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8/13/2019 Bai Giang Mon Bao Ve Role - Dung Cho in An
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BO V RLE
TRONG H THNG IN
2/7/2012 Ging vin: Nguyn Xun Tng
B mn H thng in
Trng i hc Bch Khoa H Ni
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Gii hi chng
Chng 1: Cc phn t chnh ca h thng bo v rle
Chng 2: Nguyn l bo v qu dng in
Chng 3: Nguyn l bo v so lch dng in
Chng 4: Nguyn l bo v khong cch
2
cng mn hc
ng : c nguy n o v c Chng 6: Bo v cc ng dy ti in
Chng 7: Bo v cc my bin p
Chng 8: Bo v cc h thng thanh gp Chng 9: Bo v cc h thng t b & khng b
Chng 10: Bo v cc my pht & ng c
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S cn thit phi c h thng bo v t ngCc s c din ra vi tc nh sng
Thi im & nguyn nhn gy s c khng bit trc Ging st, hng cch in, cnh cy va chm
Phn x ca con ngi khng th kp thi
Cc thao tc tron tnh hun khn c khn m bo chnh xc..
4
Khi nim chung
Cn thit phi c h thng bo v t ng cch ly cc phn t bs c h thng bo v rle (BVRL)
Rle bo v l thit b:T ng ghi nhn & phn ng:
Tnh trng lm vic khng bnh thng ca cc thit b v h thng
Cch ly cc phn t b s c (ct my ct MC)
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Thit b bo v n gin nht l cc cu ch (cu chy),aptomat
5
Khi nim chung
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Thit b bo v phc tp hn l cc rle vi cc nguyn l khcnhau:Rle qu dng, so lch, khong cch
Rle tri qua nhiu th h pht trin
6
Khi nim chung
Rle in c Rle tnh (bn dn) Rle k thut s
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Tin cyL kh nng m bo thit b lm vic ng khi c s c v khng tc
ng ch bnh thng
Chn lcPht hin loi tr ng phn t b s c
7
Cc yu cu i vi h thng BVRL
S c ti N3: ch yu cu BV3 tc ng, cc BV cn li s tr v khi s c b
loi tr S c ti N2nhnh: ch BV ti nhnh tc ng m bo chn lc
Vi cc li in c cu hnh phc tp: S dng cc nguyn l bo v phc tp hn
N1 N2 N3
I> I> I>1 2 3
nhnh
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Tc ng nhanhGim thiu tc hi ca s c
Cc bo v tc ng nhanh: thi gian tc ng 50ms
Yu cu chn lc & tc ng nhanh: mu thun vi nhau trong mt strng hp
8
Cc yu cu i vi h thng BVRL
N1 N2 N3
X l: m bo sao cho thit hi ti ph ti l nh nht
Dng cc nguyn l bo v khc
I> I> I>1 2 3
m bo chn lc: tb1> tb2> tb3
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nhyc trng cho kh nng cm nhn s c ca rle
H s nhy Kn:
Bo v chnh : K = 1 52
9
Cc yu cu i vi h thng BVRL
Kn= Gi le o c khi c
Gi khi ng ca le
Bo v d phng: Knmin= 1,21,5
Tnh kinh tLi in phn phi: s dng cc bo v qu dng n gin
Li truyn ti: s dng bo v so lch dc, bo v khong cch, haib bo v chnh
My bin p h p: bo v bng cu ch
My bin p truyn ti: bo v bng bo v so lch
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Cc phn t chnh ca h thng
Chng 01
11
bo v
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Tn gi chung: BI, CT, TI
Nhim v:Bin i t l dng in s cp th cp (5A hoc 1 A)
Cch ly mch s cp v th cp
To s phi hp dng in gia cc pha
12
My bin dng in1.1
BI cao p BI h p S nguyn l
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S tha th
13
My bin dng in
1.1
BI l tng
Sai s ca BI xut hin do tn ti ca dng t ha
in p xut hin pha th cp
Vth cp=Ith cp*(Zcun th cp+Zdy dn ph+ Zthit b ni vo)
Ti tng Vth cptng tng dng t ho Ie tng sai s caBI
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Qui c cc tnhCn thit vi : bo v lm vic da theo
hng dng in.
Cc tnh cng tn c nh du : hnhsao, chm trn, chm vung...
Trn bn v: cc tnh cng tn v cnh
14
My bin dng in1.1
nhau. Xc nh nhanh cc tnh BI:Coi chiu dng in i t pha s cp qua
rle khng i chiu
Rle
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Qui c cc tnh
15
My bin dng in1.1
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u ni BI lc thnh phn TTKDng 3 BI ring bit
16
B lc dng inth t khng (I0)1.1
Ia
Ib
Role
3I0
V rt gn
Do s dng 3 BI ring bit nn s c sai s gia cc BI
ch bnh thng, pha s cp l i xng: lun c dng in chy quarle do sai s ca BI
Ch s dng o dng chm t ln dng mng in c dng chm t ln:mng in trung tnh ni t trc tip
c
Role
Ia Ib Ic 3I0+ + =
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u ni BI lc thnh phn TTKDng BI th t khng (Flux Summation CT hoc Core Balance CT)
Bin dng c mt li t hnh xuyn
Cun dy c phn b u trn li Dy dn s cp chy xuyn qua li t (ng knh trong 1025 cm)
17
1.1 B lc dng inth t khng (I0)
u sai u ng
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BI th t khng
u ng: dy ni t v cp chy xuyn qua li t
Ngc li u sai: dng in chy qua v cp c th trit tiudng in s c (hon ton hoc mt phn) v rle c thkhng nhn c thng tin s c.
18
My bin dng in1.1
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u ni BI lc thnh phn TTKng dng ca BI th t khng
Do ch s dng mt li t sai s o lng rt nh
S dng cho cc mng in c dng chm t nh (mng in c trung tnhcch in hoc ni t qua cun dp h quang)
19
My bin dng in1.1
Do c 3 pha chy qua li t ng knh li t ln kch thc BI ln thch hp trang b vi ng cp hoc u cc my pht in
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Bo v chng chm t nhy cao (Aptomat chng git)
20
ng dng thc t ca BI th t khng1.1
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Ti danh nh & Cp chnh xc
Mt BI: c nhiu cun th cp phc v cc mc ch khcnhau.
Ti danh nh v chnh xc ca cc cun th cp ny tuthuc vo loi ti.
21
My bin dng in1.1
, , , , :Yu cu chnh xc trong ch ti bnh thng hoc nh mc.
Phm vi hot ng chnh xc trong khong 5120% ca dng in
chnh xc thng l: 0.2 hoc 0.5 vi chun IEC
Hoc 0.15 hoc 0.3 hoc 0.6 vi chun IEEE.
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BI dng cho mc ch olng
Vi mc ch o lng
Li t s bo ha nhanh bo v cc dng c o khidng in s c tng tbin.
22
My bin dng in1.1
Cpchnh
xc
Sai s gii hn
Ti % dngnh mc
Sai s ln%
Sai s gcpha
(pht)
ng dng
Phng thnghim
o mchnh xc
BI dng cho o lng ck hiu theo:
Cp chnh xc
Cng sut nh mc
o mchnh xc
o m
thng mi
o mchnh xc
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BI dng cho h thng bo v rle
Vi mc ch bo v rle & ghi s c: chnh xc thp hn so vi loi dng cho mc ch o lng
in p bo ho cao hn (kh b bo ho)
Cp chnh xc:
23
My bin dng in1.1
, ,
C100800 (IEEE).
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Tm hiu thng s ca BIVi mc ch bo v rle
C thm thng s ALF: h
s gii hn dng in theo chnh xc
V d: BI dng cho mc
24
My bin dng in1.1
Cpchnh
xc
Sai s gii hn
Ti % dngnh mc
Sai s ln%
Sai s gc pha(pht)
ng dng
Chun hadng c otrong cng
nghip
ng h o
Ti % ti nhmc
c o v c u ng h o
Bo v rle
Bo v rle
Bo v rle
5P20 30VA
Cp chnh xc 5P
P: dng cho mcch bo v rle
(Protection) H s gii hn dng: 20
Ti 20 ln dng nh mc, BI vnm bo sai s theo tiu chun
Cng sut nh mc 30VA
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Tm hiu thng s ca BIVi mc ch bo v rle
25
My bin dng in1.1
5P20 30VA
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Thc t, mi BI c th c:1 hoc 2 cun th cp Mc ch o lng
2 ti 4 cun th cp ng dng bo v rle.
