I Sucker Rod Pumping (2)
March 22, 2017 | Author: Jainik Jain | Category: N/A
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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 Pumping System
Walking Beam
Pitm an
Fulcrum Force Well Load
Counter Balance
(a) Conventional Unit
Walking Beam
Pitm an
Fulcrum Force Well Load
Counter Balance
(b) Lufkin Mark II Unit
Pitm an
Fulcrum
Force
(c) Air-Balanced Unit
Counter Balance
Walking Beam
Well Load
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
A C
P
H
G
R
I Figure 1-6: Definitions of conventional pumping unit API geometry dimensions
Table 1-1: Conventional pumping unit API geometry dimensions
API Unit Designation
A
C
I
P
H
G
R1, R2, R3
Cs
(in.)
(in.)
(in.)
(in.)
(in.)
(in.)
(in.)
(lb)
Torque 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
210 210 210 210 210 180 180 180 180 180 180 180 180 152 155 155 155 155 155 155
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
120 120 120 120 120 120 120 120 120 120 120 120 120 120 111 111 111 111 111 111
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
237.88 237.88 237.88 237.88 237.88 237.88 237.88 238.88 238.88 238.88 238.88 238.88 238.88 238.88 213 213 211 211 211 211
86.88 86.88 86.88 86.88 86.88 86.88 86.88 89.88 89.88 89.88 89.88 89.88 89.88 89.88 75 75 75 75 75 75
47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 47, 41, 35 42, 36, 30 42, 36, 30 42, 36, 30 42, 36, 30 42, 36, 30 42, 36, 30
-1500 -1500 -1500 -1500 -1500 -650 -650 -650 -520 -520 -400 -400 -400 570 -120 -120 55 55 0 0
80.32 80.32 80.32 80.32 80.32 68.82 68.82 68.82 68.45 68.45 68.45 68.45 68.45 58.12 57.02 57.02 57.05 57.05 57.05 57.05
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 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 C-114D-133-54
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 64
111 111 96 96 96 111 111 96 96 96 84 96 96 96 84 84 84 84 84 84 72 72 64 72 64
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 74.5
211 211 180 180 180 211 211 180 180 180 150.13 180 180 180 152.38 152.38 150.13 150.13 150.13 150.13 132 132 116 132 116
75 75 63 63 63 75 75 63 63 63 53.38 63 63 63 53.38 53.38 53.38 53.38 53.38 53.38 45 45 41 45 41
42, 36, 30 42, 36, 30 37, 32, 27 37, 32, 27 37, 32, 27 42, 36, 30 42, 36, 30 37, 32, 27 37, 32, 27 37, 32, 27 32, 27, 22 37, 32, 27 37, 32, 27 37, 32, 27 32, 27, 22 32, 27, 22 32, 27, 22 32, 27, 22 32, 27, 22 32, 27, 22 27, 22, 17 27, 22, 17 24, 20, 16 27, 22, 17 24, 20, 16
550 550 0 0 0 800 800 450 450 450 115 800 800 800 450 450 300 300 550 550 360 360 0 500 330
47.48 47.48 48.37 48.37 48.37 40.96 40.96 41.61 41.61 41.61 40.98 35.99 35.99 35.99 35.49 35.49 35.49 35.49 31.02 31.02 30.59 30.59 30.85 26.22 26.45
129 129 129 129 129 111 111 111 111 111 111 96 96 96 96 96 96 96 84 84 84 84 84 72 72
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 64 64 64 64 64 64 57 57 64 61 51 51 53 53 56 56 56 56 48 47 48 48 45 45
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 51 51 64 56 56 56 51 51 48 47 51 51 47 47 48 48 48 48 48 47 48 48 36 36
74.5 74.5 64 64 74.5 65.63 65.63 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
116 116 103 103 116 105 105 105 103 103 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 39 39 41 37 37 37 39 39 37 39 39 39 39 39 37 37 37 37 37 39 37 37 31 31
