252M302 Downhole Hydraulics Level I

7/16/2019 252M302 Downhole Hydraulics Level I

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Downhole Hydraulics

Level I


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2 Downhole Hydraulics Level I

Learning Objectives

After completion of this module you will be able to:

Calculate the

Tubular Volume

Hydrostatic Pressure and Pressure Differential

Tubular and Annular Flow Velocity

Buoyancy and Snubbing Forces

Hook Load

Forces on a packer


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Volume

Volume = Cross Sectional Area

Height

Where D is the internal/external

diameter in inches (for theconversion factor to apply)

ID

OD

1029

22

2

inD

ft

bblsVolume

factorconversionftHinAbblsVolume


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Why Use 1029

ft

bblsD

ft

bbls

L

V

ft

bbls

D70830.00097138ft

bbls

L

V

ft

bblsD

ft

bbls

L

V

ft

bblsD

ft

bbls

L

V

ft

bbls

in

ftinD

ft

bbls

L

V

LDV

1029

4

144

1781.0

4

1781.0

144

1

4

1781.0

1444

4

2

2

2

2

32

2

22

2

The

Diametermust be in

inches for

the

conversion

factor to

apply


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Volume Calculations (1)

(I) Calculate the volume of a 4.5in tubing, 3.959in

ID down to 9,000ft.

bblsftftbblsftVolume

ft

bblsinVolume

IDVolumeInternal

137000,9015232.0000,9@

015232.01029

959.3

10292

2

Where ID is the

internal diameter of

the tubular


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(II) Calculate the annular volume between a 4.5in

tubing 3.959in ID and a 1.75in CT string down to

TD @ 9,000ft.

bblsftft

bblsftVolume

ft

bblsininVolume

ODIDVolumeInternal

110000,9012256.0000,9@

012256.01029

75.1959.3

102922

22

Where ID is the internal

diameter of the tubular and ODis the external diameter of the

CT string


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(III) Calculate the volume displaced by a 1.75in

CT wall thickness string down to 9,000ft.

bblftft

bblsftVolume

ft

bblsinVolume

IDODVolumeInternal

6.9000,9001071.0000,9@

001071.01029

40.175.1

102922

22

Where OD is the external

diameter of the CT and ID isthe internal diameter of the CT

string


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Pressure Calculations


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Pressure

Pressure is defined as the force per unit area

exerted on a surface

Pressure exerts in all directions

When solving oilfield problems, there is two types

of pressure to consider: applied and hydrostatic

pressure


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Applied Pressure

Applied pressure is due to a pump or similar

means.

Applied pressure is felt throughout the system

equally.


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Hydrostatic Pressure

Hydrostatic pressure is fluid pressure due to the weight of

fluid above it. Both gases and liquids exert hydrostatic

pressure.

Hydrostatic pressure is present at all points below the

surface of a fluid, but unlike applied pressure it is not

constant.

The hydrostatic pressure at any point depends on thefluid density and the depth (True Vertical Depth TVD)

below the fluid surface.


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Phydrostatic = Fluid Weight True Vertical Depth

The Fluid Weight can be expressed in terms of the

Density (mass per unit volume)

Specific Gravity (a comparison to the density of water)

Hydrostatic pressure is usually defined by fluid gradient

gal

lbsWeightFluidft

psiGradientFluid 052.0


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Flow Velocity


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Flow Velocity

For Coiled Tubing Operations, it is important to

calculate the velocity of the fluid traveling in the

annular space between the CT string and the

tubular. Why?

ftbblsVolume

min

bblsRatePumpFluid

ftVelocityFlow

min


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2 OD.204 WT

10,350 ft

3.5 OD

9.3 lb/ft

EOT 8,875 ft

5.50 OD

15.5# ft

EOC 9,948 ft

1. What is the capacity of:

CT ReelAnnulus CT/Tbg

Annulus CT/Csg

2. What is the flow velocity in the annulus ?

CT/Tbg CT/Csg

1/2 bpm

1.0 bpm

3.0 bpm

3. How long does it take to circulate BU ?

Annulus CT

1/2 bpm

1.0 bpm

3.0 bpm

4. Whats the expected top of cement from a 9 bbl slurry

(from bottom up) ? CT out of slurry.

5. How many lbs of sand can there be along

the casing from the bottom up to the EOT ?

