An Introduction to Symmetrical Components, System Modeling and Fault Calculation

Transcription

1 An ntrductin t Symmetricl Cmpnents, System Mdeling nd Fult Clcultin Presented t the 35 th Annul HANDS-ON Rely Schl Mrch - 6, 8 Wshingtn Stte University Pullmn, Wshingtn By Stephen Mrx, nd Den Bender Bnneville Pwer Administrtin Symmetricl Cmpnents Mrch 6, 5

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3 ntrductin The electricl pwer system nrmlly pertes in blnced three-phse sinusidl stedy-stte mde. Hwever, there re certin situtins tht cn cuse unblnced pertins. The mst severe f these wuld be fult r shrt circuit. Exmples my include tree in cntct with cnductr, lightning strike, r dwned pwer line. n 98, Dr. C. L. Frtescue wrte pper entitled Methd f Symmetricl Crdintes Applied t the Slutin f Plyphse Netwrks. n the pper Dr. Frtescue described hw rbitrry unblnced 3-phse vltges (r currents) culd be trnsfrmed int 3 sets f blnced 3-phse cmpnents, Fig.. He clled these cmpnents symmetricl cmpnents. n the pper it is shwn tht unblnced prblems cn be slved by the reslutin f the currents nd vltges int certin symmetricl reltins. Fig. By the methd f symmetricl crdintes, set f unblnced vltges (r currents) my be reslved int systems f blnced vltges (r currents) equl in number t the number f phses invlved. The symmetricl cmpnent methd reduces the cmplexity in slving fr electricl quntities during pwer system disturbnces. These sequence cmpnents re knwn s psitive, negtive nd zer-sequence cmpnents, Fig. Fig. Symmetricl Cmpnents Pge

4 The purpse f this pper is t explin symmetricl cmpnents nd review cmplex lgebr in rder t mnipulte the cmpnents. Knwledge f symmetricl cmpnents is imprtnt in perfrming mthemticl clcultins nd understnding system fults. t is ls vluble in nlyzing fults nd hw they pply t rely pertins.. Cmplex Numbers The methd f symmetricl cmpnents uses the cmmnly used mthemticl slutins pplied in rdinry lternting current prblems. A wrking knwledge f the fundmentls f lgebr f cmplex numbers is essentil. Cnsequently this subject will be reviewed first. Any cmplex number, such s jb, my be represented by single pint p, pltted n Crtesin crdintes, in which is the bsciss n the x xis f rel quntities nd b the rdinte n the y xis f imginry quntities. This is illustrted in Fig.. Fig.. Referring t Fig.., let r represent the length f the line cnnecting the pint p t the rigin nd the ngle mesured frm the x-xis t the line r. t cn be bserved tht r cs (.) b r sin (.). Prperties f Phsrs A vectr is mthemticl quntity tht hs bth mgnitude nd directin. Mny quntities in the pwer industry re vectr quntities. The term phsr is used within the stedy stte lternting liner system. t is used t vid cnfusin with sptil vectrs: the ngulr psitin f the phsr represents psitin in time, nt spce. n this dcument, phsrs will be used t dcument vrius c vltges, currents nd impednces. A phsr quntity r phsr, prvides infrmtin but nt nly the mgnitude but ls the directin r ngle f the quntity. When using cmpss nd giving directins t huse, frm given lctin, distnce nd directin must be prvided. Fr exmple ne culd sy tht huse is miles t n ngle f 75 degrees (rtted in clckwise directin frm Nrth) frm where m stnding. Just s we dn t sy the ther huse is - miles wy, the mgnitude f Symmetricl Cmpnents Pge

5 the phsr is lwys psitive, r rther the bslute vlue f the length f the phsr. Therefre giving directins in the ppsite directin, ne culd sy tht secnd huse is miles t n ngle f 55 degrees. The quntity culd be ptentil, current, wtts, etc. Phsrs re written in plr frm s Y Y (.) Y cs jy sin (.) where Y is the phsr, Y is the mplitude, mgnitude r bslute vlue nd is the phse ngle r rgument. Plr numbers re written with the mgnitude fllwed by the symbl t indicte ngle, fllwed by the phse ngle expressed in degrees. Fr exmple 9. This wuld be red s t n ngle f 9 degrees. The rectngulr frm is esily prduced by pplying Eq. (.) The phsr cn be represented grphiclly s we hve demnstrted in Fig.., with the rel cmpnents cinciding with the x xis. When multiplying tw phsrs it is best t hve the phsr written in the plr frm. The mgnitudes re multiplied tgether nd the phse ngles re dded tgether. Divisin, which is the inverse f multiplictin, cn be ccmplished in similr mnner. n divisin the mgnitudes re divided nd the phse ngle in the denmintr is subtrcted frm the phse ngle in the numertr. Exmple. Multiply A B where A 535 nd B 345. Slutin A B Exmple. Slve D C where C 535 nd D 5 3. Slutin C 535 D Symmetricl Cmpnents Pge 3

6 3. The j nd pertr Recll the pertr j. n plr frm, j 9. Multiplying by j hs the effect f rtting phsr 9 withut ffecting the mgnitude. Tble 3. - Prperties f the vectr j. j. j 9 j 8 j 3 7 j 9 j j Exmple 3. Cmpute jr where R 6. Slutin jr 9 (6 ) 5 Ntice tht multiplictin by the j pertr rtted the Phsr R by mgnitude. Refer t Fig. 3. R 9, but did nt chnge the () R jr R (b) R j Fig. 3.. j effects Symmetricl Cmpnents Pge 4

7 n similr mnner the pertr is defined s unit vectr t n ngle f, written s. The pertr, is ls unit vectr t n ngle f 4, written 4. Exmple 3. Cmpute R where Slutin R (6 ) 8 R 6. R () A R R (b) j R Fig. 3.. effects Tble 3. - Prperties f the vectr. j j 3 j Symmetricl Cmpnents Pge 5

8 4. The three-phse System nd the reltinship f the 3 n Wye cnnected system the vltge mesured frm line t line equls the squre rt f three, 3, times the vltge frm line t neutrl. See Fig. 4. nd Eq. (4.). The line current equls the phse current, see Eq. (4.) Fig. 4. LL 3 LN (4.) L (4.) n Delt cnnected system the vltge mesured frm line t line equls the phse vltge. See Fig. 4. nd Eq. (4.3). The line current will equl the squre rt f three, 3, times the phse current, see Eq. (4.4) Fig. 4. LL (4.3) L 3 (4.4) Symmetricl Cmpnents Pge 6

