PEH:Tubing Selection, Design, and Installation - PetroWiki

08 Jul.,2024

 

PEH:Tubing Selection, Design, and Installation - PetroWiki


A design factor is the specific load rating divided by the specific anticipated load. A design factor less than 1.0 does not necessarily mean the product will fail, and neither does a design factor in excess of 1.0 mean that the product will not fail. As a result, design factors are generally selected on the basis of experience. The designer has the responsibility to select the design factors to suit particular needs and to reflect field experience. The condition of the tubing and the severity of a failure should have a significant effect on the design factors used. Design factors greater than 1.0 are recommended. Table 3.10 contains design factor guidelines.


The internal-yield pressure rating for tubing is based on an API variation of Barlow&#; s formula and incorporates a 0.875 factor that compensates for the 12.5% reduction tolerance in wall thickness allowed in manufacturing.

....................(3.2)

In general, these values should not be exceeded in operation. To be on the safe side, a minimum design factor of 1.25 based on the internal-yield pressure rating is suggested; however, some operators use different values.

In medium to high pressure wells, especially in sour service when L80, C90, and T95 API grades are used, the general stress level in the tubing should not exceed the minimum yield strength for L80 or the SSC threshold stress (generally 80% of the minimum yield strength) for C90 and T95 grades.

The joint or body yield strength for the tension design factor varies widely in practice. A simple approach is to assume a relatively high design factor of 1.6 based on the tubing weight in air and ignore other loading conditions. The calculations for loads in tension are usually for static conditions and ignore dynamic loads that may occur in running and pulling the tubing. They also may ignore collapse loads that reduce tension strengths. The pulling or drag loads are not commonly known. These may be relatively high in directional wells. Typically, the highest loads in tension occur in unsetting the packer during pulling operations. In some cases, shear pins in packers result in substantial loads in unsetting that should be accounted for in design.

The condition of the tubing after several years of service in the well is another unknown that needs to be compensated for either in design or by use of a higher tension design factor. When considering all these factors and making adjustments for drag, shear pins, and collapse pressures, a minimum design factor of 1.25 in tension for pulling is suggested. However, field experience has shown, in general, that tubing in new condition (meets API minimum requirements) can be loaded in tension to its minimum yield joint strength during pulling operations without a tension failure. Tension failures during pulling operations should be avoided because the results usually are costly. It is better to cut or back off the tubing rather than have a tension failure. Table 3.11 shows approximate setting depths for various API grades.


A collapse resistance design of 1.1 is suggested. Collapse resistance for tubing is covered in API Bull. 5C3. This standard provides conservative values for design, assuming the tubing cross section is not abnormally elliptical (oval). Any mechanical deformity in the tubing resulting in an out-of-round cross section may cause a considerable reduction in its collapse resistance. The collapse resistance value for a given tubing size, weight, and grade is based on numerous experimental tests and strength of material equations. The minimum value is designated as the API collapse resistance rating. Collapse ratings are reduced by tension loading. For example, a 23% yield stress in tension reduces the collapse resistance by approximately 14%. The biaxial effect should be used to design the tubing for critical tension and collapse conditions. Fig. 3.2 shows an ellipse of biaxial yield stress.

Tubing Design Considerations

A design factor is the specific load rating divided by the specific anticipated load. A design factor less than 1.0 does not necessarily mean the product will fail, and neither does a design factor in excess of 1.0 mean that the product will not fail. As a result, design factors are generally selected on the basis of experience. The designer has the responsibility to select the design factors to suit particular needs and to reflect field experience. The condition of the tubing and the severity of a failure should have a significant effect on the design factors used. Design factors greater than 1.0 are recommended.contains design factor guidelines.The internal-yield pressure rating for tubing is based on an API variation of Barlow&#; s formula and incorporates a 0.875 factor that compensates for the 12.5% reduction tolerance in wall thickness allowed in manufacturing. ....................(3.2)In general, these values should not be exceeded in operation. To be on the safe side, a minimum design factor of 1.25 based on the internal-yield pressure rating is suggested; however, some operators use different values.In medium to high pressure wells, especially in sour service when L80, C90, and T95 API grades are used, the general stress level in the tubing should not exceed the minimum yield strength for L80 or the SSC threshold stress (generally 80% of the minimum yield strength) for C90 and T95 grades.The joint or body yield strength for the tension design factor varies widely in practice. A simple approach is to assume a relatively high design factor of 1.6 based on the tubing weight in air and ignore other loading conditions. The calculations for loads in tension are usually for static conditions and ignore dynamic loads that may occur in running and pulling the tubing. They also may ignore collapse loads that reduce tension strengths. The pulling or drag loads are not commonly known. These may be relatively high in directional wells. Typically, the highest loads in tension occur in unsetting the packer during pulling operations. In some cases, shear pins in packers result in substantial loads in unsetting that should be accounted for in design.The condition of the tubing after several years of service in the well is another unknown that needs to be compensated for either in design or by use of a higher tension design factor. When considering all these factors and making adjustments for drag, shear pins, and collapse pressures, a minimum design factor of 1.25 in tension for pulling is suggested. However, field experience has shown, in general, that tubing in new condition (meets API minimum requirements) can be loaded in tension to its minimum yield joint strength during pulling operations without a tension failure. Tension failures during pulling operations should be avoided because the results usually are costly. It is better to cut or back off the tubing rather than have a tension failure.shows approximate setting depths for various API grades.A collapse resistance design of 1.1 is suggested. Collapse resistance for tubing is covered in APIThis standard provides conservative values for design, assuming the tubing cross section is not abnormally elliptical (oval). Any mechanical deformity in the tubing resulting in an out-of-round cross section may cause a considerable reduction in its collapse resistance. The collapse resistance value for a given tubing size, weight, and grade is based on numerous experimental tests and strength of material equations. The minimum value is designated as the API collapse resistance rating. Collapse ratings are reduced by tension loading. For example, a 23% yield stress in tension reduces the collapse resistance by approximately 14%. The biaxial effect should be used to design the tubing for critical tension and collapse conditions.shows an ellipse of biaxial yield stress.

Tubing string design must consider all reasonably anticipated loads imposed during running, producing, stimulation, workovers, and pulling operations. The design must ensure that failures will not occur under these operations; however, the designer typically selects the most economical weight and grade that meets the performance requirements. Computer software is available for tubing design, but the designer must ensure that all design conditions are met adequately.

