406 lesson 18

Petroleum Engineering 406

Lesson 18 Directional Drilling

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

Homework:

READ. “Applied Drilling Engineering”Ch. 8, pp. 351-363

REF. API Bulletin D20, “Directional Drilling Survey Calculation Methods and

Terminology”

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)

Inclination

Plane Y

or I

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


Houses

Oil-Water Contact

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 Recover

Cement 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”)

Horizontal Departure to Target

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

πθθ

θ

θ

θ

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 1Target 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


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


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 1’30” 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

Imax

From Chart,

Try = 27oImax

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


(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 1

12 500 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

221-

421

4212

1424141

max

3.26max

(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

(ii) Measured Depth of Well

265,9180

26.33,8202,500

LrDMD Holdrad11

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,

MD

= 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 start Apply weight to:

– set chisel point &– shear pin

Drill 12’-20’ Remove whipstock Enlarge hole

Jetting Bit

Fast and economical For soft formation One large - two

small nozzles Orient large nozzle Spud periodically No rotation at first

Small Jets

Jetting

Wash out pocket Return to normal

drilling Survey Repeat for more

angle if needed

Mud MotorsDrillpipe

Non-magnetic Drill Collar

Bent Sub

Mud MotorRotating Sub

Increasing Inclination

Limber assembly Near bit stabilizer Weight on bit forces

DC to bend to low side of hole.

Bit face kicks up

Hold Inclination

Packed hole assembly

Stiff assembly Control bit weight

and RPM

Decrease Inclination

Pendulum effect Gravity pulls bit

downward No near bit stabilizer

Packed Hole Assemblies

Drillpipe

HW DP

String Stabilizer

Steel DC

String Stabilizer

String Stabilizer

MonelDCSteel 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,

Angles BABA 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 BAavg

2

AAA BA

avg

Average Angle MethodVertical Plane:

A

B

IA

IB

IAVG

IAVG

2

III BAavg

avgAB

avgAB

IsinABH

IcosABV

Average Angle MethodHorizontal Plane:

avg

avgavg

avgavg

IcosABZ

AcosIsinABN

AsinIsinABE

N

B

AA

AB

AAVG

EE

N

A

avgAB IsinABH

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.

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

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

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

406 lesson 18

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