Bai Giang Bao Ve Role Trong Nha May Dien - Nha May Thuy Dien Son La

8/12/2019 Bai Giang Bao Ve Role Trong Nha May Dien - Nha May Thuy Dien Son La

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11/2/2013 Ging vin: Nguyn Xun Tng

B mn H thng ini hc Bch khoa H Ni

Email: [email protected]

H THNG HA & CP NHT KIN THCBO V RLE TRONG NH MY IN

(NH MY THY IN SNLA)


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Chng 01

3

My bin p dng in(dng cho h thng rle bo v ) Nguyn l hot ng Qui c cc tnh

S thay th - Sai s ca my bin dng in Bin dng in hot ng ch bo ha - Hin tng h mchth cp bindng

Phn bit bin dng dng cho o lng v bo v Kt cu ph bin ca BI u, nhc im

Phng thc o lng thnh phn dng in th t khng


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Nguyen Xuan Tung Department of Electric Power System Hanoi University of Science and Technology

Copyright of Nguyen Xuan Tung

Department of Electric Power System Hanoi University of Science and Technology

Do not copy or distribute without permission of the author




Tn gi chung: BI, CT, TINhim v:

Bin i t l dng in s cp th cp (5Ahoc 1 A)Cch lymch s cp v th cpTo s phi hp dng in gia cc pha

4

My bin dng in

BI cao p BI h p S nguyn l{

Is cp*ws cp = Ith cp*wth cp

Nguyn l{ hot ng

CT: Current Transformer (ting Anh)


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S thay th

Sais ca BIxut hin do tn ti ca dng t hain p xut hin pha th cpVth 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 ca BI

Xm: c gi tr t 100-10.000 (c gi tr ln vi cc BI ct s bin ln)

5

My bin dng in

V th cp

Z cun th cp

BI ltng


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Sai s ca BI

Sais gc (angle error) Sais v ln (ratio error) Sais tng hp (composite error)

6

My bin dng in

V thcp

Z cun thcp

BI ltng

Sais hay cp chnh xcca BIc cho theo sai s tng hp


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c tnh t ha ca BIQuan h gia dng in t ha cn thit (Ie) sinh ra mt inp h mch V

im gp VK:L mt im trn ng cong t haT : tng in p ln thm 10%cn tng dng t ha 50%

7

My bin dng in

in p im gp VK(Knee-point)

Vng lm vic tuyn tnh

Vng bo ha


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c tnh t ha ca BITh nghim xc nh c tnh t ha

8

My bin dng in

B to dng

BI

Bng kt qu


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Quic cc tnh (theo IEC 60044-1)Cn thit vi : bo v lm vic da theohng dng in.Cc tnh cng tn c nh du : hnhsao, chm trn, chm vung...

Trn bn v: cc tnh cng tn v cnhnhau. Xcnh nhanh cc tnh BI:

Coichiu dng in i t pha s cp qua

rle khng i chiu

9

My bin dng in

Rle


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Khai bocc tnh ca my bin dng (rle s)10

My bin dng in

im u sao v pha ng dy im u sao v pha thanh gp


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Qui c cc tnh11

My bin dng in


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Hin tng h mch th cp BIGy qu in p nguy him

12

My bin dng in

BI ltng

H mch

o H mch th cp: ton b dng s cp lm nhim v t ha lio Li t b bo hao T thng tn ra ngoi li thp


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Hin tng h mch BIDng sng in p u ra ca BI khi h mch

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My bin dng in


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S thay th khi BI b bo ha Dng khi tnh tonbo v so lch tng tr cao Khi BIb bo ha:

Dngu vo tng cao

Dngu ra c dng xung nhn gi tr hiu dng rt nhNh vy: c th coi l c dng u vo BInhng khng c dng ura dng u vo khp mch qua nhnh t ha tng ngvi vic tng tr nhnh t ha gim rt thp, coi nh xp x bng 0

14

My bin dng in

BI ltng

X

=0

Z s cp Z th cp

S thay th khi BIb bo ha


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Hin tng h mch BIC cu ni tt mch dng khi tho thit b nh th

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Rle,ng ho...

Rle,ng ho...

My bin dng in


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Ti danh nh &Cp chnh xcMt BI: cnhiu cun th cp - phc v cc mc ch khcnhau.Ti danh nh v chnh xc ca cc cun th cp ny tuzthuc vo loi ti.

