Reading Material Ch 12 Sucker Rod Pumping

Guo, Boyun / Computer Computer Assited Assited Petroleum Petroleum Production Production Engg

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Part III

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Artificial Lift Methods

Most oil reservoirs are of the volumetric type where the driving mechanism is the expansion of solution solution gas when reservoir pressure pressure declines declines because of fluid production. production. Oil reservoirs reservoirs will eventually not be able to produce fluids at economical rates unless natural driving mechanisms (e.g., aquifer and/or gas cap) or pressure maintenance mechanisms (e.g., water flooding or gas injection) are present to maintain reservoir energy. The only way to obtain high production rates of a well is to increase production pressure drawdown by reducing the bottom-hole pressure with artificial lift methods. Approximately 50% of wells worldwide need an artificial lift. The commonly used artificial lift methods include the following: . Sucker rod pumping . Gas lift . Electrical submersible pumping . Hydraulic piston pumping . Hydraulic jet pumping . Plunger lift . Progressing cavity pumping

Each method has applications for which it is the optimum installation. Proper selection of an artific artificial ial lift method method for a given given produc productio tion n system system (reserv (reservoir oir and fluid fluid propert properties ies,, wellbor wellboree configurati configuration, on, and surface surface facility facility restraints) restraints) requires a thorough thorough understandi understanding ng of the system. system. Economics analysis is always performed. Relative advantages and disadvantages of artificial lift systems are discussed in the beginning of each chapter in this part of this book. The chapters in this part provide production engineers with fundamentals of sucker rod pumping and gas lifts, as well as an introduction to other artificial lift systems. The following three chapters are included in this part of the book: Chapte Chapterr 12: Chap Chapte terr 13: 13: Chapte Chapterr 14: 14:

Sucker Sucker Rod Pumpin Pumping g Gas Gas Lift Lift Other Other Artifi Artificia ciall Lift Lift Methods Methods


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12

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Sucker Sucker Rod Pumping

Contents 12.1 12.1 Intr Introd oduc ucti tion on 12/1 12/162 62 12.2 12.2 Pump Pumpin ing g Syst System em 12/1 12/162 62 12.3 12.3 Poli Polish shed ed Rod Rod Moti Motion on 12/1 12/165 65 12.4 12.4 Load Load to the the Pump Pumping ing Unit Unit 12/1 12/168 68 12.5 12.5 Pump Pump Deliver Deliverabi abilit lity y and Power Power Requir Requireme ements nts 12/170 12/170 12.6 12.6 Procedu Procedure re for Pumpin Pumping g Unit Select Selection ion 12.7 12.7 Princip Principles les of Pump Perfor Performan mance ce Anal Analys ysis is 12/1 12/174 74 Summ Summar ary y 12/1 12/179 79 Refe Refere renc nces es 12/1 12/179 79 Prob Proble lems ms 12/1 12/179 79

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ARTIFI ARTIFICIA CIAL L LIFT LIFT METHOD METHODS S

12.1 Introduction Sucker rod pumping is also referred to as ‘‘beam pumping.’’ It provides mechanical energy to lift oil from bottom hole to surface. It is efficient, simple, and easy for field people to operate. It can pump a well down to very low pressure to maximize oil production rate. It is applicable to slim holes, multiple completions, completions, and high-tempera high-temperature ture and viscous viscous oils. The system system is also also easy easy to change to other wells with minimum cost. The major disadvantages of beam pumping pumping include excessive friction in crooked/ crooked/ deviated holes, solid-sensitive problems, low efficiency in gassy wells, limited depth due to rod capacity, and bulky in offs offsho hore re oper operat atio ions ns.. Beam Beam pump pumpin ing g tren trends ds incl includ udee improved pump-off controllers, better gas separation, gas handli handling ng pumps, pumps, and optimi optimizat zation ion using using surfac surfacee and bottom-hole cards.

12.2 Pumping Pumping System As shown shown in Fig. Fig. 12.1, 12.1, a sucker sucker rod pumping pumping system system cons consis ists ts of a pump pumpin ing g unit unit at surf surface ace and and a plun plunge gerr pump submerged in the production liquid in the well. The prime mover is either an electric motor or an internal combustion engine. The modern method is to supply each well with its own motor or engine. Electric motors are most desirable because they can easily be automated. The power from the prime mover is transmitted to the input shaft of a gear reducer by a V-belt drive. drive. The output shaft of the gear reducer drives the crank arm arm at a lower speed (4–40 revolutions per minute [rpm] depending on well characteristics and fluid properties). The rotary motion of the crank arm is converted to an oscillatory motion

Horse head

Walking beam Pitman Counter weight Gear reducer V-Belt

Bridle Polished rod Oil

by means of the walking the walking beam through beam through a pitman a pitman arm. arm. The horse’s horse’s head and and the hanger the hanger cable arrangement cable arrangement is used to ensure that the upward pull on the sucker rod string is vertical at all times (thus, no bending moment is applied to the the stuffing box). box). The The polished rod and and stuffing box combine to maintain a good liquid seal at the surface and, thus, force fluid to flow into the ‘‘T’’ connection just below the stuffing box. Conven Conventio tional nal pumpin pumping g units units are availab available le in a wide wide range range of sizes, sizes, with stroke stroke lengths lengths varyin varying g from from 12 to almost 200 in. The strokes for any pumping unit type are available in increments (unit size). Within each unit size, the stroke length can be varied within limits (about six different different lengths lengths being possible). These different different lengths lengths are achieved by varying the position of the pitman arm connection on the crank arm. Walking beam ratings are expressed in allowable polished ished rod loads loads (PRLs) (PRLs) and vary vary from from approx approxima imatel tely y 3,000 3,000 to 35,000 35,000 lb. Counte Counterba rbalan lance ce for convent conventiona ionall pumping units is accomplished by placing weights directly on the beam (in smaller units) or by attaching weights to the rotating rotating crank arm (or a combin combinati ation on of the two method methodss for larger larger units) units).. In more more recent recent design designs, s, the rotary counterbalance can be adjusted by shifting the position of the weight on the crank by a jackscrew or rack and pinion mechanism. There are two other major types of pumping units. These are are the the Lufk Lufkin in Mark Mark II and and the the AirAir-Ba Bala lance nced d Unit Unitss (Fig. 12.2). The pitman arm and horse’s head are in the same side of the walking beam in these two types of units (Class (Class III lever system). Instead of using counter-weights counter-weights in Lufkin Mark II type units, air cylinders are used in the airbalanced units to balance the torque on the crankshaft.

Stuffing box

Prime mover

Crank Sampson post

Tee

Gas

Casing Tubing Sucker rod

h t g n e l e k o r t

S

h t g n e l

e k o r t S

Downhole pump

Figure 12.1

A diagrammatic drawing of a sucker rod pumping system (used, with permission, from Golan and Whitson, 1991).


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SUCKER SUCKER ROD PUMPIN PUMPING G

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beam counterbalance units, and M is M is for Mark II units. The second field is the code for peak torque rating in thousands thousands of inch-pounds inch-pounds and gear reducer. D reducer. D stands for double-reduction gear reducer. The third field is the code for PRL rating in hundreds of pounds. The last field is the code for stroke length in inches.

The American Petroleum Institute (API) has established designations for sucker rod pumping units using a string of characters containing four fields. For example, C---228D---200---74: The first field is the code for type of pumping unit. C is is for conventional units, A is for air-balanced units, B B is for

Walking beam

Fulcrum n a m t i P

Force Well load

Counter balance

(a)

Walking beam

Fulcrum n a m t i P

Force Well load

Counter balance

(b)

Walking beam

Fulcrum

n a m t i P

Force

e c n a l a b r e t n u o C

Well load

(c) Figure 12.2 Sketch of three types of pumping units: (a) conventional unit; (b) Lufkin Mark II Unit; (c) air-balanced unit.


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Figure 12.3 illustrates the working principle of a plunger pump pump.. The The pump pump is inst instal alle led d in the the tubi tubing ng stri string ng below the dynamic liquid level. It consists of a working barrel and liner, liner, standing valve (SV), and traveling valve (TV) at the bottom of the plunger, plunger, which is connected to sucker rods. rods. As the plunger is moved downward by the sucker rod string string,, the TV is open, which allows allows the fluid fluid to pass pass

through the valve, which lets the plunger move to a position just above the SV. During this downward motion of the plunger, the SV is closed; thus, the fluid is forced to pass through the TV. When the plunger is at the bottom of the stroke and starts an upward stroke, the TV closes and the SV opens. As upward motion continues, the fluid in the well below the the SV is draw drawn n into into the the volu volume me abov abovee the the SV (flu (fluid id

Tubing Sucker rods Working barrel and liner

Travelling valve plunger Standing valve

(a)

(b)

(c)

(d)

moving down, near the bottom of the stroke; (b) plunger moving up, Figure 12.3 The pumping cycle: (a) plunger moving near the bottom of the stroke; (c) plunger moving up, near the top of the stroke; (d) plunger moving down, near the top of the stroke (used, with permission, from Nind, 1964).

(a) Tubing pump

(b) Rod pump

Figure 12.4 Two types of plunger pumps (used, with permission, from Nind, 1964).


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passing through the open SV). The fluid continues to fill the volume above the SV until the plunger reaches the top of its stroke. There There are two basic basic types types of plunge plungerr pumps: pumps: tubing pump and rod pump (Fig. 12.4). For the tubing pump, the working barrel or liner (with the SV) is made up (i.e., attached) to the bottom of the production tubing string and must be run into the well with the tubing. The plunger (with the TV) is run into the well (inside the tubing) on the sucker rod string. Once the plunger is seated in the working barrel, pumping can be initiated. A rod pump (both working barrel and plunger) is run into the well on the sucker rod string and is seated on a wedged type seat that is fixed fixed to the bottom bottom joint joint of the producti production on tubing tubing.. Plunger diameters vary from 5 ⁄ 8 4 5 ⁄ 8 in Plunger Plunger area varies from 0:307in:2 to 17:721 721 in:2 .

