KOYA UNIVERSITY
PETROLEUM ENGINEERING DEPARTMENT
THIRD STAGE
PRODUCTION TECHNOLOGY I , S5
CHAPTER THREE
COMPLETION EQUIPMENT
@DPTE 2025
Dr. Sarhad Ahmed Farkha
Sarhad.ahmed@koyauniversity.org
PhD Petroleum Engineering – Production Engineering – Koya University – October 2024
MSc Petroleum Engineering- Teesside Uni. United Kingdom October 2013
BSc Petroleum Engineering – Koya University July 2010
Member of Society of Petroleum Engineer (SPE), Onepetro access
Contents
8
➢
Production Tubing strings
a. Tubing Specifications
b. Tubing Inspections
c. Helical Buckling in Production Tubing
➢
Production packers


General consideration in packer selection
Types of production packers
◼
◼
➢
Retrievable packers
Permanent packers
Flow control system
◼
◼
◼
Subsurface control equipment
Safety systems:
◼ Subsurface controlled safety valves (SSV)
◼ Surface controlled safety valves (SCSSV)
◼ Slide Sleeve Door (SSD)
Operation considerations of Bottom hole chokes and Regulators
@KOU 2024-2025 DPTE
36
Problem Related to
Production Tubing
Helical Buckling in Tubing
Helical Buckling in Tubing
37
Introduction
• During completion tubing design process, it is necessary to calculate
variations in length for the stress applied under load conditions.
• When these have been determined it will confirm the suitability of the
selected tubing.
• Changes in temperature and in pressure inside and outside of tubing
sealed in a packer, depending on the type of packer and how it is set
will:
•
•
Increases or decreases the length of the tubing with a packer permitting free
motion of the tubing.
Unseal a packer not rigidly attached to the casing.
• Movement can only occur if the tubing is free to move.
• If the tubing is not free to move and is anchored to a packer then stress
will be subjected to the tubing string and packer.
• Tubing movement upwards (contraction) is assumed to be negative and
downwards (lengthening) is positive.
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Helical Buckling in Tubing
38
Supplied Force
• Assume there is Tubing hung freely inside the well, If tension force effect
on steel from bottom, the steel be in tensile case and remain straight as
shown is figure (3.1-A).
• Either compressive force effect on steel from bottom, the steel be in
compress case and may be buckled.
• The helix shown in figure (3.1-B) has a variable pitch as the compressive
force is progressively lowered by the weight of the pipe hanging below.
• The buckling effect is greater when pressure differential is applied across
the pipe. Unless the tubing string is shorter or the compressive force is
exceedingly high, some of the tubing will be buckled and the rest straight.
@KOU 2024-2025 DPTE
Helical Buckling in Tubing
39
Supplied Force
A
Figure 3.1: Force effect
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B
Helical Buckling in Tubing
40
Supplied Force
• The exact point between the buckled and straight section is the “Neutral
Point”.
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Low
Helical Buckling in Tubing
41
Supplied Force
• The neutral point can be calculated from the following:
𝒏 = 𝑭Τ𝑾
(3.1)
Where: n = distance from the bottom of tubing to the neutral point
F = supply force, for compression force have +ve value (Ib)
W= Weight per unit length (Ib/in)
• The pitch length can be calculated from the following:
𝒉=𝝅
𝟖𝑬𝑰
𝑭
𝟏ൗ
𝟐
(3.2)
Where: h = pitch (in)
E= young modules of elasticity, for steel = 30 × 106 psi
I = moment of inertia of tubing cross section with respect to its diameter, 𝑖𝑛4
𝐼 = 𝜋Τ64 (𝐷4 − 𝑑 4 )
Where:
D= outside diameter of tubing, in
d= inside diameter of tubing
@KOU 2024-2025 DPTE
(3.3)
High
Helical Buckling in Tubing
42
Supplied Force
Figure (3.2) shows different types of packers hold on it production tubing
Figure 4.2: Types of production packer
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Helical Buckling in Tubing
43
Supplied Force
• To analyze the effect of that predicted to occur in the production tubing assume that the
tubing contain fluid applied pressure at the packer seal, this pressure induced actual force
on tubing wall, that estimate as:
𝑭𝒂 = 𝑨𝒑 − 𝑨𝒊 𝑷𝒊
Where:
(3.4)
Fa = actual force
Ap = area based on the packer diameter, in2
Ai = inside diameter of tubing, in2
Pi = pressure at packer seal in tubing, psi
• In fact, the force that cause buckling in tubing is not the actual force, its greater that actual
force called “Fictitious Force”.