Ccdng c o (kW, KVar, A, kWh, kVArh):Yucu chnh xc trong ch ti bnh thng hoc nh mc.Phm vihot ng chnh xc trong khong 5120%ca dng in chnh xc thng l: 0.2 hoc 0.5 vi chun IECHoc 0.15 hoc 0.3 hoc 0.6 vi chun IEEE.

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Thng s ca my bin dng in


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So snh BI dng choo lng bo v rle17

Thng s ca my bin dng in

Hng mc so snh BI dng cho o lng BI dng cho bo v rlePhm vi hot ngchnh xc

(0,051,2)x Inh mc(o dng ti bnh thng hoc qu ti chophp)

ti (10-20-30)x Inh mc(m bo o c dng s c)

Lit Bo ha nhanh bo v cc dngc o khis c, dng intng cao(bo ha trong khong 520 ln Im)

in p bo ha cao hn(VK)(kh b bo ha)

chnh xc chnh xc cao0.2 hoc 0.5 vi chun IEC0.15 hoc 0.3 hoc 0.6 vi chun IEEE

chnh cc thp hn 5P hoc 10P theo chun IEC

Thit b ni ti kW, KVar, A, kWh, kVArh Rle, b ghi s c

La chn cp 5P hay 10P tythuc loi bo vBo v so lch tng tr cao 3PBo v qu dng in 10P


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BI cho bo v rleCng sut nh mcCp chnh xcC thm thng s ALF: h s giihn dng in theo chnh xc

18

V d thng s ca my bin dng in

5P20 30VA

Cp chnh xc 5P

P: dng cho mcch bo v rle

(Protection) H s gii hn dng: 20Ti 20 ln dng nh mc, BI vnm bo sai s theo tiu chun

Cng sut nh mc 30VA

BI cho o lng Cngsut nh mcCp chnh xc

30VA Cp chnh xc 0,5

Cp chnh xc 0,5

Cng sut nh mc 30VA


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BIo lng19

Qui nh sai s BI

Theo tiu chun IEC 60044-1


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BIbo v20

Qui nh sai s BI

Theo tiu chun IEC 60044-1


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So snh ng cong t ha ca BIo lng v bo v21

Qui nh sai s BI


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Theo tiu chun IEEE C57.1322

Qui nh sai s BI

BI dng chobo v

BI dng choo lng


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Dng cho mc ch c bitBo v so lch thanh gp

C rt nhiu BICc BI phi c cng c tnh lm vic gim dng khng cn bng

Bin dng cp X: thng s c cho chi tit hnDng nh mcT s binin p im gp VKDng in kch t ng vi in p im gp

in tr ln nht cho php pha mch nh th

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My bin dng in cp X (hay PX theo IEC)


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Thc t, mi BI c th c:1 hoc 2 cun th cp- Mc ch o lng2 ti 4 cun th cp-ng dng bo v rle.

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My bin dng in

Cun s cp

Cc cun th cp


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Thit k BIph bin Li t v cun dy nm trong thng cha thp gn vi t(Dead tank type ), dy th cp chy un theo hnh ch U

Lit v cun dy nm trong thng cha pha trn nh (Livetank type), dy th cp thng chy thng qua li t.Loi hn hp

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My bin dng in


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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 cch

in

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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My bin dng in


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BI loi thng cha bn trn (Live tank)Dydn s cp ngn, gim pht nhitC dng nh mc v dng n nh nhitcao hnTrng tm cao hn, km n nh v mt ckh so vi loi thng cha di (dead tank)Khi ch to vi li t ln c th gy titrng ln v mt c kh i vi s cchin.Kh lm mt cccun th cp

Loi hn hp

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My bin dng in


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Bin dng chn s (Bushing CTs)28

My bin dng in


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Nguyn l:Dng BIt ti trung tnh ca MBA

Dngin chy qua trung tnh my bin p: I A+IB+IC = 3I0 Sais php o ch do mt BI gy ra:

chnh cao Tuy nhincn c dy trung tnh t BI

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o lng dng in th t khng (3I0)

Ia+Ib+Ic=3I0 A

B

C

BItrung tnh 3I0 3I0


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Nguyn l:Dng 3 BI ringbit

Dos dng 3 BI ringbit 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 BICh s dng o dng chm t ln dng mng in c dng chm t ln:mng in trung tnh ni t trc tip

30


I a

I b

I c

RoleI a I b I c 3I 0 + + =

Role

3I 0

V rt gn

Ia+Ib+Ic=3I0


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BI th t khngu ng: dy ni t v cp chy xuyn quali t

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Rle

V kimloi ca cp

BI0


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BI th t khngNgc li - u sai: dy ni t v cp khng chy xuyn qua li t

Dng in chy qua v cp c th trit tiu dng in s c (honton hoc mt phn): rle c th khng nhn c thng tin s c.

