Talk:Siphon/Archive 7

Lets start a new section just for less scrolling
@ Tobermory - I would express siphon workings about the same as the current article does. If my posts seem confusing to you it is probably because my posts are long as it is, so I can't give detailed massively long explanations of everything, and I can't include an entire physics textbook in every post. In my very first response to you, the first four questions I asked you were questions trying to get to your understanding of suction, but you skipped my questions. When you finally answered the key question, I was able to recognize that I needed to quote some explanation to clear that up for you. I don't think you can give two exact quotes where my concepts shifted, except where I started saying that it is entirely gravity that drives the siphon instead of gravity and pressure together. But that really wasn't a shift, it was a clarification, because I always said that it was pressure that pushed the liquid up powered by the pull of gravity on the liquid on the down side. Mindbuilder (talk) 02:12, 2 May 2014 (UTC)

@ DHeyward - I'm aware the flow lowers the pressure, but do you realize that the lowering by that method can't lower the pressure anywhere near enough to account for the pressure difference needed to raise the water? If you don't believe me, calculate it yourself. Potter 1971 states that for a siphon with a frictionless fluid, the dynamic pressure would be (1/2)dv^2 lower than the static case (where d is the density of the liquid, say water, and v is the velocity). Keep it simple and consider a siphon going at 1m/s. If you haven't calculated it yourself, why would you be so convinced that it can account for the force needed to raise the liquid? That effect would also be even far less still for extremely slow siphons since the velocity is squared. Whatever theory of siphons one has must account for slow ones as well. I don't think just because something started something that it must continue, but in this case it does continue. The process involves some conversion of potential energy, but the question is what forces from where are involved in that conversion. For example in the two buckets hanging from a rope over a pulley example, there is some conversion of potential energy, but that doesn't mean there is no force exerted by the rope on the rising bucket. After sucking the liquid to the top of a siphon it starts to get easier, not because atmospheric pressure is ceasing to push the liquid up, but rather because you are getting help with your sucking by the weight of the liquid on the descending side.

Consider a siphon, say a half meter tall, starting with the two buckets at equal height, so that at first it won't flow. Imagine we put a T fitting at the top with a top extension tube going up another meter. Say we attach a vacuum pump to the top of the tube and suck the water up both sides till the water in it is a half meter up the tube coming out the top. Then we close the valve at the top of the one meter top extension. Now while the bucket water levels are still at equal heights, the water level in the top extension will be about 1m above the bucket levels, so the pressure of the air in the upper half of the top extension will be about .9atm. Since that .9atm air is a gas under pressure it will be exerting a force upward, sideways, and downward, trying to expand. Thus the water in the top extension would get pushed down if it wasn't for atmospheric pressure pushing it way up there and not just pushing the water up but also helping the water squeeze that .9atm pressurized air up into the top of the top extension against the valve. Now if we lower one of the buckets just slightly, say a tenth of a mm, the water will begin to move very slowly from the siphon effect out of the slightly higher bucket down to the slightly lower bucket. Finally we have a working siphon moving at a very slow speed. The speed is so slow that the pressure drop from velocity would be negligible. Notice what doesn't happen in this siphon. The .9atm air in the top of the top extension does not succeed in expanding and forcing its way down by any easily measurable amount. Atmospheric pressure is still pushing the water all the way up past the moving part of the siphon up into the top extension with still enough force left over to prevent the air at the top from expanding. When we lower one of the buckets slightly, you don't think that force of atmospheric pressure on the water in the up side of the siphon just suddenly disappears do you? If the pressure was lowered significantly by the movement in the water, the .9atm bubble would expand down considerably.

Now lets move the buckets back so that the levels are so close together that the movement of the siphon is almost not measurable, but the water is still going up, so it IS still a working siphon, and we still have to figure out what forces make the water go up. Lets consider a basic vector force diagram on a cubic mm of water. We can evaluate this almost as if it is static. We can just ignore the pressure forces on the sides of the cube because they're opposite and equal and for our purposes they cancel and have none of the vertical component we're interested in. The weight of the cube of water is a downward vector in our force diagram. The force on the top of the cube equals the pressure at that depth times the 1mm^2 area. Thats a downward force vector in our diagram. The force on the bottom of the cube is the slightly higher pressure at that slightly greater depth times the 1mm^2 area. Thats an upward force vector in our force diagram. The upward vector from pressure on the bottom is slightly greater in magnitude than the downward vector from the pressure on the top. The difference between the magnitudes of those two pressure vectors happens to be, not coincidentally, the magnitude of the weight vector.