9 The pwer equtin, fr three phse system, is S 3 LL L (4.5) P 3 LL L cs (4.5b) Q 3 LL L sin (4.5c) where S is the pprent pwer r cmplex pwer in vlt-mperes (A). P is the rel pwer in Wtts (W, kw, MW). Q is the rective pwer in ARS (rs, krs, Mrs). 5. The per-unit System 5. ntrductin n mny engineering situtins it is useful t scle, r nrmlize, dimensined quntities. This is cmmnly dne in pwer system nlysis. The stndrd methd used is referred t s the per-unit system. Histriclly, this ws dne t simplify numericl clcultins tht were mde by hnd. Althugh this dvntge is eliminted by the clcultr, ther dvntges remin. Device prmeters tend t fll int reltively nrrw rnge, mking errneus vlues cnspicuus. Using this methd ll quntities re expressed s rtis f sme bse vlue r vlues. The per-unit equivlent impednce f ny trnsfrmer is the sme when referred t either the primry r the secndry side. The per-unit impednce f trnsfrmer in three-phse system is the sme regrdless f the type f winding cnnectins (wye-delt, delt-wye, wye-wye, r delt-delt). The per-unit methd is independent f vltge chnges nd phse shifts thrugh trnsfrmers where the bse vltges in the winding re prprtinl t the number f turns in the windings. Mnufctures usully specify the impednce f equipment in per-unit r percent n the bse f its nmeplte rting f pwer (usully ka) nd vltge ( r k). The per-unit system is simply scling methd. The bsic per-unit scling equtin is ctul _ vlue per unit (5.) bse_ vlue The bse vlue lwys hs the sme units s the ctul vlue, frcing the per-unit vlue t be dimensinless. The bse vlue is lwys rel number, wheres the ctul vlue my Symmetricl Cmpnents Pge 7

10 be cmplex. The subscript pu will indicte per-unit vlue. The subscript bse will indicte bse vlue, nd n subscript will indicte n ctul vlue such s Amperes, Ohms, r lts. Per-unit quntities re similr t percent quntities. The rti in percent is times the rti in per-unit. Fr exmple, vltge f 7k n bse f k wuld be 7% f the bse vltge. This is equl t times the per unit vlue f.7 derived bve. The first step in using per-unit is t select the bse(s) fr the system. Sbse = pwer bse, in A. Althugh in principle Sbse my be selected rbitrrily, in prctice it is typiclly chsen t be MA. bse = vltge bse in. Althugh in principle bse is ls rbitrry, in prctice bse is equl t the nminl line-t-line vltge. The term nminl mens the vlue t which the system ws designed t perte under nrml blnced cnditins. Frm Eq. (4.5) it fllws tht the bse pwer equtin fr three-phse system is: Slving fr current: S 3 3 bse bsebse (5.) bse S 3bse 3 bse Becuse S3Φbse cn be written s ka r MA nd vltge is usully expressed in kilvlts, r k, current cn be written s: kabse bse mperes (5.3) 3kbse Slving fr bse impednce: bse bse bse S bse bse r bse bse kbsex hms (5.4) ka bse kbse hms (5.4b) MA bse Symmetricl Cmpnents Pge 8

11 Given the bse vlues, nd the ctul vlues:, then dividing by the bse we re ble t clculte the pu vlues bse bse bse pu After the bse vlues hve been selected r clculted, then the per-unit impednce vlues fr system cmpnents cn be clculted using Eq. (5.4b) pu pu r ( ) MA bse ( ) pu bse kbse (5.5) ka bse ( ) pu kbse (5.5b) t is ls cmmn prctice t express per-unit vlues s percentges (i.e. pu = %). (Trnsfrmer impednces re typiclly given in % t the trnsfrmer MA rting.) The cnversin is simple per unit Then Eq. (5.5) cn be written s percent _ vlue bse bse ka MA % (5.6) k bse kbse t is frequently necessry, prticulrly fr impednce vlues, t cnvert frm ne (ld) bse t nther (new) bse. The cnversin is ccmplished by tw successive pplictin f Eq. (5.), prducing: Substituting fr new pu ld pu ld bse nd ld bse new bse new bse nd re-rrnging the new impednce in per-unit equls: new ld new ld ka bse kbse pu pu (5.7) ld new kabse kbse n mst cses the turns rti f the trnsfrmer is equivlent t the system vltges, nd the equipment rted vltges re the sme s the system vltges. This mens tht the vltge-squred rti is unity. Then Eq. (5.7) reduces t Symmetricl Cmpnents Pge 9

12 new new ld MA bse pu pu (5.8) ld MAbse We cn quickly chnge frm ne impednce vlue in hms, t nther impednce vlue in hms by dividing by the ld bse vltge nd multiplying by the new bse vltge in hms. This is shwn in Eq. (5.9) new new ld k bse hm hm (5.9) ld kbse Exmple 5. A system hs Sbse = MA, clculte the bse current fr ) bse = 3 k b) bse = 55 k Then using this vlue, clculte the ctul line current nd phse vltge where pu, nd. 5 pu t bth 3 k nd 55 k. Slutin kabse Using Eq. (5.3) bse mperes 3kbse ) bse mperes 5A 3 3 Frm Eq. (5.) b) bse mperes. A 3 55 ctul ctul (5.9) pu pu bse (5.) bse At 3 k ctul A 4 c) A d) ctul.5 3k 5k At 55 k ctul A 544 e) A f) ctul.5 55k 63k Symmetricl Cmpnents Pge

13 Exmple 5. A 9 MA 55/4.5 uttrnsfrmer hs nmeplte impednce f.4% ) Determine the impednce in hms, referenced t the 55 k side. b) Determine the impednce in hms, referenced t the 4.5 k side Slutin First cnvert frm % t pu. % pu.4 Arrnging Eq. (5.5) nd slving fr ctul gives k ( ) pu MA ) bse.4 bse 55k 3. 5 ; therefre 55 9 b) 4.5k A check cn be mde t see if the high-side impednce t the lw-side impednce equls the turns rti squred Applictin f per-unit Appling this t rely settings, prcticl exmple cn be shwn in clcultin f the settings fr rely n trnsmissin line. Fr distnce relys cmmn setting fr zne is 85% f the line impednce. ne shuld be set nt less thn 5% f the line, with cre t nt ver rech the zne f the next line sectin. f this des then zne will need t be crdinted with the next line sectin zne. Referring t Fig. 5. the line impednce fr the 6 k line is hms. Using the bve criteri f 85% fr zne nd 5 % fr zne the relys wuld be set t Fr zne ( ) 85%(59.38 ) Symmetricl Cmpnents Pge

14 ( ) Fr zne ) 5%(59.38 ) ( ( ) k x x 6 k x x = 8 hms = 8 hms % = MA 6/5k Fig 5. Fr the relys n the 5 k side f the trnsfrmer, the impednce f the trnsfrmer needs t be clculted. Frm exmple 5. we see tht 5 5k Next the line impednce needs referenced t the 5 k side f the trnsfrmer. Using equtin 5.9 new new ld k bse hm hm (5.9) ld kbse Substituting, the line impednce equls 5 k 5 hm hms 6 Adding this t the trnsfrmer, the impednce setting fr the relys n the 5 k side f the trnsfrmer is Using the sme criteri fr zne nd zne rech. Fr zne ) 85%(34.88 ) ( ( ) Fr zne ) 5%(34.88 ) ( ( ) Given these vlues, ne cn esily see tht by ignring the bse vlues f the vltges the rely settings wuld nt be dequte. Fr exmple if the 6 k settings were pplied t Symmetricl Cmpnents Pge