A reasonable approach must be taken to prevent overdesign. The design need not prevent worst-case scenario failures but rather for all cases that have a reasonable probability of occurring. For instance, assume that there is a shallow tubing leak in which the shut-in tubing pressure is applied in the casing annulus on top of a column of heavy annulus fluid and, subsequently, that the tubing pressure at bottom is reduced quickly to a low value. This event would require tubing with a very high collapse pressure rating. If such a condition is considered to have a reasonable probability of occurring, the tubing string should be designed accordingly or adequate steps should be taken to prevent such a series of events.

The highest tensile loads normally occur at or near the top (surface) of the well. Collapse loads reduce the permitted tension loads, as shown by the biaxial graph in Fig. 3.2, and should be considered when applicable. Fortunately, the casing annulus pressure is normally low at the surface; thus, collapse pressure effects at the surface often can be ignored, but not in all cases. Buoyancy, which reduces the tensile loads, is sometimes ignored on shallow wells, but it should be considered on deeper wells. A condition that frequently determines the required tension yield strength of the tubing occurs when unsetting a partially stuck packer or using a shear-pin-release type packer in wells in which buoyancy is not applicable.

High-burst tubing loads typically occur near the surface with little or no annulus pressure under shut-in tubing conditions or during well stimulation treatments down the tubing. High-burst conditions also may occur deep in the hole with high surface pressures imposed on top of relatively high-density tubing fluid and when the annulus is empty or contains a light-density annulus fluid. Both of these conditions must be evaluated during the design of a tubing string for a specific well.

The burst resistance of the tube is increased because of tension loading up to a certain limit. In tubing- and casing-design practice, it is customary to apply the ellipse of plasticity only when a detrimental effect results. For a conservative design, this increase in burst resistance normally is ignored. Compression loads reduce burst resistance and must be considered when they occur. Such a condition can occur near the bottom of the well with a set-down packer and a relatively high internal tubing pressure and a relatively low annulus pressure. A typical design case in burst is to assume that the tubing is full of produced fluid and that the annulus is empty, which is a common situation for pumped wells.

Because tension loading reduces collapse resistance, the biaxial effect should be used to design for problem regions. A common practice in tubing design is to assume that the tubing is empty and that the annulus is full of fluid. Such conditions are common in low-pressure gas wells or oil wells that may be swabbed to bottom. Typically, the highest collapse pressures are near the bottom of the well. For combination tubing-string design, the collapse and tensile loads should be evaluated at the bottom and top of any tubing size, weight, or grade change.

In directional wells, the effect of the wellbore curvature and vertical deviation angle on the axial stress on the tubing body and couplings/joints must be considered in the tubing design. Current design practice considers the detrimental effects of tubing bending, but the favorable effect (friction while running) is neglected. Wall friction, which is unfavorable for upward pipe movement, generally is compensated for by addition of an acceptable overpull to the free-hanging axial tension. Overpull values are best obtained from field experience but can be calculated with available commercial software computer programs.

Single and Combination/Tapered Tubing Design

Many operators prefer one uniform weight (constant ID) and API grade tubing from top to bottom. Thus, it is not possible to mix different sections of the tubing during running or pulling operations throughout the life of the well. Most relatively shallow (< 9,000 ft), low-pressure (< 4,000 psi) wells have noncombination strings. As the pressures and depths increase, there comes a point at which a higher grade (stronger) or heavier weight (increased wall thickness) tubing must be used to meet load conditions and achieve acceptable design factors. For the same size diameter tubing, a higher grade normally is preferred over an increase in tubing weight. Such a choice is usually less expensive and maintains a constant internal diameter, which simplifies wireline operation inside the tubing.

Unlike casing design, which often has numerous grades and weights in a combination design, tubing design seldom has more than two different grades or weights. Such restriction may increase the cost of the tubing string but simplifies the running and pulling procedures. Deep and high-pressure wells may require more than two weights, grades, or diameters. When more than one grade or weight are used, each should be easily identifiable. To separate different weights and grades, a pup joint or different collar types may be used. For example, one section could use standard couplings and another could use beveled couplings. Painted and stenciled markings on the outside of the tubing are inadequate once the tubing is used because such markings are often obliterated.

The use of two or three different diameter sizes is sometimes advantageous. The larger tubing size may have high-joint-yield strength and permit a higher flow rate. The largest diameter is run on the top and a smaller tubing size on bottom. In such cases, the surface wellhead valves often are sized to permit wireline work in the larger tubing to prevent operational problems. A smaller tubing OD size on bottom may be necessary because of casing diameter restrictions.

Tubing Outside Diameter Limitations

The tubing OD must have adequate clearance with the casing ID. The tubing size selected should permit washover and fishing operations, in case the tubing becomes stuck and requires recovery. A wash pipe must be available that has an outside coupling dimension less than the casing drift diameter and an internal drift diameter that is greater than the tubing coupling OD plus provide a minimum of 1/8-in. clearance for adequate circulation. Also, the tubing OD should permit use of an overshot inside the casing, which limits the tubing OD size and/or the coupling OD. For example, 3 1/2-in. OD tubing with regular API EUE couplings (OD = 4.500 in.) inside 5 1/2-in., 17.00 casing (drift diameter = 4.767 in.) could not be washed over with available wash pipe. Even 3 1/2-in. specialty joint tubing with a joint OD of 3.875 in. would be an impractical, risky washover operation because the couplings would require milling. Nevertheless, special circumstances may require special proprietary tubing in close tolerance applications. Special wash-pipe sizes often can be rented from the tool service companies. The tubing designer should check the success of washover and fishing operations for their particular planned condition and the area of operation.

Multicompletions with parallel tubing strings often result in limiting the tubing and/or coupling size. If two tubing strings are to be run and pulled independently, the sum of the tubing coupling ODs should be less than the casing drift diameter. For example, inside 7-29.00 casing with a drift diameter of 6.059 in., parallel 2 3/8-in. tubing strings with EUE couplings may be planned. In such a case, beveled and special-clearance couplings with an OD of 2.910 in. typically are used. The sum of the two ODs is 5.82 in. Experience shows that if the couplings are beveled (top and bottom), these strings can be run and pulled independently. The auxiliary tubing equipment such as gas lift mandrels and safety valves often cause more clearance problems than the tubing couplings.

