Drilling Bits
Types of bits
• Drag Bits
• Roller Cone Bits
• Diamond Bits
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
Bit Selection
Minimum Cost Per Unit length $/ft
Bit cost + rig cost X (tripping time + Drilling time) C /L =
The Roller Cone Bits
Two-cone bit (Milled tooth soft formation only)
Three cone bit (milled tooth, Tungsten carbide inserts)
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
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
Fluid flow through water courses in bit
Proper bottomhole cleaning is very
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
Design Factors
Angle formed by the axis of the Journal and the axis of the bit
JOURNAL
ANGLE
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
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
Sealed,
self-lubricated roller bit journal bearing
design details
INSERTS
SILVER PLATED BUSHING RADIAL SEAL
BALL BEARING
GREASE RESERVOIR CAP
BALL RETAINING PLUG
Roller
Cone
Bearings
Bearings
• Ball Bearings (point contact)
• Roller Bearings (line contact)
• Journal bearing (area contact)
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?
Grading of Dull Bits
How do bits wear out?
• Tooth wear or loss
• Worn bearings
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 estimated3/8 of the bearing life is gone
Grading of Dull Bits
How do bits fail?
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
Roller cone
bit wear
IADC ROLLER CONE
BIT CLASSIFICATION
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
IADC Classification
• 4-Character Design/Application Code
– First 3 Characters are NUMERIC– 4th Character is ALPHABETIC
Examples
637Y
medium-hard insert bit;
friction bearing with gage protection; conical inserts
135M
soft formation Milled tooth bit; roller bearings with gage protection; motor application
447X
soft formation insert bit; friction bearings
with gage protection; chisel inserts
Sequence
• Numeric Characters are defined:
– Series 1st
– Type 2nd
– Bearing & Gage 3rd
• Alphabetic Character defined:
– Features Available 4th135M
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 Bits The higher the series number,
the harder/more abrasive the rock
1
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
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
13
Features Available
• FOURTH CHARACTER
• Features Available (Optional)
• Sixteen (16) Alphabetic Characters
• Most Significant Feature Listed
(i.e. only one alphabetic character should be selected).
135
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 Milled Tooth • T - Two-Cone Bit • W - Enhanced C/S• X - Chisel Tooth Insert • Y - Conical Tooth Insert • Z - Other Shape Inserts
135
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.
Drag Bits
Drag bits drill by physically “plowing”
or “machining” cuttings from the
Natural
Diamond
bit
junk slot cuttings radial flow high ∆p across faceSoft
Formation
Diamond bit
Larger diamonds Fewer diamonds Pointed noseHard
Formation
Diamond bit
Smaller diamonds More diamonds Flatter noseNatural 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
• 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 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.
Side view of
diamond bit
PDC
bits
Courtesy Smith Bits
At about $10,000-150,000 apiece, PDC bits cost five to 15 times more than roller cone bits
Coring
bit
PDC +
natural
diamond
Bi-Center bit
Relative Costs of Bits
Diamond WC Insert Milled Bits 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
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,
PDC Bits
• Parameters for effective use
include
weight on bit
mud pressure
flow rate
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
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
PDC Bits
– 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
PDC Bits
• 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
PDC Bits
• 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
• 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 whirlPDC 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 formationsPDC Bits
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
Less than 1/3 of cutting element is gone
BT - Broken Cutter
More than 1/3 of cutting element is broken to
Grading of Worn PDC Bits – cont’d
LT - Lost Cutter
Bit is missing one or more cutters
LN - Lost Nozzle
Bit is missing one or more nozzles
Best Penetration Rate
• Approach A
• Achieved by removing cuttings efficiently
from below the bit • Maximize the
hydraulic power available at the bit
• Approach B
• Drilling fluid hits bottom of the hole with greatest force
• 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
Nozzle Velocity
0.95
to
equal
usually
t
coefficien
discharge
Nozzle
10
074
.
8
4=
×
∆
=
− d b d nC
p
C
v
ρ
Bit Pressure Drop
2 2 2 510
33
.
