Bo v relay trong ht

Bo v relay trong htMY BIN DNG IN V MY BIN IN PNHM 8Nguyn Vn Thi41103179Bi Quang Minh 41102037i. MY BIN DNG IN

1. nh nghaMy bin dng in l kh c in c nhim v bin i dng in s cp I1 trong mch in c in p cao v dng in th cp I2 tng ng vi thit b o lng thng qua t s nBI Dng in I2 thng l 1A, 5A, i khi ln n 10A. Cc thng s nh mc: Um , Im , Zm Ngoi ra cn c thng s khc nh sai s, cp chnh xc, ph ti th cp K hiu: BI, CT,TI

2. nh du cc tnhnh du 2 mi dy s cp I v IInh du 2 mi dy th cp 1 v 2Nu u ni I 1 v II 2 th dng in qua ti l khng iThc hnh: Ni mch in nh hnh v v cc tnh ca bnh in v in k G. Khi nhn nt cng tc in k G ch theo chiu thun th nh du nh hnh v.

3.iu kin lm vic ca bin dng inBin dng in bo v lm vic cng nng n hn bin dng in o lng, ngha l khi qu ti bin dng in vn hin th ng tr s.Chn bin dng in cn c vo dng in s cp cc iTng tr ph ti th cp phi Zpt Zcp tng tr cho cho php Ph ti ca bin dng in ch c mc ni tip

Khng c cho th cp bin dng in h mch v khi ta c I0 = I1 rt ln lm t thng b bo ha bng u gy sc in ng cm ng xung lm h hng cch in. Cun th cp phi ni t. (l do an ton)Lu : Khi c ti lm vic, bin dng khng c h mch th cp, nu cn tho g th phi ni tt 2 mi th cp. Nu Zpt tng cao th cng lm cho in p th cp tng Zpt .

4. cp chnh xc ca bin dng innh ngha: Cp chnh xc ca bin dng in c gi theo sai s ln nht v tr s I%max khi n lm vic trong cc iu kin sau:Tn s nh mc f = 50HzDng in s cp I1 = (1 n 1.2) I1m Ph ti th cp Zpt = (0.25 n 1 ) Z2m

Cp chnh xc: Do cu to li thp ( ), dng in s cp I1 dy qun v, ph ti th cp ( ) lm cho I1 I2' Sai s gm: tr s I v gc pha IDa vo th vect c th xc nh biu thc tnh cc thnh phn sai s:

ABCO

gim sai s th li thp phi tt th ( ) s nh dn n sai s nh Khi I1 c gi tr ln th sai s s nh nhng li gay pht nng. Ph ti c tnh cht tr th ( ) nh dn n sai s nhTuy nhin, i vi bin dng in c cu to cho th ( ) c nh, sai s bin dng in ph thuc vo ( ) v(I1 ) m thi.

Cn c vo sai s m ngi ta chia lm cc cp chnh xc: 0.2, 0.5, 1, 3, 10. Cp chnh xc 0.2 dng cc dng c o lng mu Cp chnh xc 0.5 dng cng t in Cp chnh xc 1 dng o lng cc dng c lp bng Cp chnh xc 3, 10 dng cc b truyn ng cho CB i vi h thng bo v rle th ty chnh xc m chn.

5. Cng sut th cp ca bin dngCng sut th cp nh mc ca bin dng S2 m l cng sut max ca ph ti m n gay sai s trong gii hn cho php.Cng sut th cp nh mc: (v Z2 rt b so vi Zpt )

hay

6.S ni dy bin dngDng qua rle IR v dng trn dy pha Ip c th bng nhau v c th khc nhau ph thuc vo s ni dy.h s s :

6.1 s sao Khi bnh thng hay N(3) th

khi N(2) th In =0, v tng dng NM 2 pha = 0. Dng NM ch chy qua 2 rle ca 2 pha b s c. Khi N(1) , ch c 1 rle ca pha s c c dng NM i qua.S dao bo v mi dng NMH s s Ksd = 1 Lu : chn 3 bin dng in ging nhau trnh tnh trng mt cn bng. thc t lun tn ti dng khng cn bng khong 0.01 n 0.02 A

6.2 s sao thiuKhi bnh thng hay N(3) th

khi N(2) th In =0, nu 2 pha c bin dng th c dng NM qua 2 rle cn dng In =0; nu 2 pha trong ch c 1 pha c bin dng th dng NM qua 1 rle dy pha s c v dy chung Khi N(1) , nu ti pha khng c bin dng th khng bo v cS dao thiu bo v NM nhiu pha, khng bo v ngn mch 1 phaH s s Ksd = 1

6.3 s bin dng u tam gic cn relay u saoKhi bnh thng hay N(3) th

v vy ta c dng vo r le s ln hn dng pha ln v lch gc 30 . H s s ty thuc vo dng NM. Nu l N th

