1) Consider the power system shown in Fig. 1. Use a power base of 500 MVA...

20=0.2 T ibi e -m ar Section 4.

GOOMVA a Ст3 * Section 2 Section 2 0.1 pce. Gb. T = b. 45 41 A 500 MVA , 26k v. Kg = 0.15 500MVA 0.3pc, 261500kV.,, 600 MVA 13.8kv.

4 0 36 50e Boe. 26 @ Y G1 Anyán : Geci otis 4OOMVA 400MVA Section 1 - 4 - - - Sect Ti 43.8kv, 00,=0.2 = * - 10.1pu. K *.Techlonz OCO=0.1 pus 21-0.2 500 MVA 22 +0.2 13.8 1500 kv. V 30 =0.1 500|13.8kr. Note: Olvide the Netwoork into Gections coherever TR is there.

500MVA Formulae: - (pu) new =(Zpu) old x Sbnew & j krbone Sbo (krbn, Zpu – me Zactual - Chroja * Finding all pue reactances on the base of 500 MVA: Geettoni: Carb)n = 13.8 kr , Cb)n = 500MVA 1. Generation 1: (krbo = 13.8kr, 1660 = 500 MVA . Go 2 130 =(04) 0 = 0.2 =(33)0 =(X)0 and (CO) o=00005 0.1 2. Transformer 1: (krbo = 13.8kr, (Gbl0 = 500 MVA so (8)= 0.1 = 124). Section - 2! (korb)n = 500 kv , Sb)n = 500 MVA i Transmision Line 1 zactiial - krohn.

500) - 5002 Sbo 500 (9) pie 500 50.1 pu = (c2) pu , xo=3364 GO X0 = 0.3pus? 2. Transmicolon Line 2: (OC2)pu = (@ca)pu = 8:00 = 0.16 pue and X0 = 3001 Go 500 = 0.48 pa csscanned with CamScanner

Beetlone (kvb)n = 2687 (Gb) n = 500MVA 1. Generator2 : (krbo = 265V , (Sb)o : 600 MVA (n = (C7)n = 0.15% 5.00 petang) = 0.125 (206 = 0.4x eo (26) - 0.0835 2. Transformer 2: (krbjo = 265V, LCD 30 - 600 MVA X)n = 0.18 oO * Be J*-0.0835 Section 3: (kro) n = 13.8kr, (b)n = 500MVA 1. Generator 3: (876)0 = 13.8kw, (Gb)o -400 MVA 041)n = (%(%) n = 0.2 x 900* 23,81% - 0.25 pue.

(TO)n = 0.1* 500 (13.8.) - 0.125 que 2. Transformer To: (ko)o = 13.6 kv, (Gb)0 - 500MVA Go (x)) = (xo) = 0.4 Pu. Note: pue Impedance on primary Cide Equal to per unit impedance on Secondary Gide for Transformer. +1. pocitéve Sequence nfeo: - — т — т т —т — т— [ 0.1 0.1 30.0833 30.25 $100 0119 (Ztb)= (0,2 +0.1+0.1)|| (0.0835 +0.125) || (0.16 +0.1 +0.25) [Zth)2 = 0.411 0.208333|| 0.51) = 0.41/(0.447911832) (th) = 0.107982240.pu CSScanned with CamScanner 30.2 milioni 50.125

2.

Nega teve Sequence netwoork:- 0.16 b mm 0.1 1 0.1 30.0830 0.25 30.1 30.1 30.125. (Zt52) = (2+h) A = 0.107982914pu B. Zero Gequence netibork:- Bim mam B: 048 02 0. 1 0 .3. 3 0.125 30.0833 .

30.08333 [Zthlo - (0.1 +0.3)|| COD333 +0.9833381 (0.48+0.1) (Zth) o = 0.cell 0.166666 || 0.58. (Zth)0 = 0.097804757 Double Line ground fault coith facelt impedance 36=0 Chet on phase B and cl - * Ia =0 = Taot Tant Iaz Іь Ib +IC = If and Vf =Vbavo = If x26=or. Jal = -(IaQ+ Jaz) If = Jao + a? Ias ta. Iar + Tao + Q .IQ 2 ta tas If = 2.100 + (IQ1+Iaz) @ma) = GIAO Vb = vc ) voo + airbai tavas = vagt avait a vaz Bequence Netzoor ki- Val = vas Тая YPT (17+59) || (Z16))+ (Zth), upp & 1 Lo Ia1 1 Tag f Tao Rzth1) 32+1), 3(zh)3 Let prefault voltage vpf - 100 CSscanned with CamScanner

Tag = ILO , (10.097801757 || 0.1079822) +(0.1079822')) Iaq = 6.9++30888441-90 pu: Tao - Ialx (746) 2 (Z4h )2 + (Z4h)0 = 6.277 308841190 x 0.1079822 j 10.1079822 +0.097807457) Tag E 0-624828589680. Itcoult carrent) = OXIRO 2.8*93.

Jao = 3.293832 3551-90 pu. It (fault Current) = 3x IPO = 9.88.149706ce pe # If anice Gb = 500 x 106 13x Kvb Vax 500x203 - 577.35026 Amps Umre If Actual =(Ifo) pux (Jp) Babe If actual. = 9.88149 x 577.35 = 5.7050849 kamps, CScanned with CamScanner