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conventional pumping units. Parameters are defined in Fig. 12.6. Air-Balanced Air-Balanced Pumping Pumping Unit. Unit. For this type of unit, the maximum acceleration occurs at the top of the stroke (the acceleration at the bottom of the stroke is less than simple harmonic motion). Thus, a lower maximum stress is set up in the rod system during transfer of the fluid load to the rods. The following analyses analyses of polished polished rod motion apply to conventional units. Figure 12.7 illustrates an approximate motion of the connection point between pitman arm and walking beam. If x x denotes the distance of B of B below below its top position C position C and is measured from the instant at which the crank arm and pitman arm are in the vertical vertical position with the crank arm vertically upward, the law of cosine gives (AB )2 ¼ (OA) OA)2 þ (OB )2  2(OA 2(OA)( )(OB OB )cos ) cos AOB ,

12.3 Polished Polished Rod Motion The theory of polished rod motion has been established since since 1950s 1950s (Nind, (Nind, 1964). 1964). Figure Figure 12.5 12.5 shows shows the cyclic cyclic motion of a polished rod in its movements through the stuffing box of the conventional pumping unit and the airbalanced pumping unit. Conventiona Conventionall Pumping Pumping Unit. Unit. For this type of unit, the accele accelerat ration ion at the bottom bottom of the stroke stroke is somewh somewhat at greater than true simple harmonic acceleration. At the top of the stroke, it is less. This is a major drawback for the conventional unit. Just at the time the TV is closing and the fluid load is being transferred to the rods, the acceleration for the rods is at its maximum. These two factors combine to create a maximum stress on the rods that becomes one of the limiting factors in designing an installation. Table 12.1 shows dimensions of some API

that is, h2 ¼ c2 þ (h þ c  x)2  2c(h þ c  x)cos vt, where v is the angular velocity of the crank. The equation reduces to x2  2x[h þ c(1  cos vt)] þ 2c(h þ c)(1  cos vt) ¼ 0 so that x ¼ h þ c(1  cos vt) 

p ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi

c2 cos2 vt þ (h2  c2 ):

When vt is zero, zero, x is also also zero, zero, which which means means that the negative root sign must be taken. Therefore, x ¼ h þ c(1  cos vt)  Acceleration is

Polished rod position

p ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi

c2 cos2 vt þ (h2 þ c2 ):

Simple harmonic motion

Conventional unit

d 1

d 2

d 1

d 2

Polished rod (a)

Polished rod

Conventional unit

Air-balanced unit

Air-balanced unit

0 (b)

90

180

270

360

Crank angle, degrees

pumping ping unit and air-balanced unit (used, with permission, permission, from Figure 12.5 Polished rod motion for conventional pum Nind, 1964).

AQ1


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A

C

P

R

H

G

I Figure 12.6 Definitions of conventional pumping unit API geometry dimensions.

Dimensions Table 12.1 Conventional Pumping Unit API Geometry Dimensions API API Uni Unitt Desi Design gnat atio ion n A (in (in.) .) C (in (in.) .) I (in (in.) .) P (in (in.) .) H (in (in.) .) G (in (in.) .) R1, R1, R2, R2, R3 (in. (in.) ) Cs (lb) (lb) Torque Factor Factor C-912D-365-168 C-912D-305-168 C-640D-365-168 C-640D-305-168 C-456D-305-168 C-912D-427-144 C-912D-365-144 C-640D-365-144 C-640D-305-144 C-456D-305-144 C-640D-256-144 C-456D-256-144 C-320D-256-144 C-456D-365-120 C-640D-305-120 C-456D-305-120 C-320D-256-120 C-456D-256-120 C-456D-213-120 C-320D-213-120 C-228D-213-120 C-456D-265-100 C-320D-265-100 C-320D-305-100 C-228D-213-100 C-228D-173-100 C-160D-173-100 C-320D-246-86 C-228D-246-86 C-320D-213-86 C-228D-213-86 C-160D-173-86 C-114D-119-86 C-320D-245-74 C-228D-200-74 C-160D-200-74 C-228D-173-74 C-160D-173-74 C-160D-143-74 C-114D-143-74 C-160D-173-64 C-114D-173-64 C-160D-143-64 C-114D-143-64 C-80D-119-64 C-160D-173-54

2 10 2 10 2 10 2 10 2 10 1 80 1 80 1 80 1 80 1 80 1 80 1 80 1 80 1 52 1 55 1 55 1 55 1 55 1 55 1 55 1 55 1 29 1 29 1 29 1 29 1 29 1 29 1 11 1 11 1 11 1 11 1 11 1 11 96 96 96 96 96 96 96 84 84 84 84 84 72

120.03 120.03 120.03 120.03 120.03 120.03 120.03 120.03 120.08 120.08 120.08 120.08 120.08 120.03 111.09 111.09 111.07 111.07 111.07 111.07 111.07 111.07 111.07 111.07 96.08 96.05 96.05 111.04 111.04 96.05 96.05 96.05 84.05 96.05 96.05 96.05 84.05 84.05 84.05 84.05 84.05 84.05 72.06 72.06 64 72.06

1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 12 120 11 111 11 111 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 96 96 96 1 11 1 11 96 96 96 84 96 96 96 84 84 84 84 84 84 72 72 64 72

148.5 148.5 148.5 148.5 148.5 148.5 148.5 148.5 144.5 144.5 144.5 144.5 144.5 148.5 133.5 133.5 132 132 132 132 132 132 132 132 113 114 114 133 133 114 114 114 93.75 114 114 114 96 96 93.75 93.75 93.75 93.75 84 84 74.5 84

237.88 237.88 237.88 237.88 237.88 237.88 237.88 238.88 238.88 238.88 238.88 238.88 238.88 238.88 2 13 2 13 2 11 2 11 2 11 2 11 2 11 2 11 2 11 2 11 1 80 1 80 1 80 2 11 2 11 1 80 1 80 1 80 150.13 1 80 1 80 1 80 152.38 152.38 150.13 150.13 150.13 150.13 1 32 1 32 1 16 1 32

8 6 . 88 8 6 . 88 8 6 . 88 8 6 . 88 8 6 . 88 8 6 . 88 8 6 . 88 8 9 . 88 8 9 . 88 8 9 . 88 8 9 . 88 8 9 . 88 8 9 . 88 89 9..88 75 75 75 75 75 75 75 75 75 75 63 63 63 75 75 63 63 63 5 53 3 . 38 63 63 63 5 3 . 38 5 3 . 38 53 53.38 53 53.38 53 53.38 53 53.38 45 45 41 45

47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 37, 37, 37, 42, 42, 37, 37, 37, 32, 37, 37, 37, 32, 32, 32, 32, 32, 32, 27, 27, 24, 27,

41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 32, 32, 32, 36, 36, 32, 32, 32, 27, 32, 32, 32, 27, 27, 27, 27, 27, 27, 22, 22, 20, 22,

35 35 35 35 35 35 35 35 35 35 35 35 35 35 30 30 30 30 30 30 30 30 30 30 27 27 27 30 30 27 27 27 22 27 27 27 22 22 22 22 22 22 17 17 16 17

1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 650 650 650 520 520 400 400 400 570 120 120 55 55 0 0 0 550 550 550 0 0 0 800 800 4 50 4 50 4 50 115 8 00 8 00 8 00 450 450 300 300 550 550 360 360 0 500

80.32 80.32 80.32 80.32 80.32 80.32 80.32 80.32 80.32 80.32 6 8 .8 2 6 8 .8 2 6 8 .8 2 6 8 .4 5 6 8 .4 5 6 8 .4 5 6 8 .4 5 6 8 .4 5 5 8 .1 2 5 7 .0 2 5 7 .0 2 5 7 .0 5 5 7 .0 5 5 7 .0 5 5 7 .0 5 5 7 .0 5 4 7 .4 8 4 7 .4 8 4 7 .4 8 4 8 .3 7 4 8 .3 7 4 8 .3 7 4 0 .9 6 4 0 .9 6 4 1 .6 1 4 1 .6 1 4 1 .6 1 4 0 .9 8 3 5 .9 9 3 5 .9 9 3 5 .9 9 3 5 .4 9 3 5 .4 9 3 5 .4 9 3 5 .4 9 3 1 .0 2 3 1 .0 2 30.59 30.59 3 0 .8 5 26.22 (Continued )


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Table 12.1

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Conventional Pumping Unit API Geometry Geometry Dimensions (Continued)

API API Uni Unitt Des Desig igna natio tion n A (in (in.) .) C (in (in.) .) I (in. (in.) ) P (in (in.) .) H (in (in.) .) G (in (in.) .) R1, R1, R2, R2, R3 (in.) (in.) Cs (lb) (lb) Torque Factor C-114D-133-54 C-80D-133-54 C-80D-119-54 C-P57D-76-54 C-P57D-89-54 C-80D-133-48 C-80D-109-48 C-57D-109-48 C-57D-95-48 C-P57D-109-48 C-P57D-95-48 C-40D-76-48 C-P40D-76-48 C-P57D-89-42 C-P57D-76-42 C-P40D-89-42 C-P40D-76-42 C-57D-89-42 C-57D-76-42 C-40D-89-42 C-40D-76-42 C-40D-89-36 C-P40D-89-36 C-25D-67-36 C-25D-56-36 C-25D-67-30 C-25D-53-30

72 72 72 64 64 64 64 64 64 57 57 64 61 51 51 53 53 56 56 56 56 48 47 48 48 45 45

a¼

64 64 64 51 51 64 56.05 56.05 56.05 51 51 48.17 47 51 51 47 47 48.17 48.17 48.17 48.17 48.17 47 48.17 48.17 36.22 36.22

64 64 64 51 51 64 56 56 56 51 51 48 48 47 51 51 47 47 48 48 48 48 48 47 48 48 36 36