6.4 s ni vo hiu s dng phaKhi bnh thng hay N(3) th

Khi N(2) nu 2 pha c bin dng th IA = - IC, IR = 2IaKhi N(1) , nu ti pha khng c bin dng th khng bo v cCh bo v ngn mch nhiu pha, khng bo v ngn mch 1 phaH s s

6.5 s lc dng th t khngKhi bnh thng hay N(3) th

tuy nhin thc t tn ti dng khng cn bng nn s khc 0Khi bt i xng:

Dng qua rle la IR = 3I0 . Vy ch bo v NM mt pha chm t v 2 pha chm t.

ii. My bin in p

1. nh nghaMy bin in p l kh c in c nhim v bin i in p s cp U1 v in p th cp U2 tng ng vi thit b o lng thng qua t s nU in p U thng l 100V (my bin in p 3 pha), V (i vi my bin in p 1 pha) Bin in p c thng s nh mc: Um , Im , Sm Ngoi ra cn c thng s khc nh sai s, cp chnh xc, ph ti th cp. K hiu: BU, VT

2. nh du cc tnh(Nh trong my bin dng in)3. iu kin lm vic ca my bin in pC th dng mi mt bin p o lng cho tng bo v. Tuy nhin, do kinh t nn thng dng mt bin p o lng cho nhiu bo v. Chn bin in p theo dng c in c yu cu cp chnh xc cao nhtTng ph ti th cp VA Spt Sm tng ng vi cp chnh xcKhng c cho th cp bin dng in ngn mch.Ph ti ca bin in p ch c mc song songCun th cp phi ni t. (l do an ton)

4. cp chnh xc ca bin in pnh ngha: Cp chnh xc ca bin in p c gi theo sai s ln nht v tr s U%max khi n lm vic trong cc iu kin sau:Tn s nh mc f = 50Hzin p s cp U1 = (0.9 n 1.1) Um Ph ti th cp Spt = (0.25 n 1 )S2m H s sng sut ph ti cos = 0.8

Ta thy 2 tam gic gay sai s bin in p: AEF: do dng t ha I0 gay ra BCF: do dng ti I2 gay raLc khng ti I2 =0 th vn c sai s do I0 gay ra.

ABCEF

Cp chnh xc: Do cu to li thp ( ), dng in ti ( ) ngha l ph thuc vo cng sut v s lng dng c o mc vo th cp, cu to bin in p ( ) lm cho U1 U2' = U2.KUSai s gm: tr s U v gc pha UDa vo th vect c th xc nh biu thc tnh cc thnh phn sai s:

gim sai s th li thp phi tt th ( ) s nh dn n sai s nh Ph ti ( ) ca bin in p khng c vt qu gi tr cho php, ngoi tra cn ph thuc vo h s cng sut ca ph ti th cpPh thuc vo cu to ca bin in p ( ) nn gim sai s ngi ta chn mt dng in trong cc cun dy v t cm trong mch c gi tr nh hn so vi MBA lc nhm gim in tr cc cun dy v in p ngn mch ca bin in p, thng th in p ngn mch bin in p khong 0.4 n 1 %.

Cn c vo sai s m ngi ta chia lm cc cp chnh xc: 0.2, 0.5, 1, 3, 10. Cp chnh xc 0.2 dng cc dng c o lng mu Cp chnh xc 0.5 dng cng t in Cp chnh xc 1dng o lng cc dng c lp bng Cp chnh xc 3, 10 dng cc b truyn ng cho CB i vi h thng bo v rle th ty chnh xc m chn.

5. S ni dy bin in p5.1S NI SAOTrong s ni sao c in p pha so vi t ta phi ni trung tnh xung t. Th cp ni sao phi c dy N. Nu dy trung tnh b t th s khng c in p pha so vi t, m ch c in p pha so vi im trung tnh ca h thng. Vi s ny ta c th ly p pha hay p dy ty .C th dng 3 my bin in p o lng ri hay dng my bin in p 3 pha 5 tr (V nu dng my bin in p 3 tr th khng c ng i cho t thng th t khng, lm cho dng t ha ln khi chm t s gay pht nng.

5.2S TAM GIC THIUTrong s ni tam gic thiu c th ly p dy v p pha so vi trung im ca p p dy, dng nhiu trong mng trung th.

5.3 S B LC TH T KHNGS cp ni sao c trung tnh ni t, th cp ni tam gic h v rle ni vo 2 mi dy h ny ( ly p th t khng).

Khi vn hnh bnh thng hay ngn mch nhiu pha th UR = 0. Tuy nhin thc t th tn ti dng khng cn bng.Khi c ngn mch chm t th UR 0.Ta c th dng u sao sao tam gic h c th ly c in p ty thch: pha hay dy

C th dng my bin p o lng 2 cun th cp u sao sao tam gic h, nh th ny ta c c in p: pha, dy v th t khng.i vi MBA hay MF ta c th ly in p th t khng t trung tnh nh bin in p o lng