Dieu Chinh Dong Co

7/26/2019 Dieu Chinh Dong Co

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24

Chng 2TRANG BIN NHM MY TIN

2.1c im cng nghNhm 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 cnhiu cng nghtin khc nhau: tin trngoi, tin trtrong, tin mt u,tin cn, tin nh hnh. Trn my tin cng c ththc hin doa, khoan vtin ren bng cc dao ct, dao doa, tar renKch thc gia cng trn mytin c thtcvi mili n hng chc mt

4

Hnh 2.1 Dng bn ngoi my tin

Dng bn ngoi ca my tin nhhnh 2.1a. Trn thn my 1 t trc2, trong c trc chnh quay chi tit. Trn gtrt t bn dao 3 v sau4. Bn dao thc hin sdi chuyn dao ct dc v ngang so vi chi tit. sau t mi chng tm dng gicht chi tit di trong qu trnh gia cng,

hoc gi mi khoan, mi doa khi khoan, doa chi tit.Sgia cng tin nhhnh 2.1b. my tin, chuyn ng quay chi titvi tc gc ctl chuyn ng chnh, chuyn ng di chuyn ca dao 2 lchuyn ng n dao. Chuyn ng n dao c thl n dao dc, nu dao dichuyn dc chi tit (tin dc) hoc n dao ngang, nu dao di chuyn ngang(hng knh) chi tit. Chuyn ng phgm c xit ni x, tr, di chuynnhanh ca dao, bm nc, ht phi.


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2.2 Phti ca ccu truyn ng chnh v n dao1. Phti ca ccu truyn ng chnhQu trnh tin trn my tin c thc hin vi cc chct khc nhau

c trng bi cc thng s: su ct t, lng n dao v tc ct v.Tc phthuc vt liu gia cng, vt liu dao, kch thc dao, dng gia

cng, iu kin lm mt v.v. theo cng thc kinh nghim

stTvV yX

m

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

vi - t: chiu su ct , mms: 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 ktip,ph

Cv, x

v, y

v, m l hsv smphthuc vo vt liu chi tit, vt liu dao v

phng php gia cngm bo nng sut cao nht, sdng my trit nht th trong qu

trnh gia cng phi lun t tc ct ti u, n c xc nh bi ccthng s: su ct t, lng n dao s v tc trc chnh ng vi ngknh chi tit xc nh. Khi tin ngang chi tit c ng knh ln, trong qutrnh gia cng, ng knh chi tit gim dn, duy tr tc ct (m/s) tiu 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 hinmt lc F gm 3 thnh phn v lc ct c xc nh theo cng thc:

Fz = 9,81CF.tx

F.sy

F.vn, [N] (2-3)

Qu trnh tin xy ra vi cng sut ct Fz

V

Fz

V

Hnh 2-2 thphti catruyn ng chnh my tin

(kW) l hng s:Pz= Fz.v.10

-3, [kW] (2-4)

Bi v lc ct ln nht Fmaxsinh ra khi lngn dao v su ct ln, tng ng vi tc ct nhVmin; cn lc ct nhnht Fmin, xcnh bi t, s tng ng vi tc ct V

max,

ngha l tng ng vi hthc:Fmax.vmin= Fmin.vmax (2-5)

Sphthuc ca lc ct vo tc nhh2.2Tuy nhin nh phn tch, dng thphtithc tca truyn ng chnh my tin c dnghai vng Fz= const v Pz= const (h 1.4)


3/21

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2. Phti ca truyn ng chnh my tin ngTruyn ng chnh my tin ng c dng c th ring, khc so vi my

tin bnh thng vcu trc v kch thc. Trn my tin ng, chi tit giacng c ng knh ln v c t trn mm cp nm ngang, hay ni cchkhc trc mm cp l theo phng thng ng. Do trng lng mm cp,trng lng chi tit ln ln nn lc ma st gtrt v hp tc kh ln.V vy ph ti trn trc ng c truyn ng chnh my tin ng l tngca cc thnh phn lc ct, lc ma st gtrt, lc ma st hp tc .

