may bom khoan.pdf

7/25/2019 May bom khoan.pdf

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1

MY BM PISTON

1.1. Vai tr, cu to, nguyn l lm vic v phn loi

1.1.1. Vai tr ca my bm piston trong cng tc khoan du kh

Trong cng tc khoan khai thc v khoan thm d my bm dung

dch khoan l b phn khng th tch ri v cng nh khng th thiu

c. My bm khoan c nhim v quan trng trong qu trnh thi cng

ging khoan. Trong qu trnh thi cng ging khoan, chong khoan ph hu

t y ging khoan, mn khoan ny phi c a ln b mt nh

mt loi nc ra gi l dung dch khoan. thc hin qu trnh trn,

chng ta phi s dng mt loi thit b trong my bm khoan ng vai

tr quan trng nht. My bm khoan c cng dng bm cht lng xung

xung ging khoan lm mt, lm sch chong khoan, lm sch y

ging khoan a mn t y ging khoan ln v c bit quan trng l gip

cho qu trnh khoan c d dng.

thc hin bm dung dch khoan xung y ging khoan, my

bm khoan thng s dng l my bm piston. My bm piston c nhng

u vit ring m cc my bm khc khng c c, v c s dng rng

ri trong khoan du kh:

- C th bm cc dung dch c trng lng ring khc nhau;

- C th bm c vi p sut ln;

- p sut v lu lng khng ph thuc vo nhau. y l yu t

quan trng p ng trong v yu cu v cng ngh khoan;

- Cu to n gin, d thay th, d sa cha v bo dng;

- bn cao v d vn chuyn.

1.1.2. S cu to ca my bm piston


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2

4

5

7

8

6

3

12

9

Pa

B1B2

S

Hnh 1.1. S cu to ca my bm piston.

1. Piston 4. Hp van 7. ng ht

2. Xi lanh 5. Van ht 8. B cht lng

3. Cn piston 6. Van y 9. ng y (x)

Khong khng gian gia piston v cc van c gi l khoang

(bung) lm vic ca my bm.

Th tch ca bung lm vic thay i ty theo v tr ca piston trong

qu trnh chuyn ng.

Trong qu trnh lm vic, piston chuyn ng tnh tin qua li trong

xi lanh. Nhng im tn cng bn phi v tn cng bn tri ca piston

c gi l im cht phi v im cht tri ca piston.

1.1.3. Nguyn l lm vic ca my bm piston

Xt trn hnh v1.1, khi piston di chuyn t v tr B2n v tr B1,

th tch bung lm vic s tng dn, p sut P trong gim i v nh hn

p sut trn mt thong ca bnh cha cht lng Pa (P < Pa). Do cht

lng t b cha s dng ln i vo ng ht, i qua vanht vo khoang lm

vic ca bm, trong lc ny van y ca bm vn ang trng thi ng.


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Khi piston chuyn ng t v tr B2n B1th my bm thc hin qu trnh

ht v lc piston dng li ti v tr B1th qu trnh ht s kt thc.

Sau pison ichuyn ng v i ngc t B1n B2. Th tch

bung lm vic gim dn, p sut cht lng tng ln, van ht ng li v

van y m ra. Cht lng c p ln van y v i theo ng y ra ngoi.

Qu trnh ny gi l qu trnh y.

Qu trnh ht v y cabm c xen k nhau. Mt qu trnh ht

v y k tip nhau c gi l mt chu k lm vic ca my bm piston.

1.1.4. Kh nng t ht ca my bm piston

Khc vi my bm ly tm, my bm pison khng cn mi (in y

cht lng) trc khi khi ng, m bm c kh nng t ht.

Tht vy, nu ta gi Vo l th tch khi khng kh trong ng ht v

bung lm vic (khi piston B2). Nu pison di chuyn n B1v B1B2= S,

th khng kh gin ra vi th tch l Vo+ F.S (F.S l th tch ca xi lanh).

