MD Mot Chieu

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Bi ging my in

NextNi dungBack

Bi ging my in

Phn 1: My in mt chiu

Phn 2: My bin p

Phn m uBi ging my in

My in

My in tnh My in quay

My bin p My inmt chiu

My inxoay chiu

ng cmt chiu

My phtmt chiu

My inkhng

ng b

My inng b

My phtkhng

ng b

ng ckhng

ng bng cng b

My phtng b

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Bi ging my in

MF

HtiuthMBA MBA

1. Vai tr ca cc loi my in trong nn kinh t quc dn:

2. Khi nim, phn loi v phng php nghin cu my in:a, i cng v my in:- Nguyn l lm vic ca my in da trn c s ca nh lut cmng in t. S bin i nng lng trong my in c thc hinthng qua t trng trong n. to c nhng t trng mnh vtp trung, ngi ta dng vt liu st t lm mch t.

my bin p mch t l mt li thp ng yn. Cn trong ccmy in quay, mch t gm hai li thp ng trc: mt quay, mtng yn v cch nhau bng mt khe h. b, Phng php nghin cu my in:

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Bi ging my in

3. S lc v vt liu ch to my in:Gm c vt liu tc dng, vt liu kt cu v vt liu cch in.

Vt liu tc dng: bao gm vt liu dn in v dn t dng chto dy qun v li st.Vt liu cch in: dng cch in cc b phn dn in vi ccb phn khc ca my v cch in cc l thp ca li st.Vt liu kt cu: ch to cc chi tit my v cc b phn chu lc cgii nh trc, v my, khung my.S lc c tnh ca vt liu dn t, dn in v cch in dngtrong ch to my in.a, Vt liu dn t:b, Vt liu dn in:

c, Vt liu cch in:Cp cch in Y A E B F H C Nhit (0C) 90 105 120 130 155 180 >180

Phn 1: My in mt chiu

Chng 1 : Nguyn l lm vic - kt cu c bn

Chng 2 : Dy qun My in mt chiu

Chng 3 : Cc quan h in t trong my

Chng 4 : T trng trong my in mt chiu

Chng 5 : i chiu

Chng 6 : My pht in mt chiu

Chng 7 : ng c mt chiu

Chng 8 : My in mt chiu c bit

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Bi ging my in

Chng 1: Nguyn l lm vic- kt cu c bn

Bi ging my in

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1.1: Cu to ca my in mt chiu

1.2: Nguyn l lm vic

1-3: cc lng nh mc

1.1: Cu to ca my in mt chiu

1. Phn tnh (Stato):

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a) Cc t chnh:(L b phn sinh ra t thngkch thch)

b) Cc t ph:t gia cc cc t chnh, dng ci thin i chiu.

c) Gng t (v my):

d) Cc b phn khc:Np my: Bo v an ton cho ngi v thit b.C cu chi than: a dng in t phn quay ra mch ngoi.

Phn I: my in mt chiu

Cc t chnh

Dy qun cc t ph

Dy qun cc t chnh

Cc t ph

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my in mt chiu

phn cm ng c in mt chiu

Cc tv

Bu lng

Cun dy

phn cm ng c in mt chiu

Bu lngcc t

cun dy

v

2. Phn ng (Rto):a) Li st phn ng: Dng dn t.

+) Vi cc my cng sut va v ln ngi ta dp lthng gi dc trc. +) Vi cc my in cng sut ln cn x rnh thnggi ngang trc.b) Dy qun phn ng: L phn sinh ra scin ng v c dng in chy qua.+) Dy qun thng lm bng ng c bccch in. trnh khi quay dy qun bvng ra ming rnh thng c nm chtbng tre, g php v u dy qun thngc ai cht.+) Vi cc M cng sut nh dy qun ctit din trn, cn my c cng sut va vln dy qun c tit din hnh ch nht.

Nm

Cchinrnh

Dyqun

Li st

Rnh

L thng gi dc trc

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my in mt chiu

phn ng ng c in mt chiu

C gp

dy qun

li thp

trc

phn ng ng c in mt chiu

C gp

cun dy

li thp

trc

c) Vnh i chiu (Vnh gp):Dng bin i dng xoay

chiu thnh dng mt chiu.Phin gp

d) Cc b phn khc:Cnh qut: Dng lm mt.Trc my: gn li st phn ng, c gp, cnh qut v bi.

Trc lm bng thp cc bon tt.

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my in mt chiu

1.2: Nguyn l lm vic

Phn tnh: Gm 1 h thng t c 2 cc N v S.

1. Nguyn l lm vic ch my pht:

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my in mt chiu

Phn ng: Gm khung dy abcd(1phn t dy qun).

d

b

ac

e

I

eFt

FtI b

d

c

aU

+

-

n

Rt

S

N

A

B

Theo nh lut cm ng in t: tr s sc in ng trong tngthanh dn ab v cd c xc nh: e = B.l.vTrong : B l tr s cm ng t ni dy dn qut qua

l l chiu di thanh dn nm trong t trng.v l vn tc di ca thanh dn.

tt

Khi mch ngoi c ti th ta c: U = E - IRTrong : E l sc in ng ca my pht.

IR l st p trn khung dy abcdU l in p gia 2 u cc

Khi vng dy s chu 1 lc tc dng gi l lc t: Ft = B.I.l

Tng ng ta s c m men in t: Mt = Ft.D/2.= B.I.l.D/2T hnh v ta thy ch my pht Mt ngc vi chiu quay phn ng nn n c gi l M hm.

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my in mt chiu

Sc in ng v dng xoay chiu cm ngtrong thanh dn c chnh lu thnh scin ng v dng 1 chiu nh h thngvnh gp chi than.Ta c th biu din scin ng v dng in trong thanh dn v mch ngoi nh hnh v:

N

S

F, Mtn

Nh vy: ch ng c th U > E cn ch mypht th U < E

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my in mt chiu

ch ng c Mt cng chiu vi chiuquay phn ng gi l mmen quay.Nu in p t vo ng c l U th ta c: U = E + IR

2. Nguyn l lm vic ch ng c:

N

S

F, Mt

n

my in mt chiu

NextChng IBack

1-3: cc lng nh mc

1. Cng sut nh mc: Pm- Ti ca M ng vi tng nhit cho php ca my theo iukin lc thit k c quy nh l cng sut nh mc ca my.- Cng sut nh mc u c tnh u ra ca my.

2. Cc i lng nh mc khc:- Cc tr s in p, dng in, tc quay, h s cng sut... ngvi Pm u l cc tr s nh mc.

Chng 2: Dy qun My in mt chiu

2.1. Nhim v, cu to, phn loi

2.2. Dy qun xp n

2.3. Dy qun xp phc tp

2.4. Dy qun sng n gin

2.5. Dy qun sng phc tp

2.6. Dy cn bng in th

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my in mt chiu

2.1: Nhim v - cu to - phn loi1. Nhim v ca dy qun phn ng:- Sinh ra c 1 sc in ng cn thit, hay c th cho 1 dng innht nh chy qua m khng b nng qu 1 nhit nht nh sinh ra 1 mmen cn thit ng thi m bo i chiu tt, cchin tt, lm vic chc chn, an ton. Tit kim vt liu, kt cu n gin.

2. Cu to ca dy qun phn ng:- Dy qun phn ng gm nhiu phn t nivi nhau theo 1 quy lut nht nh.- Phn t dy qun l 1 bi dy gm 1 hay nhiu vng dy m 2 u ca n ni vo 2 phin gp. - Cc phn t ni vi nhau thng qua 2 phingp v lm thnh cc mch vng kn.

u ni

Cnh tc dng

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my in mt chiu

Chng 2

Nu trong 1 rnh phn ng (rnh thc) ch t 2 cnh tc dng (dy qun 2 lp) thrnh gi l rnh nguyn t. Nu trong 1 rnh thc c 2u cnh tc dng vi u = 1,2,3... th rnh thc chia thnh u rnh nguyn t.

u=1 u=2

u=3

Quan h gia rnh thc Z v rnh nguyn t Znt : Znt = u.ZQuan h gia s phn t ca dy qun S v s phin gp G: S = G.

Znt = S = G

3. Phn loi:- Theo cch thc hin dy qun:

+ Dy qun xp n v xp phc tp.+ Dy qun sng n v sng phc tp.

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my in mt chiu

+ Trong 1 s trng hp cn dng c dy qun hn hp: kt hpc dy qun xp v sng.

Dng xp Dng sng

- Theo kch thc cc phn t: dy qun c phn t ng u v dyqun theo cp.

