1 Petroleum Engineering 406 Floating Drilling Lesson 11b Motion Compensation.
406 Lesson 18
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Petroleum Engineering 406
Lesson 18 Directional Drilling
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Lesson 10 - Directional Drilling
When is it used? Type I Wells (build and hold) Type II Wells (build, hold and drop) Type III Wells (build) Directional Well Planning & Design Survey Calculation Methods
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Homework:
READ. “Applied Drilling Engineering”Ch. 8, pp. 351-363
REF. API Bulletin D20, “Directional Drilling Survey Calculation Methods and
Terminology”
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What is Directional Drilling?
Directional Drilling is the process of directing a wellbore along some trajectory to a predetermined target.
Basically it refers to drilling in a non-vertical direction. Even “vertical” hole sometimes require directional drilling techniques.
Examples: Slanted holes, high angle holes (far from vertical), and Horizontal holes.
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North
Direction Angle
Direction Plane X
Inclination Angle
Z Axis (True Vertical Depth)
Inclination
Plane Y
or I
or A
Non-Vertical Wellbore
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Figure 8.2 - Plan view of a typical oil and gas structure under a lake showing how directional wells could be used to develop it. Best locations? Drill from lake?
Lease Boundary
Surface Location for Well No. 1
Bottom Hole Location for Well 2
Surface Location for Well No. 2
Houses
Oil-Water Contact
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Figure 8.3 - Typical offshore development platform with directional wells.
NOTE: All the wells are directional
Top View
5 - 50 wells per platform
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Figure 8.4 - Developing a field under a city using directionally drilled wells.
Drilling Rig Inside Building
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Fig. 8.5 - Drilling of directional wells where the reservoir is beneath a major surface obstruction.
Why not drill from
top of mountain?
Maximum lateral displ.?
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Figure 8.6 - Sidetracking around a fish.
Sidetracked Hole Around Fish
Fish Lost in Hole and Unable to Recover
Cement Plug
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Figure 8.7 - Using an old
well to explore for new oil by sidetracking
out of the casing and
drilling directionally.
Possible New Oil
Sidetracked Out of Casing
Oil Producing Well Ready to Abandon
Old Oil Reservoir
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Figure 8.8 - Major types of wellbore trajectories.
Build and Hold Type
Continuous Build
Build-hold Drop and/or Hold (Modified “S” Type)
Build-hold and Drop (“S Type”)
Horizontal Departure to Target
Type I
Type III
Type II
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Figure 8.10 - Geometry of the
build section.
Build Section
Build Radius:
BUR*
,
00018r1
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Build Section:
deg rad
180*
100Lr
)cos(1rDD'dev.Horiz.
sinrD'C' depth ical Vert
rL arc,of Length
11
11
11
11
111
πθθ
θ
θ
θ
BUR*
,
00018r1
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Build-hold-and drop for the case where:
42131 xrr and xr
Target
Drop Off
End of Build
Start of Buildup
Type II
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Build-hold-and drop for the case where:
Kickoff
End of Build
Maximum Inclination Angle
Drop Off
Target
42131 xrr and xr
Type II
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Fig. 8-14. Directional well used to intersect multiple targets
Target 1Target 2
Target 3
Projected Trajectory Projected Trajectory with Left Turn to Hit Targets
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Fig. 8-15. Directional
quadrants and
compass measurements
N18E S23E
A = 157o
N55W
A = 305oS20W
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Figure 8-16: Plan View
Lead Angle
Lake
Surface Location for Well No. 2
Projected Well Path
Target at a TVD 9,659
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Example 1: Design of Directional Well
Design a directional well with the following restrictions:
Total horizontal departure = 4,500 ftTrue vertical depth (TVD) = 12,500 ftDepth to kickoff point (KOP) = 2,500 ftRate of build of hole angle = 1.5 deg/100 ft
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Example 1: Design of Directional Well
This is a Type I well (build and hold)
(i) Determine the maximum hole angle (inclination) required.
(ii) What is the total measured depth of the hole (MD)?
