Swpsc Paper on Buckling x - Viscous Drag

8/16/2019 Swpsc Paper on Buckling x - Viscous Drag

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Beam Pump Rod Buckling and Pump Leakage Considerations.

By: James C. Cox, Texas Tech Uni ersity, !. "ickens, BP, J. Lea, Texas TechUni ersity

#ntroduction

Pump slippage occurs on the upstroke and the slippage $rom a%o e the plunger, throughthe plunger&%arrel inter$ace, and to %elo' the plunger ser es to lu%ricate the plunger&

%arrel action. Usual industry $igures are (uoted to %e in the range o$ )&*+ o$ productionto pro ide lu%rication. This $igure could come under scrutiny as 'ell. #$ the slippage istoo large, then the system %ecomes ine$$icient. This can %e due to a 'orn plunger&%arrel,actually due to 'orn tra eling al e, or due to si ing the pump 'ith too large o$ aclearance.

Recent tests ha e sho'n that older e(uations used %y industry $or years may ha e o er& predicted the slippage. #$ this is true, as it appears to %e, then pumps can %e si ed 'ithlarger clearances, possi%ly eliminating some o$ the compression at the %ottom o$ the rodstring, and still not allo' excessi e leakage. #n this paper, a ery simple deri ation o$ theslippage e(uation is de eloped 'hich also pro ides an indication o$ the contri%ution toslippage due to plunger speed.

-ince %uckling considerations may arise $rom pump clearance and other $actors, rod %uckling e(uations are presented and re ie'ed. Rods %uckle due to outside $orces actingin compression at the %ottom o$ the rod string on the do'nstroke. nly negati e/e$$ecti e0 $orces excluding %uoyancy $orces contri%ute to %uckling and /true $orces0'hich include %uoyancy should not %e used in %uckling e(uations. 1xamples o$ ho' tocalculate the rod pro2ected area to %uoyancy induced pressure $orces are presented andexamples o$ calculating /true $orces0 and /e$$ecti e $orces0 are presented.

Pump $orces that contri%ute to %uckling include $orce to mo e the plunger in the %arreland also the pressure drop across the tra eling al e as the pump tra els do'n'ard.These $orces are calculated and examples are presented to the reader. The compressi e$orces predicted to initiate %uckling on the %ottom o$ the rod string range $rom a%out ))&34* l%$s $or rod si es $rom si e * to si e 35.

3


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Plunger& Barrel -lippageRelati ely recent 'ork 6Re$erences 3,), 7 89 ha e sho'n test results to indicate thetraditional expressions $or slippage 6 Re$erence 4, $or example9 greatly o er predict theslippage 6especially at larger diametral clearances o$ o er 5.5309 $or plunger&%arrelleakage on the upstroke o$ the %ottom hole pump in a %eam pump system.

-ho'n in ppendix #, is a short deri ation o$ the traditional ersion o$ the leakageexpression contrasted against the ne' leakage expression deri ed $rom recent test 'ork.

lso sho'n in ppendix #, is the traditional leakage term %ut also 'ith an additional term'hich accounts $or the plunger speed on the upstroke. This is a small e$$ect %ut one thatshould %e present. #t sho's that the plunger speed does increase the rate o$ slippage.

#n general slippage is approximately constant regardless o$ the pump speed. nother 'ayo$ saying this is that slippage is a %igger percentage o$ the production at lo' rates than itis at high pumping rates. dditional testing on plunger slippage considering a largenum%er o$ aria%les is under 'ay at the Texas Tech test 'ell in a pro2ect sponsored %yse eral interested operators 'ith donations $rom a num%er o$ endors.

Rod Buckling in Rods o er the Pump;hen studying pump slippage due to pump clearances, it comes to mind that theclearances in the pump ha e an e$$ect on 'hat compression the rods might see on thedo'nstroke. #n ppendix ##, the e(uations sho'ing 'hat $orces compressi e $orces arere(uired $or %uckling rods o er the pump are listed. Rod %uckling is not necessarilycatastrophic, %ut 'hen rods %uckle, then side $orces %et'een the rods and tu%ing occur,adding 'ear to %oth components.


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%uckling criteria 6 ppendix ##9 to determine i$ the %ottom region o$ the %ottom rod 'ill %uckle. ctually i$ a negati e Te$$ occurs at any other place in the rod string as 'ell, thenthe same criterion can %e used to determine i$ %uckling might %e occurring else'here inthe rod string.