If two tubing strings are to be run clamped together, then the sum of the smaller tubing body OD and the OD of the coupling of the second or larger string must be less than the casing drift diameter. In these cases, a full-size drawing of the cross sections of the tubulars used may be helpful. The actual clearance may depend on the clamp design. The use of parallel strings of 3 1/2-in. tubing inside 9 5/8-in. casing is another common practice, and tubing OD limitations must be considered in such installations.

API Minimum Performance Properties of Tubing

Tubing performance properties are found in API Bull. 5C2, [6] and the formulas used in the following examples are found in API Bull. 5C3.[7] API Bull. 5C2, includes tables showing minimum tubing performance properties. [Tables shown in printed volume, but removed here because API did not provide permission for their use in PetroWiki.]

Example 3.1
Design a tubing string for a 9,000-ft hydropressured vertical well that is relatively straight, that will be used to flow 500 BOPD, and that will be completed inside 4 1/2-11.60-K55 casing. The well is to be completed with compression-set type packer and 9.0 ppg inhibited salt water in annulus. An overpull to free the packer of 15,000 lbf is anticipated. A maximum surface-treating pressure of 3,000 psi is expected.

Solution
Select 2 3/8-4.70-J55 EUE tubing (see tables in API Bull. 5C2[6] ). The 2 3/8-in. size is suitable for the flowing rates (see the chapter on inflow in this section of the handbook), and larger EUE tubing sizes cannot be run and washed over inside this size and weight casing. Smaller OD sizes of tubing will save no significant investment and will complicate wireline work. Select the lightest standard weight available for the initial design and check to ensure that it meets all design conditions. The J55 grade is the most cost-effective grade available. It typically is used as a first selection for most relatively shallow, low-pressure, and low-rate design cases.

Calculate the fluid gradient, gf.

....................(3.3)

0.052 psi/ft/lbm/gal is obtained from 0.433 psi/ft/8.32 lbm/gal, which is the conversion factor from lb/gal to psi/ft.

Check design conditions for tension. Calculate the resulting hook load for a 9,000-ft length of tubing in air from

....................(3.4)

The value of wn is obtained from Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]. This calculation results in a superimposed tubing tension axial (hook) load at the surface in air of 42,300 lbf.

The weight of the tubing string in a fluid is the tubing weight in air minus the axial buoyancy load(s):

....................(3.5)

The results of the tubing cross-section metal area,

....................(3.6)

times the hydrostatic pressure at depth,

....................(3.7)

are used to calculate the axial buoyancy load,

....................(3.8)

In this example,



Eq. 3.5 can now be used to calculate the hook load in fluid at surface before setting the packer.



Compare these values to the tubing performance properties. With a joint-yield strength rating, Fj, of 71,700 lbf (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]) the design factor in tension in air is

....................(3.9)

which is an acceptable design factor in tension in air, whereas the design factor in fluid is

....................(3.10)

which is an acceptable design factor in tubing considering buoyancy.

Consider pulling conditions. With a stuck packer requiring 15,000 lbf of overpull, Fop, at packer to free, assume no buoyancy contribution because the packer is stuck.

....................(3.11)

The design factor when considering overpull is

....................(3.12)

which is an acceptable design factor in tension during pulling operations.

An overpull any greater than 15,000 lbf would not be acceptable because Dt would be less than 1.25.

Check burst and collapse loads and compare to the tubing performance properties. The maximum allowed internal pressure differential is

....................(3.13)

With an internal-yield burst-pressure rating, pyi, of 7,700 psi (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]) and a wellhead surface pressure, pwh, of 3,000 psi, calculate the design factor in burst.

....................(3.14)

which is an acceptable design factor in burst and is much higher than the 1.25 suggested.

The minimum collapse pressure without axial stress, pcr, = 8,100 psi (See Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]). Assume an annulus full of 9.0 ppg fluid and an empty tubing string. With Eq. 3.7, pbh = 9,000 ft × (9.0 × 0.052) psi/ft = 4,212 psi.

....................(3.15)

which is an acceptable design factor in collapse and is much higher than the 1.1 suggested.

Check burst at bottom of hole under pumping conditions. Assume tubing filled with 9.0 ppg salt water with 100 psi surface tubing pressure and empty annulus.

....................(3.16)

....................(3.17)

which is an acceptable design factor in burst.

Select and order tubing material. Order per API Spec. 5CT : 9,000 ft plus 300 ft of 2 3/8;-4.70-J55 EUE-8R, range 2, seamless or electric weld, and one set of pups with standard EUE couplings. In addition, order one container of API-modified thread compound and specify delivery date and shipping instructions.

Example 3.2
Design tubing for relatively deep high-pressure gas well with CO2 and H2S. Assume the following conditions: casing designation = 5 1/2-23.00-L80; measured depth, Dm, = 14,000 ft; true vertical depth, DtV, = 13,000 ft; gas rate = 15 MMcf/D, 10 bbl of condensate per MMcf, 40 ppm hydrogen sulfide resulting in a partial pressure of 0.40 psi for the H2S and a 2% (20,000 ppm) carbon dioxide; pwh = 10,000 psi during stimulation; pbh = 9,000 psi; Tbh = 250°F; Tsf = 125°F; completion fluid weight = 14.0 ppg of inhibited solids free salt water; fluid gradient = 0.728 psi/ft; anticipated drag on tubing when pulling = 5,000 lbf; and packer shear pins setting = overpull = 25,000 lbf.

Solution
Because of the anticipated rate of 15 MMcf/D, 2 7/8-in. tubing will permit flow at a significantly higher rate than 2 3/8-in. tubing. The use of 3 1/2-in. tubing is not normally recommended within 5 1/2-in. casing because fishing operations would be difficult. On the basis of experience, the use of 3 1/2-in. tubing rather than 2 7/8-in. tubing would not significantly improve the production rate in this case.

Select the tubing weight and grade. Because surface pressures of 10,000 psi are anticipated, the tubing must have a minimum internal-yield pressure greater than 10,000 psi. With a design factor of 1.25 in burst, the required minimum internal-yield pressure is 12,500 psi (1.25 × 10,000). Because the partial pressure of H2S is 0.40 psi (greater than 0.05 psi), a sour service tubing grade must be used. See NACE MR-01-75. [16]

The obvious choice in the design is 2 7/8-7.90-L80 tubing with an inside diameter (ID) of 2.323 in. (For 2 7/8-in. tubing, the lightest weight of 6.5 lbm/ft for J55 and L80 grades do not have an adequate internal-yield pressure rating.) See Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]. The 2 7/8-7.90-L80 tubing has a 13,440 psi internal-yield pressure value, which is more than adequate. Because of the high gas pressure, a proprietary connection joint with 100% joint strength and with metal-to-metal seals should be considered.

