Bits Nozzles

8/17/2019 Bits Nozzles

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Drilling Bits

And Hydraulics Calculations


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Drilling Bits


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Types of bits

• Drag Bits

• Roller Cone Bits• Diamond Bits


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a) Shearing the formation as PDC and TSPdiamond bits do

b) Ploughing / Grinding the formation, asnatural diamond do

c) Crushing; by putting the rock in compressionas a roller bit

Cutting Mechanisms


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The Ideal Bit *

1. High drilling rate

2. Long life

3. Drill full-gauge, straight hole

4. Moderate cost

* (Low cost per ft drilled)


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Bit Selection

Minimum Cost Per Unit length $/ft

Bit cost + rig cost X (tripping time + Drilling time)C /L = -----------------------------------------------------------------

Footage Progress


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$ Per Foot


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The Roller Cone Bits

Two-cone bit (Milled tooth soft formation only)

Three cone bit (milled tooth, Tungsten carbide inserts)

Four-cone bit (milled tooth, for large hole size)


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Fluid flow through jets in the bit (nozzles)


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Tungsten

Carbide Insert

Bit

Milled

Tooth

Bit


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Rotary Drill BitsRoller Cutter Bits - rock bits

• First rock bit introduced in 1909 by

Howard Hughes

• 2 - cone bit

• Not self-cleaning


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Rotary Drill Bits

• Improvements

• 3 - cone bit (straighter hole)

• Intermeshing teeth (better cleaning)

• Hard-facing on teeth and body• Change from water courses to jets

• Tungsten carbide inserts• Sealed bearings

• Journal bearings


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Rotary Drill Bits

• Advantages

• For any type of formation there is asuitable design of rock bit

• Can handle changes in formation

• Acceptable life and drilling rate

• Reasonable cost


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Fluid flow through water courses in bit

Properbottomhole

cleaning is veryimportant


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Fluid flow through jets in the bit (nozzles)


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Three Cone Bit

Three equal sized cones

Three Identical legs

Each cone is mounted on bearings run on a pin from the leg

The three legs are welded to make the pin connection

Each leg is provided with an opening ( to fit Nozzle)


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Design Factors

Dictated by the Hole size and Formation properties


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Angle formed by the axis of the

Journal and the axis of the bitJOURNALANGLE


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The Angle of the Journal influence the size of the cone

The smaller the Journal angle the greater the gouging andscrapping effect by the three cones


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Offset Cones

HardSoft

Medium


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Teeth


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Bearing

Outer & Nose

Bearings

•Support Radial Loads

Ball Bearing

•Support axial loads

•Secure the cons on

the legs


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Rotary Drill Bits

• Milled Tooth Bit (Steel Tooth)

Long teeth for soft formations

Shorter teeth for harder formations

Cone off-set in soft-formation bit results inscraping gouging action

Self-sharpening teeth by using hardfacing

on one side High drilling rates - especially in softer

rocks


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MilledTooth Bit

(Steel

Tooth)


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Rotary Bits

• Tungsten Carbide Insert Bits

• Long life cutting structure in hard rocks

• Hemispherical inserts for very hard rocks

• Larger and more pointed inserts for softer rock

• Can handle high bit weights and high RPM

• Inserts fail through breakage rather than wear

(Tungsten carbide is a very hard, brittle material)


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Tungsten

CarbideInsert

Bits


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Sealed Bearing

Lubrication System


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Sealed, self-lubricated roller bit

journal bearingdesign details

INSERTS

SILVER PLATED BUSHING

RADIAL SEAL

BALL BEARING

GREASE RESERVOIR CAP

BALL RETAINING

PLUG

BALL RACE


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Roller

Cone

Bearings


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Bearings

• Ball Bearings (point contact)

• Roller Bearings (line contact)

• Journal bearing (area contact)

• Lubrication by drilling fluid . . . or . . .


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Bearings

• Journal Bearings (area contact)

• Wear-resistant hard surface on journal

• Solid lubricant inside cone journal race• O - ring seal

• Grease

• Sealed Bearings (since 1959)• Grease lubricant (much longer life)• Pressure surges can cause seal to leak!

Compensate?


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Grading of Dull BitsHow do bits wear out?

• Tooth wear or loss

• Worn bearings

• Gauge wear


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Grading of Dull Bits

How do bits wear out?

• Steel teeth - graded in eights of original

tooth height that has worn away

e.g. T3 means that3/8 of the originaltooth height is worn

away


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Broken or Lost Teeth

• Tungsten Carbide Insert bit

e.g. T3 means that 3/8 of the inserts

are broken or lost


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How do bits fail?

• Bearings: B3 means that an estimated

3/8 of the bearing life is gone

Balled up Bit Cracked Cone

G di f D ll Bi


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How do bits fail?

Washed out Bit Lost Cone

G di f D ll Bit


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How do bits wear out?

4 Examples:

• T3 – B3 - I

• T5 – B4 - 0 1/2

• Gauge Wear:

• Bit is either in-Gauge or out-of-Gauge

• Measure wear on diameter (in inches),using a gauge ring

BIT

GAUGE RING


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Roller conebit wear

problems


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IADC ROLLER CONE

BIT CLASSIFICATIONSYSTEM


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IADC System• Operational since 1972

• Provides a Method of Categorizing Roller Cone

Rock Bits

• Design and Application related coding• Most Recent Revision

‘The IADC Roller Bit Classification System’

1992, IADC/SPE Drilling Conference

Paper # 23937


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IADC Classification

• 4-Character Design/Application Code

– First 3 Characters are NUMERIC

– 4th Character is ALPHABETIC

135M or 447X or 637Y


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Examples

637Y

medium-hard insert

bit;

frict ion bearing withgage protection;

conical inserts

135Msoft formation

Milled tooth bit ;

roller bearings withgage protection;

motor application

447Xsoft formation insert bit;

frict ion bearings

with gage protection;chisel inserts


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Sequence

• Numeric Characters are defined: – Series 1st

– Type 2nd

– Bearing & Gage 3rd

• Alphabetic Character defined:

