EXAMPLES FOR SOLIDS REMOVAL EQUIPMENT EFFICIENCY

What is the solids removal equipment efficiency?

This exercise involves the cost benefit of increasing solids removal equipment efficiency to 80% for the 4%vol drilled-solids concentration: With 80% removal efficiency and 1143 bbl of drilled solids reporting to the surface, 914 bbl would be discarded and 229 bbl returned to the pits. The drilling fluid needed to dilute the 229 bbl to 4%vol would require adding (229 bbl / 0.04), or 5725 bbl of new drilling fluid. This 5725 bbl would consist of 229 bbl of drilled solids and 5496 bbl of clean drilling fluid.

Assuming the 35%vol concentration of drilled solids in the discard, the discard volume would be 914 bbl / 0.35, or 2611 bbl. The excess volume of drilling fluid is (5725 bbl2611 bbl), or 3114 bbl. The drilling-fluid cost for 80% removal efficiency and 4%vol drilled solids would be (5725 bbl\$90 / bbl)þ[(3114 bblþ2611 bbl)\$40], or \$744,000. Increasing the SREE from 56.5 to 80% decreases the cost of the new fluid and disposal from \$1,616,000 to \$744,000 for this 6000 feet of hole. The \$872,000 difference could justify significant changes in the drilling-fluid system.

What is the volume of new drilling fluid built while diluting drilled solids remaining in the drilling fluid?

Never-Wrong Drilling Company is drilling a 12¼-inch hole from a casing seat at 4000 feet to the next casing seat at 10,000 feet. The 12-ppg waterbased drilling fluid has an 18%vol solids content with a methylene blue test of 18 lb / bbl. After drilling the well, a caliper indicates that the borehole washed out to an average diameter of 14 inches. While drilling this 5000-foot interval, a volume of 7163 bbl of clean drilling fluid was metered into the system to maintain a constant solids concentration of 18%vol.

This is a fairly common example. From these data, the SREE can be calculated. First, the hole volume, or volume of solids reaching the surface, is calculated as 1143 bbl.

hole volume=(14.0 inches)²⁄1027=1143 bbl

This assumes that the rock drilled has no porosity. The value of solids could be reduced by 20% to a value of 914 bbl, but the accuracy of the diameter measurement is insufficient to justify such a minutia.

Second, the concentration of drilled solids in the drilling fluid must be determined. For this calculation, the density (SG) of the barite will be assumed to be 4.2 g / cc and that of the low-gravity solids 2.6 g / cc.

(volume fraction of low-gravity solids) = 62.5 + 2.0 (volume fraction of solids) + 7.5 (mud weight, ppg)

This calculation indicates that the concentration of low-gravity solids is 8.5%vol. Since 2%vol is bentonite, the drilled-solids concentration in this 12-ppg drilling fluid is 6.5%vol. The total solids concentration is 18%vol, so the barite concentration must be 9.5%vol. The solidsremoval equipment will discard some drilled solids and also retain some in the system. The solids remaining in the tanks must be diluted with clean drilling fluid to produce a 6.5%vol drilled-solids concentration. The liquid added to the mud system to dilute the returned solids will be the dilution volume.

drilled solids returned = ( drilled solids concentration ) × ( dilution volume ).

This equation says that the drilled solids in the new drilling fluid added to the tanks will be diluted to the drilled-solids concentration (6.5%vol). The total fluid added to the mud system (which is the dilution volume) will consist of two components, the clean drilling fluid and the drilled solids returning to the system.

Third, calculate the volume of drilled solids returned to the system by
the solids-control equipment. The volume of clean drilling fluid added
during the drilling of this 6000 feet of hole was 7163 bbl. The drilled-solids
concentration in the drilling fluid was maintained at 6.5%vol. From the
preceding equation: 7163 bbl = drilled solids returned [(1 / 0.065) -1], or
drilled solids returned to the system = 497 bbl.

Fourth, calculate the removal efficiency of the surface equipment on this rig. Since 1143 bbl arrived at the surface during the drilling of this interval, 646 bbl were discarded. The removal efficiency would be the ratio of the solids discarded and the solids arriving 100, or 56.5%.

Without commenting on the hole problems associated with drilled solids, the cost of this inefficient removal system will perhaps be surprising. For this example, the drilling fluid will be assumed to cost \$90 / bbl and the disposal costs will be assumed to be \$40 / bbl.

How much drilling fluid was discarded during this 6000-foot interval? This would be the sum of the volume discarded with the solids-removal equipment and the excess volume needed for dilution of the drilled solids.

This system discarded 646 bbl of drilled solids. These drilled solids were not dry; they carried with them some of the drilling fluid. For 200- mesh shaker screens and other equipment, a concentration of 35%vol drilled solids in the discard is a reasonable number. This means that the volume of discard would be (646 bbl / 0.35), or 1846 bbl.

