Post on 06-Jun-2018
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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