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1
Casing Program
1000’
2000’
3000’
4000’
5000’
6000’
7000’
8000’
9000’
10 ppg 15 ppg 20 ppg
Lost circulation
10,000’
Fracture gradient
Pore pressure
Possible Kick or collapse
Mud wt.
API Grades
•Minimum yield is 80% of average yield
2
Non API Grades
Range
Range Length Average Lengthft ft
1 16 – 25 222 25 – 34 313 over 34 42
API standard lengths of casing (API RP 5B1, 1988)
Weight of casing•Nominal weight – Theoretical weight of 20 ft threaded and coupled casing joint.•Plain end weight – The weight of the casing joint excluding the threads and coupling.•Threaded and coupled weight – The average weight of the joint including the threads and coupling.
3
Casing Properties1. Tension
• Design is based on yield, i.e. “permanent deformation”• Based on nominal not minimal wall thickness• Minimum acceptable wall thickness is 87.5%
F ten A s σ yield⋅:=
A s πD2 d2−
4⋅:=
Example 7.1
Casing Design Example 7.1
Compute the body-yield strength for 20-in., 133 lb/ft, K-55 casingwith 0.635 in wall thickness. D 20in:= Ftenσyield 55000psi:=
t 0.635in:=
d D 2 t⋅−:=d 18.73 in=
As
As πD2 d2−
4⋅:=
As 38.631 in2=
Ften As σyield⋅:=
Ften 2.125 106× lbf=
4
Casing Properties2. Burst
• Minimum acceptable wall thickness is 87.5%• Barlow’s equation takes this into account.
p br0.875 2⋅ σ yield⋅ t⋅
D:=
Example 7.2
Casing Design Example 7.2
Compute the burst pressure rating for 20-in, 133 lb/ft, K-55 casingwith a wall thickness of 0.635in.
σyield 55000psi:=
D 20in:=
t 0.635in:=
pbr0.875 2⋅ σyield⋅ t⋅
D:=
pbr 3.056 103× psi=
5
Casing Properties3. Collapse
• Four modes of collapse failure• Elastic collapse
• Stress is uniform across wall thickness• Large diameter to thickness ratios only
• Transition collapse• Young’s Modulus is stress dependent• As stress increases E decreases
• Plastic collapse • Casing yields on inside diameter first.
• Yield Strength collapse• Ellipse of plasticity
• D/t determines mode of failure• Empirical coefficients
• Used to make equations fit observations
Young’s Modulus
6
Elastic Collapse
In thin walled casing with a high D/t ratio the stress is uniform across the wall thickness.
From “Casing Design Theory and Practice”by Rahman and Chilingarian
Stress from external pressureIn thick walled casing the stress is not constant across the wall thickness
7
Stress from external pressure
From “Casing Design Theory and Practice”by Rahman and Chilingarian
•The inside yields first.•As stress increases the plastic – elastic boundary moves outward.•Failure does not occur until a sufficient amount of casing has yielded.
Tangent Modulus
From “Casing Design Theory and Practice”by Rahman and Chilingarian
•The modulus of elasticity represents the slope of the stress vs. strain curve.•When Young’s modulus of elasticity is stress dependent it is referred to as the tangent modulus.
8
From “Casing Design Theory and Practice”by Rahman and Chilingarian
Young’s Modulus and temperatureYoung’s Modulus is also begins to vary at high temperatures.
From “Casing Design Theory and Practice”by Rahman and Chilingarian
API Critical Collapse
9
Table 7.5
From Applied Drilling Engineering, SPE Textbook series Vol. 2
These values only apply when axial stress is zero!
Example 7.3Casing Design Example 7.3Compute the collapse pressure rating for 20-in., 133 lb/ft, K-55casing with a wall thickness of 0.625 in.
σyield 55000psi:=
D 20in:=
t 0.635in:= Dt
31.496=
From table 7.5 the value for D/t falls in the transition range. So Eq. 7.7will be used.
From table 7.4 the "F" values are:F4 1.989:=
F5 0.036:=
pcr σyieldF4Dt
F5−
⋅:=
pcr 1.493 103× psi=
10
Collapse and tension
From “Casing Design Theory and Practice” by Rahman and Chilingarian
Ellipse of Plasticity
From “Casing Design Theory and Practice”by Rahman and Chilingarian
Tension has a detrimental effect on collapse and a positive effect on burst
The ellipse can only be used for yield strength failure mode!
11
Tension and failure mode
From “Casing Design Theory and Practice”by Rahman and Chilingarian
Example 7.5Casing DesignExample 7.5Fa 1000000 lbf:= D 20in:=
pi 1000psi:= t .635in:=Dt
31.496=
σyield 55000psi:=as 38.631in2:=
σzFaas
:=σz 2.589 104× psi=
σz pi+
σyield0.489=
σyield_eff 134
σz pi+
σyield
2
⋅−12
σz pi+
σyield
⋅−
σyield⋅:= σyield_eff 3.639 104× psi=
The effective yield stress is made dimensionless, as "Y", to allow the "F" factors to be calculated.
