PETE 411
Well Drilling
Lesson 36
Torque and Drag Calculations
1
Torque and Drag Calculations
 Friction
 Logging
 Hook Load
 Lateral Load
 Torque Requirements
 Examples
2
Assignments:
PETE 411 Design Project
due December 9, 2002, 5 p.m.
HW#18 Due Friday, Dec. 6
3
Friction - Stationary
• Horizontal surface
N
• No motion
• No applied force
S Fy = 0
N=W
W
N= Normal force = lateral load = contact force = reaction force
4
Sliding Motion
N
• Horizontal surface
• Velocity, V > 0
• V = constant
N
F
• Force along surface
N=W
W
F=N=W
5
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.
6
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.
7
Effect of Friction (no doglegs):
2. Consider Effect of Friction ( no doglegs):
8
Effect 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
T  W cos I  Ff
T  W cos I  W sin I
(3)
9
Effect of Friction (no doglegs):
(b) Raising: Friction still opposes motion,
so
T  W cos I  Ff
T  W cos I  W sin I
(4)
10
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)
11
Solution
From Equation (3) above,
T  W cos I  W sin I
(3)
When pipe is barely sliding down the wellbore,
T  0
 0  W cos I  0.4W sin I
12
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:
  cot I
13
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?
14
Problem 2 - Solution - Force
(a) What force will it take to move this pipe along
the horizontal section of the wellbore?
N
F=?
F=0
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
15
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
Then, as before,
N = W = 30 lb/ft * 8,000 ft = 240,000 lbf
F
Torque = F*d/2 = Nd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft
Torque to rotate pipe, T = 21,000 ft-lbf
16
Problem 2 - Equations Horizontal
F = N
N=W
T=F*s
( s=d/24 )
W
Force to move pipe,
F = W
Torque, T = Wd/(24 )
= 72,000 lbf
= 21,000 ft-lbf
An approximate equation, with W in lbf and d in inches
17
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 f.
Taking moments about the point P:
T
Torque, T = W * (d/2) sin f in-lbf
F
d/2 f
Where f = atan  = atan 0.3 = 16.70
o
T = 240,000 * 7/24 * 0.2873 = 20,111 ft-lbf
P
W
18
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.  = 0.3
19
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.]
20
Solution to Problem 3
(a) Hook load when rotating off bottom:
21
Solution to Problem 3 - Rotating
HL  HL2000  HL8000
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.
22
Solution to Problem 3 - lowering
2 (b) Hook load when RIH:
The hook load is decreased by friction in the
wellbore.
Ff  N
In the vertical portion,
N  30 * 2000 * sin o0  0
o
Thus,
F2000  0
23
Solution to Problem 3 - lowering
In the inclined section,
N = 30 * 8,000 * sin 60
= 207,846 lbf
24
Solution to Problem 3 - Lowering
Thus, F8000 = 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
25
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
26
Solution to Problem 3 - Summary
RIH
2,000
ROT
POH
MD
ft
10,000
0
60,000
120,000
180,000 240,000
27
Solution to Problem 3 - rotating
2(d) Torque when rotating off bottom:
In the Inclined Section:
N  W sin I
F  N
Torque
 Force * Arm
d
 Ff *
2
28
Solution to Problem 3 - rotating
(i) As a first approximation, assume the pipe
lies at lowest point of hole:
d
d
d
Torque  Ff    N   W sin I 
 2
 2
 2
7 1 
 0.3 * 30 * 8000 * sin 60 *  * 
 2 12 

Torque  18,187 ft - lbf
29
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
30
Solution to Problem 3 - Rotating
Assume “Equilibrium”
at angle f as shown.
F
Along Tangent
F
 0  Ff  W sinI sin f
…… (6)
N  W sinI sin f
Perpend . to Tangent
 0  N  W sinI cos f
N  W sinI cos f
…… (7)
31
Solution to Problem 3 - rotating
N W sin I sin f
Solving equations (6) & (7) 

N W sin I cosf
  tan f
1
f  tan (  )
(8)
32
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
f

tan
(

)

tan
(0.3)
From Eq. (8),
f  16.70

33
Solution to Problem 3 - rotating
Evaluating the problem at hand:
From Eq. (6), Ff  W sin I sin f
 30 * 8000 * sin60 * sin 16.70

Ff  59.724 lbf
34

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
35
Solution to Problem 3
2 (d) (ii) Alternate Solution:
36
Solution to Problem 3
Taking moments about tangent point,
d
T  W sin I sin 
2
7
 30 * 8000 * sin60 * sin 16.70 *
24


