d Drilling Calcs

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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

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