4. Cc bc dy qun:- Bc dy qun th nht y1 :khong cch cnh tc dng 1 & 2 ca 1 phn t.- Bc dy qun th hai y2 : khong cch cnh tc dng th haica phn t 1 v cnh tc dng 1 ca phn t th 2.- Bc dy tng hp y : khong cch gia 2 cnh tc dng thnht ca hai phn t lin k.- Bc vnh gp yG : khong cch gia 2 thanh gp ca 1 phnt.

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my in mt chiu

Dy qun c phnt ng u

Dy qun c phnt theo cp

2.2: Dy qun xp n1. Bc cc v cc bc dy qun:a) Bc cc : L chiu di phn ng di 1 cc

D l ng knh phn ng l bc ccp l s i cc

b) Cc bc dy qun: Bc dy qun th nht y1: y1 = p2Znt

Trong : l 1 s hoc phn s y1 l 1 s nguyn.

+ Nu y1 = ta c dy qun bc .

+ Nu y1 > ta c dy qun bc di.

+ Nu y1 < ta c dy qun bc ngn.p2Znt

p2Znt

p2Znt

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1 2 3

y1y y2

my in mt chiu

= [cm..]p

D. 2

pi

= [rnh ng. t]p

Znt2

2. S khai trin:

Khai trin dy qun xp n MMC c Znt = S = G = 16, 2p = 4.

- Bc dy qun tng hp (y) v bc vnh gp (yG ): c im ca dy qun xp n l 2 u ca 1 phn t

ni vo 2 phin gp k nhau nn yG = y = 1.

a) Tnh cc bc dy qun:y1 = = = 4 (Bc ) y2 = y1 - y = 4 -1 = 3.

y = yG = 1.p2Znt

416

b)Th t ni cc phn t:Cn c vo cc bc dy qun ta c th b tr cch ni cc phn t thc hin dy qun.

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- Bc dy qun th hai y2: Trong dy qun xp n: y1 = y2 + y y2 = y1 - y.

my in mt chiu

+y1

- Gi s ti thi im kho st phn t 1 nm trn ng trung tnhhnh hc ( l ng thng trn b mt phn ng m dc theo ncm ng t bng 0).- V tr ca cc cc t trn hnh v phi i xng nhau, khong cchgia chng phi u nhau. Chiu rng cc t bng 0,7 bc cc.V trca chi than trn phin i chiu cng phi i xng, khong cchgia cc chi than phi bng nhau. Chiu rng chi than ly bng 1 phin i chiu.- Yu cu chi than phi t v tr dng in trong phn t khi bchi than ngn mch l nh nht v sc in ng ly ra 2 u chithan l ln nht. Nh vy chi than phi t trn trung tnh hnh hcv trc chi than trng vi trc cc t. Khai trin

Lp trn 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1

5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4Lp di

c) Gin khai trin:

my in mt chiu

SNSN

16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

A1 +A +

A2 +B1 -

B -B2 -

n

S khai trin dy qun MMCDy qun xp n c Znt = S = G = 16, 2p = 4.

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my in mt chiu

3.Xc nh s i mch nhnh: A1

B2

A2

B1

(-)Nhn t ngoi vo dy qun phn ng cth biu th bng s sau:

- Ta thy: dy qun phn ng l 1 mch in gm 4 mch nhnhsong song hp li. (Mch nhnh song song l phn dy qun nmgia 2 chi in c cc tnh khc nhau).Nu my c 2p cc th s c 2p mch nhnh song song.

Kt lun:- Trong dy qun xp n gin th s mch nhnh song song bngs cc t hay s i mch nhnh song song bng s i cc : a = p- Nu dy qun xp tho mn 2 iu kin: chi than nm trnng trung tnh hnh hc v h thng mch t i xng th scin ng cc nhnh bng nhau v t gi tr ln nht.

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(+)

my in mt chiu

2-3: dy qun xp phc tp1. Bc dy qun:

c im ca dy qun xp phc tp l yG = m (m = 2, 3, 4...). Thng thng ch dng m = 2. Trong nhng my cng sut tht lnmi dng m > 2.Khi m = 2 = yG:- Nu s rnh nguyn t v s phn t l chn th tac 2 dy qun xp n c lp.

- Nu s rnh nguyn t v s phn t l ta c 2 dyqun xp n nhng khng c lp m ni tip nhauthnh 1 mch kn.

Nh vy c th coi dy qun xp phc tp gm m dy qun xp n lm vic song song nh chi than. V chi than phi c b rng m ln phin gp mi c th ly in ra.

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my in mt chiu

y1y y2

1 2 3 4 5

2. S khai trin:Dy qun xp phc tp c: yG = m = 2; 2p = 4; Znt = S = G = 24.

a) Cc bc dy qun: y2 = y1 - y = 6 - 2 = 4

yG = y = 2

6424

p2Zy1 ===

b) Th t ni cc phn t:

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Khp kn+y1

Khp kn

Lp trn 1 3 5 7 9 11 13 15 17 19 21 23 1

7 9 11 13 15 17 19 21 23Lp di 1 3 5

+y1

Lp trn 2 4 6 8 10 12 14 16 18 20 22 24 2

8 10 12 14 16 18 20 22 24Lp di 2 4 6

my in mt chiu

Gin khai trin dy qun MMCDy qun xp phc tp yG = m = 2; 2p = 4; Znt = S = G = 24.

N SSN

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

n

A1 + B1 -

A + B -

A2 + B2 -

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 2223 24

Cc t v chi in nh dy qun xp n. Ch khc l b rng chiin 2 ln phin gp c th ly in ng thi 2 dy qun ra.

my in mt chiu

Dy qun xp phc tp do m dy qun xp n cng u chungchi than do s i mch nhnh song song ca dy qun: a = m.p.

2-4: dy qun sng n1. Bc dy qun:

y1 = .

Dy qun sng n khc vi dy qun xp n yG.p2

Znt

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Mun cho khi qun xong vng th nht u cui ca phn t th p phi k vi u u ca phn t u tin th s phin i chiu mcc phn t vt qua phi l: p.yG = G 1 yG = (G l sphin gp).

Du (+) ng vi dy qun phi. Du (-) ng vi dy qun tri.

p1G

y1 = . y2 = y - y1 = yG - y1.

y = yG =

p2Znt

p1G

my in mt chiu

y1 y2

y

2. S khai trin:Khai trin dy qun sng n c Znt = S = G = 15; 2p = 4

a) Bc dy qun:y1 = = - = 3 (bc ngn) . y2 = y - y1 = 7 - 3 = 4.

y = yG = = = 7 (dy qun tri)

415

43

2115

p2Znt

p1G

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+y1

Lp trn 1 8 15 7 14 6 13 5 12 4 11 3 10

4 11 3 10 2 9 1 8 15Lp di 7 14 6

2 9

13 5

1

12

+y2

my in mt chiu

SNSN

Gin khai trin dy qun MMCDy qun sng n c Znt = S = G = 15; 2p = 4

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

B2 -A2 +

B -

B1 -

A +

A1 +

1 23 4 5 6 7 8 9 10 11 12 13 14 15

n

Quy lut ni dy ca dy qun sng n l ni tip tt c cc phnt di cc cc c cng cc tnh li ri ni vi cc phn t di cccc c cc tnh khc cho n ht.

my in mt chiu

Dy qun sng n ch c 1 i mch nhnh song song: a = 1.

2-5: dy qun sng phc tp1. Bc dy qun: Tng t nh vi dy qun sng n.

Ring bc vnh gp: yG pmG

=

2. S khai trin:

a) Tnh bc dy qun: y1 = = = 4 (dy qun bc ngn)

yG = 8 = y; y2 = y - y1 = 8 - 4 = 4.

p2Znt

4

2

4

18

2218

pmG

=

=


m = 2; 2p = 4; Znt = S = 18.

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Khp kn

Khp kn

+y1

Lp trn 1 9 17 7 15 5 13 3 11 1

5 13 3 11 1 9 17 7 15Lp di

+y1

Lp trn 2 10 18 8 16 6 14 4 12 2

6 14 4 12 2 10 18 8 16Lp di

+y2

+y2

my in mt chiu

Gin khai trin dy qun MMCDy qun sng phc tp c: m = 2; 2p = 4; Znt = S = 18

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N SSN

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 23 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

n

B2 -A2 +

B -

B1 -

A +

A1 +

my in mt chiu

Dy qun sng phc tp gm m dy qun sng n hp li do s i mch nhnh song song ca dy qun sng phc tp: a = m.

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2.6: Dy cn bng in th

1.iu kin dy qun i xng:

- Dy qun MMC tng ng nh 1 mch in gm 1 s nhnhsong song ghp li. Mi nhnh gm 1 s phn t ni tip nhau.

