Could you explain in a different way what is causing the overall pressure increase? I have a hard time getting unconfused with the explanation given in the video and by some comments.
@@Yehan-xt7cw The air that's down there is pressurized to the same pressure that the column of water causes. That pressure doesn't disappear once it moves up, so it's as if you put a second column of water on top of the previous one. Loosely speaking, when the bubble is down, the water pressure is fighting the air pressure. When the bubble is up, they're not fighting, they're cooperating. If you stand on a pogo stick that's on a bathroom scale, the scale will measure just your and the stick weight, but the pogo stick is compressed enough to counteract your weight. Now, lower your ceiling to the point that the ceiling is barely touching your head while you're on the pogo stick. The force on the scale is equal to your weight + stick weight, force on the ceiling is zero. If you now step on the scale yourself and wedge the pogo stick between you and the ceiling, the pogo stick will have to compress the same amount in order to fit, thus pushing the equivalent of your weight down on top of your head, and same force goes to the ceiling. The scale is now measuring twice your weight + the stick weight, the ceiling now has one time your weight on it as a force equivalent.
A good analogy is to think about standing on a spring, compressing it. You are pressing downward with exactly your weight, and the spring is holding you up. Place a ceiling directly above your head, so your head is almost touching it. Now step off the spring, and you're still pushing against the ground with exactly your weight, same as before. And now, to simulate the pressurized air being at the top of the tube, we put the spring on your head. The spring isn't compressed by your weight anymore, so it expands and presses against the ceiling, and thus pushes down on you. So now you're pushing against the ground with your weight PLUS the force of the spring pushing you downward by trying to expand against the ceiling.
do you think if you got a trampoline and then tightened the springs, so there was no spring force at all, the weight of the system would just magically change? I don't think so myself. Springs settle and they do not and cannot change the weight, no matter what. I could be wrong, but I don't think I am in this case.
@@deucedeuce1572 The point is that when the bubble is on the bottom, it's acting like a spring under your feet, while when the bubble is on top, it's acting like a spring on your head (the weight being negligible; just the springiness). You don't feel the pressure from the spring when it's under you any more than you feel the pressure from the ground. But if it's above you pushing down (off of the lid of the container), you feel the pressure of both the ground AND the spring pushing down on you.
@@stevenjones8575 For every action there's an equal and opposite reaction, is there not. If there is "X" amount of weight on the spring, then there is "X" amount of weight on the spring... and if the spring weighs "Z" amount of weight... the spring weighs "Z". If you weigh the man and the trampoline for example, it doesn't matter if there are springs on the trampoline or if the springs are replaced with an equal weight of non-stretching wire. I could just be misunderstanding what he was saying... but I almost always know what he's talking about in his videos. Have spent most of my life studying all fields of science. I'm not usually one to misunderstand, especially when it comes to scientific "laws". (As for equal and opposite reactions, if you have 10 pounds pushing on a spring, the spring is pushing back 10lbs. The spring will always compress to the exact weight to where it's pushing back on the object the exact same amount that the object is pushing on the spring. It matters not if it's a spring or a solid block of steel in that aspect. The force against each object is equal.
The force the trampoline has to exert to hold you up is equal to force it has to exert on the ground plus it's own weight. So in your example it's still equal
This is a really well established fact for any petroleum engineer because the well can receive a gas influx from the formation that we call "gas kick", this gas rises up the well which ((increases)) the pressure on the bottom formation which can cause it to fracture, so the most dangerous kick is the gas kick.
Petroleum engineer here. This is very easy to understand with mathematics, perhaps with the help of pressure gradient chart (P vs depth). But seeing it visualized in real experiment like this is a delight. Thank you.
Could you understand it using Bernoulli’s equation as well, technically? Since the total gravitational potential energy of the fluid is decreasing as the bubble goes up, it makes sense to think pressure goes up to compensate.
Old oil and gas guy here. I remember leaning this in one of the countless hours of training and education that we had to maintain to keep our field certs. I was in frac, and we didn’t have to worry so much about gas bubbles in the day to day pumping operations in the formations we were pumping into. We just needed to be aware of such phenomenon. Thanks for reminding me of it!
The thumbnail, which was at one point also part of the video (3:20) suggests, that it isn't a closed system which is confusing. Without the closed top I would even expect the pressure to decrease.
Another way to think about this: When the water is above the air, gravity is pulling the water down onto the air, which exerts a compression force which is opposite to the pressure of the air pushing outwards. Therefore, part of the air's pressure is being counteracted by gravity, and the total net force from the air's pressure is reduced by that amount. When the water is on the bottom, its gravity is no longer trying to compress the air, so there is no force to counteract the outward air pressure, so the pressure it exerts on the _container_ becomes higher. It is very counterintuitive, though..
This is actually how I thought about it in the beginning when he originally asked what would happen. That being said, his explanation was a bit to complex, although i suppose it had to be for a visual representation
@@tristanmoller9498 may be cuz air was compressed not the water, if the test done on gel or solid with liquid you might see the difference but pressure might be very small in that scenario
There was one flaw with the question at 0:02 as the right side of the pipe looks closed and the left side (highest) point looks open. Hint would be helpful to mention a closed system. With that I went in wrong direction but finally with recognizing a closed system I couldn't solve correctly as well, but it would have saved some time.
Simplest explanation by analogy: what happens if you hold your breath while you ascend from a deep dive? The air in the lungs will want to expand. The exact same thing is happening here. Air is being compressed by water and if that compression is removed without the air being allowed to expand then that air will apply higher pressure against the container that is confining it.
That is not what is happening. The feeling of air wanting to expand is due to pressure difference between air in lungs and surrounding water. The pressure applied by the gas doesnot change, that is Boyles law, it applies the same pressure when it is at the bottom as well as at the top. That is the whole puzzle! How does the pressure increase when pressure of the bubble doesnot change!? The answer is the pressure in the air at the bottom balances out the weight of the water column above but when it goes up it pushes the water column harder in to the base.
@creatureofhabit7049 Sheesh.The weight of water in both cases is what generates the equalizing pressure. It is EXACTLY the same. When the water weight is removed, the (unchanging) pressure of the air pocket exerts greater FORCE on the containment whether it is lungs or a tube filled with water. There is no paradox or mystery. Just realize that water itself can act as (part of) the containment.
@@tomszabo7350 i think you did not understand what i wrote. I never said that the weight of the water changes ever. I said when the air pocket is at the bottome it balances out the weight of the water column and exerts equal force on the bottom of the pipe. When the air pocket goes on top then it exerts the same force on top of the water column i.e. both the water column and the air pocket push the bottom of the pipe. Also can you rephrase your comment, it was hard for me to understand what youbwant to convey. As far as i know the force on the lungs when you come up is due to pressure imbalance between inside(lung air) and outside (water). Neither the air inside the lungs is exerting more pressure nor force. Pls help me better understand your point if i misunderstood.
I found the correct answer with a slight variation of the reasoning of the video: as the air goes up, it "should" undergo an expansion because it "should" be subject to a pressure decrease. But the volume being fixed, the expansion is constrained: the pressure of the air stays the same: the intuition that the pressure doesn't change is correct, but it applies to the air, not to the whole system. Therefore, with the same pressure of air, the pressure at the bottom is higher when the air is at the top because you have to include the weight of the long column, while the initial situation only had a small water column added to the air pressure.
I was so close to getting it... I thought for sure the air pressure would decrease as it moved up, and I realized that it wouldn't be able to expand against the water. But I assumed the temperature of the air would decrease with the pressure decrease in a fixed volume
@@Zaros262 yeah I almost got confused about that too, but then I realized that the temperature variation only occurs if the volume can vary: it's the work of the pressure forces integrated on the volume variation. No volume variation = no work = no variation of T :)
Yep, this was also my reasoning, thanks for putting it into words! By the way, @@renedekker9806 your Dutch is showing 😉, but I agree, this is a "top" explanation.
The pressure is kinetic as increasing with the movement of water due to the influence of gravity, so the water has to stop at the bottom using its kinetic energy to form pressure from a vacuum
The trick is to find the reference point that has the constant pressure. When the vessel is open, it's the atmospheric pressure at the opening that is constant. When it's sealed however, due to liquid being almost non-compressible, the air bubble is now the constant pressure.
The thumbnail didn't really show it as closed. But the open tube case is easy as the pressure at the bottom of an open topped tube of incompressible fluid is always proportional to the height of the fluid column (in other words it goes down if open).
I think he bent the tubing out of its natural shape, which deformed the tubing and caused it to flatten, which increased the pressure inside. I think his entire explanation is wrong also... although I can't say 100%. I just don't think it works like that. As far as I know, a spring would settle with weight on it and would not, ever in any circumstance change the total weight/mass of a sealed system. Your car doesn't weigh less, because it's on shocks. I just don't believe that. If you change the shocks/springs with a steel rod that weighs exactly the same amount, your car would still weigh exactly the same if put on top of a scale in both configurations. Then if you filled the car with water and put an air balloon in the trunk or in the cab of the car, it wouldn't magically change the weight of the car (comparing the two configurations with water and balloons)... and/or if you tied the balloon to the floor of the car or let the balloon float to the top of the car. As long as the car, the water and the air all weigh the same and contain the same mass, the weight will stay the same (as far as I know). I hope you forgive the long explanation and repetition. It's not an easy thing to explain in just written words.
