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/EX5.9/Ex5_9.sce | 37 +++++++++++++++++++++++++++++++++++++
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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.9
+
+clc;clear; // clears the console and command history 
+
+// Given data
+I_L= 80          // dc shunt motor current in A
+V_t = 220        // supply voltage in V
+N = 800          // speed in rpm
+R_a = 0.1        // armature resistance in ohm
+R_sh = 50        // shunt field resistance in ohm
+P_if = 1600      // iron and friction losses in W
+
+// caclulations 
+I_sh = V_t/R_sh        // shunt field current in A
+I_a = I_L-I_sh         // armature current
+E_b = V_t-I_a*R_a      // back emf in V
+P_in = V_t*I_L         // i/p power in W
+P_md = E_b*I_a         // mechanical power developed in the armature in W
+P_cu = P_in-P_md       // copper loss in W
+T_a = 9.55*(E_b*I_a)/N // armature torque in N-m
+P_0 = P_md-P_if        // o/p power in W
+T_sh = 9.55*(P_0/N)    // shaft torque in N-m
+n= (P_0/P_in)*100      // efficiency 
+
+// display the result 
+disp("Example 5.9 solution");
+printf(" \n Copper loss \n P_cu = %.2f W \n", P_cu);
+printf(" \n Armature torque \n T_a = %.2f N-m \n", T_a);
+printf(" \n Shaft torque \n T_sh = %.2f N-m \n", T_sh);
+printf(" \n Efficiency \n n = %.0f percent \n", n);
+
-- 
cgit