26
My bin dng in1.1
Cun s cp
Cc cun th cp
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Thit k BI ph bin
Li t v cun dy nm trong thng cha thp gn vi t(Dead ank e), dy th cp chy un theo hnh ch U
Li t v cun dy nm trong thng cha pha trn nh (Livetank type), dy th cp thng chy thng qua li t.
27
My bin dng in1.1
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BI loi thng cha bn di (Dead tank)Trng tm thp, n nh v mt c kh
C th ch to vi li t loi to m khnggy qu ti v mt c kh i vi s cchin
28
My bin dng in1.1
Dy s cp c chiu di ln nn gy phtnng nhiu hn hn ch v dng nnh nhit (ln nht 63kA/1 giy)
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BI loi thng cha bn trn (Lie tank)Dy dn s cp ngn, gim pht nhit
C dng nh mc v dng n nh nhit
cao hnTrng tm cao hn, km n nh v mt c
kh so vi loi thng cha di (dead tank)
29
My bin dng in1.1
Khi ch to vi li t ln c th gy titrng ln v mt c kh i vi s cchin.
Kh lm mt cc cun th cp
Loi hn hp
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M bi i (BU)
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in p danh nh s cp v th cpBU ngoi tri thng s dng in p pha:
in p danh nh ca cun s cp l in p danh nh ca li in.
ng dng o lng: phm vi in p lm vic: 80120%ng dng bo v rle: t 0.05 1.5 hoc 1.9 ln in p danh nh.
31
My bin in p (BU)1.2
M bi i (BU)
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BU ki t phnp BU loi cm ng in t thng thng
La chn kinh t nht i vi cp in p ti 145kV
BU kiu t phn p (CVT Coupled Voltage Transformer) La chn khi dng cp cao p
Thng c s dng kt hp vi h thng thng tin ti ba PLC
32
My bin in p (BU)1.2
u cao p
in khng b
Tphn
p Mch dp dao ng cnghng
BU cm ngthng thng
u ra
M bi i (BU)
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C trc BU ki t phnp1. Bnh gin du
2. Cc phn t t phn p
5. in khng b
7. Cun dy s cp v th cp
33
My bin in p (BU)1.2
.
M bi i (BU)1 2
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BU ki t phn p in khng b: c tnh ton trit tiu thnh phn dung
khng ca t phn pTng tr ngun nhn t pha ti l xp x 0 cng sut u ra ly ra
ln nht
B d ch ha do t hn ra
34
My bin in p (BU)1.2
Mch gim dao ng cng hng: l in tr ti, c th ni cuntam gic h
My bin in p (BU)1 2
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Cc loi BU H s gii hn in p VfKhi xy ra s c trong HT: Uhac th tng ln ti mt gi tr l Vfln
Udanh nh.Tiu chun IEC a ra cc gi tr h s Vfnh sau:
1.9 i vi cc h thng c trung tnh khng ni t trc tip
35
My bin in p (BU)1.2
1.5 i vi cc h thng c trung tnh ni t trc tipLi t ca cc bin in p khng c php bo ho khi in p tng
ti cp in p gii hn theo h s Vf.
My bin in p (BU)1 2
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Cp chnh xc theo tiu chun IEC 600442
36
My bin in p (BU)
Cpchnhxc
Sai s gii hn
Ti % in pnh mc
Sai s ln%
Sai s gc pha(pht)
ng dng
Ti % ti nhmc
Phng th nghim
o m chnh xc
1.2
Cng sut danh nh (cos=0,8)10, 15, 25, 30, 50, 75, 100, 150, 200, 300, 400, 500 VA
Chun ha dng c otrong cng nghip
ng h o
Bo v rle
Vf: H s gii hn in p
Bo v rle
o m tiu chun
My bin in p (BU)1 2
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B lc in p TTK (U0) Cch u ni BU cho mc ch bo v rle
37
My bin in p (BU)1.2
in p TTK ly ra t cun tam gic h
My bin in p (BU)1 2
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B lc in p TTK (U0) Chn t s bin p cho cun tam gic hTrung tnh cch in
38
My bin in p (BU)1.2
T s bin c th l:
My bin in p (BU)1 2
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B lc in p TTK (U0) Chn t s bin p cho cun tam gic hTrung tnh ni t trc tip
39
My bin in p (BU)
Vectin p khi xy ra scchm t mt pha trong mng ctrung tnh ni t trc tip
1.2
T s bin c th l:
Ngun thao tc1 3
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Hin nay ch yu s dng h thng acqui
40
Ngun thao tc1.3
Ngun thao tc1 3
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Cung cp nng lng cho:Cung cp nng lng trong trng hp mt in t pha h thng
Cp nng lng cho cc thit b bo v rle
Cc thit b iu khin, chiu sng khn cpNng lng thao tc ng/ct my ct...
41
Ngun thao tc1.3
c lp vi ch h thngDung lng ln, chi ph thp
H thng acqui p ng c cc yu cu ny
Ngun thao tc1.3
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Cc loi acqui ph binAcqui acid
Cc bn cc l Pb (cc dng) v PbO2 (cc m)
in th mt ngn l 2V
42
Ngun thao tc1.3
Ngun thao tc1.3
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Cc loi acqui ph binAcqui kim
Bn cc l bt Niken (+) v bt Cadimi ()
in th mt ngn: 1,2V D bo dng, tui th cao di hn (1015
ln acqui acid), chi ph ban u t hn
43
Ngun thao tc1.3
Ngun thao tc1.3
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Ch npNp duy tr
Lun duy tr dng np nh gi in p acqui khng thp qu 2,25V & 1,35 V
cho acqui acid v kim Dng np duy tr khng ln trnh to bt kh
44
Ngun thao tc
p sung Lun gi in p khng qu 2,25 & 1,35V cho mi ngn
Ngun thao tc1.3
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S dng nng lng b tCung cp nng lng ch trong khong thi gian ngn
45
Ngun thao tc
Ti im t
rle ra
Ngun choh thng
rle
Knh thng tin trong h thng in1.4
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Knh thng tin chnh OPGW: Cp quang
Microwavew link: thngtin v tuyn (viba)
46
g g g
ng in thoi ni bv chung
H thng ti ba (PLC)
Knh thng tin trong h thng in1.4
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H thng ti ba S dng chnh ng dy ti in truyn tn hiu thng tin, iu khin
Thch hp khi cn trao i thng tin gia cc trm bin p v cc ng
dy truyn ti thng c sn
47
g g g
B chn B chn
B chn: ngn tn hiu suy hao thng qua in dung thanh gp
B cng hng: c tng tr cao ti tn s dng in cng nghip v tng tr rt thp ti tn stn hiu
Thanh
gp titrm
Thanh
gp titrm
T/P T/P
Bcng
hng
Bcng
hng
B
thu/pht
B
thu/pht
Knh thng tin trong h thng in1.4
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H thng ti ba
48
g g g
B chn
T/P
Rle1.4
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49
Rle c (c phn chn ng qa) Da trn nguyn tc bin i in c
Dng in chy qua cc cun dy s sinh ra lc t tc ng
ti cc c cu, chi tit tha hnh khc. Tn gi xut pht t nguyn l lm vic
C th phn loi theo chng loi:Cun ht np t s dng ph bin nht
a chuyn ng (cc cng t)
Cm ng, rle nhit..
Rle1.4
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50
Rle c (c phn chn ng qa)
Hot ng vi dng in ac hoc dc
Tiu th cng sut ln Thi gian tc ng ko di do qun tnh cc phn quay
Rle1.4
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51
Rle c (c phn chn ng qa)
Rle qu dng vi c tnh thi gian ph thuc
Rle1.4
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52
Rle tnh (khng c phn chn ng qa)
S dng cc thit b bn dn
Tiu th cng sut t hn n chc nng, khng c kh nng kt ni
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Rle1.4
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54
H s tr ca rle Khi dng in chy qua rle tng ln n mt gi tr no rle s
ng tip im rle tc ng
Gi tr dng in ti rle tc ng: dng in khi ng (Ik) Thc hin thao tc ngc li: gim dn dn dng in qua rle n
mt gi tr no rle s nh tip im rle tr v
Gi tr dng in ti rle tr v: dng in tr v (I tv)
H s tr v K=
Rle c : Ktv=0,80,85
Rle tnh, rle s: Ktv=0,90,95
IIk
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Chng 02
55
Nguyn l bo v qu dng c thi gian2.1
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Bo v qu dng (I> hoc 50 & 51):Chng li cc dng s c qu dng mt pha, hai pha & ba pha v s c
chm t.
Bo v khi ng khi:
56
,
c ci t trc trong rle.
