Quy phạm trang bị điện

Cu 66: Theo quy phm trang b in, th no l trng thi cn bng ca

li in nhiu pha v trng thi khng cn bng ca mt li in

nhiu pha?

Trng thi cn bng ca li in nhiu pha (Balanced state of a polyphase

network): Trng thi trong in p v dng in trong cc dy dn pha to

thnh cc h thng nhiu pha cn bng.

Trng thi khng cn bng ca mt li in nhiu pha (Unbalanced state of a

polyphase network): Trng thi m trong in p v/hoc dng in trong cc

dy dn pha khng to thnh cc h thng nhiu pha cn bng.

Cu 67: Theo quy phm trang b in, th no l tin cy cung cp

in v an ton cung cp in?

tin cy cung cp in (Service reliability): Kh nng ca mt h thng in

p ng c chc nng cung cp in trong nhng iu kin n nh, theo thi

gian quy nh.

an ton cung cp in (Service security): Kh nng hon thnh chc nng cung

cp in ca h thng in ti mt thi im cho trong vn hnh khi xut hin

s c.

Cu 68: Theo quy phm trang b in, ngi ta quy nh nh th no v

mu sn cc thanh dn ca li in 3 pha 4 dy, li in 1 pha r

nhnh t h thng 3 pha, li in 1 chiu?

Mu sn thanh dn cng tn mi cng trnh in phi ging nhau.

Thanh dn phi sn mu nh sau:

1. i vi li in xoay chiu ba pha 4 dy: pha A mu vng, pha B mu

xanh l cy, pha C mu , thanh trung tnh mu trng cho li trung tnh cch ly,

thanh trung tnh mu en cho li trung tnh ni t trc tip.

2. i vi in mt pha: dy dn ni vi im u cun dy ca ngun

in mu vng, dy ni vi im cui cun dy ca ngun mu . Nu thanh dn

ca li in mt pha r nhnh t thanh dn ca h thng ba pha th phi sn theo

mu cc pha trong li ba pha.

3. i vi li in mt chiu: thanh dng (+) mu , thanh m (-) mu

xanh, thanh trung tnh mu trng.


b tr v sn mu thanh dn i vi thit b phn phi trong nh , in

xoay chiu 3 pha?

i vi thit b phn phi trong nh, in xoay chiu ba pha:

Khi thanh ci b tr thng ng: thanh trn (A) mu vng; thanh gia (B) mu

xanh l cy; thanh di (C) mu . Khi cc thanh b tr nm ngang, nm nghing hoc

theo hnh tam gic: thanh xa ngi nht (A) mu vng; thanh gia (B) mu xanh l cy;

thanh gn ngi nht (C) mu . Trng hp ngi c th tip cn c t hai pha th

thanh pha gn hng ro hoc tng ro (A) mu vng, thanh xa hng ro hoc tng ro

(C) mu .

Cc thanh r nhnh t thanh ci: nu nhn t hnh lang vn hnh, thanh tri (A)

mu vng, thanh gia (B) mu xanh l cy v thanh phi (C) mu .


b tr v sn mu thanh dn i vi thit b phn phi ngoi tri , in

xoay chiu 3 pha?

i vi thit b phn phi ngoi tri, in xoay chiu ba pha:

Thanh ci v thanh ng vng: thanh gn my bin p in lc nht (A) mu

vng, thanh gia (B) mu xanh l cy, thanh xa nht (C) mu .

Cc thanh r nhnh t h thng thanh ci: nu nhn t thit b phn phi ngoi tri

vo cc u ra ca my bin p in lc, thanh tri (A) mu vng, thanh gia (B) mu

xanh l cy, thanh phi (C) mu .

ng dy vo trm: nu nhn t ng dy vo trm, ti v tr u ni, thanh tri

(A) mu vng, thanh gia (B) mu xanh l cy, thanh phi (C) mu .

Thit b phn phi ngoi tri dng dy dn mm lm thanh ci th sn mu pha

chn s ca thit b hoc chm sn x mc thanh ci.


b tr v sn mu thanh dn i vi in mt chiu?

i vi in mt chiu:

Khi thanh ci b tr thng ng: thanh trn (thanh trung tnh) mu trng; thanh

gia (-) mu xanh; thanh di (+) mu .

Khi thanh ci b tr nm ngang: nu nhn t hnh lang vn hnh, thanh trung tnh

xa nht mu trng, thanh gia (-) mu xanh, thanh gn nht (+) mu .

Cc thanh r nhnh t thanh ci: nu nhn t pha hnh lang vn hnh, thanh tri

(thanh trung tnh) mu trng, thanh gia (-) mu xanh, thanh phi (+) mu .

Trng hp c bit, nu thc hin nh trn m gp kh khn v lp t hoc phi

xy thm tr gn cc thanh ci ca trm bin p lm nhim v o pha th cho php

thay i th t mu ca cc thanh.

Cu 72: Theo quy phm trang b in, ngi ta quy nh trong cng

trnh in phi c cc bin php m bo an ton no?

Trong cng trnh in phi c cc bin php m bo an ton sau:

Dng loi cch in thch hp. Trng hp c bit phi dng cch in tng

cng.

B tr c ly thch hp n phn dn in hoc bc kn phn dn in.

Lm ro chn.

Dng kho lin ng cho kh c in v cho ro chn ngn nga thao tc nhm.

Ct t ng tin cy v nhanh chng cch ly nhng phn thit b in b chm chp

v nhng khu vc li in b h hng.

Ni t v thit b in v mi phn t ca cng trnh in c th b chm in.

San bng th in, dng my bin p cch ly hoc dng in p 42V tr xung.

Dng h thng bo tn hiu, bin bo v bng cm.

Dng trang b phng h.

Cu 73: Theo quy phm trang b in, cc ro ngn v tm che phn

mang in trong nh , nh cng cng, ca hng v trong gian sn xut,

gian in phi p ng nhng yu cu no?

Trong nh , nh cng cng, ca hng v.v. v hoc tm che phn mang in

khng c c l. Trong gian sn xut v gian in c php dng v hoc tm che c

l hoc kiu li.

Ro ngn v tm che phi c kt cu sao cho ch tho hoc m bng c l hoc

dng c ring.

Ro ngn v tm che phi c bn c hc. i vi thit b trn 1kV, chiu

dy ca tm che bng kim loi khng c nh hn 1mm. V che dy dn nn a su

vo trong my, thit b v dng c in.

Cu 74: Theo quy phm trang b in. ngi ta quy nh cc cng trnh

in u ni vo h thng in phi p ng nhng iu kin k thut

no?

