Fluid flow and characteristics in a vacuum (Cryophorus)

[Ripcrow]

TL;DR Summary

What equations do we use for finding mass flow , velocity and boundary layer of fluid in a vacuum.

trying to calculate the mass flow of Water vapour that is produced within a vacuum of 30 inhg when the water temp is 40 degrees. How is velocity calculated in this scenario. Also what boundary layer does the water vapour have.

 

[Chestermiller]

Summary:: What equations do we use for finding mass flow , velocity and boundary layer of fluid in a vacuum.

trying to calculate the mass flow of Water vapour that is produced within a vacuum of 30 inhg when the water temp is 40 degrees. How is velocity calculated in this scenario. Also what boundary layer does the water vapour have.

So you have air in the gas phase, right?

 

[boneh3ad]

I am so incredibly confused by this. You have water in a vacuum that you have defined as essentially being a hard vacuum, but is this a sealed container? Outer space? What is there to generate a boundary layer?

I mean, the only obvious thing here is that the water is going to boil rapidly.

 

[russ_watters]

Summary:: What equations do we use for finding mass flow , velocity and boundary layer of fluid in a vacuum.

trying to calculate the mass flow of Water vapour that is produced within a vacuum of 30 inhg when the water temp is 40 degrees. How is velocity calculated in this scenario. Also what boundary layer does the water vapour have.

There's a process or source that you haven't specified and is needed to answer the question. The mass flow rate in particular is something typically specified in the process description. As thin as it is, I can only guess you are looking to find out what happens when 40F water is released into the vacuum of space.

Also, 30 in hg is a low-precision way to describe a hard vacuum, is that what you intended?

 

[Chestermiller]

Isn't 30 inches of Hg the same as 1 atmosphere? So where is the vacuum?

 

[russ_watters]

Isn't 30 inches of Hg the same as 1 atmosphere? So where is the vacuum?

"Vacuum of 30 inches" means -30 inches gauge pressure, or 0 inches absolute pressure; full vacuum.

 

[Ripcrow]

It’s called a cryoforus system. Two containers connected by a tube. One container has warm water in it at 40 degrees Celsius and the other container is kept near freezing. When you pull a -30 in hg vacuum the water will boil and the water vapour will try to fill the vacuum. As this happens the water vapour will travel the tube and enter the cold container where it will condense back to water provided the vacuum is not sufficient to boil the water in the cold container. The cycle should continue until the water is evaporated or temperatures equalise provided there is no leaks and the vacuum holds . It’s an old idea.
I have found it difficult to find how fast the evaporation rate is in a vacuum,
How to calculate how fast the water vapour travels from one container to the other in a vacuum , and what is the boundary layer of water vapour At 40 degrees The cryoforus idea could be a more efficient method of water desalination .

 

[Ripcrow]

So you have air in the gas phase, right?

No air. Water vapour and then condense back to liquid. The idea of using 40 degree water is the energy needed is much less then heating water to 100 degrees and result is the same. Liquid to vapour and then condense to liquid. At 40 degrees in a vacuum less energy is required then getting the same change in waters phases at 100 degrees.

 

[Ripcrow]

I apologise. I have been spelling it wrong. It’s called a cryophorus.

 

[boneh3ad]

No air. Water vapour and then condense back to liquid. The idea of using 40 degree water is the energy needed is much less then heating water to 100 degrees and result is the same. Liquid to vapour and then condense to liquid. At 40 degrees in a vacuum less energy is required then getting the same change in waters phases at 100 degrees.

You are neglecting the energy required to draw a hard vacuum.

 

[Ripcrow]

You are correct. The vacuum only has to created once though. Provided there are no leaks.

 

[russ_watters]

I apologise. I have been spelling it wrong. It’s called a cryophorus.

[edited after seeing the corrected spelling]
Here's the wiki on it:
https://en.wikipedia.org/wiki/Cryophorus#:~:text=A cryophorus is a glass,tube of the same material.

It contains other components you haven't yet specified, such as a heat source and sink. What should we do about that? Your description of how you want it to work is still woefully inadequate:

Two containers connected by a tube. One container has warm water in it at 40 degrees Celsius and the other container is kept near freezing. When you pull a -30 in hg vacuum the water will boil and the water vapour will try to fill the vacuum.

