CH¯ NG 6 TR€O L¯U C”NG SU¤T

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CHƯƠNG 6 TRÀO LƯU CÔNG SUẤT

Text of CH¯ NG 6 TR€O L¯U C”NG SU¤T

  • 1.GII TCH MNG Trang 77 CHNG 6 TRO LU CNG SUT 6.1. GII THIU: Nhim v ca gii tch mng l tnh ton cc thng s ch lm vic, ch yu l dng v p ti mi nt ca mng in. Vic xc nh cc thng s ch mng in rt c ngha khi thit k, vn hnh v iu khin h thng in. Mt s ln cc thut ton c xut trong 20 nm tr li y. Trong chng ny ta gii thiu cc phng php trn cc kha cnh nh: D chng trnh ha, tc gii, chnh xc.... Vic tnh ton dng cng sut phi c tin hnh tng bc v hiu chnh dn. Bn cnh mc ch xc nh trng thi tnh th vic tnh ton dng cng sut cn l mt phn ca cc chng trnh v ti u v n nh. Trc khi c s xut hin ca my tnh s, vic tnh ton dng cng sut c tin hnh bng thit b phn tch mng. T nm 1956, khi xut hin my tnh s u tin th phng php tnh dng cng sut ng dng my tnh s c xut v dn dn c thay th cc thit b phn tch mng. Ngy nay cc thit b phn tch mng khng cn c dng na. 6.2. THIT LP CNG THC GII TCH. Gi s mng truyn ti l mng 3 pha i xng v c biu din bng mng ni tip dng nh trn hnh 6.1a. Cc phn t ca mng c lin kt vi nhau nn ma trn tng dn nt YNt c th xc nh t s . Theo s 6.1a ta c: INt = YNt .VNt (6.1) 1 p . . 0 + Vp - Ip P Sp (b)(a) Hnh 6.1 : S a cng ca ng dy truyn ti YNt l mt ma trn tha v i xng. Ti cc cng ca mng c cc ngun cng sut hay in p. Chnh cc ngun ny ti cc cng lm cho p v dng lin h phi tuyn vi nhau theo (6.1) chng ta c th xc nh c cng sut tc dng v phn khng bm vo mng (quy c cng sut dng khi c chiu bm vo mng) di dng hm phi tuyn ca Vp v Ip. Ta c th hnh dung ngun cng sut bm vo mng ni ngang qua cng ti u dng ca ngun bm nh hnh 6.1b.

2. GII TCH MNG Trang 78 Phn loi cc nt: - Nt P -Q l nt m cng sut tc dng P v cng sut phn khng Q l c nh, nh nt P 6.1 chng hn )()( SP LP SP GP SP LP SP GP SP p SP ppp QQjPPjQSIV +=+= (6.2) Vi Vp = ep +jfp Ch s GP v LP ng vi cng sut ngun pht v cng sut tiu th P. S cho bit cng sut c nh (hay p t). - Nt P -V tng t l nt c cng sut tc dng P c nh v ln in p c gi khng i bng cch pht cng sut phn khng. Vi nt ny ta c: SP LP SP GP SP ppp PPPIV ==]Re[ * (6.3) SP pppp VfeV =+= )( 22 (6.4) - Nt V-q (nt h thng) r rng nt ny in p v gc pha l khng i. Vic a ra khi nim nt h thng l cn thit v tn tht I2 R trong h thng l khng xc nh trc c nn khng th c nh cng sut tc dng tt c cc nt. Nhn chung nt h thng c ngun cng sut ln nht. Do ngi ta a ra nt iu khin in p ni chung l n c cng sut pht ln nht. nt ny cng sut tc dng PS (s k hiu nt h thng) l khng c nh v c tnh ton cui cng. V chng ta cng cn mt pha lm chun trong h thng, gc pha ca nt h thng c chn lm chun thng mc zero radian. in p phc V c nh cn Ps v Qs c xc nh sau khi gii xong tro lu cng sut cc nt. 6.3. CC PHNG PHP GII QUYT TRO LU CNG SUT: Theo l thuyt th c hai phng php tn ti l phng php s dng ma trn YNt v phng php s dng ma trn ZNt. V bn cht c hai phng php u s dng cc vng lp. Xt v lch s phng php th phng php YNt a ra trc v ma trn YNt d tnh v lp trnh, thm ch ngy nay n vn s dng vi h thng khng ln lm, phng php ny gi l phng php Gauss -Seidel. ng thi phng php Newton cng c a ra phng php ny c u im hn v mt hi t. Sau khi cch loi tr trt t ti u v k thut lp trnh ma trn vevt tha lm cho tc tnh ton v s lng lu tr t hn, th phng php Newton tr nn rt ph bin. Ngy nay vi h thng ln ti 200 nt hay hn na th phng php ny lun c dng. Phng php dng ma trn ZNt vi cc vng lp Gauss - Seidel cng c tnh hi t nh phng php Newton nhng ma trn ZNt l ma trn y nn cn b nh hn ct gi chng, l hn ch chnh ca phng php ny Trong chng ny chng ta ch gii thiu nguyn l ca cc phng php, cn cc phng php c bit nh: S l ma trn tha, sp xp ti u php kh, lc , ..... khng c cp n. 3. GII TCH MNG Trang 79 6.4. LCH V TIU CHUN HI T. Php gii tro lu cng sut c coi l chnh xc khi tha mn iu kin t (6.2) n (6.4) m ch yu l phi m bo chnh xc (6.4), hai tiu chun hi t ph bin l: - Mc cng sut tnh ton nt no theo Vp v Ip bn tri ng thc (6.2) n (6.4) ph hp tng ng vi gi tr cho sn bn phi. S sai khc ny gi l lch cng sut nt. - lch in p nt gia 2 vng lp k tip nhau. Sau y ta xt tng tiu chun c th: + Tiu chun lch cng sut nt: T (6.1) v (6.2) ta c = +== n q qpqp SP p SP ppp SP pp VYVjQPIVSS 1 *** (6.5) Tch phn thc v phn o ca (6.5) ta c lch cng sut tc dng v lch cng sut phn khng thch hp cho c (6.2) v (6.3). Biu din trong ta vung gc nh sau: Ta s dng k hiu sau: ppppp VjfeV =+= qppq pqpqpq jBGY = += Vi tng nt P -V hay P - Q Dng ta vung gc: ]))(()Re[( 1 = += n q qqpqpqpp SP PP jfejBGjfePP (6.6a) Dng ta cc: += = n q qpqpqpqpqp SP pp VBGVPP 1 ||)sincos(|| (6.6b) Vi tng nt P - Q Dng ta vung gc: ]))(()Im[( 1 = += n q qqpqpqpp SP pp jfejBGjfeQQ (6.7a) Dng ta cc: = = n q qpqpqpqpqp SP pp VBGVQQ 1 ||)cossin(|| (6.7b) Tiu chun hi t chung nht c dng trong thc t l: Pp Cp cho tt c nt P -V v P -Q Qp Cq cho tt c nt P -Q Gi tr Cp v Cq c chn t 0,01 - 10 MVA hay MVAR ty theo trng hp. + Tiu chun lch in p: Gi s bc lp l k, lch in p gia hai vng lp k v k +1 l: ( ) ( )kk p VVV = +1 cho tt c cc nt P - Q Tiu chun hi t l: Vp Cv cho tt c cc nt P - Q 4. GII TCH MNG Trang 80 Gi tr Cv t 0,01 n 0,0001 6.5. PHNG PHP GAUSS - SEIDEL S DNG MA TRN YNT: d hiu phng php ny ta gi thit tt c cc nt l nt P-Q tr nt h thng V - q. V in p ca nt h thng hon ton bit nn khng c vng lp no tnh cho nt ny. Ta chn nt h thng l nt cn bng. Do Vq (q s) coi l p ca nt q so vi nt s (k hiu nt s l nt h thng). Vi tt c cc nt, tr nt th s l nt h thng ta rt ra c t (6.1) v (6.2): = === n q qpq P P P npVY V S I 1 * * ...2,1 ; p s (6.8) Tch Ypq, Vp trong ra ri chuyn v ta c: npVY V S Y V n pq q qpq P P pp p ...2,1 1 1 * * = = = ; p s (6.9) Cc vng lp ca phng trnh Gauss - Seidel c thnh lp nh sau: = + )( 11 )( 313 )( 212)( 1 11 11 )1( 1 ....... 1 k nnss kk k k VYVYVYVY V jQP Y V = + )( 22 )( 121)( 2 22 22 )1( 2 .......... 1 k nnss k k k VYVYVY V jQP Y V = ++ ++ )()( 11 )( 11 )1( 11)( )1( ................ 1 k npnsps k PPP k PPP k Pk P PP pp k p VYVYVYVYVY V jQP Y V = + ++ )1( 11 )1( 11)( )1( ....... 1 k nnnsns k nk n nn nn k n VYVYVY V jQP Y V (6.10) Hay vit di dng tng qut l: pq k p p p q n pq k qpq k qpq k p YV S VYVYV 1 .*)( 1 1 )()1()1( + = = = ++ Ma trn YNt l ma trn thu c khi ta xa i hng s v ct s ma trn YNt. V VNt, INt cng c c bng cch xa i phn t s. Ta vit li ma trn YNt bng cch gm cc phn t ng cho, ma trn gm cc phn t tam gic di ng cho, ma trn gm cc phn t tam gic trn ng cho. YNt = D - L - W (6.11) Vi: = X O X O X D = O O O X O W = O X O O O L 5. GII TCH MNG Trang 81 Vy cc vng lp c vit gn li nh sau: [ ]).