Giáo Trình Máy Điện-Khí Cụ Điện

7/29/2019 Gio Trnh My in-Kh C in

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TRNG AI HOC S PHAM KY THUAT TP. HCMKHOA IEN

BO MON: C S KY THUAT IEN------------0-----------

GVC-ThS.NGUYEN TRONG THANGGV-ThS.TRAN PHI LONG

GIAO TRNH

MAY IEN-KH CU IEN

TP. HCM Thang 12 / 2005


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LI NOI AU

Giao trnh MAY IEN-KH CU IEN nham giup sinh vien bac ai hoc hoac caoang nganh Cong nghe ien- ien t, Cong nghe ien t Vien thong cua trngai Hoc S Pham Ky Thuat TP.HCM lam tai lieu hoc tap cung co the dung lam tailieu tham khao cho sinh vien nganh ien Cong Nghiep, ien T ong va cac nganhkhac lien quan en lnh vc ien-ien t.Giao trnh May ien- Kh cu ien trnh bay nhng ly thuyet c ban ve: cau tao;

nguyen ly lam viec; cac quan he ien t; cac ac tnh cung nh cac hien tng vat lyxay ra trong: May ien mot chieu; May bien ap; May ien khong ong bo, May ienong bo va cac kh cu ien thong dung.e giup sinh vien de dang tiep thu kien thc mon hoc, giao trnh trnh bay noi dungmot cach ngan gon, c ban. moi chng co v du minh hoa, cau hoi va bai tap e

sinh vien co the hieu sau hn nhng van e mnh a hoc.

Cac tac ga


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Trang

PHN 1: MY IN ................................................................................... 1

CHNG I: MY IN MT CHIU ................................................................2

1.1. i cng v my in mt chiu..................................................................2 1.2. Cc quan h in t trong my in mt chiu.............................................7 1.3. My pht in mt chiu..............................................................................15 1.4. M my v iu chnh tc ng c in mt chiu................................32

CHNG II: MY BIN P ................................................................................43

2.1. i cng v my bin p..............................................................................43 2.2. T ni dy v mch t ca my bin p..........................................................54 2.3. Cc quan h in t trong my bin p...........................................................67 2.4. My bin p lm vic ti xc lp i xng...................................................85

CHNG III: CC VN L LUN CHUNGCA MY IN XOAY CHIU ...................................................98

3.1. Sc in ng ca dy qun my in xoay chiu..........................................98 3.2. Dy qun my in xoay chiu........................................................................106 3.3. Sc t ng ca dy qun my in xoay chiu..............................................115

CHNG IV: MY IN KHNG NG B ................................................126

4.1. i cng v my in khng ng b...........................................................126 4.2. Quan h in t trong my in khng ng b............................................ 130

4.3. M my v iu chnh tc ng c khng ng b...................................158

CHNG V: MY IN NG B ..................................................................172

5.1. i cng v my in ng b.......................................................................172 5.2. Quan h in t trong my in ng b........................................................ 178 5.3. My pht in ng b.....................................................................................189 5.4. M my v iu chnh cng sut phn khng ca ng c ng b............... 204

PHN 2: KH C IN ...........................................................................208

CHNG I: MCH T V S BIN I NNG LNG IN C ......... 209

1.1. Khi nim chung..................................................................................................209 1.2. T dn trong khe h khng kh............................................................................213 1.3. Nam chm in t mt chiu................................................................................215 1.4. Nam chm in t xoay chiu .............................................................................218 1.5. Nam chm vnh cu..............................................................................................222 1.6. Lc in ng.......................................................................................................226


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CHNG II: CC LOI KH C IN THNG DNG .........................................233

2.1. Contactor.............................................................................................................233 2.2. Cu ch bo v.....................................................................................................237 2.3. Aptomat...............................................................................................................241 2.4. Thit b chng dng in r.................................................................................248 2.5. Relay dng in...................................................................................................251 2.6. Relay in p........................................................................................................252

2.7. Relay trung gian...................................................................................................253 2.8. Relay thi gian.....................................................................................................254 2.9. Relay tc .........................................................................................................255


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1

PHN 1 :

MY IN


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2

CHNG 1 : MY IN MT CHIU

1.1 I CNG V MY IN MT CHIU

Trong nn sn xut hin i, my in mt chiu vn c coi l mt loi my quan trng. N c

th dng lm ng c in, my pht in ...

ng c in mt chiu c c tnh iu chnh tc rt tt, v vy c dng nhiu trong

nhng ngnh cng nghip c yu cu cao v iu chnh tc nh cn thp, hm m ...My pht in mt chiu dng lm ngun in cho ng c in mt chiu, lm ngun in mt

chiu kch thch t trong my in ng b, dng trong cng nghip m in...

Nhc im ca my in mt chiu : Gi thnh t, s dng nhiu kim loi mu, ch to v bo

qun c gp phc tp...

I. CU TO CA MY IN MT CHIU:

Kt cu ca my in mt chiu c th phn thnh 2 phn chnh l : phn tnh v phn quay .

1.Phn tnh (stato):

y l phn ng yn ca my gm cc b phn chnh sau:

a.Cc t chnh:L b phn sinh ra t trng gm c li st cc t v dy qun kch t lng ngoi li st cc t.

Li st cc t lm bng nhng l thp k thut in hay thp cacbon dy 0,5 n 1mm p li v tn

cht . Trong my in nh c th dng thp khi . Cc t c gn cht vo v my nh cc bulng.

Dy qun kch t c qun bng dy ng bc cch in v mi cun dy u c bc cch in

k thnh 1 khi v tm sn cch in trc khi t trn cc cc t . Cc cun dy kch t t trn cccc t ny c ni ni tip vi nhau .

Hnh 1-1. Cc t chnh

b. Cc t ph:

c t gia cc cc t chnh v dng ci thin i chiu. Li thp ca cc t ph thng

lm bng thp khi v trn thn cc t ph c t dy qun m cu to ging nh dy qun cc t

chnh. Cc t ph c gn vo v nh bulng.


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3

c. Gng t:

Dng lm mch t ni lin cc cc t, ng thi lm v my. Trong my in va v nh

thng dng thp tm dy un v hn li. Trong my in ln thng dng thp c .

d. Cc b phn khc:

- Np my : Bo v my khi b nhng vt ngoi ri vo lm h hng dy qun hay an ton

cho ngi khi chm phi in. Trong my in va v nh, np my c tc dng lm gi bi .

- C cu chi than : a dng in t phn quay ra ngoi. C cu chi than gm c chi

than t trong hp chi than v nh 1 l xo t cht ln c gp. Hp chi than c c nh trn gi

chi than v cch in vi gi. Gi chi than c th quay c iu chnh v tr chi than cho

ng ch. Sau khi iu chnh xong th dng vt c nh cht li.


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2. Phn quay ( rto ):

Gm cc b phn sau :

a. Li st phn ng: Dng dn t.Thng dng nhng tm thp k thut in dy 0,5 mm ph cch in mng 2 mt ri p

cht li gim hao tn do dng in xoy gy nn. Trn l thp c dp hnh dng rnh sau khi pli th t dy qun vo.

Trong nhng my c trung tr nn, ngi ta cn dp nhng l thng gi khi p li thnh li

st c th to c nhng l thng gi dc trc .

Trong nhng my hi ln th li st chia thnh tng on nh. Gia cc on y c 1 khe h

gi l khe thng gi ngang trc. Khi my lm vic, gi thi qua cc khe lm ngui dy qun v li

st .

Trong my in nh, li st phn ng c p trc tip vo trc. Trong my in ln, gia trc

v li st c t gi rto. Dng gi rto c th tit kim thp k thut in v gim nh trng lng

rto .

b. Dy qun phn ng:

L phn sinh ra sc in ng v c dng in chy qua. Dy qun phn ng thng lm bng

dy ng c bc cch in. Trong my in nh dng dy c tit din trn. Trong my in va v

ln dng dy tit din ch nht. Dy qun c cch in cn thn vi rnh ca li thp .

trnh khi quay b vng ra do sc ly tm, ming rnh c dng nm cht hoc phi

ai cht dy qun. Nm c th lm bng tre, g hay baklit .


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Hnh 1-7. Mt ct rnh phn ng

c. C gp: Dng i chiu dng in xoay chiu thnh 1 chiu .C gp gm nhiu phin ng c ui nhn cch in vi nhau bng lp mica dy 0,4 n 1,2

mm v hp thnh 1 tr trn. Hai u tr trn dng 2 vnh p hnh ch V p cht li. Gia vnh p v

tr trn cng cch in bng mica. ui vnh gp c cao hn ln mt t hn cc u dy ca cc

phn t dy qun vo cc phin gp c d dng .

Hnh 1- 8. C gp.

d. Cc b phn khc :

- Cnh qut : qut gi lm ngui my.- Trc my : Trn t li st phn ng, c gp, cnh qut v bi. Trc my thng lm

bng thp cacbon tt .

2. Cc tr s nh mc:

Ch lm vic nh mc ca my in l ch lm vic trong nhng iu kin m xng ch

to quy nh. Ch c c trng bng nhng i lng ghi trn nhn my v gi l nhng

i lng nh mc.

- Cng sut nh mc Pm ( KW hay W)- in p nh mc Um (V)

- Dng in nh mc Im (A)

- Tc nh mc nm ( vg/ph)

- Hiu sut nh mc m


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CU HI

1. Ti sao li st phn ng phi lm bng thp k thut in ? Ti sao v my mt chiu khng

dng gang l vt liu r tin v d c ?

2. ngha ca tr s cng sut nh mc ghi trn nhn my ? Cng sut nh mc ghi trn ng c

in l cng sut c u trc hay cng sut in a vo ng c in ?

3. Cc b phn chnh ca my in mt chiu v cng dng ca cc b phn ?


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1.2 CC QUAN H IN T TRONG MY IN MT CHIU.

I. Sc in ng cm ng trong dy qun my in mt chiu:Cho dng in kch thch vo dy qun kch thch th trong khe h s sinh ra t thng. Khi phn

ng quay vi mt tc nht nh no th trong dy qun s cm ng mt sc in ng. Sc

in ng ph thuc vo t thng di mi cc t, tc ca my, s thanh dn ca dy qun v

kiu dy qun .