74.5 74.5 74.5 64 64 74.5 6 5 . 63 6 5 . 63 65.63 64 64 57.5 56 64 64 56 56 57.5 57.5 57.5 57.5 57.5 56 57.5 57.5 49.5 49.5

116 116 116 103 103 116 105 105 10 1 05 103 103 98.5 95 103 103 95 95 98.5 98.5 98.5 98.5 98.5 95 98.5 98.5 84.5 84.5

41 41 41 39 39 41 37 37 37 39 39 37 39 39 39 39 39 37 37 37 37 37 39 37 37 31 31

2 4, 2 4, 2 4, 2 1, 2 1, 2 4, 2 1, 2 1, 2 1, 2 1, 2 1, 1 8, 1 8, 2 1, 2 1, 1 8, 1 8, 1 8, 1 8, 1 8, 1 8, 1 8, 1 8, 1 8, 1 8, 1 2, 1 2,

2 0, 2 0, 2 0, 1 6, 1 6, 2 0, 1 6, 1 6, 1 6, 1 6, 1 6, 1 4, 1 4, 1 6, 1 6, 1 4, 1 4, 1 4, 1 4, 1 4, 1 4, 1 4, 1 4, 1 4, 1 4, 8 9

16 16 16 11 11 16 11 11 11 11 11 10 10 11 11 10 10 10 10 10 10 10 10 10 10

3 30 3 30 3 30 1 05 1 05 4 40 3 20 3 20 320 1 80 1 80 0 1 90 2 80 2 80 2 80 2 80 1 50 1 50 1 50 1 50 2 75 3 75 2 75 2 75 150 150

2 6 .4 5 2 6 .4 5 2 6 .4 5 2 5 .8 2 5 .8 2 3 .5 1 2 3 .3 2 3 .3 2 3 .3 2 2 .9 8 2 2 .9 8 23.1 2 2 .9 2 2 0 .5 6 2 0 .5 6 1 9 .9 2 1 9 .9 2 2 0 .2 7 2 0 .2 7 2 0 .2 7 2 0 .2 7 1 7 .3 7 1 7 .6 6 1 7 .3 7 1 7 .3 7 1 4 .5 3 1 4 .5 3

c amax ¼ v2 c(1 þ ): (12:1) h It also appears that the minimum value of acceleration is c amin ¼ v2 c(1  ): (12:2) h If N is N is the number of pumping strokes per minute, then

d 2 x : dt2

Carryi Carrying ng out the differe differenti ntiati ation on for acceler accelerati ation, on, it is found that the maximum acceleration occurs when vt is equal to zero (or an even multiple of p radians) and that this maximum value is

C x B

h

Pitman arm

A w t t

c

Crank arm

O w

= length of pitman arm ( h ) OA = length of crank arm ( c ) AB

OB

= distance from center O to to pitman arm -walking beam connection at

B

Approximate e motion motion of connection connection point point between between pitman pitman arm and walking walking beam (used, (used, with with permission permission,, Figure 12.7 Approximat from Nind, 1964).


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v¼

(12:3)

The maximum downward acceleration of point B (which (which occurs when the crank arm is vertically upward) is cN 2 c 2 1 þ (ft=sec ) 91:2 h

or 2

amax ¼

   

cN g c 2 1 þ (ft =sec ): 2936:3 h

(12:4)

(12:5)

cN 2 g c 1  (ft =sec2 ): 2936:3 h

 

(12:6)

It follows that in a conventional pumping unit, the maximum upward acceleration of the horse’s head occurs at the bottom of the stroke (polished rod) and is equal to amax ¼

d 1 cN 2 g c 1 þ (ft =sec2 ), : d 2 2936 3 h

 

(12:7)

where d where d 1 and d 2 are shown in Fig. 12.5. However, 2cd 2 ¼ S , d 1



or cd 2 S ¼ : d 1 24

S f ¼ specific gravity of fluid in tubing D ¼ length of sucker rod string (ft) A p ¼ gross plunger cross-sectional area (in :2 ) Ar ¼ sucker rod cross-sectional area (in :2 ) specific weight of steel (490 lb=ft3 ) g s ¼ specific M ¼ Eq. (12.11). Note that for the air-balanced unit, M in M in Eq. (12.13) is replaced by 1-c/h 1-c/h.. Equation Equation (12.13) can be rewritten as

(12:9)

amin ¼

SN g c 1  (ft=sec2 ): 70471:2 h





(12:14)

g s DAr , 144

(12:15)

144W 144W r : g s D

(12:16)

Substituting Eq. (12.16) into Eq. (12.14) yields

(12:10)

Similarly,

 

g s DAr SN 2 M : 144 70471:2

PRLmax ¼ S f (62:4)

where M where M is is the machinery factor and is defined as c M ¼ (12:11) ¼ 1 þ : h 2

DA p DAr g s DAr  S f (62:4) þ 144 144 144

which can be solved for A for A r , which is Ar ¼

or we can write Eq. (12.9) as SN 2 g M (ft (ft=sec2 ), 70471:2

þ

W r ¼

So substituting Eq. (12.8) into Eq. (12.7) yields

 

(12:13)

If the weight of the rod string in air is

(12:8)

SN 2 g c 1 þ (ft=sec2 ), 70471:2 h



where

PRLmax ¼ S f (62:4)

2cd 2 S ¼ d 1 12

amax ¼

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A p  Ar ) g s DAr þ 144 144

g DAr SN 2 M , þ s 144 70471:2

where S where S is is the polished rod stroke length. So if S if S is is measured in inches, then

amax ¼

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involves only acceleration of the rods. Also, the friction term term andthe weightof weightof the plunge plungerr are neglec neglected ted.. We ignore ignore the reflective forces, which will tend to underestimate the maximum PRL. To compensate for this, we set the upthrust force to zero. Also, we assume the TV is closed at the instant at which the acceleration term reaches its maximum. With these assumptions, the PRLmax becomes PRLmax ¼ S f (62:4)D 4)D(

Likewise the minimum upward (a ( amin ) acceleration of point B B (which occurs when the crank arm is vertically downward) is amin ¼

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2p N N (rad=sec): 60

amax ¼

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(12:12)

For air-balanced units, because of the arrangements of the levers, the acceleration defined in Eq. (12.12) occurs at the bottom of the stroke, and the acceleration defined in Eq. (12.9) occurs at the top. With the lever system of an airbalanced unit, the polished rod is at the top of its stroke when the crank arm is vertically upward (Fig. 12.5b).

þ W r

DA p W r  S f (62:4) þ W r g s 144

SN 2 M : 70471:2





(12:17)

The above equation is often further further reduced by taking the flui fluid d in the the seco secondterm(thesub ndterm(thesubtr trac acti tiveterm veterm)) asan API508 with S f ¼ 0.78. Thus, Thus, Eq. (12.17) (12.17) becomes (where g s ¼ 490) PRLmax ¼ S f (62:4)

DA p SN 2 M  0:1W r þ W r þ W r 144 70471:2



or PRLmax ¼ W f þ 0:9W r þ W r DA

The load exerted to the pumping unit depends on well depth, rod size, fluid properties, properties, and system dynamics. dynamics. The maximum PRL and peak torque are major concerns for pumping unit.

12.4.1 12.4.1 Maximum Maximum PRL The PRL is the sum of weight of fluid being lifted, weight of plunger, weight of sucker rods string, dynamic load due to accele accelerat ration ion,, fricti friction on force, force, and the up-thr up-thrust ust from from below below on plunger plunger.. In practi practice, ce, no force force attrib attributa utable ble to fluid fluid accele accelerat ration ion is requir required, ed, so the accele accelerat ration ion term term

SN 2 M , 70471:2





(12:18)

where W where W f ¼ S f (62:4) 144 p and is called the fluid load (not to be confused with the actual fluid weight on the rod string). Thus, Eq. (12.18) can be rewritten as PRLmax ¼ W f þ (0:9 þ F 1 )W r ,

12.4 Load to the Pumping Pumping Unit



(12:19)

where for conventional units F 1 ¼

SN 2 (1 þ hc )

(12:20)

70471:2

and for air-balanced units F 1 ¼

SN 2 (1  hc ) 70471:2

:

(12:21)

12.4.2 12.4.2 Minimum Minimum PRL The minimum PRL occurs while the TV is open so that the fluid column weight is carried by the tubing and not


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the rods. The minimum load is at or near the top of the stroke. Neglecting the weight of the plunger and friction term, the minimum PRL is W r PRLmin ¼ S f (62:4) þ W r  W r F 2 , g s which, for API 508 oil, reduces to PRLmin ¼ 0:9W r  F 2 W r ¼ (0:9  F 2 )W r ,

(12:22)

where for the conventional units F 2 ¼

(12:23)

70471:2

and for air-balanced units 2

F 2 ¼

SN (1 þ

c ) h

70471:2

:

(12:24)

12.4.3 12.4.3 Counterwe Counterweights ights To reduce the power requirements for the prime mover, a counterbalance load is used on the walking beam (small units) or the rotary crank. The idea counterbalance load C is the average PRL. Therefore, C ¼ ¼

1 1 W f þ 0:9W r þ (F 1  F 2 )W r 2 2

(12:25)

or for conventional units 1 SN 2 c C ¼ ¼ W f þ W r 0:9 þ 2 70471:2 h





(12:26)

1 SN 2 c : W f þ W r 0:9  2 70471:2 h

(12:27)

and for air-balanced units C ¼ ¼





C ¼ ¼ C s þ W c

r d 1 , c d 2

C s ¼ structure unbalance, lb W c ¼ total weight of counterweights, lb r ¼ distanc distancee betwee between n the mass mass center center of counte counterrweights and the crank shaft center, in.

