After drilling 1000 ft of hole with a 12.5-lb / gal drilling fluid circulated at 25 bbl / min, the hole was circulated clean. This required four hole volumes to eliminate all solids in the discard. Assuming that the formation averaged about 13%vol porosity, a multiarmed caliper indicated

that a volume of 97.3 bbl of new hole was drilled. The drilling fluid was freshwater-based mud weighted with barite and contained 2%vol bentonite, no oil, and 5%vol drilled solids. While drilling this interval, 1350 sacks (sx) of barite (100 lb / sx) were added to the system, and the drilled solids remaining in the system were diluted as required to control their concentration at the targeted 5%vol. Some drilling fluid was pumped to the reserve pits, and all discards of the solids-control equipment were captured in a container to be shipped back to shore. One drilling-fluid technician reported that 200 bbl were hauled to shore, and another reported that 180 bbl were captured.

These data represent information available in most field operations. Certainly, knowledge of the volume of solids reaching the surface is necessary for any calculation. These data will be assumed to be available from a hole caliper and circulation of the hole clean after drilling a particular interval. But it is one of the most difficult parameters to determine. Solids do not report to the surface in the same order that they were drilled, nor do they report in a predictable period of time. The preceding problem was deliberately set up to remove all of the drilled interval. Next, it is needed to know

- the volume of clean drilling fluid added to the active system to dilute the cuttings from the hole.
- the volume of clean drilling fluid added to the active system to dilute the drilled solids remaining in the system.

With a weighted mud, the number of sacks of barite and an analysis of solids concentrations in the drilling fluid allow a calculation of the clean drilling fluid added. Similarly, if the liquid volume added (water, oil, or synthetics) is known, the volume of clean drilling fluid can be calculated. Finally, if all of the discard volumes are captured in a disposal tank or container, the volume of discard can be measured.

Calculation of Solids Removal Equipment Efficiency does not require knowledge of the volume of the circulating system if the other information is available. The system has reached a stable drilled solids concentration, and the changes to the system are the primary concern. In actual practice, the system is dynamic, with continuous additions of small amounts of drilling fluid ingredients and continuous discards from the solids-removal equipment. At the drilling rig, sand traps are dumped with a variety of quantities of good drilling fluid. For this reason, these calculations should involve a reasonably long interval of hole to include all of the solids reaching the surface.

## Solution

*Volume of New Drilling Fluid Built While Drilling the Interval*

Assuming a density (SG) of low-gravity solids of 2.6 g / cc and a barite density of 4.2 g / cc, the content of low-gravity solids can be calculated from the equation:

*VLG = 62.5 + 2 Vs -7.5(MW)*

where:

*VLG=content of low-gravity solids, %vol*

*Vs=total solids content, %vol*

*MW=mud weight, lb / gal.*

Rearranging to solve for total solids:

*VS=[VLG+7.5(MW)-62.5]⁄2.0*

For a 12.5-ppg drilling fluid containing 5%vol drilled solids and 2%vol

bentonite (VLG=7%vol), the total solids are 19.1%vol. Since the bentonite

and drilled solids account for 7%vol, the remaining 12.1%vol is

barite.

*A barrel of barite, SG 4.2, weighs 1470 lb:*

*(4.2)(8.34 lb/gal)(42 gal/bbl)/100%vol.*

The 1350 sx barite, or 13,500 lb, added during the drilling of this interval

are equivalent to 91.8 bbl of barite. Assume that theMWand drilled solids

were maintained at the stated levels during the drilling of the interval.

Stated in an equation:

*barite volume in the drilling fluid = (12.1%)(volume of new drilling fluid built).*

*91.8 bbl barite = (0.121)(volume of new drilling fluid built).*

The volume of new drilling fluid built is 759 bbl. The volume of drilled

solids would be 5% (759 bbl) or 38 bbl.

*Solids Removal Equipment Efficiency*

Solids Removal Equipment Efficiency is calculated from the ratio of the volume of drilled solids discarded (97.3 bbl−38.0 bbl) to the volume of drilled solids arriving at the surface (97.3 bbl).

*Solids Removal Equipment Efficiency = [59.3 bbl⁄97.3 bbl] = 61 %*

The volume of clean drilling fluid added to dilute the 38 bbl of retained drilled solids must be exactly the volume discarded if the pit levels remained constant.

The new drilling fluid built consists of the retained drilled solids and the clean drilling fluid added to dilute those solids to 5%vol. This is stated mathematically:

*new drilling fluid volume built = clean drilling fluid added **+ drilled solids remaining*

The volume of clean drilling fluid added would therefore be (759 bbl- 38 bbl), or 721 bbl.

*Volume of Drilling Fluid Created from Adding*

*the Clean Drilling Fluid*

The volume of clean drilling fluid added was 721 bbl (as calculated from the amount of barite added). The pit levels would decrease by the volume discarded. The discarded volume is 57.3 bbl of drilled solids and the associated drilling fluid. After this decrease, the pit levels increase by 721 bbl. This could have been added as water and a blend of ingredients or, as most common, as individual components during the drilling process.

The pit levels decrease only by the quantity of drilling fluid removed from the system. If nothing were removed, the pit levels would not change (except by the volume of drill pipe added to the system). The 97.3 bbl of drilled solids have been added to the system, but 97.3 bbl of new hole means that the pit levels stay constant. The pit levels will drop by the amount of fluid and solids removed from the pits and will rise by the volume of new material added to the system.

The volume of clean drilling fluid added was 721 bbl. The volume of material removed was either 200 bbl or 180 bbl, depending on which drilling-fluid technician is correct. This would indicate that a volume of 521 or 541 bbl of excess drilling fluid was built while drilling this interval.

As we previously noted in the case of unweighted drilling fluid,

- it is obviously futile to seek to prevent all drilling fluid from dripping from the end of the shale shaker, nor is the attempt needed when Solids Removal Equipment Efficiency is around 60%;
- the drilling fluid will need to be discarded eventually, and shale shakers do a better separation when effluent is wet;
- wet effluent will usually point the way to using shaker screens with smaller openings. Coarse screens allow more drilled solids to pass through.

## Inaccuracy in Calculating Discard Volumes

The volume of discard was either 200 bbl or 180 bbl. In either case the discard tanks had to contain 59.3 bbl of drilled solids that came to the surface from the drilling operation. In one case, the drilled-solids concentration of newly drilled solids in the discard would be 59.3 bbl / 200 bbl, or 29.7%vol. In the other, it would be 59.3 bbl / 180 bbl, or 33.9%vol.

In the 200-bbl case, 141 bbl of good drilling fluid were discarded; and in the 180-bbl case, 121 bbl of good drilling fluid were discarded. In other words, the total difference was only 20 bbl of good drilling fluid. So either 521 or 541 bbl of good drilling fluid would be sent to storage. The difference was relatively insignificant. Note that the 20-bbl error did not in any way affect the calculation of the Solids Removal Equipment Efficiency.

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