Hydrostatic Vacuum Distillation Idea

[jrodatus]

Hey, I came across this idea a couple of years ago. It's been a long time since I took introductory physics. But I'm curious as to whether this is feasible. If I'm crazy, just say so!

See attached image.

1. Intake tube, vacuum chamber and condenser are initially filled with water. Valve (A) remains open; valve (B) is closed. Vacuum chamber and condenser are raised to a height more than 10 meters above the intake reservoir.

2. Condenser is temporarily oriented such that its contents can drain into the chamber while the hydrostatic vacuum is forming. Water level in the intake tube settles at a height about 10 meters above the reservoir, balancing atmospheric pressure, while a vacuum is sustained in the chamber and condenser above that height. Valve (A) is closed.

3. At this point, the shelf inside the vacuum chamber has retained a significant amount of water from draining back down the intake tube. A solar collector focuses sunlight through the top of the glass container to slightly warm this water (>80˚F).

4. The warm water rapidly vaporizes in the presence of the vacuum. Water vapor diffuses into the condenser tube, condenses and accumulates at valve (B).

5. There must be enough space between the condenser and valve (B) to hold the same volume of water as does the evaporation shelf in the vacuum chamber. Once the transfer is complete, valve (B) is opened to discharge the distilled liquid. (Obviously there should be a separate valve on the chamber to let in atmosphere lest we blow the distilled water back inside).

6. In order to restart the process, both valves are opened while the displacement pump refills the vacuum chamber and condenser with water. Condenser is temporarily raised/oriented such that its contents do not flow out the open valve (B).

7. Once chamber and condenser are filled, valve (A) remains open and valve (B) is closed. Return to step 2.

It seems that the only energy supplied besides sunlight is that for refilling the vacuum chamber with water each time. Here's my attempt at estimating the work required per use.

Intake tube diameter: 1 cm
Intake tube length: 1000 cm
Intake tube volume: 785.4 cm^3 = .7854 L
Vacuum chamber volume: 2 L
Condenser volume: 1.5 L

:: Total volume of water displacement: .7854 + 2 + 1.5 = 4.2854 L

Mass of displacement: 4.2854 kg
Height of displacement: 1 m (guess)

:: Work done: 4.2854 kg * 9.81 m/s^2 * 1 m = 42.04 J

Evaporation shelf volume: 1 L
Energy input per liter distilled: ~42 J/L

Considering that it ordinarily takes a whopping 2,573,598 J to boil off a liter of water, that figure is pretty unbelievable. Maybe I need to account for the work to overcome atmospheric pressure in lifting the water? Hmm.

 

[SteamKing]

Sounds similar to a flash distillation system, where water is heated to a certain temperature, and then let into a chamber which has a reduced pressure.

If the water is heated to a temperature above the boiling point at chamber pressure, then pure water will flash into steam, which can be collected and condensed.

This is one type of water maker used aboard ship.

 

[jim hardy]

It seems that the only energy supplied besides sunlight is that for refilling the vacuum chamber with water each time. Here's my attempt at estimating the work required per use.

i think your concept is valid in principle.

in practice i expect you'll have trouble with gases other than water vapor.
Real condensers have "air ejectors" to remove noncondensibles. Most general purpose water contains dissolved air.
The 'air ejector' is a substantial vacuum pump. Mine were steam powered so it's hard for me to put a number on their power requirement.Steamking probably knows more than i about evaporators. We made our boiler water by ion exchange resin beds.

 

[jrodatus]

in practice i expect you'll have trouble with gases other than water vapor.
Real condensers have "air ejectors" to remove noncondensibles. Most general purpose water contains dissolved air. The 'air ejector' is a substantial vacuum pump. Mine were steam powered so it's hard for me to put a number on their power requirement.

Thanks, that's what I was looking for. I guess you're saying that the escaped air will increase the chamber pressure enough that the water's boiling point will rise too much and nothing will happen. These figures seem to suggest that the pressure increase wouldn't be significant. But maybe they're wrong:

Air dissolvable in water at 1 atm and 25˚C: ~0.023 g/kg
Mass of water: 1 kg
Mass of escaped air: 0.023 g = 0.000023 kg
Volume of vacuum chamber and condenser: 3.5 L = 0.0035 m^3
Density of diffused air: .000023 kg / 0.0035 m^3 = 0.0066 kg/m^3

Pressure increase due to diffused air: 0.0066 kg/m^3 * 287.058 J/kg*K * 298.15 K = 564.87 Pa = 0.1668 inHg
(http://en.wikipedia.org/wiki/Density_of_air#Temperature_and_pressure)

In theory, 0.1668 inHg shouldn't really affect the boiling point that much?

ion exchange resin beds.

But what do I know?

 

[jim hardy]

Nice calc!

23milligrams = 23/29 millimole , or about 17ml at STP.