15 the 5 k relys, zne wuld ver rech the remte terminl. Cnversely, if the 5 k settings were pplied t the 6 k relys zne wuld nt rech pst the remte terminl nd wuld thus nt prtect the full line. Fig Clculting ctul vlues frm per-unit n the fllwing sectins we will discuss symmetricl fults. The nlysis f the fults uses the per-unit. A impednce nd vltge f the system is express in per-unit. Then the fult current nd fult vltge is slved nd tht vlue will be given in per unit. Next we need t cnvert frm per-unit t ctul mps nd vlts by using the bse vlues. Using the bve equtins it is esy t prve the fllwing equtins. The MA fr three phse fult is given s MABse MAFult (5.) FultPU Or MAFult fr MABse (5. ) PU Or Fult Fult Bse _ Current (5.) PU Fult _ Current 5.3 Cnverting per-unit Fult, (5. ) PU 3k Fult Befre using the per-unit impednce f trnsfrmer frm mnufcture nmeplte yu must first cnvert it t per-unit vlue f yur system. Typiclly the three-phse pwer bse f MA is used. This is dne by first cnverting the per unit impednce t n ctul impednce (in hms) t 55k nd then cnverting the ctul impednce t per- Bse Symmetricl Cmpnents Pge 3

16 unit impednce n the new bse. Repet, this time cnverting the per unit impednce t n ctul impednce (in hms) t 4.5k nd then cnverting the ctul impednce t per-unit impednce n the new bse. n the prblem 3 t the end f this dcument, the trnsfrmer nmeplte dt is fr rti f 55/4.5k r.74, wheres BPA s ASPEN mdel uses nminl vltges f 55k nd 3k fr rti f.83. Becuse BPA used trnsfrmer rti in ASPEN mdel tht ws different thn the trnsfrmer nmeplte vlues, we hve discrepncy in the per-unit impednce vlues tht we btined. The prblem rises becuse when trnsfrmer is pplied t the BPA system the trnsfrmer tp used will ften be different thn the ne used in the nmeplte clcultins. Wht is the crrect wy t cnvert the per-unit impednce t the BPA bse? Becuse the ctul impednce f the trnsfrmer will vry when different tps re used, the mst ccurte wy t mdel the impednce wuld be t ctully mesure the impednce with the trnsfrmer n the tp tht will nrmlly be used n the BPA system. This impednce wuld then be cnverted t per-unit vlue n the BPA mdel bse. Since this isn t nrmlly pssible, clse pprximtin cn be mde by ssuming tht the per-unit impednce given n the nmeplte will remin the sme fr the different tp psitins f the trnsfrmer. Find the trnsfrmer tp psitin tht mst clsely mtches the rti f the ASPEN mdel (.83 fr 55/3k trnsfrmer), then cnvert the nmeplte per-unit impednce t n ctul vlue bsed n either the high- r lw-side vltge given fr tht tp psitin. This ctul impednce is then cnverted t per-unit vlue n the BPA mdel bse, using the high-side BPA vltge bse if the high-side vltge ws used fr the cnversin t ctul impednce, r using the lw-side BPA vltge bse if the lw-side vltge ws used fr the cnversin t ctul impednce. See prblem Sequence Netwrks Refer t the bsic three-phse system s shwn in Fig. 6.. There re fur cnductrs t be cnsidered:, b, c nd neutrl n. b c n bn cn n Fig. 6. Symmetricl Cmpnents Pge 4

17 The phse vltges, p, fr the blnced 3Φ cse with phse sequence bc re n p (6.) bn b p (6.b) 4 cn c p p (6.c) The phse-phse vltges, LL, re written s 3 b b LL (6.) bc b c LL (6.b) c c LL (6.c) 9 5 Equtin (6.) nd (6.) cn be shwn in phsr frm in Fig. 6.. Fig. 6. There re tw blnced cnfigurtins f impednce cnnectins within pwer system. Fr the wye cse, s shwn in Fig. 4., nd with n impednce cnnectin f, the current cn be clculted s P (6.3) Y Y Where is between 9 nd + 9. Fr greter thn zer degrees the ld wuld be inductive ( lgs ). Fr less thn zer degrees the ld wuld be cpcitive ( leds ). Symmetricl Cmpnents Pge 5

18 The phse currents in the blnced three-phse cse re (6.4) p (6.4b) b p 4 (6.4c) c p See Fig. 6.. fr the phsr representtin f the currents. 7. Symmetricl Cmpnents Systems The electricl pwer system pertes in blnced three-phse sinusidl pertin. When tree cntcts line, lightning blt strikes cnductr r tw cnductrs swing int ech ther we cll this fult, r fult n the line. When this ccurs the system ges frm blnced cnditin t n unblnced cnditin. n rder t prperly set the prtective relys, it is necessry t clculte currents nd vltges in the system under such unblnced perting cnditins. n Dr. C. L. Frtescue s pper he described hw symmetricl cmpnents cn trnsfrm n unblnced cnditin int symmetricl cmpnents, cmpute the system respnse by stright frwrd circuit nlysis n simple circuit mdels, nd trnsfrm the results bck int riginl phse vribles. When shrt circuit fult ccurs the result cn be set f unblnced vltges nd currents. The thery f symmetricl cmpnents reslves ny set f unblnced vltges r currents int three sets f symmetricl blnced phsrs. These re knwn s psitive, negtive nd zer-sequence cmpnents. Fig. 7. shws blnced nd unblnced systems. Fig. 7. Cnsider the symmetricl system f phsrs in Fig. 7.. Being blnced, the phsrs hve equl mplitudes nd re displced reltive t ech ther. By the definitin f symmetricl cmpnents, b lwys lgs by fixed ngle f nd lwys hs the sme mgnitude s. Similrly c leds by. t fllws then tht (7.) b ( 4 ) (7.b) Symmetricl Cmpnents Pge 6

19 c ( ) (7.c) Where the subscript () designtes the psitive-sequence cmpnent. The system f phsrs is clled psitive-sequence becuse the rder f the sequence f their mxim ccur bc. Similrly, in the negtive nd zer-sequence cmpnents, we deduce (7.) b ( ) c ( 4 ) (7.b) (7.c) (7.3) (7.3b) b (7.3c) c Where the subscript () designtes the negtive-sequence cmpnent nd subscript () designtes zer-sequence cmpnents. Fr the negtive-sequence phsrs the rder f sequence f the mxim ccur cb, which is ppsite t tht f the psitive-sequence. The mxim f the instntneus vlues fr zer-sequence ccur simultneusly. Fig.7. n ll three systems f the symmetricl cmpnents, the subscripts dente the cmpnents in the different phses. The ttl vltge f ny phse is then equl t the sum f the crrespnding cmpnents f the different sequences in tht phse. t is nw pssible t write ur symmetricl cmpnents in terms f three, nmely, thse referred t the phse (refer t sectin 3 fr refresher n the pertr). b c (7.4) (7.4b) b b b (7.4c) c c c Symmetricl Cmpnents Pge 7

20 We my further simplify the nttin s fllws; define (7.5) (7.5b) (7.5c) Substituting their equivlent vlues b c (7.6) (7.6b) (7.6c) These equtins my be mnipulted t slve fr,, nd in terms f, b, nd c. b c (7.7) 3 b c (7.7b) 3 b c (7.7c) 3 t fllws then tht the phse currents re b c (7.8) (7.8b) (7.8c) The sequence currents re given by b c (7.9) 3 b c 3 (7.9b) b c 3 (7.9c) The unblnced system is therefre defined in terms f three blnced systems. Eq. (7.6) my be used t cnvert phse vltges (r currents) t symmetricl cmpnent vltges (r currents) nd vice vers [Eq. (7.7)]. Symmetricl Cmpnents Pge 8

