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Constant energy of system doesn't imply no work done within

If this idea of work being multiplied and work done in a system of constant energy still bothers you, consider a mass hanging from a spring, oscillating up and down. As the mass goes up, the spring does work on the mass increasing its kinetic and gravitational potential energy. As the mass goes down, gravity does work on the mass, and the mass does work on the spring, increasing the spring's elastic potential energy. Assuming little or no drag or losses in the spring, this work would happen over and over again without any energy input into the system. It's like a magic infinite work multiplying system! :) Seriously though, the fact that the energy of the system remains constant, doesn't imply that no work is being done within. Mindbuilder (talk) 20:32, 4 May 2014 (UTC)

Please see the First law of thermodynamics. You are still having trouble identifying internal and external forces and energy. If you add/subtract no heat and the energy is constant, there is no work. --DHeyward (talk) 21:39, 4 May 2014 (UTC)

So you're saying that when the spring pulls the oscillating weight up, the spring does no work on the weight? Mindbuilder (talk) 21:55, 4 May 2014 (UTC)

Oh, no, I see, you're saying that there is no energy going in or out of the oscillating spring system, but if the atmosphere is lifting the liquid in the siphon then that is an input into the siphon system from the atmosphere that would change the constant energy of the siphon's constant energy equation. Is that what you're saying? Mindbuilder (talk) 22:05, 4 May 2014 (UTC)

But no, you state "If you add/subtract no heat and the energy is constant, there is no work.". There is clearly work being done within the oscillating weight on a spring system, yet no energy is being added or subtracted if the spring is lossless and there is no aerodynamic drag. Do you claim no work is being done within the weight/spring system, or do you claim there is work being done but there is energy being added/subtracted, or what? How do you think the first law of thermodynamics applies to the oscillating weight/spring system? Mindbuilder (talk) 22:28, 4 May 2014 (UTC)

@ Mindbuilder

Re your comment on moving water from the lower reservoir to the higher reservoir: "The work is done twice."

Really?. Done Twice?

Putting aside friction and other losses, when the water is moved, there is now 54000 litres at the higher reservoir that has gained potential head from being raised, and thus, the amount of stored energy would be equal to 1 kw, the energy it used to pump it. But you are claiming that there is another kw of stored energy from the atmospheric pressure? You claim the work was done twice, thus 2 kw of energy used, however there is only 1 kw of stored energy. What has happened to the other 1 kw of energy? And stay with the two reservoirs, don't head off on a boat loaded with lead, since there is a major issue with claiming the work is done twice.

@MB I'm saying the first law of thermodynamics is true and the atmosphere does no work. A siphon is explained in terms of energy entirely by gravity. From the output being lower than the reservoir. The reservoir can be made so large such that there is substantially no change. Gravity, and the internal potential energy that is freely converted from pressure to velocity to height. It's constant in a siphon and flow in the tube is realized only when the outlet is lower than the inlet. A siphon is really no different than simply draining the reservoir but is constrained in that the fluid pressure can't be negative (which is the lift limit). Since the siphon is closed, energy in the fluid is changed between the three types of energy available (pressure, gravitational potential, height). The energy at any point in the siphon tube is equal to the energy of any point in the source reservoir. In terms of energy, it is an extension of the reservoir. --DHeyward (talk) 22:48, 4 May 2014 (UTC)

What you're missing is that as work is transformed between the types of energy in the siphon, that work is being done to effect that transformation. Its easier to see this in the spring/mass system. Do you claim that no work is being done within the oscillating spring/mass system? Mindbuilder (talk) 23:14, 4 May 2014 (UTC)

That's the first law of thermodynamics. You are confusing internal and external. --DHeyward (talk) 23:28, 4 May 2014 (UTC)

@121.220.139.44 - I'll get to your questions. But I'd like to stick to this thread for a bit, as I think it illuminates the both threads.

You can base your answer on the first law of thermodynamics or whatever you want. I'm trying to find out how you interpret the first law of thermodynamics. I already know how I would answer the question. And I think I know how you would, but it is not quite clear to me how you would answer it. This gets to the core of our dispute. Is no work being done within the oscillating spring/mass system? Mindbuilder (talk) 23:34, 4 May 2014 (UTC)

No work is done. It's a free exchange of energy of an internal friction-free system. In a siphon, the atmosphere is external to the siphon system. An infinitely sized atmosphere and an infinite reservoir has a constant energy associated with the fluid in the reservoir. It's constant. The reservoir level need not change for a siphon to work. No work is done while the fluid is within the tube as it changes from height, velocity and pressure. It's a constant. --DHeyward (talk) 00:14, 5 May 2014 (UTC)

Try this: place liquid mercury in a vessel. Put liquid water on top of it. Put a siphon tube down to the liquid mercury and start a siphon of mercury. Is it the water that you think is pushing the mercury? The water is entirely supported by mercury and has potential energy from gravity. --DHeyward (talk) 00:32, 5 May 2014 (UTC)

Thank you for that answer. After all this time, I think finally we're really hitting on the key to our disagreement. In a short search of the oscillating mass on a spring I found at least two references to work being done by the spring on the mass. The one that states it most directly is this http://www.regentsprep.org/Regents/physics/phys02/spring/qtsprengy.htm where it states: "...The energy exchange is reversed as the spring does work on raising the ball to its original height...." Here is another: http://www.hippocampus.org/homework-help/Physics-B/Oscillations%20and%20Gravitation_Work%20Mass-Spring%20System.html Here is another, although its a question that asks for the work to be calculated and it doesn't quite go all the way to oscillation: http://www.physicsforums.com/showthread.php?t=276317

So it appears that at the very least, my position that conversion of energy within a system does involve work being done, is not unique. In fact I think my position that there is work being done is near universal. None of the definitions of work that I've seen in looking at this have any qualification that it is not work if the force over the distance is converting between energy types within a closed system. It seems natural to consider any force over a distance on an object to be work. I'm sure you can't find a quote that states explicitly that the conversion of energy in the mass/spring doesn't involve work being done. Nor do I think you can find a quote that says that the conversion of energy within the siphon is not work being done. Nor do I think you can find a quote about the first law of thermodynamics that explicitly says that a change of energy within a system does not involve work being done. I don't see anything in discussions of the 1st law that seems to me to imply what you state. So the question is, when you see the first law of thermodynamics described, how do you get that interpretation out of it? It would be best if you can quote here explicit words or discussion of the 1st law and how you interpret those words to get those results. Lets stick with the spring on a pendulum example for a while till we get our work definition and 1st law issues straightened out, because it's much simpler than the siphon. If we can't agree about the mass/spring system, then we don't have much hope of agreeing on the more complex siphon system.

As for your mercury siphon example I'd say the mercury is pushed up by the water and the atmosphere together. I don't see what you're trying to imply by mentioning the water being supported by the mercury. Mindbuilder (talk) 02:32, 5 May 2014 (UTC)

Please read the wiki article on the first law of thermodynamics. That should dispell your personal definition of work. Unique or not, it's simply wrong and seems to be fundamentally why you cannot grasp that gravity is the only force. The atmosphere is really just a compressible fluid in a gravitational field. The water on mercury example is to demonstrate that the fluid pressure is gravitational and even though the water weighs more than air (and mercury more than water) the conditions to siphon are unchanged - i.e. outlet lower than the mercury level and the pressure component of the static energy equation is not negative at any point in the siphon tube. Gravity, though doesn't require the weight of the object above to act on the object below. The "infinite reservoir and atmosphere" shows that as the displacement of source reservoir water by siphon flow approaches zero, there is still a siphon. The atmosphere isn't doing work on a siphon. It's constant energy system from inlet to outlet and transfers energy between gravitational energy, pressure energy, and kinetic energy. Pullies, teeter-totterrs, springs, and counterweight doors aren't as confused as your posts sound when you seem to need a mystery force to oppose gravity when the potential energy is present to exchange with gravity. Kids on the teeter-totter know how to move weight around so the heavier kid moves less distance up than the lighter kid. How they moved the lighter kid higher than the heavier kid does not need some additional force acting against gravity. Even if the heavier kid displaces the more atmosphere than the lighter kid, no one seriously suggests that air pressure moved the lighter kid to a higher height than the heavier kid. --DHeyward (talk) 03:16, 5 May 2014 (UTC)

