Title: Petroleum Engineering 406
1Petroleum Engineering 406
- Lesson 20
- Directional Drilling
- (continued)
2Lesson 17 - Directional Drilling contd
- Tool-Face Angle
- Ouija Board
- Dogleg Severity
- Reverse Torque of Mud Motor
- Examples
3Homework
- READ
- Applied Drilling Engineering, Chapter 8
- (to page 375)
4Tool Face (g)
Solution
Fig. 8. 30 Graphical Ouija Analysis.
5Over 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.
6Fig. 8.33 Basis of chart construction is a
trigonometric relationship illustrated by two
intersecting planes
b
a
aN
b dogleg angle
De
7Problem 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.
8Problem 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
9Solution to Problem 1, part 1
- I. Use Equation 8.43 to calculate .
- The dogleg severity,
10Solution to Problem 1, part 2
- 2. Use Equation 8.42 to calculate the direction
change. - New direction 3450 5.30 350.30 N9.7W
11Problem 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
12Solution to Problem 2, part 1
- 1. Find b using Equation 8.53
-
-
13Solution to Problem 2, part 2
- 2. Now calculate ? from equation 8.48.
14Solution to Problem 2, part 3
- 3. The dogleg severity,
-
- ? 4.01o / 100 ft
- Alternate solution Use Ouija Board
15Fig. 8.31 Solution to Example 8.6.
16Problem 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
17Solution to Problem 3, part 1
b 3.01o
18Solution to Problem 3, part 1
- 1. From Equation 8.43
- the dogleg severity,
19Solution 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
20Some Equations to Calculate b
Eq. 8.53
Eq. 8.54
Eq. 8.55
21Overall 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, ?.
-
22Fig. 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
25aN - a 2.8o
b 3o
26b 3o
d 3.5o/100 ft
27(aaN)/2 5.7o
aN - a 2.8o
b 3o
De 11o
d 3.5o/100 ft
28Problem 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?
29Solution to Problem 4
30Solution to Problem 4, cont.
radians
31Solution to Problem 4, cont.
- If Length of drillpipe 7,300 ft.,
- ?M 0.001043 ?15.6822278.88
- 4.77 radians
3/4 revolution!
137.2
32Example 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
33New 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.
34Dogleg Severity
- From Equation 8.43
- the dogleg severity,
35With jetting bit 325o
345o
qM 20o
307o
Fig. 8.36 Solution for Example 8.10.
228o
36Tool 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.