The movement of the gravitational potential energy was the key for me in fixing this concept in my mind. Great examples to facilitate comprehension. Thanks.
The potential/gravitational energy argument is actually completely wrong. You can tell it is wrong by considering the same problem, but replace the air with a low density incompressible liquid. The potential energy would change in the same way when the bubble moves, but the pressure in the system would remain constant, or even decrease depending on where you measure it. It is not gravitational potential energy making the change. It is a change in the internal energy of the system due to the compressive nature of the air and its interaction with the container. This is something not stressed in the video, but fundamentally all this pressure is applied by the tube/container to hold the system at fixed volume. When the air bubble is at the bottom the weight of water helps to keep it at fixed volume, at the top the water is no longer helping, so all this force comes from the container, increasing the net force of the container on the fluid. This is the change resulting in the pressure increase. It is kind of hidden because it is not a typical type of energy change, it is very specific to this situation.
@@CerealReviewer If we applied the ideal gas law, i.e. PV = nRT. Which factor do you believe that must be changed and cause the P to go up? This is the part that I don't undertand. It seems to me that neither V, n nor T should change (a lot)
@porker2008 you are right, the pressure of the air bubble is constant because the volume and temperature are fixed. But the source of the confining pressure changes, in one case the water column provides the pressure, in the other the container applies the pressure. With the container applying more pressure the internal energy of the system increases.
@@CerealReviewer Thanks for the quick response. My thought was that the pressure we messured at the bottom should be the sum of the air pressure, plus rho*g*h, where rho is the density of the water, g is the gravitational constant and h being the depth of the water above that point. So after we moved the bubble to the top, nothing changed except h increased. so the overall pressure increased. Is this a correct explaination?
I find it even more confusing and fascinating in reverse - you can create vacuum! Start with the bubble on top but at atmospheric pressure this time (i.e. seal the tube when the bubble is on top), and somehow move it down to the bottom. 1. As explained in this video, the bubble would want to stay fixed in volume and (atmospheric) pressure at the bottom. The system’s pressure decreases, but that’s not the interesting part. 2. Note the pressure at the top of the tube is (atmospheric pressure - rho*g*h). Since this pressure cannot be negative, with big enough liquid density*height, e.g. > 76 cm of mercury (new video ?😄), you’ll beat the atmospheric pressure and create vacuum on the top of the tube! Since this is a fixed-volume system, this means the bubble at the bottom will be compressed! 3. The same mechanical analogy applies (love it btw). Start with the spring on top, with an initial compression force F0 to mimic the atmospheric pressure. Invert the bottle. If mg > F0, the weight will compress the spring, creating “vacuum” on top. You got me on this one. Keep on the great content!
The same observation through off my assumption and we can reference it like Boyle's law the partial pressures of each and when the water column is above the air bubble that means the bubble is compressed equal to the weight in water column of the liquid also the water is not able to separate from the top of the vessel meaning that a partial vacuum is occurring at the top of the liquid column. Therefore when the volume of the gas is above it would if pressurized, expand like a spring to exert force on the top of the liquid column and or eliminate the partial vacuum previously at the top
And that bubble pressure increase was then probably also the reason for the oil rig explosion in some video I cannot locate anymore, where they are drilling the hole and suddenly they scream they hit an unexpected gas pocket and need to abandon the area, seconds later greyish-brown thick liquid erupts from the hole and covers the whole area, including a nearby city. Or was that one of my strange dreams many years back that I now falsely assume having been a video? I even remember clearly how some of the shots in the video looked, for example a street light next to a low, white wall in the dark under a tree in the rain.
This is the second of your videos I've watched, but the way you take the time to explain what is happening and why after the experiment made you very worth subscribing to for me.
Chemical engineer here, I think the easiest way to perceive this is to (as always) examine the extreme case. If you were to run this experiment with a column >> 10m then the weight of the water would exceed atmospheric pressure (assuming your gas bubble was injected at atmospheric pressure). You could then imagine at the top, a vacuum would be created, because the force of the water is now stronger than the gas bubble (the gas bubble would compress). This would be an area of zero pressure pushing down on the top of the column of water. Let the bubble rise and now the vacuum has been replaced with a non-zero pressure pushing down on the same column of water.
Any vacuum at the top would reduce the effect, though, and if the tube is J-shaped, it could even reverse the direction of the effect. Take a J shape where in the beginning the left part of the J has 1m of air, all the way to the bottom, with 10m of water on the right side, with vacuum on top, and 99 more meters of vacuum on top of that. In this case, the pressure at the bottom would be 10mH2O=1bar. Now let the air move to the top right of the J. With the water incompressible, the air would now fill up exactly 100m of tube, which (at constant temperature) would reduce the pressure in the air to 1bar/100 = 0.01bar. Now, since the hight of the water column is now only 9m, the total pressure at the bottom would be 0.9 bar(from the water) + 0.01bar (from the air) for a total of 0.91 bar. For comparison, if the top of the J did NOT have vacuum (the tube ended exactly where vacuum would have formed), the air would retain the 1 bar pressure, and the pressure at the bottom would be 1.9 bar.
I would say it increases. Water is incompressible so its volume is fixed, and therefore the volume of the gas is fixed too. Assuming that the temperature of the gas is fixed as well, then so will be its pressure. So when the gas is moved to the top, it pushes on the water below it with the same pressure it had at the bottom, so the extra water column remaining will increase the pressure at the bottom.
what do you mean by extra water column? the height and amount of water is the same right, so i'm still confused how there is more pressure edit: i get it now
Yeah, it's a great channel. I think he might be wrong about this experiment though. I can't say with absolute certainty. He probably knows more than me about this subject and other kinds of science... but at the same time, it don't mean he can't ever be wrong. I think the PVC systems pressure went up, because he changed the shape of the tubing and used force to do so. When the bottom was curved like in the first test, it was in it's natural resting state, which is the shape that it was manufactured to be in... but when he straightens it out, he not only has to use a certain amount of force to do that... but straightening the tubing causes it to flatten, because it's a tube being bent outside of its normal shape. For example, if you did the opposite and sealed off the ends of a perfectly straight copper tube then bend it, the pressure inside would increase in that case too.
I’m a scuba diver. And with my experience and with playing with soda bottles sealed and un sealed bottles. So I knew there would be a change in pressure. Great job. Thank you 😊
Well, if you watched the video beyond 10 seconds you would have seen that when the question was asked it was a closed system and it is mentioned it's a closed system.
Yay I got it right! I expected the gas at the lower end to counter the pressure from the water above, like a force from the opposite direction. Only 1:40 in but curious for the reason nonetheless
I think it would have been a good idea to show that in the starting state, you could take out the caps on top of the tube and it would do nothing, but after the bubble transfer if you take off the cap, it would blow out and the pressure would decrease. So at 0:05 when you see both column, it does not seem to be capped at the top so the pressure is higher in B, if it's capped, it depends on if the cap has been put after or before the bubble transfer. Great brain teaser and counterintuitive answer anyway, I love it !
For bubble on right side: Pr = pw * g * h1 = pw * g * h2 + Pb h1 = height of water measured on the left h2 = height of water measured from the right Pb = pressure from air pocket pw = density of water g = gravitational acceleration For bubble on left side: Pl = pw * g * h3 + Pb = pw * g * h4 Isolate and solve for Pb: Pb = Pl - pw*g*h3 = Pr - pw*g*h2 Pl = Pr + pw*g*(h3-h2) Since h3 > h2 (seen visually in the video), we know that the pressure at that point when the bubble is on the left side will be greater.
I came up a similar expression, except that I did not assume the pressure at top of the left are is zero at the beginning. It is more realistic to assume it is close to atmosphere. However, this assumption does not alter the result.
@@andrewliang4713 As long as there is a nanometer thin layer of air before sealing the container, the atmospherics pressure is there. Even you use a pump after sealing, it will take a few minutes to remove all gases dissolved in the water.
@DragonYang01 That makes a lot of sense when I think about filling up the tube in an actual experiment. I didn't even consider that! Just goes to show the difference between actual application and textbook questions, haha.
It has no consequence for your mathematical result, but I think the right hand side of your equation for PL is incorrect. It is not pw * g * h4. The reason is, that this equally leads to a contradiction in which Pb can become negative (solve for Pb). I think the reason is, if the fluid on the left would be compressible it's density would increase. But water isn't. So the water on the right in case of the bubble being on the left can be treated as a "wall". That also explains the observation in the experiment in the video.
Initially I thought that the system would stay the same, for the same reason as you: it's sealed. As soon he cut back at 2:45 I realized that since air is 1) not hydraulic and 2) buoyant, it would compress and exert a force on the water, effectively making it "weigh" less and decrease the pressure.
The air bubble is sealed. In the beginning this air is compressed and by changing the location of the bubble to the top this compression remains. So there are two states: A: Static pressure of water column defines overal pressure and air bubble compression. B: Static pressure of water column (this time lower) PLUS air compression from A as air bubble has no additional volume to expand. By ventilation on top at B the pressure decreases. 3:16 is missleading as top is open (= atmospheric pressure)
According to the formulas: Case 1: When the bubble is at the bottom Its P = h*d*g ; h= height of vessels or tube, d= density of liquid, g means gravity When the "air bubble" gets on top the pressure becomes P = P(atmosphere) + h*d*g
The pressure of the bubble depends on the height of the water column when the tube was sealed, so it would be greater than atmospheric pressure. Pbubble=Patm+pghi. The final pressure would be Patm+pgHi+pgHf.
