If the mechanical equipment does not remove a significant portion of the drilled solids reporting to the surface, it can become very expensive to maintain a reasonable level of undesirable drilled solids. Dilution, then, becomes a major portion of the solids-management strategy. Calculations indicate the performance of the solids-removal equipment.

This set of calculations is simply a material balance of the volumes added and the volumes discarded. The calculations will be based on a drilling-fluid processing plant system in which the solids-removal section is removing either 100, 90, 80, or 70% volume of the drilled solids arriving at the surface. Obviously 100% removal efficiency is not currently possible while retaining the liquid phase. Removal percentages listed are used simply to demonstrate the method and concepts of solids removal efficiency minimization.

The average drilled-solids concentration in the discard stream has been selected as 35%vol. The underflow from a decanting centrifuge and the mud cleaner discharge stream will contain 55–63%vol solids. The underflow from desilters and desanders will vary around 35%vol solids. The discharge stream from a shaker can vary from 45%vol, for very coarse screens, to 20%vol solids for very fine screens. The average for all of these devices is assumed to be 35%. With very coarse screens, most of the liquid is removed from the large solids as they travel down the shaker screen. So, the concentration of solids is much higher than it is with screens with smaller openings, although the total volume of solids discarded is usually smaller. Screens with smaller openings remove more solids as well as more liquid. The reason is related to the ratio between the surface area and the volume of the cuttings. For example, a golf ball would retain very little drilling fluid when removed from the fluid. Grind the ball into very small pieces and the volume of liquid would increase greatly. The volume of the solid (golf ball) would not change, but the surface area would change greatly. More liquid is required to wet the increased surface area.

To examine (1) the quantity of discards, (2) the clean drilling fluid needed to dilute the remaining drilled solids, and (3) the excess volume of drilling fluid built, four SREEs will be discussed: 100%, 90%, 80%, and 70%.

### Example 1

#### 100% Removal Efficiency

In the 100% removal efficiency case, all of the drilled solids reaching the surface are removed. In addition to drilled solids, the solids-control equipment also removes drilling fluid clinging to those solids. With a 35%vol concentration of solids, the discard will contain 65%vol drilling fluid. The pit levels will decrease by the volume of drilled solids and drilling fluid removed from the system, as shown in Figure 1. (This makes the assumption that the rock porosity and compressibility are zero.) No drilling fluid is needed for dilution because no drilled solids are retained in the system. However, 286 bbl of drilling fluid are needed to replace the material removed from the system in order to keep the pit levels constant. The drilled-solids concentration will decrease. In other words, a level of 4%vol drilled solids will not and cannot be sustained. In the preceding example, the 40 bbl of drilled solids originally in the drilling fluid will now be distributed into 1100 bbl of drilling fluid (the 1000 bbl of initial volume plus the 100 bbl of hole drilled). The drilledsolids concentration will decrease to 3.6%vol, or (40 bbl / 1100 bbl) 100. The ratio of clean drilling fluid needed per barrel of hole drilled is 2.86 [286 bbl / 100 bbl]. In this case, the 4%vol drilled solids in the system will be diluted and the concentration of drilled solids will be reduced.

### Example 2

#### 90% Removal Efficiency

In the 90% removal efficiency situation, the volume of newly drilled solids removed from the system is 90 bbl. The discards will again be assumed to contain 35%vol solids. The total volume of discard would be [90 bbl / 0.35], or 257 bbl (Figure 2). The solids returned to the system (10 bbl) must be reduced to a 4%vol concentration. The 10 bbl of newly drilled solids must become 4%vol of newly built mud. So the new mud built would be 250 bbl [10 bbl / 0.04]. The 250 bbl would be composed of 240 bbl of clean drilling fluid and 10 bbl of drilled solids. Since the volume removed by the solids-control equipment was 257 bbl, this would be the volume available in the mud pits to hold the dilution fluid. Since 240 bbl is needed to dilute the remaining 10 bbl of drilled solids, an additional 17 bbl must be added to bring the pit level. This is almost a balanced system, that is, no excess drilling fluid is needed to dilute the drilled solids returning to the system. If the discarded volume exactly matches the required volume needed for dilution, the minimum quantity of drilling fluid will be built. The optimum removal efficiency for any targeted drilled-solids level may be calculated by mathematically equating the removal volume to the dilution volume required. The ratio of clean drilling fluid needed per barrel of hole drilled is 2.57 [257 bbl / 100 bbl]. Again, the 4%vol drilled solids in the system will be decreased because more dilution was needed to keep the pit levels constant than was needed to dilute the 10 bbl of retained solids.

The concentration of drilled solids would be less than the targeted 4%vol. The volume of drilled solids would be the original 40 bbl plus the 10 bbl retained in the system by the solids-control equipment. The system volume would be the original 1000 bbl plus the 257 bbl of new hole drilled, or 1100 bbl. The drilled-solids concentration would be 3.98%vol.

### Example 3

#### 80% Removal Efficiency

For the case of the 80% removal efficiency, 229 bbl of drilled solids and drilling fluid will be discharged (Figure 3). Although this is only 21 bbl less than the 90% removal efficiency, the dilution volumes are significantly higher. The dilution of the 20 bbl of returned drilled solids to a 4%vol level requires the addition of 480 bbl [20 bbl / 0.04] to the system. The reconstituted 500 bbl of drilling fluid will contain 20 bbl of drilled solids and 480 bbl of clean drilling fluid. Since only 229 bbl of space is available, 271 bbl of drilling fluid must be discarded. The total discard will therefore be the 229 bbl from the solids-removal equipment and the 271 bbl of drilling fluid. The volume of clean drilling fluid needed per barrel of hole drilled is 4.8%. The excess drilling fluid generated is 271 bbl.

### Example 4

#### 70% Removal Efficiency

For the case of 70% removal efficiency, 70 bbl of drilled solids will be discarded in a 200-bbl slurry that contains 130 bbl of drilling fluid (Figure 4). So the pit levels will decrease by the removal of 200 bbl from the system. The 30 bbl retained in the system must be diluted to 4%vol. The new drilling fluid contains 4%vol drilled solids, or mathematically: drilled solids¼ (4%)(new drilling fluid). The volume of retained drilled solids that must be diluted is 30 bbl, so the volume of new drilling fluid would be (30 bbl / 0.04), or 750 bbl. This new drilling fluid will contain 30 bbl of drilled solids; so, 720 bbl of clean drilling fluid must be added to the system. Only 200 bbl of volume is available in the drilling fluid system, so 520 bbl of drilling fluid must be discarded from the system or stored. The volume of clean drilling fluid needed per barrel of hole drilled is 720 / 100, or 7.0. Note that the excess drilling fluid generated is 520 bbl.

## Clean Fluid Required to Maintain 4%vol Drilled Solids

The volume of clean drilling fluid required is a function of the targeted drilled-solids concentration, 4%vol for these cases, and the SREE.

The line designated by the triangles in Figure 5 indicates that the volume of clean drilling fluid required to dilute the drilled solids remaining in the drilling fluid is less than the volume of fluid required to return the pit levels back to their original values. In this case the targeted drilled-solids concentration will be decreasing. The line designated by the squares indicates that more fluid was required to dilute the solids remaining in the pits after processing through the solids-control equipment. The pit levels would increase so much that excess drilling fluid would need to be removed from the system. The intersection of these two lines would indicate the smallest quantity of drilling fluid required to maintain a concentration of 4%vol drilled solids in the system. This intersection is predictable and the calculation is presented in the next section.