tong hop he dien co

7/29/2019 tong hop he dien co

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Thi Nguyn, 3/22/2013 2

Tnghphincl mn hc chuyn ngnh ca ngnh Tngha, y c th

xem l kinthctnghpcanhiuhcphncsngnh v chuyn ngnh. Trongnhiu

nm qua, c kh nhiu ti liu trong v ngoi nccpnkinthcthuclnhvc

ny, tuy nhin mtgio trnh chun v y ph hpvichngtrnh otoksin

chuyn ngnh T ng ha xnghip cngnghip ca Trngi hcK thut Cng

nghip cn thiu.

pngyu cuoto,cbitl oto theo hthngtn ch,B mn T

nghaKhoain binson xong gio trnh hcphnTnghphinc1 v

csdngc hiuqu cho vicgingdyca gio vin v vichctpca sinh vin

ihc v hc vin cao hc. HcphnTnghphincny l hcphn th2 ca

mn hc,hcphn ny c phn bchngtrnh l 3 tn chvini dunggm 5 chng.

Vi thi gian ngn,iukinv ti liu tham khobhnchnn gio trnh ny

chcchn cn nhiuthiu st, mongbnc v cc ngnghip thngcm v cho kin

nggp chng ti hon thinni dungca gio trnh ny ctthn.

Xin chn thnh cm n!


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- Sch, gio trnh chnh:

Gio trnh Tng hp h in c do b mn T ng ho XNCN -Khoa in bin son (ang bin son).

- Sch tham kho:[1]. Trn Th, V Quang Lp (bin kho); C s iu khin t ng truyn

ng in; NXB Khoa hc v k thut, H Ni, 2004

[2]. Bi Quc Khnh, Nguyn Vn Lin, Phm Quc Hi, Dng Vn Nghi;iu chnh t ng truyn ng in; NXB Khoa hc v k thut,H Ni, 2002.

[3]. Bi nh Tiu; C s truyn ng in t ng; NXB Khoa hc v kthut, H Ni, 1984.

[4]. V Quang Lp, Trn Xun Minh; Gio trnh K thut bin i; Trngi hc K thut Cng nghip, Thi Nguyn, 1998.

[5]. Cyril W. Lander; Power Electronics; 1993.

[6]. . . , . . , . . ,. . ;

; ,, 1974.


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1.1. Khi qut v h thng tu ng

1.2. o kim tn hiu v tr

1.3. H thng tu ng v tr sensin v phng php thit k

1.4. Hiu chnh trng thi ng ca h thng ty ng v tr


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1.1. KHI QUT VHTHNGTUNG

1.1.1. ngdngcahthngtungHthng ty ng,thccht l hthngCTTthchiniukhinv tr

vilngttrcbin thin ty . Hthngtungcngdngrtrng

ri trong thctin.Nhimvcbncah l thchiniukhinccuchp

hnh bm st chnh xc ivilngtv tr, ilngiukhin(lngu

ra) thng l v tr khng gian caccusnxut, lc lngt thay i trong

qu trnh lm vic th h thng c th lm cho ilngiukhin bm st v

duy tr mtmt cch chnh xc v tr caccusnxut theo yu cu. V diu

khinccu p trc cn, trong qu trnh cn kim loi,phi lm cho khe hgia

hai trc c thtin hnh tiuchnh; iukhinqu tch gia cng ca my ctiukhins v iukhin bm ca my chp hnh; ccu li tng trn tu

thuyn; ccuiukhin anten raacacm sng pho hay knh vinvngin

tnhmngmc tiu; iukhinng tc cangi my. Nhng v d trn

yu l nhngngdngcthvhthngiukhintungv tr.


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1.1.2. Cc bphnchyucahthngtungvtr

v nguyn l lm vicca n


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(1) Bokimvtr: Do chit p RP1 v RP2 to thnhbokimv tr (gc), trong trc quay cachit p RP1 nivi bnh iukhin lm gc cho trc (gc

t), trc quay cachit p RP2 thng qua ccunivibphnph ti lm

phnhi gc quay, haibchit p uccpinnhnguninmtchiu

U, nhvy c thchuyn tham sv tr trctip thnh ilnginu ra.

(2)Bkhuchiso snh in p: Do 2bkhuchi 1A, 2A to thnh, trong b

khuchi 1A ch lm nhimvo pha, cn 2A c tc dng so snh v khuch

iin p, tn hiuu ra lm tn hiuiukhinbkhuchi cng sutcp

tip theo, ngthi c khnngnhnbitcc tnh in p (phngv m dngca gc pha).


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(3)Bkhuchicng sutochiu: cung cp cho ngcchp hnh cah

thngtungch c khuchiin p l cha, cnphikhuchi cngsut, ccbkhuchi cng sut ny thng dng chnhluiukhinhocb

bini xung p iuchrng xung nungctruynng l ngcmt

chiu,trnghp dng ngcxoay chiu thbkhuchi cng sutthng l

bintn.

(4) Ccuchp hnh: ngcbm (ngcinmtchiu,ttrngvnhcu)truynngccuchp hnh mangphti (dn anten ra a),giangcv

phtithng cbphntruynlc(hpgimtc).

(1), (2), (3), (4): L cc bphnchyu, khng th thiu to nn h thngiukhintungv tr, ch c linh kinhocthitbcth l c th khc nhau, v

d, c th dng cc bokimv tr khc nhau, dng ngcinmtchiu

hoc xoay chiu v.v...


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1.1.3. So snh hthngtungvtrvihthngiutc

Qua cc phn tch trn d dng nhn ra nhngch khc nhau v ging nhau giahthngtungv tr (sau ygitt l hthngtung) v hthngiu

tc. C hai u l hthng kn (cphnhi),tc l thng qua vic so snh lng

u ra cahthngvilng cho trc(lngt)to ra tn hiuiukhin

hthng, v vy nguyn l ca hai hthng ny l ging nhau.

ilng cho trccahthngiutc l hngs, d cho mcnhiunh

th no, u mong ilngu ra nnh, v thchtlngchngnhiucah

thng lun t ra quan trngnht. Cn trong hthngtung th tn hiutv

tr l thng xuyn thay i, l ilngthayitu, yu culngu ra

bm chnh xc theo s thay ica lng cho trc, tnh nhanh nhy, tnh linhhot, tnh chnh xc thch nghi u ra trthnh c trngchyucah thng

tung. Hay ni cch khc chtlng bm l ch tiu chyucahthng ny.


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T hnh 6.1 c ththy,hthngtung c th xy dng trn cshthngiutc ci thm mch vng v tr, mch vng v tr l ctrngcu trc chyu

cahthngtung. V vyhthngtungthngphctphnhthng

iutc.


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1.1.3. Phn loihthngtungvtr

Hnh 1.2: Cu trc in hnh ca h ty ng kiu m phng


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Hthngtungiukhin gc phakius:

S liucho trc

DA

BNhn din

pha

Biu khin

tc

ng cchp hnh

Bn my

o v tr(Thit b ng b cm ng)

Hnh 1.3: Cu trc h thng ty ng iu khin gc pha


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Hthngtungiukhin xung s:

S liucho trc

DA

BThut tono chiu

Biu khin

tc

ng cchp hnh

Bn my

o kim v tr(Thit b quang tr)

Hnh 1.4: S nguyn l h thng ty ng iu khin xung s

D*

D


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Hthngtungiukhinkium s:

S liucho trc

DA

BThut tono chiu

Biu khin

tc

ng cchp hnh

Bn my

o kim v tr(a m quang in)

Hnh 1.5: S nguyn l h thng tu ng iu khin m s

D*

D


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1.2 OKIM TN HIUV TR

1.2.1. Sensin (SS)

Hnh 1.6: Nguyn l cu to sensin

Cun dyrotor

Cun dystator Sensin l mtbcmbinchuynv gc S,

trong h thng tu ng thng dng theo

tngcpi nhau. Sensin c rotornivib

phn iu khin th c gi l sensin pht

(my pht tin), cn sensin c rotor ni vi

trcccuchp hnh th c tn l sensin thu

(my pht tin).

Sensin mt pha: Gmmtcun dy kch tmt pha v mtbcun dy chnhbc,cundy kch tlpt trn rotor, cct kch thch thng lm thnh dngccn,nhvy cth lm cho trkhng u vo khng thay i theo v tr ca rotor, cun dy chnhbc lcun dy 3 pha, thngcqunri,b tr trn stator, lch pha nhau 120o, u theo kiuY.


