Bits Nozzles
Short Description
Bits Nozzles selection...
Description
Drilling Bits And Hydraulics Calculations
Drilling Bits
Types of bits
• Drag Bits • Roller Cone Bits • Diamond Bits
Cutting Mechanisms a) Shearing the formation as PDC and TSP diamond bits do
b) Ploughing / Grinding the formation, as natural diamond do c) Crushing; by putting the rock in compression as a roller bit
The Ideal Bit * 1. High drilling rate 2. Long life 3. Drill full-gauge, straight hole 4. Moderate cost * (Low cost per ft drilled)
Bit Selection
Minimum Cost Per Unit length $/ft
Bit cost + rig cost X (tripping time + Drilling time) C /L = ----------------------------------------------------------------Footage Progress
$ Per Foot
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)
Fluid flow through jets in the bit (nozzles)
Tungsten Carbide Insert Bit
Milled Tooth Bit
Rotary Drill Bits Roller Cutter Bits - rock bits • First rock bit introduced in 1909 by Howard Hughes • 2 - cone bit • Not self-cleaning
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
Rotary Drill Bits • Advantages • For any type of formation there is a suitable design of rock bit • Can handle changes in formation • Acceptable life and drilling rate • Reasonable cost
Proper bottomhole cleaning is very important
Fluid flow through water courses in bit
Fluid flow through jets in the bit (nozzles)
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)
Design Factors
Dictated by the Hole size and Formation properties
JOURNAL ANGLE
Angle formed by the axis of the Journal and the axis of the bit
The Angle of the Journal influence the size of the cone The smaller the Journal angle the greater the gouging and scrapping effect by the three cones
Offset Cones Hard
Soft
Medium
Teeth
Bearing Outer & Nose Bearings •Support Radial Loads Ball Bearing •Support axial loads •Secure the cons on the legs
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 in scraping gouging action Self-sharpening teeth by using hardfacing on one side High drilling rates - especially in softer rocks
Milled Tooth Bit (Steel Tooth)
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)
Tungsten Carbide Insert Bits
Sealed Bearing Lubrication System
INSERTS
SILVER PLATED BUSHING RADIAL SEAL BALL RACE BALL RETAINING PLUG
BALL BEARING
Sealed, selflubricated roller bit journal bearing design details GREASE RESERVOIR CAP
Roller Cone Bearings
Bearings • Ball Bearings (point contact) • Roller Bearings (line contact) • Journal bearing (area contact) • Lubrication by drilling fluid . . . or . . .
Bearings • Sealed Bearings (since 1959) • Grease lubricant (much longer life)
• Pressure surges can cause seal to leak! Compensate?
• Journal Bearings (area contact) • Wear-resistant hard surface on journal • Solid lubricant inside cone journal race • O - ring seal • Grease
Grading of Dull Bits How do bits wear out?
• Tooth wear or loss • Worn bearings • Gauge wear
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 that 3/8 of the original tooth height is worn away
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
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
Grading of Dull Bits How do bits fail?
Washed out Bit
Lost Cone
Grading of Dull Bits How do bits wear out?
• Gauge Wear: • Bit is either in-Gauge or out-of-Gauge • Measure wear on diameter (in inches), using a gauge ring
4 Examples:
BIT
• T3 – B3 - I • T5 – B4 - 0 1/2
GAUGE RING
Roller cone bit wear problems
IADC ROLLER CONE BIT CLASSIFICATION SYSTEM
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
IADC Classification • 4-Character Design/Application Code – First 3 Characters are NUMERIC – 4th Character is ALPHABETIC
135M
or
447X
or
637Y
Examples
135M
447X
637Y
soft formation
soft formation insert bit;
Milled tooth bit;
friction bearings
medium-hard insert bit;
roller bearings with
with gage protection;
friction bearing with
gage protection;
chisel inserts
gage protection;
motor application
conical inserts
Sequence 135M
or
447X
or
• Numeric Characters are defined: – Series
1st
– Type
2nd
– Bearing & Gage
3rd
• Alphabetic Character defined: – Features Available
4th
637Y
Series 135M
or
447X
or
637Y
• FIRST CHARACTER • General Formation Characteristics • Eight (8) Series or Categories • Series 1 to 3 Milled Tooth Bits • Series 4 to 8 Tungsten Carbide Insert Bits The higher the series number, the harder/more abrasive the rock
Define Hardness Hardness
UCS (psi)
Examples
Ultra Soft
< 1,000
gumbo, clay
Very Soft
1,000 - 4,000
unconsolidated sands, chalk, salt, claystone
Soft
4,000 - 8,000
coal, siltstone, schist, sands
Medium
8,000 - 17,000
sandstone, 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
Bearing & Gage 135M
or
447X
or
637Y
• 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
Features Available 135M
or
447X
or
637Y
• FOURTH CHARACTER • Features Available (Optional) • Sixteen (16) Alphabetic Characters • Most Significant Feature Listed (i.e. only one alphabetic character should be selected).
