15. Torque And Drag Calculations 2w3i6g

PETE 661 Drilling Engineering Lesson 15 Torque and Drag Calculations Slide 1 of 62

Torque and Drag Calculations ◆ Friction ◆ Logging ◆ Hook Load ◆ Lateral Load ◆ Torque Requirements ◆ Examples Slide 2 of 62

HW #8 Well Survey due 11-04-02

Slide 3 of 62

Friction - Stationary • Horizontal surface

N

• No motion • No applied force

Σ Fy = 0 N=W W N= Normal force = lateral load = force = reaction force Slide 4 of 62

Sliding Motion N

• Horizontal surface • Velocity, V > 0 • V = constant

µ N

F

• Force along surface N=W

W

F=µ N=µ W Slide 5 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. Slide 6 of 62

Frictionless, Inclined, Straight Wellbore:

∑F = 0

along wellbore :

∆T = W cos I

(1)

∑F = 0

⊥ ar to wellbore :

N = W sin I

(2)

These equations are used for ROTATING pipe.

Slide 7 of 62

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

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Effect of Friction (no doglegs): Frictional Force, F = µ N = µ W sin I where 0 < µ < 1 (µ is the coeff. of friction) usually 0.15 < µ < 0.4 in the wellbore (a) Lowering: Friction opposes motion, so

∆T = W cos I − Ff ∆T = W cos I − µ W sin I

(3) Slide 9 of 62

Effect of Friction (no doglegs):

(b) Raising: Friction still opposes motion, so

∆T = W cos I + Ff ∆T = W cos I + µW sin I

(4)

Slide 10 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 µ =0.4)

Slide 11 of 62

Solution

From Equation (3) above, ∆T = W cos I − µW sin I

(3)

When pipe is barely sliding down the wellbore, ∆T ≅ 0

∴ 0 = W cos I − 0.4W sin I Slide 12 of 62

Solution ∴ cot I = 0.4

or tan I = 2.5 I = 68.2 

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

µ = cot I Slide 13 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. µ = 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?

Slide 14 of 62

Problem 2 - Solution - Force (a) What force will it take to move this pipe along the horizontal section of the wellbore? F=?

N

F=0

W N = W = 30 lb/ft * 8,000 ft = 240,000 lb F = µ N = 0.3 * 240,000 lb = 72,000 lb

Force to move pipe, F = 72,000 lbf Slide 15 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.

T d/2 F

Then, as before, N = W = 30 lb/ft * 8,000 ft = 240,000 lbf

Torque = F*d/2 = µ Nd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft

Torque to rotate pipe, T = 21,000 ft-lbf Slide 16 of 62

Problem 2 - Equations - Horizontal F=µ N

N=W

T=F*d

W

Force to move pipe, Torque,

F=µ W

T = µ Wd/(24 )

= 72,000 lbf

= 21,000 ft-lbf

An approximate equation, with W in lbf and d in inches Slide 17 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 φ . Taking moments about the point P: Torque, T = W * (d/2) sin φ in-lbf

T d/2 φ

Where φ

= atan µ = atan 0.3 = 16.70

o

T = 240,000 * 7/24 * 0.2873 = 20,111 ft-lbf

F P

W Slide 18 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. µ = 0.3

Slide 19 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.] Slide 20 of 62

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

Slide 21 of 62

Solution to Problem 3 - Rotating

HL = HL 2000 + HL 8000 0. 5

↓ lb lb  = 30 * 2000 ft + 30 * 8000 ft * cos 60 ft ft

= 60,000 lbf + 120,000 lbf

HL = 180,000 lbf

When rotating off bottom. Slide 22 of 62

Solution to Problem 3 - lowering 2 (b) Hook load when RIH: The hook load is decreased by friction in the wellbore.