Khi cng trnh in cn u vo h thng in, ngoi nhng th tc xy dng c

bn c quy nh cn phi c s tho thun ca c quan qun l h thng in, phi

tun theo cc vn bn php quy hin hnh v nhng iu kin k thut u ni nh

sau:

1. Lp phng n xy dng cng trnh trong h thng in.

2. Tng hp s liu ph ti in trong khu vc s xy dng cng trnh.

3. D kin im u vo h thng in (trm in, nh my in hoc ng dy

dn in), cp in p nhng im u, trang b ti im u ni.

4. Chn in p, tit din v chng loi ca ng dy trn khng hoc ng cp

v phng tin iu chnh in p, nu nhng yu cu v tuyn ng dy. i vi cng

trnh ln cn phi nu thm phng n chn s mch u.

5. Nu yu cu v s cn thit phi tng cng li in hin c do u thm cng

trnh mi (tng tit din dy dn, thay th hoc tng cng sut my bin p).

6. Nu nhng yu cu ring i vi cc trm in v thit b ca h tiu th in

c u vo h thng nh: cn c bo v t ng cc u vo, cho php cc ng

dy lm vic song song, cn c cc ngn in d phng v.v.

7. Xc nh dng in ngn mch tnh ton.

8. Nu nhng yu cu v bo v rle, t ng, cch in, bo v chng qu in

p.

9. Nu cc bin php nng cao h s cng sut.

10. Nu cc yu cu v o m in nng.

11. Xc nh nhng iu kin u trang b in c l in, thit b in cao tn..

12. Nu nhng yu cu i vi cc cng trnh ph tr v cc cng trnh khc (nh

thng tin lin lc v.v.).

Cu 75: Theo quy phm trang b in, th no l h thng nng lng,

h thng in?

H thng nng lng l tp hp cc nh my in, li in v li nhit c ni

vi nhau, c lin h mt thit, lin tc trong qu trnh sn xut, bin i v phn phi

in v nhit.

H thng in l h thng nng lng khng c li nhit.

Cu 76: Theo quy phm trang b in, th no l trm in, trm bin

p, trm ct, trm b cng sut phn khng v trm cch in kh?

Trm in l mt phn t ca h thng in, c th l trm pht in, trm bin p,

trm ct hoc trm b cng sut phn khng v.v.

Trm bin p l trm c cc my bin p lc kt ni hai hoc nhiu li in c

in p khc nhau.

Trm ct l trm gm thit b ng ct, cc thanh ci, khng c my bin p lc.

Trm b cng sut phn khng gm hai loi:

1. Trm b cng sut phn khng bng t in.

2. Trm b cng sut phn khng bng my b ng b.

Trm cch in kh (Gas insulated substation - GIS): Trm gm cc thit b in

c bc kn, c cch in bng cht kh (khng phi l khng kh).

Cu 77: Theo quy phm trang b in, ngi ta quy nh nhng li

in cp in p no c trung tnh ni t trc tip, trung tnh cch ly,

trung tnh ni t qua cun dp h quang hoc trung tnh ni t qua

in tr nh?

Li 500, 220, 110kV l loi trung tnh ni t trc tip.

Li 6, 10, 35kV l loi trung tnh cch ly c th ni t qua cun dp h quang

in, trong trng hp c bit c th ni t trc tip.

Li 15, 22kV l loi trung tnh ni trc tip, trong trng hp c bit c th

trung tnh cch ly hoc ni t qua in tr nh.

i vi li in 6 35kV c im trung tnh ni t qua cun dp h quang th

vic b dng in dung khi c chm t c thc hin trong cc trng hp sau:

1. li in 35 kV: khi dng in chm t ln hn 10A.



4. t hp khi my pht in - my bin p 6 22 kV: khi dng in chm t

ln hn 5A.

Cu 78: Trnh by tnh trng lm vic ca im trung tnh cch in

trong mng in 3 pha?

Trung tnh ni t trc tip l im trung tnh ca my bin p hoc ca my pht

in c ni trc tip vi trang b ni t hoc c ni vi t qua mt in tr nh

(th d nh my bin dng v.v.).

Trung tnh cch ly l im trung tnh ca my bin p hoc ca my pht in

khng c ni vi trang b ni t hoc c ni vi trang b ni t qua cc thit b tn

hiu, o lng, bo v, cun dp h quang c ni t hoc thit b tng t khc c

in tr ln.

Trung tnh ni t hiu qu l trung tnh ca mng in ba pha in p ln hn

1kV c h s qu in p khi ngn mch chm t khng ln hn 1,4.

H s qu in p khi ngn mch chm t trong mng in ba pha l t s gia

in p ca pha khng b s c khi c ngn mch chm t v in p pha trc khi

c ngn mch chm t.

Cu 79: Trnh by tnh trng lm vic ca im trung tnh trong mng

in 3 pha ni t qua cun dp tt h quang?

Ni t qua cun dp h quang ( cun Petersen) hay cn gi l ni t cng

hng:

Z = j.X

X=1/C

in khng ca cun dp h quang c la chn b dng in dung khin cho dng

ny trong gii hn cho php d cho di li rt ln.

u im:

Dp nhanh h quang khi chm t 1 pha, dng chm t rt nh, c khi trit

tiu hon ton.

st p khi chm t 1 pha nh.

Hn ch nh hng n ng dy in thoi.

Nhc im:

Khi chm t, cc pha lnh phi chu in p dy.

S c cch in c th gy h quang dao ng gy qu in p trn cch in

cc pha lnh.

Cun h quang phi iu chnh c thch nghi vi cu trc vn hnh thay

i ca li.

S bo v s c chm t phc tp v kh tm ch s c

Gi thnh cao, i hi bo qun.

i vi li cp bin php ny khng mang li li ch v trong li cp i a

s cc s c l do h hng vnh vin cch in.

Cu 80: Trnh by tnh trng lm vic ca im trung tnh trong mng

in 3 pha trc tip ni t?

Mng in trung tnh trc tip ni t (Z = 0):

Khi xy ra chm t mt pha s gy ngn mch v li in phi c ct ra, iu

ny cho php gim mc cch in trn ng dy v cp, ch c th p dng mc cch

in theo in p pha thay v in p dy nh trung tnh cch t do gi thnh li

in h.

Dng in ngn mch 1 pha c th rt ln gy tc hi i vi thit b trm v

ng dy, lm tng gi ha MBA ngun v cp, gy in p cm ng ln n ng

dy bn cnh v ng dy in thoi.

Khi chm t li b ct in lm gim tin cy cung cp in.

Li 15 20 kV nu cc tc hi ni trn c hn ch n mc cho php.

Cu 81: Trnh by cu to v nguyn l lm vic v ng dng ca MBA

lc 3 pha 2 dy qun?

Cu to: Ngoi v (thn) ca my, My bin p gm hai b phn chnh: Li thp

v bi dy.

