BO V CC PHN T CHNH TRONG H THNG IN

PGS.TS L KIM HNG

A. GII THIU CHUNG V MY PHT IN

My 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 MF

I.1. Cc dng h hng: - 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

kp). (2) - 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 MF Tu 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, ...

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B. CC BO V RLE CHO MY PHT IN

I. 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.

52 1BI

MF

2BI

87G

2RI

+

Ct MC

+

4Rth

-

+

5RT

MF

MC

Bo tn hiu

Rf Rf

+

3RI

Bo tn hiu t mch th

1RI +

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

a)

b)

Trong :

: dng hn ch dng in khng cn bng (I- Rf KCB), 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: = IIR 1T - I2T = ISL (1-1)

Vi I , I l dng in th cp ca cc BI hai u cun dy. 1T 2TBnh thng hoc ngn mch ngoi, dng vo rle 1RI, 2RI l dng khng cn bng

I : KCBISL = I1T - I2T = IKCB < I (dng khi ng rle) (1-2) KR

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

INnI

ISL = I1T - I2T = > I (1-3) KR

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

a)

ISL = IKCBT < IKR

I2TI1 T

b)

I1T

I2T

ISL KRI

N InI >

Trong : - 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:

I

FnI

IR = (1-4)

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. I K .IKB at KCBtt (1-5) IKCBtt = Kn.KKCK.fi .I (1-6) Nngmax

Trong : - K : h s an ton tnh n sai s ca rle v d tr cn thit. Kat at 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.

- K : h s tnh n s ng nht ca cc BI (K = 0,51). n n- f : sai s tng i ca BI, f i i 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

mFn

IISL = (1-7)

Dng khi ng ca bo v:

mFI

at In

KI = (1-8) KB

Nh vy, iu kin chn dng khi ng cho 1RI, 2RI: I = max{KKB at .IKCBtt; Kat .I } (1-9) mF

Dng in khi ng ca rle:

I

KB)3(

nI.K

I = (1-10) KR

15

Vi K(3) l h s s . Sau khi tnh c I ta s chn c loi rle cn thit. KR

Kim tra nhy Kn ca bo v:

Kn = KB

minNII

(1-11)

Vi INminV bo v c tnh chn lc tuyt i nn yu cu K

: dng in ngn mch 2 pha u cc my pht khi my pht lm vic ring l. n > 2.

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:

I = 0,2.I (1-12) KS(3RI) mF 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 > t (1-13) ct Nngoit5RT = tctNng + t (1-14)

Trong : - tc Nngt- t: bc chn lc thi gian, thng t = (0,25 0,5) sec.

: thi gian ln nht ca cc bo v ni vo thanh gp in p my pht.

Nhn xt: - Bo v s tc ng khi ngn mch nhiu pha trong cun dy stator my pht.

RI

Vng bo v

I1S

I2S

I1T

I2T

ILV

BIH IH

BILV

1BI

2BI

Hnh 1.3: Bo v so lch dng in c hm cun dy stator MF

- Bo v khng tc ng khi chm chp gia cc vng dy trong cng 1 pha hoc khi xy ra chm t 1 im trong cun dy phn tnh.

tng nhy ca bo v so lch ngi ta c th s dng rle so lch c hm.

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 I v ILV H. - Dng in vo cun lm vic I : LV

SL.

T2T1LV.

IIII == (1-15) - Dng in hm vo cun hm IH:

IH = I1T + I (1-16) 2TKhi lm vic bnh thng hay ngn mch ngoi vng bo v: Dng in I1T cng

chiu vi dng I2T: I1T I2T ISL = I = ILV 1T - I2T = I (1-17) KCBIH = I1T + I 2.I2T 1T > I (1-18) LV

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 I = ILV 1T + I2T 2.I1T > IH (1-19)

16

bo v s tc ng. Nhn xt:

- Bo v hot ng theo nguyn tc so snh dng in gia I v ILV H, 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 bnh thng, dng in th cp I

Hnh 1.4: Bo v so lch c hm tc ng nhanh cho MF cng sut ln

RL2

RL1

ILV I1S

I1TI2T

I2S

BIG IHILV

RH/2

ULV

RH/2 UH

D1 D2

RL1

RL2

n RG u ra

A

B

C RLV

ILV BIG

CL

1T v I2T ca cc nhm bin dng 1BI, 2BI chy qua in tr hm RH, to nn in p hm UH, cn hiu dng th cp (dng so lch) ISL chy qua bin dng trung gian BIG, cu chnh lu CL v in tr lm vic RLV to nn in p lm vic ULV. Gi tr in p UH > ULV, bo v khng tc ng.

Khi ngn mch trong vng bo v, in p U >> ULV H, 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 RL , RL2 2 ng tip im li, rle ct u ra s c cp ngun thao tc qua hai tip im ni tip RL v RL1 2 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 E v E bng nhau v ngc pha nhau, L1 2 1 = L2, phn b in p trong mch nh hnh 1.5b.

Hnh 1.5: Bo v so lch dng rle tng tr cao cho MFa) 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.

1BI IN 2BI N

1BI USL

USL

R1 R2

L1E1 RSL E2

E1

E2

L2 E1 L1 USL RSL L2 E2

USL E2

R1 R2

E1 USL

E2=0

R1 R2

E1 L1 USL RSL

E1

a)

b)

c)

d)

USL = 0

17

Tr s in p t ln rle so lch RU ph thuc vo quan h gia cc in tr R1 v R . in tr R , R2 1 2 gm in tr cun dy th cp v dy dn ph ni gia hai nhm bin dng 1BI v 2BI, vi R1 = R U2 SL = 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:

I

21"N

1SL n)RR.(IU += (v RSL >> R ) (1-20) 2

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

= I

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

vi: - n : t s bin dng ca BI. I- 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 cn c thm thnh phn dng in do h thng

v . Mch in ng tr v phn b in p nh hnh 1.5d. Gi tr in p t ln rle so lch RU:

"NFI

"NHI

I

21"NH

"NF

SL2 n)RR).(II( U ++= (1-21)

m bo tnh chn lc, in p khi ng ca rle so lch RU phi chn ln hn min{USL1; U }, ngha l: SL2

I

21"Nat

n)RR.(I.K +

U = K .U = (1-22) KR at SL1

Vi K = (1,15 1,2) l h s an ton. atThi 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.

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).

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

a)

XB

tI = (0,4 0,5) sec

0,7XB

XF

t

t

X

b)

BU

F

BI

I

U

RZ

BA TG

tII

jX

0

0

UF

R

ZK

RK

jXK

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:

ZI = Z + 0,7.Z (1-23) k F BThi 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

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 =I =ILV 1T - I2T = I (1-25)

RL K

R R

LV H

2BI I1S

1BI I2S ILV

I1T

I2T

BILV BIH

IH

Ct MC

4

3

2

1

0 1 2 3 4

I*LV

I*H

ILV = IH

ILV = f(IH)

a) b)

KCB

87G 87G 87G

Hnh 1.8: S bo v so lch ngang theo m s

IH > I nn bo v khng tc ng LVKhi 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

IH = I1T - I2T = I KCB 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

Ct 1MC

Lf3

+ Bo tn hiu

RI

RT Rth

+

-

C

C B A

O2

O1

1 2

T

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

87BI

a)

b)

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 O v O1 2 gia

2 nhnh song song ca cun dy. * ch bnh thng hoc ngn mch ngoi:

U12 = V - V 0 (1-27) 1 2nn 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 = V - V 0 (1-28) 1 2

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: I = (0,05 0,1).I (1-30) KB mF

I

KBn

I I = (1-31) KRt 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 V . V1 2III. 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:

2C

2q

p(1)

0Xr

.UI

+= (1-33)

Trong : - : s phn trm cun dy tnh t trung im n v tr chm t ( 1). - U : in p pha ca my pht. p- rq: in tr qu ti ch s c. - : dung khng 3 pha ng tr ca tt c cc phn t trong mng in p 0CX

my pht. 0

Nu b qua in tr qu ti ch s c (r =C C..j.3

1X0

= 0), dng chm t bng: q(1)I = 3...C .U (1-34) 0 p

Khi chm t xy ra ti u cc my pht ( = 1) dng chm t t tr s ln nht: (1)

maxI = 3..C .U (1-35) 0 pNu 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)

maxI 30 A i vi mng c U = 6 kV (1)

maxI 20 A i vi mng c U = 10 kV (1)

maxI 15 A i vi mng c U = (15 20) kV (1)

maxI 10 A i vi mng c U = 35 kV Kinh nghim cho thy rng dng in chm t 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.

(1)I

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 i

22

C31da vo dung dn gia 3 cun dy stator my pht, thng yu cu: R ()

(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.