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Rle

V kim loi ca cp

BI0


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u ni BI lc thnh phn TTKng dng ca BI th t khng

Do ch s dng mt li tsai s o lng rt nhS dng cho cc mng in c dng chm t nh (mng in c trung tncch in hoc ni t qua cun dp h quang)

Do c 3 pha chy qua li tng knh li t ln kch thc BI lnthch hp trang b vi ng cp hoc u cc my pht in

34



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Bo v chng chm t nhy cao (Aptomat chng git)35

ng dng dn dng ca BIth t khng

Nt bm th nghimin tr mch th nghim

Cun dy mch th nghim

Ti(VD:Bnh nng lnh) S c chm v(chm t)

Cun ly tn hiu dngchm t (dng so lch)

Cun ht ca Aptomat

Ngun cp


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Chng 01

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My bin in p(dng cho h thng rle bo v ) Nguyn l hot ng Qui c cc tnh S thay th - Sai s ca my bin in p BU kiu t phn p u, nhc im Phng thc o lngthnh phn dng in th t khng


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Tn gi: BU, VT, PT (potential transformer)Nhim v

Bin i t l in p s cp sang in p th cp theo tiu chun(100Vhoc 110V)Cch lymch s cp v cc thit b, ngi vn hnh bn th cp

Quic cc tnhCc tnh cng tn c nh du : hnh sao, chm trn, chmvung...Trn bn v: cc tnh cng tn v cnh nhau.

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My bin in p (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.

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My bin in p (BU)


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BU kiu tphn pBU loi cm ng in t thng thng

La chn kinh t nht i vi cp in p ti 145kVBU 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

39

My bin in p (BU)

u cao p

in khng b

Tphn

p Mch dp dao ng cng

hng

BU cm ngthng thng

u ra

K{ hiu trn s

520kV


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Cu trc BU kiu tphn p1. Bnh gin du2. Cc t phn p5. in khng b7. BU loi cm ng thng thng

(in p thp)8. u cc cao p

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My bin in p (BU)


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BU kiu t phn pin khng b: c tnh ton trit tiu thnh phn dungkhng ca t phn p

Tng tr ngun nhn t pha ti l xp x 0 cng sut u ra ly raln nhtBdch pha do t phn p gy ra

Mch gim dao ng cng hng: l in tr ti, c th ni cun

tam gic h

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My bin in p (BU)


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S nguyn lca BUt phn p Siemens42

My bin in p (BU) Tham kho


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S nguyn lca BUt phn p GE43

My bin in p (BU) Tham kho

b


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Cc loi BUH s gii hn in p Vf

Khixy ra s c trong HT: U pha c th tng ln ti mt gi tr l Vf lnUdanh nh.Tiu chun IECa ra cc gi tr h s Vf nh sau:

1.9 i vi cc h thng c trung tnh khng ni t trc tip 1.5 i vi cc h thng c trung tnh ni t trc tip

Lit ca cc bin in p khng c php bo ho khi in p tngti cp in p gii hn theo h s Vf .

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My bin in p (BU)

bi i ( )


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Cp chnh xc theo tiu chun IEC 60044-2

Cng sut danh nh (cos=0,8)10, 15, 25, 30, 50, 75, 100, 150, 200, 300, 400, 500 VA

45

My bin in p (BU)

Cpchnhxc

Sai s gii hn

Ti % in pnh mc

Sai s ln%

Sai s gc pha(pht)

ng dng

Chun ha dng c otrong cng nghip

ng h o

Ti % ti nhmc

Phng th nghim

o m chnh xc

Bo v rle

Vf: H s gii hn in p

Bo v rle

o m tiu chun

l i h


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Nguyn l UA+UB+UC=3U046

o lng in p th t khng (3U0)

l i h


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S dng BUt ti trung tnh

Dng ba BUloi mt pha

47

o lng in p th t khng (3U0)

S cp

Th cp

M bi i (BU)


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B lc in p TTK (U0)Dng BU loi 3 pha 5 tr

in p TTK ly ra t cun tam gic h

48

My bin in p (BU)

