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Downhole Hydraulics
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Page 1: Downhole Hydraulic I

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

D

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

D

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?

36

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