Hnh 1.10: Cc phng n ni t trung tnh MF

R K BA Rt a) b) c)

- 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 MF-MBA thng I < 5 (A) ch cn t bo v n gin hn bo tn hiu chm t stator m khng cn ct my pht.

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.

MF

1MC

7BI0

FCO

3RI

4RI

5R RTh G

6RT

+ +

T bo v chng nm

ngoi

+

+

Bo tn hiu

C

2BU

t

1MC

-

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:

KCBtt.

IC

.B

.A

.

IR

.I

n1)III(

n1I =++= (1-37)

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:

I >IKR KCBttKhi 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. - C , C : in dung pha i vi t ca my pht v h thng. 0F 0HT- U : in p pha ca my pht. pFDng in vo rle bng:

pF0HTD U..C.3.I = (1-39) bo v c th tc ng c cn thc hin iu kin:

KCBttD II I (1-40) KB n gin, ta gi thit dng chm t i qua bo v v dng khng cn bng tnh ton ngc pha nhau.

DIKhi s vng chm b, dng in chm t nh v bo v c th c vng cht gn trung tnh my pht.

Khi chm t mt pha ngoi vng bo v, dng in i qua bo v: pF0FD U..C.3.I = (1-41)

bo v khng tc ng trong trng hp ny, dng khi ng ca bo v phi c chn: KCBttqDKB III +> (1-42)

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 :

)IUkC3(KKI maxKCBttpFq0

tv

atRI4KB += (A) (pha s cp) (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:

I = (2 3) (A) (pha s cp) (1-45) KB4RIT 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:

MBA RU RT

MF BU

V FCO

+ +

-

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

iu kin1: UK KCBmax iu kin2: U RU > U

KRU chn theo iu kin n nh nhit ca rle v thng ly bng 15V.

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 do qu s c bn ngoi. tRT = tmax (BV ca phn t k cn) + t. (1-47)

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.

III.3.1. Phng php bin :

Hnh 1.14: Bo v chm t dy qun stator

51N

50N R BA Rt 59 BU Rt

50N a) b) c)

C0F

I(1)F I(1)H

I(1) C0H

Hnh 1.13: Chm t trong cun dy stator MF

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:

(1)I H >> I(1) vi IF F = 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 I cp in p my pht. max

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)

R

1RU

Lf3

2RU

2BU0

Z1 Z2

MF

1BU0

N F

a

b

b)

a)

c)

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 pht

UFUF

UFUF

N

N

N

N

N F

F

F

UN UN

100%

100%

100%

F

50%

50%

50% d)

N

F

F

UN UN

26

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 F N F

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).

, U l in p hi bc ba khi my pht khng ti v U , U khi my pht y ti.

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 L v s tc ng khi c chm t trong cun dy stator. f3

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 Z , Z1 2 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.

K

IH

K L I -I(1)

Vng tc ng

Vng hm

Ilv IU

b)

3I0 = I(1)D

3U0

IU

Ilv CL1

CL2

BTH1

BTH2

R1

C2

C1 R2

RI

t

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

a)

27

Rle so snh tng quan gia dng in lm vic I v dng in hm ILV H theo quan

h : I = IH - I (1-48) LV

Trong : 1 (1-49a) IH = IU + I 1ILV = IU - I (1-49b)

Vi IU l dng in ly t ngun in p U ; ly t b lc dng th t khng. 01D

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 I

I&

>I iuH LV 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 R , C vi thng s c la chn thch hp. 2 2

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 RU . Khi c chm t, in p U xut hin, RU0 0 0 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: (1)I = IA - I (1-50) D

Trong : - I dng in c to nn bi thit b to thm ti. A(1)- I D dng in chm t chy qua bo v. Trong trng hp ny gc pha gia in p th t khng U v dng in tng I0

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 = I + IA Dv gc pha gia in p th t khng U

(1) v dng in tng I0 nm trong min tc

ng ca bo v. Rle tc ng ct vi thi gian t1.

28

Ct MC

c)

R BI0N

BI0

BIG

I = IA+ I(1)

IA

I(1)

I(1)

U0

I Min hm

Min tc ng

U0

I

Min hm

Min tc ng

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.

MC

BI0

MF

BI0N R

BIG RU0 Thit b to thm

ti

A.I

)1(

.I

0.U

a

RU0 t2

t1

ng RU0 Ct RU0

I Thit b bo v

I(1)

IA R BI0N

BI0

BIG I = IA - I(1)

a)

c)

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 U t vo rle s suy gim khi in tr ch s c ln. ab- 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

1LF

IB

MBA

R C

20Hz

BIG

RB

I

RC

Hm

2LF

Lm vic

I

ILV

RL IC=IH

Ct

Hnh 1.18: S bo v 100% cun dy stator chng chm t c a thm in p hm 20Hz vo trung im MF

- 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 I chy qua in tr t RB B. Thnh phn I thng qua bin dng trung gian BIG v b lc tn s 2LF c nn thnh dng in lm vic.

- ILV a vo rle so snh vi dng in hm IH cng do ngun 20Hz to nn thng qua in tr t Rc , dng in hm c tr s khng i. ch lm vic bnh thng (R = ) dng in I c xc nh theo in dung ca cun dy i vi t C nn c tr s b do I < ILV H v rle s khng tc ng.

- Khi c chm t, dng I c xc nh ch yu theo in tr chm t R ,

I >ILV H 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, P ..P2 6 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

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).

R

C

C

R

12,5Hz

R0

I

RU0

I 12,5Hz

BUG

C

LF

BIG

R

12,5Hz R BU0

S 900

I

IM

a) R RU0

b)

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.

480

M3

M2

M1

E D

C B

ms

P6 P4P2

P5 P3 P1

0

1/ 2 chu k (+)

Tn hiu m ho

1/ 2 chu k(-)

S chu k 12,5Hz

400 320 240 160 80 2 1

560 2 1

160 80 7 6 5 4 3 3 t

t

IM

A

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: C ba phng php c s dng pht hin chng chm t mt im mch kch

t : * Phng php phn th.

64

Cun kch t R

MFkt

HNH 1.21 : Bo v chm t rotor bng 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)

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 qua rle in p, khi c mt im chm t s 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:

+

36RT

Bo tn hiu

35RI

+ +

-

37RG

47C

48CC

Ti mch kch t

Ti trc MF

34BG

52N

2RCL U~

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

DC

+

36RT

-

Bo tn hiu

35RI

+ +

-

37RG

47C

48CC

Ti mch kch t

Ti trc MF

34BG

52N

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

ph AC

32

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.

C

S2

S1 R

LF

BUG

RY

Ct MC

U(50Hz) R

Rotor ca my kch t My kch t

Cun dy rotor ca my pht in

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 S , S1 2. 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 ca chi than S

X

1 R=0

2

R CR= 0 C=

X/ 2 R=

jX

C= const R=var

R= const C= Var

R / 2

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.

1, S2 lm cho tin cy ca s khng cao v tr s ca in tr tip xc c th nh hng n tr s o ca rle. Ngoi ra bn thn h thng kch thch mt chiu cng c th nh hng n s lm vic ca bo v khi in dung ca mch kch thch i vi t C ln, in tr r R ln nht c th o c 10 k.

khc phc nhc im ny ngi ta dng s vi ngun in ph mt chiu hoc xoay chiu vi tn s thp c dng sng hnh ch nht.

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 R , R1 2 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:

RRRUI

M ++= (1 -51)

21

21RR

R.RR +=Trong : Lu rng RM

U(1Hz)

c) R= 0 R= 5K C= 2F

b)

Hnh 1.26: S nguyn l pht hin chm t trong cun dy rotor MF dng ngun in p ph 1Hz c dng sng ch nht (a), v dng sng t vo b phn o UM vi cc tr s in tr khc nhau (b v c)

+

_

M

R

R1

Uktph(1Hz)

C

RM

Cp 2 ct MF

UM

Cun dy rotor MFCp 1 cnh bo

R2

Ngun kch t

I

a)

Thanh gp t dng

in dung i vi t ca mch kch t C mc song song vi in tr R s lm tc

thi tng tr s dng in I v in p UM thi im u ca mi na chu k ca ngun in p U.

in tr R c tc dng lm suy gim tr s ca I v UM. R cng b suy gim cng nhanh, trn hnh 1.26b v 1.26c trnh by dng sng UM o c cho hai tr s ca R khc nhau.

Bo v c chnh nh tc ng bo hiu khi in tr r R tt di 80k (mc 1) v tc ng ct my pht khi R < 5k (mc 2).