Scp

Thcc p

3V 0

ABC

A B C N

o c in p th t khng:o Phi c ng dn cho t thng TTKo Cn thm 2 tr (khng qun dy)o Trung tnh cun s cp phi ni t (nu khng s ch o c thnh phn hi

N d i hi


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Auxiliary power supplyCu hnh

49

Ngun t dng in p mt chiu

Anti-paralleling diodesFor preventing high circulating currents

If one battery be faulty The fullycharged battery cannot discharge into it

Float charger Charge a battery at a similar rate as a batteryself-discharges Thus maintaining a full capacity battery. Used where the battery rarely gets discharged

Boost charger Charge a battery at a high current forshortening charging time Enables the quick charging of depletedbatteries.


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Chng 02

50

Cc nguyn lbo v trong HT Nguyn l bo v qu dng/qu dng ct nhanh/ chm t /c hng Nguyn l bo v so lchdng in c hm Nguyn l bo v tng tr thp (BV khongcch)

Cc nguyn l bo v khc


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51

Nguyn lbo v qu dng Nguyn l bo v qu dng c thi gian (51):

o

Dng in khi ng- Thi gian t o c tnh lm vic (c lp vs. ph thuc)

Nguyn l bo v qu dng ct nhanh (50)o Nguyn l m bo chn lc - Dng in khi ng

Nguyn l bo v qu dng th t khng c thi gian/ ct nhanh

(50/50N)o Dng in khi ngo nhy

Nguyn l bo v qu dng in c hng(67)

Chng02

N l{ b d thi i


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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 cchm t.

Bo v khi ng khi:Dng in ca mt pha, hai pha hoc c ba pha vt qu mt gi tr c ci t trc trong rle.

Cth 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

52

Nguyn l{ bo v qu dng c thi gian

Ng l{ b d g thi gi


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Bo v qu dng vi c tnh thi gian c lp:Thi gian lm vic (tr) ca bo v khng ph thuc vo ln dngn mch

53


+

Iqua r le

Thi gian lm vic

Ikhi ng

t lm vickhng i theo dng in

Ng l{ b d g thi gi


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Bo v qu dng vi c tnh thi gian ph thuc:Thi gian lm vic: ph thuc t l nghch vo ln ca dng ingn mchTrong thc t th thi gian tc ng t l vi t sIngn mch / Ikhi ng

54



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Tnh ton thi gian lm vic


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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 cs c c th nhiu bo v cng khi ngTuy nhin, bo v gn ch s c nht s phi tc ng trc

S c ti N2: BV2khi ng & BV1 cth cng khi ng cngm thi gianBV2phi tc ng loi tr s c, BV1 khi s tr v t thi gian

tBV2 I>1 2

N2

Tnh ton thi gian lm vic


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BV2phi tc ng loi tr s c, BV1 khi s tr v t thi giantBV2


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Nguyn tc: m bo tnh chn lcbng phn cp dng inS c ti phn on no: ch bo v ti c php khi ng

Ccbo v khng cn phi hp thi gianThi gian tc ng t xp x 0 giy (thng t 50 80ms) tn gi:bo v qu dng ct nhanh (50 hay I>>)Do cchchn lc bng dng in dng in khi ng tnh theo:

Ik=Kat*Ingn mch ngoi vng max (H s Kat=1,1 1,2)

Khngbo v c ton b i tng khng s dng lm bo vchnh

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Nguyn l{ bo v qu dng ct nhanh (50 hay



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Vngc bo v ct nhanhIk=Kat*Ingn mch ngoi vng max (H s Kat=1,1 1,2)

Dngngn mch gim dn khiim ngn mch i xa ngun ln dng ngn mch ph thuc vo ch ca h thngV ng cong biu din dng ngn mch Inmax & Inmin dc ng dy

59


L(km)

I> I>1 2

IN

Inmax

Inmax

Ik BV1

Ik BV2

Lct nhanh min

Lct nhanh max

Lct nhanh min=0

Lct nhanh max

Phn bit chc nng I> & I>> (51 & 50)


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Bo v qu dng

60

Phn bit chc nng I> & I>> (51 & 50)

Khi ng khi:Ingn mch >Ikhi ng

Bo v qu dng c thi gian(I> hay 51)

Dng khi ng tnh theo dnglm vic ln nht (Ilvmax)Khi xy ra s c c th c bov ti ch v bo v pha trncng khi ngm bo tnh chn lc: phihp phn cp thi gian (t)C th dng lm bo v chnh

