bao ve role

7/27/2019 Bao ve role

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7.1.1. Mcchcabovrle

7.1.2. Cc sc trong thitbin v li

7.1.3. Cc ch lm vic khng bnh thng

7.1.4. Cc phntcabovrle

7.1.5. Biudinrle trn bnv

7.1.6. Cc yu cucbnivibovrle

7.1.7. Cc dngc tnh cabovrle

7.1.8. Sni dy camchbovrle

7.1.9. Cc sni my bin dng v cun dy rle7.1.10. S phn b dng khi ngnmch hai pha sau my bin p

7.1.11. Ngun thao tc

XIT


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7.1.1. Mc ch ca bo v rle.

Nhnghuqu nguy him trn c thkhcphccnu pht hinkpthi cc sc, cc ch

lm vic khng bnh thng v c bin php s l nhanh chng.thchinnhimv trnngi ta sdng cc thitbbovnhcu ch, p t mt, v tinbnht l bovrle.

Bovrle l mtdngcbncatng ho, thiu n hthng cung cpin khng th lmvic bnh thng v tin cy.

Bovrlethchinvickim tra lin tc cc trng thi v cc ch lm vicca cc phnt

trong hthng c nhngphnng thch hp khi xuthinsc hay nhngch lm vickhng bnh thng:

7.1. NHNG VN C BN CA BO V RLE HTCC 2


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7.1.1. Mc ch ca bo v rle

Trong hthnginhinibovrlegnlinvi cc hthngtng ho nhmphchnhanh chng ch lm vic bnh thngca cc h tiu th.

Cc hthngtng ho thnggpnh:

Thitbtngnglpli.

Thitbtngngngund phng v cc thitbd phng.

Thitbtnggimti theo tns.

Khi xuthinsc,bovrles nhanh chng hnchsc tn ph ca n bng cch c lp

imschocloitr cc phntbsc ra khimngin sau mtkhongthi gian

nhkt khi pht hin.

Khi xuthinch lm vic khng bnh thng,tu theo tngmc m bovrletin

hnh nhng thao tc cnthitphchich lm vic bnh thnghoc bo tn hiu cho

cng nhn vn hnh bit tnh trng lm vic khng bnh thngcamnginhoc thitb

in.



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XIT

7.1.2. Cc sc trong thitbin v li

Phnln cc sc trong hthngin l ngnmch.

Nguyn nhn cbncasc l do:

Cch in lm vic lu ngy b gi ci

Khng bomngkhong cch gia cc pha

Qu in p

Cc tc ngvc l (t dy, chmchp dy dn)

Sai lmca nhn vin thao tc (ngctcu dao cch ly khi c tihocngnmchcha

cgiitrht ...).



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7.1.2. Cc s c trong thit b in v li

1. Ngn mch.

QN = 0,24.I2N.r.tN

Ngnmch l mtdngsc nguy himnht. Ngnmch c thxy ra gia cc pha, ngnmchgia cc pha km theo chmt,ngnmchmt pha. Trong my pht in v my bin

p ngoi cc dngsc trn cn c dngngnmchgia cc vng dy trong mt pha.

Huqucangnmch c th tm ttnh sau:

Khi dng ngnmchchy qua cc phntcamngin trong thi gian tN gy ra mt

lngnhit theo nhlut JunLenx:

Nhitlng ny v h quang intiv tr sc c th ph hu cc thitbti cc v tr

sc,sc ph hu cng lnnu dng ngnmch cng ln v thi gian tN tnti cng

di. Mt khc dng ngn mch ln cn gy nguy him cho cc phn t khng s c

nhng c dng ny chy qua.



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Khi ngnmchin p camnggimxung ph hus lm vic bnh thngca cc h

tiu th,phnln cc thitb dng in l cc ngc khng ngb. M men quay ca c

ngctlvi bnh phngin p cpnuccca n:

V vy khi in p gimnhius lm cho tccangcgimxungditcnh

mcthm ch c thbdngli.

M = k.U2

Cc thitbchiu sng l mtphnngkth hai ca cc h tiu thin, khi in p gim

xungcng ph hus lm vic bnh thngca n.cbitnhycmvisgimin pl cc my tnh int v cc thitbint.



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Huqu nguy himnhtcangnmch l ph huch lm vic song song nnhca

cc my pht, dnnmtnnh v tan gi hthng.

Trong ch lm vic bnh thng m men quay ca tuabin bngvi m men cnint

do tc quay cattctuc bin ca cc my pht khng thay i v ngb, khi ngn

mchtiim K trn thanh ci ca nh my in A, in p trn thanh ci gimxungbng

khng do ph tiin t v m men cnin t trn trc my pht gimxungbng

khng,

Trong thi gian ny lnghinc(hocnc) vo tuabin chagimkp do tc quay

ca tuabin tngvt ln (hin tng lng tc)dnn qu in p ucc my pht nh

my in A.



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XIT Hnh 7.1. nh hng ca s gim in p khi ngn mch

M N

Ph ti

N

A B

(a)

(b)



7 1 NHNG VN C BN CA BO V RLE


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7.1.2. Cc s c trong thit b in v l

Mt khc cc my pht inca nh my in B xa imngnmch N, v vyin p trn

thanh ci ca n gnbngnhmc. Do ttc cc my pht ca nh my in A khng mang

ti v l do trn hoc BVRL tc ngloitr n ra khili,ttcphticahthngchuynsang nh my in B, lm cho cc my pht ca nh my in B b qu ti v tc quay ca

chng gimxung. Do ngnmch lm cho tc quay ca cc my pht hai nh my in

A v B khc nhau (mtngb).

Khi ngnmch lu c th ph hus lm vicnnhca cc ngc khng ngb. Khi inp gim,tc quay cangcgim. (nuhstrttngtihstrttihn) th ngc

i vo vng lm vic khng nnh v c thtcngcgim nhanh v dnghn. Cng v

stnghstrt, cng sutphn khng m ngc tiu thcng tng ln, do c thdn

nch l sau khi ngnmchcloitr c sthiuht cng sutphn khng, lm cho in

p cahthnggimtngt, gy nhhngns lm vic bnh thngcangc.

Ngnmch l mtscrt nguy him, nn ngnmchcncgiitrmt cch nhanh chng


7 1 NHNG VN C BN CA BO V RLE


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2. Chm t mt pha trong li c trung tnh cch in vi t.

Khi chmtmt pha trong li c trung tnh cch invit th dng chmtmt pha rtnh

chbao gm thnh phn dng inin dung ca cc phn t, do vy n khng lm lch, m

in p gia cc pha. Song in p ca hai pha khng sc so vittng ln bngin p d

uy hip cch inca cc pha khng sc.

Do chmtmt pha khng cnct nhanh nhngcng khng ctnti qu lu.



7 1 NHNG VN C BN CA BO V RLE CC


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7.1.3. Cc ch lm vic khng bnh thng

1. Qu ti.

Qu ti lm cho dng inchy qu thitblnhn dng nhmc,thitbsb pht nng qu

mc cho php, lm tng hao mn cch in v c thdn tichhhng thitb. ngn

ngasccathitb khi qu ticnphi p dng cc bin php gimtihoc l ctthitb

khili.

2. Gim tn s

Nguyn nhn lm cho tnsgim l cng sutcangunnhhn cng sut tiu th do ct

ngtmts my pht ang lm vic. Khi tnsgim, lm gimtc quay caccusnxu

c thdnti ph v qu trnh cng nghcanhngt my, dy truynsnxut c yu cutc quay khng ica cc ngcin.


7 1 NHNG VN C BN CA BO V RLE HTCC 2


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7.1.3. Cc ch lm vic khng bnh thn3.in p tng cao.

in p tng cao qu gi tr cho php thngxuthin cc my pht thuin khi ctt

ngtphti. Khi ctti cc my pht thuin lm tngtc quay, tngscinngca

my pht ntrs nguy him cho cch inca n.bov my pht cnphigim dng

kch thch hoc l ct my pht

Qu p gy nguy him cho thit b c th xut hin khi ng hoc ct mt pha ng dy

truyntiinnng di c dung dnln.

Cc huqu nguy him trn c thcloitrbngbovrle.




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7.1.4. Cc phn t ca bo v rle

Hthngbovrlethnggmmts cc kh cin v mtsrle, ghp ni thnh mts

sthngnht. Cu trc camts BVRL c th chia thnh 3 khi chnh nh sau:

Khio lng: l cc thitbo lng, cc bcmbinchuyni thnh tn hiuin

nh MBD, MBAL, cc sens ... N thng xuyn theo di mt thng s no ca

mchinnh dng, p, nhit... bini thnh tn hiuphnnh ln rle.

Khi lgic: l cc rletipnhn tn hiucakhiolng, n s tc ng(tcthihoc

c duy tr thi gian) nu tn hiu theo di vt qu ngng ci t. Cc r le c thgm

nhiutng ghp ni c thm chcnngchnlc,tnghp,khuychi...

Khichp hnh: l cc kh cchuynmch v thng dngnht l my ct.




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XIT

7.1.4. Cc phn t ca bo v rle

Rle tc ng khi cc trs m n phnngtng ln gi l rlecci, cn ngc li l

rlecctiu. Rletnsphnngvitnstng ln hocgimxung.

Rlenhitphnng khi trsnhitcaitng theo di tng ln.

Cc rlephgm: Rlethi gian dng duy tr thi gian tc ngcabov,rlechth

dng tn hiu ho v xc nh tc ngcabov,rle trung gian dng truynt

tc ngca cc rlecbnn my ct v dng lin hngthigia cc phnt

cabov.




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C hai cch biudinrle trn bnv.

1. Cch thnht.

Rlecbiudindidng hnh chnht c na vng trn pha trn. Cun dy rlec

hiungm l tphndi (trong phnchnht) v thng khng v, tipimca rle

cvphn trn (phnna vng trn). Rle trong sbov cn c cc u vo. Lo

rlec k hiubngch ci uca tn rle v c ghi vo phnchnhthocc gh

bng k hiu (hnh 7-2a).

