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+// chapter 14
+// example 14.4
+// fig. 14.13
+// Determine motor torque, speed of the motor, supply power factor, firing angle and power fed back to the supply
+// page-865-866
+clear;
+clc;
+// given
+HP=10; // in HP (power of motor)
+E=210; // in V
+N=1000; // in rpm (speed)
+Ia=30; // in A (armature current)
+Ra=0.25; // in ohm (armature resistance)
+Es=230; // in V (supply voltage)
+Ka_phi=0.172; // in V/rpm (motor voltage constant)
+alpha=45; // in degree
+// calculate
+Em=Es*sqrt(2); // calculation of peak voltage
+// part- (a)
+Ka_phi_rad=Ka_phi*(60/(2*%pi)); // changing unit from V/rpm to V/rad/s
+T=Ka_phi_rad*Ia; // calculation of motor torque
+Ea=(2*Em/%pi)*cosd(alpha); // calculation of armature voltage
+Eb=Ea-Ia*Ra; // calculation of back emf
+N=Eb/Ka_phi; // calculation of speed
+EI=Es*Ia; // calculation of volt-ampere rating
+Ps=Ea*Ia; // calculation of supplied power
+Pf=Ps/EI; // calculation of power factor
+printf("\nThe motor torque is \t\t T=%.1f Nm",T);
+printf("\nThe speed of the motor is \t N=%.2f rpm",N);
+printf("\nThe supply power factor is \t Pf=%.2f",Pf);
+// part  (b)
+Ea2=-Eb+Ia*Ra; // calculation of back emf when in regenerative action Here minus sign indicates regenerative action
+// since Ea=(2*Em/%pi)*cosd(alpha), therefore we get
+alpha2=acosd(Ea2*(%pi/(2*Em))); // calculation of corresponding firing angle
+Pg=Eb*Ia; // calculation of power from dc machine
+Pr=Ia^2*Ra; // calculation of power lost in armature resistance
+Ps=Pg-Pr; // calculation of supplied power
+printf("\n\nThe firing angle to keep the motor current at its rated value is \t alpha2=%.2f degree",alpha2);
+printf("\nThe fpower fed back from the supply is \t Ps=%.1f W",Ps);
+// Note :The answers vary due to precise calculations
+\ No newline at end of file