C th lm vic vi thi gian tr m bo tnh chn lc
Thi gian tr c th l c lp so vi dng in hoc ph thucvo dng in hai loi c tnh thi gian tc ng
Nguyn l bo v qu dng c thi gian2.1
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Bo dng i c nh hi gian c l:Thi gian lm vic (tr) ca bo v khng ph thuc vo ln dng
ngn mch
57
+
Nguyn l bo v qu dng c thi gian2.1
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Bo dng i c nh hi gian h hc:Thi gian lm vic: ph thuc t l nghch vo ln ca dng in
ngn mch
Trong thc t th thi gian tc ng t l vi t s Ingn mch/ Ikhi ng
58
Tnh ton dng khi ng2.1.1
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i kin Phi khng khi ng ch bnh thng
Phi khi ng vi dng s c nh nht ( nhy)
m bo tnh chn lc
59
t n c a p n onDng lm vic gi thit ang ln nht (Ilma)
I> I>1 2
C
CTG1 TG2
Il ma
60
Tnh ton dng khi ng2.1.1
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60
I> I>1 2
C
CTG1 TG2
N2
Il ma
t
Il maS c
BV2 c MC
ng c mm li
Im m I BV1
Dng in (A)
1
i N2
.
1. Dng in tng ln Sau gim i mt cht sau giaion qu
2. in p TG2 gim i cc ng c gim tc
2. BV1 & BV2 khi ng m thi gian
3. Ti thi im t2: BV2 ct my ct, loi tr s c1. Dng in gim i do s c c loi tr
2. in p TG2 hi phc cc ng c m my tr li xut hin dng in m my ln
3. Dng in m my gim dn theo thi gian n gi trn nh BV1 phi dng m thi gian d ang cdng m my BV1 phi tr v
2Dng ngn mch b c
X hin dng m m
BV1 hi Dng m
hi gian
BV1 & BV2 khing
phi chn
61
Tnh ton dng khi ng2.1.1
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61
t
Il ma
Im m
I BV1
1 2
m bo iu kin Itv> Im myta t Itv=Kat*Im my H s an ton Kat ty chn: Kat=1,11,3
Biu din dng m my theo dng lm vic ln nht: Im my= Kmm*Ilv max H s m my K
mmty thuc vo nhiu yu t: v tr t ng c so vi v tr tbo v, s lng,
chng loi ng c...
T : Itv=Kat*Im my= Kat* Kmm*Ilv max
Theo nh ngha ca h s tr v Ktv=Itv/IkIk= Ik=ItvKtv
Kat* Kmm*Ilv maxKtv
Cng hc ci cng nh dng khi ng
62
Tnh ton thi gian lm vic2.1.2
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m bo tnh chn lc gia cc bo v bng phn cp thi gian Tn gi: bo v qu dng lm vic c thi gian (I> hay 51)
Nguyn tc:Khi c s c c th nhiu bo v cng khi ng
Tuy nhin, bo v gn ch s c nht s phi tc ng trc
62
S c ti N2: BV2 khi ng & BV1 c th cng khi ng cng
m thi gianBV2 phi tc ng loi tr s c, BV1 khi s tr v t thi gian
tBV2 I>1 2
N2
63
Tnh ton thi gian lm vic2.1.2
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BV2 phi tc ng loi tr s c, BV1 khi s tr v t thi giantBV2
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Phi hp cc bo v bng c tnh ph thucPhc tp hn
64
I> I>1 2 I> 3
31 t
C nhiu loi c tnh ph thucKhc nhau v dc (mc ph thuc)
Standard Inverse (SI): dc tiu chun
Very Inverse (VI): rt dc
Extremely Inverse (EI): cc k dc
2 L (km)
65
Tnh ton thi gian lm vic2.1.3
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Phm vi s dng ca cc c tnh ph thucStandard Inverse (SI): c tnh dc bnh thng
S dng trong hu ht cc trng hp cn phi gia cc bo v
Very Inverse (VI): c tnh rt dc S dng khi ln dng in s c dc ng dy cn bo v thay i mnh.
65
Extremely Inverse (EI): c tnh cc dc Thi gian tc ng t l nghch vi bnh phng ca dng in.
ng dy mang cc ti c dng khi ng t bin thi im u v d nh
t lnh, my bm, ng c ln... Phi hp vi cc cu ch hoc cc thit b t ng li.
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67
gun l u dng c nhnh (50 h I>>)2.2
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ng c c nhnhIk=K*Ingn mch ngi ng m ( K=1,11,2)
ng ngn mch gim dn khi im ngn mch i ngun
ln dng ngn mch h huc ch c h hng ng cng iu din dng ngn mch Inm& Inmindc ng d
L(km)
I> I>1 2
I
Inm
Inm
Ik B1
Ik B2
Lc nhnh min
Lc nhnh m
Lc nhnh min=0
Lc nhnh m
68
gun l u dng c hng (67)2.3
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li in c ngun hi h
S c i 1 c h B3 & B2 khi ng
I> 1 I> 5I>4 I>6I> 3I>21 2
N1
tBV34 I>6I> 3I>2HT1 HT2
N2
tBV3>tBV2
69
Nguyn l bo v qu dng c hng (67)2.3
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I>
+
Gii php: s dng bo v qu dng loi c nh hng Bo v qa dng c hng ch tc ng khi:Dng in chy qua bo v theo hng qui nh (hng dng
thng qui c t thanh gp ng dy)Dng in vt qua gi tr khi ng ca bo v
> = > +
I> 1 I> 5I>4 I>6I> 3I>2HT1 HT2
C th phn chia ra 2 nhm bo v
70
Nguyn l bo v qu dng c hng (67)2.3
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V phng din bo v rle: ng dy hai ngun cphaimch hnh tia
I> 1 I> 5I>4 I>6I> 3I>2HT1 HT2
I> 1 I> 5I> 3HT1
I>4 I>6I>2 HT2
tBV5tBV3=tBV5+ ttBV1=tBV3+ t 0,511,5
tBV2 tBV4=tBV2+ t tBV6=tBV4+ t0,3
0,8 1,3
71
Nguyn l bo v qu dng c hng (67)2.3
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Kim tra s lm vic ca bo v:
I> 1 I> 5I>4 I>6I> 3I>2HT1 HT2
0,511,5 0,3 0,8 1,3
N1
I> 1 I> 5I>4 I>6I> 3I>2HT1 HT2
0,511,5 0,3 0,81,3
N2
72
Bo v qu dng c kha in p thp (51&27)2.4
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Tn gi khc 51V L do s dng
ng dy di Dng ngn mch cui
ng dy nh
Bo v khng
nhy
S dng thm khu phn bit gia s cv qu tibng in p(kha in p thp)Khi s c: in p gim thp hn
Khi qu ti (nng): in p vn nm trong ngng cho php
Mang ti nng
ca bo v lnKat* Kmm*Ilv maxKtv
Ik=
73
Bo v qu dng c kha in p thp (51&27)2.4
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Tn gi khc 51V
Il ma Il ma
C MCC MC
Khng c kha in p (51) C kha in p thp (51 & 27)
*Ilv max
Kat* Kmm
KtvIk=
U hay 51N)2.5
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S dng b lc dng in th t khng Tnh ton dng khi ng ch bnh thng:
V l thuyt: dng qua rle bng 0 Thc t: do cc BI c sai s dng in qua rle khc 0
rle khng tc ng: t dng khi ng ln hn dng in sinh ra do sai
Ia
Ib
Ic
Role
Ia Ib Ic 3I0+ + =
s ny
Gi tr ci t: Ikhi ng 51N=(0,10,3)Inh mc BI
Ch s c:dng in qua rle tng gp nhiu ln bo v tcng
Do gi tr khi ng t thp bo v c nhy cao
Thi gian lm vic:Phi hp vi cc bo v qu dng th t khng khc
75
Phn bit chc nng I> & I>> (51 & 50)
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Bo q dng
Khi ng khi: Ingn mch>Ikhi ng
Bo v qu dng c thi gian
(I> hay 51) Dng khi ng tnh theo dng
lm vic ln nht (Ilvmax)
Khi xy ra s c c th c bo
v ti ch v bo v pha trncng khi ng
m bo tnh chn lc: phihp phn cp thi gian (t)
C th dng lm bo v chnh
Bo v qu dng c t nhanh
(I>> hay 50) Dng khi ng tnh theo dng
ngn mch ngoi ln nht(In ngoi max)
Khi xy ra s c: ch bo v tiphn on s c khi ng
Khng cn phi hp thi gian(ct nhanh)
Khng bo v c ton b itng ch l bo v d phng
76
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Chng 03
dng in (I)
77
Nguyn l
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Thit b
So snh tng dng in i vo & i ra ca i tng c bov: tng dng in ny khc 0 bo v tc ng.