V dy qun c 2a mch nhnh ghp song song nn sc in ng ca dy qun bng sc inng cm ng trn mt mch nhnh, ngha l bng tng s.. ca cc thanh dn ni tip trong mch

nhnh .

Sc in ng trung bnh cm ng trong thanh dn c chiu di tc dng l, chuyn ng trong t

trng vi tc v bng :

etb = Btb.l.v ( 1-1)

Tc quay : v = 60260. npnD

Btb =l.

Trong :Btb : t cm trung bnh trong khe hD : ng knh ngoi phn ng : bc cc

p : s i ccn : tc quay phn ng

: t thng khe h di mi cc

Th vo biu thc (1-1) ta c :

etb = 2p 60

n( 1-2)

Gi N l tng s thanh dn ca dy qun th mi mch nhnh song song c N/2a thanh dn ni tipnhau v nh vy s.. ca my bng :

E= nCna

Npea

Netb 60

.

2 ( V) ( 1-3)

Trong :

( Wb).

n ( vg/ph).

Ce =a

Np

60

.: h s ph thuc vo kt cu ca my v dy qun

Chiu ca E ph thuc vo chiu v n v c xc nh theo quy tc bn tay phi ( hnh 1- 9 ).


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Hnh 1-9. Xc nh s.. phn ng v mmen in ttrong my pht in mt chiu

S phn tch trn da trn gi thit dy qun bc , s.. trn cc thanh dn ca phn t ucng s hc vi nhau. Nu l bc ngn th s.. ca cc thanh dn ca mt phn t s cng vct

nn s.. ca c phn t s nh hn so vi phn t bc v nh vy s.. phn ng cng nh imt t. Nhng v trong my in 1 chiu khng cho php bc ngn ln nn nh hng t ( thng lkhng xt n khi tnh s.. ).

II. Mmen in t v cng sut:

Khi my in lm vic, trong dy qun phn ng s c dng in chy qua. Tc dng ca ttrng ln dy dn c dng in s sinh ra mmen in t trn trc my.

Gi thit mt ch lm vic no ca my in mt chiu, t trng v dng in phn ng di 1 cc nh hnh :

Theo quy tc bn tay tri mmen in t do lc in t tc dng ln cc thanh dn c chiu t

phi sang tri .

Lc din t tc dng ln tng thanh dn bng :

f = Btbl.i (1- 4)


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Nu tng s thanh dn ca dy qun bng N, dng in trong mch nhnha

Ii

2=

th mmen in t tc dng ln dy qun phn ng bng :

2.

2

DNla

IBM

tb= ( 1-5)

Trong :Btb : t cm trung bnh trong khe hI : dng din phn nga : s i mch nhnhl: chiu di tc dng ca thanh dnD : ng knh ngoi phn ng

Ta c : D =

p2, Btb =

l

=> M =

a

pN

2I= CM I ( N.m) (1-6)

=> M =

a

pN

281,9

1I ( kG.m) ( 1-7)

Trong :

: t thng di mi cc tnh bng Wb

a

NpCM .2

.

: h s ph thuc vo kt cu ca my in

Trong my pht in, khi quay my theo mt chiu nht nh trong t trng th trong dy dn

sinh ra s.. m chiu c xc nh theo quy tc bn tay phi. Khi c ti th dng in sinh ra scng chiu vi s.. nn mmen in t sinh ra s ngc chiu vi chiu quay ca my. V vy my pht in, mmen in t l mmen hm ( hnh 1-10).

Trong ng c in, khi cho dng in vo phn ng th di tc dng ca t trng, trong dyqun s sinh ra mmen in t ko my quay, v vy chiu quay ca my trng vi chiu quay cammen ( hnh 1-10).

Cng sut ng vi mmen in t ly vo ( my pht ) hay a ra ( ng c ) gi l cng sutin t v bng :

Pt = M ( 1-8 )

Trong :M : momen in t

=60

.2 n: tc gc phn ng

Thay vo cng thc ( 1-5 ) ta c :

Pt = M =a

Np

2

. I

n

a

pNn

6060

2I = E.I ( W) ( 1- 9 )


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Nh vy, trong my pht in cng sut in t chuyn cng sut c M thnh cng sutin EI. Ngc li, trong ng c in cng sut in t chuyn cng sut in E I thnh cngsut c M .III. Qu trnh nng lng v cc phng trnh cn bng:

1. Tn hao trong my in mt chiu:a. Tn hao pc: Gm tn hao bi, tn hao ma st chi than vi vnh gp, tn hao do thng gi,

tn hao ny ph thuc ch yu vo tc quay ca my v lm cho bi, vnh gp nng ln .

b.Tn hao stpFe : Do t tr v dng in xoy trong li thp gy nn. Tn hao ny ph thuc

vo vt liu, chiu dy ca tm thp, trng lng li thp, t cm v tn s f . Khi li thp nhhnh th t hao thp t l vi f(1,2 1,6) v B2 .Hai loi tn hao trn khi khng ti tn ti nn gi l tn hao khng ti :

P0 = pc + pFe ( 1-10)

Tn hao c v st sinh ra mmen hm v mmen ny tn ti khi khng ti nn gi l mmenkhng ti M0 .

M0 =

0P ( 1-11)

vi l tc gc ca rto.

c. Tn hao ngpcu : Tn hao ng trong mch phn ng pCu. v tn hao ng trong mchkch thch pCu.tTn hao ng trong mch phn ng bao gm tn hao ng trong dy qun phn ng I

2r ; tnhao ng trong dy qun cc t ph I

2rf; tn hao tip xc gia chi than v vnh gp ptx. Thngvi chi than graphit in p ging trn ch tip xc ca hai chi than 2 txU = 2 V nn ptx = 2 I .

Hin nay thng gp tt c cc tn hao ng trn phn ng li v vit di dng p = I2R

vi R = r + rf+ rtx bao gm in tr dy qun phn ng r , in tr dy qun ph rf, in tr tipxc chi than rtx .

Tn hao ng trong mch kch thch gm tn hao ng ca dy qun kch thch v tn hao ngca in tr iu chnh trong mch kch thch. V vy pCu.t = UtIt , vi Ut l in p t trn mch

kch thch; It l dng in kch thch.

d. Tn hao ph pf:Tn hao ph trong thp c th do t trng phn b khng u trn b mt phn ng, cc

bulng c vt trn phn ng lm t trng phn b khng u trong li st, nh hng ca rng, rnh

lm t trng p mch ... sinh ra .

Tn hao trong ng c th do qu trnh i chiu lm dng in trong phn t thay i, dng

in phn b khng u trn b mt chi than lm tn hao tip xc ln, t trng phn b khng u

trong rnh lm cho trong dy dn sinh ra dng in xoy, tn hao trong dy ni cn bng sinh ra.Trong my in 1 chiu pf tng i kh tnh, thng ly bng 1% cng sut nh mc .

2. Qu trnh nng lng trong my in mt chiu v cc phng trnh cn bng:

a. My pht in:

My pht in bin c nng thnh in nng nn my do mt ng c s cp bt k ko quayvi mt tc nht nh .

Gi thit cng sut kch t do mt my khc cung cp nn khng tnh vo cng sut a tng c s cp vo my pht in. Cng sut c a vo P 1, tiu hao i mt phn b vo tn haocpcv t hao st pFe, cn i b phn bin i thnh cng sut in t Pt .

Ta c :


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P1 = Pt +( pc + pFe ) = Pt + Po (1-12)

Pt = EI (1-13)

Khi c dng in chy trong dy dn th c tn hao ng nn cng sut in a ra P2 bng :P2 = Pt pcu = EI - I

2R = UI ( 1-14)

Gin nng lng :

Hnh 1-11. Gin nng lng ca my pht in mt chiu.

Chia 2 v ca phng trnh trn vi I ta c :

U = E - IR ( 1-15)

y l phng trnh cn bng s.. ca my pht in 1 chiuC th vit cng sut a vo, cng sut khng ti v cng sut in t theo dng mmen nhn

vi tc gc v nh vy cng thc ( 1- 12) vit thnh :

M1 = M0 + M

Chia 2 v cho ta c :M1 = M0 + M ( 1-16)

y l phng trnh cn bng mmen ca my pht in 1 chiuTrong :

M1 l mmen a voM l mmen in tM0 l mmen khng ti

b. ng c in:ng c in ly cng sut in vo v truyn cng sut c ra u trcCng sut in m ng c nhn c t li vo bng :

P1 = UI = U(I + It ) ( 1-17)

Trong :I = I + It l dng in t li in vo ( I l dng in vo phn ng, I t l dng in kch

thch ) .U l in p u cc my

Cng sut P1, mt phn cung cp cho mch kch thch UIt cn phn ln i vo phn ng UI ,tiu hao mt t trn dy qun ng trn mch phn ng pCu. , cn i b phn l cng sut in tPt .

P1 = pCu. + pCu.t + Pt ( 1-18)


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Cng sut in t sau khi chuyn thnh cng sut c th cn tiu hao mt t b vo tn haoc pc v tn hao st pFe ( gi chung l tn hao khng ti Po ). Phn cn li l cng sut a ra u trc P2 = M2 .

Pt = pc+ pFe + P2 = P0 + P2 ( 1- 19)

Gin nng lng nh hnh :

T cng thc ( 1-17 ) v ( 1-18 ) ta c cng sut in trong mch phn ng bng :

UI = Pt + pCu. = EI + I2R (1-20)

Chia 2 v cho I ta c :

U = E + IR ( 1-21)

y l phng trnh cn bng s.. ca ng c in 1 chiuT cng thc (1-19) ta c th vit :

M = M0 + M2 ( 1-22)

Chia 2 v cho ta c :M = M0 + M2 ( 1-23)

y l phng trnh cn bng mmen ca ng c in 1 chiu

Vi :

M2 l mmen a ra u trc my

M0 l mmen khng ti

IV. TNH CHT THUN NGHCH TRONG MY IN MT CHIU:

My in 1 chiu c th dng lm my pht in, cng c th dng lm ng c in. Trong my

pht in, chiu ca mmen in t v tc quay ngc nhau, cn dng in v s.. cng chiu .

