For the power system shown at right the impedance matrix 2 is j0.5 j0.2 3 j0.4 0.145 0.120 0.0692 Zbus=j 0.120 0.246 0.101 | 0.0692 0.101 0.164 (3) j0.2 j0.25 Pre-fault all bus voltages are 1 pu. Then a three-phase fault to ground occurs at bus 3. What are the post-fault per-unit bus voltages? (Express in polar phasor form |V|Z0° ) ~) 1pu 1pu (~ V, = 0.578 Lo° V;= 0.384 L0° V;= %3D feuetot bus -j6.098 Z33 AVI 0.o692 0.4220 AV2 jG.098 (j) 0.101 0.l6159 D.164 %3D
For the power system shown at right the impedance matrix 2 is j0.5 j0.2 3 j0.4 0.145 0.120 0.0692 Zbus=j 0.120 0.246 0.101 | 0.0692 0.101 0.164 (3) j0.2 j0.25 Pre-fault all bus voltages are 1 pu. Then a three-phase fault to ground occurs at bus 3. What are the post-fault per-unit bus voltages? (Express in polar phasor form |V|Z0° ) ~) 1pu 1pu (~ V, = 0.578 Lo° V;= 0.384 L0° V;= %3D feuetot bus -j6.098 Z33 AVI 0.o692 0.4220 AV2 jG.098 (j) 0.101 0.l6159 D.164 %3D
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 7.15P
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Can someone please explain how to convert V1=-0.4220 to poloar phasor form? (V1=0.578<0)
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New voltage at a bus = old bus voltage + change in voltage
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