// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 6 : SYNCHRONOUS MACHINES // EXAMPLE : 6.19 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA v = 440; // Operating voltage of the Synchronous Motor in Volts p = 6; // Total number of Poles m = 3; // Number of phase xs = 5; // Synchrouons reactances per phase in Ohms f = 50; // Frequency in Hertz p_hp = 10; // Motor delivering power in HP loss = 1000; // Total iron,copper and friction losses in Watts pf = 0.8; // Power factor lagging I = 10; // Motor drawing current in Amphere at 0.8 PF lagging // 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 Po = p_hp*746; // Output power in Watts Pi = Po+loss; // Input power in Watts eta = (Po/Pi)*100; // Efficiency // we have V = E*cos(delta) - I*xs*sin(theta), 254 = E*cos(delta) - 5*10*0.6, so E*cos(delta) = 254 + 30 = 284 and E*sin(delta) = I*xs*cos(theta) = 5*10*0.8 = 40 by solving these two equations we get delta = atand(40/284) = 8.01 degree delta = 8.01; // Power angle in degree E = 40/sind(delta); // Induced EMF per phase in Volts P = (3*V*E*sind(delta))/(xs*1000); // Power in Kilo-watts T = (P*1000)/ws; // Torque in Newton-meter // DISPLAY RESULTS disp("EXAMPLE : 6.19: SOLUTION :-"); printf("\n (a) Efficiency, eta = %.2f Percent \n",eta) printf("\n (b) Induced EMF, E = %.f V per phase and \n\n Power (Torque) angle = %.2f degree \n",E,delta) printf("\n (c) Power , P = %.4f kW \n",P) printf("\n (d) Torque , T = %.2f N-m \n",T)