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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.
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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)
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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]
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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 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
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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 :
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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
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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
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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)
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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.