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1.2.1. Sensin (SS)

Hnh 1.7: S nguyn l ni sensin kiu iu khin

uss = Ussm sin(t - + 900) cos(1 -2 ) (6.1)

uss = Ussm sin sin(t - + 900) (6.2)

Hnh 1.8: V tr gc ca my t chnh gc


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1.2.2. Bbin p quay (Br)

Hnh 1.9: B bin p quay

Ubr

R1

R2

S1

S2U2

U1


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1.2.2. Bbin p quay (Br)

ubr(t) = m[u1(t)cos + u2(t)sin] = mUm cos(ot +) (6.3)

Ubrm = kUfm cos( - 90o) = kUfm sin (6.4)

Hnh 1.10: Thit b o kim sai s gc do bin p quay to thnh


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1.2.3. Bngbcmng(BIS)

Hnh 1.11: B ng b cm ng kiu ng thng


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1.2.4. a m quang in

1.2.4.1.amkiu gia s

Hnh 6.12: Nguyn l lm vic a m quang kiu gia s

a) S nguyn l; b) th lng ra


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1.2.4. a m quang in

1.2.4.2.a

mkiu

tr

tuyt

i

Hnh 1.13: a m quang in tr tuyt i

a) a m ch nh phn b) a m ch tun hon


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1.2.4. a m quang in

1.2.4.2.amkiutrtuyti

Bng 1.1: Bng i chiu v tr trc a m quang in v m s


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1.3 HTHNGTUNGV TR SENSIN V PHNG PHP THITK

1.3.1. Cuto v m hnh ton hccahthngtungvtr sensin

1.3.1.1. Sensin

Hnh 1.14: H thng tu ng v tr sensin


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1.3.1.1. Sensin

a) Quan hgia bin in p u ra vim; b) Scu trc trngthi ngca sensin

Hnh 1.15: Khu o kim sai s gc bng sensin


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1.3.1.2.Khuchinhy pha


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1.3.1.3.Khuchicng sutochiu

C th lbchnhluiukhin c o dng hoc ccb xung in p

ochiu,

Vichnhluiukhin th hm truyncabKCS:

bK

s 1


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1.3.1.4.Ccuchp hnh

L ngctruynng, cn gi l ngcbm.

Thngsdngngcmtchiu, xoay chiu, cc ngccbit,

khi dng ngcmtchiu th hm truyncangcl:

ngin ha gnng l:

e

2

m e m

1/ C

T T s T s 1

e

m

1/ C

T s 1


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1.3.1.5.Bgimtc

Khu nigiangcv ccusnxut,nuccusnxutchun

ng tnh tin th n cn c nhimvbinidngchuynng. Cu

trc v hm truyncabgimtcttc gc sang gc quay l:

Hnh 1.17: S cu trc trng thi ng b gim tc


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Scu trc trng thi ngca tonbhthng ty ngv tr sensin

c m t trn hnh 1.18, trong Rp lbiuchnhv tr. Cch thit

k n cng cc tham sca n sc trnh by cthphn sau.

Hnh 1.18: S cu trc trng thi ng h thng ty ng v tr sensin

M hnh cu trc trng thi ng


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1.3.2. Phn tch sai s trng thi nnh v tnh ton tham scah

thngtungvtr sensin

1.3.2.1. Sai snguyn l (sai shthng)

Hnh 6.19: Cu trc h thng ty ngBng6.2: Phm vi saiscc loi linh kinokim


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thngtungvtr sensin

1.3.2.1. Sai snguyn l (sai shthng)

Hnh 1.20: Cc dng tn hiu vo in hnh

a) u vo l v tr b) u vo l vn tc c) u vo l gia tc


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thngtungvtr sensin1.3.2.1. Sai snguyn l (sai shthng)

Sai s h thng vi h loi I:


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Sai s h thng vi h loi II:

2

s 2 2s 0

1 s D(s)e lims 0s s D(s) KN(s)

2

sv 2 2s 0

1 s D(s)e lims 0

s s D(s) KN(s)

2

sa 3 2s 0

1 s D(s) 1e lims

s s D(s) KN(s) K


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H s phm cht tc Kvv h s phm cht gia tc Ka:

*

mv

sv

Ke

*

m

a

sa

Ke

*

mv * s 0

m

2

s 0

K lims 1 W(s)1

lims

s 1 W(s)

*

2ma * s 0

m

2s 0

K lims 1 W(s)1

limss 1 W(s)


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thngtungvtr sensin1.3.2.2. Sai snhiu

Hnh 1.21: Tc ng ca nhiu ln h thng ty ng


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Ta c:m 2

c d e 1 2

(s) W (s)

I (s)R (T s 1) 1 W (s)W (s)

Hnh 1.22: nh hng ca nhiu phti i vi h thng ty ng v tr

Hnh 1.23: S cu trc trng thi ng trng thi nhiu ph ti


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Sai s do nhiuphti Ic=const

Hthngloi I:

Hthngloi II: ec = 0

c d

c1

I R

e K


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1.4. HIUCHNHTRNG THI NGCAHTHNG TY NGV TR

1.4.1. Hiuchnhnitip(hiuchnhbngbiuchnh)

Scu trc

Hnh 1. 25: S cu trc h thng tu ng mt mch vng v tr


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Lm gnngmus hm truynitngiukhin:

TmTes2 + Tms + 1 TmTes2 + (Tm + Te)s +1 = (Tms + 1)( Tes + 1)

iukingnng l:

Te Tm / 10

Hm truynitngngin ha:

tt

p m

KW (s)

s(T s 1)(T s 1)


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a) S mch in

Hnh 1.26: B iu chnh PID

b) c tnh tn bin logarit


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Ta cphng trnh vit cho mchindng ton t Laplace:

Trong :


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Sau khibini, ta tm c hm struyncabiuchnhv tr:


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Hnh 1. 27:S cu trc h thng tu ng hiu chnh ni tip bng b PID



Do :

Trong :

Ta c cu trc hthngtung cbiuchnhv tr PID:


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1.4.2. Hiuchnh song song (phnhi)1.4.2.1.Hiuchnh song songbngphnhimtc

Ta c scu trc trng thi ng:

Hnh 1.28: H thng ng c hiu chnh mc song song


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1.4.2. Hiuchnh song song (phnhi)1.4.2.1.Hiuchnh song songbngphnhimtc

Ta c scu trc trng thi ng:

Hnh 1. 29: S c cu trc trng thi ng h thng tu ng c phn hi m tc quay


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2.1. Loi hnh c bn ca h iu tc ng c khng ng b

2.2. H thng kn iu tc bng iu chnh in p xoay chiu-h thng iu tc kiu tiu hao cng sut trt


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2.1. LOI HNH CBNCAHIUTCNGC KHNG NGB

Cc loihthngiutcngckhng ngb:

1. iutcgimin p;

2. iutcsdngb ly hptrtint;

3. iu tc in tr ni tip mch rotor ng c dy

qunngckhng ngb;

4. iu tc ni cp ng c khng ngb rotor dy

qun;5. iutc thay isicc;

6. iutcbintn v.v...


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2.1. LOI HNH CBNCAHIUTCNGC KHNG NGB

Cc hiu tc ng ckhngngb theo gc chuyn inng

lng:

1. H thng iu tc tiu hao cng sut trt - ton b cng sut trt

chuyn thnh nhitnng tiu hao mt.

2. Hthngiutckiu ti sinh - mtbphnca cng suttrtb tiu

hao i,phnln cn linhc thitbchnhluctrvmngin

hocchuyn ho thnh dngcnng dng vo vic c ch khc.

3. H thngiu tc cng sut trt khng thay i - trong h thng nykhng trnh khi tiu hao cng sut trn vng ng rotor, nhngs tiu

hao cng suttrthunh khngphthuc vo tc cao hay thp, v

thhiusut kh cao.


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2.2. HTHNG KN IUTCBNGIUCHNHIN P XOAY CHIU -

HTHNGIUTCKIU TIU HAO CNG SUTTRT

Hnh 2.1: S nguyn l iu tc bin p ng c khng ng b

BATN -Bin p t ngu; KBH-B in khng bo ho;

BXC-B iu p xoay chiu dng triac


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HTHNGIUTCKIU TIU HAO CNG SUTTRT

Cc h iu tc ng ckhng ngb theo gc chuyninng

lng:

Hnh 2.2: Mch in tng ng trng thi n nh ng c in cm ng


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HTHNGIUTCKIU TIU HAO CNG SUTTRT2.2.1. ctnh ccangckhng ngb khi thay iin p

Phng trnh c tnh c in:

Phng trnh c tnh c:

11 2

2 2 221 1 1 2

UI IR

(R ) (L L )s

2

p p 1 22m 2t 2

2 2 21 1 21 1 1 1 2

3n 3n U R / sP RM I

s R(R ) (L L )

s


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HTHNGIUTCKIU TIU HAO CNG SUTTRT2.2.1. ctnh ccangckhng ngb khi thay iin p

H s trt ti hn:

M men ti hn (m men cc i):

2t

2 2 2

1 1 1 2

RsR (L L )

2

p 1

t2 2 2

1 1 1 1 1 2

3n UM

2 R R (L L )


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2.2. HTHNG KN IUTCBNGIUCHNHIN P XOAY CHIU -HTHNGIUTCKIU TIU HAO CNG SUTTRT

2.2.1. ctnh ccangckhng ngb khi thay iin p

Hnh 2.3: ng c tnh c ca ngc in cm ng vi cc gi tr khc

nhau ca in p trn stator

Hnh 2.4: ng c tnh c ca ng cin cm ng vi rotor in tr cao khi thay

i gi tr in p trn stator


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2.2.2. H thng kn iu chnh tc bng thay i in p

v ctnh tnhcah

Hnh 2.5: H thng kn iu chnh tc bng cch thay i in p c phn hi m tc a) s nguyn l; b) c tnh tnh


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2.2.2. H thng kn iu chnh tc bng thay i in p

v ctnh tnhcah

Hnh 2.6: S cu trc trng thi tnh h thng iu tc ng ckhng ng b bng thay i in p stator


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2.2.3. Scutrc trngthi nggnng

Hnh 2. 7: S cu trc trng thi ng gn ng ca h thng iu tcng c khng ng b bng thay i in p stator


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Hm truynBB:

Hm truynphnhitc:

Hm truynngc:

Vi: v

bb

KW (s)

s 1

fn

fn

W (s)T s 1

A 1 KK 2

1 21A m

2 2

p 1A

2s K1W (s)

J RU T s 1s 1

3n U

1 AA 1K

1A 1A

22sK

U U

2

1 2m 2 2

p 1A

J RT

3n U


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Hnh 2.8: S cu trc trng thi ng tuyn tnh ho gn ng ca ng c KB


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3.1. Phng trnh iu khin c bn ca h thng truyn ngbin tn ng c khng ng b ba pha

3.2 Cc b bin tn kiu tnh

3.3. Nghch lu iu ch rng xung hnh sin (spwm)

3.4. c tnh c ca ng c khng ng b ch tnh khi iu khinphi hp tn s v in p

3.5. H thng h iu tc bin tn iu khin t s in p / tn s khng i

3.6. H thng kn iu tc bin tn iu khin tn s trt3.7. M hnh ton hc nhiu bin s ca ng c khng ng b ba pha

v php bin i to

3.8. H thng iu tc bin tn iu khin vector


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3.1. PHNG TRNH IUKHINCBNCAHTHNGTRUYNNGBINTN-NGC KHNG NGB BA PHA

Gi trhiudngcascinngcmngcami pha statorngc

khng ngb ba pha l:

Ef= 4,44f1N1kN1m

Trong :Ef l gi tr hiu dng ca sc in ng cm ng do t thng khe h

khng kh trong mi pha statorngckhng ngb ba pha gy ra,

nvo l V;

f1: l tnsmch stator, nv do l Hz;N1 l s vng quncamicun dy mi pha stator;

kN1 l hscun dy ivi sng cbn;

m l t thng khe hkhng kh micc,nv l Wb.


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3.1.1. iutcthphntnscbn

Quy lutiuchnh:

f

1

Econst

f

1

1

Uconst

f

Hnh 3.1: c tnh iu khin t s in p v tn s hng s:

a) khng b st p mch stator; b) c b st p mch stator


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3.1.2. iutccao hntnscbn

Hnh 3.2: c tnh iu khin iu tc bin tn

ng c khng ng b


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3.2 CC BBINTNKIUTNH

3.2.1. Thitbbintn gin tip(thitbbintn xoay chiu-mtchiu-xoay chiu)

Hnh 3.3: Thit b bin tn gin tip


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3.2 CC BBINTNKIUTNH

3.2.1. Thitbbintn gin tip(thitbbintn xoay chiu-mtchiu-xoay chiu)

Hnh 3.4:Bbintn gin tip c khu trung gian mtchiu

a)Bintn dngchnhluiukhinbngtiristor

b)Bintn dngchnhlu khngiukhin c thm bbini xungin p

c)Bintn dngchnhlu khngiukhinvinghchluiuchPWM


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3.2 CC BBINTNKIUTNH

3.2.2. Bbintntrctip (xoay chiu- xoay chiu)

Hnh 3.5: Thit b bin tn trc tip (xoay chiu -xoay chiu)

Hnh 3.6: S nguyn l b bin tn trc tip


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3.2 CC BBINTNKIUTNH

3.2.2. Bbintntrctip (xoay chiu- xoay chiu)

Hnh 3.7: th in p u ra ca thit b bin tn xoay chiu -xoay chiu hnh sin


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3.2 CC BBINTNKIUTNH

3.2.3. Bbintnngunin p v ngun dng in

Hnh 3.8: Cu trc bin tn xoay gin tip

a) B bin tn ngun in p b) B bin tn ngun dng in


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3.3.NGHCHLUIUCHRNG XUNG HNH SIN (SPWM)

Hnh 3.9: Cu trc bin tn gin tip bng tiristor thng s dng

Hnh 3.10: Cu trc bin tn gin tip vi nghch lu PWM


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8.3.1. Nguyn l lm viccabnghchlu SPWM

Hnh 3.11: Th tspxp cc xunghnh ch nht cng bin tngngvi sng hnh sin:

a) Sng hnh sin;b) th sng tng ng caSPWM


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Hnh 3.12a: S nguyn l mch lc khi nghch lu SPWM


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Hnh 3.12b: S khi mch khng ch nghch lu SPWM


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Hnh 3.13: Phng php iu ch rng xung v th vi trng hp iu ch mt cp

a) Sng mang tam gic v sng iu ch hnh sin; b) th sng u ra SPWM


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Hnh 3.14: th xung in p u ra ca SPWM khi iu khin mt cc


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Hnh 3.15: th sng u ra 3pha khi nghch lu SPWM kiu

hai cc:a) Sngiuch3 pha v sng titam gic;

b) uA0= f(t);

c) uB0= f(t);

d) uC0= f(t);

e) in p dy uAB= f(t)


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Nd

i 1k 1 i 1

2U 2i 1 k u(t) sin(k )sin sin k t

k 2 N 2

Nd i

1mi 1

2U 2i 1U sin( )

2 N 2

N N N2d m m m

1mi 1 i 1 i 1

d

N

mi 1

2U U 2U 2U2i 1 2i 1 2i 1 1U sin( ) sin sin ( ) 1 cos(2i 1)

2N N U 2N N 2N N 2 N

1U 1 cos(2i 1)

N N


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3.3.2. iuchngb v khng ngbcabnghchlu SPWMiuch phn onngb:

Hnh 3.16: Quan h ng cong gia fm v ftkhi iu ch phn on ng b


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3.3.3. M thciuchca SPWM v cch thchin n

Hnh 3.17: Phng php ly mu t nhin to thnh sng ca SPWM


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Hnh 3.18: Phng php ly mu quy tc to sng SPWM

a) Php ly mu quy tc I, b) Php ly mu quy tc II


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Hnh 3.19: th sng SPWM 3pha


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3.3.3.3.Phngphp loib sng hi chnh(Harmonic Elimination Method)

Hnh 3.20: th sng SPWM kiu mt ccvi 3 xung trong mt na chu k


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3.3.4. Bnghchluspwm iukhin vng tr(Hyserresis-band) dngintnsngctcao

Hnh 3.21: S khi b iu khin dng in mt pha b nghch lu SPWMiu khin vng tr dng in: DHC l b so snh vng tr


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3.3.4. Bnghchluspwm iukhin vng tr(Hyserresis-band) dngintnsngctcao

Hnh 3.22: th dng in v in p u ra ca b nghch lukhi iu khin vng tr dng in


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3.4. C TNH CCANGC KHNG NGBCHTNH KHIIUKHINPHIHPTNS V IN P

Ta c:

Khi s rtnh:

Khi s tingnn 1:

21 1 2t p 2 2 2 2

1 1 2 1 l 2

U s RM 3n ( )

(sR R ) s (L L )

21 1t p

1 2

U sM 3n ( )

R

21 1 2t p 2 2 , 2

1 1 1 l1 l2

U R 1M 3n ( )

ss R (L L )


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Hnh 3.23: c tnh c ca ng ckhng ng b ba pha khi tn s v

in p khng i

Hnh 3.24: c tnh c iu tc bintn khi iu khin t s in p v tn

s khng i


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3.4.2. ctnh ckhiphihpiukhin in p v tns

Khi U1/1 = const:

2

1 1 2c p 2 2 2 2

1 1 2 1 1 2

U s RM 3n

(sR R ) s (L L )

2 t1 2

1p

1

R Ms

U3n

2

1t p

21

21 11 2

1 1

U3 1M n

2R R

(L L )


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S thay thmt pha ngc:

Hnh 3.25: Mch in tng ng trng thi n nh cang c in cm ng v sc in ng cm ng


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Khi Ef/1 = const:

Khi s rtnh:

Khi s tingnn 1:

f2

2

221 2

EI

R( L )

s

2

2

p f 2 f 1 2t p2 2 2 2 '2

1 1 2 1 22 '221 2

3n E R E s R M 3ns R s LR

Ls

2

f 1t p

1 2

E sM 3n s

R

2

f 2t p

1 1 2

E R 1M 3n

s L s

2

ft p

1 2

E3 1M n

2 L


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Khi Er /1 = const:r

2

2

EI

R / s

2

2p r 2 r 1

t p2

1 1 22

3n E R E sM 3n

s RR

s

r 1 1 1 rmE 4,44f N kN


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Hnh 3.26: c tnh c cc ch iu ch phi hp in p - tn s khc nhau:

a) iu khin U1/1 = const; b) iu khin Ef/1 = const; c) iu khin Er/1 = const


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3.5. HTHNGHIUTCBINTNIUKHINTSIN P / TN

S KHNG I3.5.1. Hthngiutcbintn gin tipngunin p

Hnh 3.27: H thng h iu tc bin tn gin tip ngun in p -ng c khng ng b

GI- b tch phn tn hiu t; GAB - b bin i tr tuyt i


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Hnh 3.28 S nguyn l b tch phn tn hiu t


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Hnh 3.29: th in p ca b tch phn tn hiu t


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Hnh 3.30: B bin i gi tr tuyt i


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Hnh 3.31: Khu iu khin in p ca h thng

iu tc bin tn ngun p


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Hnh 3.32: B to hm TH: a) s nguyn l; b) c tnh u vo u ra


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Hnh 3.33: Khu iukhintnskhinghchlu tiristor

UFC:Bchuyniin p tns; PPX:B phnphi xung;

DPI:Bnhnbitcc tnh; KX: bkhuchi xung;

RF:Biuchnhkhiiuchnhtns


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Hnh 3.34: B nghch lu ngun p - ng c khng ng b


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3.6. HTHNG KN IUTCBINTNIUKHINTNSTRT

3.6.1. Khi nimcbnviukhintnstrtPhng trnh chuynngcbn:

Vi:

t c

p

J dM M

n dt

f f2

2 2 2

2 1 2222

2 22 2 2 22

2 1 22

2

E sEI '

R (s L )R( L )

sR / s R

cosR (s L )R

( L )s

'

t m m 2 2M C I cos


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3.6.1. Khi nimcbnviukhintnstrtTa c:

V:

ts = s1, ngthinhngha l tnstrt, th:

f 2t m m 2 2

2 1 2

sE RM C

R (s L )

f 1 1 N1 m 1 1 N m 1 1 N1 m

4, 44 1E 4,44f N K N K N K

2 2

2 1 2

t m 1 N1 m 2 2

2 1 2

s R1M C N K

R (s L )2

2 s 2t m m 2 2

2 s 2

RM K

R ( L )

2 st m m

2

M KR


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3.6.2. Quy lutiukhintnstrt

Hnh 3.35: c tnh Mt= f(s) iu khin theogi tr m = const


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f f1 2 2 2 0

2 1 m1 2

E EI I I ; I ; I

R j Lj L

s

.