IV.3. Bo v chng chm t im th hai mch kch t:

Ct 1MC

4Rth

+

Bo tn hiu hiu

-

2RT

+

1RI

3RG

V

7PA

9CN 5CC

6N

BIH

Ti mch kch t

Ti trc my pht

10CN

r3 r4

5N

a

c b

b)

RI

Ti trc MF

r1 r2

r1 r2

a)

Hnh 1.27: S bo v chng chm t th hai mch kch ta) S nguyn l b) S bo v

35

Bo v chng chm t im th hai mch kch t (hnh 1.27) c a vo lm vic sau khi c tn hiu bo chm t mt im mch kch t. Thng bo v c t trn mt bng di ng v c dng chung cho nhiu t my ca nh my. Bo v lm vic da trn nguyn tc cu bn nhnh: Khi chm t mt im mch kch t, ngi ta iu chnh cho cu cn bng nh ng h V. Khi cu cn bng ta c: 4r

3r2r1r = , do khng c dng qua

1RI, bo v khng tc ng. Khi chm t im th hai mch kch t s lm cho cu mt cn bng, c dng qua

1RI v 2RT c in, sau mt thi gian 3RG c in i bo tn hiu thng qua 4Rth, ct my ct ng thi ni tt cun dy ca 1RI trnh b h hng v t gi cho 3RG thng qua mch t gi.

Cc phn t trong s : - 3RG: rle trung gian, bao gm cc tip im: Tip im a: a tn hiu i ct my pht. Tip im b: bo v RI khng b chy (ni tt RI). Tip im c: tip im t gi.

- BI : ly thnh phn xoay chiu ca nhiu tng cng tc ng hm cho RI. H- 9CN: cu ni, dng kho bo v khi sa cha hoc khng mun bo v tc

ng. - 6N: nt n, kt hp vi ng h V iu chnh cho cu cn bng khi xy ra

chm t im th nht mch kch t. - 5N: nt n, gii t gi sau khi bo v tc ng i ct my ct. - 5CC: cun cn nhm hn ch thnh phn nhiu xoay chiu, trnh lm cho RI tc

ng nhm. - 10CN: kho bo v khng cho ct my ct.

V. Bo v chng qu in p (59) in p u cc my

pht c th tng cao qu mc cho php khi c trc trc trong h thng t ng iu chnh kch t hoc khi my pht b mt ti t ngt.

59II

59I

t

BU

Ct MC

n h thng iu chnh U(gim kch t)

MC Ct kch t

MF

Hnh 1.28: Bo v chng qu in p hai cp t MF

Khi mt ti t ngt, in p u cc cc my pht thu in c th t n 200% tr s danh nh l do h thng t ng iu chnh tc quay ca turbine nc c qun tnh ln v kh nng vt tc ca rotor my pht cao hn nhiu so vi my pht turbine hi.

cc my pht nhit in (turbine hi hoc turbine kh) cc b iu tc lm vic vi

tc cao, c qun tnh b hn nn c th khng ch mc vt tc thp hn, ngoi ra cc turbine khi hoc hi cn c trang b cc van STOP ng ngun nng lng a vo turbine trong vng vi msec khi mc vt tc cao hn mc chnh nh.

Mt khc, cc my pht thu in nm xa trung tm ph ti v bnh thng phi lm vic vi cc mc in p u cc cao hn in p danh nh b li in p ging trn h thng truyn ti, khi mt ti t ngt mc in p li cng tng cao.

Qu in p u cc my pht c th gy tc hi cho cch in ca cun dy, cc thit b u ni u cc my pht, cn i vi cc my pht lm vic hp b vi MBA s lm bo ho mch t ca MBA tng p, ko theo nhiu tc dng xu.

Bo v chng qu in p u cc my pht thng gm hai cp hnh 1.28.

36

I* Cp 1 (59 ) vi in p khi ng: UK59I = 1,1UFm (in p nh mc MF). Cp 1 lm vic c thi gian v tc ng ln h thng t ng iu chnh kch t gim kch t ca my pht.

II* Cp 2 (59 ) vi in p khi ng: UK59II = (1,31,5)UFm. Cp 2 lm vic tc thi, tc ng ct MC u cc my pht v t ng dit t trng ca my pht.

VI. Bo v chng ngn mch ngoi v qu ti Mc ch t bo v: - Chng ngn mch trn cc phn t k (thanh gp my pht, my bin p,...) nu

bo v ca cc phn t ny khng lm vic. - Chng qu ti do h thng ct gim mt s ngun cung cp. - Lm d tr cho BVSLD my pht in. thc hin bo v chng ngn mch ngoi v qu ti ta c th s dng cc phng

thc bo v sau:

VI.1. Bo v qu dng in: Vi cc my pht b v trung bnh, ngi ta thng s dng bo v qu dng in c kho in p thp (hnh 1.29). Bo v thng c 2 cp thi gian:

MC

F

BU

MBA

BI

&

2I

2II

Ct MC

Dng my pht

Hnh 1.29: Bo v qu dng in c kho in p thp

27

50

Cp 1 (2I) tc ng ct MC u cc my pht (nu ni vi thanh gp in p my pht) hoc MC ca b MF-MBA. Cp 1 c phi hp vi thi gian tc ng ca bo v d phng ca ng dy v MBA.

Cp 2 (2II) tc ng dng my pht nu sau khi ct MC u cc my pht (c thanh

gp in p my pht) hoc u hp b (MF-MBA) m dng s c vn tn ti (tc l s c xy ra bn trong hp b hoc my pht).

Kha in p thp cho php phn bit ngn mch vi qu ti v cho php bo v lm vic chc chn khi my pht c kch t bng chnh lu ly in t u cc my pht. Trong trng hp ny dng ngn mch s suy gim nhanh chng khi xy ra ngn mch ti u cc my pht. Trong mt s s ngi ta cn dng bin php m bo cho bo v tc ng chc chn l ch ly tn hiu in p thp sau khi rle dng in tr v do s suy gim dng ngn mch.

Dng in khi ng ca rle qu dng 50 (khi bo v qu dng c kho in p thp 27):

maxlvItv

at50K InK

KI = (1 -53) vi I l dng in lm vic ln nht qua cun th cp ca BI. lvmax

37

VI.2. Bo v chng ngn mch ngoi v qu ti MF: Qu ti gy pht nng cun dy stator c th do nhiu nguyn nhn nh my pht

in vn hnh vi h s cng sut thp, thnh phn cng sut phn khng vt qu mc cho php, c h hng trong h thng lm mt hoc h thng iu chnh in p lm cho my pht b qu kch thch. Cun dy rotor cmg c th b qu ti ngn hn trong qu trnh iu chnh in p khi my pht y ti cng sut tc dng.

Thi gian chu ng qu ti ca cc cun dy my pht c gii hn v ph thuc vo mc qu ti, kt cu ca my pht, h thng lm mt v cng sut ca my pht. Thng cc nh ch to cho sn quan h gia mc qu ti (I* = I/Im) vi thi gian qu ti cho php ca tng loi my pht in.

Ct

1MC

18RT

24RI 25RI

32LI2

20RT 19RT

27RI 26RI

Ct MCpd

BI

MF

Bo tn hiu Bo tn hiu

Hnh 1.30: S bo v chng qu ti v ngn mch ngoi

1MC

C nhiu nguyn l khc nhau c th c p dng thc hin bo v chng qu

ti cho cun dy ca my pht in: theo s o trc tip ca nhit cun dy, nhit ca cht lm mt hoc gin tip qua tr s dng din chy qua cun dy.

bo v chng ngn mch ngoi v qu ti cho my pht ngi ta c th s dng s hnh 1.30, thc cht y cng l mt bo v qu dng. Trong :

- 24RI, 18RT; 25RI, 20RT: chng qu ti v ngn mch i xng. - 26RI, 19RT; 27RI, 20RT: chng qu ti v ngn mch khng i xng. - 32LI2: b lc dng th t nghch ( nng cao nhy cho bo v, thng dng

cho cc my pht c cng sut ln).

VI.3.Tnh chn cc thng s ca rle:

VI.3.1. Bo v chng qu ti i xng 24RI, 18RT: Dng in khi ng ca 24RI:

Itv

mFatRI24K n.K

I.KI = (1-54)

Thi gian tc ng ca 18RT: t18RT = (7 9) sec (1-55)

VI.3.2. Bo v chng ngn mch i xng 25RI, 20RT:

Itv

mFmmatRI4K n.K

I.K.KI = (1-56) t20RT = t + t (1-57) max cc phn t ln cn

38

VI.3.3. Bo v chng qu ti khng i xng 26RI, 19RT: Dng in khi ng cho rle 26RI c chn theo hai iu kin: iu kin 1: IK26RI phi ln hn dng th t nghch lu di cho php I : 2cp

.IK I 2cpatK26RI = (1-58) - i vi my pht in turbine nc: I2cp = 5%.I mF- i vi my pht in turbine hi: I = 10%.I 2cp mF iu kin 2: Rle phi tr v sau khi ct ngn mch ngoi.