21 Exmple 7. Given 553, b 764, c 75, find the symmetricl cmpnents. The phse cmpnents re shwn in the phsr frm in Fig. 7.3 c b Unblnced cnditin Fig. 7.3 Slutin Using Eq. (7.7) Slve fr the zer-sequence cmpnent: 3 b c Frm Eq. (7.3b) nd (7.3c) 3.5 b c 3.5 Slve fr the psitive-sequence cmpnent: b c Frm Eq. (7.b) nd (7.c) 5. 3 b c 5. Slve fr the negtive-sequence cmpnent: b c Symmetricl Cmpnents Pge 9

22 Frm Eq. (7.b) nd (7.c).9 48 b c The sequence cmpnents cn be shwn in phsr frm in Fig Fig. 7.4 Using Eq. (7.6) the phse vltges cn be recnstructed frm the sequence cmpnents. Exmple 7. Given 3.5, 5.,.99 cmpnents. Shwn in the phsr frm in Fig. 7.4 Slutin Using Eq. (7.6) Slve fr the A-phse sequence cmpnent:, find the phse sequence Slve fr the B-phse sequence cmpnent: b Symmetricl Cmpnents Pge

23 Slve fr the C-phse sequence cmpnent: c This returns the riginl vlues given in Exmple 5.. This cn be shwn in phsr frm in Fig Fig. 7.5 Ntice in Fig. 7.5 tht by dding up the phsrs frm Fig. 7.4, tht the riginl phse, Fig. 7.3 quntities re recnstructed. Understnding symmetricl cmpnents cn id us in truble shting prblems. Sme relys mesure psitive sequence pwer. Mdifying equtin 7.7b fr pwer, it fllws then tht the psitive sequence pwer is P P Pb Pc (7.) 3 Fr blnced system, t unity pwer fctr, the phse sequence pwer shwn in phsr frm in Fig 7.6 Symmetricl Cmpnents Pge

24 Fig. 7.6 Frm equtin 7., the psitive sequence pwer equls, P P P P blnced system this simplifies t P P. 3 b c. Under Fig. 7.7 This is n pprximtin f the verge f the A-phse, B-phse nd C-phse pwer. See figure 7.7 Let s ssume tht the B nd C phse vltges were swpped. Frm phsr digrm we see tht the B-phse pwer nw leds A-phse by degrees, nd C-phse pwer lgs by degrees. Figure 7.8 Fig. 7.8 Symmetricl Cmpnents Pge

25 Frm equtin 7., the psitive sequence pwer equls, P P P P 3 b c. Hwever, with the B-phse nd C-phse vltge swpped, Figure 7.9 shws the phsr digrm Fig Blnced nd Unblnced Fult nlysis Let s tie it tgether. Symmetricl cmpnents re used extensively fr fult study clcultins. n these clcultins the psitive, negtive nd zer-sequence impednce netwrks re either given by the mnufcturer r re clculted by the user using bse vltges nd bse pwer fr their system. Ech f the sequence netwrks re then cnnected tgether in vrius wys t clculte fult currents nd vltges depending upn the type f fult. Symmetricl Cmpnents Pge 3

26 Given system, represented in Fig. 8., we cn cnstruct generl sequence equivlent circuits fr the system. Such circuits re indicted in Fig. 8.. Fig. 8. The psitive-sequence impednce system dt fr this exmple in per-unit is shwn in Fig. 8.. Fig. 8. Assuming the negtive-sequence equls the psitive-sequence, then the negtivesequence is shwn in Fig 8.3 Fig. 8.3 The zer-sequence impednce is greter thn the psitive nd fr ur purpse is ssumed t be three times greter. Als becuse f the wye-delt trnsfrmer, zer-sequence frm the genertr will nt pss thrugh the trnsfrmer. This will be shwn in sectin.. er-sequence is shwn in Fig 8.4 Symmetricl Cmpnents Pge 4

27 Fig. 8.4 The Thevenin equivlents fr ech circuit is reduced nd shwn in Fig. 8.5 Fig. 8.5 Ech f the individul sequence my be cnsidered independently. Since ech f the sequence netwrks invlves symmetricl currents, vltges nd impednces in the three phses, ech f the sequence netwrks my be slved by the single-phse methd. After cnverting the pwer system t the sequence netwrks, the next step is t determine the type f fult desired nd the cnnectin f the impednce sequence netwrk fr tht fult. The netwrk cnnectins re listed in Tble 8. Tble 8. - Netwrk Cnnectin Three-phse fult - The psitive-sequence impednce netwrk is nly used in three-phse fults. Fig. 8.3 Single Line-t-Grund fult - The psitive, negtive nd zer-sequence impednce netwrks re cnnected in series. Fig. 8.5 Line-t-line fult - The psitive nd negtive-sequence impednce netwrks re cnnected in prllel. Fig. 8.7 Duble Line-t-Grund fult - All three impednce netwrks re cnnected in prllel. Fig. 8.9 Symmetricl Cmpnents Pge 5

28 The system shwn in Fig. 8. nd simplified t the sequence netwrk in Fig. 8.5 nd will be used thrughut this sectin. Exmple 8. Given.99 9 pu,.75 9 pu,.759 pu, cmpute the fult current nd vltges fr Three-phse fult. Nte tht the sequence impednces re in per-unit. This mens tht the slutin fr current nd vltge will be in per-unit. Slutin The sequence netwrks re intercnnected, nd shwn + - Nte tht fr three phse fult, there re n negtive r zer-sequence vltges. The current is the vltge drp crss j.75 j The phse current is cnverted frm the sequence vlue using Eq. (7.8). - j pu b ( j5.7) () 5.75 pu c ( j5.7) () 5.73 pu Clculting the vltge drp, the sequence vltges re j.75 j5.7. pu. Symmetricl Cmpnents Pge 6

29 The phse vltges re cnverted frm the sequence vlue using Eq. (7.6)..... pu b. (.) (.). pu b c c. (.) (.). pu The per-unit vlue fr the current nd vltge wuld nw be cnverted t ctul vlues using Eq. (5.9) nd Eq. (5.) nd knwing the bse pwer nd vltge fr the given system. See exmple 5. fr reference. c b The currents nd vltges cn be shwn in phsr frm. Exmple 8. Given.99 9 pu,.75 9 pu,.759 pu, cmpute the fult current nd vltges fr Single line-t-grund fult. Nte tht the sequence impednces re in per-unit. This mens tht the results fr current nd vltge will be in per-unit. Slutin The sequence netwrks re intercnnected in series, s shwn. Becuse the sequence currents re in series, nd using hms lw. ( ) ( j.99 j. 8 pu j.75 j.75) The phse currents re cnverted frm the sequence vlue using Eq. (7.8). Substituting int Symmetricl Cmpnents Pge 7

30 Eq. (7.8) gives b 3 c Refer t Tble 3.: Nte tht 3. This is the quntity tht the rely see s fr Single Line-t- Grund fult. Substituting j. 8pu 3 3( j.8) j5. 46 pu Clculting the vltge drp, the sequence vltges re Substituting in the impednce nd current frm bve c j.99( j.8).36 j.75 j j.75 j The phse vltges re cnverted frm the sequence vlue using Eq. (7.6). b b.36 (.68) (.39).38 pu c.36 (.68) (.39). pu The per-unit vlue fr the current nd vltge wuld nw be cnverted t ctul vlues using Eq. (5.9) nd Eq. (5.) nd knwing the bse pwer nd vltge fr the given system. See exmple 5. fr reference. The currents nd vltges cn be shwn in phsr frm. Symmetricl Cmpnents Pge 8

