Transcript LESSON 11 Directional Drilling

Petroleum Engineering 406
Lesson 20
Directional Drilling
(continued)
1
Lesson 17
- Directional Drilling cont’d
Tool-Face Angle
Ouija Board
Dogleg Severity
Reverse Torque of Mud Motor
Examples
2
Homework:
READ:
Applied Drilling Engineering”, Chapter 8
(to page 375)
3
Solution
Tool Face (g)
Fig. 8. 30: Graphical Ouija Analysis.
4
Over one drilled interval (bit run)
GIVEN:
a = 16
o
Solution
Tool Face (g)
De = 12
o
aN = 12
o
De = 12o
o
Initial Inclination = 16
g=?
o
b=?
o
Fig. 8. 30: Graphical Ouija Analysis.
5
Fig. 8.33
b
a
aN
De
Basis of chart
construction is a
trigonometric
relationship
illustrated by two
intersecting
planes
b = dogleg angle
6
Problem 1
Determine the new direction (eN) for a
whipstock set at 705 m with a tool-face
setting of 450 degrees right of high side
for a course length of 10 m.
The inclination is 70 and the direction is
N15W. The curve of the whipstock will
cause a total angle change of 30/30 m.
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Problem 1
a=7
o
e = 345
o
(inclination)
(azimuth)
g = 45
o
g = 45
(tool face angle)
L = 10 m
(course length)
o
d = 3 / 30 m
(dogleg severity)
o
eN = ?
o
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Solution to Problem 1, part 1
I. Use Equation 8.43 to calculate b .
The dogleg severity,
b
d  (i)
L

d L 3  10 m
o
b 

1
i
30 m
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Solution to Problem 1, part 2
2. Use Equation 8.42 to calculate the
direction change.
De  arc tan
tan b sin g
sin a  tan b cos a cos g

De  tan
tan 1 sin 45
1

sin 7  tan 1 cos 7 cos 45
De  tan
1
0.092027  5.3

New direction =3450 +5.30 = 350.30 = N9.7W
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Problem 2
Determine where to set the tool face angle, g
for a jetting bit to go from a direction of 100
to 300 and from an inclination of 30 to 50.
Also calculate the dogleg severity, assuming
that the trajectory change takes 60 ft.

a=3
aN  5
e = 10

e
Find
g and d
e N  30
L  60 ft
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Solution to Problem 2, part 1
1. Find b using Equation 8.53
b  cos 1cos De sin aN sin a  cos aN cos a 
1

 cos cos 20 sin 5 sin 3  cos 5 cos 3
o
o
o
o
o

 cos 1 0.999116  2.4097o
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Solution to Problem 2, part 2
2. Now calculate g from equation 8.48.
g  cos
g  cos
1
1
 cos a cos b  cos aN 


sin a sin b


 cos 3o cos 2.4097o  cos 5o 


o
o
sin 3 sin 2.4097


g  cos
1
0.7052  45.15
o
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Solution to Problem 2, part 3
3. The dogleg severity,
b
2.4097
d  (i) 
 100
L
60
o
d = 4.01o / 100 ft
Alternate solution: Use Ouija Board
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Fig. 8.31: Solution to Example 8.6.
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Problem 3
Determine the dogleg severity following a
jetting run where the inclination was changed
from 4.3o to 7.1o and the direction from N89E
to S80E over a drilled interval of 85 feet.
1. Solve by calculation.
2. Solve using Ragland diagram
a  4.3o
e  89
o
aN  7.1o
L = 85 ft
Da = 7.1 - 4.3 = 2.8.
eN  100
De = 100 - 89 = 11
o
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Solution to Problem 3, part 1
1. From Equation 8.55
 2 Da
2 De
2  a  aN 
b  2 sin sin
 sin
sin 

