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Downhole Hydraulics
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Agenda
1.Basics Hydrostatic, Applied Pressure, Differential Pressure
2.Buoyancy (Archimedes’law review)
3.Hook Load and Buoyancy Factor (300.037 of field DH)
– Open ended pipe
– Plugged Pipe
4.Neutral Point (important when undoing a thread)
5.Changes in Tubing Length (TBG, DP, DC)
– Due to Temperature
– Due to Stress (own weight)
– Due to Ballooning/Reverse Ballooning (= added Tbg pressure or annulus pressure)
6. Free Point
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Basics
Pressure = Force / Area Force = Pressure x Area
Hydrostatic Pressure:
Pressure caused by a column of fluid
Phyd (psi) = Density (ppg) x Length (ft) x 0.052
Applied Pressure :
Usually associated with a pump, or pressure from the formation.
Differential Pressure:
The difference between pressures acting on different sides of a body (a pipe, a piston, etc...
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6,000 ft
10,000 ft
9 ppg brine
3,000 psi surface
Differential Pressure Example
Calculate the differential pressure
acting on the tubing just above the
packer (10,000 ft)
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Solution
P annulus = 9 ppg x 10,000 ft x 0.052 = 4,680 psi
P tubing = 3000 + [( 9 ppg x 6,000 ft ) + ( 16 ppg x 4,000 ft )] x 0.052 = 9,136 psi
P differential = P tbg - P ann = 9,136 - 4,680 = 4,456 psi
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ANSWER
Pann = 4,680 psi
Ptbg = 9,136 psi
Pdiff = 4,456 psi
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Buoyancy
Any body immersed in a fluid will receive an upward force
called buoyant force F
The buoyant force F is equal to the weight of the volume of the
fluid displaced by that body.
The bouyancy force is proportional to the weight of the fluid.
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Any body immersed in a fluid will receive an upward force
called buoyant force F.
The buoyant force F is equal to the weight of the volume of
the fluid displaced by that body.
The force F = DPhyd x Area
F
DPhyd
Buoyancy
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Hook Load
This is the actual weight supported by the hook when a string is
in the well
It combines the weight of the pipe with buoyancy due to fluid
hydrostatic pressure
Also called : effective weight
HOOK LOAD = Weight in Air - Buoyancy Force
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Given
Bull Plugged Pipe
51/2” Casing 17 lb/ft
Calculate the Hook Load
5,000 ft
10 ppg BRINE
Hook Load Example 1
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Solution
A = [ p x (5.5)2 ] / 4 A = 23.76 in2
P. hyd = 5,000 ft x 10 ppg x 0.052 = 2,600 psi
Buoy. Force = 2,600 psi x 23.76 in2 = 61,776#
Weight in Air = 5,000 ft x 17#/ft = 85,000#
Hook Load = 85,000 # - 61,776 # = 23,224#
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ANSWER
B. Force = 61,776 #
Weigh in air = 85,000 #
Hook Load = 23,224 #
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GIVEN
30” Csg / 196#/ft @ 1,000ft,
ID = 28.27”
Displace with 8.5ppg Sea.W. Calculate Hook Load at the end of cement job
1,000 ft
15.8 ppg CMT
950 ft
Sea Water
Hook Load Example 2
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Solution
Outer Area = ( p x 302 ) / 4 = 706.85 in2
Inner Area = ( p x 28.272 ) / 4 = 627.68 in2
Internal Pressure = 0.052 [ ( 950 ft x 8.5 ppg) + ( 50 ft x 15.8 ppg) ] = 461 Psi
External Pressure = 0.052 x 1,000 ft x 15.8 ppg = 822 Psi
Hyd Force (inside) = 461 psi x 627.68 in2 = 289,363 #
Hyd Force (outside) = 822 psi x 706.85 in2 = 581,030 #
Weight in air = 1,000 ft x 196 lb/ft = 196,000 #
HOOK LOAD = (196,000 + 289363) - 581,030 = - 95,667 #
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ANSWER
Hook Load = 95,667 #
THE CASING WILL FLOAT !
