Section I

Sucker Rod Pumping

1.1 Introduction

Advantages• provides mechanical energy to lift oil• efficient, simple and easy to operate• pumps a well down to very low pressure• applicable to slim holes, multiple completions, and high-temperature and viscous oils

• easy to change to other wells with minimum cost

Disadvantages• excessive friction in crooked/deviated holes• solid-sensitive problems• low efficiency in gassy wells• limited depth due to rod capacity• bulky in offshore operations.

1.2 PumpingSystem

Pitm

an

Well Load

Fulcrum

Force

CounterBalance

Walking Beam

(a) Conventional Unit

Fulcrum

Pitm

an Force

CounterBalance

Well Load

Walking Beam

(b) Lufkin Mark II Unit

Fulcrum

Pitm

an

Force

Cou

nter

Bal

ance

Walking Beam

Well Load

(c) Air-Balanced Unit

Figure 1-3: The pumping cycle: (a) plunger moving down, near bottom of stroke; (b) plunger moving up, near bottom of stroke; (c) plunger moving up, near top of stroke; (d)

plunger moving down, near top of stroke (From Nind, 1964)

Figure 1-4: Two types of plunger pumps (From Nind, 1964)

1.3 Polished Rod Motion

Figure 1-5

G

I

HR

P

AC

Figure 1-6: Definitions of conventional pumping unit API geometry dimensions

Table 1-1: Conventional pumping unit API geometry dimensions

A C I P H G R1, R2, R3 Cs

(in.) (in.) (in.) (in.) (in.) (in.) (in.) (lb)

C-912D-365-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32C-912D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32C-640D-365-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32C-640D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32C-456D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32C-912D-427-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 -650 68.82C-912D-365-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 -650 68.82C-640D-365-144 180 120.03 120 148.5 238.88 89.88 47, 41, 35 -650 68.82C-640D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -520 68.45C-456D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -520 68.45C-640D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45C-456D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45C-320D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45C-456D-365-120 152 120.03 120 148.5 238.88 89.88 47, 41, 35 570 58.12C-640D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 -120 57.02C-456D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 -120 57.02C-320D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05C-456D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05C-456D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05C-320D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05

Torque Factor

API Unit Designation

C-228D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05C-456D-265-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48

C-320D-265-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48C-320D-305-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48C-228D-213-100 129 96.08 96 113 180 63 37, 32, 27 0 48.37C-228D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37C-160D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37C-320D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96C-228D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96C-320D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61C-228D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61C-160D-173-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61C-114D-119-86 111 84.05 84 93.75 150.13 53.38 32, 27, 22 115 40.98C-320D-245-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99C-228D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99C-160D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99C-228D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49C-160D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49C-160D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49C-114D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49C-160D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02C-114D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02C-160D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59C-114D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59C-80D-119-64 84 64 64 74.5 116 41 24, 20, 16 0 30.85C-160D-173-54 72 72.06 72 84 132 45 27, 22, 17 500 26.22C-114D-133-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45

C-80D-133-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45C-80D-119-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45C-P57D-76-54 64 51 51 64 103 39 21, 16, 11 105 25.8C-P57D-89-54 64 51 51 64 103 39 21, 16, 11 105 25.8C-80D-133-48 64 64 64 74.5 116 41 24, 20, 16 440 23.51C-80D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3C-57D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3C-57D-95-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3

C-P57D-109-48 57 51 51 64 103 39 21, 16, 11 180 22.98C-P57D-95-48 57 51 51 64 103 39 21, 16, 11 180 22.98C-40D-76-48 64 48.17 48 57.5 98.5 37 18, 14, 10 0 23.1

C-P40D-76-48 61 47 47 56 95 39 18, 14, 10 190 22.92C-P57D-89-42 51 51 51 64 103 39 21, 16, 11 280 20.56C-P57D-76-42 51 51 51 64 103 39 21, 16, 11 280 20.56C-P40D-89-42 53 47 47 56 95 39 18, 14, 10 280 19.92C-P40D-76-42 53 47 47 56 95 39 18, 14, 10 280 19.92C-57D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27C-57D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27C-40D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27C-40D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27C-40D-89-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37

C-P40D-89-36 47 47 47 56 95 39 18, 14, 10 375 17.66C-25D-67-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37C-25D-56-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37C-25D-67-30 45 36.22 36 49.5 84.5 31 12, 8 150 14.53C-25D-53-30 45 36.22 36 49.5 84.5 31 12, 9 150 14.53

API Designation

C – 228D – 200 – 74.

The first field is the code for type of pumping unit. C = Conventional unitsA = Air-Balanced unitsB = Beam Counterbalance unitsM = Mark II units.

