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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 2642/CH3/EX3.9/Ex3_9.sce | 34 ++++++++++++++++++++++++++++++++++
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 create mode 100755 2642/CH3/EX3.9/Ex3_9.sce

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diff --git a/2642/CH3/EX3.9/Ex3_9.sce b/2642/CH3/EX3.9/Ex3_9.sce
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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.9
+
+
+clc;clear; // clears the console and command history 
+
+// Given data
+P_i = 350   // iron loss of transformer in W
+P_cu = 650  // copper loss of transformer in W
+kVA = 30    // kVA ratingss of transformer
+pf = 0.6    // power factor
+
+// caclulations 
+P_tloss = (P_i+P_cu)*10^-3   // total full load loss in kW
+P_out = kVA*pf               // o/p power at full load in kW
+P_in = P_out+P_tloss         // i/p power at full load
+n_1 = (P_out/P_in)*100       // efficiency at full load
+kVA_out = kVA*sqrt(P_i/P_cu) // o/p kVA corresponding to maximum efficiency 
+P_01 = kVA_out*pf            // o/p power in W
+P_tloss1 = 2*P_i             // maximum efficiency iron loss=copper loss in W
+P_in1 = P_01+P_tloss1*10^-3  // i/p power in kW
+n_2 = (P_01/P_in1)*100       // efficiency
+
+// display the result 
+disp("Example 3.9 solution");
+printf(" \n Efficiency at full load \n n_1 = %.2f percent \n", n_1);
+printf(" \n Out put power \n P_01 = %.1f kW \n", P_01);
+printf(" \n Efficiency \n n_2 = %.2f percent \n", n_2);
+
-- 
cgit