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Transcript of 4 Casing Design
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MASTER
Well Drill ing
Casing Design
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Casing Design
Why Run Casing?
Types of Casing Strings
Classification of Casing
Wellheads Burst, Collapse and Tension
Example
Effect of Axial Tension on Collapse Strength
Example
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Read Applied Drilling Engineering, Ch.7
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Casing Design
Why run casing?
1. To prevent the hole from caving in2. Onshore - to prevent contamination of
fresh water sands
3. To prevent water migration toproducing formation
What is casing? Casing
Cement
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Casing Design - Why run casing, cont’d
4. To confine production to the wellbore
5. To control pressures during drill ing
6. To provide an acceptable environment for
subsurface equipment in producing wells
7. To enhance the probability of drilling to total
depth (TD)
e.g., you need 14 ppg to control a lower zone,
but an upper zone will fracture at 12 lb/gal.
What do you do?
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Functions of CasingIndividually
Drive pipeProvides a means of
nippling up diverters
Provides a mudreturn path
Prevents erosion of ground below rig
Conductor pipeSame as Drive pipe
Supports the weight
of subsequent casingstrings
Isolates very weakformations
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Functions of Casing
IndividuallySurface casing
Provides a means of
nippling up BOPProvides a casing
seat strong enoughto safely close in a
well after a kick.Provides protection
of fresh water sands
Provides wellborestabilization
Intermediate orprotective casing
Usually set in thefirst abnormallypressured zone
Provides isolation of
potentiallytroublesome zones
Provides integrity to
withstand the highmud weightsnecessary to reach TD or next csg seat
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Functions of Casing IndividuallyProduction casing
Provides zonalisolation (prevents
migration of water toproducing zones,isolates differentproduction zones)
Confines productionto wellbore
Provides the
environment to installsubsurfacecompletionequipment
Liners
Drilling liners
Same as
Intermediate orprotective casing
Production liners
Same as production
casing
Tieback liners
Tie back drilling orproduction liner to the
surface. Convertsliner to full string of casing
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Types of Strings of Casing
1. Drive pipe or structural pile{Gulf Coast and offshore only}
150’-300’ below mudline.
2. Conductor string. 100’ - 1,600’
(BML)
3. Surface pipe. 2,000’ - 4,000’
(BML)
Diameter Example
16”-60” 30”
16”-48” 20”
8 5/8”-20” 13 3/8”
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Types of Strings of Casing
4. Intermediate String
5. Production String (Csg.)
6. Liner(s)
7. Tubing String(s)
7 5/8”-13 3/8” 9 5/8”
Diameter Example
4 1/2”-9 5/8” 7”
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Example Hole and String Sizes (in)
Structural casing
Conductor string
Surface pipe
IntermediateString
Production Liner
Hole Size
30”20”
13 3/8
9 5/8
7
Pipe Size
36”26”
17 1/2
12 1/4
8 3/4
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Example Hole and String Sizes (in)
Structural casing
Conductor string
Surface pipe
IntermediateString
Production Liner
Hole Size
30”20”
13 3/8
9 5/8
7
Pipe Size
36”26”
17 1/2
12 1/4
8 3/4
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Example Hole and String Sizes (in)
Structural casing
Conductor string
Surface pipeIntermediateString
Production Liner
250’
1,000’
4,000’
Mudline
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Classification of CSG.
1. Outside diameter of pipe (e.g. 9 5/8”)
2. Wall thickness (e.g. 1/2”)
3. Grade of material (e.g. N-80)
4. Type to threads and couplings (e.g. API LCSG)
5. Length of each joint (RANGE) (e.g. Range 3)
6. Nominal weight (Avg. wt/ft incl. Wt. Coupling)
(e.g. 47 lb/ft)
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Length of Casing Joints
RANGE 1 16-25 ft
RANGE 2 25-34 ft
RANGE 3 > 34 ft.
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Casing Threads and Couplings
API round threads - short {CSG }API round thread - long {LCSG }
Buttress {BCSG }Extreme line {XCSG }
Other …
See Halliburton Book...
