99

Chng 6

BAO VE KHOANG CACH

6.1 NGUYEN TAC HOAT ONG

Bao ve dong ien cc ai, co hng va khong hng, co thi gian lam viec chon theo nguyen tac tng cap, oi khi qua ln va trong mang vong co so nguon ln hn hai, hoac mang vong co mot nguon nhng co nhng ng cheo khong qua nguon, khong the am bao cat chon loc nhng phan t h hong. Nh vay, can phai tm cac nguyen tac BV khac va am bao tac ong nhanh, va chon loc va co o nhay tot oi vi mang phc tap bat ky. Mot trong cac BV o la bao ve khoang cach (BVKC).

Bao ve khoang cach la loai BV co bo phan c ban la bo phan o khoang cach, lam nhiem vu xac nh tong tr t cho at BV ti iem NM. Thi gian lam viec cua BV phu thuoc vao quan he gia ien ap UR, dong ien IR a vao phan o lng cua BV va goc lech pha R gia chung. Thi gian nay tang len, khi tang khoang cach t cho h hong en cho at BV. Bao ve at gan cho h hong nhat co thi gian lam viec be nhat. V the, BVKC ve nguyen tac bao am cat chon loc oan h hong trong cac mang co hnh dang bat ky vi so lng nguon cung cap tuy y vi thi gian tng oi be.

Ngi ta dung rle tong tr lam bo phan o khoang cach. No phan ng trc tiep theo tong tr, ien tr hoac khang tr cua ng day (Z& , R, X). Tuy bo phan khoang cach phan ng theo Z& , R, X ngi ta phan biet BVKC loai tong tr, ien tr hoac ien khang. Bao ve khoang cach c dung thong dung nhat la loai tong tr.

e bao am tac dung chon loc trong mang phc tap, ngi ta dung BVKC co hng, ch tac ong khi hng cong suat NM i t thanh gop en ng day. Thi gian tac ong cua cac BV theo cung mot hng c phoi hp vi nhau sao cho, khi NM ngoai pham vi ng day c BV, thi gian tac ong cua BV ln hn mot cap, so vi BV cua oan b NM.

S phoi hp chnh xac gia cac rle khoang cach tren he thong ien at c bi viec chnh nh cac vung va thi gian tac ong cua cac vung khac nhau. Thong thng, BVKC se bao gom BV vung I co hng tc thi va mot hoac nhieu vung vi thi gian tr hoan. Cac tam chnh nh va thi gian tac ong cho ba vung BVKC at tai MC hai au ng day B, C c cho tren hnh 6.1.

Thong thng, vung BV th I co thi gian tac ong tc thi chiem khoang 80% chieu dai ng day BV. Ket qua la con 20% e am bao sai so rle tranh tac ong mat chon loc oi vi phan ng day tiep theo do nhng sai so cua cac BU va BI, d lieu ve tong tr ng day cung cap khong chnh xac khi chnh nh va o lng cua rle. oi vi mot vai ng dung, trong o cac so ket hp nay cho phep tam chnh nh cua vung I co the c tang en 90% (khi ma d lieu tong tr ng day c o chnh xac). Phan con lai cua ng day khong c bao phu bi vung I th c BV bi BV co hng cap 2 co thi gian tr hoan. Tam chnh nh vung II cua BV thong thng chnh nh bao phu toan bo ng day BV, cong vi 50% cua ng day ke can ngan nhat hay dai hn 120% ng day BV. Thi gian tr hoan cua vung II phai c chnh nh e phan biet vi BV chnh

100

cua phan ng day ke tiep, bao gom BVKC cap I cong vi thi gian cat cua may cat.

B23 (a)

vung I (b) (c) (e)

(d)

B34

vung II

B32

2 3

vung III

4

Vung = 80% tong tr ng day c bao veVung = ng day c bao ve + 50% ng day th hai ngan nhat.Vung T = 1,2 (ng day c bao ve + ng day th hai dai nhat)Vung N = 20% hng ngc cua ng day c bao ve

I

II

III

III

(g)

Hnh 6.1 ac tnh thi gian/khoang cach cho ba vung bao ve khoang cach

Bao ve d tr t xa cho tat ca cac s co tren ng day ke can, thng c cung cap bi BV cap III, co thi gian tr hoan ln hn e phan biet vi BV vung II cong vi thi gian cat cua may cat. Vung III co tam chnh nh phai t nhat bang 1,2 lan tong tr ng day BV va tong tr ng day ke tiep dai nhat. cac he thong ien c ket noi vi nhau, anh hng cua nguon CS s co cac thanh cai t xa se la nguyen nhan lam cho tong tr bieu kien o c cua rle ln hn nhieu tong tr thc ti iem s co va ieu nay can phai c xem xet khi chnh nh cho vung III. Trong cac he thong phan phoi hnh tia vi mot au cung cap nguon khong b anh hng nay.

Bao ve d tr t xa cap 3 oi khi co mot vung BV ngc nho (thng khoang 20% phan ng day c BV) them vao vi tam chnh nh thuan cua no (ac tnh offset). Vung BV d tr tai cho nay c cung cap vi thi gian tr hoan e BV nhng s co thanh cai va nhng s co ba pha gan thanh cai khi cac BV khac khong tac ong c. Trong vai s o, mot tiep iem tc thi khi ong vi nhng s co ben trong ac tnh offset cua vung III c dung e cung cap BV cho s co gan hoac kiem tra ng day e BV khi ong MC vao ng day ang b s co. Nhat la trng hp s co ba pha gan do khong loai bo dao cach ly noi at an toan t viec sa cha ng day trc o. oi vi ng dung nay, thi gian tr hoan vung III c noi tat trong thi gian ngan khi ong MC bang tay.

Gian o vung BV va thi gian phoi hp ba cap cua BVKC tong quat cho hnh 6.2.

Xet v du hnh 6.2 la mang co hai nguon BV at ca hai pha au ng day va gia thiet hoat ong co hng (BV1, 2, 3..., 6). Phoi hp thi gian lam viec cua cac BVKC theo ac tuyen hnh nac thang (H.6.2b).

Khi NM tai iem N1 gia tram BC, BV3, 4 gan cho NM nhat (khoang cach l3, l4) tac ong vi thi gian nho nhat cap 1 It3 ,

It4 ; BV1 va 6 co khoang cach l1, l6 cung khi ong, nhng no ch

co the tac ong vi thi gian tr hoan IIIt1 , IIIt6 va c coi nh la BV d tr trong trng hp

oan BC khong the cach ly. BV2 va 5 cung co cung khoang cach en cho NM nhng khong khi ong v khong ung hng. Neu iem NM khong nam khoang gia ng day ma nam ve mot pha ng day (iem N2), th BV3 se tac ong vi thi gian cap II IIt3 , BV4 van lam viec vi

It4 .

101

Trong trng hp NM tai thanh gop C, th s co c co lap bang BV3 va 6 vi thi gian cap II IIt3 va

IIt6 con BV4 va BV5 khong khi ong. BVKC co ac tnh thi gian tng cap nh tren, hien nay c s dung rat rong rai, so lng vung BV va cap thi gian thng la 3. Chieu dai vung BV va thi gian cua moi vung co the chnh nh c.

~1 2 3

L1L3 L4

N1 N2B C4 5

~D

L6

t2II

t1I

t4III

t2I

t1II

t6III

t4II

t3I

t3II

t4I

t1III

t6II

t5I

t5II

t6I

t5III

6A

Hnh 6.2 Phoi hp thi gian cua bao ve khoang cach

6.2 AC TUYEN KHI ONG CUA BAO VE KHOANG CACH VA BIEU DIEN CHUNG TRONG MAT PHANG PHC TONG TR Z

Ngi ta quan sat s lam viec cua rle tong tr tren mat phang Z. Truc thc at ien tr r, con truc ao j th at khang tr x. Tong tr tren cc cua rle RRR IUZ &&& /= co the bieu dien qua

ien tr va khang tr di dang RZ& = rR + jxR hay trong toa o cc co gia tr tuyet oi 22RRR xrZ +=& va goc pha cua no la RRR rxtg /= .

ac tuyen khi ong cua BV la ng bien xac nh ieu kien tac ong cua moi BV c bieu dien trong mat phang tap Z. Mot oan bat ky cua mang ien, v du ng day AB co the c bieu dien theo cac truc r, x bang vect lABAB ZZ && = . Neu ien tr cua cac oan trong mang co goc lll rx artg /= th quy tch cac iem tren mat phang phc ac trng cho tong tr ng day la ng thang vi goc nghieng l (oi vi ng day truyen tai )8565 ol .

iem au cua ng day c BV, ni at BV A, at goc toa o (H.6.3b). Cac iem nam trong vung tac ong cua BV A co toa o dng va nam trong phan t th nhat cua mat phang. Cac iem cua mang nam ben trai A c bieu dien bang cac iem co toa o am va nam trong phan t th ba cua mat phang tap. Tong tr cua ng day AB c bieu dien bang oan AB, ng day L2 ng vi oan BC, con L3 ng vi oan AD.

N/

102

~ZNL3 L2

D AN B

C

Jrhq

I

ZDA ZAB ZBC

N~

Cjx

B B

Nrhq

A rN

D

LN

a) b)

Hnh 6.3 Bieu dien ng day tren mat phang phc Z

Tong tr NZ& t cho at BV A en cho NM (iem N tren hnh 6.3a) c bieu dien bang oan AN nghieng mot goc N = l vi truc r. Neu NM qua ho quang (tr cua ho quang la tr tac dung), th tong tr ti cho NM c bieu dien bang vect

,NA& bang tong hnh hoc cua vect AN

va tr ho quang rhq

,

AN = ZN + rhq (6.1)

Bieu thc (6.1) ung trong trng hp ch co mot nguon I cung cap cho iem NM. Khi co hai nguon cung cap, do gia cac dong NM i t nguon I va J co goc lech pha, vect

,NN nam

nghieng so vi truc r. Khi o

hqNR rKZZ .+= &&

trong o: k - he so phc; N

hqhq I

lkr =

lhq - ien tr ho quang phu thuoc vao o dai va dong IN trong ho quang.

