[ExI] Glacier Geoengineering
hkeithhenson at gmail.com
Sun Jan 31 09:52:20 UTC 2010
The object is to freeze a glacier to bedrock.
The average heat flow over the entire Earth is 87 mW · m-2
Call it 100-mW/square meter, tenth of a watt.
A square km would have a heat flux of 100,000 watts or 100 kW
Propane absorbs 428 kJ/kg evaporating. It boils at one atmosphere at
-43 deg C. (Propylene boils about 10 deg colder so might use that
So to pull 100 kW (a hundred kJ in a second) would take about 1/4 kg
of propane per second. That's 15 kg/minute. Propane has half the
density of water, so it would be in the range of 30 l/minute going
down the hole. Coming back up as vapor, a cubic meter has a mass of
about 1.9 kg/cubic meter, so 15 kg/minute would be ~eight cubic meters
per minute, or half a cubic meter per second.
Eight m/sec is an ok rate for gases so an area of 1/16 square meter is
enough. As square pipe, it would be 250 cm square. So something like
300 cm pipe would be large enough.
Air (at -50 C) will carry away heat at about 1.5 kJ/m^3 per deg K.
For a five deg temperature difference between the air and the propane,
air will carry away about 7.5 kJ/m^3. For 100 kW, that's 13 cubic
meters per second or 48,000 cubic meters per hour. If 10 km/hr is the
average air speed through the heat exchanger, then the cooling air
intake area would need to be 4.8 square meters.
I think one per square km isn't close enough to freeze the water at
the bottom of a glacier to the rock. They might be put in at 300
meter intervals. (I don't have the patience to find and run a heat
Because the heat pipe only works part of the year, the system would
have to be doubled or tripled in scale. This is only enough to take
out the heat coming out of the earth. Probably need it somewhat
larger to pull the huge masses of ice in a few decades down to a
temperature where they would flow much slower.
Glaciers cover about the same percentage of the earth as farmland. I
don't know how much of them would have to be blocked to slow them
down, perhaps 5-10 percent of the area.
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