Just curious about many things like hydrostatics

In summary, this forum is an online resource for people who are looking for help with homework questions. The person found PF through Googling and found the forum because it was the first thing that came up. They have many questions and are unsure of how to ask them. The most important thing is to be persistent and ask as many questions as possible. There is no one right way to ask a question. The first thing to do is to understand the concept and then ask questions to confirm or clarify what you understand. Although the pipe is filled with water, there are various effects that happen as a result of the atmospheric pressure. Most importantly, the vapor turns into liquid and the vapor is lost to the air above.
  • #1
ChicagoDad
14
3
How did you find PF?: I was Googling something and this forum popped up answering a completely different question. It was still very interesting. I wasn't sure what this forum was about and I was prompted to sign up while trying to figure that out. I then received an explanation and am glad I did sign up.

I have so many questions. I feel like I should have a mentor because they are all on the level of homework questions. But I'm 53 and not in school (and probably won't be going back). What is the best way to ask my questions?

Is this a place to ask a question? I have many, but I posted this one to Facebook awhile back and not a soul could help me...

We start with a platform 20 meters high at sea level with a pipe suspended from it with three open valves (one at the bottom, one just above 10 meters from the ground and one at the top) with the bottom of the pipe submerged in a pool of water. (Fig 1)

Then we close the valve at the bottom of the pipe and fill the pipe with water. (Fig 2)

After the pipe is filled with water, we close the top valve. (Fig3)

Then we open the bottom valve.
I believe atmospheric pressure leads to the water boiling to the 10 meter mark. I believe that water vapor takes up less room than liquid water. I believe that the pipe will be cooler due to liquid losing heat energy when it converts to gas. Am I correct so far? FIig 4)
How much cooler would that top chamber be (I realize that it depends on the volume, I'm looking for a formula)?
Would any impurities in the water separate from the vapor now that it's not in a liquid solution? Which? Metals and other minerals? Would it depend on their density?
Please ignore the vacuum question for the moment. I believe that I can better ask that question in reference to an upcoming diagram.

Then we close the middle valve. (Fig 5)

Then we open the top valve. (Fig 6)
What happens in that top chamber?
Is there a vacuum? If so, is there a formula to show the force of the vacuum created relative to its volume (what is it)?
Does the vapor condense into liquid?
Is it a fairly passive reaction (like a sudden mist appearing and gently forming into larger droplets and collecting on the bottom of the chamber) or violently (like with thermal shock)?
Would it be uniform with kind of a cascade forming on one end end moving towards the other (like tapping a bottle of super cooled water and watching it freeze across the bottle)? Or would pockets develop in seemingly random parts of the chamber nearly instantly pulling and pushing on each other creating chaotic looking turbulance? Or something else altogether?
Would anything that came out of solution come back into solution, or would this distill some of the impurities out of the water?
What would it look like?
If the top was a cap instead of a valve and quickly removed instead of opened, would the vacuum created displace the water vapor sending it into the air above? If so, how much of that water vapor would be lost to the air above it (is there a formula based on the dimensions of the pipe)? Would all the water thrown into the air condense into droplets or would some evaporate immediately (is there a formula to know how much would become droplets in the air or evaporate based on the volume spewed, wind speed, temperature and relative humidity?

Would any of this be different if the pipe was less than a drop of water in diameter?

If instead of there being a top chamber, it was a sealable removable section, and we did the same experiment then took off that top section full of water vapor...
How would temperature effect the vapor inside? Would it condense if it became cold enough? If at what temperature (formula for solving that question)? Because the liquid would take up more room, how much pressure would be created? Would it end up being equal to the pressure of the water before the atmospheric pressure turned the liquid into water vapor initially?
If that section was weighted and dropped into a lake (or otherwise effected by higher outside pressure), would the vapor condense? When (what formula would I use to find out)?

I have no real application in mind. Desalination and water purification seem like obvious next things to ponder based on what the answers to these questions are. And what can be achieved using this different effects in general. But, essentially, I'm just trying to wrap my mind around what happens.

I have many other questions, but this one was already prepared due to my recent Facebook post.

