🚀 go-pugleaf

John M. Cavallo wrote:
> 
> Peter Connor wrote:

[deletions here...   recall that the original posting was a description
of a "perpetual" fresh water fountain, using reverse osmosis driven
by differential density of salt vs. fresh water, at great depth.]

> > The pressure on the water at the bottom of the tube owing to the volume
> > of water above it would be equal to that on the water outside the tube
> > at the same depth, (imagine the water without the tube being there), the
> > only difference to this equilibrium is the additional pressure placed on
> > the pure water due to osmosis, this extra pressure forces the water up
> > out of the tube.

Er, this is a bit off.  The pressure inside the tube at the bottom is
always lower than the pressure outside; upward flow occurs only if the
P is lower by *more* than the osmotic pressure of the brine outside.
The fountain is driven by differential density, exactly as the original
poster described.  But only until equilibrium is reached.  

> > To answer another query on this thread the energy is actually coming
> > from a reduction in the potential energy of the ocean as a whole as
> > pushing the water out will of course reduce the overall sealevel.

No, I think the original problem had the fresh water flowing back
into the sea, at its upper surface, so the level stays the same.
However, the energy *does* come from reduction of gravitational
potential energy, as the salt concentration comes to its
equilibrium distribution, which has a lower center of mass.

... The system,
> therfore, would violate the first law of thermodynamics unless the fresh
> water level is at (or below) sea level.

That is true, at equilibrium.  The thing to realize about this
problem is that, in an ocean mixed by input of heat, the salt
concentration is *not* at equilibrium.  Cut off all such inputs,
and the salt will settle under gravitation until the concentration
at the bottom is much greater than at the top.  In such an
equilibrium ocean, the fountain will not work, because osmotic
pressure at the membrane will have increased in exact proportion
to the density difference that is the fountain's only driving force.
If we start in non-equilibrium, the fountain works by helping to
bring the system more rapidly into equilibrium, like a catalyst.
But then, without further input of energy, it will stop.  In fact,
the fresh water will eventually fall below sea level, to a
depth determined by the osmotic pressure at the sea's surface.
This must be true because, if you added an osmotic membrane at
the top, the system must finally come to equilibrium at the top
as well as at the bottom.

 The entire pressure drop has to
> be across the membrane.

I don't follow your reasoning here.  The pressure drop due to
flow can occur anywhere; though I agree, it will mostly be at
the membrane.  Of course the statement is absurd if you meant
it to apply to static pressure, so I assume you didn't.

> 
> John.