Potable Water - The Third Way.

On Sat, 29 Sep 2007 19:44:35 -0700, Keith Hughes
<keithahughes.TakeThisOut@qwest.net> wrote:

>Vic Smith wrote:
>> On Sat, 29 Sep 2007 17:43:02 -0500, Brian Whatcott
>> <betwys1.TakeThisOut@sbcglobal.net> wrote:
>>
>>> On Sat, 29 Sep 2007 20:40:57 GMT, richardcasady.TakeThisOut@earthlink.net (Richard
>>> Casady) wrote:
>>>
>>>
>>>> There is a reverse osmosis watermaker intended for liferaft use, with
>>>> a hand pump, and RO takes hundreds of psi. That is what you want, if
>>>> you actually need high pressure.
>>>>
>>>> Casady
>>>
>>> I looked up an example
>>> The Katadyn Survivor 35 hand pumped was formerly
>>> called the PUR Survivor 35 RO.
>>> At 30 strokes/minute for 1.2 gall/hr - it costs $1500.
>>>
>> Calorie expenditure by the survivor(s) could be a problem here.
>
>Oh yeah, right. Now you want to survive also. Geez, what next?
>
>> The strokes for this RO unit can probably be performed by devising a
>> simple hydraulic pump to move gears, cams, and levers.
>> The pump cylinder itself would probably need an inverted U tube with
>> legs perhaps 32' or 33' long.
>> An initial vacuum might be applied to the top of the U-tube by using a
>> fitting that can be connected to the PUR Survivor 35 RO.
>> Once the water starts flowing through the vane at one end of the U
>> tube, and the vane shaft is turning the gears, cams and levers will be
>> clacking way, running that PUR unit on auto, good as gold.
>> After that it's all gravy until you have to change the membrane.
>> In the meantime you can spend your time fishing until rescued.
>
>Sounds like perpetual motion to me, but I'm having a hard time
>envisioning what you're describing above.
>
Sorry, it was all said jokingly, but appears to be a poor joke.
I just went in a circle from the perpetual U-tube distiller to that
concept being employed to perpetually pump a purchased RO unit.
I never intended to make sense, except maybe to say it's time to go
fishing.

--Vic
 >> Stay informed about: Potable Water - The Third Way. 

In rec.boats.cruising jim.isbell <jim.isbell.DeleteThis@gmail.com> wrote:
:I give up. My Masters degree in Physics is of no value here. My

Not if you think salt water has a lower boiling point than fresh, no.
 >> Stay informed about: Potable Water - The Third Way. 

On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard"
<wilburhubbard RemoveThis @thefarm.invallid> wrote:

>When it gets full you haul it up and empty in
>into your tanks. Reverse osmosis without any energy used to get it.
>Ain't Wilbur brilliant?
You haul it up without using any energy to do it? Absolutely not/ It
will take a foot pound for each pound for each foot you haul it.
No your basis for perpetual motion will not work. And is the opposite
of brilliant.

Casady
 >> Stay informed about: Potable Water - The Third Way. 

Richard Casady brought forth on stone tablets:
> On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard"
> <wilburhubbard.TakeThisOut@thefarm.invallid> wrote:
>
>
>>When it gets full you haul it up and empty in
>>into your tanks. Reverse osmosis without any energy used to get it.
>>Ain't Wilbur brilliant?
>
> You haul it up without using any energy to do it? Absolutely not/ It
> will take a foot pound for each pound for each foot you haul it.
> No your basis for perpetual motion will not work. And is the opposite
> of brilliant.
>
> Casady

Well, not quite. The harvested fresh water is actually buoyant in the
sea water. Hauling up the water is energy free. Hauling up the
container and the rope is not, however.