So you mixed English and SI units and didn't tell us?! C'mon! If you're trying to make it difficult for us to help you, you are succeeding. Please describe your system fully!:

• Is the warm water continuously heated to keep it warm? How/how is it controlled? How big is the container and how much water? What's the starting pressure? How much head space is there?
• If the vacuum container contains a vacuum, how can it have a temperature? Are you talking about the container walls? If so, describe the container. Is the container chilled? How? What is it made of? How big is it?
• Does the tube have a valve in it? Does the vacuum container have air in it at the start of the process? Is there a vacuum pump to create and maintain the vacuum or do we assume the system starts with a hard vacuum, then the valve opens, then it goes to equilibrium?

As this happens the water vapour will travel the tube and enter the cold container where it will condense back to water provided the vacuum is not sufficient to boil the water in the cold container.

I thought you wanted a hard vacuum? So, how much vacuum do you really want? Regardless, why would lowering the pressure of an already existing 40C vapor cause it to condense? Are you actively cooling it too? How?

The cycle...

What cycle? Nothing you have described so far indicates a cycle. It looks like a single, linear, unsteady process that after some time reaches an equilibrium and stops unless it is continuously driven by outside energy sources/sinks that you haven't described.

I have found it difficult to find how fast the evaporation rate is in a vacuum,
How to calculate how fast the water vapour travels from one container to the other in a vacuum , and what is the boundary layer of water vapour At 40 degrees.

I feel like the term "boundary layer" here is misplaced and has no meaning for this problem. The rest of those are just difficult because the process appears to be unsteady, poorly defined and contradictory.

The cryoforus idea could be a more efficient method of water desalination .

Ahh, so it should be a continuous process? And we're helping you design/invent it? Have you researched how partial-vacuum desalination systems actually work?

I think, though, that the basic answer to your question is that the desalination water flow rate is a function of the heat transfer rates, so you need to design the system and specify your rates (and they have to be possible/make sense) to answer that. Or you start with the desired desalination flow rate and work backwards to find the energy requirements.

 

[boneh3ad]

You are correct. The vacuum only has to created once though. Provided there are no leaks.

I don't know how you can make that claim. Presumably you wish to remove the water somehow. To do that, you need to open it up. It's also not going to remain at vacuum when that water evaporates. I don't think you've thought this through.

 

[russ_watters]

Yeah, related to the last few post, but maybe a finer point on it: the title of this thread and thrust of the question are about fluid [flow] dynamics. But I don't think what you really should/need to evaluate has much of anything to do with fluid dynamics. This is just plain thermodynamics. It's an energy and mass transfer/conversion problem. There is overlap of course, but in many ways it is a much different way of approaching the problem.

 

[Ripcrow]

I only wanted to know how to work it out as I have searched and can’t find a correct answer so either I’m not understanding the solutions given or what I’m looking for is not available. An example of use for this system is in western Qld where artesian water exits the ground at 80 degrees. They don’t have hot water systems in Quilpie ,they have cooling tanks instead. You get in the shower and turn the hot tap on for 5 minutes and the hot water straight from the town bore arrives to shower with. How’s that for a heat source. The water is high in sulphur though. Soap won’t lather in it. So how about a cryophorus system driven by the artesian heat ( heat source ) providing clean water. Here’s the reason for wanting to know how to work out mass flow and velocity. How big does a system have to be to distill a reasonable amount of water. How long will it take to distill 1000 ltrs of water. I have other questions also like does surface area matter in relation to the amount of water vapour shifted ( will a round container with a diameter of 50 cms produce the same amount of water vapour as a container of 100 cms diameter. Does depth of container matter. The example I provided in my earlier comment was just a guide to what I wanted. 40 degree water boils at roughly -27 in hg and 7 degree water boils at -29.5 in hg( I don’t have the boiling point / pressure tables I found in front of me at the moment) therefore provided the hot side is kept at 40 degree and the cold side is kept at 7 degrees and the pressure is maintained above the cold sides boiling point and below the hot sides boiling point (-27.5 in hg ) the water vapour should continue to transfer to the cold side.

