BO V CC PHN T CHNH TRONG H THNG INPGS.TS L KIM HNG

A. GII THIU CHUNG V MY PHT INMy pht in (MF) l mt phn t rt quan trng trong h thng in (HT), s lm vic tin cy ca cc MF c nh hng quyt nh n tin cy ca HT. V vy, i vi MF c bit l cc my c cng sut ln, ngi ta t nhiu loi bo v khc nhau chng tt c cc loi s c v cc ch lm vic khng bnh thng xy ra bn trong cc cun dy cng nh bn ngoi MF. thit k tnh ton cc bo v cn thit cho my pht, chng ta phi bit cc dng h hng v cc tnh trng lm vic khng bnh thng ca MF.

I. Cc dng h hng v tnh trng lm vic khng bnh thng ca MFI.1. Cc dng h hng:kp). (2) Ngn mch nhiu pha trong cun stator. (1) Chm chp gia cc vng dy trong cng 1 pha (i vi cc MF c cun dy Chm t 1 pha trong cun dy stator. (3) Chm t mt im hoc hai im mch kch t. (4)

I.2. Cc tnh trng lm vic khng bnh thng ca MF:- Dng in tng cao do ngn mch ngoi hoc qu ti. (5) - in p u cc my pht tng cao do mt ti t ngt hoc khi ct ngn mch ngoi. (6) Ngoi ra cn c cc tnh trng lm vic khng bnh thng khc nh: Ti khng i xng, mt kch t, mt ng b, tn s thp, my pht lm vic ch ng c, ...

II. Cc bo v thng dng cho MFTu theo chng loi ca my pht (thu in, nhit in, turbine kh, thu in tch nng...), cng sut ca my pht, vai tr ca my pht v s ni dy ca nh my in vi cc phn t khc trong h thng m ngi ta la chn phng thc bo v thch hp. Hin nay khng c phng thc bo v tiu chun i vi MF cng nh i vi cc thit b in khc. Tu theo quan im ca ngi s dng i vi cc yu cu v tin cy, mc d phng, nhy... m chng ta la chn s lng v chng loi rle trong h thng bo v. i vi cc MF cng sut ln, xu th hin nay l lp t hai h thng bo v c lp nhau vi ngun in thao tc ring, mi h thng bao gm mt bo v chnh v mt s bo v d phng c th thc hin y cc chc nng bo v cho my pht. bo v cho MF chng li cc dng s c nu phn I, ngi ta thng dng cc loi bo v sau: - Bo v so lch dc pht hin v x l khi xy ra s c (1). - Bo v so lch ngang cho s c (2). - Bo v chng chm t mt im cun dy stator cho s c (3). - Bo v chng chm t mch kch t cho s c (4). - Bo v chng ngn mch ngoi v qu ti cho s c (5). - Bo v chng in p u cc my pht tng cao cho s c (6). Ngoi ra c th dng: Bo v khong cch lm bo v d phng cho bo v so lch, bo v chng qu nhit rotor do dng my pht khng cn bng, bo v chng mt ng b, ...

13

B. CC BO V RLE CHO MY PHT INI. Bo v so lch dc (87G)I.1. Nhim v v s nguyn l:Bo v so lch dc (BVSLD) c nhim v chng ngn mch nhiu pha trong cun dy stator my pht. S thc hin bo v nh hnh 1.1.Bo tn hiu+

Bo tn hiu t mch th+

MC

4Rth

Ct MC

5RT

+

+

+

3RI

1RI

2RI

Rf MF Rf 1BI

52

b) a)

MF

87G

Hnh 1.1: S bo v so lch dc cun stator MF; s tnh ton (a) v theo m s (b)

2BI

Trong : - Rf: dng hn ch dng in khng cn bng (IKCB), nhm nng cao nhy ca bo v. - 1RI, 2RI, 4Rth: pht hin s c v a tn hiu i ct my ct u cc my pht khng thi gian (thc t thng t 0,1 sec). - 3RI, 5RT: bo tn hiu khi xy ra t mch th sau mt thi gian cn thit (thng qua 5RT) trnh hin tng bo nhm khi ngn mch ngoi m tng t mch th. Vng tc ng ca bo v l vng gii hn gia cc BI ni vo mch so lch. C th y l cc cun dy stator ca MF, on thanh dn t u cc MF n my ct.

I.2. Nguyn l lm vic:BVSLD hot ng theo nguyn tc so snh lch dng in gia hai u cun dy stator, dng vo rle l dng so lch: (1-1) IR = I1T - I2T = ISL Vi I1T, I2T l dng in th cp ca cc BI hai u cun dy. Bnh thng hoc ngn mch ngoi, dng vo rle 1RI, 2RI l dng khng cn bng IKCB: ISL = I1T - I2T = IKCB < IKR (dng khi ng rle) (1-2) nn bo v khng tc ng (hnh 1.2a). Khi xy ra chm chp gia cc pha trong cun dy stator (hnh 1.2b), dng in vo cc rle 1RI, 2RI:14

I ISL = I1T - I2T = N > IKRTrong : - IN: dng in ngn mch. - nI: t s bin dng ca BI Bo v tc ng i ct 1MC ng thi a tn hiu i n b phn t ng dit t (TDT). Trng hp t mch th ca BI, dng vo rle l:

nI

(1-3)ISL = IKCBT < IKR I1 T I2T I2T I1T ISL

IN > I KR nI

a)

b)

I IR = F nI

(1-4)

Hnh 1.2: th vct ca dng in trong mch BVSLD

a) Bnh thng v khi ngn mch ngoi b) Khi ngn mch trong vng bo v

Dng in ny c th lm cho bo v tc ng nhm, lc ch c 3RI khi ng bo t mch th vi thi gian chm tr, trnh hin tng bo nhm trong qu trnh qu khi ngn mch ngoi c xung dng ln. s hnh 1.1, cc BI ni theo s sao khuyt nn bo v so lch dc s khng tc ng khi xy ra ngn mch mt pha pha khng t BI. Tuy nhin cc bo v khc s tc ng.

I.3. Tnh cc tham s v chn Rle:I.3.1. Tnh chn 1RI v 2RI:Dng in khi ng ca rle 1RI, 2RI c chn phi tho mn hai iu kin sau: iu kin 1: Bo v khng tc ng i vi dng khng cn bng cc i IKCBmax khi ngn mch ngoi vng bo v. IKB Kat.IKCBtt (1-5) IKCBtt = Kn.KKCK.fi .INngmax (1-6) Trong : - Kat: h s an ton tnh n sai s ca rle v d tr cn thit. Kat c th ly bng 1,3. - KKCK: h s tnh n s c mt ca thnh phn khng chu k ca dng ngn mch, KKCK c th ly t 1 n 2 tu theo bin php c s dng nng cao nhy ca bo v. - Kn: h s tnh n s ng nht ca cc BI (Kn = 0,51). - fi: sai s tng i ca BI, fi c th ly bng 0,1 (c k n d tr, v cc my bin dng chn theo ng cong sai s 10%). - INngmax: thnh phn chu k ca dng in chy qua BI ti thi im u khi ngn mch ngoi trc tip 3 pha u cc my pht. iu kin 2: Bo v khng c tc ng khi t mch th BI. Lc dng vo rle 1RI, 2RI: (gi s MF ang lm vic ch nh mc)

I ISL = mFDng khi ng ca bo v: IKB =

nI

(1-7)

Nh vy, iu kin chn dng khi ng cho 1RI, 2RI: IKB = max{Kat .IKCBtt; Kat .ImF } Dng in khi ng ca rle: IKR =

K at I mF nI

(1-8) (1-9) (1-10)

K (3) .I KB nI15

Vi K(3) l h s s . Sau khi tnh c IKR ta s chn c loi rle cn thit. Kim tra nhy Kn ca bo v: Kn =

Vi INmin: dng in ngn mch 2 pha u cc my pht khi my pht lm vic ring l. V bo v c tnh chn lc tuyt i nn yu cu Kn > 2.

I N min I KB

(1-11)

I.3.2. Tnh chn Rle 3RI:Dng khi ng s cp ca rle 3RI phi ln hn dng khng cn bng cc i khi ngn mch ngoi vng bo v. Nhng trong tnh ton th iu kin n nh nhit ca rle l quyt nh. Theo kinh nghim c th chn dng khi ng cho 3RI: IKS(3RI) = 0,2.ImF (1-12) Ta tnh c IKR ca 3RI v chn c loi rle tng ng.

I.3.3. Thi gian lm vic ca 5RT:Khi xy ra ngn mch ngoi vng bo v, c th xut hin nhng xung dng ln thong qua lm cho bo v tc ng nhm do vy phi chn thi gian tc ng ca 5RT tho mn iu kin: t5RT > tct Nngoi (1-13) t5RT = tctNng + t (1-14) Trong : - tctNng: thi gian ln nht ca cc bo v ni vo thanh gp in p my pht. - t: bc chn lc thi gian, thng t = (0,25 0,5) sec. Nhn xt: - Bo v s tc ng khi ngn mch nhiu pha trong cun dy stator 1BI I1S I1T my pht. - Bo v khng tc ng khi BIH BILV chm chp gia cc vng dy trong Vng bo IH cng 1 pha hoc khi xy ra chm t v 1 im trong cun dy phn tnh. ILV RI tng nhy ca bo v so I2S I2T 2BI lch ngi ta c th s dng rle so lch c hm. Hnh 1.3: Bo v so lch dng in c hm cun dy stator MF

I.4. Bo v so lch c hm:S bo v nh hnh 1.3. Rle gm c hai cun dy: Cun hm v cun lm vic. Rle lm vic trn nguyn tc so snh dng in gia ILV v IH. - Dng in vo cun lm vic ILV: Dng in hm vo cun hm IH: IH = I1T + I2T (1-16) Khi lm vic bnh thng hay ngn mch ngoi vng bo v: Dng in I1T cng chiu vi dng I2T: I1T I2T ISL = ILV = I1T - I2T = IKCB (1-17) IH = I1T + I2T 2.I1T > ILV (1-18) nn bo v khng tc ng. Khi xy ra ngn mch trong vng bo v: Dng in I1T ngc pha vi I2T: I1T = -I2T IH = I1T - I2T 0 ILV = I1T + I2T 2.I1T > IH (1-19)16

-

I LV = I 1T I 2T = I SL

.

.

(1-15)

bo v s tc ng. Nhn xt: - Bo v hot ng theo nguyn tc so snh dng in gia ILV v IH, nn nhy ca bo v rt cao v khi xy ra ngn mch th bo v tc ng mt cch chc chn vi thi gian tc ng thng t = (15 20) msec. - Bo v so lch dc dng rle c hm c th ngn chn bo v tc ng nhm do nh hng bo ho ca BI. - i vi cc my pht in c cng sut ln c th s dng s bo v so lch hm tc ng nhanh (hnh 1.4). ch lm vic A bnh thng, dng in th cp I1T v I2T ca cc nhm ILV B bin dng 1BI, 2BI chy qua I1S I2S RL1 in tr hm RH, to nn C CL RLV RL1 in p hm UH, cn hiu ULV dng th cp (dng so lch) BIG ISL chy qua bin dng trung ILV RL2 gian BIG, cu chnh lu CL I2T v in tr lm vic RLV to RL2 I1T D1 D2 BIG nn in p lm vic ULV. IH Gi tr in p UH > ULV, ILV RH/2 n RG bo v khng tc ng.RH/2 UH u ra

Hnh 1.4: Bo v so lch c hm tc ng nhanh cho MF cng sut ln Khi ngn mch trong vng bo v, in p ULV >> UH, dng in chy qua rle RL1 lm rle ny tc ng ng tip im RL1 li. Dng in lm vic sau khi nn chy qua rle RL2, RL2 ng tip im li, rle ct u ra s c cp ngun thao tc qua hai tip im ni tip RL1 v RL2 i ct my ct u cc my pht. Ngoi ra, ngi ta cn dng rle so lch tng tr cao bo v so lch my pht in (hnh 1.5). Rle so lch RU trong s c tng tr kh ln s tc ng theo in p so lch USL, ch lm vic bnh thng v khi ngn mch ngoi, cc bin dng 1BI, 2BI (c chn ging nhau) c cng dng in my pht i qua do cc sc in ng E1 v E2 bng nhau v ngc pha nhau, L1 = L2, phn b in p trong mch nh hnh 1.5b.1BI IN 2BI N E1 USL 1BI a) R1 E1 E1 b) L1 USL R2 RSL USL = 0 E2 L2 E2 E1 E1 L1 c) E1 R1 USL L1 R1 R2 E2=0

USL RSL USL R2 RSL USL L2

E2 E2

d)

Hnh 1.5: Bo v so lch dng rle tng tr cao cho MF a) S nguyn l b) Mch in ng tr v phn b in p trong ch lm vic bnh thng c) nhm 2BI b bo ho khi ngn mch ngoi v hon ton d) khi c ngn mch trong.17

Tr s in p t ln rle so lch RU ph thuc vo quan h gia cc in tr R1 v R2. in tr R1, R2 gm in tr cun dy th cp v dy dn ph ni gia hai nhm bin dng 1BI v 2BI, vi R1 = R2 USL = 0 Khi xy ra ngn mch trong vng bo v: * Trng hp my pht lm vic bit lp vi h thng: Dng in qua 1BI l dng ca my pht. Dng in qua 2BI bng khng E2 = 0. in p t ln rle so lch RU hnh 1.5c:

Trong : " - I N : tr hiu dng ca dng siu qu khi ngn mch trn u cc my pht. vi:

I "N .(R1 + R 2 ) U SL1 = nI

(v RSL >> R2)

(1-20)

I "N = I(3)Nngmax = I(3)Nu cc MF

- nI: t s bin dng ca BI. - RSL: in tr mch so lch (gm rle v dy ni). * Trng hp my pht ni vi h thng: Khi ti im ngn mch, ngoi dng in do bn thn my pht cung cp I " cn c thm thnh phn dng in do h thng NF v I " . Mch in ng tr v phn b in p nh hnh 1.5d. Gi tr in p t ln rle so NH lch RU:

m bo tnh chn lc, in p khi ng ca rle so lch RU phi chn ln hn min{USL1; USL2}, ngha l:

(I "NF + I "NH ).(R1 + R 2 ) U SL2 = nI

(1-21)

Vi Kat = (1,15 1,2) l h s an ton. Thi gian tc ng ca bo v thng: t = (15 20) msec Nhn xt: - i vi cc MF c cng sut ln, hng s thi gian tt dn ca thnh phn mt chiu trong dng in ngn mch c th t n hng trm msec, gy bo ha mch t ca cc my bin dng v lm chm tc ng ca bo v khi c ngn mch trong vng bo v. V vy cn phi s dng s bo v tc ng nhanh trc khi xy ra bo ha mch t ca my bin dng, tc l: tbh > tbv, vi tbv l thi gian ct ngn mch ca bo v; tbh thi gian bo ho mch t ca BI.

