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+// chapter 14
+// example 14.8
+// Determine no load speeds, firing angle, supply power factor and the speed regulation
+// page-868-870
+clear;
+clc;
+// given
+P=150; // in HP (power rating of motor)
+E0=650; // in V (voltgae rating of motor)
+N1=1750; // in rpm
+Eac=460; // in V (three-phase input)
+Ia=170; // in A (rated armature current)
+Ra=0.099; // in ohm (armature resistance)
+La=0.73; // in mH (armature inductance)
+Ka_phi=0.33; // in V/rpm (motor voltage constant)
+alpha1=0, alpha2=30; // in degrees (firing angle)
+neta=10; // in % (armature current at no load as a percentage of rated armature current)
+// calculate
+E=Eac/sqrt(3);
+Em=E*sqrt(2);
+Ea1=(3*sqrt(3)/%pi)*Em*cosd(alpha1);
+Eb1=Ea1-Ia*Ra;
+N_NL1=Eb1/Ka_phi;
+Ea2=(3*sqrt(3)/%pi)*Em*cosd(alpha2);
+Eb2=Ea2-Ia*Ra;
+N_NL2=Eb2/Ka_phi;
+Eb3=Ka_phi*N1;
+Ea3=Eb3+Ia*Ra;
+// since Ea=(3*sqrt(3)/%pi)*Em*cosd(alpha), therefore we get
+alpha3=acosd((Ea3/Em)*(%pi/(3*sqrt(3))));
+I_A=sqrt((1/%pi)*Ia^2*(2*%pi/3));
+VA=3*E*I_A;
+Ps=Ea3*Ia;
+PF=Ps/VA;
+Eb4=Ea3-((neta/100)*Ia*Ra);
+N_NL3=Eb4/Ka_phi;
+SR=((N_NL3-N1)/N1)*100;
+printf("\nThe no load speeds are");
+printf("\n\t\t\t For alpha=%.f degree, N_NL= %.2f rpm",alpha1,N_NL1);
+printf("\n\t\t\t For alpha=%.f degree, N_NL= %.2f rpm",alpha2,N_NL2);
+printf("\n\nThe firing angle to obtain a speed of N=%.f rpm is \t alpha= %.2f degree",N1,alpha3);
+printf("\nThe supply power factor is \t\t\t\t PF= %.2f",PF);
+printf("\nThe speed regulation is \t\t\t\t\t SR= %.2f %%",SR);
+// Note: The answer varies slightly due to precise calculations
+\ No newline at end of file