calculation equations for directional drilling

1

Directional Drilling Calculations

2

Type I Type II Type III

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

3

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

4

N18E

N55WS20W

S23E

Azimuth

Angle

5

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)

6

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)

7

(i) Maximum Inclination

Angle

r1 18 000

15

,

. r2 0

D4 1

12 500 2 500

10 000

D

ft

, ,

,

8

(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

9

(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

10

(ii) Measured Depth of Well

265,9180

26.33,8202,500

LrDMD Holdrad11

ft 518,13MD

11

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

* Important: Hit target within specified radius

12

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

L=MD1-2

Calculate = dogleg angle

DLS =L

13

Wellbore Surveying Methods

Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential

Other Topics Kicking off from Vertical Controlling Hole Angle

14

I, A, MD

15

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.

16

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

17

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

18

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

19

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

20

The Average Angle Method

21

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

22

ft 71.8350cossin19400

cossin

502

8020

2

192

2414

2

AVGAVG

DCAVG

DCAVG

AIMDNorth

AAA

III

The Average Angle Method

23

The Average Angle Method

ft

AIMDEast AVEAVG

76.9950sinsin19400

sinsin

ft

IVert AVG

21.378cos19400

cos400

24

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

25

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

26

The Balanced Tangential Method

ft 59.59

)80cos24sin20cos14(sin2

400

)AcosIsinAcosI(sin2

MDNorth DDCC

27

The Balanced Tangential Method

96.66ft

)80sin24sin20sin14(sin2

400

)AsinIsinAsinI(sin2

MDEast DDCC

28

The Balanced Tangential Method

ft77.376)14cos24(cos2

400

)IcosI(cos2

MDVert CD

29

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

30

Minimum Curvature Method

31

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)

32

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

33

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

34

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

35

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

36

The Radius of Curvature Method

ft 79.83

180

)2080)(1424(

)20sin80)(sin24cos400(cos14

180

)AA)(II(

)AsinA)(sinIcosI(cosMDNorth

2

2

CDCD

CDDC

37

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

38

The Radius of Curvature Method

ft 73.377180

1424

)14sin400(sin24

180

II

)IsinI(sinMDVert

CD

CD

39

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

40

The Tangential Method

80A 24I

20A 14I

ft 400MD D, toC fromdepth Measured

DD

CC

DD AIMDNorth cossin

ft 25.2880cos24sin400

41

The Tangential Method

ft 22.16080sinsin24400

sinsin

DD AIMDEast

ft 42.36524cos400

Icos400Vert D

42

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

43

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 1,016 1,028 3,865

44

Torque and DragCalculations

45

Friction - Stationary

• Horizontal surface

• No motion• No applied force

Fy = 0

N = W

N

W

N= Normal force = lateral load = contact force = reaction force

46

Sliding Motion

• Horizontal surface

• Velocity, V > 0

• V = constant

• Force along surface

N = W

F = N = W

N

W

F N

47

Frictionless, Inclined, Straight Wellbore:

1. Consider

a section

of pipe

in the

wellbore.

In the absence of FRICTION the forces acting on the pipe are buoyed weight, axial tension and the reaction force, N, normal to the wellbore.

48

Frictionless, Inclined, Straight Wellbore:

pipe.ROTATING for used are equations

(2) : wellbore to 0

(1) : wellborealong 0

ar

These

F

F

IcosWT

IsinWN

49

Effect of Friction (no doglegs):

2. Consider Effect of Friction ( no doglegs):

50

Effect of Friction (no doglegs):

Frictional Force, F = N = W sin I

where 0 < < 1 ( is the coeff. of friction)

usually 0.15 < < 0.4 in the wellbore

(a) Lowering: Friction opposes motion, so

(3)IsinWIcosWT

FIcosWT f

51

Effect of Friction (no doglegs):

(b) Raising: Friction still opposes motion,

so

IsinWIcosWT

FIcosWT f

(4)

52

Problem 1

What is the maximum hole angle (inclination angle) that can be logged

without the aid of drillpipe, coiled tubing or other tubulars?

(assume =0.4)

53

Solution

From Equation (3) above,

(3)

When the logging tool is barely sliding down the wellbore,

IsinWIcosWT

0T

IsinW4.0IcosW0

54

Solution

This is the maximum hole angle (inclination) that can be logged without the aid of tubulars.

Note:

68.2I

2.5Ior tan 4.0Icot

Icot

55

Problem 2

Consider a well with a long horizontal section. An 8,000-ft long string of 7” OD csg. is in the hole. Buoyed weight of pipe = 30 lbs/ft. = 0.3

(a) What force will it take to move this pipe along the horizontal section of the wellbore?

(b) What torque will it take to rotate this pipe?

56

Problem 2 - Solution - Force

(a) What force will it take to move this pipe along the horizontal section of the wellbore?

