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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 EngineeringCh. 8, pp. 351-363
REF.API Bulletin D20, Directional DrillingSurvey Calculation Methods andTerminology
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What is Directional Drilling?
Directional Drilling is the process ofdirecting a wellbore along some trajectory
to a predetermined target.
Basically it refers to drilling in a non-vertical
direction. Even vertical hole sometimesrequire 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 AngleZ Axis (True Vertical
Depth)
q, a
or I
f, e or A
Non-VerticalWellbore
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Figure 8.2 - Plan view of a typical oil and gas structure under a lake showing howdirectional 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
SurfaceLocation forWell No. 2
Houses
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Figure 8.3 - Typical offshore development platformwith directional wells.
NOTE: All thewells are
directional
Top View
5 - 50 wellsper platform
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Figure 8.4 - Developing a field under a cityusing directionally drilledwells.
Drilling Rig Inside Building
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Fig. 8.5 - Drilling of directional wells where thereservoir is beneath a major surface obstruction.
Why notdrill from
top ofmountain?
Maximum
lateraldispl.?
The BakkenPetroleumSystem: An
Example Of AContinuous
PetroleumAccumulation
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Figure 8.6 -Sidetrackingaround a fish.
Sidetracked HoleAround Fish
Fish Lost in Hole andUnable to Recover
Cement Plug
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Figure 8.7 -
Using an oldwell to explorefor new oil by
sidetrackingout of thecasing and
drilling
directionally.
PossibleNew Oil
SidetrackedOut of Casing
Oil Producing WellReady to Abandon
Old Oil Reservoir
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Figure 8.8 - Major types of wellbore trajectories.
Build and
Hold Type
ContinuousBuild
Build-hold Drop and/or Hold
(Modified S Type)
Build-hold and Drop (S Type)
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:
degrad
180*
100Lr
)cos(1rDD'dev.Horiz.
sinrD'C'depthicalVert
rLarc,ofLength
11
11
11
11
111
BUR*
,
00018r1
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Build-hold-and drop for the case where:
42131 xrrandxr
Target
Drop Off
End of Build
Start of Buildup
Type II
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Build-hold-anddrop for the case
where:
Kickoff
End of Build
Maximum
InclinationAngle
Drop Off
Target
42131 xrrandxr
Type II
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Fig. 8-14. Directional well used to intersectmultiple targets
Target 1
Target 2
Target 3
Projected Trajectory Projected Trajectory
with Left Turn to HitTargets
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Fig. 8-15.
Directionalquadrants and
compass
measurements
N18E S23E
A = 157o
N55W
A = 305oS20W
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Figure 8-16: Plan View
Lead Angle
Lake
Surface
Locationfor WellNo. 2
Projected Well Path
Target at aTVD 9,659
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Example 1: Design ofDirectional Well
Design a directional well with the following
restrictions:Total horizontal departure = 4,500 ft
True vertical depth (TVD) = 12,500 ft
Depth to kickoff point (KOP) = 2,500 ft
Rate 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 holeangle (inclination) required.
(ii) What is the total measured depthof 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 130Increase in Driftper 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
ImaxFrom Chart,Try = 27
oImax
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 of1.5 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
Imax8,266MD2= 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'
25.5cos
8,3561,7002,50025.5atMD
13,577'
27cos
266,8'800,1'500,227atMD
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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 = 1030 per 100 feet
The first part of the calculation is thesame 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 depthat completion of build up(M.D.=900 ft. and TVD = 886)
d) Measured depth, horizontal departureand TVD for 1 /100 ft from chart.
0
2
1
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Solve:
For the distances corresponding to thesides 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
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Example 1: Design of DirectionalWell
(i) Determine the maximumhole anglerequired.
(ii) What is the totalmeasured depth(MD)?
(MD = well depth measured along thewellbore,
not the vertical depth)
(i) M i
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(i) Maximum
Inclination
Angle
r1 18 000
15
,
.
