Bao Ve Cac Phan Tu Chinh Trong HTD

7/31/2019 Bao Ve Cac Phan Tu Chinh Trong HTD

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A. GII THIU CHUNG V MY PHTIN

My pht in (MF) l mt phn t rt quan trng trong h thng in (HT), slm vic tin cy ca cc MF c nh hng quyt nh n tin cy ca HT. V vy, ivi MFc 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 cccun 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 hhng v tnh trng lm vic khng bnhthng ca MF

I.1. Cc dng hhng:

- 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 chng c, ...

II. Cc bo v thng dng cho MF

Tu theo chng loi ca my pht (thuin, nhit in, turbine kh, thuin tchnng...), cng sut ca my pht, vai tr ca my pht v s ni dy ca nh my invi 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 nhi vi cc thit

bin 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 vclp 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 dngcc 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 mtng b, ...

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B. CC BO V RLE CHO MY PHTIN

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 cundy stator my pht. S thc hin bo v nh hnh 1.1.

52

1BI

MF

2BI

87G

2RI

+

C tMC

+

4Rth

-

+

5RT

MF

MC

Bo tn hiu

Rf

Rf

+

3RI

Bo tn hiu tmch th

1RI

+

Hnh 1.1: S bo v so lch dc cun statorMF; 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 thy l cc cun dy stator ca MF, on thanh dn tu cc MFn my ct.

I.2. Nguyn l lm vic:

BVSLD hot ng theo nguyn tc so snh lch dng in gia hai u cun dystator, 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 vocc rle 1RI, 2RI:

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I

NnI

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

Hnh 1.2: th vctca dngin trong mchBVSLD

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

a)

ISL = IKCBT < IKR

I2TI1 T

b)

I1T

I2T

ISL KRI

N

In

I

Trong :- IN: dng in ngn mch.- nI: t s bin dng ca BI

Bo v tc ng i ct 1MCng thi a tn hiu i n b phntng dit t (TDT).

Trng hp t mch th caBI, 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 stc 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 ngi 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, KKCKc th ly t 1 n 2 tu theo bin php c s dng nng cao nhy cabo v.

- K : h s tnh n sng nht ca cc BI (K = 0,51).n n- f : sai s tng i ca BI, fi i c th ly bng 0,1 (c kn 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 khngc tc ng khi t mch thBI.Lc dng vo rle 1RI, 2RI: (gi s MF ang lm vic chnh mc)

I

mFnI

ISL = (1-7)

Dng khi ng ca bo v:

mFI

at InK

I = (1-8)KB

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

Dng in khi ng ca rle:

I

KB)3(

nI.KI = (1-10)KR

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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

minN

II

(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 scp ca rle 3RI phi ln hn dng khng cn bng cc i khingn mch ngoi vng bo v. Nhng trong tnh ton th iu kin n nh nhit ca rle lquyt nh. Theo kinh nghim c th chn dng khi ng cho 3RI:

I = 0,2.I (1-12)KS(3RI) mFTa tnh c IKRca 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 lnthong qua lm cho bo v tc ng nhm do vy phi chn thi gian tc ng ca 5RT thomn iu kin:

t5RT > t (1-13)ct Nngoi

t5RT = tctNng + t (1-14)Trong :

- tcNngt- 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 statormy pht.

RI

Vngbov

I1S

I2S

I1T

I2T

ILV

BIH

IH

BILV

1BI

2BI

Hnh 1.3: Bo v so lch dngin c hm cundy stator MF

- Bo v khng tc ng khichm chp gia cc vng dy trong

cng 1 pha hoc khi xy ra chm t1 im trong cun dy phn tnh. tng nhy ca bo v so

lch ngi ta c th s dng rle solch 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)- Dngin 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: I1TI2TISL = 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 = -I

2T

IH = I1T - I2T 0I = ILV 1T + I2T 2.I1T > IH (1-19)

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bo v s tc ng. Nhn xt:

- Bo v hot ng theo nguyn tc so snh dng in gia I v ILV H, nn nhyca bo v rt cao v khi xy ra ngn mch th bo v tc ng mt cch chc chn vi thigian 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 lchhm tc ng nhanh (hnh 1.4).

ch lm vicbnh thng, dng in thcp I

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

RL2

RL1

ILVI1S

I1TI2T

I2S

BIG

IHILV

RH/2

ULV

RH/2UH

D1 D2

RL1

RL2

n RGu ra

A

B

C RLV

ILV

BIG

CL

1T v I2T ca cc nhmbin dng 1BI, 2BI chy quain tr hm RH, to nnin p hm UH, cn hiudng th cp (dng so lch)ISL chy qua bin dng trung

gian BIG, cu chnh lu CLv in tr lm vic RLV tonn 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 RL1lm rle ny tc ng ng tip im RL1 li. Dng in lm vic sau khi nn chy qua rleRL , RL2 2ng tip im li, rle ct u ra sc cp ngun thao tc qua hai tip imni tip RL v RL1 2i ct my ct u cc my pht. Ngoi ra, ngi ta cn dng rle solch tng trcao bo v so lch my pht in (hnh 1.5).Rle so lch RU trong s ctng trkh ln s tc ng theo in p so lch USL, ch lm vic bnh thng v khingn 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, phnbin p trong mch nh hnh 1.5b.

Hnh 1.5: Bo v so lch dng rle t ng trcao cho MFa) S nguyn l b) Mch in ng tr v phn bin p trong ch lm vic bnh thngc) nhm 2BI b bo ho khi ngn mch ngoi v hon ton d) khi c ngn mch trong.

1BI IN 2BI N

1BIUSL

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

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Tr sin p t ln rle so lch RU ph thuc vo quan h gia cc in trR1 vR . in trR , R2 1 2 gm in trcun dy th cp v dy dn ph ni gia hai nhm bindng 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 RUhnh 1.5c:

I

21"N

1SL n)RR.(I

U+

= (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 trmch 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 bin p nh hnh 1.5d. Gi trin 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 hnmin{USL1; U }, ngha l:SL2

I21

"

Natn )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 cacc 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 camy 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.

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I.5. Bo v khong cch (21):

i vi cc MF cng sut ln ngi ta thng s dng bo v khong cch lmbo v d phng cho BVSL (hnh 1.6a).

Hnh 1.6: S nguyn l (a); c tnh thi gian (b) v c tuy nkhi 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 khingn mch ngoi MBA, vng th nht ca bo v khong cch c chn bao gm inkhng ca MF v khong 70% in khng ca MBA tng p ( bo v hon ton cun hca 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 thc dng vng trn vi tm gc to hoc hnh bnh hnh vi nghing ca cnh bn

bng nghing ca vctin 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 cady qun. C th xy ra chm chp gia cc vng dy trong cng mt nhnh (cun dyn) hoc gia cc vng dy thuc hai nhnh khc nhau trong cng mt pha, dng introng cc vng dy b chm chp c tht n tr s rt ln. i vi my pht in mcun dy stator l cun dy kp, khi c mt s vng dy chm nhau sc in ng cm ngtrong hai nhnh s khc nhau to nn dng in cn bng chy qun trong cc mch vng sc v t nng cun dy c th gy ra h hng nghim trng. Trong nhiu trng hp khixy ra chm chp gia cc vng dy trong cng mt pha nhng BVSLD khng th pht hinc, v vy cn phi t bo v so lch ngang chng dng s c ny.

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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 ccvng dy trong cng mt pha thng km theo chm v, nn bo v chng chm t tcng (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, ura cc nhnh a ra ngoi nn vic bo v so lch ngang tng i d dng. Ngi ta c thdng 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 ccnhnh 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

BILVBIH

IH

CtMC

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 ngLVKhi xy ra chm chp gia cc vng dyca hai nhnh khc nhau cng mt pha,gi thit ch my pht cha mangti, ta c: I1T = -I2T

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 cundy stator, th cp ca BI ni qua b lc sng hi bc ba L3f dng gim dng khng cn

bng i vo rle.

