1. Lesson 36 Torque and Drag Calculations
PETE 411 Well Drilling
Lesson 36 Torque and Drag Calculations

2. Torque and Drag Calculations
Friction
Logging
Hook Load
Lateral Load
Torque Requirements
Examples

3. Assignments:
PETE 411 Design Project due December 9, 2002, 5 p.m. HW#18 Due Friday, Dec. 6

4. Friction - Stationary S Fy = 0 Horizontal surface N No motion
No applied force
S Fy = 0
N = W N W
N= Normal force = lateral load = contact force = reaction force

5. Sliding Motion Horizontal surface N Velocity, V > 0 V = constant
Force along surface
N = W
F = N = W
 N F W

6. 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. Frictionless, Inclined, Straight Wellbore:

8. Effect of Friction (no doglegs):
2. Consider Effect of Friction ( no doglegs):

9. 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. Effect of Friction (no doglegs):
(b) Raising: Friction still opposes motion, so
(4)

11. 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. Solution From Equation (3) above, (3)
When pipe is barely sliding down the wellbore,

13. Solution
This is the maximum hole angle (inclination) that can be logged without the aid of tubulars.
Note:

14. 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. Problem 2 - Solution - Force
(a) What force will it take to move this pipe along the horizontal section of the wellbore? N
F = ?
F = 0 W
N = W = 30 lb/ft * 8,000 ft = 240,000 lb
F = mN = 0.3 * 240,000 lb = 72,000 lb
Force to move pipe, F = 72,000 lbf

16. 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. T
d/2 F
Then, as before,
N = W = 30 lb/ft * 8,000 ft = 240,000 lbf
Torque = F*d/2 = mNd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft
Torque to rotate pipe, T = 21,000 ft-lbf

17. Problem 2 - Equations - Horizontal
N = W
T = F * s
F = mN
( s=d/24 ) W
Force to move pipe, F = mW = 72,000 lbf
Torque, T = mWd/(24 ) = 21,000 ft-lbf
An approximate equation, with W in lbf and d in inches

18. 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.
Taking moments about the point P:
Torque, T = W * (d/2) sin f in-lbf T F
d/2 f P
Where f = atan m = atan 0.3 = 16.70o
T = 240,000 * 7/24 * = 20,111 ft-lbf W

19. 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. 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. (a) Hook load when rotating off bottom:
Solution to Problem 3
(a) Hook load when rotating off bottom:

22. Solution to Problem 3 - Rotating
When rotating off bottom.

23. 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, 0o

24. Solution to Problem 3 - lowering
In the inclined section,
N = 30 * 8,000 * sin 60
= 207,846 lbf

25. Solution to Problem 3 - Lowering
Thus, F8000 = mN
= 0.3 * 207,846 = 62,352 lbf
HL = We, We, F F8000
= 60, , ,354
HL = 117,646 lbf while RIH

26. Solution to Problem 3 - Raising
2(c) Hood Load when POH:
HL = We, We, F F8000
= 60, , ,354
HL = 242,354 lbf POH

27. Solution to Problem 3 - Summary
ROT
RIH
2,000
POH MD ft
10,000
240,000
60,000
120,000
180,000

28. Solution to Problem 3 - rotating
2(d) Torque when rotating off bottom:
In the Inclined Section:

29. Solution to Problem 3 - rotating
(i) As a first approximation, assume the pipe lies at lowest point of hole:

30. 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.
The pipe will tend to climb up the side of the wellbore…as it rotates

31. Solution to Problem 3 - Rotating
Assume “Equilibrium”
at angle f as shown.
…… (6)
…… (7)

32. Solution to Problem 3 - rotating
Solving equations (6) & (7)
(8)

33. Solution to Problem 3 - rotating
(ii) continued
Taking moments about the center of the pipe:
Evaluating the problem at hand:
From Eq. (8),

34. Solution to Problem 3 - rotating
Evaluating the problem at hand:
From Eq. (6),

35. Solution to Problem 3 - rotating
Evaluating the problem at hand:
From Eq. (9),

36. 2 (d) (ii) Alternate Solution:
Solution to Problem 3
2 (d) (ii) Alternate Solution:

37. Taking moments about tangent point,
Solution to Problem 3
Taking moments about tangent point,

38. 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 =

39. Effect of Doglegs
(1) Dropoff Wellbore

40. A. Neglecting Axial Friction (e.g. pipe rotating)
Effect of Doglegs
A. Neglecting Axial Friction (e.g. pipe rotating)
W sin I + 2T

41. A. Neglecting Axial Friction
Effect of Doglegs
A. Neglecting Axial Friction

42. Effect of Doglegs B. Including Friction (Dropoff Wellbore)
While pipe is rotating
(10)&(11)

43. B. Including Friction While lowering pipe (RIH) (as above) i.e. (12)
Effect of Doglegs
B. Including Friction
While lowering pipe (RIH)
(as above)
i.e (12)

44. B. Including Friction While raising pipe (POH) (13) (14)
Effect of Doglegs
B. Including Friction
While raising pipe (POH)
(13)
(14)

45. Effect of Doglegs
(2) Buildup Wellbore

46. A. Neglecting Friction (e.g. pipe rotating)
Effect of Doglegs
A. Neglecting Friction
(e.g. pipe rotating)

47. A. Neglecting Axial Friction
Effect of Doglegs
A. Neglecting Axial Friction

48. Effect of Doglegs B. Including Friction (Buildup Wellbore)
When pipe is rotating
(15)&(16)

49. B. Including Friction While lowering pipe (RIH) (15) (17)
Effect of Doglegs
B. Including Friction
While lowering pipe (RIH)
(15)
(17)

50. While raising pipe (POH) (18) (19)
Effect of Doglegs
While raising pipe (POH)
(18)
(19)

51. 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.
Buoyed weight of pipe = 30 lbm/ft
m = 0.25

52. Problem # 4 - Curved Wellbore with Friction
T = 100,000 lbf

53. 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. 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. Solution 4(a) - Rotating
T68 = 101,315 lbf
From equation (11),
T60 = 100,000 lbf

56. Solution 4 (b)
(b) Tension in pipe 100 ft Up-Hole when Pipe is being lowered:
From equation (10):

57. From equation 10, From equation 12,
Solution 4 (b)
From equation 10,
From equation 12,

58. Solution 4(b) - Lowering
T68 = 97,153 lbf
From equation 12,
T60 = 100,000 lbf

59. Solution 4 (c)
(c) Tension in Pipe 100 ft Up-Hole when pipe is being raised:
From equation (10),

60. Solution 4 (c)
From equation 12,

61. Solution 4(c) - Raising From equation 12, T68 = 105,477 lbf

62. Solution 4(a, b and c) SUMMARY
Rot , ,315
RIH , ,153
POH , ,477

63. 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. for 40 ft distance, i.e., Lateral load on centralizer,
Solution 4 (d)
for 40 ft distance,
i.e., Lateral load on centralizer,

65. Alternate Approach
(d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated)
From above,
So, 30 ft up-hole,

66. From Eq. (10), for 40 ft centralizer spacing,
Alternate Approach
From Eq. (10),
for 40 ft centralizer spacing,

67. Centralizer