Ch bnh thng:Dng in chy qua rle nh hnh v
Dng chy qua rle: l dng chnh lch do sai s ca BI cc pha
Bnh thng
Irle= +
78
Nguyn l
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Ch s c ngoi vng:Dng in l dng s c c gi tr ln sai s BI ln hn
Phn b dng in tng t ch bnh thng
Dng in chy qua rle s ln chnh nh rle khng tc ng dng khi ng ln, gim nhy
Vn bo : ii h n bi v tr t cc BI
S c ngoi
Thit b
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80
Nguyn l
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Tng kt:Thit b
Irle= +
S c trong
Irle= +
S c ngoi
Irle= +
Bnh thng
Dng in chy qua rle l do sai s BI Dng in chy qua rle l tng
dng s c
81
Bo v so lch c hm
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Bo v rle so lch thng thngcc rle c th tc ngnhm do:Sai s ln ca cc BI khi ngn mch ngoi
Chuyn u p... Bo v so lch c hm:hot ng da theo t hp ca hai loi
S c ngoi vng: dng hm c gi tr ln dng so lch nhS c trong vng: dng hm nh Dng so lch ln.
82
Bo v so lch c hm
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T hp dng in cho bo v rle so lch c hmS dng cc bin dng trung gian (BITG)
T hp thm ra dng in hm (Ih)
83
Bo v so lch c hm
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T hp dng in cho bo v rleso lch c hm
Vit phng trnh cn bng sc t
BITG2 BITG1I1*w1I1*w1=Isl*w2
Chn w1=w2
Isl=I1 I2
I1*w3+I2*w4=Ih*w5
Chn w3= w4= w
w5=2*w
Ih=0,5*( I1 + I2)Tng qut: Ih=Kh*(I1+I2)
Kh: h hm, c h i chnh ha
i heo ng cn 5
84
Bo v so lch c hm
=( )sl
I I I
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Bng cch s dng BI trung gian, c th to ra t hp Xt s vn hnh Bo v so lch c hm
1 2
1 2
( )
( )
sl
h hI K I I = +
100% (I1+I2)
I1
100% (I1+I2)
I1
I2
Isl
Ih=Kh* (I1+I2)
I2
Isl
Ih=Kh* (I1+I2)
()> () bo khng
tc ng
()< () bo tc
ng
Ch bnh thng hoc s c ngoi Ch s c trong vng
85
Bo v so lch c hm
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La chn h s hmTng h s hm (Kh): rle
hm tt nhy tcng ca rle km i.
Hiu ng ngc li khi gimh s hm 100% (I1+I2)
I1
100% (I1+I2)
H s hm c th c iuchnh bng cch thay i svng dy cun w5.
Isl
Ih=Kh* (I1+I2)
I2
Isl
Ih=Kh* (I1+I2)
86
Bo v so lch c hm
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Rle c: h s hm l c nh Rle s: thay i
87
Bo v so lch c hm
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Rle minh ha
Rle so lch RQ4, sn xut bi hng AEG giai on19501960
Rle so lch s dng a quay (loi
1 pha)
88
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Chng 04
(bo v khong cch)
89
Nguyn l hot ng
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Bo v khong cch da trn cc gi tr dng in v in p tiim t rle xc nh tng tr s c
Nu tng tr s c ny nh hn gi tr tng tr ca ng dy ci t trong rle th rle s tc ng rle tng tr thp
Gi tr tng tr m rle xc nh c ch l mt gi tr gitng.
Do tng tr gm hai thnh phn R & X, nn thun tin s sdng mt phng tng tr biu din s lm vic ca bo vkhong cch
90
Nguyn l hot ng
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90/213
Xt s n gin:
Bnh thng:ZR(b)=ZD+Zph ti ZD
R
jX
100%ZD
ZD
Zpt
ZD+Zptim
lm viclc bnh
thng
91
Nguyn l hot ng
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91/213
Xt s n gin:
S c:v d ti 50% ng dy:ZR(c)=ZDs c=50%ZD< ZD
im s c di chuyn vo ng
tng tr ng dy
R
jX
50%ZD
ZD
Zpt
ZD+Zptim
lm viclc bnh
thng
imlm vickhi s
c
h l i l kh h
92
Nguyn l hot ng
8/13/2019 Bai Giang Mon Bao Ve Role - Dung Cho in An
92/213
c tnh lm ic ca rle khong cch im lm vic lc bnh thng v khi s c: khi s c im lm vic
lun ri vo ng tng tr ng dy c th ch cn ch to ctnh tc ng ca rle l mt ng thng trng vi ng tng trng dy
jX
c tnh tc
ng l mtng thng
R
50%ZD
ZD
Zpt
ZD+Zpt
imlm viclc bnhthng
imlm vic
khi sc
t h l i l kh h
93
Nguyn l hot ng
8/13/2019 Bai Giang Mon Bao Ve Role - Dung Cho in An
93/213
c tnh lm ic ca rle khong cch Do sai s, do s c c th xy ra qua cc tng tr trung gian nn gi
tr rle o c khi s c c th ri ra ln cn ng tng tr ngdy.
Nu ch ch to c tnh tc ng l mt ng thng th rle c ths khng lm vic trong cc trng hp ny. khc phc th cc nh
ch to thng c m rng c tnh tc ng v c hai pha ca
ng dy tr thnh vng tc ng.
R
jX
100%ZD
ZDZD+Zpt
imlm viclc bnhthng
im scri rangoi rle khng
tc ng
c tnh tcng l mtng
thng hp
im scrivo vng tc ng
c tnhtc ng
c mrng
R
jX
ZDZD+Zpt
imlm viclc bnhthng
C d t h th
94
Nguyn l hot ng
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Cc dng c tnh thng gp Thc t, c tnh lm vic c m rng theo nhiu dng khc nhau
p ng tt hn vi mi loi s c v ch vn hnh ca h thng
ZD
95
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Chng 06
Vi cc ng dy cao p (li truyn ti)
96
Cc loi bo v p dng
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Vi cc ng dy cao p (li truyn ti)Bo v chnh: bo v so lch dng in, bo v khong cch
Bo v d phng: bo v qu dng
Vi cc ng dy trung p (li phn phi)Thng s dng cc bo v qu dng (l do kinh t)
Cc bo v ng dy thng c trang b thm chc nng tng li (79)
ng dy c chiu di ln phi dng knh truyn gi tn
97
Bo v so lch dc ng dy
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ng dy c chiu di ln phi dng knh truyn gi tnhiu dng in gia cc pha.
ng dy 2
ngun cp
ng dy rnhnh
Knh n c hk ni heo mchng ng nh dhng ng
n
Hai loi bo v so lch dng in
98
c im
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Hai loi bo v so lch dng inT hp dng in 3 pha theo t l mt tn hiu so snh cc tn
hiu ny ca hai pha. nhy b thay i theo dng s c
La chn t s theo tng iu
kin c th
Ly mu dng in ca c 3 pha chuyn thnh tn hiu s chuyn n so snh cng vi u i din. Knh truyn ch yu l cp quang (PLC khng m bo bng thng)
S tun hon dng in
99
Nguyn l bo v so lch vi knh truyn tin
d b
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S tun hon dng in
Cun hm
Cun so lch
Rle so lch
ng dy cn bo v
Bnh thng: dng in chy tun hon qua knh truyn tinc im:
H mch knh truyn: tc ng nhm
Ngn mch knh truyn: kha, khng tc ng nhm
S cn bng in p
100
Nguyn l bo v so lch vi knh truyn tin
d b
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S cn bng in p
Cun hm
Rle so lch
ng dy cn bo v
Bnh thng: dng in khng chy qua knh truyn tinc im:
H mch knh truyn: kha, khng tc ng nhm
Ngn mch knh truyn: tc ng nhm
c tnh lm ic ca rle khong cch
101
Bo v khong cch (BVKC)
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jX
ZDZD+Zpt
imlm viclc bnh
c tnh lm ic ca rle khong cch c tnh lm vic ca BVKC: l mt min tc ng
jX
ZDZD+Zpt
imlm viclc bnh
R
thng
Gi tr chnh nh:Do sai s ca php o nn vng tc ng khng th t bao trm ton b
ng dy
Thng t bao trm 8085% chiu di ng dy
im scrivo vng tc ng
R
thng
L thuyt Ci t thc t
Cc dng c tnh thng gp
102
Bo v khong cch
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Cc dng c tnh thng gp Thc t, c tnh lm vic c m rng theo nhiu dng khc nhau
p ng tt hn vi mi loi s c v ch vn hnh ca h thng
ZD
Cc ng ci t ca bo khong cch
103
Bo v khong cch
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Cc ng ci t ca bo khong cch Thng c chnh nh vi 3 vng tc ng
Vng I: tc ng tc thi
Vng II & III: tc ng c tr theo nguyn tc phn cp thi gian, phi hpvi cc bo v lin k
Chi tit ci t cc ng ca bo
104
Bo v khong cch
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Chi tit ci t cc ng ca bo Vng ICc rle c: vng I thng t 80% tng tr ng dy
Cc rle s: th gi tr ny c th tng ti 85%.
hin tng bo v tc ng vt vng vi cc s c ngoi ln cncui ng dy.
