QUESTION 1 A 100 km, 400 kV transmission line is subject to an “a" phase to earth fault at 80 km. If the fault path resistance is 10 2, calculate the fault current. Data:- For generation: Za1 = Za2 = 1 + j16 N and Zao = 0.5 + j8 N For the line: ZLI = ZL2 = 1.8 + j31 N and ZLo= 10 + j88 N

QUESTION 1 A 100 km, 400 kV transmission line is subject to an “a" phase to earth fault at 80 km. If the fault path resistance is 10 2, calculate the fault current. Data:- For generation: Za1 = Za2 = 1 + j16 N and Zao = 0.5 + j8 N For the line: ZLI = ZL2 = 1.8 + j31 N and ZLo= 10 + j88 N

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

Chapter11: Transient Stability

Section: Chapter Questions

Problem 11.19P

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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

StaySafe Orangel. 36
O O 73
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WearAMask M|..
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50
University of Botswana
Faculty of Engineering and Technology
Department of Electrical Engineering
EEB 465 - Analysis of Power Systems
and
EEB 525 - Power Systems Analysis
Due Date: 2nd May 2022
Assignment 2
QUESTION 1
A 100 km, 400 kV transmission line is subject to an "a" phase to earth fault
at 80 km. If the fault path resistance is 10 2, calculate the fault current.
Data:-
For generation:
Zs1 = Zs2 = 1 + j16 N and Zs0 = 0.5 + j8 N
For the line:
ZLI = ZL2 = 1.8 + j31 2 and ZLO = 10 + j88 2
QUESTION 2
What are the fundamental differences between primary and back-up
protection?
EEB 465 - EEB 525
By: Prof. E. Matlotse
Page 1
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