12.4.4 12.4.4 Peak torque and speed speed limit limit The peak torque exerted is usually calculated on the most severe possible assumption, which is that the peak load (polished rod less counterbalance) occurs when the effective crank length is also a maximum (when the crank arm is horizontal). Thus, peak torque T torque T is is (Fig. 12.5) d 2 T ¼ ¼ c½C  (0:9  F 2 )W r  : d 1

(12:28)

Substituting Eq. (12.25) into Eq. (12.28) gives 1 S ½C  (0:9  F 2 )W r  2

or 1 1 1 S b W f þ (F 1 þ F 2 )W r c 2 2 2



(12:29)

(12-30)

Because Because the pumping unit itself is usually not perfectly perfectly balanced balanced (C s 6 peak torque torque is also also affect affected ed by ¼ 0), the peak structure unbalance. Torque factors are used for correction: 1 ½PRLmax (TF 1 ) 2

þ PRLmin(TF 2 )  , : 0 93

(12:31)

where TF 1 ¼ maximum upstroke torque factor TF 2 ¼ maximum downstroke torque factor 0.93 ¼ system efficiency. For symmetrical symmetrical conventiona conventionall and air-balanced air-balanced units, TF ¼ TF 1 ¼ TF 2 . There is a limiting relationship between stroke length and cycles per minute minute.. As given given earlier earlier,, the maximum maximum value of the downward acceleration (which occurs at the top of the stroke) is equal to SN 2 g 1  hc

 

70471:2

,

(12:32)

(the + refers to conventional units or air-balanced units, see Eqs. [12.9] and [12.12]). If this maximum acceleration divided by g by g exceeds unity, the downward acceleration of the hanger is greater than the free-fall acceleration of the rods at the top of the stroke. This leads to severe pounding when when the polish polished ed rod shoulder shoulder falls falls onto onto the hanger hanger (leadin (leading g to failur failuree of the rod at the shoulder shoulder). ). Thus, Thus, a limit of the above downward downward acceleration acceleration term divided by g by g is limited to approximately 0.5 (or where L is determined by experience in a particular field). Thus, SN 2 1  hc

 

(12:33)

#L

or N limit limit ¼

s ffiffi ffi ffi ffi ffi ffi ffi ffi ffi

70471:2L : S (1 (1  hc )

(12:34)

For L For L ¼ 0.5, N limit limit ¼

where

T ¼ ¼



70471:2

The counterbalance load should be provided by structure unbalance unbalance and counterweig counterweights hts placed at walking walking beam (small units) or the rotary crank. The counterweights can be selected from manufacturer’s catalog based on the calculated C culated C value. The relationship between the counterbalance load C load C and and the total weight of the counterweights is

T ¼ ¼

1 2SN 2 W r (in:-lb): T ¼ ¼ S W f þ 4 70471:2

amax = min ¼

1 (PRLmax þ PRLmin ): 2

Using Eqs. (12.19) and (12.22) in the above, we get C ¼ ¼

or

T ¼ ¼

SN 2 (1  hc )

12/169 12/169

187:7 (1  hc ) S (1

p ffiffi ffi ffi ffi ffi ffi ffi ffi

:

(12:35)

The minus sign is for conventional units and the plus sign for air-balanced units.

12.4.5 Tapered apered rod rod strings strings For deep well applications, it is necessary to use a tapered suckerrod suckerrod stringsto stringsto reducethe reducethe PRLat thesurface. thesurface. Thelarger diameter rod is placed at the top of the rod string, then the next largest, and then the least largest. Usually these are in sequences up to four different rod sizes. The tapered rod strings are designated by 1/8-in. (in diameter) increments. Taperedrodstringscanbeidentifiedbytheirnumberssuchas a. No. 88 is a nontapered 8/8- or 1-in. 1-in. diameter rod string string b. No. 76 is a tapered string string with 7 ⁄ 8 -in. diameter rod at the top, then a 6/8-in. diameter rod at the bottom. c. No. 75 is a three-way three-way tapered tapered string consisting consisting of 7 ⁄ 8 -in. diameter rod at top 6/8-in. 6/8-in. diameter rod at middle 5 ⁄ 8 -in. diameter rod at bottom d. No. 107 is a four-way four-way tapered string string consisting consisting of 10/8-in. (or 11 ⁄ 4 -in.) diameter rod at top 9/89/8-in in.. (or (or 11 ⁄ 8 -in.) -in.) diameterrod diameterrod below below 10/8-in 10/8-in.. diamete diameterr rod 8/8-in. (or 1-in.) diameter rod below 9/8-in. diameter rod 7 ⁄ 8 -in. diameter rod below 8/8-in. diameter rod


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Tapered rod strings are designed for static (quasi-static) lads with a sufficient factor of safety to allow for random low level dynamic dynamic loads. loads. Two criter criteria ia are used in the design of tapered rod strings: 1. Stres Stresss at the top top rod rod of each rod size is the the same same throughout the string. 2. Stress in the top rod of the smallest smallest (deepest) set of rods should be the highest (30,000 psi) and the stress progressively decreases in the top rods of the higher sets of rods.

Then the expected maximum PRL is PRLmax ¼ W f  S f (62:4)

W r þ W r þ W r F 1 g s

¼ 5,770  (0:9042)(62:4)(6,138)=(490) : þ 6,138 þ (6,138)(0:794) ¼ 16,076lbs < 21,300lb,OK AQ2

c. The peak torque torque is calculated by Eq. (12.30): T ¼ ¼

The reason for the second criterion is that it is preferable that any rod breaks occur near the bottom of the string (otherwise macaroni).

1 2SN 2 W r S W f þ 4 70471:2







!

1 2(85:52)(22)2 (6,138) 5,770 þ ¼ (85:52) 5,770 4 70471:2

Exam Exampl plee Prob Proble lem m 12.1 12.1 The The foll follow owin ing g geom geomet etry ry dimensions are for the pumping unit C – 320D – 213 – 86:

280,056 lb-in: < 320,000lb-in:OK: ¼ 280,056 d. Accurate Accurate calculation calculation of counterbalan counterbalance ce load requires requires the minimum PRL:

d 1 ¼ 96:05in. d 2 ¼ 111in. c ¼ 37 in. c/h ¼ 0.33.

F 2 ¼

If this unit is used with a 21 ⁄ 2 -in. plunger and 7 ⁄ 8 -in. rods to lift 25 8API gravity crude (formation (formation volume factor 1.2 rb/stb rb/stb)) at depth depth of 3,000 3,000 ft, answer answer the follow following ing questions: a. What is the maximum maximum allowa allowable ble pumping pumping speed speed if L ¼ 0.4 is used? b. What is the expected maximum maximum polished polished rod load? c. What is the expected expected peak torque? torque? d. What is the desired counterbalan counterbalance ce weight to be placed at the maximum position on the crank? Solution

The pumpin pumping g unit C – 320D – 213 – 86 has a peak torque of gearbox rating of 320,000 in.-lb, a polished rod rating rating of 21,300 21,300 lb, and a maximum maximum polished polished rod stroke of 86 in. a. Based Based on the config configura uratio tion n for conven conventio tional nal unit unit shown in Fig. 12.5a and Table 12.1, the polished rod stroke length can be estimated as d 2 111 S ¼ 2c ¼ (2)(37) ¼ 85:52in: d 1 96:05 The maximum allowable pumping speed is

N ¼ ¼

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s ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi s ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi 70471:2L ¼ (1  hc ) S (1

(70471:2)(0:4) ¼ 22 SPM: (85:52)(1  0:33)

b. The maximum PRL can be calculated calculated with Eq. (12.17). The 258 API gravity has an S an S f ¼ 0:9042. The area of the 21 ⁄ 2 -in. plunger is A is A p ¼ 4:91in:2 . The area of the 7 ⁄ 8 -in. rod is A is A r ¼ 0:60in:2 . Then

SN 2 (1  hc ) 70471:2

¼

(85:52)(22)2 (1  0:33) ¼ 0:4 70471:2

W r þ W r  W r F 2 g s 6,138 ¼ (0:9042)(62:4) þ 6,138  (6,138)(0:4) 490 2,976 lb ¼ 2,976

PRLmin ¼ S f (62:4)

C ¼ ¼

1 1 (PRLmax þ PRLmin ) ¼ (16,076 þ 2,976) ¼ 9,526lb 9,526lb : 2 2

A product catalog of LUFKIN Industries indicates that the structure unbalance is 450 lb and 4 No. 5ARO counterweights placed at the maximum position (c ( c in this case) on the crank will produce an effective counterbalance load of 10,160 lb, that is, W c

(37) (96:05) þ 450 ¼ 10,160, (37) (111)

which gives W gives W c ¼ 11,221lb. To generate the ideal counterbalance load of C C ¼ 9,526 lb, the counterweights should be place on the crank at r¼

(9,526)(111) (37) ¼ 36:30in: (11,221)(96:05)

The computer computer program program SuckerRodPumpingLoad.xls can be used for quickly seeking solutions to similar problems. It is availab available le from from the publishe publisherr with with this this book. book. The solution is shown in Table 12.2.

12.5 Pump Deliverability and Power Power Requirements Liquid flow rate delivered by the plunger pump can be expressed as q¼

A p S p E v (24)(60) N (bbl=day) 144 12 B o 5:615

or DA p (3,000)(4:91) W f ¼ S f (62:4) ¼ (0:9042)(62:4) 144 144

¼ 5,770lbs W r ¼ F 1 ¼

g s DAr (490)(3,000)(0:60) ¼ ¼ 6,138lbs 144 144

SN 2 1 þ hc

 

70471:2

(85:52)(22)2 (1 þ 0:33) ¼ ¼ 0:7940: 70471:2

q ¼ 0:1484

A p NS p E v (stb=day), B o

where S where S p (in.), E v is p is the effective plunger stroke length (in.), E the volumetric efficiency of the plunger, and B and B o formation volume factor of the fluid.

