Casing Design
Vamegh Rasouli
1
Casing Design - Introduction What is casing? Why run casing?
Casing
Cement
1. To preven preventt the the hole hole from from caving in, 2. Onsh Onshor ore: e: to pre preven ventt contamination of fresh water sands, 3. To prev preven entt wat water er migration to producing formation,… 2
1
Casing Design - Why run casing - cont’d 4. To confine confine prod producti uction on to the wellb wellbore ore,, 5. To contro controll pressure pressures s during during drillin drilling, g, 6. To provide provide an acceptab acceptable le environm environment ent for for subsurface equipment in producing wells, 7. To enhance enhance the probabi probability lity of drilling 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 to do? 3
Typical Sequence of Csg. Strings
4
2
Functions of Casing Individually Drive pipe • Driven & cemented to shallow depth in predrilled or pre-dug holes • Provides a mud return path to surface, • Prevents erosion of ground below rig.
Conductor pipe • Same as Drive pipe, • Supports the weight of next casing strings, • Isolates very weak formations. • Diverter installed to shale shaker • Corrosion barrier
5
Functions of Casing Individually Surface casing
Intermediate casing
• Provides a means of nippling up BOP,
• Usually set in the first abnormally pressured zone,
• Provides a casing seat strong enough to safely close in a well after a kick, • Provides protection of fresh water sands, • Provides wellbore stabilization.
• Provides isolation of potentially troublesome zones, • Provides integrity to withstand the high mud weights necessary to reach TD or the next csg. Seat. 6
3
Functions of Casing Individually – Individually – cont’d Production casing
Liners
• Provides zonal isolation (prevents migration of water to producing zones and isolates different production zones) • Confines production to wellbore • Provides the environment to install subsurface completion equipment
• Drilling liners – Same as Intermediate casing
• Production liners – Same as production casing
• Tieback liners – Tie back drilling or production liner to the surface. Converts liner to full string of casing
7
Types of Strings of Casing Diameter
Example
1. Drive Pipe or Structural Pile (Gulf Coast and offshore only) 150’150’-300’ BML
16”16”-60”
30”
2. Conductor String 100’ - 1,600’ BML
16”16”-48”
20”
3. Surface Pipe 2,000’ - 4,000’ BML
85/8”-20”
133/8”
8
4
Types of Strings of Casing – Casing – cont’d
4. Intermediate String 5. Production String
Diameter
Example
75/8”-133/8”
95/8”
4½”-9 4½”-95/8”
7”
9
Casing Programs
10
5
Casing Programs – Programs – cont’d
11
Casing Selection Chart
12
6
Example Hole and String Sizes String Sizes (in) Hole Size
Pipe Size
36”
Conductor casing
30”
26”
Surface string
20” 133/8”
17½”
Intermediate pipe
12¼”
Intermediate String
95/8”
77/8”
Production Liner
51/2” 13
Classification of CSG. • • • • • •
Outside diameter of pipe (e.g. 9 5/8”) Wall Wall thi thick ckne ness ss (e. (e.g. g. ½”) ½”) Grade of material (e.g. N-80) Type to threads and couplings (e.g. API LCSG) Length of each joint (e.g. Range III) Nominal weight (e.g. 47 lb/ft)
14
7
Most Common Grades
Minimum Yield Strength (KPSI)
Ultimate Tensile Strength (KPSI)
H-40
40
60
J-55
55
75
K-55
55
95
C-75
75
95
L-80
80
95
N-80
80
100
C-90
90
100
C-95
95
105
P-110
110
125
V-150
150
160
15
Length of Casing Joints RANGE
LENGTH (ft)
I
16 - 25
II
25 - 34
III
> 34 16
8
Casing Threads and Couplings • API round threads – threads – short
( CSG )
• API round thread - long
( LCSG )
• Buttress
( BCSG )
• Extreme line
( XCSG )
• Other … See Halliburton Book... 17
Casing Threads and Couplings – Couplings – cont’d
Rounded Threads •8 threads per inch •4½” to 20”
Square Threads •Longer •Stronger • 4½” to 20”
Integral Joint •Smaller ID, OD •Costs more •Strong • 5” to 10 103/ 3/4” 4”
18
9
19
20
10
Wellhead & Christmas Tree •Wellhead •Hang Casing Strings •Provide Seals
•Christmas Tree •Control Production
from Well
21
Wellhead & Christmas Tree – Tree – cont’d
22
11
Casing Performance - Uniaxial Loadings • Axial Tension (couplings & body) • Burst Pressure • Collapse Pressure • Bending • Buckling 23
Casing Performance - Uniaxial • Tension Strength/Failure
24
12
Tension Strength • Tension Strength – Couplings: API Tables for various couplings – Body (perm. deform.)