- Dy qun phi m bo 1 s yu cu sau:+ m bo v cm ng t: H thng mch t phi c cu to

i xng, t thng cc cc nh nhau.+ iu kin v dy qun: Tt c cc dy qun to thnh mch

nhnh phi tng ng nhau v s phn t ca cc nhnh cng phitng ng.

- iu kin bnh thng: sc in ng sinh ra trong cc mch nhnh song song bng nhau, dng in phn b u trong cc nhnh.

my in mt chiu

- Dy cn bng in th lm mt s khng i xng ca mch t trong M cn bng in th cc mch nhnh ca dyqun xp nm di cc cc t c cng cc tnh c gi l dycn bng loi 1. Bc th yt bng s phin i chiu di mii cc:

yt = a

GpG

=

- Dy cn bng lm mt s phn b khng i xng ca inp trn vnh gp gi l dy cn bng loi 2.

Bc th: yt = a

Ga

S=

2. Dy cn bng in th loi 1:

3. Dy cn bng loi 2:

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my in mt chiu

Chng 3:Cc quan h in t trong my

3.1: Sc in ng dy qun phn ng

3.2: M men in t - cng sut in t

3.3: Cn bng nng lng - tn hao- hiu sut

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my in mt chiu

3.1: Sc in ng dy qun phn ngSc in ng trung bnh cm ng trong 1 thanh dn c chiu

di l, chuyn ng vi vn tc v trong t trng bng: etb = Btb.l.v l bc ccD l ng knh phn ng. p l s i cc.n l tc quay phn ng(v/pht)

: t thng khe h di mi cc t (Wb)

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60np2

60nD

... =

pi

l

v =

Btb =

l

60np2 .

60np2 .. etb = .l. =

Nu gi N l tng s thanh dn ca dy qun th mi mch nhnh song song s c thanh dn ni tip nhau. Nh vy sc inng ca my: a2

N

E =

Trong : Ce = l h s ph thuc kt cu my v dy qun.

a2N

na60

pN..

a60pN

.etb = Hay E = Ce..n (V)

my in mt chiu

3.2. M men in t - cng sut in t1. Mmen in t:

Khi M lm vic trong dy qun phn ngs c dng in chy qua. Tc dng ca ttrng ln dy dn c dng in chy qua ssinh ra mmen in t trn trc my.

- Lc in t tc dng ln tng thanh dn: f = Btb.l.i

M = Btb. .l.N. . a2

I 2

D Thay D = v Btb = ta c:pi

p2l.

M = . .l.N. = ..I = CM. .I (Nm) l.

a2I

pi

2p2

pia2pN

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Gi N l tng s thanh dn ca dy qun v dng trong mch nhnhl: i

= I

/2a th mmen in t tc dng ln dy qun phn ng l:

Trong : CM = l h s ph thuc kt cu my.pia2pN

l t thng di mi cc t (Wb).

my in mt chiu

S

nM

B Btb

2. Cng sut in t:

Cng sut ng vi mmen in t (ly vo vi my pht v a ravi ng c) gi l cng sut in t: Pt = M.

= l tc gc phn ng.60

n2pi

pia2pN

60n2pi

a60pN

- Trong ch my pht: M ngc chiu quay vi phn ng nnng vai tr l mmen hm. My chuyn cng sut c (M.) thnh cng sut in (EI).- Trong ch ng c: M c tc dng lm quay phn ng cngchiu vi chiu quay phn ng. My chuyn cng sut in (EI) thnh cng sut c (M.)

NextChng 3Back

Pt = ..I . = .n..I = E . I

my in mt chiu

1. Tn hao trong MMC:

3.3: Cn bng nng lng - tn hao - hiu sut

a) Tn hao c( pc):b) Tn hao st (pFe):

c) Tn hao ng (pcu): Gm 2 phn: - Tn hao ng trn mch phn ng: pcu. = I

2RVi: R = r + rb + rf + rtx ; (r : in tr dy qun phn ng;

rf : in tr cc t ph; rb : in tr dy qun b; rtx: in tr tip xc chi than.)

- Tn hao ng trn mch kch thch:(Bao gm tn hao ng ca dy qun kch thch v ca in triu chnh trong mch kch thch): pcu.kt = Ukt Ikt

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0p

- Hai loi tn hao trn tn ti ngay c khi khng ti nn gi l tnhao khng ti: p0 = pc + pFe. Tn hao ny sinh ra M hm ngay ckhi khng ti nn gi l M khng ti: M0 =

Chng 3

my in mt chiu

d) Tn hao ph: (pf)Trong ng v trong thp u sinh ra tn hao ph. Tn hao phthng kh tnh. Ta ly pf = 1%Pm

2. Gin nng lng v hiu sut:

a) My pht in:

- Gin nng lng:

- Hiu sut:

( )11

cuFeco1

1

2

Pp1

PpppP

PP

=

++==

NextChng 3Back

Pt = P1 - (pc + pFe) = P1 - p0 = E IP2 = Pt - pcu = U.I

my in mt chiu

P1 Pt

pcupFepc

P2

b) ng c in:Ta c cng sut in m ng c nhn t li:

P1 = U.I = U.(I + Ikt )Vi: I = I + Ikt l dng nhn t li vo.

U l in p u cc my.Pt = P1 - (pcu. + pcu.kt)Pt = EI

Cn li l cng sut c a ra u trc: P2 = M. = Pt - (pc +pFe)

- Gin nng lng:

P1 Pt

pcu. + pcu.kt pFe pc

P2- Hiu sut:

( )11

cuFeco1

1

2

Pp1

PpppP

PP

=

++==

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my in mt chiu

4-1: t trng lc khng ti-T Trng cc t chnh-

4-2: t trng khi c ti

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Chng 4 : T trngtrong my in mt chiu

my in mt chiu

4-1: t trng lc khng ti

1.T trng chnh v t trng tn:

T thng chnh l t thng i qua khe h khng kh gia phn ng vcc t trong phm vi 1 bc cc.

0

T thng ca cc t c tnh nh sau: c = 0 + = 0(1+/ 0 ) = 0.t

NextChng 4Back

0Vi t = 1+ l h s tn t ca cc t chnh.

2. Sc t ng cn thit sinh ra t thng:

- Do mch t hon ton i xng v sc t ng cc cc t nhnhau nn ta ch cn tnh cho 1 i cc.

my in mt chiu

- c t thng chnh 0 ta cn cung cp cho dy qun kchthch 1 sc t ng F0 no . n gin cho vic tnh ton ta dngcch phn on mch t thnh 5 on: khe h khng kh (), rngphn ng (hr), lng phn ng (l), cc t (hc), gng t (lG).

NextChng 4Back

B

S

Khi sc t ng cn thit cho 1 i cc s tnh nh sau:F0 = I.W = H.l = 2H. + 2Hr.hr + H.l + 2Hc.hc + HG.lG

= F + Fr + F + Fc + FGTrong : h ch chiu cao, l ch chiu di. Trong mi on cng t trng c tnh: H = vi B = , S, l t thng, tit din, h s t thm ca cc on.

a) Sc t ng trn khe h F: F = 2H.* Khi phn ng nhn:

- Do khe h gia cc t v phn ng khng u: gia th khe hnh, 2 u mp cc t khe h ln: max = (1,5 2,5) nn phn b tcm nhng im thng gc vi b mt phn ng cng khc nhau.

my in mt chiu

hc hc

hrhr

l

lG

B

b

- n gin ta thay ng cong t cm thc tbng 1 hnh ch nht c chiu cao l B v y lb = . sao cho din tch hnh ch nht bngdin tch bao bi ng cong thc t. (b l cungtnh ton ca cc t cn l h s tnh toncung cc). Trong MMC c cc t ph th = 0,62 0,72;

MMC khng c cc t ph th = 0,7 0,8

NextBack

- Phn b t cm di 1 cc t biu din nhhnh v. T cm gia cc t c gi tr ln nhtcn 2 mp cc tr s gim dn v ng trungtnh hnh hc gia 2 cc t th bng 0.

Chng 4

my in mt chiu

Gi l l chiu di phn ng theo dc trcv lc l chiu di cc t th ta c chiu di tnhton l = . Vi l = l1- ng.bg

ng,bg l s rnh v chiu rng rnh thng gi.2

ll c+ Bl

lcl1

T cm khe h khng kh:

Sc t ng c tnh:

=

=

blSB

.

=

==

.

.....

bl2B2H2F

001

* Khi phn ng c rng: - Khi tnh ton ta phi quy i phn ngc rng v phn ng nhn bng cch tngkhe h khng kh l ' = K. vi ' cgi l tr s tnh ton ca khe h.