31 Exmple 8.3 Given.99 9 pu,.75 9 pu,.759 pu, cmpute the fult current nd vltges fr Line-t-Line fult. Nte tht the sequence impednces re in per-unit. This mens tht the slutin fr current nd vltge will be in perunit. Slutin The sequence netwrks re intercnnected, s shwn Becuse the sequence currents sum t ne nde, it fllws tht The current is the vltge drp crss in series with j.75 j.75 j. 86 pu j. 86pu The phse current is cnverted frm the sequence vlue using Eq. (7.8). j.86 j.86 pu b ( j.86) ( j.86) 4. 95pu c ( j.86) ( j.86) 4. 95pu Clculting the vltge drp, nd referring t Fig. 8.7, the sequence vltges re ( j.75)( j.86).5 pu Symmetricl Cmpnents Pge 9

32 The phse vltges re cnverted frm the sequence vlue using Eq. (7.6) pu b. (.5) (.5). 5pu c. (.5) (.5). 5pu The per-unit vlue fr the current nd vltge wuld nw be cnverted t ctul vlues using Eq. (5.9) nd Eq. (5.) nd knwing the bse pwer nd vltge fr the given system. See exmple 5. fr reference. b c b c The currents nd vltges cn be shwn in phsr frm. Exmple 8.4 Given.99 9 pu,.75 9 pu,.75 9 pu, cmpute the fult current nd vltges fr Duble Line-t-Grund fult. Nte tht the sequence impednces re in per-unit. This mens tht the slutin fr current nd vltge will be in per-unit. Slutin The sequence netwrks re intercnnected, s shwn in Fig. 8.9 Becuse the sequence currents sum t ne nde, it fllws tht + ( ) - The current is the vltge drp crss in series with the prllel cmbintin f nd Substituting in, nd,, nd, then slving fr - Symmetricl Cmpnents Pge 3

33 j3. 73pu ( ) ( ) j.75 ( ) ( ) j.99 The phse current is cnverted frm the sequence vlue using Eq. (7.8). j.75 j3.73 j.99 pu b j.75 ( j3.73) ( j.99) pu j.75 ( j3.73) ( j.99) pu c 9 Clculting the vltge drp, nd referring t Fig. 8.9, the sequence vltges re ( j.75)( j.99).348 pu The phse vltges re cnverted frm the sequence vlue using Eq. (7.6) pu b.348 (.348) (.348) pu c.348 (.348) (.348) pu Refer t Tble 3.: R The per-unit vlue fr the current nd vltge wuld nw be cnverted t ctul vlues using Eq. (5.9) nd Eq. (5.) nd knwing the bse pwer nd vltge fr the given system. See exmple 5. fr reference. b c The currents nd vltges cn be shwn in phsr frm. Symmetricl Cmpnents Pge 3

34 9. Oscillgrms nd Phsrs Attched re fur fults tht were inputted int rely nd then cptured using the rely sftwre. Three-phse fult. Cmpre t exmple (8.) Fig 9. Fig 9.b Fig 9.c Symmetricl Cmpnents Pge 3

35 Single Line-t-Grund fult. Cmpre t exmple (8.) Fig 9. Fig 9.b Fig 9.c Symmetricl Cmpnents Pge 33

36 Line-t-Line fult. Cmpre t exmple (8.3) Fig 9.3 Fig 9.3b Fig 9.3c Symmetricl Cmpnents Pge 34

37 Duble Line-t-Grund fult. Cmpre t exmple (8.4) Fig 9.4 Fig 9.4b Fig 9.4c Symmetricl Cmpnents Pge 35

38 . Additin Symmetricl Cmpnents cnsidertins. Symmetricl Cmpnents int Rely Using directinl grund distnce rely it will be demnstrted hw sequentil cmpnents re used in the line prtectin. T determine the directin f fult, directinl rely requires reference ginst which the line current cn be cmpred. This reference is knwn s the plrizing quntity. er-sequence line current cn be referenced t either zer-sequence current r zer-sequence vltge, r bth my be used. The zer-sequence line current is btined by summing the three-phse currents. See Fig.. b b c c r Fig. Frm Eq. (7.9) b c r This is knwn s the residul current r simply 3. 3 (.) The zer-sequence vltge t r ner the bus cn be used fr directinl plriztin. The plrizing zer-sequence vltge is btined by dding n uxiliry ptentil trnsfrmer t the secndry vltge. The uxiliry trnsfrmer is wired s brkendelt nd the secndry inputted t the rely. See Fig. Symmetricl Cmpnents Pge 36

39 A B C A 3 B b C c Frm Eq. (7.7) the zer-sequence vltge equls b c (.) 3 b c 3 (.) Exmple. Using the vlues btined frm exmple 8., clculte 3. Slutin b.38 pu c. pu pu The zer-sequence vltge is.8 8 pu. By cnnecting the vlue in the reverse gives 3 which equls.8 pu. Pltting this, we cn shw in phsr frm wht the rely see s, lgging 3 by the line ngle. n this cse resistnce is neglected, therefre lgs by 9. (see Fig.3). Symmetricl Cmpnents Pge 37

40 Fig.3. Symmetricl Cmpnents thrugh Trnsfrmer This sectin will lk t current flw thrugh wye-delt trnsfrmer bnk. t will be shwn in the next chpter tht fr fults tht include grund tht zer-sequence quntities will be generted. t cn be shwn using symmetricl cmpnents tht zer-sequence cmpnents cnnt pss thrugh delt-wye trnsfrmer bnks. f zer-sequence is flwing n the wye side, the currents will be reflected t the ther side, but circulte within the delt. Fig.4 The current n the left side is A n b Fig.4 Symmetricl Cmpnents Pge 38

41 Frm equtin 7. we hve A B (.3 ) A A A (.3 b) B B B Substituting n the right side f the equtin 8. gives ( ) B = ) ( ) ( ) (.4) A ( A B A B A B The zer-sequence currents re in-phse, therefre equtin.3 simplifies t ( ) B = ) ( ) (.5) A ( A B A B Where ( A B) 3 A3 nd ( A B) 3 B 3 ( 3 A3 ) ( 3 A 3 ) n 3 ( A3 A 3 ) (.6) n n blnced system where there is n negtive r zer-sequence current then equtin.6 reduces t 3 ( A 3 ) (.7) n As cn be seen the current will shift by 3 when trnsferring thrugh trnsfrmer cnnected delt-wye. The sme cn be prve when lking t the vltges. Nw cnsider the cnnectin in Fig.5. n n b n c A n c Fig.5 Symmetricl Cmpnents Pge 39

