24

Chng 2 TRANG B IN NHM MY TIN

2.1 c im cng ngh Nhm my tin rt a dng, gm cc my tin n gin, my tin vn nng, chuyn dng, my tin ngTrn my tin c th thc hin c nhiu cng ngh tin khc nhau: tin tr ngoi, tin tr trong, tin mt u, tin cn, tin nh hnh. Trn my tin cng c th thc hin doa, khoan v tin ren bng cc dao ct, dao doa, tar renKch thc gia cng trn my tin c th t c vi mili n hng chc mt

4

Hnh 2.1 Dng bn ngoi my tin Dng bn ngoi ca my tin nh hnh 2.1a. Trn thn my 1 t trc 2, trong c trc chnh quay chi tit. Trn g trt t bn dao 3 v sau 4. Bn dao thc hin s di chuyn dao ct dc v ngang so vi chi tit. sau t mi chng tm dng gi cht chi tit di trong qu trnh gia cng, hoc gi mi khoan, mi doa khi khoan, doa chi tit. S gia cng tin nh hnh 2.1b. my tin, chuyn ng quay chi tit vi tc gc ct l chuyn ng chnh, chuyn ng di chuyn ca dao 2 l chuyn ng n dao. Chuyn ng n dao c th l n dao dc, nu dao di chuyn dc chi tit (tin dc) hoc n dao ngang, nu dao di chuyn ngang (hng knh) chi tit. Chuyn ng ph gm c xit ni x, tr, di chuyn nhanh ca dao, bm nc, ht phi.

25

2.2 Ph ti ca c cu truyn ng chnh v n dao 1. Ph ti ca c cu truyn ng chnh Qu trnh tin trn my tin c thc hin vi cc ch ct khc nhau c trng bi cc thng s: su ct t, lng n dao v tc ct v. Tc ph thuc vt liu gia cng, vt liu dao, kch thc dao, dng gia cng, iu kin lm mt v.v. theo cng thc kinh nghim

stT vV yX

mvCv = , [m/ph] (2-1)

vi - t: chiu su ct , mm s: lng n dao, l dch chuyn ca dao khi chi tit quay c mt vng, mm/vg T: bn ca dao l thi gian lm vic ca dao gia hai ln mi dao k tip, ph Cv, xv, yv, m l h s v s m ph thuc vo vt liu chi tit, vt liu dao v phng php gia cng m bo nng sut cao nht, s dng my trit nht th trong qu trnh gia cng phi lun t tc ct ti u, n c xc nh bi cc thng s: su ct t, lng n dao s v tc trc chnh ng vi ng knh chi tit xc nh. Khi tin ngang chi tit c ng knh ln, trong qu trnh gia cng, ng knh chi tit gim dn, duy tr tc ct (m/s) ti u l hng s, th phi tng lin tc tc gc ca trc chnh theo quan h: v = 0,5dct.ct (2-2) vi dct: ng knh chi tit, m Trong qu trnh gia cng, ti im tip xc gia dao v chi tit xut hin mt lc F gm 3 thnh phn v lc ct c xc nh theo cng thc: Fz = 9,81CF.txF.syF.vn , [N] (2-3) Qu trnh tin xy ra vi cng sut ct Fz

V

Fz

V

Hnh 2-2 th ph ti ca truyn ng chnh my tin

(kW) l hng s: Pz = Fz.v.10-3 , [kW] (2-4) Bi v lc ct ln nht Fmax sinh ra khi lng n dao v su ct ln, tng ng vi tc ct nh Vmin; cn lc ct nh nht Fmin , xc nh bi t, s tng ng vi tc ct Vmax, ngha l tng ng vi h thc: Fmax.vmin = Fmin.vmax (2-5) S ph thuc ca lc ct vo tc nh h2.2 Tuy nhin nh phn tch, dng th ph ti thc t ca truyn ng chnh my tin c dng hai vng Fz = const v Pz = const (h 1.4)

26

2. Ph ti ca truyn ng chnh my tin ng Truyn ng chnh my tin ng c dng c th ring, khc so vi my tin bnh thng v cu trc v kch thc. Trn my tin ng, chi tit gia cng c ng knh ln v c t trn mm cp nm ngang, hay ni cch khc trc mm cp l theo phng thng ng. Do trng lng mm cp, trng lng chi tit ln ln nn lc ma st g trt v hp tc kh ln. V vy ph ti trn trc ng c truyn ng chnh my tin ng l tng ca cc thnh phn lc ct, lc ma st g trt, lc ma st hp tc .

Hnh 2.3 th ph ti ca truyn ng chnh my tin ng

Trn hnh 2.3a, l th biu din cc thnh phn cng sut ca truyn ng chnh v s ph thuc ca chng vo tc mm cp: P1 cng sut khc phc lc ct; P2 cng sut khc phc lc ma st g trt; P3 v P4 cng sut khc phc lc ma st trong hp tc tng ng do lc ct v s quay ca mm cp; P5 - tng cng sut ca truyn ng chnh. Trn hnh 2-3b, l cc thnh phn mmen tng ng vi tc ca mm cp. Thnh phn lc ma st ph thuc vo tc nh hng ln n qu trnh qu ca truyn ng chnh. Do khi lng ca mm cp v chi tit ln v s khc nhau ca h s ma st lc ng yn v chuyn ng nn mmen cn tnh khi khi ng ca truyn ng c th t ti 60 80% momen nh mc. V momen qun tnh tng qui i v trc ng c c th t ti 8 9 ln momen qun tnh ca ng c nn qu trnh khi ng ca h thng din ra chm vi momen cn tnh ln. Theo mc gia tc ca ng c, momen cn tnh s gim nhanh v khi tc tng th n t thay i. 3. Ph ti ca truyn ng n dao Lc n dao ca truyn ng n dao c xc nh theo cng thc: dmsxad FFkFF ++= , [N]

27

Cng sut n dao ca my tin c xc nh bng cng thc: , [kW] 310.. = adadad vFP Cng sut n dao thng nh hn cng sut ct 100 ln v tc n dao c xc nh bi lng n dao v tc gc chi tit: , [m/s] (2-6) 310.'. = ctad sv nh hn tc ct nhiu ln. y 2'

ss = , [mm/rad] Lc v mmen ph ti ca truyn ng n dao khng ph thuc vo tc ca n, v ph ti ca truyn ng n dao ch c xc nh bi