¼ 2 2 F y = D ¡ d YP 4
¡
¢
F y = pipe body yield strength D
= external diameter (nominal)
d
= internal diameter
Y P = yield stress
25
Tension Strength – Strength – Example Example 1 Compute the body-yield body- yield strength for a 7”, N-80, N -80, 23 lb/ft casing. Solution: Solution: From API Table (1 ( 1 & 2)
D = 7 in in d = 6:366 in in Y P = 80; 000 psi 2 ¼ 2¡ Fy = 7 6:366 £ 80; 000 = 532 ks ksi 4
¡
¢
26
13
Tension Strength Formula • Uses Nominal Diameter • API minimum Thickness Thickness 87.5% of original original (nominal) thickness • Yield Strength • Rupture much larger • May deform plastically
27
Casing Performance - Uniaxial Burst (Internal Pressure) • Yield the body
P
• Yield the coupling • Leak the coupling
28
14
Burst (Internal Pressure) • Barlow (API allows 87.5% of thickness) • Thin Wall Assumption
P br = 0:875
µ ¶ 2 YP t D
P br = pipe body burst pressure D
= nominal diameter
Y P = yield stress
29
Burst (Internal Pressure) – Pressure) – Example Example 2 Compute the body burst pressure for a 7”, NN80, 23 lb/ft casing. Solution: Solution: • From API Table D = 7 in in d = 6:366 in in
!
7 ¡ 6:366 t= = 0:317 in in 2
YP = 80; 000 ps psi P br = 0:875
µ
2 £ 80; 000 £ 0:317 7
¶
= 6; 340 ps psi 30
15
Collapse (External Pressure)
31
Collapse (External Pressure) – Pressure) – cont’d The following factors are important: • The collapse pressure resistance of a pipe depends on the axial stress (biaxial stress) • There are different regimes of collapse failure (depends on ratio D/t ) • Yield Strength Collapse (thick wall) • Plastic Collapse • Transition Collapse • Elastic Collapse
(Empirical Formulation from API)
32
16
Collapse (External Pressure) – Pressure) – cont’d • Yield Stress Collapse Pressure (thick wall) pi r 2 r2 ¡ r2 + po r2 r2 ¡ r2 ¡ o o i i ¾r (r) = 2 2 r2 (ro ¡ ri ) ¾t (r) =
pi r2 i
¡ ¢ ¡ ¢
¡ ¢ ¡ ¢
r2 + r2 ¡ po r2 r2 + ri2 o o 2 2 2 ( r r ) ¡ r o
i
33
Collapse (External Pressure) – Pressure) – cont’d • Yield Strength Collapse Pressure
PY = 2 YP P
·
(D=t =t) ¡ 1 (D=t =t)2
¸
P Yp = pipe body collapse pressure D t
= nominal diameter = wall thickness
Y P = yield yield stress stress (effective (effective for biaxial biaxial stress) stress)
34
17
Collapse (External Pressure) – Pressure) – cont’d • Plastic Collapse Pressure
PP = YP
µ
A ¡ D=t =t B
¶
¡ C
P P = pipe body collapse pressure D
= nominal diameter
t = wall thickness Y P = yield stress (effective for biaxial stress) A, A, B, C , F , G, Material (and stress) dependent coefficients 35
Coef.’s Formulas (API Bull. 5C3)
36
18
Collapse (External Pressure) – Pressure) – cont’d • Transition Collapse Pressure
PT = YP
µ
¶
F ¡ G = D t
P P = pipe body collapse pressure D
= nominal diameter
t = wall thickness Y P = yield stress (effective for biaxial stress) A, B, C , F , G, Material (and stress) dependent coefficients coefficients 37
Collapse (External Pressure) – Pressure) – cont’d • Elastic Collapse Pressure
PE =
