1. Drill Dtring Course

Drilling Technology

By

Dr. Khaled Abdel Fattah

CONTENTSCONTENTS

1. Introduction

2. Drill String Design

3. Formation Pressure Prediction

4. Fracture Pressure Determination

5. Casing Setting Depth Selection

6. Casing Design

7. Cementing Design

8. Borehole Problems

INTRODUCTIONINTRODUCTION

Drill String Design

PURPOSE OF DRILL STRINGPURPOSE OF DRILL STRING

Provide Fluid Conduit from Rig to Bit

Impart Rotary Motion to the Bit

Provide Weight on Bit

Lower and Raise Bit in the Well

Allow Formation Evaluation & Testing

DRILL STRING COMPONENTSDRILL STRING COMPONENTSDrill PipeBottom Hole Assembly

– Drill Collars– Stabilizers– Jars– Reamers– Shock Subs– Bit

DRILL PIPE DRILL PIPE

Consists of pipe Body and Tool joint

Available in three Length Ranges

Range Length, ft

1 18-22

2 27-30

3 38-45

DRILL PIPE DESCRIPTIONDRILL PIPE DESCRIPTION Pipe OD

Nominal Weight

Pipe Grade

Type of Upset

Connection Thread

Premium Classification

DRILL PIPE GRADESDRILL PIPE GRADES

GradeYield Strength, psi

E75,000

X95,000

G105,000

S135,000

API SEAMLESS INTERNAL UPSET API SEAMLESS INTERNAL UPSET DRILL PIPE DRILL PIPE STRENGTHSTRENGTH

PIPE SIZEPIPE SIZE Pipe OD is Outside Diameter of Plain End pipe Available Sizes are :

– 2-3/8”, – 2-7/8”,– 3-1/2”, – 4”, – 4-1/2”,– 5”, – 5-1/2”,– 6-5/8”

DRILL PIPE WEIGHTDRILL PIPE WEIGHT

Nominal Weight of pipe per foot

including wt of regular API connection

Adjusted Weight is Actual Weight of pipe

and Connection

Depends on Grads and Connection Type

DRILL PIPE DRILL PIPE CLASSIFICATIONCLASSIFICATION

New No wear and has never

Premium Uniform wear and a minimum wall thickness of 80% of original wall

Class 2 Allows drill pipe with a minimum wall thickness of 65 % with all wear on one side so long as the cross-sectional area is the same as premium class, i.e. based on not more than 20 uniform wall reduction

Class 3 Allows drill pipe with a minimum wall thickness of 55% with all wear on one side

SINGLE DRILL PIPESINGLE DRILL PIPE

fWBfBHWHLfBcWcLMOPY

L

9.0

MOPBWLBWLLWBfHHfccf

0.9 × Yield Strength = Weight of DP + Weight of DC +

Weight of HWDP+MOP

DRILL COLLARSDRILL COLLARS

Large OD Small ID Great Stiffness Thirty- Foot Joints

FUNCTIONS OF DRILL FUNCTIONS OF DRILL COLLARSCOLLARS

Provide Weight on Bit

Stabilize Bit

Provide stiffness to BHA and

Minimize Directional Control Problems

COLLAR SIZE SELECTIONCOLLAR SIZE SELECTION

Minimum Collar OD =

2(Casing Coupling OD) – Bit OD

Select Largest Drill Collar

That can be fished out of hole

WEIGHT ON BIT CALCULATIONSWEIGHT ON BIT CALCULATIONSBUOYANCY FACTOR METHODBUOYANCY FACTOR METHOD

CoscWBF

SFWOBcL

WOB = Weight on Bit, IbSF = Safety Factor (1.1-1.15)BF = Buoyancy Factor Wc = Drill Collar Weight in Air, Ib/ft

= Maximum Hole Angle at BHA, degrees

HEAVY WEIGHT DRILL PIPEHEAVY WEIGHT DRILL PIPE

Reduces failures at transition zone

Reduces downhole torque and drag in

directional drilling

Reduces differential sticking

DRILL STRING DESIGNDRILL STRING DESIGN Drill Collar DesignDrill Collar Design

Bit weight (lbs) Air Weight (DRILL COLLAR) = _____________

B. F. x S. F.

Where: B.F. = Buoyancy Factor S.F. = Safety Factor

(i.e., 0.90 = 10% S.F.; 0.70 – 30% S.F.)

Correcting S.F. after Air Weight Design is complete

=100 - (desired bit weight) x (100)

(Air Weight Calculated) x (B.F.)

