L THNH BC

GIO TRNH

THIT B IN (Ti bn c sa cha v b xung)

NH XUT BN KHOA HC V K THUT H NI - 2003

MC LC

Mc lc Phn th nht C S L THUYT THIT B IN Khi nim chung v thit b in Chng 1 H quang in 1.1. i cng v h quang in 1.2. H quang in mt chiu 1. 3. H quang in xoay chiu 1. 4. Qa trnh phc hi in p ca h quang in 1. 5. Cc bin php v trang b dp h quang trong thit b in Chng 2 Tip xc in 2. 1. i cng v tip xc in 2. 2. Tip im ca thit b in Chng 3 Pht nng 3. 1. i cng 3. 2. Ch lm vic di hn ca vt th ng nht 3. 3. Ch lm vic ngn hn ca vt th ng nht 3. 4. Ch lm vic ngn hn lp li ca vt th ng nht 3. 5. S pht nng khi ngn mch Chng 4 Lc in ng 4. 1. Khi nim chung 4. 2. Cc phng php tnh lc in ng 4. 3. Tnh lc in ng ca vt dn 4. 4. Lc in ng trong mch in xoay chiu 4. 5. Cng hng c kh v n nh lc in ng Chng 5 C cu in t v nam chm in 5. 1 Khi nim chung v mch t 5. 2 Tnh t dn khe h khng kh ca mch t 5. 3 Tnh ton mch t 5. 4 i cng v nam chm in 5. 5. Tnh lc ht in t nam chm in mt chiu 5. 6. Nam chm in xoay chiu v vng chng rung 5. 7 Nam chm in 3 pha 5. 8. C cu in t chp hnh Phn th hai THIT B IN H P Chng 6 Rle 6. 1. Khi nim chung v rle 6. 2. Rle in t 6. 3. Rle in ng 6.4. Rle t in 6. 5. Rle cm ng 6. 6. Rle thi gian - Rle nhit -Rle tc -- Rle iu khin 6.7 R le tnh Chng 7 Cm bin 7. 1. Khi nim chung 7. 2. Cm bin in tr 7. 3. Cm bin in cm 7. 4. Cm bin cm ng - Cm bin in dung - Cm bin im 7.5. Cm bin quang Chng 8 Cng tc t-khi ng t-cu ch-ptmt

Trang

5

7 9

11 12 14

17 20

25 26 27 28 30

31 31 32 36 38

40 41 44 48 49 52 54 54

58 60 62 63 64 66

78 81 86 87

8.1. Cng tc t 8.2. Khi ng t

8.3. Cu chy(cu ch) 8.4. ptomat Chng 9 Cc b n nh in 9. 1. Khi nim chung v cc b n nh in 9. 2. n p st t khng t 9. 3. n p st t c t 9. 4. n p khuch i t 9. 5. n p bin tr than 9.6. n p Servomotor 9.7. n p kiu b 9.8. n p in t Phn th ba THIT B IN TRUNG V CAO P Chng 10 Dao ngt 10. 1. Cc nh ngha v c tnh ca thit b ng ct 10. 2. Dao cch li 10. 3. Cu dao ni t mt tr 10. 4. C cu thao tc tc ca dao cch li v cu dao ni t 10.5. Cu dao cao p 10. 6. Dao cch li v cu dao ph ti li trung p Chng 11 My ngt in 11.1. Chc nng-phn loi-cch la chn v cu trc 11. 2. Nguyn l ct v cc iu kin ng ct khc nghit 11. 3. Mi trng dp h quang v nguyn l tc ng 11.4. C cu tc ng v iu khin 11.5. Mt s loi my ngt cao v siu cao p Chng 12 Thit b chng st 12. 1. Khi nim chung 12. 2. Thit b chng st ng 12. 3. Chng st van 12. 4. Chng st van t 12. 5. Chng set xit kim loi 12. 6. Chng st VariSTAR UitraSIL Chng 13 Khng in 13.1. Khi nim chung 13.2. La chn v kim tra khng in Chng 14 Bin p o lng 14.1. Bin in p o lng 14.2. Bin dng in Chng 15 H thit b SCADA 15.1. Cng dng v chc nng ca h SCADA 15.2 T chc SCADA trong h thng in lc 15.3. Phn mm RUNTIME thng l ca SCADA 15.4. H phn mm thng phm ca SCADA cng nghip 15.5. Cc mng truyn tin ca h SCADA

15.6 Truyn tin trong h SCADA

Ph lc Ti liu tham kho

88

91 95 98

101

105 105 106 108 109 110 111 112

113 115 119 119 120 121

123 137 140 152 155

171 172 173 176 177 181

186 186

189 192

196 199 204 204 205 207

Li ni u

"Gio trnh Thit b in c bin son trn c s cng chi tit mn hc "Thit b in" cho cc ngnh K thut in, T ng ha, K thut Nhit-in lnh. Trong qu trnh bin son, tc gi c tham kho cc gio trnh "C s l thuyt kh c in", "Phn t t ng", "Kh c in h p ", "Kh c in cao p",... c trng i hc Bch khoa H Ni xut bn. Gio trnh ny dng lm ti liu ging dy v hc tp cho sinh vin ngnh in, in t, Cng ngh Nhit -in lnh cng nh lm ti liu tham kho cho ki s v cn b ki thut ngnh in cng nh cc chuyn ngnh lin quan.

Ni dung ca gio trnh cp n cc vn l thuyt c bn ca thit b in v gii thiu mt s thit b in thng dng hin nay.

Gio trnh ny c chia lm ba phn: + Phn th nht: L thuyt c s. + Phn th hai: Thit b in h p. + Phn th ba: Thit b in trung - cao p. Trong qu trnh bin son, Tc gi nhn c s gip v cung cp ti liu ca: - Cc Thy, C gio trong b mn Thit b in -in t, trng i hc Bch khoa H

Ni. - Cc ng nghip trong nhm Thit b in trng i hc K thut Nng nh GVC.

L Vn Quyn, ThS.V Nh Tin. - Cc Ki s cng tc ti c quan i din cc hng thit b nh ABB, SIEMENS,

COOPER,... v cc Ki s ca trung tm iu in Quc gia. c bit l s gip tn tnh ca TS.Trn Vn Chnh trong vic hiu nh v ng gp

thm nhiu kin cho ni dung Gio trnh. Mc d, tc gi c nhiu c gng trong vic bin son gio trnh nht l cp n

nhng thit b in hin i nhm p ng yu cu nng cao cht lng o to, phc v nhu cu cng nghip ha - hin i ha hin nay nhng vi kh nng v kinh nghim c hn, chc chn khng trnh khi thiu st. Sch sau khi c nh xut bn Khoa Hc v K Thut pht hnh, tc gi cng nhn c nhiu kin ng gp v khch l ng vin ca cc Thy C gio v nhiu k s, cn b k thut ang cng tc ti cc trng i hc cng nh cc cng ty, x nghip ca

ngnh in. Tc gi xin chn thnh cm n v rt mong tip tc nhn c s ng gp kin ca ng o bn c gio trnh c hon thin hn na trong cc ln ti bn sau.

Mi th t, gp xin gi v ban bin tp nh xut bn Khoa Hc v K Thut - H Ni v b mn Thit b in - trng i hc K thut Nng. Tc gi xin chn thnh cm n.

Tc gi

TI LIU THAM KHO 1. Teo ex A -. H. A -MOCKBA Bca -1985. 2. C s l thuyt kh c in - B mn My in- Kh c in - i hc Bch khoa H Ni - 1978. 3. Kh c in, kt cu s dng v sa cha - Nguyn Xun Ph, T ng -Nh xut bn Khoa hc v k

thut - 1997. 4. Cm nang thit b ng ct - Nh xut bn Khoa hc v K thut -H Ni - 1998. 5. Gio trnh K thut in cao p - V Vit n - i hc Bch khoa H Ni - 1972. 6. Phn t t ng - Nguyn Tin Tn, Phm Vn Chi - B mn My in - Kh c in - i hc Bch

khoa H Ni - 1980. 7. Gio trnh Kh c in, dng cho ngnh in kh ha- i hc Bch khoa H Ni - 1979. 7. Gio trnh Kh c in- i hc Bch khoa H Ni - 1985. 8. Static Relays - ABB. 9. Low Oil Content Circuit - Breakers for Outdoor Stations 10...72.5 kV. E.I.B. 10. SF6 Circuit - Breakers with Spring Operating Mechanism 72.5... 170 kV. AEG. 11. Gas - Insulated Switchgear 72.5... 525 kV. AEG. 12. Metal - Enclosed, SF6 - Gas Insulated High Voltage Switchgear (V.I.S.). series B3 up to 420kV.

AEG. 13. Gio trnh Cm bin- Phan Quc Ph, Nguyn c Chin-Nh xut bn Khoa hc v K thut - H

Ni -2000. 14. Quy trnh vn hnh v bo dng cc loi my ct du- Nh xut bn Khoa hc v K thut-H Ni

-1996. 15. Quy trnh vn hnh v bo dng my ngt SF6-Tng cng ti in lc Vit Nam- H Ni -1998. 16. Nh my in v trm bin p - Trnh Hng Thm, Nguyn Hu Khai, o Quang Thch, L Vn t,

Phm Vn Ha, o Kim Hoa -Nh xut bn Khoa hc v K thut -H Ni -1996.

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PHN TH NHT L THUYT C S

KHI NIM CHUNG V THIT B IN

Thit b in c cp y l cc loi thit b lm cc nhim v: ng ct, iu khin, iu

chnh, bo v, chuyn i, khng ch v kim tra mi s hot ng ca h thng li in v cc loi my in. Ngoi ra thit b in cn c s dng kim tra, iu chnh v bin i o lng nhiu qu trnh khng in khc.

Thit b in l mt loi thit b ang c s dng rt ph bin c mt trong hu ht cc lnh vc sn xut ca nn kinh t, t cc nh my in, trm bin p, h thng truyn ti in, n cc my pht v ng c in trong cc x nghip cng nghip, nng nghip, giao thng,... v trong c lnh vc an ninh quc phng.

Thit b in s dng nc ta hin nay c nhp t rt nhiu nc, rt nhiu hng sn xut khc nhau v cc th h. C c cc thit b c thi gian s dng 40 n 50 nm, rt lc hu v cc thit b rt hin i mi nhp. Chnh v vy cc quy cch khng thng nht, gy kh khn cho vn hnh, bo dng v sa cha. Do qa nhiu chng loi thit b in vi cc tiu chun ki thut rt khc nhau, nn trong s dng hin nay nhiu khi khng s dng ht tnh nng v cng sut ca thit b hoc s dng khng ng gy h hng nhiu, lm thit hi khng nh cho nn kinh t. Chnh v vy vic o to v cp nhp nng cao kin thc v thit b in c bit l cc thit b mi cho cc cn b ki thut qun l v vn hnh thit b in l mt i hi rt cp thit. Gio trnh ny nhm trang b nhng l lun c bn, hiu nguyn l lm vic, c im cu to cc loi thit b in thng dng trong t ng truyn ng, trong h thng in v trong cc linh vc iu khin my in,...nhm gip sinh vin cc ngnh nng lng khi ra trng c th la chn, vn hnh, sa cha, ci tin thit b in hoc mt s b phn ca thit b in, c bit cung cp nhng kin thc lm c s tip cn cc thit b hin i. 1. Phn loi thit b in

thun li cho vic nghin cu, vn hnh s dng v sa cha thit b in ngi ta thng phn loi nh sau: a) Phn theo cng dng + Thit b in khng ch: dng ng ct, iu chnh tc chiu quay ca cc my pht in, ng c in (nh cu dao, p t mt, cng tc t,...). + Thit b in bo v: lm nhim v bo v cc ng c, my pht in, li in khi c qu ti, ngn mch, st p,...( nh rle, cu ch, my ct,...). + Thit b in t ng iu khin t xa: lm nhim v thu nhn phn tch v khng ch s hot ng ca cc mch in nh khi ng t,...

+ Thit b in hn ch dng ngn mch (nh in tr ph, cun khng,...). + Thit b in lm nhim v duy tr n nh cc tham s in (nh n p, b t ng iu chnh

in p my pht,...) + Thit b in lm nhim v o lng (nh my bin dng in, bin p o lng,...).

b) Phn theo tnh cht dng in + Thit b in dng trong mch mt chiu. + Thit b in dng trong mch xoay chiu. c) Phn theo nguyn l lm vic Thit b in loi in t, in ng, cm ng, c tip im, khng c tip im,... d) Phn theo iu kin lm vic

+ Loi lm vic vng nhit i kh hu nng m, loi vng n i, c loi chng c kh chy n, loi chu rung ng,... e) Phn theo cp in p c

+ Thit b in h p c in p di 3kV. + Thit b in trung p c in p t 3kV n 36 kV. + Thit b in cao p c in p t 36kV n nh hn 400 kV.

+ Thit b in siu cao p c in p t 400 kV tr ln.

6

2. Cc yu cu c bn ca thit b in - Phi m bo s dng c lu di ng tui th thit k khi lm vic vi cc thng s k

thut nh mc. - Thit b in phi m bo n nh lc in ng v n nh nhit khi lm vic bnh

thng, c bit khi s c trong gii hn cho php ca dng in v in p. - Vt liu cch in chu c qu p cho php.