8
t d bA
C
q
p
ρ
−×
=
∆
Hydraulic Power
HP
P
pq
P
H H8
.
272
1714
400
1169
1714
=
×
=
∆
=
Hydraulic Impact Force
lbs
F
F
p
q
C
F
j j b d j5
.
820
1169
12
400
95
.
01823
.
0
01823
.
0
=
×
×
×
=
∆
=
ρ
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
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 highenough 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 themaximum allowable pressure rating of the pump and the surface equipment.
Maximum Nozzle Velocity
Nozzle Velocity
i.e.
so the bit pressure drop should be maximized in order to obtain the maximum nozzle velocity
ρ
4 b d n10
*
074
.
8
P
C
v
=
∆
− b nP
v
∝
∆
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.
pressure. surface allowable maximum the and rate n circulatio minimum the at satisfied, are above 2 & 1 when maximized is vn ∴Maximum Bit Hydraulic Horsepower
The hydraulic horsepower at the bit is
maximized when is maximized.
(
∆
p
bitq)
d pump
bit
p
p
p
=
∆
−
∆
∆
where may be called the
parasitic
pressure
loss in the system (friction).
dp
∆
bit d pumpp
p
p
=
∆
+
∆
∆
Maximum Bit Hydraulic Horsepower
.
turbulent
is
flow
the
if
cq
p
p
p
p
p
p
d=
∆
s+
∆
dp+
∆
dc+
∆
dca+
∆
dpa=
1.75∆
In general, where
∆
p
d=
cq
m0
≤
m
≤
2
Maximum Bit Hydraulic Horsepower
0
)
1
(
p
when
0
1714
1714
pump 1
=
+
−
∆
=
∴
−
∆
=
∆
=
∴
+ m Hbit m pump bit Hbitq
m
c
dq
dP
cq
q
p
q
p
P
d pump bitp
p
p
=
∆
−
∆
∆
m dcq
p
=
∆
Maximum Bit Hydraulic Horsepower
when
maximum
is
1
1
p
when
.,
.
)
1
(
p
when
.,
.
d pump Hbit pump dP
p
m
e
i
p
m
e
i
∴
∆
⎟
⎠
⎞
⎜
⎝
⎛
+
=
∆
∆
+
=
∆
pump dp
m
p
⎟
∆
⎠
⎞
⎜
⎝
⎛
+
=
∆
1
1
0
)
1
(
p
pump−
+
=
∆
mq
m
c
Maximum Bit Hydraulic Horsepower
Examples
-In turbulent flow, m = 1.75
pump bit pump pump dp
of
%
64
p
p
of
36%
%
100
*
p
1
75
.
1
1
p
∆
=
∆
∴
∆
=
∆
⎟
⎠
⎞
⎜
⎝
⎛
+
=
∆
∴
p d p 1 m 1 p ∆ + = ∆In laminar flow, for Newtonian fluids, m = 1
pump b pump pump dp
of
%
50
p
p
of
50%
%
100
*
p
1
1
1
p
∆
=
∆
∴
∆
=
∆
⎟
⎠
⎞
⎜
⎝
⎛
+
=
∆
∴
Maximum Bit Hydraulic Horsepower
Examples - cont’d
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
Maximum Jet Impact Force
The jet impact force is given by Eq. 4.37:
)
(
c
0.01823
01823
.
0
d pump d bit d jp
p
q
p
q
c
F
∆
−
∆
=
∆
=
ρ
ρ
Maximum Jet Impact Force
But parasitic pressure drop,
2 2
01823
.
