In April 2010, the worst oil spill ever recorded occurred when an explosion and fire on the Deepwater Horizon offshore oil-drilling rig left 11 workers dead and began releasing oil into the Gulf of Mexico. One of the attempts to contain the spill involved pumping drilling mud into the well to balance the pressure of escaping oil against a column of fluid (the mud) having a density significantly higher than those of seawater and oil. In the following problems, you may assume that seawater has a specific gravity of 1.03 and that the subsea wellhead was 5053 ft below the surface of the Gulf.(a) Estimate the gauge pressure (psig) in the Gulf at a depth of 5053 ft.(b) Measurements indicate that the pressure inside the wellhead is 4400 psig. Suppose a pipe between the surface of the Gulf and the wellhead is filled with drilling mud and balances that pressure. Estimate the specific gravity of the drilling mud.(c) The drilling mud is a stable slurry of seawater and barite SG 4:37. What is the mass fraction of barite in the slurry?(d) What would you expect to happen if the barite weight fraction were significantly less than that estimated in Part (c)? Explain your reasoning.Felder, Richard M.; Rousseau, Ronald W.; Bullard, Lisa G.. Elementary Principles of Chemical Processes, 4th Edition (Page 71). Wiley. Kindle Edition.

Respuesta :

Explanation:

A.

P = Ds * g * h

Height, h = 5053 ft

In m,

1541.165 m

g = 9.81 m/s^2

SG = density of substance, Ds/density of water

1.03 = Ds/1000

= 1030 kg/m^3

P = 1030 * 9.81 * 1541.165

= 1.56 x 10^7 N/m^2

1 Psi = 6894.76 Pa = 6894

76 N/m^2

= 2258.6 psig.

B.

Pressure = 4400 psig

Since,

1 Psi = 6894.76 Pa = 6894

76 N/m^2

= 3.03 x 10^7 Pa

Density = 3.03 x 10^7/(1541.165 * 9 81)

= 2006.6 kg/m^3

SG = density of substance, Ds/density of water

= 2006.6/1000

= 2.01.

C.

Density of barite = 4.370 * 1000

= 4370 kg/m^3

Total mass of slurry = mass of barite + mass of seawater

= 1030 + 4370

= 5400 kg in 1 m3 of slurry.

Mass fraction = mass of barite/total mass of slurry

= 4370/5400

= 0.81.

D.

The mud weight is a key factor. Too low mud weight may result in collapse problems, hence spill of the well will not be contained. This is because of the pressure of escaping oil against a column of fluid (the mud) having a density significantly higher than those of seawater and oil.