The second field is the code for peak torque rating in 1000 in.-lb. D stands for Double Reduction Gear Reducer.

The third field is the code for polished rod load rating in 100 lb.

The last field is the code for stroke length in inches.

Approximate Motion

Figure 1-7

If x denotes the distance of B below its top position C and is measured from the instant at which the crank arm and pitman arm are in the vertical position with the crank arm vertically upward, the law of cosine gives

( ) ( ) ( ) ( )( ) AOBOBOAOBOAAB cos2222 −+=

( ) ( ) txchcxchch ωcos2222 −+−−++=

i.e.,

( )[ ] ( )( ) 0cos12cos122 =−++−+− tchctchxx ωω

( ) ( )2222 coscos1 chtctchx −+±−+= ωω

( ) ( )2222 coscos1 chtctchx ++−−+= ωω

where ω is the angular velocity of the crank. The equation reduces to

so that

When ωt is zero, x is also zero, which means that the negative root sign must be taken. Therefore,

2

2

dtxda =

( )hcca += 12

max ω

Acceleration is

Carrying out the differentiation for acceleration, it is found that the maximum acceleration occurs when ωt is equal to zero (or an even multiple of π radians) and that this maximum value is

(1-1)

( )hcca −= 12

min ω

602 Nπω =

It also appears that the minimum value of acceleration is

(1-2)

If N is the number of pumping strokes per minute then

(1-3)(rad/sec)

⎟⎠⎞

⎜⎝⎛ +=

hccNa 1

2.91

2

max

⎟⎠⎞

⎜⎝⎛ +=

hcgcNa 1

3.2936

2

max

The maximum downward acceleration of point B(which occurs when the crank arm is vertically upward) is

or

(ft/sec2) (1-4)

(ft/sec2) (1-5)

⎟⎠⎞

⎜⎝⎛ −=

hcgcNa 1

3.2936

2

min

⎟⎠⎞

⎜⎝⎛ +=

hcgcN

dd

a 13.2936

2

2

1max (ft/sec2)

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

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

(1-6)

(1-7)

(ft/sec2)

Sdcd

=1

22

122

1

2 Sdcd

=

241

2 Sd

cd=

where d1 and d2 are shown in Figure 1-5. But

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

or

(1-8)

⎟⎠⎞

⎜⎝⎛ +=

hcgSNa 1

2.70471

2

max

MgSNa2.70471

2

max =

So substituting Eq (1-8) into Eq (1-7) yields

or we can write Eq (1-9) as

(ft/sec2)

(1-9)(ft/sec2)

(1-10)

hcM +=1

⎟⎠⎞

⎜⎝⎛ −=

hcgSNa 1

2.70471

2

min (ft/sec2)

where M is the machinery factor and is defined as

Similarly,

(1-11)

(1-12)

1.4.1 Maximum PRL

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

2.704711441441444.62PRL

2

maxMSNDADAAA

DS rsrsrpf

γγ (1-13)

Equation (1-13) can be rewritten as

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛++−=

2.704711441441444.62

1444.62PRL

2

maxMSNDADADAS

DAS rsrsr

fp

fγγ

(1-14)

144rs

rDA

Wγ

=

DW

As

rr γ

144=

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛++−=

2.704714.62

1444.62PRL

2

maxMSNWW

WS

DAS rr

s

rf

pf γ

If the weight of the rod string in air is

which can be solved for Ar which is

Substituting Eq (1-16) into Eq (1-14) yields

(1-17)

(1-15)

(1-16)

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛++−=

2.704711.0

1444.62PRL

2

maxMSNWWW

DAS rrr

pf

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

2.704719.0PRL

2

maxMSNWWW rrf

( )144

4.62 pff

DASW =where

and is called the fluid load (not to be confused with the actual fluid weight on the rod string).

(1-18)

The above equation is often further reduced by taking the fluid in the second term (the subtractive term) as an API 50° with Sf = 0.78. Thus, Eq (1-17) becomes (where γs= 490)

or

( ) rf WFW 1max 9.0PRL ++=

( )2.70471

12

1hcSN

F+

=

( )2.70471

12

1hcSN

F−

=

Thus, Eq (1-18) can be rewritten as

where for conventional units

and for air-balanced units

(1-21)

(1-19)

(1-20)

1.4.2 Minimum PRL

( ) 2min 4.62PRL FWWWS rrs

rf −+−=

γ

( ) rrr WFWFW 22min 9.09.0PRL −=−=

which, for API 50° oil, reduces to

(1-22)

( )2.70471

12

2hcSN

F−

=

( )2.70471

12

2hcSN

F+

=

where for the conventional units

and for air-balanced units

(1-24)

(1-23)

1.4.3 Counterweights

( )minmax21 PRL PRL +=C

( ) rrf WFFWWC 2121

21 9.0 −++= (1-25)

The idea counter-balance load C is the average PRL. Therefore,

Using Eqs (1-19) and (1-22) in the above we get

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

hcSNWWC rf 2.70471

9.02

21

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

hcSNWWC rf 2.70471

9.02

21

or for conventional units

and for air-balanced units

(1-27)

(1-26)

2

1

dd

crWCC cs +=

where

Cs = structure unbalance, lbsWc = total weight of counterweights, lbsr = distance between the mass center of

counterweights and the crank shaft center, in.

