gt - may ep thuy luc - (142).pdf

8/14/2019 GT - May ep thuy luc - (142).pdf

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Gio trnh c khMy p thy

lc


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Chng 1

Cc khi nim c bn

1.1. Nguyn l hot ng v phn loi

My p thu lc l mt my cng c s dng ngun lc l h thng thulc, da trn nguyn l nh lut Pascal. Nu ta c 2 xilanh-pittng c ni vi

nhau bng ng dn, nhhnh 1-1a, bn trong cha y cht lng. Di tc dng

ca ngoi lc ln pittong 1,P1, s to ra mt p sut trong cht lngp, gi l p

sut thu tnh. Theo nh lut Pascal, p lc pc truyn cho ton b khi cht

lng nm trong 2 xilanh v lun c hng vung gc vi mi thnh ng. p sut

cht lng c to ra c gi tr1

1

f

Pp = . Nhvy, do p sut cht lng lun c

chiu vung gc vi pittng ln 2, nn chng to ra p lc tc dng ln pittng 2

c gi tr P2= p.f2. Chnh lc ny s to ra cng nng bin dng vt liu 3.T , ta c:

1

2

12

f

fPP =

(1.1)

C ngha l, lc P2 lun bng tch ca lc P1vi t s gia din tch f2ca

pittng 2 trn din tch f1ca pittng 1. Nhvy, t s f2/f1cng ln, p lc dng

gia cng vt liu cng ln.

Hnh 1-1.My p thu lc

a. nguyn l hot ng; b. s kt cu; c. s my p c dm di ng

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Theo hnh 1.1.b, kt cu my p thu lc gm cc cm chnh sau:

Thn khung my;

H thng thu lc;

H thng iu khin.

Nguyn l hot ng ca my p: Xi lanh cng tc 4 c c nh trn dm

ngang trn 6 v lin kt vi dm ngang c nh di 9 qua cc tr dn hng 7,

to thnh thn khung my. Pittng 5 chuyn ng trong xi lanh 4, c gn vidm di ng 8, c trt theo tr dn hng. Trn dm di ng c bn my trn

vi cc rnh lp bulng lp khun trn. DM di ng c chuyn ng i

xung nh pittng cng tc v chuyn ng i ln nh pittng tr v 11. Trn

dm c nh di c lp bn my (di) dng lp khun di. Do s dng

ngun cht lng p sut cao, nn gia xi lanh v pittng thng dng cc loi

m kn (giong) trnh r r lm gim p lc cht lng.

Cc thng s c bn ca my p thu lc:

Lc p nh mc di tc ng ca p sut ti a ca cht lng gi l lc p

danh ngha PH, c xc nh bng tch s gia p sut danh ngha ca khi chtlngpvi din tch tit din ngang ca pittng cng tcf.

Chiu cao kn ca my H - khong cch gia hai bn my.

Khong lm vic S qung ng dm di ng c kh nng trt t do.

Tc dm di ng i xung l tc khng ti, thng s dng tc

nhanh.

Tc p l tc khi p to hnh bin dng kim loi, thng chm.

Tc tr v ca bn my l tc khng ti, ph thuc xilanh-pittng i

ln.

Kch thc bn ca dm ngang dng lp bn my.

Kch thc bn my di, dng lp khun.

My p thu lc rt a dng. Vi mt cm to lc bm - pittng- xi lanh c

th lp thnh nhiu dng my khc nhau, phc v cc dng cng ngh khc nhau.

Tu theo chc nng cng ngh, my p thu lc c phn thnh my p

gia cng vt liu kim loi (hnh 1-2.a), my p gia cng vt liu phi kim loi

(hnh 1-.2.b) v cc my cng dng khc. Trong ti liu ny ch yu gii thiu

my p kim loi. My p kim loi c chia thnh 5 nhm: my p rn t do

dp th tch, my p chy kim loi, my p dp tm, my p dng trong lp rp

v my p p kim loi ph thi. Cng vi s pht trin ca cng ngh gia cng p

lc, cc dng my mi dn xut hin nhm p ng yu cu ca tng cng nghring bit.



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Hnh 1-2.Phn loi my p theo chc nng cng ngh

a. my gia cng kim loi; b. my gia cng phi kim loi

My p thu lc gia cng kim loi

Rn

Myctphi

Bngui

Dp

Ctbavia

Dn

i-Dt

inh

t

Chut

Thanh-ng

Thanhnhhnh

pchyngui

M rn v d th tch M ch

My p thu l c dng gia cng v t li u phi kim lo i

My giacng ccbt phi

kim loi

My giacng cht

do

My pcc tmphi kim

loi

My tng gia

cng nngcht do

My giacng bng

pp dng incc

My ngvin v

ng bnh

My dp tm My nn sa v lp

rp

My x l ph

liu kim loi

M

yctvtliutm

Dp

tmtcdngn,kp

D

bncaosu

Unv

m

Kovdpdn

D

tmd

U

nvgpmptmdy

Nnsa,tinhchnh

plprp

Chn

ngbnh

nggi

pbtkimloi



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My p nhm th nht l my p dng rn t do v dp th tch. My

dng rn t do v rn trong khun n gin c lc p danh ngha PH= 5 120

MN(500 12000 T). My p dp th tch dng dp nng cc chi tit lm t

thp hoc hp kim nhm v hp kim magi, PH= 10 700 MN(1000 70000 T).

My p t l, dng t l su phi thp trng thi nng trong khun kn, PH

= 1,5 30 MN(150 3000 T). My p chut ko cc phi rn bng thp, PH= 0,75 15 MN(75 1500 T).

Nhm th hai gm cc my p dng p chy hay p n cc sn phm

dng ng - thanh nh hnh t hp kim mu v thp, c p lc PH = 0,4 120 MN.

Nhm th 3 bao gm: my p dp tm tc ng n, ch c xilanh cng

tc p vi PH = 0,5 10 MN (50 - 1000 T). My p vut su cc chi tit hnh tr,

vi tc ng kp c xilanh cng tc to lc p v xilanh to lc p bin,

PH = 0,3 40 MN (30 - 4000 T). My p cao su PH = 10 200 MN (1000 - 20000

T). My p gp mp, to mt bch, un v dp cc vt liu dng tm dy, PH = 3

45 MN (300 - 4500 T). My p lc, un lc cc cc vt liu dng tm dy trng thi nng, PH = 3 200 MN (300 - 20000 T).

Nhm th 5 thuc cc loi my p ng gi v ng bnh, dng p ph

liu dng nhphoi kim loi v cc ph liu kim loi, PH = 1 6 MN (100 - 600

T). My p vt liu phi kim loi bao gm: my p vt liu bt, my p cht do

v my p p cc dng tm, phin.

Ngy nay nhiu dng my mi xut hin: my p vt liu bn lng kiu

ng, c h thng kp cht khun bng c kh, nhng xilanh p vt liu bn lng

c thit k p 2 n 3 cp p lc. Cc my p chy chi tit dng cc di c bn

my di ng, my p un cc profin dng un v tu thu...Tnh cng ngh ca my p thy lc ph thuc kt cu ca thn my (kiu

ct, kiu hai tr, kiu mt tr v kiu chuyn dng) v kiu dng v s lng xi

lanh (pittng, pittng nhiu bc...). My bn tr c nh c s dng rng ri,

cc dm ng di chuyn theo phng thng ng (hnh 1-b). i khi khung my

c lm theo kiu chuyn ng (hnh 1-c). Cc my p n cc chi tit dng

thanh thng c kt cu dng nm ngang gim chiu cao nh xng.

Trn hnh 1-3 trnh by cc dng xi lanh ca my p, xi lanh kiu tr, kiu

pittng tr nhiu bc l loi xi lanh tc dng n. Xi lanh cng tc kiu pittng

nhiu bc

c s dng trong tr

ng hp khi phn d

i pittng i qua xi lanhcng tc (v d: my p thanh - ng).



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Cc xi lanh kiu pittng c s

dng rng ri khi dng du nhn lm

cht lng cng tc. Trong trng hp

ny, chi tit bt kn cho pittng thng

dng dng vng gng (xcmng).

Xilanh kiu pittng l xilanh tc dng

hai chiu, c th to lc nn gia cngkhi p sut cht lng tc dng mt

trn, v c th tr v khi p sut cht

lng tc ng pha di pittng. Trong

my p c xilanh cng tc t pha

di khung my, c th khng c

xilanh y v, trong trng hp ny,

xilanh chuyn ng tr v nh trng

lng ca phn chuyn ng ca my

p. Xilanh cng tc c ni vi thngcha cht lng qua cc ng dn.

1.2. truyn dn thu lc v thit b thu lc ca my p

Cc thnh phn ca h thng my p thu lc bao gm: my p, b phn

truyn dn, phn thu hi cht lng, cc thng cha, cc b phn iu khin - b

phn phi, cc van, cc ng ng ni, ct ni lin kt tt c cc phn t k

trn thnh mt h thng thng nht, h thng in v b iu khin.

Ngun cung cp cht lng p sut cao cho my p quyt nh dng dn

ng ca my p. H thng cung cp cht lng quyt nh s thu lc v tcng ca my p (hnh 1-4).

C 2 dng cht lng s dng trong my p thu lc l du khong v nh

tng.

Cc dng dn ng gm dn ng kiu dng bm v dn ng kiu dng b

tng p. Trong dng dn ng dng bm c phn lm 2 loi, c dng bnh tch

p v khng dng bnh tch p. Khi dng loi dn ng khng c bnh tch p,

ngun cp cht lng p sut cao cho my p ch thc hin t cc bm.

H thng dn ng c bnh tch p l h thng cht lng cng tc c cp

ng thi t bnh tch p v t bm hnh trnh cng tc.

Hnh 1-3. Cc dng xilanh ca my pthu lc

a. kiu Plud; b. kiu nhiu bcc. kiu pittng



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i vi cc h thng dn ng kiu tng p, cht lng c cp cho my p

trong hnh trnh cng tc nh b tng p, cht lng cng tc c cp theo tng

lng nht nh. B tng p l bm xilanh p sut cao.

Hnh 1-4.Phn loi trm p thu lc

Cht lng cng tc cng l mt c trng ca my p thu lc, chng quyt

nh c im kt cu ca my p. Du khong l dng cht lng cng tc dng

trong my p thu lc c p lc khng ln. Chng c nhiu u im nh nht

cao, h s bin dng th tch nh, khng gy n mn chi tit. Nhng du khong

t. Ngc li, dng nh tng lm cht lng cng tc c gi thnh h thngdng cho cc my p thu lc c lc danh ngha ln, lng cht lng ln, tnh

kinh t tt.

Khi s dng kiu dn ng kiu bm c bnh tch p, bnh tch p c nhim

v tch tr nng lng trong ton b chu trnh cng tc ca my p thc hin

hnh trnh cng tc. Nh , lm u ti cho bm v ng c in. Nh c im

ca dn ng kiu bm c bnh tch p l nng lng tiu hao ph thuc vo tr

lc bin dng ca phi v dung lng ca bnh tch p.

i vi loi dn ng kiu bm khng c bnh tch p, th cng sut nh

mc ca ng c v bm c xc nh thng qua cng sut ln nht do my p

to ra. B dn ng s tiu th nng lng to ra cng c ch ca my p.



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S dn ng t b tng p dng cho hi hoc kh nn, nng lng tiu th

cng khng ph thuc vo tr lc bin dng ca phi. N c th m bo thc

hin c mt s ln cc hnh trnh ngn v lp li. Dn ng t b tng p c

kh s m bo tiu th nng lng khng ph thuc vo cng thc hin, n cng

m bo s lng ln cc hnh trnh lp li v lng bin dng ng u ca u

ba vo phi kim loi.

1.3. Cht lng cng tc v p sut s dng

Trong cc my p thu lc, cht lng cng tc thng dng l nh tng

hoc du khong. trnh g cho cc chi tit nhxilanh, pittng, cc chi tit iu

khin v ng ng, nc c cho thm 2 3 % cht nh tng. Thnh phn

ca cht nh tng bao gm: 83 87% du khong, 12 14% axit oleic v 2,5%

xt nng 40%. Du khong thng c dng l du my, du cng nghip,

du tuabin...