- Thit b in phi m bo lm vic tin cy, chnh xc an ton, gn nh, d lp rp, d kim tra, sa cha.

- Ngoi ra cn yu cu phi lm vic n nh iu kin kh hu mi trng m khi thit k cho php. Chng 1. H QUANG IN

1.1. I CNG V H QUANG IN

1. Khi nim chung

H quang in thc s c ch khi c s dng trong cc lnh vc nh hn in, luyn thp,...nhng lc ny h quang cn c duy tr chy n nh.

Nhng trong cc thit b in nh cu ch, cu dao, my ct,...h quang li c hi cn phi nhanh chng c loi tr. Khi thit b in ng, ct (c bit l khi ct) h quang pht sinh gia cc cp tip im ca thit b in khin mch in khng c ngt dt khot. H quang chy lu sau khi thit b in ng ct s lm h hi cc tip im v bn thn thit b in. Trong trng hp ny m bo lm vic tin cy ca thit b in yu cu phi tin hnh dp tt h quang cng nhanh cng tt.

7

Bn cht ca h quang in l hin tng phng in vi mt dng in rt ln (ti khong 104 n 105 A/cm2), c nhit rt cao (ti khong 5000 60000C) v in p ri trn cc m b (ch khong 10 20V) v thng km theo hin tng pht sng. S phn b ca in p v cng in trng dc theo chiu di h quang c biu din trn hnh 1-1a.

Dc theo chiu di h quang c chia lm ba vng l: vng xung quanh cc m (cch cc m khong 10-4 n 10-5cm) vng ny tuy in p nh ch 8 n 10V nhng khong cch cng rt b nn cng in trng rt ln c 105 n 106 V/cm. Cn vng c chiu di gn ht h quang l vng thn, vng ny c cng in trng ch khong 10 n 50 V/cm. Vng cn li cn c gi l vng cc dng c cng in trng ln hn vng thn nhng cc yu t xy ra y theo cc l thuyt hin i th t nh hng n qu trnh pht sinh v dp h quang nn khng c cp.

c tnh u(i) ca h quang mt chiu c th biu in theo cng thc Kapzow c dng:

uhq = a+ bl + ni

dlc+

Vi: a, b, c, d l cc hng s ph thuc vt liu lm tip im v cc yu t bn ngoi (v d tip im ng c a= 30; b=17; c=41; d=33). C n l s m, ph thuc vo nhit vt liu dng cc, theo thc nghim thng ly n = 2,62.T.10-4, trong T l nhit ca vt liu dng cc.

c tnh u(i) vi l l chiu di h quang c dng hypcbn nh hnh 1-1b. 2. Qa trnh pht sinh v dp tt h quang a) Qu trnh pht sinh

H quang in pht sinh l do mi trng gia cc in cc (hoc gia cc cp tip im) b ion ha (xut hin cc ht dn in). Ion ha c th xy ra bng cc con ng khc nhau di tc dng ca nh sng, nhit , in trng mnh,.... Trong thc t qu trnh pht sinh h quang in c nhng dng ion ha sau: - Qu trnh pht x in t nhit; Qu trnh t pht x in t. - Qu trnh ion ha do va chm. - Qu trnh ion ha do nhit . a.1) S pht x in t nhit in cc v tip im ch to t kim loi, m trong cu trc kim loi lun tn ti cc in t t do chuyn ng v mi hng trong qu o ca cu trc ht nhn nguyn t. Khi tip im bt u m ra lc nn vo tip im gim dn khin in tr tip xc tng ln ch tip xc dng in b tht li mt dng tng rt ln lm nng cc in cc (nht l cc m nhiu e). B t nng, ng nng ca cc in t tng nhanh n khi cng nhn c ln hn cng thot lin kt ht nhn th in t s thot ra khi b mt cc m tr thnh in t t do. Qu trnh ny c gi l pht x in t nhit. a.2) S t pht x in t

a) b)

Hnh 1-1: a) H quang mt chiu; b) c tnh

K A

UA

UTh

UK

E[V]EK

Eth EA

Vng Vng thn Vng

lhq[m]

I

Uhql

50mm20 0

0 2 4 6 8 10 12

50100

150200

8

Khi tip im hay in cc va m ra lc u khong cch cn rt b di tc dng ca in p ngun ngoi th cng in trng rt ln, nht l vng cc m c khong cch nh c th ti hng triu V/ cm. Vi cng in trng ln cc m mt s in t c lin kt yu vi ht nhn trong cu trc s b ko bt ra khi b mt ca tt tr thnh cc in t t do, hin tng ny gi l t pht x in t. Khi c in t t pht x v pht x in t nhit nng lng c gii phng rt ln lm nhit khu vc h quang tng cao v pht sng, c bit khi ct mch in p cao v c dng ti ln th h quang chy v pht sng rt mnh lit. a.3) Ion ha do va chm

Sau khi tip im m ra, di tc dng ca nhit cao hoc ca in trng ln (m thng thng l c hai) th cc in t t do s pht sinh chuyn ng t cc dng sang cc m. Do in trng rt ln nn cc in t chuyn ng vi tc rt cao. Trn ng i cc in t ny bn ph cc nguyn t v phn t kh s lm bt ra cc in t v cc ion dng. Cc phn t mang in ny li tip tc tham gia chuyn ng v bn ph tip lm xut hin cc phn t mang in khc. Do vy m s lng cc phn t mang in tng ln khng ngng, lm mt in tch trong khong khng gian gia cc tip im rt ln, l qu trnh ion ha do va chm. a.4) Ion ha do nhit

Do c cc qu trnh pht x in t v ion ha do va chm, mt lng ln nng lng c gii phng lm nhit vng h quang tng cao v thng km theo hin tng pht sng. Nhit kh cng tng th tc chuyn ng ca cc phn t kh cng tng v s ln va chm do cng cng tng ln. Khi tham gia chuyn ng cng c mt s phn t gp nhau s kt hp li phn li thnh cc nguyn t. Cc nguyn t khuch tn vo mi trng xung quanh, gp nhit thp s kt hp li thnh phn t, hin tng ny gi l hin tng phn li (phn ng phn li thu nhit lm gim nhit ca h quang, to iu kin cho kh ion). Cn lng cc ion ha tng ln do va chm khi nhit tng th gi l lng ion ha do nhit. Nhit c hin tng ion ha do nhit cao hn nhiu so vi nhit c hin tng phn li. V d khng kh c nhit phn li khoang 40000K cn nhit ion ha khong 80000K.

Tm li, h quang in pht sinh l do tc dng ca nhit cao v cng in trng ln sinh ra hin tng pht x in t nhit v t pht x in t v tip theo l qu trnh ion ha do va chm v ion ha do nhit. Khi cng in trng cng tng (khi tng in p ngun), nhit cng cao v mt dng cng ln th h quang chy cng mnh lit. Qu trnh c thot nng lng ht nhn nn thng km theo hin tng pht sng chi la. Nu tng p lc ln mi trng h quang th s gim c tc chuyn ng ca cc phn t v do vy hin tng ion ha s gim. b) Qu trnh h quang tt

H quang in s b dp tt khi mi trng gia cc in cc khng cn dn in hay ni cch khc h quang in s tt khi c qu trnh phn ion ha xy ra mnh hn qu trnh ion ha. Ngoi qu trnh phn li ni trn, song song vi qu trnh ion ha cn c cc qu trnh phn ion gm hai hin tng sau: b.1) Hin tng ti hp

Trong qu trnh chuyn ng cc ht mang in l ion dng v in t gp c cc ht tch in khc du l in t hoc ion dng tr thnh cc ht trung ha (hoc t dng hn). Trong l thuyt chng minh tc ti hp t l nghch vi bnh phng ng knh h quang, v nu cho h quang tip xc vi in mi hin tng ti hp s tng ln. Nhit h quang cng thp tc ti hp cng tng. b.2) Hin tng khuch tn

Hin tng cc ht tch in di chuyn t vng c mt in tch cao(vng h quang) ra vng xung quanh c mt in tch thp l hin tng khuch tn. Cc in t v ion dng khuch tn dc theo thn h quang, in t khuch tn nhanh hn ion dng. Qu trnh khuch tn c trng bng tc khuch tn. S khuch tn cng nhanh h quang cng nhanh b tt. tng qu trnh khuch tn ngi ta thng tm cch ko di ngn la h quang.

1.2. H QUANG IN MT CHIU 1. Khi nim chung

9

Chng ta kho st y mt qu trnh xut hin h quang gia hai in cc trong mt mch in mt chiu nh hnh 1-2.

Gi in p ngun l U0 ,in tr mch l R, in cm mch l L v rhq c trng cho in tr h quang vi in p trn h quang l uhq. Theo nh lut Kic khp II, ta c phng trnh cn bng in p trong mch khi m tip im v h quang bt u chy nh sau:

U0 = i.R + uhq + L dtdi

(1.1)

Khi h quang chy n nh th dng in khng i i=I v c dtdi

= 0 phng trnh cn bng p s l :

U0 = uR+ uhq = I.R+ I.rhq (1.2) Cc thnh phn in p trong phng trnh (1.1) c th hin trn hnh 1-2. Vi: ng 1-l

in p ngun; ng 2- l in p ri trn in tr R v ng 3- l c tnh u(i) ca h quang. Theo th cc ng c tnh 2 v 3 giao nhau hai im A v B. Ti A v B phng trnh

(1.2) c tha mn, cc im A, B c gi l hai im chy ca h quang . -Xt ti B: H quang ang chy nu v mt l do no lm dng in i tng ln hn IB th theo th ta

nhn thy sc in ng t cm trn L l Ldtdi

< 0 (ngc chiu dng tng) s lm dng in i gim

xung li IB. Cn ngc li nu i gim nh hn IB th L dtdi

> 0 s lm i tng tr li gi tr IB, do vy im

B c gi l im h quang chy n nh. -Nu cng tng t ta xt ti im A, khi h quang ang chy n nh vi i= IA nu v mt l do no i

gim nh hn IA th L dtdi

< 0 nn dng tip tc gim n 0 v h quang tt. Cn nu i tng ln hn IA th

trn c tnh ta thy Ldtdi

> 0 nn dng tip tc tng n IB v h quang chy n nh ti im B, vy

im A gi l im h quang chy khng n nh. 2. iu kin dp tt h quang in mt chiu

10

c th dp tt c h quang in mt chiu cn loi b c im h quang chy n nh (im B). Trn c tnh ta nhn thy s khng c im chy n nh khi ng c tnh 3(in p trn h quang) cao hn ng c tnh 2 (l c tnh in p ri trn in tr R) nh hnh 1-2b (tc l h quang s tt khi Uhq> U0- UR). nng cao ng c tnh 3 thng thc hin hai bin php l tng di h quang(tng l) v gim nhit vng h quang xung, c tnh nh hnh 1-3.

3. Qu in p trong mch in mt chiu

Khi ct mch in mt chiu thng xy ra qu in p, khi mch c in cm ln nu tc ct cng nhanh th qu in p cng ln.

Nu ti thi im ct c I= 0 th : U0 = L dtdi

+ uhq , hay ta c:

uhq - U0 = - L dtdi

= U (1.3) U l tr s qu in p xoay chiu. Trong mch mt chiu lm vic vi cng sut ln li c nhiu vng dy khi dp h quang in qu in p s xy ra rt ln c th gy nh thng cch in v h hng thit b. hn ch hin tng qu in p ngi ta thng dng thm mt mch in ph mc song song vi ph ti. Mch ny c th l in tr, in tr v t ni tip hoc mt chnh lu mc ngc.

Hnh 1-2: c tnh h quang mt chiu v iu kin tt

I[A]

U[V]

U01

2 3

UR

Uhq Ldi/dt>0 Ldi/dt

< 0

Ldi/dt< 0

I[A]

U [V]

2

3

c)

a)

+ - Uo

R

rhqL

I

b)

T1 T2L1

a) I[A]

Hnh 1-3: c tnh khi ko di v gim nhit h quang

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1.2. H QUANG IN XOAY CHIU

1. Khi nim chung

c im ca mch xoay chiu l trong mt chu k bin thin dng in c hai ln qua tr s i= 0. Khi c h quang th ti thi im khi i= 0 qu trnh phn ion ha xy ra mnh hn qu trnh ion ha. Khi i= 0 h quang khng dn in v y l thi im tt dp tt h quang in xoay chiu.

Khi h quang in xoay chiu ang chy ta a dng in v in p ca h quang vo dao ng k ta s c dng sng ca dng in v in p h quang nh hnh 1-4.

Dng in c dng sng gn ging sng hnh sin cn in p th trong mt na chu k c hai nh nhn tng ng vi hai gi tr in p chy ( Uch) v in p tt (Ut) ca h quang in. T dng sng thu c trn mn hnh dao ng k ta xy dng c c tnh Vn -Am pe (V-A) ca h quang in xoay chiu nh hnh 1-4.

Ta nhn thy thi im dng in qua tr s 0 nu in p ngun nh hn tr s in p chy (Uch) th h quang s tt. Do vy qu trnh dp h quang in xoay chiu ph thuc rt nhiu vo tnh cht ca ph ti.

