This document presents the casing design and drilling engineering project for an oil and gas well. It includes:
1) Casing design calculations for 5 casing strings considering loading conditions like tension, compression, burst/collapse. Grade and dimensions are selected.
2) Drill string design calculations to determine minimum drill collar lengths at different depths to avoid compression on drill pipe.
3) Equations for axial loading, hydraulic calculations, and hoisting system are presented. Stress analysis shows second intermediate casing material needs upgrading to withstand loads.
4) Tables and figures show results like casing design, burst pressures, stress combinations, drill string specifications and calculated collar lengths.
IRJET- Geotechnical Investigation of Different Soil Samples using Regression ...
project in drilling engineering
1. Eastern Macedonia and Thrace Institute of Technology
MSc in Oil & Gas Technology
“Drilling Engineering”
“Project in Drilling Engineering”
Iliopoulos P.
Tsiknopoulos K.
Ballis. Th.
Supervisor: Gaganis B.
June 2015
2. EasternMacedoniaand Thrace Institute of Technology
MSc inOil & Gas Technology
DrillingEngineering
2
Table of Contents
Casing design.........................................................................................................................3
Axial Loading......................................................................................................................4
Tension..........................................................................................................................5
Compression ..................................................................................................................6
Burst/Collapse....................................................................................................................7
Drill string dimensions............................................................................................................9
HYDRAULICS ........................................................................................................................ 14
Hoisting system.................................................................................................................... 16
Appendix............................................................................................................................. 18
3. EasternMacedoniaand Thrace Institute of Technology
MSc inOil & Gas Technology
DrillingEngineering
3
Casing design
The answer to the question (5a) is included at the excel spreadsheet that was given as
an input for this project and is illustrated at the next table.
For the next question we have created the appropriate chart using the table in the
appendix which clearly shows the minimum number of casing strings needs to be placed
from top to bottom of our wellbore. In order to do so we have determine the EMD
while we have taken into account the minimum safety margin. (EMD pore + 0.5 = EMD
margin pore) and (EMD fracture -0.5 = EMD margin)
Figure 1 Mud densityline
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According to the previous chart we are able to decide about three casing strings taking
into account the initial casings (conductor, surface). So we created a model which has
five casings. Their dimensions were selected from the SPE textbook and the sample
charts that were given. Also we must take into account the corresponding mud density
and the buoyancy factor because we further want to calculate the stresses with the wet
weight.
Table 1 Casingdesign
Casings Depth(ft)
OD
Bit
CASING
OD (in)
Mud
Density
Buoyancy
Factor
Weight Grade Casing
ID (in)
Conductor 0 - 100 - 30 9,6 0,853 234,29 X-52 28,5
Surface
casing
0 - 2000 26 20 9,6 0,853 94 H-40 19,124
1st
intermediate
0- 5800 17,5 13,375 9,6 0,853 68 C-90 12,415
2nd
intermediate
0 - 10100 12,25 9,625 15 0,771 40 C-90 8,835
Production
liner
9900 -
13000
7,875 5,5 16,4 0,749 23 C-90 4,67
Axial Loading
F = ((D-L)*(Wa * BF) + Fpull)*2
Where, safety factor = 2
Wa = Casing Weight
Fpull = overpull force and BF = (1-
𝜌 𝑚
𝜌 𝑠
)
Ften =
π
4
σyield (OD2
– ID2
)
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Tension
Conductor
F = ((100 * 234.29 * 0.853) + 10000) * 2 = 59969.9 lbf
Ften = π/4 * 52000*(302
– 28.5 2
) = 3583771.8 lbf
Ften > F Can withstandthe stress
Surface casing
F = ((2000 * 94 * 0.853) + 10000) * 2 = 340728 lbf
Ften = π/4 * 40000*(202
– 19.1242
) = 1076706.2 lbf
Ften > F Can withstandthe stress
1st
intermediate
F = ((5800 * 68 * 0.853) + 10000) * 2 = 692846.4 lbf
Ften = π/4 * 90000*(13,3752
– 12.4152
) = 1750068.17 lbf
Ften > F Can withstandthe stress
2st
intermediate
With H-40
F = ((10100 * 32.3 * 0.771) + 10000) * 2 = 523046.7 lbf
