+r_si = 2000 ; // specific insulation resistance is 2000 MΩ/1000ft . From Table 5.2
+
+// CALCULATIONS
+// For case (a)
+r_si0 = r_si * l/1000 ;
+R_i = r_si0 * log10 (D/d) ; // Total Insulation resistance in MΩ
+
+// For case (b)
+P = kv^2/R_i ; // Power loss due to leakage current in W
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.3 : SOLUTION :-") ;
+printf("\n (a) Total insulation resistance at 60 degree F , R_i= %.2f MΩ \n",R_i) ;
+printf("\n (b) Power loss due to leakage current , V^2/R_i = %.4f W \n",P) ;
+
+printf("\n NOTE : ERROR : Mistake in textbook case (a) \n") ;
-- 
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