6 £ 46:95 10 2
(D=t =t) [(D=t =t) ¡ 1]
P E = pipe body collapse pressure D t
= nominal diameter = wall thickness
38
19
Collapse (External Pressure) – Pressure) – cont’d • A, A, B, C , F , G … • These values are for the uniaxial stress • Different values for effective yield stress • For Biaxial calculate the effective Yield Stress and interpolate the coefficients (Coef.’s depend on Yield Stress) 39
Collapse (External Pressure) – Pressure) – cont’d
(D= D=tt)Y P =
p
(A ¡ 2)2 + 8(B + C=Y = Y P ) + (A (A ¡ 2) 2 (B (B + C= Y P )
40
20
Collapse (External Pressure) – Pressure) – cont’d • Upper Limit for Plastic Collapse
(D= D=tt)P T
Y P (A ¡ F ) = C + Y P (B ¡ G)
41
Collapse (External Pressure) – Pressure) – cont’d • Upper Limit for Transition Collapse
(D= D=tt)T E =
2 + B= B=A A 3 B= B=A A
42
21
Collapse (External Pressure) – Pressure) – cont’d • Boundaries for Axial Stress = 0
43
Collapse (External Pressure) – Pressure) – Example Example 3 Calculate the Collapse pressure rating for a 7 in, N -80, 23 lb/ft casing. Solution: Solution: 7 in, N-80, 23 lb/ft
t = 0.317 in
Grade
A
B
C
F
G
N-80
3.071
0.0667
1,955
1.988
0.0434
(D= D=tt) =
7 = 22:08 ! 0:317
Plastic co collapse for N8 N80
44
22
Collapse (External Pressure) – Pressure) – Example Example 3 P P = YP
PP = 80; 000 £
µ
µ
A ¡ B D=tt D=
¶
¡C
¶
3:071 ¡ 0:0667 ¡ 1; 955 = 3; 3; 836 ps psi 22:08
45
Triaxial Collapse •
Effect of Axial Axial Stress in the Collapse Resistance – Resistance – Effective Effective Yield Stress
•
Von Mises Criteria (Distortion Energy) • Material fails (ductile – (ductile – yield yield failure) when total distortion energy equals uniaxial test distortion energy 2
(¾a ¡ ¾t ) + (¾t ¡ ¾r ) + (¾r ¡ ¾a ) = 2 Y 2
2
2
P
46
23
Triaxial Collapse – Collapse – cont’d • Triaxial
Ye =
s
Y
• Biaxial
Ye =
2
P
¡3
r
2
µ
¾a + pi 2
Y ¡ 3 ¾a P 2
¶
2
³ ´
2
¾a ¡ pi ¡ 2
¾a ¡ 2
Y e = effective yield stress Y P = uniaxial stress t stress t s = axial stress a pi = internal pressure ( p pi << p << po)
47
Triaxial Collapse – Collapse – cont’d Linear Interpolation for the coefficients • Coef.’ Coef.’s s depe depend nd on on Yiel Yield d Stre Stress ss • For Y L < Y e < Y U interpolate using (linear)
Ye ¡ YL X e = X + (X U ¡ X L ) YU ¡ YL
48
24
Coef.’s Formulas (API Bull. 5C3)
A = 2: 2:8762 + 0: 0:10679 £ 10¡5 YP + 0:21301 £ 10¡10 Y 2 ¡ 0:53132 £ 10¡16 Y 3 P
P
B = 0: 0:026233 + 0:50609 £ 10¡6 YP C = ¡465:93 + 0: 0:030867 YP ¡ 0:10483 £ 10¡7 Y 2 + 0:36989 £ 10¡13 Y 3 P
46:95 £ 106 F = YP
h
¡ 3B=A 2 + B=A
h
3 B=A 2 + B=A
B=A =A
ih
1
i
P
3
¡ 3B=A 2 + B=A
2
i
G = F B =A 49
Triaxial Collapse – Collapse – Example Example 4 • For the casing of Example 3, calculate the corrected critical collapse pressure if a section of 2,000 ft, ft, 7 in, N-80, 23 lbm/ft casing is suspended below it (assume linear weight of 23 lbf/ft and empty borehole - no buoyancy effect). What is the corrected collapse pressure if the internal pressure is 1,000 psi?