25 bbl @ .0.00246 bbls/ft43 bbl @ .004813 bbls/ft

21 bbls @ .01992 bbls/ft

104 25.1

208 50.2

624 150.6

128 50.9

64.1 25.5

21.4 8.5

378 ft. @ 42.0168 ft/bbl. (0.0238 bbls/ft) , = 9570ft

14.3 lbs/ft x 1,073 = 15,344 lbs

CT Calculations (1)


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1.75 OD0.156 WT

10,350 ft

3.50 OD

9.3 lb/ft

EOT 8,875 ft

7.00 OD

23.0# ft

EOC 9,948 ft

1. What is the capacity of:

CT ReelAnnulus CT/Tbg

Annulus CT Csg

2. What is the flow velocity in the annulus ?

CT/Tbg CT/Csg

1/2 bpm

1.0 bpm

3.0 bpm

3. How long to circulate BU ?

Annulus CT

1/2 bpm

1.0 bpm

3.0 bpm

4. How high would a 9 bbl slurry of cement

subtend from bottom up ? CT out of slurry.

5. How many lbs of sand can there be along

the casing from the bottom up to the EOT ?

20.8 bbls @ 0.00201 bbls/ft50.8 bbls @ .00572 bbls/ft

39.06 bbls @ .03641 bbls/ft

87.4 13.7

175 27.5

524.5 82.4

179.7 41.6

89.9 21

29.9 7

228 ft. @ 25.38 ft/bbl.

23.645 lbs/ft x 1073 = 25,371 lbs

CT Calculations (2)


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Hook Load

Definitions CT Area, BuoyancyForce, Snubbing Load


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External and Internal Areas in theCT Downhole End

External Area: Based on the

pipe OD, Wellbore pressure acts

here upwards.

Metal Area: The differencebetween the external and the

internal areas. Wellborepressure acts here.

Internal Area: Based on the pipe

ID. CT internal pressure acts

here downwards.

ID

OD


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Buoyancy Force

The upward force acting on an object placed in a

fluid

The buoyancy force is equal to the weight of fluid

displaced by the object

Fbuoyancy

= Weight of the fluid displaced by the CT (lbs)

= Volume of Fluid displaced by the CT Density of Fluid in the Well


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Buoyancy Force (2)

Calculate the Buoyancy Force for the following

cases:

(I) 1.5in CT string, 0.109in wall with no check

valve full of water @ 3,490ft in a well full ofwater.

lbsf

gal

lbsgals

gal

lbs

bbl

galf t

in.in.

720

34.84.86

34.842490,31029

2821

1029

5122

Fbuoyancy=Weight of the fluid displaced by the CT


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Buoyancy Force (3)

(II) 1.5in CT string, 0.109in wall with check valve

full of water @ 3,490ft in a well full of water.

Buoyancy Force acting on the full cross sectional

area of the CT It does not matter what fluid isinside

lbs

gal

lbsgals

gal

lbs

bbl

galf t

in.

673,2

34.85.320

34.842490,31029

512

Fbuoyancy=Weight of the fluid displaced by the entire CT st


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Snubbing Force

CT going into a well with pressure

The pressure pushes against the tubingproducing an upward force.

The Snubbing Capacity is the force required to

push the CT into a pressurized wellhead,

overcoming this upward push.

Snubbing Load = WHP External Cross Sectional Area of the CT + Stripper Friction Lo


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Snubbing Force

05000

1000015000200002500030000350004000045000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000WHP (psi)

Snubbing

load

(lbs)

1.25 1.5 1.75 2 2 3/8

Snubbing capacity HR440:

Snubbing capacity HR480:


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Hook Load

Hook load is the actual weight

supported by the Injector Head when

the CT string is in the well

The Reading of the Weight Indicatorequals the Hook Load plus the effect

of the reel tension and the stripperload

Note: The following slides refer to the Hook Load,

the Reading of the Weight Indicator will be dealt

closer during the TFM Training Session


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Hook Load in a Vertical Well (1)Well and CT full of water. No added forces and zero pressure.