– Features Available 4th

135M135M or or 447X447X or or 637Y637Y


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Series

• FIRST CHARACTER

• General Formation Characteristics

• Eight (8) Series or Categories

• Series 1 to 3 Milled Tooth Bits

• Series 4 to 8 Tungsten Carbide Insert BitsThe higher the series number,

the harder/more abrasive the rock

1135M35M or or 4447X47X or or 6637Y37Y


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Define Hardness

Hardness UCS (psi) Examples

Ultra Soft < 1,000 gumbo, clay

Very Soft 1,000 - 4,000unconsolidated sands, chalk,

salt, claystone

Soft 4,000 - 8,000 coal, siltstone, schist, sands

Medium 8,000 - 17,000sandstone, slate, shale,

limestone, dolomite

Hard 17,000 - 27,000

quartzite, basalt, gabbro,

limestone, dolomite

Very Hard > 27,000 marble, granite, gneiss

UCS = Uniaxial Unconfined Compressive Strength

B i & G


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Bearing & Gage

• THIRD CHARACTER• Bearing Design and Gage Protection

• Seven (7) Categories

– 1. Non-Sealed (Open) Roller Bearing – 2. Roller Bearing Air Cooled

– 3. Non-Sealed (Open) Roller Bearing Gage Protected

– 4. Sealed Roller Bearing

– 5. Sealed Roller Bearing Gage Protected

– 6. Sealed Friction Bearing

– 7. Sealed Friction Bearing Gage Protected

131355MM or or 444477XX or or 636377YY

F t A il bl


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Features Available

• FOURTH CHARACTER

• Features Available (Optional)• Sixteen (16) Alphabetic Characters

• Most Significant Feature Listed(i.e. only one alphabetic character should be selected).

135135MM or or 447447XX or or 637637YY

IADC F t A il bl


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IADC Features Available

• A - Air Application

• B - Special Bearing/Seal

• C - Center Jet

• D - Deviation Control

• E - Extended Nozzles

• G - Gage/Body Protection• H - Horizontal Application

• J - Jet Deflection

• L - Lug Pads

• M - Motor Application

• S - Standard MilledTooth

• T - Two-Cone Bit

• W - Enhanced C/S• X - Chisel Tooth Insert

• Y - Conical Tooth Insert

• Z - Other Shape Inserts

135135MM or or 447447XX or or 637637YY


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Drag Bits

Cutter may be made from:

Steel

Tungsten carbide

Natural diamonds

Polycrystalline diamonds (PDC)

Drag bits have no moving parts, so it is less likelythat junk will be left in the hole.


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Fishtail type drag bit

D Bit


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Drag Bits

Drag bits drill by physically “plowing”

or “machining” cuttings from the

bottom of the hole.


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Natural Diamond Bits PDC Bits

Nat ral


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Natural

Diamondbit

junk slotcuttings

radial flow

high pacross face


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SoftFormation

Diamond bit

Larger diamonds

Fewer diamonds

Pointed nose


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HardFormation

Diamond bit

Smaller diamonds

More diamonds

Flatter nose

Natural Diamonds


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Natural Diamonds

The size and spacing of diamonds on a

bit determine its use.

NOTE: One carat = 200 mg precious stones

What is 14 carat gold?

Natural Diamonds


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Natural Diamonds

• 2-5 carats - widely spaced diamondsare used for drilling soft formations such as

soft sand and shale

• 1/4 - 1 carat - diamonds are used for drillingsand, shale and limestone formations of

varying (intermediate) hardness.

•1/8 - 1/4 carat - diamonds, closely spaced, are

used in hard and abrasive formations.

When to Consider Using a Natural


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g

Diamond Bit?

1. Penetration rate of rock bit < 10 ft/hr.

2. Hole diameter < 6 inches.3. When it is important to keep the bit and

pipe in the hole.4. When bad weather precludes making trips.

5. When starting a side-tracked hole.

6. When coring.

* 7. When a lower cost/ft would result


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Top view of diamond bit


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Side view ofdiamond bit


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PDCbits

CourtesySmith Bits

PDC Cutter


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At about $10,000-150,000 apiece, PDC bits cost five to 15 times more

than roller cone bits

PDC Bits


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The Rise in Diamond Bit Market Share

Coring


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Coring

bit

PDC +

naturaldiamond

Bi-Center bit


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Courtesy Smith Bits

Relative Costs of Bits


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Diamond WC Insert MilledBits Bits Tooth Bits

$/Bit

• Diamond bits typically cost several times as much as tri-

cone bits with tungsten carbide inserts (same bit diam.)

• A TCI bit may cost several times as much as a

milled tooth bit.