K l h s khe h: t1 l bc rng;br1 l chiu rng nh rng

++

=10b

10tK

1r

1

Ta c sc t ng phn ng khi c rng :

F = 2.H.' F = 2.H. K. = 2.F1. K = 2. .K' = K.

... bl0

NextChng 4Back

my in mt chiu

br1

t1

b) Sc t ng trn rng phn ng:

Hrtb

Hr1t1

br1

x

br2

t2

Hr2

T thng i qua 1 bc rng t1 l t = B.l.t1Xt 1 tit din ng tm vi mt phn ng,

cch nh rng 1 khong x th t thng i qua tit din gm 2 thnh phn:

xrrxt +=Chia 2 v ca (1) cho Srx (tit din rng) ta c:

(1)

rx

xr

rx

rx

rx

t

SSS

+

=

rx

rx

t BS

=

rx

rx

rx BS

=

l tr s t cm tnh ton v thc t ca rngv

NextChng 4Back

(1)

Vi phn ng c rng v rnh khi t thng i qua khe hkhng kh th phn lm 2 mch song song i vo rng v rnh phnng. Do t dn ca thp ln hn khng kh nhiu nn i b phn tthng i vo rng. t1 l bc nh rng

t2 l bc chn rng

my in mt chiu

*) ngha ca Brx:Coi ton b t thng u i qua rng phn ng (mch t cha

bo ho). Khi Btx> 1,8 Tesla th mch t bt u bo ho, t tr tngdn, ta khng th b qua t cm trn rnh. Khi thnh phn th 2 ca biu thc (1') biu din nh sau:

rxxr0rxxrrx

xr

xr

xr

rx

xr KHKBSS

SS....

=== (2)- Gi thit rng nhng mt ct hnh tr ngang rng v rnh cc cao x u l nhng mt ng tr ca t trng th khi c th xemnh Hrx = Hrx Thay vo (1'): rxxrrxrx KHBB .. '0+= (3)

Gi tr biu thc ny c th tm c t ng cong t ho B = f(H) v qua cc bc tnh c th sau:

+ V ng cong t ho ca li st phn ngKhi bit kch thc ca rng v rnh ta c: 1

klblt

SSK

crx

x

rx

xrrx ==

Vi: Kc l h s p cht li st; Stx, tx l tit din rng v bc rng cao x; l, l l chiu di thc v tnh ton ca li st.

my in mt chiu

B(2)

(1)

0Hrx H

rxrx0 KH ..

Brx

+ T ng cong t ho cho cc tr s ca tcm Brx ta tm c Hrx tng ng Brx theo (3).Sau v ng biu din (ng 2).

- Ngoi ra tr s t cm tnh ton ca rng cng c

th xc nh theo biu thc: crx

1

rx

trx klb

ltBS

B

=

= (4)

* Thc t khi tnh ton sc t ng rng ch cn tnh H 3 im theochiu cao rng: nh rng, chn rng v gia chiu cao rng. Khi trs tnh ton ca cng t trng trung bnh bng:

( )2rrtb1rr HH4H61H ++= (5)

Gi hr l chiu cao rng th sc t ng rng i vi 1 i cc l:Fr = 2. Hr. hr (6)

* n gin ta ch cn xc nh t cm v t trng tit din cchchn rng 1/3 lm tr s tnh ton. Khi ta c: Fr = 2. Hr1/3. hr (7)Trong my in 1 chiu t cm ni hp nht ca rng = 1,8 2,3 T.

my in mt chiu

c) Sc t ng trn lng phn ng:

T thng sau khi i qua khe h khng kh vo phn ng qua rng v rnh s phn b khng u: gn rng c t cm ln hnnhng s khc bit khng ln lm nn c th ly t cm trung bnh lng phn ng tnh ton.

T cm lng phn ng:c

00

klh2S2SB

....

=

=

=

Trong : l t thng phn ng;

l tit din lng phn ng.

20

=

cklhS .. =

Theo ng cong t ho t B H

Khi sc t ng trn lng phn ng c tnh: F = H.l

NextChng 4Back

my in mt chiu

d) Sc t ng trn thn cc t v gng t:

Tnh ton sc t ng cc t ta phi xt n nh hng ca tthng tn. Khi t thng cc t ln hn t thng chnh: c = 0.tvi t = 1,15 1,25 l h s tn t.

Thc t do tn ra khp cc t nn t thng cc phn trncc t v gng t cng khc nhau. Nhng n gin ho tnh ton tacoi nh trn cc t v gng t c t thng khng i. (G = 1/2c)

Ta c: v

Vi Sc v SG l tit din cc t v gng tc

cc S

B

=

G

cG S2

B

=

Sc t ng cc t v gng t: Fc = 2.Hc.hc hc: chiu cao cc tFG = HG.lG lG: chiu di trung bnh ca gng t.

NextChng 4Back

my in mt chiu

3. ng cong t ho ca MMC:

ng biu din quan h gia 0 v F0gi l ng cong t ho ca MMC.

F0

a b cmNu ko di on tuyn tnh ta c quan

h = f(F). Khi t thng t gi tr nh mcth on ab c trng cho sc t ng khe hcn on bc c trng cho sc t ng ri trncc phn st ca mch t.

ab = F F0 = ac = F + Fr + F + Fc + FGbc = Fr + F + Fc + FGt th k l h s bo ho ca mch t.

ab

ack

F

F0==

Trong cc M trit li dng vt liu v cng sut khi in

p l nh mc ta chn im lm vic l im chm bo ho (im c: im m ng cong t ho bt u cong vi k = 1,1 1,35).

NextChng 4Back

my in mt chiu

4-2: t trng ca my in khi c ti

1. T trng cc t chnh: N

S

T hnh v phn b t trng ca cc t chnhca my 2 cc ta thy t trng chnh nhn trccc lm trc i xng v khng thay i v trtrong khng gian.

2. T trng phn ng:

a) Chiu ca t trng phn ng:

* Khi chi than t trn trung tnh hnh hc:

Chng 4Back Next

- Trc sc t ng tng ca c dy qun sinhra lun lun trng vi trc chi than.

Trung tnhhnh hcN

S

N

n

S

my in mt chiu

* Nu dch chi than khi trung tnh hnhhc ng vi 1 on b trn phn ng:

- Phn tch sc t ng phn ng F thnh 2 thnhphn: + Sc t ng ngang trc Fq.

+ Sc t ng dc trc: Fd.

b) S phn b t trng trn b mt phn ng:

* Khi chi than trn ng trung tnh hnh hc: (*)Ta xt 1 mch vng i xng vi im gia ca 2 chi than th

1 im cch gc l x sc t ng c tnh nh sau:Fx = A.2x (A/i cc)

A = (A/cm): l ph ti ng ca phn ng.

i = l dng trong thanh dn.

.

.

DiN

pi

a2I

NextBack

Sc t ng s ln nht khi x = . Khi : F = A.2. = A..2

2

Chng 4

my in mt chiu

N

n

S

Trung tnhvt l

Trung tnhhnh hc

N

CDn

S

A B

b

b

(*)

Fd

Fq

F

NextChng 4Back

- Nu b qua t tr ca li thp tht tr ca mch phn ng ch cn l 2 khe h khng kh nn t cm ca phnng di mt cc t c biu din: Bx = 0.Hx = 0. = 0.

= 0. .x 2F x

2x2A.

A

Di mi bc cc trong phm vi 2b dng in sinh ra sc t ngdc trc cn trong phm vi ( - 2b) dng in sinh ra sc t ngngang trc: Fd = A.2b (A/i cc)

Fq = A. ( - 2b)

FxA/2

Bx

N S

F

* Khi chi than dch khi trung tnhhnh hc 1 khong b no :

Tm li: t trng phn ng ph thuc vo v tr ca chi than vmc ca ti.

my in mt chiu

(*)

NextChng 4Back

3. Phn ng phn ng:a) Khi chi than t trn ng trung tnh hnh hc:

mN S

F (4)

(3)(1)

(2)

m

- S phn b ca t trng do ttrng cc t chnh v t trngphn ng hp li nh sau:

(4): T trng tng khi mch tbo ho.

(1): T cm ca cc t chnh.(2): T trng phn ng.(3): T trng tng khi mch t

cha bo ho.

*)Tm li:

Khi chi than t trn trung tnh hnh hc ch c phn ng ngang trclm mo dng t trng khe h v xut hin ng trung tnh vt l

my in mt chiu

m

m

NextChng 4Back

b) Khi chi than dch khi trung tnh hnh hc: Phn tch sc t ng phn ng thnh 2 phn: Fq v Fd.- Thnh phn ngang trc Fq c tc dng lm mo dng t trngcc t chnh v kh t 1 t nu mch t bo ho.- Thnh phn dc trc Fd nh hng trc tip n t trng cc tchnh. N c tc dng kh t hoc tr t tu theo chiu x dch cachi than. (*)

Do yu cu ca i chiu ch cho php quay chi than theochiu quay ca phn ng trong trng hp my pht cn ng c thngc li.

my in mt chiu

4. T trng cc t ph: Sf N Nf S Sf

NextBack

F

(2)

(3)

(1)

Hb-1

Hb-2

Hb-3

(1): Sc t ng cc t chnh.