1- Li thp My bin p:

Dng dn t thng chnh ca my, c ch to t vt liu dn t tt, thng l

thp k thut in mng ghp li. gim dng in xoy trong li thp, ngi ta dng

l thp k thut in, hai mt c sn cch in ghp li vi nhau thnh li thp.

2- Bi dy ca My bin p:

Bi dy c ch to bng dy ng hoc nhm c tit din trn hoc ch nht,

bn ngoi dy dn c bc cch in (bng giy hoc may). My bin p c cng sut

nh th lm mt bng khng kh. My c cng sut ln th lm mt bng du, v thng c

cnh tn nhit.

- i vi dy nhm c u vit cho nh ch to c bn l gi thnh r, nhng kh

thi cng nht l khi hn ni, kh sa cha, d s c, tn hao cao.

- i vi dy ng th ngc li vi dy nhm.

Cc hng nc ngoi ngi ta dng dy nhm v dy ng, ty theo kt cu loi

My bin p, lc ny kt cu bi dy v mi hn rt vng chc. Cc hng trong nc c

thng hiu ngi ta dng dy ng m khng s dng dy nhm v uy tn v thng

hiu.

Nguyn l lm vic:

Khi ta ni dy qun s cp vo ngun in xoay chiu in p U1 s c dng in

s cp I1 (hnh 1).

Dng in I1 sinh ra t thng fi bin thin chy trong li thp. T thng ny mc

vng ng thi vi c hai dy qun s cp v th cp c gi l t thng chnh.

Theo nh lut cm ng in t:

e1 = - W1 dfi/dt; e2 = - W2 dfi/dt

W1, W2 l s vng dy qun s cp v th cp.

Hnh 1

Khi My bin p c ti, di tc ng ca sc in ng e2, c dng in th cp

I2 cung cp in cho ti.

T thng fi bin thin hnh sin fi = fiMax sinWt. Ta c:

k = E1/ E2= W1/ W2 , k c gi l h s bin p.

B qua in tr dy qun v t thng tn ra ngoi khng kh ta c:

U1/ U2 xp x E1/ E2 = W1/ W2 = k

B qua mi tn hao trong My bin p, ta c:

U2 I2 xp x U1 I1 suy ra U1/U2 xp x I2/I1 = W1/W2 = k

ng dng:

My bin p c th chuyn i hiu in th (in p) ng vi gi tr mong

mun, v d t ng dy trung th 10 kV sang mc h th 220 V hay 400 V dng trong

sinh hot dn c. Ti cc nh my pht in, my bin p thng chuyn hiu in th

mc trung th t my pht in (10 kV n 50 kV) sang mc cao th (110 kV n 500

kV hay cao hn) trc khi truyn ti ln ng dy in cao th. Trong truyn ti in

nng vi khong cch xa, hiu in th cng cao th hao ht cng t.

Cu 82: Nu iu kin ha 2 my bin p lm vic song song, phn tch

mt trong cc iu kin y?

iu kin cho cc My bin p lm vic song song:

in p nh mc s cp v th cp ca cc my phi bng nhau tng ng.

Cc my phi c cng t ni dy.

in p ngn mch ca cc my phi bng nhau: UnI% = UnII% =.....UnN%

Phn tch iu kin in p ngn mch ca cc my phi bng nhau:

UN kt Phn b ti

Bng nhau T l theo cng sut

Khc nhau Ln MBA cng sut ln Nh My non ti

Nh MBA cng sut nh Ln My qu ti

Theo quy nh, UN% khng khc nhau qu 10%

Cu 83: Trnh by cu to, nguyn l lm vic v ng dng ca my bin

p o lng?

My bin in p c dy qun s cp ni vi li in v dy qun th cp ni vi

Volt mt hay vi cun dy song song ca Watt mt hoc vi cun dy ca role bo v.

Tng tr Z ca loi my ny rt ln nn my bin p lm vic trng thi gn nh khng

ti, in p ri trong my rt nh, do sai s v tr s U% v v gc u gia U1 v U2

u nh.

Ch khi s dng my bin in p khng c ni tt mch th cp, v nh th

s tng ng vi ni tt mch s cp v dn n gy ra s c ngn mch li in.

Cu 84: Trnh by cu to, nguyn l lm vic v ng dng ca my bin

dng in?

My bin dng in c dy qun s cp v ni ni tip vi mch cn o dng

in, dy qun th cp gm nhiu vng dy c ni vi Ampe mt hoc cc cun dy

ni tip ca Watt mt hay rle bo v.

Tng tr Z ca nhng dng c ny rt nh v trng thi lm vic ca my bin

dng l trng thi ngn mch, li thp khng bo ho ( = 0,8 1 wb) v I0 0, do

cc sai s o lng v tr s:

v sai s v gc i cng nh.

S kt ni v th vecto ca my bin dng in.

Ch khi s dng my bin dng khng c dy qun th cp h mch v

nh vy I0 = I1 rt ln, li thp bo ho nghim trng ( = 1,4 1,8 wb) s nng v lm

chy dy qun. Hn na khi bo ho s lm cho sc in ng tng vt n in p u

th cp ln rt cao khng an ton cho ngi s dng.

m bo an ton dy qun th cp c ni t mt u.

Cu 85: Nu tc dng ca cc thit b ng ct: aptomat, cng tc t,

khi ng t, chng c s dng vo u trong li in?

Tc dng ca cc thit b ng ct:

Aptomat: l mt loi kh c in ng ct v bo v chnh trong mch in

h p. N c s dng ng ct t xa v t ng ct mch khi thit b in hoc

ng dy pha sau n b ngn mch hoc qu ti, qu p, km p, chm t

Cng tc t: l loi kh c in h p c s dng iu khin ng ct

mch t xa t ng hoc bng nt n cc mch in lc c ph ti in p n 500V,

dng in n 600A.

Khi ng t: dng iu khin t xa vic ng ct, o chiu v bo v

qu ti cho cc ng c roto lng sc.

Ni t chng trong li in:

Thit b bo v phi t ni thun tin cho vn hnh, bo dng trnh b

h hng do c hc. Vic t chng phi m bo khi vn hnh hoc khi tc ng khng

gy nguy him cho ngi v khng gy h hng cc vt xung quanh.

Vic vn hnh v bo dng thit b bo v c phn dn in h phi do

ngi c chuyn mn m nhim.

Cn t thit b bo v ti cc v tr trong mch in m tit din dy

dn gim nh (v pha ph ti in) hoc ti cc v tr cn m bo nhy v tnh chn

lc.