Thanks.
Happy to be here.
- Ed
 

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  • #2
ChicagoDad said:
What is the best way to ask my questions?
Welcome to PF. :smile:

I've moved your New Member Introduction thread to the Mechanical Engineering forum, which is probably the best match for now. :smile:
 
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  • #3
Thank you
 
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  • #4
ChicagoDad said:
I have so many questions.
Your stream of questions regarding the operation of a water-filled barometer can be answered, and each step explained by the wonders-of-science.
Most of your commentary is reasonable, but there are a couple of misunderstandings, probably because you were working by yourself.
How do you want to proceed?
 
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  • #5
Thank you for getting back to me, Baluncore! I didn't see the notification that anyone had. I'll be sure to check and not wait from now on.

I'm open. Whichever method seems best. Maybe where I am wrong in my assumptions. I am so happy to be on the verge of having a better understanding!
 
  • #6
Start by turning on email notifications to your watched threads.
 
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  • #7
ChicagoDad said:
I didn't see the notification that anyone had. I'll be sure to check and not wait from now on.
Baluncore said:
Start by turning on email notifications to your watched threads.

@ChicagoDad -- Click on USER at the top right of the page, and select Preferences. I don't get e-mails for replies in my threads, but you can enable that:

1678466103275.png
 
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  • #8
I'm having trouble locating that. Is it because I'm using the app and Chrome on my phone?
I can post screenshots of all the places that I've looked. Buy I thought that I should just probably ask the question first.
 
  • #9
If you're having trouble seeing the USER link at the top of your screen, try turning your phone sideways to get more width in. Does that help?
 
  • #10
Yes (feeling foolish). Thanks. That worked perfectly.
 
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  • #11
ChicagoDad said:
Yes (feeling foolish).
No need to feel foolish! It's a non-obvious trick that we have to use semi-often on phones. :smile:
 
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  • #12
It is difficult to answer, in an orderly way, the complex braided questions in the long original post.

The first thing to do is reduce the complexity of the analysis. Perform the analysis slowly, so everything has time to reach a state of thermal equilibrium. You can then ignore time and temperature, to consider only the “hydrostatic” relationship between height, pressure, mass, and volume.

Later, once you understand the static relationships between variables, you can go back to analyse time, and the rate of temperature change, due to dynamic changes in pressure resulting from the individual transitions between states, that make up the essay long process.
 
  • #13
Thank You Baluncore,

I can see how I could be asking too much at once. I was trying to set up the conditions for one question, but was also trying to leave the window open to be corrected about any misconceptions that I may have. I am completely self taught (aside from YouTube) and have no one to correct me if I'm mistaken about a concept along the way.

So let's begin (if it's alright with you) with my primary question. In the last picture that I posted, a 20 meter(-ish) pipe was previously filled with water with the bottom of that pipe submerged while a valve at the bottom was shut (preventing the water from falling straight through). Then the valve at the top of the filled pipe was closed. Then the valve at the bottom of the submerged pipe was opened. At which time, all of the water within the pipe above 10 meters(-ish, assuming the bottom of the pipe is at sea level) boils into water vapor due to atmospheric pressure). Then a valve just above that line of water vapor (roughly just above 10 meters) is closed trapping that negatively pressurized (compared to the outside) water vapor. Then the top valve is opened quickly.

What does that look and sound like?

Assuming there's no wind, does the vapor turn to liquid in the pipe? Is it a cascade effect? Does the pressure difference cause the vapor to form above the pipe? Is it a violent or elegant reaction (like tapping a supercooled bottle of water and watching the ice form through it progressively from one side to the other? Is it cold due to the heat energy lost in the reaction?

Yes, many questions. But really just a single thorough description of the event will answer most of them. I can't imagine the event lasting longer than a few seconds. What are those seconds like?

I do have more questions. But this is the one that inspired the others.

If you would walk me through those 3-5? seconds, I would really appreciate it.

Thank You Again,
Ed (ChicagoDad)
 
  • #14
ChicagoDad said:
At which time, all of the water within the pipe above 10 meters(-ish, assuming the bottom of the pipe is at sea level) boils into water vapor due to atmospheric pressure).
Not exactly.
Boiling is typical of a liquid heated from below, where bubbles of gas move up through the liquid.

ALL the water within the pipe does NOT boil, nor evaporate. The entire column of water fell about half way, leaving water vapour in its place above 10 metres.
The steam tables show the vapour pressure of water at 20°C is 2.336 kPa.