1.4.4 Peak Torque and Speed Limit

( )[ ]1

229.0

ddWFCcT r−−=

( )[ ]rWFCST 221 9.0 −−=

Peak torque T is (see Figure 1.5)

Substituting Eq. (1-25) into Eq. (1-28) gives

(1-29)

(1-28)

or

( )[ ]rf WFFWST 2121

21

21 ++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

2.704712

2

41 r

fWSN

WST

or

(1-30)(in-lbs)

Torque factors and efficiency are used in practice:

[ ]93.0

)()( 2min1max21 TFPRLTFPRL

T+

=(1-31)

Torque factor is defined:

Load Rod PolishedCrankshaft the toExerted Torque

=TF

=1TF Maximum upstroke torque factor occurred when the crank is in the horizontal position

Maximum downstroke torque factor occurred when the crank is in the horizontal position

=2TF

Approximate Maximum Torque FactorConventional and Air Balance Units

Stroke (in.) TF1 (in.) TF2 (in.)16 8.5 8.524 13 1330 16 1636 19 1942 22 2248 26 2654 29 2964 34 3474 39 3986 45 45100 52 52120 63 63144 75 75168 87 87

Approximate Maximum Torque FactorMark II Units

Stroke (in.) TF1 (in.) TF2 (in.)

64 29 37

74 34 43

86 39 51

100 47 57

120 55 71

144 66 88

168 79 102

As given earlier the maximum value of the downward acceleration is equal to

( )2.70471

12

minmax/hcgSN

a±

= (1-32)

( )L

SN hc

≤±

2.7047112

(1-33)

Maximum Permissible Pumping Speed

( )hcSLN

m12.70471

limit =

( )hcS

Nm17.187

limit =

or

For L = 0.5

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

(1-34)

(1-35)

1.4.5 Tapered Rod Strings

Tapered rod strings can be identified by their numbers such as:

a. No. 88 is a non-tapered 8/8” or 1” diameter rod stringb. No. 76 is a tapered string with 7/8” diameter rod at the top,

then a 6/8” diameter rod at the bottom.c. No. 75 is a 3 way tapered string consisting of

7/8” diameter rod at top6/8” diameter rod at middle5/8” diameter rod at bottom

d. No. 107 is a 4 way tapered string consisting of10/8” (or 1 1/4”) diameter rod at top9/8” (or 1 1/8”) diameter rod below 10/8” diameter rod8/8” (or 1”) diameter rod below 9/8” diameter rod7/8” diameter rod below 8/8” diameter rod

There are two criteria used in the design of tapered rod strings:

1.Stress at the top rod of each rod size is the same throughout the string

2. Stress in the top rod of the 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.

The following geometry dimensions are for the pumping unit C – 320D – 213 – 86:

d1 = 96.05 in.d2 = 111 in.c = 37 in.c/h = 0.33

Example Problem 1-1:

If this unit is used with a 2 1/2” plunger and 7/8 in.rods to lift 25 °API gravity crude (formation volumefactor 1.2 rb/stb) at depth of 3,000 ft, answer thefollowing questions:

a) What is the maximum allowable pumping speed if L = 0.4 is used?

b) What is the expected maximum polished rod load?c) What is the expected peak torque?d) What is the desired counter-balance weight to be

placed at the maximum position on the crank?

The pumping unit C – 320D – 213 – 86 has a peak torque of gearbox rating of 320,000 in-lbs, a polished rod rating of 21,300 lbs, and a maximum polished rod stroke of 86 in.