Bng 1-1

Quan h c tnh cc b phn ca my p thu lc v cht lngc tnh ca cc b phn ph thuc vo cht lng

cng tcB phn my p thulc

Nc - Nh tng Du khongm kn cc pittngng knh t 60-70mm khi p sut cao

Vng bt kn hoc xcmng

Khe h ng knh giapittng v xilanh l nh

B phn phi chtlng p sut cao Kiu van

Kiu van trt, c th sdng cc van loi khc

Bm Tc chm c kchthc tng i ln Tc nhanh v kch th

cnh

m kn ca thit bthu lc

Kiu mmR kn b mt hoc dngvng xcmng

Xi lanh Kiu pittngKiu pittng vi xi lanhng knh ln

Bnh tch pKhng c cc phn tngn cch v c ccphn t ny

Ch c cc phn t phncch gia du v hi

Cc tnh cht c bn ca cht lng cng tc l tnh chu nn v nht. H

s nn th tch i vi nc (nh tng) 5.10-6 cm2/N, vi du khong

l 6.10-6cm2/N, tng ng vi vng p sut lm vic ca cc my p. p sut chtlng cng ln, th h s nn th tch cng gim.



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nht ca du khong dng trong cc my p thu lc vo

khong 1,5 60E (BY), nhit khong 500C v p sut kh quyn. Trong khi

nht ca nh tng khng chu nh hng ca p sut, th nht ca du

thay i t ngt khi p sut du tng. Khi p sut ca du tng n gn 30MPa,

nht ca du tng ln gp i. Nhvy, cn phi tnh n quan h gia p sut

cht lng v nht trong cc kt cu c th tch cht lng ln v chuyn ng

vi p sut cao, nhtrong my p rn. Nhit bc chy ca hi du dao ng

trong khong t 160 2100C, v vy, cn phi ch khi rn p cc phi nng.

Cc loi du c nht nh, th nhit t bc chy cng thp.

Nhvy, cht lng cng tc s dng trong h thng thu lc ng vai tr

quan trng c tnh quyt nh i vi cc c im kt cu ca c cu dn ng,

h thng iu khin v ca c my p (bng 1-1).

p sut danh ngha ca cht lng cng tc trong my thu lc c tiuchun ho theo OCT 356 - 80. Cc p sut thng thng l 20, 30 v 40 MPa.

1.4. Chu trnh cng tcChu trnh cng tc l thi gian cc bc trong qu trnh gia cng bin dng.

Thi gian ca chu trnh ca my p dng tng qut c th biu din nhsau:

chkhgacgctTaKtTcht ttttttttT +++++++= (1.2)

trong :

tT- thi gian dm ngang treo, l thi gian tin hnh cp phi hoc lp dng

c,

tKt- thi gian chy khng ti, l thi gian dm ng dch chuyn xung n

lc dng c tip xc vi phi,

tTa- thi gian tng p sut cc xi lanh cng tc,tgct - thi gian hnh trnh cng tc, l thi gian tin hnh gia cng theo cng

ngh cn thit,

tc- thi gian nn p chi tit di p lc,

tga- thi gian gim p sut cc xi lanh cng tc,

tkh- thi gian hnh trnh tr v ca dm ngang,

tch- thi gian chuyn v tr ca c cu iu khin.

Cc my p c cc cng dng khc nhau, thi gian chu trnh c th khc

nhau, do s lng thnh phn v gi tr ca cc thnh phn cng khc nhau.

Th d, nh khi l phi trn my p rn, khng c cc thnh phn tT

v tgtc.



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Gi tr ca tng chu k ring bit do dng dn ng quyt nh. Th d, nu

Tchtvt qu tcrt nhiu, nn s dng kiu dn ng bng bm c bnh tch p,

cho nn ta c:

Kt

KtKt

v

St = ;

ct

ctct

v

St = ;

Kh

KhKh

v

St =

trong :

SKt, Sct, SKh- hnh trnh khng ti, hnh trnh cng tc, v hnh trnh tr v,

vKt, vct, vKh- tc trung bnh ca hnh trnh khng ti, ca hnh trnh cng

tc v ca hnh tnh tr v.

Tc chuyn ng ca dm ngang ca cc my p hin i c trnh by

bng 1-2.

Bng 1-2

c tnh tc ca my p thu lc (mm/s)

Lc p (MN)

>5


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+ Cc xi lanh cng tc c ni vi thng cht lng p sut cao, cn xi lanh

y v c ni vi thng cha,

+ Cc xi lanh cng tc v xi lanh kh hi c ni vi nhau v vi ngun

cht lng p sut cao,

+ Cc xi lanh cng tc c ni vi ngun cht lng p sut cao, cn cc xi

lanh y v c ni vi bnh tch p.

Cc bin php thao tc hnh trnh y v:

+ Cc xi lanh cng tc c ni vi thng cha, cn cc xi lanh y v c

ni vi ngun cht lng p sut cao,

+ Khi xi lanh cng tc c ni vi thng cha, cc phn chuyn ng i

xung c thc hin di tc ng ca chnh trng lng ca n. Trong trng

hp ny nn b tr cc xi lanh tr v trn my p v hnh trnh i xung thc hin

di tc ng ca cht lng p sut cao.



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Chng 2

my p thy lc dn ng kiu bm khngc bnh tch p

2.1. Chc nng v hot ng ca cc cm chi tit

Hnh 2-1 biu din s my p dn ng bng nh tng nc kiu bm

khng c bnh tch p. Nguyn l hot ng nhsau:

Dm 1 gn vi pittng dng

to lc p khi gia cng kim loi.

Van 2 m cp cht lng cng tc

p sut thp, lu lng ln, cho my

p khi hnh trnh khng ti, bo m

tc chuyn dch ca dm ngangnhanh. Khi hnh trnh cng tc, van

2 ngn cch bnh cha vi ng

cht lng p sut cao.

ng c thy lc chp hnh 3 t

ng nng van cp 2 v m bo

cho cht lng t xilanh cng tc tr

v thng cha, khi dm ngang

hnh trnh y v. Thng cha 4

cung cp cht lng cho my p, khi

hnh trnh khng ti, p sut

khng kh trong bnh thng bng

0,4 0,8 MPa (4 - 8 kG/cm2). B

van t ng 5, b trit ti, s chuyn

bm sang lm vic ch khng

ti sau khi t p sut nh.

Van tun hon 6, van 1 chiu 7, bm pittng 8 dng cp cht lng cng

tc p sut cao trong hnh trnh cng tc. B phn phi 9 c chc nng iu khin

my p. Van mt chiu 10 ca cc xilanh y v dng cp v x cht lng

cng tc ra khi xilanh. Van 11 l van cp, cn van 12 l van x ca xilanh cngtc.

Hnh 2-1.S my p dn ng bngbm nc khng c bnh tch p



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pha di bn tri hnh 2-1, trnh by biu pha ng m cc van cab phn phi. Trn trc ng biu th hnh trnh ca van trn van, trn trc

ngang biu th gc quay ca cn iu khin ca b phn phi.

Dng cht lng cng tc tng ng vi cc v tr cc van trong b iu

khin:

- v tr III ("Dng") cc van 11 v 12 c m, cn van 10 ng. Cht

lng t bm qua b van t ng trit ti 5 n b phn phi, qua van 11 v van 12

n ng dn (ng ht) v bm. Nhvy, dm ca my p, khng chu p

lc ca cht lng nn gi nguyn v tr do van 10 ng.

- v tr II (hnh trnh khng ti) cc van 10 v 12 m. Cht lng t thng

cha qua van cp 2 t ng nng ln, chy ti xilanh cng tc, dm ngang i

xung. Di tc dng ca trng lng cc phn chuyn ng v p sut cht lng

t thng cha ln pittng cng tc, cht lng t xilanh y v c nn v chy

qua van 10 v thng cha v xilanh cng tc.

- v tr I (hnh trnh cng tc) cc van 10 v 11 m. Cht lng t bm qua

van 11 v phn trn ca v van 2 ti xilanh cng tc. T xilanh y v, cht

lng c nn qua cc van 10 v 11 ang m ti xilanh cng tc ca my p.

Van in y cng bng p sut cht lng cp t bm ti v lc ca b tr

du 3 khng m van. Khi p sut trn ng cng tc t tr s t trc

b t ng trit ti th van tun hon s nng ln v bm s bt u lm vic

khng ti. My p thc hin vic p phi rn.

- v tr IV (hnh trnh y v) van 12 m, cht lng di p sut t bm s

m van 10 v in y xilanh y v, cn t xilanh cng tc cht lng c y

qua van 12 c tit din nh ti thng cha. p sut trong xilanh cng tc gim

xung, b tr dn 3 nng van cp ln v cht lng t xilanh cng tc qua van cp

s c nn t do tr v thng cha, dm ngang ng c nng ln.

Khi thit k b phn phi ca my p c dn ng kiu bm khng c bnh

tch p, cn ch bt k v tr no ca cn iu khin th tt c cc van khng

c ng ng thi. Nu ng ng thi, c th dn ti va p thy lc trn

ng ng v lm cho b t ng chu ti lm vic t ngt v gy hng hc.

2.2. Cc loi bm

Trong h thng truyn ng ca my p thy lc, bm tc dng tnh c

s dng rng ri nht. Cc bm ny, theo loi cht lng s dng, c chia ra loibm nh tng v bm nc.



14/142


Trong cc loi bm nc (bm nh tng nc) dn ng my p thy

lc, ngi ta thng s dng bm kiu pittng c trc khuu, b tr nm ngang,

c cng sut ti 1500kW.

p sut cht lng cng tcthng dng l 20 hoc

32MPa (200 hoc 320

kG/cm2). Cc bm c tc

ng n v tc dng kp.

S ca bm 1 pittng

tc dng n c trnh

by trn hnh 2-2.

K hiu f l din

tch ca pittng, v l vn

tc, q l lng cp ca mt

pittng, r l bn knh

khuu, l gc quay catrc khuu v l vn tc gc quay ca trc khuu.

n gin ha cho cc tnh ton, ta coi chiu di thanh truyn bng v

cng. Khi :

vfq .= v sinrv= , vy ta c sinfrq=

Hnh 2-3.Biu lu lng ca bm

a. Mt pittng; b. Hai pittng; c. Ba pittng

Hnh 2-2.S bm mt pittng tc dng ngin

1. trc khuu; 2. tay bin; 3. con trt;4. pittng; 5. m kn; 6. van y; 7. thng cha;8. ng ng p sut cao; 9. van ht; 10. v bm



15/142


C ngha l, lu lng cht lng cng tc l mt hm Sin ca gc quay ca

trc khuu. Nhvy, trong trng hp bm 1 xilanh, lu lng cht lng khng

ng u theo thi gian. iu nh hng n qu trnh nn p kim loi bin

dng.

Trn hnh 2-3 trnh by biu lu lng ca bm ph thuc vo s

pittng, lu lng cc i c coi bng mt n v. Bm ba pittng tc dng

n, trn trc c ba khuu c b tr lch nhau 1200, cho lu lng v p sutcht lng tng i u theo thi gian.

Bm 5 pittng khng c s dng, v khi kt cu bm phc tp, chiu

rng bm tng, s ng u ca lu lng t c ci thin.

Cc van t ng dng phn phi cht lng trong h thng thu lc. Khi

trc khuu t mt gi tr vng quay nht nh, c gi l s vng quay ti hn,

van lm vic km theo ting g khi van ng m, lu lng cht lng tr nn

khng u do s ng v m ca van khng tng ng vi hnh trnh pittng.

Tc trung bnh vcca pittng, theo cc s liu vn hnh ca bm, thng

ly bng 0,5 1,5 m/s, nhng khng qu 3m/s, v, nu khng s xy ra s mimn nhanh cc m kn ca pittng v xut hin ting g ca van.

30

n.SvC = (2.1)

trong :

S - hnh trnh ca pittng (m);

n - s vng quay ca trc khuu trong mt pht, gii hn do ting g

ca van, thng ly n = 125 180 v/ph.

Lu lng thc t, hoc c ch ca bm, m3/s

061 = fsznQe (2.2)

trong :

f - din tch nh pittng, m2,

z - s pittng,

n s vng quay ca trc khuu, v/ph;

0- h s tn hao th tch ca bm.

Cng sut trn trc khuu ca bm c xc nh theo cng thc:

M

epQN

=0

1000 (2.3)

trong :

M- h s tn hao c kh ca bm, khong 0,80 0,85,



16/142


p - p sut ca cht lng do bm to ra, MPa.

Tc trung bnh ca cht lng trn ng ng ht khng c qu 0,3m/s.

Mc nc cao nht ca nc trong thng cha phi cao hn trc ca pittng

khong 4 - 5m. Tc trung bnh ca nc ti cc van ht thng khng qu

3m/s v cc van y khng qu 6m/s.

