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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 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.8 :
+// Page number 215
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 33.0 // Line Voltage(kV)
+D_2 = 2.0 // Conductor diameter(cm)
+D_1 = 3.0 // Sheath diameter(cm)
+
+// Calculations
+R_2 = D_2/2 // Conductor radius(cm)
+R_1 = D_1/2 // Sheath radius(cm)
+g_max = V/(R_2*log(R_1/R_2)) // RMS value of maximum stress in the insulation(kV/cm)
+g_min = V/(R_1*log(R_1/R_2)) // RMS value of minimum stress in the insulation(kV/cm)
+
+// Results
+disp("PART II - EXAMPLE : 7.8 : SOLUTION :-")
+printf("\nMaximum stress in the insulation, g_max = %.2f kV/cm (rms)", g_max)
+printf("\nMinimum stress in the insulation, g_min = %.2f kV/cm (rms)", g_min)