Khi p sut khng kh trong xi lanh l:

P = Pa..

o

o

V

V F S (1.1)

T (1.1) ta thy P < Pa. Do cht lng t b cha s i vo ng ht

v dng ln theo cao c xc nh nh sau:

aP Ph

(1.2)

Nu pison tip tc lm vic, cht lng t b cha s dng ln dn

theo ng ht v in y khoang lm vic ca bm. Khi xem nh my

bm t mi xong.

1.2. u v nhc im ca my bm piston

1.2.1. u im

- Cu to n gin, d thay th, bo dng.


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- C th to ra p sut ln.

- C th bm c cc dung dch c trng lng ring khc nhau.

- p sut v lu lng khng ph thuc vo nhau.

- My bm c bn cao.

1.2.2. Nhc im

- Chuyn ng ca cht lng qua bm khng n nh, do lu

lng ca bm b dao ng.

- Kt cu ca bm cng knh.

1.3. Phn loi my bm piston

1.3.1. Theo phng php truyn lc

- My bm truyn ng bng tay.

- My bm truyn c truyn ng gin tip.

- My bm c truyn ng trc tip.

1.3.2. Theo cch b tr xi lanh

- My bm thng ng.

- My bm nm ngang.

1.3.3. Theo cu to ca pison

- My bm c piston dng a.

- My bm c piston dng trc.

1.3.4. Theo cht lng cn bm

- My bm dng bm nc l.- My bm dng bm axit.

- My bm dng bm dung dch.

1.3.5. Theo cch tc dng (s ln ht y sau 1 vng quay ca trc)

- My bm tc dng n.


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- My bm tc dng kp.

- My bm tc dng ba.

- My bm tc dng bn

1.3.6. Theo p sut

- My bm p sut thp (P < 10 at).

- My bm p sut trung bnh (P = 10 20 at).

- My bm p sut cao (P > 20 at).

1.3.7. Theo lu lng

- My bm lu lng thp (Q < 15 3 /m h ).

- My bm lu lng trung bnh (Q = 15 60 3 /m h ).

- My bm lu lng cao (Q > 60 3 /m h ).

1.4. Lu lng ca my bm piston

1.4.1. Lu lng l thuyt trung bnh

Th tch lm vic ca my bm tc dng n l:

V = F.S (1.3)

Th tch lm vic ca my bm tc dng kp l:

V = (2F - f).S (1.4)

F: din tch b mt lm vic ca mt piston,

F: din tch mt ct cn piston.

Gi n l s vng quay ca trc bm quay c trong mt pht.

a. Lu lng l thuyt trung bnh ca my bm tc dng n l:

. .

60

F S nQ (1.5)

b. Lu lng l thuyt trung bnh ca my bm tc dng kp l:


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(2 ). .

60

F f S nQ

(1.6)

c. Lu lng l thuyt trung bnh ca my bm tc dng ba l:

3. . .60F S nQ (1.7)

d. Lu lng trung bnh ca my bm tc dng bn l:

2.(2 ). .

60

F f S nQ

(1.8)

1.4.2. Lu lng trung bnh thc

Lu lng trung bnh thc ca my bm piston bao gi cng nh

hn lu lng l thuyt tnh trn v nhng l do sau:

- C khng kh lt vo bm.

- B phn lt kn ca bm v cc van khng th m bo tuyt i

kn khi bm lm vic dn n r r cht lng.

- S ng - m chm ca van ht v van y trong qu trnh ht v

y k tip nhau lm tht thot cht lng.

V vy, lu lng trung bnh thc ca my bm piston c xc

nh theo cng thc sau:

.t

Q Q (1.9)

l hiu sut lu lng ca my bm,

= 0,85 0,90 ng vi my bm nh (D < 150 mm),

= 0,90 0,95 ng vi my bm va (D 150 300 mm),

= 0,95 0,98 ng vi my bm ln (D > 300 mm),


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1.5. Ct p ca bm piston

Hnh 1.2. S tnh ton ct p

Kh nng truyn nng lng ca bm vi dng dung dch c th

hin bng s chnh lch nng lng n v ca dng dung dch hai mt

ct trc sau ca my bm.

