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7/30/2019 N IN T CNG SUT trung
1/49
NIN TCNG SUT
GVHD : TRNG TH BCH THANH
SVTH : L XUN TRUNG.
MSSV : 105110393
NHM : 34
7/30/2019 N IN T CNG SUT trung
2/49SVTH: L XUN TRUNG 11D1CLC
CHNG I
TNG QUAN VCC PHNG PHP IU CHNH TC
A. NG CIN MT CHIU.
1.1. Khi qut chung:
ng c in mt chiu cho php iu chnh tc quay lin tc trng mt phmvi rng v trong nhiu trng hp cn c c tnh c c bit, thit bn gin hnv r tin hn cc thit biu khin ca ng cba pha.V mt su im nh
vy cho nn ng c in mt chiu c s dng rt ph bin trong cng nghip,trong giao thng vn ti.
2.1. Cu to ca n c in mt chiu.
ng c in mt chiu chia thnh 2 thnh phn chin:
- phn tnh( stato).
Gm cc b phn nh sau:
Cc chnh t: l b phn sinh ra ttrng, gm li st cc t v dy qun kch t.
+ Li st cc t lm bng l thp k thut in dy( 0,5-1)mm p li v tn cht.
+ Dy qun kch tc qun bng dy ng bc cch in.
Trong cc my c cng sut nh cc t chnh l mt nam chm vinh cu. Cntrong my c cng sut ln cc tl nam chm in.
Cc t pht gia cc tchnh dng ci thin trnh trng lm vic cu myin v i chiu.
+ li thp cc t ph c th l mt khi hoc c gp bng cc l thp ty theoch lm vic.
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Xung quanh cc t phc t giy qun cc t ph, dy qun cc t phc ni vi dy qun phn ng.
Gng t: dng lm mch t ni lin cc cc t, ng thi lm v my.
- Phn quay(roto)Bao gm cc b phn sau:+ li thp phn ng: dng dn t, thng dng nhng tm thp k thut indy 0.5mm phcch in mng hai mt ri p cht li gim tn hao do dngin xoy gy ln.
Trong my in nh li thp phn ng c p trc tip vo trc cn tong myin ln gia trc v li st c gi roto.
Dy qun phn ng l phn sinh ra sc in ng v c dng in chy qua, n
thng c lm bng ng bc cch in.Cgp: dng i chiu dng in xoay chiu thnh mt chiu.
C cu ni than: dng a dng in t phn quay ra ngoi.2.2. phn loi ng c in mt chiu.
Cng nh my pht, ng c in cng c phn loi theo cch kch thch tthnh cc ng c sau:- ng c in kch tc lp:ng c in mt chiu kch t c lp c qun kch tc cp in t mtngun in ngoai c lp vi ngun in cp cho mch phn ng.- ng c kch t ni tip:ng c kch t ni tip c cun kch t mc ni tip vi cun dy phn ng.- ng c kch t hn hp:Gm 2 dy qun kch t: dy qun kch t song song v dy qun kch t ni tiptrong dy qun kch t song song l ch yu.2.3. Nguyn l lm vic ca ng cin mt chiu
Khi cho in p mt chiu U vo hai chi in, trong dy qun phn ng c
dng in I. Cc thanh dn c dng in nm trong t trng, s chu lc Ft
tc dng lm cho rto quay.
Khi phn ng quay c na vng, v tr cc thanh dn i ch
cho nhau, do c phin gp i chiu dng in, gi cho chiu
lc tc dng khng i, m bo ng cc chiu quay khng i.
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Khi ng c quay, cc thanh dn ct t trng, s cm ng sc in ng E..
ng cin mt chiu sc in ng E ngc chiu vi dng in I nn
sc in ng Ecn c gi l sc phn din.
Phng trnh in l:
3.phng trnh c tnh c:
iu khin c tc ng c in mt chiu kch tc lp th ta phi phntch, tm cc mi quan h gia tc vi cc thng s khc ca ng c ta ra phng php iu khin. ng c in mt chiu kch tc lp th dngkch tc lp vi dng phn ng.V c nui bi hai ngun mt chiu c lpvi nhau.
Hnh 1. S ni dy ca ng cmt chiu kch tc lp
Theo hnh 1 ta vit phng trnh cn bng in p ca mch phn ng :
U=E+ ( R+ Rf)I (1)
Trong : U: in p t ln phn ng ng c(V).
E: Sc in ng phn ng (V)
R: in trca mch phn ng ()
Rf: in trph trong mch phn ng ()
I: Dng in mch phn ng ()
Vi R= r+ rcf+rb+ rct
Trong : r in trcun dy phn ng ()
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Rcf: in trcun cc t ph()
Rb: in trcun b ( )
Rct: in trtip xc chi in ()
Sc in ng Eca phn ng ng c c xc nh theo biu thc sau:
trong : p : Si cc tchnh
N : S thanh dn tc dng ca cun dy phn ng
a : Si mch nhnh song song ca cun dy phn ng: T thng kch tdi mi cc t(wb)
: Tc gc (rad/s)
K =
: H s cu to ng c.Nu biu din sc in ng theo tc quay n (vng/ pht) th
= =
V vy : Vi : : H s sc in ng ca ng c. .
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Thay (1) v (2) v bin i ta c : ; (4)Biu thc (4) l phng trnh c tnh c in ca ng c.Mt khc, mmen in
t: Mt= . (5)Nu b qua tn tht trong cc trc, tn tht t qut mt v tn tht trong thp thmmen ctrn trc ca ng cbng mmen in t, ta k hiu l M, tc l :
Mt= Mc=M
Vy phng trnh c tnh cca ng cl :
( 6)Biu thc (6) l phng trnh c tnh cca ng c in mt chiu kch tc
lp .
Khng xt n nh hng ca phn ng phn ng ngang trc lm gim t thng ca ng ctc l xem =const th quan h=f(M,I) l tuyn tnh.
