PETE 7212 Well Completion Design Casing Design
Casing Design
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Why Run Casing? Types of Casing Strings Classif lassificati ication on of Casing Casing Wellheads Burst, Collapse and Tension Example Effect of Axial Tension on Collapse Strength Example Loui ouis sianaStateUni nive versi rsity ty College of Engineering
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Casing Design
• Why run casing? –
To prevent the hole from collapsing.
–
Onshore - to prevent contamination of fresh water sands
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To prevent water migration to producing formation
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To confine production to the wellbore
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To control pressures during drilling
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To provide an acceptable environment for subsurface equipment in producing wells LouisianaStateUniversity
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Casing Design –
To confine production to the wellbore.
–
To control pressures during drilling.
–
To provide an acceptable environment for subsurface equipment in producing wells.
• You need 14 ppg to control a lower zone, but an upper zone will fracture at 12 ppg. What do you do?
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Types of Casing Strings Depth BML (ft)
Diameter Range
Example
Drive pipe or structural pile (Gulf coast and offshore only)
150 – 300
16" – 60"
30"
Conductor casing
100 –1600
16" – 48"
20"
Surface casing
2000 –4000
8-5/8" – 20"
13-3/8"
Intermediate casing
As needed
7-5/8" – 13-3/8"
9-5/8"
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Types of Strings of Casing Depth BML
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(ft)
Diameter Range
Example
Production casing
As needed
4-1/2" – 9-5/8"
7"
Liner(s)
As needed
4-1/2" – 9-5/8"
Tubing string(s)
As needed
Up to 4"
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Example Hole and Casing Sizes Hole Size
Pipe Size
36"
Structural casing
30"
26"
Conductor casing
20"
17-1/2"
Surface casing
13-3/8"
12-1/4"
Intermediate casing
9-5/8"
8-3/4"
Production liner
7" LouisianaStateUniversity
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Example Hole and String Sizes (in) Structural casing Conductor casing 250' 1,000' 4,000' Surface casing Intermediate casing Production Liner 8
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Classification of Casing • Outside diameter of pipe • Wall thickness • Grade of material • Type to threads and couplings • Length of each joint (RANGE) • Nominal weight per foot
(e.g. 9 5/8”) (e.g. 1/2”) (e.g. N-80) (e.g. API LCSG) (e.g. Range 3) (e.g. 47 lb/ft)
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Casing Grades and Properties Minimum Ultimate Most Common Yield Strength Tensile Grades (psi) Strength (psi)
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H-40
40,000
60,000
J-55
55,000
75,000
K-55
55,000
95,000
C-75
75,000
95,000
L-80
80,000
95,000
N-80
80,000
100,000
C-90
90,000
100,000
C-95
95,000
105,000
P-110
110,000
125,000
V-150
150,000
160,000
σ
σu
σy
ε
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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
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API round threads (short)
CSG
API round threads (long)
LCSG
Buttress Extreme line
BCSG XCSG
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Typical API Design Factors Required
Design
10,000 psi
Collapse 1.125
11,250 psi
100,000 lbf
Tension 1.8
180,000 lbf
10,000 psi
Burst
11,000 psi
1.1
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Pore Pressure Conductor Pipe
Abnormal
Surface Casing Production Casing Production Tubing Liner
Normal Pore Pressure 0.433 – 0.465 psi/ft 14
Abnormal Pore Pressure > 0.465 psi/ft LouisianaStateUniversity College of Engineering
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Louisfrom ianaStatebottom University Design
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Wellhead Assembly
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Wellhead Assembly
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Casing Design Tension
Tension Depth Burst Collapse
Collapse STRESS Burst: Collapse: Tension: 18
Assume full reservoir pressure all along the wellbore. Hydrostatic pressure increases with depth Tensile stress due to weight of string is highest at top LouisianaStateUniversity College of Engineering