Yσyield_eff
1psi:=
Y 3.639 104×=
12
7.5
F1 2.8762 0.10679 10 5−⋅ Y⋅+ 0.21301 10 10−⋅ Y2⋅+ 0.53132 10 16−⋅ Y⋅−:=
F2 0.026233 0.50609 10 6−⋅ Y⋅+:=
F3 465.93− 0.030867 Y⋅+ 0.10483 10 7−⋅ Y2⋅− 0.36989 10 13−⋅ Y3⋅+:=
F4
46.95 106⋅
3F2F1
⋅
2F2F1
+
3
⋅
Y
3F2F1
⋅
2F2F1
+
F2F1
−
⋅ 1
3F2F1
⋅
2F2F1
+
−
2
⋅
:=
F5 F4F2F1
⋅:=
7.5
F1 2.941= F2 0.045= F3 645.079= F4 2.102= F5 0.032=
pcr σyield_effF4Dt
F5−
⋅:=
pcr 1.267 103× psi=
pcr pi+ 2.267 103× psi=
13
Example 7.6Casing Design Example 7.6Determine the maximum axial stress for a joint of the following casing when it is subjectedto 400 kips of axial tension across a dogleg severity of 4 deg/100ft. Find the stress usingboth the soft string and stiff stirng models.
σyield 80000psi:= Fa 400000 lbf:= α 4:= Lj 36ft:=
w 39lbfft
:= D 7.625in:= d 6.625in:= From Table 7.6
As πD2 d2−
4⋅:=
As 11.192 in2=
The first step is to find the axial stress due to tension in the string .
σzFaAs
:=σz 3.574 104× psi=
7.6
The next step is to find the axial stress due to bending.
(The units that come out in the derivation in fig 7.14 are lbf/in3.)σzbending 218lbf
in3α D⋅:=
σzbending 6.649 103× psi=σtotal σz σzbending+:=
σtotal 4.239 104× psi=
To find the stress from bending using the stiff string model, eq. 7.22, the moment ofinertia must be calculated.
Iπ64
D4 d4−( )⋅:=I 71.37 in4=
KcFa
30 106⋅ psi I⋅:=
Kc 0.164 ft-1=
Note that Lj is divided by 12. The numberused in equation 7.22 a must be in inches notfeet in order for the units to work out right.σzbendstiff 218
lbf
in3α⋅ D⋅
6 Kc⋅Lj12
⋅
tanh 6 Kc⋅Lj12
⋅
⋅:=
σzbendstiff 1.974 104× psi=
14
7.6
Using the bending force instead of bending stress equation,
Fab 63 α⋅ D⋅ w⋅6 Kc⋅ Lj⋅
tanh 6 Kc⋅ Lj⋅( )⋅:=
Fab 2.212 105× lbf=
σzbFabAs
:=σzb 1.977 104× psi=
σTotalstiff σzb σz+:=σTotalstiff 5.551 104× psi=
Example 7.7
Casing Design Ex. 7.7 Using the Pore-Pressure gradient in fig 7.21 pick casing points.
From “Applied Drilling Engineering” by Bourgoyne, Chenevert, Milheim & Young
15
7.7
Step 1. Plot the design curvesAdd trip margin of .5 lb/gal to the pore pressure curveSubtract a 0.5 lb/gal kick margin from the Fracture gradient curve.
Step 2. Choose the casing points.
Draw a vertical line that intersects the design mud wt curve at total depth, and the kicktolerance curve farther up the well. The depth where the line intersects the kick tolerancecurve is the next casing point.
Using a scale measure this depth to be 11,400 ftDraw a horizontal line from this point to where it intersects the trip margin curve. Thendraw a vertical line from this intersection up to where it intersects the kick tolerance curveagain. This is another casing point.Using a scale measure this depth to be 4,000 ftThe casing program calls for surface casing to be set to 4,000 ft and intermediate casing to11,400 ft. The 7in production casing will be run to the total depth of 15000 ft
Example 7.8
Casing Design Example 7.8Select casing sizes.
Step 1. Find the commonly used bit size to run for casing from Table 7.7
Starting from the bottom with the 7 incasing table 7.7 sugests an 8 5/8 inbit will be needed.
Step 2. Find the next smallest sizecasing that will accomodate this bitfrom table 7.8
From table 7.8 it is seen that 9 5/8 incasing is required for an 8 1/2 in bit.
This process is repeated for the 9 5/8in casing to find that a 12 1/4" bit isneeded, and 13 3/8 in casing will berequired for the 12 1/4" bit.
So 13 3/8 in surface casing will be runto 4,000 ft, and 9 5/8 in intermediatecasing will be run to 11,400 ft.
From “Applied Drilling Engineering” by Bourgoyne, Chenevert, Milheim & Young
16
7.8
From “Applied Drilling Engineering” by Bourgoyne, Chenevert, Milheim & Young
Surface CasingBurst Collapse
17
Example7.9Casing Design Example 7.9
Design the surface casing for the last 2 examples.
