DIFFERENTIAL STICKING FORCE
The differential sticking force is given by:
Differential sticking force (DSF) = (Hs - Pf) x effective contact area x friction factor (12.1)
whereHs = hydrostatic pressure of mud
Pf = formation pressure
In Equation (12.1), the most difficult terms to determine are the effective contact area and
the friction factor between the mud cake and the pipe steel. To a first approximation the
Differential sticking force (DSF) = (Hs - Pf) x effective contact area x friction factor (12.1)
whereHs = hydrostatic pressure of mud
Pf = formation pressure
In Equation (12.1), the most difficult terms to determine are the effective contact area and
the friction factor between the mud cake and the pipe steel. To a first approximation the
effective area may be calculated as the product of the height of the exposed permeable
formation times 20% of the perimeter of the drillpipe or drillcollars.
Another equation for estimating the contact area is given by
formation times 20% of the perimeter of the drillpipe or drillcollars.
Another equation for estimating the contact area is given by
completely valid as the contact area is affected by a number of variables including the
friction factor (time-dependent), the amount of bend in the drillpipe or collars, hole angle
and thickness of the filter cake.
The surface estimate of the thickness of the filter cake can be very different from that
occurring downhole.
Example : Differential Sticking Force
Determine the magnitude of the differential sticking force across a permeable zone of 30 ft
in thickness using the following data:
Differential pressure = 500 psi
Area of contact is 20% of effective drillpipe perimeter
Filter cake = 1/2 in (12.7mm); friction factor = 0.1.
Drillpipe OD: 5"
Solution
Perimeter of drillpipe = π x OD = π x 5 = 15.71 in
DSF = (Hs - Pf) x h x 20% x 15.71
= 500psi x (30ft x 12 in) x 20% x 15.71
= 565,560 lb
FREEING DIFFERENTIALLY STUCK PIPE .
There are basically two ways in which a differentially stuck pipe can be released:
• reduction of hydrostatic pressure
• spotting pipe release agents
• reduction of hydrostatic pressure
• spotting pipe release agents
REDUCTION OF HYDROSTATIC PRESSURE
The reduction of hydrostatic pressure is the obvious and most successful method of freeing a differentially stuck pipe. The lowering of the hydrostatic pressure reduces the side loading
forces on the pipe and therefore reduces the force required to free the pipe from the filter
cake. There are several methods by which this may be achieved. However prior to
implementing this action the following factors should be seriously considered:
1. Are there other pressured zones in the open hole section?
2. Will these exposed zones kick if the hydrostatic pressure is reduced?
3. The confidence level in the accuracy of pore pressure estimates made while drilling
and the pressure control equipment.
4. The effects of a reduction in hydrostatic pressure on the mechanical stability of all
exposed formations.
5. The volumes of base oil or water required to achieve the required reduction in
hydrostatic pressure. (This may well influence the method chosen).
All the above factors need to be carefully considered prior to reducing the hydrostatic
pressure as the potential for inducing a well control problem or formation instability are
considerably increased. The following methods for reducing hydrostatic pressure can be
used:
• circulation & reducing mud weight
• displacing the choke
• the ‘U’ tube method
forces on the pipe and therefore reduces the force required to free the pipe from the filter
cake. There are several methods by which this may be achieved. However prior to
implementing this action the following factors should be seriously considered:
1. Are there other pressured zones in the open hole section?
2. Will these exposed zones kick if the hydrostatic pressure is reduced?
3. The confidence level in the accuracy of pore pressure estimates made while drilling
and the pressure control equipment.
4. The effects of a reduction in hydrostatic pressure on the mechanical stability of all
exposed formations.
5. The volumes of base oil or water required to achieve the required reduction in
hydrostatic pressure. (This may well influence the method chosen).
All the above factors need to be carefully considered prior to reducing the hydrostatic
pressure as the potential for inducing a well control problem or formation instability are
considerably increased. The following methods for reducing hydrostatic pressure can be
used:
• circulation & reducing mud weight
• displacing the choke
• the ‘U’ tube method
CIRCULATION & REDUCING MUD WEIGHT
In this method, the drilling mud is circulated and its weight is gradually reduced. The
minimum mud weight required to balance the highest pore pressure in open hole should be
determined and the mud weight cut back in small stages. Close attention must be made to all
kick indicators whilst circulating down (reducing) the mud weight, frequent flow checks
should also be made. Whilst reducing the mud weight, tension should be held on the pipe.
minimum mud weight required to balance the highest pore pressure in open hole should be
determined and the mud weight cut back in small stages. Close attention must be made to all
kick indicators whilst circulating down (reducing) the mud weight, frequent flow checks
should also be made. Whilst reducing the mud weight, tension should be held on the pipe.