Now what is responsible for this water pressure force holding these cubes of water up against the pull of gravity? Well, if we put the buckets back at the same level so the siphon goes static again, the vector diagram will be almost identical. There will still be that large upward pressure force vector at the bottom of each cube. Since that vacuum pump we used to fill the siphon can't actually apply a pulling force to the liquid, we realize that the upward pressure force in the vector diagram ultimately comes from atmospheric pressure. Do you deny that there exists this upward pressure force vector countering the downward gravity vector in a static siphon, in an almost unmeasurably slow siphon, and even in a normally moving siphon? Mindbuilder (talk) 09:30, 2 May 2014 (UTC)


 * Mindbuilder, I am well aware of the parameters of suction. Please consider this, there are several concepts that are relevant to the understanding of siphons.


 * NEGATIVE PRESSURE GENERATED BY OCTOPUS SUCKERS: A STUDY OF THE TENSILE STRENGTH OF WATER IN NATURE
 * Understanding suction attachment mechanisms requires knowledge of the physics of water under negative pressure (pressures below 0 MPa) (see Kier and Smith, 1990). If water is contained in a leak-proof, expandable container, such as a sucker, exerting force to expand the container decreases the pressure of the water with little actual increase in volume. This is because water's cohesiveness resists expansion, and the decrease in pressure balances the expansive force (see Denny, 1988; Kier and Smith, 1990). Water at decreased pressure behaves like a solid in tension because of its low expansibility (high bulk modulus). Water is thus sometimes described as having tensile strength. Water breaks when the pressure falls to the cavitation threshold for that particular sample.


 * Your answer to my question; 'I would express siphon workings about the same as the current article does', doesn't actually offer anything as the current article is a dogs breakfast of conflicting and contradictory statements. I am interested in understanding how you concieve and directly explain a siphon as your comments are not clear and are contradictory. Tobermory  conferre 12:23, 2 May 2014 (UTC)


 * Do you realize that those suckers will completely lose all ability to exert any negative pressure if a small air bubble gets under the sucker? Likewise if a siphon had any negative pressure in it, all negative pressure would immediately cease to pull if even a tiny air bubble got in, as siphons in vacuum do completely empty upon the entry of an air bubble. But typical siphons hardly notice an air bubble going through. Of course, even with a little air bubble underneath, those suckers can still "suck" in the way a normal suction cup does, by lowering the pressure underneath and letting the external ambient pressure push the object and sucker together. Which isn't really suction at all, it's pushing together. I doubt you can cite exact quotes of either my statements or statements in the article that are contradictory.  Mindbuilder (talk) 13:31, 2 May 2014 (UTC)


 * Your example of the vacuum pump and tee can be replaced with valves in the bucket and a fill port instead of the vacuum pump. Two buckets at same level, close off tube in the bucket with valve.  Fill from the top.  Close off top valve when apparatus is full (same level as vacuum).  Lower one bucket.  Open bucket valves.  Siphon flows.  No air pressure difference needed.  You are confusing the method to extract the potential energy of the height difference with the mechanism with which it is released.  It doesn't matter if you push the ball up the hill or pull it.  You have convinced yourself that since you can pull the ball, the "pull" force must be necessary when in fact, the only necessary component is that the ball is at the top of the hill.  How it got there is not part of the mechanism for how it would roll down.  Nor is the incline of the hill relevant, just the height.  Once you have filled the tube, by whatever mechanism, and created a difference in height, the siphon will flow.  If you don't see how filling from the top negates your whole argument then you don't understand how siphons work.  Even the fact that you can turn the vacuum pump off is a clue that the atm pressure is not relevant.  Further, you conflate headspace requirements with atmospheric pressure.  --DHeyward (talk) 16:22, 2 May 2014 (UTC)


 * My example didn't rest on how the siphon was filled. I considered writing it basically as you stated, closed valves on the bottom ends and adding liquid from the top. But once you close the valve at the top of the top extension and open the ones at the bottom, and if the buckets start level, and if you added the right amount of water so it stabilizes with the heights the same as in my example, then the rest of my example is the same. The key point is that the force vectors on that cubic mm of water are the same either way, and contain an upward force vector countering gravity. When the experiment is done by filling from the top, do you agree that the force diagram for that cubic mm would still have an upward vector countering gravity? Forget the ultimate source or cause of that vector for a while if you want. Is that upward force vector there? Its not clear to me if you even think that force vector is there or not. I don't want to move on to further discussions only to find out that I'm wasting my time because I never convinced you that vector was there. So I need you to answer one way or the other - is that upward force vector there? Or are you saying that upward force vector is only there while you're filling the siphon by suction, but after it's filled from the top, or after the siphon starts flowing extremely slowly, then it vanishes? Mindbuilder (talk) 20:19, 2 May 2014 (UTC)