Well if you'd go read the entire physics section of your local university library, then you would understand I'm right. See that doesn't work. You can't just refer to a large body of writing and say that something in there supports your position. I've studied the laws of thermodynamics, and I understood them, and I skimmed the Wikipedia article to see if there was anything in there that I missed that would support this idea that its not work if its just an energy conversion within the system. I don't see anything that implies what you're saying. My definition of work is not just my personal definition, it's the same as all those I've seen in physics textbooks and on the net. Not one of those definitions say that it's not work if it is energy conversion within a closed system. Here's one from the Boston University physics department "Whenever a force is applied to an object, causing the object to move, work is done by the force." Where did you get your personal definition from? Just quote some of the words from that 1st law article or anywhere on the net. You should be able to narrow it down somewhat. Mindbuilder (talk) 04:33, 5 May 2014 (UTC)

Geez. Work is measured as energy. Please tell me how different forms of the same energy in a lossless systems do work? The sum of energy change is zero. You already have agreed the energy per unit volume of an incompressible fluid is constant throughout the fluid. The whole point of the siphon is no external energy is needed for operation. The atmosphere does not lose energy to a siphon. A siphon does not gain energy from the atmosphere. All of a siphon's energy is derived from (and can be calculated by) the height difference between output and input. There is simply no other source of energy required to describe it and whence no other component of work is present. --DHeyward (talk) 05:04, 5 May 2014 (UTC)

Also, don't confuse the energy for priming it with energy for operation. Just like your spring oscillator, a siphon condition's state isn't naturally a siphon. The oscillator must be primed to start oscillation. The same is true for the siphon. Whether the priming is done by evacuation (i.e sucking on the tube) or by filling from the top is not relevant to the operation. --DHeyward (talk) 05:37, 5 May 2014 (UTC)

@ Mindbuilder Re your earlier comments 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....."

then you add....."But that doesn't imply that no work happens as we move it up and down."

You contradict yourself.

Then you add

"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."

In a siphon, there is no swapping of water around. All the water on the up side moves up. You can add a constant steam of air bubbles, and the water between each air bubble doesn't swap places with any other water. And if you say it swaps place once, why stop at once, why not say twice, why not 5 times. That way you can claim to get 5 times the work.

If a siphon can work in a vacuum without atmospheric pressure, what is doing the 2nd lot of work then? — Preceding unsigned comment added by 124.182.170.250 (talk) 05:49, 5 May 2014 (UTC)

@ 124.182.170.250 - I still intend to reply to your two previous posts later today. And you're right, as my statement didn't clearly express what I meant, it did technically contradict my other statement, as you cite in your last comment. What I meant was there is very little need for the person moving the cube around to do work to move it. I didn't mean to imply that the ambient water would do very little work on the cube as the cube moved around.

@ DHeyward - This is not all of my response to your previous posts, but I thought I'd let you chew on this while I work on the rest.

If you won't quote the 1st law I'll quote it for you. Consider this statement of the 1st law from MIT:

"The change in energy of a system is equal to the difference between the heat added to the system and the work done by the system."

It seems you've misinterpreted this in two ways. You seem to have concluded that if there is no change in the internal energy of the system and no heat added to the system, then there is no work done by the system. You almost can't be blamed for interpreting it that way because that IS what it literally says. But the statement is simplified as all statements in physics are, and you're supposed to realize that what it is actually saying is that there is no work done by the system on the world outside the system. It says nothing about work done inside the system in converting one type of energy to an equivalent amount of another type of energy inside the system.

An odd consequence of defining work inside a system to not be work, would be that if you set the boundaries of your system to include the entire earth, then there would be no work taking place anywhere on earth, except for what work went out to or came in from space. Then you could change whether something was really work or not just by specifying different boundaries for your system. Although you cant change whether a particular example of work is work by changing how you define the boundaries of your system, you can change whether that example of work is work going in or out of your system by changing the boundaries of your system.

Second, when the 1st law talks about work done by the system on the world outside the system, it is talking about the net work done by the system on the world outside. If the system does one joule of mechanical work to the world outside the system and the outside world does one joule work back into the system, then the net work done by the system to the outside world is zero. But a net work of zero doesn't mean that the work done in, and the work done out, are not work or that the work didn't happen. It just means those bits of work add to zero. The work happens, it just adds up with all the other work that happens, to zero. Mindbuilder (talk) 22:16, 5 May 2014 (UTC)

Again, no. Not the way it works but you don't seem to understand. Try this, look at Pascal's siphon picture. Once mercury fills the tube, it siphons. Do you not see that once the mercury vessels are completley covered with water that the pressure on the lower vessel's mercury surface is higher than the upper and the flow is against the static pressure? Do you not see that once the fluid connection is made is when flow starts? Please tell me how you think atmospheric pressure is related to this problem, other than the initial filling of the siphon tube? How is it doing work if the level of interface between air and fluid is not moving while the siphon is running? --DHeyward (talk) 23:42, 5 May 2014 (UTC)

Step back from the siphon for a minute. Do you still think that work done within the system that doesn't enter or exit, is not work? If you define the borders of a system to include the whole earth is there no work going on inside? How small do you have to make the system so that work contained inside it ceases to be considered work? If the system does work on the outside world and the world does equal work back in, for a net of zero, does that mean no work went in or out, because the net was zero? We cant settle the siphon if we can't even agree on what the first law of thermodynamics means. Mindbuilder (talk) 01:25, 6 May 2014 (UTC)

Processes on the entire earth are lossy. It's a silly question that only shows you are not comfortable with where energy can be extracted and where it cannot. A lossless oscillator is clearly not doing work. It's certainly not doing twice as much work as you claimed earlier. The static atmosphere is not doing work either as it is simply a gas in a gravitational potential. A siphon can do work when it's operating and that work is defined by density, gravity and height (note that atmospheric pressure is not in the equation because it's external and static). It's that simple and you've said as much yet obstinately fail to understand that the static atmosphere is not operating on a siphon. You've hinted a slight grasp with motion through a distance but then can't see that the atmosphere doesn't need to move at all for a siphon to operate. Even in your narrow view, the atmosphere would need to move to do work. When potential energy, linetic energy and pressure energy in the fluid is changed from one form to another in the siphon, there is no loss. It is contained completely by the liquid system, whence no work is needed or done by the atmosphere. Work would be needed if energy were lost in the conversion. That is not the case. The energy at any point in the siphon is the same as any other point. In the up leg, the pressure drops but it is exactly equal to the energy gained by height and velocity. There is no room for work done by the atmosphere because there is no change in total energy. You are not grasping this. --DHeyward (talk) 01:49, 6 May 2014 (UTC)

We seem to have a much different definition of the word work. Would you say that gravity does work on a pendulum on the way down? Mindbuilder (talk) 06:38, 6 May 2014 (UTC)

Nope. But your avoiding the issue. A pendulum's energy is completely converted from potential to kinetic and back. 0 work. Now please describe how the non-moving atmosphere is doing work on the siphon. --DHeyward (talk) 06:51, 6 May 2014 (UTC)

The atmosphere does work on the siphon by pushing the liquid down. But the atmosphere doesn't do net work on the siphon system because then the descending liquid does equal work on the atmosphere as it gets pulled out by gravity. In Pascals siphon, the relevant atmosphere is the water. The water does work pushing down the mercury in the upper jar. The mercury runs over the siphon and into the lower jar where it does equal work pushing up the water. It is no coincidence that the water pushed up by the bottom jar has the same volume as the amount of mercury pushed down in the top jar. No NET work into the siphon by the atmosphere, but there is work into the siphon by the atmosphere. Do you see how it is possible there can be work into a system even when there is no net work into the system? Mindbuilder (talk) 08:10, 6 May 2014 (UTC)

No, because the water doesn't move from potentials in Pascals siphon. Incompressible fluids have the same energy throughout the fluid as depth pressure is traded for gravity. We've been through that. The water displacing the mercury on the top reservoir is no different than water over the bottom reservoir. There isn't even a transfer of energy. So if the water is not moving at its surface, how is it doing work on the internal mercury? Why isn't the weight of the mercury sufficient to explain it that you need a less dense and non-potential energy source to move it? Bernoulli's principle has all the components to move the mercury without needing external pressure. Pressure is lower with fluid height. Pressure is lower with fluid velocity. If all the energy and forces are explained without external pressure, why do you need your admittedley net zero force to move it? The math and physics don't need it. Do you see the problem with creating a criteria that you admit has no effect? --DHeyward (talk) 08:51, 6 May 2014 (UTC)

@ Mindbuilder

Re your comment: "No NET work into the siphon by the atmosphere, but there is work into the siphon by the atmosphere"

A siphon operates in a vacuum without atmospheric pressure, what is doing the 2nd lot of work then? — Preceding unsigned comment added by 124.182.141.21 (talk) 08:36, 6 May 2014 (UTC)

As I've stated before, siphons in vacuum do rely on liquid cohesion. And of course if there is no atmosphere then the atmosphere will do no net work and no work at all. I'm sorry I didn't reply to your previous posts.

