Question

25] Problem 4 Part a Fig represent the saturation curve at 1200 rpm of a series de motor .The motor has Ka 40 and wound with 8 turns per pole. Total series field and armature-circuit resistance are 25m Ω and m 2, respectively. Draw the equivalent circuit. Part b For a certain load the same motor runs at 200v while taking 325 A in current. If core loss is 220 W, F& W loss is 40 W find Determine Flux per pole at E-200 (part a) (A) (B) Determine Flux per pole at an excitation at 2600 at per pole (part a) (C) Torque developed (part b) (D) Motor speed (Part b) (E) Efficiency (part b) CiiticaT esist 350 250 200 150 100 S0 000 1500 2000 2500 3000 Excitation, At per pole 3500

Answer 1

Part B (A) E = ka phi w_m 200 = 40 times phi times 2 pi tim 1200/60 phi = 200/40 times 40 pi = 0.039 weber/pole (B) excitation emf = 230 at excitation at 2600 per 230 = 40 times phi times 2 pi times 7200/60 phi = 230/40 times 40 pi = 0.045 weber/pole (C) v_t = E_f + I_a(r)_a + r_f) 200 = E_f + 325(75 times 10^- 3) E_f = 175.625 vell Power developed = E_f I_a = 57.07 times 10^3 w \r\n (D) E_1/E_2 = w_1/W_2 200/175.625 1200/N_2 implies N_2 = 1053.75rpm \r\n Torque developed = E_f I_a/w = 57.07 times 1000/2 pi times 1053.75 times 60 = 517.18w (E) eta = 57.07 times 1000/325 times 200 times 100 = 87.8%