effect of opening the barrier between a tube with one side pressurized and the other side a vacuum

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?

• I'd expect the answer to depend to some extent on tube diameter and some properties of the tube's inner surface. (I expect that this is an inefficient way to pump water). – RedGrittyBrick Apr 12 '16 at 13:25
• This is actually an attempt to design a rocket and propulsion system using the water pressure at 5 km below the ocean with a tube 20 m in diameter. I expect the water and air to be expelled quite rapidly from 500 atm of pressure into a vacuum. – waychow Apr 14 '16 at 5:19
• I did a few sums and concluded a Saturn V produces 50 times as much initial thrust as this vacuum tube. As the water reaches the top, a 5 km Saturn-V sized tube would be lifting 375,000,000 Kg of water. The speed of the column as it neared the surface might be less than many people might think. – RedGrittyBrick Apr 14 '16 at 9:26
• Sorry, 375,000,000 tonnes of water. – RedGrittyBrick Apr 14 '16 at 9:52
• I got a rough figure of 300,000,000 kg of water traveling at the integral of the gravity equation from 0 to 5 km with a force of 9.8 m/s to equal roughly 3 billion newtons. Equal to about the thrust of both solid rocket boosters for the shuttle. – waychow May 8 '16 at 6:01

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.

• Would an object with a mass of 100,000 then be ejected from the end of the tube at about 700 mph or 1,100 kph? What if a De Laval nozzle was used to reduce the diameter to 10 m? Would that only double the velocity? Or much more? – waychow May 8 '16 at 4:24
• If the vacuum tube was extended above sea level, how high would the inertia carry the water level to? Would the object continue beyond the water level? Or would energy have been transferred to the object as soon as the water level began slowing down and cause the object to continue independently because the effect of earth's gravity is less on the object than on the water column? – waychow May 8 '16 at 4:45
• No! The initial velocity of 1027 ft/s would be the initial velocity as the water enters the tube 5 km below sea level. At that depth the pressure is 500 atm forcing the flow. However, as the water enters the tube at that velocity and travels in a vertical direction upward, the velocity will decrease as a result of the changing water level in the tube. That is, as the tube is filled, the initial 500 atm pressure will continue to be reduced as the result of the weight of the water. The reduction in pressure will be for each 2.31 feet the depth is reduced the 500 atm will reduce by 1 psi. – Gerard De Santis May 12 '16 at 14:02
• With no Bernouli effect from the De Laval nozzle, would the integral of the gravity equation from 0 m to 5,000 m provide an approximate velocity?...Given ideal gas. How much would friction and turbulence affect the velocity? – waychow May 14 '16 at 4:04