T hai iu kin trn v theo kinh nghim ngi ta chn:

I

mFRI26K n

I.1,0I = (1-59)

Thi gian tc ng ca 19RT thng c chn: t19RT = (7 9) sec (1-60)

VI.3.4. Bo v chng ngn mch khng i xng 27RI, 20RT: Dng khi ng ca 27RI chn theo cc iu kin sau: iu kin 1: Bo v khng c tc ng khi t mt pha trong h thng ni vi

nh my. iu kin 2: Bo v phi phi hp nhy vi cc bo v ln cn.

Trn thc t tnh ton dng th t nghch kh phc tp, theo kinh nghim ngi ta chn:

I

mFRI26K n

I)6,05,0(I = (1-61)

t gi tr dng khi ng tnh c ta c th chn c rle thch hp. Thi gian tc ng ca rle 20RT phi phi hp vi cc bo v ln cn:

t20RT = t + t (1-62) max cc phn t ln cnVI.3.5. Kim tra nhy ca bo v: nhy K ca bo v c tnh theo cng thc sau: n

KB

minNn I

IK = (1-63)

Tu vo nhim v ca bo v m gi tr nhy ca bo v phi t yu cu. Khi lm bo v chnh K 1,5 v khi ng vai tr lm bo v d tr K 1,2. n n

VI.4. Bo v dng th t nghch: (hnh 1.31) Dng in th t nghch c th xut hin trong cun dy stator my pht khi xy ra

t dy (hoc h mch mt pha), khi ph ti khng i xng hoc ngn mch khng i xng trong h thng.

Qu ti khng i xng nguy him hn qu ti i xng rt nhiu v n to nn t thng th t nghch 2 bin thin vi vn tc 2 gp hai ln tc ca rotor, lm cm ng trn thn rotor dng in ln t nng rotor v my pht. Dng th t nghch I2 cng ln th thi gian cho php tn ti cng b, v vy bo v chng dng in th t nghch c thi gian tc ng t ph thuc t l nghch vi dng I : 2

22

2

mF

2

1

KII

Kt

= (1-64)

39

LI2

30 2

51 51

t1 t1

Ct MC Cnh bo

52

HNH 1.31: Bo v dng in TTN cho my pht

10

I*2

0,6 0,5 0,4 0,3 0,2 0,1

t (sec) 10 10 t (sec)

0,4 0,3 10 0,1

I*2

10 8 6 4 2 0

IK1

IK2

t1

t2

HNH 1.32: C TNH THI GIAN PH THUC (A) V C LP C HAI

b) a)

Trong :

mF

2cp

II

-K , K l h s t l, K1 2 2 =

vi: - l hng s i vi tng loi rle c th. - I2cp: dng th t nghch cho php vn hnh lu di, n ph thuc vo chng loi

my pht, cng sut v h thng lm mt ca cun dy rotor. - I : dng in nh mc ca my pht. mF

- I*2: dng th t nghch tng i, I*2 = mF

Bo v c th c c tnh thi gian ph thuc t l nghch theo quan h t = f(I

2

II

2) (hnh 1.32a) hoc c tnh thi gian c lp 2 cp (hnh 1.32b): cp 1 cnh bo v cp 2 i ct my ct.

40

VII. BO V CHNG MT KCH T Trong qu trnh vn hnh my pht in c th xy ra mt kch t do h hng trong

mch kch thch (do ngn mch hoc h mch), h hng trong h thng t ng iu chnh in p, thao tc sai ca nhn vin vn hnh... Khi my pht b mt kch t thng dn n mt ng b stator v rotor. Nu h mch kch thch c th gy qu in p trn cun rotor nguy him cho cch in cun dy.

ch vn hnh bnh thng, my pht in ng b lm vic vi sc in ng E cao hn in p u cc my pht UF (ch qu kch thch, a cng sut phn khng Q vo h thng, Q > 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).

Gii hn pht nng mp li thp stator

Gii hn pht nng cun rotor

Gii hn pht nng cun stator

Min lm vic bnh thng P (MW)

- Q

+ Q

R

- jX

ZF

A

T

Zpt + jX

(IV)

(I) (II)

(III)

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 U v U l 180D BC

in p a vo cc b bin i dng sng (hnh sin sang hnh ch nht) S0 (hnh 1.34).

1 v S2 tng ng bng:

D.

BC.

1.

UU.aU = (1-65) D

.BC

.2

.UU.bU = (1-66)

Xd

0,5Xd 0

Min qu kch thch (E > 0, Q > 0)

Min thiu kch thch (E < 0, Q < 0)

E

I, Q H thng

U

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.

0

a)

b) c)

B

41

Gc lch pha gia 1 v 2 s c kim tra. ch bnh thng = 0.

U.

U 0, rle khng lm vic. Khi b mt kch t = 1800, rle s tc ng. Gc khi ng c chn khong 900. 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:

BC.

D.

BC.

U.aUU.b >> (1-67)

A.U

B.U C

.U

B.I

BC.U

BC.U BC

.Ub D

.U

BC.Ua

C.I

A.I BC

.I

BC.I

A B C BUG

BIG

U2

U1

aUBC

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 B Bo 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

BU

~

~ S1

S1

& RL S3 S4

-1

UD

IB

IC

bUBC

Ct MF

a)

U1

U2

-U1

t

S1

S2

S3 = - S1.S2

S4 = S3

k Tn hiu ct

t

t

t

t

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

b) c)

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 +jX

B (ngn mch) C (ct ngn mch)

A (bnh thng)

Z

0 R

1

2

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.

F1BI

BA1BU

P & Ct

MC

IC

2BI

IU U 1 2 3

Z< X<

4

2BU

HNH 1.37: S nguyn l ca bo v chng trt cc t (dao

ng in)

+jX

R

-ZF

N

X

M

XH

2

1

HNH 1.36: c tnh khi ng hnh thu knh pht hin dao

ng in

43

0 10 300 t (sec)

I*2,5

2

1,5

1

2 (cun dy stator)

1 (cun dy rotor)

HNH 1.39: Quan h gia mc qu ti v thi gian qu ti cho php

ca cc cun dy my pht

+jX

R 0

XdF

0,9XB

HNH 1.38: c tnh khi ng hnh ch nht pht hin dao

ng in

IX. bo v chng lung cng sut ngc Cng sut s i chiu t h thng vo my pht nu vic cung cp nng lng cho

Turbine (du, kh, hi nc hoc dng nc...) b gin on. Khi my pht in s lm vic nh mt ng c tiu th cng sut t h thng. Nguy him ca ch ny i vi cc my pht nhit in l Turbine s lm vic ch my nn, nn lng hi tha trong Turbine lm cho cnh Turbine c th pht nng qu mc cho php. i vi cc my pht diezen ch ny c th lm n my.

bo v chng ch cng sut ngc, ngi ta kim tra hng cng sut tc dng ca my pht. Yu cu rle hng cng sut phi c nhy cao pht hin c lung cng sut ngc vi tr s kh b (thng ch b p li tn tht c ca my pht trong ch ny). Vi cc my pht in Turbine hi, cng sut khi ng P bng: k

P = (0,01 0,03)Pk m (1-68) Vi cc my pht thu in v Turbine kh:

P = (0,03 0,05)Pk m bo nhy ca bo v cho cc my pht cng sut ln, mch dng in ca bo v thng c u vo li o lng ca my bin dng (thay cho li bo v thng dng cho cc thit b khc). Bo v chng cng sut ngc thng c hai cp tc ng: cp 1 vi thi gian khong (2 5) sec sau khi van STOP khn cp lm vic v cp th 2 vi thi gian ct my khong vi chc giy khng qua tip im ca van STOP (hnh 1.40).

m (1-69)

F

BI

BU

92 2II

& 2I

Cp 2

Cp 1 Ct

Van STOP

HNH 1.40: S nguyn l ca bo v chng cng sut ngc

44

X. Mt s s bo v my pht in dng rle s

X.1.S bo v my pht in cng sut trung bnh ( 1MW): Phng n 1:

S s dng cc bo v sau: - 51: bo v qu dng c thi gian. - 51N: bo v qu dng chng chm t c thi

gian. - 46: bo v dng th t nghch. - 49: rle nhit . Phng n 2: hnh 1.42 - 51: bo v qu dng c thi gian. - 51N: bo v qu dng chng chm t c thi

gian. - 46: bo v dng th t nghch. - 64: bo v chng chm t cun dy

rotor. - 32: rle nh hng cng sut. - 40: rle pht hin mt kch t my

pht in.

X..2.S bo v my pht in cng sut ln (> 1MW): (hnh 1.43)

S s dng cc bo v sau: - 51: bo v qu dng c thi gian. - 51N: bo v qu dng chng chm t

c thi gian. - 46: bo v dng th t nghch. - 32: rle nh hng cng sut. - 40: rle pht hin mt kch t my

pht in. - 49: rle nhit . - 87,87N: rle so lch chng chm pha

v chm t.