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Ct 1MC

Lf3

+Bo tn

hiu

RI

RT Rth

+

-

C

CBA

O2

O1

1

2

T

Hnh 1.9: S bo v so lchngang cho cc pha MF, s tnhton (a) v theo m s(b)

87

BI

a)

b)

CN: cu ni, bnh thng CN v tr 1 v bo v tc ng khng thi gian. Khi mypht 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 th2 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 gia2 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 hnhnhng phi chuyn cu ni sang v tr 2 trnh trng hp bo v tc ng nhm khi ngnmch 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 vc 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, nnthng xc nh theo cng thc kinh nghim:

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

I

KB

nI

I = (1-31)KR

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 tim th nht mch kch t th cu ni CN c chuyn sang v tr 2. Thi gian tc ng carle RT c xc nh nh sau:

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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 kcht.

- 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 cundy stator. Tuy nhin n khng th thay th hon ton cho BVSLD c v khi ngn mchtrn 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 chmt im th hai mch kch t khng tc ng) do s khng i xng ca t trng lm choV .V1 2

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 tqua 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 racc cun dy bn cnh gy ngn mch nhiu pha.V vy, cn phi t bo v chng chmt 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 trqu ti ch s c.- : dung khng 3 pha ng tr ca tt c cc phn t trong mng in p0CX

my pht.0

Nu b qua in trqu ti ch s c (r

=C C..j.3

1X

0

= 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 lain ti ch chm t lm hng cun dy v li thp ti ch s c, v vy bo v cn phi tcng ct my pht. Phn ln s c cun dy stator l chm t mt pha v cc cun dy cchin nm trong cc rnh li thp. gii hn dng chm t trung tnh my pht thng nit qua mt tng tr. Cc phng php ni t trung tnh c trnh by trong hnh 1.10.

(1)I

Nu tng trtrung 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 hndng. Nu tng tr trung tnh qu cao, dng chm t b lm cho rle khng tc ng.Ngoi ra in trqu ln s xut hin hin tng cng hng qu gia cc cun dy vit v ng dy kt ni. trnh hin tng ny khi tnh chn in trtrung tnh cc i

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C31

da 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 chomy 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 trni t dovy gi trin th ny phi ln m bo nhy ca rle.Hnh 1.10gii thiu mt s phng n p dng ni t trung tnh my pht.

Phng n a: Trung tnh ni t qua in trcao Rt (hnh1.10a) gii hn dngchm t nh hn 25A. Mt phng n khc cng ni t qua in trthp cho php dngchm t c tht n 1500A.

Phng n b: Trung tnh ni t qua in khng c khng trb (hnh 1.10b), viphng n ny cho php dng chm t ln hn khi dngphng n a, gi tr dng chmt khong (25100)% dng ngn mch 3 pha.

Phng n c: Trung tnh nit qua my bin p BA hnh 1.10c,in p ca cun sMBA bng inp my pht, in p ca cun thMBA khong 120V hay 240V.

Hnh 1.10: Cc phng n n i t trung tnh MF

R K BA Rta) b) c)

- i vi s c thanhgp 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 vn gin hn bo tn hiuchm t stator m khng cn ctmy 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 tngun xoay chiu ly t 2BU.

MF

1MC

7BI0

FCO

3RI 4RI 5R RThG

6RT

+ +

T bo vchng nm

ngoi

+

+

Bo tn hiu

C

2BU

t1MC

-

Hnh 1.11: S bo v ch ng chm t 1 i m cun stator MF

-

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- 3RI: rle chng chm t 2 pha ti hai im khi dng bo v so lch dc t 2pha (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 mypht.

- 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.

In1

)III(n1

I (1-37)

Dng in khng cn bng do cc pha pha scp ca 7BI0t khng i xng vicun th cp v do thnh phn kch t ph gy nn. Dng in khi ng ca rle cn phichn ln hn dng in khng cn bng trong tnh trng bnh thng ny:

I >IKR KCBtt

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 tim 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 chtgn 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 chmt 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 bov. Bo v s tc ng ct my pht nhrle 3RI. Trong trng hp ny rle 4RI cng khing 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 vkhi 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 scp) (1-43)Dng khi ng ca rle 4RI:* Dng khi ng ca 4RI c chn theo 2 iu

kin:Bo v khngc tc ng khi ngn mch ngoi vng bo v, khi :

)IUkC3(KKI maxKCBttpFq0tvatRI4KB (A) (pha scp) (1-44)

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Theo gi tr dngin s cp b nht tngng vi dngin khi ng cctiu ca 4RI (gi tr ny ph thuc vo cu to v nhy ca rle 4RI). i vi cc rlethng gp gi tr ny khong:

I = (2 3) (A) (pha scp) (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 trnh hng ca nhng gi tr qu

ca dng in dung khi chm t mt pha trong mng in p my pht, ngi ta thngchn:

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 vybo v ch cn bo tn hiu, y ch cn dng s bo vn gin, lm vic theo in pth 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 mtim cun stator b MF-MBA

iu kin1: UK KCBmaxiu kin2: URU > UKRU chn theo iu

kin n nh nhit ca rle v thng lybng 15V.

Thng chn theo iu kin 2 l thoiu kin 1.

Rle thi gian dng to thi giantr trnh trng hp bo v tc ng nhmdo 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 pthng c cng sut b v s bo vthng 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

50NR

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 tpha in dung h thng I(1)H ln hn nhiu so vi thnh phn chm t t pha in dungmy pht I(1)F ngha l:

(1)

I H >> I

(1)

vi IF F = 3.j..C.U

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V dng chm t I(1) (hnh 1.13) ph thuc vo v tr ca im chm t, nnnu xy ra chm t gn trung tnh ( 0) bo v s khng nhy, v vy phng

php ny ch bo vc khong 70% cun dy stator my pht k tu 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 trcao R : (hnh 1.14a)

My bin dng t dy ni trung tnh MF qua in trni t R, cun th cpni vo rle dng ct nhanh (c m s 50N). Tr s dng in t ca rle ly bng 10% gitr dng in chm t cc i cp in p my pht. y l tr st nh nht c tnh n an ton khi thnh phn dng in th t khng t h thng cao p truyn qua in dungcun dy MBA ti my pht. nng cao hiu qu ca bo v ngi ta c tht thm bov dng cc i (51N) c c tnh thi gian ph thuc c tr s dng in t khong 5% gitr 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 virle qu in p (59) song song vi ti trRt nhm n nh s lm vic cho MBA v to gitrin p t ln rle qu in p. Tr sin p t khong (5,4 20) V. S ch c th

bo vc khong 90% cun stator tnh tu cc my pht. Ngi ta cng c th sdng phng n hnh 1.14c bo v chng chm t cun stator my pht. Cun th cp

ca MBA c mc thm ti trRt, in trny lm tng thnh phn tc dng chm t lnkhong 10A v trn mch th cp ny t bin dng ni vo rle dng cc i (50N). Gi trt 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 scp ca BI chn bng hoc nhhn dng in i qua cun scp ca MBA ni t.