24, 20, 16 24, 20, 16 21, 16, 11 21, 16, 11 24, 20, 16 21, 16, 11 21, 16, 11 21, 16, 11 21, 16, 11 21, 16, 11 18, 14, 10 18, 14, 10 21, 16, 11 21, 16, 11 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 18, 14, 10 12, 8 12, 9
330 330 105 105 440 320 320 320 180 180 0 190 280 280 280 280 150 150 150 150 275 375 275 275 150 150
26.45 26.45 25.8 25.8 23.51 23.3 23.3 23.3 22.98 22.98 23.1 22.92 20.56 20.56 19.92 19.92 20.27 20.27 20.27 20.27 17.37 17.66 17.37 17.37 14.53 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
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
( AB )2 = (OA)2 + (OB )2 − 2(OA)(OB ) cos AOB i.e.,
h = c + (h + c − x ) − 2c(h + c − x ) cos ωt 2
2
2
where ω is the angular velocity of the crank. The equation reduces to x 2 − 2 x[h + c(1 − cos ωt )] + 2c(h + c )(1 − cos ωt ) = 0
so that
(
x = h + c(1 − cos ωt ) ± c cos ωt + h − c 2
2
2
2
)
When ωt is zero, x is also zero, which means that the negative root sign must be taken. Therefore,
(
x = h + c(1 − cos ωt ) − c cos ωt + h + c 2
2
2
2
)
Acceleration is
a=
2
d x dt
2
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
a max = ω c (1 + 2
c h
)
(1-1)
It also appears that the minimum value of acceleration is
a min = ω c (1 − 2
c h
)
(1-2)
If N is the number of pumping strokes per minute then
2πN ω= 60
(rad/sec)
(1-3)
The maximum downward acceleration of point B (which occurs when the crank arm is vertically upward) is
a max
cN 2 ⎛ c⎞ = ⎜1 + ⎟ h⎠ 91.2 ⎝
(ft/sec2)
(1-4)
cN g ⎛ c ⎞ (ft/sec2) = ⎜1 + ⎟ 2936.3 ⎝ h⎠
(1-5)
or 2
a max
Likewise the minimum upward (amin) acceleration of point B (which occurs when the crank arm is vertically downward) is
cN g ⎛ c⎞ = ⎜1 − ⎟ 2936.3 ⎝ h⎠ 2
a min
(ft/sec2)
(1-6)
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
a max
d 1 cN 2 g ⎛ c⎞ = ⎜1 + ⎟ (ft/sec2) d 2 2936.3 ⎝ h⎠
(1-7)
where d1 and d2 are shown in Figure 1-5. But
2cd 2 =S d1 where S is the polished rod stroke length. So if S is measured in inches, then 2cd 2 S = 12 d1
or
cd 2 S = d1 24
(1-8)
So substituting Eq (1-8) into Eq (1-7) yields
a max
SN 2 g ⎛ c⎞ = ⎜1 + ⎟ (ft/sec2) 70471.2 ⎝ h⎠
(1-9)
or we can write Eq (1-9) as 2
a max
SN g = M (ft/sec2) 70471.2
(1-10)
where M is the machinery factor and is defined as
M =1+
c h
(1-11)
Similarly,
a min
SN 2 g ⎛ = ⎜1 − 70471.2 ⎝
c⎞ ⎟ (ft/sec2) h⎠
(1-12)
1.4.1 Maximum PRL
PRL max = S f (62.4)D
(A
− Ar ) γ s DAr γ s DAr ⎛ SN 2 M ⎞ ⎜⎜ ⎟⎟ + + 144 144 144 ⎝ 70471.2 ⎠
p
(1-13)
Equation (1-13) can be rewritten as PRL max
DAr γ s DAr γ s DAr ⎛ SN 2 M ⎞ ⎜⎜ ⎟⎟ = S f (62.4) − S f (62.4 ) + + 144 144 144 144 ⎝ 70471.2 ⎠ DAp
(1-14)
If the weight of the rod string in air is
Wr =
γ s DAr
(1-15)
144
which can be solved for Ar which is
144Wr Ar = γ sD
(1-16)
Substituting Eq (1-16) into Eq (1-14) yields PRL max
⎛ SN 2 M ⎞ ⎟⎟ = S f (62.4) − S f (62.4) + Wr + Wr ⎜⎜ γs 144 ⎝ 70471.2 ⎠ DAp
Wr
(1-17)
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 (117) becomes (where γs= 490) PRL max
⎛ SN 2 M ⎞ ⎟⎟ = S f (62.4) − 0.1Wr + Wr + Wr ⎜⎜ 144 ⎝ 70471.2 ⎠
PRL max
⎛ SN 2 M ⎞ ⎟⎟ = W f + 0.9Wr + Wr ⎜⎜ ⎝ 70471.2 ⎠
DAp
or
where
W f = S f (62.4 )
(1-18)
DA p 144
and is called the fluid load (not to be confused with the actual fluid weight on the rod string).
