bom piston

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

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 v 1.1, khi piston di chuyn t v tr B2 n 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 van ht vo khoang lm vic ca bm, trong lc ny van y ca bm vn ang trng thi ng.

3

Khi piston chuyn ng t v tr B2 n B1 th my bm thc hin qu trnh ht v lc piston dng li ti v tr B1 th qu trnh ht s kt thc.

Sau pison i chuyn ng v i ngc t B1 n 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 ca bm 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 B1 v 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

VV 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:

6

(2 ). .

60F 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 ). .

60F 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:

.tQ 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

vA A A

A AP Vl Z

g

(1.10)

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

.2

vB B B

B BP 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 VB 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 v thi gian c gi l cng sut thy lc hay cng sut c ch (Ntl);

9

. .75tlQ HN (1.17)

Cng sut thy lc chnh l c nng m cht lng trao i vi bm trong 1 n v thi gian.

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

tn hao c kh, c nh gi bng h s tl, 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 b do thay i tc dng chy khi cht lng chuyn ng t b cha n ng y. Ngoi ra cn thng lc qun tnh ca van.

ttll

HH

(1.18)

Ht, Hl: Ct p thc t v ct p l thuyt.

+ Tn hao th tch V: c xc nh bng h s ht y:

tVl

QQ

(1.19)

Qt, Ql: Lu lng thc t v lu lng l thuyt.

Nh vy, cng sut trn trc ca piston l cng sut lm vic hay cng sut ch bo (Nlv):

.

tllv

tl V

NN

(1.20)

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

Nh vy, cng sut ca ng c s l:

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:

10

LS

BA O

xS

B2B1

Pa

tldc

NN

. .tl V c (1.22)

Thng thng, = 0,67 0,85.

1.8. Phng trnh chuyn ng ca my bm piston (phng trnh vn tc)

Ta xt hnh v sau:

Hnh 1.3. Qu trnh chuyn ng ca piston.

L: chiu di thanh truyn,

R: bn knh ca tay quay,

S: hnh trnh ca piston,

X: qung ng chuyn ng ca piston t v tr im cht tri.

: tc gc ca trc khuu,

: gc quay ca tay quay t im cht tri.

T hnh v (1.2) ta thy:

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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. .

60r 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 tr rdt dt dt

=> . .sinc r (1.26)

* Gia tc tc thi ca piston l:

2sin. . .cos . .cosdc d dj r r r

dt 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 my bm 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:

12

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

QmQ

(1.30)

1.9.1. Lu lng ca my bm tc dng n

Hnh 1.4. th lu lng my bm tc dng n

13

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. . ..30

F r nQ F c

. . .2 . . ..60 60 30tb tb

F s n F r n F r nQ F c

max1tb

QmQ

1.9.2. My bm tc dng kp

Hnh 1.5. th lu lng my bm tc dng kp

14

O360300O180O60O OO

Qtb

2401200

Qmax

Trong my bm tc dng kp gm 2 xi lanh tc dng n, cc cht

khuu c t lch nhau 1 gc 180 o . Nh vy, nu mt piston thc hin bc y th piston kia thc hin bc ht. iu ny cng hon ton ng vi bm 1 xi lanh tc dng kp.

max max. . ..30

F r nQ F c

2. . . 2. .2 . . ..

60 60 15tb tbF s n F r n F r nQ F c

max2tb

QmQ

1.9.3. th lu lng ca my bm tc dng ba

Trong my bm tc dng ba, ba xi lanh tc dng n lm vic hon ton nh nhau, nhng tay quay ca chng c t lch nhau 1 gc l 120 o .

Hnh 1.6. th lu lng ca my bm tc dng ba.

15

Qtb

O360270O180OO

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. .


max3tb

QmQ

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 bm tc 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.

16

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

max1, 41. . . .2. . . .sin

30F r nQ db dc F r

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


max 1,41. 1,114tb

QmQ

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.

18

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 trong bnh 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

19

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 ha ht.

- 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 lm thay 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 PP

(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 VP VP

V

(1.32)

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

max min min max minmin

.( )2 2.tb

P P P V VPV

(1.33)

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

max min max min max minmax min

2.( )

tb tb tb

P P V V V V VP V V V V

(1.34)

V l lng cht lng c tch li trong bnh 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:

. .. . . .sin30tb

F r nd V dQ Q dt F r dt dt

. . 1. .sin . . .(sin ).30

30

F r n dF r d F r dn

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:

22

x

bb

Hh

dH

a aPa

- 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:

tbVV

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 cht lng 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 qtP 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 vHg g g

(1.36)

: h s tn tht cc b;

m : h s tn tht do ma st;

.

hh

F cvF

: vn tc ti ng ht;

F: tit din b mt piston;

Fh: tit din ng ht.

b. Tn tht Hv:

.vG RH

f

(1.37)

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

R: lc cng ca l xo (N);

f: tit din a van.

c. Tn tht Hqt:

24

. ..qt h

L FH jg F

(1.38)

2. .cosj r

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

2 .1 .( ). .

