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I HC QUC GIA THNH PH H CH MINH
TRNG I HC BCH KHOA
KHOA IN - IN T
-----o0o-----
TIU LUN MN HC
BO V RLE NNG CAO
GVHD: PGS. TS. NGUYN HONG VITHVTH: NG XUN QUANG
MSHV: 12180119
TP. H CH MINH, 5/2014
TP HCM
BK
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Bo v Rle nng cao
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Ni dung
1. Khi nim chung ................................................................................................... 2
1.1 Nhim v ca bo v Rle .................................................................................... 2
1.2 Yu cu c bn ca mch bo v: .......................................................................... 21.2.1 Tnh chn lc .................................................................................................... 2
1.2.2 Tc ng nhanh ................................................................................................. 3
1.2.3 nhy ............................................................................................................ 3
1.2.4 Tnh bo m .................................................................................................... 3
2. Mt s loi Rle bo v ........................................................................................ 3
2.1 Bo v qu dng.................................................................................................... 4
2.1.1 Nguyn l hot ng ......................................................................................... 4
2.1.2 Bi tp p dng ................................................................................................. 4
2.2 Bo v khong cch .............................................................................................. 9
2.2.1 Nguyn l hot ng ......................................................................................... 9
2.2.2 Bi tp p dng ............................................................................................... 10
3. So snh t u dy my bin p tam gic/sao ni t v sao ni t/sao ni t .. 16
3.1 Trng hp ngn mch 3 pha (T s MBA bng 1): ............................................ 16
3.2 Trng hp ngn mch 2 pha (T s MBA bng 1): ............................................ 16
3.3 Trng hp ngn mch 1 pha chm t (T s MBA bng 1): ............................. 17
3.4 Kt lun .............................................................................................................. 17
4. Ti liu tham kho .............................................................................................. 18
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1. Khi nim chung
1.1Nhim v ca bo v Rle
Khi thit k v vn hnh bt k mt h thng in no cn phi k n kh nng phtsinh ngn mch v cc tnh trng lm vic khng bnh thng trong h thng in y.
Ngn mch l loi s c c th xy ra v nguy him nht trong h thng in. Hu quca ngn mch l:
St in p mt phn ln ca h thng in; Ph hy cc phn t b s c bng tia la in; Ph hy cc phn t c dng ngn mch chy qua do tc ng nhit v c; Ph hy n nh ca h thng in.
Ngoi cc loi ngn mch, trong h thng in cn c cc tnh trng vic khng bnh
thng. Mt trong nhng tnh trng vic khng bnh thng l qu ti. Dng in quti lm tng nhit cc phn dn in qu gii hn cho php lm cch in cachng b gi ci hoc i khi b ph hy.
ngn nga s pht sinh s c v s pht trin ca chng c th thc hin cc binphp ct nhanh phn t b ngn mch ra khi mng in, loi tr nhng tnhtrng lm vic khng bnh thng c kh nng gy nguy him cho thit b v h dngin.
m bo s lm vic lin tc ca cc phn khng ngn mch trong h thng in
cn c nhng thit b ghi nhn s pht sinh ca ngn mch vi thi gian b nht, phthin ra phn t b ngn mch v ct phn t b ngn mch ra khi h thng in. Thit
b ny c thc hin nh nhng kh c t ng c tn gi l rle. Thit b bo vc thc hin nh nhng rle c gi l thit b bo v rle (BVRL).
Nh vy nhim v chnh ca thit b BVRL l t ng ct phn t ngn mch ra khih thng in. Ngoi ra thit b BVRL cn ghi nhn v pht hin nhng tnh trng lm
vic khng bnh thng ca cc phn t trong h thng in, ty mc m BVRL cth tc ng i bo tn hiu hoc i ct my ct. Nhng thit b BVRL phn ng vi
tnh trng lm vic khng bnh thng thng thc hin tc ng sau mt thi gianduy tr nht nh (khng cn phi c tnh tc ng nhanh nh cc thit b BVRLchng ngn mch).
1.2Yu cu c bn ca mch bo v:1.2.1 Tnh chn lc
Tc ng ca bo v m bo ch ct phn t b ngn mch ra khi h thng in cgi l tc ng chn lc. Khi c ngun cung cp d tr cho h tiu th, tc ng nh
vy to kh nng cho h tiu th tip tc c cung cp in.