from a comment above regarding trying to have heat in a vacuum it is totally achievable. The transfer of heat thru the walls of the container will still continue up till the water level. There is still a medium for heat to transfer to.
Regarding the maintenance of a vacuum I assume that as the mass is only transferring from one container to the other and volume in the closed system will not change ( possibly smaller volume due to cooler temperature once water has transferred to the cold side ) the vacuum should not be affected but this is only true if dissolved air in the water is not accounted for. I don’t see that having a vacuum pump and a microcontroller monitoring the amount of vacuum as an issue and can easily remove the Dissolved air from the system.
If we used the artesian water at 80 degrees and the correct vacuum it’s plausible that the cold side only needs to be slightly colder to make the system work( would require precise control of vacuum though).
Also I don’t see why a continuous system can’t be developed by having a piston pump located at the bottom of the cold side. Example could be that the pump starts to operate (draw piston back ) and a vale between the pump and cold tank opens and provided the water is above the pump inlet gravity should allow water to fill the pump cylinder. the valve closes and the pump continues to pump the water to a container because it is now free of the vacuum. Same with filling the hot side with water. If the outlet pump is removing 100 ml of distilled water then another pump and valve system could be filling the hot side With 100 mls of water.Timed right and the volume in the cryophorus would not alter that much during water inlet and off take.
the reason I wanted to know about boundary layer of water vapour and how to calculate it for differing temps is the fact that I believe the bigger the temp difference means more water could be distilled over the same time period. The cooling tanks are shaded but still subject to high ambient temps. It’s drinkable temp but only just in the middle of summer. I would probably need a bigger temp difference. So I thought with a endless source of heat why not utilise that heat to cool the cold side via the use of peltier plates My idea was to use the heat to produce electricity via peltier plates then feed that electricity to another peltier plate that acts as a cooler. The heat sinks that are generally used with peltier plates are about 30 to 40 mm high. As Pelletier are very inefficient I thought I might be able to make my own heat sink for the cold side To get more efficient cooling with them . Build something like a thin set of plates that the water vapour has to pass through.A plate top and bottom that the vapour passes through. I thought if the spacing was really close and the vapour had to squeeze through some efficiency would be gained over using readily available heat sinks. But the question is how small can the gap be. Is .4 mm enough. Do I need 2 mm. The mass flow and velocity and boundary layer are needed to properly work out how much area Of cooling plate would be required.
I’m not asking for anyone to invent my idea. As you can see I have plenty of ideas. Some may not work and some will need modifying to make them work. I asked for assistance in the maths to understand how to work it out and any questions I may have. For instance in one set of evaporation temp vs pressure tables I located it gave a microns scale. But no where in the article or the table did it describe what the microns represented.

 

[Ripcrow]

I don't know how you can make that claim. Presumably you wish to remove the water somehow. To do that, you need to open it up. It's also not going to remain at vacuum when that water evaporates. I don't think you've thought this through.

Why won’t it remain in a vacuum. The mass is only transferring from one container to another. The cryophorus is a very old physics apparatus

 

[boneh3ad]

Why won’t it remain in a vacuum. The mass is only transferring from one container to another. The cryophorus is a very old physics apparatus

If your liquid is turning to a gas, you now have molecules occupying the space where previously there were none, ergo, no longer a vacuum.

 

[russ_watters]

I’m not asking for anyone to invent my idea. As you can see I have plenty of ideas. Some may not work and some will need modifying to make them work.

'Inventing an idea' isn't really a thing. Ideas are free, exist only in your head, and don't take any work or money to create. An invention is a physical thing. Turning the idea into a working device is what "inventing" is.

Here’s the reason for wanting to know how to work out mass flow and velocity. How big does a system have to be to distill a reasonable amount of water. How long will it take to distill 1000 ltrs of water. I have other questions also like does surface area matter in relation to the amount of water vapour shifted ( will a round container with a diameter of 50 cms produce the same amount of water vapour as a container of 100 cms diameter. Does depth of container matter. The example I provided in my earlier comment was just a guide to what I wanted. 40 degree water boils at roughly -27 in hg and 7 degree water boils at -29.5 in hg( I don’t have the boiling point / pressure tables I found in front of me at the moment) therefore provided the hot side is kept at 40 degree and the cold side is kept at 7 degrees and the pressure is maintained above the cold sides boiling point and below the hot sides boiling point (-27.5 in hg ) the water vapour should continue to transfer to the cold side.

We can walk you through this, but it isn't an easy thing if you don't have any background in thermodynamics. I'll give a rough start:

The basic design process goes like this:
1. Specify the desired outputs. (1000 liters...let's say, per day)
2. Define the known inputs. (80C water, electricity)
3. Describe the process to turn the inputs into the outputs.
4. Calculate the process requirements.
5. If the process doesn't work, re-define #1 or #2, if possible, and re-do the calculations.