Cc chi tit chnh ca bm (hnh 2-4) l thn v c trc khuu v b

dn hng, trc khuu, thanh truyn, con trt, pittng, thn van v cc van.Thn v bm 1 c c bng gang, c khi lng ln gim rung cho

bm do tc ng ca cc khi lng chuyn ng tnh tin. Thn v c tnh

ton kh nng chu ko v chu un di tc dng ca hp lc ca tt c cc

pittng. i vi bm ba pittng, hp lc ca cc pittng thng ly l bng 2.f.p

(f- din tch nh pittng, p - p sut nh mc ca bm).

ng sut tng hp cho php do un v ko khi tnh ton thng ly

bng [] = 7 8 MPa.

Hnh 2-4.Bm kiu ngang ba pittng

Trc khuu 2 c ch to bng thp 45 v qua cng ngh rn. ca trc

khuu thng lm theo kiu trt hoc ln. Trc khuu thng b tr hai hoc

bn . bn ca trc khuu v p sut c tnh ton theo hai phng

php trnh by trong phn my p trc khuu.

Thanh truyn 3 v con trt 4 c ch to bng cng ngh rn hoc c.

Chiu di thanh truyn thng ly L 5r (r bn knh khuu). B hng con

trt c dng hnh tr hoc phng. cc b dn hng hnh tr, nu b mn

c th thay cc ng lt bng gang. i vi cc b dn hng kiu phng th v tr

ca con trt c iu chnh bng cc m. p lc n v cho php trn b dn



17/142


hng thng ly l 0,2 0,3 MPa. i vi cht ca con trt, p lc n v cho

php 9MPa, cn i vi m khuu cho php p lc n v 6 7 MPa.

Pittng 5 ca bm c ch to bng thp c bn cao v tnh chng g tt

20X13 v 3X13. B mt pittng c ti v gia cng mi bng. Lin kt ca

pitng vi con trt thng l theo kiu t do.

bt kn pittng ngi ta dng vng m trn nn vi cao su. m bo

thm t bng nc hoc nh tng cho m, chiu cao ca m phi nh hn

hnh trnh ca pittng.

Thn van 6 c rn t thp 25 hoc 35.

Cc van ht v van y c b tr cnh nhau hoc xp thnh dy theo chiu

ng. Cc van, van, l xo v chi tit lin kt c ch to t thp khng g

hoc brnz (ng thanh) cht lng cao. Van 7 thng c ch to c 8 dng

cn hng theo phn hnh tr ca tit din lu thng.

Hnh 2-5.Bm pittng kiu hng knh

Khi xc nh tit din lu thng ca cc van, thng xut pht t tr s

nng ca van (4 - 5mm). Cc bm nc c th s dng bm du, nhng khng

hp l v tnh cht ca du cho php thc hin kt cu ca bm du n gin hn

v gn hn.

Cc bm pittng lm vic vi du khong, c phn lm 2 loi: bm

pittng kiu hng knh v kiu chiu trc.

S bm c cc pittng b tr kiu hng knh c trnh by hnh 2-5.a, kt cu ca bm c trnh by hnh 2-5.b. Rto 1 ca bm quay quanh

trc phn phi c nh 3. Cc pittng 2 c b tr kiu hng knh v chuyn



18/142


ng tnh tin. Cc pittng c t trn con ln 5 v ln theo vng nm trong bloc

di ng 6. Trc phn phi c cc van 4 ht v y du. v tr c trnh by

trn hnh 2.5.a, ht du c thc hin qua na pha trn ca trc phn phi, cn

vic y du - qua na pha di. Khi dch chuyn bloc di ng, c th thay i

lu lng v hng ca du. Cc khe h gia pittng v cc l tng ng khng

c qu 40m (i vi pittng ng knh ti 40mm).

p sut ca cc bm pittng kiu hng knh c ch to t ti 20MPa.

S r r gia trc phn phi v rto, s un trc do ti trng tc ng lm hn ch

vic tng p sut. ng thi c cu phn phi cn hn ch s vng quay ca rto

bm pittng kiu hng knh, khng c vt qu 1000v/ph. Thng thng

tng lu lng, ngi ta s dng bm nhiu xilanh.

Th tch du do pittng y ra sau 1 vng quay ca rto l (m 3/ph):

hzd

q4

2

=

trong :

d - ng knh pittng, (m),

h - hnh trnh pittng, (m),

z - s pittng, thng l s l.

T hnh 2-5.a ta thy, hnh trnh ca pittng bng 2e, vy lu lng l

thuyt ca bm trong 1 pht l (m3/s):

eznd120

n.qQ 2

== (2.4)

trong : e - lch tm ca trc khuu (m).

Lu lng thc t hay c ch l:

QQe 0= trong : 0- h s tn hao th tch ca bm, ly bng 0,8 0,95.

Cng sut trn trc rto ca bm (kW):

MpQ

N e

=

0

1000 (2.5)

trong :

p - p sut, (MPa),

M- h s tn hao c kh ca bm, thng ly bng 0,94 0,96.

Cc bm thng c ch to theo tiu chun lu lng nh hn 1000

lt/ph. Kt cu ca bm pittng kiu hng knh c pittng dch chuyn t do

c trnh by hnh 2-5.b.



19/142


C th thay i lu lng ca bm pittng kiu hng knh bng cch thay

i v tr ca bloc bm so vi roto nh c cu trc vt.

Cc bm c iu chnh t ng lu lng theo ng cong p sut cng tc

cng c s dng rng ri (hnh 2-6). Di tc dng ca l xo 2, bloc di ng

ca bm 1 s dch chuyn trong rnh dn hng 4 ti vt chn 5, vt dng iu

chnh lu lng cc i ca bm.

Khi p sut trong ng ra 7 tng, mt phn cht lng i v tc ng lnpittng 6, lm dch chuyn bloc bm sang phi. Lu lng ca bm s gi khng

i, khi p sut nh hn pH, tng ng vi lc gi ca l xo do vt 3 iu chnh.

Khi p sut nh hn PH, cng sut ca bm s tng t l thun vi p sut trong

ng cng tc. Khi p sut ln hn pH, lu lng ca bm bt u gim theo qui

lut tuyn tnh, do cng ca l xo quyt nh.

Vic s dng nhiu l xo lp ni tip s cho php nhn c cc quy lut

ph thuc khc nhau gia lu lng v p sut cht lng.

C cu c trnh by trn hnh 2-6 m bo gi p sut nh trc ch

lu lng nh nht v h thng lm vic khng c va p khi c thay i tit ngt cui hnh trnh cng tc.

Hnh 2-6.Bm c iu chnh t ng lu lng theo p sut

gi cho cng sut ca bm lun n nh, cn phi thay i lu lng

ph hp vi p sutptheo quy lut:

p.

K

NQ

1= (2.6)

trong :

N - cng sut ca bm,

K - hng s ph thuc vo th nguyn ca p v Q.

Quan h gia lu lng v p sut ca bm c th xc nh nh iu khinlc l xo kt hp vi c cu cam. Cc thng s ca bm m bo theo iu kin

ca biu thc (2.6) c gi l bm cng sut n nh.



20/142


Bm cng sut n nh s gi nguyn tr s cng sut tiu th, bt u t p

sut nh, cho ti p sut gii hn ln nht.

Bm pittng kiu chiu trc gm c a phn phi di ng 1 (hnh 2-7.a),

bloc xilanh quay 2, cc pittng 3, cn pittng 4, a nghing 6 lin kt vi cc cn

4 theo kiu bn l.

a phn phi 1 c cc ca hnh vng cung 7, du c pittng ht v yi qua cc ca ny. Bloc xilanh 2 v a nghing 6 quay c l nh trc 5.

Lng du c mt pittng cp sau 1 vng quay ca a nghing l:

hd

V4

2=

trong :

d - ng knh nh pittng,

h - hnh trnh pittng.

Nu s lng pittng l z, sau mt vng quay ca bloc xilianh, th tch du

c cp l:

hzd

V4

2=

Lu lng l thuyt trung bnh

hznd

Q4

2= (2.7)

trong :

n- s vng quay trong mt pht,

h = Dsin.

Khi

= sinznDd

Q4

2 (2.8)

Cc bm kiu chiu trc c 2 loi: dng lu lng c nh v v dng lu

lng thay i, h s hu ch ca chng cao hn so vi bm pittng kiu hng

knh. Do r r t hn cc c cu phn phi, v khi cht lng phn b pha u

mt th khe h gia a phn phi v bloc xilanh s thng xuyn c t ng

loi b. Cc bm kiu chiu trc c ch to vi p sut cng tc l 20 - 35MPa.

Cc bm pittng kiu van bm du, c ch to pittng b tr theo kiu

hng knh v kiu mt hng. Nh c kn kht cao ca b phn phi kiu van,

cc bm kiu ny c dng trong trng hp cn du p sut cao trn 30MPa,40MPa v cao hn. Cc bm ny c ch to dng c cng sut khng i,



21/142


nhng cng c kt cu bm vi cng sut c th iu chnh. Tc quay trc

lch tm thng l 1500v/ph.

Bm c cc kt cu c bn khc nhau v lch tm ca trc khuu - thanh

truyn, c con trt dng pittng v b dn hng dng xilanh.

Hnh 2-7.S bm

Trn hnh 2-7.b trnh by mt ngn ca bm kiu van lch tm c pittng b

tr mt hng. Bnh lch tm 3 tc dng ln pittng 2 qua kiu ln v y

pittng ln, khi cht lng c nn qua van y 5 v i ra.

Di tc dng ca l xo 6 Pittng h xung, ng thi vic ht cht lngqua van 1 vo bung cha. Trc lch tm v cc pittng c b tr trong thn

van 4.

Lu lng c ch ca bm l: (m3/ph)

eznd

Q e2

2

0

= (2.8)

trong : d - ng knh pittng (m),

e - lch tm (m),

n - s vng quay ca trc lch tm (v/ph),

0- h s tn hao th tch, vi p sut 30MPa 0 = 0,9.Khe h gia xilanh v pittng c ng knh 20 - 30 mm thng dng

l 15 30m.



22/142


2.3. Cng sut ca bm v ng c ca my p thy lcCng ca my p c xc nh trong thi gian tcthc hin hnh trnh cng

tc, khi bin dng to hnh kim loi. Gi thit trong h thng dn ng khng c

tn hao nng lng.

Ta a vo cc k hiu sau:

P - lc ca my p ti im cho trc ca hnh trnh pittng,

p - p sut cht lng trong xilanh my p,

S - hnh trnh ca pittng,

Pdn- lc p danh ngha ca my p,

Sc- hnh trnh cng tc.

Nu b qua cc tn hao trong h thng thy lc, th quan h gia cc cng

sut ca b dn ng bm khng c bnh tch p khng c bnh trn trc dn

ng bm, thi im bt k ca hnh trnh cng tc tun theo ng thc:

dcbp NNN == (2.9)

trong :

Np- cng sut ca my p hnh trnh cng tc,Nb- cng sut ca bm,

Nc- cng sut ca ng c in.

Khi my nn lm vic, c thi im Nbt gi tr cc i. T biu thc (2.9)

suy ra, cng sut ca bm cng c gi tr cc i. Ngha l bm phi c tnh

theo cng sut cc i ca my p v c xc nh bng lc ln nht v tc

cho trc ca chuyn ng con trt.

V thi gian hnh trnh cng tc l nh nht khi s dng ton b cng sut

ca bm, nn nhn c tCnh nht, bm phi lm vic vi cng sut nh

mc trong sut ton b hnh trnh cng tc:bNKpQ = (2.10)

trong :

Q - lu lng ca bm,

p - p sut ca bm to ra,

K - h s ph thuc vo th nguyn ca Q, p v N.

Gi bm l l tng, nu m bo c iu kin (2.10) trong sut hnh

trnh cng tc v c h s c ch bng 1. cc iu kin thc t th cc bm ca

my p thy lc khng lm vic vi cng sut khng i, c bit thi im bt

u hnh trnh cng tc v i vi nhiu qu trnh cng ngh khng cn c psut cao m ch cn nng sut cao.



23/142


Hnh 2-8a biu din th lc p vi s n gin nht ca my p mt

xilanh dn ng khng c bnh tch p t bm c lu lng khng i. Phn din

tch c gch Oab t l vi cng khng c bm s dng v c trng cho vic

s dng cng sut ca bm.

Hnh 2-8. th lc p

Tr s ca tung pi vi im a s tng ng vi p sut bm khng s

dng ti thi im v t l vi phn cng sut khng s dng ca bm.

QpN ''= , Q= const.

Trng hp bm ckt cu n gin nht, b dn ng ca bm khng cbnh tch p thng ch s dng mt phn nh cng sut nh mc ca my.

th lc dn ng t bm l tng c trnh by hnh 2-8.b. ycng sut ca bm c s dng ht. Mc hon thin ca cc dn ng thc t

cn c nh gi bng cch so snh vi dn ng t bm l tng, lm vic vi

cng sut khng i.