0
+
−
∆
=
∴
=
∆
m d p d j m dq
c
q
p
c
F
cq
p
ρ
ρ
)
(
c
0.01823
d pump d jq
p
p
F
=
ρ
∆
−
∆
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,
p d
p
2
m
2
p
∆
+
=
∆
Maximum Jet
Impact Force
Examples
-p b p dp
of
%
47
p
and
p
of
%
53
p
1.75,
m
if
,
∆
=
∆
∆
=
∆
=
Thus
p b p dp
of
33%
p
and
p
of
%
67
p
1.00
m
if
,
∆
=
∆
∆
=
∆
=
Also
p d p 2 m 2 p ∆ + = ∆Nozzle Size Selection
-1. Show opt. hydraulic path
2. Plot ∆p
dvs q
3. From Plot, determine
optimum q and ∆p
d4. Calculate
5. Calculate
Total Nozzle Area:
(TFA)
6. Calculate Nozzle Diameter
d pump bit
p
p
p
=
∆
−
∆
∆
opt b d opt opt tp
C
q
A
)
(
10
*
311
.
8
)
(
2 2 5∆
=
−ρ
With 3 nozzles:
π
3
A
4
d
tot N=
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, P
pump= 2,800 psi
At 247 gal/min, P
pump= 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
Example 4.31 - 1(a), 485 gpm
Calculate pressure drop through bit nozzles:
2 2 2 5
10
*
311
.
8
:
)
34
.
4
.(
t d b
A
c
q
p
Eq
ρ
−=
∆
psi 906 1,894 -2,800 loss pressure parasitic psi 1,894 32 12 4 3 (0.95) ) 485 )( 6 . 9 )( 8.311(10 p 2 2 2 2 -5 b = = ∴ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π = ∆Example 4.31 - 1(b), 247 gpm
psi
p
b491
32
12
4
3
)
95
.
0
(
)
247
)(
6
.
9
)(
10
(
311
.
8
2 2 2 2 5=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
=
∆
−π
psi
409
491
-900
loss
pressure
parasitic
=
=
∴
Plot these two
points in Fig. 4.36
(q
1, p
1) = (485, 906)
Example 4.31 - cont’d
2. For optimum hydraulics:
gal/min
650
000
,
3
)
91
.
0
)(
250
,
1
(
714
,
1
714
,
1
q
max max=
=
=
P
E
P
Hp1,
Interval
)
a
(
gal/min 225 q psi 875 , 1 ) 000 , 3 ( 2 2 . 1 2 2 2 p min max d = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ∆ P m2,
Interval
(b)
3,
Interval
(c)
3 2 1Example 4.31
3. From graph, optimum point is at
) ( 10 * 311 . 8 ) ( 2 2 5 opt b d opt opt t p C q A ∆ = ∴ − ρ ) 700 , 1 ( * ) 95 . 0 ( ) 650 ( * 6 . 9 * 10 * 8.311 2 2 -5 =
( )
d
N opt ndsin
2 opt0.47
in
14
32
A
=
⇒
=
psi
p
psi
gal
q
,
p
1
,
300
b1
,
700
min
650
∆
d=
⇒
∆
=
=
psi
p
psi
gal
q
,
p
1
,
300
b1
,
700
min
650
∆
d=
⇒
∆
=
=
Example 4.32
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
Mud Program
Mud Plastic Yield
Depth Density Viscosity Point
(ft) (lbm/gal) (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
gal/min.
681
423
,
3
)
85
.
0
)(
600
,
1
(
714
,
1
p
E
P
714
,
1
q
max Hp max=
=
=
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,
(
)
psia
1,245
423
,
3
1
75
.
1
1
1
1
max=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
∆
p
m
p
dSolution
Interval 3
For a minimum annular velocity of
120 ft/min opposite the drillpipe,
(
)
gal/min
395
60
120
5
.
4
05
.
10
448
.
2
2 2 min=
⎟
⎠
⎞
⎜
⎝
⎛
−
=
q
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
Table
5,000 38 490 320 20 20 888
6,000 38 601 320 20 25 1,004
7,000 38 713 320 20 29 1,120
8,000 51 1,116 433 28 75* 1,703
9,000 57 1,407 482 27* 111* 2,084
* Laminar flow pattern indicated byHedstrom number criteria.
d dpa dca dc dp s
p
p
p
p
p
p
Depth
∆
∆
∆
∆
∆
∆
Table
The proper pump operating conditions
and nozzle areas, are as follows:
5,000 600 1,245 2,178 0.380
6,000 570 1,245 2,178 0.361
7,000 533 1,245 2,178 0.338
8,000 420 1,245 2,178 0.299
9,000 395 1,370 2,053 0.302
in.)
