DKBVHTD - Cong Thuc Bao Ve Role

7/30/2019 DKBVHTD - Cong Thuc Bao Ve Role

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CHNG 5: BO V QU DNG

1. Tnh ton bo vdng in cc i cho ng dy U (kV)

Tnh dng lm vic cc i IlvM

Chn my bin dng v tnh t s bin dng,

Chn s ni my bin dng,

Chn kiu r-le

(Dng khi ng ca bo v: ktc.IlvM)

Dng khi ng role bo v cc i

itv

lvMmmsdtc

kdR nk

Ikkk

I .

....

Chn dng t ca role

kdRdR II

Dng khi ng thc t ca bo v:

sd

idR

kdI

k

nII

.

nhy ca bo v:

kdI

k

nhI

Ik

)3(

87,0

Kim tra my bin dng theo dk lm vic tin cy ca cun ct.

Nh vy my bin dng chn m bo yu cu lm vic tin cy cho cun ct.

(Tnh ton ln lt cho cc bo v)

1.1 Tnh ton thi gian bo v theo c tnh c lp (thi gian tc ng khngph thuc dng in) (cho mng in nhiu cp)

Cp thi gian bo v gia bv1 v bv2:

dtqttMC ttstt 111 .

Thi gian tc ng bv2:


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112 ttt

Nu c bo v 3 th tnh t3= t2+ t2tng t

1.2. Tnh ton thi gian theo c tnh ph thuc.(thi gian tc ng ph thucdng in) (cho mng in nhiu cp)

Bi s dng ca rle bv1 ti N1:

11

111

. dR

kBV

In

Im (

1kI dng ngn mch tiN1)

Thi gian t ca ca rle bv1

14,0

)1( 02.0 1111

BId

mtt

Chn thi gian

11 dcd tt

Bi sdng in ca role bv1 ti N2:

1.1

211'

dRi

kBV

In

Im

Thi gian tc ng thc t bv1 khi c nhn mch N2

1

14.0.

02.0'.

11

1

'

2.1

BV

cdnm

tt , ( 11 dcd tt )

phn cp bv1 v bv2:

dtqtntMC ttstt '

2111

Thi gian tc ng ca bv2 khi ngn mch N2. 1' 2.12 ttt n

(Tnh ton Tng ti vi cc bo v khc)

2. Tnh ton bo v ct nhanh cho ng dy Ucb(KV)

in trh thng:HTk

cb

HT

S

Ux

.

2

in trng dy: lrRd

.0

, lxXd

.0


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Tng trng dy : 22 ddd XRZ

Tng trngn mch: 22 )( dHTdk XXRZ

Dng ngn mch 3 pha ti N1(im trn thanh ci gn ngun):cb

HTkk

U

SI

.3

.1

Dng ngn mch 3 pha ti N2:k

cbk

Z

UI

.32

Chn my bin dng ( I1BI> Ilv Max ), chn s ni bin dng.

(1) Dng khi ng role ct nhanh:i

ksdtcRkdCN

n

IkkI 2..

.

Chn dng t IdRCN>IkdCNR

Dng khi ng thc t ca bvcn:sd

idRCNkdCN

k

nII

..

nhykdCN

k

nhI

Ik

)3(

2.87.0

t c nhy cn thit bo v ct nhanh chtc ng vi dng ngn mch

l:

Ik=2.IkdCN

T lvng tc ng ct nhanh: )(100% HTkdCN

XI

E

Zdm

(i vi ng dy 1 ngun

cp)

Kim tra bin dng theo iu kin lm vic tin cy ca cun ct: Icc = 0,05Ik1

Nh vy my bin dng chn m bo yu cu lm vic tin cy cho cun ct.