Trn hnh 2-8.c trnh by th lc p khi bm kiu pittng c trc khuu

lm vic vi ba mc p sut v lu lng cp cho my p mt xilanh:

bccbbaa NQpQpQp === (2.11)

Cc im a, b, c ca th lc l cc im cng sut khng i.

pc > pb > pa v Qc< Qb< Qa

Phn cng bm khng dng c gim i, nn thi gian hnh trnh cngtc, cng sut ca bm v ng c cng c th gim, trong trng hp dn ng

t bm lu lng khng i.



24/142


Trn hnh 2-8d trnh by th lc dn ng my p mt xilanh t hai bm

c c tnh khc nhau v mt ng c in lm vic. Cc bm c chn t iu

kin:

P1(Q1+ Q2) = p2Q2= Nb (2.12)

Bt u t p sut p1bm c ngt vi cc thng s p1v Q1. Sau my

p nhn c cht lng t bm vi cc thng s p2v Q2

Cc im a v d trn ng cong Oad (hnh 2-8.d) tng ng cng sutkhng i ca bm.

Trn hnh 2-8.e trnh by th lc ca my p mt xilanh khi lm vic t

bm cng sut khng i. Trong trng hp ny, mi quan h pQ = C t c

bng cch s dng pittng c bit chuyn dch bloc di ng ca bm pittng

hng knh, pittng cn bng nh l xo v iu chnh lng dch chuyn ca bloc

di ng nh c cu cam. n im a, bm lm vic vi cng sut khng i

nhng khng s dng ton b cng sut ca chng. Sau khi i qua im a, cng

sut c gi khng i cho n khi kt thc hnh trnh cng tc.

Vic a cc thng s lm vic ca b dn ng gn hn vi cc thng slm vic ca bm l tng trong thi gian tc thc t c thc hin bng cc

phng php sau: s dng cc bm c iu chnh kiu bc thang lu lng theo

p sut; s dng cc bm c c tnh khc nhau; s dng mt lot cc bm ging

nhau; s dng cc bm c thay i t ng cng sut; s dng my p nhiu

xilanh lm vic vi cc p sut khc nhau v c bng cch s dng t hp cc

phng php k trn.

ci thin vic s dng cng sut ca cc ng c in trong thi gian ton

b chu trnh cng tc Tchtca my p, c th a cc thng s ca b dn ng

gn cc thng s ca ng c l tng (l ng c c th cp trong sut khong

thi gian Tchtmt cng sut khng i) c tnh bng t s:cht

cht

T

A (Acht- cng

c ch ca my p).

Khi khng c bnh tch p, vic a s dn ng ti gn vi s lm vic ca

ng c l tng c th t c bng cc phng php k trn gim cng

sut ca bm. Thng thng, thi gian dnh cho cc cng on ph li nhiu hn

thi gian ca hnh trnh cng tc ca my p. Nhvy, b dn ng bt buc

phi lm vic khng ti trong thi gian di.

Cng sut ca ng c in khi bm chy khng ti thng vo khong

10 15% cng sut cc i. V vy, trng trng hp cc yu t cng ngh chophp, th hp l nht l gim mt phn tc ca hnh trnh cng tc hoc tng



25/142


t scht

p

T

t. Trong trng hp ny, thi gian ton b ca chu trnh c th thay i

khng ng k do vic gim thi gian cho hnh trnh tip cn, hnh trnh y v

v chuyn ch .

Cc qu trnh cng ngh ring, th d khi t l, s yu cu tc ca hnh

trnh cng tc lc cao, lc thp khc nhau, do cc tnh cht cng ngh ca vt liu

phi quyt nh. Trong trng hp ny, bng cch ngt mt s bm cng ng

c, c th t c cng sut ca bm v ng c in ti u.

Khi chn cng sut ca ng c in cho my p thu lc, ngi ta thng

phn bit hai ch : Ch lm vic lu di v ch lm vic ngn - lp li.

Theo ch lm vic lu di ngi ta thng chn ng c in cho dn

ng kiu bm c bnh tch p, v ngi ta thng chn ng c in theo ch

lm vic ngn - lp li cho dn ng kiu bm khng c bnh tch p. Thng c

th chn cng sut nh mc ca ng c in bng mt na cng sut ca bm.

C th gim tip theo cng sut ca ng c in dn ng khng c bnh

p bng cch t mt bnh trn trc ni gia bm v ng c in.c cu dn ng c bnh , vic thay i s vng quay v s nh nng

lng ca bnh ph thuc vo c tnh c hc ca ng c in. Trong trng

hp ny biu thc (2.9) s c dng:

Np= NbNc (2.13)

Yu t gii hn y l cng sut ca bm - cng sut ng c c th gim

2 - 3 ln v ph thuc vo c tnh ti P = f(S) v t scht

p

T

t.

Trong mt s trng hp, vic s dng trit cng sut ca ng c in

v bm c th t c bng cch lin kt nhiu my p c cng lc p vo mtmy p ln.

Khi lc p cng tc ca my p tng ng vi p sut gii hn ca cht lng

trong bm th bm c ngt ra khi my p v chuyn lm vic cho my p

khc. Phng php trn s gim hoc trit tiu ton b s lm vic khng ti ca

bm v v vy n l phng php kinh t.

La chn dn ng bm khng c bnh tch p, c th c xc nh bng

ch lc tc dng ca my p.

Cc qu trnh cng ngh trong gia cng bng p sut, theo c tnh ca ch

lc, c th chia ra lm 6 nhm chnh (hnh 2-9).



26/142


Hnh 2-9. th c trng ca cc lc cng ngh

Nhm I (cc qu trnh p chy, vut...) - lc to ra trong thi gian hnh trnh

ca pittng my p c gi gn nhkhng i, ngha l 0dS

dP. Mc in

y th lc l = 70 - 80% (c xt n lc cc i cui hnh trnh).

Nhm II (cc qu trnh chn, vut...) lc tng u theo hnh trnh ca

pittng my p, mi quan h P = f(S) gn nh tuyn tnh, ngha l

%;const

dS

dP7060= .

Nhm III (un, dp khi nng, ng bnh, dp tm, ng gi...) th lc

c th chia lm hai on: on u lc tng t t theo hnh trnh ca pittng

(dPdS const), on sau lc tng mnh, ngha l dS

dP; = 10 25%.

Nhm IV (dp vut su vt liu tm, dp bng cao su...) - Lc thay i u

theo hnh trnh ca pittng:dPdS const; - 40 70%.

Nhm V (Qu trnh cht, t...) - lc tng t t vi hnh trnh cng tc

t

ng i ngn v sau gim cn nhanh hn, dS

dP

= 0; = 25 60%.



27/142


Nhm VI (cc qu trnh dp ni, dp tinh...) - lc tng t ngt theo hnh

trnh ca pittng, ngha ldS

dPvi hnh trnh cng tc ngn, thng c

xc nh theo bin dng n hi ca my p: = 40 45%.

Cc th tnh ton lc cn biu th khng nhng cng bin dng do cho

phi, m cn c cng bin dng n hi ca cc b phn ca my p. i vi cc

tnh ton c lin quan n vic chn la b dn ng, cc th thc t, nn c

thay bng cc th c n gin ho gm cc on thng ni nhau. Vic

xt cc th lc c trng cho php a ra cc kt lun v vic s dng loi dn

ng loi ny hoc loi khc.

Th d vi cc qu trnh ca nhm

III, my p c p lc khng ln, nn khng

cn phi trang b loi dn ng kiu bm

c bnh tch p. H s thu lc khng ng

k do mc in y th lc nh.

y, khng nn s dng loi dn ng kiubm khng c bnh tch p kt cu n

gin nht (bm lu lng khng i v

my p mt xilanh). Cng sut ca b dn

ng thu lc khng c s dng mc

cn thit. V vy, nn s dng b dn ng

t cc bm c c tnh cng tc khc nhau.

Xt tnh ton cc thng s h dn

ng thu lc ca my p mt xi lanh dp nng t hai bm c c tnh khc

nhau. th tnh ton my p dp nng

c trnh by trn hnh 2-10. Theo th, nhn thy qu trnh p c phn thnh hai on, v vym trong trng hp

ny, ch cn s dng hai bm c c tnh khc nhau l hp l.

S chuyn i t phn nm ngang ca th n phn dc c biu th ti

im B (Sp - aSp; P1). K hiu Q1v Q2l lu lng ca cc bm p sut thp v

p sut cao; p1v p2l cc p sut ca cc bm cc im B v C tng ng. Ta

xc nh mi quan h gia lu lng v p sut ca cc bm chn:

bNpKQpKQ =+ 1211 (2.14)

trong :

K- h s;

Nb- cng sut ca bm l tng:

maxb pkQN 2=

Hnh 2-10. th tnh ton lc dp nng



28/142


K hiu p1/pmax= b; ta c:

1121

QQb

bQ =

=

vib

b

=1

.

C th dng cc tc trung bnh ca hnh trnh cng tc xc nh Q2.

Gi F l din tch pittng cng tc ca my p, xc nh tc trung bnh theocng thc:

p

ptb

t

Sv = (2.15)

Ch : tP= t1+ t2, t hnh 2.10 ta c:( )

21

1

Q

FbSat

p= (2.16)

2

2

Q

FaSt

p= (2.17)

Thay th cc cng thc (2.15) v (2.16) vo (2.14) ta nhn c:

( )[ ]abaFQ

vtb+

=1

2 (2.18)

Vi a = 1/4 v b = 1/4 th

vtb2,3 v2 (2.19)

m bo ch c mt tc trung bnh th lu lng ca bm p sut cao

c th ly mt cch gn ng bng mt na so vi lu lng yu cu ca mt

bm.

Cng sut ca cc bm cp hai:22

vPN dnQ = (2.20)

dp nng, nu xt phng trnh (2.18) ta c:

322 ,

vPN tbdnQ = (2.21)

2.4. Cc loi thit b thy lc lm vic vi du khong

thay i hng chuyn ng ca dng du trong h thng thu lc ca

my p ngi ta s dng cc b phn phi kiu van trt (hnh 2-11). Khe hgia pittng v xilanh ca van trt thng vo khong 5 30 m. V ca b

phn phn phi c kch thc tng i ln, nn bc v van trt c ghp cng.



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trnh hin tng van trt b p vo mt pha, ng dn du vo van s

dng kt cu rnh vng. Trn b mt ca xi lanh c lm cc rnh c b rng 0,5

0,75 mm v su 0,3 0,5 mm. V vy, lc lm dch chuyn pittng s khng

ln v khng vt qu 10 20N.

Hnh 2-11.B phn phi kiu van trt

Cc lc khng cn bng theo chiu trc, gy ra do tc dng phn lc ca

dng du, thng c xt khi thit k cc b phn phi kiu van trt.

trng ( che) ca vnh gi pittng i vi rnh trn xilanh thng ly

khong 2 3 mm.

Kch thc cc ng dn trong b phn phi, ph thuc vo kh nng

thng qua v c tnh ton vi vn tc cho php ca dng du l v = 3 6 m/s.Din tch tit din ca ng dn (dm2):

v

Qf

600=

vi Q - tnh bng lt/ph; v - m/s.

kn ti cc v tr mi ghp gia cc chi tit ca b dn ng thu lc,

nh kn ca cc van kiu van trt, ph thuc vo khe h, vn tc, p sut v

tnh cht vt l ca cht lng cng tc (ph thuc ch yu vo nht).

Cc b phn phi kiu van trt c ch to vi p sut lm vic

ti 50 MPa (500 kG/cm2) v lu lng t 8 3000 lt/ph. Cc b phn phi ny

thng khng m bo kn tuyt i, nhng lng r r qua n thng khng

ng k v gim nhiu khi gim khe h gia pittng v xi lanh ca van.



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Pittng 1 c t ng v tr gia nh cc l xo 4 v ng lt 3, khi bm

c ni vi thng cha du qua cc ng dn trong pittng, cc ng thot t

xi lanh c ng kn. Khi pittng dch chuyn sang tri, bm cp du vo ng

S2, ng S1thng vi ng x. Cn khi pittng dch chuyn sang phi, ng

S1c ni thng vi bm, ng S2thng vi ng x.

Tu theo s lng cc l dn du ti v thot du, cc b phn phi kiu

van trt c chia ra cc loi: hai ng, ba ng, bn ng dn v.v... Cntheo s lng cc v tr lm vic ca pittng th cc b phn phi c chia ra:

loi hai v tr, ba v tr v.v... Theo s cc ng dn du t bm v tr gia ca

pittng th cc b phn phi c chia ra lm b phn phn phi kiu ng gia

hoc m gia.

