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## Transients in Power Systems

by: Fredy Okuneva

95

0

49

# Transients in Power Systems ECE 524

Fredy Okuneva
UI
GPA 3.81

Brian Johnson

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COURSE
PROF.
Brian Johnson
TYPE
Class Notes
PAGES
49
WORDS
KARMA
25 ?

## Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 49 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 95 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
IZ 39OU NOISSEIS SIAIELLSAS HEIMOd NI SiNEIISNVELL 1729 333 ECE524 Transients in Power Systems Session 18 Page 18 Spring 2008 Multiphase Transmission Line Modeling Basic Theory The process of modeling lines with more than one conductor begins with the partial differential equations relating voltage and current on an incremental line segment with length A13 The basic circuit elements for one of these conductors is shown below The partial differential equations representing this line segment in the limit as A2 gt 0 for the single conductor case now become matrix equations where 86 at 39gij Z 1 IIlg 1 z e quot8 Cla 2 The analysis presented below can be applied to lines with any number of conductors but we will continue the development for a line with 3 conductors such that 39 R b R lRl Rl Rllc Rh and Liza Lizb Liic lLll Lllm L21 Ll L Lin 39 C 39b 0 M C a C b 0t l Orin lz l Session 18 Page 28 Spring 2008 ECE524 Transients in Power Systems These matrices are generally symmetric If the line is transposed the elements on the main diagonal of each matrix are all equal and the offdiagonal terms of each matrix are also equal If the system is excited at a particular frequency u then the equations become z 2 lt3 Y lE lt4 Where Z29 jo ij2j and iji j Note that this will also result if a Fourier Transformation is applied to the equations Equations 3 and 4 can be rewritten in the same manner used for the single phase case to giVe magi Ls 5 A modal transformation can be used to decouple these matrix equations We can define two modal matrices T6 and T where TE is the eigenvector matrix for Z Y and T1 is the eigenvector matrix Y Z The most effective way to calculate these is to use a dedicated eigenvalue routine such as the BIG function in Matlab These are constant matrices independent of time and position We can then define vectors Em and m to represent the variables transformed into the modal domain such that E TEETH 1 2 mm 8 A 1 Next substitute Em and m into equations 5 and 6 resulting in 2 Em C 8292 821m Ml il 8132 10 27 V1 9219365223 5 5 3 ECE524 Transients in Power Systems Session 18 Page 48 Spring 2008 aEm 1 I I r 39 Z I m 03 lTil1lYIl lTelEm 377I1lEm A Tel 1lZ llY llTel lTel llZ lTiXYTIlY llTel I ernllY tl 20 We can also see that A l l llY39llZ39lml lTilquot1Y39lTeXTJllZ39llTiD lY12llZInl 21 A diag12 m YAHZLL implies that and are diagonal This holds for both the transposed and untransposed cases With the diagonalized form of the matrix differential equations in 20 and 21 the decoupled equations can now be solved as scalar equations using the methods discussed for single phase lines The resulting equations can be more easily solved if the resistance terms are lumped at the ends of the lines and lines are treated as lossless Then for Em E1E2E3 Enquot and 1m 111213 Inlt we can find modal propagation velocities and characteristic impedances for representing the traveling waves on the