(sq
(psi)
(psi)
(gal/min)
)
ft
(
(5)A
p
(4)
p
(3)
Rate
(2)Flow
Depth
)
l
(
∆
d∆
b tTable
The first three columns were read directly
from Fig. 4.37.
(depth, flow rate and ∆p
d)
Col. 4
(∆p
b)
was obtained by subtracting
shown in Col.3 from the maximum pump
pressure of
3,423 psi
.
Col.5
(A
tot)
was obtained using Eq. 4.85
d
p
∆
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.
swab surgeP
P
,
v
fixed
For
pipe=
Figure 4.40B
- Velocity profile for laminar flow pattern when closed
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
Pump volumetric output and circulating pressure
Q= K.L(2D2-d2).spm.η
v/100 for double acting pump Q= K.L.D2.spm.η
v/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)
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 = --- ft/min IDp2
Annular Average Velocity
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 Assuming the total string is DP;
24.51 x Q
Annular Velocity Vann = --- ft/min Dh2 - OD
Nozzle Jet Velocity
V
n= 0.321 (Q/A)
ft/s
Fluid Flow
Laminar Flow Re < 2000
Turbulent Flow Re > 4000 Re = 15.46 ρ DV / µ
Newtonian fluid
Non Newtonian fluid
Bingham Plastic Fluid Power-Law Fluid
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
Vc
D YP
D
=
97
+
97
+
8 2
2 2pv
pv
.
ρ
ρ
ft/min V > Vc Turbulent flow V < Vc Laminar flowPressure Loss in the System
Pressure losses in the surface equipment Pressure loses in the drilling string
Pressure drop in the surface
equipment
Pressure Drop in Drillpipe
P2 = c . QN 8.91 x 10-5 ρN-1 PV2-N . L c = ---IDpN+3 8.91 x 10-5 ρN-1 QN PV2-N . L P2 = ---IDpN+3 P2 = f ρ V2 L / 25.8 dPressure Drop in annulus
P3 = c . QN c = 8.91 x 10-5 ρN-1 PV2-N L / (D h - ODp)3 (Dh + ODp)N+3 8.91 x 10-5 ρN-1 QN PV2-N . L P3 = ---(Dh - ODp)3 (D h + ODp)N+3 P3 = f ρ V2 L / 21.1 (D h - ODp)Pressure drop across the bit
Pb = Pstandpipe - (P1+P2+P3) ρ Q2 Pb = ---12,032 Cn2 A T2 Cn = Nozzle Coefficient (~ 0.95)Nozzle Velocity Vn ft/s
V
n
P
b
= 33 36
.
Best Penetration Rate
• Approach A
• Achieved by removing cuttings efficiently
from below the bit • Maximize the
hydraulic power available at the bit
• Approach B
• Drilling fluid hits bottom of the hole with greatest force
• 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
Max. Hydraulic Power at the bit
Pb . Q / 1714 hp
Jet Impact Force below the bit
IF = Q/58 (ρ Pb)0.5 Max IF when P b = [N/(N+2)] Psp 61.6 x 10-3 ρ Q2 / A T Pb = (Psp - PCS) Pcs = c QN HHPb = (Psp Q - c QN+1 )/1714 Differentiate wrt Q = 0 Pb = (N/N+1) Psp
Nozzle Selection
A
T= 0.0096 Q (ρ /P
b)
0.5= .32 Q/Vn
d
n= 32 (4 A
T/3π)
0.5Total 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
Types of flow
Laminar
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
We often assume that fluid flow is
turbulent if N
re> 2100
cp.
fluid,
of
viscosity
µ
in
I.D.,
pipe
d
ft/s
velocity,
fluid
avg.
