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# Transients in Power Systems ECE 524

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This 120 page Class Notes was uploaded by Fredy Okuneva on Thursday October 22, 2015. The Class Notes belongs to ECE 524 at University of Idaho taught by Brian Johnson in Fall. Since its upload, it has received 32 views. For similar materials see /class/227723/ece-524-university-of-idaho in ELECTRICAL AND COMPUTER ENGINEERING at University of Idaho.

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Date Created: 10/22/15

as 39ou NOISSHS SNELLSAS HHMOd NI ISLNEIISNVHJ 1729 333 m Universityofldaho osmntcbmmggmca E m Univearsityofldaho m Universityofldaho m Universityofldaho Universityofldaho Swarm GAP RafgCVbg Seasz CTM awex M GAP z 3 4 fink i quot1 Yam L Bffass breaker 2a39l ria Q 00 4 00 4535C M Universityofldaho D DAL GAP V YECfJE SCHEZ L E 27 AP 3 CterL 1 H 1N a G L S S G39an 31 ask S4 Qedksef c brmk er GM 4 J GAP 3 quio V Lilo 3d r BVPASS brcuk v Qev iny 4 4493 Qquot 88quot 73900 Msec m Universityofldaho ZNC OXICSE ROTECVVz 5057 sgt Mm 3739 Pass Erecake E 35 g osmvtob hmzcn z 2 m Universityofldaha Vein LAer Abraan c9J 2 KW V406 gm 3 285090 w Universityofldaho GOLbsweaemv39 McDDEL 9 c 3w if IT 0 5 n r km v Strl s CAP 2013 11quot MOV R Wm me A wJ I 3 Wm Universityofldaho M C T C LIMACTEB Sift 19va LP 49 Universityofldaho EQUIVALewr Clem CS VW H W 2kN 11 Xvw 11 WW TObbst 7VY M0 p Z AL V Universityofldaho Z 56 QFAW reReg 3 I fg AZ7 pleuwws jr CAquotP3gt ECE 52439 Transients in Power Systems Session 44 Page 111 Spring 2008 Lightning Examples PSCAD Solution Consider a transmission line with towers that are 40m tall and spaced 280m apart Assume that there is a single shield wire with a characteristic impedance of 520 ohm and assume that the tower ground strap has a characteristic impedance of 135 ohm and the tower has a footing resistance of 30 ohm Assume a propagation velocity of the speed of light for the ground wires and 085 times the speed of light for the tower ground strap Assume that the phase conductors are 75 of the way up the tower and that the ground wire is segmented and open at the top of each of the adjacent towers Consider the case of a lightning strike where the current rises to 40kA in 2 us and falls to 20kA after 40 us De ne parameters De ne speed of light Rfoot 300hm Z07gw 5200hm Htower 40m Zeitower 135ohm Hconductor 3 075 Ht0wer Hconductor 30m DSpan 280m 7 F Mo 4n10 80 885410 12 m 1 5 km 0 c 3 X 10 080 sec vgw c 5 km vtower 085c vtower 25483 X 10 Q ysec 107 6sec ECE 524 Session 44 Page 211 Transients in Power Systems Spring 2008 a Find the peak voltage at the tower top the bottom of the tower and at the conductor height if the lightning strike the top of the tower PSCAD Schematic from le Lightninglpsc m It Surge Simple Lightning Surge x 39 40kAlT1t C1t T1 i ll i lt3quot i ll i a an rite 4WltI IW iFiI E GWL GWL GWR GWR 3 39 J g El 3 ltltT TovTL39er1 T J gtgt i LJeMOi ill Vtop CrossA g Vfoot H 0 El K5 Notes 3 0 Note the transmission line models used for the Tow er2 ltltI ground straps and static wires 0 PSCAD requires aterminating impedance at the end aemoi of a line it can t be open circuited A very large ltT resistance has been placed at the end of the line 6 o i 30ohm EOE 52o Sessmn M Page 3m Tvanswems m Paw svmms Spnng 2m Current surge source dulog box Wm mms m nnmatemmunsmne mmmmmnwm mum mam shin Q mmmmm mm Cnmmem am New P10 oflxghtmng current W has n a BMW nnm nmm nmm nnsm Voltages at towertop cross arm and footing resistance Nam aams u u 1 W v up aussA vmd u m umnm nu um I E m unwm unEnm ECE 52439 Session 44 Page 411 Transients in Power Systems Spring 2008 Zoom in on peak of voltage waveform M39n ais y W ot39ID odrm odm OdIB39n 0dEn 003910n oo39izn 003914n Od 39n 00391871 As one would expect the top shows the biggest peaks in voltage Note also the time delay in the voltage at the footing resistor Vpeakitop l5609103kV at twp 25075ysec Vpeakicrossarm l4876103kV at tcross 25400ysec 3 Note that this is actually the Vpeakifoot 3 132943910 kV at tfoot 3 43275Hsec second relative peak in this waveform Session 44 Page 511 ECE 52439 Spring 2008 Transients in Power Systems b Repeat of the lightning strike occurs 100m away from the tower top in either direction PSCAD Circuit Diagram from le Lightning2psc Ia f x i Simple Lightning Surge 40kNT1 t C1t T1 ollv VH3 is gtW Wltlt IgtWl IWltlt hgtgtW GWL 2 GWL 2 GWL 4Wltlt DU wuo OLQCIi ii L gi ltltll Tow er1 52 we OLQOL LJQNO Vt CrossA 4 RE Vfoot