 * Sure. Weight, counterweight = 0 -> there's gravity and no motion.  Fluid isn't moving.  Now replace your vertical "up" tube with an 30 degree angled tube that contains twice as much water per cm in elevation with the same diameter tube. Down tube is vertical, same diameter so that the up tube has twice the volume, same diameter and same static level of buckets. Does the maximum height a siphon change?  Do you agree that the height of the liquid column supported in that static situation is not dependent on the mass in the column or the cross-sectional area of the tube?  --DHeyward (talk) 00:11, 3 May 2014 (UTC)


 * A 30deg up tube doesn't change the maximum height the siphon can support. I agree the height is not dependent on the mass in the column or the cross sectional area. I'm completely missing how those things are relevant to our discussion though. Does the fact that if you had a barometer at a 30deg angle, atmospheric pressure would still be considered the force pushing the liquid up, bear on the 30deg angle issue? It seems simpler to just stick to vertical tube siphons so we don't have to factor in the tube walls supporting half the weight though.


 * So if we agree that there is this upward pressure force vector on our cubic mm of water, then next is what shall we attribute that upward force to. In the static siphon with the buckets at the same level and air in the top half of the top extension pushing down, we have a situation that is pretty much the same as in a drinking straw, except the lower part is split in two, and the valve at top is closed rather than having someone sucking. It is also like a barometer but with some gas pressure at top. If atmospheric pressure suddenly disappeared, the upward "pressure" force on the cubes of water in the siphon would disappear, but there would still be a down gravity vector and a downward force originating from the pressurized bubble at top. With only two downward force vectors on the water, it would rapidly accelerate downward and empty into the bucket very quickly. It seems pretty clear that most physicists would say that it is atmospheric pressure supplying the force so those little force vectors can hold up those water cubes, just as in a barometer or drinking straw. Would you agree most physicists would say that?


 * Now if we change the levels of the buckets very slightly so the siphon starts flowing extremely slowly, those upward force vectors are only going to change very slightly, right? And even while the siphon is now flowing extremely slowly, those upward force vectors still have to do the job of pushing hard enough to support the weight of the liquid and keep that .9atm bubble at top from pushing its way down, right? Or do you suggest that at the moment the bucket levels start to become slightly uneven and the water just starts to flow extremely slowly, that suddenly those force vectors which were formerly attributed entirely to atmospheric pressure, suddenly have nothing or very little to do with atmospheric pressure, and suddenly those upward force force vectors should be attributed to gravity, a Bernoulli pressure reduction at such an extremely slow velocity, the flow of the siphon, the conversion of potential energy, or some other cause?


 * There are at least two ways you can make something move, you can apply a force to it, or if there are already two opposing forces already acting on it, you can remove or weaken one of the opposing forces. This is not precisely accurate but it helps to look at it like this: In the static siphon you've got the force of atmospheric pressure pushing the water up both sides. At the top of the bend you've got the atmospheric pressure from each side battling each other and coming to a stalemate. When you lower one of the buckets slightly, that makes the column of water on the down side of the siphon slightly taller than on the up side. Atmospheric pressure on the down side now has a bit of a harder time getting to the top against the added weight/pressure. So now there is an imbalance of force from atmospheric pressure on each side at the top of the siphon, and the atmospheric pressure going up the up side wins. Every time a chunk of liquid rolls over the top like a train car, so to speak, it goes from being an impediment to siphon flow, to opposing the atmospheric pressure on the down side and thereby reducing the resistance the atmospheric pressure faces on the up side to lifting up the next chunk entering the siphon.


 * The fact that it is gravity powering the siphon doesn't change the fact that atmospheric pressure pushes the liquid up. What matters is that the pressure on the down side is reduced, thereby reducing the impediment to atmospheric pressure pushing the liquid up. That pressure on the down side can be reduced by gravity action on the column of water as if it is a big syringe plunger, or the entire down side of the siphon could be replaced by an actual syringe plunger, powered by gravity acting on the water of an alpine lake turning a turbine, generating electricity, powering an electric motor to pull on the syringe plunger. Or the syringe plunger could be pulled by an electric motor powered by solar panels, never involving gravity as the ultimate power source. In all those situations and in the real siphon, it is atmospheric pressure pushing the liquid up, regardless of the power source. Mindbuilder (talk) 22:41, 3 May 2014 (UTC)


