Some explanation of the pressures involved with the blown Deepwater Horizon well

[gristy]

Blow-out preventer (Schlumberger)



Pictures of the oil flows (Drillingahead)



...
If a person on either side of you pushed you with equal force at the same time, you don't move, because the two forces balance out. It is only if there is one force, or if one of the two pushes harder, that you move. In other words, where there are a number of forces acting on a body, it is the size of the difference in pressures, and the direction of that difference, that controls the movement.

Consider, here we are drilling merrily away (and have cased the well near the surface, and hit no more fluids on the way down) and at 6,000 ft. we penetrate the rock that is capping the well, and enter the rock with the oil in it. The oil (in the rock) is at some fraction of the overburden pressure, since it is trapped in the rock, and for the sake of this example I am going to say that it is at 5,000 psi , the fluid in the well is at 3,000 psi, the height of the mud column.

There is a difference of 2,000 psi. We are drilling a hole some 6-5/8th inches in diameter. That has an area of about 34.5 square inches. The total force we have suddenly applied to the bottom of the well (bit and fluid) is thus (area x pressure difference) 34.5 x 2,000 = 69,000 lb (or 35 tons). Oops!


more: http://www.theoildrum.com/node/6421#more