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Yu cu tc ng chn lc cng khng loi tr kh nng bo v tc ng nh l bo vd tr trong trng hp hng hc bo v hoc my ct ca cc phn t ln cn.
Cn phn bit 2 khi nim chn lc:
Chn lc tng i: theo nguyn tc tc ng ca mnh, bo v c th lm vicnh l bo v d tr khi ngn mch phn t ln cn.
Chn lc tuyt i: bo v ch lm vic trong trng hp ngn mch chnhphn t c bo v.
1.2.2 Tc ng nhanh
Cng ct nhanh phn t b ngn mch s cng hn ch c mc ph hoi phn t, cng gim c thi gian trt thp in p cc h tiu th v cng c kh nng
gi c n nh ca h thng in.
gim thi gian ct ngn mch cn phi gim thi gian tc ng ca thit b bo vrle. Tuy nhin trong mt s trng hp thc hin yu cu tc ng nhanh thkhng th tha mn yu cu chn lc. Hai yu cu ny i khi mu thun nhau, v vy
ty iu kin c th cn xem xt k cng hn v 2 yu cu ny.
1.2.3 nhy
Bo v rle cn phi nhy i vi nhng ngn mch v tnh trng lm vickhng bnh thng c th xut hin nhng phn t c bo v trong h thng in.
Thng nhy c c trng bng h s nhy Knh. i vi cc bo v lm victheo cc i lng tng khi ngn mch (v d, theo dng), h s nhy c xc nh
bng t s gia i lng tc ng ti thiu (tc dng ngn mch b nht) khi ngn
mch trc tip cui vng bo v v i lng t (tc dng khi ng).
1.2.4 Tnh bo m
Bo v phi lun lun sn sng khi ng v tc ng mt cch chc chn trong tt ccc trng hp ngn mch trong vng bo v v cc tnh trng lm vic khng bnhthng nh trc.
Mc khc bo v khng c tc ng khi ngn mch ngoi. Nu bo v c nhim vd tr cho cc bo v sau n th khi ngn mch trong vng d tr bo v ny phi khing nhng khng c tc ng khi bo v chnh t gn ch ngn mch hn chatc ng. tng tnh m bo ca bo v cn:
Dng nhng rle cht lng cao. Chn s bo v n gin nht (s lng rle, tip im t). Cc b phn ph (cc ni, dy dn) dng trong s phi chc chn, m bo.
Thng xuyn kim tra s bo v.2. Mt s loi Rle bo v
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2.1Bo v qu dng
2.1.1 Nguyn l hot ng
Bo v dng in cc i l loi bo v phn ng vi dng trong phn t c bo v.Bo v s tc ng khi dng in qua ch t thit b bo v tng qu mt gi tr nhtrc no .
Dng khi ng ca bo v IK, tc l dng nh nht i qua phn t c bo v mc th lm cho bo v khi ng, cn phi ln hn dng ph ti cc i ca phn tc bo v ngn nga vic ct phn t khi khng c ngn mch.
C th m bo kh nng tc ng chn lc ca cc bo v bng 2 phng php khc
nhau v nguyn tc:
Phng php th nht - bo v c thc hin c thi gian lm vic cng ln
khi bo v cng t gn v pha ngun cung cp. Bo v c thc hin nhvy c gi l BV dng in cc i lm vic c thi gian.
Phng php th hai- da vo tnh cht: dng ngn mch i qua ch ni bo vs gim xung khi ngn mch cng cch xa ngun cung cp. Dng khi ngca bo v IK c chn ln hn tr s ln nht ca dng trn on c bov khi xy ra ngn mch on k (cch xa ngun hn). Nh vy bo v c thtc ng chn lc khng thi gian. Chng c gi l bo v dng in ct
nhanh.