2. Cch th hai.

Rlecbiudindidng khai trin (hnh 7-2b). Cun dy rle v tipimcv ring

bit trong nhngphn khc nhau cas v c k hiubng cc ch ci tngng. Cchbiudin ny c uim l dc,nht l ivinhngsphctp.




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Hnh 7.2. Bieu dien rle tren s o nguyen ly




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1. Tc ng nhanh

Ctngnmchcnphic tin hnh vikhnng nhanh nhthnch tc hi do d

ngnmch gy ra:

nng cao hiuqungtngng dy v thanh ci d phng

gimnhthi gian cc h tiu thphi lm vicviin p gimthp

bo ton s lm vicnnhcahthngin.

Tronghthnginhinibo ton tnh nnh yu cuthigian ctngnmchrtnh

Th d vi ng dy truyn ti in nng (300500) kV cnphi ct s c sau khon

(0,10,2)s k tkhixuthinngnmch,vingdy (110220)kV sau khong (0,150,3)vi liphn phi (610)kVxa ngun thigian ctngnmch cho php khong (1,53)s

chng khngnhhngn tnh nnhcahthng




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Tuy nhin ivich lm vic khng bnh thng yu cuvthi gian tc ngcabov

khng c qu nhanh. Bi v cc ch lm vic khng bnh thng c tnh chttmthi vc tnh t loitr, th dnh qu tingnhn khi khingngc khng ngb. Ct qu

nhanh trong trnghp ny sdntichctnhm cc h tiu th.

V vyct cc thitb khi xuthin cc ch khng bnh thng gy nguy him cho thit

phictin hnh vithi gian chmcnthit.




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7.1.6. Cc yu cucbnivibovrle2. Tc ngchnlc

bom tin cy cung cpin cho cc h tiu th th bovrlephi tc ngchnlc

Bovrlecgi l c chnlcnu n chctphntbsc ra khiliin v bo to

s lm vic bnh thngca cc h tiu th khc.

Vd:

Trn (hnh 7.3), khi ngnmch ti N1 bovcnphictng dy bscbng my c

MC7, ngha l bng my ctgnv tr scnht.

Khi ngnmchti N2 chcho php my ct MC4 v my ct MC6 tc ng,ctng dy s

c 1, cn ng dy 2 tiptc lm vic cung cpin cho h tiu th.



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Hnh 7.3. S biu din im s c ngn mch



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7.1.6. Cc yu cucbnivibovrl3. Phi c nhy cao.

Bovphi tc ng khi xuthinsc trong vng tc ngca n v vng bovca

bov ngay sau n nubov ny v mt l do no khng tc ng.

bom tin cy cung cpin cho cc h tiu th th bo

V d: (hnh v 7.4)

Hnh 7.4. Vng tc ng ca bo v


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XIT


INmin l dng ngnmchnhnhtchy qua bov khi ngnmchcui vng bov

Ikdbv l dng khingcabov.

nhycabovcc trngbnghsnhy. ivibovphnngvi dng

ngnmch.

Trong :

Bovcgi l c nhynu knh tnh ton lnhnhocbng knh yu cuivitng

loibov.

kdbv

minNnh

I

Ik

nh.tt nh.yck k



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Bovrle khng c tchi tc ng khi ngnmch trong vng bovca n v khng

c tc ngnhm ln. Yu cuv tin cyrt quan trngbi v tchi lm vic hay tcng sai camtbov no udntichct sai mtsh tiu th.

Khi ngnmchti N (hnh 7.5) m bov I khng lm vic th bov III s lm vic v dnti

chct sai trm B v trm C. Nu v mt l do no m bov IV tc ngnhmln th dn

tichngng cung cpin hon ton cho h tiu th.

tin cycabovphthuc vo chtlngrle, cng tct v cc thitb khc trong s v vo ktcucas.

XIT

7.1.6. Cc yu cucbnivibovrl4. Phi tc ng tin cy.

Hnh 7.5. S bo v ng dy nhiu cp



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Trong qu trnh vn hnh cc thng scaliin lun lun thay i do ta nghocct

cc my pht, cc my bin p, cc ng dy vo hoc ra khimngin, do dng ngn

mchcng thay i,nhngbovrlephi tc ngmt cch chcchn khi c ngnmch.

XIT


5. Phi tc ng c lp vi iu kin vn hnh ca li in



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7.1.7. Cc dng c tnh ca bo v rle

XIT

1. c tnh c lp (Hnh 7.6)

Bovloi ny c thi gian tc ng kh chnh xc, nhng c nhcim l bomiukin tc ngchn lc th thi gian tc ngcabovcp trn qu ln, khng bom yu

cu tc ng nhanh.ivibov c c tnh clp th thi gian tc ngcabovc

thchin trn rlethi gian, khng phthuc vo dng chy qua rle dng.

Nu dng inchy qua bov

chavt qu dng chnhnh

Ikd th bo v khng tc ng.

Khi dng inchy qua bov

ln hn hoc bng dng khi

ng th bovs tc ngvi

thi gian cnh ttd khng ph

thuc vo dng in chy qua

n.Hnh 7.6. c tnh tc ng c lp



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2. c tnh ph thuc. (Hnh 7.7)

uimcabov ny l c thbovc qu ti, cho php tngcngct nhanh kh

ngnmchuonng dy cbovtc l sc cng nngn th thi gian tc ng

cng nhanh, sbovngin v khng dng cc rlethi gian, trung gian v tn hiu.

Nhcimcabov l phihpthi gian tc ngchnlccabov ny phctp v vy

n chc dng trong nhngtrnghpphctp

Thi gian tc ng ca bo vph thuc vo dng in chy

qua n. Dng in chy qua

bov cng ln th thi gian tc

ng cng nh.

Hnh 7.7. c tnh tc ng ph thuc



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3. c tnh ph thuc hn ch. (Hnh 7.8)

Va c unhcimcabov c c tnh phthuc,va c unhcimcabov cc tnh clp.

Hnh 7.8. c tnh ph thuc hn ch



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Rlescp l rle dng in hay in p thng thngutrctip vo dng in hay in

mchscpcaphntcbov.

uimcarlescp l khng cn my bin dng v bin p olng, khng cnngun tha

tc v cp khngch.

Rlethcp l rlecmc vo cunthcpca my bin dng hay my bin p olng.

Rlethcpcsdngrng ri v n c uimhnrlescp l cch ly vimngi

p cao, t xa phntcbov,thuntin trong vn hnh, c th tiu chun ho vi dn

nhmc 1A hoc 5A hay in p nhmc 100V, khng phthuc vo dng v p mchscp

XIT

7.1.8. S ni dy ca mch bo v rle

1. Theo cch ni rle c chia ra:



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7.1.8. S ni dy ca mch bo v rle

2. Theo phng php tc ng n c cu ngt ca thit b ng ct mch in chia ra:

Rle tcngtrctip:Cng l rle dng in hay in p thng thng, tc ngtrctip vo

ccungtcathitbctmchin. Rle tc ngtrctip c th l rlescphocth

cp.

Rle tc ngtrctip khng cnngun thao tc nhngrle ny cchto c lc tc ng

ln. V vyrle tc ngtrctip c chnh xc khng cao v tiu th cng sutln.

Rle tcnggin tip:L rle dng tipimca mnh khngchmchiukhin my ct

hay bo tn hiu.

Bov c rle tc ng gin tipcnphi c ngun thao tc, nhnglc tc ngcarleloiny khng cnlnlm v vy chng c chnh xc cao hn v cng sut tiu thnhhnrle

tc ngtrctip. Ngy nay bovrle c rlethcp tc ng gin tipcsdngrng

ri nht



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7.1.9. Cc s ni my bin dng v cun dy rle

XIT

Cch biudin vc t dng thcp

Chiuca vc t dng thcp

I2 trn biu vc tphthuc

vo chiu dng quy c ca

dng inclyivicun

dy scp.

Phng php biudinth hai nginthuntinhnphng php u, n cho php khi

xy dngbiu vc t,chiuca vc t dng s v thcpcv trng nhau. V vy

phng php th hai csdngvbiu vc t cc phntip theo.y, khi xy dngbiu vc t ta b qua sai sv gc pha ca my bin dng

Hnh 7.10. Biu din vc t dng th cp.


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7.1.9. Cc s ni my bin dng v cun dy rl

XIT

1.S ni my bin dng v cun dy rle theo hnh sao .

My bin dng ct trn ttc cc pha. Cc cun dy thcpca my bin dng v cccun dy carlecni theo hnh sao v cc im trung tnh ca chng cni linv

nhau bngmt dy dngi l dy trung tnh (dy dntrv).

Hnh 7.11. S u my bin dng theo s sao



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Rle R4 khng tc ng v dng inchy qua rle 4 l tng dng in pha:

0I3IIII0cba4R

Trnghp ny cc rleu c ks =1

BI

A

a1Rn

III

BI

B

b2Rn

III

BI

C

c3Rn

III

Trong ch lm vic bnh thng v khi ngnmch ba pha dng inchy qua rle 1,

2, 3 l dng in pha.

Kho st cc ch:7.1.9. Cc s ni my bin dng v cun dy rle

Trong thct do sai sca cc my bin dng khc nhau nn trong rle 4 vn c dng chy

qua gi l dng khng cn bng: IN4 = Ikcb 0.

Hnh 7.12. th vc t dng in khi lm vicbnh thng v khi ngn mch 3 pha



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Khi ngnmch hai pha.Dng ngnmchchchy trong hai pha bsc v tngng c dng chy trong cc rlen

vo my bin dng t trong pha sc, cn pha khng sc khng c dng ngnmchchy

qua.

Gis khi ngnmch hai pha A v B.

Hai pha A v B c dng ngnmch, cn pha C khng c:

IA =I(N)A ; IB = I(N)B = - IA ; IC = 0

tngng c: Ia = - Ib v Ic= 0

BI

Aa1R

n

III

BI

B

b2Rn

III IR3 = Ic = 0

Hai rle R1 v R2 c Ks =1

Rle R3 c Ks = 0

Dng chy qua rle 4:

IR4 = IR1 + IR2 = Ia + Ib = 0

Nhvyrle 4 mc vo dy dntrvs khng tc ng khi ngnmchgia hai pha.