(.. )()()1(1)1( S k nutNut k nut kk VVYVWVLDV ++= ++ nutnut Vi : = snsk n nn spsk p pp sSk S k NutNut VY V jQP VY V jQP VY V jQP VVY )*( )*( 1)*( 1 11 )( ),( (6.12) k : = 1 Chn tr s in p ban u Vp (0) , p = 1, 2,... n Xc nh s liu vo Tnh Vp (k+1) theo (6.10) P = 1, 2,.... n Xc nh thay i cc i ca in p Max|Vp (k+1) | = |Vp (k+1) - Vp (k) | p = 1, 2,... Hnh 6.2 : S khi phng php Gauss _ Seidel Kim tra |Vp (k+1) | max < Cv In kt qu Vp = Vp (k+1) + V0 Tnh dng cng sut, Tnh dng cng sut, in p...... Vp = Vp (k+1) + V0 k : =1 END BEGIN 6. GII TCH MNG Trang 82 Kim tra hi t nh sau: V k p k p CVVMax max p cal p QQ = min p cal p QQ < min p cal p QQ = Tnh nh tnh vi nt P - Q v khng iu chnh in p. Nu trong tnh ton tip theo gim xung trong phm vi gii hn th tnh ton nh nt P - Vcal pQ 6.5.2. Tnh ton dng chy trn ng dy v cng sut nt h thng: Sau khi cc php tnh v vng lp hi t. Dng chy trn ng dy v cng sut nt h thng c tnh nh sau: Ipq Ypq Ipq qp 0 + Vq - + Vp - Ypq/2 Ypq/2 0 Hnh 6.3 : S ca ng dy truyn ti Xt ng dy ni t nt p n nt q c tng dn ni tip v Ypq v tng dn r l Y pq, dng in ng dy c xc nh: 2/)( ' pqppqqppq YVYVVI += Dng cng sut chy t p n q l: ]2/)[( '**** pqPpqqpppqpq YVYVVVjQP +=+ (6.17) Dng cng sut chy t q n p l: ]2/)[( '**** pqqpqpqqqpqp YVYVVVjQP +=+ (6.18) Tn tht cng sut ng dy s bng tng i s ca Ppq +jQpq v Pqp +jQqp Cng sut nt h thng c tnh bng tng cc dng cng sut chy trn cc ng dy c u ni vi nt h thng: 6.5.3. Tng tc hi t: Phng php s dng vng lp YNt hi t chm bi v trong h thng ln mi nt thng c dy ni n 3 hay 4 nt khc. Kt qu l lm cho tin trnh lp yu i vic ci thin in p mt nt s nh hng n cc nt ni trc tip vo n. V vy k thut tng tc c s dng nng cao tc hi t. Phng php ph bin nht l SOR (Successive - over - relaxation) phng php gim d qu hn lin tip. Ni dung phng php l c sau mi vng lp th s hiu chnh in p trn cc nt P - Q bng cch sau: )( )()1( )( )1( k p k tnhp k p VVV = ++ (6.19) V Vp (k+1) l: )1()()1( ++ += k p k p k p VVV (6.20) H s a gi l h s tng tc c xc nh theo kinh nghim gia 1 v 2, thng (1 < a < 2). 8. GII TCH MNG Trang 84 Nu a chn hp l th tc hi t tng mnh, nhn chung gi tr thc ca a l t 1,4 n 1,6. Nu a l s phc th phn thc v phn o ca in p c tng tc ring bit: [ ] [ ])()1( )( )()1( )( )1( ImRe k p k tnhp k p k tnhp k p VVjVVV += +++ (2.21) V (6.22))1()()1( ++ += k p k p k p VVV Vi a v b u l s thc: 6.5.4. u v nhc im ca phng php dng YNt: Ma trn YNt kh d thnh lp v phng php gii l trc tip nn lp trnh tr nn n gin. B nh c dng lu tr cc phn t khc khng nm trn ng cho chnh. Sau khi s dng tnh i xng ca YNt th vic tnh ton v lu tr cng gn hn. V trong h thng mi nt ni n 3 hay 4 nt khc nn mi vng lp cho tng nt s dng n s lu tr cc nt ny, do php tnh s tng ln rt nhiu. S php tnh trong mi bc lp t l vi s nt n, nu s nt l n th s php tnh l n2 . Vi h thng c 200 nt hay hn na phng php ny t ra km hiu qu v rt kh hi t nu c nh hng ca iu kin no chng hn c mt ca t ni tip (t b dc) so vi phng php Newton. 6.6. PHNG PHP S DNG MA TRN Z NT: gii thch v phng php ny u tin ta gi thit khng c nt P-V cc nt u l P - Q (gm n nt) v mt nt cn bng (chn nt cn bng l nt h thng). Trng hp c tn ti nt P - V s xt phn 6.6.3: Gi thit cc thng s ca mng tuyn tnh khi c th xem ngun dng nt th p l Jp l t hp tuyn tnh ca dng in gy ra bi in p Vp v in p cc nt khc Vq (q = 1... n, q p). y l nguyn l xp chng ca mng in. YNt .VNt = INt YNt, VNt , INt c ngha nh (6.1) Nhim v ca chng ta l tm VNt. tm VNt c th dng phng php kh lin tip hay phng php Crame nhng cc phng php ny rt cng knh khi n ln. y ta cp n phng php ma trn. Do YNt l ma trn vung, i xng v khng suy bin nn ta c: VNt = YNt -1 . INt YNt -1 = ZNt : Gi l ma trn tng tr nt ca mng in. Do