1. 1
PETE 411
Well Drilling
Lesson 37
Coiled Tubing
2. 2
Coiled Tubing
What is Coiled Tubing?
Uses of Coiled Tubing
Properties of Coiled Tubing
Drilling with Coiled Tubing
Buckling
3. 3
Buckling of Coiled Tubing
Buckling Modes
Sinusoidal and Helical Buckling
Buckling in Horizontal or Inclined Sections
Buckling in Vertical Section
Buckling in Curved Wellbores
Prediction of Buckling Loads
“Lockup” of Tubulars
16. From
OGJ
July 8,
2002
p.62
Coil
Tubing
Drilling on
the North
Slope
16
17. 17
Coil Tubing Drilling on the North Slope
Drilling Rates routinely in excess of
250 ft/hr - drilling in sandstone
Laterals longer than 2,500 ft
Good incremental oil production
Used electrical umbilical for MWD
Used mud motors and 3 ¾-in PDC bits
18. 18
Advantages
No rig required
No connections - fast tripping
Disadvantages
Fatigue life limit (cycles)
Pressure and tension
Diameter and ovality
19. 19
Reference:
“Coiled Tubing Buckling Implication in
Drilling and Completing Horizontal Wells”
by Jiang Wu and H.C. Juvkam-Wold, SPE
Drilling and Completion, March, 1995.
24. Sinusoidal Buckling in a Horizontal Wellbore
When the axial compressive load along the coiled
tubing reaches the following sinusoidal buckling
load Fcr, the intial (sinusoidal or critical) buckling
of the coiled tube will occur in the horizontal
wellbore.
24
2 0.5
cr e F = (EIW / r)
r
25. 25
Consider:
= p - =
(2 1.688 )12*65.45
4
2.67 lb
W 3.07 1 8.6
ö çè= - æ -
E 30,000,000 psi (steel)
3.07 lb
0.2225 lb
ft
ft
65.45
ft
231
W
gal mud
2" OD, 1.688" ID; 8.6 #
e
2 2
=
= = ÷ø
in
*
2 0.5
Fcr = (EIWe / r)
26. F E IWe
cr = 2
3.317 lbf
26
= p - = p - =
= - = 3.875 - 2
-
r D HOLE
OD
Then, F 2 30*10 *0.3869*0.2225
0.9375
0.9375 in
2
2
(2 1.688 ) 0.3869 in
64
(OD ID )
64
I
6 0.5
cr
4 4 4 4 4
ö
= ÷ ÷ø
ç çè æ
=
Consider:
,
=
r
27. 27
Sinusoidal Buckling Load
A more general Sinusoidal Buckling Load
equation for highly inclined wellbores (including
the horizontal wellbore) is:
F EIWe sin
cr
= 2 q
r q
28. 28
Sinusoidal Buckling Load
For the same 2” OD coiled tubing, at q = 45o
F EIWe sin
cr
= 2 q
r
6 0 5
F * * . * . sin ÷ ÷ø
2 30 10 0 3869 0 2225 45
0 9375
o .
cr .
ö
ç çè æ
=
Fcr = 2,789 lbf
30. Helical Buckling in a Horizontal Wellbore
When the axial compressive load reaches the
following helical buckling load Fhel in the horizontal
wellbore, the helical buckling of coiled tubing then
occurs: ( ) r
30
F E IWe
hel = 2 2 2 -1
( ) 0 9375
6
F = -
* * . * . hel 2 2 2 1 30 10 0 3869 0 2225
.
Fhel = 6,065 lbf
31. 31
General Equation
A more general helical buckling load
equation for highly inclined wellbores
(including the horizontal wellbore) is:
( ) r
F EIWe sin
hel
= 2 2 2 -1 q
33. 33
Buckling in Vertical Wellbores:
In a vertical wellbore, the buckling of coiled tubing
will occur if the coiled tubing becomes axially
compressed and the axial compressive load
exceeds the buckling load in the vertical section.