K at .I "N .(R1 + R 2 ) UKR = Kat.USL1 = nI

(1-22)

18

I.5. Bo v khong cch (21):i vi cc MF cng sut ln ngi ta thng s dng bo v khong cch lm bo v d phng cho BVSL (hnh 1.6a). TG BA BUU tII jXK R 0 tI = (0,4 0,5) sec t a) b) 0 RK X jX UF ZK

F BI

RZI

XB

t0,7XB

XF

Hnh 1.6: S nguyn l (a); c tnh thi gian (b) v c tuyn khi ng (c) ca bo v khong cch cho MF V khong cch t MBA n my ct cao p kh ngn, trnh tc ng nhm khi ngn mch ngoi MBA, vng th nht ca bo v khong cch c chn bao gm in khng ca MF v khong 70% in khng ca MBA tng p ( bo v hon ton cun h ca MBA), ngha l: I Z k = ZF + 0,7.ZB (1-23) Thi gian lm vic ca vng th nht thng chn tI = (0,4 0,5) sec (hnh 1.6b). Vng th hai thng bao gm phn cn li ca cun dy MBA, thanh dn v ng dy truyn ti ni vi thanh gp lin k. c tuyn khi ng ca rle khong cch c th c dng vng trn vi tm gc to hoc hnh bnh hnh vi nghing ca cnh bn bng nghing ca vct in p UF hnh 1.6c.

II. Bo v so lch ngang (87G)Cc vng dy ca MF chp nhau thng do nguyn nhn h hng cch in ca dy qun. C th xy ra chm chp gia cc vng dy trong cng mt nhnh (cun dy n) hoc gia cc vng dy thuc hai nhnh khc nhau trong cng mt pha, dng in trong cc vng dy b chm chp c th t n tr s rt ln. i vi my pht in m cun dy stator l cun dy kp, khi c mt s vng dy chm nhau sc in ng cm ng trong hai nhnh s khc nhau to nn dng in cn bng chy qun trong cc mch vng s c v t nng cun dy c th gy ra h hng nghim trng. Trong nhiu trng hp khi xy ra chm chp gia cc vng dy trong cng mt pha nhng BVSLD khng th pht hin c, v vy cn phi t bo v so lch ngang chng dng s c ny.

19

K

RL R I1SLV

I*LV Ct MC

R H IH BIH

4 3 ILV = IH

2BI1BI I2S ILV

I2T I1T b)

2 1 0 1 2 3 ILV = f(IH) 4 I*H

BILV a)

Hnh 1.7: Bo v so lch ngang c hm (a) v c tnh khi ng (b) i vi MF cng sut va v nh ch c cun dy n, lc chm chp gia cc vng dy trong cng mt pha thng km theo chm v, nn bo v chng chm t tc ng (trng hp ny khng cn t bo v so lch ngang). Vi MF cng sut ln, cun dy stator lm bng thanh dn v c qun kp, u ra cc nhnh a ra ngoi nn vic bo v so lch ngang tng i d dng. Ngi ta c th dng s bo v ring hoc chung cho cc pha.

II.1. S bo v ring cho tng pha: (hnh 1.7, 1.8)Trong ch lm vic bnh thng hoc ngn mch ngoi, sc in ng trong cc nhnh cun dy stator bng nhau nn I1T = I2T. Khi : IH = I1T + I2T = 2.I1T (1-24) ISL =ILV=I1T - I2T = IKCB (1-25) IH > ILV nn bo v khng tc ng 87G 87G 87G Khi xy ra chm chp gia cc vng dy ca hai nhnh khc nhau cng mt pha, gi thit ch my pht cha mang ti, ta c: I1T = -I2T Hnh 1.8: S bo v so lch ngang theo m s IH = I1T - I2T = IKCB ILV = I1T + I2T = 2.I1T (1-26) ILV > IH nn rle tc ng ct my ct u cc my pht.

II.2. S bo v chung cho cc pha: (hnh 1.9)Trong s BI c t gia hai im ni trung tnh ca 2 nhm nhnh ca cun dy stator, th cp ca BI ni qua b lc sng hi bc ba L3f dng gim dng khng cn bng i vo rle.

20

A

B

C

+ RI Lf3

T 1 2 + RT

Bo tn hiu

a)

Rth Ct 1MC

O1 O2

C -

Hnh 1.9: S bo v so lch ngang cho cc pha MF, s tnh ton (a) v theo m s (b)

b)

BI

87

CN: cu ni, bnh thng CN v tr 1 v bo v tc ng khng thi gian. Khi my pht chm t 1 im mch kch t (khng nguy him), CN c chuyn sang v tr 2 lc bo v s tc ng c thi gian trnh tc ng nhm khi chm t thong qua im th 2 mch kch t.

II.2.1. Nguyn l hot ng:Bo v hot ng trn nguyn l so snh th V1 v V2 ca trung im O1 v O2 gia 2 nhnh song song ca cun dy. * ch bnh thng hoc ngn mch ngoi: U12 = V1 - V2 0 (1-27) nn khng c dng qua BI do bo v khng tc ng (cu ni v tr 1). * Khi xy ra chm chp 1 im mch kch t, my pht vn c duy tr vn hnh nhng phi chuyn cu ni sang v tr 2 trnh trng hp bo v tc ng nhm khi ngn mch thong qua im th 2 mch kch t. * Khi s c (chm chp gia cc vng dy): U12 = V1 - V2 0 (1-28) nn c dng qua BI bo v tc ng ct my ct.

II.2.2. Dng khi ng ca rle:Dng in khi ng ca bo v c xc nh theo cng thc: IKB Kat.IKCBtt (1-29) Thc t vic xc nh dng khng cn bng tnh ton IKCBtt tng i kh, nn thng xc nh theo cng thc kinh nghim: IKB = (0,05 0,1).ImF (1-30) IKR =

I KB nI

(1-31)

t c th chn c loi rle cn thit.

II.2.3. Thi gian tc ng ca bo v:Bnh thng bo v tc ng khng thi gian (cu ni CN v tr 1). Khi chm t im th nht mch kch t th cu ni CN c chuyn sang v tr 2. Thi gian tc ng ca rle RT c xc nh nh sau:

21

tRT = tBV 2 im kt + t (1-32) Trong : - tBV 2 im kt: thi gian tc ng ca bo v chng chm t im th hai mch kch t. - t: bc chn thi gian, thng ly t = 0,5 sec. Nhn xt: - Bo v so lch ngang cng c th lm vic khi ngn mch nhiu pha trong cun dy stator. Tuy nhin n khng th thay th hon ton cho BVSLD c v khi ngn mch trn u cc my pht bo v so lch ngang khng lm vic. - Bo v tc ng khi chm t im th hai mch kch t (nu bo v chng chm t im th hai mch kch t khng tc ng) do s khng i xng ca t trng lm cho V1 V2.

III. Bo v chng chm t trong cun dy stator (50/51n)Mng in p my pht thng lm vic vi trung tnh cch in vi t hoc ni t qua cun dp h quang nn dng chm t khng ln lm. Tuy vy, s c mt im cun dy stator chm li t li thng xy ra, dn n t chy cch in cun dy v lan rng ra cc cun dy bn cnh gy ngn mch nhiu pha.V vy, cn phi t bo v chng chm t mt im cun dy stator. Dng in ti ch chm t khi trung im ca cun dy my pht khng ni t l:

I (1) = Trong : -

.U p

2 2 rq + X C0

(1-33)

: s phn trm cun dy tnh t trung im n v tr chm t ( 1). Up: in p pha ca my pht. rq: in tr qu ti ch s c. X C0 : dung khng 3 pha ng tr ca tt c cc phn t trong mng in p

my pht. X C0 =

Nu b qua in tr qu ti ch s c (rq = 0), dng chm t bng: (1-34) Khi chm t xy ra ti u cc my pht ( = 1) dng chm t t tr s ln nht: (1-35) Nu dng chm t ln cn phi t cun dp h quang (CDHQ), theo quy nh ca mt s nc, CDHQ cn phi t khi:

1 3. j..C0

I (1) = 3...C0.Up

I (1) max = 3..C0.Up

I (1)max 30 A i vi mng c U = 6 kV I (1)max 20 A i vi mng c U = 10 kV

I (1)max 15 A i vi mng c U = (15 20) kV I (1)max 10 A i vi mng c U = 35 kV (1) Kinh nghim cho thy rng dng in chm t I 5A c kh nng duy tr tia la

in ti ch chm t lm hng cun dy v li thp ti ch s c, v vy bo v cn phi tc ng ct my pht. Phn ln s c cun dy stator l chm t mt pha v cc cun dy cch in nm trong cc rnh li thp. gii hn dng chm t trung tnh my pht thng ni t qua mt tng tr. Cc phng php ni t trung tnh c trnh by trong hnh 1.10. Nu tng tr trung tnh ln dng chm t c th gii hn nh hn dng in nh mc my pht. Khng c cng thc tng qut no cho gi tr ti u ca tng tr gii hn dng. Nu tng tr trung tnh qu cao, dng chm t b lm cho rle khng tc ng. Ngoi ra in tr qu ln s xut hin hin tng cng hng qu gia cc cun dy vi t v ng dy kt ni. trnh hin tng ny khi tnh chn in tr trung tnh cc i22

1 da vo dung dn gia 3 cun dy stator my pht, thng yu cu: R () 3C (1-36) vi C l in dung ca mi cun dy stator my pht. Nu in tr trung tnh thp, dng in chm t s cao v s gy nguy him cho my pht. Khi in tr trung tnh gim nhy ca rle chng chm t gim do in th th t khng nh. Rle chng chm t s cm nhn in th ging trn in tr ni t do vy gi tr in th ny phi ln m bo nhy ca rle. Hnh 1.10 gii thiu mt s phng n p dng ni t trung tnh my pht. Phng n a: Trung tnh ni t qua in tr cao Rt (hnh1.10a) gii hn dng chm t nh hn 25A. Mt phng n khc cng ni t qua in tr thp cho php dng chm t c th t n 1500A. Phng n b: Trung tnh ni t qua in khng c khng tr b (hnh 1.10b), vi phng n ny cho php dng chm t ln hn khi dng phng n a, gi tr dng chm t khong (25100)% dng ngn mch 3 pha. Phng n c: Trung tnh ni t qua my bin p BA hnh 1.10c, in p ca cun s MBA bng in p my pht, in p ca cun th MBA khong 120V hay 240V. - i vi s c thanh gp cp in p my pht khi I > 5 (A) cn phi ct my pht. - i vi s ni b MFMBA thng I < 5 (A) ch cn t R K BA Rt bo v n gin hn bo tn hiu chm t stator m khng cn ct a) b) c) my pht.Hnh 1.10: Cc phng n ni t trung tnh MF

III.1. i vi s thanh gp in p my pht:S hnh 1.11 c dng bo v cun dy stator my pht khi xy ra chm t. Bo v lm vic theo dng th t khng qua bin dng th t khng 7BI0 c kch t ph t ngun xoay chiu ly t 2BU.1MC + 3RI + 4RI5RG

Bo tn hiu + RTh + Ct 1MC6RT

FCO 2BU

T bo v chng nm ngoi

7BI0

MF

Hnh 1.11: S bo v chng chm t 1 im cun stator MF

23

- 3RI: rle chng chm t 2 pha ti hai im khi dng bo v so lch dc t 2 pha (s sao khuyt). - 4RI: rle chng chm t 1 pha cun dy stator. - 5RG: kho bo v khi ngn mch ngoi. - 6RT: to thi gian lm vic cn thit bo v khng tc ng i vi nhng gi tr qu ca dng in dung i qua my pht khi chm t 1 pha trong mng in p my pht. - Rth: rle bo tn hiu.

III.1.1. Nguyn l hot ng:Tnh trng lm vic bnh thng, dng in qua rle 3RI, 4RI: . . . 1 . 1 . I R = (I A + I B + I C ) = I KCBtt (1-37) nI nI Dng in khng cn bng do cc pha pha s cp ca 7BI0 t khng i xng vi cun th cp v do thnh phn kch t ph gy nn. Dng in khi ng ca rle cn phi chn ln hn dng in khng cn bng trong tnh trng bnh thng ny: IKR >IKCBtt Khi xy ra chm t 1 pha trong vng bo v: Dng qua ch chm t bng: ID = (3...C0HT + 3...C0F).UpF (1-38) Trong : - : phn s vng dy b chc thng k t im trung tnh cun dy stator. - C0F, C0HT: in dung pha i vi t ca my pht v h thng. - UpF: in p pha ca my pht. Dng in vo rle bng: (1-39) I D = 3...C0HT .U pF bo v c th tc ng c cn thc hin iu kin: IKB I I KCBtt (1-40) D n gin, ta gi thit dng chm t i qua bo v v dng khng cn bng tnh ton ngc pha nhau. Khi s vng chm b, dng in chm t I nh v bo v c th c vng cht D gn trung tnh my pht. Khi chm t mt pha ngoi vng bo v, dng in i qua bo v: (1-41) I D = 3...C0F .U pF bo v khng tc ng trong trng hp ny, dng khi ng ca bo v phi c chn: (1-42) I KB > I Dq + I KCBtt y chng ta chn iu kin nng n nht l khi dng in chm t qua bo v v dng khng cn bng c chiu trng nhau, ng thi phi chn gi tr ca dng in chm t bng gi tr qu ln nht v chm t thng l khng n nh. Khi xy ra chm t 2 pha ti hai im, trong c mt im nm trong vng bo v. Bo v s tc ng ct my pht nh rle 3RI. Trong trng hp ny rle 4RI cng khi ng nhng tn hiu t 4RI b tr do 6RT.

III.1.2. Tnh chn Rle:* Dng khi ng ca rle 3RI: Vic xc nh dng khng cn bng i qua bo v khi ngn mch ngoi vng bo v rt phc tp v th ngi ta thng chnh nh vi mt d tr kh ln, theo kinh nghim vn hnh thng chn: IKB3RI = (100 200) (A) (pha s cp) (1-43) * Dng khi ng ca rle 4RI: Dng khi ng ca 4RI c chn theo 2 iu kin: Bo v khng c tc ng khi ngn mch ngoi vng bo v, khi :

I KB 4RI =

K at (3C0k qU pF + I KCBtt max ) (A) (pha s cp) K tv

(1-44)

24

Theo gi tr dng in s cp b nht tng ng vi dng in khi ng cc tiu ca 4RI (gi tr ny ph thuc vo cu to v nhy ca rle 4RI). i vi cc rle thng gp gi tr ny khong: IKB4RI = (2 3) (A) (pha s cp) (1-45) T hai iu kin trn chng ta s chn c dng in ln hn lm dng in tnh ton. * Thi gian lm vic ca rle 6RT: loi tr nh hng ca nhng gi tr qu ca dng in dung khi chm t mt pha trong mng in p my pht, ngi ta thng chn: t6RT = (1 2) sec (1-46)

III.2. i vi s ni b MF-MBA:Vi s ni b, khi xy ra chm t mt im cun dy stator dng chm t b v vy bo v ch cn bo tn hiu, y ch cn dng s bo v n gin, lm vic theo in p th t khng nh hnh 1.12. Gi tr khi ng ca RU (UKRU) thng chn theo hai iu kin sau: + iu kin1: UKRU > UKCBmax iu kin2: UKRU chn theo iu + kin n nh nhit ca rle v thng ly RT bng 15V. RU MBA Thng chn theo iu kin 2 l tho iu kin 1. Rle thi gian dng to thi gian tr trnh trng hp bo v tc ng nhm FCO do qu s c bn ngoi. V tRT = tmax (BV ca phn t k cn) + t. (1-47)BU

III.3. Mt s s khc:MF ni vi thanh gp in p thng c cng sut b v s bo v thng da trn nguyn l lm vic theo bin hoc hng dng in chm t.