Step 1. Eliminate casing that won’t meet burst requirements during a well control operation.Find and plot internal and external pressure, differential pressure, and a design line.Use similar triangles to determine the minimum depths of the casing sections.
Step 2. Prepare a graph to find the collapse design load.Find the mud level inside the pipe if circulation is lost at the next casing point.Determine the minimal internal pressure anticipated from the lost circulation.Find and plot the external pressure from surface to shoe.Find and plot the differential pressure with a design factor included.
Step 3. Find the cheapest casing that will meet the collapse requirements at the shoe.
Step 4. Use similar triangles to find the maximum setting depth of the next cheapest casing section. Draw a free body diagram to determine the axial forces at the section change.
Step 5. Compute the effective yield stress.Using iterative approach find the corrected collapse pressure and new max setting depth.
Step 6. Repeat this process with the next cheapest string until surface or burst limitations are met.
Step 7. Check that the casing meets the tension requirements while it is being run.
Summary. Start with burst at surface during well control.Then design for collapse with lost circulation at next casing point.Finally check for tension while casing is being run.
7.9Step 1. Eliminate casing that wont meet burst requirements during a well control operation.
D 13.375in:= Dfburst 1.1:= Dfcollapse 1.1:= Dftension 1.6:= gp 0.465psift
:=
Depthshoe 4000ft:= Depthfinal 11400ft:=
M 16g
mole:=Temp 520R 0.012
Rft
Depthshoe+:= Temp 568R=
The fracture gradient is read from fig 7.21 to be 14.1 lb/gal. The injection pressure is .3 lb/gal so thedesign gradient will be equivalent to 14.4 lb/gal.
ρf 14.4lbfgal
:=
The pressue inside the casing at the shoe is..
Units must be used with conversion factors with mathcad. Or the conversion factor can be left out entirely.
piburst Depthshoe ρf⋅ 0.052⋅gal
ft in2⋅:=
piburst 2995psi=
The pressure inside the casing at surface is..Cheating mathcad out of units for ideal gas equation
ρg
2995 15+( ) 16⋅ 1⋅lbfgal
80.3 1⋅ 521 0.012 4000⋅+( )⋅:= ρg 1
lbfgal
= ρg 0.052⋅gal
ft in2⋅⋅ 0.055
psift
=
psurface piburst Depthshoe ρg⋅ 0.052gal
ft in2⋅⋅
⋅−:=
psurface 2776psi=
18
7.9
Fig. 7.25 Burst design
0
500
1000
1500
2000
2500
3000
3500
4000
0 1,000 2,000 3,000 4,000
Pressure
Dep
th
PburstPoPi
The external pressure at the shoe is..
po Depthshoe 0.465⋅psift
:=
po 1860psi=
The design differential pressure is..
∆pshoe piburst po−:=
∆pshoe 1135psi=
∆pdsgnsurf psurface Dfburst⋅:=
∆pdsgnsurf 3054psi=
∆pdsgnshoe ∆pshoe Dfburst⋅:=
∆pdsgnshoe 1249psi=
The rated burst pressures of the two weakest casing types are..
BurstH40 1730psi:= BurstJ55 2730psi:=
7.9Using similar triangles find the minimum depth the cheapest casings can be run to.
X Depthshoe:=
Y ∆pdsgnsurf ∆pdsgnshoe−:=
X 4000ft=
Y 1805psi=
b1 ∆pdsgnsurf BurstH40−:= b1 1324psi=
b2 ∆pdsgnsurf BurstJ55−:= b2 324psi=
a1XY
b1⋅:=
a2XY
b2⋅:=
The minimum depth the H-40 casing can be set is..
a1 2933ft=
The minimum depth the J-55 casing can be set is..
a2 717ft=
19
7.9Step 2. Prepare a graph to find the collapse deign load.
From fig. 2.1 the mud wieght in use when the casing is run will be 9.3 lbf/gal, and 13.7 lbf/gal at TVD.
ρmud1 9.3lbfgal
:=
ρmud2 13.7lbfgal
:=
The pressure outside the casing at the time is is set is..
po4000 Depthshoe ρmud1⋅ 0.052⋅gal
ft in2⋅⋅:=
po4000 1934psi=
When circulation is lost at the next casing point the mud level will fall to..
Dm
ρmud2gp
.052gal
ft in2⋅
−
ρmud2Depthfinal⋅:=
Dm 3959ft=
The pressure inside the casing at the shoe when circ. is lost is..
pishoe Depthshoe Dm−( ) ρmud2⋅ 0.052gal
ft in2⋅⋅
⋅:=
pishoe 29psi=
7.9Fig 7.26 Collapse design line
0
500
1000
1500
2000
2500
3000
3500
4000
0 500 1000 1500 2000 2500
Pressure
The differential pressure that tends to collapse the casing is..