LJQVO CrggsA 4 ZJQMQL lgt Twig g 394 S 8 30oh m ECE 524 Session 44 Page 611 Transients in Power Systems Spring 2008 Voltages at tower top cross arm and footing resistance Main Gapi s 2w IVtop i O39ossA iVfoot 2 that the peak E 1ges are higher in this 003900 003910m 00 0m 005 001pm 0050m i Wn a s Zoom in on peak of voltage waveform E gt 06a 00i12m 0d14m 0an 00le 003910m 003912m 003914m 003916m Againthe top shows the biggest peaks in voltage Note also the time delay in the voltage at the footing resist0 Vpeakitop 20707103kV at twp 28275ysec Note that the time to the peak is delayed due to travel time along the ground wire Vpeakicrossarm l9251103kV at tcross 28650ysec Vpeakifoot l6244103kV at Ifoot 40525ysec Note that this is actually the second relative peak in this waveform but it is earlier than the last case ECE 524 Session 44 Page 711 Transients in Power Systems Spring 2008 c Repeat part a if the adjacent towers are each grounded with the same ground strap characteristics and the same footing resistances o For this case we will just model the adj actent towers in detail since we aren39t concerned with waiting for further re ections to come back We do model both ground wires at the top of each though The model includes the static wires beyond the next tower as well terminated with a very large resistance 0 From le Lightning3psc TIME lSimple Lighming Surge V 40kNT1t C1 tT1 a 3 i AIWltltamp WH LIW i mi m wwi ulna mam GWL GWL GWR GWR GWR2 GWR2 VJSMOi LJSMQL QJSMOi 4 Li 4 Vlp J E39 g 339 g I 3 ii 5 a El CrosA A j Vfool K E E a U a o o m g m T 5 TowerZ 3 i T 4 Transients in Power Systems vain Graphs l OrossA Session 44 Page 811 Spring 2008 oltages at tower top ross arm and footing sistance o Zoomed voltages at 0 Note that the peak tower top cross arm and footing resistance voltage is significantly lower than the other case s oltages at tops of e next towers to th eft and ri ht 5 5 gt MainzGrapi s i m IVtop ilCrossA IVfoot i gt 0000 00me 0004m 0006m 0008m 00i0m 003912m 00i4m 003916m i NaintGraphs IVT L IVT R 35000k op i i I 3000 M 25000k ii 20000k 5 5 15000k gt 10000k 5000k 000 0000 0005m 00i0m 00i5m 39 003925m 39 003930m 0035m ECE 524 Session 44 Page 911 Transients in Power Systems Spring 2008 Direct comparison of PSCAD and ATP Results for the First Simulation Case 0 Import PSCAD output data le PSCADout Column information retrieved from Cnoname01out Jnf le lt2 Time vector H PS C ADoutlt0gt Phase A source voltage Vtop PSCADout Surge iS PSCADoutlt1gt Phase A load voltage Vfoot PSCADoutlt3gt Phase A switch voltage VcrossA PSCADoutlt4gt 0 Import ATP Output Datafile exported by PlotXY data I CatpLAADF Time vector ta datalt0gt Top VtopA datalt1gt VcrossA datalt2gt Foot Vfoota datalt3gt lt4 Isurge data 0 Compare the lightning surge currents ECE 524 Session 44 Page 1011 Transients in Power Systems Spring 2008 0 Compare tower top voltages I I I I 0 1105 2 105 3 105 4 105 5 105 ttta 0 From this View they look identical now zoom in 15 106 0 210 410 610 810 6110 5 ttta 1210 51410 5 Again a very good match ECE 524 Session 44 Page 1111 Transients in Power Systems Spring 2008 0 Compare voltages at the crossarm 15106 Vcros SA VcrossA 0 110 210 5 310 5 410 5 ttta 0 Compare voltage at footing 15 39106 I I 139106 Vfoot Vfoota 5105 0 l 5 l 5 l 5 l 5 5 0 13910 210 310 410 510 ttta ECE 52439 Transients in Power Systems Session 44 Page 19 Spring 2008 Lightning Example Consider a transmission line with towers that are 40m tall and spaced 280m apart Assume that there is a single shield wire with a characteristic impedance of 520 ohm and assume that the tower ground strap has a characteristic impedance of 135 ohm and the tower has a footing resistance of 30 ohm Assume a propagation velocity of the speed of light for the ground wires and 085 times the speed of light for the tower ground strap Assume that the phase conductors are 75 of the way up the tower and that the ground wire is segmented and open at the top of each of the adjacent towers Consider the case of a lightning strike where the current rises to 40kA in 2 its and falls to 20kA a er 40 Ms De ne parameters De ne speed of light Rfoot 300hm Z07gw 5200hm Htower 40m Zeitower 1350hm Hconductor 3 07539Htower Hconductor 30m Dspan 280m H F Mo 441710 7 80 885410 12 m m l 5km c c3gtlt10 No80 sec ngc 5 km vtower 085gtc vtower 25483X 10 Q usec 10 6sec gogau Sessmn u P392215 TvanswemsmPuwevSv ems s mg Inna I top L L ugmmng sinks the top ofthe tower ATPDraW Schemauc moxg GndWwe Gm Wwe WARR VFAR39L Envy Paste enhvedalagnd mm m Lahe