Mc

V

V1 V2 V3

Mc

V

V1 V2 V3

Hnh 2.4 th ph ti ca truyn ng n dao

khi lng b phn di chuyn ca my v lc ma st g trt v hp tc . Trn th ph ti ca truyn ng n dao hnh 2.4, di tc rng v1< v v2 momen ph ti s thay i tuyn tnh theo tc 3) Thi gian my Thi gian my (thi gian gia cng) ca my tin c xc nh:

ad

M vlt

310.= , [s] (2-7) Trong : l l chiu di gia cng , mm ct l tc gc chi tit, rad/s s lng n dao, mm/vg Kt hp (2-6) v (2-7) ta c cng thc tnh thi gian my:

'.slt

ctNM = , [s] (2-8)

Nh vy gim thi gian gia cng, ta phi tng tc ct v lng n dao v nng sut s tng. 2.3 Phng php chn cng sut ng c truyn dng chnh ca my tin Truyn ng chnh my tin thng lm vic ch di hn. Tuy nhin, khi gia cng cc chi tit ngn, cc my trung bnh v nh, do qu trnh thay i nguyn cng v chi tit chim thi gian qu ln nn truyn ng chnh phi tin hnh tnh ton mt ch nng n nht. Gi thit trn my tin thc hin gia cng chi tit nh hnh 2-5. Cc nguyn cng khi gia cng gm 4 giai on: 1 v 3 - tin ct hoc tin ngang; 2 v 4 - tin tr (tin dc). Ph ti ca ng c trong tng nguyn cng ph thuc vo cc thng s ch ct, vt liu chi tit dao v.v

28

Qu trnh tnh ton nh sau: a) T cc yu t ch ct gt, theo

1cc cng thc (2-1), (2-3), (2-4) v (2-8) xc nh tc ct, lc ct, cng sut ct v thi gian gia cng d

ng vi tng nguyn cng. Nu tc ct tnh c khng ph hp tc ca my (theo s liu k thut c

2 d 1d 0

234

l4 l2l3 l1

1

2 d 1d d0

234

l4 l2l3 l1

Hnh 2-5 Chi tit c gia cng trn my tin

kh) th chn ly tr s c sn trong my gn ging vi tc ct tnh ton. Dng tr s ny tnh li Pz, tm, theo (2-4) v (2-8). Tr s V, Pz, tm ny c dng chnh thc trong ton b bi ton. b) Chn nguyn cng nng n nht v gi thit nguyn cng y my lm vic ch nh mc. T xc inh hiu sut ca my ng vi ph ti ca tng nguyn cng theo cng thc:

btaMM

M

mshi

hi

++=+= 1

1

a, b - h s tn hao khng bin i v bin i.

Cng sut trn trc ng c ng vi tng nguyn cng : i

ziDi

PP =

Gi thit trong thi gian g lp, tho g chi tit, chuyn i t nguyn cng ny sang nguyn cng khc, ng c quay khng ti (m khng ct in ng c) th cng sut trn trc ng c lc ny l cng sut khng ti ca my, tc l bng lng mt mt khng i: Po= a.Pcm (2-9) ng vi cng sut ny l thi gian ph ca my, chng c xc nh theo tiu chun vn hnh ca my t0c) ng c c th chn theo cng sut trung bnh hoc cng sut ng tr:

==

==

+

+= n

jj

imi

n

jj

ici

tb

tt

PPP

10

4

1

10

4

1

hoc ==

==

+

+= n

jj

imi

j

n

jjmi

ici

dt

tt

tPtPP

10

4

1

01

20

4

1

2 ..

trong :

29

Pci, ti cng sut trn trc ng c, thi gian my ca nguyn cng th i P0j, t0j- cng sut khng ti trn trc ng c, thi gian lm vic khng ti ca my, P0j = P0n - s khong thi gian lm vic khng ti Chn ng c c cng sut nh mc ln hn 20 30% cng sut trung bnh hay ng tr:

Pc

Tck

t

P0 P0 P0 P0

Pc1

Pc2=Pc m

Pc3

Pc4

t01 t02 t03 t04 tm4tm3tm2tm1

Pc

Tck

t

P0 P0 P0 P0

Pc1

Pc2=Pc m

Pc3

Pc4

t01 t02 t03 t04 tm4tm3tm2tm1

Hnh 2-6 th ph ti ca ng c

Pm (1,2 1,3) Ptb hoc Pm= (1,2 1,3)Pt (2-12) d) ng c truyn ng chnh my tin cn phi c kim nghim theo iu kin pht nng v qu ti 2.4 Nhng yu cu v c im i vi truyn ng in v trang b in ca my tin 1. Nhng yu cu v c im chung a. Truyn ng chnh: Truyn ng chnh cn phi c o chiu quay m bo quay chi tit c hai chiu, v d khi ren tri hoc ren phi. Phm vi iu chnh tc trc chnh D< (40125)/1 vi trn iu chnh = 1,06 v 1,21 v cng sut l hng s (Pc = const). ch xc lp, h thng truyn ng in cn m bo cng c tnh c trong phm vi iu chnh tc vi sai s tnh nh hn 10% khi ph ti thay i t khng n nh mc. Qu trnh khi ng , hm yu cu phi trn, trnh va p trong b truyn lc. i vi my tin c nng v my tin ng dng gia cng chi tit c ng knh ln, m bo tc ct ti u

30

v khng i (v = const) khi ng knh chi tit thay i, th phm vi iu chnh tc c xc nh bi phm vi thay i tc di v phm vi thay i ng knh:

min

max

min

max

min

max

min

max

min

max ..ct

dct

ct DD

vv

vD

Dv

D === (2-13)

nhng my tin c nh v trung bnh, h thng truyn ng in chnh thng l ng c khng ng b roto lng sc v hp tc c vi cp tc . cc my tin c nng, my

