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ece524-s18 - mnm mNA g afizamzd Hz woémm m<m m2m...

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Unformatted text preview: mnm mNA g. afizamzd Hz woémm m<m+m2m mmmmSZ so. 5 H. c3m<mfimeO\.—Qm_do : _ _ Guzmanqozamro _ ECE 524: ' Session 18; Page THO Transients in Power Systems Spring 2012 ECE 524: Lecture 18 Three phase TRV transients MVAR :m iOOOkW My; 10‘ 6m 6M.» Consider the three phase system shown below: ' 12mm _‘ N51121: 2"”: 34.5w @ '- * = “a. 2 ma Sm" VLL I: 34.5kV Vm I: E-VLL Vm *"—’” 28.1710] ZS 2m 3-10—50hm 2 VLL QL := 12MVAR XL 2: XL a 99.199 QL XL 263 l H j: W E . m Lph 2-w-60Hz LP“ “ deMCIfiIL‘ CW 2: 2119 [Mei-M 1 1 1 fE1 :m fn = 6938.13“ 2m llfihcph 5 TH z: i Tn = 144.13us fn M Choose simuiation time step: T“ 48 --—- = . s 30 y” ca<mwmfi<oiam30 .\ .. ECE 524: Session 18; Page 2/10 Transients in Power Systems Spring 2012 Case 1: Zn :m ooohm Breaker currents 100 400 -200 0.00 0.02 0.04 0,06 0.08 [5] 0.10 (Hie Examplepw; xnvarn CIVSWA -LOAOA ctVSWf-l ALOADE! CIVSWC —LDADC Neutrai to ground voltage: 0.0 -'?’.5 -15.0 ~22.5 .N M '30}; 49.5 50.5 51.5 52.5 53.5 {ITS} 54.5 i (1 iie Example.plfi; xwv ar 1) v:N!EU ~ 509 W] 37.5 25.0 12.5 0.0 42.5 -25.0 -37.5 I500 0.00 0.02 0.0‘3 0.06 . [3? 0,10 (1 He Exampfiegfid: x—var I) v:LOA€)A ~NEU v:LOAOB NEU v:$._OADC -NEU Session 18; Page 6110 Spring 2012 ECE 524: Transients in Power Systems Case 3: Repeat with CT! :2 3-01,}. (In : 60F Iv WS'kcv Breaker currents: 300 25;“ 100 0 400 ~200 4500 0.00 0.02 0 .04 0.06 0.08 is; 0.10 (fife Exampiepmmwam) 01V5WA vLOADA c:V5WB -LOADB c:vswc -LOADC Neutral current afier first phase clears: 5.00 [A] 3.75 2.50 1.25 0.00 4.25 ~25!) 33.75 -5.00 49.0 50.6 50.2 50.4 50.6 50.8 51.0 51.2 ins; 51.4 We ExampteL17.pl4; x—var l) c:NEU - Neutrai current when ali phases clear: 5.00 [A] 3.75 2.50 ‘ 1.25 0.00 4.25 «2.50 -3.75 -5.00 53.5 53.8 54.1 54.4 54.7 imsi 55.0 (Ma ExampleL1?.p!4: xvvar l) c:NEU - ECE 524: Session 18; Page 8110 Transients in Power Systems Spring 2012 Breaker voltages: 20 —70 0.0330 0.0464 0.0598 0.0732 0.0866 {8} 01000 (file ExampéeL17.;)l4;x-vart)V:VSA -LOADA V:VSB -LOADB v:VSC -LGADC Arm: 78.846kV Case 4: high resistance grounci: Rgr I ph 1 chh m 1.33x106n chh := W 2-1T-60HZ'Cph W kg. :m 3.xcph Rgr ; 3.98MB 90 $le Wm 84.459kV 7° 50 Vmax Vm 2 3 30 10 ~10 -30 «50 -7‘0 - 0.0330 0.0464 0.0598 0.0732 0.0866 [3] 01000 (file Example£17.pl4; x-var l) ViVSA -LOADA v :VSB ~LOA08 v:VSC -LOADC ECE 524: Session 18; Page 9/10 Transients in Power Systems Spring 2012 Case 5: Add grounding resistance (high resistance ground——limit In to 20A): VLL Inmax :2 20A "Ram: Wfié Kg? 2 995.93 a use 1000 ohm nmax Breaker voltages: 18 -34 —60 0.0330 0.0464 0.0598 0.0732 0.0866 [5] 0.1000 (file ExampleL17.pi4;x-var£)VIVSA —LOA{)A v:VSB -L.0Af)B v:VSC -LOADC Case 6: Now try a low resistance ground (In 2 500A) 2mm? 500A 53%: “k Kg1r = 39.849 use 400m ' nmax 60 W: 56.771kV {:1 Vj: = 2.02 20 -2{} an} -60 0.0330 0.0464 0.0598 0.0732 0.0866 [5] 0.1000 (file ExampleL17.pM; x—var :} v:VSA -LOADA v:VSB ‘LOADB vzvsc -LOADC ...
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