Q2) The single-line diagram of a three-phase power system is shown in the figurebelow. The inductor connected to Generator 3 neutral has a reactance of 0.05per unit using generator 3 ratings as base. Use a 1000-MVA, 765-kV base inthe zone of line 1-2. Equipment ratings are given as follows:Synchronous generators:G1 1000 MVA15 kVG2 1000 MVA15 kVG3 500 MVA13.8 kVG4 750 MVA13.8 kVTransformers:T1 1000 MVAT2 1000 MVAT3 500 MVAT4 750 MVATransmission lines:1-2 765 kV1-3 765 kV2-3 765 kV15 kV A/765 kV Y15 kV A/765 kV Y15 kV Y/765 kV Y15 kV Y/765 kV YX = X₂ = 0.18, Xo = 0.07 per unitX=X₂=0.20, Xo = 0.10 per unitX=X₂ = 0.15, Xo = 0.05 per unitX=0.30, X₂ = 0.40, Xo = 0.10 per unitX₁ = 50 2, Xo = 150 2X₁ = 40 2, Xo = 100 2X₁ = 40 2, Xo = 100 2atLine 1-3Line 1-22X = 0.10 per unitX = 0.10 per unitX = 0.12 per unitX = 0.11 per unitLine 2-33youtly a) Draw the zero, positive, and negative sequence reactance diagrams.b) For a fault at bus 3, determine the thevenin equivalent of each sequencenetwork as viewed from the fault bus. Prefault voltage is 1 per unit. (use thewye-delta transformation)AсBZAB=ZBCZCA =ZCAZAZB+ Z₂Zc+ ZcZAZcZAZB+ ZBZc+ ZcZAZAZAZB+ ZBZc+ZCZAZBBZ₁ =Z₁ =Zc =ZABZCAZAB + ZBC + ZCAZABZBCZAB + ZBC + ZCAZCAZBCZAB + ZBC + ZCAc) Determine the subtransient fault current in per unit and in kA, as well as theper unit line-to-ground voltages at the fault bus for a bolted single line-to-ground fault at bus 3.

Solution:Given data:Synchronous generators3 rating = 1000 KVA , 750 kVA , 500 kVA and give…

Q2) The single-line diagram of a three-phase power system is shown in the figure below. The inductor connected to Generator 3 neutral has a reactance of 0.05 per unit using generator 3 ratings as base. Use a 1000-MVA, 765-kV base in the zone of line 1-2. Equipment ratings are given as follows: Synchronous generators: G1 1000 MVA 15 kV G2 1000 MVA 15 kV G3 500 MVA G4 750 MVA Transformers: 13.8 kV 13.8 kV T1 1000 MVA T2 1000 MVA T3 500 MVA T4 750 MVA Transmission lines: 1-2 765 kV 1-3 765 kV 2-3 765 kV 15 kV A/765 kV Y 15 kV A/765 kV Y 15 kV Y/765 kV Y 15 kV Y/765 kV Y X=X₂ = 0.18, Xo = 0.07 per unit X=X₂=0.20, Xo = 0.10 per unit X = X₂ = 0.15, Xo = 0.05 per unit X=0.30, X₂ = 0.40, Xo = 0.10 per unit X₁ = 50 2, Xo = 150 2 X₁ = 40 2, Xo = 100 2 X = 40 Ω, Χ = 100 Ω X = 0.10 per unit X = 0.10 per unit X = 0.12 per unit X = 0.11 per unit

Q2) The single-line diagram of a three-phase power system is shown in the figure below. The inductor connected to Generator 3 neutral has a reactance of 0.05 per unit using generator 3 ratings as base. Use a 1000-MVA, 765-kV base in the zone of line 1-2. Equipment ratings are given as follows: Synchronous generators: G1 1000 MVA 15 kV G2 1000 MVA 15 kV G3 500 MVA G4 750 MVA Transformers: 13.8 kV 13.8 kV T1 1000 MVA T2 1000 MVA T3 500 MVA T4 750 MVA Transmission lines: 1-2 765 kV 1-3 765 kV 2-3 765 kV 15 kV A/765 kV Y 15 kV A/765 kV Y 15 kV Y/765 kV Y 15 kV Y/765 kV Y X=X₂ = 0.18, Xo = 0.07 per unit X=X₂=0.20, Xo = 0.10 per unit X = X₂ = 0.15, Xo = 0.05 per unit X=0.30, X₂ = 0.40, Xo = 0.10 per unit X₁ = 50 2, Xo = 150 2 X₁ = 40 2, Xo = 100 2 X = 40 Ω, Χ = 100 Ω X = 0.10 per unit X = 0.10 per unit X = 0.12 per unit X = 0.11 per unit

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter9: Unsymmetrical Faults

Section: Chapter Questions

Problem 9.1P

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