Slide 1

PETE 661

Drilling Engineering

Lesson 13

Directional Drilling

Slide 2

Directional Drilling

When is it used?

Type I Wells

Type II Wells

Type III Wells

Directional Well Planning & Design

Survey Calculation Methods

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

I II III

Slide 3

Read ADE Ch.8 (Reference)

HW #7Cementingdue 10-25-02

Slide 4

Inclination Angle, I

Direction Angle, A

Slide 5

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Max.Horiz.

Depart.?

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Type I Type II Type III

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

Slide 13

x

y

I

I

r

rL

In the BUILDSection

x = r (1 - cos I)

y = r sin I

L = r rad

degI r180

= L

BUR*

000,18r

Slide 14

Slide 15

Fig. 8.11

42131 xrr and xr

Slide 1642131 xrr and xr 3D Wells

Slide 17

N18E

N55WS20W

S23E

Azimuth

Angle

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Slide 19

Example 1: Design of Directional 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

• Type I well (build and hold)

Slide 20

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)

Slide 21

(i) Maximum Inclination

Angle

r1 18 000

15

,

. r2 0

D4 1

12 500 2 500

10 000

D

ft

, ,

,

Slide 22

(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

Slide 23

(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

Slide 24

(ii) Measured Depth of Well

265,9180

26.33,8202,500

LrDMD Holdrad11

ft 518,13MD

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* The actual well path hardly ever coincides with the planned trajectory

* Important: Hit target within specified radius

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What is known?I1 , I2 , A1 , A2 ,

L=MD1-2

Calculate = dogleg angle

DLS =L

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(20)

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Wellbore Surveying Methods

Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential

Other Topics Kicking off from Vertical Controlling Hole Angle

Slide 30

I, A, MD

Slide 31

Example - Wellbore Survey Calculations

The table below gives data from a directional survey.

Survey Point Measured Depth Inclination Azimuth along the wellbore Angle Angle

ft I, deg A, deg

A 3,000 0 20 B 3,200 6 6 C 3,600 14 20 D 4,000 24 80

Based on known coordinates for point C we’ll calculate the coordinates of point D using the above information.

Slide 32

Example - Wellbore Survey Calculations

Point C has coordinates:

x = 1,000 (ft) positive towards the east

y = 1,000 (ft) positive towards the north

z = 3,500 (ft) TVD, positive downwards

Z

E (x)

N (y)C

Dz

N

D

C

yx

Slide 33

Example - Wellbore Survey Calculations

I. Calculate the x, y, and z coordinates of points D using:

(i) The Average Angle method

(ii) The Balanced Tangential method

(iii) The Minimum Curvature method

(iv) The Radius of Curvature method

(v) The Tangential method

Slide 34

The Average Angle Method

Find the coordinates of point D using the Average Angle Method

At point C, X = 1,000 ft

Y = 1,000 ft

Z = 3,500 ft

80A 24I

20A 14I

ft 400MD D, toC fromdepth Measured

DD

CC

Slide 35

The Average Angle Method

80A 24I

20A 14I

ft 400MD D, toC fromdepth Measured

DD

CC

Z

E (x)

N (y)

C

Dz

N

D

C

yx

Slide 36

The Average Angle Method

Slide 37

The Average Angle Method

This method utilizes the average of I1 and I2 as an inclination, the average of A1 and A2 as a direction, and assumes all of the survey interval (MD) to be tangent to the average angle.

From: API Bulletin D20. Dec. 31, 1985

Slide 38

ft 71.8350cossin19400

cossin

502

8020

2

192

2414

2

AVGAVG

DCAVG

DCAVG

AIMDNorth

AAA

III

The Average Angle Method

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The Average Angle Method

ft

AIMDEast AVEAVG

76.9950sinsin19400

sinsin

ft

IVert AVG

21.378cos19400

cos400

Slide 40

The Average Angle Method

At Point D,

X = 1,000 + 99.76 = 1,099.76 ft

Y = 1,000 + 83.71 = 1,083.71 ft

Z = 3,500 + 378.21 = 3,878.21 ft

Slide 41

The Balanced Tangential Method

This method treats half the measured distance (MD/2) as being tangent to I1 and A1 and the remainder of the measured distance (MD/2) as being tangent to I2 and A2.