Eammem 7mg ar smr a want r We r Ltd r Vahage M EK W M ECE 524 Session 44 Page 39 Transients in Power Systems Spring 2008 A TP Data File BEGIN NEW DATA CASE C i i i i i i i i i i i i i i i i i i i i i i i i i i i C Generated by ATPDRAW January Wednesday 4 2006 C A Bonneville Power Administration program C Programmed by H K Heidalen at SEfAS 7 NORWAY 199472003 C C dT gtlt TmaX gtlt Xopt gtlt Copt gt 2 5E79 5E75 500 1 o o o o o 1 o c i 2 3 4 5 6 7 8 c 345678901234567890123456789012345678901234567890123456789012345678901234567890 BRANCH C lt n lgtlt n 2gtltreflgtltref2gtlt R gtlt L gtlt C gt C lt n lgtlt n 2gtltreflgtltref2gtlt R gtlt A gtlt B gtltLenggtltgtltgt0 lCROSSARFOOT 135255E8 30 l 0 0 RFOOT 30 lVFARL TTOP 520 3E8 280 lTTOP VFARR 520 3E8 280 lTTOP CROSSA 1352 55E8 lO gt gt gt 000 o gtlt n 2gtlt Tclose gtltTopTde gtlt le gtltVfCLOP gtlt STROKETTOP SOURCE type gt MEASURING l C lt n lgtltgtlt Ampl gtlt Freq gtltPhaseT0gtlt Al gtlt Tl gtlt TSTART gtlt TSTOP l3STROKEil 4E4 OUTPUT TTOP CROSSA BLANK BRANCH gt E76 2E4 4E75 5E77 1 Plot of lightning current I I 10 20 30 40 1039B 50 Me probZa pi4 w ar 1 i STROKETFC E ECE 524 Session 44 Page 49 Transients in Power Systems Spring 2008 Voltages at tower top cross arm and footing resistance 1 10 14 30 4o 10 6 50 10 20 file probZa pill erart v TRAP v CROSSA v RFOOTV Zoom in on peak of voltage waveform 0 i 2 4 6 8 10 12 14 1039B 1B fiie probZa pill wart v Win v CROSSA v RFOOTV As one would expect the top shows the biggest peaks in voltage Note also the time delay in the voltage at the footing resistor vpeakitop 15609r103kV at twp 25075Msec Vpeakicrossarm 14876gt103kV at tcross 25400Msec 3 Note that this is actually the Vpeakifoot 3 132943910 kV at tfoot 3 43275Hsec second relative peak in this waveform ECE 524 Session 44 Page 59 Transients in Power Systems Spring 2008 b Repeat of the lightning strike occurs 100m away from the tower top in either direction T sTROKE ATP Draw C1rcu1t D1agra1n n Gnd Wre Gn d V re VFARL E Grd V re VFARR E STRIKE CROSSA Tower Voltages at tower top cross arm and footing resistance I u u u u u u u u 30 40 quot10B 50 fiie probe piA wart v TTOP v CROSSA v RFOOTV Note that the peak voltages are higher in this case ECE 524 Session 44 Page 69 Transients in Power Systems Spring 2008 Zoom in on peak of voltage waveform 3 6 Me probe pill wart v min v CROSSA v RFOOTV Againthe top shows the biggest peaks in voltage Note also the time delay in the voltage at the footing resistor vpeakitop 20707r103kV at twp 28275usec Note that the time to the peak is delayed due to travel time along the ground wire Vpeakicrossarm 19251gt103kV at tcross 28650usec Vpeakifoot 16244gt103kV at tfoot 40525usec Note that this is actually the second relative peak in this waveform but it is earlier than the last case c Repeat part a if the adjacent towers are each grounded with the same ground strap characterist and the same footing resistances 0 For this case we will just model the adjactent towers in detail since we aren39t concerned with waiting for further re ections to come back 0 We do model both ground wires at the top of each though ECE 524 Session 44 Page 79 Transients in Power Systems Spring 2008 Gnd iWely nd iWe EH E E9 CROSSA Tower 30 ohm I I 30 40 10396 50 l 10 20 fiie proch pi4 erar t v 77 iF v CROSSA v RFOOTV I 8 10 10396 12 2 4 6 Me proch pi4 wan v TWP v CROSSA v RFOOTV ECE 524 Session 44 Page 89 Transients in Power Systems Spring 2008 103 300 250 200 150 100 I 30 40 1039B 50 10 20 Me prob2c pi4 wan w vr APP v VFARL ECE 524 Transients in Power Systems Session 44 Page 99 Spring 2008 Typical cable data 15 kV underground pipe type cable below at 60 Hz Assume the sheaths are grounded and the pipe are grounded Inner Pipe Diameter 348 in Outer Pipe Diameter 358 in Outer Jacket Diameter 41 in Resistivity ofpipe 2828E8 ohmm Relative permeability of pipe 10 Relative permittivity inside pipe 10 Relative permittivity of jacket 35 Angle to phase 1 0 degrees Radius to center phase 1 081 in Angle to phase 2 120 degrees Radius to center phase 2 081 in Angle to phase 3 120 degrees Radius to center phase 3 081 in Depth ofcenter ofpipe 1 meter SC cable inner radius 0 in SC core diameter 08 in SC Sheath inner diameter 1324 in SC Sheath outer diameter 14 in Resistivity SC Core 1724E8 ohmm Relative permeability of core 10 Relative permeability of insulator 1 10 Relative permittivity of insulator 1 35 Resistivity sheath 1724E8 ohmm Relative permeability of sheath 10 1717 39OU NOISSEIS SIAEIJSAS EEIMOd NI SiNElISNVELL 1725 332 