S sdng 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 hibc 3 (a); th vcttrong ch vn hnh bnh thng (b); khichm ttrung tnh (c) v khi chm tu cc im my pht

UF

UF

UF

UF

N

N

N

N

NF

F

F

UN

UN

100%

100%

100%

F

50%

50%

50%d)

N

F

F

UN

UN

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Cc s bo v m t trn khng bo vc hon ton cun stator my pht khixy ra chm t mt pha. Vi cc my pht cng sut ln hin i, yu cu phi bo v100% cun dy stator khi xy ra s c trn, ngha l bo v phi tc ng khi xy ra chmt 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 thnhphn sng hi bc ba, gi tr ca thnh phn in p ny ph thuc vo tr sin khng cathit b ni vi trung tnh my pht, in dung vi t ca cun stator, in dung ni t cacc 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 ccim khc nhau trn cun dy stator ta c phn bin p nh trn hnh 1.15b. y khiu UN F N F

Khi xy chm t u cc hoc trung tnh my pht, in p sng hi u cckhng chm t tng ln gn gp hai ln so vi ch tng ng trc khi chm t (hnh1.15c,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 sin p hi bc ba trung tnhmy pht v tr sin p hi bc ba ly cun tam gic hca 2BU. Rle le in p 2RUni 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 vc khong 90% cunstator tnh tu cc my pht, y rle 2RU cng bo vc khong (70 80) % cunstator tnh tim trung tnh. Nh vy s phi hp lm vic gia 1RU v 2RU c th bo vc ton b cun stator my pht khi xy ra chm t mt pha.

Cc tng trZ , Z1 2c chn sao cho ch lm vic bnh thng in p t ln2RU bng khng, khi xy ra chm t cun statorin p t ln rle s ln hn nhiu sovi 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 chmt khong 90% cun dy k tu cc my pht.

K

IH

KL

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)

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

I = IH - I (1-48)LVTrong :

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 vcthnh 1.16b ta c th thy rng, iu kin lm vic ca bo vcxc nh theo du ca I, bo v s tc ng ct MC khi I > 0, ngha l I

I&

>I iuH LV nyc tho mn nu chm t xy ra trong vng bo v. ng K-L trn th vcthnh1.16b 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 trR1 thay chotin C1 th s c th s dng bo v cho cc my pht c trung tnh ni t qua intrln. Khi y thnh phn tc dng ca dng in tc dng sc 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 sangmch R , C vi thng sc la chn thch hp.2 2

Mt phng n khc thc hin bo v chng chm t cun dy stator my phtc trung tnh khng ni t hoc ni t qua in tr ln lm vic trc tip vi thanh gpin 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 nyc a vo lm vic khi pht hin c chm t v lm tng thnh phn tc dng ca dngin s c ln khong 10A, to iu kin thun li cho vic xc nh hng dng in. Thit

b to thm ti bao gm BI0Nu vo trung tnh ca my pht, ti R ca BI ny c ngmbng tip im ca rle in p RU . Khi c chm t, in p U xut hin, RU0 0 0ngtc thi tip im ca mnh v duy tr mt khong thi gian t2 cho s lm vic chcchn.

T s bin i ca BIG trong mch thit b to thm ti c chn sao cho thnhphn 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 chmt 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 Ia 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.

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CtMC

c)

RBI0N

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 ch ng chm t cun dy stator MF c thi t b to thmti (a) th vctkhi c chm t ngoi (b) v trong (c) vng bo v.

MC

BI0

MF

BI0N R

BIGRU0 Thit bto thm

ti

A.I

)1(

.I

0.U

a

RU0 t2

t1

ngRU0

CtRU0

I

Thit b bo v

I(1)

IA

RBI0N

BI0

BIGI= IA - I

(1)

a)

c)

Shnh 1.17c th bo vc 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 cacun dy khng ln (khng vt qu 10% Up) nn xc xut xy ra hng hc vin (chnghn do cch in bnh thng) rt thp nn cc my pht cng sut b ngi ta thngkhng i hi bo v ton b cun dy.

i vi cc MF ni b vi MBA, thng thng cun dy MBA pha my pht utam 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 U0u cc tam gic h ca BU t u cc MF, hoc u ra caMBA 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 docc nguyn nhn chc .

Ngy nay bo v 100% cun dy stator chng chm t, ngi ta thng dng haiphng php sau y:

- Theo di s bin thin ca hi bc ba ca sng in p trung im v u ccMF.

- 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 trch 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.

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khc phc nhng nhc im ny ngi ta dng phng php a thm mtin 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 dyMF (hnh 1.18): MF

1LF

IB

MBA

RC

20Hz

BIG

RB

I

RC

Hm

2LF

Lm vic

I

ILV

RL

IC=IHCt

Hnh 1.18: S bo v 100% cun dy statorchng chm t c a thm in p hm 20Hzvo trungim MF

- Dng in I t ngun 20Hz saukhi qua b lc 1LF c phn thnh hai

thnh phn I chy qua BU0 ni vi trungtnh MF v I chy qua in trt RB B.Thnh phn I thng qua bin dng trunggian BIG v b lc tn s 2LF c nnthnh dng in lm vic.

- ILV a vo rle so snh vidng in hm IH cng do ngun 20Hz tonn thng qua in trt Rc , dng inhm c tr s khng i. ch lmvic bnh thng (R = ) dng in I c xc nh theo in dung ca cundy 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 nghipkhi c chm t xy ra u cc MF.

Mt phng n khc thc hin bo v 100% cun dy stator chng chm t ldng ngun ph 12,5Hz (vi tn s cng nghip l 60Hz ngi ta dng 15Hz) c tn hiuc m ha a vo mch s cp thng qua BU0u vo mch trung tnh ca MF(hnh 1.19a).

Trong ch lm vic bnh thng, dng in I chy qua im trung tnh MFc xc nh theo tr sin dung ng tr ca MF l C (hnh 1.19b).

Khi xy ra chm t, in tr chm t Rc ghp song song vi C lm tngdng in n tr s I > I (hnh 1.19c). Rle u ra s phn ng theo s tng dng inv 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 tnhiu 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 trc nh hng ca nhiu do dng in pha scp v

php o c tin hnh ring cho tng na chu k dng v m s trnh c nh hngca nhiu c tn s bi ca 12,5Hz.

30


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20/177

IV. Bo v chng chm t mch kch tca MF (64)

i vi MF, do ngun kch t l ngun mt chiu nn khi chm t mt immch kch t cc thng s lm vic ca my pht hu nh thay i khng ng k. Khichm t im th hai mch kch t, mt phn cun dy kch t s b ni tt, dng in quach cch in bnh thng c th rt ln s lm hng cun dy v phn thn rotor. Ngoira dng in trong cun rotor tng cao c th lm mch t b bo ho, t trng trong mypht 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 tbo v bo tn hiu khi c mt im chm t trong mch kch t v tc ng ct my phtkhi xy ra chm t im th hai.

i vi MF cng sut ln (my pht thuin), hu qu ca vic chm t imth hai trong mch kch t c th rt nghim trng, v vy khi chm t mt im trongcun 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 kcht :

* Phng php phn th.

64

Cunkcht R

MFkt

HNH 1.21 : Bo v chm t rotorbn hn h hn 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 cundy rotor, ngi ta dng in tr mc song songvi cun dy kch t, im gia ca in tr niqua rle in p, khi c mt im chm t sxut hin mt in thrle in p, in th nyln nht khi im chm t u cun dy. trnh vng cht khi im chm t gn trungtnh 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 tnhiu

35RI

+ +

-

37RG

47C

48CC

Ti mch kch t

Ti trc MF

34BG

52N

2RCL

U~

HNH 1.23: S bo v chng chm t1 im cun rotor dng ngun in phDC

+

36RT

-

Bo tnhiu

35RI

+ +

-

37RG

47C

48CC

Ti mch kch t

Ti trc MF

34BG

52N

HNH 1.22: S bo v chng chmt 1 im cun rotor dng ngun inph AC

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S bo vc trnh by hnh 1.22. in p ngun ph xoay chiu thng bngin 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: tin 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 hmch 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 kim no trong

mch kch t bo vu c th tc ng. Tuy nhin do dng ngun xoay chiu nn phichng 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, nhb 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 vin giathit 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 MFhin i:

i vi cc MF c h thng kch t khng chi than vi cc it chnh lu lptrc tip trn thn rotor ca my pht, in dung ca h thng kch ti vi t s tng ln

ng k v h thng bo v chng chm t ca cun dy rotor cng trnn phc tp .Cc s bo v chng chm t mt im trong cun dy rotor ca cc MF hini thng tc ng ct my pht ( loi tr xy ra chm t im th hai) v da trn mttrong nhng nguyn l sau:

- o in dn trong mch kch t (i vi t) bng tn hiu in p xoay chiu tns 50Hz.