lines as shown by 1 L V0 and Z0 4 22 L505 00 1 1 V1 and Z1 m 23 tL le V 01 1 Lg U2 and Z2 2 w 24 tLgog 05 H 25 1 L 1 2 and Zn L 26 tLgog V on However solving the voltage equation using the modal transformation matrix Te and the current equation using T7 will result in different values for the characteristic impedances Therefore it will be necessary to include the transformation to get useful information The propagation velocities will be the same in any case since they come from the eigenvalues for the set of equations 2 7 e L 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 017 39OU NOISSEIS SIAEJJSAS HHMOd NI SiNEIISNVHJ i725 EDS wmo 1 Ww H O 1114I11T4J 71V YV 90 7 rirJilziIilrWJJ qnumueoau W JIMV 8250893xi111111151H QUAQUOLQUQQ Zunw AwaUQunur W 2 I J QqRQGAroo 14Irr 425221 ib K w mwo c 1iriar 01 V 0 M F F JIf MmmMAHAHDmrprr 1 1111111 aihv AHWIIII m Midway Bus VOW gs Voltage Revergcl at Remy Locchom Time m8 20f8 600000 600000 2mm00 Vo age ACross Se es Copcchor v No Gap ashing Max156848 MWM 138565 150000 Avg 301567 Abs156848 RMSz44254 TOOOOO CF 354422 i FF 146749 3 50000 Z w M me 74W W mwwrquot wwmv Wwwu 2 Oi fp gt A x isoooogj EOOOOO39 AWSOOOO 0 W00 ZOO Tk e n18 30f8 N dway Bus Vo ege VOHOge Revera FoHowed by 000 ashmg 000000 7 7 L EMQX484BOO i v 0 4000OO r39w A A Abg 521029 7 HMS 271707 00107702 200000 f I I I y F 1221 E 39W H WW 3 0 H I a I f K A NMAMWV J a I quotHNVV HH i f J I I I A 4 0 O O O O V V V K V Vi V 600000 7 a s e 1 z 2 E 1 1 1 1 e O 100 200 Time ms 4of8 Vo age V Vo age Acmx5 Se es Capachor w Cop HGSth 200000 waoooogju wooooog 50000 200000 T O Max168933 MMm 160479 Avg112336L3 Abs68933 ARMSz26552 C 636231 FF 215235 C Q l D 50f8 N dwoy End FQUH C0N 0t 6000 0 4000 rw 2000 3 53 0 5 2000 V E 5 l t 1 0 100 200 Time m8 U C CD 6of8 5000 deay End Cuwe t 4000 w4wmwwwrn w 3000 2000 u H woooiE I Max42229 MUM 23678 Avg 734592 AbS 42229 m RMS406282 CF 397329 FF 44682 CuwentA A A A A 2V V V 4WOOO 22000 300035 A A M w WVVVW TMne OAS 70f8 MCon Subsystem 1ISCHPSISCAP1 WA c 691 IMUM B IMU U39C a EH Um 3910 ng am ENE 36 2E M I I 663333 61666 625 633333 64166 15 10f5 Subsystem 1ISCAPSISCEP1 PM 153 400 2EID GOD 3GB EDI 100 2E r f I 010m 311044 0119 012895 31382 U1446 U1561 20f5 Subsystem 1 ISCEPS ISCHP I quot56 NJ I 300 200 100 kh 100 200 300 lh UH k i F5UTE E AUG E k 30 20 WW 0 M 0 003333 01666 025 033333 04166 06 3 of5 Subsystem 1ISCAPSISCRP1 300 200 100 REE 100 200 300 k 20 E w KER b 0 003333 01000 025 033333 04160 05 4of5 Subsystem 1ISCAP5ISCRP1 STKTE 108333 11 D 25 033333 04136 50f5 osmEEEEMSED E osvaQBEmSED 5 Universityofldaho M Q T C HMACTEU Vx b L49 1 44 ECE 52439 Session 6 Page 17 Transients in Power Systems Spring 2008 Transient RL Circuit Example Given a 230kV34kV A Ygrounded 100MVA transformer supplied supplied by a source with a short circuit MVA of 2000 The transformer has a per unit reactance of 01pu on its ratings base Both the transformer and the source impedance have an X R ratio of 12 Calculate the complete phase A fault current if a fault occurs on the LV side of the transformer at 70 degrees past the voltage peak Define Units MVA 1000kW SBASE 100MVA pu l Transformer parameters Srated 100MVA XoverR 12 th 230kV Vlv 345kV Xtrans 3 Olpu Xtrans XoverR Rtrans 3 Rtrans 001pu Find the transformer impedances referred to the LV side since the fault occurs on that side V1 2 ZBLV V ZBLV 1199 Srated RtransLV 3 Rtrans ZBLV RtransLV 009929 XtransLV 