m Universityofldaho m Universityofldaho QSMEKOEEEZS i universityofldaho m Universityofldaho m Universityofldaho m Universityofldaho Tmnsw osmEkDEmSED a 34 217 39OU NOISSES SIAEIJSAS HEIMOd NI SiNElISNVHJ 1725 EDEI Session 41 Page 118 Spring 2008 ECE524 Transients in Power Systems Numerical Oscillations in EMTPLike Programs 1 Causes of Numerical Oscillations The Electromagnetic transients program and its variants all use the the trapezoidal rule numer ical integration method The trapezoidal rule is a second order numerical integration technique that is simple to implement astable numerically stable in stiff systems and fast However it is also susceptible to numerical oscillations when differentiating step changes in voltage or current The program user should be aware of the potential for these oscillations when simulating circuits and systems using EMTP The most common events leading to numerical oscillations are a step change in current through and inductor or a step change in voltage across a capacitor The cause if this problem can be seen by looking at the differential equations for the inductor and the capacitor r i Inductor Voltage 1 L L g 1 dt at do CL paC LtOT39 Current t C H t 2 One can think of the numerical oscillations as resulting from forcing an in nite 171 It across and inductor or an infinite dvdt on a capacitor One possible solution is to produce a more accurate circuit model by including parasitic capacitances internal resistances in a capacitor 11 Mathematical Representation The problem can be represented by modeling the equation ll it 3 using the trapezoidal rule We can integrate both sides of the equation resulting in the following e V ewe t 1t tm mlt 1175 At 4 Next we rewrite the equation using the trapezoidal rule Remember we are finding an approxi mation of the area of the trapezoid bounded by 3 t and t At ya 3105 Ar 9 5ltmlttgt are Am 5 1 17 was 933 msi Session 41 Page 218 Spring 2008 ECE524 Transients in Power Systems Now we can solve this equation for 1t which could be either the current through a capacitor or the voltage across an inductor 2 we mltt m mom ye At 6 Start out with rt At 0 We will now have y undergo a step change from yt At 0 to 10 The function will stay at 10 for the rest of the time period of interest Ve would expect yt and xt to behave as shown in Figure 1 Note that we would expect to be an impulse Dirac Delta Function at at 1 z t and zero everywhere else no L0 Time t At t At Figure 1 Plots showing yt and expected It for example case Plug the know values into the equation for 361 resulting in 2 1 t 1lt gt At 7 Now step ahead and nd xt At 2 2 2 i m Ar N Ain 0 0 At 8 if we continue ahead another time step we nd that 2 A I 9 tt 2 t N tan Universityofldaho Luz 31w ECE524 Transients in Power Systems Session 41 Page 618 Spring 2008 122 Capacitor Voltage The second example is a single phase system with a sinusoidal voltage source supplying a pi circuit A switch connecting the source to the pi circuit closes at 16667 msec causing a step changing in the capacitor voltage on the capacitor near the source Figure 5 shows EMTP simulation results The rst plot shows the current through the capacitor connected to the switch Notice the large amplitude of the oscillations in this case The second plot zooms in on the oscillations themselves and the third plot shows that the voltage across the capacitor is not impacted The EMTP data le is provided below aw x104 I 0 2 quotI l I l L I I 0002 0004 0006 0008 001 0012 0014 0016 0018 002 p A lca l N l 1 l I l l 1 l I 0017 0017 0017 0017 0017 00171 00171 00171 00171 00171 00171 10 X 15 1 l I l l l I I 10 a I 5 0 4L 0 l l l l 1 0002 0004 0006 0008 001 0012 0014 0016 0018 002 Time sec 4 l Va1V Figure 5 Capacitor example BEGIN NEW DATA CASE C EMTP Datafile showing numerical oscillations when changing the C voltage across a capacitor C C Miscellaneous data C DeltaTlt TMaxlt X0pt C0ptlt EpsilnltTolMatlt T8tart 10E5 002 C IDutltIPlotltIDoublltKSSOutlt Max0utlt IPunlt MemSavlt 1Catlt NEnergltIPrSup 111 1 1 C C Circuit data C Bus1gtBu52gtBu53 gtBus4gtlt Rlt Llt C 67139 2h ECE524 Transients in Power Systems Session 41 Page 1018 Spring 2008 it it Figure 8 Snubber circuit provides a bypass path for the switch the numerical oscillations are avoided and a more accurate model of the power electronic circuit is used However some circuit topologies done need snubbers diode recti ers for example in addi tion switches with large safe operating areas SOA such as IGBT s and MOSFET s don t need snubbers