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PETE 628 Lesson 6

Build Curve Design

Petroleum Engineering Department Texas A&M University

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Outline  Dogleg severity  Single radius build curve -

- No tangent section  Complex tangent build curve -

- Unequal build rates)  The “Ideal” build curve  Tangent section inclination...

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Outline - cont’d   Casing and drilling fluid programs   Positive displacement mud motors

(PDM’s)   Bit side forces - change in trajectory

READ: to page 46 in the TEXTBOOK

HW #3 - Max. Tool Length - due 02-10-04

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Dogleg Angle

β

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Dogleg Angle  Total (3D) change in hole angle

between two survey points in a borehole:

β  = dogleg angle

Ref. API Bulletin D20

21221 IsinIsin2Asin

2Isinsin2 ⎟

⎠

⎞⎜⎝

⎛ Δ+⎟⎠

⎞⎜⎝

⎛ Δ=β −

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Dogleg Angle...

 Example: Point 1 Point 2 Change I1 = 20o I2 = 30o ΔI = 10o A1 = 70o A2 = 130o ΔA = 60o

MD1 = 10,200’ MD2 = 10,350’ ΔMD = 150’

1

2

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Calculation of DL Angle

β = dogleg angle

21221 IsinIsin2Asin

2Isinsin2 ⎟

⎠

⎞⎜⎝

⎛ Δ+⎟⎠

⎞⎜⎝

⎛ Δ=β −

30sin20sin260sin

210sinsin2 221 ⎟

⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛=β −

( ) deg93.252244.0sin2 1 ==β −

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Dogleg Severity, δ

 Dogleg severity, δ = β (100/ΔMD) = 25.93 (100/150)

= 17.29 deg/100 ft

( The dogleg severity is the dogleg angle

over a measured depth of 100 ft )

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  In general, the dogleg severity,

Dogleg Severity - cont’d

MD

100 ⎟⎠

⎞⎜⎝

⎛Δ

β=δ

2122 IsinIsin2Asin

2Isinsina

MD200

⎟⎠

⎞⎜⎝

⎛ Δ+⎟⎠

⎞⎜⎝

⎛ Δ⎟⎠

⎞⎜⎝

⎛Δ

=δ

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠

⎞⎜⎝

⎛ Δ+⎟⎠

⎞⎜⎝

⎛ Δ=β −

21221 IsinIsin2Asin

2Isinsin2

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Single Radius Build Curve

 Problem: Where should the KOP be located if the TVD at EOC is 10,000 ft, and there is no tangent section ?

BUR = 15 deg./100 ft Hole angle at start of build = 5o Hole angle at end of build = 92o

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Single Radius Build Curve, cont’d

ΔT

I1

I2

I1

I2

R

R

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Single Radius Build Curve - cont’d

 R = 18,000/(π BUR) = 382 ft (BUR = 15)

 ΔT = R (sin I2 - sin I1)

 TKOP = Tb - ΔT = 10,000 - 382 (sin 92 - sin 5)

TKOP = 9,652 ft

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Complex Tangent Build Curve

 Problem: Determine the depth of the kickoff point for a complex tangent build curve if:

 Depth of target = 9,200 ft  First BUR = 15 deg/100 ft  Second BUR = 12 deg/100 ft  Tangent angle = 45 deg.  Length of tangent = 120 ft   I1 = 0 deg; I2 = 90 deg

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Complex Tangent Build Curve, cont’d

KOP R1

R2

Ltan

45

EOC

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Complex Tangent Build Curve, cont’d

0o R1

R2

Ltan

EOC 90o

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Complex Build Curve - Solution

 R1 = 18,000/(π BUR1) = 18,000/15 π = 382 ft  R2 = 18,000/(π BUR2) = 18,000/12 π = 477 ft  First build section, ΔTB1 = R1 (sin I2 - sin I1) = 382 (sin 45 - sin 0 ) = 270 ft  Tangent section, ΔTtan = L cos I = 120 cos 45 = 85 ft