MF

Hnh 1.12: S bo v chm t mt im cun stator b MF-MBA

III.3.1. Phng php bin : I(1)F C0F I(1)H51N BA Rt 59 BU 50N Rt

I(1)

C0H

R

50N

a)

b)

c)

Hnh 1.13: Chm t trong cun dy stator MF

Hnh 1.14: Bo v chm t dy qun stator

Phng php bin thng c s dng khi thnh phn dng in chm t t pha in dung h thng I(1)H ln hn nhiu so vi thnh phn chm t t pha in dung my pht I(1)F ngha l: I(1)H >> I(1)F vi IF = 3.j..C.U

25

V dng chm t I(1) (hnh 1.13) ph thuc vo v tr ca im chm t, nn nu xy ra chm t gn trung tnh ( 0) bo v s khng nhy, v vy phng php ny ch bo v c khong 70% cun dy stator my pht k t u cc my pht. Ngoi s nu phn III.1, sau y chng ta s xt thm mt s s bo v theo phng php bin khc sau: Trung tnh my pht ni t qua in tr cao R: (hnh 1.14a) My bin dng t dy ni trung tnh MF qua in tr ni t R, cun th cp ni vo rle dng ct nhanh (c m s 50N). Tr s dng in t ca rle ly bng 10% gi tr dng in chm t cc i cp in p my pht. y l tr s t nh nht c tnh n an ton khi thnh phn dng in th t khng t h thng cao p truyn qua in dung cun dy MBA ti my pht. nng cao hiu qu ca bo v ngi ta c th t thm bo v dng cc i (51N) c c tnh thi gian ph thuc c tr s dng in t khong 5% gi tr dng chm t cc i Imax cp in p my pht. My pht ni t trung tnh qua MBA: (hnh 1.14b) MBA ni t t trung tnh my pht in, va c chc nng nh mt khng in ni t ca my pht va cung cp ngun cho bo v. Cun th cp ca MBA c ni vi rle qu in p (59) song song vi ti tr Rt nhm n nh s lm vic cho MBA v to gi tr in p t ln rle qu in p. Tr s in p t khong (5,4 20) V. S ch c th bo v c khong 90% cun stator tnh t u cc my pht. Ngi ta cng c th s dng phng n hnh 1.14c bo v chng chm t cun stator my pht. Cun th cp ca MBA c mc thm ti tr Rt, in tr ny lm tng thnh phn tc dng chm t ln khong 10A v trn mch th cp ny t bin dng ni vo rle dng cc i (50N). Gi tr t ca rle ny khong 5% gi tr dng in chm t cc i cp in p my pht. Dng in th cp ca BI chn 1A cn dng in pha s cp ca BI chn bng hoc nh hn dng in i qua cun s cp ca MBA ni t. S s dng in p sng hi bc 3: (hnh 1.15)N MF F

2BU0 1RU R Z1 a Z2 Lf3 1BU0 a) b) N F 2RU b N UN UN

50%

UF UF F 100% UF UF

c)

N

F

N

50% 50%

F 100% F 100%

d)

N

F

N UN UN

Hnh 1.15: S bo v chm t 100% cun stator theo in p hi bc 3 (a); th vct trong ch vn hnh bnh thng (b); khi chm t trung tnh (c) v khi chm t u cc im my pht26

Cc s bo v m t trn khng bo v c hon ton cun stator my pht khi xy ra chm t mt pha. Vi cc my pht cng sut ln hin i, yu cu phi bo v 100% cun dy stator khi xy ra s c trn, ngha l bo v phi tc ng khi xy ra chm t mt pha bt k v tr no cun dy stator my pht. Mt trong nhng phng php la chn y l s dng in p sng hi bc ba. Do tnh phi tuyn ca mch t my pht nn in p cun dy stator lun cha thnh phn sng hi bc ba, gi tr ca thnh phn in p ny ph thuc vo tr s in khng ca thit b ni vi trung tnh my pht, in dung vi t ca cun stator, in dung ni t ca cc dy dn, thanh dn mch my pht v in dung cun dy MBA ni vi my pht in. Trong iu kin vn hnh bnh thng, nu o in p sng hi bc ba vi t cc im khc nhau trn cun dy stator ta c phn b in p nh trn hnh 1.15b. y k hiu UN, UF l in p hi bc ba khi my pht khng ti v UN, UF khi my pht y ti. Khi xy chm t u cc hoc trung tnh my pht, in p sng hi u cc khng chm t tng ln gn gp hai ln so vi ch tng ng trc khi chm t (hnh 1.15c,d). Nguyn l lm vic ca s bo v l so snh tr s in p hi bc ba trung tnh my pht v tr s in p hi bc ba ly cun tam gic h ca 2BU. Rle le in p 2RU ni qua b lc tn s hi bc ba Lf3 v s tc ng khi c chm t trong cun dy stator. Nh phn tch phn trc, rle in p 1RU ch bo v c khong 90% cun stator tnh t u cc my pht, y rle 2RU cng bo v c khong (70 80) % cun stator tnh t im trung tnh. Nh vy s phi hp lm vic gia 1RU v 2RU c th bo v c ton b cun stator my pht khi xy ra chm t mt pha. Cc tng tr Z1, Z2 c chn sao cho ch lm vic bnh thng in p t ln 2RU bng khng, khi xy ra chm t cun stator in p t ln rle s ln hn nhiu so vi in p t ca 2RU.

III.3.2. Phng php hng dng in chm t: (hnh1.16)Phng php hng dng in chm t c th m rng vng bo v chng chm t khong 90% cun dy k t u cc my pht.

3U0 IU

K Vng tc ng

R1 BTH1 Ilv

C1 C2 CL1

R2

K

I

-I(1)

L Vng hm

tRI IU IH Ilv

3I0 = I

(1)

D

BTH2 a)

CL2

b)

HNH 1.16 : bo v c hng chng chm t cun dy stator thanh gp in p mf

27

h :

Rle so snh tng quan gia dng in lm vic ILV v dng in hm IH theo quan I = IH - ILV (1-48)

Trong :

IH = IU + I1 (1-49a) ILV = IU - I1 (1-49b) &1 ly t b lc dng th t khng. Vi IU l dng in ly t ngun in p U0; I D T th vct hnh 1.16b ta c th thy rng, iu kin lm vic ca bo v c xc nh theo du ca I, bo v s tc ng ct MC khi I > 0, ngha l IH >ILV iu ny c tho mn nu chm t xy ra trong vng bo v. ng K-L trn th vct hnh 1.16b l ranh gii gia min tc ng v min hm ca bo v. Nu chuyn mch kho K (hnh 1.16a) u vo in p U0 qua in tr R1 thay cho t in C1 th s c th s dng bo v cho cc my pht c trung tnh ni t qua in tr ln. Khi y thnh phn tc dng ca dng in tc dng s c so snh vi thnh phn phn khng ca dng in khi trung im cun dy my pht khng ni t. Nu thnh phn tc dng v thnh phn phn khng ca dng in chm t gn bng nhau, ngi ta s dng s c tn gi l s 450 khi y kho K s chuyn sang mch R2, C2 vi thng s c la chn thch hp. Mt phng n khc thc hin bo v chng chm t cun dy stator my pht c trung tnh khng ni t hoc ni t qua in tr ln lm vic trc tip vi thanh gp in p my pht trnh by trn hnh 1.17. Trong phng n ny ngi ta s dng thit b to thm ti th t khng. Ti ny c a vo lm vic khi pht hin c chm t v lm tng thnh phn tc dng ca dng in s c ln khong 10A, to iu kin thun li cho vic xc nh hng dng in. Thit b to thm ti bao gm BI0N u vo trung tnh ca my pht, ti R ca BI ny c ng m bng tip im ca rle in p RU0. Khi c chm t, in p U0 xut hin, RU0 ng tc thi tip im ca mnh v duy tr mt khong thi gian t2 cho s lm vic chc chn. T s bin i ca BIG trong mch thit b to thm ti c chn sao cho thnh phn tc dng ca dng in a vo b so snh pha xc nh ng hng s c. Hnh 1.17b,c trnh by s nguyn l v th vct xc nh hng s c khi chm t xy ra bn trong (hnh 1.17b) v bn ngoi (hnh 1.17c) cun dy stator my pht. Khi chm t ngoi vng bo v, dng in tng I a vo b so snh pha: I = IA - I(1)D (1-50) Trong : - IA dng in c to nn bi thit b to thm ti. - I(1)D dng in chm t chy qua bo v. Trong trng hp ny gc pha gia in p th t khng U0 v dng in tng I vt qua tr s gc lm vic gii hn nn s khng c tn hiu ct . Khi chm t trong cun dy stator MF ta c: I = IA + I(1)D v gc pha gia in p th t khng U0 v dng in tng I nm trong min tc ng ca bo v. Rle tc ng ct vi thi gian t1.

28

MC

.

U0. (1) I

RU0 a

t2 t1

ng RU0 Ct RU0 Ct MC BI0N

BI0 IA R BIG U0 I(1)

I(1)

BI0.

I

Thit b bo v MF RU0 BI0N R

I = IA+ I(1)

IABIG Thit b to thm ti

c)

Min hm

Min tc ng I

a)

BI0

I(1)I

U0Min hm

IA

BI0N

R

I = IA - I(1)

BIG

Min tc ng

c) Hinh 1.17 : S bo v chng chm t cun dy stator MF c thit b to thm ti (a) th vct khi c chm t ngoi (b) v trong (c) vng bo v. S hnh 1.17c th bo v c 90% cun dy. Khi chm t trong vng 10% cn li (gn trung im) bo v khng nhy. Tuy nhin, do in p phn ny ca cun dy khng ln (khng vt qu 10% Up) nn xc xut xy ra hng hc v in (chng hn do cch in b nh thng) rt thp nn cc my pht cng sut b ngi ta thng khng i hi bo v ton b cun dy. i vi cc MF ni b vi MBA, thng thng cun dy MBA pha my pht u tam gic nn chm t pha co p dng th t khng khng nh hng n MF. Vi cc im chm t xy ra trong mng cp in p my pht c th pht hin bng s xut hin U0 u cc tam gic h ca BU t u cc MF, hoc u ra ca MBA u vi trung im ca MF. Vi cc MF cng sut ln, ngi ta yu cu phi bo v 100% cun dy stator chng chm t ngn nga kh nng chm t vng gn trung im ca cun dy do cc nguyn nhn c hc . Ngy nay bo v 100% cun dy stator chng chm t, ngi ta thng dng hai phng php sau y: - Theo di s bin thin ca hi bc ba ca sng in p trung im v u cc MF. - a thm mt in p hm tn s thp vo trung im ca cun dy MF. * Phng theo di s bin thin ca sng hi bc ba (xem mc III.3.1) c mt s nhc im: - Khi chm t vng gn gia cun dy, bo v c th khng lm vic v thnh phn sng hi bc ba trong in p qu b. - in p Uab t vo rle s suy gim khi in tr ch s c ln. - S khng pht hin c chm t khi MF khng lm vic.Trong mt s MF, thnh hi bc ba khng ln bo v c th pht hin c.29

khc phc nhng nhc im ny ngi ta dng phng php a thm mt in p hm tn s thp vo mch trung tnh ca MF. * Phng php a thm mt in p hm tn s thp vo trung im ca cun dy MF (hnh 1.18): MF MBA - Dng in I t ngun 20Hz sau khi qua b lc 1LF c phn thnh hai thnh phn I chy qua BU0 ni vi trung tnh MF v IB chy qua in tr t RB. Thnh phn I thng qua bin dng trung R C gian BIG v b lc tn s 2LF c nn thnh dng in lm vic. 20Hz 1LF I I - ILV a vo rle so snh vi dng in hm IH cng do ngun 20Hz to IB RB nn thng qua in tr t Rc , dng in BIG hm c tr s khng i. ch lm RC vic bnh thng (R = ) dng in I c xc nh theo in dung ca cun 2LF dy i vi t C nn c tr s b do ILV < IH v rle s khng tc ng. Hm RL Lm vic ILVCt

IC=IH

Hnh 1.18: S bo v 100% cun dy stator chng chm t c a thm in p hm 20Hz vo trung im MF - Khi c chm t, dng I c xc nh ch yu theo in tr chm t R , ILV>IH rle s tc ng ct my pht. - Cc b lc tn s 1LF, 2LF m bo cho s ch lm vic vi thnh phn 20Hz, ngoi ra b lc 1LF bo v cho my pht 20Hz khi b qu ti bi dng in cng nghip khi c chm t xy ra u cc MF. Mt phng n khc thc hin bo v 100% cun dy stator chng chm t l dng ngun ph 12,5Hz (vi tn s cng nghip l 60Hz ngi ta dng 15Hz) c tn hiu c m ha a vo mch s cp thng qua BU0 u vo mch trung tnh ca MF (hnh 1.19a). Trong ch lm vic bnh thng, dng in I chy qua im trung tnh MF c xc nh theo tr s in dung ng tr ca MF l C (hnh 1.19b). Khi xy ra chm t, in tr chm t R c ghp song song vi C lm tng dng in n tr s I > I (hnh 1.19c). Rle u ra s phn ng theo s tng dng in v theo tn hiu phn hi c m ha. Trn hnh 1.20 trnh by vic m ha tn hiu bng cch thay i thi gian pht tn hiu v thi gian dng .Trong cc khong thi gian ny nhiu php o c tin hnh: M1, M2 v M3 cho khong thi gian truyn tn hiu v P1, P2..P6 cho khong thi gian dng. Phng php ny cho php loi tr c nh hng ca nhiu do dng in pha s cp v php o c tin hnh ring cho tng na chu k dng v m s trnh c nh hng ca nhiu c tn s bi ca 12,5Hz.

30

R C b) I RU0 R S 900 a) R BUGRU0

C 12,5Hz

R C R0 LF

IBU0 12,5Hz BIG

R

IM c)

I

12,5Hz

Hnh 1.19 : S nguyn l (a) ca bo v 100% cun dy stato MF chng chm t dng bin php bm tn hiu 12,5Hz c m ho v s xc nh dng in chm t I khi lm vic bnh thng (b) v khi chm t (c). Cc s bo v 100% cun dy stator chng chm t thng c s dng kt hp vi s bo v 90% tng tin cy cho h thng chm t.A Tn hiu m ho IM 1/ 2 chu k (+) DM1 M3

B

C

t

EP1 P3 P5

1/ 2 chu P2 P4 P6 M2 k(-) S chu k 4 3 1 2 3 6 7 1 2 5 12,5Hz ms 0 80 160 240 320 400 480 560 80 160

t

Hnh 1.20: Biu bm tn hiu 12,5Hz c m ho thc hin bo v 100% cun dy stator chng chm t. A- chu k hoy ng; B- thi gian pht tn hiu; C- thi gian dng; Thi gian o; E- thi gian kim tra tn hiu phn hi

31

IV. Bo v chng chm t mch kch t ca MF (64)i vi MF, do ngun kch t l ngun mt chiu nn khi chm t mt im mch kch t cc thng s lm vic ca my pht hu nh thay i khng ng k. Khi chm t im th hai mch kch t, mt phn cun dy kch t s b ni tt, dng in qua ch cch in b nh thng c th rt ln s lm hng cun dy v phn thn rotor. Ngoi ra dng in trong cun rotor tng cao c th lm mch t b bo ho, t trng trong my pht b mo lm cho my pht b rung, ...gy h hng nghim trng my pht. i vi MF cng sut b v trung bnh (my pht nhit in), thng ngi ta t bo v bo tn hiu khi c mt im chm t trong mch kch t v tc ng ct my pht khi xy ra chm t im th hai. i vi MF cng sut ln (my pht thu in), hu qu ca vic chm t im th hai trong mch kch t c th rt nghim trng, v vy khi chm t mt im trong cun dy rotor bo v phi tc ng ct my pht ra khi h thng.