Hnh 1.41

52

64

51 32 46 40

51N

Hnh 1.42 - 27: rle in p thp. - 59: rle qu in p. - 81: rle tn s. - 64F: chng chm t cun dy rotor.

X.3. S bo v b MF-MBA: Phng n 1: hnh 1.44 - 87U: bo v so lch dc chung cho my pht v MBA tng p v MBA t dng. - 87T: bo v so lch dc MBA tng p v MBA t dng. - 51: bo v qu dng c chnh nh thi gian. - 51N: bo v qu dng chng chm t c thi gian. - 63: rle p sut dng cho MBA. - 71: rle hi dng cho MBA. - 64R, 64R2: bo v chng chm t 1 im v 2 im mch kch t. - 51N, 59N: bo v chng chm t cun dy rotor. - 87G: bo v so lch chng chm pha trong my pht. - 49S: bo v qu nhit cun dy stator. - 59: rle qu in p. - 81N: rle tn s. - 24: rle qu t. 78: rle kim tra ng b. - 40: rle pht hin mt kch t my pht in. - 21: rle khong cch - 32: rle nh hng cng sut..

45

Phng n 2: hnh 1.45

87

27

81

64F

27

59

51N

51

87N

32 46 40 49

52

Hnh 1.43

51N

59N

64R2 64R

E

46 21 71

63 51N

87T 87G

32

40

78 49S 81N

59 24 51

87U

51N

71 63

87T

HNH 1.44: S bo v b my pht v my bin p .

46

51N

63 87T

Kim tra cch in li

50 51

T1

87T

63

6,3kV

50 51

64

40

21 59

46 50 51 81

G

TE1

TU

220kV CSV

MC

MC

MC

Mch t ng kch thch

ng h o lng

HNH 1.45: S BO V B MY PHT V MY

87G

47

48

A. GII THIU CHUNG I. MC CH T BO V

Trong h thng in, my bin p l mt trong nhng phn t quan trng nht lin kt h thng sn xut, truyn ti v phn phi. V vy, vic nghin cu cc tnh trng lm vic khng bnh thng, s c... xy ra vi MBA l rt cn thit.

bo v cho MBA lm vic an ton cn phi tnh y cc h hng bn trong MBA v cc yu t bn ngoi nh hng n s lm vic bnh thng ca my bin p. T ra cc phng n bo v tt nht, loi tr cc h hng v ngn nga cc yu t bn ngoi nh hng n s lm vic ca MBA.

II. CC H HNG V TNH TRNG LM VIC KHNG BNH THNG XY RA VI MBA

II.1. S c bn trong MBA: S c bn trong c chia lm hai nhm s c trc tip v s c gin tip.

1. S c trc tip l ngn mch cc cun dy, h hng cch in lm thay i t ngt cc thng s in.

2. S c gin tip din ra t t nhng s tr thnh s c trc tip nu khng pht hin v x l kp thi (nh qu nhit bn trong MBA, p sut du tng cao...).

V vy yu cu bo v s c trc tip phi nhanh chng cch ly MBA b s c ra khi h thng in gim nh hng n h thng. S c gin tip khng i hi phi cch ly MBA nhng phi c pht hin, c tn hiu bo cho nhn vin vn hnh bit x l. Sau y phn tch mt s s c bn trong thng gp.

Hnh 2.1: Ngan mch nhieu pha trong cuon day MBA

c/ b/ a/

A C B A B C A C

II.1.1. Ngn mch gia cc pha trong MBA ba pha:

Dng ngn mch ny (hnh 2.1) rt him khi xy ra, nhng nu xy ra dng ngn mch s rt ln so vi dng mt pha.

53

II.1.2. Ngn mch mt pha:

Khong cach t trung tnh en iem chm (%

cuon day)

Dong s cap

Hnh 2.3: Dong ien chm at mot pha ca MBA noi at qua tong tr

100

I

IS

% ca dong 1xmaxI100

80

60

40

20

80 60 40 20 0

Dong chm

Ix IS

Z

Hnh 2.2: Ngan mch mot pha chm at

C th l chm v hoc chm li thp MBA. Dng ngn mch mt pha ln

hay nh ph thuc ch lm vic ca im trung tnh MBA i vi t v t l vo khong cch t im chm t n im trung tnh.

Di y l th quan h dng in s c theo v tr im ngn mch (hnh 2.3). T th ta thy khi im s c dch chuyn xa im trung tnh ti u cc MBA, dng in s c cng tng.

II.1.3. Ngn mch gia cc vng dy ca cng mt pha: Khong (7080)% h hng MBA l t

chm chp gia cc vng dy cng 1 pha bn trong MBA (hnh 2.4).

Hnh 2.4: Ngan mch gia cac vong day trong cung mot pha

Trng hp ny dng in ti ch ngn mch rt ln v mt s vng dy b ni ngn mch, dng in ny pht nng t chy cch in cun dy v du bin p, nhng dng in t ngun ti my bin p IS c th vn nh (v t s MBA rt ln so vi s t vng dy b ngn mch) khng cho bo v rle tc ng.

Ngoi ra cn c cc s c nh h thng du, h s dn, h b phn iu chnh u phn p ...

II.2. Dng in t ho tng vt khi ng MBA khng ti: Hin tng dng in t ho tng vt c th xut hin vo thi im ng

MBA khng ti. Dng in ny ch xut hin trong cun s cp MBA. Nhng y khng phi l dng in ngn mch do yu cu bo v khng c tc ng.

II.3. S c bn ngoi nh hng n tnh trng lm vic ca MBA: 3. Dng in tng cao do ngn mch ngoi v qu ti. 4. Mc du b h thp do nhit khng kh xung quanh MBA gim t ngt. 5. Qu in p khi ngn mch mt pha trong h thng in...

54

B. CC LOI BO V THNG S DNG BO V MBA

I. BO V CHNG S C TRC TIP BN TRONG MBA

I.1. Bo v qu dng in: I.1.1. Cu ch:

Vi MBA phn phi nh thng c bo v ch bng cu ch (hnh2.5). Trong trng hp my ct khng c dng th cu ch lm nhim v ct s c t ng, cu ch l phn t bo v qu dng in v chu c dng in lm vic cc i ca MBA. Cu ch khng c t trong thi gian qu ti ngn nh ng c khi ng, dng t ho nhy vt khi ng MBA khng ti...

I.1.2. Rle qu dng in: My bin p ln vi cng sut (1000-1600)KVA hai dy

qun, in p n 35KV, c trang b my ct, bo v qu dng in c dng lm bo v chnh, MBA c cng sut ln hn bo v qu dng c dng lm bo v d tr. nng cao nhy cho bo v ngi ta dng bo v qu dng c kim tra p (BVQIKU). i khi bo v ct nhanh c th c thm vo v to thnh bo v qu dng c hai cp (hnh 2.6). Vi MBA 2 cun dy dng mt b bo v t pha ngun cung cp. Vi MBA nhiu cun dy thng mi pha t mt b.

Hnh 2.5

CC

I.2. Bo v so lch dc: i vi MBA cng sut ln lm

vic li cao p, bo v so lch (87T) c dng lm bo v chnh. Nhim v chng ngn mch trong cc cun dy v u ra ca MBA.

IS

Hnh 2.6: S nguyn l bo v qu dng ct nhanh v c thi gian

+en rle tha hanh chung

-

+ RI RI RT

87T

Bo v lm vic da trn nguyn tc so snh trc tip dng in hai u phn t c bo v. Bo v s tc ng a tn hiu i ct my ct khi s c xy ra trong vng bo v (vng bo v l vng gii hn gia cc BI mc vo mch so lch).

55

RI RI RI

Hnh 2.7: S o nguyen l bo ve so lech MBA 2 cuon day

Th

en rle tha hanh chung

+

+

Rth

Khc vi bo v so lch cc phn t khc (nh my pht...), dng in s cp hai (hoc nhiu) pha ca MBA thng khc nhau v tr s (theo t s bin p) v v gc pha (theo t u dy). V vy t s, s BI c chn phi thch hp cn bng dng th cp v b s lch pha gia cc dng in cc pha MBA.

Dng khng cn bng chy trong bo v so lch MBA khi xy ra ngn mch ngoi ln hn nhiu ln i vi bo v so lch cc phn t khc.

Cc yu t nh hng nhiu n dng khng cn bng trong bo v so lch MBA khi ngn mch ngoi l:

6. Do s thay i u phn p MBA. 7. S khc nhau gia t s MBA, t s BI, nc chnh rle. 8. Sai s khc nhau gia cc BI

cc pha MBA. V vy, bo v so lch MBA thng

dng rle thng qua my bin dng bo ho trung gian (loi rle in c in hnh nh rle PHT ca Lin X) hoc rle so lch tc ng c hm (nh loi ZT ca Lin X).