Do vng I khng cn phi phi hp vi bt c bo v no nn thi
gian tc ng c th t xp x 0 giy.
Cc ng ca bo khong cchV II
105
Bo v khong cch
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Cc ng ca bo khong cch Vng IIm bo bo v 100% chiu di ng dy
t t nht 120% tng tr ng dy cn bo v.
Thng thng vng II c ci t bng 100% tng tr ng dycn bo v + 50% tng tr ca ng dy ngn nht lin k
Thi gian lm vic ca vng II c phi hp vi vng I vi bc phncp thi gian t nh trnh by trong phn bo v qu dng.
Cc ng ca bo khong cchV III
106
Bo v khong cch
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g g Vng IIIl vng bo v d phng chng li tt c cc s c trn ng dy
lin k
Do gi tr khi ng thng t ln hn 20% ca tng tr tnh tv tr t rle ti cui ng dy di nht lin k.
Thi gian tc ng ca vng III c phi hp vi thi gian tc ngvng II.
Minh ha
107
Bo v khong cch
ng dy 1 ng dy 2 ng dy 3
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107/213
Ngn
ng dy 1 g y g y
1015% 1015% 1015%Vng I Bo v 2 Vng I Bo v 3
A B C D
Vng I Bo v 1
Thanh gp B
Thanh gp A
Vng It = 0 giy Tng tr
ng dy
Vng IIt = t giy
Vng IIIt = 2t giy
Cc t nh hng n BVKC nh hng ca ti
108
Bo v khong cch
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g nh hng ca tiTi ca ng dy cng c th biu din di dng tng tr
Trn mt phng tng tr: vng ti c rng hay co hp ty theoh s cng sut ca ti
mang ti nng: vng ti c th
chng ln vo c tnh tc ng
Vic chng ln ti nh hng
n vng 3 ca BVKC
Vng 3
Vng ti
Cc t nh hng n BVKC nh hng ca ti Cch x l
109
Bo v khong cch
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109/213
nh hng ca ti Cch x lVng 3 m rng c gii hn
S dng cc c tnh a gic
B nh hng chng ln ti Khng b nh hng chng ln ti
Cc t nh hng n BVKC nh hng ca in tr h quang ti im s c
110
Bo v khong cch
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110/213
nh hng ca in tr h quang ti im s cS c trn ng dy thng km theo h quang
H quang c tnh cht in tr (Rhq)in tr h quang ny lm php o tng tr ng dy b sai lch
R
jX
ZD
Zpt
ZD+Zptim
lm viclc bnhthng
imlm vickhi s
c
Rhq=0
R
jX
ZD
Zpt
ZD+Zpt
im lm vic khi
scnm ngoivng tc ng
Rhq>0
Rhq
Cc t nh hng n BVKC nh hng ca in tr h quang ti im s c
111
Bo v khong cch
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111/213
nh hng ca in tr h quang ti im s cKhc phc: S dng c tnh t gic c min tc ng m rng v
pha trc R
jX jX
R
ZD
Zpt
ZD+Zpt
im lm vic khiscnm ngoi
vng tc ng
c tnh MHO
Rhq
R
ZD
Zpt
ZD+Zpt
im lm vic khiscnm trongvng tc ng
c tnh t gic
Rhq
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Phng thc phi hpM rng vng 1
113
Phi hp s lm vic ca cc BVKC
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g g
Truyn tn hiu cho php ct Truyn tn hiu ct trc tip
Truyn tn hiu cho php ct vi vng 1 khng m rng
Truyn tn hiu cho php ct vi vng tc ng m rng
Truyn tn hiu kha
Truyn tn hiu ct trc tip
V d phng thc truyn tn hiu cho php khng m rngvng 1:
114
Phi hp s lm vic ca cc BVKC
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o
Khi vng 1 khi ngo Ct my ct ti ch
o Truyn tn hiu cho php ti
Gi tnhiu i
Ct my
u i din
o Ti u i din, bo v s tcng tc thi khi:
o khi ng
o Nhn c tn hiu cho php
ca u i dino S ch yu cu mt knh truyn
duy nht do ch c mt Vng 1 cngi tn hiu qua knh truyn
Tn hiu n tu i din
c t
V d da theo khuyn co ca hng SIEMENSLi hnh tia (li phn phi)
115
Cc phng thc bo v c bn
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( p p )
o
Bo v chnh l bo v qu dng(51&51N)
o Km theo chc nng t ng li (79)
o Bo v qu dng th t nghch (46):tng nhy vi cc s c khngi xng.
V d da theo khuyn co ca hng SIEMENSng dy song song
116
Cc phng thc bo v c bn
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o
Bo v chnh l bo v qu dng(51&51N)
o Chc nng bo v chng qu ti (49)
o Bo v qu dng th t nghch (46):tng nhy vi cc s c khngi xng.
o S dng bo v qu dng loi chng (67&67N) do dng cng sutkhi ngn mch c th chy theo c
hai chiu
V d da theo khuyn co ca hng SIEMENSng dy truyn ti
117
Cc phng thc bo v c bn
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o Bo v chnh l bo v khong cch (21&21N) c lin ng
o Bo v d phng l cc bo v qu dng
V dng dy truyn ti
118
Cc phng thc bo v c bn
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C ng truyn tin
bng thng rng
o Gm hai b bo v chnh
o Bo v so lch dc (87L)o Bo v khong cch (21&21N)
c lin ng
o C th s dng bo v qu
dng lm d phng
Ch 07
119
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Chng 07
Phng in s xuyn S c phapha, phat i vi cun dy cao v h p
120
Cc loi s c
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S c gia cc vng dy trn cng cun dy.
S xm m ca hi nc vo du cch in cng l mt yu tgy nn s c.
Vi cc my bi n p ln n i ti ng dy truy n ti th khi c
st nh vo ng dy, sng vi dc u sng ln s ivo trong my bin p v c th gy thng cch in cuicun dy my bin p.
S c chm chp gia cc vng dy: dng in trong cc vngdy b s c ln nhng dng in ti hai u ca my bin ph i kh k ( h d )
121
Cc loi s c
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thay i khng ng k (theo t s vng dy).Khi mi xy ra s c th ch mt s t vng dy b nh hng, nhng
nu khng loi tr nhanh th c th gy s c lan trn.
S c li t: cc trng hp s c vi dng in ln c th gy
x lch li t, v lm tng ln dng in xoy, gy pht nhit
v c th dn ti s c ln hn. S c thng du chnh my bin p: c th xy ra hin tng r
r du lm mc du b h thp gy ngu him cho cch in v
lm mt my bin p. Hng b chuyn i u phn p (OLTC)
Dng t ha khi ng khng ti Khi ng my bin p khng ti c th xut hin dng t ha,
122
Cc loi s c
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qu t thng li thp hoc khi bo ha BI...u lm tng dngso lch v bo v c th s tc ng.
T thng trong li t c th tng ti 280% ln ty theo:
T d trong li th th t thng
Thit k ca my bin p
S lng my bin p lm vic song song
Vi gi tr tng vt li t chc chn s b bo ha dngin b mo dng sng trm trng
Dng in ny gi l dng in t ha xung kch khi ng mybin p.
Dng t ha khi ng khng ti Hnh dng dng in t ha xung kch khi ng my bin p
123
Cc loi s c
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: dng in t ha xung kch cha ch yul thnh phn sng hi bc 2 v bc 3 s dng hmbo vkhi ng my bin p khng ti.
124
Cc loi bo v cho my bin p
Loi sc Loi bo v
Bo vso lch
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Scpha-pha v pha-t cun dy
Bo vqu dng
Bo vchng chm t hn ch
Bo v so lch
c g a c c v ng yRle kh (Buchholz)
Scli tBo v so lch
Rle kh (Buchholz)
Scthng du my bin p
Bo v so lch
Rle kh (Buchholz)
Bo vchng chm t thng my bin p
Qu tthng Bo vchng qu tthng
Qu nhit Bo vchng qu ti
Bo so lch c hm I (87) Dng lm bo v chnh cho my bin p
125
Cc loi bo v my bin p
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Phm vi bo v c gii hn bi v tr t BI
Cc nh hng cn lu :T u dy my bin p
Loi b thnh phn dng in TTK
Hiu chnh t s bin dngLa chn h s hm trong cc tnh trng lm vic
Hm bo v khi ng my bin p khng ti
Hin tng qu t thng
nh hng ca t d m bin p MBA t u dy hai pha khc nhau dng in cc pha b lch
h
126
Cc loi bo v my bin p
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gc nhauT u dy Y0/11 th dng s cp v th cp lch nhau 11x300=3300.