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GIVEN
5 1/2” Csg / 17#/ft @ 5,000ft
10 ppg MUD ; Open End
Calculate the Hook Load
5 1/2” Csg
17 lb/ft
5,000 ft
10 p
pg
MU
D
Hook Load Example 3
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Solution
Area = p / 4 ( OD2 - ID2 )
Area = 0.7854 x [(5.5in)2 – (4.89in)2] = 4.962 in2
P hyd = 5,000 ft x 10 ppg x 0.052 = 2,600 Psi
Buoy. Force = 2,600 psi x 4.962in2 = 12,900 #
Weight in Air = 5,000 ft x 17 lb/ft = 85,000 #
HOOK LOAD = 85,000 lb - 12,900 lb = 72,100 #
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ANSWER
Hook Load = 72,100 #
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-->
Buoyancy Factor = 1 - ( Mud Weight / 231 x density of pipe )
with steel density = 0.2833 lb/in3
BF = 1 - ( 0.01528 x Mud Weight )
Note 1:
The buoyancy factor for different mud weights can be found in
the handbook, page 300.037.
Note 2:
The buoyancy factor can only be applied when using the same
fluid inside and outside the pipe, so there is no differential
pressure between annulus and tubing.
Buoyancy Factor
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-->
Given
5000.ft of 17 #/ft Casing
10.ppg Mud
Calculate the Hook Load
Solution
B.F. =1- (0.01528x10) = 0.8472
Eff Weight = 17#/ft x 0.8472 = 14.4 #/ft.
Hook Load = 5000' x 14.4#/ft= 72,000#
Buoyancy Factor on Example 1
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In deviated well we have to take into account the fact that the pipe is in contact with the wellbore
This will generate Drag Forces (Friction)
True Hook Load in Deviated Well
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q
T
W
R
W = Bouyant weight of the string
R = Reaction against wellbore
T = Tension in the string = HL
True Hook Load in Deviated Well
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Static Condition
Tension T = W cos q
R.I.H
Tension T = W cos q - Friction
P.O.H
Tension T = W cos q + Friction
Only a pull test RIH can confirm the
true Friction drag force
q
T
W
R
W = Bouyant weight of the string
R = Reaction against wellbore
T = Tension in the string
True Hook Load in Deviated Well
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Hook load of a static string is equal to:
Weight in air -- buoyancy -- weight supported by the hole
Hook load of a dynamic string is equal to:
Static hook load + drag forces ( + while POH / - while RIH )
Drag = Total of normal forces x Friction Coefficient
Drag will change when buckling/helical buckling occurs in the
well
Confirmation of the exact drag can be done only by doing
RIH/POH tests prior to the job
True Hook Load in Deviated Well
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Neutral Point:
It is the the point in a string which
is not under tension nor under
compression.
Hook LoadNeutral Point
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Neutral Point:
It is the point in a string which is
nor under tension nor under
compression.
Hook Load
Tension
NEUTRAL POINT
(off bottom because of
bouyancy force)
Neutral Point
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Neutral Point:
It is the UNIQUE point in a string
which is not under tension nor
under compression.
If we slack off 10,000lb to set the
packer the neutral point will move
up
Hook Load
Tension
Neutral Point
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Neutral Point:
Is the point in a string which is not
under tension nor under
compression
If we slack off 10,000lb to set the
packer the neutral point will move
up
NEUTRAL POINT ??
Hook Load
Tension
10,000lbNeutral Point
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Neutral Point:
Is the point in a string which is not
under tension nor under
compression
If we slack 10,000lb to set the
packer the neutral point will move
up
Hook Load
Tension
Compression
10,000lb
NEUTRAL POINT
Neutral Point
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Neutral Point Calculation
Calculate the effective weight of the pipe (lbf/ft effective
using the bouyancy factor table)
Divide the weight required on the packer by the effective
weight of the pipe (lbf/ft)
That result is : the length of pipe required to effectively have
the required weight on the packer.