IV.1 Bo v chng chm t mt im mch kch t:t : C ba phng php c s dng pht hin chng chm t mt im mch kch * Phng php phn th. * Phng php dng ngun ph AC. * Phng php dng ngun ph DC.

IV.1.1 Phng php phn th: (hnh1.21)

Cun kch t

MFkt R 64

Trong s bo v chng chm t cun dy rotor, ngi ta dng in tr mc song song vi cun dy kch t, im gia ca in tr ni HNH 1.21 : Bo v chm t rotor qua rle in p, khi c mt im chm t s bng phng php phn th xut hin mt in th rle in p, in th ny ln nht khi im chm t u cun dy. trnh vng cht khi im chm t gn trung tnh cun dy kch t, ngi ta chuyn nc thay i in u vo rle tc ng.

IV.1.2. Phng php dng ngun in p ph AC:Bo tn hiu Bo tn hiu

+ +35RI

+ + +35RI36RT

+36RT 37RG

52N

37RG

52N

34BG

47C 48CC 34BG

47CTi trc MF

Ti trc MF Ti mch kch t

U~

48CC

CL

2R

Ti mch kch t

HNH 1.22: S bo v chng chm t 1 im cun rotor dng ngun in ph AC32

HNH 1.23: S bo v chng chm t 1 im cun rotor dng ngun in ph DC

S bo v c trnh by hnh 1.22. in p ngun ph xoay chiu thng bng in p cun kch t. - 34BG: bin p trung gian, ly in t thanh gp t dng. - 35 RI: rle dng in, pht hin s c. - 36RT: rle thi gian, to thi gian tr trnh trng hp bo v tc ng nhm khi ngn mch thong qua. - 37RG: rle trung gian. - 52N: nt n gii tr t gi. - 47CC: cu ch bo v. - 48C: t in dng cch ly mch kch t mt chiu vi mch xoay chiu. Nguyn l lm vic ca s nh sau: - Bnh thng, pha th cp ca bin p trung gian 34RG h mch do khng c dng qua rle 35RI, bo v khng tc ng. - Khi xy ra chm t mt im mch kch t, th cp ca bin p trung gian khp mch, c dng chy qua rle 35RI lm cho bo v tc ng i bo tn hiu. S c u im l khng c vng cht ngha l chm t bt k im no trong mch kch t bo v u c th tc ng. Tuy nhin do dng ngun xoay chiu nn phi chng s xm nhp in p xoay chiu vo ngun kch t mt chiu.

IV.1.3 . Phng php dng ngun in p ph DC:Phng php ny khc phc c nhc im ca phng php trn bng s hnh 1.23, nh b chnh lu it m ta c th cch li ngun mt chiu v ngun xoay chiu. Ngun in ph mt chiu cho php loi tr vng cht v thc hin bo v 100% cun dy rotor chng chm t. S c nhc im l s lin h trc tip v in gia thit b bo v v in p kch t UKT c tr s kh ln i vi cc MF c cng sut ln.

IV.2. Mt s s bo v chng chm t mt im trong cc MF hin i:i vi cc MF c h thng kch t khng chi than vi cc it chnh lu lp trc tip trn thn rotor ca my pht, in dung ca h thng kch t i vi t s tng ln ng k v h thng bo v chng chm t ca cun dy rotor cng tr nn phc tp . Cc s bo v chng chm t mt im trong cun dy rotor ca cc MF hin i thng tc ng ct my pht ( loi tr xy ra chm t im th hai) v da trn mt trong nhng nguyn l sau: - o in dn trong mch kch t (i vi t) bng tn hiu in p xoay chiu tn s 50Hz. - o in tr ca mch kch t (i vi t) bng tn hiu in p mt chiu hoc tn hiu sng ch nht tn s thp. Nguyn l o in dn ca mch kch t i vi t ca MF c h thng kch t khng chi than trnh by trn hnh 1.24.My kch t Rotor ca my kch t R C S1 LF Ct MC

Cun dy rotor ca my pht in U(50Hz) BUG

S2 R RY

HNH 1.24: bo v chng chm t cun rotor MF c h thng kch t khng chi than vi it chnh lu lp trc tip trn thn rotor theo nguyn l o in dn.

33

Ngun in p ph xoay chiu tn s 50Hz c t vo mch trung tnh ca cun dy my kch thch xoay chiu ba pha v thn rotor ca MF thng qua cc vnh gp v chi than S1, S2. B lc tn s LF ch cho tn s cng nghip chy qua rle o in dn RY loi tr nh hng ca hi bc cao trong php o. in dn m rle RY o c ch yu xc nh theo in tr R v in dung C i vi t ca mch kch t. Trn hnh 1.25 trnh by qu o ca nt vct tng tr Z m rle o c cho hai trng hp: Khi R = const, C = var v khi C = const, R = var. Rle RY c chnh nh vi hai mc tc ng: mc cnh bo vi c tnh khi ng 2 v mc tc ng ct my pht vi c tnh khi ng 1. c tnh 1 bc ly mt phn ca gc phn t th hai v th ba trn mt phng ta m bo cho bo v tc ng mt cch chc chn khi c chm t trc tip (R 0). S bo v hnh 1.24 c mt s nhc im l: s c mt 2 ca chi than S1, S2 lm cho tin cy ca s khng cao v tr 1 s ca in tr tip xc c th R=0 CR= 0 R nh hng n tr s o ca rle. Ngoi ra bn thn h thng kch C= thch mt chiu cng c th nh hng n s lm vic ca bo C= const v khi in dung ca mch kch R=var R / 2 thch i vi t C ln, in tr r R ln nht c th o c 10 k. khc phc nhc im X/ 2 R= const ny ngi ta dng s vi R= X C= Var ngun in ph mt chiu hoc xoay chiu vi tn s thp c dng sng hnh ch nht. jX Hnh 1.25: c tnh bin thin ca tng tr i vi t ca mch kch t v c tnh tc ng ca Rle o in dn chng chm t mch roto MF ng b. 1- c tnh ct; 2- c tnh cnh bo. Trn hnh 1.26 trnh by nguyn l pht hin chm t trong cun dy rotor ca MF c kch thch t ngun in t dng qua b chnh lu Thyristor dng ngun tn hiu sng ch nht c tn s 1Hz. Cc in tr ph R1, R2 c chn c ch s kh ln so vi in tr RM to in p UM t vo b phn o lng M. Dng in do ngun in ph U to ra bng: U I= (1 -51) R + RM + R Trong :

R1.R 2 R1 + R 2 Lu rng RM 0). Khi my pht lm vic ch thiu kch thch hoc mt kch thch, sc in ng E thp hn in p UF, my pht nhn cng sut phn khng t h thng (Q < 0) (hnh 1.33a,c). Nh vy khi mt kch t, tng tr o c u cc my pht s thay i t Zpt (tng tr ph ti nhn t pha my pht) nm gc phn t th nht trn mt phng tng tr phc sang ZF (tng tr ca my pht nhn t u cc ca n trong ch Q < 0) nm gc phn t th t trn mt phng tng tr phc (hnh 1.33b).(II) + jX 0 (IV) (I) Zpt 0,5Xd B Xd (III) - jX A T Min qu kch thch (E > 0, Q > 0) R Min thiu kch thch (E < 0, Q < 0) +Q E I, Q U

H thng

a)

ZFGii hn pht nng cun rotor Gii hn pht nng cun stator P (MW)

b)

c) 0Min lm vic bnh thng

Hnh 1.33: Mt kch t MF a) thay i hng cng sut Q. b) thay i tng tr o c cc my pht. c) gii hn thay i ca cng sut my pht.-Q

Gii hn pht nng mp li thp stator

Khi xy ra mt kch t, in khng ca my pht s thay i t tr s Xd (in khng ng b) n tr s Xd (in khng qu ) v c tnh cht dung khng. V vy pht hin mt kch t my pht in, chng ta c th s dng mt rle in khng cc tiu c Xd < Xk < Xd vi c tnh vng trn c tm nm trn trc -jX ca mt phng tng tr phc. c tnh khi ng ca rle in khng cc tiu hnh 1.33b c th nhn c t s nguyn l hnh 1.34a. Tn hiu u vo ca rle l in p dy Ubc ly u cc my pht v dng in pha Ib, Ic ly cc pha tng ng. in p s cp UBC c a qua bin p trung gian BUG sao cho in th cp c th ly ra cc i lng a.UBC v b.UBC (vi b > a) tng ng vi cc im A v B trn c tnh in khng khi ng hnh 1.33b. Khi mt kch t, dng in chy vo my pht mang tnh cht dung v vt trc in p pha tng ng mt gc 900. Hiu dng in cc pha B v C thng qua bin dng cm khng BIG to nn in p pha th cp UD vt trc dng in IBC mt gc 900. Nh vy gc lch pha gia hai vct in p UD v UBC l 1800 (hnh 1.34). in p a vo cc b bin i dng sng (hnh sin sang hnh ch nht) S1 v S2 tng ng bng:

U1 = a. U BC U D U 2 = b. U BC U D. . .

.

.

.

(1-65) (1-66)

41

Gc lch pha gia U 1 v U 2 s c kim tra. ch bnh thng = 00, rle khng lm0 vic. Khi b mt kch t = 1800, rle s tc ng. Gc khi ng c chn khong 90 . Cc h s a, b c chn (bng cch thay i u phn p ca BUG) sao cho cc im A v B trn hnh 1.34b tho mn iu kin:

.

.

b. U BC > U D > a. U BCA B C BU U1 aUBC IB IC BIG.

.

.

.

(1-67)

BUG

U2

~ ~bUBC

S1 & S1 -1 S3

S4

Ct RL MF

UD U1

-U1 t U2

a)UA. .

IB.

S1I BC

IA.

t.

UC

.

U BC.

UB.

S2 t S3 = - S1.S2 t S4 = S3.

IC

I BC

t

.

.

UD

b U BC.

U BCc)

b)

a U BC

k

Tn hiu ct t

HNH 1.34: S bo v chng mt kch t my pht in dng rle in khng cc tiu a) s nguyn l; b) th vct; c) dng sng ca cc i lng Khi mt kch thch, gc pha dng in thay i, gc lch pha c kim tra thng qua di ca tn hiu S3 = - S1.S2. Nu > k (hnh 1.34c) bo v s tc ng i ct my pht trong khong thi gian t (1 2) sec.

VIII. BO V CHNG MT NG BBo v chng mt ng b i khi cn c tn gi l bo v chng trt cc t. Khi my pht in ng b b mt kch t, rotor my pht c th b mt ng b vi t trng quay. Vic mt ng b cng c th xy ra khi c dao ng cng sut trng h thng in do s c ko di hoc do ct mt s ng dy trong h thng. Hu qu ca vic mt ng b gy nn s dao ng cng sut trong h thng c th lm mt n nh ko theo s tan r h

42

thng in, ngoi ra n cn to ra cc ng sut c nguy him trn mt s phn t ca my pht. pht hin s c ny c th s dng nguyn l o tng tr u cc my pht. Trn hnh 1.35 trnh by c tnh bin thin ca mt vct tng tr o c trn u cc my pht trong qu trnh s c v xy ra dao ng in trong h thng. ch vn hnh bnh thng, mt vct tng tr nm v tr im A. khi xy ra ngn mch mt vct dch chuyn t A n B, sau khi bo v ct ngn mch vct tng tr nhy t B sang C v nu xy ra dao ng, mt vct chu k u tin s dch chuyn theo qu o 2... Hnh vi ny ca vct tng tr khi c dao ng in c th c pht hin bng mt rle vi c tnh khi ng nh trn hnh 1.36. c tnh khi ng c dng hnh elp hoc thu knh 1 v dng in khng 2 kt hp vi nhau theo nguyn l v. Khi c dao ng nu qu o ca mt vct Z i vo min khi ong im M v ra khi min khi ng im N di c tuyn 2 (hnh 1.37) c ngha l tm dao ng (tm in) nm trong min tng tr ca b MF-MBA, bo v s tc ng ct my pht ngay trong chu k dao ng u tin.Dao ng in 1 2 Z 0 +jX B (ngn mch) C (ct ngn mch) A (bnh thng) R

HNH 1.35: Hnh trnh ca vct tng tr Z khi xy ra s c v dao ng Nu tm dao ng nm pha h thng qu o ca mt vct Z s nm cao hn c tuyn 2, khi y bo v s tc ng ct sau mt s chu k nh trc. Trn hnh 1.37 trnh by s nguyn l ca bo v chng trt cc t, bo v gm b phn o khong cch vi c tuyn thu knh1 kt hp vi b phnhn ch theo in khng 2 gii hn min tc ng t pha h thng, b phn m chu k dao ng 3 ct my pht khi s chu k t tr s t trc. pha cao p ca MBA tng c t thm b phn nh hng cng sut 4 thc hin chc nng ging nh b phn 2 v lm nhim v d phng cho b phn ny. Thay v c tuyn tng tr kt hp 1 v2 trn hnh 1.36 ngi ta c th s dng c tuyn hnh ch nht nh trn hnh 1.38 pht hin dao ng in.2BU +jX X XH 2 N

4 2BI

PCt

BAR 1BU

MC

&

1 -ZF

M

F1BI

I

1

U Z I kcbmax (2-23) ngoi: Theo iu kin phi hp v nhy vi cc bo v ng dy ni vo thanh gp ca trm: I k K at .3I ott (2-24)Trong : K at : h s an ton khi phi hp c th chn Kat = (1,1-1,2). I ott : dng th t khng (TTK) ti ch t bo v, ng vi dng ngn mch no gy ra dng TTK ln nht.

Khi chn I k theo iu kin (2-24) th iu kin (2-23) cng c tho mn, v vy thng ch tnh theo iu kin (2-24).

70

nhy ca bo v: Khi lm bo v chnh: 3I (2-25) K n = 0min 1,5 I K Ly (3I0min ) khi ngn mch trn thanh gp ca trm. Khi lm bo v d tr: Kn 1,2. Lc dng 3I0min l dng b nht khi ngn mch cui vng d tr. in khng TTK ca MBA Vi MBA hai dy qun in khng th t thun (TTT) bng in khng th t nghch (TTN) bng in khng th t khng X1B = X2B = X0B . MBA ba pha ba dy qun ni /Yo/Y loi ny thng c s dng vi cun ni vi my pht in, cun Yo ni vi thanh ci cao p, cun Y l trung p 35KV thng trung tnh khng ni t. Do vy tng tr TTK ca loi ny bng tng tr TTT ca cun Yo. Nu t ni dy /Yo/Yo, vi cun c ti, in khng TTK ca mi cun chnh bng TTT, MBA t ngu in khng TTK ca mi cun chnh bng in khng TTT. Up (n) I1 = (2-26) (x1 + z(n) ) Dng ngn mch (n)n Z

In 0

NM 1 pha(A) 1 NM 2 pha chm t (B,C) 1,1

x2 + x0 x 2 .x 0 x2 + x0

I1 x2 I1 x2 + x 0

Trong : - n: dng ngn mch. x1 x1 - Io: dng in th t khng. x2 x2 x0 - z(n) : Tng tr s c thm vo. - x1: in khng th t thun ti im ngn mch. x0 - x2: in khng th t nghch ti im ngn mch. (1,1) (1) N1 N1 - x0: in khng th t khng ti im ngn mch. V d ta c s thay th tnh ton MBA hai cun dy ca hnh 2.20. Xc nh dng th t khng khi ngn mch mt pha v dng th t khng khi ngn mch hai pha chm t trn thanh gp (im N1 khi bo v lm nhim v bo v chnh). Chn gi tr ln hn lm gi tr tnh ton dng khi ng, gi tr nh hn dng kim tra nhy ca bo v. Khi bo v lm nhim v d tr dng 3I0min ly cui vng bo v (cui ng dy di nht ni n thanh ci MBA t bo v).

