torque and drag

7/28/2019 Torque and Drag

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© 2001 PetroSkills LLC, All Rights Reserved

TORQUE AND DRAG

Drag is the increase in string

weight when pulling out of the

hole or the reduction in string

weight while tripping in the hole



© 2001 PetroSkills LLC, All Rights Reserved2

Torque and Drag

Torque is the force required toturn the drill string

In a perfectly vertical well, thetorque and drag in a well arenegligible

In directional wells, torque anddrag can be significant

In horizontal or extended reachwells, torque and drag arecrucial




Torque and Drag

The magnitude of the torque anddrag is determined by the force

with which the pipe contacts the

hole wall and the frictioncoefficient between the wall and

pipe

The contact force is called the

normal force




Torque and Drag

First, consider simple friction

F

W

The force F

required to movethe block of weight

W (normal force) is

F = μW Where μ is the

friction coefficient




Torque and Drag

If the plane isinclined, the

normal force is a

function of thecosine of θ

T=-W sin θ +

μ W cos θ θ

T




Torque and Drag

At 0 degrees, thenormal force

equals the

weight of theblock

At 90 degrees,

the normal forceis zero

θ


http://slidepdf.com/reader/full/torque-and-drag 7/66© 2001 PetroSkills LLC, All Rights Reserved7

Torque and Drag

The same is true for a drill stringin the hole

However, inclination is measured

from the vertical rather than thehorizontal

If the inclination is zero, the normal

force is zero

If the inclination is 90o, the normal

force equals the weight of the pipe



Torque and Drag

Class problem:Given the two blocks below

F1

W = 100 lbs

F2

W = 100 lbs



Torque and Drag

Multiple choice – which is true

A. F1 = F2

B. F1 < F2

C. F1 > F2

D. None of the above



Torque and Drag

Pipe weight is not the onlysource of drag

When hole curvature is

considered, an additional force

is added to the normal force

That force is a function of pipetension and dogleg severity



Torque and Drag

Most of the torque and drag in adirectional well comes from pipe

tension (or compression) in a

dogleg

Horizontal and other high angle

wells do have a significant drag

component from the pipe weight



Torque and Drag

The normalforce is a

function of

tension anddogleg severity

Consider a pipe

segment bentaround a corner



Torque and Drag

Summing forces in the x and ydirection yields

avg x I W I T F sinsin

avg y I AT F sinsin



Torque and Drag

The normal force is the vectorialsum of the x and y forces

Once the normal force is

calculated, the drag for the pipesegment can be calculated

22

sinsinsinsin avg avg N I AT I W I T F



Torque and Drag

The drill string isbroken intosegments starting atthe bit where the

tension is knownThe normal force iscalculated and thetension at the top of

the segment iscalculated based ondrag and pipe weight



Torque and Drag

The tension at the top of the

segment becomes the tension at

the bottom of the next segment

The drag and pipe weight for that

segment are used to calculatethe tension at the top

N avg

F I W T T cos12



Torque and Drag

The process is repeated until thecalculations reach the surface ,

which is the hook load

Tripping in the hole, the drag

acts in the opposite direction

N avg F I W T T cos12



Torque and Drag

While tripping in the hole, thereis an additional normal force that

can be imposed on the pipe

If there is sufficient

compression, the pipe can be

buckle

In a vertical well, the pipe

buckles almost immediately



Torque and Drag

In a directional well, it takes afinite amount of compression to

cause buckling

A very simple buckling equation

is as follows:



Torque and Drag

The critical buckling load is a

function of the sine of theinclination, the radial clearance

and the moment of inertia

At zero degrees, the pipebuckles immediately

F E Ag I

r crit 2

I sin



Torque and Drag

As the pipe gets stiffer (larger OD and smaller ID), it takes more

force to cause buckling

As the hole gets bigger, the pipe

buckles easier

In oil field terms, the equation isas follows



Torque and Drag

If the pipe is buckled, an

additional normal force isapplied according to the

following equation:

)(sin)/)((1082.9

445

OD D I B ft Wt IDOD F

h

crit

W rF

E N

f

2

4 I



Torque and Drag

Note that the normal forcechanges with the square of the

force that exceeds the buckling

load




Torque and Drag

Calculate the critical buckling load of 4 ½” OD (3.826” ID), 16.6#/ft (nominal)

drill pipe at 90o in an 8 ½” hole. The

mud weight is 10 ppg. The actual wt/ftof the drill pipe is 18.3#/ft with tool

joints.