Cc bo v dng in cc i lmvic c thi gian chia lm hai loi tng ng vi ctnh thi gian c lp v c tnh thi gian ph thuc c gii hn. Bo v c c tnh
thi gian c lp l loi bo v c thi gian tc ng khng i, khng ph thuc votr s ca dng in qua bo v. Thi gian tc ng ca bo v c c tnh thi gian
ph thuc gii hn, ph thuc vo dng in qua bo v khi bi s ca dng so vidng IK tng i nh v t ph thuc hoc khng ph thuc khi bi s ny ln.
2.1.2 Bi tp p dng Bi tp 1: Tnh ton bo v dng cc i 51
Dng ti cc i:
3
3
9 10I 167.35A
0.9 34.5 3
x
x x
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3
2 3
9x10I I + 325.89A
0.9x34.5x 3
3
1 29 10I I 484.43A
0.9 34.5 3x
x x
Bt u tnh ton cho rle xa ngun nht. Xt ti v tr BV3:
Tr t dng ti rle BV3 l 100% ImCT3, t s CT3=200/5, nh vy dng khi ngrle bn s cp l 200A hay Ik3= 200A.
Chn rle c dc U2:2
5.950.18 .TD
1t
m
Vi: m: Bi s dng ngn mchTD: H s chn thi gian
Chn thi gian ct cho rle BV3 ng vi trng hp ngn mch ti 3,
Inm=Inm3max=2000A l t=0.1s.
Vi m=10, t=0.1s, thay vo phng trnh tnh TD:
2
5.950.18 .TD
1t
m
=>TD=0.42 => Chn: TD=1
ng cong c tnh ca rle BV3 (Ikd=200A, TD=1), kt qu bng sau:
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Xt ti v tr BV2:
Tr t dng ti rle BV2 l 100% ImCT2, t s CT2=400/5, nh vy dng khi ngrle bn s cp l 400A hay Ik2= 400A.
Chn rle c dc U2:2
5.950.18 .TD
1t
m
Vi: m: Bi s dng ngn mch
TD: H s chn thi gian
Thi gian ct cho rle BV2 ng vi trng hp ngn mch ti 2, Inm=Inm2max=3000A lt=t + 0.21s=0.3 + 0.21=0.51s.
Vi m=7.5, t=0.51s, thay vo phng trnh tnh TD:
25.950.18 .TD
1t
m
=>TD=1.77 => Chn: TD=2
ng cong c tnh ca rle BV2 (Ikd=400A, TD=2), kt qu bng sau:
Xt ti v tr BV1:
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Tr t dng ti rle BV1 l 100% ImCT1, t s CT1=600/5, nh vy dng khi ngrle bn s cp l 600A hay Ik1= 600A.
Chn rle c dc U2:2
5.950.18 .TD
1t
m
Vi: m: Bi s dng ngn mch
TD: H s chn thi gian
Thi gian ct cho rle BV1 ng vi trng hp ngn mch ti 2, Inm=Inm1max=5000A lt=t + 0.22s=0.3 + 0.22=0.52s.
Vi m=8.33, t=0.52s, thay vo phng trnh tnh TD:
2
5.950.18 .TD
1t
m
=>TD=1.77 => Chn: TD=2
ng cong c tnh ca rle BV1 (Ikd=600A, TD=2), kt qu bng sau:
ng cong phi hp gia cc rle:
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Kim tra nhy:
33 min
nhay
3
I 1500k 7.5
I 200
BV nm
kd
(t)3
2 minnhay
2
I 1500k 3.75
I 400
BV nm
kd
(t)3
1 minnhay
1
I 1500k 2.5
I 600
BV nm
kd
(t) Bi tp 2: Tnh ton bo v ct nhanh 50
Xt ti v tr BV3:
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Tr dng ct nhanh cho rle BV3: chn Kat=1.25
3 3
maxI K .I 1.25 2000 2500ABV
CN at nm x
Dng ngn mch ch min sau my ct BV3 c th xem bng gi tr ngn mch
min ti thanh ci s 2:
2 3
minI 2250A
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thi gian ny t ng tng ln khi khong cch t ch ni bo v n im ngn mchtng ln. Bo v t gn ch ngn mch nht c thi gian lm vic b nht.
L quan h gia thi gian tc ng ca bo v vi khong cch hay tng tr n v tr
ngn mch.