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Rounded Threads
* 8 per inch
~ Square Threads
* Longer* Stronger
Integral Joint
* Smaller ID, OD
* Costs more
* Strong
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23 lb/ft
26 lb/ft
N-80
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API Design Factors (typical)
Collapse 1.125
Tension 1.8
Burst 1.1
Required
10,000 psi
100,000 lbf
10,000 psi
Design
11,250 psi
180,000 lbf
11,000 psi
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Normal Pore Pressure Abnormal Pore Pressure
0.433 - 0.465 psi/ft gp
> normal
Abnormal
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23Design from bottom
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X-mas TreeWing Valve
Choke Box
Master
Valves
Wellhead
• Hang Csg. Strings
• Provide Seals
• Control Productionfrom Well
Press. Gauge
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Casing Design
Burst: Assume full reservoir pressure all along the wellbore.
Collapse: Hydrostatic pressure increases with depth
Tension: Tensile stress due to weight of string is highest at top
STRESS
Tension
Burst
Collapse
Collapse
Tension
Depth
Burst
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Casing Design - Tension
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Tensile force balance on pipe body
A*Fsyieldten
σ=
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Tensile force balance on pipe body
Example 7.1:
Compute the body-yield strength for 20-
in., K-55 casing with anominal wall thicknessof 0.635 in. and a
nominal weight perfoot of 133 lbf/ft.
A*Fsyieldten
σ=
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Tensile force balance on pipe body
Solution: This pipe has a minimum
yield strength of 55,000 psiand an ID of:
.in730.18)635.0(200.20d =−=
K55
A*F syieldten σ=
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Tensile force balance on pipe body
Thus, the cross-sectional area of steel is
and a minimum pipe-body yield
is predicted by Eq. 7.1 atan axial force of:
.in.sq 63.38)73.1820(4
A 22s =−=
π
lbf 000,125,2)63.38(000,55Ften ==
A*F syieldten σ=
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Pipe Body Yield Strength
where
p
22
y Y)dD(4P −=
inpipe,of diameter insided
inpipe,of diameter outsideD
psistrength,yieldminimumspecifiedY
lbf strength,yieldbodypipeP
p
y
=
=
=
=
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Casing Design - Burst
(from internal pressure)
4 Internal Yield Pressure for pipe
4 Internal Yield Pressure for couplings
4 Internal pressure leak resistance
pp
Internal
Pressure
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Internal Yield Pressure for Pipe (Burst)
where
⎥⎦
⎤
⎢⎣
⎡
= D
t Y2
875.0Pp
inpipe,of O.D.D
inthickness,wallnominalt
psistrength,yieldminimumY
psipressure,yieldinternalP
p
=
=
=
=
FP = DLP
F T = 2tLYP
DLP = 2tLYP
⎥⎦
⎤⎢⎣
⎡=
D
t2YP
p
FP
FT
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Example
For 7”, 26 #/ft P-110 pipe
9,955
7*26.276)-(7*110,000*2*0.875
=
=
psi960,9P = (to the nearest 10 psi)
…agrees with Halliburton Tables.
⎥⎦
⎤⎢⎣
⎡=
D
tY2875.0P
p
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Casing Design - Burst
Example
Design a 7” Csg. String to 10,000 ft.
Pore pressure gradient = 0.5 psi/ft
Design factor, Ni=1.1
Design for burst only.