21 m 2

1 f f2 21 m

1 2 1 m 1 2

21 m 2

2 s m 20 0

2 2 s 21 2

R

j (L L )1 1 sI E ER Rj L

j L j L ( j L )s s

Rj (L L )

R j (L L )sI IR

R ' j Lj Ls

2 2 2

2 s m 21 0 2 2 2

2 s 2

R (L L )I I

R L


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Hnh 3.36: ng cong hm s I1 = f(s), khi duy tr m= const


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3.6.1. Khi nimcbnviukhintnstrt(1) Sdngbbin tnngun dng, lm cho i tngiukhin c kh

nng thch nghi nhanh trng thi ng v tin cho vic hm ti sinh,

l cs nng cao chtlngngcahthng.

(2) Cnggingnhhthngiutc hai mch vng kn cangcin

mt chiu, mch vng ngoi l vng tc quay, mch vng trong l

vng dng in. u ra cabiuchnhtc quay R l gi tr chotrccatnstrtidin cho m men t.


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3.6.1. Khi nimcbnviukhintnstrt


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Hnh 3.37: S cu trc h thng iu tc bin tn iu khin tn s trt


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Hnh 3.38: c tnh lm vic trn bn gc phn t ca h thng iu khin tn s trt


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3.7. M HNH TON HCNHIUBINSCANGC KHNG NGB

BA PHA V PHP BINITO

3.7.1. Tnh cht ca m hnh ton hc trng thi ng ca ng c

khng ngb

Hnh 3.39: S cu trc iu khinnhiu bin ca ng c khng ng b

Hnh 3.40: S cu trc iu khin h thng iutc bin tn ca ng c khng ng b nhiu bin


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BA PHA V PHP BINITO

3.7.1. Tnh cht ca m hnh ton hc trng thi ng ca ng c

khng ngb

Hnh 3.41: M hnh vt l ng c khng ng b 3 pha


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BA PHA V PHP BINITO

3.7.2. M hnh ton hcnhiubin cangckhng ngb ba pha

Hnh 3.41: M hnh vt l ng ckhng ng b 3 pha

Phng trnh i xng in p ca nhm cun dy

mch stator 3 pha l:

Phng trnh i xng in p ca nhm cun dy

mch rotor 3 pha sau khi tnh chuynivmch stator

l:

AA A 1

BB B 1

CC C 1

du i R

dt

du i Rdt

du i R

dt

aa a 2

bb b 2

cc c 2

du i R

dt

du i R

dt

du i R

dt


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3.7.2.1.Phngtrnh cn bngin pPhng trnh in p cvitdng ma trn

Hoc c thvit: u = Ri + p

A A A

B B B

C C C

a a a

b b b

c c c

u iR 0 0 0 0 0

u i0 R 0 0 0 0

u i0 0 R 0 0 0p

u i0 0 0 R 0 0

u i0 0 0 0 R 0

u i0 0 0 0 0 R


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3.7.2.2.Phngtrnh tthngPhng trnh t thng cvitdng ma trn:

Hoc c thvit: = Li

A AA AB AC Aa Ab Ac A

B BA BB BC Ba Bb Bc B

C CA CB CC Ca Cb Cc C

a aA aB aC aa ab ac a

b bA bB bC ba bb bc b

c cA cB cC ca cb cc c

L L L L L L i

L L L L L L i

L L L L L L i

L L L L L L i

L L L L L L i

L L L L L L i


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3.7.2.2.Phngtrnh tthng


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3.7.2.2.Phngtrnh tthngPhng trnh t thng hon chnhvitdidng ma trnkhi:

Trong :


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3.7.2.2.Phngtrnh tthng

Thayphng trnh t thng vophng trnh in p, ta c:

m1 t1 m1 m1

ss m1 m1 t1 m1

m1 m1 m1 t1

1 1L L L L

2 2

1 1L L L L L

2 2

1 1L L L L

2 2

m1 t2 m1 m1

rr m1 m1 t2 m1

m1 m1 m1 t2

1 1L L L L

2 2

1 1L L L L L

2 2

1 1L L L L

2 2

0 0

T 0 0

rs sr m1

0 0

cos cos( 120 ) cos( 120 )

L L L cos( 120 ) cos cos( 120 )

cos( 120 ) cos( 120 ) cos

di dL di dLu Ri p(Li) Ri L i Ri L i

dt dt dt d


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3.7.2.3.Phngtrnh chuynngTrong trnghptng qut,phng trnh chuynngcahthngtruynng

in c dng:

Trong : Mc l m menphti (m men cn);

J l m men qun tnh cahthng;

D l hscn m men cntlvitc quay;

K l hsnhi m men quay;

np

l sicctngc.

iviphti m men khng i, D = 0, K = 0, th:

t c

p p p

J d D K M M

n dt n n

t c

p

J dM M

n dt


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3.7.2.4.Phngtrnh m men

T T

m

1 1W i i Li

2 2

m mt p

i constm i const

W WM n

srT T

t p p

rs

0 L1 L 1

M n i i n i i2 2

L 0

T Trs sr t p r s s r

0 0

p m1 A a B b C c A b B c C a A c B a C b

L L1M n i i i i

2

n L [(i i i i i i ) sin (i i i i i i )sin( 120 ) (i i i i i i )sin( 120 )]


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3.7.2.5. M hnh ton hcngckhngngb ba phaM hnh ton hcnhiubinscangckhng ngb 3 pha khi chuti m

men khng i:

M hnh ngcbiudinbng khng gian trng thi:

Tp c

p

di Lu Ri L i

dt

1 L J dn i i M2 n dt

d

dt

1 1

TpT0

c

di LL (R )i L u

dtnnd L

i i Mdt 2J J

d

dt


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3.7.3. Php binito v ma trnchuyni

3.7.3.1. Khi nimcbn v nguyn tcca phpbinito

Hnh 3.42: M hnh vt l ng c in mt chiu hai cc:

F- cun dy kch t, A - cun dy rotor, C- cun dy b


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3.7.3.1. Khi nimcbn v nguyn tcca phpbinito

Hnh 3.43: M hnh vt l cc cun dy ng c in xoay chiu,

m hnh tng ng v m hnh ng c in mt chiu

a) M hnh cc cun dy xoay chiu ba pha

b) M hnh tngngxoay chiu hai pha

c) M hnh cun dy ngcmtchiu quay trn


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3.7.3.2. Ma trnchuynito trongiukin cngsutbtbin

1

2

n

u

uu

u

1

2

n

i

ii

i

1

2

n

u

uu

u

1

2

n

i

ii

i


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3.7.3.3. Phpchuyni3 pha/2 pha (phpchuyni3/2)

Hnh 3.44: V tr vector khng gian ca h to 3 pha v 2 pha cng vi sc t ng cun dy


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33/ 2

2

1 11

2 2

N 3 3C 0

N 2 2

K K K

1 T 33/ 2 3/ 2

2

1 0 K

N 1 3C C K

N 2 2

1 3 K2 2

2 2

1 23 3 33/ 2 3 /2

2 2 2 22

1 1 31 0 01 0 K2 2 2 1 0 0N N N3 3 1 3 3 3

C C ( ) 0 K 0 0 0 1 0 IN 2 2 2 2 N 2 2 N

0 0 2K K K K 0 0 3K 1 3

K2 2


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V vy:

V: nn:

2

3

2

N31

2 N

3

2

N 2

N 3

22K 11

K2

3/ 2

1 11

2 2

2 3 3C 0

3 2 21 1 1

2 2 2

1

2/3 3/ 2

11 0

2

2 1 3 1C C

3 2 2 2

1 3 1

2 2 2


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A

B

C

1 1i1

i 2 2 2i

i 3 3 3i0

2 2

A

B

C

1 0i

i2 1 3i

i3 2 2i

1 3

2 2

A

B

30

i i2

i i12

2

A

B

30

ii 2

ii 1 1

6 2


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3.7.3.3. Phpchuyni2 pha/2 pha

Hnh 3.45: H to c nh v h to quay 2 pha

v vector khng gian sc t ng


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3.7.3.3. Phpchuyni2 pha/2 pha

M M

2r/2s

T T

i i icos sin

Ci i isin cos

2r/2s

cos sinC

sin cos

1

M

T

i ii cos sin cos sin

i ii sin cos sin cos


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3.7.3.5. Phpchuynito vunggc/tacc(php chuyni K/P)

C thsdng cng thc khc tm :

2 2

1 M Ti i i

1p 1p 1p

1p 1p T

1p 1p 1p 1p T M

sin sin (2cos ) sin i2 2 2tg2 1 cos i i

cos cos (2cos )

2 2 2

1 T1p

T M

i2tg

i i


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3.7.3.6. Php chuynithtocnh 3 pha btk sang htoquay 2 pha (phpchuyni3s/2r)

Trc tin chuynthta ba pha ABC sang hta hai pha cnh,

sau thchinchuynth hai phai cnh sang h hai pha quay dq.

