a) In Three Phase Faults that will occur separately on A, B, C and D Busbars, Calculate the currents.b) Maximum fault currents to pass through KA, KB, KC1, KC and KD Protection devices determine.Help me please... Thank you so much. TransformerSTRF = 6 MVAXTRF = % 3Generator AB UB / Uc = 36 kV/6 kVDKAXHCD = j 0,5 Ohm00KciEКсКвXHAB = j 0,5 OhmKDSGA = 4 MVAXGA = % 4ScD = 2 MVASload =5 MVAUload = 6 kVUGA = 36 kVXGD = % 2UGD = 6 kVGeneratorCosp=1

Consider the Base MVA =4 MVA Base KV=36 KV

STRF = 6 MVA XTRF = % 3 UB / Uc = 36 kV/6 kV Generator A B C D KA XHCD = j 0,5 Ohm Kci Kc KB XHAB = j 0,5 Ohm KD SGA = 4 MVA XGA = % 4 UGA = 36 kV SGD = 2 MVA Sload =5 MVA Uload = 6 kV XGD = % 2 UGD = 6 kV Generator Cosp=1

STRF = 6 MVA XTRF = % 3 UB / Uc = 36 kV/6 kV Generator A B C D KA XHCD = j 0,5 Ohm Kci Kc KB XHAB = j 0,5 Ohm KD SGA = 4 MVA XGA = % 4 UGA = 36 kV SGD = 2 MVA Sload =5 MVA Uload = 6 kV XGD = % 2 UGD = 6 kV Generator Cosp=1

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

100%

a) In Three Phase Faults that will occur separately on A, B, C and D Busbars, Calculate the currents.

b) Maximum fault currents to pass through KA, KB, KC1, KC and KD Protection devices determine.

Help me please... Thank you so much.

Transformer
STRF = 6 MVA
XTRF = % 3
Generator A
B UB / Uc = 36 kV/6 kV
D
KA
XHCD = j 0,5 Ohm
00
KciE
Кс
Кв
XHAB = j 0,5 Ohm
KD
SGA = 4 MVA
XGA = % 4
ScD = 2 MVA
Sload =5 MVA
Uload = 6 kV
UGA = 36 kV
XGD = % 2
UGD = 6 kV
Generator
Cosp=1

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