M,P

VVgh Vmax

P

M

Vmin

2-7 Biu momen v cng sut ng c trong truyn ng chnh

M,P

VVgh Vmax

P

M

Vmin

2-7 Biu momen v cng sut ng c trong truyn ng chnh

tin ng, h thng truyn ng chnh iu chnh 2 vng, s dng b bin i ng c in mt chiu (BB ) v hp tc : khi v< vgh m bo M = const; khi v> vgh th P= const. B Bin i c th l my pht mt chiu hoc b chnh lu dng Thyristor. b. Truyn ng n dao: Truyn ng n dao cn phi o chiu quay m bo n dao hai chiu. o chiu bn dao c th thc hin bng o chiu ng c in hoc dng khp ly hp in t. Phm vi iu chnh tc ca truyn ng in hoc dng khp ly hp in t. Phm vi iu chnh tc ca truyn ng n dao thng l D = (50 300)/1 vi trn iu chnh = 1,06 v 1,21 v momen khng i (M = const). ch lm vic xc lp, sai lch tnh yu cu nh hn 5% khi ph ti thay i t khng n nh mc. ng c cn khi ng v hm m. Tc di chuyn bn dao ca my tin c nng v my tin ng cn lin h vi tc quay chi tit m bo nguyn lng n dao. my tin c nh thng truyn ng n dao c thc hin t ng c truyn ng chnh, cn nhng my tin nng th truyn ng n dao c thc hin t mt ng c ring l ng c mt chiu cp in t khuch i my in hoc b chnh lu c iu khin. c. Truyn ng ph: Truyn ng ph ca my tin khng yu cu iu chnh tc v khng yu cu g c bit nn thng s dng ng c khng ng b rto lng sc kt hp vi hp tc . 2.Cc s iu khin in hnh my tin ng v my tin c nng Cc my tin ng v my tin c nng c mt trong cc ch lm vic c bn l tin mt u. t c nng sut ln nht ng vi cc thng s ca ch ct ti u, yu cu phi duy tr tc ct khng i. t c iu , khi ng knh D ca chi tit gim dn, cn phi iu chnh tc

31

gc ca chi tit ct theo lut hyperbol: ct.D = const. Sau y ta xt mt s s iu khin in hnh.

RTr3

RTr1

RVRTr2

RTr3

RTr3

RD FT1UV UD

Bn dao

PRTr2(T) 1BK

RT

RTr2(N) 2BKRN

RTr1

+ -

-

+

KTRTr1

RT

+ -

KN

RN

RT

RNX

-

+

FT2RC

BB CUc

-

attric ng knh chi tit gia cng khi tin mt u l bin tr DD. Con trt ca n lin h vi bn dao qua b iu tc P. Phm vi di chuyn ln nht ca con trt s tng ng vi ng knh ln nht ca chi tit gia cng trn mt my. in p t ln bin tr RD c ly t my pht tc FT1 t l vi tc gc ca chi tit, v vy UD~ ctD. in p t ln bin tr RV l in p n nh. in p ly con trt ca RV s t l vi tc ct.

+

Rv

BB CUc

Bn dao

PFT2

RD

Uph

(a)

(b)

FTC

(c)

X32C1

C2U~ CL2

CL1

X31 CKFT

Uc Uph BB

RTr3

RTr1

RVRTr2

RTr3

RTr3

RD FT1UV UD

Bn dao

PRTr2(T) 1BK

RT

RTr2(N) 2BKRN

RTr1

+ -

+

-

KTRTr1

RT

+ -

KN

RN

RT

RNX

-

+

FT2RC

BB CUc

-

+

Rv

BB CUc

Bn dao

PFT2

RD

Uph

(a)

(b)

FTC

(c)

X32C1

C2U~ CL2

CL1

X31 CKFT

Uc Uph BB

Hnh 2-8 Cc s iu khin duy tr tc ct l hng s (v = const)

32

Hiu in p cc u con trt ca bin tr RV v RD l UV-UD c t vo rle 3 v tr RTr2. R le ny s iu khin ng c X t tc quay ca ng c chnh C. Khi khi ng, bin tr Rc v tr tng ng vi tc gc mm cp nh nht, cn UD = 0. Sau khi khi ng, ng c chnh (rle KT hoc KN tc ng), do tip im RTr2(T) kn nn rle RT tc ng, ng c X quay theo chiu thun ng vi s tng tc ca ng c chnh v in p my pht tc FT1. Khi in p UD=Uv, rle RTr2 mt in nn RT ngt nn ng c X dng c hm ng nng. Tc ca ng c chnh s tng ng vi tc ct t trc v v tr bn dao khi bt u gia cng. Khi gia cng, bn dao di chuyn ti tm, con trt ca bin tr di chuyn v hng gim UD, do rle RTr2, RT li tc ng; ng c X li quay theo chiu tng tc ng c trc chnh, nh vy duy tr c in p UD~ct.D l hng s. Khi tc gc ng c chnh t gi tr ln nht, cng tc hnh trnh 1BK tc ng, ng c X dng quay. Khi dng mm cp, rle RTr2 tc ng tng ng vi tip im RTr2(N) ng v ng c X quay theo chiu gim tc ng c chnh, con trt bin tr Rc c di chuyn v v tr ban u, cng tc hnh trnh 2BK s b tc ng dng ng c X. Tc ct c duy tr khng i vi chnh xc ph thuc chnh xc ch to b phn lin h gia bn dao v bin tr RD, mc tuyn tnh ca c tnh bin tr RD v pht tc, nhy im khng ca rle cc tnh RTr2, v n nh ca cc thng s ca s khi nhit v in p li thay i. Trn hnh 2-8b l s iu khin tc quay ca ng c C theo hm ca ng knh chi tit gia cng theo nguyn l Uc Uph D. in p ch o Uc t l vi tc ct c t bng bin tr RV. in p phn hi Uph D . Nu h thng iu chnh c b iu chnh PI th lun lun c: Uc = Uph D ngha l Vz = D Trn hnh 2-8c l s iu khin duy tr tc ct l hng s thc hin bng cc attric ng knh v tc kiu khng tip im. in p pht ra ca attric X31 t l vi tc di Vz. in p phn hi ly t my pht tc FT, cun dy kch t pht tc c cp t attric X32 qua cu chnh lu CL2 t l vi ng knh ca chi tit UCL2 = K1D; nh vy in p pht tc UFT = K2D. S iu khin m bo Uc= Uph = K2D v iu khin .D = const chnh xc duy tr tc ct ph thuc vo nhng yu t: c tnh phi tuyn ca attric X32 v pht tc, ng cong t tr ca pht tc.