With suitable flotation, the container could be made neutral-buoyant,
and so hauling it up could be free also, Finally, if the rope were HD
polyethylene or something else with about 1.0 density, the rope could be
free to hoist too. It would be necessary to attach a weight greater
than the weight of water to be harvested to the container in order to
get it to sink. This weight would then be disconnected/abandoned before
hoisting the recovered water. From an energy standpoint, the investment
would be that necessary to cover the friction in the hauling apparatus,
and the the invested energy content of the abandoned weight (steel:
high, concrete: medium, rock: free).

Venting the container to the surface would be impractical. Evacuate it
instead.

With Wilbur, one must be careful to not discard the wheat with the chaff...

bob
s/v Eolian
Seattle
 >> Stay informed about: Potable Water - The Third Way. 

On Tue, 02 Oct 2007 09:59:46 -0700, RW Salnick <salnick.TakeThisOut@no.spam.org>
wrote:

>Richard Casady brought forth on stone tablets:
>> On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard"
>> <wilburhubbard.TakeThisOut@thefarm.invallid> wrote:
>>
>>
>>>When it gets full you haul it up and empty in
>>>into your tanks. Reverse osmosis without any energy used to get it.
>>>Ain't Wilbur brilliant?
>>
>> You haul it up without using any energy to do it? Absolutely not/ It
>> will take a foot pound for each pound for each foot you haul it.
>> No your basis for perpetual motion will not work. And is the opposite
>> of brilliant.
>>
>> Casady
>
>Well, not quite. The harvested fresh water is actually buoyant in the
>sea water. Hauling up the water is energy free. Hauling up the
>container and the rope is not, however.
>
>With suitable flotation, the container could be made neutral-buoyant,
>and so hauling it up could be free also, Finally, if the rope were HD
>polyethylene or something else with about 1.0 density, the rope could be
>free to hoist too. It would be necessary to attach a weight greater
>than the weight of water to be harvested to the container in order to
>get it to sink. This weight would then be disconnected/abandoned before
>hoisting the recovered water. From an energy standpoint, the investment
>would be that necessary to cover the friction in the hauling apparatus,
>and the the invested energy content of the abandoned weight (steel:
>high, concrete: medium, rock: free).
>
>Venting the container to the surface would be impractical. Evacuate it
>instead.
>
>With Wilbur, one must be careful to not discard the wheat with the chaff...
>
>bob
>s/v Eolian
>Seattle


And how much of the time are you sailing in 500 ft deep water, which
was the original specification?

Bruce in Bangkok
(brucepaigeATgmailDOTcom)
 >> Stay informed about: Potable Water - The Third Way. 

brucedpaige.DeleteThis@gmail.com brought forth on stone tablets:
> On Tue, 02 Oct 2007 09:59:46 -0700, RW Salnick <salnick.DeleteThis@no.spam.org>
> wrote:
>
>
>>Richard Casady brought forth on stone tablets:
>>
>>>On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard"
>>><wilburhubbard.DeleteThis@thefarm.invallid> wrote:
>>>
>>>
>>>
>>>>When it gets full you haul it up and empty in
>>>>into your tanks. Reverse osmosis without any energy used to get it.
>>>>Ain't Wilbur brilliant?
>>>
>>>You haul it up without using any energy to do it? Absolutely not/ It
>>>will take a foot pound for each pound for each foot you haul it.
>>>No your basis for perpetual motion will not work. And is the opposite
>>>of brilliant.
>>>
>>>Casady
>>
>>Well, not quite. The harvested fresh water is actually buoyant in the
>>sea water. Hauling up the water is energy free. Hauling up the
>>container and the rope is not, however.
>>
>>With suitable flotation, the container could be made neutral-buoyant,
>>and so hauling it up could be free also, Finally, if the rope were HD
>>polyethylene or something else with about 1.0 density, the rope could be
>>free to hoist too. It would be necessary to attach a weight greater
>>than the weight of water to be harvested to the container in order to
>>get it to sink. This weight would then be disconnected/abandoned before
>>hoisting the recovered water. From an energy standpoint, the investment
>>would be that necessary to cover the friction in the hauling apparatus,
>>and the the invested energy content of the abandoned weight (steel:
>>high, concrete: medium, rock: free).
>>
>>Venting the container to the surface would be impractical. Evacuate it
>>instead.
>>
>>With Wilbur, one must be careful to not discard the wheat with the chaff...
>>
>>bob
>>s/v Eolian
>>Seattle
>
>
>
> And how much of the time are you sailing in 500 ft deep water, which
> was the original specification?
>
> Bruce in Bangkok
> (brucepaigeATgmailDOTcom)