Ta nhn thy trong mch c ph ti in tr thun d dp h quang hn trong mch c ti in cm, bi mch thun tr khi dng in qua tr s khng (thi gian i=0 thc t ko di khong 0,1s ) th in p ngun cng bng khng (trng pha), cn mch thun cm khi dng bng khng th in p ngun ang c gi tr cc i (in p vt trc dng in mt gc 900). 2. Dp tt h quang in xoay chiu

H quang in xoay chiu khi dng in qua tr s 0 th khng c cung cp nng lng. Mi trng h quang mt dn tnh dn in v tr thnh cch in. Nu cch in ny ln v in p ngun khng duy tr phng in li th h quang s tt hn. nh gi mc cch in ca in mi vng h quang l ln hay b ngi ta dng khi nim in p chc thng. in p chc thng ( Uch.t ) cng ln th mc cch in ca in mi cng cao.

Qu trnh dp tt h quang in xoay chiu khng nhng ty thuc vo tng quan gia ln ca in p chc thng vi ln ca in p h quang m cn ph thuc tng quan gia tc tng ca chng. Nu tc tng in p chc thng ln hn tc phc hi in p ngun (hnh 1-5: ng 1 v ng 2 khng giao nhau im no) th h quang s tt hon ton. Trong cc thit b in khi tip im m ra khong cch tng dn lm cch in in mi tng dn (ng 1), na chu k sau cng dc hn na chu k trc.

Hnh 1-4: c tnh ca h quang xoay chiu

i(t )

12

UchUt

U[V]

t

a)

U

I

Uch

Ut

b)

12

Ngc li, tc phc hi in p m nhanh hn tc tng ca in p chc thng ( lm ng 1 v ng 2 giao nhau) th h quang s chy li. Tm li : dp tt h quang in xoay chiu hon ton th ta phi lm sao tng in p chc thng (ng 1) vt cao hn nh ca ng biu din in p phc hi h quang (ng 2). Khi in p ngun l1000V th trong lc dng in qua tr s 0 sau khong 0,1s mc cch in khu vc ny t n gi tr xuyn thng tc thi khong 150 n 250V.

1.4. QU TRNH PHC HI IN P CA H QUANG IN

1. Khi nim

Gi tr tc thi ca in p ngun xut hin gia cc tip im sau khi ngt mch trong qu trnh qu c gi l in p phc hi. a) Trong mch in mt chiu

Ty thuc tnh cht ca ti l in tr, in cm hay in dung m in p phc hi cng khc nhau. Thc t tn ti in dung gia cc dy dn khc nhau, dy dn vi t hay gia cc bi dy vi nhau. Trong mch khi c c R, L, C th in p phc hi ty theo gi tr in tr R m c th dao ng tun hon hay khng. Khi mch R, L, C m c mc thm t in song song vi h quang th trc khi dng in trit tiu t c np v phng in tr li, in p phc hi s dao ng tun hon khi R nh.

Nhng nu tr s in tr R ln s khng th c dao ng tun hon c. b) Trong mch in xoay chiu

Nu h quang c dp tt vnh vin th qu trnh phc hi in p c dng bin thin vi tn s nh dn v bng 0. Nu h quang xut hin li th qu trnh phc hi b ngt v in p gim nhanh t gi tr Uch n gi tr b nht ng vi in p ri trn h quang.

Nu mch in c in tr ln th in p phc hi trn tip im khi c h quang s khng cn xut hin li (c dng khng tun hon). mch in xoay chiu th tn s in p ngun fngun thng thng rt thp so vi tn s dao ng ring ca mch c L v C.

f ngun

13

Trn hnh 1-6d biu din in p phc hi khi ngt mch ng dy khng ti. 2. Nng lng h quang a) Dng mt chiu c tnh dp tt h quang ph thuc vo nng lng h quang. Nng lng h quang dng mt chiu tnh theo :

+t

0

2

hq dt.i).i.RU(2IL=W (1.5)

T phng trnh thy rng ton b nng lng 2I.L 2

tch ly trong mch trc lc ngt cng

vi nng lng ngun sau khi bt phn nng lng tn hao trn in tr R nm trong mch chnh l nng lng h quang (Whq).

Do vy mch mt chiu, in cm ca mch cng ln th nng lng h quang s cng ln, khi h quang s kh dp tt. b) Dng in xoay chiu

H quang xoay chiu dp tt lc i = 0, do nng lng in t xem nh bng 0 v ta c :

n.=t

0hq R.i)i.dt.-(u=W (1.6)

Vi n l s lng bn chu k trong khong thi gian chy ca h quang. Kt qu l dng xoay chiu th nng lng h quang l nng lng ngun tr bt i phn tn hao tc dng. Khc vi dng mt chiu ton b nng lng c a tr v ngun. Nu dng in c ngt trc lc i qua tr s 0 th mt phn ca nng lng t s khng a v ngun m cung cp cho h quang. Do ng trn quan im nng lng m xt th ngt mch dng xoay chiu d dng hn ngt mch dng mt chiu cng mt cng sut.

ng thi ta cn thy mun gim nng lng h quang (mt chiu v xoay chiu) th phi cn gim thi gian t chy ca h quang.

Hnh 1-6: Cc ng c tnh in p phc hi sau khi ct mch trong cc trng hp: a,b) ph ti in cm, c)ph ti in tr , d)ph ti dung

e(t)i(t

) ut t

Uphm Em =90

L

e(t)i(t

) ut t

Um=2Em

Em =90

L

Uphm =0

e(t)i(t

)ut

t

R e(t)i(t

)ut

t

C

i=0

=90 d)c)

b)a)

14

3. Cng thc qui c v cng sut ngt

c trng cho kh nng ngt ln nht ca thit b ng m mch, ngi ta a vo khi nim cng sut ngt (Sngt) c xc nh theo qui c theo cng thc sau :

)MVA(I.CS mngt ngt= (1.7) Trong: C=m.Um=3.Umfa= pha bacho trngc:mdyU.3 .

[kA].mach, m ong thit bcua mcnh t in ng dong dung hiu trgia lamngt I

dung). hiu tr(gia dy mcnh apin lamdyU

dung). hiu tr(gia phamcnh apin la mfaU

Ingtm l dng in ln nht ng vi lc u tin cc tip im ri xa nhau in p nh mc ca thit b ng m mch. Trong cc cng thc trn xt tr s ca cc thng s c bn khi ngt gi tr thit b in khng b xy ra h hng.

1.5. BIN PHP V TRANG B DP H QUANG TRONG THIT B IN 1. Cc bin php v trang b dp h quang trong thit b in cn phi m bo yu cu

-Trong thi gian ngn phi dp tt c h quang, hn ch phm vi chy h quang l nh nht. -Tc ng m tip im phi ln. -Nng lng h quang sinh ra phi b, in tr h quang phi tng nhanh. -Trnh hin tng qu in p khi dp h quang.

2. Cc nguyn tc c bn dp h quang in

-Ko di ngn la h quang. -Dng nng lng h quang sinh ra t dp. -Dng nng lng ngun ngoi dp. -Chia h quang thnh nhiu phn ngn dp. -Mc thm in tr song song dp.

3. Trong thit b in h p thng dng cc bin php v trang b sau a) Ko di h quang in bng c kh

y l bin php n gin thng dng cu dao cng sut nh hoc rle. Ko di h quang lm cho ng knh h quang gim, in tr h quang s tng dn n tng qu trnh phn ion dp h quang. Tuy nhin bin php ny ch thng c dng mng h p c in p nh hn hoc bng 220V v dng in ti 150 A. b) Dng cun dy thi t kt hp bung dp h quang

Ngi ta dng mt cun dy mc ni tip vi tip im chnh to ra mt t trng tc dng ln h quang sinh ra mt lc in t ko di h quang. Thng thng bin php ny kt hp vi trang b thm bung dp bng aming. Lc in t ca cun thi t s thi h quang vo tip gip aming lm tng qu trnh phn ion. c) Dng bung dp h quang c khe h quanh co

Bung c dng bng aming c hai na li lm v ghp li hp thnh nhng khe h quanh co (khi ng knh h quang ln hn b rng khe th gi l khe hp).

Khi ct tip im lc in ng sinh ra s y h quang vo khe quanh co s lm ko di v gim nhit h quang. d) Phn chia h quang ra lm nhiu on ngn

Trong bung h quang pha trn ngi ta ngi ta t thm nhiu tm thp non. Khi h quang xut hin, do lc in ng h quang b y vo gia cc tm thp v b chia ra lm nhiu on ngn. Loi ny thng c dng li mt chiu di 220 V v xoay chiu di 500 V. e) Tng tc chuyn ng ca tip im ng

15

Ngi ta b tr cc l dao ng, c mt l chnh v mt l ph (thng l cu dao) hai l ny ni vi nhau bng mt l xo, l dao ph ct nhanh do l xo n hi(l xo s lm tng tc ct dao ph) khi ko dao chnh ra trc . f) Kt cu tip im kiu bc cu

Mt im ct c chia ra lm hai tip im song song nhau, khi ct mch h quang c phn chia lm hai on v ng thi do lc in ng ngn la h quang s b ko di ra lm tng hiu qu dp. 4. Cc bin php v trang b dp h quang thit b in trung v cao p a) Dp h quang trong du bin p kt hp phn chia h quang

cc my ct trung p cc tip im ct c ngm trong du bin p, khi ct h quang xut hin s t chy du sinh ra hn hp kh (ch yu l H) lm tng p sut vng h quang, ng thi gim nhit h quang. Cc my ct in p cao mi pha thng c phn ra lm nhiu ch ngt. b) Dp h quang bng kh nn

Dng kh nn trong bnh c sn hoc h thng ng dn kh nn khi h quang xut hin (tip im khi m) s lm m van ca bnh kh nn, kh nn s thi dc hoc ngang thn h quang lm gim nhit v ko di h quang. c) Dp h quang bng cch dng vt liu t sinh kh

Thng dng trong cu ch trung p, khi h quang xut hin s t chy mt phn vt liu sinh kh(nh thy tinh hu c,...) sinh ra hn hp kh lm tng p sut vng h quang. d) Dp h quang trong chn khng Ngi ta t tip im ct trong mi trng p sut ch khong 10-6 n 10-8 N/ cm2. mi trng ny th bn in cao hn rt nhiu bn in ca khng kh nn h quang nhanh chng b dp tt. e) Dp h quang trong kh p sut cao

Kh c nn p sut ti khong 200 N/cm2 hoc cao hn s tng bn in gp nhiu ln khng kh. Trong cc my ct in p cao v siu cao p hin nay thng s dng kh SF6 c nn trong cc bnh kh nn dp h quang. H quang dp trong mi trng SF6 rt m bo(bi v ngay c iu kin p sut thng h quang cng tt nhanh trong mi trng kh SF6).

16

Chng 2. TIP XC IN

2.1. I CNG V TIP XC IN

17

1. Khi nim

Ch tip gip gia hai vt dn in cho dng in chy t vt dn ny sang vt dn kia gi l tip xc in. B mt ch tip gip ca cc vt dn in gi l b mt tip xc in.

Tip xc in chia ra lm ba dng chnh: -Tip xc c nh: l hai vt dn tip xc lin kt cht cng bng bulng, inh vit, inh riv,... -Tip xc ng m: l tip xc m c th lm cho dng in chy hoc ngng chy t vt ny sang vt khc (nh cc tip im trong thit b ng ct). -Tip xc trt: l vt dn in ny c th trt trn b mt ca vt dn in kia (v d nh chi than trt trn vnh gp my in).

Tip xc ng m v tip xc trt u c hai phn, phn ng (gi l tip im ng) v phn tnh (gi l tip im tnh).

Ba dng tip xc trn u c th tin hnh tip xc di ba hnh thc: -Tip xc im: l hai vt tip xc vi nhau ch mt im hoc trn b mt din tch vi ng knh rt nh (nh tip xc hai hnh cu vi nhau, hnh cu vi mt phng, hnh nn vi mt phng,...) -Tip xc ng: l hai vt dn tip xc vi nhau theo mt ng thng hoc trn b mt rt hp (nh tip xc hnh tr vi mt phng, hnh tr vi tr,...) -Tip xc mt: l hai vt dn in tip xc vi nhau trn b mt rng(v d tip xc mt phng vi mt phng,...).

Cc yu cu i vi tip xc in ty thuc cng dng, iu kin lm vic, tui th yu cu ca thit b v cc yu t khc. Mt yu t ch yu nh hng ti tin cy lm vic v nhit pht nng ca tip xc in l in tr tip xc Rtx. 2. in tr tip xc

Xt khi t hai vt dn tip xc nhau(hnh 2-1) , ta s c din tch b mt tip xc : Sbk= a . l.

Nhng trn thc t din tch b mt tip xc thc nh hn nhiu a.l v gia hai b mt tip xc d gia cng th no th vn c nhp nh, khi cho tip xc hai vt vi nhau th ch c mt s im trn tip gip tip xc. Do din tch tip xc thc nh hn nhiu din tch tip xc biu kin Sbk= a.l.

Din tch tip xc cn ph thuc vo lc p ln trn tip im v vt liu lm tip im, lc p cng ln th din tch tip xc cng ln.