III.2. Bo v I0 my bin p c hai pha ni t dng rle qu dng in: MBA c hai dy qun ni t trc tip (hnh 2.21), dng 3I0 i nh hnh v. Trong : IoN2(1-2): dng 3Io do ngun I cung cp khi ngn mch chm t ti N2. IoN1(2-1): dng 3Io do ngun II cung cp khi ngn mch chm t ti N1. IoN1(1-1): dng th t khng tng cung cp n im ngn mch N1. IoN2(2-2): dng th t khng tng cung cp n im ngn mch N2. V th, cn t BVI0 c hng, thng c 2-3 cp tc ng.

71

Cp I: L BVI0 ct nhanh, phi hp vi BVI0 ng dy ni n thanh ci pha t bo v: (2-27) I k I = K atK fm I k Iz max Trong : Kat: h s an ton, Kat = 1,1. I Kfm: h s phn mch I0, K fm = 0 bve. I 0 day I0 bv: dng I0 qua ch t bo v. I0 dy: dng I0 qua ng dy c Ik Iz max. IkIz max: dng chnh nh cp 1 ca BVI0 ng dy c tr s ln nht trong tt c cc ng dy ni n thanh ci MBA c bo v. Thi gian chnh nh: (2-28) tI = tIzmax+t tIzmax: thi gian tc ng ca bo v ng dy c Ik Iz max. Cp II: Chn phi hp vi cp 2 ca BVI0 ng dy, tnh tng t nh cp I trn, thay k hiu I bng k hiu II. nhy cp I v cp II: N1 N2 I 3I 0min II (2-29) KnI = 1,5 I K I I0N1(1-1) I0N1(2-1) 3I 0min (2-30) K n II = 1,5 I K II I0N2(2-2) trong 3I0min ly vi ngn mch ngay trn thanh gp ca trm. Cp III: L bo v qu dng in v III hng, tnh nh BVI0 ca MBA c mt pha ni Hnh 2.21: Dong ngan mch vi at t.MBA co hai day quan noi at

IV. TNH TON CC BO V SO LCHIV.1. Bin dng cho bo v so lch:Nh ni trn vi bo v so lch MBA s u dy BI c chn c th b s lch pha gia dng in cc pha MBA do t u dy MBA gy ra. V d MBA c t u dy /Y-11 th dng th cp lch 300 so vi dng s cp. dng in th cp MBA khng lch pha nhau, ngi ta ni mch th cp ca BI ngc li, ngha l pha ni sao ca MBA ngi ta ni BI theo kiu v ngc li. Mc ch l trnh dng khng cn bng qu ln chy qua bo v so lch trong trng thi lm vic bnh thng cng nh khi ngn mch ngoi c th lm cho bo v tc ng nhm. S u dy BI theo cc cch u cc cun dy MBA khc nhau nh hnh 2.22. * V d cch chn my bin dng: my bin p hai cun dy Sm= 20 MVA, Um =110 Kv/ 6 Kv, t ni dy MBA Y/ -11. + Chn my bin dng cp in p 110 Kv, mch th cp BI ni nn dng in cun dy bng dng in pha. Do vy dng in tnh ton chn BI pha cao p bng: 3 3.S m 3.20.10 I sC = = = 181,8A 3.U m 3.110 Chn loi bin dng 200/5 A.

72

+ Chn bin dng pha h 6 kV. Mch th cp BI ni sao. Dng in tnh ton chn BI pha h p bng: Sm 20.103 I sH = = = 1937A 3.U m 3.6 Chn bin dng loi 2000/5 A. Dng in th cp BI hai pha tng ng bng: I I tC = sC . 3 = 4,55A nI I sH = 4,84A nI chnh lch dng in th cp hai pha bng : I tC I tH 4,55 4,84 .100 = .100 = 6.4% I tC 4,55 I tH =

Gn y, trong rle so lch hin i ngi ta thc hin vic cn bng pha v tr s dng in th cp cc pha ca MBA ngay trong rle so lch./Y-1AB

/Y-11AB

a b c

a b c

C A B C N A

C A B C N 11

B

a b cB

A

a b cbB

a/

cC

a b

b/

aC

c

Y/-11 Y/-1AB

A

a b c n a b

a b cA B C

B

C

C

c

A

B C

A

1

n

a b c

d/

11

A

c/C

c

aB

aC

b c

b

B

73

Y/Y-0AB

Y/Y-6 a b cA C A B C

a b cA

C A B C

a b cB

a b c cB

e/

C

c

a b

f/C

b a

Hnh 2.22: S o noi day mch th cap may bien dong phu hp vi to noi day MBA: A, B, C: vect dong ien s cap pha A, B, C ca MBA. a, b, c: vect dong ien th cap pha a, b, c ca MBA.

IV.2. Bo v so lch dng in c s dng bin dng bo ha trung gian:(Loi PHT)Trnh t tnh ton: 1. Xc nh dng s cp tt c cc pha ca MBA hoc bin p t ngu c bo v. Dng ny c xc nh tng ng cng sut nh mc (cng sut nh mc ca cun dy khe nht) cn i vi MBA t ngu th tng ng vi cng sut truyn qua ca n. Xc nh t s bin dng da vo dng in s cp va tnh trn. Theo cc t s bin i ca t my bin dng tnh cc dng th cp tng ng trong cc nhnh ca bo v : I IT , I IIT , I IIIT i khi ngi ta chn t s bin dng ln hn gi tr tnh ton ca n c th chn s vng dy ca BIG gn vi gi tr tnh ton ca n hn, do lm tng nhy ca bo v. Lp bng gi cc tr tnh ton trn: STT 1 2 3 4 Tn gi cc i lng Dng s cp cc pha ca MBA tng ng vi cng sut nh mc H s bin i ca BI T ni dy ca BI Dng th cp trong cc nhnh ca bo v tng ng vi cng sut nh mc Gi tr bng s cho pha UC UT UH

Chn pha c gi tr dng in s cp ln nht lm pha c bn. 2. Xc nh dng ngn mch s cp cc i chy qua MBA khi ngn mch ngoi trong ch lm vic cc i tt c cc pha ca MBA. 3. Tnh ton dng in khng cn bng s cp cha k n thnh phn I "' kcbtt do chn s vng dy khng chnh xc gy ra.

74

Dng khng cn bng s cp ton phn tnh theo cng thc sau: ' I kcbtt = I 'kcbtt + I 'kcbtt + I "' kcbttVi:

(2-31)

- I 'kcbtt : thnh phn do sai s ca my bin dng gy nn:I 'kcbtt = K kck .K n.fi .I Nng max

(2-32)

Kkck: h s k n thnh phn khng chu k trong qu trnh qu . i vi rle PHT c my bin dng bo ha vi cun dy ngn mch, h s ny ly bng 1. Kn: h s ng nht ca cc my bin dng, i vi bo v MBA thng ly bng 1. fI: sai s cc i cho php ca BI, fI max= 10%. I Nngmax : thnh phn chu k ca dng ngn mch chy qua MBA khi ngn mch ba pha trc tip ngoi vng bo v. I" kcbtt : thnh phn do vic iu chnh in p ca MBA c bo v gy nn.' I 'kcbtt = U I Nng max + U I Nng max

(2-33)

Trong o : U, U: sai so tng oi do viec ieu chnh ien ap cac pha ca MBA c bo ve lay bang na khong ieu chnh cho tng pha tng ng. ong thi khi tnh so vong day ca may bien dong bao hoa trung gian phi lay gia tr trung bnh ca ien ap pha co ieu chnh. I Nng max va INng max: thanh phan chu ky ca dong chy qua pha co ieu chnh ien ap ca MBA khi ngan mch ngoai tnh toan. I "' kcbtt : thanh phan do viec chn so vong day cac pha khong c bn khong phu hp vi gia tr tnh toan ca chung gay nen: WItt WI W WII (2-34) I "' .I INng max + IItt .I IINng max kcbtt = WItt WIItt WItt, WIItt: s vng tnh ton ca cc cun dy my bin dng bo ha trung gian i vi cc pha khng c bn xc nh theo yu cu cn bng sc t ng khi ngn mch ngoi v lm vic bnh thng. I cbT.Wcb = I IT .WItt = I IIT .WIItt (2-35) WI ,WII: cc s vng c chp nhn (s nguyn) ca cun dy my bin dng bo ha trung gian cc pha khng c bn tng ng . Biu thc (2-33) v (2-34) vit cho MBA ba pha v MBA t ngu. i vi MBA hai cun dy cn b bt s hng th hai v phi ca cc biu thc ny. 4. Xc nh s b gi tr dng khi ng ca bo v Ik cha k n thnh "' phn I kcbtt . Theo iu kin chnh nh khi gi tr tnh ton ln nht ca dng khng cn bng tnh ton:Trong :

I k K at .I kcbtt(2-36) Vi Kat: h s an ton k n sai s ca rle v d tr, c th ly bng 1,3. Theo iu kin chnh nh khi gi tr nhy vt ca dng in t ho khi ng MBA khng ti :

75

I k K.I mB (2-37) Trong : ImB l dng in nh mc tng ng vi cng sut nh mc ca MBA (ca cun dy c cng sut ln nht) v vi cng sut mu ca MBA t ngu cha k n h s nhit i ho, ly theo pha c bn. K: l h s chnh nh chn trong khong 1,0 - 1,3 khi tnh ton bo v my bin dng bo ha trung gian. Theo hai iu kin (a) v (b) ta chn gi tr ln nht lm gi tr tnh ton. 5. S b kim tra nhy c th xc nh xem c th dng rle PHT c hay khng hay phi dng rle c c tnh hm loi ZT. s b kim tra nhy cn xc nh dng ngn mch trc tip khi h hng xy ra trn cc cc MBA trong tnh trng tnh ton. Tnh trng tnh ton y cn cp n c ch lm vic ca MBA v c ch lm vic ca h thng. H s nhy ca bo v xc nh theo cng thc: I (2-38) K n = R I kR Trong IR l dng trong cun dy rle. Dng ny ph thuc vo dng ngn mch v s ni dy ca my bin dng. Trn hnh 2.23 v s phn b dng in trong mch bo v so lch ca MBA 3 cun dy i vi mt s trng hp ngn mch khc nhau. n gin, h s nhy c xc nh vi gi thit l ton b dng ngn mch ch chy t mt pha n. IkR: dng khi ng ca rle tng ng vi s vng pha c dng IR chy qua. Nu h s nhy tnh c ln hn 2 th s tip tc tnh ton cho rle PHT theo trnh t tip theo di y cn khng th c th khng cn tnh thnh phn Ikcbtt do iu chnh in p gy nn vi gi thit l khi thay i u phn p ta cng s thay i i lng t ca bo v. Trong nhng trng hp khi khng tnh n thnh phn Ikcbtt m bo v vn khng m bo c nhy cn thit hoc l phi bt buc k n thnh phn khng cn bng th nn dng cc bo v c c tnh hm loi ZT (xem mc 3). i vi nhng trng hp ng th MBA vo mt pha in p no hoc khi MBA ba cun dy (hay t ngu) lm vic trong tnh trng mt my ct pha no ct ra th c th cho php ta h thp yu cu v nhy ca bo v so lch. Trong nhng trng hp ny nu bo v khng nhy th cc bo v khc nh bo v rle hi, hay bo v d tr ca MBA s tc ng ct MBA. 6. Xc nh s vng cun c bn ca bin dng bo ha trung gian, tng ng vi dng khi ng ca bo v (pha c bn l pha c dng in th cp BI ln nht). F (2-39) Wcbtt = kR I kRcbTrong : IkRcb l dng khi ng ca rle tnh qui i v pha c bn. N bng t s gia dng khi ng s cp vi h s bin i ca BI pha c bn c tnh n s ni dy, IkRcb = Ks.(Ik / nI). Gi tr nI theo pha c bn. FkR: Sc t ng (A-Vng) khi ng ca rle

Loi PHT-562 FkR = 60AV Loi PHT-565 FkR = 100AV V dng in pha c bn l ln nht nn s vng cun c bn ca bin dng bo ha trung gian l b nht. Ch thch hnh 2.23: S phn b dng trong MBA ba cun dy v rle so lch khi: a: ngn mch 2 pha pha ngun.76

b: ngn mch 1 pha pha ngun. c: ngn mch 2 pha pha cun sao. d: ngn mch 3 pha cun tam gic. e: ngn mch 2 pha pha cun tam gic.Nguon An BI(2) IN

B(2) IN

C(2) IN

nBI

n BI(2) IN

(2) IN

wIk=3w I

n BI(2) IN(2) IN

w III

=1

w II

2

n BI

n BI

w III

nBII

nBIIIa bNguon

c

a

b

c

a/

A

BI (1) N nBI

C

I (1) N

wIwI

= w IIw III nBIIIa b c

w II nBII

(1) IN

w III = 3w I

nBI

a

b

c

b/

77

Nguon

A

B(2) IN

Cn BI(2) IN

nBIwIwI

n BI(2) IN

IN

(2)

(2) IN

n BI(2) IN

= w II3w I

w III =

w II

(2) IN

(2) IN

2

(2) IN

n BI

n BI

w III

nBII

nBIIIa bNguon

c

a

b

c

c/

An BI(3) IN

BnBI(3) IN

C(3) IN

nBI

n BI(3) 3I N

wIk=3w I

(3) IN

(3) IN

(3) IN

n BI(3) 3I N

w III

=1

(3) IN

3

(3) IN

(3) IN

3

w II

n BI

(3) 3I N

3

n BI

w III

nBII

nBIIIa b c a b c d/

78

Nguon AIN(2)

BI 2.N(2)

CIN(2)

n BI =

3

nBI 3

nBI 3

nBI 3

wI

(2) IN

2I N

(2)

3(2) IN

(2) IN

(2) IN

3(2)

3

k=

3w I

w III

=1w III

2 .I N

3(2) IN

3

(2) IN

w II

(2) IN

3

nBII

nBIII = 1a b c a b ce/

Hnh 2.23: S o phan bo dong trong MBA va mch th cap BI khi xy ra ngan mch trong vung bo ve

7. S vng ca cc cun dy cc pha khc, xc nh t iu kin cn bng sc t ng trong my bin dng bo ha trung gian khi MBA lm vic bnh thng v khi c ngn mch ngoi theo biu thc (2-35): I cbT .Wcb = I IT .WItt = I IIT .WIItt S vng tnh ton pha I: I (2-40) WItt = Wcb. cbT I IT S vng tnh ton pha II: I (2-41) WII tt = Wcb. cbT I IIT Wcb: s vng cun c bn ca BIG sau khi ly trn (v pha s nguyn gn nht) tng ng vi s vng tnh ton thc t c c ca BIG. Nu s vng tnh ton WItt , WIItt tnh theo (2-40) v (2-41) ra nhng s l th ly v s nguyn gn nht pha ln hn hoc b hn, sao cho dng in khng cn bng tng Ikcb c k n thnh phn I "' (do vic chn s vng dy WI v WII khng kcb ph hp vi gi tr tnh ton ca chng gy nn) trong mi trng hp ngn mch ngoi s khng lm cho bo v tc ng nhm. Cn ch rng i vi loi rle FkR= 100A th sai s do vic ly trn s vng dy gy nn (thnh phn I "' ) ni chung s b hn l i vi loi rle c FkR= 60A. kcb