Thus, Eq (1-18) can be rewritten as
PRL max = W f + (0.9 + F1 )Wr
(1-19)
where for conventional units
SN (1 + 2
F1 =
c h
)
70471.2
(1-20)
and for air-balanced units F1 =
SN 2 (1 −
c h
70471.2
)
(1-21)
1.4.2 Minimum PRL
PRL min = − S f (62.4)
Wr
γs
+ Wr − Wr F2
which, for API 50° oil, reduces to
PRL min = 0.9Wr − F2Wr = (0.9 − F2 )Wr
(1-22)
where for the conventional units
SN (1 − ) F2 = 70471.2 2
c h
(1-23)
and for air-balanced units
SN (1 + ) F2 = 70471.2 2
c h
(1-24)
1.4.3 Counterweights The idea counter-balance load C is the average PRL. Therefore,
C=
1 2
(PRL max + PRL min )
Using Eqs (1-19) and (1-22) in the above we get
C = 12 W f + 0.9Wr +
1 2
(F1 − F2 )Wr
(1-25)
or for conventional units 2 ⎛ SN c⎞ 1 ⎟⎟ C = 2 W f + Wr ⎜⎜ 0.9 + 70471.2 h ⎠ ⎝
(1-26)
and for air-balanced units
⎛ SN c⎞ ⎟⎟ C = W f + Wr ⎜⎜ 0.9 − 70471.2 h ⎠ ⎝ 2
1 2
(1-27)
r d1 C = C s + Wc c d2 where Cs = structure unbalance, lbs Wc = 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 Peak torque T is (see Figure 1.5)
d2 T = c[C − (0.9 − F2 )Wr ] d1
(1-28)
Substituting Eq. (1-25) into Eq. (1-28) gives
T = 12 S [C − (0.9 − F2 )Wr ] or
(1-29)
[
T = 12 S 12 W f +
1 2
(F1 + F2 )Wr ]
or 2 ⎛ 2 SN Wr ⎞ 1 ⎟⎟ T = 4 S ⎜⎜W f + 70471.2 ⎠ ⎝
(in-lbs)
(1-30)
Torque factors and efficiency are used in practice:
T=
1 2
[PRLmax (TF1 ) + PRLmin (TF2 )] 0.93
(1-31)
Torque factor is defined:
Torque Exerted to the Crankshaft TF = Polished Rod Load TF1 =
Maximum upstroke torque factor occurred when the crank is in the horizontal position
TF2 =
Maximum downstroke torque factor occurred when the crank is in the horizontal position
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
Maximum Permissible Pumping Speed As given earlier the maximum value of the downward acceleration is equal to
a max/ min =
SN (1 ± 2
SN 2 g (1 ± 70471.2 c h
70471.2
)
≤L
c h
)
(1-32)
(1-33)
or
N limit
70471.2 L = S (1 m hc )
(1-34)
For L = 0.5
N limit =
187.7 S (1 m
c h
)
(1-35)
The minus sign is for conventional units and the plus sign for air-balanced units.
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 7/8” 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.
Example Problem 1-1:
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
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?
Solution: 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:
d2 111 S = 2c = (2)(37) = 85.52 in. d1 96.05 The maximum allowable pumping speed is:
N=
70471.2 L = c S (1 − h )
= 22 SPM
(70471.2)(0.4) (85.52)(1 − 0.33)
(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 W f = S f (62.4 )
Wr =
γ s DAr 144
DA p 144
=
= (0.9042)(62.4 )
(3,000)(4.91) = 5,770 lbs 144
(490)(3,000)(0.60) = 6,138 lbs 144
SN (1 + hc ) (85.52)(22) 2 (1 + 0.33) = = 0.7940 F1 = 70471.2 70471.2 2
Then the expected maximum PRL is:
PRL max = W f − S f (62.4)
Wr
γs
+ Wr + Wr F1
= 5,770 − (0.9042)(62.4)(6,138) /(490) + 6,138 + (6,138)(0.794)
= 16,076 lbs
< 21,300 lbs, OK
(c) The peak torque is calculated by Eq (1-30):
2 2 ⎛ ⎞ ⎛ SN W 2 2 ( 85 . 52 )( 22 ) (6,138) ⎞ r ⎟⎟ = 14 (85.52) ⎜⎜ 5,770 + ⎟⎟ T = 14 S ⎜⎜W f + 70471.2 70471.2 ⎠ ⎝ ⎠ ⎝
= 280,056 lb-in. < 320,000 lb-in. OK
Accurate calculation of counter-balance load requires the minimum PRL:
SN (1 − hc ) (85.52)(22) 2 (1 − 0.33) F2 = = = 0.4 70471.2 70471.2 2
PRL min = − S f (62.4)
Wr
γs
+ Wr − Wr F2
6,138 = −(0.9042)(62.4 ) + 6,138 − (6,138)(0.4) 490 = 2,976 lbs
C=
1 2
(PRL max + PRL min ) = 12 (16,076 + 2,976) = 9,526
lbs
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,
(37) (96.05) + 450 = 10,160 Wc (37) (111)
which gives Wc = 11,221 lbs. In order to generate the ideal counter-balance load of C = 9,526 lbs, the counterweights should be place on the crank at
(9,526)(111) r= (37) = 36.30 (11,221)(96.05)
in.