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
135M
or
• L - Lug Pads • M - Motor Application • S - Standard Milled Tooth • T - Two-Cone Bit • W - Enhanced C/S • X - Chisel Tooth Insert • Y - Conical Tooth Insert • Z - Other Shape Inserts
447X
or
637Y
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 likely that junk will be left in the hole.
Fishtail type drag bit
Drag Bits Drag bits drill by physically “plowing” or “machining” cuttings from the bottom of the hole.
Natural Diamond Bits
PDC Bits
Natural Diamond bit
junk slot cuttings radial flow high ∆p across face
Soft Formation Diamond bit
Larger diamonds Fewer diamonds Pointed nose
Hard Formation Diamond bit
Smaller diamonds More diamonds Flatter nose
Natural Diamonds
The size and spacing of diamonds on a bit determine its use.
NOTE: One carat = 200 mg What is 14 carat gold?
precious stones
Natural Diamonds • 2-5 carats - widely spaced diamonds are used for drilling soft formations such as soft sand and shale
• 1/4 - 1 carat - diamonds are used for drilling sand, 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 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
Top view of diamond bit
Side view of diamond bit
PDC bits
Courtesy Smith Bits
PDC Cutter
PDC Bits
At about $10,000-150,000 apiece, PDC bits cost five to 15 times more than roller cone bits
The Rise in Diamond Bit Market Share
Coring bit
PDC + natural diamond
Bi-Center bit
Courtesy Smith Bits
Relative Costs of Bits
$/Bit Diamond Bits
WC Insert Bits
Milled Tooth Bits
• Diamond bits typically cost several times as much as tricone bits with tungsten carbide inserts (same bit diam.) • A TCI bit may cost several times as much as a milled tooth bit.
PDC Bits Ref: Oil & Gas Journal, Aug. 14, 1995, p.12
• Increase penetration rates in oil and gas wells • 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 • Parameters for effective use include weight on bit
mud pressure flow rate rotational speed
PDC Bits • Economics • Cost per foot drilled measures Bit performance 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 ¾ Bit Demand U.S Companies sell > 4,000 diamond drill bits/year Diamond bit Market is about $200 million/year Market is large and difficult to reform When bit design improves, bit drills longer
PDC Bits ¾ Bit Demand, cont’d – Improvements in bit stability, hydraulics, and cutter design => increased footage per bit – Now, bits can drill both harder and softer formations
PDC Bits ¾ 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 ¾ Bit design, cont’d • Factors that limit operating range and economics: – Lower life from cutter fractures – Slower ROP from bad cleaning
PDC Bits • Cutters • Consist of thin layer of bonded diamond particles + a thicker layer of tungsten carbide • Diamond • 10x harder than steel • 2x harder than tungsten carbide • Most wear resistant material but is brittle and susceptible to damage
PDC Bits ¾ Cutters, cont’d • Diamond/Tungsten Interface • Bond between two layers on cutter is critical • Consider difference in thermal expansion coefficients and avoid overheating • Made with various geometric shapes to reduce stress on diamond
PDC Bits 4 Cutters, cont’d • Various Sizes • Experimental dome shape • Round with a buttress edge for high impact loads • Polished with lower coefficient of friction
PDC Bits • 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 sometimes reinforces whirl • Can cause PDC cutters to chip and break
PDC Bits Preventing Bit Whirl • Cutter force balancing • Bit asymmetry • Gauge design • Bit profile • Cutter configuration • Cutter layout
PDC Bits Applications • PDC bits are used primarily in • Deep and/or expensive wells • Soft-medium hard formations
PDC Bits 4 Application, cont’d Advances in metallurgy, hydraulics and cutter geometry • Have not cut cost of individual bits • Have allowed PDC bits to drill longer and more effectively • Allowed bits to withstand harder formations
PDC Bits • Application, cont’d • PDC bits advantageous for high rotational speed drilling and in deviated hole section drillings • Most effective: very weak, brittle formations (sands, silty claystone, siliceous shales) • Least effective: cemented abrasive sandstone, granites
Grading of Worn PDC Bits
CT - Chipped Cutter
BT - Broken Cutter
Less than 1/3 of cutting element is gone
More than 1/3 of cutting element is broken to the substrate
Grading of Worn PDC Bits – cont’d
LT - Lost Cutter
LN - Lost Nozzle
Bit is missing one or more cutters
Bit is missing one or more nozzles
Diamond bit wear problems
Best Penetration Rate • Approach B • Approach A • Achieved by removing • Drilling fluid hits bottom of the hole cuttings efficiently with greatest force from below the bit • Maximize the hydraulic power available at the bit
• Maximize Jet Impact Force
Optimum bit hydraulics 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 Pb Calculate optimum AT (TFA) and select jets
Bit Nozzles
Bit Nozzles
Nozzle Velocity vn = C d
∆pb 8.074 × 10−4 ρ
Cd = Nozzle discharge coefficient usually equal to 0.95
Bit Pressure Drop 8.33 ×10 ρq ∆pb = 2 2 Cd At −5
2
Hydraulic Power ∆pq PH = 1714 1169 × 400 PH = = 272.8 HP 1714
Hydraulic Impact Force F j = 0.01823Cd q ρ∆pb F j = 0.01823 × .95 × 400 12 ×1169 F j = 820.5lbs
Jet Bit Nozzle Size Selection • Proper bottom-hole cleaning • will eliminate excessive regrinding of drilled solids, and • will result in improved penetration rates
¾ Bottom-hole cleaning efficiency • is achieved through proper selection of bit nozzle sizes
Total Pump Pressure • 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 drill collars and the hole wall • Pressure loss in the annulus between the drill pipe and the hole wall • Hydrostatic pressure difference
(ρ varies)
Jet Bit Nozzle Size Selection - Optimization Through nozzle size selection, optimization may be based on maximizing one of the following: ¾ Bit Nozzle Velocity ¾ Bit Hydraulic Horsepower ¾ Jet impact force
• There is no general agreement on which of these three parameters should be maximized.