Hnh 2.3 thphti ca truyn ng chnh my tin ng

Trn hnh 2.3a, l thbiu din cc thnh phn cng sut ca truynng chnh v sphthuc ca chng vo tc mm cp: P

1 cng sut

khc phc lc ct; P2 cng sut khc phc lc ma st gtrt; P3v P4cng sut khc phc lc ma st trong hp tc tng ng do lc ct v squay 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 phthuc vo tc nh hng ln n qu trnhqu ca truyn ng chnh. Do khi lng ca mm cp v chi tit ln vskhc nhau ca hsma st lc ng yn v chuyn ng nn mmen cntnh khi khi ng ca truyn ng c tht ti 60 80% momen nhmc. V momen qun tnh tng qui i vtrc ng cc tht ti 8 9

ln momen qun tnh ca ng cnn qu trnh khi ng ca hthng dinra chm vi momen cn tnh ln. Theo mc gia tc ca ng c, momencn tnh sgim nhanh v khi tc tng th n t thay i.

3. Phti ca truyn ng n daoLc n dao ca truyn ng n dao c xc nh theo cng thc:

dmsxad FFkFF ++= , [N]


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Cng sut n dao ca my tin c xc nh bng cng thc:, [kW]310.. = adadad vFP

Cng sut n dao thng nhhn cng sut ct 100 ln v tc n daoc xc nh bi lng n dao v tc gc chi tit:

, [m/s] (2-6)3

10.'.

= ctad sv nhhn tc ct nhiu ln.

y2

's

s = , [mm/rad]

Lc v mmen phti ca truyn ng ndao khng phthuc vo tc ca n, v phti ca truyn ng n dao chc xc nh bi

Mc

V

V1 V2 V3

Mc

V

V1 V2 V3

Hnh 2.4 thphti catruyn ng n dao

khi lng bphn di chuyn ca my v lcma st gtrt v hp tc .Trn thphti ca truyn ng n dao hnh2.4, di tc rng v1< v v2 momen

phti sthay i tuyn tnh theo tc 3) Thi gian my

Thi gian my (thi gian gia cng) ca my tin c xc nh:

ad

Mv

lt

310.= , [s] (2-7)

Trong : l l chiu di gia cng , mmctl tc gc chi tit, rad/s

s lng n dao, mm/vgKt hp (2-6) v (2-7) ta c cng thc tnh thi gian my:

'.s

lt

ct

NM

= , [s] (2-8)

Nhvy gim thi gian gia cng, ta phi tng tc ct v lng ndao v nng sut stng.

2.3 Phng php chn cng sut ng ctruyn dng chnh ca my tinTruyn 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 qutrnh thay i nguyn cng v chi tit chim thi gian qu ln nn truynng chnh phi tin hnh tnh ton mt chnng nnht.

Gi thit trn my tin thc hin gia cng chi tit nhhnh 2-5. Ccnguyn cng khi gia cng gm 4 giai on: 1 v 3 - tin ct hoc tin ngang;2 v 4 - tin tr(tin dc). Phti ca ng ctrong tng nguyn cng phthuc vo cc thng schct, vt liu chi tit dao v.v


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Qu trnh tnh ton nhsau:a) Tcc yu tchct 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 dng vi tng nguyn cng. Nu tcct tnh c khng ph hp tcca my (theo sliu kthut c

2 d1

d0

234

l4 l2

l3 l1

1

2 d1

d d0

234

l4 l2

l3 l1

Hnh 2-5 Chi tit c gia cng trnmy tin

kh) th chn ly trsc sn trongmy gn ging vi tc ct tnhton.

Dng trsny tnh li Pz, tm,theo (2-4) v (2-8). TrsV, Pz, tmny c dng chnh thc trong ton bbi ton.

b) Chn nguyn cng nng nnht v githit nguyn cng y my lmvic chnh mc. T xc inh hiu sut ca my ng vi phtica tng nguyn cng theo cng thc:

bt

aMM

M

mshi

hi

++

=+

=

1

1

a, b - hstn hao khng bin i v bin i.

