Thank you for visiting A well is being drilled at a vertical depth of 12 200 ft while circulating a 12 lbm gal mud at a rate of 9. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The density of the kick material, assuming it entered as a slug, is 5.28 lbm/gal.
To calculate the density of the kick material, we need to consider the mud gained in the pit and the volume of the annular space. The mud gained in the pit over a 5-minute period is 15 barrels. The annular capacity opposite the 5-in. drillpipe is 0.0775 bbl/ft. The vertical depth of the well is 12,200 ft. Using these values, we can calculate the annular volume as 0.0775 bbl/ft * 12,200 ft = 945.5 bbl.
Next, we calculate the total volume of the kick material by adding the mud gained in the pit to the annular volume: 15 bbl + 945.5 bbl = 960.5 bbl.
Now, we can calculate the density of the kick material by dividing the weight of the kick material by its volume. Since we know that 1 gallon of mud weighs 12 lbm, we can convert the volume from barrels to gallons: 960.5 bbl * 42 gal/bbl = 40,381 gal.
Finally, we divide the weight of the kick material by its volume to get the density: 12 lbm/gal * 40,381 gal / 40,381 gal = 5.28 lbm/gal.
Therefore, the density of the kick material, assuming it entered as a slug, is 5.28 lbm/gal.
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