Din tch tip xc thc mt im(nh mt cu tip xc vi mt phng) xc nh bi:

S = d

F (2.1)

Trong : F l lc p vo tip im [kg].

d l ng sut chng dp nt ca vt liu lm tip im [kg/cm2].

Bng 2.1: ng sut chng dp nt ca mt s kim loi thng dng

Kim loi ng sut d

[N/cm2] Kim loi ng sut d

[N/cm2] bc 30.400 ng cng

(hp kim) 51.000

ng mm 38.200 nhm 88.300 Nu tip xc n im th din tch s ln ln n ln so vi biu thc (2.1).

Hnh 2-1: Tip xc ca hai vt dn

21 a

l 1

2

a

18

Dng in chy t vt ny sang vt khc ch qua nhng im tip xc, nh vy dng in cc ch tip xc s b tht hp li, dn ti in tr nhng ch ny tng ln.

in tr tip xc ca tip im kiu bt k tnh theo cng thc:

Rtx = mF

K [ ] ( 2.2)

K: h s ph thuc vt liu v tnh trng b mt tip im ( theo bng tra). m: h s ph thuc s im tip xc v kiu tip xc vi: +Tip xc mt m = 1 +Tip xc ng m = 0,7 +Tip xc im m = 0,5

Bng 2.2: Tra tr s K trong cng thc (2.2) Kim loi tip xc Tr s K [ .N] Kim loi tip xc Tr s K [ .N] ng - ng ( 0,08 n 0,14).10-2 st - ng ( 3,1).10-2 bc - bc ( 0,06)10-2 nhm - ng ( 0,38).10-2 nhm - nhm ( 0,127).10-2

Ngoi cng thc (2.2) l cng thc kinh nghim, ngi ta cn dng phng php gii tch dn gii rt ra cng thc tnh in tr tip xc im:

Rtx =

..2

d

nF (2.3)

: in tr sut ca vt dn [ .cm]. n: s im tip xc.

F: lc nn [kg]. Do vy r rng in tr tip xc ca tip im nh hng n cht lng ca thit b in, in

tr tip xc ln lm cho tip im pht nng. Nu pht nng qu mc cho php th tip im s b nng chy, thm ch b hn dnh. Trong cc tip im thit b in mong mun in tr tip xc c gi tr cng nh cng tt, nhng do thc t c nhiu yu t nh hng n Rtx nn khng th gim Rtx cc nh c nh mong mun. 3.Cc yu t nh hng n in tr tip xc (Rtx)

in tr tip xc b nh hng ca nhiu yu t vi mc khc nhau, ta xt y mt s yu t ch yu sau: a) Vt liu lm tip im

T (2.3) ta thy h s chng dp nt d b th Rtx b. V vy ng v mt yu cu c in tr tip xc b nn dng cc vt liu mm lm tip im. Nhng thc t cn phi kt hp cc yu t khc(nh bn c) nn vt liu thng dng l ng, ng thau m thic, thp m thic,...

19

b) Lc p ln tip im Cng t cng thc (2.2) v (2.3) lc F

cng ln th Rtx cng nh (hnh 2-2) ng 1 biu din in tr tip xc gim theo chiu lc tng, nu gim lc nn ln tip im in tr tip xc Rtx thay i theo ng 2. Ta c th gii thch l v khi tng lc nn b ln mt tip xc th khng nhng b mt tip xc b bin dng n hi m cn b ph hy cc b. Khi ta gim lc p th mt s im tip xc vn cn gi nguyn nh khi lc p ln tc dng. Tng lc p ch c tc dng gim Rtx giai on u in tr ln v trung bnh. Khi lc p ln th d c tng lc p ln na th in tr tip xc vn khng thay i. c) Hnh dng ca tip im

Hnh dng ca tip im cng nh hng n Rtx. Cng mt lc nhng kiu tip xc khc nhau th Rtx cng khc nhau. T cc cng thc trn ta thy Rtx ca tip xc mt nh nht v c h s m ln nht (tra t cng thc 2.2). d) Nhit ca tip im

Nhit ca tip im thay i s lm Rtx thay i theo kt qu th nghim vi nhit nh hn 2000C c th tnh Rtx qua cng thc:

R tx( ) = Rtx (0)(1+23 ) [ ] (2.4)

Trong : Rtx(0): in tr tip xc 00C, : h s nhit in tr [1/0C]. : Nhit ca tip im [0C].

e) Tnh trng b mt tip xc B mt tip xc khi b bn hoc khi b oxit ha c Rtx ln hn nhiu Rtx ca tip im sch (do

c nhiu im khng c tip xc trc tip bng vt liu lm tip im). Khi b oxy ha cng nhiu th nhit pht nng trn b mt tip xc cng cao. Tip im b oxy ha c in tr tip xc tng hng chc ln(v oxit ca phn ln kim loi dn in km hn nhiu kim loi nguyn cht). f) Mt dng in Din tch tip xc c xc nh ty theo mt dng in cho php. Theo kinh nghim i vi thanh dn bng ng cho tip xc nhau khi ngun tn s 50 Hz th mt dng in cho php l:

Jcp = SI

[( 0,31 - 1,05 .10-4 (I-200)] [A/mm2] ( 2.5)

Trong : I l gi tr dng hiu dng ; S=Sbk din tch tip xc biu kin. Biu thc (2.5) trn ch ng khi dng in bin thin trong khong t 200 n 2000A. Nu ngoi tr s th c th ly: I < 200 A ly Jcp = 0,31 [A/ mm2] I > 2000 A ly Jcp = 0.12 [A/ mm2]. Khi vt dn tip xc khng phi l ng th mt dng cho phep i vi vt liu y c th ly theo cng thc sau:

Jcp vt liux = Jcp.ng x liu)vt (

ng)(txRR

(2.6)

2.2. TIP IM THIT B IN

1. Vt liu lm tip im

Rtx[106]

F[kg]0 5 10 15 20 25 100

200

300

400

12

Hnh 2-2: in tr tip xc khi lc nn tng

20

tha mn tt cc iu kin lm vic khc nhau ca tip im thit b in th vt liu lm tip im phi c c nhng yu cu c bn sau:

-C dn in cao(gim Rtx v chnh in tr ca tip im). -Dn nhit tt (gim pht nng cc b ca nhng im tip xc). -Khng b oxy ha (gim Rtx tng n nh ca tip im). -C kt tinh v nng chy cao (gim mi mn v in v gim s nng chy hn dnh

tip im ng thi tng tui th tip im). -C bn c cao (gim mi mn c kh gi nguyn dng b mt tip xc v tng tui th

ca tip im). -C do ( gim in tr tip xc). -D gia cng khi ch to v gi thnh r. Thc t t vt liu no p ng c y cc yu cu trn. Trong thit k s dng ty tng

iu kin c th m trng nhiu n yu cu ny hay yu cu khc. Nhng vt liu thng dng gm: a) ng ki thut in: ng nguyn cht thu c bng in phn. N p ng hu ht cc yu cu trn. Nhc im chnh ca ng ki thut in l rt d b oxit ha. b) ng caimi: ng ki thut in pha thm caimi c tnh cht c cao chng mi mn tt, kh nng chu c h quang tt hn ng ki thut in thng thng. c) Bc: l vt liu lm tip im rt tt do c dn in cao v c in tr tip xc n nh. Nhc im ch yu l chu h quang km nn s dng b hn ch. d) ng thau: hp kim ng vi km c s dng lm tip im dp h quang. e) Cc hp kim ng khc: hp kim ng vi nhm, ng vi mangan, ng vi niken, ng vi silic v cc hp kim ng khc c s dng lm tip im, ng thi lm l xo p (v d tip im tnh ca cu ch). Nhng tip im nh vy khi b t nng d b mt tnh n hi. f) Thp c in tr sut ln: thp thng b oxy ha cao nhng l vt liu r nn vn c s dng lm tip xc c nh dn dng in ln, trong cc thit b thp thng c m. g) Nhm: c dn in cao, r nhng rt d b oxy ha lm tng in tr sut. Nhc im na l hn nhm rt phc tp, bn c li km. h) Vonfram v hp kim vonfram: c mi mn v in tt v chu c h quang tt nhng c in tr tip xc rt ln. Hp kim vonfram vi vng s dng cho tip im c dng nh. Hp kim vi molipen dng lm tip im cho nhng thit b in thng xuyn ng m, khi dng in ln th vonfram v hp kim vonfram s dng lm tip im dp h quang. i) Vng v platin: khng b oxy ha do c in tr tip xc nh v n nh, c s dng lm tip im trong thit b in h p c dng in b v quan trng. Vng nguyn cht v platin nguyn cht c bn c thp nn thng c s dng dng hp kim vi mlipen hoc vi irii tng bn c. j) Than v graphit: c in tr tip xc v in tr sut ln nhng chu c h quang rt tt. Thng dng lm cc tip im m khi lm vic phi chu tia la in, i khi lm tip im dp h quamg. k) Hp kim gm: hn hp v mt c hc ca hai vt liu khng nu chy m thu c bng phng php thiu kt hn hp bt hoc bng cch tm vt liu ny ln vt liu kia. Thng vt liu th nht c tnh cht k thut in tt, in tr sut v in tr tip xc nh, t b oxy ha.Vt liu th hai c tnh cht c cao v chu c h quang. Nh vy, cht lng kim loi gm l do tnh cht ca hn hp quyt nh. Kim loi gm s dng rng ri nht thng c gc bc nh : bc-niken, bc- oxit caimi, bc- vonfram, bc-mlipen. Ngoi ra i khi ngi ta s dng kim loi gm c gc ng nh: ng -vonfram, ng -mlipen, ng caimi lm tip im chnh v tip im dp h quang. Ch +Vi tip xc c nh thng dng vt liu l ng, nhm, thp. +Vi tip xc ng/m ty theo dng dn, nu : -Dng in b dng bc, ng, platin, vonfram, i khi vng, mlipen, niken. -Dng va n ln dng ng thau, kim loi hoc hp kim t nng chy nh vonfram, molipen,... -Dng in ln th thng dng hp kim gm (sn phm hai kim loi dng bt p li p lc ln, nhit cao. Hp kim gm rt cng chu c dng ln, khuyt im l dn in km, nn thng c ch to dng tm mng hn trn b mt tip im ca thit b).

21

2. Mt s kt cu tip im a) Phn ra lm cc loi theo cu to Tip xc c nh c cc dng -Ni hai thanh tit din ch nht. -Ni hai thanh tit din trn (thanh trn ni vi nhau thng trong cc thit b in nh my ngt in, my bin dng,...). Loi tip xc ng m v tip xc trt phn theo dng in

-Dng b : I 10 [mA]. -Dng va: I 100 [A]. -Dng ln: I > 100 [A].

b) Tip im rle Thng dng bc, platin tn hn g vo tip im, kch thc tip im do dng in cho php

quyt nh (theo bng c trong cc s tay thit k). c) Tip im thit b in khng ch

Cc thit b nh cng tc t, ptmt v thit b cao p thng c dng in ln. Th nhng tip im chnh mc song song vi tip im h quang khi tip im v tr ng dng in s qua tip im chnh (tip im) lm vic, khi m hoc bt u ng tip im h quang s chu h quang. Do bo v c tip im lm vic. Ta thng thy tip im c cc dng nh hnh 2-3. +Hnh ngn: dng trong cng tc t, tip im ng va trt va ln trn tip im tnh do vy c th t lm bc lp oxit trn b mt tip xc. +Tip im bc cu: dng trong rle v cng tc t. +Tip im i din: dng my ngt in p cao. +Tip im hoa hu: gm mt cnh hnh thang ging cnh hoa hu hay ch z, tip im ng l mt thanh dn trn. +Tip im vut ma: tip im ng kiu sng dao c th trt gia hai vut trn (lm tip im tnh) l xo v dy c ni cht vi vut. +Tip im chi: tip im ng hnh chi gm nhng l ng mng 0,1 0,2 mm xp li trt ln sng dao tip im tnh. tng lc p trn tip im hnh chi th thng c thm bn n hi. Loi ny khi chi b chy s lm in tr tng nhanh do t dng lm tip im h quang.

22

3. Nguyn nhn h hng tip xc v bin php khc phc a) Nguyn nhn h hng

Nguyn nhn h hng tip xc c rt nhiu, ta xt mt s nguyn nhn chnh sau: a.1) n mn kim loi

Trong thc t ch to d gia cng th no th b mt tip xc tip im vn cn nhng l nh li ti. Trong vn hnh hi nc v cc cht c hot tnh ha hc cao thm vo v ng li trong nhng l nh s gy ra cc phn ng ha hc to ra mt lp mng mng rt gin. Khi va chm trong qu trnh ng lp mng ny d b bong ra. Do b mt tip xc s b mn dn, hin tng ny gi l hin tng n mn kim loi. a.2) Oxy ha

Mi trng xung quanh lm b mt tip xc b oxy ha to thnh lp oxit mng trn b mt tip xc, in tr sut ca lp oxit rt ln nn lm tng Rtx dn n gy pht nng tip im. Mc gia tng Rtx do b mt tip xc b oxy ha cn ty nhit . 20-30oC c lp oxt dy khong 25.10-6mm. Theo th nghim tip im ng ngoi tri sau mt thng Rtx tng ln khong 10%. nhit ln hn 700C s oxit ha rt nhanh. Theo th nghim 1000C sau ch mt gi Rtx ca tip im ng tng khong 50 ln. Ngoi ra vic lun phin b t nng v lm ngui cng tng qu trnh xit ha. a.3) in th ha hc ca vt liu tip im

Mi cht c mt in th ha hc nht nh. Ly H lm gc c in th m (-) th ta c bng mt s kim loi c in th ha hc nh bng sau:

Bng 2.3: in th ha hc ca mt s kim loi

Kim loi Ag Cu H Sn Ni Co Fe Al in th ha hc [ V].