 * We don't say that because it's not true. When the outlet is lowered slightly water flows slightly and creates a pressure drop that is propagated back to the source at the speed of sound in the liquid.  What's important to note is that the source doesn't change.  Flow starts and creates the pressure differential at the outlet.  How much is pressure energy and how much is velocity energy of the fluid is governed by Bernoulli's principle.  Energy throughout the system is conserved and is not changing at the inlet.  The atmospheric pressure is irrelevant.  It does no work and the pressure difference that moves the fluid starts at the outlet and moves to the inlet as an energy equation.  At any point in the siphon, the energy is the same (either as velocity, pressure or height).  In the static case where the inlet and outlet are the same height and there is no flow, the fluid pressure is the same.  It doesn't matter what the pressure is and it doesn't contribute to the siphon.  If the atmosphere was a force that actually propelled liquid, it would matter.  --DHeyward (talk) 23:13, 3 May 2014 (UTC)

So if the down side of the siphon is disconnected and replaced by a suction pump, would the atmosphere be doing any work pushing the liquid up the up side? Mindbuilder (talk) 23:32, 3 May 2014 (UTC)


 * Nope. Energy is conserved.  A manometer does no work.  Any work that is done would have to be by the pump.  It, too, would have a pressure difference that starts on the pump end and propagates to the atmosphere side at the speed of sound.  The fluid is moved before the atmosphere can act on it.   --DHeyward (talk) 23:57, 3 May 2014 (UTC)

@ DHeyward

Re your comment "At any point in the siphon, the energy is the same (either as velocity, pressure or height)." This is not correct for a working siphon, as Bernoulli's equation forgets to include friction. For a frictionless pipe, the equation is correct, but in a siphon, we are losing most of the potential head at the inlet (what you refer to as height) to friction in the pipe. There is also a little of it lost to entrance and exit losses, i.e. velocity head is lost on exit of the pipe. It doesn't change what I believe to be the purpose of your general argument, more just a note that pipe friction is also there and the energy is being lost along the way, so at any point in the working siphon, the energy is not the same. — Preceding unsigned comment added by 58.169.253.149 (talk) 23:39, 3 May 2014 (UTC)


 * That's correct. I am simplifying to ideal liquid and flow.  It's not really necessary for the fundamental processes and the principal that flow lines have equal energy.  There are lots of idealities that we know contribute to practical limits of a siphon.  --DHeyward (talk) 23:57, 3 May 2014 (UTC)


 * I agree we should stick to the frictionless model till we can agree on that simpler situation.


 * So lets say we just start with a vertical tube with air in it and we attach a vacuum cleaner to the top and start sucking up some water. Since the typical vacuum cleaner won't get anywhere close to complete vacuum, there will still be some air at the top of the tube pushing down on the water as we suck it up, so the air above can't be doing the work to lift the water, because it's applying a downward force not a lifting force. Then the atmosphere would be doing the work lifting the water right? Mindbuilder (talk) 00:13, 4 May 2014 (UTC)


 * Again, no.  If the fluid moves, the liquid closest to the vacuum pump moves first.  That is where the pressure difference is created and that is when flow starts and why.  That movement and pressure change propagates through the fluid at the speed of sound in the fluid. The atmospheric pressure is still at equilibrium with the reservoir and  that doesn't change.  --DHeyward (talk) 01:06, 4 May 2014 (UTC)


 * After the liquid is sucked up it has a higher gravitational potential energy. That energy had to come from somewhere. That energy couldn't have come from the air above the liquid because that air was pushing down. So what did work on the liquid to raise it up the tube? Mindbuilder (talk) 01:21, 4 May 2014 (UTC)


 * The machine that moved the air out of the tube did work. That's it.  It's quite plain that the fluid closest to the vacuum moved first.  It's somewhat obvious that this movement traverses the column at the speed of sound in the fluid.  That's from the pressure in the fluid which starts with the same energy as any other unit volume in reservoir.  Potential energy is exchanged between liquid pressure and height.  There is no distinction in energy between liquid at the surface at maximum height and liquid at the bottom and minimum height.  One has maximum gravitational energy and minimum liquid pressure, the other minimum height and maximum pressure.  The is also obvious in that the depth of the source tube inlet is not a factor. --DHeyward (talk) 01:54, 4 May 2014 (UTC)

@ Mindbuilder Re your comment "After the liquid is sucked up it has a higher gravitational potential energy"

Putting aside friction, but it also has less pressure energy by the same increase in potential head as DHeyward points out. In any static pipeline over undulating terrain, potential head plus pressure head will always be equal at any point. You are only considering potential head, and forgetting that it no longer has the pressure head — Preceding unsigned comment added by 124.182.57.9 (talk) 02:09, 4 May 2014 (UTC)