42 Substituting equtin 7. nd reducing gives ( C ) = ) ( ) ( ) (.8) A ( A C A C A C n( 3 A 3 ) ( 3 A3 ) n ( 3 3 ) (.9) 3 A A As seen frm the prir exmple equtin.9 will reduce t n 3( A 3 ) if there is n negtive r zer-sequence current, which is the cse fr blnced system. By inspectin f the equtins bve fr ANS stndrd cnnected delt-wye trnsfrmer bnks if the psitive-sequence current n ne side leds the psitive current n the ther side by 3, the negtive-sequence current crrespndingly will lg by 3. Similrly if the psitive-sequence current lgs in pssing thrugh the bnk, the negtive-sequence quntities will led 3. The directin f the phse shifts between the delt-cnnected winding nd the wyecnnected winding depends n the winding cnnectins f the trnsfrmer. The winding cnfigurtins f trnsfrmer will determine whether r nt zer-sequence currents cn be trnsfrmed between windings. Becuse zer-sequence currents d nt dd up t zer t neutrl pint, they cnnt flw in neutrl withut neutrl cnductr r grund cnnectin. f the neutrl hs neutrl cnductr r if it is grunded, the zer-sequence currents frm the phses will dd tgether t equl 3 t the neutrl pint nd then flw thrugh the neutrl cnductr r grund t mke cmplete pth..3 Sequence mpednces: Trnsfrmer. Trnsfrmers with t lest tw grunded wye windings When trnsfrmer hs t lest tw grunded-wye windings, zer-sequence current cn be trnsfrmed between the grunded-wye windings. The currents will dd up t 3 in the neutrl nd return thrugh grund r the neutrl cnductr. The currents will be trnsfrmed int the secndry windings nd flw in the secndry circuit. Any impednce between the trnsfrmer neutrl pints nd grund must be represented in the zer-sequence netwrk s three times its vlue t crrectly ccunt fr the zer-sequence vltge drp crss it. Symmetricl Cmpnents Pge 4

43 Belw n the left is three-phse digrm f grunded-wye, grunded-wye trnsfrmer cnnectin with its zer-sequence netwrk mdel n the right. Ntice the resistnce in the neutrl f the secndry winding is mdeled by 3R in the zer-sequence netwrk mdel. P 3R S P 3 3 S R Reference Bus. Trnsfrmers with grunded-wye winding nd delt winding When trnsfrmer hs grunded-wye winding nd delt winding, zersequence currents will be ble t flw thrugh the grunded-wye winding f the trnsfrmer. The zer-sequence currents will be trnsfrmed int the delt winding where they will circulte in the delt withut leving the terminls f the trnsfrmer. Becuse the zer-sequence current in ech phse f the delt winding is equl nd in phse, current des nt need t enter r exit the delt winding. Belw n the left is three-phse digrm f grunded-wye-delt trnsfrmer cnnectin with its zer-sequence netwrk mdel n the right. 3. Auttrnsfrmers with grunded neutrl Auttrnsfrmers cn trnsfrm zer-sequence currents between the primry nd secndry windings if the neutrl is grunded. er-sequence current will flw thrugh bth windings nd the neutrl grund cnnectin. Belw n the left is three-phse digrm f grunded neutrl uttrnsfrmer with its zer-sequence netwrk mdel n the right. Symmetricl Cmpnents Pge 4

44 4. Auttrnsfrmers with delt tertiry f n uttrnsfrmer hs delt tertiry, zer-sequence current cn flw thrugh either the primry r secndry winding even if the ther winding is pen circuited in the sme mnner tht zer-sequence current cn flw in grundedwye-delt trnsfrmer. f the grund is remved frm the neutrl, zer-sequence current cn still flw between the primry nd secndry windings, lthugh there will nt be ny trnsfrmtin f currents between the primry nd secndry windings nly between the prtil winding between the primry nd secndry terminls nd the delt tertiry. This is nt nrml cnditin thugh, s it will nt be nlyzed here. Nte tht when mdeling three-winding trnsfrmers the impednce needs t be brken int the impednce f the individul windings. Symmetricl Cmpnents Pge 4

45 5. Other trnsfrmers Other trnsfrmer cnfigurtins, such s ungrunded wye-ungrunded wye, grunded wye-ungrunded wye, ungrunded wye-delt, nd delt-delt will nt llw zer-sequence currents t flw nd will hve n pen pth in the zersequence netwrk mdel. Sme f these cnfigurtins re shwn belw with their zer-sequence netwrk mdels. P S Reference Bus n the preceding trnsfrmer cnnectin digrms the vlues f t the terminls f the primry nd secndry windings will be equl n per-unit bsis. They will ls hve the sme per-unit vlues within the wye nd delt windings; hwever, the per-unit vlues f current within the windings f n uttrnsfrmer re smewht mre difficult t determine becuse prt f the winding crries bth primry nd secndry currents. f the mgnitude f current within the winding f n uttrnsfrmer needs t be knwn, it cn be determined by equting the mpere turns f the primry winding t thse f the secndry winding nd slving. f tertiry is invlved, it will need t be included in the equtin ls. Symmetricl Cmpnents Pge 43

46 Mgnitude f trnsfrmer zer-sequence impednce The zer-sequence impednce f single-phse trnsfrmer is equl t the psitivesequence impednce. When three single-phse units re cnnected s three-phse unit in cnfigurtin tht will trnsfrm zer-sequence currents (grunded wye-grunded wye, grunded wye-delt, etc.), the zer-sequence impednce f the three-phse unit will nrmlly be equl t the psitive-sequence impednce. n trnsfrmers built s three-phse units, i.e. with three-phse cre, in cnfigurtin cpble f trnsfrming zer-sequence currents, the zer-sequence impednce will be the sme s the psitive-sequence impednce if the trnsfrmer cre is f the shell type. f the cre is f the cre type, the zer-sequence impednce will be different thn the psitive-sequence impednce. This is becuse the zer-sequence excittin flux des nt sum t zer where the three legs f the cre cme tgether nd is frced t trvel utside f the irn cre, thrugh the il r the trnsfrmer tnk where the mgnetic permebility is much less thn the irn cre. This results in lw impednce (high cnductnce) in the mgnetizing brnch f the trnsfrmer mdel. The lrger zer-sequence mgnetizing current results in lwer pprent zer-sequence impednce. Using lwer vlue f zer-sequence impednce in the trnsfrmer zer-sequence mdel is sufficient fr mst fult studies, but t btin highly ccurte zer-sequence mdel f three-phse crefrm trnsfrmer, the mgnetizing brnch cnnt be neglected. Mgnitude f trnsfrmer zer-sequence impednce Typicl bnks re shwn in the reference sectin f this dcument. The winding lbels re H, M, nd L, r high, medium nd lw-vltge windings respectively. The L winding is ls clled the tertiry winding. The mnufcture prvides the lekge impednces between the windings s HM, HL, AND ML, usully n different ka r MA rtings t the rted winding vltges. The equivlent wye lekge impednces re btined frm the fllwing equtins: H ( HM HL ML) M ( HM ML HL) H ( HL ML MM ) As check,, etc. HM H M t is pssible fr ne f the vlues t be negtive. The junctin pint f the wye hs n physicl significnce. The wye is mthemticl equivlent vlid fr current nd vltge clcultins t the trnsfrmer terminls. Symmetricl Cmpnents Pge 44

47 . System Mdeling. System Mdeling: Trnsmissin Lines Trnsmissin lines re represented n ne-line digrm s simple line cnnecting busses r ther circuit elements such s genertrs, trnsfrmers etc. Trnsmissin lines re ls represented by simple line n impednce digrms, but the digrm will include the impednce f the line, in either hm r per-unit vlues. Smetimes the resistive element f the impednce is mitted becuse it is smll cmpred t the rective element. Here is n exmple f hw trnsmissin line wuld be represented n n impednce digrm with impednces shwn in hms: n blnced three-phse system the impednce f the lines nd lds re the sme, nd the surce vltges re equl in mgnitude. We cn clculte the single-phse current, but must tke int ccunt the vltge drp crss the mutul impednce cused by the ther phse currents. Frm Fig., the vltge drp in A-phse is AΦ BΦ s s m m CΦ s m S A m B m C Fig. (.) Fr the cse f blnced three-phse current ( ). Therefre: B C A S m A (.b) Dividing by A shws the psitive-sequence impednce f the line equls the self impednce minus the mutul impednce. A S m (.) A The negtive-sequence current encunters negtive-sequence impednce which is equl t the psitive-sequence impednce Symmetricl Cmpnents Pge 45