Plunger ?iscous @rag


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assem%ly $lo' through area should %e as large as possi%le, especially 'hen pumping athigher speeds.


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-chlichting th 1dition, page , $lo' through t'o $lat plates, top place mo ing at U tothe right 6or up9 representing the plunger, top plate distance h $rom lo'er plate.

u = the local elocity at any y, $tKsecU = the elocity o$ the top plate, $tKsech = the distance %et'een plates, $tx = the distance along streamline %et'een plates, $tµ = the iscosity o$ the $luid

= ?iscosity, cp x .5555)5E l%$&secK$t )

dpKdx = l%$K$t8

τ = shear stress, l%$K$t)

93696)

)

h y

h y

dxdph

U h y

u −−= µ

9M6)

N

9O6)

Q

dxdph

hU

dydu

dxdph

hU

dydu

h y

h y

µ µ µ τ

µ

+−=−=

+=

=

=

#n the a%o e $igure, the dra'ing is relati e to %eing on the Plunger. Then on the upstrokeas the plunger mo es up'ard, the %arrel appears and is mo ing do'n'ard relati e to the

plunger as a%o e. The leakage is relati e to the plunger, or 'hat passes %eneath the plunger is leakage $rom a%o e the plunger to %elo' the plunger.

Upstroke: Leakage Analysis:

*


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upstrokeonly$oraccountto)Ka%o eleakage,BP@*. 3*K8 55x)4x$lo'xu upstroke9on6rateleakageBP@,

$t, 94344K696$lo'

$tKsecinter$ace, %arrel plunger throughelocitya erageu

$tKsec , 963))

93696)

96

)))

)5

)

5

≅

=−=

−=

−=−−

==∫ ∫

xdido

dxdphU

h

dyh y

h y

dxdph

U h y

h

dyuu

hh

π

µ

µ

The pressure decreases in the direction o$ the positi e U and , so the pressure gradientis a negati e alue. There$ore the sign o$ the second term %ecomes a positi e term 'hen∆ pKL is inserted %elo' $or dpKdx:

di9K) = @C = diameter clearance o$ %arrel #@ plunger @, in, C= 6do&di9P = pressure di$$erence across pump, psi 6 ∆ p9µ = iscosity o$ $luid, cp

L = length o$ plunger, $t

Leakage 0 a ple Cal-+lations:@ = ).55 inchesC=).55F ).55 = .55F inches∆ p = P = )555 psiL = 4 $tµ = 8 cpU = * $tKsec


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ctually the elocity U is the instantaneous pump elocity during the upstroke. napproximation might %e the a erage elocity on the upstroke at the pump %ut %etter

'ould %e to %reak up any leakage calculation into time increments, each 'ith the correctinstantaneous pump elocity.

BP@,leakage= 643.E 9 U@C > 6F8 4*9 @C8PKµl= 43.E x * x ).55 x 5.55F > F8 4* x ).55 x .55F 8 x )555 K 8 x 4 = 8.8 > 34.8= 3 . * %pd 6note: plunger elocity contri%utes 3E+ o$ leakage $or * $tKsec9

Varco = F 5 x ).55inches x )555psi x 6.55F9 3.*) K 64F x 8 cp9 = 3*. %pd6note: Varco does not determine leakage $rom plunger elocity9

6F8 4*9 @C8PKµl= 43.E x * x ).55 x 5.53 > F8 4* x ).55 x .53 8 x )555 K 8 x 4 = 4.3E > ) .E


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= 8).3 %pd 6note: plunger elocity contri%utes 38+ o$ leakage $or * $tKsec9

Varco = F 5 x ).55 inches x )555 psi x 65.539 3.*) K 64F x 8 cp9 = )).5 %pd

6F8 4*9 @C8PKµl= 43.E x * x ).55 x 5.53* > F8 4* x ).55 x .53* 8 x )555 K 3) = .)E > E4.)= 355.* %pd 6note: plunger elocity contri%utes + o$ leakage $or * $tKsec9

Varco = F 5 x ).55inches x )555psi x 6).53* ).9 3.*) K 64F x 8 cp9 = 45.F %pd

The rco leakage still much less at ery large clearances compared to the theoreticale(uation.