Bo v qu dng ct nhanh(I>> hay 50)

Dng khi ng tnh theo dngngn mch ngoi ln nht(In ngoi max)

Khi xy ra s c: ch bo v tiphn on s c khi ngKhng cn phi hp thi gian(ct nhanh)Khng bo v c ton b i

tng ch l bo v d phng

Bo v qu dng th t khng (I> hay 51N)


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S dng b lc dng in th t khng

Tnh ton dng khi ngch bnh thng:

Lthuyt: dng qua rle bng 0Thc t : do cc BI c sais dng in qua rle khc 0 rle khng tc ng: t dng khi ng ln hn dng in sinh ra do sais ny

Gitr ci t: Ikhi ng 51N =(0,10,3)I nh mc BI

Ch s c: dng in qua rle tng gp nhiu ln bo v tcngDo gitr khi ng t thp bo v c nhy cao

61

Bo v qu dng th t khng (I0> hay 51N)

I a I b I c 3I 0 + + =


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V d: bo v qu dng cho mt ngn l


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64

V d: bo v qu dng cho mt ngn l

A

B

C

I> I> I>

I>> I>> I>>

I0>

I0>>

S c hai pha N(2)


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Bo vqu dng c hng(67)


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66

Bo v qu dng c hng (67)

I> = I> + WI>

+

-

I> 1 I> 5I>4 I>6I> 3I>2HT1 HT2

Gii php: s dng bo v qu dng loi c nh hngBo 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

C th phn chia ra 2 nhm bo v



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V phng din bo v rle: ng dy hai ngun cphaimch hnh tia

67


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



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Kim tra s lm vic ca bo v:68


I> 1 I> 5I>4 I>6I> 3I>2HT1 HT20,511,5 0,3 0,8 1,3

N1

I> 1 I> 5I>4 I>6I> 3I>2HT1 HT20,511,5 0,3 0,81,3

N2

Phn t nh hngcng sut


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Phn t nh hng cng sut:Vecto in p thng c dng lm i lng tham chiu

69

Phn t nh hng cng sut

S u ni b phn nh hng


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B nh hng cng sut:S dng in p lm i lng tham chiuc u ni m bo: rle c nhyv tc ng ng trong mi trng hp.S u ni tiu chun i vi cc rle

s v rle tnh l s 900

Chitit phng thc u ni: Dngin t mt phain p dy ca hai pha cn li

Gi thit cos=1 hay =00 th in p thamchiu v dng in to vi nhau gc 900chnh l tn gi ca s

70

S u ni b phn nh hng

S u ni b phn nh hng


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L dochn in p l i lng tham chiu:Khixy ra s c ba pha: in p gim thp, nu s dng in p phath rle nh hng c th khng nhy, s dng in p dys tng c gi tr in p a vo rle.

Khixy ra s c pha-pha v d gia pha 1 & 2: in p U12 c th rtthp (c th bng 0 nu s c gn bo v) rle nh hng khng nhy, trong khi in p U23 vn cn ln phi s dngin p dy vi pha khng s c cn li lm in p tham chiu.

71

S u ni b phn nh hng


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72

Nguyn lbo v so lch Nguyn l bo v so lch dng in (87) Nguyn l bo v so lch dng in c hm

o Nguyn l hot ngo La chn h s hmo Ni t BI

Nguyn l bo v so lch tng tr cao (87H)

Nguyn l bo v so snh pha dng in

Chng02

Nguyn l{ bo v so lch dng in


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Thit b

So snh tng dng in i vo &i ra ca i tng c bo

v: tng dng in ny khc 0bo v tc ng.

Ch bnh thng:Dng in chy qua rle nh hnh vDng chy qua rle: l dng chnh lch do sai s ca BI cc pha

73


Bnh thng

Irle= +



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Ch s c ngoi vng:Dng in l dng s c c gi tr lnsai s BI ln hnPhn b dng in tng t ch bnh thngDng in chy qua rle s lnchnh nh rle khng tc ng

dng khi ng ln, gim nhy

Vng bo v:gii hn bi v tr t cc BI

74


S c ngoi

Thit b



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Ch s c trong vng:Dng in chy qua rle bng tng dng hai phac gi tr lnrle s tc ng ngay

75


S c trong vng

Thit b

Irle= +



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Tng kt:76


Irle= +

S c trong

Thit b

Irle= +

S c ngoi

Irle= +

Bnh thng

Dng in chy qua rle l do sai s BIDng in chy qua rle l tndng s c

Nguyn l{ bo v so lch dng in c hm


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Bo v rle so lch thng thngcc rle c th tc ng

nhm do:Sai s ln ca cc BI khi ngn mch ngoiChuyn u p...

