Dilution refers to the process of adding a liquid phase to a drilling fluid to decrease the drilled-solids concentration. Dilution is used in several ways. If no solids-control equipment is used or if the equipment is used ineffectively, dilution may be the principal method of keeping drilled solids to a reasonably low level. This is an expensive solution to the problem. For example, to decrease drilled solids by 50% requires that 50% of the system be discarded and replaced with clean drilling fluid. Usually dilution is used after processing by solids-removal equipment to dilute drilled solids remaining in the drilling fluid. Dilution may be added as a clean drilling fluid or as the liquid phase of a drilling fluid with the other necessary drilling fluid ingredients, usually through a chemical barrel and a mud hopper. In this discussion, dilution will refer specifically to the clean drilling fluid necessary to decrease drilled-solids concentration. Clean drilling fluid is the liquid phase with all necessary additives such as barite, polymers, clay, etc.
As an example of dilution, consider a well in which the drilling-fluid specifications suggest that the volume percentage of (%vol) drilled-solids concentration should be, and is, 6%vol. Assume that 10 bbl of formation solids are brought to the surface and that no solids-removal equipment is used. All 10 bbl would be retained in the drilling-fluid system. These solids would require dilution to maintain the 6% volume concentration in the new drilling fluid.
The new drilling-fluid volume would consist of drilled solids (10 bbl) and clean drilling fluid (dilution). If the drilled solids (10 bbl) are 6% volume of the new drilling fluid built, the volume of new drilling fluid built would be 167 bbl (or 10 bbl / 0.06). The volume of clean drilling fluid (dilution) required may be calculated from the statement that the new drilling-fluid volume, 167 bbl, would consist of 10 bbl of drilled solids plus the clean drilling fluid. Obviously the dilution, or the clean drilling fluid, needed would be 157 bbl. This dilution would increase the pit volume by 157 bbl. When nothing is removed from the system, the pits would overflow if they were originally full. The only volume available to accept dilution would be the volume (liquid and solids) removed from the pit system. The volume removed would include all fluid and solids removed by the solids-removal equipment and any drilling fluid removed from the system to be stored or discarded. This is an important concept and is not trivial. Dilution calculations are based on simple material balances—addition and subtraction of volumes added and removed.
Before the dilution (clean drilling fluid) is added, the pit levels would remain the same as they were before the formation was drilled. The solids that were drilled occupy the same volume as they did before they were pulverized by the drill bit. The new hole volume is exactly the volume of material added to the drilling fluid system; consequently, drilling new hole does not change the pit levels. This, too, is an important concept and is the basis of determining the liquid levels in a drilling fluid system before and after using the solids-removal equipment.
Material added to the drilling-fluid system during drilling consists of solids and fluid contained within the formation. (Actually the pit levels would increase slightly by the volume of drill pipe steel added during the drilling of that interval and would decrease by the volume of gas released at the surface. The volume of gas released would be calculated on the basis of the bottom-hole volume that entered the drilling-fluid system. This factor will be ignored in the calculations; however, for purposes of this calculation, the drill pipe will be returned to its original position.)
For example, consider a 13.9-ppg freshwater-based drilling fluid with 2%vol bentonite and 25%vol total solids. Given the specific gravity (SG) of low-gravity solids as 2.6 and of barite as 4.2, the drilled solids would be 8%vol. This may be calculated from the equations presented in Chapter 3 on solids calculations. In the preceding problem, the new drilling fluid built would include 10 bbl of drilled solids consisting by volume of 17% barite, 2% bentonite, and 81% water. The dilution liquid would thus be 81% of the volume of the 167 bbl, or 135 bbl of water.
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