I've got to pull myself away from this debate. I'll have to leave it up to others to settle this. I'll probably not be back by here for quite a while. Bye. Mindbuilder (talk) 08:51, 6 May 2014 (UTC)

Liquids in a vacuum are still under the effects of gravity. Gravity is responsible for the way liquids flow to fit their container, not atmospheric pressure. The weight of the liquid in the Inlet container has positive pressure on its container, even in a vacuum. Falling liquid on the outlet side creates negative pressure(Strong vacuum or suction force)in the siphon tube. This pressure difference between the Inlet and Outlet sides of the siphon is what drives the fluid flow. Gravity is the only force needed to make a siphon work. It is responsible for the weight of the liquid(positive pressure) on the inlet side and the falling liquid(negative pressure) in the tube on the outlet side. 69.170.106.237 (talk) 03:10, 1 September 2014 (UTC)

@DHeyward - Do you still think gravity does no work on a falling object? I think I finally figured out where you were getting that idea. It seems your idea comes from the way potential energy calculations are done. When doing potential energy calculations it is important NOT to include the work done by gravity in the work term, but rather let the work done by gravity be accounted for implicitly in the potential energy term. But it is important to realize that this is just a convention to simplify the calculations. It doesn't mean gravity is doing no work just because you don't include it in the work term. You are accounting for the work done by gravity in the potential energy term. If you included it in the work term as well as the potential energy term, you would be counting the work done by gravity twice. Here is an MIT physics homework problem calculating the work done by gravity on a falling object: http://ocw.mit.edu/high-school/physics/exam-prep/work-energy-power/forces-potential-energy/8_01t_fall_2004_ic_sol_w06d3_1.pdf There is also a problem from MIT to calculate the work done by gravity on a pendulum (problem 2c): http://ocw.mit.edu/high-school/physics/exam-prep/work-energy-power/forces-potential-energy/8_01_fall_1999_final.pdf and the answer is given here: http://ocw.mit.edu/high-school/physics/exam-prep/work-energy-power/forces-potential-energy/8_01_fall_1999_finalsol.pdf as mgl rather than zero. Now I've cited an MIT physics instructor showing that gravity does do work on falling objects. And gravity is a force that often acts over a distance in the direction of motion, so it fits the definition of a force doing work. Can you cite a single comment from anyone, anywhere, that supports the idea that gravity does no work on falling objects? Mindbuilder (talk) 18:02, 5 November 2014 (UTC)

No, you are not correct in forming only a part of the problem. You cannot simply look at the falling side. When a siphon is static balance (outlet and inlet surface level are equal), there is no flow. The upper container has potential energy. You can measure the potential energy at any point in the balanced, non flowing siphon and it is constant. No work is being done in the static case since nothing is moving. When the outlet is lowered so the flow starts, there is a conversion of gravitational energy to kinetic. That's work done by gravity. All of it done by gravity, none done by atmospheric pressure. The atmospheric forces don't push/pull liquid, it's all gravity and explained completely by gravity. The energy at any point in the siphon is constant as it is a fluid (ideal, non-compressible fluids are at equal energy meaning the gravitational potential energy of water at the surface exactly equals the energy at the bottom of the bucket (gravitational height energy is traded for pressure energy). This is why the siphon tube depth in the source bucket is irrelevant (it can be just below the surface or the bottom of the bucket, all that matters is the outlet height compared to the source surface (i.e. the source energy is higher than the outlet energy so it must flow as the energy equation must be equal so kinetic energy is the difference). --DHeyward (talk) 07:10, 6 November 2014 (UTC)

I thought we needed to clear up the basic physics before we could make progress on the details of the siphon. Do you agree now that an object in free fall has work done on it by gravity? Do you agree that a swinging pendulum has work done on it by gravity? Do you agree that a weight hanging from a lossless spring and bobing up and down, has work done on the weight by gravity and by the spring as it goes up and down? Mindbuilder (talk) 08:34, 6 November 2014 (UTC)

Object in freefall does work and it's from gravity. Change in height is net work. Pendulum and spring -> no work as the velocity and potential energy are the same at the end of the cycle. It could do work but that would mean the each cycle, the distance traveled is less. That distance loss between cycles is work but the ideal case there is no work, simply a transfer of energy from potential to kinetic and back. This isn't a physics class so I will not debate you endlessly. Note that potential energy in a fluid system depends only on the surface height. The work in a siphon is limited to the difference in height between surface of the top reservoir and outlet height. The liquid in the top of the tube has the exact same energy as the liquid in the bottom of the source container. pressure energy + kinetic energy + potential energy = constant (there is no analogue to pressure energy in your simple spring and pendulum). Again, this is why the location of the inlet opening of the U-Tube doesn't matter but the outlet does and the outlet matters only relative to the surface of the source (the source surface pressure energy and kinetic energy are 0 and it's only energy is gravitational). --DHeyward (talk) 15:24, 6 November 2014 (UTC)

If you don't need me to give you a physics class, you can just walk away and let me restore the siphon article. Did you see my second MIT cite (problem 2c): http://ocw.mit.edu/high-school/physics/exam-prep/work-energy-power/forces-potential-energy/8_01_fall_1999_final.pdf where the work done by gravity on the pendulum as the arm swings from horizontal down to vertical equals mgl? There is no principle of physics that says that in a lossless cycle there is no work done. Where did you get that idea? Can you cite any comment from anywhere to back up that idea? Mindbuilder (talk) 16:22, 6 November 2014 (UTC)

The laws of thermodynamics. See Work (physics)#Work by gravity. It is true that that at the bottom of the pendulum, the potential energy has been transformed to kinetic but it returns to the same height and zero velocity. Returning to the same height results in a net zero change in height. "work" would be any losses in the system such that it didn't return to the same height. Since it does return to the same height, the work done by the pendulum is 0 by definition. --DHeyward (talk) 18:18, 6 November 2014 (UTC)

You're talking net work of the lossless system. But it is important to recognize that work can have taken place even if in the end there is no net work. The work by gravity formula you linked to is this: W = F(y2 - y1). So when a pendulum swings from horizontal down to vertical, the bob goes down an amount (y2 - y1) = L where L is the length of the arm. Gravity does work on the bob. Work is done according to that formula. Of course the formula also says that if the bob is allowed to swing back up on the other side, then y3-y1 = 0 and therefore in the end the net work is zero. But that doesn't mean the work on the way down wasn't done, it just means work was done on the way down and then more (negative)work was done on the way up, and the total adds to zero. As the bob swings down to the bottom, work is done, and that work done becomes a fact of history that can't be changed. More (negative)work can be done on the way up and added to cancel out the work done on the way down, but the new work done doesn't go back in time to undo the fact that work had been done previously. It's like if you add three marbles to a cup and then take three out. In the end there was net zero marbles added to the cup, but that doesn't change the fact that there were three marbles added to the cup. Notice that in the MIT swinging pendulum problem there was no mention of what happend after the pendulum reached the bottom. The work done on the bob by gravity on the way down was mgL regardless of whether the bob swung back up the other side or hit a sand bag. If the work done on the pendulum on the way down depended on what happened after reaching bottom, then the correct answer to the problem wouldn't have been mgL, but rather the correct answer would be: "Not enough info given to solve the problem". One of the primary advantages of the potential energy formulas is that if all you care about is the result at the endpoint, you can sort of ignore all or much of the work done in between. But that doesn't mean all the work in between didn't happen. The net work after swinging back up is zero, but don't you agree that work is done by gravity on the bob on the way down, regardless of whether the bob hits a sand bag at the bottom or is allowed to swing back up? (and I'm not talking about the work done by the bob on the sand bag, I'm talking about the work done by gravity on the bob on the way down, whether it hits a sandbag or not) Mindbuilder (talk) 19:51, 6 November 2014 (UTC)