Trong ng c in th mmen v tc quay cng chiu cn dng in v s.. ngc chiu nhau.

V vy ch cn c iu kin khch quan khc nhau th my s c tnh cht lm vic khc nhau.

Gi s my ang lm vic trng thi my pht :

R

UEI

ngha l E > U . My sinh ra in t hm .

Nu ta gim t thng hoc tc n gim E xung mt cch thch ng th E s nh hn

U v dng in I s i chiu, E v I ngc chiu nhau. Do chiu ca t thng khng i nn


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mmen in t ( M = CM I ) s i du, ngha l M v n cng chiu v mmen in t t

mmen hm thnh mmen quay. My chuyn t my pht in sang ng c in. y l tnh

cht thun nghch ca my in. Tch ng c s cp ra ta s c 1 ng c in mt chiu thng

thng.

V D

V d 1: Mt my pht in 1 chiu lc quay khng ti tc n = 1000 vg/ph th s.. pht rabng E0 = 222 V . Hi lc khng ti mun pht ra s.. nh mcE0.m = 220 V th tc no.m bng bao nhiu khi gi dng in kch t khng i ?

Gii

Gi dng in kch t khng i ngha l t thng khng iTheo cng thc ( 1-3) ta c :

momoe

e

mo

o

n

n

nC

nC

E

E

...

==

Khi Eo.m = 220 V th :

no.m = n.o

mo

E

E .= 1000. 990

222

220 vg/ph

V d 2 : Mt ng c in 1 chiu kch thch song song, cng sut nh mc Pm = 5,5 kW,Um = 110 V, Im = 58 A ( tng dng in a vo bao gm dng in phn ng I v kch t It,nm = 1470 vg/ph. in tr phn ng R = 0,15 , in tr mch kch t rt = 137 , in p gingtrn chi than 2 txU = 2 V. Hi s.. phn ng, dng in phn ng v mmen in t ?

GiiDng in kch t :

It =tr

U= 8,0

137

110 A

Dng in phn ng :I = Im It = 58 0,8 = 57,2 A

Sc in ng phn ng :

E = U IR - 2 txU = 110 ( 57,2 . 0,15 ) 2 = 99,4 V

Mmen in t :

M = 9,36

60

1470.2

2,57.4,99

60

2

n

IEIE N.m

Nu tnh ra kG.m th :

M = 76,381,9

9,36 kG.m


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CU HI

1. Sc in ng trong my in 1 chiu ph thuc vo nhng yu t no ?

2. Phn tch gin nng lng ca my pht v ng c in 1 chiu , t a ra cc quan h

v cng sut, mmen, dng in v s.. ?

BI TP

1. Mt ng c in 1 chiu kch thch song song c cc s liu sau : Um = 220 V, R = 0,4 ,

Im = 52 A , rt = 110 v tc khng ti no = 1100 vg/ph . Hy tm :

- Sc in ng phn ng lc ti nh mc

- Tc lc ti nh mc

- Cng sut in t v mmen in t lc ti nh mc

Khi phn tch b qua dng in khng ti .

p s : E.m= 200 V

nm = 1000 vg/ph

Pt = 10 kW , Mt = 95,5 N.m2. Mt ng c in 1 chiu kch thch song song c cc s liu sau :

Pm = 96 kW, Um = 440 V, Im = 5 A, It = 5 A, nm = 500 vg/ph, R = 0,078 . Tnh :

-Mmen nh mc u trc M2

-Mmen in t khi dng in nh mc

-Tc quay lc khng ti k tng ( I = 0)

p s : M2 = 1833,5 N.m

Mt = 2007,7 N.mno = 523 vg/ph

3. Mt my pht in kch thch ngoi c Um = 220 V , nm = 1000vg/ph . Bit rng tc

n = 750 vg/ph th s.. lc khng ti Eo = 176 V. Hi s.. v dng in phn ng lc ti nh mc

ca my l bao nhiu, bit in tr phn ng R = 0,4 .

p s : E.m = 234,6 V

I.m= 36,5 A


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1.3 MY PHT IN MT CHIU

I. I CNG:

Trong nn kinh t quc dn, nhiu ngnh sn xut nh luyn kim, ho cht, giao thng vn ti ...

i hi phi dng ngun in mt chiu, v ngy nay vn khng th thay th c dng in mt

chiu mc d vic dng dng in xoay chiu trong cng nghip rt ph bin. Thng thng

c ngun in mt chiu c th dng cc my pht in mt chiu quay bng cc ng c s cp nh

ng c in xoay chiu, ng c t trong, tuabin ...

Hnh 1-13. S nguyn l ca my pht in mt chiu

a. My pht in mt chiu kch thch c lpb.My pht in mt chiu kch thch song songc. My pht in mt chiu kch thch ni tipd. My pht in mt chiu kch thch hn hp

Ty theo cch kch thch cc t chnh, my pht in mt chiu c phn loi nh sau :

1. My pht in mt chiu kch thch c lp : Bao gm my pht kch thch bng nam chm

vnh cu v my pht kch thch in t. Loi u ch c ch to vi cng sut nh. Loi th hai

c dy qun kch thch ly dng in t cquy, li in mt chiu hoc my pht in mt chiu

ph v c dng nhiu trong cc trng hp cn iu chnh in p trong phm vi rng, cng sut

ln, in p thp (4 24)V hoc in p cao trn 600 V.

2.My pht in mt chiu t kch thch : C dng in kch thch ly t bn thn my pht in.

Tu theo cch ni cc dy qun kch thch, ta c :

My pht in mt chiu kch thch song song

My pht in mt chiu kch thch ni tip

My pht in mt chiu kch thch hn hp

Trong mi trng hp, cng sut kch thch chim (0,3 5) % cng sut nh mc ca my.

Theo hnh (1-13) ta thy rng cc my pht kch thch song song v kch thch hn hp I= I It

cn my pht kch thch ni tip I = I = It .


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II. CC C TNH CA MY PHT IN MT CHIU

My pht in mt chiu c 4 i lng c trng l U, I , It v n . Tr tc n c ng cs cp gi khng i, ba i lng cn li : U, I , It l nhng i lng bin thin c lin h chtch vi nhau . Vi 3 i lng c th thnh lp 3 mi quan h c bn : U = f(I) khi It = C

te ; U =f(It) khi I = C

te v I = f(It) khi U = Cte .

Da vo , khi nghin cu my pht in mt chiu ta c cc c tnh sau y :

1) c tnh khng ti Uo = Eo = f(It) khi I= 0, n= Cte

.2) c tnh ngn mch In = f(It) khi U= 0 , n= Cte .

3) c tnh ngoi U = f(I) khi It = Cte , n= Cte .

4) c tnh ti U = f(It) khi I = Cte , n= Cte .

5) c tnh iu chnh It = f(I) khi U = Cte , n= Cte .

Trong nm c tnh trn th c tnh khng ti l trng hp c bit ca c tnh ti khi I = 0 vc tnh ngn mch l trng hp c bit ca c tnh iu chnh khi U=0 . Tt c 5 c tnh uc th thnh lp c bng th nghim trc tip trn my pht in. Tuy nhin trong mt s trnghp, n gin ch cn lm 2 th nghim khng ti v ngn mch, sau da vo tam gic c tnh suy ra 3 c tnh cn li.

Di y ta s nghin cu chung cc c tnh khng ti, c tnh ngn mch v cch thnh lp tamgic c tnh ca cc loi my pht in. V v c bn chng khng c g khc nhau. Cn cc ctnh lm vic bao gm c tnh ngoi v c tnh iu chnh s c nghin cu ring bit i vitng loi my .

c tnh khng ti Uo = Eo = f(It) khi I = 0, n = Cte .

Lm th nghim cho my pht in lm vic tc n khng i , cu dao h mch khngni vi ti bn ngoi ( I = 0 ), o cc tr s I t v U tng ng ta s c c tnh khng ti .

Cn ch rng, i vi my pht in kch thch c lp, do c th i chiu dng in kchthch nn ta c thv c ton b chu trnh tr i xng ABABA gia hai tr s gii hn ca dngin kch thch Itm ng vi in p c ( 1,15 1,25)Um .

Hnh 1-14. c tnh khng ti ca my pht in mt chiu.

i vi my pht in t kch thch, do cc tnh u my ( chi than ) l c nh v khng

th thc hin c It nn ta ch c th v c chu trnh ph ABA gia + Itm v 0 .


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on OB trn hnh (1-14) l s.. ng vi t d trong mch t ca my. S.. ny rt nh,

thng bng 2 3% Um nn ta c th b qua, v vy c tnh khng ti ca my pht in mt chiu

l ng trung bnh i qua gc to AOA biu th bng ng t nt. cng chnh l ng

cong t ho ca my pht in suy ra c khi tnh ton mch t ca my lc khng ti .

c tnh ngn mch In = f(It) khi U= 0 , n= Cte .

Hnh 1-15. c tnh ngn mchca my pht in mt chiu

Trc ht cn ch rng c c tnh ngn mch tt c cc loi my pht in mt chiu u

phi c kch thch c lp. Nu em ni ngn mch cc chi than v cho my pht in lm vic

tc khng i ri o cc tr s It v I tng ng ta c c tnh ngn mch. Khi ngn mch :

U = E - IR = 0 => E = IR

Do R rt nh, mt khc phi gi cho dng in I khi ln qu tr s (1,25 1,5)Im nn E rt nhv dng in kch thch It tng ng cng rt nh . V It nh nn mch t ca my khng bo ho( = Cte ) , E It do I It v c tnh ngn mch l mt ng thng. Nu my c kht d th ng thng ny i qua gc to ( ng 1 trn hnh 1-15 ). Nu my cha c kh td ta s c ng 2 . c c tnh ngn mch tiu chun ta v ng thng song song vi ng 2qua gc ta .