3 Xtrans ZBLV XtransLV 1199 XtransLV LtransLV LtransLV 3157mH 21t60Hz ECE 524 Session 6 Page 27 Transients in Power Systems Spring 2008 Source impedance MVAsc 2000MVA Vpu 10pu MVAsc MVAscju MVAsc 1m 20pu SBASE 2 z Vpu z 0 05 m source 3 m source Pu MVAsc 1m 11 atanXoverR 11 8524deg Zsource stourceej39n Zsource 00042 00498ipu Source impedance in Ohms referred to LV side ZsourceiOhm 3 Zsource ZBLV ZsourceiOhm 0049 0593 1mZsourceiOhm LsourceiLV z LsourceiLV 157mH If we wanted to find source impedance referred to LV side in 1 step 2 VIV atanXoverR MVASC 0049 0593iQ Equivalent circuit since this is a three phase fault we can use per phase analysis RsrC Lsrc Rtrans Ltrans Fault VLNLV Session 6 Page 37 Spring 2008 ECE 524 Transients in Power Systems Transient Current Solution Requiv 3 Re ZsourceiOhm RtransLV Requiv 0159 Lequiv 3 LsourceiLV LtransLV Lequiv 473 mH Decay constant R xz x3142Hz Lequiv l 1 1 0032s 7 Driving point voltage 2 VIII 2 V1V Vm 2817kV Fault inception angle If we de ne our driving voltage as vt Vmsinoa t 1 rad n 21t60Hz n 37699 s Since the fault occurs 70 degrees phase the peak of the voltage for analytical solution let fault occur at time t 0 I 90deg 70deg I l60deg Initial condition i0 0A Switch is open Equiavlent impedance 2 2 ZequiviLV 3 Requiv m Lequiv ZequiviLV 1799 ECE 524 Transients in Power Systems Session 6 Page 47 Spring 2008 Create time vector 4 39 60Hz t 011074sec Two components to the current For a more complete derivation see Vt Vmsinnt 1 httpwwweceuidahoedueepowerECE423LecturesL18ece423518pdf Vm 155t mj51nmt n Vm 7m 1transt Sln 11 e ZequiviLV 439104 4 Vt 2 10 1550 0 itrans 4 210 4 I 4 10 0 002 004 0 06 239104 15 39104 mama 1104 5000 ECE 524 Session 6 Page 57 Transients in Power Systems Spring 2008 i0 1 i550 itrans Other items of interest peak current Peak current occurs when the derivative of the current equals 0 the second time in this case t39 39 LNmcosmt I 11 Asin ne77Lt ZequiviLV Plot this for c0mparis0n dit 3 w39COSDt I 11 7vsin 1067 kit equlvi 1 39107 I I I 5 39106 300 it Note that the second T0 0 zero crossing is at the peak 1n th1s case 0 5 106 I I I 1 10 0 001 002 003 Session 6 Page 67 Spring 2008 ECE 524 Transients in Power Systems sec Now we want to nd the values of t where the derivative is zero millisecond 1000 tzero 001sec Given oac0sntzer0 I 11 Asin ne7 knew 39 0 Vm ZequiviLV tpeak Findtzer0 tpeak 888millisecond ditpeak 106 X 10 9 Simulation results lt I I ulo o Settings for simulation 7 5 At 510 sec tmax 0lsec Lequiv 473 mH vIn 28169V Requiv 0159 Note ATP uses peak voltage PSCADEMTDC uses RMS linetoline Since the source is a cosine set phase angle for sine 9 90deg Ecgau Sessmn Page77 mnscm m PuWEv Sv ems Spnng 2m Txme forfault to occur let run 1 full cycle andthen 70 degrees past posmvc peak 1 160 t 1 1 0 02407 Thxs is umc that mc sthch closes m son K 360 m S 391u u u uuz u mvhvuxuavocw E sPlntXannt u 5 ma 2 25 so 35 WEN a We mm m x var c w an o I39 nuaza 727271 IYIEEI I S 39OU NOISSEIS SIAELLSAS EEIMOd NI SiNEIISNV tIJ 1729 333 m4 5m mum OLGUIQEELQED i 331033522 5 5m mum oxmvtob azp E ocmutc gmzcs E a 4 3 3m ace 5 L5 53 m Universityofldaho m Universityofldaho Wwwsw at 3 law mom exam ROBEEZCD E 062 a EEEQED 2 a 3 3 3m Baggage s as 5294 L5 m Universityofldaho m Universityofldaho Z N ozmtisbm mzc E N 3 doom 3 oxmvtobm ng s 5 3 yum mU 5 5254 LS niversityofidaho

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