Role-Dang Xuan Quang-12180119.pdf

5/25/2018 Role-Dang Xuan Quang-12180119.pdf

1/19

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


2/19

Bo v Rle nng cao

1

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


3/19

Bo v Rle nng cao

2

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.


4/19

Bo v Rle nng cao

3

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


5/19

Bo v Rle nng cao

4

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


6/19

Bo v Rle nng cao

5

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:


7/19

Bo v Rle nng cao

6

Xt ti v tr BV2:


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


Xt ti v tr BV1:


8/19

Bo v Rle nng cao

7


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 phi hp gia cc rle:


9/19

Bo v Rle nng cao

8

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:


10/19

Bo v Rle nng cao

9

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


11/19

Bo v Rle nng cao

10

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


12/19

Bo v Rle nng cao

11

'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:


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:


13/19

Bo v Rle nng cao

12

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:


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:


14/19

Bo v Rle nng cao

13

Bi tp 4: Tnh ton bo v khong cch 21 ti A

Bo v vng I:


I

1Z K . 0.8 27.9 65 22.32 65o o

A at LZ x

Bo v vng II:


II

1 22

0.5 0.5Z min ;

k kA L L Tpd L pd TZ xZ xZ


15/19

Bo v Rle nng cao

14

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


16/19

Bo v Rle nng cao

15

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:


17/19

Bo v Rle nng cao

16

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):



18/19

Bo v Rle nng cao

17


3.3Trng hp ngn mch 1 pha chm t (T s MBA bng 1):



3.4Kt lun

Trong trng hp ngn mch bt i xng pha th cp/ti pha th cp mt ixng:


19/19

Bo v Rle nng cao

18

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.

Documents

Role-Dang Xuan Quang-12180119.pdf