ece524-s13

ece524-s13 - ma.0 ZOHmmmm m>_m.m>m mm>>On ZH...

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Unformatted text preview: ma .0: ZOHmmmm m_>_m:.m>m mm>>On_ ZH mFZmHmzék wmm mum. ECE 524 Session 13; Page 1/1 Transients in Power Systems Spring 2012 ECE 524: Homework #2 Due Session 17 (February 22 on campus, March 7 for Engineering Outreach) Solve each of these problems analytically and then simulate them using a transient simulation program. Compare your results. wt Problem in text book. Neglect load in motor loads in analytical calculations and reduce circuit to simple RLC circuit. Do the simulations first without the loas and t n with th loads modelled as a series R—L circuit in a more complete circuit model. H % Problem in textbook: Neglect load in motor loads in analytical calculations and reduce circuit to simple RLC circuit. Do the simulations first without the loads and then with the loads modelled s a series R—L circuit. 3 in textbook: Neglect the loads for analytical calculations. Problem ® Problem in text book: Neglect the loads for analytical calculations. eg Problem 5 in text book: Neglect the loads for analytical calculations. w, , Gnu—utobigmza ozmu_kofi_£m>m:., Emth ozmutcbgmz: ECE 524: Session 11; Page 13/17 Transients in Power Systems Spring 2012 o 10 20 (file Exampl1R.pl4; x-var t) t: VMAX vzct - v:CZ ~ Analytical Solution: (31 = 40.114111: C2 = 22.291111: L1 := 9.2uH V]m = 28.17 -kV Actual value with C1 in the system Undamped resonant frequencies: l l f := —--—--— f = 439.35— 01 C1 + 01 S l l f02 = 9596.15 — Likely to be damped foz := —— C1 'C2 S severly by a resistance 2'7" L1' C1 + C2 that can damp the lower * frequency, so neglect this resonant circuit. Driving voltage for fm: From charge balance: C1~VC1(0) = (C1 + C2) -V(oo) Assume near voltage peak when switch 0587633.? OsmExoktwLmZCD ECE 524: Session 11; Page 14/17 Transients in Power Systems Spring 2012 C1 Vinfinity I: Vm' C1 + C2 Vinfinity = 18.11.kv VdfiVingPOint :: Vm _ Vinfinity Vdrivingpoint : 10-06‘kV Damped resonant voltage will be: Vfoi - _Vdrivingpoint’e ‘Sm'wdt 1 Where: OL = CI'CZ 2-R- Find maximum voltage that can be added to source: $1 vmaxAdded := 33.5w —— vm VmemAdded = 5.33-kV % P; So we want the positive peak of the damped sine wave to be: OL-t VmaxAdded = —Vdrivingpoint'e— 'Sin'wdt Or with the numbers added: 5.33kV = (10.06kV)-e‘0”.sin-wdt Positive maximum occurs when: sin. wdt = —1 This implies that: tpk- (2.7:~f01)2 — a2 = — ECE 524: Session 11; Page 16/17 Transients in Power Systems Spring 2012 Then, taking a square root .4.(2.7T.f01)2 0L :2 OL = 368.71--L 9-112 sec2 + Exp_atpk2-4 sec R 11 - ———1-—--— eca 0L _ + C2) R ._ __.__1___ ’_ 2.OL.(C1+ C2) R = 21.73-Ohm r H1 w '- 2 f 2—— 2 wd 1 d-— ( '7" 01) CL 2—“ =435.42—— - s However, there is a problem with this result. This result assumes that the transient happens so fast that the 60Hz waveform is not varying. It turns out that the transient will be worse the switching transient occurs prior to the 60 Hz peak with a tradeoff between the driving point voltage and the sum of the resonant response plus the 60 Hz response. Solve again, and use the undamped resonant frequency from above, assuming that R won't change it too much: ms := 10— 3sec Then solving for the time to the peak (3/4 cycle at natural frequency). Estimate the initial switching time: 1 k1: — . ms k = . -ms tp 6 15 61 tp 106 0H2 Using the time from the simulation... ...
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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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ece524-s13 - ma.0 ZOHmmmm m>_m.m>m mm>>On ZH...

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