To bring it back to the siphon, the relevant net height difference between source reservoir surface and outlet is the only work done in the siphon system and only gravity is doing work. You are confusing yourself with "net work" and double the work and all sorts of other stuff. The difference in a fluid system is that pressure is another way to store energy (in addition to gravity potential and kinetic). --DHeyward (talk) 20:45, 6 November 2014 (UTC)

The pendulum is much simpler than the siphon. We seem to have a fundamental disconnect about some important basic physics here. It would be very much more difficult to come to an understanding about the basic physics in the much more complicated context of the siphon. Lets settle the easy stuff first before moving on to the siphon. Do you agree that gravity does work on the falling pedulum bob on the way down, regardless of whether or not the bob swings back up to give zero net work? Mindbuilder (talk) 22:08, 6 November 2014 (UTC)

Nope. You cannot extract it in a closed system. First law of thermodynamics states it very concisely. An ideal pendulum does no work. A pendulum with friction does mgh amount of work but only after it has completely stopped is all the work accomplished. The work is the same regardless of whether it swung once or one thousand times and it needs to be at rest to have done mgl amount of work. Only then does work = heat which is the fundamental principle of the First law of thermodynamics. In no case does an ideal spring or pendulum do work as all the energy remains internal to the system and there is no addition or subtraction. --DHeyward (talk) 23:56, 6 November 2014 (UTC)

You're right that there is no work extracted from a closed lossless system. It's true that an ideal pendulum does no significant work on anything outside its system But I'm not talking about extracting work from the system. I'm talking about whether any work takes place within the pendulum system. The work done by gravity on the bob on the way down converts the gravitational potential energy of the bob to kinetic energy, but the total energy remains constant. But that is still work done by gravity on the bob on the way down, right? Mindbuilder (talk) 00:35, 7 November 2014 (UTC)

First law of thermodynamics states it very concisely. Work within a closed system that isn't doing work on its surroundings or taking heat in is a meaningless construct. It's just a restatement of tradeoffs in energy and that's all a siphon is. A static balance siphon (output and input surface height are equal, no flow, tube is filled), trades pressure energy for gravitational energy (no flow means no kinetic energy). Lift max is achieved when pressure energy reaches zero and it's all gravitational potential energy. A fundamental property of incompressible fluids is that energy in the fluid system is equal throughout. A swimming pool surface has the same same energy as the swimming pool bottom - it trades high gravitational energy at the surface for pressure at the bottom. It's the reason why water seeks it's own level. It's the reason why the siphon flow can only be started when the outlet is lower than the source level (and why the source surface matters and not the depth at which the "up tube" is submerged). This is fundamental fluid properties and understanding conservation of energy. Incidentally, the same concepts apply to water distribution systems - "up" is irrelevant as long as the lower drain is below the source. It's why water can flow "up" to the second story of a house - a siphon still exhibits the same physics as that case with additional feature that pressure energy can still be traded off for potential gravitational energy above the surface. You can do a non-siphon experiment. Take two identical jugs at the same height with one full and one empty - connect the bottom of the jugs with a tube that sags so there is a down and up side. The full jug will fill the empty jug until they are level. The is still an "up" tube and a "down" tube. There is no need for anything other than an understanding of the energy tradeoffs. A siphon is no different and the energy tradeoffs are identical. --DHeyward (talk) 01:25, 7 November 2014 (UTC)

Lets say you put a pair of brake pads at the bottom of the pendulum so that when it swings down it immediately stops at the bottom. By the formula you cited above, the work done by gravity on the falling pendulum would be WorkDoneByGravity = Weight(finalHeight - initialHeight) right? But that's a lossy system. Now lets say instead of brake pads at the bottom you have a device at the bottom which when when hit by the pendulum it sends a frictionless flywheel spinning and transfers all the kinetic energy of the pendulum bob at the bottom of the swing into kinetic energy in the flywheel. Now since the pendulum is at a different height at the start and end, then gravity does work. But the system of the pendulum and flywheel together is lossless. Sometime in the far future we could apply a friction brake to that flywheel and dissipate the energy. But whether that lossiness happens or not in the future wouldn't go back in history and change whether gravity did work on the pendulum on the way down. Likewise, if we make the decision to stop the pendulum at the bottom with a friction brake or let it continue up the other side, that decision doesn't go back to the past and change whether gravity did work on the pendulum on the way down. Gravity did work on the pendulum either way, right?

A simpler example would be if we just drop a ball onto a ramp in the shape of a quarter circle so that the ball goes from falling vertically to rolling along a horizontal plane. The ball has a different height at the end from the beginning so gravity does work on the falling ball, and the energy is conserved as the horizontal kinetic energy of the ball rolling along the horizontal surface. Whether we let the ball roll forever or stop it with some friction or let it roll up another ramp back to the same initial height, doesn't go back in time and change whether gravity did work on the ball on the way down, right?

When you calculate the work done by gravity as the pendulum swings down to the bottom and then back up, You take the integral of the work done by gravity over the pendulum path. That roughly means you add up all the work done over each tiny bit of movement. When you add the work done on the way up to the work done on the way down, you get zero, but that doesn't mean the work on the way down equals zero and didn't exist. It just means you added more work and the sum was zero. That's the convention of physics, that's what the work integral formulas say, that the work done by gravity on the way down is non-zero, right? Mindbuilder (talk) 16:28, 7 November 2014 (UTC)

You are clearly confused about the difference in work definitions and the physics behind it. Please state how you think anything about this relates to siphons. WP:NOTFORUM. --17:32, 7 November 2014 (UTC)

Because you've stated that the atmosphere does no work pushing the liquid up the siphon, and I'm trying to get you to realize that just because work in from the atmosphere minus work out to the atmosphere happens to add up to zero in a particular case, doesn't imply that no work went in. In other words just becaue x + y = 0 doesn't imply that both x = 0 and y = 0. The atmosphere does work into the siphon system pushing the liquid up, then the liquid on the down side does work out to the atmosphere on the way down, adding up to zero. But just because the sum is zero doesn't mean the atmosphere did no work in. It did no NET work in, but it still did some in. And the fact that the atmosphere did some work pushing the liquid up is important, even if that work was later undone to zero.

To see the atmosphere doing work consider this device.

Imagine you use a vacuum pump where the arrow is to remove all the air from within the cylinder leaving a pure vacuum. Then you disconnect the vacuum pump, leaving the opening sealed, haul the vacuum pump 1000km away, melt down the vacuum pump, and let the whole contraption sit around for a month with the pure vacuum inside it. When you pull down on the latch to release it today, then the atmosphere pushes the piston and car up, doing work today. The vacuum pump did work a month ago but it won't do any today because it's a thousand km away and was melted down a month ago. Something that was melted down a month ago can't do work today. Something does work today lifting the car, and it's not the vacuum pump. Now you might say that no work is done because the energy of the system is the same after the atmosphere pushes the car up, due to the car having higher potential energy, but then we could just change the example to remove the car and put some friction brakes rubbing on the piston rod on the way up to dissipate the energy. Then it would be lossy.

Now you could also remove the car and connect a 100% efficient generator and battery to the piston rod and store all the work the atmosphere does pushing the piston up. Then you could use the energy in the 100% efficient battery to drive a 100% efficient motor to drive the piston back down and do an equal amount of work back out to the atmosphere. The net work done by the atmosphere would then be zero, but just because the net ended up being zero doesn't mean the atmosphere didn't do an important bit of non-zero work pushing the piston up and charging the battery, right?