In this case, the surface of the liquid evaporates to fill the space above the falling water column. It is the surface that evaporates because water below the surface is under higher hydrostatic pressure than the surface water, so the surface evaporates first.

The 10-metre level is decided by the Atmospheric pressure pushing up, balanced against the density of water pushing down.

ChicagoDad said:
Then a valve just above that line of water vapor (roughly just above 10 meters) is closed trapping that negatively pressurized (compared to the outside) water vapor. Then the top valve is opened quickly.
What does that look and sound like?
The upper section contains water vapour at an absolute pressure of about 2.336 kPa. When the top valve is opened, a bolt of air accelerates into and down the upper tube, hitting the closed mid-valve and being reflected back up the tube. It will sound like a double pop. The first will be a depression wave, as the column of air begins to move, followed by the pressure wave reflection that will be about 20 metres delayed, 60 ms later. What happens with multiple reflections is the sound of an organ-pipe, and needs to be modelled using transmission line theory or numerical simulation. The sound will have a distinct note, I guess somewhere around 1/0.06 = 16.5 Hz, more like a flutter or fart than a sound.

ChicagoDad said:
Assuming there's no wind, does the vapor turn to liquid in the pipe? Is it a cascade effect? Does the pressure difference cause the vapor to form above the pipe? Is it a violent or elegant reaction (like tapping a supercooled bottle of water and watching the ice form through it progressively from one side to the other? Is it cold due to the heat energy lost in the reaction?
As the bolt of air accelerates down the tube, the pressure of the water vapour in the tube rises, so the water vapour dissolves in the wavefront of the air. A mist or cloud will NOT form ahead of the accelerating wavefront. Mist may appear just behind the wavefront and in the upper tube during the acceleration, from humidity in the air and the pressure drop at the entrance. The visual effect will be determined by the relative humidity at the time, and the shape of the open-end of the tube.
 
  • #15
It may interest you to know that this is not just an abstract physics problem, but a real world problem that tripped up at least one engineer. The diagram below shows the system. A tank containing ##TiO_2## slurry with specific gravity about 2, had a pump that supplied a distribution pipe located about 40 feet above the tank liquid level. The line had a valve located just after the pump that opened when the pump was on, and closed when the pump turned off. The discharge end of the line was below the surface of the slurry.
TiO2 Loop.jpg

Every time the pump started, there was a KABLAAMM that shook the pipes enough that the pipes were in danger of breaking.

The flow rate and viscosity were such that the entire line ran full when the pump was running. When the pump stopped and the valve closed, the slurry continued to flow until the liquid level in the downward portion of the pipe was about 15 feet above the tank liquid level. There was a vacuum* from that point up to the distribution pipe at 40 feet up. When the pump started up, the flow was helped by the vacuum until the new supply of slurry met the slurry standing in the downpipe with a loud crash.

The solution was to add another valve in the downpipe just above the tank to keep the slurry from running down and creating a vacuum gap. Problem solved.

*More correctly water vapor as discussed by @Baluncore above. But for engineering purposes, it's a vacuum.
 
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  • #16
When a vacuum is gradually pulled on a water column, two things happen. Firstly, the gasses dissolved in the water come out of solution. They stick to the wall, or rise to the surface, which could be wrongly interpreted as the water beginning to boil. Then, at the boiling point, 2.336 kPa at 20°C, the water is free of previously dissolved gasses, so a liquid-gas exchange equilibrium is established at the surface.

The “empty” space at the top of a water barometer immediately contains water vapour, and maybe half of the gasses previously dissolved in the remaining 10 metres of suspended water column. In the longer term, it is necessary to (vacuum) pump that space, to remove atmospheric gasses that diffuse through the water into that space. The water-barometer column, is really a diffusion column, for atmospheric gasses.
 
  • #17
Baluncore said:
When a vacuum is gradually pulled on a water column, two things happen. Firstly, the gasses dissolved in the water come out of solution. They stick to the wall, or rise to the surface, which could be wrongly interpreted as the water beginning to boil. Then, at the boiling point, 2.336 kPa at 20°C, the water is free of previously dissolved gasses, so a liquid-gas exchange equilibrium is established at the surface.