Ta c nng lng n v ti mt ct (A - A):

.

2

v

A A AA A

P Vl Z

g

(1.10)

Nng lng n v ti mt ct ( B - B) :

.

2

v

B B BB B

P Vl Z

g

(1.11)

Trong

: p sut dng chy ti mt ct ( A - A ), (B - B);

ZA ZB

PB, VB

PA, VA

B B

A A


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VA; VB : Vn tc dng chy ti mt ct ( A - A), (B - B);

: H s iu chnh ng nng.

: Trng lng ring ca cht lng bm.

=> chnh lch nng lng hai mt ct ( A - A), (B - B ) l:

l = lA lB= + + (ZB - ZA) (1.12)

- Nu l > 0: Th cht lng c my cung cp cho nng lng. Hay

l my thc hin qu trnh bm, my thu lc gi l my bm.

- Nu l < 0: cht lng truyn nng lng cho my thu lc, my

gi l ng c thu lc.

Gi H = lB - lA l ct p ca my thu lc (my bm). Ta c nh

ngha: Ct p H ca my bm l nng lng n v (tc nng lng) trng

lng cht lng ca cht lng trao i c vi my thu lc.

H =V V

B AP P

+ 2g

B B AV V + (ZB - ZA) (m ) (1.13)

Gi thnh phn th nng n v l ct p tnh:

Ht = + Z (m) (1.14)

Gi thnh phn ng nng n v l ct p ng:

H= (m) (1.15)

=> H = Ht + H (mt ct nc) (1.16)

1.6. Cng sut (N)

Cng sut ca ng c (Nc) chi ph cho qu trnh bm lm vic bao

gm cc thnh phn sau:

- Chi ph cng sut nng mt lu lng Q ln cao H trong 1

n vthi gian c gi l cng sut thy lc hay cng sut c ch (Ntl);


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. .

75tlQ H

N (1.17)

Cng sut thy lc chnh l c nng m cht lng trao i vi bm

trong 1 n vthi gian.

- Chi ph cng sut thng cc tn hao thy lc, tn hao thtch,

tn hao c kh, c nh gi bng hstl, V v c.

+ Tn hao thy lc tl: bao gm chi ph thng cc sc cn thy

lc do ma st vi thnh ng v cc tn hao cc bdo thay i tc dng

chy khi cht lng chuyn ng tbcha n ng y. Ngoi ra cn

thng lc qun tnh ca van.

t

tl

l

HH

(1.18)

Ht, Hl: Ct p thc tv ct p l thuyt.

+ Tn hao thtch V: c xc nh bng hsht y:

tV

l

Q

Q (1.19)

Qt, Ql: Lu lng thc tv lu lng l thuyt.

Nh vy, cng sut trn trc ca piston l cng sut lm vic hay

cng sut chbo (Nlv):

.tl

lv

tl V

NN

(1.20)

+ Tn hao c kh (c): l cc tn hao tng c n trc ca piston.

Nh vy, cng sut ca ng c sl:

Nc=. .

. . 75. . .tl

tl V c tl V c

N Q H

(1.21)

1.7. Hiu sut ()

Hiu sut ton phn ca my bm c xc nh theo cng thc:


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- Mt vng quay ca trc khuu th piston di chuyn c qung

ng l 2S.

Vn tc trung bnh ca piston s l:

Ctb= 2 .60S n = 2.2. .60

r n = .15r n (1.23)

Gi thit chiu di thanh truyn ln hn tay quay nhiu ln, lc ny

khi tay quay quay c 1 gc th piston di chuyn c mt on l x.

Ta c:

X = AB = Ctb.t = AO BO (1.24)

AO = r, BO = r.cos

Suy ra: x = r r.cos= r.(1 - cos) (1.25)

* Phng trnh vn tc tc thi ca piston l:

C = ( ).sin .sindx d d t r rdt dt dt

=> . . sinc r (1.26)

* Gia tc tc thi ca piston l:

2sin. . .cos . .cosdc d d

j r r rdt dt dt

(1.27)

1.9. Lu lng tc thi ca my bm piston

Ta c cng thc xc nh lu lng ca my bm piston l:

.Q F c (1.28)

Do F l tit din ca ng ht (vi qu trnh ht) hoc ng y (vi

qu trnh y) l i lng khng i v c l vn tc tc thi ca my bm

ti mt thi im bt k, nn lu lng ca mybm piston theo cng thc

(1.21) l lu lng tc thi.

