calculation equations for directional drilling
DESCRIPTION
Calculations and examples for directional drilling engineeringTRANSCRIPT
1
Directional Drilling Calculations
2
Type I Type II Type III
Build and Hold
Build-Hold and Drop
ContinuousBuild
KOP
EOC
3
x
y
I
I
r
rL
In the BUILDSection
x = r (1 - cos I)
y = r sin I
L = r rad
degI r180
= L
BUR*
000,18r
4
N18E
N55WS20W
S23E
Azimuth
Angle
5
Example 1: Design of Directional Well
Design a directional well with the following restrictions:
• Total horizontal departure = 4,500 ft
• True vertical depth (TVD) = 12,500 ft
• Depth to kickoff point (KOP) = 2,500 ft
• Rate of build of hole angle = 1.5 deg/100 ft
• Type I well (build and hold)
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Example 1: Design of Directional Well
(i) Determine the maximum hole angle required.
(ii) What is the total measured depth (MD)?
(MD = well depth measured along the wellbore,
not the vertical depth)
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(i) Maximum Inclination
Angle
r1 18 000
15
,
. r2 0
D4 1
12 500 2 500
10 000
D
ft
, ,
,
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(i) Maximum Inclination Angle
500,4)820,3(2
500,4)820,3(2000,10500,4000,10 tan2
x)rr(2
x)rr(2)DD(xDDtan2
221-
421
4212
1424141
max
3.26max
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(ii) Measured Depth of Well
ft 265,9L
105,4sinL
ft 4,105
395500,4x
ft 395
)26.3 cos-3,820(1
)cos1(rx
Hold
Hold
Hold
1Build
10
(ii) Measured Depth of Well
265,9180
26.33,8202,500
LrDMD Holdrad11
ft 518,13MD
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* The actual well path hardly ever coincides with the planned trajectory
* Important: Hit target within specified radius
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What is known?I1 , I2 , A1 , A2 ,
L=MD1-2
Calculate = dogleg angle
DLS =L
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Wellbore Surveying Methods
Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential
Other Topics Kicking off from Vertical Controlling Hole Angle
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I, A, MD
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Example - Wellbore Survey Calculations
The table below gives data from a directional survey.
Survey Point Measured Depth Inclination Azimuth along the wellbore Angle Angle
ft I, deg A, deg
A 3,000 0 20 B 3,200 6 6 C 3,600 14 20 D 4,000 24 80
Based on known coordinates for point C we’ll calculate the coordinates of point D using the above information.
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Example - Wellbore Survey Calculations
Point C has coordinates:
x = 1,000 (ft) positive towards the east
y = 1,000 (ft) positive towards the north
z = 3,500 (ft) TVD, positive downwards
Z
E (x)
N (y)C
Dz
N
D
C
yx
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Example - Wellbore Survey Calculations
I. Calculate the x, y, and z coordinates of points D using:
(i) The Average Angle method
(ii) The Balanced Tangential method
(iii) The Minimum Curvature method
(iv) The Radius of Curvature method
(v) The Tangential method
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The Average Angle Method
Find the coordinates of point D using the Average Angle Method
At point C, X = 1,000 ft
Y = 1,000 ft
Z = 3,500 ft
80A 24I
20A 14I
ft 400MD D, toC fromdepth Measured
DD
CC
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The Average Angle Method
80A 24I
20A 14I
ft 400MD D, toC fromdepth Measured
DD
CC
Z
E (x)
N (y)
C
Dz
N
D
C
yx
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The Average Angle Method
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The Average Angle Method
This method utilizes the average of I1 and I2 as an inclination, the average of A1 and A2 as a direction, and assumes all of the survey interval (MD) to be tangent to the average angle.
