Answered: Q1: Calculate, the fault level for a…

Q1: Calculate, the fault level for a three phase fault at the midpoint of transmission line, and the fault current supplied by each generator (Assume SB = 100 MVA).

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

Chapter7: Symmetrical Faults

Section: Chapter Questions

Problem 7.2MCQ: Even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical...

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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.

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Q1: Calculate, the fault level for a three phase fault at the midpoint of transmission line, and the fault current
supplied by each generator (Assume SB = 100 MVA).
Ti
Ö of a Ⓒo
100 MVA
110 MVA
(11/132) KV
11 kV
X=12%
X=25%
T.L
Š
100 km
(0.085+j 0.145) 52/km
O
+
T₂
· ¹ · Ö
90 MVA
(132/11) kV
X-10%
80 MVA
11 kV
X-18%

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