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authorSiddharth Agarwal2019-09-03 18:27:40 +0530
committerSiddharth Agarwal2019-09-03 18:27:40 +0530
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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.15
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+// Refer figure 3.17 page no. 101
+
+S = 500 * 10 ^ 6; // Rating of power transformer in VA
+V1 = 400 * 10^3; // HV side rating of the power transformer in Volts
+V2 = 131 * 10^3; // LV side rating of the power transformer in Volts
+pcu = 5; // Rated Copper loss in Percentage
+pi = 1; // Rated Core loss in Percentage
+
+
+// CALCULATIONS
+
+Pcu = S*(pcu/100); // Rated Copper loss in Watts
+Pi = S*(pi/100); // Rated Core loss in Watts
+kt = 0.25*3 + 0.75*3 + 1*3 + 0.5*3 + 1.0*3 + 0.25*6 + 1.0*3; // From graph figure 3.17 page no. 101
+out = S*kt; // Output energy in kilo-watt-hour
+kt2 = 0.54375; // From graph figure 3.17 page no. 101
+eloss = 24*Pi + S*kt2; // Energy required in losses in kilo-watt-hour { Energy required in losses = 24*Pi + sigma(copper loss * duration)}
+eta = 100*(out/(out+eloss)); // All day efficiency
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.15: SOLUTION :-");
+printf("\n (a) All day efficiency = %.2f percent \n",eta)