// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 6 : SYNCHRONOUS MACHINES // EXAMPLE : 6.33 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA v = 6.6*10^3; // Operating voltage of the Synchronous motor in Volts P = 5*10^6; // Operating power of the Synchronous motor in Watts pf = 1.0; // Power factor xd = 3.0; // Direct axis reactances in Ohms xq = 1.0; // Quadrature axis reactances in Ohms eta = 0.98; // OPerating efficiency // CALCULATIONS V = v/sqrt(3); // Per phase voltage in Volts I = P/(eta*v*sqrt(3)); // Line current in Amphere delta = atand((xq*I)/v); // power angle in degree E = v*cosd(delta) + xd*I*sind(delta); // Induced EMF in Volts Tmax = ((3*E*V*sind(90))/xd) + ((3*V^2*sind(180))/2)*((1/xq)-(1/xd)); // Maximum electromagnetic torque in N-m T = ((3*E*V*sind(delta))/xd) + ((3*V^2*sind(2*delta))/2)*((1/xq)-(1/xd)); // Actual electromagnetic torque in N-m Ratio = Tmax/T; // Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque // DISPLAY RESULTS disp("EXAMPLE : 6.33: SOLUTION :-"); printf("\n (a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is %.2f \n",Ratio) printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" ); printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) delta = 2.41 instead of %.2f degree \n ",delta); printf("\n (b) E = 6379 instead of %.2f V \n ",E); printf("\n (c) Ratio = 10.84 instead of %.2f \n ",Ratio); printf("\n From Calculation of the Power angle (delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle (delta) value is WRONGLY calculated and the same used for the further Calculation part \n")