So a sipon is not powered by the atmosphere, but the atmosphere does important work pushing the liquid up the siphon, even if that work in from the atmosphere goes right back out on the way down, right? Mindbuilder (talk) 20:06, 7 November 2014 (UTC)

None of what you say has anything to do with a siphon. Forget a siphon, take two vessels and one with a hole in the bottom and a tube to the lower vessel. Gravity drains it from upper to lower. No priming necessary. Now move the tube around: up, down, wherever. Lift it above the surface of the upper vessel like a siphon. The flow is not changed. The flow is completely described by gravity and the height difference between upper and lower liquid surface heights (whether it's a siphon or city water system, it does not matter). Atmospheric pressure plays only one role: it sets the pressure head for the liquid start point which creates a boundary condition for lift but is not a force in the analysis of the system or flow. The only force is gravity just like letting the pendulum go it only has gravity. Since the force driving the whole apparatus is described as gravity, we don't need to create any more forces. In addition, atmospheric pressure pushes the liquid from both sides. There is no atmospheric forces necessary to sustain a siphon or describe any portion of it's operation since the atmospheric pressure cancels out. When you figure out how the upstairs bathroom gets water, when it is higher than the downstairs water source, you will understand. It's not atmospheric pressure pushing it upstairs. --DHeyward (talk) 22:20, 7 November 2014 (UTC)

Gravity is a downward force. It can't make anything go up directly. There has to be some other intemediate force to make the liquid go up. Like in a seesaw where a downward force on one end leads to upward forces in the beam and seat on the other side that make the other end go up. In a siphon in vacuum there are attractive forces due to the cohesion of the liquid molecules that pull the liquid up. But if a siphon gets a bubble in it then cohesion can't be the upward force because gasses can't pull up. So what is the upward force in a typical siphon or in the Figure 4 siphon? You cite Bernouli's description of the flow, but don't you realize that atmospheric pressure is one of the forces that join together in Bernouli's equation to describe the flow? If you leave the forces of atmospheric pressure out of the Bernouli's equation for a sea level siphon, the equation would be a massively incorrect model of the siphon. The atmospheric pressure cancels in the end but it does important things in between. Do you remember my picture of the two trucks being pushed up a hill in talk archive 6 ? Mindbuilder (talk) 23:33, 7 November 2014 (UTC)

I might as well post it again here. Note that even though the push from the guy on the left is canceled by the guy on the right, it is still the canceled push that is responsible for pushing the little truck up the hill. It's kind of odd that a canceled push can still do significant object moving, but it can. Just as in the siphon, the atmospheric pressure cancels, but still does significant liquid moving. Mindbuilder (talk) 23:48, 7 November 2014 (UTC) Of course the reason these seemingly canceled forces can still push things up is because they're not really canceled. In this diagram, the weight of the heavier truck prevents the force from the guy on the right from getting all the way to the top to cancel the guy on the left. In a siphon the greater weight of the liquid on the taller down side prevents the atmospheric pressure at the exit from getting all the way to the top or the entrance to cancel atmospheric pressure at the entrance. Mindbuilder (talk) 02:43, 8 November 2014 (UTC)

Once again, no. You just talked about a pendulum. It goes up as it trades kinetic energy for gravitational energy. There is no need for an "up" force. Same with a siphon. It doesn't matter in an apparatus whether the lower reservoir is filled by a direct pipe from the upper that never goes "up" or whether the "down" is followed by an "up" side (i.e. a bathtub in a two-story house) or whether it's a classic siphon with an up leg and a down leg. As long as the height difference is the same, it works. Bournoulli trades pressure energy in the reservoir for gravitational potential energy. Atmospheric pressure is equal on both sides. It is irrelevant to the flow. Its the same physics as the tap in your kitchen faucet. Anything else is pure nonsense. It's easily seen by changing pipe diameters where pressure and velocity head is exchanged. Siphons work just fine whether the pressure change in a tube is negative or positive since the only concern is height difference. Draw a siphon with a narrow up-tube and wide horizontal tube. Works just fine even though the pressure energy is higher in the horizontal tube than the up-tube. You even drew the picture where the liquid in the up-leg weighs more than the down leg. It still works because it doesn't matter.--DHeyward (talk) 05:07, 8 November 2014 (UTC)

I was afraid that if I described partially canceled atmospheric pressure at the exit as due to the greater "weight" of the liquid on the taller side, that it might cause confusion, but I couldn't immediately think of a better way to describe it. Yes, I drew the diagram of the fat up tube siphon, so I know and understand that it is not the "weight" of the water in the down tube that matters but rather the height. But I had hoped you would still get the idea that the water in the taller down side was counteracting the pressure at the exit somewhat. Do you see from the car diagram how a force that seems to be canceled by an equal and opposite force can actually end up not being canceled because of the intervening effect of gravity on some of the mass?

As for the liquid going up because of kinetic energy, the speed of the liquid entering the siphon is often much to low to carry it up to the top by momentum without an additional force. Consider a tall siphon, say 5 meters tall. How fast would the liqud have to be going at the start to carry it five meters up by momentum? About the same speed as if it had just fallen 5 meters, right? How fast would the liquid be going if the difference between the surface of the source and discharge reservoirs was only one mm. The speed would be very low, much much to low to even get close to 5 meters without additional force to push or pull it up, right? But it would still work, even though the kinetic energy or momentum alone couldn't explain the molecules going up against gravity.

Also imagine you have a running siphon and you close a valve at the exit so the flow stops. Say you also let a bubble into the top just big enough to disconnect the up and down columns of liquid so the liquid on the down side can't pull up the liquid on the up side. Now when you open the valve and let flow resume, the liquid on the up side will go from standing still to accelerating upward until the steady state speed is reached. Since it is starting from a standstill it has no momentum, and it isn't being pulled up because gas bubbles can't pull, so there must be a force other than gravity to accelerate those molecules upwards right? F=ma The liquid molecules have mass, they start accelerating upwards, there must be an upward force. Where does it come from? It can't be hydrostatic pressure of the reservoir liquid alone, because that is only enough pressure to get the liquid up to the height of the surface of the reservoir. To go up the tube farther requires more pressure - atmospheric pressure, right? Just like we describe it in a barometer, the atmospheric pressure pushes the liquid up. Mindbuilder (talk) 06:34, 8 November 2014 (UTC)

You're confusing how a siphon starts with a running siphon. A siphon can be started a number of ways. Atmospheric pressure and a vacuum can start a siphon (which you use to explain a running siphon, which is wrong). Filling a siphon from the top can start a siphon. Neither of those methods require an atmosphere. Filling your tub with water has an "up" tube and no atmospheric pressure is needed. It's the same physics. Siphons used to drain lakes don't use a vacuum. The have valves at the bottom of the tubes and the inlet and outlet. They fill the tubes with water from the top, seal the top and open the outlet valves. Another method is a submersible pump in the source that fills the tube. The only requirement is that the tube is filled. It does not matter how the tube is filled. It's a key point that the siphon behaves just as your indoor plumbing once the condition that it's a continuous fluid system is met. Like the pendulum, it swings whether it's pushed while at rest or whether it's lifted and dropped. That condition is not necessary to describe the motion and has no bearing on its operation. Whether the liquid system is down, over and up or up over and down, - once primed the physics is the same. Velocity of fluid is only one of three energies in the system: pressure and gravity are the other two. Velocity can be traded for pressure by simply changing pipe size. Pressure energy changes with depth (gravitational energy). --DHeyward (talk) 15:21, 8 November 2014 (UTC)

Consider the 5m siphon with a 1mm source to discharge reservoir altitude difference. After the siphon has run for a little bit, assuming the tube cross section is constant, the speed of the liquid up the siphon will settle to a constant value for the whole trip up the up side. If the liquid is moving up the siphon at a constant speed there must be an upward force countering the downward force of gravity, because if gravity were the only force on the molecules, the molecules would have a downward acceleration of g (of course a downward acceleration doesn't necessarily imply downward movement if they were launched with a sufficient upward velocity). Ignoring friction effects, the flow speed of a siphon with a source to discharge altitude difference of one millimeter happens to be equal to the speed something would have after a freefall of one mm. For momentum to carry the liquid up 5m would require an initial speed equal to the speed of an object after being in freefall for 5m. So the speed of the liquid entering a tall siphon is not nearly enough to get it over the top. There has to be some other force to help it overcome gravity.

When you fill a bathtub, atmospheric pressure isn't needed because an alternative source of pressure is typically provided by a pressure pump.

