Question

A three phase power transformer of 100 (KVA), 22 / 12.7 (kV) and Δ-Y has a resistance of 0.15 (pu) and a reactance of 0.6 (pu). The elements of the excitation branch are R_m = 100 (pu) and X_m = 20 (pu)

a) If this transformer supplies a load of 80 (KVA) with an FP = 0.8 delay, draw the phasor diagram of one phase of the transformer.

b) What is the voltage regulation of the transformer bank in these circumstances?

c) Draw the equivalent circuit referred to the low voltage side of a phase of this transformer. Calculate all the transformer impedances relative to the low voltage side.

d) Determine the losses in the transformer and the efficiency of the transformer under the conditions of part b).

Answer 1

100 KVA 22/12.7 (kW) B-1 R=0,ispu X = 0.6P. u Rm = 100po km = 200.0 SL = 80 KVA Pf = 0.8 Lay 32KV Rakv 7 paid fokua PL 80 X103 80 x3 : 6.29. KA V3 VL V2 Y(12.2 xinp3 In -6.ar at 26.86 Lag Rm= 100 Ro Roy U-) = Rp.wix Base I6= krni Rmpu = Rmact X IVA), 100 Rimact X Ol A Line current - Phase corrent Rm = 484kn (22j2 Xpact - Xp.ux (Kb)? 20x 222 = 96.8k Do = 22X162 V 484003 0.0445A Il = -0.22 A 96.8 I, EL =) I 27*6:29 N2 N 22 111 = 3.63A

TEEV 3.631 Here we taken I'=8, only To H B = 3 su 20.224 Iza 6.29A VEN b Voltage regulation J. R Cos« 3 Xsinga = 0.15 X0.870,6806 Ves 0.48.1. с Stuivalent circuit de fered to secondary 12/12-7 ku Roq Xog Mmm 22 leue Roz COIS X mUtb) =0:15X (12.79 = 241.93 0 6 x en = 967.7471 Boz = Roluxor 997.san I'= PL = 6.294 I2 = P2+ Po' = 6:23 +02

core loss Copperlass=2 U2 Rm = 1kw 484 1 Rey = 8.68 (0.15.2 =%6kw 6-29 24 241-93 9.6kw Esticiency = BA VB Pl coster VLE2.C0542 + Lolby 80x0,8 80x0.8 +1+9-6 97.5")