PDC Bits


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PDC BitsRef: Oil & Gas Journal, Aug. 14, 1995, p.12

• Increase penetration rates in oil and gaswells

• Reduce drilling time and costs

• Cost 5-15 times more than roller cone bits• 1.5 times faster than those 2 years earlier

• Work better in oil based muds; however,these areas are strictly regulated

PDC Bits


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PDC Bits

• Parameters for effective use

include

weight on bit

mud pressure

flow rate

rotational speed

PDC Bits


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• Economics

• Cost per foot drilled measures Bitperformance economics

• Bit Cost varies from 2%-3% of total cost, but

bit affects up to 75% of total cost

• Advantage comes when

- the No. of trips is reduced, and when- the penetration rate increases

PDC Bits


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Bit Demand

U.S Companies sell > 4,000 diamond drillbits/year

Diamond bit Market is about $200million/year

Market is large and difficult to reform

When bit design improves, bit drills longer

PDC Bits


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– Improvements in bit stability, hydraulics,and cutter design => increased footage per bit

– Now, bits can drill both harder and softerformations

Bit Demand, cont’d

PDC Bits


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Bit Design,

PDC bit diameter varies from 3.5 in to 17.5 in

• Goals of hydraulics:

– clean bit without eroding it

– clean cuttings from bottom of hole

PDC Bits


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• Factors that limit operating rangeand economics:

– Lower life from cutter fractures

– Slower ROP from bad cleaning

Bit design, cont’d

PDC Bits


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• Cutters

• Consist of thin layer of bonded diamondparticles + a thicker layer of tungsten carbide

• Diamond• 10x harder than steel

• 2x harder than tungsten carbide

• Most wear resistant materialbut is brittle and susceptible to damage

PDC Bits


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• Diamond/Tungsten Interface

• Bond between two layers on cutter iscritical

• Consider difference in thermalexpansion coefficients and avoid

overheating• Made with various geometric shapes to

reduce stress on diamond

Cutters, cont’d

PDC Bits


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• Various Sizes

• Experimental dome shape

• Round with a buttress edge for highimpact loads

• Polished with lower coefficient of friction

4 Cutters, cont’d

PDC Bits


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• Bit Whirl (bit instability)

• Bit whirl = “any deviation of bit rotation

from the bit’s geometric center”

• Caused by cutter/rock interaction forces

• PBC bit technology sometimesreinforces whirl

• Can cause PDC cutters to chip and break

PDC Bits


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Preventing Bit Whirl

• Cutter force balancing• Bit asymmetry

• Gauge design• Bit profile

• Cutter configuration

• Cutter layout

PDC Bits


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Applications

• PDC bits are used primarily in

• Deep and/or expensive wells

• Soft-medium hard formations

PDC Bits


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Advances in metallurgy, hydraulicsand cutter geometry

• Have not cut cost of individual bits

• Have allowed PDC bits to drill longerand more effectively

• Allowed bits to withstand harderformations

4 Application, cont’d

PDC Bits


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• Application, cont’d

• PDC bits advantageous for high rotationalspeed drilling and in deviated hole sectiondrillings

• Most effective: very weak, brittle formations(sands, silty claystone, siliceous shales)

• Least effective: cemented abrasive sandstone,granites

Grading of Worn PDC Bits


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CT - Chipped Cutter Less than 1/3 of cuttingelement is gone

BT - Broken Cutter More than 1/3 of cuttingelement is broken to

the substrate

Grading of Worn PDC Bits – cont’d


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LT - Lost Cutter Bit is missing one ormore cutters

LN - Lost NozzleBit is missing one ormore nozzles

Diamond bit wear problems


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Best Penetration Rate


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• Approach A

• Achieved by removingcuttings efficientlyfrom below the bit

• Maximize thehydraulic power

available at the bit

• Approach B

• Drilling fluid hitsbottom of the holewith greatest force

• Maximize Jet ImpactForce

Optimum bit hydraulics


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Find the flow rates for different pump pressures (before POOH)

Use the values to calculate C and N

Get the expression for optimum flow rate

Establish optimum flow rate Q

Find the system pressure drop

Get the optimum system pressure drop (from either approach A or

Establish optimum Stand pipe pressure and check with pump capacity

Calculate optimum P b

Calculate optimum AT (TFA) and select jets

Bit Nozzles


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Bit Nozzles


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Nozzle Velocity


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0.95toequalusually

tcoefficiendischarge Nozzle

10074.8 4

=

×∆= −

d

bd n

C

pC v ρ

Bit Pressure Drop


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22

251033.8

t d

b AC

q p ρ −×=∆


97/197

Hydraulic Power


98/197

HPP

pqP

H

H

8.2721714

4001169

1714

=×

=

∆=

Hydraulic Impact Force


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lbsF

F

pqC F

j

j

bd j

5.820

11691240095.01823.0

01823.0

=

×××=∆= ρ

Jet Bit Nozzle Size Selection


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• Proper bottom-hole cleaning

• will eliminate excessive regrinding of drilledsolids, and

• will result in improved penetration rates

Bottom-hole cleaning efficiency

• is achieved through proper selection of bitnozzle sizes

Total Pump Pressure


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• Pressure loss in surf. equipment

• Pressure loss in drill pipe

• Pressure loss in drill collars

• Pressure drop across the bit nozzles

• Pressure loss in the annulus between the drillcollars and the hole wall

• Pressure loss in the annulus between the drillpipe and the hole wall

• Hydrostatic pressure difference ( varies)

Jet Bit Nozzle Size Selection

- Optimization -


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- Optimization -

Through nozzle size selection,

optimization may be based onmaximizing one of the following:

Bit Nozzle Velocity

Bit Hydraulic Horsepower

Jet impact force

• There is no general agreement on which ofthese three parameters should be maximized.

Maximum Nozzle Velocity


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Nozzle velocity may be maximized consistent withthe following two constraints:

• 1. The annular fluid velocity needs to be high

enough to lift the drill cuttings out of the hole.

- This requirement sets the minimumfluid circulation rate.

• 2. The surface pump pressure must stay within the

maximum allowable pressure rating of thepump and the surface equipment.


