// Electric Machinery and Transformers
// Irving L kosow 
// Prentice Hall of India
// 2nd editiom

// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
// Example 8-17

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 );
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
PF_SM = 1 ; // power factor  of the synchronous motor

// 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

kvar_final = kvar_original - kvar_IM ; // 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 8-17 Solution : ");
printf(" \n The synchronous motor operates at the same efficiency as the IM");
printf(" \n that has been replaced, and therefore the total power of the system");
printf(" \n is unchanged. The solution involves construction of table that shows ")
printf(" \n the original condition of the system, the change, and the final condition.\n");
printf(" \n Original kVA = %d kVA \n ", kVA_original );
printf(" \n Original kvar = \n" );disp(%i*kvar_original);

printf(" \n Induction motor kW = %d kW \n ", kW_IM );
printf(" \n Induction motor kVA = %.f kVA \n ", kVA_IM );
printf(" \n Induction motor kvar = ");disp(%i*kvar_IM)

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   0     \t%.1f \t\t 1.0 ",kW_IM,kW_IM);
printf(" \n Final    : \t%d \t\tj%.f \t\t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
printf(" \n __________________________________________________________________________");