From: API Bulletin D20. Dec. 31, 1985
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ft 71.8350cossin19400
cossin
502
8020
2
192
2414
2
AVGAVG
DCAVG
DCAVG
AIMDNorth
AAA
III
The Average Angle Method
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The Average Angle Method
ft
AIMDEast AVEAVG
76.9950sinsin19400
sinsin
ft
IVert AVG
21.378cos19400
cos400
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The Average Angle Method
At Point D,
X = 1,000 + 99.76 = 1,099.76 ft
Y = 1,000 + 83.71 = 1,083.71 ft
Z = 3,500 + 378.21 = 3,878.21 ft
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The Balanced Tangential Method
This method treats half the measured distance (MD/2) as being tangent to I1 and A1 and the remainder of the measured distance (MD/2) as being tangent to I2 and A2.
From: API Bulletin D20. Dec. 31, 1985
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The Balanced Tangential Method
ft 59.59
)80cos24sin20cos14(sin2
400
)AcosIsinAcosI(sin2
MDNorth DDCC
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The Balanced Tangential Method
96.66ft
)80sin24sin20sin14(sin2
400
)AsinIsinAsinI(sin2
MDEast DDCC
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The Balanced Tangential Method
ft77.376)14cos24(cos2
400
)IcosI(cos2
MDVert CD
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The Balanced Tangential Method
At Point D,
X = 1,000 + 96.66 = 1,096.66 ft
Y = 1,000 + 59.59 = 1,059.59 ft
Z = 3,500 + 376.77 = 3,876.77 ft
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Minimum Curvature Method
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Minimum Curvature Method
This method smooths the two straight-line segments of the Balanced Tangential Method using the Ratio Factor RF.
RF = (2/DL) * tan(DL/2) (DL= and must be in radians)
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Minimum Curvature Method
The dogleg angle, , is given by:
radians 36082.020.67
0.935609
))2080cos(1(24sinsin14-14)-cos(24
))AAcos(1(IsinIsin)IIcos(Cos CDDCCD
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Minimum Curvature Method
The Ratio Factor,
ft 25.6001099.1*59.59
RF)IcosIsinAcosI(sin2
MDNorth
01099.12
67.20tan*
3608.0
2RF
Z
tan2
RF
DDCC
2
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Minimum Curvature Method
ft 380.911.01099*376.77
RF)IcosI(cos2
MDVert
ft 97.721.01099*96.66
RF)AsinIsinAsinI(sin2
MDEast
DC
DDCC
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Minimum Curvature Method
At Point D,
X = 1,000 + 97.72 = 1,097.72 ft
Y = 1,000 + 60.25 = 1,060.25 ft
Z = 3,500 + 380.91 =3,888.91 ft
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The Radius of Curvature Method
ft 79.83
180
)2080)(1424(
)20sin80)(sin24cos400(cos14
180
)AA)(II(
)AsinA)(sinIcosI(cosMDNorth
2
2
CDCD
CDDC
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The Radius of Curvature Method
ft 95.14
180
)2080)(1424(
)80cos20)(cos24cos14(cos400
180
)AA)(II(
)AA)(cosIcosI(cosMDEast
2
2
CDCD
DCDC
2180
CDCD
DCDC
AAII
AcosAcosIcosIcosMDEast
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The Radius of Curvature Method
ft 73.377180
1424
)14sin400(sin24
180
II
)IsinI(sinMDVert
CD
CD
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The Radius of Curvature Method
At Point D,
X = 1,000 + 95.14 = 1,095.14 ft
Y = 1,000 + 79.83 = 1,079.83 ft
Z = 3,500 + 377.73 = 3,877.73 ft
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The Tangential Method
80A 24I
20A 14I
ft 400MD D, toC fromdepth Measured
DD
CC
DD AIMDNorth cossin
ft 25.2880cos24sin400
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The Tangential Method
ft 22.16080sinsin24400
sinsin
DD AIMDEast
ft 42.36524cos400
Icos400Vert D
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The Tangential Method
ft 3,865.42365.423,500 Z
ft 1,028.2528.251,000Y
ft 1,160.22160.221,000X
D,Point At
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Summary of Results (to the nearest ft)
X Y Z
Average Angle 1,100 1,084 3,878
Balanced Tangential 1,097 1,060 3,877
Minimum Curvature 1,098 1,060 3,881
Radius of Curvature 1,095 1,080 3,878
Tangential Method 1,016 1,028 3,865
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Torque and DragCalculations
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Friction - Stationary
• Horizontal surface
• No motion• No applied force
Fy = 0
N = W
N
W
N= Normal force = lateral load = contact force = reaction force
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Sliding Motion
• Horizontal surface
• Velocity, V > 0
• V = constant
• Force along surface
N = W
F = N = W
N
W
F N
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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.