Your idea is already a thing, but it doesn't seem like you've really researched it, and you really should. Just google "Vacuum distillation". Here's an example:

https://sites.google.com/site/kjdesalination/vacuum-distillation

The steps in the process are:
1. Feed water.
2. Spray water into low pressure chamber. Some water flashes to steam, the rest falls into a pool at the bottom of the chamber (falling through a heat exhchanger where more boils).
3. Steam is collected, pressurized to atmospheric pressure(and heated), then run through the heat exchanger in the chamber to condense it into water.
4. Hot output water preheats feedwater.
5. (Not shown) Your water starts warm, so your fresh water product is pretty hot, and should go through an additional heat exchanger to cool it.

Every step in the process needs the conditions to be specifically defined.

from a comment above regarding trying to have heat in a vacuum it is totally achievable. The transfer of heat thru the walls of the container will still continue up till the water level. There is still a medium for heat to transfer to.

Two problems with that:
1. A real device doesn't use the walls of the container to transfer heat, it uses a purpose-made heat exchanger. The container itself will be insulated to prevent heat transfer.
2. You need to stop saying "vacuum" when diving into the specifics. You need to specify the *actual* pressure that you are looking for, because it's not a hard vacuum/-30". If it's -27", that's fine.

That's enough for now -- I think we're going to need to do a complete walkthrough/design of the process for you, if we're up for it...

Again; this has basically nothing to do with fluid dynamics/"boundary layer", so please forget about that part. You can't analyze a boundary layer until you define its properties, and in reality you don't even need to do that because your heat exchanger will likely be something that's already been modeled for you, and you just need to plug your requirements into the model. I design (select) heat exchangers frequently and I've never done one from scratch as a professional -- I haven't done such calculations since college. Real heat exchangers are likely too complicated to manually model anyway (and definitely too cumbersome to bother).

 

[Ripcrow]

But when that gas is cooled back to a liquid the volume is the same. As I stated. The cryophorus works and is used in many physics labs

 

[Ripcrow]

I only asked for the equations and perhaps an explanation of any concepts I didn’t fully understand. My background is year 11 physics and maths 2 while studying to get entry to the airforce. After my eyesight was deemed not good enough I quit school and took a very different career path. That was 30 years ago. In the last 5 years I have learned to calculate the thrust output from a turbo based jet engine. Build it and test it to find the equations to be true. That’s the best education I’ve ever had I think. The people that walked me through the processes and assisted by explaining concepts I had forgotten about actually explained things better then the physics teachers in school. The fact that the boundary layer has to be taken into account when designing flow paths in a jet to allow proper mass flow and thermal properties to be maintained makes me believe that the equation to define the boundary layer in a cryophorus system such as I want to build is important. I only asked for the correct equation. I’ll continue on my own thanks.

 

[Ripcrow]

I don’t really care what a real device does. I have a scenario that I have described that I’d like the system to work with. Not everything can be bought off a shelf.

 

[256bits]

You are asking for a custom design for your scenario.

What you need is a comparison of how similar systems work to get a feel of where to start.
Even batch processes.
The thing to look for is vacuum distillation as mentioned above by Russ.
Or look for how to make moonshine, with a still - same thing.
It's what is used in a chemistry lab for distillation.
You can research those items, and apply for a continuous flow, and find some thermodynamic equations regarding vapour pressure of water at temperature, heat input, cooling load, vacuum, ...

Fluid boundary layer is of little concern.

 

[boneh3ad]

I only asked for the equations and perhaps an explanation of any concepts I didn’t fully understand.

The problem is that there is clearly quite a bit you don't yet understand and you get defensive about any pushback.

 

[Ripcrow]

My original question. Have you read it. Where does it say I need help designing a system or device. How many here have commented without fully understanding what a cryophorus is and how it works. I asked for the proper equations and have stated that I have been unable to find them or I have failed to understand them.

 

[Ripcrow]

Here’s the link to what a cryophorus system is. It’s been around since 1813. https://en.wikipedia.org/wiki/Cryophorus.

