Torque and Drag Calculations x
Friction
x
Logging
x
Hook Load
x
Lateral Load
x
Torque Requirements
x
Examples
HW #8 Well Survey due 11-04-02
Friction - Stationary • Horizontal surface
N
• No motion • No applied force
Σ Fy = 0 N=W W N= Normal force = lateral load = contact force = reaction force
Sliding Motion N
• Horizontal surface • Velocity, V > 0
F
N
• V = constant • Force along surface
W
N = W F = µ N = µ
W
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.
Frictionless, Inclined, Straight Wellbore:
∑
F
=0
along wellbore :
∆T = W cos I
(1)
∑
F
=0
⊥ar to wellbore :
N = W s in I
(2)
These
equations are used for ROTATING pipe.
Effect of Friction (no doglegs): 2. Consider Effect of Friction ( no doglegs):
Effect of Friction (no doglegs): Frictional Force, F = µ N = µ W 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
∆T = W co s I − Ff ∆T = W co s I − µ W sin I
(3)
Effect of Friction (no doglegs):
(b) Raising: Friction still opposes motion,
so
∆T = W cos I + Ff ∆T = W cos I + µ W sin I
(4)
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 µ =0.4)
Solution
From Equation (3) above,
∆T = W cos I − µ W sisinn I
(3)
When pipe is barely sliding down the wellbore,
∆T ≅ 0
∴ 0 = W cos I − 0.4W sin I
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:
µ = co cott I
Problem 2 Consider a well with a long horizontal section. section. An 8,000-ft long string of 7” OD csg. is in the hole. µ = 0.3 Buoyed weight of pipe = 30 lbs/ft. (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?
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
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 Torque = F*d/2 =
F
Nd/2 = 0.3 * 240,000 lbf * 7/(2 * 12) ft
Torque to rotate pipe, T = 21,000 ft-lbf
Problem 2 - Equations - Horizontal F=
N=W
N
T=F*d
W
Force to move pipe,
Torque,
T=
F=
Wd/(24 )
W
= 72,000 lbf
= 21,000 ft-lbf
An approximate equation, with W in lbf and d in inches
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
T
F
in-lbf d/2
Where
= atan
= atan 0.3 = 16.70
P
o
T = 240,000 * 7/24 * 0.2873 = 20,111 ft-lbf
W
Problem 3 A well with a measured depth of 10,000 ft. may be approximated as being perfectly vertical to the th e kick-off point at 2,000 ft. A string of 7” OD 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
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.]
Solution to Problem 3 (a) Hook load when rotating off bottom:
Solution to Problem 3 - Rotating
HL = HL 2000
+ HL 8000 0. 5
= 30
lb ft
* 2000 ft + 30
lb ft
↓
* 8000 ft * cos 60
= 60,000 lbf + 120,000 lbf H L = 1 8 0,0 0 0 lb lbf f
When rotating off bottom.
Solution to Problem 3 - lowering 2 (b) Hook load when RIH:
The hook load is decreased by friction in the wellbore.
Ff = µ N
In the vertical portion,
N = 3 0 * 2000 * sin o0
o o
Thus,
F2000
=0
=0
Solution to Problem 3 - lowering
In the inclined section,
N = 30 * 8,000 * sin 60 = 207,846 lbf
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
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
Solution to Problem 3 - Summary
RIH
ROT POH
MD ft
Solution to Problem 3 - rotating 2(d) Torque when rotating off bottom: In the Inclined Section:
N = W s in I F = µ N
Torque
= Force * Arm = Ff *
d 2
Solution to Problem 3 - rotating (i) As a first approximation, assume the pipe lies at lowest point of hole: Torque
d d d = Ff = µ N = µ W sin I 2 2 2
= 0.3 * 30 * 8000 * sin 60 Torque
7 1 * * 2 12
= 18,187 ft - lbf
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 F f . The pipe will tend to climb up the side of the wellbore…as it rotates
Solution to Problem 3 - Rotating Assume “Equilibrium” at angle φ as shown.
