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+//Chapter 5:Dc Motor Drives
+//Example 23
+clc;
+
+//Variable Initialization
+
+//The motor is operated using dynamic braking method
+N=600     //speed in rpm
+Vs=220    //source voltage in v
+Ra=0.12   // armature resistance in ohms
+delta_min=0.1   //manimum value of duty ratio
+delta_max=0.9   //maximum value of duty ratio
+
+//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) Maximum braking speed is 800rpm with armature current of 70 A
+N1=800   //maximum braking speed in rpm 
+Ia=70    //armature current in A
+E1=N1/N*E(7)   //at 70A motor back emf is 202V 
+Rbe=E1/Ia-Ra   //effective value of braking resistance
+Rb=Rbe/(1-delta_min)   //required braking resistance
+
+//(ii)When the speed of the motor is 87 rpm
+//now torque is maximum when the duty ratio is maximum
+N1=87   //speed in rpm
+R=Rb*(1-delta_max)+Ra
+
+Ia=If(5)   //value of armature current for the given value of E=184V 
+Ke_phi=E(5)/(2*%pi*N)*60
+T=Ke_phi*Ia   //required torque
+
+
+//Results
+mprintf("(i)Hence braking resistance is:%.2f ohm",Rb)
+mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)