G FOTO 100 MVA 13.8 kV X₁ = 0.15 p.u. X₂ = 0.17 p.u. Xo = 0.05 p.u. AY 100 MVA A:Y 13.8:138 kV X = 0.1 p.u. Line X1 X₂ = 20 Χο = 60 Ω M T2 2 01040 OYI #A 100 MVA 100 MVA 13.8 KV Y:A-138:13.8 kV X₁ = 0.20 p.u. X = 0.1 p.u. X₂ = 0.21 p.u. X = 0.10 p.u. X = 0.05 p.u. n
G FOTO 100 MVA 13.8 kV X₁ = 0.15 p.u. X₂ = 0.17 p.u. Xo = 0.05 p.u. AY 100 MVA A:Y 13.8:138 kV X = 0.1 p.u. Line X1 X₂ = 20 Χο = 60 Ω M T2 2 01040 OYI #A 100 MVA 100 MVA 13.8 KV Y:A-138:13.8 kV X₁ = 0.20 p.u. X = 0.1 p.u. X₂ = 0.21 p.u. X = 0.10 p.u. X = 0.05 p.u. n
Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
10th Edition
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Russell E. Smith
Chapter7: Alternating Current, Power Distribution, And Voltage Systems
Section: Chapter Questions
Problem 7RQ
Related questions
Question
A single line diagram of a power system is shown in Figure Q2.1. The neutrals for
the generators and transformers are solidly grounded. The motor neutral is grounded
through reactance ?? = 0.05 per unit on the motor base.
i）Calculate the phase currents in kA following a three-phase fault at busbar 2.
ii) Calculate the phase currents in kA when a phase-alpha-to-ground fault occurs
at busbar 2.
Transcribed Image Text:YÖ
100 MVA
13.8 kV
X₁ = 0.15 p.u.
X₂ = 0.17 p.u.
Xo = 0.05 p.u.
AY
100 MVA
A:Y 13.8:138 kV
X = 0.1 p.u.
Line
X₁ = X₂ = 20
Χο = 60 Ω
FYA
он
M
100 MVA
13.8 kV
Y:A-138:13.8 kV X₁ = 0.20 p.u.
100 MVA
X = 0.1 p.u.
X₂ = 0.21 p.u.
X = 0.10 p.u.
= 0.05 p.u.
X₂ =
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Step 1: Define the given variable
VIEWStep 2: Per unit value for line reactance
VIEWStep 3: Draw the positive, negative and zero sequence network
VIEWStep 4: Determine the Thevenin's equivalent reactance for each sequence when fault occurs at bus bar 2
VIEWStep 5: Determine the current value for three phase fault
VIEWStep 6: Determine the current value for single line to ground fault
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