You mention the two other energies as pressure energy and gravitational potential energy. The pressure energy at the entrace of the siphon is what supplies the force to push the liquid up against gravity. You do realize that pressure is defined as the force per unit area right? Hopefully you can see that the whole idea of pressure energy is that the pressure exerts a force on the fluid pushing from a zone of higher pressure, e.g. the siphon entrance, to a zone of lower pressure, e.g. the top of the siphon. That's the force pushing the liquid up the siphon at a steady speed against gravity. The pressure at the entrance is the sum of two components, the hydrostatic pressure and the atmospheric pressure. The hydrostatic pressure can only supply enough force to push the liquid up to the level of the surface of the reservoir. It's the contribution of the atmospheric pressure to the pressure energy that is needed to push the liquid up the rest of the way to the top, right? Mindbuilder (talk) 17:07, 8 November 2014 (UTC)

No, there doesn't need to be another force just as there is no other force in a swinging pendulum. It goes up. Evergy is constant in the fluid. It trades pressure head for gravitational head. The bathtub filling is identical to a syphon. Imagine your 5 mm diameter tube attached to the top reservoir and draining directly into a lower reservoir (no siphon). As it's doing that, raise the middle tube above the top reservoir. Nothing changes. It flows the same. Fills at the same rate. Whether your flowing tube starts with simple gravity and the tube is raised to create the siphon condition doesn't change any of the equations. Water to fill a tub flows just fine as it exchanges pressure and gravity energies. Same with the siphon. Fill rates are the same as long as the supply surface and the outlet height maintain the same difference Get it?

The pressure energy of the source increase from surface to bottom while its gravitational energy decreases. It's the sum of pressure and gravitational energy that is constant through the fluid. The pressure at the bottom of the source (i.e. the bottom of a swimming pool) is higher. But a syphon (or drain) doesn't care where the tube is placed in the source reservoir because the potential energy in the fluid is constant (gravity head + pressure head = constant). You can place the intake tube just below the surface or all the way to the bottom. You can check the pressure of a pool by swimming to the bottom (it increases). You can check that the bottom pressure has no more energy than the surface since putting a tube to the bottom doesn't cause water to fly out the tube. Water levels itself precisely to make the energy equivalent. A deep ocean has a lot more pressure at the bottom than a shallow one but the energy available is the same so the surface is level. A fluid has a constant energy which is what Bernoulli's equation is. --DHeyward (talk) 19:28, 8 November 2014 (UTC)

I get that the flow rate stays the same if the entrance and exit levels stay the same, even when you move the tube up and down. But that doesn't imply that all the forces within the tube stay the same, only that the NET sum effect of the forces ends up the same.

Consider the force diagram on an object like a liquid molecule moving upward in a gravitational field. If the only force acting on the object is gravity, then wouldn't the object be accelerating downward at g? (moving upward but accelerating downward) If the object is moving upward at a steady speed then its acceleration is zero, because its speed is steady, right? If its acceleration is zero, then that means the net force on the object is zero, right? Since its in a gravitational field, it has a downward force on it, so there must also be an upward force for the net force to add up to zero, right? Mindbuilder (talk) 20:05, 8 November 2014 (UTC)

A incompressible fluid has three sources of potential energy (not forces). An object in space has two sources of energy. A ball thrown in the air has gravity as the force. Initial velocity is the energy and at the point it received it's kinetic energy, the force that created it is irrelevant and unnecessary in the solution. Solving the problem for all points requires knowing the initial velocity and mass. No extra forces are needed to explain it. An incompressible fluid, though, has other properties. One of these properties in a non-flowing liquid is that the sum of gravitational energy and pressure energy is constant at any point. Then, when it's flowing through a fixed diameter tube, the free energy tradeoff from gravity is pressure, not velocity. No force is necessary to create the tradeoff and it happens whether the liquid flows up or down. Starting the flow from the tube on the ground and then raising should show that there is no other forces necessary for understanding. It's also too complicated to try to visualize a single molecule in an incompressible fluid (i.e constant density but variable pressure). These are macro properties of a state of matter, in this case an ideal incompressible liquid. --DHeyward (talk) 23:28, 8 November 2014 (UTC)

suggestion for Article structure

Because every Wikipedia article of significant complexity should be divided according to probable audiences (ie: primary school, secondary school, advanced), it appears to me that the article should be divided into something like 'levels of understanding', so that a brief coherent and general description of the use, structure, and simplest principles of a siphon are generally apprehended, perhaps with an acknowledgement that there are at least several more subtle phenomena at work.
Then, a description of the limits of the initial description could introduce those more subtle components and their effects, followed by more rigorous descriptions of those components.
It might also help if written analyses of these siphon components are first on a static system, before dynamics are addressed.

If my contribution is not productive, perhaps I will return with a re-write attempt of my own, for the surprisingly (and delightfully) complex and subtle issues in this 'obvious' and everyday phenomenon.

Initial siphon.doc: only the essential structural requirements & fluid weight imbalance, which is the effect of the greatest magnitude.

List of other effects, in order(?) of significance: atmospheric pressure (gauge,absolute, on fluid & tube surfaces, lift height limit, etc.), viscosity, bubbles-tube diameter, bernoulli, cohesion(surface tension), ...
--Wikidity (talk) 18:59, 15 September 2014 (UTC)

The first, easiest and most basic concept is energy is constant in the fluid system. Siphons exchange pressure, kinetic and gravitational energy between all three. Though most people can visualize solid objects exchanging kinetic and potential energy (i.e. ball rolling down a hill), a siphon adds pressure energy of the fluid. The nature of ideal fluids is that energy is constant throughout it's depth. The water at the surface of a lake has the same potential energy as the water at the bottom. The bottom has less gravitational potential energy but more pressure energy. It's static and not moving so the analogy to the ball and hill can be confusing at this point. This concept of equal energy is not intuitive but necessary to understanding. Once that is understood, the balanced siphon can be explained (outlet and surface are at same height, no flow). The energy of the fluid in the balanced siphon is constant as the same pressure energy and gravity potential are traded off. From the top reservoir, down to the intake, up the tube, down the tube to the balanced output of non-flowing fluid, - the fluid energy is constant. It converts pressure energy to gravitational potential energy. The next step is the limiting case of the ideal fluid. This is when the all the pressure is converted to gravitational energy. Zero pressure is the lowest an ideal liquid can support and is the theoretical lift limit. The next step is to lower the output and unbalance the siphon so it flows. This adds kinetic energy in the tradeoff. Pressure, velocity and all the other goodness that happens when liquids flow start to come into play. It starts with Bernoulli's equation mathematically but non-ideal fluids also add their quirks. "weight" of the fluid is an incomplete concept in describing the flow and, I believe, the ball/hill scenario is why people incorrectly state it as a weight problem. Siphons with very wide up tubes and very narrow down tubes still work the same even though the absolute weight and weight ratios are different. The energy is the same though and to make that work, fluid velocity and pressure adapt to satisfy conservation of energy. The working force is gravity but the energy at each point can take the three forms (pressure, gravitational, kinetic) stated above as long as they sum to the constant. --DHeyward (talk) 02:55, 16 October 2014 (UTC)

This statement in the article, "Another simple demonstration that liquid tensile strength isn't needed in the siphon is to simply introduce a bubble into the siphon during operation. The bubble can be large enough to entirely disconnect the liquids in the tube before and after the bubble, defeating any liquid tensile strength, and yet if the bubble isn't too big, the siphon will continue to operate with little change." is incorrect.

The short version of the reason is, "You can't siphon sand," because sand has no significant tensile force between grains.

The liquid on the long leg downstream from the bubble (mentioned in the quote) holds together and acts as a piston pulling down and lowering the pressure in the bubble. As long as that pressure is lower than the hydrostatic pressure of the fluid just upstream of the bubble, the siphon will run.

When the weight of the downstream liquid is decreased, the pressure drop in the bubble is smaller. If the pressure in the bubble is too high, the siphon stops. This has been called "vapor lock." [See Siphoning—a weighty topic James McNeill, Phys. Today 64, 8, 10 (2011)] 72.95.57.92 (talk) 19:22, 26 October 2014 (UTC)john dooley

John,

Just to clarify. You can have the following Air bubble Air lock Vapor bubble Vapor lock

Re your comment: "....If the pressure in the bubble is too high, the siphon stops. This has been called "vapor lock." "

This is incorrect. A vapor bubble is caused when the pressure is too low, i.e. it drops below the vapor pressure of the liquid, not when the pressure is too high.