- o in trca mch kch t (i vi t) bng tn hiu in p mt chiu hoctn hiu sng ch nht tn s thp. Nguyn l o in dn ca mch kch t i vi t caMF c h thng kch t khng chi than trnh by trn hnh 1.24.

C

S2

S1R

LF

BUG

RY

C tMC

U(50Hz) R

Rotor ca my kch tMy kch t

Cun dy rotor camy pht in

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

dn.

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Ngun in p ph xoay chiu tn s 50Hz c t vo mch trung tnh ca cundy my kch thch xoay chiu ba pha v thn rotor ca MF thng qua cc vnh gp vchi 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 Ci vi t ca mch kch t.

Trn hnh 1.25 trnh by quo ca nt vct tng trZ 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 khing 2 v mc tc ng ct my pht vi c tnh khi ng 1. c tnh 1 bc ly mt phnca gc phn t th hai v th ba trn mt phng ta m bo cho bo v tc ngmt cch chc chn khi c chm t trc tip (R 0).

S bo v hnh 1.24 cmt s nhc im l: s c mtca chi than S

X

1

R=0

2

RCR= 0

C=

X/ 2

R=

jX

C= constR=var

R= const

C= Var

R/ 2

Hnh 1.25:c tnh bi n thin ca t ng tr i vit ca mch kch tv c tnh tc ng ca Rleo in dn chng chm t mch roto MFng

b. 1- c tnh ct; 2- c tnh cnh bo.

1, S2 lm cho tin cy ca s khng cao v trs ca in tr tip xc c thnh hng n tr so ca rle.

Ngoi ra bn thn h thng kchthch mt chiu cng c th nhhng n s lm vic ca bov khi in dung ca mch kchthch i vi t C ln, in trr R ln nht c tho c 10k.

khc phc nhc imny ngi ta dng s vingun in ph mt chiu hocxoay chiu vi tn s thp cdng sng hnh ch nht.

Trn hnh 1.26 trnh by nguyn l pht hin chm t trong cun dy rotor caMFc kch thch t ngun in t dng qua b chnh lu Thyristor dng ngun tn hiusng ch nht c tn s 1Hz.

Cc in trph R , R1 2c chn c ch s kh ln so vi in trRM to inp UMt vo b phn o lng M.

Dng in do ngun in ph U to ra bng:

RRRU

IM

(1 -51)

21

21

RR

R.R

RTrong :

Lu rng RM


23/177

U(1Hz)

c)R= 0R= 5K

C= 2F

b)

Hnh 1.26: S nguyn l pht hin chm t trongcun dy rotor MF dng ngun in p ph 1Hz cdng sng chnht (a), v dng sngt vo b

phn o UMvi cc tr sin trkhc 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 gpt dng

in dung i vi t ca mch kch t C mc song song vi in trR s lm tcthi tng tr s dng in I v in p UM thi im u ca mi na chu k ca ngunin p U.

in trR c tc dng lm suy gim tr s ca I v UM. R cng b suy gimcng nhanh, trn hnh 1.26b v 1.26c trnh by dng sng UMo c cho hai tr s ca Rkhc 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 thhai mch kch t:

Ct 1MC

4Rth

+

Bo tnhiuhiu

-

2RT

+

1RI

3RG

V

7PA

9CN5CC

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 ch ng chm t thhai mch kch t

a) S nguyn l b) S bo v

35


24/177

Bo v chng chm t im th hai mch kch t (hnh 1.27) c a vo lm vicsau khi c tn hiu bo chm t mt im mch kch t. Thng bo vc t trn mt

bng di ng v c dng chung cho nhiu t my ca nh my. Bo v lm vic da trnnguyn tc cu bn nhnh: Khi chm t mt im mch kch t, ngi ta iu chnh cho

cu cn bng nhng h V. Khi cu cn bng ta c:4r3r

2r1r , 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 qua1RI 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 quamch 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 tgi.

- 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 mypht c th tng cao qu mccho php khi c trc trc trongh thng t ng iu chnhkch t hoc khi my pht bmt ti t ngt.

59II

59I

t

BU

Ct MC

n h thng iuchnh U(gim kch t)

MC

Ct kch t

MF

Hnh 1.28: Bo v ch ng qu in p hai c p tMF

Khi mt ti t ngt,in p u cc cc my phtthu in c th t n 200%tr s danh nh l do h thngtng iu chnh tc quayca turbine nc c qun tnhln v kh nng vt tc carotor my pht cao hn nhiu sovi my pht turbine hi.

cc my pht nhit in (turbine hi hoc turbine kh) cc biu tc lm vic vitc cao, c qun tnh b hn nn c th khng ch mc vt tc thp hn, ngoi ra ccturbine khi hoc hi cn c trang b cc van STOP ng ngun nng lng a voturbine trong vng vi msec khi mc vt tc cao hn mc chnh nh.

Mt khc, cc my pht thuin nm xa trung tm ph ti v bnh thng phi lmvic vi cc mc in p u cc cao hn in p danh nh b li in p ging trn hthng 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, ccthit bu ni u cc my pht, cn i vi cc my pht lm vic hp b vi MBA slm 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.

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25/177

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 tng iu chnh kch t gimkch 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 tng 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 trungbnh, ngi ta thng s dngbo v qu dng in c khoin p thp (hnh 1.29). Bo vthng c 2 cp thi gian:

MC

F

BU

MBA

BI

&

2I

2II

CtMC

Dngmy pht

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

27

50

Cp 1 (2I) tc ng ctMC u cc my pht (nu nivi thanh gp in p my pht)hoc MC ca b MF-MBA. Cp1 c phi hp vi thi gian tcng ca bo v d phng cang dy v MBA.

Cp 2 (2II) tc ng dng my pht nu sau khi ct MC u cc my pht (c thanhgp in p my pht) hoc u hp b (MF-MBA) m dng s c vn tn ti (tc l s cxy 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 lmvic 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 tiu cc my pht. Trong mt s s ngi ta cn dng bin php m bo cho bo v tcng chc chn l ch ly tn hiu in p thp sau khi rle dng in trv do s suygim dng ngn mch.

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

maxlvItv

at50K InK

KI = (1 -53)

vi I l dng in lm vic ln nht qua cun th cp ca BI.lvmax

37


26/177

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 phtin 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 mypht b qu kch thch. Cun dy rotor cmg c th b qu ti ngn hn trong qu trnh iuchnh 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 vomc qu ti, kt cu ca my pht, h thng lm mt v cng sut ca my pht. Thngcc nh ch to cho sn quan h gia mc qu ti (I* = I/Im) vi thi gian qu ti cho phpca tng loi my pht in.

