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+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt
+E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt
+
+theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees
+theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees
+
+R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm
+
+// writing given voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt
+E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt
+
+// writing given impedance(in ohm)in exponential form as follows
+Z_1 = 2.01 * expm(%i * 84.3*(%pi/180) ); // %pi/180 for degrees to radians conversion
+Z_2 = Z_1 ;
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// Calculations
+E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case a
+I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b
+E_gp1 = E_1_mag;
+P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W
+
+// case c
+E_gp2 = E_2_mag;
+P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W
+
+// case d
+// but consider +ve vlaue for P_2 for finding losses, so
+P2 = abs(P_2);
+losses = P_1 - P2 ; // Losses in the armature in W
+
+// E_r_a yields -80 degrees which is equivalent to 100 degrees, so
+theta = 100 - I_s_a ; // Phase difference between E_r and I_a in degrees
+
+check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7
+R_aT = R_a1 + R_a2 ; // total armature resistance of alternator 1 and 2 in ohm
+double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7
+
+// Display the results
+disp("Example 7-9 Solution : ");
+printf(" \n a: I_s = ");disp(I_s);
+printf(" \n    I_s = %.2f <%.2f A \n ",I_s_m, I_s_a );
+
+printf(" \n b: P_1 = %.f W (power delivered to bus)",P_1);
+printf(" \n    Slight variation in P_1 is due slight variations in ")
+printf(" \n    magnitude of I_s,& angle btw (E_gp1,I_s)\n")
+printf(" \n    P_2 = %.f W (power received from bus)\n",P_2);
+
+printf(" \n c: Losses: P_1 - P_2 = %d",losses);
+printf(" \n    Check: E_a*I_s*cos(theta) = %d W ",check );
+printf(" \n    Double check : (I_s)^2*(R_a1+R_a2) = %d W ",double_check );