Downhole Hydraulic I

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(2)

S chlu mb er g er P riv ate

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)

(3)

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

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S chlu mb er g er P riv ate 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)

(5)

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

(6)

ANSWER

P

_ann

=

4,680 psi

P

_tbg

=

9,136 psi

P

_diff

=

4,456 psi

(7)

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.

(8)

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 =

D

P

x Area

F

DP_hyd

(9)

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

(10)

S chlu mb er g er P riv ate Given

Bull Plugged Pipe 51/₂” Casing 17 lb/ft Calculate the Hook Load

5,000 ft

10 ppg BRINE

(11)

Solution

A = [

p

x (5.5)

2

_{] / 4}

_{A = 23.76 in}

2

P. hyd

= 5,000 ft x 10 ppg x 0.052 = 2,600 psi

Buoy. Force = 2,600 psi x 23.76 in

2

_{= 61,776#}



Weight in Air = 5,000 ft x 17#/ft

= 85,000#



Hook Load

= 85,000 # - 61,776 #

= 23,224#



(12)

ANSWER

B. Force =

61,776 #



Weigh in air =

85,000 #



Hook Load =

23,224 #



(13)

S chlu mb er g er P riv ate 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

(14)

Solution

Outer Area = ( p x 302 _{) / 4 = 706.85 in}2 Inner Area = ( p x 28.272 _{) / 4 = 627.68 in}2

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 # 

(15)

ANSWER

Hook Load =

95,667 #



(16)

S chlu mb er g er P riv ate 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 ppg MU D

(17)

Solution

Area =

p

/ 4 ( OD

2

_{- ID}

2

₎

Area = 0.7854 x [(5.5in)

2

_{– (4.89in)}

2

_]

_{= 4.962 in}

2

P hyd = 5,000 ft x 10 ppg x 0.052

= 2,600 Psi

Buoy. Force = 2,600 psi x 4.962in

2

_{= 12,900 #}



Weight in Air = 5,000 ft x 17 lb/ft

= 85,000 #



HOOK LOAD = 85,000 lb - 12,900 lb

= 72,100 #



(18)

ANSWER

Hook Load =

72,100 #



(19)

S chlu mb er g er P riv ate -->

Buoyancy Factor = 1 - ( Mud Weight / 231 x density of pipe )

with steel density = 0.2833 lb/in

3

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.

(20)

S chlu mb er g er P riv ate --> 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#

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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)

(22)

S chlu mb er g er P riv ate q T W R

W

=

Bouyant weight of the string

R

=

Reaction against wellbore

T =

Tension in the string = HL

(23)

S chlu mb er g er P riv ate 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

True Hook Load in Deviated Well

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S chlu mb er g er P riv ate Neutral Point:

It is the the point in a string which is not under tension nor under compression.

Hook Load

(26)

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)

(27)

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

(28)

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 10,000lb

Neutral Point

(29)

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

(30)

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.

(31)

S chlu mb er g er P riv ate 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 ppg MUD 15,000 lb

(32)

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

(33)

Answer

NP @

9,092 ft from surface

NP @

908 ft from Packer

NP depth = 9,092 ft

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S chlu mb er g er P riv ate 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 ppg MU D 15,000 lb 6” DC 79.4 lb/ft

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S chlu mb er g er P riv ate 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 ppg MUD 15,000 lb 6” DC 79.4 lb/ft

(36)

Example 3

Due 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” DP 19.5 lb/ft 10 ppg MUD 6” DC 79.4 lb/ft 600 ft lenght DLT + SV At 1000 ft

(37)

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

(38)

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

(39)

Solution

Q5

(40)

Factors that can affect tubing length:



Temperature



Stress



Ballooning / Reverse Ballooning

Changes in Tubing Length ΔL

(41)

Temperature will change due to :

Production

Injection

If Temperature

Increases

=>

Pipe Expands

Decreases

=>

Pipe Contracts

Changes in Temperature

(42)

Temperature Effect:

D

L = Lo x ß x

D

T

where:

–

Lo = original length of pipe

–

ß = temperature elongation factor (6.9 x 10

-6

/°F)

–

D

T = change in average temperature

If both end of the tubing are fixed a force F will be

generated

F = 207 x A x

D

T

(43)