Hnh 2 thc tnh cca ng c in mt chiu kch tc lp
Tth ta c : khi I= 0 hoc M =0 ta c:
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(7)0: c gi l tc khng ti l tng ca ng c, khi =0 ta c:
V M = k (9)Inm, Mnm: c gi l dng in ngn mch v mmen ngn mch.
Mt khc :phng thnh c tnh (4) v (6) cng c th vit di dng:
V : ta suy ra t (5)Trong : R = R
+ R
f,
: c gi l st tc ng vi gi trca M (hay I) , ta c th biu din c
tnh c in v c tnh ctrong hn vtng i vi iu kin tthng l nh
mc ( =m)
( Rcb=Um/ Imc gi l in trcbn )
T(4) v (6) ta vit c tnh c in v c tnh c n vtng i
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*=1- R*I* (13)
= 1- R*M(14) cng ca c tnh c:
Cng sut (nng lng in)
Tphng trnh l tng : IU =(E+IR)I (16)
Ta c Pin=Pt+
Trong Pt = IE cng sut in t.
P =I2R tn hao cng sut ctrn in trphn ng .
Thc t :Pin=Pt+ P+ P0 (17)
Vi P0 tn hao ma st do s quay
T biu thc (4) hoc (6) ta thy l mt hm ph thuc R, ,U : =f(R,,U) do iu chnh tc ng c in mt chiu kch tc lpc ba phng php iu khin sau :
- iu khin in trph phn ng
- iu khin t thng kch t
- iu khin in p phn ng
Sau y ta xem xt tng phng php iu khin mt
2. CC PHNG PHP IU KHIN NG C IN MT CHIUKCH TC LP
2.1 Phng php iu khin bng in trph phn ng (Rf):
Nguyn l iu chnh:
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Ni thm in trph Rfvo mch phn ng
Ta phn tch trn nn ta c = fRf, kt, U, gi thit rng :Nu gi=m=const ; U= Um= const; R=const th =f(Rf)
Mun thay i c gi tr Rfca mch phn ng bng cch ni tip mt in trph(Rf) thay i c gi trvo mch phn ng. Lc ny ta c : R = R+ Rf
Tphng trnh c tnh c :
Tphng trnh trn ta thy : khi tng gi tr ca Rf th tc ca ng cgim,khi gim gi tr ca Rf th tc ca ng ctng .
Lc ny ta c tc khng ti l tng:
Nh vy khi thay i Rf cho ta mt hc tnh nh sau:
Hnh 1-3 : c tnh cca ng c in mt chiu kch tc lp nhng intrph khc nhau
Nhn xt: Nu Rfcng ln th tc ng ccng gim, ng thi Inm v Mnmcng gim. Phng php ny c dng hn chdng in ng ckhi khing.
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- u im : n gin , d thc hin.
-Nhc im : + cng c tnh cthp.
+ Tn tht nng lng trn in trln .
+ Phm vi iu chnh hp.
2.2. Phng php iu chnh bng tthng kch t:
Nguyn l iu chnh:
iu chnh t thng kch t ca ng c in mt chiu l iu chnh m men int ca ng cM = kIv sc in ng quay ca ng c
E=k
Hnh 1-4 :S ni dy iu chnh kch t ca ng cin mt chiu kch tc lp.
T biu thc (4) v (6) ta thy = f(U, kt, Rf), nu gi U=Um=const v in trphn ng R= const (Rf=0 ) lc ny = f(kt). thay i c tc ta cn
thay i kt, m t thng kch t do dng kch t sinh ra.
Vy iu chnh kt ta mc thm bin trRv vo mch kch t, khi iu chnhkt ta phi tun theo iu kin sau.
Khng thtng dng kch t Ikt ln hn dng nh mc ca cun dy kch t v nph hng cun kch t v khi kt = m bo ha ri, nu mun tng Ikt th ktcng khng tng ng knn ta iu chnh bng cch gim kt.
Trong trng hp ny ta c:
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- Tc khng ti l tng
- cng ca c tnh c
Do cu to ca ng c in, thc thng iu chnh gim t thng. Nn khi tthng gim th x tng, cn s gim. Ta c thc tnh cvi x tng dn v cng ca c tnh gim dn khi gim t thng.
Hnh 1-5 : c tnh cv c in ca ng c in mt chiu kch tc lp khigim t thng.
T th ta nhn thy khi tthng thay i vi m>1 >2ta c:
- Dng in ngn mch : = - M men ngn mch : Mnm = kxInm = var (Mnm > Mnm1 > Mnm2)
T th c tnh ta thy 0
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- Tc khng ti l tng: tng.- cng c tnh c :
gim.
- Di iu chnh khng rng. : D = 1 : 2+ Tc nh nht b chn bi c tnh t nhin (=m)
+ Tc ln nht maxb gii hn bi bn ckh v iu kin chuyn mch cang c
- u im : Cng sut mch iu chnh nh, tn tht nng lng nh
- Cng c th sn xut nhng ng cgii hn iu chnh 1:5 thm ch n 1:8nhng phi dng nhng phng php khng chc bit, do cu to v cngngh ch to phc tp khin cho gi thnh ca my tng ln.
2.3. Phng php iu khin ng c in mt chiu kch t c lp bngthay i in p phn ng:
Gi thit = m=const; R=const,(Rf=0), M = const. Lc ny = f(U)
Khi thay i in p phn ng ta c:
Nh vy khi thay i in p phn ng cho ta mt hcc c tnh sau.
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Vi Um>U1>U2>U3>>Ui
Hnh 1-6 : thc tnh cca ng c in mt chiu kch tc lp khi inp phn ng thay i.
Nh vy khi thay i in p phn ng ng cta c mt hc tnh csongsong vi c tnh t nhin.
Nhn xt :
- u im : + khng gy n
+ khng gy tn hao phtrong ng c
+ di iu chnh rng D 10 : 1+ cng c tnh ckhng i trong tn di iu chnh
+ d tng ha.