(a)Based on the configuration for conventional unit shown in Figure 1-5(a) and Table 1-1, thepolished rod stroke length can be estimated as:

Solution:

in. 52.8505.96

111)37)(2(21

2 ===ddcS

( ) ( )33.01)52.85()4.0)(2.70471(

12.70471

−=

−=

hcS

LN

The maximum allowable pumping speed is:

= 22 SPM

( ) ( ) lbs 770,5144

)91.4)(000,3(4.62)9042.0(144

4.62 === pff

DASW

lbs 138,6144

)60.0)(000,3)(490(144

=== rsr

DAW

γ

(b) The maximum PRL can be calculated with Eq(12-17). The 25° API gravity has an Sf = 0.9042. The area of the 2 ½” plunger is Ap = 4.91 in.2 The area of the 7/8” rod is Ar = 0.60 in.2 Then

( ) ( ) 7940.02.70471

33.01)22)(52.85(2.70471

1 22

1 =+

=+

= hcSN

F

( ) 1max 4.62PRL FWWWSW rrs

rff ++−=

γ( ) )794.0)(138,6(138,6)490/()138,6(4.62)9042.0(770,5 ++−=

lbs 076,16=

Then the expected maximum PRL is:

< 21,300 lbs, OK

(c) The peak torque is calculated by Eq (1-30):

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

2.70471)138,6()22)(52.85(2770,5 )52.58(

2.704712

2

41

2

41 r

fWSNWST

= 280,056 lb-in. < 320,000 lb-in. OK

( ) ( ) 4.02.70471

33.01)22)(52.85(2.70471

1 22

2 =−

=−

= hcSN

F

( ) 2min 4.62PRL FWWWS rrs

rf −+−=

γ( ) )4.0)(138,6(138,6

490138,64.62)9042.0( −+−=

Accurate calculation of counter-balance load requires the minimum PRL:

= 2,976 lbs

( ) ( ) lbs 526,92,976 076,61PRL PRL 21

minmax21 =+=+=C

160,10450)111()05.96(

)37()37(

=+cW

A product catalog of LUFKIN Industries indicates that the structure unbalance is 450 lbs and 4 No. 5ARO Counterweights placed at the maximum position (c in this case) on the crank will produce an effective counter-balance load of 10,160 lbs. That is,

221,11=cW 526,9=C

30.36)37()05.96)(221,11(

)111)(526,9(==r

which gives lbs. In order to generate the ideal counter-balance load of lbs, the counterweights should be place on the crank at

The computer program SuckerRodPumpingLoad.xlscan be used for quickly seeking solutions to similar problems. It is available from the publisher with this book. Solution is shown in Table 1-2.

in.

SuckerRodPumpingLoad.xlsDescription: 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 section; and 2) view result in the Solution section.Input Data:

Pump setting depth (D): 3,000 ftPlunger diameter (dp): 2.5 in.Rod section 1, diameter (dr1): 1 in.

length (L1): 0 ftRod section 2, diameter (dr2): 0.875 in.

length (L2): 3,000 ftRod section 3, diameter (dr3): 0.75 in.

length (L3): 0 ftRod section 4, diameter (dr4): 0.5 in.

length (L4): 0 ftType of pumping unit (1 = conventional; -1 = Mark II or Air-balanced): 1Beam dimension 1 (d1) 96.05 in.Beam dimension 2 (d2) 111 in.

Crank length (c): 37 in.Crank to pitman ratio (c/h): 0.33Oil gravity (API): 25 oAPIMaximum allowable acceleration factor (L): 0.4

Solution:

= 85.52 in.

= 22 SPM

= 4.91 in.2

= 0.60 in.

= 5,770 lbs

= 6,138 lbs

1

22ddcS =

( )hcS

LN−

=1

2.70471

4

2p

p

dA

π=

( )144

4.62 pff

DASW =

4

2r

rdA π

=

144rs

rDA

Wγ

=

= 0.7940

= 16,076 lbs

= 280,056 lbs

= 0.40

= 2,976 lbs

= 9,526 lbs

( )2.70471

12

1hcSN

F±

=

( ) 1max 4.62PRL FWWWSW rrs

rff ++−=

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

2.704712

2

41 r

fWSN

WST

( )2.70471

12

2hcSN

Fm

=

( ) 2min 4.62PRL FWWW

S rrs

rf −+−=

γ

( )minmax21 PRL PRL +=C

615.5)60)(24(

12144 o

vpp

BES

NA

q =

o

vpp

BENSA

q 1484.0=

Liquid flow rate delivered by the plunger pump can be expressed as

or

(stb/day)

1.5 Pump Deliverability and Power Requirements

(bbl/day)

1.5.1 Effective Plunger Stroke Length

The magnitude of the rod stretch is

EADW

lr

rfr =δ (1-36)

Tubing stretch can be expressed by a similar equation. That is

EADW

lt

tft =δ (1-37)

EADW

nlr

rro =δ

( )2.70471

12hcSN

n±

=

The magnitude of the rod stretch due to acceleration is called plunger over travel:

But the maximum acceleration term n can be written as

(ft) (1-38)

so that Eq (1-38) becomes

( )2.70471

12hc

r

rro

SNEADW

l±

=δ (ft) (1-39)

