Page 193 - Shale Shakers Drilling Fluid Systems
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176   SHALE SHAKERS AND DRILLING FLUID SYSTEMS





          Experimental and calculation procedure:

          A sample of discard is placed in a 40.10 gm crucible and weighed:

            Crucible + Sample Weight = 114.94 gm

          The wet sample weight is 74.84 gm. Since the wet discard density is 1.77 gm/cc, the wet sample has a volume of
          74.84 gm/1.77 gm/cc = 42.19 cc.

          After heating overnight at 250°F, the crucible and sample weight is 91.08 gm. The dry solids weight in the sample is:
          91.8 gm - 40.10 gm = 50.98 gm

          The dry solids weight percent in the discard is the weight of dry solids divided by the wet sample weight times 100 or:

            [(50.98 gm)/74.84 gml x 100 = 68.11 wt%

          The volume of the dry sample is calculated by subtracting the volume of water lost from the volume of the wet sample:
          the 42.19 cc wet sample lost 114.94 cc - 91.08 cc or 23.86 cc of water. The volume of the dry sample is:

            42.19 CC - 23.86 CC = 18.33 CC

          The density of the dry solids is the weight of dry solids divided by the volume of dry solids, or 50.98 gm/18.33 cc or
          2.78 gm/cc.







           CALCULATION OF BARITE DISCARD                        2. The volume of low-gravity solids is the total
                                                                   volume of dry solids minus the volume of
          Assuming that all of the drilled and other low-          barite.
        gravity solids in the drilling fluid have a dried       3. The volume of low-gravity solids in one cc
        density of 2.47 gm/cc and the barite has a den-            of solids = 1 cc minus the volume of barite
        sity of 4.2 gm/cc, the weight percent of barite in         in 1 cc of solids.
        the dry sample may be calculated from the follow-
        ing mass-balance equation:                              This can be expressed as follows:






        or












          To determine the terms on the right side of
        the equation:
                                                                This equation may be reduced to the expression:
           1. The volume of barite is the density (4.2 gm/cc)
             divided by the weight of barite                    D = 0.4119W R + 2.47
                                                                            D
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