- Nhc im : + phng php iu chnh ny cn mt b ngun c th thay itrn in p ra.
+ iu khin phc tp
Kt lun : Vi s pht trin ca khoa hc k thut nh hin nay th phng phpny cng c s dng ph bin trong sn xut cng nhtrong cc lnh vc khc.
2.4. Nuyn l iu chnh:
c tnh iu chnh :
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iu chnh in p phn ng, ta phi s dng mt b bin i, iu chnh cin p u ra cp cho mach phn ng ca ng c.
Hnh 1-7 : S ngyn l iu chnh ng c
B bin i dng bin i in p xoay chiu ca li in thnh mt chiuv iu chnh c gi trin p u ra theo yu cu. in trtrong ca b bini. Rbph thuc vo loi thit b, v thng thng cng sut ca b bn i vng cxp x bng nhau nn Rbcng c gi trng k so vi R ca ng c.
Ts nguyn l ta c s thay th
Hnh 1-8 : S thay thnguyn l iu chnh ng c
Ts thay thta c phng trnh cn bng in p.
E= Ub(Rb+ R)I (20)
Sc in ng ca ng c
E= km (21)
T biu thc (20) v (21) ta c
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km= Ub( Rb+ R)I(22)
Tphng trnh (22) ta c phng trnh c tnh c in.
Phng trnh c tnh c:
Vi : M =kmI
Trong ml tthng nh mc ca ng c, m= const.Tc khng ti l tng
cng ca c tnh c
Ta thy tc khng ti l tng khng ph thuc vo M, I m ph thuc vo Ub.
Ub=kaUk
Trong : ka l h s khuych i ca b bin i
Ukl in p iu khin
Tphng trnh (23) v (24) ta c thc tnh c biu din nh sau.
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Hnh 1-9: thc tnh c in khi Ub thay i
T th ta thy, khi Ubthay i th ta c nhng tc khng ti l tng khc
nhau, cn cng c tnh ckhng i v c tnh iu chnh dc hn c tnh tnhin.
Nhn xt :
- iu chnh tc trong bt k vng ti no, k c khng ti l tng , c tnhc iu chnh tuy mm hn c tnh tnhin nhng cng hn phngphp dngbin trv thay i kt
- iu khin phc tp, vn u t ln nhng n l phng php tt hn so vihai phng php trn
Kt lun : Sau khi phn tch ba phng php iu khin nu trn th phngphp iu khin bng thay i in p phn ng l tt hn c. Cho nn em chnphng php lm phng php nguyn cu cho ti ca em.
3. CC B BIN I IU CHNH TC NG C MT CHIUKCH TC LP BNG THAY I IN P PHN NG .
- Hin nay trong cng nghip ngi ta thng s dng ba loi b bin i sau:
+ B bin i my pht in mt chiu
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+ B bin i xung p mt chiu
+ B bin i chnh lu c iu khin
- Tng ng vi vic s dng cc b bin i ta c cc h truyn ng sau:
+ H my phtng c( F )
+ Hiu chnh xung png c( XA )
+ H chnh lu Tiristo ng c( T )
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Chng II
TNG QUAN V B CHNH LU THYRISTOR HNH CU 3 PHA.
THIT KS N NGHUYN L H THNG CHNH LUNG C IN 1 CHIU(h T- ) KCH TC LP.
A. TNG QUAN V B CHNH LU THYRISTOR HNH CU 3 PHA.I. TNG QUAN V B CHNH LU THYRISTOR HINH CU 3
PHA.1. 1 gii thiu v thyristor.
Thyristor l linh kin gm 4 lp bn dn P1-N
1-P
2-N
2lin tip to nn
ba cc: anot A, catot K v cc iu khin G(gate) Ti ba v tr tip xcnhau ca cc lp P1-N1-P2-N2 to nn cc lp tip gip J1,J2,J3.V l thuyt c hai loi thyristor:
- Thyristor kiu N hay thyristor c cc iu khin G ni vi vng N gn anot.
- Thyristor kiu P hay thyristor c cc G ni vi vng P gn catot.- Hot ng ca thyristor:- Thysristor lm vic ba trng thi: ng mkha
UBR: in p ngc nh thng
UBO: in p t mca thyristor
UTO: in p ri trn Thyristor
IL latching
IH dng duy tr.
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+ thysristor kha nu UAK< 0 v s vn kho nu ta cho UAK> 0.+ thysristor chuyn qua trng thi t kha sang dn nu ng thi m bo 2iu kin: UAK> 0, v c dng iu khin IG mnh( v c cng sut vthi gian). Khi thyristor dn nu ngt dng iu khin i( cho IG=0) n svn dn chng nao dng in qua n ln hn mt gi tr gi l dng duy tr.
- Trong thc tngi ta s dng thyristor kiu N nhiu hn.II. CHNH LU THYRISTOR CU 3 PHA.
1.1 s chnh lu cu thyristor cu 3 pha.
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1.2cc tham s chnh ca mch chnh lu cu 3 pha:- gi trtrung bnh in p chnh lu.
Khi th Ud0 = 2,34.U2- Tr strung bnh dng in qua van:
- in p ngc ln nht m van phi chu khi lm vic:
- Cng sut tnh ton mch MBA ngun: Sba = 1,05.Pd- Hs mch cu in p chn lu Km = 0,057 s chnh lu 3 pha( H
gm 6 thyristor chia than hai nhm)+ Nhm catot chung: T1,T3,T5+ Nhm anot chung: T2,T4,T6.
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- in p pha:= =
= G m c tnh tgiao im ca cc na hnh sin ( hay thi gian chuynmch t nhin).
- Hot ng ca s .Gi thit T5 v T6 ang cho dng chy qua VF = U2C ,VG = U2b khi cho xung iu khin m T1. Thyristor mv U2a >0. S mcaT1 lm cho T5 b kha li mt cch t nhin v U2a > U2c, lc ny T6 v T1
cho dng chy qua. in p trn ti Ud = Uab =U2a = U2b.Khi cho sung iu khin mT2. Thyristor ny mv khiT6 dn dng, n t U2b nn anot T2 m U2b > U2c. s mca T2 lm cho T6b kha li mt cch t nhin U2b > U2c.