 * @ DHeyward - There are plenty of physicists and there are published papers and references to those papers stating that the liquid is pushed up a siphon. There are some that deny it and claim that the liquid is pulled up. I have seen none that deny both theories. If you have a reference that states that the liquid is not pushed up by atmospheric pressure as has been the long consensus in this article, then quote the particular words that state it is not pushed up by atmospheric pressure. You can't just cite a physics textbook and claim that all those formulas tell you by your odd interpretation that you are right and the other published articles that explicitly contradict your odd interpretation is wrong. If you want to advocate the liquid cohesion theory and present the controversy between the two theories, then we can probably do that. But if you want to remove the only theories specifically endorsed in the literature, you need to present quotes that specifically deny the theory you want to remove, not just a bunch of equations that you alone derive that conclusion from. This will be my third revert of your attempt to remove the theory from the literature that has been long established here. I'll have more discussion soon. Mindbuilder (talk) 05:09, 4 May 2014 (UTC)


 * I had a mistaken memory of what the three revert rule was. I thought it was that when three reverts had been made back to the previous consensus, that further edits were prohibited pending some kind of dispute resolution. Thats why I mentioned that in my comment and edit note. I also didn't remember anything about the 24 hour limit. I'll get straightened out on that right away. Mindbuilder (talk) 05:26, 4 May 2014 (UTC)


 * The machine that moved the air out of the tube certainly did work, but it didn't do any work on the water because it never exerted any force on the water. It removed some of the force of the air on the water, but the air above the water was doing negative work on the water and after its pressure was reduced, it still ended up doing negative work on the water, so what was doing the positive work? Are you claiming that while it gained gravitational potential energy that it lost some other kind? See my next response to 124.182.57.9.


 * @124.182.57.9 - You don't have to sign in, but it would be nice if you'd pick a nickname and just type that at the end of your posts so we can keep you separate from other commenters and know what to call you.


 * Say you start with a large tank of water at the bottom, maybe a ton of water. Then you have a pipe going up to another tank of the same size but sealed against air entry. Where the pipe goes into the bottom of the upper tank there is a valve. At the top of the upper tank you attach an air suction pump. You suck all the water from the full lower tank up to the upper tank by removing the air(not pumping the water directly), and as it gets full, you slow down the suction pump until the water stops rising and you close off the valve on the bottom of the upper tank and turn off the suction pump. You let the air pressure equalize to atmospheric at the top of the upper tank. If that's a small volume of air space to come back up to atmospheric pressure, then that is a small amount of energy, but nothing near the kind of energy to raise a ton of water a few meters. You had a big tank of water at one altitude, and now you have another tank of water at a higher gravitational potential energy a few meters higher. The water in both tanks is not moving at start and end, so there's no kinetic energy from Bernoulli's equation to worry about here. Where is this pressure energy? A pressure gauge will give the same readings any place in the top tank as it did in the bottom tank(if they're the same shape) except for a tiny difference in the atmospheric pressure a few meters up, but that can't account for anywhere near the work needed to lift a ton of water a few meters. Since the air the suction pump was evacuating was pressing down on the water, it couldn't be the suction pump doing upward work on the water to raise its gravitational potential energy. So where did the work come from to lift the water if not from atmospheric pressure? Mindbuilder (talk) 06:36, 4 May 2014 (UTC)


