I Sucker Rod Pumping (2)

Section I

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 Pumping

System

Pit ma n Well Load Fulcrum Force Counter Balance Walking Beam

Fulcrum Pit ma n Force Counter Balance Well Load Walking Beam

Fulcrum Pit ma n Force Counter Balance 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)

1.3 Polished

Rod Motion

G I H R P A C

Table 1-1: Conventional pumping unit API geometry dimensions

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

(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 80.32 C-912D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32 C-640D-365-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32 C-640D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32 C-456D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 -1500 80.32 C-912D-427-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 -650 68.82 C-912D-365-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 -650 68.82 C-640D-365-144 180 120.03 120 148.5 238.88 89.88 47, 41, 35 -650 68.82 C-640D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -520 68.45 C-456D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -520 68.45 C-640D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45 C-456D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45 C-320D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 -400 68.45 C-456D-365-120 152 120.03 120 148.5 238.88 89.88 47, 41, 35 570 58.12 C-640D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 -120 57.02 C-456D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 -120 57.02 C-320D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05 C-456D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05 C-456D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05 C-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 57.05 C-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 47.48 C-320D-305-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48 C-228D-213-100 129 96.08 96 113 180 63 37, 32, 27 0 48.37 C-228D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37 C-160D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37 C-320D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96 C-228D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96 C-320D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-228D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-160D-173-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-114D-119-86 111 84.05 84 93.75 150.13 53.38 32, 27, 22 115 40.98 C-320D-245-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-228D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-160D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-228D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49 C-160D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49 C-160D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49 C-114D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49 C-160D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02 C-114D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02 C-160D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59 C-114D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59 C-80D-119-64 84 64 64 74.5 116 41 24, 20, 16 0 30.85 C-160D-173-54 72 72.06 72 84 132 45 27, 22, 17 500 26.22 C-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 26.45 C-80D-119-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45 C-P57D-76-54 64 51 51 64 103 39 21, 16, 11 105 25.8 C-P57D-89-54 64 51 51 64 103 39 21, 16, 11 105 25.8 C-80D-133-48 64 64 64 74.5 116 41 24, 20, 16 440 23.51 C-80D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3 C-57D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3 C-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 22.98 C-P57D-95-48 57 51 51 64 103 39 21, 16, 11 180 22.98 C-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 22.92 C-P57D-89-42 51 51 51 64 103 39 21, 16, 11 280 20.56 C-P57D-76-42 51 51 51 64 103 39 21, 16, 11 280 20.56 C-P40D-89-42 53 47 47 56 95 39 18, 14, 10 280 19.92 C-P40D-76-42 53 47 47 56 95 39 18, 14, 10 280 19.92 C-57D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-57D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-40D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-40D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-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 17.66 C-25D-67-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37 C-25D-56-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37 C-25D-67-30 45 36.22 36 49.5 84.5 31 12, 8 150 14.53 C-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 units

A = Air-Balanced units

B = Beam Counterbalance units M = 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

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

( ) ( ) ( )

AB

2

=

OA

2

+

OB

2

−

2

( )( )

OA

OB

cos

AOB

(

h

c

x

)

c

(

h

c

x

)

t

c

h

2

=

2

+

+

−

2

−

2

+

−

cos

ω

(

)

[

1

cos

]

2

(

)(

1

cos

)

0

2

2

−

+

−

+

+

−

=

t

c

h

c

t

c

h

x

x

ω

ω

(

)

2 2

(

2 2

)

cos

cos

1

t

c

t

h

c

c

h

x

=

+

−

ω

±

ω

+

−

(

)

2 2

(

2 2

)

cos

cos

1

t

c

t

h

c

c

h

x

=

+

−

ω

−

ω

+

+

where ω is the angular velocity of the crank.

The equation reduces to

so that

When

ω

t is zero, x is also zero, which means that

the negative root sign must be taken.

Therefore,

2

2

dt

x

d

a

=

( )

_h

c

c

a

_max

=

ω

2

1

+

Acceleration is

Carrying out the differentiation for acceleration,

it is found that the maximum acceleration

occurs when

ω

t is equal to zero (or an even

multiple of

π

radians) and that this maximum

value is

( )

_hc

c

a

_min

=

ω

2

1

−

60

2 N

π

ω

=

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)

⎟

⎠

⎞

⎜

⎝

⎛ +

=

h

c

cN

a

1

2

.

91

2 max

⎟

⎠

⎞

⎜

⎝

⎛ +

=

h

c

g

cN

a

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)

⎟

⎠

⎞

⎜

⎝

⎛ −

=

h

c

g

cN

a

1

3

.