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 (dp): 2.5 in. Rod section 1, diameter (dr1): 1 in. length (L1): 0 ft Rod section 2, diameter (dr2): 0.875 in. length (L2): 3,000 ft Rod section 3, diameter (dr3): 0.75 in. length (L3): 0 ft Rod section 4, diameter (dr4): 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 oAPI
Maximum allowable acceleration factor (L):
0.4
Solution:
S = 2c N=
=
πd p2
85.52 in.
22 SPM
=
4.91 in.2
=
0.60 in.
=
5,770 lbs
=
6,138 lbs
4
πd r2 4
W f = S f (62.4 ) Wr =
=
70471.2 L S (1 − hc )
Ap = Ar =
d2 d1
γ s DAr 144
DA p 144
SN 2 (1 ± hc ) F1 = 70471.2 PRL max = W f − S f (62.4)
Wr
γs
+ Wr + Wr F1
2 ⎛ SN Wr ⎞ 2 ⎟⎟ T = 14 S ⎜⎜W f + 70471.2 ⎠ ⎝
SN 2 (1 m hc ) F2 = 70471.2 PRL min = − S f (62.4)
C=
1 2
Wr
γs
+ Wr − Wr F2
(PRL max + PRL min )
=
0.7940
=
16,076
lbs
=
280,056
lbs
=
0.40
=
2,976
lbs
=
9,526
lbs
1.5 Pump Deliverability and Power Requirements Liquid flow rate delivered by the plunger pump can be expressed as
Ap
S p E v (24)(60) q= N 144 12 Bo 5.615
(bbl/day)
or
q = 0.1484
A p NS p E v Bo
(stb/day)
1.5.1 Effective Plunger Stroke Length The magnitude of the rod stretch is
δl r =
W f Dr Ar E
(1-36)
Tubing stretch can be expressed by a similar equation. That is
δl t =
W f Dt At E
(1-37)
The magnitude of the rod stretch due to acceleration is called plunger over travel:
Wr Dr δl o = n (ft) Ar E
(1-38)
But the maximum acceleration term n can be written as
n=
SN 2 (1 ±
c h
70471.2
)
so that Eq (1-38) becomes 2 c ( ) 1 SN ± Wr Dr h (ft) δl o = Ar E 70471.2
(1-39)
Let us restrict our discussion to conventional units. Then Eq (1-39) becomes 2
W r D r SN M (ft) δl o = Ar E 70471.2
(1-40)
Eq (1-40) can be rewritten to yield δlo in inches. Wr is
W r = γ s Ar D r and γS = 490 lb/ft3 with E = 30 x 106 lb/m2 Eq (1-40) becomes 2 δl o = 1.93 × 10 −11 Dr2 SN(in) M
(1-41)
Plunger stroke is approximated using the above expressions as
S p = S − δl r − δl t + δl o or
12 D ⎡ ⎛ 1 1 ⎞ SN 2 M Wr ⎤ + ⎟⎟ − Sp = S − ⎢W f ⎜⎜ ⎥ E ⎣ ⎝ Ar At ⎠ 70471.2 Ar ⎦
(in)
(1-42)
If pumping is carried out at the maximum permissible speed limited by Eq (1-34), the plunger stroke becomes 12 D ⎡ ⎛ 1 1 ⎞ 1+ Sp = S − + ⎟⎟ − ⎢W f ⎜⎜ E ⎢⎣ ⎝ Ar At ⎠ 1 −
c h c h
LW r ⎤ ⎥ Ar ⎥⎦
(in)
For the air-balanced unit the term
1+ 1−
c h c h
is replaced by its reciprocal.
