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

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.

Question

100%

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
13.8 kV
G4 750 MVA
13.8 kV
Transformers:
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
X₁ = 40 2, Xo = 100 2
at
Line 1-3
Line 1-2
2
X = 0.10 per unit
X = 0.10 per unit
X = 0.12 per unit
X = 0.11 per unit
Line 2-3
3
youtly

a) Draw the zero, positive, and negative sequence reactance diagrams.
b) For a fault at bus 3, determine the thevenin equivalent of each sequence
network as viewed from the fault bus. Prefault voltage is 1 per unit. (use the
wye-delta transformation)
A
с
B
ZAB=
ZBC
ZCA =
ZCA
ZAZB+ Z₂Zc+ ZcZA
Zc
ZAZB+ ZBZc+ ZcZA
ZA
ZAZB+ ZBZc+ZCZA
ZB
B
Z₁ =
Z₁ =
Zc =
ZABZCA
ZAB + ZBC + ZCA
ZABZBC
ZAB + ZBC + ZCA
ZCAZBC
ZAB + ZBC + ZCA
c) Determine the subtransient fault current in per unit and in kA, as well as the
per unit line-to-ground voltages at the fault bus for a bolted single line-to-
ground fault at bus 3.

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