V:


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A A

3/ 2 B B

0 C C

1 11

2 2i i i

2 3 3i C i 0 i

3 2 2i i i

1 1 1

2 2 2

3s/2r

0 0

1 1 3 1 3 11 cos sin cos sin cos

2 2 2 2 2 2cos sin 0

2 3 3 2 1 3 1 3C sin cos 0 0 s in sin cos sin

3 2 2 3 2 2 2 20 0 1

1 1 1 1 1 1

2 2 2 2 2 2

cos cos 120 cos 120

2s

3

0 0in sin 120 sin 1201 1 1

2 2 2


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Ma trnchuyningc(t 2 pha quay sang 3 cnh)ca n l:

1 T 0 0

2r /3s 3s / 2r 3s / 2r

0 0

1cos sin

2

2 1C C C cos 120 sin 120

3 2

1cos 120 sin 120

2

0 0

0 0

3s/2r

cos cos 120 cos 120

2C sin sin 120 sin 120

3 1 1 1

2 2 2


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3.7.4. M hnh ton hcngckhng ngb trn hto quay 2

pha btkcangckhng ngb

3.7.4.1.Phngtrnh in p trnhto dq0Ta c quan hchuyniin p stator l:

Trong hto ABC,phng trnh in p pha A l:

A d1

0 0

B q1

C 010 0

1cos sin

2u u

2 1u cos 120 sin 120 u

3 2u u

1cos 120 sin 120

2

A d1 q1 01

A d1 q1 01

A d1 q1 01

2 1u (u cos u sin u )

3 2

2 1i (i cos i sin i )

3 2

2 1( cos sin )

3 2

A A 1 Au i R p


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3.7.4.1.Phngtrnh in p trnhto dq0Ly cc cng thcchuynica uA, iA, A thay vobiuthc vbini ta c:

t p =11 l tc gc cah to quay tngivi stator; v l btk, nn 3cng thc sau ylnltc thnh lp:

1 1 1 1 1 1 1 1d 1 d d q q 1 q q d 01 1 01 01

1u R i p p cos u R i p p sin (u R i p ) 0

2

d1 1 d1 d1 11 q1

q1 1 q1 q1 11 q1

01 1 01 01

u R i p

u R i p

u R i p

d2 2 d2 d2 12 q2

q2 2 q2 q2 12 q2

02 2 02 02

u R i p

u R i p

u R i p


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3.7.4.2.Phngtrnh tthng trn hto dqMa trnchuyni t thng t ta 3 pha quay sang h ta 2 pha quay vi tc

quay khc nhau, ma trnchuyni l C3r/2r, da theo tc quay tngigia hai h

ta xem xt, v hnh thc C3r/2rgingvi C3s/2r, chcni gc 1 thnh gc gia

trc d v trc l 2.

0 0

1 1 1

0 0

3s/2r 1 1 1

cos cos( 120 ) cos( 120 )2

C sin sin( 120 ) sin( 120 )3

1 1 1

2 2 2

0 0

2 2 2

0 0

3r /2r 2 2 2

cos cos( 120 ) cos( 120 )2

C sin sin( 120 ) sin( 120 )3

1 1 1

2 2 2


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3.7.4.2.Phngtrnh tthng trn hto dqV th ta c:

d1 A

q1 B

01 3s / 2r C

d2 3r / 2r a

q2 b

02 c

C 00 C

d1 d1

q1 q1

101 3s / 2r ss sr 013s/2r

1d2 3r / 2r rs rr d23r/2r

q2 q2

02 02

i

iC 0 L L iC 0

0 C L L i0 C

i

i

d1 d1

q1 q1

101 3s / 2r ss sr 013s/2r


q2 q2

02 02

i

iC 0 L L iC 0

0 C L L i0 C

i

i

d1 d1

q1 q1

101 3s / 2r ss sr 013s/2r


q2 q2

02 02

i

iC 0 L L iC 0

0 C L L i0 C

i

i


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3.7.4.2.Phngtrnh tthng trn hto dq

1m1 t1 m1 m10 0

1 1 1

1 0 0

3s / 2r ss 3s / 2r 1 1 1 m1 m1 t1 m1 1

m1 m1 m1 t1

11 1 cos sinL L L L22 2cos cos( 120 ) cos( 120 )

2 1 1

C L C sin sin( 120 ) sin( 120 ) L L L L cos( 1203 2 21 1 1

1 1L L L L2 2 2 2 2

0 0

1

0 0

1 1

1

) sin( 120 ) 2

1cos( 120 ) sin( 120 )

2

t1 m1

1

3s / 2r ss 3s / 2r t1 m1

t1

3L L 0 0

2

3

C L C 0 L L 02

0 0 L


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3.7.4.2.Phngtrnh tthng trn hto dqt2 m1

1

3r / 2r rr 3r / 2r t2 m1

t2

m1

1

3s/ 2r sr 3r / 2r m1

m1

1

3r / 2r rs 3s / 2r m1

3L L 0 0

2

3C L C 0 L L 0

2

0 0 L

3L 0 0

2

3C L C 0 L 0

2

0 0 0

3L 0 0

2

3C L C 0 L 0

2

0 0 0


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3.7.4.2.Phngtrnh tthng trn hto dq

Trong :

Lm=(3/2)Lm1 - Hcmgia cc cun dy stator v rotorngtr trn htrcto dq0;

Ls=Lt1+(3/2)Lm1 - T cm cc cun dy stator 2 pha tngng trn hto dq0;

Lr=Lt2+(3/2)Lm1 - T cm cc cun dy stator 2 pha tngng trn hto dq0;

d1 s m d1

q1 s m q1

01 t1 01

d2 m r d2

q2 m r q2

02 t2 02

L 0 0 L 0 0 i

0 L 0 0 L 0 i

0 0 L 0 0 0 i

L 0 0 L 0 0 i

0 L 0 0 L 0 i

0 0 0 0 0 L i


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3.7.4.2.Phngtrnh tthng trn hto dqCn ch , hcm Lm c trsbng 3/2 ln gi trccihcm Lm1gia haipha tu trong nhm cc cun dy 3 pha ban u.

T hai hng th ba v th su trong cng thc (3.102) c ththy thnhphn trn

trc 0 cat thng l:

T cng thc tnh t thng c th thy:

01 = Lt1i01 v 02 = Lt2i02 ,chng c lp vi nhau, khng c nh h-ng g i vi trc d, q sau ny trongm hnh ton hc s khng xt ti. V

vy:


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3.7.4.2.Phngtrnh tthng trn hto dqCn ch , hcm Lm c trsbng 3/2 ln gi trccihcm Lm1gia haipha tu trong nhm cc cun dy 3 pha ban u.

T hai hng th ba v th su trong cng thc (3.102) c ththy: 01 = Lt11i01

v 02 = Lt21i02 chng clpvi nhau, khng c nhhng g ivitrc d, qsau ny trong m hnh ton hcs khng xt ti. V vy:

d1 s d1 m d2d1 d1s m

q1 s q1 m q2q1 q1s m

d2 d2m r d2 m d1 r d2

q2 q2m r q2 m q1 r q2

L i L iiL 0 L 0

L i L ii0 L 0 L

iL 0 L 0 L i L ii0 L 0 L L i L i

Hoc:


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3.7.4.2.Phngtrnh tthng trn hto dq0

Hnh 3.46: M hnh vt l ca ng c khng ng b chuyn i sang trc dq


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3.7.4.2.Phngtrnh tthng trn hto dq0

Phng trnh in p catrc d, q:

d1 d11 s 11 s m 11 m

q1 q111 s 1 s 11 m m

d2 d2m 12 m 2 r 12 r

q2 q212 m m 12 r 2 r

u iR L p L L p L

u iL R L p L L pu iL p L R L p L

u iL L p L R L p


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3.7.4.3.Phngtrnh m men vphngtrnh chuynngtrn hta dq0

0 0

t p m1 A a B b C c A b B c C a A c B a C bM n L [(i i i i i i )sin (i i i i i i )sin( 120 ) (i i i i i i )sin( 120 )]

S dng cc ma trn C-13s/2rv C-13r/2rca php bin i ngc, ta -

c:

1 1

A d1

0 0

B 1 1 q1

C 010 0

1

2 2

a

0 0

b 2 2

c0 0

2 2

1cos sin2

i i2 1

i cos 120 sin 120 i3 2

i i1

cos 1 120 sin 1202

1cos sin

2i

2 1i cos 120 sin 120

3 2i

1cos 120 sin 120

2

d2

q2

02

i

i

i


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3.7.4.3.Phngtrnh m men vphngtrnh chuynngtrn hta dq0