From: API Bulletin D20. Dec. 31, 1985

Slide 42

The Balanced Tangential Method

ft 59.59

)80cos24sin20cos14(sin2

400

)AcosIsinAcosI(sin2

MDNorth DDCC

Slide 43

The Balanced Tangential Method

96.66ft

)80sin24sin20sin14(sin2

400

)AsinIsinAsinI(sin2

MDEast DDCC

Slide 44

The Balanced Tangential Method

ft77.376)14cos24(cos2

400

)IcosI(cos2

MDVert CD

Slide 45

The Balanced Tangential Method

At Point D,

X = 1,000 + 96.66 = 1,096.66 ft

Y = 1,000 + 59.59 = 1,059.59 ft

Z = 3,500 + 376.77 = 3,876.77 ft

Slide 46

Minimum Curvature Method

Slide 47

Minimum Curvature Method

This method smooths the two straight-line segments of the Balanced Tangential Method using the Ratio Factor RF.

RF = (2/DL) * tan(DL/2) (DL= and must be in radians)

Slide 48

Minimum Curvature Method

The dogleg angle, , is given by:

radians 36082.020.67

0.935609

))2080cos(1(24sinsin14-14)-cos(24

))AAcos(1(IsinIsin)IIcos(Cos CDDCCD

Slide 49

Minimum Curvature Method

The Ratio Factor,

ft 25.6001099.1*59.59

RF)IcosIsinAcosI(sin2

MDNorth

01099.12

67.20tan*

3608.0

2RF

Z

tan2

RF

DDCC

2

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Minimum Curvature Method

ft 380.911.01099*376.77

RF)IcosI(cos2

MDVert

ft 97.721.01099*96.66

RF)AsinIsinAsinI(sin2

MDEast

DC

DDCC

Slide 51

Minimum Curvature Method

At Point D,

X = 1,000 + 97.72 = 1,097.72 ft

Y = 1,000 + 60.25 = 1,060.25 ft

Z = 3,500 + 380.91 =3,888.91 ft

Slide 52

The Radius of Curvature Method

ft 79.83

180

)2080)(1424(

)20sin80)(sin24cos400(cos14

180

)AA)(II(

)AsinA)(sinIcosI(cosMDNorth

2

2

CDCD

CDDC

Slide 53

The Radius of Curvature Method

ft 95.14

180

)2080)(1424(

)80cos20)(cos24cos14(cos400

180

)AA)(II(

)AA)(cosIcosI(cosMDEast

2

2

CDCD

DCDC

2180

CDCD

DCDC

AAII

AcosAcosIcosIcosMDEast

Slide 54

The Radius of Curvature Method

ft 73.377180

1424

)14sin400(sin24

180

II

)IsinI(sinMDVert

CD

CD

Slide 55

The Radius of Curvature Method

At Point D,

X = 1,000 + 95.14 = 1,095.14 ft

Y = 1,000 + 79.83 = 1,079.83 ft

Z = 3,500 + 377.73 = 3,877.73 ft

Slide 56

The Tangential Method

80A 24I

20A 14I

ft 400MD D, toC fromdepth Measured

DD

CC

DD AIMDNorth cossin

ft 25.2880cos24sin400

Slide 57

The Tangential Method

ft 22.16080sinsin24400

sinsin

DD AIMDEast

ft 42.36524cos400

Icos400Vert D

Slide 58

The Tangential Method

ft 3,865.42365.423,500 Z

ft 1,028.2528.251,000Y

ft 1,160.22160.221,000X

D,Point At

Slide 59

Summary of Results (to the nearest ft)

X Y Z

Average Angle 1,100 1,084 3,878

Balanced Tangential 1,097 1,060 3,877

Minimum Curvature 1,098 1,060 3,881

Radius of Curvature 1,095 1,080 3,878

Tangential Method 1,160 1,028 3,865

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Building Hole Angle

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Holding Hole Angle

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CLOSURE

LEAD ANGLE

(HORIZONTAL) DEPARTURE

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Tool Face Angle