Question

03 Consider the following one line diagram. The generator is rated at 25 MVA, IN has subtransient reactance of 20 %. The transmission line has a reactance of 10 The subtransient reactance of motorlis 0.345 pu where is that ol second more pu. Each motor has the following ratings. 10 MVA and 11 kV. Both transformers are rated at 25 MVA and have a reactance of 8.05% each; the transformers 1 and 2 have voltage ratio of 11/121 KV and 121/1KV respectively. All the pu/percentage reactances given here are on a base appropriate for analysis without any conversions: More over the above values constitute positive sequence reactances of the nelWoin. The negative sequence reactances are same as positive sequence reactances. lhe zero sequence reactance of gen is 6%, and a current limiting reactancer of 2 ohms connected in the neutral to ground path of generator. The zero sequence reactances or motors 1 and 2 are 8.2 % and 16.4% respectively. The zero sequence reactance or transformers are 8.05% each. The zero sequence reactance of transmission line is JUU ohms. a) Draw labeled positive negative and zero sequence network diagrams of the system. b) Redraw the sequence network diagram for the case when a single line to ground fault occurs at the primary side of transformer 1 (pre fault voltage is 1 pu), clearly showing fault current paths. c) Also find out total fault current mete v Gen TL PAY YA Motos

Answer 1

a) the seg diagram X20-2005 T.L. T OROS ,-04107 X01 & Gon All the values mentioned in the questions were gree on base appropriate scale as mentioned. Mas we didn't have to do any conversions to find them. Calculation for Xpe of transmission line- M = Xr MUA BOX = 0.1707 pu Hare Mure 25 MUA KVA = 121 kV

- re seg diagram XTQ=0.0805 mm 3 X q = 0.2 XT2 = 0.0805 rom- IXTL = 0.1707 om 0.69 = 0.3453 - ne seq diagram un same as the sea- diag gero seq. diagram XTLO T 8-0-0305 3x20 x=0.080S 3X0 0.5122 to 0.050s 0720020 Aside X = 0.062 closed A side shunt closed X = 0.082 3X n = 1023 2 € 491-00% 3X=0 Xho - 300 x 25 = 0.5122pu.

Ti primary La the fault on seq. diag 0.0805 0.1707 soso.o elleer twee recer I If 30.69 If wlower If = If tI f 0.345 منفرد کام mQueen fault impedance: X = 0.2 11(00805 +0:1707+0.0805 + (0.345 11 0.69)] & 00th 859 pa X, = 0.1474pu - ve seq diag 0.080s 0.1707 som to soso.o reele to Ist If 30.69 0.23 YIf 0.345 n fault impedance X = 0.211 (0.0805 +0.1707 + 0.0805 + (0.345/10.69) A = 0. 1474 pm

zero seq 0.080s 22150 trecerea freue frecreuer 0 10.080SF 0.0823 نور wewe mwen 0.06 س تكون الم Xo = 0.06 (see that to the right of the fault, the switch is open to those elements to the right of the fault will not aich affeit the zeno seg) 3E – 3 (1) If la 8. 1474 to 1474 +0.06. + x x + x Il = 8.45spa