∆pcshoe po4000 pishoe−:=∆pcshoe 1905psi=
∆pcDm Dm ρmud1⋅ 0.052⋅gal
ft in2⋅⋅:=
∆pcDm 1915psi=
∆pcshoedsgn ∆pcshoe Dfcollapse⋅:=∆pcshoedsgn 2096psi=
∆pcDmdsgn ∆pcDm Dfcollapse⋅:=∆pcDmdsgn 2106psi=
20
7.9
Step 3. Find the cheapest casing that will meet the collapse requirements at the shoe.
From table 7.6 C-75 68 lb/ft casing is the weakest casing that meets the collapse pressure at the shoe.
Csg wt collapse D d AsH40 48.00 740 13.375 J-55 54.50 1130 13.375 12.615 15.51K-55 61.00 1540 13.375 12.515 17.49K-55 68.00 1950 13.375 12.415 19.45C-75 68.00 2220 13.375 12.415 19.45
Step 4. Use similar triangles to find the maximum settingdepth of the next cheapest casing section.
Pcr 1950ft:=
y3959ft2106ft
Pcr⋅:= Plug the Pcr values in here y 3666ft=
7.9Draw a free body diagram to determine the axial forces at the section change.
w 68lbfft
:=
L Depthshoe y−:= L 334ft=
p1 pishoe:= p2 0psi:=
Asa 19.45in2:=
Asb 19.45in2:=
Fa1top w L⋅ p1 Asa⋅− p2 Asb⋅+:= Fa1top 22161lbf=
Fa2btm Fa1top:=
21
7.9Use the axial force to compute the effective yield stress.
σyield 55000:= Fa Fa2btm:= As Asb:= d 12.415in:= pi p2:= pi 0psi=
tD d−
2:= t 0.480 in=
σzFaAs
:= σz 1139psi=
σz pi+
σyield0.0207 psi=
σyield_eff 1psi234
σz pi+
σyield
2
⋅−12
σz pi+
σyield
⋅−
σyield⋅:= σyield_eff 54421psi=
The effective yield stress is made dimensionless to allow the "F"factors to be calculated.Yeff
σyield_eff1psi
:=
7.9
F1 2.8762 0.10679 10 5−⋅ Yeff⋅+ 0.21301 10 10−⋅ Yeff2⋅+ 0.53132 10 16−⋅ Yeff
3⋅−:=
F2 0.026233 0.50609 10 6−⋅ Yeff⋅+:=
F3 465.93− 0.030867 Yeff⋅+ 0.10483 10 7−⋅ Yeff2⋅− 0.36989 10 13−⋅ Yeff
3⋅+:=
F4
46.95 106⋅
3F2F1
⋅
2F2F1
+
3
⋅
Yeff
3F2F1
⋅
2F2F1
+
F2F1
−
⋅ 1
3F2F1
⋅
2F2F1
+
−
2
⋅
:=
F5 F4F2F1
⋅:=
F1 3= F2 0= F3 1188.8= F4 2= F5 0=
22
7.9Step 5. Compute the effective yield stress.
Using itterative approach find the corrected collapse pressure and new max setting depth.
pcr σyield_effF4Dt
F5−
⋅:=pcr 1939psi= pcr pi+ 1939psi=
This new value is plugged back into the similar triangles calculation (Pcr) to determine a new L1 and pcr
Here are the results with the K-55 68 lb/ftcasing.
Pcr L Fa Stress axial (σz + pi/σ) eq. 7.11 σeff
1,950 334 22,166 1,140 0.0207 0.9895 54,4211,939 356 23,615 1,214 0.0221 0.9888 54,3831,938 357 23,699 1,219 0.0222 0.9887 54,3801,938 357 23,704 1,219 0.0222 0.9887 54,380
7.9Since this results in less than 500 feet of C-75 the K-55 68 lb/ft pipe will not be used. Instead the nextweakest casing is selected.
The process is repeated with K-55 61 lb/ft casing. The results are tabulated below.
Pcr L Fa Stress axial (σz + pi/σ) eq. 7.11 σe
1,540 1,106 74,670 4,270 0.0776 0.9589 52,7411,516 1,151 77,718 4,444 0.0808 0.9571 52,6431,515 1,153 77,852 4,452 0.0809 0.9571 52,6391,515 1,153 77,858 4,452 0.0810 0.9571 52,639
Depthk_55_61lb Depthshoe 1153ft−:= Plug in L valuehere
This is the maximum depth this casing can be runto.
Depthk_55_61lb 2847ft=
Step 6. Repeat this process with the next cheapest string until surface or burst limitations are met.
The process is repeated with K-55 54.5 lb/ft casing. The long vertical length of the triangle is replacedwith Shoe depth less L1, and the large horizontal value is replaced with the last calculated (pcr + pi). Thecollapse rating of the casing plus any internal pressure if there is any is used for Pcr
The results are tabulated below.
23
7.9Pcr L Fa Stress axial (σ z + p i/σ) eq. 7.11 Pcr 1,130 723 121,991 7,864 0.1430 0.9208 1,1301,114 753 123,813 7,981 0.1451 0.9195 1,1141,114 754 123,851 7,983 0.1452 0.9195 1,1141,114 754 123,852 7,983 0.1452 0.9195 1,114
Depthk_55_54.5lb Depthk_55_61lb 754ft−:=
Plug in L valuehere
This is the maximum depth this casing can be runto.