Phi t thit b bo v ngay ti ch u phn t c bo v vi ng

dy cung cp. Khi cn thit, cho php chiu di ca on dy r nhnh gia thit b bo v

v ng dy cung cp n 6m. Tit din ca on dy ny c th nh hn tit din ca

ng dy cung cp nhng khng nh hn tit din ca dy dn sau thit b bo v.

Cu 86: Trnh by cu to, nguyn l lm vic ca my ct in du?

Du lm nhim v g trong my ct nhiu du v my ct t du?

Cu to:

1. Tip xc ng

2. Tip xc tnh

3. X xuyn

4. V my ct

5. H quang

6. L xo tch nng

7. Cc bt dy ngun t my ct

8. Cc bt dy ti ra

9. Thanh truyn ng

Nguyn l lm vic:

Nu my ct ang v tr ng, tip xc ng ng cht vo tip xc tnh, l xo

tch nng trng thi nn, n tn hiu ch mu , dng in t ngun bt qua cc bt

dy v ti.

Khi c tn hiu ct t role hoc t kho iu khin th b truyn ng c gii

phng khi v tr ng, l xo tch nng y thanh truyn ng sp xung, tip xc ng

ri khi tip xc tnh, mch in c ct, h quang pht sinh gia 2 u tip xc ng

v tnh khi ct t nng cc b lm du b phn tch thnh hi l hn hp kh cacbon

hydro nh, trong hydro chim ti 70%, p sut c th t ti (100-140 N/cm2 ) lm

cho du b xo trn mnh, y tia h quang vo su trong du. Mt khc lc in t do

dng in chy ngc chiu cng lm tia h quang chy su trong lp du ngoi. V vy

h quang b lm ngui v dp tt, tuy vy tc lung kh khng mnh dp tt h

quang nhanh, nn loi my ct ny thng b ko di.

Du c nhim v:

My ct nhiu du: va l cht cch in ng thi sinh khi dp tt h

quang.

My ct t du: Lng du t ch sinh kh dp tt h quang cn cch in

l cht rn.

Cu 87: Trnh by cu to v nguyn l lm vic v ng dng ca chng

st ng?

Cu to:

ng cch in c kh nng sinh cht kh khi tip xc vi h quang in

(viniplash, phibro bakelit)

u trn ng bt kn bng in cc kim loi, cch dy dn khong cch Sn.

u di ng h (in cc xuyn) v ni t.

Nguyn l lm vic:

Sn; St l khe h bn ngoi v bn trong chng st ng.

Khi c qu in p truyn ti ch t chng st ng, c bin ln hn kh

nng cch in ca (Sn + St), gy phng in cp (Sn + St) tn qu in p vo t.

Dng in xung gy ra in p d t ln cch in nn chng st ng phi

ni t tt.

Dng in k tc xut hin cng dng in xung, t nng thnh ng, sinh

lng kh p sut ln thi qua u h ca ng v dp tt h quang khi dng in qua tr

s 0.

Chng st ng c gii hn trn v gii hn di cho dng in k tc c th

dp tt: gii hn di: gii hn kh ln dp h quang; gii hn trn: bn ca

ng.

Khe h Sn: iu chnh c theo iu kin phi hp cch in ca thit b

cn bo v. Khe h St khng iu chnh c, chn theo iu kin dp h quang.

ng dng:

H tr bo v cho TBA cng sut nh, t quan trng bng cch t trn

khong vt trc trm.

Bo v DK khng treo dy chng st.

Bo v cc im cch in yu trong li.

Cu 88: Trnh by cu to v nguyn l lm vic v ng dng ca chng

st van?

Cu to ca chng st van gm hai phn:

Bn ngoi l mt ng s hay cht do cch in c hnh dng v kch thc ty

thuc cp in p nh mc s dng.

Bn trong ng cha hai phn t chnh l khe h phng in v in tr phi

tuyn.

Khe h phng in bao gm nhiu cp khe h ghp ni tip. Mi cp khe h

c ch to bi 2 a ng mng dp nh hnh. gia l mt tm m mica hoc ba

cch in dy khong 1mm to khe h phng in. Mi chng st van c s cp khe

h phng in ty theo nh ch to thit k.

in tr phi tuyn gm cc tm hnh tr trn ghp ni tip. in tr phi tuyn

c th l Vilit hoc Tirit hoc ZnO... (thng l Vilit).

Nguyn l lm vic:

Ch yu ph thuc vo tnh cht ca in tr Vilit. Khi in p t ln Vilit tng

cao th gi tr in tr ca n gim v ngc li khi in p gim xung th in tr s

tng ln nhanh chng.

Khi c qu in p t ln chng st van, in tr ca chng st van nhanh chng

h thp xung to iu kin tho ht sng st qua n xung t, n khi t ln chng

st van ch cn l in p mng th in tr ca chng st van li tng ln rt ln chm

dt dng k tc vo thi im thch hp nht.

ng thi trong khi tho st, in p d trn chng st van cng c gi tr nh,

iu ny s m bo an ton cho thit b c bo v.

Chng st van c c tnh tc ng tng i bng phng nn chng st van

khng nhng c tc dng h thp bin m cn lm gim dc ca sng st. V th,

n c th bo v chng c hin tng xuyn kch gia cc vng dy trong cng mt

pha ca cc my in.

in tr Vilit d b nhim m s thay i c tnh in v lm mt tc dng ca

chng st van, do cn c bin php chng nhim m cho in tr Vilit.

ng dng: thng dng bo v trm bin p; dng bo v my in.

Cu 89: Trnh by cc loi chng st cho ng dy ti in trn khng?

Tc dng ca tng loi?

DK in p 110kV tr ln phi c bo v khi st nh trc tip bng dy

chng st trn sut chiu di ng dy, tr mt s on tuyn c bit khng th b tr

c dy chng st. on tuyn ny phi c bin php chng st khc b sung.

DK in p t 22kV tr xung khng yu cu bo v khi st nh bng dy chng

st trn sut chiu di.

DK in p 35kV khng phi bo v bng dy chng st (tr on vo trm

35/0,4kV) nhng cc ct phi ni t ng vi yu cu.

on DK i vo trm bin p phi c bo v trnh qu in p kh quyn ph

hp vi yu cu bo v trm.

Cu 90: Trnh by cc loi chng st cho trm bin p? Tc dng ca

tng loi?

TBA v TBPP ngoi tri in p 22 - 500kV phi -c bo v chng st nh trc

tip. Khng cn bo v chng st nh trc tip i vi TBA in p 22 - 35kV ngoi tri

c MBA cng sut mi my n 1600kVA v khng ph thuc vo s gi st trong nm.

Nh t TBPP v TBA nn -c bo v chng st nh trc tip. Mi cc nh t

TBPP v TBA bng kim loi phi -c ni t.

i vi h thng x l du, trm my b ng b, nh in phn, kho cha cc bnh

kh hydro, b tr trong khu vc TBA vic bo v chng st nh trc tip phi thc hin theo

tiu chun chng st cho cc cng trnh xy dng hin hnh.