Thay gi tr c t cng thc (1.19) vo (1.21) ta c c:


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a1

OOO

QtbQmax

360180900 a2

. . . .sinQ F c F r (1.29)

T cng thc (1.22) ta nhn thy lu lng tc thi ca my bm

piston l mt hm s ph thuc vo i lng sin . Do th lu lng

ca my bm s c dng th ca hm sin.Gi m l h s lu lng ca my bm piston, m c xc nh theo

cng thc:

max

tb

Qm

Q (1.30)

1.9.1. Lu lng ca my bm tc dng n

Hnh 1.4. th lu lng my bm tc dng n


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Qtb

Qtb2

Qtb1

OO

Qmax

3601800

Trong bc y, na u ca bc kp lu lng cht lng thay i

theo hnh sin oa1a2. bc ht khng c cht lng b y ra v c biu

din bng on thng a2a3.

max max . . .. 30F r nQ F c

. . .2 . . ..

60 60 30tb tbF s n F r n F r n

Q F c

max

1tb

Qm

Q

1.9.2. My bm tc dng kp

Hnh 1.5. th lu lng my bm tc dng kp


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Qtb

O

360270O

180OO

Qmax

900

nhng phn m hai xi lanh c cng bc y th ta phi dng cch

cng th tm Q tng cng trn ng y. Cui cng ta c ng

cong lu lng tng cng, c 6 ln cc i sau mi vng quay ca trc

khuu.

max max

. . ..

30

F r nQ F c

3. . . 3. .2 . . .3. .


Q F c

max

3tb

Qm

Q

1.9.4. th lu lng ca my bm tc dng bn

Ngi ta ch to my bm tc dng bn gm 4 xi lanh tc dng n,

trc khuu ca chng t lch nhau 1 gc 90 o , hoc gm 2 xi lanh tc

dng kp v tay quay ca chng t lch nhau 90 o .

Hnh 1.7. th lu lng ca my bmtc dng bn.

ng biu din lu lng tng cng 4 bung lm vic l tng

tung ca th lu lng do 4 bung lm vic sinh ra. N t 4 ln cc

i sau mi vng quay ca trc khuu.


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Lu lng tc thi ln nht ca my bm th hin bng tung dc

v n c ln bng tng hai on db v dc (db = dc), tng ng vi =

45 o .

. . 1,41. . . .. . .sin . .sin 4530 60

or n F r nQ db F r F

max

1, 41. . . .2. . . .sin

30

F r nQ db dc F r

4. . . 4. .2 . 8 . .4. .


Q F c

max 1,41. 1,114tb

Qm

Q

i vi my bm tc dng 4 gm 2 xi lanh tc dng kp th mc

khng ng u ca lu lng s ln hn 1,11 ty theo t l ng knh

cn v ng knh piston.

* Nhn xt:

- Cc my bm c s xi lanh l th h s n nh lu lng nh.

- Cc my bm c s xi lanh chn th h s n nh lu lng ln.

- V mt kinh t k thut th my bm gm nhiu xi lanh tc dng

n s cng knh, phc tp nn vi my bm hin nay thng s dng

nhng loi sau:

+ My bm tc dng 3 gm 3 xi lanh tc dng n.

+ My bm 2 xi lanh tc dng kp (tc dng bn).

+ My bm tc dng kp.

1.10. Khc phc hin tng chuyn ng khng n nh ca cht lng

trong my bm piston

1.10.1. Tc hi ca chuyn ng khng n nh ca cht lng trong my

bm piston

- Lm tng tn tht thy lc.


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- Xy ra s rung ng khi lm vic, gy ra va p thy lc lm h

hng cc b phn ca my bm.