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2500’
10,000’
Imax
Imax
TVD1
4,500’
12,500’
Type I: Build-and-Hold
HD1
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Uniform 1’30” Increase in Drift per 100 ft of hole
drilled
10,000’
Vert.
Depth
4,500’ Horizontal Deviation
0’
Try Imax = 27o ??
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Solution
Type I Well 1.5 deg/100’
2500’ Available depth
= 12,500-2,500
= 10,000’
10,000’
Imax
Imax
From Chart,
Try = 27oImax
TVD1
HD1
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Build Section
Imax
Imax
TVD1
HD1
MD1 = 1,800’ (27/1.5)
TVD1 = 1,734’
HD1 = 416’
Remaining vertical height
= 10,000 - 1,734 = 8,266’
From chart of 1.5 deg/100’, with Imax = 27o
In the BUILD Section:
8,266’
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Solution
Horizontally:
416 + 8,266 tan 27o = 4,628
We need 4,500’ only:
Next try Imax = 25’ 30 min
Imax8,266’ MD2 = 1,700’ (25.5/1.5)
TVD2 = 1,644’
HD2 = 372’
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Solution:
Remaining vertical depth = 10,000-1644
= 8,356 ft.
Horizontal deviation = 372+8,356 tan 25.5
= 4,358 ft. { 4500 }
Approx. maximum angle = 26
What is the size of target?
4
10
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MD = MDvert + MDbuild + MDhold
13,500'MD
13,458' 25.5 cos
8,3561,7002,50025.5 at MD
13,577' 27 cos
266,8'800,1'500,2 27 at MD
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Type II Pattern
Given: KOP = 2,000 feet
TVD = 10,000 feet
Horiz. Depart. = 2,258 feet
Build Rate = 20 per 100 feet
Drop Rate = 10 30’ per 100 feet
The first part of the calculation is the same as previously described.
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Procedure - Find:
a) The usable depth (8,000 feet) b) Maximum angle at completion of
buildup (180) c) Measured depth and vertical depth at
completion of build up (M.D.=900 ft. and TVD = 886)
d) Measured depth, horizontal departure and TVD for 1 /100 ft from chart.
0
2
1
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Solve:
For the distances corresponding to the sides of the triangle in the middle.
Add up the results.
If not close enough, try a different value for the maximum inclination angle, Imax
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Example 1: Design of Directional Well
(i) Determine the maximum hole angle required.
(ii) What is the total measured depth (MD)?
(MD = well depth measured along the wellbore,
not the vertical depth)
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(i) Maximum Inclination
Angle
r1 18 000
15
,
.
0r2
D4 1
12 500 2 500
10 000
D
ft
, ,
,
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(i) Maximum Inclination Angle
500,4)820,3(2
500,4)820,3(2000,10500,4000,10 tan2
x)rr(2
x)rr(2)DD(xDDtan2
221-
421
4212
1424141
max
3.26max
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(ii) Measured Depth of Well
ft 265,9L
105,4sinL
ft 4,105
395500,4x
ft 395
)26.3 cos-3,820(1
)cos1(rx
Hold
Hold
Hold
1Build
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(ii) Measured Depth of Well
265,9180
26.33,8202,500
LrDMD Holdrad11
ft 518,13MD
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We may plan a 2-D well, but we always get a 3D well (not all in one plane)
Horizontal
Vertical
ViewN
View
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Fig. 8-22. A curve representing a wellbore between survey stations A1 and A2
MD,
MD
= dogleg
angle
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Directional Drilling
1. Drill the vertical (upper) section of the hole.
2. Select the proper tools for kicking off to a non-vertical direction
3. Build angle gradually
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Directional Tools
(i) Whipstock (ii) Jet Bits (iii) Downhole motor and bent sub
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Whipstocks
Standard retreivable Circulating Permanent Casing
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Setting a Whipstock
Small bit used to start Apply weight to:
– set chisel point &– shear pin
Drill 12’-20’ Remove whipstock Enlarge hole
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Jetting Bit
Fast and economical For soft formation One large - two
small nozzles Orient large nozzle Spud periodically No rotation at first
Small Jets
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Jetting
Wash out pocket Return to normal
drilling Survey Repeat for more
angle if needed
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Mud MotorsDrillpipe
Non-magnetic Drill Collar
Bent Sub
Mud MotorRotating Sub
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Increasing Inclination
Limber assembly Near bit stabilizer Weight on bit forces
DC to bend to low side of hole.