I got confused because I assumed that the tube was open. the first diagram he shows also has open containers. with an open tube the pressure would decrease, because when the air would reach the top it would be able to disperse.
i somehow got it right. im no pro but used basic high school physics. At 0:10 , PA = Po + pgh1 At 0:30 PB = Po + pgh2 (h2>h1) so PB>PA thus pressure at bottom increases Similarly for top At 2:00 PA = Po + pgh1 -pgh2 (h2>h1) so overall Po - pg(h2-h1) At 2:30 PB = Po + pgh3 (although h3 is very small) overall Pb>Po and Po>PA thus i concluded Pb>Pa and thus pressure at the top also increases Hope im correct even though i used basic fluid mechanics and the new perpective of COG->P and squeezy air was new and interesting to me. and it does make sense if i write bernoullis equation for the system. Feel free to correct me if im wrong.
It will increase as the pressure of the air column remains same if we assume the temperature is constant thus the pressure at bottom will be pressure of air + dgh where d is the density of water, since h increasesz pressure. Increases
I know this from working in the automotive brake industry. A small amount of air in a brake line could dramatically affect the stopping distance ( by several feet). So your stopping force is proportional to the pressure applied, (for tmoc system actually a lot more force than the applied pressure, but at a certain point you get to the knee point and it eventually becomes 1-1 input to output force). Having to bleed brake systems for basically a cubic millimeter of air is very annoying depending on the type/design of the calipers. Some types seem to be prone to getting air stuck. Thats why straight off the assembly line usually has the best brake performance due to manufacturers using a push/pull method to fill the brake line when the system started off dry. That is usually better at getting any air out of the system.
Conservation of energy - when the water is at the top, the gravitational potential energy is higher so the pressure potential energy must be lower, and vice versa
3:15 I want to say that the plexiglass setup and the tube setup aren't the same. In the tube setup, there's initially more water above the height of the point at which the pressure is measured. However, in the plexiglass setup, there's exactly as much water below that point's height. Assuming the volume stays the same, this means there's also exactly as much water above that height.
I got this from the thumbnail, admittedly by assuming the anti-intuitive answer was correct and then thinking about why. Here is how I thought about it. Water is incompressible (or at least very nearly so) when the gas bubble is at the bottom, its pressure is equal to that of a water column from the lower surface to the upper one, and the pressure at the bottom is exactly that of a water column equal to the height of the tank. Now we release the gas bubble to the top. As the system is constant volume, the gas cannot expand. Its pressure when at the top is exactly the same as it was in the lower position. So the pressure at the bottom is now equal to that of the height of the water column, PLUS that of the former difference between the two surfaces.
I was confused when he said the energy turned into pressure, It's like saying that force of the magnet sticking to my fridge comes from magnetic energy@@deinauge7894
@@rainaldkoch9093this is the only explanation that makes sense to me. The pressure increases when the air bubble rises to the top because the gravitational potential energy of water molecules is getting converted to kinetic energy. Which implies that the temperature of the system must increase at least slightly to account for the increase in pressure. Since his set up is not truly "closed", in the sense that heat exchange will still occur between the contents of the tube and the surroundings, does this also mean that the pressure will eventually return to its original value once thermal equilibrium is achieved??
Before this goes to far im going to say it increases becasue the air acts like a shock absorber. Edit: i was a plumber for a very long yime and also dabnle in hydrologics
I get you said this before you watched the vid, but if the pressure on the air is increased then it puts more pressure on its container as well. It’s why a box container has weight pushing against its sides. So it would work as a shock absorber but it would apply some of that pressure onto its container. But part of it would just compress and not push against the side. Idk though that’s my guess. I’m just 15 so I don’t know a whole lot unlike you.
Indeed, for a vertical pipe, the gravitational potential energy of the water changes. However, for the U-shaped tube, it is not true! The water also raises on the other side! The problem can not be treated by neglecting the process and just looking at the initial and final states. The process of the bubble rising is quite rapid and far from being quasi-static (not only the gas doesn't travel in one piece, but the rise of the tiny bubbles it splits into is accelerated), so basic thermodynamics doesn't work well either (i.e. the law pV/T= const for the gas is valid only for quasi-static processes). The gas, however, heats up while being pushed up by the liquid's pressure (work is definitely being done on it), and this increases its pressure (just not proportionally, since the process is not quasi-static). Why does the liquid remain at higher pressure after the thermal equilibrium with the room (if that even happens, since, as you mentioned, the pipe could even burst from the high pressure) is probably somewhat ansewrable by non-equilibrium thermodynamics. One other factor to take into account is the work you yourself are doing to turn the tube back and forth.
3:21 the bubble is working with its weight when it’s above, when it’s below it’s against the weight of the water that’s why the pressure changes with local.
There is a fault in the premiss. You in no way stated that the air bubble introduced into the system, was under pressure. You created the assumption that the system was first created with the bubble at the top, where it was not under pressure, and caused the bubble to move to the lower position. Under that circumstance, there is no pressure when the bubble is at the top, thus you failed to provide the exact circumstances under which the bubble was created.
It doesn't matter. You can create the system by starting with the bubble at the top. You will reduce the pressure when you try to move it to the bottom. Potential Energy of the system is a state function.
We know that there is enough pressure to counter the weight of liquid, otherwise he couldn't have kept the bubble entirely on one side (we would have observed a bubble of vacuum on top in all cases).
I'm not sure why it would need to be stated that the air bubble is under pressure. It's a closed system; the air bubble will always be under some amount of pressure. If it wasn't, it would be a vacuum.
This video is what made me subscribe. There are so many counterintuitive things answered by science. Trying to stick with intuitive explanations doesn't always lead to the right conclusion.
It's pretty easy if you think about it like a spring. At the bottom, the spring is like a cushion acting in opposition to gravity at the reference point. At the top, the spring goes from acting against gravity to acting in tandem with gravity at the reference point.
Except its wrong to think of it like a spring, since the water doesnt put pressure on the bubble, since its closed at the top. Easy to demonstrate by opening the bottom, the water wont flow down, right? What the explanation in the video overlooks is that the pressure does increase at the end of the bottom where the bubble was i believe.
@@barrygillis the water would flow down if you opened the bottom... I'm not sure if that's what you're trying to describe but that's what I'm interpreting. If you opened the bottom, whether the bubble was at the top or bottom, the water would tend to flow out of the tube toward hydraulic equilibrium. Depending on the amount of surface tension present in the tube, the water may or may not stop flowing due to a vacuum forming at the closed top end. This is a separate force, that also acts like a spring, but is not present in the first set of conditions.
Reading through the comments, someone mentioned the Bernoulli equation and that was spot on. As a ChE who reviews various subjects in my field periodically I was chagrined I didn't make that connection. In a fluid system, the sum of the changes in velocity, height, and pressure (scaled by appropriate constants) are constant. If you change one something else needs to change. The center of gravity of they system decreased and the decrease in potential energy resulted in an increase in pressure.
0:40 I think it's something to do with the compression of the air bubble. Maybe the air will expand because it's not being compressed by the liquid, causing the air to compress the liquid instead as it has less space. Something about that doesn't seem right, but I'm expecting a counterintuitive solution so that's my guess.
Increase. The volume of air is constant since the liquid is incompressible, so the pressure of the air remains the same. The pressure of the air matches the pressure of the water on the interface, and water increases on pressure with depth, so pressure increases with depth from the air, no matter where the air is. More intuitively, when the water is pushing down it squeezes the air, giving it pressure. If you move the air to the top, still at the same pressure, now the air is pushing down on the water, increasing the starting pressure of the water. (written before the explanation)
In the initial image though it doesn't look as though it is a sealed system it looks to me like it's open at the top because you can clearly see a top piece on the lower end but not on the upper end.
Another way to look at it, when we consider the air pressure to be neutral with the bubble on top: if the air bubble is down, it wants to be "compressed" by the weight of the water above it, so the water is "pulled apart", creating negative pressure similar to capillary forces in trees.
This makes perfect sense. The water is acting sort of like a spring and decreasing pressure. As the water goes up, the force of the compression is going from bottom to top, to top to bottom.
8:40 I got it right reasoning like that - when the air is at the top there is more potential energy. But I think the illustrated diagram where A or B is presented may be wrong, though... The potential energy for the systems illustrated as A and B is equal, I think? The bubble in the illustration isn't big enough. Right?
my guess: I first tend towards saying, with the air bubble further up, pressure should decrease. However, since air is compressible, and the bubble e.g. has been compressed when being down, i guess: pressure doesnt change.
Well.. water is imcompressible.. but air is compressible. The air is compressed and will remain so where ever it is. So ading the compressed air that is compressed the the full level to the top the adding all the water minus the top, now you have one full level pf pressure in the gas, then about 3/4 in level. So the relative pressure should go up... say about 75%
Very educational! I love the fact that you asked very intelligent questions, one in particular that I didn’t think of but realized that it was an excellent question as soon as you said it. Where did the extra energy come from? Excellent observation when he mentioned the center of gravity lowered
I've seen a few of your videos/shorts and thought they were interesting, but this one made me subscribe. Great job explaining a very non intuitive concept!