2936

2 min

⎟

⎠

⎞

⎜

⎝

⎛ +

=

h

c

g

cN

d

d

a

1

3

.

2936

2 2 1 max (ft/sec2)

Likewise the minimum upward (a_min) 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)

S

d

cd =

1 2

2

12

2

1 2

S

d

cd

=

24

1 2

S

d

cd =

where d

₁

and d

₂

are shown in Figure 1-5. But

where S is the polished rod stroke length.

So if S is measured in inches, then

or

⎟

⎠

⎞

⎜

⎝

⎛ +

=

h

c

g

SN

a

1

2

.

70471

2 max

M

g

SN

a

2

.

70471

2 max

=

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

or we can write Eq (1-9) as

(ft/sec

2

₎

(1-9)

(ft/sec

2

₎

h c

M

=1

+

⎟

⎠

⎞

⎜

⎝

⎛ −

=

h

c

g

SN

a

1

2

.

70471

2 min

(ft/sec

2

)

where M is the machinery factor and is defined as

Similarly,

(1-11)

1.4.1 Maximum PRL

(

)

(

)

_⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − = 2 . 70471 144 144 144 4 . 62 PRL 2 max M SN DA DA A A D S_f p r γs r γs r

_(1-13)

Equation (1-13) can be rewritten as

(

)

(

)

_⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − = 2 . 70471 144 144 144 4 . 62 144 4 . 62 PRL 2 max M SN DA DA DA S DA S _f p _f r γ s r γ s r

(1-14)

144

r s r

DA

W

=

γ

D

W

A

s r r

γ

144

=

(

)

(

)

_⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − = 2 . 70471 4 . 62 144 4 . 62 PRL 2 max M SN W W W S DA S _{r r} s r f p f 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)

(

)

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+

−

=

2

.

70471

1

.

0

144

4

.

62

PRL

2 max

M

SN

W

W

W

DA

S

_f p _{r r r}

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+

=

2

.

70471

9

.

0

PRL

2 max

M

SN

W

W

W

_{f r r}

(

)

144 4 . 62 p f f DA S W =

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 S

_f

= 0.78. Thus, Eq

(1-17) becomes (where

γ

_s

= 490)

(

)

_r f

F

W

W

₁ max

0

.

9

PRL

=

+

+

( )

2

.

70471

1

2 1 h c

SN

F

=

+

( )

2

.

70471

1

2 1 h c

SN

F

=

−

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

where for conventional units

and for air-balanced units

(1-21)

(1-19)

1.4.2 Minimum PRL

(

)

₂ min

62

.

4

PRL

S

W

W

_r

W

_r

F

s r f

+

−

−

=

γ

(

)

_r r r

F

W

F

W

W

_{2 2} min

0

.

9

0

.

9

PRL

=

−

=

−

which, for API 50° oil, reduces to

(

)

2

.

70471

1

2 2 h c

SN

F

=

−

(

)

2

.

70471

1

2 2 h c

SN

F

=

+

where for the conventional units

and for air-balanced units

(1-24)

(1-23)

1.4.3 Counterweights

(

_{max min}

)

2 1

_PRL

+

_PRL

=

C

(

)

_r r f

W

F

F

W

W

C

=

1₂

+

0

.

9

+

1_{2 1}

−

₂

(1-25)

The idea counter-balance load C is the

average PRL. Therefore,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+

=

h

c

SN

W

W

C

_{f r}

2

.

70471

9

.

0

2 2 1

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

+

=

h

c

SN

W

W

C

_{f r}

2

.

70471

9

.

0

2 2 1

or for conventional units

and for air-balanced units

(1-27)

(1-26)

2

1

d

d

c

r

W

C

C

=

_s

+

_c

where

C

_s

= structure unbalance, lbs

W

_c

= total weight of counterweights, lbs

r = distance between the mass center of

counterweights and the crank shaft

center, in.

1.4.4 Peak Torque and Speed Limit

(

)

[

]

1 2 2

9

.

0

d

d

W

F

C

c

T

=

−

−

_r

(

)

[

C

F

W

_r

]

S

T

=

1₂

−

0

.

9

−

₂

Peak torque T is (see Figure 1.5)

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

(1-29)

(1-28)

(

)

[

W

f

F

F

W

r

]

S

T

=

1_{2 2}1

+

1_{2 1}

+

₂

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

2

.

70471

2

2 4 1 r f

W

SN

W

S

T

or

(1-30)

(in-lbs)

Torque factors and efficiency are used in practice:

[

]

93

.