Maximum Nozzle Velocity Nozzle velocity may be maximized consistent with the 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 minimum fluid circulation rate. • 2. The surface pump pressure must stay within the maximum allowable pressure rating of the pump and the surface equipment.
Maximum Nozzle Velocity
Nozzle Velocity
i.e.
vn = Cd
∆ Pb −4 8 . 074 * 10 ρ
v n ∝ ∆Pb
so the bit pressure drop should be maximized in order to obtain the maximum nozzle velocity
Maximum Nozzle Velocity This (maximization) will be achieved when the surface pressure is maximized and the frictional pressure loss everywhere is minimized, i.e., when the flow rate is minimized. ∴ v n is maximized when 1& 2 above are satisfied, at the minimum circulation rate and the maximum allowable surface pressure.
Maximum Bit Hydraulic Horsepower The hydraulic horsepower at the bit is maximized when (∆p bit q) is maximized.
∆ p pump = ∆ p d + ∆ p bit
∆ p bit = ∆ p pump − ∆ p d where ∆p d may be called the parasitic pressure loss in the system (friction).
Maximum Bit Hydraulic Horsepower The parasitic pressure loss in the system,
∆p d = ∆p s + ∆p dp + ∆p dc + ∆p dca + ∆p dpa = cq
1.75
if the flow is turbulent. In general,
∆ p d = cq
m
where 0 ≤ m ≤ 2
Maximum Bit Hydraulic Horsepower
∆ p bit = ∆ p pump − ∆ p d ∴ PHbit
∆ p d = cq
∆pbit q ∆p pump q − cq = = 1714 1714
dPHbit ∴ = 0 when dq
m
m +1
∆p pump − c(m + 1)q = 0 m
Maximum Bit Hydraulic Horsepower ∆p pump − c(m + 1)q = 0 m
i . e ., when i . e ., when
∆ p pump = ( m + 1 ) ∆ p d ∆pd
⎛ 1 ⎞ = ⎜ ⎟∆p ⎝ m +1⎠
∴ P Hbit is maximum
∆pd
pump
when
⎛ 1 ⎞ =⎜ ⎟ ∆ p pump ⎝ m +1⎠
Maximum Bit Hydraulic Horsepower - Examples In turbulent flow, m = 1.75
1 ∆p d = ∆p p m +1
1 ⎛ ⎞ ∴ ∆p d = ⎜ ⎟ ∆ p pump * 100 % ⎝ 1 . 75 + 1 ⎠ = 36% of ∆ p pump ∴ ∆ p bit = 64 % of ∆ p pump
Maximum Bit Hydraulic Horsepower Examples - cont’d In laminar flow, for Newtonian fluids, ∴ ∆pd
⎛ 1 ⎞ = ⎜ ⎟ ∆ p pump * 100 % ⎝1+1⎠ = 50% of ∆ p pump
∴ ∆ p b = 50 % of ∆ p pump
m=1
Maximum Bit Hydraulic Horsepower • In general, the hydraulic horsepower is not optimized at all times • It is usually more convenient to select a pump 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 minimum required for proper cuttings removal
Maximum Jet Impact Force The jet impact force is given by Eq. 4.37:
F j = 0.01823 cd q ρ ∆pbit = 0.01823 c d q ρ ( ∆p pump − ∆p d )
Maximum Jet Impact Force F j = 0.01823 c d q ρ ( ∆p pump − ∆pd ) But parasitic pressure drop,
∆ p d = cq ∴ F j = 0 .01823 c d
m
ρ ∆p p q − ρ cd q 2
m+2
Maximum Jet Impact Force Upon differentiating, setting the first derivative to zero, and solving the resulting quadratic equation, it may be seen that the impact force is maximized when,
2 ∆p d = ∆p p m+2
Maximum Jet Impact Force - Examples Thus, if m = 1.75,
2 ∆p d = ∆p p m+2
∆pd = 53% of ∆pp and ∆pb = 47% of ∆pp
Also, if m = 1.00
∆pd = 67% of ∆pp and ∆pb = 33% of ∆pp
Nozzle Size Selection - Graphical Approach -
1. Show opt. hydraulic path 2. Plot ∆pd vs q 3. From Plot, determine optimum q and ∆pd ∆ p bit = ∆ p pump − ∆ p d 4. Calculate 5. Calculate 2 −5 8 .311 * 10 ρ q opt Total Nozzle Area: ( At ) opt = 2 C d ( ∆ p b ) opt (TFA)
