ece524-s12

ece524-s12 - Nfl .0: ZOHmmmm m_>_m_._.m>m...

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Unformatted text preview: Nfl .0: ZOHmmmm m_>_m_._.m>m Mm>>On_ ZH mFZmHmZEH vmm mUm ECE 524: Session ll; Page 5/17 Transients in Power Systems Spring 2012 Repeat if capacitor is storing charge: Veto :2 —-Vm Charged to negative peak voltage . Vm _ Vc1_0 Imaxl '= 20 “films 1m} = 8192.97A ‘ VcapPK I= x/E-Ipms-Xci + (Vm —— VOLO) VcalpPK = 84.85 .kV ATP Model (M): W set time step near 5% of period of resonant freq usec := 10— 6sec: 1 tn can use deltaT=l 00 tn .— f—n tn — 1829.47-lLsec 2—0 —— 91.47-usec microseconds ATPDraw schematic “ ’11» VS 01 i 102 40.11%? 22.29%“; Peak and RMS steady—state current (Cl inserted for steady—state initialization) * ‘‘‘‘‘ “ 1m: 304.89A ' ka := 43 1 . 14A A ’0‘!“ 500 375 250 125 0 425 ~250 E. Z 3 ~375 -5oo , , x x u I I r I I 000 0.02 0.04 0.06 0,08 [s] 0.10 (file Exampl1.pl4; x—vart) 0151 El t (CWRMS ECE 524: Session 11; Page 6/17 Transients in Power Systems Spring 2012 Now suppose capacitor C1 had trapped charge. Use the added circuit to precharge the capacitor. The resistor has a very small resistance and is used to prevent a loop of switches. ie 2229uFilf «b T 40.11%? 9000 [A] 5000— 3000— Current response: «3000— ka := 8172A -6000— 45" ‘90001——-—-—r————1—v—1—r—-——1—-——1—t 30 35 4O 45 50 55 60 65 [rm] 70 (file Examp|1.pl4; x~vart) o:B'I ~CSW1 0:81 -C1 Voltage response: 90 [WV] 60 Vpk := 84.462kV 30 -30 —60 —90 30 35 4O 45 50 55 60 65 [ms] 70 (file Exampl1.pl4; x-var t) v (:1 / v:VS L/i ECE 524: Session 11; Page 7/17 Transients in Power Systems Spring 2012 Suppose that X/R of the source is not infinite, and instead X/R = 10. % : +onv‘C lo ) E}: 3 2 Recall 28:: J— 2: R2+X2= ZS +X2= X2_i+1 sc 5 10 100 R: Es) “5 (95) Xs XS = 0.793 S2 LS :2 — Ls = 2.103 ~mH w Xs RS 2: T6 RS = 0.0793 9 2 Rs 7 1 . _ = —1.19 x 10 — Negative result, so 2‘Ls Ls'Cl 52 underdamped circuit. _RS OL := 2-LS Particular solution: 1 — w-Ls U)‘ 1 n := atan —-R— n = 89.93-deg (av/2 2W ECE 524: Transients in Power Systems iL_p(t) I= -——-—-———-——- -sin(w-t + d) + n) Homogeneous Solution: ‘3 iL_h(t) = eu't-[kl-(cos(wd-t) + kZ-sin(wd-t) ] ~ ~ Boundary conditions: iL(O) = 0 Switch was open. 0 = eo.(k1.1 + k2.()) Therefore k1 := 0 Vm = LS-kz-ea'O-(cx-sifiwdfl) + wd-cos(wd-O)) Vm = -——- k = 3890.61 A LS-l-wd-l 2 k2: iL_h(t) := kZ-ea't-(sin(wd-t)) iLU) 3: iL_h(‘[) + iL ~13“) Session 11; Page 8/17 Spring 2012 ECE 524: Transients in Power Systems 4000 [A] 2500 1000 -500 -2000 -3500 6000 , 1‘ k2.ea.t-(sin(wd-t)) (meé‘iggp.4;“335:3: ~01 6?“ 66"” 72:22 7778 ["5] 8333 Positive peak: 3800.7A Negative peak: —4086.3A Capacitor voltage vat) = mt) dt + we) t ._ k2 eOL-t Vc_wd() -— C1 2 2 0L +UJd Vc“) Ix Ve_w(t) + Vc_wd(t) -(o¢-sin(wd- ) -cos(w-t + <1) + 11) Session 11; Page 10/17 Spring 2012 t —wd-cos(wd-t)) Go,“ ECE 524: Session 11; Page 11/17 Transients in Power Systems Spring 2012 0 0.01 0.02 0.03 0 0.01 0.02 003 Max: 54.6kV Min: ~51.0 kV ECE 524: Transients in Power Systems ATP results: 60 [W] 40 20 ~20 -40 ~60 50.000 55.556 61.111 66666 72.222 77.777 {ms} 83.333 (file Exampipl4; x-vart) v:C1 ~ Max: 54.1kV Min: -51.2kV G) ‘V' wit" \ @f. :‘L, 3; Switch in capacitor 2 through a resistance: Session 11; Page 12/17 Spring 2012 Eliminate the source resistance again. Suppose the bank 2 is energized through a resistor with bank 1 already in the system. Determine the resistance needed to limit the peak line to ground voltage on either bank to 33.5kV C1'C2 C :=—————— “1 C1+C2 Ceq = 14.33%]? Simulation Results With R = 25.01 ohm, switching at t = 15.6lrns. Results identical with L1 ignored. ...
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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-s12 - Nfl .0: ZOHmmmm m_&amp;gt;_m_._.m&amp;gt;m...

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