Zbus 1.0 Introduction The Zbus is the inverse of the Ybus, i.e., (1) Since we know that

Transcription

1 us. Introducton he us s the nverse of the us,.e., () Snce we now tht nd therefore then I V () V I () V I (4) So us reltes the nodl current njectons to the nodl voltges, s seen n (4). In developng the power flow prolem, we choose to wor wth us. he reson for ths s tht the power flow prolem requres n tertve soluton tht cn e mde very effcent when we use us, due to the sprsty (lots of zeros) n the mtrx used n performng the terton (the Jcon mtrx we wll dscuss ths more lter).

2 However, n developng fult nlyss methods (done n EE 457), we wll choose to wor wth us. he mn reson for choosng to wor wth us n fult nlyss s tht, s we wll see, us qunttes chrcterze condtons when ll current njectons re zero except one, correspondng to the fulted us. We cn use some cretve thnng to express tht one current njecton (the fult current). Once we hve tht one current njecton, eq. (4) s very esy to evlute to otn ll us voltges n the networ, nd once we hve us voltges, we cn get ll currents everywhere. hese currents re the currents under the fult condtons nd re used to desgn protecton systems. he us s not sprse (no zeros). But fortuntely, fult nlyss does not requre tertve solutons, nd so computtonl eneft of sprsty s not sgnfcnt n fult nlyss.

3 . he menng of us elements We cn wrte the -us relton for the sme networ s V V V I I I () We understnd y eq. () tht the ndependent sources re ll current sources, nd eq. () llows us to compute the voltges resultng from those current sources eng njected nto the networ. hese current sources re the equvlent representton of the genertor voltge sources. Let s nspect more closely one of the equtons n (). Artrrly choose the second equton. V I I I ()

4 Now solve for drvng pont mpednce of us,. V I I I () But wht f we set I nd I to,.e., wht f we open-crcut uses nd? In other words, let s dle ll sources. hs mens tht we wll open ny current sources t nodes nd so tht there re no sources there (ut there my e lod mpednces). hen eq. () s: V I I I (4) Equton (4) sys tht s the rto of us voltge to the us current njecton when ll sources re dled. hs s the defnton of the hevenn mpednce!!!! 4

5 Concluson: he dgonl elements of the - us re the hevenn equvlent mpednces seen loong nto the networ t tht us. hs s useful ecuse, s seen n the lst set of notes, the current nto the fult my e computed s I f f V f hev he remnder of these notes comes from chpter 9 of Bergen & Vttl. I often cover t n EE 456 (ut other nstructors do not). hey show you how to construct the -us for lrge networ wthout performng mtrx nverson. One mportnt ttrute to uldng the - us for fult nlyss s tht the genertor sutrnsent rectnces should e ncluded. hs mens tht t s necessry to nclude n ddtonl us for every genertor n order to enle dstngushng etween the hgh sde of the genertor nternl voltge from the 5

6 networ sde of the genertor sutrnsent rectnce. he fgure elow llustrtes the dfference etween genertor representton for power flow nlyss (the focus of EE 456) nd genertor representton for fult nlyss (the focus of EE 457). jx d Genertor representton for power flow nlyss Genertor representton for fult nlyss. Self dmttnce nd drvng pont mpednce ou should recll tht t s esy to develop the -us. From tht, one cn nvert t to otn the -us. However, n spte of the fct tht Mtl s qute cple of mtrx nverson for smll dmenson, you re NO 6

7 ALLOWED to thn out just nvertng us snce we must, eventully, lve n the rel world of 5+ us models. As result, we must turn to the so-clled - us uldng lgorthm, sed on modfcton to n lredy exstng us. o understnd t, we wll frst loong t modfcton to n lredy exstng us. 4. us modfctons (See Appendx 7) Let s ssume tht we hve -us system wth us gven y (7) Assume the rnch etween uses nd s numered s rnch. hen let s modfy rnch y ddng nother crcut etween us nd us hvng n dmttnce of Δy. How wll the us chnge? 7

8 Add Δy to dgonl elements n postons (,) nd (,). Sutrct Δy from off-dgonl elements n postons (,) nd (,). he resultng mtrx s n y y y y (8) So the new -us s just the old -us wth the ddton of Δy n four postons, s ndcted y eq. (9): n y y y y y (9) he mtrx on the rght of eq. (9) cn ctully e wrtten s product of two vectors. Defne vector correspondng to 8

9 9 the modfcton of rnch (connected etween us nd us ) s us us or us us () hen notce tht: () Susttuton of () nto (9) yelds: n y () In generl, nytme we modfy -us, then n y () where s constructed ccordng to: t poston, - t poston j f we dd or remove rnch etween uses nd j. t us f we dd or remove shunt t us. Why s eq. () of nterest to us? Reson s tht there s nce wy to nvert n expresson n the form of eq. ().