Hnh 2-11 trnh by kt cu b phn phi bn ng dn, ba v tr v m

gia.

Theo kiu iu khin s dch chuyn ca van trt, cc b phn phi c cc

loi: iu khin bng tay, bng cam, bng thu lc, bng in v bng in thy

lc.Trn hnh 2-11 pha di c trnh by s b phn phi bn ng dn,

ba v tr c iu khin bng in t, s dch chuyn ca pittng v y ng c li

bng cc l xo v v tr gia.

Pittng van trt c ch to t thp cc bon cao (Y8A, Y10) hoc thp

thm cc bon (20X), c ti v ram thp t cng HRC 58 - 62. B mt ngoi

ca pittng van trt c mi bng, o van thng c lm t hp kim ng

bronz, t thp cht lng cao v c th t thp thm cc bon. Mt trong ca o

van

c r bng v

c c nh vo thn v ca b phn phn phi bng cchp cng.

Thn v b phn phn phi c ch to bng cng ngh rn t thp 45.

Cc van trong cc h thng thu lc ca my p thc hin cc chc nng

sau y: bo v khi b qu ti, m bo mt p sut nht nh khng i cc

phn khc nhau ca h thng, lm gim p sut ca dng cht lng, m bo th

t thc hin hot ng ca cc xi lanh cng tc v p sut bin i.

Van an ton c kt cu n gin nht l kt cu kiu van bi, nhng kt cu

ny khng m bo c p sut khng i, c bit l khi lng cht lng tiu

th ln v p sut cao, v khng c nh hng v gim chn cho bi.



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cc h thng thy lc ca my p

ngi ta s dng nhiu cc van c s

nguyn l c trnh by hnh 2-12. iu

chnh hoc t p sut cho van bng cch

thay i lc p ca l xo 7 qua nt vn 9.

Cn nt vn 9 c bt kn bng m 8. Tit

din lu thng ca van bi v l 5 trong thn

v lm tng tit din ca l 3 trong van 1. L

xo 2 p van 1 vo van. Khi t p sut t

trc, van 1 s b nng ln trn v p l xo 2,

do c chnh p sut cc khoang C v D

(v c du cp qua l 3). L A c thng

vi bm, l B - thng vi h thng v nh

du c a v thng cha. Sau khi p sut

khoang C gim xung, l xo 2 s y van 1 tr v ng li v gim l ng

du v thng cha. Van s ng chm do pittng c l thng nh. Nu p sut

tng ln th qu trnh li c lp li.

u di ca van 1 c chi tit 4 c chc nng lm trit tiu nng lng

ca tia du phun qua khe dng cn ca van khi m.

Cc van k trn c cc loi khc nhau:

- Van an ton hoc van trn.

- Van gim ti: c tc dng lm gim p sut trong h thng hoc trong mt

phn no ca h thng, bng cch ni thng khoang D vi ng x, qua b

phn phi dng van trt c iu khin bng cam hoc nam chm in.

- Van th t tc dng: m bo th t hot ng ca hai xi lanh N01 v N02.

Trong trng hp ny, khoang A c thng vi bm, khoang B ni thng vi xi

lanh N01, cn khoang C thng vi xi lanh N02. Du i qua van bi c dn tr v

thng cha bng ng ring, cn l 10 van khi khng lm vic. Sau khi p

sut xi anh N01 t mc nh, van 1 c nng ln v du c a ti xi

lanh N02.

Hnh 2-12. Van an ton



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- Van c tc dng thay i lin tc p sut ca h thng: s dung trong my

p nn. Trong trng hp ny, khoang D c ng thot du ti b phn tit lu

lm thay i lng du qua l tit lu 3. Khoang sau b phn tit lu c

ni thng vi thng cha.

Hnh 2-13.Kt cu v k hiu van

Trn hnh 2-13.a biu din cu to ca van an ton. L xo 3 c tnh vi lc

tng ng ng knh ca pittng tr dn 2.

Nu p sut h thng cao hn p sut nh, pittng 2 s nng li vantrt 1 ln v cht lng t h thng s quay v thng cha. Khi li van trt h

xung, qu trnh iu chnh c lp li. p sut thay i u, v l xo 3 c lcp khng ln. i khi trn ng dn ti pittng tr dn 2, ngi ta t b tit

lu. Van c th dng vi chc nng nhvan th t tc dng. Khi l B c

ni vi xi lanh N02. Sau khi t p sut nh h thng th cht lng s i ti

xi lanh N02.

t ng cp cc xung vo mch iu khin in - thu lc ca my p

khi t c p sut nh trc, ngi ta thng dng r le p sut. Trong r le p

sut, chi tit lm vic (pittng) s p l xo v lm ng mch in ca cng tc.



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Bng 2.1Cc k hiu trong h thng thu lc

Tn gi K hiu Tn gi K hiu

ng p sut caoBm lu lng khng

i

ng iu khin phBm lu lng iu

chnh

ng x

Hng dng chyng c in

Ch ni Trc quay

Thng cha L xo

Bnh tch p B lc

Rle p sut ng h p sut

Xilanh kiu pittng tr

Ch cc ng ng

ct nhau khng c

mi ni

B tr thit b thy lc Xi lanh kiu pittng

K hiu ca cc thit b khc trn s nguyn l ca h thng thu lc,

c trnh by bng 2-1.2.5. B tr cc thit b thu lc ca my p dn ngbng bm du

Cc thit b thy lc dng dn ng kiu bm du c sn xut theo n

t hng. Trong h thng thy lc ca my p, cc thit b thy lc phi m bo

nhng yu cu sau: thc hin hnh trnh khng ti vi tc cao v thc hin

hnh trnh cng tc vi tc ph hp vi cng sut nh nht ca bm v ng

c in; trong trng hp cn thit phi gi c chi tit di p lc nht nh;

gim p sut ng u trong cc xi lanh cho n khi bt u hnh trnh y v;trit ti cho bm trong thi gian ngng ngh; c kh nng iu chnh p sut v.v...



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Tt c nhng yu cu trn khng phi bt buc i vi tt c cc my p

thu lc, m n ph thuc vo chc nng ca my.

Hnh 2-14.a biu din s dn ng my p t bm lu lng khng i,

khng c hnh trnh khng ti tc cao. H thng dn ng bao gm: bu lc 1,

bm 2 (Q = const) v van an ton 3. B phn phi kiu van trt 4 iu khin

bng tay, cho php a lng du c bm cp ti cc khoang S1v S2ca xilanh pittng 5.

Hnh 2-14.S dn ng kiu bm ca my p

Tc ca hnh trnh khng ti bng tc hnh trnh cng tc. Tc

hnh trnh y v ca pittng ln hn nhiu so vi tc ca hnh trnh cng tc,

v S1> S2. Lng du a ti khoang S1trong thi gian hnh trnh y v s ln

hn lng du do bm cp. Cn xt n yu t ny khi xc nh tit din lu

thng ca cc van ca b phn phi v ca cc phn ng ng tng ng. S

dn ng khng cho php thc hin hnh trnh khng ti tc cao, chng c

dng my p c lc p ti 0,1 - 0,15 MN (10 - 15 tn) hoc my p c hnh

trnh khng ti nh.



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Tit din lu thng ca cc ng trn ng tng p c ly theo tc cho

php ca dng du l 3 6 m/s. i vi ng ng ht v ng ng x th tc

cho php ca dng du l 0,75 1,5 m/s.

S dn ng my p t bm c cng sut khng i, vi hnh trnh

khng ti nhanh c trnh by hnh 2-14.b. y xi lanh 5 c thm pittng

nh 6 dng cho hnh trnh khng ti.

Din tch khoang S3nh hn mt t so vi khoang S1. Trong thi gian hnh

trnh khng ti th bm y du vo khoang S3, khoang S1 c in y cht

lng cng tc qua van mt chiu c iu khin 9 (van cp) t thng cha 8 bng

cch t chy. Khi p sut khoang S3tng ln, van 7 lm vic: du t bm i vo

cc khoang S1 v S3, hnh trnh cng tc c thc hin. Trong thi gian hnh

trnh y v th du t khoang S3tr v khoang cha qua b phn phi 4. Du t

khoang S1tr v thng 8 qua van 9 hin ang m do c p sut khoang S2. Bi

v thng 8 du a n nhiu hn du i ly, nn phn du tha s c chy v

khoang cha ca bm. Thng pittng 6 c thay bng hai xi lanh pittng. Bdn ng kiu ny c s dng cho cc my nn nm ngang v my nn c lc

p nh.

Khi khi lng ca cc phn chuyn ng tng i ln v khung my p

c kiu ng, th ngi ta s dng xi lanh pittng bnh thng.

Trong thi gian hnh trnh khng ti, cn m bo cp du cho xi lanh,

ngi ta lp van cp ngay trc tip n xi lanh ca my p v b tr bn trong

thng cp du. i khi du trong thng trng thi c p sut d ca khng kh

hoc nit (p = 0,4 MPa). Van cp c tnh ton vi tc lu thng ca dng

du khng qu 1,5 m/s.Tng khi lng ca cc phn chuyn ng s tng tc ca hnh trnh

khng ti lm cn tr vic in y cht lng cng tc cho xi lanh. V vy, ngi

ta a thm van hm 10 vo h thng (hnh 2-14.b), to thnh t hp ca van mt

chiu v van an ton. Van an ton y c iu chnh sao cho du khng i

qua khi p sut ca du do khi lng ca cc phn chuyn ng v p sut ca

du trong thng 8 to ra. Nhvy, c th gi dm ngang bt k v tr no, nu

s dng b phn phi 4 c m gia.

ng nng tch tr c ca cc phn chuyn ng, dch chuyn vi tc

hnh trnh khng ti, khng bin thnh cc dng va p truyn cho phi khi

cc phn chuyn ng tip xc vi phi, cn phi gi tc cng tc n trc

thi im p kim loi.



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Trong h thng (hnh 2-14.c), van 7 dng iu chnh th t chuyn ng,

c iu khin khng phi bng p sut, m bng c cu cam. Cam K c gn

trn dm ngang di ng. Khoang S1c ni vi ngun t bm. Khi cam tip xc

vi van 7 th tc ca dm di ng gim xung. Trc thi im ny th khoang

S1c cp du qua van 9 t thng 8.

s dng mt cch tt hn cng sut ca ng c in, ngi ta dng

cch kt hp hai bm: mt bm c lu lng ln v p sut nh c dng cho

hnh trnh khng ti v phn ti nh ca hnh trnh cng tc, cn bm th hai

c dng cho phn ti ln ca hnh trnh cng tc (hnh 2-14.d). Khi sc cn

chuyn ng ca x ngang nh, th lu lng ca cc bm H1v H2c kt hp

vi nhau. Khi tng lc v p sut ti tr s cao hn p sut t trc van 7,

th van ny s chuyn du t bm H2v thng. Van ngc 11 c ng li, cn

do c p sut cng tc nh bm H1to ra, van 7 vn m. Du do bm H2cp s

quay li thng khi p sut d bng 0. Nu p sut trong h thng gim xung

di gi tr t van, lu lng ca bm H2c kt hp vi lu ca bm H1.

B phn phi 4 c m gia. B tr van an ton 3, trnh qu ti cho h

thng.

C th dng bm p sut cao lu lng khng i v thay i. Trong trng

hp bm c lu lng thay i, th trong thi gian di c th to lc ln gii hn

m du khng b nng. i khi kh nng ny c thc hin bm c lu lng

rt nh khng i.



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Chng 3

My p thu lc dn ng kiu bmc bnh tch p

3.1. Thnh phn ca my v cng dngMy p s dng dn ng t trm bm c bnh tch p c trnh by

hnh 3-1.

Hnh 3-1.My p thu lc c dn ng t trm bm c bnh tch p

Khi ng b tng p, van mt chiu 7 trong b phn phi s ngt xi lanh

cng tc vi bnh tch p 15. B tng p trung gian 13 c nhim v iu tit p

sut ca cht lng cng tc cp cho my p. Bnh tch p 15 gm c bnh thy lc

v bnh kh. Van mc ti thiu 14 c b tr nh mc p sut cht lng trong

bnh thu lc ca b tch p khng c gim qu mc. Bnh tch p khng

pittng 15 lm nhim v tr cht lng c p sut t bm a n trong nhngkhong thi gian ngh ca my p v cp cht lng cho my nhng khong thi

gian lm vic. My nn kh p sut cao 16 dng cp kh nn cho cc bnh kh



38/142


ca bnh tch p, c tnh ton thit k cng sut , trong thi gian lm

vic ca my p, lng tiu th khng kh khng nhiu. Thng ca bm 17 m

bo cp cht lng cho bm v cha cht lng tha t thng cp du tr v. Bm

18 p sut cao dng np cho bnh tch p, thng ngi ta s dng kiu bm

trc khuu ba pittng. Van gim ti 20 ca bm c nhim v chuyn bm sang

lm vic ch khng ti khi bnh tch p y.

