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#1
August 15th 03, 02:24 AM
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Effects of possible hydrothermal explosion at Yellowstone Lake
Grant wrote:
Bob Harrington wrote: Anybody have a decent estimate on the volume of water in Yellowstone lake as compared to the volume of water in a decent mid latitude cyclonic system, or a tropical storm? Thinking the answer might be surprising... Don't know about Yellowstone lake, but here are some back-of-the-envelope values for 1) a midlatitude cyclone: average 30 mm of water (liquid + vapor) over an area of, say, 1000x1000 km is equivalent to 30 cubic km of liquid water, or as much as you would find in a 17x17 km by 100 meter deep (on average) lake. 2) a typical isolated thunderstorm cell: average of 60 mm of water in the column over an area of approximately 100 sq. km, or 6e6 cubic meters of water substance, equal to a 780 x 780 x 10 meter lake. That's still a lot of water to vaporize in a single explosion (you'd need 1.5e16 joules of energy as a minimum, equivalent to a 3.5 megaton hydrogen bomb). Having said that, just because a convective cell *contains* that much water mass doesn't mean you'd need to vaporize nearly that much to *initiate* convection. Simply adding a few thousand joules/kg of latent heat energy to the lowest km of the atmosphere over a 10x10 km area would probably do the trick. Vaporizing a mere 100 tons of water (requiring minimum energy input equivalent to around 60 tons of TNT) would probably get you there unless the atmosphere was particularly stable. Furthermore, if that 100 tons of liquid water is already superheated (under pressure) to a sufficiently high temperature -- around 400-500 Celsius would do the trick I think, then no added energy is needed .. all you have to do is relieve the pressure abruptly and you've got both your hydrothermal explosion and, probably, a convective rainshower if not a thunderstorm. Fun with simple math.... - Grant Thanks! The brain cell is on vacation this week... |
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