10 5. Mtrx nverson lemm (Appendx 6) he mtrx nverson lemm (MIL), otherwse nown s the Shermn-Morrson formul, s s follows. Suppose we hve n n n symmetrc mtrx whose nverse s nown nd we wsh to fnd the nverse of +μ, where μ s sclr nd s n n vector. hen the MIL sys tht: (4) where s n n vector gven y (5) nd γ s sclr gven y (6) here s proof of MIL n the text, pge 6. It s lso dscussed n most oos on lner lger.

11 hs formul s useful for gettng modfed -us, snce = -. In other words, eq. (4), (5), nd (6) ecome: n (7) where s n n vector gven y (8) nd γ s sclr gven y (9) he mplcton s huge. Consder tht somehow, we re le to get n ntl us, denoted, for some suporton of our networ. hen we cn chnge us to reflect the ddton of n element to the networ, y usng eqs. (8) nd (9) to compute the nd γ, whch re n turn used n eq. (7) to get the new -us, denoted n. Repeted pplcton of ths wll eventully provde us wth the entre -us. he lgorthm mplementng ths pproch s clled the -us uldng lgorthm.

12 6. Exmple (Exmple A7. n text) We re gven the followng us for - node networ. j he dmttnce of rnch, locted etween nodes nd, s chnged from j5 to j5. Fnd the new us, n. Soluton: he chnge n dmttnce of rnch, locted etween uses nd, s y y, j Note tht the fct tht the dmttnce ecme more negtve mens tht the mpednce got less postve,.e., the mpednce decresed. hs s cused y the ddton of nother crcut, s shown elow. x=j. y=-j5 x dd =j. y dd =-j x=j. y=-j5 x new =j.667 Orgnl sunetwor Add n crcut y new =-j5 Resultng sunetwor

13 Followng the exmple of eq. (), we hve: us us Wrtng out eq. (), we hve: n y y j Now, f we hd, we could esly compute n. But tht s not our ojectve. Our ojectve s to compute n =( n ) -. By eq. (7), (8), nd (9), ths s n where μ=δy. So frst let s compute. ht s

14 .... j j Wht s? More generclly, we oserve: In other words, fnds the dfference etween the columns of correspondng to the uses termntng the rnch eng modfed, n ths cse, uses nd,.e.,, where. Now let s compute γ. y ( j) j. j.5 ( j.667). j.4. j.75 4

15 5 Fnlly, we cn compute n ccordng to: ) ( j j j j j j n j j j j n

16 6 7. A closer loo t γ Recll tht y But. Susttuton yelds: y Consder the term where rnch connects uses nd j. hen we cn wrte: jj j jj j jj j j jj j j jn jj j j j nn nj n n jn jj j j n j n j ) ( j us us.

17 he expresson for γ ecomes y y j jj If there s no crcut etween uses nd j, z j jj where z =/Δy s the mpednce of the new crcut dded. 8. Some specl cses he MIL s used to uld us when you hve two exstng uses nd you wnt to dd n lne. But how do you get those exstng uses n the frst plce? hs requres some specl cses for modfyng us. We wll consder three specl cses, nd then we wll gve the -us uldng lgorthm. 8. Addng us connected to ground he stuton s s llustrted n Fg. 5, where we re just ddng n us wth shunt, ut we re not (yet) connectng the us to the networ. 7

18 he networ y =/z Fg. 5 Although ths stuton does not me much sense, t corresponds to step tht we must te n uldng the -us. o understnd the pproch to ths stuton, we return to the us. How wll us e modfed? We smply ncrese the dmenson of us y, wth only the new dgonl element eng non-zero, nd t wll hve vlue y. he result of ths s shown elow, where s the -us efore the ddton of the new node. n y () Here, s n n vector of s. Invertng eq. (), we get: n n y / y z () 8