0

)

(

)

(

_{1 min 2} max 2 1

_PRL

_TF

_PRL

_TF

T

=

+

(1-31)

Torque factor is defined:

Load

Rod

Polished

Crankshaft

the

to

Exerted

Torque

=

TF

= 1

TF

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

=

2

Approximate Maximum Torque Factor

Conventional and Air Balance Units

Stroke (in.) TF₁ (in.) TF₂ (in.)

16 8.5 8.5 24 13 13 30 16 16 36 19 19 42 22 22 48 26 26 54 29 29 64 34 34 74 39 39 86 45 45 100 52 52 120 63 63 144 75 75 168 87 87

Approximate Maximum Torque Factor

Mark 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

1

2 min max/ h c

g

SN

a

=

±

(1-32)

( )

L

SN

_hc

≤

±

2

.

70471

1

2

(1-33)

Maximum Permissible Pumping Speed

(

_hc

)

S

L

N

m

1

2

.

70471

limit

=

(

_hc

)

S

N

m

1

7

.

187

limit

=

or

For L = 0.5

The minus sign is for conventional units

and the plus sign for air-balanced units.

(1-34)

1.4.5 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 string

b. 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 top

6/8” diameter rod at middle 5/8” diameter rod at bottom

d. No. 107 is a 4 way tapered string consisting of 10/8” (or 1 1/4”) diameter rod at top

9/8” (or 1 1/8”) diameter rod below 10/8” diameter rod 8/8” (or 1”) diameter rod below 9/8” 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

The following geometry dimensions are for

the pumping unit C – 320D – 213 – 86:

d

₁

= 96.05 in.

d

₂

= 111 in.

c

= 37 in.

c/h = 0.33

If this unit is used with a 2 1/2” plunger and 7/8 in.

rods to lift 25 °API gravity crude (formation volume

factor 1.2 rb/stb) at depth of 3,000 ft, answer the

following 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:

in.

52

.

85

05

.

96

111

)

37

)(

2

(

2

1 2

=

=

=

d

d

c

S

(

)

(

85

.

52

)

(

1

0

.

33

)

)

4

.

0

)(

2

.

70471

(

1

2

.

70471

−

=

−

=

h c

S

L

N

The maximum allowable pumping speed is:

(

)

(

)

5,770 lbs 144 ) 91 . 4 )( 000 , 3 ( 4 . 62 ) 9042 . 0 ( 144 4 . 62 = = = p f f DA S W lbs 138 , 6 144 ) 60 . 0 )( 000 , 3 )( 490 ( 144 = = = s r r DA W γ

(b) The maximum PRL can be calculated with Eq

(12-17). The 25° API gravity has an S

_f

=

0.9042. The area of the 2 ½” plunger is A

_p

=

4.91 in.

2

_{The area of the 7/8” rod is A}

r

= 0.60

(

)

(

)

7940

.

0

2

.

70471

33

.

0

1

)

22

)(

52

.

85

(

2

.

70471

1

2 2 1

=

+

=

+

=

h c

SN

F

(

)

₁ max

62

.

4

PRL

W

S

W

W

_r

W

_r

F

s r f f

−

+

+

=

γ

(

62.4

)

(6,138)/(490) 6,138 (6,138)(0.794) ) 9042 . 0 ( 770 , 5 − + + =

lbs

076

,

16

=

Then the expected maximum PRL is:

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

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = 2 . 70471 ) 138 , 6 ( ) 22 )( 52 . 85 ( 2 770 , 5 ) 52 . 5 8 ( 2 . 70471 2 2 4 1 2 4 1 r f W SN W S T

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

(

)

(

)

4

.

0

2

.

70471

33

.

0

1

)

22

)(

52

.

85

(

2

.

70471

1

2 2 2

=

−

=

−

=

h c

SN

F

(

)

₂ min

62

.

4

PRL

S

W

W

_r

W

_r

F

s r f

+

−

−

=

γ

(

)

6

,

138

(

6

,

138

)(

0

.

4

)

490

138

,

6

4

.

62

)

9042

.

0

(

+

−

−

=

Accurate calculation of counter-balance load

requires the minimum PRL:

(

PRL

PRL

) (

₂1

1

6

,

076

2,976

)

9

,

526

lbs

min max 2 1

+

=

+

=

=

C

160

,

10

450

)

111

(

)

05

.

96

(

)

37

(

)

37

(

₊

₌

c

W

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 = c W 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.xls

can 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.