12.5.1 12.5.1 Effective Effective plunger plunger stroke stroke length The motion of plunger at the pump-setting depth and the motion of the polished rod do not coincide in time and in magnitu magnitude de becaus becausee sucker sucker rods rods and tubing tubing string stringss are

AQ3


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Table 12.2 Solution Given by Computer Program Sucker Sucker RodPumpingLoad.xls SuckerRodPumpingLoad.xls Description: This spreadsheet calculates the maximum allowable pumping speed, the maximum PRL, the minimum PRL, peak torque, and counterbalance load. Instruction: (1) Update parameter values in the Input the Input section; section; and (2) view result in the Solution the Solution section. section. Input Data: Pump setting depth (D (D): Plunger diameter (d (d p ): Rod section 1, diameter (d ( d r1 ): length (L (L1 ): Rod section 2, diameter (d ( d r2 ): length (L (L2 ): Rod section 3, diameter (d ( d r3 ): length (L (L3 ): Rod section 4, diameter (d ( d r4 ): length (L (L4 ): Type of pumping unit (1 ¼ conventional; 1 ¼ Mark II or Air-balance d) d): Beam dimension 1 (d (d 1 ) Beam dimension 2 (d (d 2 ) Crank length (c (c): Crank to pitman ratio (c/h (c/h)): Oil gravity (API): Maximum allowable acceleration factor (L ( L): Solution: 2 S ¼ 2c d d 1

N ¼ ¼ A p ¼

3,000 ft 2.5 in. 1 in. 0 ft 0.875 in. 3,000 ft 0.75 in. 0 ft 0.5 in. 0 ft 1 96.05 in. 111 in. 37 i n . 0 .3 3 25 8 API 0 .4

¼ 85.52 in.

q ffiffi ffi ffi ffi ffi ffi ffi 70471:2L S (1 (1hc) d p 2 p d 4

¼ 22 SPM

Ar ¼ DA W f ¼ S f (62:4) 144 p g s DAr W r ¼ 144

¼ 4.91 in:2 ¼ 0.60 in. ¼ 5,770 lb ¼ 6,138 lb

F 1 ¼ 70471:2h r PRLmax ¼ W f  S f (62:4) W þ W r þ W r F 1 g

¼ 0.7940 8 ¼ 16,076 lb

d r2 p d 4

SN 2 (1c)



2

SN W r T ¼ ¼ 14 S W f þ 270471 :2



s

SN 2 (1c) F 2 ¼ 70471:2h r PRLmin ¼ S f (62:4) W þ W r g s C ¼ ¼ 12 (PRLmax þ PRLmin )

¼ 280,056 lb ¼ 0.40 ¼ 2,976 lb ¼ 9,526 lb

 W r F 2

elastic. Plunger motion depends on a number of factors including including polished rod motion, motion, sucker rod stretch, stretch, and tubing stretch. The theory in this subject has been well established (Nind, 1964). Two major sources of difference in the motion of the polishedrodandtheplungerareelasticstretch(elongation)ofthe rod string and overtravel. Stretch is caused by the periodic transfer of the fluid load from the SV to the TV and back again. The result is a function of the stretch of the rod string and the tubing string. Rod string stretch is caused by the weight of the fluid column in the tubing coming on to the rodstringat rodstringat the the botto bottom m ofthestroke ofthestroke when when theTV closes(th closes(this is load is removed from the rod string at the top of the stroke whenthe TVopens).It is apparentthatthe apparentthatthe plungerstrokewill plungerstrokewill be less than the polished rod stroke length S length S by by an amount equal equal to therod stret stretch.The ch.The magni magnitud tudee of the the rodstretc rodstretch h is

dl r ¼

W f Dr , Ar E

(12:36)

W f ¼ weight of fluid (lb) Dr ¼ length of rod string (ft) Ar ¼ cross-sectional area of rods (in:2 ) E ¼ modulus of elasticity of steel (30  106 lb=in:2 ). Tubing stretch can be expressed by a similar equation: W f Dt At E

W r Dr (ft): (12:38) Ar E But the maximum acceleration term n can be written as

dl o ¼ n

n¼

where

dl t ¼

But because the tubing cross-sectional area At is greater than the rod cross-sectional cross-sectional area Ar , the stretch of the tubing tubing is small small and is usuall usually y neglec neglected ted.. Howeve However, r, the tubing stretch can cause problems with wear on the casing. Thus, for this reason a tubing anchor is almost always used. Plunger overtravel at the bottom of the stroke is a result of the upward acceleration acceleration imposed on the downwarddownwardmoving sucker rod elastic system. An approximation to the extent of the overtravel may be obtained by considering a sucker rod string being accelerated vertically upward at a rate n times the acceleration of gravity. The vertical force force require required d to supply supply this this acceler accelerati ation on is nW r . The The magnitude of the rod stretch due to this force is

(12:37)

SN 2 1  hc

 

70471:2 so that Eq. (12.38) becomes

W r Dr SN 2 1  hc (ft ) , (12:39) dl o ¼ Ar E 70471:2 where again the plus sign applies to conventional units and the minus sign to air-balanced air-balanced units. Let us restric restrictt our discus discussio sion n to conven conventio tional nal units. units. Then Eq. (12.39) (12.39) becomes

 


12/172 12/172

dl o ¼

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W r Dr SN 2 M (ft): Ar E 70471:2

(12:40)

Equation (12.40) can be rewritten to yield dl o in inches. W r is W r ¼ g s Ar Dr 3

and g S ¼ 490lb=ft with E with E ¼ 30  106 lb=m2 . Eq. (12.40) becomes

dl o ¼ 1:93  1011 D2r SN 2 M (in:),

(12:41)

which is the familiar Coberly expression expression for overtravel overtravel (Coberly, 1938). Plunger stroke is approximate approximated d using the above expresexpressions as S p ¼ S  dl r  dl t þ dl o or S p ¼ S 

12D 12D E

 

 W f

1 1 þ Ar At



SN 2 M W r (in :):  70471:2 Ar



(12:42)

If pumpin pumping g is carried carried out at the maximum maximum permiss permissibl iblee speed limited by Eq. (12.34), the plunger stroke becomes S p ¼ S 

12D 12D E

 

 W f

1 1 þ Ar At





1 þ hc LW r (in :): 1  hc Ar



For the air-balanced unit, the term reciprocal.

1þch 1ch

(12:43)

is replaced by its

12.5.2 12.5.2 Volumet Volumetric ric efficiency efficiency Volumetric efficiency of the plunger mainly depends on the rate of slippage of oil past the pump plunger and the solution–gas ratio under pump condition. Metal-to-meta Metal-to-metall plungers plungers are commonly commonly available available with plunger-toplunger-to-barrel barrel clearance on the diameter diameter of 0.001, 0.002,  0.003,  0.004, and  0.005 in. Such fits are referred to as 1, 2, 3, 4, and 5, meaning the plunger outside outside diameter is 0.001 in. smaller than the barrel inside diameter. In selecting a plunger, one must consider the viscos viscosity ity of the oil to be pumped. pumped. A loose loose fit may be accept acceptabl ablee for a well well with with high high viscos viscosity ity oil (low (low API8 gravity). But such a loose fit in a well with low viscosity oil may be very very ineffi inefficien cient. t. Guideli Guidelines nes are as follow follows: s: a. Low-vi Low-visco scosit sity y oils (1–20 cps) cps) can be pumped pumped with a plunger to barrel fit of 0.001 in. b. High-viscosi High-viscosity ty oils (7,400 cps) will probably carry sand in suspen suspensio sion n so a plunger plunger-to -to-ba -barrel rrel fit or approx approxiimately 0.005 in. can be used. An empirical formula has been developed that can be used to calculate the slippage rate, qs (bbl/day), through the annulus between the plunger and the barrel:



2:9





d b þ d p D p k p d b  d p qs ¼ , : 0 1 L p m d b

(12:44)

where k p ¼ a constant d p ¼ plunger outside diameter (in.) d b ¼ barrel inside diameter (in.) D p ¼ differential pressure drop across plunger (psi) L p ¼ length of plunger (in.) viscosity of oil (cp). m ¼ viscosity The value of k k p is 2:77  106 to 6 :36  106 depending on field conditions. An average value is 4 :17  106 . The value of  p may p may be estimated on the basis of well productivity

index and production rate. A reasonable estimate may be a value that is twice the production drawdown. Volume Volumetric tric effici efficienc ency y can decrea decrease se signif significa icantl ntly y due to the presen presence ce of free free gas below the plunger. plunger. As the fluid is elevated and gas breaks out of solution, there is a significant difference between the volumetric displacement of the bottom bottom-ho -hole le pump and the volume volume of the fluid delive delivered red to the surface. surface. This This effect effect is denote denoted d by the shrinkage factor greater than 1.0, indicating that the bottom-hole pump must displace more fluid by some additio itiona nall perce percent ntag agee than than the the volum volumee deliv delivere ered d to the the surface surface (Brown, 1980). The effect of gas on volumetric efficiency depends on solution–gas ratio and bottom-hole pressure. pressure. Down-hole devices, called ‘‘gas anchors,’’ anchors,’’ are usuall usually y instal installed led on pumps pumps to separa separate te the gas from from the liquid. In summary, volumetric efficiency is mainly affected by the slippage slippage of oil and free gas volume volume below plunger. plunger. Both effects are difficult to quantify. Pump efficiency can vary over a wide range but are commonly 70–80%.

12.5.3 12.5.3 Power requiremen requirements ts The prime mover should should be proper properly ly sized sized to provid providee adequate power to lift the production fluid, to overcome friction loss in the pump, in the rod string and polished rod, rod, and in the pumping pumping unit. unit. The power required required for lifting fluid is called ‘‘hydraulic power.’’ It is usually expressed in terms of net lift: P h ¼ 7:36  106 qg l LN ,

(12:45)

where P h ¼ hydraulic hydraulic power, hp q ¼ liquid production rate, bbl/day g l ¼ liquid specific gravity, water ¼ 1 LN ¼ net lift, ft, and LN ¼ H þ

ptf , 0:433g l

(12:46)

where H ¼ depth to the average fluid level in the annulus, ft ptf ¼ flowing tubing head pressure, psig. The power requir required ed to overco overcome me fricti friction on losses losses can be empirically estimated as P f ¼ 6:31  107 W r SN :

(12:47)

Thus, the required prime mover power can be expressed as P pm ¼ F s (P h þ P f ),

(12:48)

where F where F s is a safety factor of 1.25–1.50. Exampl Examplee Proble Problem m 12.2 12.2 A well well is pump pumped ed off off (flu (fluid id level is the pump depth) with a rod pump described in Example Problem 12.1. A 3-in. tubing string (3.5-in. OD, 2.99 2.995 5 ID) ID) in the the well well is not not anch anchor ored. ed. Calcul Calculat atee (a) (a) expect expected ed liquid liquid produc production tion rate rate (use (use pump volume volumetri tricc efficiency 0.8), and (b) required prime mover power (use safety factor 1.35).