Hnh 7-13: th vc t dngin khi ngn mch 2 pha



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Khi ngnmch hai pha ngthichmt trong hthng c trung tnh nit

Gis khi ngnmch hai pha A v B ngthichmt. Chhai pha A v B c dng ngn

mchchy qua v lch pha vi nhau 1200, tngt dng chy qua rle 1 v rle 2 cng

lch pha nhau 1200.

IR1 = Ia, IR2 = Ib, IR3 = 0

Ba rle R1 , R2 v R4 c ks =1

Rle R3 c ks =0

60j

a e.I

IR4 = IR1 + IR2 = Ia + Ib =

Hnh 7.14. th vc t dng in khi

ngn mch 2 pha chmt



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Ch ring pha A c dng ngnmchchy qua:

IA = IN ; IB =0; IC = 0

Dng trong rle:

IR1 = IR4 = Ia

Ba rle R1 v R4 c ks =1

Rle R2, R3 c ks =0

Nhn xt chung: Ta thymidngngnmchu c dng inchy qua t nht l 2 r

viks = 1. S ny csdngbov trong hthng c trung tnh nit

Khi ngnmchmt pha, gis pha A.

Hnh 7.15. th vc t dng inkhi ngn mch 1 pha


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7.1.9. Cc s ni my bin dngv cun dy rle

XIT

Kho st cc ch :

Trong ch lm vic bnh thng v khi ngnmch ba pha

BI

Aa1R

nIII

BI

C

c2Rn

III

IR4 = Ia + Ic = - Ib

Cc rle c ks = 1

Hnh 7-16b: th vct in p khi lm vic bnh thnv khi ngn mch 3 pha

Khi ngnmch hai pha c t my bin dng (A v C)

IA = - IC, IB = 0 nn:

IR1 = Ia, IR2 = Ic= - Ia

IR4 = Ia + Ic = 0

Cc rle R1 , R2 c ks =1

Rle R4 c ks = 0Hnh 7.16c. th vc t dng in

khi ngn mch 2 pha A v C


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7 1 9 Cc s ni my bin dng v cun dy rle



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XIT

Khi ngnmch hai pha gia pha c t my bin dng v pha khng t my bin dng.

Gis hai pha A v B.

IA= IN; IB= IN = - IA ;

IC = 0; IR= Ic - Ia = - Ia

C ks = 1

Khi ngn mch mt pha pha khng t

my bin dng (pha B):IB = IN ; IA = 0 ; IC = 0

Tngng: Ia = Ic = 0

IR= Ic - Ia = 0

Nhvy khi ngnmchmt pha pha khng t my bin dng, khng c dng chy quarle,bovmcli. Do s ny c dng bovngnmchgia cc pha trong

hthng c trung tnh khng nit(hthng khng c dngngnmchmt pha).

Hnh 7.17d. th vc t dng in khingn mch 2 pha t my bin dng

v pha khng t my bin dng


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7 1 10 S phn b dng khi ngn mch hai pha sau my bin p



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7.1.10. S phn b dng khi ngn mch hai pha sau my bin p

My bin p c tdu dy /-11 csdngrng ri nht, nn ta nghin cu my bin

loi ny.

ngin ta coi hsbin p ca my bin p bng 1. Ts dng in pha:

Ngnmch hai pha sau my bin p c tu dy /hoc/ v khi ngnmchmt pha

sau my bin p ni

/

0.

1. S phn b dng khi ngn mch hai pha sau my bin p ni /hoc/ .

I.3I

II

IIY

TC.d

YSC.d3

W

W

I

I

Y

Y

Khi ngnmch hai pha pha , gisgia pha b v pha c, dng trong pha khng sc Ia = 0cn dng trong pha sc b v c bng dng ngnmch.

Ic = - Ib = IN




55/159

XIT

p g g p y p

Trong tam gic dng ngnmch IN chia thnh hai phn: Mtphnchy qua cun b, mtphn

chy qua hai cunni tip a v c. Tng tr hai cun dy a v c ni tipgp hai ln tng tr

cun dy b nn dng trong cun dy pha b bng2/3IN

cn dng trong cun dy a v c bng

1/3IN.

Hnh 7.23.S nguyn l v s phn b dng khingn mch 2 pha sau my bin p u/-11




56/159

XIT

Dng in trong cc cun dy ni hnh lnhn ln dng inchy trong cun dy pha

cng tn pha tam gic.

3

I

3.3

I

3.IINN

aA

3

I3.

3

I3.II NNbC

NNcB I.3

23.I.

3

23.II

Khi ngnmchgia cc pha a v b hoc a v c th s phn b dng cngtngt.

Nh vy khi ngnmch hai pha pha , trong ttc cc pha hnh sao u c dng ngn

mch. Dng trong hai pha c trsbng nhau v cng chiu, dng trong pha cn libngtng

dng hai pha trn v c chiungcli.

Tng t khi ngnmch hai pha pha hnh sao trong s/cngnhngnmch hai

pha pha tam gic phn tch trn.

By gi ta phn tch cc iukin lm viccabovni theo s sao , sao thiu,hius

dng hai pha trong iukinkho st trn (c ngha l ngnmch sau my bin p).




57/159

XIT

Trong mt pha casxuthin dng (nBI l hsbinica my bin dng), cn trong ha

pha cn lixuthin dng . Tng dng in trong dy dn khng bng khng. Rle 1, 2, 3 u t

ngnhng c nhy khc nhau, mt trong ba rle c nhygpinhyca hai rle cn lv c dng lngpichy qua.

Dng inchy qua c hai pha v dy dnngc. Dng trong dy dnngcbngtng hnh h

dng hai pha. Nu my bin dng t trn hai pha c dng ngnmchnhhnchy qua th

nhynhhn hai lnnhycas sao .khcphcnhcim ny ta t thm mtrlena vo dy dnngc. Dng chy qua rle ny l tng hnh hcca cc pha kia

Trong trnghp ny s hai bin dng ba rle (hai rlemc vo dng pha, mtrlemc vo

tng dng hai pha) c nhybngs ba pha.

Nu my bin dng t trn hai pha c dng ngnmchnhhn th IR = 0. V vys ny khng

c dng bovngnmch sau my bin p ni/hoc/.

Trong s sao .

Trong s sao thiu.

S hiu s.


58/159




59/159

XIT

ngin ta cng coi hsbinica my bin p l 1. Ts dng in pha:

1W

W

I

I

00 Y

Y

Y

Y

Khi ngnmchmt pha pha cun dy0, gisngnmchmt pha b:

Ib = IN, Ia = Ic = 0

Dngngnmchmt pha l dngngnmch khng ixng.gii cc bi ton khng ixng ta

c th dng phng php cc thnh phnixng.

Phng trnh sccvitnh sau (y ta ly pha b lm chun):

0III

0III

IIII

0c2c1c

0a2a1a

N0b2b1b

Dng ton t a, c thbiudin cc vc t thnh phnixng qua cc vc tcamt pha c

ly lm cs, v d pha b.




60/159

XIT

Nhn vo (hnh 7-25) ta vitc cc cng thc sau:

0022

1 bbb IIaaI

1b1a I.aI 1b2

1c I.aI 2b2

2a I.aI 2b2c I.aI

N0b2b1b IIII

0IaIIa 0b2b1b2

120jea 240j2 ea

Hphng trnh 1 cvitlinh sau:

Hnh 7.25. Gin vc t dng in khi lm vic bnh thnghoc khi ngn mch i xng

0I).aa(I).aa(2b

2

1b

2

2b2

1b2 I).aa(I).aa(

Ly phng trnh (3) tr i phng

trnh (4) ta c:

(5)


61/159

V ta gi thit h s bin i ca my bin p l 1 nn ta c:7.1.10. S phn b dng khi ngn mch hai pha sau my bin p



62/159

XIT

3

I.2III N2B1BB

V ta githithsbinica my bin p l 1 nn ta c:

Ib1 = IB1 = Ib2 = IB2 = Ia0 = IA0 = Ib0 = IB0 = Ic0 = IC0

Ia1 = IA1, Ic1 = IC1

Do :

Tngt dng trong cc pha khc:

N1B2A1AA I3

1

IIII

N1B2C1CC I3

1IIII

Nhvy khi ngnmchmt pha pha 0 ca my bin p, trong ttc cc pha pha hnh sao

u c dng ngnmchchy qua. Dng trong hai pha c trsbng nhau v cng chiu, dng

trong pha cn li c trsgpi v c chiungcli. Ngnmchmt pha sau my bin p

ni/0 iukin lm vicca cc bovni theo s sao , sao thiu,hiuscng

gingnhtrnghpngnmch hai pha sau my bin p ni/hoc/. Nubov dng

s sao thiu, hai bin dng hai rlenhyca n s km hns ba pha.khcphc

cngphi dng thm rleth ba ni vo tng dng hai pha, khi ny nhycas hai bin

dng ba rlebngnhycas ba pha.

7 1 11 Ngun thao tc



63/159


1.nhngha v cc yu cu chung

Ngunin thao tc l ngunc dng cung cp dng in cho mchiukhin c khong

cch ca cc my ct, cho mch thao tc cabovrle,thitbtng,iukhin xa v cc

dng tn hiu ho khc.

Ngun cung cp cho mch thao tc ca cc phnt dng ctng dy, thitb v ctsc

phicbit tin cy. V vy yu cuchyucangun dng thao tc l trong chsc v

trong ch lm vic khng bnh thngcaliphimbotrsin p v cng sutcn

thitbovrle v cc thitbtng tc ng v phing v ct tin cy my c

tngng.

Ngun thao tc c thmtchiuhoc xoay chiu.

XIT

7.1.11. Ngun thao tc2 Ngun thao tc mt chiu



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XIT

2. Ngun thao tc mtchiu

Ngun thao tc mtchiu l cc b c quy in p (110220)V, trmbin p cng sutnh

th dng (2448)V, cttp chung cung cp cho ttc cc mch thao tc.