79

i vi MBA ba cun dy khi s vng chp nhn ca ba pha khc nhau th pha c bn s ni vo cun dy so lch ca BIG cn cc pha kia s ni vo cun dy cn bng. i vi MBA ba cun dy khi s vng chp nhn hai pha no ging nhau v vi MBA hai cun dy th pha c bn s ni vo mt cun dy cn bng no (hoc mt phn ca cun dy cn bng v mt phn ca cun dy so lch) cn pha kia s ni vo cun dy cn bng cn li. Cch ni ny cho php ta chn s vng dy pha c bn gn vi gi tr tnh ton ca n hn. 8. Xc nh gi tr chnh xc ca dng khng cn bng s cp ca bo v c k n thnh phn I " ' theo cng thc (2-31) v (2-34): kcbttI "' kcbtt = W WII WItt WI .I IINng max .I INng max + IItt WIItt WItt

' I kcbtt = I 'kcbtt + I 'kcbtt + I "' kcbtt

Cuon can bang I 21 1 28 14 7 00 1234 0 1234 3 2 4

Cuon so lech3 2 1 0 8 16 24 12 2011 12 6

Cuon can bang II5

RI

RI

7

Hnh 2.24.a: S o noi day rle PHT 565 e thc hien bo ve so lech MBACuon can bang 21 1 28 14 7 001 234 01234 3 2 4

Cuon so lech3 2 1 0 8 16 24 12 2011 12 6

Cuon can bang II5

RI

RI

7

Hnh 2.24.b: S o noi day rle PHT 565 e thc hien bo ve so lech MBA

9. Xc nh gi tr chnh xc ca dng khi ng s cp ca bo v theo gi tr chnh xc ca dng khng cn bng, tnh theo biu thc (2-36). Tnh i dng khi ng ny sang pha th cp ca BI pha c bn. Nu nh dng tnh i ny ln hn dng khi ng ca rle tng ng vi s vng c chp nhn pha c bn th phi chn li s vng pha c bn v80

pha b hn gn nht. Sau li tnh li s vng ca cc cun dy ca BIG cc pha cn li tng ng vi s vng pha c bn va c chp nhn Wcb theo cc cng thc (2-40), (2-41). C nh vy tnh cho n khi no dng khi ng ca bo v c k n thnh phn I " ' kcbtt bng hoc b hn dng khi ng ca bo v c chp nhn mi thi. 10. Xc nh cc dng ngn mch s cp v nhng dng th cp tng ng ca BI, ri tnh h s nhy theo cng thc (2-38) hoc cng thc sau: I .W Kn = R (2-42) Fk R IR: dng th cp mi pha ca bo v c k n du ca n dng ngn mch ang kho st. W: s vng ca cun dy BIG pha tng ng. H s nhy Kn phi tnh cho tt c cc trng hp lm vic khc nhau ca MBA c bo v cng nh ca h thng v mng in cung cp, khi c nhng dng ngn mch khc nhau xy ra trn cc MBA c bo v. Khi dng cng thc (2-42) cn phi tnh s phn b ca dng s c v dng cc pha ca mch bo v. Khi dng cng thc (2-38) cn xc nh dng trong my bin dng BIG ca rle vi gi thit l ton b dng ngn mch ch chy mt pha cung cp dng s c ln nht dng ngn mch ang kho st. Nu tnh ton s b th dng cng thc (2-38) tin hn. Nu nh nhy ti thiu nhn c trong tnh trng tnh ton b hn gi tr cho php, ng thi gi tr tnh ton ca dng khi ng ca bo v li chn theo iu kin chnh nh khi dng khng cn bng khi ngn mch ngoi th phi tnh ton li bo v vi rle c c tnh hm loi ZT.Bng tng hp gi tr tnh ton:

STT Tn i lng tnh ton 1

K hiu v cng thc tnh

2 3 4 5 6 7 8 9

Dng khi ng ca rle pha I I kRcb = K s k (Ik khi cha c bn nI "' k n I kcb ) S vng tnh ton ca cun dy F my bin dng bo ha trung Wcbtt = kR . I kRcb gian pha c bn S vng pha c bn s b Wcb c chp nhn FkR Dng khi ng ca rle pha I kRcb = c bn tng ng Wcb S vng tnh ton ca cun dy I my bin dng bo ha trung WItt = Wcb cbT . I IT gian pha I S b chn s vng pha I WI S vng tnh ton ca cun dy I my bin dng bo ha trung WIItt = Wcb cbT . I IIT gian pha II S b chn s vng pha II WII Thnh phn dng khng cn WItt WI .I INngmax bng s cp do vic chn s I "' kcbtt = WItt vng pha I khc vi gi tr tnh ton ca n gy nn trong trng81

Gi tr bng s

10 11 12 13

hp ngn mch tnh ton Dng khng cn bng tnh ton pha s cp c k n thnh "' phn I kcbtt Gi tr chnh xc dng khi ng ca s cp ca bo v Gi tr chnh xc dng khi ng ca s cp ca rle pha c bn S vng ca cc cun dy ca my bin dng bo ha cui cng c chp nhn

' I kcbtt = I 'kcbtt + I 'kcbtt

+ I "' kcbtt

I kRcb K at .I kcbttI k nI Pha c bn Wcb Pha I WI Pha II WII I kRcb = K s

IV.3. Bo v so lch dng BIG c c tnh hm ZT:Ch dn chung: 1. Rle loi ZT c mt cun hm nm trong my bin dng bo ho trung gian cho php ta hm cc bo v bng dng in ly t my bin dng t pha no . c tnh khi ng ca rle khi c hm Flv = f(Fh) ph thuc vo gc gia dng lm vic Ilv v dng hm Ih. Trn (hnh 2.25) v c tnh gii hn tng ng ca cc gi tr hm ln nht v b nht. 2. S ni cc cun dy trong rle v s bo v my bin p hai v ba cun dy bng rle loi ZT trnh by trn hnh 2.26a, 2.26b trong cun hm c ni ti pha c dng ngn mch ln hn.

1300 1100 900 800 600 400 200 0

Flv (A-V)

Vung khi ong

Vung ham

200

400

600 800

Fh (A-V)

Hnh 2.25: ac tnh khi ong ca rle loi ZT

3. Dng rle so lch c cun hm cho php ta khng cn chnh nh dng khi ng theo dng khng cn bng khi ngn mch ngoi. Khi sc t ng do cun hm sinh ra s m bo cho bo v rle khng tc ng. V th bo v dng rle c cun hm thng c nhy cao hn so vi cc bo v khng c cun hm (loi PHT). Bo v dng rle ZT -1 thng dng trong trng hp sau y: - My bin p hai hoc ba cun dy c my ct trong mch t dng. - My bin p hai hoc ba cun dy (hay bin p t ngu), ni vo h thng c cng sut ln qua hai my ct. - My bin p ba cun dy trong mt cun ni vi pha khng c ngun.

82

Cuon ham2

6 1 3 5 7 9 11

Cuon can bang 4I0 1 2 3 0 1 2 3

Cuon so lech9 8 8 16 24 12 20 1112

32103

1

Cuon can bang II5

RI

RI

7

Cuon th cap

Hnh 2.26a: S o noi day rle ZT - 1

4. bo v khng tc ng khi ngn mch ngoi trong trng hp cun hm khng lm vic (v d my ct ni vo cun hm ct ra) hoc khi ng MBA khng ti, cn phi chn dng in khi ng theo cc iu kin nh i vi bo v PHT: * Theo iu kin chnh nh khi dng khng cn bng tnh ton ln nht khi ngn mch ngoi:I k K at .I kcbtttrong : h s an ton Kat = 1,5.

(2-43)

Dng khng cn bng tnh ton Ikcbtt chn dng khi ng Ikmin c xc nh nh rle PHT.' I kcbtt = I 'kcbtt + I 'kcbtt + I " ' kcbtt

I 'kcbtt = K kck .K n.fi .I Nng max' I 'kcbtt = U I Nng max + U I Nng max

ti:

WItt WI W WII .I INng max + IItt .I IINng max WItt WIItt * Theo iu kin chnh nh khi dng t ho nhy vt khi ng MBA khng I "' kcbtt =I k K.I mTtrong : h s K = 1,2 -1,5

(2-44)

* Theo iu kin chnh nh khi dng ngn mch ln nht khi ngn mch sau MBA t dng (hoc khng in t dng). Nu trong mch t dng khng t bin dng cho bo v so lch. Thng thng iu kin ny khng phi l iu kin tnh ton.

83

Trong hai iu kin (a), (b) h s an ton ly cao hn i vi loi rle PHT v trong rle so lch loi ZT khng c cun dy ngn mch, nh hng ca dng qu s ln hn. 5. S vng ca cun dy lm vic ca my bin dng bo ha (cun cn bng v cun so lch) tng ng vi dng Ikmin xc nh tng t nh i vi bo v dng rle PHT. Theo cng thc (2-39), (2-40), v (2-41): F I I Wcbtt = kR ; WItt = Wcb . cbT ; WIItt = Wcb . cbT I kRcb I IT I IIT 6. Cun hm ca rle ZT nn ni vo t my bin dng no m bo c nhy ca bo v ni chung cao nht. Mun vy phi lm sao cho: Dng khi ng ca bo v cng b cng tt. Khi ngn mch trong vng bo v (trong nhng trng hp kim tra nhy) tc ng hm cng b cng tt. Thc t chn c pha ni vi cun hm mt cch hp l cn phi xem xt mt cch c th. V d i vi cc s ni vi my tng p (hnh 2.27) c th chn t my bin dng ni vi cun hm nh sau:I II I II

a/ en cuo n ham

b/ en cuo n ham

III

~

F

TD

III

~

F

TD

I

II I en cuo n ham

II

en cuo n ham

c/

d/

III

~

F

TD

III

~

F

TD

Hnh 2.27: S o noi day cac MBA tang ap 3 cuon day (t ngau) dung rle so lech co cuon ham

84

* i vi my bin p tng p hai v ba cun dy (hoc MBA t ngu) cc cp in p u c ngun, trong mch t dng c t my ct cn trong mch my pht khng t my ct th cun hm ca my bin dng nn ni vo t my bin dng t t dng (hnh 2.27a). * i vi MBA tng p ba cun dy c ba cp in p u c ngun cung cp, trong mch my pht v mch t dng u c t my ct th cun hm ca rle so lch nn ni vo tng dng in ca hai nhnh my pht v t dng (hai t my bin dng ca hai nhnh ni song song) khi phi chnh nh dng khi ng ti thiu ca bo v theo dng khng cn bng khi h hng trong my pht in (hnh 2.27b). * i vi my bin p tng p hai hoc ba cun dy ni vi h thng cng sut ln qua hai my ct th cun hm ca rle so lch nn ni vo t my bin dng t trong mch ca mt trong hai my ct ( hnh 2.27c). * i vi MBA tng p ba cun y lm vic theo s b, pha in p trung khng c ngun cung cp, trong mch t dng v mch my pht khng t my ct th cun hm ca rle so lch nn ni vo t my bin dng t pha cp in p trung (hnh 2.27d). Sau y chng ta kho st mt s v d v cch ni cun hm trong bo v ca my bin p gim p (hnh 2.28):I II I en cuo n ham b/ II

en cuo n ham a/

III I II II

III

I

I en cuo n ham

c/

d/

e/ en cuo n ham

III

III

II

Hnh 2.28: S o noi day cac MBA gim ap 2, 3 cuon day (hoac t ngau) dung rle so lech co cuon ham

85

* i vi cc MBA gim ni vi h thng c cng sut ln qua hai my ct th cun hm ca rle so lch nn ni vo t my bin dng t trong mch ca mt trong hai my ct (hnh 2.28a). * i vi cc my bin p gim, ba cun dy ch ni vi mt ngun cung cp (hnh 2.28b) khi c iu p di ti v dng khi ng ca bo v tnh theo iu kin chnh nh khi dng khng cn bng khi ngn mch ngoi ( c hai pha khng c ngun) ln hn dng khi ng tnh theo iu kin chnh nh khi dng t ho nhy vt th cun hm ca rle so lch nn ni vo t my bin dng t trong mch ni vi ngun. Nu khi ngn mch ngoi mt pha no dng khng cn bng rt ln v dng khi ng ca bo v tnh theo iu kin chnh nh khi dng khng cn bng pha ny ln hn l dng khi ng ca bo v tnh theo iu kin chnh nh khi dng t ho nhy vt, cn khi ngn mch pha kia dng khi ng ca bo v tnh theo iu kin chnh nh khi dng khng cn bng b hn dng khi ng ca bo v tnh theo iu kin chnh nh khi dng t ho nhy vt th cun hm ca rle so lch nn ni vo t my bin dng t pha c dng khng cn bng ln hn. Cch ni ny cho php ta chn dng khi ng ca bo v theo iu kin chnh nh khi dng t ho nhy vt. * i vi MBA t ngu gim p ni vi hai ngun cung cp (hnh 2.28c) cun hm ca rle so lch nn ni vo t my bin dng t mt trong hai mch ni vi ngun. * i vi MBA gim ba cun dy c hai ngun cung cp (hnh 2.28d) khi cun dy in p trung hoc cun dy in p thp c in tr bng khng (r = 0) th cun hm ca rle so lch nn ni vo t my bin dng t pha c in tr bng khng. Tuy nhin trong trng hp khi cng sut ca ngun pha no b hn so vi pha kia th nn chn cch ni cun hm theo nhng ch dn mc trn. * i vi cc MBA gim hai cun dy (hnh 2.28e) cun hm ca rle so lch nn ni vo t my bin dng t pha khng ngun ca MBA. Cch ni ny lm tng nhy ca bo v khi c ngn mch trong MBA v khi y cun hm s khng lm vic. Trong trng hp chung c th chn cch ni cun hm mt cch hp l c th lm nh sau: Ln lt t cun hm tt c cc pha ca MBA c bo v vi gi thit l my ct t pha ny ct ra, xc nh dng khng cn bng tnh ton cc i Ikcbttmax khi cun hm khng lm vic v dng khi ng ti thiu tng ng Ikmin ca bo v i vi tt c cc phng n t cun hm khc nhau. Phng n no c dng khi ng ti thiu Ikmin b nht s c dng v khi y nhy ca bo v s cao nht. Nu nh hai phng n no cho gi tr Ikmin gn nh nhau hoc hon ton ging nhau (khi iu kin tnh ton l iu kin chnh nh khi dng t ho nhy vt) th nn chn phng n c tc ng hm t hn khi ngn mch trong vng bo v tng nhy ca bo v. Cn ch rng c th c trng hp nu ni cun hm vi pha m khi ngn mch ngoi pha dng khng cn bng tnh ton khng phi l ln nht th phi chn s vng ca cun hm qu ln v khi ngn mch trong vng bo v nhy ca bo v qu thp, trong nhng trng hp nh vy phi tnh ton li vi phng n ni cun hm vo pha khc. 7. Khi cun hm lm vic nu c ngn mch ngoi bo v s khng tc ng. kim tra tnh khng tc ng ca bo v (c hm) phi dng c tnh khi ng tng ng vi tc ng hm b nht (ng cong II trn hnh 2.25) tnh ton c n gin hn ngi ta c th thay th ng cong bng tip tuyn ca n i qua gc to , khi ta thay th nh vy th d tr khng tc ng ca bo v khi ngn mch ngoi s cao hn. R rng rng bo v chc chn s khng tc ng nu nh khi ngn mch ngoi, im biu din quan h Flv, Fh nm thp hn ng tip tuyn ny, ngha l: I .W K at lvT lvtt tg (2-45) I hT .Wh