33

thc hin php nhn cc tn hiu t l vi v D, c th dng b nhn bng in t thay cho my pht tc. u im ca n l iu chnh trn, tin cy cao. Nhc im l kh chnh nh mch sao cho qu trnh qu ti u trong ton b iu chnh. Mt yu cu c bit i vi my tin c nng v my tin ng l duy tr lng n dao khng i. iu c th thc hin bng s 2-9. in p ch o ca h thng truyn ng n dao c ly t my pht tc FT1 ni cng vi trc ng c truyn ng chnh C. Khi UcdD= K1D = K2C v D/ c= const. Chit p RD s t lng n dao

FT2RD

BB2 DUcdFT1BB1 C

FT2RD

BB2 DUcdFT1BB1 C

Hnh 2-9 S duy tr lng n dao l hng s

2.5 Mt s s iu khin my tin in hnh 1. S iu khin truyn ng chnh my tin nng 1A660 My tin nng 1A660 c dng gia cng chi tit bng gang hoc thp c trng lng 250N, ng knh chi tit ln nht c th gia cng trn my l 1,25m. ng c truyn ng chnh c cng sut 55kW. Tc trc chnh c iu chnh trong phm vi 125/1 vi cng sut khng i, trong phm vi iu chnh tc ng c l 5/1 nh thay i t thng ng c. Tc trc chnh ng vi 3 cp ca hp tc c gi tr nh sau: cp 1: ntc = 1,6 8 vng / pht cp 2: ntc = 8 40 vng/ pht cp 3: ntc = 40 200 vng/ pht Truyn ng n dao c thc hin t ng c truyn ng chnh. Lng n dao c iu chnh trong phm vi 0,064 26,08 mm/vg Truyn ng chnh c thc hin t h thng F-. iu chnh tc ng c bng cch thay i dng in kch t ca ng c, cn sc in ng ca my pht gi khng i. a/ Mch ng lc ng c quay truyn ng chnh c cp in t my pht F. ng c s cp quay my pht F khng th hin trn s . Kch t ca ng c l cun CK(2). Kch t ca my pht l cun CKF(9). ng c lm vic c cn G(l) = 1, ni in p my pht vi ng c ng thi K2 (l) = 0, gii phng mch hm ng nng. Cun kch t

34

CK(2) c cp in m bo t thng v cun kch t my pht CKF(9) c in to t thng F lm cho my pht F to ra in p UF . Rle RC(l) bo v qu dng c tip im l RC(27). Khi dng in qua ng c ln hn gi tr cho php, RC(l) = 1, RC(9) = 0, ct in mch iu khin ( dng 27) Rle RH(l) v RCB(l) c gi tr tc ng khc nhau. Ga tr tc ng ca RCB bng gi tr nh mc ca in p my pht; cn gi tr tc ng ca RH bng 10% gi tr nh mc ca in p my pht. RG1 v RD1 l hai cun dng ca rle RG v RD. Hai cun p tng ng l RG2(9) v RD2(8). Hai cun dng v p ni ngc cc tnh nhau. Bnh thng khi cun p c in s lm cho tip im ca rle tng ng ng li. Nu dng in trong ng c ln hn gi tr cho php th cun dng s to ra lc y ln hn lc ht ca cun p lm cho tip im ca n m ra. C th khi: RG(9) = 1, RG(8) = 1; nu I> Icf1 Fy RG1> FhtRG2 RG(8) = 0; RD(8) = 1, RD(4) = 1, nu I> Icf2 Fy RD>Fht RD2 RD(4) = 0, b/ Mch kch t ng c Cun CK(2) l cun kch t ca ng c c cp t ngun mt chiu cng ngun vi cun CKF(9) v l ngun cp cho mch khng ch. Bin tr KT(2) ni tip vi cun CK thay i dng in chy qua n, lm thay i t thng thay i tc ng c trn tc c bn. Khi RKT(2) v R(2) b ni tt th dng CK bng nh mc. Rle dng RT(2) c gi tr tc ng bng dng nh mc ca CK. Rle dng RTT(2) l rle bo v thiu t thng . Gi tr tc ng ca n nh thua dng CK nh nht to ra tc ln nht ca ng c. c/Mch kch t my pht Cun CKF(9) l cun kch t my pht c cp in bi cu tip im T,N(6) v N,T(10). Khi T(6) = 1, v T(10) = 1, tng ng vi chiu quay thun ca ng c. Khi N(6) = 1, v N(10) = 1, tng ng vi chiu quay ngc ca ng c. in tr Rf ni tip vi cun CKF(9) nhm gim dng qua n, kt qu in p ca my pht gim nhm lm gim dng trong ng c. d/Cc iu kin lm vic ca my

1. Phi dng kch t cho ng c RTT(1) = 1, 2. Phi dng bi trn DBT(36) = 1, K4(36) = 1, K4(29) = 1, 3. Cc bnh rng n khp: 1KBR(39) = 1, 2KBR(39) = 1,

3KBR(39) = 1, 4KBR(39) = 1, 4RL(39) = 1, 4RL(29) = 1, 4. Tr s tc c chn T(29) = 1,

