Post on 13-Apr-2015
description
1
Petroleum Engineering 406
Lesson 20
Directional Drilling
(continued)
2
Lesson 17- Directional Drilling cont’d
Tool-Face Angle Ouija Board Dogleg Severity Reverse Torque of Mud Motor Examples
3
Homework:
READ: Applied Drilling Engineering”, Chapter 8
(to page 375)
4
Fig. 8. 30: Graphical Ouija Analysis.
Solution Tool Face ()
5
Fig. 8. 30: Graphical Ouija Analysis.
Solution Tool Face ()
Initial Inclination = 16o
GIVEN:
= 16o
= 12o
= 12o
= ? o
= ? o
New Inclination - 12o
= 12o
Over one drilled interval (bit run)
6
Fig. 8.33
Basis of chart construction is a trigonometric relationship illustrated by two intersecting planes
= dogleg angle
7
Problem 1
Determine the new direction () 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
= 7o (inclination)
= 345o (azimuth)
= 45o (tool face angle)
L = 10 m (course length)
= 3o/ 30 m (dogleg severity)
= ? o
= 45o
9
Solution to Problem 1, part 1
I. Use Equation 8.43 to calculate .
The dogleg severity,
o1m30
m103
i
L
)i(L
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
cos cos tansin
sintantan arc
45cos7cos1tan7sin
45sin1tantan 1
3.5092027.0tan 1
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.
= 3
= 10
Find ft 60 L
30
5
N
N
and
12
Solution to Problem 2, part 1
1. Find using Equation 8.53
o1
ooooo1
NN1
4097.2999116.0cos
3cos5cos3sin5sin20coscos
coscossinsincoscos
13
Solution to Problem 2, part 2
2. Now calculate from equation 8.48.
o1 15.457052.0cos
oo
ooo1
4097.2sin3sin
5cos4097.2cos3coscos
sinsin
coscoscoscos N1
14
Solution to Problem 2, part 3
3. The dogleg severity,
= 4.01o / 100 ft
Alternate solution: Use Ouija Board
10060
4097.2(i)
L
o
15
Fig. 8.31: Solution to Example 8.6.
16
Problem 3Determine 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 oN
o
oN
o
100 89
7.1 3.4
= 7.1 - 4.3 = 2.8. = 100 - 89 = 11
17
Solution to Problem 3, part 1
1. From Equation 8.55
2/1
2221
2
1.73.4sin
2
11sin
2
8.2sinsin2
2/1
N2221
2sin
2sin
2sinsin2
= 3.01o
18
Solution to Problem 3, part 1
1. From Equation 8.43
the dogleg severity,
10085
01.3(i)
L
feet 100/5.3 o
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
7.111o
Ragland Diagram
20
Some Equations to Calculate
)coscossinsincos(cos arc NN
NN sinsincossinsincoscos
Eq. 8.53
Eq. 8.54
2
sin2
sin2
sinsin arc 2 N222
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
= 11o
= 5.7o
= 2.8o
= 3o
= 3.5o/100 ft
24
= 11o
= 5.7o
25
= 2.8o
= 3o
26
= 3o
= 3.5o/100 ft
27
= 11o
= 5.7o
= 2.8o
= 3o
= 3.5o/100 ft
28
Problem 4 - Torque and Twist1.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 = 11.5*106 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,
gdrillstrinBHAmotorM GJ
ML
GJ
ML
GJ
ML
44444 in22.19826.35.432
IDOD32
J Pipe, for
444 in 6.229813.2732
J Collars, for
30
Solution to Problem 4, cont.
o
44
2
pipecollarsM
2.137rad
deg180rad 394.2
radians 394.288.278,268.15001043.0
in22.19
in12650,3
in6.229
in12300
inlbf
105.11
ftin
12lbs.ft000,1
J
L
J
L
G
M
radians
31
Solution to Problem 4, cont.
If Length of drillpipe = 7,300 ft.,
M = 0.001043 15.68+2*2278.88]
= 4.77 radians * rad
deg180
oM 3.273 ~ 3/4 revolution!
137.2
32
Example 8.10
Design a kickoff for the wellbore in Fig. 8.35.
From Ouija Board,
= S48W = 228o N = N53W = 307o
= 2o L = 150 ft N = 6o
= 79o Find and
33
Fig. 8.36: Solution for Example 8.10.
Where to Set the Tool Face
High Side
Present Direction
High Side
New Direction
34
Dogleg Severity
From Equation 8.43
the dogleg severity,
100150
8.5(i)
L
feet 100/87.3 o
35
Fig. 8.36: Solution
for Example
8.10.
307o
With jetting bit: 325o345o
228o
M = 20o
36
Fig. 8.36: Solution for Example 8.10.
Where to Set the Tool Face
High Side
Present Direction
New Direction
Tool Face Setting Compensating for Reverse
Torque of the Motor
High Side