+0.8 +0.345 0 -0.14 - 0.2 -0.255 -0.44 - 1.34

23

Hai kim loi c in th ha hc khc nhau khi tip xc s to nn mt cp hiu in th ha hc, gia chng c mt hiu in th. Nu b mt tip xc c nc xm nhp s c dng in chy qua, v kim loi c in th hc m hn s b n mn trc lm nhanh hng tip im. a.4) H hng do in

Thit b in vn hnh lu ngy hoc khng c bo qun tt l xo tip im b hoen r yu i s khng lc p vo tip im. Khi c dng in chy qua, tip im d b pht nng gy nng chy, thm ch hn dnh vo nhau. Nu lc p tip im qu yu c th pht sinh tia la lm chy tip im. Ngoi ra, tip im b bn, r s tng in tr tip xc, gy pht nng dn n hao mn nhanh tip im. b) Cc bin php khc phc

bo v tip im khi b r v lm gim nh in tr tip xc c th thc hin cc bin php sau: b.1) i vi nhng tip xc c nh nn bi mt lp m chng r hoc qut sn chng m. b.2) Khi thit k ta nn chn nhng vt liu c in th ha hc ging nhau hoc gn bng nhau cho tng cp. b.3) Nn s dng cc vt liu khng b oxy ha lm tip im. b.4) M in cc tip im: vi tip im ng, ng thau thng c m thic, m bc, m km cn tip im thp thng c m caini, niken, km,... b.5) Thay l xo tip im: nhng l xo r, yu lm gim lc p s lm tng in tr tip xc, cn lau sch tip im bng vi mm v thay th l xo nn khi lc nn cn qu yu. b.6) Kim tra sa cha ci tin: ci tin thit b dp h quang rt ngn thi gian dp h quang nu iu kin cho php. 4. Tnh trng lm vic ca tip im khi ngn mch

Khi c ngn mch, nhit ch tip xc tng cao lm gim tnh n hi v cng c kh ca tip im. Nhit cho php khi ngn mch quy nh: -Vi ng, ng thau: [ ] = (200 300)0C -Nhm: [ ] = (150 200)0C Ty thi gian ngn mch c mt dng in cho php khc nhau nh bng 2-4.

Bng 2.4: Mt dng in cho php Vt liu tip xc Mt dng in cho php jcp [A/mm2 ] Thi gian ngn mch [s] 1s 5s 10s ng 152 67 48 ng thau 75 38 27 Nhm 89 40 28 Ngoi ra cn ty tnh trng lm vic ca tip im khi ngn mch xy ra nh: -Tip im v tr ng khi ngn mch

Theo cng thc kinh nghim Butkvich: Im = K. F (2.7) Vi: Im: dng in bin lm tip im nng chy hn dnh. K: h s ty vt liu lm tip im v s im tip xc. F: lc nn ln tip im, F = (20 50) kg. H s K trong mt s trng hp c th sau: + Tip im chi ng, ng thau: K= 3000 n 4000 + Tip im hnh ngn bng ng: K= 4100. + Tip im kiu cm ng, ng thau: K= 6000. -Tip im trong qu trnh ng b ngn mch

Lc ny sinh lc in ng ko di tip im, tip im ng c tc ln d sinh hin tng hn dnh v c chn ng.

Khi dng chy trong vt dn t tit din ln sang tit din nh thng b un cong sinh lc in ng theo cng thc:

24

F = 1,02.10-8.i2lndD

(2.8)

D,d: ng knh tit din ln v nh [cm]. -Tip im trong qu trnh ngt b ngn mch

Pht sinh h quang c th lm chy tip im. Ty kim loi c tr cc tiu p v cc tiu dng c th pht sinh h quang.

Bng 2.5: Tr s dng, p cc tiu Kim loi tip im W Ag Cu Al Fe

Imin [A] 0,8 0,75 0,42 0,5 0,55 Umin [V] 11,5 12 14 12,5 12,5

+Khi ct dng b Nu I Imin , U Umin : Gia hai tip im hnh thnh mt cu kim loi nng chy, cu b t kim loi s chy t ant sang catt. V vy tip im l ant b mn. Nu I Imin , U Umin : Hnh thnh cc ion n bn ph pha catt, kim loi s chuyn t catt sang ant. +Trng hp ct dng trung bnh v dng in ln

H quang ln c catt v ant u b mn. Cn ch tip im ng khi ng c khi b hao mn nhiu hn khi m. S hao mn t l vi dng in, s ln ng m v lng in tch qua tip im v thi gian chy ca h quang, l cc hao mn v in (do dng in gy ra). Ngoi ra cn hao mn v c, thng thng hao mn v c bng (1 3)% hao mn in. Chng 3. PHT NNG 3.1. I CNG 1. Khi nim chung Nhit lng sinh ra do dng in chy qua trong cun dy hay vt dn in khi thit b in lm vic s gy pht nng. Ngoi ra trong thit b in xoay chiu cn do tn hao dng xoy v t tr trong li st t cng sinh ra nhit. Nu nhit pht nng ca thit b in vt qu tr s cho php th thit b in s nhanh b h hng, vt liu cch in nhanh b gi ha, bn c kh ca kim loi b gim st. Nhit cho php ca cc b phn ca thit b in tham kho theo bng cho sn. Trong tnh ton pht nng thit b in thng dng khi nim chnh nhit l hiu s gia nhit pht nng v nhit mi trng xung quanh thit b in 0. vng n i cho php =350C, vng nhit i =500C. S pht nng thit b in cn ty thuc vo ch lm vic. Thit b in c ba ch lm vic: di hn, ngn hn v ngn hn lp li. 2. Cc ngun nhit trong thit b in-Cc phng php truyn nhit

25

Trong thit b in mt chiu s pht nng ch yu l do tn hao ng. i vi thit b in xoay chiu, s pht nng sinh ra ch yu l do tn hao ng trong dy qun v tn hao st t trong li thp, ngoi ra cn tn hao do hiu ng b mt. Song song vi qu trnh pht nng c qu trnh ta nhit gm: dn nhit, bc x nhit v i lu nhit. Qu trnh dn nhit, nhit lng dn tnh theo cng thc

dQ = - .

X

Q .dS.dt

Trong : dQ: nhit lng c dn theo phng x.

X

Q: graien nhit lu theo phng x; dS: din tch nhit lu i qua, dt: thi gian; :

h s dn nhit [W/0C.cm]. Bc x nhit: ph thuc b mt ta nhit i lu nhit: phn lm i lu t nhin v i lu cng bc, i lu ph thuc vo v tr phn b ca vt th, kch thc b mt, tnh cht mi trng xung quanh vt v nhit mi trng.

Nu xt c ng thi ba hnh thc trn th c cng thc Niutn sau:

P = .S. hay = SP

Trong : P: nhit lng ta ra; S: din tch ta nhit. : chnh nhit ca vt dn vi mi trng. : h s ta nhit [N/0C.cm2]. Dng cng thc trn rt tin nhng sai s c (15 25)% H s tra trong ti liu thit k: +Vi cun dy truyn nhit tt trong phm vi nhit 750C 1200C h s l: = 11.10-4 n 12,98.10-4 [W/0C cm2] +Vi cun dy truyn nhit km: = 9,84.10-4 n 11,52.10-4 [W/0C. cm2].

3. Nhit pht nng v cp cch in Nhit mi trng xung quanh quy nh cho cc nc vng n i

0 = 350C, nc vng nhit

i 0 = 400C. Nhit pht nng chnh lch = - 0 quy nh vng n i th: =350C, vng nhit i =500C. Cp cch in: cn c vo kh nng chu nhit pht nng ln nht ca vt liu cch in m khng lm ph hy tnh cht c ca n, ngi ta chia vt liu cch in ra cc cp cch in gm cp: A : [T0] = (90 105)0C E : [T0] = (105 120)0C B : [T0] = (120 140)0C Cc b phn thit b in quy nh + Vt liu khng bc cch in xa vt cch in [T0] =110. + Dy ni tip xc c nh [T0] = 750C + Tip xc m bc [T0] =1200C + Vt liu dn in c bc cch in th: -Cp O: [T0] 800C -Cp A : [T0] 950C -Cp B: [T0] 1100C + Vt liu khng dn in khng bc cch in [T] 1100C Ngoi ra ch lm vic khc nhau c nhit ln nht cho php khc nhau.

3.2. CH LM VIC DI HN CA VT TH NG NHT

26

Thit b in lm vic di hn tc l thit b in c th lm vic lin tc lu di nhng thi gian lm vic phi khng nh hn thi gian cn thit thit b pht nng n nhit n nh. Khi c dng in I chy trong vt dn s gy ra tn hao mt cng sut P v trong thi gian dt s gy ra mt nhit lng:

P.dt = RI2dt (3.1) Nhit lng hao tn ny bao gm hai phn: -t nng vt dn G.C.d -Ta ra mi trng xung quanh S ..dt. Ta c phng trnh cn bng nhit ca qu trnh pht nng:

P.dt = G.C.d + S ..dt (3.2) Trong : G l khi lng vt dn [g] C l t nhit vt dn ta nhit [J/g]. l chnh nhit [00C]. l h s ta nhit [W/cm2]. T (3.2) ta c phng trnh :

C.G

P =dtd

+C.G

.S . (3.3) Gii phng trnh vi phn (3.3) vi iu kin ti t = 0 th chnh nhit ban u l 0, ta c:

= .SP (1 -

tGC

S

e

) + 0 t

GCS

e

(3.4)

t T = .SC.G l hng s thi gian pht nng.

.SP = : chnh nhit n nh. Ta c:

= ( 1- Tt

e

) + 0 Tt

e

(3.5)

Khi t = 0 m 0 = 0 th:

= .(1- Tt

e

) (3.6) Khi ngt dng in (I = 0), qu trnh pht nng chm dt v qu trnh ngui lnh bt u xy ra, ngha l P.dt = 0, ta c phng trnh ngui lnh:

I2R.dt = 0 (3.7) V: G.C.d + S +dt = 0 nn c:

dtd + .S

C.G = 0 (3.8)

Vi iu kin khi ngt dng in chnh lch nhit bng chnh lch nhit n nh. Gii phng trnh vi phn (3.8) ta c biu thc th hin qu trnh ngui lnh:

= .e tT

Hng s thi gian pht nng T l khong thi gian cn thit t nng vt ln ti chnh nhit n nh nu khng c s ta nhit ra mi trng xung quanh . Xc nh hng s T bng gii tch, ta c: P dt = G.C.d

dtd

= C.G

P th = C.G

P .t + 0

0 0.632

3

t[s]

1 2

0

T

AB

Hnh 3-1: Pht nng di hn

27

Nu 0 = 0 th: = C.GP .t

Khi 0 = th t = T. T = C.G

P.T v theo cng thc Niutn =

S.

P

.

Ta c: T= .SC.G

(3.9)

Dng phng php v cng c th xc nh c gi tr T. T gc ta gc ta v ng tip tuyn vi ng cong 1 v ng cong 2. Ta nhn c AB = T.

ddt

t =0 = T

= tg = ABBC

Trong BC = vy AB = T. Qu trnh pht nng c ta nhit ra mi trng xung quanh th sau thi gian T chnh lch nhit ch t ti gi tr 0,632 .

3.4. CH LM VIC NGN HN CA VT TH NG NHT

ch lm vic ngn hn chnh lch nhit ca thit b in sau thi gian lm vic cha t ti tr s n nh th thit b in ngng lm vic. Nhit pht nng ch ny l nh nht. Khi ngng lm vic (I= 0) th qu trnh ngui lnh li bt u. Gi s lm vic di hn ng cong pht nng l ng 1 trong hnh 3-2. Ph ti lc ny l Pf :

Pf = S.f (3.10) Sau thi gian tlv (thi gian lm vic ngn hn) chnh nhit mi t ti tr 1 < f, nn thit b in lm vic non ti v cha li dng ht kh nng chu nhit. T ta thy rng c th nng ph ti ln sau thi gian lm vic ngn hn tlv chnh nhit va t ti tr s cho php f, ph ti lc ny l Pn:

Pn = S. max (3.11) ng cong pht nng trng hp ny l ng 2. im M trn ng 2 tha mn phng trnh chnh nhit ca qu trnh pht nng.

f = max (1- e Ttlv

) (3.12)

Sau thi gian lm vic tlv dng in ngng chy vo vt dn do vt dn ngui lnh theo quy lut nh khi lm vic di hn (ng 3).