- Trong trng hp dng nhiu bm lm vic, c th xy ra hin

tng cng hng bin dao ng ca p sut dn n gim bn ca

thit b.

1.10.2. Bin php khc phc chuyn ng khng n nh ca cht lng

trong my bm piston

1.10.2.1. Khi thit k

- Dng my bm tc dng kp.

- Dng my bm tc dng ba (gm 3 my bm tc dng n ghp

vi nhau).

1.10.2.2. Khi s dng

Dng bnh iu ha iu ha lu lng v p sut ca my bm

khi lm vic. C hai loi bnh iu ha gm:

- Bnh iu ha ht (lp trn ng ht).

- Bnh iu ha y (lp trn ng y).

1. Bnh iu ha ht

Trong qu trnh lm vic ca my bm, mt phn cht lng c

tch ly trong bnh iu ha. Trn mt thong ca cht lng trong bnh lun

c mng ngn cht lng vi khng kh c p sut chn khng. V th cht

lng t trong ng ht ln bnh c th xem nh l mt dng chy n nh..

Do s gim c tn tht nng lng trong ng ht.

Khi lp bnh iu ha ht, ta c th:- Tng chiu cao ht ca my bm.

- Gim c dao ng p sut ca my bm trong qu trnh ht.


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PxPx

x

(Px < Pa)

Pa

Hnh 1.8. Bnh iu ha ht

Trong qu trnh lm vic ca my bm, mt phn cht lng c

tch ly trong bnh iu ha. Trn mt thong ca cht lng trong bnh lun

c mng ngn cht lng vi khng kh c p sut chn khng. V th cht

lng t trong ng ht ln bnh c th xem nh l mt dng chy n nh..

Do s gim c tn tht nng lng trong ng ht.Khi lp bnh iu ha ht, ta c th:

- Tng chiu cao ht ca my bm.

- Gim c dao ng p sut ca my bm trong qu trnh ht.

2. Bnh iu ha y

Trong qu trnh y, mt phn lu lng ca my bm (phn ln

hn lu lng trung bnh) c tch ly trong bnh, mc cht lng trongbnh dng ln, nn khng kh phn trn ca bnh v to ra p sut ln.

Khi van y ng, nh p sut ca khi khng kh b nn trong bnh th

cht lng tip tc c y ra ng y, dao ng lu lng v p sut


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s

Pxx

P > Pa

trong ng y s gim, dng chy ca cht lng trong ng y s iu ha

v n nh hn.

Hnh 1.9. Bnh iu ha y

S dng bnh iu ha y c tc dng lm gim lc qun tnh trong

ng y ca my bm, do s gim tn tht lu lng v p sut trn ng

y.

* Ch :

- Bnh iu ha cng t st my bm th cng c li.

- Trong trng hp ng ht ngn th khng cn lp bnh iu haht.

- Bnh iu ha y dng trong mi trng hp.

3. Tnh chn bnh iu ha

Kch thc ca bnh iu ha c xc nh da trn gi tr ln nht

ca mc khng n nh p sut. Trong bnh iu ha v c s thay i

mc cht lng lm cho khng kh trong bnh b nn v n ng thi s lmthay i p sut ca khng kh.


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V t b , P t b

V m a x , P m i n

V m i n , P m a x

V, P: th tch v p sut cht kh trong bnh 3 mc trong 1 vng

quay ca trc my bm.

Thc t, nu mc khng n nh ca p sut trong bnh nh th ta

coi nh l m bo c n nh:

max min 0,025 0,05tb

P P

P

(1.31)

Ta cho rng trong qu trnh thay i th tch, khng kh trong bnh

gin n ng nhit. Theo Booilo Mariot ta c:max max min min

min minmax

max

. . .

.tb tbP V P V P V

P VP

V

(1.32)

p sut trung bnh ca khng kh trong bnh iu ha l:

max min min max min

min

.( )

2 2.tbP P P V V

PV

(1.33)

Thay (1.25) v (1.26) vo cng thc (1.24) ta c c:

max min max min max min

max min

2.( )

tb tb tb

P P V V V V V

P V V V V

(1.34)

V l lng cht lng c tch li trong b nh iu ha.