6. Calculate Nozzle Diameter With 3 nozzles:
4Atot dN = 3π
Example 4.31 Determine the proper pump operating conditions 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 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: Eq .( 4 . 34 ) : ∆ p b = ∆p b =
8.311(10
-5
8 . 311 * 10 2
c d At
)( 9 .6 )( 485 )2
2 ⎡ π ⎛ 12 ⎞ ⎤ 2 (0.95) ⎢3 ⎜ ⎟ ⎥ ⎢⎣ 4 ⎝ 32 ⎠ ⎥⎦
−5
2
ρ q
2
2
= 1,894 psi
∴ parasitic pressure loss = 2,800 - 1,894 = 906 psi
Example 4.31 - 1(b), 247 gpm
∆ pb =
8 . 311 (10
−5
)( 9 . 6 )( 247 )
⎡ π 2 ( 0 . 95 ) ⎢ 3 ⎢⎣ 4
⎛ 12 ⎞ ⎤ ⎜ ⎟ ⎥ ⎝ 32 ⎠ ⎥⎦ 2
2
2
= 491 psi
∴ parasitic pressure loss = 900 - 491 = 409 psi (q1, p1) = (485, 906) (q2, p2) = (247, 409)
Plot these two points in Fig. 4.36
Example 4.31 - cont’d
3 2
2. For optimum hydraulics:
(a ) Interval 1, q max =
1
1,714 PHp E Pmax
(b) Interval 2,
1,714(1,250)(0.91) = = 650 gal/min 3,000
⎛ 2 ⎞ ⎛ 2 ⎞ ∆p d = ⎜ ⎟ Pmax = ⎜ ⎟ ( 3, 000 ) ⎝m+2⎠ ⎝ 1 .2 + 2 ⎠ = 1,875 psi
(c) Interval 3,
q min = 225 gal/min
Example 4.31 3. From graph, optimum point is at gal q = 650 , ∆ p d = 1,300 psi ⇒ ∆ p b = 1,700 psi min 8 .311 * 10 ρ q opt −5
∴ ( At ) opt =
2
C d ( ∆ p b ) opt
Aopt = 0.47 in
2
⇒
2
-5
8.311*10 * 9.6 * (650) = 2 (0.95) * (1,700)
(d N )opt = 14
nds
32
in
2
gal q = 650 , ∆ p d = 1,300 psi ⇒ ∆ p b = 1,700 psi min
Example 4.32 Well Planning It is desired to estimate the proper pump operating conditions and bit nozzle sizes for maximum bit horsepower at 1,000-ft increments for an interval of the well between surface casing at 4,000 ft and intermediate casing at 9,000 ft. The well plan calls for the following conditions:
Example 4.32 Pump: 3,423 psi maximum surface pressure 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 ft. 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 Depth (ft)
Mud Density (lbm/gal)
Plastic Yield Viscosity Point (cp) (lbf/100 sq ft)
5,000
9.5
15
5
6,000
9.5
15
5
7,000
9.5
15
5
8,000
12.0
25
9
9,000
13.0
30
12
Solution The path of optimum hydraulics is as follows: Interval 1
q max =
1,714 PHp E p max
1,714(1,600)(0.85) = 3,423
= 681 gal/min.
Solution Interval 2 Since measured pump pressure data are not available and a simplified solution technique is desired, a theoretical m value of 1.75 is used. For maximum bit horsepower,
⎛ 1 ⎞ ⎛ 1 ⎞ ∆p d = ⎜ ⎟ (3,423 ) ⎟ pmax = ⎜ ⎝ 1.75 + 1 ⎠ ⎝ m +1⎠ = 1,245 psia
Solution Interval 3 For a minimum annular velocity of 120 ft/min opposite the drillpipe,
(
q min = 2.448 10 .05 − 4.5 = 395 gal/min
2
2
)
⎛ 120 ⎞ ⎜ ⎟ ⎝ 60 ⎠
Table The frictional pressure loss in other sections is computed following a procedure similar to that outlined above for the sections of drillpipe. The entire procedure then can be repeated to determine the total parasitic losses at depths of 6,000, 7,000, 8,000 and 9,000 ft. The results of these computations are summarized in the following table:
Table
Depth ∆ps ∆pdp ∆pdc ∆pdca ∆pdpa ∆pd 5,000 6,000 7,000 8,000 9,000
38 38 38 51 57
490 601 713 1,116 1,407
320 320 320 433 482
20 20 20 25 20 29 28 75* 27* 111*
* Laminar flow pattern indicated by Hedstrom number criteria.
