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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.17
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+
+V = 440;                       // Operating voltage of the Synchronous Motor in pu
+E = 200;                       // Induced voltage in Volts
+xs = 8.0;                      // Synchronous reactance in Ohms
+f = 50;                        // Frequency in Hertz
+pa = 36;                       // Power angle in degree
+
+
+// CALCULATIONS
+
+v = V/sqrt(3);                 // Rated phase voltage in Volts
+ws = 2*%pi*f;                  // Synchronous speed in Radians per second
+// To calculate the power factor angle refer page no 438 n figure 6.50
+// Since E*cos(delta) < v so Power factor is lagging, let power factor angle be theta from ohasor diagram figure 6.50:- page no. 438
+// v = E*cos(delta) + I*xs*sin(theta), I*sin(theta) = (254-0.809*200)/8 = 11.525
+// Similarly, E*sin(delta) = I*xs*cos(theta), I*cos(theta) = (200*0.59)8 = 14.70
+// From above two equations, tan(theta) = 0.784
+theta = -38.1;                                  // Power factor angle in degree (minus sign because of lagging)
+pf = cosd(theta);                               // Power factor lagging
+I = 14.7/cosd(theta);                           // Line current in Amphere (I*cos(theta) = 14.7) 
+p = 3 * v * 14.7;                               // Input to motor in watts ( p = 3*V*I*cos(theta), I*cos(theta) = 14.7) 
+P = (3*E*v*sind(pa))/(xs*1000);                 // Power in Kilo-watts
+T = (P*1000)/ws;                                // Torque in Newton-meter
+// For Power factor unity
+// let the current will be I2, thus 3*v*I2 = 3*v*I*cos(theta) , I2 = I*cos(theta) = 14.10 A
+// let ecitation will be E2, thus v = E2*cos(delta2) and E2*sin(delta2) = I2*xs, E2*cos(delta2) = 254 and E2*sin(delta2) = 117.60, by solving these two equations we get E2 = sqrt(254^2+117.6^2) = 279.90 V and delta2 = atand(117.6/254) = 24.84 degree
+E2 = 279.90;
+delta2 = 24.84;
+P_1 = (3*v*E2*sind(delta2))/(xs*1000);                     // Power in kilo-watts
+T_1 = (P_1*1000)/ws;                                       // Torque in Newton-meter
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.17: SOLUTION :-");
+printf("\n (a) Line current, I = %.2f A \n",I)
+printf("\n (b) Power factor angle = %.1f degree \n",theta)
+printf("\n (c) Power , P = %.3f kW \n",P)
+printf("\n (d) Torque , T = %.2f N-m \n",T)
+printf("\n (e) Power factor = %.2f lagging \n",pf)
+printf("\n To make the Power factor to UNITY requirements are:- \n (a) Excitation EMF, E = %.2f V \n",E2)
+printf("\n (b) Power angle = %.2f degree \n",delta2)
+printf("\n (c) Power , P = %.3f kW \n",P_1)
+printf("\n (d) Torque , T = %.2f N-m \n",T_1)