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Derrick Calculations
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Derrick Calculations
Use 3-1/2drill pipe
5 drill collars
Drill pipe (3/1/2, 9.5 lbf/ft)
(From Applied Drilling Engineering, page 19, Chapter 1)
ID = 2.992
OD = 3.5
Drill collars (5, 59 lbf/ft)
(From Fundamentals of Drilling Engineering, page 587, Table 9.1, Mitchell, R. F. and Miska, S. Z)
ID = 1-3/4
OD = 5
Drillstring weight at depth 5250
Assume 30 length for the drill collars
Type Amount Length
(ft)
Total Length
(ft)
Weight/ft Total Weight
(lbf)DC 4 30 120 59 7,080
DP 1 5130 5,130 9.5 48,735
Total drill string weight 55,815 lbf
With a safety factor (SF) of 1.5,
Or
Maximum load on derrick will be seen when running 7, 32 lbf/ft casing of 283,574.42 lbf
(determined from the casing design calculation)
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Tension in fast line can be calculated from the following equation:
Where W is the load supported by the travelling block
E is the efficiency
n is the number of lines
With W = 283,583.5 lbf,
n E En Ff
6 0.874 5.244 54,077.7
8 0.841 6.728 42,149.7
10 0.81 8.1 35,010.3
12 0.77 9.24 30,690.9
Tension in the dead line can be determined from the following equation:
n fs
6 47,263.9
8 35,447.9
10 28,358.4
12 23,632.0
The load of the derrick is determined from the following equation:
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n Ff Fs W Fd Assuming
SF=1.5
6 54,077.7 47,263.9 283,583.5 384,925.1 577,387.7
8 42,149.7 35,447.9 283,583.5 361,181.2 541,771.8
10 35,010.3 28,358.4 283,583.5 346,952.2 520,428.2
12 30,690.9 23,632.0 283,583.5 337,906.3 506,859.5
If 6 lines were to be used , the rig needs to be capable of handling an anticiapted load of 580,000
lbf.
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Cement Calculations
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Cement Calculations
Starting with the surface casing
Excess = 150%
Hole size = 15
Frac = Max Weight =
From the Red Book (Halliburton)
Looking at page 107, Section 122, under Vol. & Hgt. Between: Tbg., Csg., D.P. & Hole Tab,
For 15 (hole) 10-3/4 (casing), gives
Looking at page 25, Section 210, under Capacity Tab,
For 10-3/4 (casing), 40.5 lbf/ft, gives
Under Technical Data, Section IV: Class G Cement
Use Class G Cement w/10% Bentonite, density = 12.8 PPG
Pump at a rate of 2-3 bbls to maintain 100 psi surface pressure
Class G w/ 10% Bentonite,
Volume of cement in shoe joint
100'
950'40'
16'
10-3/4''
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40 ft shoe joint
Go to Capacity Tab, Section 210
Number of sacks =
Volume of cement in 16, 65 lbf/ft (casing) 10-3/4, 40.5 lbf/ft annulusLooking at Section 221, page 129, under Vol. & Hgt. Between; Tbgs., Tbg. & Csg., Csgs., D.P.
& Csg. Tab, gives
So, conductor casing is set until 100
Number of sacks =
Volume of cement in the open hole 15 10-3/4 casing annulus
With an excess of 150%, so
Number of sacks =
Total volume = Total number of sacks = Thus, use approximately 410 sacks.
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10-3/4, 40.5 lbf/ft, K-55, to 950
OD = 11.75
ID = 10.05
Drift = 9.894
Collapse resistance = 1580 psi
Burst resistance = 3130 psi
Capacity = 0.0981
7, 32 lbf/ft, C-75 to 5250
OD = 7.656
ID = 6.094
Drift = 5.969
Collapse resistance = 8,230 psi
Burst resistance = 8,490 psi
100'
950'40'
16''
10-3/4''
5250'
15'' hole
7''
8-5/8'' hole
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Capacity = 0.036
For the 7 (casing) 10-3/4 (casing), from Vol. & Hgt. Between; Tbgs., Tbg. & Csg., Csgs.,D.P. & Csg. Tab, it is found that the flow is 0.0505
As for the 7 (casing) 8-5/8 (hole), from Vol. &Hgt. Between: Tbg., Csg., D.P. & HoleTab, it is found that the flow is 0.0247
Displacement
10-3/4, 40.5 lbf/ft to bump plug
Bump plug at 3 bpm (or barrels per minute)
Time =
Bump plug w/
Bump plug w/ 1,000 psi
Now for the production casing
Referring to the previous diagram,
Class C, Econolite 3%, yield is
Class G, Bentonite 8%, yield is
For the 7 (casing) 8-5/8 (hole), from Vol. & Hgt. Between: Tbg., Csg., D.P. & Hole Tab,it is found that the flow is 0.0247
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Class C
Total volume Total number of sacks = Thus, use approximately 120 sacks
Class G
Total volume = Total number of sacks = Thus, use approximately 450 sacks
Pressure
Total =
Displacement
7, 32 lbf/ft to bump plug
Bump plug at 5 bpm (or barrels per minute)
Time =
Bump plug w/
Bump plug w/ 5,000 psi
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Casign Design Calculations
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Casing Design Calculations
Assume:
Trip margin= 0.5 lbm/gal
Kick margin= 0.5 lbm/gal
Fracture gradient= 0.8 psi/ft
Pore pressure gradient= 0.439 psi/ft (Applied Drilling Engineering, Table 6.1, page 247, for
California Area)
Safety Factor (SF)= 2
Pore Pressure + Trip margin= 8.44 + 0.5 = 8.94 PPG
Fracture PressureKick margin= 15.380.5 = 14.88 PPG
Surface Casing
The surface casing design follow the guidelines that are outlined in Example 7.9 (From Applied
Drilling Engineering).
