L11 - ECE 524: Transients in Power Systems Session 11; Page...

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ECE 524: Transients in Power Systems Session 11; Page 1/17 Spring 2012 ECE 524: Lecture 11 Define units: MW 1000kW  MVA MW  MVAr MVA  f 60Hz  ω 2 π f  ω 376.99 rad s t 0 0.000001sec 2 60 sec   Example 1: For the system below find the following: Worst case inrush current when closing into C 1 with C 2 open 1. jX s C 2 C 1 Assume source and capacitors are Y connected. V LL 34.5kV  RMS Capacitor Banks: Q 1 18MVAr  at 34.5kV Q 2 10MVAr  at 34.5kV L 1 19.2 μ H  between caps Source I sc 25kA  Determine circuit parameters: Find capacitance using: Q V LL  2 X c = X c1 V LL 2 Q 1  X c1 66.13 Ω X c2 V LL 2 Q 2  X c2 119.03 Ω C 1 1 ω X c1  C 1 40.11 μ F C 2 1 ω X c2  C 2 22.29 μ F
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ECE 524: Transients in Power Systems Session 11; Page 2/17 Spring 2012 Find source impedance using: I sc V ln X s = X s V LL 3 I sc  X s 0.797 Ω L s X s ω  L s 2.11 mH R s 0ohm  no source resistance Equivalent circuit: L s C 2 C 1 L 1 Load Neglect load for now. 1. Worst case inrush current when closing into C 1 with C 2 open V m V LL 2 3  V m 28.17 kV ϕ 90deg  v s t () V m sin ω t ϕ  Z 0 L s C 1  Z 0 7.26 Ω ω n 1 L s C 1  f n ω n 2 π  f n 546.61 1 s Initial conditons: V c1_0 0V  capacitor is discharged i Ls_0 0A  no load current
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ECE 524: Transients in Power Systems Session 11; Page 3/17 Spring 2012 Resonant frequency f n 1 2 π 1 L s C 1  f n 546.61 1 s approx. 9 times source frequency ω n 2 π f n  Use biggest possible voltage difference for given Vc1(0) Homogeneous solution: i h_1 t () V m V c1_0 Z 0 sin ω n t   peak current at resonant frequency V m V c1_0 Z 0 3880.89A R s 10 10 ohm  Set to a very small number more numerically stable Particular solution: η atan 1 ω C 1 ω L s R s  η 90 deg i p_1 t 2 3 V LL 1 ω C 1 ω L s sin ω t ϕ η  I Prms 1 3 V LL 1 ω C 1 ω L s  I Prms 304.9A i 1 t i p_1 t () i h_1 t 
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ECE 524: Transients in Power Systems Session 11; Page 4/17 Spring 2012 Worst possible current will be when both the 60Hz current and the resonant current peak at the same time. Since there is no resistance, the resonant response will continue forever. So at some point in time they will peak at the same time. Realistically, we would look at the peak in the first 60Hz cycle. I max1 V m V c1_0 Z 0 2 I Prms  I max1 4312.08A Capacitor voltage (assume switch at voltage peak) v cap1 t () 2 I Prms sin ω t ϕ X c1 V m V c1_0  cos ω n t  V capPK 2 I Prms X c1 V m V c1_0  V capPK 56.68 kV
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This note was uploaded on 02/28/2012 for the course ECE 524 taught by Professor Staff during the Spring '08 term at Idaho.

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L11 - ECE 524: Transients in Power Systems Session 11; Page...

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