The one-line diagram of a power system is shown in the figure below. A single-line-to-ground (SLG) fault occurs at bus 2. Assume the fault impedance Z,= j0.08 pu. Calculatethe fault current (in per unit).T,2MLineX, = X2 = 40 NXo =1200%3D100 MVA13.8 kV100 MVA100 MVAX, = 0.15 13.8-kVA/138-kVYX = 0.10 per unit138-kV Y/13.8-kV A100 MVAX2 = 0.17X = 0.10 per unit13.8 kVX, = 0.20X2 = 0.21Xo = 0.10X, = 0.05 per unitXo = 0.05 per unit

A single line to ground fault occurs at bus 2. So the fault current will be- Ea=Genetaror terminal…

The one-line diagram of a power system is shown in the figure below. A single-line-to- ground (SLG) fault occurs at bus 2. Assume the fault impedance Zj=j0.08 pu. Calculate the fault current (in per unit). 2 M Line X, = X2 = 40 N 1 Xo = 1200 100 MVA 13.8 kV 100 MVA 100 MVA X, = 0.15 X2 = 0.17 X, = 0.05 per unit 138-kV Y/13.8-kV A X = 0.10 per unit 13.8-kVA/138-kVY 100 MVA X = 0.10 per unit 13.8 kV X, = 0.20 X2 = 0.21 0 10

The one-line diagram of a power system is shown in the figure below. A single-line-to- ground (SLG) fault occurs at bus 2. Assume the fault impedance Zj=j0.08 pu. Calculate the fault current (in per unit). 2 M Line X, = X2 = 40 N 1 Xo = 1200 100 MVA 13.8 kV 100 MVA 100 MVA X, = 0.15 X2 = 0.17 X, = 0.05 per unit 138-kV Y/13.8-kV A X = 0.10 per unit 13.8-kVA/138-kVY 100 MVA X = 0.10 per unit 13.8 kV X, = 0.20 X2 = 0.21 0 10

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

The one-line diagram of a power system is shown in the figure below. A single-line-to-
ground (SLG) fault occurs at bus 2. Assume the fault impedance Z,= j0.08 pu. Calculate
the fault current (in per unit).
T,
2
M
Line
X, = X2 = 40 N
Xo =1200
%3D
100 MVA
13.8 kV
100 MVA
100 MVA
X, = 0.15 13.8-kVA/138-kVY
X = 0.10 per unit
138-kV Y/13.8-kV A
100 MVA
X2 = 0.17
X = 0.10 per unit
13.8 kV
X, = 0.20
X2 = 0.21
Xo = 0.10
X, = 0.05 per unit
Xo = 0.05 per unit

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