Investigate tension load conditions. Use Eq. 3.4 to calculate Fa = Lp × wn = 14,000 ft × 7.9 lbm/ft = 110,600 lbf. Use Eq. 3.7 to find the hydrostatic pressure at depth, pbh = 13,000 × 14 × 0.052 = 9,464 psi. Use Eq. 3.8 to calculate the buoyancy effect in 14 ppg fluid, Fb = Am × pbh = 2.254 in. 2 × 9,464 psi = 21,332 lbf. Use Eq. 3.5 to calculate Ff = Fa &#; Fb = 110,600 lbf &#; 21,332 lbf = 89,269 lbf. With Fj = 180,300 lbf, use Eqs. 3.9 and 3.10 to calculate Dt = Fj/Fa = 180,300/110,600 = 1.63, which is an acceptable design factor in tension in air, and Dt = Fj/Ff = 180,300/89,269 = 2.02, which is an acceptable design factor in tension considering buoyancy.

Consider pulling conditions. Buoyancy is neglected because the packer is set.

....................(3.18)

This is the required hook load to unset the packer. Use Eq. 3.12 to calculate Df = Fj/Ps = 180,300/140,600 = 1.28, which is an acceptable design factor in tension.

Check collapse conditions. pcr = 13,890 psi for 2 7/8-7.9-L80 tubing (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]). Assume the casing annulus is filled with 14 ppg fluid with no surface pressure and the tubing pressure is bled off after a plug was set in the bottom of the tubing or a tubing safety valve at bottom is closed, which is a reasonable possibility over the life of the well. Use Eq. 3.7 to calculate pbh = DtV × gf = 13,000 ft × (14.0 × 0.052) psi/ft = 9,464 psi, and use Eq. 3.15 to calculate Df = pcr / pbh = 13,890/9,464 = 1.47, which is an acceptable value. Ensure that the surface annulus pressure is kept less than 3,163 psi [ (13,890/1.1) &#; 9,464] in the event that the tubing pressure is bled off.

Select and order the tubing material. Request that the tubing meet API Spec. 5CT. Order 14,500 ft of 2 7/8-7.90-L80 Type 13 Cr, Range 2, seamless tubing with a proprietary connection and one set of pup joints with same type connections as tubing. In addition, order all accessories with the same connection and an appropriate thread lubricant. State the required delivery and follow API RP 5C1 on tubing handling.

Example 3.3
Design tubing for a relatively deep sweet-oil well. Make a dual grade tubing-string design (to reduce cost). Assume the following conditions: casing designation = 7-26.00-N80; Dm = 11,000 ft; DtV = 11,000 ft; desired flow rate under gas lift conditions = 1,500 B/D from 10,000 ft; pww = 5,000 psi; pwh = 5,000 psi; pbh = 6,200 psi; Tbh = 200°F; Tsf = 125°F; completion fluid in annulus, wf, = 11.0 ppg of inhibited solids free salt water; fluid gradient, gf, = 0.052× wf = 0.572 psi/ft; packer shear pins setting = overpull (Fop) = 50,000 lbf. The well is relatively straight with small drag forces while pulling, and it is to be circulated with salt water before pulling tubing. Assume Fd = 0.

Solution
Select tubing size. Because of the anticipated flow rate, 3 1/2-in. tubing was selected. There is no clearance problem with 3 1/2-in. tubing inside the 7-in. casing. Smaller tubing sizes would result in high friction losses and loss in production rate. Larger tubing sizes would not increase production rates sufficiently and would result in clearance problems inside the 7-in. casing. EUE tubing with modified couplings (see API SR13 seal ring) are selected to provide adequate leak resistance.

Select tubing weight and grade. 3 1/2-9.30-J55 tubing is checked to determine if all design conditions are met.

Check collapse on bottom. pcr = 7,400 psi (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6]) for the selected tubing. Assume worst/maximum collapse design condition occurs at bottom where annulus is full of 11 ppg fluid and tubing pressure is zero (possible under completion conditions if well is swabbed down.) Use Eq. 3.7 to calculate pbh = DtV × gf = 11,000 ft × 0.572 = 6,292, and use Eq. 3.15 to calculate Dc = pcr / pbh = 7,400/6,292 = 1.176, which is adequate because 1.1 is acceptable.

Check burst at bottom. Assume casing annulus is empty and tubing is full of produced water. This is possible under gas lift conditions if the annulus injection pressure is bled off with tubing full of produced fluid plus surface wellhead pressure. Use Eq. 3.16 to calculate the burst pressure on bottom, 11,000 × 0.465 + 100 = 5,115 + 100 = 5,215 psi. With an internal-yield pressure for 3 1/2-9.30-J55 of 6,980 psi, use Eq. 3.17 to calculate 6,980/5,215 = 1.34, which is adequate because 1.25 is acceptable. With a maximum stimulation burst pressure at surface of 5,000 psi, use Eq. 3.14 to calculate Db = 6,980/5,000 = 1.396, which is adequate for burst.

Check tension loads at surface. For 3 1/2-9.30-J55 or N80 tubing, use Eq. 3.4 to calculate 11,000 ft × 9.3 lb/ft = 102,300 lbf. Use Eqs. 3.7 and 3.8 to calculate the axial buoyancy load, Fb = 2.590 in.2 × (11,000 × 11.0 × 0.052) psi = 16,296 lbf. Use Eq. 3.5 to calculate the weight in 11 ppg fluid, 102,300 &#; 16,296 = 86,004 lbf. For 3 1/2-9.30-J55 EUE tubing (100% joint efficiency), Fj = 142,500 lbf. Use Eq. 3.9 to calculate the design factor in tension, Dt, for 3 1/2-9.30-J55 EUE tubing in air, 142,500/102,300 = 1.39, which does not account for necessary overpull. The recommended design factor for weight in air is 1.6; therefore, the design factor is not adequate. A higher grade at top must be used for adequate tension design conditions.

