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+//Chapter 5:Dc Motor Drives
+//Example 21
+clc;
+
+//Variable Initialization
+
+//Ratings of the series motor
+N=600     //speed in rpm
+Vs=220    //source voltage in V
+Ra_Rf=0.12    //combine armature resistance field resistance
+
+//Magnetisation curve at N
+If=[10, 20,30, 40, 50,  60, 70, 80]    //field current in A
+E =[64,118,150,170,184,194,202,210]    //terminal voltage in V
+
+//Solution
+//(i)When the duty ratio is 0.6 and motor current is 60 A
+delta=0.6    //duty ratio
+Ia1=60       //motor current in A
+Va1=delta*Vs //terminal voltage for the given duty ratio
+E1=Va1-Ia1*Ra_Rf      //back emf for the given duty ratio
+
+//For Ia1=60 A the terminal voltage is 194 V as given in the magnetization curve
+N1=E1/E(6)*N          //motor speed for the given duty ratio
+
+//(ii)When the speed is 400rpm and the duty ratio is 0.65
+delta=0.65    //duty ratio
+N2=400        //speed in rpm
+Va1=delta*Vs //terminal voltage for the given duty ratio
+
+//From the magnetization characteristic for the speed of 400rpm the current Ia=70 A
+E1=Va1-If(7)*Ra_Rf      //back emf for the given duty ratio
+T=(E1*If(7))/N2/(2*%pi/60)    //required torque
+
+
+//Results
+mprintf("(i)Hence the motor speed is :%.1f rpm",N1)
+mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)