Ften = π/4 * 40000*(9.625 2 – 9.0012) = 365.135 lbf
Ften < F Cannot withstand the stress
With C-90
F = ((10100 * 40 * 0.771) + 10000) * 2 = 642.968 lbf
Ften = π/4 * 90000*(9.625 2 – 8.835 2) = 1030839.8 lbf
Ften > F Can withstand the stress
Production liner
F = ((3100 * 23 * 0.749) + 5000) * 2 = 116807.4 lbf
Ften = π/4 * 90000*(5.52
– 4,672
) = 596666.2 lbf
Ften > F Can withstandthe stress
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Compression
Conductor
F = 100 * 234.29 = 23429 lbf
Ften = π/4 * 52000*(302
– 28.5 2
) = 3583771.8 lbf
Ften > F Can withstandthe stress
Surface casing
F = 2000 * 94 = 188000 lbf
Ften = π/4 * 40000*(202
– 19.1242
) = 1076706.2 lbf
Ften > F Can withstandthe stress
1st
intermediate
F = 5800 * 68 = 394400 lbf
Ften = π/4 * 90000*(13,3752
– 12.4152
) = 1750068.17 lbf
Ften > F Can withstandthe stress
2st
intermediate
With H-40
F = 10100 * 32.3 = 326230 lbf
Ften = π/4 * 40000*(9.625 2 – 9.0012) = 365.135 lbf
Ften < F Can withstand the stress
With C-90
F = 10100 * 40 = 404000 lbf
Ften = π/4 * 90000*(9.625 2 – 8.835 2) = 1030839.8 lbf
Ften > F Can withstand the stress
Production liner
F = 3100 * 23 = 71300 lbf
Ften = π/4 * 90000*(5.52
– 4.672
) = 596666.2 lbf
Ften > F Can withstandthe stress
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After the above calculations we conclude that instead of using H-40 type to the 2st
intermediate casing, we use C-90.The worst case scenario is obviously on tension
loadings.
Burst/Collapse
We assume a gas density of 3 ppg in case of a kick.
Internal pressure: Pi = 0.052 * ρgas * L
Maximum burst pressure: Pbr = 0.875
2 𝑡 𝜎 𝑦𝑖𝑒𝑙𝑑
𝑑𝑛
External pressure: Pf = 0.465 * Depth
Table 2 Burst calculations
Depth
(ft)
Pe (psi) Pff (psi) Pi (psi)
Burst
Pressure
(psi)
Max
Pburst
(psi)
Result
Surface casing
0 0 998,4 686,4 686,4
1530 Accept
2000 930 998,4 998,4 68,4
1st Intermidiate
casing
2000 930 4575,272 3982,472 3052,472
5652 Accept4000 1860 4575,272 4294,472 2434,472
5800 2697 4575,272 4575,272 1878,272
2st Intermidiate
casing
5800 2697 8734,076 8063,276 5366,276
6464 Accept8000 3720 8734,076 8406,476 4686,476
10100 4696,5 8734,076 8734,076 4037,576
Production Liner
9900 4603,5 11573,12 11089,52 6486,02
11884 Accept10100 4696,5 11573,12 11120,72 6424,22
13000 6045 11573,12 11573,12 5528,12
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Table 3 Stress combination
After completing the previous calculations we found out that our second intermediate
casing cannot withstand the total load according to the elliptic equation.
According to the given formula: σz term2 + σt term2 - σt term2* σz term2 >1 So we
conclude to change the dimensions of that casing from C-90 to C-95.
Table 4 New casing design
Casings Depth(ft) OD
Bit
CASING
OD (in)
Mud
Density
Buoyancy
Factor
Weight Grade Casing
ID (in)
Conductor 0 - 100 - 30 9,6 0,853 234,29 X-52 28,5
Surface
casing
0 - 2000 26 20 9,6 0,853 94 H-40 19,124
1st
intermediate
0- 5800 17,5 13,375 9,6 0,853 68 C-90 12,415
2nd
intermediate
0 -
10100
12,25 9,625 15 0,771 40 C-95 8,835
Production
liner
9900 -
13000
7,875 5,5 16,4 0,749 23 C-90 4,67
Depth
(ft)
Inner P Outer P
Axial
Load
Area
Axial
stress
σt term
σz
term
ellipse result
0 14,7 0 188000 26,91766 6984,263 0,008578 0,175 0,0292
2000 14,7 998,4 0 26,91766 0 -0,57404 0,0004 0,3297
2000 14,7 998,4 258400 19,4452 13288,63 -0,15795 0,1478 0,0701
4000 14,7 1996,8 122400 19,4452 6294,612 -0,31826 0,0701 0,1285
5800 14,7 2895,36 0 19,4452 0 -0,46253 0,0002 0,214
5800 14,7 4524 172000 11,45378 15016,88 -0,63656 0,167 0,5394
8000 14,7 6240 84000 11,45378 7333,826 -0,8788 0,0817 0,8507
10100 14,7 7878 0 11,45378 0 -1,11003 0,0002 1,2323
9900 14,7 8442,72 71300 6,629624 10754,76 -0,67118 0,1197 0,5451
10100 14,7 8613,28 66700 6,629624 10060,9 -0,68476 0,112 0,5581
13000 14,7 11086,4 0 6,629624 0 -0,88171 0,0002 0,7776
Production
Liner
Accept
Accept
Unaccept
Accept
1st
Intermidiat
e casing
2st
Intermidiat
e casing
Surface
casing
Stress Combination
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Drill string dimensions
Table 5 Given data
Specifications OD ID weight Steel
density
Drill pipe 5’’ 4,276’’ 19,5lbf/ft 490lbm/𝑓𝑡3
Drill collars 5’’ 2,5’’
Having into account all the given specifications for drill pipe and drill collars, it is
absolutely necessary to determine the suitable lengths. These lengths should ensure
that during the drilling process any point of drill pipe is not under compression and any
point of drill collars is not under tension. Different WOB at three stages, give us detailed
information in order to locate exactly the point (neutral point) above which there is not
tendency to buckling. The correct position which satisfies the previous restrictions is the
top of the collars.