Example of Drill Collar DesignExample of Drill Collar DesignFind the correct drill collar Air Weight to pick

up. if desired bit weight is 70,000 Ibs.. and the Mud Weight is 9.1 Ib/gal. (68 pcf). The hole size is 171/2", and a tapered string with 15% S.F. is desired. Consider using;

11" x 3" (8970 Ib/single)9-3/4" x 3" (6900 Ib/single) 8 1/2" x 3" (5070 Ib/single)

70,000 lbsAir Weight = 0.86 x 0.85 = 95,800 lbs.

B.F. = ( 65.44 – 9.1) / 65.44 = 0.86*S.F. (15%) = 100 - 15 = 85% (0.85)Try (1):(3) 11" drill collar = 26,910 Ibs,(3) 9 3/4" drill collar = 20,700 Ibs,(3) 8 1/2" drill collar = 15,210 Ibs,Total = 62,820 Ibs,Required = 95,800 Ibs,Remaining = 32,980 Ibs,

Example of Drill Collar DesignExample of Drill Collar Design

Second, Try: (3) Additional 9-3/4" = 20,700 Ibs

(3) Add. 81/2" = 15,210 Ibs.

Total 35,910 Ibs.

So With:

(3) 11" drill collar = 26,910 Ibs

(6) 9 3/4" drill collar = 41,400 Ibs

(6) 81/2" drill collar = 30.420 |bs.

98,730 Ibs.


Actual Air Weight = 98.730 Ibs Calculated Air Weight = 95,800 Ibs.

Difference. = 2,930 Ibs.

Corrected S.F. = 100 - (70,000) x (100)

(98,730) x (0.86)

100 - 82.4 = 17.56% (New S.F.)


DRILL STRING DESIGNDRILL STRING DESIGNSINGLE-GRADE DRILL PIPESINGLE-GRADE DRILL PIPE

MOPBwLBwLLwB fHHfccf

0.9 × Yield Strength = Weight of DP + Weight of DC + Weight of HWDP + MOP

L = Length of the Drill Pipe, ft




Lc = Length of the Drill Collar, ft




LH = Length of the Heavy Weight Drill Pipe, ft




wH = Nominal weight of the H/Weight Drill Pipe, lb/ft




wc = Nominal weight of the Drill Collar, lb/ft




w = Nominal weight of the Drill Pipe, lb/ft




Bf = Buoyancy Factor, fraction

= (1 – m/steel)




MOP = Max Overpull on the drill string by the Drawworks, lb




f

fHHfcct

wB

BwLBwLMOPPL

9.0Pt = Yield Strength of pipe, lb

DRILL STRING DESIGNDRILL STRING DESIGNDUAL-GRADE DRILL PIPEDUAL-GRADE DRILL PIPE

f

fHHfccf

Bw

BwLBwLBwLMOPYL

2

112

9.0

L1 = Length of the Drill Pipe—Grade-1, ft

w1 = Nominal weight of the Drill Pipe—Grade-1, lb/ft

L2 = Length of the Drill Pipe—Grade-2, ft

w2 = Nominal weight of the Drill Pipe—Grade-2, lb/ft

MAXIMUM OVERPULLMAXIMUM OVERPULL(MOP)(MOP)

Maximum Overpull should not exceed 90% of Tensile Strength of Weakest

Drill Pipe

Pa = Pt x 0.9

Where:Pa = maximum allowable design load in tension, lb.Pt = theoretical tension load from hand book, Ib.

MAXIMUM OVERPULLMAXIMUM OVERPULL(MOP)(MOP)

M.O.P. = Pa / P

Where:P submerged load hanging below

the top joint of drill pipe, lb.

fdcdcdpdp BwLwLP )(

dp

dcdc

fdp

t

W

LW

BSFW

PL

**

9.0

COLLAPSE PRESSURECOLLAPSE PRESSURE

. .

Collapse RatingAllowable Collapse

S F

Collapse P = External P – Internal PCollapse P = External P – Internal P

BURST PRESSUREBURST PRESSURE

. .

Burst RatingAllowable Burst

S F

Burst P = Internal P – External PBurst P = Internal P – External P

Torsional StrengthTorsional Strength

The actual torque applied to the pipe during drilling is difficult to measure, but may be approximated by the following equation.

