|
Potable Water - The Third Way.
On Wed, 03 Oct 2007 08:13:20 -0700, RW Salnick
wrote: brought forth on stone tablets: On Tue, 02 Oct 2007 09:59:46 -0700, RW Salnick wrote: Richard Casady brought forth on stone tablets: On Sat, 22 Sep 2007 17:51:56 -0400, "Wilbur Hubbard" 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) |
Potable Water - The Third Way.
|
Potable Water - The Third Way.
|
Potable Water - The Third Way.
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 |
Potable Water - The Third Way.
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 |
Potable Water - The Third Way.
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 |
Potable Water - The Third Way.
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 |
Potable Water - The Third Way.
|
Potable Water - The Third Way.
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 |
| All times are GMT +1. The time now is 02:35 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
Copyright ©2004 - 2014 BoatBanter.com