 * No, they all agree it's gravity. So do you when asked what the fundamental requirement is (height difference - even knowing that the atmospheric pressure at the outlet is higher than the inlet, it's not logical to sa the atmosphere works against itself to "push" or "pull").  All sources agree that it's governed by Bernoulli principle.  It is true that the atmosphere at the surface of the reservoir is at equilibrium with the atmosphere.  They all agree that the energy of a static reservoir is constant per unit volume and is the sum of liquid pressure and gravity.  At the surface gravitational PE dominates and pressure is low (atmospheric equilibrium).  At depth, pressure dominates and gravity PE is low.  Everyone agrees that it is irrelevant where that PE comes from, just that the sum is constant.  I provided many sources yet you continue to misinterpret even your own.  You obstinately stick to your own misunderstanding and use the talk page for your theories about atmospheric pressure driving a syphon which is simply incorrect as you even acknowledge that a) the actual pressure doesn't matter, b) the atmospheric pressure difference between inlet and outlet doesn't matter and even agree that outlet atmospheric pressure is higher at the outlet as it is lower, c) the cross sectional area of the "up" tube doesn't matter d) the "up" tube intake can be at the reservoir surface or bottom and it doesn't matter.  Now given all those things that you acknowledge don't matter, how can you still say it's pressure? How can you claim that it's atmospheric pressure when the net change in atmospheric pressure from inlet to outlet opposes flow in the siphon direction?   Seriously, it's like looking at someone stand on a table, conclude that the table supports their weight.  Then, as they jump off the table, conclude the table must have thrown them in the air because it statically held them so how could they jump without the table pushing them in the air?  It's such a fundamental misinterpretation of force, work and energy as to defy reason.  So, no, the article should not postulate "pulled" or "pushed", rather it is a conversion between three types energy: pressure, gravitational and kinetic.  It's governed by Bernoulli's Principle and the Newton's second Law.  Look at this picture: VenturiFlow.png. The kinetic energy increases at the expense of the fluid pressure, as shown by the difference in height of the two columns of water.]]  You can assume that the air flow through the tube behaves as a fluid.  There is no work being done on the U-Tube manometer.  Can the direction of flow be determined by the manometer reading?  There is simply an exchange between different types of energy and the tube statically reflects the difference.  It's like measuring the height of a ball at rest at the top of the hill vs. the rolling ball at the bottom.  The ball has kinetic energy at the bottom but it doesn't change the fact that you can still measure the hill.  The absolute pressure of each side is not relevant.  Because there is flow, there is a pressure gradient in the horizontal tube, and secondly because the flow is faster in the narrow section, there is an abrupt conversion of pressure energy to kinetic energy (or kinetic to pressure.  The venturi meter measures the pressure head change only, not work.  A siphon converts these three forms of energy through it's entire apparatus not unlike the way a ball rolling down a hill converts gravitation energy to kinetic energy.  An ideal lossless system can oscillate between gravitational potential energy and kinetic energy indefinitely.  No force is needed to oppose gravity to climb back up the hill when the ball has that energy (m g h) already in the form of kinetic energy.  The same is true for a siphon.  "lift" is the conversion of pressure head to height in the gravitational field. Flow converts gravitational and pressure energy into kinetic energy.  --DHeyward (talk) 07:19, 4 May 2014 (UTC)


 * Your latest thought experiment is great but until you can explain the energy you add and take away it's meaningless. The pump does the work so the energy required to move water from the lower tank to the upper will be reflected in the energy of the vacuum pump (doesn't matter if you pump the water to the higher tub using vacuum pump or submersible in lower tank).  There is no free lunch.  You can't add net potential energy to the fluid with just static tools.  You are verging on perpetual motion if you think you can move water for free.  A siphon doesn't do that.  Neither does a manometer.  --DHeyward (talk) 07:19, 4 May 2014 (UTC)

@ mindbuilder Re: "So where did the work come from to lift the water if not from atmospheric pressure?"

To move water, you can calculate the kw required via the formula: l/s x m/head divided by 102.04 x 100 divided by pump efficiency (example 75) x 100 divided by motor efficiency (example 95). To move water from your lower reservoir to the upper is going to require some energy, whether that is done with a vacuum pump or a normal water pump. The end result of raising the water is the same. When you finish moving the water, at the end of the month or the end of the quarter, you are going to get an account from the electricity supply company for the energy you used. So the energy comes from the power point on the wall that is supplying the electricity, the original source is unknown, could be a wind generator, could be coal fired power station, could be a hydro plant. You don't get another account from nature for using atmospheric pressure. If atmospheric pressure did the work, why are you paying the electricity company? Has it taken twice the amount of kw to move the water, one for atmospheric pressure, the other for the energy the electricity company supplied you and which you paid for?

Re two current theories. One is gravity and atmospheric pressure (water pushed up), the other is gravity and cohesion (water pulled up). There is plenty of examples of scenarios that question one or the other of these. Bubbles in a siphon question cohesion & tensile strength. Water dripping out of a tap or overfilling a glass clearly show a cohesion/surface tension, tensile strength limit that it can support about 3 to 4mm of it's own weight, again creating issues with cohesion and chain analogy theories. Fat up leg siphons also question the chain analogy. Siphons operating in a vacuum question the atmospheric pressure theory. The fact that both current major theories have issues might be because both might not be the correct theory, that maybe there is an alternative theory, that may utilize some existing knowledge, yet can have an explanation that each of the tests that are thrown up by each side to justify there view.

The issue is further compounded by outside theories that don't really relate to siphons, example tree height and the chain bead siphon, and just provide a distraction.

The fact that you are using one of the current know theories doesn't make you right. And the fact that DHeyward is putting forward a theory that is not one of the main two doesn't make him wrong. Regardless of past understanding, the fact that the issue has not been resolved gives even more reason to listen to alternative views, since that is where the answer should lie.