C t1MC

18RT

24RI 25RI

32LI2

20RT19RT

27RI26RI

C t

MCpd

BI

MF

Bo tn hiuBo tn

hiu

Hnh 1.30: S bo v ch ng qu ti v ng n mch ngoi

1MC

C nhiu nguyn l khc nhau c thc p dng thc hin bo v chng quti cho cun dy ca my pht in: theo so trc tip ca nhit cun dy, nhit cacht 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 dngs 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


27/177

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: IK26RIphi ln hn dng thtnghch lu di cho php I :2cp

.IKI 2cpatK26RI = (1-58)- i vi my pht in turbine nc: I2cp = 5%.ImF- i vi my pht in turbine hi: I = 10%.I2cp mFiu kin 2: Rle phi trvsau 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 khngc 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 cn

VI.3.5. Kim tra nhy ca bo v:

nhy K ca bo vc 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. Khilm 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 thtnghch: (hnh 1.31)

Dng in th t nghch c th xut hin trong cun dy stator my pht khi xy rat dy (hoc h mch mt pha), khi ph ti khng i xng hoc ngn mch khng ixng trong h thng.

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

ph thuc t l nghch vi dng I :2

22

2

mF2

1

KI I

Kt

(1-64)

39


28/177

LI2

30 2

51 51

t1 t1

Ct MCCnh bo

52

HNH 1.31: Bo v dngin TTN cho mypht

10

I*2

0,6

0,5

0,4

0,3

0,2

0,1

t (sec)1010 t (sec)

0,4

0,3

10

0,1

I*2

1086420

IK1

IK2

t1

t2

HNH 1.32: C TNH THI GIAN PH THU C A V C L P C HAI

b)a)

Trong :

mF

2cp

I

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

vi:- l hng si 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) (hnh1.32a) hoc c tnh thi gian c lp 2 cp (hnh 1.32b): cp 1 cnh bo v cp 2 i ct myct.

40


29/177

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 trongmch kch thch (do ngn mch hoc hmch), h hng trong h thng tng iu chnhin p, thao tc sai ca nhn vin vn hnh... Khi my pht b mt kch t thng dn nmt ng bstator v rotor. Nu hmch kch thch c th gy qu in p trn cun rotornguy him cho cch in cun dy. ch vn hnh bnh thng, my pht in ng b lm vic vi sc in ngE cao hn in p u cc my pht UF (ch qu kch thch, a cng sut phn khng Qvo 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, Q > 0)

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

E I, QH thng

U

Hnh 1.33: Mt kch tMFa) thay i hng cng sut Q.b) thay i tng tro c cc my pht.c) gii hn thay i ca cng sut my pht.

0

a)

b)c)

B

41


30/177

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

U.

U 0, rlekhng lm vic. Khi b mt kch t = 1800, rle s tc ng. Gc khi ng c chnkhong 900. Cc h s a, b c chn (bng cch thay i u phn p ca BUG) sao chocc 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 thngqua 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 bi khi cn c tn gi l bo v chng trt cc t. Khimy pht in ng b b mt kch t, rotor my pht c th b mt ng b vi t trngquay. 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 sng dy trong h thng. Hu qu ca vic mt ng bgy nn s dao ng cng sut trong h thng c th lm mt n nh ko theo s tan r h

BU~

~S1

S1

& RLS3 S4

-1

UD

IB

IC

bUBC

CtMF

a)

U1

U2

-U1

t

S1

S2

S3 = - S1.S2

S4 = S3

kTn hiu ct

t

t

t

t

t

HNH 1.34: S bo v chng mt kch tmy phtin dng rle in khngcc tiu a) s nguyn l; b) th vct; c) dng sng ca cc i lng

b) c)

42


31/177

thng in, ngoi ra n cn to ra cc ng sut cnguy him trn mt s phn t ca mypht. pht hin s c ny c th s dng nguyn l o tng tru cc my pht.

Trn hnh 1.35 trnh by c tnh bin thin ca mt vcttng tro c trn ucc my pht trong qu trnh s c v xy ra dao ng in trong h thng. ch vnhnh bnh thng, mt vcttng trnm v tr im A. khi xy ra ngn mch mt vctdch chuyn t A n B, sau khi bo v ct ngn mch vct tng trnhy t B sang C vnu xy ra dao ng, mt vctchu k u tin s dch chuyn theo quo 2... Hnh vi

ny ca vct tng tr khi c dao ng in c thc pht hin bng mt rle vi ctnh khi ng nh trn hnh 1.36. c tnh khi ng c dng hnh elp hoc thu knh 1 vdng in khng 2 kt hp vi nhau theo nguyn l v. Khi c dao ng nu quo camt vctZ i vo min khi ong im M v ra khi min khi ng im N dic tuyn 2 (hnh 1.37) c ngha l tm dao ng (tm in) nm trong min tng trca bMF-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 vcttng trZ khi xy ra scv daong

Nu tm dao ng nm pha h thng quo ca mt vctZ s nm cao hn ctuyn 2, khi y bo v s tc ng ct sau mt s chu k nh trc. Trn hnh 1.37 trnh bys 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 tpha h thng, b phn m chu k dao ng 3 ct my pht khi s chu k t tr sttrc. pha cao p ca MBA tng c t thm b phn nh hng cng sut 4 thc hinchc nng ging nh b phn 2 v lm nhim v d phng cho b phn ny. Thay v ctuyn tng trkt hp 1 v2 trn hnh 1.36 ngi ta c th s dng c tuyn hnh ch nhtnh trn hnh 1.38 pht hin dao ng in.

F

1BI

BA

1BU

P

&CtMC

IC

2BI

I

U U1 2 3

Z


32/177

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 quti v thi gian qu ti cho php

ca cc cun dy my pht

+jX

R0

XdF

0,9XB

HNH 1.38:c tnh khi nghnh chnhtpht hin dao

ngin

IX. bo v chng lung cng sut ngc

Cng sut si chiu t h thng vo my pht nu vic cung cp nng lng choTurbine (du, kh, hi nc hoc dng nc...) b gin on. Khi my pht in s lmvic nh mt ng ctiu th cng sut t h thng. Nguy him ca ch ny i vi ccmy pht nhit in l Turbine s lm vic ch my nn, nn lng hi tha trongTurbine lm cho cnh Turbine c th pht nng qu mc cho php. i vi cc my phtdiezen ch ny c th lm n my.

bo v chng ch cng sut ngc, ngi ta kim tra hng cng sut tcdng 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 phttrong ch ny). Vi cc my pht in Turbine hi, cng sut khi ng P bng:kP = (0,01 0,03)Pk m (1-68)

Vi cc my pht thuin v Turbine kh:P = (0,03 0,05)Pk

m bo nhy ca bo v cho ccmy pht cng sut ln,mch dng in ca bo vthng c u vo li olng ca my bin dng(thay cho li bo v thngdng cho cc thit b khc).Bo v chng cng sutngc thng c hai cptc ng: cp 1 vi thigian khong (2 5) sec saukhi van STOP khn cp lmvic v cp th 2 vi thi gianct my khong vi chc giykhng qua tip im ca vanSTOP (hnh 1.40).

m (1-69)

F

BI

BU

92 2II

& 2I

Cp 2

Cp 1Ct

Van STOP

HNH 1.40: S nguyn l ca bo v chng cng sutngc

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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 mypht in.

X..2.S bo v my pht incng 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 mypht 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..

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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 2171

6351N

87T

87G

32

40

7849S81N

59 2451

87U

51N

7163

87T

HNH 1.44: S bo v b my pht v my binp .

46


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51N

63 87T

Kim tra cch inli

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 tng kch thch

ng ho lng

HNH 1.45: SBO V B MY PHT V MY

87G

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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 nhtlin kt h thng sn xut, truyn ti v phn phi. V vy, vic nghin cu cc tnhtrng 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 bntrong 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 ngnnga cc yu t bn ngoi nh hng n s lm vic ca MBA.

II. CC HHNG V TNH TRNG LM VICKHNG BNH THNG XY RA VI MBA

II.1. Sc 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 it ngt cc thng sin.