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Casing Design – Tension
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Casing Design
• Collapse (from external pressure) –
Yield Strength Collapse
–
Plastic Collapse
–
Transition Collapse
–
Elastic Collapse
Collapse pressure is affected by axial stress 20
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Casing Design – Collapse
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Casing Design – Burst
• • •
Internal yield pressure for pipe Internal yield pressure for couplings Internal pressure leak resistance
P
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P
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Casing Design Example – Burst
• Design a 7” casing 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
• Calculate probable reservoir pressure. p res
= 0.5
psi ft
(10,000 ft ) = 5,000 psi
• Calculate required pipe internal yield pressure rating pi
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= pres N i = (5,000)(1.1) = 5,500 psi LouisianaStateUniversity College of Engineering
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Example
• Select the appropriate casing grade and weight from tables: –
Burst pressure required = 5,500 psi l
7", J-55, 26 lb/ft has BURST Rating of 4,980 psi
l
7", N-80, 23 lb/ft has BURST Rating of 6,340 psi
l
7", N-80, 26 lb/ft has BURST Rating of 7,240 psi
• Use N-80 casing, 23 lb/ft
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Collapse Pressure
• The following factors are important: –
The collapse pressure resistance of a pipe depends on the axial stress
• There are different types of collapse failure
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–
Yield strength collapse
–
Plastic collapse
–
Transition collapse
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Elastic collapse
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Casing Design
• Collapse pressure correction for axial stress –
(Neglects internal pressure) PC
= PCR
2 σ σ A A 1 − 0.75 − 0.5 σ σ Y Y
PC = Adjusted collapse rating PCR = Collapse rating for zero axial stress σ A = Axial stress in pipe (psi) σy = Pipe material minimum yield strength (psi)
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Casing Design – Collapse
• Calculate D/t to determine the proper equation to use for calculating the collapse rating –
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Yield strength collapse
Plastic collapse
Pyp
Pp
= 2 σ y (D / t ) −2 1 (D / t )
F = σ y 1 − F2 − F3 D / t
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Casing Design – Collapse –
–
Transition collapse:
Elastic Collapse:
Pt
F = σ y 4 − F5 D / t
Pe
=
46.95 × 106 D D − 1 t t
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Casing Design – Collapse If axial tension is zero: Grade
Yield
Plastic
Transition
Elastic
J-55
14.81
25.01
37.31
N-80
13.38
22.47
31.02
P-110
12.44
20.41
26.22
D / t
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Example 2
• Determine the collapse rack rating of 5 1/2" (J55, 14 lb/ft) casing under zero axial load. –
Calculate D/t: D t
=
5.500 1 (5.500 − 5.012 ) 2
= 22.54
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Example 2
• Check the mode of collapse –
Table 7.4 on p. 309 of Bourgoyne et al. shows that, for J-55 pipe with 14.81 < D/t < 25.01, the mode of failure is plastic collapse.
• The plastic collapse is calculated from: Pp
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A = Y p − B − C D / t 2.991 = (55,000 ) − 0 .0541 − 1206 22.54 ≈ 3 ,120 psi LouisianaStateUniversity College of Engineering
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Example 3
• Determine the collapse rating for a 5 1/2" (J-55, 14 lb/ft) casing under axial load of 100,000 lbf –
The axial tension will reduce the collapse pressure as follows: PCC
σz =
= PCR
Tension Area
=
3 1− 4
2 1 σ σ z z − σ y 2 σ y
100 ,000
π (5.5 2 − 5.0122 )
= 24820 psi
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Example 3 (cont.) –
The axial tension will reduce the yield stress to an apparent value:
σ y, e
2 24820 24820 = (55000 ) 1 − 0 .75 − 0 .5 55000 55000 = 38,216 psi
Here the axial load decr eased the J-55 rating to an apparent “ J-38.2” rating
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Example 3 (cont.) F = σ y ,e 1 − F2 − F3 D / t 2. 945 = (38 ,216 ) − 4 .557 x10 − 2 − 700 .43 22 .54 ≈ 2,550 psi Pp
…compared to 3,117 psi with no axial stress! 35
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