Thanks. First positive reply accepting that the system works. Lots of ways to heat and cool. Just need the maths to calculate it out.Tom.G said:The cyrophorus approach looks, and acts, like a souped-up version of a Solar Still as used in camping. See part 4, Solar Stills, in this link: (not very detailed but a place to start)
https://www.wikihow.com/Get-Water-While-Camping#
The difference in your case is you already have the heat source (the hot water). What you seem to be searching for is a way to increase the efficient usage of using that heat to obtain distilled water.
Two straight forward approaches are:
1) use a heat sink of some sort that is colder than the water to condense the humidity. If a separate chamber or container is used, you have to ensure air/moisture circulation between the hot and cold parts.
2) use reduced air pressure to increase the evaporation rate from the hot water.
It sounds like you are trying to use both approaches, which would surely increase the collection rate. You still need to ensure moisture circulation between the hot and cold parts though. Perhaps a solar electric panel driving a small fan (computer fan?) would be adequate for a small to medium installation. I'm thinking to rather large passages between two chambers, one for hot-to-cold and the other for cold-to-hot.
The biggest problem may be maintaining a temperature difference though. Can the cold side be immersed in a local body of water, like a pond, lake or ocean? The advantage of that is their temperature remains close to the daily average temperature, not peak temperature. And don't forget a Sun shade for the cold side. Another cooling approach is evaporative cooling, keep a wet blanket wrapped around the cold side if nothing else is available.
I would suggest you try a prototype without the vacuum first to get all the details worked out. Remember that a good vacuum will easily collapse most common containers. Perhaps a beer keg, if you can find one, would withstand a mild to moderate vacuum. Those 20 and 55 gallon (80 and 200 litre) drums are not near strong enough.
Have fun! and let us know what you come up with.
Cheers,
Tom
Nobody has suggested that the idea won't work, and I gave you an approach and offered to walk you through it, but you didn't seem interested. If you really want help, you'll need to put some more focused effort into it and lose the attitude/actually accept the help that is being offered. Your current tack getting you nowhere.Ripcrow said:Thanks. First positive reply accepting that the system works. Lots of ways to heat and cool. Just need the maths to calculate it out.
The attitude comes from constantly being told the system doesn’t work or that I can’t build a heat exchanger or that the current solution is the best. I asked for the equations calculate the mass flow rate ,calculate the velocity and how to find the boundary layer. Sure I have gave examples of where such a system could be used but I have never stated I was going to build one. All I asked was for some guidance on the maths.russ_watters said:Nobody has suggested that the idea won't work, and I gave you an approach and offered to walk you through it, but you didn't seem interested. If you really want help, you'll need to put some more focused effort into it and lose the attitude/actually accept the help that is being offered. Your current tack getting you nowhere.
The equations/math of this is really easy. You're just looking for the heat of vaporization/condensation (read from a steam table) on each end, times the flow rate, and the pumping power is volumetric flow rate times pressure (more complicated if you are pumping/compressing a vapor though). The work here is in defining/designing the process steps. I could pick them for you, but then this would be my design, not yours.
So, step/process 1: You said you have/want to use 40 C water and want to distil 1000 liters. Look up the pressure and the heat of vaporization in a steam table. Multiple the heat of vaporization by the mass of water. Also, specify the time. Then divide the energy by time to get the rate at which you need to add heat to boil your 1,000 L.
Ok. I’ll set a few parameters so we can work through it.russ_watters said:Nobody has suggested that the idea won't work, and I gave you an approach and offered to walk you through it, but you didn't seem interested. If you really want help, you'll need to put some more focused effort into it and lose the attitude/actually accept the help that is being offered. Your current tack getting you nowhere.
The equations/math of this is really easy. You're just looking for the heat of vaporization/condensation (read from a steam table) on each end, times the flow rate, and the pumping power is volumetric flow rate times pressure (more complicated if you are pumping/compressing a vapor though). The work here is in defining/designing the process steps. I could pick them for you, but then this would be my design, not yours.
So, step/process 1: You said you have/want to use 40 C water and want to distil 1000 liters. Look up the pressure and the heat of vaporization in a steam table. Multiple the heat of vaporization by the mass of water. Also, specify the time. Then divide the energy by time to get the rate at which you need to add heat to boil your 1,000 L.
Yes the vaporisation rate has me lost. I’m thinking that will depend on how quickly the vapour above the fluid will be moved away allowing more vaporisation to occur. As others have said as the water turns to vapour the pressure immediately above the water rises and stops the vaporisation process.Tom.G said:I think your limiting factors will be the evaporation rate and the condensation rate that you can reach, not the boundary layer of the water vapor to the walls of the plumbing; after all, you can use larger diameter plumbing. I understand there are tables for pipe size versus pressure versus flow rate, but I don't have any details.
Counterflow shell-and-tube heat exchangers seem the best bet.
For a cold source, there is always a cooling tower, but it needs a pump; and moderately clean makeup water or frequent cleaning/maintenance. Perhaps there is enough pressure & flow from the artesian well to supply that pumping energy.
Cheers,
Tom
Ok, I didn't know about the 20 C to 40 C part of the process, but that is the correct answer.Ripcrow said:Ok. I’ll set a few parameters so we can work through it.
1000 ltrs at 20 degrees celcius needs 23.244 kilowatts to reach 40 degrees celcius in 1 hour.
Your value in torr is right, the value in inches hg is expressed wrong. It's 27.7473 inches of vacuum, and at this point it would be better to just keep it in absolute pressure; it is 2.18". Also, bar is a more common unit and it is weird to mix American and SI units.Pressure requirement is 55.3 torr Or 27.7473 in hg.
