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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 |
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