Cc sung iu khin lch nhauc ln lt a n cc iu khin ca
cc thyristor theo th t1,2,3,4,5,6,1,2..Trong mi nhm khi 1 Thyristor m n s kha ngay thyristor dn dng
trc n.Tm tt bng:
Gi trtrung bnh in p ti:
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Nhn xt: Chnh lu cu thyristor cu 3 pha l loi c s dng nhiu nht trongthc t v c nhiu u im hn c. N cho php u thng vo li in 3 pha, p mch rt nh(5).Nu c dng bin p th gy mo li it hn cc loi trn ng thi cng sutMBA cng ch xp x bng cng sut ti. cng sut mch chnh lu c th rt lnln ti hng trm kw.
Nhc im ca n l st p trong mch van gp i s hnh tia nn khng phhp vi in p ra ti di 10V.
2.2.2. Hin tng trng dn .
Gi s T1 , T2ang dn dng . Khi = 1 cho xung iu khin mT3 , do Lc0 nn iT3 khng thtng t ngt t0 Id v dng iT1cng khng th gim tngt t Id 0 . C ba thyristor T1 , T3 , T5u dn dng , hai ngun ea , eb ningn mch .
Nu chuyn gc to t0 n 1 ta c :
ea = 2 . U2sin( +6
5 + ) (2-2)
eb = 2 . U2sin( +6
+ ) (2-3)
in p ngn mch Uc = UbUa = 6 . U2sin( + )Dng in ngn mch icc xc nh bi phng trnh :
Uc = 6 . U2sin( + ) = 2Xc. ddi
c ; ic =cX
U.2
.62 [ cos cos( + ) ] ; (2-5)
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Hnh 2-6: S chnh lu cu ba pha c hin tng trng dn .
Hnh 2-7 : S dng sng chnh lu cu ba pha c trng dn .
Dng in chy trong T1 l iT1 = Td - Ic ;
Dng in chy trong T3 l iT3 = ic ;Gi thit qu trnh trng dn kt thc khi = 2 v k hiu = 21 l gc trngdn .
Khi = , iT1 = 0 , ta c biu thc sau :
2
ac ee
Ud
0
1 3 5
6 2 4
5;1 1;3 3;5
6;2 2;4 4;6
2
cb ee
2
ba ee
u2a u2b u2c
VG
VF
FG
T1T4
T3
T6
T5T2
Lc
I
eaebec
LcLc
ic
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cos cos( + ) =2.6
..2
U
IX dc (2-6)
Hnh dng in p ti Udtrong giai on trng dn trong khong 1 2 . T2 dn
dng , T1 , T3 trng dn dng . Vy c th vit cc phng trnh sau :
ea - ec - 2Lcdt
diT1 = Ud (2-7)
ebec - 2Lc.dt
diT3 = Ud (2-8)
iT3 + iT1 = iT2 = Id =const (2-9)
T 3phng trnh ta rt ra :
Ud =2
ba ee - ec (2-10)
Do trng dn ( Lc 0) nn tr trung bnh ca in p ti b gim i mt U ,v c xc nh nh sau :
U =.2
6
0 0
).sin(.63
)2
( ddee
eba
b
=.2
63[cos - cos(+ ) ] (2-11)
T ta c thxc nh c : U =
cLX ...3 ; (2-12)
2.2.3. Nghch lu ph thuc.
Ta c ch chnh lu dng in trung bnh trn ti Idv in p trung bnh
Ud lun cng chiu . Cng sut tiu th trn ti P = Ud.Id lun dng v chiuca cng sut lun t pha ngun xoay chiu chuyn qua ti mt chiu , ta ni bbin i lm vic ch chnh lu.
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Hnh 2-8 : a ) Ch chnh lu ; b ) Ch nghch lu .Khi tng gc m , gi tr trung bnh ca in p chnh lu Ud gim i nhng
vn cn dng . Tip tc tng > 2
, in p chnh lu trung bnh i du . Cc
M tr thnh m ( - ) v N thnh dng ( + ) , in pUd tng dn n khi = .V in p Udi chiu trong khi Id c chiu khng i nn cng sut P i du .iu ny c ngha l ti mt chiu trthnh ngun pht v cng sut truynngc t pha ti v ngun . Ta ni b bin i chuyn sang lm vic chnghch lu . Lu rng ch nghch lu , li xoay chiu nhn cng sut tcdng t pha ti nhng vn tip tc cung cp cng sut phn khng v nh hngn dng sng , tn s ca in p xoay chiu ta ni rng s ny lm vic ch
nghch lu ph thuc .
li c th nhn nng lng t pha ti th ti phi l mt ngun pht vin p ca ti phi ln hn in p ca ngun m bo cho van bn dn phncc thun
Hnh 2-9 : S dng sng ca nghch lu ph thuc .
Ti
a )
M
N
_
+
TiUd
Id
BB
Cng sut P = Ud.Id > 0
b )
M
N
+
_
Ud
Id
BB
Cng sut P = Ud.Id < 0
u2a u2b u2c
2 30
U VG
VF
t
1 3 5
6 2 4 6 2 4
1 3
Utb
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Trong trng hp nghch lu ph thuc th ta phi c gc iu khin > /2ta c in p trung bnh ca nghch lu ph thuc l Utb< 0 nh hnh 2-9 .
B.S NGHUN L H H THNG CHNH LU NG C IN 1CHIU ( H T) KCH TC LP.