F = ? F = 0N

W

N = W = 30 lb/ft * 8,000 ft = 240,000 lb

F = N = 0.3 * 240,000 lb = 72,000 lb

Force to move pipe, F = 72,000 lbf

57

Problem 2 - Solution - Force

(b) What torque will it take to rotate this pipe?

As an approximation, let us

assume that the pipe lies on

the bottom of the wellbore.

Then, as before, N = W = 30 lb/ft * 8,000 ft = 240,000 lbf

Torque = F*d/2 = Nd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft

Torque to rotate pipe, T = 21,000 ft-lbf

F

T

d/2

58

Problem 2 - Equations - Horizontal

Torque, T = Wd/(24 ) = 21,000 ft-lbf

F = N T = F * dN = W

W

Force to move pipe, F = W = 72,000 lbf

An approximate equation, with W in lbf and d in inches

59

Horizontal - Torque

A more accurate equation for torque in a horizontal wellbore may be obtained by taking into consideration the fact that a rotating pipe will ride up the side of the wellbore to some angle .

Taking moments about the point P:

Torque, T = W * (d/2) sin in-lbf

Where = atan = atan 0.3 = 16.70o

T = 240,000 * 7/24 * 0.2873 = 20,111 ft-lbf

FT

d/2 P

W

60

Problem 3

A well with a measured depth of 10,000 ft. may be approximated as being perfectly vertical to the kick-off point at 2,000 ft. A string of 7” OD csg. is in the hole; total length is 10,000 ft. The 8,000-ft segment is inclined at 60 deg. Buoyed weight of pipe = 30 lbs/ft. = 0.3

61

Problem 3

Please determine the following:

(a) Hook load when rotating off bottom

(b) Hook load when RIH

(c) Hook load when POH

(d) Torque when rotating off bottom

[ ignore effects of dogleg at 2000 ft.]

62

Solution to Problem 3

(a) Hook load when rotating off bottom:

63

Solution to Problem 3 - Rotating

When rotating off bottom.

lbf 120,000lbf 000,60

60cos*ft 8000*ft

lb30ft 2000*

ft

lb30

HLHLHL

5.0

80002000

lbf 000,180HL

64

Solution to Problem 3 - lowering

2 (b) Hook load when RIH:

The hook load is decreased by friction in the wellbore.

In the vertical portion,

Thus, 0F

0osin*2000*30N

2000

o

NFf

0o

65

Solution to Problem 3 - lowering

In the inclined section,

N = 30 * 8,000 * sin 60

= 207,846 lbf

66

Solution to Problem 3 - Lowering

HL = We,2000 + We,8000 - F2000 - F8000

= 60,000 + 120,000 - 0 - 62,354

Thus, F8000 = N = 0.3 * 207,846 = 62,352 lbf

HL = 117,646 lbf while RIH

67

Solution to Problem 3 - Raising

2(c) Hood Load when POH:

HL = We,2000 + We,8000 + F2000 + F8000

= 60,000 + 120,000 + 0 + 62,354

HL = 242,354 lbf POH

68

Solution to Problem 3 - Summary

MDft

RIHROT

POH

Axial Tension

HL

69

Solution to Problem 3 - rotating

2(d) Torque when rotating off bottom:In the Inclined Section:

NF

IsinWN

2

d*F

Arm*Force

Torque

f

70

Solution to Problem 3 - rotating

(i) As a first approximation, assume the pipe lies at lowest point of hole:

12

1*

2

7*60sin*8000*30*3.0

2

dIsinW

2

dN

2

dFTorque f

lbf-ft 187,18Torque

71

Solution to Problem 3 - rotating

(ii) More accurate evaluation:Note that, in the above figure, forces are not balanced; there is no force to balance the friction force Ff.

The pipe will tend to climb up the side of the wellbore…as it rotates

72

Solution to Problem 3 - Rotating

Assume “Equilibrium”

at angle as shown.

sinIsinWFF fTangentAlong 0

cosIsinWNF Tangentto.Perpend 0

…… (7)

sinIsinWN …… (6)

cosIsinWN

73

Solution to Problem 3 - rotating

Solving equations (6) & (7)

(8))(tan

tan

cosIsinW

sinIsinW

N

N

1

74

Solution to Problem 3 - rotating

(ii) Continued

Evaluating the problem at hand:

From Eq. (8),

Taking moments about the center of the pipe:

2

d*FT f

70.16

)3.0(tan)(tan 11

75

Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (6),

lbf 724.59F

70.16sin*sin60*8000*30

sinIsinWF

f

f

76

Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (9),

lbf-ft 420,17Torque

12

1*

2

7*59,724

2

d*FT f

77

Solution to Problem 3

2 (d) (ii) Alternate Solution:

78

Solution to Problem 3

Taking moments about tangent point,

24

7*70.16sin*sin60*8000*30

2

dsinIsinWT

lbf-ft 420,17T

79

Solution to Problem 3

Note that the answers in parts (i) & (ii) differ by a factor of cos

(i) T = 18,187

(ii) T = 17,420

cos = cos 16.70 = 0.9578