Bo v chng st nh trc tip vo TBA v TBPP ngoi tri -c dng kim thu st

b tr trn cc kt cu xy dng hoc dy thu st. C th s dng cc ct cao (ct DK,

ct lp n pha v.v.) lm ct thu st. Cho php b tr cc kim chng st trn ct cng gn

MBA hoc in khng phn mch.

Ct cng MBA, ct cng ca in khng phn mch v kt cu ca TBPP ngoi

tri cch xa MBA hoc in khng theo mch ni t chung nh hn 15m th c th lp

kim thu st khi in tr sut t-ng -ng ca t vo ma st nh hn 350m v tun

theo cc iu kin sau:

1. B tr chng st van (CSV) ngay trn cc u ra ca cun dy MBA 6 - 35kV hoc

cch cc u ra khng qu 5m theo chiu di dy dn.

2. Phi m bo ni t t ct t kim thu st n mch ni t chung bng 3 - 4 tia.

3. Trn mch ni t chung cch ct c kim thu st 3 - 5m phi ng 2 - 3 cc ni t di

t 3 n 5m.

4. cc TBA n 35kV c b tr kim thu st trn ct cng MBA, in tr ca trang b ni

t khng -c ln hn 4 (khng tnh n cc b phn ni t bn ngoi mch vng ni

t ca TBPP ngoi tri).

5. Dy ni t ca CSV v MBA -c u vo mch ni t nn b tr sao cho im ni

t ca CSV nm gia im ni t ca dy ni t ct cng c kim thu st v im ni

t ca MBA.

Cu 91 : So snh u nhc im ca s cu trong v cu ngoi ? ng

dng ?

S cu trong c s dng khi hai ng dy lm vic song song, c chiu di

ln (thng l 70 km), hay c s c trn ng dy nhng li t phi ng m cc mch

my bin p.

S cu ngoi c s dng khi ng dy c chiu di ngn (l 70 km), t s

c trn ng dy nhng li thng xuyn ng ct cc my bin p khi ph ti cc tiu

gim tn tht in nng trong cc my bin p.

Cu 92 : Nu cc c cu o c bn ? Phn tch mt trong cc c cu ?

Cc c cu o c bn l :

o dng in.

o in p v kim tra cch in.

o cng sut.

o tn s.

o lng khi ha ng b.

Phn tch o cng sut :

Phi o cng sut theo cc yu cu sau:

1. o cng sut tc dng v cng sut phn khng cho tng my pht in. i vi my

pht in 100MW tr ln, phi dng ng h c cp chnh xc 1,0.

2. o cng sut tc dng i vi tng my bin p v ng dy 6kV tr ln cp in t

dng cho nh my in.

3. o cng sut phn khng ca my b ng b v t b 25MVAr tr ln.

4. o cng sut tc dng v cng sut phn khng ca MBA tng p hai cun dy ca

nh my in.

o cng sut tc dng v cng sut phn khng pha h p v trung p ca MBA tng

p ba cun dy (k c my bin p t ngu c s dng cun dy th ba) ca nh my

in. i vi MBA lm vic trong khi vi my pht in, vic o cng sut pha h p

thc hin mch my pht in.

5. Ti cc trm bin p gim p, o cng sut tc dng v cng sut phn khng cho tng

my bin p 110kV tr ln, i vi cc my bin p khc ch cn o cng sut tc dng.

i vi my bin p ba cun dy gim p - o pha trung p v h p.

i vi my bin p hai cun dy gim p - o pha h p.

Khng cn o cng sut cc my bin p phn phi h p. Ti nhng my bin p khng

ch cng sut theo ma th phi c ch u ng h di ng.

6. ng dy 110kV tr ln c dng cng sut t hai pha, k c my ct mch vng - o

cng sut tc dng v phn khng.

7. Phi d kin v tr u dng c di ng o cng sut ti cc im cn kim tra nh k

dng cng sut tc dng v phn khng.

Phi dng dng c o c thang o hai pha i vi mch c hng cng sut thay

i.

Cu 93 : V cc s o m in nng tc dng mt pha v pha pha

(trc tip v gin tip) ?

Cu 94 : Nu phng php tnh tn tht in p trn ng dy ca

mng in a phng c ph ti tp trung ?

Xt mng PP cung cp cho 3 ph ti tp trung nh- (HV).

S thay th ca mng s c dng:

Cng sut trn cc on:

S01 = S3 + S2 + S1 = (p1 + p2 + p3) + j(q1 + q2 + q3)

S12 = S2 + S3 = (p2 + p3) + j(q2 + q3)

S23 = S3 = p3 + jq3

Tnh U theo cng sut chy trn cc on:

dm

23232323

dm

12121212

dm

010101012312013

U

xQrP

U

xQrP

U

xQrPUUUU

2 1 0

S1 = p1 + jq1 S2 = p2 + jq2

3

S3 = p3 + jq3

2

S2

1 0

S1

S01 S12

3

S23

S3

2 1 0 r01 + jx01 3 r12 + jx12

r23 + jx23

P01 + jQ01 P12 + jQ12 P23 + jQ23

Tng qut cho mng c n ph ti:

dm

ijijijij

U

xQrPU

)xQrP(.U.1000

100100.

U

U%U ijijijij2

dmdm

Trong : U - [V].

Pij ; Qij - [kW] ; [kVAr].

Udm - [kV].

rij ; xij - [].

Tnh U theo cng sut ca tng ph ti:

dm

3333

dm

2222

dm

11110302013

U

XqRp

U

XqRp

U

XqR.pUUUU

Tng qut:

dm

iiii

U

XqRpU

)XqRp(.U.1000

100100.

U

U%U iiii2

dmdm

2 1 0

p1 + jq1 p2 + jq2

3

R3 + jX3

R1 + jX1

R2 + jX2

p3 + jq3

+ V coi mng l tuyn tnh nn chng ta c th

s dng nguyn tc xp chng. Tc l tn tht

in p n im cui cng ca mng (im 3)

bng tng tn tht in p gy ra bi 3 ph ti

trn cc on t ph ti n u ngun:

Trong : pi ; qi - ph ti tc dng v phn khng [kW]; [kVAr].

Ri ; Xi - in tr v in khng t ph ti i v ngun [].

Ch : Biu thc tng qut trn ch -c dng tnh tn tht in p t ngun n im

cui cng cu l-i. Khi p dng tnh U t ngun n mt im bt k s dn n sai

(khng s dng -c).