Bo v so lch c hm:hot ng da theo t hp ca hai loi

dng in so lch (Isl ) & hm (Ih):S c ngoi vng: dng hm c gi tr ln dng so lch nhS c trong vng: dng hm nh- Dng so lch ln.

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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)

78




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T hp dng in cho bo v rle

so lch c hm

Vit phng trnh cn bng sc tng cho cc BITG

79


BITG2 BITG1I1*w1-I2*w1=Isl*w2Chn w

1=w

2Isl=I1 - I2

I1*w3+I2*w4=Ih*w5Chn w

3= w

4= w

w5=2*w Ih=0,5*( I1 + I2)

Tng qut:Ih=K h*(I1+I2)K h: h s hm, c th iu chnhthayi theo s vng cun w5



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Bng cch s dng BI trung gian, cth to ra t hp Xts vn hnh Bo v so lch c hm

80

guy { bo v so c d g c 1 2

1 2

=( )

( ) sl

h h

I I I

I K I I

I 1 I 2

I sl

I h=K h* (I 1+I 2 )

100% (I 1+I 2 )

I 1

I 2

I sl

I h=K h* (I 1+I 2 )

100% (I 1+I 2 )

(I h)> (I sl ) bo v khng

tc ng

(I h)< (I sl ) bo v tc

ng

Ch bnh thng hoc s c ngoi Ch s c trong vng



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La chn h s hmTng h s hm (Kh): rlehm tt nhy tcng ca rle km i.Hiu ng ngc li khi gimh s hm

H s hm c th c iuchnh bng cch thay i svng dy cun w5.

81

g y { g

I sl

I h=K h* (I 1+I 2 )

100% (I 1+I 2 )

I 1

I 2

I sl

I h=K h* (I 1+I 2 )

100% (I 1+I 2 )


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Bo v so lch c hm


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Ni t BI

83

Ni t ti mt im

Ccrle c: c lin h trc tipv in gia cc pha

Phi ni t ring tng mch

Ccrle s: mch dng ca ccBI cch lyv mt in

Vn ni t BI vi bo v so lch


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Ni t BI

84

Bo v so lch MF

Bo v so lch MBA

BI trunggian

BI trung gian Rle c

Rle s

Bo v so lch c hm


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Ni t BI

85

Bo v so lch tng tr cao

Rle

Bo v so lch c hm


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Ni t BI

86

Bo v so lch tng tr cao

Nguyn l{ bo v so lch tng tr cao (87H)


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Bo v so lch c th tc ng nhm khi BI b bo ha

V d hin tng bo ha ca cc BI vi bo v thanh gp

87

g y g ( )

Vng bo v

S c trong vng bo v

Vng bo v

S c ngoi vng bo v

Bo ha


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Xts n gin: thanh gp ch gm hai ngn l90

g y g ( )

IBI1

BI2

Bo ha

IBI1Ltng

Z s cp Z th cp

X

>> 0 X =0

Z s cpZ th cp

BI2Ltng

Gi tr rt ln(coi lhmch)

Gitr xp x 0(ni tt)



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Xts n gin: thanh gpch gm hai ngn l91

g y g ( )

IBI1Ltng

Z s cp Z th cp Z s cpZ th cp

BI2Ltng

Nhnh song song

BIb bo ha: khng c dng th cp BI cnli: dng cp ra khp mch qua rle v tng tr th cp BIb boha c dng qua rle rle s tc ng (tc ng sai)Gii php: hn ch dng qua rle bng cch ghp ni tip mt in trc gi tr ln rle so lch tng tr cao

I

Z th cp Z th cp

R>>

Rle so lch tng tr cao



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Yucu i vi BICc BI c cngt s binNn s dng bin dng cp XBI chns c th s dng nu cc cun dy BIthuc loi qun phnb u

Cc BI nn c cngthit k in tr cun th cp BI nnchn nh nht c thKhng nnni cc thit b khc ti li dng cho 87Hin p im gp Vk phi ln hn t nht 2 ln in p khi ng carle:

