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/CH8/EX8.6/Ex8_6.sce | 34 ++++++++++++++++++++++++++++++++++
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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.5
+
+clc;clear; // clears the console and command history 
+
+// Given data
+V = 210      // supply voltage in V
+f = 50       // supply frequency in Hz
+P = 4        // number of poles
+P_0 = 400    // i/p power in W
+I_0 = 1.2    // line current in A
+V_0 = 210    // line voltage
+P_fw = 150   // total friction and windage losses in W
+R = 2.2      // stator resistance between two terminals in ohm
+
+// caclulations 
+R_1 = R/2                               // per phase stator resistance in ohm
+P_scu = 3*I_0^2*R_1                     // copper loss in W
+P_core = P_0-P_fw-P_scu                 // stator core loss in W
+R_0 = (V_0/sqrt(3))^2/(P_core/3)        // no-load resistance in ohm
+// alternate approach
+phi_0 = acosd(P_core/(sqrt(3)*V_0*I_0)) // power factor angle
+X_0 = (V_0/sqrt(3))/(I_0*sind(phi_0))   // magnetizing reactance per phase in ohm
+
+// display the result 
+disp("Example 8.6 solution"); 
+printf(" \n No-load resistance \n R_0 = %.1f ohm \n", R_0 );
+printf(" \n Magnetizing reactance per phase \n X_0 = %.0f ohm \n", X_0 );
+
-- 
cgit