Bits Nozzles

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)

ƒ Long teeth for soft formations

ƒ Shorter teeth for harder formations

ƒ Cone off-set in soft-formation bit results in scraping gouging action

ƒ Self-sharpening teeth by using hardfacing on one side

ƒ 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 estimated

3/8 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

ƒ ‘The IADC Roller Bit Classification System’ ƒ 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 4th

135M

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:

ƒ

Steel

ƒ

Tungsten carbide

ƒ

Natural diamonds

ƒ

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 face

Soft

Formation

Diamond bit

ƒ Larger diamonds ƒ Fewer diamonds ƒ Pointed nose

Hard

Formation

Diamond bit

ƒ Smaller diamonds ƒ More diamonds ƒ Flatter nose

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

• 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.

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

ƒ

weight on bit

ƒ

mud pressure

ƒ

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

ƒ U.S Companies sell > 4,000 diamond drill bits/year

ƒ Diamond bit Market is about $200 million/year

ƒ 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,

ƒPDC 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 whirl

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

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 P_b

Nozzle Velocity

0.95

to

equal

usually

t

coefficien

discharge

Nozzle

10

074

.

8

4

=

×

∆

=

₋ d b d n

C

p

C

v

ρ

Bit Pressure Drop

2 2 2 5

10

33

.

8

t d b

A

C

q

p

ρ

−

×

=

∆

Hydraulic Power

HP

P

pq

P

H H

8

.

272

1714

400

1169

1714

=

×

=

∆

=

Hydraulic Impact Force

lbs

F

F

p

q

C

F

j j b d j

5

.

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 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.

so the bit pressure drop should be maximized in order to obtain the maximum nozzle velocity

ρ

4 b d n

10

*

074

.

8

P

C

v

=

∆

₋ b n

P

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 v_n ∴

Maximum Bit Hydraulic Horsepower

The hydraulic horsepower at the bit is

maximized when is maximized.

(

∆

p

_bit

q)

d pump

bit

p

p

p

=

∆

−

∆

∆

where may be called the

parasitic

pressure

loss in the system (friction).

d

p

∆

bit d pump

p

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

m

0

≤

m

≤

2

Maximum Bit Hydraulic Horsepower

0

)

1

(

p

when

0

1714

1714

pump 1

=

+

−

∆

=

∴

−

∆

=

∆

=

∴

+ m Hbit m pump bit Hbit

q

m

c

dq

dP

cq

q

p

q

p

P

d pump bit

p

p

p

=

∆

−

∆

∆

m d

cq

p

=

∆

Maximum Bit Hydraulic Horsepower

when

maximum

is

1

1

p

when

.,

.

)

1

(

p

when

.,

.

d pump Hbit pump d

P

p

m

e

i

p

m

e

i

∴

∆

⎟

⎠

⎞

⎜

⎝

⎛

+

=

∆

∆

+

=

∆

pump d

p

m

p

⎟

∆

⎠

⎞

⎜

⎝

⎛

+

=

∆

1

1

0

)

1

(

p

_pump

−

+

=

∆

m

q

m

c

Maximum Bit Hydraulic Horsepower

Examples

-In turbulent flow, m = 1.75

pump bit pump pump d

p

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 d

p

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 j

p

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 d

q

c

q

p

c

F

cq

p

ρ

ρ

)

(

c

0.01823

_{d pump d} j

q

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 d

p

of

%

47

p

and

p

of

%

53

p

1.75,

m

if

,

∆

=

∆

∆

=

∆

=

Thus

p b p d

p

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

_d

vs q

3. From Plot, determine

optimum q and ∆p

_d

4. Calculate

5. Calculate

Total Nozzle Area:

(TFA)

6. Calculate Nozzle Diameter

d pump bit

p

p

p

=

∆

−

∆

∆

opt b d opt opt t

p

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

_b

491

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

_Hp

1,

Interval

)

a

(

gal/min 225 q psi 875 , 1 ) 000 , 3 ( 2 2 . 1 2 2 2 p min max d = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ∆ P m

2,

Interval

(b)

3,

Interval

(c)

3 2 1

Example 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} nds

in

2 opt

0.47

in

14

32

A

=

⇒

=

psi

p

psi

gal

q

,

p

1

,

300

_b

1

,

700

min

650

∆

_d

=

⇒

∆

=

=

psi

p

psi

gal

q

,

p

1

,

300

_b

1

,

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

_d

Solution

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 by

Hedstrom 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 t}

Table

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 surge

P

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 P_t 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 V_dp = --- ft/min ID_p2

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 V_ann = --- ft/min D_h2 _{- 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 2

pv

pv

.