48 Symmetricl Cmpnents Pge 46 m S A A (.3) Fr the zer-sequence impednce, becuse, b nd c re in phse with ech ther, C B A then zer-sequence vltge drp is given in equtin.4 A m m A S C m B m A S (.4) A m S (.4b) Dividing ech side by A give the zer-sequence impednce: m S A A (.5) The result gives the zer-sequence impednce s functin f the self nd mutul impednce f the line. The zer-sequence impednce is lwys lrger thn the psitivesequence becuse we re dding tw times the mutul impednce t the self impednce, insted f subtrcting the mutul impednce frm the self impednce.

49 . System Mdeling: Subtrnsient, Trnsient, nd Synchrnus Rectnce f Synchrnus Genertrs A synchrnus genertr is mdeled by n internl vltge surce in series with n internl impednce. Belw is typicl ne-line digrm symbl fr genertr. The circle represents the internl vltge surce. The symbl t the left f the circle indictes tht the three phses f the genertr re wye-cnnected nd grunded thrugh rectnce. The symbl fr synchrnus mtr is the sme s synchrnus genertr. A typicl impednce digrm representtin f synchrnus genertr is shwn in Fig... Xg Rg Eg + t - Fig.. When mdeling the impednce f synchrnus genertr (r mtr), the resistive cmpnent is usully mitted becuse it is smll cmpred t the rective cmpnent. When fult is pplied t pwer system supplied by synchrnus genertr, the initil current supplied by the genertr will strt t lrger vlue, nd ver perid f severl cycles it will decrese frm its initil vlue t stedy stte vlue. The initil vlue f current is clled the subtrnsient current r the initil symmetricl rms current. Subtrnsient current decreses rpidly during the first few cycles fter fult is initited, but its vlue is defined s the mximum vlue tht ccurs t fult inceptin. After the first few cycles f subtrnsient current, the current will cntinue t decrese fr severl cycles, but t slwer rte. This current is clled the trnsient current. Althugh, like the subtrnsient current, it is cntinully chnging, the trnsient current is defined s its mximum vlue, which ccurs fter the first few cycles f subtrnsient current. After severl cycles f trnsient current, the current will rech finl stedy stte vlue. This is clled the stedy stte current r the synchrnus current. Symmetricl Cmpnents Pge 47

50 The resn why the current supplied by the synchrnus genertr is chnging fter fult is becuse the incresed current thrugh the rmture f the genertr cretes flux tht cuntercts the flux prduced by the rtr. This results in reduced flux thrugh the rmture nd therefre reduced generted vltge. Hwever, becuse the decrese in flux tkes time, the genertr vltge will be initilly higher nd decrese ver time. We ccunt fr the chnging genertr vltge in ur mdel by using different vlues f rectnce in series with the internl genertr vltge. We use three vlues f rectnce t mdel the genertr during the perid fter fult inceptin: the subtrnsient rectnce (Xd ) is used during the initil few cycles; the trnsient rectnce (Xd ) is used fr the perid fllwing the initil few cycles until stedy stte vlue is reched; the synchrnus rectnce (Xd) is used fr the stedy stte perid. The impednce digrms fr synchrnus genertr (r mtr) during the subtrnsient, trnsient, nd synchrnus perids re shwn in Fig..3. Fig..3 The rectnce f synchrnus mtrs re the sme s fr synchrnus genertrs. f the line t synchrnus mtr develps three-phse fult, the mtr will n lnger receive electricl energy frm the system, but its field remins energized nd the inerti f its rtr nd cnnected ld will keep the rtr turning fr sme time. The mtr is then cting like genertr nd cntributes current t the fult Symmetricl Cmpnents Pge 48

51 .3 System Mdeling: Trnsfrmers Trnsfrmers re represented in ne-line digrms by severl symbls. Belw re sme typicl nes. The first is tw-winding trnsfrmer cnnected delt- grunded wye, nd the secnd is three-winding trnsfrmer cnnected grunded wye-delt-grunded wye. An impednce mdel f prcticl tw-winding trnsfrmer is shwn in Fig..4. Fig..4 n the mdel, : represents the winding rti f the idel trnsfrmer shwn by the tw cupled cils, BL in prllel with G represents the mgnetizing susceptnce nd cnductnce which mke up the mgnetizing brnch, E represents the excittin current, r nd x represent the lekge impednce f winding,r nd x represent the lekge impednce f winding, nd represents the primry vltge nd current respectively, nd nd represent the secndry vltge nd current respectively. Becuse nrml fult nd ld currents re very much lrger thn the mgnetizing current, E, we cn mit the mgnetizing brnch frm ur mdel. We cn ls mit the idel trnsfrmer if we refer the lekge impednces t either the primry- r secndry-side f the trnsfrmer. The lekge impednce f ne side f the trnsfrmer cn be referred t the ther side f the trnsfrmer by multiplying it by the squre f the turns rti. Belw is the simplified impednce digrm with the mgnetizing brnch remved nd the lekge impednce f the secndry winding referred t the primry side f the trnsfrmer. Symmetricl Cmpnents Pge 49

52 Our impednce mdel cn be further simplified by letting R r r X x x When using this simplified mdel, ny impednces nd vltges cnnected t the secndry side f the circuit must nw be referred t the primry side. As n exmple, the fllwing trnsfrmer mdel will be cnverted t the simplified impednce mdel. The mgnetizing brnch nd the lekge resistnces hve been mitted t simplify the prblem. The secndry-side impednce is multiplied by the squre f the turns rti befre being trnsferred t the primry side. j6. * 8.33 = j46.3ω Symmetricl Cmpnents Pge 5

53 This is dded t the high side t get n impednce f j5ω + j46.3ω = j466.3ω The simplified mdel is shwn in Fig..5.4 Sme dditinl pints DC Offset Fig..5 n trnsmissin netwrk, the sudden ccurrence f shrt circuit will result in sinusidl current tht is initilly lrger nd decreses due t the chnging ir gp flux in the synchrnus genertrs. We ve seen tht this is mdeled by subtrnsient, trnsient, nd synchrnus rectnce in ur genertr mdel. n circuit cntining resistnce nd inductnce (RL circuit), such s in trnsmissin netwrk, the sudden ccurrence f shrt circuit will ls result in DC ffset in the current tht ccurs fter fult is pplied. Cnsider the RL circuit belw: f the switch is clsed t time t=, the vltge rund the circuit is mxsin(ωt+φ) = Ri + Ldi/dt Slving this differentil equtin fr the instntneus current, i, gives i = mx [sin(ωt+φ-θ) e-rt/lsin(φ-θ)] / Where = (R + (ωl) nd θ = tn-(ωl/r) Symmetricl Cmpnents Pge 5

