sổ tay cdt chuong 20(4)-cc chap hanh

8/6/2019 s tay cdt Chuong 20(4)-Cc Chap Hanh

1/27

20C cu chp hnhMassimo Sorli Politecnico di Torino

Stanley S. IpsonUniversity of Bradford

1.4 H thng chp hnh thu lc v kh nn .......................1

1.4 H thng chp hnh thu lc v kh nn

Massimo sorli and Stefano Pastorelli

Gii thiuChc nng chnh ca mt h thng chp hnh l tc ng ln i tng cn iu khin t c mc ch m

Chc nng c th thc hin c nh h thng c cu chp hnh c kh nng chuyn nng lng s cp sang lng c hc cui cng.

Ba dng nng lng chnh m h thng c cu chp hnh s dng l: in, thu nng, v kh nn. Dng nng lnht tng ng vi c cu chp hnh in nh: ng c in, cun cm, v cc nam chm in. Hai dng nng ltng ng vi cc xi lanh (ng c tuyn tnh) hoc ng c quay, c bn tng t v hnh dng v kch thc. Sng tng ng ca chng c iu khin nh tc ng ca lu cht khng nn c (du thu lc, du th, hocht lng vi nht thp hn) hoc lu cht (fluid) nn c (kh nn hoc kh gas ni chung).

Ngoi ra cn c cc dng nng lng khc nhng t thng dng hn trong cc h thng t ng. V d nh, nnho hc v nhit nng gy ra s chuyn pha ca vt liu hay s gin n nhit ng lc ca h thng thnh chuyhc.

Di y ta s xt cc c tnh ca cc h servo thy lc v kh nn cho php iu khin lin tc mt trong hai vt l c trng cho nng lng ca dng lu cht l p sut v lu lng. Thng thng, vic iu khin p sudng trong nhng trng hp cn phi to ra quy lut ca lc hoc mmen, trong khi vic iu khin lu ldng iu khin ln cc i lng ng hc nh v tr, tc v gia tc.

iu khin lin tc lc hoc tc c th c thc hin mt cch hiu qu nh c c cu chp hnh lu chtu im d thy so vi cc c cu chp hnh in l kh nng duy tr h thng di ti khng gii hn vi cc thkhin ph hp; kh nng thc hin cc di chuyn tuyn tnh trc tip tc cao khng cn cc thit b bin chuquay thnh tuyn tnh v c di hot ng rng, c bit cc h thng thu lc c kch thc gii hn v qun tnh th

H thng chp hnh thu lc v kh nnH thng chp hnh l mt phn trong cc my t ng, bao gm phn cng sut v phn iu khin c minh

hnh 20.86. Phn cng sut gm cc thit b tc ng n hot ng di chuyn ca h thng. Phn iu khin thtrnh x l thng tin theo vng kn vi lut iu khin c thit lp da trn tn hiu mu v tn hiu t cm bin phn vn hnh. Cc tn hiu lnh t phn iu khin c gi ti phn vn hnh nh cc thit b trung gian, cc tc nhim v chuyn v khuch i cc tn hiu (khi cn thit) sao cho cc c cu chp hnh c th s dng trc tihiu ny. Cc phn trung gian c th l b truyn ng tc , phn tip xc ca cc ng c in hay cc van ptrong cc c cu chp hnh thu lc hoc kh nn. Hnh 20.87 m t hot ng ca mt h thng chp hnh dng cng sut l b phn chp hnh - trong hnh v l xilanh kp - phn u v phn sau khoang xilanh c ni vi v phi 4/2, c chc nng iu chnh cng sut dng lu cht

Lnh chuyn van l lnh t phn iu khin. Lnh ny c a ra cn c theo k hoch chuyn ng v vngmong mun ca xilanh, da trn nhng tn hiu phn hi t cc cm bin trong xilanh - c m t trong hnh v d

cc cng tc gii hn.Cc h thng chp hnh c th c dng khng lin tc v lin tc ty thuc vo loi hnh t ng ho nhng phn iu khin v phn c cu chp hnh. H thng chp hnh khng lin tc s rt hiu qu khi chng c ngloi hnh t ng ho khng lin tc, in hnh nh dy chuyn lp rp v dy chuyn x l cc b phn my mc;

1


2/27

S tay C in t

cch khc, cc h thng c cu chp hnh lin tc c ng dng trong cc qu trnh x l lin tc, cc thit b tng t cho cc h servo.

HNH 20.86H thng chp hnh

HNH 20.87H thng chp hnh dng nng lng lu cht

HNH 20.88S ca mt h thng servo dng nng lng lu cht

Cc c cu chp hnh thu lc hay kh nn, hoc c dng tnh tin (xilanh) hoc dng quay (ng c) l cc h ttc bi chng c th xc nh v tr ca cc phn di ng (ca tay n ng vi trng hp xilanh; ca trc ng vi trl ng c) mi im trong qu trnh chuyn ng. S hot ng ca xilanh v ng c chu nh hng ln c(tnh v ng) ti ni tip xc gia cc phn di ng. nh hng ca ma st do vic thay i cc iu kin ma st vtrong chuyn ng ca cc phn di ng trong c cu chp hnh lm tng kh nng xy ra hin tng trt c bih thng kh nn, hoc cc chuyn ng khng lin tc cc tc di chuyn rt thp.

Khi tnh n nh hng ca ma st, s c mt ca cc thit b lm im ta cho cc phn di ng ca c cu chduy tr iu kin p sut chun nh gi , vng m s lm tng phi tuyn ti v tr cn bng ca c cu chp hcho vic nh v h thng rt kh t c chnh xc cao. khc phc vn ny trong mt s ng dng c dng c cu chp hnh khng c vng bt kn v d nh trng hp cc trc thu tnh v thu ng ( fluid static, fluid dynamic bearing).

Cc phn t trung gian nh van phn phi trong hnh v ng vai tr quan trng trong vic xc nh ch hot c cu chp hnh. Trong trng hp thc t, khi n ch chuyn ng qua li, vi v tr ca c cu chp hnh imtrnh, thng dng van phn phi dng s c t 2 n 3 v tr, hnh. 20.87.

Ni cch khc, khi cn iu khin lin tc v tr v lc truyn, cn dng cc thit b lin tc nh cc van t l v vhoc dng cc thit b s kt hp vi iu bin tn hiu, v d cc b iu khin rng xung PWM (PulsModulation).

Do vy, h thng chp hnh thu lc hoc kh nn l mt h thng c phn hi (h thng servo) vi s hothnh 20.88. Cu trc thc t ca c cu chp hnh servo thu lc dng tnh tin c s hot ng nh hnh 20.88ra trong hnh 20.89. C cu chp hnh ny l mt thit b c lp gm c xilanh, van v b chuyn i v tr.

H thng chp hnh lu cht c iu khin l mt h c in t in hnh. N l s kt hp ca cc phn c khcc thit b iu khin v cm bin; v n thng i hi phi c qu trnh m phng xc nh kch thc v cc phn t khc nhau sao cho chng ph hp vi cc c tnh mong mun.

K hiu chun ca cc thnh phn khc nhau trong h thng thu lc v kh nn cng nh cc nh ngha v c ph tr c xc nh theo chun ISO 1219 Cc thnh phn ca h thng nng lng thu lc, kh nn - k hiu hncc s mch; Phn 1: K hiu hnh hc, Phn 2: S mch

H thng servo thu lc v kh nn

2


3/27

C cu chp hnh

H thng servo thu lc hay kh nn l cc h iu khin cng sut c u ra, chng c th l cc h thng sekhin v tr, tc hoc h servo iu khin lc, mmen, p sut.

iu khin cc i lng u ra bao gm vic iu khin lu lng dng lu cht c nh ngha l lu lchy hoc dng kh qua cc thnh phn ca h thng servo.