2.1 Tnh ton bo v ct nhanh cho trm bin p U1/U2(kV)

Dng nh mc ca bin p: IBA = SBA/sqrt(3).U1

Dng ngn mch quy vs cp: Ik3)

=Ik3)

.U2/U1 (ICN1> Ik3)

)

Theo dng tha t bin: ICN2> Ib = kb.IBA

Chn trong Ik(3)

v Ib dng ln hn tnh dng khi ng cu bo v


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(Dng khi ng ca bo v: ktc.Ib hay ktc.Ik(3)

),

chn bin dng c I1BI> IBA, tnh ni

Tnh ton cc bc tng tnh (1) vi Ik2=Ik(3)

hay Ib

Ch : tnh ton bo v qu dng in l tnh ton c bo v cc i v bo v ctnhanh

CHNG 6:BO VC HNG

1. Tnh ton bo vc hng cho mng in U(kV)

Tnh dng in lm vic cc i chy qua bo v (IlvM1), chn bin dng

(I1BA>IlvM1)

Dng khi ng ca role:itv

lvMsdmmtckdR

nk

IkkkI 11

...

Chn dng t cho role: 1kdRdRII

Dng khi ng thc t bv:sd

idRkdBV

k

nII

.

Tnh in tr trn on dy c bo v (v d bo v 1-gn ngun A nht, choon dy 12): 1201212012 ., lxXlrR

in trh thng:Ak

HTA

S

UX

.

2

in trngn mch: 2122

12 )( HTk XXRZ

Dng ngn mch 3 pha:k

kAZ

UI

.3

)3(

Thi gian tc ng: t2 = t5

Mot cach gan dung co the coi phn cp thi gian t= const:

dtqtntMc ttstt 1

Thoi gan tac dong cua bao ve 3,4


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tttt 243

Thoi giant ac dong bao ve 1,6

tttt 361

CHNG 7: BO V SO LCH

1. Bo v so lch cho thanh ci (sl dc)

Dng khi ng ca bo vc tnh n tin cy: Ikdbv = ktc.Ikcb.max

Chn bin dng, tnh ni, ksd

Tnh dng khng cn bng cc i : = (3)Dng khi ng ca role : = . Chn dng t ca role l IR>IkdR

Dng khi ng thc t ca bo v so lch: = .

nhy ca bo v: = .

=0.87.

(3)

2. Bo v so lch cho MBA (sl dc)

Dng nh mc hai pha ca mba:

1 = 3.1 ; 2 =

3.2

Gi trdng in th cp bin dng hai pha ca mba thc t l:

21 = 1 .1 22 =?Sai s do chnh lch dng in pha th cp:

2 = 2 22

Nu s chnh lch gia cc dng in th cp l qu ln nn chn li bin dng.V d chn li bd pha th cp


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2 = 12 ; 22 =2. 2

Xc nh dng khng cn bng: = ( + + 2) . Vi IkMax.ng l dng ngn mch ti thanh ci th cp

Dng khi ng ca bo v: = . (2) Dngkhi ng ca role: = .2 Chon dng t cho role: IdR>IkdR

Dng khi ng thc t ca bv so lch: . = .2

nhy = . = (2) / .=0,87.Ik3)/IkdSLTng nhy bng role c cun hm: =n.((I2I/I2II)-1)

3. Bo v so lch cho my pht(sl dc)

Dng nh mc ca my pht: = 3. t chn bin dng, s ni bdDng ngn mch 3 pha:

(3)

=.

Dng khng cn bng: = (3)Chn trong Ikcb v InF dng no ln hn tnh dng khi ng ca bo v. Cc bctip theo tnh nh (2)

CHNG 8: BO V KHONG CCH

in trgitng trn cc role: ZR= lk.z0.ni/nu

Bo v 3 cp:

in trkhi ng ca bo v vng 1 phi nhhn in trng dy bo v:

ZA1= k1.ZAB

Dien tro khoi dong vung 2: ZAII=k2(ZAB + k1ZBC)

ZBC in trng dy BC lin ngay sau ABDien tro khoi dong vung 3: ZA

III= k2[(ZAB + k2(ZBC + k1ZCD)]


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Bo v khong cch cho ng dy(c 2 bo v BVA v BVB)

*Chn my bin p v my bin dng theo bi (cp in p ng dy v dnglm vic trn ng dy). Suy ra

= 1

2

= 12 *xc nh in tron dy.

= 0 trong in tr on dy AB,0 sut in tr ng dy(/km). Tng tcho cc on dy cn li (on BC).