0r2
D4 112 500 2 500
10 000
D
ft
, ,
,
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(i) Maximum Inclination Angle
500,4)820,3(2
500,4)820,3(2000,10500,4000,10tan2
x)rr(2
x)rr(2)DD(xDDtan2
22
1-
421
421
2
14
2
4141
maxq
3.26max
q
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(ii) Measured Depth of Well
ft265,9L
105,4sinL
ft4,105
395500,4x
ft395
)26.3cos-3,820(1
)cos1(rx
Hold
Hold
Hold
1Build
q
q
(ii) M d D th f
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(ii) Measured Depth of
Well
265,9180
26.33,8202,500
LrDMD Holdrad11
q
ft518,13MD
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We may plan a 2-D well, but we alwaysget a 3D well (not all in one plane)
Horizontal
Vertical
ViewN
View
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Fig. 8-22. A curve representing a wellborebetween survey stations A1 and A2
MD, a1, e1DMD
a2, e2b = doglegangle
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Directional Drilling
1. Drill the vertical (upper) section ofthe hole.
2. Select the proper tools for kicking offto 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 whipstockEnlarge hole
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Jetting Bit
Fast andeconomical
For soft formationOne 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
drillingSurvey
Repeat for moreangle if needed
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Mud Motors
DrillpipeNon-magneticDrill Collar
Bent SubMud Motor
RotatingSub
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Increasing Inclination
Limber assembly
Near bit stabilizer
Weight on bit forcesDC to bend to lowside of hole.
Bit face kicks up
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Hold Inclination
Packed holeassembly
Stiff assemblyControl bit weight
and RPM
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Decrease Inclination
Pendulum effect
Gravity pulls bit
downwardNo near bit stabilizer
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Packed Hole Assemblies
Drill
pipe
HW DP
String
Stabilizer
Steel DC
String
StabilizerString
Stabilizer
MonelDCSteel DC
NBStab
Vertical Calculation Horizontal Calculation
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3D View Dog Leg Angle
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3D View Dog Leg Angle
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Deflecting Wellbore Trajectory
0
90
180
270
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Bottom Hole Location
10,000:TVD
ft2,550:Distance
E53N:Direction
o1-
22
53N
EtanDirectionClosure
NE2,550Closureft1,535
53cos2,550N
ft2,037
53sin550,2E
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Survey Calculation Methods
1. Tangential Method
= Backward Station Method= Terminal Angle Method
Assumption: Hole will maintainconstant inclination and azimuth
angles between survey points
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BAB
BAB
BA
BA
IsinABHIcosABV:nCalculatio
A,AAngles
I,IesAngl
ABDistance
AofLocation:Known
Poor accuracy!!
A
B
IA
IB
IB
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Average Angle Method= Angle Averaging Method
Assumption: Borehole is parallel to thesimple average drift and bearing anglesbetween 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
2AAA BAavg
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Average Angle Method
Vertical Plane:
A
B
IA
IB
IAVG
IAVG
2
III BAavg
avgAB
avgAB
IsinABH
IcosABV
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Average Angle Method
Horizontal Plane:
avg
avgavg
avgavg
IcosABZ
AcosIsinABNAsinIsinABE
D
DD
N
B
AA
AB
AAVG
EE
DN
A
avgAB IsinABH
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Change in position towards the east:
Change in position towards the north:)1..(2
AA
sin2
II
sinLEx
BABA
DD
)2..(2
AAcos
2
IIsinLNy BABA
DD
)3..(2
II
cosLZ
BA
D
Change in depth:
Where L is the measured distancebetween the two stations A & B.
Example
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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 showsthat the inclination is 15 degrees from vertical,
and the direction is 45 degrees east of north. Themeasured distance between this station and thenext is 300 ft.
Example
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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
Solution
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Solution
H12 = L12 sin Iavg = 300 sin 20 = 103 ft
DE = H12 sin Aavg = 103 sin 55 = 84 ftDN = H12 cos Aavg = 103 cos 55 = 59 ft
DZ = L12 cos Iavg = 300 cos 20 = 282 ft
202
25152
III 21av g
552
6545
2
AAA 21
avg
Solution contd
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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 ftz2 = z1 + DZ = 3,000 + 282 ft = 3,282 ft