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How quickly would a 600 pound block of ice float to the surface from a depth of 200 feet?
Wiki User
∙ 12y ago
To make this roughly answerable, I will assume the following:
1) The block of ice is roughly spherical.
2) The water is sea water with a salinity of about 3.5%
3) Both the water and the ice are at about -1.9°C (the melting point of sea water with a salinity of 3.5%).
4) There is no heat transfer between the water and the ice. I.e. there are no phase changes going on at the interface and the mass of the ice does not change.
5) The acceleration of the block is assumed to be quick compared to the time taken to travel 200ft. Thus I will calculate terminal velocity only.
6) The density of sea water at -1.9°C is about 1020kg/m3. The density of sea ice is about 935kg/m3 at -1.9°C
The volume of a sphere of sea ice weighing 600lbs (272kg) would be 291L and have a diameter of 822mm. The mass of sea water displaced would be 297kg. Thus the bouyancy force would be about 245N. This upwards force would be an initial upwards force. As the block accelerates, the drag of the water will increase to the point that it cancels the bouyancy force at a terminal velocity.
As the drag coefficient of a sphere changes with the speed, an iterative solution to find the final velocity will be required. Thus, assuming the speed is 2m/s, the Reynolds number would be about 1020 x 0.822 x 2 / (1.88 x 10E-3) = 892000. At this Re, the drag coefficient of a sphere is about 0.12. The drag would be: D = 1/2 x 1020 x 2^2 x 0.53 x 0.12 = 130N.
This is less than the bouancy force, so the speed would be higher. The drag coefficienct would be roughly the same (maybe a little higher), so the speed will be higher in proportion to the square root of the force difference. Thus sqrt(245/130) x 2 = 2.74 m/s.
200ft is 60.96m. This takes about 22.2 seconds at this speed. It will take slightly longer if the ice block starts from rest, so an answer of 25 seconds is probably in the right ball park.
Considering the assumptions, there is one major issue. A sphere has a pretty low drag coefficient at this speed. A typical block of ice may have a drag coefficient significantly higher (say 4 times). This will reduce the square root of the speed proportionally. Thus I would expect a randomly shaped ice block to rise at half the speed and take twice as long, about 50 seconds. An optimally shaped block (teardrop with a length to diameter ratio of 10) might rise as quickly as 15 seconds while a flat plate would take more than a minute.
Concerning the other assumptions: On the way up, friction with the water will tend to wear the corners, so I would expect the drag coefficient to decrease, the speed to increase and the mass to decrease.
Wiki User
∙ 12y ago
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