🚀 go-pugleaf

Robert Virtue wrote:

> O.K.
> Using reference material within arm range.
> osmotic pressure of a salt solution of 1000 mg/l = 78.6 kPa
> therefore 35,000 mg/l approx. = 2751 kPa (not quite true for electrolytes but it
> will do)
> = 280 m water head
> Density fresh water = 1 g/cm3
> Density sea water = 1.025 g/cm3
>  Column of water required to obtain pressure differential of 280 m =
> 280/0.025 = 11,200 m.

Thanks for giving us some actual numbers.  Puts some well-needed
perspective on all this.  Note, in addition, that this figure requires
the average salinity to be a constant function of depth; but that surely
is not the case.  The bottom should be saltier than the top (sorry,
don't
know how much it actually is) and certainly *must* be so if the ocean
were in equilibrium.  So, the required depth would actually be greater;
indeed, at equilibrium, there is *no* depth sufficient to cause flow.

[...remainder of most enlightening article deleted...]

I would only add that the energy input required to keep this running
forever, consists in *lifting* the bypass water above its equilibrium
level, not just moving it out of the way.  There's only so much you
can move it out of the way before it has to start piling up, so to
speak.  The thing that keeps this from being a p.m. of the 2nd Kind,
is that the equilibrium concentration gradient in a gravitational
field is *not* zero; you have to get a TdS for diffusion upward that
exactly equals the nonzero dE due to gravity.

IMO this is the *really* "kewl" thing about this system (if I may
borrow the Ur-original poster's word):  the fact that, by balancing
diffusion against gravitational settling, the salt ions will always
conspire to turn the machine off, eventually.  And that's true for
any system, regardless of the actual concentrations, densities, or
even the value of g.

> By the way OSMOSIS is  what you get when you turn off the power at an RO plant :-)

:-)  Nice to hear from an expert who knows whereof he speaks.