T bieu thc tren ta thay, khi NM au ng day rhq nho hn khi NM cuoi ng day, v dong khi NM au ng day ln hn. Moi oan ng day, ta co the xay dng t giac ABB'N', bieu th t giac s co (H.6.3b) cua oan c BV. Muon BV tac ong bat ky s co nao nam trong oan c BV AB (NM trc tiep hay qua ien tr trung gian) th toan bo hnh t giac s co phai nam ben trong vung tac ong cua ac tuyen khi ong cua BV. Tuy nhien, e at c tnh tac ong chon loc cao nhat cua BV oi vi nhieu s co khac nhau trong oan c BV (v du: cac che o NM ben ngoai, che o qua tai, D...) th dien tch chenh lech gia dien tch t giac s co va vung tac ong cua ac tuyen khi ong cua BV phai be nhat, chang han ac tuyen khi ong co dang hnh elip (H.6.4b).

103

jx jx jx

R R

Zkmax

Zkmax

L R

Zkjx

R

RZk

R

jx jx jxZkmax

R X = kkR

R

a) b) c)

g)f)e)d) Hnh 6.4 Cac dang ac tuyen khi ong cua bao ve khoang cach

Hnh 6.4 gii thieu cac loai ac tuyen khi ong cua BVKC thong dung. ac tuyen cua moi rle la quy tch cua nhng iem thoa ieu kien kRR ZZ && = . Phan gach nghieng cua ac tuyen ma

trong o kRR ZZ && < ng vi vung tac ong cua rle. Ngoai vung nay kRR ZZ && > rle khong tac

ong. ac tuyen khi ong c bieu dien bang phng trnh ac tnh ),,( RRRk IUfZ =& trong mat phang phc Z.

Hnh 6.4a: ac tuyen khi ong la vong tron tam 0 (OHM).Zk = k = hang so. Bao ve tac ong khi ZR < ZkR khong phu thuoc vao goc R . ac tuyen nay goi la tong tr

khong co hng. Loai ac tuyen nay thng c ap dung cho mang ien di 35kV.

Hnh 6.4b: ac tuyen la vong tron i qua goc 0 (tong dan MHO). ien tr khi ong cua BV phu thuoc vao goc R . Tong tr khi ong co o nhay cc ai khi RnR = hay = l . Bao ve khong khi ong oi vi ZR nam trong phan t th ba cua mat phang phc nen c goi la BV tong tr co hng. ac tuyen thc te khong i qua goc toa o 0 do ban than bo phan so sanh khong u nhay. V the, NM au ng day gan cho at BV co the BV khong lam viec, oan nay goi la vung chet cua BV.

Hnh 6.4c: ac tuyen khi ong la ng thang song song vi truc R; ac tuyen khi ong c bieu dien bang phng trnh xR = k = hang so, goi la ac tuyen ien khang. Bao ve vi ac tuyen nay lam viec khong phu thuoc vao ien tr NM trung gian.

Hnh 6.4d: ac tuyen co dang hnh ellip. ac tuyen nay i qua goc toa o va co o nhay cc ai khi maxRnhayl = . ay la ac tuyen cua BV tong tr co hng, co tnh chon loc cao.

Hnh 6.4e: ac tuyen co dang hnh a giac. ac tuyen nay co the trung kht vi t giac s co, nhng cau truc cac bo phan cua BV co ac tuyen nay rat phc tap, nen ch c dung cho ng day quan trong, ien the cao, cong suat ln chong cham at.

Mot vai dang ac biet: e nang cao o nhay cua BV cap III co the s dung ac tuyen vi dang hai ng tron cat nhau (H.6.4f) hay vong tron lech (H.6.4g) offset MHO.

104

6.3 NGUYEN TAC THC HIEN RLE KHOANG CACH

Bang 6.1 Bo phan so sanh co cac dang ac tuyen khac nhau

ac tuyen Dang ac tuyen So sanh suat So sanh pha

Co hng

A& B& A& B&

RR IZU&&& + RR IZU

&&& R IZ

&& RU&

Tong tr khong hng

R IZ&& RU

& RR IZU&&&

RR IZU&&& +

Tong tr

co hng

R IZ&& RR IZU

&&& + RR UIZ&&&

RU&

Tong tr

co hng

ac biet

( ) Ro IZZ &&& ( )oR ZZU2 &&& RR UIZ &&& RoR IZU &&&

Khang tr

RRR UIR2IZ2&&&&&

RU& R IZ

&& RR UIZ&&&

trong o: Z& = R + jx - tong tr khi ong cua ng day BV

oZ& - tong tr xac nh vong tron lech tam.

Nguyen tac tac ong cua tat ca cac bo phan o khoang cach eu da tren s so sanh gia tr tuyet oi hay goc pha cua hai ai lng vect A& va B& . Chung la ham cua dong IR va UR a vao rle. Trng hp tong quat

RR UkUkA &&& 21 += ; RR UkUkB &&& 43 += (6.2)

trong o k1, k2... k4 la cac so phc.

Bang cach thay oi k1, k2 k4 trong (6.2), ta co the nhan c bo phan khoang cach vi cac dang ac tuyen khac nhau (xem muc 2 phan 1.3). Sau ay gii thieu bang 6.1 cho mot vai bo phan so sanh cua bo phan khoang cach co cac ac tuyen c ban.

R

R

R

R

R

jx

jx

jx

jx

jx

Zo

Z

R

jx

nhmax

105

6.4 CACH CHON UR, IR A VAO BO PHAN KHOANG CACH E PHAN ANH NGAN MACH GIA CAC PHA

Dong ien IR va ap UR nhan t BI va BU phai c chon a vao rle tong tr e cho tong tr ZR o c hai au cc rle khoang cach phai ty le vi khoang cach t cho at BV en cho NM va khong phu thuoc vao dang NM. e thoa man c yeu cau o, ien ap RU& phai bang ien

ap giang tnh en iem NM RNNZI && , con dong IR bang dong NM IN, khi o

RNN

RNN

R

RR ZI

ZIIU

Z &&

&&

&

&&

===

Nh vay, bo phan khoang cach phai noi vao ap va dong cua vong NM. e cho cac bo phan khoang cach lam viec chnh xac khi co NM hai pha, can phai at ba rle ng vi s co gia cac pha AB, BC va AC. Cung co loai s o ch dung mot bo phan khoang cach nhng can co thiet b oi noi truyen qua ap va dong pha tng ng tuy theo s co xay ra gia cac pha nao. Cac s o noi day bao am c ty le ZR va IN khi co NM gia cac pha khong dung c cho trng hp mot pha cham at. V vay, rle khoang cach chong cham at se co s o noi day khac.

Sau ay gii thieu mot s o noi day cua cac phan o lng BVKC chong NM nhieu pha c s dung rong rai.

6.4.1 S o noi rle tong tr noi vao ien ap day va hieu so dong pha

Cach chon cac dong ien va ien ap noi vao rle cho trong bang 6.2 va hnh 6.5.

Khi NM ba pha ca ba bo phan khoang cach eu nam trong cung ieu kien nh nhau

hZIUUII npRNR 1)3()3()3()3( 33...;3 ===

vi: )3(NI - dong NM ba pha i tren cac pha

Z1 - tong tr th t thuan tren 1km ng day

h - chieu dai t cho at BV en cho NM.

Nh vay hZIU

ZR

RR 1)3(

)3()3( &

&

&&

==

Trong trng hp NM hai pha, v du pha B va C (H.6.6b), ch co rle B lam viec ung. oi vi rle nay

)2()2( 2 NcbR IIII &&& == ; hZIUU NBCR ..2 1)2()2( &&

==

va hZIU

ZR

RR .1)2(

)2()2( &

&

&&

==

Rle RI& RU&

A ba II && abU&

B cb II && bcU&

C ac II && caU&

Bang 6.2 S o noi day rle tong tr

106

RZCRZBA B C

ABC

Hnh 6.5 S o noi rle khoang cach chong NM nhieu pha

RZA

RI

IN(3)

IN(3)

IN(3)

URUR

UR

h

A

B

C

RZ

IN(2)

UR

A

B

C

a) b)

Hnh 6.6 Ngan mach nhieu pha tren ng day bao ve

h

Tren cac cc cua rle A va C co dong )2(NI va ien ap ln hn )2(

bcU , v the tong tr tren cac cc cua nhng rle nay ln hn Z1 va nhng rle nay khong tac ong. Bao ve ch can mot rle B tac ong e m may cat. Khi NM hai pha AB hoac AC th rle A hoac rle C se lam viec ung. Khao sat tng t khi NM hai pha cham at BC hay AB hay CA tng ng rle B, A hoac C lam viec.

Nh vay, s o nay bao am tong tr at tren rle ZR nh nhau tat ca cac dang NM nhieu pha tai cung mot iem. o la u iem ln cua s o. e hieu ro rang hn, chung ta tnh tong tr o c cua rle t v tr rle ti cho NM.

Khao sat tong tr bieu kien o c cua rle khoang cach at tren ng day truyen tai: tnh tong tr bieu kien o c bi rle can thiet phai tnh NM tren ng day truyen tai. Qua trnh tnh toan NM chi tiet c trnh bay trong quyen Cac bai toan tnh NM va BV rle cung tac gia, ay ch gii thieu cac bc chnh va ket qua at c. Mang tnh toan tieu bieu cho hnh 6.7.