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Burst Example
1. Calculate probable reservoir pressure.
psi000,5ft000,10*ft
psi5.0pres ==
2. Calculate required pipe internal yieldpressure rating
psi500,51.1*000,5N*pp iresi ===
Ni = API Design Factor for BURST = 1.1
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Example
3. Select the appropriate csg. grade and wt.
from the Halliburton Cementing tables:
Burst Pressure required = 5,500 psi
7”, J -55, 26 lb/ft has BURST Rating of 4,980 psi7”, N-80, 23 lb/ft has BURST Rating of 6,340 psi
7”, N-80, 26 lb/ft has BURST Rating of 7,249 psi
Use N-80 Csg., 23 lb/ft
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23 lb/ft
26 lb/ft
N-80
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Casing Design - Collapse
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Collapse Pressure
The following factors are important:
4 The collapse pressure resistance of a pipedepends on the axial stress
4 There are different types of collapsefailure
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Collapse Pressure
There are four different types of collapse
pressure, each with its own equation forcalculating the collapse resistance:
4 Yield strength collapse
4 Plastic collapse
4 Transition collapse4 Elastic collapse
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If Axial Tension is Zero:
Yield Strength Plastic Transition Elastic
→ )t/D(
J -55 14.81 25.01 37.31
N-80 13.38 22.47 31.02
P-110 12.44 20.41 26.22
Casing Design - Collapse
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Casing Design
Collapse pressure - with axial stress
1.
⎪
⎭
⎪⎬⎫
⎪
⎩
⎪⎨⎧
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ −
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ −=
P
A
2/12
P
APPA
Y
S5.0
Y
S75.01 Y Y
YPA = yield strength of axial stressequivalent grade, psi
YP = minimum yield strength of pipe, psi
SA = Axial stress, psi (tension is positive)
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Casing Design - Collapse
2. Calculate D/t to determine proper equation
to use for calculating the collapse pressure
Yield Strength
Collapse :
Plastic Collapse:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟ ⎠ ⎞⎜
⎝ ⎛
−⎟ ⎠
⎞⎜⎝
⎛
=2p YP
tD
1t
D
Y2P
CB
t
DA YP pp −
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎟ ⎠
⎞⎜⎝
⎛ =
C C ’
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Transition
Collapse:
Elastic
Collapse:
⎥⎥⎥
⎥
⎦
⎤
⎢⎢⎢
⎢
⎣
⎡
−⎟ ⎠
⎞⎜⎝
⎛ = G
t
D
F YP p T
2
6
E
1t
D
t
D
10X95.46
P⎥⎦
⎤⎢⎣
⎡−⎟
⎠
⎞⎜⎝
⎛ ⎟ ⎠
⎞⎜⎝
⎛ =
Casing Design - Collapse, cont’d
E l 2
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Example 2
Determine the collapse strength of 5 1/2”O.D., 14.00 #/ft J -55 casing under zero
axial load.
1. Calculatethe D/t ratio: ( )
booknHalliburtoFrom
54.22
012.5500.52
1500.5
tD
↑
=−
=
C C
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If Axial Tension is Zero:
Yield Strength Plastic Transition Elastic
→ )t/D(
J -55 14.81 25.01 37.31
N-80 13.38 22.47 31.02
P-110 12.44 20.41 26.22
Casing Design - Collapse
Example 2
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Example 2
2. Check the mode of collapse
Table (above) shows that,
for J -55 pipe,with 14.81 < D/t < 25.01
the mode of failure is plastic collapse.