This whole time I was thinking "why don't you turn it on its side and measure, since vertical orientation you must also factor in gravity, whereas horizontal you can largely ignore it."
I found it easier to understand with something squishy like an orange. Let's say you have a tube with a screw top. Put a spring in the tube, followed by an orange on top. The spring will compress due to the weight of the orange. Screw the top tight to the orange, then flip the tube over. The spring will no longer be compressed by the weight, so it will attempt to expand, squishing the orange. Sub the orange for liquid, and the spring for gas. When you try to squish a liquid you get pressure on the container.
Pausing at 4:17 the only thing i can thing of is the nature of air how its more compressable than water so that when the water is above the bubble its actually compressing the air to allow for more soace in the chamber giving a false preassure reading once the trapped air is released the air isnt able to be compressed by the weight of the warer reducing the available volume for the water.
A simple thing to consider is that gravity has an influence on everything in our lives. If we can one day gain better knowledge and mastery over gravity then we can do much greater things.
Two ways i can see someone easily understanding it. 1. Fill a waterbottle with air at the bottom of a pool seal it and then take it to the top of the pool. Open the bottle fast and you can see the pressure that was stored as the weight of the water in the pool was being used to compress the air bubbles. 2. Imagine a large amount of sand over an explosive to absorb the energy, then imagine that same explosive at the top instead of the bottom. Without the sand to compress it its free to show the entirety of energy stored. Idk feel like it makes sense when viewed like that
Pressure should decrease because it's only the height of the water column that affects pressure of water in a vessel, barring external forces being applied (which would include atmospheric pressure or lack thereof). Thumbnail showed it being an open vessel which would decrease for sure. This is a closed vessel, curious to see what happens
The fact that the air bubble moved upwards on its own when the "tail" was deformed has clued me in that the potential energy change might be the solution here. Great demonstration!
I immediately intuited this as being part way to a liquid barometer. If the column of liquid kept going higher, eventually the pressure would drop to the point that there would be a vacuum and a bubble would form at the top. So as long as there's no air in the column of liquid and the tube is rigid enough, the pressure will continuously drop the higher you go.
7:25 Just to be more accurate, stating water is incompressible is incorrect. It IS compressible but takes an absurd amount of pressure to do it, such as industrial water jets. Normally I would not even bring this up but I hold your channel in the highest regards and felt it should be pointed out.
Water is compressible on the micron scale. After such it's functionally solid. Also water is what fluids are based off of due to the stability of water physics.
For the compressible air pocket to have equal volume in the two cases, the liquid-air-interface must maintain equal pressure. Adding the contribution by gravity acting on the liquid from the interface down to the bottom, it is clear that the pressure is higher at the bottom when the air pocket is in the upper position, so increase when the pocket is moved to the top.
Since the liquid is incompressible and the tank is fixed, the volume of the gas remains the same, and the temperature is also constant so the gas pressure is the same, but the pressure at the same height of the liquid is the same so in the low gas scenario the pressure is lower everywhere in the tube.
I love your video sir but This is wrong interpertation as in 7:59 the presure in whole tube will not inc as the botom end will still have same presure as befor, it only changes how presure is distributed in starting demostration he measure presure between ends which inc because high presure gas goes up and replaced by low presure liquid. [To get analogy thick two block of same area one more denser below and less dense above presure at ground will be same even if we reverse the order. but presure in intersection or within lower block (low dense) after filp will have high presure then intial case.] So presure change “inside tube” not at botom because order switches not due to dec in potential energy. But presure at top inc because property of gases to have same presure throughout and same on all walls, that causes expotion in mining sites. Thats my idea thought
As a physics professor, I kept starting to write a comment, but then you explained exactly what I was about to say. Like four times in a row. So.... great job!
I am analyzing and commenting at the same time trying to understand this puzzle. Someone please correct me if I am wrong. I assume there is no extra pressure pumped into the closed system. I wish you had just laid the tubes down horizontally and showed the pressure in psi for the starting position. 6:47 the 350 grams is the same when you flipped it. Were you trying to confuse us with the way you presented this demonstration? Note: 0.8 bar is 12. psi. Using psi makes more sense. 3:08 you show water traveling up the tube when air should be traveling up and the water down. I think you are showing that raising a water column increases psi. I will revisit this puzzle later maybe after someone corrects me.
Fascinating. I’m not sure if you answered this in the video or not (I’m still confused), but if I understand it correctly, the water being on top pushes the air below down to compress it and the air acts as a cushion effectively taking one for the team to reduce overall tube pressure, but is not as capable of cushioning the pressure when on top because it can then only utilize the tube pressure and not the water’s weight to compress. Since the air is less compressed in the top position, the tube pressure increases. I initially thought pressure would decrease because the water went to the lower portion and thus closer to the ground/the gravitational pull becomes (minisculy so) stronger.
i got it right immediately. the air bubble invokes a force upon the water which counteracts the pressure of the water column you created, so when the air bubble goes to its most natural location at the top of the column it’s then adding to the total amount of pressure instead of subtracting from it. air is a fluid too, & as a sailor it’s really important to have a good grasp of fluid dynamic since that’s the very mechanic that sailing exploits to function. in the world we live in, anything that seems counterintuitive could very well be the exact answer your looking for. sailing downwind wing on wing won’t be any faster than tacking upwind. to the untrained eye it’s confusing because when you’re with the wind, sails all the way out, completely level with the water you’d think you’d be going the maximum speed you can, but you literally have to sail against the wind, making frequent & regular turns with your sails basically all the way in, with your entire body hanging off the boat as a counterweight (hiking) to prevent it from tipping, since your simultaneously using the boat as a wing in the air & a paddle in the water, all while surfing on numerous forms of tension
0:23 guess at this point. I would say there is a slight potential energy change that would slightly increase the pressure on the vessel because the denser fluid is being elevated. That said I could easily see how I would be wrong here.
I am a drilling engineer, we have to get certified in well control before doing any operations. This one is a huge topic.
Could you explain in a different way what is causing the overall pressure increase?
I have a hard time getting unconfused with the explanation given in the video and by some comments.
@@Yehan-xt7cw The air that's down there is pressurized to the same pressure that the column of water causes. That pressure doesn't disappear once it moves up, so it's as if you put a second column of water on top of the previous one.
Loosely speaking, when the bubble is down, the water pressure is fighting the air pressure. When the bubble is up, they're not fighting, they're cooperating.
If you stand on a pogo stick that's on a bathroom scale, the scale will measure just your and the stick weight, but the pogo stick is compressed enough to counteract your weight.
Now, lower your ceiling to the point that the ceiling is barely touching your head while you're on the pogo stick.
The force on the scale is equal to your weight + stick weight, force on the ceiling is zero.
If you now step on the scale yourself and wedge the pogo stick between you and the ceiling, the pogo stick will have to compress the same amount in order to fit, thus pushing the equivalent of your weight down on top of your head, and same force goes to the ceiling.
The scale is now measuring twice your weight + the stick weight, the ceiling now has one time your weight on it as a force equivalent.
@@therflash _" the water pressure is fighting the air pressure"_
That was the missing link. Now I'm getting it. 👍
@@therflash That was a really good explanation, thanks!
@Oldasianguy It's a similar concept, yes.
A good analogy is to think about standing on a spring, compressing it. You are pressing downward with exactly your weight, and the spring is holding you up. Place a ceiling directly above your head, so your head is almost touching it. Now step off the spring, and you're still pushing against the ground with exactly your weight, same as before. And now, to simulate the pressurized air being at the top of the tube, we put the spring on your head. The spring isn't compressed by your weight anymore, so it expands and presses against the ceiling, and thus pushes down on you. So now you're pushing against the ground with your weight PLUS the force of the spring pushing you downward by trying to expand against the ceiling.
????
do you think if you got a trampoline and then tightened the springs, so there was no spring force at all, the weight of the system would just magically change? I don't think so myself. Springs settle and they do not and cannot change the weight, no matter what. I could be wrong, but I don't think I am in this case.
@@deucedeuce1572 The point is that when the bubble is on the bottom, it's acting like a spring under your feet, while when the bubble is on top, it's acting like a spring on your head (the weight being negligible; just the springiness). You don't feel the pressure from the spring when it's under you any more than you feel the pressure from the ground. But if it's above you pushing down (off of the lid of the container), you feel the pressure of both the ground AND the spring pushing down on you.
@@stevenjones8575 For every action there's an equal and opposite reaction, is there not. If there is "X" amount of weight on the spring, then there is "X" amount of weight on the spring... and if the spring weighs "Z" amount of weight... the spring weighs "Z". If you weigh the man and the trampoline for example, it doesn't matter if there are springs on the trampoline or if the springs are replaced with an equal weight of non-stretching wire.
I could just be misunderstanding what he was saying... but I almost always know what he's talking about in his videos. Have spent most of my life studying all fields of science. I'm not usually one to misunderstand, especially when it comes to scientific "laws".
(As for equal and opposite reactions, if you have 10 pounds pushing on a spring, the spring is pushing back 10lbs. The spring will always compress to the exact weight to where it's pushing back on the object the exact same amount that the object is pushing on the spring. It matters not if it's a spring or a solid block of steel in that aspect. The force against each object is equal.