In this cases the program user can add numerical snubbers These numerical snubbers can also be added for conventional switches as well although they have less of a basis in reality If a capacitor alone is used a capacitance value of roughly 1 2 nF will suf ce If a RC snubber is used as shown in Figure 8 the time constant for the RC snubber must still be greater than the simulation time step to avoid problems The resistance value should be chosen so the RC time constant is a minimum of 2 3 times the simulation time step At Performance of the RC snubber will vary with the circuit and the user may need to vary R and C values for best performance The inductor current example from earlier was simulated with a RC snubber across the switch as shown in Figure 9 The first plot shows the resulting voltage Notice that there are no oscillations The second plot zooms in on the voltage waveform Note that the voltage sees a decaying exponential from overdamped RLC response The third plot shows the snubber current Notice also that the time step used in the data le below is smaller to capture the LC resonance BEGIN NEW DATA CASE C EMTP Datacase showing numerical oscillations resulting from opening C a switch in series with an inductor C C Add a snubber across the switch C C C Miscellaneous data C DeltaTlt TMaxlt X0ptlt COptlt EpsilnltTolMatlt TStart 4E6 003 C IOutlt IPlotltquotIDoubllt KSSOutltMax0utlt IPunltMemSavlt ICatlt NEnerglt IPrSup 11 1 1 5 2h ECE524 Session 41 Page 1118 Transients in Power Systems Spring 2008 x16 1 r e mww V bus V c I l i 0005 001 0015 002 0025 003 l snubber A m ac ca Figure 9 Inductor current example with snubber added C C Circuit data C Bus1 gtBu52 gtBus3 gtBus4gtlt Rlt Llt C GENl B081 20 1 BUS2 25 C Snubber circuit RC 3DELTAT GENl 8081 1500 010 1 BLANK ends circuit data C C Switch data C Bus gtBus gtlt Tcloselt Topenlt Ie 0 BUSl BU32 1 1E 3 BLANK ends switch data C C Source data C Bus gtltIltAmplitudeltFrequencylt TOlPhiOlt OPhi0 lt Tstartlt Tstop 14GEN1 10000E3 60 0 1 111 BLANK ends Source data C C Output Request Data C Bus gtBus gtBusgtBusgtBus gtBus gtBus gtBus gtBusgtBus gtBusgtBusgtBus gt BUSl BUSZ GENl BLANK ends output requests BLANK ends plot request BEGIN NEW DATA CASE BLANK ends all cases 23 Reducing the Time step In some cases reducing the simulation time step can also eliminate numerical oscillations How ever7 this is not a general purpose solution It depends on the presence of resistances in the circuit preferably in parallel with inductances or in series with the capacitances In many cases the required time step is also far too small for practical simulation Also since the amplitude of the oscillation varies with lAt the smaller time step may also make the oscillations larger of St P 395 Um 7W W Universityofldaho Cox 1 meow a mw ma a w ww m acukohm Ohm mg gmr db 1 4 mam Scum 1W 07 QM skunk E C3mltmwmm olamro Universityofldaho Universityof Idaho lolw LE 390 NOISSEIS SIAELLSAS EEIMOd NI SiNEIISNVELL 1729 EDEI 665 52 as Universityofldaho quot a a h mm sm w w 43 39 gags 0 ng 3 m Universityofldaho m Universityofldaho WWW m Universityofldaho Universityofldaho Universityofldaho 1 m ocmvtobimSED i m Universityofldaho m Universityofldaho OLMEEEEMSED 5 ocwttogmgmzcs E U Department nglgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 1 Trapped Charge 0 Circuit breakers open near natural current zero 0 Current through a capacitance 900 out of phase with voltage 0 Results in charge trapped on capacitance 0 Easy to visualize with shunt capacitor banks 0 Will also occur with transmission lines and cables 0 Important concern when energizing circuits 0 Simple emtp line models won t allow discharge Trapped Charge Spring 2008 U Department ofElgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 2 Energizing a Line 0 We saw earlier that energizing a line with no trapped charge can result in up to a 2 pu voltage at receiving end 0 Distributed capacitance stores charge 0 Big currents can arise when reclose unless match voltages 0 Worst case when very small source impedance 0 Can t match all three phases well with three phase line Trapped Charge Spring 2008 ECE 524 Transients in Power Systems U Department of E lectrical Session 22 Page 3 I Engineering Receiving End Voltage When EnergizingReclosing Open Line I I vonage Von H n as n ma me see Trapped Charge Spring 2008 ECE 524 Transients in Power Systems U Department of E lectrical Session 22 Page 4 I Engineering Sending End Voltage When EnergizingReclosing Open Line vowage Von Trapped Charge