Hnh 2.8 cho s nguyn l mt pha ca bo v so lch c dng my bin dng bo ha trung gian. Trong my bin dng bo ha trung gian c hai nhim v chnh:

9. Cn bng cc sc t ng do dng in trong cc nhnh gy nn tnh trng bnh thng v ngn mch ngoi theo phng trnh:

WN

IIIT

IIT

IIIS

IIS

RI

Hnh 2.8: S o nguyen li bo ve so lech co dung may bien dong bao hoa trung gian

WlvTWlvSWcbI WcbII

WN

IIT(WcbI + WlvS) + IIIT(WcbII + WlvS) = 0 10. Nh hin tng bo ha ca

mch t lm gim nh hng ca dng in khng cn bng Ikcb (c cha phn ln dng khng chu k).

56

I.3. Bo v MBA ba cun dy dng rle so lch c hm: Nu MBA ba cun dy ch c cung cp ngun t mt pha, hai pha kia ni

vi ti c cc cp in p khc nhau, rle so lch c dng nh bo v MBA hai cun dy (hnh 2.9a). Tng dng in th cp hai BI pha ti s cn bng vi dng in th cp BI pha ngun trong iu kin lm vic bnh thng. Khi MBA c hn mt ngun cung cp, rle so lch dng hai cun hm ring bit b tr nh hnh 2.9b.

Nguon

c ham

b/ c lviec 87

co the co nguon

ti

c lviec

a/ c ham

87

Nguon

Hinh 2.9: S o bo ve so lech co ham

MBA ba cuon day

I.4. Bo v chng chm t cun dy MBA: i vi MBA c trung tnh ni t, bo v chng chm t mt im trong

cun dy MBA c th c thc hin bi rle qu dng in hay so lch th t khng. Phng n c chn tu thuc vo loi, c, t u dy MBA.

Khi dng bo v qu dng th t khng bo v ni vo BI t trung tnh MBA, hoc b lc dng th t khng gm ba BI t pha in p c trung tnh ni t trc tip (hnh 2.10). i vi trng hp trung tnh cun dy ni sao ni qua tng tr ni t bo v qu dng in thng khng nhy, khi ngi ta dng rle so lch nh hnh 2.12a. Bo v ny so snh dng chy dy ni t IN v tng dng in 3 pha (IO). Chn IN l thnh phn lm vic v n xut hin khi c chm t trong vng bo v. Khi chm t ngoi vng bo v dng th t khng (IO tng dng cc pha) c tr s bng nhng ngc pha vi dng qua dy trung tnh IN.

+ RI

IN

RT RI

+ +

Hnh 2.10: S nguyn l bo v chng chm t MBA bng bo v qu dng in

57

Cc i lng lm vic v hm nh sau: NI I lv &= (2-1) (2-2) ; III oh1 N &&& += III oh2 N &&& = Cc dng in hm c phi hp vi nhau v ln to nn tc dng hm theo quan h: )IIIIk(I 0N0Nh &&&& += (2-3)

Vi : dng dy ni t; k: hng s t l. NKho st cch lm vic ca rle so lch th t khng:

I& ;IIII CBAo &&&& ++ Khi chm t bn ngoi:

ngc pha vi v bng nhau v tr s: .

oI& NI&N

Gi thit chn k=1, lc IIo && =

,I2IIIII ,II NNNNNN hlv &&&&&& =+== .2II lvh =

Hnh 2.11: S o nguyen ly bo ve so lechth t khong co ham

lvI&

h2I&

h1I& H2

H1

Cuon lviec

I

NI& oI&

Khi chm t bn trong, ch c thnh phn qua trung tnh: ; 0I0 =&

;II Nlv && = 0.0I0IIh =+= &&

&&& =

NNQua phn tch trn ta thy, khi

chm t bn trong thnh phn hm khng xut hin. Nh th ch cn dng

chm t nh xut hin khi chm t trong vng bo v (vng gii hn gia cc BI), bo v s cho tn hiu tc ng. Ngc li khi chm t bn ngoi tc ng hm rt mnh.

Nu cun sao MBA ni t qua tng tr cao, rle so lch 87N c th khng nhy tc ng, ngi ta c th thay bng rle so lch chng chm t tng tr cao 64N (hnh 2.12b). Rle so lch tng tr cao c mc song song vi in tr R c tr s kh ln.

Trong ch lm vic bnh thng hay ngn mch ngoi vng bo v (vng gii hn gia cc BI), ta c:

(2-4) NooNu b qua sai s ca BI, ta c dng in th cp chy qua in tr R bng

khng v in p t ln rle cng bng khng, rle s khng tc ng.

III

Khi chm t trong vng bo v, lc I0 = 0 nn I0 = IN ton b dng chm t s chy qua in tr R to nn in p rt ln t trn rle, rle s tc ng.

a/

IC

IB

IA

Z

IO

IN 87N Rle so lech th t khong

b/

64N

R RL

Z

IO

IN

Hnh 2.12: S o nguyen ly bo ve so lech th t khong

58

I.5. Bo v MBA t ngu: Bo v chnh MBA t ngu cng l bo v so lch. Bo v da trn c s nh

lut Kirchoff, l tng vect dng in vo ra cc nhnh ca i tng bo v bng khng (ngoi tr trng hp s s).

b/ c bT

a

87 87 87

C B

A

87

a/

Hnh 2.13: Bo ve so lech MBA t ngau Bo v so snh dng in thuc hai nhm: nhm BI ni vo u cc MBA v

nhm BI ni vo trung tnh MBA. Nu bo v ch dng mt bin dng t trung tnh MBA, cc BI t u cc MBA c ni thnh b lc th t khng v ni n mt rle, khi to thnh bo v so lch chng chm t bn trong MBA t ngu (hnh 2.13a).

Trong trng hp cun th ba (cun tam gic) khng ni vi ti, my bin p t ngu dng lin kt h thng siu cao p v cao p. S bo v c th thc hin nh hnh 13b, cc BI c phi hp trn mi pha gn trung tnh (im cui ca cun dy MBA) v dng 3 rle, lc bo v p ng chng ngn mch nhiu pha v mt pha bn trong cun dy chnh MBA t ngu. S ny khng p ng khi s c cun dy th ba, bo v cho cun dy th ba trong trng hp ny ngi ta thng dng bo v qu dng in.

87T

Hnh 2.14: S o nguyen ly bo ve so lech MBA t ngau

Bo v tt c cc cun dy MBA t ngu tng t nh bo v cho MBA ba cun dy (hnh 2.14).

II. BO V CHNG S C GIN TIP BN TRONG MBA

C cc loi bo v sau: Rle kh (BUCHHOLZ). Bo v qu nhit. Rle pht hin tc tng, gim p sut du. Bo v dng du b iu p. S dng loi no l tu quan im ca nh sn xut v tu tng c my.

Thng c dng ph bin l rle kh (hnh 2.15).

59

II.1. Rle kh Buchholz (96B): Rle hot ng da vo s bc hi ca du my bin p khi b s c v mc

h thp du qu mc cho php.

a)

n bnh du ph

T thng du MBA

Phao 1

Phao 2

Bnh du ph

Thng MBA

96B

b)

Hnh 2.15: Nguyn l cu to (a) v v tr b tr trn MBA ca rle hi Rle kh c t trn on ng ni t thng du n bnh dn du ca

MBA. Rle c hai cp tc ng gm c hai phao bng kim loi mang bu thu tinh c tip im thu ngn hay tip im t. ch lm vic bnh thng trong bnh y du, cc phao ni l lng trong du, tip im rle trng thi h. Khi kh bc ra yu (v d v du nng do qu ti), kh tp trung ln pha trn ca bnh rle y phao s 1 xung, rle gi tn hiu cp 1 cnh bo. Nu kh bc ra mnh (chng hn do ngn mch cun dy MBA t trong thng du) lung kh di chuyn t thng du ln bnh dn du y phao s 2 xung gi tn hiu i ct my ct ca MBA.

Mt van th c lp trn rle: Khi th nghim rle, lp my bm khng kh nn vo u van th. M kha van, khng kh nn bn trong rle cho n khi phao h xung ng tip im.

Mt nt nhn th kim tra s lm vic ca 2 phao. Khi nhn nt th n na hnh trnh, s tc ng c kh cho phao trn h xung (lc ny c 2 phao ang nng ln v rle cha y du) ng tip im bo hiu (cp 1) ca phao trn. Tip tc nhn nt th n cui hnh trnh, s tc ng c kh cho phao di cng b h xung (do phao trn h xung ri) ng tip im m my ct (cp 2) ca phao di.

Da vo thnh phn v khi lng hi sinh ra ngi ta c th xc nh c tnh cht v mc s c. Do trn rle hi cn c thm van ly hn hp kh sinh ra nhm phc v cho vic phn tch s c. Rle hi tc ng chm thi gian lm vic ti thiu l 0,1s; trung bnh l 0,2s.