TD - That length of pipe = Neutral point
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GIVEN
5” DP - 19.5#/ft, in 10ppg fluid
PKR @ 10,000ft set with 15,000#
CALCULATE the position of the Neutral Point
5” DP
19.5 lb/ft
10,000 ft
10 p
pg
MU
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15,000 lb
Neutral Point - Example 1
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Solution
Buoyancy Factor = 1 - ( 0.01528 x 10 ) = 0.8472
DP effective weight = 19.5 x 0.8472 = 16.52 lb/ft
DP total Weight in Fluid = 10,000’ x 16.52 #/ft = 165,200 lb
Hook Load = 165,200lb - 15,000lb (on Packer) = 150,200lb
Neutral Point Depth = 150,200ft / 16.52#/ft = 9,092 ft
We can also calculated the Neutral Point position from the Packer:
Neutral Point (from Packer) = 15,000 / 16.52 = 908 ft
Neutral Point Depth= 10,000ft - 908ft = 9,092 ft
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Answer
NP @ 9,092 ft from surfaceNP @ 908 ft from Packer
NP depth = 9,092 ft
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GIVEN
3000ft of 5” DP - 19.5 lb/ft
500ft of 6” DC - 79.4 lb/ft
PKR @ 3,500ft set with 15,000#
CALCULATE the position of the Neutral
Point
5” DP
19.5 lb/ft
10 p
pg
MU
D
15,000 lb
6” DC
79.4
lb/ft
Neutral Point - Example 2
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SOLUTION
DP effective weight = 16.52 lb/ft
DC effective weight = 67.27 lb/ft
DP total weight = 49,560 lb
DC total weight = 33,635 lb
Hook Load = 68,195 lb
As the Hook Load is > than DP weight, the neutral point is In the drill collars section
Neutral Point depth = 3,277 ft
5” DP
19.5 lb/ft
10 p
pg
MU
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15,000 lb
6” DC
79.4
lb/ft
Neutral Point - Example 2
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Example 3Due to emergency situation in off shore , the well has
to be shut down temporarily. 9-5/8 in DLT Packer + 6-
1/8 in Storm Valve planned to be set around 1000 ft
depth. At the same time client wants to have the bit
500 ft off bottom when the packer is set.
Questions :
1. What is the total hook load before you set the
Packer?
2. Is the 6-1/8 in Storm Valve able to perform this job?
Why?
3. What will be the Hook Load you need to have before
unscrewing the Storm Valve (after the packer set)?
4. What will be the total tensile load supported by the
DLT Packer?
5. Is the 9-5/8 in DLT packer able to support this load?
Why?
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5” DP
19.5 lb/ft
10 p
pg
MU
D
10,000 ft
6” DC
79.4 lb/ft
600 ft lenght
DLT + SV
At 1000 ft
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Solution
Q1
Bouyancy Factor = 1 – (0.01528 x 10 ppg) = 0.8472
Total DC length = 600 ft
Total DP length = 10,000 ft – 500 ft – 600 ft = 8,900 ft
Total DC eff. wt = 0.8472 x 79.4 lb/ft x 600 ft = 40,360.6 lbs
Total DP eff. wt = 0.8472 x 19.5 lb/ft x 8900 ft = 147,031.6 lbs
Total Hook Load = 40,360.6 lbs + 147,031.6 lbs = 187,392.2 lbs
Q2
Yes, because tensile load max of 6-1/8 in Storm Valve is 363 klbs
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Solution
Q3
Total DP length = 1,000 ft (from surface to SV depth)
Total DP eff. wt = 0.8472 x 19.5 lb/ft x 1,000 ft = 16,520.4 lbs
Q4
Total DC length = 600 ft
Total DP length = 9,500 ft – 600 ft – 1000 ft = 7,900 ft
Total DC eff. wt = 0.8472 x 79.4 lb/ft x 600 ft = 40,360.6 lbs
Total DP eff. wt = 0.8472 x 19.5 lb/ft x 7900 ft = 130,511.2 lbs
Total Hook Load = 40,360.6 lbs + 147,031.6 lbs = 171,141.8 lbs
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Solution
Q5
Yes, because hang off weight max of 9-5/8 in DLT Packer is 375 klbs
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Factors that can affect tubing length:
Temperature
Stress
Ballooning / Reverse Ballooning
Changes in Tubing Length ΔL
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Temperature will change due to :
Production
Injection
If Temperature
Increases => Pipe Expands
Decreases => Pipe Contracts
Changes in Temperature
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Temperature Effect:
DL = Lo x ß x DT
where:
– Lo = original length of pipe
– ß = temperature elongation factor (6.9 x 10-6 /°F)
– DT = change in average temperature
If both end of the tubing are fixed a force F will be generated
F = 207 x A x DT
where A = cross section area of pipe (in2).