This problem problem can be quickly solved using using the computer program SuckerRodPumpingFlowrate&Power.xls.. The SuckerRodPumpingFlowrate&Power.xls The solut solutio ion n is shown in Table 12.3. Solution

12.6 Procedure for Pumping Unit Unit Selection The following procedure can be used for selecting a pumping unit:


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SuckerRodPumpingFlowrate&Power.xls er.xls Table 12.3 Solution Given by SuckerRodPumpingFlowrate&Pow SuckerRodPumpingFlowRate&Power.xls Description: This spreadsheet calculates expected deliverability and required prime mover power for a given sucker rod pumping system. Instruction: (1) Update parameter values in the Input the Input section; section; and (2) view result in the Solution the Solution section. section. Input Data: Pump setting depth (D (D): Depth to the liquid level in annulus (H ( H ): Flowing Flowing tubing head pressure ( p ( ptf ): Tubing outer diameter (d ( d to to ): Tubing inner diameter (d (d ti ti ): Tubing anchor (1 ¼ yes; 0 ¼ no): Plunger diameter (d (d p ): Rod section 1, diameter (d ( d r1 ): length (L (L1 ): Rod section 2, diameter (d ( d r2 ): length (L (L2 ): Rod section 3, diameter (d ( d r3 ): length (L (L3 ): Rod section 4, diameter (d ( d r4 ): length (L (L4 ): Type of pumping unit (1 ¼ conventional; 1 ¼ Mark II or Air-balance d) d): Polished rod stroke length (S (S ) Pumping Pumping speed (N (N ) Crank to pitman ratio (c/h (c/h)): Oil gravity (API): Fluid formation formation volume factor (B (B o ): Pump volumetric efficiency (E (E v ): Safety factor to prime mover power (F ( F s ): Solution:

At ¼ A p ¼

d t2 p d 42 d p p d 4 d r2 p d 4

Ar ¼ DA W f ¼ S f (62:4) 144 p g s DAr W r ¼ 144 M ¼ ¼ 1  hc S p ¼ S  12E D W f

h  

A NS E

1 Ar

2

SN M þ A1t  70471 :2

q ¼ 0:1484 p B o p v ptf LN ¼ H þ 0:433g 433g l P h ¼ 7:36  106 qg l LN P f ¼ 6:31  107 W r SN P pm ¼ F s (P h þ P f )

W r Ar

i

1. From From the maximu maximum m anticip anticipate ated d fluid fluid produc productio tion n (based on IPR) and estimated volumetric efficiency, calculate required pump displacement. 2. Based on well depth and pump displacemen displacement, t, determine determine API rating and stroke length of the pumping unit to be used.This used.This canbe done done usingeithe usingeitherr Fig. Fig. 12.8 12.8 orTable12.4. 3. Select tubing size, plunger size, rod sizes, and pumping speed from Table 12.4. 4. Calculate Calculate the fractional length length of each section section of the rod string. 5. Calculate the length of each section of the rod string to the nearest 25 ft. 6. Calculate Calculate the acceleration acceleration factor. factor. 7. Determine Determine the effective effective plunger stroke length. 8. Using the estimated estimated volumetric efficiency, efficiency, determine determine the probable production rate and check it against the desired production rate. 9. Calculate Calculate the dead weight of the rod string. 10. Calculate Calculate the fluid load. 11. Determine peak polished rod load and check it against the maximum beam load for the unit selected. 12. Calculate Calculate the maximum stress at the top of each rod size size and check check it agains againstt the maximu maximum m permiss permissibl iblee working stress for the rods to be used.

4 ,0 0 0 f t 4 ,0 0 0 f t 10 0 f t 3.5 in. 2.995 in. 0 2.5 in. 1 in. 0 ft 0.875 in. 0 ft 0 .7 5 i n . 4 ,0 0 0 f t 0.5 in. 0 ft 1 86 i n . 22 spm 0.33 8 25 8 API 1.2 rb/stb 0 .8 1 .3 5

¼ 2.58 in:2 ¼ 4.91 in:2 ¼ 0.44 in. ¼ 7,693 lb ¼ 6,013 lb ¼ 1.33 ¼ 70 in. ¼ 753 sbt/day ¼ 4,255 ft ¼ 25.58 hp ¼ 7.2 hp ¼ 44.2 hp 13. Calculate the ideal counterbalance counterbalance effect and check it againstthe againstthe counterb counterbalan alanceavailab ceavailable le forthe unitselected. unitselected. 14. From the manufa manufactu cturer rer’s ’s literat literature ure,, determ determine ine the position of the counterweight to obtain the ideal counterbalance effect. 15. On the assumption assumption that the unit will be no more than 5% out of counterbalance, calculate the peak torque on the gear reducer and check it against the API rating of the unit selected. 16. Calculate Calculate hydraulic hydraulic horsepower, friction horsepower, horsepower, and brake horsepower of the prime mover. Select the prime mover. 17. From the manufacturer manufacturer’s ’s literature, obtain the gear redu reduct ctio ion n rati ratio o and and unit unit shea sheave ve size size for for the the unit unit select selected, ed, and the speed of the prime mover. From this, determine the engine sheave size to obtain the desired pumping speed. Example Problem 12.3 A well is to be be put on a sucker sucker rod pump. The proposed pump setting depth is 3,500 ft. The anticipated production rate is 600 bbl/day oil of 0.8 specific gravity against wellhead pressure 100 psig. It is assu assumed med that the the work workin ing g liqu liquid id level level is low, low, and and a sucker rod string having a working stress of 30,000 psi is


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to be used. Select surface and subsurface equipment for the instal installat lation ion.. Use a safety safety factor factor of 1.35 for the prime prime mover power. Solution

1. Assumi Assuming ng volumet volumetric ric efficie efficiency ncy of 0.8, 0.8, the requir required ed pump displacement is

Therefore, the selected pumping unit and rod meet well load and volume requirements. 7. If a LUFKIN LUFKIN Indust Industrie riess C - 320D – 213 – 86 unit is chosen, the structure unbalance is 450 lb and 4 No. 5 ARO counterweights placed at the maximum position (c in this case) on the crank will produce an effective counterbalance load of 12,630 lb, that is,

(600)=(0:8) ¼ 750bbl=day:

W c

2. Based on well depth 3,500 ft and pump displacement displacement 750 bbl/day, Fig. 12.8 suggests API pump size 320 unit with 84 in. stroke, that is, a pump is selected with the following designation:

(37) (96:05) þ 450 ¼ 12,630lb, (37) (111)

which gives W gives W c ¼ 14,075lb. To generate the ideal counterbalance load of C C ¼ 10,327 lb, the counterweights should be placed on the crank at

C-320D C-3 20D---213-86 -21 3-86

r¼

3. Table 12.4 g suggests the following: following: Tubing size: 3 in. OD, 2.992 in. ID Plunger size: 21 ⁄ 2 in. Rod size: 7 ⁄ 8 in. Pumping speed: 18 spm 4. Table 12.1 12.1 gives d gives d 1 ¼ 96:05 in., in., d 2 ¼ 111in., c 111in., c ¼ 37 in., and h ¼ 114 in., thus c/h ¼ 0.3246. The spreadsheet spreadsheet program SuckerRodPumpingFlowRate&Power.xls program SuckerRodPumpingFlowRate&Power.xls gives gives AQ4

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(10,327)(111) (37) ¼ 31:4 in: (14,076)(96:05)

8. The LUFKIN LUFKIN Industries Industries C – 320D – 213 – 86 unit has a gear ratio of 30.12 and unit sheave sizes of 24, 30, and 44 in. are available. If a 24-in. unit sheave and a 750rpm electric electric motor motor are chosen, chosen, the diameter diameter of the motor sheave is d e ¼

(18)(30:12)(24) : ¼ 17:3 in: (750)

qo ¼ 687bbl=day > 600 bbl/day, OK P pm ¼ 30:2 hp 5. The spreadsheet spreadsheet program SuckerRodPumpingLoad.xls program SuckerRodPumpingLoad.xls gives PRLmax ¼ 16,121lb 4,533 lb PRLmin ¼ 4,533 T ¼ 247,755 lb < 320,000 in.-lb, OK C ¼ 10,327 lb

AQ5

6. The cross-section cross-sectional al area of the 7 ⁄ 8 -in. rod is 0.60 in.2. Thus, the maximum possible stress in the sucker rod is 30,000 psi: s max max ¼ (16,121)=(0:60) ¼ 26,809psi < 30,000 AQ6

OK

12.7 Principles of Pump Performance Performance Analysis The efficiency efficiency of sucker rod pumping units is usually usually analyzed using the information from pump dynagraph and polisher polisher rod dynamometer dynamometer cards. Figure Figure 12.9 shows a schematic of a pump dynagraph. This instrument is installed immediately immediately above above the plunger plunger to recordthe plunger stroke and the loads carried carried by the plunger plunger during the pump cycle. cycle. The relative motion between the cover tube (which is attached to the pump barrel and hence anchored to the tubing) and the calibrated rod (which is an integral part of the sucker rod string) is recorded as a horizontal line on the recording tube. This is achieved by having the recording tube mounted mounted on a winged winged nut thread threaded ed onto onto the calibrated rod and prevented from rotating by means of

2,500 Curve

) y a 2,000 d / l b b ( t n 1,500 e m e c a l p 1,000 s i D p m u 500 P

AP I s iz e

Stroke

A

40

34

B

57

42

C D E

80 114 160

48 54 64

F

228

74

G H

320 640

84 144

0 0

2,000

4,000

6,000

8,000

10,000

12,000

Pump Setting Depth (ft) Figure 12.8 Sucker rod pumping unit selection chart (used, with permission, from Kelley and Willis, 1954).


