giao trinh thiet ke dien tu cong suat

8/14/2019 Giao Trinh Thiet Ke Dien Tu Cong Suat

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TRNG I HC BCH KHOA H NI

B MN THIT B IN - IN T

------------------------------

ti liu hng dnthit k thit b in t cng sut(Dnh cho sinh vin chuyn ngnh Thit b in - in t)


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Mc ch yu cu:

Trong nhng nm gn y cng vi vic pht trin ngy cng mnh mca cc lnh vc khoa hc, ng dng ca chng vo cng nghip ni chung vcng nghip in t ni ring, cc thit b in t c cng sut ln c chto ngy cng nhiu, c bit l ng dng ca n vo cc ngnh kinh t qucdn v i sng, lm cho yu cu v s hiu bit v thit k cc loi thit bny l ht sc cn thit i vi cc k s ngnh in.

gip cho sinh vin mt k nng ng dng nhng kin thc l thuyt hc v mn hc thit b in t cng sut vo vic thit k nhng b nguncng sut hon chnh, thit k thit b in t cng sut (TK), t mc chhon thin l thuyt v nng cao k nng ng dng lm mc ch chnh.

Mi sinh vin c nhn mt ti thit k c lp cho mnh, c trchnhim hon thnh ni dung c ra theo nhim v TK, vi ni dung nysinh vin phi thit k thnh nhng thit b hon chnh c th ng dng

trong thc t sn xut.Ni dung TK

Thuyt minh s qua v cng ngh ca ti m b ngun cp in Chn s mch ng lc. Thit k, tnh chn cc thit b c bn ca mch ng lc (bao gm

chn van bn dn, tnh ton cc thng s nh mc c bn, tnh tonmy bin p hay cun khng nu c, tnh chn cc thit b ng ct vbo v.........).

Thit k tnh chn mch iu khin. Thit k kt cu (t in).8.1 Tm tt l thuyt. cp ngun cho ti mt chiu, chng ta cn thit k cc b chnh lu

vi mc ch bin i nng lng in xoay chiu thnh mt chiu. Cc loi bbin i ny c th l chnh lu khng iu khin v chnh lu c iu khin.Vi mc ch gim cng sut v cng, ngi ta thng mc song song ngc


3/122

mt hay ba pha. Cc thng s quan trng ca s chnh lu l: dng in vin p ti; dng in chy trong cun dy th cp bin p; s ln p mchtrong mt chu k. Dng in chy trong cun dy th cp bin p c th l mtchiu, hay xoay chiu, c th phn loi thnh s c dng in bin p mtchiu hay, xoay chiu. S ln p mch trong mt chu k l quan h ca tn ssng hi thp nht ca in p chnh lu vi tn s in p xoay chiu.

Theo hnh dng cc s chnh lu, vi chuyn mch t nhin chng tac th phn loi chnh lu thnh cc loi s sau.

1. Chnh lu mt na chu k.

Hnh 8.1. S chnh lu mt na chu k.

s chnh lu mt na chu k hnh 8.1 sng in p ra mt chiu sb gin on trong mt na chu k khi in p anod ca van bn dn m, dovy khi s dng s chnh lu mt na chu k, chng ta c cht lng inp xu, tr s in p ti trung bnh ln nht c tnh:

Udo = 0,45.U2 (8 -1)Vi cht lng in p rt xu v cng cho ta h s s dng bin p xu:

Sba = 3,09.Ud.Id. (8 -2)

nh gi chung v loi chnh lu ny chng ta c th nhn thy, y lloi chnh lu c bn, s nguyn l mch n gin. Tuy vy cc cht lngk thut nh: cht lng in p mt chiu; hiu sut s dng bin p qu xu.

Do loi chnh lu ny t c ng dng trong thc t.Khi cn cht lng in p kh hn, ngi ta thng s dng s chnh

lu c chu k theo cc phng n sau.

2. Chnh lu c chu k vi bin p c trung tnh.

U2

R L

T

U1


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s ging ht nhau, mi na chu k c mt van dn cho dng in chy qua.Cho nn c hai na chu k sng in p ti trng vi in p cun dy c vandn. Trong s ny in p ti p mch trong c hai na chu k, vi tn sp mch bng hai ln tn s in p xoay chiu. Hnh dng cc ng congin p, dng in ti (Ud, Id), dng in cc van bn dn I1, I2 v in p cavan T1 m t trn hnh 8.3a khi ti thun tr v trn hnh 8.3b khi ti in cmln.

Hnh 8.3. Cc ng cong in p, dng in ti, dng in cc

van v in p ca Tiristo T1

in p trung bnh trn ti, khi ti thun tr dng in gin on ctnh:

Ud = Udo.(1+cos)/2. (8 -3).

vi: - Udo: in p chnh lu khi khng iu khin v bng Udo = 0,9.U2

Gc m ca cc Tiristo.

Khi ti in cm ln dng in, in p ti lin tc, lc ny in p mtchiu c tnh:

0t1 t2 t3

Ud Id

I1

I2

t

t

t

t

0t1 t2 t3

Ud

Id

I1

I2

t

t

t

t

c.b

p1 p2 p3

UT1 UT1


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tt hn. Dng in chy qua van khng qu ln, tng in p ri trn van nh.i vi chnh lu c iu khin, th s hnh 8.2 ni chung v vic iukhin cc van bn dn y tng i n gin. Tuy vy vic ch to bin pc hai cun dy th cp ging nhau, m mi cun ch lm vic c mt na chuk, lm cho vic ch to bin p phc tp hn v hiu sut s dng bin p xuhn, mt khc in p ngc ca cc van bn dn phi chu c tr s ln nht.

3. Chnh lu cu mt pha.

Hnh 8.4. S chnh lu cu mt pha iu khin i xng.

Hot ng ca s ny khi qut c th m t nh sau. Trong na bnk in p anod ca Tiristo T1 dng (+) (lc catod T2 m (-)), nu c xungiu khin cho c hai van T1,T2 ng thi, th cc van ny s c m thng

t in p li ln ti, in p ti mt chiu cn bng in p xoay chiuchng no cc Tiristo cn dn (khong dn ca cc Tiristo ph thuc vo tnhcht ca ti). n na bn k sau, in p i du, anod ca Tiristo T3 dng(+) (catod T4 m (-)), nu c xung iu khin cho c hai van T3,T4 ng thi,th cc van ny s c m thng, t in p li ln ti, vi in p mtchiu trn ti c chiu trng vi na bn k trc.

Chnh l

u cu mt pha hnh 8.4 c cht l

ng in p ra hon ton gingnh chnh lu c chu k vi bin p c trung tnh, nh s hnh 8.2. Hnhdng cc ng cong in p, dng in ti, dng in cc van bn dn v inp ca mt van tiu biu gn tng t nh trn hnh 8.3a.b. Trong s nydng in chy qua van ging nh s hnh 8.2, nhng in p ngc vanphi chu nh hn U = 2 U

T4 T1U2

T3

L

T2

R


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Hinh 8.5. Phng n cp xung chnh lu cu mt pha

Chnh lu cu mt pha iu khin khng i xng c th thc hin bnghai phng n khc nhau nh hnh 8.6. Ging nhau hai s ny l: chngu c hai Tiristo v hai it; mi ln cp xung iu khin ch cn mt xung;in p mt chiu trn ti c hnh dng ( xem hnh 8.7a,b) v tr s ging nhau;

ng cong in p ti ch c phn in p dng nn s khng lm vicvi ti c nghch lu tr nng lng v li. S khc nhau gia hai s trnc th hin r rt khi lm vic vi ti in cm ln, lc ny dng in chyqua cc van iu khin v khng iu khin s khac nhau.

b

Hnh 8.6. S chnh lu cu mt pha iu khin khng i xng.

Trn s hnh8.6a (vi minh ho bng cc ng cong hnh 8.7a) khiin p anod T1 dng v catod D1 m c dng in ti chy qua T1, D1 nkhi in p i du (vi anod T2 dng) m cha c xung m T2, nng lngca cun dy ti L c x ra qua D2, T1. Nh vy vic chuyn mch ca ccvan khng iu khin D1 D2 xy ra khi in p bt u i du Tiristo T1 s

dieu khienMach

T1 (T3)

T2 (T4)

D

D

U

R

T1

T2

L

D2

D1 D1 D2

T2 T1

RL


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Khi in p i du nng lng ca cun dy L c x ra qua cc it D1,D2, cc van ny ng vai tr ca it ngc. Chnh do m cc Tiristo s t

ng kho khi in p i du. T ng cong dng in cc van trn hnh8.7b c th thy rng, s ny dng in qua Tiristo nh hn dng in quacc it.

Hnh 8.7. Gin cc ng cong in p, dng in ti(Ud, Id), dng in cc van bn dn ca cc s a- hnh 8.6a; b- hnh 8.6b.

Nhn chung cc loi chnh l

u cu mt pha c cht l

ng in p t

ngng nh chnh lu c chu k vi bin p c trung tnh, cht lng in mtchiu nh nhau, dng in lm vic ca van bng nhau, nn vic ng dngchng cng tng ng nhau. Mc d vy chnh lu cu mt pha c uim hn ch: in p ngc trn van b hn; bin p d ch to v c hiusut cao hn. Th nhng chnh lu cu mt pha c s lng van nhiu gp hailn, lm gi thanh cao hn, st p trn van ln gp hai ln, chnh lu cu iu

khin i xng th vic iu khin phc tp hn.Cc s chnh lu mt pha cho ta in p vi cht lng cha cao, bin

p mch in p qu ln, thnh phn hi bc cao ln iu ny khng p ngc cho nhiu loi ti. Mun c cht lng in p tt hn chng ta phi sdng cc s c s pha nhiu hn.

0

t1 t2 t3

Ud

Id

IT1

0

t1 t2 t3

Ud

Id

t

IT2

ID1

ID2

IT1

IT2

ID1

ID2

t

t

t

t

t

t

t

t t

t

t

a. b.


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giao nhau c coi l gc thng t nhin ca cc van bn dn. Cc Tiristior chc m thng vi gc m nh nht ti thi im gc thng t nhin (nh vy

trong chnh lu ba pha, gc m nh nht = 00 s dch pha so vi in p phamt gc l 300).

H h 8 8 Ch h l i b h

T1

B T2

C T3

A

RL

Ud

Id

UT1

t1 t2 t3 t4I1

I2

I3

Ud

t

Id

t1 t2 t3 t4

I1

I2

I3

t

t

t t

t

t

tt

t

b.

0

c.


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V d trong khong t2 t3 van T1 kho cn T2 dn do van T1 phi chumt in p dy U

AB

, n khong t3 t4 cc van T1, T2 kho, cn T3 dn lcny T1 chu in p dy UAC.

Khi ti thun tr dng in v in p ti lin tc hay gin on phthuc gc m ca cc Tiristo. Nu gc m Tiristo nh hn 300, cc ngcong Ud, Id lin tc, khi gc m ln hn > 300 in p v dng in ti ginon (ng cong Ud, Id trn hnh 8.8c).

Hnh 8.9. ng cong in p ti khi gc m = 600vi a.- ti thun tr, b.- ti in cm.

Khi ti in cm (nht l in cm ln) dng in, in p ti l ccng cong lin tc, nh nng lng d tr trong cun dy ln duy trdng in khi in p i du, nh ng cong nt m trn hnh 8.9b (tngt nh vy l ng cong Ud trn hnh 8.8b). Trn hnh 8.9 m t mt v d sosnh cc ng cong in p ti khi gc m= 600 ti thun tr hnh 8.9a vti in cm hnh 8.9b

Tr s in p trung bnh ca ti s c tnh nh cng thc (1 - 4) nuin p ti lin tc, khi in p ti gin on (in hnh khi ti thun tr v gc

m ln) in p ti c tnh:

Trong ; Udo = 1,17.U2f. in p chnh lu tia ba pha khi van la it.

t t

a. b.

A B C A A B C A

)51(3

sin13

+= Udo

Ud


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trung tnh chu dng in ti.

5. Chnh lu tia su pha:

Hnh 8.10. Chnh lu tia su pha.a.- S ng lc; b.- ng cong in p ti.

S chnh lu tia ba pha trn c cht lng in p ti cha tht ttlm. Khi cn cht lng in p tt hn chng ta s dng s nhiu pha hn.

Mt trong nhng s l chnh l

u tia su pha. S ng lc m t trnhnh 8.10a.

S chnh lu tia su pha c cu to bi su van bn dn ni ti binp ba pha vi su cun dy th cp, trn mi tr bin p c hai cun gingnhau v ngc pha. in p cc pha dch nhau mt gc l 60 0 nh m t trnhnh 8.10b. Dng sng in p ti y l phn dng hn ca cc in p phavi p mch bc su. Vi dng sng in p nh trn, ta thy cht lng in

p mt chiu c coi l tt nht.

Theo dng sng in p ra (phn nt m trn gin hnh 8.10b) chngta thy rng mi van bn dn dn thng trong khong 1/6 chu k. So vi cc s khc, th chnh lu tia su pha dng in chy qua van bn dn b nht.Do s chnh lu tia su pha rt c ngha khi dng ti ln Trong trng

A

*

*

R L

*

T1

B T2

C T3

T4A*

T5B*

T6C*

t

A B CA* B*C*


11/122

R

T1

T3

T5

L

T6

T4

T2

A B C A

X1

X2

X3

X4

X5

X6

t1 t2 t3 t4 t5 t6 t7

Ud

Uf

I1

I3

I5

I2

I4

I6

0

Theo hot ng ca chnh lu cu ba pha iu khin i xng, dng in

Ud

Uf

A B C A

Ud

UfA B C A


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chy qua ti l dng in chy t pha ny v pha kia, do ti mi thi imcn m Tiristo chng ta cn cp hai xung iu khin ng thi (mt xung nhm anod (+), mt xung nhm catod (-)). V d ti thi im t1 trn hnh8.11b cn m Tiristo T1 ca pha A pha anod, chng ta cp xung X1, ng thiti chng ta cp thm xung X4 cho Tiristo T4 ca pha B pha catod cc thiim tip theo cng tng t. Cn ch rng th t cp xung iu khin cngcn tun th theo ng th t pha.