𝑭𝒇 = 𝑨𝒑 𝑷𝒊
(3.5)
• To simulate the real condition in oil well, it will have completion fluid in both the tubing
and the annulus, this fluid cause outer and inner pressure; therefore; Fictitious Force is
equal:
𝑭𝒇 = 𝑨𝒑 (𝑷𝒊 − 𝑷𝒐 )
@KOU 2024-2025 DPTE
(3.6)
Helical Buckling in Tubing
44
Supplied Force
• There are three methods in which tubing is connected to the packer,:
• Packers permitting free movement: tubing is fully free to move either way
• Packers not-permitting downward movement: the tubing is positioned
where it is fully free to move upward but its downward movement is
restricted, and stress applied to the packer
• Anchored tubing: the tubing is connected to the packer by being threaded to
or latched to the packer.
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Helical Buckling in Tubing
45
Supplied Force
• Several effects must be evaluated to accurately determine the tubing
movement or stress situation as shown in figure (3.3).
1
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2
3
4
Figure (3.3) effect of varies force on tubing movement
Helical Buckling in Tubing
46
Packers permitting free movement
• These are three pressure induce effects which produce forces that
move the tubing and effect of temperature.
• These effects are:
1. Piston effect (Hooke’s law)
2. Buckling effect
3. Ballooning effect
4. Temperature effect
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Helical Buckling in Tubing
47
Piston effect (Hooke’s law)
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Helical Buckling in Tubing
48
Packers permitting free movement
1-Piston Effect (Hooke’s law)
• The force that effect on cross section area of tubing called Stress and the relative
change in tubing length called Strain.
• The ratio of stress to strain called young Module of elasticity, therefore; Hooke’s
law link between stress and strain within the elastic range as:
𝑭𝒂
ൗ𝑨𝒔
𝑬 = ∆𝑳𝟏
ൗ𝑳
∆𝑳𝟏 =
@KOU 2024-2025 DPTE
𝑳𝑭𝒂
𝑬𝑨𝒔
(3.7)
(3.8)
Helical Buckling in Tubing
49
Packers permitting free movement
1-Piston Effect (Hooke’s law)
• The tubing is run into a completion fluid with equivalent fluid density inside and
outside the tubing which results is a reduction of the load due to buoyancy,
𝐹𝑎 = 𝐴𝑝 − 𝐴𝑖 𝑃𝑖 − (𝐴𝑝 − 𝐴𝑜 )𝑃𝑜
• Where:
∆𝐿1 = change in length due to hook’s law effect, in
L = Length of the tubing string to the packer depth, in
𝐹𝑎 = Actual force acting on bottom of tubing, Ib
𝐴𝑠 = Cross sectional area of tubing, in2
𝑃𝑜 = pressure at the packer set in annulus, psi
𝐴𝑜 = area based on the outside diameter of tubing, in2
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(3.9)
Helical Buckling in Tubing
50
Packers permitting free movement
1-Piston Effect (Hooke’s law)
• If there is an alteration from this initial condition causing a change in pressure
forces across the packer seal unit then a piston effect is cased.
• This will alter the tensile load on the top and bottom of the tubing, therefore; in
case of variation operation conditions:
𝐹𝑎 = 𝐴𝑝 − 𝐴𝑖 ∆𝑃𝑖 − (𝐴𝑝 − 𝐴𝑜 )∆𝑃𝑜
(3.10)
• Where:
∆𝑃𝑖 = (𝑃𝑖𝑓 − 𝑃𝑖𝑖 ) = change in tubing pressure at the packer, psi
∆𝑃𝑜 = (𝑃𝑜𝑓 − 𝑃𝑜𝑖 ) = change in annulus pressure at the packer, psi
Where, the subscription f refers to final condition, and i to initial condition.