~ ~

h

ZN

R

CEs

A NZL

ZED

QB

EU

RSEs

+Zs ZN ZU

+EU

hZL N (1 - h)ZLQ

Hnh 6.7 Mang ien tieu bieu tnh bao ve khoang cach cho ng day

Hnh 6.8 S o thay the mang ien (H.6.7 )

107

Trong hnh 6.7: RQ - ng day co at BVKC EZ& - tong tr lien ket mang ngoai ng day

LZ& - tong tr ng day BV RQ; NZ& - tong tr cham trung gian h - khoang cach t cho at rle ti iem cham.

Hnh 6.8 la s o thay the mang hai ca Thevenin cua mang ien hnh 6.7. Mach th t tng ng cua mang tren, sau khi bien oi /Y, c ve hnh 6.9, (chon pha chuan la pha a).

hZLo (1-h)ZLo

ZEo

ZUoZSo

VRo

+

QoRoVo

+

IRo

NoIo

N2I2

(1-h)ZLohZLo

Zjo

ZTo

ZKo

ZP

+

E1

ZI2

O2

Vo+ Zo

Io

Jo Ko

Oo Oo

No

R

hZL2 (1-h)ZL1

ZE2

ZS2

VR2

+

Q2R2V2

+

IR2

N2I2

N2I2

(1-h)ZL2hZL2

Zj2

ZT2

ZK2

ZP2

+

E1

ZI2

O2

V2+ Z2

I2

J2 K2

O2 O2

N2

R

hZL1 (1-h)ZL1

ZE1

ZU1ZS1+

ENO1

VR2

+

Q2R1V1

+

IR1

N1I1

N1 I1

(1-h)ZL1hZL1

Zj1

ZT1

ZK1

Zp1

+

E1

ZI1

O1

V1

+ Z1

+

N1

I1

EN

J1

O1

R K1

a)

b)

c)

ZU2

Hnh 6.9 Mang th t tng ng cua hnh 6.8

108

Tren s o vi cac gia tr tnh toan

++=

UEi

EiSi

SiUi

UiEiSiKi

Ji

Ii

ZZ

ZZ

ZZ

ZZZZ

Z

Z

&&

&&

&&

&&&

&

&

&

1

vi i = 1, 2, 0.

1- Mang th t thuan

NE& - ien the tai N trc luc s co.

111 LJT ZhZZ &&& += - tong tr nhanh rle th t thuan (nhanh trai)

111 )( LKP ZhlZZ &&& += - tong tr nhanh song song th t thuan (nhanh phai)

11

1111

PT

PTI ZZ

ZZZZ&&

&&&&

++= - tong tr tng ng th t thuan

111 aR ICI &&& = - thanh phan dong qua rle th t thuan

vi 11

11

PT

P

ZZZC

&&

&&

+= - he so phan bo dong th t thuan

nen 1

1

1

11 K

CZZEC

IN

NR &

&

&&

&&&

=

+= hay 111 CIK R &&& =

vi N

N

EZZ

K&

&&& +

=1

1 ; 1

111 K

CZhZV LNR &

&&&& +

=

hay 1111 CZhZVK LNR &&&&& +=

2- Mang th t nghch

222 LJT ZhZZ &&& += - tong tr nhanh rle th t nghch

222 )( LKP ZhlZZ &&& ++= - tong tr nhanh song song th t nghch

T2 P22 I2

T2 P2

Z ZZ Z

Z Z= +

+

& && &

& & - tong tr tng ng th t nghch

R2 2 a2I C I=&& & - thanh phan dong qua rle th t nghch

vi P22T2 P2

ZC

Z Z=

+

&&

& & - he so phan bo dong th t nghch.

3- Mang th t khong

TOPO

POo

POTO

POTOIOo ZZ

ZC

ZZZZ

ZZ&&

&&

&&

&&&&

+=

++= ;

T cac mach s o th t nay, chung ta tnh c cac ai lng ien khi NM vi cac dang khac nhau. Trnh t tnh toan cung nh ket qua c trnh bay quyen Cac bai toan tnh ngan mach

109

va bao ve rle. Trong phan nay, se s dung cac ket qua tnh toan o e tnh tong tr bieu kien o c bi rle khi co NM tren ng day.

6.4.2 Phan tch tong tr bieu kien o c cua rle

Gia s s o noi rle la ien ap day va hieu hai dong pha.

1- Tong tr o c cua rle khi ngan mach ba pha

1

11

1

)( CZ

ZhCZ

IIkUK

ZZ NLNba

ababA &

&&

&

&

&&

&&& +==

== (6.3)

1

11

1

)( CZ

ZhCZ

IIkUK

ZZ NLNcb

bcbcB &

&&

&

&

&&

&&& +==

== (6.4)

1

11

1

)( CZ

ZhCZ

IIkUK

ZZ NLNac

cacaC &

&&

&

&

&&

&&& +==

== (6.5)

vi

1

111

11

1

1

P

TTP

P

ZZZZ

ZC

&

&&&

&&

+

=

+= (6.6)

(he so phan bo dong th t thuan la ham cua ),, UES ZZZ &&&

trong o: 1TZ& - tong tr nhanh rle (pha trai)

1PZ& - tong tr nhanh song song (pha phai).

T (6.3), (6.4), (6.5), thay rang tong tr o c bi moi rle la tong tr cua tong tr ng day, t v tr at rle ti iem NM cong vi mot phan cua tong tr s co c tnh bang hang so C1. Khi NM gan iem at rle

ZP1 >> ZT1 nen C1 1 (6.7)

Khi NM gan vi iem cuoi ng day at BV

ZP1

110

Nhan xet: Tong tr bieu kien o c cua rle ln hn tong tr thc t iem at rle ti iem NM, nen vung BV thc se b thu nho lai. Tnh trang nay c goi la di tam.

2- Tong tr o c cua rle khi ngan mach hai pha B, C

21

22

211

)1()1(

3

CaCaZ

CaCZ

ZhZZ NLabA &&&

&&

&&&&

+++

++== (6.9)

21

1 CCZ

ZhZZ NLbcB &&&

&&&

++== (6.10)

2

21

2

22

11

)1()1(

3

CaCaZ

CaCZ

ZhZZ NLcaC &&&

&&

&&&&

++

++== (6.11)

vi 22

22

TP

P

ZZZC

&&

&&

+= - he so phan bo mang th t nghch

Z2 - tong tr tng ng th t nghch.

Trng hp nay, rle B o tong tr la tong tr ng day cong vi tong tr NM, con hai rle A va C tong tr o c gom tong tr NM ng day cong vi cac tong tr phc khac. Thng trong hau het cac trng hp C1 gan bang C2.

Bang 6.3 Cho ket qua tnh toan tong tr o c khi NM nhieu pha (C1 C2)

Rle N3 N2

A 1

N1L

CZZh&

&& +

211o

2

21

N1L

CdCdZ3

CaCZZh

&&

&

&&

&&

++

++

B 1

N1L

CZZh&

&& +

21

N1L

CCZZh

&&

&&

++

C 1

N1L

CZZh&

&& +

2o11

2

22

1

N1L

CdCdZ3

CaCZZh

&&

&

&&

&&

+

++

6.4.3 Tong tr o c cua rle khi ngan mach cham at (C1 C2)

Bang 6.4 Ket qua tnh toan tong tr bieu kien o c bi rle khi NM cham at

Rle N1 N1,1

A 21

22oN1L

aCZaZZ3Zh

++

&

&&&&

( )( ) ox211ox21o

ox2oxx2No1L

ZCdCdZCdZZ3Z2ZZdZh&&&&&

&&&&&&

++

+++

B 21

2oN1L

CCZ2ZZ3Zh

&&

&&&&

+++ ( )( ) ox21x21

oxx2N1L

ZCCZCZ2ZZZh

&&&&&

&&&&

++

++

C 221

2oN1L

CaCZaZZ3Zh

&&

&&&&

++

( )( ) ox2o11x211

ox2oxx2N11L

ZCdCdZCdZZ3Z2ZZdZh&&&&&

&&&&&&

++

++

vi oo ad 30312 == ; oad 150311 == ;

oaad 90322 ==

22 ZZZ Nx &&& += ; GoNox ZZZZ &&&& 3++=

Nhan xet: Cac tong tr o c nay gom ca hai thanh phan th t nghch va zero cong vi tong tr s co. Lu y rang, tong tr rle B (rle b-c) khi cham at mot pha a (gia thiet khi C1 C2)

111

tong tr o c gom, tong tr cham thc te cong vi tong tr rat ln, co ngha la rle nay se khong tac ong trong trng hp s co nay. oi vi NM hai pha cham at, tong tr o c bi tat ca ba rle gom tong tr thc ti iem NM cong vi mot ham phc tong tr s co trong pha. Lu y trong s o nay khi co cham at, tong tr rle o c sai so rat nhieu so vi tong tr thc, nen e chong BV cham at ngi ta se chon s o khac co bu thanh phan th t khong.

Trong cac trng hp ac biet C1 = C2, cac bieu thc tnh tong tr bieu kien o c bi rle c tom tat bang 6.5 va 6.6. Thc te C1 va C2 gan bang nhau nen bieu thc cac bang nay c dung rong rai e tnh toan.