Disadvantages of this methods are:
• It is slow, and remember the force required to free pipe is time dependent.
• The volume increase required may overload the surface pit handling capability.
This may be a serious problem when OBM is used.
• The active volume will be increasing during the reduction in mud weight,
making kick detection difficult.
• It is slow, and remember the force required to free pipe is time dependent.
• The volume increase required may overload the surface pit handling capability.
This may be a serious problem when OBM is used.
• The active volume will be increasing during the reduction in mud weight,
making kick detection difficult.
DISPLACING THE CHOKE
This method is applicable to floating rigs where BOPS are placed on the seabed. The
hydrostatic pressure can be quickly and effectively reduced by displacing the choke line to
base oil or water. The well is shut in using the annular preventer and the displaced choke line
opened thereby reducing the overbalance.Note that the annular preventer isolates the
wellbore from the hydrostatic head of mud in the riser from rig floor to the annular preventer.
The advantage of this method is that if any influx is taken, the well can be immediately
killed by closing the choke and opening the annular. This action again exposes the well to the
active hydrostatic pressure from rig floor to TD. The disadvantage of this method is that the
amount of reduction in hydrostatic pressure is limited to the water depth. This may well
result in a limited reduction in shallow water, or in the case of deep water, an excessive
reduction in hydrostatic pressure
hydrostatic pressure can be quickly and effectively reduced by displacing the choke line to
base oil or water. The well is shut in using the annular preventer and the displaced choke line
opened thereby reducing the overbalance.Note that the annular preventer isolates the
wellbore from the hydrostatic head of mud in the riser from rig floor to the annular preventer.
The advantage of this method is that if any influx is taken, the well can be immediately
killed by closing the choke and opening the annular. This action again exposes the well to the
active hydrostatic pressure from rig floor to TD. The disadvantage of this method is that the
amount of reduction in hydrostatic pressure is limited to the water depth. This may well
result in a limited reduction in shallow water, or in the case of deep water, an excessive
reduction in hydrostatic pressure
THE ‘U’ TUBE METHOD
The U-tube method is used to reduce the hydrostatic pressure of mud to a level equal or
slightly higher than the formation pressure of the zone across which the pipe got
differentially stuck.Clearly, the objective is to free the differentially stuck pipe safely without
losing control of the well by inadvertently inducing underbalanced conditions. A pipe free
agent should be spotted across the permeable zone prior to adopting the ‘U’ tube method.
The mathematics required for the full method is laborious, however,
slightly higher than the formation pressure of the zone across which the pipe got
differentially stuck.Clearly, the objective is to free the differentially stuck pipe safely without
losing control of the well by inadvertently inducing underbalanced conditions. A pipe free
agent should be spotted across the permeable zone prior to adopting the ‘U’ tube method.
The mathematics required for the full method is laborious, however,
SPOTTING PIPE RELEASE AGENTS
The severity of differentially stuck pipe can be reduced by the spotting of pipe release
agents. Pipe release agents are basically a blend of surfactants and emulsifiers mixed with
base oil or diesel oil and water to form a stable emulsion. They function by penetrating the
filter cake, therefore making it easier to remove and at the same time, reduce the surface
tension between the pipe and the filter cake.
Due to the time dependency of the severity of differential sticking, the pipe release agent
should be spotted as soon as possible after differential sticking is diagnosed. Typically the
pill will be prepared whilst initially attempting to mechanically free the pipe; ie by pulling
and rotating.
agents. Pipe release agents are basically a blend of surfactants and emulsifiers mixed with
base oil or diesel oil and water to form a stable emulsion. They function by penetrating the
filter cake, therefore making it easier to remove and at the same time, reduce the surface
tension between the pipe and the filter cake.
Due to the time dependency of the severity of differential sticking, the pipe release agent
should be spotted as soon as possible after differential sticking is diagnosed. Typically the
pill will be prepared whilst initially attempting to mechanically free the pipe; ie by pulling
and rotating.