𝐹𝑎 = +𝑣𝑒 Compression force → ∆𝐿1 = - ve, decrease in tubing length
𝐹𝑎 = −𝑣𝑒 Tension force → ∆𝐿1 = + ve, increase in tubing length
• Substituting equation (10) in equation (8):
∆𝐿1 = −
@KOU 2024-2025 DPTE
𝐿 𝐴𝑝 −𝐴𝑖 ∆𝑃𝑖 −(𝐴𝑝 −𝐴𝑜 )∆𝑃𝑜
𝐸𝐴𝑠
(3.11)
Helical Buckling in Tubing
51
2- Helical Buckling effect
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Helical Buckling in Tubing
52
Packers permitting free movement
2-Helical Buckling effect,
• Helical buckling is initiated by compressive force acting on the bottom of the
tubing and it is the formation of helical spirals in the tubing string.
• The helix shown in figure (3.4) has a variable pitch as the compressive force is
progressively lowered by differential is applied across the pipe.
• Unless the tubing string is shorter or the compressive force is exceedingly
high, some of the tubing will be buckled and the rest straight. The exact point
between the buckled and straight section is the “Neutral Point”.
• However, in order to complete the understanding of the effect which leads to
variations in length due to buckling, we must also consider the effect caused by
pressure differential across the pipe.
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Helical Buckling in Tubing
53
Packers permitting free movement
2- Helical Buckling effect,cont.,
• Helical buckling is caused by the effect of the pressure which acts on the
lateral surface of the pipe wall as the convex surface of the bend in a greater
force is larger than the concave surface, as shown in figure (3.4)
@KOU 2024-2025 DPTE
Figure (3.4): Pressure induced helical buckling effect
Helical Buckling in Tubing
54
Packers permitting free movement
2- Helical Buckling effect, cont.,
• The internal pressure will therefore exert a greater force on the convex side of
the helix, then that exerted on the concave section of the same bend.
• The resulting force will, therefore, created the helical buckling configuration.
The same occurs when the stable external pressure is greater than the internal
pressure also resulting in helical buckling.
• When the natural point is within the tubing length (and so the helix can fully
develop), the length reduction due to helical buckling (ΔL2) can be calculated
by the following formula:
L > n:
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𝑟 2 𝐹𝑡2
∆𝐿2 = −
8𝐸𝐼𝑊
(3.12)
Helical Buckling in Tubing
55
Packers permitting free movement
2- Helical Buckling effect, cont.,
• If the tubing is very short (as happens for example on selective type
completion between two packers) all the string may be affected by buckling
and there is no natural point.
• In this case, the length reduction due to the buckling effect is dependent upon
the entire length of the string and can be calculated by the following formula:
L ≤ n:
𝑟 2 𝐹𝑡2
𝐿𝑊
∆𝐿2 = −
−
−
8𝐸𝐼𝑊
𝐹𝑓
@KOU 2024-2025 DPTE
𝐿𝑊
2−
𝐹𝑓
(3.13)
Helical Buckling in Tubing
56
Packers permitting free movement
2- Helical Buckling effect, cont.,
• Where:
𝐹𝑓 = 𝐴𝑝 𝑃𝑖 − 𝑃𝑜
𝑟 = (𝑰𝐷𝑪 −𝐷)ൗ2
𝑊 = 𝑊𝑠 + 𝑊𝑖 − 𝑊𝑜
•
•
•
•
•
•
𝐹𝒕 = 𝐴𝒑 (∆𝑃𝑖 − ∆𝑃𝑜 )
(3.14)
(3.15)
(3.16)
𝑊𝑖 = 𝐴𝑖 × 𝑤𝑖𝑒𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡𝑢𝑏𝑖𝑛𝑔
𝑊𝑜 = 𝐴𝑜 × 𝑤𝑖𝑒𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡𝑢𝑏𝑖𝑛𝑔
𝐼𝐷𝑐 = inside diameter of casing, in
W = Weight per unit length, lb/in
𝑊𝑠
=Weight of tubing per unit length
𝑊𝑖
=Weight of fluid contained inside tubing per unit length (based on ID of
tubing).