35

F

RCG RG1 RD1

RHK2

RhRCB

391KBR 2KBR 3KBR 4KBR

4RL

CTC22RL1RL 38

CTC11RL2RL 37

DBTK4 36

K3 34RCBK1

RT

K2 32GRH

G 31NT

RH

RNT RC

K1 29

LN

LT

K2K1

K44RLT

KN 27DM3 2KX KT

KT 261KX 3RLKNLN 25

M2

3RL

20

RCB

3RLLN

LT

19

M1

LT

N

18

T2RLLNK3

T 17N1RLLTK1TT

TN

2RL

1RL

T

N

K1

K4

K4

C

H1

H2

16

15

14

K2

N

N

T

T

RD2

RG2

K1 GRG2C

Rf

T N

CKF

6

8

910

CKRT RTT

R

K2KT

K3 K3 G

K1 RD1C

1

2

3

4

5

+ -

KT

KT

KN

KNCK11

7

11

1213

21

22

23

24

30

33

35

28

F

RCG RG1 RD1

RHK2

RhRCB

391KBR 2KBR 3KBR 4KBR

4RL

CTC22RL1RL 38

CTC11RL2RL 37

DBTK4 36

K3 34RCBK1

RT

K2 32GRH

G 31NT

RH

RNT RC

K1 29

LN

LT

K2K1

K44RLT

KN 27DM3 2KX KT

KT 261KX 3RLKNLN 25

M2

3RL

20

RCB

3RLLN

LT

19

M1

LT

N

18

T2RLLNK3

T 17N1RLLTK1TT

TN

2RL

1RL

T

N

K1

K4

K4

C

H1

H2

16

15

14

K2

N

N

T

T

RD2

RG2

K1 GRG2C

Rf

T N

CKF

6

8

910

CKRT RTT

R

K2KT

K3 K3 G

K1 RD1C

1

2

3

4

5

+ -

KT

KT

KN

KNCK11

7

11

1213

21

22

23

24

30

33

35

28

Hnh 2-10. S truyn ng chnh my tin h F- (1660)

36

5. Chiu quay c chn: chn ng c quay thun CTC1(37) = 1, 1RL(37) = 1, 1RL(17) = 1 v 1RL(19) = 1; chn quay ngc CTC2(38) = 1, 2RL(38) = 1, 2RL(18) = 1 v 2RL(20) = 1,

e/ Khi ng (khi ng thun) Cc iu kin lm vic . Chiu quay c chn. n nt M1(22) LT(22) = 1, LT(17) = 1, + LT(22,23) = 1, + LT(29) = 1, K1(29) = 1, K1(30) = 1, + K1(34) = 1, + K1(17) = 1, T(17) = 1, T(16) = 1, + T(20) = 0, + T((30) = 1, G(31) = 1, G(32) = 1, K2(32) = 1, K2(30) = 1, ni vi K1(30) to ra mch duy tr cho K1(29). Kt qu khi n nt M1, cc phn t sau y c in: K1, T, G v K2. Trn mch ng lc, G(l) = 1, ni F vi ; K2(l) = 1, gii phng mch hm ng nng. K2(1) = 1, R(2) b ni tt; G(3) = 1, KT(2) b ni tt; ICK = m = m. K2(8) = 1, + T(6) = 1, + T(10) = 1, RG2(9) = 1, RG(8) = 1, Rf b ni tt nn ICKF = m UF nhanh chng tng n gi tr nh mc. ng c khi ng cng bc lm cho tc tng nhanh nhng dng in c th vt qu gi tr cho php. Nu I>Icf1 FRG1>FhRG2 RG(8)= 0, Rf+CKF ICKF UF I Khi I

37

Khi dng in trong cun kch t ICK = m th rle RT(2) = 1, RT(35) = 0, K3(34) = 0, K3(20) = 0, T(17) = 0, T(6) = 0, + T(10) = 0, ICKF = 0, UF gim v Ud ng c hm ti sinh gim tc. Khi UF Ud RH(l) = 0, RH(29) = 0, + T(30) = 0, G(31) = 0, G(32) = 0, + RH(33) = 0, K2(32) = 0. Trn mch ng lc G(l) = 0, K2(l) = 1, ng c hm ti sinh gim tc v khng. Hm my khi ng c ang quay ngc - (ngi c t nghin cu). g/ Th my Cc iu kin lm vic , chiu quay c chn; gi s chn chiu quay thun. n TT(18) hoc TN(19) T(17) = 1, T(30) = 1, G(31) = 1, G(32) = 1, K2(32) = 1. Kt qu ta c T, G, K2 c in. Vic khi ng din ra tng t nh m t nh khi n nt M1 nhng khng c duy tr (do khng c K1). Dng ICK= m RT(2) = 1, RT(35) = 1 nn K3 khng th c in KT lun lun b ni tt ng c ch tng tc n tc c bn. Khi th nt n, ng c thc hin vic hm ti sinh do gim in p my pht v hm ng nng. Th ngc - (ngi c t nghin cu). h/ iu khin tc t xa S dng ng c xec v (servomotor) 1(12) quay bin tr KT(2). Mun tng tc, n M1(22) hoc M2(25) LT(22) = 1, hoc LN(25) = 1, LT(22,23) = 1, hoc LN(23,24) = 1, KT(26) = 1, KT(11) = 1 v KT(13) = 1, 1(12) = 1, quay KT v pha phi tng tc ng c v 1KX(26) l cng tc gii hn hnh trnh ca KT bn phi. Mun gim tc, n M3(27) KN(27) = 1, KN(11) = 1, + KN(13) = 1, 1(12) = 1, quay KT(2) v pha tri lm gim tc ng c v 2KX(27) l cng tc gii hn hnh trnh ca KT bn tri. j/ Mch tn hiu n H1(14) sng bo hiu du bi trn. n H2(15) sng bo hiu thiu du bi trn Ci C(16) ku bo hiu thiu du bi trn khi ang lm vic.