Sdng 2phng trnh ma trn trn i iA ,iB ,iC ,ia ,ib ic trong cng thc m men

thnh cc thnhphnca d, q, 0, ngthi ch tiv tr tngica stator v

rotor: = 1 - 2

Sau khi ginc,cui cng snhnc cng thc m men rtngin trn h

to dq0: Mt = npLm(iq1id2id1iq2)

V: = 11 - 12

t p m q1 d 2 d1 q 2 c

p

J dM n L (i i i i ) M

n dt

ddt

3 h h kh 2


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3.7.4.4.Scu trc trngthi ngvmchintngngtrngthi ng

cangcin khngngb trnhto dq

t:

d1 d1 d1 d1s m1 11

q1 q1 q1s m1 11

d2 d2 d2m r2 12

q2 q2 q2s r2 12

u i iL p 0 L p 0R 0 0 0 0 0 0

u i i0 L p 0 L p0 R 0 0 0 0 0

u i iL p 0 L p 00 0 R 0 0 0 0

u i i0 L p 0 L p0 0 0 R 0 0 0

q1

d2

q2

T

d1 q1 d2 q2

T

d1 q1 d2 q2

T

d1 q1 d2 q2

u [u u u u ]

i [i i i i ] [ ]

s m1

s m1

m r2

s r2

L 0 L 0R 0 0 0

0 L 0 L0 R 0 0R , L

L 0 L 00 0 R 0

0 L 0 L0 0 0 R

3 74 M h h t h kh b t h 2


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Vectorscinng quay:

Phng trnh cnbngin p:

d1 11 d111

q1 11 q111

r

d2 12 d212

q2 12 q212

0 0 00 0 0

e0 0 0

0 0 0

ru Ri Lpi e

3 74 M h h t h kh b t h t 2


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Hnh 3.47: S cu trc trng thi ng nhiu bin s ca ng c khng ng b



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Hnh 3.48: S thay th trng thi ng ca ng c khng ng b trn h to dqa) S thay th trc d; b) S thay th trc q



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3.7.5. M hnh ton hcngcin khngngb trnhtocnh 2 pha

Phng trnh in p:

Phng trnh t thng:

1 1 1 1

1 1 1 1

2 2 2 2 2

2 2 2 2 2

u Ri p

u R i p

u R i p

u R i p

1 s 1 m 2

1 s 1 m 2

2 m 1 r 2

2 m 1 r 2

L i L i

L i L i

L i L i

L i L i

3 74 M hnh ton h khng b trn h t q a 2


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1 1 2 2t p m

M n L i i i i

1 11 s m

1 11 s m

2 2m m 2 r r

2 2m m r 2 r

u iR L p 0 L p 0

u i0 R L p 0 L p

u iL p L R L p L

u iL L p L R L p

d1 1 1

q1 1 1

d2 2 2

q2 2 2

i i cos i sin

i i sin i cos

i i cos i sini i sin i cos

3 74 M hnh ton hc ng c khng ng b trn h to quay 2


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Phng trnh in p:

Phng trnh m men:

Phng trnh chuynng v cng thc vi phn gc quay tng tnh trnghpivihtacnh.

d1 d11 s 1 s m 1 m

q1 q11 s 1 s 1 m m

d2 d2m m 2 r s r

q2 q2s m m s r 2 r

u iR L p L L p L

u iL R L p L L pu iL p L R L p L

u iL L p L R L p

1 1 2 2t p mM n L i i i i



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3.7.7. M hnh ton hccangckhng ngb theo nhhngttrng

trnhta quayngb 2 pha - m hnh ton hchta M, T

V:

1 s 1 s m 1 mM1 M1

1 s 1 s 1 m mT1 T1

m m 2 r s r M2 M2

s m m s r 2 r T2 T2

R L p L L p Lu i

L R L p L L pu iL p L R L p Lu i

L L p L R L pu i

t p m T1 M 2 M1 T 2M n L (i i i i )



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3.7.7. M hnh ton hccangckhng ngb theo nhhngttrng

trnhta quayngb 2 pha - m hnh ton hchta M, T

1 s 1 s m 1 mM1 M1

1 s 1 s 1 m mT1 T1

m 2 rM2 M2

s m s r 2 r T2 T2

R L p L L p Lu i

L R L p L L pu iL p 0 R L p 0u i

L 0 L R L pu i

2 r M2 mt p m T1 M 2 M1 T 2 p m T1 M2 T1

m r

2 mp m T1 M2 T1 T1 M2 p T1 2

r r

L i L

M n L (i i i i ) n L i i ( i )L L

Ln L i i i i i ) n i .

L L

3 8 H THNG IU TC BIN TN IU KHIN VECTOR


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3.8. HTHNGIUTCBINTN IUKHIN VECTOR

3.8.1. Scutrc php binita v m hnh ngcmtchiu

tngngcangckhng ngb

Hnh 3.49: Scu trc binitangckhngngb3/2)Bini 3 pha /2 pha;

VR)Bini quay ngb;

) Gcgiatrc M v trc(trc A)



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Hnh 3.50: tng cu trc h thng iu khin vector


3.8.2. tngvcutrc hthngiukhin vector



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3.8.3. Phngtrnh cbniukhin vector

1 s 1 s m 1 m M1M1

1 s 1 s 1 m m T1T1

m 2 r M2

s m s r 2 T2

R L p L L p L iu

L R L p L L p iu

L p 0 R L p 0 i0

L 0 L R i0

2M1 2

m

T p 1i

L

m2 M1

2

Li

T p 1

r

2

2

LT

R

mT2 T1

r

Li i

L

mt p T1 2

r

LM n i

L



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3.8.3. Phngtrnh cbniukhin vector

Giiquytvniukhin tnsphihpviiukhin dng in, c th

thy:

Phng trnh iukhintnstrt:

2s T2

2

Ri

m T1s

2 2

L iT

3.8. H THNG IU TC BIN TN IU KHIN VECTOR


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3.8.4. Hthngiutcbintn gin tipngun dng iniukhin

m men dngtnstrtmch vng htthng

Hnh 3.51: H thng iu tc bin tn gin tip ngun dng

R- b iu chnh tc quay; RI - b iu chnh dng in;

K/P - b bin i ta vung gc / to cc



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3.8.5. M hnh quan st tthng rotor

T ma trnin p hto, ngthit u2 = 0, u2 = 0, ta c:

2 m 1 r 2

2 m 1 r 2

L i L i

L i L i

2 2 m 1

r

2 2 m 1

r

1i ( L i )

L

1i ( L i )

L

1 2 1 2 2

1 2 1 2 2

m r m m 2

m r m m 2

L pi L pi L i L i R i 0

L pi L pi L i L i R i 0



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Sau khibinisc m hnh quan st t thng rotor:

2 2 2 m 1

2

2 2 2 m 1

2

1p ( L i ) 0

T

1p ( L i ) 0

T

2 2 2 m 1

2

2 2 2 m 1

2

1p ( L i ) 0

T

1p ( L i ) 0T

3 8 5 M hnh quan st t thng rotor


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3.8.5.1. M hnh quan st tthng rotor trnhtacnh hai pha

Hnh 3.52: M hnh quan st thng rotor trn h to c nh hai pha

3.8.5. M hnh quan st t thng rotor


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3.8.5.1. M hnh quan st tthng rotor trnhtacnh hai pha

Hnh 3.53: M hnh quan st t thng trn h to quay hai phatheo nh hng t trng

3.8.6. H thng iu tc bin tn PWM vi mch vng kn t thng v tc


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3.8.6. Hthngiutcbintn PWM vimch vng kn tthng v tc

quay kiumch vng trdng in

Hnh 3.54: H thng iu tc bin tn PWM kiu vng tr dng in vi mchvng kn t thng v tc quay

RBiuchnhtc quay; RMBiuchnh mmen;

RBiuchnhtthng; FTBcmbintc quay


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4.1. Nguyn l iu tc ni cp v cc dng c bn ca n

4.2. Cht lng h thng iu tc ni cp

4.3. c tnh c ca ng c khng ng b trong h iu tc ni cp

4.4. H thng kn iu tc ni cp vi hai mch vng

4.5. H thng iu tc ni cp siu ng b

4.6. Mt s vn c bit ca h thng iu tc ni cp

4.1. NGUYN L IUTCNICP V CC DNGCBNCA N


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4.1.1. S lm viccangckhng ngb khi rotor c thm sc

inng

Lc ngckhng ngb lm vic,scinng pha mch rotorca

n l:

E2 = sE20 (9.1)

Trong : s l hs trtcangckhng ngb; E20 l scin

ng pha cangckhng ngb rotor dy qun khi rotorng yn,

hay gi l scinngmchh, s... nhmcmch rotornut

vo mch statorin p xoay chiuvitns v gi trin pbngnh

mc.



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inng

Cng thc (9.1) chng t, tr s ca sc in ng E2 ca rotor t l

thunvihstrt s, ngthitns f2 ca n cngtlthunvi s:

f2 = sf1. Lc rotorcni dy bnh thng,phng trnh dng in pha

rotor l:

Trong R2 l intrmi pha cacun dy rotor; X20 l in khng

tncami pha cun dy rotor khi s = 1.

