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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 6 : SYNCHRONOUS MACHINES
+
+// EXAMPLE : 6.21
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+v = 440;                       // Operating voltage of the Synchronous Motor in Volts
+f = 50;                        // Operating Frequency of the Synchronous Motor in Hertz
+xd = 10;                       // Direct axis reactances in Ohms
+xq = 7.0;                      // Quadrature axis reactances in Ohms
+p = 6;                         // Total number of Poles
+pf = 0.8;                      // Power factor lagging
+i = 10;                        // Motor drawing current in Amphere 
+
+
+// CALCULATIONS
+
+V = v/sqrt(3);                                  // Phase voltage in Volts
+ws = (4*%pi*f)/p;                               // Synchronous speed in Radians per second
+theta = acosd(pf);                              // Power factor angle in degree
+I = 10*(cosd(theta)+(%i*sind(theta)));          // Motor drawing current in Amphere at 0.8 PF leading
+delta = atand((i*xq*cosd(theta))/(V+i*xq*sind(theta)));             // Power angle for motoring mode in degree
+Iq = i*cosd(theta+delta);                                           // Current in Amphere
+Id = i*sind(theta+delta);                                           // Current in Amphere
+Eo = V*cosd(delta) + Id*xd;                                         // Induced EMF in Volts
+P = ((3*V*Eo*sind(delta))/xd)+(3*V^2*((1/xq)-(1/xd))*sind(2*delta))/2;                      // Power in Watts
+T = ((3*V*Eo*sind(delta))/(xd*ws))+(3*V^2*((1/xq)-(1/xd))*sind(2*delta))/(2*ws);            // Torque in Newton-meter 
+
+// when the machine is running as alternator, the magnitude of induced EMF = 323.38V. Let the new current will be Inew at lagging power factor thetanew. Now torque angle is 10.71 deg from phasor diagram Figure 6.51 and page no. 444 we get V+Id*xd*cos(delta)-Iq*xq*sin(delta) = Eo*cos(delta), 254+9.825*Id-1.3Iq = 317.75, 9.825*Id-1.3*Iq = 63.75, 7.56*Id-Iq = 49 and we have Id*xd*sin(delta)+Iq*xq*cosdelta) = Eo*sin(delta), 1.85*Id+6.88*Iq = 60.1, 0.27*Id+Iq = 8.74 by solving these two equations we get Idnew = 123.85/10.095 = 12.27A and Iqnew = 5.43A
+Iqnew = 5.43;                                   // New current in Amphere
+Idnew = 12.27;                                  // New current in Amphere
+Inew = sqrt(Idnew^2 + Iqnew^2);                 // New total Current in Amphere
+// We know that torque angle, tan(delta) = (I*xd*cos(theta))/(V+I*Xq*sin(theta)) so by calutaion for new power factor angle thetanew we get, tan(10.17) = (13.42*7*cos(thetanew))/(254+13.42*7*sin(thetanew)), 0.189(254+13.42*7*sin(thetanew) = 13.42*7*cos(thetanew), 48-93.94cos(thetanew)+17.75*sin(thetanew) = 0 by solving this equatuon we gwt thetanew = 49.5 lagging
+thetanew = 49.5;                                    // New power factor angle in degree
+pfnew = cosd(thetanew);                             // Power factor lagging
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 6.21: SOLUTION :-");
+printf("\n (a) Induced EMF, E = %.2f V \n",Eo)
+printf("\n (b) Power (Torque) angle = %.2f degree \n",delta)
+printf("\n     Power , P = %.2f W \n",P)
+printf("\n     Torque , T = %.2f N-m \n",T)
+printf("\n (c) when the machine is running as alternator requirements are:- \n\n     New Current = %.2f A\n",Inew)
+printf("\n     Power factor = %.3f lagging \n",pfnew)