Weighted Drilling Fluid Comparison with CaCl₂

I asked a question previously, and the answer was quite illuminating. The answer ushered me forth in a direction that finally made sense. I ended up finding this website, chem.latech.edu/~ramu/cmen522/homework/hw06\_04a.pdf, that further explained the dynamic as well. I gathered what was necessary to be done, but wasn't quite sure how to apply it. I am currently a chemist for the oil field and am trying to find an accurate representation of percent water and CaCl2\*subscript\*, as well as the correct specific gravity of the complete solution. Given a 10.5 Mud weight (10.5 lb/gal), I have struggled to find volumes necessary to add and partial volume actually present.


Further explained - I applied the formula designated $V /cm^3 = 1001.97 + 30.2207(m^.5) − 36.7889m + 37.0838(m^1.5) − 5.7447(m^2)$ for the corrected solution volume and was really close to the correct "loss" of volume from the packing that the answer described. The two following formulas for partial volumes$ −36.7889 + 15.1104(m^−.5) + 55.6257(m^.5) − 11.4894m $(for CaCl2\*subscript\*) and $18.048 + 0.2722(m^.5) − 0.3341(m^1.5) + 0.1035(m^2) $(for water), don't add up to the total solution loss. I am a bit baffled as to why this is the case. Granted my physical chemistry knowledge is a bit dated, I would assume my excel document is correct. At the 10.5 weighted mud, my molality was 3.586. This resulted in a volume of 1105 with fluid loss and a 1183 of theoretical volume. The excess volume then should be 78.27. The formula for the solute given that molality gave me a volume of 17.63 and the solution was 35.33. The volume together is 52.95. This doesn't equal the excess volume which would suggest there was an error in one of the formulas, or that I have made a significant error on my part.


