Tổng hợp bài tập về nhà - Thủy lực.pdf

7/25/2019 Tng hp bi tp v nh - Thy lc.pdf

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I HC QUC GIA THNH PHHCH MINH

TRNG I HC BCH KHOA

BO CO BI TP THY LC

MN: K THUT THY LC-KH NN

GVHD: TS. Trn Nguyn Duy Phng

Nhm: HBo

Sinh vin:

Nguyn Tt Nam 21202269Trng Trng Giang 21200889Phm Vn Trng 21204194Nguyn Phc Hng 21201503Nguyn K Nht 21202639Bi Cng Nam 21202221Hunh Anh V 21204600Nguyn Tin Vit 21204501LXun Vinh 21204534Nguyn Vn S 21203202

Nguyn Ph Vn 21204461Phm Qu L i 21201871

Tp.HCM, thng 12/2014


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BI TP SCH POWER HYDRAULICS

A.1 CIRCUITRY QUESTIONS

A.1.1

Sa li sai trong s

Sa: Thay i ng ra van 2 trng thi

A.1.2 Sa li sai trong s


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Sa: i chiu van mt chiu

A.1.3

Trong mch biu din hnh A.3, cho rng p lc do ti trng trn hnh trnh mrng l 150 barv trn hnh trnh thu hp l 70 bar, v p sut ca van an ton c ci t l 165 bar, vmt


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thp sut - thi gian cho mt chu kmrng v thu hp hon thnh. thnn bt u v ktthc vi vtr gia ca van iu khin hng. Gishnh trnh mrng mt 12 giy v hnhtrnh thu hp 9 giy.

TL:


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A1.4:

Nu c tnh ca mch? ng dng nhiu nht l my no?

Bi lm:

-c tnh:

Du c bm ln nhmotor in, thot u p sut ca du khng thng c lc l xo cavan iu khin hng v van an ton nn du schy dn ht ln xy lanh, to p sut y

pittong sang tri p vt. Khi p sut du ln, thng c lc l xo ca van iu khin


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hng th lm van chuyn trng thi, du sdn c vbn tra xy lanh, to p lc y pittongvphi. Nhng du khng chy vbngay m khi p sut vt qu gii hn ca van an ton thmi chy vb.

-ng dng ca van:

+c sdng chyu cho cc loi my p, my nng.

A.1.5

Sa li sai trong hnh . Nhng li thc trong mch hai bm hn hthng 1 bm. N thngc sdng u ?

Trli :

y, chiu ca van mt chiu bngc .

Cch sa : i ngc chiu li .

u im : C thto ra c lu lng ln p sut thp v lu lng nhp sut cao.

ng dng : Dng nhiu trong cc hthng nn .

A.1.6

u im ca bm b p sut trong ng dng ny ? Hiu sut ca mch sc nng cao nhthno ?


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Trli:

Van iu khin 3 trng thi iu khin bng in: (ttri sang phi)1: Thu vo

2: Giti3: y raBm c lu lng ring thay i c iu khin bng hmt hthy lc (c mt van iukhin bng l xo 2 trng thi) qua a lc (c 2 ng vo iu khin trn 2 u a lc: 1 u lnv 1 u nh, skhc nhau ny nhm to ra mt ngu lc thay i vtr a )Khi van iu khin 3 trng thi trng thi 3: y ra. Lu cht stbm y trc tip vo

pha bn tri ca xy lanh. Lu cht pha bn phi ca xy lanh c xvb. Trong qu trnhny, lu lng ca bm stng dn (do schnh lch lc gia 2 u a lc u di > utrn). Khi n trng thi p vt, p sut trong hthng tng ln v van iu khin bng l xo 2trng thi c kch hot, lu lng bm gim xung (do lu cht u di c thng vi

b, u trn>u di) p sut ca hthng tip tc tng n gi trci t van iu khin psut.

Li thca phng php iu khin ny l to ra c mt lc p ln nhng t tn hi ti bmdo lu lng khi p thp.Van iu khin p sut trc tip cng t bh hi hn v lu lng nh.Thi gian y ra nhanh do xy lanh c cung cp mt lu lng ln. Hthng ny gn nhhnhthng sdng nhiu bm.


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Van tit lu trong hthng c mc ch chng sc cho h.

i vi vtr 1: Thu vo . Trnh tging nh y ra nhng ngc li.

i vi vtr 2: Giti. Ging nh cc hthng khc tc l ngt dng lu cht vo xy lanh vcho xqua van iu khin p sut trc tip.

A.1.7

Trong hthng c 1 van 2 trng thi vtr thng ng nhvo lc l xo. Trng thi kia ckch hot bng in v c thng vi b.

Khi hthng hot ng bnh thng. van 2 trng thi trng thi ng.

Khi xy ra sc, nh p sut tng t ngt do van iu khin trc tip hng. Ta

skch hot trng thi th2 ca van. Tc l thng lu cht bm tbm vb

nhm hng chh hng do sc.

A.1.8

Xylanh trong mch di y bsai v khng ththc hin hnh trnh i ra.


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Sdng nhng van mu sa li mch xylanh c ththc hin c hnh trnh

i ra v lui vkhi tc ng vo van iu khin bng tay. Vi chc nng i ra, ban

u phi di chuyn vi tc cao ri chm li khi p sut gy ra bi ti t n

gi trthit lp.

A.1.9.iu c bit ca van mt chiu trong mch v mc ch ca n l g?


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Mc ch ca van mt chiu: khng cho du bht ngc ln v trnh va p iu c bit cacheck valve: N c t vtr thot du


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A.1.10. Ti sao mch thy lc di y khng hot ng? Hy xut cch sa i n gin sa li thit ksai st ny.

Do p sut iu khin hai nhnh ca van phn phi c cng gi trnn van sng im khngdch chuyn do mch khng hot ng.

Gii php: c thbhai ng iu khin thy lc ca van phn phi

A.1.11.

Cun dy A c kch trong sut hnh trnh i ra nhng ngt trong qu trnh li v. Chc nngny c tc ng nh thno trong mch?


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TL: mch c thci t p sut an ton 2 cp 300 bar hoc 100 bar. Khi khng kch cun A thmch c gii hn p sut 100 bar. Khi kch cun A th mch c ci t p sut an tonl 300 bar. Khi lui vp sut gii hn 100 bar t ta c hnh trnh i ra ti ln hn nhiu so

vi khi lui v, vic sdng hthng nh vy sm bo mt mt nng lng l nhnht.A.1.12. Sa li sai trong mch v xc nh ththot ng ca cc xylanha) Khi cun 1 c kch

b) Khi cun 2 c kchMc ch ca van C l g?


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TL: sai cm van tun t. Thay thbng cm van nh hnh sau:

a) Khi cun 1 c kch in: pittng ca xylanh A lui vtrc sau n cui hnh trnh thpittng ca xylanh B tin ra. Lc ny van tun tc tc dng.

b) Khi cun 2 c kch in: pittng ca xylanh B lui vtrc sau n cui hnh trnh thpittng ca xylanh A tin ra. Lc ny van tun tkhng c tc dng.

Tc dng ca van gim p C: do nhnh B c p sut hot ng nhhn p sut trong mchchnh nn phi dng van gim p ci t p sut hot ng cho nhnh B.


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A.1.13

Cho mch thy lc, hy gii thch hot ng ca mch v tnh ton p sut ci t. Bit ti 10tn, ng knh xylanh l 100mm

Trli:

Nu van phn phi c kch cun dy bn tri th ti c nng ln, kch cun bn phi th hti nhng dng du cn qua van tit lu nn tc hti ty thuc vo vic ci t van tit lu.

Cn khi trng thi gia th ti c gi cn bng nh hthng van mt chiu.Tnh ton psut ci t cho van xti ca hthng:

Din tch tc ng ca dng du khi nng ti:

=4 = 7,85.10

M


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p sut ci t van xti

>= 12,7 A.1.14

Cho mch thy lc nh hnh. Gii thch tnh nng ca mch trn. Ti sao dng van phn phi 2cun dy li tt hn dng van phn phi 1 cun dy thit lp trng thi cho van A? Kiuchuyn i trng thi ca van trn l g? Nu lu lng ring l 25 ml/vng v van chnh luc ci t gi tr thp tc l thuyt cc i ca ng c l bao nhiu

a) Chai cun dy 1 v 2 c kch

b) Cun dy 1 c kch

c) Cun dy 2 c kch

Trli:


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Tnh nng ca mch:lm thay i tc quay ca m t thy lc cng nh i chiu ca m tDng van phn phi 2 cun dy tt hn dung van phn phi 1 cun dy v: Gip van chuyntrng thi ddng hn

a) Chai cun dy c kch:40 rev/min

b) Cun dy 1 c kch: 840 rev/min

c) cun dy 2 c kch : 240 rev/min

A.1.15

Trong my kp thy lc thcng ny, mc ch ca 0.5 lt bnh cha l g? Nu van gii hn psut c ci t l 200bar, hy tnh p sut kh trong bnh cha.

Tc dng ca bnh tch p: Khi van iu khin trc tip chuyn vtr sang bn tri lu lngc bm qua van 1 chiu, t, mt phn du tch trong bnh tch p, mt phn y xilanh qua

phi. Nu xi lanh qua gii hn phi( kp cht) th p sut hthng stng ln, khi p sutvt qu p sut c t van gii hn p sut( 200bar) , du chchy qua van gii hn p


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sut. Van 1 chiu khng cho du chy ngc li, lc ny chnh lu lng do bnh tch p cungcp cho xi lanh sgicho xi lanh kp cht.

p sut kh trong bnh cha c ci t khong 90% p sut hthng

P= 0.9x200=180 bar.

A.1.16

Xi lanh trong mch c yu cu mt hnh trnh n tc cao nhng khong thi gian btthng. Hy nhn xt v sa i nu cn thit. Chc nng ca van iu khin lu lng l g?Tnh thit kp sut ci t cho van A.

Lc p sut hthng cha t ti p sut ci t ca van gii hn p sut th cht lng va i voxilanh va vo cha trong bnh tch. Khi p sut trong bnh tch t ti p sut ci t ca van Alm van A mcho 1 lng du nhtbm chy vbcha. Khi p sut trong hthng s


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gim , bnh tch hot ng lm tng p t ngt. Cng lc va c du vo xi lanh do bm cungcp va c du do bnh tch cung cp, lm cho xilanh co tc hnh trnh cao.Khi p sut hthng n v vt qu 200 bar, lm cho van gii hn p sut ( ci t 200bar) mlng du schy vbcha.

Van iu khin lu lng A cn ci t p sut bng i a vi p sut trong bnh tch v phinhhn gi tr200 bar.

A.1.17

Xylanh yu cu bt u mra vi tc cao di mt ti thp. V hon thnh gii hn chuyni tc thp v ti cao. Gii thch mch ny lm iu nh thno? Ti sao c 2 bnh tchp? Mc ch ca cng tc p sut P l g?

Lc u khi van phn phi i trng thi p sut hthng tng, cht lng c bm ln hthng, n np vo bnh tch p A v B. Khi p sut hthng vt qu 50 bar th bnh tchp A lm vic xlu lng vo xilanh lm n chuyn ng nhanh nhng ti nh. gncui hnh trnh, cn pittong schm vo switch in lm kn mch iu khin van phn

phi ca bnh tch p B, van phn phi i trng thi, bnh B xlu lng lm tng p

?

?

?

??

A

B

P

220 bar

200 bar


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trong hthng t ngt, trong khi lu lng ra xilanh cha kp lc ny pittong schuynng chm li v vi ti cao.

Mch c 2 bnh tch p:+ Bnh A: dng gicho p sut trong ng du chnh c n nh.+ Bnh B: dng kt hp vi cng tc p sut pha trn tc ng lm thay i tc chuyn ng ca piston.

Cng tc P c iu khin bi p sut trong hthng, dng iu khin ng hoc ngtmch xlu lng khi p sut hthng tng cao.

A.1.18

Hthng truyn ng thy lc khng o chiu ny c mt ti qun tnh cao v n c nhnthy chy qu tc cng nh cn c iu chnh gim tc . Hy nu ra mt sa i phhp.

V ti c qun tnh cao nn gim tc cn t thm mt van over-center ni tip ving c thy lc trong hthng trnh ng c lm vic qu ti, giti vtr cn

bng hoc phanh ng c mt cch tt.

A.1.19

Hnh 19

?

LOAD


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Trong mch truyn thy lc ny, mc ch ca phn tvan iu khin trc tip 3 cng, 3 vtrl

g? Ci t ph hp cho van gii hn lin quan l g?

Mc ch ca van iu khin trc tip 3 cng, 3 vtr l khi lu lng nhnh bn no ln hnth lm van i trng thi vpha bn kia v ng du kia thng qua van gii hn p sut vb

bm cung cp lu lng gm 1 bm 2 chiu lu lng thay i cung cp vo mch chnh v 1bm 1 chiu lu lng cnh cung cp lu lng nhhn hoc dng x lu lng khi psut vt qu 250 bar trong mch chnh

Ci t p sut cho cc van gi hn lin quan: Van

A.1.20

Truyn ng thulc ny chu stng p khi van iu khin trc tip chuyn trng thi. Hynu sa i gim bt stng p .

=> gim p lc khi van iu khin chuyn trng thi th ta nn thay bm 1 chu c lu lngcnh iu khin van iu khin trc tibng bm 1 chiu c lu lng thay i v gim lulng cung cp cho mch iu khin van iu khin trc tip

Hnh 20


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A.2 CIRCUITRY QUESTIONS

A.2.1.

Mt bm c lu lng ring l 1.7 cm3 trn vng quay vi tc 1500 vg/ph. Nu bm c hiu sut thtch l 87% v hiu sut tng l 76%. Hy tnh:

a) Lu lng thc ca bm (l/ph)

Ta c: = Vy = 0.87 1.7 101500 = 2.2185 /b) Cng sut vo khi p l 150 bar

Ta c: = = .= .. = 0.73 A.2.2.

Mt bm bnh rng c lu lng thc 15 l/ph, p sut l 200 bar v tc quay l 1430 vg/ph. Nu pvo l 6.8 kW v hiu sut ng c l 87%, hy tnhlu lng ring ca bm (cm3/vg)

Ta c: = = = 6.8 = .= 0.735M = = .. = 0.845Vy

=

=

.= 0.0124

=12.4 .A.2.3.

Mt hthng thy lc bm 32 l/ph lu cht ti p sut 260 bar. Bm c sdng l bm piston hngtrc c lu lng lu lng ring l 28 cm3/vg. Tc quay l 1430 vg/ph v hiu sut tng l 0.85, hiusut thtch l 0.9. Hy tnh:

a) Phn trm lu lng ring ln nht so vi lu lng thc

% = = 320 . 9 2 8 1 0 1430 100% = 88.8%

b) Cng sut cn cung cp cho bm

= = 600 = 16.3 A.2.4.


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My bm c lu lng ring 8.8 ml/vng, tc 2880 vng/pht. Hiu sut thtch v hiu sutc ln lt l 0.93 v 0.91.

TL:

a) Tnh lu lng thc ca bm

3 310 0.93 8.8 10 2880 23.6 /p v pQ p D n l ph

b) Tnh cng sut vo ca bm khi p sut l 350bar

0

23.6 35016.27

600 600 0.93 0.92 600

p p

in

v t

Q p Q pP kW

p p p

A.2.5.

Mt mch thy lc sdng 25 lt cht lng trong mt pht v c cung cp bi mt my bm c

lu lng ring l 12.5 cm3/vng v tc 2880 vng/pht. Bm c hiu sut thtch l 0.85 vhiu sut tng l 0.75, p sut ca hthng l 180bar.

TL:

a) Lu lng thc ca bm

3 310 0.85 12.5 10 2880 30.6 /p v pQ p D n l ph

b) Cng sut cp cho bm

0

30.6 180

12.24600 0.75 600

pin

Q pP kW

p

c) Cng sut do dng chy chy qua van an ton

30.6 25 180' 1.68

600P kW

A.2.6.Mt mch thy lc gm c 1 bm bnh rng lu lng cnh cung cp thy lc n 1 xylanh cng knh xylanh l 100 mm, ng knh trc 56 mm, chiu di hnh trnh l 400 mm. Lu lng cabm tng tng bc 1ml/vg bt u t5ml. Hiu sut thtch v hiu sut tng l 88% v 80%. Bm

c truyn ng trc tip tng c in vi tc c ti l 1430 vng/ph. Chn bm thch hp xylanh hon thnh 1 chu ktrong 12s.

TL:

Thtch cht lng bm vo xylanh trong lc piston y:


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2 2 3 33.14

0.1 0.4 3.14 104 4

dV A L D L m

Thtch cht lng bm vo xylanh trong lc piston li v:

2 2 2 2 3 33.14

0.1 0.056 0.4 2.1553 104 4

lV A a L D d L m

Lu lng m bm phi cung cp c ththa chu klm vic trong 12 giy:

3

4 33.14 2.1553 10

4.413 10 / 26476.5 /12 12

d lp

V VQ m s ml ph

M p v p pQ D n

26476.5

22 /0.88 1430

p

p

v p

QD ml ph

n

A.2.7.

Mt my bm c lu lng ring 25 ml/vng, tc 1440 vng/pht, cng sut my bm 10kW. Hiusut tng v hiu sut c ln lt la 85% v 90%. Tnh:

= 0.85

= 0.9

a) Lu lng thc ca bm

Qp = pv Dp10-3n =..2510-31440 = 34 (lit/pht)

b) p sut ln nht bm c thcung cp nu motor khng qu ti

= 600=> = 600 =10 0.85 60034 = 150( )A.2.8.

Bm c lu lng theo l thuyt l 35 lt/pht v hiu sut thtch l 90%, ng knh xylanh l110mm, ng knh trc thanh truyn l 65mm, hnh trnh pittng l 700mm. Tnh:

a) Vn tc i v vca xylanh

Q = VA


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3

1 2

35 10 0.93.31( / min)

0.11

4

QV m

A

21

3

2 2

35 10 0.9

5.09( / min)0.11 0.65

4

Q

V A m

b) Thi gian hon thnh 1 chu k

t= = 1 2

0.7 0.7

3.31 5.09

S S

V V = 0.349 = 20.9

A2.9.

Mt bm thulc c c dng chy c yu cu theo hthng thulc nh hnh hnh A2.1. Hthng

c p sut lm vic l 125 bar v p sut ln nht l 200 bar. p sut c tch trtrc l 90% p sutlm vic ln nht. Tnh:

a.Lu lng thc ca bm theo yu cu:

Theo hnh vta c : Q =.+.+. = 15(lit/pht) = 0,25 (lit/s)

b.Thtch ln nht c thcha ca bnh tch p:

Giai onthi gian(s) Bm(l/s) Yu cu(l/s)

Lu lng cungcp ca bnh tchp

Thtch bnh tch p(l)

0-5 0,25 0,75 0,5 0,5.5=2,5

5-10 0,25 0 0,25 0,25.5=1,25

10-15 0,25 0,5 0,25 0,25.5=1,25

15-25 0,25 0 0,25 0,25.10=2,5

25-30 0,25 0,25 0 0

Nh vy thtch bnh tch p c thtch ti a l 2,5 lit.


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c.Thtch bnh tch p ginh qu trnh xv np u u ng nhit:

P1=90%.Plv= 90%.125 = 112,5 bar => Pltd= 113,5 bar

P2 = Pmax= 200 bar => P2td= 201 bar

P3 = Plv = 125 => P3td= 126 barA.2.10.

Mt xylanh p c ng knh 140 mm v ng knh thanh y l 100 mm c vn tc ban u l 5 m/phtv vn tc p cui cng l 0,5 m/pht. p sut hlc di chuyn l 40 bar v p sut nn cui cng l 350bar. Githit 2 bm c hiu sut thtch v hiu sut tng ln lt l 0,95 v 0,85. Xc nh:a) Lu lng n xylanh lc di chuyn v p cui cng.b) Lu lng thch hp cho mi bm.c) Lu lng ring ca mi bm nu vn tc l 1720 vng/pht.d) Cng sut yu cu ca motor bm trong lc di chuyn v nn.e) Vn tc li vnu p sut yu cu ti a l 25 bar.

Giia) Lu lng n xylanh lc di chuyn:

Q1= V1A1= 501.423.14/4 = 76.9 (l/min)Lu lng n xylanh lc p:Q2= V2A1= 51.423.14/4 = 7.69 (l/min)

b) Lu lng thch hp cho bm 1:q1= Q1-Q2 = 76.9-7.69= 69.21 (l/min)Lu lng thch hp cho bm 2:q2= Q2= 7.69 (l/min)

c) Lu lng ring ca bm 1:

Dp =3

10

p

v

Q

p n

= 69.21/(0.95172010-3) = 42.36 (cm3/rev)

Lu lng ring n bm 2:

Dp =3

10

p

v

Q

p n

= 7.69/ (0.95172010-3) =4.7 (cm3/rev)

d) Cng sut yu cu ca motor bm trong lc di chuyn:

Pin=0 600

pQ p

p

= 7740/(0.85600) =6.04 KW

Cng sut yu cu ca motor bm trong lc nn:Pin=

0 600

pQ p

p

= 7.7350/(0.85600) =5.3 KW

e) Tit dinA=3.14.(D2-d2)/4=3.14.(1402-1002)/4=7536mm2

Vn tc li v:V= Q/A = 77/ (7536.10-3)= 10.2 ( m/min)


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A.2.11.

Mt bm c quay 1440v/p c lu lng ring l 12.5ml/vng v hiu sut thtch 87% c dng cp lu cht cho mt chu trnh vi hai xilanh. Nu kch thc xilanh l: ng knh xilanh 63mm x cnxilanh 35mm x hnh trnh 250mm v 80mm x 55mm x 150mm, tm thi gian nhnht ca chu kcho chai xilanh khi ko di v rt ngn hon ton.Gii:Lu lng thc tca bm: Qp = p.Dp.np= 0,87.12,5.10-3.1440=15,66(l/p)

Xy lanh 1:

Tit din piston: A1= =., =3,117.10-3 m2Din tch vnh khn: a1 =

() = (,,) =2,155.10-3 m2

Xy lanh 2:

Tit din pitton: A2= =., =5,027.10-3 m2Din tch vnh khn: a2 =

()

=(,,)

=2,651.10-3 m2

Thi gian xilanh ko di v rt ngn

T1=. . = ,.,.. ,.... = 0.084=5.04s

T2= . = ,.... ,.... = 0.074 = 4.44

Vy thi gian chu k ngn nht c hai xy lanh m rng v thu li y : Tmin=T1 + T2= 5,04+4.44=9,48s

A.2.12.

Mt xilanh thy lc c yu cu tc ng mt lc y dc trc ti thiu l 25 tn v lc ko vti thiul 15 tn. Xc nh kch thc tiu chun hmt thch hp ca xilanh nu p sut cc i trong hthngl 200 bar. Githit p lc hng trc ng gp 9 ln p lc hng trc tnh. Cn p sut bao nhiu trongxilanh to lc cn thit nu nh nh hng ca i p khng ng k.Bi gii:

Ta c: = 0.9. . Suy ra: = ,=..,...= 131.7Chn theo tiu chun ta c: = 140, = 90

Xc nh p sut cn thit cho mch:

= ,= 177

= ,()= 181A.2.13.

Mt xi lanh thy lc biu din nh hnh A.22, gia tc cho 1 ti 50 tn trong 50mm theo phngngang ttrng thi nghn khi t vn tc 10m/pht. Hsma st gia ti v mt dn hng l

=0.1, githuyt i p bng 0. Xc nh:


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a) Kch thc tiu chun ca xilanh nu p sut cho php cc i ca xilanh la 180 barb) Lu lng yu cu y piston vi vn tc 3m/pht.

Bi lm

a) Ta c:

2

2

2

0 2* *

0.28( / )2*

v a s

va m s

s

Lc qun tnh Fqt=ma= 50000x0.28=13.89kNLc ma st: Fms= .N= 0.1x50000x9.81=49.05kN.

Suy ra p sut trong xi lanh l:3

5

2

(13.89 49.05).10180.10

*4

qt msF FF

PDA A

Suy ra: D = 67mmTheo tiu chun chn D=80mm va d=45mm

b) Khi piston tin, c lu lng l

Q1=V.A=20.8

3.10 . 15.08( / )4

l p

Khi piston lui, c lu lng l

Q2=V.a=2 20.8 0.45

3.10 . 10.31( / )4

l p

Vy lu lng yu cu l: 15.08+10.31=25.39( l/p)

A.2.14Xilanh c ng knh l 50mm v ng knh piston l 32mm, c dng nng hti 3 tn.S

mch nh hnh A.23. Lu lng thc c cabm l 8 l/p.Van gii hn p sut c ci t 180 barv lu lng ra xilanh qua van tit lu l 4 l/p.Tnh:

a) Tc ib) p sut c P1khi vn tc xilanh n nh.c) p sut ghi P1khi xilanh hot ng khng tid) t van tit lu c hp lu lng bao nhiu nu tc c bng tc m

Bi lm

a) Tc y piston:2 2

43.45( / )

0.05 0.032.

4

qv m p

A a

b) Trng lng ca ti : P= 3000x9.81=29430N

Lc tc n vo y piston l: F1=180x105xA=180x105x2.0,05

4

=35342N

Lc tc ng vo din tch vnh khan: F2= F1P=5912N


28/203

Trang 28

Suy ra P2= 2 2 25912

50,990.05 0.032

4

Fbar

A a

c) Nu lm vic khng ti th:2

F =1

F

p sut P= 1 2 235342 304,870.05 0.0324

Fbar

A a

d) Lu lng ci t cho van lu lng tit lu vn tc i bng vn tc v

v 3.45( / )m p =q Q

A a A

Suy ra Q= 6,77( l/p)

A.2.15

Mt xilanh yu cu nng mt ti 6000kg ln cao 4m.p sut lm vic khng c qu 100 bar ngvo xilanh.Bqua nh hng ca p.Xilanh c nh vtheo mt bch, ti y c dn hng nivi trc biston xoay c.Xc nh kch thc cho cn piston va xilanh.

Bi lm

Ti gii hn:2

2

EJK

L

Vi K= 6000kg

E= 6 22,1.10 ( / )kg cm

4

J .64

d ( d : ng knh piston)

L=4m

Suy ra: d=5,5cm

Theo tieu chun ta chn d=56mm

nng 6000kg ln cn p sut l 100bar, nn :

2

56000.9,81 58,86100.10

F FP A cmA P

58,86.486,57D mm

Theo tieu chun ta chn D= 100mm


29/203

Trang 29

Vy D=100mm,d=56mm

A2.16.Mt xylanh thy lc c ng knh xylanh l 125mm v ng knh thanh truyn l 90mm csdng di chuyn mt ti nng 20 tn vi tc nhanh nht l 5m/pht. ti trng phi c a nnghti im cui ca xylanh trong sau mt hnh trnh di 50mm. p sut hthng c tc ng ln

Xylanh trong khong thi gia tr. Xc nh:a) p sut c yu cu nng ti? giskhng c p lc trli pha bn hng ca

xylanh.b) Tc dng chy cht lng nng cao ti vi tc 5m/phtc) p sut trung bnh ca hi m trong xylanh trong khong thi gian tr( lc gim tc

xylanh)Gii

a) p sut yu cu nng ti l

= 2 0 1 0

1254 10= 163 A.2.17.trung tm trc ca mt my tin c iu khin bi mt cng sut khng i c chuynng bi hthng thy lc. ng c thy lc lu lng bin i khng o chiu c phm vi bin itc t300 n 2500 vng/pht. Cng sut ti a c yu cu ti u ra ca ng c thy lc l6KW. p sut ti a cho php ca bm vn chuyn l 125 bar v p sut tn tht gia bm v ng cthy lc l 5 bar. Hiu sut momen v hiu sut thtch ca cbm v ng c l 0,85. Gisqu trnhtruyn vng lp h. Xc nh :

a) Lu lng l thuyt ca ng c v lu lng thc ca bm c cung cp?

b) Cng sut u vo ca bm ?

Giia) p sut tc ng ln ng c l :

pdc125-5=120 barLu lng ti u ra ca ng c l:

=600 =600 6120 = 30 /Lu lng cp vo ng c( lu lng l thuyt) l

= ,= 41,52 / Lu lng thc QBmthucm bm cn cung cp cho ng c l 41,52 lt/phtGi q l lu lng 1 vng quay ca bm, n l tc quay ca bm (vng/pht)

=


30/203

Trang 30

= ,(),= 0,023 0,191 = 23 191 b) Cng sut u vo ca bm l

= 600=

125 41,520,85 600= 11,97

A.2.18)Mt xy lanh thy lc di chuyn 1 ti i ngang 1khong cch l 3m. Ti c gn cnh trn

piston v c dn hng. Lc y tc dng ln xy lanh khi c ti l 1,6 tn, lc ko vkhi khngti l 0,7 tn. Hiu sut p lc ng bng 0,9 ln p lc tnh.

Nu p sut ti a ca hthng l 150bar.Hy xc nh li kch thc xylanh v tnh ton li p sut vn hnh ca hthng.Gii.

- Xc nh ng knh trc xylanh:

Ti gii hn: 1.6 1600K t Kg . Hsan ton:S = 3.5

Modun n hi: 6 22,1.10 ( / )E kg cm

Moment qun tnh ca ti: 2 / 64J d

Ta c :

42

2

42 3 6

E64 1600 3.5 15064 3.336( )

2.1 10

d

K S d cmL

Tiu chun chn d=36mm

- Xc nh ng knh xylanh:

Lc F = 1600*9.81=15696(N); p sut ln nht: p=150 bar; Hiu sut xy lanh: 0.9

Din tch piston c tnh nh sau: / .A F p

5 3 2 2(1600 9.81) / (150 10 0.9) 1.16 10 11.6A m cm

Vy4

3.8A

D cm

. Chn D=50mm

Vy xylanh c chn c kch thc : D=50mm, d=36mm

Xc nh li p sut lm vic gii hn.

2

1600 490.6( ar)

0.05 0.9

Fp b

A

A.2.19


31/203

Trang 31

Cho 1 my ct dng xylanh thy lc c phn hi, khi chu trnh i ra ca xy lanh l piston inhanh vi tc 10m/min ng vi qung ng 1m vi lc ct lc l 2500kg. Khi cng tcp sut chuyn trng thi th tc di chuyn ca piston khi l 0,25m/min ng vi qungng l 0,5m vi lc ct khi l 10 tn. p sut ln nht ca xy lanh l 200bar.a. chn kch thc xy lanh theo tiu chun .

b. Tnh ton lu lng bm.GiiChn kch thc xylanh tiu chun.

ng knh cn pittong : Khi pittong ang thc hin vic ct vi lc ct l 10 tn.

Chiu di quy c:1 0.5

0.75 752 2

lL m cm

Ti trng gii hn: K=10 tn. hsan ton S=3-4, chn S=3

Modun n hi: 6 22,1.10 ( / )E kg cm

Moment qun tnh ca cn:4

64

dJ

Ta c:

42

2

42 3 6

E64 10000 3 7564 3.588( )

2.1 10

d

K S d cmL

Theo tiu chun chn d=45mm

- ng knh xilanh l :

Khi piton i thc hin vic ct l lc p sut t cc i v hthng lc ny l nguy him nht nntnh ng xylanh sao cho m bo cho trng hp ny.

Lc F = 10000*9.81=98100(N); p sut ln nht: p=200 bar

Nn din tch yu cu ca xylanh l: 3 25

981005 10

200 10

FA m

P

Vy3

4 5 1079D mm

. Chn D=80mm v d= 45mm

Tnh lu lng bm lc ra nhanh:

20.08

0.25 1.256 /4Q vA l ph

Lc ra chm:2

0.082.5 12.56 /

4Q vA l ph

A.2.20)ng knh piston = 125mm


32/203

Trang 32

ng knh cn piston = 80mm

Hnh trnh = 350mm

Thi gian 1 chu k= 15s

Ti trng y = 20 tn

Ti trng thu v= 10 tn

Din tch ton phn:2

2 2125/ 4 12271.8

4D mm

Din tch bn c cn y:2 2

2 2 2(125 80 )( d ) / 4 7245.34

D mm

a) p sut l thuyt ca hthng khi xylanh y ra:3

6

20 10 9.81160 ar

12271.8 10EP b

b) p sut l thuyt ca hthng khi xylanh thu vo:3

6

10 10 9.81135.4 ar

7245.3 10R

P b

c) Lu lng cn cung cp bm hot ng:6 3

1 2( ) L (12271.8 7245.3) 10 350 10 60 1000 27.3 /15 15

A AVQ l ph

t

d) Lu lng ring ca bm27.3 1000 0.9 1440 21( )p p v p pQ D n D D ml

e) Nng lng ln nht cn cung cp cho bm:

27.234 1609.5(KW)

600 0.85 0.9 600 0.85 0.9

p pQ P

P

A.2.21

Trong hthng thy lc vng h, motor c lu lung ring l 0.5l/rev, tc quay l 65rev/min. Bmthy lc cung cp lu lung cho motor c tc quay l 1440rev/min. Hiu sut c v HS tng ca bmv motor ln lut l 95%v 85%. Tnh:a) Tnh lu lung ring ca bmb) p sut ca motor nu momen u ra ca motor l 1000Nmc) Cng sut u vo ca Bm nu chnh lch p sut trn dung ng gia Bmv motor l 5barGii:

a) Lu lung vo ca motor0, 5 65

36,310,895

p

v

D nQ

(l/ph)

Lu lung l thuyt ca bm:


33/203

Trang 33

36,3140,57

0,895o

v

QQ

(l/ph)

Lu lung ring ca bm:40,57

0,0282 28, 21440

o

p

Q

D n

(ml/vng)

b) Momen cn thit cung cp cho ng c:

10001052,6

0,95m

t

TT

(Nm)

p sut ng c l:

3

2 2 1052, 6132

0,5 10

m

m

m

Tp

D

(bar)

c/ Cng sut thy lc u ra ca bm

36, 344(132 5)8,3

600 600p

Q PP

kW

Cng sut u vo ca bm:

8,39,76

0,85

p

o

PP

kW

A2.22Mt ng c thy lc c yu cu cung cp moment l 100Nm tc ti a 600 vng/pht. p sutrt khi qua ng c l 150bar. Hiu sut moment v hiu sut thtch u l 0.9. Xc nh:

a/ Lu lng ring ph hpb/ Lu lng cn cp vo ng c

Gii

a/ Lu lng ring ca ng c

5

2 100 2

4.654150 10 0.9

m

m

m m t

T

D P

10-5

m3

/rev = 46.54 cm3

/rev

b/ Lu lng thc cung cp cho motor:

Q= m m

n v

D n

=46.54 600

310260.9

cm3/min = 31.026 l/min

A.2.23


34/203

Trang 34

Mt my tin c truyn ng trc tip bng ng c thy lc. My tin c dng gia cng 1 trcc ng knh ti a l 60mm. Lc ct ti a trn ao tin l 2KN, tc quay ti a ca u my tin l700vg/ph. p sut ci t van gii hn p sut l 200bar, tng chnh lch p sut gia van gii hn psut v motor l 10bar, p sut ngc ti motor la 5bar. Hiu sut tng v hiu sut thtch ln lt l0.85 v 0.9. Xc nh:

a/ Lu lng ring ti thiu tnh theo cm3/rad

b/ Lu lng n motor ti tc quay ti a

c/ Cng sut u vo ti a ca motor

Gii

a/ Momen ln nht do lc ct gy ra

33 60.10

2.10 60( )2 2

m

dT F Nm

Hiu sut mment

0,850,94

0,9

m om t

m v

p sut lm vic ca ng c

200 10 5 185mp bar

Lu lng ring ti thiu cn cp cho ng c

2

5 4

60 103.45

0.94 185 10 10m

m t m

TD

P

cm3/rad

b/ Lu lng qua ng c

3.45 2700

100016.86

0.9

mm

m v

D nQ

l/min

c/ Cng sut ln nht cung cp cho ng c

32 16,86.10 1

200 10 6,6(kW)60 0,85

m m

m o

p QP

A.2.24


35/203

Trang 35

Lu lng ring ca mt m t thy lc l 475cm3/vng c dng quay trc tip tang trng bng tic ng knh l 0,7m. p sut tc ng trn m t l 140bar v lu lng thc t vo m t 48 l/ph.Hiu sut tng v hiu sut c ca m t tng ng l 0,9 v 0,94. Xc nh:

a) Moment ca tang trng bng ti

b)

Cng sut (kW) cung cp cho tang trng bng tic) Tc tuyn tnh ca vnh ai bng ti

Gii:

a) Moment ca tang trng bng ti:6 5. 475.10 .140.10

. 0,94. 995( )2 2

m mt

D PT Nm

b) Cng sut cung cp cho bng ti:3

5

0

48.10. . .140.10 .0,9 10, 08( )

60.1000m mP Q P kW

c) Tc tuyn tnh cu vnh ai bng ti:

3

6

0,948.10 .

. . 0,9496,75(vg/ ph)

475.10

m m m vm m

v m

D n QQ n

D

0, 72 . . 2 .96, 75. 213( / )

2t m t

v n R m ph

A.2.25

Mt chic xe nng 2 tn chy ln 1 con dc c dc l 1 trong 10 (cao l 1 trn chiu nghing ca

dc l 10 o dc theo sn) tc 20km/h. Hslc ma st ln l 0,1. Chic xe c y bng mt hthng thy lc vi 2 bm c lu lng ring cnh ca ng c c lp ti cc bnh xe c ng knh850mm. Hiu sut thtch v hiu sut c ca m t u l 0.95. p sut ln nht ngang qua ng c l250 bar. Xc nh:

a) Lu lng ring ng c yu cub) Lu lng ln nht ca dng cht lng c cp tbm

Gii:

a) GisT l momen m mi m t thy lc sinh ra, khi tng momen gy ra tc dng lm

quay cc bnh xe l 2T.Lc do momen sinh ra l: 2F = 2T/R (Tnh cho 2 bnh xe).p dng phng trnh cn bng lc cho chic xe, ta c:

mg.sin+ mgf.cos= 2F

vi: sin=cos= 1 =


36/203

Trang 36

F =mg.si+mgf.cs =

.,. +.,.,. = 1957(N)T = F.R = 1957.. = 832 (Nm)

Mt khc: T = tDmPm

M theo phng trnh p dng cho m t thy lc th:

t=.TDPDm=

.TP=...,= 22.10-5(m3/vg) = 220 (ml/vg)

b) Svng quay ca bnh xe cng chnh l svng quay ca m t do :

nm=.vR=

.., = 124,8 (vg/ph)v=

D.Q Qm= D. = ..,, = 289.10-4(m3/ph) = 28,9 (l/pht)Do tng lu lng ln nht do bm cp cho m t l

Qp= 2.28,9 = 57,8 (l/ph)

A.2.26

Mt thng trn thc n c khi lng 50 kg v bn knh quay trn l 0,4 m c iu khin bi mtng c thy lc. Thng ny c thtng tc ttrng thi nghln tc ti a l 120 vng / pht trong0,5 giy. Ap sut ti a trn ng c l 200 bar. Ly hiu sut thtch v hiu sut m-men xon tngng l 0,96 v 0,95. Xc nh:

a. Lu lng ring ph hp ca ng c.

b. Lu lng thc tyu cu ca bm iu khin ng c tc ln nht nu hiusut bm c xem nh ging hiu sut ca ng c.

Bi gii:

a. Lc tc ng vo thng: . 50.10 500F m g (N).

M-men ti: . 500.0, 4 200m

T F r (N.m).

Ta c:. .

2

m t m m

m

D PT

Lu lng ring ca ng c: 55

2 . 2 .2006,65.10

. 0, 95.200.10

m

m

m t m

TD

P

3

m

rev

= 66,5 mlrev

b. Lu lng thc t:

c.5

3. 6,65.10 .1208,3.10

0,96

m mm

m v

D nQ

3

mph


37/203

Trang 37

A.2.27.

Mt h thng vn ti bng thy lc dng ko mt xe goong ( ging tu ha nhng khng c u ko )trong khu m . Trng lng xe ti l 5 tn . dc ca on ng l 1:10 . Dy ti c qun theo lptrong 1 ci ng ( ging nh 1 ci trng ), nhng n gin cho vic tnh ton gi s ng knh ng

lun l 1 m . Qung ng t h thng ti cui ng ray l 1500m .Dy ti c khi lng ring l4kg/m .ng dy c khi lng 1,5 tn khi khng c dy ti qun vo v c bn knh qun tnh l 0,5m.

Xe c th c ko t trng thi ng yn di chn dc ti vn tc ln nht 5km/h trong 10s. p sutcc i trong moto ti l 300 bar . Hiu sut c v th tch u l 95% .

a. Tnh dung tch trn 1 vng ca motob. Xe c th dng t ngt gia ng , gia tc ln nht ca xe lc khi ng li l?

Tnh c momen qun tnh ca ng dyc. Nu bm cung cp cho moto ti c hiu sut th tch l 92% v hiu sut ton phn l

87%. Tnh lu lng ring cn thit ca bm khi c ni vi ng c c tc l 2200

v/pd. Tnh cng sut ca ng c cung cp cho bm , bit chnh p do mt nng gia bm

v ti l 30bar.

Bi lm

- Gia tc ca xe lc khi ng :25 5

18.10 36

va

t

(m/s2)

- Phng trnh ng lc hc i vi xe :

.F m a

5 9,81

(5000 1500 1500.4).( ) 13998, 6136 10

F (N)

T suy ra : . 13998,61.0,5 6999,3mT F R (N.m)

M.

2

m m t

m

D PT

5

2 . 2 .6999, 31,54

. 300.10 .0, 95

m

m

m t

TD

P

(l/rev)

b. Momen qun tnh ca ng dy : I = mR2= 1500.0,52= 375 kg.m2Phng trnh ng hc lc xe khi ng li :

M(a+g/10) + I.a/R2= FSuy ra a = 0,425 m/s2Ta c lu lng ring ca bm l : .Dp v p pQ pn n

M lu lng ring ca motor l :


38/203

Trang 38

.m mm

v

D nQ

mn

M lu lng do bm cung cp b mt mt do hiu sut ca bm mi bng lu lng camotor nn ta c :

0

.

. . . m mv p pv

D npn pn D n

mn

Suy ra0

. 1,54.26,530,0254

. . . 0, 95.0, 87.0, 92.2200

m mp

v v p

D nD

mn pn n pn (l/v)

Ta c mt mt t bm qua motor bng 30 bar nn p sut ln nht bm l 330 bar.Do cng sut cn thit cung cp cho bm l :

3

0

. . 25, 4.10 .2200.0, 92.330. 32, 5

.600 0,87.600

p p v

p

D n pnP P

pn

kw

A.2.28.Mt h thng thy lc s dng 1 bm c lu lng ring thay i c tc dng lm quay mt motor clu lng ring c nh .c trnh by s A24

a. Cho bit :( ) 1

( )( ) 1

pm

m m

Ks

Y s d s

Trong : s l bin laplace , Kpl hng s lu lng bm cho tc m , dml lu

lng ring ca motor v l hng s thi gian .( Xem lu cht l khng nn c ) . Hs r r chung l hng s .

b. Xc nh t nguyn tc u tin ca h thng mt n v nhp vo sau thi gian l nhnhau l 3 . Hy tnh gi tr ca . Trong h thng c cc thng s gi tr ring : Ti trng tnh I = 100Nms2H s r r : = 12.10-3l/pht/barLu lng ca motor , dm = 25 ml/radHng s lu lng bm : Kp= 5.10-3m2.s-1

c. Hy v th cc cho tn s phn ng ca h thng hm vo sint cho gi tr ca gia 1 v 10 rad/s

Bi lm

a. R r ca bm = .pp

P

Lu lng thc t bm = Kp.Y - .pp

P = q

R r motor = .m mP


39/203

Trang 39

Lu lng thc s dng trong motor = q - .m mP = m .dm

Do :m .dm = Kp.Y - .

ppP - .m mP

B qua s gim p trong h thng : P = Pm= Pp

V nu

l h s r r ca bm v motor th m

.dm = Kp.Y -.P

Nu momen ra ca motor l Tmth

Tm= dm.P = I. md

dt

Trong I l ti trng tnh ca u ra .Thay P vo ta c :

I. m

d

dt = dm.(Kp.Y -

m .dm )/ .

m

S dng bin i laplace :

Is m

(s) =

.( . ( ) ( ).d )m P m m

d K Y s s

m (s) .(Is +

2

md

) = .Y(s)m p

dK

2(s) (d . / ) / / (1/ (1 ))( )

pmm p m

m

KK Is d s

Y s d

(*)

b. Ta c :3

2 5 6 2

12.10 .1000,32

60.10 .1000.(25.10 )m

d

T (*) suy ra3

/ 3

6

5.10(1 ) (1 ) 190

25.10

p t

m

m

Ke e

d

(rad/s)

c. Ta c Y = sint suy ra Y(s) =2 2

s

M( ) = [(t2+ 2)1/2]-1.0,2.10-3= [(1+0,322)1/2]-1.0,2.10-3 = tan-1(0,32)


40/203

Trang 40

A.2.29.

Xe ti duc dn ng bng hthng thy lc bng lu lung. Cc thng sv yu cu ca xe nh sau: Tng trng lung ca xe l 7 tn Sbnh dn ng l 2 Lc cn l 100kg/tn Bn knh bnh xe l 0,4 m

Tc ln nht ca xe l 15km/h. Thi gian t duc vn tc cc i l 5 giy. Xe c khnng leo lndc ti a l 1:4, vn tc ti a v gia tc ti a l khng quan trng.Bm c thtch ring thay i vi lu lung cung cp ti a theo l thuyt bt u t20 l/min v tng lnmi ln l 10 l/min khi ng c tng tc. Mt thy lc c dung tch khng i vi dung tch lc khng l 0,1 l/vng v tng thm 0,05 l/vng.p sut vn hnh hthng ti a l 300 bar. Bm v mto u c hao ph thtch l 0,95; hao ph momentca mto l 0,94; hao ph tng ca Bm l 0,9.a) Bqua mt mt trn dung ng. Chn ng c v Bm ph hp, xc nh cng sut u vo ca bmb) Nu 70% khi lung ca xe tp trung bnh sau, 30% khi lung bnh truc v hsko gia bnhxe v mt ng l 0,95 xc nh dc ln nht xe c thleo.

Giia)

Svng quay ca motor:

1560 99, 47

3, 6 (2 0, 4)m

n

vng/pht


41/203

Trang 41

Gia tc ca motor:

15 5

3, 6 5 6a

m/s2

Lc ma st.100 7 9, 81 6867

msF (N)

Lc qun tnh:

57000 5833, 3

6qt

F (N)

Lc cn thit ln dc khi dc ln nht l:

70009,81 17167,5

4F (N)

Tng lc motor cn pht ng khi tc quay ln nht l:

6867 5833,3 12700,3o

F (N)

Tng lc motor cn pht ng khi leo dc ln nht l:

6867 17167,5 24034,5oF (N)

Momen mi motor cn c l:

T=24034,5.0,4/2 = 4806,9 Nm

Lu lng ring ca motor l:

3

5

4806,9 2.210 1, 07

0, 94 0, 94 300 10m

TD

P

(l/vng)

Nn ta chn lu lng ring ca bm l 1,1 l/vng.

Lu lng ca motor:

99, 47.1,12 2 230, 36

0,95 0, 95

m m

m

n DQ

(l/ph)

Lu lng ca bm:

242,10,95

mp

QQ (l/ph)

p sut ca hthng cn l: 300/2 = 150 bar.


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Cng sut vo ca bm:

67,25600.0,9

p p

p

Q PN kW

b) Lc cn xe c thleo dc:7000.0,7.9,81.0, 95 45665,55oF (N)


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BI TP THY LC FLUID POWER ENGINEERING

CHNG 1 SCH FLUID POWER ENGINEERING

F.1.1

Nu r chc nng ca hthng in

Hthng in c dng kim sot hthng iu khinTn hiu in i voc xl v chuyn i a ra hthng iu khinhthngng iu khin hoc hthng iu khin quay- Hthng in c dng trong:

+ Hthng in c+ Hthng in kh nn+ Hthng in thy ng lc+ Hthng in thy tnh

F.1.2

Tho lun ngn gn cc nguyn tc hot ng ca cc hthng in khc nhau v a ra ccchng trnh hot ng cn thit

- Cc hthng nng lng c hc sdng cc yu tc hc truyn ti v kim saots nnglng c hc

VD: Hthng in kh nn l hthng in sdng kh nn nh l ngun hot ng. Nguyntc hot ng tng tvi hthng in gip chuyn i nng lng in thnh nng lng plc ca kh nn.

Qu trnh chun bkhng kh l cn thit sdng. vic chun bkhng kh bao gm vic lc,sy kh v bi trn bng du. khng kh nn c lu trtrong cc bnh cha. Nng lngca kh nn li c iu khin bi cc van p lc, van lu lng, van iu khin hng. sau mi c chuyn thnh nng lng c hc bng xy lanh kh nn.

- Hthng thy ng lc in truyn ti l hthng dng nng lng in truyn ti nnglng bng cch tng chyu ng nng ca cht lng. ni chung, cc hthng bao gmmt my bm thy lc, mt tua bin.Cc ng dng ca hthng in c gii hn trong chuyn ng quay

C hai loi chnh :+ Khp ni thy lc : vc bn l da trn cht lng ly hp, n bao gm mt my bm iukhin bi cc trc u vo v mt tuabin, cng vi trc u ra. Khi cnh qut my bmquay, du chy vo tuabin tc cao. Sau du tc ng ln cc cnh tua bin.+ Hthng chuyn i m men xon. Kh cn kim sot m men xon u ra v phi datrn ts truyn.


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- Hthng in thy tnh c truyn bng cc tng chyu l nng lng p lc cht lng. cchthng c sdng rng ri trong thit bdi ng, my bay, kim sot cc con tu

F.1.3

Vmch cu mt hthng thy lc n gin cc biu tng chun v gii thch cc yu tcbn ca n mt cch tm lc

- Mch sdng mt van iu khin hng. du c m t bm ln i vo pha bn tri caxylanh tc ng kp, y piston di chuyn sang phi v p vt. khi p sut dng chy thngc lc ci t l xo ca van iu khin hng lm van chuyn trng thi lm piston chyntrng tha sang chiu ngc li, nhng du khng lp tc trvbc, do van an ton, nhvy hthng c tc dng givt ng yn cho n khi no p sut vt qu p sut ci t van an ton th lc ny chuyn trng thi cho du vb.

F.1.4


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Nu u nhc im ca cc hthng thy lc:

u im:- Tlsc mnh thy lc so vi trng lng- Tbi trn

- Khng c hin tng bo ha trong cc hthng thy lc so vi bo ha trong cc myin. ti a m men xon ca ng c in l tlthun vi in hin ti, nhng n bhnchbi bo ha t.

- Tlm men xon trn khi lng ln, gip tng tc v p ng nhanh ca hthng ngc thy lc.

- cng cao ca xy lanh thy lc cho php dng ti ti bt lvtr trung gian- n gin bo vchng qu ti- Khnng lu ginng lng trong hthng c quy thy lc- Tnh linh hot ca hthng truyn ng so vi cc hthng c kh

- Sn c chai chuyn ng quay v chuyn ng thng- An ton lin quan n nguy c dnHthng thy lc c nhng nhc im sau:- in thy lc khng c sn , khng ging nh inDo o my pht in thy lc phc yu cu- Chi ph sn xut cao do cc yu cu nhthng qua quy trnh cn t chnh xc cao- Qun tnh cao cu ng dy truyn ti lm nhanh hng hthng theo thi gian- Gii hn nhit hot ng ti a v ti thiu- Chy nguy him khi dng cc loi du khong sn

- Vn lc du

F.1.5

Vmch mt vthng thy lc n gin, bao gm mt my bm hng van iu khin , xylanh thy lc, van xv o p sut. nu chc nng ca nhng thnh phn v tho lun chi tit vhthng truyn ti in trong mch.

- Cng tc p sut skhi ng lm n bt sng khi lc du c du hiu bngt- Khi van iu khin iu khin hng trng thi gia th du c cp vo ng P1 xtheo

ng T- Khi van iu khin trng thi trn cng, motor squay, van iu khin p c ci t mc p sut ln hn mc p sut lm motor quay mt cht, khi motor bngt, p sut dutng u ln thm v thot ra khi bng ng van iu khin p, mch c bo v.- Khi van iu khin c kch trng thi di cng , n cng lm cho motor quay nhngtheo chiu ngc li . mch vn c bo vtheo nguyn tc trn.


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F.1.6

Din tch piston :2

3 2

1

.0,17, 857.10 ( )

4A m

Din tch vnh khn:2 2

3 21

.(0,1 0,07 ) 4.10 ( )4

A m

Tnh ton lc ti, F: (do u ra ca xi lanh thng vibnn coi p2=0)

3 5

1 2 1 1 2 2 7,857.10 .200.10 0 157,14(KN)F F F A p A p

Tc dng chy trli, QT:1 2

20, 37( / min)p

T T

Q QQ lA A

Tc piston, v:3

1 2 3

2

20,37.100,0849( / )

7,857.10

TQ

v v m sA

Nng lng in bm thy lc:1 1

40.200. 13.33( )

600Q p KW

Nng lng xy lanh: F.v= 0.0849*157,14*10^3=13.33(KW)

F.1.7

Din tch piston :2

3 2

1

.0,17,857.10 ( )

4A m

Din tch vnh khn:2 2

3 2

2

.(0,1 0,07 )4.10 ( )

4A m

V hthng 2 u ca xy lanh thng nhau nn: p1=p2

1 2 1 1 2 1 1 1 2

25

(A )

0.07200 10 77( )

4

F F F A p A p p A

KN

1 1 2

Qin Qout Qp Qout

v A A A

Xt1 2

Qout Qp Qout

A A

ta c:

2 2 2

40 4041.63(l/ ph)

1 2 0.1 0.1 0.07

Qout Qout Qout Qout Qout

A A

Qin=Qout+Qp=41.63+40=81.63


47/203

Trang 47

3

3

41.63 100.17345( / )

2 60 4 10

Qoutv m s

A

Nh vy ta tnh cNng lng ca cn y ca piston: F*v=77*10^3*0.17345=13.33(KW)

nng lng in bm thy lc:1

40.200. 13.33( )

600pQ p KW

Nhn xt: so vi F. hthy lc ca F. 1.6 th F. 1.7 hthng u cung cp cho cn y piston 1nng lng bng nhau. Tuy nhin h1.7 th cn y xy lanh c lc nhhn (77KN 0.0849 m/s).V chai hthng th nng lng bm lun bng nng lng ca cn y piston l 13.33KW

F.1.85 2.A . 2 / 0.611. . 2.139.10 / 870 2.4674( )out d t t t Q C p A A mm

V u ra ca van tit lu l bn cha nn : 139( )c

p p bar 5 2.A . 2 / 0.611. . 2.139.10 / 870 2.4674( )out d t t t Q C p A A mm

F.1.9Vli mch F. 1.8 chnng ti. Vi bm lu lng, van an ton, cc kch thc khng ihy tnh khi lng ti ln nht m hthng c thnng ln. Tnh ton tt ccc thng svnhnh chny. Bqua mt mt thulc trong cc bphn, trvan tit lu (bm ga).Cc thng snh c.

Gii


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Trang 48

Khi lng ti ln nht m hthng c thnng:= 350 ( ) = 134500 = 134.5 Ta c lu lng sau khi qua van tit lu l:

= = 0.07 2.61 10

= 1.827 10

=

Do cng sut bm khng i nn:

= 0.027 = 600 = 1477.8 Lu lng Qout: = = 0.55 /CHNG 2 SCH FLUID POWER ENGINEERING

F.2.1

Trnh by mi quan hgia p sut v nng lng mt i trong ng truyn thy lc ving knh ng l hng strong dng chy tng

Gii

Tn tht p sut trong ng ng thy lc c ng knh l hng svi dng chy tng.2

2

L vP

D

Vi

2

4

64/ Re

Re

Q

v D

vD

Trong :

D: ng knh ng (m)

L: chiu di ng (m)

Re: hsreynolds

V: vn tc (m/s)

P : p sut hao ht (Pa)

Tn tht nng lng

2

4

128 LQN

D


49/203

Trang 49

F.2.2

Trnh by smt mt nng lng thy lc ca 1 ng truyn thy lc v suy ra kt quchong ni song song v ng ni tip.

Trli

L LP R Q

Trong RL: khnng chng r rNs/m5

QL: lu lng r rm3/s

2

L LN R Q

F.2.3

Trnh by qun tnh trong ng ng thy lcTrli

Nng lng hot ng trong ng ng thy lc bnh hng bi qun tnh lu cht.nh hng bi qun tnh c xy dng:

2

4

dvF ma P A AL

dt

Q dQv P I

A dt

L LI

A D

I: qun tnh cht lng (kg/m4)

F.2.4

Trnh by qun tnh thy lc trong hai ng ng ni tip hoc song song.Gii

1 2

1 2

L LI

A A

(ni tip)

1 2

LIA A

(song song)

F.2.5

nh ngha mun n hi cht lng v trnh by mun n hi ca hn hp du-kh.Gii


50/203

Trang 50

Khnng nn c l khnng thay i thtch khi p sut thay i. Vi cht lng tinh khit,mi lin hgia thtch v p sut l:

/

PB

V V

Mun n hi ca lu cht c xc nh bi nhit v p sut.Mun n hi cho hn hp du-kh l:

1enP BB

Pn B B

: Tlthtch kh.n: Sm a hng.

F.2.6

Trnh by dung khng ca hai ng ng ni tip hoc song song.

GiiLu lng u ng ng l

1Q , p sut P; cui ng ng ta c lu lng l

2Q :

Q in out V Q Q dt Trong :

QV l skhc nhau gia lu lng vo v ra

nh hng ca snn lu cht:

C

VV P

B

CdV V dP

dt B dt

VyQ L CV V V

LV phthuc vo cht liu ng ng, dy thnh ng, ng knh thnh ng v p sut h

thng.Vi 2 ng ng khc nhau th 0LV

Q C L

L

V V V

V

in out

d d d

dt dt dt

dV dP

Q Q B dt dt

Mt khc: CV dP dP

Q VB dt dt

2

4

D LC

B

Trong : C l dung khng ca ng ng.


51/203

Trang 51

F.2.7

Trnh by stng p sut vi 1 lng cht lng bmc kt trong ng khi chu stng nhit T .

Gii

Thtch cht lng thay i bi nhit :T

V V T

Trong : V: thtch lc u, 3m

: hsgia nhit. 10,0007K

T : tng nhit C

VT : bin i thtch khi cht lng c gia nhit

Trong trng hp cht lng bmc kt, thtch sl hng s, vy P stng ln

0

. 0

T C

C

V V V

VV P

B

VT V P

B

P B T

F.2.8

Trnh by ngn gn tc ng ca p sut hi bo ha trong hthng thy lc.

Gii

p sut hi bo ha c xc nh nh p sut m cht lng ssi. N tng ln vi stngca nhit du.Cht lng ssi khi khi p sut cht lng thp hn p sut hi bo ha. iu ny sgy ra hintng si bt kh ni m p sut ng ng gim t ngt. Si bt kh gy ph hoi ngng nh n mn,

F.2.9

Tnh phn trm skhc nhau vthtch ca cht lng nu B=1,4 Gpa. p sut cht tng l10Mpa.

Gii

Tlthay i thtch ca cht lng:/

/

P PB V V

V V B

6

9

10.10/ .100%

1, 4.10V V -0,714%

=> Thtch gim 0,714%


52/203

Trang 52

F.2.10

ng ng thy lc c ng knh D, chiu di L, dy h, modun n hi E, modun n hidu B. Chng minh hsma st ca ng ng cho bi

a/

2

4

D LC

B

bqua cc bin dng thnh ng

b/2

1

4

D L DC

B B Eh

xem xt cc bin dng xuyn tm ca ng

c/2

1 5

4 4

D L DC

B B Eh

xem xt cbin dng xuyn tm v trc

Trli:

(a)

(b+c)


53/203

Trang 53

T(2.58) (2.59), (2.77) v (2.61) suy ra

F.2.11

Tnh skhc nhau gia lu lng ng vo v ng ra nu dP/dt = 13.4 Mpa/s, D=10mm, L=3m,B=1.3, h=210 Gpa.

Gii


54/203

Trang 54

c

dPQ C

dt

M2

2 26

9 9

1 5( )4 4

.(10.10 ) .30 1 5.10( ).60.13, 4.10 0,157( / )

4 1, 3.10 4.210.10 .1C

D L DC B Eh

Q l ph

F.2.12

(a) Vit cng thc tnh v tnh nht cht lng trong ng ng di y: L1=L2=10mm,D=8mm, c=2m, = 0.02 Ns/m2(b) Nu ng ng c chuyn ng xoay, vit cng thc tnh nht v tnh gi trca chng.

GiiHsma st nht ca cht lng l:

.0,02.8.102

2v

DLf

c

5.02

F.2.13

Tnh lu lng r rqua hng knhL

Q v hscn r r LR . Cho bit cc thng s:

D=12mm, 7c m , L=20mm, 20, 018 /Ns m , 21pP MPa , 0TP .

Tnh ton li khi 100c m . Biu lin h LQ c .

Gii

a/

33 63

6 8 3

3

12.10 . 7.10.21.10 6, 29.10 /

12 12.0, 018. 20.10L

DcQ P m s

L

33,77.10 / minl

3

14 5

33 3 6

12.0,018. 20.10123,34.10 /

12.10 . 7.10L

LR Ns m

Dc

b/

33 63

6 4 3

3

12.10 . 100.10.21.10 1,83.10 /

12 12.0, 018. 20.10L

DcQ P m s

L

11 / minl


55/203

Trang 55

3

11 5

333 6

12.0,018. 20.10121.15.10 /

12.10 . 100.10L

LR Ns m

Dc

c/ Biu lin h LQ c .

F.2.14

a/ Tm cng thc lin hX(s)/ P (s)1 2

P P P .

- p lc:2

. .4

P

DF P

- Lc ma st: 2. .t

DL dXF

c dt

- Lc ko l xo: 2. .lx

F k x

- Phng trnh cn bng lc:2

. . 2. . 2. .4

D DL dXP k x

c dt

b/ Tnh v vthp ng khi 4P bar


56/203

Trang 56

2

32 2

5

3

8.101 1. . 2. . . . . . .4.10 .

4 4 2. 4 2.20.10

D DP k x x P

k

45,03.10 0,503m mm

Ta c:2

2 2 .4

D DL dxP kxc dt

2

12 . .

24

DLdt dx

DcP kx

2

5 312 . .ln 2 6, 28.10 .ln 20,11 40.10

2 4

DL Dt P kx x

c k

+C

5 3 56,28.10 .ln 20,11 40.10 6,28.10 .ln 20,11t x

F.2.15

Tnh p sut trc van tit lu,t

A , cho l dng chy tng. Bit:

Lu lng bm 14 / minQ l

Hsnht ng lc hc 20, 0185 /Ns m

Hstrao i 0,611D

C

Khi lng ring 2866 /p Kg m

Din tch van tit lu 23A mm

Cc thng sng 1 2 314 , 8 , 10D mm D mm D mm

1 2 31,3 , 2 , 1,5L m L m L m

GiiCt p sau van tit lu:

2 2

1 1 2 2

1 1 2 2

64 64. . . .

Re 2 Re 2l

L V L Vh

D g D g

Trong : 4 31 14 / min 2,33.10 /Q l m s


57/203

Trang 57

4

1

1 2 23

1

2,33.101, 52 /

14.10..4

4

Qv m s

D

3

1 11 866.1,52.14.10Re 996,13

0,0185pv D

1

2

3

64 1,3 1,52. . 0,7

996,13 14.10 2.9,81lh m

Ta c:2 3

2 2

2 2 3 3

2 2 3 3

64 64. . . .

Re 2 Re 2l l

L V L Vh h

D g D g

2 2

2 2 3 3

2 2 3 32 3

64 64. . . .

2 2

L V L V

pv D pv DD g D g

2 2 3 3 2 2 3 3

2 22 2 2 2

2 32 2 3 3 2 3

. . . .

4 4

L Q L Q L Q L Q

D DD A D A D D

2 3

2 34 4

2 3

. .L L

Q QD D

4 4

2 3

3 2 2 2

3 2

3 2

2 10. . . . 3, 26.

1,5 8

14 / min

L DQ Q Q Q

L DQ Q l

5 3214

3, 29 / min 5, 48.10 /3,26 1

Q l m s

5

22 2 2

32

5,48.101, 09 /

8.10..4

4

Qv m s

D

2

3

2 2

2

2

3

866.1,09.8.10

Re 408,190,0185

64 2 1,09. . 2,37

408,19 8.10 2.9,81l

pv D

h m

1 20,7 2,37 3,07

l l lh h h m

p sut sau van tit lu:2

. . 3,07.866.9,81 26081,06 0,26lP h p g Pa bar


58/203

Trang 58

Ta c:2

. .d t

PQ C A

p

p rt qua van tit lu:2

32

3

114.10 . 86660. . 7016435,5 70,16

. 2 0,611.3.100 2d t

Q pP Pa bar

C A

p trc van tit lu l:2

70,16 0,26 70,42P P P bar

F. 2.16

Tnh p sut u vo bm 60L/min. Cho bit bmv kch thc ng ng ht l di 1m,ng knh l 15mm, hsnht ng hc l 0,2cm2/s, trng lng ring du l 900kg/m3

Gii

Ta c p sut u ng ng ni vi bbng p sut kh quyn = 0.981 barp sut ri trong ng ng: eP gh

Trong 2

64Re * 2

eLVh

D g

3 360 / min 10 /Q L m s

3 3

3 2

10 105.66 /

* 0.25 * * 0.25 *(15 10 )

QV m s

A D

3 3

2

900 5 (15 10 ) 900 5 (15 10 )Re 4245

900 0.2 100

VD

2

3

64 1 5.661.64( )

4245 15 10 2 9.81e

h m

900 9.81 1.64 14479.56( ) 0.14( )P Pa bar Vy p sut trc bm = 0.981-0.14=0.84 (bar)

F. 2.17

Tnh ton p sut mt mt bi ng ng sau:


59/203

Trang 59

trong hsnht ng hc =20 cSt=20.10-6m2/sGii:

p dng cng thc trang 42 sch

ta c

3

1 3 2

1

10*10 / 60

0.25* *(13*10 )

QV

A

=1.256 m/s

1

1

1.256*13*10 ^ 3Re 816.4

20 *10 ^ 6

V D

3

2 3 2

2

10 *10 / 60

0.25* *(8*10 )

QV

A

= 3.315 m/s

3

2

2 6

3.315*8*10

Re 132620*10

V D

2 2

1 2

1 2

1 1 2 2

64 6416171.93 112710 128881.93

Re * 2 Re * 2

L V L VP P P

D D

=1.3 bar

F.2.18


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Trang 60

Tnh ton sthay i ca pittong choo bi hthng sau

Gii

Bin i thtch di sthay i p lc nn:2

1 2 0

1 5 1 5( ) ( )

4 4 4c p

D d DV C p C p p A x L

B Et B Et

Vi :3100 10

56.8( )0.0176p

Fp bar

A

3 2 3

9 9 3 9 6

45

1 (13 10 ) 1 5 13 1020.0176 0.5 51.4 10 4.2 10 10 6 10 4 1.4 10 4 210 1 10

c

A

V

= 12 5 36.909 10 3.92 10p m

12

36.909 102.23 10

0.0176

c

p

Vx p m

A

2.23( )x mm

CHNG 3 SCH FLUID POWER ENGINEERING


61/203

Trang 61

F.3.1

Cc dng tn tht nng lng trong hthng thy lc:

Trong phng trnh Becnuly vit cho dng chy thc, shng hW l nng lng ca n vtrng lng cht lng btn tht, khc phc sc cn ca dng chy trong on dng ang xt,ta chia tn tht lm 2 dng:

+ Tn tht dc c sinh ra trn ton bbdi dng chy u hoc khng u l tn thtxy ra dc theo ng di chuyn ca dng chy do sma st ca cht lng vi thnh rn tipxc

+ Tn tht cc bsinh ra ti nhng ni c bit dng chy bbin dng t ngt

+ Nguyn nhn ca tn tht ct nc, d di dng no cng do ma st gia cc phn tcht lng do ma st trong sinh ra. Cng ti nn bi lc ma st ny bin thnh nhit nng mkhng ly li c

Vi githit l cc dng tn tht xy ra c lp vi nhau th tn tht nng lng hw ca dngchy c thvit: hw = shdshc

trong : shd: tng cng cc tn tht dc ng ca dng chyshc: tng cng cc tn tht cc bca dng chy

F.3.2

Gii thch chi tit cc cch tnh ton tn tht nng lng do ma st trong hthng thy lc

Cc tn tht trong ng ng phthuc vo hnh hc, nhm bmt, tnh cht cht lng v h

sRenoyds

Ba loi chnh ca dng chy chy tng, chy ri, chy na ri

Cng thc Renoyds: D vDRev

Trong : D: ng knh trong

: vn tc dng chy2

4Q

D

: nht (Pa.s)

v : nht ng hc

Cng thc xc nh tn tht nng lng:2

(*)PL Pv

D z

hsma st phthuc chyu vo hsReynolds (Re). vic chuyn tchy tng sang chy ri


62/203

Trang 62

Re>2300. Trong rng hp dng chy tng, bng cch thay thv v Re trong phng trnh (*) .Ta c biu thc tnh tn tht p sut

4

128

P

LQ RQ

D

F.3.3

Gii thch chi tit cc m hnh ca dng truyn ng thy lc. gii thch r rng githit cabn v tnh ton ma trn ng vo v ra p sut v lu lng

Dng truyn ng thy lc tht sl mt hthng tham sphn phi. Cc chuyn ng cacht lng trong iu kin thong qua xy ra di tc ng ca cht lng qun tnh, ma st Tc du, p sut v nhit thay i tim ny sang im khc dc theo chiu di ng v

bn knh ng,

Cc tnh ton trnn phc tp khi cn nhc tt ccc thng sdu v dng chy+ Cc dng chy tng mt chiu+ p lc cht lng v tc c xem xt nh l gi trtrung bnh v c coi l lin

tc theo phng ng cho

Trong phn u tin, du di chuyn di tc ng ca lc ma st. Do , chuyn ng ca nc m tbi phng trnh lin tc p lc, P, tc dng chy, Q

P0- P1 = RQ1Q0= Q1

p dng bin i Laplace, sau khi sp xp li ta c

0 1 1

0 1 1

1

0 1

P s P s P sRR

Q s Q Q s

Trong 5

4

128(Ns/ m )

LR

D

R: khng ma trn

F.3.4

Phng trnh 3.19 m tmi lin hgia p sut v tc dng chy trong ng ng thy lc.

Gii thch I, R ,C

* Ta c: P0- P1= RQ1=> Q0= Q1


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p dng bin i Laplace ra c

0 1 1

0 1 1

1

0 1

P s P s P sR

s s

R

Q Q Q s

Trong 5

4

128(Ns/ m )

LR

D

R: khng ma trn

* Phng trnh m tschuyn ng ca du

2

1 2

1 2

d

dt

QP P I

Q Q

p dng bin i Laplace ra c

2 2

2 2

0

0

11

0 1

sP s P s P s

Q s Q s

I

Q s

Trong : I tng cc dng qun tnh 254 (Ns/m )PLI

D

* Hiu qunn du trong phn cui cng, cc mi quan hc thsuy ra

1

2

2

dP

dtL

L

Q C

P P

Q

p dng bin i Laplace ta c

2

2 1

1

0L L

s L LC s

P s P s P sC

Q s Q Q s

Trong : C mc np ma trn3

2

(m / Pa)4

D LI

P

Chuyn ma trn lin quan ti cc thng s: P0, Q0, P1, QL

10 2

10 2

L

L

P s P s P s P sR RI RIC

Q s Q s Q s Q s

Hoc

2

0

0

1

1

Ls s s

Ls

P s P s

Q

IC RC I R

Cs Q s

F.3.5


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(a) Rt ra cc chc nng chuyn giao ca mt dng kt thc m, QL/Po v tm thy nhng biuhin cho cc thng sca n.

- Tnh cc thng sc trng.

+ Dng khng: 34

128/

LR Pa m

D

+ Dng qun tnh: 42

4/

L LI kg m

A D

+ Dng in dung:2

4

D LC

B

3 /m Pa

(b) Mt dng thy lc m cui cng l mca cho khng kh c nhng iu sau ythng s:ng chiu di L = 3,5 m, ng knh ng D = 8 mm, mt du = 850 kg / mnht = 0,018 Ns / m v slng ln du mun B = 1,3 GPa.Tnh ton cc hsca hm truyn ca n p ng vi mt p lc bc u vo l 0,5 bar.

+ Dng khng: 9 34 4

128 128.0,018.3,50,63.10 /

.0,008

LR Pas m

D

+ Dng qun tnh: 8 42 24 4.850.3,5 0, 59.10 /.0,008

L LI kg mA D

+ Dng in dung:2 2

13 3

9

.0,008 .3,51,35.10 /

4 4.1,3.10

D LC m Pa

B

F.3.6


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Gii thch ngn gn hot ng chnh ca my bm thtch.

Trli

Hot ng ca bm thtch c tm tt nh sau:- Trong qu trnh gin n, bung bm ni vi ng ht. Qu trnh gin nny to ra p

sut chn khng trong bung bm lm lu cht c ht vo trong bung bm.- Khi thtch bung bm t n gi trti a th bung c ngn cch vi ng ht.- Trong giai on co li bung bm kt ni vi ng y sau lu cht by ra khi

bm v t p sut cn thit thng c lc cn ng ra.- Chu trnh cp dng li khi thtch bung bm t gi trti thiu. sau bung bm c

phn cch vi ng y v qu trnh ctip din.

F.4.2

Vit biu thc m tbm l tng v bm thc t.Trli:

Bm l tngLu lng bm c nh ngha l thtch cht lng cung cp bi cc my bm trn mi

vng quay, giskhng c r rv bqua snn ca du. N phthuc vo gi trti a v tithiu ca thtch bung bm, slng bung bm v shnh trnh bm trn mi vng quay. Thtch ny phthuc vo dng hnh hc ca my bm, v vy n cng c gi l thtch hnh hcVg. c cho bi phng trnh sau y

Vg= (VmaxVmin)zii = shnh trnh bm trn mi vng quayVg= lu lng bm, m3/vngVmax= thtch ti a ca bung, m3Vmin = thtch ti thiu ca bung, m3z = sbung bm

Gismy bm l tng, khng c r rni b, khng c ma st, v khng c tn tht p suttc dng chy ca bm c cho bi:

Qt = Vg.nQt= lu lng bm l thuyt, m3/sn = tc bm, vng/s

Bm thc tNng lng thy lc c truyn bi cht lng bng cc my bm thc tt hn so vi nng

lng c hc c cung cp da vo lng thtch, ma st v tn tht. Lu lng dng chy bmthc tQpt hn so vi dng chy l thuyt Qt, chyu l do:

- R rni b- Bm to bt v kh- Khnng chu nn ca cht lng


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- Lm y mt phn ca my bm do qun tnh cht lngNguyn nhn thnht ca tn tht nng lng l r rni b. Trn thc tkhi hot ng theo

cc iu kin thit kchnh xc, tn tht dng chy chyu l do r rni bQL. Dng r rquakhe hhp v thay i tuyn tnh da trn chnh lch p sut. Khnng chng r rni bRL, tlthun vi nht ca du, , v tlnghch vi ly tha ba ca khe htrung bnh c.

QL= P/RLQ = QtQL

Vi RL= K /c3Cho mc p sut cao v tng khe hhng tm, tng tc dng chy tht thot v dng

r trnn hn lon.

nh hng ca hin tng r rc thhin bng hiu sut thtch, vc quy nhnh sau:

t L Lv

t t t g L

Q Q QQ P1 1 1Q Q Q V R n

Hiu sut thtch trong phm vi 0.80.99. Bm piston c hiu sut thtch cao, trong khibm cnh gt v bm bnh rng ni chung hiu sut thtch thp hn. Ma st l nguyn nhn thhai gy tn tht nng lng. Ma st nht v ma st c gia cc phn tbm phn tn nng lngMt phn ca mmen xon c tiu thkhc phc lc ma st. Phn ny l mmen xon mast TF. N phthuc vo tc bm, p lc dng v nht ca du. V vy, to ra p lcnh yu cu, nn to ra mmen xon cao hn. Tn tht ma st trong cc my bm c nh gi

bng hiu sut c, m:

F Fm

(T T ) T T

T T

T = mmen xon thc t, NmTF= mmen xon ma st NmTTF= mmen xon chuyn i sang p lc, Nm = tc quay trc bm, rad/s


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Nguyn nhn thba ca tn tht nng lng l tn tht p sut bn trong my bm. p lcc to ra tbn trong bung bm, Pc, ln hn p lc ra ti a, P. Tn tht chyu l do tntht cc b. Nhng tn tht c gi trkhng ng ki vi my bm c tc di 50 vng/sv c ngha l tc du thp hn 5m/s. i vi tc ln hn, tn tht p sut tlthun v

bnh phng ca tc dng chy. Nhng tn tht p sut c nh gi bng hiu sut thy lch

h

C C

QP P

QP P

PC: p sut trong bung bm (Pa)P: p sut ra (Pa)

Biu thc cho hiu sut tng T, c cho bi:

t CT v m h

F

Q P

(T T )

Nng lng c hc (T TF) c chuyn i thnh nng lng thy lc bng QtPC, sau:

T= vmhTrong hot ng trng thi n nh, bm chuyn thc c m t

Q = Vg.n.vNh= Nm. T

Hoc QP = 2 nTTSau

g

m h

V

T P2 Nh= nng lng thy lc WNm= nng lng c WP = chnh p u ra v u vo ca bm P = P - PiPa

Nu p lc bm du vo, Pi, l qu nhso vi p lc dng, P, sau n c thc bqua v chnh p, P, bng vi p sut bm li ra, P th:

g

m h

VT P

2

F.4.3Bm c thtch hnh hc l 23cm3, c vn hnh tc 1450vg/ph, cung cp p sut 20MPaGithit bm l tng, tnh lu lng ca bm Qt, tng cng sut N, cng sut ra Noutvmmen dn ng Ttnu p sut vo l 100kPa.

Qt = Vgn = 2310-61450

60= 5.5610-4m3/s = 33.36 l/ph


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N = QtP = 5.5610-4(20106-100103) = 11064.4 WNout= QtP = 5.5610-420106= 11120 W

Tt=6

6 323 10 (20 10 100 10 )2 2

gV

P

= 72.8 N.m

F.4.4Hnh bn di gii thch struyn ti in trong hthng thy lc, nh mi quan hnhn quca hthng bin thin. Hy tho lun chi tit viu ny.

Trli:- M t1 mch khp kn- Hthng in cung cp 1 in thkch lm motor quay vi tc n rad/ s lm quay trc

bm sinh ra 1 lu lng gQ V .n tc ng vo xylanh vi vn tcQ

vA

vi A l din tch ca

xylanh nng ti.Ti c nng bi 1 lc F bng trng lng ti sinh ra p sut trn mt

xylanhF

p

A

. p sut P n bm v sinh ra 1 moment xon gPV

T

2

tc dng vo motor in

sinh ra dng cp cho hthng in. Hthng ctip tc tun hon.

F.4.5

Tho lun vnhng loi mt nng khc trong bm thtch.Trli:

- R r trong chnh bm - Ma st- S tt p trong ng i ca du

F.4.6

Tho lun vhin tng cavitation trong bm thtch.Trli:

Cc c tnh to bt trong mt my bm m ttc ng ca p lc u vo ca lu lngmy bm. Gim p lc bm u vo vi cc gi trthp hn so vi p sut hi dn n sbay hhoc si ca du. Cc dng chy ca cc ca bm to thnh mt hn hp lng, kh c gii

phng v hi. mc khng hoc p lc li ra rt thp, th bm c bqua v d, hi khng


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ngng tv hc hi khng st p. Nhng trong iu kin hot ng bnh thng, my bm cti bi p lc ti trng ln. Khoang hi st p do sngng tqu nhanh ca hi khi chuyn ncc vng p sut cao hn. Do , tc dng chy rng ca bm gim. Ni chung, tng 1% trongthtch hi trong dng chy hi lm gim hiu sut thtch khong 1%.

Ngoi vic gim hiu sut thtch, cc phn tbm ang phi chu p lc tc ng ln tcc cht lng o t lp y khng gian hi bst p. p lc tc ng ln n gi trrt cao, lnn 7000 bar. Khi to bt kh, slm tng mc v cng ting n my bm. Bmt cacc phn tbn trong my bm bh hng bi cc lthng do p lc tc ng ln bmt. Do p lc bm u vo phi cao hn p sut hi bo ha trong du nhit hot ng ti a bnggi trthch hp. Gi trny c gi l gi trd to bt vo khong 0,3-0,4 bar.

Hin tng khng mong mun ny c thtrnh c bng cch thc hin cc hot ng saubt ckhi no c th:

- Gim tht thot p lc dng bm u vo bng cch tng ng knh ng ng ht vgim chiu di ca n.

- Trnh sdng cc blc ng dy u vo v cc phn tgy mt nng cc bkhc- Tng p lc bm ht bng cch lm mt trong cc cch sau:- Sdng bm tng p.- Sdng mt bkn.- Gn my bm di bnm cch mt khong cch thch hp.

F.4.7

Trnh by ngn gn vphn loi bm thy lc.Trli: Cc loi bm trong cng nghip

a) Bm vi lu lng cnh- bm bnh rng n khp ngoi- bm bnh rng n khp trong- bm pittng hng trc- bm trc vt- bm pittng dy- bm cnh gt kp- bm rto

b) Bm vi lu lng thay i

- bm pittng hng tm- bm pittng hng trc (truyn bng a nghing)- bm pittng hng trc (truyn bng khp cu)- bm cnh gt n

Cc loi bm thtcha) Bm bnh rng

- bm bnh rng n khp ngoi


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- bm bnh rng n khp trongb) Bm trc vt

- Bm trc vt loi p sut thp- Bm trc vt loi p sut trung bnh- Bm trc vt loi p sut cao

c) Bm cnh gt- bm cnh gt n- bm cnh gt kp

d) Bm pittng- Bm pittng hng tm- Bm pittng hng trc

F.4.8

Vs ca trc cong bm pittng hng trc cong. Gii thch ngn gn chc nng ca n v

vit biu thc tnh thtch hnh hc ca n.

Trli:Hnh 4.39 cho thy mt cu to in hnh ca trc cong bm piston hng trc. Bm bao gm

mt trc a (1), khi xi lanh (3), piston (4), v mt tm cng (5). Kt thc cu ca cc pistonc gn vo a (2), cng vi trc bm. Khi trc bm quay, khi xi lanh cng quay theo. Khxi lanh trt trn tm cng, trong bao gm hai lkim sot dng qung (xem hnh .4.13) .

Trc truyn ng quay quanh trc Ox nm ngang trong khi trc quay ca khi xi lanh nghingmt gc . Khi xi lanh nghing gn vi piston lun phin chuyn ng so vi khi xilanh. Do, mi piston thc hin mt chuyn ng qua li gia im cht trn v di ca n. Chuynng piston tim cht di n im cht trn to ra mt hnh trnh ht. Cht lng c htthng qua vic mca kim sot pha bn ht ca tm cng vo khi xi lanh. Khi trc bm tiptc xoay v piston di chuyn tim cht trn xung im cht di, cht lng c di chuynra ngoi thng qua vic mca kim sot khc ( bn chu p lc). Trong hnh trnh phn phi


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trc bm hot ng trn a bng m-men xon cn thit to cc lc quay trc piston chng lip lc ti.Biu thc thtch hnh hc c cho bi biu thc sau y:

sinh D 2

sin4

g

dV zAh D z

Trong A = din tch piston m2D = ng knh vng trn pitch md = ng knh piston mh = hnh trinh i ca piston mz = spiston= gc nghing rad

F.4.9

Hy trnh by vbin dng chy ca bm pittng hng trc cng vi sdao ng p sut cabm khi chu ti bi vi phun.Trli:

Sdao ng lu lng v p ca bm ni chung:Vmt l thuyt, lu lng bm c tnh l Q =Vgn. Biu thc ny a ra lu lng

trung bnh ca bm. Trn thc t, lu lng bm l khng phi l hng s. Mi bung bm cungcp lu lng bng vi gim thtch ca n. Lu lng thc ca bm ti mt khonh khcno ci m l tng ca lu lng c cung cp bi cc bung bm th c ni vi ng ra t

lc .Lu lng bm c cp bi bung bm c gi trl 0 ngay ti lc bt u hnh trnh bm. Ntng dn cho n khi t gi trti a ti im gia ca hnh trnh.


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Skhng lin tc ca lu lng bm thtchSau sgim u cho n gi trnhnht lc kt thc hnh trnh. Do , lu lng thc t

dao ng (hnh v). Cng dao ng th c nh gi bng hsdao ng, v c xc nh

=

.100%

Trong ,: l hsdao ng ca lu lng.: lu lng ln nht, m3/s: lu lng nhnht, m3/s: lu lng trung bnh, m3/sSdao ng lu lng dn ti sdao ng p sut v schuyn ng khng tng ng ca xlanh thy lc v ng c.Xt trng hp 1 bmtit lu u ra(hnh bn di) v bqua snn cht lng. p sut ti ura l:

2

2 2

2

2 2

2

ax ax2 2

2

min min2 2

2

2

2

2

d t

m m

d t

m m

d t

d t

P QC A

P QC A

P QC A

P QC A


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Trong : l hsdao ng ca p sut: p sut ln nht u ra, Pa: p sut nhnht, Pa: p sut trung bnh, PaNu lu lng dao ng t0,9n 1,04th =14%, =24,16%. Trn thc t, cho

rng c snh hng nn du, sdao ng p sgim c bit l stng thtch u ra.

Sdao ng lu lng v p bm piston hng trci vi bm piston hng trc, cc piston chuyn ng n gin v hi ha theo quy lut

hnh sin. Lu lng to bi mi piston th bng tc ca n nhn vi din tch ca mi pistonBqua nh hng ca r rdu, qun tnh du v snn du, lu lng to bi mipiston cngc dng hnh sin. Hnh bn di thhin lu lng to bi tng piston ring rtrong 5 piston.

Lu lng ca bm th bng tng lu lng to bi mi piston. C thc n 2 hoc 3piston to lu lng cng 1 lc. Tng lu lng my bm c thhin trong hnh trn. Bmc ti bi 1 van tit lu. p sut v lu lng c vtng i gn vi gi trPmv Qmtrnhnh trn.

Sdao ng lu lng ca bm piston th c tnh ton khc nhau ty vo spiston. p

sut c tnh ton khi bqua snn ca du. Kt qutnh cho thy l sdao ng dng cabm gim khi slng piston l sl, v vy i vi bm piston hng trc th spiston nn lsl.

F.4.10

Vbiu vsthay i gc nghing ca a nghing trong bm piston hng trc. Gii thchngn gn chc nng ca n v vit biu thc vthtch hnh hc ca n


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Trli:Hnh 4.16 cho thy cu to v hot ng ca mt my bm a lc rung. Trc a (1) quay

v dn ng cho khi xi lanh (5). Ca v khi xi lanh c cng trc quay. Khi xi lanh (5) vpiston ca n (6), lun phin vi cc trc. Mi piston c gn vo mt bc lt lp m trt(3). Cc piston v bc lt lp m trt ca chng c chn vo cc lca tm chn (4). Do tm chn quay cng vi cc piston v khi xi lanh. N c dn hng xoay trong mt mt

phng song song vi a lc (2) bi mt hng cnh (8). Quo ca bc lt lp m trtc xc nh bi a lc v tm chn. Trong khi quay, mi piston thc hin mt chuyn ngxoay chiu. Trong qu trnh ny, mt lng nc, tng ng vi din tch piston v hnh trnhht hoc cung cp qua chai lkim sot trong cc tm cng (7).

Khi xi lanh c y chng li cc tm cng bng l xo (9), gim thiu sr rthng quakhe h, phn chia u hot ng bm. Khi c p lc tch t, n hot ng trn cc khi xi lanhvi mt lc xit cht bng {0.25 (d d) P}. Lc ny hot ng bn phi, chng li lc ydo sphn bp lc trong khe hgia cc khi xi lanh v tm cng. Lc hot ng ny l ktqugim nh hng ca khe hv gim thiu sr rthng qua n. Thtch hnh hc my

bm c cho bi biu thc

= 4tan

l gc nghing (rad)

F.4.11

Vbiu ca a lc rung bm pittng hng trc vi pittng nghing. Gii thch ngn gn chcnng ca n v vit biu thc cho thtch hnh hc ca n.Trli:


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Nguyn l hot ng: khi trc nghing quay th slm thay i thtch bn trong xilanh

lm p sut trong thay i, dn n vic chnh lch p sut v to p sut y cht lng thylc ra bn ngoi, truyn p sut cho mch lm vic.Gc nghing cng ln th lu lng ra cng ln, khi gc nghing v0 th khng c lu lng bmra ngoi.

F.4.12

Trnh by ngn gn nhng g cn thit cho bm c lu lng ring thay i.a) Tng qut:

Bm lu lng ring thay i th t hn v thit kphc tp hn nhiu so vi bm lulng ring cnh. Tuy nhin c 2 l do cc nh thit ksdng bm lu lng ring thay

i l tit kim v iu khin.b) Bm b p kiu cnh gt

Kt cu v hot ngKt cu ca bm cnh gt thay i lu lng vi roto khng cn bng p c thhin

trong hnh 4.35 ti 4.37. Bm bao gm: vhp (1), roto (2), cnh gt (3), vng stato (cam) (4)vt iu chnh (6). Vtr ca cam (stato) (4) c iu khin theo chiu ngang bng 2 pittng: mtci nh(10) v mt ci ln (11). Vtr thng ng ca cam c iu khin bng vt iu chnh(7). Bung bm (8) c hnh thnh bi cnh (3) v roto (2) vng stato (4), v mt bn (9). m bo chc nng bm trong thi gian khi ng, stato (4) c t vtr lch tm bi l xo(5). Stato c t lch tm cng vi roto. Sau do roto (2) quay, thtch ca bung (8) tng vcht lng c in y thng qua 1 cng ht (S). khi thtch bung c in y, bung skhng cn ni vi cng ht. Sau roto (2) tip tc quay, thtch ca bung gim khi n thngvi pha c p. Du sau sc y ra khi bm qua cng (P). Pittng iu khin nh(10)lun ni vi ng bm c p. Pittng ln c ti bi l xo (5). Sau khi bm hot ng vp sut ca bm thot ra, siu p ca l xo xc nh vtr ca stato v bm tip tc quay.


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Hnh 4.35

Hnh 4.36


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Hnh 4.37

c)Bm pittng hng trc cong iu khin cng sutM tv hot ng

Hnh 4.39 thhin bm pittng hng trc vi trc cong iu khin cng sut c ththayi lu lng. bm gm 2 nhm chnh. Nhm u tin l bdn ng cha trc, tm cng, v 7

pittng trong mt khi hnh tr. Nhm th2 l nhm iu khin, trong c cc pittng cm binp sut, pittng servo, thanh dn hng, van ng, l xo iu chnh v hai l xo iu khin. hai

nhm c kt ni bng cht. khu vc nhhn pha trn ca pittng servo lun kt ni vi cngra ca bm, thng qua nhng lgim xc. Bung pittng servo thp hn c ni vi p hoc b

pittng cm bin p sut c ni vi ng p thng qua cc lgim xc thhai. Pittng ny dichuyn xung di ng xuyn qua thanh dn hng v iu chnh l xo theo hng khc.

F.4.13

Gii thch ngn gn hot ng chnh ca bm nhin liu loi dao ng.Trli:

bm piston hng trc vi a xoay (xem hnh trn), a c xoay bng trc truyn

ng. Thn xilanh c c nh. Piston di chuyn vo trong bng a xoay, trong khi chuyn ngpiston theo hng ngc li c bo m bng l xo. Trong hot ng bm, bung bm c

ni vi ng du vo v ra qua van 1 chiu. Van 1 chiu ng vo nn p lc thp trnh to

lhng trong bm.


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F.4.14

Phn loi v gii thch ngn gn hot ng chnh ca bm piston hng knh. Sdng hnh 4.194.20, 4.21Trli:


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Bm piston lm vic theo nguyn l thtch, bm piston c truyn ng bi ng cchuyn ng quay ca trc ng c c bin i thnh chuyn ng tnh tin ca piston 1 trong

xi lanh 2. Nu tay quay tvtr im C2qua theo chiu mi tn n im C1th piston di chuynvpha tri, thtch bung lm vic 5 tng dn, p sut p trong gim i v b hn p sut mt thong bcha pa (p < pa). Do cht lng tbht qua van ht 6 vo bung lm vic 5trong khi van y 4 ng. l qu trnh ht ca bm. Sau , tay quay tip tc quay timC n im C2, piston i chiu chuyn ng sang phi, thtch bung lm vic gim dn lm psut tng ln, van ht 6 bng, van y 4 mcht lng chy vo ng y thc hin qu trnhy ca bm. Qu trnh ht v y ca bm piston din ra gin on v xen kln nhau, to nnqu trnh lm vic lin tc ca bm. Mt qu trnh ht v y ktip nhau gi l mt chu trnhlm vic ca bm.a) Bm tc dng n (bm tc dng mt chiu)

Trong loi bm ny, cht lng lm vic vmt pha ca piston, mt chu k lm vic capiston chc mt qu trnh ht v y ni tip. Cu to v nguyn l lm vic ca bm tc dngn c thhin cthtrong hnh 4.2.b) Bm tc dng kp (bm tc dng hai chiu)

Trong loi bm ny, piston lm vic chai pha, do c hai bung lm vic A v B, haivan ht 1,4 v hai van y 2,3. Trong mt chu trnh lm vic ca bm c hai qu trnh ht v haiqu trnh y (khi bung A ht th bung B y v ngc li).

