A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. 10.av
A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. 10.av
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
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.52P
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Question
![b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal
voltages throughout the network are of 1 p.u. and the impedance of the electric
arc is neglected. Sequence impedance parameters of the generator,
transmission lines, and transformer are given in Figure Q3, where X and Y are
the last two digits of your student number.
jX(1) j0.1Y p.u.
jX2)= j0.1Y p.u.
jXko) = j0.1X p.u.
V₁ = 120° p.u. V₂ = 120° p.u.
(i)
(ii)
0
jX(1) = j0.2 p.u.
1 jx(2) j0.2 p.u. 2
jX1(0) = j0.25 p.u.
jXT(1)
jXT(2)
종 3
j0.1X p.u. JX3(1)
j0.1Y p.u.
j0.1X p.u. JX3(2)
j0.1Y p.u.
jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u.
0
=
x = 1,
jX2(1) j0.2Y p.u. V₁=1/0° p.u.
jX(2(2) = j0.2Y p.u.
jX2(0) = j0.3X p.u.
=
V3 = 120° p.u.
Figure Q3. Circuit for problem 3b).
For example, if your student number is c1700123, then:
y = 7
=
=
jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u.
Assuming a balanced excitation, draw the positive, negative and zero
sequence Thévenin equivalent circuits as seen from bus 4.
4
Determine the positive sequence fault current for the case when a three-
phase-to-ground fault occurs at bus 4 of the network.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48d63296-e4c0-4d86-9f5e-2017d17269b2%2F1701a7a0-4acd-4e75-b348-98530ea9fb58%2Fvb7me0s_processed.png&w=3840&q=75)
Transcribed Image Text:b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal
voltages throughout the network are of 1 p.u. and the impedance of the electric
arc is neglected. Sequence impedance parameters of the generator,
transmission lines, and transformer are given in Figure Q3, where X and Y are
the last two digits of your student number.
jX(1) j0.1Y p.u.
jX2)= j0.1Y p.u.
jXko) = j0.1X p.u.
V₁ = 120° p.u. V₂ = 120° p.u.
(i)
(ii)
0
jX(1) = j0.2 p.u.
1 jx(2) j0.2 p.u. 2
jX1(0) = j0.25 p.u.
jXT(1)
jXT(2)
종 3
j0.1X p.u. JX3(1)
j0.1Y p.u.
j0.1X p.u. JX3(2)
j0.1Y p.u.
jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u.
0
=
x = 1,
jX2(1) j0.2Y p.u. V₁=1/0° p.u.
jX(2(2) = j0.2Y p.u.
jX2(0) = j0.3X p.u.
=
V3 = 120° p.u.
Figure Q3. Circuit for problem 3b).
For example, if your student number is c1700123, then:
y = 7
=
=
jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u.
Assuming a balanced excitation, draw the positive, negative and zero
sequence Thévenin equivalent circuits as seen from bus 4.
4
Determine the positive sequence fault current for the case when a three-
phase-to-ground fault occurs at bus 4 of the network.
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