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1 2

0 90 180 360

QmaxQtb

O O O

a

* By gi ta s i xc nh gi tr V.

Ta xt cho trng hp my bm piston tc dng n 1 xi lanh.

th lu lng ca n c dng nh sau:

Phn din tch 1a2 l phn lu lng d.

Khi phn t th tch cht lng i vo bnh nh sau:

. .. . . .sin

30tbF r n

d V dQ Q dt F r dt dt

. . 1. .sin . . .(sin )

.3030

F r n d F r d F r d

n

Ton b khi lng cht lng vo bnh iu ha s l:

2

1

1. . (sin ) . .( cos )V F r d F r

T iu kin V = 0 ta c: 1 11sin 0,323rad

2 1 2,817rad

=> V = 1,1.F.r = 0,55.F.S (S = 2r)

* Vi phng php lm tng t ta c c cc kt qu sau:


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x

bb

Hh

dH

a a

Pa

- Vi my bm tc dng kp 1 xi lanh: V = 0,21.FS.

- Vi my bm tc dng kp 2 xi lanh tc dng n: V = 0,042.FS.

- Vi my bm tc dng ba gm 3 xi lanh tc dng n: V =

0,009.FS.Da vo kt qu V tnh c trn v gi tr trong khong gii

hn 0,025 0,05 ta s xc nh c lng kh cn np vo bnh cho tng

loi my bm c th:

tb

VV

Kch thc bnh iu ha c chn trn c s lng kh cha trong

bnh m ta xc nh cng thc trn ch chim khong 2/3 th tch bnh.

1.11. Qu trnh ht ca my bm piston

hP

: l ct p trong xi lanh khi ht (m);

Pa: p sut mt thong (at);

Hh: chiu cao ht ca my bm (m);


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C: tc chuyn ng ca piston (m/s);

Hoh: tn tht ct p trn ng ht (m);

Hv: tn tht ct p ti van ht (m);

Hqt: tn tht ct p do qun tnh chuyn ng ca khi chtlng trn ng ht (m).

Vit phng trnh becnuli cho 2 mt ct (a-a) v (b-b) ta c:

2

2a h

h oh v qt

P P cH H H H

g (1.35)

Trong cc i lng tn tht c xc nh nh sau:

a. Tn tht Hoh:2 2 2

. . .2 2 2

h h hoh m

v v vH

g g g (1.36)

: h s tn tht cc b;

m : h s tn tht do ma st;

.h

h

F cv

F

: vn tc ti ng ht;

F: tit din b mt piston;

Fh: tit din ng ht.

b. Tn tht Hv:

.vG R

Hf

(1.37)

G: trng lng van v l xo (N);

R: lccng ca l xo (N);

f: tit din a van.

c. Tn tht Hqt:


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..

.qt h

L FH j

g F (1.38)

2. .cosj r

* Thay cc tn tht vo trong (1.28) ta c c:

2 .1 .( ). .

2 . .h a

h

h h

P P F c G R L F H j

F g f g F

(1.39)

*Nhn xt:

- hP

ph thuc chuyn ng ca piston (c) v gia tc ca n (j).

- Gi tr hP

cng nh th Hhcng ln.

*Bin php tng kh nng ht ca my bm piston :

- Tng p sut mt thong bng cch cho cht lng vo bnh kn.

- Gim chiu cao ht.

- Gim cc yu t gy tn tht (chiu di ng ht, van, ct ni,).

1.12. Qu trnh y ca my bm piston

Ta gi Pl p sut trong xi lanh qu trnh y. Khi P

l ct p

tng ng vi Pv n bng tng cc ct p sau:

- Ctp cht lng i qua van y (Hv).

- Ct p nng cht lng ln ti chiu cao y (H).

- Ct p thng tt c cc lc cn trong ng y (Hod).

- Ct p cn bng lc qun tnh (Hqt).

- Ct p thng p sut trc ca ra ca ng y (Pod).


