Grace Robert. Advanced Blowout and Well Control
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286
Advanced
Blowout
and
Well
Control
s
-
LK>o
psi
-0p
Figure
610
-
Snubbing
Drillpip
Through
the
Annular
Solution:
1.
When
the
annulat
is
closed
on
the tube, the force
associated
with the pressure
can
be determined
using
Equation
6.2:
Fwp
=
0.7854Dp2P,
Fwp
=
0.7854(4.52)(1500)
Fv
=
23,857
Ibf
Special Services in Well Control
287
s
-
1.500
psi
e
Figure
6
11
-
Snubbing the Tool
Joint
Through the Annular
2.
When the annular is closed on a tool
joint,
the force
is
calculated using the diameter of the connection:
Fwp
=
0.7854(
6.5’)(
1500)
In
addition
to
the pressure area force, the friction force must be
considered. Friction is that force which
is
tangent to
the
surface
of
contact between
two
bodies
and
resisting movement. Static friction is the
force that resists the initiation of movement. Kinetic friction is the force
resisting movement when one
body
is
in
motion relative to the other. The
force required to overcome static friction is always greater
than
that
required to maintain movement (kinetic friction). Since friction is a
288
resistance
to
motion,
it
acts
in the direction opposite the pipe movement.
Friction acts upward when snubbing or stripping into a well and
downward when snubbing or stripping out of
a
well. The magnitude of
the force required
to
overcome
fiiction
is
a
function
of
the roughness of
the surface
areas
in
contact,
total
surfixe
area,
the lubricant being used
and the closing force applied to the BOP.
Additional friction or
drag
may result between the snubbing string
and the wall of the hole.
In
general, the larger the dogleg severity,
inclination and tension (or compression) in the snubbing string, the greater
the friction due to
drag.
In
addition
to
the forces associated with pressure and friction, the
buoyant force
affects
the snubbing operation. Buoyancy is the force
exerted by a fluid (either gas or liquid)
on
a body wholly or partly
immersed and is
equal
to
the weight
of
the fluid displaced by the body.
Advanced
Blowout
and
Well
Control
M
D
uI
P
iM
_...
.....'
+
'm
Fb
Fb
-
P,
L(E
Dp2
1
-pi
Li(
$Di2)
4
~~~~~~
Figure
6.12
-
The
Buoyant
Force
Special
Services in Well Control
289
As
illustrated in Figure
6.12,
the buoyant force,
FB,
is given by
Equation 6.3:
F’
=
O.7854(p,,,Dp2L-piDi2Li)
(6.3)
Where:
Pm
=
Mud gradient
in
annulus, psi/ft
Pi
=
Fluid gradient inside pipe, psi/ft
Dp
=
Outside diameter
of
pipe, inches
Di
=
Inside diameter
of
pipe, inches
L
=
Length
of pipe below
BOP,
feet
Li
=
Length
of
column inside pipe, feet
If the pipe
is
being snubbed into the hole dry, the density
of
the air
is negligible and the
piDi2
Li
term is negligible.
If
the inside of the pipe is
full or partially full, the
piDi2
Li
term
cannot be ignored.
If
the annulus
is
partially filled with gas, the
pmDi
L
term must be broken into its
component parts.
If
the annulus contains muds
of
different densities, each
must
be considered. The determination
of
the buoyant force is illustrated
in Example
6.2,
and
Equation
6.3
becomes
Where:
Ll
=
Column length
of
fluid having
a
density gradient
pml
L,
=
Column length
of
fluid having
a
density gradient
p,,,,
&
=
Column length of fluid having
a
density gradient
pm3
L,
=
Column length
of
fluid having
a
density gradient p,
ExamuIe
6.2
Given:
Schematic
=
Figure 6.12
Mud gradient,
Pm
=
0.624pdft
Length
of
pipe,
L
=
2,OOOfeet
290
Advanced
Blowout
and
Well
Control
Tubular
=
4
%-inch
16.6
#/f€
drillpipe
Tubular
is
dry.
Required:
The buoyant force
Solution:
The buoyant force
is
given by Equation
6.3:
304
With
dry
pipe, Equation
6.3
reduces to
FB
=
0.7854pmD;
L
F,
=
0.7854( 0.624)( 4.
52)(
2000)
FB
=
19,849
Zbf
In
this
example, the buoyant force is calculated to be
19,849
Zbf
.
The buoyant force acts across the exposed cross-sectional
area
which is
the end
of
the drillpipe and reduces the effective weight
of
the pipe.
Without the well pressure force,
Fwp
,
and the fiiction force,
Ff
,
the
effective weight of the
2,000
feet
of drillpipe would be given by
Equation
6.4:
Special Services
in
Well
Control
291
Example
6.3
Given:
Example
6.2
Required:
Determine the effective weight of the
4
%-inch drillpipe.
