7.3 Installation loads
7.3.8 Temperature load
Increased temperature causes the casing to increase in length, and the ability of the casing move to accommodate this change determines the resulting stresses. In the installation phase, for uncemented casing which is free to elongate, no additional stress will result. The amount of elongation is derived below.
FIGURE G-10 VARIATION WITH DEPTH OF AXIAL FORCE DUE TO A POINT LOAD APPLIED ABOVE A PACKER
If an element, of casing length dz, is subjected to a temperature increase ∆T, the element will expand by an amount given by;
α∆T dz
where α is the coefficient of linear expansion of the material. A value of 6.9 x 10-6 per degree Fahrenheit (1.24 x 10-5 per degree Centigrade) is appropriate for steel.
Since the change in temperature, ∆T , will generally be different for each element, dz, the above expression must be integrated over the entire length of the casing to find the total change in length.
If we consider a situation in a vertical well where the surface temperature is Ts, then the change in temperature at depth z will be given by;
∆T = (Ts + giz) - Ts = giz (G-20)
where gi is the thermal gradient of the hole into which the casing is run.
Integrating over the whole length L to find the length change ∆L
Example
For a 10,000 ft (3,048 m) string run into a hole with a thermal gradient of 0.02°F/ft, (0.036°C/m) then the change in length is (from Eq. G-21), in field units;
6.9 x 10-6 x 0.02 x 10,0002 7.3.9 Maximum installation load
For each point along the string, it must be confirmed that the casing can accommodate the maximum force that the string at that point will have seen during installation. This is best achieved by considering self-weight, buoyancy, and bending first, and then adding shock or drag loads later by the principle of superposition.
Consider a point X in a vertical casing string, an arbitrary distance y from the shoe. As the string is run, the axial force at point X is illustrated in Figure G-11. As described earlier, the axial force at any point is given by the weight in air of the casing below that point corrected for the pressure (buoyancy) load. As a string is run deeper into a well, and the hydrostatic pressure at the shoe increases, the pressure (buoyancy) load will increase. As a result, since the weight in air below X is constant, the axial force at point X will decrease. The maximum static axial force experienced by any point in the string during installation, therefore, is the force present when that point is at surface. If the maximum static axial force experienced by any point in the string during installation is plotted against its final as landed depth, line 4 results.
This holds for a vertical well. For deviated wells, the true vertical projection of the casing length should be used in all calculations.
In a deviated well, or vertical well with localised doglegs, all casing that has to pass through a dog leg must be designed to withstand the bending loads imposed. For each point in the string it is necessary to calculated the sum of the buoyant axial load and the bending load as that point passes through the dogleg. Since the bending load will be constant through the dog leg, the maximum combined load will coincide with the maximum buoyant axial load over the dogleg interval. This latter load will always be at the top of the dogleg interval, when the string is landed.
In Figure G-12, for a well which has a constant dog leg of 5°/100 ft (1.64°/10 m) below the kick-off point, line 5 represents the maximum installation axial force, i.e. buoyant weight plus bending experienced by each point along the casing referenced to its final depth. It can be seen that adding the bending force to line 4 overestimates the maximum load by a constant amount.
Similarly, adding the bending force to line 3, gives a line that corresponds to the axial force at each point in the casing once installed, but underestimates the maximum axial force experienced by each point in the casing during installation- except at the top of the build-up section.
FIGURE G-11: MAXIMUM STATIC INSTALLATION LOAD AS A FUNCTION OF DEPTH FOR A VERTICAL WELL
For a combination string, which will experience different bending loads and will have different tensile capacities along its length, it will be necessary to plot the maximum experienced load line when installing the string to ensure sufficient capacity is present at all depths. This is specially important for the running of liners through high build-up sections into straight inclined sections.
Example
Consider the following example as an illustration of the difference between installation axial loads and as landed axial loads.
A well has the profile shown in Figure G-13. It is necessary to examine the axial loads in a string of 9 5/8 in (0.2445 m) 47 lb/ft (69.9 kg/m) casing run in 0.65 psi/ft (14.7 kPa/m) mud to 8,000 ft TVD(2,438 m).
FIGURE G-12: MAXIMUM STATIC INSTALLATION LOAD AS A FUNCTION OF DEPTH FOR A DEVIATED WELL
FIGURE G-13: WELL PROFILE CORRESPONDING TO ACCOMPANYING EXAMPLE
The force due to bending is given, in field units, by Eq. G-10;
For φ = 5°/100 ft Fb = 218 x 9.625 x 5 x 13.57 = 142,366 lb For φ = 2°/100 ft Fb = 218 x 9.625 x 2 x 13.57 = 56,947 lb In SI units, using Eq. G-11;
For φ = 1.64°/10 m Fb = 183 x 106 x 0.2445 x 1.64 x 8.754 x 10-3 = 642,363 N For φ = 0.66°/10 m Fb = 183 x 106 x 0.2445 x 0.66 x 8.754 x 10-3 = 258,512 N
All casing that will eventually sit below 2000 ft TVD (610m) will have to pass through the 5°/100 ft (1.64°/10 m) dog leg and must be designed accordingly. It can be shown that the static forces experienced during installation (excluding drag and shock loads) are greater than those as landed, particularly for those sections that will eventually be below 4,000 ft TVD (1,219 m).
Consider such a point in the casing which, when landed, will be 2000 ft TV (610 m) above the shoe. As the casing is run, the maximum force experienced by that point will be as the point passes the kick-off depth. For the purpose of this example, assume that the point of interest, when at the kick-off depth, is 2400 ft TV (731 m) above the shoe. At that time, the axial force at that point is given by the weight in air of the string below that point, minus the buoyancy force. In field units, from Eq. App. 6-18;
Fa = (2400 x 47) - (4400 x 0.65 x 13.57) = 73,990 lb
The bending force due to the 5°/100 ft dogleg must be added to give a total force of 216,356 lb.
When that point is in the as landed condition the axial force due to the buoyant weight will be, from Eq. App. 6-18;
Fa = (2000 x 47) - (8000 x 0.65 x 13.57) = 23,436 lb
The bending force due to the 2°/100 ft dogleg must be added to give a total of 80,383 lb.
In SI units;
With the point of interest at the kick-off depth:
Fa = (731 x 9.8 x 69.9) - (1341 x 14.7 x 103 x 8.754 x 10-3 ) = 328,185 N
The bending force due to the 1.64°/10 in dogleg must be added to give a total force of 970,548 N.
When that point is in the as landed condition:
Fa = (610 x 9.8 x 69.9) - (2438 x 14.7 x l03 x 8.754 x 10-3 ) = 104,131 N
The bending force due to the 0.66°/10 in dogleg must be added to give a total force of 362,643 N.