Depthk_55_54.5lb 2093ft=
The H-40 48 lb/ft casing can not be run above 2,933 ft because of burst limitations. The K-55 54.5 lb/ftcasing cannot be run above 716 ft because of burst limitation.
Therefore:Section 4 - K-55 61 lb/ft will be run to 716 ft Length = 716 ftSection 3 - K-55 54.5 lb/ft will be run to 2,093 ft Length = 1,376 ftSection 2 - K-55 61 lb/ft will be run to 2,847 ft Length = 754 ftSection 1 - C-75 68 lb/ft will be run to 4,000 ft. Length = 1,153 ft
7.9Step 7. Check that the casing meets the tension requirements while it is being run.
Find the axial force at each sectionchange.W1 68
lbfft
:= L1 1153ft:= P1 Depthshoe ρmud1⋅ 0.052gal
ft in2⋅:=
W2 61lbfft
:= L2 754ft:= P2 Depthk_55_61lb ρmud1⋅ 0.052gal
ft in2⋅:=
W3 54.5lbfft
:= L3 1376ft:= P3 Depthk_55_54.5lb ρmud1⋅ 0.052gal
ft in2⋅:=
W4 61lbfft
:= L4 716ft:= P4 L4 ρmud1⋅ 0.052gal
ft in2⋅:=
P1 1934psi= P2 1377psi= P3 1012psi= P4 346psi=
dC_75_68lb 12.415in:= dK_55_61lb 12.515in:= dK_55_54.5lb 12.615in:=
As1π4
D2 dC_75_68lb2−( )⋅:= As1 19in2=
As2π4
D2 dK_55_61lb2−( )⋅:= As2 17in2=
As3π4
D2 dK_55_54.5lb2−( )⋅:= As3 16in2=
24
7.9∆As2 As1 As2−:= W1 L1⋅ 78404lbf=∆As2 2in2=
W2 L2⋅ 45994lbf=∆As3 As2 As3−:= ∆As3 2in2=
W3 L3⋅ 74992lbf=
∆As4 As2 As3−:= ∆As4 2in2= W4 L4⋅ 43676lbf=
Fabtm1 P1− As1⋅:= Fabtm1 37615− lbf=
Fatop1 Fabtm1 W1 L1⋅+:= Fatop1 40789lbf=
Fabtm2 Fatop1 P2 ∆As2⋅+:= Fabtm2 43485lbf=
Fatop2 Fabtm2 W2 L2⋅+:= Fatop2 89479lbf=
Fabtm3 Fatop2 P3 ∆As3⋅+:= Fabtm3 91477lbf=
Fatop3 Fabtm3 W3 L3⋅+:= Fatop3 166469lbf=
Fabtm4 Fatop3 P4 ∆As4⋅−:= Fabtm4 165785lbf=
Fatop4 Fabtm4 W4 L4⋅+:=Fatop4 209461lbf=
7.9Fabtm1dsgn Fabtm1 Dftension⋅:= Fabtm1dsgn 60184− lbf=
Fabtm1dsgn2 Fabtm1 100000lbf+:= Fabtm1dsgn2 62385lbf=
Fatop1dsgn Fatop1 Dftension⋅:= Fatop1dsgn 65263 lbf=
Fatop1dsgn2 Fatop1 100000lbf+:= Fatop1dsgn2 140789lbf=
Fabtm2dsgn Fabtm2 Dftension⋅:= Fabtm2dsgn 69576lbf=
Fabtm2dsgn2 Fabtm2 100000lbf+:= Fabtm2dsgn2 143485lbf=
Fatop2dsgn Fatop2 Dftension⋅:= Fatop2dsgn 143166lbf=
Fatop2dsgn2 Fatop2 100000lbf+:= Fatop2dsgn2 189479lbf=
Fabtm3dsgn Fabtm3 Dftension⋅:= Fabtm3dsgn 146363lbf=
Fabtm3dsgn2 Fabtm3 100000lbf+:= Fabtm3dsgn2 191477lbf=
Fatop3dsgn Fatop3 Dftension⋅:= Fatop3dsgn 266350lbf=
Fatop3dsgn2 Fatop3 100000lbf+:= Fatop3dsgn2 266469lbf=
Fabtm4dsgn Fabtm4 Dftension⋅:= Fabtm4dsgn 265257lbf=
Fabtm4dsgn2 Fabtm4 100000lbf+:= Fabtm4dsgn2 265785lbf=
Fatop4dsgn Fatop4 Dftension⋅:= Fatop4dsgn 335138lbf=
Fatop4dsgn2 Fatop4 100000lbf+:= Fatop4dsgn2 309461lbf=
25
7.9Plot the design tension curve
Fig 7.29 A xial te n s io n
0
500
1,000
1,500
2,000
2,500
3,000
3,500
4,000
-100,000 -50,000 0 50,000 100,000 150,000 200,000 250,000 300,000 350,000 400,000
T e n s io n , lb sD
epth
, ft
Fa ( lbs ) Fa * 1.6 Fa + 100K
Casing Design Example 7.10
How much surface pressure can be put on casing to displace cement?