Spring 2008 U Department nglgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 5 Sending End Current When EnergizingReclosing Open Line Lme Current m Hr current Amp 72m nu n me n ma me see Trapped Charge Spring 2008 U Department ofElgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 6 Reducing Energization Transient 0 Can also reduce energization transient by using a pre insertion resistor Z S Line 0 Recall that TV iii so choose R ZC 0 Close switch in series with R rst now Vapphed l 2V1 o No re ection of when wave returns from far end of the line 0 Can close rnain switch after a few travel times will be small transient o Resistor must be sized for current it will carry and energy dissipation Trapped Charge Spring 2008 U Department nglgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 7 Preinsertion Resistors for Three Phase Line 0 How do you choose R for a three phase or higher line 0 Set it equal to one of the line modes 0 Won t eliminate transient but will cut it down 0 Often sized at a typical value for line mode characteristic impedance 5009 Trapped Charge Spring 2008 U Department ofElgctricgl ECE 524 Transients in Power Systems I Engineering Session 22 Page 8 Receiving End Voltage with PreInsertion Resistors Voltage Volt um um n n1 n12 n4 as n ma me see Trapped Charge Spring 2008 U Department of E lectrical I Engineering ECE 524 Transients in Power Systems Session 22 Page 9 Sending End Voltage with PreInsertion Resistors Vonage Von Trapped Charge Spring 2008 U Department of E lectrical I Engineering E CE 524 Transients in Power Systems Session 22 Page 10 Sending End Current with PreInsertion Resistors 0mm Amp 74nn nus an a me see Trapped Charge Spring 2008 AZ 39OU NOISSEIS SWELLSAS HEIMOd NI SiNElISNVHJ i725 EDEI L17 3 ECE 524 Session 27 Page 611 Transients in Power Systems Spring 2008 Matrix of eigenvectors Ti for current equation 06 071 041 346gtlt 10 ml 1 eigenvecswz Ti 054 0 081 Ii2 eigenvalsYZ 1 64 x 10 12 l 06 o71 041 1 39 12 m2 L59x10 J o71 06 o41 2 298x10 1ll Ti 1 eigenvecsCL Ti 2 0 054 081 eigenvalsCL 6 3 X 10 11 071 O6 041 nnz 29 x 10 H j Note that LC and ZY have a similar matrix of eigenvectors but the last two columns are swapped but different eigenvalues Also note that ZY and Y2 have the same eigenvalues Likewise LC and CL have the same eigenvalues different order 1 0 717gtlt 10 T Em 9 4 a L EL 3i L17 m Lac us Gk m SW WM ATPbmu mM adtm m 5 y h Smut ampa M 3 4L9 m Universiwofldaho 1 mmdana CawcAmu m MOM w 27 a w LIB am ccmvabmm gtc2 E ECE 524 TRANSIENTS IN POWER SYSTEMS SESSION no 35 m Universityofldaho L Ne 93 branch wa Q wm a 0231033522 E m Universityofldaho g U40 HMS ozmvtob ggmcn i osmUIoDEEgtED i tyofldaho O IVEI39SI O mUn VJ ocmvtob aSED i U1 ECE524 Lecture 33 Limitations 0 Limited to two winding transformers 0 It is very easy to create numerical problems in the simulation with the ideal transformer Transformers Spring 2008 E CE524 U I Saturable Transformer Lecture 33 0 Model has builtin circuit elements Winding resistance Leakage inductance can t enter 0 Core loss resistance tauHM W Magnetizing branch not entered as an L in mH Can set all except leakage to O to simplify Enter winding to winding ratios awed Transformers Spring 2008 lE S m5 is at 9E 39OU NOISSEIS SIAEJJSAS HEMOd NI SiNEIISNVHi 1725 EDEI Session 36 Page 46 Spring 2008 ECE 524 Transients in Power Systems 0 Note that the 00 point was added to get the appropriate plot but the program will add it internally when running the case 1328 U Wm 996 664 332 039 I A ms 0 o 1 7 33 5 o 6 6 Autotransformer case AUTOTRANS 100 200 4300 quot400 I I I I I I I I I 10 20 30 40 n39s 50 file lectBSApl4 xvart vHV vHV vLV vLV 9 na ECE 524 Session 36 Page 56 Transients in Power Systems Spring 2008 Ground Transformer Example Ground Transformer Resistor Current for SLG fault 1 E A 9 10 v O II I II n r a F A V r A a39 H 3 15 I I I I 45 50 55 60 ms 65 W 9 37 35 40 file lect368pl4 xvart chVA RNEG ECE 524 Session 36 Page 66 Transients in Power Systems Spring 2008 Phase currents to the transformer for SLG fault not all in phase 20 A 15 1 I r I I I 48 50 52 54 56 58 60 file lec1368pi4 xvart CIVSENDA GNDPA CIVSENDB GNDPB CZVSENDCGNDF C Ground transformer resistor current for a three phase fault 30 PA 08 39839 r I I I 17 r I I I I 000 002 004 006 008 s 010 file IectSBBpM xv ar t chVA RNEG E C3mltmwmm o1amro U What is Needed in ECE524 I Transformer Models Lemmas 0 Amount of detail depends on frequency of desired response 0 Power flow stability often just model leakage perhaps winding