Khi chng ta cp ng cc xung iu khin, dng in s c chy tpha c in p dng hn v pha c in p m hn. V d trong khong t1 t2 pha A c in p dng hn, pha B c in p m hn, vi vic m thngT1, T4 dng in dc chy t A v B.

Khi gc m van nh hoc in cm ln, trong mi khong dn ca mtvan ca nhm ny (anod hay catod) th s c hai van ca nhm kia i ch cho

nhau. iu ny c th thy r trong khong t1 t3 nh trn hnh 8.11b TiristoT1 nhm anod dn, nhng trong nhm catod T4 dn trong khong t1 t2 cnT6 dn tip trong khong t2 t3.

in p ngc cc van phi chu chnh lu cu ba pha s bng 0 khivan dn v bng in p dy khi van kho. Ta c th ly v d cho van T1(ng cong cui cng ca hnh 8.11b) trong khong t1 t3 van T1 dn in

p bng 0, trong khong t3 t5 van T3 dn lc ny T1 chu in p ngc UBA,n khong t5 t7 van T5 dn T1 s chu in p ngc UCA.

Khi in p ti lin tc, nh ng cong Ud trn hnh 8.11b tr s in pti c tnh theo cng thc (8 -4).

Khi gc m cc Tiristo ln ln ti gc > 600 v thnh phn in cmca ti qu nh, in p ti s b gin on nh cc ng nt m trn hnh8.11d (khi gc m cc Tiristo =900 vi ti thun tr). Trong cc trng hp

ny dng in chy t pha ny v pha kia, l do cc van bn dn c phn ccthun theo in p dy t ln chng (cc ng nt mnh trn gin Ud cacc hnh v 8.11b, c, d), cho ti khi in p dy i du, cc van bn dn s ch h kh


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cc van bn dn T1,T2,T3,D1,D2,D3. Cc Tiristo c dn thng t thi imc xung m cho n khi m Tiristo ca pha k tip. V d T1 m thng t t1

(thi im pht xung m T1) ti t3 (thi im pht xung m T2). Trong trnghp in p ti gin on Tiristo c dn t thi im c xung m cho nkhi in p dy i du. Cc it t ng dn thng khi in p t ln chngthun chiu. V d D1 phn cc thun trong khong t4 t6 v n s m chodng in chy t pha B v pha A trong khong t4 t5 v t pha C v pha Atrong khong t5 t6.

D1

D2

D3

T1

T2

T3

LR

A B C A

X1

X2

X3

t1 t2 t3 t4 t5 t6 t7

Ud

Uf

T1

T2

T3

D1

D2

D3

0


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Vic kch m cc van iu khin trong chnh lu cu ba pha c iukhin d dng hn, nhng cc iu ho bc cao ca ti v ca ngun ln hn.

So vi chnh lu cu ba pha iu khin i xng, th trong s ny viciu khin cc van bn dn c thc hin n gin hn. Ta c th coi mchiu khin ca b chnh lu ny nh iu khin mt chnh lu tia ba pha.

Chnh lu cu ba pha hin nay l s c cht lng in p tt nht,hiu sut s dng bin p tt nht. Tuy vy y cng l s phc tp nht.

7. Chnh lu khi c it ngc

Nh nu trn, khi chnh lu lm vic vi ti in cm ln, nnglng ca cun dy tch lu s c x ra khi in p ngun i du. Trongtrng hp ny nh m t trn hnh 8. 13 khi in p ngun i du do it Dt ngc in p ln cc tiristo (trong cc khong 0t1, p1t2, p2t3), nncc tiristo b kho in p ti bng 0. Dng in chy qua cc tiristo I1, I2 ch

U1

T1

U2

T2

D

R

U2

L

t

0t1 t2 t3

UdId

I1

I2

t

t

t

t

ID

p1 p2 p3

( ) ( ) )61(cos1(max)2

3cos1(max)

2

33+=+=

UdayUfUtb

Hnh 8.13. Chnh lu mtpha vi bin p trung tnh

a- s ng lc b- gin cc ng cong.


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15

Cc h s c bn ca cc s chnh lu Bng 8.1

in p ti c tnh Ud H s in p

Ti thun tr chnh lu so vi in pti

so vi inp xoaychiu

Tn s chnh lu

S chnh lu

dng inlin tc

dng in ginon

Ti incm(dng lin

tc)

ku= Udo/U2f kn1= Un/Udo kn2= Un/U2f

CL na chuk Udo.1+cos

2Udo.cos

2

0,45

2 1,41

CL c chuk vi binp c trung

tnh Udo.1+cos2

Udo.cos

22

0,9

3,14

2 2 2,82

T1

L

T2

U2R

U1

U2

U2

RL

T

U1


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16


Ti thun tr chnh l

u so vi in pti so vi inp xoaychiu

Tn s

chnh lu

S chnh lu

dng inlin tc

dng in ginon

Ti incm(dng lintc)


CL cu mtpha

iukhinX Udo.1+cos2

Udo.cos 22

0,9

2

1,57

2

1,41

CL cu mtpha

iukhin

KX Udo.1+cos2

22

0,9

2

1,57

2 1,41

D2

A?D1

T1

T2

RL

T4

T2

R

T1

T T3

L

D1 D2

T2 T1

RL


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17


Ti thun tr chnh l


Tn s

chnh lu

S chnh lu

dng inlin tc

dng in ginon

Ti incm(dng lintc)


CL tia bapha

Udocos

6

khi

+

3sin1

3

0dU

6

khi

Udocos

2

63

1,17

3

2

2,09

6 2,45

CLtia supha

Udocos

3

khi

+

6

sin10dU

3

khi

Udocos

23

1,35

2 2 2,82

T1

B T2

C T3

A

RL

A*

*

*

R L

*

T1

B*

T2

C* T3

TA?A

A?B

A?C


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18


Ti thun tr chnh l


Tn s

chnh lu

S chnh lu

dng inlin tc

dng in ginon

Ti incm(dng lintc)


CL cu bapha iu

khin X Udocos

3

khi

+

6sin1

0d

U

3

khi

Udocos

63

2,34

3

1,45

6 2,45

CL cu bapha iu

khin KX

Udo.1+cos2 63

2,34

3

1,45

6 2,45

R

T1

T3

T5

L

T6

T4

T2

D1

D2

D3

T1

T2

T3

LR


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19

Bng cc h s dng in v bin p ca cc s chnh lu Bng 8.2

H s dng in HSCS bin p

Van bn dn Th cp bin p S cp bin pTn s chnh lu

Hnh dngdng in

TBktb=Iv/Id

Hiu dng khd=Ihd/Id

Hnh dngdng in

k2= I2/Id Hnh dngdng in

k1=I1/kba.Id

ks=Sba/

Pdmax

ks1=S1ba/Pdmax

ks2=S2ba/Pdmax

1 2 3 4 5 6 7 8 9 10 11Mt nachu k

c chu kvi BA ctrung tnh

2

1

0,5

2

1

0,71

2

1

0,71

1,0 1,48 1,23 1,74

Cu mtpha

KX

1,0

Cu mtphaKKX

2

1

0,52

1

0,71

1,23 1,23 1,23


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20



Hnh dngdng in

TBktb=Iv/Id

Hiu dng khd=Ihd/Id

Hnh dngdng in


k1=I1/kba.Id

ks=Sba/

Pdmax

ks1=S1ba/Pdmax

ks2=S2ba/Pdmax

2

2

Cu mtpha

KKX

2

+

2

+

2

2

Tia ba pha3

1

0,333

1

0,583

1

0,58 3

2

0,82

1,345

1,209

1,48

Tia supha 6

1

0,176

1

32

1

0,29

1/20,71

1,26 1,05 1,48

Cu ba

pha X 3

1

0,333

1

0,583

2

0,82

3

2

0,82

1,05 1,05 1,05


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21



Hnh dngdng in

TBktb=Iv/Id

Hiu dng khd=Ihd/Id

Hnh dngdng in


k1=I1/kba.Id

ks=Sba/

Pdmax

ks1=S1ba/Pdmax

ks2=S2ba/Pdmax

/3

2

2

3

2

0,82

khi


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8.2. M t khi qut yu cu ca ti

Sau khi nhn c nhim v v ni dung thit k, ngi thit k cn tmhiu ti liu v loi ti, trong cc ti liu chuyn su ca loi ti m cn phi thitk b ngun cp in.

iu cn thit nht ca l phi c mt s hiu bit v loi ti m mnh cpin, nhng c im c bn, nhng yu cu ca ti i vi ngun in (chnghn nh tr s hay hnh dng dng in c g c bit, n nh v vng iuchnh in p trn ti...).

V d: Thit k ngun chnh lu cho ti m in. Ngi thit k cn tm hiucc gio trnh chuyn nghnh in ho, chuyn su m in, c hiu bit cbn cn thit v m in. Nhng kin thc cn bit ti thiu m ngi thit k cnbit l c nhng phng php m no hin nay ang dng, m in l g? Cu

trc mt b m nh th no? Dng in cn chy qua b m ph thuc nhng yut no? in p cn cp cho b m l in mt chiu. Ti m in thuc loi ti

R-C-E. Tuy nhin in tr trong ca b m nh do hng s thi gian np, x trt nh. C th coi nh hng ca t l khng ng k. Sc in ng E trong bm thng nh nn chng ta cng c th b qua.

8.3. La chn s thit k.

Trn c s ca nhng hiu bit thu c phn tm hiu cng ngh, tinhnh chn s thit k. y l khu quan trng nht c tnh cht quyt nh sng n ca ton b TK. c th lm tt c vic ny, trc tin cn xc nhloi ngun thit k l loi ngun no, trong cc loi ngun c bn: chnh lu; iukhin in p xoay chiu; bm p mt chiu hay thit b bin tn.

Sau khi xc nh c ngun c bn th tin hnh nghin cu mt cch

t

ng i chi tit v loi ngun .V d: Thit k b ngun cho ti m in, th sau khi tm hiu v cng ngh

m, ta bit rng loi ngun c bn cho m in l in mt chiu. Cc loi ngunmt chiu c th cp in cho b m bao gm my pht in mt chiu, chnh lu.

My pht in mt chiu vi nhc im: c ghp mau hng; thit b cng knh;


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ch : cc thng s c bn, c im ca s khi c v khng iu khin, hotng, u nhc im ca s .

Qua vic phn tch u, nhc im ca cc s chng ta tin hnh chnmt s hp l cho ti.

Cn c chn s chnh lu thit k l:

Yu cu ca ti v cht l

ng ngun cp, di iu khin, n nh dngin v in p ti.

Cc thng s c bn ca ti: cng sut; in p ; dng in; p mchcho php.

Loi ngun cp- mt chiu hay xoay chiu, mt pha hay ba pha, tr sin p v tn s.

Kh nng c th c v ngun vt t linh kin, nht l cc van bn dnng lc. Gi thnh, kch thc. Cc cn c ph khc.Mt s gi v cch la chn s nh sau:

Chnh lu mt pha thng c chn khi ngun cp l li in mt pha,hoc cng sut khng qu ln so vi cng sut li (lm mt i xng in pli) v ti khng c yu cu qu cao v cht lng in p mt chiu.

Trong chnh lu mt pha, nu ti c dng in ln v in p thp, th s chnh lu mt pha c chu k vi bin p c trung tnh c u im hn. Bi vtrong s ny tn hao trn van bn dn t hn, nn cng sut tn hao trn van sovi cng sut ti nh hn, hiu sutt thit b cao hn, in p ngc ca van ln(nu in p cao m chn s ny c th khng chn c van bn dn). Nu tic in p cao v dng in nh, th vic chn s cu chnh lu mt pha hp lhn, bi v h s in p ngc ca van trong s cu nh hn, do chng tad chn van hn.


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ti c lm vic ch nghch lu hon tr nng lng v li, nh ng c inmt chiu chng hn.

i vi cc loi ti c in cm ln (v d nh cun dy kch t ca myin), li dng nng lng ca cun dy x ra v bo v van khi mt in tngt, ngi ta hay chn phng n mc thm mt it ngc song song vi ti.

Cc s chnh lu ba pha thng c chn, khi ngun cp l li ba pha

cng nghip v khi ti c yu cu cao v cht lng in p mt chiu.Chnh lu tia ba pha thng c la chn, khi cng sut ti khng qu ln

so vi bin p ngun cp ( trnh gy mt i xng cho ngun li), v khi tic yu cu khng qu cao v cht lng in p mt chiu. i vi cc loi ti cin p mt chiu nh mc l 220V, s tia ba pha c u im hn tt c. Biv theo s ny, khi chnh lu trc tip t li chng ta c in p mt chiu l220V.1,17 =257,4V. c in p 220V khng nht thit phi ch to bin p,m ch cn ch to ba cun khng anod ca van l .