∑F
Along Tangent
∑F
= 0 = Ff − W sinI sin φ µN = W sinI sin φ
Perpend. to Tangent
…… (6)
= 0 = N − W sinI cos φ
N = W sin I cos φ
…… (7)
Solution to Problem 3 - rotating
Solving equations (6) & (7)
⇒
µ N N
=
W s in I si sin n φ W sin I co sφ
µ = ta tan n φ −1
φ = ta tan n ( µ )
(8)
Solution to Problem 3 - rotating (ii) continued Taking moments about the center of the pipe: T
=
Ff *
d 2
Evaluating the problem at hand: −1
−1
tan n ( µ ) = ta tan n (0.3) From Eq. (8), φ = ta
φ = 16 .70
Solution to Problem 3 - rotating Evaluating the problem at hand: From Eq. (6), Ff
= W sisinn I sisinn φ = 30 * 8000 * sin60
Ff = 5 9.7 2 4 lb lbf f
* si sin n 1 6.7 0
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 ,42 420 0 ft - lb lbf f
Solution to Problem 3 2 (d) (ii) Alternate Solution:
Solution to Problem 3 Taking moments about tangent point,
T = W sin I sin Ο
d 2
= 30 * 8000 * sin60
* sin 16.70 *
T = 1 7,4 2 0 ft - lb lbf f
7 24
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
Effect of Doglegs (1) Dropoff Wellbore
δ = dogleg angle
Effect of Doglegs A. Neglecting Axial Friction (e.g. pipe rotating)
∑F
along normal
+
: W sin I + (T + ∆T) sin
sinII + sT sin WWsin 2T
δ 2
δ 2
δ
+ T sin − N = 0 2
δ
+ ∆T sin − N = 0 2
δ N ≅ W si sin n I + 2T si sin n 2
(10)
∆≅
Effect of Doglegs
A. Neglecting Axial Friction
∑
δ δ Falong tangent : (T + ∆ T ) co coss − W co coss I − T co coss = 0 2 2 δ ∆T co coss = W co coss I 2
δ coss → 1 ⇒ co 2
s s o c I I T W
(11)
Effect of Doglegs B. Including Friction (Dropoff Wellbore) While pipe is rotating
δ N = W si sin n I + 2T sin sin 2 ∆T = WcosI
(10)&(11)
Effect of Doglegs B. Including Friction While lowering pipe (RIH) δ N = W sin I + 2T sin 2
(as above)
∆T = W cos I − µ N i.e.
δ (12) ∆T = W co coss I − µ ( W si sin n I + 2T si sin n ) 2
Effect of Doglegs B. Including Friction While raising pipe (POH)
∆T = W cos I + µ N ∆T =
δ
W c o sI + µ ( W s in I + 2T s in ) 2
d d δ Torque = µ N ≅ µ ( W si sin n I + 2T si sin n ) 2 2 2
(13)
(14)
Effect of Doglegs (2) Buildup Wellbore
δ = dogleg angle
Effect of Doglegs A. Neglecting Friction (e.g. pipe rotating)
∑F
along normal
:
W sin I − ( T + ∆T ) sin W sin I − 2T sin
δ 2
δ 2
δ
− T sin − N = 0 2
δ
− ∆T sin − N = 0
N ≅ W sin I − 2T sin
2
δ 2
∆≅
Effect of Doglegs
A. Neglecting Axial Friction
∑
δ δ Falong tangent : (T + ∆ T ) co coss − W co coss I − T co coss = 0 2 2 δ coss = W co coss I ∆T co 2
δ coss → 1 ⇒ co 2
s s o c I I T W
(16)
Effect of Doglegs
B. Including Friction (Buildup Wellbore) When pipe is rotating
δ N = W sin I − 2T sin 2
∆T = WcosI
(15)&(16)
Effect of Doglegs B. Including Friction While lowering pipe (RIH) δ N = W sin I − 2T si sin n 2
(15)
coss I − µ N ∆T = W co
δ coss I − µ W si sin n I − 2T si sin n ∆T = W co 2
(17)
Effect of Doglegs While raising pipe (POH) coss I + µ N ∆T = W co i.e.
∆T = WcosI + µ WsinI - 2Tsin
δ 2
d d δ Torque = µ N ≅ µ W si sin n I − 2T si sin n 2 2 2
(18)
(19)
Problem #4 - Curved Wellbore with w ith Friction
In a section of our well, hole angle drops at the rate of 8 of 8 degrees per 100 per 100 ft. ft. The axial tension is 100,000 lbf at lbf at the location where the hole angle is 60 degrees. degrees.