The “empty” space at the top of a water barometer immediately contains water vapour, and maybe half of the gasses previously dissolved in the remaining 10 metres of suspended water column. In the longer term, it is necessary to (vacuum) pump that space, to remove atmospheric gasses that diffuse through the water into that space. The water-barometer column, is really a diffusion column, for atmospheric gasses.
Thank you Baluncore.

I've been hung up on your explanation for almost 3 weeks now. I thought that I should figure this out before asking other questions, but I'm still struggling. So I'm just going to ask:

So when water is in a container in a vacuum chamber and the pressure is low enough that the vapor pressure reaches equilibrium, it boils at room temperature. But in my imaginary pipe set up, that top part will bring the other gasses (impurities in the water) out of solution, but will not boil?

I have so many other questions. But if I'm understanding the basic premise incorrectly, the rest are mute.

Or are you saying that the impurities who's vapor pressure is less than water will come out first, then it will boil?

If you stop up a syringe that has warm water in it and pull the syringe with enough pressure, the water in the syringe will boil, right? The classic school demonstration of boiling water at room temperature. Isn't that what's happening in my imaginary 20 meter water filled capped pipe above 10 meters (sea level) when I open the bottom valve submerged in the pool?

Or am I just getting hung up on "boiling" being the wrong term being used when people refer to water boiling at room temperature?

You said, "the surface of the liquid evaporates to fill the space above the falling water column. It is the surface that evaporates because water below the surface is under higher hydrostatic pressure than the surface water, so the surface evaporates first."

That evaporation isn't "boiling"?

I think that I'm getting hung up on terminology and am kicking myself for not just asking this sooner.
 
  • #18
ChicagoDad said:
So when water is in a container in a vacuum chamber and the pressure is low enough that the vapor pressure reaches equilibrium, it boils at room temperature. But in my imaginary pipe set up, that top part will bring the other gasses (impurities in the water) out of solution, but will not boil?
Water boils when it changes from liquid to a gas, (steam). Atmospheric gasses dissolve, as a gas, in cold water. The solvent is water. The dissolved gasses cannot be said to boil, when the solvent becomes saturated, and some of the dissolved gasses are released from the solvent.

If you warm the water, or reduce the pressure, those dissolved gasses gradually come out of solution in the water. All dissolved gasses have left the water, as the water reaches its boiling point. You boil water, to remove the dissolved oxygen, before making a nice cup of tea.

We know the space above the water will be saturated with water vapour at the vapour pressure of water at that temperature. The atmospheric gasses that came out of solution also remain in the space above the water column.

If the supported barometric column of liquid water has a 10 m height, then at the mid-point of the column, the pressure will be half-atmospheric. The water there will not boil, and there will still be some atmospheric gasses dissolved in the water. The gas exchange between the liquid and the space above, takes place at the surface, where the hydrostatic pressure is a minimum.
 
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  • #19
Baluncore said:
Not exactly.
Boiling is typical of a liquid heated from below, where bubbles of gas move up through the liquid.

ALL the water within the pipe does NOT boil, nor evaporate. The entire column of water fell about half way, leaving water vapour in its place above 10 metres.
The steam tables show the vapour pressure of water at 20°C is 2.336 kPa.

In this case, the surface of the liquid evaporates to fill the space above the falling water column. It is the surface that evaporates because water below the surface is under higher hydrostatic pressure than the surface water, so the surface evaporates first.

The 10-metre level is decided by the Atmospheric pressure pushing up, balanced against the density of water pushing down.The upper section contains water vapour at an absolute pressure of about 2.336 kPa. When the top valve is opened, a bolt of air accelerates into and down the upper tube, hitting the closed mid-valve and being reflected back up the tube. It will sound like a double pop. The first will be a depression wave, as the column of air begins to move, followed by the pressure wave reflection that will be about 20 metres delayed, 60 ms later. What happens with multiple reflections is the sound of an organ-pipe, and needs to be modelled using transmission line theory or numerical simulation. The sound will have a distinct note, I guess somewhere around 1/0.06 = 16.5 Hz, more like a flutter or fart than a sound.As the bolt of air accelerates down the tube, the pressure of the water vapour in the tube rises, so the water vapour dissolves in the wavefront of the air. A mist or cloud will NOT form ahead of the accelerating wavefront. Mist may appear just behind the wavefront and in the upper tube during the acceleration, from humidity in the air and the pressure drop at the entrance. The visual effect will be determined by the relative humidity at the time, and the shape of the open-end of the tube.
I really appreciate this description.