54.22=t
D
C i D i C ll
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Casing Design - Collapse
Calculate D/t to determine proper equation touse for calculating the collapse pressure
Plastic Collapse: CB
t
DA YP pp −
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎟ ⎠
⎞⎜⎝
⎛ =
Example 2
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Example 2
The plastic collapse is calculated from:
206,10541.054.22
991.2000,55
CBt/D
A YP pp
−⎥⎦
⎤⎢⎣
⎡−=
−⎟ ⎠ ⎞⎜
⎝ ⎛ −=
psi117,3P p =Halliburton Tablesrounds off to 3,120 psi
Example 3
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Example 3
Determine the collapse strength for a 5 1/2” O.D.,14.00 #/ft, J -55 casing under axial load of 100,000lbs
The axial tension will reduce the collapse pressureas follows:
P
p
A2
p
APA Y
Y
S5.0
Y
S75.01 Y
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⎟⎟
⎠
⎞⎜⎜
⎝
⎛ −⎟⎟
⎠
⎞⎜⎜
⎝
⎛ −=
( ) psi
Area
F S A A 820,24
012.55.54
000,100
22
=−
==π
Example 3 cont’d AA YS S
Y ⎥⎤
⎢⎡
⎟ ⎞
⎜⎛
⎟ ⎞
⎜⎛
507501
2
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Example 3 cont d
The axial tension will reduce the collapsepressure rating to:
psi216,38
000,55000,55
820,245.0
000,55
820,2475.01 Y
2
PA
=
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡⎟ ⎠
⎞⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ −=
Here the axial load decreased the J-55
rating to an equivalent “J-38.2” rating
P
p
A
p
APA Y
Y Y Y
⎥⎥⎥
⎦⎢⎢⎢
⎣
⎟⎟
⎠⎜⎜
⎝ −
⎟⎟
⎠⎜⎜
⎝ −= 5.075.01
Example 3 cont’d
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Example 3 - cont d
551,243.70010x557.4
54.22
945.2216,38
CBt/D
A YP
2
PAp
=−⎥⎦
⎤⎢⎣
⎡−=
−⎟ ⎠
⎞⎜⎝
⎛ −=∴
−
psi550,2P p ≈
…compared to 3,117 psi with no axial stress!
Example 3 cont’d
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Example 3 - cont d
We shall be using API Tables to correct for theeffect of axial tension on collapse strength of casing.
The Halliburton Cementing Tables list thecollapse resistance of 5 ½ -in, 14.00 lb/ft J -55casing at 3,120 psi.
The axial tension in this case would derate the
collapse strength to about 2,550 psi.
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Combined Loading
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Linear Interpolation
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p
(iii) CmSP
(ii) CmSP
(i) CmSP
cmxy
22
11
+=
+=+=
+=
Linear Interpolation
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ea te po at o
⇒−=−− )SS(mPP )ii()iii( 1212
12
12
S S PPm
−−=
)()(P )()(1
12
12
11
S S S S
PPS S mPiii −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−=−=−−
Linear Interpolation
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p
( )12
12
11 PP
SSSSPP −⎟⎟
⎠ ⎞⎜⎜
⎝ ⎛
−−+=∴
With design factor:
( )⎥⎦
⎤⎢⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛
−
−−= 21
12
11cc PP
SS
SSP
.F.D
1P
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α = dogleg severity, deg/100 ft
= angle build rate, deg/100 ft
απ= 00018,RadiusBuild
Length of arc, L = R∆θRL
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∆L = (R + r)∆θ - R∆θ
θ∆=θ∆=∆ 2drL n
( ) 180100122
d
L2
d
L
L nn πα=
θ∆=
∆=ε∆
nn
6
d218d
180400,2
10*30E α=α
π=ε∆=σ∆
nd218α=σ∆
∆θR + rR
g R
snAd218F α= (7.14a)
Figure 7.14 - Incremental stress caused by
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bending of casing in a directional well
The area of steel, As, can be expressed
conveniently as the weight per foot of pipe divided by the density of steel.For common field units, Eq. 7.14a
becomes
)14.7.........(....................64 bwd F nab α =
ly.respectivelbf/ft,and in.,ft,0degrees/10lbf,
of unitshave and ,,, where wd F nab α
Example
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)14.7.........(....................64 bwd F nab α =
/53 in7
100deg/5
,
ft lbf wd
ft
n
=
=
=α
Fab
= 64 * 5 * 7 * 35 = 74,400 lbf
Fab = 74,400 lbf
Casing Design Example
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g g p
Example Problem
API Design Factors
“Worst Possible Conditions”
Effect of Axial Tension on Collapse Strength
Iteration and Interpolation
Design for Burst, Collapse and Tension
Casing Design Example
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Design a 9 5/8-in., 8,000-ft combinationcasing string for a well where the mud wt.will be 12.5 ppg and the formation porepressure is expected to be 6,000 psi.
Only the grades and weights shown areavailable (N-80, all weights). Use API
design factors.
