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Slide 2 of 62. Torque and Drag Calculations. Friction Logging Hook Load Lateral Load Torque Requirements Examples. . Slide 3 of 62. HW
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1. Slide 1 of 62 PETE 661 Drilling Engineering
2. Slide 2 of 62 Torque and Drag Calculations Friction
Logging
Hook Load
Lateral Load
Torque Requirements
Examples
3. Slide 3 of 62
4. Slide 4 of 62 Friction - Stationary Horizontal surface
No motion
No applied force
S Fy = 0
N = W
5. Slide 5 of 62 Sliding Motion Horizontal surface
Velocity, V > 0
V = constant
Force along surface
N = W
F = ??N = ??W
6. Slide 6 of 62 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.
7. Slide 7 of 62 Frictionless, Inclined, Straight Wellbore:
8. Slide 8 of 62 Effect of Friction (no doglegs): 2. Consider Effect of Friction ( no doglegs):
9. Slide 9 of 62 Effect of Friction (no doglegs): Frictional Force, F = mN = mW sin I
where 0 < m < 1 (m is the coeff. of friction)
usually 0.15 < m < 0.4 in the wellbore
(a) Lowering: Friction opposes motion, so
(3)
10. Slide 10 of 62 Effect of Friction (no doglegs): (b) Raising: Friction still opposes motion,
so
11. Slide 11 of 62 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)
12. Slide 12 of 62 Solution From Equation (3) above,
(3)
When pipe is barely sliding down the wellbore,
13. Slide 13 of 62 Solution This is the maximum hole angle (inclination) that can be logged without the aid of tubulars.
Note:
14. Slide 14 of 62 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. m = 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?
15. Slide 15 of 62 Problem 2 - Solution - Force (a) What force will it take to move this pipe along the horizontal section of the wellbore?
16. Slide 16 of 62 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.
17. Slide 17 of 62 Problem 2 - Equations - Horizontal
18. Slide 18 of 62 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 f.
19. Slide 19 of 62 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. m = 0.3
20. Slide 20 of 62 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.]
21. Slide 21 of 62 Solution to Problem 3 (a) Hook load when rotating off bottom:
22. Slide 22 of 62 Solution to Problem 3 - Rotating When rotating off bottom.
23. Slide 23 of 62 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,
24. Slide 24 of 62 Solution to Problem 3 - lowering In the inclined section,
N = 30 * 8,000 * sin 60
= 207,846 lbf
25. Slide 25 of 62 Solution to Problem 3 - Lowering
26. Slide 26 of 62 Solution to Problem 3 - Raising 2(c) Hood Load when POH:
27. Slide 27 of 62 Solution to Problem 3 - Summary
28. Slide 28 of 62 Solution to Problem 3 - rotating 2(d) Torque when rotating off bottom:
In the Inclined Section:
29. Slide 29 of 62 Solution to Problem 3 - rotating (i) As a first approximation, assume the pipe lies at lowest point of hole:
30. Slide 30 of 62 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.
31. Slide 31 of 62 Solution to Problem 3 - Rotating Assume “Equilibrium”
at angle f as shown.
32. Slide 32 of 62 Solution to Problem 3 - rotating Solving equations (6) & (7)
(8)
33. Slide 33 of 62 Solution to Problem 3 - rotating (ii) continued
Taking moments about the center of the pipe:
Evaluating the problem at hand:
From Eq. (8),
34. Slide 34 of 62 Solution to Problem 3 - rotating Evaluating the problem at hand:
From Eq. (6),
35. Slide 35 of 62 Solution to Problem 3 - rotating Evaluating the problem at hand:
From Eq. (9),
36. Slide 36 of 62 Solution to Problem 3 2 (d) (ii) Alternate Solution:
37. Slide 37 of 62 Solution to Problem 3 Taking moments about tangent point,
38. Slide 38 of 62 Solution to Problem 3 Note that the answers in parts (i) & (ii) differ by a factor of cos f
(i) T = 18,187
(ii) T = 17,420
cos f = cos 16.70 = 0.9578
39. Slide 39 of 62 Effect of Doglegs (1) Dropoff Wellbore
40. Slide 40 of 62 Effect of Doglegs A. Neglecting Axial Friction (e.g. pipe rotating)
41. Slide 41 of 62 Effect of Doglegs A. Neglecting Axial Friction
42. Slide 42 of 62 Effect of Doglegs B. Including Friction (Dropoff Wellbore)
While pipe is rotating
(10)&(11)
43. Slide 43 of 62 Effect of Doglegs B. Including Friction
While lowering pipe (RIH)
(as above)
i.e. (12)
44. Slide 44 of 62 Effect of Doglegs B. Including Friction
While raising pipe (POH)
(13)
(14)
45. Slide 45 of 62 Effect of Doglegs (2) Buildup Wellbore
46. Slide 46 of 62 Effect of Doglegs A. Neglecting Friction
(e.g. pipe rotating)
47. Slide 47 of 62 Effect of Doglegs A. Neglecting Axial Friction
48. Slide 48 of 62 Effect of Doglegs B. Including Friction (Buildup Wellbore)
When pipe is rotating
(15)&(16)
49. Slide 49 of 62 Effect of Doglegs B. Including Friction
While lowering pipe (RIH)
(15)
(17)
50. Slide 50 of 62 Effect of Doglegs While raising pipe (POH)
(18)
(19)
51. Slide 51 of 62 Problem #4 - Curved Wellbore with Friction In a section of our well, hole angle drops at the rate of 8 degrees per 100 ft.
The axial tension is 100,000 lbf at the location where the hole angle is 60 degrees.
52. Slide 52 of 62 Problem # 4 - Curved Wellbore with Friction
53. Slide 53 of 62 Evaluate the Following: (a) What is the axial tension in the pipe 100 ft. up the hole if the pipe is rotating?
(b) What is the axial tension in the pipe 100 ft up the hole if the pipe is being lowered into the hole?
(c) What is the axial tension in the pipe 100 ft up the hole if the pipe is being pulled out of the hole?
(d) What is the lateral load on a centralizer at incl.=64 if the centralizer spacing is 40 ft?
54. Slide 54 of 62 Solution 4(a) - Rotating Axial tension 100 ft up hole when pipe is rotating :
Pipe is rotating so frictional effect on axial load may be neglected.
55. Slide 55 of 62 Solution 4(a) - Rotating From equation (11),
56. Slide 56 of 62 Solution 4 (b) (b) Tension in pipe 100 ft Up-Hole when Pipe is being lowered:
From equation (10):
57. Slide 57 of 62 Solution 4 (b) From equation 10,
From equation 12,
58. Slide 58 of 62 Solution 4(b) - Lowering From equation 12,
59. Slide 59 of 62 Solution 4 (c) (c) Tension in Pipe 100 ft Up-Hole when pipe is being raised:
From equation (10),
60. Slide 60 of 62 Solution 4 (c)
61. Slide 61 of 62 Solution 4(c) - Raising From equation 12,
62. Slide 62 of 62 Solution 4(a, b and c) SUMMARY
Rot 100,000 101,315
RIH 100,000 97,153
POH 100,000 104,477
63. Slide 63 of 62 Solution 4 (d) (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated):
From above,
This is for 100 ft distance
64. Slide 64 of 62 Solution 4 (d) for 40 ft distance,
i.e., Lateral load on centralizer,
65. Slide 65 of 62 Alternate Approach (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated)
From above,
From above,
So, 30 ft up-hole,
66. Slide 66 of 62 Alternate Approach
From Eq. (10),
for 40 ft centralizer spacing,
67. Slide 67 of 62 Centralizer