3.1 Tng quan v h truyn ng T- khng o chiu.
Khi dng cc b chnh lu c iu khin hay l b chnh lu dng thyristor lmb ngun mt chiu cung cp phn ng ng c in mt chiu, ta c in p chnhlu ca h:
Vy ta c phng trnh c in v phng trnh c tnh c ca h chnh lu T
khng o chiu.
l tc khng ti l tng, v lc vng dng in ginon h s c thm mt lng st p nn ng c tnh biu din dc hn. Tc khng ti l tng thc s ln hn tc khng ti gitng Vy khi thay i gc iu khin th thay i tn (- v
ta sc mt hc tnh song song vi nhau nm bn phi mt phng to [I] hoc [ M].
Trong : XBA: in khng my bin p. fl: Tn sli.
: in cm tng mch phn ng. m: s pha chnh lu
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a. Ch dng lien tc.Dng chnh lu Id chnh l dng in phn ng ng c in. Da vo s thayth ta vit c phng trnh c tnh:
2.3. Tng quan h truyn ng T c o chiu.2.3.1. Nguyn tc c bn xy dng h truyn ng T-o chiu :
- Gi nguyn chiu dng in phn ng v o chiu dng kch t .
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- Gi nguyn dng kch tv o chiu dng phn ng nhng c phn rabn s chnh :
+ Truyn ng dng mt b bin i cp cho phn ng v o chiu quay
bng cch o chiu dng kch t .+ Truyn ng dng mt b bin i cp cho phn ng v o chiu quay
bng cng tc t chuyn mch phn ng ( t thng gikhng i ) .
+ Truyn ng dng hai b bin i cp cho phn ng iu khin ring .
+ Truyn ng dng hai b bin i ni song song iu khin chung .
Tuy nhin , mi loi s u c u nhc im ring v thch hp vi tngloi ti , trong phn ny ta chn b truyn ng dng hai b bin i ni song songngc iu khin chung , bi n dng cho di cng sut va v ln c tn sochiu cao vthc hin o chiu m hn .Trongs ny ng c khng nhngo chiu c m cn c
2.3.2.Phng php iu khin chung :
S gm hai b bin i G1 v G2, u song song ngc vi nhau v cccun khng cn bng Lc . Tng b bin i c th lm vic ch chnh luhoc nghch lu .
Phng php iu khin kiu tuyn tnh : 1 + 2 =
Lc ny c hai mch chnh lu cng c pht xung iu khin , nhng lunkhc ch nhau : mt mch ch chnh lu ( xc nh du ca in p mtchiu ra ti cng l chiu quay ang cn c ) cn mch kia ch nghch lu . Vhai mch cng du cho mt ti nn gi tr trung bnh ca chng phi bng nhau .
Ut = Ud1 = U d2 ; (2-13)
Nu dng in lin tc ta c : Ud1 = Ud0.cos1 ;
Ud2 = Ud0.cos2 ;
Vy : Ud0.cos1 = Ud0.cos2 ;
Hay : cos1+ cos2 = 0 ; suy ra 1 + 2 = 1800 ;
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Nu 1 l gc mi vi G1, 2 l gc mi vi G2 th s phi hp gi tr1 v 2 phi c thc hin theo quan h :
1+ 2 = 1800 ;
S phi hp ny gi l phi hp iu khin tuyn tnh (hnh 2-11) .
Hnh 2-11 : S phi hp tuyn tnh ca 1v 2 .
Gi s cn ng c quay thun , ta cho G1 lm vic ch chnh lu ,1= 0 90
0 , Ud1 > 0 , by gi2 > 900 , G2 lm vic ch nghch lu , Ui2 0 ;
Ui2 = U0cos 2 < 0 ;
Chai in p Ud1 v Ui2u c ln phn ng ca ng c M . ng c chc thnghe theo Ud1 v quay thun . ng c t chi Ui2 v cc thyristor khngth cho dng chy tcatt n ant .
Khi 1= 1 = 900
, th Ud1 = Ui2= 0 , ng c trng thi dng.th hm tisinh .
2 = 0 1 = 0
1=1800
2=1800
300
600
900
1500
1800
1200
0
ucuc1
uc2
uc4 uc3
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Hnh 2-12 : S chnh lu cu ba pha dng phng php iu khin chung .
Gi s ucl in p iu khin biu khin cn khi ng M quay thunta cho uc = uc1 (hnh 2-11) .
1 = 300, 2 = 180
0 - 1 = 1500 , Ud1 =
2
3U0 , Ud2 = -
2
3U0 ;
G1 lm vic ch chnh lu cn G2 chun b sn sng lm vic chnghch lu . Nu by gicn gim tc ng c , ta cho uc = uc2 , cc gc m:
1 = 600, 2 = 180
0 - 1 = 1200 , Ud1 =
2
1 U0, Ud2 = -2
1 U0 .
T1T4
T3
T6
T5T2
Lc
FG
T4T1
T6
T3
T2T5
+ -Ckt
M
Ikt
L
Lc
Lc
Lc
u2au2bu2c
G1
G2
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Lc ny , do qun tnh nn sc in ng E ca ng c vn cn gi nguyntr s ng vi trng thi trc , E > Ud1 b bin i G1 b kho li .
Mt khc E > |Ud2| nn b bin i G2 lm vic ch nghch lu ph
thuc , trnng lng tch lutrong ng c v ngun in xoay chiu . Dng inphn ng i du , chy t M vo G2ng c b hm ti sinh , tc gim xungn gi trng vi Ud1 .
Nu cho in p iu khin uc < 0 th G2 s lm vic ch chnh lu , cnG1 s lm vic ch nghch lu ph thuc .
Vy bng cch thay i in p iu khin uc ( uc > 0 hoc uc < 0 ) ta s thayi c gc m1v 2 :
+ Nu uc> 0 th 1 < 900, 2 > 90
0 dn n b chnh lu G1 lm vic chchnh lu , cn b bin i G2 lm vic ch nghch lu ph thuc kt qu llm cho ng c quay thun T .