Rle N1 N1,1

A 12

11

oN1L

C3Zj

CdZZ3Zh

&

&

&

&&& +

+ ( )( )ox2x2o1

ox2oxx2No1L

ZdZdCZZ3Z2ZZdZh

&&&

&&&&&&

+++

B 1

N1L

CZZh&

&& +

C 12

11

oN1L

C3Zj

CdZZ3Zh

&

&

&

&&&

++ ( )( )ox2x211

ox2oxx2N11L

ZdZdCZZ3Z2ZZdZh

&&&

&&&&&&

++

V du: ng day 161kV dai 160km c BV bang rle khoang cach dung s o ien ap day va hieu dong hai pha. Tong tr th t thuan va khong cua ng day la

1LZ& = 27,99 + j66,71

0LZ& = 56,57 + j48,50

Mot s co NM hai pha bc xay ra tai N cach v tr at rle 70% ng day. thi iem xay ra NM tong tr nguon nh sau 21 SS ZZ && = = 0,002 + j0,005 vt; 0SZ& = 0,004 + j0,010 vt

21 UU ZZ && = = 0,003 + j0,006 vt; 0UZ& = 0,006 + j0,012 vt

21 EE ZZ && = = 0,2 + j0,5 vt; 0EZ& = 0,4 + j2 vt

Cac gia tr tren tnh vi c ban: Scb = 100MVA; Zcb = 259,21 . ien tr NM trung gian ho quang la ZN = 0,04 vt, vi cung c ban tren. Tnh

- Tong tr (vt) t iem rle ti iem NM 1LZh &

- Tong tr 1TZ& va 1PZ&

- Hang so 1C& va 2C&

- Tong tr bieu kien o c bi rle.

Ve cac iem o c tren mat phang tong tr.

Rle N3 N2

A 1

N1L

CZZh&

&& +

1

2

1

N1L

CZ3j

CZaZh

&

&

&

&&

B 1

N1L

CZZh&

&& +

1

N1L

C2ZZh&

&& +

C 1

N1L

CZZh&

&& +

1

2

1

N2

1LC

Z3jC

ZaZh&

&

&

&&

Bang 6.5 Tong tr rle o c khi NM nhieu pha (C1 = C2)

Bang 6.6 Tong tr o lng khi ngan mach cham at (C1 = C2)

112

Giai: Tnh tong tr ng day BV trong vt (Scb = 100MVA; Zcb = 259,21 )

21,259

71,6699,2721

jZZ LL+

==&& = 0,1080 + j0,2574 vt

Theo au bai, tai v tr cham h = 0,7 nen

1LZh & = 0,7 (0,1080 + j0,2574) = 0,0756 + j0,1802 vt

va NZ& = 0,04 + j0 vt

T cac bieu thc tnh cac gia tr cua s o thay the

111

111

UES

SUI ZZZ

ZZZ

&&&

&&&

++= = 0,00017 o1986,68

111

111

UES

ESJ ZZZ

ZZZ&&&

&&&

++= = 0,00542 o1986,68

111

111

UES

EUK ZZZ

ZZZ

&&&

&&&

++= = 0,00381 o1986,68

Tnh cac tong tr cac nhanh trong s o thay the

111 JLT ZZhZ &&& += = 0,0776 + j0,1852 vt

111 )1( KLP ZZhZ &&& += = 0,0363 + j0,0896 vt

11 PT ZZ && + = 0,1139 + j0,2721 vt

Tong tr tng ng th t

111

111 L

PT

PT ZZZZZZ &&&

&&& +

+= = 0,0248 + j0,0592 vt

11

11

PT

P

ZZZC

&&

&&

+= = 0,3192 + 0,0028 vt

V mach th t thuan va nghch giong nhau, nen 12 ZZ && = ; 12 CC && =

Dung ket qua cua bang tnh tong tr bieu kien o c cua rle (bang 6.5):

Rle A: 221

13 ZCj

CaZZhZ NLab &&&

&&&=

= 0,1954 67,24 + 0,1253 59,95 + 0,3481 22,78

Rle B: NLbc ZCZhZ &

&&&

11

2

1+= , , , ,0 1954 67 24 0 0627 0 05= +

Rle: 211

2

13 ZCjZ

CaZhZ NLca &&

&&

&& +=

= ooo 27,1573481,022,601288,024,671957,0 ++

113

Tong tr bieu kien o c cua ba rle c ve tren hnh 6.10; cach ve vect tong tr o c la tong cua vect 1LZh & va vect bieu dien

cac so hang ke tiep tnh tong tr o c.

Gia thiet s dung rle khoang cach co ac tuyen khi ong MHO la vong tron co ng knh bang ung vi chieu dai ng day ZL. Nhan thay rang, rle B se tac ong v tong tr o c nam trong vong tron ac tnh con cac rle A va C khong tac ong c. Thc te, ch can mot rle B tac ong se i m may cat au ng day.

6.5 CACH CHON UR, IR A VAO BO PHAN KHOANG CACH E PHAN ANH CHAM AT MOT PHA

Hau het he thong ng day truyen tai cao va sieu cao ap c noi at trung tnh trc tiep hoac thong qua ien tr hoac ien khang. Trong cac loai s co xuat hien tren ng day truyen tai ien, s co cham at chiem hn 90% so trng hp. Gia thiet he thong ng day c noi at nhieu iem (tai cac trung tnh cua MBA). Khi o, dong th t khong o qua ien tr ho quang gia pha dan va at c tao ra t trung tnh noi at, dong ien th t khong nay co the rat ln. Cac ien ap th t khong va dong th t khong c o tai v tr rle, c s dung e thc hien BV chong cham at, v ay la tn hieu am bao chac chan la he thong ang khong bnh thng. Khi he thong van hanh bnh thng, dong th t khong cung co the co do s khong can bang gia cac pha, nhng dong nay nho hn nhieu khi so sanh vi dong s co. Tong tr th t cua he thong phu thuoc vao cau truc cua he thong. Cau truc nay khong phu thuoc vao tnh trang tai ma ch phu thuoc vao viec ong cat mach. ieu nay, co ngha la tong tr th t khong la hang so khi he thong khong co s ong cat mach. ay la tnh chat cho thay s khac biet cua dong th t khong va dong th t thuan, luon b thay oi tuy theo tai cua ng day tng ng vi he thong tai va nguon thay oi. Mot ac tnh khac cua mang th t khong la tong tr th t khong co o ln vao khoang t hai en sau lan ln hn thanh phan tong tr th t thuan cua ng day. Do o oi vi ng day dai, tong tr th t khong se thay oi ln khi s co di chuyen t au ng day en cuoi ng day.

Rle khoang cach chong cham at cung co the c cung cap tn hieu ien ap pha va dong ien bu t cac dong ien tren ng day.

Xem xet s co mot pha cham at (pha a) cac thanh phan th t c noi vi nhau nh hnh 6.11. Tong cac thanh phan ien ap tai v tr N, ta co

aoNaaao IZVVV &&&&& 321 =++

hay aoNRLRRLRRoLoRo IZIZhVIZhVIZhV &&&&&&&&&&& 3)()()( 212111 =++

jXZCaZNa2

C1

+ j 3C1

Z2

ZL1

Zbc

hZL1

Zab

ZNaC1

-j 3C1Z2

-

R

ZN2c1

O

Hnh 6.10 Tong tr bieu kien o c cua rele khi ngan mach hai pha b - c

114

3ZN

hZL1 (1 - h)ZL1

ZE1

ZS1 ZU1

R1

IRo

I I Io 1 2 = = hZLo (1 - h)ZLo

ZEo

ZSo ZUo

Ro

IRo

Io

Vo

Oo

VRo

+

E+

O1

V1

I1

hZL2 (1 - h)ZL2

ZE2

ZS2 ZU2

R2

IR2

I2

V2

O2

VR2

+

N2

VRo

+

Hnh 6.11 Mang noi ket th t khi chon at mot pha

Tra bang cac gia tr tng ng vi s co mot pha cham at c trnh bay trong phan tnh toan NM cua quyen Cac bai toan tnh NM va BV rle cung tac gia, ta co

22111 ;; aoRaRaooRo ICIICIICI &&&&&&&&& === (6.12)

do o aoNaLRaLRaoLooRo IZIZhCVIZhCVIZhCV &&&&&&&&&&& 3)()()( 21221111 =++ (6.13)

ien ap tai R pha s co (a)

aoNaLaLLooRRRoaR IZIZhCIZhCZhCVVVV &&&&&&&&&&&&&& 321211121 +++=++=

= aoNLLoooaRL IZZhZhICIZh &&&&&&& 3)( 11 ++

= aoNLLoooaRL IZZhZhICIZh &&&&&&&& 3)( 11 ++ (6.14)

115

= [ ])1/( 11 += LLoaooaRLaR ZZICIZhV &&&&&& (6.15) 1LZh & =

+ aoo

L

LoaR

aR

ICZZI

V

&&

&

&&

&

131

(6.16)

Bieu thc tren cho thay, rle chong cham at pha a c cap tn hieu ien ap pha a va dong ien pha a se cho ra ket qua co sai so (tng ng vi trong dau ngoac). Do o, dong ien cung cap cho rle phai c hieu chnh goi la dong bu. Co the thc hien thong qua bien dong ien nh hnh 6.12.

3Co aoI

IaR

IbR

IcR

N1 N2N1 aRI AG

BG

CG

AB

BC

CA

rle chongcham pha

rle chongcham at

Hnh 6.12 S o au BI co bu dong th t khong

Luc o dong ien cung cap vao rle

;3,

;3,

1

2

1

2

NNICII

NNICII aoobRbaooaRa &&&&&&&& +=+=

1

23,

NNICII aoocRc &&&& += (6.17)

hay buaRa III &&& +=,' ; bubRb III &&& +=

,'

bucRc III &&& +=,' ; ( )12 /3 NNICI aoobu &&& = (6.18)

Neu chon ty so MBI nh sau

= 1

3

1

11

2

L

Lo

ZZ

NN

&

&

Khi o rle se xac nh chnh xac tong tr t iem at rle en iem s co

GRCaR

aR

aooCaR

aRL IkI

VICkI

VZh

&&&

&

&&&&

&&

+=

+=

31 (6.19)

trong o: kC - he so bu; IGR - dong ien ve th t khong.

vi 1

1

3 L

LLoC Z

ZZk&

&&&

= ; aooGR ICI &&& 3= (6.20)

Phan tch tren ay c thc hien vi gia thiet la ng day c chuyen v hoan toan. oi vi trng hp ng day khong chuyen v, phan dong bu se co khac i, va c xac nh da tren ien ap ri tren ng day.

