Velocity of water column 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?
 A: 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.