Let us restrict our discussion to conventional units. Then Eq (1-39) becomes

2.70471

2MSNEADW

lr

rro =δ (ft) (1-40)

rrsr DAW γ=

MSNDl ro22111093.1 −×=δ (in) (1-41)

Eq (1-40) can be rewritten to yield δlo in inches. Wr is

and γS = 490 lb/ft3 with E = 30 x 106 lb/m2

Eq (1-40) becomes

otrp lllSS δδδ +−−=

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

r

r

trfp A

WMSNAA

WEDSS

2.704711112 2

Plunger stroke is approximated using the above expressions as

or

(in)

(1-42)

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

r

r

hchc

trfp A

LWAA

WEDSS

111112

hchc

−

+

11

If pumping is carried out at the maximum permissible speed limited by Eq (1-34), the plunger stroke becomes

For the air-balanced unit the term

is replaced by its reciprocal.

(in) (1-43)

1.5.2 Volumetric Efficiency

Guidelines are

a. Low viscosity oils (1 to 20 cps) can be pumped with a plunger to barrel fit of -0.001”.

b. High viscosity oils (7400 cps) will probably carry sand in suspension so a plunger to barrel fit or ~0.005” 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.

( ) ( )pb

pbpbps L

pd

ddddkq Δ+−

= 1.0

9.2

μ(1-44)

1.5.3 Power Requirements

The power required for lifting fluid is called hydraulic power. It is usually expressed in terms of net lift:

Nlh LqP γ61036.7 −×= (1-45)

and

l

tfN

pHL

γ433.0+=

(1-46)

SNWP rf71031.6 −×=

)( fhspm PPFP +=

The power required to overcome friction losses can be empirically estimated as

Thus the required prime mover power can be expressed as

(1-47)

(1-48)

A well is pumped off (fluid level is the pump depth) with a rod pump described in Example Problem 1-1. A 3” tubing string (3.5” OD, 2.995 ID) in the well is not anchored. Calculate (a) expected liquid production rate (use pump volumetric efficiency 0.8), and (b) required prime mover power (use safety factor 1.35).

Example Problem 1-2:

This problem can be quickly solved using the computer programSuckerRodPumpingFlowrate&Power.xls. Solution is shown in Table 1-3.

Solution:

Table 1-3: Solution given by SuckerRodPumpingFlowrate&Power.xls

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 section; and 2) view result in the Solution section.

Input Data:

Pump setting depth (D): 4,000 ftDepth to the liquid level in annulus (H): 4,000 ftFlowing tubing head pressure (ptf): 100 ftTubing outer diameter (dto): 3.5 in.

Tubing inner diameter (dti): 2.995 in.Tubing anchor (1 = yes; 0 = no): 0Plunger diameter (dp): 2.5 in.Rod section 1, diameter (dr1): 1 in.

length (L1): 0 ftRod section 2, diameter (dr2): 0.875 in.

length (L2): 0 ftRod section 3, diameter (dr3): 0.75 in.

length (L3): 4,000 ftRod section 4, diameter (dr4): 0.5 in.

length (L4): 0 ft

Type of pumping unit (1 = conventional; -1 = Mark II or Air-balanced): 1

Polished rod stroke length (S) 86 in.

Pumping speed (N) 22 spm

Crank to pitman ratio (c/h): 0.33

Oil gravity (API): 25 oAPI

Fluid formation volume factor (Bo): 1.2 rb/stb

Pump volumetric efficiency (Ev): 0.8

Safety factor to prime mover power (Fs): 1.35

Solution:

= 2.58 in.2

= 4.91 in.2

= 0.44 in.

4

2t

td

Aπ

=

4

2p

p

dA

π=

4

2r

rd

Aπ

=

= 7,693 lbs

= 6,013 lbs

= 1.33

= 70 in.

= 753 sbt/day

= 4,255 ft

= 25.58 hp

= 7.2 hp

= 44.2 hp

( )144

4.62 pff

DASW =

144rs

rDA

Wγ

=

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

r

r

trfp A

WMSNAA

WEDSS

2.704711112 2

hcM ±= 1

o

vpp

BENSA

q 1484.0=

l

tfN

pHL

γ433.0+=

Nlh LqP γ61036.7 −×=

SNWP rf71031.6 −×=

)( fhspm PPFP +=

1.6 Procedure for Pumping Unit Selection

The following procedure can be used for selecting a pumping unit:

1) From the maximum anticipated fluid production (based on IPR) and estimated volumetric efficiency, calculate requiredpump displacement.

2) Based on well depth and pump displacement, determine API rating and stroke length of the pumping unit to be used. This can be done using either Figure 1-8 or Table 1-4.

3) Select tubing size, plunger size, rod sizes, and pumping speed from Table 1-4.

4) Calculate the fractional length of each section of the rod string.

