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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 IV : UTILIZATION AND TRACTION
+// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS
+
+// EXAMPLE : 1.23 :
+// Page number 706
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+hp = 25.0 // Motor rating(hp)
+T = 100.0/60 // Heating time constant(hour)
+theta = 40.0 // Temperature rise(°C)
+t = 0.5 // Time(hour)
+n = 0.85 // Motor maximum efficiency
+
+// Calculations
+output = hp*735.5/1000 // Output of motor(kW)
+output_max = output*n // Power at maximum efficiency(kW)
+theta_f2 = theta/(1-exp(-t/T)) // θ_f2(°C)
+loss = 1+(output/output_max)**2 // Losses at 18.4 kW output in terms of W
+P = ((theta_f2/theta*loss)-1)**0.5*output_max // Half-hour rating of motor(kW)
+P_hp = P*1000/735.5 // Half-hour rating of motor(hp)
+
+// Results
+disp("PART IV - EXAMPLE : 1.23 : SOLUTION :-")
+printf("\nHalf-hour rating of motor, P = %.f kW = %.1f hp (metric)\n", P,P_hp)
+printf("\nNOTE: ERROR: Calculation mistake from final temperature rise onwards in textbook")