You said above that you wanted to process 1000 liters, and used the mass correctly in a calculation; 1000 liters is 1000 kg.is mass of water equal to 40,000?..
We'll come back to that later. You haven't done the calculation I asked/described in post #34, and we need it.further parameter will be set at cold side temp of 1 degree so pressure will be 4.9 torr or 29.7318 in hg.
You've specified the vaporization rate: it is 1000 kg/hr. Now you need to calculate the energy/power input requirement to make that happen, per my description in post #34. Once you know how much energy/power you need, then you can decide how to do the heating.Ripcrow said:Yes the vaporisation rate has me lost.
You're trying to design a continuous, steady process here. The pressure is whatever you want it to be, and it's constant -- you have to design the device physically to make what you want to happen, happen. But first, you need to design/describe the device/process mathematically. I'm trying to walk you though the calculations. Please do the calculation I asked/described in post #34.I’m thinking that will depend on how quickly the vapour above the fluid will be moved away allowing more vaporisation to occur. As others have said as the water turns to vapour the pressure immediately above the water rises and stops the vaporisation process.
You asked me to calculate the mass of water multiplied by the heat of vaporisation. 1000 litres of water equals 1000 kg multiplied by 40 degrees. I calculated energy input required to reach 40 degrees celcius assuming that the water would have a starting temp Therefore 20 degrees celcius.russ_watters said:Ok, I didn't know about the 20 C to 40 C part of the process, but that is the correct answer.
Your value in torr is right, the value in inches hg is expressed wrong. It's 27.7473 inches of vacuum, and at this point it would be better to just keep it in absolute pressure; it is 2.18". Also, bar is a more common unit and it is weird to mix American and SI units.
You said above that you wanted to process 1000 liters, and used the mass correctly in a calculation; 1000 liters is 1000 kg.
We'll come back to that later. You haven't done the calculation I asked/described in post #34, and we need it.
You've specified the vaporization rate: it is 1000 kg/hr. Now you need to calculate the energy/power input requirement to make that happen, per my description in post #34. Once you know how much energy/power you need, then you can decide how to do the heating.
You're trying to design a continuous, steady process here. The pressure is whatever you want it to be, and it's constant -- you have to design the device physically to make what you want to happen, happen. But first, you need to design/describe the device/process mathematically. I'm trying to walk you though the calculations. Please do the calculation I asked/described in post #34.
Also; I'm not sure, but the vibe I'm getting from these last few sentences is that you may have a misunderstanding of how boiling works. The process in your device is basically the same as boiling a pot of water on your stove. You apply heat, and the water heats up and boils. The faster you apply heat, the faster it boils. You need to calculate how fast you need to apply heat to boil 1000 kg in in hour. That's the calculation I described for you in post #34.
No, "heat of vaporization" is the amount of energy (heat) required to turn liquid water into vapor, at a specific temperature and pressure. You take water at 40 C and 55 torr and turn it into vapor at 40 C and 55 torr. It's the "enthalpy -- evaporated" number in a steam table like this one:Ripcrow said:You asked me to calculate the mass of water multiplied by the heat of vaporisation. 1000 litres of water equals 1000 kg multiplied by 40 degrees. I calculated energy input required to reach 40 degrees celcius assuming that the water would have a starting temp Therefore 20 degrees celcius.
Hopefully I have this right now. 2,406,100 kJ per hour to vaporise 1000 kg of waterruss_watters said:No, "heat of vaporization" is the amount of energy (heat) required to turn liquid water into vapor, at a specific temperature and pressure. You take water at 40 C and 55 torr and turn it into vapor at 40 C and 55 torr. It's the "enthalpy -- evaporated" number in a steam table like this one:
https://www.efunda.com/materials/water/steamtable_sat.cfm
Yes. And in terms of power, that's 668 kW.Ripcrow said:Hopefully I have this right now. 2,406,100 kJ per hour to vaporise 1000 kg of water
I assumed you would want a valve between the tanks. It's a good idea because it is adjustable and you don't have to be as precise about sizing the pipe (which is very difficult).If you imagine a hot tank and a cold tank connected by a pipe with a tap between them. There can be two different pressures... I only realized last night I hadn’t described this system needing a valve...
So, if the condenser operates at 30C, repeat the exercise you just did to find the pressure and the heat removal requirement. Then you'll have the process fully defined and we can start figuring out how to create a device to make it happen.[previous post] ...using a cooling tank to supply the cold side at about 30 degrees Celsius the tables suggest the example is valid Provided the vacuum is kept above the boiling point of 30 degrees Celsius water.
Tried the cooling calculation. Used the efunda link and found the evaporation energy and pressure required but something is wrong. It worked out at 675 kw which is more then what is required to evaporate at higher temps. I assumed that the heat of vaporisation had to be removed from the vapour to cause condensation so the maths would have been the heat of vaporisation of water at 40 degrees celcius minus the heat of vaporisation of water at 30 degrees Celsius and the answer would have represented the amount of heat needed to be removed. At this stage I have 668 -675= -7 kw. I adjusted the vaporisation pressure in the calculator so the pressure was the same as the 40 degree water and it still worked out at 668 kw which would indicate that the Heat values are the same.russ_watters said:Yes. And in terms of power, that's 668 kW.
I assumed you would want a valve between the tanks. It's a good idea because it is adjustable and you don't have to be as precise about sizing the pipe (which is very difficult).
So, if the condenser operates at 30C, repeat the exercise you just did to find the pressure and the heat removal requirement. Then you'll have the process fully defined and we can start figuring out how to create a device to make it happen.
A preview:
Hopefully you recognize at this point that the heating and cooling requirements of this process are enormous, and have put some thought into how you want to do the heating and cooling...