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Maximum Nozzle Velocity


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This (maximization) will be achieved whenthe surface pressure is maximized and thefrictional pressure loss everywhere isminimized, i.e., when the flow rate is

minimized.

pressure.surfaceallowablemaximumtheandratencirculatiominimumtheat

satisfied,areabove2&1whenmaximizedisvn∴

Maximum Bit Hydraulic Horsepower


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The hydraulic horsepower at the bit ismaximized when is maximized.q) p( bit∆

d pump bit p p p ∆−∆=∆

where may be called the parasitic pressureloss in the system (friction).

d p∆

bitdpump ppp ∆+∆=∆



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.turbulentisflowtheif

cq p p p p p p 75.1dpadcadcdpsd =∆+∆+∆+∆+∆=∆

In general, wherem

d cq p =∆ 2m0 ≤≤

The parasitic pressure loss in the system,



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0)1( pwhen0

17141714

pump

1

=+−∆=∴

−∆=

∆=∴

+

m Hbit

m

pumpbit Hbit

qmcdq

dP

cqq pq pP

d pump bit p p p ∆−∆=∆m

d cq p =∆


0)1(p =+−∆ mqmc


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whenmaximumis

1

1 pwhen.,.

)1( pwhen.,.

d

pump

Hbit

pump

d

P

pm

ei

pmei

∴

∆⎟

⎠

⎞⎜

⎝

⎛ +

=∆

∆+=∆

pumpd pm

p ∆⎟ ⎠

⎞⎜⎝

⎛ +

=∆1

1

0)1( p pump =+∆ qmc


- Examples -


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p

In turbulent flow, m = 1.75

pump bit

pump

pumpd

pof %64 p

pof 36%

%100* p175.1

1 p

∆=∆∴

∆=

∆⎟ ⎠

⎞⎜⎝

⎛ +

=∆∴

pd p1m

1p ∆

+=∆


Examples - cont’d


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In laminar flow, for Newtonian fluids, m = 1

pump b

pump

pumpd

pof %50 p

pof 50%

%100* p11

1 p

∆=∆∴

∆=

∆⎟ ⎠

⎞⎜⎝

⎛ +

=∆∴

p



112/197

• In general, the hydraulic horsepower is notoptimized at all times

• It is usually more convenient to select apump liner size that will be suitable for

the entire well

• Note that at no time should the flow rate be

allowed to drop below the minimumrequired for proper cuttings removal

Maximum Jet Impact Force


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The jet impact force is given by Eq. 4.37:

)(c0.01823

01823.0

d d pump

bit d j

p pq

pqcF

∆−∆=

∆=

ρ

ρ



114/197

But parasitic pressure drop,

2201823.0

+−∆=∴

=∆

md pd j

md

qcq pcF

cq p

ρ ρ

)(c0.01823 d d pump j p pqF ∆−∆= ρ


Upon differentiating setting the first derivative


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Upon differentiating, setting the first derivativeto zero, and solving the resulting quadratic

equation, it may be seen that the impactforce is maximized when,

pd p

2m

2 p ∆

+

=∆

Maximum Jet

Impact Force- Examples -

pd p2m

2 p ∆

+=∆


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Examples

p b

pd

pof %47 p and

pof %53 p 1.75,mif ,

∆=∆

∆=∆=Thus

p b

pd

pof 33% pand

pof %67 p 1.00mif ,

∆=∆

∆=∆= Also


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Nozzle Size Selection

- Graphical Approach -


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1. Show opt. hydraulic path

2. Plot ∆ pd

vs q

3. From Plot, determine


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optimum q and ∆ pd

4. Calculate

5. Calculate

Total Nozzle Area:(TFA)

6. Calculate Nozzle Diameter

d pump bit p p p ∆−∆=∆

opt bd

opt opt t

pC

q A

)(

10*311.8)( 2

25

∆=

− ρ

With 3 nozzles:π 3

A4d tot N =

Example 4.31

Determine the proper pump operating


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Determine the proper pump operatingconditions and bit nozzle sizes for max.

jet impact force for the next bit run.

Current nozzle sizes: 3 EA 12/32”

Mud Density = 9.6 lbm.gal

At 485 gal/min, Ppump = 2,800 psi

At 247 gal/min, Ppump = 900 psi

Example 4.31 - given data:

Max pump HP (Mech ) = 1 250 hp


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Max pump HP (Mech.) = 1,250 hp

Pump Efficiency = 0.91

Max pump pressure = 3,000 psig

Minimum flow rate

to lift cuttings = 225 gal/min

Example 4.31 - 1(a), 485 gpm

Calculate pressure drop through bit nozzles:2


123/197

22

2510*311.8:)34.4.(

t d b Ac

q p Eq

ρ −

=∆

psi9061,894-2,800losspressureparasitic

psi1,894

32

12

43(0.95)

)485)(6.9)(8.311(10

p 222

2-5

b

==∴

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ π=∆

Example 4.31 - 1(b), 247 gpm

)247)(69)(10(3118 25−


124/197

psi p b 491

32

12

43)95.0(

)247)(6.9)(10(311.82

22

25

=

⎥⎥⎦⎤

⎢⎢⎣⎡ ⎟

⎠ ⎞⎜

⎝ ⎛

=∆π

psi 409491-900losspressureparasitic ==∴

Plot these twopoints in Fig. 4.36

(q1, p1) = (485, 906)(q2, p2) = (247, 409)


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Example 4.31 - cont’d

2. For optimum hydraulics:1Interval)a(

32

1


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gal/min650000,3

)91.0)(250,1(714,1714,1qmax

max ===P

E P Hp

1, Interval)a(

gal/min225q

psi875,1

)000,3(22.1

22

2 p

min

maxd

==

⎟ ⎠ ⎞⎜

⎝ ⎛

+=⎟

⎠ ⎞⎜

⎝ ⎛

+=∆ P

m 2,Interval(b)

3,Interval(c)