888 1,004 1,120 1,703 2,084
Table The proper pump operating conditions and nozzle areas, are as follows: ( l) Depth (2)Flow Rate (3) ∆ p d (4) ∆ p b ( ft )
(gal/min)
5,000 6,000 7,000 8,000 9,000
600 570 533 420 395
(psi)
1,245 1,245 1,245 1,245 1,370
(5)A t
(psi) (sq in.)
2,178 2,178 2,178 2,178 2,053
0.380 0.361 0.338 0.299 0.302
Table The first three columns were read directly from Fig. 4.37. (depth, flow rate and ∆pd) Col. 4 (∆pb) was obtained by subtracting ∆pd shown in Col.3 from the maximum pump pressure of 3,423 psi. Col.5 (Atot) was obtained using Eq. 4.85
Surge Pressure due to Pipe Movement When a string of pipe is being lowered into the wellbore, drilling fluid is being displaced and forced out of the wellbore. The pressure required to force the displaced fluid out of the wellbore is called the surge pressure.
Surge Pressure due to Pipe Movement An excessively high surge pressure can result in breakdown of a formation. When pipe is being withdrawn a similar reduction is pressure is experienced. This is called a swab pressure, and may be high enough to suck fluids into the wellbore, resulting in a kick.
For fixed
v pipe ,
Psurge = Pswab
Figure 4.40B
- Velocity profile for laminar flow pattern when closed pipe is being run into hole
The Hydraulics Parameters Pump Volumetric output and circulation pressure Pt Flow rate Bit nozzle jet velocity Annular velocity Pressure losses in the system Pump Hydraulic power output Pressure drop across the bit nozzles Hydraulic Power at the bit Jet impact force
Pump volumetric output and circulating pressure
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
Can be measure directly (flow-meter) Can be calculated
Average Velocity in Drillpipe
Assuming the total string is DP; 24.51 x Q Velocity Vdp = -----------------IDp2
ft/min
Annular Average Velocity Assuming the total string is DP; 24.51 x Q Annular Velocity Vann = -----------------Dh2 - ODp2
ft/min
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 required to efficiently remove cuttings from the hole
Nozzle Jet Velocity
Vn = 0.321 (Q/A)
ft/s
Minimum 350 ft/s or 100 m/s
Fluid Flow Newtonian fluid Non Newtonian fluid Bingham Plastic Fluid Power-Law Fluid Re = 15.46 ρ DV / µ Laminar Flow
Re < 2000
Turbulent Flow
Re > 4000
Bingham Plastic Model At the wall zero Fluid velocity Viscosity independent of time Particles travel parallel to the pipe axe (max. velocity at the center).
Critical Velocity Vc
97 pv + 97 pv + 8.2 ρD YP Vc = ρD 2
V > Vc
Turbulent flow
V < Vc
Laminar flow
2
ft/min
Pressure Drop
Pressure Loss in the System
Pressure losses in the surface equipment Pressure loses in the drilling string Pressure loses in the annulus
Pressure drop in the surface equipment
P1 = E ρN-1 (PV)2-N QN
N=1.8 or can be measures
Pressure Drop in Drillpipe P2 = f ρ V2 L / 25.8 d f
is a friction factor depends on the type of flow
P2 = c . QN c
P2
8.91 x 10-5 ρN-1 PV2-N . L = ------------------------------------IDpN+3 8.91 x 10-5 ρN-1 QN PV2-N . L = ----------------------------------------IDpN+3
Pressure Drop in annulus P3 = f ρ V2 L / 21.1 (Dh - ODp) f is a friction factor depends on the type of flow P3 = c . QN c = 8.91 x 10-5 ρN-1 PV2-N L / (Dh - ODp)3 (Dh + ODp)N+3 P3
8.91 x 10-5 ρN-1 QN PV2-N . L = ------------------------------------(Dh - ODp)3 (Dh + ODp)N+3
Pressure drop across the bit
Pb = Pstandpipe - (P1+P2+P3) ρ Q2 Pb = --------------------12,032 Cn2 AT2 Cn = Nozzle Coefficient (~ 0.95)
Nozzle Velocity Vn ft/s
Vn = 33.36
Pb
ρ
Best Penetration Rate • Approach B • Approach A • Achieved by removing • Drilling fluid hits bottom of the hole cuttings efficiently with greatest force from below the bit • Maximize the hydraulic power available at the bit
• Maximize Jet Impact Force
Optimum bit hydraulics 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 Pb Calculate optimum AT (TFA) and select jets
Max. Hydraulic Power at the bit Pb . Q / 1714 Pb = (Psp - PCS)
hp
Pcs = c QN
HHPb = (Psp Q - c QN+1 )/1714
Differentiate wrt Q = 0
Pb = (N/N+1) Psp
Jet Impact Force below the bit IF = Q/58 (ρ Pb)0.5 Max IF when Pb = [N/(N+2)] Psp 61.6 x 10-3 ρ Q2 / AT
Nozzle Selection
AT = 0.0096 Q (ρ /Pb)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 • 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 drill pipe and the hole wall • Hydrostatic pressure difference
(ρ varies)
Types of flow Laminar
Turbulent
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
_
N Re =
928 ρ v d
ρ = fluid density, lbm/gal
where _
v = avg. fluid velocity, ft/s d = pipe I.D., in µ = viscosity of fluid, cp.