Nu in p Vk chn thp hn: khi cs c, BIs sinh ra in p c th khng ln lm rle tc ng (chn ln hn 2 ln m bo nhy ti thiu l2)

92

in p im gp VK(Knee-point)



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Hot ng vi s c ngoi93

BI ltng: khng bo ha BIb bo ha


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Nguyn l{ bo v so snh pha dng in


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Bo v so snh pha dng in (87PC)

Rle ch trao i thng tin v pha dng inCh trao i cho nhau hai trng thi: dng in ang chu kz m haydng (logic 0 v 1)Vi cc rle hin i: so snh ring bit ba pha (cp quang)

Khis dng knh truyn bng thng thp (v d: ti ba PLC):S dng cc thnh phn i xngHoc t hp cc thnh phn i xng theo t l nht nh

95



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Nguyn l so snh pha96

S c trong vng bo vS c ngoi vng bo v

khng trng khp khng trng khp0

0 01

1 trng hon ton trng hon ton



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Ccyu t nh hng

ch non ti hoc khng ti: dng in dung c th lm bo v tcng nhmGii php: s dng b phn pht hin s c (phn t qu dng) khi ngtruyn tin.

Vi cc rle k thut s: sai s v gc pha do vic ly mu tn hiuDo nh hng ca dng in dung: gc ci t cho BV solch phaphi tng ln

BV solch pha ch nn p dng cho ng dy ti 400km.Tnhti dao ng gc pha do nh hng ca dao ng in

97


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Nguyn l{ hot ng


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Bo v khong cch da trn cc gi tr dng in v in p ti

im t rle xcnh tng tr s c

Nu tng tr s c ny nh hn gi tr tng tr ci t trongrle th rle s tc ng rle tng tr thp Z< (hoc 21)

Tng tr gm hai thnh phn R & X: thun tin phn tch ss dng mt phng tng tr biu din s lm vic ca bov khong cch

99

R

jX

ZD

Rle khong cch loi in c n gin


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Nguyn l{ hot ng


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Xts n gin:

Tnh ton tng tr rle o c trong cc ch S c : v d ti 50%ng dy:ZR(sc)=ZDs c=50%ZD < ZD

im s c di chuyn vo ngtng tr ng dy

101

R

jX

50%Z D

ZD

ZD+Zpt

imlm viclc bnhthng

imlm vickhi s

c

Nguyn l{ hot ng


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c tnh lm vic ca rle khong cch

im lm vic lc bnh thng v khi s c: khi s c im lm vlun ri vo ng tng tr ng dyc th ch cn ch to ctnh tc ng ca rle l mt ng thng trng vi ng tng trng dy

102

c tnh tcng l mtng thng

R

jX

ZD

ZD+Zpt

imlm vic

lc bnhthng imlm vickhi s

c

Nguyn l{ hot ng


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c tnh lm vic ca rle khong cch

Do sai s, do s c c th xy ra qua cc tng tr trung gian nn gtr rle o c khi s c c th ri ra ln cn ng tng tr dy.Nu ch ch to c tnh tc ng l mt ng thng th rle c ts khng lm vic trong cc trng hp ny. khc phc th ccch to thng c { m rng c tnh tc ng v c hai pha cang dy tr thnh vng tc ng.

103

R

jX

100%Z D

ZDZD+Zpt

imlm viclc bnhthng

im s c ri rangoi rle khng

tc ng

c tnh tcng l mt

ngthng hp

im s c rivo vng tc ng

c tnhtc ngc m

rng

R

jX

ZDZD+Zpt

imlm viclc bnhthng

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 nhaup ng tt hn vi mi loi s c v ch vn hnh ca h thng

104

ZD


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Ci t cc vng bo v


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Cc vng ci t ca bo v khong cch

Thng c chnh nh vi 3 vng tc ngVng I: tc ng tc thiVng II & III: tc ng c tr theo nguyn tc phn cp thi gian, phi hvi cc bo v lin k

106

Ci t cc vng bo v


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Vng 1 (Z1) : bo v khong 80-90%ng dy AB Khngth ci t bo v 100%ng dy:

Sais ca BI, BU: BI cth b bo ha v nh hng ti chnh xcca tng tro c.nh hng ca h s phn b dng inTnh ton tng tr da trn gi thuyt: b qua in dung ng dy, cc pha opha tun t...tuy nhin thc t iu ny khng th hon ton chnh xc.