38

2.S iu khin truyn ng chnh my tin ng 1540

AT

AT2

BA2

CL2 CL3

BA6

BB1

BA3 BA5 Lk

BB2

BA4

K2

R1

R2

CK

RTT

RCH

O2

r2

K

CKFT O1

r1

Uk

Tr

R8 R8

1 1521 23

+ -

19

R12

R11

R

R913

R11

R9

R9

R9 25

17

R7

R7

R3

11

9

5

7

3

RV

R1027 29

31

R9

R10

RTr1

O3

33 35R11

R9 R9

R11

R10

43

RTr2

R10 R9

RD

37 39

35

41

R9

45 49FT

R4

R3

R3

R4

47

RVD

R9

R951

CL1

BA1

(1)K1D1 M1

RA

(3)K2D2 M2

K2

(2)

K1

RTh (4)

R11 R5R1 (5)

R6 R3 R4

R11(6)

(7)

(8)R2

R5(9)

R3 R4 (10)

R3

R6

R3 R4 (12)

R4 (11)

K1

BK1 BK2 BK3 BK4 D3MT

R8 R7

RBTR5

MN

R5

R6

R6

R5R6

(13)

(14)

(15)(16)

R9LV HC

R7 (17)

AT1

R8 (18)R5

R6

RC RTT

BK1R9R7

(19)

D3 D4 D5 D6

(20)

RA

RA

(21)

RTh

RAK RAL RBT

(22)

RBT R8 (24)C

R10R5RTr2

R6

RTr1 R11

R12R1 RTT

R2

BK1 BK2

K2

RBT

H1

H2

H3

(23)

RH

K1

n cc truyn ng

ph

AT

AT2

BA2

CL2 CL3

BA6

BB1

BA3 BA5 Lk

BB2

BA4

K2

R1

R2

CK

RTT

RCH

O2

r2

K

CKFT O1

r1

Uk

Tr

R8 R8

1 1521 23

+ -

19

R12

R11

R

R913

R11

R9

R9

R9 25

17

R7

R7

R3

11

9

5

7

3

RV

R1027 29

31

R9

R10

RTr1

O3

33 35R11

R9 R9

R11

R10

43

RTr2

R10 R9

RD

37 39

35

41

R9

45 49FT

R4

R3

R3

R4

47

RVD

R9

R951

CL1

BA1

(1)K1D1 M1

RA

(3)K2D2 M2

K2

(2)

K1

RTh (4)

R11 R5R1 (5)

R6 R3 R4

R11(6)

(7)

(8)R2

R5(9)

R3 R4 (10)

R3

R6

R3 R4 (12)

R4 (11)

K1

BK1 BK2 BK3 BK4 D3MT

R8 R7

RBTR5

MN

R5

R6

R6

R5R6

(13)

(14)

(15)(16)

R9LV HC

R7 (17)

AT1

R8 (18)R5

R6

RC RTT

BK1R9R7

(19)

D3 D4 D5 D6

(20)

RA

RA

(21)

RTh

RAK RAL RBT

(22)

RBT R8 (24)C

R10R5RTr2

R6

RTr1 R11

R12R1 RTT

R2

BK1 BK2

K2

RBT

H1

H2

H3

(23)

RH

K1

n cc truyn ng

ph

Hnh 2-11. S truyn ng chnh my tin h T- (1540)

39

ng c 1 l ng c truyn ng chnh c cng sut 70kW; in p phn ng 440V. Phm vi iu chnh tc bng iu chnh in p phn ng l Du = 6,7/1 v iu chnh t thng l D= 3/1. a/ Mch ng lc: ng c quay truyn ng chnh c cp in t b bin i BB1. BB1 gm b chnh lu cu 3 pha dng Thyristor, khng c my bin p nn phi s dng cun khng Lk chng tc tng dng ant v h thng pht xung iu khin cho Thyristor. in p Uk c t vo khu so snh ca h thng pht xung iu khin. Khi Uk thay i s lm cho gc m thay i thay i in p ra ca b BB1 nhm thay i tc ng c di tc c bn. in p Uk l u ra ca b khuch i mt chiu K; u vo ca K gm c hai knh: - knh 1: t vo chn 21-23 ca K l hiu s ca 2 gi tr in p: in p ch o Uc ly trn in tr R(5-9) v in p phn hi m tc ly trn my pht tc FT(45- 49). Do Uk = k(Uc UFT) vi k l h s khuch i ca b khuch i K - knh 2: l khu hn ch dng in trong ng c gm 3 bin p BA3, BA4, BA5 c cun s cp ni song song vi cun khng Lk; cun th cp ni vi chnh lu CL3 c in p u ra t ln in tr r1, ni vi it O1 v transistor Tr. Khi dng in trong ng c ln hn gi tr cho php th in p ri trn Lk ln in p trn CL1 cng nh trn r1 ln cho O1 thng lm cho transistor Tr m. Kt qu l in p ra ca b khuch i mt chiu gim nhm lm gim in p ra ca BB1 gim dng trong ng c khng vt qu gi tr cho php. b/ Mch kch t CK l cun kch t ca ng c c cp t b bin i BB2. BB2 gm b chnh lu 3 pha hnh tia ni song song ngc v hai h thng pht xung iu khin cho hai nhm Thyristor ni anot chung v catot chung iu khin theo phng php c lp. Khi R1 = 1, nhm chnh lu pha trn ( nhm catot chung) lm vic, cun CK c dng to ra t thng ng vi chiu quay thun ca ng c. Khi R2 = 1, nhm chnh lu pha di (nhm anot chung) lm vic, cun CK c dng to ra t thng ng vi chiu quay ngc ca ng c. Rle RTT l rle bo v thiu t thng . Khi dng qua n, RTT = 1. c/ Phi hp iu khin gia in p phn ng v t thng ca ng c in p phn ng ca ng c l 440V. Khi UBB < 420V th in p do khu o lng H t ln in tr r2 cha O2 thng; h thng pht