The force the trampoline has to exert to hold you up is equal to force it has to exert on the ground plus it's own weight. So in your example it's still equal
This is a really well established fact for any petroleum engineer because the well can receive a gas influx from the formation that we call "gas kick", this gas rises up the well which ((increases)) the pressure on the bottom formation which can cause it to fracture, so the most dangerous kick is the gas kick.
Well I didn't think this would be true even for such small height
I like that you put several seconds of black screen at the end so that the automatic video suggestions don't block anything important.
I hate those kind of pop-ups. Mamy times It does cover value info or images
Those aren't automatic, you set them up as an uploader. You can choose not to have any or you can move them around wherever you want on screen
@@theclipreaper huh
@@isaacorlich *valuable
@@YunxiaoChu it's true!!
Petroleum engineer here. This is very easy to understand with mathematics, perhaps with the help of pressure gradient chart (P vs depth). But seeing it visualized in real experiment like this is a delight. Thank you.
Could you understand it using Bernoulli’s equation as well, technically? Since the total gravitational potential energy of the fluid is decreasing as the bubble goes up, it makes sense to think pressure goes up to compensate.
Shame most of that petroleum gets used to fund a car dependent suburban hellscape instead of idk science and stuff
@@HUEHUEUHEPony world run on money tbh I don't even blame him
@@c4r4dd1ct6More like a differential equation if you think about a spring
@@HUEHUEUHEPony you mean civilization?
Old oil and gas guy here. I remember leaning this in one of the countless hours of training and education that we had to maintain to keep our field certs. I was in frac, and we didn’t have to worry so much about gas bubbles in the day to day pumping operations in the formations we were pumping into. We just needed to be aware of such phenomenon. Thanks for reminding me of it!
The thumbnail, which was at one point also part of the video (3:20) suggests, that it isn't a closed system which is confusing. Without the closed top I would even expect the pressure to decrease.
Me too
Exactly, the moment he tried it with the closed tube I was like "woooh, that's completely different"
Yes, with an open top it's a simple hydrostatic pressure dependent only on depth.
This movie isn't acceptable
Posters who don't pay attention to such aspects should be condemned as scientists who are not good enough.
Another way to think about this: When the water is above the air, gravity is pulling the water down onto the air, which exerts a compression force which is opposite to the pressure of the air pushing outwards. Therefore, part of the air's pressure is being counteracted by gravity, and the total net force from the air's pressure is reduced by that amount.
When the water is on the bottom, its gravity is no longer trying to compress the air, so there is no force to counteract the outward air pressure, so the pressure it exerts on the _container_ becomes higher.
It is very counterintuitive, though..
Best explanation
This is actually how I thought about it in the beginning when he originally asked what would happen.
That being said, his explanation was a bit to complex, although i suppose it had to be for a visual representation
How come the water doesn’t act as a spring as much as the air does?
because liquids are incomprehensible @@tristanmoller9498
@@tristanmoller9498 may be cuz air was compressed not the water, if the test done on gel or solid with liquid you might see the difference but pressure might be very small in that scenario
There was one flaw with the question at 0:02 as the right side of the pipe looks closed and the left side (highest) point looks open. Hint would be helpful to mention a closed system. With that I went in wrong direction but finally with recognizing a closed system I couldn't solve correctly as well, but it would have saved some time.
He said it was a closed system. Listening comprehension please.
Simplest explanation by analogy: what happens if you hold your breath while you ascend from a deep dive? The air in the lungs will want to expand. The exact same thing is happening here. Air is being compressed by water and if that compression is removed without the air being allowed to expand then that air will apply higher pressure against the container that is confining it.
Beautiful explanation, thank you.
🙏
Perfect explanation.
That is not what is happening. The feeling of air wanting to expand is due to pressure difference between air in lungs and surrounding water. The pressure applied by the gas doesnot change, that is Boyles law, it applies the same pressure when it is at the bottom as well as at the top. That is the whole puzzle! How does the pressure increase when pressure of the bubble doesnot change!? The answer is the pressure in the air at the bottom balances out the weight of the water column above but when it goes up it pushes the water column harder in to the base.
@creatureofhabit7049 Sheesh.The weight of water in both cases is what generates the equalizing pressure. It is EXACTLY the same.
When the water weight is removed, the (unchanging) pressure of the air pocket exerts greater FORCE on the containment whether it is lungs or a tube filled with water. There is no paradox or mystery. Just realize that water itself can act as (part of) the containment.
@@tomszabo7350 i think you did not understand what i wrote. I never said that the weight of the water changes ever. I said when the air pocket is at the bottome it balances out the weight of the water column and exerts equal force on the bottom of the pipe. When the air pocket goes on top then it exerts the same force on top of the water column i.e. both the water column and the air pocket push the bottom of the pipe.
Also can you rephrase your comment, it was hard for me to understand what youbwant to convey.
As far as i know the force on the lungs when you come up is due to pressure imbalance between inside(lung air) and outside (water). Neither the air inside the lungs is exerting more pressure nor force.
Pls help me better understand your point if i misunderstood.
I found the correct answer with a slight variation of the reasoning of the video: as the air goes up, it "should" undergo an expansion because it "should" be subject to a pressure decrease.
But the volume being fixed, the expansion is constrained: the pressure of the air stays the same: the intuition that the pressure doesn't change is correct, but it applies to the air, not to the whole system.
Therefore, with the same pressure of air, the pressure at the bottom is higher when the air is at the top because you have to include the weight of the long column, while the initial situation only had a small water column added to the air pressure.
Well explained! That's how I reasoned as well.
That's a more intuitive explanation, I think. At least it clicks better with my brain. Top.
I was so close to getting it... I thought for sure the air pressure would decrease as it moved up, and I realized that it wouldn't be able to expand against the water. But I assumed the temperature of the air would decrease with the pressure decrease in a fixed volume
@@Zaros262 yeah I almost got confused about that too, but then I realized that the temperature variation only occurs if the volume can vary: it's the work of the pressure forces integrated on the volume variation. No volume variation = no work = no variation of T :)
Yep, this was also my reasoning, thanks for putting it into words!
By the way, @@renedekker9806 your Dutch is showing 😉, but I agree, this is a "top" explanation.
Yup. I got it wrong.
Got it wrong too but I assumed the top was open to the atmosphere -so no spring.
No, WE got it wrong
@@ryanjohnson3615 ye dang
@Nulley0 implying that everyone got it wrong? I didn't.
The pressure is kinetic as increasing with the movement of water due to the influence of gravity, so the water has to stop at the bottom using its kinetic energy to form pressure from a vacuum
Glad to have got it wrong in exactly the same way you did haha.
Me too
28/4/2024 Sunday 8:56PM
I was really confused by this. And then the moment you introduced the spring "Because the air is springy" it instantly clicked. Perfect illustration!
The trick is to find the reference point that has the constant pressure. When the vessel is open, it's the atmospheric pressure at the opening that is constant. When it's sealed however, due to liquid being almost non-compressible, the air bubble is now the constant pressure.
My brain wasn't registering that the tube was completely sealed. Which means now we need to see this with the top end open.
The thumbnail didn't really show it as closed. But the open tube case is easy as the pressure at the bottom of an open topped tube of incompressible fluid is always proportional to the height of the fluid column (in other words it goes down if open).
and if it were then the pressure would decrease instead - the top is at atmospheric pressure.
With sealed top, it's almost pipette, where bubble gives some freedom of movement...
I suspect the average pressure throughout the vessel does not change.
I think that his experiment doesn't match the question too. The jar in the question is not sealed.
I was very surprised. I thought the pressure would stay the same since it's a sealed system.
You also have to take into account gravity. If you removed gravity I wonder if it would stay the same 🤔
This is why education is important 🤦🏻♂️
@@ACatLoversHandle if you remove gravity, there's no such thing as top and bottom 😁
I think he bent the tubing out of its natural shape, which deformed the tubing and caused it to flatten, which increased the pressure inside. I think his entire explanation is wrong also... although I can't say 100%. I just don't think it works like that. As far as I know, a spring would settle with weight on it and would not, ever in any circumstance change the total weight/mass of a sealed system. Your car doesn't weigh less, because it's on shocks. I just don't believe that. If you change the shocks/springs with a steel rod that weighs exactly the same amount, your car would still weigh exactly the same if put on top of a scale in both configurations. Then if you filled the car with water and put an air balloon in the trunk or in the cab of the car, it wouldn't magically change the weight of the car (comparing the two configurations with water and balloons)... and/or if you tied the balloon to the floor of the car or let the balloon float to the top of the car. As long as the car, the water and the air all weigh the same and contain the same mass, the weight will stay the same (as far as I know). I hope you forgive the long explanation and repetition. It's not an easy thing to explain in just written words.
@@alexrvolt662 No gravity, no weight.
That is very counterintuitive but with the spring example, it makes sense!
The movement of the gravitational potential energy was the key for me in fixing this concept in my mind. Great examples to facilitate comprehension. Thanks.
The potential/gravitational energy argument is actually completely wrong. You can tell it is wrong by considering the same problem, but replace the air with a low density incompressible liquid. The potential energy would change in the same way when the bubble moves, but the pressure in the system would remain constant, or even decrease depending on where you measure it. It is not gravitational potential energy making the change. It is a change in the internal energy of the system due to the compressive nature of the air and its interaction with the container. This is something not stressed in the video, but fundamentally all this pressure is applied by the tube/container to hold the system at fixed volume. When the air bubble is at the bottom the weight of water helps to keep it at fixed volume, at the top the water is no longer helping, so all this force comes from the container, increasing the net force of the container on the fluid. This is the change resulting in the pressure increase. It is kind of hidden because it is not a typical type of energy change, it is very specific to this situation.