The concept works. Since asking for the equations to find out if this could be built to be a better way of desalination I have been informed of everything but the equations. Some have hinted that the cryophorus doesn’t even work. Current desalination of seawater in Australia costs between $5-$10 a kilolitre and is very energy intensive. As a thought experiment, could the cryophorus be used to deliver fresh water and be less energy intensive?

when building a jet engine the equations are straight forward and easy to understand. Pressure drop and temperature drops are easy concepts. In a cryophorus there is no pressure drop between containers. It relies on the simple facts that hot moves to cold and waters vapour pressure depends on the air pressure acting on it. Calculating air velocities in a jet involves temperature and pressure drop. How do you calculate the rate at which hot moves from cold with no pressure drop? Building a jet involves calculating boundary layer so that the mass can flow at the right velocities. Googling the boundary layer of water vapour in a vacuum doesn’t give much. If the boundary layer is accounted for the cryophorus could be optimised for performance and efficiency.

I asked for help with the equations and understanding them better. I didn’t ask about industrial vacuum distillation systems that use a lot of energy.

The cost of emergency desalination of sea water for disaster areas rus at about $14 per kilolitre and requires lots of energy. Sometimes that energy is not available. As a thought experiment could a sun driven cryophorus system be built to distill water in emergency situations using less energy and still provide a viable volume of fresh water. We will never know unless the right equations and understanding of those equations is used .

 

[berkeman]

The cost of emergency desalination of sea water for disaster areas rus at about $14 per kilolitre and requires lots of energy. Sometimes that energy is not available.

Here’s the link to what a cryophorus system is. It’s been around since 1813. https://en.wikipedia.org/wiki/Cryophorus.

Sorry if this has been addressed in the thread already, but you are saying that current desalination methods are very energy intensive. How much energy does it take to make the liquid Nitrogen used in this device?

 

[Ripcrow]

Sorry if this has been addressed in the thread already, but you are saying that current desalination methods are very energy intensive. How much energy does it take to make the liquid Nitrogen used in this device?

View attachment 276719

I don’t know how much it costs. I didn’t intend to use liquid nitrogen. Given that heat moves to cold provided there is a temperature difference the cryophorus will work. The greater the difference the greater the volume of flow of course. Theoretically if you built a sealed pipeline ( cryophorus) from the equator to one of the poles and created the Right amount of vacuum the water vapour would travel from the equator to the pole and condense as fresh water.

 

[berkeman]

I believe the "cryo" part of the name of that apparatus refers to cryogenic temperatures. If you just use a -10C refrigeration unit, does it still work? With what efficiency and condensation rate?

Say it takes 1kW to heat the water to flow through a distillation column versus to make the liquid nitrogen and keep it cryogenic -- which apparatus would have the highest rate of generating desalinated water?

 

[Ripcrow]

I believe the "cryo" part of the name of that apparatus refers to cryogenic temperatures. If you just use a -10C refrigeration unit, does it still work? With what efficiency and condensation rate?

Say it takes 1kW to heat the water to flow through a distillation column versus to make the liquid nitrogen and keep it cryogenic -- which apparatus would have the highest rate of generating desalinated water?

If you look at the water vapour temp vs pressure scales you can see that the water will condense at the same pressure it boils at if the vapour is cooled even slightly. In my above example using artesian heated water at 80 degrees Celsius and 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. I’m not fighting the science on this apparatus. The maths would answer the question of how best to build a viable unit ( if it was possible ). Without the maths it’s a very expensive testing regime.

 

[Tom.G]

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

 

[Ripcrow]

Look at a Newcomen steam engine. Steam fillsthe cylinder on the down stroke and cold water then floods the cylinder walls causing the steam to condense to water reducing volume and therefore creating a vacuum that lifts the cylinder up.

 

[Ripcrow]

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

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]

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

 

[russ_watters]

Thanks. First positive reply accepting that the system works. Lots of ways to heat and cool. Just need the maths to calculate it out.

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.

 

[Ripcrow]

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.

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.
for your equation you supplied does that apply to fluid flow in a vacuum. The great thing is I can substitute numbers and and mathematically determine whether a system could be built viably.I have no set parameters
I have seen a functioning cryophorus distillation system without the high tech heat exchanges and the build looks really simple. Is it viable or economical. I don’t know. Don’t have access to any more data on it.

 

[Ripcrow]

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.
1000 ltrs at 20 degrees celcius needs 23.244 kilowatts to reach 40 degrees celcius in 1 hour. Pressure requirement is 55.3 torr Or 27.7473 in hg .
is mass of water equal to 40,000?..

further parameter will be set at cold side temp of 1 degree so pressure will be 4.9 torr or 29.7318 in hg.

