- Question 2 20 A Simple Three Bus Power System Is Shown In Fig Q2 The System Data Is Given In Table Q2 For A Phase T 1 (47.97 KiB) Viewed 98 times
Question 2 [20] A simple three-bus power system is shown in Fig. Q2. The system data is given in Table Q2. For a phase-to-phase fault at busbar 3, calculate the following: (15) The positive phase sequence network. (4) (16) The negative phase sequence network. (3) (17) The reduced (single) positive phase sequence impedance (x₁). (3) (18) The reduced (single) negative phase sequence impedance (x2). (3) (19) The negative phase sequence current component (1₂). (2) (2) (20) The positive phase sequence current component (1₁). (21) The fault current (p.u.). (3)
T3 ني Line 13 Line23 2 t G1 G2 T1 Line 12 T2
Table Q2: Item Sequence reactance (in p.u. referred to same base) X₁-0.18, X₂-0.18, xo=0.10 p.u. G1 G2 X₁=0.18, X₂=0.18, xo=0.10 p.u., earthing reactance = j0.05 X₁-0.08, X2=0.08, xo=0.16 p.u G3 T1 X₁-0.15, X₂-0.15 p.u., Xo = 0.18 T2 T3 X₁ = 0.12, X₂ = 0.12, xo = 0.12p.u. X₁ = 0.10, X₂ = 0.10, Xo = 0.10p.u. X₁ X2 X1, Xo=0.70 p.u Line 12 Line 13 X1 X2 X1, Xo=0.70 p.u Line 23 X₁ X₂=X1, Xo=0.70 p.u The following formula may be useful : BH IR [1 1 1 TIRO ly = 1 a² a IR1 2 IB 1 a a IR2