+ Nu uc< 0 th 1 > 900, 2 < 90
0 dn n b chnh lu G1 lm vic chnghch lu ph thuc , cn b bin i G2 lm vic ch chnh lu , kt qu llm cho ng c quay theo chiu ngc N .
c im ca cho dng ang xt l c mt dng in lc th chy tG1 vo G2 , lc th chy t G2 vo G1 m khng qua mch ti . Ngi ta gi dng
in ny l dng in tun hon .Dng in tun hon lm cho my bin p v cc thyristor lm vic nng n
hn . hn chdng in tun hon ngi ta dng bn in cm Lc(nh hnh 2-12) . Nh th slm tng cng sut t v gi thnh h thng . Tuy nhin phng
php iu khin chung cho php iu chnh nhanh ti a .
- Xc nh dng in tun hon icc .
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s
Xt trung hp 1 = 30
0
, 2 = 180
0
- 1 = 150
0
, nh hnh 2-13 . Trong khong1n 3 : c T1 v T6 , T2 v T5 dn dng ,nhng ant T5 v catt T2 c cng mtin th , khng c dng chy t T5 sang T2 . Ch c dng tun hon chy t G1vo G2 qua T1 v T6. in p tun hon trong khong ny l :
ucc12 = u2au2b = 6 U2sin( + /6) ;
Nu chuyn to t O sang O2 , ta c :
ucc12 = -6
U2sin = 2.Xc ddicc12
;
icc12 =c
X
U
2
.6 2 .cos + C ;
Khi = 1 , icc12 = 0, ta c :
6 2 4
1 3 5
= 300
= 1500
1
2
2
u u u
1 2 3 4
6 2
3 5
1-1 0
0
0
0
icc21 icc12
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icc12 = -cX
U
2
.62 .(cos - cos1) ; (2-14)
Tip tc xt cc khong khc , kt qu nhn c cho php ta kt lun :
hon icc12 chy t G1 vo G2, v ba xung dng in tun hon icc12 chy t G2 voG1 .
Tr trung bnh ca dng in tun hon :
Icc =2
3
1
1.2
.6 2
cX
U(cos - cos1)d =
cX
U
2
.63 2 .(sin11cos1) (2-15)
Phng php iu khin kiu phi tuyn : 1 + 2 = + .
y l kiu iu khin phi hp khng hon ton th lc ny s c thm h sphi tuyn v ta c :
1 + 2 = + ;
Gc ph thuc vo cc gi tr ca 1v 2 mt cch phi tuyn .
Hnh 2-14 : a ) S iu khin chung phi hp kiu tuyn tnh .
Id
0dm
1min
2max
Ic
1= /2
-0dm
1max
2min
a )
Id
1min
2max
0dm
-0dm
1max
2min
1= /2
Ic
b )
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b ) S iu khin chung phi hp kiu phi tuyn .
2.3.2.Phng php iu khin ring :
Hai mch chnh lu hot ng ring bit . Mch ny hot ng (c pht
xung iu khin ) th mch kia hon ton ngh ( b ngt xung iu khin ) . V vyloi trc dng in tun hon v khng cn cun khng cn bng Lc . Songtrong qu trnh o chiu cn c thi gian cht ( nh nht l vi ms ) cho vanca mch phi ngng hot ng kp phc hi tnh cht kho ri mi bt u phtxung cho mch kia hot ng . V vy cn mt khi logic iu khin o chiu tincy v phc tp .
thay i trng thi lm vic ca cc b chnh lu th phi dng thit bcbit chuyn cc tn hiu iu khin t b chnh lu ny sang b chnh lu kia .Bi vy , khi iu khin ring , cc dc tnh c s bgin on ti trc tung . Nhvy , khi thc hin thay i ch lm vic ca h skh khn hn v h c tnhlinh hot km hn khi iu chnh tc .
Trong phng php iu khin ring cng c phi hp iu khin kiu tuytnh v phi tuyn .
T
T
T
T
T
T
FG
T4T1
T6
T3
T2T5
+ -Ckt
M
Ikt
L
u2
u2
u2
G1
G2
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Hnh 2-15: S cu chnh lu cu ba pha c dng phng php iu khinring .
Do s ny dng phng php iu khin ring nn trong mch khng c
dng in tun hon dn n trong mch ta khng dng bn cun khng cn bngLc .
S nghuyn l h truyn ng T-:
B bin i van thyristor l mt loi ngun in p mt chiu, n trc tip bin idng soay chiu thnh dng mt chiu. vic iu chnh in p u ra ca b bini, c thc hin bng cch iu chnh gc m
ca van.
in p chnh lu Ud0( in p khng ti u ra) c dng p mch vi tn sp mch l n trong mt chu k ca in p s cp ca my bin p lc.
S chnh lu cu n= 2m.(vi m l spha).s hnh tia n=m.
Gi sin p cp cho b bin i van c dng.
K
n
Uk
C0
T1
T3
T5
T4
T6
T2
u2a u2b u2c
BA
U1, f1
+ -CK RK
CCSX
BKC
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tTa bit sau mt chu kdng in v in p lp li nn ta ch cn xt cho mtchu kl , coi in trvan Rv=0.
S thay th mch.Mt b bin i van c th bao gm : My bin p lc, t van, khng lc, thit bbo v v h thng iu khin.
S thay th c dng nh sau:
S thay th chnh lu thyristor ng c in mt chiu
Khi van dn ta c phng trnh:
- Vi : =
= Nhn xt:- u im : H( T-) tc ng nhanh, tn tht nng lng t, kch thc v
trng lng nh, khng gy n v d tng ha do cc van bn dn c h
s khch ai ln, iu rt thun tin cho vic thit lp cc h thng tng iu chnh nhiu vng nng cao cht lng cc c tnh tnh v ccc tnh ng ca h thng.
- Nhc im: Do cc van bn dn c tnh phi tuyn, dng in p chnh lura c p mch cao, khnng linh hot v chuyn trng thi lm vickhng cao, gy tn tht ph v lm su iu kin chuyn mch trn c gp.