An air bubble will not stop a siphon working if the vertical height of the air bubble is less than the vertical height of the siphon outlet below the siphon inlet.

An air lock in a pipeline is where an air bubble moves to the downward leg, and in the case of a gravity pipeline (which a siphon is), the vertical height of the bubble equals or exceeds the vertical height of the inlet and outlet difference. In the case of a pumped pipeline with undulations, an air bubble that moves to the downward leg of the undulations will increase the pressure required at the pump equal to the vertical height of the air bubble. If the increased pressure required exceeds the pressure that the pump is capable of, then the situation that has arisen is referred to as an air lock.

As noted, a vapor bubble is when the pressure drops below the vapor pressure.

A vapor lock refers generally to fuel pumps, that can pump a liquid but not a vapor. If the fuel turns to vapor and gets into the pump, the fuel pump will not work.

In a siphon, if the vertical height of the top of the siphon gets too high, above 10 metres at sea level, then the water will turn to vapor and the pressure gradient is affected since the vapor does not have the density of water.

Pressure in water is due to vertical height and density, not volume or the weight of water. Friction losses aside, changing the tube size so the volume and weight of water changes does not change the pressure. — Preceding unsigned comment added by 124.187.6.131 (talk) 22:50, 10 November 2014 (UTC)

It actually does. Reducing tube size increases velocity. Kinetic energy is gained at the expense of pressure energy so bubbles can form in narrower tubes. Moving from a static condition to flowing flowing can also reduce pressure but the limiting pressure remains the same. --DHeyward (talk) 02:00, 11 November 2014 (UTC)

Too many words

Both the article and this talk page are far too wordy. The article is written in long, rambling paragraphs of long, detailed sentences: the very first sentence is 91 words long! Start with the simple cases and go into greater detail later in the article. Try to organize the article more logically. Cut back on the "siphon terminology" section and move some of the content to independent articles.

Also, the discussion on this page does not seem to be converging. I would recommend that you find some reliable sources specifically on siphons rather than trying to re-derive the physics from first principles. --Macrakis (talk) 04:03, 11 November 2014 (UTC)

Unfortunately there is no final word in the reliable sources because articles are still being published supporting the liquid cohesion theory. Apparently the siphon with a bubble at top doesn't occur to the authors of those articles until after they've published, and then they're too embarrased to admit their error so they stubbornly come up with the most bizare theories to avoid admitting their mistakes. DHeyward is another problem. He is very persistent and has an odd and possibly unique misunderstanding of physics. He thinks that "pressure energy" can cause the liquid to go up without exerting an upward force on the liquid against gravity. He turned the first section of the article into a highly technical and incorrect reference to Bernoulli's equation (Bernoulli's equation is good, just his interpretation is mistaken). Bernoulli's equation is fine for further down the article, but is inappropriate for the standard Wikipedia audience, and thus shouldn't be near the top even if DHeyward's theory was correct. I'm planning soon to remove his uniqe theories (and bring in backers to make it stick if necessary), since progress is hard to make in our discussions because it's hard to get him to state clearly when he agrees or even disagrees with my points. But that kind of thing is to be expected from somebody stubborn and confused like that. Take a look at the state of the article back in March, and see if you think we should revert or partially revert. It was rather messed up back then as well, but it was better. Would you be with me on that? Mindbuilder (talk) 04:55, 11 November 2014 (UTC)

Deviating from the well-understood and well-supported position of Bernoulli is a non-starter. It's well referenced, self consistent and accepted. Fringe theories need not apply. It's not difficult. Simplifying the language is fine but stating incorrect and fallacious physics is not. Siphons start without suction and are prevalent. Saying the opposite is just wrong. Likewise, liquids flow from a lower point to a higher point without siphons (i.e. a tub on the second floor). Wording that expresses it as physics instead of magic is fine. The physics is gravity. --DHeyward (talk) 05:44, 11 November 2014 (UTC)

The article has been in bad shape for at least a year -- I didn't look further back.

This is Wikipedia, not a physics journal. We are looking for the views in reliable sources, not our own thinking. If there are differing views in sources which are equally respectable, we present both views and present the debate. I do not know what the state of the modern literature is on syphons, but whatever it is, is what we should be reflecting. Arguing on this page is a waste of time.

Besides the technical content, the general organization and writing style on this page is abominable. There is lots of work to do. Don't waste your energy trying to teach each other physics from the ground up. --Macrakis (talk) 05:59, 11 November 2014 (UTC)

I have given up on fixing the style of the siphon article. With the physics issues you can discuss it and give clear examples to prove nearly any siphon issue, and come to agreement. But with style, it's just one person's opinion against another. I made some improvements a few years ago, but too many people came along and insisted on messing it up with a wide variety of odd fluid physics conceptual explanations. So I finally settled on just trying to make sure that the important physics examples and concepts are there for clear thinking readers to sort out from the garbage. Actually, as far as I know, there is only one modern author still publishing articles advocating the cohesion theory - Stephen Hughes. Another author published an article a few years ago, but has since backed off to a more neutral position on the cohesion theory. There are a few articles from decades ago also advocating the cohesion theory, but that was before the liquid separating bubble example, the air start siphon of figure 4 in this article, and the carbon dioxide gas siphon became widely known. But regardless which of the two common theories are right, DHeyward has made huge changes to the article based on his unique theory, and I was trying to get him to come around by discussion rather than just forcing the issue. Mindbuilder (talk) 21:44, 11 November 2014 (UTC)

@ DHeyward. re change in tube size changing pressure. Yes you are right, in a flowing pipe, a change in tube part way along a pipeline will have a small change to the pressure due to a change in velocity and thus velocity head. In a static situation, there is no change. I was trying to correct comments made by John Dooley that spoke of the weight of the water, and referring to the bigger picture that pressure relates to vertical height and not volume. — Preceding unsigned comment added by 121.222.59.63 (talk) 10:28, 11 November 2014 (UTC)

Re the opening section that discusses the Bernoulli's principle: " ......that the sum of all energy at any given point in the siphon flow is constant. Because the reservoir generally has a constant energy per unit volume (the sum of pressure energy and gravitational potential energy is constant), the outlet must also have the same energy....."

In a static situation, before water flows in a pipeline, the sum of all energy at any individual points is the same as a selected 2nd point. However, once the water starts to flow, friction loss is incurred and Bernoulli's principle or formula is incorrect since it makes no allowance for friction and from a practical point of view, that energy is lost as heat. So for a flowing siphon, the total energy at the start and end are entirely different. At the siphon outlet, there is no gravitational potential head or pressure head, only velocity head. The whole Bernoulli's principle section at the start needs cleaning up and moved further down, it is not helping to explain siphons.

Re Pascals Siphon. The main article states: "Specifically, Pascal demonstrated that siphons work via gravity...." Yet an article with link below claims that "Pascal demonstrated that the siphon worked by atmospheric pressure, not by horror vacui, ...." Does anyone know exactly what Pascal claimed was the result of his experiment?

• http://mysite.du.edu/~jcalvert/tech/fluids/hydstat.htm

Bernoulli's principle is applied as an ideal inviscous fluid much like how we simplify motion to Newtons equations. All energy (including friction loss) comes from the potential energy of the source liquid and can be modeled as increased head requirement. Pascal statement is that mercury did not need a vacuum (i.e. atmospheric pressure difference) and worked under normal atmospheric conditions. The water is level in the tank that contains mercury so the atmospheric pressure is constant. Note the description in the source is identical to the article. No change in atmospheric pressure, the mercury rises in both tubes as water head is added to the contraption (the water adds energy and the response of mercury is directly related to the addition of water. The height of mercury is equal in both tubes as the water height/head determines it). Without any change in atmospheric pressure, the fluid in the lower reservoir tube reverses direction when a fluid connection is made. That is Pascal's claim. He does not claim the atmosphere does anything (except as the normal setting of baseline energy at the surface of the water) and discounts explicitly the notion that a vacuum is needed to cause the mercury to rise and siphon (i.e. the mercury barometer is a vacuum device). --DHeyward (talk) 03:27, 12 November 2014 (UTC)

@DHeyward. re "ideal inviscous fluid"?? Why include an imaginary fairytale fluid that doesn't exist in reality? Does it help the cause? No. In practice, the energy at the outlet for a flowing siphon is not the same as at the start. That Bernoulli stuff at the start does not help explain siphons given it totally neglects friction. The Bernoulli principle is more appropriate to explain that energy in a pipeline can change between gravitational, pressure and velocity head. I agree with you on the Pascal siphon, that he was pointing out issues related to vacuum. It appears that others are then taking it upon themselves to claim he was advocating the importance of gravity or atmospheric pressure. I wonder if the Pascal siphon is actually a siphon or just what is called the siphon effect. i.e. to get the mercury to move and to continue to move, the water pressure needs to be high enough that it would flow anyway, even if the 2nd half of the siphon didn't exist.

