Question

Q-1 A 60Hz, 480V/240V,960V A transformer is supplying a 960W passive load at 240V. The resistances of the primary side and secondary side of the transformer are 1.092 and 0.252 respectively. If the no load losses of the transfomer is given by P-Vã 4800 W where Vn is the input voltage of the transformer in Volt. Answer the following: a) Calculate transformer efficiency b) Do you allow this transformer to be connected to 240V 50Hz distribution system? Explain. c What should be the adjustment in the input voltage of the transformer, if the operating frequency changes to 40Hz? Q-2 A 15 kVA, 240V/120 V, 60 Hz,two-winding transformer is to be reconnected as a 240V/360 V step- up autotransformer. From test work on the two-winding transfomer, it is known that its rated voltage core losses and coil losses are 280 W and 300 W, respectively. For this autotransformer, (a) determine the apparent power rating, (b) the full-load efficiency if supplying 360 V to a 0.8 PF lagging load, and (c) the full-load efficiency if supplying 360 V to a unity power factor load. Q-3 A 25 kVA, 2400V/240 V, 60 Hz, two-winding tran sfomer is to be applied in a distribution system with service at 2000V/200V, 50 Hz. (a) Is there any fundamental problem with this application? (b) Determine the apparent power rating in the 50 Hz application

Answer 1

V_1 = 480 V: V_2 = 240 V: f = 60 HZ: s = 960 V A output power P_2 = 960 w i.e cos_2 = 1 {because S = P} Transformation Ratio K = v_2/v_1 = 246/480 = 0.5 R_1 = ohm and R_2 = 0.25 ohm By shifting R_2 to Primary code R^1_2 = R_2/K^r = 0.25/0.5^2 = 1 ohm therefore Total Resistance T/F = R_T = R_1 + R^1_2 = 2 ohm Secondary current I_2 = P_2/V_2 cos _2 = 690/240 times 1 = a A Promany current I_1 = K I_2 = 0.5 times 4 = 2 A (P_cu) copper losses of T/F = I^2_r R_T = 2^2 times 2 = 8 W No load losses P_NL = V_in^2/4800 = 480^2/4800 = 48 W therefore Efficiency eta = P_2/P_2 + P_c4 + P_NL times 100 = 960/690 + 8 + 48 times 100 eta = 64.48 % Actual Rating V_1 = 480 V Actual