We often assume that fluid flow is turbulent if Nre > 2100
µ
Turbulent Flow Newtonian Fluid
Turbulent Flow Bingham Plastic Fluid In Pipe
_ 1 . 75
dp f ρ v µ = 1 . 25 dL 1800 d 0 . 75
0 . 25
_ 1 . 75
ρ v µp dp f = dL 1800 d 1 . 25 0 . 75
0 . 25
In Annulus _ 1 . 75
dp f ρ v µ = 1 . 25 dL 1,396 (d 2 − d 1 ) 0 . 75
0 . 25
_ 1 . 75
ρ µp v dp f = 1 . 25 dL 1,396 (d 2 − d 1 ) 0 . 75
0 . 25
API Power Law Model K = consistency index n = flow behaviour index
API RP 13D
τ=K γn
SHEAR STRESS τ psi 0
SHEAR RATE, γ , sec-1
Rotating Sleeve Viscometer (RPM * 1.703) VISCOMETER RPM
SHEAR RATE
3 100
ANNULUS
5.11 170.3
300 600
DRILL STRING
511 1022
sec -1
BOB
SLEEVE
Pressure Drop Calculations • Example
Calculate the pump pressure in the wellbore shown on the next page, using the API method.
• The relevant rotational viscometer readings are as follows:
• R3 = 3 • R100 = 20 • R300 = 39 • R600 = 65
(at 3 RPM) (at 100 RPM) (at 300 RPM) (at 600 RPM)
Pressure Drop Calculations Q = 280 gal/min
ρ = 12.5 lb/gal PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES + ∆PDC/ANN + ∆PDP/ANN + ∆PHYD
PPUMP
Pressure Drop In Drill Pipe Power-Law Constant (n): ⎛ R 600 n = 3 . 32 log ⎜⎜ ⎝ R 300
⎞ ⎟⎟ ⎠
⎛ 65 ⎞ = 3 . 32 log ⎜ ⎟ = 0 . 737 ⎝ 39 ⎠
Fluid Consistency Index (K):
K =
5.11 R600 n
1,022
5.11 * 65 dyne sec n = = 2.017 0.737 1,022 cm 2
Average Bulk Velocity in Pipe (V): 0 . 408 Q V = D2
0 . 408 * 280 ft = = 8 . 00 2 3 . 78 sec
OD = 4.5 in ID = 3.78 in L = 11,400 ft
OD = 4.5 in ID = 3.78 in L = 11,400 ft
Pressure Drop In Drill Pipe Effective Viscosity in Pipe (µe):
⎛ 96V µe = 100 K ⎜⎜ ⎝ D
⎞ ⎟⎟ ⎠
⎛ 96 * 8 ⎞ µe = 100 * 2.017⎜ ⎟ ⎝ 3.78 ⎠
n −1
⎛ 3n + 1⎞ ⎜ ⎟ ⎜ 4n ⎟ ⎝ ⎠
n
0.737−1
0.737
⎛ 3 * 0.737 + 1⎞ ⎜ ⎟ ⎝ 4 * 0.737 ⎠
= 53 cP
Reynolds Number in Pipe (NRe): NRe
928 D Vρ = µe
928 * 3.78 * 8.00 * 12 .5 = = 6,616 53
Pressure Drop In Drill Pipe NOTE: NRe > 2,100, so Friction Factor in Pipe (f):
a=
b=
So,
log n + 3.93
=
50
1.75 − log n
a NRe
a NRe
b
log 0.737 + 3.93 = 0.0759 50
1.75 − log 0.737 = = 0.2690 7
7
f =
f =
OD = 4.5 in ID = 3.78 in L = 11,400 ft
b
0 .0759 = = 0 .007126 0 .2690 6,616
Pressure Drop In Drill Pipe
OD = 4.5 in ID = 3.78 in L = 11,400 ft
Friction Pressure Gradient (dP/dL) : f V ρ ⎛ dP ⎞ ⎜ ⎟ = 25.81 D ⎝ dL ⎠ 2
psi 0.007126 * 8 2 * 12 .5 = = 0.05837 25.81 * 3.78 ft
Friction Pressure Drop in Drill Pipe : ⎛ dP ⎞ ∆P = ⎜ ⎟ ∆L ⎝ dL ⎠
= 0.05837* 11,400
∆Pdp = 665 psi
Pressure Drop In Drill Collars
OD = 6.5 in ID = 2.5 in L = 600 ft
Power-Law Constant (n): ⎛ R 600 n = 3 . 32 log ⎜⎜ ⎝ R 300
⎞ ⎛ 65 ⎞ ⎟⎟ = 3 . 32 log ⎜ ⎟ = 0 . 737 ⎝ 39 ⎠ ⎠