Bang 6.7

Rle URa IRo

A aU& aI& + 3KcIo

B bU& bI& + 3KcIo

C cU& cI& + 3KcIo

116

Nh vay dung s o bu dong ien, tong tr hai au rle ty le vi khoang cach t cho at BV en cho cham at, khong phu thuoc vao dang cham at. S o nay c s dung rat rong rai. Tong tr tren cac cc cua rle B va C cua pha khong s co ZR > Zl va BV khong tac ong nham.

Tom lai, ap va dong a vao rle khoang cach chong cham at cho bang 6.7. Vi KC theo cong thc (6.20) khi BV ng day n. Trong trng hp ng day song song he so bu se thay oi va c khao sat chng 8.

6.6 CHON CAC THAM SO CUA BAO VE Chon cac tham so cua BVKC la chon thi gian tac ong va ac tuyen lam viec tong tr khi

ong cua cac vung BV khac nhau. Quan sat mang ien cho nh hnh 6.13. Gia thiet tai cac tram A B C D co at BVKC co hng tac ong len cac may cat MC1, MC2,... MC6. Nguyen tac c ban e tnh toan BV la tnh chon loc, v du NM tai N1 th ch co MC1 va MC2 c cat ra e co lap s co.

Cap I cua bao ve

- ac tnh lam viec cua rle tong tr thng c chon dang co hng, co the la ac tuyen MHO hoac t giac hoac ellip (H.6.4) tuy theo nhiem vu va oi tng BV.

- Thi gian lam viec cua BV cap I la thi gian tac ong rieng cua BV (tac ong tc thi khong can bo phan thi gian).

- Tong tr khi ong cap I IkZ c chon theo ieu kien sao cho bo phan khoang cach cap I

khong tac ong khi co NM ngoai pham vi ng day BV.

~

~ ~

FB T1

INT1

N3

N2N1MC1

AB N4 C

MC4 MC5 MC6MC3

MC2 FC

ZIIkaZIka

tIIA

tIA

tIIIA

FA

Hnh 6.13 Chon tham so cua bao ve khoang cach

Do o IkZ c chon nho hn tong tr ng day BV vi sai so ln nhat van am bao

Tong tr khi ong BVKC cap I chong NM nhieu pha

lZkZkZ LIk 111 1

&&==

trong o: Z1 - tong tr th t thuan tren moi km ng day BV l - chieu dai ng day BV

1LZ - tong tr th t thuan ng day BV

117

k1 = 0,8 0,9 - he so xet en sai so BI, BU, ien tr cham trung gian... lZkZkZ oLo

Ik 11

&&==

vi ko = 0,75 0,85, ko thng c chon nho hn k1 v ien tr cham trung gian mot pha vi at thng ln hn cham gia cac pha.

He so bu cho s o BV chong cham at 1

1

3 L

LoLc Z

ZZk

=

vi oLL ZZ

&& ,1

la tong tr th t thuan va khong cua ng day BV.

He so bu cua bieu thc (6.23) c dung cho ng day n. Trong trng hp BV ng day song song, neu van dung he so bu nh (6.23) do co hien tng ho cam, vung BV cap I chong cham at mot pha se b tnh trang di tam, trong nhieu trng hp can thiet co the hieu chnh he so bu. Van e nay se c khao sat chng 8.

Cap II cua bao ve

Tong tr khi ong cap II ( IIkZ ) c chon theo ieu kien sao cho vung BV cap II phai bao phu toan bo ng day mot cach chac chan e BV phan con lai cua ng day ma BV cap II phai vt ra pham vi ng day BV (theo yeu cau ve o nhay vung cap II phai bao phu t nhat 120% ng day BV 2,1/

1= L

IIknh ZZk ).

e am bao tnh chon loc va thi gian tac ong nhanh, o dai vung II va thi gian lam viec cua no phai c tnh toan phoi hp vi cac BV cua cac phan t noi vi thanh cai cua cac ng day (chang han ng BC, MBA T1 tren s o hnh 6.13).

Thong thng, thi gian lam viec cua BV cap II cua cac may cat lan can c chon bang nhau ttt III += (6.24)

vi tI la thi gian tac ong nhanh cap I cua phan t tiep theo; t = 0,3 0,5s. e thoa man chon loc trong ieu kien tren th yeu cau o dai cua vung II khong c vt qua pham vi BVCN hoac cap I cua phan t noi vao thanh cai cuoi ng day BV.

Tong tr khi ong cap II c tnh toan phoi hp theo ieu kien tren vi phan t noi vao thanh cai cuoi ng day co tong tr nho nhat.

T cac ieu kien tren tong tr khi ong cap II c tnh

)( 1111IBAB

IIA ZklZkZ += (6.25)

vi: k1 = 0,85 0,9 - he so phoi hp vi BV cap I tiep sau khi ke en sai so cua BV k11 = 0,8 1 - he so tnh en sai so BV cap II lAB - chieu dai ng day BV Z1 - tong tr ng day BV tren moi km

IBZ - tong tr khi ong cap I nho nhat cua phan t xuat phat t thanh cai cuoi ng day BV.

Khi tnh toan chon gia tr IkZ can lu y cac trng hp dong NM tai cho NM khac vi dong

NM qua v tr at rle se a en sai so o lng tong tr cua rle. Trong cac trng hp nay, can hieu chnh gia tr khi ong bang cac he so c goi la he so phan dong (kpd).

118

V du, he thong ien (H.6.13) co nguon tai thanh cai B nen dong NM tai BV1 khac vi dong tai iem NM N2. Trong trng hp nay, gia tr tong tr khi ong cap II tai v tr MC1 can c tnh theo

)( 1111IB

pdAB

IIA Zk

klZkZ && += (6.26)

vi 2NBN

NADpd I

Ik = la he so phan dong la ty so dong NM qua iem at rle INAB so vi dong NM tai

iem NM 2NBNI la iem cuoi vung II cua BV A(N2).

Trng hp nay c nhac lai hnh 6.19 ro hn.

Tng t, can hieu chnh gia tr tong tr khi ong cap II theo ieu kien NM sau MBA T1

)(1

1

1111 T

TAB

IIA Zk

klZkZ && += (6.27)

vi: 1TZ

& - tong tr MBA co cong suat ln nhat noi vi tram B

1

/ NITNABT IIK = - he so phan dong la ty so dong NM qua v tr at BV INAB va dong NM

qua MBA - T1 khi NM ngay sau MBA.

Gia tr tong tr khi ong cap II cua MC1 tram A c chon la gia tr nho nhat cua (6.25), (6.26), (6.27).

Gia tr c chon nay phai c kiem tra ve o nhay khi co NM tai thanh cai tram B, (iem NH).

Vi cac ng day ngan, tong tr ng day nho (5 10) yeu cau ve o nhay cao hn 5,1IInhk v BV ng day NM xay ra ho quang.

Trong trng hp o nhay vung II khong am bao ( 2,1

119

oi vi BV cap II chong cham at mot pha cho ng day n, ieu kien tnh toan chon cac gia tr at cung tng t nh chong NM nhieu pha va he so bu c tnh theo (6.23). Trong trng hp BV cho ng day co ho cam, can lu y hien tng vung BV hieu qua cap II c ni rong hn gia tr chnh nh (hien tng qua tam) khi van hanh ng day vi cac che o khac nhau. Can thiet co the hieu chnh he so bu, van e nay se c khao sat ky hn chng 8.

Cap III cua bao ve

Muc ch cua vung BV cap II la d tr cho ng day BV va cac phan t noi vao thanh cai cuoi ng day BV. Tong tr khi ong cap III co the c chon mot trong hai ieu kien: - Vung BV cap III bao phu toan bo ng day BV noi t thanh cai cuoi ng day BV. V du, theo s o hnh 6.13 BCAB

IIIA ZZZ && +> ; 1TAB

IIIA ZZZ && +>

Tong tr c chon ...};max{{2,11TBCAB

IIIA ZZZZ &&& += (6.30)

Thi gian lam viec cap III ttt IIBIIIA +=

ng lien net hnh 6.15 bieu dien minh hoa vung BV va thi gian lam viec cap III cua BV A c chon theo ieu kien nay.

- Bao ve cap III khong c tac ong khi tai lam viec cc ai

minZZIIIA & (6.31)

vi Zmin la gia tr tong tr nho nhat nhn t v tr rle khi phu tai he thong che o cc ai

max

maxmin

ptpt I

UZ =

trong o: Umin - ien ap nho nhat khi phu tai cc ai ( mU95,09,0 )

maxptI - dong ien lam viec ln nhat qua BV.

Vung BV cap III chon theo ieu kien nay thng rat rong nen thi gian cap III phai c phoi hp vi thi gian cap III cua phan t tiep sau ttt IIIB

IIIA +='

ng t net hnh 6.15 bieu dien vung BV va thi gian lam viec cap III chon theo ieu kien tai cc ai.

oi vi BVKC cap III chong cham at mot pha cua ng day n, ieu kien tnh toan cung tng t nh chong NM nhieu pha va vi he so bu dong theo bieu thc (6.23).

t

A B C D

cachkhoangI

AZ

IIAZ

IBZ

t

IIIAZ

IIBZ

IIBt

IIIBt

t IIIBZ

Hnh 6.15 Vung bao ve va thi gian lam viec cua BVKC cap III

120

Trong trng hp BV ng day co ho cam, can lu y hien tng qua tam e chon thi gian tac ong cap III hp ly.