2
2
 2 

1/ 2
 2 2.8
2 11
2  4.3  7.1
b  2 sin sin
 sin
sin 

2
2
2



1/ 2
1
1
b = 3.01
o
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Solution to Problem 3, part 1
1. From Equation 8.43
the dogleg severity,
b
3.01
d 
(i) 
 100
L
85
d  3.5 / 100 feet
o
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Solution to Problem 3, part 2
2. Construct line of length a (4.3o)
Measure angle De
(11o )
Construct line of length aN (7.1o)
Measure length b
4.3
(Measure angle g)
o
b
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Ragland Diagram
7.1
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Some Equations to Calculate b
Eq. 8.53
b  arc cos(cos De sin aN sin a  cos a cos aN )
Eq. 8.54
cosb  cosDe sinaN sin a  cos Da  sin a sin aN
Eq. 8.55
 Da 
2  De 
2  a  aN 
b  2 arc sin sin 
  sin 
 sin 

 2 
 2 
 2 
2
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Overall Angle Change and
Dogleg Severity
Equation 8.51 derived by Lubinski is used
to construct Figure 8.32,
a nomograph for determining
the total angle change b and
the dogleg severity, d.
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Fig. 8.32: Chart
for determining
dogleg severity
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(aaN)/2 = 5.7
o
aN  a = 2.8
o
b=3
o
De = 11
o
d = 3.5 /100 ft
o
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(aaN)/2
o
= 5.7
De = 11
o
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aN  a = 2.8
o
b=3
o
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b=3
o
d = 3.5 /100 ft
o
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(aaN)/2 = 5.7
o
aN  a = 2.8
o
b=3
o
De = 11
o
d = 3.5 /100 ft
o
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Problem 4 - Torque and Twist
1.Calculate the total angle change of
3,650 ft. of 4 1/2 inches (3.826 ” ID)
Grade E 16.60 #/ft drill pipe and 300 ft.
of 7” drill collars (2 13/16” ID) for a bitgenerated torque of 1,000 ft-lbf. Assume
that the motor has the same properties
as the 7” drill collars. Shear modules of
steel, G = 11.5*106 psi.
2. What would be the total angle change if
7,300 ft. of drill pipe were used?
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Solution to Problem 4
From Equation 8.56,
 ML 
 ML 
 ML 
M  





 GJ motor  GJ BHA  GJ  drillstring
(
)
(
)


4
4
for Pipe, J 
OD  ID 
4.5 4  3.826 4  19.22 in4
32
32
(
)
 4
for Collars, J 
7  2.813 4  229.6 in4
32
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Solution to Problem 4, cont.
M  L 
L  
 M   
  
G  J  collars  J pipe 
in
1,000 ft.lbs  12
ft

lbf
11.5  10 2
in
radians
 300  12in 3,650  12 in 



4
4
19.22 in 
 229.6 in
 0.001043 15.68  2,278.88  2.394 radians
180 deg
   2.394 rad 
 137.2o
 rad
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Solution to Problem 4, cont.
If Length of drillpipe = 7,300 ft.,
M = 0.001043 15.68+2*2278.88]
= 4.77 radians *
M  273.3
o
180 deg
 rad
~ 3/4 revolution!
137.2
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Example 8.10
Design a kickoff for the wellbore in Fig. 8.35.
e = S48W = 228o
eN = N53W = 307o
a = 2o
L = 150 ft
aN = 6o
De = 79o
Find b, g and d
o
From Ouija Board, b  5.8 , g  97
o
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New
Direction
Where to Set the Tool Face
o
b  5.8
o
g  97
High
Side
Present Direction
High
Side
Fig. 8.36: Solution for Example 8.10.
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Dogleg Severity
From Equation 8.43
the dogleg severity,
b
5 .8
d
(i) 
 100
L
150
d  3.87 / 100 feet
o
34
o
o
With jetting bit: 325
345
o
M = 20
o
307
Fig. 8.36:
Solution
for
Example
8.10.
o
228
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Tool Face
Setting
New
Direction
High
Side
Where to Set the Tool Face
Compensating for Reverse
Torque of the Motor
Present
Direction
High
Side
Fig. 8.36: Solution for Example 8.10.
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