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/CH5/EX5.6/Ex5_6.sce | 32 ++++++++++++++++++++++++++++++++
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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.6
+
+clc;clear; // clears the console and command history 
+
+// Given data
+I_L1 = 5      // dc shunt motor current
+V_t = 230     // terminal voltage in V
+N_1 = 1000    // speed in rpm
+R_a = 0.2     // armature resistance in ohm
+R_F = 230     // field resistance in ohm
+I_L2 = 30     // dc shunt motor current
+
+// caclulations 
+// at noload condition
+I_sh = V_t/R_F         // shunt field current in A
+I_a1 = I_L1-I_sh       // armature current in A
+E_b1 = V_t-I_a1*R_a    // back emf in V
+// under load condition
+I_a2 = I_L2-I_sh       // armature current in A
+E_b2 = V_t-I_a2*R_a    // back emf in V
+N_2 = (E_b2/E_b1)*N_1   // motor speed under load condtion in rpm
+
+// display the result 
+disp("Example 5.6 solution");
+printf(" \n Speed under load condition \n N_2 = %.1f rpm \n", N_2 );
+
-- 
cgit