Check worst possible tension design case. Pull at surface to overcome drag and shear pins in packer with no buoyancy effect on tubing above packer. Use Eq. 3.4 to calculate Fa, and use Eq. 3.18 to calculate Ft = 11,000 × 9.3 + 50,000 + 0 = 102,300 + 50,000 = 152,300 lbf. Use

....................(3.19)

to calculate 152,300 × 1.25 = 190,375 lbf. Use Minimum Performance Properties of Tubing tables in API Bull. 5C2 [6] to find Fj = 207,200 lbf for 3 1/2-9.30-N80 tubing, which is acceptable. Suggest the use of as much J55 as feasible to reduce tubing string cost. For maximum pull load on 3 1/2-9.3-J55, applying the acceptable design factor = 142,500/1.25 = 114,000 lbf. Calculate the maximum feet of 3 1/2-9.30-J55 from

....................(3.20)

Assume Lp = 6,800 ft for 3 1/2-9.30-J55, and Lp = 11,000 &#; 6,800 = 4,200 ft for 3 1/2-9.30-N80 tubing. Use Eq. 3.12 to calculate the design factor in tension, Dt, for 3 1/2-9.30-N80, 207,200/152,300 = 1.36, which is acceptable. For Fa = 152,300 lbf, the design factor for 3 1/2-9.30-J55 can be calculated as 142,500/(152,300 &#; 4,200 × 9.3) = 1.26. Do not exceed the 50,000-lbf overpull load, because this would over load the top of the J55 tubing.

Select and order tubing material. Request that tubing meet API Spec. 5CT. Order 4,400 ft of 3 1/2-9.30-N80 with EUE modified API SR13 beveled couplings and S or EW, range 2; one set of pup joints for 3 1/2-9.30-N80 EUE modified API SR 13 standard couplings; 7,000 ft of 3 1/2-9.30-J55 with EUE modified API SR 13 standard couplings and S or EW, range 2; and one container of API modified thread compound as per API RP 5A3. Specify delivery date and shipping instructions. Some operators might prefer to use L80 rather than N80 3 1/2 tubing and to heat-treat the J55 after upsetting. Both these options increase the cost of the tubing string but may increase the operating life.

Example 3.4
Design tubing for a deep high-pressure gas well. Complete the well with 7-29.00-P110 casing to 13,900 ft and a 5-in. liner (4.031 in. ID) from 13,800 to 16,650 ft. Perforations are to be from 16,530 to 16,570 ft with a permanent packer at 16,500 ft. The bottomhole pressure is estimated to be 14,850 psi with a bottomhole temperature of 340°F and a surface-flowing temperature of 150°F. The well has a surface shut-in pressure of 12,445 psi with a gas gradient, gg, of 0.146 psi/ft. The well initially will produce approximately 10 MMcf/D of gas with a 10 BC/MMcf and 10 BW/MMcf into a 1,000-psia sales system. The gas gravity is 0.7 and contains 1% of nitrogen and 1% carbon dioxide, but the H2S is only 1 ppm. The formation may require acid stimulation with a maximum surface-treating pressure of 10,000 psi. Before perforating, the 17.4 ppg mud will be circulated out and replaced with 10 ppg clean inhibited salt water. After perforating, the well will be killed, the packer and tubing installed, and the annulus filled with 10 ppg clean inhibited salt water. If needed, batch inhibition is planned to protect the tubing from erosion/corrosion.

Solution
Select tubing sizes. The type of completion and the size of the tubing string must be selected before making the tubing design. Fig 3.3 shows an inflow performance and outflow performance graph comparing the production with 2 3/8-, 2 7/8-, and 3 1/2-in. tubing strings. This graph shows that a full string of 2 3/8-in. tubing would restrict production significantly; thus, the amount of 2 3/8-in. tubing should be limited. The 2 7/8-in. tubing produces the well near its maximum rate, whereas the use of 3 1/2-in. tubing results in only a small production rate increase and will cost substantially more. The 5-in. liner (4.031 in. ID) will make washover and fishing 2 7/8-in. tubing difficult; therefore, 2 3/8-in. tubing will be used in the liner section of the well. Thus, the top portion of the tubing string will be 2 7/8-in. tubing and the lower portion inside the liner will be 2 3/8-in. tubing.


Now that the approximate sizes of tubing have been determined, the tubing design can be made for tension, collapse, and burst conditions. In general, select the lowest weight per foot and grade that is acceptable. This will normally result in the most economical design.

Select weights and grades. The most common approach in casing and tubing design is to start at the bottom and work your way back to the surface; however, in this high-pressure well, burst is a major consideration. Draw a pressure-depth graph as shown in Fig. 3.4.


To control the shut-in surface tubing pressure of 12,445 psi with a design factor of 1.25, calculate the suggested minimum internal-yield pressure rating required, pyi = 12,445 × 1.25 or 15,556 psi. As the Minimum Performance Properties of Tubing tables in API Bull. 5C2 show, 2 7/8-7.90-P110 is suitable, which has an internal yield of 18,480 psi. API grades C90 and T95 could also be used, but these grades are usually more costly than P110. Because the H2S partial pressure is less than 0.05 psi, the nonsour service grade N80 and P110 can be used.

Because tension reduces the collapse rating and collapse reduces the tension rating, start at the bottom where tension is small and collapse is normally high. Actually, at the bottom (because of buoyancy forces), the tubing is in compression when run in fluid. Draw a schematic tubing depth chart as shown in Fig. 3.5.


Check collapse and tension stresses. Start at the bottom of the hole and work to the surface-checking tension and collapse at any size, weight, or grade change. The tensile load increases moving upward, but the collapse differential pressure decreases.

To calculate the collapse differential, use

....................(3.21)

With the annulus full of 10 ppg salt water and assuming that the tubing pressure bled to zero, a 0.52 × 16,500 = 8,580 psi collapse differential would result on the bottom of the hole. The 23/8;-4.70-N80 tubing has a collapse of 11,780 psi, resulting in a design factor 11,780/8,580 = 1.37, which is acceptable. Keep the annulus pressure at the surface to a maximum of 1,500 psi in normal operations to avoid possible collapse if the tubing pressure at bottom is bled down to zero.