19 So modfyng us to ccommodte new us connected to ground s esy, nd t results n Modfcton # nd Rule # n Secton 9.5. Modfcton #: Add rnch wth mpednce z from new us (numered n+) to the reference (ground) node. Rule #: n s gven y n z () 8. Add rnch from new to exstng us We ssume the new us s numered n+ nd the exstng us s numered. he stuton s s llustrted n Fg. 6. Bus y =/z Exstng networ Fg. 6 9

20 he dervton of wht to do here s sed on the sme pproch ten n secton 8.,.e., we frst see wht hppens to the us, nd then consder the nverson of the us. In ths cse, the dervton s lttle tedous nd I wll not go through t. ou cn refer to the text, pp he result s Modfcton # nd Rule # n Secton 9.5. Modfcton #: Add rnch wth mpednce z from new us (numered n+) to n exstng node. Rule #: Denote the th column of s, nd the th element of s. n s then gven y n z () Exmple : A networ conssts of sngle us connected to the reference node through n mpednce of j.5 ohms. Gve us.

21 Soluton: hs requres pplcton of rule. n z But does not exst,.e., t s mtrx of dmenson. herefore, n j.5 Exmple : A us s dded to the networ of exmple through rnch hvng rectnce of j.5. Gve new us. Soluton: hs requres pplcton of rule. n z Here, ecuse conssts of just sngle us (us ), = = =,.e., = = =. herefore n j.5 j.5 j.5 j.5 j.5 z j.5 j.5 j.

22 8. Add rnch etween exstng uses hs s the orgnl stuton we wored on to motvte MIL. It s llustrted n Fg. 8. Exstng networ Bus y =/z Bus j Fg. 8 We now wht to do wth ths. he result s summrzed s Modfcton #4 nd Rule #4 n Secton 9.5. Modfcton # 4: Add rnch z etween exstng th nd j th nodes. Rule #4: Denote the th column of s, nd the j th column of s j, nd denote the th, jj th, nd j th elements of s, jj, j. hen n s gven y n j z j jj

23 We hve lredy gven one exmple of ths rule (Exmple A7., pge ove), nd nother one s gven n the text s Exmple 9.. So we wll not llustrte further. here s one more modfcton necessry, tht we wll cll modfcton #, nd tht s to dd rnch wth mpednce etween exstng node nd reference (ground). hs dffers from Modfcton # ecuse n Modfcton #, the node dd not exst prevously. Now t does. he ssue s t trcy, nd we wll ddress t n secton 8.5. Before we do tht, however, t my help to te loo t the lgorthm used to uld - us, s we hve enough nformton to do tht now us uldng lgorthm Step : Numer the nodes of the networ strtng wth those nodes t the ends of rnches connected to the reference node.

24 Step : Develop the -us for ll uses wth connecton to the reference node. hs s Modfcton # usng Rule #, whch mens tht the resultng mtrx wll e dgonl mtrx consstng of the vlues of the shunt mpednces long the dgonl (ll off-dgonls wll e zero). Recll tht ths step s sed on: he networ y n =/z z Note: uldng the us n ths wy vods the necessty of rule, snce ths step conssts only of ddng shunt mpednces smultneous wth nodes (modfcton #, rule #), nd fter we re done, we wll not hve ny more shunt mpednces to dd, nd therefore t wll not e possle to dd shunt mpednce to n exstng us. Lovely. 4

25 Step : Add new node to the th (exstng) node of the networ v new rnch hvng mpednce z. Contnue untl ll nodes of the networ hve een dded. hs s modfcton #, Rule #, sed on: Exstng n networ z Step : Add rnch etween the th nd j th nodes. Contnue untl ll remnng lnes hve een dded. hs s modfcton #4, Rule #4, whch s sed on: n j z j jj Bus y =/z Exstng networ Bus y =/z Bus j 5

26 8.5 Addng shunt to exstng us: motvton he us uldng lgorthm provdes wy to uld the us usng only Rules #,, nd 4. he reson why ths wors s tht the frst step of the lgorthm s to uld the us for ll uses tht hve shunt elements. hs step elmntes the posslty tht t some lter step of the lgorthm, we wll hve to dd shunt to n exstng us. hus we vod Rule #, whch ddresses ddng shunt to n exstng us. Unfortuntely, there re stutons tht requre Rule # whch we cnnot vod. Consder, for exmple, tht you wnt to develop progrm whch wll compute fult currents for ech us, one fulted us t tme. An mportnt queston s, Wht does fulted us men? 6