EQUATIONS
Minimum drill collars length: Ldc= Fb/Wdc*BF
Stability force: Fs(depth)= AiPi-AoPo
Wdc = π/4(OD2-ID2)
Ft=Wdp*Ldp+Wdc*Ldc+0,052*ρmud*(D-Ldc)*(Adc-Adp)-0,052*ρmud*D*Adc-WOB
BF = 1 –
𝑀𝑢𝑑 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
65,5𝑙𝑏𝑚/𝑔𝑎𝑙
Where:
OD,ID= external and internal diameter
Pi= internal pressure
Po=external pressure
Fb=WOB
65,5= weigth og a gallon of steel
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Calculations
Wdc=π/4*((52
− 2,52
)/0,2945)=50,1lbf/ft
490lbm/𝑓𝑡3
=65,5lb/gal
For 5800ft
BF = 0,853
L=
𝐹𝑏
𝑊𝑑𝑐(1−
𝑃𝑓
𝑃𝑠
)
=
𝐹𝑏
𝑊𝑑𝑐∗𝐵𝐹
=
25000𝑙𝑏𝑓
50,1∗0,85
= 587ft = 590ft
For 10100ft
BF=0,771
L =
𝐹𝑏
𝑊𝑑𝑐(1−
𝑃𝑓
𝑃𝑠
)
=
𝐹𝑏
𝑊𝑑𝑐∗𝐵𝐹
=
50000𝑙𝑏𝑓
50,1𝑙𝑏𝑓
𝑓𝑡
∗0,77
= 769ft=770ft
For 13000ft
BF=0,749
L =
𝐹𝑏
𝑊𝑑𝑐(1−
𝑃𝑓
𝑃𝑠
)
=
𝐹𝑏
𝑊𝑑𝑐∗𝐵𝐹
=
75000𝑙𝑏𝑓
50,1𝑙𝑏𝑓
𝑓𝑡
∗0,749
= 1999ft=2000ft
The previous calculations were conducted using an outer drill collar diameterequal to
5’’. Using Wdc=50,1lbf/ft and the above minimum drill collars lengths at the excel file
we observed that a larger OD collar is required, in order to avoid any segment of our
drill pipe to be under compression. So we selected a collar with OD equal to 8’’.
Unfortunately the suitable ID diameter of our production liner is equal to 4.67’’, but
using this ID we can’t use any of the previous collar diameters. In order to proceed
with tis project we assume an OD of the drill collars equal to 8’’.
New Wdc = 155 lbf/ft
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For 5800 buckling
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
-200000 -100000 0 100000 200000
Depth,ft
Axial forces, lbf
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For 10100 buckling
0
2,000
4,000
6,000
8,000
10,000
12,000
-600000 -400000 -200000 0 200000 400000
Depth,ft
Axial forces , lbf
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For 13000 buckling
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
-800000-600000-400000-200000 0 200000 400000
Depth,ft
Axial forces, lbf
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Grade of the drill pipes
The following formula give us the opportunity to select the suitable grades of drill pipes
while includes the wet weight of drill string an over pulling force at all depths
F’t= (Wcd*Ldc+Wdp*Ldp)*BF+ pulling force
σi= F’t/A
Where : σι > σyield
Table 6 Drill stringdimensions and specifications
DEPTH WOB Ldc Ldp Ft(lbf) GRADES
5800 25000 590 5210 264667 D-55
10100 50000 770 9330 332290 E-75
13000 75000 2000 11000 492850 X-95
HYDRAULICS
EQUATIONS
For the calculations we assumed a sphericity ψ = 0.801, the mean diameter of the
cuttings equal to 0.0025”.