From above the top of the liner at 13,800 ft to the permanent packer at 16,500 ft, 2,700 ft of 2 3/8-4.70-N80 tubing is tentatively selected. Use 2,800 ft of 2 3/8-in. tubing to avoid interference with the liner top. At 13,700 ft, the tubing size can be increased safely to 2 7/8 in., which will allow a higher flow rate. To simplify wireline operations, the tubing weight for all 2 7/8-in. tubing is the same.

For burst considerations, the design requires a minimum of 7.9 lbm/ft tubing. There is a &#;11,188 buoyancy force because of the fluid acting on the bottom tubing area. At 13,700 ft, the 2 3/8-in. tubing will have a load of 4.7 lbm/ft × 2,800 ft = 13,160 lbf; however, the tensile load on the tubing is altered slightly because of the tubing area change at 13,700 ft. This results in an axial load at 13,700 ft of 13,160 &#; 11,188 + 7,929 &#; 14,690 = &#;4,789 lbf; thus, the effect of tension on collapse can be neglected because the tubing is in compression.

The maximum burst pressure on bottom may occur during stimulation. Calculate the burst differential from

....................(3.22)

Assuming a surface-treating pressure of 10,000 psi, the tubing full of acid (gradient = 0.45 psi/ft), and the annulus full of 10 ppg (gradient = 0.52 psi/ft) salt water, use Eq. 3.22 to calculate a burst differential on bottom of 10,000 + 0.45 × 16,500 &#; 0.52 × 16,500 = 8,845 psi. The use of a design factor of 1.25 in burst will require an internal-yield pressure of 8,845 × 1.25 = 11,056 psi. The 2 3/8-4.70-N80 tubing has an API internal yield of 11,200 psi (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 ), which is acceptable.

Burst and collapse conditions now need to be checked at all depths where tubing size, weight, or grade changes are planned. Burst is of primary importance. Check burst at the changed over from 2 3/8 in. to 2 7/8 in. at 13,700 ft. Use Eq. 3.22 to calculate the tubing burst pressure differential during stimulation: 10,000 psi + 0.45 psi/ft × 13,700 ft &#; 0.52 psi/ft × 13,700 ft = 9,041 psi. The use of a design factor in burst, Db, of 1.25 would require a burst resistance rating of 9,041 psi × 1.25 = 11,301 psi. Thus, at this depth, the 2 3/8, 4.7, N80 tubing is acceptable and 2 7/8, 7.9, N80 tubing is acceptable because it has an API internal pressure rating of 13,440 psi (see Minimum Performance Properties of Tubing tables in API Bull. 5C2 .) and a collapse resistance of 13,890 psi. Using a design factor in burst, Db, of 1.25, the maximum burst differential for 2 7/8, 7.9, N80 should not exceed 13,440/1.25 = 10,752 psi.

At the surface during stimulation, 2 7/8-in., 7.9, P110 is required as shown previously. The depth of the crossover from P110 to N80 needs to be calculated. This depth is where the burst pressure differential is equal to 10,752 psi for 2 7/8-in., 7.9, N80 tubing. The worst case condition is during shut-in when a surface pressure of 12,445 psi occurs and the tubing is full of 0.146-psi/ft gas.

....................(3.23)

where ga is the annulus fluid gradient and gt is the tubing fluid gradient.

Using Eq. 3.23, Lp = (12,445 psi &#; 13,440/1.25 psi)/(0.52 psi/ft &#; 0.146 psi/ft) = 4,527 ft. Thus, 2 7/8-in., 7.9, P110 tubing is to be used from the surface to 4,527 ft and 2 7/8;-in., 7.9, N80 tubing is to be used from 4,527 to 13,700 ft. Table 3.13 summarizes the sizes, weights and grades selected.


Calculate the hook load of the tubing string in air and in fluid for the various tubing sizes, weights, and grades.