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

zi

Di

PP

=

Gi thit trong thi gian g lp, tho g chi tit, chuyn i tnguyncng ny sang nguyn cng khc, ng c quay khng ti (m khng ctin ng c) th cng sut trn trc ng clc ny l cng sut khng tica 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 nhtheo tiu chun vn hnh ca my t0c) ng cc thchn theo cng sut trung bnh hoc cng sut ng tr:

==

==

+

+

=n

jj

imi

n

j

j

i

ci

tb

tt

PP

P

10

4

1

1

0

4

1

hoc

==

==

+

+

=n

j

j

i

mi

j

n

j

jmi

i

ci

dt

tt

tPtP

P

1

0

4

1

0

1

2

0

4

1

2 ..

trong :


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Pci, ti cng sut trn trc ng c, thi gian my ca nguyn cng thiP0j, t0j- cng sut khng ti trn trc ng c, thi gian lm vic khng tica my, P0j= P0n - skhong thi gian lm vic khng ti

Chn ng c c cng sut nh mc ln hn 20 30% cng sut trungbnh hay ng tr:

Pc

Tck

t

P0 P0 P0 P0

Pc1

Pc2=Pcm

Pc3

Pc4

t01 t02 t03 t04 tm4tm3tm2tm1

Pc

Tck

t

P0 P0 P0 P0

Pc1

Pc2=Pcm

Pc3

Pc4

t01 t02 t03 t04 tm4tm3tm2tm1

Hnh 2-6 thphti ca ng c

Pm (1,2 1,3) Ptbhoc Pm= (1,2 1,3)Pt (2-12)d) ng c truyn ng chnh my tin cn phi c kim nghim theoiu kin pht nng v qu ti

2.4 Nhng yu cu v c im i vi truyn ng in v trang binca my tin1. Nhng yu cu v c im chung

a. Truyn ng chnh: Truyn ng chnh cn phi c o chiu quaym bo quay chi tit chai chiu, v dkhi 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).

chxc lp, h thng truyn ng in cn m bo cng ctnh ctrong phm vi iu chnh tc vi sai stnh nhhn 10% khi phti thay i tkhng n nh mc. Qu trnh khi ng , hm yu cu phitrn, trnh va p trong btruyn lc. i vi my tin cnng v my tinng dng gia cng chi tit c ng knh ln, m bo tc ct ti u


7/21

30

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

min

max

min

max

min

max

min

max

min

max ..

ct

dct

ctD

D

v

v

v

D

D

vD ===

(2-13)

nhng my tin cnhv trungbnh, hthng truyn ng inchnh thng l ng ckhng ng

broto lng sc v hp tc c vicp tc . cc my tin cnng, my

M,P

VVgh Vmax

P

M

Vmin

2-7 Biu momen v cng sutng ctrong truyn ng chnh

M,P

VVgh Vmax

P

M

Vmin

2-7 Biu momen v cng sutng ctrong truyn ng chnh

tin ng, hthng truyn ng chnhiu chnh 2 vng, sdng bbin ing cin mt chiu (BB ) vhp tc : khi v< vghm boM = const; khi v> vghth P= const. BBin i c thl my pht mt chiuhoc bchnh 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 ochiu ng cin hoc dng khp ly hp in t. Phm vi iu chnh tcca truyn ng in hoc dng khp ly hp in t. Phm vi iu chnhtc ca truyn ng n dao thng l D = (50 300)/1 vi trn iuchnh = 1,06 v 1,21 v momen khng i (M = const).

chlm vic xc lp, sai lch tnh yu cu nhhn 5% khi phti thay i tkhng n nh mc. ng ccn khi ng v hm m. Tcdi chuyn bn dao ca my tin cnng v my tin ng cn lin hvitc quay chi tit m bo nguyn lng n dao.

my tin cnhthng truyn ng n dao c thc hin tng ctruyn ng chnh, cn nhng my tin nng th truyn ng n dao cthc hin tmt ng cring l ng cmt chiu cp in tkhuch imy in hoc bchnh lu c iu khin.

c. Truyn ng ph:Truyn ng phca my tin khng yu cu iuchnh tc v khng yu cu g c bit nn thng s dng ng ckhng ng brto lng sc kt hp vi hp tc .

2.Cc siu khin in hnh my tin ng v my tin cnngCc my tin ng v my tin cnng c mt trong cc chlm vic

cbn l tin mt u. t c nng sut ln nht ng vi cc thng sca chct ti u, yu cu phi duy tr tc ct khng i. t ciu , khi ng knh D ca chi tit gim dn, cn phi iu chnh tc


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

RTr3

RTr1

RVRTr2

RTr3

RTr3

RD FT1UV UD

Bn dao

P

RTr2(T) 1BKRT

RTr2(N) 2BKRN

RTr1

+ -

-

+

KTRTr1

RT

+ -

KN

RN

RT

RN

X

-

+

FT2

RC

BB CUc

-

attric ng knh chi tit gia cng khi tin mt u l bin trDD. Contrt ca n lin hvi bn dao qua biu tc P. Phm vi di chuyn lnnht ca con trt s tng ng vi ng knh ln nht ca chi tit giacng trn mt my. in p t ln bin trRDc ly tmy pht tcFT1 tlvi tc gc ca chi tit, v vy UD~ ctD. in p t ln bintrRVl in p n nh. in p ly con trt ca RVstlvi tc ct.

