Assume that the surface system contains 1000 bbl of drilling fluid, the targeted drilled-solids level is 4%vol, and 100 bbl of drilled solids report to the surface. For reference, 100 bbl is the volume of 1029 feet of a 10-inch-diameter hole.

If this volume of 100 bbl of drilled solids remains in the drilling fluid, the pit levels remain constant. The drilling-fluid system has increased to 1100 bbl because new hole of 100 bbl was drilled. The volume of the rock represented by the new hole is virtually the same whether it is ground into cuttings or is solid rock. The 4%vol drilled-solids concentration before drilling means that the drilling fluid contained 0.041000 bbl, or 40 bbl of drilled solids when drilling started. After drilling, these 40 bbl plus the 100 bbl of new drilled solids would be in the 1100-bbl drillingfluid system, or (140 bbl / 1100 bbl)100, or 12.7%vol drilled solids.

To reduce the concentration of drilled solids to 4%vol by only adding clean drilling fluid would require adding enough clean drilling fluid to reduce the 100 bbl of drilled solids to a volume concentration of 4%.

100 bbl = (0.04) new drilling fluid built

new drilling fluid built = 2500 bbl.

This newly built drilling fluid would consist of drilled solids plus clean drilling fluid, or

2500 bbl = clean drilling fluid þ 100 bbl’

clean drilling fluid = 2400 bbl.

Observe that this volume is independent of the original volume of the system.

If the tanks are full when the drilling starts, the 2400 bbl of clean drilling fluid cannot be contained in the mud tanks. The only volume available for the clean drilling fluid is the volume of discard removed from the system. Since nothing is removed from the system, the volume added must be removed to return the pit levels to the original level. The excess drilling fluid (2400 bbl) would need to be removed from the drilling-fluid system to keep the pits from overflowing. Not only would the cost of the clean drilling fluid be prohibitive, but this fluid must also eventually go to a disposal site.

The intent of this analysis is to build the basis for the concept of an appropriate removal efficiency to build the minimum quantity of new drilling fluid and also minimize the volume of discarded fluid. The assumption is that the drill pipe will be returned to the position at which it started drilling the interval. Although the rock does not change in volume, the pit levels will rise because of the volume of the steel added as the drill string enters the well.

Evaluation of SREE is very difficult to do with only short-term tests. Capturing equipment discards for only 15 minutes to 2 hours will not provide sufficient data to calculate SREE. The quantity of solids entering the drilled solids removal equipment is usually unknown and impossible or very difficult to determine. Drilled solids do not arrive at the surface in the same order in which they were drilled. Fluid flow in the annulus is usually laminar. The center part of the annular flow moves faster than the portion of the flow adjacent to the formation or the drill pipe. A 30-minute sample from the shale shaker discards may contain samples that were drilled 2 to 4 hours apart, even in a well bore drilled to gauge. With rugosity and borehole enlargements, lag times are extended. Mud loggers often observe formation cuttings that were drilled several days prior to the sample’s being taken from the end of the shale shaker. This is the reason that material balances are difficult to obtain with only snapshots. Circulating a borehole clean before drilling an interval, drilling a known volume of solids, and circulating all of the cuttings from the well bore will permit estimating a known volume of solids arriving to the surface. Another variable in this analysis is the porosity of the formation. Solids analysis requires being informed of the initial condition of the quantity, or volume, of solids that arrive at the surface.