25/33

25

> n121

nn

RG

Q

H DB

CA0

p dng phng trnh becnuli cho 2 mt ct (a-a) v (b-b) nh hnh

v mc 1.8 ta c:

22 2 .' ( ) ( ) 1 . .

2 . od od

v d

p L F PF F cH H j

F F g g F

(1.40)

*Nhn xt:

- p sut Pc gi tr ln nht khi piston bt u chuyn ng (x =

S)v nh nht khi piston cui hnh trnh y(x = 0).

1.13. ng c tnh ca my bm piston

1.13.1. ng c tnh c bn H = f(Q) vi hai s vng quay lm vic

khc nhau n1v n2(n1> n2)

Hnh 1.9. ng c tnh l thuyt

Theo l thuyt th ng c tnh l thuyt ca my bm l AB (n1)

v CD (n2) do H v Q khng ph thuc vo nhau.

Thc t khi p sut tng th s c tn tht lu lng do cht lng rr qua b phn lm kn hoc van an ton m x bt cht lng v b ht

khi p sut bung lm vic qu cao, iu ny lm cho lu lng thc ca

my bm b gim. V vy ng c tnh c on AG v CR.


26/33

26

Q = f(H)

N = f(H)

H

N, Q,

0

1.13.2. ng c tnh lm vic Q = f(H), N = f(H), = f(H) ng vi n =

const

Hnh 1.10. ng c tnh lm vic.

- iu chnh Q bng cch thay i H, vi n = const.

- Khi H = const th Q, N, s t l thun vi n.

1.13.3.ng c tnh xm thc ca my bm

Hin tng xm thc my bm l hin tng xut hin bt kh

trong cht lng c bm. Nguyn nhn chnh gy ra hin tng xm

thc l do sxut hin cc bt kh, xy ra khi:

- Chiu cao ht qu ln lm gim nhit si.

- Nhit cht lng qu cao.

- Trong cht lng c kh ng hnh.

- ng ng ht qu nh, qu di lm tng tn tht thy lc.

ng c tnh xm thc cho thy khnng lm vic bnh thng

ca my bm ng vi svng quay khng i v nhit lm vic nht

nh phthuc chn khng ca my bm.


27/33

27

Qo_2

_2o 1(nt)

2(//)

a

No

N1

P

Qo

P1

Po

O

P

Q

Hnh 1.11. ng c tnh xm thc ca my bm

K1, K2l im gii hn phm vi lm vic an ton ca bm ng vi

trsp sut chn khng gii hn. Nu chn khng vt qu cc trs

gii hn th bm slm vic trong tnh trng bxm thc.

1.14. Ghp my bm

Trn hnh v1.12 biu din sphthuc gia p sut trong htun

hon v lu lng trong qu trnh khoan (biu din qua ng Oa).

Hnh 1.12. c tnh khi ghp my bm

Q

Q2

Q1

K2 K1Kgh

n2

n1

n2>> n1


28/33


29/33

29

. . .

60

i F S nQ

Tcng thc trn ta thy thay i gi trlu lng Q, ta c th

thc hin nhng phng php sau:

- Thay i s cp piston-xi lanh (i): s cp piston xi lanh t l

thun vi lu lng ca my bm, c thtng thm 1 hay 2 cp ty theo

thit kca my bm ang sdng.

- Thay i cp piston xi lanh: phng php ny chnh l thay i

ng knh cp piston xi lanh (thay i tit din F ca piston). Mi my

bm u c thit ksao cho ph hp mt vi bpiston xi lanh (trong

cng tc khoan du kh thng c t 6 n 12 b) vi ng knh cppiston xi lanh thay i trong khong d = 10 12 mm.

- Tng chiu di hnh trnh piston (S), phng php ny thc hin

trong qu trnh thit k my bm cho t hp hay nhim v nht nh.

Trong cng tc khoan ti khoan trng, phng php ny khng sdng

c.

- Thay i t trng ca cht lng cn bm: khi t trng cht lng

gim th lu lng ca my bm stng ln v ngc li.