Cc b c quy bom cung cp cho mch thao tc trong mithiimviin p v cng

sutcnthit, khng phthuc vo mngin chnh. V vy n l ngun thao tc tin cynht.

Nhng n c nhcim l gi thnh t,phi c my np c quy v nh cha ring, phi

thng xuyn boqun,mngmtchiu ko di phctp.

nng cao tin cy, limtchiuc phn on thnh mtsonmchc cung

cpclpt cc thanh ci ring cab c quy.

Tnh khng hon hoca cc cu ch ckim tra bngrle TH (hnh 7-27a). Mchct CC

v khitipimthngm trong btruynng my ct MC ckim tra bngrle KT, cho

tn hiu khi hmch (khi my ctangngtipim lin ng MC ng).

Trong li dng mtchiu c thxuthin dng chmt. Gischmtti hai im N1,

N2 (hnh 7-27c), cc tipimcarlebov BV bnitt trong cunct CC xuthin dng,

my ctsct ra.ngnngas tc ngnhvy c thsdng vn mt V1, V2 v rle tn

hiu TH (hnh 7-27a) kim tra vhintngchmt trong li dng mtchiu.




65/159

XIT

Hnh 1.27. S nguyn l mng cung cpin mt chiu


66/159




67/159

XIT

2. Ngun thao tc xoay chiu

(tham kho)


68/159

7.2.1. Nguyn l tc ng ca bo v qu dng in

7.2.2. Bo v qu dng in cc i cho ng dy

7.2.3 S bo v dng cc i cho ng dy

7.2.4. Bo v dng cc i khi chm t kp

7.2.5. Dng khi ng ca bo v dng cc i

7.2.6. Thi gian duy tr ca bo v dng cc i

7.2.7. nhy ca bo v dng cc i

7.2.8. Bo v dng cc i c kho in p cc tiu

XIT


69/159

7.2.1. Nguyn l tc ngcabov qu dng in

7.2. B V QU D CC I CHO NG DY HTCC 2


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XIT

Nguyn tcphihp tnh tc ngchn lcca cc bovct nhanh cthchinbng

cch chnhnh gi tr dng khing.

Bov qu dng incci l mtdngbovcbncang dy c ngun cung cp

tmt pha. Trong li c ktcuphctphn,i khi bov dng ccicsdng

nhmtbovph.

Trong li c ngun cung cp tmt pha, bov dng ccisct trn umi

onng dy v pha c ngun,nhvymionng dy c mtbov ring, khi

xuthinsconng dy no th bovung dy phi tc ngct

ng dy sc.

7.2.2. Bo v qu dng in cc i cho ng dyKhi h t i t i t h th d t i N4 (h h 7 34) d h



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XIT

Khi ngn mch ti N4 th tt

c cc bovukhing,nhngbov 4 c thi gian

duy tr nh nht s tc ng

trc, ng dy s c

cct ra. Khi scc

giitr,bov 1, 2, 3 c thi

gian duy tr lnhnchakptc ng,strvtrng thi

ban u.

Khi ngnmch timtim no trong h thng, v d ti N4 (hnh 7.34), dng ngnmch

chy qua ttc cc onng dy, c thdnti lm cho ttc cc bov 1, 2, 3, 4 u tc

ng. Song theo iukinchnlcchcho php bov 4 tuonng dy sc tc

ng v chonng dy bscbct ra.

mboiukin tc ngchnlc th phichnhnhthi gian duy tr cabovcp trn

philnhnthi gian duy tr cabovcpdignmtkhongthi gian t.

Hnh 7.34.c tnh thi gian ca bo v


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Khi ngn mch trong on AB bo v ct

7.2.2. Bo v qu dng in cc i cho ng dy7.2. B V QU D CC I CHO NG DY HTCC 2


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XIT

nhanh 1 s tc ngvi thi gian tCN1, cn

bov dng cci 1 khng kp tc ng. Khi

ngnmch trong on BC bovct nhanh 1khng tc ng, cn bov dng cci 1 s

tc ng vi thi gian tCD1. Khi ngn mch

trong on CD, bov dng cci 1 mc d

khingnhng khng kp tc ng,bov

ct nhanh 2 vi thi gian tc ngngnhn

s tc ngtrc.

bomiukin tc ngchnlc th:Hnh 3.35.S phi hp bo v ct nhanh

v bo v dng cc itCD1 = tCD2 + t (7.1)Ik.CN1 > Ik.CN2 (7.2)

Trong :

- tCD1

, tCD2

: l thi gian duy tr cabovcci 1 v 2.

- Ik.CN1 , Ik.CN2: l dng khingcabovct nhanh 1 v 2.

bov cho cc ng dy ta sdng cc bov dng cci sau:

7.2.3 S bo v dng cc i cho ng dy



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XIT

Bov dng cci c ththchin ba pha hoc hai pha, tc ngtrctiphoc gin tip.

- Theo ngun thao tc chia ra: Bov c ngun thao tc mtchiu v bov c ngun thao tc

xoay chiu.

- Theo c tnh thi gian tc ng chia ra: Bov c c tnh thi gian tc ngclp v ph

thuc.

1. Sbovt trn ba pha, ngun thao tc mtchiu

2. Sbovt trn hai pha, ngun thao tc mtchiu

1. Sbovt trn ba pha, ngun thao tc mtchiu.7.2.3 S bo v dng cc i cho ng dy



75/159

XIT

a)Bov c ctnhthig ian tcngclp.

Trong sbov ba pha, my bin dng v rlecni theo s sao (hnh 7-36).

Nhngphntcbncabov dngcci

Nguyn l hotng.

Hnh 7.36. S ni dy ca bo v dngcc i c c tnh tc ng c lp

Thi gian tc ngcabovcthchin

trn rle thi gian ThG, khng ph thuc vo

tr s dng ngn mch, v vy bo v nyc gi l bo v c c tnh thi gian tc

ngclp.

unhcim:

-ngin,vn hnh d dng.- D dng phihpthi gian tc ng vi ccbov khc.- Khi dng ngnmchlnbov khng thctnhanh hn.

b )Bov c ctnhthig ian tcngphthuc.7.2.3 S bo v dng cc i cho ng dy



76/159

XIT

Trong sbo

Trong s khng sdngrle thi gian v thi gian tc ngcabov do rle dng thc

hin, khng c rle trung gian v tipimcarle dng c khnngtiln, khng c rle tn

hiu v rle dng c tn hiuc (con bi), n c khi rle tc ng.

Nhvyrle dng cmng PT -80 c th lm thay chcnngcachpb cc rle dng,

rlethi gian, rle tn hiu, v rle trung gian.

Hnh 7-37. S ni dy ca bo v dncc i c c tnh tc ng ph thuc

Bov c c tnh thi gian tc ngph thuc l

bov c thi gian tc ng(thi gian duy tr) ph

thuc vo trs dng ngnmchchy qua bov tbv

= f(I).

Bovdng ny cthchinbngrle tc ng

khng tcthi, c thi gian tc ngphthuc vo

dng chy qua rle,thngsdngrle dng cmngloi PT -80. Sbov c c tnh phthuc

cbiudin trn (hnh 7-37).

Khc vibov c c tnh thi gian tc ngclp(ng 1 hnh 7-38), bov c c tnh

7.2.3 S bo v dng cc i cho ng dy7.2. B V QU D CC I CHO NG DY HTCC 2


77/159

XIT

thi gian tc ngphthuc(ng 2) v phthuchnch(ng 3) tc ng khi

IR = (12)Ik

V vybov khng tc ng khi qu tingnhn(Iqtnh). Ngoi ra bov c c tnh thi gian

tc ngphthuc cho php tngcngct khi scung dy, nu dng ngnmch

ti N1 lnhnnhiu dng ngnmchcuing dy.

Hnh 7.38. Cc dc tnh tc ngca bo v qu dngC

Song vic duy tr thi gian bov

c c tnh thi gian tc ngc

lpnginhnrtnhiubov

c c tnh thi gian tc ngph

thuc, v vy n chcsdng

trong nhngtrnghpthunli.



78/159

XIT

c) unhcimsbovt trn ba pha, ngun thao tc mtchiu.

Sbov dng cci ba pha phnngvi ttc cc dngngnmch, v vy n

csdng trong li c trung tnh nittrctip.

Trong li c trung tnh cch inhocnit qua cun Petecxen sbov ba pha

khng csdng v:

- S ba pha thns hai pha, v cnnhiuthitb v dy nihn.

- Bov ba pha c trnghp lm vic khng chnlcbngbov hai pha khi chmtng

thi ti hai im khc pha ca nhng l ng dy khc nhau cng ni chung mt ngun

(chmt kp).


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80/159

Rle dng khi ng mc vo hiu s dngSmtrle



81/159

XIT

hai pha, phn ng vi tt c cc dng ngn

mchgia cc pha.

IR = Ic Ia

Hnh 7.41. S bo v 2 pha 1 rle

uimcasmtrle:

sthitb v dy ni l t nht.

Nhcim:

-nhy km hns hai bin dng hai rle khi ngnmchgia pha c t my bin dng

v pha khng t my bin dng. Nhcim ny c ngha khi dng ngnmchlnhn dng

phti khng nhiulm.

- Khng bovc khi ngnmch hai pha sau my bin p ni/hoc/hocngnmch

mt pha sau my bin p ni0/0.

- Khi rle v dy ni khng hon ho,bov khng lm vic

Vy,smtrlechcsdng trong li phn phi c trung tnh cch invi

(610) kV v bovngcin.


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83/159


84/159

Nubov 1 v 2 c thi gian duy tr ging nhau, khi chmtxuthin trn hai pha A v C t

hai im N1 v N2 (hnh 7-44b) trn pha c t my bin dng th c hai bov 1 v 2 u t

7.2.4. Bo v dng cc i khi chm t kp7.2. B V QU D CC I CHO NG DY HTCC 2


85/159

XIT

( ) p y g

ng c ngha l khng chnlc. Bov 3 pha trong iukin ny cng tc ng khng chnlc.