• 𝑊𝑜
=weight of annulus fluid displaced by bulk volume of tubing per unit
length
(based on OD of tubing).
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Helical Buckling in Tubing
57
Packers permitting free movement
2- Helical Buckling effect, cont.,
• The fictitious force equal zero in initial condition for most operations where
(𝑃𝑖 = 𝑃𝑜 ) . But in final condition, may be pressure change occurs, therefore
fictitious force calculated by use equation (3-14).
• Equations (3- 12) and (3-13) used to calculate the reduction in tubing length
only not lengthening. That means the fictitious force is compression force (+
ve).
• It should be remember that, to calculate the variations in length, the variations
of the forces compared to initial conditions must be calculated. Therefore, to
sum up:
•
•
In the ∆𝑳𝟏 (Hooke’s law), the variation of the piston force (Actual force, 𝑭𝒂 ) must be
used.
In the ∆𝑳𝟐 (Buckling), the variation of the fictitious force (𝑭𝒇 ) must be used when
this is positive, otherwise, being a tensile force, it cannot buckle the string and
∆𝑳𝟐 = 0
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Helical Buckling in Tubing
58
3- Ballooning effect
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Helical Buckling in Tubing
59
Packers permitting free movement
3- Ballooning Effect:
• The third element which changes the length of a string, due to the changes to
internal and external pressure, is caused by ballooning.
• This effect occurs when (Δ𝑃 = 𝑃𝑖 − 𝑃𝑜 ) is positive and tend to swell the
tubing which, contracts axially or shortens.
• On the other hand, when (Δ𝑃 = 𝑃𝑖 − 𝑃𝑜 ) is negative, the tubing is squeezed
and expands axially or elongates. This is termed reverse ballooning, as shown
in figure (3.5).
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Helical Buckling in Tubing
60
Packers permitting free movement
3- Ballooning Effect, comt.,:
Figure (3.5): Ballooning effect
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Helical Buckling in Tubing
61
Packers permitting free movement
3- Ballooning Effect, cont.,:
• The normally used simplified formula to calculate ballooning or reverse
ballooning effect in tubing length (∆𝐿3 ) is:
∆𝐿3 =
𝑉
−
𝐸
∆𝜌𝑖 −𝑅2 ∆𝜌𝑜 −(𝛿 1+2𝑉ൗ2𝑉 ) 𝐿2
(𝑅2 −1)
−
2𝑉
𝐸
∆𝑃𝑖𝑠 −𝑅2 ∆𝑃𝑜𝑠 𝐿
(𝑅2 −1)
(3-17)
• Where:
V = Poisson’s ratio (0.3 for steel)
R = tubing OD / tubing ID
∆𝛒𝐢 = change in density of fluid inside tubing, lb/in3
∆𝛒𝐨 = change in density of fluid outside tubing, Ib/in3
∆𝐏𝐢𝐬 = change in tubing pressure at the surface.
∆𝐏𝐨𝐬 = change in annulus pressure at the surface.
𝛅
= pressure drop in tubing due to flow, psi/in. (usually considered as 𝛅 = 0, i.e. in
steady and 𝛅 = +ve. In flow rate).
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Helical Buckling in Tubing
62
4- Temperature effect
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Helical Buckling in Tubing
63
Packers permitting free movement
4- Temperature Effect:
• The final effect considered when calculating tubing length variations, is the
temperature effect which usually induces the largest movement.
• During a well operation, stimulation the temperature of the tubing may be
much less than that in, either the initial or flow rate conditions.
• During well stimulation, significant quantities of fluids are pumped through
the tubing at ambient surface temperature which may change the temperature
of the tubing by several degrees.