1

Example 4.31

3. From graph, optimum point is at

l


127/197

)(

10*311.8

)( 2

25

opt bd

opt

opt t pC

q

A ∆=∴

− ρ

)700,1(*)95.0(

)650(*6.9*10*8.311

2

2-5

=

( ) ind opt N nds2opt 3214in0.47A =⇒=

psi p psigal

q b 700,1300,1 p ,

min

650 d =∆⇒=∆=


128/197

psi p psigal

q b 700,1300,1 p ,min

650 d =∆⇒=∆=

Example 4.32

Well Planning


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It is desired to estimate the proper pumpoperating conditions and bit nozzle sizes formaximum bit horsepower at 1,000-ftincrements for an interval of the wellbetween surface casing at 4,000 ft and

intermediate casing at 9,000 ft. The wellplan calls for the following conditions:

Example 4.32

Pump: 3,423 psi maximum surface pressure


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1,600 hp maximum input

0.85 pump efficiency

Drillstring: 4.5-in., 16.6-lbm/ft drillpipe(3.826-in. I.D.)

600 ft of 7.5-in.-O.D. x 2.75-in.-I.D. drill collars

Example 4.32

Surface Equipment: Equivalent to 340


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Surface Equipment: Equivalent to 340ft. of drillpipe

Hole Size: 9.857 in. washed out to 10.05 in.

10.05-in.-I.D. casing

Minimum Annular Velocity: 120 ft/min

Mud Program

Mud Plastic YieldDepth Density Viscosity Point


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Depth Density Viscosity Point

(ft) (lbm/gal) (cp) (lbf/100 sq ft)

5,000 9.5 15 5

6,000 9.5 15 57,000 9.5 15 5

8,000 12.0 25 99,000 13.0 30 12

Solution

The path of optimum hydraulics is asfollows:


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follows:

Interval 1

gal/min.681

423,3

)85.0)(600,1(714,1

p

EP714,1

qmax

Hp

max

=

==

Solution

Interval 2

Si d d t t


134/197

Since measured pump pressure data are not

available and a simplified solution techniqueis desired, a theoretical m value of 1.75 isused. For maximum bit horsepower,

( )

psia1,245

423,3175.1

1

1

1max

=

⎟ ⎠

⎞⎜⎝

⎛ +

=⎟ ⎠

⎞⎜⎝

⎛ +

=∆ pm

pd

Solution

Interval 3


135/197

For a minimum annular velocity of120 ft/min opposite the drillpipe,

( )

gal/min395

601205.405.10448.2 22min

=

⎟ ⎠ ⎞⎜

⎝ ⎛ −=q

Table

The frictional pressure loss in othersections is computed following a


136/197

sections is computed following a

procedure similar to that outlined above forthe sections of drillpipe. The entireprocedure then can be repeated to

determine the total parasitic losses atdepths of 6,000, 7,000, 8,000 and 9,000 ft.The results of these computations aresummarized in the following table:

Table

ddpadcadcdps p p p p p p Depth ∆∆∆∆∆∆


137/197

5,000 38 490 320 20 20 8886,000 38 601 320 20 25 1,0047,000 38 713 320 20 29 1,120

8,000 51 1,116 433 28 75* 1,7039,000 57 1,407 482 27* 111* 2,084

* Laminar flow pattern indicated byHedstrom number criteria.

pp

Table

The proper pump operating conditionsand nozzle areas, are as follows:


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5,000 600 1,245 2,178 0.3806,000 570 1,245 2,178 0.361

7,000 533 1,245 2,178 0.338

8,000 420 1,245 2,178 0.2999,000 395 1,370 2,053 0.302

in.)(sq (psi) (psi) (gal/min) )ft(

(5)A p(4) p(3) Rate(2)Flow Depth)l( t bd ∆∆

Table

The first three columns were read directlyfrom Fig 4 37 (depth flow rate and ∆p )


139/197

from Fig. 4.37. (depth, flow rate and ∆pd)

Col. 4 (∆pb) was obtained by subtracting

shown in Col.3 from the maximum pumppressure of 3,423 psi.

Col.5 (Atot) was obtained using Eq. 4.85

d p∆


140/197

Surge Pressure due to Pipe Movement

When a string of pipe isbeing lowered into the


141/197

being lowered into the

wellbore, drilling fluid isbeing displaced and forcedout of the wellbore.

The pressure required toforce the displaced fluid out

of the wellbore is called thesurge pressure.

Surge Pressure due to Pipe Movement

An excessively high surge pressure canresult in breakdown of a formation


142/197

result in breakdown of a formation.

When pipe is being withdrawn a similarreduction is pressure is experienced. Thisis called a swab pressure, and may behigh enough to suck fluids into the wellbore,resulting in a kick.

swabsurge PP ,vfixedFor pipe =

Figure 4.40B


143/197

- Velocity profile for laminar flow pattern when closedpipe is being run into hole

The Hydraulics Parameters

Pump Volumetric output and circulation pressure Pt


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Flow rateBit nozzle jet velocity

Annular velocity

Pressure losses in the system

Pump Hydraulic power output

Pressure drop across the bit nozzlesHydraulic Power at the bit

Jet impact force

Pump volumetric output and circulating pressure

Q= K L(2D2 d2) spm η /100 for double acting pump


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Q= K.L(2D2-d2).spm.ηv/100 for double acting pump

Q= K.L.D2.spm.ηv/100 for single acting pump

Q in GPM if K=.00679

Q in BPM if K=.000126

Circulating Pressure = Total Pressure loss (except at the bit)

+ Pressure drop across the bit nozzle

Flow rate Q


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Can be measure directly (flow-meter)

Can be calculated

Average Velocity in Drillpipe


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Assuming the total string is DP;

24.51 x Q

Velocity Vdp = ------------------ ft/minID p2

Annular Average Velocity

Assuming the total string is DP;


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Minimum velocity govern by the lifting capacity of the drilling fluid

Maximum velocity in sensitive formation 100 ft/min.

