Lecture 19

# Lecture 19 - Announcements Be reading Chapter 8 HW 8 is 7.6...

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1 Announcements ! Be reading Chapter 8 ! HW 8 is 7.6, 7.13, 7.19, 7.28; due Nov 3 in class. ! Start working on Design Project. Tentatively due Nov 17 in class. This page was for UI ECE 476 course. Please refer to OSU ECE 5042 syllabus for relationship between chapters and lectures/ handout references. dgk au2013

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2 In the News Earlier this week the Illinois House and Senate overrode the Governor’s veto of the Smart Grid Bill; it is now law On Tuesday Quinn called it a “smart greed” plan Law authorizes a ten year, \$3 billion project to “modernization” the electric grid in Illinois All ComEd customers and most of Ameren will get “smart” meters Effort will be paid for through rate increases over the period; ComEd estimated the average increase of \$36 per year would be offset by electricity savings
3 Network Fault Analysis Simplifications ! To simplify analysis of fault currents in networks we'll make several simplifications: 1. Transmission lines are represented by their series reactance 2. Transformers are represented by their leakage reactances 3. Synchronous machines are modeled as a constant voltage behind direct-axis subtransient reactance 4. Induction motors are ignored or treated as synchronous machines 5. Other (nonspinning) loads are ignored

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4 Network Fault Example For the following network assume a fault on the terminal of the generator; all data is per unit except for the transmission line reactance 2 19.5 Convert to per unit: 0.1 per unit 138 100 line X = = generator has 1.05 terminal voltage & supplies 100 MVA with 0.95 lag pf
5 Network Fault Example, cont'd Faulted network per unit diagram * ' a To determine the fault current we need to first estimate the internal voltages for the generator and motor For the generator 1.05, 1.0 18.2 1.0 18.2 0.952 18.2 E 1.103 7.1 1.05 T G Gen V S I = = ° " # = = ∠ − ° = ° % & ' (

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6 Network Fault Example, cont'd f The motor's terminal voltage is then 1.05 0 - (0.9044 - 0.2973) 0.3 1.00 15.8 The motor's internal voltage is 1.00 15.8 (0.9044 - 0.2973) 0.2 1.008 26.6 We can then solve as a linear circuit: 1 I j j j j × = ∠ − ° ∠ − ° × = ∠ − ° = .103 7.1 1.008 26.6 0.15 0.5 7.353 82.9 2.016 116.6 9.09 j j j ° ∠ − ° + = ∠ − ° + ∠ − ° =
7 Fault Analysis Solution Techniques !

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