ii.A fault now occurs as shown in figure Q3a500 MVAG₁G213.8 kV20 %FT1GND500 MVA13.8A/500 Y kVX=12%Line 1jo.5 2750 MVA500 Y/18A kvmX = 10%G313 سلب1000 MVA500 Y/204 kVX = 10%,Line 2j0.592750 MVA18 kV18%1000 MVA20 kV17%T2Figure Q3aModify the reactance circuit in (i) so that it can be used to calculate the faultcurrent.Assuming V₁ =1 p.u. calculate the fault current in p.u. ampere.

The three phase system is here voltage at fault point isNow we need to calculate the fault current.

Answered: iii. Figure Q3a Modify the reactance…

iii. Figure Q3a Modify the reactance circuit in (i) so that it can be used to calculate the fault current. Assuming V₁ =1 p.u. calculate the fault current in p.u. ampere.

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

Chapter5: Transmission Lines: Steady-state Operation

Section: Chapter Questions

Problem 5.25MCQ

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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

ii.
A fault now occurs as shown in figure Q3a
500 MVA
G₁
G2
13.8 kV
20 %
F
T1
GND
500 MVA
13.8A/500 Y kV
X=12%
Line 1
jo.5 2
750 MVA
500 Y/18A kvm
X = 10%
G3
13 سلب
1000 MVA
500 Y/204 kV
X = 10%,
Line 2
j0.592
750 MVA
18 kV
18%
1000 MVA
20 kV
17%
T2
Figure Q3a
Modify the reactance circuit in (i) so that it can be used to calculate the fault
current.
Assuming V₁ =1 p.u. calculate the fault current in p.u. ampere.

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