+

Rv

BB CUc

Bn dao

P

FT2

RD

Uph

(a)

(b)

FT

C

(c)

X32C1

C2U~

CL2

CL1

X31 CKFT

Uc Uph BB

RTr3

RTr1

RVRTr2

RTr3

RTr3

RD FT1UV UD

Bn dao

P

RTr2(T) 1BKRT

RTr2(N) 2BKRN

RTr1

+ -

+

-

KTRTr1

RT

+ -

KN

RN

RT

RN

X

-

+

FT2

RC

BB CUc

-

+

Rv

BB CUc

Bn dao

P

FT2

RD

Uph

(a)

(b)

FT

C

(c)

X32C1

C2U~

CL2

CL1

X31 CKFT

Uc Uph BB

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


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32

Hiu in p cc u con trt ca bin trRVv RD l UV-UDct vo rle 3 v tr RTr2. R le ny siu khin ng cX t tc quay ca ng cchnh C.

Khi khi ng, bin trRc v tr tng ng vi tc gc mm cpnhnht, cn U

D= 0. Sau khi khi ng, ng cchnh (rle KT hoc KN

tc ng), do tip im RTr2(T) kn nn rle RT tc ng, ng cX quaytheo chiu thun ng vi stng tc ca ng cchnh v in p my phttc FT1. Khi in p UD=Uv, rle RTr2 mt in nn RT ngt nn ng cX dng c hm ng nng.

Tc ca ng cchnh stng ng vi tc ct t trc v vtrbn dao khi bt u gia cng.

Khi gia cng, bn dao di chuyn ti tm, con trt ca bin trdi chuynvhng gim UD, do rle RTr2, RT li tc ng; ng c X li quaytheo chiu tng tc ng c trc chnh, nh vy duy tr c in p

UD~ct.D l hng s. Khi tc gc ng cchnh t gi trln nht, cngtc hnh trnh 1BK tc ng, ng cX dng quay.

Khi dng mm cp, rle RTr2 tc ng tng ng vi tip im RTr2(N)ng v ng cX quay theo chiu gim tc ng cchnh, con trt

bin trRc c di chuyn vvtr ban u, cng tc hnh trnh 2BK sbtc ng dng ng cX.

Tc ct c duy tr khng i vi chnh xc phthuc chnhxc chto bphn lin hgia bn dao v bin trRD, mc tuyn tnhca c tnh bin trRDv pht tc, nhy im khng ca rle cc tnhRTr2, v n nh ca cc thng s ca skhi nhit v in p lithay i.

Trn hnh 2-8b l siu khin tc quay ca ng cC theo hmca ng knh chi tit gia cng theo nguyn l UcUphD. in pcho Uctlvi tc ct c t bng bin trRV. in p phn hiUphD . Nu hthng iu chnh c biu chnh PI th lun lun c:Uc = UphD ngha l Vz= D

Trn hnh 2-8c l siu khin duy tr tc ct l hng sthc hinbng cc attric ng knh v tc kiu khng tip im. in p pht raca attric X31 tlvi tc di Vz. in p phn hi ly tmy pht tc

FT, cun dy kch t pht tc c cp tattric X32 qua cu chnh luCL2 tlvi ng knh ca chi tit UCL2= K1D; nhvy in p pht tcUFT = K2D.

Siu khin m bo Uc= Uph= K2D v iu khin .D = constchnh xc duy tr tc ct phthuc vo nhng yu t: c tnh phi

tuyn ca attric X32 v pht tc, ng cong ttrca pht tc.


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thc hin php nhn cc tn hiu tlvi v D, c thdng bnhnbng in tthay 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 tiu trong ton biu chnh.