40

xung m cc Thyristor phi m vi gc m nh nht in p ra ca BB2 l ln nht tng ng vi dng kch t ca ng c l ln nht. Khi UBB 420V, in p trn r2 cho O2 thng, h thng pht xung ca BB2 thay i c gc m (tu gi tr t) lm thay i in p ra ca BB2 lm thay i dng kch t ca ng c lm tng tc ng c trn tc c bn. d/ iu kin lm vic ca my - n M1 K1(1) = 1, ng in cho cc truyn ng ph; K1(3) = 1, v K1(12) = 1, cp in cho cc dng t (12) ( 24). Nu in p li RA(21) = 1, RA(2) = 1, duy tr cho cun K1; - du bi trn v p lc du: RAK(23) = 1, RAL = 1, RBT(23) = 1, RBT(13) = 1, - Cc bnh rng c n khp: BK1(13) = 1, BK2(13) = 1, - X ngang c kp cht : BK3(13) = 1, - Truyn ng nng h x thi lm vic: BK4 = 1, e/ Khi ng n M2(3) K2(3) = 1, K2(4) = 1, v K2(l) = 1, lm cho BB1 v BB2 c in chun b cho mch ng lc lm vic. Mun khi ng thun, n MT(13) R5(13) R5(14) = 1, + R5(18) = 1, + R5(5) = 1, R1(5) = 1, v R5(9) = 1, R3(9) = 1. Do R1 c in nn h thng pht xung ca BB2 lm vic dng CK tng ln gi tr nh mc. Khi dng CK t n gi tr chnh nh (nh thua dng nh mc) th rle bo v thiu t thng RTT tc ng RTT(17) = 1, R12(17) = 1, [R1(17) ng)] v RTT(18) = 1, R8(18) = 1 R8(15) tomch duy tr cho R5 (gm R8(15) + R7(15) + R5(14). Kt qu khi n MT ta c c R5, R1, R3, R8 v R12 c in. R8(15-13) = 1, + R8(1-3) = 1, R(5-9) c t in p Uc do ngun CL2 cp; R12(19-21) = 1, + R3(41- 45) = 1, + R3(45- 49) = 1, s ni Uc vi UFT qua cc im (t dng ngun sang m ngun) sau: 15, 13, 17, 19, 21, 23, 35, 41, 45, 49, 47, 7, 5, 3, 1. Vi gi tr Uc - UFT ny t vo b khuch i mt chiu K lm cho Uk 0, UBB1 0 ng c khi ng. Trong qu trnh khi ng, nu dng in trong ng c ln hn gi tr cho php th khu hn ch dng tham gia vo lm vic. Khi thay i bin tr R(5-9), Uk thay i lm thay i gc m lm thay i tc ng c di tc c bn. Khi UBB 420V th O2 thng, cho php h thng pht xung ca BB2 thay i gc m thay i dng trong cun CK lm thay i tc trn tc c bn. Lu l th ti im 45 dng hn so vi im 49 v im 17 dng hn so vi im 35. Do it O3 (33-35) thng RTr1 = 0. Khi ng ngc, n MN(15) - t nghin cu

41

f/ Hm my Gi s ng c ang quay thun nh trnh by mc e/. Cc phn t ang c in l R5, R1, R3, R8, R12. n nt dng D3(13) R5(13) = 0, R5(5) = 0, R1(5) = 0, + R5(9) = 1, nhng R3(9) = 1, + R5(18) = 0, R8(18) = 0, R8(1-3) = 0, + R8(15-13) = 0, Uc t ln trn R(5-9) bng 0 Uk UFT ngha l t l vi tc ca ng c. Lc ny, th ti im 35 ln hn th ti im 17 (do Uc =0) nn iot O3 kho, RTr1(33-35) = 1, RTr1(15) = 1, R11(15) = 1, R11(17-23) = 1, + R11(19-35) = 1, + R11(17-19) = 0, + R11(23-35) = 0, cc tnh dng ca FT c t vo im 21 cho ph hp vi cc tnh u vo ca b K. R11(5) = 0, + R11(7) = 1, R2(8) = 1. Trn b BB2, nhm chnh lu phi trn dng lm vic, nhm chnh lu pha di lm vic. Tc ng c gim tc o chiu quay. Trong giai on gim tc ny, in p Uk do t l vi tc nn cng gim theo lm cho in p ra ca b BB1 cng gim nn tc gim cng nhanh. Qu trnh gim tc lm cho th ti im 35 cng gim; n lc th ti im 35 gn bng th ti im 33 th RTr1(33-35) thi tc ng R11(15) = 0, R11(19-35) = 0, + R11(17 -23) = 0, ct in p t vo b K(21-23) Uk= 0 UBB1= 0 ng c dng quay. Nu n mt trong cc nt D3 D6 RA(21) = 0, RA(2) = 0, K1(1) = 0; iu ny cng nh n vo D1(1). Khi K1(12) = 0, R5(13) = 0, v R8(18) = 0, qu trnh hm xy ra tng t nh n D3. Nu n vo D2(3) K2(3) = 0, K2(l) = 0, cc b bin i BB1 v BB2 mt in, ng c dng t do. Hm khi ng c ang quay ngc- t nghin cu g/Th my Quay b khng ch KC(17) v v tr HC R7(17) = 1, R7(15) = 0, mt duy tr cho R5 ch th my. h/Tin ct hay tin mt u Khi tin ct, lc dao ct i dn vo tm chi tit th tc quay ca chi tit cn phi tng tng ng m bo cho lng ct l khng i nhm gi vng nng sut ca my. Lc tin ct, chn ch tin ct trn mt my cho BK5(20) = 1, R9(20) = 1. Ch tin ct tng t nh ch tin thng, ch thm c R9 tc ng, ngha l khi ta chn ch tin ct quay thun chng hn th cc phn t c in l R5, R1, R3, R8, R12, R9. Lc ny in p Uc t ln bin tr Rv do R9(3-5) = 0, + R9(9-110 = 0, R9(13-25) = 1, R9(17-29) = 1;

