What happens if you connect the ocean floor and the surface with a pipe?

[abrek]

TL;DR

if you take a pipe with one end closed and lower it into the water to a great depth, for example, 1OOO meters, after which we open the closed end, the pressure difference will drive the water through the pipe and then to the surface will you get an endless fountain?

 

[erobz]

No, the pipe will fill to the surface of the water. It will not shoot out of the top of the pipe in an endless geyser.

 

[kuruman]

What drives the water inside to go up the pipe is the pressure difference between the level of the water inside and outside. When these levels are equalized, there will be no such pressure difference.

 

[Lnewqban]

Welcome, @abrek !

You will get a temporary fountain, not an endless one.
You can't get more pumping work out than you put in by forcing the pipe down against its natural buoyancy.

 

[abrek]

No, the pipe will fill to the surface of the water. It will not shoot out of the top of the pipe in an endless geyser.

What drives the water inside to go up the pipe is the pressure difference between the level of the water inside and outside. When these levels are equalized, there will be no such pressure difference.

thanks for the reply. I thought this was possible due to the large difference in pressure at the bottom of the ocean and the usual pressure at the surface

What drives the water inside to go up the pipe is the pressure difference between the level of the water inside and outside. When these levels are equalized, there will be no such pressure difference.

 

[erobz]

You probably get some oscillation in which the water could spill out if the top momentarily, but it will not be constant, and will die out from friction/viscous effects.

 

[erobz]

Out curiosity I come to this ODE,

$$ g ( H - h ) - \frac{v^2}{2} = h v \frac{dv}{dh} $$

or with $u = \frac{v^2}{2}$

$$ h \frac{du}{dh} + u -g (H-h) = 0 $$

Can we tell if this equation will produce a reasonable solution. I would like to say at $h= H$ we get:

$$ h \frac{du}{dh} = - u $$

implying that $v^2$ is still decreasing there (the beginning of oscillation), but can $v^2 = 0$ @ $h = H$ be ruled out?

I'm also concerned that there is no mass at $h = 0$ implying infinite acceleration there. Perhaps some mass must be included around the inlet if this is too fishy. But then my question is; if so, how much?

 

[A.T.]

TL;DR Summary: if you take a pipe with one end closed and lower it into the water to a great depth, for example, 1OOO meters, after which we open the closed end, the pressure difference will drive the water through the pipe and then to the surface will you get an endless fountain?

Have you never used a drinking straw? If it doesn't work there, why should it work with 1000m? The pressure difference increases at the same rate as the height the water has to climb.

 

[Lnewqban]

It is also important to note that the water flowing up inside the pipe will be colder and denser than the water outside it.
@abrek,
Would that fact make your fountain effect stronger or weaker?

 

[jrmichler]

I'm also concerned that there is no mass at h=0 implying infinite acceleration there. Perhaps some mass must be included around the inlet if this is too fishy. But then my question is; if so, how much?

Fixed that for you. For a round pipe inlet, a rule of thumb is that the velocity at one inlet diameter from from the inlet is 0.1 times the velocity in the inlet. That includes a volume three times the diameter of the inlet, and one inlet diameter think. The velocity through that volume varies by a full order of magnitude, which makes the problem computationally challenging. But it gives you a starting point. Just make a guess at the volume of fluid moving at peak velocity with kinetic energy equal to that of the actual volume, which is moving at a wide range of velocities. It's a negligibly small source of error for such a long pipe, so your estimate needs to be only good enough to make a numerically stable solution.

 

[erobz]

Fixed that for you. For a round pipe inlet, a rule of thumb is that the velocity at one inlet diameter from from the inlet is 0.1 times the velocity in the inlet. That includes a volume three times the diameter of the inlet, and one inlet diameter think. The velocity through that volume varies by a full order of magnitude, which makes the problem computationally challenging. But it gives you a starting point. Just make a guess at the volume of fluid moving at peak velocity with kinetic energy equal to that of the actual volume, which is moving at a wide range of velocities. It's a negligibly small source of error for such a long pipe, so your estimate needs to be only good enough to make a numerically stable solution.

Thank you for confirming my suspicion. I think if I include any mass outside the pipe (even just moving at $v$ ) without being realistic about the velocity distribution it should be at least a stable solution as you say, unfortunately it becomes a more challenging ODE. I'll try to see what happens!

 

[OCR]

. . .and one inlet diameter think thick.

Fixed that for you. . . .

.

 

[hmmm27]

I thought this was possible due to the large difference in pressure at the bottom of the ocean and the usual pressure at the surface

It is, to start off with but, when the water level in the tube reaches the surface level, the weight pushing down is equal to the buoyancy pushing up.

But, don't despair : fresh water is lighter than sea water, so...

If you stick a seawater-RO filter assembly onto the bottom end of your 11km long pipe and mosey over to the Challenger Deep, you could make a 90ft high fountain of fresh water in the middle of the Pacific Ocean.

 

[erobz]

Updating the EOM to include some mass $m$ accelerating at $\ddot h$ outside of the pipe:

$$ \left( \frac{m}{\rho A} + h \right) \frac{du}{dh} + u = g( H-h) $$

Where $u=\frac{v^2}{2}$

This transformed linear ODE can be solved by integration factor:

$$ \frac{du}{dh} + \overbrace{\frac{1}{\frac{m}{\rho A} + h }}^{P(x)} ~u = \overbrace{\frac{g( H-h)}{ \frac{m}{\rho A} + h }}^{f(x)} $$

Integrating factor:

$$ e^{\int P(x) dx} = e^{\int \frac{dh}{\frac{m}{\rho A} + h }}= e^{\ln \left| \frac{m}{\rho A} + h \right|} = \frac{m}{\rho A} + h $$

and

$$ \frac{d}{dh} \left[ \left( \frac{m}{\rho A} + h \right) ~u \right] = g( H-h) $$

$$ \implies v(h) = \sqrt{ \frac{g}{\frac{m}{\rho A} + h } \left( Hh- \frac{h^2}{2} \right)} $$

This gives a peak fountain height ( $v= 0$ ) of $h = 2H$.

This ( obviously) ignores friction (viscosity) which would be more than significant here, but I did small scale test with a clear large diameter straw ( $D \approx 8 ~\rm{mm}$) dipped $H \approx 100 \rm{mm}$ and qualitatively it peaked at about a half of $H$ above the water surface.

Thoughts, concerns/criticism welcome as always?