Trn hnh 3-1 gc di bn phi c biu m cc van. Bng 3-1 lit kv tr cc van phn phi cc giai on khc nhau ph hp vi biu trn.

Bng 3.1

V tr cc van ca b phn phi

Cc vanGiai

on 4 5 6 7 8 9 10

I M M M

II M M MT

III M

IV M M

V M M M

Ch thch: M - m; - ng ; MT - m t ng

H thng hot ng theo cc giai on sau: I - hnh trnh cng tc, p sut

t b tng p trung gian; II - hnh trnh cng tc, p sut t bnh tch p; III -

hnh trnh khng ti (cp du); IV - gi dm ngang; V - hnh trnh y v.

3.2. Phn loi v kt cu bnh tch p

Bnh tch p ca cc my p thu lc c hai loi chnh: loi bnh tch p ti

trng v bnh kh thu lc (dng hi).

Theo kiu ca c cu phn phi gia khng kh v cht lng th tch p kiu

bnh kh thu lc cn chia ra lm cc loi khng c pittng, loi c pittng v loi

mng.

3.2.1. Bnh tch p ti trng

Bnh tch p ti trng c kt cu dng ng di, tc dng nhmt xi lanh,trong ng c pittong c gn thm mt khi ti trng lm bng gang, to p

lc nn cho cht lng.



39/142


Trng lng khi ti trng c tnh nhsau G:p.f.aG=

trong :

a - h s tnh ton xt n nh hng ca ma st do s bt kn ca pittng,

thng dng a = 1,1;

f - din tch tit din ngang ca pittng;

p - p sut cht lng yu cu.

bo m an ton ca c cu v kh nng lm vic theo ng yu cu

ca bnh tch p, ngi ta s dng cc dng c cu:

- C cu chng tro cht lng c bm n ca bnh tch p ti trng. C

cu gm t hp cc van kh hi v van gim ti ca bm, bm ny dn ng nh

c kh t khi ti trng ca bnh tch p v kt ni gia mch t ca bm vi

thng cha ca bm.

- C cu hiu chnh an ton khi v ng bao gm khoang cha van bi, c

nn bng trng lc ca khi ti vo l ni ng dn vi bnh tch p. Khi ng dn

n my b v, di p sut ca nc van bi o chiu sang v tr khc v ng

kn l dn cht lng n my p.

- C cu nn t t khi ti ln b , l mt van tit lu c lp trn

ng ng dn n my. Khi khi ti trng ca bnh ti h xung c khng ch

chiu cao, t ngay trn khi ti qua h thng tay gt nng cc van tit l u, van

lm iu tit lu lng nc chy ra t bnh tch p v nh lm gim ph tn

ca cht lng.

S lm vic ca bnh ti ko theo cc lc va p trong h thng thu lc do

chuyn ng nng ca khi ti thnh nng l

ng ca p sut cht lng.u im ca bnh ti trng l bo m n nh p sut c to khi qu

trnh gim ti ca bnh khc nhau. Nhc im ca bnh ti trng l c chiu cao

ln, cn mng ln v khi ti nng, c lc va p thu lc trong h thng v kh

tng dung tch cng tc bnh tch p.

Ngy nay, bnh tch p ti trng c s dng khi cn bo m tiu hao

cht lng khng ln v p sut cht lng cng tc khng i, khng ph thuc

vo loi bnh tch p.

3.2.2. Bnh tch p kh - thu lc kiu pittng

Bnh tch p kh - thu lc kiu pittng (hnh 3-2) gm pittng 1, xi lanh kh2, xi lanh thu lc 3, bnh kh 4 v my nn kh 5.



40/142


T s gia din tch F v din tch f c gi l h s tng p K, thng

chn t 1 n 100. Khi s dng kh nn c p sut 0,6 - 0,7 MPa, kh nn c

ly t h thng kh nn chung ca nh my.

Th tch kh ca b tch p VB, gm th tch cc bnh kh, th tch ccng

ng dn kh v th tch phn khoang nm trn pttng khi pittng nm v tr trn

cng, c tnh ton khi bit th

tch VP (VP - th tch cng tcca bnh tch p, ngha l th

tch cht lng y ra khi xi

lanh thu lc khi pittng dch

chuyn t v tr trn cng ti v

tr di cng).

Th tch VB thng c

chn xut pht t iu kin sao

cho h s chnh lch p sut cho

phpmax

minmax

pppm = khng

vt qu 10 20%.Khi khng kh trong bnh tch p dn n, th tch do kh chim s tng ln

mt lng l:

KVf

FVSF pp ==

trong :f

FK=

Trng thi ca kh

c biu din bng ph

ng trnh:( )nPBmin

nBmax

KVVpVp += (3.1)

trong : pmax, pmin- p sut khng kh trn pittng khi pittng v tr trn

cng v di cng.

n - ch s a bin vi p sut 20 MPa v bng 1,29 n 1,30.

minmax pm

p

=1

1 (3.2)

Thay pmaxt biu thc (3.3) vo (3.2) ta nhn c

( )nPBminnBmin

KVVpV

m

p +=

1

1 (3.3)

Kt qu ta c cng thc tnh VB nhsau:

Hnh 3-2.Bnh tch p kh - Thu lc kiupittng



41/142


n

nP

n

PB

m

mKV

m

KVV

=

=11

1

11

1 (3.4)

Nu n = 1, m = 0,1 v K = 10, ta c VB= 100 VP.

Cc bnh tch p kiu pittng c kt cu tng i phc tp, kch thc ln

v thng c s dng cht lng l nh tng v du khong. u im ca kt

cu l cht lng cng tc c kh nng to p sut ln, ti 60 100 MPa.Bnh tch p kiu pittng khng nn s dng khi th tch cng tc tng i

nh v p sut cht lng cao.

3.2.3. Bnh tch p kh thu lc kiu khng c pittng

Bnh tch p gm c bnh thu lc, trong khng kh trc tip p ln b

mt ca cht lng v cc bnh kh nn c ni vi nhau. H s k trong cng thc

(3.4) vi bnh tch p khng c pittng c gi tr bng 1 v th tch VB c tnh

bng:

VB= 9.VP (3.5)

Hnh 3-3.S thu lc iu khin bnh tch p kh thu lc kiu khng c

pittng. 1. bnh thu lc; 2. bnh kh; 3. b phn phi iu khin van mc ti

thiu; 4. van mc ti thiu; 5. van gim ti; 6. b phn phi iu khin van gim

ti; 7. t thit b in; 8. b iu chnh thu ngnTrn hnh 3.3 trnh by s b iu khin bnh tch p kh thu lc kiu

khng c pittng. Bn trong b iu khin 8, cha mt lng thu ngn. Khi mc



42/142


cht lng trong bnh 1 tng ln, thu ngn s ln lt ng cc tip im platin

na bn tri ca b iu khin 8. Cc nam chm in E1v E2dng ng cc

tip im. Khi cht lng trong bnh thu lc 1 t mc trn th nam chm in E 2

ng v b phn phi 6 s chuyn cc bm sang lm vic ch khng ti.

Tng t nhvy, khi cht lng t mc thp th ng nam chm in E1v b

phn phi 3 s thc hin vic ng van mc thp 4.

Hot ng ca b iu khin mc cht lng kiu thu ngn da trn c s

nh lut bnh thng nhau. Phng trnh cn bng cht lng i vi c hai khoang

ca b iu chnh c dng:

(H + h)= 2 hp+ (H - h) B (3.6)

trong :

H- chiu cao mc cht lng cng tc bnh thu lc;

h - lng dch chuyn ca thu ngn trong b iu khin;

- trng lng ring ca cht lng cng tc;

p- trng lng ring ca thu ngn;B- trng lng ring ca kh quyn p sut p.

Ngoi b iu khin kiu thu ngn ngi ta cn s dng cc b iu khin

c cm bin kiu cm ng t kiu in tip xc.

Nhc im ca bnh tch p kiu khng c pittng l nc bo ho khng

kh, iu ny lm nh hng n bn ca cc van, b phn phi v cc thit b

khc, v vy p sut s dng bnh tch p kiu ny rt t khi qu 30 MPa.

Cc u im ca cc bnh tch p kiu khng c pittng l th tch cng tc

ln ca cht lng rt ln v gim nguy c xut hin va p thy lc - do khng c

cc chi tit trung gian gia khng kh v cht lng, tn tht khng kh tng i

nh, c kh nng d dng tng th tch cng tc ca my.

Cc bnh tch p kiu khng c pittng c s dng cc my cn c th

tch cng tc ca nc ln v p sut 32 MPa.

Khi s dng du khong lm cht lng cng tc, du thng b oxy ho do

oxy ca khng kh v s b mt cc tnh cht ca du v c th to nn cc hn

hp. Ngi ta thng s dng cc bnh tch p du - kh c cc mng cao su

phn chia du v kh, ph bin dng kh nit. Trong trng hp ny c th sdng bnh tch p c pittng.



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3.3. Tnh ton th tch cng tc ca bnh tch p v lulng ca bm

i vi my p c ng kiu bm - bnh tch p c lp, th Vp c tnh

theo lng tiu th cc i c th ca cht lng p sut cao.

70,

FsFS

VBi Pmaxp

p

+

= (3.7)

trong :

SPmax- hnh trnh cng tc cc i ca my nn kh;

FP - din tch mt u ca cc pittng cng tc;

s - hnh trnh ca dm ngang;

FB - din tch mt u ca cc pittng tr v;

0,7 - h s xt n tn tht v lng d tr cht lng cn thit.

Khi my p lm vic vi b tng p trung gian, Vp

c tnh nh

sau:

70,

FsFSF'S

V

BMMiPmaxp

p

++

=

(3.8)

trong :

S'Pmax- hnh trnh cng tc cc i n khi ng b tng p;

SM- hnh trnh ton b ca pittng b tng p;

FM- din tch ca pittng ln ca b tng p.

Phng php tnh VPnu trn c s dng rng ri cho cc my p tc

cao, trong t s gia thi gian hnh trnh cng tc tp chia cho thi gian caton chu trnh Tchtl nh.

Nu tp gn bng Tcht th khi tnh th tch cng tc cn phi tnh n lu

lng QHca bm:

( )opHHp ttQ,

VV +=

70 (3.9)

trong :

VH- l lng tiu th nc p sut cao ca my p trong thi gian mt chu

trnh;

t0- thi gian hnh trnh y v.Lu lng ca bm QH (l/ph) c xc nh theo cng thc:



44/142


cht

HH

t

.VQ

0

60

= (3.10)

trong :

VH- lng tiu th nc ca my p, dm3(lt);

0- h s tn hao th tch ca bm;Tcht- thi gian mt chu trnh (giy);

Bng 3-2

Tc ca pittng my p, khi p chy cc ng bng ng thau v ng

thanh

Tc pittng (mm/s)

Pittng chnh hnh tr Pittng hnh tr bcLc p

(MN) Hnh trnh

cng tc

Hnh trnh

khng ti

Hnh trnh

y v

Hnh trnh

cng tc

Hnh trnh

khng ti

v kh hi

25 80 200 300 130 350

15 100 250 400 180 400

6 120 350 500 - -

C th tnh ton th tch cng tc ca bnh tch p dng chung cho mt

nhm cc my p, theo phng php ca P.S.Istomin, da trn c s cc s liu

v tc chuyn ng ca pittng my p khi cc hnh trnh cng tc, khng

ti v kh hi, v chiu di ca phi c gia cng v cng sut ca bm. Th d,

cc s liu v tc chuyn ng ca pittng my p, khi p chy cc ng bng

bronz v latng clit k trong bng 3-2. Vi mi loi my p, khi bit tc

yu cu, kch thc ca pittng v phi, ta xc nh thi gian chu trnh l lngtiu th nc p sut cao trong mt chu trnh.

Sau dng th tiu th nc p sut cao ca mi my p theo thi gian,

trn trc honh l trc thi gian, cn trc tung l trc lu lng Q (hnh 3-4).

Th tch cng tc ca bnh tch p cn phi c tr s sao cho lng tiu th

nc cc i ca nc trong khong thi gian tnh ton, th mc nc trong th

tch ny khng b gim qu mc cho php. Vi gi thit, lng nc t c

lng tiu th ln nht khi cc cng on p ca nhiu my p mt lc.