Ff = µN

In the vertical portion, oo

N = 30 * 2000 * sin o0 = 0 Thus,

F2000 = 0 Slide 23 of 62

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

Slide 24 of 62

Solution to Problem 3 - Lowering Thus, F8000 = µ N = 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 Slide 25 of 62

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 Slide 26 of 62

Solution to Problem 3 - Summary

2,000

RIH

ROT POH

MD ft

10,000 0

60,000

120,000 180,000 240,000 Slide 27 of 62

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

N = W sin I F = µN

Torque = Force * Arm d = Ff * 2

Slide 28 of 62

Solution to Problem 3 - rotating (i) As a first approximation, assume the pipe lies at lowest point of hole: d  d  d  Torque = Ff   = µN  = µW sin I  2 2 2 7 1  = 0.3 * 30 * 8000 * sin 60 *  *   2 12  

Torque = 18,187 ft - lbf Slide 29 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. The pipe will tend to climb up the side of the wellbore…as it rotates

Slide 30 of 62

Solution to Problem 3 - Rotating Assume “Equilibrium” at angle φ

∑F

as shown.

Along Tangent

∑F

= 0 = Ff − W sin I sin φ …… (6) µN = W sin I sin φ

Perpend. to Tangent

= 0 = N − W sin I cos φ

N = W sin I cos φ

…… (7) Slide 31 of 62

Solution to Problem 3 - rotating

µN W sin I sin φ Solving equations (6) & (7) ⇒ = N W sin I cosφ µ = tan φ −1

φ = tan ( µ )

(8)

Slide 32 of 62

Solution to Problem 3 - rotating (ii) continued Taking moments about the center of the pipe: d T = Ff * 2 Evaluating the problem at hand: −1 −1 φ = tan ( µ ) = tan (0.3) From Eq. (8),

φ = 16.70

 Slide 33 of 62

Solution to Problem 3 - rotating Evaluating the problem at hand: From Eq. (6), Ff = W sin I sin φ

= 30 * 8000 * sin60 * sin16.70 Ff = 59.724 lbf

Slide 34 of 62

Solution to Problem 3 - rotating Evaluating the problem at hand: d From Eq. (9), T = Ff * 2 7 1  = 59,724 *  *   2 12  Torque = 17,420 ft - lbf Slide 35 of 62

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

Slide 36 of 62

Solution to Problem 3 Taking moments about tangent point,

d T = W sin I sin Ο 2 7 = 30 * 8000 * sin60 * sin 16.70 * 24 



T = 17,420 ft - lbf Slide 37 of 62

Solution to Problem 3

Note that the answers in parts (i) & (ii) differ by a factor of cos φ (i) T = 18,187 (ii) T = 17,420 cos φ = cos 16.70 = 0.9578 Slide 38 of 62

Effect of Doglegs (1) Dropoff Wellbore

δ = dogleg angle

Slide 39 of 62

Effect of Doglegs A. Neglecting Axial Friction (e.g. pipe rotating) δ δ ∑ Falong normal : W sin I + (T + ∆T) sin 2 + T sin 2 − N = 0 δ δ sinII + + sT sin + ∆T sin − N = 0 WWsin 2 2 2T

δ N ≅ W sin I + 2T sin 2

(10) Slide 40 of 62

Effect of Doglegs ≅

∆

A. Neglecting Axial Friction

δ δ ∑ Falong tangent : (T + ∆T) cos 2 − W cos I − T cos 2 = 0

W

T

δ cos → 1 ⇒ 2

I s o c

δ ∆T cos = W cos I 2 (11) Slide 41 of 62

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

δ N = W sin I + 2T sin 2

(10)&(11)

∆T = WcosI Slide 42 of 62

Effect of Doglegs B. Including Friction While lowering pipe (RIH) δ N = W sin I + 2T sin 2

(as above)

∆T = W cos I − µN

i.e. ∆T = W cos I − µ ( W sin I + 2T sin δ ) (12) 2

Slide 43 of 62

Effect of Doglegs B. Including Friction While raising pipe (POH) ∆T = W cos I + µN

δ ∆ T = W cosI + µ ( W sin I + 2T sin ) 2 δ d d Torque = µN  ≅ µ  ( W sin I + 2T sin ) 2  2  2

(13)

(14) Slide 44 of 62

Effect of Doglegs (2) Buildup Wellbore

δ = dogleg angle

Slide 45 of 62

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

∑F

along normal

δ δ : W sin I − ( T + ∆T ) sin − T sin − N = 0 2 2 δ δ W sin I − 2T sin − ∆T sin − N = 0 2 2