54 The imprtnt thing t nte frm the slutin is tht there is sinusidl cmpnent tht represents the stedy-stte slutin fr the current (mx sin(ωt+φ-θ) / ) nd expnentilly decying cmpnent (-mx e-rt/lsin(φ-θ) / ). Sme pints t nte but the expnentilly decying r DC ffset cmpnent: The initil vlue f the DC ffset is determined by wht pint in the cycle the vltge wvefrm is t when the fult ccurs (the vlue f φ) nd will rnge frm up t the vlue f the stedy stte cmpnent. The dc cmpnent will decrese with time cnstnt f L/R. The lrger the rti f inductnce t resistnce in the circuit, the lrger the time cnstnt, nd the slwer the dc cmpnent will decy. Three time cnstnts fter the switch is clsed, the dc ffset will hve decyed t 5% f its initil vlue. DC ffset is n imprtnt cnsidertin in sizing brekers. Mst mdern micrprcessr-bsed relys re immune t DC ffset becuse fter the nlg signls re cnverted t digitl signls, they cn be mthemticlly filtered t remve the DC cmpnent. Therefre the DC cmpnent desn t need t be cnsidered in the rely settings. Sme electrmechnicl relys re immune t DC ffset, nd sme ren t. Clpper nd plunger type units re generlly nt immune, nd DC ffset will hve t be llwed fr in the rely settings (ne guideline is t set pickup t 6% f the desired c pickup current). Cylinder type units, used in distnce relys, re immune t DC ffset. The different vlues f the AC fult current shuld be cnsidered in the rely settings. The subtrnsient fult current shuld be used in setting instntneus current elements, wheres the synchrnus fult current shuld be used in current elements with lng time delys. Symmetricl Cmpnents Pge 5

55 Prblem Prblems BPA s system mdel uses three-phse pwer bse f MA. The line-t-line vltge bse is 55k fr the 5 system, 3k fr the 3 system, nd 5k fr the 5 system. ) An undervltge rely n the 5 system is set t pick up t.85 pu (per unit) f the phse-t-grund vltge. Wht is the phse-t-grund vltge tht the undervltge rely will pick up t? b) A three-phse fult n the 5 system results in fult current f 75A. Wht is the per unit vlue f this current? c) Wht is the bse impednce fr the 5 system? d) Wht is the bse impednce fr the 3 system? e) Wht is the bse impednce fr the 5 system? Prblem Frm ur exmple 5., the percent impednce f 55/4.5k uttrnsfrmer is.4% bsed n its nmeplte vlue f 9MA. Suppse we need t mdel this trnsfrmer in BPA s ASPEN mdel which uses MA pwer bse. Wht wuld the per-unit impednce be? Prblem 3 Frm ur exmple in 5., cnvert the per-unit impednce t per-unit vlue in threephse pwer bse f MA. ) First cnvert the per unit impednce t n ctul impednce (in hms) t 55k nd then cnvert the ctul impednce t per-unit impednce n the new bse. b) Repet, this time cnverting the per unit impednce t n ctul impednce (in hms) t 4.5k nd then cnverting the ctul impednce t per-unit impednce n the new bse Prblem 4 Cnvert the per-unit impednce f the trnsfrmer in the exmple t per-unit vlue in the BPA mdel with three-phse pwer bse f MA by first cnverting the per unit impednce t n ctul impednce (in hms) t 3 k nd then cnverting the ctul impednce t per-unit impednce n the new bse. Symmetricl Cmpnents Pge 53

56 Prblem 5 j5.o j6.o = 5O 8.33: Using the trnsfrmer mdel cnvert frm hms t per-unit. The vltge bse fr the primry side will be 5k, nd the vltge bse fr the secndry side will be 3.8k. The pwer bse fr bth sides is MA. Prblem 6 Belw is ne line digrm f prtil pwer system. The tw genertrs re identicl, ech rted 3.8k nd 5MA with subtrnsient rectnce f Xd = 5%. The tw genertrs re tied t cmmn bus which is cnnected t trnsmissin line with delt-grunded wye trnsfrmer rted t 5MA, 3.8k/5k nd n impednce f 9.7%. The trnsmissin line is 3 miles lng nd hs n impednce f j.5ω. At the end f the trnsmissin line is grunded wye-grunded wye trnsfrmer, rted 5MA, 5k/3k with n impednce f 7.4% tht cnnects the line t 3k bus. The remining pwer system cnnected t the 3k bus is nt shwn. Frm the bve infrmtin, drw the impednce digrm with impednces shwn in their per-unit vlues. Use vltge bses f 3.8k, 5k, nd 3k fr the crrespnding prts f the system, nd use pwer bse f MA fr the whle system. Symmetricl Cmpnents Pge 54

57 Prblem 7 Frm the impednce digrm, determine the per-unit nd mpere vlues f subtrnsient current in ech genertr nd t the fult fr three-phse fult pplied n the 3k bus with bth genertrs perting t.pu vltge. The genertrs cn be cmbined int their Thevenin equivlent s shwn belw. j j.7 j.389 j.5 F +.pu 3-phse fult - Prblem 8 Frm the ne line digrm f prtil pwer system tht we used in prblem 6. Frm the bve infrmtin, we drew the psitive-sequence impednce digrm using subtrnsient impednces fr the genertrs nd with impednces shwn in their per-unit vlues. Nrmlly the psitive-sequence netwrk is drwn with the reference bus (which is the neutrl pint) shwn t the tp insted f the bttm. The negtive-sequence rectnce f the genertrs is equl t their psitive-sequence subtrnsient rectnce. Drw the psitive nd negtive-sequence netwrks fr the pwer system with impednces shwn in their per-unit vlues. Prblem 9 Ech genertr hs zer-sequence rectnce f 5% nd is grunded thrugh rectnce f Ω. The trnsmissin line hs zer-sequence impednce f.9 + j75.9ω. The grunded wye-grunded wye trnsfrmer hs zer-sequence rectnce f 4.8%. Drw the zer-sequence impednce digrm. Symmetricl Cmpnents Pge 55

58 Prblem Frm the bk Prtective Relying Principles nd Applictins; Furth editin, by J. Lewis Blckburn nd Thms J. Dmin. Prblem 3.4 The pwer trnsfrmer cnnectins shwn in the figure belw re nnstndrd nd quite unusul with tdy s stndrdiztin. Hwever, this cnnectin prvides n excellent exercise in understnding phsrs, plrity, nd directinl sensing rely cnnectins. Cnnect the three-directinl phse relys A, B, C t line-side current trnsfrmers (CT s) nd bus-side vltge trnsfrmers (Ts) fr prper pertin fr phse fults ut n the line. Use the 9 6 cnnectin. Ech directinl rely hs mximum trque when the pplied current leds the pplied vltge by 3. The uxiliry Ts shuld be cnnected t prvide the relys with equivlent line-side vltges. The currents re cnnected s tht when, b, nd c re flwing in the trip directin indicted by the trip directin rrw, the secndry currents flw thrugh the directinl units frm the plrity f the rely. The uxiliry trnsfrmer is wired such tht the secndry vltge reflects the primry vltge. Delt Wye n the primry, is reflected Delt Wye n the secndry. With the trip directin f the currents estblished in the directinl unit current cils, the vltges bc, n unit A, c n unit B, nd b n unit C must be cnnected frm plrity t nn-plrity n the directinl unit vltge cils. Symmetricl Cmpnents Pge 56