II.2. Rle bo v qu nhit cun dy MBA (26W): Nhit nh mc my bin p ph thuc ch yu vo dng in ti chy qua

cun dy MBA v nhit ca mi trng xung quanh. Tu theo tng loi cng nh cng sut nh mc ca MBA m di nhit cho php ca chng c th thay i, thng thng nhit ca cun dy di 95oC c xem l bnh thng.

Thit b ch th nhit cun dy c trnh by nh hnh 2.39 (tng t thit b ch th nhit du).

60

o nhit cun dy MBA ngi ta thng dng thit b loi AKM 35, y l thit b s dng in tr nhit c phn t t nng c cp in t bin dng pha cao v h my bin p. Rle nhit cun dy gm bn b tip im (mi b c mt tip im thng m, mt tip im thng ng vi cc chung) lp bn trong mt nhit k c kim ch th.

Hnh 2.40: Thit b ch th nhit cun dy C cu rle gm: ch th quay ghi s o, mt b phn cm bin nhit, mt

ng mao dn ni b phn cm bin nhit vi c cu ch th. Bn trong ng mao dn l cht lng c nn li. S co gin ca cht lng trong ng mao dn thay i theo nhit m b cm bin nhn c, tc ng ln c cu ch th v bn b tip im. ng thi, tc ng ln c cu ch th v cc tip im, cn c mt in tr t nng. Cun dy th cp ca mt my bin dng in t ti chn s my bin p c ni vi in tr t nng. chnh nh cho phn t t nng, ngi ta s dng mt bin tr t t iu khin cnh my bin p. Tc dng ca in tr t nng (ty theo dng in qua cun dy my bin p) v b cm bin nhit ln c cu o cng cc b tip im s tng ng vi nhit im nng, nhit ca cun y.

Thit b ch th nhit cun dy

C 4 vt iu chnh nhit t tr s tc ng cho 4 b tip im. Ty theo thit k, cc tip im rle nhit c th c ni vo cc mch, bo hiu s c nhit cun dy cao, mch t ng m my ct c lp my bin p, mch t ng khi ng v ngng cc qut lm mt my bin p.

Rle nhit cun dy hot ng 2 cp: Cp 1: Khi nhit cun dy MBA 115oC s bo ng bng tn hiu n

ci. Cp 2: Khi nhit cun dy MBA l 120oC th bo ng bng tn hiu n

ci v tc ng i ct my ct c lp my bin p ra khi li. Ngoi ra, rle nhit cun dy MBA cn c tc dng a cc tn hiu i

iu khin h thng lm mt cho MBA. V d i vi MBA lm mt bng qut thi th h thng qut mt s lm vic khi nhit cun dy MBA t n mt trong cc gi tr 750C cun cao, 800C cun h v 600C i vi nhit du. H thng ny s dng khi nhit cun dy v du MBA gim 100C di cc gi tr khi ng trn.

II.3. Rle nhit du (26Q): o nhit lp du trn s dng hai ng h. Mt ng h nhit du

bo tn hiu 800C v mt ng h nhit du tc ng ct my ct 900C. Cc ng h ny s dng nguyn l cm ng nhit . Phn t cm ng nhit c b trong hp nh v c t gn nh ca thng du ca my bin p.

61

Tn hiu ra

Dng ti

Phn t cm ng nhit Phn t sinh nhit

nh my bin p

Hnh 2.38: Cch lp rle nhit trong my bin p

Rle nhit du gm c c cu ch th quay ghi s o, mt b phn cm

bin nhit, mt ng mao dn ni b phn cm bin nhit vi c cu ch th. Bn trong ng mao dn l cht lng (dung dch hu c) c nn li. S co gin ca cht lng (trong ng mao dn) thay i theo nhit m b phn cm bin nhit nhn c, s tc ng c cu ch th v cc tip im. Cc tip im s i trng thi m thnh ng, ng thnh m khi nhit cao hn tr s t trc. B phn cm bin nhit c lp trong l tr bc kn, pha trn np my bin p, bao quanh l tr l du, o nhit lp du trn cng ca my bin p. Thng dng nhit k c 2 (hoc 4) vt iu chnh nhit c th t sn 2 (hoc 4) tr s tc ng cho 2 (hoc 4) b tip im ring r lp trong nhit k. Khi nhit cao hn tr s lp t cp 1, rle s ng tip im cp 1 bo tn hiu s c nhit du cao ca my bin p. Khi nhit tip tc cao hn tr s cp 2, rle s ng thm tip im cp 2 t ng ct my ct, ct in my bin p, ng thi cng c mch i bo hiu s c ct do nhit du cao (B phn ch th nhit nh hnh 2.39). Trong :

1. B phn cm bin nhit. 2. ng mao dn (capillary tubo). 3. Kim ch th nhit . 4. Hai vt iu chnh nhit hai b tip im . 5. Hai b tip im rle nhit du . Nhit mi trng s dng : -100C n 700 C. Thang o : -200C 0 +1300C. Thang iu chnh : -200C 0 +1300C. Sai s ca tr s o c : + 30C. Khong sai bit tc ng ca tip im : 10-14.

II.4. Cu to rle mc du ti my bin p (33):

Thit b ch th mc du thn my

du

ng du ni n thnmy

ng th c bnh silicagel

Thit b ch th mc du b i nc

ng du ni n b i nc

Hnh 2.41: V tr lp rle mc du ti my bin p

62

Rle mc du gm hai b tip im lp bn trong thit b ch th mc du, my bin p c b i nc in p c ti (b iu p di ti) th thng gin n du c chia lm hai ngn (hnh 2.41). Ngn c th tch chim phn ln thng gin n, c ni ng lin thng du qua rle hi n thng chnh my bin p ( c th tch gin n du cho my bin p). Ngn c th tch chim phn nh hn nhiu ca thng gin n, s c ni ng lin du n thng cha b iu p di ti. Thng chnh my bin p v thng b i nc c thit k ring r, khng c lin thng du vi nhau. V vy, c hai thit b ch mc du lp ti hai u thng gin n o mc du ca hai ngn thit b ch th mc du my bin p v thit b ch th mc du b iu p di ti.

7

8

9

4

1 2

5

6

3

Hnh 2.42: Cu to ca thit b ch th mc du 1. V my. 6. Kim ch th. 2. Vng m . 7. Mt ch th.

3. Phao. 8. Thanh quay. 4. Nam chm vnh cu. 9. Trc quay.

5. Nam chm vnh cu. C cu ca thit b ch th mc du gm hai b phn (hnh 2.42): B phn iu

khin v b phn ch th. B phn iu khin c mt phao (3), thanh quay (8) trc quay (9) c lp nam chm vnh cu (4). B phn iu khin lp trn v my (u thng gin n) c vng m. B phn ch th gm kim ch (6) lp trn trc mang mt nam chm vnh cu (5). B phn ch th c lm bng nhm trnh b nh hng t trng nam chm v chng nh hng ca nc.

Khi mc du nng h th phao (3) nng h theo. Chuyn ng nng h ca phao c chuyn thnh chuyn ng quay ca trc (9) nh thanh quay (8). Khi quay t trng do nam chm (4) s iu khin cho nam chm (5) quay sao cho hai cc khc tn (N v S) ca hai nam chm i din nhau (hai cc cng tn c lc y, hai cc khc tn c lc ht nhau). Do vy kim ch th quay theo nam chm (5), ghi c mc du trn mt ch th. B phn ch th cng tc ng ng m cc tip im rle mc du a tn hiu vo mch bo ng hoc mch ct ty theo tng thit k.

II.5. Bo v p sut tng cao trong my bin p (63): Rle bo v d phng cho my bin th lc, ch danh vn hnh l R.63. Khi

c s c trong my bin p, h quang in lm du si v bc hi ngay, to nn p sut rt ln trong my bin p. Thit b an ton p sut lp trn np thng chnh my bin p s m rt nhanh (m ht van khong 2ms) thot kh du t thng chnh MBA ra mi trng ngoi, p sut trong thng chnh s gim. Trong thit b an ton p sut c gn rle p sut.

63

S khi ca bo v R.63 ti trm:

Ct my ct Tn hiu t BI

Hnh 2.43: S khi bo v R.63 tnh trng lm vic bnh thng, van a b nn bi l xo nn lm kn thng

chnh my bin p. Khi c s c bn trong thng chnh my bin p th p sut trong thng chnh tng cao s ln hn p lc nn ca l xo, van a s chuyn ng thng ln, lm h thnh khe h xung quanh chu vi van a. Kh s thot ra ti khe h vng m, lm gim p sut trong thng. Khi van a di chuyn ln th cng tc ng ln ci ch th c kh bung ln, ng thi tc ng tip im rle p sut gi tn hiu ti mch bo ng v t ng ct my ct c lp my bin p ra khi li in. Khi p sut tr li bnh thng, mun ti lp li MBA th phi nhn ci ch th c kh ( b bung ln) v v tr c, ng thi t li rle p sut bng nt nhn.