Leakage plots $or plunger elocity = 3 $tKsec, 'ith a%o e data same

F


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Leakage plots $or plunger elocity = * $tKsec, 'ith a%o e data same

%o e data 'ith plunger elocity ariation $rom 3 to * $ps

E


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Conclusions:

The $lat plate theory is a good $it to the rco e(uation up to a%out .53 in clearance. Thetheory here sho's idea o$ 'hat portion o$ leakage is due to plunger elocity using

iscous data. !o'e er this classical theory predicts that the $aster the plunger elocity onthe upstroke, then the more leakage.

lso leakage is predicted to ary slightly 'ith plunger elocityW ho'e er it is practicallynear constant $or di$$erent production rates. -o in $ield trials, a high production rate

'ould %e predicted to ha e a%out the same leakage rate as a lo' production rate usingthis theory. r in other 'ords, leakage 'ould %e a much %igger percentage o$ a lo' production rate 6gi en the same P and C and iscosity, and plunger length9 than it 'ould %e o$ a high production rate.

Pre ious $ield trials seemed to indicate that slippage decreased as the rate increased,ho'e er could it %e that $or larger production rates, the slippage 'as 2ust a smaller

percent o$ the 'hole and not in experimental accuracyX lso there could %e a largeentrance loss as $luids enter the plunger&%arrel inter$ace and this could a$$ect the results,

%ut this is not predicted here.

APP02D3 33 Rod Buckling 1(uations

35


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05+ations or o* B+-kling

33


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w is weight per foot in liquid E, steel, = 30 x 0 ! psi " = #o#ent of inertia of $ross se$tion, π d % &!%Rourk, R. J., and Young, ;. C., 'or#ulas for (tress and (train , DcHra'&!ill, "Y,"Y,3EF), P. *8E

3)

t

w E" L

ft in

w E" 'eff

psi x E

ind "

ft l)ww air

$ 83. *O344 ).*)FF

.54E5F85x35E*.5

OKE*.

l%$ F5.5*

OK344

).*)FF 5.54E5F 85x35 T. E*

O96E*.

3585

54E5F.4K4K

K*)FF.)93)F..36E.)93)F.36 :Rods#nchne$ornCalculatio1xample

8K3)

8K3)

8K3))

))

8K3)

44

==

=−=

−

−==

====−=−=

π

π

π

π π

γ


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Criti-al 7or-e an* Lengt' or o* B+-kling

38


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APPENDIX THREE: Fluid Pressure Forces, Archimedes Principle,True/Effective ForcesDepth: ft Assume ! rod "ith upsets#ne inch rods: $%&' l(/ft in !ir, Are! in )*section: '%+ -. s in%0ei1ht of )* rod len1th:π 2)2)/. 2 ft 2 Ac 2 ρ /).. 3 $%&' 2

ρ 3. +%- l(m/cu ft , from 0rf30r!4)%5%)$ γ6Ac 3 $%&'./ 4 π 2 ) /. 2 . +%-/)..6 3 )%'&)& 4)*rods6Ac 3 0tpft / d $ 2 '%7+8Ac 4+/ 9s6 3 $%$$7 / % +-$ 2 %7+8 3 )%'&)+Ac 47/.9s6 3 )%877 / %+- $ 2 %7+8 3 )%'&)8Ac 4-/ 9s6 3 )%)7 / %8$- $ 2 %7+8 3 )%' ++0here Ac multiplies the !re! of the (od tocorrect pro;ected !re! to !ccount for the rodupsets%

T"o st!tements ofArchimedes Principle

Archemedes Principle:Depth: $- feet0t! 3 $- feet 2 $%& for one inch rod 3 +$%- l(f 0tf 3 --%- 4)%5%)$ γ6 3 +$%- 2 % +$ 3 87%$$ l(f

or:P F 3 --%- > &%$+ 8 3 87%$$ l(f This s! s ?0ei1ht in !ir > sum of vertic!llpro;ected pressure 2 !re! forces 3 0ei1ht in fluid*

Also s!me !s 0tf 3 0t! 5 "t of fluid displ!ced3 +$%- 5 $-2%+ -.2)%'&'828$%./).. 3 87%$$ l(f


F

0t!