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Complex Curve - Solution - cont’d

 Second build section, ΔTB2 = R2 (sin I2 - sin I1) = 477 (sin 90 - sin 45) = 140 ft

 TKOP = TVD - ΔTB1 - ΔTtan - ΔTB2 = 9,200 - 270 - 85 - 140

TKOP = 8,705 ft

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Complex Tangent Build Curve, cont’d

0o R1 = 382

R2= 477

Ltan

EOC 90o

8,705

270

85

140

Depth of target = 9,200 ft

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The “Ideal” Build Curve

R1

R2

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The “Ideal” Build Curve - cont’d

 First build section   Design for maximum build rate based on

probable range   Determine ACTUAL build rate

 Second build Section   Reduce build rate as needed to hit target

by changing tool face angle

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Tool Face Angle (γ)

  Example:   20o build motor   When tool face angle

is zero,   Angle build rate

= 20 deg./100 ft

γ = 0

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Tool Face Angle - cont’d

 When tool face angle = + 30 deg.

 Angle build rate = 20 cos 30 = 17.3 deg/100 ft

γ = 30 γ = -30

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Optimum Tangent Angle

Itan = sin-1[sin If - BURMaxBURMin (Tb - Tt)K1 (BURMax - BURMin) ]

where K1 = 18,000/π

for BUR in deg/100 ft

REF: TEXT p.26

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Opt. Tangent Angle - cont’d

  Itan = inclination of tangent, deg.   If = final well inclination, deg.  BURMax = maximum expected BUR  BURMin = minimum expected BUR  Tb = base of target  Tt = top of target

 NOTE: K1 = 1,719 if BUR deg./30 m

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Target TVD at EOC

where,  Ttgt = target TVD at EOC  BURExp = expected or planned BUR

Ttgt = Tt + K1 1BURExp

- 1BURMax

sin If - sin Itan

REF: TEXT p.26

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Tangent Section - Example

  Assumptions: If = 86 deg.

BURMax = 12 deg./100 ft

BURMin = 10 deg./100 ft

BURExp = 11 deg./100 ft

K1 = 18,000/ π = 5,729.58

Base of target at TVD, Tb = 10,020 ft

Top of target at TVD, Tt = 10,000 ft

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Optimum Tangent Angle

Inclination of tangent section,

Itan = 52.01 deg.

Itan = sin-1[sin If - BURMaxBURMin (Tb - Tt)K1 (BURMax - BURMin) ]

I tan = sin -1 [sin 86 - 12 * 10 (10,020 - 10,000) 5,730 (12 - 10) ]

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Target TVD at EOC

Ttgt = 10,009.09 ft

Ttgt = Tt + K1 1BURExp

- 1BURMax

sin If - sin Itan

T tgt = 10,000 + 5,730 1 11

- 1 12

sin 86 - sin 52.01

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Basic horizontal well casing programs

 No intermediate casing

  Intermediate casing to KOP

  Intermediate casing to EOC

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No Intermediate Casing

245 mm

SALTS

Bakken Shale

North Dakota

TVD - 3,250 m

Lateral - 762 m

14o/30 m 90o

140 mm

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Intermediate Casing to KOP

245 mm

Sloughing Shale Lost Circulation Differential Sticking

Swan Hills

Alberta

Jean Marie

N.E. B.C. 18o/30 m

178 mm

114 mm

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Intermediate Casing to EOC

273 mm

Sloughing Shales

Alberta

- Deep Basin

TVD - 2,780 m

Lateral - 275 m

20o/30 m

90o

114 mm

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Casing and Drilling Fluid Program

9 5/8” Flowing Salt Zones

Bakken Oil Well

Use weighted invert oil mud (oil is continuous phase)

Casing cemented up from EOC

14o/100 ft

Pf = 0.62 psi/ft

5 1/2”

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Casing and Drilling Fluid Program

10 3/4”

Sloughing Shales Overpressured gas

Cadomin Gas Well

Use weighted invert mud to csg. point.

