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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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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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Chn in trt ca role l ZdR1A