v
lbm/gal
density,
fluid
ρ
where
_=
=
=
=
µ
d
v
ρ
928
N
_ Re=
Turbulent Flow -Newtonian Fluid 25 . 1 25 . 0 75 . 1 _ 75 . 0 f d 1800 v dL dp = ρ µ Turbulent Flow -Bingham Plastic Fluid
25 . 1 25 . 0 p 75 . 1 _ 75 . 0 f d 1800 v dL dp ρ µ =
(
)
1.25 1 2 25 . 0 p 75 . 1 _ 75 . 0 f d d 396 , 1 v dL dp − µ ρ =(
)
1.25 1 2 25 . 0 75 . 1 _ 75 . 0 f d d 396 , 1 v dL dp − µ ρ = In Annulus In PipeAPI Power Law Model
K = consistency index n = flow behaviour index
SHEAR STRESSτ
psi
τ = K γ
nSHEAR RATE, γ , sec-1
0
Rotating Sleeve Viscometer
VISCOMETER RPM 3 100 300 600 (RPM * 1.703) SHEAR RATE sec -1 5.11 170.3 511 1022 BOB SLEEVE ANNULUS DRILL STRINGPressure 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:
• R
3= 3
(at 3 RPM)• R
100= 20
(at 100 RPM)• R
300= 39
(at 300 RPM)• R
600= 65
(at 600 RPM)PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES + ∆PDC/ANN + ∆PDP/ANN + ∆PHYD Q = 280 gal/min
ρ
= 12.5 lb/galPressure Drop
Calculations
PPUMPPower-Law Constant (n):
Pressure Drop In Drill Pipe
Fluid Consistency Index (K):
Average Bulk Velocity in Pipe (V):
OD = 4.5 in ID = 3.78 in L = 11,400 ft 737 . 0 39 65 log 32 . 3 R R log 32 . 3 n 300 600 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 737 . 0 600 2.017 sec 022 , 1 65 * 11 . 5 022 , 1 11 . 5 cm dyne R K n n = = = sec ft 00 . 8 78 . 3 280 * 408 . 0 D Q 408 . 0 V = 2 = 2 =
Effective Viscosity in Pipe (µe):
Pressure Drop In Drill Pipe
Reynolds Number in Pipe (NRe):
OD = 4.5 in ID = 3.78 in L = 11,400 ft n 1 n e n 4 1 n 3 D V 96 K 100 ⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = µ − cP 53 737 . 0 * 4 1 737 . 0 * 3 78 . 3 8 * 96 017 . 2 * 100 737 . 0 1 737 . 0 e ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = µ − 616 , 6 53 5 . 12 * 00 . 8 * 78 . 3 * 928 V D 928 N e Re µ = = ρ =
NOTE: NRe > 2,100, so
Friction Factor in Pipe (f):
Pressure Drop In Drill Pipe
OD = 4.5 in ID = 3.78 inL = 11,400 ft So, b Re N a f = 0759 . 0 50 93 . 3 737 . 0 log 50 93 . 3 n log a = + = + = 2690 . 0 7 737 . 0 log 75 . 1 7 n log 75 . 1 b = − = − = 007126 . 0 616 , 6 0759 . 0 N a f b 0.2690 Re = = =
Friction Pressure Gradient (dP/dL) :
Pressure Drop In Drill Pipe
OD = 4.5 in ID = 3.78 inL = 11,400 ft
Friction Pressure Drop in Drill Pipe :
400 , 11 * 05837 . 0 L dL dP P ⎟ ∆ = ⎠ ⎞ ⎜ ⎝ ⎛ = ∆
∆P
dp= 665 psi
ft psi 05837 . 0 78 . 3 * 81 . 25 5 . 12 * 8 * 007126 . 0 D 81 . 25 V f dL dP = 2ρ = 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛Power-Law Constant (n):
Pressure Drop In Drill Collars
Fluid Consistency Index (K):
Average Bulk Velocity inside Drill Collars (V):
OD = 6.5 in ID = 2.5 in L = 600 ft 737 . 0 39 65 log 32 . 3 R R log 32 . 3 n 300 600 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 n 737 . 0 n 600 cm sec dyne 017 . 2 022 , 1 65 * 11 . 5 022 , 1 R 11 . 5 K = = = sec ft 28 . 18 5 . 2 280 * 408 . 0 D Q 408 . 0 V = 2 = 2 =