Tam gic c tnh

Nu thnh lp tam gic c tnh trn h ta chung c trc honh It , ta v cc c tnh khng

ti ( ng 1 ) v c tnh ngn mch (ng 2) nh hnh (1-16) .

Hnh 1-16. Xy dng tam gic c tnhtrong trng hp phn ng phn ng kh t

I

It

1

2


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Gi s rng khi ngn mch trong phn ng c dng in Im tng ng vi dng in kchthch It = OC. Dng in kch thch dnh mt phn OD sinh ra s.. khc phc in p ri trnin tr phn ng ImR = AD = BC ; phn cn li DC = AB dng khc phc phn ng phn nglc ngn mch. Tam gic ABC gi l tam gic c tnh c cnh BC t l vi dng in phn ng I vcnh AB trong iu kin mch t khng bo ho t l vi phn ng phn ng ( ngha l cng t l vidng in I ) .

ln ca cnh AB ph thuc vo loi my v ln nht my in 1 chiu khng c dy

qun b v cc t ph. my c dy qun b v cc t ph th phn ng phn ng hu nh b trit

tiu ( cnh AB 0 ). my in 1 chiu kch thch hn hp, dy qun ni tip c tc dng tr t v

nu s.t. ca n ln hn AB , ngha l ngoi phn s.t. trit tiu nh hng ca phn ng cn s.t.

tr t th cnh AB s nm v pha bn phi ca BC ( hnh 1-17 ) .

Hnh 1-17. Xy dng tam gic c tnhtrong trng hp phn ng phn ng tr t

1. c tnh lm vic ca my pht in mt chiu kch thch c lp

a. c tnh ngoi U = f(I) khi It = Cte , n= Cte .

Khi I tng, in p ri trn dy qun phn ng tng. Mt khc, do phn ng phn ng tng theo

I nn s.. E gim. Kt qu l in p U u my pht in gim xung.

Dng ca c tnh ngoi ca my pht in kch thch c lp c trnh by trn hnh (1-18) .

Hnh 1-18. c tnh ngoi ca my pht inmt chiu kch thch c lp


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Hiu s in p lc khng ti v lc ti nh mc vi iu kin dng in kch t bng dng

in kch t nh mc c quy nh l bin i in p nh mc :

100%m

mo

mU

UU

U =

my pht in mt chiu kch thch c lp Um = 5 15% .Nh ni trn, c tnh ngoi c th c c bng th nghim trc tip hoc bng phng

php gin tip da vo c tnh khng ti v tam gic c tnh nh trnh by trn hnh (1-19). Hycho c tnh khng ti ca my v on OP = It = C te, on PP ng vi It cho biu th in pU = E lc khng ti Eo ( I = 0 ) v xc nh im xut pht D ca c tnh ngoi. t tam gic ABCc cnh AB, BC theo t l ng vi I= Im sao cho im A nm trn c tnh khng ti v cnh BCtrn ng thng ng PP th on PC s l in p khi I = Im v tng ng ta c im D v gc

phn t th hai . chng minh ta thy rng, nu U = PC th E = U + ImR = PC+ CB = BP = AQ . Lc khng

ti c E = AQ cn c dng in kch thch t I t (o) = OQ ; khi c ti nh mc phi tng dng in

kch thch ln mt lng It = QP = AB b li s kh t ca phn ng phn ng. Ton dng in

kch thch lc ny ng l :

It = It (o) + It = OQ + QP = OP nh cho trc .

Nu I =2

1Im th tam gic c tnh c cc cnh bng mt na cnh ca tam gic ABC. Cng

lm nh trn ta s xc nh c im D. Tip tc theo trnh t trn, ta xc nh c mt s im

khc ng vi cc tr s khc nhau ca dng in I. Ni cc im D, D, D ... li ta c c tnh

ngoi U = f(I) khi I t = Cte , n = Cte . Trn thc t do nh hng ca bo ho, khi I tng v U gim,

cnh AB ca tam gic c tnh khng cn t l vi I nn ng c tnh ngoi thu c bng th

nghim trc tip hi lch i theo ng t nt trn hnh (1-19) .

Hnh 1-19. Xy dng c tnh ngoi ca my pht in kch thchc lp t c tnh khng ti v tam gic c tnh.


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im ng vi U = 0 ca c tnh ngoi cho ta tr s ca dng in ngn mch khi kch thch honton y . V R rt b, dng in ngn mch In = (5 15)Im v rt nguy him c th gy vng latrn vnh gp v ng lc in ng rt ln do phi trang b my ct t ng cc nhanh tch my

pht in ra khi li khi xy ra ngn mch t nhin. Ch rng bin php ny khng bo v ckhi xy ra ngn mch bn trong my

b. c tnh iu chnh It = f(I) khi U = Cte , n= Cte .

c tnh iu chnh cho ta bit cn iu chnh dng in kch thch th no gi cho in pu ra ca my pht khng i khi thay i ti. ng biu din c tnh iu chnh trn hnh (1-20)

cho thy khi ti tng cn phi tng dng in kch thch sao cho b c in p ri trn I v nh

hng ca phn ng phn ng . T khng ti ( U= Um ) tng n ti nh mc (I = Im ) thng

phi tng dng in kch thch ln 15 25% .

Phng php dng c tnh iu chnh bng c tnh khng ti v tam gic c tnh c trnh bytrn hnh 1-21 .

Hnh 1-21.Xy dng c tnh iu chnh ca my pht inkch thch c lp t c tnh khng ti v tam gic c tnh


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Vi tr s xc nh ca Uo = Um = MP , khi I = 0 ta c im M ng vi dng in kch thch

It = OM. Nu t tam gic c tnh ABC ng vi ti nh mc Im sao cho nh A nm trn c tnh

khng ti v nh C nm trn ng thng FC ( ng vi U = Um = Cte ) v h ng thng BN th

on ON cho ta tr s dng in kch thch ti nh mc. Vic chng minh c tin hnh tng t

nh trng hp dng c tnh ngoi mc a. tm nhng im khc ta k nhng on thng AC,

AC ,...song song vi cnh AC, nm gia c tnh khng ti v ng thng CF sau h nhng

ng thng ng ct trc honh ti N , N ...Cc on ON, ON, ...s biu th cc dng in kch

thch ng vi cc tr s ca dng in i xc nh bng t s gia cc on AC, AC, ... vi cnh

huyn AC .

Do nh hng ca bo ho, ng c tnh iu chnh thu c bng th nghim trc tip c

dng theo ng t nt trn hnh (1-21) .

2. c tnh ca my pht in mt chiu kch thch song songMy pht in mt chiu kch thch song song c dy qun kch thch c ni song song vi

dy qun phn ng c th t sinh ra dng in kch thch cn thit m khng cn ngun in bn

ngoi.Ta bit, khi my ngng hot ng, trong li thp cc t, gng t cn li t d. Nu h

mch kch thch ( It =0 ) v quay my pht in n tc nh mc, do c t thng d trong dyqun phn ng s cm ng c sut in ng E v trn cc my s to ra mt in p

U = (2 3)%Um. Nu ni kn mch kch thch th trong n s c dng in It =tr

Uvi rt l in tr

ca mch kch thch. Kt qu l sinh ra s.. Itwt. Nu sut in ng ny sinh ra t thng c chiutrng vi t thng d th my s uc tng kch t, in p u cc s tng v c tip tc nh vymy s tip tc t kch thch c. Nu t thng sinh ra ngc chiu vi t d th my s b kh t,khng th t kch v to ra in p c .

thy r qu trnh to ra in p ca my pht in kch thch song song ta vit phngtrnh in p cho mch vng kn bao gm dy qun kch thch v dy qun phn ng. B qua phnng phn ng rt nh sinh ra bi dng in It khi chy qua dy qun phn ng v gi s rng h st cm ca dy qun kch thch I t = C

te , c th b qua R v n rt nh so vi rt , ta c :

Edt

diLir tttt ( 1-24)

vi iu kin ban u t = 0 , it = 0 .

Theo biu thc trn, s.. cm ng sinh ra trong dy qun phn ng ph thuc vo dng inkch thch it v tc quay n ca my . Nu gi thit n = C

te th E = f(It) , y chnh l c tnh khngti ca my pht in .

Phng trnh vi phn (1-24) c th gii trn my tnh hoc tnh bng phng php gn ng .mun tnh bng phng php gn ng th phi biu th c tnh khng ti cho bit ca my bng

biu thc gii tch c dng :

ntnto iaiaaE ...

1

1 (1-25)

vi 0oa , chnh l s.. Ed do t thng d sinh ra


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Gii h phng trnh (1-24), (1-25) ta c s bin i ca dng in i t t it = 0 n tr s xc

lp it = It . Khi dng in t n tr s xc lp It th 0dt

dit v in p to ra u my bng :

rtIt = E = Uo (1-26)T phng trnh (1-24), (1-25) ta thy iu kin c nghim (hay iu kin my t kch

v to ra c in p) l :

- Khi it = 0 tr s 0oa ngha l my phi c t d v nu khng phng trnh (1-24) khngc li gii no khc it = 0, ni khc i l my khng th t kch thch c .

- Chiu quay ca my pht phi theo chiu nht nh sinh ra dng in it > 0 v nu it < 0

s lm cho E = 0, phng trnh (1-24) khng c li gii v my khng th t kch c .

- Nu rt qu ln, dng in It xc lp s rt nh, in p xc lp bng E d ca my .

in p Uo thnh lp c u my theo biu thc (1-26) cng c th tm trn hnh (1-22)

v chnh l giao im M ng vi tr s it = It ca c tnh khng ti v ng thng 1 biu th

quan h U= itrt .

Hnh 1-22. in p xc lp ca my pht kch thch song songng vi cc tr s khc nhau ca rt

T hnh (1-22) ta thy rng khi it < It th o hm :

tt

t

t

irEdt

diL > 0

V dng in it tng, kt qu l my s t kch thch c n in p ng vi giao im M

0dt

diL tt v dng in kch thch c tr s xc lp bng It .