Solution:
The effective weight,
yff
,
is given by Equation
6.4:
yf
=
16.62000)
-
19849
yf
=
13,351
lb.
As
illustrated
in
this
example, the weight of drillpipe is reduced
from
33,200
pounds to
13,351
pounds by the buoyant force.
The
maximum
snubbing or stripping force required occurs when
the string is
first
started,
provided the pressure remains constant. At
this
point, the weight of the string and the buoyant force are
minimal
and
may
generally be ignored. Therefore,
the
maximum
snubbing force,
F,,
,
can
be calculated
from
Equation
6.5:
Where:
F,,
=
Maximum
snubbing force,
Zbr
Fwp
=
Well pressure force,
Zbf
Ff
=
Frictional pressure force,
Zbf
As
additional pipe
is
run
in
the
hole,
the downward force
attributable to the buoyed weight of the string increases until it is equal to
the well pressure force,
Fv
.
This
is generally referred to
as
the balance
292
Advanced Blmt and Well Control
point and is the point at which the snubbing string will
no
longer be forced
out of the hole by well pressure. That is,
as
illustrated
in
Figure 6.13,
at
the balance point the well force,
F,,
is exactly
equal
to
the weight of the
tubular being snubbed
into
the hole. The length
of
empty pipe at the
balance point is given by
Equation
6.6:
En-
4p
=
Wp
-
0.0408pDp
Where:
hP
=
Length
at
balance point, feet
5-
=
Maximumsnubbingforce,
Ibr
Wp
=
Nominal
pipe weight,
p
=
Muddensity,ppg
Dp
=
Outside
diameter
of tubular, inches
After the pipe is filled, the net downward force is
a
positive
snubbing force
as
given by Equation 6.1.
In
a
nod snubbing situation, the work string is
run
to
a
point
just above the balance point without Wg the work string. While
snubbing, the well force must
be
sufficiently greater
than
the weight of the
pipe to cause the slips to grip the pipe firmly. It
is
intended that, after the
pipe
is filled, the weight of the pipe is sufficient
to
cause the slips
to
grip
the pipe firmly.
This
practice increases the string weight and reduces the
risk
of dropping the work string near the balance point.
The determination of the balance point is illustrated in
Example 6.4:
Eramule
6.4
Given:
4 %-inch 16.6
#/ft
drillpipe is to be snubbed
ram
to
ram
into a
well containing 12-ppg mud with
a
shut-in wellhead pressure
of
2500
psi. The friction contributable
to
the
BOP
ram
is
3,000
Zbf
. The internal diameter of the drillpipe is 3.826 inches.
Required:
1.
The
maximum
snubbing force required
Special Services in
Well
Control
293
2.
3.
Length
of
empty
pipe
to
reach
the
balance
point
The
net
downward
force
after
the pipe
is
filled
at the
balance
point
'.....
.-
4
Fw
MUD
FW
-
Ff
+
FWP
+
Fb
-
WpL
u
Figure
113
-
Balance
point
Solution:
1.
The
maximum
snubbing
force
is
given
by Equation
6.5:
Em
=
Fwp
+
Ff
Combining
Equations
6.5
and
6.2:
294
Advanced
Blowout
and
Well
Control
6-
=
0.7854Dp2P,
+
Ff
F,,
=
0.7854(4S2)(2500)
+3000
F,,
=42,761
Ibf
2.
The
length
of
empty pipe at the balance point is given by
Equation
6.6:
42,76
1
"
=
16.60- 0.0408(12)(4.5*)
hP
=
6,396
feet
3.
The net
force after
the
pipe is
filled
is given by
Equation
6.1
:
F,
=
W~L
-
(
F/
+
F~
+
F,)
Since
cnm
=
Ff
+F,,
En
+
Fv
=
85,958
Ibf
The buoyant
force,
FB
,
is given by Equation
6.3:
Special
Services
in
Well
Control
295
FB
=
0.7S54(pmDiL
-pmDi2Lj)
FB
=
0.7854[( 0.624)( 4.52)(6396)
-
(0.624)( 3.S262)( 6396)]
FB
=
17,591
lbs
Therefore,
F,
=
6,396(16.6) -42,761- 17,591
en
=
45,822
lbs
EQUIPMENT SPECIFICATIONS
In
hydraulic snubbing operations, the hoisting power required is
produced by pressure applied
to
a multi-cylinder hydraulic jack. The
jack cylinder
is
represented
in
Figure
6.14.
Pressure is applied to different
sides
of
the jack cylinder depending on whether snubbing or stripping is
being done.
During
snubbing, the jack cylinders are pressurized on
the
piston rod side and on the opposite side for lifting or stripping.
Once the effective
area
of
the jack
is
known,
the force required
to
lift or snub
a
work string can
be
calculated using Equations
6.7
and
6.8:
l&
=
0.7854PwNcDpd
2
Where:
<nub
=
Snubbing force,
Zbf