202

2 2

2 20

sEI

R (sX )



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inng

20 ph

22 22 20

sE EI

R (sX )

Hnh 4.1: S u s... ph (Eph) trongmch rotor ng c khng ng b

Khi a vo mch rotormt sc in ng

ph Ephiukhinc v mcnitipvi

sc in ng E2 ca mch rotor. Eph c

cng tn s, nhng c th cng pha hoc

ngc pha vi E2, nh trn hnh 9.1. Trong

trng hp ny, phng trnh dng in s

l:



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inngKhi m men ph ti Mc l hng s, c th coi dng in rotor I2 cng l

hng. Gi thittrc khi c scinngph,ngcang lm vicnnh

vi gi trhstrt s = s1. Sau khi ascinngphngcdu vo, do

m men ph ti l hngs,v tri cng thc (9.2) l hngs (I2), v thhs

trtcangcbucphitng ln v nnhvi s=s2 (s2>s1), quan hgia s1,

s2 v s...phtha mnbiuthc:

2 20 ph 1 202

2 2 2 2

2 2 20 2 1 20

s E E s EI

R (s X ) R (s X )



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4.1.2. Tolpscinngphv hthngtruynngiutcni

cpascinngph vo mch rotorngckhng ngb rotor dy qun r

rng l c th lm thay i tc quay cangc,nhng do tnscasc

inngcmng E2camch rotorngcin thay i theo hstrt, nn

tnscascinngph Eph

cngbucphi thay i theo tc quay ca

ngc. giiquytvn ny mt cch nginnht, trong hthngthc

t,u tin in p xoay chiu trong mchcun dy rotorngccbini

thnh scinngmtchiu, sau so snh n vi scinngphmt

chiu,iukhin gi tr bin scinngphmtchiu l c thiuchnh

ctc quay cangc.Nhvychuynvn thay ictnsv gi trca s...ph xoay chiu sang vniuchnh gi trca s...ph

mtchiu khng lin quan g ntns, lm cho vic phn tch v iukhind

dng irtnhiu.



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cp

Hnh 4.2: S nguyn l h thng truyn ng ni cp ng c khng ng b



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cpTa cphng trnh cnbngscinng trong mchinmtchiuchnhlu

nh sau:

Trong:K1, K2 - cc hsph thuc vo loisbbiniB1 v B2, khi B1 v

B2 u dng chnhlucu 3 pha th: K1 = K2 = 2,34

Ud1 - in p chnhlu trung bnh u ra caB1;

U2 - gi trhiudngin p pha thcpca mybin p nghchlu BA;Ud2 - in p nghchlu trung bnh u ra caB2;

- gc iukhinnghchlucaB2;

R - intrtngngcamch rotor quy iv pha mtchiumtchiu.



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4.1.3. Cc dngkhc cahthngiutcnicp

Hnh 4.3: S nguyn l h thng truyn ng ni cp in lcvi B2 mc theo s tia ba pha



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Hnh 4.4: H thng iu thng ni cp in lc thi k u



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Hnh 4.5: S nguyn l h thng iu tc ni cp c kh

4.2. CHTLNGHTHNGIUTCNICP


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4.2.1. ctnh ccahthngiutcnicp

Phng trnh cnbngscinngcamchinmtchiu pha rotor khi lmvic khng ti l tng:

Trong : s0 l hs trt khng ti l tng; U2 l in p hiudng thcp

mybin p nghchlu BA.

C ththy,vi cc gc khc nhau, ng cong Mt=f(s) khi iutcnicp

ngckhng ngb l gnnh song song, tng tnhngc tnh ccaiutciu p ngcinmtchiu.

0 20 2

20

20

s E U cos

U coss

E

4.2. CHTLNGHTHNGIUTCNICP h


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4.2.1. ctnh ccahthngiutcnicp

Hnh 4.6: c tnh c ca ng c khng ng b khi iu tc ni cp

a) ng c ln b) ng c b

4.2. CHTLNGHTHNGIUTCNICP4 2 2 M


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4.2.2. My bin p nghchlu

ngindthy, c thda vo trng thi lm vic khng ti l tng tnhchnin p pha pha thcp U2 ca mybin p nghchlu:

Trong :

s0max l hstrt khng ti l tngccingvitc quay cctiu

ch khng ti l tngcangcc xc nh theophm vi iutc

cahthng;

min l gc nghch lu khi lm vic vi h s trt cc i, lc ny gcnghchlunhnhtthngchn: min = 300.

0 20 0max 202

min

s E s EU

cos cos

4.2. CHTLNGHTHNGIUTCNICP4 2 3 D l ( t) thit b i t i


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4.2.3. Dung lng(cng sut)thitbiutcnicp

Do :

m 2m 2mS 3U I

0max 20 0max 202m 0 max 20 200

min

s E s E 1U 1,15s E 1,15E (1 )

cos30 D

m 20 2m

1S 3,45E I (1 )

D

4.2. CHTLNGHTHNGIUTCNICP4 2 4 Hi t h th i t i


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4.2.4. Hiusutcahthngiutcnicp:

Hnh 4.7: Phn tch hiu sut h thng iu tc ni cp

a) Hng i ca cng sut h thng iu tc ni cp; b) Biu nng lng h thng



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co co t co2nc

v 1 F t 1 s 2 s

t co t

t 1 2 s t s

P P P (1 s) PP100% 100% 100%

P P P (P P ) (P P P )

P (1 s) P P (1 s)100% 100%

P (1 s) P P P P (1 s) P

t co2

R

v 1

t co

t 1

P (1 s) PP100% 100%

P P

P (1 s) P(1 s) 100%

P P

Hnh 4.8: ng cong= f(s) ca h thngiu tc ni cp in lc v iu tc in tr

ph trong mch rotor



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Hiusutca haiphng php iutc khiphti c tnh cht khc nhau cch ra bngdiy:

Bng 4.1: Hiu sut iu tc ni cp in lc v iu tc mc ni tip in tr mch rotor

4.3. C TNH CCANGCKB TRONG HIUTCNICP4 3 1 i p v dng i h i h h l rotor


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4.3.1. in p v dng in ca mch in chnh lu rotor ng c

khng ngb:a) Trngthi lmvicthnht:

Hnh 4.9: Mch in chnh lu rotor

2a 20 1e 2sE sin(s t )6

1 Do d

20

1 Do d

20

2X sIcos (1 )

6E s

2X Icos (1 )6E

20d

Do

6EI (1 cos )

2X

D0d d

3X sU I

D0d1 20 d

3X sU 2,34sE I

4.3. C TNH CCANGCKB TRONG HIUTCNICP4 3 1 in p v dng in ca mch in chnh lu rotor ng c


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khng ngb:

Hnh 4.10: th in p v dng in s chnh lu rotorng vi cc gc chuyn mch khc nhau:

a) < 60o; b) = 60o;



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khng ngb:

Hnh 4.10: th in p v dng in s chnh lu rotorng vi cc gc chuyn mch khc nhau:

c) > 60o



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khng ngb:b) Trngthi lmvicthhai:

- Dng inchnhlu trung bnh:

- in p chnhlu trung bnh:

Trong hai cng thc trn, tngngvi: p 0, = 600.

20 20d p p pD0 D0

6E 6EI cos cos sin

2X 2X 6

p p D0d1 20 20 p d

cos cos 3sXU 2,34sE 2,34sE cos I

2



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khng ngb:b) Trngthi lmvicthhai:

Hnh 4.11: B chnh lu rotor: Id= f() , Id=f(p)

4.3. C TNH CCANGCKB TRONG HIUTCNICP4 32 M men in t ca ng c khng ng b khi iu tc ni cp


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4.3.2. M men intcangckhng ngb khi iutcnicp

Hnh 4.12: S thay th tng ng ca h thng iu tc ni cp in lc



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- in p u ra cabchnhlu rotorvng lm victh II:

- in p pha dng mtchiubnghchlu:

- V c: Ud1 = Ud2 + IdRCK

d1 d0 p D d D0 D

3U sU cos 2 U I ( X .s 2R )

d2 2 T d BA BA

3U 2,34U cos 2 U I ( X 2R )



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Ta cphng trnhbiuthhstrt s:

Ta c c tnh iutccahthngiutcnicp:

Trong cng thc trn b qua nhhngcaUD, UT v R1'.

V khi cho p = 00, ta c cng thcc tnh tc quay cahthngiutcnicpvng lm victh I.

20 p 2 d D0 BA BA D CK

0

20 p D0 d

3 32,34(E cos U cos ) I ( X X 2R 2R R )

n n3

2,34E cos X .I

2 d BA BA D CK

20 p D0 d

32,34U cos I ( X 2R 2R R )

s3

2,34E cos X .I

0

0

n ns

n

Vi:



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Githit:

Ta c c tnh tc rt gncangctrong hthngiutcnicp:

d

0 de e

U I R 1n n ( ) (U I R )

C' C



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tm m men intcangckhng ngb trong hthngiutcnicp, c thbtut quan h cng sut Pscamchinchnhlu rotor.

tm m men intcangckhng ngb khibchnhlu rotortrng

thi lm victh I, ta cho p=0.

Gi trccica m men int trong cng thc trn c th tm c khi t:

D0s d0 p d d

3X sP (sU cos I )I

s D0

t d 0 p d d

0 0

P 3X1M (U cos I )Is

D0 D0t d 0 d d d0 d d

0 0 d0

3X 3X1 1M (U I )I U (1 I )IU

t

d

dM0

dI

4.3. C TNH CCANGCKB TRONG HIUTCNICP4.3.2. M men in t ca ng c khng ng b khi iu tc ni cp


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Khi : 20d d.1m

D0

6EI I

2X

2

20t t.1m

0 D0

27EM M

6 X

v:

Quan hgia hai ilng l Ud0 = 2,34E20.

