Submission Date: June 03, 2022 PROBLEM 2. A single-line diagram of the power system shown in the Figure, where negative-

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Submission Date June 03 2022 Problem 2 A Single Line Diagram Of The Power System Shown In The Figure Where Negative 1 (293.89 KiB) Viewed 37 times

Submission Date: June 03, 2022 PROBLEM 2. A single-line diagram of the power system shown in the Figure, where negative- and zero-sequence reactances are also given. The neutrals of the generator and A-Y transformers are solidly grounded. The motor neutral is grounded through a reactance X₂ = 0.05 per unit on the motor base. (a) Draw the per-unit zero-, positive-, and negative sequence networks on a 100-MVA, 13.8-kV base in the zone of the generator. [15pts] (b) Reduce the sequence networks to their Thévenin equivalents, as viewed from bus 2. Pre-fault voltage is VF = 1.05L0° per unit. Pre-fault load current and A -Y transformer phase shift are neglected. [10pts] (c) Calculate the per-unit subtransient fault currents in phases a, b, and c for a bolted three-phase-to-ground short circuit at bus 2. [10pts] (d) Calculate the subtransient fault current in per-unit and in kA for a bolted single line- to-ground short circuit from phase a to ground at bus 2. Also calculate the per-unit line-to-ground voltages at faulted bus 2. [15pts] (e) Calculate the subtransient fault current in per-unit and in kA for a bolted line-to-line fault from phase b to c at bus 2. [10pts] G 1 2 M Line 어머 X₁ 100 MVA 138-kV Y/13.8-KV A X = 0.10 per unit Xo = X₂ = 100 MVA 13.8 kV X" = 0.15 X₂ = 0.17 = 0.05 per unit Xo 100 MVA 13.8-kVA/138-kVY X = 0.10 per unit 20 Ω 60 Ω 100 MVA 13.8 kV X" = 0.20 X₂ = 0.21 Xo = 0.10 X₁ = 0.05 per unit