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From: Walter Banks on 28 Jun 2010 07:37 io_x wrote: > what about this? > __________________________ >   >  weitght==?  >   >  __________  >   tap >     >       > ^^^^^^^   ^^^^^^^^ surface submarine > ^ > /\ >  > point where oil exit > > there is someone of you know what is the weight > for the above structure (**in the water**) for > controbilance oil pressure? For a rough estimate you have about 15000 feet of granite less 5000 feet of water. Granite is about 2.6 times the weight of sea water more or less 40,000 feet of water equivalent The pressure on exit pipe is about 400005000 equals 35000 feet of water. How much weight? Surface area of the leak with a weight equal to a column of water 35000 feet high. Regards, w..  Walter Banks Byte Craft Limited http://www.bytecraft.com
From: Geoff on 28 Jun 2010 08:22 On Mon, 28 Jun 2010 13:04:03 +0200, "io_x" <a(a)b.c.invalid> wrote: >what about this? >__________________________ >  > weitght==?  >  > __________  >  tap >    >      >^^^^^^^   ^^^^^^^^ surface submarine > ^ > /\ >  >point where oil exit > >there is someone of you know what is the weight >for the above structure (**in the water**) for >controbilance oil pressure? > > W = P/A where W = weight in pounds, P = pressure in PSI, A = area of contact If the exit point is a cylinder then A = 4/3*PI*R^2 where R is the radius of the cylinder. Estimated pressure of the BP well is in excess of 6800 PSI. Diameter of the bore of the LMRP exit flange is 21 inches. Pressure at 5000 fsw is 151 ATA. (2220 PSI)
From: Geoff on 28 Jun 2010 08:43 On Mon, 28 Jun 2010 05:22:39 0700, Geoff <geoff(a)invalid.invalid> wrote: >On Mon, 28 Jun 2010 13:04:03 +0200, "io_x" <a(a)b.c.invalid> wrote: > >>what about this? >>__________________________ >>  >> weitght==?  >>  >> __________  >>  tap >>    >>      >>^^^^^^^   ^^^^^^^^ surface submarine >> ^ >> /\ >>  >>point where oil exit >> >>there is someone of you know what is the weight >>for the above structure (**in the water**) for >>controbilance oil pressure? >> >> > Correction: W = P * A where >W = weight in pounds, >P = pressure in PSI, >A = area of contact in square inches >If the exit point is a cylinder then A = 4/3*PI*R^2 >where R is the radius of the cylinder. > >Estimated pressure of the BP well is in excess of 6800 PSI. >Diameter of the bore of the LMRP exit flange is 21 inches. > >Pressure at 5000 fsw is 151 ATA. (2220 PSI) (It's early here.) :)
From: Frank Kotler on 28 Jun 2010 09:31 io_x wrote: > what about this? > __________________________ >   >  weitght==?  >   >  __________  >   tap >     >       > ^^^^^^^   ^^^^^^^^ surface submarine > ^ > /\ >  > point where oil exit Cool ASCIIart! (Annie was prettier :) > there is someone of you know what is the weight > for the above structure (**in the water**) for > controbilance oil pressure? Doesn't matter, we can pile rocks on it until it's "enough". Similar to the first solution Bee Pee tried (as I understand it). The problem appears to be that "icelike crystals" (methane clathrates, I guess) plug up your "tap". (Why they don't plug up the leak is a mystery to me!) Seems to me the easiest thing would be to let it flow to the surface. Quit pumping "dispersants" into it. Currents and turbulence will keep it from being a "point source", but it ought to hit the surface in a relatively small area. Surround it with barges and booms and collect it from there. We can go a mile under the ocean, and drill another couple miles to get this oil, but we can't scoop it up off the surface of the water? C'mon! Alternatively, they make plastic out of the stuff. Turn it into plastic and just roll it up. :) We can't plug the leak, but we can broadcast hires pictures of it around the world. Something a little "unbalanced" about our abilities, you think? Best, Frank
From: Walter Banks on 28 Jun 2010 10:08 Walter Banks wrote: > io_x wrote: > > > what about this? > > __________________________ > >   > >  weitght==?  > >   > >  __________  > >   tap > >     > >       > > ^^^^^^^   ^^^^^^^^ surface submarine > > ^ > > /\ > >  > > point where oil exit > > > > there is someone of you know what is the weight > > for the above structure (**in the water**) for > > controbilance oil pressure? > > For a rough estimate you have about 15000 feet of granite > less 5000 feet of water. > > Granite is about 2.6 times the weight of sea water more or less > 40,000 feet of water equivalent > > The pressure on exit pipe is about 400005000 equals > 35000 feet of water. > > How much weight? > > Surface area of the leak with a weight equal to > a column of water 35000 feet high. > To add some clarity and a missing back of envelope information . The above doesn't account for the weight of the oil in the well. 15000 feet at 0.88 the weight of sea water. Equivalent to 13000 feet of water also pushing back against the oil outflow This reduces the pressure at leak to about 23000 feet of water. The overburden of rock can have different densities from about 1.75 times water to 2.7 times water. This changes the differential pressure at the leak to a range of 8000 feet of water to 23000 feet of water or 3700 PSI to 10700 PSI This quick calculations is consistent with published numbers of up to 10000 psi at the well head and is within the range of the 6800PSI quote elsewhere in this thread. Regards, w..  Walter Banks Byte Craft Limited http://www.bytecraft.com

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