50
25
Triaxial Collapse – Collapse – Example Example 4 Solution: Solution: Weight of Casing Below Point in Question
F = 2; 000 £ 23 = 46 46; 000 lb lbf Cross Section Area
2 ¼ 2¡ Ac = 7 6:366 = 6: 6:6555 in2 4
¡
Axial Stress
¾a =
¢
46; 000 = 6; 6; 912 psi 6:6555 51
Triaxial Collapse – Collapse – Example Example 4 Effective Yield Stress (biaxial)
Ye =
Ye =
s
80; 000
2
r
¡3
¾a 2
³ ´ µ ¶
Y2¡3 P
6; 912 2
2
2¡
¡
¾a 2
6; 912 = 76 76; 320 psi 2
(an equivalent “N“N -76.32”) 52
26
Triaxial Collapse – Collapse – Example Example 4 Interpolated Coef.’s Grade
A
B
C
F
G
C-75
3.054
0.0642
1,806
1.990
0.0418
“N“N-76.32”
3.059
0.0649
1,845
1.992
0.0422
N-80
3.071
0.0667
1,955
1.998
0.0434
API F’s Formulas (MsExcel Spreadsheet) Y p p
76320
A
B
C
F
G
3. 0 58
0. 0649
1 84 5
1. 992
0. 0422
53
Triaxial Collapse – Collapse – Example Example 4 (D=t = t) =
Collapse Regime
7 = 22:08 0:317
• Yield Regime U-Limit:
(D=t) t)Y P =
q
(A ¡ 2)2 + 8(B + C =Y P ) + (A (A ¡ 2) 2 (B (B + C = Y P )
p
(3:058 ¡ 2)2 + 8 £ (0:0649 + 2 £ (0:0649 +
1;845 ) 76;320 1; 845 ) 76;32 0
+ (3 (3:058 ¡ 2) = 13 13:54 < 22:08
54
27
Triaxial Collapse – Collapse – Example Example 4 Plastic Regime U-Limit:
(D= D=tt)P T =
YP (A ¡ F ) C + YP (B ¡ G)
76; 320 £ (3:058 ¡ 1:992) 2 2 :0 8 < = 22:79 1; 845 + 7 6; 320 £ (0:0649 ¡ 0:0422) 2)
Collapse occurs in the Plastic Regime
55
Triaxial Collapse – Collapse – Example Example 4
Plastic Collapse Strength P P = YP
PP = 76; 320 £
µ
µ
A ¡ B D =t
¶
¡C
¶
3:058 ¡ 0:0649 ¡ 1; 845 = 3; 3; 772 ps psi 22:08
(compare with 3,830 psi for the unstressed casing) 56
28
Triaxial Collapse – Collapse – Example Example 4 Effect of Internal Pressure Critical pressure expressions are for pressure differential. However, the effective yield stress should account for the internal pressure, since the yield will start at the internal wall. The triaxial expression must be used:
Y e =
s
Y 2 ¡ P
3
µ
¾a + pi 2
¶
2
¾a ¡ pi ¡ 2 57
Triaxial Collapse – Collapse – Example Example 4 Y e =
s
80; 0002 ¡ 3
µ
6; 912 + 1; 1; 000 2
¶
2
¡
6; 912 ¡ 1; 000 2
Y e = 76 76; 750 ps psi sy ield
76750
A 3. 060
PP = 76; 750 £
B
C
F
G
0. 0651
1858
1. 992
0. 0424
µ
3:060 ¡ 0:0651 22:08
¶
¡ 1; 858
= 3 ; 782 psi
(Po )max = 3; 3; 782 + 1 ; 000 = 4; 4 ; 782 psi 58
29
Pressure Collapse Table
59
Casing Design Criteria Biaxial Method or Uniaxial Method • Burst – Conductor – Surface and Intermediate Casing – Production Casing
• Collapse • Tension 60
30
Casing Design Criteria – Criteria – cont’d Burst Conductor: – External pressure is zero – The maximum internal pressure is the formation fracture pressure at the depth of the conductor set s et depth. If the fracture pressure is unknown, assume pff =1 psi/ft – F.S.=1.1 – Neglect the gas density inside the conductor
61
Burst of Conductor
62
31
Casing Design Criteria – Criteria – cont’d Burst of Surface & Intermediate Csg.: • •
External pressure: hydrostatic pressure of the heaviest mud used to drill the hole and set the casing Internal pressure: based on pore pressure at the final depth of the next casing. If the pore pressure at the bottom of the next casing is not known, assume the following:
½
r p r p p p
= 0:564 p pssi=ft depth < 8,000 ft ft = 0:650 ps psi=ft depth ¸ 8,000 ft ft
• Assume that a fraction f (usually not less than 40%) of the length is evacuated by gas and (1-f (1-f ) fraction of the length remains filled with drilling fluid. • Neglect the gas density inside the casing. • F.S.=1.1 • Worse scenario at the top of string 63
Burst of Surf. & Interm. Csg.