2.2.2 Bi tp p dng Bi tp 3: Tnh ton bo v khong cch 21 ti ABC
H s qui i tng tr v pha th cp:
n 1000 / 1k 0.4783
n 230000 /110CT
qd
VT
Tng tr ng dy L1:
1
1 1 1Z . 100.(0.089 0.476) 48.42 79.41o
L L Z j 0
1 1 0Z . 100.(0.426 1.576) 163.26 74.87o
L L Z j Tng tr ca ng dy L2:
1
2 2 1Z . 60.(0.089 0.476) 29.05 79.41o
L L Z j 0
2 2 0Z . 60.(0.426 1.576) 97.95 74.87
o
L L Z j
Tng tr ti:
2 2U 230Z 26.45 36.87
2000 0.8 2000 0.6
o
tSL x i x
Tng tr cc ng dy v ti qui v pha th cp:
'1 1
1 1Z k . 0.4783 48.42 79.41 23.16 79.41o o
L qd LZ x
'0 0
1 1Z k . 0.4783 163.26 74.87 78.09 74.87o o
L qd LZ x
'1 1
2 2Z k . 0.4783 29.05 79.41 13.89 79.41o o
L qd LZ x
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'0 0
2 2Z k . 0.4783 97.95 74.87 46.85 74.87o o
L qd LZ x
'Z k 0.4783 26.45 36.87 12.65 36.87o ot qd t xZ x
Tnh bo v khong cch 3 cp ti ABC:
Bo v vng I:
Chng ngn mch nhiu pha, gi tr khi ng l:
I '1
1Z K . 0.8 23.16 79.41 18.53 79.41o o
ABC at LZ x
Chng chm t, gi tr khi ng l:
IG IZ Z 18.53 79.41oABC ABC
H s b th t khng:
'0 '11 1
'1
1
78.09 74.87 23.16 79.41k 0.79 6.453 3 23.16 79.41
o o oL Lbu o
L
Z ZZ x
Bo v vng II:
Chng ngn mch nhiu pha, gi tr khi ng l:
II '1 '1
1 2Z 0.5 23.16 79.41 0.5 13.89 79.41 30.11 79.41o o o
ABC L LZ xZ x
Kim tra iu khin nhy:
II
'1
1
Z 30.11k 1.3 1.2
23.16
ABCnhay
LZ
(t)Chng chm t, gi tr khi ng l:
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IIG IIZ Z 30.11 79.41oABC ABC H s b th t khng:
'0 '1
1 1
'11
78.09 74.87 23.16 79.41k 0.79 6.45
3 3 23.16 79.41
o ooL L
bu oL
Z Z
Z x
Bo v vng III:
Chng ngn mch nhiu pha, gi tr khi ng l:
III '1 '1
1 2Z 1.2 ( ) 1.2 (23.16 79.41 13.89 79.41 ) 44.46 79.41o o o
ABC L Lx Z Z x
Chng chm t, gi tr khi ng l:
IIIG IIIZ Z 44.46 79.41oABC ABC
H s b th t khng:
'0 '1
1 1
'1
1
78.09 74.87 23.16 79.41k 0.79 6.45
3 3 23.16 79.41
o ooL L
bu o
L
Z Z
Z x
Chn c tnh bo v ti ABC:
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Bi tp 4: Tnh ton bo v khong cch 21 ti A
Bo v vng I:
Chng ngn mch nhiu pha, gi tr khi ng l:
I
1Z K . 0.8 27.9 65 22.32 65o o
A at LZ x
Bo v vng II:
Chng ngn mch nhiu pha, gi tr khi ng l:
II
1 22
0.5 0.5Z min ;
k kA L L Tpd L pd TZ xZ xZ
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H s phn dng khi ngn mch cui L2:
Trng hp ch cc i ti A, cc i ti B:
12
2
I 907.66k 0.371
I 2435.14
Lpd L
L
Trng hp ch cc i ti A, cc tiu ti B:
12
2
I 675.6k 0.434
I 1556.96L
pd L
L
Trng hp ch cc tiu ti A, cc i ti B:
12
2
I 862.36k 0.358
I 2411.12L
pd L
L
Trng hp ch cc tiu ti A, cc tiu ti B:
12
2
I 933.43k 0.42
I 2222.13L
pd L
L
Chn trng hp kpd=0.358 (nh nht) tnh gi tr khi ng cho trng hp ngnmch cui ng dy L2:
II