Optimum Annular Velocity is at the minimum flow rate requiredto efficiently remove cuttings from the hole

24.51 x QAnnular Velocity Vann = ------------------ ft/min

Dh2 - OD p

2

Nozzle Jet Velocity


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Vn = 0.321 (Q/A) ft/s

Minimum 350 ft/s or 100 m/s

Fluid Flow

Newtonian fluid

Non Newtonian fluid


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Laminar Flow Re < 2000

Turbulent Flow Re > 4000

Re = 15.46 ρ DV / µ

Non Newtonian fluid

Bingham Plastic Fluid

Power-Law Fluid

Bingham Plastic Model

At the wall zero Fluid velocity


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Viscosity independent of time

Particles travel parallel to the

pipe axe (max. velocity at the

center).

Critical Velocity Vc


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Vc D YP

D

= + +97 97 8 22 2 pv pv . ρ

ρ

ft/min

V > Vc Turbulent flow

V < Vc Laminar flow


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Pressure Drop

Pressure Loss in the System


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Pressure losses in the surface equipment

Pressure loses in the drilling string

Pressure loses in the annulus

Pressure drop in the surface

equipment


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P1 = E ρ N-1 (PV)2-N Q N N=1.8 or can be measures

Pressure Drop in Drillpipe

P2 = f ρ V2 L / 25.8 d


156/197

P2 = c . Q N

8.91 x 10-5 ρ N-1 PV2-N . Lc = -------------------------------------

ID p N+3

8.91 x 10-5 ρ N-1 Q N PV2-N . LP2 = -----------------------------------------ID p

N+3

f is a friction factor depends on the type of flow

Pressure Drop in annulus

P3 = f ρ V2 L / 21.1 (Dh - OD p)


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P3 = c . Q N

c = 8.91 x 10-5 ρ N-1 PV2-N L / (Dh - OD p)3 (Dh + OD p) N+3

8.91 x 10-5

ρ N-1

Q N

PV2-N

. LP3 = -------------------------------------(Dh - OD p)

3 (Dh + OD p) N+3

f is a friction factor depends on the type of flow

Pressure drop across the bit


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P b = Pstandpipe - (P1+P2+P3)

ρ Q2

P b = ---------------------12,032 Cn2 AT

2

Cn = Nozzle Coefficient (~ 0.95)

Nozzle Velocity Vn ft/s


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V P

nb

= 3336. ρ

Best Penetration Rate

• Approach A

• Achieved by removing

• Approach B

• Drilling fluid hits


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cuttings efficientlyfrom below the bit

• Maximize thehydraulic poweravailable at the bit

bottom of the holewith greatest force

• Maximize Jet ImpactForce

Optimum bit hydraulics

Find the flow rates for different pump pressures (before POOH)



161/197


Get the expression for optimum flow rate

Establish optimum flow rate Q

Find the system pressure drop

Get the optimum system pressure drop (from either approach A or

Establish optimum Stand pipe pressure and check with pump capacity

Calculate optimum P b

Calculate optimum AT (TFA) and select jets

Max. Hydraulic Power at the bit

P b . Q / 1714 hp

N


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Jet Impact Force below the bitIF = Q/58 (ρ P b)0.5 Max IF when P b = [N/(N+2)] Psp

61.6 x 10-3 ρ Q2 / AT

P b = (Psp - PCS) Pcs = c QHHP b = (Psp Q - c Q

N+1 )/1714 Differentiate wrt Q = 0

P b = (N/N+1) Psp

Nozzle Selection


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AT = 0.0096 Q (ρ /P b)0.5 = .32 Q/Vn

dn = 32 (4 AT /3π)0.5

Total Pump Pressure

• Pressure loss in surf. equipment

• Pressure loss in drill pipe


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• Pressure loss in drill collars• Pressure drop across the bit nozzles

• Pressure loss in the annulus between the drill

collars and the hole wall

• Pressure loss in the annulus between the drillpipe and the hole wall

• Hydrostatic pressure difference ( varies)

Types of flow

Laminar Turbulent


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Fig. 4-30. Laminar and turbulent flow patterns in a circular pipe: (a) laminar

flow, (b) transition between laminar and turbulent flow and (c) turbulent flow

Turbulent Flow -Newtonian Fluid

lbm/galdensity,fluidρ where =

µ

dvρ928 N

_

Re =


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We often assume that fluid flow is

turbulent if Nre > 2100

cp.fluid,of viscosityµ

inI.D., piped

ft/svelocity,fluidavg. v _

=

=

=

Turbulent Flow -Newtonian Fluid

75.1 _

Turbulent Flow -Bingham Plastic Fluid

75.1 _

In Pipe


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25.1

25.0_

75.0f

d1800

v

dL

dp µρ=25.1

25.0p

75.0f

d1800

v

dL

dp µρ=

( ) 25.112

25.0

p

75.1 _75.0

f

dd396,1v

dLdp

−µρ=( ) 25.112

25.075.1 _

75.0

f

dd396,1v

dLdp

−µρ=

In Annulus

API Power Law Model

K = consistency indexn = flow behaviour index

τ = K γ n

API RP 13D


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SHEARSTRESS

τ psi

SHEAR RATE, γ , sec-10

Rotating Sleeve Viscometer

VISCOMETERRPM

(RPM * 1.703)

SHEAR RATE

sec -1


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3100

300600

5.11170.3

5111022

BOB

SLEEVE

ANNULUS

DRILLSTRING

Pressure Drop Calculations

• Example Calculate the pump pressure inthe wellbore shown on the next page, using the API method.