Step 1. Determine how much additional axial force is available.
d 12.459in:=
Ftenmax 853000lbf:=
Fa 300000lbf:=
DFtension 1.3:=
∆FaFtenmax
DFtensionFa−:=
∆Fa 356154lbf=
Step 2. Find how much pressure generates this much tension using equation 7.27.
pi∆Fa 4⋅
π d2⋅
:=
pi 2921psi=
Example 7.10
26
Casing Design Example 7.11
Compute the maximum change in axial force resulting from degradation of mud in the annulus aftercementing.
Step 1. Find the average pressure change after degradation with equation 7.32
ρ 1 14lbfgal
:=
ρ 2 9lbfgal
:=
Depth 8000ft:=
D 10.75in:=
∆poavgDepth
2ρ 2 ρ 1−( )⋅ .052⋅
gal
ft in2⋅
:=
∆poavg 1040− psi=
Step 2. Find the change in axial stress caused by this pressure change using equation 7.36
∆Fa 0.471− D2⋅ ∆poavg⋅:=
∆Fa 56607lbf=
Example 7.11
Example 7.12Casing Design Example 7.12
Perform a stability analysis on the intermediate string after cement placement and while drilling to 15K.
Step 1. Find and plot the axial forces in the casing.
Step 2. Find and plot the stability forces.
Step 3. Locate the intersection of the stability and force curves.
Step 4. Find the force and length changes in all sections while drilling ahead.
Step 5. Determine what portion of length change is allowed for by bending due to buckling.
Step 6. Plot the new Fa and Fs curves.
Step 7. Locate the intersection of the new Fa and Fs curves.
Step 8. Determine how much tension is needed at landing to lower intersection to top of cement.
27
7.12Given
Casing weight w1 40.0lbfft
:= w2 43.5lbfft
:= w3 47.0lbfft
:=
Inside diameter d1 8.835in:= d2 8.755in:= d3 8.681in:=
Internal area Ai1 61.306in2:= Ai2 60.201in2:= Ai3 59.187in2:=
External area Ao1 72.760in2:= Ao2 72.760in2:= Ao3 72.760in2:=
Steel area As1 11.454in2:= As2 12.559in2:= As3 13.573in2:=
Depthshoe 10000ft:=ρm1 10.0
lbfgal
:= gm1 ρm1 0.052⋅gal
ft in2⋅:=
TVD 15000:=ρm2 16.0
lbfgal
:=gm2 ρm2 0.052⋅
gal
ft in2⋅:=
∆T 30F:=ρcmt 15.7
lbfgal
:=gcmt ρcmt 0.052⋅
gal
ft in2⋅:=
7.12Step 1. Find and plot the axial forces in the casing.
Find the forces when the cement was placed.
F1 8000ft gm1⋅ 2000ft gcmt⋅+( )− Ao1⋅:=
F1 421.484− 103× lbf=
F2 10000ft gm1⋅ 593psi+( ) Ai1⋅:=
F2 355.146 103× lbf=
F3 6200ft gm1⋅ 593psi+( )− Ai1 Ai2−( )⋅:=
F3 4.218− 103× lbf=
F4 1800ft gm1⋅ 593psi+( )− Ai2 Ai3−( )⋅:=
F4 1.550− 103× lbf=
W1 w1 10000ft 6200ft−( )⋅:=
W1 152.000 103× lbf=
W2 w2 6200ft 1800ft−( )⋅:=
W2 191.400 103× lbf=
W3 w3 1800⋅ ft:= W3 84.600 103× lbf=
F5 W1 W2+ W3+ F1+ F2+ F3+ F4+:=
F5 355.893 103× lbf=
28
7.12
Fig 7.34 Fa & Fs vs. Depth
0
2000
4000
6000
8000
10000
-100 0 100 200 300 400
Force 1000 lbf
Dep
th
Fa_btm_sec1 F1 F2+:=
Fa_btm_sec1 66.338− 103× lbf=
Fa_top_sec1 Fa_btm_sec1 W1+:=
Fa_top_sec1 85.662 103× lbf=
Fa_btm_sec2 Fa_top_sec1 F3+:=
Fa_btm_sec2 81.444 103× lbf=
Fa_top_sec2 Fa_btm_sec2 W2+:=
Fa_top_sec2 272.844 103× lbf=
Fa_btm_sec3 Fa_top_sec2 F4+:=
Fa_btm_sec3 271.293 103× lbf=
Fa_top_sec3 Fa_btm_sec3 W3+:=
Fa_top_sec3 355.893 103× lbf=
7.12Find the stability forces
The stability force = piAi-poAo
Fs_btm_sec1 593psi 10000ft gm1⋅+( ) Ai1⋅ 8000ft gm1⋅ 2000ft gcmt⋅+( ) Ao1⋅−:=
Fs_btm_sec1 66.338− 103× lbf=
Fs_top_cmt 593psi 8000ft gm1⋅+( ) Ai1⋅ 8000ft gm1⋅ Ao1⋅−:=
Fs_top_cmt 11.294− 103× lbf=
Fs_top_sec1 593psi 6200ft gm1⋅+( ) Ai1⋅ 6200ft gm1⋅ Ao1⋅−:=
Fs_top_sec1 573.238− 100× lbf=
Fs_btm_sec2 593psi 6200ft gm1⋅+( ) Ai2⋅ 6200ft gm1⋅ Ao2⋅−:=
Fs_btm_sec2 4.791− 103× lbf=
Fs_top_sec2 593psi 1800ft gm1⋅+( ) Ai2⋅ 1800ft gm1⋅ Ao2⋅−:=
Fs_top_sec2 23.944 103× lbf=
Fs_btm_sec3 593psi 1800ft gm1⋅+( ) Ai3⋅ 1800ft gm1⋅ Ao3⋅−:=
Fs_btm_sec3 22.394 103× lbf=
Fs_top_sec3 593psi Ai3⋅:=Fs_top_sec3 35.098 103× lbf=
29
7.12Step 2. Find and plot the stability force after cementing on Fig 7.34.
