13. directional drilling

PETE 661

Drilling Engineering

Lesson 13

Directional Drilling

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

Read ADE Ch.8 (Reference)

HW #7Cementingdue 10-25-02

Inclination Angle, I

Direction Angle, A

Max.Horiz.

Depart.?

Type I Type II Type III

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

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

Fig. 8.11

42131 xrr and xr

xrr and xr 3D Wells

N18E

N55WS20W

S23E

Azimuth

Angle

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)

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

,

. r2 0

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

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

* Important: Hit target within specified radius

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

L=MD1-2

Calculate = dogleg angle

DLS =L

Wellbore Surveying Methods

Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential

Other Topics Kicking off from Vertical Controlling Hole Angle

I, A, MD

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.


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


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

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


80A 24I

20A 14I


DD

CC

Z

E (x)

N (y)

C

Dz

N

D

C

yx


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

ft 71.8350cossin19400

cossin

502

8020

2

192

2414

2

AVGAVG

DCAVG

DCAVG

AIMDNorth

AAA

III



ft

AIMDEast AVEAVG

76.9950sinsin19400

sinsin

ft

IVert AVG

21.378cos19400

cos400


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

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


ft 59.59

)80cos24sin20cos14(sin2

400

)AcosIsinAcosI(sin2

MDNorth DDCC


96.66ft

)80sin24sin20sin14(sin2

400

)AsinIsinAsinI(sin2

MDEast DDCC


ft77.376)14cos24(cos2

400

)IcosI(cos2

MDVert CD


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

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)


The dogleg angle, , is given by:

radians 36082.020.67

0.935609

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

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


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


ft 380.911.01099*376.77

RF)IcosI(cos2

MDVert

ft 97.721.01099*96.66

RF)AsinIsinAsinI(sin2

MDEast

DC

DDCC


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

The Radius of Curvature Method

ft 79.83

180

)2080)(1424(

)20sin80)(sin24cos400(cos14

180

)AA)(II(

)AsinA)(sinIcosI(cosMDNorth

2

2

CDCD

CDDC


ft 95.14

180

)2080)(1424(

)80cos20)(cos24cos14(cos400

180

)AA)(II(

)AA)(cosIcosI(cosMDEast

2

2

CDCD

DCDC

2180

CDCD

DCDC

AAII

AcosAcosIcosIcosMDEast


ft 73.377180

1424

)14sin400(sin24

180

II

)IsinI(sinMDVert

CD

CD


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

The Tangential Method

80A 24I

20A 14I


DD

CC

DD AIMDNorth cossin

ft 25.2880cos24sin400


ft 22.16080sinsin24400

sinsin

DD AIMDEast

ft 42.36524cos400

Icos400Vert D


ft 3,865.42365.423,500 Z

ft 1,028.2528.251,000Y

ft 1,160.22160.221,000X

D,Point At

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

Building Hole Angle

Holding Hole Angle

CLOSURE

LEAD ANGLE

(HORIZONTAL) DEPARTURE

Tool Face Angle

13. directional drilling

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