I have so many more questions about it (like if the top of the sealed pipe full of water vapor is opened at the top and there's a screen or some other porous obstruction hindering the mist from leaving quickly, will that water vapor then just become liquid water again? Was some of the mass of the water lost when that bottom valve was opened and before the middle one shut? If so, a lot?), but I need to make sure that I understand the initial concept first. I don't want to add misunderstanding to a false premise.

Thanks again for your help and patience with me.
 
  • #20
Baluncore said:
Water boils when it changes from liquid to a gas, (steam). Atmospheric gasses dissolve, as a gas, in cold water. The solvent is water. The dissolved gasses cannot be said to boil, when the solvent becomes saturated, and some of the dissolved gasses are released from the solvent.

If you warm the water, or reduce the pressure, those dissolved gasses gradually come out of solution in the water. All dissolved gasses have left the water, as the water reaches its boiling point. You boil water, to remove the dissolved oxygen, before making a nice cup of tea.

We know the space above the water will be saturated with water vapour at the vapour pressure of water at that temperature. The atmospheric gasses that came out of solution also remain in the space above the water column.

If the supported barometric column of liquid water has a 10 m height, then at the mid-point of the column, the pressure will be half-atmospheric. The water there will not boil, and there will still be some atmospheric gasses dissolved in the water. The gas exchange between the liquid and the space above, takes place at the surface, where the hydrostatic pressure is a minimum.
I should have just asked much sooner. Thank you.
 
  • #21
Baluncore said:
Not exactly.
Boiling is typical of a liquid heated from below, where bubbles of gas move up through the liquid.

ALL the water within the pipe does NOT boil, nor evaporate. The entire column of water fell about half way, leaving water vapour in its place above 10 metres.
The steam tables show the vapour pressure of water at 20°C is 2.336 kPa.

In this case, the surface of the liquid evaporates to fill the space above the falling water column. It is the surface that evaporates because water below the surface is under higher hydrostatic pressure than the surface water, so the surface evaporates first.

The 10-metre level is decided by the Atmospheric pressure pushing up, balanced against the density of water pushing down.The upper section contains water vapour at an absolute pressure of about 2.336 kPa. When the top valve is opened, a bolt of air accelerates into and down the upper tube, hitting the closed mid-valve and being reflected back up the tube. It will sound like a double pop. The first will be a depression wave, as the column of air begins to move, followed by the pressure wave reflection that will be about 20 metres delayed, 60 ms later. What happens with multiple reflections is the sound of an organ-pipe, and needs to be modelled using transmission line theory or numerical simulation. The sound will have a distinct note, I guess somewhere around 1/0.06 = 16.5 Hz, more like a flutter or fart than a sound.As the bolt of air accelerates down the tube, the pressure of the water vapour in the tube rises, so the water vapour dissolves in the wavefront of the air. A mist or cloud will NOT form ahead of the accelerating wavefront. Mist may appear just behind the wavefront and in the upper tube during the acceleration, from humidity in the air and the pressure drop at the entrance. The visual effect will be determined by the relative humidity at the time, and the shape of the open-end of the tube.
I think I understand the difference between boiling and the gasses coming out of solution. So I'll move on to those next questions...

When that full vertical pipe with the bottom submerged in a pool and top valve closed has it's bottom valve opened, how much water is lost from the pipe?

I realize that the width of the inside of the pipe matters in terms of estimating an exact quantity. But I just mean:
sea level, 20 meter pipe, 75°F (water, pipe and air)..
If the pipe is full of water and the middle valve open while the bottom and top are shut. Then the bottom valve is opened to effect the pressure inside. Then closed. Then the top valve is opened just enough to let air in. How full of water will that pipe end up being? 2/3? 9/10? Just over half? My guess is somewhere between 2/3 - 9/10, but that's just a guess.
 
  • #22
ChicagoDad said:
When that full vertical pipe with the bottom submerged in a pool and top valve closed has it's bottom valve opened, how much water is lost from the pipe?
The loss is not relevant. The water remaining in the column will reach to about 10 metres above the surface of the pool. It has little to do with proportions, and much to do with atmospheric pressure and the density of water.
The sectional profile of the tube is also irrelevant to the question.