Bao ve khoang cach cap III, ngoai chc nang d tr cho ng day cung co the d tr cho NM tren thanh cai va NM gan v tr at BV. e thc hien chc nang nay, ngi ta dung at tnh tong tr co hng ngc (offset-MHO). Gia tr tong tr at hng ngc khoang 10 25% tong tr khi ong cap I cap IV cua BV. Trong trng hp khong s dung ac tnh off-set MHO, ngi ta dung them cap BV th t BV mot phan ngc ng day. Tong tr khi ong hng ngc la 25% tong tr khi ong cap I neu ng day BV ngan (< 30km) hay ZII = 10% ZI neu ng day BV dai.

Bao ve cap IV hng ngc con dung cho cac s o BVKC pilot khoa hay cho phep (se c khao sat chng 8). Vung BV cap IV trong trng hp nay c tnh toan ln hn vung BV cap II cua BV au kia ng day, ngha la

ABIIB

IVA ZZZ %.120 (6.32)

Trong thc te ap dung tnh toan, neu kiem tra cac yeu to anh hng en sai so do tong tr nam trong gii han cho phep, ngi ta thng chon tong tr ba cap nh sau:

Cap I: 85 90% tong tr ng day BV

Cap II: tong tr ng day BV +50% tong tr phan t tiep sau ngan nhat

Cap III: 1,2 (tong tr ng day BV + tong tr phan t tiep sau dai nhat).

Tren hnh 6.16 gii thieu cac dang ac tuyen cua BVKC cap 3 thng gap:

Trng hp a: dung ba rle tong tr khong co hng phoi hp cung rle nh hng cong suat e xac nh vung lam viec cua ba cap.

Trng hp b: dung ba rle tong tr co hng cho ba cap khac nhau.

Trng hp c: dung ac tnh t giac.

CB III

II

IAR

jx

B

CIII

II

I

CIII

B III

AR

IV

jx jx

R

a) b) c)

A

Hnh 6.16 Cac dang ac tuyen cua bao ve

Bao ve cap III, ngoai chc nang d tr cho ng day, cung co the d tr cho thanh cai va cac iem NM gan cho at BV. Ngi ta thng dung ac tnh off-set-MHO (H.6.16b) hoac cap IV (H.6.16c). Tong tr bieu kien th cap cua rle c tnh

BU

BIkR n

nZZ .=

121

6.7 NHNG YEU TO LAM SAI LECH S LAM VIEC CUA RLE KHOANG CACH 6.7.1 Anh hng cua ien tr qua o tai cho ngan mach, tong tr nguon va tong tr lien ket mach ngoai

Khao sat s lam viec cua rle khoang cach at tai tram A BV ng day RQ hnh 6.17. ng CD tng trng tong tr tng ng lien ket cua mach khac gia hai tram A va B. Chung ta muon rle R cat NM R oi vi cac s co trong khoang hs cua oan RQ (hs tnh bang n v tng oi va c goi la he so at cua rle khoang cach). ieu nay co ngha la rle ch tac ong cat cho en khi nao NM xay ra trc iem at nay, sau iem at nay th tac ong cat co thi gian cap II. iem at nay tr thanh iem can bang cua rle R, hay ngng tac ong cua bo phan khoang cach. Khao sat N3 trc tiep tai iem can bang, tnh c

RSLsR IZhV &&& = (6.33)

vi: VR - ien the tai rle; IRS - dong ien qua rle; ZL - tong tr ng day RQ hs - he so at (ty so tong tr t R en iem can bang so vi tong tr ng day RQ).

V tong tr o c bi rle la ty so ien ap a vao rle vi dong qua rle, nen ta tnh tong tr o c la

LsRS

RRS ZhI

VZ &

&

&&

== (6.34)

Gia tr RSZ& la ranh gii gia tac ong va khong tac ong goi la ngng tac ong, neu NM xay ra gan R, ien the tai rle nho hn va dong qua rle tang len, nen tong tr o c ZR hsZL (6.35)

Con neu s co xay ra t R ln hn he so hs cua (6.34) th rle se khong tac ong. Tieu chuan tac ong cua BVKC theo bieu thc (6.35) oc lap vi tnh trang lam viec cua he thong nguon phat he thong cc ai hay cc tieu, va ieu kien van hanh ong ngat ng day khac nhau. Tieu chuan anh gia khoang cach theo (6.35) ch tai v tr c o lng rle se o, nen co the bo phan o lng se o khong chnh xac tong tr t v tr at rle en iem NM.

~ ~A

C

B

D

QN

R ZL

ZEES EU

IRS

hSES

+

R N Q

EU+ZN

iS

iR

hZL

ZE

(1-h)hZL ZU

iQ

Hnh 6.17 Bao ve khoang cach ng day RQ Hnh 6.18 Mang tng ng ng day

e hieu ro cac yeu to anh hng chung ta khao sat tren mang ien hnh 6.17. Theo tieu

chuan tac ong (6.35), ch so at hs la hang so. oi vi ng day, hs c chon mot khoang 0,8 0,9, ngha la vung BV cua ng day khoang 80 90% ng day, oi vi s co xay ra trong khoang t 20 10% con lai cua ng day, rle ng day au B se cat, va sau o rle au A se cat may cat R co thi gian nh mang hnh tia mot nguon.

122

Mang tng ng cho hnh 6.18 vi ZL la tong tr ng day c BV RQ, NM ba pha xay ra tren ng day tai v tr N cach R mot khoang h, rle tai R se xac nh v tr NM bang cach o tong tr ng day nhn t R, chieu dai ng day a biet, gia thiet chon tr at la hs = 0,9. Khoang cach o lng tai rle R la ham so cua bien so he thong, chang han nh tong tr cua mang va goc pha cac nguon trc khi s co, v tat ca cac bien so nay anh hng en dong va ap tai R.

Tnh va ve cac ac tnh mach cho ng day BV va tong tr o c bi rle tai R tren cac che o lam viec khac nhau cua he thong tren mat phang tong tr cho hnh 6.19.

Nhan xet: Tong tr ng day c bieu dien la ng thang ve t goc O ky hieu: ZL

- ac tnh khi ong rle khoang cach MHO la ng tron i qua goc O, ng knh bang 0,9ZL va co cung goc vi goc ng day RQ.

- ng 1 nam ben phai gan ng ZL la ng tong tr o c bi rle tai R khi h thay oi t 0 1, khi NM qua ien tr trung gian la 0,01 vt. Ta thay, khoang cach o c tai ngng tac ong cua rle la hc = 0,85 thay v 0,9 (90% ng day) nh mong muon (tr so at hs = 0,9), ieu nay xay ra la do cac yeu to nh tong tr nguon, va tr so cua tong tr tng ng mach ngoai lam mot phan dong ien NM i qua mach khac. ay ta cung lu y, NM tai iem 0,9 ng day khong lam khi ong c rle. Tnh trang nay ngi ta goi la di tam (vung BV b thu nho lai).

- ng 2 c ve khi ien tr cham trung gian la 0,05 vt (Zn ln hn nam lan ng 1), ng nay cat ac tuyen khi ong tai h = 0,76. Nhan thay sai so o lng nhieu hn khi co ien tr cham nho hn (ng 1). Trng hp nay cung a en di tam.

- ng 3 va 4 bieu dien tong tr o c bi rle, khi tai he thong trc khi s co thay oi. ng 3 cat ac tuyen khi ong tai h = 0,96 la ng ma trc khi s co co nguon ES = 1 vt va goc ien ap sm trc goc ien ap EU mot goc 60 (trng hp nay hc > hs a en vung BV m rong them, c goi la hien tng qua tam). ng 4 bieu dien khi tnh trang nguon trc s co la EU = 1,0 va sm trc nguon ES mot goc 60, ng nay cat ac tuyen khi ong h = 0,7 (co sai so ln nhat).

h = 0,85

ZL

h = 0,96

1 2

3

h = 0,76h = 0,9s

X (vt)0,9

0,8

0,7

0,6

0,5

0,4

0,3

0,2

0,1

0,0

-0,1-0,2 0,0 0,2 0,4 0,6 0,8

4

R (vt)

h = 0,070

1: = 0 ; Z = 0,01 o N2: = 0 ; Z = 0,05 o N3: = 60 ; Z = 0,05 o N4: = -60 ; Z = 0,05 o N

ZL

Hnh 6.19 Tong tr ng day bao ve va tong tr rle khoang cach

o c tren mat phang tong tr - goc lech gia ES va EU

123

Sau ay gii thieu bang 6.8 ket qua tnh toan vi cac thong so he thong thay oi cua mang (H.6.17). Tr so hc (ngng tac ong rle) cot cuoi cung la giao iem cua ng ac tuyen khi ong rle va ng tong tr bieu kien o c bi rle. Tnh toan vi gia thiet tong tr nguon la thuan khang va tong tr cham trung gian la thuan tr.

Bang 6.8 S lam viec cua rle khoang cach trong cac che o khac nhau cua he thong

Trng hp S XS U XU ZE ZN hC Hien tng sai so

a 0 0,1 0 0,1 0,2 + j 1,0 0,01 + j 0 0,85 Di tam

b 0 0,1 0 0,1 0,2 + j 1,0 0,05 + j 0 0,76 Di tam

c 60o 0,1 0 0,1 0,2 + j 1,0 0,05 + j 0 0,96 Qua tam

d 0 0,1 60o 0,1 0,2 + j 1,0 0,05 + j 0 0,70 Di tam

e 0 1,0 60o 0,1 0,2 + j 1,0 0,05 + j 0 0,62 Di tam

f 0 0,1 60o 1,0 0,2 + j 1,0 0,05 + j 0 0,79 Di tam

g 0 1,0 60o 0,1 0,4 + j 2,0 0,05 + j 0 0,60 Di tam

h 0 1,0 60o 0,1 0,1 + j 5,0 0,05 + j 0 0,65 Di tam

T bang ket qua nay ta nhan xet:

- Tr so ien tr cham trung gian cang ln cang co khuynh hng lam giam tr so hc v tong tr sang pha phai cang nhieu, a en tnh trang di tam cang nhieu

- Tong tr nguon pha at rle cang cao lam rle cang b di tam (trng hp e) - Tong tr lien lac mach ngoai cang ln cang lam giam hc (trng hp g). Neu ca ba trng hp tren xay ra cung mot luc th sai so o lng rat ln (trng hp g). Tai

cua he thong cung anh hng quan trong, sai so cang ln neu goc pha cua nguon EU sm hn nguon ES. Khi nguon pha trai (ES) sm hn nguon EU rle co khuynh hng b qua tam (trng hp c).