2. S c gin tip din ra t t nhng s tr thnh s c trc tip nu khngpht 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 trctip phi nhanh chng cch ly MBA b sc ra khi h thng in gim nhhng n h thng. S c gin tipkhng i hi phi cch ly MBA nhng

phi c pht hin, c tn hiu bo chonhn vin vn hnh bit x l. Sau y

phn tch mt s s c bn trong thnggp.

Hnh 2.1: Ngan mch nhieu phatrong cuon day MBA

c/b/a/

A CB A B C A C

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

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

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II.1.2. Ngn mch mt pha:

Khong cacht trung tnhen iemchm (%

cuon day)Dong s cap

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

100

I

IS

% ca dong 1xmaxI100

80

60

40

20

806040200

Dong chm

IxIS

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 lnhay nh ph thuc ch lm vic ca im trung tnh MBA i vi t v t l vokhong cch tim chm t n im trung tnh.

Di y l th quan h dng in s c theo v tr im ngn mch (hnh2.3). T th ta thy khi im s c dch chuyn xa im trung tnh ti u ccMBA, 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 tchm chp gia cc vng dy cng 1 pha bntrong MBA (hnh 2.4).

Hnh 2.4: Ngan mch gia cac vongday trong cung mot pha

Trng hp ny dng in ti chngn mch rt ln v mt s vng dy b ningn mch, dng in ny pht nng t chycch in cun dy v du bin p, nhngdng in t ngun ti my bin p IS c thvn nh (v t s MBA rt ln so vi s tvng dy b ngn mch) khng cho bo vrle tc ng.

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

II.2. Dng in tho 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 scp MBA. Nhng ykhng phi l dng in ngn mch do yu cu bo v khng c tc ng.

II.3. Sc 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...

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B. CC LOI BO V THNG SDNG BO V MBA

I. BO V CHNG SC TRC TIP BN TRONGMBA

I.1. Bo v qu dng in:

I.1.1. Cu ch:

Vi MBA phn phi nh thng c bo v ch bng cuch (hnh2.5). Trong trng hp my ct khng c dng th cuch lm nhim v ct s c tng, cu ch l phn t bo v qudng in v chu c dng in lm vic cc i ca MBA. Cuch khng c t trong thi gian qu ti ngn nhng ckhi

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 dyqun, in p n 35KV, c trang b my ct, bo v qu dng inc dng lm bo v chnh, MBA c cng sut ln hn bo vqu 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 ctnhanh c thc thm vo v to thnh bo v qu dng c hai cp (hnh 2.6). ViMBA 2 cun dy dng mt b bo vt pha ngun cung cp. Vi MBA nhiucun 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 vchng ngn mch trong cc cun dy vu ra ca MBA.

IS

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

+en rle thahanh chung

-

+RI RI RT

87T

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

55


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RI RIRI

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

Th

en rle thahanh chung

+

+

Rth

Khc vi bo v so lch cc phn t khc (nh my pht...), dng in scphai (hoc nhiu) pha ca MBA thng khc nhau v tr s (theo t s bin p) vv gc pha (theo tu 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 tnh 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 boho trung gian (loi rle in cin hnhnh rle PHT ca Lin X) hoc rle solch tc ng c hm (nh loi ZT caLin X).

Hnh 2.8 cho s nguyn l mtpha ca bo v so lch c dng my bindng bo ha trung gian. Trong my

bin dng bo ha trung gian c hai nhimv chnh:

9. Cn bng cc sc t ng do

dng in trong cc nhnh gy nn tnhtrng bnh thng v ngn mch ngoitheo phng trnh:

WN

IIIT

IIT

IIIS

IIS

RI

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

WlvTWlvSWcbIWcbII

WN

IIT(WcbI + WlvS) + IIIT(WcbII + WlvS) = 010.Nh hin tng bo ha ca

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

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I.3. Bo v MBA ba cun dy dng rle so lch c hm:Nu MBA ba cun dy chc 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 haicun dy (hnh 2.9a). Tng dng in th cp hai BI pha ti s cn bng vi dngin 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 theconguon

ti

c lviec

a/ c ham

87

Nguon

Hinh 2.9: S o bo ve so lech co hamMBA 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 thc thc hin bi rle qu dng in hay so lch th t

khng. Phng n c chn tu thuc vo loi, c, tu dy MBA.Khi dng bo v qu dng th t khng bo v ni vo BI t trung tnhMBA, hoc b lc dng th t khng gm ba BI t pha in p c trung tnh nit trc tip (hnh 2.10). i vi trng hp trung tnh cun dy ni sao ni qua tngtrni t bo v qu dng in thng khng nhy, khi ngi ta dng rleso lch nh hnh 2.12a. Bo v ny so snh dng chy dy ni t IN v tng dngin 3 pha (IO). Chn IN l thnh phn lm vic v n xut hin khi c chm t trongvng 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 ch ng chm t MBAbng bo v qu dngin

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Cc i lng lm vic v hm nh sau:

NII 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 tkhong co ham

lvI&

h2I&

h1I&

H2

H1

Cuon lviec

I

NI&

oI&

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

;II Nlv&& =

0.0I0IIh =+= &&

&&& =

NN

Qua phn tch trn ta thy, khichm t bn trong thnh phn hmkhng xut hin. Nh th ch cndngchm 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 trcao, rle so lch 87N c th khng nhy tc ng, ngi ta c th thay bng rle so lch chng chm t tng trcao64N (hnh 2.12b). Rle so lch tng trcao c mc song song vi in trR c trs kh ln.

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

(2-4)NooNu b qua sai s ca BI, ta c dng in th cp chy qua in trR 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 chmt s chy qua in trR to nn in p rt ln t trn rle, rle s tc ng.

a/

IC

IB

IA

Z

IO

IN 87NRle so lechth t khong

b/

64N

R RL

Z

IO

IN

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

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I.5. Bo v MBA tngu:Bo v chnh MBA t ngu cng l bo v so lch. Bo v da trn csnh

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

b/c

ba

87 87 87

C

BA

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 vnhm BI ni vo trung tnh MBA. Nu bo v ch dng mt bin dng t trungtnh MBA, cc BI t u cc MBA c ni thnh b lc th t khng v ni nmt 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 tamgic) khng ni vi ti, my bin p t ngudng lin kt h thng siu cao p v cao p.S bo v c th thc hin nh hnh 13b, ccBI c phi hp trn mi pha gn trung tnh

(im cui ca cun dy MBA) v dng 3 rle,lc bo vp ng chng ngn mch nhiupha v mt pha bn trong cun dy chnh MBAt ngu. S ny khng p ng khi s ccun dy th ba, bo v cho cun dy th batrong trng hp ny ngi ta thng dng bov qu dng in.

87T

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

Bo v tt c cc cun dy MBA t ngutng t nh bo v cho MBA ba cun dy (hnh2.14).

II. BO V CHNG SC GIN TIP BN TRONGMBA

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 biu p.

S dng loi no l tu quan im ca nh sn xut v tu tng cmy.Thng c dng ph bin l rle kh (hnh 2.15).

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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 bnhdu ph

T thng d uMBA

Phao 1

Phao 2

Bnh d u ph

ThngMBA

96B

b)

Hnh 2.15: Nguyn l cu to (a) v v tr btr trn MBA ca rle hi

Rle kh c t trn on ng ni t thng du n bnh dn du caMBA. Rle c hai cp tc ng gm c hai phao bng kim loi mang bu thu tinhc tip im thu ngn hay tip im t. ch lm vic bnh thng trong bnhy du, cc phao ni llng 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 yphao s 1 xung, rle gi tn hiu cp 1 cnh bo. Nu kh bc ra mnh (chng hndo ngn mch cun dy MBA t trong thng du) lung kh di chuyn t thng duln bnh dn du y phao s 2 xung gi tn hiu i ct my ct ca MBA.