ChicagoDad said:
If the pipe is full of water and the middle valve open while the bottom and top are shut. Then the bottom valve is opened to effect the pressure inside. Then closed. Then the top valve is opened just enough to let air in. How full of water will that pipe end up being?
1. When the top valve is closed, and you open the bottom valve, you have made a water barometer. The top of the water column will settle about 10 metres above the surface of the pool. That 10 metre water column is supported by the atmospheric pressure on the surface of the pool.
2. When you close the bottom valve, you fix the water level and volume in the barometer.
3. When you open the top valve and so let in the air, you have made a tall bucket, with water remaining 10 metres deep. (That is, assuming the bottom valve was at the same level as the surface of the pool).
 
  • #23
Baluncore said:
The loss is not relevant. The water remaining in the column will reach to about 10 metres above the surface of the pool. It has little to do with proportions, and much to do with atmospheric pressure and the density of water.
The sectional profile of the tube is also irrelevant to the question.1. When the top valve is closed, and you open the bottom valve, you have made a water barometer. The top of the water column will settle about 10 metres above the surface of the pool. That 10 metre water column is supported by the atmospheric pressure on the surface of the pool.
2. When you close the bottom valve, you fix the water level and volume in the barometer.
3. When you open the top valve and so let in the air, you have made a tall bucket, with water remaining 10 metres deep. (That is, assuming the bottom valve was at the same level as the surface of the pool).
So once the liquid water becomes water vapor, re-pressurizing won't turn it into liquid water again?

I didn't see that one coming.

Yes. A big water barometer. Exactly. I just assumed that the water vapor would become liquid again when air was reintroduced minus the volume lost through the bottom of the pipe early in the process and the other gases (impurities) that came out of solution.

My next set of questions WERE going to be about how much of a decrease in pressure would it take for the water vapor to become water again. My next proposition WAS going to be... actually still is:

Instead of a single 20m pipe with a valve at the top, middle and bottom, if there were two 10m pipes with valves on either end that could be connected. And the rest was essentially the same leaving the bottom pipe full of water who's impurities that would be lost to gasses freed from solution under those conditions already freed and trapped in the top section. Then doing it again with another pair of pipes. Then taking the two bottom pipes filled with water and disconnecting them from their top pipes with the trapped water vapor. Then connecting the water filled bottom pipes together and sticking the bottom of one end in a pool. Then opening the bottom and middle valves.

Do you get the same result? My instinct says yes. But my instinct also thought the water vapor in the top section would form into liquid again when re-presurrized.

Why doesn't that happen? Is it that evaporation pulls the gasses apart, but it takes some sort of catalyst to get the hydrogen and oxygen molecules to bond again? If so, what sort of catalyst?

I don't want to get ahead of myself, but I'm really curious about something else...

If I started by taking a very large (100m to make it a non-issue) semi ridged hose (bendable, but not one that collapes on itself) and submerged it completely. Then capped off one end. Then pulled that capped end up over a wall 11m high wall and back down over the other side back into the water with a weight to drag it down. Would atmospheric pressure cause there to be nothing but water water vapor in the entirety of the hose that's above that 10m threshold a meter before going over the wall?

What if I pulled the hose through until only the open end was submerged while nearly 90% of the hose had been pulled over and dropped back down the other side of the wall and left to fall 78 meters back under the water?

Because everything over 10m and hanging over the other side into the water has become water vapor on its journey, does it just remain as vapor? What if I then capped the other end and tossed it over the wall and held that end at sea level allowing the rest to sink. Would I have a 100m hose that was still only 10m full of liquid although it's 100m vertically under water?
 
  • #24
jrmichler said:
It may interest you to know that this is not just an abstract physics problem, but a real world problem that tripped up at least one engineer. The diagram below shows the system. A tank containing ##TiO_2## slurry with specific gravity about 2, had a pump that supplied a distribution pipe located about 40 feet above the tank liquid level. The line had a valve located just after the pump that opened when the pump was on, and closed when the pump turned off. The discharge end of the line was below the surface of the slurry.
View attachment 323799
Every time the pump started, there was a KABLAAMM that shook the pipes enough that the pipes were in danger of breaking.