Mt yu cu c bit i vi my tin cnng v my tin ng l duytr lng n dao khng i. iu c ththc hin bng s2-9. in pcho ca hthng truyn ng n dao c ly tmy pht tc FT1 nicng vi trc ng ctruyn ng chnh C. Khi UcdD= K1D= K2CvD/ c= const. Chit p RDst lng n dao

FT2

RD

BB2 D

Ucd

FT1BB1 C

FT2

RD

BB2 D

Ucd

FT1BB1 C

Hnh 2-9 Sduy tr lng n dao l hng s

2.5 Mt ssiu khin my tin in hnh1. Siu khin truyn ng chnh my tin nng 1A660

My tin nng 1A660 c dng gia cng chi tit bng gang hocthp c trng lng 250N, ng knh chi tit ln nht c thgia cng trnmy l 1,25m. ng ctruyn ng chnh c cng sut 55kW. Tc trcchnh c iu chnh trong phm vi 125/1 vi cng sut khng i, trong phm vi iu chnh tc ng cl 5/1 nhthay i tthng ng c.Tc trc chnh ng vi 3 cp ca hp tc c gi trnhsau:

cp 1: ntc = 1,6 8 vng / phtcp 2: ntc = 8 40 vng/ phtcp 3: ntc = 40 200 vng/ phtTruyn ng n dao c thc hin tng ctruyn ng chnh. Lng

n dao c iu chnh trong phm vi 0,064 26,08 mm/vgTruyn ng chnh c thc hin th thng F-. iu chnh tc

ng c bng cch thay i dng in kch t ca ng c, cn sc inng ca my pht gikhng i.a/Mch ng lcng cquay truyn ng chnh c cp in tmy

pht F. ng cscp quay my pht F khng thhin trn s. Kch tca ng c l cun CK(2). Kch tca my pht l cun CKF(9).ng clm vic c cn G(l) = 1, ni in p my pht vi ng cng thi K2 (l) = 0, gii phng mch hm ng nng. Cun kch t


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34

CK(2) c cp in m bo tthng v cun kch tmy phtCKF(9) c in to tthng F lm cho my pht F to ra in p UF.

Rle RC(l) bo vqu dng c tip im l RC(27). Khi dng in quang c ln hn gi trcho php, RC(l) = 1, RC(9) = 0, ct inmch iu khin ( dng 27)

Rle RH(l) v RCB(l) c gi trtc ng khc nhau. Ga tr tc ngca RCB bng gi trnh mc ca in p my pht; cn gi tr tc ngca RH bng 10% gi trnh mc ca in p my pht.

RG1 v RD1 l hai cun dng ca rle RG v RD. Hai cun p tngng 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 ngng li. Nu dng in trong ng c ln hn gi tr cho php th cundng sto ra lc y ln hn lc ht ca cun p lm cho tip im ca nmra. Cthkhi:

RG(9) = 1, RG(8) = 1; nu I> Icf1Fy RG1> FhtRG2RG(8) = 0;RD(8) = 1, RD(4) = 1, nu I> Icf2Fy RD>FhtRD2RD(4) = 0,b/ Mch kch tng c

Cun CK(2) l cun kch t ca ng cc cp t ngun mtchiu cng ngun vi cun CKF(9) v l ngun cp cho mch khng ch.Bin trKT(2) ni tip vi cun CKthay i dng in chy qua n,lm thay i tthng thay i tc ng ctrn tc cbn. KhiRKT(2) v R(2) bni tt th dng CKbng nh mc.

Rle dng RT(2) c gi trtc ng bng dng nh mc ca CK.Rle dng RTT(2) l rle bo vthiu tthng

. Gi trtc ng ca

n nhthua dng CKnhnht to ra tc ln nht ca ng c.c/Mch kch tmy pht

Cun CKF(9) l cun kch tmy pht c cp in bi cu tip imT,N(6) v N,T(10). Khi T(6) = 1, v T(10) = 1, tng ng vi chiu quaythun ca ng c. Khi N(6) = 1, v N(10) = 1, tng ng vi chiu quayngc ca ng c. in trRf ni tip vi cun CKF(9) nhm gim dngqua n, kt quin p ca my pht gim nhm lm gim dng trong ngc.d/Cc iu kin lm vic ca my

1.

Phi dng kch tcho ng cRTT(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. Trstc c chn T(29) = 1,


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36

5. Chiu quay c chn: chn ng cquay thun CTC1(37) = 1,1RL(37) = 1, 1RL(17) = 1 v 1RL(19) = 1; chn quay ngcCTC2(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 duytr cho K1(29). Kt qukhi n nt M1, cc phn tsau y c in: K1, T,G v K2.

