7. Determine the fault current for a 3-phase symmetrical fault and L-G fault for thedistribution system shown. The sequence impedance are (i) 11-kV line Z₁ = Z₂ = 0.8+j3.5 , Zo = 0.5 + j3.0 (ii) 100 kVA transformer on its own base Z₁ = Z₂ = j0.05,Zo = j0.08.11 KV/400V100kVAY+/A

Types of open circuit fault- 1) Symmetrical fault- Three phase line to ground fault Three phase…

Answered: the fault current for a

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the fault current for a

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 7.7P

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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

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7. Determine the fault current for a 3-phase symmetrical fault and L-G fault for the
distribution system shown. The sequence impedance are (i) 11-kV line Z₁ = Z₂ = 0.8
+j3.5 , Zo = 0.5 + j3.0 (ii) 100 kVA transformer on its own base Z₁ = Z₂ = j0.05,
Zo = j0.08.
11 KV/400V
100kVA
Y+/A

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Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows: Generator G1:Â Â Â Â 50Â MVA,Â 12kV,Â X=0.2 per unit Generator G2: 100Â MVA, 15 kV, X=0.2 per unit Transformer T1: 50 MVA, 10 kV Y/138kVY,X=0.10 per unit Transformer T2: 100 MVA, 15 kV /138kVY,X=0.10 per unit Each 138-kV line: X1=40 A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV. Prefault load current is neglected. (a) Draw the positive-sequence reactance diagram in unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine (b) the ThĂ©venin equivalent at the fault, (c) the subtransient fault current in per unit and in kA rms, and (d) contributions to the fault from generator G2 and from transformer T2.
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