PETE 411 Well Drilling

PETE 411Well Drilling Lesson 36 Torque and Drag Calculations

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

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

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

Sliding Motion N • Horizontal surface • Velocity, V > 0 • V = constant • Force along surface N = W F = N = W F  N 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 N F = ? F = 0 (a) What force will it take to move this pipe along the horizontal section of the wellbore? 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 T d/2 (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. 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 Taking moments about the point P: Torque, T = W * (d/2) sin f in-lbf 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. T F d/2 f P Where f = atan m = atan 0.3 = 16.70o W T = 240,000 * 7/24 * 0.2873 = 20,111 ft-lbf

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.]

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 HL = We,2000 + We,8000 + F2000 + F8000 2(c) Hood Load when POH: = 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 0 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 The pipe will tend to climb up the side of the wellbore…as it rotates (ii) More accurate evaluation: Note that, in the above figure, forces are not balanced; there is no force to balance the friction force Ff.

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),

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

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

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

Effect of Doglegs A. Neglecting Axial Friction

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

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

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

Effect of Doglegs (2) Buildup Wellbore

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

Effect of Doglegs A. Neglecting Axial Friction

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

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

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

PETE 411 Well Drilling