Every thinks that it has to be pushed or pulled, that they are the only two options. And neither are providing an undisputed answer. So the answer must lie elsewhere and I think that answer is in better understanding pressure gradients.

It would be good to see further vacuum tests done with ionic liquid, to see it's maximum siphon height, to drill a hole in the siphon crest to see what happens, if the liquid had tensile strength, it would continue to operate.

Maybe instead of continuing to find new ways to explain your current understanding, maybe a little effort should be put into giving genuine consideration to the information that is being presented. Dr Hughes is also locked into an understanding, and is unable to consider alternatives.

Atmospheric pressure definitely plays the part in setting the maximum height of a practical siphon and keeping the water as a liquid. However I believe the pressure part of atmospheric pressure may not be fully playing the part that is claimed, and it is the different density of water and air that is a key component that is overlooked. — Preceding unsigned comment added by 60.231.229.245 (talk) 07:35, 4 May 2014 (UTC)


 * Don't mix up the idea of doing work on something and doing work to remove a force that then enables another force to do work. Take for example if you have a boat loaded with a heavy lead weight. You use a crane based on shore to slowly lift the weight just until there is a tiny air gap between it and the boat, so that the boat is no longer supporting the weight, the weight is entirely supported by the crane. In the process of lifting the weight, the hull of the boat will rise up in the water somewhat. The water will do some work pushing the hull of the boat up. And the crane will do work equal to the work done by the water. But the crane won't do any work on the boat hull. The crane is not touching the hull. The weight is not glued to the hull. Neither the crane nor the weight being lifted will apply an upward pull or force on the hull.


 * I realize of course that the suction pump does work in my two tank example. And it is no coincidence that the work the suction pump does is the same as the work done in raising the water, plus inefficiencies. But in order for one object to do work on another object, it has to apply a force, and there has to be movement, and the force has to be in the same direction as the movement, or else the amount of work done is multiplied by the cosine of the angle between the direction of the force and the direction of the movement. That suction pump can't do work on the water telepathically, and there are no net magnetic or electric fields reaching out directly from the pump to the water, so the suction pump isn't doing work on the water directly. Now you might say that the suction pump does work on the air it is sucking on. But if the suction pump does work on the air in the tube between the suction pump and the tank, and that pump work is being transferred through the air to the water, then that air would have to somehow do work on the water that is being sucked up. But although the air is at reduced pressure, it is not applying an upward pulling force, it is applying a downward directed force. The water is moving up, so the angle difference between the direction of the work and the direction of the movement is 180deg. The cos(180deg) is -1. Thus the air above the water is not making a positive contribution to the work done on the water. Since the reduced pressure air being sucked by the suction pump is not doing the positive work on the water, then neither can the suction pump be doing the positive work on the water through that reduced pressure air.


 * @DHeyward - When you claim that I have a unique perspective, you are mixing up two distinct claims that I am making. My claim that atmospheric pressure pushes the liquid up a siphon is stated explicitly by many physicists from hundreds of years ago to present. Even those sources who deny it have acknowledged that it is a common position. On the other hand, I don't recall seeing an explicit statement of my claim that atmospheric pressure does work on the liquid as it goes up the siphon. But I didn't put that in the article. I didn't leave it out because I doubt it, I just don't see value to put it in. Mindbuilder (talk) 09:17, 4 May 2014 (UTC)

@ Mindbuilder

Re your comment: "I realize of course that the suction pump does work in my two tank example. And it is no coincidence that the work the suction pump does is the same as the work done in raising the water, plus inefficiencies"

To move say 54000 litres from a lower reservoir to a reservoir 5 metres higher within 1 hour at a flow rate of 15 l/s will take approx. 1 kw of energy for 1 hour, ignoring any pipe friction 15 x 5 / 102.04 x 100 / 75 x 100/95.

If atmospheric pressure is supplying 1kw of energy, and the pump is using 1 kw of energy, then is it? 1: using 2 kw to move the water, 1kw from atmospheric pressure and 1 from the electricity company? 2: using 1km from the electricity company only? 3: 1 kw from atmospheric pressure only?