Example : Reduction of Hydrostatic Pressure
Calculate the volume of oil required to reduce the hydrostatic pressure in a well by 500 psi,
using the following data:
mud weight = 10 ppg
hole depth = 9,843 ft
drillpipe = OD/ID = 5 in/4.276 in
hole size = 12.25 in
specific gravity oil = 0.8 (6.7 ppg)
Solution
Initial hydrostatic pressure = 0.052 x10x 9843 = 5,118 psi
Required hydrostatic pressure = 5,118 - 500 = 4,618 psi
Thus,
New hydrostatic pressure = pressure due to (mud and oil) in drillpipe
4618 = 0.052x 10xY (mud) +0.052x (6.7) x (9843-Y) (oil)
Calculate the volume of oil required to reduce the hydrostatic pressure in a well by 500 psi,
using the following data:
mud weight = 10 ppg
hole depth = 9,843 ft
drillpipe = OD/ID = 5 in/4.276 in
hole size = 12.25 in
specific gravity oil = 0.8 (6.7 ppg)
Solution
Initial hydrostatic pressure = 0.052 x10x 9843 = 5,118 psi
Required hydrostatic pressure = 5,118 - 500 = 4,618 psi
Thus,
New hydrostatic pressure = pressure due to (mud and oil) in drillpipe
4618 = 0.052x 10xY (mud) +0.052x (6.7) x (9843-Y) (oil)
where Y = height of mud in drillpipe.
Therefore,Y = 6,927 ft
Hence,
height of oil = 9,843 - 6,927 = 2,916 ft
volume of oil = capacity of drillpipe x height
= 290.79 ft3
= 51.7 bbl
Note that when the required volume of diesel oil is pumped inside the drillpipe, the
hydrostatic pressure at the drillpipe shoe becomes 4,618 psi, while the hydrostatic pressure
in the annulus is still 5,118 psi. This difference in the pressure of the two limbs of the well
causes a back-pressure on the drillpipe which is the driving force for removing the diesel oil
from the drillpipe and reducing the level of mud in the annulus. It is only when the annulus
level decreases that the hydrostatic pressure against the formation is reduced and the stuck
pipe may be freed.
When the formation pressure is unknown, it is customary to reduce the hydrostatic pressure
of mud in small increments by the U-tube technique until the pipe is free.
A variation of the U-tube method is to pump water into both the annulus and the drillpipe to
reduce hydrostatic pressure to a value equal to or just greater than the formation pressure.
This method is best illustrated by an example.
Therefore,Y = 6,927 ft
Hence,
height of oil = 9,843 - 6,927 = 2,916 ft
volume of oil = capacity of drillpipe x height
= 290.79 ft3
= 51.7 bbl
Note that when the required volume of diesel oil is pumped inside the drillpipe, the
hydrostatic pressure at the drillpipe shoe becomes 4,618 psi, while the hydrostatic pressure
in the annulus is still 5,118 psi. This difference in the pressure of the two limbs of the well
causes a back-pressure on the drillpipe which is the driving force for removing the diesel oil
from the drillpipe and reducing the level of mud in the annulus. It is only when the annulus
level decreases that the hydrostatic pressure against the formation is reduced and the stuck
pipe may be freed.
When the formation pressure is unknown, it is customary to reduce the hydrostatic pressure
of mud in small increments by the U-tube technique until the pipe is free.
A variation of the U-tube method is to pump water into both the annulus and the drillpipe to
reduce hydrostatic pressure to a value equal to or just greater than the formation pressure.
This method is best illustrated by an example.
Example : : Simplified U-Tube Method
The following data refer to a differentially stuck pipe at 11,400 ft:
Formation pressure = 5,840 psi
Intermediate casing = 9.625 in, 40# at 10,600 ft
Drillpipe = OD /ID = 5/4.276 in
The following data refer to a differentially stuck pipe at 11,400 ft:
Formation pressure = 5,840 psi
Intermediate casing = 9.625 in, 40# at 10,600 ft
Drillpipe = OD /ID = 5/4.276 in
Mud density = 12.3 ppg
It is required to reduce the hydrostatic pressures in the drillpipe and the annulus so that both
are equal to the formation pressure.