Tuy nhin bov hai pha c uimnibthns ba pha, nuchmtxy ra trn pha ct my bin dng N2 v trn pha khng t my bin dng N1 (hnh 7-44c). Bov hai pha trong

trnghp ny chctmtng dy sc cn ng dy khc c chmtmt pha vnc

gilitiptc lm vic. Bov ba pha trong trnghp ny tc ng khng chnlc,ctc ha

ng dy vittc cc dngchmt kp khc nhau.

Hnh 7.44. S biu din chm t kp.

Cc trng hp bo v tc ng khng chn lc


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7.2.5. Dng khi ng ca bo v dng cc i7.2. B V QU D CC I CHO NG DY HTCC 2


87/159

XIT

V d, khi ngnmch ti N2 (hnh 7.45), rle dng cabov 1 v 2 ukhing. Sau khi

ngnmchc loitrbngbov 2 th rle dng cabov 1 cnphitrv. Munvy

dng trvcarle dng trong bov 1 phi lnhn dng ph ticang dy chy qua

bov sau khi ct dng ngnmch.

Sau khi ct dng ngnmch, in p phchi, cc ng c b hm li trong qu trnh ngn

mchttngtc ln tc ban u. Dng inchy trn ng dy vnlnhn dng lm vic

nnh v sgimdnv gi trnnh.

Trnghpxunht dng lm vicnnh l dng phticci Ilv max cac hai ng

dy hocnh nhng hnch l I Ilv max ca ring anng dy 1. Trong iukin ny

mbo cho rle dng cabov 1 trvmt cch tin cy th:

Hnh 7.45. Phi hp cc bo v dng cc i

Itv > kk. Ilv max (7.4)

Trong : kk l hskhing tnh tistng dng in do cc ngctkhing. Thng lykk = (1,52).

Dng trvcarle cn c th xc nh theo biuthc sau:

I = k k I (7 5)

7.2.5. Dng khi ng ca bo v dng cc i7.2. B V QU D CC I CHO NG DY HTCC 2


88/159

XIT

Itv = kdt. kk. Ilv max (7.5)

Ta bitrng:

Trong: kdt l hsdtrtnh tisais dngtrvcarlev pha cao hnlybng(1,11,2).

bv.kd

tv

tvI

Ik

Trong: ktv l hstrvcarle dng.

Vy:maxlv

tv

kddt

tv

tv

bv.kdI.

k

k.k

k

II (7.6)

Nuiukin (7.4) c tho mn th iukin(7.3) cngctho mn v dng khingca

bov bao gicnglnhn dng trvca n.

Dng khingcarle dng:BI

bv.kdsdR.kd

n

I.kI

Trong: ksd

l hss; nBI

l tsbinicamybin dng.

- vis sao v sao thiuksd = 1- vishius ksd = 3

Hnh 7.46. S biu din dng ngn mch


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7.2.6. Thi gian duy tr ca bo v dng cc



90/159

XIT

Trong: tdt l thigian dtr,thnglybng0,1 s.

bov 1 khng kp tc ng khi ngnmch trn ng dyD2 th thi gian duy tr cabo

v 1 philnhnthi gian tntingnmch khi ngnmch trn ng dyD2:

t1 > tNM = t2 + tSS2 + tMC2

C th do sai scarlethi gian cabov 1 m thi gian duy tr cabov 1 c thgimi

mtlng tSS1 l thi gian sai sv pha nhanh hncarlethi gian cabov 1.

V vychn:t1 = t2 + tSS2 + tMC2 + tSS1 + tdt

t = t1 - t2 = tSS2 + tMC2 + tSS1 + tdt (7.7)

Biuthc (7.7) dng tnh phn cpthi gian tc ngcabov c c tnh clp

Hnh 7.47. Phi hp thi gian tc ngca bo v dng cc i


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2. Chnthi gian tc ngcabov7.2.6. Thi gian duy tr ca bo v dng cc



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XIT

Thi gian duy tr cabov c c tnh tc ngclpc xc nhxut pht t phn cp

thi gian cho theo biuthc:

t1 = t2 + t (7.8)

Thi gian duy tr cabov c c tnh tc ngphthuchocphthuchnchcngphi

tho mn iukin (7.8), nhngthi gian tc ngcabov ny ph thuc vo dng in

chy qua n, cnphibitgiihn dng m iukin ny (7.8) cnphithchin. Githit

rngng dyD2 (hnh 7.47) c trang bbov c c tnh phthuchnch. Yu cu

chnc tnh cabov 1 tho mn iukin tc ngchnlc.

Bov 1 cnphi c thi gian duy tr lnhn thi gian duy tr cabov 2 mtcpt khi

ngnmch trong vng ko theo skhingcac hai bov c ngha l ngnmch trn

ng dy 2. Khi ngnmch trn ng dyD1 th thi gian duy tr cabov 1 khng cn

philnhnthi gian duy tr cabov 2. Nungnmchtiim N2 (u vng bovca

bov 2)

Dng chy qua hai bov 1 v 2 l IN2 th vittc cc imngnmch sau N2 c ngha l trong

vng bovcabov 2 th dng ngnmchsnhhn.iukin tc ngchnlcphic

7.2.6. Thi gian duy tr ca bo v dng cc 7.2. B V QU D CC I CHO NG DY HTCC 2


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XIT

g g g g p

thchinvi ttc cc dng ngnmchnhhnhocbng dng ngnmchcci ti N2

IN2.max . Trnghpngnmch trn ng dyD1, thi gian tc ngcabov 1 khng cnphiphihpvibov 2, thi gian tc ng cng nh cng c li. V khi ngnmch trn ng

dyD1 dng ngnmchchy qua bov 1 slnhn IN2.max.

Cc bctin hnh:

- Xy dng cc c tnh t = f(I) cabov 2 v bov 1 cn phihpvi nhau.

- Xc nh dng ngnmchcci IN2.maxchy qua bov 1 v 2 khi ngnmchu vng bovcabov 2.

- Cnc vo c tnh cabov 2 xc nhthi gian tc ngcabov 2 ngvi IN2.max l t2.

- Theo iukin tc ngchnlc,thi gian tc ngcabov 1 ngvi IN2.max cnphiln

hnthi gian t2cabov 2, mtcpthi gian tc ngt:

ttt21

(7.9)

iukin (7.9) phitho mn vittc cc dng ngnmchnhhn IN2.max.c tnh cabov

tho mn iukin (7.9) cthchin trong thitkbng cch chnrle,cthchin trong

7.2.6. Thi gian duy tr ca bo v dng cc 7.2. B V QU D CC I CHO NG DY HTCC 2


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XIT

( ) g g g

vn hnh bng cch iuchnhtrstthi gian carle.

-c tnh cabov 1 cchn,c xy dng chung cng mthtrcvic tnh cabov 2 d dng kim tra iukin tc ngchnlc.

ttt21

Ch :thuntin, khi xy dng chung mth

trc,c tnh camtsbovcqui i v dng pha s cp, nhngkhi ny phi ch tisnimchdng ca bov c ngha l phiknks.

Nu cc bovcnphihpc tnh t cc pha khc nhau ca my binp th c tnh ca n cnphic

qui i v dng ca cng mt cpin p.

Hnh 7.48. Phi hp thi gian tc ng ca bo vdng cc i s dng r le dng c c tnh

tc ng chn lc

7.2.7. nhy ca bo v dng cc i

nhy ca bo v c c trng bi h s nhy:



95/159

XIT

5,1I

I

kbv.kd

min.Nnh

nhycabovcctrngbihsnhy:

(7-10)

Trong : IN.min l dng ngnmchnhnhtcui vng bov.

Vng bovcabov dng cci bao gmng dy cbov v phnng dy tip

theo hay my bin p.

nhycabov dng ccibov cho ng dy philnhnhocbng 1,5. V dng

thc tchy trong rle khi ngnmch c thnhhn trs tnh ton IN.min do tnh dng ngn

mch khng chnh xc, nhhngcaintrtiv tr sc v sai sca my bin dng lm

gim dng thcp.

Nunhycabovcci khng mbo c thkhcphc:

- Thay bngbov qu dng incci c kho in p cctiu- B sung thm bovtht khng.

Hnh 7.49. Vng bo v cabo v dng cc i


96/159


97/159

2. Dng khingcarle dng

Dng khingcarle dng cchnhnh theo dng ph ti lu di lnnht. Dng ny

trong thc t c th nh hn 1 5 n 2 ln dng ph ti cc i:

7.2.8. Bo v dng cc i c kho in p cc tiu7.2. B V QU D CC I CHO NG DY HTCC 2


98/159

XIT

trong thct c thnhhn 1,5 n 2 ln dng phticci:

maxlv

tv

dt

kdbv I.k

k

I (7.11)

V vynhycabovctng ln.

3. in p khi ng ca rle in p cc tiu RU1

in p khingcarlein p RU1cchn sao cho:

- Khng tc ngngvimcin p lm vicnhnht cho php.- Tc ng khi ngnmch ngoi vng bov v phitrv khi sccgiitr.

Nubomciukin 2 th iukin 1 stho mn. V vy Ukdcchn theo iukin

trvcarle.bomiukin ny th in p trvcarle Utvphinhhnin p

lm vicnhnht Ulv min:

BU

minlv

tvn

UU

dtBU

minlvtv

k.n

UU

Trong: nBU l hsbin p ca mybinin p olng,Vi: kdt l hsdtrlybng(1,11,2).


99/159

4. nhy ca bo v

nhycarlein p ckim tra theo in p tn dlnnht khi ngnmchcui vng

bov (UN max). U

7.2.8. Bo v dng cc i c kho in p cc tiu7.2. B V QU D CC I CHO NG DY HTCC 2


100/159

XIT

( N.max)5,1

U

Uk

min.N

kdnh (7.12)

Thc t thy rngng dy di cung cp tngun cng sut lnhoc trn ng dy c

khng in,nhycarlein p thng khng (v ving dy ny in p tn d kh

ngnmch cui vng bovln). V vybov c kho in p chsdng cho ng dy

ngn v trung bnh.