Trn c s cc th ring l v tiu th nc cc my p, ta dng th

tng hp cho cc thi im bt u p trng nhau.T th tng hp ca lng tiu th, ta xc nh lng tiu th nc cc

i Vmax, trong chu k nh tmax, bng thi gian np y bnh tch p.



45/142


Hnh 3-4. Cc th lng tiu th nc p sut cao

Lu lng cn thit ca cc bm c th c xc nh nh sau. Nu vi

my p 1, lng tiu th nc p sut cao cho 1 chu trnh Tcht1l V1v nu c n1

my p, th lu lng cn thit ca cc bm l.

1

111

60

chtt

.nVQ = (lt/pht) (3.11)

Tng t, ta tnh c lu lng ca cc bm cho cc my p khc:

- Tng lu lng ca cc bm: Q = Qi.- Th tch cng tc ca bnh tch p:

VP= Vmax- Qtmax (3.12)

Nn sao cho vic np th tch cng tc ca bnh tch p c thc hin

trong thi gian khng hn 20 - 30 giy v khng vt qu 90 - 120 giy. iu

ny cho ta kh nng ngt cc bm nu nh

cc aptomat khng lm vic.S dng l thuyt xc sut trong tnh ton th tch cng tc ca bnh tch p

i vi mt nhm cc my p. Phng php ny s hp l nht khi p dng cho



46/142


trng hp cp cht lng cho mt nhm cc my p nhnhau t mt bnh tch p.

Theo phng php ca P.S.Istomin, th tch VPnhn c s tng ln, v xc sut

ton hc l t s gia s lng cc trng hp xut hin cc s kin theo mun,

chia cho tng s cc trng hp.

Xc sut s kin hn hp c c trng bi du hiu chung cho mt s s

kin ring (nhng khng trng nhau), bng tng cc xc sut ca cc s kin ny

(nh l v cng xc sut). Xc sut sao cho s kin 1 v s kin 2 xy ra ngthi bng tch ca cc xc sut ring (nh l nhn xc sut).

Gi thit tng qut ta c M my p (cc my p khng ging nhau), mi

my p s ly nc p sut cao trong khong thi gian t iv c thi gian ton b

chu trnh l Ti.

Xc sut ly nc ca my p th nht:

1

11

T

tP =

Theo nh l v tch xc sut, ta c xc sut ly nc ng thi ca tt c

cc my p:P = P1. P2...Pm (3.13)

Xt trng hp khi trong xng c M my p ging nhau lm vic, thi

gian ly nc cho mi my p l t v thi gian ton b chu trnh ca mi my l

T giy.

Ta xc nh xc sut ly nc ca K my p trong tng s M my p ang

lm vic.

Xc sut bt k thi im no c K my p cng ly nc:

PMK= PK (3.14)

Xc sut bt k thi im no c K my p trong s M cng ly nc,

cn (M - K) my khng ly nc l:

PT

tT=

1 ; ( ) KMK PP 1

i vi bi ton y th K my no trong s M my s ly n c. Ta nhn

c s lng cc nhm c K my t M my s l s cc t hp ca M theo K

ngha l CMK. Xc sut chung sao cho trong bt k thi im no cng c K

my p no trong s M my, cng ly nc p sut cao, cn M - K my s

khng ly nc:

( ) KMKKM

PPP = 1 (3.15)

Th tch cng tc ca bnh tch p c th nh hn lng tiu th c th vcht lng ca tt c cc my p:



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VP=Vmax (3.16)trong :

- h s s dng ng thi hoc trng nhau (cho bit khong bao nhiumy p ta cn d kin khi tnh ton);

Vmax- th tch nc p sut cao c cc my p ly.

Ta xc nh thi gian xc sut lm vic ng thi ca 1,2,3... cc my ptrong tng s M my, trong mt ca lm vic (420 pht):

( ) KMKKM

PPP = 1420 (3.17)

trong : P - thi gian hot ng tng i trung bnh ca mi my p:

T

ttP p 0+

=

Tnh bnh tch p m bo cho s lng K my p no , trong tng s

M my, sao cho khi c s c ngt cung cp cht lng p sut cao c th xy ra,

s khng lm nh hng nhiu ti cng vic ca xng.

Th d: Trong xng c lp 6 my p c c tnh gn ging nhau

P =

+T

ttP 0 = 0,15

Xc nh h s trng nhau. Cng thc tnh ton v kt qu ca n c a

trong bng 3.3.

Trong trng hp cho th tch bnh tch p c tnh cho 4 my p.

67,06

4==

Bi v, thi gian lm vic ng thi ca 5 my (0,164 pht) thc t l khng th

gy ra s gin on hot ng ca xng.

Trong trng hp cho th th tch bnh tch p c tnh cho 4 my p, dothi gian lm vic ng thi ca 5 my l 0,164 pht, thc t s khng th gy ra

s gin on hot ng ca xng. V vy = 4/6 = 0,67.Th tch phn hnh tr ca bnh thu lc ca bnh tch p kiu khng c

pittng ca th tch cng tc l Vp, th tch an ton pha di l VH.a.o v th tch

d tr trn l VB.P.O.

Th tch d tr trn (lt) c xc nh t iu kin lm vic ca van gim

ti ca bm:

== 3360 pa

H

panBPO

tQ

FtvV (3.18)

trong :

vn- vn tc tng mc cht lng trong bnh tch p (dm/s);



48/142


F - din tch tit din cht lng trong bnh (dm2),

Tpa3 - thi gian lm vic ca van gim ti ca bm, thng ly bng 3-5

giy;

QH- lu lng ca bm (lt/ph);

Bng 3-3

Xc nh s my p cho php

S lng cc my p lm

vic ng thi( ) KMKK

M PPC

1 KM

, Pht

0 ( ) 3771501 6006

,PPC = 159,0

1 ( ) 39933301 5116 ,PPC = 167,7

2 ( ) 1761801 4226

,PPC = 73,8

3 ( ) 0414501 3336

,PPC = 17,4

4 ( ) 0549,01

244

6

= PPC 2,3

5 ( ) 0003901 1556

,PPC = 0,164

6 ( ) 0000101 0666

,PPC = 0,004

Cn c th tch an ton di khng xy ra ht kit cht lng trong bnh

tch p v gy ra s lt kh p sut cao vo h thng ca my p.

Vic xc nh th tch ny c tin hnh t iu kin lm vic ca van mc

ti thiu.

VH.a.o= v0FtM (3.19)

trong :

v0 - vn tc gim mc cht lng trong bnh tch p, theo cc s liu th

nghim khng c vt qu 2,5 m/s.

tM- thi gian ng van mc ti thiu, ly bng 3-5 giy.

Thng cha ca bm (hnh 3.5) cn c kch thc tip nhn cht lng

cha trong cc xi lanh cng tc v xi lanh y v (V1+ V2), cha trong bnh tch

p (V3) v thng np (V4).

Th tch thng cha ca bm:

VHao= 1,5 (V1+ V2+ V3+ V4) (3.20)

p sut cht lng trong thng cha ca bm thng bng p sut kh quyn.Cc thng cha ca bm thng c lm t thp 10, 20, theo kiu hn. Cc



49/142


thng ny c t v tr cao nht

theo kt cu khung nh cha trm

bm - bnh tch p cho php, nhm

m bo ct p cn thit trn

ng ng ht ca bm.

Cc thnh dc ca cc thngcha kch thc ln (hnh 3-5), bn

trong c lin kt bng cc tm

ngang c khot l cc v tr i

nhau. Cc ng trong thng cha

ca bm, ng t thng np v ng

dn n cc bm, c b tr cc pha i din nhau ca thng. Nhvy, cht

lng phi i con ng zc zc v s m bo kh nng loi b lng khng kh

ho tan, cng cc cn bn ln trong cht lng.

3.4. Tnh ton ng lc hc my p dn ng bng bmc bnh tch p

Hnh trnh cng tc ca dm ngang c ngha ln trong vic quyt nh

tnh thch hp v mt cng ngh ca my p.

Hot ng ca my p thu lc trong thi gian hnh trnh cng tc c

thy r trn hnh 3-6a.

H s cn quy dn ca ng ng xi lanh cng tc c xc nh t cng

thc:

+=

== 2

12

1

2 11

i

n

ii

ii

in

ipp

FFd

lF (3.21)

trong :

Fp- din tch ca pittng cng tc;

- h s tn tht ma st ng ng, i vi nc nh tng thng lybng 0,03;

li- chiu di on ng ng;

di- ng knh trong ca on ng ng;

Fi- din tch tit din ngang on ng ng;I - s th t ca on ng ng hoc l sc cn cc b;

n - l s on ng ng hoc v tr cc b;

Hnh 3-5.Thng cha ca bm



50/142


i- h s sc cn cc b.Chiu di quy dn ng ng ca xi lanh cng tc c xc nh t biu

thc:

=

=n

ii

i

pp l

F

FL

1

Tng t, ta c th tm cc h sc cn quy dn B, yv chiu di ng ngLB, Lyca xi lanh tr v v xi lanh cn bng.

Hnh 3-6.Hot ng ca my p theo thi gian

a- hnh trnh cng tc; b - hnh trnh khng ti; c- hnh trnh y v; P- xi

lanh cng tc; B- xi lanh kh hi; H - thng np; A - bnh tch p

Trn cc phn chuyn ng ca my p trong hnh trnh cng tc c cc lctc dng sau y: khi lng G, p lc nc ln pittng cng tc ppFp, lc ma st

R1trong cc m kn ca xi lanh v phn dn hng ca dm ngang di ng, lc

cn ca phi R(x) = Kx + R0(R0- lc cn ban u ca phi); p sut nc ln

pittng cn bng (pyFy) v ln pittng y v (pBFB).

Gc ta tng ng vi im bt u hnh trnh cng tc (im O), chiu

dng ca trc 0x ta chn theo chiu chuyn ng ca dm ngang my p. Chp

nhn cc gi thit: cc van iu khin c m vi tit din khng i, cht lng

b nn, my p v ng ng l cng tuyt i, ct p trong bnh tch p v

thng cha chnh l khng i, gi tr tc ct p v v tr ct p l v cngnh.



51/142


S dng cc k hiu c, ta vit phng trnh chuyn ng dm ngang

my p nhsau:

( ) BByypp FpFpxRRGFpdt

xdm += 12

2

(3.22)

xc nh p sut ca cht lng trong xi lanh tng ng, s dng phng

trnh Becnuly i vi chuyn ng khng n nh. Vi

ng dn ca xi lanhcng tc, ta chn mt tit din trn b mt cht lng trong bnh tch p, cn tit

din th hai l trn b mt ca pittng cng tc, khi :

2

22

2 dt

xd

g

L

g

vpp

pnpap

= (3.23)

trong : Pa- p sut trong bnh tch p.

Tng t, nh phng trnh (3.23), ta vit phng trnh cho chuyn ng

ca cht lng cc xilanh kh hi v xi lanh cn bng:

2

22

2 dt

xd

g

L

g

vpp BnBHB

++= (3.24)

2

22

2 dt

xd

g

L

g

vpp

ynyBay

++= (3.25)

Thay cc phng trnh (3.23) (3.25) vo phng trnh (3.22) v sau khi

chia c hai v cho FP. Ta nhn c:

+

pppp

pnp

a

F

Rx

F

K

F

R

F

G

dt

xd

g

L

g

vp 012

22

2-

p

BH

p

yyn

p

yy

p

ya

F

Fp

dt

xd

F

F

g

L

g

v

F

F

F

Fp

2

22

2

2

2

2

22

2 dt

xd

F

m

dt

xd

F

F

g

L

g

v

F

F

pp

BBn

p

BB

= (3.26)

lm n gin cho cc tnh ton tip theo biu thc cui cng, ta dng

cc k hiu sau:

p

BB

b

yyyp

p F

F

g

L

F

F

g

F

g

L

g

L

F

mM +++

= (3.27)

n= p+ yp

BB

p

y

F

F

F

F+ (3.27')



52/142


pPp

BH

p

ya

pa

F

R

F

R

F

Fp

F

FP

F

Gpp 01 += (3.28)

H s ca x trong biu thc (3.26), k hiu l:

=pF

KC

Thay vo biu thc (3.26) cc gi tr M, n, p v C nhn c:

02

2

=

++

pCx

g

v

dt

dvM nn

n (3.29)

trong :

M - c trng cho qun tnh ca h thng;

n- h s tn hao ma st khi cht lng chuyn ng;

p - p sut to tc cho dm ngang.