86

IlvT v IhT: dng lm vic v hm th cp tnh i v pha th cp ca t my bin dng ni vi cun hm ca rle so lch. Wh: s vng ca cun hm c s dng. Wlvtt: s vng tnh ton ca cun lm vic pha c ni vi cun hm. tg: tg ca gc ca ng tip tuyn vi c tnh khi ng thp nht qua rle (theo s liu ca nh ch to tg 0,83) Kat: h s an ton Kat = 1,5. Mt khc ta li c : I lvT I lv = I hT Ih V khi ngn mch ngoi I lv = I kcbtt nn t (45) c th tm c s vng cn thit ca cun hm:Wh K at .I kcbtt Wlvtt . Ih tg

Trong :

(2-46)

Dng ngn mch tnh ton chn s vng cun hm theo biu thc (2-46) l ngn mch ba pha xy ra trong ch khi I ln nht: K .I (2-47) I = at kcbtt Ih xc nh gi tr ln nht ca I cn phi kho st nhng trng hp ngn mch pha do tc ng ca cun hm bo v s khng lm vic. ng thi trong trng hp chung tin hnh tnh ton cho nhng phng n lm vic khc nhau ca MBA v ca h thng in. Khi sc t ng lm vic do dng khng cn bng gy nn ln hn 100A c th dng biu thc (2-46) v thay ng c tuyn khi ng bng tip tuyn ca n (tg = 0,85). Khi sc t ng b hn 100A c tuyn khi ng thc t khc nhiu so vi ng tip tuyn ca n, v vy nu tnh theo ng tip tuyn th s vng cun hm nhn c s cao hn gi tr cn thit ca n. Mt khc vic xc nh tnh trng tnh ton s dng ng cong thc t s gp kh khn v khi tng t s I theo (2-47) i vi c tnh thc t th tg cng tng theo. 8. nhy ca bo v khi h hng xy ra trong vng bo v trong nhng ch khi cun hm khng lm vic c th c trng bng h s nhy tnh theo cc cng thc: IR Kn = I kRFk R Cc ch lm vic v dng ngn mch tnh ton cng nh h s nhy ti thiu cho php cng chn tng t nh l i vi cc bo v rle PHT. 9. nhy ca bo v trong nhng trng hp cun hm lm vic (c dng chy qua cun hm) h s nhy trong trng hp ny tnh theo cng thc: Flv (2-48) Kn = Flvk.R

Trong : I kcbtt v Ih : dng khng cn bng s cp tnh ton xc nh theo (2-31) v dng hm s cp dng ngn mch tnh ton chn s vng ca cun hm ca rle so lch.

hoc

Kn =

I R .W

87

Flv: sc t ng lm vic ca rle khi ngn mch trc tip. FlvkR: sc t ng lm vic khi ng ca rle trong iu kin khi bo v gii hn ca min khi ng trong trng hp ngn mch kho st, nhng khng phi ngn mch trc tip m qua in tr qu . Sc t ng lm vic ca rle so lch xc nh theo biu thc : (2-49) Flv = I lvT .Wlv IlvT: dng th cp cc pha khc nhau c k n du ca chng dng ngn mch trc tip kho st. Wlv: s vng ca cun lm vic tng pha tng ng. Khi c mt ngun cung cp, hoc tnh ton n gin ngay c khi c nhiu ngun cung cp c th xc nh sc t ng lm vic theo biu thc sau: Flv = I lvT .Wlvc (2-50) Trong : - IlvT: dng trong cun lm vic ca rle khi ngn mch trc tip (nu c nhiu ngun cung cp th gi thit dng ch chy theo pha cung cp ch yu). - Wlvc: s vng cun lm vic ca rle so lch pha c ngun cung cp (khi c nhiu ngun cung cp ly pha c ngun cung cp ch yu). Cc dng trong cc cun lm vic IlvT v IlvT trong cc biu thc (249) v (2-50) cn phi xc nh c k n dng ngn mch v s ni my bin dng ca bo v. Sc t ng lm vic khi ng ca rle FlvkR trong biu thc (248) xc nh theo c tnh khi ng ca rle khi tc ng hm ln nht theo trnh t sau:Xc nh sc t ng hm ca rle so lch Fh khi ngn mch theo biu thc: Fh = I hT .Wh (2-51)Trong :

Trong :

IhT: dng trong cun hm ca rle so lch dng ngn mch ang kho st c k n s ni dy ca BI. Wh: s vng dy ca cun hm. Trn mt phng Flv, Fh v im biu din tng ng vi Flv, tnh theo (2-49) v (2-50) v Fh tnh theo (2-51) cho trng hp ngn mch kho st. V ng thng OA, ng thng ny l qu tch nhng im tng ng vi dng Flv (A-V) 1300 ngn mch kho st qua cc in tr trung gian khc nhau, bi v 1100 khi ngn vi nhng in tr trung Vung tac gian khc nhau s phn b dng ong 900 in v quan h gia cc dng lm vic v hm s khng thay 800 i. im ct nhau gia ng Flvtt thng ny vi ng cong khi 600 ng ca rle tng ng vi tc A ng hm ln nht (im A trn 400 hnh 2.29) s l im biu din Vung ham gii hn tc ng ca rle, do 200 sc t ng lm vic tng ng vi im ny s l im lm vic Flh FlvkR khi ng FlvkR dng ngn 0 200 400 600 800 (A-V) mch ang kho st c in tr trung gian.Hnh 2.29: Xac nh sc t ong lam viec khi ong ca rle theo ac tnh khi ong khi o ham ln nhat.88

Trong :

xc nh nhy ca bo v cn phi chn ng tnh trng lm vic ca h thng, ca nh my in hoc trm v ca bn thn my bin p c bo v sao cho dng in qua cun hm l ln nht. H s nhy ti thiu Kn(hmmin = 2. i vi trng hp ng th MBA hoc khi MBA ba cun dy (hay t ngu) lm vic trong tnh trng mt my ct pha no ct ra th h s nhy ti thiu c th cho php ly thp hn. Cn ch rng khi xc nh nhy ca bo v nn k n sai s ca rle v bo v c th tc ng chc chn im lm vic tng ng vi dng ngn mch kho st phi nm cao hn ng c tnh khi ng t nht l 10% so vi tung ca n.

IV.4. My bin dng ph cho bo v so lch:Ngoi BI chnh ngi ta c th dng BI ph b lch pha, v hiu chnh dng khng cn bng vo rle do cc u phn p do khc t s bin i gia MBA v BI chnh. i vi MBA hai cun dy my bin dng ph c phi hp vi iu kin dng ph ti ln nht ca MBA (hnh 2.30). Vi MBA ba cun dy, chn BI theo cun dy c cng sut nh mc ln nht (hnh 2.31). V d: Cho bng vng dy theo u ni v s u dy my bin dng ph ca rle so lch MBA ca hng GEC. Hy chn t s BI ph, s vng dy ca cun BI ph trong bo v so lch MBA. My bin p 3 pha: S = 30MV, 11/66KV, /Y. 1-2 2-3 3-4 4-5 5-6 x-7 7-8 8-9 S1-S2 u ni S T s bin 5 5 5 5 125 25 25 25 215 vng i 1 /1 A

Bien dong ph C.ham

87 87C.lam viec

Hnh 2.30: Bo ve so lech co dung BI ph

89

S cap BI ph P1 P2 5 Bien dong ph

3Bien dong ph

4

x

C.ham

Gii:S1

87

C.lam viec S2 87 au ra ti rle

87

Hnh 2.31: Bo ve so lech MBA ba cuon day co dung BI ph

30.103 = 1574,6A Dng nh mc pha 11 KV: 3.11 Cun dy MBA pha 11KV ni, BI ni Y, dng th cp vo rle (chn BI chnh 1600 /1): 1574,6.1 = 0,984A I TI = 1600 30.103 = 262,4 A 3.66 Cun dy MBA pha 66KV ni Y, BI ni Y, dng th cp a vo rle (chn BI chnh 300 /1): 262,43.1 I TII = = 0,875A 300 b lch pha BU ph u Y/. T s (66KV / 11KV) = 0,875 / 0,984 = 0,899. Cun dy th cp S1-S2 l 215 vng. S vng ca cun s cp BI ph nI c tnh: 0,984.215 0,984.215 0,875.n = ; n1 = = 139,6 140 von 1 3 3.0,875 Chn hai u ni cun s cp BI ph l 2 v 6 (5 + 5 + 5 + 125) = 140 vng.

Dng nh mc pha 66 KV:

90

Kt hp bo v so lch v bo v chm t cun dy MBA:

C th lin kt bo v so lch th t khng cun dy MBA v bo v so lch dc MBA. T hnh 2.32 ta thy nu trung tnh cun sao ni t qua in tr 1 n v tng i, h thng bo v so lch dc c tr s t 20% s pht hin chm ch 42% cun dy tnh t u ng dy. Yu cu s ca hai b BI cho hai h thng so lch khc nhau, BI ca bo v so lch TTK ni sao, trong khi BI ca h thng bo v so lch dc th ni pha cun dy ni sao MBA, dng hai b BI th tn km, c th dng mt b BI cho hai h thng so lch theo cc cch sau: Dng my bin dng ph tng. Dng my bin dng ph Y/. Phng php sau c dng rng ri hn v c th iu chnh dng khng cn bng cho bo v so lch dc.

Im 100 80 60 40 20 0 100 80 60

(%)

Bo ve chm at

B.ve so lech 40 20 0

% Cuon day c bo ve

Hnh 2.32: Vung bo ve cuon day MBA theo dong khi ong s cap

a) S BI ph tng: BI ph tong co bon cuon day s cap giong nhau noi ti cac BI chnh nh hnh 2.33. Cac BI noi oi vi he thong so lech dc, dong a vao rle so lech dc la hieu dong hai pha. oi vi BI ph tong dong vao rle van la dong tng pha, tong cac dong la 0 khi ieu kien can bang cuon th t ca BI ph la dong t trung tnh MBA, rle at cuon th BI ph se bo ve chm at cuon day noi sao MBA, nh the s o nay bo ve so lech dc va bo ve so lech th t khong lam viec theo ac tnh rieng ca mnh. b) S BI ph Y/: Khi BI chnh ni Y pha cun cao MBA th t BI ph ni Y/ hiu chnh gc pha nh hnh 2.34. Rle chng chm t c ni t pha s BI ph. Lu trong cc s kt hp trn rle lm vic ng, khi chn BI chnh phi tnh n ti ca BI chnh, BI ph v cc rle.V. BO V SO LCH KHI C DNG T HO NHY VT, HIN TNG QU KCH T MBA

Khi ng MBA khng ti dng in t ho nhy vt pha ngun, tng cc dng ny khng phn bit vi dng ngn mch bn trong MBA. trnh tc ng nhm trong trng hp ny c cc phng php sau: Tc ng chm: Dng t ho l dng qu , tt nhanh nn c th trnh bng cch cho rle tc ng c thi gian. Vi cc bo v hin i ngi ta thc hic bin php hm ho tn bc 2. Dng in t ho c th phn tch ra cc thnh phn bc 2, 3, 4, ... nhng trong thnh phn bc 2 ln hn c. Hn na trong dng in ngn mch dng in bc 2 khng c nn thnh phn bc 2 c s dng n nh bo v chng li hin tng qu xung kch t ha khi ng MBA khng ti, khi thnh phn bc 2 ln hn gi tr t, bo v s b kho. Cn khi xut hin qu kch t MBA, c thnh phn sng hi bc 5 chim phn ln, thnh phn bc 5 ny c dng cho mc ch n nh bo v. Bo v s b kho khi thnh phn sng hi bc 5 ln hn gi tr t.

91

C.ham 87 C.lam viec 87

64

Hnh 2.33: S o so lech dung BI ph tong VI. MT S S BO V TIU BIU CHO MBACc k hiu trn s :50/51: Bo v qu dng in 2 cp 50/51N: Bo v qu dng th t khng 2 cp 51N : Bo v qu dng th t khng 27 : Bo v in p thp 81 : Rle t ng sa thi ph ti 96B: Bo v hi ca dng du MBA 74 : Rle kim tra cun ct my ct 96P : Bo v hi ca dng du ca b CA 26Q: Bo v nhit du MBA tng cao 2 cp 66 : Bo v p sut tng cao trong MBA 26W : Bo v nhit cun dy MBA - 2 cp 66 OLTC: Bo v p sut tng cao trong b A 33 : Bo v mc du MBA gim thp - 2 A, V : Ampemet, Vnmet cp Wh : My m in nng tc dng W: Oatmet Varh : My m in nng phn khng Var :Varmet

92

DK 110kV N-01 TBA 500Kv Nng

74

50/51

WH

VARH 96P1

A 87 63 63 OLTC

T196B 1 2 26Q 1 2 51N 331 2

26W 1 21 2

A 90 WH VARH VAR W V

F-24kV 50A

74

81

27

LI IN PHN PHI 22KV

Hnh 2.34: S bo v MBA 2 cun dy tiu biu

93

A. GII THIU CHUNGI. T VN S c xy ra vi thanh gp rt t, nhng v thanh gp l u mi lin h ca nhiu phn t trong h thng nn khi xy ra ngn mch trn thanh gp nu khng c loi tr mt cch nhanh chng v tin cy th c th gy ra nhng hu qu nghim trng v lm tan r h thng. Vi thanh gp c th khng cn xt n bo v qu ti v kh nng qu ti ca thanh gp l rt ln. Bo v thanh gp cn tho mn nhng i hi rt cao v chn lc, kh nng tc ng nhanh v tin cy.

II. NGUYN NHN GY S C TRN THANH GPCc nguyn nhn gy ra s c trn thanh gp c th l: H hng cch in do gi ci vt liu. Qu in p. My ct h do s c ngoi thanh gp. Thao tc nhm. S c ngu nhin do vt dng ri chm thanh gp. i vi h thng thanh gp phn on hay h thng nhiu thanh gp cn cch ly thanh gp b s c ra khi h thng cng nhanh cng tt. Cc dng h thng thanh gp thng gp nh hnh 3.1. Mi s h thng thanh gp c chc nng v tnh linh hot lm vic khc nhau i hi h thng bo v rle phi tho mn c cc yu cu . Cc dng h thng bo v thanh gp nh sau: Kt hp bo v thanh gp vi bo v cc phn t ni vi thanh gp. Bo v so lch thanh gp. Bo v so snh pha. Bo v c kho c hng. Trong loi 1, 2 ph hp cho cc trm va v nh 3, 4 dng cho cc trm ln.

a) S mt thanh gp

b) S mt thanh gp phn on bng MC

d/ He thong hai thanh gop co thanh gop vong c/ He thong hai thanh gop

95

e) He thong hai thanh gop li f) S mt ri

B. CC DNG BO V THANH GPI. BO V THANH GP BNG CC PHN T NI KT VI THANH GPH thng bo v ny bao gm bo v qu dng in hoc bo v khong cch ca cc phn t ni vo thanh gp, n c vng bo v bao ph c thanh gp. Khi ngn mch trn thanh gp s c c cch ly bng bo v ca cc phn t lin kt qua thi gian ca cp th hai.

