36_ Torque and Drag Calculations

Slide 1

PETE 411Well Drilling

Lesson 36 Torque and Drag Calculations

Slide 2

Torque and Drag Calculations

u Frictionu Loggingu Hook Loadu Lateral Loadu Torque Requirementsu Examples

Slide 3

Assignments:

PETE 411 Design Projectdue December 3, 2001, 5 p.m.

HW#17 Due Monday, Nov. 26

Slide 4

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 5

Frictionless, Inclined, Straight Wellbore:

pipe.ROTATING for used are equations

(2) : wellbore to 0

(1) : wellborealong 0

ar

These

F

F

∑

∑

⊥=

= IcosWT =∆

IsinWN =

Slide 6

Effect of Friction (no doglegs):

2. Consider Effect of Friction ( no doglegs):

Slide 7


Frictional Force,

where

(a) Lowering: Friction opposes motion, so

(3)

0.4}0.2{usually friction] of coeff.[ 10

IsinWNF f

≤≤=≤≤

==

µµµ

µµ

IsinWIcosWT

FIcosWT f

µ−=∆

−=∆

Slide 8


(b) Raising: Friction still opposes motion, so

IsinWIcosWT

FIcosWT f

µ+=∆

+=∆

Slide 9

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

Slide 10

Solution

From Equation (3) above,(3)

When pipe is barely sliding down the wellbore,

IsinWIcosWT µ−=∆

0T ≅∆

IsinW4.0IcosW0 −=∴

Slide 11

Solution

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

Note:

o68.2I

2.5Ior tan 4.0Icot

=

==∴

Icot=µ

Slide 12

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. = 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?

≅ µ

Slide 13

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 = µN = 0.3 * 240,000 lb = 72,000 lb

Force to move pipe, F = 72,000 lbf

Slide 14

Problem 2 - Solution - Force

(b) What torque will it take to rotate this pipe?

As an approximation, let us assume that the pipelies on the bottom of the wellbore.

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

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

F

T

d/2

Slide 15

Problem 2 - Equations - Horizontal

Torque, T = µWd/(24 ) = 21,000 ft-lbf

F = µN T = F * dN = W

W

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

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

Slide 16

Horizontal - Torque

A slightly 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

Where φ = atan µ = atan 0.3 = 16.70o

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

FT

d/2 φP

W

Slide 17

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

Slide 18

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 19

Solution to Problem 3

(a) Hook load when rotating off bottom:

Slide 20

Solution to Problem 3 - Rotating

When rotating off bottom.

lbf 120,000lbf 000,60

60cos*ft 8000*ftlb30ft 2000*

ftlb30

HLHLHL5.0

80002000

+=

+=

+=

↓o

lbf 000,180HL =

Slide 21

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

Slide 22


In the inclined section,

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

Slide 23


62,354-0-120,00060,000

FFHLFLHL

lbf 354,62846,207*3.0NF ,Thus

8000200080002000

8000

+=

−−+=

=== µ

lbf 646,117HL = RIH while

Slide 24

Solution to Problem 3 - raising

2(c) Hood Load when POH:

62,3540120,00060,000

FFHLHLHL 8000200080002000

+++=

+++=

POH lbf 354,242HL ←=

Slide 25

Solution to Problem 3 - Summary

2,000

10,000

MDft

60,000 120,000 180,000 240,000

RIH ROT

POH

0

Slide 26

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=

=

Slide 27


(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

o

µµ

lbf-ft 187,18Torque =

Slide 28


(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 29


∑

∑

−==

=

⇒−==

⊥ (7) cosIsinWN0F

(6) sinIsinWN

sinIsinWF0F

shown. as angleat m"equilibriu" Assume

tangentar to

f tangentalong

φ

ϕµ

φ

φ

Slide 30


Solving equations (6) & (7)

(8))(tan

tan

cosIsinWsinIsinW

NN

1 µφ

φµ

φφµ

−=

=

=⇒

Slide 31


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

Evaluating the problem at hand:

From Eq. (8),

2d*FT f=

o70.16

)3.0(tan)(tan 11

=

== −−

φ

µφ

Slide 32



From Eq. (6),

lbf 724.59F

70.16sin*sin60*8000*30

sinIsinWF

f

f

=

=

=

oo

φ

Slide 33



From Eq. (9),

lbf-ft 420,17Torque

121*

27*59,724

2d*FT f

=

=

=

Slide 34


2 (d) (ii) Alternate Solution:

Slide 35


Taking moments about tangent point,

247*70.16sin*sin60*8000*30

2dsinIsinWT

oo=

Ο=

lbf-ft 420,17T =

Slide 36


Note that the answers in parts (i) & (ii) differ by a factor of cos φ

9578.070.16coscos == φφ

Slide 37

Effect of Doglegs

(1) Dropoff Wellbore angle dogleg=δ

Slide 38

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+≅

Slide 39

Effect of Doglegs

A. Neglecting Axial Friction

(11) 12

cos

IcosW2

cosT

02

cosTIcosW2

cos)TT(:F tangentalong

⇒→

=∆

=−−∆+∑

δ

δ

δδ

Slide 40

Effect of Doglegs

B. Including Friction (Dropoff Wellbore)

While pipe is rotating

(10)&(11)

WcosIT

2sinT2IsinWN

=∆

+=δ

Slide 41

Effect of Doglegs

B. Including FrictionWhile lowering pipe (RIH)

(as above)

i.e. (12)

2sinT2IsinWN δ

+=

NIcosWT µ−=∆

)2

sinT2IsinW(IcosWT δµ +−=∆

Slide 42

Effect of Doglegs

B. Including FrictionWhile raising pipe (POH)

(13)

(14)

NIcosWT µ+=∆

)2

sinT2IsinW(IcosWT δµ ++=∆

)2

sinT2IsinW(2d

2dNTorque δµµ +

≅

=

Slide 43

Effect of Doglegs

(2) Buildup Wellbore angle dogleg=δ

Slide 44

Effect of Doglegs

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

0N2

sinT2

sinsTIsinW

0N2

sinT2

sin)TT(IsinW:F normal along

=−∆++

=−+∆++∑δδ

δδ

(15) 2

sinT2IsinWN δ+≅

−

− 2Τ

−

−

−

Slide 45

Effect of Doglegs

A. Neglecting Axial Friction

(16) 12

cos

IcosW2

cosT

02

cosTIcosW2

cos)TT(:F tangentalong

⇒→

=∆

=−−∆+∑

δ

δ

δδ

Slide 46

Effect of Doglegs

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

(15)&(16)

WcosIT

2sinT2IsinWN

=∆

−=δ

Slide 47

Effect of Doglegs

B. Including FrictionWhile lowering pipe (RIH)

(15)

(17)2sinT2IsinWIcosWT

NIcosWT

2sinT2IsinWN

δµ

µ

δ

−−=∆

−=∆

−=

Slide 48

Effect of Doglegs

While raising pipe (POH)

(18)

(19)2sinT2IsinW

2d

2dNTorque

22Tsin-WsinIWcosIT .e.i

NIcosWT

δµµ

δµ

µ

−

≅

=

+=∆

+=∆

Slide 49

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.

0.25

lb/ft 30pipe of wt.Buoyed

=

=

µ

Slide 50

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? o

Slide 51

Solution 4 (a)

(a) Axial tension 100 ft up hole when pipe is rotating :

Pipe is rotating so frictional effect on axial load may be neglected.

o64I2

6860I

AVG

AVG

=

+=

Slide 52

Solution 4 (a)

From equation (11),

315,1000,100T

lbf 1,315

64cos*ft100*ftlb30

IcosWT

68 +=∴

=

=

=∆

o

o

rotating lbf 315,101T68 ←=o

Slide 53

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

=

+=

+=

+=

oo

δ

Slide 54

Solution 4 (b)

From equation 10,

From equation 12,

NIcosWT µ−=∆

lbf 162,4F

16,648*0.25NForceFriction

f =

== µ

Slide 55

Solution 4 (b)

From equation 12,

T)(T 867,2000,100T68 ∆+−=∴ o

lbf 153,97T68 =o

-2,847

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

=

−=∆ o

Slide 56

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

=

+=

+=

+=

oo

δ

Slide 57

Solution 4 (c)

lbf 162,4F

16,648*0.25NForceFriction

f =

== µ

From equation 12,

NIcosWT µ+=∆

Slide 58

Solution 4 (c)

From equation 12,

T)(T 5477000,100T

lbf 5477

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

68 ∆++=∴

=

+=∆

o

o

lbf 477,105T68 =o

Slide 59

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

=

= oθ

Slide 60

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

=

=

Slide 61

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

====

o

o

θθ

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

=+=

Slide 62

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

=

+=

+=

+=oo

δ

lbf 685,6N .centr =

∴

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36_ Torque and Drag Calculations