* Ch :

My bm dung dch khoan l b phn quan trng nht trong qu

trnh tun hon dung dch khoan. My bm cn cung cp lu lng dung

dch cn thit trong qu trnh khoan. Lu lng ca my bm khoan c

la chn da vo cc thng s tiu chun sau:

- Vn tc nng dung dch khoan trong khong khng vnh xuyngia ging khoan v ct cn khoan;

- Ra sch dng c khoan;

- Thi gian ti a nng ht mn ln mt;

- Dng dng chy trong khong khng;


30/33

30

3

1

4

2

H

v1

d

v

f1

v1

d1

- n nh thnh ging khoan;

- Khoan bng ng c y.

p lc y ca my bm lin quan trc tip n tn tht p sut

trong h thng tun hon dung dch, tn tht vi cc vi phun dng c ph, vi s st p ng c y, vi lu lng v cc tnh cht vt l ca

dung dch.

1.16. Slm vic ca van trong my bm piston v tnh ton van

1.16.1. Cu to van (van ht v van y tng tnhau)

Hnh 1.13. Cu to van

1. a van

2. van

3. L xo

4.Thnh bung lm vic

V1: vn tc qua van;

F1: din tch thot ca van;

L: chu vi a van;

H: chiu cao nng ca a van;

V: tc qua khe hvan;


31/33

31

: hsdng chy khi cht lng chuyn ng trong khe h

van.

1.16.2. Tnh ton van

Trong my bm piston, van ht v van y c cu to tng tnhnhau v dng ngn cch cc bung lm vic ca my bm vi ng ht

v ng y. Van tmdi p sut cht lng, l xo lc ny sbnn v

a van tm. Van ng li nhtrng lng ca a van v sc nng ca

l xo.

Do dng chy ca cht lng l u v lin tc nn ta c:

1 1. .v f F c (1.41)

Khi van nng ti cao H th cht lng khi lng cht lng do

piston y ra v i qua khe hvan l:

. . . .

. . . .sin

. . . .

F c v L H

F c F r H

v L v L

(1.42)

Khi 2

th Hmax=

. .

. .

F r

v L

Cng tc thit kvan gm cc vic sau:

- Xc nh ng knh lvan.

- ng knh a van.

- Chiu cao nng ln nht.

- Kch thc l xo.

- Thit kcu to v tnh ton bn cc phn tca van.

iu kin tnh ton thit k van: L van phi lm vic m v

khng c ting g p khi ng van. Spht ra ting g p c lin quan

mt thit n tc khi a van hxung . Mun khng sinh ra ting


32/33

32

g p th theo Kykoleb sphthuc gia Hmaxv svng quay ca trcuj

khuu ca bm sl:

max. (800 1000)n H

Khi xc nh c Hmaxth ta sxc nh c chu vi a van L:

max

. .

. .

F rL

v H

(1.43)

Tc cht lng qua khe hvan c xc nh da vo loi cht

lng v chiu cao ht ca my bm:

-

i vi my bm piston c cng sut nhth v = 6 8 m/s.

-

i vi my bm piston c cng sut ln th v = 12 14 m/s.Gi tr Hmin c ly ph thuc vo kch thc ht mn ln nht,

thng th chn Hmin 2,5 mm.

Khi bit L th ta s tnh c ng knh a van t cng thc:

Ld

Khi cht lng i qua van c chnh p, v tc v lin quan n H

theo cng thc sau:

2. .v g H (1.44)

Gi tr H trong cng thc trn cn phi ln thng trng lng

van (G) v sc cng ca l xo khi m van (R):

.

G RH

f

(1.45)

T ta c c:

2

2 . . .. 2

G R vv g R f G

f g

(1.46)

Ta s c Rmaxtng ng vi vmax.

Khi ng van th sc cng ca l xo s l Rovi:


33/33

max

1 1( ).2 3o

R R

Cn c vo Rmax, Ro, v cc thong s d1, s vng l xo I, ta s chn

c l xo cho van ca my bm.

may bom khoan.pdf

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