Post on 04-Nov-2015
description
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 .
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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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