Fluid Consistency Index (K): K=
5.11R 600 1,022
n
5.11 * 65 dyne sec n = = 2.017 0.737 1,022 cm 2
Average Bulk Velocity inside Drill Collars (V): 0 . 408 Q V= D2
0 . 408 * 280 ft = = 18 . 28 2 2 .5 sec
Pressure Drop In Drill Collars Effective Viscosity in Collars(µe):
⎛ 96V ⎞ µe = 100 K ⎜ ⎟ ⎝ D ⎠
n −1
⎛ 3n + 1⎞ ⎜ ⎟ ⎝ 4n ⎠
⎛ 96 * 18.28 ⎞ µe = 100 * 2.017⎜ ⎟ 2.5 ⎠ ⎝
OD = 6.5 in ID = 2.5 in L = 600 ft
n
0.737−1
⎛ 3 * 0.737 + 1⎞ ⎜ ⎟ ⎝ 4 * 0.737 ⎠
0.737
= 38.21cP
Reynolds Number in Collars (NRe): NRe
928 D V ρ = µe
928 * 2.5 * 18 .28 * 12 .5 = = 13,870 38 .21
Pressure Drop In Drill Collars NOTE: NRe > 2,100, so Friction Factor in DC (f):
f =
a NRe
b
log n + 3.93 a= 50
log 0.737 + 3.93 = = 0.0759 50
1.75 − log n
1.75 − log 0.737 = = 0.2690 7
b=
So,
7
a f = b NRe
OD = 6.5 in ID = 2.5 in L = 600 ft
0.0759 = = 0.005840 0 .2690 13,870
Pressure Drop In Drill Collars Friction Pressure Gradient (dP/dL) : f V ρ ⎛ dP ⎞ ⎟ = ⎜ 25.81 D ⎝ dL ⎠ 2
0.005840 * 18 .28 2 * 12 .5 psi = = 0.3780 25 .81 * 2.5 ft
Friction Pressure Drop in Drill Collars : ⎛ dP ⎞ ∆P = ⎜ ⎟ ∆L ⎝ dL ⎠
= 0.3780 * 600
∆Pdc = 227 psi
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop across Nozzles ∆P =
156ρ Q
2
(D
2
N1
∆P =
+ DN2 + DN3 2
2
)
2
156 * 12.5 * 280 2
(11
2
+ 11 + 12 2
)
2 2
∆PNozzles = 1,026 psi
DN1 = 11 32nds (in) DN2 = 11 32nds (in) DN3 = 12 32nds (in)
Pressure Drop in DC/HOLE Annulus Q = 280 gal/min
ρ = 12.5 lb/gal DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
8.5 in
Pressure Drop in DC/HOLE Annulus
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
Power-Law Constant (n): ⎛R n = 0 . 657 log ⎜⎜ 100 ⎝ R3
⎞ ⎛ 20 ⎞ ⎟⎟ = 0 . 657 log ⎜ ⎟ = 0 . 5413 ⎝ 3 ⎠ ⎠
Fluid Consistency Index (K): K =
5.11R100 170 .2
n
5.11 * 20 dyne sec n = = 6.336 0.5413 170 .2 cm 2
Average Bulk Velocity in DC/HOLE Annulus (V): ft 0 . 408 Q 0 . 408 * 280 V = = = 3 . 808 2 2 2 2 8 .5 − 6 .5 sec D 2 − D1
Pressure Drop in DC/HOLE Annulus
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
Effective Viscosity in Annulus (µe): ⎛ 144V ⎞ ⎟⎟ µ e = 100 K ⎜⎜ ⎝ D2 − D1 ⎠
n −1
⎛ 2n + 1⎞ ⎟ ⎜ ⎜ 3n ⎟ ⎠ ⎝
⎛ 144 * 3.808 ⎞ µ e = 100 * 6.336⎜ ⎟ ⎝ 8 .5 − 6 .5 ⎠
n
0.5413 −1
⎛ 2 * 0.5413 + 1⎞ ⎟ ⎜ ⎝ 3 * 0.5413 ⎠
0.5413
= 55.20 cP
Reynolds Number in Annulus (NRe): NRe =
928 (D2 − D1 ) V ρ µe
928 (8.5 − 6.5) * 3.808 * 12.5 = = 1,600 55.20
Pressure Drop in DC/HOLE Annulus
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
NOTE: NRe < 2,100 Friction Factor in Annulus (f): 24 f = NRe
24 = = 0 .01500 1,600 f V ρ 2
⎛ dP ⎞ ⎜ ⎟= ⎝ dL ⎠ 25.81(D 2 − D1 )
0.01500 * 3.808 2 * 12.5 psi = = 0.05266 25.81 (8.5 − 6.5 ) ft