85.0)10)(015.0(1))(015.0(1

B MW B




Torque and Drag

It takes 27,343 lbs to buckle 4 ½” drill

pipe in an 8 ½” horizontal well

lbs F

F

crit

crit

343,27

)5.45.8(

90sin)85.0)(3.18)(826.35.4(1082.9445

)(

sin)/)((1082.9 445

OD D

I B ft Wt IDOD F

h

crit




Torque and Drag

Class ProblemCalculate the critical buckling load for 1.5”

coiled tubing in an 8 ½” hole

1.5” OD by 1.31” ID 1.42 lbs per foot

Fluid is water at 8.34 ppg

Inclination is 90

o

Answer 607 lbs




Torque and Drag

In 4 ½” casing it takes 1012 lbs

In 2 7/8” tubing it takes 1652 lbs

lbs F

F

crit

crit

607)5.15.8(

90sin)87.0)(42.1)(31.15.1(1082.9 445

87.0)34.8)(015.0(1))(015.0(1

B MW B

T d D




Torque and Drag

Torque can be calculated in avery similar manner

However, the drag is generally

ignored while calculating the

tension in a simplified torque

and drag model

T d D




Torque and Drag

Weight Indicator

Hook Load Down

Hook Load UpRotating Weight

Martin Decker

T d D




Torque and Drag

Drag influence is alleviated whenthe pipe is rotated so the tension

is simply

The normal force is calculated

based on the pipe weight and thetorque can be calculated from

the normal force

avg I W T T cos12

T d D




Torque and Drag

Torque is in foot-poundsThe normal force supplies the pounds

The radius of the pipe supplies the

moment arm or foot

The bit torque is entered and thetorque is summed up the holethe same as the drag

M M F R N 2 1

T d D




Torque and Drag

Compare the drag from twodifferent wells

The average inclination is the same

in both wells

Case A has an average dogleg

severity of 4o/100’

Case B has an average doglegseverity of 2o/100’

T d D




Torque and Drag

Also assume that each case has

50,000 lbs or 200,000 lbs tensionThe tension would change depending

upon how much hole was drilled

below the dogleg

MD I A DLS

2000

2100

2200

2300

40

44

48

52

135

135

135

135

4o/100

4o/100

4o/100

4o/100

MD I A DLS

2000

2100

2200

2300

43

45

47

49

135

135

135

135

2o/100

2o/100

2o/100

2o/100

Case A Case B

T d D




Torque and Drag

The air weight of the 4 1/2 inchdrill pipe is 18.1 lbs/ft including

tool joints (can be found in API

RP7G)

Calculate the buoyant weight per

100 feet

W Length wt ft Bouyancy Factor /

W 100 18 1 1 0 015 10. .

T d D




Torque and Drag

The drag is equal to the normal

force times the friction

coefficient

W lbs 1538 50 100. / '

T F N

22 sinsinsinsin avg avg I AT I W I T T

T d D




Torque and Drag

Make the calculations at 100 footintervals

Case A: 50,000 pound tension

502/4852 avg I

I 52 48 4

A 135 135 0

2250sin0sin5000050sin50.15384sin5000040.0 T

T lbs drag 92371.

T d D




Torque and Drag

In order to calculate the drag inthe next interval, the tension at

2,200 feet must be calculated.

From Equation 6:


T

250000 1538 50 50 923 71 . cos .

T lbs2 51912 64 .



Torque and Drag




Torque and Drag

Now calculate the drag from2,000 feet to 2,100 feet

2246sin0sin64.5191246sin50.15384sin64.5191240.0 T

T lbs drag 100582.


82.100546cos50.153864.519122 T

T lbs2 53987 19 .

Torque and Drag




Torque and Drag

I avg 44 40 2 42/

I 44 40 4

A 135 135 0

T F N

2s0sin19.5398742sin50.15384sin19.5398740.0 T

T lbs drag 1094 60.


60.109442cos50.153819.539872 T

T lbs2 5622512 .

Torque and Drag




Torque and Drag

Total drag for the 300 footinterval in Case A with 50,000

pound tension is:

Drag 923 71 1005 82 1094 60. . .

Drag lbs 3024 13.

Torque and Drag




Torque and Drag

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

50,000.00

51,912.64

53,987.19

56,225.12

TOTAL

923.71

1005.82

1094.60

3024.13

4o/100’ with 50,000 lbs

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds) 2300

2200

2100

2000

200,000.00

206,098.02

212,474.74

219,134.88

TOTAL

5109.09

5307.99

5516.81

15,933.89

4o/100’ with 200,000 lbs

At 4o/100’

The

tension istimes 4

The drag is

times 5.3

Torque and Drag




Torque and Drag

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

50,000.00

51,270.12

52,611.89

54,025.55

TOTAL

240.66

273.04

306.96

820.66

2o/100’ with 50,000 lbs

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

200,000.00

203,364.09

206,829.06

210,395.56

TOTAL

2334.63

2396.24

2459.80

7190.67

2o/100’ with 200,000 lbs

At 2o/100’

The

tension istimes 4

The drag is

times 8.8

Torque and Drag




Torque and Drag

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

50,000.00

51,912.64

53,987.19

56,225.12

TOTAL

923.71

1005.82

1094.60

3024.13

4o/100’ with 50,000 lbs

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

50,000.00

51,270.12

52,611.89

54,025.55

TOTAL

240.66

273.04

306.96

820.66

2o/100’ with 50,000 lbs

At 50,000

lbs tension

The doglegseverity is

times 2

The drag is

times 3.7

Torque and Drag




Torque and Drag

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

200,000.00

206,098.02

212,474.74

219,134.88

TOTAL

5109.09

5307.99

5516.81

15,933.89

4o/100’ with 200,000 lbs

DEPTH

(feet)

