Fig. 1. I a. V ag I c. I n. V cg. Z n Z Y. I b. V bg

Transcription

1 ymmetricl Compoets equece impedces Although the followig focuses o lods, the results pply eqully well to lies, or lies d lods. Red these otes together with sectios.6 d.9 of text. Cosider the -coected lced lod, Fig.. t is grouded through impedce. Becuse of this, the eutrl poit my ot e t the sme potetil s the groud. g c cg g Fig.

2 Let s use KL to write the voltge equtios for the three phse-to-groud voltges g, g, d cg s fuctio of the lie currets d the lod impedces. Assume o mutul couplig etwee phses. g g () c cg We c replce if we pply KCL t the juctio ode t the ceter of the. c () ustitutio of () ito () yields: c g c g () c c cg Expdig the through d the collectig terms with like currets yields: g c g (4) c cg

3 Now let s write these voltge equtios i mtrix form. g g cg compct ottio, eq. (5) is: c (5) c c c (6) Now pose the followig questio. f we c somehow fid wy to trsform eq. (5) ito equtio tht reltes sequece voltges o the LH to sequece currets o the RH, wht will the impedce mtrix look like? To swer this questio, we eed to derive eq. (7). (7)

4 We refer to the impedce mtrix,, tht reltes sequece voltges to sequece currets, s the sequece impedce mtrix. To derive eq. (7), cosider wht we hve: eq. (6), repeted here for coveiece. c c c (6) Recll tht c =A d c =A. ustitutig ito eq. (6) yields: A ca (8) Now pre-multiply oth sides y A -. This is: A A A ca (9) The left hd side is just. A ca () Compriso of () with (7) idictes tht the sequece impedce mtrix,, is give y A ca () 4

5 5 o wht does look like? We kow ll three elemets of eq. () so why do t we do the mtrix mth d fid out c A A Multiplyig the two right-hd mtrices: Now multiply the remiig mtrices: 9 Pluggig this expressio ito eq. () c A A

6 g g g () Now this is mzig thig ll off-digol terms re zero! Wht does this me? t mes tht the oly curret tht determies the zero sequece voltge is the zero sequece curret. the oly curret tht determies the positive sequece voltge is the positive sequece curret. the oly curret tht determies the egtive sequece voltge is the egtive sequece curret. This is the cse wheever the impedce mtrix is digol, with off-digols ll. 6

7 We sy tht the three equtios represeted y the mtrix reltio re ucoupled i tht o vrile (curret) ppers i more th oe equtio. o these ucoupled equtios re: g g g () The relly ice thig out these equtios is tht they represet seprte d distict NGLE PHAE CRCUT!!!! Therefore we c just pply EE perphse lysis to lyze them. Fig. illustrtes the sigle phse circuits. 7

8 = + ero sequece etwork Positive sequece etwork Negtive sequece etwork Fig. 8

9 ome questios:.why does t the eutrl impedce pper i the positive & egtive sequece etworks? Becuse the positive d egtive sequece etworks coti lced currets oly, d lced currets sum to d therefore do ot cotriute to flow i the eutrl..why do we hve i the zero sequece etwork isted of just? Recll A + B + C =. We defied. o ctully flows i the. But our zero sequece etwork hs oly flowig. Therefore, to oti the correct voltge drop see i the eutrl coductor with flow of oly, we model the zerosequece impedce s. The the voltge drop is, s it should e. 9

10 .Wht do these three etworks look like if the eutrl is solidly grouded (o eutrl impedce)? Positive d egtive sequece etworks re the sme. ero sequece is the sme except =. 4.Wht do these three etworks look like if the eutrl is ugrouded (flotig)? Positive d egtive sequece etworks re the sme. ero sequece hs ope circuit, which mes. 5.Wht is eefit of the C trsformtio? Aswer: f the lod (or lie, or lod d lie) is symmetric (so tht is digol), the the three etworks will decouple d we c lyze ulced situtio with three seprte per-phse lyses.

11 6. Wht if the lod (or lie, or lod d lie) is ot symmetric? Look t this cse closely. Cosider geerl --c impedce mtrix s give elow (cosistet with the mtrix of eq..4, pg. 47, i text) c c c c c cc (4) This is geerl impedce mtrix i tht The digols,, d cc my differ. The off-digols, c, c my differ. The ove occurs whe the phse geometry is ot equilterl & trspositio is ot used. Recll eq. (), repeted here for coveiece: A ca () The we compute usig the --c impedce mtrix of eq. (4):

12 A c A c c c c cc We will ot go through the detiled mtrix multiplictio here ut isted will just provide the expressios for ech of the 9 terms i the mtrix, s follows: cc c c () cc c c (4) cc c c (5) cc c c (6) cc c c (7) cc c c (8)

13 For our +- circuits to e decoupled (d thus oti the dvtge of symmetricl compoet decompositio see questio 5 ove), the off-digol elemets of must e. o wht re the coditios for the offdigol elemets of to e? We oti these coditios y settig (5)- (8) to (mkig off-digol elemets ) d solvig them simulteously. We will ot go through this pi here. Rther, you should e le to ispect eqs. (5)-(8) d otice tht for them to e, it must e true tht cc (9) (Digol phse impedces must e equl) c c () (Offdigol phse impedces must e equl).

14 4 Uder the coditios imposed y eqs. (9) d (), it will e the cse tht the +- impedces re give y () () () Check it for the exmple we worked ove correspodig to Fig. : c

15 ome dditiol oservtios from this work: By eq. (4), repeted here for coveiece, cc c c (4) the positive d egtive sequece impedces re lwys equl, idepedet of whether the lod is symmetric or ot. This is true for trsmissio lies, cles, d trsformers. By eq. (), repeted here for coveiece, cc c c () is ot equl the -sequece impedce to the positive d egtive sequece impedces of symmetricl lod uless the off-digol phse impedces = c = c re. t is ot true for rottig mchies ecuse positive sequece currets, rottig i the sme directio s the rotor, produce fluxes i the rotor iro differetly th the egtive sequece currets which rotte i the opposite directio s the rotor. 5