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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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Grading of Dull Bits
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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Grading of Dull Bits
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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Grading of Dull Bits
How do bits fail?
Washed out Bit Lost Cone
G di f D ll Bit
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Grading of Dull Bits
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 −×=∆
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Hydraulic Power
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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 ∆+∆=∆
Maximum Bit Hydraulic Horsepower
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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,
Maximum Bit Hydraulic Horsepower
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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 =∆
Maximum Bit Hydraulic Horsepower
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
Maximum Bit Hydraulic Horsepower
- 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 ∆
+=∆
Maximum Bit Hydraulic Horsepower
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
Maximum Bit Hydraulic Horsepower
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• 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
Maximum Jet Impact Force
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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
Maximum Jet Impact Force
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
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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−
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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
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)(
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 =∆⇒=∆=
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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
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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
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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
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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 ∆∆∆∆∆∆
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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 )
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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∆
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Surge Pressure due to Pipe Movement
When a string of pipe isbeing lowered into the
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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
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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
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- 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
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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)
Use the values to calculate C and N
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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
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):
Pressure Drop In Drill Pipe OD = 4.5 inID = 3.78 inL = 11,400 ft
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):
Pressure Drop In Drill Pipe OD = 4.5 inID = 3.78 inL = 11,400 ft
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) :
Pressure Drop In Drill Pipe OD = 4.5 inID = 3.78 inL = 11,400 ft
ft
psi05837.0
783*8125
5.12*8*007126.0
D8125
Vf
dL
dP 22
==ρ
=⎟⎠
⎞⎜⎝
⎛
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Friction Pressure Drop in Dril l Pipe :
400,11*05837.0LdL
dPP =∆⎟
⎠
⎞⎜⎝
⎛ =∆
Pdp = 665 psi
ft78.3*81.25D81.25dL ⎠⎝
Power-Law Constant (n):
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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Fluid Consistency Index (K):
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
−
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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
Pressure Drop In Drill Collars
NOTE: NRe > 2,100, soFriction Factor in DC (f): b
ReN
af =
9337370log933nlog
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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
Pressure Drop In Drill Collars
ft
psi3780.0
52*8125
5.12*28.18*005840.0
D8125
Vf
dL
dP 22
==ρ
=⎟
⎠
⎞⎜
⎝
⎛
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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=∆
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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
Power-Law Constant (n):
DHOLE = 8.5 inODDC = 6.5 inL = 600 ft
Pressure Dropin DC/HOLE Annulus
5413.03
20log657.0
R
Rlog657.0n
3
100 =⎟ ⎠
⎞⎜⎝
⎛ =⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
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Fluid Consistency Index (K):
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):
DHOLE = 8.5 inODDC = 6.5 inL = 600 ft
n1n
12e n3
1n2
DD
V144
K100 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−=µ
−
Pressure Dropin DC/HOLE Annulus
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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
DHOLE = 8.5 inODDC = 6.5 inL = 600 ft
NOTE: NRe < 2,100
Friction Factor in Annulus (f):
01500.06001
24
N
24f
R===
Pressure Dropin DC/HOLE Annulus
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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
Power-Law Constant (n):
Pressure Dropin DP/HOLE Annulus
DHOLE = 8.5 inODDP = 4.5 inL = 11,400 ft
5413.03
20log657.0
R
Rlog657.0n
3
100 =⎟ ⎠
⎞⎜⎝
⎛ =⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
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Fluid Consistency Index (K):
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):
Pressure Dropin DP/HOLE Annulus
n1n
12
e
n3
1n2
DD
V144K100 ⎟⎟
⎠
⎞⎜⎜
⎝
⎛ +⎟⎟
⎠
⎞⎜⎜
⎝
⎛
−
=µ−
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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−=
Pressure Dropin DP/HOLE Annulus
NOTE: NRe < 2,100
Friction Factor in Annulus (f):
02299.0044,1
24
N
24
f Re ===
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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
Cumulative Distance from Standpipe, ft
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
Cumulative Distance from Standpipe, ft
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
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n = 1.0
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d8.25
vf
dL
dp _
2ρ=