*xc nh dng t cho phn t khi ng ca bv ADng khi ng ca RI xc nh ging bv dng cc i

> = *chn dng t role >*dng khi ng thc t ca bo v.

> = > +,bo v A

in trkhi ng vng 1 ca bv A: = 1 -in trkhi ng ca role: . = -chn in trt vng 1ca role

.

-in trkhi ng thc t ca BV khong cch vng 1:: . = . - nhy ca bv vng 1: = .

* in trkhi ng vng 2 ca bv A: = 2( + 1)-in trkhi ng ca role vng 2: . = -chn in trt vng 2 ca role . (


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-in trkhi ng thc ca role: . = . +, bo v B

* in trkhi ng vng 1 ca bv B: = 1 -in trkhi ng vng 1ca role: . = -chn in trt vng 1ca role . (


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KBV HT

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nhy ca bv vng 1: = . vng 2 ca bv A:trong scc on dy sau AB on BC c in trnh nht vvy in trkhi ng vng 2 ca bv A l:

* in trkhi ng vng 2 ca bv A tnh n in trqu :

= 2( + 1( + 0.5 ))

= 28700( + . ) .31.4

in trkhi ng ca role: ZR.AII= ZA

II.ni/nu

* chn in trt vng 2 ca role . * in trkhi ng thc ca role: . = . * thi gian tc ng vng 2: 2 = 1 + * thi gian tc ng vng 3: 3 = 2 +

Tnh ton bv khong cch A ng dy c thm ngun cung cp (2 ngun A lngun chnh c cho chiu di v B ngun ph)

Trc ht xc nh tng trca cc on dy ( ZAC, ZCD)

*dng chy trn on CD: = + (l dng chy trn ngun chnhA, l dng chy trn ngun ph B)

*chn mba

* = 12trong 1(> )

*h s phn dng = 1 + * in trca ng dy AD c tnh n h s phn dng:

= + * in trkhi ng tnh ton ca role:

= 1 *chn in trt ca role .( nhhn cht


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* in trkhi ng thc ca role: = . * nhy ca bv : =

Bo vcho ng dy 1 ngun cung cp(AB) phn ra thnh 2 nhnh (BM v BN).

Xc nh tng trcc on dy ZAB , ZBN

*dng tng chy trn on dy AB: = + *chn mba =( un1/ un2)* = 12

* h s phn dng

= 1

* in trca ng dy AN c tnh n h s phn dng:

= + trong l tng tron dy BN)* in trkhi ng tnh ton ca role:

= 1

*chn in trt ca role .( nhhn cht

1) in trkhi ng vng 1 ca bo v A:(zA1 < ZAB )

=> zA1 = K1 ZAB

2) in trkhi ng vng 2 ca bo v A:=> zAII = k2(ZAB + K1 ZBC )

HAY => zAII =

zmin

1++

3) in trkhi ng vng 3 ca bo v A:=> zAIII = k2(ZAB + k2(ZBC + K1

ZCD )


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HAY => zAIII = ktc ZAB + ktc (1)zBIIkp

4) Gi tr khi ng ca rle vng 1 ca bo v A:

zR.AI = zA

Ini

nu

= k1zABni

nu

***cn c vo dng khi ng ca rle zR.AI ta chn zd.A

I nc chnh nh gn nht

vpha di zR.AI (zd.A

I zR.AI )***in trkhi ng thc t ca bo v khong cch A vng 1:

zkd.A=zd.A

I nu

n i

I

***H s nhy ca rle in trvng 1 c xc nh nh sau:

knh =zAB

zkd .aI

5) in trkhi ng c tnh n qu trnh qu :

zd = zd + 0.5Rqd

vd: zAI = k1(zAB + 0.5RqdAB )

RqdAB =28700(a + v tk)Ik1.4

CHNG 10 BO V MY PHT V NG C IN

Dng in nh mc ca my pht: IF=

SF

3.Un

Da vo dng in nh mc chn my bin dng: ni =InI

In2

Dng in ngn mch 3 pha : I(3)k=Ex . IF

E

* : sut in ng trong hn vtng i

X

: in trsiu qu dc trc

a/ bo v dng cc i :


