What would happen when a tube with an air space pressurized to 500 atm by the water exerted at 5 km below the surface was opened into a vacuum chamber rising up 5 km to the surface? The bottom of the tube would be opened to allow the water to rush in and up the tube simultaneously as the barrier between the pressurized section and the vacuum section was opened. At what optimal time, would the top of the tube be opened at sea level to allow the rising column of water escape the tube?
Don't let the size and length of the tube confuse you. What you described can be viewed as a big nozzle with a diameter of 20 meters with a water pressure of 500 atm. entering the back side of the nozzle. The formulas that can be used are:
V = sq. root of 2gh where: V = velocity in ft/s. g = acceleration const = 32.2 ft/s^2 h = head in feet of liquid (16,402.2 ft.) in 5 km.
In doing the math., V = sq. root of (2)(32.2)(16,402.2) = 1,027 ft/s
Another formula to use is: V = (gpm * .321)/A
where: V = velocity in ft/s A = nozzle area in in^2 .321 = conversion constant.