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Mark Borgerson

In article ,
says...
Mark Borgerson wrote:
In article ,
says...
snip

Yes, and this "migration" is simple diffusion. *And* you have (in the
example above) 33' of column it has to diffuse through on the seawater
side, and however many feet of column on the freshwater side it has to
traverse prior to condensation. If both columns (fresh and sea) are
referenced to the same height, then the evacuated column height on both
sides will be the same, and that diffusion path will be up to 66'. That
does not happen quickly.

How do you get 33' as 1/2 of the diffusion path.

A quick thumbnail guesstimation at where equilibrium would likely be
reached. I didn't take the time to calculate the exact heights.

I think there will be
about 33 feet of water in the column on each side

Then I think you would be wrong, unless your columns are significantly
longer than that, probably more like 50+ feet.

---to provide the
weigth that pulls the pressure down. That would leave only about
7 feet of water vapor path on each side of the column.

There is no vacuum to hold the water up - the vacuum is what you are
trying to *create*. The water columns will drop until there is an
equilibrium point reached between the external atmospheric pressure, the
height (weight as you state) of the water column, and the pressure in
the headspace (the U-tube). The water columns *must* retreat, or the
headspace stays at atmospheric pressure. If the tubes are long enough,
and the initial column heights are high enough, then when you reach
equilibrium, you'll have close to a vacuum and close to 33' water column
heights. And a lot more empty headspace than you started with.

I see the problem. I am assuming that you completely fill a 40 foot
tube with water using a pump capable of providing about 16-20 PSIG.
That fills the tube completely with water--at which point you
close the tube (with a one-way valve). When you release the
pressure at the bottom end, the water falls to the point where the
weight of the water column is one atm (about 14.7PSIA) minus the
vapor pressure of water at 20deg C. The vapor pressure of water
at 20C is about 17.5mmHg, or about 2.3% of the 760mmHg standard
atmosphere.

Since a mercury has a density 13.6, the column of water will
be 13.6 * (760- 17.6)mm high. That's 10.1m high, or
about 33.12 feet high. In a 40-foot tube, that would leave
about 7 feet of water vapor at the top of the tube and 33 feet
of water below the vapor.

Use the ideal gas law: PV=nRT

For our evacuation purposes, nRT is a constant (#moles is constant, R
doesn't change, and assume constant temperature), so if you start with a
volume of 1 liter, and a pressure of 14.7 psia, and you want to reduce
that pressure to 1.47psia, then you need a 10-fold volume increase. You
want to reduce it to 0.147psia? then you need a 100-fold initial-volume
increase.

What is the 1 liter to which you refer?

This is not a closed system---the tube is open to a reservoir at
atmospheric pressure at the bottom.

I'm assuming that you start with a head space (or initial volume)
of zero. You then simply have to evaporate enough water to fill
the top of the tube with water vapor to the point where vapor
pressure + water weight = 1ATM.

I'm not sure that 'diffusion' is the proper term for the motion
of the water vapor. After all, the heat engine is providing
water vapor on one side and condensing it on the other---so there
is a net mass flow and probably a small pressure differential to
move the vapor.

Well, diffusion is the primary mechanism. What happens when your 'heat
engine' creates water vapor? It doesn't just immediately condense on
the other side. It creates pressure on the heating side, which does two
things. One, it drives both the water columns *downward*, and it raises
the boiling point on the seawater side (it does, however, make
condensation on the fresh side more efficient as well). You can't look
at this as a static system where the pressure stays the same or the
column heights stay the same. It's a dynamic system, and will reach an
equilibrium point with the columns much lower than the initial starting
point, and the headspace pressure much higher.

Well, not much higher----only about 17.5 mmHG higher. But that IS
a lot higher than zero! ;-)

And don't forget, there will also be significant evaporation (due to low
partial pressures) on the freshwater side that will be in equilibrium
with (and in opposition to) the condensation process. It's not as simple
a system as it seems.

That's why this system *will* work, but it must work very slowly.

Still (pun intended), you need a lot of heat to provide the energy
to evaporate the water or it will soon cool to the point where
its vapor pressure is reduced and the process slows drastically.

My 'guess' would be that the system would end up operating around
4-5psia when equilibrium is reached, which would require a temp of about
60°C (140°F) to maintain boiling.

AHA!, you're assuming a much higher operating temperature than me.
I was assuming something on the order of 20 to 25C. You're going
to have to add to your energy budget the heat necessary to raise
the water temperature from 20C to 60C, then.

If the equilibrium pressure is really 1/3ATM, then there will be
about 20 feet of water in the 40-foot tube and 20 feet of vapor.
If you're going to work at those temperatures and pressures, you
probably need only a 22-foot tube.

Here in my neck of the woods, our energy from the sun ranges from about
220-360 BTU/ft^2/Hr measured at normal incidence, depending on the time
of year. A couple of decades ago I worked at a solar test lab and we
tested all kinds of collectors, including swimming pool collectors which
are unglazed (i.e. no cover over them to exclude wind). Bare copper
tubes, painted black, with no wind, are about 15% efficient at solar
absorption (#'s are from my old memory, so...) when the tubings'
longitudinal surface is perpendicular to the incident angle. However,
with a 3 mph wind (per ASHRAE 95-1981 which we used for indoor system
simulations) that efficiency drops to the low single digits. When you
factor in off-angle response (i.e. since the tubes won't be on a
tracking mount to keep them 'aimed at the sun") the basic efficiency
drops from ~15% to probably ~8%, and with the wind, between -3% to 3%.
So, using only the tube as a collector is a real challenge. Probably be
better using a flat-plate collector as the primary heater, but that's
another major addition to the complexity.

Of course, too much heat would kill the system with over pressurization.

The fact that the water 'boils' near room temperature does not
reduce the amount of heat required to change the water from
liquid to vapor.

No, in fact the lower pressure raises it a bit. Latent Heat of
Vaporization for water is inversely proportional to the pressure, albeit
the change is less than 10% IIRC.

As has been discussed, the simple idea does not address the problems
of salt buildup in the seawater side, or the addition of dissolved
gasses to the vacuum part of the loop.

Non-condensables are a rate limiter for the process, unless you want to
spend more energy for vacuum deaeration.

With a large enough (or double) sal****er tube you might get a
convection cell going with the cold, saltier water sinking and
pulling up warmer seawater to the top.

Certainly possible, but not easily doable.

You could solve the dissolved gas problem by periodically pumping
both tubes up enough to displace the accumulated gases.

Well, if you added a convection cell as above (another system that
requires time to reach an equilibrium condition to work), then the
periodic headspace purging would quench both the distillation and the
seawater convection systems. In reality, the purging would be likely be
very frequent given the size of tubes that would be practical.

Now the project is getting complex enough that an RO system
starts to look attractive!

Yep, sure does.

Keith Hughes

Mark Borgerson

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