Hai dng ph bin thng dng trong cc ng dng hin nay l: h thng servo thu lc trong lu cht tndng cht lng v h thng servo kh nn, trong lu cht tn ti di dng kh nn. p sut lm vic ca h ththu lc thng c gi tr trong khong t 150 n 300 bar, trong khi vi cc h thng kh nn p sut ny l di

Nhm cc h thng servo thu lc s dng cc loi du - cc loi lu cht c nht cao cho cc h thng servp sut iu khin cao v cc cht d chy nh xng du cho ngnh t v hng khng (JPA, JPB,), c ng dcc mch nhin liu ca cc ng c t chy. Cc h thng servo khc s dng cht lng truyn lc c dngnghip v hng hi.

HNH 20.89H thng servo thy lc dng xi lanhCc h thng servo kh nn c ng dng trong cng nghip, t ng ho sn sut, t ng ho qu trnh, gia

hng khng, ng bin, ng b, ng st. Kh nn trong nhng ng dng nu trn c to ra bi my nn lkh t mi trng. Trong cc ng dng khc, lu cht lm vic khng phi l kh nn m l kh gas c bit. Hin ch thng servo dng lu cht ng lnh c dng trong cc phng tin i li v cc h thng lm lnh cng ngcnh cn c nhng h thng servo dng kh t (LPG. mtan, prban) trong cc ng dng gia dng v dng Nit trng dng p sut cao. T nhng phn tch s b ny, chng ta nhn thy rng cc h thng servo thy lc v kh nngia vo qu trnh sn xut sn phm v cng vi c cu servo in tr thnh b phn khng th thiu trong qu trnht ng ho. Ngoi ra, chng cn ng vai tr nh cc thit b pht ng iu khin c, tch hp vo trong bn sn phm. Sau y chng ta s xem xt cc c cu chp hnh servo c lp t trong my bay v ngy cng ph bcc phng tin giao thng ng b hin nay.

H thng chp hnh thy lcThnh phn ca mt h thng thy lc gm:

Bm: l h thng to cng sut thy lc;C cu chp hnh: l cc phn t chuyn nng lng thy lc sang nng lng c hc;Van: l b iu chnh nng lng thy lc;ng ni: ni cc thnh phn ca h thng chp hnh;B lc, cquy, ngun cung cp;Cht lng: l cc cht truyn nng lng gia cc phn t khc nhau trong mch;Cc cm bin v b chuyn i;

H thng hin th, o lng v cc thit b iu khin.

3


4/27

S tay C in t

HNH 20.90Phn loi bm

BmBm chuyn nng lng in v nng lng c hc sang dng nng lng thu lc. Chng to ra dng cht ln

thng thu lc vi p sut c xc nh bi cn thu lc theo hng xui dng t pha pht. Cc dng chnhc gii thiu trn hnh 20.90.

Cc my bm ly tm c cng sut cao p sut thp. Chng khng c cc van trong nhng c mt khong trng l phn quay v phn tnh bo m dng chy tnh ph hp. Ngc li, nhng loi bm c ng dng ph bithy tnh hay bm c dch chuyn dng li c p sut cao vi cng sut hn ch. Chng gm cc phn t nh np phn chia khu vc cng sut vi khu vc u vo, chng c th a cc xung vo dng chy trn u ra v thdng cc cht lng c c tnh bi trn v kh nng chu ti cao gim ma st gia cc phn trt ca bm. C bm c dch chuyn khng i v thay i. Dng chnh ca bm thy tnh hay bm c dch chuyn dngrng, van quay v dng piston.

Bm bnh rng Bm bnh rng c cc dng bnh rng ngoi, bnh rng trong, trc vt. tt c cc dng , bm u gm 2 b

t trong v sao cho khe h gia chng l nh nht.Hnh 20.91 l s ca bm vi cc bnh rng ngoi. Cc bnh rng quay ngc nhau s khin du nm trong

khng gian gia rng v thnh ca bnh rng c truyn t ch tht ti li ra. Ty theo dng rng, bm bnh rng cc dng: bnh rng tr, bnh rng xon c v bnh rng cam.Cc dng bm vi bnh rng trong c chc nng tng t nh trn nhng cc bnh rng quay theo cng mt chi

20.92 m t cu to ca bm 2 tng(two-stage). Vi bnh rng trc vt, c th c mt hay nhiu rotor, cc phn t c rxon c tng t nh ng ren xoy trn c ca trc vt. Cht lng c truyn trong trc theo chiu quay ca trdng bm ny m bo dng chy m, gim ting ng, mc nhiu thp.

Tc quay thng thng trong khong gia 1000 v 3000 vng/pht, cng sut t 1 v 100 kW. p sut c th 250 bar, cc bm bnh rng ngoi gi tr ny cn cao hn. Dng chy l hm ca khong cch dch chuyn ca bm gc u vo, vi cc gi tr t 0.1 ti 1000 cm2/vng quay. tng cc gi tr ny c th dng bm kp. Bm bnh rngc hiu sut cao, thng c gi tr trn di 90%.

Bm dng van quay

Bm van (Hnh. 20.93) thng gm 1 stator v 1 rotor, c th quay c lp i vi cc phn khc. Cc van cchuyn trong cc rnh ring t quanh stator hoc rotor v phn cch bi cc th tch thay i tng ng.

4


5/27

C cu chp hnh

HNH 20.91Bm bnh rng tr ngoi (hng Casappa)

HNH 20.92Bm bnh rng trong (hng Truninger)

Trong hnh 20.93, cng nh trong hu ht cc cu trc, cc van c bi rotor c th quay pha bn trong statoquay dn n s di chuyn khi cht lng gia 2 van lin tip nhau t bn trong ti u vo ca phn pht. Dngcho php di p sut lm vic ti 100 bar v dng chy dao ng thp hn, tnh ln hn nhiu so vi cc bm drng.

HNH 20.93Bm dng van quay

HNH 20.94Bm pittng dng trc (Bosch Rexroth)5


6/27

S tay C in t

Bm pittng Cc bm pittng c th c mt hoc nhiu xilanh, mi xilanh c 1 pittng trt trong n. S di chuyn ca khi c

t trong ra ngoi xc nh bi khong di chuyn ca pittng trong xilanh, cung cp bi cc van vo ra hoc cc ca. Tdng sp xp hnh hc ca cc xilanh tng ng vi trc quay ca ng c, cc bm piston c chia ra bm dng bent axis v swash plate) v bm hng tm. Hnh 20.94 m t s ca 1 bm piston dng trc di chuyn khongdng swash plate. Di p sut lm vic tng ng ca bm pittng ln hn cc dng trc, c th t ti p su400-500 bar nhng nhc im l c nhiu dng chy khng u hn.

C cu chp hnh truyn ngCc c cu chp hnh ny chuyn nng lng thy lc ca cht lng b nn sang dng nng lng c. Cc c

hnh l cc ng c thu lc, v c bn, dng chuyn ng sinh ra, tng t nh trng hp bm, trong ccquay, na quay phn to ra cc chuyn ng quay l cc trc, v trong cc ng c tuyn tnh chuyn ng qua li, ra chuyn ng quay l cc xi lanh thu lc.

Cc ng c quay v na quayTrong cc thut ng gii thch, cc ng c quay gn ging nh bm quay. C cc ng c dng bnh rng, van

dng trn hoc ng. Tuy nhin, nguyn l hot ng th ngc li vi bm. Hnh 20.95 l k hiu ng c thy quay. Cc ng c dng na quay to cc dao ng hoc chuyn ng tuyn tnh, chuyn ng quay ca mt vantip vi u ra ca trc hoc kt hp vi 1 c cu thanh rng truyn ng bi 1 piston, vi mt bnh rng ni vi trc nh v d trong hnh 20.96. Cc ng c van na quay sinh ra mmen xon tc thi trn trc u ra; V vy chgi l cc ng c mmen thy lc.

HNH 20.95K hiu cc ng c thu lc quay

HNH 20.96C cu chp hnh thu lc quay (Parker Hannin)

ng c tuyn tnhCc ng c thu lc tuyn tnh tip tc c s dng ph thng nht trong cc dng c cu chp hnh. Chn

chuyn ng thng bi hnh trnh ca thanh truyn c ni vi 1 piston trt trong xilanh. C s khc bit gia xng n v tc ng kp. Trc y cc ng c ch c duy nht mt hnh trnh lm vic do vy p sut ca cht6


7/27

C cu chp hnh

tc dng ln b mt ca piston theo mt chiu duy nht; hnh trnh ko li c thc hin nh lc tc ng ln mtcn xilanh to ra, hoc nh lc l xo xon kt hp vi c cu chp hnh trong 1 khoang. Cui cng c hai hnh php cht lng tc ng ln lt ln c 2 mt ca piston, nhm to ra c 2 hnh trnh tin hoc li. Cc xi lanh tcc th c thanh truyn n hoc kp. N bao gm mt ng bt c 2 u v mt piston di ng trong trng cha 1 hotruyn c ni ra bn ngoi to chuyn ng. Khi n c c nh vi ming m hn kn, piston chia xilankhoang. Vic a du b nn vi p sut cao vo mt trong cc khoang thng qua cc ng dn c bit pha u, lch p sut gia 2 mt piston truyn 1 sc p ra pha ngoi nh thanh truyn. Hnh 20.97 gii thch hot ng ca 1 xilanhthy lc tc ng kp vi 1 thanh truyn n. C cu chp hnh thanh truyn n c coi nh cc xilanh khng bi vng lm vic pha thanh truyn th nh hn vng ca pittng.

HNH 20.97C cu chp hnh pittng tc ng kp, thanh truyn n (Atos)

HNH 20.98K hiu cc c cu chp hnh

loi ny cc lc tc ng v tc theo 2 hai chiu khc nhau, vi cng p sut trong 2 khoang.Cc c cu chp hnh thy lc c th chng c qu ti, nu ti vt qu lc y c th, thanh truyn s d

chuyn ng ngc li, nhng khng gy hng. Tuy nhin, cc xilanh c th b hng hoc t nht b gim hiu sut dti khng dc theo trc ca thanh truyn s sinh ra phn lc t thanh truyn v trc pittng dn ti hin tmn, gim kht v gy ra s r r du.

Nhng thng s chnh ca ng c tuyn tnh l kch thc, hnh trnh pittng, p sut lm vic ti a, dng chtcch lin kt. K hiu cc dng c cu chp hnh khc nhau c nu trong hnh 20.98.

VanVan l b phn trong h thy lc c nhim v iu chnh cng sut thy lc truyn ti c cu chp hnh. Va

chng l ng/m dng chy ca du hoc phn hng n theo yu cu, do cho php iu chnh hai i lng vca vic truyn cht lng l p sut v lu lng dng chy. Phn loi van theo ch hot ng, c cc dng:

Van nh hngVan ng-m

Van iu chnh p sutVan iu chnh lu lng dng chy

Ngi ta thng s dng van t l lu lng, van servo, van t l p sut trong cc ng dng c cu servo lin tc

Van nh hng Van nh hng quyt nh lng v chiu dng du qua n bng s dch chuyn ca cc phn di ng tng

chng hoc tc ng t bn ngoi. Van nh hng cng c xem nh van phn phi, c phn bit nh dng phn t di ng, do chng c phn loi theo cu trc bn trong ca chng, bng s lng kt ni c th vi cngoi v s v tr chuyn mch.

7


8/27

S tay C in t

HNH 20.99S ca van bn ngn, hai v tr

Cc phn t di ng c th c dng poppet hoc dng cun. Cc van poppet khng phn bit dng cht lng vchu tc ng bi cc tp cht trong cht lng , nhng yu cu lc tc ng ln khi khng th b vi p lc ducun cho php kt ni ng thi theo mt vi dng khc nhau vi cc s chuyn khc nhau, do thng dng hlinh hot ca chng. S kt ni c th c xc nh bi s ng chy hoc cch th hin trn b ngoi thn vanchuyn tng ng vi s s kt ni c th ca van, vi cc di chuyn thch hp ca cc phn di ng.

Hnh 20.99 m t s hot ng ca van cun 4 ngn, 2 v tr (k hiu l 4/2) ni vi mt ng c tuyn tnh hkp. v tr 1 (Hnh 20.99(a)) ngun cung cp lin h vi khoang sau ca xilanh qua u ra A, cn khoang trc x qB. Trong s ny, pittng thc hin mt hnh trnh tin v pha trc khi thanh truyn hng ra ngoi. v tr 20.99(b)), s di chuyn ca van trt khin vic cp v x 2 khoang ngc li, pittng thc hin hnh trnh ngc

Van nh hng nhiu v tr c biu din bi cc hnh vung t cnh nhau m t cc kt ni tng v tr. Hnh biu din mt s k hiu van nh hng theo chun ISO. V tr gia ca cc van 3 v tr thng l v tr dng van cun v cc v tr ph. Cc van nh hng c th iu khin theo nhiu cch (Hnh. 20.100): bng tay; c kh: b nh cam, n by,...; thy lc v kh nn: cc lu cht b nn; in t, trc tip hoc gin tip ph thuc vo lc nam chm in t thng hng vi van trt hoc bng dng cht lng, hng ca chng c iu khin bi hng nh hn van iu khin chnh.

Cc van ng/ngt Cc van ng ngt l cc van n hng, ch cho php dng cht lng chy theo mt hng. Do chng cn d

chiu ngc li nn chng cn c gi l van khng phn hi (nonreturn) hay cc van kim tra. Cc van ng/ngc t trong mch thy lc gia bm v c cu chp hnh nn khi my pht dng, dng cht lng trong h thc a vo b cha m lu li trong ng ng. Vic ny s ngn s lng ph nng lng cho vic lp y ngm bo v tr ca c cu chp hnh di ti.

V mt cu trc, cc van kim tra bao gm mt c cu chp hnh km theo mt qu cu hoc pittng c dutr ca n nh lc y ca l xo (van khng phn hi), hoc bng p sut chnh lch gia bn trong hoc bn ngohng).

Van iu chnh p sut C 2 dng van iu chnh p sut c bn: cc van hn ch p sut hay van an ton v van gim p.

8


9/27

C cu chp hnh

Hnh 20.100K hiu cc van

Cc van hn ch p sut m bo hot ng chnh xc ca h thng, ngn p sut vt qu cc mc nguy hithng. Thng c mt van p sut ti a trong cc mch thy lc tho bt k dng chy d tha no m h thcn n vo b cha. iu ny l do cc my pht hoc cc bm dch chuyn dng, cung cp dng cht lng lin tn khng c s dng hoc khng c van hn ch p hoc van p sut ti a th n s lm cho p sut trong h thn cc gi tr khng cho php. Cc van hn ch p sut c th tc ng trc tip hoc gin tip. Trng hp trc cp mt lc l so vi ti trng c nh t trc, bng vi lc chng li p sut ca ca chn, hoc ca iu chnh m bcho p sut m ti a. Trng hp gin tip thay th tc ng ca l so bng tc ng ca dng cht lng c iu klc bi mt van dn hng

Chc nng ca cc van iu chnh p sut l gi 1 p sut khng i cui dng, khng ph thuc s thay i p dng. Vic iu chnh cc mc p sut c th t bng tay bng tn hiu dn hng hoc bng tn hiu in tncc trng hp sau, cc van iu chnh p sut c th hot ng trong mch in vng kn, khi chng c b chuyn o p sut iu khin.

Cc van iu chnh lu lng Cc van dng ny dng iu chnh lu lng dng cht lng i qua n. N hot ng nh van tit lu n gi

t mt vi phun c din tch thay i. Dng qua van tit lu l hm ca din tch qua v hiu gia p sut u v cuDo , van tit lu n gin rt nhy cm vi ti, bi lu lng dng chy cng ph thuc s gim p sut in v chu s chi phi ca cc thnh phn khc trong mch.

Trong trng hp van iu chnh lu lng b p sut, lu lng dng chy c duy tr khng i trn mc thiu (thng bng 10 bar) nh mt hm ring ca cc im t ngoi bng tay hoc in. Trong trng hp ny, vvan tit lu ni tip, mt ci c nh v ci cn li t ng thay i, duy tr s gim p sut khng i van tnh v bo m lu lng dng chy khng i.

K hiu ca cc van iu chnh lu lng theo chun ISO c gii thiu trong hnh 20.101.

HNH 20.101K hiu ca cc van iu khin dng

Van t l v van servoCc van servo bt u xut hin t cui nhng nm 1930 v c ng dng ch yu trong lnh vc qun s

khng. Sn phm thng mi u tin xut hin gia nhng nm 50. Ngy nay, cc van servo v van t l c s dri trong lnh vc dn s, hng khng, v tr, t ng ha, cng nghip. Thng thng, chng c dng cho iu khtc v tr, tc v lc ca mt c cu chp hnh thy lc, yu cu chnh xc v v tr hoc chnh xc trong iuv di tn s lm vic, trong c hai cu trc iu khin vng m v vng kn.

Van servo hay van t l l mt b phn ca h thy lc c th sinh ra u ra iu khin c v l hm ca uin. Thit b chuyn i tn hiu in thnh tc ng van cun hoc van a (s-pp hnh nm) l nam chm in c hm (ng c mmen ln) hoc cun cm t l. ng c hm chuyn dng in mt chiu nh sang mmen tcmt rotor ch lng cc. Cun cm t l sinh ra lc n hng trn hm phn ng di ng ca dng introng cun dy, vi c tnh duy tr lc ny xp x khng i trong di dch chuyn con chy. ng c hm, vi cdng in v in cm thp, c thi gian p ng ngn hn cun cm servo, c bit l hot ng vi dng in csinh ra cng sut c thp hn. Do , ng c hm to thnh tng dn hng thng thy trong cc van servo, cn cservo c dng trong cc van t l tc ng trc tip ln cc van cun.

Lu lng dng chy hoc chnh p sut c th c iu khin trc tip bi cc van servo hoc van t l,Van servo hoc van t l thng c cc c im c bn sau:

Cc tn hiu u vo chnh xcTrTuyn tnh u vo v ra.Di cht (dead band)

9


10/27

S tay C in t

rng diCc tn hiu u vo c c trng bi dng tn hiu v di bin thin. Ph bin l cc tn hiu dng in (10

4-20 mA) in p (0-10 V). chnh xc c nh gi bi hiu gia gi tr t v gi tr thc. N c tnh theo pht l. c tnh tr nhn c t chnh vi im t ln v xung tng ng. Cc gi tr ca n biu din t l gti a v gi tr ton thang o. Tuyn tnh l 1 c tnh c th c lng trn ton di lm vic. Trong di tng i, c biu din bng t l phn trm lch ti a ca quan h vo/ra ca phn hi quy tuyn tnh ca n. Thng thkhin v tr yu cu tuyn tnh cao hn so vi iu khin tc , p sut hoc lc. Di cht (dead band) xc nh

vo ti thiu m bo s bin thin u ra vn c chp nhn. Khng ging nh trn, rng di l mt c tBNG 20.11Nhng im khc nhau chnh gia cc van servo v van t l thu lc

Van servo Van t l

Chuyn i c in Mmen ng c hai chiu vi nozzle-flapper hoc vi phun

Cun cm servo nhiu hng (1020W)

Dng in u vo 100 200 mA < 3A

Lu lng dng chy 2 200 l/pht (dng mt tng) vi vanc gim p sut = 70 bar

10 500 l/pht (dng mt tng) vivan c gim p sut = 10 bar

Tr


11/27

C cu chp hnh

HNH 20.102Vi phun ng servo nozzle-apper, (Moog)

HNH 20.103S van servo dng ng phun jet-pipe

S kch thch ny gy ra s khng cn bng cc lc phn cui ca phn ng ng c, sinh ra mmen lm bn thng v ng phun quay. S dch chuyn ca ming nozzle i hi s phn phi khc nhau gia 2 ng di n v kt qra chnh p sut im cui ca van cun vi 1 lc tt yu trn n l nguyn nhn gy ra s dch chuyn. Phnvan cun v ng phun jet-pipe, sinh ra phn hi v tr. dch chuyn ca van v ng phun jet-pipe lm bin dng lhi, sinh ra mt lc tng t l vi n, c th cn bng mmen cp cho ng phun jet-pipe v lc sinh ra im cui Theo cch ny, h thng tm c mt v tr cn bng, t l vi in p u vo. L so phn hi nh tm van trt c nh, do khng cn thit nh tm l so. Hnh 20.104 l s khi m t cc thnh phn van servo jet-pipe vch thch cc i lng vt l a ra trong hnh 20.103.

HNH 20.104S khi ca van servo dng ng phun jet-pipe

11


12/27

S tay C in t

HNH 20.105Cun cm servo t l (a) s cun cm, (b) c tnh cun cm

Van t l Cc van t l c th c chia nhm thnh van t l dng, v nhm van t l p sut (hn ch v gim p sut

trng hp u, tc ng ca cun cm servo phn ng (Hnh. 20.105(a)) thay th ng cun chnh ca van, c gil so trn thn van.

Cc c tnh lc ca cun cm servo, trong ton b hnh trnh c th ca ng cun, l hm ca dng in uminh ha trong hnh 20.105(b). ng vi cc gi tr c th ca dng in vo, c mt v tr nht nh ti lc icun cm cn bng vi lc phn hi ca l so.

Vi trng hp yu cu v tr chnh xc ca van trt, c th dng phn hi v tr. Hnh 20.106 minh ha mt dng ny. Tn hiu vo cun cm servo, t modul phn hi l sai s c b bi mng PID gia tn hiu t v tn hi t b chuyn i v tr LVDT. chnh xc v kh nng lp li ca van s tng nh phn hi v tr, khi sai s trdo ma st gia cc phn di ng c b mt phn.

Cc van lu lng t l c th c 2 tng. Trong trng hp ny, u ra ca van t l dn hng a ti khoangmt van cun c kch c ln hn, cho php t c lu lng iu khin ln hn ng thi gim c cc c Hnh 20.107 l van lu lng t l 2 tng.

Trong cc van t l iu chnh p sut, hot ng ca cun cm servo s tc ng ln nh chp nn sao cho c chnh c p sut trong khoang u t pha nh ca n. Hnh 20.108 minh ha hot ng ca van t l tng p.

HNH 20.106Van t l lu lng (Bosch Rexroth)

12


13/27

C cu chp hnh

HNH 20.107Van t l lu lng 2 tng: (1) tng van cun chnh, (2) tng dn hng, (3) LVDT ca tng dn hnLVDT ca tng chnh (Atos)

HNH 20.108Van t l tng p (Bosch Rexroth)

M hnh h thng servo thy lc iu khin v tr Hnh 20.109 l s mt h thng servo thy lc iu khin v tr tc ng kp, c cu chp hnh 2 im cui

khin bi mch vng kn vi van servo 4 ng. V tr x ca piston xc nh khi lc tc ng ln n cn bng: ngoy do p sut P1, P2 tc ng ln khoang ca pittng, lc ma st v lc qun tnh. p sut P1 v P2 c xc nh bi lulng dng QC1 v QC2 vo ra cc khoang. T l lu lng QFI tng ng vi lu lng du r gia pittng v barrel. Van t llu lng iu khin lu lng du da trn tn hiu tref t c cu b GC. u vo ca b b l sai s eV gia tn hiuVSET, tng ng vi v tr tay n mong mun x o bi chuyn i v tr LVDT. C cu chp hnh c m hnh b phng trnh lu lng lin tc trong khoang v phng trnh cn bng ng lc hc cho tay n. Cc phng trnh

c dng:( ) ( )IN OUT

d V dV d Q Q V

dt dt dt r r

r r - = = + (20.39)

INQ = tng lu lng voOUT Q = tng lu lng ra

r = t trngV = th tcht = thi gian

dV d dP

V

r

r b= - = - (20.40)

Chng bao gm

13


14/27

S tay C in t

IN OUT dV V dP

Q QV dt b

- = + (20.41)Phng trnh lin tc cho khoang 1 l:

0 1c smCI FI c

V A x V dPQ Q A x

dt b+ +- = +& (20.42)

HNH 20.109S ca h thng servo thu lc iu khin v tr

Phng trnh lin tc cho khoang 1 l:

0 22

c smFI C c

V A x V dPQ Q A x

dt b

- +- = +& (20.43)

Trong :QC1 = lu lng vo khoang 1QC2 = lu lng vo khoang 2QFI = lu lng r gia pittng v barrelAc = phn y = (Dal2 - Dst2) /4Dal = ng knh xilanhDst = ng knh tay nV0 = th tch khoang vi pittng gia = Ac L/2L = di hnh trnh chuyn ng ca pittng

X = dch chuyn ca pittng (x=0 v tr trung tm)Vsm = dead bank volume

Phng trnh cn bng ca piston l

1 2 0c c e AP A P A F Mx F - - - - =&& (20.44)

Trong M: khi lng dch chuynFe: Lc suy rngFA: lc ma st = ( ) ATT x F sign x g +& &

: h s ma st nht

FA: Lc ma st inVic r c th c m hnh nh cc in tr tng v cc iu kin trng thi n nh:P

RQ

D= (20.45)

Trong :R: in tr Q: lu lng dng chyP: chnh p sut

Trong trng hp ca mt ng hnh khuyn, ta c:3 2(1 1.5 )12

DhQ Pl

p em+

= D (20.46)

Trong :

14


15/27

C cu chp hnh

D: ng knh mth: su ngch = (D - d)/2d: ng knh cun : lch tm = 2e/(D h)e: khong cch gia mt trc v trc cun : nht ngl: di ngch

HNH 20.110S khi ca van

HNH 20.111S tng ng biu din lu lng dng chy qua van t l

c tnh ng ca b chuyn i in c v ca van cun c th nhn dng bi mt m hnh tuyn tnh bc 2 gt v v tr van trt xV, biu th nh hnh 20.110, trong K S (m/V) l h s tnh ca van, n (rad/s) l tn s t nhin cavan, v l h s tt dn ca van.

Lu lng iu chnh bi cc van t l c minh ha chi tit trong s hnh 20.111. Ta c, cc din tch xc 1,A2, A3, A4, hm ca dch chuyn van xV theo khi nim hnh hc c bn, lu lng dch chuyn qua van l hm ca p suc tnh theo:

11 1 1 12( )d s SQ AC sign P P P Pr

= - - (20.47)

4 4 1

2( )d s T T Q A C sign P P P Pr = - - (20.48)

2 2 2 2

2( )d s SQ A C sign P P P Pr = - - (20.49)

3 3 2

2( )d s T T Q A C sign P P P Pr = - - (20.50)

trong PS: p sut ngun, PT: p sut x ca vng cha, Cd l h s lu lng ca cc cng o. Do , chng ta c:1 1 4CQ Q Q= - (20.51)

2 3 2CQ Q Q= - (20.52)

H cc phng trnh a ra trn c th tuyn tnh ln cn im lm vic, xc nh bi vng chuyn ca vanV0, psut ti gim PL0 = P1 - P2, v v tr x0.

HNH 20.112S khi m hnh h servo thu lc tuyn tnh iu khin v trVi gi thit d khng ng k, s ca h tuyn tnh c ch ra trn hnh 20.112 vi:

15


16/27

S tay C in t

GC hm truyn ca phn b [V/V]'

2S Q C

OLV C PQ

K K AK

A K g=

- h s khuch i tc tnh/m s

V

T P RK K = khuch i tnh ca chuyn i v tr [V/m]

2

PQOLF

C PQ

K K

A K g

=-

hng s lcm

sN

ns tn s t nhin ca van t l [rad/s] z h s gim chn ca van t l

( )20 1 / A PQ CC

K AM

s g = - tn s cng hng thu lc [rad/s]

( )( )

20

20

/

2 1 /PQ C

A

PQ C

MK C A

C M K A

g z

g

-=

- h s gim chn thy lc

2

0

C

PQ

AC K

t = Hng s thi gian ca nhiu lc [s]

trong :' max

max

V S

AK

V = h s tnh van t l

2mV

0 0,V L

QV A P

QK

A= h s khuch i lu lng van t l

3

2

msm

C A din tch pittng [m2]

0 0,V L

QL A P

QK

P= H s khuch i lu lng-p sut van t l

3msPa

g h s ma st nht/N

s m

2

0

4 CT

AC

V b= cng trong nc v tr gia [N/m]

T V tng th tch ca hai khoang [m3]M: khi lng cc phn di chuyn [kg]Hm truyn vng h:

2 2

2 2 2 2

12 2

RET n AOL C OLV TP

n n A A A

V G G K K

e s s s s ss s

zs s z s s= =

+ + + + (20.53)

'

0 2

C S Q C T PC OLV TP

C PQ

G K K A K K G K K

A K g= =

- (20.54)

2 2

2 2 2 2 2 2

5 4 3 25 4 3 2 1

( 2 )( 2 )1

1

R C OLV n ACL

set n n A A A C OLV n A TP

K G K x G

X s s s s s G K K

a s a s a s a s a s

s s zs s z s s s s

= =+ + + + +

=+ + + + +

(20.55)

trong :

5 42 2

0 0

1 2, An A n A n A

a aK K

z z

s s s s s s

= = +

16


17/27

C cu chp hnh

3 2 12 20 0 0

41 1 1 2 1, , A An n A A n A

a a aK K K

z z z z s s s s s s

= + + = + =

Hm truyn gia u ra v nhiu, theo cn bng ng hc l:2 2 2

2 2 2 2 2 2

( 1)( 2 )( 2 )( 2 )

OLF n n AFCL

e n n A A A C OLV n A TP

K s s s x G

F s s s s s G K K t zs s s

zs s z s s s s- + += =

+ + + + + (20.56)

p ng tnh l:

' '0

1PQOLF e C OLV TP C S Q C TP C S P C TPs

K K x F G K K G K K A K G K K A K =

= = = (20.57)

t:

0 0,

,V L

Q LPQ P

P V A P

K PK K

K A= = h s p sut 2

Pam

Cui cng, cng tnh xc nh bi

'

0

eC S P C TP

s

F G K K A K

x == (20.58)

D suy ra hng s thi gian, t hm truyn vng kn l h s a1 = 1/K 0.Kt lun:

H s khuch i tc K OLV, h s khuch i tnh vng h K 0 ph thuc ng k vo h s khuch i t l lulng K Q v tng khi K Q tng. K Q tng khi PS tng, gim khi PL0 tng, v khng bin thin vi AV0. Vi gi thit di 1000 Ns/m, tc ng ca K PQ khng ln, thc t khng ng k.

Hng s lc K OLF ph thuc h s khuch i lu lng-p sut K PQ v tng theo K PQ. |K PQ| tng vi AV0 v viPL0, gim khi PS tng; do vy |K OLF| gim khi PS tng. Tn hao dn ti tng K OLF.

cng tnh ph thuc nhiu vo khuch i p sut ca van v tng theo thng s ny. Da vo K P gim t lvi d dn ti gim cng tnh. Hn na, da vo |K PQ| tng vi AV0 trong khi K Q khng bin thin theo AV0, khuch i p sut gim khi AV0 tng, do cng tnh gim nu van lm vic vi cc iu kin m h

Hn na, da vo |K PQ| gim khi PS tng, trong khi K Q tng khi PS tng, h s khuch i p sut tng khi PS tng,do cng tnh tng vi PS. Tn s cng hng cht lng tng vi c ca cht lng C0 v gim khi khi lng M tng. Thc t n khng

b nh hng bi h s khuch i lu lng-p sut K PQnu


18/27

S tay C in t

HNH 20.113Phn loi cc my nn kh

HNH 20.114Cc my nn pittng: (a) c cu n, (b) c cu i

My nnCc loi my nn s dng to ra kh nn c tng kt trong hnh 20.113. Trong cc my nn th tch, kh

c ht bng van trong bung nn ni m th tch b gim do vic nn kh ga. Khi t n mt p sut nh trc sut m dn khi kh n ni s dng.

Ngc li, cc my nn kh ng hay cc my tuabin nn kh, nng lng ng lc c bin i thnh p lc truga lm quay bnh cng tc.

Cc my nn pittng iu chnh vic nn kh ga nh ca chuyn ng ca pittng, chuyn ng nay gy ra bi m

ni pittng v c cu trc khuu thanh truyn, bn trong mt xylanh khng r r kh. Chng c th l dng c cu i, vi mt hay nhiu pittng v mt hay nhiu k. Chng c th t ti p sut hng trm bar, trong trng hp nv tc dng kh l nghn khi kh mi gi, trong trng hp nhiu xylanh. Cc my nn dng cnh c mt rto, tm vi trc quay ca xylanh, mt s cnh c th chuyn ng xung quanh trc ca n.

18


19/27

C cu chp hnh

HNH 20.115My nn cnh quay (Preumofore)

Trong chuyn ng quay lin tc ca rto, cc cnh c quay ly tm vi im ta ca stato, th tch cc bung i theo hnh trnh gc, m bo kh nn ht vo mt u v y ra u cn li. p sut nn l di 15 bar, vdng ln nht l 500m3/h. So vi cc my nn pittng chuyn ng qua li, cc my nn loi ny c n, rung thpnh gn hn.

Cc my nn dng xon c 2 rto quay ngc chiu nhau pha trong stato, mt cng chiu vi cc vu (cam) lcng chiu vi cc vu (cam) lm. S kt hp cc mt nghing ca hai rto lm gim th tch trong sut quay v nnp sut ph bin l di 15 bar, dng kh lin tc c cung cp, t gi tr ln ti 3000 m3/gi.

Tng t nh vy, cc my nnRoots,cng c xem nh cc my siu np, c lp 2 rto hnh s 8 quay ng pha trong stato sao cho kh ga c truyn t u ht vo ti u x ra. Hiu sut ca cc my nn dng ny thp gia bn thn hai rto, gia cc vu v v bc, v vy chng c s dng cho cc p sut nn thp, di 2 bar. Tuchng cho php hot ng m khng phi bi trn, ging nh cc my nn xon, so thu c kh khng c du.

Hai dng my nn hng trc v hng knh c s dng to ra cc lung kh nn c tc t vi nghn m3/h.

Cc b x l kh nn Ngun cung cp kh nn n h thng tu ng (servosystem) thng c mt b phn x l kh bn trong, bao g

b lc, c ni vi mt mng li cung cp v phn phi kh ga, mt b iu chnh p sut, i khi km theo mtm L. Hnh 20.116 l mt v d v mt b tch hp lc v iu chnh p sut. u tin, kh c a qua b lc v bi mt my lm lch dng, cc tp cht lng v rn s lng xung y, cng nh tc dng ca mng y hnh nn bnung hoc vi lc, c t di khi hnh tr thng t ong. Sau , kh c lc lan truyn n u vo cchnh p sut, ghp vi mt ca sp trong trng thi cn bng v p lc nn. Vic iu khin p sut cui dng bi v tr ca ca sp chnh, ni iu khin dng ra. Khe h c ng khi p lc do p sut cui dng, tc ng lngn v pittng (tnh tin), cn bng vi p lc pha u l so, ti trng cho trc c thit lp bng cch quay khin. Ngc li, nu p sut kh nn thp hn mc yu cu, lung kh s c iu khin theo hng b p sut ca sp (np) c ng tr li, khi t p sut yu cu. Trng hp nu p sut iu tit cao hn mc yu cu th s c m gia vng s dng v ca x.

19


20/27

S tay C in t

HNH 20.116B lc kh nn/My gim p (Metal Work)

Cc van kh nnCc van kh nn c chc nng tng t vi cc van c s dng trong cc h thng thu lc. c bit, iu

ph hp i vi cc van nh hng dng s v dng t l. Thm ch trong cc h thng kh nn, c ng cun hay c phi 4, 3, 2 ng vi 2 hay 3 v tr lm vic v c thao tc bng tay, c kh, kh nn hoc in.

Cc van t l v thc cht l tng t nh cc van thu lc v c th dng c vi c b bin i c in mhoc c cu ph tc ng trc tip trn ng. Cng nh cc thit b iu khin dng kh, cc van s bng in iu ba lung cng c s dng, cc tn hiu iu khin c iu ch bng phng php iu ch rng xung PWwidth modulation), iu bin tn s xung PFM (pulse frequency modulation), iu bin m xung PCM (pulsemodulation), iu ch s lng xung PNM (pulse number modulation) hoc kt hp cc phng php trn.

Cho n khi cc van iu p c quan tm, cc van t l ba ng c th dng c i vi c cu chp hncho php bin i tn hiu in chun ho u vo thnh p sut iu khin u ra vi chnh xc cao.

Cc van iu ch rng xung PWM Cu to ca cc van PWM tng t nh cc van s iu khin bng in, nhng s dng k thut iu ch

xung gi ti cun dy iu khin t l dng kh cung cp dng iu khin ny, tn hiu t ca in p u vtn hiu tng t VREF (v d 0-10V) c bin i thnh tn hiu s VPWM (ON/OFF) bng mt chuyn i ring. Thngthng (tu s la chn), tn hiu iu bin c to trc tip t mt b iu chnh s, ging nh PLC.

Hnh 20.117 m t nguyn l hot ng ca PWM. Tn hiu in p s a n cun dy c iu chnh bi mxung, vi bin , chu k T khng i, nhng c rng t ca mi xung l mt hm tuyn tnh ca gi tr in chiu. Gi tr m trung bnh ca van l mt hm ca rng xung t, c th l ca cng sut c ch t/T. Cc xung PWchung khng c hi tip, nn van p lc xui dng v tc lung kh ph thuc loi mch kh nn hin thi.

Hnh 20.118 ch ra hai s m t nguyn l hot ng ca cc van 2 ng, 2 v tr, vi PWM, c s dng n b iu chnh lu lng kh (Hnh 20.118 (a)) v b iu chnh p sut (Hnh 20.118 (b)) . Trong m hnh a, van iul lu lng kh i qua hai im, im cung cp c p sut PS (feed pressure) v im cui dng c p sut PV c duy trkhng i (n nh). Trong trng hp ny, lu lng ch l hm ca m van v do l mt hm c dng tuyTrong m hnh b, cc van c t theo hnh ch thp l v d v iu khin p sut PR trong mt bnh c th tch kh V,tng ng vi vic iu chnh lu lng kh vo G1 v ra G2. Gradien theo thi gian ca p sut iu khin tng ng vi l

lng G vo bnh cha.

20


21/27

C cu chp hnh

HNH 20.117Cc tn hiu u vo v u ra iu ch rng xung PWM

HNH 20.118Cc van dng s iu ch rng xung PWM: (a) iu chnh lu lng, (b) iu chnh p sut

HNH 20.119Van a dng s, hai ng

Cc tham s nh hng n vic iu chnh tt/ bt nh van iu ch rng xung PWM bao gm:Thi gian van ng v m Ph thuc s ln m/ng ca p sut xui dng v ngc dngKch c vanChu k T hay tn s sng mang iu ch f=1/TTui th ca van

Dng van ng/m ny c nhng c tnh tri ngc nhau nh s ln ng/m nh nhng lu lng kh qua vando khi thit k h thng lun lun cn c s dung ho gia mt bn l phn gii iu khin v tuyn tnh tt vl p ng ng hc nhanh.

21


22/27

S tay C in t

Trong cc ng dng c cu servo kh nn, di tn s sng mang f=1/T thng nm trong khong 20 n 100 Hzthi gian ng/ m ca cc van l t 1 5 ms.

Hnh 20.119 m t van thng ng 2/2, van ny m bo thi gian m ca van l nh nht (di 1ms). Cc cc do gim cc phn chuyn ng, vi vic s dng mt van a (poppet) ni vi mt thanh dao ng nh, thc ma st gia cc phn c chuyn ng tng i.

Cc van iu chnh p sut t l

Cc van ny thng l van ba ng, vi a i hoc ng cun. Nguyn l hot ng ca cc van a ging nh iu chnh p sut. Tng t nh cc b iu chnh p sut, van a s phn bit mi trng p sut cao v mi sut iu chnh l nh trng thi cn bng gia p lc to bi p sut iu chnh v p lc do khi iu khin to ny c th l lc ca phn ng cun cm servo hoc do p sut iu khin bi khi iu khin tc ng ln pittnmng chn ni vi van a poppet.

HNH 20.120Van p lc theo t l (i xng) (Parker)

HNH 20.121S iu khin ca van p lc theo t l (i xng)

Hnh 20.120 ch ra mt v d v van p lc i xng vi a i. p sut cung cp (ti cc cng), p sut iu csut ra tng ng l P, A v E. cc v tr minh ho trong hnh v, van a 2 l im cui hnh trnh ca n, im 1 din vi mt c nh. Tng t, van a iu chnh 3 lin h vi van a 2 bi phn gia 5. m gia cng P vc xc nh bi s cn bng lc tc ng ln pittng ca van a 3, c bit lc FR do p sut iu chnh PR trong khoang ph 4 hng thng xung, lc FC do tc ng ca p sutiu chnh PC pha ngoi, hng ln. Nu FR =FC, cc thn dichuyn trn van cc v tr nh trn hnh v, do p sut iu khin PC ca khoang c lp vi phn ngun v phn x. NuFR > FC, sau 2 van a di chuyn hng xung v khe h m ra v chuyn khi kh ti u ra v cn bng li PC mt gitr xc nh. Trong trng hp ngc li, nu FR < FC, van a iu chnh di chuyn hng ln, nhung khi gi nguyn v trim cui hnh trnh ca n, phn 5 m v cho php khi kh chuyn t cng A ti ca x.

Trong hnh 20.120, 6 v 7 l cc van PWM m /ng, iu chnh p sut PR ca khoang ph 4. S iu khin kh nnca van c ch ra trong hnh 20.121. Cc van PWM, 2 ng nhn tn hiu iu bin t khi iu chnh. Chng g

ng iu khin lu lng vo khoang 4 (Hnh. 20.120), mt ng iu khin lu lng x ra. Bng tc ng ttn hiu iu khin c chuyn sang tn hiu p sut t l.

22


23/27

C cu chp hnh

Hnh 20.122S mt h thng servo kh nn vi cc van s 2 ng

Hnh 20.123S mt h thng servo kh nn vi cc van t l 3 ng

M hnh ho h thng servo kh nnS ca mt h thng servo kh nn c kh nng iu khin v tr, tc , lc c th tng t nh h thng c

thu lc m t trong hnh 20.109. Tn hiu ca b chuyn i cc i lng xc nh, cn phi c hi tip theovng kn, ph thuc i lng c iu khin.

Trong s hnh 20.122, v tr trc ca pittng, hi tip bng b bin i v tr, c xc nh bng cch iu sut bung nn 1 v 2 do cc tip xc i xng dng. Vic lin h v tr c so snh vi tn hiu hi tip v sai s bng mt b iu chnh. Da trn vic m van, tn hiu c gi n cc van iu chnh, a v cc bung ca p

thit l i xng. p lc tc dng ln b mt ca pittng s chng li ngoi lc. Gi thit s in c bn vanng ng, iu khin bng in. Gii php ny cho php s dng cc van kch thc nh, c th ph hp vi toyu cu ca c cu ph kh nn. phng php ny, t l m ca cc van c iu chnh bng phng php irng xung tn hiu s. Mi cp van V11, V12 v V21,V22 to thnh mt van ba ng u ra kt ni vi khoang ca pittndo s hnh 20.122 c th c xem nh tng ng vi s hnh 20.123 vi cc van ba ng iu khinV1 v V2.

M hnh hnh tr m phng mt h thng vi cc phng trnh vi phn cp ba, hai bc ca khi khng kh tr bung t v mt bc ca trng thi cn bng truyn ng.

Cc ln sau c ly vi phn (ch s di 1 v 2 tng ng vi cc bung t 1 v 2 ca cc pittng):A phn p ln pittngFe nhiu ca ngoi lc ln cn ca pittngG tc khi khng kh vo bung tM khi lng cc phn di ng ca pistonn h s kh a hngP p sut trong khoang xilanhPi p sut ban u trong khoang xilanhPamb p sut mi trngR hng s khng khTi nhit ban u trong khoang xilanhx v tr tay n xo vi gi tr ban u l x0x0 na hnh trnh pistonxm di cht h s ma st nht

1 1 1 1(1 ) /

1 0 1 1 1 0 1( )( / ) ( )i

n nm i m

dP G nRT P n dx dt A x x x P P x x x dt -

= -+ + + + (20.58)

23


24/27

S tay C in t

2 1 2 2(1 ) /

2 0 2 1 2 0 2( )( / ) ( )i

n nm i m

dP G nRT P n dx dt A x x x P P x x x dt -

= -+ - + - (20.59)

21 1 2 2

2

( ) ( ) /amb amb eP P A P P A F dx dt d x dt M

g- - - - -= (20.60)

Van i xng dng V1 c m hnh ho nh mt in tr kh nn bin i. Cc phng trnh c s dng tntc khi kh G qua in tr kh nn, c c tnh ho bi mt dn C v tiu chun ti hn b, ph hp vi c6358, lin kt vi hai mi trng A v B, vi cc p sut tng ng PA v PB, theo chiu dng t A n B, l:

Lung tc m thanh:

0 0 B A A

PG P C b

Pr = < (20.61)

Lung h m:

2

0

/1 11

B A B A

A

P P b PG P C b

b Pr

- = - < -(20.62)

Lung tc m thanh:

0 BG P Cr = - vi 0 B A

P bP

< (20.63)

Lung h m:2

0

/11

A BB

P P bG P C

br

- = - - -vi 1 A

B

Pb

P< (20.64)

Gi s c mt tn hiu lng cc, gi thit l Vref >0 tng ng vng ni ngun cung cp v vng lm vic, Vref


25/27

C cu chp hnh

g) 1 2( , , , )e x x F P P x =&& && & bc 2 Cn bng pittng (xem (20.60)

h) 1 1 ( , )ref ref set ret V V x x = bc 0 iu khin van V1

i) 2 2 ( , )ref ref set ret V V x x = bc 0 iu khin van V2 Nu chng ta mun thc hin mt phn tch tuyn tnh, c th gi thit rng cc phng trnh a), b), g), h), i)

di dng tuyn tnh.

Khi xem xt tc dng ca cc van c), d) chng c gi thit l tc dng chy u vo ti mi im thunm trong vng h m tc l Vref >0, v tc m thanh vng x vi Vref 0 vi php gii tch tng ng.

1 1 1nG C Pr = (20.73)

Ln cn P1 = P1r

1 1 1 12 1n r G P C K Cr = = (20.74)

Cng thc (20.74) c ngha vi Vref1< 0, C1 < 0 vi php gii tch tng ng.Tnh trung bnh K 11 v K 12 ta c h s K meanxc nh bi:

41 111 1241

( )2 2(1 )

n s r

mean

P b PK K K

b

r -+= =-

(20.75)

Hnh 20.124S khi ca m hnh tuyn tnh ca h servo kh nn vi iu khin v tr

Lu lng t l c tuyn tnh l hm ca C1, do ta c:41 1

1 1 1 141

( )2(1 )

n s r P b PG C K Cb

r -= =-

(20.76)

Tng t vi van 2, ta c:25


26/27

S tay C in t

42 2

2 2 2 242

( )2(1 )

n s r P b PG C K C

b

r -= =

-(20.77)

Phng trnh lin tc ca khi lng trong bung t pittng e) v f) , c tuyn tnh quanh im t, xc nh b

1 1 2 2

1 1 2 2

1 2

, ,

0, 0, 00, 1

n r r

r r r

m m

x x P P P P

P P P P x x

x x n

= = =

= = = = = == = =

& & & & & &

1 1 01 1 1

r r P A x x G x A PRT RT

+= + && (20.78)

2 2 02 2 2

r r P A x x G x A PRT RT

-= + && (20.79)

S khi ca m hnh tuyn tnh ch ra trong hnh 20.124.Bng cch dng bin i Laplace ca h thng ca cc phng trnh tuyn tnh, gi thit cc van ging nhau, ta c

2 2 2

2 2 2 2 2 2 . .( 2 ) ( 2 ) ( 2 )n A A

c OLV OLF e

n n A A A A A A

x G K e K sF C I

s s s s s s

s s s

zs s z s s z s s

= - ++ + + + + +

&

(20.80)

trong C.I. l cc iu kin u, K OLV h s khuch i tc tnh, K OLF h s khuch i ca nhiu lc, As v A x tngng l tn s t nhin ca c cu chp hnh v h s gim chn, v Gc lkhi b.

Kt qu ny c ch ra trn s khi hnh 20.125. Hnh 20.126, l so vng kn vi phn hi v tr. C thtng t ca s ny so vi h thng thu lc. c bit, gi thit rng:

1 2r r sP P Pd = = vi 0.6 0.9d -

1 2 A A A= = c cu chp hnh c 2 im cui1n = bin i ng nhit

HNH 20.125S khi ca m hnh mch h

HNH 20.126S khi ca m hnh vng kn

Chng ta c th m t cc thng s chnh ca h thng servo kh nn:*

* *

(1/ )1 12 (1 ) (1 )

nOLV V

RT bK K

b A br d -=

- - (20.81)

20 0[1 ( / ) ]

2r

OLF s

x x x K P Ad

-= (20.82)

26


27/27

C cu chp hnh

20 0

2

[1 ( / ) ]s

Ar

P A

x m x x

d s =

- (20.83)

20 0[1 ( / ) ]

8r

As

x x x P Am

z gd

-= (20.84)

Mt vn c bn khi thit k l c th chn kch c v cc c tnh ca cc thnh phn ph, u tin l hot m hnh tuyn tnh sau kim tra trn mt m hnh h phi tuyn hon ton.

Ti liu tham kho[1] Andersen, B. W.,The analysis and design of pneumatic systems, Wiley, New York, 1967. Bouteille, D., Belforte, G.,

Automazione essibile, elettropneumatica e pneumatica, Tecniche Nuove, Milano, 1987.[2] Belforte, G., D Alo, N., Applicazioni e prove dellautomazione a uido, Levrotto & Bella, Torino, 1997.[3] Belforte, G., Manuello Bertetto, A., Mazza, L., Pneumatica: corso completo, Tecniche Nuove, Milano, 1998.[4] , J. F., Reethof, G., Shearer, J. L., Fluid power control , MIT Press, Cambridge, 1960.[5] Dranseld, P, Hydraulic control systemsdesign and analysis of their dynamics, Springer, Berlin, 1981.[6] Esposito, A., Fluid power with applications, 5th ed., Prentice-Hall, Upper Saddle River, NJ, 2000.

[7] Gotz, W., Hydraulics. Theory and applications, Robert Bosch Automation Technology Division Training, Ditzingen,1998.[8] ,A. H., Fluid power troubleshooting, 2nd ed., Dekker, New York, 1995.[9] Introduction to hydraulic circuits and components,The University of Bath, Bath, 2000.[10] Introduction to control for electrohydraulic systems, The University of Bath, Bath, 1999.[11] Jacazio, G., Piombo, B.,Meccanica applicata alle macchine 3: Regolazione e servomeccanismi, Levrotto & Bella,

Torino, 1994.[12] , J. E.,Electrohydraulic servo systems, 2nd ed., Penton IPC, Cleveland, 1977.[13] Johnson, J. L., Design of electrohydraulic systems for industrial motion control , Penton IPC, Cleveland, 1991.[14] Johnson, J. L., Basic electronics for hydraulic motion control , Penton IPC, Cleveland, 1992.[15] Lewis, E. E., Stern, H., Design of hydraulic control systems, McGraw-Hill, New York, 1962.

[16] Mang, T., Dresel, W., Lubricants and lubrications, Wiley-VCH, Weinheim, 2001.[17] McCloy, D., Martin, H. R.,The control of uid power , Longman, London, 1973.[18] Merritt, H. E., Hydraulic control systems, Wiley, New York, 1967. Moog, Technical Bulletins, 101-152, Moog, New

York.[19] Muller, R., Pneumatics. Theory and applications, Robert Bosch Automation Technology Division Training,

Ditzingen, 1998.[20] Nervegna, N.,Oleodinamica e Pneumatica, Politeko, Torino, 1999.[21] Parr, A., Hydraulics and pneumatics: a technicians and engineers guide, 2nd ed., Butterworth Heinemann, Oxford,

1998.[22] , D., Kolk, R. A.,Mechatronics system design, PWS publishing company, Boston, 1997.[23] Tonyan, M. J.,Electronically controlled proportional valves: selection and application, Dekker, New York, 1985.[24] Viersma, T. J., Analysis, synthesis and design of hydraulic servosystems and pipelines, Elsevier, Amsterdam, 1980.[25] Yeaple, F.,Fluid power design handbook , 3rd ed., Dekker, New York,

sổ tay cdt chuong 20(4)-cc chap hanh

Documents