T  17,420 ft - lbf
37
Solution to Problem 3
Note that the answers in parts (i) & (ii) differ
by a factor of cos f
(i) T = 18,187
(ii) T = 17,420
cos f = cos 16.70 = 0.9578
38
Effect of Doglegs
(1) Dropoff Wellbore
  dogleg angle
39
Effect of Doglegs
A. Neglecting Axial Friction
(e.g. pipe rotating)
F
along normal
: W sin I  (T  T) sin
WWsin
sinI I+2T
sT sin

2
 T sin

2

2
 T sin

2
N0
N0

N  W sin I  2T sin
2
(10)
40
Effect of Doglegs
A. Neglecting Axial Friction
F
along tangent
: (T  T) cos
T cos
cos

2
1

2

2
 W cos I  T cos

2
0
 W cos I
T  W cos I
(11)
41
Effect of Doglegs
B. Including Friction (Dropoff Wellbore)
While pipe is rotating
N  W sin I  2T sin

2
(10)&(11)
T  WcosI
42
Effect of Doglegs
B. Including Friction
While lowering pipe (RIH)
N  W sin I  2T sin

(as above)
2
T  W cos I  N
i.e. T  W cos I   (W sin I  2T sin  ) (12)
2
43
Effect of Doglegs
B. Including Friction
While raising pipe (POH)
T  W cos I  N

T  W cos I   (W sin I  2T sin )
2
(13)

d
d
Torque  N     ( W sin I  2T sin )
2
 2
 2
(14)
44
Effect of Doglegs
(2) Buildup Wellbore
  dogleg angle
45
Effect of Doglegs
A. Neglecting Friction
(e.g. pipe rotating)
F
along normal


: W sinI  T  T  sin  T sin  N  0
2
2


W sin I  2T sin  T sin  N  0
2
2

N  W sin I  2T sin
2
46
Effect of Doglegs
A. Neglecting Axial Friction
F
along tangent
: (T  T) cos
T cos
cos

2
1

2

2
 W cos I  T cos

2
0
 W cos I
T  W cos I
(16)
47
Effect of Doglegs
B. Including Friction (Buildup Wellbore)
When pipe is rotating
N  W sin I  2T sin

2
(15)&(16)
T  WcosI
48
Effect of Doglegs
B. Including Friction
While lowering pipe (RIH)
N  W sin I  2T sin

2
(15)
T  W cos I   N
T  W cos I   W sin I  2T sin

2
(17)
49
Effect of Doglegs
While raising pipe (POH)
T  W cos I  N
i.e. T  WcosI   WsinI - 2Tsin

2

d
d
Torque   N       W sin I  2T sin
2
 2
 2
(18)
(19)
50
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
 = 0.25
51
Problem
#4
- Curved
Wellbore
with
Friction
T = 100,000 lbf
52
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?
53
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.
54
Solution 4(a) - Rotating
From equation (11),
T68 = 101,315 lbf
T  W cos I
lb

 30 *100 ft * cos 64
ft
 1,315 lbf
 T68  100 ,000  1,315
T68  101,315 lbf
T60 = 100,000 lbf
 rotating
55
Solution 4 (b)
(b) Tension in pipe 100 ft Up-Hole when Pipe is being
lowered:
From equation (10):
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
56
Solution 4 (b)
From equation 10,
Friction Force  N  0.25 *16,648
Ff  4,162 lbf
From equation 12,
T  W cos I  N
57
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
58
Solution 4 (c)
(c) Tension in Pipe 100 ft Up-Hole when pipe is being
raised:
From equation (10),
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
59
Solution 4 (c)
Friction Force  N  0.25 *16,648
Ff  4,162 lbf
From equation 12,
T  W cos I  N
60
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
61
Solution 4(a, b and c)
SUMMARY
T60
T68
Rot
100,000
101,315
RIH
100,000
97,153
POH
100,000
104,477
62
Solution 4 (d)
(d) Lateral load on centralizer if spacing = 40 ft. (after
pipe has been rotated):
From above,
at   64

N  16,648 lbf
This is for 100 ft distance
63
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
64
Alternate Approach
(d) Lateral load on centralizer if spacing = 40 ft.
(after pipe has been rotated)
From above, at   60  , T  100,000 lbf
From above, at   68 , T  101,315 lbf
So, 30 ft up-hole,
T  100,000  1,315 * (30 / 100 ) lbf
T  100,395 lbf
65
Alternate Approach
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
66
Centralizer
67