64
32
Casing Design Criteria – Criteria – cont’d Burst of Production Casing: • •
External pressure: hydrostatic pressure due to formation saltwater (SG ( SGsw = 1.1542) Internal pressure: based on pore pressure at the final depth Dc (production depth). If the pore pressure at the bottom of the casing is not known assume the foll owing:
½ • •
r p r p p p
= 0:564 p pssi=ft depth < 8,000 ft ft ¸ = 0:650 ps psi=ft depth 8,000 ft ft
Assumed the whole internal casing filled with gas (gas lift production) Pressure inside the casing determined as follows: D¡D g (D ¡D) ¡M ¼ c ¹ T R
pi = pb e • •
pb e
c
40;000
F.S.=1.1 Worse scenario at the top of string
65
Burst of Production Casing
66
33
Casing Design Criteria – Criteria – cont’d Collapse: • Collapse due to fluid in the annulus between the casing and the borehole • Considered the heaviest drilling fluid used to drill the hole and set the casing • Assume casing empty • No buoyancy • F.S.=1.0 (neglect the strengthening effect of cement; most of the casing will not be empty) • Worse scenario at the bottom of string 67
Collapse of Casing
68
34
Casing Design Criteria – Criteria – cont’d Tension: • Corresponds to the weight of the casing measured in the air (no buoyancy effect) • F.S.: – 1.6 for couplings – 1.8 for casing body • Worse scenario at the top of string
69
Casing Design Example Evaluate the burst and collapse pressure loadings and design an appropriate surface casing using the biaxial method. Check for axial load. – – – – –
Setting depth of the casing string: 4000 ft Mud density as setting the string: 10.0 lb/gal Setting depth of the next csg. string: 11000 ft Mud density of the next phase: 10.5 lb/gal Casing size and coupling: 10 3/4” Buttress threads, minimum grade K-55 – Assume f = 40%.
70
35
Casing Design Example cont’d Burst Loading (this is a surface csg.) • External Pressure: po(psi) = 0.052 x 10 lb/gal x D(ft) po = 0.52 x D
71
Casing Design Example cont’d • Internal Pressure: pp=0.650
psi/ft (Dnc>8,000 ft)
pp = 11,000 x 0.650 = 7,150 psi (1-0,4)xDnc = 6,600 ft p6600= 7,150-0.052x10.5x6,600=3,546 psi pi= 3,546 psi
72
36
Casing Design Example cont’d Burst Pressure – Pressure – cont’d: F.S. = 1.1 pab = (pi-po) pab= 3,546 – 3,546 – 0.52D 0.52D
73
Casing Design Example cont’d Collapse Loading • External Pressure: po = 0.52 x D • Internal pressure = 0 psi • F.S. = 1.0 pac = 0.52 x D 74
37
Casing Design Example cont’d Design for Burst • Start at bottom (minimum burst pressure) pab,4000 = 3,546 – 3,546 – 0.52 0.52 x 4000 = 1,466 psi Cheapest casing: (p.320-321) K-55, 40.50 lb/ft, Burst Strength 3,130 psi Minimum depth that can go: pab,D = 3,546 – 3,546 – 0.52 0.52 x D = 3130 psi / 1.1 Dmin = 1347 ft 75
Casing Design Example cont’d • Continue with next cheapest Casing K-55, 45.50 lb/ft, Burst Strength 3,580 psi Minimum depth that can go: pab,D = 3,546 – 3,546 – 0.52 0.52 x D = 3,580 psi / 1.1 Dmin = 561 ft
76
38
Casing Design Example cont’d • Continue with next cheapest Casing K-55, 51.00 lb/ft, Burst Strength 4,030 psi Minimum depth that can go: pab,D = 3,546 – 3,546 – 0.52 0.52 x D = 4,030 psi / 1.1 Dmin = -226 ft (above surface)
77
Casing Design Example cont’d Burst Diagram 0 ft 103 /4 K-55 51.00 lb/ft 561 ft 103 /4 K-55 45.50 lb/ft 1347 ft
103 /4 K-55 40.50 lb/ft
4000 ft
78
39
Casing Design Example cont’d Design for Collapse (uniaxial) • Start at top (minimum collapse pressure) pac = 0.52 x D Cheapest casing: K-55, 40.50 lb/ft, Collapse Strength 1,580 psi Maximum depth that can go: pac,D = 0.52 x D = 1,580 psi / 1.0 Dmax = 3,038 ft 79
Casing Design Example cont’d • Continue with next cheapest Casing K-55, 45.50 lb/ft, Collapse Strength 2,090 psi
Maximum depth that can go: pac,D = 0.52 x D = 2,090 psi / 1.0 Dmax = 4.019 ft
80
40
Casing Design Example cont’d Collapse Diagram 0 ft
103 /4 K-55 40.50 lb/ft
3038 ft 103 /4 K-55 45.50 lb/ft 81
4000 ft
Casing Design Example cont’d Combine Two Diagrams 0 ft
103 /4 K-55 51.00 lb/ft
561 ft 103 /4 K-55 45.50 lb/ft 1347 ft
+ 3038 ft
=
burst 103 /4 K-55 40.50 lb/ft collapse 103 /4 K-55 45.50 lb/ft
4000 ft
82
41