@@CerealReviewer If we applied the ideal gas law, i.e. PV = nRT. Which factor do you believe that must be changed and cause the P to go up? This is the part that I don't undertand. It seems to me that neither V, n nor T should change (a lot)
@porker2008 you are right, the pressure of the air bubble is constant because the volume and temperature are fixed. But the source of the confining pressure changes, in one case the water column provides the pressure, in the other the container applies the pressure. With the container applying more pressure the internal energy of the system increases.
@@CerealReviewer Thanks for the quick response. My thought was that the pressure we messured at the bottom should be the sum of the air pressure, plus rho*g*h, where rho is the density of the water, g is the gravitational constant and h being the depth of the water above that point. So after we moved the bubble to the top, nothing changed except h increased. so the overall pressure increased. Is this a correct explaination?
@@porker2008 hard to say without a more detailed conversation, but it sounds like you get the gist of it.
I find it even more confusing and fascinating in reverse - you can create vacuum!
Start with the bubble on top but at atmospheric pressure this time (i.e. seal the tube when the bubble is on top), and somehow move it down to the bottom.
1. As explained in this video, the bubble would want to stay fixed in volume and (atmospheric) pressure at the bottom. The system’s pressure decreases, but that’s not the interesting part.
2. Note the pressure at the top of the tube is (atmospheric pressure - rho*g*h). Since this pressure cannot be negative, with big enough liquid density*height, e.g. > 76 cm of mercury (new video ?😄), you’ll beat the atmospheric pressure and create vacuum on the top of the tube! Since this is a fixed-volume system, this means the bubble at the bottom will be compressed!
3. The same mechanical analogy applies (love it btw). Start with the spring on top, with an initial compression force F0 to mimic the atmospheric pressure. Invert the bottle. If mg > F0, the weight will compress the spring, creating “vacuum” on top.
You got me on this one. Keep on the great content!
BTW there are vacuum gauges manufactured to do exactly this.
That L shaped thing in the thumbnail doesn't look like a sealed system.
Exactly. It looks like it is only sealed at the lower end.
That was my observation as well. If it is open to the air, the pressure is directly related to the height of the water column.
It looks like the blue stuff isn't fluid. probably just a solid model made for the visual
@@DarthCalculushuh
The same observation through off my assumption and we can reference it like Boyle's law the partial pressures of each and when the water column is above the air bubble that means the bubble is compressed equal to the weight in water column of the liquid also the water is not able to separate from the top of the vessel meaning that a partial vacuum is occurring at the top of the liquid column. Therefore when the volume of the gas is above it would if pressurized, expand like a spring to exert force on the top of the liquid column and or eliminate the partial vacuum previously at the top
And that bubble pressure increase was then probably also the reason for the oil rig explosion in some video I cannot locate anymore, where they are drilling the hole and suddenly they scream they hit an unexpected gas pocket and need to abandon the area, seconds later greyish-brown thick liquid erupts from the hole and covers the whole area, including a nearby city.
Or was that one of my strange dreams many years back that I now falsely assume having been a video?
I even remember clearly how some of the shots in the video looked, for example a street light next to a low, white wall in the dark under a tree in the rain.
It's one of those problems where once you know the answer, it seems so obvious but until then, it's anything but. Great content!
This is the second of your videos I've watched, but the way you take the time to explain what is happening and why after the experiment made you very worth subscribing to for me.
Chemical engineer here, I think the easiest way to perceive this is to (as always) examine the extreme case. If you were to run this experiment with a column >> 10m then the weight of the water would exceed atmospheric pressure (assuming your gas bubble was injected at atmospheric pressure). You could then imagine at the top, a vacuum would be created, because the force of the water is now stronger than the gas bubble (the gas bubble would compress). This would be an area of zero pressure pushing down on the top of the column of water. Let the bubble rise and now the vacuum has been replaced with a non-zero pressure pushing down on the same column of water.
Any vacuum at the top would reduce the effect, though, and if the tube is J-shaped, it could even reverse the direction of the effect.
Take a J shape where in the beginning the left part of the J has 1m of air, all the way to the bottom, with 10m of water on the right side, with vacuum on top, and 99 more meters of vacuum on top of that. In this case, the pressure at the bottom would be 10mH2O=1bar.
Now let the air move to the top right of the J. With the water incompressible, the air would now fill up exactly 100m of tube, which (at constant temperature) would reduce the pressure in the air to 1bar/100 = 0.01bar.
Now, since the hight of the water column is now only 9m, the total pressure at the bottom would be 0.9 bar(from the water) + 0.01bar (from the air) for a total of 0.91 bar.
For comparison, if the top of the J did NOT have vacuum (the tube ended exactly where vacuum would have formed), the air would retain the 1 bar pressure, and the pressure at the bottom would be 1.9 bar.
I would say it increases. Water is incompressible so its volume is fixed, and therefore the volume of the gas is fixed too. Assuming that the temperature of the gas is fixed as well, then so will be its pressure. So when the gas is moved to the top, it pushes on the water below it with the same pressure it had at the bottom, so the extra water column remaining will increase the pressure at the bottom.
nice, you nailed it.
Perfect!
what do you mean by extra water column? the height and amount of water is the same right, so i'm still confused how there is more pressure
edit: i get it now
Yup
From this result, I think it can be said that the bubble acts as a constant pressure regulator for the water which have the same height level.
I love how this channel has grown over the years and now is my favorite go-to place for science related videos on youtube.
Truee
Yeah, it's a great channel. I think he might be wrong about this experiment though. I can't say with absolute certainty. He probably knows more than me about this subject and other kinds of science... but at the same time, it don't mean he can't ever be wrong. I think the PVC systems pressure went up, because he changed the shape of the tubing and used force to do so. When the bottom was curved like in the first test, it was in it's natural resting state, which is the shape that it was manufactured to be in... but when he straightens it out, he not only has to use a certain amount of force to do that... but straightening the tubing causes it to flatten, because it's a tube being bent outside of its normal shape. For example, if you did the opposite and sealed off the ends of a perfectly straight copper tube then bend it, the pressure inside would increase in that case too.
@@deucedeuce1572 He returned the shape back. But pressure remained at higher value.
@@d4slaimless Good point. Appreciate the reply.
69-th like!
5:47 Jail Speedrun Any% 💀
"Hey Guuuys, look at this ODD little contraption I build."
?
😂 Def not 💯%
I’m a scuba diver. And with my experience and with playing with soda bottles sealed and un sealed bottles. So I knew there would be a change in pressure. Great job. Thank you 😊
When everyone gets wrong, it's probably because something is wrong with question. The picture is confusing because it doesn't look like sealed system.
This.
Well, if you watched the video beyond 10 seconds you would have seen that when the question was asked it was a closed system and it is mentioned it's a closed system.
I was still telling my TV that the pressure shouldn't change, until you pointed out the shifted center of gravity.
Wow that’s so interesting! It was completely wrong with what I thought. This type of physics is truly incredible!
Yay I got it right! I expected the gas at the lower end to counter the pressure from the water above, like a force from the opposite direction. Only 1:40 in but curious for the reason nonetheless
Turns out it was exactly the reason behind it. 🥳
Good one, as an oilfield drilling supervisor we all should know this. It's one basic knowledge on well killing method --> volumetric method.
I think it would have been a good idea to show that in the starting state, you could take out the caps on top of the tube and it would do nothing, but after the bubble transfer if you take off the cap, it would blow out and the pressure would decrease.
So at 0:05 when you see both column, it does not seem to be capped at the top so the pressure is higher in B, if it's capped, it depends on if the cap has been put after or before the bubble transfer.
Great brain teaser and counterintuitive answer anyway, I love it !
For bubble on right side:
Pr = pw * g * h1 = pw * g * h2 + Pb
h1 = height of water measured on the left
h2 = height of water measured from the right
Pb = pressure from air pocket
pw = density of water
g = gravitational acceleration
For bubble on left side:
Pl = pw * g * h3 + Pb = pw * g * h4
Isolate and solve for Pb:
Pb = Pl - pw*g*h3 = Pr - pw*g*h2
Pl = Pr + pw*g*(h3-h2)
Since h3 > h2 (seen visually in the video), we know that the pressure at that point when the bubble is on the left side will be greater.
I came up a similar expression, except that I did not assume the pressure at top of the left are is zero at the beginning. It is more realistic to assume it is close to atmosphere. However, this assumption does not alter the result.
@DragonYang01 My line of thinking was that since the vessel is completely enclosed, the atmospheric pressure wouldn't act on the surface of the water!
@@andrewliang4713 As long as there is a nanometer thin layer of air before sealing the container, the atmospherics pressure is there. Even you use a pump after sealing, it will take a few minutes to remove all gases dissolved in the water.
@DragonYang01 That makes a lot of sense when I think about filling up the tube in an actual experiment. I didn't even consider that! Just goes to show the difference between actual application and textbook questions, haha.
It has no consequence for your mathematical result, but I think the right hand side of your equation for PL is incorrect. It is not pw * g * h4. The reason is, that this equally leads to a contradiction in which Pb can become negative (solve for Pb). I think the reason is, if the fluid on the left would be compressible it's density would increase. But water isn't. So the water on the right in case of the bubble being on the left can be treated as a "wall". That also explains the observation in the experiment in the video.