S chlu mb er g er P riv ate 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 = D Temp = Force applied = BRINE 70o.F T T 150o.F 15,000 lb 3.1/2” Tbg 12.8 lb/ft

Changes in Temperature - Example

(44)

A =

P/4 ( 3.5

2

_{- 2.764}

2

₎

_{A = 3.62 in}

2

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 3.62in

2

_{x (- 40) deg F}

(45)

S chlu mb er g er P riv ate 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.F T T 150o.F 15,000 lb 3.1/2” Tbg 12.8 lb/ft

Changes in Temperature - Example

(46)

The stretch caused by stress is calculated with the

Hooke's law:

F x L

S =

---E x A

Where: S = Stretch (= elongation) (ft.)

– F = Force pulling on tubing (lbf) – L = Original length of tubing (ft.) – E = Young’s Modulus (30 x 106 psi)

ΔL Due to Stress

(47)

Hook Load is Maxi at the top

of the string and zero at the

bottom

Hook Load

?

(48)

Hook Load is Maxi at the top

of the string and nil at the

bottom

Hook Load

(49)

Hook Load is Maxi at the top

of the string and nil at the

bottom

We can average the stress to

calculate the stretch

D

L.

Hook Load

(50)

Hook Load is Maxi at the top

of the string and nil at the

bottom

We can average the stress to

calculate the stretch

D

L

10,000 ft

Hook Load

Average Stress

ΔL Due to Stress

(51)

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

(52)

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 in

2

Stretch = ( 54,000 lb x 10,000 ft) / (30 x 10

6

_{psi x 3.62 in}

2

₎

(53)

S chlu mb er g er P riv ate 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 W_{in mud} =10.84 #/ft Hook Load =108,400 # Stretch DL =4.99 ft 10,000 ft Hook Load Average Stress

(54)

Internal Tubing Pressure will create

Ballooning

=>

Shorten the Tubing

External Tubing Pressure ( Annulus ) will create

Reverse Ballooning

=>

Elongates the Tubing

Ballooning / Reverse Ballooning

(55)

S chlu mb er g er P riv ate Depth

(56)

Ballooning

Pres sure

(57)

S chlu mb er g er P riv ate Depth ???? ft

Ballooning

Reverse Ballooning

Pres

sure

Pres

sure

???? ft

(58)

If tubing is free to expand or shorten we will have

to deal with

Ballooning Stretch:

D

P

_tb

- R

2

D

_P

an

D

L = 2L x 10

-8

_x

---R

2

_{- 1}

Where :



D

_P

_tb

_{= change in tubing pressure}



D

_P

_an

_{= change in annulus pressure}

–

R

= Ratio = tubing OD / tubing ID

(59)

If the tubing is not free to expand or shorten we will

have to deal with

Ballooning Force:

F = 0.6 [ (

D

P

_tb

x Ai ) - (

D

P

_an

x Ao ) ]

Where :

–

Ai

= Internal Section Area

–

Ao

= External Section Area

(60)

GIVEN

3.1/2” Tbg / 12.8 #/ft

Mud = 10 #/gal

Calculate the change in length

or force due to Ballooning

SOLUTION

10,000 ft ???? ft

3000psi

(61)

Solution

If the string is allowed to shorten :

ΔL = 2L x 10

-8

[ ( ΔP

tb

- R

2

ΔP

an

) / ( R

2

- 1 ) ]

R = 3.5 / 2.764 = 1.2663

R

2

- 1 = 0.6035

L = 10,000 ft

ΔP

tb

= 3,000 psi

ΔP

an

= 0

ΔL = 2 x 10,000 ft x 10

-8

[ ( 3,000 ) / 0.6035 ]

ΔL = 0.994 ft = 12 in ( shorter )

(62)

S chlu mb er g er P riv ate GIVEN 3.1/2” Tbg / 12.8 #/ft Mud = 10 #/gal

Calculate the change in length or force due to Ballooning

SOLUTION

If pipe Free

DL = 12 in shorter If pipe not Free

F = 10,800 # tension

3000psi

(63)

S chlu mb er g er P riv ate 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.

(64)

The free point can be calculated from Hooke's Law as:

E A S L = ---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)

Free Point Calculations

(65)

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.

(66)

S chlu mb er g er P riv ate 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

(67)

S chlu mb er g er P riv ate 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