I.1. S bo v dng in:H thng bo v dng cc bo v dng in ca MBA, ng dy v bo v dng in t thanh gp (hnh 3.2). Khi ngn mch trn thanh gp cn thc hin ct my ct phn on trc sau mt thi gian tr cc my ct ngun ni vi thanh gp s c c ct ra. Bo v t trn thanh gp cn phi hp vi thi gian ca bo v ng dy ni vi thanh gp. Phi hp vi bo v ng dy: t I = t Iz+ t MCI vi t z l thi gian ct nhanh ng dy. Cp thi gian th hai d tr cho cp th hai ca ng dy: II t II = t z + t, MC Thi gian ca bo v dng cc i ca phn t c ngun phi ln hn thi gian ca my ct: t MBA = t II + t. MC gim thi gian loi tr s c trn thanh gp xung mc thp nht, cn kho bo v ca phn t ni vi ngun

51Bo ve ng day

51

51

Hnh 3.2: Bo v dng in thanh ci

96

bng cc rle ca cc l ra cp in cho ph ti.51

tTG tH&Khoa

t151

t251

Hnh 3.3: Bo ve dong ien thanh cai co tac ong lien hp

I.2. Nguyn tc thc hin kho rle dng (hnh 3.3):Cc phn t ngun c bo v dng cc i c hai cp thi gian tc ng tH v tTG. Cp thi gian tH c chn phi hp vi bo v cc phn t khc trong h thng, cn cp thi gian tTG loi tr s c trn thanh gp, b hn nhiu so vi tH. Khi s c trn ng dy ra, bo v qu dng ca cc l ny gi tn hiu kho mch ct vi thi gian tTG ca my ct ngun, ng thi a tn hiu tc ng ct my ct thuc ng dy b s c. Thng thng s c trn ng dy ra s c ct vi thi gian t1, t2 tu theo v tr im ngn mch. Nu cc bo v hoc my ct tng ng t chi tc ng th sau thi gian tH bo v qu dng phn t pha ngun s tc ng ct my ct pha ngun. Khi ngn mch trn thanh gp bo v cc xut tuyn ra khng khi ng nn khng gi tn hiu kho my ct pha ngun v thanh gp s c c ct ra vi thi gian tTG.

I.3. Dng rle nh hng cng sut kho bo v nhnh c ngun ni vi thanh ci:Nguyn tc thc hin kho bng rle nh hng cng sut khi cc phn t ni vi thanh gp c ngun cung cp t hai pha. Rle kho tc ng khi hng cng sut ngn mch ra khi thanh gp. Khi ngn mch trn mt nhnh c ngun phn t nh hng cng sut trn nhnh khi ng. Khi ngn mch trn thanh gp rle nh hng cng sut khng khi ng v thanh gp c ct ra khi ngun.1 21RI1 2RI1 1RI2 2RI2 1RW 1RI 2RW 2RI 1RW 2RW RG RG

Hnh 3.4: Bo ve dong ien thanh gop dung RW khoa cac tac ong97

II. BO V SO LCH THANH GPII.1. Cc yu cu khi bo v so lch thanh gp:S s lch thanh gp cn tho mn cc yu t sau: Phn bit vng tc ng (tnh chn lc). Kim tra tnh lm vic tin cy. Kim tra mch nh th BI.Vung I Vung II

II.1.1. Phn bit vng tc ng: Mt h thng thanh gp gm c hai hay nhiu Vung III thanh gp khc nhau, khi c s c trn thanh gp no h thng bo v rle phi ct tt c cc my ct ni ti thanh gp . thc hin yu cu ny, mch th cp ca tt c cc BI ca mt thanh gp ni song Hnh 3.5: Vung bo ve he thong hai thanh gop song v ni vi dy dn ph, t a vo rle bo v thanh gp , khi nhnh no c ni vi thanh gp no th BI ca n s c ni vi dy dn ph ca thanh gp bng tip im ph ca dao cch ly. m bo, tt c cc im trn thanh gp nm trong vng bo v c gii hn bi cc BI. II.1.2. Kim tra mch th cp BI: Khi dy dn mch BI b t hay chm chp s gy ra dng khng cn bng chy vo rle so lch c th rle hiu nhm a tn hiu i ct cc my ct. i vi bo v thanh gp trong thc t vn hnh xc sut xy ra h hng mch th cp ln nn h thng bo v thanh gp cn c b phn pht hin h hng mch th cp BI.87B

87B

87B

95

Hnh 3.6: S o phat hien t mchMt trong nhng mch n gin pht hin t mch th cp l dng rle pht hin t mch th BI (rle 95 hnh 3.7) t ni tip hay song song vi mch bo v thanh gp (87B).

98

87B

87B

87B

95

Hnh 3.7: S o phat hien t mch th dung rle noi song songII.1.3. Kim tra tnh lm vic tin cy: Bo v thanh gp lm vic nhm s gy thit hi to ln nn hot ng ca s phi lun c kim tra. H thng kim tra phi tho mn cc yu cu sau: - H thng kim tra phi thc hin bng rle khc lm vic c lp vi rle chnh (rle K hnh 3.8a) - Tc ng nhanh nh bo v chnh. - Ngun cung cp ca rle kim tra phi khc vi ngun cung cp cho bo v chnh. - N cho tc ng khi ngn mch trong vng bo v v khng tc ng khi c ngn mch ngoi.I

C1

E

II

C2

AC1 C2

D

B Day dan phC1

C

V

C1

C2

C2

V I II

Kiem tra V I II K

I II V

+

Hnh 3.8a: Bo ve so lech he thong 2 thanh gop co thanh gop vong99

Trong s trn c 3 vng bo v ring bit. Mi mch ni vi 1 b bin dng to thnh vng bo v I, II v V. Mch iu khin my ct gm cc tip im ca rle phn bit vng bo v ghp ni tip vi tip im ca rle kim tra.V d khi xy ra ngn mch trn thanh gp I, lc ng thi tip im ca rle bo v cho thanh gp I v tip im ca rle kim tra ng mi a ngun iu khin ct cc my ct ni vi thanh gp I.

I

C1

K Cat C

II

C1 C2 C2

Cat B

V

Cat E Cat A Cat D

Hnh 3.8b: S o mch ieu khienII.2. Bo v so lch thanh gp dng rle dng in:Nguyn l so lch cn bng dng hay p thng c dng bo v thanh gp. Bo v loi cn bng p (hnh 3.9): Cc cun th cp BI c ni sao cho khi ngn mch ngoi v lm vic bnh thng, sc in ng ca chng ngc chiu nhau trong mch, rle c mc ni tip trong mch dy dn ph. - Khi ngn mch ngoi, cng nh khi lm vic bnh thng c dng ph ti & & & & chy qua, cc s ETI , ETII bng nhau. V d I TI = I TII v n I = n II nn: & & & R = ETI ETII I Z trong Z l tng tr ton mch vng. & & - Khi ngn mch trong vng bo v cc s ETI , ETII cng nhau v to thnh dng trong rle lm bo v tc ng.IR=0 IR0

ETI

ETII

ETI

ETII

N a/ b/

Hnh 3.9: S o so lech loi can bang ap100

S nguyn l bo v so lch dng thanh gp c hai mch nh hnh 3.10. RI RI RI Vng bo v c gii hn gia cc BI. Dng in khng cn bng khi ngn mch ngoi trong s ny thng rt ln do: Dng t ho BI khc nhau. Ti mch th cp BI khc nhau. Hnh 3.10: S o bo ve so lech dung rle dong Mc bo ho ca BI do ien thnh phn khng chu k ca dng ngn mch gy ra khc nhau. Thi gian suy gim ca thnh phn khng chu k c nh gi bng hng s thi gian tu thuc vo loi phn t ni kt vi thanh gp b s c. Mt vi tr s tiu biu nh sau: My pht cc li c cun cm: 0,15sec. My pht cc li khng c cun cm: 0,3sec. My bin p: 0,04sec. ng dy: 0,04sec. T cc s liu trn ta nhn thy nu c my pht ni vi thanh gp, thnh phn khng chu k ca dng ngn mch s tn ti lu hn v BI b bo ho nhiu hn. Vi bo v so lch dng rle dng in nn s dng c tnh thi gian ph thuc phi hp vi thi gian gim dn ca thnh phn khng chu k dng ngn mch. khng b nh hng bi hin tng bo ho li thp ca BI khi ngn mch ngoi, ngi ta dng BI vi li khng phi l st t (BI tuyn tnh, li khng kh). u im ca BI ny l: - Khng b bo ho. - p ng nhanh v khng b qu . - Tin cy, d chnh nh. R - Khng nguy him khi h mch th cp. Tuy nhin khuyt im ca loi ny l cng sut u ra th cp thp v gi thnh rt t. S dng BI tuyn tnh thng l s so lch cn bng p (hnh 3.11). Khi ngn mch ngoi tng dng bng khng v in th a vo rle bng khng. Khi ngn mch trong vng Hnh 3.11: S o so lech can bo v, hiu in th sut hin qua rle tng tr v lm rle tc ng. bang ap

II.3. Bo v so lch thanh gp dng rle dng in c hm: khc phc dng khng cn bng ln ca bo v so lch thanh gp khi dng rle dng in ngi ta cng c th dng rle so lch c hm. Loi rle ny cung cp mt i lng hm thch hp khng ch dng khng cn bng khi ngn mch ngoi c dng khng cn bng ln. Dng in so lch Isl (dng lm vic) : &sl = & lv = & TI & TII I I I I (3-1) Dng in hm IH: & I H = K( & TI + & TII ) I I (3-2) ITI ITII Vi K l h s hm, K < 1. Trong ch lm vic bnh thng, hay 87B Cuon lviec khi ngn mch ngoi vng bo v, dng in Cuon hamHnh 3.12: S o nguyen ly bo ve so lech co ham101

lm vic s b hn nhiu so vi dng in hm nn rle so lch khng lm vic. Khi ngn mch trong vng bo v (v d ch c mt ngun cung cp n thanh gp), lc ny: & lv = & TI > & H I I I (3-3) nn rle so lch s lm vic.

II.4. Bo v so lch thanh gp dng rle tng tr cao (khng hm):Rle so lch tng tr cao c mc song song vi in tr R c tr s kh N1 ln. Trong ch lm vic bnh thng v khi ngn mch ngoi vng bo v (im Rle tong tr cao N2), ta c: & = & TI & TII = 0 I I I (3-4) ITI RL R ITII Nu b qua sai s ca my bin dng, th dng in th cp ca BI chy qua in tr R c th xem bng khng. N2 Khi ngn mch trong vng bo v Hnh 3.13: Bo ve thanh gop bang (im N1) ton b dng ngn mch s rle so lech tong tr cao chy qua in tr R to nn in p t trn rle rt ln, rle s tc ng. S (hnh 3.14) trnh by phng n thc hin bo v rle tng tr cao i vi thanh gp. n gin, ta xt trng hp s thanh gp ch c hai phn t (G, H) v my bin dng c H G thng s ging nhau. Rle c mc ni RBIH RlH RlG tip vi mt in tr n nh RR, vic mc ni tip mt in tr n nh RR s INMT RBIG xH RR lm tng tng tr mch rle nn phn xG ln dng khng cn bng (do s bo ho RL khng ging nhau gia cc BI khi ngn mch ngoi) s chy trong mch BI b bo ha c tng tr thp hn, ngha l N2 RR c tc dng phn dng qua rle. Hnh 3.14: S o thay the mch th cap BI Nu xem cc my bin dnghon ton ging nhau th RBIG = RBIH (in tr th cp BI), dy dn ph c c trng bi R1H v R1G (hnh 3.14) v in khng mch t ha xH, xG. ch ngn mch ngoi, nu cc my bin dng khng b bo ha th xH v xG c tr s kh ln nn dng in t ha c th b qua, dng in ra vo nt cn bng nhau (nh lut 1 Kirchoff) do pha th cp BI khng c dng chy qua rle, rle khng tc ng. Trng hp ti t nht l my bin dng t trn phn t c s c bo ha hon ton, gi thit ngn mch ngoi nhnh H lm BI nhnh H b bo ha hon ton (xH = 0) ngha l bin dng H khng c tn hiu u ra, tnh trng ny c biu th bng cch ni tt xH (hnh 3.14). My bin dng G cho tn hiu u ra ln hn, khng b bo ha. Dng in ngn mch pha th cp ( INMT) phn b qua cc tng tr nhnh gm RlH, RBIH v nhnh rle: Dng in qua rle: I NMT (RlH + RBIH) (3-5) IR = RR + RlH + RBIH

102

Nu RR c gi tr nh, IR s gn bng INMT, iu ny l khng cho php. Mt khc, nu RR ln khi IR gim. Phng trnh (3-5) c th vit gn ng vi sai s cho php nh sau: I NMT (RlH + RBIH ) IR = (3-6) RR (3-7) U R = I R .RR = I NMT .(RlH + RBIH ) Mun tng nhy ca bo v cn chn BI c in tr cun th RBI b v gim n mc thp nht in tr ca dy dn ni t BI n rle. Khi ngn mch trn thanh gp tt c cc dng in pha s cp u chy vo thanh gp, pha th cp tt c cc dng in u chy vo rle, c th gy qu in p trn cc ca rle. chng qu p cho rle c th mc song song 1 in tr phi tuyn vi rle. Nhng yu cu c bn khi s dng s ny l: - T s BI ca tt c cc nhnh ging nhau. - in th th cp BI ln. - in tr cun dy th cp BI nh. - Ti dy dn ph nh.

III. BO V SO SNH PHABo v so snh pha dng in i vo v i ra khi phn t c bo v, v vy nn c tn l bo v so snh pha. Pha ca dng in c truyn qua knh truyn so snh vi nhau (hnh 3.14). lch pha: (3-8) = 1 + 2 = trong : 1, 2 l gc A B ISI ISII pha tng ng ca dng ~ ~ in i vo v i ra khi phn t c bo v. ITI 1 ITII 2 ch lm vic bnh thng v khi ngn T T mch ngoi gc pha ca Kenh F F dng in hai u gn nh nhau nn 00. Khi ngn mch trong vng T.hieu cat bo v, dng in hai T.hieu cat pha ngc nhau nn Hnh 15: S o nguyen ly bo ve so sanh pha dong ien 1200. Trn thc t do nh hng ca in dung phn b ca phn t c bo v nn trong ch lm vic bnh thng v khi ngn mch ngoi 1800, trnh bo v tc ng nhm phi chn gc khi ng k ln hn mt gii hn no , thng khong (300-600). TG N1 S nguyn l bo v so snh pha dng in ca MC bo v thanh gp hnh 3.16. IS1 IT1 Khi ngn mch trn thanh IR1 gp (im N1) dng in s IT2 IR2 & tc Ct MC cp v th cp BI tt c cc IT3 N2 IS2 phn t c pha ging nhau IS3 IR3 (hnh 3.17a), thi gian trng D1 D2 D3 hp tn hiu tc cho na chu k103

Hnh 3.16: S nguyn l so snh pha dng in thc hin bo v thanh gp

(dng hoc m) ln (i vi h thng c f=50 Hz), thi gian tCmax = 10ms) cho bo v tc ng (tC t).a) iT1 iT2 iT3 iR1 iR2 iR3Tn hieu cat

b) iT1 iT2 iT3 t iR1 iR2 iR3Tn hieu cat

t

tC = 0

t t tC Hnh 17: Pha dong ien khi ngan mch ben trong (a) va ngan mch ben ngoai (b)

Khi ngn mch ngoi vng bo v thanh gp (im N2), dng in chy qua BI ca phn t b s c c pha ngc vi dng in trong cc my bin dngca phn t khng b s c, thi gian trng tn hiu bng khng, bo v s khng lm vic (hnh 3.17b).

IV. BO V D PHNG MY CT HNGMy ct l phn t tha hnh cui cng trong