Nu rt tng th ng U = itrt c dc ln hn ( ng 2), in p thnh lp s nh hn. Tr s

rt ng vi ng 3 trng vi on thng ca c tnh khng ti gi l in tr ti hn rt(th), khi


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in p u my pht s khng n nh. Nu rt > rt(th) ta c ng thng 4 v in p u my bng

s.. Ed , hay my khng th t kch c .

Do tnh cht bo ho ca mch t , bng cch tng rt ta c th iu chnh c in p xc lp

u my n tr s nh nht Umin = (0,65 0,75)Um . Trong trng hp cn iu chnh trong phm vi

rng ng vi Um:Umin = 5:1 ( hoc 10:1) cn phi un cong on u ca c tnh khng ti

(hnh 1-23). Mun vy, phi lm cho mch t sm bo ho bng cch x rnh cc t .

Hnh 1-23. Un cong phn thng ca c tnh khng tica my pht in mt chiu

a. c tnh ngoi U = f(I) khi rt = Cte , n= Cte .

Hnh 1-24. c tnh ngoi ca my pht in mt chiukch thch song song (2) v kch thch c lp (1)

Khi tng ti, in p ca my pht in kch thch song song gim nhiu hn in p ca my

pht in kch th ch c lp v ngoi nh hng ca phn ng phn ng v in p ri trn R nh

trong my pht in kch thch c lp, trong my pht in kch thch song song vi s.. E cn

gim theo dng in kch t It . V vy m thay i in p kch thch song song cng ln hn,

thng %1210mU .

im c bit ca my pht kch thch song song l dng in ti ch tng n mt tr s nht

nh I = Ith, sau nu tip tc gim rt ca ti mch ngoi th I khng tng m gim nhanh n tr

s Io xc nh bi t d ca my in v ng vi im P .


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S d nh vy l do my lm vic trong tnh trng khng bo ho ng vi on thng ca ng

cong t ho, dng It gim lm cho E, U gim rt nhanh. in p U gim nhanh hn I t lm cho dng

in ti I gim n tr s Io . Nh vy, s c ngn mch u my pht kch thch song song khng

gy nguy him nh trng hp my pht kch thch c lp.

Cch thnh lp c tnh ngoi t c tnh khng ti v tam gic c tnh tin hnh nh my

pht kch thch c lp . iu khc c bn l my pht kch thch c lp It = const, cn y It

ph thuc vo U v ng U = rtIt l ng OP qua gc ta (hnh 1-25) .

Hnh 1-25. Xy dng c tnh ngoi ca my pht in mt chiukch thch song song bng c tnh khng ti v tam gic c tnh

Tam gic c tnh ABC y s tnh tin trong vng gii hn gia c tnh khng ti v ng

OP, trong khi my pht kch thch c lp, trong vng gii hn gia c tnh khng ti v ng

thng ng PP ( hnh 1-19) .


c tnh iu chnh ca my pht kch thch song song ging nh my pht kch thch c lp.

Bi v, i vi bn thn my pht vic iu chnh dng in kch thch gi in p khng i khi

ti thay i khng ph thuc vo vic dng in kch thch c ly t u - t mt ngun khc bn

ngoi hay t u cc ca my. iu cn ch l my pht kch thch song song khi tng ti, in

p st nhu hn nn mc tng dng in kch thch phi nhiu hn, do c tnh iu chnh s

dc hn

3. c tnh ca my pht in mt chiu kch thch ni tip:

Trong my pht kch thch ni tip, dy qun kch thch c mc ni tip vi dy qun phn

ng . Cng v vy m s vng dy ca dy qun kch thch t hn nhiu so vi s vng dy ca dy

qun kch thch ca my pht kch thch song song, nhng tit din ca dy ln hn mt cch tng

ng .

My pht kch thch ni tip thuc loi t kch thch, cn phi c t d v phi c quay theo

chiu quy nh t thng ban u trng vi t d, hn na mch ngoi phi khp kn qua 1 in

tr, ni khc i l my ch c kch thch khi c ti. V It = Ir= I nn khi n= Cte ch cn 2 i lng


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bin i l U v I. Do my pht in ny ch c 1 c tnh ngoi U = f(I) , cn cc c tnh khc

ch c th thnh lp c theo s kch thch c lp .

c tnh ngoi (ng cong 2) v phng php suy t tam gic c tnh c trnh by trn

hnh 1-26.

Hnh 1-26. c tnh ngoi camy pht in mt chiu kch thch ni tip

Trn hnh, ng cong 1 l c tnh khng ti, ng 3 l quan h IR = f(I) .Tnh tin tam gic c tnh ABC ng vi Im n v tr ABC sao cho A nm trn c tnh khng

ti th C s nm trn c tnh ngoi. Thay i cc cnh ca tam gic c tnh ng vi tr s dngin I v tin hnh tng t nh trn ta c ton b ng c tnh ngoi .

T c tnh ngoi ta thy in p ca my pht kch thch ni tip thay i rt nhiu theo ti nnthc t loi my ny rt t c s dng .4. c tnh ca my pht in mt chiu kch thch hn hp

My pht in kch thch hn hp c ng thi hai dy qun kch thch song song v ni tipcho nn trong n tp hp cc tnh cht ca 2 loi my ny. Tu theo cch ni, s.. ca hai dy qunkch thch c th cng chiu hoc ngc chiu nhau .

Trng hp sau rt t gp v ch p dng vi nhng mc ch c bit (nh my pht hn).Khi ni thun hai dy qun hai dy qun kch thch, dy qun song song ng vai tr chnh cn dyqun ni tip ng vai tr b li tc dng ca phn ng phn ng v in p ri trn R, nh mmy c kh nng iu chnh t ng c in p trong 1 phm vi ti nht nh .

a. c tnh ngoi U = f(I) khi n= Cte

Hnh 1-27. c tnh ngoi ca my pht inmt chiu kch thch hn hp


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Hnh 1-27 trnh by c tnh ngoi ca my pht kch thch hn hp. Khi ni thun, in p

u cc c gi hu nh khng i (ng 2) . Trng hp b tha in p s tng khi ti tng

(ng 1 ). iu ny c ngha c bit khi cn b hao ht in p trn ng dy ti in gi

cho in p h tiu th in khng i .

Nu ni ngc 2 dy qun kch thch, khi ti tng, in p s gim nhanh hn my pht kch

thch song song (ng 3 v 4) .

Phng php dng c tnh ngoi t c tnh khng ti v tam gic c tnh ging trng hp

my pht kch thch song song.

Hnh 1-28. Xy dng c tnh ngoi ca my pht in mt chiukch thch hn hp bng c tnh khng ti v tam gic c tnh

Trn hnh (1-28), ng cong 1 biu th c tnh khng ti, ng 2 quan h U = I trt v

ng 3 - in p ri trn in tr phn ng IR . Tam gic c tnh trn hnh ng vi trng hp

b tha (xem hnh 1-17). Cho A1B1C1 l tam gic c tnh ng vi2

mI

I . Tnh tin A1B1C1 theo

ng thng 2 sao cho nh C1 chim v tr C1 trn ng thng 2 v nh A1 chim v tr A1 trn

ng 1 th on CG1 = D1E1 l in p ng vi dng in ti bng2

mI

. Nu ABC l tam gic c

tnh ng vi I = Im th tng t ta c CG = DE l in p ng vi ti nh mc . Nu nh C trng

vi M giao im ca ng cong 1 v ng thng 2 th Um = Uo . Khi cn b in p ri trn

ng dy ti in gi cho h dng in nhn c in p nh mc phi tng cng dy qun

kch thch ni tip sao cho in p u cc my pht bng on DE ng vi c tnh ngoi c dng

theo ng nt t .


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c tnh iu chnh ca my pht kch thch hn hp trnh by trn hnh 1-29, trongh ng

cong 1 l c tnh iu chnh khi ni thun hai dy qun kch thch v b bnh thng, ng cong

2 b tha v ung cong 3 khi ni ngc .

Hnh 1-29. c tnh iu chnh ca my pht in mt chiukch thch hn hp

III. MY PHT IN MT CHIU LM VIC SONG SONG

Nhng my pht in c ghp lm vic cng mt ng dy phn phi in nng l mt

i hi thc t nhm bo m an ton cung cp in v s dng kinh t nht cc my pht. Vi mt

h thng my pht nh vy, v d khi ti gim, mt hoc mt s my pht s ngng hot ng cc

my pht cn li lm vic vi cng sut nh mc, do hiu sut s cao. l s lm vic song

song ca cc my pht.

Di y ta s xt cc iu kin cn thit ghp cc my pht lm vic song song v s

phn phi cng nh chuyn cng sut gia cc my pht .

1. iu kin lm vic song song ca my pht in mt chiu

Gi s c hai my pht in mt chiu I v II, trong my I lm vic vi mt ti I no .

Mun ghp my pht II vo lm vic song song vi my pht I cn phi gi ng nhng iu kin

sau :

- iu kin cng cc tnh, ngha l phi ni cc dng ca my II vo cc dng ca thanh gpv cc m vo cc m ca thanh gp .

- S.. ca my pht II phi bng in p U ca thanh gp.

- Nu nhng my lm vic song song thuc loi my pht kch thch hn hp th cn c iu

kin th 3 : ni dy cn bng gia cc im m v n nh hnh (1-30) .


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Hnh 1-30. My pht in mt chiu lm vc song song :

a) my pht kch thch song song; b) my pht kch thch hn hpiu kin th nht cn phi m bo tht cht ch, v nu khng, sau khi ng cu dao ghp

sonh song hai my pht I v II s b ni ni tip nhau thnh mch kn khng qau in tr ca ti, gynn tnh trng ngn mch ca c 2 my. Nu iu kin th 2 khng tho mn th sau khi ghp vo,my II hoc phi nhn ti t ngt (nu E > U) v lm cho in p ca li in thay i hoc lmvic theo ch ng c (nu E < U). S cn thit ca iu kin th 3 c th gii thch nh sau : nukhng c dy cn bng th sau khi ghp song song nu t nhin v mt l do no tc ca 1trong 2 my , v d ca my I tng th s.. E 1 tng do I1 tng. V cc dy qun kch thch songsong v ni tip ca my pht kch thch hn hp thng c ni thun nn khi I1 tng, E1 cngtng v c tip tc nh vy khin cho my I s ginh ly ht ti v b qu ti, ng thi buc my IIgim dn ti v chuyn t ch my pht sang ch ng c. Nu c dy cn bng th s trnhc hin tng trn, v dng in phn ng ca my I tng s c phn phi cho dy qun kchthch ni tip ca c 2 my khin cho s.. ca c 2 my u tng .