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for API Sucker Rod Rod Pumping Units Table 12.4 Design Data for (a) Size 40 unit with 34-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)

1,000–1,100 1,100–1,250 1,250–1,650 1,650–1,900 1,900–2,150 2,150–3,000 3,000–3,700 3,700–4,000

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 1

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

(b) Size 57 unit with 42-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)

1,150–1,300 1,300–1,450 1,450–1,850 1,850–2,200 2,200–2,500 2,500–3,400 3,400–4,200 4,200–5,000

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 1

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

(c) Size 80 unit with 48-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)

1,400–1,500 1,550–1,700 1,700–2,200 2,200–2,600 2,600–3,000 3,000–4,100 4,100–5,000 5,000–6,000

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 1

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

(d) Size 114 unit with 54-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)

1,700–1,900 1,900–2,100 2,100–2,700 2,700–3,300 3,300–3,900 3,900–5,100 5,100–6,300 6,300–7,000

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 1

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

(e) Size 160 unit with 64-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)

Pumping Speed (stroke/min)

2,000–2,200 2,200–2,400 2,400–3,000 3,000–3,600 3,600–4,200 4,200–5,400 5,400–6,700 6,700–7,700

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 1

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

24–19 24–19 24–19 23–18 22–17 21–17 19–15 17–15

⁄ 8 ⁄ 8 3 ⁄ 4 3 ⁄ 4 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 5 5 ⁄ 8 – 3 ⁄ 6 7

⁄ 8 ⁄ 8 3 ⁄ 4 3 ⁄ 4 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 5 5 ⁄ 8 – 3 ⁄ 6 7

⁄ 8 ⁄ 8 3 ⁄ 4 3 ⁄ 4 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 5 5 ⁄ 8 – 3 ⁄ 6 7

⁄ 8 ⁄ 8 3 ⁄ 4 3 ⁄ 4 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 4 5 ⁄ 8 – 3 ⁄ 5 5 ⁄ 8 – 3 ⁄ 6 7



24–19 24–19 24–19 24–19 24–19 23–18 22–17 21–17


24–19 24–19 24–19 24–19 23–18 23–19 21–17 19–17


⁄ 8 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 5 ⁄ 8 – 3 ⁄ 4 – 7 ⁄ 8 5 ⁄ 8 – 3 ⁄ 4 – 7 ⁄ 8 5 ⁄ 8 – 3 ⁄ 4 – 7 ⁄ 8 7

24–19 24–19 24–19 24–19 24–19 24–19 22–18 21–18

24–19 24–19 24–19 23–18 22–17 21–17 19–16 17–16


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(f) Size 228 unit with 74-in. stroke Pump Depth (ft)

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)


2,400–2,600 2,600–3,000 3,000–3,700 3,700–4,500 4,500–5,200 5,200–6,800 6,800–8,000 8,000–8,500

23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 11=16

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 2 2 2

7

⁄ 8 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 5=8---3 ⁄ 4 – 7 ⁄ 8 5=8---3 ⁄ 4 – 7 ⁄ 8 5=8---3 ⁄ 4 – 7 ⁄ 8

24–20 23–18 22–17 21–16 19–15 18–14 16–13 14–13

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)


23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 11=16

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2

7

Plunger Size (in.)

Tubing Size (in.)

Rod Sizes (in.)


23 ⁄ 4 21 ⁄ 2 21 ⁄ 4 2 13 ⁄ 4 11 ⁄ 2 11 ⁄ 4 11=16

3 3 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2 21 ⁄ 2

7

18–14 17–13 16–13 15–12 14–12 14–11 13–10 12–10

(g) Size 320 unit with 84-in. stroke Pump Depth (ft)

2,800–3,200 3,200–3,600 3,600–4,100 4,100–4,800 4,800–5,600 5,600–6,700 6,700–8,000 8,000–9,500 (h) Size 640 unit with 144-in. stroke Pump Depth (ft)

3,200–3,500 3,500–4,000 4,000–4,700 4,700–5,700 5,700–6,600 6,600–8,000 8,000–9,600 9,600–11,000

7

⁄ 8 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8

7

⁄ 8 –1 ⁄ 8 –1 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8 3 ⁄ 4 – 7 ⁄ 8

--1 --1 --1 --1 --1 --1

7

--1 --1 --1 --1 --1 --1

23–18 21–17 21–17 20–16 19–16 18–15 17–13 14–11

Sucker rod string Self-aligning bear Lugs

Tubing

Rotating tube with spiral grooves Cover tube with vertical grooves Recording tube

Calibrated rod

winged nut Stylus

Lugs Plunger assembly Pump liner

Figure 12.9 A sketch of pump dynagraph (used, with permission, from Nind, 1964).


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two lugs, which are attached to the winged nut, which run in vertical grooves in the cover tube. The stylus is mounted on a third tube, which is free to rotate and is connected connected by a self-aligning bearing to the upper end of the calibrated rod. Lugs attached to the cover tube run in spiral grooves cut in the outer surface of the rotating tube. Consequently, vertic vertical al motion motion of the plunge plungerr assemb assembly ly relati relative ve to the barrel results in rotation of the third tube, and the stylus cuts a horizontal line on a recording tube. Any change in plunger loading causes a change in length of the section of the calibrated rod between the winged nut supporting the recording tube and the self-aligning bearing supporting the rotating tube (so that a vertical line is cut on the recording tube by the stylus). When the pump is in operation, the stylus traces a series of cards, one on top of the other. To obtain a new series of cards, the polished rod at the well head is rotated. This rotation is transmitted to the plunger in a few pump strokes. Because the recording tube is prevented from rotating by the winged nut lugs that run in the cover tube grooves, the rotation of the sucker rod string causes causes the winged nut to travel travel—up —upwar ward d or downward downward depending depending on the direction of rotation—on rotation—on the thread threaded ed calibr calibrate ated d rod. rod. Upon the comple completio tion n of a series of tests, the recording tube (which is 36 in. long) is removed. It is important to note that although the bottom-hole dynagraph records the plunger stroke and variations in plunger loading, no zero line is obtained. Thus, quantitative interpretation interpretation of the cards becomes somewhat somewhat speculative unless a pressure element is run with the dynagraph. Figure 12.10 shows some typical dynagraph card results. Card (a) shows an ideal case where instantaneous valve actions at the top and bottom of the stroke are indicated. In general, however, some free gas is drawn into the pump on the upstroke, so a period of gas compression can occur on the down-stroke before before the TV opens. This is shown in card (b). Card (c) shows gas expansion during the upstroke giving a rounding of the card just as the upstroke begins. Card (d) shows fluid pounding that occurs when the well is almost pumped off (the pump displacement rate is higher than the formation formation is potential potential liquid production production rate). This fluid pounding results in a rapid fall off in stress in the rod string and the sudden imposed shock to the system. Card (e) shows that the fluid pounding has progressed so that the mechanical shock causes oscillations in the system. Card (f) shows that the pump is operating at a very low volumetric efficiency where almost all the pump stroke is being lost in gas compression and expansion (no liquid is being pumped). This results in no valve action and the area between the card nearly disappears (thus, is gas locked). Usually, Usually, this gas-locked condition condition is only temporary, and as liquid leaks past the plunger, the volume of liquid in the pump barrel increases until the TV opens and pumping recommences.

12/177 12/177

The use of the pump dynagr dynagraph aph involves involves pulling pulling the rods and pump from the well bath to install the instrument and to recover the recording tube. Also, the dynagraph cannot be used in a well equipped with a tubing pump. Thus, the dynagraph is more a research instrument than an operational operational device. Once there is knowledge from a dynagraph, surface dynamometer cards can be interpreted. The surface, or polished rod, dynamometer is a device that records the motion of (and its history) the polished rod during the pumping cycle. The rod string is forced by the pumping unit to follow a regular time versus position pattern. However, the polished rod reacts with the loadings (on the rod string) that are imposed by the well. The surface dynamometer cards record the history of the variatio variations ns in loading loading on the polished polished rod during during a cycle. The cards have three principal uses: a. To obtain information information that can be used to determine load, torque, and horsepower changes required of the pump equipment b. To improve pump operating operating conditions conditions such as pump speed and stroke length c. To check well conditions after installation installation of equipment to prevent or diagnose various operating problems (like pounding, etc.) Surface Surface instruments instruments can be mechanical, mechanical, hydraulic, and electrical. electrical. One of the most common mechanical mechanical instruinstruments is a ring dynamometer installed between the hanger bar and the polished rod clamp in such a manner as the ring may carry the entire well load. The deflection of the ring is propor proportio tional nal to the load, and this this deflec deflectio tion n is amplified amplified and transmitted transmitted to the recording recording arm by a series of levers. A stylus on the recording arm traces a record of the imposed loads on a waxed (or via an ink pen) paper card located on a drum. The loads are obtained in terms of polished rod displacements by having the drum oscillate back and forth to reflect the polished rod motion. Correct interpretation of surface dynamometer card leads to estimate of various parameter values. . Maximu Maximum m and minimum minimum PRLs PRLs can be read read direct directly ly

from the surface card (with the use of instrument calibration). These data then allow for the determination of the torque, torque, counterbalan counterbalance, ce, and horsepower horsepower requirements for the surface unit. . Rod stretc stretch h and contract contraction ion is shown shown on the surface surface dynamometer dynamometer card. This phenomenon phenomenon is reflected reflected in the surface surface unit unit dynamo dynamomet meter er card card and is shown shown in Fig. 12.11a for an ideal case. . Acceleration Acceleration forces cause the ideal card to rotate rotate clockwise. The PRL is higher at the bottom of the stroke and lower at the top of the stroke. Thus, in Fig. 12.11b, Point A is at the bottom of the stroke.

down-stroke, (c) gas expansion on Figure 12.10 Pump dynagraph cards: (a) ideal card, (b) gas compression on down-stroke, upstroke, (d) fluid pound, (e) vibration due to fluid pound, (f) gas lock (used, with permission, from Nind, 1964).