2 . .h a

hh h

P P F c G R L FH jF 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 Hh cng 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 P l p sut trong xi lanh qu trnh y. Khi P

l ct p

tng ng vi P v n bng tng cc ct p sau:

- Ct p 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

> 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:

2

2 2 .' ( ) ( ) 1 . .2 .

od odv

d

p L F PF F cH H jF F g g F

(1.40)

* Nhn xt:

- p sut P c 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 n1 v 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 r r 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

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 s xut 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 kh nng lm vic bnh thng ca my bm ng vi s vng quay khng i v nhit lm vic nht nh ph thuc chn khng ca my bm.

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, K2 l im gii hn phm vi lm vic an ton ca bm ng vi tr s p sut chn khng gii hn. Nu chn khng vt qu cc tr s gii hn th bm s lm vic trong tnh trng b xm thc.

1.14. Ghp my bm

Trn hnh v 1.12 biu din s ph thuc gia p sut trong h tun 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 K1 Kgh

n2

n1

n2 >> n1

28

Gi s trong qu trnh khoan, chng ta cn mt lu lng l Qo, khi p sut v cng sut thy lc tng ng s phi l Po v No. Nhng ti

thi im lm vic, ch c my bm c N1 no , gi s l N1 = 2oN .

Nh vy, c th t c gi tr cng sut No th ta phi tin hnh kt hp lm vic nhng my bm c N1.

1. Ghp ni tip hai my bm

Khi tin hnh ghp ni tip hai my bm, mi my bm s c lu

lng l Qo v p sut l 2oP . Khi s lm vic ca mi my bm s

c xc nh ti im 1 trn ng N1.

Tuy nhin, khi cho hai my bm ghp ni tip nhau lm vic, nu mt my b s c khng hot ng c th lu lng ca my kia khng i nhng my bm ch c p sut bng mt na p sut yu cu cho nn my bm khng lm vic bnh thng c.

khc phc hin tng trn ta cn lp thm my bm trn ng ht ca my bm piston (c th lp my bm ly tm).

2. Ghp song song hai my bm

Khi tin hnh ghp song song hai my bm, mi my bm s c

lu lng l 2

oQ v p sut l Po. Khi s lm vic ca mi my bm s

c xc nh ti im 2 trn ng N1.

Khi mt my bm gp s c m khng hot ng c th lu lng s gim i mt na, cn p sut s gim i 4 ln (im P1). Khi qu trnh khoan khng m bo nhng s khng xy ra hin tng phc tp g i vi my bm.

1.15. iu chnh lu lng ca my bm piston

Ta c cng thc:

29

. . .

60i F S nQ

T cng thc trn ta thy thay i gi tr lu 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 th tng thm 1 hay 2 cp ty theo thit k ca my bm ang s dng.

- 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 k sao cho ph hp mt vi b piston xi lanh (trong cng tc khoan du kh thng c t 6 n 12 b) vi ng knh cp piston xi lanh thay i trong khong d = 1012 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 s dng c.

- Thay i t trng ca cht lng cn bm: khi t trng cht lng gim th lu lng ca my bm s tng 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 xuyn gia 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

3

1

4

2

H

v1

d

v

f1v1

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. S lm vic ca van trong my bm piston v tnh ton van

1.16.1. Cu to van (van ht v van y tng t nhau)

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 h van;

31

: h s dng 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 t nh nhau v dng ngn cch cc bung lm vic ca my bm vi ng ht v ng y. Van t m di p sut cht lng, l xo lc ny s b nn v a van t m. Van ng li nh trng 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 h van l:

. . . .. . . .sin

. . . .

F c v L HF c F rHv L v L

(1.42)

Khi 2

th Hmax = . .. .

F rv L

Cng tc thit k van gm cc vic sau:

- Xc nh ng knh l van.

- ng knh a van.

- Chiu cao nng ln nht.

- Kch thc l xo.

- Thit k cu to v tnh ton bn cc phn t ca van.

iu kin tnh ton thit k van: L van phi lm vic m v khng c ting g p khi ng van. S pht ra ting g p c lin quan mt thit n tc khi a van h xung . Mun khng sinh ra ting

32

g p th theo Kykoleb s ph thuc gia Hmax v s vng quay ca trcuj khuu ca bm s l:

max. (800 1000)n H

Khi xc nh c Hmax th ta s xc nh c chu vi a van L:

max

. .. .F rLv H

(1.43)

Tc cht lng qua khe h van c xc nh da vo loi cht lng v chiu cao ht ca my bm:

- i vi my bm piston c cng sut nh th 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 Gf g

(1.46)

Ta s c Rmax tng ng vi vmax.

Khi ng van th sc cng ca l xo s l Ro vi:

33

max1 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.

bom piston

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