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If an air filled tube 5 km long with a diameter of 20 m was submerged vertically under water one end 5 km underwater and the other end at the surface, how fast would the water column rise if the ends were opened quickly?

Would the water rise at nearly the rate of an equal mass of water falling 5 km? Would it be the integral of the distance?

How high would the water spurt out the top?

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  • $\begingroup$ Now assume the air filled tube has a pressure of 500 atm. before opening. Such as if the tube bottom was opened first to allow the air in the tube to be compressed by the water pressure at 5 km below sea level first before opening the top of the tube. $\endgroup$ – waychow Apr 2 '16 at 6:29
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If you neglect the viscosity of air and water and friction along the tube, the compressibility, the temperature gradient and change, and assume the water can come freely to the pipe, and air/water can go out freely directly at pipe exit, then it's a direct application of Archimedes principle: the rising force in the column corresponds to the weight of the part "replaced" by air.

Initially it is thus $f(0) = \int_0^{50km}\rho g h$ $\approx 5.10^7$ .

But once the filling started, with level at y(t) (0 at sea level, positive with depth) we get $f(t)= \int_0^{y(t)}\rho g h$ $\approx 5.10^3 y^2$, i.e. the force decrease to 0 as the filling complete.

Then, acceleration is $a(t)=f/m$, with m(t) the filled part $\int_{y(t)}^{50km}\rho g h$ $\approx 5.10^7-5.10^3 y^2$, so $a(t)\approx \frac{5.10^7}{5.10^7-5.10^3 y^2}-1 = \frac{1}{1-10^{-4} y(t)^2}-1$

So you have to solve the differential equation $y"= \frac{1}{1-10^{-4} y(t)^2}-1$. The velocity is then y'.

For the "how high the column goes in air", I would try 2 very approximative ways:

  • the column keeps it's pipe shape (then it should raise up to some altitude, then falls with the whole process reverse in huge oscillation)
  • the column explode in big parcels not weighting on the sub-sea column, but big enough to not suffer to much air drag. Then it's a classic balistic problem from velocity reached at sea level.
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  • $\begingroup$ Sorry my calculus is a little rusty. So around 700mph? $\endgroup$ – waychow Apr 2 '16 at 5:23
  • $\begingroup$ If the pipe extended above sea level, is inertia of the column of water the reason it rises above sea level? $\endgroup$ – waychow Apr 2 '16 at 5:42
  • $\begingroup$ I've done this experiment on a small scale and the water rises above the water level. $\endgroup$ – waychow Apr 2 '16 at 5:44
  • $\begingroup$ inertia: yes. 700mph: well, I'm not sure how to solve analytically the differential equation :-/ $\endgroup$ – Fabrice NEYRET Apr 3 '16 at 18:34
  • $\begingroup$ If the tube was pressurized by the water trying to reach equilibrium at sea level by a gate then opened to max instantly at mid-depth, and the upper half of the tube was connected via a De Laval nozzle that also reduced the diameter of the tube to 10 m and was a vacuum tube rising another 2,500 m above sea level, would the water column be traveling over 1,000 mph with a force of about 3 billion newtons? $\endgroup$ – waychow May 9 '16 at 6:31

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