42

in p UFT t ln bin tr RD do R9(35- 41) = 0, R9(37-35) = 1, R9(39- 41) = 1, R9(47-51) = 1, in p t vo b khuch ai K lc ny l URV - URDChn bin tr RD ni vi chuyn ng n dao theo chiu hng tm. Khi dao i vo tm chi tit th chn bin tr RD dch chuyn theo hng gim nh URD lm cho in p t vo K tng nn tc ng c s tng tng ng. Dao cng i su vo tm chi tit th th ti im 43 cng gim n mc chnh lch th ti im 31 vi 43 ln cho RTr2 tc ng RTr2(13) = 1, R10(13) = 1, R10(29-31) = 0, R10(37- 43) = 0, R10(27-29) = 1, R10(37-39) = 1, in p t vo b khuch i m bo tc ng c c gi tr khng i khng ph thuc vo s dch chuyn ca chn bin tr RD trong sut thi gian gia cng cn li. j/ Mch tn hiu: - n H1(20) sng BB1 v BB2 ang c in, sn sng lm vic. - n H2(21) sng du bi trn - n H3(22) sng cc bnh rng n khp - Ci C(24) ku ln thiu du bi trn khi ang lm vic. 3.S iu khin truyn ng n dao my tin ng 1540 truyn ng my tin c nng v my tin ng, thng dng h thng truyn ng ring cho bn dao. V h thng ny c cng sut khng ln v phm vi iu chnh tc rng nn thng s dng h thng KM- v ngy nay l h thng T- H thng truyn ng n dao m bo iu chnh tc n dao lm vic trong phm vi 0,059 470 m/ph. H thng truyn ng n dao l h thng T- khng o chiu thc hin trong h thng kn c phn hi m tc nh my pht tc FT2. Phm vi iu chnh ng c l 200/1 bng cch thay i in p phn ng, m bo M= const. Phn ng ng c 1 c cung cp t b bin i dng Thyristor khng o chiu c cung cp t bin p BA1. Cun kch t ca my pht tc FT2 c cung cp t b chnh lu BB. in p iu khin t vo b bin i l hiu ca in p ch o v in p phn hi tc : Uk = Uc Uft = Vc Trong Uc : in p ch o ly trn bin tr RD1 hoc RD2 Uft : in p my pht tc FT2 ni cng vi ng c truyn ng n dao 1

43

CL1

BA1

AT1

(1)

BK1

(2)

K

AT1

BA1

AT1

BB

1CK1 CKFT2

FT1

FT2

R10R10

RD2

R10

R3

R1

RD1

2

D M

R1

RXR1

R2

R3

KC112 1 RL KT

R1 C1 R4C2

R5

R2 C3 R6C4

R7KC112 1R2 R8R9

R9R8

R4NC1

(3)

BK2

(4)

R5NC2

(5)

BK3(6)

R6NC3BK4

(7)

BK5(8)

R7NC4

BK1 BK2

(9)

(10)NC5KC2

BK3 BK5

(11)

(12)NC6

R4 R5

R6 R7

(13)

(14)

R8NC7

R9NC8

R2 KR4

R5

R6

R7

H1

H2

H3

H4

(15)

(16)

CL1

BA1

AT1

(1)

BK1

(2)

K

AT1

BA1

AT1

BB

1CK1 CKFT2

FT1

FT2

R10R10

RD2

R10

R3

R1

RD1

2

D M

R1

RXR1

R2

R3

KC112 1 RL KT

R1 C1 R4C2

R5

R2 C3 R6C4

R7KC112 1R2 R8R9

R9R8

R4NC1

(3)

BK2

(4)

R5NC2

(5)

BK3(6)

R6NC3BK4

(7)

BK5(8)

R7NC4

BK1 BK2

(9)

(10)NC5KC2

BK3 BK5

(11)

(12)NC6

R4 R5

R6 R7

(13)

(14)

R8NC7

R9NC8

R2 KR4

R5

R6

R7

H1

H2

H3

H4

(15)

(16)

Hnh 2-12. S iu khin truyn ng n dao my tin h T- (1540) ch gia cng tin ct, rle R10 (khng v trong s ) khng c in, tip im thng kn ca n ng nn in p ch o ly trn bin tr RD1. ch mi mt u, rle R10 c in, in p ch o c ly trn bin tr RD2 t l vi in p my pht tc FT1 v do my pht tc ni cng vi trc ng c truyn ng chnh nn tc ng c n dao s t l vi tc ng c truyn ng chnh. Nh vy tc di chuyn bn dao s thay i nhp nhng vi tc quay chi tit gi lng n doa s l hng s trong qu trnh gia cng. La chn ch di chuyn ca dao hay bn dao c thc hin bng cc cng tc chuyn i C1 C4, cc rle tng ng R4 R7 s c in

44

v ng ngun cho cc nam chm in ca cc khp ly hp in t NC1 NC4 - Di chuyn ln ca dao: ng C1, rle R4 c in, NC1 c in - Di chuyn xung ca dao: ng C2; rle R5 c in, NC2 c in - Di chuyn ti tm ca bn dao: ng C3. rle R6 c in, NC3 c in - Di chuyn xa tm ca bn dao: ng C4, rle R7 c in, NC4 c in. Thc hin hm cc dao v bn dao bng cc khp ly hp in t NC5 v NC6. Khi hai khp NC5 v NC6 c in do cc rle tng ng R4 n R7 mt in, dao v bn dao c hm dng. Khi cn dng dao v bn dao m khng cn hm cng bc th t KC2 v tr 1(bn tri). Lc ny cc khp in t NC5 v NC6 khng c in. S m bo s lm vic ca truyn ng n dao ba ch : n dao lm vic, di chuyn nhanh v chm bng s dng b khng ch KC1. ch n dao lm vic, t b khng ch KC1 v tr 0; n nt M, rle R1 c in (nu truyn ng chnh lm vic th tip im RL kn), in p ch o c ly trn bin tr RD1 t vo b bin i qua tip im R1. Dng my bng cch n nt D. Mun di chuyn nhanh dao hoc bn dao, t KC1 v tr 2 bn tri, n nt M, rle R2 c in, v tip ng cng tc t K, ng c 2 c in khng duy tr, bn dao s di chuyn nhanh. di chuyn chm bn dao hoc dao, t KC1 v tr 1 bn tri, n nt M, rle R3 c in, in p ch o c ly trn RD1 qua tip im R3 s c tr s b tng ng vi tc nh. S c cc bo v sau: Bo v dng in cc i v ngn mch nh aptmat AT1, AT2 v bo v gii hn chuyn ng ca v bn dao bng cc cng tc hnh trnh cui BK1 BK5 S n dao ch lm vic khi:

- Truyn ng chnh lm vic: tip im L kn. - ng c bm du lm vic: tip im KT2 kn - X my c kp cht: tip im RX kn - dao c di chuyn khi c ni: tip im R1 kn - Bn dao ch di chuyn khi bn dao c ni: tip im R2 kn

Cc n tn hiu 1 4 bo hiu ch di chuyn ca dao v bn dao tng ng.