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