Situation: NoProfit Drilling Company is drilling 100 bbl of hole daily in a formation with 15% porosity. For four consecutive days, 400 bbl of discards and fluid were captured each day in discard tanks. The pit levels remained constant, but some drilling fluid was jetted to the reserve pits daily to keep the pits from overflowing. The unweighted drilling fluid weighed 9.4 ppg daily and contained 2%vol bentonite

Since no barite is contained in the drilling fluid, the volume of lowgravity solids (VLG) is the same as the volume of total suspended solids (Vs). Assume SG of the low-gravity solids to be 2.6.

The equation for determining VLG is:

*VLG = 62.5 + 2.0×Vs-7.5(MW)*

where MW=mud weight. Or, Vs=VLG=7.5(MW)−62.5=8%vol.

Part of this 8%vol low-gravity solids concentration was 2%vol bentonite, so the content of low-gravity drilled solids was 6%vol.

The pit levels decrease by the quantity of material removed from them. If no fluid or solids are removed from the system, the pit levels remain constant (except for the increase in volume of the drill pipe entering the hole). In this case, the volume decrease is 400 bbl daily, or 1600 bbl. This must also be the total volume of clean mud added to the system if the pit levels are returned to their original position.

The %vol drilled solids in the mud remained at 6% daily. So, the drilled solids retained must be 6% of the volume of the newly built mud, or volume of retained solids = (0.06)(volume of new mud built).

The new drilling fluid built daily comprises the clean mud added and the drilled solids that remain in the drilling fluid after it has been circulated through the solids-removal equipment.

*volume of new drilling fluid built in the drilling fluid system*

*= drilled solids retained + clean drilling fluid added.*

The quantity of drilled solids retained can be substituted into that equation, resulting in:

*volume drilling fluid mud built*

*=(0:06)(volume new drilling fluid built) + clean drilling fluid.*

The volume of clean drilling fluid must be exactly the volume that was discarded, or 1600 bbl. This gives an equation with one unknown:

*volume of new mud built=0.06×volume of new mud built= 1600 /(1-0.06) = 1702 bbl.*

Since the drilled solids retained are 6% of the volume of the new mud built, the drilled solids retained=0.08×1702 bbl=136 bbl.

With 15% porosity, the 400 bbl drilled resulted in the addition of 340 bbl of solids to the system. If 136 bbl of drilled solids were retained, 204 bbl were discarded. This gives a ratio of 204 bbl / 340 bbl, or 0.60. The SOLIDS REMOVAL EQUIPMENT EFFICIENCY is 60%.

For comparison, the procedure in API RP 13C used to calculate the same SOLIDS REMOVAL EQUIPMENT EFFICIENCY is presented in the Appendix.

## Excess Drilling Fluid Built

Normal discards from fine screens on linear motion shale shakers and from hydrocyclones contain about 35%vol solids, as shown in Figure 6. If all of these are drilled solids, the volume of drilling fluid discarded with the drilled solids can be calculated. The statement that the volume of discarded solids is equal to 35% of the discarded volume can be written:

*discarded drilled solids=0.35×(volume of fluid discarded with drilled solids)*

*204 bbl=0.35×(volume of tatol fluid discarded with drilled solids)*

volume discarded with drilled solids= 583 bbl.

The 583 bbl of waste discard would contain 204 bbl of drilled solids and

379 bbl of drilling fluid.

Since a total of 1600 bbl were discarded, 1017 bbl of good drilling fluid were removed from the system along with the 583 bbl of waste drilled solids and drilling fluid. The 1017 bbl of good drilling fluid was the excess clean fluid added to dilute the retained drilled solids and could be pumped to a storage pit. However, eventually this excess drilling fluid must go to disposal. Ideally, the amount of clean drilling fluid added to the system will be exactly the volume discarded with the drilled solids (583 bbl).

Consider a case in which the discard from the hydrocyclones and the shale shaker was very wet (meaning that a large volume of liquid was discarded with the drilled solids), so that the discard contained 20%vol drilled solids instead of 35%vol.

*discarded drilled solids=0.20(volume of fluid discarded with drilled solids)*

*50.5 bbl = (0:20)(volume of fluid discarded with drilled solids)*

or volume of fluid discarded with drilled solids=253 bbl. Again, since a total of 400 bbl were discarded daily, 400 bbl – 253 bbl, or 147 bbl, of good drilling fluid were also discarded.

A conclusion should be obvious at this point. Efforts to eliminate all dripping of drilling fluid from the end of a shale shaker are futile and not needed when the SOLIDS REMOVAL EQUIPMENT EFFICIENCY is around 60%. This drilling fluid will need to be discarded eventually, and shale shakers do a better separation when the effluent is still wet. This will usually point the way toward using shaker screens with smaller openings. Coarse screens allow more drilled solids to pass through.

## One Reply to “SOLIDS REMOVAL EQUIPMENT EFFICIENCY IN AN UNWEIGHTED DRILLING FLUID”

Comments are closed.