This is one of your best videos. Surprising result, great series of experiments, thorough explanation of what's going on.
Damn, that really was surprising, but very cool
Initially I thought that the system would stay the same, for the same reason as you: it's sealed. As soon he cut back at 2:45 I realized that since air is 1) not hydraulic and 2) buoyant, it would compress and exert a force on the water, effectively making it "weigh" less and decrease the pressure.
The air bubble is sealed. In the beginning this air is compressed and by changing the location of the bubble to the top this compression remains. So there are two states:
A: Static pressure of water column defines overal pressure and air bubble compression.
B: Static pressure of water column (this time lower) PLUS air compression from A as air bubble has no additional volume to expand.
By ventilation on top at B the pressure decreases.
3:16 is missleading as top is open (= atmospheric pressure)
According to the formulas:
Case 1: When the bubble is at the bottom
Its P = h*d*g ; h= height of vessels or tube, d= density of liquid, g means gravity
When the "air bubble" gets on top the pressure becomes
P = P(atmosphere) + h*d*g
Yes me too I also did the same!!! Are you a JEE aspirant?
Why p(atmosphere) if it is a sealed tube ?
The pressure of the bubble depends on the height of the water column when the tube was sealed, so it would be greater than atmospheric pressure. Pbubble=Patm+pghi.
The final pressure would be Patm+pgHi+pgHf.
So... P*2?
I got confused because I assumed that the tube was open. the first diagram he shows also has open containers.
with an open tube the pressure would decrease, because when the air would reach the top it would be able to disperse.
Same here
10 seconds in the video and it's both mentioned and shown to be a closed system
Pressure in a sealed vessel is potential energy, so you’re converting types of potential energy
No, pressure can only do work if the volume changes.
i somehow got it right. im no pro but used basic high school physics.
At 0:10 , PA = Po + pgh1
At 0:30 PB = Po + pgh2 (h2>h1) so PB>PA thus pressure at bottom increases
Similarly for top
At 2:00 PA = Po + pgh1 -pgh2 (h2>h1) so overall Po - pg(h2-h1)
At 2:30 PB = Po + pgh3 (although h3 is very small) overall Pb>Po and Po>PA thus i concluded Pb>Pa and thus pressure at the top also increases
Hope im correct even though i used basic fluid mechanics and the new perpective of COG->P and squeezy air was new and interesting to me. and it does make sense if i write bernoullis equation for the system.
Feel free to correct me if im wrong.
It will increase as the pressure of the air column remains same if we assume the temperature is constant thus the pressure at bottom will be pressure of air + dgh where d is the density of water, since h increasesz pressure. Increases
I know this from working in the automotive brake industry. A small amount of air in a brake line could dramatically affect the stopping distance ( by several feet). So your stopping force is proportional to the pressure applied, (for tmoc system actually a lot more force than the applied pressure, but at a certain point you get to the knee point and it eventually becomes 1-1 input to output force). Having to bleed brake systems for basically a cubic millimeter of air is very annoying depending on the type/design of the calipers. Some types seem to be prone to getting air stuck. Thats why straight off the assembly line usually has the best brake performance due to manufacturers using a push/pull method to fill the brake line when the system started off dry. That is usually better at getting any air out of the system.
I'm going to guess "B". Currently at 0:11.
what?
Conservation of energy - when the water is at the top, the gravitational potential energy is higher so the pressure potential energy must be lower, and vice versa
But it could take or free energy to move the air from one place to the other...
3:15 I want to say that the plexiglass setup and the tube setup aren't the same. In the tube setup, there's initially more water above the height of the point at which the pressure is measured. However, in the plexiglass setup, there's exactly as much water below that point's height. Assuming the volume stays the same, this means there's also exactly as much water above that height.
I got this from the thumbnail, admittedly by assuming the anti-intuitive answer was correct and then thinking about why.
Here is how I thought about it. Water is incompressible (or at least very nearly so) when the gas bubble is at the bottom, its pressure is equal to that of a water column from the lower surface to the upper one, and the pressure at the bottom is exactly that of a water column equal to the height of the tank.
Now we release the gas bubble to the top. As the system is constant volume, the gas cannot expand. Its pressure when at the top is exactly the same as it was in the lower position. So the pressure at the bottom is now equal to that of the height of the water column, PLUS that of the former difference between the two surfaces.
When the air goes up the potential gravitational energy decreases so the energy stored by the pressure differential increases.
that's what he says in the video, but it's wrong. The pressure in the bubble stays the same, and the pressure change in water doesn't take energy.
@@deinauge7894 This here.
The potential energy goes into kinetic energy of the water passing the bubble, then into thermal energy of the water by viscous damping.
I was confused when he said the energy turned into pressure,
It's like saying that force of the magnet sticking to my fridge comes from magnetic energy@@deinauge7894
@@rainaldkoch9093this is the only explanation that makes sense to me. The pressure increases when the air bubble rises to the top because the gravitational potential energy of water molecules is getting converted to kinetic energy. Which implies that the temperature of the system must increase at least slightly to account for the increase in pressure. Since his set up is not truly "closed", in the sense that heat exchange will still occur between the contents of the tube and the surroundings, does this also mean that the pressure will eventually return to its original value once thermal equilibrium is achieved??
Before this goes to far im going to say it increases becasue the air acts like a shock absorber.
Edit: i was a plumber for a very long yime and also dabnle in hydrologics
I get you said this before you watched the vid, but if the pressure on the air is increased then it puts more pressure on its container as well. It’s why a box container has weight pushing against its sides. So it would work as a shock absorber but it would apply some of that pressure onto its container. But part of it would just compress and not push against the side. Idk though that’s my guess. I’m just 15 so I don’t know a whole lot unlike you.
Indeed, for a vertical pipe, the gravitational potential energy of the water changes. However, for the U-shaped tube, it is not true! The water also raises on the other side!
The problem can not be treated by neglecting the process and just looking at the initial and final states. The process of the bubble rising is quite rapid and far from being quasi-static (not only the gas doesn't travel in one piece, but the rise of the tiny bubbles it splits into is accelerated), so basic thermodynamics doesn't work well either (i.e. the law pV/T= const for the gas is valid only for quasi-static processes). The gas, however, heats up while being pushed up by the liquid's pressure (work is definitely being done on it), and this increases its pressure (just not proportionally, since the process is not quasi-static). Why does the liquid remain at higher pressure after the thermal equilibrium with the room (if that even happens, since, as you mentioned, the pipe could even burst from the high pressure) is probably somewhat ansewrable by non-equilibrium thermodynamics.
One other factor to take into account is the work you yourself are doing to turn the tube back and forth.
3:21 the bubble is working with its weight when it’s above, when it’s below it’s against the weight of the water that’s why the pressure changes with local.
Who else laughed when he said "It looks like a bomb or something" haha
There is a fault in the premiss. You in no way stated that the air bubble introduced into the system, was under pressure. You created the assumption that the system was first created with the bubble at the top, where it was not under pressure, and caused the bubble to move to the lower position. Under that circumstance, there is no pressure when the bubble is at the top, thus you failed to provide the exact circumstances under which the bubble was created.
It doesn't matter. You can create the system by starting with the bubble at the top. You will reduce the pressure when you try to move it to the bottom. Potential Energy of the system is a state function.
We know that there is enough pressure to counter the weight of liquid, otherwise he couldn't have kept the bubble entirely on one side (we would have observed a bubble of vacuum on top in all cases).
I'm not sure why it would need to be stated that the air bubble is under pressure.
It's a closed system; the air bubble will always be under some amount of pressure. If it wasn't, it would be a vacuum.
You could even do it with oil instead of air.
This video is what made me subscribe. There are so many counterintuitive things answered by science. Trying to stick with intuitive explanations doesn't always lead to the right conclusion.
It's pretty easy if you think about it like a spring. At the bottom, the spring is like a cushion acting in opposition to gravity at the reference point. At the top, the spring goes from acting against gravity to acting in tandem with gravity at the reference point.
Except its wrong to think of it like a spring, since the water doesnt put pressure on the bubble, since its closed at the top. Easy to demonstrate by opening the bottom, the water wont flow down, right? What the explanation in the video overlooks is that the pressure does increase at the end of the bottom where the bubble was i believe.
@@barrygillis the water would flow down if you opened the bottom... I'm not sure if that's what you're trying to describe but that's what I'm interpreting.
If you opened the bottom, whether the bubble was at the top or bottom, the water would tend to flow out of the tube toward hydraulic equilibrium. Depending on the amount of surface tension present in the tube, the water may or may not stop flowing due to a vacuum forming at the closed top end. This is a separate force, that also acts like a spring, but is not present in the first set of conditions.
The example with the spring immediately made me understand why it works the opposite way I naively thought it would.
Best person to practically and simply explain this if your struggling is your plumber.
I understood it, when you explained it with the transformation of potential energy into pressure @8:20 . A very interesting problem. Thank you.
1:05 and 2:29 Pressure doesn't change much, it's 0.8something measured right next to the bubble. If no gravity it would be absolutely the same.
