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.

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) 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

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

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

Boiling Point Elevation
The boiling point of a solution is higher than that of the pure
solvent. Accordingly, the use of a solution, rather than a pure
liquid, in antifreeze serves to keep the mixture from boiling in a hot
automobile engine. As with freezing point depression, the effect
depends on the number of solute particles present in a given amount of
solvent, but not the identity of those particles. If 10 grams (0.35
ounces) of sodium chloride are dissolved in 100 grams (3.5 ounces) of
water, the boiling point of the solution is 101.7°C (215.1°F; which is
1.7°C (3.1°F) higher than the boiling point of pure water). The
formula used to calculate the change in boiling point ( Tb) relative
to the pure solvent is similar to that used for freezing point
depression:

Tb = i Kb m,

where Kb is the boiling point elevation constant for the solvent
(0.52°C·kg/mol for water), and m and i have the same meanings as in
the freezing point depression formula. Note that Tb represents an
increase in the boiling point, whereas Tf represents a decrease in
the freezing point. As with the freezing point depression formula,
this one is most accurate at low solute concentrations.

From:
http://www.chemistryexplained.com/Ce...roperties.html

In article ,
says...
SNIP

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.


You seem to have missed the fact that I proposed filling the tubes
completely with water so that the initial head space would be zero.
At that point you release the pressure on the water and it falls
to the point where water weight plus vapor pressure equals 1ATm.

At that point, you essentially have two water barometers,
interconnected at the top. One is salty and warm, and
one is fresh and cold. Neither need be too much longer
than 33 feet. The actual height of the water will be
less than 32 feet by a factor dependent on the temperature
of the water in the warm side.


The real practical problem lies in the addition of the dissolved
gases in the seawater to the water vapor in the headspace.
What we have here is a rather inefficient degassing column.
I spent a lot of time degassing seawater while working on
my MS in chemical oceanography. I was trying to measure
the dissolved hydrogen in seawater, and the oxygen, nitrogen,
methane, and other gases kept getting in the way!

Getting rid of the disssolved gases in the headspace and
as bubbles forming on the sides of the tube is going to
be a major headache. As soon as you release the pressure
and start warming the seawater side, bubbles are going
to form all along the tube as the temperature rises and
the pressure is less than 1ATM except at the bottom
of the tube.


Mark Borgerson

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


You seem to have missed the fact that I proposed filling the tubes
completely with water so that the initial head space would be zero.

No, it won't be zero. It can't be. If it is, then you have a solid
liquid stream, and it's just a siphon. You have to have headspace. And
it has to be sufficient to maintain separation of the seawater and
freshwater to prevent contamination when filling the tubes. And it has
to be large enough to prevent percolation carryover when boiling is
initiated.

At that point you release the pressure on the water and it falls
to the point where water weight plus vapor pressure equals 1ATm.

A solid liquid loop will not separate into two separate columns. They
have to be separated by a headspace. You can heat the seawater side and
create a headspace by liberating dissolved gases, then let the columns
drop to create vacuum, but you will have contaminated the freshwater side.


At that point, you essentially have two water barometers,
interconnected at the top. One is salty and warm, and
one is fresh and cold. Neither need be too much longer
than 33 feet. The actual height of the water will be
less than 32 feet by a factor dependent on the temperature
of the water in the warm side.

The real practical problem lies in the addition of the dissolved
gases in the seawater to the water vapor in the headspace.
What we have here is a rather inefficient degassing column.
I spent a lot of time degassing seawater while working on
my MS in chemical oceanography. I was trying to measure
the dissolved hydrogen in seawater, and the oxygen, nitrogen,
methane, and other gases kept getting in the way!

Getting rid of the disssolved gases in the headspace and
as bubbles forming on the sides of the tube is going to
be a major headache.

Not a headache, an impossibility (they're not really dissolved at that
point though) :-) That, and the increase in pressure due to water vapor
will make this an oscillating, self-quenching system. It'll require
more and more heat as the partial pressures of the non-condensables
increases, and the column heights will drop as the pressure goes up,
with the diffusion path increasing the whole time.

Keith Hughes

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


You seem to have missed the fact that I proposed filling the tubes
completely with water so that the initial head space would be zero.