Design tubing for a deep high-pressure gas well. Complete the well with 7-29.00-P110 casing to 13,900 ft and a 5-in. liner (4.031 in. ID) from 13,800 to 16,650 ft. Perforations are to be from 16,530 to 16,570 ft with a permanent packer at 16,500 ft. The bottomhole pressure is estimated to be 14,850 psi with a bottomhole temperature of 340°F and a surface-flowing temperature of 150°F. The well has a surface shut-in pressure of 12,445 psi with a gas gradient,, of 0.146 psi/ft. The well initially will produce approximately 10 MMcf/D of gas with a 10 BC/MMcf and 10 BW/MMcf into a 1,000-psia sales system. The gas gravity is 0.7 and contains 1% of nitrogen and 1% carbon dioxide, but the HS is only 1 ppm. The formation may require acid stimulation with a maximum surface-treating pressure of 10,000 psi. Before perforating, the 17.4 ppg mud will be circulated out and replaced with 10 ppg clean inhibited salt water. After perforating, the well will be killed, the packer and tubing installed, and the annulus filled with 10 ppg clean inhibited salt water. If needed, batch inhibition is planned to protect the tubing from erosion/corrosion.Select tubing sizes. The type of completion and the size of the tubing string must be selected before making the tubing design.shows an inflow performance and outflow performance graph comparing the production with 2 3/8-, 2 7/8-, and 3 1/2-in. tubing strings. This graph shows that a full string of 2 3/8-in. tubing would restrict production significantly; thus, the amount of 2 3/8-in. tubing should be limited. The 2 7/8-in. tubing produces the well near its maximum rate, whereas the use of 3 1/2-in. tubing results in only a small production rate increase and will cost substantially more. The 5-in. liner (4.031 in. ID) will make washover and fishing 2 7/8-in. tubing difficult; therefore, 2 3/8-in. tubing will be used in the liner section of the well. Thus, the top portion of the tubing string will be 2 7/8-in. tubing and the lower portion inside the liner will be 2 3/8-in. tubing.Now that the approximate sizes of tubing have been determined, the tubing design can be made for tension, collapse, and burst conditions. In general, select the lowest weight per foot and grade that is acceptable. This will normally result in the most economical design.Select weights and grades. The most common approach in casing and tubing design is to start at the bottom and work your way back to the surface; however, in this high-pressure well, burst is a major consideration. Draw a pressure-depth graph as shown inTo control the shut-in surface tubing pressure of 12,445 psi with a design factor of 1.25, calculate the suggested minimum internal-yield pressure rating required,= 12,445 × 1.25 or 15,556 psi. As the Minimum Performance Properties of Tubing tables in APIshow, 2 7/8-7.90-P110 is suitable, which has an internal yield of 18,480 psi. API grades C90 and T95 could also be used, but these grades are usually more costly than P110. Because the HS partial pressure is less than 0.05 psi, the nonsour service grade N80 and P110 can be used.Because tension reduces the collapse rating and collapse reduces the tension rating, start at the bottom where tension is small and collapse is normally high. Actually, at the bottom (because of buoyancy forces), the tubing is in compression when run in fluid. Draw a schematic tubing depth chart as shown inCheck collapse and tension stresses. Start at the bottom of the hole and work to the surface-checking tension and collapse at any size, weight, or grade change. The tensile load increases moving upward, but the collapse differential pressure decreases.To calculate the collapse differential, use ....................(3.21)With the annulus full of 10 ppg salt water and assuming that the tubing pressure bled to zero, a 0.52 × 16,500 = 8,580 psi collapse differential would result on the bottom of the hole. The 2;-4.70-N80 tubing has a collapse of 11,780 psi, resulting in a design factor 11,780/8,580 = 1.37, which is acceptable. Keep the annulus pressure at the surface to a maximum of 1,500 psi in normal operations to avoid possible collapse if the tubing pressure at bottom is bled down to zero.From above the top of the liner at 13,800 ft to the permanent packer at 16,500 ft, 2,700 ft of 2 3/8-4.70-N80 tubing is tentatively selected. Use 2,800 ft of 2 3/8-in. tubing to avoid interference with the liner top. At 13,700 ft, the tubing size can be increased safely to 2 7/8 in., which will allow a higher flow rate. To simplify wireline operations, the tubing weight for all 2 7/8-in. tubing is the same.For burst considerations, the design requires a minimum of 7.9 lbm/ft tubing. There is a &#;11,188 buoyancy force because of the fluid acting on the bottom tubing area. At 13,700 ft, the 2 3/8-in. tubing will have a load of 4.7 lbm/ft × 2,800 ft = 13,160 lbf; however, the tensile load on the tubing is altered slightly because of the tubing area change at 13,700 ft. This results in an axial load at 13,700 ft of 13,160 &#; 11,188 + 7,929 &#; 14,690 = &#;4,789 lbf; thus, the effect of tension on collapse can be neglected because the tubing is in compression.The maximum burst pressure on bottom may occur during stimulation. Calculate the burst differential from ....................(3.22)Assuming a surface-treating pressure of 10,000 psi, the tubing full of acid (gradient = 0.45 psi/ft), and the annulus full of 10 ppg (gradient = 0.52 psi/ft) salt water, useto calculate a burst differential on bottom of 10,000 + 0.45 × 16,500 &#; 0.52 × 16,500 = 8,845 psi. The use of a design factor of 1.25 in burst will require an internal-yield pressure of 8,845 × 1.25 = 11,056 psi. The 2 3/8-4.70-N80 tubing has an API internal yield of 11,200 psi (see Minimum Performance Properties of Tubing tables in API), which is acceptable.Burst and collapse conditions now need to be checked at all depths where tubing size, weight, or grade changes are planned. Burst is of primary importance. Check burst at the changed over from 2 3/8 in. to 2 7/8 in. at 13,700 ft. Useto calculate the tubing burst pressure differential during stimulation: 10,000 psi + 0.45 psi/ft × 13,700 ft &#; 0.52 psi/ft × 13,700 ft = 9,041 psi. The use of a design factor in burst,, of 1.25 would require a burst resistance rating of 9,041 psi × 1.25 = 11,301 psi. Thus, at this depth, the 2 3/8, 4.7, N80 tubing is acceptable and 2 7/8, 7.9, N80 tubing is acceptable because it has an API internal pressure rating of 13,440 psi (see Minimum Performance Properties of Tubing tables in API.) and a collapse resistance of 13,890 psi. Using a design factor in burst,, of 1.25, the maximum burst differential for 2 7/8, 7.9, N80 should not exceed 13,440/1.25 = 10,752 psi.At the surface during stimulation, 2 7/8-in., 7.9, P110 is required as shown previously. The depth of the crossover from P110 to N80 needs to be calculated. This depth is where the burst pressure differential is equal to 10,752 psi for 2 7/8-in., 7.9, N80 tubing. The worst case condition is during shut-in when a surface pressure of 12,445 psi occurs and the tubing is full of 0.146-psi/ft gas. ....................(3.23)whereis the annulus fluid gradient andis the tubing fluid gradient.Using= (12,445 psi &#; 13,440/1.25 psi)/(0.52 psi/ft &#; 0.146 psi/ft) = 4,527 ft. Thus, 2 7/8-in., 7.9, P110 tubing is to be used from the surface to 4,527 ft and 2 7/8;-in., 7.9, N80 tubing is to be used from 4,527 to 13,700 ft.summarizes the sizes, weights and grades selected.Calculate the hook load of the tubing string in air and in fluid for the various tubing sizes, weights, and grades.

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Hook load in air = 13,700 ft × 7.9 lbm/ft + 2,800 ft × 4.7 lbm/ft = 121,390 lbf.

Hook load in fluid = 121,390 + 7,929 &#; &#; 11,188 = 103,441 lbf.

Body/joint yield strength for 2 7/8, 7.9, P110 = 247,900 lbf.

Body/joint yield strength for 2 7/8, 7.9, N80 = 180,300 lbf.

Body/joint yield strength for 2 3/8, 4.7, N80 = 104,300 lbf.

Maximum allowed hook load at surface for 2 7/8, 7.9, P110 tubing = 247,900/1.25 = 198,320 lbf.

Maximum allowed hook load at surface for 2 7/8, 7.9, N80 tubing = 4,527 × 7.9 + 180,300/1.25=180,003 lbf.

Maximum allowed hook load at surface for 2 3/8, 4.7, N80 tubing = 13,700 × 7.9 + 104,300/1.25= 191,670 lbf.

Thus, the limiting condition is for pulling on the 2 7/8, 7.9, N80 tubing, which allows a hook load at the surface of 180,003 lbf. For this string design, an overpull over the weight in fluid would be 180,003 lbf &#; 103,441 lbf = 76,562 lbf.