T khao sat tren, chung ta co khai niem rle khoang cach c goi la di tam khi tong tr ng day nho hn tong tr bieu kien o c ti iem NM. Phan tram di tam c nh ngha

%100.R

BVR

ZZZ

&

&&

vi: RZ& - tong tr at rle

BVZ& - tong tr khoang ng day c BV thc s (vung BV hieu qua).

Mot rle khoang cach c goi la qua tam khi tong tr thc ng day ti iem cham ln hn tong tr bieu kien o c cua rle. Phan tram qua tam c nh ngha

%100.R

NR

ZZZ

&

&&

e khac phuc cac sai so o lng neu tren, can thiet a vao rle nhieu tn hieu o lng cac v tr khac nhau nhieu hn, nh vay rle lam viec chon loc hn, nhng cau tao rle se phc tap cung nh gia thanh se tang cao.

e lam ro them anh hng cua ien tr trung gian ho quang va s khac nhau cua dong ien trong nhanh at BV va cho NM ta xet cac trng hp cu the sau:

124

1- Anh hng cua ien tr ho quang len thi gian tac ong cac cap cua BVKC

Viec vung BV b thu nho co the c chap nhan vung BV cap I, tuy nhien vung II viec thu nho vung BV (hoac vung BV cap II b di tam do anh hng cua ien tr ho quang) co the anh hng en cac iem NM cuoi cung ng day BV, vung cap II b cat vi thi gian cap III. Do o, khi chon tr at cho thi gian cap II can tnh toan kiem tra khoang b thu nho. Viec tnh toan co the thc hien cho rle co ac tnh MHO BV cap II ng day truyen tai nh sau

Tong tr iem cuoi vung II

dkZZ ZL =+= &&&2

jXd

aKz

1L

bZL

Rhq

R

jX

KzRhq

b

RL 1a) b)

ZL

d a

Hnh 6.20 Tnh toan vung b thu nho do ien tr trung gian ho quang oi vi ac tnh MHO

iem nay tng ng vi ng knh d cua vong tron MHO. T hnh 6.20 ;222 bad += ;cos2222 += ZhqZhq kRkRa && 222 )sin)cos( ++= LhqL ZRZb &&

Giai cac phng trnh tren, ta c

+

= cos1

4cos1

2 2

22

2

L

Z

L

Z

L

ZLhq Z

kZk

ZkZR

&

&

&

&

&

&

Tr so Rhq nay la tr so ln nhat e BV cap II cat bang thi gian cap II oi vi tat ca cac iem NM tren ng day BV (tong tr bieu kien o c cua rle nam trong vong tron ac tnh cap II).

V du: Khi kZ = 0,2 va = 60o th Rhq = 0,29ZL hay ZL = 3,45Rhq Mot cach e giam anh hng cua ien tr ho quang la ng knh cua vong tron ac tnh

khong trung vi ac tnh ng day va lech theo chieu kim ong ho nh hnh 6.20b.

oi vi ng day co hai nguon, viec xac nh ien tr trung gian ho quang khong n gian nh mang hnh tia. Trng hp tong quat, khao sat ng day co hai nguon pha trai va pha phai cua ng day s co. Xac nh tong tr tng ng pha trai va phai Z1, Z2. Khi co NM tren ng day, dong ien NM gom hai thanh phan I1 t pha nguon trai va I2 t pha nguon phai. Ngi ta co the tnh tong tr tong, nhn bi rle pha trai la

hqLR

RR RZ

ZZIVZ

++== 1

2

1&

&&

&

&&

vi LZ& la tong tr thc t cho at rle ti iem NM; SL ZZZ &&& +=1

125

Nh the, tong tr ho quang bieu kien o c tang t gia tr thc cua no ti gia tr nh so hang th hai cua ZR. V tong tr nguon phu thuoc vao cong suat phat, nen ty so cua tong tr bieu kien tren se phu thuoc vao tnh trang he thong, va ien tr ho quang bieu kien se ln nhat oi vi cac iem cham gan rle. Lu y, ien tr ho quang bieu kien co the la so ao v ty so tong tr trong so hang th hai co the co goc pha khac khong.

2- Anh hng do s khac nhau cua dong ien trong cac nhanh gia cho at bao ve va cho ngan mach

Khi khao sat vung th II cua BVKC can phai ke en anh hng cua dong ien phu vao cua cac nguon nam gia cho NM va cho at BV. V du mang cho hnh 6.21, khi NM tai iem N, tong tr tren cc rle at tai tram A:

lZk

lZlZII

lZI

lZIlZIIUZ

pdAB

AB

BNAB

AB

BNABAB

AB

ARA ..

1....

....1111

11 &&&

&&

&

&&&&

&

&& +=+=

+==

trong o 1+=

&&&&

Co ngha la, tong tr at vao rle ln hn tong tr thc te cua ng day t cho at BV en cho NM va rle se lam viec nh trng hp NM b di ra xa mot oan (a en tnh trang di tam).

~

~

~

IAB

ICB

IAB

IBN

LLAB

AB

N

N

C

RZ

RZ

Hnh 6.21 Anh hng cua he so phan nhanh dong k1 len s lam viec cua rle tong tr

Trng hp b th ngc lai BNAB II && > nen kpd > 1 va

)(..1. 111 llZlZklZZ AB

pdABRA +

126

cng cac ac tnh co hng cua ac tnh MHO. S phan cc bu khong co hieu qua trong trng hp nhng s co ba pha gan v tr at BV, trong o tat ca ien ap cua cac pha co hien tng sup o ien ap. e khac phuc trng hp nay, ngi ta s dung phng phap nh ien ap trc s co. Rle khoang cach the he ky thuat so co ac tnh Micromho va Quadramho co mach phan cc ong bo so.

Cac mach phan cc ong bo nh vai chu ky ien ap trc s co vi cac thong tin chnh xac ve tan so va pha. Khi ng day c ong ien, mot vai chu ky can phai c phan cc ong bo e t ong khoa tan so va pha cua nguon cung cap chnh, nham muc ch cho cac n v offset MHO cua vung III tc thi nhng s co trong luc ong ien ng day ang b s co. Cac mach phan cc bu va nh cho phep tac ong nhanh oi vi nhng s co gan v tr at rle theo chieu thuan cua rle va am bao on nh oi vi nhng s co gan nhng theo hng ngc. Cac ac tnh Micromho va Quaramho se am bao viec thc thi nay ngay ca trng hp co xet en nhng sai so qua o trong MBA kieu tu ien. Cac phng an ien ap bu c cho trong quyen bai tap.

6.7.3 Anh hng cua sai so o lng

Sai so BI, BU lam cho tr so tong tr ZR at tren cc cua rle thay oi. V the, cac BI dung trong BV phai c kiem tra theo ng cong sai so 10% khi NM trc tiep ba pha cuoi vung BV th nhat. e giam ton hao ien ap trong cac day dan phu mach ien ap, oi khi phai dung day co tiet dien ln hoac dung MBA o lng co ien the th cap 200V thay v 100V.

6.7.4 Anh hng cua cach noi day may bien ap ong lc at gia cho bao ve va cho ngan mach

Khi NM sau MBA co to noi day Y/Y th rle tong tr se lam viec nh trng hp NM tren ng day, tong tr at vao rle se bang tong so cac tong tr cua ng day va MBA. Neu to noi day cua MBA la Y/ hoac /Y th rle tong tr se lam viec khac i, v khi NM hai pha sau MBA dong ien pha s cap va th cap cua MBA khong nhng khac nhau ve tr so ma con khac nhau ve goc pha.

6.7.5 Anh hng cua dao ong ien

Dao ong ien (D) phat sinh khi cac may phat ien trong he thong (HT) lam viec mat ong bo. Khi co D, th dong tang len, ap giam va BV se phan ng giong nh luc co NM oi xng. Trong trng hp co D, yeu cau khong cat MC ma phai tm bien phap e khoi phuc tnh ong bo cua he thong. Quan sat he thong ien tng ng hai may phat (MF) co suat ien ong Es, ER va ng day tai ien AB (H.6.22a). Goc lech pha gia hai suat ien ong nay la , khi cong suat truyen tai tren oan AB cang ln th goc cang tang. Nhng tang co gii han, e bao am tnh on nh cua he thong. T hnh d dong ien dao ong do chenh lech gia Es va ER:

ZEEI Rsdd &&&

& =

vi Z& la tong tr tng ng cua mach co dong dao ong chay qua.

Hnh 6.22 o th vect cua dong va ap khi co dao ong

~ ~~A B

A BES

-ES

UA

C

C

UB

ER

ER

SA Id

RES

UA 180o

UB

ER

127

Neu EEE RS == && th 2

sin.2

=

ZEIdd

T hnh 6.22d ta cung thay, ien the giam dan khi tien dan en trung tam ng day, ien the nay cang giam khi tang len, va ien the nay se tien ti 0 khi = 180o (khi o vect ES va ER ngc chieu nhau, hnh c). iem co ien the 0 nay goi la tam ien cua HT hay tam cua D. Trong qua trnh dao ong thay oi keo theo s thay oi ien ap tat ca cac iem mang ien. Khi = 0, ien ap tat ca cac iem trong mang bang nhau va co gia tr cc ai Um ax = E. Khi tang, ien ap trong mang ien giam xuong va co gia tr cc tieu tam ien. Khi = 180o ien ap tam ien giam xuong en khong, con tat ca cac iem khac cua he thong U = Id.Z. Ta quan sat thay s thay oi tong tr at tren au cc rle trong qua trnh dao ong tren mat phang tong tr Z (H.6.23).