If you remove that 17 ml between cycles you'll be okay
just be aware it doesn't condense so will have to be removed mechanically.

At 1/.017 = 56, so at 1/56 atmosphere it'll occupy a liter

is that 1809 pascals?

In theory, 0.1668 inHg shouldn't really affect the boiling point that much?

My steam tables are packed away, so can't check. But you're right, not a lot. Just don't let it accumulate in your chamber.

Lastly - remember that air is denser than steam, molecular weight is 29 vs 18, so it'll accumulate in the bottom.

old jim

 

[jrodatus]

My steam tables are packed away, so can't check. But you're right, not a lot. Just don't let it accumulate in your chamber.

http://www.trimen.pl/witek/calculators/wrzenie.html
Looks like going from 0.5 to 0.6668 inHg it should rise from 15 to 20˚C (59-68˚F).

But there has to be other factors hindering the vacuum formation. Otherwise I might as well expect to reach near 0.1668 inHg, boiling point -2˚C. And that's ridiculous!

If you remove that 17 ml between cycles you'll be okay just be aware it doesn't condense so will have to be removed mechanically.

Maybe I'm missing something. In step 5 atmosphere is let back into the chamber before repriming with water. (Via the displacement pump). It seems to me that the entire 3.5 L of air (including the original 17mL) is just forced out of the chamber because it's lighter than the incoming water.

I'm failing to see why a mechanical vacuum pump is needed between cycles. But my education is an A.S. Math, so that could be the problem here :uhh:

Thanks for your time Mr. Jim.

 

[jrodatus]

Sounds similar to a flash distillation system, where water is heated to a certain temperature, and then let into a chamber which has a reduced pressure.

If the water is heated to a temperature above the boiling point at chamber pressure, then pure water will flash into steam, which can be collected and condensed.

This is one type of water maker used aboard ship.

Wow thanks- I just realized you indirectly answered my big question! A flash distillation system uses multiple stages because once the pressure of the vapor reaches a certain point, no more water can evaporate and we have to restart the cycle. So this system probably won't produce 1 L for every 42 J. The rate could be much lower. Hmm...

temperature easily reachable with solar collector: 39˚C (102˚F)
maximum pressure permissible to boil water at 39˚C: 6772.78 Pa
(http://www.trimen.pl/witek/calculators/wrzenie.html)

R = 8.3144621 J/mol*K
T = 312.15 K (39˚C)
P = 6772.78 Pa
V = 3.5 L = 0.0035 m^3
n = unknown
PV/RT = n
(6772.78 Pa * 0.0035 m^3) / (8.3144621 J/mol*K * 312.15 K) = 0.009133506 mol

molar mass of water: 18.0152 g/mol
:: amt of water that can vaporize B4 boiling point exceeds solar warmth: 0.009133506 * 18.0152 = 0.17 g

So vaporizing 0.17 mL of water "saturates" the vacuum chamber and condenser. If we stopped the cycle there, our production would be 247044 J/L.

But as the vapor condenses in the cooler coil, I suppose the chamber pressure will drop again to continue the evaporation process. So there may not be a problem after all?

 

[jim hardy]

But there has to be other factors hindering the vacuum formation. Otherwise I might as well expect to reach near 0.1668 inHg, boiling point -2˚C. And that's ridiculous!

it's not ridiculous.

IF secondhand information is allowable -
a colleague used to operate flash evaporators on a submarine. They evaporate water at low pressure to chill it and that's the refrigeration for shipboard airconditioning. It had steam eductors to generate the vacuum, i suppose that'd be quieter than reciprocating compressors.
He said it would freeze solid if you let it.Here's a link to vapor pressure of ice so you won't think I'm crazy:
http://www.its.caltech.edu/~atomic/snowcrystals/ice/ice.htm

I'm failing to see why a mechanical vacuum pump is needed between cycles. But my education is an A.S. Math, so that could be the problem here

My mistake not yours, i just didnt realize you were removing all the air in step 5.
I gather you completely fill the chamber with your displacement pump at bottom?
That'd displace the air provided your vent is at top. Displacing air with water is mechanically removing it, just i didn't get that from your word picture. I have a touch of Asperger's and tend to miss subtleties...

Thanks for your time Mr. Jim.

No, the thanks go to you PF members. I am an old retired guy grateful to you who let me play in your sandbox. Thanks.

Not to hijack - just a thought... if you want to go further start another thread ..
Thought Experiment:
Some 'green' buildings use water vapor cycle for air conditioning.
I have for years longed to make a residential solar water cycle airconditioning booster for the sunbelt.
What's missing is conventional thinking always leaps to continuous instead of batch mode, when sunshine is itself a batch process useful maybe ten hours a day.
If you could evaporate a 55 gallon drum of water over course of a day you'd be producing about a ton of airconditioning , almost enough for a modest house. At night it could drain back to reset for next day.

old jim