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. Rod vibration causes a serious complication in the in-

terpret terpretati ation on of the surface surface card. card. This This is result result of the closing of the TV and the ‘‘pickup’’ of the fluid load by the rod string. This is, of course, the fluid pounding. This phenomenon sets up damped oscillation (longitudinal and bending) in the rod string. These oscillations result in waves moving from one end of the rod string to the other. Because the polished rod moves slower near the top and bottom of the strokes, these stress (or load) fluctua fluctuatio tions ns due to vibrat vibration ionss tend tend to show show up more promine prominentl ntly y at those those locati locations ons on the cards. cards. Figure Figure 12.11c shows typical dynamometer card with vibrations of the rod string.

Figure 12.12 presents a typical chart from a strain-gage type of dynamometer measured for a conventional unit operated with a 74-in. stroke at 15.4 strokes per minute. It shows the history of the load on the polished rod as a function of time (this is for a well 825 ft in depth with a No. 86 three-tapere three-tapered d rod string). Figure 12.13 reproduces reproduces the data in Fig. 12.12 in a load versus displacement diagram. In the surface chart, we can see the peak load of 22,649 22,649 lb (which is 28,800 psi at the top of the 1-in. rod) in Fig. Fig. 12.1 12.13a 3a.. In Fig. Fig. 12.1 12.13b 3b,, we see see the the peak peak load load of 17,800 17,800 lb lb (which (which is 29,600 29,600 psi at the top of the 7 ⁄ 8 -in. rod). In Fig. 12.13c, we see the peak load of 13,400 lb (whic (which h is 30,3 30,300 00 psi psi at the the top top of the the 3 ⁄ 4 -in. -in. rod). rod). In

Rod stretch Rod contraction

Zero line

W r-W rb+W f W r-W rb

(a)

Zero line

Zero line A B C

F

(b)

W r-W rb+W f

W r-W rb

Zero line

(c)

Zero line

Dynamometer Card: (a) ideal card (stretch and contraction), contraction), (b) ideal card (acceleration), Figure 12.11 Surface Dynamometer (c) three typical cards (used, with permission, from Nind, 1964). 24,000 21,000 b l , d 18,000 a o 15,000 l d o 12,000 r d e 9,000 h s i l 6,000 o P 3,000 0 0

t f , d t 2 o r n e d m e 4 h e c s a i l l o p s 6 P i d

8 0

1

2

3

4 5 Time, sec

6

7

dynamometer chart. Figure 12.12 Strain-gage–type dynamometer

8

9


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12/179 12/179

28,000 24,000

(a)

20,000

(b)

16,000 b l , d a o L

(c)

12,000 8,000 4,000

(d)

0

−4,000

8

7

6

5

4

3

2

1

0

−1

−2

−3

−4

Displacement, ft Figure 12.13 Surface to down hole cards derived from surface dynamometer card. Fig. 12.13d is the dynagraph card at the plunger itself. This card indicates gross pump stroke of 7.1 ft, a net liquid stroke of 4.6 ft, and a fluid load of W f ¼ 3,200lb. 3,200lb. The shape of the pump card, Fig. 12.13d, indicates some downhole gas compression. The shape also indicates that the tubing anchor is holding properly. A liquid displacement rate rate of 200 bbl/ bbl/day day is calcula calculated ted and, and, compar compared ed to the surface surface measured measured production production of 184 bbl/day, bbl/day, indicated indicated no seri seriou ouss tubi tubing ng flow flowin ing g leak leak.. The The nega negati tive ve in Fig. 12.13d is the buoyancy of the rod string. The information derived from the dynamometer card (dynagraph) can be used for evaluation of pump performance and troubleshooting of pumping systems. This sub ject is thoroughly addressed by Brown (1980).\

Summary This chapter presents the principles of sucker rod pumping systems systems and illustrates a procedure procedure for selecting selecting components of rod pumping systems. Major tasks include calculations of polished rod load, peak torque, stresses in the rod string, pump deliverability, and counterweight placement. Optimization Optimization of existing pumping systems is left to Chapter 18.

References brown, k.e. The

Technology of Artificial Artificial Lift Methods Methods,, Vol. 2a. Tulsa, OK: Petroleum Publishing Publishing Co., 1980. coberly, c.j. Problems in modern deep-well pumping. Oil pumping. Oil Gas J . May 12, 1938. golan, m. and whitson, c.h. Well Performance, Performance, 2nd edition. Englewood Cliffs: Prentice Hall, 1991. nind, t.e.w. Principles of Oil Well Production. Production. New York: McGraw-Hill McGraw-Hill Book Co., New York, 1964.

Problems 12.1 If the dimensions dimensions d d 1 , d 2 , and c and c take the same values for both conventional conventional unit (Class I lever system) and air-balanced unit (Class III lever system), how different will their polished rod strokes length be? 12.2 What are the advantages advantages of the Lufkin Mark II and air-balanced units in comparison with conventional units?

12.3 Use your knowledge knowledge of kinematics kinematics to proof that for Class I lever systems, a. the polished rod will travel travel faster in down stroke than in upstroke if the distance between crankshaft and the center of Sampson post is less than dimension d dimension d 1 . b. the polished polished rod will travel faster faster in up stroke than in down stroke if the distance between crankshaft and the center of Sampson post is greater than dimension d dimension d 1 . 12.4 Derive a formula formula for calculating calculating the effective effective diameter of a tapered rod string. 12.5 Derive formulas formulas for calculating calculating length fractions fractions of equal-top-r equal-top-rod-str od-stress ess tapered rod strings strings for (a) twosized sized rod string strings, s, (b) threethree-siz sized ed rod string strings, s, and (c) four-s four-size ized d rod string strings. s. Plot Plot size size fracti fractions ons for each case as a function of plunger area. 12.6 A tapered rod string consists of sections of 5 ⁄ 8 - and 1 ⁄ 2 -in. rods and a 2-in. plunger. Use the formulas from Problem 12.5 to calculate length fraction of each size of rod. 12.7 A tapered rod string consists consists of sections of 3 ⁄ 4 -, 5 ⁄ 8 -, 1 and ⁄ 2 -in. rods and a 13 ⁄ 4 -in. plunger. plunger. Use the formulas from Problem 12.5 to calculate length fraction of each size of rod. 12.8 12.8 The follow following ing geomet geometry ry dimensi dimensions ons are for the pumping unit C – 80D – 133 – 48: d 1 ¼ 64in. d 2 ¼ 64in. c ¼ 24 in. h ¼ 74.5 in. Can this unit be used with a 2-in. plunger and 3 ⁄ 4 -in. rods to lift 308 API gravity crude (formation (formation volume factor factor 1.25 rb/stb) at depth of 2,000 ft? If yes, what is the required counter-balance load? 12.9 12.9 The follow following ing geomet geometry ry dimensi dimensions ons are for the pumping unit C – 320D – 256 – 120: d 1 ¼ 111:07in. d 2 ¼ 155in. c ¼ 42in. h ¼ 132in. Can this unit be used with a 2 1 ⁄ 2 -in. plunger and 3 ⁄ 4 -, 7 ⁄ 8 -, 1-in. 1-in. tapere tapered d rod string string to lift lift 22 8API gravit gravity y crude crude (for (forma mati tion on volum volumee fact factor or 1.22 1.22 rb/s rb/stb tb)) at a depth depth of 3,000 ft? If yes, what is the required counter-balance load?


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12.10 A well is pumped off with a rod pump described described in Problem Problem 12.8. 12.8. A 21 ⁄ 2 -in. tubing string (2.875-in. (2.875-in. OD, 2.441 ID) in the well is not anchored. Calculate (a) expected liquid production rate (use pump volumetric efficiency 0.80) and (b) required prime mover power (use safety factor 1.3). 12.11 A well is pumped with a rod pump described in Problem 12.9 to a liquid level of 2,800 ft. A 3-in. tubing string (31 ⁄ 2 -in. OD, 2.995-in. ID) in the well is anchored. anchored. Calculate Calculate (a) expected expected liquidproduction rate (use pump volumetric efficiency 0.85) and (b) required prime mover power (use safety factor 1.4). 12.1 12.12 2 A well well is to be put put on a suck sucker er rod pump. pump. The The proposed pump setting depth is 4,500 ft. The anticipated production rate is 500 bbl/day oil of 40 8 API

gravity against wellhead pressure 150 psig. It is assumed that the working liquid level is low, and a suck sucker er rod rod stri string ng havi having ng a work workin ing g stres stresss of 30,000 psi is to be used. Select surface and subsurface face equipme equipment nt for the instal installat lation ion.. Use a safety safety factor of 1.40 for prime mover power. 12.13 A well is to be put on a sucker rod pump. The proposed pump setting depth is 4,000 ft. The anticipated production rate is 550 bbl/day oil of 35 8 API gravity against wellhead pressure 120 psig. It is assume sumed d that that work workin ing g liqui liquid d leve levell will will be abou aboutt 3,000 ft, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsur subsurfac facee equipm equipment ent for the instal installat lation ion.. Use a safety factor of 1.30 for prime mover power.

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- Pls. verify figure number number change to 12.7 - Pls. check ‘‘ok’’ ‘‘ok’’ in Eq. 12.30 - Pls. check ‘‘ok’’ ‘‘ok’’ in Eq. 12.31 - Pls. check ‘‘ok’’ ‘‘ok’’ - Pls. check ‘‘ok’’ ‘‘ok’’ - Pls. check ‘‘ok’’ ‘‘ok’’

Reading Material Ch 12 Sucker Rod Pumping

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