KCl-Water-Polymer mud in horizontal

20o/100 ft

0.35 psi/ft 4 1/2”

7 5/8”

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Csg. and drilling fluid program - cont’d

 Bakken oil well - N. Dakota  Lateral hole = 3,000 ft  Production casing cemented up from EOC

through a stage collar

 Cadomin gas well - Alberta  No cemented liner  Two stage cement job on intermediate

casing

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Marker Beds

 Some marker beds will be continuous to a control well, and some will be discontinuous.

 The continuous marker may change depth relative to the target.

 Markers with isopachs constant to the zone are the most desirable ...

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Continuous vs. discontinuous marker beds.

Horizontal Well

Horizontal Well

Vertical Control Well

SHALE MARKER (continuous)

COAL - (discontinuous)

TARGET

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Marker bed dip angle is unequal to target dip angle

Horizontal Well

Horizontal Well

Vertical Control Well

t1

TARGET

Itarget = Imarker

SHALE MARKER

t1 = t2

t2

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Faults in uphole marker beds

Horizontal Well

Horizontal Well

Vertical Control Well

C

tAM tAM ≠ tAV tAV

C’

A’ A

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Discontinuous reservoirs

LATERAL VARIATION

VERTICAL VARIATION

FAULTS

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Target Zone is at Expected TVD

BUR = 5o/100 ft

TARGET ZONE

BUR = 20o/100 ft

EOC AT 90o

TOP OF TARGET ZONE (from samples)

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Target Zone at Lower TVD due to Formation Dip

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Cross Section of Mud Motors showing different Lobe Patterns

STATOR

ROTOR

High Speed - Low Torque Motors

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Cross Section of Mud Motors showing different Lobe Patterns

HIGH TORQUE - LOW SPEED MOTORS

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Conventional single-lobe downhole Mud Motor

DUMP VALVE

ROTOR UNIVERSIAL JOINT SHAFT

BIT

THRUST and

RADIAL BEARINGS

DRIVE SHAFT

STATOR

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Medium Radius Angle Build Assembly

BEARING ASSEMBLY CONNEC.-ROD HOUSING

MOTOR NON-MAGNETIC DRILL COLLAR

3.2’

6.3’

BENT SUB

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Combined Effect of Bent Sub and Bent

Motor

BENT MOTOR

θ1 = 1.5o

BENT SUB

θ2 = 2.5o

θ = θ1 + θ2 = 4.0o

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Bending Stresses at Kickoff

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Bending stresses at equilibrium

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Lateral forces on bit from kickoff to equilibrium

FEET DRILLED (MD)

FOR

CE

ON

BIT

1,

000’

s o

f lb

f

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Kickoff in a soft homogeneous formation

Bottomhole Assembly will reduce Curvature

BUREXP & REXP

High Initial Dogleg BUR >> BUREXP

R1

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Kickoff in a soft formation underlying a hard formation

Bottomhole Assembly will NOT wipe out Dogleg

Very Hard Siltstone

Very Soft Coal

High Dogleg and Contact Point

Sloughing

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Bit shank wear in

abrasive formation

Bit Contact Force

Abrasive, Hard Formation

Gauge Cutting

Structure

Shank

Shank Wear

Slow Bit Recipro.

Bit Cone

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Dogleg Angle Equations

21221 IsinIsin2Asin

2Isinsin2 ⎟

⎠

⎞⎜⎝

⎛ Δ+⎟⎠

⎞⎜⎝

⎛ Δ=β −

( ) ( )[ ])AAcos(1IsinIsinIIcos(cos 1221121 −−−−=β −

( )[ ]2121121 IcosIcosIsinIsinAAcoscos +−=β −

⎟⎠

⎞⎜⎝

⎛ +⎟⎠

⎞⎜⎝

⎛ −+⎟⎠

⎞⎜⎝

⎛ −=β −

2IIsin

2AAsin

2IIsinsin2 2121221221