ece524-s14

Ece524-s14 - 3.0 ZOHmmmm m>_m.m>m mm>>On ZH m.Zm_Hm2</n vmm mom ECE 524 Session 14 Page 1/4 Transients in Power Systems Spring 2012 ECE

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Unformatted text preview: 3 .0: ZOHmmmm m_>_m_._.m>m mm>>On_ ZH m._.Zm_Hm2</n_._. vmm mom ECE 524: Session 14; Page 1/4 Transients in Power Systems Spring 2012 ECE 524: Lecture 14 MVA :2 lOOOkW MVAR :2 MVA pu := 1 w := 2-7T-(60HZ) Example 6-1: 0 Source impendance (convert to ohms on 13.8kV side of xfmr) VB ;= 13.8kV W V32 ZB :2 m ZSLLV I: (0.002pu +j-0.O286pu)-ZB ZSLLV = (0.0038 + 0.05450 9 Rs1_Lv 3: R€(ZSI_LV) RS]_LV = 000389 1m Z51 LV LSl_LV :: *L—T—‘l L51_LV = 1444751qu U.) Not needed at the moment, but for completeness: zso_LV :2 (0.0038pu +j-O.O398pu)-ZB ZSUV = (0.0072 + 0.0758i) o 0 Transformer Impedance Referred to 13.8kV): fomr 2: 7.5% XoRxfmr := 18.6 0xfmr ;= atan(XoRxfm) exfmr = 86,9225.ng V132 ‘Zz‘v 3OMVA 90$ j'exfmr ZT1_LV I= fomr'e - ZTLLV = (0.0256 + 0.4754i) Q RTlfiLV 1= R€(ZT1_LV) RT1_LV = 0.02560 1m er LV LTLLV 3: —-—(—-—:—l LTLLV = 1261.0732-HH o) ECE 524: Session 14; Page 2/4 Transients in Power Systems Spring 2012 0 Generator Impedance Referred to 13.8kV): XORgen :: 48 X” X” gen '_ gem” 35.3MVA X"gen : 0.80929 XII gen R 1: R = 0.01699 gen XoRgen gen L .— X'lgen gen " w Lgen = 2.1466-mH 0 Capacitor Qcap := 6MVAR 2 VB X := X = 31.749 0 Qcap C Y Cannth Qfluu 1 CbankY := Cbanky = 83.5mm? W'Xc 0 Load Model S V32 . Zloadflseries : “WW Zload—Series : + (25.5 +j-15.8)MVA 1m Zload series Lloadwseries 3: if) Lload_series : 8-8694‘mH R V82 R 7 4682 9 Load '“ ZSISMVA Load ’ - ECE 524: Session 14; Page 4/4 Transients in Power Systems Spring 2012 w 1 2 fem :2 C C fem = 6155.23-Hz 2% 2 Vm := -3—-13.8kV Vm = 11.2677-kv Vm kawdcg :2 ——-————- kaiclcg = 18.2091-kA LCICZ / Cseries 30 ,‘ , .. ozmvtobfihgmta ECE 524: Session l4; Page 4/10 Transients in Power Systems Spring 2012 : l l , Compare to: f02 = 438.26— f01 = 9596.15— “ea 5 o So the approximation used above is close, it helps that there is a wide separation in frequencies. Initial Voltage Conditions: é I _ . r ‘7 g ' J- .L Vcl_0 :: Vm Vc2fi0 3: 0V N H \ {to L2 2012: ‘C— 201 = 8‘] Peak current: Vc1_0 “ Vc2_0 ———————- lpkl = 24.33- A Very large current I '2 pkl - Z01 at a high frequency! Note: This is circulating between the capacitors, not through the source..... For comparison, the steady-state current would be: VLL 158:: -————-————‘[3————-——— llssl = 477.53A , 1 1 ‘1 J-XS-i- , + -—,——-—-,——- —J‘Xcl _J'X02+J'w'L2 Or with just 1 capacitor (as we saw a few lectures ago): VLL fl I: ~-—————— 2 304.9A SS—lcap ’Xs + _.l 'Xcl I i Iss__lcapI Now look at the voltage at the bus: First look at the term at the above resonant frequency: ECE 524: Session 14; Page 5/l0 Transients in Power Systems Spring 2012 Vcl_O'Cl + Vc2_0‘C2 C1 + C2 Vmfinjty = 18.11 -kV There will be an oscillating voltage based on this: 0301 :2 2-7T-f01 V020) :2 Vinfinity —— Vinfinity'COS(W01't) «a This represents the increase in voltage at that capacrtor From charge balance: Vmfinity 2: This represents the decrease Vol“) : Vinfinity + (Vin _ Vinfinity) 'Cos(w01't> in voltage at this capacitor Note that: (Vm — vinfinity) = 10.06.1(V Note that these are only correct if the high freq resonant term is considered in isolation The other terms correct the offsets 4x104 3><104 4 t 2><lO_ ‘L N Session 14; Page 6/10 Spring 2012 ECE 524: Transients in Power Systems There will be 2 other terms in the response: 1. The 60 Hz response-—-just the steady—state solution 2. A term that is equivalent to the two capacitors adding in parallel, resonating with the source impedance. 1 l f02 = f ;:—————————-— wzw 2'7T'1’LS'iC1-i-C2i S This is how the circuit would behave when the high frequency transient dies out. We looked at this part of the response last time with adding the resistor. This will add to the above response since the source circuit will respond to the change in voltage across C1. LS z := ————- z = 5.829 02 C1+C2 02 Vm _ Vinfinity 1210k = 1728.69A Z02 Izpk :2 The voltage at this frequency will be the same on both capacitors and it will have an amplitude of: (vm — vinfimty) = 10.06-kV Capacitor Cl voltage (unzoomed)-~note multiple frequencies: ECE 524: Session 14; Page 7/10 Transients in Power Systems Spring 2012 50.0 *103 37.5 % 25.0 12.5 0.0 -12.5 -25.0 -37.5 -50.0 0.00 5.55 11.10 16.65 22.20 27.75 *10'3 33.30 (file capsvvit.pl4; x-var t) v:VS v:C1 — 15.5 16.4 17.3 18.2 19.1 *10-3 20.0 (filecapswit.p|4;x«vart)v:VS V’Ci u v:CZ « Two frequencies, one is 180 degrees out of phase on the two caps and the other is in phase Zoom in to first few cycles at high frequency: ECE 524: Session 14; Page 8/10 Transients in Power Systems Spring 20 l 2 40 *103 35 /\ /\ First peak of blue line is 36.59kV, compare ’ j to 36.22kV the analytical 25 for green cl) is 28.5 V compared to 28.169 and the ' e peak is 8.26kV compared to 8.21kV 16.650 16.688 16.726 16.764 16.802 ’“10'3 16840 (file capswinM; x»var i) v.C1 « v:CZ ~ Source current (compared to Vs/ 1 O) 3000 *10 2000 1000 —1000 -2000 —3000 a -r- l “r u T'“ I ‘1 | 0 10 20 30 4O *10’3 50 (file capswitpl4; x~var t) c:81 -C1 .1 1V3 factors: 1 1 U 1 offsets: O 0 i) Note that the high frequency term is barely noticable ECE 524: Session 14; Page lO/lO Transients in Power Systems Spring 2012 Zoom in on the current between caps: Calculated 24.3 3l<A with high freq term alone 30 I ’10 -10 —20 -30 16.60 16.55 16.70 16.75 16.80 16.85 *10‘3 16.90 (file capswit‘pl4; x—v ar 1) (3:82 {320 l 1 T1 2: 0.016693sec T2 :2 0.016797sec —— = 9615.38— T2 - T1 S ...
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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-s14 - 3.0 ZOHmmmm m>_m.m>m mm>>On ZH m.Zm_Hm2</n vmm mom ECE 524 Session 14 Page 1/4 Transients in Power Systems Spring 2012 ECE

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