(3): Sc t ng cc t ph.(2): Sc t ng phn ng.

Hnh b-3: Phn b t cm.Hnh b-2: Sc t ng tng.

my in mt chiu

Chng 4

Tc dng ca cc t phl sinh ra 1 sc t ng trittiu t trng phn ngngang trc v to ra t trngngc chiu vi t trngphn ng khu vc ichiu.

NextChng 4Back

Hnh c3: Cc t ph ctc dng tr t.

Hc3

Hc2

Hc1

SN NfSfS

F

Hnh c1: Cc t phkhng nh hng ti ttrng tng.Hnh c2: Cc t ph ctc dng kh t.

my in mt chiu- Khi chi than t trn trung tnh hnh hc th cc t ph

khng nh hng n cc t chnh.

Cch ni cc t ph:

F

NfSf

N

S

5. T trng dy qun b:

(1)

(2)

F

NextChng 4Back

- Tc dng ca dy qun b lsinh ra t trng trit tiu phnng phn ng lm cho t trngkhe h v cn bn khng b modng na.

ng (1): Sc t ng caphn ng phn ng ngang.

ng (2): Sc t ng cady qun b.

ng (4): Phn b t trng tngkhi c c dy qun b v cc t ph.

Cch ni dy qun b:

ng (3): Sc t ng khi khng ti.

N NfS Sf S

(3)

my in mt chiu

(4)

Chng 5 : i chiu

5.1: Nguyn nhn sinh ra tia latrn vnh gp

5.2: Qu trnh i chiu

5.3: Cc phng php ci thini chiu

NextPhn IBack

my in mt chiu

5.1: Nguyn nhn gy tia la trn vnh gp

NextChng 5Back

1. Nguyn nhn v c kh:- Vnh gp khng ng tm vi trc.- S cn bng b phn quay khng tt.- B mt vnh gp khng phng do nhng phin i chiu hoc mi ca cch in gia cc phin i chiu nh ln.- Lc p chi than khng thch hp (mnh qu c th lm mn chiv vnh gp), kt chi trong hp chi, hp chi than khng c gicht hay t khng ng v tr.

2. Nguyn nhn v in:- Do sc in ng phn khng khng trit tiu ht sc in ngi chiu.- Do s phn b khng u mt dng in trn mt tip xc vquan h phi tuyn ca in tr tip xc: rtx = f(t,) vi l thng sc trng cho tc dng nhit v hin tng in phn di chi than.

my in mt chiu

3. Cc cp tia la in:

Bng cc cp tia la in (*)

- Ta thy tia la mnh gy hao mn nhanh chng chi than v vnhgp. Do tia la cp 2 ch cho php vi nhng ti xung ngn hn, tia la cp 3 ni chung l khng cho php.

- Ch lm vic lu di vi cp tia la 1.

NextChng 5Back

my in mt chiu

1.Mt s khi nim:

5.2: Qu trnh i chiu

a) Qu trnh i chiu:

NextChng 5Back

b) Chu k i chiu:

2i 2i 2it = 0 0 < t < Tc t = Tc

Qu trnh i chiu ca dng in khi phnt di ng trong vng trung tnh hnh hc v bchi than ni ngn mch gi l s i chiu.

i i i i i i i ii i i i

S NA2

B1

A1 S

B2

my in mt chiu

N

- Qu trnh i chiu ca dng in trong mi phn t tn ti trong 1 khong thi gian rt ngn.

NextChng 5Back

- Khong thi gian dng in hon thnh vic i chiu gi l chuk i chiu (Tc). l thi gian cn thit vnh gp quay i 1 gcng vi chiu rng ca chi, ngha l Tc = (1)

G

c

v

b

bc l chiu rng ca chi gp.vG l vn tc di ca vnh gp.

Gi DG l ng knh vnh gp Ta c: bG = pi. bG l bc vnh gpG : s phin gp vG = DG. pi. n = bG.G.n

GDG

Gi G l h s trng khp th: G = Thay vo (1):

Tc = = G. (2)G

c

bb

nGbb

G

c

.. nG1.

(y l chu k i chiuca dy qun xp n)

my in mt chiu

2. Cc sc in ng trong mch vng i chiu:- Sc in ng t cm eL:

dc

ccL

T

i2L

dt

diLe ==

- Sc in ng h cm eM:

===n

1n

dc

uMn

nn

1n

n

1MnM M.

T

i2e

dt

di.Mee

Mn l h s h cm gia phn t ang xt v phn t th n.in l dng in trong phn t th n

NextChng 5Back

my in mt chiu

- Sc in ng i chiu ec: sinh ra khi phn t i chiu chuynng trong t trng tng hp ti vng trung tnh:

ec = 2.W.Bc.lc.v.W: s vng ca phn t i chiu.Bc: T cm i chiu

- Sc in ng phn khng epk: epk = eL + eM m bo i chiu c tt th sc in ng phn khng

phi lun lun ngc chiu vi sc in ng i chiu.

3. Phng trnh i chiu:

Theo nh lut Kishop 2 vit cho mch vng phn t b:i.rpt + i1.(rtx1 + rd) - i2.(rtx2 + rd) = e (1)Trong :rpt : in tr ca phn t i chiu.rd : in tr dy ni.rtx1,2: in tr tip xc ca chi than vi phin gp 1 v 2.e : Tng cc sc in ng sinh ra trong phn t i chiu:e = eM + eL + ec = epk + ecTheo nh lut Kishop 1 vit cho cc nt 1 v 2:Nt 1: i + i - i1 = 0 (2)Nt 2: - i + i - i2 = 0 (3)

i: dng ngn mch chy trong phn t i chiu.i1, i2: dng chy qua dy ni vi cc phin i chiu 1 v 2.

NextChng 5Back

i i i i

1 2i1 i2

a b c

1 2

my in mt chiu

Cc dng i, i1, i2 c th tnh t cc phng trnh (1), (2) v (3) nu cc i lng khc bit.

i.rpt + i1.(rtx1 + rd) - i2.(rtx2 + rd) = ei + i - i1 = 0

- i + i - i2 = 0

- mc gn ng gi thit rpt 0, rd 0. Ta c: i1.rtx1 - i2.rtx2 = e i + i - i1 = 0- i + i - i2 = 0 1tx2tx

1tx2tx

1tx2tx

rr

ei.

rr

rri

+

+

+

=

V gi thit rpt 0, rd 0 nn (rtx1+ rtx2) l tng tr ca phn ti chiu khi b chi than ngn mch v dng if chnh l dng ngn mch trong phn t gy bi e

NextChng 5Back

(4)

my in mt chiu

if

Gi thit rng rtx1, rtx2 t l nghch vi b mt tip xc ca chiin vi phin i chiu 1 v 2. Trong qu trnh i chiu t 0 nTc vi iu kin bc = bG th cc b mt tip xc c tnh nh sau:

ST

tS

dctx =2 S

T

tTS

dc

dc1tx

=

Vi S l din tch tip xc ton phn cachi vi phin gp.

txdc

dctx

1tx1tx r

tT

Tr

S

Sr

== txdc

tx2tx

2tx rt

Tr

S

Sr ==

Thay (5), (6) vo (4) ta c:

(5)

nm

dc r

ei.

T

t21i

+

=

)1T(t

T.rr

dc

2dc

txnm

=

(6)

(7)

Vi

NextChng 5Back

K hiu rtx l in tr tip xc ton phnng vi mt tip xc ton phn S ta c:

(*)

(7): L phng trnh i chiutng qut.

my in mt chiu

SStx2

Stx1

4. Cc loi i chiu:a) i chiu ng thng:

e = 0 v dng in i chiu l

dc

i.T

t21i

=

Mt dng in 2 pha i vo v i ra bngnhau: J1 = J2 nn khng c tia la xut hin. (*)

b) i chiu ng cong:

if, rnm

0

(1)

(2)

(3)

tm e 0 nn if 0.

Dng in ph ny s cng vi dng c bnlm cho dng in i chiu tr nn phi tuynv ta c i chiu ng cong.

nmf

r

ei

=

(1): rnm(t)(2): if(t) khi e > 0(3): if(t) khi e < 0

Ta biu din if di dng sau: (e = const):

NextChng 5Back

my in mt chiu

Tc

1

t

i1t

-i

0

i22+i

i

* Gi s epk > ec hay e > 0 v githit rtx = const th dng i chiuc dng:

* Gi s epk < ec hay e < 0 th if idu theo ng (3). Khi dng ichiu c dng:

Ta thy 1 < 2 nn J1 < J2 => i chiuvt trc v c tia la u vo cachi.

