In April 2010, the worst oil spill ever recorded occurred when an explosion and
ID: 2003480 • Letter: I
Question
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. Estimate the gauge pressure (psig) in the Gulf at a depth of 5053 ft. 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. The drilling mud is a stable slurry of seawater and barite (SG = 4.37). What is the mass fraction of barite in the slurry? What would you expect to happen if the barite weight fraction were significantly less than that estimated in Part (c)? Explain your reasoning.Explanation / Answer
(a) Pressure due to seawater on the wellhead = gh
Since water density =1000 kg/m3
Specific gravity = seawater/pure-water
So sea water density seawater = 1000 x 1.03 = 1030 kg/m3
g=9.8 m/s2
h = 5053 ft = 1540 meter
Pressure = 1030 x9.8x1540 = 155.44 X105 pascal
Total pressure on wellhead = Atmospheric preessure + pressure due to seawater = 155.44 X105 + 1.01 X105 = 156.45 x 105 Pa
Since 1 psig is equal to 6894.75728 pascal,
So Total pressure on wellhead = 2269.12 psig
(b) Here seawater is replaced by mud.
Total pressure due to mud = 4400x6894.75728 -101325 = 302.36 x 105 Pascal
mudgh = 302.36 x 105
mud = 302.36 x 105 /(9.8x1540)=2003.45 Kg/m3
Specific gravity = mud/pure-water = 2003.45 /1000 = 2
(c) As the mud is mixture of seawater and barite,
mud can be written as = (Mseawater + Mbarite)/Total Volume
mud = (Mseawater + Mbarite)/(Mseawater/seawater + Mbarite/bartie)
On putting values of all densities we get Mseawater=0.57 x Mbarite
So mass fraction of barite in slurry = Mbarite/(Mseawater + Mbarite)= 1/(1+0.57)= 0.64
(d)
If barite weight fraction, whose density is higher than seawater is reduced, then required pressure 0f 4400psig will not build up on wellhead. And the oil will be able to escape.