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- Khc phc: thit k truyn ng van c gng lm hp vng dng gin onbng cch ni khng lc ln, tng s ln p mch, ni van m.
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Chng III
TNH TON MCH NG LC
Vi cc s liu:
- Ngun in li xoay chiu 3 pha: 220V/380V.- Ti l ng c in mt chiu kch tc lp c:+ Pm = 10KW ; Um = 220V ; Im = 63A; R = 0,3.+ H s d trin p Ku =1,5 1,8+ H s d trdng in Kd = 1,1 1,4
I. Tnh ton cc thng s cn li c bn ca ng c.ta c:
=
= 0,86 =0,145
ta chn = 0,86 l hiu sut cu ng c in 1 chiu.II. Tnh ton chn van ng lc.
Van n lc c la chn da vo cc yu tc bn: dng ti, s chn,iu kin tn nhit, in p lm vic.
1 . tnh ton v chn thyristor.a.in p ngc ca van.
(1.1)Vi = thay vo (1.1) lc c th tnh:
= = = 230.3 (V).Trong : in p ti, ngun xoay chiu, ngc ca van.
Ud=220, = cc h sin p ngc v in pti.
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c th chn van theo in p hp l, in p ngc ca van cn chn phi lnhn in p lm vic Ulv qua h s d trin p Kdtu = ( 1,5 ta chn :
Kdtu = 1,8:
= 1,8. 230,3 =414,45 (v).b. Dng in lm vic ca van.
Chn dng in qua van.Vi s cu 3 pha ta c:
= 36,4 (A).Chn h s d tr dng l ( 1,1 ) ta chn =1,4
= 1,4 . 36,4 = 50,96 (A).Van chu c dng in nh mc l: = 50,96 (A).Sau khi tnh ton ta phi chn thyristor vi cc thng s: Idmv = 50,96 (A),
Unv = 414,54 (V).ta chn c 6 thyritor loi : T60N600BOC.Vi cc thng s:+ Dng in nh mc ca van : =60 (A)+ in p ngc t ln van: = 600 (V).+ st p trn van: + Dng in r : 25 (mA).+ Dng in nh cc i : .+Dng in xung iu khin: = 150(mA).+ in p xung iu khi: = 1,4 (V).+ Thi gian chuyn mch ( mv kha) : + Nhit lm vic cc i 125 (0C ).
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+ Dng duy tr: + o hm in p:
= 400( V/s).
2. Tnh ton my bin p( MBA).T cc thng sc bn :
+ in p trung bnh trn ti: Ud = 220 (V).
+ Dng lm vic trung bnh trn ti( A- DC) : Id = 63( A).
Ta chn my bin p 3 pha hnh tr.
im p pha s cp MBA: in p chnh lu khng ti.
Ta c: = 235 (V).Ud = 220 (V).
in p st p trn van.
.St p bn trong my bin p khi c ti. Bao gm cc st p
trn in tr v st p trn in cm cc ai lng ny khi chon s bvo khong ( 5 chn = 13,2 ( V).st p trn dy ni: . = . = 0 Cng sut biu kin cu my bin p.
. = 1,05. = 12209,3 ( W).
- cng sut biu kin my bin p.
- h s cng sut ca s mach ng lc cu 3 pha iukhin i xng.
in p cudy s cp : in p cu cc cun dy th cp.
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= = 100,43 ( V).in p chnh lu khng ti.
H sin p chnh lu mch cu 3 pha iu khin i xng.
+ Vi in p cun s cp : U1 = 380(V).+ T s bin p: n = = 100,6/ 380 = 0,26.
Dng in ca cun dy:- Dng in chy trong mi pha cun th cp:
= 63 . = 51,44 (A)- Dng in chy trong mi pha cun s cp.:
= . n = 51,44 . (A).a. Tnh ton s b mch t. Tit din s b tr.
Ta c : . = 6. = 54,13 (cm2)
Cng sut my bin p tnh bng W.
- H s ph thuc vo hnh thc lm mt. Ta chon lm mt my bin pkh .ms tr ca my bin ap( v my bin p 3 pha chon m= 3).f - tn s ngun in xoay chiu. vi li in Vit Nam ta chn f= 50(H).
ng knh tr: = 8,3 ( cm).Chn ha ng knh tr theo tiu chun: d = 8(cm).( theo bng 41a trang213 sch thit k my bin p in lcPHAN TTH).
Chn loi thp 330 cc l thp c dy 0,5mm chn s b mt t cmtrong tr BT = 1 T.
Chn t s m = h/d = 2,3. Suy ra h = 2,3 . d = 18,4 cm.( thng thng m =2 2,5 ).
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- Chn chiu cao tr h = 18 cm. S vng dy ca cun s cp:
W1 =
0,1.10.1,61.50.44,4
380
...44,44
1
BQf
U
Fe
= 280 ( vng)
Chn W1= 280 (vng)
S vn dy ca cun th cp:W2 =W1 .
U
U.
1
2= 280.
380
43,100=72,8 ( vng).
Chn W2 = 73 (vng).
- Vi f = 50 nn ta chn B= 1T( Tesla). Tit din dy dn cun s cp my bin p:
Ban u chn : ( 2 ). .
- Chn dy dn c tit din hnh ch nht chun ha theo tiu chun cchin l :
.
- Kch thc ca dy c kn cch in l: .- Tnh li mt dng in trong cun s cp:
= 2,76 A/. Tit din dy dn qun th cp my bin p:
Ban u chn : ( 2 ).
.
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- Chn dy dn c tit din hnh ch nht chun ha theo tiu chun cchin l : .
- Kch thc ca dy c kn cch in l:
.
- Tnh li mt dng in trong cun scp: = 2,722 A/
Kt cu dy qun qun s cpThc hin dy qun kiu ng tm b tr theo chiu dc trc.
- S vng dy trn mt lp ca qun s cp: = 39,1 (vng)Ly = 39 ( vng).