F.4.15

Vbiu ca bm bnh rng n khp ngoi. Gii thch ngn gn chc nng v a ra biu thc

vthtch hnh hc ca n.Trli:

Bm bnh rng l loi c nhiu roto. C 4 loi bm bnh rng l: n khp trong, n khpngoi, bm trc vt v gerotors.


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Khoang bm c hnh thnh bi bmt ca 2 rng lin k, bn trong ca vhp v 2 tm

phng. Trong chuyn ng ca bnh rng, cc bnh rng cha n khp to thnh khoang bmKt qucha c p, cng vi lc ht cht lng buc phi chy vo cng ht (7). Cht lng iny khoang sau c di chuyn bi bnh rng tpha ht vpha c p. Ti y bnh rng nkhp mt ln na v y cht lng ra khi khoang bm v ngn khng cho chy ngc li vnght.

Trong trng hp bnh rng n khp ngoi vi 2 bnh rng tc ng, thtch hnh hc cabm c cho bi:

2 2

2 ( sin )gV bm z Trong b = chiu dy ca bnh rng, m

m = mun bnh rng, mz = srng ca 1 bnh rng = gc p lc ca rng, rad

F.4.16

Tho lun vtc gii hn cabm bnh rng v suy ra mi quan hcn thit.Trli:

Trong bm bnh rng, du i vo bung bm dc theo chu vi bnh. Khi vo my bm, ch tlng bt u quay vi cc bnh rng v c chu cc lc ly tm. Nhng lc c xu hng y ni v ra khi bungbm. V vy, tc bm ti a nn c hn chv p lc u vo l caotrnh hin tng ny. Mt biu hin cho tc ti a l suy lun trong nhng iu sau y:


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So vi cc lc p lc v ly tm hot ng trn mt phn tca cht lng. (Xem hnh 4.24), ccmi quan hsau y c suy lun bqua thut ng(drd) so vi (rd): rd =Pbrd rP dP b F hoc drF br dP

Lc ly tm2 2

2

2

0 0

2

2

r . .d . . r. .

. r.

r

P2

c

r

p r

c

F m r b d r

dP r d

dP r d

r

p lc u vophi ln hn cc lc p lc ly tm . V vy, tc bm ti a nn chn chnh sau:

= 2n> < 12 2

Bi trn l mt yu tp t tc ti thiu c nghtrong mt smy bm. Cc bnhrng c np bi cc lc p, FP, v thit blc lng lin lc, FC, nh thhin trong hnh 4.25Cc cnh qut l nonpressure bi thng v cc lc p lc l khng cn bng. Nhng lc ctruyn ti trc . V vy, trong trng hp trt mt mang bi trn thy ng, tc bm tithiu, nmin, c khuyn khch m bo bi trn theo yu cu.

F.4.17


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Tho lun ngn gn sbin ng dng chy trong bm bnh rng n khp ngoi.Trli:

Theo l thuyt, tc dng chy c tnh theo Q = Vgn. Biu thc ny cung cp gi trtrung bnh ca tc dng chy. Thc ttc dng chy khng phi hng s. Mi bm cungcp tc dng chy bng vi tc gim thtch ca n. Tc dng chy ca bm ti mtthi im no l tng ca tc dng chy cung cp tc thi bi khoang kt ni vi ca cungcp

Tc dng chy cung cp bi khoang ca bm bt u ti gi tr0 khi vo hnh trnh cungcp. N tng dn cho ti khi t gi trti a ti im gia ca hnh trnh cp. Sau n gimdn vgi trnull ti cui hnh trnh cp. Do lu lng thc sbin ng nh trong hnh 4.9.

Hnh 4.9ln ca sbin ng dng chy c nh gi bi hsbin ng c nh ngha l

ax min100%

mQ

m

Q Q

Q

Trong Q: hsbin ng dng chyQmin: gi trnhnht ca lu lng m3/sQmax: gi trln nht ca lu lng m3/sQm= Vg: lu lng dng chy trung bnh m3/s

Kt quca bin ng dng chy l p sut dao ng v chuyn ng khng u ca xylanhv ng c.

Xem xt trng hp ca mt ng ng tit lu (hnh 4.10) v bqua snn ca cht lng

p sut ra cabm c cho bi


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Hnh 4.10

Trong P= hsbin ng p sutPmin= gi trnhnht ca p ra ca bm, PaPmax= gi trln nht ca p ra ca bm, PaPm = gi trtrung bnh ca p ra, Pa


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Nu tc dng chy dao ng gia 0.9Qmv 1.04 Qm, sau Q = 14% v P= 27.16%Trn thc t, xem xt n snn ca du, p sut dao ng c bit tng thtch ca ng ra.

F.4.18

Hy ni vvn nn du trong bm bnh rng n khp ngoi.

Trli:Trong hot ng bnh thng ca bm, khi rng n khp, thtch du c gili trong khng

gian gia hai rng lin tip. vng quay ca bnh rng lm gim thtch du bkt v p sut can tng ln ti gi trrt ln. (hnh 4.23). gim 1% thtch du lm tng p sut t100 ti 200

bar. Sgia tng qu mc p sut du c thtrnh c bng cch sdng mt trong nhng kthut sau:

- Ct rnh trn cc tm mt bn truyn qua khng gian lin kt rng pha c p.- Thit kbnh rng vi srng nhvi khe h0.4 ti 0.5mm- Sdng btruyn ng bnh rng nghing.

Hnh 4.23

F.4.19

Gii thch ngn gn hot ng chnh ca bm bnh rng n khp trong v tho lun nhng cim vlu lng v hiu sut ca bm.Trli:

Trong na vng quay u, bnh v vnh nhkhp, khong trng gia cc rng tng. Trongtrng hp ny, du sau khi c ht ttqua lS, khng thay i lu lng v c a sangcc khong trng gia cc rng ca bnh rng 2, vnh 3 v cc chi tit trung gian hnh li limLu cht sau tip tc c y qua lP.Bm bnh rng n khp trong thng c dng trong nhng trng hp yu cu cng vngcao, n nh.


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Nguyn l hot ng bm bnh rng n khp trongBnh ch ng v bnh b ng lun t lch tm. Khi bnh ch ng quay ko theo bnh

b ng quay cngchiu trong Stato. Cht lng trong cc rnh rng theo chiu quay ca ccbnh rng vn chuyn t khoang ht n khoang y vng theo v bm. Khoang ht v khoangy c ngn cch vi nhau bi li chn.Lu lng ring: 0.4 250 cm3/vng. p sut: n 300 bar. Vn tc quay: 500 n trn 3000vng/pht. Lm vic t gy n nht trong cc loi bm.

F.4.20

Gii thch ngn gn hot ng chnh ca bm bnh rng (hnh 4.28) v bm trc vt.Bm bnh rng

Hnh bn minh hot mt ct ngangmt loi bm bnh rng in hnh, gmmt cp bnh rng c to hnh snBnh rng bn trong (hay rotor) dn ng

bnh rng bn ngoi.Bnh rng dn ng c t rng hn

bnh rng bdn. Cc bung bm c

to ra nhcc cp rng lin knhau. Khirotor quay, cc rng trn n c giacng chnh xc nn n sdn dng vomt trong ca b phn u ra. Bungcha smrng ra khi bnh rng ra khiv n st thtch ln nht khi rng ca

rotor bdn chm vo gia im cht trn. Trong na vng quay cn li, khong trng nhdnthay thlu cht u ra. Thtch hnh hc ca bm bnh rng c cho bi:

g max minV bz(A A ) trong b l chiu cao rng, z l srng rotor.

Bm trc vis
http://4.bp.blogspot.com/_b_eyIHkBHqA/SjdbXm14HII/AAAAAAAAAeQ/w5KX-Yu_zXY/s1600-h/bom+banh+rang+an+khop+trong.jpg


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i vi bm trc vis, bung cha thtchc to thnh gia ren v v.

Cng thc tnh thtch hnh hc bm trcvis 2 cp l:

2 2 2g sin2V (D d )s D s

4 2 2

ViD d

cos2D

F.4.21

Gii thch ngn gn cu to v hot ng ca bm cnh gt cn bng p sut.Trli:

1. Trc 2. Phn quay 3. Vnh khung tnh 4. Cnh gt 5. a c nh mt bn6. V 7. B 8. Cnh gt trong

Cu to v hot ng: Hnh 4.30 cho thy cu to mt my bm cnh gt lu lng cnhRotor (2) c iu khin bi trc (1) v quay trong vng stato (cam vng) (3). Cc cnh gt (4)


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c gn vo khe cm xuyn trong rotor. N c p xuyn tm ra pha ngoi c tip xcvi cc bmt bn trong ca vng cam di tc ng ca lc ly tm. Khi ci t p lc trong ccdng cung cp, mc ch ca mt scnh gt c p lc m bo vic kn kht phi btrchnh xc. Bung bm c cu to bi hai cnh lin tip (cnh qut), bmt bn trong ca vngcam, bmt ngoi ca rotor, v hai bmt ca cc tm bn (5). Thtch ca nhng thay i trong

bung bm (tng hoc gim) trong vng xoay do hnh dng hnh bu dc ca vng cam. Trongvic mrng ca n, l bung c kt ni vi ng ht thng qua mt ltrong tm bn. Do, du bht vo trong bung. Sau , trong qu trnh thu hp ca n, bung bm chim chduvo dng u ra thng qua mt ltrn tm bn. Bng cch ny, mi bung thc hin hai hnhtrnh bm trn mi vng quay.Do hnh dng hnh bu dc, cu to bm l i xng. C hai p lc v hai bung ht i dinnhau. Do , cc cnh qut l p lc cn bng. N c np chyu do mmen xon bm truynng. Bng cch ny, ti trng mang c gim thiu, lm gim ma st v mi mn v lm tnghiu qubm v tui th.

F.4.22Gii thch ngn gn chc nng ca cu trc intra-vane trong bm cnh gt cn bng p sut (hnh4.33 v 4.34)Trli:

Hnh 4.16 cho thy cu to v hot ng ca my bm cnh gt lu lng cnh cnhRotor (2) c iu khin bi trc (1) quay trong vng tnh (vng cam) (3). Cc cnh qut (4)c gn vo khe cm trong bnh cnh qut. Chng c t hng tm ra pha ngoi c thtip xc vi bmt bn trong ca vng cam di tc ng ca lc ly tm. Khi p lc c sinhra trong dng phn phi, cc gc ca mt scnh qut c p lc c m bo ng kn


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nhng vtr thch hp. Bung bm c bao bc bi hai l cnh qut lin tip (cnh qut), bmtbn trong ca vng cam, bmt ngoi ca cnh qut, v hai bmt ca tng bn (5). Nhngthay i thtch trong bung bm (tng hoc gim) trong vng quay do hnh dng hnh bu dcca vng cam. Trong vic mrng ca n, bung c kt ni vi ng ht thng qua mt ltrn tm bn. Do , du bht vo trong bung. Sau , trong qu trnh thu hp ca n, bung

bm chim chdu vo dng u ra thng qua mt ltrn tm bn. Bng cch ny, mi bungthc hin hai hnh trnh bm trn mi vng quay.

F.4.23

Gii thch ngn gn cu to, hot ng, iu khin sthay i lu lng ring ca bm cnh gt(hnh 4.35 ti 4.38)Trli:

Cu to my bm cnh gt chuyn bin vi mt cnh qut p lc khng cn bng c thhin trong hnh 4.35 v 4.37. Bm bao gm: hp v(1), bnh cnh qut (2), cc cnh qut (3)

vng tnh (vng cam) (4) v mt vt iu chnh (6). Vtr ca vng cam (stator) (4) c iukhin theo chiu ngang bng hai piston: mt piston iu khin nh(10) v mt ci ln (11). Vtrthng ng ca cam c xc nh bi cc c vt iu chnh (7). Cc bung bm (8) c hnhthnh bi cc cnh qut (3), bnh cnh qut (2), vng tnh (4), v cc tng bn (iu khin) (9)m bo chc nng bm trong thi gian khi ng, vng tnh (4) c tchc vtr c lp(vtr dch chuyn ) bng l xo (5). Vng cam c t lch tm i vi cc cnh qut. Sau chuyn ng quay ca rotor (2), thtch ca cc bung bm (8) tng v lp y bi cht lng


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thng qua mt knh ht (S). Khi thtch bung t n ti a, bung (8) bngt kt ni tphang ht. Sau , khi cnh qut (2) tip tc xoay, thtch ca bung gim khi chng c kt nthng vi cc vng p lc. Du sau c di chuyn n dng thot bm qua cc knh p lc(P). Piston iu khin nh(10) kt ni thng trc vi dng p lc bm. Piston ln c np bimt l xo (5), trong t cc cam vtr c lp, lc khi ng ca my bm. Sau , khi bmhot ng v p lc thot bm c sinh ra, p lc l xo ca bung xc nh vtr ca vng tnh(vng cam) v do to nn lu lng bm.

F.4.24Gii thch ngn gn cu to, hot ng, iu khin lu lng ring ca bm pittng hng trcvi biu khin cng sut l hng s(hnh 4.39 v 4.40)Trli:


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Bm piston hng trc l loi bm c bittong t song song vi trc ca roto v ctruyn bng khp hoc hoc bng a nghing

Bm pittng hng trc (Axial piston pump): thit krt a dng, lu lng thay i ddng, thun tin cho vic iu khin tng cng sut theo ti, c dng phbin trong ccmy thulc hin nay.

Cu to- Thn my thng lm bng gang, gia cng bn tinh- Hthng a nghing: bao gm mt nghing c c thquay quanh 1 trc ngang, u

trn c thgng them bphn iu chnh gc nghin- Hpiston: bao gm nhiu piston ging nhau,

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