Changes in Temperature
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GIVEN
15,000 lb weight on PackerPumping Fluid @ 70o.F
CALCULATE
Force left on Packer when string Temperature is down to 70o F
SOLUTION
Area =D Temp = Force applied =
BRINE
70o.FT
T 150o.F
15,000 lb
3.1/2” Tbg
12.8 lb/ft
Changes in Temperature - Example
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A = P/4 ( 3.52 - 2.7642 ) A = 3.62 in2
Temp. Average = ( 150 deg F + 70 deg F ) / 2 = 110 deg F
DT = 70 deg F – 110 deg F - 40 deg F
F = 207 x A x DT = 207 x 3.62in2 x (- 40) deg F
F = - 29,974 lb
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GIVEN
15,000 lb weight on Packer
Pumping Fluid @ 70o.F
CALCULATE
Force left on Packer when string Temperature
is down to 70o F
SOLUTION
Area = 3.62 in2
D Temp. = 40o F
Force applied = 29974 lbf - 15000 lbf =
14974 lbf
THE PACKER IS UNSET !!
BRINE
70o.FT
T 150o.F
15,000 lb
3.1/2” Tbg
12.8 lb/ft
Changes in Temperature - Example
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The stretch caused by stress is calculated with the Hooke's law:
F x LS = -----------
E x AWhere: S = Stretch (= elongation) (ft.)
– F = Force pulling on tubing (lbf)
– L = Original length of tubing (ft.)
– E = Young’s Modulus (30 x 106 psi)
– A = Cross sectional area (in2)
ΔL Due to Stress
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Hook Load is Maxi at the top of the string and zero at the bottom
Hook Load
?
ΔL Due to Stress
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Hook Load is Maxi at the top of the string and nil at the bottom
Hook LoadΔL Due to Stress
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Hook Load is Maxi at the top of the string and nil at the bottom
We can average the stress to calculate the stretch DL.
Hook LoadΔL Due to Stress
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Hook Load is Maxi at the top of the string and nil at the bottom
We can average the stress to calculate the stretch DL
10,000 ft
Hook Load
Average
Stress
ΔL Due to Stress
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GIVEN
3.1/2” tbg / 12.8 #/ft
Mud = 10 #/gal
Calculate the change in length
caused by stress
SOLUTION
10,000 ft
Hook Load
Average
Stress
DL Due to Stress - Example
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Buoyancy factor = 0.8472
Pipe Weight in mud = 12.8 #/ft x 0.8472 = 10,84 #/ft
Hook Load = 10.84 #/ft x 10,000 ft = 108,400 #
Average Stress Hook Load / 2= 54,000 #
Cross Sectional Area already calculated = 3.62 in2
Stretch = ( 54,000 lb x 10,000 ft) / (30 x 106 psi x 3.62 in2 )
= 4.99 ft
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GIVEN
3.1/2” Tbg / 12.8 #/ft
Mud = 10 #/gal
Calculate the change in length caused
by stress
SOLUTION
B.F. (from handbook) =0.8472
Pipe Win mud =10.84 #/ft
Hook Load =108,400 #
Stretch DL =4.99 ft 10,000 ft
Hook Load
Average
Stress
DL Due to Stress - Example
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Internal Tubing Pressure will create
Ballooning
=>
Shorten the Tubing
External Tubing Pressure ( Annulus ) will create
Reverse Ballooning
=>
Elongates the Tubing
Ballooning / Reverse Ballooning
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Depth
Ballooning / Reverse Ballooning
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Ballooning
Depth
Pre
ssu
re
???? ft
Ballooning / Reverse Ballooning
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Depth ???? ft