Mt van thc lp trn rle: Khi th nghim rle, lp my bm khng khnn vo u van th. Mkha van, khng kh nn bn trong rle cho n khi phaoh xung ng tip im.

Mt nt nhn th kim tra s lm vic ca 2 phao. Khi nhn nt thnna hnh trnh, s tc ng ckh cho phao trn h xung (lc ny c 2 phao angnng ln v rle cha y du) ng tip im bo hiu (cp 1) ca phao trn. Tiptc nhn nt thn cui hnh trnh, s tc ng ckh cho phao di cng b hxung (do phao trn h xung ri) ng tip im mmy ct (cp 2) ca phao

di. Da vo thnh phn v khi lng hi sinh ra ngi ta c th xc nh ctnh cht v mc s c. Do trn rle hi cn c thm van ly hn hp khsinh ra nhm phc v cho vic phn tch s c. Rle hi tc ng chm thi gian lmvic 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 nhcng 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).

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o nhit cun dy MBA ngi ta thng dng thit b loi AKM 35,y l thit b s dng in trnhit c phn tt 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 cmt tip im thng m, mt tip im thng ng vi cc chung) lp bn trongmt nhit k c kim ch th.

Hnh 2.40: Thit b chth nhit cun dy

Ccu rle gm: ch th quay ghi so, mt b phn cm bin nhit, mtng mao dn ni b phn cm bin nhit vi ccu ch th. Bn trong ng mao dnl cht lng c nn li. S co gin ca cht lng trong ng mao dn thay i theonhit m b cm bin nhn c, tc ng ln ccu ch th v bn b tip im.ng thi, tc ng ln ccu ch th v cc tip im, cn c mt in trt nng.Cun dy th cp ca mt my bin dng in t ti chn s my bin p c nivi in trt nng. chnh nh cho phn tt nng, ngi ta s dng mt

bin trt tiu khin cnh my bin p. Tc dng ca in trt nng (tytheo dng in qua cun dy my bin p) v b cm bin nhit ln ccu o cngcc b tip im s tng ng vi nhit im nng, nhit ca cun y.

Thi t b ch th nhit cun dy

C 4 vt iu chnh nhit t tr s tc ng cho 4 b tip im. Tytheo thit k, cc tip im rle nhit c thc ni vo cc mch, bo hiu s

c nhit cun dy cao, mch tng mmy ct c lp my bin p, mchtng 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 nci.

Cp 2: Khi nhit cun dy MBA l 120oC th bo ng bng tn hiu nci 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 iiu khin h thng lm mt cho MBA. V di vi MBA lm mt bng qut thith h thng qut mt s lm vic khi nhit cun dy MBA t n mt trong ccgi tr 750C cun cao, 800C cun h v 600C i vi nhit du. H thng nys 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 dubo tn hiu 800C v mt ng h nhit du tc ng ct my ct 900C. Ccng h ny s dng nguyn l cm ng nhit . Phn t cm ng nhit c btrong hp nh v c t gn nh ca thng du ca my bin p.

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Rle mc du gm hai b tip im lp bn trong thit b ch th mc du, my bin p c bi nc in p c ti (biu p di ti) th thng gin nduc 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 tchgin ndu cho my bin p). Ngn c th tch chim phn nh hn nhiu ca thnggin n, sc ni ng lin du n thng cha biu p di ti. Thng chnh

my bin p v thng bi nc c thit k ring r, khng c lin thng du vinhau. V vy, c hai thit b ch mc du lp ti hai u thng gin no mcdu ca hai ngn thit b ch th mc du my bin p v thit b ch th mc du biu p di ti.

7

8

9

4

12

5

6

3

Hnh 2.42: Cu to ca thit b chth mc du1. V my. 6. Kim chth.2. Vngm . 7. Mt chth.3. Phao. 8. Thanh quay.

4. Nam chm vnh cu. 9. Trc quay.5. Nam chm vnh cu.

Ccu ca thit b ch th mc du gm hai b phn (hnh 2.42): B phn iukhin v b phn ch th. B phn iu khin c mt phao (3), thanh quay (8) trcquay (9) c lp nam chm vnh cu (4). B phn iu khin lp trn v my (uthng gin n) c vng m. B phn ch th gm kim ch (6) lp trn trc mang mtnam chm vnh cu (5). B phn ch thc lm bng nhm trnh bnh hngt trng nam chm v chng nh hng ca nc.

Khi mc du nng h th phao (3) nng h theo. Chuyn ng nng h caphao c chuyn thnh chuyn ng quay ca trc (9) nh thanh quay (8). Khiquay t trng do nam chm (4) siu khin cho nam chm (5) quay sao cho haicc 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), ghic mc du trn mt ch th. B phn ch th cng tc ng ng mcc tip imrle mc du a tn hiu vo mch bo ng hoc mch ct ty theo tng thitk.

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 psut rt ln trong my bin p. Thit b an ton p sut lp trn np thng chnh my

bin p s m rt nhanh (mht van khong 2ms) thot kh du t thng chnhMBA ra mi trng ngoi, p sut trong thng chnh s gim. Trong thit b an ton

p sut c gn rle p sut.

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S khi ca bo v R.63 ti trm:

Ct my ctTn 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 thngchnh my bin p. Khi c s c bn trong thng chnh my bin p th p sut trongthng chnh tng cao s ln hn p lc nn ca l xo, van a s chuyn ng thngln, lm hthnh khe hxung quanh chu vi van a. Kh s thot ra ti khe hvngm, lm gim p sut trong thng. Khi van a di chuyn ln th cng tc ng lnci ch th ckh bung ln, ng thi tc ng tip im rle p sut gi tn hiu timch bo ng v tng ct my ct c lp my bin p ra khi li in. Khi p

sut trli bnh thng, mun ti lp li MBA th phi nhn ci ch th ckh ( bbung ln) v v tr c, ng thi t li rle p sut bng nt nhn.

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

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

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

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

R.63Ct my ctTn 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 gianR86.

II.7. Rle kha trung gian (86):Rle kha trung gian R.86 thng c dng l loi kiu MVAJ-21 nh chto 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 hiuc gi ti t cc rle khc. Rle ny c th hot ng hai ch tc thi hocc 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

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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 tli 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 QUTI

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

Tuy nhin rle dng in khng th phn nh c ch mang ti caMBA trc khi xy ra qu ti. V vy i vi MBA cng sut ln ngi ta s dngnguyn 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 khcnhau trong my bin p v tu theo mc tng nhit m c nhiu cp tc ng khcnhau: cnh bo, khi ng cc mc lm mt bng tng tc tun hon ca khngkh hoc du, gim ti my bin p.

Nu cc cp tc ng ny khng mang li hiu qu v nhit my bin pvn vt qu gii hn cho php v ko di qu thi gian quy nh th my bin p sc 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 cpin p mt b.

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Vi MBA ba cun dy v MBA t ngu mt trong cc b bo v dng incc 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 mtrle dng in c kim tra p. Cc bo v chng dng in tng cao do ngn mchngoi dng lm bov d tr cho bo v

chnh ca MBA khingn mch nhiu phaMBA, n cn lm

bo v d tr chobo v ca cc phnt ln cn nu iukin nhy cho

php.

LI2

RURT

+

T BU noi vaothanh gop TA

-

+

++

LI2

RIRI

TA

++

RWRI

LU2

+

RU

RT

HA

CA

T BU noivao thanhgop TA

T BU noi vaothanh gop CA

+

RI RW

-

Hnh 2.18 chos nguyn l bov chng ngn mchngoi cho my bin

p t ngu. Trong rle nh hngcng sut (RW) chtc ng khi hngcng sut ngn mchtruyn t my binp n thanh gp caop, cn theo chiungc li th khngtc ng.

