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PETE 411Well Drilling

Lesson 36

Torque and Drag Calculations

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Torque and Drag Calculations

◆ Friction◆ Logging◆ Hook Load◆ Lateral Load◆ Torque Requirements◆ Examples

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Assignments:

PETE 411 Design Projectdue December 9, 2002, 5 p.m.

HW#18 Due Friday, Dec. 6

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Friction - Stationary

• Horizontal surface• No motion• No applied force

Σ Fy = 0

N = W

N

W

N= Normal force = lateral load = contact force = reaction force

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Sliding Motion

• Horizontal surface

• Velocity, V > 0

• V = constant

• Force along surface

N = W

F = µ N = µ W

N

W

Fµµµµ N

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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.

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Frictionless, Inclined, Straight Wellbore:

pipe.ROTATING for used are equations

(2) : wellbore to 0

(1) : wellborealong 0

ar

These

F

F

∑

∑

⊥=

= IcosWT =∆

IsinWN =

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Effect of Friction (no doglegs):

2. Consider Effect of Friction ( no doglegs):

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Effect of Friction (no doglegs):

Frictional Force, F = µN = µW 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

(3)IsinWIcosWT

FIcosWT f

µ−=∆

−=∆

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Effect of Friction (no doglegs):

(b) Raising: Friction still opposes motion,

so

IsinWIcosWT

FIcosWT f

µ+=∆

+=∆

(4)

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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)µ

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Solution

From Equation (3) above,(3)

When pipe is barely sliding down the wellbore,

IsinWIcosWT µ−=∆

0T ≅∆

IsinW4.0IcosW0 −=∴

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Solution

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

Note:

�68.2I

2.5Ior tan 4.0Icot

=

==∴

Icot=µ

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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?

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Problem 2 - Solution - Force

(a) What force will it take to move this pipe along the horizontal section of the wellbore?

F = ? F = 0N

W

N = W = 30 lb/ft * 8,000 ft = 240,000 lb

F = µµµµN = 0.3 * 240,000 lb = 72,000 lb

Force to move pipe, F = 72,000 lbf

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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.

Then, as before, N = W = 30 lb/ft * 8,000 ft = 240,000 lbfTorque = F*d/2 = µµµµNd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft

Torque to rotate pipe, T = 21,000 ft-lbf

F

T

d/2

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Problem 2 - Equations -Horizontal

Torque, T = µµµµWd/(24 ) = 21,000 ft-lbf

F = µµµµN T = F * sN = W

W

Force to move pipe, F = µµµµW = 72,000 lbf

An approximate equation, with W in lbf and d in inches

( s=d/24 )

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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 φφφφ

in-lbf

Where φφφφ

= atan µµµµ = atan 0.3 = 16.70o

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

FT

d/2 φφφφP

W

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Problem 3A 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

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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.]

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Solution to Problem 3

(a) Hook load when rotating off bottom:

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Solution to Problem 3 - Rotating

When rotating off bottom.

lbf 120,000lbf 000,60

60cos*ft 8000*ftlb30ft 2000*

ftlb30

HLHLHL5.0

80002000

+=

+=

+=

↓�

lbf 000,180HL =

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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, 0F

0osin*2000*30N

2000

o

=

==

NFf µ=

0o

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Solution to Problem 3 - lowering

In the inclined section,

N = 30 * 8,000 * sin 60= 207,846 lbf

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Solution to Problem 3 - Lowering

HL = We,2000 + We,8000 - F2000 - F8000

= 60,000 + 120,000 - 0 - 62,354

Thus, F8000 = µµµµN = 0.3 * 207,846 = 62,352 lbf

HL = 117,646 lbf while RIH

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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

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Solution to Problem 3 - Summary

2,0002,0002,0002,000

10,00010,00010,00010,000

MDft

60,00060,00060,00060,000 120,000120,000120,000120,000 180,000180,000180,000180,000 240,000240,000240,000240,000

RIH ROT

POH

0000

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Solution to Problem 3 - rotating

2(d) Torque when rotating off bottom:In the Inclined Section:

NF

IsinWN

µ=

=

2d*F

Arm*Force

Torque

f=

=

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Solution to Problem 3 - rotating

(i) As a first approximation, assume the pipe lies at lowest point of hole:

=

=

=

=

121*

27*60sin*8000*30*3.0

2dIsinW

2dN

2dFTorque f

�

µµ

lbf-ft 187,18Torque =

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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

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Solution to Problem 3 - Rotating

Assume “Equilibrium” at angle φ as shown.

φ−==∑ sinIsinWFF fTangentAlong 0

φ−==∑ cosIsinWNF Tangentto.Perpend 0

…… (7)

φ=µ sinIsinWN …… (6)

φ= cosIsinWN

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Solution to Problem 3 - rotating

Solving equations (6) & (7)

(8))(tan

tan

cosIsinWsinIsinW

NN

1 µφ

φµ

φφµ

−=

=

=⇒

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Solution to Problem 3 - rotating

(ii) continuedTaking moments about the center of the pipe:

Evaluating the problem at hand:

From Eq. (8),

2d*FT f=

�70.16

)3.0(tan)(tan 11

=

== −−

φ

µφ

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Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (6),

lbf 724.59F

70.16sin*sin60*8000*30

sinIsinWF

f

f

=

=

=

��

φ

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Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (9),

lbf-ft 420,17Torque

121*

27*59,724

2d*FT f

=

=

=

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Solution to Problem 3

2 (d) (ii) Alternate Solution:

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Solution to Problem 3

Taking moments about tangent point,

247*70.16sin*sin60*8000*30

2dsinIsinWT

��=

Ο=

lbf-ft 420,17T =

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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

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Effect of Doglegs

(1) Dropoff Wellbore angle dogleg=δ

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Effect of Doglegs

A. Neglecting Axial Friction (e.g. pipe rotating)

0N2

sinT2

sinsTIsinW

0N2

sinT2

sin)TT(IsinW:F normal along

=−∆++

=−+∆++∑

δδ

δδ

(10) 2

sinT2IsinWN δ+≅

W sin I + 2T

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Effect of Doglegs

A. Neglecting Axial Friction

(11) 12

cos

IcosW2

cosT

02

cosTIcosW2

cos)TT(:F tangentalong

⇒→

=∆

=−−∆+∑

δ

δ

δδ

IcosWT ≅∆

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Effect of Doglegs

B. Including Friction (Dropoff Wellbore)

While pipe is rotating

(10)&(11)

WcosIT

2sinT2IsinWN

=∆

+= δ

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Effect of Doglegs

B. Including FrictionWhile lowering pipe (RIH)

(as above)

i.e. (12)

2sinT2IsinWN δ+=

NIcosWT µ−=∆

)2

sinT2IsinW(IcosWT δµ +−=∆

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Effect of Doglegs

B. Including FrictionWhile raising pipe (POH)

(13)

(14)

NIcosWT µ+=∆

)2

sinT2IsinW(IcosWT δµ ++=∆

)2

sinT2IsinW(2d

2dNTorque δµµ +

≅

=

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Effect of Doglegs

(2) Buildup Wellbore angle dogleg=δ

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Effect of Doglegs

A. Neglecting Friction (e.g. pipe rotating)

( ) 0N2

sinT2

sinTTIsinW:F normalalong =−δ−δ∆+−∑

2sinT2IsinWN δ−≅

0N2

sinT2

sinT2IsinW =−δ∆−δ−

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Effect of Doglegs

A. Neglecting Axial Friction

(16) 12

cos

IcosW2

cosT

02

cosTIcosW2

cos)TT(:F tangentalong

⇒→

=∆

=−−∆+∑

δ

δ

δδ

IcosWT ≅∆

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Effect of Doglegs

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

(15)&(16)

WcosIT

2sinT2IsinWN

=∆

−= δ

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Effect of Doglegs

B. Including FrictionWhile lowering pipe (RIH)

(15)

(17)2sinT2IsinWIcosWT

NIcosWT

2sinT2IsinWN

δµ

µ

δ

−−=∆

−=∆

−=

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Effect of Doglegs

While raising pipe (POH)

(18)

(19)2sinT2IsinW

2d

2dNTorque

22Tsin-WsinIWcosIT .e.i

NIcosWT

δµµ

δµ

µ

−

≅

=

+=∆

+=∆

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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

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Problem# 4

- Curved Wellbore

with Friction

T = 100,000 lbf

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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? �

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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.

26860IAVG

+=

oAVG 64I =

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Solution 4(a) - Rotating

From equation (11),

315,1000,100T

lbf 1,315

64cos*ft100*ftlb30

IcosWT

68 +=∴

=

=

=∆

�

�

rotating lbf 315,101T68 ←=�

T60 = 100,000 lbf

T68 = 101,315 lbf

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Solution 4 (b)

(b) Tension in pipe 100 ft Up-Hole when Pipe is being lowered:

From equation (10):

lbf 16,648N

13,9512,696

4sin*000,100*264sin*100*30N

2sinT2IsinWN

=

+=

+=

+=

��

δ

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Solution 4 (b)

From equation 10,

From equation 12,

NIcosWT µ−=∆

lbf 162,4F

16,648*0.25NForceFriction

f =

== µ

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Solution 4(b) - Lowering

From equation 12,

T)(T 867,2000,100T68 ∆+−=∴�

lbf 153,97T68 =�

-2,847

162,4)64cos*100*30(T

=

−=∆ �

T60 = 100,000 lbf

T68 = 97,153 lbf

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Solution 4 (c)

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

lbf 16,648N

13,9512,696

4sin*000,100*264sin*100*30N

2sinT2IsinWN

=

+=

+=

+=

��

δ

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Solution 4 (c)

lbf 162,4F

16,648*0.25NForceFriction

f =

== µ

From equation 12,

NIcosWT µ+=∆

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Solution 4(c) - Raising

From equation 12,

T)(T 5477000,100T

lbf 5477

162,4)64cos*100*30(T

68 ∆++=∴

=

+=∆

�

�

lbf 477,105T68 =�

T60 = 100,000 lbf

T68 = 105,477 lbf

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Solution 4(a, b and c)SUMMARY

Rot 100,000 101,315

RIH 100,000 97,153

POH 100,000 104,477

T60 T68

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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

lbf 648,16N

64 at

=

= �θ

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Solution 4 (d)

for 40 ft distance,

i.e., Lateral load on centralizer,

∴

lbf 6,659 10040*648,16N .centr

=

=

lbf 1200ftlb30*pipe offt 40 :Note

lbf 659,6N .centr

=

=

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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,

lbf 101,315T ,68at lbf 100,000T ,60 at

====

�

�

θθ

lbf 395,100Tlbf )100/30(*315,1000,100T

=+=

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Alternate Approach

From Eq. (10),

for 40 ft centralizer spacing,

lbf 6,685N

5,6061,079 40/100}*{4

)6.1sin(*395,100*264sin*40*30N2

sinT2IsinWN

=

+=

+=

+=��

δ

lbf 685,6N .centr =

∴

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Centralizer