T cc biu thc (3.10), (3.11), (3.12) v gi Kp = fn

PP

l h s qu ti cng sut ta rt ra:

Kp = fn

PP

= f

max

=

Ttlv

e1

1

> 1 (3.13)

V cng sut t l vi bnh phng dng in nn:

KI = fn

II

= PK =

Ttlv

e1

1

(3.14)

KI : h s qu ti v dng in.

max f 1

0 t[s] tlv

1

2 3

Hnh 3-2: Pht nng khi ngn hn

M

28

V du: Mt thit b in c T = 180s nu lm vic di hn th dng in cho php If = 100 A nhng nu lm vic ngn hn trong thi gian tlv = 5 s th c th tng dng din ln bao nhiu ?. Gii:

KI = 1

1 etTlv

= 1

15

180 e = 6

Vy dng cho php ln nht l: In = KI. If = 6.100 = 600 [A].

3.4. CH LM VIC NGN HN LP LI CA VT TH NG NHT

y l ch m thit b in lm vic trong mt thi gian tlv m nhit pht nng cha t

ti bo ha v sau ngh mt thi gian tng m nhit cha gim v nhit ban u ri li tip tc lm vic v ngh xen k. Qu trnh lm vic v ngh c lp li tun hon nh vy. th hin mc lm vic lp, ngi ta dng khi nim h s lm vic (cn gi h s ng in):

L% = ngtlvt

lvt

+ .100% (3.15)

Trong thc t L% thng bng 25%, 40%, 60%. Trong ch lm vic ngn hn lp li, nhit pht nng nh hn ch lm vic di hn nhng ln hn ch ngn hn. Tng thi gian lm vic tlv v thi gian ngh tng gi l thi gian chu k tck.

tck = tlv + tng Ta gi thit ti thi im ban u chnh nhit ca vt dn l 0 sau thi gian lm vic tlv

vt dn c t nng n chnh nhit l:

1= (1-e Ttlv

) + 0 e Ttlv

(3.16) Sau thi gian ngh tng vt dn ngui xung nhit

:

2 = 1 e Ttng

(3.17) Chu k tip theo vt dn li b t nng ti chnh nhit :

3= (1- e Ttlv

) + 2 e Ttlv

(3.18) Sau mt s chu k nhit chnh lch nhit

t n chnh nhit cc i max v chnh lch nhit cc tiu min khng thay i, ta gi l thi k n nh. Tng t nh trn, ta vit:

Qu trnh pht nng max = (1- e Ttlv

) + min e T

tlv (3.19)

Qu trnh ngui lnh: min = max . e Ttng

(3.20) Gii hai phng trnh ny ta c:

t[s]

tlvtngtcK

34

1

2

maf mmi

Hnh 3-3: Pht nng khi ngn hn lp li

29

max = T

tt

Tt

nglv

lv

e1

e1

+

(3.21)

Vi: : chnh nhit n nh bng chnh nhit cho php f [0C]. max: chnh nhit ln nht khi lm vic ngn hn lp li [0C]. C: max< f = nn c th cho tng ti thm ln lm vic nh ng cong pht nng 2(ng vi nl> f) hnh 3-3, sau thi gian lm vic = f. Ta c:

f = nl T

ttTt

nglv

lv

e1

e1+

(3.22)

H s qu ti cng sut: Kp = cf

nl

=

TtTt

lv

CK

e1

e1

(3.23)

H s qu ti dng in:

KI = fnlII

= PK =

TtT

t

lv

CK

e1

e1

(3.24)

Hnh 3-3 so snh c tnh pht nng khi lm vic trong ch ngn hn lp li (ng 3) vi c tnh pht nng khi lm vic di hn (ng 1) ta thy khi lm vic ngn hn lp li li c th tng thm ph ti (ng 4).

3.5. S PHT NNG KHI NGN MCH

Thi gian xy ra ngn mch rt ngn nn nhit cung cp cho vt th hon ton dng t

nng vt dn v gn ng ta coi khng c nhit lng ta ra mi trng xung quanh. Trong thi gian dt dng in ngn mch sinh ra nhit lng l:

dQ = K2m. I2 .R.dt = K2m .I2 . sl .dt (3.25)

Trong : Km = II nm , vi Inm l tr s dng ngn mch qua vt dn; I l dng in nh mc qua

vt dn; S l tit din vt th. Ton b nhit lng do dng in ngn mch sinh ra dng t nng vt dn ln chnh nhit l dnm . Ta c phng trnh:

dQ = C.G.dmn = C.S.l. .dmn (3.26) Vi l khi lng ring ca vt dn. C l nhit dung ring ca vt dn. So snh biu thc (3.25) v (3.26) ta c: dmn = c.

K2m

2

F

I

.dt. Ly tch phn ta c:

30

nm = cm

K

.

.

2

.

t0

dt.2

S

I (3.27)

-Khi I = const th: dt2

S

It

0

= 2

S

I

t = J2t. C: nm= cm

K

.

.

2

J2t (3.28)

Nu chnh nhit lc bt u ngn mch l th khi kt thc ngn mch chnh nhit s l: ,nm = + nm. Trong thc t , C thay i theo nhit : C = C0 [ 1+ b0 ( + nm )], = 0 [ 1+ 0 ( + nm )]. Trong : C0: nhit dung ring khi = 0; b0: h s nhit t nhit. 0: in tr sut khi = 0; 0: h s nhit in tr. Thay vo (3.28) ta c:

nm =

2m

Kdt.

2

S

I.

)]nm(0b1[0c

)]nm(01[0

++++

(3.29)

Chng 4. LC IN NG

4.1. KHI NIM CHUNG

Mt vt dn t trong t trng, c dng in I chy qua s chu tc ng ca mt lc. Lc c hc ny c xu hng lm bin dng hoc chuyn di vt dn t thng xuyn qua n l ln nht. Lc chuyn di gi l lc in ng. Chiu ca lc in ng c xc nh theo quy tc bn tay tri.

trng thi lm vic bnh thng, thit b in c ch to lc in ng khng lm nh hng g n bn vng kt cu. Khi ngn mch dng tng ln rt ln (c lc ti hng chc ln Im) do lc in ng s rt ln. Trong mt s trng hp dng ln, lc c th ti hng chc tn. Lc lm bin dng i khi c th lm ph v kt cu thit b. Do cn phi nghin cu lc in ng ngn nga tc hi ca n khi la chn, tnh ton v thit k thit b in.

Ngoi ra ngi ta cn nghin cu ng dng lc in ng ch to cc thit b in nh rle in ng, c cu o in ng,...

4.2. CC PHNG PHP TNH TON LC IN NG

1. Phng php s dng nh lut Bio-Xavar-Laplax Theo quan im ca phng php ny lc in ng l kt qu tng tc ln nhau ca dy dn l

mang dng in I v t trng do dy dn khc to nn.

- Lc in ng tc dng ln chiu di l khi c dng in I t trong t trng c t cm B l: == sin.l.B.IFhayBxl.IF Vi gc l gc hp bi l v B ( l cng chiu I ). l gc xc nh theo chiu quay nh nht.

- Dng vi phn l BxldIFd .= = sin.dl.B.IFd (4-1)

C : dl

trng chiu dng in i. T ta c lc in ng :

==l

0

l

0

sinBdl.IFdF = I.B. l. sin (4-2)

B

dl l

M

I

Hnh 4-1: Lc in ng

31

- Nu hai dy dn cng trong mt mt phng = 900 th =l

0

IBdlF =I.B.l.

Mun xc nh c F ta phi tm c quan h B = B(l), cm ng t ph thuc kch thc dy dn.

- Theo Bio-Xavar-Laplax th cng t cm ti mt im M B c tr s l : == 202 00 rsin.dlI4B hay, r rxldI4B (4-3)

Trong :

=

ln. tao 0r vald do

gmt ph vigoc thng tcam ng t vec: B

0r vald bi hpgoc :

M.n dl cach t khoang la :r

1 0r co Mn dl chon t n v vectla 0r

2. Phng php cn bng nng lng

Xt mt dy dn c dng in chy qua nh hnh 4-2. Khi dy dn dch chuyn theo hng x mt on dx th lc in ng c xc nh bi :

dw = F.dx dxdwF = (4-4)

Trong : + dw : bin thin nng lng t trng ca vt dn

mang dng in khi di chuyn mt on dx. + x : phng chuyn di c th c ca dy dn di tc

dng ca lc F.

+ Chiu F trng vi chiu dx. V d: xt h hai vt dn mang hai dng in i1 ; i2 nh hnh 4-3 t song song cch nhau mt khong x. Nng lng t trng ca h l:

=

=

++==

++=

]cm/J[dx

dL.i

21F

]cm/J[dx

dL.i

21F

: la se re ringdung taclc co Tadx

iMiiL21iL

21d

dxdw

F

: la dung taclc vaiMiiL21iL

21W

2222

1211

21222

211M

21222

211M

Khi vt th bin dng hoc chuyn di ta gi thit cc dng in bng hng s. Theo phng php ny mun tnh lc ta phi bit c biu thc ton hc ca h s t cm L v h cm M theo x. Cc phng php tnh L v M nu trong gio trnh l thuyt trng in t.

4.3. TNH LC IN NG TC DNG LN VT DN

x A A

B B

F

dx

l

Hnh 4-2: Lc tc dng vo thanh dn.

x dx I2

I1

Hnh 4-3: Lc gia hai vng dy

32

1. ng dng phng php cn bng nng lng Ta xt lc in ng trong mt s trng hp vt dn ng nht nm trong t trng u. Cc

trng hp khc c th tham kho ti liu chuyn ngnh ch to thit b. a) Lc in ng tc dng ln mt vng dy c dng i nm trong mt t trng

Gi thit bn knh vng dy R, bn knh dy dn r (hnh 4-4). Lc in ng c xu hng ko cng vng dy dn bung ra. Gi thit lc phn b u trn chu vi vng dy. Gi fR l lc tc dng ln mt

n v di chu vi theo hng knh, lc tc dng tng: dRdL.I

21f.R.2F 2R == (4-6)

Theo Kic khp c:

= 75,1rR8lnRL 0 .

V ta gi thit 1Rr2 >r th:

al.I.10.04,2F 28= [kg] (4.11)

Nu dng trong hai dy cng chiu th hai dy dn s ht nhau v ngc chiu th y nhau. 2. ng dng nh lut Bio-Xavar-Laplax a) Lc in ng tc dng ln hai dy dn t trong cng mt mt phng

Hnh 4-5: Hai thanh t song song

l

I

2r

a

dF B l

FTd

I

2r

Hnh 4-4: Lc cng vng dy

33

Trn hnh 4-6 l hai dy dn l1 v l2 cng t trong mt mt phng. Dy dn l1 mang dng I1 dy dn l2 mang dng I2.

Ta tm s phn b lc ln dy dn l2. Ta chn trc tung oy trng vi dy l1 (chn h xoy hnh 4-6). Dng I1 n v dy trong dy l1 to

ra on dl c cng t cm l :

20

10

r

rxydI

4Bd

GGG

= hay:

210

r

)sin(dyI4

dB =

V c: sin( )= sin nn: 21

0r

sindyI4

dB =

Lc tc dng ln on dl2 do I1dy gy ra l: Bdxld.IFd 22

GGG = Hay:

02221

0 90sin.r

sindl.dyII4

dB =

T hnh 4-6 ta c :

y=cotg ==

sinxr;d

sin

xdy;2

Vy:

= d.sin.dl.x.4

II.dF 2

210 (4.12)

Lc tc dng ln on dl2 v tr x trn do dng I1 chy trong l1 gy ra l :

=

2

1

d.sinx.4

I.I.dF 210x (4-13)

Lc tc dng ln mt n v di ca dy l2 ti v tr xi do 11 ltrongI gy ln l :

i

i1i22102

xx x

coscos.

4I.I.

dl

dFF i

i

== (4-14) Ch y : khi chn cc im tnh x dc chiu di l2 gc v di x bin thin dn n cc lc Fx bin

thin khng u dc chiu di l2 ca dy 2. im tc dng ca lc tng F s qua trng tm dy l2. Bng phng php v ta c th bit s phn b ca lc dc chiu di dy l2. b) Lc in ng gia hai dy dn t song song trong mt dy di v tn

Hnh 4-7, xt khi dy l1 = ; dy l2 = l khong cch gia hai dy x = a. p dng biu thc (4.14) ta thay 1 = ; 2 = 0; x = a vo ta c : consta.4

I.I.2F 210xi =

= Lc in ng tc dng ln dy dn l2 l :

a

l.

