Question

b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 +j0 p.u. and the impedance of the electric arc is neglected (Zf-0+)0p.u). The positive, negative and zero sequence impedance parameters of the generator, transmission lines and transformer are given in Figure Q3 x,(1) 30. 15 p.u. 1岚12,-j0.15 p.u. 2 T2)0.15 Figure Q3. Circuit for problem 3b) (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits [4 marks] Determine the positive sequence fault current for the case when a three- (ii) phase-to-ground fault occurs at bus 4 of the network [3 marks] (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 4. Recall that phasors can be expressed in terms of their symmetrical components as where F stands for any three-phase quantity (e.g., current, voltage) [5 marks] (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 4 [5 marks]

Answer 1

The sequence N/W Thevenin equivalent -ve sequence N/W. Thevenin equivalent \r\n Zero sequence N/W Thevenin equivalent For three phi Fault at but (4) taking the positive sequence CRt only. Vertical-bar I_f(1) Vertical-bar fault current = Vertical-bar 1 < 0 degrees/j 0.6 Vertical-bar = 1.67 pre For 1 - phi to graund fault at bus (4) we take all the sequence ckts & connect them in series. \r\n Vertical-bar I_f(1) Vertical-bar = Vertical-bar 1 < 0 degrees Vertical-bar /(0.65 + 0.6 + 0.4) j Vertical-bar Vertical-bar I_f(1) Vertical-bar = 0.61 per Now fault current is given by = 3 I_f(1) = 3 times 0.61 per = 1.82 per For phase to phase fault at bus (4) we take only the & -ve sequence ckts. & connect them in parallel. Vertical-bar I_f(1) Vertical-bar = Vertical-bar 1 < 0 degrees/j 0.6 + j 0.65 Vertical-bar = 0.8per. the fault current is given by = squareroot 3 I_f(1) = squareroot 3 times 0.8 = 1.38 per