bai giang mon bao ve role - dung cho in an

8/13/2019 Bai Giang Mon Bao Ve Role - Dung Cho in An

1/213

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

[email protected]


2/213

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


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


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)



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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



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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



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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

+



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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



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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



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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



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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



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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



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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



71/213

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


73/213

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


74/213

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)


75/213

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


76/213

Chng 03

dng in (I)

77

Nguyn l


77/213

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


78/213

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


79/213

80

Nguyn l


80/213

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


81/213

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


82/213

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


83/213

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


84/213

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


85/213

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


86/213

Rle c: h s hm l c nh Rle s: thay i

87

Bo v so lch c hm


87/213

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


88/213

Chng 04

(bo v khong cch)

89

Nguyn l hot ng


89/213

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


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


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


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


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


94/213

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


95/213

Chng 06

Vi cc ng dy cao p (li truyn ti)

96

Cc loi bo v p dng


96/213

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


97/213

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


98/213

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


99/213

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


100/213

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)


101/213

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


102/213

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


103/213

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


104/213

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


105/213

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


106/213

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


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


108/213

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


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


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


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


112/213

Phng thc phi hpM rng vng 1

113

Phi hp s lm vic ca cc BVKC


113/213

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


114/213

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


115/213

( 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



116/213

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



117/213

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



118/213

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


119/213

Chng 07

Phng in s xuyn S c phapha, phat i vi cun dy cao v h p

120

Cc loi s c


120/213

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


121/213

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


122/213

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


123/213

: 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


124/213

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


125/213

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



126/213

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).


S c chm t ngoi vng

127



127/213

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


S dng BI trung gian c cun tam gic: loi tr thnh phn I0h b

128



128/213

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



129/213

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



130/213

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


1 2w 3 3,813 3 2, 202 3 06i


131/213

BI u tam gic th dngpha & dng dy khc

nhau

1 2

2 1

, ,3,06

w 0, 719 0, 719i= = = =


132/213

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



133/213

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



134/213

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



135/213

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


Hi bc 2


136/213

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



137/213

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



138/213

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



139/213

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



140/213

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



141/213

MBA cng st nh

142

S phng thc bo v my bin p

Bo v chnh: bo v qu dng in phacao p


142/213

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


Bo v chnh: bo v so lch dng in(87T)


143/213

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



144/213

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



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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



146/213

MBA ba cn d

147



147/213

MBA t ng

148



148/213

MBA t ng ln qan trng

149



149/213

Chng 08

150


150/213

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


151/213

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



152/213

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



153/213

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



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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


BUS 1

+

BUS 1


155/213

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



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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



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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



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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


159/213

Phng thc bo v khuyn co cho cc MF

160

Bo v my pht in


160/213

(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


161/213

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


162/213

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)


163/213

t dng khi ng thp)


165/213

g g

S c ngn mchkhng i xng

Nguyn tc thc hin

166

Bo v so lch (87G)


166/213

Bo v so lch

dc

ngang


167/213

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


168/213

Dng in trong cun dy stato(Amatre Crrent HeatingLimit): khng c vt qumc pht nng cho php


169



169/213

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


170



170/213

+


171/213

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)

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Khi nhn Q (thiu kch t): in khng o c nh hn 0 (X


173/213

Bin i trc quan

2. S dng gi tr tng dn (ring Siemens p dng) c tnh lm ic

174



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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)


175/213

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)


176/213

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)


177/213

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.

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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



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Cun tam gic h ca MBA

to trung tnh gi

Phng php o dng chm t

180



180/213

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



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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%


182/213

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



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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


Bnh thng


184/213

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


S dng cng i s bo 90%


185/213

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



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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



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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


188/213

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)


189/213

g

y rung ng mn u c p c n u rung v qungng cho php

Phng thc bo v n gin

190



190/213

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



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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)


192/213

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)


193/213

Hng dn bi tp di

194


194/213

bi

195Ni dung cn lm


195/213

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


196/213

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


197/213

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


198/213

Inmax N9

Inmin N9

I0max N9

I0min N9

Inmax N5

Inmin N5

I0max N5

I0min N5

199Tnh ton chnh nh cc bo v


199/213

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



200/213

L(km)

Inmax N9

Inmax N5

Ik50/BV1Ik50/BV2

Lct nhanh min

Lct nhanh max

Lct nhanh min=0

Lct nhanh max



201/213

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


Ik50N/BV1I


202/213

L(km)

3I0max N9

3I0max N5

Ik50N/BV2

Lct nhanh min

Lct nhanh max

Lct nhanh min=0

Lct nhanh max


203/213


204/213


N9

N5


205/213

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)


N9

N5

i t t


206/213

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)


N9

N5

Cn li T cha bit (T : bi s thi gian ca bo v): cn tm


207/213

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


N9N5

cn li


208/213

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


N9N5

ca BV2 (vi dng in I I )


209/213

ca BV2 (vi dng in InmaxN5InmaxN9)

L(km)D1 D2 tBV2/N9(Inmax)

N8N7N6N5 N9

Tnh ton tip vi BV1:


N5

Ti N5: nu BV2 khng tc ng th BV1 phi tc ng phi chnh


210/213

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


N5

Phng trnh lm vic ca BV1


211/213

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


N9N5

lm vic tng ng ca BV1 (vi dng in InmaxN1InmaxN5)


212/213

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


tBV1/Inmin tBV2/Inmin


213/213

L(km)D1 D2 tBV2/Inmax

tBV1/Inmax

t

KT THC

bai giang mon bao ve role - dung cho in an

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