• date post

05-Jan-2016
• Category

## Documents

• view

28

8

Embed Size (px)

description

Cal

### Transcript of 36 Torque and Drag Calculations

• 1PETE 411Well Drilling

Lesson 36

Torque and Drag Calculations

• 2Torque and Drag Calculations

• 3Assignments:

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

HW#18 Due Friday, Dec. 6

• 4Friction - Stationary

Horizontal surface No motion No applied force

Fy = 0N = W

N

W

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

• 5Sliding Motion

Horizontal surface

Velocity, V > 0

V = constant

Force along surface

N = W

F = N = W

N

W

F N

• 6Frictionless, 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.

• 7Frictionless, Inclined, Straight Wellbore:

pipe.ROTATING for used are equations

(2) : wellbore to 0

(1) : wellborealong 0

ar

These

F

F

=

= IcosWT =

IsinWN=

• 8Effect of Friction (no doglegs):

2. Consider Effect of Friction ( no doglegs):

• 9Effect of Friction (no doglegs):

Frictional Force, F = N = W sin I

where 0 < < 1 ( is the coeff. of friction)usually 0.15 < < 0.4 in the wellbore

(a) Lowering: Friction opposes motion, so

(3)IsinWIcosWT

FIcosWT f

=

=

• 10

Effect of Friction (no doglegs):

(b) Raising: Friction still opposes motion,

so

IsinWIcosWT

FIcosWT f

+=

+=

(4)

• 11

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)

• 12

Solution

From Equation (3) above,(3)

When pipe is barely sliding down the wellbore,

IsinWIcosWT =

0T

IsinW4.0IcosW0 =

• 13

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=

• 14

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?

• 15

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 = N = 0.3 * 240,000 lb = 72,000 lb

Force to move pipe, F = 72,000 lbf

• 16

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 = Nd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft

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

F

T

d/2

• 17

Problem 2 - Equations -Horizontal

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

F = N T = F * sN = W

W

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

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

( s=d/24 )

• 18

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 .

Taking moments about the point P:Torque, T = W * (d/2) sin in-lbf

Where = atan = atan 0.3 = 16.70o

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

FT

d/2 P

W

• 19

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. = 0.3

• 20

Problem 3

(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.]

• 21

Solution to Problem 3

(a) Hook load when rotating off bottom:

• 22

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=

• 23

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 =

0o

• 24

Solution to Problem 3 - lowering

In the inclined section,

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

• 25

Solution to Problem 3 - Lowering

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

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

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

HL = 117,646 lbf while RIH

• 26

Solution to Problem 3 - Raising

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

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

HL = 242,354 lbf POH

• 27

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

• 28

Solution to Problem 3 - rotating

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

NF

IsinWN

=

=

2d*F

Arm*Force

Torque

f=

=

• 29

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

lbf-ft 187,18Torque =

• 30

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 wellboreas it rotates

• 31

Solution to Problem 3 - Rotating

Assume Equilibrium at angle as shown.

== sinIsinWFF fTangentAlong 0

== cosIsinWNF Tangentto.Perpend 0 (7)

= sinIsinWN (6)

= cosIsinWN

• 32

Solution to Problem 3 - rotating

Solving equations (6) & (7)

(8))(tan

tan

cosIsinWsinIsinW

NN

1

=

=

=

• 33

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

=

==

• 34

Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (6),

lbf 724.59F

70.16sin*sin60*8000*30

sinIsinWF

f

f

=

=

=

• 35

Solution to Problem 3 - rotating

Evaluating the problem at hand:

From Eq. (9),

lbf-ft 420,17Torque

121*

27*59,724

2d*FT f

=

=

=

• 36

Solution to Problem 3

2 (d) (ii) Alternate Solution:

• 37

Solution to Problem 3

247*70.16sin*sin60*8000*30

2dsinIsinWT

=

=

lbf-ft 420,17T =

• 38

Solution to Problem 3

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

(i) T = 18,187(ii) T = 17,420

cos = cos 16.70 = 0.9578

• 39

Effect of Doglegs

(1) Dropoff Wellbore angle dogleg=

• 40

Effect of Doglegs

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

0N2

sinT2

sinsTIsinW

0N2

sinT2

sin)TT(IsinW:F normal along

=++

=+++

(10) 2

sinT2IsinWN +

W sin I + 2T

• 41

Effect of Doglegs

A. Neglecting Axial Friction

(11) 12

cos

IcosW2

cosT

02

cosTIcosW2

cos)TT(:F tangentalong

=

=+

IcosWT

• 42

Effect of Doglegs

B. Including Friction (Dropoff Wellbore)

While pipe is rotating

(10)&(11)

WcosIT

2sinT2IsinWN

=

+=

• 43

Effect of Doglegs

B. Including FrictionWhile lowering pipe (RIH)

(as above)

i.e. (12)

2sinT2IsinWN +=

NIcosWT =

)2

sinT2IsinW(IcosWT +=

• 44

Effect of Doglegs

B. Including FrictionWhile raising pipe (POH)

(13)

(14)

NIcosWT +=

)2

sinT2IsinW(IcosWT ++=

)2

sinT2IsinW(2d

2dNTorque +

=

• 45

Effect of Doglegs

(2) Buildu