Khi tr lc bin dng ca phi l khng i, th phng trnh chuyn ng

ca dm (3.29) c th vit l:

02

2

=

+

p

g

v

dt

dvM nn

n (3.30)

Thc t l tr lc bin dng khng i trn phn ln di ca hnh trnh

cng tc ca cc qu trnh kiu p, chut, t ...

Nu nhchuyn ng pittng my p vi tr lc bin dng khng i ca

phi l chuyn ng n nh th 0=dt

dvn v vn tc chuyn ng l:

n

gp

U

=

2

(3.31)

Thay gi tr p t biu thc (3.31) vo cng thc (3.30) nhn c:

022

22

=+g

u

g

v

dt

dvM n

nn

n (3.32)

Sau khi bin i v ly tch phn biu thc (58) ta c:

Cvu

vuln.

u.gM

tn

n +

+

=

2

12 (3.33)

trong : C = 0; v khi t = 0 th Vn= 0.

K hiu:

u

Mg

n= (3.34)



53/142


V thay vo biu thc (3.33) nhn c :

1

1

+

=

t

t

n

e

euv (3.36)

Thc t, ngi ta chp nhn tc vn t c tr s n nh khi t = 4.Thi gian t1tng ng vi vn= 0,96 u, thng tr s t1ca cc my p thy lcrt nh (nh hn 0,1 thm ch c khi lc tng t ngt t 0 ti 0,5 PH).

Cng thc (3.31) ca chuyn ng n nh c th s dng tnh ton tc

ca my p trong hnh trnh cng tc.

Xt s thay i ca u ph thuc vo cc tr lc bin dng khc nhau ca

phi.

K hiu:

pp

Bn

p

ya

pa

F

R

F

Fp

F

Fp

F

Gpp 10 += (3.37)

Nu 00 =pF

R th max

n

ug

.P

u =

=

200

Nu 00

50 p.,F

R

p

= th maxn

, u,g

.P

u 7020

50 =

=

Nu 00

750 p.,F

R

p

= th maxn

, u,g

.P

u 502

4

0750 =

=

Trong trng hp, khi 00 pF

R

p

= , c ngha l vi cc iu kin gn iu

kin nh mc, th tc dm ngang my p bng 0.

Ta xc nh t l gia u v pp(pp- p sut trong xi lanh cng tc, tng ng

hnh trnh cng tc), khi m cng sut ca my p t cc i:2Kupp ap

Nu coi nh 0ppa = , iu m thc t hon ton cho php cng sut ca

my p l:

papp FKupuFupN 2==

( ) 03 2

== pa FKupdudN

Nhvy: Ku2= Pa/3



54/142


aap pKupp3

22 == (3.38)

My p c th t cng sut cc i, khi lc my phi thng gn bng tr

lc bin dng ca phi, bng 2/3 lc p danh ngha ca my p.

3.4.1. Hnh trnh khng ti ca my p

S tnh ton ca my p thy lc khi hnh trnh khng ti c trnh by hnh 3-6.b. Chiu dng ca trc x l chiu hng xung di. im 0 l im

u tnh ton, l v tr bt u ca dm ngang khi hnh trnh khng ti. Trn phn

chuyn ng ca my p c cc lc sau y tc dng: trng lng G ca phn

chuyn ng, p sut nc trong xi lanh cng tc ( ppFp ), p sut trong xi lanh

kh hi ( BBFp ), trong xi lanh cn bng ( yyFp ) v cc lc ma st.

msH RRR +=1

trong :

RH- lc cn ma st phn dn hng ca dm ngang chuyn ng khi hnh

trnh khng ti;

Rms: lc cn ma st phn m kn ca xi lanh

i

ims

D

KPR =

trong :

K - 0,1 i vi m vng t lm kn, K 0,6 0,7 i vi m t bung;Pi- lc ca pittng to ra (dN);

Di- ng knh pittng (cm).

Phng trnh chuyn ng ca dm ngang c dng:

12

2

RFpFpGFpdt

xdm yyBBpp += (3.39)

c th s dng c phng trnh Becnuli vi chuyn ng khng n

nh ng ng dn t thng np n xi lanh cng tc, ly mt mt ct trn b

mt cht lng thng np, cn mt ct kia trn pittng cng tc:

= ppn

pHp zdt

xd

g

L

g

vpp

2

22

2 (3.40)

trong : PH - p sut thng np.

B qua ct p do tc to ra, tng t nhtrn nhn c:



55/142


+++= BBn

BHB zdt

xd

g

L

g

vpp

2

22

2 (3.41)

2

22

2 dt

xd

g

L

g

vpp

ynyay ++= (3.42)

Thay cc biu thc (3.39) - (3.41) vo (3.38) chia c hai v cho Fp v k

hiu:

p

yy

p

BBp

p F

F

g

L

F

F

g

L

g

L

F

mM +++

=

P

yy

P

BBpn

F

F

F

F++=

pp

BB

P

ya

P

BH

pH

F

R

F

FZ

F

FP

F

Fp

F

Gpp 1+=

p = const

ta c:

02

2

=

+

p

g

v

dt

dvM nn

n (3.43)

Phng trnh (3.42) cho hnh trnh khng ti, v mt cu trc trng vi

phng trnh cho hnh trnh cng tc khi sc cn ca phi khng i, nh ng cc

tr s ca M, n, p y c ngha khc.Nhtnh ton cho thy, thi gian tc tc t1khng vt qu 0,01s. V thi

gian tng tc nh, nn khi tnh ton tc ca hnh trnh khng ti c th s

dng cng thc (3.31).

Tc u c th tng bng cch tng p sut thng np, tng khi lngcc phn chuyn ng ca my p, gim lc cn trn ng dn thng np - xi

lanh cng tc.

C th gim tc u bng cch tng lc cn ng dn xi lanh y v v

bng cch s dng cc xi lanh cn bng.

my p hot ng bnh thng, khi tnh ton hnh trnh khng ti cn

ch ti c tnh thay i p sut xi lanh cng tc. p sut cht lng gim n

0 dn n s gin on dng cht lng v gy s ht kh ngoi mi trng vo

xi lanh.

iu kin khng b gin on dng cht lng:pp0 (3.44)

trong : pp- p sut dtrong xi lanh cng tc.



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Tc cc i ca hnh trnh khng ti t c khi ch n nh:

02

2

= ppHp zg

upp (3.45)

Trong tc u c xc nh theo cng thc (3.31). C th ngn s gin

on ca dng chy bng cch t thng np trung gian gn my p hoc thng

np chnh gn my p, bng cch tng

ng knh cc xi lanh cn bng, tng psut thng np, tng tit din ca ng ng np nh t b van tit lu mt

chiu trn tuyn ca cc xi lanh y v.

3.4.2. Hnh trnh y v ca my p

Khi dm ngang y v, nc t trong xi lanh cng tc tr li thng np qua

van x ca b phn phi v qua van np ca xi lanh cng tc. Lng nc i qua

van x ca b phn phi nc, thng chim khong t hn 7% tng hp nc

c y ra khi xi lanh cng tc. V vy, iu kin tnh ton l nc c y ra

ch qua van np. S tnh ton my p thu lc i vi hnh trnh y v c

trnh by trn hnh 3-6.c. Chiu dng ca trc x ta ly l chiu hng trn.

Trn cc phn chuyn ng ca my p khi hnh trnh y v c cc lc tc

dng sau y: lc cn c ch R0, trng lng G, p sut nc ln xi lanh cng tc

( ppFp ), ln pittng kh hi ( BBFp ) v ln pittng cn bng ( yyFp ), lc ma st

cc m kn ca xi lanh Rmsv phn dn hng RHm ta coi bng 0.

Trn c s nh lut 2 ca ng lc hc ta c:

msoppyyBB RRGFpFpFpdt

xdm +=

2

2

(3.46)

Khi hnh trnh y v, cng nhkhi hnh trnh cng tc ca my p, ta bqua ct p tc v ct p v tr i vi cc xi lanh kh hi v cn bng. S dng

phng trnh Becnuli i vi chuyn khng n nh, ta tm cc p sut pp , Bp , v

yp . Sau khi thay tr s ca cc p sut ny vo phng trnh (3.46) v k hiu:

p

yy

p

BBp

p F

F

g

L

F

F.

g

L

g

L

F

mM +++

=

B

yy

p

BBP

F

F

F

F++=

p

mspH

p

ya

p

Ba

FRRGzp

FFp

FFpp ++++= 0



57/142


p = const v qua mt vi php bin i ta nhn c phng trnh

02

2

=

+

p

g

v

dt

dvM nn

n (3.47)

Nu khi chuyn ng l n nh ( 0=dt

dvn ) khi tc chuyn ng ca

dm ngang c biu th bng phng trnh (3.31).Bi v, cc qu trnh ca dm ngang xy ra rt nhanh, cho nn xc nh

tc ca dm ngang hnh trnh y v c th s dng cng thc (3.31). Tc

u c th tng bng cch tng p sut trong bnh tch p, tng din tch ca cc

xilanh y v v xilanh cn bng hoc l gim sc cn tuyn xilanh cng tc -

thng np.

3.5. Kt cu v tnh ton cc chi tit ca h thng np

Cc thit b np bao gm: thng np, c b tr gn my p hoc trn

xilanh cng tc. to p sut ln cht lng, thng ngi ta s dng h thngkh nn ca nh my, p sut 0,6 0,8 MPa. Van np ni xilanh cng tc vi

thng np.

Th tch ton b ca nc trong thng np Vc xc nh, xut pht t

th tch nc Vpdo cc pittng cng tc y ra sau mt hnh trnh ton b. Nhm

ngn nga s lt kh t thng np vo ng ng dn n van np thng ly

V= (2-2,25)Vp.

Gi s, p sut khng kh trong thng np khi tiu th cht lng c thay

i theo quy lut ng nhit:pVB= const

trong :

VB- th tch ban u ca khng kh trong thng cha.

p sut khng kh sau khi loi tr th tch Vps b gim xung v ngi ta

ly pmin= 0,75max.

S thay i trng thi khng kh c biu din bng cng thc:

pmin(VB +Vp) = pmaxVB (3.48)

T , thu c: VB= 3Vp.

Th tch ton b VHca thng npVH = (22,5)Vp+ 3Vp = (55,5)Vp (3.49)



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Cc thng np c ch to theo kiu hn t thp tm: dy ca phn hnh

tr v phn hnh cu thng khng nh hn 8mm.

dy ca thnh c tnh theo cc thng s nhtnh chiu dy ni hi.

kh va p khi u ba chm phi, trn ng np trc my p, ngi

ta lp b b va p thu lc, c thng np c lp van bo him l xo, ng o

mc n

c, ng h p sut v ba van (van x, van dn n

c t

ng ng n

cv van dn kh nn).

ngt thng np v tr ni vi my p, ngi ta t van chn mt chiu

trn ng ng np, dng ngt thng np ra khi my p v nu cn thit s

chuyn nc t xilanh cng tc v thng np.

a lng nc tha v thng

bm, thng np c van thot du c

iu chnh p sut nh hn 0,2MPa so

vi p sut m van an ton ca thng

np s lm vic.S ca van thot du c trnh

by hnh 3-7. Cht lng t thng np

theo ng ng A n khoang di

van 5 v l tit lu 1 - vo khoang trn

van 5. Vt nng 3 s gi p sut nh

thng qua van 2 c kch thc nh,

c ni vi pittng 4. Khi p sut vt qua mc chnh p trn van 5 v di van

5, n s lm nng van v chuyn cht lng t thng np v thng bm qua ng

ng B. Khi gim p sut trong thng van np, van 2 c ng li v van chnh 5

cng s ng li trn van. Van chuyn c iu khin bng s thay i mc

nc trong thng np, c th c s dng thay th van c iu khin bng

p sut.

S nguyn l ca van c trnh by trn hnh 2-1. Van gm c van np

2 v dn ng thy lc 3 nng van. Dn ng c ni vi ng ca xilanh

y v. h tr m van np, i khi ngi ta cn lm van np c thm van ph

gim ti c tit din nh. Tit din lu thng ca van np c tnh theo tc

hnh trnh khng ti vxxcho tr

c v theo tc chn tr

c vH.Kca dng chtlng trong tit din lu thng ca van np khi n c m hon ton

Hnh 3-7. Van thot du



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K.H

x.x

v

v.Ff= (3.50)

trong :

F - din tch ca pittng cng tc,

vx.x- thng khong 300 400mm/s,

vH.K- khng qu 6-7 m/s.V vy, giai on thit k ban u thng ngi ta ly:

f = 0,05F (3.51)

Sau khi chn cc kch thc ca ng np, ngi ta tin hnh tnh ton

kim tra ng

gt - may ep thuy luc - (142).pdf

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