500 feet? That's only 83 fathoms. My sailing area is Puget Sound, much
of which is 150 fathoms or more. Why? Is Thailand in a skinny water zone?

bob
s/v Eolian
Seattle
 >> Stay informed about: Potable Water - The Third Way. 

On Wed, 03 Oct 2007 08:13:20 -0700, RW Salnick <salnick.TakeThisOut@no.spam.org>
wrote:

>brucedpaige@gmail.com brought forth on stone tablets:
>> On Tue, 02 Oct 2007 09:59:46 -0700, RW Salnick <salnick.TakeThisOut@no.spam.org>
>> wrote:
>>
>>
>>>Richard Casady brought forth on stone tablets:
>>>
>>>>On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard"
>>>><wilburhubbard.TakeThisOut@thefarm.invallid> wrote:
>>>>
>>>>
>>>>
>>>>>When it gets full you haul it up and empty in
>>>>>into your tanks. Reverse osmosis without any energy used to get it.
>>>>>Ain't Wilbur brilliant?
>>>>
>>>>You haul it up without using any energy to do it? Absolutely not/ It
>>>>will take a foot pound for each pound for each foot you haul it.
>>>>No your basis for perpetual motion will not work. And is the opposite
>>>>of brilliant.
>>>>
>>>>Casady
>>>
>>>Well, not quite. The harvested fresh water is actually buoyant in the
>>>sea water. Hauling up the water is energy free. Hauling up the
>>>container and the rope is not, however.
>>>
>>>With suitable flotation, the container could be made neutral-buoyant,
>>>and so hauling it up could be free also, Finally, if the rope were HD
>>>polyethylene or something else with about 1.0 density, the rope could be
>>>free to hoist too. It would be necessary to attach a weight greater
>>>than the weight of water to be harvested to the container in order to
>>>get it to sink. This weight would then be disconnected/abandoned before
>>>hoisting the recovered water. From an energy standpoint, the investment
>>>would be that necessary to cover the friction in the hauling apparatus,
>>>and the the invested energy content of the abandoned weight (steel:
>>>high, concrete: medium, rock: free).
>>>
>>>Venting the container to the surface would be impractical. Evacuate it
>>>instead.
>>>
>>>With Wilbur, one must be careful to not discard the wheat with the chaff...
>>>
>>>bob
>>>s/v Eolian
>>>Seattle
>>
>>
>>
>> And how much of the time are you sailing in 500 ft deep water, which
>> was the original specification?
>>
>> Bruce in Bangkok
>> (brucepaigeATgmailDOTcom)
>
>500 feet? That's only 83 fathoms. My sailing area is Puget Sound, much
>of which is 150 fathoms or more. Why? Is Thailand in a skinny water zone?
>
>bob
>s/v Eolian
>Seattle

From Singapore north through either the Gulf of Thailand or up the
west coats of Malaysia or most of the western part of Indonesia 150
ft. of water would be deep water. Wilbur's invention isn;t going to
work very well over here.


Bruce in Bangkok
(brucepaigeATgmailDOTcom)
 >> Stay informed about: Potable Water - The Third Way. 