34


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Neutr!l Point:Depth: $- feet0t! 3 $- feet 2 $%&l(f/ft for one inch rod 3 +$%- l(f 0tf 3 --%- 4)%5%)$ γ6 3 +$%- 2 % +$ 3 87%$$ l(f #r:P 3 %.777 2 $- 3 )'% 7$- psiAc 4to !ccount for !re! of upsets6 3 )%'&'8F 3 P 2 A 2 Ac 3 )'% 7$- 2 %+ -. 2 )%'&'8 3 &%$+ 8 l(f 0t! 3 0t! > F 3 --%- > &%$+ 8 3 87%$$ l(f At (ottom, &%$+ 8 l(f/%+ -. s in 3 ))% ) psicompression 4true compression, &%$+ 8 l(f istrue compression force6Neutr!l point: &%$+ 8/$%& 3 7%)&& ft from (ottom4$- > 7%)&& ft6 4$%&6 3 87%$$ tension !t surf!ce, or"ei1ht in fluid, 0tf, "hich is true force !tsurf!ce, l(f%


F

Wta

F35&%$+ 8 l(f P3)'% 7$- psiA3%+ -. s in

0t!

$- ft

7%)&&ft

Neutr!lPoint

True Force Effective ForceTeff 3 Ttrue @ PoAo< !n point in rod

Teff 3 Ttrue @ Po4Ao2Ac635&%$+ 8 l(f @ )'% 7$-4%+ -.64)%'&'863 ero

0tf 3 87%$$ l(f

Ftrue3Bero Teff 3 ' @PoAoAc3 &%..8 psiAoAc3 %'& l(f

0tf 3 87%$$ l(f

TrueCompression

Teff 3 0tf @ PoAo30tf @ Bero3 0tf

3*


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F30tf !t surf >en1th 2 "t/ft

387%$$ > $%&4$-635&%$

0t!

$- ft

7%)&&ft

Neutr!lPoint

True Force Effective ForceTeff 3 Ttrue @ PoAo< !n point in rod

Teff 3 Ttrue @ Po4Ao2Ac635&%$+ 8 l(f @ )'% 7$-4%+ -.64)%'&'863 ero

0tf 3 87%$$ l(f

Ftrue3Bero Teff 3 ' @PoAoAc3 &%..8 psiAoAc3 %'& l(f

0tf 3 87%$$ l(f

TrueCompression

8 ft

True 30tf > 82$%&3.-% $ l(f < 8 ft

Teff30tf @PoAoAc3.-% $ @$%-&&2)%'&'82%+ -.3. %'.8 l(f < 8 ft

P3%.777283$%-&& psi

3


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IIENDIX THREE:r the fluid pound effect% the su(;ect,(ut not contr!r to most d n!mometer evidence

usu!ll presented for fluidPlot of Ttrue !nd Teff vs%Depth: No e2tern!l forces !pplied%

3


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Pt%=355 psi

-i e F. *F4in )

c=3.5E5).E54 GK$tL=3*8EZ

-i e . 53 in )

c=3.5E5)).))8 GK$tL=3*E Z

-i e .44) in )

c=3.5E*

3. 88 GK$tL=3F *Z

APPENDIX I0 a ple o 8tati- r+e%0 e-ti e Loa*s on apere* o* 8tring

F3P4A)* >A+/ 9s6

F3P4A7/.9s*6

F3P4A+/ 9s* >A7/.9s*6

P< ;unction

P< (ottom

P< ;unction

0rf 3 0r! 5 F9s

3 0r!4)5 %)$ γ6A9s !re pro;ected !re!s includin1 effects of upsets4correct "ith >Psurf 2 Ao if Psurf

si1nific!nt6

Teff = True+PoAo at anypoint in string

Distributionsof True and

Teff shown onnext page

3F


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t %ottom, no external loads: Po = 355 > .488[*555 = )) .* psi=.44) s( in, c = 3.5E* P =)) . *[.44)[3.5E*=35E psi

Ttrue = &35E l%$l Te$$=True>Po = &35E > 35E = 5.55 l%$s

Pt%=355 psi

-i e F. *F4in )

c=3.5E5).E54 GK$tL=3*8EZ

-i e . 53 in )

c=3.5E5)).))8 GK$tL=3*E Z

-i e .44) in )

c=3.5E*3. 88 GK$tL=3F *Z

Po = 355 > .488[838* = 34* .* psi Te$$ = True > Po oTrue = 3F *[.44) 35E = 3E4F.*4 l%$ Te$$=3E4F.*4 >34* [.44)[3.5E*=) *4l%$

∆ =6. 53[3.5E)&.44)[3.5E*9=.3 )8 in )

Po o=34* .*[3 )8 = )*3.38 l%$ Te$$ = Ttrue > Po oTtrue = 3E4F.*4 &)*3.38 = 3 E l%$ Te$$ = 3 E >34* .*[. 53[3.5E5)=) *).4 l%$