Nguyn l bo v so lch yu cu dng in hai pha cn so snh
qua bo v s tc ng nhm phi hiu chnh gc pha.
Rle c & Rle nh:hiu chnh gc pha bng BI trung gian.
Rle :hiu chnh gc pha c thc hin bng phn mm:BI c th u hnh Y cho mi cun dy
Khai bo vo rle cc t du dy ca my bin p v my bin dng(nu cn thit).
Loi b thnh phn dng in TTK
S c chm t ngoi vng
127
Cc loi bo v my bin p
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Nu khng loi b: tc ng nhm
Dng qua rle ln hn 0
Rle c th tc ng nhm
BI trung gian khngc cun tam gic
Loi b thnh phn dng in TTK
S dng BI trung gian c cun tam gic: loi tr thnh phn I0h b
128
Cc loi bo v my bin p
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chy vo bo v
BI trung gian ccun tam gic
Dng qua rle bng 0
Hi chnh t s bin dng
Dng cn bng c th sinh ra khi:BI h bi kh bi
129
Cc loi bo v my bin p
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BI cc pha c t s bin khc t s bin p
Hoc khi dng in th cp ca cc BI khng ging nhau
Hi chnh t s bin dng
Dng cn bng c th sinh ra khi:BI h t bi kh t bi
130
Cc loi bo v my bin p
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BI cc pha c t s bin khc t s bin p
Hoc khi dng in th cp ca cc BI khng ging nhau
Rle so
lch
Hi chnh t s bin dng
Chn BI trung gianBI t i th d
131
Cc loi bo v my bin p
1 2w 3 3,813 3 2, 202 3 06i
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BI u tam gic th dngpha & dng dy khc
nhau
1 2
2 1
, ,3,06
w 0, 719 0, 719i= = = =
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Bo chng chm t hn ch 87N (Restricted Earth Falt REF)
Bo v qu dng TTK (50N & 51N) c th khng nhy bo v cho cun dy my bin p S c ti im gn trung tnh
133
Cc loi bo v my bin p
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bo v cho cun dy my bin p S c ti im gn trung tnhcun dy u hnh sao, ni t qua tng tr: dng s c rt b.
Bo v chng chm t hn ch (87N): so lch TTK
:(phm vi bo v b hn ch).
Bo chng chm t hn ch 87N (Restricted Earth Falt REF)
Bo v c th l dng so lch tng tr cao hoc so lch c hm
C nhy cao v dng khi ng c th t thp
134
Cc loi bo v my bin p
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C nhy cao v dng khi ng c th t thp
Dng in a vo rle l ton b dng s c ch khng ch lmt thnh phn c bin i qua t s bin chy trn pha
s vng dy b s c bn cun h p).
Dng t ha ng kch khi ng m bin p khng ti
Bo v s tc ng nhm khi ng my bin p khng ti
135
Cc loi bo v my bin p
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Hi bc 2
Vi rle c: cho bo v lm vic vi thi gian tr khng tin cydo lc ng my bin p c th gp s c.
Dng t ha ng kch khi ng m bin p khng ti
Rle s: hm theo sng hi bc 2
136
Cc loi bo v my bin p
Hi bc 2
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L do dng sng hi bc 2 (bc chn):
Dng t ha xung kch (qu ) lun cha thnh phn sng hi bc 2
Dng s c khng cha thnh phn sng hi bc 2 v cc sng hi bcchn
Dng t ha ch xc lp c th b mo sng do li t bo ha, tuynhin dng ny cng khng cha cc thnh phn sng hi bc chn.
Sng hi bc 2: c trng ring bit ca dng t ha xung kch s dng thnh phn sng hi bc 2 ny t ng hm bov so lch khi ng khng ti my bin p.
Bo chng q t thng li t
Qu t thng li t xy ra khi:in p h thng b tng cao
137
Cc loi bo v my bin p
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in p h thng b tng cao
Tn s h thng b gim thp
Qu t thng qu khng gy
tr
Bo v vi c tnh thi gian phthuc khi ng khi t s V/fvt ngng ci t.
Bo chng chm t thng m bin p
Thng my bin p c t cch in
S dng chc nng bo v rle chng hin tng chm t v
138
Cc loi bo v my bin p
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S dng chc nng bo v rle chng hin tng chm t vthng
Thc hin: ni mt BI vo dy ni gia v thng v t.
Rle kh (Bchhol)
V tr: trng ng ng ni t thng du chnh my bin p lnthng du ph.
139
Cc loi bo v my bin p
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g p
Rle Buchhloz: hai t hp phao nm l lng.
Rle kh (Bchhol)
Qu ti:kh ga t thng du tch t ln trn theo ng dn duy mc du trn np rle Buchholz xung phao s 1 (bn
140
Cc loi bo v my bin p
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trn) chm xung, ng tip im khi ng cnh bo qa ti
thc hin qu trnh san ti cho my bin p.
nhanh, kh tch t mnh v i ln trn x y vo rle cp hai
khi ng i ct cc pha ni vi ngun ca my bin p.
Rle kh (Bchhol)
141
Cc loi bo v my bin p
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MBA cng st nh
142
S phng thc bo v my bin p
Bo v chnh: bo v qu dng in phacao p
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Bo v d phng: bo v chng chm t
hn ch (87N) v bo v qu dng th tkhng t ti trung tnh
MBA cng st ln
143
S phng thc bo v my bin p
Bo v chnh: bo v so lch dng in(87T)
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Bo v d phng:
Bo v qu dng pha cao p & h p
Bo v chng chm t hn ch 87N
Ch dng hai b rle: p dng cho cc mybin p khng quan trng
MBA nhn in t hai l
144
S phng thc bo v my bin p
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Bo v chnh: bo v so lch dng in(87T)
Bo v d phng:
Bo v qu dng pha cao p & h p
Bo v chng chm t hn ch 87N Dng 3 b rle ring bit: tng tin
cy
MBA lm ic song song
145
S phng thc bo v my bin p
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Tng t cc phng thc bo v trc
Do my bin p vn hnh song song: cc bo v qu dng l loi c nh
hng (67 & 67N)
MBA lm ic song song c thanh gp phn on
146
S phng thc bo v my bin p
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MBA ba cn d
147
S phng thc bo v my bin p
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MBA t ng
148
S phng thc bo v my bin p
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MBA t ng ln qan trng
149
S phng thc bo v my bin p
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Chng 08
150
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Thanh gp: kt ni nhiu phn tH thng bo v c vai tr quan trng
Thi gian tc ng cc ngn m bo n nh ca h thng
Rle hin i: tc ng ch trong vng 1 chu k
151
Bo v cc h thng thanh gp
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Rle hin i: tc ng ch trong vng 1 chu k
H thng bo v: phi c tin cy & an ton cao
nhiu khu c lp
Nguyn l "2 trong 3" m bo an ton
c im:S lng ngn l ln s lng tn hiu, dy dn ln s dng cu
trc phn tn
C ch kim tra chng tc ng nhm: check zone
152
Bo v cc h thng thanh gp
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C ch kim tra chng tc ng nhm: check zone
D gp hin tng bo ha my bin dng
Bo ha
S c trong vng bo v
S c ngoi vng bo v
Gii thiu cu trc bo v thanh gp kiu tp trungS dng nguyn l bo v so lch
Tn hiu dng in t tt c cc ngn l c a v rle trung tm:s lng dy dn nhiu.