TENSION

(pounds)

INTERVAL

DRAG

(pounds)

2300

2200

2100

2000

200,000.00

203,364.09

206,829.06

210,395.56

TOTAL

2334.63

2396.24

2459.80

7190.67

2o/100’ with 200,000 lbs

At 200,000

lbs tension

The doglegseverity is

times 2

The drag is

times 2.2

Torque and Drag




Torque and Drag

Increasing both tension anddogleg severity will increase the

torque and drag

However, for a 2o /100’ dogleg, ittakes twice as much hole to

achieve the same inclination as a

4o /100’ dogleg

Torque and Drag




Torque and Drag

Class Problem

The buoyant weight of a pipe segment is-2000 lbs (100 foot long) and there is100,000 lbs tension at the bottom of thesegment

The average inclination for the pipesegment is 200 and the friction coefficient is0.30

Two cases:

Case 1 Case 2

ΔI = 00 ΔI = 00

Δ A = 00 Δ A = 180

What is the tension at the top of the pipesegment

Torque and Drag




Torque and Drag

Drilling torque increasing withdepth

45

40

35

30

25

20

1510,000 10,500 11,000 11,500 12,000 12,500 13,000 13,500 14,000 14,500 15,000

PDC Torque

Trend

Roller Cone Torque Trend

Increased Torque Caused by Cuttings

Buildup

Depth, ft

Erratic Torques Caused by Torsional Dynamics

T

o r q u e ,

1 0 0 0 l b f - f t

Torque and Drag




Torque and Drag

Three ways to change the drag ina well

Change the friction coefficient

Change the directional profile

Change the string weight or tension

Torque and Drag




Torque and Drag

Reducing the friction coefficientcan have the most dramaticaffect on torque and drag

The friction coefficient can beaffected by mud type, bentonitecontent, solids content and variousadditives

Generally, oil base muds will havethe lowest friction coefficient but notalways

Torque and Drag




Torque and Drag

Friction coefficients will depend uponthe torque and drag model used

Below are some ranges for friction

coefficients from Sperry SunDrilling Fluid μ in Casing μ in Formation

Oil based 0.16 to 0.20 0.17 to 0.25

Water based 0.25 to 0.35 0.25 to 0.40

Brine 0.30 to 0.40 0.30 to 0.40

Torque and Drag




Torque and Drag

With a top drive, pipe can be runin the hole even if it will not fall

by its own weight because

rotation reduces the drag

Torque and Drag



© 2001 PetroSkills LLC, All Rights Reserved

Direction of

Friction Drag

V A

= Axial Velocity

Component

V C

= Circumferential

Velocity Component

V R

= Resultant Velocity Component

Axial Drag

RotationV C

DownwardTravel

V R

V A

F N = Normal Force

q g

Resultant Drag = μFN = Constant

Circumferential Drag

Axial

Drag

Axial drag is reduced when rotational speed is increased.

Torque and Drag




Torque and Drag

Changing the directional profilecan affect the torque and drag

Consider a well with a target TVD

of 15,500 feet and departure of 8,800 feet

There are any number of directional profiles that can be

used to hit the target

Torque and Drag




Torque and Drag

Torque and Drag




Torque and Drag

The directional profile does notmake a significant difference

Getting from point A to point B

takes a certain amount of drag

Generally, lower dogleg

severities will yield slightly lower

drag values

Torque and Drag




Torque and Drag

When drilling a directional well,the actual dogleg severity will begreater than the planned dogleg

severity; therefore, the actualdrag values will be greater thanthe calculated values

To compensate, you can use aslightly higher friction coefficientwhen planning

Torque and Drag




Torque and Drag

Reducing the pipe weight will

reduce the tension and therefore

drag values

Replace drill collars with HWDPand drill pipe

Drill pipe can be run in

compression in a directional wellif required but should not exceed

the critical buckling load

Torque and Drag




Torque and Drag

Torque and Drag




Torque and Drag

In horizontal wells, the drill pipewill not fall in the hole under its

own weight

The pipe must be pushed intothe hole using the weight of the

pipe above the critical inclination

Torque and Drag




Torque and Drag

The critical inclination is werethe drag equals the weight

component along the axis of the

holeThe critical inclination can be

calculated based on the friction

coefficient

1tan 1 I

Torque and Drag




Torque and Drag

Anything above the criticalinclination must be pushed

Keep the pipe in the horizontal

section as light as possibleDrill collars require more weight

to push them along the wellbore

Torque and Drag




Torque and Drag

The pipe in the horizontalsection should be as light as

practical so limit the BHA

HWDP is used in and above thecurve

Drill collars may or may not be

required in the vertical portionabove the HWDP

Torque and Drag



Torque and Drag

Typical drill string design for

horizontal well

Higher Weight

or StandardDrillpipe

Steerable Motor

Bit

MWD/LWD

Non-magCompressive

Service Pipe

Premium Drillpipe

0

2000

4000

6000

8000

10,000

12,000

14,000

16,0000 1000 2000 3000 4000 5000

HWDP

D e p t h ,

f t

torque and drag

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