The flow rate is 400 gal/min
Slip velocity using stokes model: Us =
138(𝑝𝑠−𝑝𝑓)𝑑
2
𝜇
Annular velocity: Ua =
𝑞
2.448∗(𝑑2
−𝑑2
)
pipe velocity: Udp =
𝑞
2.448∗𝑑2
total nozzle area: At = 3*π/4*(13/32)
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Δpbit =
8.311∗10−5∗𝑝∗𝑞2
𝑐 𝑑
2∗𝐴 𝑡𝑜𝑡
2
Neutonian friction pressure loss
There exist several rheological fluids models such as Bingham Plastic Model, Power Law
Mode, Robertson-Stuff Model and Herschel-Bulkley Model using fluid hydrodynamics.
Some of them are utilized to characterize drilling fluids while some are not applicable to
drilling fluids. In this assignment we assume that the drilling mud is a Newtonian fluid.
Pipe:
𝒅𝑷𝒇
𝒅𝑳
=
𝝁 𝒗̅
𝟏𝟓𝟎𝟎 𝒅 𝟐
Annulus:
𝒅𝑷𝒇
𝒅𝑳
=
𝝁 𝒗̅
𝟏𝟓𝟎𝟎 (𝒅 𝟐−𝒅 𝟏) 𝟐
Mud pump pressure: Ppwp = Ps + Pd + Pa + Pd, where Ps is the Surface equipment
pressure loss and we assume that is equal to zero.
Table 7 Required pump pressure
All the above calculations have been conducted with a viscosity of 5 cp, so that the mud
velocity is always greater than the slip velocity at all different depths.
Depth
U
annulus
pipe
U
anullu
s
collar
Udp Udc
At
nozzle
Initial
slip U
F Nre New F
Final
slip U
Pressur
e loss
DPbit
Pressur
e loss
Dppipe
Pressur
e loss
Dpanull
us
Recuired
pump
pressure
Ppwp
0-2000 0.48 0.54 8.94 26.14 0.33 0.21 2.59 9.27 170.00 0.03 1289.71 6.34 0.17 1296.21
2000-
5800
1.27 1.81 8.94 26.14 0.33 0.21 2.59 9.27 170.00 0.03 1289.71 16.71 7.63 1314.05
5800-
10100
3.08 11.62 8.94 26.14 0.33 0.11 3.00 7.99 200.00 0.01 2015.16 25.94 1141.54 3182.64
10100-
13000
4.41 17.56 8.94 26.14 0.33 0.09 3.48 6.90 230.00 0.01 2203.25 45.81 1300.00 3549.05
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Hoisting system
At this stage we should compute the maximum expected load which stresses the
hoisting system during the drilling process. We consider two different scenarios for the
drill string and casing strings, while we assume an extra pull load of 20000 psi and 10000
psi respectively.
Table 8 Maximum load from drill string
Table 9 Maximum load from casings
The red boxes in the above tables indicate the maximum load in each case that the
hoisting system is able to withstand.
Minimum number of lines
Ff=(W*1,6)/(n*E)<=Fmax E*n=> (W*1,6)/(Fmax)
Where: n: number of lines between the crown and the traveling block
E: efficiency of hoisting system
W: maximum hoisting load of 413174.81lbf.
For E=0,874 and n=6 the inequality is satisfied.
Depth
(ft)
Pipe (ft)
Weight
(lbf/ft)
Collars
(ft)
Weight
(lbf/ft)
Total
pipe (lbf)
Total
collar
(lbf)
Mud
(ppg)
Extra Pull
load (lbf)
Wet
weight
(lbf)
Total
weight (lbf)
2000 1750 250 155 34125 38750 9.6 20000 62194.08 82194.08
5800 5210 590 155 101595 91450 9.6 20000 164751.38 184751.38
10100 9930 770 155 193635 119350 15 20000 241309.05 261309.05
13000 11000 2000 155 214500 310000 16.4 20000 393174.81 413174.81
19.5
Depth
(ft)
Weight
(lbf/ft)
Weight
(lbf/ft)
Weight
(lbf/ft)
Total
casing
(lbf)
Mud
(ppg)
Extra Pull
load (lbf)
Wet
weight
(lbf)
Total
weight
(lbf)
2000 19.5 155 94 188000 9.6 10,000 160446 170446
5800 19.5 155 68 394400 9.6 10,000 336595 346595
10100 19.5 155 40 404000 15 10,000 311481 321481
13000 19.5 155 23 299000 16.4 10,000 224136 234136
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Time required to pull 90 ft
Max hook power: Ph = Pd * E = 500 hp *0.874 = 437 hp
Max hoisting speed: v = Ph / W = (437 / 413174.81) * 33000 = 34.9 ft/min
Time required: t = s / v = 90 ft / 34.9 ft/min = 2.57 min