2. Phn phi v chuyn ti gia cc my pht in:Sau khi ghp my pht in II lm vic song song vi my pht in I , do E2 = U nn my II

cha tham gia pht in v ton b ti vn do my I m nhim (I1 = I , I2 = 0). Lc c tnhngoi ca 2 my pht in c trnh by nh cc ng 1 v 2 trn hnh (1-31) .

Hnh 1-31. Phn phi ti gia cc my pht in


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Mun my II nhn ti phi tng E2 > U v nh vy c tnh ngoi ca n s tnh tin ln trn(ng nt t 2 ). V dng in ti tng I bn ngoi khng i nn mun gi cho in p U camng in khng i th cng vi vic tng E2 phi ng thi gim thch ng E1 sao cho c tnhngoi ca my I tnh tin xung di n v tr thch ng (ng t nt 1 ), sao cho in p U =Cte ta c I1 + I2 = I. Vic thay i E1 v E2 c thc hin bng cch thay i cc dng in kchthch t It1 v It2 ca mi my hoc bng cch thay i tc quay ca cc ng c s cp ko ccmy pht . Trong thc t vn hnh, ngi ta thng dng phng php thay i dng in kchthch phn phi li ti gia cc my pht. Tuy nhin c 2 phng php trn u khin cho cngsut ca ng c thay i v lc b iu chnh ca chng s tc ng lm thay i lng nhinliu a vo cc ng c s cp .

Nh vy nu mun chuyn ti hon ton t my pht I sang my pht II ch vic tip tc tngE2 v gim E1 ng thi cho n khi E1 = U. lc my pht II hon ton m nhim ti (I2 = I) vc th tch my I ra khi li in. Ch rng nu gim It1 qu nhiu th E1 < U v my I s lmvic ch ng c in tiu th cng sut in ly t my pht II. Nu ng c s cp l ng cnhit hoc ng c thu lc th khng cho php lm vic ch , v s gy h hng ng c scp. Vic iu chnh cc dng in kch thch It1 v It2 phi tin hnh rt chm v lin tc v mt sthay i nh ca cc dng in s lm cho dng in I1 v I2 thay i rt nhiu .

T hnh (1-31) ta thy rng khi lm vic song song trong iu kin lc khng ti s.. E bngnhau v kch thch t khng i, th lc c ti my pht in no c c tnh ngoi cng ( dcnh) s nhn nhiu ti (trng hp my I). Ngc li, my pht in c c tnh ngoi mm ( dc

ln) s nhn t ti (trng hp my II). Tnh trng lm vic nh vy khng li, v vy li dng ttcng sut my cn phi m bo cho c tnh ngoi ca cc my pht in mt chiu lm vic songsong biu th trong h n v tng i hon ton trng nhau. Trong trng hp ti s lun lunt ng phn phi gia cc my pht theo t l cng sut .

V D

V d 1 : cho mt my pht in mt chiu kch thch song song 25 kW, 1800 vg/ph,R = 0,09 , in p ging trn chi than Utx = 2 V, phn ng phn ng lc ti y (I = Im, bqua It ) tng ng vi dng in It = 0,05 A . ng cong t ho ng vi tc nh mc nh sau

It (A) 1 1,5 2 3 4 5 6Uo(V) 134 180 209 237 256 268 279

Tnh :a.in tr ca mch kch t rt .

b.in p khng ti (in tr mch kch t gi khng i) .

Giia. Khi ti y :

Im = I = 230

2500= 108,7 A

E = U + IR = 230 + 108,7. 0,09 +2 = 241,8 V

T ng cong t ho suy ra : It = 3,25 A.Tuy nhin khc phc phn ng phn ng, tn thc t ta phi c :

It = 3,25 + 0,05 = 3,3 A

Vy : rt = 6,693,3

230' tI

U


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30

b. in p lc khng ti Uo l giao im ca ng thng Uo = rtIt = 69,6 It v c tnh khng ti.Bng phng php v ta suy ra giao im ng vi It = 3,56 A v Uo = 247,6 V.

V d 2 : cho mt my pht in kch thch c lp c cc s liu lc ti y U = 220 V, It = 2,5

A = Cte , I = 10 A, n = 1000 vg/ph. s vng dy ca dy qun kch thch wt = 850 .

ng cong t ho 750 vg/ph c cc tr s :

It (A) 1,0 1,6 2 2,5 2,6 3 3,6 4,4

Uo(V) 78 120 150 176 180 193,5 206 225

Tnh :a. in p khng ti n = 1000 vg/ph

b. S ampe-vng kh t ca phn ng phn ng khi ti y .c. in p u cc khi qu ti 25%.

Gii

a. V s.. t l vi tc nn :

750

1000

)750(

)1000(

E

E

E(1000) = 176 235750

1000 V

b. Sc in ng ca my pht khi ti y tc 1000 vg/ph :E = U + IR = 220 + 10. 0,4 = 224 V

tc 750 vg/ph :

E(750) = 168

1000

224750 V

T ng cng t ho ta tm c ng in kch t tng ng It = 2,35 A.

Vy s ampe-vng kh t bng :850.(2,5 - 2,35) = 127,5 A.vg

c. Khi qu ti 25 % phn ng phn ng s tng 25% tng ng :It = (2,5 2,35).1,25 = 0,1875 A

Dng in kch thch c hiu qu bng :

It = 2,5 - 0,1875 = 2,315 A

T ng cong t ho suy ra E(750) = 165 V .Do :

E(1000) = 220750

1651000 V

in p u cc s bng :U = E IR = 220 (10.1,25).0,4 = 215 V


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CU HI :

1. Khi ly c tnh khng ti, trong qu trnh tng in p, c nn gim dng in kch t ri tngtip tc khng ? Ti sao?

2. Vi mt in tr nh hn in tr ti hn rt(th) nu n < nm th trong qa trnh t kch thch camy pht in kch thch song song, in p u cc ca my s ra sao. Trong trng hp nh thno my s khng th t kch c ?

3. Gii thch v sao cc c tnh suy ra t tam gic c tnh hi khc cc c tnh c c t thnghim trc tip v c dng nh cc ng nt t trn hnh v ?

4. Tm cc nguyn nhn khin my pht in kch thch song song khng t kch v to ra cin p ?5. Nu my pht in kch thch song song khng t kch thch c do mt t d th gii quyt

nh th no to ra c in p .6. Khi ti chung khng i nu tng kch thch ca my pht in I m khng gim kch thch

ca my pht in II lm vic song song vi my pht in I th ti s phn phi li gia 2 my nhth no ? in p ca li lc ra sao ?BI TP1. Cho 1 my pht in mt chiu c Pm = 215 kW, Um = 115 V v n = 450 vg/ph. in tr cady qun phn ng v cc t ph bng 0,002 , 2 2 txU V . Cc s liu ca c tnh khng tiv c tnh ngn mch nh sau :

It (A) 5 10 15 20 25 30 35Uo(V) 49 87 108 119,3 125,2 129,5 135

It (A) 0 6Inm(A) 0 Im

a. V tam gic ngn mchb. Dng kch thch ngoi sao cho my y ti U = Um , I = Im , n = nm . Nu b ti i, tnh %U ?

c. Dng kch thch ngoi khin cho khi khng ti U = Um gi It = C

te

th khi I = Im, %U bng baonhiu ?p s : b. 10 %

c. 18,2 %.

2. Hai my pht in kch thch song song c cc s liu sau :

My Pm (kW) Uo (V) N (vg/ph) UmI 20 1000 230 210II 15 1200 240 210

Gi s quan h U = f(I) l ng thng . Tnh :a. Cng sut ca mi my khi ti tng l 20kW v in p lc ?

b. Ti tng ln nht vi iu kin khng my no b qu ti ?p s : a. P1 = P2 = 10 kW ; U = 220 V

b. 53 P kW

3. Cho hai my pht in mt chiu lm vic song song vi ti tng P = 100 kW . Sc in ng cachng E1 = 250 V, E2 = 248 V, dng kch t It1 = It2 = 21 A . in tr ca dy qun phn ng R 1 =R 2 = 0,05 . Hy xc nh in p chung v cng sut ca mi my .

p s :U= 237,42

P1 = 45,25 kWP2 = 54,75 kW


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1.4 M MY V IU CHNH TC NG C IN MT CHIU

I. M my ng c in mt chiu m my ng c in mt chiu c tt, phi thc hin c nhng yu cu sau:

+ Mmen m my (hay khi ng) Mkphi c tr s cao nht c th c hon thnh qu trnhm my, ngha l t c tc quy nh trong thi gian ngn nht.

+ Dng m my (hay khi ng ) Ik phi c hn ch n mc nh nht trnh cho dyqun khi b chy hoc nh hng n i chiu.

Trong khun kh nhng yu cu trn, ngi ta p dng 3 phng php m my sau y :

a. M my trc tip (U= Um)b. M my nh bin trc. M my bng in p thp (U < Um)

Trong tt c mi trng hp, khi m my bao gi cng phi m bo c max ngha l trckhi ng ng c vo ngun in, bin tr iu chnh dng in kch thch phi v tr ng vi tr snh nht sau khi ng cu dao ng c c kch thch ti mc ti a v theo biu thc (1-6)mmen ng vi mi tr s ca dng in I lun lun ln nht. Hn na phi m bo khng xy

ra t mch kch thch v trong trng hp 0 , M = 0, ng c khng quay c, do E = 0 v theo biu thc (1-21) dng in I s rt ln lm chy vnh gp v dy qun .