Gia thiet tai tram A cua ng day AB, co at BVKC, co ac tuyen khi ong la tong tr co hng. Khi he thong lam viec bnh thng, khong vt qua 15 20o va cac nh cua tong tr ma BVKC o c luc o nam ngoai vung tac ong cua ac tuyen (v du iem P1 ng vi trng hp RS EE && = , P2 ng vi trng hp

RS EE && > , P3 ng vi trng hp RS EE && < ). Khi co dao ong, thay oi neu RS EE && th quy ao cua cac nh tong tr nhn bi rle khoang cach la ho ng cong (ng 1 va 2 nam trong ho o) va la ng thang 3 khi RS EE && = .

Khi co NM, rle khoang cach o tong tr ng day AB, con trong trng hp co dao ong no se o tong tr AP. Khi tang th iem P se di chuyen va iem P co the nam trong vung khi ong cua BV. Can phan biet hai trng hp dao ong ong bo va khong ong bo.

Trng hp dao ong ong bo: khong lam cho cac MF mat on nh, va BV khong c phep lam viec e he thong nhanh chong khoi phuc on nh.

Trng hp dao ong khong ong bo: cac MF ang lam viec mat on nh, cac ket qua nghien cu nhng nam gan ay cho thay vi nhng ieu kien nhat nh th trong trng hp nay MF van co the tr lai lam viec ong bo vi nhau. Noi tom lai, yeu cau rat quan trong oi vi BV la khong c tac ong khi co dao ong.

oi vi BVKC, e tranh tac ong nham khi co dao ong, ngi ta chon ac tuyen khi ong khong co cha tam D. Tren thc te, bien phap nay ch co the ap dung cho vung I cua BV, cach th hai la lam cho BV tac ong vi thi gian tr hoan khoang 1 2s. Bien phap nay ch c ap dung khi s tr hoan khong lam anh hng en on nh cua he thong. Cach th 3, hien nay ang dung rong rai cho cac ng day cao the la dung bo khoa t ong khi co xuat hien D, da vao nguyen tac o toc o tong tr thay oi. Khi dao ong toc o thay oi cham nhieu so vi khi NM.

Hnh 6.23 S thay oi tong tr au cc rle khi co dao ong

1

2

3

A

B

S

P2

123 RR

R

P3

Z (E < E )R S R

Z (E < E )R S R

jX

128

6.7.6 Anh hng cua tu ien bu doc len bao ve khoang cach

Tren cac ng day dai, sieu cao ap, ngi ta thng lap noi tiep vao ng day cac bo tu ien c goi la cac tu bu doc. Cac bo tu nay lam giam khang tr cua ng day, tang gii han truyen tai cong suat theo ieu kien on nh cua he thong, giam ton that va cai thien ieu kien phan bo ien ap doc theo chieu dai ng day nhng che o cong suat khac nhau.

Z

NX = 0,6XC L

XL

X = 0,6XC L

X 0,5XL

L2XC

0,4XL0,4XL

XN

-XC-0,6XL

-0,1XL

XC = 0,6XL

L3

L3

XL

0,4XL

XC

L

X

L

X

L0,4XL

0,4XLXC

2XC

2

XC2

XL3

XC2

= 0,3X1XC

2XC

2= 0,3X1

XC2

a) b) c)

d) e) g)

XC2

L

XL

Hnh 6.24 Tong tr bieu kien o c cua rle at tai au ng day co tu noi tiep

Cac bo tu ien bu doc co the at tap trung hoac phan tan theo chieu dai ng day. Tuy theo mc o bu va v tr at cua bo tu ma tong tr o c au ng day se khac nhau. Mc o bu doc c ac trng bang he so bu doc kC bieu dien gia ty so chung tr XC cua bo tu doc va khang tr XL cua ng day. oi vi ng day truyen tai sieu cao ap thng chon he so bu doc kC = 0,25 0,7 (ng day 500kV Bac - Nam cua Viet Nam co kC = 0,6). Hnh 6.24 trnh bay quan he bien thien cua tong tr o c au ng day co he so bu kC = 0,6 vi cac phng an at tu bu khac nhau. Thong thng, e de dang viec lap at, van hanh va bao quan, ngi ta at tu bu doc gan thanh cai cac tram ng au ng day (H.6.24b,d,g). oi vi cac ng day co at tu bu doc, s lam viec cua BVKC co the b sai lech, khi NM xay ra nhng v tr khac nhau tren ng day. Chang han, rle khoang cach RZ at au A (H.6.24b) khong the nhn thay NM tai N1 sau bo tu XCA (H6.24a), va ca mot phan ng day gan o v tong tr cua tu va phan ng day nay nam pha sau lng ngoai vung tac ong cua rle khoang cach RZ1. Ngc lai, rle khoang cach RZ3 lam nhiem vu BV oan ng day trc o co the phan ng sai, tac ong mat chon loc v iem N1 lai ri vao mien tac ong cua RZ3 (H.6.25). e ngan chan tac ong sai cua rle khoang cach, trong trng hp nay co the dung phng phap noi tat bo tu khi xay ra NM va cho vung 1 cua cac rle khoang cach tac ong vi o tre chng 0,1 0,15s.

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Khi bo tu b noi tat, ac tnh tong tr cua ng day c BV tr lai giong nh nhng ng day bnh thng khong co tu bu doc va rle khoang cach co the am bao tnh chon loc bnh thng (H.6.25b).

3 1A

XCA N1 XCBB

2 4XLRL

jX

R

XL

A

XCA

Z3

RL

ZLZ1

Z1

B

B

jXZ1

Z1

Z3

A

XCBXCB

RL

BXL

R

RZ3 RZ1 RZ1RZ2

Hnh 6.25 Bao ve khoang cach tren ng day co at bu doc

a) Khi tu lam viec bnh thng; b) Khi tu b noi tat

cac bo tu ien bu doc hien ai, ngi ta s dung HTBV bang ien tr phi tuyen (Variston ZnO), khe phong ien va may cat ien au song song vi bo tu. Khi co NM tuy theo iem s co (tr so cua dong NM) va thi gian ton tai s co ma cac thiet b nay se lam viec va noi tat bo tu lai. Tat nhien, noi tat bo tu va cho vung I cua rle khoang cach lam viec tre se anh hng en tnh tac ong nhanh cua BV.

Cac nguyen ly BV khac nh so lech dong ien, so sanh pha... khong b anh hng bi bu doc co the c dung e BV chnh song song vi BVKC.

Anh hng cua tu noi tiep tren ng day se c khao sat ky hn chng BV ng day.

6.7.7 Anh hng cua ho cam ng day song song

Trng hp nay se c trnh bay trong chng BV ng day.

6.8 ANH GIA LNH VC NG DUNG CUA BAO VE KHOANG CACH Nh co mot so u iem nhat nh nen BVKC c ng dung rong rai trong cac mang ien cao ap.

1- Cac u iem chnh cua bao ve khoang cach - am bao tnh chon loc trong mang co cau truc bat ky va co so nguon cung cap tuy y. - Vung I cua BV chiem gan 80 90% o dai phan t c BV va co thi gian lam viec rat

be. ieu nay rat quan trong oi vi ieu kien on nh he thong la phai cat nhanh phan t s co gan thanh gop nha may ien va cac tram iem nut cong suat ln.

- Co o nhay cao oi vi NM.

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2- Cac nhc iem cua bao ve khoang cach - Phc tap ve mat s o cung nh ban than cua cac rle thuoc ve s o BV. BVKC dung rle

ien c la loai BV co nhieu rle v can nhieu tiep iem nhat. - Cac s o BV khong tiep iem dung ban dan, vi mach th phc tap ve mat s o logic va

co rat nhieu phan t chc nang. - Khong am bao cat tc thi NM tren toan bo ng day c BV. - Phan ng theo dao ong va phu tai, v vay viec tranh phu tai lam giam o nhay BV, giam

tac dung d tr cho phan t ke. e chong dao ong phai dung thiet b khoa va BV cang tr nen phc tap.

Tuy con nhng khuyet iem nh tren, nhng hien nay BVKC van la BV thong dung nhat dung e BV chnh hay d tr cho cac ng day cao ap hay trung ap.

Cau hoi chng 6: 1-Xac nh khoang cach t cho at r le en iem ngan mach thong qua cac gia tr o nao? 2-Giai thch tai sao cap I cua bao ve khoang cach tnh toan 80% en 90% tong tr ng day

bao ve. 3-Gia thch tai sao gia tr tnh toan cap II cua bao ve khoang cach theo ieu kien tong tr khi

ong phai ln hn 120% tong tr ng day bao ve va vung bao ve cua no nho hn vung bao ve cap I cua bao ve ng day ngan nhat tiep sau.

4-Trnh bay cac cach tnh toan tong tr khi ong cap III cua bao ve khoang cach. 5-Tai sao can co bao ve vung ngc cap IV? 6-Tai sau phai dung hai bo bao ve khoang cach chong ngan mach nhieu pha va mot pha rieng

biet. 7-Cach kiem tra o nhay bao ve khoang cach. 8-Hien tng di tam,qua tam la g ? cho v du khi dung bao ve khoang cach. 9-Cac yeu to anh hng en sai so cua bao ve khoang cach,trng hp nao se a en di

tam hay qua tam. Websides tham khao

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