High School

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 bbl/min when the well begins to flow. Fifteen barrels of mud are gained in the pit over a 5-minute period before the pump is stopped and the blowout preventers are closed.

After the pressures stabilized, an initial drillpipe pressure of 400 psia and an initial casing pressure of 550 psia were recorded. The annular capacity opposite the 5-in., 19.5-lbf/ft drillpipe is 0.0775 bbl/ft. The annular capacity opposite the 600 ft of 3-in. ID drill collars is 0.035 bbl/ft.

Tasks:

1. Compute the density of the kick material assuming the kick entered as a slug.
- **Answer:** 5.28 lbm/gal.

2. Compute the density of the kick material assuming the kick mixed with the mud pumped during the detection time.
- **Answer:** 1.54 lbm/gal.

3. Do you think that the kick is a liquid or a gas?

4. Compute the pressure that will be observed at the casing depth of 4,000 ft when the top of the kick zone reaches the casing if the kick is circulated from the well before increasing the mud density.
- **Answer:** 3,198 psia.

5. Compute the annular pressure that will be observed at the surface when the top of the kick zone reaches the surface if the kick is circulated to the surface before increasing the mud density. The annular capacity inside the casing is also 0.0775 bbl/ft.
- **Answer:** 1,204 psia.

6. Compute the surface annular pressure that would be observed at the surface when the top of the kick zone reaches the surface if the mud density is increased to the kill mud density before circulation of the well.
- **Answer:** 1,029 psia.

7. Using the data from Exercise 4.10, compute the pit gain that will be observed when the kick reaches the surface if the kick is circulated to the surface before increasing the mud density. Assume that the kick remains as a slug and that any gas present behaves as an ideal gas.
- **Answer:** 99.8 bbl.

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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