Ballooning Reverse Ballooning
Pre
ssu
re
Pre
ssu
re
???? ft
Ballooning / Reverse Ballooning
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If tubing is free to expand or shorten we will have to deal with Ballooning Stretch:
DPtb - R2 DPanDL = 2L x 10-8 x -----------------------------
R2 - 1
Where :
DPtb = change in tubing pressure
DPan = change in annulus pressure
– R = Ratio = tubing OD / tubing ID
Ballooning / Reverse Ballooning
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If the tubing is not free to expand or shorten we will have to deal with Ballooning Force:
F = 0.6 [ ( DPtb x Ai ) - ( DPan x Ao ) ]
Where :
– Ai = Internal Section Area
– Ao = External Section Area
Ballooning / Reverse Ballooning
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GIVEN
3.1/2” Tbg / 12.8 #/ftMud = 10 #/gal
Calculate the change in length or force due to Ballooning
SOLUTION
10,000 ft ???? ft
3000psi
Ballooning - Example
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Solution
If the string is allowed to shorten :
ΔL = 2L x 10-8 [ ( ΔPtb - R2 ΔPan ) / ( R2 - 1 ) ]
R = 3.5 / 2.764 = 1.2663 R2 - 1 = 0.6035
L = 10,000 ft
ΔPtb = 3,000 psi
ΔPan = 0
ΔL = 2 x 10,000 ft x 10-8 [ ( 3,000 ) / 0.6035 ]
ΔL = 0.994 ft = 12 in ( shorter )
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GIVEN
3.1/2” Tbg / 12.8 #/ftMud = 10 #/gal
Calculate the change in length or force due to Ballooning
SOLUTION
If pipe FreeDL = 12 in shorter
If pipe not FreeF = 10,800 # tension
10,000 ft ???? ft
3000psi
Ballooning - Example
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Definition:
Free point is the point in the string above which a stuck pipe is free (drilling incident)
Determination:
Apply an upward force, F1, to ensure that all the string is in tension.
Mark a reference point on the pipe.
Apply more upward force, F2, ( below the yield strength of the pipe ).
Measure the stretch S in inches.
Calculate the Free Point from Hooke's Law.
Free Point
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The free point can be calculated from Hooke's Law as:
E A SL = -----------------
12 DF
Where:– S = Pipe Stretch ( in )– DF = F2 - F1 ( lb )– L = Free Point (ft)– E = Young's Modulus ( 30 x 106 psi )– A = Cross sectional area ( in2 )
For steel pipes of linear weight = W (lb/ft)
L = 735 x 103 W S / DF
Free Point Calculations
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10,000 ft of 3.1/2" Grade “E” D.P. ( 13.3 #/ft ) are stuck in a hole.
The driller obtained the following data, after pulling on the pipe:
F1 = 140,000 lb
F2 = 200,000 lb
S = 4 ft
QUESTIONS :
1. Check that F1 is above the string weight.
2. Check that F2 is less than the yield strength of the pipe.
3. Calculate the Free Point position.
Free Point - Example
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SOLUTION
1. String weight = 13.3 #/ft x 10,000 ft = 133,000 lb
(Should be less with buoyancy effect)
2. Yield strength of 3 1/2in, Grade “E” Drill Pipe > 240,000 lb
3. Free point:
L = ( 735 x 103 x 13.3 x 4 x 12 ) / 60,000
Depth = 7820 ft
Free Point - Example
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1. Review
- Hydrostatic
- Applied Pressure
- Differential Pressure
2. Buoyancy
3. Hook Load and Buoyancy Factor
- Open Ended Pipe
- Plugged Pipe
4. Neutral Point
5. Changes in Tubing Length
- Due to Temperature
- Due to Stress
- Due to Ballooning/Reverse Ballooning
6. Free Point
Module Summary
Recommended