Design for “worst possible conditions.”
Casing Design - Solution
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g g
Before solving this problem is it necessary tounderstand what we mean by “Design Factors”
and “worst possible conditions”.
API Design Factors
Design factors are essentially “safety factors”that allow us to design safe, reliable casingstrings. Each operator may have his own set
of design factors, based on his experience,and the condition of the pipe.
Casing Design
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Casing Design
In PETE 661, we’ll use the design factorsrecommended by the API unless otherwise
specified.
These are the API design Factors:
Tension and J oint Strength: N T = 1.8
Collapse (from external pressure): Nc= 1.125Burst (from internal pressure): Ni = 1.1
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g g
What this means is that, for example, if we
need to design a string where the maximumtensile force is expected to be 100,000 lbf ,we select pipe that can handle 100,000 * 1.8
= 180,000 lbf in tension.
Note that the Halliburton Cementing Tables
list actual pipe strengths, without safetyfactors built in.
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Unless otherwise specified in a particularproblem, we shall also assume the following:
Worst Possible Conditions
1. For Collapse design, assume that thecasing is empty on the inside (p = 0 psig)
2. For Burst design, assume no “backup”fluid on the outside of the casing (p = 0 psig)
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Worst Possible Conditions, cont’d
3. For Tension design,assume no buoyancy effect
4. For Collapse design,
assume no buoyancy effect
The casing string must be designed to stand up to the
expected conditions in burst, collapse and tension.
Above conditions are quite conservative. They are also
simplified for easier understanding of the basic concepts.
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Burst Requirements (based on the expected pore
pressure)
The whole casing string must be capable of withstanding this internal pressure without failing inburst.
psi600,6P
1.1* psi000,6
Factor Design* pressure poreP
B
B
=
=
=
D e p t h
Pressure
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Collapse Requirements
For collapse design, we start at the bottomof the string and work our way up.
Our design criteria will be based onhydrostatic pressure resulting from the 12.5ppg mud that will be in the hole when the
casing string is run, prior to cementing.
Casing Designe
p t h
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Collapse Requirements, cont’d
severelessare
tsrequiremencollapsetheholetheupFurther
.bottomtheatd'reqpsi850,5P125.1*000,8*5.12*052.0
factordesign*depth*weightmud*052.0P
c
c
←=
=
=
D e
Pressure
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Req’d: Burst: 6,600 psi Collapse: 5,850 psi
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Note that two of the weights of N-80 casingmeet the burst requirements, but only the
53.5 #/ft pipe can handle the collapserequirement at the bottom of the hole (5,850psi).
The 53.5 #/ft pipe could probably run all theway to the surface (would still have to checktension), but there may be a lower costalternative.
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To what depth might webe able to run N-80, 47#/ft? The maximumannular pressure that this
pipe may be exposed to,is:
psi231,4125.1
760,4
factor design
pipeof pressureCollapse
Pc===
D e p t h
Pressure
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First Iteration
At what depth do we see this pressure (4,231psig) in a column of 12.5 #/gal mud?
ft509,65.12*052.0
231,45.12*052.0
Ph
h*5.12*052.0P
c1
1c
===∴
=
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Weight, W1 = 53.5 #/ft * 1,491 ft= 79,769 lbf
This weight results in an axialstress in the 47 #/ft pipe
psi877,5
in13.572
lbf 769,79
areaend
weightSof
21 ===
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The API tables show that the above
stress will reduce the collapse resistancefrom4,760 to somewhere between
4,680 psi (with 5,000 psi stress)
and 4,600 psi (with 10,000 psi stress)
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Interpolation between these values showsthat the collapse resistance at 5,877 psi
axial stress is:
psi148,4125.1
666,4P
psi666,4)600,4680,4(*)000,5000,10(
)000,5877,5(680,4P
cc1
1c
==
=−−
−−=
With the design factor,
( )21
12
11c1P PP
S S
S S P −⎟⎟
⎠
⎞⎜⎜⎝
⎛
−
−−=
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This (4,148 psig) is the pressure at a
depth
Which differs considerably from the
initial depth of 6,509 ft, so a seconditeration is required.