This could happen when we “slack-off” weight at
the surface to apply bit weight for drilling and
pushing the coiled tubing through the build section
and into the horizontal section.
34. 34
Buckling in Vertical Wellbores:
Lubinski derived in the 1950’s the following
buckling load equation for the initial buckling
of tubulars in vertical wellbores:
=
F 1.94(EIW )
=
F 1.94(30*10 *0.3869*0.2225 )
F 161 lbf
cr,b
6 2 1/3
cr,b
2 1/3
cr,b e
=
35. 35
Buckling in Vertical Wellbores:
Another intitial buckling load equation for
tubulars in vertical wellbores was also
derived recently through an energy analysis:
=
F 2.55(EIW )
F 2.55(30*10 *0.3869*0.2225 )
Alternate F 212 lbf (Table 1)
cr,b
6 2 1/3
cr,b
2 1/3
cr,b e
=
=
36. 36
Helical Buckling in Vertical Wellbores:
A helical buckling load for weighty tubulars in
vertical wellbores was also derived recently through
an energy analysis to predict the occurrence of the
helical buckling:
F 5.55(EIW 2 )1/3
hel,b e
461 lbf
=
=
37. Helical Buckling in Vertical Wellbores:
37
This helical buckling load predicts the first
occurrence of helical buckling of the weighty
tubulars in the vertical wellbore. The first
occurrence of helical buckling in the vertical
wellbore will be a one-pitch helical buckle at
the bottom portion of the tubular.
38. Helical Buckling in Vertical Wellbores:
The upper portion of the tubular in the vertical
wellbore will be in tension and remain straight.
When more tubular weight is slacked-off at the
surface, and the helical buckling becomes more
than one helical pitch, the above helical buckling
load equation may be used for the top helical
pitch of the helically buckled tubular
38
39. 39
Helical Buckling in Vertical Wellbores:
The top helical buckling load Fhel,t is calculated by
simply subtracting the tubular weight of the initial
one-pitch of helically buckled pipe from the
helical buckling load Fhel,b, which is defined at the
bottom of the one-pitch helically buckled tubular:
= -
F 5.55(EIW ) W L
2 1/3
e
e hel
2 1/3
hel,t e
0.14(EIW )
=
40. Helical Buckling in Vertical Wellbores:
40
Where the length of the initial one-pitch of
helical buckling or the first order helical
buckling is:
L (16 EI /W )1/3 (10)
e
2
hel = p
41. Helical Buckling in Vertical Wellbores:
From Table 1, it is also amazing to find out that
the top helical buckling load, Fhel,t, is very close to
zero. This indicates that the “neutral point”, which
is defined as the place of zero axial load (effective
axial load exclusive from the hydrostatic pressure
force), could be approximately used to define the
top of the helical buckling for these coiled tubings.
41
42. Helical Buckling in Vertical Wellbores:
42
2 1/3
=
F 0.14(EIW )
0.14(30*10 *0.3869*0.2225 )
=
F 12 lbf
hel,t
6 2 1/ 2
hel,t e
=
43. Sinusoidal: cr = 2 = 3,317 lbf
43
Buckling of 2” x 1.688” CT
Horizontal
F E IWe
r
hel ( ) cr F = 2 2 -1 F
Helical: = 6,065 lbf
44. 44
Buckling of 2” x 1.688” CT
Vertical
Sinusoidal, bottom:
or
F = 1 . 94 (E IW 2 ) 1 /
3 =
cr,b e 161 lbf F = 2 . 55 (E IW 2 ) 1 /
3 =
cr,b e 212 lbf
45. 45
Buckling of 2” x 1.688” CT
Vertical
Helical, bottom:
F . (E IW ) lbf /
Helical, top:
hel,b 5 55 e 461 2 1 3 = =
F = 0 . 14 (E IW 2 ) 1 /
3 =
hel,b e 12 lbf
