// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN // TURAN GONEN // CRC PRESS // SECOND EDITION // CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION // EXAMPLE : 6.10 : clear ; clc ; close ; // Clear the work space and console // GIVEN DATA X_C = 6.2292 ; // commutating reactance when all 3 breakers are closed I_db = 1600 ; // dc current base in A V_db = 125 * 10^3 ; // dc voltage base in V I_d = I_db ; // Max continuous current in A V_d = 100 * 10^3 ; // dc voltage in V alpha = 0 ; // Firing angle α = 0 degree // CALCULATIONS // For case (a) R_c = (3/%pi) * X_C ; R_cb = V_db/I_db ; // Resistance base in Ω V_d_pu = V_d/V_db ; // per unit voltage I_d_pu = I_d/I_db ; // per unit current R_c_pu = R_c/R_cb ; // per unit Ω E_pu = (V_d_pu + R_c_pu * I_d_pu)/cosd(alpha) ; // Open ckt dc voltage in pu V_d0 = E_pu * V_db ; // Open ckt dc voltage in V // For case (b) E = V_d0/2.34; // Open ckt ac voltage on wye side of transformer in V // For case (c) E_1LN = 92.95 * 10^3 ; // voltage in V E_1B = E_1LN ; E_LN = 53.44 * 10^3 ; // voltage in V a = E_1LN/E_LN ; n = a ; // when LTC on neutral X_c_pu = 2 * R_c_pu ; E_1_pu = E_1LN / E_1B ; // per unit voltage cos_delta = cosd(alpha) - ( (X_c_pu * I_d_pu)/( (a/n) *E_1_pu) ) ; delta = acosd(cos_delta) ; u = delta - alpha ; // For case (d) cos_theta = V_d/V_d0 ; // pf of rectifier theta = acosd(cos_theta) ; // For case (e) Q_r = V_d*I_d*tand(theta) ; // magnetizing var I/P // For case (f) d_V = E_LN - E ; // necessary change in voltage in V p_E_LN = 0.00625 * E_LN ; // one buck step can change in V/step no_buck = d_V / p_E_LN ; // No. of steps of buck // DISPLAY RESULTS disp("EXAMPLE : 6.10 : SOLUTION :-") ; printf("\n (a) Open circuit dc Voltage , V_d0 = %.2f V \n",V_d0); printf("\n (b) Open circuit ac voltage on wye side of transformer , E = %.2f V \n",E); printf("\n (c) Overlap angle , u = %.2f degree \n",u) printf("\n (d) Power factor , cosθ = %.3f \n",cos_theta); printf("\n and θ = %.2f degree \n ",theta); printf("\n (e) Magnetizing var input to rectifier , Q_r = %.4e var \n",Q_r); printf("\n (f) Number of 0.625 percent steps of buck required , No. of buck = %.f steps \n",no_buck);