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

Input Data:

Pump setting depth (D): 3,000 ft Plunger diameter (d_p): 2.5 in. Rod section 1, diameter (d_r1): 1 in. length (L₁): 0 ft Rod section 2, diameter (d_r2): 0.875 in.

length (L₂): 3,000 ft Rod section 3, diameter (d_r3): 0.75 in.

length (L₃): 0 ft Rod section 4, diameter (d_r4): 0.5 in.

length (L₄): 0 ft Type of pumping unit (1 = conventional; -1 = Mark II or

Air-balanced): 1

Beam dimension 1 (d₁) 96.05 in. Beam dimension 2 (d₂) 111 in.

Crank length (c): 37 in. Crank to pitman ratio (c/h): 0.33

Oil gravity (API): 25 o_API

Maximum 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 2 2 d d c S =

(

_hc

)

S L N − = 1 2 . 70471 4 2 p p d A = π

(

)

144 4 . 62 p f f DA S W = 4 2 r r d A = π 144 r s r DA W = γ

= 0.7940 = 16,076 lbs = 280,056 lbs = 0.40 = 2,976 lbs = 9,526 lbs

(

)

2 . 70471 1 2 1 h c SN F = ± ( ) 1 max 62.4 PRL W S W W_r W_rF s r f f − + + = γ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = 2 . 70471 2 2 4 1 r f W SN W S T

(

)

2 . 70471 1 2 2 h c SN F = m ( ) 2 min 62.4 PRL S W W_r W_rF s r f + − − = γ

(

_{max min}

)

2 1

_PRL

+

_PRL

=

C

615

.

5

)

60

)(

24

(

12

144

_o v p p

B

E

S

N

A

q

=

o v p p

B

E

NS

A

q

=

0

.

1484

Liquid flow rate delivered by the plunger pump

can be expressed as

or

(stb/day)

1.5 Pump Deliverability and

Power Requirements

1.5.1 Effective Plunger Stroke Length

The magnitude of the rod stretch is

E

A

D

W

l

r r f r

=

δ

_(1-36)

Tubing stretch can be expressed by a

similar equation. That is

E

A

D

W

l

t t f t

=

δ

_(1-37)

E

A

D

W

n

l

r r r o

=

δ

( )

2

.

70471

1

2 h c

SN

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

so that Eq (1-38) becomes

( )

2

.

70471

1

2 h c r r r o

SN

E

A

D

W

l

=

±

δ

(ft) (1-39)

Let us restrict our discussion to conventional

units. Then Eq (1-39) becomes

2

.

70471

2

M

SN

E

A

D

W

l

r r r o

=

δ

(ft) (1-40)

r r s r

A

D

W

=

γ

M

SN

D

l

_o

=

1

.

93

×

10

−11 _r2 2

δ

(in) (1-41)

Eq (1-40) can be rewritten to yield

δ

l

_o

in

inches. W

_r

is

and

γ

_S

= 490 lb/ft

3

with E = 30 x 10

6

lb/m

2

Eq (1-40) becomes

o t r p

S

l

l

l

S

=

−

δ

−

δ

+

δ

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

−

=

r r t r f p

A

W

M

SN

A

A

W

E

D

S

S

2

.

70471

1

1

12

2

Plunger stroke is approximated using the above

expressions as

or

(in)

⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = r r h c h c t r f p A LW A A W E D S S 1 1 1 1 12 h c h c

−

+

1

1

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.

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, q

_s

(bbl/day), through the annulus between the

plunger and the barrel.

(

) (

)

p b p b p b p s

L

p

d

d

d

d

d

k

q

=

−

_0.1

+

Δ

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:

N l h

q

L

P

=

7

.

36

×

10

−6

γ

(1-45)

and

l tf N

p

H

L

γ

433

.

0

+

=

(1-46)

SN

W

P

_f

=

6

.

31

×

10

−7 _r

)

(

_{h f} s pm

F

P

P

P

=

+

The power required to overcome friction losses

can be empirically estimated as

Thus the required prime mover power can

be expressed as

(1-47)

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).

This problem can be quickly solved using the

computer program

SuckerRodPumpingFlowrate&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 ft Depth to the liquid level in annulus (H): 4,000 ft Flowing tubing head pressure (p_tf): 100 ft Tubing outer diameter (d_to): 3.5 in. Tubing inner diameter (d_ti): 2.995 in. Tubing anchor (1 = yes; 0 = no): 0 Plunger diameter (d_p): 2.5 in. Rod section 1, diameter (d_r1): 1 in.

length (L₁): 0 ft

Rod section 2, diameter (d_r2): 0.875 in.

length (L₂): 0 ft

Rod section 3, diameter (d_r3): 0.75 in.

length (L₃): 4,000 ft

Rod section 4, diameter (d_r4): 0.5 in.