Chnh lu cu ba pha nn chn, khi cn cht lng in p mt chiu tt, vy l s c cht lng in p ra tt nht, trong cc s chnh lu thnggp. S chnh lu cu ba pha iu khin khng i xng c mch iu khinn gin hn, nn trong a s cc trng hp ngi ta hay chn phng n cu bapha iu khin khng i xng. V d lm ngun cho my hn mt chiu, iu

khin kch t my pht xoay chiu cng sut nh, cc b ngun cho cc thit bin ho nh m in, in phn....

S chnh lu cu ba pha iu khin i xng c dng nhiu trong cctrng hp ti c yu cu v vic hon tr nng lng v li, v d nh iukhin ng c in mt chiu.

gim tit din dy qun th cp bin p, cc cun dy th cp bin p c

th u tam gic ().S tia su pha, vi vic ch to bin p phc tp v phi lm thm cun

khng cn bng, nn thng c la chn khi ti c dng in qu ln m theos cu ba pha chng ta khng chn c van theo dng in.

Cng mt tr s in p v dng in ti nh nhau s dng s cng nhiu


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thng s c bn ca s thit k.Cc thng s c bn ca mch ng lc cn c tnh chn l: cc van bn

dn ng lc, my bin p ng lc (hoc cun khng trong mch ng lc),aptomat, cu ch, dy ni......

8.4.1 Tnh chn van ng lc:

Hai thng s cn quan tm nht khi chn van bn dn cho chnh lu l inp v dng in, cc thng s cn li l nhng thng s tham kho khi la chn.

Khi p ng c hai thng s c bn trn cc thng s cn li c ththam kho theo gi sau:

- Loi van no c st p U nh hn s c tn hao nhit t hn.- Dng in r ca loi van no nh hn th cht lng tt hn.- Nhit cho php ca loi van no cao hn th kh nng chu nhit tthn.- in p v dng in iu khin ca loi van no nh hn, cng sut

iu khin thp hn.

- Loi van no c thi gian chuyn mch b hn s nhy hn. Tuy nhintrong a s cc van bn dn thi gian chuyn mch thng t l nghch vi tn

hao cng sut.Cc van ng lc c la chn da vo cc yu t c bn l: dng ti, s

chn, iu kin to nhit, in p lm vic.

Cc thng s c bn ca van ng lc c tnh nh sau:

in p ngc ca van c tnh:

Ulv

= knv

. U2

(8 -1)

vi U2 = Ud/ku thay vo (8 -1) lc Ulv c th tnh

U

d

nvlv k

U.k=U (8 -2)


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Unv = kdtU.Ulv. (8 -3)kdtUthng c chn ln hn 1,6.

Tnh dng in ca van.

Dng in lm vic ca van c chn theo dng in hiu dng chy quavan theo s chn (Ilv = Ihd). Dng in hiu dng c tnh:

Ihd = khd. Id (8 - 4)Trong : Ihd , Id- Dng in hiu dng ca van v dng in ti;

khd- H s xc nh dng in hiu dng (tra bng 8.2).

van bn dn c th lm vic an ton, khng b chc thng v nhit, cnphi chn v thit k h thng to nhit hp l. Theo iu kin to nhit cchn tin hnh tnh thng s dng in nh mc ca van cn c.

Dng in nh mc ca van (Imv) c th chn theo gi sau: khi khngcnh to nhit v tn hao trn van P10 Ilv), khi c cnh to nhit vi din tch b mt cho php vanlm vic ti 40%Imv (Imv > 2,5.Ilv), khi c cnh to nhit din tch b mt vc qut thng gi c th cho php van lm vic ti 60 %Imv (Imv > 1,6.Ilv ), khi ciu kin lm mt bng nc c th cho php lm vic gn ti 100% Imv .

V qu trnh thng gi t nhin khng c tt lm, do khi tn hao trnvan PV = UV.Ilv c khong 100 W/van tr ln, vic i lu khng kh t nhinxung quanh cnh to nhit xy ra chm, nhit to ra mi trng khng kp. Vvy theo kinh nghim, khi PV > 100 W/van cn c qut lm mt cng bc. Chitit v cch chn van tham kho trong phn bo v qu dng van trong ti liuny.

V d: Cn chn van ng lc cho mt b chnh lu cu mt pha vi thngs c bn ca s chnh lu: U= 100 V, I= 100 A.

Van ng lc cn chn c thng s:

in p ngc ca van


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Ilv = Ihd = khd .Ithay s vo vi khd tra t bng 2 ta c

Trong : Ulv - in p cc i khi lm vic [V];

Ilv , Ivhd- dng in lm vic v dng in hiu dng ca van [A];kU- h s in p ca s ;

khd- h s dng in hiu dng;

(cc h s khd ,kU tra trong Bng 8.1, 8.2 ca ti liu ny)

Vi cc thng s lm vic ca van trn, chn iu kin lm vic ca van l

c cnh to nhit vi y din tch to nhit, khng qut i lu khng kh(iu kin lm vic ca van do ngi thit k t chn).

Thng s cn c ca van ng lc l:

Unv = kdtU . Ulv = 2 . 157 = 314 V (chn kdtU = 1.6 2)

Imv = ki . Ilv = 4 . 70 = 280 A (vi iu kin lm mt chn Ilv = (10 30)% Imv y chn Ilv = 25% Imv )

c th chn c van cho lm vic vi cc thng s nh mc c bntrn, chng ta tra bng thng s cc van ( it, tiristo) chn cc van c thng sin p ngc (Unv), dng in nh mc(Imv) ln hn gn nht vi thng s tnh c trn.

Theo cch c th chn v d (tra t bng 8.4, 8.5):

it loi HD310/04-6 vi cc thng s nh mc :

- Dng in nh mc ca van Imv = 300 A,- in p ngc cc i ca van Unv = 400 V,- st p trn van U = 1,6 V,

AIlv 712

100==


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- Dng in d Ir = 50 mA,- in p iu khin Uk = 3 V,- Dng in iu khin Ik = 0,2 A.8.4.2 Tnh ton my bin p:

Cc i lng cn c cho tnh ton mt bin p chnh lu:1. in p chnh lu khng ti

Udo = Ud + U v + Uba + Udn (8 - 5)

Trong : Ud- in p chnh lu;

Uv- st p trn cc van (tr s ny c ly t cc thng s ca

cc van chn trn) ;Uba = Ur + Ul - st p bn trong bin p khi c ti, bao gm

st p trn in trUr v st p trn in cm Ul nhng i lng ny khi

chn s b vo khong (5 10)%;

Udn - st p trn dy ni;

Udn = Rdn.Id =(.l/S).Id . (8 6)2. Xc nh cng sut ti a ca ti v d vi ti chnh lu xc nh

Pdmax = Udo . Id (8 - 7)

3. Cng sut bin p ngun cp c tnh

Sba = ks . Pdmax (8 - 8)

Trong : Sba - cng sut biu kin ca bin p [W];ks - h s cng sut theo s mch ng lc (c th tra bng

2)

Pdmax - cng sut cc i ca ti [W].


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Sba - cng sut bin p tnh bng [W];kQ - h s ph thuc phng thc lm mt;

kQ= 4 5 nu l bin p du;

kQ = 5 6 nu l bin p kh;

m - s tr ca my bin p (bin p ba pha c m=3, mt pha c

m-1);f - tn s ngun in xoay chiu f=50 Hz.

Tit din ca tr gn ng c th tnh theo cng thc kinh nghim

5. Tnh ton dy qun bin p.

Thng s cc cun dy cn tnh bao gm s vng v kch thc dy.

Thng s cc cun dy qun s cp v cc cun dy th cp, ni chung cchtnh dy s cp v th cp nh nhau nn y ch gii thiu cch tnh chung chocc cun dy.

S vng dy ca mi cun c tnh

( )vongBQf

UW

Fe ...44,4

10. 4= (8 - 11)

Trong : W - s vng dy ca cun dy cn tnh

U - in p ca cun dy cn tnh [V];

B - t cm (thng chn trong khong (1,0 1,8) Tesla tu thuccht lng tn).

QFe - tit din li thp [cm2].

Nu coi f = 50 Hz chn B = 1Tsla lc gn ng c th tnh

[ ] )108(cm

m

S0,1Q 2baFe =


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- in p cun dy th cp c tnh:

U

d

k

UU 02 = (8 -13)

Trong : Ud0 - tnh t (8 -5);

kU - tra t h s in p chnh lu bng 8.1.

- in p cun dy s cp U1 bng in p ngun cp.Tnh dng in ca cc cun dy.

Cch th nht:

Xc nh dng in cc cun dy bng cch tra dng in s v th cp theobng 8.2

Cch th hai:

i vi nhng chnh lu c dng in xoay chiu i xng nh cc chnhlu cu, dng in c tnh gin tip qua cng sut pha s v th cp.

vi:

S1ba = ks1. Pdmax. (8 -16)

S2ba = ks2. Pdmax. (8 -17)

Trong : S1ba , S2ba - cng sut pha s,th cp bin p.ks1 , ks2 - cc h s cng sut pha s, th cp ca bin p. Cc

h s ny c th tra theo bng 8.2.

Tnh tit din dy dn:

( )148U

SI

1

ba11 =

( )158U

SI2

ba21

21 =


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Trong : d - ng kng dy qun.

SCu tit din dy qun.

Nu chn dy qun ch nht, cn tra bng kch thc dy (bng 8.3a)

chn kiu v kch thc dy6. Tnh kch thc mch t

Chn s b cc kch thc c bn ca mch t

Chn hnh dng ca tr

Nu cng sut nh (di 10 KVA) ngi ta thng chn tr ch nht (hnh

8.1) vi cc kch thc QFe = a . b. Trong a - b rng tr, b - b dy trNu cng sut ln ngi ta chn tr nhiu bc [......]

Chn l thp: thng l thp c cc dy 0,35 mm v 0,5 mm

Din tch ca s cn c:

Qcs = Qcs1 + Qcs2 (8 -20)

vi:Qcs1 =kl.W1.SCu1

Qcs2 =kl.W2.SCu2

Trong : Qcs,- din tch ca s [mm2];

Qcs1,Qcs2 - phn do cun s cp v th cp chim ch [mm2];

W1, W2 - s vng dy s, th cp;SCu1, SCu2 - tit din dy qun s, th cp [mm

2];

kl - h s lp y thng chn 2,0 3,0

Chn kch thc ca s

= Cu

S4d


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s ph cho tnh li.Chiu rng ton b mch t C = 2c + x.a (x =2 nu l bin p mt pha, x = 3

nu l bin p ba pha), chiu cao mch t H = h + z.a (z=1 nu l bin p mt pha,z = 2 nu l bin p ba pha)

Hnh dng kt cu mch t th hin nhhnh 8.1

7.Kt cu dy qun:

Dy qun c b tr theo chiu dc tr, mi cun dy c qun thnhnhiu lp dy. Mi lp dy c qun lin tc, cc vng dy st nhau. Cc lpdy cch in vi nhau bng cc ba cch in. Cch tnh cc thng s ny nhsau:

S vng dy trn mi lp W1l:

Khi dy qun tit din trn c tnh

n

g

ld

hhW

=1

Trong : h - chiu cao ca s,

dn - ng knh dy qun k c cch in;

b

hH

c

C

a c

Hnh 8.1 S kt culi thp bin p

(8 21)


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hg- khong cch cch In, khi dy qun ch nht c kch thcln, thng chn trong khong (2 - 10)mm.

S lp dy Sld trong ca s c tnh bng t s, s vng dy W ca cundy W1 hoc W2 cn tnh, trn s vng dy trn mt lp W1l

l

ldW

WS

1

=

B dy ca mi cun dy bng tng b dy ca cc lp dy d. s ld cng cchin cc lp dy trong cun dy cn tnh lp cd.sld.

Bdct = d. sld + cd.sld (8 -23)

Trong :

Bdct- b dy ca cun dy cn tnh,

cd - b dy ca ba cch in.Ba cch in c cc dy: o,1; 0,3; 0,5;1,0; 2,0; 3,0 mm.

dn - ng knh ngoi ca dy (nu dy qun tit din hnh chnht th thay dn bng an.

Tng b dy cc cun dy Bd

Bd = Bd1 + Bd2 + ......+ cdt + cdn (8 -24)Trong : Bd1, Bd2 - b dy cun dy s v th cp;

cdt, cdn - b dy cch in trong cng v ngoi cng.

Trc khi tnh kh lng st v ng cn kim tra xem ca s chn hp l cha. Kch thc ca s c,h ch ng khi b dy cc cun dy phi nh hnchiu rng ca s (Bd < c i vi bin p mt pha v 2Bd < c nu l bin p bapha). Kch thc hp l gia cun dy v tr c = c -Bd vi bin p mt pha v c= c - 2.Bd vi bin p ba pha trong khong (0,5 - 2) cm. Khong cch ny cnthit m bo cch in v lm mt. Trong trng hp ngc li, b dy Bd cccun dy ln hn chiu rng c ca ca s, chn li cc kch thc ca s c,h.