Trn mch ng lc, G(l) = 1, ni F vi ; K2(l) = 1, gii phngmch hm ng nng.

K2(1) = 1, R(2) bni tt; G(3) = 1, KT(2) bni tt; ICK

= m = m.K2(8) = 1, + T(6) = 1, + T(10) = 1, RG2(9) = 1, RG(8) = 1, Rf

bni tt nn ICKF= m UFnhanh chng tng n gi trnh mc.ng ckhi ng cng bc lm cho tc tng nhanh nhng dng

in c thvt qu gi trcho php.Nu I>Icf1FRG1>FhRG2RG(8)= 0, Rf+CKF ICKFUFIKhi I


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Khi dng in trong cun kch tICK= 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, UFgim vUdng c hm ti sinh gim tc.

Khi UFUdRH(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 chm ti sinh gim tc vkhng.

Hm my khi ng cang quay ngc - (ngi c tnghin cu).g/ Thmy

Cc iu kin lm vic , chiu quay c chn; gischnchiu quay thun.

n TT(18) hoc TN(19) T(17) = 1, T(30) = 1, G(31) = 1, G(32) = 1, K2(32) = 1. Kt quta c T, G, K2 c in.

Vic khi ng din ra tng tnh m tnhkhi n nt M1 nhngkhng c duy tr (do khng c K1). Dng ICK= m RT(2) = 1,

RT(35) = 1 nn K3 khng thc in KT lun lun bni tt ngcchtng tc n tc cbn.

Khi thnt n, ng cthc hin vic hm ti sinh do gim in p mypht v hm ng nng.Thngc - (ngi c tnghin cu).h/iu khin tc txa

Sdng ng cxec v (servomotor) 1(12) quay bin trKT(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 vKT(13) = 1, 1(12) = 1, quay KT vpha phi tng tc ng cv 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) vpha tri lm gim tc ng cv 2KX(27)l cng tc gii hn hnh trnh ca KT bn tri.

j/ Mch tn hiun H1(14) sng bo hiu du bi trn.n H2(15) sng bo hiu thiu du bi trnCi C(16) ku bo hiu thiu du bi trn khi ang lm vic.


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ng c1 l ng c truyn ng chnh c cng sut 70kW; in pphn ng 440V. Phm vi iu chnh tc bng iu chnh in p phnng l Du= 6,7/1 v iu chnh tthng l D= 3/1.

a/ Mch ng lc:ng cquay truyn ng chnh c cp in tbbin i BB1.

BB1 gm bchnh lu cu 3 pha dng Thyristor, khng c my bin pnn phi sdng cun khng Lk chng tc tng dng ant v hthng

pht xung iu khin cho Thyristor. in p Uk c t vo khu so snhca hthng pht xung iu khin. Khi Uk thay i slm cho gc mthay i thay i in p ra ca bBB1 nhm thay i tc ng cdi tc cbn.

in p Uk l u ra ca bkhuch i mt chiu K; u vo ca Kgm c hai knh:

- knh 1: t vo chn 21-23 ca Kl hiu sca 2 gi trin p: in pcho Ucly trn in trR(5-9) v in p phn hi m tc ly trnmy pht tc FT(45- 49). Do

Uk= k(Uc UFT)

vi k l hskhuch i ca bkhuch i K- knh 2: l khu hn chdng in trong ng cgm 3 bin p BA3, BA4,BA5 c cun scp ni song song vi cun khng Lk; cun thcp ni vichnh lu CL3 c in p u ra t ln in tr r1, ni vi it O1 vtransistor Tr. Khi dng in trong ng c ln hn gi tr cho php thin p ri trn Lk ln in p trn CL1 cng nhtrn r1 ln choO1 thng lm cho transistor Tr m. Kt qu l in p ra ca bkhuchi mt chiu gim nhm lm gim in p ra ca BB1 gim dng trongng ckhng vt qu gi trcho php.

b/ Mch kch tCK l cun kch t ca ng c c cp t b bin i BB2.

BB2 gm bchnh lu 3 pha hnh tia ni song song ngc v hai hthngpht xung iu khin cho hai nhm Thyristor ni anot chung v catot chungiu khin theo phng php c lp.