For the same reason we neglect friction for demonstrating kinetic energy derived from a force. It's easier to understand. Many people have solved cannonball trajectories by neglecting friction and terminal velocities. We use conic sections all the time which invariably neglect friction. Bernoulli's equation is correct for a siphon and shows that all the energy transfer is from the height difference (friction would be an added term). The Pascal siphon is an excellent example as the flow is completely determined by the gravitational height difference between the two reservoirs. The flow is the same whether it's a siphon inverted U-tube or whether it's a an upright u-tube. Once it's a connected fluid the sum of all energies must be equal. The lower elevation head of the bottom reservoir requires a velocity head component to compensate. The ideal case is solvable and makes it easier to see. See page 22 here [1] where a siphon is solved using ideal fluid and Bernoulli's equation. --DHeyward (talk) 23:19, 12 November 2014 (UTC)

I very much agree with DHeyward on this one. I think it is legitimate to stick to the idealized model on this page, and I wouldn't recommend adding in the complication of flow friction. That's best left for a general advanced pipe flow article. Mindbuilder (talk) 23:24, 12 November 2014 (UTC) It might be a good idea though to make it very clear in the article that real siphons often differ very significantly from the idealized model. Mindbuilder (talk) 23:27, 12 November 2014 (UTC)

I thought the purpose of the Wikipedia article was to explain REAL siphons!

There is no way we could ever get a perfect description of a real siphon. We have to settle for a useful approximation. Someone can easily visit the pipe flow page to find out how to refine their approximation. Mindbuilder (talk) 01:15, 13 November 2014 (UTC)

It's easy to add it. It's just "+ Energy lost" on the right side of the equation. It's a complicated construction using surface roughness, Reynolds number and a host of other factors that create heat loss. That could be added to the detail section. --DHeyward (talk) 01:36, 13 November 2014 (UTC)

REAL siphons include friction loss. It shouldn't be added as an after thought. Seriously. There is the inclusion of Z tubes, and trees and chains, yet when it comes to friction loss it's like "Oh, we can't go there, that doesn't fit into our perfect fairytale world". It's not complicated: Siphon gravitational head less outlet velocity head = friction loss.

That's fundamentally correct. Calculating friction in a pipe done by design requires surface roughness, velocity, Reynolds number and more than just a measurement difference. It's the difference between prediction and measurement. --DHeyward (talk) 04:41, 13 November 2014 (UTC)

Calculating friction in a pipe does not have to include a Reynolds number or velocity. There are a variety of formulas that can be used and the more popular ones don't use a Reynolds number or velocity. For siphons however, there is not a need to discuss friction formulas. However gravity pipelines, including a siphon have friction as an essential component of the flow rate achieved. Rather than conserving energy, gravity pipelines and siphons, when started, keep increasing their flow rate until the energy at the outlet is only velocity head, i.e. no pressure head. And thus the energy at the outlet of a flowing siphon does NOT equal the energy at the start. The 2nd paragraph in the article should reflect this.

While I feel strongly that an idealized model is perfectly ok, it might be a good idea to add some more info about non-ideal flow. I wouldn't object. Mindbuilder (talk) 06:02, 13 November 2014 (UTC)

Pascal's Siphon error

"......The mercury rises in both sides of the tube (upper and lower) and each tube is level with each other once the lower tube reaches the upper reservoir surface....."

This clearly is not correct.

If the mercury is rising in both tubes, then how can they be level with each other once the lower tube reaches the upper reservoir surface. It has to be one or the other, it can't be both. They can't start off at different levels, both rise and then be level when the lower level reaches where the upper level started.

A further review of Pascal's siphon reveals the following:

Mercury has a density 13.6 times that of water.

If you were to build a Pascal's siphon, with the upper reservoir 2 cm above the lower reservoir, and the cross tube 1 cm above the upper reservoir, and then started to fill the outside container with water, what would happen?. If you added 13.6 cm of water above the lower reservoir, the lower tube would have Mercury rise 1 cm, the upper tube would have Mercury rise by 0.85 cm (11/13.6).

If no further water is added, the Mercury levels would be static as per above, with air in the cross tube and there would be no flow of Mercury.

If you added a further 2 cm of water, so the water level is 15.6 cm of water above the lower reservoir or 13.6 cm above the higher reservoir, then the Mercury would rise 1.15 cm in the lower tube and 1 cm in the upper tube. This would see the Mercury in the upper tube reach the base of the cross tube.

If you added a further 2 cm of water, so the water level is 17.6 cm of water above the lower reservoir or 15.6 cm above the higher reservoir, then the Mercury would rise 1.3 cm in the lower tube and 1.15 cm in the upper tube. This would see the Mercury start to flow across the cross tube and cascade down the lower tube. The flow would not be consider a siphon, rather it is being moved via the additional pressure the water is creating on top of the Mercury, compared to only air on the upper reservoir tube. This is similar to someone placing a straw in a container, sealing the top of the container so the straw pokes out of it but the container is otherwise air tight, then increasing the air pressure inside the top of the container.

As the level of the Mercury in the top container drops by 0.15 cm, the Mercury flow will cease. So Mercury flow will only occur at this point if there is a pressure difference cause by the water level and the outside air that is great enough to get the Mercury above the cross tube.

If the water level was 27.2 cm above the lower reservoir, the Mercury would rise 2 cm in the lower tube, and be equal with the level of the upper reservoir. At this point, the level of the Mercury in the upper tube would have risen in theory 1.85 cm, and be 0.85 cm above the cross tube. However flow across the cross tube is going to occur and the Mercury will cascade down into the lower tube.

If no further water was added, then Mercury would continue to flow across and cascade down until the upper level dropped back to the cross tube level, 1 cm above the upper reservoir and the lower reservoir level would have risen by a similar amount to the fall in the upper reservoir if the containers were the same size. The lower tube level would be just below the cross bar.

If it was raised more than 40.8 cm (13.6 x 3), then the Mercury level in the lower tube would have risen all the way to the cross bar, and the upper tube level would have risen in theory 2.85 cm above the upper level, with flow occurring across the cross bar. Since the Mercury pressure from the upper tube is greater the Mercury pressure in the lower tube, the upper tube pressure will force Mercury down into the lower tube, and flow will continue until the Mercury levels are equal.

Whilst it has the appearance of a siphon, whether it flows is related to the pressure from the water on the upper reservoir. It is not a siphon as we known it at all.

Note that an alternative to the Pascal Siphon can be built that performs in a similar manner yet does not utilize Mercury. Build the a device similar to Pascal's siphon. Put water in the two containers rather than Mercury. Now seal the top of the container so it is airtight, with the single riser from the siphon apparatus poking out the top. Have a 2nd hole in the top seal of the container, and start pumping air in. Water will flow across the connecting bar when the air pressure is high enough to raise the water. — Preceding unsigned comment added by 121.223.41.81 (talk • contribs) 21:42, November 13, 2014‎ (time of last change)

Your numbers look right. The mercury level in the two tubes will clearly not be at equal heights as water is added. At least not until the mercury reaches the horizontal section. Go ahead and fix it if you want, or I will eventually. We'll see if DHeyward wants to chime in about it. Mindbuilder (talk) 00:31, 29 December 2014 (UTC)

Hatnote

The hatnote states "This article is about the domestic device." It clearly isn't. It's about the physical principles of the siphon, and a number of practical applications (domestic and otherwise). I can't immediately think of a single-word summation of that, but I suggest somebody changes it. GrindtXX (talk) 12:39, 6 February 2015 (UTC)

Somebody changed the disambiguation description of the siphon. A bit of an improvement I'd say. Mindbuilder (talk) 22:49, 28 February 2015 (UTC)