(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.
k p (d b − d p )
2.9
qs =
μ
d
(d
0.1 b
b
+ d p ) Δp Lp
(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:
Ph = 7.36 × 10 qγ l LN −6
(1-45)
and
LN = H +
ptf 0.433γ l
(1-46)
The power required to overcome friction losses can be empirically estimated as −7
Pf = 6.31×10 Wr SN
(1-47)
Thus the required prime mover power can be expressed as
Ppm = Fs ( Ph + Pf )
(1-48)
Example Problem 1-2:
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).
Solution: This problem can be quickly solved using the computer program SuckerRodPumpingFlowrate&Power.xls. Solution is shown in Table 1-3. 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): Depth to the liquid level in annulus (H): Flowing tubing head pressure (ptf): Tubing outer diameter (dto):
4,000 ft 4,000 ft 100 ft 3.5 in.
Tubing inner diameter (dti): Tubing anchor (1 = yes; 0 = no): Plunger diameter (dp): Rod section 1, diameter (dr1): length (L1): Rod section 2, diameter (dr2): length (L2): Rod section 3, diameter (dr3): length (L3): Rod section 4, diameter (dr4): length (L4):
2.995 in. 0 2.5 in. 1 in. 0 ft 0.875 in. 0 ft 0.75 in. 4,000 ft 0.5 in. 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):
Solution:
At =
Ap = Ar =
1.35
π d t2 4
πd p2
=
2.58 in.2
=
4.91 in.2
=
0.44 in.
4 πd r2 4
W f = S f (62.4 )
DA p =
7,693
lbs
=
6,013
lbs
=
1.33
=
70
=
753
=
4,255
ft
Ph = 7.36 × 10 −6 qγ l LN
=
25.58
hp
Pf = 6.31×10 −7 Wr SN
=
7.2
hp
Ppm = Fs ( Ph + Pf )
=
44.2
hp
Wr =
144
γ s DAr 144
M = 1±
c h
12 D ⎡ ⎛ 1 1 ⎞ SN 2 M Wr ⎤ Sp = S − + ⎟− ⎥ ⎢W f ⎜ E ⎣ ⎜⎝ Ar At ⎟⎠ 70471.2 Ar ⎦
q = 0.1484
LN = H +
A p NS p E v
in. sbt/day
Bo
ptf
0.433γ l
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 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.
Example Problem 1-3:
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: 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 gives qo = 687 bbl/day > 600 bbl/day, OK Ppm = 30.2 hp
(5) Spreadsheet program SuckerRodPumpingLoad.xls gives PRLmax = 16,121 lbs PRLmin = 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
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.
(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,
(37) (96.05) Wc + 450 (37) (111)
= 12,630 lbs
which gives Wc = 14,075 lbs. In order to generate the ideal counterbalance load of C = 10,327 lbs, the counterweights should be place on the crank at
(10,327)(111) r= (37) = 31.4 (14,076)(96.05)
in.
(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
(18)(30.12)(24) de = (750)
= 17.3 in.
2,500
Pump Displacement (bbl/day)
Curve 2,000
1,500
1,000
API Size
Stroke
A B C D E F
40 57 80 114 160 228
34 42 48 54 64 74
G H
320 640
84 144
500
0 0
2,000
4,000
6,000
8,000
10,000
Pump Setting Depth (ft)
Figure 1-8: Sucker rod pumping unit selection chart (After Kelley and Willis, 1954)
12,000
Table 1-4: Design data for API sucker rod pumping units
Pump Depth (ft) 1000-1100 1100-1250 1250-1650 1650-1900 1900-2150 2150-3000 3000-3700 3700-4000
Pump Depth (ft) 1150-1300 1300-1450 1450-1850 1850-2200 2200-2500 2500-3400 3400-4200 4200-5000
