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ECE4762011_Lect3 - ECE 476 POWER SYSTEM ANALYSIS Lecture 3...

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Lecture 3 Three Phase, Power System Operation Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS
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2 Reading and Homework For lecture 3 please be reading Chapters 1 and 2 For lectures 4 through 6 please be reading Chapter 4 we will not be covering sections 4.7, 4.11, and 4.12 in detail though you should still at least skim those sections. HW 1 is 2.9, 22, 28, 32, 48; due Thursday 9/8 For Problem 2.32 you need to use the PowerWorld Software. You can download the software and cases at the below link; get version 15. http://www.powerworld.com/gloversarma.asp
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3 Three Phase Transmission Line
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4 Per Phase Analysis Per phase analysis allows analysis of balanced 3 φ systems with the same effort as for a single phase system Balanced 3 φ Theorem: For a balanced 3 φ system with All loads and sources Y connected No mutual Inductance between phases
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5 Per Phase Analysis, cont’d Then All neutrals are at the same potential All phases are COMPLETELY decoupled All system values are the same sequence as sources. The sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence; later in the course we’ll discuss negative and zero sequence systems.
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6 Per Phase Analysis Procedure To do per phase analysis 1. Convert all load/sources to equivalent Y’s 2. Solve phase “a” independent of the other phases 3. Total system power S = 3 V a I a * 4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts) 5. If necessary, go back to original circuit to determine line-line values or internal values.
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7 Per Phase Example Assume a 3 φ , Y-connected generator with V an = 1 0 ° volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with V a”b” = 1 0 ° through a second transmission line which also has an impedance of j0.1 per phase. Find 1. The load voltage V a’b’ 2. The total power supplied by each generator, S Y and S
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8 Per Phase Example, cont’d First convert the delta load and source to equivalent Y values and draw just the "a" phase circuit
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9 Per Phase Example, cont’d ' ' ' a a a To solve the circuit, write the KCL equation at a' 1 (V 1 0)( 10 ) V (3 ) (V j 3 j j - ∠ - + + - ∠ - 30° 29(-10 29 = 0
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