Question

A 400-MVA, 200-kV/20-kV, three phase Y-Y transformer has an equivalent series impedance of ZTL = 0.012 + 20.0162 per phase referred to the low-voltage side (secondary side). The transformer is supplying a three-phase load of 300-MVA, 0.8 power factor lagging at terminal voltage of 20 kV (line-to-line) on its low-voltage side (secondary side) as shown in Fig. 22. The primary is supplied from a transmission line with an impedance of ZL = 0.5 + j1.0 per phase. The line-to-neutral load terminal voltage can be chosen as a reference 400 MVA V = 20 kV V.? Z = 0.5+ j1.09 200/20 kV Z = 0.012 + 10.016 2 GC Y 300 MVA 0.8 pf (lag) Vier = 20 kV Fig. 22

Answer 1

(a) S: Load MVA = 300 MVA Load voltage, V = 20 kV (line-to-line) Load pf = 0,8 (lagging). Iload - L o ad] L-fos (1) p. 5) Considering line to central - reference :- 623920x0,8) Z-C5'(0.8) T10.8253_-36.87° k4] Lv3v losy (6) I Lead is the current an LV side of the transformer, Ip is the current on the tv side of the trausformer. I = Tkazad na vool) - (10.8253_-36.87°) x 2) = 12000 dosi 1108253 286,87) KA (C) Vload on = 20 LB LV Ilond = 10,82532-36.87° kA. Zload = (oad pu) =(293) 20 10182534 6)= 1,0667_36.87's = 16.85333 +30:64)

& Zrl= (0.012 + 30.016). ZH = (N) 2 ZIL = 200)? [0012 +30.016)] = (1.2+j16).21 el Mo Load impedance referred to the primary (Hu) side' of the transformer, Zload = (200) zwad = (85-33 +364.00) The per phase equivalent ckt, referred to the His side of the transformer is shown in fig. ū 2=0.54j10us ZH = (85.33+)64.00). ELO =(1-24; 116) figl v=Tg Road [Applying potential divider sule] Line-to-line load terminal voltage = 20kr (on the LV side of the transformers " wad 6:57}}0)+(1:1203503) - 20LO) (1+265 (85.33+j64.00) = 118.637/0.55417° kV. Generates terminal roltage, line-to-oneutral) = 118.637kV lVg 1 (line-to-line) = 131 Vgpul 205.4853 kV