34/122

VFe = 3a.b.h + 2C.a.b = QFe.(3h + 2C)- nu l bin p ba pha;VFe = 2a.b.h + C.a.b = QFe.(2h + C)- nu l bin p mt pha;

Vi: QFe;a;b;c;h;C - l cc kch thc ca li thp c i thnh dm.

mFe = 7,85 kg/dm3

9. Khi lng ng.

Khi lng ng bng tch ca th tch VCu cun dy ng cn tnh nhn vitrng lng ring ca ng mCu:

MCu = VCu.mCu (kg) (8 - 26)

Trong :

VCu - th tch khi ng ca cc cun dy v c tnh [dm3];

VCu = SCu.lTrong : SCu - tit din dy dn [dm

2];

l - chiu di ca cc vng dy [dm];

mCu = 8,9kg/dm3

Chiu di dy qun c tnh bng cch ly chiu di mi vng nhn vi s

vng dy trong cun. Cc vng trong cun dy c chu vi khc nhau cho nnchng ta hay ly chu vi trung bnh tnh. Chiu di trung bnh ca cc vng dyc th tnh gn ng . Dtb khi coi Dtb l ng knh trung bnh ca cun dy trn,

l = W.. Dtb ( 8 - 27)

Trong :

Dtb - ng knh trung bnh ca cun dy v c tnh:

Dtb = (Dt+ Dn)/2

Trong : Dt,Dn - ng knh trong v ngoi ca cun dy.

ng knh trong ca cun dy trong cng c tnh:


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Dn = Dt + 2.(d + cd).sldCh : vi cc cun dy bn ngoi, th D t ca cun ngoi s bng Dn ca

cun trong.

Nu coi cun dy l khi hp ch nht th a tb+ btb l chu vi trung bnh cavng dy ch nht. Nh vy chiu di dy ng tnh theo cng thc:

l = W.( atb+ btb ). ( 8 - 28)

Trong :

atb = at+ Bd/2 chiu rng trung bnh ca vng dy;

btb = bt+ Bd/2 chiu di trung bnh ca vng dy;

at, bt - cc kch thc trong ca cun dy.

10. Tnh tng st p bn trong bin p.

in p ri trn in tr:

dr IW

WRRU

+=

2

1

212 (8 - 29)

Trong :

R1, R2 - in tr thun ca cc cun dy s v th cp c tnh:R = .l/S

Vi: = 0,0000172 mm - in tr sut ca ng;

l, S - chiu di v tit din ca dy dn [mm, mm2];

Id- dng in ti mt chiu [A].

in p ri trn in khng:Ux = mf. X.Id/ (8 - 30)

Trong :


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Rbk- Bn knh trong cun dy th cp [m2

].h - Chiu cao ca s li thp [m].

cd - B dy cc cch in cc cun dy vi nhau (nu l bin p dng nh,

gia cc cun dy c lt bng ba cch in dy (0,3 1) mm, cn i vinhng bin p dng ln, cn phi cch ly bng cc a php c cc dy lnhn) [m].

Bd1, Bd2 - B dy cun dy s v th cp [m].

= 314 rad.

11. in tr ngn mch my bin p

12. Tng tr ngn mch my bin p:

13. iin p ngn mch phn trm ca my bin p:

14. Dng in ngn mch my bin p:

)328(r*

w

wrr 1

2

1

2

2nm

+=

)338(xrz2

nm

2

nmnm +=

)358(z

UI

nm

dm2

nm =

)348(100*U

z*IU

dm2

nmdm0

0nm =


37/122

8.4.3 Tnh chn cc thit b bo v:1. Bo v qu dng in

Bo v ngn mch v qu ti v dng in: dng aptomat hoc cu ch.Nguyn tc chn cc thit b ny, l chn theo dng in, vi Ibv = (1,11,3) Ilv.Dng bo v ngn mch ca aptomat khng vt qu dng ngn mch ca mybin p.

Khi lm vic van bn dn c st p, do c tn hao cng sut P = U.Ilv.Tn hao cng sut ny sinh nhit. Mt khc van ch c lm vic ti nhit tia cho php Tcp no (cc tr s thng gp vo khong 125

0C - xem ct 8 bng8.4 hay ct 12 bng 8.5). Do chng ta phi tm cch bo v qu nhit cho vanbn dn.

Mun bo v qu nhit cho cc van bn dn l phi chn ng dng in

van theo ch lm mt. Lm mt van hin nay ph bin ngi ta thng dngcnh to nhit. Din tch b mt to nhit c th c tnh gn ng theo cngthc:

tntn

k

PS

= (8 - 36)

Trong : Stn - din tch b mt to nhit [cm2 ];

P - tn hao cng sut [ W ];

- chnh nhit so vi mi trng = Tlv -Tmt

Tlv -Tmt - nhit lm vic v nhit mi trng [oC ];

ktn - h s c xt ti iu kin ta nhit (trong iu kin lm mt

t nhin khng qut cng bc, thng chn ktn =(6 10).10-4 [W/cm2oC ]).

Sau khi tnh xong din tch b mt to nhit, tin hnh thit k (chn) cnhto nhit, sao cho b mt tnh. Cnh to nhit phi m bo din tch bmt tip xc vi khng kh, dy cnh, khong cch gia cc cnh.Nhng loi cnh to nhit hin nay c th bn rn trn th trng, hoc thit kmi theo hnh dng tng t nh hnh 8 2


38/122

a. Lm mt ch bng v van bn dn

Nu cng sut to nhit khi van lm vic P = U.Ihd< 20W, cho php van

lm vic vi dng in ti a ti 10% Im m khng cn cnh to nhit. Cch chnny c th hiu l v van bn dn khng to nhit khi cho lm vic vi dngin ln hn 10% Im. V d: c loi van vi Im =100A, U=1V, van ny chophp lm vic khng cnh to nhit vi dng in ti a ti 10A, nhng loi vanvi Im = 500A, U=1V cho php lm vic ti a ti 20A (P = 20W) mc d tnht s phn trm c th l 10%. Im = 50A.

b. Lm mt bng cch gn van bn dn ln cnh to nhit

Khi van bn dn c mc vo cnh to nhit bng ng hay nhm, nhit ca van c to ra mi trng xung quanh nh b mt ca cnh to nhit. S tonhit nh th ny l nh vo s chnh nhit gia cnh to nhit vi mi trngxung quanh. Khi cnh to nhit nng ln, nhit xung quanh cnh to nhit tng

h

h

b

a

c z

ho

1

Cc kch thc c bnnn chn:a < 200 mm;b < 200 mm;h < 180 mm;ho = 5-15 mm;c = 3 5 mm;z = 5 12mm.

Hnh 8.2 Hnh dng v kch thc gii hn cho cnhto nhit mt van bn dn


39/122

c. Lm mt cng bc bng qut.Khi c qut i lu khng kh thi dc theo khe ca cnh to nhit, nhit

xung quanh cnh to nhit thp hn tc dn nhit ra mi trng xung quanh tthn, hiu sut to nhit cao hn. Do cho php lm vic vi dng in ti a ti70%Im (Ilv< 70%Im).

d. Lm mt bng nc.

Khi thit k h thng lm mt bng nc hiu sut trao i nhit tt hn,cho php lm vic vi dng in ti a ti 90% Im. Qu trnh lm mt bng ncphi m bo x l nc khng dn in. Bng cch kh Ion trong nc, hocgim dn in ca nc (tng in tr nc) theo nguyn tc tng chiu dihay gim tit din ng ng dn nc ta c th coi dn in ca nc khngng k.

Th

ng th

ng i vi cc loi ngun cng sut, gi an ton cho cc vanbn dn ngi ta thng hay thit k qut lm mt cng bc ngay c trongtrng hp dng ti khng qu ln so vi dng nh mc ca van. Vi cng sutto nhit c vo khong 100W/ van l cn c qut lm mt cng bc.

Trong trng hp dng in lm vic qu ln (so vi dng cho php lm

D2 D1 D2D2 D1 D2D1

R R L L

I1 I2

D1

a b c d

Hnh 8.3 Cc s mc song song van bn dna- mc song song trc tip; b- mc qua in tr; c- mc quacun cm; d- mc qua h cm.


40/122

tnhTrong : U - hiu st p ca cc van khi cng tr s dng in ln I1;

R2dg- in tr ng ca van D2 ti im lm vic I1.

gim s phn b khng u trn, ngi ta c th mc ni tip vi cc vancc in tr (hnh 8. 3b) vic s dng in tr ch c ngha khi in p ri trnin tr l khng ng k, nu in p ri trn in tr ln, tn hao cng sut ln,

lm cho hiu sut ca chnh lu thp. khc phc nhc im ny chng ta cth thay th in tr bng cc cun dy in cm (hnh 8.3c). Thng cc cuncm ny c ch to c li khng kh.

s hnh 8.3d, cn bng dng in cc van c thc hin tt hn khicun khng c ch to c li thp, vi cc cun dy mc ngc u nhau. S ny cn c bit c ngha, khi s dng cho trng hp cc van iu khin mkhng ng thi.

2. Bo v qu in p cho thit b bn dn:

Linh kin bn dn ni chung v bn dn cng sut ni ring, rt nhy cmvi s thay i ca in p. Nhng yu t nh hng ln nht ti van bn dn mchng ta cn c phng thc bo v l:

in p t vo van ln qu thng s ca van. Xung in p do chuyn mch van. Xung in p t pha li xoay chiu, nguyn nhn thng gp l do ct

ti c in cm ln trn ng dy.

Xung in p do ct t ngt bin p non ti. bo v van khi lm vic di hn m khng b qu in p, chng ta cn

chn ng cc van bn dn theo in p ngc.a. Mc ni tip van bn dn.

Sau khi tnh c tr s in p lm vic ca van theo (8 - 2) v (8 - 3) tinhnh chn van theo in p, tr s in p van c chn phi ln hn tr s tnh


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R

V2 V2V2

V1 V1V1

R

R

R

R

R

C

C

C

C

V1

V2

D

D

D

D

a. b. c. d.

Thng gp nht trong thc t, phn b u in p khi mc ni tipngi ta hay mc theo s hnh 8.4a. S ny n gin d thc hin. Ngoi s ny ra, chng ta c th phn b in p bng t nh hnh 8.4b,c. hoc s dngcc it n p phn b in p.

b. Bo v xung in p khi chuyn mch van bn dn.

Hnh 8.5 Bo v thit b in t khi chc thng do xung in p.

Bo v xung in p do qu trnh ng ct cc van c dng bng ccmch R - C mc song song vi cc van bn dn. S n gin ca loi mch

R

T C C C

CC

D2

D4

D6

D1

D3

D5

R

R R

R2

R1

a. b. c.

Hnh 8.4 S mc ni tip cc van


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bo v xung in p t li in, chng ta mc song song vi ti uvo mt mch R - C, nhm lc xung nh m t trn hnh 8.5b. Khi xut hin xungin p trn ng dy, nh c mch lc ny m nh xung gn nh nm li honton trn in tr ng dy. Tr s R,C ph thuc nhiu vo ti. Thng s thamkho theo [..] R = (5 20), C = 4F.

c. Bo v van bn dn khi nh thng do ct bin p non ti.

bo v van do ct t ngt bin p non ti, trong a s cc b bin ingi ta thng mc mt mch R-C u ra mt chnh lu cu ba pha ph bngcc it cng sut b, nh m t trn hnh 8.5c. Tr s t C trong trng hp nyc th c tnh:

Trong : I - Dng in t ho bin p %;

I2;U2 - Dng in, in p th cp bin p;

KTU - Kh nng tng in p cho php ca van, thng c

chn KTU= 1,25 1,5.

Thng thng tr s t thng chn trong khong 10 200 F.

Bin in p xung khi ng bin p nh hn nhiu so vi khi ct do mch trn cho php bo v qu in p trong c hai trng hp ny.

8.5 Tnh ton cun khng lc dng in p mch.

8.5.1 Khi qut v dng in p mch.

S p mch ca in p chnh lu lm cho dng in ti cng p mch

theo, lm xu i cht lng dng in mt chiu, nu ti l ng c in mtchiu lm xu qu trnh chuyn mch c gp ca ng c, lm tng pht nng cati do cc thnh phn sng hi.

Thng thng chng ta nh gi nh hng ca p mch dng in theo trhiu dng ca sng hi bc nht bi v sng hi bc nht chim mt t l vo

)388(U

I

1K

I30C

2

2

2

Tu

=


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Trong :LL - tr s in cm lc p mch cn thit[Henry];

Id.m - dng in nh mc ca b chnh lu [A];

= 314 - tn s gc [1/s];

K = 1,2,3... - bi s sng hi;

m - s ln p mch trong mt chu k;

Ud.max- bin thnh phn sng hi ca in p chnh lu [V];

I1*% - tr hiu dng ca dng in sng hi c bn ly t s theo dng

in nh mc ca chnh lu. Tr s ny cho php I1*%


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Trong s chnh lu cu v tia ba pha, thnh phn sng hi bc nht (K=1)c bin ln nht. Bin sng hi bc cng cao s cng gim, tc dng cacun khng lc vi cc thnh phn hi bc cao ny cng c hiu qu hn. V vytnh in cm ca cun khng i vi cc s chnh lu ch cn tnh theo thnhphn songs hi bc nht l .

Quan h gia t l ca bin sng hi theo tr trung bnh in p chnh l

u

0

max.

d

dn

U

U vi gc m van bn dn ca cc s chnh lu cu v tia ba pha xy

dng theo cng thc (8 - 40) m t trn hnh 8 -6.

Cc ng cong 1,2 trn hnh 8 -6 m t quan h0

max.

d

dn

U

U = f() vi K=1 cho

cc s cu ba pha iu khin i xng v tia ba pha.

i vi s chnh lu iu khin khng i xng, khi gc iu khin nh, thnh phn sng hi vi K = 2 v K.m = 6 (ng 4 trn hnh 8 -6), cn khigc iu khin bt u t = 250 thnh phn hi bc nht K = 1 (ng 3 trnhnh 8 -6) c bin ln hn.