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dng khi ng role: = . . . . chn dng t ca role IdRI>

dng khi ng thc t : > = >.

nhy ca bo v : Knh =0.87.(3) >

b/ Bao ve cat nhanh :

+ Dong ien k cua bao ve cat nhanh: IkdCN.R = (3)

. chon dong kd+ Dong k thc te cua bao ve cat nhanh: Ikd.CN =

.

c/ bao ve so lech :

+ Dong ien khong can bang :

Ikcb = ka.kcl.si.I(3)

k

Neu ikcb


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Thi gian tac dong bv vung 2: t2 = t1 + 1t

CHNG 11: BO V MY BIN P

Bo v role cho my bin p:Chn in p c bn: Ucb=Un2=37kv.

in trh thng : XHT =2 .

in trca my bin p : XBA=2

100

Tng trngn mch: XK= XHT + XBA

Dng ngn mch ba pha ti im trn thanh ci th cp quy vpha s cp

(3) = 3

a) Bo v so lch cho my bin p:Dng in nh mc hai pha s cp ca my bin p

1 = 31 2 = 32Gi trdng in th cp hai pha ca my bin dng chnh mc:

I2I =1.1

1 I2II =2.2

2 Sai s do s chnh lch dng in pha th cp:

2 = 2 2

2

Dng in khng cn bng:

= (ka.kcl si + +s2i )(3)Dng in khng cn bng -> = ?Dng khi ng ca bo v: Ikd= ktv Dng in khi ng ca role IkdRSL=

22

Dng khi ng thc t ca bo v so lch : Ikd.Sl=IkdSL=2 . .

2

nhy ca bo v l :


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= 0,87. (3)

.

Vi s vng dy ca cun cn bng l , chn s vng dy san bng dng inth cp:

= 22 1Chn s vng dy cn mc thm pha s cp l = 2

b, bo v ct nhanh:

dng khi ng ca bv ct nhanh c chn t mt trong hai iu kin

-Ln hn dng ngn mch ngoi (3)-Ln hn dng t bin t ha ca mba: Ib =Kb .In2Chn ci no c dng ln hn

ICN= ktv .Ib

Dng khi ng ca role bo v ct nhanh c xc nh IkdCN.R= 1

Dngkhi ng thc t ca bo v ct nhanh : IkdCN = .

1T l vng ct nhanh

m% =100

( )

c, bo v chng qu ti

Dng khi ng ca role bo v chn qu ti :

Ikd.R.qt = .1 .1

1 Dng khi ng thc t ca bo v chng qu ti:

Ikd.qt = 1

1

CHNG 12: BO VTHANH CI V NG DY

1. Tnh ton bo vcho ng dy in p U(kV)Dng in chy trn cc on dy nhnh: = 3Dng in chy trn on dy chnh bng tng dng in ca cc on dy nhnh.


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in trca cc on dy: R=l.r0; X=l.x0 = 2 +2in trh thng: = 2 in trngn mch ti thanh ci B tnh t ngun A: = 2 + ( + )2Dng ngn mch trn thanh ci B: (3) = 3

1.1Bo vdng in cc iDng khi ng ca role bo v A:

1 = .1 . .

Chn role s vi dng t ca role l: IdR>IkdR1

Dng khi ng thc t ca bo v: = . 1 nhy ca bo v: nh = .0.87

1.2Bo v ct nhanhDng khi ng ca bo v ct nhanh:

=

1

.

Chn role s vi dng t ca role l: IdRCN>IkdRCN

Dng khi ng thc t ca bo v ct nhanh: = .1 T lvng tc ng ct nhanh: % = 100 (

)

1.3Bo v khong cch- Vng 1 ca bo v Ain trkhi ng vng 1 ca bo v A: Zkd1A=k1(ZAB+0.5RqdAB)

Trong : k1=0.8; RqdAB =28700(+.1)

(3)

Chn bin p h s bin p nu=Un1/Un2

in trkhi ng ca role A: Z1 = Zkd1A . 1


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Trang 17

Chn in trt ca role l ZdR1A

DKBVHTD - Cong Thuc Bao Ve Role

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Transcript of DKBVHTD - Cong Thuc Bao Ve Role