Bit face kicks up
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Hold Inclination
Packed hole assembly
Stiff assembly Control bit weight
and RPM
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Decrease Inclination
Pendulum effect Gravity pulls bit
downward No near bit stabilizer
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Packed Hole Assemblies
Drillpipe
HW DP
String Stabilizer
Steel DC
String Stabilizer
String Stabilizer
MonelDCSteel DC
NB Stab
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Vertical Calculation Horizontal Calculation
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3D View Dog Leg Angle
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Deflecting Wellbore Trajectory0
90
180
270
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Bottom Hole Location
10,000 :TVD
ft 2,550 :Distance
E 53 N :Direction
o1-
22
53N
EtanDirection Closure
NE 2,550Closure
ft 1,535
53 cos 2,550N
ft 2,037
53 sin 550,2E
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Survey Calculation Methods
1. Tangential Method
= Backward Station Method
= Terminal Angle Method
Assumption: Hole will maintain constant inclination and azimuth angles between survey points
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BAB
BAB
BA
BA
IsinABH
IcosABV:nCalculatio
A ,A Angles
I ,I es Angl
ABDistance
Aof Location :Known
Poor accuracy!!
A
B
IA
IB
IB
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Average Angle Method = Angle Averaging Method
Assumption: Borehole is parallel to the simple average drift and bearing angles between any two stations.
Known: Location of A, Distance AB,
Angles BABA A ,A ,I ,I
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(i) Simple enough for field use
(ii) Much more accurate than
“Tangential” Method
A
B
IA
IB
IAVG
IAVG
2
III BAavg
2
AAA BA
avg
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Average Angle MethodVertical Plane:
A
B
IA
IB
IAVG
IAVG
2
III BAavg
avgAB
avgAB
IsinABH
IcosABV
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Average Angle MethodHorizontal Plane:
avg
avgavg
avgavg
IcosABZ
AcosIsinABN
AsinIsinABE
N
B
AA
AB
AAVG
EE
N
A
avgAB IsinABH
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Change in position towards the east:
Change in position towards the north:
)1..(2
AAsin
2
IIsinLEx BABA
)2..(2
AAcos
2
IIsinLNy BABA
)3..(2
IIcosLZ BA
Change in depth:
Where L is the measured distance between the two stations A & B.
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Example
The coordinates of a point in a wellbore are:
x = 1000 ft (easting)
y = 2000 ft (northing)
z = 3000 ft (depth)
At this point (station) a wellbore survey shows that the inclination is 15 degrees from vertical, and the direction is 45 degrees east of north. The measured distance between this station and the next is 300 ft….
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Example
The coordinates of point 1 are:
x1 = 1000 ft (easting)
y1 = 2000 ft (northing) I1 = 15o
z1 = 3000 ft (depth) A1 = 45o
L12 = 300 ft
At point 2, I2 = 25o and A2 = 65o
Find x2 , y2 and z2
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Solution
H12 = L12 sin Iavg = 300 sin 20 = 103 ft
E = H12 sin Aavg = 103 sin 55 = 84 ft
N = H12 cos Aavg = 103 cos 55 = 59 ft
Z = L12 cos Iavg = 300 cos 20 = 282 ft
202
2515
2
III 21avg
552
6545
2
AAA 21
avg
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Solution - cont’d
E = 84 ft
N = 59 ft
Z = 282 ft
x2 = x1 + E = 1,000 + 84 ft = 1,084 ft
y2 = y1 + N = 2,000 + 59 ft = 2,059 ft
z2 = z1 + Z = 3,000 + 282 ft = 3,282 ft