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Frictionless, Inclined, Straight Wellbore:
pipe.ROTATING for used are equations
(2) : wellbore to 0
(1) : wellborealong 0
ar
These
F
F
IcosWT
IsinWN
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Effect of Friction (no doglegs):
2. Consider Effect of Friction ( no doglegs):
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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
(3)IsinWIcosWT
FIcosWT f
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Effect of Friction (no doglegs):
(b) Raising: Friction still opposes motion,
so
IsinWIcosWT
FIcosWT f
(4)
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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)
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Solution
From Equation (3) above,
(3)
When the logging tool is barely sliding down the wellbore,
IsinWIcosWT
0T
IsinW4.0IcosW0
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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
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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?
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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
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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 lbf
Torque = 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
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Problem 2 - Equations - Horizontal
Torque, T = Wd/(24 ) = 21,000 ft-lbf
F = N T = F * dN = W
W
Force to move pipe, F = W = 72,000 lbf
An approximate equation, with W in lbf and d in inches
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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
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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
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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.]
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Solution to Problem 3
(a) Hook load when rotating off bottom:
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Solution to Problem 3 - Rotating
When rotating off bottom.
lbf 120,000lbf 000,60
60cos*ft 8000*ft
lb30ft 2000*
ft
lb30
HLHLHL
5.0
80002000
lbf 000,180HL
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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
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Solution to Problem 3 - lowering
In the inclined section,
N = 30 * 8,000 * sin 60
= 207,846 lbf
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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
67
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
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Solution to Problem 3 - Summary
MDft
RIHROT
POH
Axial Tension
HL
69
Solution to Problem 3 - rotating
2(d) Torque when rotating off bottom:In the Inclined Section:
NF
IsinWN
2
d*F
Arm*Force
Torque
f
70
Solution to Problem 3 - rotating
(i) As a first approximation, assume the pipe lies at lowest point of hole:
12
1*
2
7*60sin*8000*30*3.0
2
dIsinW
2
dN
2
dFTorque f
lbf-ft 187,18Torque
71
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
72
Solution to Problem 3 - Rotating
Assume “Equilibrium”
at angle as shown.
sinIsinWFF fTangentAlong 0
cosIsinWNF Tangentto.Perpend 0
…… (7)
sinIsinWN …… (6)
cosIsinWN
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Solution to Problem 3 - rotating
Solving equations (6) & (7)
(8))(tan
tan
cosIsinW
sinIsinW
N
N
1
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Solution to Problem 3 - rotating
(ii) Continued
Evaluating the problem at hand:
From Eq. (8),
Taking moments about the center of the pipe:
2
d*FT f
70.16
)3.0(tan)(tan 11
75
Solution to Problem 3 - rotating
Evaluating the problem at hand:
From Eq. (6),
lbf 724.59F
70.16sin*sin60*8000*30
sinIsinWF
f
f
76
Solution to Problem 3 - rotating
Evaluating the problem at hand:
From Eq. (9),
lbf-ft 420,17Torque
12
1*
2
7*59,724
2
d*FT f
77
Solution to Problem 3
2 (d) (ii) Alternate Solution:
78
Solution to Problem 3
Taking moments about tangent point,
24
7*70.16sin*sin60*8000*30
2
dsinIsinWT
lbf-ft 420,17T
79
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