• The formula used to calculate the change of length due to temperature effect
(∆𝐿4 ) is:
∆𝐿4 = 𝐶𝐿 ∆𝑇
@KOU 2024-2025 DPTE
(3.18)
Helical Buckling in Tubing
64
Packers permitting free movement
4- Temperature Effect:
∆𝐿4 = 𝐶𝐿 ∆𝑇
(3.18)
• Where:
C = materials coefficient of thermal expansion, for steel = 69 𝑥 10−6 𝑝𝑒𝑟 ˚𝐹
L = length of tubing string, feet
ΔT = change in temperature, ˚F
∆𝐿4 = change in length, feet
•
Where the average temperature variation in the string can be calculated as
follows:
∆𝑇 =
@KOU 2024-2025 DPTE
(𝑇𝑓𝑖𝑛𝑎𝑙 −𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 )𝑡𝑜𝑝 ℎ𝑜𝑙𝑒 +(𝑇𝑓𝑖𝑛𝑎𝑙 −𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 )𝑏𝑜𝑡𝑡𝑜𝑚 ℎ𝑜𝑙𝑒
2
(3.19)
Helical Buckling in Tubing
65
Total
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Helical Buckling in Tubing
66
Packers permitting free movement
• The sum of the length changes (ΔL) obtained from the change in pressure
induced force and temperature effects, gives the total shift of the bottom end of
the string at the packer depth where it is free to move in the packer-bore. This
sum is calculated:
∆𝐿 = ∆𝐿1 + ∆𝐿2 + ∆𝐿3 + ∆𝐿4
• ∆𝐿1 = −
• L > n:
• ∆𝐿3 =
𝐿 𝐴𝑝 −𝐴𝑖 ∆𝑃𝑖 −(𝐴𝑝 −𝐴𝑜 )∆𝑃𝑜
𝐸𝐴𝑠
𝑟 2 𝐹𝑡2
∆𝐿2 = −
,
8𝐸𝐼𝑊
𝑉
−
𝐸
∆𝜌𝑖 −𝑅2 ∆𝜌𝑜 −(𝛿 1+2𝑉ൗ2𝑉 ) 𝐿2
• ∆𝐿4 = 𝐶𝐿 ∆𝑇
@KOU 2024-2025 DPTE
L ≤ n:
𝑟 2 𝐹𝑡2
𝐿𝑊
∆𝐿2 = −
−
−
8𝐸𝐼𝑊
𝐹𝑓
(𝑅2 −1)
−
2𝑉
𝐸
2−
∆𝑃𝑖𝑠 −𝑅2 ∆𝑃𝑜𝑠 𝐿
(𝑅2 −1)
𝐿𝑊
𝐹𝑓
Helical Buckling in Tubing
67
Example:
The following well data from work-over operations by cementing formation produce water
(Aquifer
Casing 7”, 32 Ib/ft, Tubing 2 7/8 ”, 6.5Ib/ft
Production packer 3.25”, set @ 10000 ft.
Temperature change -20°F (Cooling)
Initial condition
Tubing & Annulus contain oil have API = 30°. Pressure inside & outside tubing at
surface (Pth, Pah) = zero.
Final condition
Annulus contain same oil & Tubing contain cement have 𝜌 = 15 lbgal.
Pressure inside tubing at surface (Pth) = 5000 psi, pressure outside tubing at surface
(Pah) = 1000 psi
Determine:
1. The change in tubing length with respect to all effects pressure and temperature
effects.