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• The relevant rotational viscometer readingsare as follows:

• R3 = 3 (at 3 RPM)• R100 = 20 (at 100 RPM)

• R300 = 39 (at 300 RPM)• R600 = 65 (at 600 RPM)

Q = 280 gal/min

Pressure Drop

Calculations

PPUMP


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PPUMP = ∆PDP + ∆PDC

+ ∆PBIT NOZZLES

+ ∆PDC/ANN + ∆PDP/ANN

+ ∆PHYD

= 12.5 lb/gal

Power-Law Constant (n):

Pressure Drop In Drill Pipe OD = 4.5 inID = 3.78 inL = 11,400 ft

737.039

65log32.3

R

Rlog32.3n

300

600 =⎟ ⎠

⎞⎜⎝

⎛ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =


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Fluid Consistency Index (K):

Average Bulk Velocity in Pipe (V):

2737.0

600 sec017.2

022,1

65*11.5

022,1

11.5

cm

dyne RK

n

n ===

secft00.8

78.3280*408.0

DQ408.0V

22 ===

Effective Viscosity in Pipe ( e):


n1n

e

4

1n3

D

V96K100 ⎟

⎟

⎠

⎞⎜⎜

⎝

⎛ +⎟⎟

⎠

⎞⎜⎜

⎝

⎛ =µ

−


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Reynolds Number in Pipe (NRe):

n4D ⎠⎝ ⎠⎝

cP53

737.0*4

1737.0*3

78.3

8*96017.2*100

737.01737.0

e =⎟

⎠

⎞⎜

⎝

⎛ +⎟

⎠

⎞⎜

⎝

⎛ =µ

−

616,653

5.12*00.8*78.3*928VD928Ne

Re ==µρ=

NOTE: NRe > 2,100, soFriction Factor in Pipe (f):


b

ReN

af =

933l


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So,

0759.050

93.3737.0log

50

93.3nloga =

+=

+=

2690.07

737.0log75.17

nlog75.1b =−=−=

007126.0616,6

0759.0N

af 2690.0bRe

===

Friction Pressure Gradient (dP/dL) :


ft

psi05837.0

783*8125

5.12*8*007126.0

D8125

Vf

dL

dP 22

==ρ

=⎟⎠

⎞⎜⎝

⎛


175/197

Friction Pressure Drop in Dril l Pipe :

400,11*05837.0LdL

dPP =∆⎟

⎠

⎞⎜⎝

⎛ =∆

Pdp = 665 psi

ft78.3*81.25D81.25dL ⎠⎝


Pressure Drop In Drill Collars OD = 6.5 inID = 2.5 inL = 600 ft

737.039

65log32.3

R

Rlog32.3n

300

600 =⎟ ⎠

⎞⎜⎝

⎛ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =


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Average Bulk Velocity inside Drill Collars (V):

2

n

737.0n

600

cm

secdyne017.2

022,1

65*11.5

022,1

R11.5K ===

secft28.18

5.2280*408.0

DQ408.0V

22 ===

Effective Viscosity in Collars( e):

OD = 6.5 inID = 2.5 inL = 600 ft

Pressure Drop In Drill Collars

n1n

e

n4

1n3

D

V96K100 ⎟

⎠

⎞⎜

⎝

⎛ +⎟

⎠

⎞⎜

⎝

⎛ =µ

−


177/197

Reynolds Number in Collars (NRe):

n4D ⎠⎝ ⎠⎝

cP21.38737.0*4

1737.0*3

5.2

28.18*96017.2*100

737.01737.0

e =⎟

⎠

⎞⎜

⎝

⎛ +⎟

⎠

⎞⎜

⎝

⎛ =µ

−

870,1321.38

5.12*28.18*5.2*928VD928Ne

Re ==µρ=

OD = 6.5 inID = 2.5 inL = 600 ft


NOTE: NRe > 2,100, soFriction Factor in DC (f): b

ReN

af =

9337370log933nlog


178/197

So,

0759.050

93.3737.0log

50

93.3nloga =+=+=

2690.07

737.0log75.17

nlog75.1b =−=−=

005840.0870,130759.0

Naf 2690.0bRe

===

Friction Pressure Gradient (dP/dL) :

OD = 6.5 inID = 2.5 inL = 600 ft


ft

psi3780.0

52*8125

5.12*28.18*005840.0

D8125

Vf

dL

dP 22

==ρ

=⎟

⎠

⎞⎜

⎝

⎛


179/197

Friction Pressure Drop in Dril l Collars :

ft5.281.25D81.25dL ⎠⎝

600*3780.0LdL

dPP =∆⎟

⎠

⎞⎜⎝

⎛ =∆

Pdc = 227 psi

Pressure Drop across Nozzles

DN1 = 11 32nds (in)DN2 = 11 32nds (in)

DN3 = 12 32nds (in)( )2

2

3N

2

2N

2

1N

2

DDD

Q156P

++

ρ=∆


180/197

DN3 = 12 32nds (in)

( )2222

2

121111

280*5.12*156P

++=∆

PNozzles = 1,026 psi

( )DDD ++

Pressure Dropin DC/HOLE

Annulus

Q = 280 gal/min


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DHOLE = 8.5 inODDC = 6.5 in

L = 600 ft

Q 280 gal/min

= 12.5 lb/gal 8.5 in


DHOLE = 8.5 inODDC = 6.5 inL = 600 ft

Pressure Dropin DC/HOLE Annulus

5413.03

20log657.0

R

Rlog657.0n

3

100 =⎟ ⎠

⎞⎜⎝

⎛ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =


182/197


Average Bulk Velocity in DC/HOLE Annulus (V):

