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

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

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

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

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

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.

Frictionless, Inclined, Straight Wellbore:

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

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)

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

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)

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

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

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?

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

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

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

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 * 0.2873 = 20,111 ft-lbf W

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

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.]

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

Solution to Problem 3 - Rotating When rotating off bottom.

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

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

Solution to Problem 3 - Lowering Thus, F8000 = mN = 0.3 * 207,846 = 62,352 lbf HL = We,2000 + We,8000 - F2000 - F8000 = 60,000 + 120,000 - 0 - 62,354 HL = 117,646 lbf while RIH

Solution to Problem 3 - Raising 2(c) Hood Load when POH: HL = We,2000 + We,8000 + F2000 + F8000 = 60,000 + 120,000 + 0 + 62,354 HL = 242,354 lbf POH

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

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

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

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

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

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

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

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

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

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

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

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

Effect of Doglegs (1) Dropoff Wellbore

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

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

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

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)

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

Effect of Doglegs (2) Buildup Wellbore

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

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

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

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

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

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

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

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?

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.

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

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

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

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

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

Solution 4 (c) From equation 12,

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

Solution 4(a, b and c) SUMMARY Rot 100,000 101,315 RIH 100,000 97,153 POH 100,000 104,477

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

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

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

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

Centralizer