27 It mens tht short-crcut s plced on the us. In relty, there s no perfect short,.e., ny short wll lwys hve some mpednce. In effect, then, fulted us mens connectng very smll shunt mpednce from us to the reference node. So, ssumng fult nlyss requres the us for ech fulted condton, how would you do the study to get the short crcut current for ech fulted us? Usng our lgorthm descred ove, wht we hve to do s to re-uld the us for ech seprte fulted us we wnt to nlyze. Buldng us s not s computtonl s nvertng the mtrx, ut t does requre some computtonl effort. 7

28 Wht we would le to do, from progrmmng perspectve, s to uld the networ us once, nd then perform more effcent mnpulton for gettng ech fult-specfc us. hs, then, s where Rule # comes n very hndy. Addng shunt mpednce from us to the reference node my e thought of n prelmnry wy s ddng rnch etween n exstng node nd new node. hs s Rule # ssue. How do we hndle Rule #? Repetng eq. (), we develop the new us s: n z () where the new us relton s V I V ref z Iref (4) 8

29 But n eq. (4), V ref =, nd so V I z Iref (5) hs fct tht the left-hnd-sde of eq. (5) s for the lst equton s sgnfcnt. It mens tht we my elmnte vrle from our soluton vector. In prtculr, we would le to elmnte I ref snce t corresponds to current tht s not useful to us (t s the sum of ll currents njected from the reference node nto the networ). here exsts certn nlytcl technque for elmntng I ref. It s clled Kron Reducton fter the fmous power system engneer Grel Kron. 9

30 8.5 Kron Reducton (secton 9.) Consder the followng mtrx equton: x c y e d z f (6) We cn wrte ths s seprte equtons: x y z e (7) cy dz f (8) Let s elmnte the vrle z from the top equton. hs s ccomplshed y multplyng the ottom equton y -d - x y z e d cy d dz d nd ddng t to the top equton. hs results n x y d f cy z d dz e d (9) Fctorng out y from the frst two terms nd notng the thrd nd fourth terms go to, we hve: x d ( c) y e d f (4) Concluson: we cn elmnte the second vrle (z) from our equton set f we force to zero the element n the frst row, second column (). f

31 hs wors f the ottom element n the lefthnd-sde s zero. In ths cse, the operton to ccomplsh our purpose wll not chnge the top element n the left-hnd-sde (x). Let s see f we cn do ths sme thng to eq. (5), repeted here for convenence: V I z Iref (5) One cn smply pply the sme pttern tht we used n the ove cse, whch ws: x d ( c) y e d f (4) to otn (note tht e nd f re zero): V z I (4) Alterntvely, one cn thn n terms of forcng the element n the second column, frst equton, to zero, through n operton where we dd multple of the second equton to the frst.

32 he gol here s to zero out the element n poston (,) of the eq. (5) mtrx. he reson we wnt to do ths s ecuse then, the frst equton ecomes ndependent of the vrle I ref. o do ths, we wll multply the second row y ( +z ) -. hen we dd ths multpled second row to the frst row. hs results n: ref I I z z z z z V ) ( ) ( * ) ( * ) ( (4) Notng n eq. (4) the terms tht go to zero, ref I I z z V * ) ( (4) And the frst equton n (4) s ndependent of I ref, so tht we my extrct t s: I z V (44) whch s the sme s eq. (4).

33 hs s Kron reducton. It cn e ppled to reduce the dmensonlty of set of lner equtons whenever the left-hnd-sde of one of the equtons s zero. Asde: he text, n Secton 9., motvtes Kron reducton n totlly dfferent, ut eqully legtmte wy. It rses the ssue of wht you cn do to the us when you hve us for whch there s nether generton or lod modeled t t. hs ssue comes often n power flow nlyss. here re, n fct, some susttons tht do not hve generton, nd they do not serve lod. Such susttons mght rse, for exmple, t the juncton of severl trnsmsson crcuts, or t trnsmsson-level trnsformer pplcton (e.g., 69V:45 V). In ths cse, the lefthnd sde of the us relton wll e zero for the correspondng crcut. he text does some good exmples of ths on pp. 9-.

34 8.6 Addng shunt to exstng us: method We re now (fnlly) n poston to stte Modfcton # nd Rule #. It s: Modfcton #: Add rnch wth mpednce z etween (exstng) th node nd the reference node. Rule #: Denote the th column of s, nd the th element of s. n,e s then gven y n e, z () he new us s then otned y performng Kron reducton on n,e to elmnte the lst vrle, tht s, z n (45) 4