Question

A three-phase line with an impedance of (0.25+j1.25) ohms/phase supplies power to three balanced 3-phase parallel connected loads: • Load 1 absorbs a total of 125 kW and 100 KVAR • Load 2 is a delta-connected impedance load with (125+j 40)ohms/phase • Load 3 absorbs 120 kVA at 0.8 power factor (lagging) Given that the line-neutral voltage at the load end of the line is 2400 V, determine the magnitude of the line-to- line voltage at the source end of the line.

Answer 1

Zline = (0.25 +j 1.25)~/ph. Load 1 Po = 125kw, Bi - 100 KVAR S, = {60.078/38-66°) kV4; 0,238.66* Londa o Land Za = (25+)40)upos zs = (3.-244.725 hulph. Convest the s into stas Lond 131,244117-25 Zy = 20 3 2 - (3-4 [n-gº-h. Load 3 S3 = 120KVA , Cosos3 = 0,8 lag 43 – 36.876 = 2400V. Load phase voltage vlohy The single phase Equiment circuit in I. {ZLinef I3 다. (b). b)

3 Load, Silpha 160.078/28.66 ) xr03 silph = 53-359 kVA. silph Il v(ph) I, = 22.233A, Ī, - 11/01 Ij =(22.2331-38-66°)4- 53.359 x103 2.400 2400 43.740 17.75 Loada I Velph) Zy Is =(54.8591-12.78°) 4 4. Load 3 120x103 40 kVA S3/ph= 3 3 40X10-3 = 16.67A Iz s3/ph Vlph) 2400 Iz z 134€ - (1667/-36-87).

II,tiIz 22.2331-38-66*+54.8591-17.75+16:671-36-87' I 17-36-313-89+ 52-247-j16-125+ 13.336-j10 Ī =(22-3534-26.09) 4 A: Vs (ph) a 4 (ph) + 1xZrine 2400+30 +692.353–25.09) (0.25+j1-25) = 2400+jo+71-505 +j93.524. vs 699 2473.274 V. Source line to Line voltage is. V (2) VB X Vs Iph) = BX 24 73.274 yle 4283.836.V.