2. The actual force, fictitious force & neutral point.
3. Mention the conclusion
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Helical Buckling in Tubing
68
Example:
Solution:
Initial Condition:
𝐴𝑃𝐼 = 141.5Τ𝛾 − 131.5
Oil specific gravity, 𝛾=0.8762
Oil density, 𝜌 = 𝜌𝑖 = 𝜌𝑜 = 62.4 × 0.8762 = 54.675 𝐼𝑏/𝑓𝑡 3
Pressure gradient inside & outside tubing:
54.675
Pressure gradient = 123 = 0.0317 Ib/in2/in
Pressure inside & outside tubing at packer depth:
𝑃𝑖 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 + 𝑃𝑡ℎ
𝑃0 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 + 𝑃𝑎𝑡𝑚
𝑃𝑖 = 𝑃𝑜 = 0.0317 × 120000 + 𝑧𝑒𝑟𝑜 = 3800 𝑝𝑠𝑖
Calculate weight per unit length, W
6.5
𝑊𝑖 = 𝑊𝑠 + 𝑊𝑖 − 𝑊𝑜 = 𝑊𝑠 + 𝐴𝑖 𝜌𝑖 − 𝐴𝑜 𝜌𝑜 =
+ 4.68 × 0.0317 − 6.49 × 0.0317
12
𝑊𝑖 = 0.4843 𝐼𝑏/𝑖𝑛
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Helical Buckling in Tubing
69
Example:
Solution:
Final Condition:
Pressure gradient inside tubing contain cement.
Pressure gradient= 15/231 = 0.065 Ib/in2/in
Pressure gradient in annulus contain oil:
As in initial condition, pressure gradient = 0.0317 Ib/in2/in
𝑃𝑖 = 0.065 × 120000 + 5000 = 12800 𝑝𝑠𝑖
𝑃𝑜 = 0.0317 × 120000 + 1000 = 4800 𝑝𝑠𝑖
6.5
𝑊𝑓 = 𝑊𝑠 + 𝐴𝑖 𝜌𝑖 − 𝐴𝑜 𝜌𝑜 = 12 + 4.68 × 0.065 − 0.0317 × 6.49
𝑊𝑓 = 0.6401 Ib/in
Pressure & density changing between initial and final condition
∆𝑃𝑖 = 12800 − 3800 = 9000 𝑝𝑠𝑖
∆𝑃𝑜 = 4800 − 3800 = 1000 𝑝𝑠𝑖
∆𝜌𝑖 = 0.065 − 0.0317 = 0.0333 Ib/in3
∆𝜌𝑜 = 0.0317 − 0.0317 = 0 Ib/in3
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Helical Buckling in Tubing
70
Example:
Solution:
• Now Calculate ∆𝐿1 using equation (3.11):
∆𝐿1 = −
𝐿 𝐴𝑝 −𝐴𝑖 ∆𝑃𝑖 −(𝐴𝑝 −𝐴𝑜 )∆𝑃𝑜
𝐸𝐴𝑠
120000 8.30−4.68 9000−(8.30−6.49)1000
∆𝐿1 = −
30×106 ×1.81
∆𝐿1 = −68 in
• Calculate ∆𝐿2 using equation (3.12 ana 3.13):
First, Calculate neutral point n by:
𝐹
𝑛 = 𝑓ൗ𝑊
𝐹𝑓 = 𝐴𝑝 𝑝𝑖 − 𝑝𝑝 = 8.30 × 12800 − 4800 = 66400 𝑝𝑠𝑖
𝑛 = 66400Τ0.64 = 103750 𝑖𝑛 = 8646 𝑓𝑡
n(8646ft)<L(10000 ft, therefore:
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Helical Buckling in Tubing
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Example:
Solution:
• Calculate ∆𝐿2 using equation (3.12)
𝑟 2 𝐹𝑡2
1.612 × (8.30(9000 − 1000)2
∆𝐿2 = −
→ ∆𝐿2 = −
→ ∆𝐿2 = −46.2 𝑖𝑛
8𝐸𝐼𝑊
8 × 30 × 106 × 1.61 × 0.64
• Calculate ∆𝐿3 using equation (3.17)
𝑽 ∆𝝆𝒊 − 𝑹𝟐 ∆𝝆𝒐 − (𝜹 𝟏 + 𝟐𝑽ൗ𝟐𝑽 ) 𝑳𝟐
𝟐𝑽 ∆𝑷𝒊𝒔 − 𝑹𝟐 ∆𝑷𝒐𝒔 𝑳
