Power Flow Analysis 0.01 +0.03 Slack Bus V₁ = 1.05/0° 3 0.02 + j0.04 0.0125+j0.025 138.6 45.2 MW Mvar 2 256.6 MW 110.2 Mvar Figure shows the one-line diagram of a simple three-bus power system with generation at bus 1. The magnitude of voltage at bus 1 is adjusted to 1.05 pu. The scheduled loads at buses 2 and 3 are as marked on the diagram. Line impedances are marked in pu on a 100 MVA base and the line charging susceptances are neglected. 1- Using the Gauss-Seidel method, determine the phasor values of the voltage at the load buses 2 and 3, accurate to four decimal places.

Power Flow Analysis 0.01 +0.03 Slack Bus V₁ = 1.05/0° 3 0.02 + j0.04 0.0125+j0.025 138.6 45.2 MW Mvar 2 256.6 MW 110.2 Mvar Figure shows the one-line diagram of a simple three-bus power system with generation at bus 1. The magnitude of voltage at bus 1 is adjusted to 1.05 pu. The scheduled loads at buses 2 and 3 are as marked on the diagram. Line impedances are marked in pu on a 100 MVA base and the line charging susceptances are neglected. 1- Using the Gauss-Seidel method, determine the phasor values of the voltage at the load buses 2 and 3, accurate to four decimal places.

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

Chapter9: Unsymmetrical Faults

Section: Chapter Questions

Problem 9.53P

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Most buses in a typical power-flow program are a.voltage controlled bus b.PQ bus c.PV bus d.Slack bus 2-The power-flow problem is the computation of voltage magnitude and phase angle at each bus in a power system under balanced three-phase steady-state conditions. False True 3-The following sequence currents were recorded in a power system under a fault condition Ipositive=j1.5pu, Inegative=–j0pu, Izero=– j0pu. The fault is a.Three-phase b.Double Line to ground c.Line to ground d.Line to line 4- he following sequence currents were recorded in a power system under a fault condition Ipositive=j1.5pu, Inegative=j0pu, Izero=j0pu. The fault is a.Three-phase b.Double Line to ground c.Line to ground d.Line to line i uploded the houndots (book )here :https://seomidea.com/hct/nu.pdf
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