P = 355 > .488[3*FE = .8F psi Te$$ = True > Po oTtrue=3 E .4 > 3*E [).))8 = *)4*.8 l%$ Te$$= *)4*.8 >

.8F [. 53[3.5E5)=* 4 .4 l%$

∆ . F*4[3.5E5 &. 53[3.5E5) = .)53 in )

Po o = .8F [.)33=3*4 l%$ Te$$ = True > Po oTtrue = *)4*.8&3*4 = *5E3 l%$ Te$$ = *5E3> .8F [. F*4[3.5E5 =* 4 .4 l%$

Ttrue = *5E3 > 3*8E[).E54 = E* 5.8 l%$ Te$$ = True > Po oTe$$ = E* 5 > 355[. F*4[3.5E5 = E 4* l%$

;r$ = ;ra63.&.3)F γ 9 & Po o = 335 ). 6 3.&.3)F9 &355[. F*4[3.5E5 = E 4 . &F*. * =E* 5.E* l%$

#$ PR 3KF larger thenPRL = E* 5 355[6.EE4&. F*493.5E=E*8 l%$

"ot adding 'tK$t x length up to loadcell

8tati- r+e%0 e-ti e Loa*s on apere* o* 8tring

3E


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)5


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Stress should be calculated by dividing load by the actual area and not thepro ected area that is used to deter!ine buoyant forces fro! pressure ti!espro ected areas"

Plotting Dyno #ards$ The top card would be when you add the truestatic forces to dyna!ic calculated forces" The botto! card is when

you efffective static forces to dyna!ic forces for! a wave e%uation!odel"

Adding trueforces todyna!ic forcesto plot

Adding effectiveforces to

dyna!ic forcesto plot

&

&

PoAo !t(ottom of(ottom rod

Adding effective static forces to dyna!ic forces gives the botto! plot whichshows 'ero negative co!pression forces to indicate buc(ling unless outsideforces such as pu!p friction appears when dyna!ic forces are calculated" Thebotto! plot !a(es it !uch easier to see co!pressive forces contributing tobuc(ling if they are present and it is suggested to plot cards li(e this at thepu!p in the future"

Pump friction if !n

Pump friction if !n

)3


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& &

)oad* lbf )oad* lbf

P o s i t i o n

ax)oads

inloads

Addin1 true lo!d to d n!mic lo!ds Addin1 effective lo!ds to d n!mic lo!ds

=m!ll !mount of compression !t thepump is !l"! s there for re!l d!t! thisis indic!tive of forces tendin1 to (uc lethe rods%% Gsu!ll !(out )''5$'' l(fs%% If !lot more or hi1h in rod strin1, then poordesi1n or pump friction%

This l!r1e ne1!tive lo!d

"ith true forces is notrel!ted to (uc lin1%%it isdue to (ou !nc forces%=ee ne2t p!1e of notes

))


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,- & - . / 0 1& 1- 1. 1/ 10 -& --

&

.&&

0&&

1-&&

1/&&

-&&&

-.&&

-0&&

2-&&

2/&&

.&&&

..&&

.0&&

3-&&

Depth 4ft6

Rod o!d 4)''' l(f6 A)F456+"A#P$ 0"&& SP+* 1.."&&& in Stro(e* Production = .07"2 bbl8d

5od - Top

5od 2 Top

Pu!p

Computer Dodel utput: *555 deep, )0 pump, F rod string, )0 pump, F -PD, 355 psi sur$ pressure, \3sp.gr.,

)8


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Plunger ?iscous @rag nalysis on @o'nstroke:

Ldi *drag 9)K6) π = is the area that $luid iscous drag acts on, $t )

5.539 = & 5.E* l%$K$t)


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Plgr el=* $tKsec

Plgr el=3. $tKsec

Conclusion:


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Belo' is e$$ecti e load: -ome measurements like this sho' ery little pump resistance.