ρ

ρ

ft/min V > Vc Turbulent flow V < Vc Laminar flow

Pressure 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 = ---ID_pN+3 8.91 x 10-5 ρN-1 _QN _PV2-N _{. L} P₂ = ---ID_pN+3 P₂ = f ρ V2 _{L / 25.8 d}

Pressure 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 = ---(D_h - OD_p)3 _(D h + ODp)N+3 P₃ = f ρ V2 _{L / 21.1 (D} h - ODp)

Pressure drop across the bit

P_b = P_standpipe - (P₁+P₂+P₃) ρ Q2 P_b = ---12,032 C_n2 _A T2 C_n = 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 P_b

Max. Hydraulic Power at the bit

P_b . Q / 1714 hp

Jet Impact Force below the bit

IF = Q/58 (ρ P_b)0.5 _{Max IF when P} b = [N/(N+2)] Psp 61.6 x 10-3 ρ Q2 _{/ A} T P_b = (P_sp - P_CS) Pcs = c QN HHP_b = (P_sp Q - c QN+1 _{)/1714 Differentiate wrt Q = 0} P_b = (N/N+1) P_sp

Nozzle Selection

A

_T

= 0.0096 Q (ρ /P

_b

)

0.5

_{= .32 Q/Vn}

d

_n

= 32 (4 A

_T

/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

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 Pipe

API Power Law Model

K = consistency index n = flow behaviour index

SHEAR STRESS_τ

psi

τ = K γ

n

SHEAR 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 STRING

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:

• R

₃

= 3

(at 3 RPM)

• R

₁₀₀

= 20

(at 100 RPM)

• R

₃₀₀

= 39

(at 300 RPM)

• R

₆₀₀

= 65

(at 600 RPM)

P_PUMP = ∆P_DP + ∆P_DC + ∆P_{BIT NOZZLES} + ∆P_DC/ANN + ∆P_DP/ANN + ∆P_HYD Q = 280 gal/min

ρ

= 12.5 lb/gal

Pressure Drop

Calculations

P_PUMP

Power-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 (µ_e):

Pressure Drop In Drill Pipe

Reynolds Number in Pipe (N_Re):

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: N_Re > 2,100, so

Friction Factor in Pipe (f):

Pressure Drop In Drill Pipe

OD = 4.5 in _{ID = 3.78 in}

L = 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 in}

L = 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(µ_e):

Reynolds Number in Collars (N_Re):

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: N_Re > 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

D_N1 = 11 32nds (in) D_N2 = 11 32nds (in) D_N3 = 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

D_HOLE = 8.5 in OD_DC = 6.5 in L = 600 ft Q = 280 gal/min

ρ

= 12.5 lb/gal _{8.5 in}

Power-Law Constant (n):

Fluid Consistency Index (K):

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

D_HOLE = 8.5 in OD_DC = 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 (µ_e):

Reynolds Number in Annulus (N_Re):

D_HOLE = 8.5 in OD_DC = 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,

D_HOLE = 8.5 in OD_DC = 6.5 in L = 600 ft

NOTE: N_Re < 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/gal

Pressure Drop

in DP/HOLE Annulus

D_HOLE = 8.5 in OD_DP = 4.5 in L = 11,400 ft

Power-Law Constant (n):

Fluid Consistency Index (K):

Average Bulk Velocity in Annulus (V_a):

Pressure Drop

in DP/HOLE Annulus

D_HOLE = 8.5 in OD_DP = 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 (µ_e):

Reynolds Number in Annulus (N_Re):

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: N_Re < 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

-P_PUMP = ∆P_DP + ∆P_DC + ∆P_{BIT NOZZLES} + ∆P_DC/ANN + ∆P_DP/ANN + ∆P_HYD P_PUMP = 665 + 227 + 1,026

+ 32 + 153 + 0

P

_PUMP

= 1,918 + 185

= 2,103 psi

∆P_HYD = 0

P_PUMP = ∆P_DS + ∆P_ANN + ∆P_HYD

∆P_DS = ∆P_DP + ∆P_DC + ∆P_{BIT NOZZLES}

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

∆P_ANN = ∆P_DC/ANN + ∆P_DP/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