Answers
Soln Taking 10.5kV; 250 MUA base at generator side. (i) calculating line reactance Xine = 0.2 x 50=100 Z Bare = (VBase) ² Ubase = 1 10 ku SBCL&L = 2 6 TUA (110) co = 4.8.42 ZB = Xune pus = Xine Cactual) ZB 10.
48.4 (line pu) = 0. 2066 pu) (11) Calculating line current 90 Peu = Kepul lipul coso p=40mw V₂ = 110kv 250 = (110) - lipul. (0-8) Cos$ = 0.6 Ipu 1 = 0.2 pu lagging
IL (p = 0-2 /-cos' (0.8) (IL (pu) = 0.2 [-36.87 pu) (iii) Calculating pre fault voltage L j0.1 pu | j0.1033 j0-10331. Ji (pu 81 млн т ізден a = 1 LO PU jXJ jxline jx line tin Vf= prefault Voltage at fault point applying KUL Up = keine). Cai' + V VA = (10.1033).(0-2 2-3687) + 160 V = 1.0125 Lo.g4 pu)
(10) calculating sub transient thevenin reactance cet fault point (replacing woltage source with short ect) jo.1 jo 1033 jo.1033 JXT j xeline I jxline jo.25 Éjx": PU nmnom XTH XTH = (xd"+xt + Xline) || ( Meine) XTH = 0.25 +0:1 + 0.1033) || (0-1033) XTH = (0.4533).(0.1033) 10.4533) + (0-1033) XTH = 0.0841 pu ( calculating 30 subtransient fault current VF = 1.0125 (0.94 pu If you XTH = 0.0841 pu
I = 1.012520.94 jo.0841 If = 12.0352 L-89.06 pu Fault current If I = 12.0352 pu Bare current SB IB = 3.
.
VB fault current in amp. If = 1 If lpu X IB If = (12.0352)x(13)2.16) If = 15.79 KA IB = IB² 250x106 JBX 1102 3 x 110x103 IB a 1312.16 Amp
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