Fig 7.34 Fa & Fs vs. Depth
0
2000
4000
6000
8000
10000
-100 0 100 200 300 400
Fa Fs
Step 3. Locate the intersection of the stability and forcecurves.
Since the intersection occurs at the shoe there is no tendency tobuckle, as is expected when the casing is suspended off bottomwhile cementing.
7.12Step 4. Find the force and length changes in all sections while drilling ahead.
An increase in internal pressure causes an increase in tension stress, a tendency to shorten the casing,and an increase in radial and tangential stress.
An increase in internal pressure causes an increase in the hydrostatic forces.
∆F3 6200− ft gm2 gm1−( )⋅ Ai1 Ai2−( )⋅:=∆F3 2.138− 103× lbf=
∆F4 1800− ft gm2 gm1−( )⋅ Ai2 Ai3−( )⋅:=∆F4 569.462− 100× lbf=
The change in hydrostatic forces F3 & F4 tends to shorten sections 2 and 3.
∆L2 ∆F36200ft 1800ft−
As2 30⋅ 106psi⋅⋅:=
∆L2 0.0250− ft=
∆L3 ∆F4 ∆F3+( ) 1800ft
As3 30⋅ 106psi⋅⋅:=
∆L3 0.0120− ft=
The change in length of section 1 is not considered because the ∆L approach assumes the bottom isfree to move.
30
7.12The average change in internal pressure is..
∆pi_avg 8000ft2
gm2 gm1−( )⋅:=∆pi_avg 1.248 103× psi=
The average ratio of internal area to cross sectional area is..
Aavg_ratio8000 6200−
8000
Ai1
As1⋅
6200 1800−8000
Ai2
As2⋅+
18008000
Ai3
As3⋅+:=
Aavg_ratio 4.8218=
From the derivation of equation 7.3..
∆Lavg2− 8000⋅ ft .3⋅ Aavg_ratio⋅ ∆pi_avg⋅
30 106psi⋅:=
∆Lavg 0.9628− ft=
The increase in temperature causes the casing to lengthen according to eq. 7.37
∆Lt 6.667 10 6−⋅ F 1− 8000⋅ ft ∆T⋅:= ∆Lt 1.600 ft=
∆Ltotal ∆Lt ∆Lavg+ ∆L2+ ∆L3+:= ∆Ltotal 0.6003 ft=
7.12Step 5. Determine what portion of length change is allowed for by bending due to buckling.