NextChng 5Back

Ta thy 1 > 2 nn J1 > J2 => i chiumang tnh tr hon. Tia la xut hin u ra ca chi in khi phn t ri khiv tr ngn mch.

my in mt chiu

+i2

i2

0

-it

i1

Tc

1

t

i

-i

Tc

1t i1

i2

2+i

0 t

i

1.Phng php dng cc t ph:

5.3. Cc phng php ci thin i chiu

Bin php c bn ci thin i chiu trong MMC l to ra ttrng i chiu ti vng trung tnh hnh hc bng cch t nhngcc t ph gia nhng cc t chnh. Mun vy sc t ng ca cct ph Ff phi ngc chiu vi sc t ng phn ng ngang trc. V tr s ngoi vic trung ho phn ng phn ng ngang trc cn phito ra 1 t trng ph sinh ra ec lm trit tiu epk.

Mt khc : Fq v epk t l vi i do sc t ng cc t ph v ttrng i chiu cng phi bin i t l vi ph ti. Mun vy dyqun cc t ph phi ni tip vi dy qun phn ng ng thi mch t cc t ph khng bo ho. Thng thng khe h gia cc t phvi phn ng ln hn so vi cc t chnh (t 1,5 2 ln).

NextChng 5Back

my in mt chiu

2. Phng php x dch chi than khi trung tnh hnh hc:

my pht: Chi than ang c t trn trungtnh hnh hc. Ta xt 1 phn t dy qun:

-i ec

epk+ieq

e

NextChng 5Back

my in mt chiu

N

n

S

*) Nhn xt: Khi M lm vic ch my pht ci thin ichiu ta phi dch chi in khi trung tnh hnh hc 1 gc: = + theo chiu quay phn ng. Cn ch ng c th dch chi in i 1 gc ngc chiu quay phn ng.(: l gc gia trung tnh hnh hc v trung tnh vt l.

: l gc ng vi iu kin t trng tng bng t trng i chiu)

4. Nhng bin php khc:Chn chi than ph hp, gim sc in ng phn khng ...

NextChng 5Back

my in mt chiu

3. Dng dy qun b: Trong cc M cng sut ln v ti thay i t ngt ngi ta

thng dng dy qun b h tr cho cc cc t ph. Dy qunb c t di mt cc t chnh v s trit tiu t trng phnng di phm vi mt cc t chnh v lm cho t trng chnh hunh khng thay i. Dy qun b ni tip vi dy qun phn ngnn c th b bt c ti no.

Chng 6 : My pht mt chiu

6.1: Nhng khi nim c bn

6.2: c tnh c bn ca my pht mt chiu

6.3: My pht mt chiu kch t song song

NextPhn IBack

6.4: My pht mt chiu kch t ni tip

6.5: My pht mt chiu kch t hn hp

6.6: My pht mt chiu lm vic song song

my in mt chiu

NextChng 6Back


1. Phn loi: Tu theo phng php kch thch cc t chnh MFMC c phn thnh 2 loi:

a) My pht in mt chiu kch t c lp:b) My pht 1 chiu t kch:

UI

Ikt

I

Hnh a

U

I I

I

Hnh b

Trong mi trng hp cng sut kch thch chim 0,3 0,5% Pm.

+ My pht mt chiu kch thch song song: I = I + Ikt (hnh a).+ My pht mt chiu kch thch ni tip: I = Ikt = I (hnh b).+ My pht mt chiu kch thch hn hp: I = I + Iktss (hnh c).

I

IU

Ikt

Hnh c

U

I

Iktss

Iktnt

I

my in mt chiu

NextChng 6Back

2. Phng trnh cn bng m men:

Ta c: P1 = pc + pFe + PtChia 2 v cho : hay: Mq = M0 + Mt

Trong : Mq l mmen ca my pht in.M0 l mmen cn khng ti.Mt l mmen in t.

+

+=

dtFeco1 PPPP

Nu t M0 + Mt = MCT (mmen cn tnh) th phng trnh cn bngmmen s l: Mq = MCT

3. Phng trnh cn bng in p:

U.I = E.I - (I2rdq + UtxI)

+=

txdq

I

UrIEU

t l in tr mch phn ng

txdq

I

Urr

+= U = E - Ir

my in mt chiu

P1

pc pFe

Pt

pcu

P2

P2 = Pt - (pcu + pf)

4. Cc c tnh ca my pht 1 chiu:

C 5 dng c tnh:+ c tnh khng ti: U0 = E = f(Ikt) khi I = 0, n = const.+ c tnh ngn mch: In = f(Ikt) khi U = 0, n = const.+ c tnh ngoi: U = f(I) khi Ikt = const, n = const.+ c tnh ph ti: U = f(Ikt) khi I = const, n = const.+ c tnh iu chnh: Ikt = f(I) khi U = const, n = const.

NextChng 6Back

my in mt chiu

1. c tnh khng ti: U = f(Ikt) khi I = 0, n = const.

6.2: Nhng c tnh c bn camy pht mt chiu

c tnh c xc nh bng thc nghim theo s th nghim

Khi I = 0 U = E = Ce..n = Ce.. c tnh lp li dng ng cong t ho ring ca my in.

Ikt

V

A

I

A'

-IktmIktm

AU

B

0Ikt

B

my in mt chiu

NextChng 6Back

2. c tnh ngn mch: In = f(Ikt) Khi U = 0, n = const.

0

Ikt

(1)

(2)In

(1): My c kh t d.(2): My cha c kh t d.

- Do U = 0 ta c E = IR ngha lton b sc in ng sinh ra bp cho st p trn mch phn ng.

- Mt khc: dng ngn mch c hn ch bng (1,25 1,5)Im vR rt nh v vy E nh Ikt tng ng nh mch t khngbo ho. Do E t l tuyn tnh vi Ikt nn I cng t l vi Ikt ctnh c dng ng thng.

NextChng 6Back

my in mt chiu

Tam gic c tnh:

Dng tam gic c tnh

(1): c tnh khng ti(2): c tnh ngn mch.

ln ca AB ph thuc vo loi my, ln nht MMC khng c cc t phv dy qun b. my c cc t ph v dy qun b phn ng phn ng hu nhb trit tiu, cnh AB 0. MMC kch t hn hp, dy qun ni tip c tc dngtr t v nu sc t ng ca n ln hn AB, ngha l ngoi phn sc t ng trit

tiu nh hng ca phn ng phn ng cn sc t ng tr t th cnh AB snm v bn phi ca BC.

Gi s khi ngn mch trong phn ng c dng Im tng ng vi dng kch thchIt = OC: 1 phn OD sinh ra sc in ng khc phc in p ri trn in trphn ng Im.R = AD = BC; Phn cn li DC = AB dng khc phc phn ngphn ng lc ngn mch.

0 C D It

Enm

Inm=Im B A

U (1)

I (2)E,I

A B

U (1)I (2)

Inm=Im

Enm

0 D C It

E,I

my in mt chiu

NextChng 6Back

ABC c cnh BC t l vi dng in phn ng v cnh AB trong iu kin mch t khng bo ho t l vi phn ng phn ng (t l vi I) gi l tam gic c tnh.

3. c tnh ph ti: U = f(Ikt) khi I = const, n = const.

NextChng 6Back

Dng c tnh:(1): c tnh khng ti.(2): c tnh ph ti.

ng (2) c th xc nh khi bit ng (1) v tam gic c tnh: Gi s bit tam gic c tnh 1 ch ti no . VD ti nhmc l tam gic ABC. Ta t tam gic sao cho nh A nm trn ctnh khng ti, cc cnh AB v BC song song vi trc honh v trctung ng thi t l vi ph ti, khi tam gic dch chuyn song songvi chnh n nh C s v nn c tnh ph ti.

0

A1 B1

C1

(2)C

BA(1)U

IktIkt

Rt

I

A

V

my in mt chiu

4. c tnh ngoi: U = f(I) Khi Ikt = const, n = const.

Um = U0 - Um vi iu kin Ikt =Iktmgi l bin i in p nh mc:

( )%155%100U

UU%U

dm

dm0=

=

0 ImI

U

E

Um

UU0

Um

* C th dng c tnh ngoi t ctnh khng ti v tam gic c tnh:Cho OP = Ikt = constPP' = UI = 0 = E im Dt tam gic ABC c AB v BC theot l ng vi I = Im sao cho A nmtrn c tnh khng ti cn BC nmtrn ng thng ng PP' PC lin p khi I = Im ta c im D' v gc phn t th 2.