T = HP x 5,250 RPM

Where: T = torque delivered to drill pipe, ft.-lbs.HP = horse power used to produce rotation of pipeRPM = revolutions per minute

Examples of Drill Pipe DesignExamples of Drill Pipe DesignExample 1Anticipated T.D. 12,000 ft.Hole Size 12 1/4 inches.Expected Mud Wt. 13.4 Ib/gal. (100 pcf).Desired S.F. or Safety (Margin of over pull) 50, 000 Ibs. (collapse S.F. 1.125).Length and Size Drill Collars (3) 10" drill collar, I.D. = 2 13/16" (12) 9" drill collar, I.D. = 2 13/16" Total Length 450 ft. Total Air Wt. 92, 340 Ibs.HWDP (21 joints) 5" x 3" x 50 Ib/ft. Total Length 630 ft. Total Air Wt. 31, 500 Ibs.Drill Pipe 5", 19.5 Ib/ft., XH, Class-2. Adj. Wt. 20.99 Ib/ft. Grade "E" and /or Grade"95“ (21.09 Ib/ft.)Tensile Strength Grade "E" API Class 2 = 311,540 Ibs. Grade "95" API Class 2 = 394,600 Ibs.Buoyancy Factor 0.795 w/13.4 Ib/gal. mud.

Examples of Drill Pipe DesignExamples of Drill Pipe Design

Solution Length of drill pipe (5", 19.5 Grade "E") to go above Hevi-wt. and drill

collar's. Pt x 0.9 - M.O.P. – WcLc = Ldp

Wdp x Bf Wdp

Ldp = ( 311,540 x 0.9) - (50,000) _ 123,840 lbs. 20.99 x 0.795 20.99 = 13,806 - 5900 = 7,906 ft. (will use 7900 ft.)*123,840 - Total Air Weight of drill collar's and Heavy weight drill pipe

(2) Calculate Air Weight hanging below this top joint of Grade "E" drill pipe

P = (ldp x Wdp ) + (Lc x Wc ) (not corrected for buoyancy) P = [(7900 ft.) x (20.99) + (123,840)]= 165,821 Ibs. + 123,840 Ibs. = 289,661 Ibs.

Examples of Drill Pipe DesignExamples of Drill Pipe Design

Solution

(3) Accumulate Length of Drilling AssemblyTapered drill collar 450 ft. Hevi-Wt. 5" 630 ft. Grade "E" 5" 7900 ft. Accum. Total 8980 ft. Require 12,000 ft.Remaining Grade 3,020 ft.(4) Length of drill pipe No.2 5” Grade "95”Ldp No2 = ( 394,600 x 0.9) - (50,000) - 289,661 lbs.

(21.09 x 0.795) 21.0918,199 - 13.735 = 4465 ft. (will use only 3020 ft.) *•289,661 Ibs. = Total Air Wt.at DRILL COLLAR Hevi Wt.,

Grade "E".

Examples of Drill Pipe DesignExamples of Drill Pipe DesignSolution (5) Calculate Total Air Weight of stringP = {(3020 ft. x 21.09) + 289,661 Ibs.} = 63.692 + 289,661 = 353,353 Ibs.

(6) Corrected for BuoyancyWeight in 13.41 Ib/gal. mud = 353,353x 0.795= 280,916 Ibs.( 7) Summary of Design Calculation Weight in

Items Length 13.4 Ib/gal.mud

-----------------------------------------------------------------------------------------------------

DRILL COLLAR 10" and 9" tapered 450 ft. 73,410 Ibs. Hevi-Wt. drill pipe (5" - 50 Ib/ft.) 630 ft. 25,043 Ibs. DRILL PIPE Grade "E" (5". 19.5 - 20.99) 7,900 ft 131,828 Ibs. DRILL PIPE Grade '•95" (5". 19.5 - 21.09) 3,020 ft. 50,635 Ibs.

Totals 12,000 ft. 280,916lbs.

Examples of Drill Pipe DesignExamples of Drill Pipe DesignSolution (8) Check on Margin of OverpullGrade “ E “ (311,540 x 0.9) - (230,281) = 50,105 IbsGrade "95" (394,600 x 0.9) - (280,916) = 74,224 Ibs.(9) This design Could Be Used Down To 13, 445 f t.Note: Only 3020 ft. of Grade"95" was used and this allowed another

1466 ft. of Grade "95" for use.(10) AIIowable Collapse Pressure For 5", 19.5, Grade "E", API Class - 2 = 4760 psi. Btm 5" @10,920'Pac = ( 4760) = 4231 psi ( 1.125) (11) Net Collapse Pressure Pc = (10,920 ft )(13.4 Ib/gal.) = 7600psi 19.251 (12) Tension Safety Factor Grade “ E” = 280,386 = 1.22 230,281 Grade "95" = 355,140 = 1.26

280,916

Drill Pipe Design Using Two Drill Pipe Design Using Two Different Grades of PipeDifferent Grades of Pipe

Example:Anticipated Total Depth with this String...................... 1 5, 000 feet

Hole Size (9-5/8" Casing (@ 12,000 feet) ....................... 81/2 inches

Expected Mud Weight......................................... 1 3. 4 Ib/gal.