Trong : H s p cht.h- Chiu cao tr.
khong cch tgng n cun dy s cp. chn s b 1,5 cm- S lp dy qun s cp: 2,7
39
280
11
111
W
Wn (lp).
- Chn s lp = 7 ( lp).- Nh vy c 280 vng chia thnh 7 lp. mt lp c 40 vng.- Chiu cao thc t ca cun s cp:
= = 18,41(cm).- Chn ng qun dy lm bng vt liu cch in c b dy:
- Khong cch t tr ti cum s cp .- ng knh trong ca ng cch in: 2 = 9 + 2.1 2. 0,1 = 10,8 (cm).
- ng knh trong ca ca qun s cp: + 2 = 10,8 + 2.0,1 = 11 (cm).
- Chn bdy cch in gia cc lp dy cu qun s cp l:
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.- B dy cun s cp:
= ( ). = ( 1,23 + 0,1).7 = 9,31 (mm).= 0,931(cm).- ng knh ngoi ca cun s cp:
+ 2 = 11 + 0,931. 2 = 12,862(cm).- ng knh trung bnh cun s cp: = = = 11,931 (cm).
- Chiu di dy qun s cp: =. = 11,931 . . 280 =10495,1 (cm) = 104,951 (m).
- Chn bdy cch in gia cun s cp v th cp: .
Kt cu dy qun th cp:- Chn s b chiu cao cun th cp: h1 = h2 = 18,41 (cm).Tnh s b vng dy trn mi lp :
= 19( vng).- Tnh ton s b lp dy qun th cp: = = = 3,8 ( lp).- Chn s lp dy qun th cp:
= 4 (lp).Vy 73 vng chia thnh 4 lp, 3 lp u mi lp 24 vng, 1 lp sau c 1vng.
- Chiu cao thc tc ca cun th cp: = (cm).
- ng knh trong ca cun th cp: 1 = 14, 862 (cm).Vi a12 : l bdy cch in cc cun dy vi nhau.(vi my bin p cdng in nh th a12 thc(0,3
chn = 1mm.)
- Chn bdy cch in gia cc lp dy cun th cp: - B dy cun th cp:- = ( ). = ( 0,224 + 0,01).4 = 0,936 (cm).- ng knh ngoi ca cun th cp:
+ 2 = 14,628 + 2.0,936 = 16,5 (cm)
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- ng knh trung bnh ca cun th cp: = = = 15,653 (cm).
- Chiu di dy qun th cp: =. = 15,653. . 73 = 3569(cm) = 35,69 (m).- ng knh trung bnh cc cun dy: = = = 13,75 (cm).
- Chn khong cch gia hai cun th cp:
= 2( cm).
Tnh kch thc mch t.Vng knh tr l d = 9 (cm) ta c s bc l 5 trong na titdin tr.( theo bng 41a trang 213 sch thit k my bin p in lc PHAN TTH).
- Tnh ton b tit din bc thang ca tr: 56,7(). ( theo bng 41a trang 213 sch thit k my bin
p in lcPHAN TTH).
- Tit din hiu qu cu tr: = 56,7 . 0,891 = 55,63( cm2).- Tng chiu dy ca cc bc thang hnh tr:
- S l thp trong cc bc:
+ Bc 1:
+ Bc 2: + Bc 3:
+ Bc 4:
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+ Bc 5: - n gin trong ch to to ta chn gng c tit din hnh ch nht
c cc kch thc sau:+Chiu dy ca gng bng chiu dy cu tr: b = dt =8 (cm).+Chiu cao ca gng bng chiu rng tp l thp th nht cu tr: a =8,5 (cm).+Tit din gng: Qbg = a
- Tit din hiu qu ca gng: .
- S l thp dng trong mt gng: Tnh chnh xc mt in cm trong tr:
---
- Mt t cm trong gng:
- Chiu rng ca s:c = 2( + + + ) + = 2( 1 +0,931+ 1 + 0,936) + 2=9,734 (cm).
- Khong cch gia hai tm truc:
- Chiu rng mch t:L = 2c + 3d = 9,734. 2 + 3. 9 = 44,468 (cm).
- Chiu cao mch t: H = h + 2a = 21 + 2 .8,5 = 38 ( cm). Tnh khi lng cu st:
- Th tch ca tr:
- Th tch ca gng: 5388
)(099,110.63,55.280.50.44,4
380
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WB
Tg
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63,55..
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g
T
g
7/30/2019 N IN T CNG SUT trung
47/49SVTH: L XUN TRUNG 11D1CLC
- Khi lng ca tr: = 3,541 . 7,85 = 27,797 (kg)
- Khi lng ca tr: = 3,541 . 7,85 = 27,797 (kg)
- Khi lng ca gng:
- Khi lng cu st:
- Th tch ca ng:
104,951.10. .+35,69.10.18,9. )= 3,573 (dm3)
- Khi lng cu ng: 8,9 . 3,573
Tnh ton cc thng s ca my bin p:- in trtng ng cun s cp my bin ap 75 o C:
. 0,02133 =0, 455().Trong ).
- in trtng ng cun s cp my bin ap 75 o C: . 0,02133 = 0,04().
Trong ).- in tr my bin p quy i v th cp:
= 0,04 + 0,455( = 0,071().- St p trn in trmy bin p:
7/30/2019 N IN T CNG SUT trung
48/49SVTH: L XUN TRUNG 11D1CLC
- in khng my bin p quy i v th cp: (
)
0,087().Trong : +
+ r = = = 7,483 (cm) bn knh trong dy qun th cp.- in cm my bin p quy i v th cp:
- St p trn in khng my bin p:
- St p trn my bin ap: =
- Tng trngn mch quy i v th cp: =
- Tn hao ngn mch trong my bin ap: . = 3.0,071. = 564(W).
-
7/30/2019 N IN T CNG SUT trung
49/49
+ Attomat: dng ng ct mch ng lc, tng bo v khi qu ti v ngnmch.
Chn attomat: