Downhole Hydraulic I

Downhole Hydraulics

Agenda 1.Basics Hydrostatic, Applied Pressure, Differential Pressure 2.Buoyancy (Archimedes’law review) 3.Hook Load and Buoyancy Factor (300.037 of field DH) 4.Neutral Point (important when undoing a thread) 5.Changes in Tubing Length (TBG, DP, DC)

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– Open ended pipe – Plugged Pipe

– 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

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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Hydrostatic Pressure:  Pressure caused by a column of fluid  Phyd (psi) = Density (ppg) x Length (ft) x 0.052

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Differential Pressure Example 3,000 psi surface 9 ppg brine

6,000 ft

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Calculate the differential pressure acting on the tubing just above the packer (10,000 ft)

10,000 ft

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Solution

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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 Pdiff = 4,456 psi

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Ptbg = 9,136 psi

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Buoyancy

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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 bouyancy force is proportional to the weight of the fluid.

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Buoyancy

DPhyd

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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.

F

The force F = DPhyd x Area 8

Hook Load

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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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Hook Load Example 1 Given Bull Plugged Pipe

10 ppg BRINE Schlumberger Private

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

5,000 ft

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Solution A = [ p x (5.5)2 ] / 4

= 5,000 ft x 10 ppg x 0.052 = 2,600 psi x 23.76 in2 = 5,000 ft x 17#/ft = 85,000 # - 61,776 #

= 2,600 psi = 61,776#  = 85,000#  = 23,224# 

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P. hyd Buoy. Force Weight in Air Hook Load

A = 23.76 in2

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ANSWER B. Force

= 61,776 #

Weigh in air = 85,000 # Schlumberger Private

Hook Load = 23,224 #

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Hook Load Example 2 GIVEN

30” Csg / 196#/ft @ 1,000ft,

15.8 ppg CMT

Displace with 8.5ppg Sea.W. Calculate Hook Load at the end of cement job

Sea Water

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ID = 28.27”

950 ft 1,000 ft

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Solution Outer Area = ( p x 302 ) / 4 = 706.85 in2 Inner Area = ( p x 28.272 ) / 4 = 627.68 in2 = 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

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Internal Pressure

= 1,000 ft x 196 lb/ft = 196,000 # 

HOOK LOAD = (196,000 + 289363) - 581,030 = - 95,667 # 

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ANSWER

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Hook Load = 95,667 #

THE CASING WILL FLOAT !

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Hook Load Example 3 GIVEN

5 1/2” Csg 17 lb/ft

Calculate the Hook Load

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10 ppg MUD ; Open End

10 ppg MUD

5 1/2” Csg / 17#/ft @ 5,000ft

5,000 ft

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Solution

= 4.962 in2 = 2,600 Psi

Buoy. Force = 2,600 psi x 4.962in2 Weight in Air = 5,000 ft x 17 lb/ft HOOK LOAD = 85,000 lb - 12,900 lb

= 12,900 #  = 85,000 #  = 72,100 # 

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Area = p / 4 ( OD2 - ID2 ) Area = 0.7854 x [(5.5in)2 – (4.89in)2] P hyd = 5,000 ft x 10 ppg x 0.052

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ANSWER Hook Load = 72,100 #  Schlumberger Private

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Buoyancy Factor Buoyancy Factor = 1 - ( Mud Weight / 231 x density of pipe ) with steel density = 0.2833 lb/in3

BF = 1 - ( 0.01528 x Mud Weight ) -->

The buoyancy factor for different mud weights can be found in the handbook, page 300.037.

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Note 1:

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. 19

Buoyancy Factor on Example 1 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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-->

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True Hook Load in Deviated Well

This will generate Drag Forces (Friction)

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In deviated well we have to take into account the fact that the pipe is in contact with the wellbore

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True Hook Load in Deviated Well q

T

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R

W

W = Bouyant weight of the string R = Reaction against wellbore T = Tension in the string = HL

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True Hook Load in Deviated Well Static Condition Tension T = W cos q

Tension T = W cos q - Friction P.O.H Tension T = W cos q + Friction

R T W

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R.I.H

q

 W = Bouyant weight of the string

Only a pull test RIH can confirm the  R = Reaction against wellbore  T = Tension in the string true Friction drag force 23

True Hook Load in Deviated Well Hook load of a static string is equal to: Weight in air -- buoyancy -- weight supported by the hole

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

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Hook load of a dynamic string is equal to:

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Neutral Point Hook Load

Neutral Point: Schlumberger Private

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

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Neutral Point Neutral Point:

NEUTRAL POINT (off bottom because of bouyancy force)

Tension

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It is the point in a string which is nor under tension nor under compression.

Hook Load

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Neutral Point Neutral Point:

If we slack off 10,000lb to set the packer the neutral point will move up

Tension

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

Hook Load

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Neutral Point Neutral Point:

If we slack off 10,000lb to set the packer the neutral point will move up

Hook Load Tension

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

10,000lb

NEUTRAL POINT ??

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Neutral Point Neutral Point:

Hook Load

Tension

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

10,000lb

NEUTRAL POINT If we slack 10,000lb to set the packer the neutral point will move up

Compression

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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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

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Neutral Point - Example 1 GIVEN

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CALCULATE the position of the Neutral Point

10 ppg MUD

5” DP - 19.5#/ft, in 10ppg fluid PKR @ 10,000ft set with 15,000#

5” DP 19.5 lb/ft

15,000 lb 10,000 ft

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Solution = 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

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Buoyancy Factor

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

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NP @ 9,092 ft from surface NP @ 908 ft from Packer NP depth = 9,092 ft

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Neutral Point - Example 2 GIVEN

5” DP 19.5 lb/ft

PKR @ 3,500ft set with 15,000#

CALCULATE the position of the Neutral Point

6” DC 79.4 lb/ft

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500ft of 6” DC - 79.4 lb/ft

10 ppg MUD

3000ft of 5” DP - 19.5 lb/ft

15,000 lb

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Neutral Point - Example 2 SOLUTION = 16.52 lb/ft

DC effective weight

= 67.27 lb/ft

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

6” DC 79.4 lb/ft

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DP total weight = 49,560 lb

10 ppg MUD

DP effective weight

5” DP 19.5 lb/ft

15,000 lb

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Example 3 Due to emergency situation in off shore , the well has to be shut down temporarily. 9-5/8 in DLT Packer + 61/8 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.

10 ppg MUD

5” DP 19.5 lb/ft

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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?

DLT + SV At 1000 ft

6” DC 79.4 lb/ft 600 ft lenght

10,000 ft

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Solution

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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 Schlumberger Private

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 Schlumberger Private

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Changes in Tubing Length ΔL Factors that can affect tubing length:

 Stress

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 Temperature

 Ballooning / Reverse Ballooning

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Changes in Temperature

If Temperature Increases => Decreases =>

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Temperature will change due to : Production Injection

Pipe Expands Pipe Contracts

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Changes in Temperature Temperature Effect:

DL = Lo x ß x DT 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 –

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where:

where A = cross section area of pipe (in2). 42

Changes in Temperature - Example GIVEN 15,000 lb weight on Packer Pumping Fluid @ 70o.F

T

o

70 .F

CALCULATE Force left on Packer when string Temperature is down to 70o F SOLUTION Area = D Temp = Force applied =

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3.1/2” Tbg 12.8 lb/ft

15,000 lb

BRINE

T

o

150 .F

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Schlumberger Private

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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Changes in Temperature - Example GIVEN 15,000 lb weight on Packer Pumping Fluid @ 70o.F

T

o

70 .F

CALCULATE Force left on Packer when string Temperature is down to 70o F SOLUTION Area D Temp. Force applied

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3.1/2” Tbg 12.8 lb/ft

15,000 lb

= 3.62 in2 = 40o F = 29974 lbf - 15000 lbf = 14974 lbf  BRINE

T

o

150 .F

THE PACKER IS UNSET !! 45

ΔL Due to Stress The stretch caused by stress is calculated with the Hooke's law: Schlumberger Private

FxL S = ----------ExA 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)

–

A = Cross sectional area (in2)

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ΔL Due to Stress Hook Load is Maxi at the top of the string and zero at the bottom

Hook Load

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?

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ΔL Due to Stress

Hook Load

Hook Load is Maxi at the top of the string and nil at the bottom Schlumberger Private

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ΔL Due to Stress

Hook Load

Hook Load is Maxi at the top of the string and nil at the bottom Schlumberger Private

We can average the stress to calculate the stretch DL.

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ΔL Due to Stress

Hook Load

Hook Load is Maxi at the top of the string and nil at the bottom Schlumberger Private

We can average the stress to calculate the stretch DL Average Stress

10,000 ft

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DL Due to Stress - Example Hook Load

GIVEN 3.1/2” tbg / 12.8 #/ft

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Mud = 10 #/gal

Calculate the change in length caused by stress SOLUTION Average Stress

10,000 ft

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Buoyancy factor

= 0.8472

Hook Load

= 10.84 #/ft x 10,000 ft = 108,400 # 

Average Stress

Hook Load / 2= 54,000 #

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Pipe Weight in mud = 12.8 #/ft x 0.8472 = 10,84 #/ft

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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DL Due to Stress - Example Hook Load

GIVEN 3.1/2” Tbg / 12.8 #/ft Mud = 10 #/gal

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Calculate the change in length caused by stress

SOLUTION B.F. (from handbook) =0.8472 Average Stress

Pipe Win mud =10.84 #/ft

Hook Load

=108,400 #

Stretch DL

=4.99 ft

10,000 ft

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Ballooning / Reverse Ballooning Internal Tubing Pressure will create Schlumberger Private

Ballooning => Shorten the Tubing External Tubing Pressure ( Annulus ) will create

Reverse Ballooning => Elongates the Tubing 54

Ballooning / Reverse Ballooning

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Depth 55

Ballooning / Reverse Ballooning Ballooning

Pressure

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???? ft

Depth 56

Ballooning / Reverse Ballooning Reverse Ballooning

???? ft

Pressure

Depth

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Pressure

Ballooning

???? ft 57

Ballooning / Reverse Ballooning

 DPtb

= change in tubing pressure

 DPan

= change in annulus pressure

–

R

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If tubing is free to expand or shorten we will have to deal with Ballooning Stretch: DPtb - R2 DPan DL = 2L x 10-8 x ----------------------------R2 - 1 Where :

= Ratio = tubing OD / tubing ID

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Ballooning / Reverse Ballooning If the tubing is not free to expand or shorten we will have to deal with Ballooning Force:

– –

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F = 0.6 [ ( DPtb x Ai ) - ( DPan x Ao ) ] Where : Ai = Internal Section Area Ao = External Section Area

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Ballooning - Example GIVEN 3.1/2” Tbg / 12.8 #/ft Mud = 10 #/gal 3000psi

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Calculate the change in length or force due to Ballooning

SOLUTION

10,000 ft

???? ft

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Solution

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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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Ballooning - Example GIVEN 3.1/2” Tbg / 12.8 #/ft Mud = 10 #/gal 3000psi

SOLUTION

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Calculate the change in length or force due to Ballooning

If pipe Free DL = 12 in shorter If pipe not Free F = 10,800 # tension 10,000 ft

???? ft

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Free Point Definition:  Free point is the point in the string above which a stuck pipe is free (drilling incident) Schlumberger Private

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.

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Free Point Calculations

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The free point can be calculated from Hooke's Law as: EAS 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) L = 735 x 103 W S / DF

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Free Point - Example

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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.

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Free Point - Example SOLUTION 1. String weight = 13.3 #/ft x 10,000 ft = 133,000 lb

2. Yield strength of 3 1/2in, Grade “E” Drill Pipe > 240,000 lb

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(Should be less with buoyancy effect)

3. Free point: 

L = ( 735 x 103 x 13.3 x 4 x 12 ) / 60,000



Depth = 7820 ft

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Module Summary 1.

Review

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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

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