 

[Ripcrow]

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

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.
The condensation process has me intrigued. I have heard many times that things lose heat in a vacuum. Does this mean that the cold side will continue getting colder if there is no outside source of heat.
the boundary layer seems to be sticking point here. Calculating jet engine mass flows means calculations allow for the boundary layer Otherwise you end up with stalled flow or expansion and therefore cooling of gases in the wrong areas. I’m thinking it’s the same here. If the boundary layer is not known and the calcutions are done just on mass flow and velocity the pipes may be too small and the vapour won’t move quick enough to allow the most efficient vaporisation in the hot side and if the pipes are to big the vapour will have room to expand and condense too early.

 

[russ_watters]

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.

Ok, I didn't know about the 20 C to 40 C part of the process, but that is the correct answer.

Pressure requirement is 55.3 torr Or 27.7473 in hg.

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.

is mass of water equal to 40,000?..

You said above that you wanted to process 1000 liters, and used the mass correctly in a calculation; 1000 liters is 1000 kg.

further parameter will be set at cold side temp of 1 degree so pressure will be 4.9 torr or 29.7318 in hg.

We'll come back to that later. You haven't done the calculation I asked/described in post #34, and we need it.

Yes the vaporisation rate has me lost.

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.

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'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.

 

[Ripcrow]

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.

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.
The cold side parameter is defined for this equation now. As you said I needed parameters to Work the answer out. Using the right units is definitely vital and its Something I wondered about. When I was taught the right way to calculate jet flows it was an older person who swapped from imperial to metric which didn’t bother me as I grew up with my father and grandfather doing the same.
The system will still be continuous. 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. One of my quirks is to assume that most people think outside the box. I assumed that when people viewed my question they would think about such a system and work out it needed more then temperature to drive the flow from the hot side to cold side. It needs pressure change also therefore a tap or valve to control flow. I only realized last night I hadn’t described this system needing a valve and that it maybe the cause of the negative comments and misunderstanding.

An example of starting the system would be draw a vacuum in the hot side and draw a greater amount of vacuum in the cold side. Heat the water and then open the valve slowly. Provided the rate of evaporation is equal to the amount of condensation. The valve will maintain the pressure difference between the sides. The pressure difference is what causes velocity which is why I asked how to calculate the velocity in a vacuum. The greater the speed at which the water vapour can be moved from the hot to the cold side the less vapour there is affecting the pressure in the hot side which affects vaporisation rates

 

[russ_watters]

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.

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

 

[Ripcrow]

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

Hopefully I have this right now. 2,406,100 kJ per hour to vaporise 1000 kg of water

 

[russ_watters]

Hopefully I have this right now. 2,406,100 kJ per hour to vaporise 1000 kg of water

Yes. And in terms of power, that's 668 kW.

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...

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).

[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.

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...

 

[Ripcrow]

Yes the Energy requirement surprised me. Is total energy required equal to the heating energy plus the enthalpy energy therefore total energy to heat and vaporise 1000 litres of 40 degree celcius water is 691 kw. Or does the enthalpy value equal the total energy.
To heat 1000 litres of water from 20 degrees to 100 degrees celcius in 1 hour at atmospheric pressure takes 93.33 kw and the enthalpy value is 626 kw.
The energy requirements between the two don’t seem logical given the 60 degree temperature difference and the pressure difference. From what I read atmospheric pressure dictated how much heat was needed to cause phase changes in water. But it doesn’t seem linear with these numbers.

 

[Ripcrow]

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...

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.

 

[Ripcrow]

Ok I found the right equations. Bernoulli gives flow rates and I found the reynolds number and calculated the boundary layer of the fluid.
Some Numbers I have are a mass flow rate 0.0129 kg per second at velocity of 512 metres per second. This is using a pressure difference of 6719 pascal and a pipe diameter of 25 mm ignoring boundary layer. Thanks to those who clarified the power requirement to vaporise the water per second in the other thread. I have a kinetic energy output of 1690 joules and require 31kw to vaporise the water per second.
This is what got me interested in the equations. It’s a long video (30 mins ) but it was an interesting concept. According to the maths here this might work but there is very little efficiency in the system. Am I missing something

 

[Ripcrow]

Wrong video.