Khi tnh in cm cun khng lc dng in p mch, cn phi cn c vomc cho php ca p mch dng in chnh lu i vi ti tr s in p

nh mc v in p cc tiu.Cc b chnh lu cu iu khin i xng, ch dng in v in p

nh mc thng c gc iu khin 300. Tr s gc ny cn c p ng khnng b st p khi in p ngun li gim v st p do tng ti ca b chnh lu.V vy gc m ban u ca cc van bn dn c th coi 0 = 30

0.

C th tnh c mc in p chnh lu nh mc ca s chnh lu iu

khin i xng c ti 0 = 300, trong s chnh lu iu khin khng i xnggc ny c tr s0 = 43

0.

Trong cc s chnh lu c iu khin, khi gc m van bn dn cng tngbin sng hi cng cao. Do vy khi tnh in cm theo (8 - 39) v xc ng


45/122

cn thit LL.Tr s in cm ca cun khng lc LckL cn mc thm lc thnh phn

dng in p mch c tnh theo cng thc:

LckL = LL - Ld -LBA. (8 - 41)

Trong :

LckL - in cm cun khng lc cn mc thm;

LL - in cm cn thit lc thnh phn sng hi dng in I1*%


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8.5.2 Thit k cun khng lc (CKL) dng in p mch.

Cc thng s cn thit cho thit k:

-in cm ca cun khng lc tnh theo (8 - 39) khi ti khng in cm(thun tr), theo (8 - 41) khi ti c in cm.

-Dng in nh mc chy qua cun khng Im (dng in ny l dng inchnh lu nh mc).

-Thnh phn dng in xoay chiu cho php ca sng hi bc nht (thngcho php I~(1) < 10%.Im.

Thng thng dy qun cun khng lai ny c tit din kh ln, do vyin tr thun ca cun khng nh c th b qua. V vy

ZCKL = XCKL = . LCKL = 2f.m.LCKL. ( 8 - 44).Trong :

f - tn s in p ngun cp f = 50 Hz;

m - s ln p mch ca s chnh lu.

Cc bc tnh ton.

1. Tnh in p ri trn cun khng:UCKL = ZCKL . I~(1). (8 -45)

2. Tnh cng sut cun khng lc:

PCKL = UCKL. I~(1). (8 -46)

3. Tnh ton li thp cun khng lc.

Tit din li thp cun khng lc:(8 -47)

T

'.

f

PkQ CKLFe =


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bin p kh.4. Tnh ton dy qun cun khng:

Khi c thnh phn dng in xoay chiu bc nht I~(1) chy qua cun khnglc th trong cun khng xut hin mt sc in ng t cm, tr s sc in ngny c tnh:

ECKL = 4,44.kdq.W.f..

= 4,44.kdq.W.f.B.QFe. (8 -48)

Trong :

kdq - h s dy qun, c th chn kdq = 1,1 1,3;

W - s vng dy cun khng lc;

f` - tn s dng in sau chnh lu f` = 50.m;

B - mt t cm ca li thp, vi B = 1,1 1,8;

Q Fe - tit din hiu qu li thp.

Vi gi thit, b qua st p trn in tr, sc in ng ECKL xp x st ptrn cun khng UCKL tnh trn (ECKL = UCKL ). T c th tnh c svng dy W ca cun khng lc:

(8 -49)

Dy qun cun khng c tit din:

T tit din SCu tra bng kch thc dy qun chn c dy qun cn thit.Vic tnh ton cc thng s, kch thc cn li ca cun khng tng t nh

tnh ton my bin p [..].

8 6 Tnh ton cun khng hn ch dng in gin on

Fedq

CKL

QBfk

UW

'.'...44,4

=

J

IS dmCu =


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ca dng in ca ng c in mt chiu cng nh

cc loi ti khc l cn thit.Hin tng gin on dng in chnh lu xy ra do nng lng in t tch

lu trong mch khng ln. ch dng in gin on gc dn ca van trnn nh hn 2m, do in p xoay chiu i du nn dng in chy qua van bndn v 0 trc khi kch m van k tip.

Nu van m ti thi im t0 no tng ng vi gc m van 0=t0 tnh

t gc to ca ng cong in p hnh sin, hoc l gc m van tnh t thiim in p bt u dng.

T s thay th ca h thng chnh lu - ng c (CL - C) [ .. ] ta cphng trnh vi phn:

(8 -50)

t:

T = L/R - hng s thi gian in t ca mch;

= arctg(T) - gc pha ca mch.

Gii phng trnh (8 -50) ta c nghim ca phng trnh vi phn:

Trong :

I0 - gi tr ban u ca dng in trong mi khong van dn A ;

Edc

- sc in ng ca ng c V .

Dng in phn ng ng c i

c dng p mch, nn ta c th phn tchthnh phn mt chiu v xoay chiu. Thnh phn mt chiu ca dng in chnhlu chnh l thnh phn tc dng v c xc nh bng gi tri trung bnh ca i

trong mt chu k

( )dt

diLiRUEtU UUVdcm ..sin 02 +++=+

( )[ ] ( ) ( )[ ] ++= 02020 .sin.cos.expsin.cos. tEEtUEIRi mdcmdcU (8 51)


49/122

khng mc ni tip vi ti mt chiu hn ch vng gin on dng in.ng c in in mt chiu c gin on dng in di nht khi ng c

lm vic tc cui da iu khin (khi m in p ca b chnh lu l thpnht). Nh vy khi gc m van bn dn ln nht max th dng in ti b ginon di nht.

in cm cn thit hn ch vng dng in gin onc tnh theo cng

thc [4]:

(8 -53)

Trong :

Lg- in cm cn thit hajn ch dng in gin on [H]. = 2.f.m - tn s gc ca dng in, vi f = 50 Hz;

Udo - in p khng ti ca chnh lu;

Idgh - dng in gii hn nh nht, dng in ny trong tnh ton nn chn

xp x dng in khng ti. C th chn Idgh0,05Im.

kgh - h s ph thuc gc m van bn dn.

= BAgh

dgh

dgd xk

I

UL .

1 0

0,15

0,2

0,25

0,3

0,35

0,4

0,45 kgh

m=3


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n gin trong vic tnh biu thc (8 -54), quan h gia k gh v gc mvan bn dn theo (8 -54) c th xy dng thnh th nh biu din trn hnh 8.7.

Tm li, tnh c in cm ca cun khng hn ch dng in gin

on chng ta cn:

T di iu khin in p (hay di iu khin tc ng c) xc nh cgc m van ln nht max

Thay vo (8 -54) (hoc tra theo hnh 8 -7) ta c h s kgh.

Thay h s kgh vo (8 - 53) ta tnh c tr s in cm cn thit Lg hn

ch vng dng in gin on.Hiu s gia in cm cn thit Lg v in cm ti Ld s l in cm ca

cun khng mc thm vo mch hn ch dng in gin on.

LCKg = Lg - Ld (8 - 55)

8.6.2 Thit k cun khng hn ch dng in gin on.

1. Thng s cn c:

in cm ca cun khng tnh theo (8 -55 ).

Dng in nh mc chy qua cum khng. Dng in ny bng dng inchnh lu Im.

Gi tr dng in gin on gii hn.

Trnh t tnh ton:

1.Tnh tng tr ca cun khng:

ZCKg = RCK + 2.fLCKg. (8 -56)

sincot1

=m

gm

kgh (8 54)


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Vi : UCK = Igh. ZCKgh.3. Tnh tit din li thp cun khng.

(8 -59)

vi: k = 5 6, f = 50.m.4. Tnh s vng dy cn c ca cun khng.

Dng in gin on c dng l cc xung dng in. Do khi chy trongcun khng lm xut hin mt sc in ng t cm Eg, sc in ng ny cxc nh theo cng thc:

Eg = 4,44.kdq.W.f.B.QFe. (8 -60)

T (8 -60) ta c:

5. Xc inh in cm mch chnh lu nhm gii hn vng dng in gin

on.

S pha S ln pmch

in cm mch chnh lu Ld [Henry]

Tia ba pha 3 3

'.

f

PkQ

CKgd

Fe =

Fedq

CK

Fedq

gd

QBfk

U

QBfk

EW

.'...44,4.'...44,4

==

= fBA

dgh

fd x

IUL

sin.46,01 2

(8 61)


52/122

Ni dy song songHai

tia ba phangc nhauc khngcn bng.Theo s tngng 12pha

3 12

Ni dy ni tip

= fBA

dgh

f

d xI

UL

sin.02684,0

1 2

Cuba pha iukhin ixng

3 6

Bng 8.3 Thng s mt s dy ng trn

ngha cc ct:

d- ng knh thc ca li ng;

SCu - tit din tnh ton ca li ng;

mCu

- trng lng ring mt mt;

R/m- in tr mt mt;

Dn- ng knh ngoi k c cch in trong khong,

=

4sin.02684,0

1 2 fBA

dgh

f

d

x

I

UL

= fBA

dgh

dd x

I

UL 2sin.126,0

1 2


53/122

dmm

SCumm2

mCugam/m

R/m/m

Dnmm

0,12 0,01131 0,101 1,59 0,14 0,15

0,13 0,01327 0,118 1,256 0,15 0,16

0,14 0,01539 0,137 1,169 0,16 0,17

0,15 0,01767 0,157 1,018 0,17 0,180,16 0,02011 0,179 0,895 0,18 0,2

0,17 0,0227 0,202 0,793 0,19 0,21

0,18 0,02545 0,226 0,707 0,2 0,22

0,19 0,02835 0,252 0,635 0,21 0,23

0,2 0,03142 0,279 0,572 0,225 0,24

0,21 0,03464 0,308 0,52 0,235 0,25

0,23 0,04155 0,369 0,433 0,255

0,280,25 0,04909 0,436 0,366 0,275 0,3

0,27 0,05726 0,509 0,315 0,31 0,32

0,29 0,06605 0,587 0,296 0,33 0,34

0,31 0,07548 0,671 0,239 0,35 0,360,33 0,08553 0,76 0,21 0,35 0,38

0,35 0,09621 0,855 0,187 0,39 0,41


54/122

dmm

SCumm2

mCugam/m

R/m/m

Dnmm

0,51 0,2043 1,82 0,084 0,56 0,58

0,53 0,2206 1,96 0,0781 0,58 0,60

0,55 0,2376 2,11 0,0725 0,60 0,62

0,57 0,2552 2,27 0,0675 0,62 0,640,59 0,2734 2,43 0,063 0,64 0,66

0,62 0,3019 2,68 0,0571 0,67 0,69

0,64 0,3217 2,86 0,0538 0,69 0,72

0,67 0,3526 3,13 0,0488 0,72 0,75

0,69 0,3729 3,32 0,0461 0,74 0,77

0,72 0,4072 3,6 0,0423 0,78 0,8

0,74 0,4301 3,82 0,04 0,8 0,83

0,77 0,4657 4,14 0,037 0,83 0,86

0,8 0,5027 4,47 0,0342 0,86 0,890,86 0,5809 5,16 0,0297 0,92 0,95

0,9 0,6362 5,66 0,027 0,96 0,99

0,93 0,6793 6,04 0,0253 0,99 1,02

0,96 0,7238 6,44 0,0238 1,02 1,05

1,00 0,7854 6,98 0,0219 1,08 1,11

1,04 0,8495 7,55 0,0202 1,12 1,15

1 08 0 9161 8 14 0 0188 1 16 1 19


55/122

dmm

SCumm2

mCugam/m

R/m/m

Dnmm

1,30 1,327 11,8 0,0132 1,38 1,41

1,35 1,4314 12,7 0,0123 1,43 1,46

1,40 1,5394 13,7 0,0113 1,48 1,51

1,45 1,6513 14,7 0,0106 1,53 1,561,5 1,7672 15,7 0,00993 1,58 1,61

1,56 1,9113 17 0,00917 1,64 1,67

1,62 2,0612 18,3 0,0085 1,71 1,73

1,68 2,217 19,7 0,00791 1,77 1,79

1,74 2,378 21,1 0,00737 1,83 1,85

1,81 2,573 22,9 0,00681 1,90 1,93

1,88 2,776 24,7 0,00631 1,97 2,00

1,95 2,987 26,5 0,00587 2,04 2,07

2,02 3,205 28,5 0,00547 2,12 2,142,10 3,464 30,8 0,00506 2,20v2,23

2,26 4,012 35,7 0,00437 2,36 2,39

2,44 4,676 41,6 0,00375 2,54 2,57

2,83 6,29 55,9 0,00278

3,05 7,306 65, 0,0024

3,28 8,45 75,1 0,00207

4,1 13,2 11,7 0,00123


56/122

8.7. Thit k mch iu khin

8. Nguyn l thit k mach iu khin.

iu khin Tiristo trong s chnh lu hin nay thng gp l iu khintheo nguyn tc thng ng tuyn tnh. Ni dung ca nguyn tc ny c th m ttheo gin hnh 8.8 nh sau.

Khi in p xoay chiu hnh sin t vo anod ca Tiristo, c th iukhin c gc m ca Tiristo trong vng in p + anod, ta cn to mt inp ta dng tam gic, ta thng gi l in p ta l in p rng ca Urc. Nhvy in p ta cn c trong vng in p dng anod.