Desired S.F. of Safety Margin - Collapse - 1.125, M.O.P. . 50, 000 Ibs.

Length and Size of Drill Collars 61/2 " x 2 13/16 x 950 ft.

wt. Ib/ft. 91 Total Wt. (air) 86,450 lbs.

Drill Pipe; 5", 19.5 x H, Grade "E" 20.99 Ib/ft. API Class No. 2

5” , 19.5, x H, Grade "95" 21.09 Ib/ft. APl Class No.2

Tensile Strength; Grade "E". API Class No. 2................ 311, 540 Ibs.

Grade "95". API Class No. 2 …………………............... 394,600 Ibs.

Buoyancy Factor w/13.4 Ib/gal............................................... 0.795


Solution1. Length of drill pipe (5" 19.5 Ib/ft., Grade "E") to go above drill collar

Ldp = (311,540 x 0.9) - 50,000 _ 86,450 Ibs.

20.99 x 0.795 20.99

= 13,806 - 4119 = 9688 ft.

2. Calculate Air Weight Hanging Below This Top Joint of Grade "E" d.

p = (9688 x 20.99) + 86,450 Ibs. = 203,351 + 86,450 = 289,801 Ibs.

3. Accumulated Length of Drilling Assembly

DRILL COLLAR ........................................... 950 feet

DRILL PIPE Grade "E“ .............................. 9,688 feet

Total ........................................ 10,638 feet

Require ...................................... 15, 000 feet

Remaining Crade "95“ ....................... 4, 362 feet


Solution

4. Length of drill pipe No. 2 (Grade "95")

Ldp No2 = (394,600 x 0.9) - 50,000 - 289,801

21.09 x 0.795 21.09

= 18,199 - 13,741 = 4,458 ft. (will use 4,362 ft.)

5. Calculate Total Air Weight of String

P = [4362 (21.09) + 289,801] = 91.995 + 289,801 = 381,796 Ibs.

6. Corrected For Buoyancy

Weight in 13.4 Ib/gal. mud =

0.795 x 381,796 Ibs. = 303, 528 Ibs.


Solution7. Summary of Design Calculation

Item Length (ft.) Weight in13.4 Ib/gal. mud

--------------------------------------------------------------------------------------------------------------

DRILL COLLAR (61/2 x 2-13/16 x 9501) 950 68,728 Ibs.

DRILL PIPE (19.5 20.99 Grade "E") 9,688 1,6b4

DRILL PIPE (19.5 - 21.09 - Grade "95") 4,362 73,136

Total 15,000 ft. 303,528 Ibs.

8. Check on Margin of Overpull (M.O.P.)

5", (19.5 - 20.99) Grade "E" -- (311,540 x 0.9) - 230.392 = 50.000 Ibs.

5". (19.5 21.09) Grade “95" - - (394,600 x 0.9) - 303.528 = 51,612 Ibs.

9. Tension Safety Factor

Grade -E" = ( 311,540 x 0.9) = 1.22

230,392

Grade- “95” = (394,600 x 0.9) = 1.17

303,528

Many engineers feel that the design S.F. should be approximately 1.25 - 1.33

Drill Pipe Design Using Tapered Drill Pipe Design Using Tapered Drill Pipe StringDrill Pipe String

Example:Anticipated Total Depth.......................................... 18,000 feet

Hole Size (7" liner - 18,000 - 11, 500 ft.)........................ 6-1 / 8 inches

Expected Mud Weight (0.833 psi/ft.)..... ...................... 16. 0 Ib/gal.

Designed Safety Factor - M.O.P. ................................ 50,000 Ibs.

Length and Size of Drill Collars ................................. 950 ft.

Weight drill collar (Air) 4 3/4” x 2 x 950 ft.) 50 Ib/ft.. ................ 47, 500 Ibs.

Drill Pipe – 31/2. 13.30 - 14.06. Grade "E", API Class-2, XH

5", 19.50 - 20.99, Grade "E", API Class-2, XH

5", 19.50 - 21.09, Grade "95", API Class-2, XH

Tensile Strength

31/2", 13.30. Grade "E", API Class-2, 212,250 Ibs.