∆r13in 9.625in−
2
:=∆r 1.688 in=
Iπ64
9.625in( )4 d14−
⋅:=
I 122.196 in4.000=
As_avg18008000
As3⋅6200 1800−
8000
As2⋅+8000 6200−
8000
As1⋅+:=As_avg 12.539 in2.000=
∆Lbu∆r2 Fbu
2⋅
8 30⋅ 106psi⋅ I⋅ w1⋅:=
Fbu Eq. 7.33
∆Fa30− 106psi⋅ As_avg⋅
8000ft∆Ltotal ∆Lbu−( )⋅:= ∆Lbu Hooks law
Fa_new F1 F2+ 2000ft w1⋅+( ) ∆Fa+:= ∆Fa
Fs_new 8000ft gm2⋅ Ai1⋅ 8000ft gm1⋅ Ao1⋅−:=
Fbu Fs_new Fa_new−:= Fa_newFs_new 105.371 103× lbf=
31
7.12∆r2
1lbf( )2⋅
8 30⋅ 106psi⋅ I⋅ w1⋅2.428 10 12−× ft=
∆Lbu 2.428 10 12−⋅ft
lbf 2Fbu
2⋅:= Fbu simplifying
30− 106psi⋅ As_avg⋅
8000ft47.019− 103×
lbfft
=
∆Fa 47.019− 103×lbfft
∆Ltotal ∆Lbu−( )⋅:= ∆Lbu
∆Ltotal 600.329 10 3−× ft=
47.019− 103×lbfft
.6003⋅ ft 28.226− 103× lbf=
∆Fa 28.226− 103× lbf 47.019 103×lbfft
∆Lbu⋅+:= ∆Lbu substituting
∆Fa 28.226− 103× lbf 47.019 103×lbfft
2.428 10 12−⋅ft
lbf 2Fbu
2⋅
⋅+:= Fbu
47.019 103⋅lbfft
⋅ 2.428⋅ 10 12−⋅ft
lbf 2⋅ 114.162 10 9−× lbf 1.000 100×−
=
7.12∆Fa 28.226− 103× lbf 114.162 10 9−× lbf 1− Fbu
2⋅( )+:= Fbu
Fa_new F1 F2+ 2000ft w1⋅+( ) ∆Fa+:= ∆Fa
F1 F2+ 2000ft w1⋅+ 13.662 103× lbf=
Fa_new 13.662 103× lbf ∆Fa+:= ∆Fasubstituting
Fa_new 13.662 103× lbf 28.226− 103× lbf 114.162 10 9−× lbf 1− Fbu2⋅( )+
+:= Fbu
13.662 103× lbf 28.226 103× lbf− 14.564− 103× lbf=
Fa_new 14.564− 103× lbf 114.162 10 9−× lbf 1− Fbu2⋅( )+:= Fbu
Fs_new 8000ft gm2⋅ Ai1⋅ 8000ft gm1⋅ Ao1⋅−:=
Fbu Fs_new Fa_new−:= Fa_new
Fs_new 105.371 103× lbf=
32
7.12Fbu 105.371 103× lbf 14.564− 103× lbf 114.162 10 9−× lbf 1− Fbu
2⋅( )+
−:= Fbu substituting
105.371 103× lbf 14.564 103× lbf+ 119.935 103× lbf=
Fbu 119.935 103× lbf 114.162 10 9−× lbf 1− Fbu2⋅−:= Fbu
a 114.162 10 9−×:=
b 1:=
cq 119.935− 103×:=
solving the quadratic..Fbu b−
b2 4 a⋅ cq⋅−
2 a⋅+:=
Fbu 2.148 103×=
Fa_new 14.564− 103× lbf 114.162 10 9−× lbf 1− 2.148 103lbf( )2⋅
+:=
Fa_new 14.563− 103× lbf= From the top of the cement..
Fs_new 105.371 103× lbf=
7.12Step 6. Plot the new Fa and Fs curves.
Fa2_top_cmt Fa_new:= Fa2_top_cmt 14.563− 103× lbf=
Fa2_top_sec1 Fa2_top_cmt 8000ft 6200ft−( ) w1⋅+:= Fa2_top_sec1 57.437 103× lbf=
Fa2_btm_sec2 Fa2_top_sec1 F3+ ∆F3+:= Fa2_btm_sec2 51.081 103× lbf=
Fa2_top_sec2 Fa2_btm_sec2 W2+:= Fa2_top_sec2 242.481 103× lbf=
Fa2_btm_sec3 Fa2_top_sec2 F4+ ∆F4+:= Fa2_btm_sec3 240.361 103× lbf=
Fa2_top_sec3 Fa2_btm_sec3 W3+:= Fa2_top_sec3 324.961 103× lbf=
Fs2_top_cmt Fs_new:= Fs2_top_cmt 105.371 103× lbf=
Fs2_top_sec1 6200ft gm2 Ai1⋅ gm1 Ao1⋅−( )⋅:= Fs2_top_sec1 81.663 103× lbf=
Fs2_btm_sec2 6200ft gm2 Ai2⋅ gm1 Ao2⋅−( )⋅:= Fs2_btm_sec2 75.963 103× lbf=
Fs2_top_sec2 1800ft gm2 Ai2⋅ gm1 Ao2⋅−( )⋅:= Fs2_top_sec2 22.054 103× lbf=
Fs2_btm_sec3 1800ft gm2 Ai3⋅ gm1 Ao3⋅−( )⋅:= Fs2_btm_sec3 20.535 103× lbf=
Fs2_top_sec3 0.0lbf:=
33
7.12
Fig 7.34 Fa & Fs vs. Depth
0
2000
4000
6000
8000
10000
-100 0 100 200 300 400
Force 1000 lbf
Dep
th
Fa_new Fs_new
Fa Fs
Step 7. Locate the intersection of the new Fa and Fs curves.
The dashed lines on the graph intersect at approximately 5,800 ft.Thus the 2,200 feet just above the top of the cement is in helicalbuckling.
Step 8. Determine how much tension is needed at landing tolower intersection to top of cement.
Fs2_top_cmt Fa2_top_cmt− 119.935 103× lbf=
An additional 119,935 lbs of tension could be added at landing toprevent the casing from buckling.
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