In article <kunbf353h0oq5kqg01mnlkuoe0cdpmr90f.RemoveThis@4ax.com>,
nsremovable.RemoveThis@iinet.net.au says...
> On Sat, 22 Sep 2007 10:55:52 -0500, Brian Whatcott
> <betwys1.RemoveThis@sbcglobal.net> wrote stuff
> and I replied:
>
> But what is the cheap source of getting the vacuum? I figured there
> had to be a vacuum, although it was not said. But how do you get it?
>
> >>Well no, he obviously hadn't figured that out. Nor can anybody figure
> >>out what is going to hold a column of water 40 ft high as was stated in
> >>the original post. The tubes may be 40 feet but the column of water will
> >>be considerably less. How much less will depend on how much energy is
> >>heating on the hot side and how much energy is cooling on the cool side.
> >>The total amount of energy needed is not going to be any different than
> >>any other distilling method.
> >> Unless you have the free or cheap sources of cooling and heating at
> >>specific temperatures this isn't going to work any better either.
> >>
> >>-jim
> >
> ><<SNIP>>

Simply put a one-way (out only) valve at the top where the two tubes
are joined. Pump water up both tubes to about 3" from the top,
displacing the air in the tubes. That will only require about
16 PSI from the pumps.

The major problem would seem to be that vigorous boiling is going
to carry over salt and contaminants from the boiling salt water unless
the tubes are large or there is some sort of debubbler on the
salt water side.

For heating and cooling, I suppose that you could use the sunny and
shady sides of a sailboat mast.

Mark Borgerson
 >> Stay informed about: Potable Water - The Third Way. 

In article <46fe85ab$0$503$815e3792@news.qwest.net>,
keithahughes.RemoveThis@qwest.net says...
>
>
> jim wrote:
> >
> > "jim.isbell" wrote:
> >>> Ah well, another great idea skuppered by dat old devil science
> >>>
> >>> Bruce in Bangkok
> >>> (brucepaigeATgmailDOTcom)
> >> A 32' column of water is a continuous vacuum pump.
>
> This is just plain wrong. As a *unit of measure* 32 feet of water
> column equals about 13.9 psi. Meaning, if you pumped a 40' column up to
> a 39' height with water, equalized the headspace to atmospheric pressure
> (assuming 14.7psia), sealed it, then allowed gravity to *drain* the
> water column to a height of 2', the resulting pressure in the headspace
> will be about 0.8psia. Now you also have 33' of empty evacuated column.
>
> >> As long as you put
> >> water (salt water) into the column it will pull down and keep a vacuum
> >> in the top of the column.
>
> Sorry, this makes no sense. Putting water in does not cause it to "pull
> down". Yes, you have supply makeup water to maintain column height lost
> to evaporation.
>
> >> The fresh water distills off the top of the
> >> saltwater column then migrates
>
> 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. I think there will be
about 33 feet of water in the column on each side---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.

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.

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.
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.

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.

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

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

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


Mark Borgerson
 >> Stay informed about: Potable Water - The Third Way. 

In article <1191179636.430473.28350 DeleteThis @y42g2000hsy.googlegroups.com>,
jim.isbell DeleteThis @gmail.com says...
> I give up. My Masters degree in Physics is of no value here. My
> Bachelors degree in Math is of no value here. My 20 years with the
> university (retired) means nothing. Someone with an opinion (however
> false) instead of facts of physical science, seems to be more able to
> swing the belief of the uninformed.
>
> I will try to explain it again.
>
> The vacuum will hold the column of water in the tube.
>
> Dont believe it?
> Test this statement, take a simple soda straw stick in in a glass of
> water put your finger over the end and lift it out. The 8" column of
> water stays in the straw because of the vacuum in the top of the
> straw. Now remove your finger and the water drops out. So, it
> doesn't TAKE a 32' column of water, but that is the tallest column of
> water that will be suspended, a simple law of physics.
>
> Thats why a lift pump like the old rocker handle pitcher pumps have to
> be replaced with either submerged or Jet pumps in deeper wells. A
> lesser column WILL work however. At the top is a vacuum. If its 32
> feet, thats the greatest vacuum you can create. There is salt water
> on one side and fresh water on the other. The salt water will boil
> earlier because of the salt content.
>
> Now test that statement.
> Put a pot on the stove and then before it comes to a boil add salt.
> Voila, it begins to boil.