2

2 220t p p

0 D0

27EM [cos cos ( )]6 X

T ta c:

tm m men in tcci, t cng thc trn ta lyo hm bcnhtca m

men theo p v chobng khng ( ), ngthi thay = 60o, c th tm c

khi p = 15o th m men cci Mt.2mvng lm victh II:

t

p

dM0

d



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Cng dng incamchmtchiutngng l:

Giao tipgia vng lm victh I v th II chnh l im lm vicgiihnca

schnhlu rotor, khibtu pht sinh hintngcngbcmchm cc

van. M men tiim ny gi l m men chuyn tip Mt.1-2. Chcn lyiu

kin= 60o, p= 0o thay vo cng thc (9.25) v (9.14) , l c th tm c:

20d.2m

D0

3EI

2X

2

20t.1 2

0 D0

27EM 8 X

20d.1 2

D0

6EI

4X



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Da vo cch chng minh trn, c th rt ra chthc:

M men ccicac tnh vn c ban ucangckhng ngb khib

qua intrca rotor:

C th tm c:

t.2m

t.1m

M0,866

M

t.1 2

t.1m

M0,75

M

2

20t.m

0 D0

3E1M

2 X

t.1m

t.m

M0,955

M t.2m

t.m

M0,826

M t.1 2

t.m

M0,716

M



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Hnh 4.13 ng c tnh c ca ng c khng ng b khi iu tc ni cp in lc

4.3.3. Phngtrnh ctnh ccangcKB khi iutcnicp

4331 Phng trnh ng c tnh c vng lm vic th I ( 600 = 00)


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4.3.3.1.Phngtrnh ngc tnh cvng lm victh I ( 600, p = 00)

Trong : s1m= s1m- s10: l s gia cahstrtcangckhng ngbt khng ti

l tngn m menbng m men cci tnh ton;s1 = s - s10: l s gia cahstrt dophti gy ra;

s10: l hstrt khng ti l tngngvitrs;

s1m: l hstrttihn khi m men cci tnh ton M t.1mtngng:

t

1m 1t.1m

1 1m

M 4

s sM2

s s

BAD BA CK

1m 10

D0

3X2R 2R R

s 2s 3X /


4332 Phng trnh c tnh c vng lm vic th I I ( = 600 = 00300)


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4.3.3.2. Phng trnh c tnh c vng lm vic th I I ( = 600, p= 00300)

Trong : s2m= s2m- s20 l s gia cahstrtccitngngvimttrsp

no xt tihintngcngbcmchm;s2 = s - s20: l s gia cahstrt dophti gy ra trong vng lm victh II

ngvi gc v p;

s20 l hstrt khng ti l tngngvi vi gc v p.

2

pt

2m 2t.1m

2 2m

4cosM

s sM2

s s

220

20 p

U coss

E cos

BA

D BA CK

2m 10

D0

3X2R 2R R

s 2s3X /


4332 Phng trnh c tnh c vnglmvic th I I ( = 600 = 00300)


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4.3.3.2.Phngtrnh c tnh cvng lmvicthI I ( = 60 , p= 0 30 )

Do vimt gi tr cho trcca th s20 thay i theo p, nn s2mcng thay itheo p. Gi tr s2m khngphibiuthhstrtthct khi m men tcciM t.2m, n ch l mtilng dng tnh ton xuthin trong qu trnhbini ton hc m thi. Biu thc thng dngvngc tnh c tinhn chotnh ton:

Khi cho p = 00, cng thc trn ph hp cho vic tnh ton c tnh cvng lm

victh I, lc sm= s1m, s = s1.Nu cho p = 00 300, cng thc trn ph hp cho vic tnh ton c tnh cvng lm victh II, khi sm = s2m, s = s2.

2

t t.1m p

m2

t t.1m p

1 1 (M / M ) / coss s

1 1 (M / M ) / cos

4.4 HTHNG KN IUTCNICPVI HAI MCH VNG


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Do hstrtc tnh tnhcahthngiutcnicp l kh ln, cho nn h

thngiukhinmchhchc dng vo trnghp chnh xc iutc

yu cu khng cao. nng cao chnh xc iutctrng thi tnh v nhn

cc tnh ngtthn, c th dng iukhinphnhi. Cnggingnhh

thngiu tcmtchiu, thng dng phng thciukhin hai mch vngkn lphnhi dng in vphnhitc quay.


4 4 1 Cu trc h thng iu khin mch vng kn


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4.4.1. Cutrc hthngiukhinmch vng kn

Hnh 4.14: H thng iu tc ni cp iu khin hai mch vng kn


4 4 1 Cu trc h thng iu khin mch vng kn


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Trong hthng trn hnh 9.14,bchnhlu c iukhin,biuchnh v s

cu trc trng thi ngca khuphnhigingnh trong hthngiutcmt

chiu, v thy khng trnh by li. Nhngmtphnmchinmtchiu

rotorcangcv c lin quan tihstrt nnphix l ring.

4.4.2.1. Hmstruyncamchmtchiu rotor

Da vo s tngngnh trn hnh 9.12 c tha ra phng trnh cn

bngin p trng thi ngcamchinmtchiu rotor trong hthngiu

tcnicp:

dd0 d2 d

dIsU U L R I

dt


4421 Hm t h t hi rotor


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4421 Hms truyn ca mch mt chiu rotor


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Bini Laplace cho hai vcabiuthc (*), c th tm c hm struyn cho

mchinmtchiu rotor:

Vi:

V:

d Lr

d0 Lrd0 d2

0

I (s) K

U T s 1U n(s) U (s)

n

Lr

LT

R

Lr

1K

R


4421 Hms truyn ca mch mt chiu rotor


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Cnphich ra rnghngsthi gian TLr

v hskhuchai KLr

trong hm s

truyncamchinmtchiu rotor trong hthngiutcnicpu l hm

scatc quay n, chng khngphi l hngs xc nh.

Hnh 4.15: S cu trc mch in mt chiu rotor


4422 Hms truyn ca ng c khngng b


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4.4.2.2. Hmstruyncangckhngngb

M men intcangckhng ngb l:

Trong :

Phng trnh chuynngcahthngtruynngin:

Hoc:

trong Ic l dng inphtitngngvi m menphti Mc.

D0t d0 d d M d

0

3X1M U I I C I

D0

M d0 d0

3X1

C U I

2

t c

GD dnM M

375 dt

2

M d c

GD dnC (I I )

375 dt


4422 Hms truyn ca ng c khngng b


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4.4.2.2. Hmstruyncangckhngngb

Ta c:

Trong :

Cn ch , hs CM l hm sca dng in Id, cho nn KMcng l hm sca Id.

M2

d c

M

Kn(s) 1

GD 1I (s) I (s) ss

375 C

M 2

M

1K

GD 1

375 C


4423 S cu trc trng thi ng ca h thng


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4.4.2.3.Scu trc trngthi ngcahthng

Hnh 4.16: S cu trc trng thi ng ca h thng iu tc ni cp

4.4.3. Xc nh cc tham sbiuchnhcahthngiutc

4.4.3.1.Thit k b iu chnh dngin


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4. 4.3.1.Thitkbiuchnh dngin

Biuchnh dng in c ththitk theo hthngin hnh loi I hochthnginhnh loi II. Nu thitk theo h thng loi I s c cim l tc ng nhanh cao,

lng qu iukhinnh, cn theo hthngloi II th c khnngchngnhiutt. Da

vo yu cu chung, mch vng dng in thng thitk theo h thngin hnh loi I.

Theo s hnh 9.16 c thvit ra hm struynmchhcamch vng dng in l:

i i i Lr

i i Lr

K ( s 1) K K W(s)

s T s 1 T s 1




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g

Hm s truyn c biu din bng cng thc trn, nu chn ng cc

tham s (chng hn ly i =TLr) chnh l cu trc ca h thng in

hnh loi I. Nhng bi v tnh bt nh ca TLr, nn cch lm thng thng

y l khng th chp nhn c.

Trong cng thc trn, nu tr s TLrkh ln, v tho mn iu kin TLr >hTi (h l rng trung tn), th c th chn i=hTi, khi c th thit

k mch vng dng in theo h thng in hnh loi II. C th chng

minh nh sau:

Do TLr l kh ln, c th coi gn ng hm s truyn ca mch in

chnh mt chiu nh l mt khu tch phn, cng thc hm truyn mchvng dng in c th vit thnh:

i i i Lr I ik 2

i i Lr i

K ( s 1) K K K ( s 1)W (s)

s T s 1 T s s (T s 1)




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g

Trong :

l h s K mch vng dng.

Nu khng tho mn iu kin TLr> hTi, th thc hin thit k h theo

h thng in hnh loi I. Da vo vic phn tch cc t liu v s liu

thc nghim lin quan cho thy lc c th da vo cn di ca

phm vi iu tc, tc l xc nh gi tr ca KLr v TLr theo s0max, hoctheo s0max/2, sau coi mch vng dng in l h thng xc lp i

thit k cc tham s b iu chnh dng in.

i i Lr i iI

i Lr i

K K K K K K

T hT L


4.4.3.2.Thitkbiuchnhtc


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c c khnng khng nhiu tt, mch vng tc u thitk theo h

thng

tong hop he dien co

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