resistance Tap changing Perhaps wyedelta phase shift Transformers Spring 2008 What is Needed in mm I Transformer Models Lemmas 0 Fault studies require more information Connection info wye delta zigzag autotransformer etc Grounding Possible impact of tertiary Transformers Spring 2008 U Low Frequency mm I Lecture 33 0 Similar modeling info to fault programs Connection information more important 0 Magnetizing branch Saturation 0 Core loss term 0 Not using per unit Need to include turns ratio Divide leakage L winding R between windings Transformers Spring 2008 Single Phase mm U I Equlvalent C1rcu1t Lecture33 Wifw o Winding resistance 0 Leakage inductance 0 Core losstota losses o Nonlinear inductor model for magnetizing branch Transformers Spring 2008 U I ATP Options ECE524 Lecture 33 0 Ideal transformer component 0 Saturable transformer component 0 BCTran preprocessor that converts description of transformer to coupled RL 0 Can also create manually using coupled RL branches Transformers Spring 2008 U Ideal transformer mm Lecture 33 o Combines ideal transformer with ideal source Simply enter transformation ratio Can be used to implement floating source too Uses frequency from basic ATPDraw settings Need to make sure this matches system frequency Setting Branch 0 forces ATP to use this frequency Branch 1 can avoid this Vm1E20 Transformers Spring 2008 Models U Accessing Transformer 0 Note that three phase and single phase options Lecture 33 Praises s Erphase Branch Linear r Branch Nnniinear r LinEsC bias r Swithes r Snurces i MEEHiHES V a 3 phase Moms E 52 User Speci ed Frequency camp at Sat vv 343a Standard Cnmpnnent scram HWrid mudei Transformers Spring 2008 U ECE524 I Lecture 33 nmnnnent IRAranuP 7 51 Amines om uw mug n mm D Envy Pasie eniiiedaiagnd Eammeni my lu Lahei Q I Hi e 1 Lack M QK W M Transformers Spring 2008 U ECE524 I Text Lecture 33 C Source data w u l 7k kt t C H l4NOdeJ C NodeLgtlt7 i77in777gtltNodeKltNodeMltNodeX 18 o The 18 card need to follow the 14 0 Node X can be used for current measurement Transformers Spring 2008 I R L Terms Lecture33 o The balance of the regulartransformer model can be created by Adding external R L for series terms Shunt resistor for core loss term Saturable inductor component for Lm o Create winding connections externally 0 Can also add capacitance Transformers Spring 2008 l ECE524 I Lecture 33 0 Limited to two winding transformers o It is very easy to create numerical problems in the simulation with the ideal transformer Transformers S pri mg 2008 ECE524 I Saturable Transformer Lecture33 0 Model has builtin circuit elements Winding resistance Leakage inductance can t enter 0 Core loss resistance Magnetizing branch not entered as an L in mH Can set all except leakage to O to simplify Enter winding to winding ratios Transformers Spring 2008 U Single Phase Saturable ECE524 Base Attributes Lamas 1 o lo F0 are steadystate point on saturation characteristic for initial Lm o RMS 0 or 1 determines how the saturation characteristic is entered LI 0 Output is information about in mangetization branch Emmi umpm Me I Na 39 I Lack gait deiimiians EK Eancei eip Transformers Spring 2008 ECE524 I Saturatlon Characterlstlc Lemmas o f RMS 0 this is current A Attributes Chazalrleristicl versus flux 39 39 39f RMS 1v quot8 393 RMS i imm 143253 current versus R MS Mariam oaamzssom were voltage at frequency of m5927i288n23 567955393n529 inquot first source in the system 232ng 223233 0 The 00 point is assumed Mm by the program ii 0 Up to 10 points can be entered cue incitiecraiacim Better to limit to 35 ave Envy Essie liew Q Transformers Spring 2008 U Viewing The Saturation mm 1 Characteristic Lemmas If RMS 0 this is current versus flux If RMS 1 this is RMS current versus RMS voltage at 579 l frequency of first source in the system Q View Nonlinearity 522 7 Fluxlinkedl W n 535 5 The 00 point is assumed by the program Up to 10 points can be entered 651i 9 8 HM Betterto limitto 35 for better 35 57 m3 i552 2191 Dane numerical behavior Does not include hysteresis Transformers Spring 2008 U ECE524 1 Three Phase Model Lamas 0 Icon changes with the connection type Here is three winding with all VWE Here is 2 winding with deltawye Q Note that there is a point to connect to measure magnetizing branch voltage Three leg core option Transformers Spring 2008 U ECE524 I Three Phase Model Lamas Enter data for each Ll A nbL ES lonamctansiic Winding Select Y D or Zigzag Three leg option here but better to use specific case from pull down menu Labet Evmmenl Uulpm 39 Hg urNa v i L i BK R Emu elv Transformers spring 2008 U Three Phase Model mm I Three Leg Core Lamas Option to model t TRAWHJSUD T v 1 homopolar rel uctance quot1quotan A ributes lchazalrteristicl gtgt SaturableThree Leg Much additional information needed See help menu EK Eancel Help Transformers spring 2008 ECE524 U I BCTRAN Interface Lemma 11 o Produces coupled RL model 0 Based on SC and 00 test data quot9 3 v r is Eleggedwaundcme v 5 Linear Lm internally Need to add external nonlinear Transformers Spring 2008 ECE524 U I Add1t10na1 BC Tran Data Lecture men Mcull Shun maul I gusi vs ssguancs I s a znn 25 s Transformers S W ng 2008 1 ECE524 I Nonlinear Devices Lemma 0 Models for resistors and Ronn29 inductors Differ model implementation L0 The Type 98 and Type 93 W Mov Type 92 inductors do not include hysteresis 313232 L 7 Type 93 rut Same user data as saturation in saturable xfmr Set with initial conditions Transformers spring 2008 U Type 96 reactor with mm I hysteresis Lamas vg o x 0 Option to enter E J residual flux along 3 a With steadystate iei Label Eammenll Q Transformers spring 2008 U ECE524 I E Xample Lecture 33 o Transformer with saturation 1i ii i r La m mm 5 i v I M i x i Transformers Spring 2008 U Results from increasing M524 voltage Lecture 33 a M Mn um quot um nns an n n n Transformers Spring 2008 ZE 39OU NOISSHS SNELLSAS HHMOd NI SiNEJISNVELL i729 EDS CV 7 KSOKV Vss V1 i B 3911 CZ ah79 Q30 cfCV Ci 1 C1 5180 xIO z r M C2 3 4C386x0 l WNW WP jalrch C wn cw 5p M E H C9 ram 4 taxi8 H C A quot E om lt3 aanx 2 6 w is n j C tC EFZI KO 32 5 SO 0 A S c Csa 6190 M QM Agoquot H 3 4W 2 c 839 CF 2 06 quot670 94 Era we quotquot1quot Mm NP r S360 74km AT 3 88 M if rquotKO 25quot 0 00 51 8F 40 1 MN lug 39 Awe6 zr 39 2 IIiDO Ms 6 71am J 3 If SAC CD 14 2i 6 69 0 Rae SCquot 00 m a mm RE 2 MJZ W Lu 4MH Typicai excitation curve 1000 00 100 D 30 1 10 So 100 1000 MmA customer MahkotaBukit farek Core i Class Burden I Current Ratio NA Tum Ratioi Ram I Project Maiaisyaitem1 1amp2 5253o VA I 309011 3000 J 63 Bur Reference MP 73 63650 yp IQSK 553 Frequency Hz 50 30001 5P25 30VA 1I r 63 V 850 1 3000 N2 500 f 50 Em Ohmswmm N2 V2 2 V N Ohmstum 21 x 10 3 0001 105 0002 235 0005 70 001 150 002 310 005 750 006 840 007 870 01 900 10 965 quotr1 039 N F25z F lt0 Mr F lt9 5 2 NZ N F2 2 augmentF25 F2r lt1 quot7 100 0 13910 001 01 1 1X10393 F lt0 712 lt0gt 1 v 39 444NBsmr A 709 x 10 4 This is the area in square meters From A We can deduce that the dimensions of the CT is height 2 01 width 004171 Therefore mean ooii length ooil length 2 2height 2width coilleng1h 0283 Mean ienth of Core Resistance of the CT windings Moore 4w1dth n Rm 2 Ohmstum N Rzmt Ohms tum NZ Moore 0524 Riot 2 63 mm 2 105 F 1 lt gt 0 444fN H L 3 A Moore 7 ffff 7 BH 2 augmentB H 0 1 0022 5724 005 11447 0148 28618 0318 57236 0656 114473 1588 286181 1779 343418 1842 400654 1906 572363 2044 5724103 BH OOOVOU39lbWNAO OLI OI OOI 30001 5P25 30VA V 2 850 2 V2 2 V9 N N 2 3000 0001 105 0002 235 0005 70 001 150 002 310 005 750 006 840 007 870 01 900 10 965 F251 F ltOgt 111 N N2 2500 f I 50 Bsatzz 18 lt1 32 F2r F N 112 augmentF25F2r 3 Ohmstum 1 6 Ohmsvtum 21 X 10 3 V A2 444NB5mf A 709 x 104 This is the area in square meters From A We can deduce that the dimensions of the CT is height 2 01 width 004171 Therefore mean coil length coil1ength 2height 2width coi1length 0283 Mean Ienth of Core Resistance of the CT windings Moore 4w1dth7I Rtot c Ohmstum N Rzmt Ohms tum NZ Moore 2 0524 Rm 63 RZW 525 F 1 M NF 0 444fN H W B 1 A Moore BB augmentB H 0 1 0022 5724 005 11447 0148 28618 0318 57236 0656 114473 1588 286181 1779 343418 1842 400654 1906 572363 2044 5724108 BH OWNODCHAUJN ko gOIgtlt17ZL39S H VOII OLI OOI OI VZL39S I 0 0 non VVO39Z OI 30001 SPZS 30VA V 2 850 V2 VEE N 0001 0002 0005 001 002 005 006 007 01 10 F251 F lt gt N 1 3000 105 235 150 310 750 840 870 900 965 Ii N2 F2r F 1 N 2 3900 63 2 23 2 Ohmstum I Y Ohmstum 21 x 10 3 N2 N F2 augmentF28 F2r Y 444NBsatf A 709 x 10 4 This is the area in square meters From A We can deduce that the dimensions of the CT is height 01 width 004171 Therefore mean coil length co length 2height 2width eoi1length 0283 Mean lenth of Core Resistance of the CT windings Moore 1 4width7r Rtot z OhmsjumN 2 641 o gt M Moore 524 Rm 2 63 F 1 NF 0 444 fN H 2 A Mcore BH augmenKBJI 1 10 100 13910 110 62 quotCu NOISSEIS SIAEIJSAS HEMOd NI SiNEIISNVELL 1725 333 m Universityofldaho rm 3 Q rm M Wanst mmo f e cm as QM 4ng 3 11 me sh aids

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