BZZZZT!!! FALSE!!!


Salt water DOES NOT boil at a lower temperature than fresh water. It
boils at a higher temperature. When you add salt to heated water,
it appears to boil because the water is superheated near the bottom of
the pan and the salt crystals provide "nuclei" that start the boiling
process. You can easily verify this by starting with two pans
of water, one salty and one fresh and heating them to their respective
boiling points and measuring the temperature. More explicit
instructions, aimed at middle school students are at:

http://aquarius.nasa.gov/pdfs/prop_fresh_sea.pdf


<snip>


Mark Borgerson






<<SNIP>>
 >> Stay informed about: Potable Water - The Third Way. 

Mark Borgerson wrote:
> In article <46fe85ab$0$503$815e3792@news.qwest.net>,
> keithahughes.RemoveThis@qwest.net 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.

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.

>
> 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.

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.

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) saltwater 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
 >> Stay informed about: Potable Water - The Third Way. 

In article <470932a8$0$493$815e3792@news.qwest.net>,
keithahughes.TakeThisOut@qwest.net says...
> Mark Borgerson wrote:
> > In article <46fe85ab$0$503$815e3792@news.qwest.net>,
> > keithahughes.TakeThisOut@qwest.net 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) saltwater 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
 >> Stay informed about: Potable Water - The Third Way. 

Mark Borgerson wrote:
> In article <470932a8$0$493$815e3792@news.qwest.net>,
> keithahughes DeleteThis @qwest.net says...
>> Mark Borgerson wrote:
>>> In article <46fe85ab$0$503$815e3792@news.qwest.net>,
>>> keithahughes DeleteThis @qwest.net says...
>> <snip>
>>
<snip>

>>> 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).

Uhmm, a manual valve is a manual valve. A "one-way" valve is a
checkvalve, and you wouldn't need to close it.

> 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.

Ahh, no. See below...

>> 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?

It is an example, for illustration purposes. It's the headspace (i.e.
the amount of volume *not* filled with water, prior to closing the valve
and letting the water columns 'fall').

The point is, whatever your starting headspace volume is, to get
anywhere near a vacuum, the *VOLUME* of the headspace must increase 100
fold. That is the relevance of the ideal gas law. If you start with a
100ml headspace, then to get a decent vacuum, the water columns have to
drop to a point where the headspace is 10L. *AT THAT POINT* you have
sufficient vacuum to support a water column of around 30'. But the
columns have dropped significantly to achieve that vacuum, and thus the
columns must be much higher, as must the initial water column height.

The *only* way the headspace volume increases is if the water columns
drop significantly, and the only way significant vacuum is created is if
the sealed volume increases tremendously.
>
> 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!

No, a lot higher. You're confusing vapor pressure with the "steam"
pressure during distillation. Huge difference. Vapor pressure is the
countervailing force (fighting condensation as it were) on the FRESH
water side of the system (and the seawater side). Vapor pressure in
both columns will be about the same, so you have to boil the seawater
column to get any significant vapor transfer. This results in *much*
higher pressure, which lowers the columns and increases the vapor path
and....

>> 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.

Yes, because you're mistaking the amount of "vacuum" you'll have
available when the system reaches equilibrium.

> I was assuming something on the order of 20 to 25C.

Then you are assuming an almost perfect vacuum, which can't happen since
the boiling Must significantly raise the headspace pressure.

> 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.

You need to get back to the gas law to see where this error lies. You
have to *create* the vacuum. That requires a HUGE increase in volume
for whatever the initial headspace is. For this to happen you need a
much longer tube to start with.

Keith Hughes
 >> Stay informed about: Potable Water - The Third Way.