# The Question and Answer of Solids Control

### Given:

An interval of 7000 feet of 12¼-inch hole is to be drilled, through Miocene sediments, below 13 3/8-inch casing set at 2000 feet with waterbased mud weighing 9.0 lb/gal. It is expected that the actual average hole diameter at the end of the interval will be 13.5 inches and that the interval will require six bits and 10 days to drill.

### Assume the following:

. The only solids present in appreciable concentrations are drilled solids and commercial bentonite.
. Commercial bentonite content is maintained at 2% by volume.
. Mud cost is \$10/bbl and liquid disposal costs are \$5/bbl.
. Formation porosity can be disregarded.
. Two hundred rotating hours will be required to drill the interval.
. Prior experience in the area indicates that the installed solids-control system (shale shakers, desanders, and desilters) can be expected to remove only 60% of the drilled solids.
. A centrifuge capable of processing 100 gpm of this mud is available at a rental cost of \$600/day plus \$1500 (total) for mobilization and demobilization.
. The decanter centrifuge, if used, will be operated only while drilling.

### Useful relationships:

1. Drilled volume, bbl/1000 ft¼0.97 (D, in.)²
2. Where %=the desired drilled solids content, the barrels of new drilling fluid that must be prepared to compensate for each barrel of incorporated drilled solids is given by the following: bbl/bbl=(100-%)/%
3. In slurries consisting of low-gravity solids and freshwater, the solids content in % by volume=7.5 (fluid density, lb/gal–8.33)

1. What volume of new drilling fluid will have to be prepared while drilling this interval if no additional solids are removed?

Drilled solids volume=70×97×(13.5)²=1237 bbl. Incorporated drilled solids are 40% of this, or 495 bbl. The solids content in this 9.0 lb/gal fluid is 7.5 (9.0 – 8.33)=5%. Of this, commercial bentonite composes 2%, leaving 3% drilled solids. The second useful relationship in the preceding list shows that [(1003)/3]¼32.33 bbl of new mud are required to compensate for each barrel of incorporated solids. Therefore, the answer to this first question is that (32.33495)¼16,003 bbl of new fluid must be prepared to compensate for the incorporated solids and maintain the desired density of 9.0 lb/gal. Note that the total new drilling fluid volume will be 16,498 bbl, the volume of the newly prepared clean drilling fluid plus the volume of the incorporated solids.

2. What is the cost of this new fluid?

At \$10/bbl preparation cost and \$5/bbl disposal cost, the total cost is \$242,520: \$160,030 for the preparation of 16,003 bbl of new fluid and \$82,490 for the ultimate disposal of 16,498 bbl.

3. What volume of solids would the centrifuge have to remove from the drilling fluid during the drilling of the interval to make its rental economically beneficial?

Rental, mobilization, and demobilization costs for 10 days’ use totals \$7500. At a total of \$15/bbl for each new barrel of fluid mixed, the breakeven point on the use of the centrifuge is reached when the new mud prepared is reduced by 500 bbl (\$7500 @ \$15/bbl). Inasmuch as each bbl of incorporated solids requires the preparation of 32.33 bbl of new fluid to be added to each bbl of incorporated solids, the removal of each bbl of drilled solids reduces the new mud volume by 33.33 bbl. To reduce the volume of new drilling fluid prepared by 500 bbl, we must separate 15 bbl (500/33.33) of drilled solids during the 10 drilling days.

4. Assuming that no dilution is required at the centrifuge, how much mud will be centrifuged during the drilling of the interval?

In 200 rotating hours, processing 100 gpm, the centrifuge will process 28,570 bbl of drilling fluid [(100 gal/min60 min/hr200 hr)/42 gal/ bbl].

5. Assuming that the mud being centrifuged weighs 9.1 lb/gal, what is the weight of drilled solids entering the centrifuge/hr?

At 9.1 lb/gal, the processed drilling fluid contains 5.8% low-gravity solids and 3.8% drilled solids. At 100 gpm, the centrifuge processes 2.38 bpm, 142.8 bph. The drilling fluid processed in an hour contains 4.28 bbl of drilled solids.

6. What percentage of the drilled solids entering the centrifuge has to be removed to offset the cost of its use?

At 3.8% drilled solids, the 28,570 bbl of drilling fluid processed in 200 operating hours contain a total of 1086 bbl of drilled solids. In question 3 we calculated that the breakeven removal volume is 15 bbl. This is 1.4% of the drilled solids in the processed fluid.

7. Assuming that one third of the drilled solids are removed from the processed drilling fluid, what is the net financial benefit or loss arising from the use of the centrifuge?

The removal of one third of the drilled solids that were not removed by the upstream solids-removal equipment, 165 bbl, eliminates the need for the preparation of (16532.33)=5334 bbl of new fluid at a cost of \$53,340, and the disposal of (16533.33)=5499 bbl at a cost of \$27,495. Use of the centrifuge reduced net operating cost by (\$53,340+27,495 7500)=\$73,335.

8. Making the same assumptions, what would the net financial benefit or loss be if the processing capacity were doubled by providing two decanter centrifuges of the same 100-gpm capacity?

Doubling the removal of drilled solids that are not separated upstream would double the total cost savings, increasing it to \$146,670.

9. Comment on the cost of not using a centrifuge for this interval.

The cost of not using a centrifuge is very significant: \$146,670 when compared with the use of two centrifuges; \$73,335 when compared with the use of a decanter centrifuge.

10. What effect, if any, will running the single centrifuge for 200 hours have upon bentonite requirements under these conditions?

In this application, the underflow—consisting of the larger solids particles and the liquid wetting them—is discarded. The bentonite is expected to remain in the overflow and be returned to the drilling-fluid system. Bentonite consumption should be unaffected.