Calculate the volumes of water required in both the annulus and the drillpipe, assuming that
the density of saltwater = 8.65 ppg.
Solution
Annulus Side
Assume the height of water in the annulus to be Y.
Required hydrostatic pressure at stuck point = 5,840 psi or
5,840 = 0.0.52x 8.65x Y + 0.052x 12.3x (11,400-Y)
Y = 7,647 ft (length of water column)
Required volume of water in annulus
= annular capacity between drillpipe and 9.625" casing x height of water
= 0.0515 (bbl/ft) x 7,6476
= 393.8 bbl
Hence, pump 393.8 bbl of water into the annulus to reduce the hydrostatic pressure in the
annulus to 5,840 psi at the stuck point. When 393.8 bbl of water is pumped into the annulus,
the drillpipe is still filled with the original mud of 12.3 ppg having a hydrostatic pressure at
the stuck point of (0.052x12.3 x 11,400) = 7,291 psi. Thus, a back-pressure equivalent to
7,291 – 5840= 1,451 psi will be acting on the annulus and will be attempting to equalise
pressures by back-flowing water from the annulus.
It is required to reduce the hydrostatic pressures in the drillpipe and the annulus so that both
are equal to the formation pressure.
Calculate the volumes of water required in both the annulus and the drillpipe, assuming that
the density of saltwater = 8.65 ppg.
Solution
Annulus Side
Assume the height of water in the annulus to be Y.
Required hydrostatic pressure at stuck point = 5,840 psi or
5,840 = 0.0.52x 8.65x Y + 0.052x 12.3x (11,400-Y)
Y = 7,647 ft (length of water column)
Required volume of water in annulus
= annular capacity between drillpipe and 9.625" casing x height of water
= 0.0515 (bbl/ft) x 7,6476
= 393.8 bbl
Hence, pump 393.8 bbl of water into the annulus to reduce the hydrostatic pressure in the
annulus to 5,840 psi at the stuck point. When 393.8 bbl of water is pumped into the annulus,
the drillpipe is still filled with the original mud of 12.3 ppg having a hydrostatic pressure at
the stuck point of (0.052x12.3 x 11,400) = 7,291 psi. Thus, a back-pressure equivalent to
7,291 – 5840= 1,451 psi will be acting on the annulus and will be attempting to equalise
pressures by back-flowing water from the annulus.
In order to contain the 393.8 bbl of water in the annulus, the drillpipe must also contain a
column of water equal in height to that in the annulus.
Thus,
volume of water required in drillpipe to prevent back-flow from annulus
= capacity of drillpipe x height of water = 0.0178 (bbl/ft) x 7,647 ft = 136 bbl
Balancing of the columns of water in the drillpipe and in the annulus can be achieved as
follows: (a) circulate 393.8 bbl of water down the annulus; (b) circulate 136 bbl of water
down the annulus; (c) circulate 136 bbl of water in the drillpipe to remove 136 bbl of water
from the annulus and to reduce the hydrostatic pressure in the drillpipe to 5,840 psi. At this
stage the hydrostatic pressure in the well is equal to the formation pressure of 5,840 psi.
If the well should kick during the operation, reverse-circulate down the annulus using the
12.3 ppg (i.e. original density) mud to recover all the water from the drillpipe. Then circulate
in the normal way through the drillpipe using 12.3 ppg mud until all the water is removed
from the annulus.
column of water equal in height to that in the annulus.
Thus,
volume of water required in drillpipe to prevent back-flow from annulus
= capacity of drillpipe x height of water = 0.0178 (bbl/ft) x 7,647 ft = 136 bbl
Balancing of the columns of water in the drillpipe and in the annulus can be achieved as
follows: (a) circulate 393.8 bbl of water down the annulus; (b) circulate 136 bbl of water
down the annulus; (c) circulate 136 bbl of water in the drillpipe to remove 136 bbl of water
from the annulus and to reduce the hydrostatic pressure in the drillpipe to 5,840 psi. At this
stage the hydrostatic pressure in the well is equal to the formation pressure of 5,840 psi.
If the well should kick during the operation, reverse-circulate down the annulus using the
12.3 ppg (i.e. original density) mud to recover all the water from the drillpipe. Then circulate
in the normal way through the drillpipe using 12.3 ppg mud until all the water is removed
from the annulus.