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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 8-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 10000 ; // kVA rating of a system
+cos_theta = 0.65 ; // power factor of the system
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+cos_theta_b = 0.85 ; // Raised PF
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 );
+cost = 60 ; // cost of the synchronous capacitor to improve the PF in dollars/kVA
+// neglect the losses in the synchronous capacitor
+
+// Calculations
+// case a : For unity PF
+// at the original load
+kW_a = kVA * cos_theta ; //
+theta = acosd(cos_theta) ; // Power factor angle of the system in degrees
+kvar = kVA * sind(theta) ; // Reactive power in kvar
+kVA_a = kvar ;
+cost_cap_a = kvar * cost ; // Cost of raising the PF to unity PF in dollars
+
+// case b
+theta_b = acosd(cos_theta_b) ; // Power factor angle of the system in degrees
+kVA_b = kW_a / cos_theta_b ; // kVA value reduction
+kvar_b = kVA_b * sind(theta_b) ; // final kvar value reduced
+kvar_add = kvar - kvar_b ; // kvar of correction added
+
+cost_cap_b = kvar_add * cost ; // Cost of raising the PF to 0.85 PF in dollars
+
+// Display the results
+
+disp("Example 8-10 Solution : ");
+printf(" \n    Note : Slight variations in the kvar and cost values are due to  ");
+printf(" \n    non-approximation of theta values while calculating in scilab.\n");
+printf(" \n a: At the original load,\n");
+printf(" \n    kW = %d kW at theta = %.1f degrees \n", kW_a , theta );
+printf(" \n    kvar = %.3f kvar\n\n    For unity PF,",kvar);
+printf(" \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kVA_a);
+printf(" \n    Cost of synchronous capacitor = $%.f \n\n",cost_cap_a );
+
+printf(" \n b: For %.2f, PF = cos(%.1f), the total power,",cos_theta_b, theta_b);
+printf(" \n    %.f kW,remains the same. Therefore,\n ",kW_a);
+printf(" \n    kVA of final system reduced to = %.f kVA \n",kVA_b);
+printf(" \n    kvar of final system reduced to = %.f kvar \n    Therefore,",kvar_b);
+
+printf(" \n    kvar of correction added = %.3f kvar\n ",kvar_add);
+printf(" \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kvar_add);
+printf(" \n    Cost of synchronous capacitor = $%.f",cost_cap_b );
+printf(" \n    or less than half the cost in part(a)");