Cu 95 : Nu phng php tnh tn tht in p trn ng dy c ph

ti u ?

ii02dm

LprU.1000

100%U (-ng dy th-ng cng 1 tit din)

Gi p0 Cng sut phn b u trn 1 n v chiu di dl. Ti im x cch ngun

1 khong lx . Trn vi phn chiu di dl c mt l-ng cng sut l dp = p0 .dl. Cng sut

ny gy ra trn on lx mt tn tht in p l dU = r0.lx.dp/Udm

dm

x00

U

dllprUd

Tn tht trn ton b on dy:

2

ll.

U

prdl

U

lprUdU

201

202

dm

00l

l

l

ldm

x0012

02

01

02

01

= )ll.(2

ll.

U

pr0102

0102

dm

00

l01

0

lx

l02

1 2 x + -ng dy b qua in khng: nhng

tr-ng hp sau: (-ng dy CC cho ph

ti c cos = 1.)

- mng h p r0 > > > x0

Ta c: p0(l02 l01) = l12.p0 = P

'2

0201 l2

ll

2 l im gia on 1-2

dm

'2

dm

'20

12U

R.P

U

l.PrU

S thay th t-ng -ng (HV)

Trong l12 = l12 /2

T s thay th t-ng -ng cch tnh nh- mt ph ti tp chung vi P = pi

t cch xa ngun 1 khong l2 = l01 + 1/2. l12

Cu 96 : Nu phng php chn aptomat, my ct, cu dao cc cp

in p khc nhau ?

1. Chn theo in p nh mc: in p nh mc ca TB. cho trn nhn my ph

hp vi mc cch in ca n v c mt mc d- no v bn, cho php TB lm vic

lu di in p cao hn nh mc 1015 % (gi l in p lm vic cc i ca TB.). V

lch in p trong iu kin lm vic bnh th-ng khng v-t qu 1015 % nh mc

nn khi la chn cc TB. theo iu kin in p cn phi tho mn iu kin sau:

dmtbdmm UU (1)

Udm m - din p nh mc ca mng m thit b mc vo.

Udmtb - in p nh mc ca thit b do nh my ch to cho trong l lch, hoc

ghi trn nhn my. Thc t vn hnh in p l-i dao ng nn ta c:

mdmmdmtbdmtb UUUU (2)

l2

0 1 2

Udmtb - tng in p cho php ca TB.

Um - lch in p c th c ca mng khi lm vic so vi nh mc trong iu

kin vn hnh.

Mc tng in p cho php ca mt s thit b:

2. Chn theo dng in nh mc: Idm l dng in c th chy qua TB. trong thi

gian lu di nhit nh mc ca mi tr-ng. Lc nhit ca phn t b t nng

nht ca TB. khng v-t qu gi tr cho php lu di.

Vic chn ng theo dng nh mc m bo khng xy ra qu t nng nguy

him cho cc phn ca TB. khi lm vic lu di ch nh mc. Dng in lm vic

cc i ca mng Ilvmax trong thi gian t 3T khng -c v-t qu dng nh mc ca

TB.

dmtblv II max

Dng in lm vic cc i xut hin khi:

+ Mch cc -ng dy lm vic song song khi ct i 1 -ng dy.

+ Mch my BA khi s dng kh nng qu ti ca chng.

+ Cc -ng cp khng d tr khi s dng kh nng qu ti ca chng.

Vic tng cao lp t TB . so vi mt n-c bin dn ti s gim in p cho php.

Mc tng in p so vi in p nh mc va nu trn ch cho php khi TB -c lp

t cao d-i 1000 m so vi mt n-c

bin. Nu cao ni lp t cao hn phi

gim bt khng -c qu Udm .

+ Cp in 1,1 Udmtb

+ S 1,15

+ Dao cch ly 1,15

+ My ct in 1,15

+ Cc my pht in, khi lm vic vi cng sut nh mc v in p gim 5% so

vi nh mc.

Nhit mi tr-ng xung quanh TB th-ng ly 35 0C . Khi nhit ni lp t

ln hn khi cn hiu chnh li dng nh mc.

0

cf

kkcfdmtb

35II

cf nhit ln nht cho php ca TB.

kk nhit khng kh ni lp t.

Tr-ng hp kk < 350C th dng cho php c th ln Idm . C mi gim ca mi

tr-ng xung quanh so vi 350C th cho php tng dng in ln hn l 0,005 Idm nh-ng

tng cng khng -c v-t qu 0,2 Idm .

Cu 97 : Nu phng php tnh tn tht cng sut, in nng ca li

in ? Cho v d c th ?

Tn tht cng sut :

Trong tnh ton -ng dy ti in, ng-i ta s dng s thay th hnh (i vi

mng 110 kV, i khi ngay c vi mng 220 kV ng-i ta th-ng b qua phn in dn

tc dng ca -ng dy. Tc l trn s ch cn li thnh phn in dn phn khng Y

= jB do dung dn ca -ng dy v th-ng -c thay th bng ph ti phn khng jQc.

Z

1 2 S1 S2

S2 = P2 + jQ2

S1

Ch : S = 3.I2dm.Z (m U3

SI )

Z.U

SS

2

2

Cng sut cui -ng dy:

)2

QQ(jP

2

QjSS 2c22

2c2

."2

Tn tht cng sut c th xc nh theo cng sut cui -ng dy:

X.U

S.jR.

U

SZ.

U

SQjPS

22

2"2

22

2"2

2

2

"2

.

Cng sut cui -ng dy:

."2

'1 SSS

Tn tht cng sut c th xc nh theo cng sut chy u -ng dy:

X.U

S.jR.

U

SZ.

U

SQjPS

21

2'1

21

121

2

1

'1

.

Khi cng sut chy cui -ng dy s l:

.

1

.

2

.

S'S"S

Cng sut i vo -ng dy s l:

2

Qj'SS 1c1

.

1

.

Trong ph ti phn khng ca -ng dy c th tnh theo in dn phn khng

theo cng thc sau:

2

B.U

2

Q 21

1c 2

B.U

2

Q 22

2c

Tn tht in nng :

Nu trong thi gian t ph ti in khng thay i, th cng sut l hng s

v tn tht in nng s -c tnh nh- sau:

A = P.t

Thc t ph ti li bin thin lin tc theo thi gian nn A phi ly tch

phn hm P trong sut thi gian kho st.

t

0

2t

0dt).t(I.R.3dt.PA

V I(t) khng tun theo mt dng hm no khng th xc nh -c tn

tht in nng theo cng thc trn. tnh tn tht in nng ng-i ta -a ra khi

nim Tmax v .

max

maxP

AT

Khi nim v : tnh in nng ng-i ta cng -a ra mt khi nim t-ng t nh-

Tmax.