U

D'

I Im 0 P Ikt

P'A B

C

D

Im/2

my in mt chiu

Ikt

Rt

I

A

V

Ikt

0I

5. c tnh iu chnh:Ikt = f(I) Khi U = const, n = const

c tnh iu chnh cho ta bit cn phi iu chnh dng kch thch nh th no gi cho in p u ra ca my pht khng thay i khi ti thay i.

NextChng 6Back

my in mt chiu

1. iu kin t kch ca my:

6.3: My pht mt chiu kch t song song

m bo my t kch c cn c cc iu kin sau:- Trong my phi tn ti 1 lng t d d = (2 3)% m- Cun dy kch thch phi u ng chiu hoc my quay ng chiu sinh ra dng ikt > 0- Nu tc quay bng hng s th in tr mch kch thch phi nhhn 1 in tr ti hn no . Hoc nu in tr mch kch thch bng hng s th tc quay phi ln hn 1 tc ti hn no .

UI

Ikt

I

Nu my pht thomn 3 iu kin trn th qu trnh t kch xy ra nh sau:

NextChng 6Back

rkt2 rkt1

U = Iktrkt

Ikt0

U rthA

my in mt chiu

h mch kch thch v quay my pht n nm. Do trong my tn ti d nn trong dy qun s cm ng 1 sc in ng E v trn 2 cc my s c 1 in p U = (2 3)% Um.

Ni kn mch kch thch trong mch kch thch s c: no . Dng ny sinh ra t thng d v tng (d + d ) > d ssinh ra dng kch thch ln hn. C nh vy my s tng kch t in p u cc tng ln v my tip tc t kch cho n khi n lm vic n nh im A.Nu d ngc chiu vi d th my s khng t kch c.

2. c tnh ngoi:U = f(I) khi Ikt = const, n = const

U

Um Um(1)

(2)

I

IthImI00

(1): c tnh ngoi ca MF kch t c lp.(2): c tnh ngoi ca MF kch t song song.

NextChng 6Back

kt

'kt

r

UI =

my in mt chiu

UB

0 Iktm Ikt

3. c tnh khng ti:

4. c tnh iu chnh:Ikt = f(I) khi U = const, n = const(1): ca my pht kch thch song song(2): ca my pht kch thch c lp

NextChng 6Back

Ikt

0I

(1)(2)

my in mt chiu

MFMC kch thch ni tip cng thuc loi t kch.

6.4: My pht mt chiu kch t ni tip

I Ikt

I

Rt

Do dng ti I = I = Ikt nn trong my ch c 2 lng ph thuc nhau l U v I nn ta ch xy dng c tnh ngoi: U = f(I) khi n = const.

- V khi ti tng in p 2 u cc my pht thay i nhiu nn thc t t dng loi my ny.

NextChng 6Back

- Khi I = Ith mch t bo ho in p bt u gim

0

U

Ud

Ith I

my in mt chiu

My pht mt chiu kch thch hn hp c 2 cun dy kch thch. Tu theo cch u dy m ta c s ni thun v ni ngc.

6.5. My pht mt chiu kch t hn hpU

I

Iktss

Iktnt

I1. c tnh ngoi: U = f(I) khi n = const

0 I

U0

U(1)

(2)

(3)(4)

Khi ni thun in p c gihu nh khng i (ng (2))

Khi b tha (ng 1) in p s tng khi ti tng. iu ny c ngha quan trng trong vic truyn ti in nng i xa.

Nu ni ngc 2 dy qun kch thch (ng (4)) khi ti tng p sgim nhanh hn so vi my pht kch thch song song (ng (3))

NextBack Chng 6

my in mt chiu

U

I

Iktss

Iktnt

I

2. c tnh iu chnh:

0 I

(2)

(1)

(3)Iktng (1) : Khi ni thun 2 dy qun kch thch v b bnh thng.

(2) : Khi b tha.

(3) : Khi ni ngc 2 dy qun kch thch.

NextChng 6Back

my in mt chiu

1. iu kin ghp cc my pht lm vic song song:

6.6: My pht mt chiu lm vic song song

Cng cc tnh: Sc in ng ca my pht II phi bng in p U ca thanh gp.Nu ghp cc my pht kch thch hn hp lm vic song song thcn c iu kin th 3: Ni dy cn bng gia cc im m v n nhhnh b.

A A

V

F2F1

Hnh a

A A

mI n

F1 F2

Hnh b

NextBack

my in mt chiu

Chng 6

2. Phn phi v chuyn ti gia cc my pht in:

IIIII

(2)

(2')

U

(1')

(1)EII=U

I = III = II+III

Ghp my pht II lm vic song song vi my pht I.

Vic thay i EI v EII bng cch bin i dng kch t IktI v IktIIhoc bng cch thay i tc quay ca cc ng c s cp.

NextChng 6Back

my in mt chiu

Do E2=U nn my II cha tham gia pht in v ton b ti vn do my I m nhn. Lc ny c tnh ngoi ca 2 my l ng (1) v (2).

Chng 7 : ng c mt chiu


7.2: M my ng c in mt chiu

7.3: ng c in mt chiu kch thch song song hoc c lp

NextPhn IBack

7.4: ng c mt chiu kch thch ni tip

7.5: ng c mt chiu kch thch hn hp

my in mt chiu

1. Phn loi:


UI

Ikt

I

Hnh b

U

I I

I

Hnh c

U

I

Iktss

Iktnt

Hnh d

II

IU

IktUktHnh a

+ ng c mt chiu kch thch hn hp: I = I + Ikt (hnh d).

+ ng c mt chiu kch thch c lp: I = I (hnh a).

+ ng c mt chiu kch thch song song: I = I + Ikt (hnh b).

+ ng c mt chiu kch thch ni tip: I = I = Ikt (hnh c).

NextChng 7Back

my in mt chiu

2. Phng trnh cn bng p:Cng sut in a vo u ng c kch thch song song l:

P1 = U.(I + Ikt)Pt = P1- (pcu.kt + pcu.)

EI = U.(I + Ikt) - (U.Ikt + I2.R) E = U - I.R

3. Phng trnh cn bng mmen:

Pt = pFe + pc + P2

+

+=

2Fecodt PPPP

Mt = M0 + M2

t: M0 + M2 = MCT (Mmen cn tnh) Mt = MCT

Trong : M0: mmen cn khng ti.M2: mmen ph ti.

NextChng 7Back

my in mt chiu

P1 Pt

pcu. + pcu.kt pFe pc

P2

1. Yu cu khi m my:

7.2. M my ng c mt chiu

Dng m my phi c hn ch n mc nh nht trnh cho dy qun khi b chy hoc nh hng xu n i chiu.

M men m my phi c tr s cao nht c th c hon thnh qu trnh m ma ngha l t c tc quy nh trong 1 thi gian ngn nht.

2. Cc phng php m my:

- Khi m my trong mi trng hp u phi m bo c maxngha l trc khi ng ng c vo ngun in, bin tr iu chnh dng kch thch phi t v tr sao cho in tr kch thch nh nht mmen t gi tr ln nht ng vi mi gi tr ca dng phn ng.

NextChng 7Back

my in mt chiu

a. M my trc tip:

Ti thi im u: n = 0 E = 0 U = I.R I = Imm = I = (5 10)Im.

V dng m my ln nn phng php ny t c s dng. Ch yu dng cho ng c cng sut vi trm ot (v R tng i ln nn Imm (4 6)Im).

RU

b. M my nh bin tr:

CU

Ikt

0 T

12 3 4

5M

rmm

- Khi m my nh bin tr dng c tnh:

mmi

i

RR

EUI

+

=

Rmmi l in tr m my th i.

NextChng 7Back

my in mt chiu

- Bin tr m my c tnh sao cho: Imm = (1,4 1,7)Im i vi cc ng c cng sut ln Imm = (2 2,5)Im vi ng c cng sut nh.

Qu trnh m my c biu din nh hnh v: n

1 2 3 4 5

0 MMc

M

I

nI1

I2M1

M2

NextChng 7Back

Thng dng m my cho nhng ng c cng sut ln kt hp c vic iu chnh tc bng cch thay i in p.

c) M my bng in p thp (Umm < Um):

Phi dng 1 ngun c lp c th iu chnh in p c cung cp cho phn ng ng c. Mt ngun khc U = Um cung cp cho mch kch thch.

my in mt chiu

1. c tnh c:

7.3. ng c in mt chiu kch thch song song hoc c lp

T phng trnh: E = Ce..n

Thay M = CM..I ta c: (1)

=

=

e

e C

R.IU

C

En

2Me

e .C.C

R.M

C

Un

=

Vi iu kin: U = const, Ikt = const khi M (hoc I) thay i th t thng cng hu nhkhng i.

ng c ny c dng trong trng hp tc hu nh khng i khi ti thay i (my ct kim loi...)

