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+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF; 
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+
+PF_SM = 0.8 ; // power factor leading of the synchronous motor
+cos_theta_SM = PF_SM; 
+sin_theta_SM = sqrt( 1 - (cos_theta_SM)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM; 
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+
+eta = 91*(1/100) ; // Efficiency of IM
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA 
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+// case a
+kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW
+kVA_SM = kW_SM / PF_SM ; // Synchronous motor kVA
+kvar_SM = kVA_SM * sin_theta_SM ; // Synchronous motor kvar
+
+kvar_final = kvar_original - kvar_IM - kvar_SM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-18 Solution : ");
+
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+printf(" \n a:");
+printf(" \n Synchronous motor kW = %d kW \n ", kW_SM );
+printf(" \n Synchronous motor kVA = %.f kVA \n ", kVA_SM );
+printf(" \n Synchronous motor kvar = ");disp(-%i*kvar_SM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed  : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added    : \t+%.f  \t\t-j%.2f     \t%.1f \t\t %.1f lead",kW_SM,abs(kvar_SM),kVA_SM,PF_SM);
+printf(" \n Final    : \t%d \t\tj%.2f \t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________\n\n");
+
+printf(" \n b: ");
+printf(" \n In Ex.8-17, a 6148 kVA, unity PF, 7500 hp synchronous motor is needed.");
+printf(" \n In Ex.8-18, a 7685 kVA, 0.8 PF leading, 7500 hp synchronous motor is needed.\n");
+printf(" \n \t Ex.8-18b shows that a 0.8 PF leading,7500 hp synchronous motor ");
+printf(" \n must be physically larger than a unity PF,7500 hp synchronous motor ");
+printf(" \n because of its higher kVA rating.");
+