Dng mt in p mt chiu Uk so snh vi in p ta. Ti thi im(t1,t4) in p ta bng in p iu khin (Urc = Uk), trong vng in p dng

anod, th pht xung iu khin Xk. Tiristo c m t thi im c xung iukhin (t1,t4) cho ti cui bn k (hoc ti khi dng in bng 0)

Udf

Urc

Udk

Ud

Xdk

t1 t2 t3 t4 t5


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Hnh 8.9. S khi mch iu khin

Nhim v ca cc khu trong s khi hnh 8.9 nh

sau:Khu ng pha c nhim v to in p ta Urc (thng gp l in p dngrng ca tuyn tnh) trng pha vi in p anod ca Tiristo

Khu so snh c nhim v so snh gia in p ta vi in p iu khinUk, tm thi im hai in p ny bng nhau (Uk = Urc). Ti thi im hai inp ny bng nhau, th pht xung u ra gi sang tng khuych i.

Khu to xung c nhim v to xung ph hp m Tiristo. Xung mTiristor c yu cu: sn trc dc thng ng, m bo yu cu Tiristo mtc thi khi c xung iu khin (thng gp loi xung ny l xung kim hoc xungch nht); rng vi rng xung ln hn thi gian m ca Tiristo; cngsut; cch ly gia mch iu khin vi mch ng lc (nu in p ng lc quln)

Vi nhim v ca cc khu nh vy tin hnh thit k, tnh chn cc khu

c bn ca ba khi trn. Chi tit v cc mch ny s gii thiu chi tit phn sau.8.7.1 Thit k s nguyn l

Hin nay mch iu khin chnh lu thng c thit k theo nguyn tcthng ng tuyn tnh nh gii thiu trn.

Theo nhim v ca cc khu nh gii thiu, tin hnh thit k, tnh chncc khu c bn ca ba khi trn, tt nht l nn chn trong cc s c trongcc gio trnh v ti liu. Trong ti liu ny ch gii thiu mt s s v d chongi thit k lm t liu tham kho la chn.

Trn hnh 8.10; 8.11; 8.12 gii thiu mt s khu ng pha, so snh, toxung in hnh

ng pha So snh To xung


58/122

A R1

R2

-E

U2U1

C

D

Tr

Ura

dng s tao in p ta bng s hnh 8.10b. Theo s ny, in p ta cphn bin thin tuyn tnh ph ht na chu k in p. Do vy khi cn iu khinin p t 0 ti cc i l hon ton c th p ng c.

Vi s ra i ca cc linh kin ghp quang, chng ta c th s dng s

to in p ta bng b ghp quang nh hnh 8.10c. Nguyn l v cht lng inp ta ca hai s hnh 8.10b,c tng i ging nhau. u im ca s hnh8.10c ch khng cn bin p ng pha , do c th n gin hn trong vicch to v lp t.

C h h i l i kh T i

U1 U2

R1A

-E

R2

D2

D

B

UrC

C

A1 A2

R1A

R2

Ur

R3

C1

CD1B

Tr

U1

GHEP QUANG

C

R2R1

DUra

+E

c.

Uv

Hnh 1.10: Mt s khu ng pha in hnh.

a- dng it v t; b- dng tranzitor v t; c- dng b ghpquang; d- dng khuych i thut ton.

a. b.

d.


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xc nh c thi im cn m Tiristo chng ta cn so snh hai tn hiuUk v Urc. Vic so snh cc tn hiu c th c thc hin bng Tranzitor(Tr) nh trn hnh 8.11a. Ti thi im Uk = Urc, u vo Tr lt trng thi tkho sang m (hay ngc li t m sang kho), lm cho in p ra cng b lttrng thi, ti chng ta nh du c thi im cn m Tiristo.

Vi mc m bo ho ca Tr ph thuc vo hiu Uk Urc = Ub, hiuny c mt vng in p nh hng mV, lm cho Tr khng lm vic ch ngct nh ta mong mun, do nhiu khi lm thi im m Tiristo b lch kh xa sovi im cn m ti Uk = Urc.

KTT c h s khuych i v cng ln, ch cn mt tn hiu rt nh (cV) u vo, u ra c in p ngun nui, nn vic ng dng KTT lmkhu so snh l hp l. Cc s so snh dng KTT trn hnh 8.11b,c rtthng gp trong cc s mch hin nay. u im hn hn ca cc s ny lc th pht xung iu khin chnh xc ti Uk = Urc.

Vi nhim v to xung ph hp m Tiristo nh nu trn tng

A3

Ura

R2Udk

R1Urc

b.

A3

Ura

R1Urc

R2Udk

c.

R1Urc

R2Udk

-E

R3

a.

Tr

Ura

Hnh 8.11: S cc khu so snh thng gp

a- bng tranzitor; b- cng mt cng o ca KTT; c- haicng KTT.


60/122

Tng khuych i cui cng bng s darlington nh trn hnh 8.12bthng hay c dng trong thc t. s ny hon ton c th p ng cyu cu v khuych i cng sut, khi h s khuych i c nhn ln theothng s ca cc tranzitor.

Trong thc t xung iu khin ch cn c rng b (c khong (10 200)s), m thi gian m thng cc tranzitor cng sut di (ti a ti mt na chu k -0.01s), lm cho cng sut to nhit d ca Tr qu ln v kch thc dy qun scp bin p d ln. gim nh cng sut to nhit Tr v kch thc dy s cpBAX chng ta c th thm t ni tng nh hnh 8.12c. Theo s ny, Tr ch mcho dng in chy qua trong khong thi gian np t, nn dng hiu dng ca

R

Uv

Tr

BAX

+E

D

a.

R

Uv

Tr

BAX

+E

D

Tr1

b.

R

Uv

Tr

BAX

+E

D

Tr1C

D

c.

Hnh 8.12: S cc khu khuych i.a- bng tranzitor cng sut; b- bng s darlington;

c- s c t ni tng.


61/122

8

6

7

4

2 1

3

5

555

+U

R1

R2

C1 C2

Ura

a.

Hnh 8.13: S phi hp to xung chm.

Cc s to chm xung c th tham kho trong [ 7 ]. Mt s s khuto chm xung m t trn hnh 8.14.

Vi mch 555 to xung ng h hnh 8.14a cho ta cht lng xung kh tt vs cng tng i n gin. S ny thng hay gp trong cc mch tochm xung.

Trong thit k mch iu khin, th

ng hay s dng KTT. Do ngdng v linh kin, khu to chm xung cng c th s dng KTT, nh cc s trn hnh 8.14b,c. Tuy nhin, y s dao ng a hi hnh 8.14b c u imhn v mc n gin, do c s dng kh rng ri trong cc mcg toxung ch nht.

T so snh &Ti khuych i

T chm xung

A

R3

C

R3

R4

+15V


62/122

62

D4

Uv

R1

R2

A

A1

+

- B

R3

R4D3

Tr1

CA2

-

+

R4

Uk

R5

R6

A3-

+

D

T1

+15V

VA

A6+

CX

Hnh 8.15 Mch iu khgin Tiristora. mch to xung iu khin n, b. mch to chm xung iu khin

D4

Uv

R1

R2

A

A1

+

- B

R3

R4D3

Tr1

CA2

-

+

R4

Uk

R5

R6

A3-

+

D T

a.

b.

Sau khi chn xong cc khu c bn, v mch hon chnh .


63/122

UA

UrcUdk

Ud

Xdk

UB

UD

UE

UF

t

t

t

t

t

t

g

V d: s ng pha chn hnh 8.10d, so snh chn hnh 8.11b, s

khuych i chn hnh 8.12c, s to chm xung hnh 8.14b, ta c s nguyn l mch iu khin Tiristo m t trn hnh 8.15

Khi c mch iu khin tin hnh gii thch hot ng ca ton bmch v hiu chnh nhng ch cha hp l.

Hot ng ca mch iu khin hnh 8.15 c th gii thch theo gin cc ng cong trn hnh 8 16 nh sau:


64/122

cc ng cong trn hnh 8.16 nh sau:

in p vo ti im A (UA) c dng hnh sin, trng pha vi in panod ca Tiristo T, qua khuych i thut ton (KTT) A1 cho ta chui xungch nht i xng UB. Phn p dng ca in p ch nht UB qua it D1 tiA2 tch phn thnh in p ta Urc. in p m ca in p UB lm m thngtranzitor Tr1, kt qa l A2 b ngn mch (vi Urc = 0) trong vng UB m.Trn u ra ca A2 chng ta c chui in p rng ca Urc gin on.

in p Urc c so snh vi in p iu khin Uk ti u vo ca A3.Tng i s Urc + Uk quyt nh du in p u ra ca KTT A3. Trongkhong 0t1 vi Uk > Urc in p UD c in p m. Trong khong t1t2in pUk v U rc i ngc li, lm cho UD lt ln dng. Cc khong thigian tip theo gii thch in p UD tng t.

Mch a hi to chm xung A4 cho ta chui xung tn s cao, vi in pUE trn hnh 8.16. Dao ng da hi cn c tn s hng chc kHz y ch m

t nh tnh.Hai tn hiu UD, UE cng c a ti khu AND hai cng vo. Khi

ng thi c c hai tn hiu dng UD, UE (trong cc khong t1t2, t4t5)chng ta s c xung ra UF. Cc xung ra UF lm m thng cc tranzitor, ktqu l chng ta nhn c chui xung nhn Xdk trn bin p xung, a tim Tiristo T.

in p Ud s xut hin trn ti t thi im c xung iu khin u tin,ti cc thi im t2, t4 trong chui xung iu khin, ca mi chu k in pngun cp, cho ti cui bn k in p dng anod.

Hin nay c nhiu hng ch to ra cc vi x l chuyn dng iukhin tiristo, c th tm hiu cc loi vi x l ny trong [8], ni chung cc vi xl iu khin tiristo rt tin li, tuy nhin nhng linh kin loi ny cha c

ph bin lm trn th trng.

1.8.2 Tnh ton cc thng s ca s mch iu khin.

rng xung iu khin tx;


65/122

Mc st bin xung sx mt i xng cho php ; in p ngun nui mch iu khin U.1.Tnh bin p xung

- Chn vt liu lm li (li bin p y c th bng l thp k thutin, hoc thng lm bng li pherit)

-.Tnh th tch li thp cn c:

Trong : tb - t thm trung bnh

o = 4 . 10-6 (H/m);

Q - tit din li st;

l - chiu di trung bnh ng sc t;

T th tch li st, tra bng chn c li c cc kch thc cn thit.

- Tnh thng s dy qun bin p xung:

S vng dy qun s cp

S vng dy qun th cp

)628(B

I.U.s.t..l.QV

2

2xx0tb

==

)638(H.

B

0tb

=

)648(B.Q

t.UW x1

=

)658(k

WW

ba

1

2 =

Sau khi la chn xong cc linh kin ca tng khuych i cui cngchng ta c cc thng s c bn ca mch ny


66/122

chng ta c cc thng s c bn ca mch ny.

in p ngun nui xung Ux;Dng in colector Ic;

H s khuych i dng in ca Tranzitor ;

in tr vo ca tng khuych i c chn:

Trong : Uv - in p vo c ly t tng so snh a sang;

Ib - dng in baz ca Tranzitor khuych i

vi Ic l dng in s cp bin p xung Ic = Ik/kbax.

3. Tnh chn tng so snh.

Tng so snh cng c tnh trn c s nhng linh kin trong s

dc chn.Nu tng so snh c chn l s hnh 8.8a, th cc in tr hn ch

colector R3 v baz R1 c chn:

Nu tng so snh c chn l cc s hnh 8.8b;c, th cc in tr hn

)688(I

UR

b

vb

( )698II 0b =

)708(I

ER

0

3

)718(I

URR

b

V

21

=

U

cc linh kin bn dn.

T i h (h h 8 8 b d) i t i t


67/122

Trong i a s cc s ng pha (hnh 8.8a;b;c;d), vic to in p ta

c tin hnh bng cch np t theo mch R - C. Thng s ca cc mch R -C c tnh theo

Trc = R2.C ( 8 - 73 )

trong : R2,C - cc thng s in tr v t in trong mch np t (trncc s hnh 8.8).

cho cc ng rng ca c nh nhn ti cui cc bn k, th Trc nn

chn c khong Trc = (0,003 0,005) s. T tr s Trc va chn, ta tin hnh tnhchn R2 v C bng cch chn trc C sau tnh R2.

Cc thng s in p xoay chiu ng pha c chn khong (5 10)V

8.9. Thit k t in.

T in thit k phi m bo cht lng cao v k thut, m thut cngnghip.

Cht lng k thut ca t in c th hin cc yu cu sau:

Kch thc hp l so vi cc thit b cn lp.

B tr linh kin hp l v khng gian.

Cc linh kin b tr theo nguyn tc trng lng ngha l nhng thit bnng b tr di thp, nhng thit b nh b tr trn cao.

Cc thit b b tr theo nguyn tc to nhit, ngha l nhng thit b tonhit t c b tr di thp cn nhng thit b to nhit nhiu c b tr trncao.

- C cc l thng gi cn thit: a s cc thit b in t cng sut cn

to nhit nhiu nn thng phi b tr qut lm mt. Thit b cn lm mt nhttrong trng hp ny l cc van bn dn, bi v cc van bn dn to nhit lnv rt nhy vi nhit .

Mch iu khin cn c bo v tt trnh nhit cao ngi ta thng

Mu sc hi ho khng qu sc s, khng qu ti, thng gp mu ghisng, trng ng, xanh nht.......


68/122

g, g g ,

Thit b b tr ngay ngn c hng, c ct.Dy ni phi t trong mng dy hoc c b thnh b gn gh.

8.9_V d tnh ton b ngun chnh lu .

Thit k b ngun chnh lu mt chiu cp in cho ng c in mt

chiu kch t c lp .