Burst
For burst considerations, use an injection pressure that is equivalent to a mud density 0.3 lbm/gal
greater than the fracture gradient. Also assume that any gas kick is composed of methane, which
has a molecular weight (M) of 16. Ideal gas behavior is also assumed for simplicity.
1. Determine potential internal pressure, pi
2. Determine potential gas pressure (Still following Example 7.9, Applied DrillingEngineering)
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Formation temperature, T is
Using Equation 4.5 from
Gradient for methane is
3. Surface Casing Pressure for design loading condition is (pressure diferential that tends toburst casing at the surface)
External pressure at surface= 0 psi
External pressure at casing seat=
Pressure diferential that tends to burst casing at the casing seat
Using a safety factor of 2,
At the surface: At the casing seat:
For 10-3/4 casing, J-55/K-55 casing grades meet the burst requirements.
The planned mud density when running the 10-3/4 casing is 8.94 PPG
Collapse
The external pressure of collapse-design load
The internal pressure for the collapse-design load is controlled by the maximum loss in
fluid level that could occur if a severe lost-circulation problem is encountered. To
determine the depth to which the mud level will fall, assume a normal-pressure and so
lost-circulation zone will be encountered near the depth of the next casing seat (5250),
and also if no permeable zones are exposed above this depth, then
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Using Equation7.24b from Applied Drilling Engineer
Depth to which the mud level will fall, Dm
( )
For these conditions, the mud level would fall to within 293 of the casing seat at 950.
The pressure differential that leads to collapse of the casing is zero at the surface.
Collapse of the casing seat at 950
External pressureinternal pressure w/ mud level at 627 ft
Using a safety factor of 2,
At657 ft: At 950 ft:
10-3/4 casing of all grades, except F-25 meet these design requirements.
Axial Tension
For 10-3/4 casing, using K-55, 40.5 lbf/ft,
(From Red Book, Halliburton)
OD= 11.75 (with coupling)
ID= 10.05
Casing weight=
Cross sectional area of steel
Using a safety factor of 2,
Or
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(overpull)
10-3/4, K-55 has a joint strength of 450,000 lbfand this is greater than the calculated
lbfat axial tension.Thus, 10-3/4, K-55, 40.5 lbf/ft surface casing will be run.
Production Casing
The production casing design will also follow the guidelines that are outlined in Example 7.9
(From Applied Drilling Engineering).
TVD= 5250
Burst
Potential internal pressure, pi
Potential gas pressure
To find the pressure differential that tends to burst casing at the surface,
External pressure at the surface= 0 psi
External pressure at the casing seat=
With a safety factor of 2,
At the surafce: At the casing seat of 5250:
Hence, 7, C-75, 32 lbf/ft casing would meet the burst requirement.
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Collapse
Earlier, it has already been determined that lost circulation was encountered, the fluidlevel would drop to 657 (950-293) from the surface.
Assume collapse if all the fluid in the wellbore was lost.
Gas gradient=
To collapse casing seat of 5250,
Using a safety factor of 2,
7, C-75, 32 lbf/ft has a collapse resistance of 8,230 psi which is greater than the
estimated collapse resistance of 4,080.2 psi.
Axial Tension
For 7 casing, using C-75, 32 lbf/ft,
(From Red Book, Halliburton)
OD= 7.656 (with coupling)
ID= 6.094
Casing weight=
Cross sectional area of steel
Using a safety factor of 2,
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Or
(overpull)
7, C-75, 32 lbf/ft, LT&C has a joint strength of 633,000 lbf, which is greater than the
worst case of 283,583.5 lbf.
Conclusion
Surface casing: 10-3/4, K-55, 40.5 lbf/ft, set at 900 in a 15 hole(Based on Table 7.7,Chapter 7, page 331, Applied Drilling Engineering)
Production casing: 7, 32 lbf/ft, C-75 set at 5250 in a 8-5/8 hole (Based on Table 7.7,
Chapter 7, page 331, Applied Drilling Engineering)
Chapter 7, page 332, Table 7.8 illustrated that the largest bit through 10-3/4, 40.5 lbf/ft
is 9-7/8 bit
15bit will pass through 16 conductor pipe
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Mud Information and Annular Velocity
Calculations
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Mud Information and Annular Velocity Calculations
There is a need to determine if there is enough rate at 375 GPM to keep the hole clean.
Assume Bingham Plastic Model (Based on the sample drilling reports)
15 hole
q= 375 GPM
Drill pipe ID= 2.992
Using Equation from Table 4.6,
Drill pipe mean velocity,
=8.94 ppg
d= 2.992
= 16
Turbulence criteria
(Turbulent flow through the drill pipe)
7 9-7/8hole annulus
Turbulence criteria
Right at turbulent criteria for the annulus.
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10-3/4 15hole annulus
Turbulence criteria
Right at turbulent criteria for the annulus.
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Directional Plan
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Directional Plan
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Drilling Program Outline
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Figure 1: Schematic arrangement of solids control equioment to be used
Shale Shaker
Desander
Mud Cleaner
Centrifuge
Reserve mud pits
Return line from well
Solids to pit
Solids to pit
Solids to pit
Solids to pit
Liquified returnto mud system
Liquids
Liquids
Liquids
Liquids
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Plan not to scale
Figure 2: Illustration of well trajectory
0
200
400
600
800
0 200 400 600 800 1000 1200 1400
South
(-)
/North(+)
West(-)/East(+)
Surface Location Reference Point Bottom Hole Location
Bottom Hole
Location- 600' North,
1200' East of "AA"
Surface Location- 300'
North, 600' East of
"AA"
"AA"