1 2
0.5 0.5Z 27.9 65 13.5 65 46.75 65
k 0.358
o o o
A L L
pd
Z Z x
H s phn dng khi ngn mch sau my bin p T:
Tng t trn, chn trng hp ch cc tiu ti A, cc i ti B:
1I 598.91
k 0.349I 1714.59L
pd T
T
Tnh gi tr khi ng cho trng hp ngn mch cui MBA T:
II
1
0.5 0.5Z 27.9 65 24.5 65 63 65
k 0.349
o o o
A L T
pd
Z Z x
Vy gi tr khi ng vng II l:
II
1 2
0.5 0.5Z 27.9 65 13.5 65 46.75 65
k 0.358
o o o
A L L
pd
Z Z x
Kim tra nhy:
II
1
46.75 65k 1.68 1.2
27.9 65
o
Anhay o
L
Z
Z
(t)
Bo v vng III:
12
2
Ik
IL
pd L
L
1I
kIL
pd T
T
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Chng ngn mch nhiu pha, so snh cc gi tr:
III1 1 22
1 1Z 1.2 ( max ; ) max 65.6 65 ;98.1 65 98.1 65
k k
o o o
A L L T
pd L pd T
x Z xZ xZ
V gi tr tng tr khi lm vic vi ti cc i:
III min2
max dd
U 0.9 110000Z 83.3
3 K K K xI cos( ) 3 1.2 1.5 1.1 400 cos(65 35 )
lvA o o
at mm tv lv lv
x
x x x x x x x x x
Vi: Kat=1.2: H s an ton;Kmm=1.5: H s m my trng hp c ng c;
Ktv=1.05-1.1: H s tr v.
Vy chn gi tr khi ng vng III l:III III
2Z 83.3A AZ
c tnh bo v ti A:
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3.So snh t u dy my bin p tam gic/sao ni t v sao ni t/sao ni t
Phn b dng trong MBA hai cun dy khi c ngn mch cc dng khc nhau pha th
cp MBA:
3.1Trng hp ngn mch 3 pha (T s MBA bng 1):
MBA vi t y dy tam gic/ sao ni t
MBA vi t y dy sao ni t/ sao ni t
3.2Trng hp ngn mch 2 pha (T s MBA bng 1):
MBA vi t y dy tam gic/ sao ni t
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MBA vi t y dy sao ni t/ sao ni t
3.3Trng hp ngn mch 1 pha chm t (T s MBA bng 1):
MBA vi t y dy tam gic/ sao ni t
MBA vi t y dy sao ni t/ sao ni t
3.4Kt lun
Trong trng hp ngn mch bt i xng pha th cp/ti pha th cp mt ixng:
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MBA c t u dy tam gic/sao ni t khng xut hin dng th t khngtrn li pha s cp.
MBA c t u dy sao ni t/sao ni t xut hin dng th t khng trn
li pha s cp.
Cc my bin p thng c X1= X2, cn Xoph thuc vo t ni dy. T ni dy ch cho dng th th khng chy qun trong cun dy m khng ra ngoi li in.T ni dy Y cho dng th th khng i qua cun dy ch khi trung tnh ni t vs chy ra li in.
Hai my bin p vi cng cng sut, t l my bin p th dng lm vic cng nhdng ngn mch qua cun dy pha s cp ca MBA c t u dy tam gic/ sao
ni t lun thp MBA c t y dy sao ni t/ sao ni t (Xt trng hp ngnmch pha th cp).
4. Ti liu tham kho[1] PGS. TS. Nguyn Hong Vit, Cc bi ton tnh Ngn mch v Bo v Rle trong
H thng in,Nh xut bn i hc Quc gia TP HCM, 2012.
[2] PGS. TS. Nguyn Hong Vit, Bo v rle v T ng ha trong h thng in,Nh xut bn i hc Quc gia TP HCM, 2012.
[3] PGS. TS. Nguyn Kim Hng, Bo v rle v T ng ha,Nh xut bn ihc Nng, 2008.
[4] Ngun ti liu trn internet.