(a) Size 40 Unit with 34-inch Stroke Plunger Size Rod Sizes (in) Tubing Size (in) (in) 2 3/4 3 7/8 2 1/2 3 7/8 2 1/4 2 1/2 3/4 2 2 1/2 3/4 1 3/4 2 1/2 3/4 1 1/2 2 5/8-3/4 1 1/4 2 5/8-3/5 1 2 5/8-3/6
Pumping Speed (stroke/min) 24-19 24-19 24-19 24-19 24-19 24-19 22-18 21-18
(b) Size 57 Unit with 42-inch Stroke Tubing Plunger Size (in) Size (in) Rod Sizes (in) 2 3/4 3 7/8 2 1/2 3 7/8 2 1/4 2 1/2 3/4 2 2 1/2 3/4 1 3/4 2 1/2 3/4 1 1/2 2 5/8-3/4 1 1/4 2 5/8-3/5 1 2 5/8-3/6
Pumping Speed (stroke/min) 24-19 24-19 24-19 24-19 24-19 23-18 22-17 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 5000-6000
1 1/4 1
2 2
5/8-3/5 5/8-3/6
21-17 19-17
Pump Depth (ft) 1700-1900 1900-2100 2100-2700 2700-3300 3300-3900 3900-5100 5100-6300 6300-7000
(d) Size 114 Unit with 54-inch Stroke Plunger Size (in) Tubing Size (in) Rod Sizes (in) 2 3/4 3 7/8 2 1/2 3 7/8 2 1/4 2 1/2 3/4 2 2 1/2 3/4 1 3/4 2 1/2 3/4 1 1/2 2 5/8-3/4 1 1/4 2 5/8-3/5 1 2 5/8-3/6
Pumping Speed (stroke/min) 24-19 24-19 24-19 23-18 22-17 21-17 19-16 17-16
Pump Depth (ft) 2000-2200 2200-2400 2400-3000 3000-3600 3600-4200 4200-5400 5400-6700 6700-7700
(e) Size 160 Unit with 64-inch Stroke Rod Sizes Plunger Size (in) Tubing Size (in) (in) 2 3/4 3 7/8 2 1/2 3 7/8 2 1/4 2 1/2 3/4-7/8 2 2 1/2 3/4-7/8 1 3/4 2 1/2 3/4-7/8 1 1/2 2 5/8-3/4-7/8 1 1/4 2 5/8-3/4-7/8 1 2 5/8-3/4-7/8
(f) Size 228 Unit with 74-inch Stroke Rod Sizes Pump Depth (ft) Plunger Size (in) Tubing Size (in) (in) 2400-2600 2 3/4 3 7/8 2600-3000 2 1/2 3 7/8 3000-3700 2 1/4 2 1/2 3/4-7/8 3700-4500 2 2 1/2 3/4-7/8 4500-5200 1 3/4 2 1/2 3/4-7/8 5200-6800 1 1/2 2 5/8-3/4-7/8 6800-8000 1 1/4 2 5/8-3/4-7/8 8000-8500 1 1/16 2 5/8-3/4-7/8
Pumping Speed (stroke/min) 24-19 24-19 24-19 23-18 22-17 21-17 19-15 17-15
Pumping Speed (stroke/min) 24-20 23-18 22-17 21-16 19-15 18-14 16-13 14-13
(g) Size 320 Unit with 84-inch Stroke Rod Sizes Pump Depth (ft) Plunger Size (in) Tubing Size (in) (in) 2800-3200 2 3/4 3 7/8 3200-3600 2 1/2 3 7/8 3600-4100 2 1/4 2 1/2 3/4-7/8-1 4100-4800 2 2 1/2 3/4-7/8-1 4800-5600 1 3/4 2 1/2 3/4-7/8-1 5600-6700 1 1/2 2 1/2 3/4-7/8-1 6700-8000 1 1/4 2 1/2 3/4-7/8-1 8000-9500 1 1/16 2 1/2 3/4-7/8-1
Pumping Speed (stroke/min) 23-18 21-17 21-17 20-16 19-16 18-15 17-13 14-11
(h) Size 640 Unit with 144-inch Stroke Pump Depth (ft) 3200-3500 3500-4000 4000-4700 4700-5700 5700-6600 6600-8000 8000-9600 9600-11000
Plunger Size (in) 2 3/4 2 1/2 2 1/4 2 1 3/4 1 1/2 1 1/4 1 1/16
Tubing Size (in) 3 3 2 1/2 2 1/2 2 1/2 2 1/2 2 1/2 2 1/2
Pumping Speed Rod Sizes (in) (stroke/min) 7/8-1 18-14 7/8-1 17-13 3/4-7/8-1 16-13 3/4-7/8-1 15-12 3/4-7/8-1 14-12 3/4-7/8-1 14-11 3/4-7/8-1 13-10 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
Problems 1-1. If the dimensions d1, d2 and c take the same values for both conventional unit (Class I lever system) and airbalanced 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 airbalanced 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 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.
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