5", 19.50, Grade "E", API Class 2, 311,500 Ibs.

5", 19.50, Grade "95", API Class-2, 394,600 Ibs.

Buoyancy Factor w/16.0 Ib/gal. ................................. 0.756


Solution1. DRILL PIPE Design ( 31/2, 13.30, Grade "E") for Below Top of liner to

T.D. 6600 ft. required.)

Ldp = (212,250 x 0.9) - 50,000 - 47,500

14.06 x 0.756 14.06

= 13,267 - 3378 = 9889 ft. (will use 5650 feet) .

2. Air Weight of String Supported by Top Joints of 31/2" drill pipe

P1 = (5650 ft.x 14.06) + (950 ft. x 50 Ib/ft.)

= 79,439 + 47,500 = 126,939 Ibs. (31/2" drill pipe + 4-3/4" drill collar)

3. drill pipe Design for 5" 19.5. Grade "E" (above 31/2", 13.30)

Ldp No.2 = (311,500 x 0.9) - 50,000 - 126,939 = 8469 ft.

20.99 x 0.756 20.99

4. Air Weight of String Supported by Top Joint of 5", 19.5, "E"

P = (8469 x 20.99) + 126,939 = 177,764 + 126,939 =

= 304,703 Ibs. (5", 19.5, "E" +31/2 + 4-3/4drill collar)


Solution

5. drill pipe Design for 5", 19.5, Grade "95", (above 5", Grade "E")

Ldp No.3 = (394,600 x 0.9) - 50,000 - 304.703

21.09 x 0756 21.09

= 4690 feet (will use 2931 ft.)

6. Air Weight of String Supported by Top Joint of 5", 19.5, Grade ’95’

P3 = (2931 ft. x 21.09) + 304,703 - 61,815 + 304,703 = 366, 518 Ibs.

(complete string weight in air)

7. Buoyancy Weight of String

366,518 x 0.756 = 277,087 lbs.


Solution

8. Summary of Drill Strinq Design

Weight in 16.0

Item Length (ft.) Ib/gal. mud (Ibs.)

Drill collars (4-3/4 x 2 x 950) 950 35,910

Drill Pipe (31/2, 13.30, Grade "E") 5,650 60,056

Drill Pipe (5", 19.5, Grade "E") 8,469 134,390

Drill Pipe (5", 19.5. Grade "95") 2,931 46,732

Totals. ............................ 18,000 ft. 277,0881bs.


Solution9. Margin of Overpull Check

31/2. 13.30, Grade "E" = (212,250 x 0.9) - 95,966 = 95,059 Ibs.

5", 19.50, Grade "E" = (311,500 x 0.9) – 230,056 = 50,294 Ibs.

5", 19.50, Grade "95" = (394,600 x 0.9) - 277,088 = 78,052 Ibs

10. Torsional Yield

31/2", 13.30. Grade "E", API CIass No. 2 =11,710 ft. Ib.

5". 19.50, Grade "E", API Class No. 2 = 26,320 ft. Ib.

11. Collapse Pressure

31/2”, 13.30, Grade "E".............10,250 psi

Pc = 17,050 ( 16.0lb/gal.) = 14,170 psi

19.251

Therefore, this pipe must be protected against collapse and a D.S.T. may result in failure.

BOTTOM HOLE ASSEMBLYBOTTOM HOLE ASSEMBLY

Purpose:

– Drill usable hole economically

– Maximize penetration rate

– Minimize formation of dog legs

– Prolong service life of bit and drill pipe

BHA COMPONENTSBHA COMPONENTSHeavy weight drill pipe

Drill Collars

Stabilizers

Shock Absorbers

Jars

Reamers

Bit

FUNCTIONS OF STABILZERSFUNCTIONS OF STABILZERS

Centralize Drill Collars in hole and increase

Stiffness

Increase ability of Drill Collars to drill

smooth straight hole

Wipe wall of hole to ensure full gage

Spiral

Stabilizers

SHOCK ABSORBERSSHOCK ABSORBERSReduce or Eliminate vertical oscillations

Maintain uniform bit load

Increase bit life

Increase ROP

Reduce drill collar failures

JARRING DEVICESJARRING DEVICES

There are sloughing formations There are sensitive shales Mud system does not have good suspension

properties There is expensive equipment in BHA

Generate upward or downward load to free stuck pipe or release fish

Jars are needed when :

Hydraulic JarsHydraulic Jars

TYPES OF BOTTOM HOLE TYPES OF BOTTOM HOLE ASSEMBLIESASSEMBLIES

1. Drill Dtring Course

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