153
Bo v cc h thng thanh gp
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s g dy d u
Gii thiu cu trc bo v thanh gp kiu phn tnKhi iu khin trung tm (Central Unit)
Khi iu khin cp ngn l (Bay Unit)
S dng nguyn l bo v so lch
154
Bo v cc h thng thanh gp
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S dng nguyn l bo v so lch
H thng chuyn mch dngMt ngn l c th ni ti thanh gp 1 hoc 2 cn chuyn mch
dng ti bo v tng ng
155
Bo v cc h thng thanh gp
BUS 1
+
BUS 1
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BUS 2
ISOLATO
R1
ISOLATOR 1 OPEN
7B 7A
CB 1
ISO 1 ISO 2
ISO 3
BYPASS
-
F1aF1c
Contact Input F1a OnContact Input F1c On
F1b
-
+
F1aF1c
Contact Input F1a OnContact Input F1c On
F1b
ISOLATOR1
ISOLATOR 1 CLOSED
7B 7A
BUS 1
Cc phng thc bo v thanh gp
S kha lin ng Thng dng li phn phi (mt ng cp ti TG)
Yu cu mt khong phn cp thi gian ngn
156
Bo v cc h thng thanh gp
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Yu cu mt khong phn cp thi gian ngn
Tn hiu kha c th ni trc tip gia cc rle (dy ng)
50
50 50 50 50 50
BLOCK
Cc phng thc bo v thanh gp
Bo v so lch S dng nguyn l hm
Dng hm: tng dng, dng ln nht, mt phn tng dng..
157
Bo v cc h thng thanh gp
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g g g, g , p g g
5187
Chc nng kim tra vng bo v (check zone)Khi mch dng t mt ngn l hng dng so lch bng chnh dng
ti ngn l rle tc ng nhm.
Gii php:
158
Bo v cc h thng thanh gp
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Dng khi ng ln hn Ii ma: gim nhy S dng chc nng check zone: ly tng tt cc cc dng in vo h thng
an g p
Ch tc ng khi: bo v vng & bo v check zone cng tc ng
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Phng thc bo v khuyn co cho cc MF
160
Bo v my pht in
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(O: ty chn, X: nn dng, Y: thy in tch nng)
Nguyn tc d phng: nguyn tc n1 hng hc 1 phn t
khng gy gin on h thng D phng mt phn:Hai h thng rle bo v
161
Bo v my pht in
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g
C th s dng chung cc bin p o lng
My ct c th ch c mt cun ct
Nguyn tc d phng: nguyn tc n1 hng hc 1 phn t
khng gy gin on h thng D phng y :Bin p o lng ring bit
162
Bo v my pht in
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Hai h thng bo v ton phn
Tn hiu ct i theo cc h thng khc nhau
My pht in ly in p kch t ca u cc:Ngn mch gn in p u cc st gim dng in ngn mch
b gim i bo v khng nhy
Gii php:
163
Chc nng bo v qu dng (50, 51)
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t dng khi ng thp)
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g g
S c ngn mchkhng i xng
Nguyn tc thc hin
166
Bo v so lch (87G)
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Bo v so lch
dc
ngang
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Generator Capability Curve: cng sut phn khng b gii hn
168
Bo v mt kch t (Loss of Field 40)
Gii hn dng in trong cunkch t (Field Crrent HeatingLimit): gii hn bi pht nngtrong cun dy roto
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Dng in trong cun dy stato(Amatre Crrent HeatingLimit): khng c vt qumc pht nng cho php
Generator Capability Curve: cng sut phn khng b gii hn
169
Bo v mt kch t (Loss of Field 40)
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Gii hn pht nng li thp ti cnh ca
stato(End Region Heating Limit): khnng nhn cng sut phn khng ca mypht ch thiu kch t
Generator Capability Curve: cng sut phn khng b gii hn
170
Bo v mt kch t (Loss of Field 40)
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+
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Da theo c tnh c hai phng php bo v chng mt
kch t (thiu kch t)1. S dng tng tr: rle tng tr nhn vo MF
Khi pht Q: in khng o c ln hn 0 (X>0)
172
Bo v mt kch t (Loss of Field 40)
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Khi nhn Q (thiu kch t): in khng o c nh hn 0 (X
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Bin i trc quan
2. S dng gi tr tng dn (ring Siemens p dng) c tnh lm ic
174
Bo v mt kch t (Loss of Field 40)
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o Khi im lm vic ca my pht vt qu c tnh 1 & 2: bo v lm vic tr
sau 10 giy
o m bo b kch t nng in p tr li mc cn thito Nu b iu chnh hng rt ngn thi gian ct Thc hin bng cch
gim st in p kch t (chc nng )o Khi vt c tnh 3: ct tc thi
Reverse Power Protection
Nng lng s cp b mtMF hot ng ch ng cNu cn h thng kch t: ng c ng b
Ngc li: nh ng c khng ng b.
h h h b
175
Bo v chng lung cng sut ngc (32R)
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Ch ng c gy nguy him cho tuabin:
n ng qu m c c n ua n o ng u c uy n clm mt
Nguy him cho hp s ca cc tuabin kh do cc hp s ny khngc thit k ch quay ngc.
Nguyn nhn:
Li vn hnhTrc trc my ct u cc khng ct khi ngng t my
Hng hc c kh
Bo vBo v c kh v theo tn hiu in
ln lung cng sut ngc ty thuc:Ma st, tn hao do tuabin hot ng nh my nn
T h i t ht
176
Bo v chng lung cng sut ngc (32R)
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Tn hao in trong my pht
n ng c ng su ng c r p p o p c n x cCh s dng thnh phn TTT ca dng v p
Sai s gc ca BU & BI gy sai s php o phi a vo trong tnhton
Bo v thng l loi c tr
Trnh cc bin ng ngn hnTrong khi ha ng b hoc dao ng inc th c lung cng sut
ngc
L doMy pht in:
Trung tnh cch in
Ni t qua tng tr
Hn ch dng chm t
177
Bo v chng chm t 90% (59N, 64G, 67G)
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Hn ch dng chm tChm t cun stato:
Cch in b ha than ti li thp
H quang ti li thp.
Thc nghim cho thy: Chm t c pht sinh h quang dng in 5A c th ph hy cch in l
thp stato s c tip theo
Khng c mt tiu chun c th v gi tr dng in chm tThng c gii hn trong khong 515A
Nguyn l bo v trong rle 7UM62
Da theo in p im trung tnh cun dy statoKhi chm t vecto in p 3 pha mt cn bng im trung tnh
b dch chuyn in th tng ln khc 0.
178
Bo v chng chm t 90% (59N, 64G, 67G)
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Da theo dng in chm t
Vi cc MF c ph ti a phng (cp) dng chm t c th xc nh s c
tng nhy: s dng BI th t khng (core balance CT)
S dng b nh hng cng sut: s c trong & ngoi my pht
Bnh thng
Cch in Ni qua tng tr
Chm t
Phng php o in p im trung tnhMy bin in p ni ti trung tnh cun dyMy bin p trung tnh
179
Bo v chng chm t 90% (59N, 64G, 67G)
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Cun tam gic h ca MBA
to trung tnh gi
Phng php o dng chm t
180
Bo v chng chm t 90% (59N, 64G, 67G)
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Logic bo v chng chm tin p im trung tnh vt qu ngng ci tDng in chm t vt qu ngng
Hng vo trong my pht
Ukhi > Ukh i h h t 510% U h
181
Bo v chng chm t 90% (59N, 64G, 67G)
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Ukhi ng> Ukhng i ng ong n hnh t510% Uha
Bo v c 9095%cun dy stato tnh t u cc
S dng sng hi bc 3 (27/59TN 3rd Harm.)in p do MF sinh ra c c thnh phn bc 3Thnh phn bc 3 c tnh cht nh thnh phn TTK
182
Bo v chng chm t 100%
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Phn b in p bc 3 lc bnh thng
S dng sng hi bc 3 (27/59TN 3rd Harm.)Chm t gn trung tnh: gi tr in p bc 3 ti u cc ln nht
s dng rle in p cao (59)
183
Bo v chng chm t 100%
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Chm t ti trung tnh
S dng sng hi bc 3 (27/59TN 3rd Harm.)Bnh thng c in p bc 3 ti trung tnhS c: in p ny v xp x 0 dng rle in p thp (27)
184
Bo v chng chm t 100%
Bnh thng
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g
Chm t ti trung tnh
S dng sng hi bc 3 (27/59TN 3rd Harm.)Dng rle in p cao (59)Hoc dng rle in p thp (27)
185
Bo v chng chm t 100%
S dng cng i s bo 90%
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S dng cng i s bo 90%
ln in p bc 3 ph thuc ti rle 7UM62x t ng thay igi tr chnh nh theo dng ti.
S dng ngun ph tn s thp
L doCc phng php trn ph thuc c tnh MF
S lng ph ti, s lng ng dy, cp
S theo in p bc 3 c im cht trong vng bo v
186
Bo v chng chm t 100%
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S theo in p bc 3 c im cht trong vng bo v
p pPht ngun ph c lp tn s thp vo cun dy my pht
Gim st dng in do in p tn s thp ny gy ra
S dng ngun ph tn s thp
Nguyn lPht in p tn s thp vo trung
tnh
in p sinh ra dng in
187
Bo v chng chm t 100%
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p g
ngun pht & in dung cun statovi t (Xc)
Khi chm t: in dng b ni ttdng in tng ln
Tn s thp : Dung khng (Xc) c gi tr ln dng
in nh dng khi ng thp tng nhy
Trnh nhiu do in p ca MF gy ra,
d lc.