Khi m my, chiu quay ca ng c in mt chiu ph thuc vo chiu ca mmen. thayi chiu ca mmen c th dng 2 phng php l : i chiu dng in trong phn ng hoc ichiu t thng, c th l dng in kch thch. Vic c th thc hin c bng cch trao i cchni cc u dy qun phn ng hoc cc u dy qun kch thch trc lc m my. Vn ichiu quay ca ng c in lc ang quay v nguyn tc cng c th thc hin c bng c 2

phng php trn.

Tuy nhin, trn thc t ch c dng phng php i chiu dng in phn ng I v dyqun kch thch c nhiu vng dy do h s t cm L t rt ln v vic thay i chiu dng inkch thch dn n s xut hin s.. t cm rt cao, gy ra qu in p nh thng cch in ca dyqun kch thch .

Sau y ta xt cc phng php m my ng c in mt chiu :

1. M my trc tip:Phng php ny c thc hin bng cch ng thng ng c in vo ngun. Nh vy lc

rto cha quay s.. E = 0 v dng in qua phn ng bng :

I =

R

U

R

EU

V trong thc t R* = 0,02 0,1 nn vi in p nh mc U* = 1 dng in I s rt ln vbng (5 10) Im cho nn phng php m my trc tip ch p dng c cho cc ng c in ccng sut vi trm ot. c my ny R tng i ln do khi m my I (4 6) Im . Trongnhng trng hp c bit mi cho php m my trc tip i vi nhng ng c c cng sut vikW .

2. M my nh bin tr:

trnh nguy him cho ng c v dng in m my qu ln, ngi ta dng bin tr mmy Rk, gm c 1 s in tr ni tip khc nhau v t trn mch phn ng (hnh 1-32).


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Hnh 1-32. S m my ng c in mt chiukch thch song song bng bin tr.

Nh vy trong qu trnh m my ta c :

ki

i

RR

EUI

vi i l ch s ng vi th t cc bc ca in tr .

Bin tr m my c tnh sao cho dng in m my Ik = (1,4 1,7)Im i vi cc ng cln v Ik = (2 2,5)Im i vi cc ng c nh. Trc lc m my tip im T nm ti v tr O vcon chy ca bin tr mch kch thch v tr b ( rc = 0). Khi bt u m my, gt T v v tr 1.

Nh cung ng M, dy qun kch thch c t di ton b in p v t thng c tr s cc i

max . Nu mmen do ng c in sinh ra ln hn mmen cn (M > M c ) rto bt u quay vs.. s tng t l vi tc quay n . Do s xut hin v tng ln ca E, dng in phn ng I sgim theo, M gim khin n tng chm hn (hnh 1-33).

Hnh 1-33. Cc quan h I, M v n i vi thi giankhi m my ng c

Khi I gim n tr s (1,1 1,3)Im ta gt T n v tr 2. V mt bc in tr b loi tr, I lptc tng n gii hn trn ca n ko theo M, n, E tng. Sau I, M li gim theo quy lut trn. Lnlt chuyn T n cc v tr 3, 4, 5. Qu trnh trn c lp li cho n khi my t n tc n = n mth Rkcng c loi tr hon ton v ng c lm vic vi ton b in p.

S bin thin ca M, I, n trong qu trnh m my trnh by trn hnh (1-33) cho thy mi khi

loi mt bc in tr, I v M tng vi hng s thi gian T 0 v h s t cm ca phn ng rt b.

Tri li s gim dn ca I v M xy ra chm chp v ph thuc vo s tng s.. E hay tc n,

ngha l ph thuc vo hng s thi gian Tcrt ln ca c khi quay .


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S bc ca in tr m my v in tr ca mi bc c thit k sao cho dng in m my

cc i v cc tiu mi bc u nh nhau m bo cho qu trnh m my c tt nht .

3. M my bng in p thp (Uk< Um)Phng php ny i hi phi dng 1 ngun in c lp c th iu chnh in p c

cung cp cho phn ng ca ng c, trong khi mch kch thch phi c t di in pU = Um ca 1 ngun khc .

y l phng php thng dng hn c trong vic m my cc ng c in cng sut ln ngoi ra cn kt hp vi vic iu chnh tc bng cch thay i in p.

II. iu chnh tc ng c in mt chiu

T biu thc E = Ce n => n =

e

eC

RIU

C

E(1-27)

m M = CM I nn ta c :

n =2

eM

eCC

MR

C

U(1-28)

T biu thc (1-28) ta thy rng vic iu chnh tc ca ng c in mt chiu c th thc hinc bng cch thay i , R v U .

- Phng php iu chnh tc bng cch thay i c p dng tng i ph bin, c

th thay i c tc lin tc v kinh t. Trong qu trnh iu chnh hiu sut = Cte v s iu

chnh da trn vic tc dng ln mch kch thch c cng sut rt nh so vi cng sut ng c. Ch

rng bnh thng ng c lm vic ch nh mc vi kch thch ti a ( = max ) nn ch

c th iu chnh theo chiu hng gim , tc l iu chnh tc trong vng trn tc nh

mc v gii hn iu chnh tc b hn ch bi cc iu kin c kh v i chiu ca my .

- Phng php iu chnh tc bng cch thm in tr ph vo mch phn ng tng R

ch cho php iu chnh tc quay trong vng di tc quay nh mc v lun km theo tn hao

nng lng trn in tr ph, lm gim hiu sut ca ng c in. V vy phng php ny ch p

dng ng c in c cng sut nh v trn thc t thng dng ng c in trong cn trc .

- Phng php iu chnh tc bng cch thay i in p cng ch cho php iu chnh tc

quay di tc nh mc v khng th nng cao in p hn in p nh mc ca ng c in.

Phng php ny khng gy thm tn hao trong ng c in nhng i hi phi c ngun ring c

in p iu chnh c .

Sau y ta s xt c tnh c v cch iu chnh tc ca tng loi ng c in:

1. ng c in kch thch song song hoc kch thch c lp:Vi nhng iu kin U = Cte , It = C

te khi M (hoc I) thay i, t thng ca ng c in

hu nh khng i, v thc ra nh hng lm gim bt t thng ca phn ng ngang trc ca phn

ng rt nh cho nn biu thc (1-28) c th vit di dng :


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35

k

MRnn

o

. (1-29)

V c tnh c ca ng c in kch thch song song l 1 ng thng nh hnh (1- 34). ngc tnh c ng vi trng hp trn mch ca phn ng khng c in tr ph v c gi l ctnh c t nhin .

Hnh 1- 34. c tnh c (v c tnh tc ) t nhin ca

ng c in mt chiu kch thch song song

Do R rt nh nn khi ti thay i t thng n nh mc, tc gim rt t (khong 2 3% tc nh mc ) cho nn c tnh c t nhin ca ng c in kch thch song song rt cng. Vi ctnh c nh vy, ng c in kch thch song song c dng trong trng hp tc hu nhkhng i khi ti thay i (my ct kim loi ...).

a. iu chnh tc bng cch thay i t thng :Nu tng in tr rc trn mch kch thch t (hnh 1-32) ng vi cc tr s khc nhau ca in

3 kch thch ta c cc c tnh c tng ng nh trn hnh (1-35) .

Hnh 1-35. c tnh c (v c tnh tc ) ca ng c in mt chiukch thch song song vi nhng dng in kch thch khc nhau

Cc ng c no ln hn nom v c nghing khc nhau , giao nhau trn trc honh ti

in ng vi dng in rt ln I =R

Utheo iu kin n = 0 ca cc biu thc (1-29) hoc (1-27).

ng thp nht trn hnh ng vi t thngm

. Giao im ca ng mmen cn ca ti

Mc = f(n) vi cc ng trn cho bit tc xc lp ng vi cc tr s khc nhau ca t thng .


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Do iu kin i chiu, cc ng c thng dng hin nay c th iu chnh tc quay bng

phng php ny trong gii hn 1 : 2 . Cng c th sn xut nhng ng c gii hn iu chnh 1 : 5

thm ch n 1 : 8 nhng phi dng nhng phng php khng ch c bit. Do cu to v cng

ngh ch to phc tp khin cho gi thnh ca my tng ln .

b. iu chnh tc bng cch thay i in tr ph Rftrn mch phn ng:

Nu ni thm in tr ph Rfvo mch phn ng th biu thc (1-29) tr thnh :

k MRRnnfo .

(1-30)

Hnh (1- 36) trnh by cc c tnh c ng vi cc tr s khc nhau ca Rf, trong ng vi Rf= 0 lc tnh c t nhin.

Hnh 1-36. c tnh c (v c tnh tc ) ca ng c inmt chiu kch thch song song nhng in tr ph khc nhau

Nu Rf cng ln c tnh c s c dc cng cao v do cng mm hn, ngha l tc s

thay i nhiu khi ti thay i. Cng nh trn, giao im ca nhng ng vi ng Mo = f(n)

cho bit tr s tc xc lp khi iu chnh tc bng thay i in tr ph Rf.

c. iu chnh tc bng cch thay i in p:Phng php ny ch p dng c i vi ng c in mt chiu kch thch c lp hoc

ng c in kch thch song song lm vic ch kch thch c lp. Vic cung cp in p c thiu chnh c cho ng c t 1 ngun c lp c thc hin trong k thut bng cch ghp thnht my pht - ng c c s nguyn l trnh by trn hnh 1-37 .

Hnh 1-37. S t my pht - ng c dng iu chnh tc bng thay i in p ng c in mt chiu kch thch c lp.


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Khi thay i U ta c mt h c tnh c c cng 1 dc (hnh 1-38) : ng 1 ng vi U m,ng 2, 3 ng vi U m > U2 > U3 v ng 4 ng vi U4 > U m .

Hnh 1-38. c tnh c (v c tnh tc ) ng c in mt chiukch thch c lp nhng in p trn phn ng khc nhau.

Ni chung v khng cho php vt qu in p nh mc nn vic iu chnh tc trn tc nh mc khng c p dng hoc ch c thc hin trong 1 phm vi hp. c im ca phng

php ny l lc iu chnh tc , mmen khng i v v I u khng i . S d I khng il v khi gim U, tc n gim lm E cng gim, nn :

te

C

R

EUI

Ngy nay, t my pht- ng c thng dng trong cc my ct kim loi v my cn thp ln a tc ng c vi hiu sut cao trong gii hn rng 1:10 hoc hn na .