5) Calculate the length of each section of the rod string to thenearest 25 ft.

6) Calculate the acceleration factor.7) Determine the effective plunger stroke length.8) Using the estimated volumetric efficiency, determine the

probable production rate and check it against the desired production rate.

9) Calculate the dead weight of the rod string.10) Calculate the fluid load.11) Determine peak polished rod load and check it

against the maximum beam load for the unit selected.

12) Calculate the maximum stress at the top of each rod size and check it against the maximum permissible working stress for the rods to be used.

13) Calculate the ideal counterbalance effect and check it against the counterbalance available for the unit selected.

14) From the manufacturer's literature, determine the position of the counterweight to obtain the ideal counterbalance effect.

15) On the assumption that the unit will be no more than five per cent out of counterbalance, calculate the peak torque on the gear reducer and check it against the API rating of the unit selected.

16) Calculate hydraulic horsepower, friction horsepower, and brake horsepower of the prime mover. Select the prime mover.

17) From the manufacturer's literature obtain the gear reduction ratio and unit sheave size for the unit selected, and the speed of the prime mover. From this determine the engine sheave size to obtain the desired pumping speed.

A well is to be put on a 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 assumed that working liquid level is low, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsurface equipment for the installation. Use safety factor of 1.35 for prime mover power.

Example Problem 1-3:

Solution:

(1) Assuming volumetric efficiency of 0.8, the required pump displacement is (600)/(0.8) = 750 bbl/day.

(2) Based on well depth 3,500 ft and pump displacement 750 bbl/day, Figure 1-8 suggests API pump size 320 unit with 84 in. stroke, i.e., a pump is selected with the following designation:C - 320D – 213 - 86

(3) Table 1-4 (g) suggests:Tubing size: 3 in. O.D., 2.992 in. I.D.Plunger size: 2 ½ in.Rod size: 7/8 in.Pumping speed: 18 spm

(4) Table 1-1 gives d1 = 96.05 in., d2 = 111 in., c = 37 in. and h = 114 in., thus c/h = 0.3246. Spreadsheet program

SuckerRodPumpingFlowRate&Power.xls givesqo = 687 bbl/day > 600 bbl/day, OKPpm = 30.2 hp

(5) Spreadsheet program SuckerRodPumpingLoad.xlsgives

PRLmax = 16,121 lbsPRLmin = 4,533 lbsT = 247,755 lbs < 320,000 in.-lbs, OKC = 10,327 lbs

The cross-sectional area of the 7/8 in. rod is 0.60 in.2 Thus the maximum possible stress in the sucker rod is

σmax = (16,121)/(0.60) = 26,809 psi < 30,000 psi, OK

Therefore, the selected pumping unit and rod meet well load and volume requirements.

450)111()05.96(

)37()37(

+cW

075,14=cW 327,10=C

4.31)37()05.96)(076,14(

)111)(327,10(==r

lbs. In order to generate the ideal counter-balance load of lbs, the counterweights should be place on the crank at

in.

(6) If a LUFKIN Industries C - 320D – 213 – 86 unit is chosen, the structure unbalance is 450 lbs and 4 No. 5ARO Counterweights placed at the maximum position (c in this case) on the crank will produce an effective counter-balance load of 12,630 lbs. That is,

= 12,630 lbs

which gives

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

)750()24)(12.30)(18(

=ed = 17.3 in.

0

500

1,000

1,500

2,000

2,500

0 2,000 4,000 6,000 8,000 10,000 12,000

Pump Setting Depth (ft)

Pum

p D

ispl

acem

ent (

bbl/d

ay) A 40 34

B 57 42C 80 48D 114 54E 160 64F 228 74G 320 84H 640 144

Curve API Size Stroke

Figure 1-8: Sucker rod pumping unit selection chart (After Kelley and Willis, 1954)

Table 1-4: Design data for API sucker rod pumping units

(a) Size 40 Unit with 34-inch Stroke

Pump Depth (ft)Plunger Size

(in) Tubing Size (in)Rod Sizes

(in)Pumping Speed

(stroke/min)1000-1100 2 3/4 3 7/8 24-191100-1250 2 1/2 3 7/8 24-191250-1650 2 1/4 2 1/2 3/4 24-191650-1900 2 2 1/2 3/4 24-191900-2150 1 3/4 2 1/2 3/4 24-192150-3000 1 1/2 2 5/8-3/4 24-193000-3700 1 1/4 2 5/8-3/5 22-183700-4000 1 2 5/8-3/6 21-18