Dng s c ny kh xy ra
Vi MF cun dy c nhiu vng (MF thy in): nn t Kh pht hin bng cc bo v thng thng
Phng php bo v vi MF c cun dy phn chia
188
Bo v chng chm chp gia cc vng dy
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Phng php bo v vi MF c cun dy phn chia
Chm t mt im: khng gy nguy him cnh bo
L tin cho chm t im th hai Chm t im th hai:Mt s vng dy b ni tt
T trng b lch
189
Bo v chng chm t cun dy roto (64R)
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g
y rung ng mn u c p c n u rung v qungng cho php
Phng thc bo v n gin
190
Bo v chng chm t cun dy roto (64R)
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S dng hai n
Bnh thng: hai n sngbng nhau
Chm t mt nhnh: n tihn
Khng pht hin chm t titrung im
S dng ng h o in
Chm t mt nhnh: kimng h lch v pha tng
ng Khng pht hin chm t ti
trung im
Phng php bm ngn ph
Phng thc bm ngun ph xoay chiu
in p xoay chiu bm vo mch rotoqua rle qu dng (64F)
T C:Hn ch dng khi c s c chm t
191
Bo v chng chm t cun dy roto (64R)
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Dng in qua rle 64F:IC: dng in dung (in dung roto)IG: dng r qua cch in ca roto (rt
nh)
Dng khi ng: (+)
Khi s c chm t in dung b ni tt dng in quarle tng ln rle khi ng.
Phng thc bm ngun ph xoay chiu
Nhc im:Hot ng ph thuc ch ni t roto
Nu ni t qua bi trc quay
Mng du dn in kmR l kh h
192Bo v chng chm t cun dy roto (64R)
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Rle khn nh
Nu tng in p bm vo chc thngmng du dn in tt
Nhc: n mn in ha ti bi trc quay
Gii php khc: chi than ni t
Phng thc bm ngun ph mt chiu
u im:Trnh c dng in dung qua t
193Bo v chng chm t cun dy roto (64R)
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Hng dn bi tp di
194
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bi
195Ni dung cn lm
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Chn s liu: La chn = S d ca php chia {S th t/n}
196Ni dung cn lm
n=3n=4
n=6
n=3 n=3 n=4
n=4
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Nu php chia khng c s d: ly s cui cng ca dy s liu
n=5 n=3
n=6 n=4
V d: s th t trong danh sch lp l 32
197Ni dung cn lm
n=3 n=4
32/3 cn d 2
32/4 cn d 0
S d bng 2: chn s th 2 ca dy s liu
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S d bng 0: chn s cui cng ca dy s liu
Tham kho ti liu km theo
Kt qu tnh ngn mch
198Chn bin dng in (BI) & Tnh ton ngn mch
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Inmax N9
Inmin N9
I0max N9
I0min N9
Inmax N5
Inmin N5
I0max N5
I0min N5
199Tnh ton chnh nh cc bo v
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Chnh nh bo v qu dng ct nhanh I>> (50)Thi gian:
t50/BV1= 0 (giy)
t50/BV2= 0 (giy)
Dng khi ng: Ik50/BV1=Kat*Ingn mch ngoi max= Kat*Inmax N5 Ik50/BV2=Kat*Ingn mch ngoi max= Kat*Inmax N9
Vi Kat=1,11,2 (ty chn)
Xc nh vng bo v ct nhanh
Da theo dng ngn mch v dng khi ng tnh xc nhvng c bo v ct nhanh Ik50/BV1= Kat*Inmax N5 Ik50/BV2=Kat*Inmax N9
200Tnh ton chnh nh cc bo v
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L(km)
Inmax N9
Inmax N5
Ik50/BV1Ik50/BV2
Lct nhanh min
Lct nhanh max
Lct nhanh min=0
Lct nhanh max
201Tnh ton chnh nh cc bo v
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Chnh nh bo v qu dng ct nhanh th t khng I0>> (50N)Thi gian:
t50N/BV1= 0 (giy)
t50N/BV2= 0 (giy)
Dng khi ng: Ik50N/BV1=Kat*3I0ngn mch ngoi max= Kat*3I0max N5 Ik50N/BV2=Kat*3I0 ngn mch ngoi max= Kat*3I0max N9
Vi Kat=1,11,2 (ty chn)
Xc nh vng bo v ct nhanh th t khng
Tng t vi bo v ct nhanh
202Tnh ton chnh nh cc bo v
Ik50N/BV1I
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L(km)
3I0max N9
3I0max N5
Ik50N/BV2
Lct nhanh min
Lct nhanh max
Lct nhanh min=0
Lct nhanh max
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205Tnh ton chnh nh cc bo v
N9
N5
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Bo v qu dng c thi gian I> (51)Thi gian: s dng c tnh ph thuc & phi hp vi cc bo v tht khng lin k t51BV2/ngn mch ti N9= tpt2+ t (giy)
Chn bc phn cp thi gian t=0,30,5 giy (ty ) T tnh ra c gi tr t51BV2/ngn mch ti N9
Bo v qu dng c thi gian I> (51)
206Tnh ton chnh nh cc bo v
N9
N5
i t t
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c gi tr t51BV2/ngn mch ti N9 bit: bo v 2 lm vic vi c tnh nh sau:
Trong cng thc ny: t: thi gian lm vic ca bo v khi s c ti im no
(khi s c ti N9 th thi gian l t51BV2/ngn mch ti N9 bit)
Ik: dng khi ng ca bo v 2 ( bit l Ik51/BV2=K*I lm vic max qua D2) I: dng ngn mch ti im ang tnh (N9): bit l Inmax N9 Cn li Tpcha bit (Tp: bi s thi gian ca bo v): cn tm
Bo v qu dng c thi gian I> (51)
207Tnh ton chnh nh cc bo v
N9
N5
Cn li T cha bit (T : bi s thi gian ca bo v): cn tm
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Cn li Tpcha bit (Tp: bi s thi gian ca bo v): cn tm
Thay tt c cc gi tr bit vo cng thc
tm ra gi tr Tp ca bo v 2: Tp/BV2
?
c tnh lm vic ca BV 2
Tm thi gian lm vic ca bo v 2 (BV2) vi cc im s c
208Tnh ton chnh nh cc bo v
N9N5
cn li
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cn li Bit dng in ti N5N9 ch max
Ln lt thay vo cng thc ca BV2 bit tnh ra thi gian tng ng V d:
Bit dng inInmax N7
Thay vo y
bit
bit
Tnh ra ctBV2/N7
Ti mi im t N5N9: tnh c thi gian lm vic tng ng
209Tnh ton chnh nh cc bo v
N9N5
ca BV2 (vi dng in I I )
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ca BV2 (vi dng in InmaxN5InmaxN9)
L(km)D1 D2 tBV2/N9(Inmax)
N8N7N6N5 N9
Tnh ton tip vi BV1:
210Tnh ton chnh nh cc bo v
N5
Ti N5: nu BV2 khng tc ng th BV1 phi tc ng phi chnh
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Ti N5: nu BV2 khng tc ng th BV1 phi tc ng phi chnhnh thi gian ca BV1:
t51BV1/ngn mch ti N5= t51BV2/ngn mch ti N5+ t (giy)
bit 0,3 giyTnh c ragi tr ny
Vi gi tr thi gian lm vic ca BV1 ti N5 bit
211Tnh ton chnh nh cc bo v
N5
Phng trnh lm vic ca BV1
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Phng trnh lm vic ca BV1
?
c tnh lm vic ca BV1
t51BV1/ngn mch ti N5 Ik51/BV1
bit InmaxN5 bit
Tnh ra c
Tng t vi BV2: ti mi im t N1N5: tnh c thi gian
212Tnh ton chnh nh cc bo v
N9N5
lm vic tng ng ca BV1 (vi dng in InmaxN1InmaxN5)
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lm vic tng ng ca BV1 (vi dng in InmaxN1 InmaxN5)
L(km)D1 D2 tBV2/N9(Inmax)
N8N7N6N5 N9N3N2N1 N4
tBV1/Inmax
t
C y phng trnh lm vic ca BV1 & BV2
Bit ln dng ngn mch c ch min: tnh ton tipthi gian v ra ng c tnh thi gian ng vi Inmin
213Tnh ton chnh nh cc bo v
tBV1/Inmin tBV2/Inmin
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L(km)D1 D2 tBV2/Inmax
tBV1/Inmax
t
KT THC