δ N ≅ W sin I − 2T sin 2 Slide 46 of 62

Effect of Doglegs ≅

∆

A. Neglecting Axial Friction

δ δ ∑ Falong tangent : (T + ∆T) cos 2 − W cos I − T cos 2 = 0 I s o c

T

δ cos → 1 ⇒ 2

W

δ ∆T cos = W cos I 2 (16) Slide 47 of 62

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

δ N = W sin I − 2T sin 2

(15)&(16)

∆T = WcosI

Slide 48 of 62

Effect of Doglegs B. Including Friction While lowering pipe (RIH) δ N = W sin I − 2T sin 2

(15)

∆T = W cos I − µ N

δ ∆T = W cos I − µ W sin I − 2T sin 2

(17) Slide 49 of 62

Effect of Doglegs While raising pipe (POH) ∆T = W cos I + µN

δ i.e. ∆T = WcosI + µ WsinI - 2Tsin 2

(18)

δ d d Torque = µ N   ≅ µ   W sin I − 2T sin 2  2  2

(19)

Slide 50 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. Buoyed weight of pipe = 30 lbm/ft

µ = 0.25 Slide 51 of 62

Problem #4 - Curved Wellbore with Friction T = 100,000 lbf Slide 52 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? Slide 53 of 62

Solution 4(a) - Rotating Axial tension 100 ft up hole when pipe is rotating :

IAVG

60 + 68 = 2

IAVG = 64

o

Pipe is rotating so frictional effect on axial load may be neglected. Slide 54 of 62

Solution 4(a) - Rotating From equation (11),

T68 = 101,315 lbf

∆T = W cos I lb  = 30 *100ft * cos 64 ft = 1,315 lbf ∴ T68 = 100,000 + 1,315

T68 = 101,315 lbf

T60 = 100,000 lbf

← rotating Slide 55 of 62

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

δ N = W sin I + 2T sin 2 N = 30 *100 * sin 64 + 2 *100,000 * sin 4 = 2,696 + 13,951 N = 16,648 lbf Slide 56 of 62

Solution 4 (b) From equation 10,

Friction Force = µN = 0.25 *16,648 Ff = 4,162 lbf From equation 12,

∆T = W cos I − µN Slide 57 of 62

Solution 4(b) - Lowering From equation 12,

T68 = 97,153 lbf

∆T = (30 *100 * cos 64  ) − 4,162 = -2,847

∴ T68 = 100,000 − 2,867

(T + ∆T) T60 = 100,000 lbf

T68  = 97 ,153 lbf Slide 58 of 62

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

δ N = W sin I + 2T sin 2 N = 30 *100 * sin 64 + 2 *100,000 * sin 4 = 2,696 + 13,951 N = 16,648 lbf Slide 59 of 62

Solution 4 (c)

Friction Force = µN = 0.25 *16,648 Ff = 4,162 lbf From equation 12,

∆T = W cos I + µN

Slide 60 of 62

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

From equation 12,

∆T = (30 *100 * cos 64 ) + 4,162 

= 5477 lbf ∴T68 = 100 ,000 + 5477

(T + ∆T)

T60 = 100,000 lbf

T68 = 105,477 lbf Slide 61 of 62

Solution 4(a, b and c) SUMMARY

T60

T68

Rot

100,000

101,315

RIH

100,000

97,153

POH

100,000

104,477 Slide 62 of 62

Solution 4 (d) (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated): From above,

at θ = 64  N = 16,648 lbf

This is for 100 ft distance Slide 63 of 62

Solution 4 (d)

∴

for 40 ft distance,

N centr .

 40  = 16,648 *    100  = 6,659 lbf

i.e., Lateral load on centralizer,

N centr. = 6,659 lbf lb Note : 40 ft of pipe * 30 = 1200 lbf ft Slide 64 of 62

Alternate Approach (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated) From above, at θ = 60 , T = 100,000 lbf From above, at θ = 68 , T = 101,315 lbf So, 30 ft up-hole,

T = 100 ,000 + 1,315 * (30 / 100 ) lbf T = 100 ,395 lbf Slide 65 of 62

Alternate Approach δ From Eq. (10), N = W sin I + 2T sin 2 N = 30 * 40 * sin 64 + 2 *100,395 * sin(1.6 ) {4 * 40/100} = 1,079 + 5,606 N = 6,685 lbf

∴ for 40 ft centralizer spacing,

N centr . = 6,685 lbf Slide 66 of 62

Centralizer

Slide 67 of 62