CT Data: 14,000 ft of 1.5, 0.109 (1.623 lbs/ft) HS90

Well Data: 3.5in Tubing, 7in Casing. Vertical @ 3,490 ft

Hook Load = Tension on cross section right below the stripper

HL = Wair BF + Wfi

where

Wairis the weight of the CT string in air (lbs)

Wfi is the weight of the fluid inside the CT (lbs)

BF is the Buoyancy Force (lbs)

BF

Wair


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Hook Load in a Vertical Well (2)CT string full of 10ppg brine and wellbore full of water. WHP=1,000psi

BF+Snubbing Lo

Wair+Wfi

HL = Wair+ Wfi BF Snubbing Load

where

Wairis the weight of the CT string in air (lbs)

Wfi is the weight of the fluid inside the CT (lbs)


Snubbing Load = WHP External Cross Sectional Area of the CT


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Hook Load in a Horizontal Well (3)

HL Waircosa + Wfi BF Snubbing Load Drag

where

Drag is the friction between the CT and the tubular = m N

N is the normal force = Wair sena - BF sena


BF

Weight

N mN

a


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Hook Load

Affected by:

CT size, WHP, density of fluids inside and outsideCT, shape of the well (inclination, doglegs, etc)

Decreases:

Lighter fluids inside CT

Increasing WHP

Increases: Heavier fluids inside CT


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1. What would the WHP be with hole full of

10 ppg fluid in annulus & 9 ppg in CT &

pump off & no CVs ?

2. What would the minimum CP be with hole full of

10 ppg fluid in annulus & 9 ppg in CT &

pump off & no CVs ?

3. What would the WI read assuming that

the CT weighed 3.923 lbs/ft ?

4. What would the WI read if 9 bbls of 15.6 ppg

were placed in the annulus from CT endfollowed by 2 bbls Fresh Water and remaining CT

volume with 9 ppg ?

2.00 O.D.

.204 W.T.10,350 ft.

3.50 O.D.

9.3 lb/ft

EOT 8,875 ft.

7.00 O.D.

23.0# ft.

EOC 9,948 ft.

0 psi

517 psi

32,048 lbf

31,760 lbf

Pressure/Force Calculations (1)


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Forces on a Packer


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Once a packer is set, the

pressure above the packer will

be applied on the surface

between the CT OD and the

wellbore ID

The pressure belowthe packeris the same as the pressure

inside the CT. So the pressure

belowthe packer produces a

net force only in the area

between the CT ID and thewellbore ID.

Any weight slacked offon top of

the packer will push the packer

down.

Differential force on a packer:

Weight slacked off

pressure

above

pressure below

pressure

inside

DF = (Pb * Ab) - (Pa * Aa) - Sla

Forces on a Packer


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1. Assuming the formation pressure were

7,700 psi & the injection pressure were 440psi,what would the CP be if the reel is full of

15% HCl & Friction Pressure @ 1 bpm = 950

psi ?

2. What would the D P across the packer be

with 10 ppg fluid in the CT/Tbg annulus and

0 WHP injecting under conditions of #1 ?

3. What would the direction and D F value be

on the packer be while set and injecting @ 1

bpm ?

4. You can slackoff 10000 lbs on top of the

packer. How much WHP is required to

neutralize the differential force?

7700 + 440 = 8140 psi

8140 - (9948 x 8.962 x .052) + 950 = 4454 psi

3025 psi

20558 lbsf

WHP = 2715 psi

2.00 O.D..204 W.T.

10,350 ft.

3.50 O.D.

9.3 lb/ft

EOT 8,875 ft.

7.00 O.D.

23.0# ft.

EOC 9,948 ft.

Through tubing Packer set at 8,875

Packer Calculations (1)


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Questions


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TOTAL PRESSURE (psi)

= HYDROSTATIC PRESS (psi) +APPLIED PRESS(psi)

HYDROSTATIC PRESSURE (psi)

= FLUID GRADIENT (psi/ft) x TRUE VERTICALDEPTH (ft)

APPLIED PRESSURE is usually either: The pump (treating) pressure

A part of the formation pressure

Zero


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Phyd = Density (ppg) x Depth (TVD) x .052 (consta

Density = 10.0 ppg Depth = 10,000 ft .052 psi/ft/ppg

Phyd = 10ppg x 10,000ft x .052

Phyd = 5,200 psi

1) Density = 9.6 ppg Depth = 14,100 ft .052 psi/ft/ppg





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Why 0.052?

1 ft

1 ft

1 ft

1 ppg x 7.48 gals/ft3

144 in / ft22

psi/ft/ppg.052

252M302 Downhole Hydraulics Level I

Documents

fluid pressure

surface pressure

typesof pressure

pressure differential

annular volume

id isthe internal diameter

annular flow velocity

inct wall thickness