Remember also that the absolute pressure of the water in the two containers, after atmospheric pressure is considered, just below the surface is essentially the same. — Preceding unsigned comment added by 121.220.216.75 (talk) 10:01, 4 May 2014 (UTC)


 * The work is done twice. We live at the bottom of an ocean of air. So imagine you have a chamber bolted to the bottom of a large tank of water. Say you pump the water out of the chamber leaving vacuum. Your pump uses a certain amount of power for a certain time to pump out the chamber. Then you can close the valve, turn off the pump and have your evacuated chamber as a sort of energy storage device. When you decide to, you can let water flow back into your chamber and extract power from the water going back into the chamber. The energy you extract as it goes back in is the same as the energy you used to evacuate it, ignoring inefficiencies. Also, in my crane lifting a weight off a boat example, would you deny that the water does work lifting the boat hull, thus the work done by the crane is doubled? Mindbuilder (talk) 11:07, 4 May 2014 (UTC)


 * Really? Stop imagining and start math.  The work is not done twice.  If the atmosphere could actually do work, we would use it.  Here's a hint: Atmospheric pressure at a surface height is based on gravity on a compressible fluid.   --DHeyward (talk) 19:09, 4 May 2014 (UTC)


 * Do you think that you can not extract energy from the water re-entering the evacuated underwater chamber? So you do some work with your pump to evacuate the chamber. You carry the pump far away so it cant do any more work in the system. Then more work is done as the water re-enters, the same amount of work done by the pump. The amount of work of the pump done twice. But during the re-entry, the pump is not doing the work because it is far away from the system. It's not a magical work multiplying scheme. You start by expending work, only after then you get work back.


 * OK, I get where the mix up is coming in here. It would seem that if the energy of the system is constant that implies no work is done to the system. Actually, work is done as the energy is transformed from one type to the other. Consider a 10cm cube of water with its upper surface level with the surface of the water in a tank. If we move this cube of water slowly so that drag effects are minimal we can move the cube of water up and down with very little need to do work, other than a tiny amount of one time work to accelerate it to a slow speed as we start to move it(and again if we change direction). But that doesn't imply that no work happens as we move it up and down. If we move the cube straight down 10cm from its original position where its top was level with the surface, gravity does work on our cube as it moves down. You might say, no, the energy of the system is constant. True, but that work by gravity still happens. The reason the energy can stay constant even though this work is being done in the system is because in order to move the cube down, another cube worth of water is displaced and raised up to fill the space our cube started out in. The cube that was displaced had a higher pressure energy down below, it lost that energy, but work had to be done on it to raise it up to a higher gravitational potential energy. Lots of work can be going on inside a system even if the total system energy remains constant.


 * In a venturi, as the fluid accelerates work is done on it by the higher pressure fluid behind. It's kinetic energy increases, but its pressure energy decreases. As the fluid slows down, it does work increasing the pressure energy of the fluid it "runs into" at high speed, and loses its kinetic energy. Work is done whenever the energy is transformed, though the total energy remains constant.


 * In the siphon, atmospheric pressure does work on the water going up, increasing its gravitational potential energy, but it looses the pressure energy it had at the bottom. Though the total may remain constant, work happens during the transformation from one form of energy to the other. Mindbuilder (talk) 19:28, 4 May 2014 (UTC)

Maths vs. Language
Why is there so much discussion regarding a physical concept? Surely maths is clearer and more concise. Bernoulli's principal is clearly in favour of gravity. Unless someone can come up with a formula describing fluid flow in terms of atmospheric pressure, we should just accept the conclusions of the formula we have.

And regarding 'atmospheric pressure is what pushes the liquid up into the low pressure zone' the pressure gradient due to atmospheric pressure on one side of the pipe will be the same as on the other side. Atmospheric pressure pushes the water back along the pipe with as much force as it pushes it through. Atmospheric pressure cancels. Brocerius (talk) 14:59, 1 June 2014 (UTC)


 * Yes, you are correct the maths is very concise. However the point of debate centres on the value of P in Bernoulli's equation at the top of the siphon. If P must always positive then the siphon is just a barometer with flow. If P can be negative then the siphon is not a special case of the barometer, but rather the barometer is a special case of the siphon. Using ionic liquids it has been shown that P can be negative. The question now revolves around the generality of this i.e. can you do it with a liquid like water? If you can use water then the chain model is applicable (to a degree). The nub of the problem is can water exist for long enough at negative P to flow over the top?--143.210.144.159 (talk) 13:19, 2 June 2014 (UTC)

Correct use of "Talk" tab
I'm not sure why, but the rules at the top of the page seem to indicate that the organizers expect this page to discuss ways to improve the nature and structure and effectiveness of the Article, rather than debate (at horrific length (I see 'why'.)) the nature of the topic itself. I don't want to inhibit the energetic (although somewhat redundant) discussion, but this probably should not be the basis for a course in the relatively settled but complex physics of siphons. One repeated flaw in the above discussion is the convenient fallacy of 'suction' or 'negative pressure', and misuse of 'gauge pressure'. --Wikidity (talk) 18:59, 15 September 2014 (UTC)