42I.1I.02

2F = (4-14)

v c ][....,2Fhay [J/cm] ...., kga

lII

a

lIIF 81021042

81021202== .

c) Lc in ng gia hai dy dn song song c chiu di bng nhau p dng cng thc (4.12) phn trc v thay x = a; dl2 = dy ta c :

l1l2

ldl

dyy

I2

I1

1

2x

Hnh 4-6: Hai thanh trong cng mt phng

34

)1cos2(cosdy.a.4

2I.1I.0dF

= (4-15)

Trn hnh 4-7 c : 22

11222ay

y)cos(coscon , a)yl(

ylcos+

==+

=

Vy :

+ +

+

= l

0 2a2y

ydyl

0 2a2)yl(

dy)yl(

a.42I.1I.0F (4-16)

Tnh tng tch phn ring r c :

+

=l

0 2a2y

ydyA

Nu t z2= y2+a2 2zdz = 2ydy v: + khi y= 0 th z= a

+khi y=1 th z= l a2 2+ i cn ta c :

.a2l2a2a2l

adz

l

0

2a2y

ydyA +=

+=

+=

i cn ta c:

a2a2l0

l 2a2u

udu- l

0 2a2)yl(

dy)yl( +=+

=+

T thay vo (4.16) ta c :

+

=

+

=la

2l

2a12a

l.2

2I.1I.0a2l2a2.a.4

2I.1I.0F

t : co 1)l

a( thal khichnh u ham higoicon hay

l

a2l

2a1)

l

a( >>+=

][).(...., :hay]/[).(..., kgl

a

a

lIIFcmJ

l

a

a

lIIF 81021042

8102120==

Khi hai thanh dn c tit din ch nht vi kch thc rng b, cao h v di l + Nu c b h, b a th :

]cm/J[810.2a

2hlln

a

harctg

a

h22h

1.l2I1I2,0F

+=

. C th vit di dng :

]cm/J[)f(810.a

l2I1I.2,0F = hay ]kg[)f(810.a

l2I1I.04,2F =

=====

=+

=

0 u l ykhil u 0 ykhi

dy- du

y;- l t u t tngl

0 2a2)yl(

dy)yl(B

2

1

al

y

d

lI1 I2

Hnh 4-7: Hai thanh song song

35

c (f) gi l hm Dwight ph thuc theo bh

ba;

a

h

+

+ Nu h

36

Theo th nghim sau t = th c i t cc i imax = 1,8 2 I v lc: FMax = CI2 = C.6,48I2.

2. Lc in ng trong mch xoay chiu ba pha Gi s dng in trong cc pha A, B, C ln lt l :

+=

=

=

)3

2t.sin(.2I2i

)3

2t.sin(.2I2i

t.sin.2I1i

a) Khi b tr ba dy trn mt mt phng (hnh 4-10a) Gi C1 hng s lc gia dy A v B, C2 dy B v C, C3 dy A v C . Ta c: + Lc tc dng ln dy pha A l:

++=+=

)3

2tsin(.tsin3C)3

2tsin(.tsin1C

2I2

3i1i3C2i1i1CF

Chn chiu tng theo thi gian t: du (+) vi lc ko v hai dy kia v (-) vi lc y ra. Tin hnh thay s ta tnh ton v tm c cc tr s lc y v lc ko cc i ca pha A

l:

++= )3C1C(3C1C23C21C222I

1kF .

Chn sin2t v cos2t du (-) lc ngc li l lc y nhau:

+++= )3C1C(3C1C23C21C222I

1F .

+ Vi dy pha C ging dy A. + Dy pha B : tng t ta c Fk2 v F2 l :

++= )2C1C(2C1C22C21C222I

2kF .

++= )2C1C(2C1C22C21C222I

2F .

Nu chn C1 = C2, C3 = 0,5C1 th ta c pha A:

=

=2I1C.615,11

F

2I1C.115,01kF

C ngha l pha A lc y gp khong 14 ln lc ko. Cn pha B th:

=

=2I1C73,12

F

2I1C73,12kF

b) Trng hp ba dy dn b tr trn ba nh tam gic u Ta gi thit ln lt ba dng in i1, i2, i3 cho trn i vo dy dn cc pha A, B, C c b tr trn ba nh tam gic u nh hnh 4-10b. Ta c h s C1=C2=C3=C

+ Lc tc dng ln dy pha A sau khi thay s v tnh ton ta c:

a) b) Hnh 4-10: Lc in ng trong mch xoay

chiu ba pha

F1 Fk1

F2 Fk2

C

Y

X A

B

C FAC

FAB6

37

tsin.C2I3t2cos.222CI2

3F == + Lc tc dng ln dy B v dy C tng t nh dy A ch c gc pha thay i.

3. Lc in ng trong ba pha khi ngn mch Dng trong cc pha khi ngn mch l :

[ ]

+++=

+=

+=

)3

2tcos()

3

2cos(teI23i

)3

2tcos()

3

2cos(teI22i

)tcos(costeI21i

Trong : :gc pha ca dng in trong pha th nht khi bt u xy ra s c; : h s cn. Nu gi thit khng xt n thnh phn khng tun hon vi e-t = 1 ta c : + Lc tc ng dy A l : F = C1i1i2 + C3i1i3 + Lc tc dng ln dy B l : F = C1i1i2 + C2i2i3

Khi xt ba dy cng nm trong mt mt phng, lc in ng khng ch ph thuc thi gian t m ph thuc c thi im xy ra ngn mch . Xt : +) Khi = - 150 m xy ra ngn mch th

= tcos

2

3

2

t2sin2I21C31F

Nu t = th

==

01k

F

2I1C46,6max1F

+) Khi = 750 m ngn mch th

+= tcos2

3

2

t2sin2I21C31kF

t = th Fk1max = 0,16C1I2, F1max = -1,5C1I2.

4.5. CNG HNG C KH V N NH LC IN NG THIT B IN

1. Cng hng c kh Khi dng in xoay chiu i qua thanh dn (thanh ci) lc in ng s gy chn ng v c th

pht sinh hin tng cng hng c kh. iu kin trnh cng hng c kh

Mun khng xy ra cng hng th tn s dao ng ring ca thanh ci phi b hn tn s sng c bn ca lc. Trong thc t ngi ta thng thay i khong cch gi thanh ci iu chnh tr s tn s dao ng ring ca thanh ci.

Tn s dao ng ring thanh ci tnh theo biu thc :

1g

J.E2l

112Z =

Trong : l : khong cch gi cch in; E : m un n hi [kg/cm2]. J : m men qun tnh (ly trc thng gc vi hng un lm chun)

38

g1 : trng lng n v di thanh ci [kg]. Nu khng thc hin c iu kin trn th c th phi gii quyt bng iu chnh tn s ring

ca thanh ci z ln hn tn s sng c bn. Ch tn s lc in ng gp hai ln tn s dng in f1 = 2fI > z. 2. n nh lc in ng

Trong thit b in phi tnh lc in ng kim tra xem thit b in c t bn c hay khng. n nh lc in ng l kh nng chu ng tc ng c kh do lc in ng sinh ra khi ngn mch.

m bo cn iu kin cn th: Im > Ixk vi : +Im : dng cho php ln nht ca thit b in, ixk : dng xung kch tnh ton khi ngn mch ba

pha. C th dng bi s cho php (Km) ln nht kim tra lc in ng.

xkmm iKI2 , trong : Km l bi s dng cho php ln nht. Ch : theo tnh ton ngn mch trong mng ba pha, lc in ng khi ngn mch mt pha (Fmax

= CI12 = C.6,48Im2) ln hn lc in ng khi ngn mch ba pha (F1max = C1.6,46Im2), nhng do khi ngn mch ba pha chiu lc thay i trong khng gian nn phi dng kim tra kh nng chu lc cc im.

- Nu thit b in khng ghi gi tr Im th c th xc nh theo cng thc :

]kA[mU3

ngS55,2xkimI =

Vi : Sng : cng sut ngt mch [MVA]; Um : in p nh mc hiu dng [kV].

40

Chng 5. C CU IN T V NAM CHM IN 5.1. KHI NIM CHUNG V MCH T

1. Khi nim Cc thit b in nh rle, cng tc t, khi ng t, p t mt,...u c b phn lm nhim v bin i t in nng ra c nng. B phn ny gm c cun dy v mch t gi chung l c cu in t, chia lm hai loi xoay chiu v mt chiu. nm c nhng quy lut in t ta xt mch t v phng php tnh ton mch t.

Mch t c chia lm cc phn: - Thn mch t. - Np mch t. - Khe h khng kh chnh v khe h ph p. - Khi cho dng in chy vo cun dy th trong cun dy c t thng i qua, t thng ny cng chia lm ba phn : a) T thng chnh l thnh phn qua khe h khng kh gi l t thng lm vic lv. b) T thng tn t l thnh phn i ra ngoi khe h khng kh xung quanh c) T thng r r l thnh phn khng i qua khe h khng kh chnh m khp kn trong khng gian gia li v thn mch t. 2. Tnh ton mch t Tnh ton mch t thc cht l gii hai bi ton: a) Bi ton thun : bit t thng tnh sc t ng F = IW loi ny gp khi thit k mt c cu in t mi. b) Bi ton nghch : bit sc t ng F = IW cn tm t thng (gp khi kim nghim cc c cu in t c sn). gii quyt c hai bi ton trn cn phi da vo cc c s l thuyt sau:

- Bit ng cong t ha ca vt liu st t. - Nm vng cc nh lut c bn v mch t. - Bit c t dn khe h.

2. Cc l thuyt c s a) ng cong t ha B = f(H) hnh 5-2 b) Cc nh lut c bn mch t

+ nh ton dng in F IW Hdll

= = + nh lut Ohm trong mch t:

MM RIW

RF ==

+nh lut Kic Khp 1 cho mch t : 0i = +nh lut Kic Khp 2 cho mch t: = iMii FR (tng i s st t p trn mt mch t kn bng tng i s cc sc t ng tc dng trong mch t ). c) T dn ca khe h

Hnh 5-1: Kt cu mch t 1.Thn mch t; 2. Np mch t ;3. Cun dy

Hnh 5-2: ng cong t ha

3 1

2

r

t

B

H

41

V mch t c t thm (h s dn t) ln hn khng kh nhiu nn t tr ton b mch t hu

nh ch ph thuc vo t tr khe h khng kh. Trong tnh ton thng dng t dn MR1G = . Tng t

nh mch in th trong mch t dn G t l thun vi tit din mch t, t l nghch vi chiu di khe h khng kh.

C : lS.gin mach ng tngS.G 0 == vi:

+

A

WbG :t dn khe h khng kh.

+cm.A

Wb10.25,1 80= : h s t thm khng kh.

+ [cm]: chiu di khe h. +S [ cm2]: din tch t thng i qua ( tit din).

Cng thc ny dng trn c s gi thit : t thng qua khe h khng kh phn b u n ( cc ng sc t song song vi nhau), cng thc ch ng khi khe h rt b, (khe h ln th cng ra mp cng khng song song). Thc t tnh t dn rt phc tp, ty yu cu chnh xc m c cc phng php tnh t dn khc nhau.

5.2. TNH T DN KHE H KHNG KH CA MCH T

1. Tnh t dn bng phng php phn chia t trng Xt v du : C mt cc t tit din

ch nht t song song vi mt phng. Gi thit chiu i t cc t xung mt phng (hnh 5-3). Nu tnh t dn khe h bng phng php phn chia t trng ta s phn t trng thnh nhiu phn nh sao cho mi phn t trng phn b u(c cc ng sc t song song vi nhau) p dng cng thc c bn tnh t dn c trn. y ta chia lm 17 phn gm : +) 1 hnh hp ch nht th tch: a. b. +) 4 hnh 1/4 tr trn c ng knh 2 chiu cao a v b +) 4 hnh tr 1/4 rng c ng knh trong 2 ng knh ngoi 2 + 2m

Bng 5.1: Cng thc tnh t dn ca cc phn Hnh dng Tn gi Cng thc tnh t dn

Hp ch nht

G = 0 ab/

b a

l

a b

m

Hnh 5-3: Phn chia t trng

42

1/4 Hnh tr c

G = 0,52.0 l. (l=a hoc b)

1/4 tr rng

-Nu >3m th: G = 0(1,28.m.l) /(2+m) -Nu

43

( )2n.m

31,0m1n.m

58,01K +++= , vi ==an;

abm

d) T dn gia mt phng v cc t t u mt phng(hnh 5-4d)

G = K .G0 = K .0 . S/

Vi ( )2n.m

31,0m5,11n.m

58,01K +++=

e) T dn gia mt phng v cc t t gia mt phng

00 G.KS..KG ==

( ) 2n.m31,0m21

n.m58,01K +++=

3. Tnh t dn bng phng php gii tch Nguyn tc ca phng php ny l da vo tnh cht tng ng gia s phn b t trng xung quanh vt dn t vi in trng xung quanh vt dn in. iu kin b ging nhau th cng gii tng t. V du: hai vt dn t t song song vi nhau, nu in trng th c cng thc:

2. va1dn vt cua in th la2,1

dung.in laCdn. trn vtin tchlaQ

)21C(=Q

Vi in tch Q, in dung C, in th . t trng c : ( )21 uuG = vi:

2. va1dn vt cua th tla:2U,1Udn. vt haigiadn tla G

dn. vt haigia thng tla

d

a

b

b)

c)

b

a

Hnh 5-4: Mt s hnh dng phn b khe h

d)

a)

44

c: C.KG = v K: h s ph thuc n v chn. y:

00K

= vi ]cm/F[10.9.4

1;Acm

b.W10.25,1110

80

== Vy vi m hnh ton hc ging nhau khi tm ra in dung C th s tm ra t dn G.