0

nn0

Mm M(I)

NextChng 7Back

my in mt chiu

k

R.Mnn 0 =

Vi v k = Ce.CM.

=

e0

C

Un

2

(1)

2. iu kin n nh ca ng c:Xt c tnh M = f(n) ca ng c in v Mc = f(n) ca ti nh hnh v:

Ta c:

vi:

l qun tnh phn quay.

dt

djMM c

+=

g4

GDj

2

=

Trng hp a: P l im lm vic ca h thng c M = Mchay 0

dn

dM=

Nu v l do no tc tng: n = nlv + n th Mc > M ng c b ghm, tc gim dn v im P n = nlv. Ngc li: nu tc gim Mc < M ng c c gia tc v t tc lm vic.

Nh vy: iu kin lm vic n nh ca h thng l: dn

dMdndM c Mc lm tc tip tc tng mi hoc s gim tc s dn n hu qu l tc gim mi.

dn

dM

dn

dM c

dndM

dndM c>

NextChng 7Back

my in mt chiu

M

0

M McP

Hnh a

nlv

nn

Hnh b

0

MM

McP

n

nnlv

3. iu chnh tc : 2

Me

e .C.C

R.M

C

Un

=

a. Phng php thay i t thng:

1

n

0

n0m

n01

n02

2m

M

Bng cch thay i tr s ca bin tr trong mch kch thch. Cc ng ny c n0 > n0mv giao nhau ti 1 im trn trc honh ng vi (n = 0, I = )

R

U

b. Thay i in tr ph trn mch phn ng:

0

nn0 Rf = 0

Rf1Rf2

Rf3 M

( )k

RR.Mnn f0

+=

NextChng 7Back

my in mt chiu

c) Phng php thay i in p:

0 M

UUm n02

n01

n03n

Vic cung cp in p cho ng c c thc hin bng 1 ngun c lp bng cch ghp thnh t hp my pht - ng c.

4. c tnh lm vic:

a) c tnh tc : n = f(I) ging c tnh c:

=

.C

R.I

C

Un

e

e

b) c tnh mmen: M = f(I) khi U = Um = const M = CM..I

Do Ikt = const khi U = const = const M = f(I) l ng thng.

M

0Ic) c tnh hiu sut: = f(I)

Khi U = Um = const.max c tnh vi dng in ti I = 0,75Im. Khi tn hao khng i trong ng c(pc + pFe) bng tn hao bin i trong mch phn ng (ph thuc rdq v t l I

2 ) 0

0,75Im

I

max

NextChng 7Back

my in mt chiu

1. c tnh c ca ng c 1 chiu kch thch ni tip:

7.4: ng c mt chiu kch thch ni tip

V I = I = Ikt trong phm vi rng c th biu th = K.I (1)Trong : K = const khi I < 0,8Im v gim i 1 cht khi I > 0,8Im

do nh hng bo ho ca mch t.

M = CM..I. Thay phng trnh (1) vo ta c:

M

22

MC

K.M

KCM

==

MC

K.M =

2Me

e .C.C

R.M

C

Un

=

=

K.C

R

K.M.C

CU

e

e

M (2)

Nu b qua R th n t l vi hay . MU

2

2

n

CM =

Khi mch t cha bo ho c tnh c c dng hypecbol bc 2.

NextChng 7Back

my in mt chiu

0 M

n

ng c mt chiu kch thch ni tip vi c tnh c rt mm c ng dng trong nhng ni cn iu kin m my nng n v cn thay i tc trong 1 vng rng (cu trc, xe in...)

Khi n gim th M tng v ngc li. Trong trng hp mt ti (I = 0, M = 0) th n c tr s rt ln v th loi ng c ny khng cho php lm vic trong iu kin c th mt ti (ai truyn...).

2. iu chnh tc :a) iu chnh tc bng cch thay i t thng:

Vic thay i t thng trong ng c kch t ni tip c th thc hin theo 3 phng php:

NextChng 7Back

my in mt chiu

Mc in tr sun vo dy qun kch thch. (Hnh a)

U

RktRS

C

I

(a)

U

C

Wkt

W'kt

I

(b)

U

RS

Rkt

I

(c)

Mc in tr sun vo mch phn ng. (Hnh c)Thay i s vng dy ca dy qun kch thch. (Hnh b)

NextChng 7Back

Hai bin php u cho cng 1 kt qu: nu dng kch thch ban u l Ikt thdng sau khi iu chnh l I'kt = k.IktVi k l h s gim.

+ Trng hp a: + Trng hp b: 1RR

RkSkt

S Fd Kb > 1: b tha - ng (2 )

- Khi FCB < Fd Kb < 1: b thiu - ng (3).

ng dng:

NextChng 8Back

my in mt chiu

8.2: My pht hn mt chiuMy pht hn phi c c tnh ngoi U = f(I) c dc cao nhhnh v. My pht hn c sn xut vi:

U = 35V (U0 = 80) v I = 500A

0

U

Rt1

Rt2Rt3 I

Thc t ch to c loi my pht c bit c s nh sau:

Khi I tng t thng ca cc cc ln d gim nhiu cn t thng ca cc cc b n khng thay i (do li thp bo ho) t thng tng (d + n) gim rt nhanh khin cho UAB h thp rt nhiu nn c tnh ngoi rt dc. Ch : Khi I tng UBC cung cp cho cc dy qun kch thch vn gi khng i v n khng i.

NextChng 8Back

+ -Rt

INnNd

C

B

SdSn

A

I"I'

F1

F2

my in mt chiu

My pht 1 cc l loi my c bit khng vnh gp cho php t c dng in ln (n 50000A) in p thp (1 50V).

8.3: My pht mt cc

Cu to nh hnh v:Hai cc t hnh tr lng vo nhau. Thanh dn t trn hnh tr trong (Rto) (hay c thdng chnh bn thn rto thay cho thanh dn) hai u ni cht vi 2 vnh C 1 v C2.

Khi rto quay trong cc thanh dn s sinh ra sc in ng vdng in ly ra t cc chi t ln 2 vnh C1 v C2. V dng in rt ln, trnh tn hao ngi ta dng chi than bng kim loi lng (thu ngn Natri) dn dng ra ngoi.My pht 1 cc c dng cho in phn, cp in cho cc nam chm in ca thit b tng tc NextChng 8

C1C2

2

S 1

N

my in mt chiu

L my pht in dng bin i chuyn ng quay thnh tn hiu in (in p). Yu cu i vi loi my ny l phi c quan h U = f(n) l ng thng v chnh xc 0,2 0,5 %.

8.4: My pht o tc

- Khi khng ti ta c: U = E = Ce..n = Ke.n (v = const).

t

R

U

t

R

U- Khi c ti : I = U = E - .R

U = E - I.R E = U.(1 + )t

R

U

t

e

t

R

R 1

n.K

R

R 1

EU

+=

+

NextChng 8Back

my in mt chiu

n

URt =

Rt1

Rt2

Hnh a

U

n

Hnh b

Nu b qua tc dng ca phn ng phn ng v st p do tip xc gia chi than vnh gp th c tnh u ra l tuyn tnh. (h a).

gim qun tnh ca phn quay v s p mch ca t thng vin p v s tn ti ca rng trn mt phn ng ta dng rto rng.

Nhim v: bin i tn hiu in (in p iu khin) nhn c thnh di chuyn c hc ca trc tc ng ln cc b phn iu khin hoc iu chnh khc.

8.5: ng c tha hnh

Yu cu: tc ng nhanh, chnh xc, mmen v tc quay phi ph thuc vo in p iu khin theo quan h ng thng.Cu to: tng t nh 1 ng c kch t c lp. N c th c iu khin trn phn ng hoc trn cc t:

ng c tha hnh tc ng nhanh ngi ta ch to phn ng c qun tnh nh di dng rto rng hoc dt hnh a c mch in.

NextChng 8Back

my in mt chiu

+ Khi iu khin cc t: in p iu khin c a vo dy qun kch thch. Nh vy cng sut iu khin s nh nhng quan hn = f(Uk) khng l ng thng.

+ Khi iu khin trn phn ng in p kch thch t thng trc trn dy qun kch thch, ng c trng thi chun b tha hnh. Khi c Uk t ln dy qun phn ng lp tc ng c hot ng. Vi phng php iu khin ny: M = f(Uk) v n = f(Uk) l nhng ng thng.

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MD Mot Chieu