V₁ = 120° p.u. jXj0.1 p.u. JX2)/0.1 p.u. JX) 10.15 p.u. V₂ - 120° p.u. j0.2 p.u. JX(1) 1 JX2) 0.2 p.u. 2 V₂ - 120° p.u. V₁- 120° p.u. JX20-j0.15 p.u. jXj0.2 p.u. 4 JX20) j0.2 p.u. jXj0.25 p.u. JX) 0.1 p.u. JXT2)-j0.15 p.u. JXT(0) JO.1 p.u. Figure Q4. Circuit for problem 4b). JX) 10.1 p.u. JX)/0.1 p.u. JXL) 10.1 p.u. (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 3.

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 3.26P: A bank of three single-phase transformers, each rated 30MVA,38.1/3.81kV, are connected in Y- with a...
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b) A fault occurs at bus 3 of the network shown in Figure Q4. 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, transformer and load are given in Figure Q4.
V₁ = 120° p.u.
V₂ = 120° p.u. V₂ = 1/0° p.u.
V₂= 120° p.u.
jXj0.1 p.u.
JX2) 0.1 p.u.
jX0j0.15 p.u.
jXn-j0.2 p.u.
1 JX(2)-j0.2 p.u. 2
jX)=j0.25 p.u.
JX20-10.15 p.u.
jXa(z)-j0.2 p.u. 4
jX2(0)=j0.2 p.u.
jXT(1) j0.1 p.u.
jXT(2)=j0.15 p.u.
jXT(0)=j0.1 p.u.
Figure Q4. Circuit for problem 4b).
=
jXj0.1 p.u.
j0.1 p.u.
-
JX(2)
JXL(0) 10.1 p.u.
=
(i) Assuming a balanced excitation, draw the positive, negative and zero
sequence Thévenin equivalent circuits as seen from bus 3.
(ii) Determine the positive sequence fault current for the case when a three-
phase-to-ground fault occurs at bus 3 of the network.
(iii) Determine the short-circuit fault current for the case when a one-phase-
to-ground fault occurs at bus 3.
Transcribed Image Text:b) A fault occurs at bus 3 of the network shown in Figure Q4. 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, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 3.
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