2. ng c in mt chiu kch thch ni tip: ng c in mt chiu kch thch ni tip, dng in kch thch chnh l dng in phn ng

It = I = I. V vy trong mt phm vi kh rng c th biu th :

= Ik (1-31)

Trong k l h s t l, l hng s trong vng I < 0,8 I m , khi I > (0,8 0,9)I m th hi gim

xung do nh hng bo ho ca mch t .

Biu thc mmen c dng :

kCICMMM

2

(1-32)

Kt hp vi biu thc (1-28) ta c :

kC

R

MkC

UCn

e

e

M

.

.(1-33)

Nu b qua R th :

M

Un hay

2

2

n

CM (1-34)

Nh vy khi mch t cha bo ho c tnh c ca ng c in mt chiu kch thch ni tipc dng ca ng hypecbn bc hai nh hnh 1-39 (ng 1 ).


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Hnh 1-39. c tnh c ca ng c in mt chiu kch thch ni tip cc trng hp iu chnh tc khc nhau

Ta thy rmg ng c in mt chiu kch thch ni tip, tc quay n gim rt nhanh khi Mtng v khi mt ti (I = 0, M = 0) c tr s rt ln. Do khng c cho loi ng c in ny lmvic nhng iu kin c th xy ra mt ti nh dng ai truyn, v khi xy ra t hoc trt ai

truyn tc quay tng ln rt cao. Thng thng ch cho php ng c lm vic vi ti ti thiuP2 = (0,2 0,25)P m .

Trn thc t do nh hng ca bo ho khi ti tng, tc ca ng c gim t hn theo ng

nt t (hnh 1-39) .

Vi c tnh c rt mm nh vy, ng c in kch thch ni tip rt u vit trong nhng ni

cn iu kin m my nng n v cn tc thay i trong 1 vng rng nh cc u my ko ti (xe

in, u my in ...) .

a. iu chnh tc bng cch thay i t thngT thng ca ng c kch thch ni tip c th thay i bng nhng bin php sau y : mc

sun dy qun kch thch bng mt in tr, thay i s vng dy ca dy qun kch thch, mc sundy qun phn ng, theo cc s hnh (1-40) .

Hnh 1- 40. Cc s iu chnh tc ng c in mt chiu kch thch ni tip

a) mc sundy qun kch thch; b) thay i s vng dy ca dy qun kch thch;c) mc sun phn ng; d) thm in tr vo mch phn ng


43/46

39

Hai bin php u dn n cng mt kt qu. Nu dng in kch thch lc u l It = I th

dng in kch thch sau khi p dng cc bin php trn s gim xung It = kI, trong k l h s

gim :

stt

st

RR

Rk

< 1 nu mc sun dy qun kch thch .

trong Rst l in tr sun .

t

t

w

wk

'

< 1 nu thay i s vng dy ca dy qun kch thch .

Nh vy trong cng thc (1-33) h s k c thay bng kk. . R rng vi cc phng php

trn ch iu chnh cm

v tc s thay i c trong vng

trn nh mc v ng c tnh s nm v pha trn ca c tnh t nhin (ng 2 hnh 1-39) .

Nu dng bin php th ba, mc sun phn ng, th in tr tng ca ton mch s b i, dngin I = It v tng ln v tc quay gim xung. Nh vy phng php ny ch iu chnh c

tc di vng nh mc v ng c tnh c tng ng nm pha di ca c tnh c t nhin(ng 3 trn hnh 1-39). V Rt rt b nn Rs. hu nh c t di ton b in p ca mng incho nn tn hao rt ln v hiu sut ca ng c gim i nhiu. Mt khc hiu qu ca iu chnh tc bng cch tng t thng cn b hn ch bi s bo ho ca mch t nn phng php ny rt ts dng .

b. iu chnh tc bng cch thm in tr vo mch phn ng:

S trn hnh 1-40 ch iu chnh c tc di tc nh mc v km theo tn hao trn

in tr ph lm gim hiu sut ca ng c nn t c ng dng . c tnh c ng vi trng hp

ny c trnh by trn hnh 1-39 ( ung 4 v 5 ).

c. iu chnh tc bng cch thay i in p:

Phng php ny ch iu chnh c tc di tc nh mc v khng cho php tng in

p qu nh mc nhng li gi c hiu sut cao do khng gy thm tn hao khi iu chnh.

Phng php ny c p dng rng ri trong giao thng vn ti v c thc hin bng cch i

ni song song thnh ni tip hai ng c. Nh vy khi lm vic song song, cc ng c s lm vic

in p U = 2

1

U m . c tnh c ca ng c in trong trng hp ny c dng ca ng 6 trn

hnh (1-39) .

3. ng c in mt chiu kch thch hn hp:

ng c in mt chiu kch thch hn hp c ch to sao cho tc dng ca cc dy qun kch

thch song song v ni tip hoc b nhau hoc ngc nhau. Trn thc t , ngi ta ch s dng loi

ng c in kch thch hn hp b v ng c in kch thch hn hp ngc khng m bo c

iu kin lm vic n nh. ng c in kch thch hn hp b c c tnh c mang tnh cht trung


44/46

40

gian gia hai loi ng c kch thch song song v kch thch ni tip . Khi ti tng t thng tng,

do c tnh c ca ng c in kch thch hn hp b mm hn so vi c tnh c ca ng c

in kch thch song song. Tuy nhin mc tng ca khng mnh nh ng c in kch thch

ni tip, cho nn c tnh c ca ng c in kch thch hn hp b cng hn so vi c tnh c ca

ng c in kch thch ni tip.

tin so snh, c tnh c ca cc loi ng c in ni trn c trnh by trn hnh (1-41).

Hnh 1-41. c tnh c ca ng c in mt chiu kch thch hn hpso snh vi cc loi ng c in mt chiu khc

ng 1 : kch thch hn hp bng 2 : kch thch hn hp ngcng 3 : kch thch song songng 4 : kch thch ni tip

Tc ca ng c in mt chiu kch thch hn hp c iu chnh nh trng hp ngc kch thch song song, d rng v nguyn tc c th p dng nhng phng php iu chnh tc

dng cho ng c in kch thch ni t ip .

ng c in mt chiu kch thch hn hp c dng trong nhng ni cn cc iu kin

mmen m my ln, gia tc quay khi m my ln, tc bin i theo ti trong mt vng rng nh

my p, my bo, my nng ti ... Thi gian gn y, ng c kch thch hn hp cn c dng

trong giao thng vn ti v c u im hn so vi ng c kch thch ni tip ch d hm bng ch

pht in tr nng lng v li in .

V D :Cho 1 my pht in kch thch song song c P m =27 kW, U m = 115 V,

n m = 1150 vg/ph, It = 5 A, hiu sut m = 86 % . in tr trong mch phn ng R = 0,02 ,

2txU = 2 V .

a. Nu em dng nh ng c in (b qua tc dng phn ng phn ng ) vi U m = 110 V,Pm = 25kW, = 0,86, tnh tc n ?

b. S bin i ca tc khi ti y n khng ti .


45/46

41

Gii

a. Khi lm vic nh my pht in th :EF = U + I(F)R + 2 txU

vi : I(F) = tm

mI

U

P = 2405

115

27000 A

=> EF = 115 + 240.0,02 + 2 = 121,8 VKhi lm vic nh ng c in :

E = U - I()R - 2 txU

vi : I() = tm

mI

U

P

= 3,259

115

1105

110.86,0

25000 A

=> E = 110 259,3.0,02 2 = 102,3 V

V : 8,121

3,102

FFe

e

F

nC

nC

E

E

V gi thit rng :

5

7,4

tF

t

F

I

I

Ta c : 10308,4

5.

8,121

3,1021150

n vg/ph

b. Khi ng c lm vic khng ti : I 0 nn Eo = U = 110 V = Ce

no v ta c th vit :

3,102

110

O

O

n

n

E

E

=> 11053,102

1101030

3,102

110

Onn vg/ph

CU HI :

1. iu kin lm vic n nh ca cc ng c in . So snh cc loi ng c in v phng

din ny ?2.Phn loi ng c in mt chiu ?

3. So snh cc c tnh ca ng c in mt chiu ?

4. Hin tg xy ra khi m my ng c kch thch song song trong trng hp mch kch thch

b t. Cng nh vy trong trng hp in tr iu chnh trn mt kch thch rc qu ln ?

5. Nu chi in t khng ng v tr m b x dch ngc chiu quay ca roto th tc ca

ng c in s nh th no ?


46/46

BI TP

1.Cho mt ng c in kch thch song song vi cc s liu sau : P m =95 kW, U m = 220 V,

Im = 470 A, Itm = 4,25 A, R = 0,025 , nm = 500 vg/ph .

Hy tnh :

a. Hiu sut ca ng c.

b. Tn hao ng trong my, tn hao khng ti v dng in khng ti .

c. Mmen ca ng c .

d. Tr s ca dng in ti hiu sut cc i .

e. in tr iu chnh Rfcn thit ng c quay vi n = nm, I = Im v t thng gim i 40% .

f. in tr Rf c n = nm, I = 0,85 Im v t thng gim i 25%.

p s : a. 91,8 %

b. P = 5,42 kW

Po = 2,04 kW, Io = 13,5 Ac. M = 1814 N.m

d. I = 349 A

e. Rf= 0,18

f. Rf= 0,136

2. Cho ng c in kch thch song song Pm =17 kW, U m = 220 V, nm = 1150 vg/ph , R =

0,1 , Mt = 12 kG.m. Hy tnh :

a. Cng sut in t Pt v dng I .

b. in tr iu chnh Rc mc trn mch phn ng ng c quay vi tc 500 vg/pt khi

mmen ti khng i (b qua phn ng ca phn ng).

p s : a. Pt = 14,15 kW , I = 66,3 A

b. Rc = 1,82 .

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