(b) Size 57 Unit with 42-inch Stroke

Pump Depth (ft) Plunger Size (in)Tubing

Size (in) Rod Sizes (in)Pumping Speed

(stroke/min)1150-1300 2 3/4 3 7/8 24-191300-1450 2 1/2 3 7/8 24-191450-1850 2 1/4 2 1/2 3/4 24-191850-2200 2 2 1/2 3/4 24-192200-2500 1 3/4 2 1/2 3/4 24-192500-3400 1 1/2 2 5/8-3/4 23-183400-4200 1 1/4 2 5/8-3/5 22-174200-5000 1 2 5/8-3/6 21-17

(c) Size 80 Unit with 48-inch Stroke

Pump Depth (ft)Plunger Size

(in) Tubing Size (in)Rod Sizes

(in)Pumping Speed

(stroke/min)

1400-1500 2 3/4 3 7/8 24-19

1550-1700 2 1/2 3 7/8 24-19

1700-2200 2 1/4 2 1/2 3/4 24-19

2200-2600 2 2 1/2 3/4 24-19

2600-3000 1 3/4 2 1/2 3/4 23-18

3000-4100 1 1/2 2 5/8-3/4 23-19

4100-5000 1 1/4 2 5/8-3/5 21-175000-6000 1 2 5/8-3/6 19-17

(d) Size 114 Unit with 54-inch Stroke

Pump Depth (ft)Plunger

Size (in) Tubing Size (in) Rod Sizes (in)Pumping Speed

(stroke/min)1700-1900 2 3/4 3 7/8 24-191900-2100 2 1/2 3 7/8 24-192100-2700 2 1/4 2 1/2 3/4 24-192700-3300 2 2 1/2 3/4 23-183300-3900 1 3/4 2 1/2 3/4 22-173900-5100 1 1/2 2 5/8-3/4 21-175100-6300 1 1/4 2 5/8-3/5 19-166300-7000 1 2 5/8-3/6 17-16

(e) Size 160 Unit with 64-inch Stroke

Pump Depth (ft) Plunger Size (in) Tubing Size (in)Rod Sizes

(in)Pumping Speed

(stroke/min)2000-2200 2 3/4 3 7/8 24-192200-2400 2 1/2 3 7/8 24-192400-3000 2 1/4 2 1/2 3/4-7/8 24-193000-3600 2 2 1/2 3/4-7/8 23-183600-4200 1 3/4 2 1/2 3/4-7/8 22-174200-5400 1 1/2 2 5/8-3/4-7/8 21-175400-6700 1 1/4 2 5/8-3/4-7/8 19-156700-7700 1 2 5/8-3/4-7/8 17-15

(f) Size 228 Unit with 74-inch Stroke

Pump Depth (ft) Plunger Size (in) Tubing Size (in)Rod Sizes

(in)Pumping Speed

(stroke/min)2400-2600 2 3/4 3 7/8 24-202600-3000 2 1/2 3 7/8 23-183000-3700 2 1/4 2 1/2 3/4-7/8 22-173700-4500 2 2 1/2 3/4-7/8 21-164500-5200 1 3/4 2 1/2 3/4-7/8 19-155200-6800 1 1/2 2 5/8-3/4-7/8 18-146800-8000 1 1/4 2 5/8-3/4-7/8 16-138000-8500 1 1/16 2 5/8-3/4-7/8 14-13

(g) Size 320 Unit with 84-inch Stroke

Pump Depth (ft) Plunger Size (in) Tubing Size (in)Rod Sizes

(in)Pumping Speed

(stroke/min)2800-3200 2 3/4 3 7/8 23-183200-3600 2 1/2 3 7/8 21-173600-4100 2 1/4 2 1/2 3/4-7/8-1 21-174100-4800 2 2 1/2 3/4-7/8-1 20-164800-5600 1 3/4 2 1/2 3/4-7/8-1 19-165600-6700 1 1/2 2 1/2 3/4-7/8-1 18-156700-8000 1 1/4 2 1/2 3/4-7/8-1 17-138000-9500 1 1/16 2 1/2 3/4-7/8-1 14-11

(h) Size 640 Unit with 144-inch Stroke

Pump Depth (ft) Plunger Size (in) Tubing Size (in) Rod Sizes (in)Pumping Speed

(stroke/min)3200-3500 2 3/4 3 7/8-1 18-143500-4000 2 1/2 3 7/8-1 17-134000-4700 2 1/4 2 1/2 3/4-7/8-1 16-134700-5700 2 2 1/2 3/4-7/8-1 15-125700-6600 1 3/4 2 1/2 3/4-7/8-1 14-126600-8000 1 1/2 2 1/2 3/4-7/8-1 14-118000-9600 1 1/4 2 1/2 3/4-7/8-1 13-109600-11000 1 1/16 2 1/2 3/4-7/8-1 12-10

1.7 Principles of Pump Performance Analysis

Figure 1-9: A sketch of pump dynagraph (From Nind, 1964)

Figure 1-10: Pump dynagraph cards: (a) ideal card, (b) gas compression on down stroke, (c) gas expansion on upstroke, (d) fluid pound, (e) vibration due to fluid pound, (f) gas lock (From Nind, 1964)

The surface dynamometer cards record the history of the variations in loading on the polished rod during a cycle. The cards have three principal uses:

a. To obtain information that can be used to determine load, torque and horsepower changes required of the pump equipment.

b. To improve pump operating conditions such as pump speed and stroke length.

c. To check well conditions after installation of equipment to prevent or diagnose various operating problems (like pounding etc).

