Question

Use power base of 100MVA to change to per unit

Consider the transmission system in figure 1 below Gen 1 Тx 2 Line 1 Gen 2 Tx3 Line 4 Line 3 R F Tx 1 Gen 3 Line 2 Figure 1 Impedance and winding details are:- Generators: Gen 1 132kV, Z=0+j12.5% on 25MVA Gen 2 132kV, Z= 0+j12.5% on 25 MVA 20KV, Z 0+j10% on 25 MVA A- connected windings Gen 3 Transformers Earthed Y windings (both sides) Earthed Y windings (both sides) Windings A-connected 66kV side; Тx 1 20/66kV, Z 0+j10% on 25MVA 132/66KV, Z 0+j12% on 50MVA 66/11KV, Z= 0+j10% on 25MVA Тx 2 Tx 3 earthed Y on the 11 kV side Transmission Lines: 132KV, Z 0+j1 N/phase 66KV, Z 0+jl O/phase 66kV, Z 0+j2 2/phase 1kV, Z 0+j O/phase Line 1 Line 2 Line 3 Line 4 Calculate the steady-state, three-phase short-circuit current in Amperes that will flow (a) if a fault is applied at point 'F' (b) Calculate the steady-state, single-phase to earth current in Amperes that will flow if a single-phase to earth fault is applied at point 'F' Ignore resistances in your calculations. Generator and transformer impedances are shown in per cent, on each machine's own rating as power base. Line impedances are shown in ohms. Assume zero sequence impedances of all lines are 2.5 x the positive sequence values A protection relay is located at point 'R'. The relay is set to trip if the current passing through it reaches 0.5 A. The current transformer ratio is 1000:5 Amps. Will the relay detect the fault at point 'F? с)

Answer 1

132KV 25 MVA 500VA 132/66 kv 25 MVA 29 2SMVA 66/11 KV -122 250VA 2oku 20/66 kv 25 25 YA Shase = 100MVA (hiver) IX MVA new a P.uolat plnen = MVA old z N bore kr- MUA Kuha = 132kU You hy al 3 1321 - 174.24 2, O = 0.5 pon 1125 /1 G - 2Bu = ی lo ی تا G + 2003 2526317 ospu 6Kupa 2020 202 203 y Ti ?u= 1X ( 2 ) x 100 = 0.4 pin. 1, 2012 () p Ty + P * O pen Similante for lines; م

L, 2pn 14 (132) x 100 = Up.m. 1 > 2pu pu. L₂ + 2pn : 2 (bez upu. Ly - Zpu 2140 eur your (a) (kt during fault looks like; 0.4 Tº u ở 1 24 = ((o-s //0.5 +4+.24) || (+0-3+4)] + " to = [(-25 + 9 +-24) |/ 8 43 + 8-4 = (y. 49/18 43 +8.4 = 11-12 = 0.0899 pu 7.12 slot%3 10 5248.79 Thanx = - in Juilt ana, I = 100 x 10 xxo 3xlixio - Tre Ijas (A) = 0.0899 x 524 8.79 = 471.86 A

The cht. unde line to grord Juult looks like: (no seg. network) Lim un seg a 2.5x the sea 2) lo don to the "Clearly no to sog. cunt will you A conted winding of T3 (c) As clear from part (6) solution, No aut will flow and hum Relay at R will not be able to detect the fault at point F.