// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 6 : SYNCHRONOUS MACHINES // EXAMPLE : 6.22 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA E1 = 1100 + (%i*0); // EMFs of two identicel synchronous Generators in Volts per phase E2 = 1100*(cosd(5)-(%i*sind(5))); // EMF in Volts per phase Zl = 1.0 + (%i*1.0); // Load impedance in Ohms per phase Zs1 = 0.15 + (%i*2.1); // Synchronous impedance in Ohms per phase Zs2 = 0.2 + (%i*3.3); // Synchronous impedance in Ohms per phase f = 50; // Frequency in Hertz // CALCULATONS Ys1 = 1/Zs1; // Synchronous Admittance in Ohms per phase Ys2 = 1/Zs2; // Synchronous Admittance in Ohms per ohase Yl = 1/Zl; // Load Admittance in Ohms per ohase V = ((E1*Ys1)+(E2*Ys2))/(Yl+Ys2+Ys1); // Terminal Voltage in Volts per phase (From Millman's Theorem) I1 = (E1-V)/Zs1; // Individual current in Amphere per phase I2 = (E2-V)/Zs2; // Individual current in Amphere per phase P1 = abs(V)*abs(I1)*cosd(atand(imag(V),real(V))-atand(imag(I1),real(I1))); // Per phase actice power in Watts P2 = abs(V)*abs(I2)*cosd(atand(imag(V),real(V))-atand(imag(I2),real(I2))); // Per phase actice power in Watts Ic = (E2-E1)/(Zs1+Zs2); // No-load circulating current in Amphere per phase // DISPLAY RESULTS disp("EXAMPLE : 6.22 : SOLUTION :-"); printf("\n (a) Terminal Voltage per phase, V = %.2f < %.1f V \n",abs(V),atand(imag(V),real(V))) printf("\n (b) Individual currents per phase, I1 = %.f < %.1f A \n\n I2 = %.1f < %.1f A \n",abs(I1),atand(imag(I1),real(I1)),abs(I2),atand(imag(I2),real(I2))) printf("\n (c) Per phase Active Power , P1 = %.f W \n\n P2 = %.1f W \n ",P1,P2) printf("\n (d) No-load current per phase, Ic = %.2f < %.2f A \n",abs(Ic),atand(imag(Ic),real(Ic)))