Rlein p ccitht khng RU0 tc ng khi xuthinngnmch 1 pha v ngnmch ha

pha chmt.

Trong ch lm vic bnh thngin p tht khng U0 = 0, nhng do sai scablc cung

cp cho rle m trn uccca n xuthinin p gi l in p khng cn bng Ukcb.

rle RU0 khng tc ng khi c in p khng cn bng, ta chn: Ukd.0 > Ukcb.

Xut pht tiukin ny Ukd.0carle RU0(1520)% Umax trn ucccab lc khi ngn

mchmt pha. Khi ngnmchmt pha in p cci trn ucccun dy tam gic hcamy bin p olng l 100V. Do :

Ukd.0 = (1520)%.U0max = (1520)%.100 = (1520) V.


101/159


102/159

7.3.1. Nguyn l tc ngcabovct nhanh

7.3.2. Scabovct nhanh

7.3.3. Thng scabovct nhanh tcthibov trn ng dyc ngun

7.3.4. Thng scabovct nhanh c duy tr thi gian bov trnng dy c ngun cung cptmt pha7.3.5.nh gi bovct nhanh

7.3.6. Phihpbovct nhanh v bov dng cci thnh bovqu dng in ba cpXIT

7.3. BV QU D CT NHANH CHO NG DY HTCC 2

Bov dngccic uim l ngin, tin cy, gi thnhr nhng c nhc im l thigian duy tr ln. c bit l ngnmch gn ngun cung cp dng

7.3.1. Nguyn l tc ngcabovct nhanh

Bovct nhanh l mtdngcabov qu dng in

cho php ct nhanh dng ngn mch.


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XIT

mchgnngun cungcp, dngngnmch l lnnht,nhngthi

gian tcnglidinht. khcphcnhcim nyngita sdngbovctnhanh.

cho php ct nhanh dng ngnmch.

Bov dng inct nhanh chia ra:- Bovctnhanh khngthigian.

- Bovctnhanh c duy trthigian.

Tc ngchnlccabovct nhanh khng thi gian cthchinbng cch hnch

btdi lm vicca n, sao cho bovct nhanh khng tc ng khi ngnmchngdy ln cn, m bovcang dy c thi gian duy tr bnghoclnhnthi gian tc

ngcabovct nhanh.

Munvyphichn dng khingcabovct nhanh lnhn dng ngnmchcci

cuing dy m n bov.

Phng php hnch vng tc ngcabovct nhanh da trn cs l trs dng ngnmchphthuc vo tngtrtrcimngnmch.

7.3.1. Nguyn l tc ng ca bo v ct nhanh

Dng ngnmchtibtkim no trn onng dy kho st c tnh nh sau:

htht

NM

EEI (7-14)



104/159

XIT

K0htk.ddht

NM

.XXXX

Trong :- Eht l scinngcahthng quyivin p tnh ngnmch.- Xht, XD.K l in khngcahthngv ngdytrcimngnmchv pha c ngun.- X0 l in khngca 1 km chiu diz,/km.- l chiu diztnguntiimngnmch

Nu theo iu kin tc ng chn lc th

bovct nhanh khng tc ng khi ngn

mch ti im M (hnh 7.60), mun vychn:

Hnh 7.60. Vng tc ng ca bo v ct nhanh

Ikd > IN.max (M)Trong:

- Ikdl dngkhingcabovctnhanh.

- IN.max (M) l dng ngn mch cc i vi imngnmchcuingdycbov.

Nhvybovct nhanh chtc ng khi ngnmch trong vng AN, khng tc ng khi c

ngnmch trong on NM. Vng NM gi l vng chtcabov, n cbovbngbov

dtr khc nhbov dng cci.


105/159

7.3.3. Thng s ca bv ct nhanh tc thi trn ng dy c ngun

1. Dng khingcabovct nhanh

Theo iu kin tc ng chn lc dng khi ng ca bo v ct nhanh khng thi gian phi



106/159

XIT

Theo iukin tc ngchnlc, dng khingcabovct nhanh khng thi gian phi

lnhn dng ngnmchccicuing dy cbov(im M). Munvy ta chn:

Ik = kdt.IN. max (M) (7-15)

Trong:

- IN. max (M) l dngngnmchlnnhtcuingdycbov.- kdt l hsdtrktisais trongvictnh IN. max (M) v saisv dngkhingcarlev cchn sao cho bovctnhanh khng tcngkhi c ngnmchtiM.

V thi gian tc ng ring cabovct nhanh bngkhong (0,020,01)s nn dng IN. max (M)c tnh tithiim ban u (t = 0) lybng dng siu qa ban u I N. max (M)Trong sbovct nhanh nu rle dng tc ng trc tipn my ct khng qua rle

trung gian, thi gian tc ngcabovct nhanh c thtti 0,02s, th khi phi tnh ti

thnh phn khng chu k ca dng ngn mch bng cch nhn IN. max (M) vi h s ka =

(1,61,8).

ivi bo v ct nhanh s dng rlein t PT ta ly kdt = (1,21,3). i vibo vct

nhanh sdngrle PT -80 (rle dng cmng),loiRleloi ny c sai slnv dng khi

ng (2025)% th kdt cly cao hn l: kdt = 1,5.

7.3.3. Thng s ca bv ct nhanh tc thi trn ng dy c ngu

Vng tc ngcabovct nhanh trong hai trnghpc xc nh trn th. Thng

ngi ta xy dng ng cong dng ngn mch ph thuc vo khong cch t u ng

2. Vng tc ngcabov



107/159

XIT

ngi ta xy dngng cong dng ngnmchph thuc vo khong cch tung

dy nimngnmchchvn hnh cci v cc tiucah thngin (ng

cong 1 v 2 hnh 7.62), imctgia chng v ng thng Ik l im cui ca vng ct

nhanh trong chcci v cctiu (AN1 v AN2).

Vng tc ngcabovct nhanh c

xc nh theo cng thc:

Hnh 7.62.Vng bo v ca bo v ct nhanh

htkd

ht

ddCN

XI

E.

X

100%X (7.16)

Trong:- XCN% l vng tc ngcabovctnhanh

tnh theophntrm v khngcnhhn20% ngdycbov.

- Ik l dngkhingcabovctnhanh.

- Xht l in khngcahthng.- XD l in khngcangdycbov.- Eht l scinngcahthng quyiv

cpin p cangdy.


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109/159


110/159


111/159

7.3.6. Phi hp bo v ct nhanh v bo v dng cc i thnh bo v qudng in ba cpPhihpbov dng cci,bovct nhanh khng duy tr thi gian v bovct nhanh

c duy tr thi gian ta cbov qu dng in ba cp. Bov qu dng in ba cp cho



112/159

XIT

c duy g a a c bo qu d g ba cp o qu d g ba cp c o

php ct nhanh ngnmchgnngun cung cp v bovd phng cho ng dy pha

sau.

c tnh thi gian tc ngcabov ba cpcbiudin trn (hnh 7.64).

Trong : Bov I l bovct nhanh khng

thi gian; Bov II l bovct nhanh c duy

tr thi gian; Bov III l bov dng cci.

Hnh 7.64. c tnh thi gian ca bo vqu dng in ba cp


113/159

7.4.1. Dng in v in p khi chm t mt pha

7.4.2. Nhng yu cu c bn i vi bo v chm t

7.4.3. Cc loi bo v chm t

XIT

7.4. BV CHM T TRONG LI C DNG CHM T NH HTCC 2

7.4.1. Dng in v in p khi chm t mt pha

Li c dng chmtnh l li c trung tnh cch invithocnit qua cundph

quang (Petecxen). Chmt trong li ny khng gy ra sgimin p gia cc pha v tng

dng in trong li n gin cho vic kho st s thay i dng v p khi mt pha chm t


114/159

XIT

dng in trong li.ngin cho vickho st s thay i dng v p khi mt pha chmt

ta coi ng dy khng ti.

Hnh 7.65. S biu din dng v p khi chm t 1 phatrong li c dng chm t nh

7.4.1. Dng in v in p khi chm t mt ph

Trong ch bnh thngin p ca dy dn pha A, B, C so vitbngin p pha: , ,

Khi khng ti c gi trbngscinng pha , , . Cc vc tin p to thnh hthng

hnh sao i xng tng hnh hc ca chng bng khng do in p trung tnh bng khng:

AU

BU

AE

BE CE



115/159

XIT

hnh sao ixng,tng hnh hcca chng bng khng, do in p trung tnh bng khng:

Utt = 0. Giin dung gia dy dn pha A, B, C vit l CA, CB, CC. Dng in dung cxc nh theo cng thc:

C

A

)C(AJX

UI

C

B

)C(BJX

UI

C

C)C(C

JX

UI

Tng dng in dung d bng khng do : I0 = 0

1.in p cc pha khi chmt

Khi chmt kim loica 1 pha no ,gis pha A. in p ca pha ny so vitgim

xungbng khng: UA = 0. V nit,im N mang in thbng khng (in thcat).

in p im trung tnh so vitbngin p giaim N v O bngvtrsnhngngc

duviscinng pha chmt.

A0Ntt EUU (7-18)

in p gia cc pha khng sc pha B v C so vitbngin p gia dy dn pha B v C

so viim N v bngin p dy:

BAB U'U CAC U'U




116/159

XIT

Biudin v qua :B'U

C'U

ttU

BttB EU'U

CttC EU'U

Hnh 7.66. Gin vc t biu dindng v p khi chm t trong li c

dng chm t nh

2. Dng in qua imchmt 1 pha

V: Nn:0UA 0I )C(A C

BA

C

B)C(B

jX

U

jX

'U'I

C

CA

C

C)C(C

jX

U

jX

'U'I




117/159

XIT

Dng inchmttiv tr chmtbngtng hnh hc dng in trong pha B v C nhng c

chiungcli:

C

CA

C

BA)C(C)C(Bcd

X

U

X

Uj)'I'I(I

C

f

C

A

C

Acd

X

U.3j

X

E.3j

X

E.3jI

Nhvy dng inchmtbng ba ln dng in dung trong ch bnh thng v chm sau

Uttmt gc 900.