4. Mt s cng thc c c bng phng php gii tch a) T dn gia mt tr song song vi mt mt phng ln (khong cch a>4r)

rraaln

l..2G22 +

=

b) T dn hai mt tr trn song song (khong cch b>4d)

l.

rbln

.G 0 =

c) T dn gia hai mt tr ng tm bn knh r1 v r2

12

0

rr

ln

l2.G =

d) T dn gia hai mt cc t ch nht t song song trong cng mt mt phng

+= 1mm21m2ln.2l.G 220 , vi d

db2m += Ngoi ra cn phng php v nhng ch dng khi cc t hnh dng phc tp khng th dng

biu din ton hc c.

5.3. TNH TON MCH T 1. Tnh mch t mt chiu + Mch t mt chiu khi lm vic, trong mch c dng khng i I, t thng =const nn khng c tn hao dng xoy, li c lm bng vt liu st t khi d gia cng c kh. Trnh t tnh ton mch t: * V mch t ng tr. * Tnh t dn G ca khe h khng kh v ton mch. * Gii mch t, tm cc tham s cha bit.

Trong qu trnh lm vic khe h khng kh thay i lm t thng bin thin do vy ta chia c ra cc trng hp: a) Tnh mch t mt chiu khi khng xt t thng r

Vi mch t khe h khng kh b, cun dy phn b u trn mch t th c th b qua t thng r.

V du: xt mch t hnh xuyn hnh 5-5; phn st t chiu di l, tit din S, khe h c t thng r ro=0. Gii: a.1) Bit cn tm F=IW (do ro=0 nn = do IW sinh ra. S

BB = , theo nh lut ton dng in c: +== .HHlIWF (*), t tr s B ta tra ra H, vi S l tit din mch t [m2]

S

l

Hnh 5-5: Mch t hnh xuyn

45

Vi tr s t cm l B th:0

BH =

, thay gi tr H vo (*) ta c F=IW.

Hoc dng phng trnh:

+=

G1RIW M

a.2) Bit IW cn tm C : += .Hl.HIW

Vi:

====

GS:n SG

BBH

..

;

00

00

con

+=

GBSHlIW . Chia hai v cho l.

Ta c: lGS.BH

lIW

+=

Trn ng cong t ha st t t oa lIW=

- Chn t l xch trc honh mH (A.vng/khong). - Chn t l xch trc tung mB (Gauss/khong).

Vi: BH

mm

.l.G

tg = (ct ng cong t ha ti b) t b h OHbc nh vy c

Bm.bc;l.GBSm.ca;Hm.Oc BHH ===

do == S.B . Rt ra trng hp tng qut

* i vi nhng bi ton sc t ng IW ging nhau, nhng khe h khng kh khc nhau (tit din S khc nhau) th c th gii c nhanh chng bng cch k t a cc on ab, ab,... to vi trc honh cc gc , ,... tung cc im b, b l tr s B cn tm.

* Khi khe h khng kh v tit din S bng nhau nhng sc t ng IW khc nhau th trn trc honh ta t nhng on thng oa, oa,...c gi tr bng

111

lWI

; lWI 22 v k ab//ab tung b, b

l tr t cm B cn tm. b) Tnh mch t mt chiu khi xt t thng r Khi np mch t m th lng t thng r ln ng k nn khi tnh phi xt n. Tnh h s t thng r : Xt mch t hnh 5-7, ta xt s phn b t thng r dc theo chiu cao mch t li.

Sc t ng trn mt on x l lx.IWFX = theo vi phn dx l dx.g.Fd xrx = (g: t dn r

trn n v chiu di x).

===x

0

x

0rx

2xrx 2

x.l

g.IWdx.g.Fd

Khi x = 0 th lx;0co rx == nn :

Hnh 5-6 Tm t thng t ng cong t ha

H

B

Oa c

b

46

2l.g.IWrrx ==

C th xem t thng r r chy qua mt t dn tp trung c gi tr bng g

l.2 , t dn r tp trung c gi

l t dn r quy i. - nh gi t thng r nhiu hay t ta dng h s t thng r :

++=++=

= trtr 1 Vi: : t thng tng do cun dy sinh ra : t thng khe h r: t thng r v t l vi t dn nn:

++=G

GGG tr

Trong : - Khi np m r ln th ly =(1,8 3). - Khi np ng r nh th ly =(1,05 1,1). Ch y:

- Khi np m c th b qua t tr ca mch t nhng phi xt n t thng r, nn c mch t ng tr nh hnh 5-8. - Khi np ng c th b qua t thng r v b nhng phi k n t tr. 2. Tnh mch t xoay chiu Mch t xoay chiu khc mch t mt chiu v nhng c im sau: a) Trong mch t xoay chiu: i=i(t) nn =msint dng bin thin c hin tng t tr, dng xoy, dng in chy trong cun dy ph thuc vo in khng ca cun dy, m in khng ph thuc t dn mch t nn t tr ton mch t cng ln (khe h khng kh cng ln) th in khng cng b v dng in trong cun dy cng ln. Khi np mch t m dng in khong I= (415)Im. Ch y: khi ng in c cu in t, phi kim tra np xem ng cha, nu np m c th lm cun dy b chy. b) Lc ht in t F bin thin F=F(t) c thi im F=0 c thi im F=Fmax dn n mch t khi lm vic b rung, hn ch rung ngi ta t vng ngn mch. T thng bin thin lm xut hin sc in ng trong vng ngn mch, trong vng c dng in mc vng khp kn, lm vng ngn mch nng ln. Gi Wnm l s vng ngn mch (thng Wnm=1). Theo nh lut ton dng in c:

( )( )

dtd.

rW

RRIW

RRWIIW

nm

2nm

t

tnmnm

:conn ++=

+=+

( )

++=

nm

2nm

tm

rWJRR

2IW , gi

nmnm

t rWx = l t khng ca vng ngn mch th c:

( )[ ]ttm JxRRIW2 ++= :JxRRZ ttt ++= vi Rt: t tr mch t. c im: t khng trong mch xoay chiu tiu th cng sut tc dng.

IW r

t

G

Gt

Gr

Hnh 5-8 Mch t ng tr khi c t thng r

rx

t

x dx

Hnh 5-7 Mch t khi c t thng

r

47

c) Trong mch t xoay chiu c tn hao dng xoy t tr lm nng mch t, c th xem nh tn hao trong vng ngn mch. Nu gi Pxt l cng sut hao tn do dng xoy v t tr th c th biu din di dng tng ng nh mt vng ngn mch.

P I rxt nm nm= 2 . hay 2mnm

2nm

2

nm

2nm

xt .r.2W.

rB

P ==

C: nm2m

xtnm

2nm X

P2rW. =

= gi l t khng thay th tng ng c trng cho tiu hao cng

sut tc dng do dng xoy v t tr. d) T dn r quy i Khc vi mch mt chiu v: - Sc t ng tng F=IW sc t

ng on X l lx.IWFX =

lxWWx = t thng mc vng on

x l rx x rxW= . Cui cng c :

3l.gGr = l t dn r

trong mch xoay chiu. V phng php tnh ton mch t xoay chiu cng ging mch t mt chiu nhng phi lu bn c im trn. V d mch t xoay chiu nh hnh 5-9:

- Khi v mch t ng tr phi xt n tc dng ca vng ngn mch, tn hao dng xoy v t tr.

- Khi np ng, b qua t thng r nhng phi k n t tr v t khng mch t nn dng nh hnh 5-10a. - Khi np mch t m, c th b qua t tr v t khng ca mch t, nhng phi xt n t thng r cho nn mch t ng tr c dng nh hnh 5-10b.

5.4. I CNG V NAM CHM IN

1. Khi nim

a) b)

3 1

2

r

t 4 i(t

)

Hnh 5-9: Mch t xoay chiu 1.Thn mch t; 2. Np mch t; 3. Cun dy;4. Vng ngn mch

Hnh 5-10: Mch t ng tr a) Khi np ng ;

IW IW

Xnm

R2 R2

XnmR1 R1

Rt

RrXt

48

Dng in chy trong cun dy s sinh ra t trng. Vt liu st t t trong t trng ny s b t ha v c cc tnh ngc li vi cc tnh ca cun dy, cho nn s b ht v pha cun dy hnh 5-11.

Nu i chiu dng in trong cun dy th t trng trong cun dy cng i chiu v vt liu st t b t ha c cc tnh ngc vi cc tnh cun dy, cho nn chiu lc ht khng i.

Vt liu st t c t thm ln hn rt nhiu ca khng kh nn t tr ton b mch t hu nh ch ph thuc vo t tr khe h khng kh. Ta thng dng khi nim t dn:

= R1G

(5.1) Do tnh cht tng ng gia mch t v mch in nn trong mch t, t dn t l thun vi

tit din mch t v t l nghch vi chiu di khe h khng kh.

= A

WbS.G 0 (5.2)

Trong : +0 t thm khng kh bng 1,25.10-8[Wb/A.cm] +S[cm2] tit din t thng i qua. + [cm] chiu di khe khng kh. Ch y: cng thc trn ch ng vi gi thit t thng trong khe khng kh phn b u (cc ng sc t phi song song) khi khe h b. Khi khe h ln tnh ton phc tp ty yu cu c th vic tnh ton c cc phng php khc nhau. Mt s cng thc dng trong tnh ton mch t

SB =

2cm

Wb

H : Cng t trng [ A/cm]=1,25 [Osted]

F;HB= = IW :l sc t ng [A.vng]

+ nh lut ton dng in FWIHdll

= =

+ nh lut m cho mch t: MR

IWG.IW ==

+ nh lut Kic khp I cho mch t: ==n

1ii 0 ti mt im.

+ nh lut Kic khp II cho mch t: trong mt mch t khp kn c:

== =n

0i

n

0iiii FR

2. Phn loi c cu in t Phn theo tnh cht ca ngun in - C cu in t mt chiu. - C cu in t xoay chiu. Theo cch ni cun dy vo ngun in - Ni ni tip. - Ni song song.

i N

Si

N

S

N

S

Hnh 5-11 Hai dng nam chm

in

49

Theo hnh dng mch t - Mch t ht chp (thng). - Mch t ht xoay (quanh mt trc hay mt cnh), mch t ht kiu pt tng.

Trong qu trnh lm vic np mch t chuyn ng khe h khng kh gia np v li thay i nn lc ht in t cng thay i. Thng tnh ton mch t nam chm in ngi ta dng hai phng php (s nu sau).

5.5. TNH LC HT IN T NAM CHM IN MT CHIU

1. Tnh lc ht in t bng phng php cn bng nng lng Nng lng t trng v in cm

Xt mch t nh hnh 5-12. Khi cho dng in i vo cun dy w c:

(5.3) dtdtdidt.i.Ruidt

dtdi.Ru

2

hay

+=

+=

Ly tch phn hai v phng trnh trn ta c :

+=t

0

t

0

t

0

2 dtdtdiRdtiuidt (5.4)

Trong ta c:

t

0uidt l nng lng ngun cung cp.

t

0

2dtRi l nng lng tiu hao trn in tr cun dy w

=t

0tWdtdt

di l nng lng tch ly trong t trng c:

=0

t idW (5.5)

Biu din bi hnh 5-13 chnh l din tch phn tam gic cong oab c quan h v i l phi tuyn.

Theo nh ngha th in cm: IL =

Trong : l t thng mc vng ca cun dy w. I :l dng in trong cun dy.

=== I tt IWL n ILiLdiw 0 22 22

con (5.6)

Hnh 5-12: Nam chm in ht chp

ik

50

Tnh lc ht in t Khi cung cp nng lng

cho c cu in t th np ca mch t c ht v pha li, khe h khng kh gia np v li gim dn.

ng vi v tr ban u ca np mch t c: = = =1 1 1; ;I I

ng vi v tr cui c: = = =2 2 2; ;I I Nng lng t trng khi v tr u s l:

=1

10

t idW = din tch oa1b1 Nng lng t trng khi v tr cui s l:

= 2

20

t idW = din tch oa2b2 (hnh 5-14) Vy nng lng ly thm t ngoi vo np mch

t chuyn ng l:

w idt =

1

2

= din tch hnh thang b1a1a2b2

(nh hnh 5-14). Theo nh lut cn bng nng lng c:

AWWW 21 ttt +=+ Trong A l nng lng lm np chuyn ng t v tr 1 n v tr 2.

21 tt WWWA += = din tch tam gic cong oa1a2

Nu gi thit mch t cha bo ha ng c tnh = f(i) ch xt on tuyn (hnh 5-15). Ta c:

V c: =I.L ( hnh 5-

16a).

( )1221 II21A = (5.8)

t: += 12 , III 12 += ( )II2

1A 11 = (5.9)

Dng vi phn :

( )dIId21dA = (5.10)

Vy lc ht in t s l:

Hnh 5-13 Hnh 5-14

sHnh 5-15

Hnh 5-16

)(2

IIW

;2

IW;

2I

W

1221

t

22t

11t 21

+=

==

1

a1

a2

I1 I[A

2

I2 0

b2

b1

I2

1=2

a1 a2

I1 I[A

0

1

a1

a2

I1= I[A

2

a)

b)

01

a1

a2

I1 I[A

2

I2

b2

b1I[A]

0 I

b a

51

== ddI

ddI

21

ddAF (5.11)

Ta xt hai trng hp sau:

a) Trng hp khi I = const th 0ddI = (nh hnh 5-16a).

LI];kg[ddI.1,5F ==