Buoyed weight of pipe = 30 lbm/ft
µ = 0.25
Problem #4 - Curved Wellbore with Friction
T = 100,000 lbf
Evaluate the Following: (a) What is the axial tension in the pipe 100 ft. 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? ft?
Solution 4(a) - Rotating Axial tension 100 ft up hole when pipe is rotating :
60 + 68 2
IAVG
=
IAVG
= 64
o
Pipe is rotating so frictional effect on axial load may be neglected.
Solution 4(a) - Rotating T68 = 101,315 lbf
From equation (11),
∆T = W co coss I lb 100 0ft * co coss 64 = 30 * 10 ft
lbf f = 1,315 lb 100 0,00 000 0 + 1,31 315 5 ∴ T68 = 10
T68
T60 = 100,000 lbf
101 1,31 315 5 lb lbf f ← rotating = 10
Solution 4 (b) (b) Tension in pipe 100 ft Up-Hole when Pipe is being lowered: From equation (10):
δ N = W si sin n I + 2T si sin n 2 N = 30 * 10 100 0 * si sin n 64
= 2,696 + 13,951 N = 16,648 lb lbf f
100 0,00 000 0 * si sin n4 + 2 * 10
Solution 4 (b) From equation 10,
Friction Force Ff
N = 0.25 = µ = 4,162
* 16,648
lbf lb f
From equation 12,
coss I − µ N ∆T = W co
Solution 4(b) - Lowering T68 = 97,153 lbf
From equation 12,
∆T = (30 *100 * cos 64 ) − 4,162
= -2,847
∴ T68 = 10 100 0,00 000 0 − 2,86 867 7
T68
153 3 = 97 ,15
lbf lb f
(T + ∆ T) T60 = 100,000 lbf
Solution 4 (c) (c) Tension in Pipe 100 ft Up-Hole when pipe is being raised: From equation (10),
δ N = W si sin n I + 2T si sin n 2 N = 30 * 10 100 0 * si sin n 64
= 2,696 + 13,951 N = 16,648 lb lbf f
100 0,00 000 0 * si sin n4 + 2 * 10
Solution 4 (c)
Friction Force Ff
= µ N = 0.25 = 4,162
* 16,648
lbf lb f
From equation 12,
∆T = W co coss I + µ N
Solution 4(c) - Raising T68 = 105,477 lbf
From equation 12,
100 0 * cos 64 ) + 4,16 162 2 ∆T = (30 *10
= 5477 lbf 100 0 ,00 000 0 + 5477 ∴ T68 = 10
T68
(T + ∆T)
= 105,477 lbf
T60 = 100,000 lbf
Solution 4(a, b and c) SUMMARY
T60
T68
Ro t
100,000
101,315
R IH
100,000
97,153
POH PO H
100,00 100, 000 0
104, 10 4,47 477 7
Solution 4 (d) (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated): From above,
at θ = 64
N = 16,648 lbf
This is for 100 ft distance
Solution 4 (d)
∴
for 40 ft distance,
40 N centr . = 16,64 648 8 * 100 0 10 lbf f = 6,659 lb i.e., Lateral load on centralizer,
N centr .
= 6,65 659 9 lb lbf f
lb Note : 40 ft of pipe * 30 ft
= 1200 lb lbf f
Alternate Approach (d) Lateral load on centralizer if spacing = 40 ft. (after pipe has been rotated) From above, at θ = 60 , T = 100,000 lbf
From above, at θ = 68 , T = 101,315 lbf
So, 30 ft up-hole,
T = 10 100 0 ,00 000 0 + 1,31 315 5 * (30 / 10 100 0 ) lbf T = 10 100 0 ,395 lbf
Alternate Approach δ sin n I + 2T si sin n From Eq. (10), N = W si 2 N = 30 * 40 * si sin n 64 + 2 * 10 100 0,39 395 5 * sin(1.6 ) {4 * 40/100} = 1,079 + 5,606
N = 6,685 lb lbf f
∴ for 40 ft centralizer spacing, N centr .
= 6,685 lbf
Centralizer