ft382,65.12*052.0
148,4h2 ==
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Second Iteration
Now consider running the 47 #/ftpipe to the new depth of 6,382 ft.
psi378,6in572.13lbf 563,86S
lbf 563,865.53*)382,6000,8(W
22
2
==
=−=
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Interpolating again,
This is the pressure at a depth of
( ) psi pcc 140,4600,4680,4*5000
5000378,6
680,4125.1
1
2=
⎭⎬
⎫
⎩⎨
⎧
⎥⎦
⎤
⎢⎣
⎡−
−−=
ft369,65.12*052.0
140,4h 3 ==
( )⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−−=
2112
1
1c1 D.F.
1P PP
S S
S S P
Casing Design
Thi i ithi 13 ft f th d l If
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This is within 13 ft of the assumed value. If more accuracy is desired (generally not
needed), proceed with the:Third Iteration
psi429,6572.13
259,87S
lbf 259,875.53*)369,6000,8(W'369,6h
3
3
3
==
=−==
Pcc3 = ?
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Third Iteration, cont’d
2
3
140,4
)600,4680,4(*
000,5
000,5429,6680,4
125.1
1
cc
cc
P psi
Pthus
==
⎭
⎬⎫
⎩
⎨⎧
−−
−=
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Third Iteration, cont’d
This is the answer we are looking for, i.e.,we can run 47 #/ftN-80 pipe to a depth of 6,369 ft, and 53.5 #/ft pipe between 6,369
and 8,000 ft.Perhaps this string will run all the way to thesurface (check tension), or perhaps an evenmore economical string would include some43.5 #/ft pipe?
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At some depth the 43.5 #/ft pipe would be
able to handle the collapse requirements,but we have already determined that it willnot meet burst requirements.
! NO∴
N-80
43 5 #/ft?
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N-8053.5 #/ft
N-8047.0 #/ft
43.5 #/ft?Depth = 5,057?
5,066?5,210?
Depth = 6,3696,3696,382
6,509
8,000
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The weight on the top joint of casing
would be
With a design factor of 1.8 for tension, a
pipe strength of
weightactual602,386
)/#5.53*631,1()/#0.47*369,6(
lbs
ft ft ft ft
=
+
required islbf 080,695602,386*8.1 =
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The Halliburton cementing tables give a
yield strength of 1,086,000 lbf for the pipebody and a joint strength of 905,000 lbf forLT & C.
surfacetoOK isft/#0.47∴
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We have 4 different weights of casingavailable to us in this case:
1. Two of the four weights are unacceptableto us everywhere in the string becausethey do not satisfy the burstrequirements.
2. Only the N-80, 53.5 #/ft pipe is capable of withstanding the collapse requirementsat the bottom of the string
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3. Since the 53.5 #/ft pipe is the mostexpensive, we want to use as little of it
as possible, so we want to use asmuch 47.0 #/ft pipe as possible.
4. Don’t forget to check to make sure thetension requirements are met; both for
pipe body, and for threads andcouplings (T&C).
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The collapse resistance of N-80, 47 #/ft willdetermine to what depth it can be run. Twofactors will reduce this depth:
Design Factor
Axial Stress (tension)
“Halliburton” collapse resistance: 4,760 psi Apply design factor: psi231,4
125.1
760,4=
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To determine the effect of axial stressrequires an iterative process:
1. Determine the depth capability without
axial stress
2. Determine axial stress at this point
ft509,65.12*052.0
231,4depth ==
Casing Design Review
3 Determine corresponding collapse resistance
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3. Determine corresponding collapse resistance
4. Determine depth where this pressure exists
5. Compare with previous depth estimate
6. Repeat steps 2-6 using the new depthestimate
7. When depths agree, accept answer(typically 2-4 iterations) (agreement towithin 30 ft will be satisfactory)