Ta c thvit:L.C..U3C.U3I 0ffcd

Trong:

- L l chiu dingdy, km.- Co l in dung dydntca 1 km ngdy.


118/159

a) Chmt qua tngtr qu b) Chm t trong li c b




119/159

XIT

Hnh 7.67. S biu din dng v p khichm t qua in tr qu

Hnh 7.68. S biu din dng v p th tkhng khi chm t trong li c trung tnh ni

t qua cun dp h quang

7.4.2. Nhng yu cucbnivibovchmt

Chm t khng gy ra dng siu qu v khng lm lch mo tr s in p gia cc pha

Yu cuivibovchmt trong li c dng chmtnh khc vinhng yu cuivbovngnmch.



120/159

XIT

Chmt khng gy ra dng siu qu v khng lm lch mo trsin p gia cc pha,

chng khng gy nhhngtis cung cpin cho cc h tiu th v khng ko theo s quti cho thitb. V vy khc vingnmch,chmt khng cnloitr nhanh.

Ctchmt l cnthit, do hiungnhitca dng chmttiv tr sc c th ph hu

cch ingia cc pha v chuyntngnmchmt pha sang ngnmchgia cc pha. Ngo

ra qu in p do chmt c th gy ra chcthnghoc phng inbmtsca cc pha

khng sc, gy ra chmtngthitinhiuim khc nhau trong li.

Theo kinh nghim,chmt trong li c b hoc trong li c dng chmtnh (2030) A

li (610) kV c thtnti trong thi gian khong 2 gi m khng lm tngsc v khng ph

hus lm vic bnh thngca cc h tiu th.

V vybovchmt trong li c dng chmtnhchcn tc ng bo tn hiu.


121/159


122/159

7.4.3. Cc loi bo v chm t2. Bovchmt c chnlc

a) Dng blcthtkhng

S ny c nhcim:



123/159

XIT

Hnh 7.70.S bo v chm tdng b lc th t khng


124/159


125/159

c) Dng khingcabovchmt

Nuli c trung tnh nit qua cundph quang:

3f)CC(

1.U3

7.4.3. Cc loi bo v chm t



126/159

XIT

BI

3

L

f

R

)(

XI

Trong li c b, dng inin dung

chy qua bov khi chmt cc

ng dy khc c tr s c th so

snh c vi dng s c. V thtrong li ny bov dng chmt

t csdng.

Trong li c b dng 3I0(L) chkhp

mch qua bo v ca ng dy b

scchmt (hnh 7-73).Hnh 7.73. S biu din dng in in dung

trong li c trung tnh ni t qua cun dp h quang

Cc my bin p
http://localhost/var/www/apps/conversion/tmp/scratch_3/MBA%20luc.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_3/MBA%20luc.ppt


127/159

Mybin p lmtphntrtquantrng t rongliin,numybin pbscslmngngcungcpin tonb haymtphnh t iuth,nhhngtisnxut, c

th gy ng uyhim chothitbv conngi. Mtkhc g i thnhca mybin p so

viccthitbkhc t ro nghthng cungcpin caohnrtnhiu, ch i phbov

sachacngln,thigiansacha lu . Do mybin pphicbovmt

cchchcchn.

7.5.1. Cc dngsc v cc loibov

7.5.2. Cc ch lm vic khng bnh thng & cc loibov

7.5.3. Bov qu dng in cho my bin pXIT

7.5. BO V MY BIN P IN LC HTCC 2

1. Cc dngsc

Ngn mch gia cc pha bn trong thng du my bin p v trn u ra cc cun dy

7.5.1. Cc dngsc v cc loibov


128/159

- Ngnmchgia cc pha bn trong thng du my bin p v trn u ra cc cun dy.

- Ngnmchgia cc vng dy trong mt pha.

- Chmtcun dy hocu ra cun dy.

- Du trong my bin dng bcn,dub phn hu.

- Vsu vo v u ra my bin p.

XIT Kinh nghimvn hnh cho thy cc dngngnmchu ra v ngnmchgia cc vng dy

trong cng mtcun dy pha l xuthin kh nhiu.

Scgia cc pha bn trong thng duxy ra t, v khong cch gia cc pha ln.

7.5.1. Cc dng s c v cc loi bo v

Khi ngnmchgia cc vng dy cng pha, dng

chy t ngun cung cp ti v tr sc khng ln

lm. V d khi ngnmch W vng dy (hnh 7.74).

Dng ngnmchbng IN, th dng scchy t



129/159

XIT

Hnh 7.74.S c ngn mch

gia cc vng dy

g g c b g N, t d g s c c y t

ngun ti c th tm c bng phng trnh cn

bngsctng:

INW1 = IN . W

Suy ra:

V ts do : IN


130/159

XIT

2. Cc loibovsc bn trong my bin p

Bovct nhanh.

Bov so lch.

Bovchmv thng my bin p.

Bovrlehi.

7.5.2. Cc ch lm vic khng bnh thng v cc loi bo v

1. Cc ch lm vic khng bnh thng

Ccch lmvickhng bnhthnghaygpnht l dnginchy t rongcun dy



131/159

Ngnmch ngoi

Ngnmch ngoi xy ra khi ngnmch trn thanh ci u ra my bin p hocngnmchtrn cc lng dy i ra t thanh ci my bin p m bov cc lng dy ny tchitc ng.

Dng ngnmch ngoi thnglnhn dng nhmcnhiuln, lm nhitcacun dymy bin p lnhn gi tr cho php, c thdntisc my bin p. V vy my bin pcnphicbovngnmch ngoi. Ngnmch ngoi cn ko theo sgimin ptrong li, do bovcnphi tc ngvithi gian nhnhtnhngphibom tnh

tc ngchnlc

XIT

mybin ptngqu g itrnhmc,xuthin t rongnhngtrnghp:Ngnmch ngoiv Quti

Qu ti

Qu tixuthin khi cc ngctkhing,tngphti do ctbt my bin p lm vic

song song, ngtngphti...




132/159

Qu tithng khng gy st p ln trong li. V vythi gian tc ngcabov qu ti

c xc nhxut pht tbn cch incacun dy my bin p.Q

Qu tithnggp l qu tingnhn v t loitr khng nguy him cho my bin p v

thi gian tntingn. V d, qu ti do ngctkhing,phtinhnhn, cc mybin p khi qu tinhvy khng yu cubov tc ng. Qu ti lu di xy ra khi tng

ngphti, do thitbtngngngund phng tc ngctmts my bin p lm

vic song song... cc trnghp qu ti ny c thcloitrbng tay.

V vybov qu ti cho my bin p chtc ng khi khng thtin hnh ct qu tibng

tay. Nhngtrnghp cn libov qu tichcn tc ng cho tn hiu. Bov qu tithchin theo nguyn tcphnng theo dng in cho tn hiuhoc tc ngn my ct

tu theo tnh chtcatngtrm.

XIT

2. Cc loibov




133/159

Bovngnmch ngoi, bao gm:

- Bov dng cci.

- Bov dng cci c kho in p cctiu.

- Bovthtnghch,tht khng...

Bov qu ti: Bov qu tithng dng l bov qu ti theo dng.

XIT


134/159


135/159

Bovct nhanh khng c tc ng khi xuthin dng t ho nhyvt khi ng

my bin p khng ti vo li th phitho mn iukin:

I = (35) I (7.29)

7.5.3. Bo v qu dng in cho my bin p



136/159

XIT

kdbv( )

dm.BA

Dng khingcabovcchnphitho mn c hai iukin trn. Khng

tc ng khi xuthinngnmch trn thanh ci u ra my bin p v khi xuthin

dng t ho nhyvt. Dng khingcabovcchnbng gi trlnnht

tnh theo hai biuthc (7.27) v (7.29).

uimcabovct nhanh l sngin, tc ng nhanh, tngcngct

scu vo my bin p.

Nhcimcabovct nhanh l chbovctphnucacun dy s

cp my bin p trvnv tr t my bin dng


137/159

2. Bov dng cci cho my bin p

Sbov dng cci cho my binp ba pha hai dy qun c ngun cung




138/159

XIT

cptmt pha cthhin trn (hnh7.76). Bo v c t pha nguncung cp my bin p nm trong vngbov.

m rng vng bo v ca bo vdng cci th my bin dng cnt

gn my ct. my bin p hai dy qunc ngun cung cpmt pha bovcntc ngn my ct MC1. Song tng tin cy khi ngn mch ngoi th cntc ng n c hai my ct MC1 vMC2 vimcchmt my ctcd

phng bngmt my ct khc.Hnh 7.76.S nguyn l

ca bo v dng cc i

Sni my bin dng v rlecabov dng ccicnphibom tc ng trong tt

c cc trnghpngnmchxy ra.

T h b h h kh




139/159

XIT

Trong li c trung tnh nitbov thchin ba pha, cn trong li c trung tnh khngtrctipnit,bovcthchin theo s sao thiu. S sao thiu c nhy km

hns ba pha khi ngnmch hai pha sau my bin p ni Y /hocngnmchmt pha

sau my bin p ni Y /Y0.

tngnhy ta t thm rleth ba vo dy dnngcca cc my bin dng (ni vo

tng dng hai pha). Khi nhycas hai bin dng ba rle c nhybngnhycas ba bin dng ba rle m li c uimhns ba pha l titkimcmtbin

dng, titkimc dy ni, trong trnghpchmt kp th tc ngchnlchns

ba pha.

Shius khng c dng bov cho my bin p ni Y /hoc Y /Y0 v c trng

hpngnmchbov khng tc ng.


140/159

nhycabov:

3,1I

Ik

kdBV

minNnh




141/159

XIT

Trong :

- IN.min: l dngngnmchnhnht ticuivngbov quyivscp. ivimybin p niY / l dngngnmch 2 pha sau mybin p, vimybin p niY /Y0 l dngngnmch 1 pha sau mybin p, vidngsck trn dngchyqua bovphi l nhnht.

- Ikd.bv: l dngkhingcabov.

Trnghpkim tra bov khng nhy th cch khcphcnh sau:

+ Nubov dngs haibin dng hairle thchuyn sang dngs haibin dng barle,

+ p dngbov dngccic kho in p cctiu.)