No, it won't be zero. It can't be. If it is, then you have a solid
liquid stream, and it's just a siphon. You have to have headspace. And
it has to be sufficient to maintain separation of the seawater and
freshwater to prevent contamination when filling the tubes. And it has
to be large enough to prevent percolation carryover when boiling is
initiated.

At that point you release the pressure on the water and it falls
to the point where water weight plus vapor pressure equals 1ATm.

A solid liquid loop will not separate into two separate columns. They
have to be separated by a headspace. You can heat the seawater side and
create a headspace by liberating dissolved gases, then let the columns
drop to create vacuum, but you will have contaminated the freshwater side.


The head space is generated by the evaporation (or boiling) of some of
the water in a column. It's exactly the same principle that you get it
you fill a closed tube full of mercury and then invert it, placing the
end in a reservoir of mercury. (We call these things barometers.)
You start with no head space, but when you invert it, VOILA!
head space appears as the mercury sinks to a level where the weight
of the mercury equals the atmospheric pressure. You get a much
better vacuum with mercury, since it has a much lower vapor pressure
at room temperature.

A column of water will behave the same way. The column just has
to be much taller.

Some of the historical references on water barometers mention that,
despite precautions, the water in the barometer eventually got
contaminated with dissolved gases and they lost their accuracy.

At that point, you essentially have two water barometers,
interconnected at the top. One is salty and warm, and
one is fresh and cold. Neither need be too much longer
than 33 feet. The actual height of the water will be
less than 32 feet by a factor dependent on the temperature
of the water in the warm side.

The real practical problem lies in the addition of the dissolved
gases in the seawater to the water vapor in the headspace.
What we have here is a rather inefficient degassing column.
I spent a lot of time degassing seawater while working on
my MS in chemical oceanography. I was trying to measure
the dissolved hydrogen in seawater, and the oxygen, nitrogen,
methane, and other gases kept getting in the way!

Getting rid of the disssolved gases in the headspace and
as bubbles forming on the sides of the tube is going to
be a major headache.

Not a headache, an impossibility (they're not really dissolved at that
point though) :-) That, and the increase in pressure due to water vapor
will make this an oscillating, self-quenching system. It'll require
more and more heat as the partial pressures of the non-condensables
increases, and the column heights will drop as the pressure goes up,
with the diffusion path increasing the whole time.\

I agree with that part---except for the oscillation part. I think
the processes are slow enough and the thermal and physical masses
are high enough that the oscillations will be damped out and you
will see a slow change to equilibrium with little or no overshoot.


Mark Borgerson

N:dlzc D:aol T:com (dlzc) brought forth on stone tablets:
Dear Richard Casady:

"Richard Casady" wrote in message
...

On Sat, 29 Sep 2007 19:44:35 -0700, Keith Hughes
wrote:


Sounds like perpetual motion to me, but I'm
having a hard time envisioning what you're
describing above.

Of course you are, since it is basically nonsense.
No mention of where the energy comes from.


Links were provided. "waste heat" (from what process?) and / or
"solar heat" have been cited so far. All the vacuum does is move
boiling temperature closer to ambient. Making more common
materials suitable for this application.

David A. Smith




.... and much more importantly, making the process viable with far
lower quality heat.

bob
s/v Eolian
Seattle

On Mon, 08 Oct 2007 10:13:17 -0700, Keith Hughes
wrote:

...I proposed filling the tubes
completely with water so that the initial head space would be zero.

No, it won't be zero. It can't be. If it is, then you have a solid
liquid stream, and it's just a siphon.
....
A solid liquid loop will not separate into two separate columns.
....
Keith Hughes

Perhaps it would be better for you to check what is the maximal rise
(head) of a syphon. Can you guess what it might be?

Brian Whatcott Altus OK

On Mon, 08 Oct 2007 05:33:39 -0700, Keith
wrote:

Boiling Point Elevation
The boiling point of a solution is higher than that of the pure
solvent. Accordingly, the use of a solution, rather than a pure
liquid, in antifreeze serves to keep the mixture from boiling in a hot
automobile engine.....
From:
http://www.chemistryexplained.com/Ce...roperties.html


Actually, no. Ethylene glycol in its pure liquid state boils near
200 degC
http://www.dow.com/ethyleneglycol/about/properties.htm

It is usually cut to 50% dilution for use as an antifreeze.

Brian Whatcott Altus OK