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## Petroleum Engineering 406

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### Petroleum Engineering 406 Lesson 20 Directional Drilling (continued) Lesson 17 - Directional Drilling cont d Tool-Face Angle Ouija Board Dogleg Severity Reverse ... – PowerPoint PPT presentation

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Title: Petroleum Engineering 406

1
Petroleum Engineering 406
• Lesson 20
• Directional Drilling
• (continued)

2
Lesson 17 - Directional Drilling contd
• Tool-Face Angle
• Ouija Board
• Dogleg Severity
• Reverse Torque of Mud Motor
• Examples

3
Homework
• Applied Drilling Engineering, Chapter 8
• (to page 375)

4
Tool Face (g)
Solution
Fig. 8. 30 Graphical Ouija Analysis.
5
Over one drilled interval (bit run)
GIVEN a 16o De 12o aN 12o
Tool Face (g)
Solution
New Inclination - 12o
De 12o
g ? o b ? o
Initial Inclination 16o
Fig. 8. 30 Graphical Ouija Analysis.
6
Fig. 8.33 Basis of chart construction is a
trigonometric relationship illustrated by two
intersecting planes
b
a
aN
b dogleg angle
De
7
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.

8
Problem 1
• a 7o (inclination)
• e 345o (azimuth)
• g 45o (tool face angle)
• L 10 m (course length)
• d 3o/ 30 m (dogleg severity)
• eN ? o

g 45o
9
Solution to Problem 1, part 1
• I. Use Equation 8.43 to calculate .
• The dogleg severity,

10
Solution to Problem 1, part 2
• 2. Use Equation 8.42 to calculate the direction
change.
• New direction 3450 5.30 350.30 N9.7W

11
Problem 2
• Determine where to set the tool face angle,
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
• e 10
• Find

12
Solution to Problem 2, part 1
• 1. Find b using Equation 8.53

13
Solution to Problem 2, part 2
• 2. Now calculate ? from equation 8.48.

14
Solution to Problem 2, part 3
• 3. The dogleg severity,
• ? 4.01o / 100 ft
• Alternate solution Use Ouija Board

15
Fig. 8.31 Solution to Example 8.6.
16
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
• L 85 ft

Da 7.1 - 4.3 2.8. De 100 - 89 11
17
Solution to Problem 3, part 1
• 1. From Equation 8.55

b 3.01o
18
Solution to Problem 3, part 1
• 1. From Equation 8.43
• the dogleg severity,

19
Solution to Problem 3, part 2
• 2. Construct line of length ? (4.3o)
• Measure angle ?? (11o )
• Construct line of length ?N (7.1o)
• Measure length ?
• (Measure angle ?)

4.3
11o
b
7.1
Ragland Diagram
20
Some Equations to Calculate b
Eq. 8.53

Eq. 8.54
Eq. 8.55
21
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 ? and
• the dogleg severity, ?.

22
Fig. 8.32 Chart for determining dogleg severity
23
(aaN)/2 5.7o
aN - a 2.8o
b 3o
De 11o
d 3.5o/100 ft
24
(aaN)/2 5.7o
De 11o
25
aN - a 2.8o
b 3o
26
b 3o
d 3.5o/100 ft
27
(aaN)/2 5.7o
aN - a 2.8o
b 3o
De 11o
d 3.5o/100 ft
28
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 bit-generated torque of 1,000
ft-lbf. Assume that the motor has the same
properties as the 7 drill collars. Shear modules
of steel, G 11.5106 psi.
• 2. What would be the total angle change if
7,300 ft. of drill pipe were used?

29
Solution to Problem 4
• From Equation 8.56,

30
Solution to Problem 4, cont.
31
Solution to Problem 4, cont.
• If Length of drillpipe 7,300 ft.,
• ?M 0.001043 ?15.6822278.88

3/4 revolution!
137.2
32
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
From Ouija Board, b 5.8o, g 97o
33
New Direction
Where to Set the Tool Face
b 5.8o g 97o
High Side
High Side
Present Direction
Fig. 8.36 Solution for Example 8.10.
34
Dogleg Severity
• From Equation 8.43
• the dogleg severity,

35
With jetting bit 325o
345o
qM 20o
307o
Fig. 8.36 Solution for Example 8.10.
228o
36
Tool Face Setting
Where to Set the Tool Face
Compensating for Reverse Torque of the Motor
New Direction
Present Direction
High Side
High Side
Fig. 8.36 Solution for Example 8.10.