Khi R1 = 1, nhm chnh lu pha trn ( nhm catot chung) lm vic, cun

CKc dng to ra tthng ng vi chiu quay thun ca ng c. KhiR2 = 1, nhm chnh lu pha di (nhm anot chung) lm vic, cun CKc dng to ra tthng ng vi chiu quay ngc ca ng c.

Rle RTT l rle bo vthiu tthng . Khi dng qua n, RTT = 1.c/ Phi hp iu khin gia in p phn ng v tthng ca ng cin p phn ng ca ng cl 440V. Khi UBB< 420V th in p do

khu o lng H t ln in trr2 cha O2 thng; hthng pht


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xung m cc Thyristor phi m vi gc m nh nht in p ra caBB2 l ln nht tng ng vi dng kch tca ng cl ln nht. KhiUBB420V, in p trn r2 cho O2 thng, hthng pht xung caBB2 thay i c gc m(tugi trt) lm thay i in p ra caBB2 lm thay i dng kch tca ng clm tng tc ng ctrntc cbn.

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 liRA(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 hx thi lm vic: BK4 = 1,e/ Khi ngn M2(3) K2(3) = 1, K2(4) = 1, v K2(l) = 1, lm cho BB1 v

BB2 c in chun bcho 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 nnh thng pht xung ca BB2 lm vic dng CK tng ln gi trnhmc. Khi dng CKt n gi trchnh nh (nhthua dng nh mc) thrle bo v thiu tthng RTT tc ng RTT(17) = 1, R12(17) = 1,[R1(17) ng)] v RTT(18) = 1, R8(18) = 1 R8(15) tomch duy trcho R5 (gm R8(15) + R7(15) + R5(14).

Kt qukhi 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 Ucdo ngun

CL2 cp; R12(19-21) = 1, + R3(41- 45) = 1, + R3(45- 49) = 1, sni UcviUFTqua cc im (tdng ngun sang m ngun) sau: 15, 13, 17, 19, 21,23, 35, 41, 45, 49, 47, 7, 5, 3, 1. Vi gi trUc- UFTny t vo bkhuchi mt chiu Klm cho Uk0, UBB10 ng ckhi ng.

Trong qu trnh khi ng, nu dng in trong ng c ln hn gi trcho php th khu hn chdng tham gia vo lm vic. Khi thay i bin tr

R(5-9), Ukthay i lm thay i gc mlm thay i tc ng cdi tc c bn. Khi UBB 420V th O2 thng, cho php h thng

pht xung ca BB2 thay i gc mthay i dng trong cun CKlmthay i tc trn tc cbn.

Lu l th ti im 45 dng hn so vi im 49 v im 17 dnghn so vi im 35. Do it O3 (33-35) thng RTr1 = 0.

Khi ng ngc, n MN(15) - tnghin cu


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f/ Hm my

Gi sng cang quay thun nh trnh by mc e/. Cc phn tang 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, Uct ln trn R(5-9) bng 0 UkUFTngha l tlvi tc ca ng c.

Lc ny, th ti im 35 ln hn th ti im 17 (do Uc=0) nn iotO3 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 tnhdng ca FT c t vo im 21 cho ph hp vi cc tnh u vo ca

bK.R11(5) = 0, + R11(7) = 1, R2(8) = 1. Trn bBB2, nhm chnh lu

phi trn dng lm vic, nhm chnh lu pha di lm vic. Tc ng cgim tc o chiu quay. Trong giai on gim tc ny, in p Ukdo tlvi tc nn cng gim theo lm cho in p ra ca bBB1 cng gimnn tc gim cng nhanh.

Qu trnh gim tc lm cho th ti im 35 cng gim; n lc th tiim 35 gn bng thti 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 bK(21-23) Uk= 0 UBB1= 0 ng cdng quay.

Nu n mt trong cc nt D3 D6 RA(21) = 0, RA(2) = 0, K1(1) = 0; iu ny cng nhn vo D1(1). Khi K1(12) = 0, R5(13) = 0,v R8(18) = 0, qu trnh hm xy ra tng tnhn D3.

Nu n vo D2(3) K2(3) = 0, K2(l) = 0, cc bbin i BB1 vBB2 mt in, ng cdng tdo.