Select and order tubing material. Order the tubing to API 5CT specifications, adding a few hundred feet of each type: seamless, range 2, and a proprietary connection integral joint or threaded and coupled with metal-to-metal seals. Also, order a set of grade P110 pup joints for the 2 7/8-in. tubing with the same proprietary connection integral joint. Order an appropriate thread compound. In addition, one special crossover 2 7/8-7.90 to 2 3/8-4.70 in grade N80 is required. (If an MTC connection is used, the crossover can be a pin × pin with a 2 7/8-N80 coupling.)

All auxiliary well equipment should have the same proprietary connection. Tubing should be hydrostatically tested to 80% of yield pressure. Ensure that proper running procedures are used.

Check with the manufacturer on ways to distinguish between the two grades of 2.875-in. OD tubing. Some operators would select 2 7/8-7.90-P110 and no 2 7/8-7.90-N80 tubing to ensure that accidental mixing of the 2.875-in. OD different grade tubing could not occur and to allow a slightly higher overpull value.

If pressures are greater than 7,000 psi and the depth is greater than 13,000 ft, a pipe-body load analysis should be performed. In sour service for L80, C90, and T95, triaxial stress intensity should be checked and a design factor greater than 1.25 maintained. See ISO Sec. B.5.2. [15]

Stretch in Tubing

When tubing is subject to an axial load, either in tension or compression, that does not exceed the elastic limit of the material, the stretch or contraction may be determined from

....................(3.24)

where ΔLt = total axial stretch or contraction, in.; F = superimposed tension or compression axial load, lbf; Lp = length of pipe, ft; E = Young &#; s modulus of elasticity for steel = 30 million psi, which is not affected significantly by tubing grade; and Am = cross-section metal area of pipe, in.2 = 0. × (do2 &#; di2).

For multiple sizes or weights, calculate stretch for each section and sum the results. This formula also can be used to determine the length of free pipe by applying a load, F, and measuring the stretch, ΔLt.

....................(3.25)

Example 3.5
Find free point for a stuck string of 2 7/8-6.50 API steel tubing string in an 11,000-ft well.

Solution
With a block-hook load of 60,000 lbf, mark the tubing at the top of rotary table. An additional 10,000-lbf load was picked up and the measured increase in length (stretch) is 20.0 in. Calculate the tubing cross-section area with Eq. 3.6. Am = π × (2. &#; 2.)/4 = 1.812 in.2 Use Eq. 3.25 to calculate Lp = ΔLt × E × Am / (12 × F) = 20.0 in. × 30,000,000 psi × 1.812 in.2 /(12 in./ft × 10,000 lbf) = 9,060 ft.

Tubing Buckling

Tubing buckling must be considered in design. See the chapter on completion design in this section of the Handbook.

Corrosion Considerations

Tubing selection for corrosive environments is a critical design responsibility. Both the inside and outside of the tubing can be damaged by corrosion. Weight-loss corrosion may be a serious problem with conventional tubing strings in wells producing salt water, especially when the water becomes the wetting phase. Acidity caused by the presence of acid gases (CO2 and H2S) normally increases the corrosion rate. When corrosion is minor, the common practice is to use standard API grades and to start batch inhibition when corrosion becomes a problem.

Corrosion/erosion, a major problem with steel tubing, occurs in most high-rate gas-condensate wells in which the gas contains CO2. The CO2 attacks the steel tubing, which creates an iron carbonate film (corrosion product); it is removed from the wall by erosion (impingement of well fluids). Rapid deep pit failure may occur from corrosion/erosion. Increasing fluid velocities and CO2 partial pressure are highly detrimental, as are increasing temperature or increasing brine production. There may be a region of conditions in which frequent batch or continuous inhibition is necessary. Gas wells with CO2 contents higher than 30 psi partial pressure and gas velocities greater than 40 fps normally require continuous or frequent batch inhibition to protect the steel tubing. CRA material is often the most cost-effective means of combatting erosion/corrosion. Some CRA material is subject to failure in brine water environments.

A different type of tubing design problem is SSC. SSC and/or hydrogen embrittlement causes a brittle-type failure in susceptible materials at stresses less than the tubing yield strength. SSC is a cracking phenomenon encountered with high-strength steels in sour (H2S) aqueous environment. Cracking also occurs in austenitic stainless steels in caustic or chloride solutions and mild steel in caustic or nitrate solutions. Susceptibility to attack of most low-alloy steels is roughly proportional to its strength. In terms of hardness, most steels are not subject to SSC failure if the hardness is less than 241 Brinell Hardness number or 23 Hardness-Rockwell C. The potential harmful level of H2S for susceptible materials has been defined as 0.05 psi partial pressure of the H2S gas phase. Carbonate-induced cracking of mild steel can occur in freshwater environments.

Use of inhibition to prevent SSC is not completely reliable because 100% effective coverage of metal surface generally is not achieved. The best solution for tubulars subject to SSC is to use materials that are not subject to SSC failures. In general, follow NACE guidelines. [16]

Dissimilar metals close to each other can influence corrosion. Because corrosion can result from many causes and influences and can take different forms, no simple or universal remedy exists for its control. Each tubing well problem must be treated individually, and the solution must be attempted in light of known factors and operating conditions.

Internal Coatings

Plastic internal coating of a tubing string is sometimes used to deter corrosion or erosion/corrosion in oil and gas wells and may increase tubing life significantly. Such cases may be in high-water-cut oil wells or gas wells with high CO2 partial pressures. These coatings are usually thin wall film applications ( < 0.01 in. thick) that are baked (bonded) onto the inside walls of the tubing string. The film thickness is small enough to allow normal wireline operations. The key to plastic coatings is selecting the correct material and its proper application. Even if the specifications call for "100% holiday free," eventually the coating comes off and holidays occur because of poor application or handling practices, wireline work, caliper surveys, blisters caused by the environment, or other reasons. Coating should not be expected to stop all weight-loss corrosion over the life of the well. Typically, a few holes may develop in time but the bulk of the tubing stays intact. In such cases, workover costs are usually lowered because the tubing often can be retrieved without major fishing operations. Because such coatings increase the smoothness, they reduce pressure drop slightly in high-rate wells and, in some cases, may be helpful in reducing paraffin and scale problems. Besides thin wall film coatings, there are other kinds of interior coating or liners for tubing that have special application. Plastic liners and cement lining have been used successfully when the reduction in ID is not a major problem, primarily for water and carbon dioxide injection tubing or for sour service production.

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