⎛ dP ⎞ ∆P = ⎜ ⎟ ∆L ⎝ dL ⎠
So,
= 0 .05266 * 600
∆Pdc/hole = 31.6 psi
Pressure Drop in DP/HOLE Annulus q = 280 gal/min
ρ = 12.5 lb/gal DHOLE = 8.5 in ODDP = 4.5 in L = 11,400 ft
Pressure Drop in DP/HOLE Annulus
DHOLE = 8.5 in ODDP = 4.5 in L = 11,400 ft
Power-Law Constant (n): ⎛ R 100 n = 0 .657 log ⎜⎜ ⎝ R3
⎞ ⎟⎟ ⎠
⎛ 20 ⎞ = 0 .657 log ⎜ ⎟ = 0 .5413 ⎝ 3 ⎠
Fluid Consistency Index (K):
K =
5.11R100 n
170.2
5.11* 20 dyne secn = = 6.336 0.5413 170.2 cm2
Average Bulk Velocity in Annulus (Va):
0.408 Q V = 2 2 D2 − D1
ft 0.408* 280 = = 2.197 2 2 8.5 − 4.5 sec
Pressure Drop in DP/HOLE Annulus Effective Viscosity in Annulus (µe): ⎛ 144V ⎞ ⎟⎟ µ e = 100 K ⎜⎜ ⎝ D2 − D1 ⎠ ⎛ 144 * 2.197 ⎞ µ e = 100 * 6.336⎜ ⎟ ⎝ 8 . 5 − 4 .5 ⎠
n−1
0.5413 −1
⎛ 2n + 1⎞ ⎜⎜ ⎟⎟ ⎝ 3n ⎠
n
⎛ 2 * 0.5413 + 1 ⎞ ⎜ ⎟ ⎝ 3 * 0.5413 ⎠
0.5413
= 97.64 cP
Reynolds Number in Annulus (NRe): NRe =
928 (D2 − D1 ) V ρ µe
=
928 (8.5 − 4.5) * 2.197 * 12.5 = 1,044 97.64
Pressure Drop in DP/HOLE Annulus NOTE: NRe < 2,100 Friction Factor in Annulus (f): f=
24 NRe
=
24 = 0 .02299 1,044
fV ρ ⎛ dP ⎞ = ⎜ ⎟ ⎝ dL ⎠ 25.81(D2 − D1 ) 2
0.02299 * 2.1972 * 12.5 psi = = 0.01343 25.81(8.5 − 4.5) ft
⎛ dP ⎞ ∆P = ⎜ ⎟ ∆L ⎝ dL ⎠
So,
= 0 . 01343 * 11,400
psi psi ∆Pdp/hole = 153.2
Pressure Drop Calcs. - SUMMARY PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES + ∆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 = 32 + 153 = 185 ∆PHYD = 0
PPUMP = 1,918 + 185 = 2,103 psi
2,103 psi
P = 0
"Friction" Pressures
"Friction" Pre ssure , psi
2,500 DRILLPIPE
2,000 1,500
DRILL COLLARS
1,000
BIT NOZZLES
500
ANNULUS
0 0
5,000
10,000
15,000
20,000
Cumulative Distance from Standpipe, ft
25,000
H ydrostatic Pre ssure , psi
Hydrostatic Pressures in the Wellbore 9,000 8,000
BHP
7,000 6,000 5,000
DRILLSTRING
ANNULUS
4,000 3,000 2,000 1,000 0 0
5,000
10,000
15,000
20,000
Cumulative Distance from Standpipe, ft
25,000
Pressures, psi
Pressures in the Wellbore 10,000 9,000 8,000 7,000 6,000 5,000 4,000 3,000 2,000 1,000 0
CIRCULATING
STATIC
0
5,000
10,000
15,000
20,000
Cumulative Distance from Standpipe, ft
25,000
Wellbore Pressure Profile 0 2,000
DRILLSTRING
Depth, ft
4,000 6,000
ANNULUS 8,000 10,000
(Static)
12,000
BIT 14,000 0
2,000
4,000
6,000
Pressure, psi
8,000
10,000
Pipe Flow - Laminar In the above example the flow down the drillpipe was turbulent. Under conditions of very high viscosity, the flow may very well be laminar. NOTE: if NRe < 2,100, then Friction Factor in Pipe (f): Then
f =
16 N Re
f V ρ 2
and
⎛ dP ⎞ ⎜ ⎟ = 25 .81 D ⎝ dL ⎠
n = 1.0
_ 2
dp fρ v = dL 25 .8 d
View more...
Comments