Effective Viscosity in Collars(µe):
Reynolds Number in Collars (NRe):
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
n 1 n e n 4 1 n 3 D V 96 K 100 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = µ − cP 21 . 38 737 . 0 * 4 1 737 . 0 * 3 5 . 2 28 . 18 * 96 017 . 2 * 100 737 . 0 1 737 . 0 e ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = µ − 870 , 13 21 . 38 5 . 12 * 28 . 18 * 5 . 2 * 928 V D 928 N e Re µ = = ρ =
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
NOTE: NRe > 2,100, so Friction Factor in DC (f): b Re N a f = So, 0759 . 0 50 93 . 3 737 . 0 log 50 93 . 3 n log a = + = + = 2690 . 0 7 737 . 0 log 75 . 1 7 n log 75 . 1 b = − = − = 005840 . 0 870 , 13 0759 . 0 N a f b 0.2690 Re = = =
Friction Pressure Gradient (dP/dL) :
Friction Pressure Drop in Drill Collars :
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
ft psi 3780 . 0 5 . 2 * 81 . 25 5 . 12 * 28 . 18 * 005840 . 0 D 81 . 25 V f dL dP = 2ρ = 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 600 * 3780 . 0 L dL dP P ⎟ ∆ = ⎠ ⎞ ⎜ ⎝ ⎛ = ∆
∆P
dc= 227 psi
Pressure Drop across Nozzles
DN1 = 11 32nds (in) DN2 = 11 32nds (in) DN3 = 12 32nds (in)(
2 2 2)
2 2 12 11 11 280 * 5 . 12 * 156 P + + = ∆∆P
Nozzles= 1,026 psi
(
)
2 2 3 N 2 2 N 2 1 N 2 D D D Q 156 P + + ρ = ∆Pressure Drop
in DC/HOLE
Annulus
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft Q = 280 gal/minρ
= 12.5 lb/gal 8.5 inPower-Law Constant (n):
Fluid Consistency Index (K):
Average Bulk Velocity in DC/HOLE Annulus (V):
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
Pressure Drop
in DC/HOLE Annulus
5413 . 0 3 20 log 657 . 0 R R log 657 . 0 n 3 100 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 n 5413 . 0 n 100 cm sec dyne 336 . 6 2 . 170 20 * 11 . 5 2 . 170 R 11 . 5 K = = = sec ft 808 . 3 5 . 6 5 . 8 280 * 408 . 0 D D Q 408 . 0 V 2 2 2 1 2 2 = − = − =Effective Viscosity in Annulus (µe):
Reynolds Number in Annulus (NRe):
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft cP 20 . 55 5413 . 0 * 3 1 5413 . 0 * 2 5 . 6 5 . 8 808 . 3 * 144 336 . 6 * 100 5413 . 0 1 5413 . 0 e ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = µ −
(
)
(
)
600 , 1 20 . 55 5 . 12 * 808 . 3 * 5 . 6 5 . 8 928 V D D 928 N e 1 2 Re = − = µ ρ − = n 1 n 1 2 e n 3 1 n 2 D D V 144 K 100 ⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = µ −Pressure Drop
in DC/HOLE Annulus
So,
DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft
NOTE: NRe < 2,100
Friction Factor in Annulus (f):