II.6. Bo v p sut tng cao trong b i nc my bin p (R.63 OLTC):

Rle bo v tc ng theo p sut thng iu p di ti my bin p lc, l bo v d phng cho my bin p. Ch danh vn hnh trn s bo v l R.63 OLTC (On Load Tap Changer).

Cu to v nguyn l vn hnh ca rle tng t nh R.63 ni trn. Khi c s c bn trong thng i nc my bin p th rle s tc ng v t ng ct my ct c lp MBA ra khi li in.

S khi ca bo v R.63 OLTC ti trm:

R.63 Ct my ct Tn hiu t BI

Hnh 2.44: S khi bo v R63 OLTC

Mun ti lp li MBA sau khi rle tc ng phi t li Rle kha trung gian R86.

II.7. Rle kha trung gian (86):

Rle kha trung gian R.86 thng c dng l loi kiu MVAJ-21 nh ch to GEC ALSTOM.

c im v ng dng ca rle nh sau: Thit b ny dng ngt mch in vi an ton cao, c bit chng c

th dng ngt mch in hoc iu khin cc hot ng ng ngt do tn hiu c gi ti t cc rle khc. Rle ny c th hot ng hai ch tc thi hoc c thi gian tr hon.

Rle MVAJ c kh nng dp tt c s phng in do in dung. Rle MVAJ l loi thit b bo v dng gim st s hot ng ca cc

loi rle bo v khc. Nguyn tc hot ng: Rle MVAJ-21 ch hot ng khi cc rle khc (c lin quan) lm vic.

Khi rle bo v chnh ca thit b hot ng th cng ng thi tc ng rle R.86

64

lm vic. R.86 hot ng s c lp ngun iu khin ca cc rle iu khin khc. Mun ti lp li s lm vic bnh thng ca mch iu khin cc thit b th phi t li R.86.

Hnh 2.17: S o nguyen

ly bo ve qua ti

Nguon

- RT

Th + +

RI

III. BO V CHNG NGN MCH NGOI V QU TI

III.1. Bo v qu ti (BVQT): C chc nng bo tn hiu qu ti MBA. Dng bo v qu dng in. MBA

hai dy qun bo v c b tr pha ngun (hnh 2.17), my bin p ba dy qun bo v qu ti c th b tr hai hoc c ba dy qun. Bo v qu ti ch b tr mt pha v i bo tn hiu sau mt thi gian nh trc.

Tuy nhin rle dng in khng th phn nh c ch mang ti ca MBA trc khi xy ra qu ti. V vy i vi MBA cng sut ln ngi ta s dng nguyn l hnh nh nhit thc hin bo v chng qu ti.

Bo v loi ny phn nh mc tng nhit nhng thi im kim tra khc nhau trong my bin p v tu theo mc tng nhit m c nhiu cp tc ng khc nhau: cnh bo, khi ng cc mc lm mt bng tng tc tun hon ca khng kh hoc du, gim ti my bin p.

Nu cc cp tc ng ny khng mang li hiu qu v nhit my bin p vn vt qu gii hn cho php v ko di qu thi gian quy nh th my bin p s c ct ra khi h thng.

III.2. Bo v dng in tng cao do ngn mch ngoi: Thng thng ngi ta dng bo v qu dng in. V nguyn tc vi MBA

ba cun dy khi c ba cp in p u c th c ngun cung cp nn t mi cp in p mt b.

65

Vi MBA ba cun dy v MBA t ngu mt trong cc b bo v dng in cc i thng l bo v c hng ( m bo tnh chn lc gia cc bo v). nng cao nhy ngi ta dng bo v dng in th t nghch (BVI2) km theo mt rle dng in c kim tra p. Cc bo v chng dng in tng cao do ngn mch ngoi dng lm bo v d tr cho bo v chnh ca MBA khi ngn mch nhiu pha MBA, n cn lm bo v d tr cho bo v ca cc phn t ln cn nu iu kin nhy cho php.

LI2

RU RT

+

T BU noi vao thanh gop TA

-

+ + +

LI2

RI RI

TA

+ +

RW RI

LU2

+

RU

RT

HA

CA

T BU noi vao thanh gop TA

T BU noi vao thanh gop CA

+

RI RW

-

Hnh 2.18 cho s nguyn l bo v chng ngn mch ngoi cho my bin p t ngu. Trong rle nh hng cng sut (RW) ch tc ng khi hng cng sut ngn mch truyn t my bin p n thanh gp cao p, cn theo chiu ngc li th khng tc ng.

Hnh 2.18: S o nguyen ly bo ve chong ngan mch ngoai

C. TNH TON BO V RLE CHO MBA C s tnh chn bo v rle cho MBA: Cn phi bit cc thng s ca MBA do nh ch to cung cp trn nhn

my hoc trong cc catalogue: V d vi MBA ba pha hai cun dy:

Thng s sn xut Um cun

dy

Loi MBA

C iu chnh in p

SBm

Uc Uh

Un(%) Pn Po Io(%)

Dng ngn mch ln nht, nh nht xut hin trong cc dng ngn mch. Cc thng s, c tnh ca my bin dng in, bin in p. Cc yu cu bo v rle ca MBA.

66

SHT

N1

51 50

U1

U2

N2

I. BO V QU DNG IN I.1. Cu ch:

Cu ch c chn theo iu kin sau: Icc Kat.Im (2-5) Vi Im: dng lm vic nh mc pha t cu ch; Kat h s an ton ly bng 1,2.

S liu tham kho t cu ch cho MBA cp in p 11 Kv Cng sut MBA Cu ch

S (KVA) I (A) Im tct (s) 100 5,25 16 3 200 10,5 25 3 300 15,8 36 10 500 26,2 50 20 1000 52,5 90 30

I.2. Bo v qu dng in: Chn my bin dng in cho bo v. nh mc th cp ca BI c tiu chun ho l 5A hoc 1A. BI c chn c dng nh mc s cp bng hay ln hn dng nh mc

cun dy MBA m n c t. i vi MBA hai cun dy dng nh mc s cp v th cp MBA ph thuc cng sut nh mc ca MBA v t l nghch vi in p. i vi MBA ba cun dy dng nh mc ph thuc vo cun dy tng ng.

m

mm lv

B

B

U3SI = (2-6)

Vi SBm: cng sut nh mc ca my bin p. UBm: in p nh mc ca MBA.

67

I.2.1. Bo v ct nhanh: Xc nh dng ngn mch s cp cc i chy qua ch t bo v khi ngn

mch ngoi (INngmax) ti im N1 trong hnh.

)x(x3

UIIht

(3)ngmax

B

1N1N +== (2-7)

xB

xht

N1(3)

Trong : xB: in khng ca MBA, m

mB

2BN

B 100.S%.UUx =

x : in khng ca h thng. ht Dng in khi ng bo v:

Nngmaxatk .IKI = (2-8) vi Kat l h s an ton, K = (1,3-1,4) at Dng khi ng th cp ca rle :

I

Nngmax(3)sat

kR n.I.KK

I = (2-9) (3)sK : h s k n s ni dy ca BI.

Kim tra nhy ca bo v ng vi tnh trng ngn mch hai pha trn cc MBA pha ni vi ngun trong ch lm vic cc tiu ca h thng (imN2).

2I

IKK

Nminn = (2-10)

Thi gian bo v: t = 0sec. I.2.2. Bo v qu dng c thi gian: Xc nh dng khi ng ca bo v:

max lvtv

mmatk .IK

.KKI = (2-11) y dng Ilv max dng lm vic max qua ch t bo v. Trong trng hp

khng bit c th ly Ilv max = IBm . Vi MBA ba cun dy dng Ilv max ly tng ng ca tng cun. Kat: h s an ton (1,1 - 1,2). Kmm: h s m my (1,3 - 1,8). Ktv: h s tr v (0,85 - 0,9).

Dng khi ng ca rle: I

k(3)s

k n.IK

I R = (2-12)

Kim tra nhy ca bo v: I

IKk

minn

N1= (2-13) Yu cu khi lm bo v chnh. :1,5Kn y IN1min dng ngn mch nh nht qua bo v khi ngn mch trc tip

cui vng bo v (im N1). Dng ngn mch tinh ton l dng ngn mch hai pha nn:

)x.(x3

UI21

1(2)N1 +

=

68

Trong : - x1 1 1B 1ht - x

:in khng th t thun tng n im ngn mch, x = x + x . 2 : in khng th t nghch tng n im ngn mch, x2 = x2B + x2ht.

Yu cu : khi lm bo v d