107

Ci t cc vng bo v


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Vng 2 (Z2) : t i thiu t 120-150% ZAB(gi tr ti thiu m bo s cst thanh gp B (F2, F3, F4) khng b ri vo vng 1 ca BV1)

Vng 2 ca BV1 l bo v d phng cho BV5 do :Vng 2 ca BV1 khng th bo v vt qu Vng 1 ca cc bo v lin k (BV5; BV7)Phi hp vi ng dy ngn nht (bo v ti 50% ca ng dy ngn nh)

108

Ci t cc vng bo v


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Vng 3 (Z3) : bao trm ng dy di nht t thanh gp B (ng dyBD) Tuy nhin khngc qu vng 2 ca cc bo v BV3; BV5; BV7

Vng 3: tnh ti kh nng nh hng ca ti nng v trng hp xy radao ng cng sut.

109

Ci t cc vng bo v


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Cc vng ca bo v khong cch

Vng hng ngc (Z3R)Vng bo v gia tng hiu qu bo v.Bo v d phng cho thanh gpt khong 25% tng tr vng I

Vng III hng ngc (Z3KR) c th dng c tnh tc ng v hng(hnh trn).

110


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Ci t cc vng bo v


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Minh ha112

Ngun

ng dy 1ng dy 2 ng dy 3

1015% 1015% 1015%Vng I Bo v 2 Vng I Bo v 3A B C D

Thanh gp B

Thanh gp A

Vng It = 0 giy Tng tr

ng dy

Vng IIt = t giy

Vng IIIt = 2t giy

Vng I Bo v 1

Thanh gp B

Thanh gp A

Vng It = 0 giy Tng trng dy

Vng IIt = tgiy

Vng IIIt = 2t

giy

Vnghngngc

f f

Thut ton tnh ton tng tr


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Cc mch vng tnh ton tng tr

S c pha-pha: 3 mch vng tnh tonPht hin s c pha-pha: ch mch vng pha-pha c nh gi

S c pha-t: 3 mch vng tnh ton

Pht hin s c pha-t: ch mch vng pha-t c nh gi

113


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N X T D f El i P S H i U i i f S i d T h l115



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Cc mch vng tnh ton tng tr

H s b th t khng K0

Rle o ng gi tr tng tr th t thun nu mt lng dng ibng (K-1)/3 c a thm vo dng in pha.L h s b dng in th t khng cn ci t

115

H s bth tkhng

Ng X T g D t t f El t i P S t H i U i it f S i d T h l g116



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Vi cc pha khng s c

Tng tr tnh ton c ln hn tng tr s cCh mang mt phn dng s cin p thng gim t hn

116

Fault Loop Equation

Single phase to groundAG, BG, CG

ZAG= VA / (1+k 0) IAZBG= V B / (1+k 0) IBZCG= V C / (1+k 0) IC

Three phase

ABC

ZABC = VA/B/C / IA/B/C

Double phase

AB, BC, CA

ZAB= (VA- VB) / I A- IBZBC= (V B- VC) / I B- ICZCA= (V C- VA) / I C- IA

Nguyen Xuan Tung Department of Electric Power System Hanoi University of Science and Technology117

Phi hp lin ng cc bo v khong cch


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Nu khng c s lin h phi hp gia cc bo v hai u

ng dy:S c ti 1015%cui ng dy mi pha s c loi tr vi thigian ca vng 2 (tr mt khong t)nh hng n n nh ca h thng

Khng th s dng h thng t ng li nhanh:My ct hai u khng c ct cng lcs c thong qua dthnh duy tr.

Khc phc: thc hin lin ng gia cc BVKC thng qua knhtruyn tng tc loi tr s c

117

HT11015%

A B

HT21015%

N1N2


Phng thc lin ng bo v (Teleprotection)


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Cc phng php

Zone 1 Extension Scheme Transfer Tripping Schemes

Direct Under-reach Transfer Tripping SchemePermissive Under-reach Transfer Tripping (PUTT) Scheme

Permissive Over-Reach Transfer Tripping (POTT) SchemeNgun yu (Weak Infeed)

Blocking Over-reaching schemeNgun yu

So snh s truyn tn hiu cho php v truyn tn hiu kh

118


Phng thc lin ng bo v (Teleprotection)


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Zone 1 Extensio

Bai Giang Bao Ve Role Trong Nha May Dien - Nha May Thuy Dien Son La

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Transcript of Bai Giang Bao Ve Role Trong Nha May Dien - Nha May Thuy Dien Son La