Hm khi ng cang quay ngc- tnghin cug/ThmyQuay bkhng chKC(17) vvtr HC R7(17) = 1, R7(15) = 0,

mt duy tr cho R5chthmy.h/Tin ct hay tin mt uKhi 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 givng nng sut ca my.

Lc tin ct, chn chtin ct trn mt my cho BK5(20) = 1, R9(20) = 1. Ch tin ct tng tnhch tin thng, ch thm cR9 tc ng, ngha l khi ta chn ch tin ct quay thun chng hn thcc phn tc in l R5, R1, R3, R8, R12, R9. Lc ny in p Uc t ln

bin trRv do R9(3-5) = 0, + R9(9-110 = 0, R9(13-25) = 1, R9(17-29) = 1;


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in p UFTt ln bin trRDdo R9(35- 41) = 0, R9(37-35) = 1, R9(39-41) = 1, R9(47-51) = 1, in p t vo bkhuch ai Klc ny l

URV- URDChn bin trRD ni vi chuyn ng n dao theo chiu hng tm. Khidao i vo tm chi tit th chn bin trRD dch chuyn theo hng gimnhURDlm cho in p t vo Ktng nn tc ng cstng tngng.

Dao cng i su vo tm chi tit th thti im 43 cng gim n mcchnh lch thti 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 bkhuch i m bo tc ng ccgi trkhng i khng phthuc vo sdch chuyn ca chn bin trRDtrong 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.Siu khin truyn ng n dao my tin ng 1540

truyn ng my tin cnng v my tin ng, thng dng hthngtruyn ng ring cho bn dao. V hthng ny c cng sut khng ln v

phm vi iu chnh tc rng nn thng sdng hthng KM-vngy nay l hthng T-

Hthng truyn ng n dao m bo iu chnh tc n dao lm victrong phm vi 0,059 470 m/ph. Hthng truyn ng n dao l hthngT-khng o chiu thc hin trong h thng kn c phn hi m tc nhmy pht tc FT2. Phm vi iu chnh ng cl 200/1 bng cch thayi in p phn ng, m bo M= const.

Phn ng ng c 1 c cung cp t b bin i dng Thyristorkhng o chiu c cung cp tbin p BA1. Cun kch tca my phttc FT2 c cung cp tbchnh lu BB. in p iu khin t vo b

bin i l hiu ca in p cho v in p phn hi tc :Uk= Uc Uft= Vc

Trong Uc: in p cho ly trn bin trRD1 hoc RD2Uft: in p my pht tc FT2 ni cng vi ng ctruyn ng n dao 1


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v ng ngun cho cc nam chm in ca cc khp ly hp in tNC1NC4

- 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 tNC5 v

NC6. Khi hai khp NC5 v NC6 c in do cc rle tng ng R4 n R7mt in, dao v bn dao c hm dng. Khi cn dng dao v bn daom khng cn hm cng bc th t KC2 vtr 1(bn tri). Lc ny cckhp in tNC5 v NC6 khng c in.

Sm bo slm vic ca truyn ng n dao ba ch: n dao lmvic, di chuyn nhanh v chm bng sdng bkhng chKC1. chn dao lm vic, t bkhng chKC1 vtr 0; n nt M, rle R1 c in

(nu truyn ng chnh lm vic th tip im RLkn), in p choc ly trn bin trRD1 t vo bbin i qua tip im R1.

Dng my bng cch n nt D. Mun di chuyn nhanh dao hoc bn dao,t KC1 vtr 2 bn tri, n nt M, rle R2 c in, v tip ng cngtc tK, ng c2 c in khng duy tr, bn dao sdi chuyn nhanh. di chuyn chm bn dao hoc dao, t KC1 vtr 1 bn tri, n nt M,rle R3 c in, in p cho c ly trn RD1 qua tip im R3 sctrsb tng ng vi tc nh.

S c cc bo v sau: Bo v dng in cc i v ngn mch nhaptmat AT1, AT2 v bo vgii hn chuyn ng ca v bn dao bngcc cng tc hnh trnh cui BK1 BK5

Sn dao chlm vic khi:-

Truyn ng chnh lm vic: tip im Lkn.- ng cbm 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 chdi chuyn khi bn dao c ni: tip im R2 knCc n tn hiu 1 4 bo hiu chdi chuyn ca dao v bn dao

tng ng.

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Dieu Chinh Dong Co