APP02D3 3


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Pressure @rop across the Tra eling ?al e on the @o'n&-troke:

typical ori$ice e(uation that could predict this pressure drop is:

sec,sec

,

,

,),

8)

8

)) ft l)f

ft l)# g

ft l)#

ft

l)f p

ft +*o,$

−

−∆

=

ρ

;here:V, $t 8Ksec

p = π 6@plgr9) K 64 x 3449, $t)

@plgr = ).5 inches ρ =density o$ $luid, l%mK$t8

gc = 8).) l%m&$tK6l%$&sec) 9

∆ p = pressure drop across T?, ps$ C = discharge coe$$icient, use C ≈ 5. $or calculations?el = elocity o$ the plunger, $tKsec 6$rom 'a e e(uation9

-ol ing $or a calculated pressure drop:

))

8

K,9)6

96

secK9,966

inl)f g *o+

, p

ft *o-el ,

$

ρ =∆

=

1xample calculation:

#nstantaneous plunger elocity = .5 $tKsec@ori$ice = 5.F40ρ = ).4 l%mK $t8

C = 5.o = π x .F4 x .F4 K 6344 x 49 = 5.558F4 $t )

V = ?el x plgr = x 8.34 x ) x ) K6344 x 49 = 5.385F $t 8Ksec∆ p, ps$ = 65.385FK65. x .558F49) ).4 K6) x 8).)9 = ))E4 l%$K$t) or 3*.E psi


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@iameter ?= $ps ?=3 $ps

less?than the flow area through the seat" Courtesy: Bennie Williams, HF ? Typical seat inside dia!eters for traveling valves$1"3 @ pu!p "//& dia!eter 1":3@ pu!p "03& dia!eter -"&&@ pu!p "7.& dia!eter -"-3 @ pu!p 1"&:& dia!eter

-ummary:%o e i$ the pump 'as a ).550 pump, and the $lo' area is )5+ less than 5.F*5 diameter

'ould calculate, then the e$$ecti e diameter 'ould %e 5.F40. #n the a%o e plot, theup'ard $orce 'ould %e a%out *) l%$s, $or $ps.

Conclusion: This could %e o$ concern $or %uckling $or higher plunger elocities on thedo'nstroke and small ori$ices. The $lo' through the T? can create enough pressure dropacross the plunger area to %e o$ concern $or %uckling.

)E


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] / The dynamic loads during $luid pounding can cause se eral detrimental e$$ectson the do'nhole e(uipment.: The rod string can experience %uckling that leads torod %reaks,W rod&to&tu%ing 'ear is increasedW shock loads contri%ute to coupling$ailure due to unscre'ingW and pump parts can %e damaged 6as 'ell as tu%ing9, i$unanchored. n the sur$ace , shock loads can damage pumping unit %earings, andcan lead to instantaneous tor(ues that o erload the speed reducer. / ./aka$s

APPENDIX IIFluid Pound vs" 5od #o!pression

85


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#s it possi%le that $luid pound does not lead to compression or %uckling in the lo'er rodsX

1 en in $luid pound, there is gas compression. ssume 3550 %ottom hole stroke. ssume^ o$ the %ottom hole stroke is $illed 'ith $luid and the other hal$ 'ith gas at 355 psi.

ssume )555 psi o er the T?. Then:

P3KP) = 355K)555 = ?)K?3 = 6 x 9 K 6*50 x 9 =-ol e $or and itZs e(ual to ).* inches or the T? 'ill open 'hen the T? is ) ^ / $romthe $luid.

#$ the initial pressure in the %arrel is *5 psi, then the T? opens 'hen the T? is 3 _0 $romthe $luid

-o the T? al'ays opens due to gas compression %e$ore it hits the $luid. #t 'ould ne er hitthe $luid 'ithout gas ha ing enough pressure to open the T? $irst.-till the plunger hits the $luid 'ith the T? open and goes $rom tra el in the gas toimpacting the $luid and tra eling do'n'ard through the $luid column.

#n %elo' $igure, $luid pound is usually sho'n in /cartoon0 $ashion as sho'ing no unusualcompression as a result o$ $luid pound, although usually in the text descri%ing $luid

pound, it usually says rod compression is a result o$ $luid pound.

83


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!o'e er in the %elo' $igure $rom Re$erence , compression is sho'n $or a $luid poundcard. This 'ould sho' compression due to $luid pound, although this type o$ %ottom holecard is usually not sho'n 'ith a $luid pound situation. Usually the do'n'ard spike,indicating compression to the rods o er the pump, is not present.

Conclusions:The conclusion might %e that $luid pound may or may not cause rod compression. #n most

%ottom hole dynagraphs 6measured or calculated9, no excess compression due to $luid pound is sho'n. lso it might generate such a short duration o$ a compression spike inthe rods, that it might not %e recorded in many cases.

areReal =alcul

ingress

Documents

Swpsc Paper on Buckling x - Viscous Drag