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Directional Drilling
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
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.
North
Direction
Angle
Direction Plane X
Inclination Angle
Z Axis (True Vertical
Depth)
q, a or I
f, e or A
Non-Vertical
Wellbore
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
Figure 8.3 - Typical offshore development platform
with directional wells.
NOTE: All the
wells are
directional
Top View
5 - 50 wells
per platform
Figure 8.4 - Developing a field under a city
using directionally drilled wells.
Drilling Rig Inside Building
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.?
Figure 8.6 -
Sidetracking
around a fish.
Sidetracked Hole
Around Fish
Fish Lost in Hole and
Unable to RecoverCement Plug
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
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)
Type I
Type III
Type II
Figure 8.10 -
Geometry of the
build section.
Build Section
Build Radius:
BUR*
,
00018r1
Build Section:
deg rad
180*
100Lr
)cos(1rDD'dev.Horiz.
sinrD'C' depth ical Vert
rL arc,of Length
11
11
11
11
111
BUR*
,
00018r1
Build-hold-and drop for the case where:
42131 xrr and xr
Target
Drop Off
End of Build
Start of Buildup
Type II
Build-hold-and
drop for the case
where:
Kickoff
End of Build
Maximum
Inclination
Angle
Drop Off
Target
42131 xrr and xr
Type II
Fig. 8-14. Directional well used to intersect
multiple targets
Target 1
Target 2
Target 3
Projected Trajectory Projected Trajectory
with Left Turn to Hit
Targets
Fig. 8-15.
Directional
quadrants and
compass
measurements
N18E S23E
A = 157o
N55W
A = 305oS20W
Figure 8-16: Plan View
Lead Angle
Lake
Surface
Location
for Well
No. 2
Projected Well Path
Target at a
TVD 9,659
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
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)?
2500
10,000
Imax
Imax
TVD1
4,500
12,500
Type I: Build-and-Hold
HD1
Uniform 130 Increase in Drift
per 100 ft of hole
drilled
10,000
Vert.
Depth
4,500 Horizontal Deviation
0
Try Imax = 27o
??
Solution
Type I Well 1.5 deg/100
2500 Available depth
= 12,500-2,500
= 10,000
10,000
Imax
ImaxFrom Chart,
Try = 27o
Imax
TVD1
HD1
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
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
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
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
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.
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
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
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)
(i) Maximum
Inclination
Angle
r1 18 000
15
,
.
0r2
D4 112 500 2 500
10 000
D
ft
, ,
,
(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
22
1-
421
421
2
14
2
4141
maxq
3.26max q
(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
q
q
(ii) Measured Depth of
Well
265,9180
26.33,8202,500
LrDMD Holdrad11
q
ft 518,13MD
We may plan a 2-D well, but we always
get a 3D well (not all in one plane)
Horizontal
Vertical
ViewN
View
Fig. 8-22. A curve representing a wellbore
between survey stations A1 and A2
MD, a1, e1
DMD
a2, e2
b = dogleg
angle
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
Directional Tools
(i) Whipstock(ii) Jet Bits(iii) Downhole motor and bent sub
Whipstocks
Standard retreivable Circulating Permanent Casing
Setting a Whipstock
Small bit used to startApply weight to:
set chisel point &
shear pin
Drill 12-20Remove whipstockEnlarge hole
Jetting Bit
Fast and economical
For soft formationOne large - two
small nozzles
Orient large nozzleSpud periodicallyNo rotation at first
Small Jets
Jetting
Wash out pocketReturn to normal
drilling
SurveyRepeat for more
angle if needed
Mud Motors
DrillpipeNon-magnetic
Drill CollarBent Sub
Mud Motor
Rotating
Sub
Increasing Inclination
Limber assemblyNear bit stabilizerWeight on bit forces
DC to bend to low
side of hole.
Bit face kicks up
Hold Inclination
Packed hole assembly
Stiff assemblyControl bit weight
and RPM
Decrease Inclination
Pendulum effectGravity pulls bit
downward
No near bit stabilizer
Packed Hole Assemblies
Drill
pipe
HW DP
String
Stabilizer
Steel DC
String
StabilizerString
Stabilizer
Monel
DCSteel DC
NB
Stab
Vertical Calculation Horizontal Calculation
3D View Dog Leg Angle
Deflecting Wellbore Trajectory0
90
180
270
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
Survey Calculation Methods
1. Tangential Method
= Backward Station Method
= Terminal Angle Method
Assumption: Hole will maintain constant inclination and azimuth
angles between survey points
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
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,
AnglesBABA A ,A ,I ,I
(i) Simple enough for field use
(ii) Much more accurate than
Tangential Method
A
B
IA
IB
IAVG
IAVG
2
III BAav g
2
AAA BAavg
Average Angle Method
Vertical Plane:
A
B
IA
IB
IAVG
IAVG
2
III BAav g
avgAB
avgAB
IsinABH
IcosABV
Average Angle Method
Horizontal Plane:
avg
avgavg
avgavg
IcosABZ
AcosIsinABN
AsinIsinABE
D
D
D
N
B
AA
AB
AAVG
EDE
DN
A
avgAB IsinABH
Change in position towards the east:
Change in position towards the north:
)1..(2
AAsin
2
IIsinLEx BABA
DD
)2..(2
AAcos
2
IIsinLNy BABA
DD
)3..(2
IIcosLZ BA
D
Change in depth:
Where L is the measured distance
between the two stations A & B.
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.
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
Solution
H12 = L12 sin Iavg = 300 sin 20 = 103 ft
DE = H12 sin Aavg = 103 sin 55 = 84 ft
DN = H12 cos Aavg = 103 cos 55 = 59 ft
DZ = L12 cos Iavg = 300 cos 20 = 282 ft
202
2515
2
III 21avg
552
6545
2
AAA 21avg
Solution - contd
DE = 84 ft
DN = 59 ft
DZ = 282 ft
x2 = x1 + DE = 1,000 + 84 ft = 1,084 ft
y2 = y1 + DN = 2,000 + 59 ft = 2,059 ft
z2 = z1 + DZ = 3,000 + 282 ft = 3,282 ft