2

n

5413.0n100 cm

secdyne336.62.170

20*11.5

2.170

R11.5K ===

secft808.3

5.65.8280*408.0

DDQ408.0V

222

1

2

2

=−

=−

=

Effective Viscosity in Annulus ( e):


n1n

12e n3

1n2

DD

V144

K100 ⎟⎟⎠

⎞

⎜⎜⎝

⎛ +

⎟⎟⎠

⎞

⎜⎜⎝

⎛

−=µ

−



183/197

Reynolds Number in Annulus (NRe):

cP20.555413.0*3

15413.0*2

5.65.8

808.3*144336.6*100

5413.015413.0

e

=⎟ ⎠

⎞

⎜⎝

⎛ +

⎟ ⎠

⎞

⎜⎝

⎛

−=µ

−

( ) ( ) 600,120.55

5.12*808.3*5.65.8928VDD928Ne

12

Re =−=µρ−=

12 ⎟ ⎠⎜⎝ ⎟⎟ ⎠⎜⎜⎝ µ


NOTE: NRe < 2,100

Friction Factor in Annulus (f):

01500.06001

24

N

24f

R===



184/197

So,

600,1NRe

( ) ( ) ft

psi05266.0

5.65.881.25

5.12*808.3*01500.0

DD81.25

Vf

dL

dP 2

12

2

=

−

=

−

ρ=⎟

⎠

⎞⎜

⎝

⎛

600*05266.0LdL

dPP =∆⎟

⎠

⎞⎜⎝

⎛ =∆

Pdc/hole = 31.6 psi

q = 280 gal/min

Pressure Dropin DP/HOLE Annulus


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= 12.5 lb/gal

DHOLE = 8.5 inODDP = 4.5 inL = 11,400 ft



DHOLE = 8.5 inODDP = 4.5 inL = 11,400 ft

5413.03

20log657.0

R

Rlog657.0n

3

100 =⎟ ⎠

⎞⎜⎝

⎛ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =


186/197


Average Bulk Velocity in Annulus (Va):

2

n

5413.0n100

cm

secdyne336.62.170

20*11.5

2.170

R11.5K ===

secft197.2

5.45.8280*408.0

DDQ408.0V

222

1

2

2

=−

=−

=

Effective Viscosity in Annulus ( e):


n1n

12

e

n3

1n2

DD

V144K100 ⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +⎟⎟

⎠

⎞⎜⎜

⎝

⎛

−

=µ−


187/197

Reynolds Number in Annulus (NRe):

⎠⎝ ⎠⎝

cP64.975413.0*3

15413.0*2

5.45.8

197.2*144336.6*100

5413.015413.0

e =⎟ ⎠

⎞⎜⎝

⎛ +⎟ ⎠

⎞⎜⎝

⎛ −

=µ−

( ) ( ) 044,164.97

5.12*197.2*5.45.8928VDD928Ne

12Re =−=µ

ρ−=


NOTE: NRe < 2,100

Friction Factor in Annulus (f):

02299.0044,1

24

N

24

f Re ===


188/197

So, psi

e

( ) ( ) ft

psi01343.0

5.45.881.25

5.12*197.2*02299.0

DD81.25

Vf

dL

dP 2

12

2

=

−

=

−

ρ=⎟

⎠

⎞⎜

⎝

⎛

400,11*01343.0LdL

dPP =∆⎟

⎠

⎞⎜⎝

⎛ =∆

Pdp/hole = 153.2 psi

Pressure Drop Calcs.

- SUMMARY -

PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES

+ ∆P + ∆P + ∆P


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+ ∆PDC/ANN + ∆PDP/ANN + ∆PHYD

PPUMP = 665 + 227 + 1,026

+ 32 + 153 + 0

PPUMP = 1,918 + 185 = 2,103 psi

PPUMP = ∆PDS + ∆PANN + ∆PHYD

∆PDS = ∆PDP + ∆PDC + ∆PBIT NOZZLES

= 665 + 227 + 1,026 = 1,918 psi

∆PANN = ∆PDC/ANN + ∆PDP/ANN

2,103 psi

P=0


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PPUMP = 1,918 + 185= 2,103 psi

∆PHYD = 0

ANN DC/ANN DP/ANN

= 32 + 153 = 185

" Friction" Pressures

1,500

2,000

2,500

e s s

u r e ,

p s i DRILLPIPE

DRILL COLLARS


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0

500

1,000

0 5,000 10,000 15,000 20,000 25,000

Cumulative Distance from Standpipe, ft

" F r i c t

i o n "

P r e

BIT NOZZLES

ANNULUS

Hydrostatic Pressures in the Wellbore

5,000

6,000

7,000

8,000

9,000

r e s

s u r e ,

p s i BHP

DRILLSTRING ANNULUS


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0

1,000

2,000

3,000

4,000

0 5,000 10,000 15,000 20,000 25,000


H

y d r o s t a t i c

P

Pressures in the Wellbore

5 000

6,000

7,000

8,000

9,000

10,000

e s ,

p s i CIRCULATING


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0

1,000

2,0003,000

4,000

5,000

0 5,000 10,000 15,000 20,000 25,000


P

r e s s u r

STATIC

Wellbore Pressure Profile

0

2,000

4,000

6,000h ,

f t

DRILLSTRING


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8,000

10,000

12,000

14,000

0 2,000 4,000 6,000 8,000 10,000

Pressure, psi

D e p t h ANNULUS

(Static)

BIT

Pipe Flow - Laminar

In the above example the flow down thedrillpipe was turbulent.

Under conditions of very high viscosity,the flow may very well be laminar


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y g ythe flow may very well be laminar.

NOTE: if NRe < 2,100, thenFriction Factor in Pipe (f):

ReN

16

f = D81.25

Vf

dL

dP2ρ

=⎟ ⎠

⎞

⎜⎝

⎛

Then and


196/197

n = 1.0


197/197

d8.25

vf

dL

dp _

2ρ=

Bits Nozzles

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