P

Pmax

Tmax 0 8760 t [h]

N Tmax: Thi gian trong nu gi thit l tt c cc h

dng in u s dng cng sut ln nht Pmax nng l-ng

in chuyn ch trong mng in bng vi l-ng in nng

thc t m mng chuyn ch trong thi gian t.

(t = 8760 gi = thi gian lm vic 1 nm).

N : L thi gian m trong nu mng lun chuyn tr vi mc tn tht cng

sut ln nht th sau mt thi gian l-ng tn tht bng l-ng tn tht thc t trong

mng sau 1 nm vn hnh

Cu 98 : Nu phng php tnh tn tht cng sut, in nng ca my

bin p ? Cho v d c th ?

Tn tht cng sut :

My bin p 2 cun dy:

Tn tht cng sut trn 2 cun dy (tc trn tng tr ZB).

B

2

2

"

B

2

2

"

cucucu X.U

SjR.

U

SQjPS

Trong : S = S2 - Cng sut ca ph ti.

Ton b tn tht cng sut trong my bin p s l:

..

B

2

2

"

feB

2

2

"

fecufe

.

B X.U

SQjR.

U

SPSSS

T y ta thy rng cng sut u vo my bin p l:

S1 = Sfe + S = SB + S2

Trong thc t ng-i ta c th xc nh tn tht cng sut trn cun dy ca my

BA bng nhng thng s cho tr-c ca my BA. Xut pht t cng thc tnh RB v XB ta

c:

Sfe = Pfe + jQfe

S1 1 2 S S

2dm

2dmN

BS

U.PR

;

2B

2BB RZX trong :

dm

2dmN

BS

U%.uZ

2

2dm

2dmN

2

dm

2dmN

BS

UP

S

U%.uX

= 2N

2

dmN2dm

2dm PS%.u

S

U

Thay RB ; XB v coi U2 = Udm (ly gn ng):

PCu = N

2

dm

PS

"S

N

2

dm

2N

2dm

2N

2

dm

Cu Q.S

"SPS%.u.

S

"SQ

2

dm

Nfe

2

dm

Nfe

.

BS

"SQQj

S

"S.PPS

Vi my bin p 3 cun dy: vic tnh ton hon ton t-ng t nh- my bin p

2 cun dy (phn tn tht trong dy cun cu tng cun dy).

Cng thc tng qut cho vic xc nh tn tht cng sut trn cc cun dy:

)jXR.(U

SS ii

2

dmi

"i

i

.

Z3

1 2 Z1 Z2

SFe = Pfe + jQfe

S2 S2 S1 S1

S3

S3

3

S2

S3

S1

Tn tht cng sut ton b my BA:

..3

1iiFeB SSS

Cng sut u vo:

..

.

B32

.

fe

.'11

.

SSSSSS

Tn tht in nng: gm 2 phn:

Tn tht khng ph thuc vo ph ti xc nh theo thi gian lm vic ca

my bin p.

Phn ph thuc vo ti xc nh theo th ph ti, nu cng sut MBA c

th nh ph ti th dng Tmax tnh .

Tn tht in nng 1 nm tnh theo l:

AB = P0.Tb + Pmax. = P0.Tb + PN.

.

Trong : Tb l thi gian vn hnh nm ca MBA 8500 8760h

Smax l ph ti cc i nm ca MBA.

Cu 99 : Nu nguyn l ca bo v qu dng cc i ? ng dng ?

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 nh

trc no .

ng dng:

V d kho st tc ng ca cc bo v dng in cc i t trong mng hnh tia

c 1 ngun cung cp, cc thit b bo v c b tr v pha ngun cung cp ca tt c

cc ng dy. Mi ng dy c 1 bo v ring ct h hng trn chnh n v trn

thanh gp ca trm cui ng dy.

Dng khi ng ca bo v IK, tc l dng nh nht i qua phn t c bo v

m c th lm cho bo v khi ng, cn phi ln hn dng ph ti cc i ca phn t

c bo v ngn nga vic ct phn t khi khng c h hng.

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 nh vy 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 v s

gim xung khi h hng cng cch xa ngun cung cp. Dng khi ng ca bo v IK

c chn ln hn tr s ln nht ca dng trn on c bo v khi xy ra ngn mch

on k (cch xa ngun hn). Nh vy bo v c th tc ng chn lc khng thi

gian. Chng c gi l bo v dng in ct nhanh.

Cc bo v dng in cc i lm vic c thi gian chia lm hai loi tng ng

vi c tnh 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 vo

tr 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 vi dng

IK tng i nh v t ph thuc hoc khng ph thuc khi bi s ny ln.

Cu 100 : Nu nguyn l ca bo v ct nhanh dng in ? ng dng ?

Bo v dng ct nhanh (BVCN) l loi bo v m bo tnh chn lc bng cch

chn dng khi ng ln hn dng ngn mch ln nht qua ch t bo v khi h hng

ngoi phn t c bo v, BVCN thng lm vic khng thi gian hoc c thi gian rt

b nng cao nhy v m rng vng BV.

ng dng:

Xt s mng, BVCN t ti u ng dy AB v pha trm A. bo v

khng khi ng khi ngn mch ngoi (trn cc phn t ni vo thanh gp trm B), dng

in khi ng IK ca bo v cn chn ln hn dng in ln nht i qua on AB khi

ngn mch ngoi. im ngn mch tnh ton l N nm gn thanh gp trm B pha sau

my ct.

IK = kat. INngmax

Trong :

INngmax: L dng ngn mch ln nht khi ngn mch ngoi vng bo v

(thng l dng N(3) )

kat: h s an ton; xt ti nh hng ca thnh phn khng chu k, vic

tnh ton khng chnh xc dng ngn mch v sai s ca rle. Thng kat= 1,2 1,3.

Khng k n ktv v khi ngn mch ngoi bo v khng khi ng.

Cu 101 : Nu nguyn l ca bo v ct nhanh c hng ? ng dng ?

Bo v qu dng in vi thi gian lm vic chn theo nguyn tc bc thang

khng m bo c tnh chn lc hoc thi gian tc ng ca cc bo v gn ngun qu

ln khng cho php. khc phc ngi ta dng bo v qu dng c hng. Thc cht

y cng l mt bo v qu dng thng thng nhng c thm b phn nh hng cng

sut pht hin chiu cng sut qua i tng c bo v. Bo v s tc ng khi

dng in qua bo v ln hn dng in khi ng IK v hng cng sut ngn mch

i t thanh gp vo ng dy.

Ngy nay hu ht cc rle qu dng c hng s c tch hp thm nhiu chc

nng nh: chc nng ct nhanh, qu dng vi c tuyn thi gian c lp v ph thuc,

nh mt s rle qu dng c hng c c tnh chn lc tuyt i v tng i, ngha

l c th va m bo chc nng ct nhanh va ng vai tr nh mt bo v d tr. Mt

trong nhng rle va nu trn l rle qu dng c hng ba cp tc ng. hiu r hn

v loi rle ny chng ta s i phn tch chn thi gian lm vic v dng in khi ng

ca bo v qu dng c hng ba cp tc ng cho mt s mng in in hnh trong h

thng in.