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 o_API

Fluid formation volume factor (B_o): 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 2 t t d A =

π

4

2 p p

d

A

=

π

4

2 r r

d

A

=

π

= 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 p f f DA S W = 144 r s r DA W = γ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = r r t r f p A W M SN A A W E D S S 2 . 70471 1 1 12 2 h c M = 1 ± o v p p B E NS A q = 0.1484 l tf N p H L γ 433 . 0 + = N l h q L P = 7.36×10−6

γ

SN W P_f = 6.31×10−7 _r ) ( _{h f} s pm F P P P = +

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 required pump 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 the nearest 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

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.

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:

(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 gives

q_o = 687 bbl/day > 600 bbl/day, OK

(5) Spreadsheet program SuckerRodPumpingLoad.xls gives PRL_max = 16,121 lbs PRL_min = 4,533 lbs T = 247,755 lbs < 320,000 in.-lbs, OK C = 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

σ

_max = (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

(

₊

c

W

075 , 14 = c W 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

(

=

e

d

_{= 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)

Pump Displacement (bbl/day

) A 40 34 B 57 42 C 80 48 D 114 54 E 160 64 F 228 74 G 320 84 H 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 24-19 1100-1250 2 1/2 3 7/8 24-19 1250-1650 2 1/4 2 1/2 3/4 24-19 1650-1900 2 2 1/2 3/4 24-19 1900-2150 1 3/4 2 1/2 3/4 24-19 2150-3000 1 1/2 2 5/8-3/4 24-19 3000-3700 1 1/4 2 5/8-3/5 22-18 3700-4000 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 24-19 1300-1450 2 1/2 3 7/8 24-19 1450-1850 2 1/4 2 1/2 3/4 24-19 1850-2200 2 2 1/2 3/4 24-19 2200-2500 1 3/4 2 1/2 3/4 24-19 2500-3400 1 1/2 2 5/8-3/4 23-18 3400-4200 1 1/4 2 5/8-3/5 22-17 4200-5000 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 21-17 5000-6000 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 24-19 1900-2100 2 1/2 3 7/8 24-19 2100-2700 2 1/4 2 1/2 3/4 24-19 2700-3300 2 2 1/2 3/4 23-18 3300-3900 1 3/4 2 1/2 3/4 22-17 3900-5100 1 1/2 2 5/8-3/4 21-17 5100-6300 1 1/4 2 5/8-3/5 19-16 6300-7000 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 24-19 2200-2400 2 1/2 3 7/8 24-19 2400-3000 2 1/4 2 1/2 3/4-7/8 24-19 3000-3600 2 2 1/2 3/4-7/8 23-18 3600-4200 1 3/4 2 1/2 3/4-7/8 22-17 4200-5400 1 1/2 2 5/8-3/4-7/8 21-17 5400-6700 1 1/4 2 5/8-3/4-7/8 19-15 6700-7700 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 24-20 2600-3000 2 1/2 3 7/8 23-18 3000-3700 2 1/4 2 1/2 3/4-7/8 22-17 3700-4500 2 2 1/2 3/4-7/8 21-16 4500-5200 1 3/4 2 1/2 3/4-7/8 19-15 5200-6800 1 1/2 2 5/8-3/4-7/8 18-14 6800-8000 1 1/4 2 5/8-3/4-7/8 16-13 8000-8500 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 23-18 3200-3600 2 1/2 3 7/8 21-17 3600-4100 2 1/4 2 1/2 3/4-7/8-1 21-17 4100-4800 2 2 1/2 3/4-7/8-1 20-16 4800-5600 1 3/4 2 1/2 3/4-7/8-1 19-16 5600-6700 1 1/2 2 1/2 3/4-7/8-1 18-15 6700-8000 1 1/4 2 1/2 3/4-7/8-1 17-13 8000-9500 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 18-14 3500-4000 2 1/2 3 7/8-1 17-13 4000-4700 2 1/4 2 1/2 3/4-7/8-1 16-13 4700-5700 2 2 1/2 3/4-7/8-1 15-12 5700-6600 1 3/4 2 1/2 3/4-7/8-1 14-12 6600-8000 1 1/2 2 1/2 3/4-7/8-1 14-11 8000-9600 1 1/4 2 1/2 3/4-7/8-1 13-10 9600-11000 1 1/16 2 1/2 3/4-7/8-1 12-10

1.7 Principles of Pump

Performance Analysis

Figure 1-10: Pump dynagraph cards: (a) ideal card,

(b) gas compression on down stroke, (c) gas expansion on upstroke,

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

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

Problems

1-1. 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 their polished 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 in down 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

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

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 o_{API 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 o_{API 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.