Thng s ng c : Um=400 (V) ;nm=980 (vng/pht) ;P=27(Kw);=0,85 ;s i cc p=2.8.9.1_La chn s thit k.

Sau khi phn tch nh gi v chnh lu ,t cc u nhc im ca cc s

chnh lu ,vi ti v ng c in mt chiu vi cng sut va phi nh trn

th s chnh l

u cu 3 pha iu khin i xng l hp l hn c ,bi l cng sut ny trnh lch ti bin p ,khng th thit k theo s mt pha

,s tia 3 pha s lm mt i xng in p ngun .Nn s thit k ta chn

l s cu 3 pha c iu khin i xng .

S c biu din trn hnh 8 -17 di y :

ba c T1T2

CA B

Cc thng s c bn cn li ca ng c c tnh .

P


69/122

Im=

dmU

P

.

=79,41(A) .

U2a ;U2b ;U2c - Sc in ng th cp my bin p ngun .

E - Sc in ng ca ng c .

R ,L - in tr v in cm trong mch .

R= 2.Rba+R+Rk+Rdt .

L= 2.Lba+L+Lk .Rk,Lk l in tr v in khng ca my bin p qui i v th cp :

Rba=R2+R1.

2

1

2

W

W

Lba=L2+L1.

2

1

2

WW .

Rk ,Lk l in tr v in khng cun lc .

Rdt in tr mch phn ng ng c c tnh gn ng nh sau :

R=0,5 .(1-).

udm

udm

I

U()=0,5 .(1-0,85) .

476,53

220=0,31 () .

L

l in cm mch phn ng ng c c tnh theo cng thc

Umanxki_Lindvit :

L

= .dmdm

dm

Inp

U

....2

60.

=0,25 .

476,53.950.3..2

60.220

=0,00345 (H) =3,45 (mH)

Trong =0,25 l h s ly cho ng c c cun b .1.10.2_Tnh chn Thyristor:

Tnh chn da vo cc yu t c bn dng ti ,iu kin to nhit ,in p lm

vic ,cc thng s c bn ca van c tnh nh sau :

Trong :

KdtU - h s d tr in p ,chn KdtU =1,8 .


70/122

dtU p , dtU ,

+) Dng lm vic ca van

c tnh theo dng hiu dng :Ilv = Ihd = Khd .Id =

3dI =

3

41,79=45,847 (A)

(Do trong s cu 3 pha ,h s dng hiu dng :Khd =3

1) .

Chn iu kin lm vic ca van l c cnh to nhit v y din tch

to nhit ; Khng c qut i lu khng kh ,vi iu kin dng nh mcca van cn chn :

Im =Ki . Ilv =3,2 . 45,847 = 147 (A)

(Ki l h s d tr dng in v chn Ki =3,2)

t cc thng s Unv ,Imv ta chn 6 Thysistor loi SCI50C80 do M sn xut c

cc thng s sau :in p ngc cc i ca van : Un = 800 (V)

Dng in nh mc ca van : Im =150 (A)

nh xung dng in :Ipik =2800 (A)

Dng in ca xung iu khin : Ik =0,1 (A)

in p ca xung iu khin : Uk =3,0 (V)

Dng in r : Ir =15 (mA)St p ln nht ca Thyristor trng thi dn l : U = 1,6 (V)

Tc bin thin in p :dt

dU=200 (V/s)

Tc bin thin dng in :

dt

dI=180 (A/s)

Thi gian chuyn mch : tcm = 80 (s)Nhit lm vic cc i cho php :Tmax=125

oC

8 9 3 Tnh ton my bin p chnh lu

Up =380 (V)

3-in p pha th cp ca my bin p


71/122

p p p y p

Ph

ng trnh cn bng in p khi c ti :Udo .cos min =Ud +2. Uv +Udn + UbaTrong :

min =100 l gc d tr khi c s suy gim in li

Uv =1,6 (V) l st p trn Thyristor

Udn 0 l st p trn dy ni

Uba = Ur + Ux l st p trn in tr v in khng my bin p .Chn s b :

Uba =6% .Ud =6% .400 = 24 (V)T phng trnh cn bng in p khi c ti ta c :

Ud0

=mincos

.2.2

badnvd UUU +++ =o10cos

2406,1.2400 +++=433,79 (V)

in p pha th cp pha my bin p :

U2=u

d

k

U=

6.3

79,433=185,45 (V)

4-Dng in hiu dng s cp ca my bin p :

I2 = dI32 = 41,79

32 = 64,84 (A)

5-Dng in hiu dng s cp my bin p :

I1 = KbaI2 =1

2

U

U.I2 =

380

45,185. 64,84 = 34,64 (A)

*)Tnh s b mch t(Xc nh kch thc bn mch t)

6-Tit din s b tr .

S

QFe=6 .50.3

33353


72/122

50.3

7-ng knh tr :

d =

eFQ.4 =

469,89.4= 10,67 (cm)

Chun on ng knh tr theo tiu chun d = 11 (cm)

8-Chn loi thp 330 cc l thp c dy 0,5 mm

Chn mt t cm trong tr Bt = (T)

9-Chn t s m=d

h= 2,3 , suy ra h = 2,3 . d = 2,3.11 = 25,3 (cm)

Ta chn chiu cao tr l 25 cm

*)Tnh ton dy qun .

10- S vng dy mi pha s cp my bin p .

W1=TFe BQf

U

...44,41 =

0,1.10.469,89.50.44,4

3804

= 191,3 (vng)

Ly W1= 191 vng

11- S vng dy mi pha th cp my bin p :

W2

=1

2

U

U.W

1=

380

45,185.191 = 93,2 (vng)

Ly W2= 93 vng

12- Chn s b mt dng in trong my bin p .

Vi dy dn bng ng ,my bin p kh ,chn J1= J2= 2,75 (A/mm2)

13- Tit din dy dn s cp my bin p .

S1 =1

1JI =

75,264,31 = 11,5 (mm2)

Chn dy dn tit din hnh ch nht ,cch in cp B .

Chun ho tit din theo tiu chun : S = 12 30 (mm2)

S2 =2

2

J

I=

75,2

84,64= 23,58 (mm2)


73/122

2

Chn dy dn tit din hnh ch nht ,cch in cp B .Chun ho tit din theo tiu chun : S2= 23,6 (mm

2)

Kch thc dy dn c k cch in : S2c = a2.b2 = 3,28.7,4 (mm x mm)

16- Tnh li mt dng in trong cun th cp .

J2=2

2

S

I= 2,74 (A/mm2)

*)Kt cu dy dn s cp :

Thc hin dy qun kiu ng tm b tr theo chiu dc trc

18- Tnh s b s vng dy tren mt lp ca cun s cp .

W11=1

.2

b

hh g. kc=

69,0

5,1.225 .0,95 = 30 (vng)

Trong :kc= 0,95 l h s p cht .

h l chiu cao tr .

hg l khong cch t gng n cun dy s cp .

Chn s b khong cch cch in gng l 1,5 cm .

19- Tnh s b s lp dy cun s cp :n11=

11

1

W

W=

30

191= 6,3 (lp)

20- Chn s lp n11=7 lp .Nh vy c 191 vng chia thnh 7 lp ,Chn 6 lp

u vo c 28 vng ,lp th 7 c 191 6.28 = 23 (vng)

21- Chiu cao thc t ca cun s cp :

h1=ck

bW .11 =95,0

69,0.28= 20,34 (cm)

22- Chn ng qun dy lm bng vt liu cch in c b dy : S01= 0,1 cm.

28- ng knh ngoi ca cun s cp .

Dn1= Dt1+2.Bd1=13 + 2.1,337= 15,467 (cm)


74/122

29- ng knh trung bnh ca cun s cp .

Dtb1=2

11 nt DD + =2

674,1513 += 14,337 (cm)

30- Chiu di dy qun s cp .

l1 = W1..Dtb= .191.14,337 = 86,02 (m)31- Chn b dy cch in gia cun s cp v th cp :cd01= 1,0 cm

*)Kt cu dy qun th cp .32- Chn s b chiu cao cun th cp .

h1= h2 = 20,34 (cm)

33- Tnh s b s vng dy trn mt lp .

W12= ck

b

h

2

2 = 95,0

74,0

34,20= 26 (vng)

34- Tnh s b s lp dy qun th cp .

n12=12

2

W

W=

26

93= 3,6 (lp)

35- Chn s lp dy qun th cp n12= 4 lp .Chn 6 lp u c 24 vng ,lp

th 4 c 93 4.24 = 21 (vng)

36- Chiu cao thc t ca cun th cp

h2= bk

W

c

.12 = 74,0.95,0

24= 18,69 (cm)

37- ng knh trong ca cun th cp.

Dt2 = Dn1+ 2.a12 = 15,674 + 2.1 = 17,674 (cm)

38- Chn b dy cch in gia cc lp dy cun th cp : cd22= 0,1 (mm)39- B dy cun s cp .

Bd2 = (a2+cd22).n12 = (0,328 + 0,01).4= 1,352 (cm)

40 k h i th

D12=2

21 nt DD +=

2

378,2013 +=16,689 (cm)


75/122

r12= 212D = 8,344 (cm)

44- Chn khong cch gia hai cun th cp :a22= 2 (cm)

*)Tnh kch thc mch t .

45- Vi ng knh tr d= 11 cm ,ta c s bc l 6 trong na tit din tr .

W2 W1

.

.

a01 hg

a12

Bd2Bd1

Hnh 1.29 Cc bc thang ghp thnh tr

1

2

345

6

46- Ton b tit din bc thang ca tr .

Qbt= 2.(1,6.10,5+1,1.9,5+0,7.8,5+0,6.7,5+0,4.6,5+0,7.4) = 86,2 (cm2)

47- Tit din hiu qu ca tr .

QT= khq.Qbt = 0,95.86,2 = 81,89 (cm2)

48- Tng chiu dy cc bc thang ca tr .

dt = 2.(1,6+1,1+0,7+0,6+0,4+0,7)= 10,2 (cm)

49- S l thp dng trong cc bc .

Bc 1 n1= 2.5,0

16 = 64 l

Bc 2 n2= 2.11

= 44 l

Bc 6 n6= 2.5,0

7= 28 l


76/122

*) n gin trong vic ch to gng t,ta chn gng c tit din hnh chnht c cc kch thc sau .

Chiu dy ca gng bng chiu dy ca tr : b = dt = 10,2 cm

Chiu cao ca gng bng chiu rng tp l thp th nht ca tr : a = 10,5 cm

Tit din gng Qbg= a x b = 107,1 (cm2)

50- Tit din hiu qu ca gng .

Qg= khq.Qbg = 0,95 .107,1 = 101,7 (cm2)51- S l thp dng trong mt gng .

hg =5,0

b=

5,0

102= 204 (l)

52- Tnh chnh xc mt t cm trong tr .

BT=TQWf

U...44,4 1

1 = 410.89,81.191.50.44,4380 = 1,094 (T)

53- Mt t cm trong gng .

Bg = BT.g

T

Q

Q= 1,094.

7,101

89,81= 0,88 (T)

54- Chiu rng ca s .c= 2.(a01+Bd1+a12+Bd2) + a22 = 2.(1+1,337+1+1,352) + 2 = 11,378 (cm)

55- Tnh khong cch gia hai tm trc .

c = c+d = 11,378 + 11 = 22,378 (cm)

56- Chiu rng mch t .

L= 2.c +3.d = 2.11,378 + 3.11 = 55,756 (cm)

57- Chiu cao mch t .H = h + 2.a = 25 + 2.10,5 = 46 (cm)

*)Tnh khi lng ca st v ng.

58 Th h


77/122

H

nh1.30

Sktc

um

yb

inp

62- Khi lng ca st .

M M +M 48 21 + 93 752 141 962 (Kg)


78/122

MFe

= MT

+Mg

= 48,21 + 93,752 = 141,962 (Kg)

63- Th tch ng .

VCu = 3.(S1.L1 + S2.L2) = 3.(12,3.10-4.86,02.10+ 23,6.10-4.55,5597.10)

= 7,1077 (dm3)

64- Khi lng ca ng .

MCu = VCu . mCu = 7,1077.8,9 =63,26 (Kg)

*)Tnh cc thng s ca my bin p .65- in tr ca cun s cp my bin p 75 0 C .

R1= .1

1

S

l= 0,02133.

3,12

02,86= 0,149 ()

Trong 75 = 0,02133 ()

66- in tr cun th cp my bin p 750

C .R2= .

2

2

S

l= 0,02133.

6,23

5597,55= 0,0502 ()

67- in tr ca my bin p qui i v th cp .

RBA = R2 + R1

2

1

2

W

W = 0,0502 + 1,492

191

93

=0,085 ()

68- St p trn in tr my bin p .

Ur = RBA.Id = 0,085 . 79,41 = 6,75 (V)69- in khng my bin p qui i v th cp .

XBA= 8 .2.(W)2.

++

3

. 2112dd

qd

BBa

h

r..10-7

= 8 .2.932.

++

3

352.1337,101,0.

34,20

837,8.314.10-7

Ux =

3XBA.Id =

30,176.79,41 = 13,35 (V)


79/122

Rdt =3 .XBA = 0,168 ()

72- St p trn my bin p .

UBA=22

xr UU + =22 67,1275,6 + = 13,45 (V)

73- in p tr ng c khi c gc mmin= 100

U= Ud0.Cosmin - 2.UV - UBA= 433,79.cos100 2.1,6 19,46 =405 (V)

74- Tng tr ngn mch qui i v th cp .