Cu 102 : Nu nguyn l ca bo v so lch ngang c hng ? ng

dng ?

Nguyn tc tc ng bo v so lch ngang da vo vic so snh dng trn 2 ng

dy song song, trong ch lm vic bnh thng hoc khi ngn mch ngoi cc dng

ny c tr s bng nhau v cng hng, cn khi pht sinh h hng trn mt ng dy th

chng s khc nhau.

ng dng:

Bo v c dng cho 2 ng dy song song ni vo thanh gp qua my ct

ring. Khi h hng trn mt ng dy, bo v cn phi ct ch ng dy v gi

nguyn ng dy khng h hng li lm vic. Mun vy bo v phi c t c 2

u ng dy v c thm b phn nh hng cng sut xc nh ng dy b h

hng.

S nguyn l 1 pha ca bo v trn hnh v. Cc my bin dng t trn 2

ng dy c t s bin i nI nh nhau, cun th ca chng ni vi nhau th no

nhn c hiu cc dng pha cng tn. Rle dng 5RI lm nhim v ca b phn khi

ng, rle 6RW tc ng 2 pha l b phn nh hng cng sut. Khi chiu dng in

quy c nh trn hnh 5.9, ta c dng a vo cc rle ny l IR = IIT - IIIT .

p a vo 6RW c ly t BU ni vo thanh gp trm. Rle 6RW s tc ng

i ct ng dy c cng sut ngn mch hng t thanh gp vo ng dy v khi c

2 ng dy u c cng sut ngn mch hng t thanh gp vo ng dy th 6RW s

tc ng v pha ng dy c cng sut ln hn.

Trong ch lm vic bnh thng hoc khi ngn mch ngoi, dng IIT , IIIT

bng nhau v trng pha. Dng vo rle IR = IIT - IIIT gn bng 0 (IR = IKCB), nh hn

dng khi ng IKR ca b phn khi ng 5RI v bo v s khng tc ng.

Khi ngn mch trn ng dy I im N, dng II > III . V pha trm A c IR =

IIT - IIIT ; cn pha trm B c IR = 2IIIT. Rle 5RI c 2 pha u khi ng. Cng sut

ngn mch trn ng dy I pha A ln hn trn ng dy II; do vy 6RW khi ng

v pha ng dy I v bo v ct my ct 1MC. V pha trm B, cng sut ngn mch

trn ng dy I c du dng (hng t thanh gp vo ng dy), cn trn ng dy

II - m. Do 6RW cng khi ng v pha ng dy I v ct my ct 1MC. Nh vy

bo v m bo ct 2 pha ca ng dy h hng I.

Khi ngn mch trn ng dy gn thanh gp (im N), dng vo rle pha

trm B l IR 0 v lc u n khng khi ng. Tuy nhin bo v pha trm A tc ng

do dng vo rle kh ln. Sau khi ct my ct 2MC, phn b dng trn ng dy c

thay i v ch n lc ny bo v pha trm B mi tc ng ct 2MC. Hin tng khi

ng khng ng thi va nu l khng mong mun v lm tng thi gian loi tr h

hng ra khi mng in.

Ngun thao tc c a vo bo v qua cc tip im ph ca 1MC v 2MC. Khi

ct mt my ct th tip im ph ca n m v tch bo v ra. Cn thc hin nh vy v

2 l do sau:

Sau khi ct 1 ng dy bo v tr thnh bo v dng cc i khng thi gian.

Nu khng tch bo v ra, n c th ct khng ng ng dy cn li khi xy ra ngn

mch ngoi.

Bo v c th ct ng dy b h hng khng ng thi. Khi ngn mch ti im

N, my ct 2MC ct trc, sau ton b dng h hng s i n ch ngn mch qua

ng dy I. Nu khng tch bo v pha trm A ra, n c th ct khng ng 1MC ca

ng dy I khng h hng.

Cu 103 : Nu nguyn l ca bo v so lch dc ? ng dng ?

Cng l loi bo v ct nhanh, da trn s chnh lch v c-ng v pha ca dng

in u v cui (-ng dy) vng bo v. Th-ng -c dng bo v my bin p,

my pht in, ng c cng sut ln, t dng bo v -ng dy.

HV S nguyn l

Vng bo v l my bin p.

BI -c t 2 u ca BA. phn nh dng in chy trong bo v.

Cc cun dy ca BI -c ni sao cho dng in trong rle bng hiu dng in

chy trong my bin dng.

....

21R III

Trong tr-ng hp bnh th-ng hoc ngn mch ngoi vng bo v (HV-1). Ta c I1

= I2 v cng pha nn hiu ca chng bng khng IR = 0. Rle khng tc ng.

Khi xy ra ngn mch, trong vng bo v (HV-2) do dng in trong 2 bin dng

ng-c chiu nhau nn dng in chy trong rle - I R = I1 + I2 > 0 (khi 2 pha BA u c

I1

I2

N

MC

MC

BA

HV-1

I1

I2

HV-2

N

I1

I2 I2

I1

ngun). Hoc IR = I1 >0 (khi BA ch c 1 ngun). Lc ny rle tc ng MC ct BA ra

khi ngun.

Trng thi lm vic bnh th-ng hoc khi ngm, xy ra ngoi vng bo v yu

cu bo v khng lm vic. C ngha l phi m bo dng th cp ca cc my bin

dng I1 = I2 c v gi tr v gc pha. Cc dng s cp I1 v I2 khc nhau v c-ng (do

t s bin i ca BA gy ra). Vy c I1 = I2 I R = 0 Phi chn my bin dng ph

hp vi t u dy ca BA. Vic cn bng dng in th cp ca my bin dng nh- trn

th-ng gp nhiu kh khn do cc my BI th-ng -c ch to sn theo tiu chun nht

nh. Nn rt kh khn m bo -c I R = 0. Trong nhiu tr-ng hp ng-i ta phi dng

thm cc BA t ngu hoc BI bo ho mc vo pha tr-c rle dng in lm s

phc tp.

So vi bo v ct nhanh. bo v so lch c -u im l khng li vng cht

nh-ng s phc tp, tn nhiu thit b nn ch -c dng ni quan trng.

i vi nhng -ng dy in p cao lm vic song song, hoc BA v ng c c

cng sut ln (quan trng), ng-i ta cn dng bo v so lch ngang da trn c s so snh

dng in gia cc -ng dy lm vic song song vi nhau.

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