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

C. TNH TON BO V RLE CHO MBACs tnh chn bo v rle cho MBA:

Cn phi bit cc thng s ca MBA do nh ch to cung cp trn nhnmy hoc trong cc catalogue:

V d vi MBA ba pha hai cun dy:

Thng s sn xutUm cun

dy

LoiMBA

C iuchnhin 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.

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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 chS (KVA) I (A) Im tct (s)

100 5,25 16 3200 10,5 25 3300 15,8 36 10500 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 scp bng hay ln hn dng nh mccun dy MBA m n c t. i vi MBA hai cun dy dng nh mc scpv 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

mmlv

B

B

U3

SI = (2-6)

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

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I.2.1. Bo v ct nhanh:

Xc nh dng ngn mch scp cc i chy qua cht bo v khi ngnmch ngoi (INngmax) ti im N1 trong hnh.

)x(x3

UII

ht

(3)ngmax

B

1N1N

+

== (2-7)

xB

xht

N1(3)

Trong : xB: in khng ca MBA,m

m

B

2BN

B100.S

%.UUx =

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

Nngmaxatk .IKI = (2-8)

vi Katl h s an ton, K = (1,3-1,4)at Dng khi ng th cp ca rle :

I

Nngmax

(3)

satkRn

.I.KKI = (2-9)

(3)sK : h s kn s ni dy ca BI.

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

2I

IK

K

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:

maxlvtv

mmatk .I

K

.KKI = (2-11)

y dng Ilv max dng lm vic max qua cht bo v. Trong trng hpkhng bit c th ly Ilv max = IBm . Vi MBA ba cun dy dng Ilv max ly tng ngca tng cun.

Kat: h s an ton (1,1 - 1,2).Kmm: h s mmy (1,3 - 1,8).Ktv: h s trv (0,85 - 0,9).

Dng khi ng ca rle:I

k(3)s

kn

.IKI R = (2-12)

Kim tra nhy ca bo v:I

IK

k

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 phann:

)x.(x3

UI

21

1(2)N1

+

=

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53/177

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 tr (ngn mch cui vng d tr).Nu nhy khng t yu cu, phi dng bo v qu dng c kim tra p

(BVQIKU). Lc dng khi ng ca bo vc tnh:

1,2Kn

maxlvtv

atk .I

K

KI = (2-14)

Khng kn Kmm v sau khi ct ngn mch ngoi cc ng ct khi ngnhng khng lm in p gim nhiu v bo v khng th tc ng.

in p khi ng ca RU< :

tvat

minlvk

.KK

UU = (2-15)

Kat =1,2, Ktv =1,15, Ulv min: in p ti cht bo v trong iu kin t khi ngca ng csau khi ct ngn mch ngoi. Thng thng c th ly (0,7-0,75) Um.

Thi gian lm vic thng c phn thnh 2 cp:Cp thnht ct my ct thcp:

tc1 = t(2) + t (2-16)vi - t(2): thi gian tc ng ln nht ca bo v k n.

- t: bc chn lc v thi gian (0,3 - 0,5)sec.Cp thi gian thhai ct tt c cc my ct ca MBA:

tc2= tc1 + t (2-17)

I.3. Bo v dng thtnghch:

tng nhy cho BVQIKU, ngi ta s dng kt hp vi BVI2 (hnh 2.19).Khi , bo v qu dng ch b tr mt pha chng ngn mch ba pha v nhyc kim tra theo dng ngn mch ba pha th cp:

1.5I

IK

k

(3)N1min

n = (2-18)

Dng khi ng ca BVI2:

mtv

atk B2 .I

K

KI = . Vi Kat = 1,2; Ktv = 0,85 (2-19)

2MC

RTRIRURI

t BU thanhgop

-

N11MC

+ ++

LI2

Ct 1 v 2 MCCt 2MC

Hnh 2.19: S o nguyen l bo ve qua dong co kiem tra ap ket hpBVI2 tac ong co thi gian

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U2(n)1N

51N

U1

Hnh 2.20: S o nguyen ly bove chong chm at MBAbang bo ve qua dong ien

II. Bo v qu ti

Dng khi ng ca bo v qu ti :tv

Bmatk

K

.IKI = (2-20)

Dng khi ng ca rle :I

k(3)s

kRn

.IKI = (2-21)

Kat= 1,05; Ktv= 0,85IBm: dng nh mc pha t bo v tnh theo cng sut nh mc MBA.

Thi gian t ca bo v:t = tbv max + t (2-22)

tbv max : thi gian ln nht ca bo v ln cn.

III. BO V DNG THTKHNG (BVI0) CAMBA TRONG MNG C DNG CHM T LN

III.1. Bo v I0 MBA mt pha ni t:Dng khi ng scp BVIO c chn theo hai iu kin : Theo iu kin chnh nh khi dng khng cn bng khi ngn mch

ngoi: kcbmaxk II > (2-23)

Theo iu kin phi hp v nhy vi cc bo vng dy ni vothanh gp ca trm:

(2-24)ottatk .3IKI Trong :

: h s an ton khi phi hp c th chn KatK at = (1,1-1,2).

: dng th t khng (TTK) ti cht bo v, ng vi dng ngn mch no gy

ra dng TTK ln nht.ottI

Khi chn 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).I

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nhy ca bo v: Khi lm bo v chnh:

5,1=K

0minn

I

3IK (2-25)

Ly (3I0min ) khi ngn mch trn thanh gp ca trm.Khi lm bo v d tr: Kn 1,2. Lc dng 3I 0min l dng b nht khi ngnmch 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 X = X = X1B 2B 0B . 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 p35KV thng trung tnh khng ni t. Do vy tng trTTK ca loi ny bng tngtrTTT ca cun Yo. Nu t ni dy /Yo/Yo, vi cun c ti, in khng TTKca mi cun chnh bng TTT,

MBA t ngu in khng TTK ca mi cun chnh bng in khng TTT.

)z(x

UI

(n)1

p(n)1

+= (2-26)

Dng ngn mch (n) n

Z n0I

NM 1 pha(A) 1 02 xx + 1I

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

02

02

xx

.xx

+ 1

02

2 Ixx

x

+

Trong :

- n: dng ngn mch.- Io: dng in th t khng.- : Tng trs c thm vo.(n)

z

- x : in khng th t thun ti im ngn mch.1- x2: in khng th t nghch ti im ngn

mch.- 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 hndng kim tra nhy ca bo v. Khi bo v lm nhim v d tr dng 3I0minly cui vng bo v (cui ng dy di nht ni n thanh ci MBA t bo v).

x0

x2

x1

(1)N1 (1,1)N1

x2

x1

x0

III.2. Bo v I0 my bin p c hai pha ni t dng rle qu dngin:

MBA c hai dy qun ni t trc tip (hnh 2.21), dng 3I0i nh hnhv. Trong :

IoN2(1-2): dng 3Io do ngun I cung cp khi ngn mch chm t ti N .2 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.

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Cp I: L BVI0 ct nhanh, phi hp vi BVI0ng dy ni n thanh cipha t bo v: (2-27)maxIzkfmatIk IKKI =Trong :

Kat: h s an ton, Kat = 1,1.

Kfm: h s phn mch I0, da0

bve0fm I

I

K = .

I0 bv: dng I0 qua cht bo v.I0 dy: dng I0 qua ng dy c Ik Iz max.IkIz max: dng chnh nh cp 1 ca BVI0ng dy c tr s ln nht trong tt

c cc ng dy ni n thanh ci MBA c bo v.Thi gian chnh nh:

tI = tIzmax+t (2-28)tIzmax: thi gian tc ng ca

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