ZBA =22

BABA XR + =22 176,0085,0 + = 0,195 ()

75- Tn hao ngn mch trong my bin p .

Pn = 3.RBA .I2 = 3.0,085.64,842 = 1072,1 (W)

P% =S

Pn .100 =33353

1,1072.100 = 3,12 %

76- Tn hao c ti c k n 15% tn hao ph .

P0 = 1,3.nf . (MT.BT2+Mg.Bg

2) = 1,3.1,15.(48,21.1,0942+93,752.0,882)

Po = 194,8 (W)P % =

S

Po .100 = 0,58 %

77- in p ngn mch tc dng .

Unr=2

2.

U

IRBA .100 =45,185

085,0.84,64.100 = 2,97 %

78- in p ngn mch phn khng .

Unx =2

2.

U

IxBA .100 =45,185

176,0.84,64.100 = 6,15 %

Imax =

+ nxnr

u

u

m eI

.

2 1.2

=

+ 0584,00297,0.

1.1,991.2

e =1686,5 (A)


80/122

< Ipik = 2800 (A)Ipik :nh xung max ca Thyristor .

81- Kim tra my bin p thit k c in khng hn ch tc bin

thin ca dng chuyn mch .

Gi s chuyn t mch T1 sang T3 ta c phng trnh .

2.LBA .dt

dic = U23 U2a = )(..6 2 SinU

dt

dicmax=

BAL

U

.2

.6 2 =310.53,0.2

45,185.6

= 428545,2 (A/s)

dt

dicmax = 0,43 (A/s) < dt

diccp = 100 (A/s)

Vy my bin p thit k s dng tt .

83- Hiu sut thit b chnh lu .

=S

IU dd. =33353

41,79.400= 95 %

8.9.4 Thit k cun khng lc .1.Xc nh gc m cc tiu v cc i .

Chn gc m cc tiu min= 10o .Vi gc mmin l d tr ta c th b

c s gim in p li .

Khi gc m nh nht =min th in p trn ti l ln nht .

Ud max = Udo .Cos min =Ud m v tng ng tc ng c s ln nht nmax=nm.

Khi gc m ln nht = max th in p trn ti l nh nht .

Ud min = Udo.Cos max v tng ng tc ng c s nh nht nmin .Ta c :

Ud min = ( ) ( )[ ]dtBAuudm RRRIDUD

+++ ..1cos..34,2.1

min2


81/122

= ( )

+++

BABAuudm XRRIU .3..120cos..34,2.201 min2

Thay s

Ud min =

( )

+++ 176,0.3

085,0378,0.412,79.120cos.45,185.34,2.20

1min

Ud min= 64,8 (V)

Thay s vo (4.3) ta c

max = arcosdo

d

U

U min = arcos45,185.34,2

648= 80,90

2.Xc nh cc thnh phn sng hi .

thun tin cho vic khai trin chui Furier ta chuyn gc to sang im

1(Hnh 1.1 b) ,khi in p tc thi trn ti khi Thyristor T1 v T4 dn

Ud = Uab =

+

6

..6 2 CosU

Vi = .tin p tc thi trn ti in Ud khng sin v tun hon vi chu k

=P

2=

6

2=

3

Trong P =6 : L s xung p mch trong mt chu k in p li .

Khai trin chui Furier ca in p Ud :

Ud = ( ) kbkaa

nno

k.6sin..6cos.

2 1++

=

= ).6(.2 1

kkSinUnma

k

o ++ =

*) bn=

0

6cos.2

dkUd =

dkU 6cos.)6

cos(.66

02 +


82/122

=

sin6

3sin.2.1)6(

12..6.322 k

U =

cos.1)6(

12..6.322 kkU

Ta c2oa = 2.

6.3U

. cos

Vy ta c bin ca in p :

Uk.n = 22 nn ba +

Uk.n = 2. 2.6.3

U

. 2222

)6(cos1)6(

1Sink

k+

Uk.n = doU.6.3

. 222

)6(1

1)6(

1tgk

k

+

Ud )6(..6.3

1

+ SinUCos kmn

3.Xc nh in cm cun khng lc .

T phn tch trn ta thy rng khi gc m cng tng th bin thnh

phn sng hi bc cao cng ln ,c ngha l p mch ca in p ,dng incng tng ln. S p mch ny lm xu ch chuyn mch ca vnh gp

,ng thi gy ra tn hao ph di dng nhit trong ng c . hn ch s

p mch ny ta phi mc ni tip vi ng c mt cun khng lc ln

Im 0,1.Im .Ngoi tc dng hn ch thnh phn sng hi bc cao ,cun khng lc

cn c tc dng hn ch vng dng in gin on .in khng lc cn c tnh khi gc m =max

Ta c

U~ = U1m.Sin(6 +) nn

I = dtU~1

= )6( 11 +Cos

Um = Im.Cos(6+1)


83/122

L ...2. LfVy Im =

Lf

Um...2.6

1

0,1 I

m

Suy ra

L

dm

m

If

U

.1,0...2.61

= 6 l s xung p mch trong mt chu k in p .Trong

U1m = 2. max22

2max 61

16

cos

tg

Udo +

U1m = 2.o

tg 9,80361136

9,80cos.45,184.34,2 20

+ = 146,95

Thay s :

L =41,79.1,0..50.2.6

78,147

= 0,00986 (H) = 9,8 (mH)

in cm mch phn ng c :

Lc = L+ 2.LBA = 6,1 + 2.0,53 = 7,16 (mH)in cm cun khng lc .

Lk = L Lc = 9,86 7,16 = 2,70 (mH)

4.Thit k kt cu cun khng lc .

Cc thng s ban u :

in cm yu cu ca cun khng lc Lk= 2,7 (nH)

Dng in nh mc chy qua cun khng Im = 79,41 (A)

Bin dng in xoay chiu bc 1 I1m = 10% Im= 7,94 (A)

1- Do dng in cun khng ln v in tr b do ta c th coi tng tr

S = U.2

1dmI = 28,58 .2

94,7= 160,46 (VA)


84/122

4- Tit din cc t chnh ca cun khng lc .Q = kQ .

50.6

46,160.5=

f

s= 3,656 (cm2)

KQ l h s ph thuc phng thc l mt ,khi lm mt bng khng kh t

nhin kQ = 5 .

Chun ho tit din tr theo kch thc c sn :

Chn Q = 4,25 (cm2)

5- Vi tit din tr Q = 4,25 (cm2)

b

L/2

L

hH c a/2

a

Hnh 1.31 Kt cu mch t cun khng

Chn loi thp 330A ,tm thp dy 0,35 mma= 20 (mm)

b= 25 (mm)

6- Chn mt t cm trong tr : BT= 0,8 T

7- Khi c thnh phn dng in xoay chiu chy qua cun cm th trong

cun cm s xut hin mt sc in ng Fk :Fk= 4,44.w.f.BT.Q

Gn ng ta c th vit : Ek = U = 28,58 (V)

Ik=2

22

2

2

94,741,79

2

1

+=

+mI

Id =79,61 (A)


85/122

9- Chn mt dng in qua cun khng .

S1=75,2

61,79=

J

Ik = 28,95 (mm2)

Chn dy dn tit din hnh ch nht ,cch in cp B ,chn Sk =29,90 mm2

Vi kch thc dy :

ak x bk =3,80 x 8,00 (mm x mm)Tnh li mt dng :

j = 66,29,29

61,79==

k

k

S

I(A/mm2)

11-Chn t s lp y :

Kl = cs

k

Q

Sw.

= 0,7

12- Din tch ca s

Qcs=7,0

9,29.63.=

ld

k

k

Sw= 26,91 (cm2)

13- Tnh kch thc mch t .

Qcs = c x h

Chn m = 3=a

h

Suy ra h= 3.a = 3.20 =60 (mm)

c =

0,6

91,26=

h

Qcs = 4,5 cm = 45 mm

14- Chiu cao mch t .

H = h + a = 60 +20 = 80 (mm)

15 Chi di h t

n1 = 97

63

1

==w

w(lp)


86/122

19- Chn khong cch cch in gia dy qun vi tr: a01 = 3mmCch in gia cc lp: cd1 = 0,1mm

20- B dy cun dy:

Bd =(ak + cd1 ).n1 =(3,8+ 0,1).9=35,1(mm)

21- Tng b dy cun dy:

Bd =Bd +a01 =35,1+3=38,1(mm)

22- Chiu di ca vng dy trong cng:l1 = 2(a+b)+2.a01 = 2(20+25)+2..3 = 108,8 (mm)23- Chiu di ca vng dy ngoi cng:

l2 = 2(a+b) + 2.(a01 + Bd ) = 2.(20+25) + 2(3+35,1)=329,4(mm)24- Chiu di trung bnh ca mt vng dy

ltb = (l1 + l2 )/2 =(108,8+329,4)/2 =219,1(mm)25- in tr ca dy qun 75o

R=75 . ltb w/sk =0,02133.219,1.10-3 .63/29,9 = 0,0098()

vi 75 =0,02133 (.mm2/m) in tr sut ca ng 75o c

ta thy in tr rt b nn gi thit ban u b qua in tr l ng

26- Th tch st

vfe = 2.a.b.h+ 2. a/2.b.l = a.b.(2h+1) = 0,125( dm3 )27- Th tch st

Mfe = Vfe . mfe = 0,944 (kg)

trong mfe l khi lng ring ca st mfe =7,85 (kg/dm3 )

28- Khi lng ng:

M cu = V cu . m cu = s k ltb.. w. m cu =3,7 (kg).Trong : mcu =8,9 ( kg/dm3 ).

8.10.5 .Tnh chn cc thit b bo v mch ng lc

1 S mch ng lc c cc thit b bo v

CB

CD

A


87/122

1 2

C

2CC

1CC

1CC

R

R1 2

2CC

1 2

1 2

R

1

2

1 2

1 2

c

R

2CC

2CC1 2

1 2

1

2

1 2

1CC

Ap

R

T1

T4

T6

1 2

1

2

R1 2

C

T2

1 2

a

2CC

3CC

1 2

DC1

2

1 2T3

2CC

T5

3CC

1

2

L

C

1 2

C

C

b

C

Hnh 8.32 .Mch ng lc c cc thit b bo v .

2.Bo v qu nhit cho cc van bn dn:

p = U. Ilv =73,4 (w)+ Din tch b mt to nhit:

S =p/k .


88/122

m mTrong :

p - tn hao cng sut (w)

- chnh lch so vi mi trng.chn nhit mi trng Tmt =40

0 c. Nhit lm vic cho php ca

Tiristo

Tcp =1250 c. Chn nhit trn cnh to nhit Tlv =800 c = Tlv - Tmt = 40

0 c

Km h s to nhit bng i lu v bc x. Chn Km = 8 [ w/m2 . 0 C ]

vy sm = 0,2294 (m2 )

chn loi cnh to nhit c 12 cnh, kch thc mi cnh a x b =10 x 10

(cm x cm).Tng din tch to nhit ca cnh S = 12.2.10.10=2400(cm2 )

3.Bo v qu dng in cho van:

+Aptomat dng ng ct mch ng lc, t ng ng mch khi qu

ti v ngn mch tiristo, ngn mch u ra bin i, ngn mch th cp my

bin p ngn mch ch nghch lu.

+ Chn 1 apomat c:Idm = 1,1 Ild = 69,8 ( A ) =70 ( A )

Udm =220 (v )

c 3 tip im chnh, c th ng ct bng tay hoc bng nam chm in.

Chnh nh dng ngn mch

Inm =2,5 Ild = 137 (A)

Dng qu ti

Iqt =1,5 Ild = 82,5 ( A )

Chn cu giao c dng nh mc

Nhm 2 cc:

dng in nh mc dy chy nhm 2cc

I2cc =1,1. Ihd = 1,1 . 45,847= 50,43 (A)

Nh 3


89/122

Nhm 3 cc:dng in nh mc dy chy nhm 3cc

I3cc =1,1. Id = 1,1 . 79,41= 87,35 (A)

vy chn cu nhy nhm: 1cc loi 80 A

2cc loi 50 A

3cc loi 100 A4.Bo v qu in p cho van:

Bo v qu in p do qu trnh ng ct Tiristo c thc hin bng

cch mc R- C song song vi Tiristo. Khi c s chuyn mch cc in tch tch

t trong cc lp bn dn phng ra ngoi to ra dng in ngc trong khong

thi gian ngn, s bin thin nhanh chng ca dng in ngc gy ra sc in

ng cm ng rt ln trong cc in cm lm cho qu in p gia Anod vcatod ca Tiristo. Khi c mch R- C mc song song vi Tiristo to ra mch

vng phng in tch trong qu trnh chuyn mch nn Tiristo khng b qu

in p

1 2

R2 C2

Hnh 8.34 .Mch R_C bo v qu in p do chuyn mch .

Theo kinh nghim R1 = (5 30 ) ; C1 = (0,25 4 ) F

Chn ti liu [4] : R1 = 5,1 ; C1= 0,25 F

cba

+Bo v xung in p t li in ta mc mch R-C nh hnh 8.35 nh

c mch lc ny m nh xung gn nh nm li hon ton trn in tr ng

dy .

T RC h th ti li [4] R 12 5 C 4 F


90/122

Tr s RC c chn theo ti liu [4] :R2= 12,5 ;C2 = 4 F+ bo v van do ct t bin p non ti ,ngi ta mc mt mch R C

u ra ca mc

giao trinh thiet ke dien tu cong suat

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