Correct interpretation of surface dynamometer card leads to estimate of various parameter values.

• Maximum and minimum PRLs can be read directly from the surface card (with the use of instrument calibration). This data then allows for the determination of the torque,counter balance, and horsepower requirements for the surface unit.

• Rod stretch and contraction is shown on the surface dynamometer card. This phenomena is reflected in the surface unit dynamometer card and is shown in Figure 1-11 (a) for an ideal case.

Figure 1-11: Surface Dynamometer Card: (a) ideal card (stretch and contraction), (b) ideal card (acceleration), (c) 3 typical cards (From Nind, 1964)

• Acceleration forces cause the ideal card to rotate clockwise. The PRL is higher at the bottom of the stroke and lower at the top of the stroke. Thus, in Figure 1-11 (b), point A is at the bottom of the stroke.

• Rod vibration causes a serious complication in the interpretation of the surface card. This is 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.

Figure 1-12: Strain-gage-type dynamometer chart

Figure 1-13: Surface to down hole cards derived from surface dynamometer card

Problems1-1. If the dimensions d1, d2 and c take the same values for both conventional unit (Class I lever system) and air-balanced unit (Class III lever system), how different will theirpolished rod strokes length be?

1-2. What are the advantages of the Lufkin Mark II and air-balanced units in comparison with conventional units?

1-3. Use your knowledge of kinematics to proof that for Class I lever systems,

(a) the polished rod will travel faster in down stroke than in upstroke if the distance between crankshaft and the center of Sampson post is less than dimension d1.

(b) the polished rod will travel faster in up stroke than indown stroke if the distance between crankshaft and the center of Sampson post is greater than dimension d1.

1-4. Derive a formula for calculating the effective diameter of a tapered rod string.

1-5. Derive formulae for calculating length fractions of equal-top-rod-stress tapered rod strings for (a) two-size rod strings, (b) three-sized rod strings, and (c) four-sized rod strings. Plot size fractions for each case as a function of plunger area.

1-6. A tapered rod string consists of sections of 5/8” and ½” rods and a 2” plunger. Use the formulae from problem 1-5 to calculate length fraction of each size of rod.

1-7. A tapered rod string consists of sections of ¾”, 5/8”and ½” rods and a 1 ¾” plunger. Use the formulae from problem 1-5 to calculate length fraction of each size of rod.

1-8. The following geometry dimensions are for the pumping unit C – 80D – 133 – 48:

d1 = 64 in.d2 = 64 in.c = 24 in.h = 74.5 in.

Can this unit be used with a 2” plunger and ¾”rods to lift 30 °API gravity crude (formation volume factor 1.25 rb/stb) at depth of 2,000 ft? If yes, what is the required counter balance load?

1-9. The following geometry dimensions are for the pumping unit C – 320D – 256 – 120:

d1 = 111.07 in.d2 = 155 in.c = 42 in.h = 132 in.

Can this unit be used with a 2 1/2” plunger and ¾”-7/8”-1”taperd rod string to lift 22 °API gravity crude (formation volume factor 1.22 rb/stb) at depth of 3,000 ft? If yes, what is the required counter balance load?

1-10. A well is pumped off with a rod pump described in Problem 12-8. A 2 ½” tubing string (2.875” 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).

1-11. A well is pumped with a rod pump described in Problem 1-9 to a liquid level of 2,800 ft. A 3” tubing string (3.5” OD, 2.995” ID) in the well is anchored. Calculate (a) expected liquid production rate (use pump volumetric efficiency 0.85), and (b) required prime mover power (use safety factor 1.4).

1-12. A well is to be put on a sucker rod pump. The proposed pump setting depth is 4,500 ft. The anticipated production rate is 500 bbl/day oil of 40 oAPI gravity against wellhead pressure 150 psig. It is assumed that working liquid level is low, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsurface equipment for the installation. Use safety factor of 1.40 for prime mover power.

1-13. 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 oAPI gravity against wellhead pressure 120 psig. It is assumed that working liquid level will be about 3,000 ft, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsurface equipment for the installation. Use safety factor of 1.30 for prime mover power.