🚀 go-pugleaf

remove-this wrote:
> 
> On Thu, 23 Oct 1997 23:10:32 +0800, Lozang Trinlae
> <lozang@mindless.com> wrote:
> 
> >John Cook wrote:
> >
> >> Here is the plan...
> >> >
> >> > Imagine a long tube reaching from the surface to the bottom of the japan
> >> > trench.  It's bottom end is a reverse osmosis filter unit.  It is filled with
> >> > fresh water.
> >> > The water outside the tube is salty and weighs around 1028 kg/cu mt - the
> >> > water inside the tube is fresh and weighs 1000 kg/cu mt.  This causes a
> >> > pressure diferential across the osmotic membrane of about 2 to 3% of the
> >> > external water pressure.
> >> >
> >> > Thats it.  Sufficient depth of water WILL cause a flow through the membrane
> >> > and up the tube.
> >> >
> >> > Such a device will sit there for'ever' emiting a fountain of fresh water.
> >> >
> >> > Kewl?
> >> >
> >> > I think the energy comes from slightly lowering the potential energy of the
> >> > salt in the ocean. Am I right about this?
> >> >
> >
> >I don't know what you mean by the 'potential energy of the salt'.
> >Basically what you have here is a type of barometer. By reducing the
> >density of the water (by filtering out the salt) in the tube that is
> >placed in the ocean, and atmospheric pressure would push the liquid up
> >into the reduced-pressure region, like the way a drinking straw works.
> >That is mechanical in nature; i.e., a mechanical potential difference
> >would occur across the filter membrane boundary. But if you look at
> >Boyle's law, P.1*V.1=P.2*V.2, you can see that V.2 (the volume of water
> >in the tube) = (P.1*V.1)/P.2. Any differences in P.1/P.2 look like they
> >will get washed out (no pun intended), by the relatively infinite value
> >of V.1 (the volume of the water in the ocean).
> >
> >However this kind of thing looks like it could be experimentally tried
> >easily at home, given the suitable filter. Go for it and let us know
> >your results!
> >
> >-Lozang Trinlae
> >
> >
> I think you have the wrong law and the wrong reasoning here.  

I don't have a very clear picture of the problem so this could be so. 

> Boyle's
> Law does not apply to water and solutions.  

Ok, I accept that Boyle's law applies to ideal gases, which the ocean
definitely is not. 

>The volume of the ocean
> has no bearing on the issue, only the depth of the membrane below the
> surface.  The pressure at this depth is a function of the depth and
> the density of the sea water.  That pressure will support a greater
> depth of fresh water inside the tube, therefore the water would tend
> to rise above the level of the ocean.

So the liquid pressure of lesser density and at lesser depth fresh water
will be lower than the water at higher density and further depth sea
water on the other side of the membrane. Then the difference in
pressures at the depth of the membrane is proportional to the difference
in water densities. But the power needed to get a flow going in a full
pipe is proportional not only to the pressure differential, but also the
force of gravity over the distance of the flow (not to mention other
factors such as average velocity). So even though there is an upward
pressure due to the density difference, wouldn't there still need to be
a differential greater than 9800 Newtons per meter^3 x the depth the
fresh water would have to travel to reach the exit of the tube, in order
to force the water out? (Density of fresh water is 1000 kg/m^3.) 

-lozang trinlae