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Reading through the comments, someone mentioned the Bernoulli equation and that was spot on. As a ChE who reviews various subjects in my field periodically I was chagrined I didn't make that connection. In a fluid system, the sum of the changes in velocity, height, and pressure (scaled by appropriate constants) are constant. If you change one something else needs to change. The center of gravity of they system decreased and the decrease in potential energy resulted in an increase in pressure.
0:40 I think it's something to do with the compression of the air bubble. Maybe the air will expand because it's not being compressed by the liquid, causing the air to compress the liquid instead as it has less space. Something about that doesn't seem right, but I'm expecting a counterintuitive solution so that's my guess.
Hey, I was (kinda) right!
The “sealed system” part of this question makes a huge difference regarding intuitive reasoning.
Just imagine the bubble as your boss. Once he gets on top of your head, you will feel more pressure.
Increase. The volume of air is constant since the liquid is incompressible, so the pressure of the air remains the same. The pressure of the air matches the pressure of the water on the interface, and water increases on pressure with depth, so pressure increases with depth from the air, no matter where the air is.
More intuitively, when the water is pushing down it squeezes the air, giving it pressure. If you move the air to the top, still at the same pressure, now the air is pushing down on the water, increasing the starting pressure of the water.
(written before the explanation)
First part of the explanation correct, the second part not entirely.
This shows again that analyzing a system by looking at its energy can help quite a lot in providing the right answer.
In the initial image though it doesn't look as though it is a sealed system it looks to me like it's open at the top because you can clearly see a top piece on the lower end but not on the upper end.
Another way to look at it, when we consider the air pressure to be neutral with the bubble on top: if the air bubble is down, it wants to be "compressed" by the weight of the water above it, so the water is "pulled apart", creating negative pressure similar to capillary forces in trees.
This makes perfect sense. The water is acting sort of like a spring and decreasing pressure. As the water goes up, the force of the compression is going from bottom to top, to top to bottom.
8:40 I got it right reasoning like that - when the air is at the top there is more potential energy.
But I think the illustrated diagram where A or B is presented may be wrong, though... The potential energy for the systems illustrated as A and B is equal, I think? The bubble in the illustration isn't big enough. Right?
my guess: I first tend towards saying, with the air bubble further up, pressure should decrease. However, since air is compressible, and the bubble e.g. has been compressed when being down, i guess: pressure doesnt change.
ok, right idea, wrong conclusion
Well.. water is imcompressible.. but air is compressible. The air is compressed and will remain so where ever it is.
So ading the compressed air that is compressed the the full level to the top the adding all the water minus the top, now you have one full level pf pressure in the gas, then about 3/4 in level. So the relative pressure should go up... say about 75%
Very educational! I love the fact that you asked very intelligent questions, one in particular that I didn’t think of but realized that it was an excellent question as soon as you said it. Where did the extra energy come from?
Excellent observation when he mentioned the center of gravity lowered
I've seen a few of your videos/shorts and thought they were interesting, but this one made me subscribe. Great job explaining a very non intuitive concept!
This whole time I was thinking "why don't you turn it on its side and measure, since vertical orientation you must also factor in gravity, whereas horizontal you can largely ignore it."
I found it easier to understand with something squishy like an orange. Let's say you have a tube with a screw top. Put a spring in the tube, followed by an orange on top. The spring will compress due to the weight of the orange. Screw the top tight to the orange, then flip the tube over.
The spring will no longer be compressed by the weight, so it will attempt to expand, squishing the orange.
Sub the orange for liquid, and the spring for gas. When you try to squish a liquid you get pressure on the container.
When you asked the question...you said "or...will the pressure increases somehow" 😂😂 and i knew it was going to increase
Pausing at 4:17 the only thing i can thing of is the nature of air how its more compressable than water so that when the water is above the bubble its actually compressing the air to allow for more soace in the chamber giving a false preassure reading once the trapped air is released the air isnt able to be compressed by the weight of the warer reducing the available volume for the water.
A simple thing to consider is that gravity has an influence on everything in our lives. If we can one day gain better knowledge and mastery over gravity then we can do much greater things.
Two ways i can see someone easily understanding it.
1. Fill a waterbottle with air at the bottom of a pool seal it and then take it to the top of the pool. Open the bottle fast and you can see the pressure that was stored as the weight of the water in the pool was being used to compress the air bubbles.
2. Imagine a large amount of sand over an explosive to absorb the energy, then imagine that same explosive at the top instead of the bottom. Without the sand to compress it its free to show the entirety of energy stored.
Idk feel like it makes sense when viewed like that
Pressure should decrease because it's only the height of the water column that affects pressure of water in a vessel, barring external forces being applied (which would include atmospheric pressure or lack thereof). Thumbnail showed it being an open vessel which would decrease for sure. This is a closed vessel, curious to see what happens
Very interesting that it increased in a closed system
The fact that the air bubble moved upwards on its own when the "tail" was deformed has clued me in that the potential energy change might be the solution here.
Great demonstration!
Only was able to understand this after he talked about the potential energy, I could not wrap my head around it before.
This sounds like a problem created by the round chicken in a vacuum mentality, which I also fell for.
3:17 - it's not increasing in pressure when the bubble is at the bottom. You just showed it do the opposite in the experiment!
I immediately intuited this as being part way to a liquid barometer. If the column of liquid kept going higher, eventually the pressure would drop to the point that there would be a vacuum and a bubble would form at the top. So as long as there's no air in the column of liquid and the tube is rigid enough, the pressure will continuously drop the higher you go.
7:25 Just to be more accurate, stating water is incompressible is incorrect. It IS compressible but takes an absurd amount of pressure to do it, such as industrial water jets. Normally I would not even bring this up but I hold your channel in the highest regards and felt it should be pointed out.
Water is compressible on the micron scale. After such it's functionally solid.
Also water is what fluids are based off of due to the stability of water physics.
For the compressible air pocket to have equal volume in the two cases, the liquid-air-interface must maintain equal pressure. Adding the contribution by gravity acting on the liquid from the interface down to the bottom, it is clear that the pressure is higher at the bottom when the air pocket is in the upper position, so increase when the pocket is moved to the top.
Since the liquid is incompressible and the tank is fixed, the volume of the gas remains the same, and the temperature is also constant so the gas pressure is the same, but the pressure at the same height of the liquid is the same so in the low gas scenario the pressure is lower everywhere in the tube.
I love your video sir but This is wrong interpertation as in 7:59 the presure in whole tube will not inc as the botom end will still have same presure as befor, it only changes how presure is distributed in starting demostration he measure presure between ends which inc because high presure gas goes up and replaced by low presure liquid. [To get analogy thick two block of same area one more denser below and less dense above presure at ground will be same even if we reverse the order. but presure in intersection or within lower block (low dense) after filp will have high presure then intial case.] So presure change “inside tube” not at botom because order switches not due to dec in potential energy. But presure at top inc because property of gases to have same presure throughout and same on all walls, that causes expotion in mining sites. Thats my idea thought
As a physics professor, I kept starting to write a comment, but then you explained exactly what I was about to say. Like four times in a row. So.... great job!
I am analyzing and commenting at the same time trying to understand this puzzle. Someone please correct me if I am wrong. I assume there is no extra pressure pumped into the closed system. I wish you had just laid the tubes down horizontally and showed the pressure in psi for the starting position. 6:47 the 350 grams is the same when you flipped it. Were you trying to confuse us with the way you presented this demonstration? Note: 0.8 bar is 12. psi. Using psi makes more sense. 3:08 you show water traveling up the tube when air should be traveling up and the water down. I think you are showing that raising a water column increases psi. I will revisit this puzzle later maybe after someone corrects me.
Loved what you said in the end. Those counterintuitive phenomena let you learn and understand the fundamentals better
Fascinating. I’m not sure if you answered this in the video or not (I’m still confused), but if I understand it correctly, the water being on top pushes the air below down to compress it and the air acts as a cushion effectively taking one for the team to reduce overall tube pressure, but is not as capable of cushioning the pressure when on top because it can then only utilize the tube pressure and not the water’s weight to compress. Since the air is less compressed in the top position, the tube pressure increases.
I initially thought pressure would decrease because the water went to the lower portion and thus closer to the ground/the gravitational pull becomes (minisculy so) stronger.
i got it right immediately. the air bubble invokes a force upon the water which counteracts the pressure of the water column you created, so when the air bubble goes to its most natural location at the top of the column it’s then adding to the total amount of pressure instead of subtracting from it.
air is a fluid too, & as a sailor it’s really important to have a good grasp of fluid dynamic since that’s the very mechanic that sailing exploits to function. in the world we live in, anything that seems counterintuitive could very well be the exact answer your looking for.
sailing downwind wing on wing won’t be any faster than tacking upwind. to the untrained eye it’s confusing because when you’re with the wind, sails all the way out, completely level with the water you’d think you’d be going the maximum speed you can, but you literally have to sail against the wind, making frequent & regular turns with your sails basically all the way in, with your entire body hanging off the boat as a counterweight (hiking) to prevent it from tipping, since your simultaneously using the boat as a wing in the air & a paddle in the water, all while surfing on numerous forms of tension
Remind me to never let you change my brakes if you didn't already understand this about hydraulic systems.
0:23 guess at this point. I would say there is a slight potential energy change that would slightly increase the pressure on the vessel because the denser fluid is being elevated. That said I could easily see how I would be wrong here.
I am at 00:40
Decrease because air is at top of it and now liquid is not pushing it ❤❤❤❤
Btw love your videoes!!