Question

A synchronous generator is connected to an infinite bus via a transformer and a network of transmission lines, as illustrated2) A fault occurs at the receiving end of line 1-3, effectively at bus 3, and after the prescribed breaker coordination interBus 1 Bus 2 Line 1-2 Xp=0.1 8p.u. inf Line 1-3 Line 2-3 다D- x0.27p.u. X23-0.15p.u. x-0.10p.u. Bus 3 Figure 1. Power system fo

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Pe fault Conditions single Line diagram Jo18 in q1utus Jo 27 jo-is Jo. Spower an generator delievers (o1py *eul power So geneduping Pefanlt Condition PImas Vint Ep =rosxl:3 O.474 Pmax=.8761 ndf.tauj 22.486 So CuTrent opersting point input= 1pu So imediagram after duning fault me difid Single line So become Canver sion A-Y jo-o418 jo.L jo 062 jo.31E Ja.03438 3LS Jo.06219 koduing fault Condifion So =1-128 mat Pm= 1.1pu Noo when the fault is cleared at S=q0° opening 1 e tinel-3 beuomes Single linehow Now Let the machine dotor behaves See us during fault Condituons all the k after fautt Condition P 4 Az Pm PsT(mat12609 Pfrom the above Curve S1 Sin o.4 581 rad 6 30 2.48181 8, sin = 77.187°= 1.36 Fad. 2801 Ema 180- s,= 1g0- 26, 30- IS3,1z to 6SNows fos decele ating qrea Az Szq00 Smat 123+1 Sing --Por) dst (2.48191: S2 IS3.7 A2- $1 Sing- 11) ds (l-12B09 Sins-1)ds+ 17-

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