The flow rate and viscosity were such that the entire line ran full when the pump was running. When the pump stopped and the valve closed, the slurry continued to flow until the liquid level in the downward portion of the pipe was about 15 feet above the tank liquid level. There was a vacuum* from that point up to the distribution pipe at 40 feet up. When the pump started up, the flow was helped by the vacuum until the new supply of slurry met the slurry standing in the downpipe with a loud crash.

The solution was to add another valve in the downpipe just above the tank to keep the slurry from running down and creating a vacuum gap. Problem solved.

*More correctly water vapor as discussed by @Baluncore above. But for engineering purposes, it's a vacuum.
Thank you for sharing that. It is neat to know that the same thing happened in a real life scenario.
 
  • #25
ChicagoDad said:
So once the liquid water becomes water vapor, re-pressurizing won't turn it into liquid water again?

I didn't see that one coming.

Yes. A big water barometer. Exactly. I just assumed that the water vapor would become liquid again when air was reintroduced minus the volume lost through the bottom of the pipe early in the process and the other gases (impurities) that came out of solution.
I had assumed that dry air was introduced from above.

At 20 °C the ppH2O = 2.336 kPa.
If the air introduced was 100% water saturated, assuming there is 10 m of vapour above the liquid, (in a cylindrical tube);
10 m * 2.336 kPa / 100 kPa = 0.230 m.

So yes, some of that vapour may condense, and it will make somewhere between zero and +230 millimetres of liquid level difference.

But, what will the Relative Humidity, RH%, be on the day of the experiment? More water may be introduced from above, as vapour in the air, that you allowed to fill the partial vacuum. If the RH% was zero, then there would be no change in the liquid level, as all the vapour present could dissolve in the dry air that entered.

Now we have an important counterintuitive back step, and about face. In effect, the rise in the liquid level is totally due to the water content of the air that you allowed to enter from above. You need to start with a simple and well-defined model and analyse that, before you introduce environmental vapour pressure corrections and thermal refinements to the model.

A sensible and reasoned discussion cannot take place at many different levels of abstraction simultaneously, it is clear that just confuses things.
 
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  • #26
Yes.
Sorry.
You're right.
I'm jumping the gun trying to wrap my head around many concepts at once. My approachch to figuring things out is far more artist than scientist (which makes sense because I am an artist and I'm not a scientist). I like to satisfy my curiosity and add things to my engineering tool box by building and trying.

I like to experiment just to understand the effects of something. Play with it. Maybe build some cool piece that incorporates the concept. But I'm often satisfied just understanding a reaction/ set of related reactions and moving on.

But I am far too reckless. I'm 53. In my 40s, I could have blinded myself playing with laserss. I could have killed us in an unsupported trench I was digging to cancel out the city lights to observe the night sky. In my 30s, I'm sure I did some damage to my eyes while playing with solar effects (cooking, hydraulics, reflection/ refraction, etc.). I blew up a projector that I was trying to invent (I never thought "BOOM" and fire was remotely within the realm of possibility. I was playing with color compensation.). A lava lamp type of design that I was building would have killed me (if I hadn't have come across an article about someone who died trying the same thing). Tripped the circuit breaker so many times in so many places. And my 20s? The only proof of God that I need is to not have died as many times as I have not died.

Because I observe something cool and I immediately want to know, "What if you heat it up? What if you cool it? What if you hook up three in series and heat one, cool the other and play with the third? What if you shake it? Build it out of aluminum? Acrylic? Hit it with a hammer? Add water? Add alcohol? Does a color filter change anything? Maybe it just needs a little more power...".

No data. No scientific rigor. No discipline.
I'm not proud of that.

So my first thought when approaching my original question was to grab a bunch of water, pvc pipe sections & fittings, a couple of thick glass 1.5 liter kombucha bottles, a bucket, a kiddie pool (in case the bucket was too small) and a roll of duct tape. And grab my friend and his van. Then drive up a largely vacant parking structure during rush hour (so we can get out if need be before the cops can reach us through bumper to bumper traffic). Then run a version of what I have pictured in the diagrams.

Then I realized that it may not be the smartest idea to play with depressurizing and repressurizing a large glass bottle an arms length away while having no real concept of what sort of pressures I would even be dealing with. So I decided to try to find my answers another way.