01500 . 0 600 , 1 24 N 24 f Re = = = ( ) ( ) ft psi 05266 . 0 5 . 6 5 . 8 81 . 25 5 . 12 * 808 . 3 * 01500 . 0 D D 81 . 25 V f dL dP 2 1 2 2 = − = − ρ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 600 * 05266 . 0 L dL dP P ⎟ ∆ = ⎠ ⎞ ⎜ ⎝ ⎛ = ∆
∆P
dc/hole= 31.6 psi
Pressure Drop
in DC/HOLE Annulus
q = 280 gal/min
ρ
= 12.5 lb/galPressure Drop
in DP/HOLE Annulus
DHOLE = 8.5 in ODDP = 4.5 in L = 11,400 ftPower-Law Constant (n):
Fluid Consistency Index (K):
Average Bulk Velocity in Annulus (Va):
Pressure Drop
in DP/HOLE Annulus
DHOLE = 8.5 in ODDP = 4.5 in L = 11,400 ft 5413 . 0 3 20 log 657 . 0 R R log 657 . 0 n 3 100 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 n 5413 . 0 n 100 cm sec dyne 336 . 6 2 . 170 20 * 11 . 5 2 . 170 R 11 . 5 K = = = sec ft 197 . 2 5 . 4 5 . 8 280 * 408 . 0 D D Q 408 . 0 V 2 2 2 1 2 2 = − = − =Effective Viscosity in Annulus (µe):
Reynolds Number in Annulus (NRe):
Pressure Drop
in DP/HOLE Annulus
n 1 n 1 2 e n 3 1 n 2 D D V 144 K 100 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = µ − cP 64 . 97 5413 . 0 * 3 1 5413 . 0 * 2 5 . 4 5 . 8 197 . 2 * 144 336 . 6 * 100 5413 . 0 1 5413 . 0 e ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = µ −(
)
(
)
044 , 1 64 . 97 5 . 12 * 197 . 2 * 5 . 4 5 . 8 928 V D D 928 N e 1 2 Re = − = µ ρ − =So, psi
Pressure Drop
in DP/HOLE Annulus
NOTE: NRe < 2,100
Friction Factor in Annulus (f):
02299 . 0 044 , 1 24 N 24 f Re = = = ( ) ( ) ft psi 01343 . 0 5 . 4 5 . 8 81 . 25 5 . 12 * 197 . 2 * 02299 . 0 D D 81 . 25 V f dL dP 2 1 2 2 = − = − ρ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 400 , 11 * 01343 . 0 L dL dP P ⎟ ∆ = ⎠ ⎞ ⎜ ⎝ ⎛ = ∆
∆P
dp/hole= 153.2 psi
Pressure Drop Calcs.
SUMMARY
-PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES + ∆PDC/ANN + ∆PDP/ANN + ∆PHYD PPUMP = 665 + 227 + 1,026
+ 32 + 153 + 0
P
PUMP= 1,918 + 185
= 2,103 psi
∆PHYD = 0
PPUMP = ∆PDS + ∆PANN + ∆PHYD
∆PDS = ∆PDP + ∆PDC + ∆PBIT NOZZLES
= 665 + 227 + 1,026 = 1,918 psi
∆PANN = ∆PDC/ANN + ∆PDP/ANN = 32 + 153 = 185
2,103 psi
P = 0
"Friction" Pressures 0 500 1,000 1,500 2,000 2,500 0 5,000 10,000 15,000 20,000 25,000
Cumulative Distance from Standpipe, ft
"F ri ct io n " P re ssu re , p s i DRILLPIPE DRILL COLLARS BIT NOZZLES ANNULUS
Hydrostatic Pressures in the Wellbore 0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 0 5,000 10,000 15,000 20,000 25,000
Cumulative Distance from Standpipe, ft
H y dr os ta ti c P re s s u re , ps i BHP DRILLSTRING ANNULUS
Pressures in the Wellbore 0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 0 5,000 10,000 15,000 20,000 25,000 Cumulative Distance from Standpipe, ft
Pr e s s u re s , p s i STATIC CIRCULATING
Wellbore Pressure Profile 0 2,000 4,000 6,000 8,000 10,000 12,000 14,000 0 2,000 4,000 6,000 8,000 10,000 Pressure, psi D e pt h, f t DRILLSTRING ANNULUS (Static) BIT