∆𝑳𝟑 = −
−
𝑬
(𝑹𝟐 − 𝟏)
𝑬
(𝑹𝟐 − 𝟏)
∆𝑳𝟑
𝟎. 𝟎𝟑𝟑𝟑 − ((𝟐. 𝟖𝟕𝟓)/(𝟐. 𝟒𝟒𝟏))𝟐 × 𝟎 − (𝟎 𝟏 + 𝟐 × 𝟎. 𝟑ൗ𝟐 × 𝟎. 𝟑 ) 𝟏𝟐𝟎𝟎𝟎𝟎𝟐
𝟎. 𝟑
= −
𝟑𝟎 × 𝟏𝟎𝟔
(((𝟐. 𝟖𝟕𝟓)/(𝟐. 𝟒𝟒𝟏))𝟐 −𝟏)
𝟐 × 𝟎. 𝟑
−
𝟑𝟎 × 𝟏𝟎𝟔
𝟏𝟎𝟎𝟎 − ((𝟐. 𝟖𝟕𝟓)/(𝟐. 𝟒𝟒𝟏))𝟐 × 𝟏𝟎𝟎𝟎 𝟏𝟐𝟎𝟎𝟎𝟎
((𝟐. 𝟖𝟕𝟓)/(𝟐. 𝟒𝟒))𝟐 −𝟏
∆𝑳𝟑 = −𝟏𝟐. 𝟑𝟖 + 𝟐. 𝟒 = −𝟗. 𝟗𝟖 𝒊𝒏
@KOU 2024-2025 DPTE
Helical Buckling in Tubing
72
Example:
Solution:
• Calculate ∆𝐿4 using equation (3.18),
∆𝐿4 = 𝐶𝐿 ∆𝑇
∆𝐿4 = 6.9 × 10−6 × 120000 × (−20)
∆𝐿4 = −16.56 𝑖𝑛
• Then ∆𝐿𝑡𝑜𝑡𝑎𝑙 equal:
∆𝐿 = ∆𝐿1 + ∆𝐿2 + ∆𝐿3 + ∆𝐿4
∆𝐿 = −68 + −46.2 + −9.98 + −16.56 = -140.74 inch
@KOU 2024-2025 DPTE
Helical Buckling in Tubing
73
Example:
Solution:
Conclusion
• Ff have +ve, value that means it is compression force, i.e., helical buckling effect occur on
tubing, and Ff (66400 lb) value is equal double value of Fa (37648 lb) value.
•
Neutral point locate through tubing length (n<L, for this used Eq.(4.12) to calculate ΔL2, and
buckling occur in most tubing (86% from original length).
• The change in tubing length result (ΔL2) from buckling helical effect have high ratio (35%)
from total length (ΔL). When neglect this change (ΔL2), it will cause moving of tubing up,
outside the seal plug and it is danger
• The total change in length (ΔL 140.74 in) is the maximum expected change. Generally, the
actual change is less than the calculated ΔL change, due to friction between tubing and packer
that ignore here.
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Helical Buckling in Tubing
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(Assignment)
In vertical well, packer is set at 10,000 ft, which is free to move. Its outside diameter is 4.5” Tubing
(15.1 lb/ft.) with ID of tubing = 3.826”
• Packer seal bore outside diameter = 5.0”
• Weight per length = 17.7 lb/ft.
• E (Young’s modulus) = 30 × 106
• µ (Poisson’s ratio) = 0.3
• Production casing size = 7” with ID = 6.049”
Initial Condition
Fluid in annulus = 10.0 ppg
Fluid in tubing = 10.0 ppg
Tubing pressure = 0 psi
Annulus pressure = 0 psi
Final Condition
Fluid in annulus = 10.0 ppg
Fluid in tubing = 8.0 ppg
Tubing pressure = 1,500 psi
Annulus pressure = 0 psi
Determine: The change in tubing length
@KOU 2024-2025 DPTE
Helical Buckling in Tubing
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End of Lecture
@KOU 2024-2025 DPTE