That was a few years ago and my questions have had time to marinate. Now I have question on top of questions on top of questions. After years of social media posting, trying to get answers from teachers in community colleges I'm not enrolled in, you tube, trying to decipher physics websites and text books that I don't understand, I found Physicsforums and you. I am extremely grateful. But I seem to be aching for answers to what happens at step 20, 23 & 32 of my mental experiments while being wrong about what happens on step 3 & 5 (figuratively speaking).

Yes, I was also thinking about introducing air through the top. I hadn't considered relative humidity (in my mind, it was a beautiful sunny 80°F day with no to little breeze).

I'll try again in a moment.

Again, thank you very much for your patience and help.
 
  • #27
Baluncore said:
You need to start with a simple and well-defined model and analyse that, before you introduce environmental vapour pressure corrections and thermal refinements to the model.
I'm trying to get a feel for what happens to the water after it comes out of its liquid state due to atmospheric depressurization.

Conditions:
Everything is 80°F (except me, who's in an environmental suit which is 80°F on the outside)

Materials: A 100m pool whose surface is at sea level. A 111m wall separating the pool (with holes so the water pressure, volume and temperature are the same on both sides) so it's top is 11m above the surface of the water. A 100m hose with an internal structure allowing it to bend in at least one direction while not collapsing (in my mind, it has a really large old wooden toy snake that only bends in two directions due to its hinges with holes drilled through it to let the water through, but it's hypothetical internal support structure shouldn't be relevant). The hose has one end capped off. The other end has a valve to open and close it. Both ends are weighted.

I submerge the hose so it is completely filled with water. I leave the open end in the water while I pull the capped end to the top of the wall and start feeding it over the other side into the other side of the pool.

At 10m (before it even reaches the top of the wall) the water on the inside of the hose turns to vapor. As I feed the hose over the other side of the wall to the point where the open end is barely still submerged, does the vapor ever become liquid again in the hose dropped over the other side of the wall? If so, at what point (height/ depth)?

If not, if I shut the valve while it was barely submerged, then just threw the whole thing over the other side and let it sink to the bottom, would it still be mostly full of water vapor?

My guess is that as the capped end gets further than a meter past the other side of the wall on its way down into the other side of the pool (so descending past 10m above sea level) water becomes liquid again.

This is the first suppositori that I'd like to confirm or be corrected about.
 
  • #28
Firstly, stick to SI units. Temperature is measured in °C or K, not °F.

Vacuum hoses are made that contain a long helix of wire, like a coil spring. That leaves the hose flexible, and prevents collapse when the atmosphere presses on it from outside.

How come your questions always make me feel like I am being dragged through a prickle bush backwards?

ChicagoDad said:
does the vapor ever become liquid again in the hose dropped over the other side of the wall?
The vapour in a barometer rests on the liquid, so the mass of water supported by the atmosphere is independent of the temperature and vapour pressure.

There is a hydrostatic pressure gradient in the water, and in the vapour. The lowest pressure is at the top of the vapour, the highest point in the 'siphon tube' that passes over the wall.
If you measure hydrostatic pressure as it falls, as you rise, from the liquid surface, through the vapour to the highest point in the tube, that pressure should rise again as you go down the other side. At some point, the pressure will be high enough, that the vapour will begin to condense. I would expect that to be at the same level as the water on the barometer side of the wall.

ChicagoDad said:
If not, ...
I may as well stop reading there.
 
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  • #29
ChicagoDad said:
I'm trying to get a feel for what happens to the water after
This is a case where a few experiments are worth 1000 words of PF posts. Buy a 50 foot (or longer) length of 1/2" ID 3/4" OD vinyl tubing, one or two rubber stoppers, and one or two C-clamps. Total cost about $75.00 from our local hardware store based on my visit there today. Add a bucket of water and a tall building with openable windows or balconies, and just get started.

That tubing appears to be thick enough to withstand vacuum without collapsing, and flexible enough that a C-clamp can clamp it shut without breaking it. MTA: I found a piece of that size tubing in my shop, clamped it, and showed that it can be clamped shut without breaking.

Do not expect to see visible boiling at the surface, but look close because who knows. Air coming out of solution should show up as tiny bubbles along the inside surface of the tube.

Keep in mind that barometric pressure varies, and that the vapor pressure of water is a function of temperature. If you know the temperature, you can use it as a water barometer. If you know the barometric pressure, it is a thermometer. All good experiments. And please let us know what you find because we can help interpret your results.
 

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