The pit levels would have decreased by 1846 bbl during the drilling of this interval. During this part of the well, a volume of 7163 bbl of clean fluid was added to the 497 bbl of drilled solids returned to the pits by the solids-control equipment. The total drilling-fluid volume built would be 7660 bbl. Since only 1846 bbl of fluid were removed from the pits, an excess volume of (7660 – 1846), or 5814 bbl, was built.

The new drilling fluid would cost (\$90 / bbl7660 bbl), or \$689,000. The discarded volume would be the sum of the discarded volume from the solids-control equipment and the excess volume generated to dilute the returned drilled solids, or 1846 bblþ5814 bbl, or 7660 bbl. This cost would be \$40 / bbl7660, or \$306,000. Total drilling-fluid cost for operating this system for 6000 feet of drilling would be \$995,000, or \$166 / ft of hole.

What volume of drilled solids was discarded?

In this inefficient drilled-solids removal system, the decision is made to drill with a polymer drilling-fluid system that required a 4%vol drilled solids concentration. The cost of reducing the drilled-solids concentration
by only 2.5% is startling.

With 56.5% removal efficiency and 1143 bbl of drilled solids reporting to the surface, 646 bbl will be discarded and 497 bbl will be returned to the pits (as before). The returned drilled solids must be reduced to 4%vol in the pits, so the dilution volume will be (497 bbl / 0.04), or 12,430 bbl. Assuming the same volume concentration of drilled solids in the equipment discard of 35%, as before, the equipment discard (and the decrease in pit volume) will be 646 bbl / 0.35, or 1850 bbl. The excess drilling fluid generated is (12,430 bbl – 1850 bbl), or 10,580 bbl.

Now the costs:

• . New drilling fluid added to the system¼12,430 bbl\$90 / bbl, or \$1,119,000
•  Disposal for this interval¼(1850 bblþ10,580 bbl)\$40 / bbl, or \$497,000
• Total cost for this 6000-foot interval¼\$1,616,000, or \$269 / ft of hole.

Decreasing the drilled-solids concentration from 6.5 to 4%vol with this inefficient removal system increases the cost by \$621,000.

What volume of drilling fluid resulted from the clean drilling fluid being added?

This exercise demonstrates the effect of a slight increase in the drilled solids concentration:
If the 80% removal efficiency were achieved and a 6%vol drilled-solids level would not create hole problems, another significant cost reduction is possible. With 80% removal efficiency and 1143 bbl of drilled solids reporting to the surface, 914 bbl would be discarded and 229 bbl returned to the pits.
The drilling fluid needed to dilute the 229 bbl to 6%vol would require adding (229 bbl / 0.06), or 3817 bbl, of new drilling fluid. This volume would consist of 229 bbl of drilled solids and 3588 bbl of clean drilling fluid. Assuming 35%vol drilled solids in the discard, the discard volume would be (914 bbl / 0.35), or 2611 bbl. The excess volume of drilling fluid is (3817 bbl2611 bbl), or 1206 bbl.
The drilling-fluid cost for 80% removal efficiency and 6%vol drilled solids would be (3817 bbl\$90 / bbl)þ[(1206 bblþ2611 bbl)\$40], or \$496,000. With an 80% removal efficiency, increasing the drilled-solids target concentration from 4 to 6%vol decreases the cost of the new fluid and disposal from \$744,000 to \$496,000 (or by \$248,000) for this 6000 feet of hole.

What is the cost of the error in the mud engineer’s estimate of material captured for discard? (i.e., does this number have to be very accurate?)

Assumed that the clean drilling fluid volume was known. Frequently, drilling fluid is not metered into the pits, but the new fluid is built by adding liquid and solids to the tanks to keep the pit levels constant and maintain the drilled-solids concentration at some target value.
Since the barite concentration in the drilling fluid was also kept at the same value, 9.5%vol, it has been used as a tracer to determine how much clean drilling fluid was added. In Exercise 1, the dilution fluid contains 9.5%vol barite, or 680 bbl. Barite weighs 1471 lb / bbl, so this would be equivalent to 1,000,000 lb of barite used (or 100,000 sx). The new drilling fluid built contained 7163 bbl of clean fluid and some volume of drilled solids that was returned to the pits by the solids-removal equipment. Barite may be used as a tracer to determine the volume of clean drilling fluid added to the system. But using measured volumes of barite in the discard is fraught with problems. The liquid removed by fine-mesh shale shaker screens does not contain the same concentration of barite as does the drilling fluid in the pits. In one field test, between 100 and 300 lb / hr of excess barite was discarded from a continuous-cloth screen. During that field test, a deficiency of as much as 100 lb / hr of barite was observed in the discard from panel screens. In this case, barite went through the screen, which decreased the barite in the discard below the concentration values of the drilling fluid in the pits.