142/159

3. Bo v dng cc i c kho in p cc tiu

Khi ngnmch trong vng bov dng chy qua bov tng ln, ng thiin p mng

gimxungrle dng RI1 v cc rlein p cc tiu RU u tc ng. Rle thi gian c

in, sau mtkhong thi gian duy tr, tipim thngmngchmng li rle trung

gian c i tc ti i th l i t c i my t t




143/159

XIT

gian c in, tc ng,tipimthngmngli,cunct c in, my ctct.

Khi qu ti khng km theo sgimin p

qu mc rle in p cc tiu khng tc

ng,rlethi gian ThG1 khng c inbo

v khng tc ngnct my ct

unhcim:Nhvybov khng tcng khi qu ti v chtc ngnct my

ct khi ngn mch. Do dng khi ng

cabovchcnchnhnh theo dng nh

mcca my bin p, nn nhycabo

v cao hn bo v dng cc i n gin.

Nhng cng c nhc im l dng nhiu

rlehn v phi c my bin p olng.

Hnh 7.77. S nguyn l ca bo vdng cc i c kho in p cc tiu

Dng khi ng ca bo v:

Dng khingcabov v dng khingcar le xc nh theo biuthc sau:

dmBA

tv

dtkdbv I.

k

kI




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BI

bv.kdsdkdR

n

I.kI

kdtl h s d tr ly bng (1,11,2)

in p khi ng: c chn xut pht t iu kin sau:

Rlein p phitrvngvimcin p lm vicnhnht:

dmminlvtvU95,0UU

dt

minlv

tvk

UU

kdt l hsdtrlybng(1,11,2)

Ta c:

tvdt

lvmin

tv

tvkd

kd

tvtv

k.k

U

k

UU

U

Uk

ktvl hstrvcarlein p cctiu: ktv= 1,15.

dm U750U950U




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dmdm

kd U75,015,1.2,1

U95,0U

Khi ngctkhingin p gimxungn (0,650,7) Udm th rlein p cctiu

cng khng c tc ng, do ta chn:

Ukd = (0,50,6) Udm

in p khingcarle:

BU

kdBVkdR

UU

BU : l h s bin i ca my bin in p o lng.

nhycabov:

Theo dng in:

31Ik minN




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3,1IIkkdBV

min.Nnh

5,1U

Uk

max.N

kdnh

INMmin l dng ngn mch nh nht cui vng bo v quy i v cp in p t bo v

Theo in p:

UN.maxl in p tn dlnnhttrn thanh cinitbov khingnmchcuivngbov

4. Bo v qu ti cho my bin p nhng my bin p c nhn vin trc:

Bo v qu ti ch cn tc ng cho tn

hiu (hnh 7.78). Rle RI chcn t trn

mt pha v qu tithngxy ra ngthi

c ba pha loi tr trng hp cho tn




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Hnh 7.78. S nguyn l ca bo v qu ti

c ba pha.loitrtrnghp cho tn

hiu khi ngn mch v qu ti ngn hn,

sdng thm rlethi gian ThG, cun dy

carle ny c tnh ton lm vicch

di hn

Dng khingcacabov qu ti:c chn xut pht t iu kin tr v

carle dng khi my bin p lm vicvi

thng snhmcnhmc:

Itv = kdt . IdmBA, dmBAtv

dt

tv

tvBV.kd I.

k

k

k

II

kdtl h s d tr ly bng 1,05

Thi gian tc ng:

cchnlnhnthi gian duy tr cabov dng cci.

tqt = tCD.max + t

- tqtl thigian duy trcabov qu ti.

t max l thi gian duy tr ca bo v dng cc i cho my bin p




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- tCD.max l thigian duy trcabov dngccicho mybin p.

5. Bov qu ti cho my bin p nhng my bin p khngc nhn vin trc:

Bov qu tithchin theo 3 cp:

Cpthnht: Lm vic khi qu tinh, tc ng cho tn hiu qua ng dy iukhin xantrmiukhin c nhn vin trc. Thi gian duy tr t1 = tCD.max + t.

Cpth hai:Duy tr mtthi gian t2: t2 = t1 + t < tcp

tcp l thigian qu ticho php ca mybin p

Cpth ba: L cpbohim, tc ngnct my ctnucpth hai v mt l do no khng lm vic. t3 = t2 + t < tcp

6. Bovrlehi cho my bin p

Bovrlehicsdngrng ri bovsc bn trong my bin p v n c

nhy cao.

S c bn trong my bin p l s c xut hin bn trong v thng my bin p c km theo




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Sc bn trong my bin p l scxuthin bn trong v thng my bin p c km theopht nhitt nng cc chi tit,dntidub phn hu(bCrc kinh) v ph huvtliu cch

in,to ra hi. Hii ln bnh dn duct pha trn my bin p v thng vi kh

quyn. Tn hiuhi v chuynngcaduv pha bnh dnducsdngthchin

bovgi l bovrlehi.

Cutocarlehi:

Rle bao gmv gang 1 c dngng ba chc c mt bch nivingdnti bnh dn du

(hnh 7.79).

Trong vrlet hai phao di ng 2a v 2b c dng hnh tr kn tni trong du. Cc phao

u quay ct do xung quanh trccnh 4. Trn u phao ttipimthu ngn 3, lmtngthu tinh trong c gn cc tipim v ngthu ngn.

v tr xc nhca phao, thu ngn sni kn cc tipim. Cc

tipimcnivi cc dy dn cch inhoc cp mma

ra bn ngoi vrle. Dy dn v cp mm khng lm nhhngti

s quay ca phao. Tipimca phao trn dng bo tn hiu, tip

imdi dng ct my ct. Phao trn ctphn trn ca

v rle phao di c ngang mc ng ni t my bin p n bnh




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vrle, phao dic ngang mcngnit my bin p n bnh

dndu, sao cho khi duchuynngs lm nhhngn n.

Hnh 7.79.V tr v cu to ca r le hi

Nguyn l tc ngcabov:

Rlect thphnmcdu trong bnh

dndu, v vy trong rle lun lun ydu.

Cc phao lun lun cy ln pha trn ti




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p c y p tiv tr cao nht ph thuc vo cch kp vo

trc,v tr ny cc tipimum.

Khi scnhhi sinh ra chmto ra bt kh

nh chuyn ng v pha bnh dn du, i

qua rle, bt kh in kn phn trn ca v

rle, y du tt xung. Tu theo s gim

mcdu, tip im trn khing, sau mt

khong thi gian, ph thuc vo cng

to ra bt kh, phao ttiv tr m tipim

carlengli, bo tn hiu.

Hnh 7.80. S ni dy ca r le hi

Khi scnng trong my bin p, hi sinh ra mnh,yduchuynng nhanh v pha

bnh dn du,y phao di chm xung,tipimdingli,cpngun cho rle trung

gian TrG (hnh 7.80) qua rle tn hiu Th, TrG tc ngict my ct.




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Schuynngcadu c tnh gin on, nn tipimca phao dingchpchn.

bomct tin cy my ctsdngrle trung gian c tipimt duy tr K hoc dng

rle trung gian c tipimtrchm.

Bovrlehicng tc ng khi gimmcdu trong my bin p, numcdugim tcho tn hiu,gimnhiu cho lnhct my ct.

Khi rlehi cho tn hiu, bo cho nhn vin vn hnh bit trong my c scnh,cnphi

chuynphti sang my bin p khc. My bin p chcct ra sau khi cthtphti

v tin hnh kim tra.

7. Bov qu dng intht khng v chmv my bin p, cun dy thcp

c trung tnh trctipnit

S v nguyn l hotng:

Vi my bin p c tu dy Y /Y0, nubov dng cci khng nhy khi ngn

mch mt pha th cp ta t thm bo v th t khng (hnh 7-81 v hnh 7-82)




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mchmt pha thcp ta t thm bovtht khng (hnh 7-81 v hnh 7-82)

Hnh 7.81.S nguyn l ca bo v th t khng Hnh 7.82. S bo v chm v MBA

Khi mt pha pha Y0 chmvhoc khi ngnmchmt pha u ra my bin p, bov c

dng ngnmchchy qua, rle dng RI tc ngnct my ct.bomiukin tc

ngchnlc,bovtht khng tc ng c thi gian.

Vibov th t khng c c tnh c lp c th dng rle thi gian ring, nhng tit




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p gkim v vng bovcabovtht khng trng vi vng bov chnh cabov dng

cci nn ngi ta thngsdngmtrlethi gian cho hai bov ny.

Bov dng ccitht khng dng ctngnmchmt pha xy ra trong bn thn my

bin p, trn thanh gp 0,4 kV cngnh trn cc phntni vo thanh gp 0,4kV vichm

MBA. Bov ny t trung tnh nittrctipca my bin p pha 0,4 kV.

Nu cc phntcpin p 0,4 kV cbovbngcu ch c dng nhmc ln(t

300 A tr ln) bov dng th t khng thng dng rle dng c c tnh ph thuc loi

RT-80 hoc PT-81. Sdnhvy l v cnphiphihpc tnh cabovvic tnh ca

cu ch ni trn. Cc trnghp cn li c th dng rle dng in t tc ngtc thi vrlethi gian


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8. Phn cpbov my bin p:

Phn cpbov my bin p cnc vo gi thnh my bin p, mc quan trngca n

trong hthng cung cpin, yu cucaphti (tnh lin tc cung cpin), tnh cht mi

trngnit my bin p... c th phn cpbov my bin p nh sau:




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- Vi my bin p gim c cng sutnhmcn 560 kVA dng cu ch bov.

- Vi my bin p gim c cng sutt (7505600) kVA trang b cc loibov sau:

+ Bovct nhanh.

+ Bov dng cci.

+ Bov qu tinucn.

+ Bovrlehi.

+ Bovtht khng nucn(ivi my bin p ni Y /Y0)

- Vi my bin p gim c cng sutt 5600 kVA tr ln ngoi cc loibovrle trn trang

b thm bov so lchdc dng in.


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