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Fundamental equations of the siphon

Imagine a simple siphon consisting of a reservoir containing the working fluid; and a tube completely filled with the working fluid. One end of the tube is open and the other end has a tap. The open end of the tube is immersed in the reservoir of working fluid and the end with the tap is outside the reservoir. If the end with the tap is below the surface of the working fluid and the tap is opened, fluid flows out through the tap and falls to the ground.

Atmospheric pressure at the surface of the working fluid is Pa. Density of the working fluid is ρf. Density of the atmosphere is ρa. The densities of the working fluid and the atmosphere are considered not to vary with height.

Whether working fluid will flow out through the tap when the tap is opened, or atmosphere will flow in and the tube will empty, is dependent on the pressures on either side of the tap before the tap is opened. The rate at which the fluid, or the atmosphere, will initially flow past the tap is also dependent on the pressures on either side of the tap before the tap is opened. Using only the information available so far, we cannot determine the equilibrium rate of flow because we don’t know the length and internal diameter of the tube, its surface roughness or the viscosity of whatever is flowing through it, but we can certainly determine the two pressures on either side of the tap before the tap is opened.

The surface of the working fluid in the reservoir is the datum. Points below the datum will be given positive distances from the datum, and points above the datum will be given negative distances. The crest of the siphon is at a height hc above the datum, so it will be a negative number. If the absolute pressure of the working fluid at the crest of the siphon is Pmin it can be calculated:

where g is the acceleration due to gravity. hc is negative, so Pmin will be less than Pa.

The tap is at a distance hx below the crest of the siphon. The absolute pressure of the working fluid adjacent to the tap is Pt and can also be calculated:

This equation can be re-arranged as follows:

Pmin is common to both equations so it can be eliminated, leaving a single equation:

Re-arranging, we find an expression for the absolute pressure of the working fluid adjacent to the tap:

Observe that hx + hc is simply the distance of the tap below the datum. Let’s call it ht. The above expression then becomes:

Observe that when the tap is closed, the absolute pressure of the working fluid at the tap is dependent only on the height of the tap below the surface of the working fluid in the reservoir. The absolute pressure at the crest of the siphon is no longer significant because Pmin has been eliminated from the expression for Pt.

On the other side of the tap is atmospheric pressure Pat. This is related to atmospheric pressure at the datum by the following expression:

When the tap is opened, the initial acceleration of the working fluid through the siphon will be dependent on the difference in pressure from one side of the tap to the other. On one side of the tap is working fluid at absolute pressure Pt. On the other side of the tap is the atmosphere with absolute pressure Pat. The difference in pressure can be called ΔP. The expression for ΔP is:

Pa disappears from the expression, leaving the following:

If ΔP is positive, when the tap is opened working fluid will leave the tube through the tap and the apparatus will operate as a siphon. If ΔP is negative, atmosphere will enter the tube through the tap and working fluid will flow down the tube back into the reservoir. The magnitude of ΔP will indicate the strength of the siphon when the tap is opened.

Of great significance is the fact that the absolute pressure of the atmosphere at the surface of the working fluid in the reservoir, Pa, is not present in the expression for ΔP. The strength of the siphon, and even whether the apparatus will function as a siphon, is independent of the atmospheric pressure.

Now let’s consider a practical siphon in which the working fluid is water, and the atmosphere is air at sea level. The density of water, ρf is 1000 kg.m-3. The density of atmospheric air at sea level and 15°C, ρa is 1.2 kg.m-3. So in the expression for ΔP, the following term:

has the value (1000 – 1.2) or 998.8. This means that the pressure difference at the tap, before the tap is opened, is mostly due to the density of the water. At sea level the density of the air reduces ΔP by about 0.12% from what it would be if the apparatus was surrounded by vacuum.

The driving force behind a common siphon is the density of the water. The effect of the density of air is very small and it serves to reduce the strength of the siphon by about 0.1%. The absolute pressure of the air at the datum does not influence the driving force. Dolphin (t) 12:53, 29 January 2011 (UTC)

Your analysis all looks good, except for your initial equation, which is based on a simplifying assumption that does not hold true under all important ranges. This equation
for Pmin (the pressure at the top of the siphon), is a linear function of hc (the height of the siphon above the upper reservoir). But a more accurate equation for Pmin (in siphons that cannot rely on liquid tensile strength like almost all practical siphons) would be a piecewise function, where the equation you gave would apply over the domain where hc does not exceed the barometric height. For the domain of the function where hc exceeded the barometric height, Pmin would be equal to approximately a constant zero(neglecting the vapor pressure of the liquid). With this more correct piecewise function, your algebraic manipulations that follow are no longer valid over their entire range, and Pa does not always cancel. It is true of course that the flow rate does not generally vary with atmospheric pressure, but only if there is enough or more than enough atmospheric pressure to push the liquid up the siphon. If there is not enough pressure to push the liquid up, then the flow rate does depend on atmospheric pressure, and the flow rate will be zero. Mindbuilder (talk) 19:56, 29 January 2011 (UTC)
@Mindbuilder: Thanks for your comment that my analysis all looks good. I agree that if hc exceeds the barometric height the equation is not valid because a siphon will not operate. Would your objection disappear if I re-wrote my first equation as follows?
for hc less than the maximum height of the column of working fluid that can be supported by the atmospheric pressure at the surface of the reservoir.
Dolphin (t) 05:09, 30 January 2011 (UTC)
As far as your thesis about the importance of a difference between the density of the atmosphere and the density of the siphon's working fluid, I agree that such a density difference is important. There is an interesting description of something similar to what you describe called Pascal's Siphon, at a page by Dr James Calvert, professor of engineering at the University of Denver, here: http://mysite.du.edu/~jcalvert/tech/fluids/hydstat.htm#Siph
Note that here we have a professor stating that in a siphon "...pressure [of the atmosphere of water in this case] on the free surfaces of the mercury in the beakers pushes mercury into[up] the tubes..." again demonstrating that this is the normal way of explaining this phenomenon. Mentioning the hydrostatic pressure of the fluid at the entrance of the tube is a distraction from the fact that the liquid is pushed up by the atmosphere. Mindbuilder (talk) 23:27, 29 January 2011 (UTC)
I accept that language like the atmosphere pushes liquid up the tube might be appropriate for use in textbooks, guide books, instruction manuals. Our task here is to find language that will be most appropriate in an encyclopedia. As you know, at WP:NOT it explains all the things that Wikipedia is not. For example, Wikipedia is not a textbook or a guide book or an instruction manual. We can use Dr Calvert's site as a source for our information but we can't simply plagiarise his exact words, and we must find suitably encyclopedic language. Dolphin (t) 05:09, 30 January 2011 (UTC)


Hydraulics details for a siphon.

Pipe 25mm PE100 Polyethylene PN 12.5, Inside Diameter 21.1mm

8 metre siphon pipe length. Siphon, 2 metre siphon rise, then 6 metre fall.

Top of siphon fitted with tee piece with gauge, and 90 degree elbow either side.

Both siphon ends placed in seperate water reservoirs.

Siphon ends just below surface, assume top reservoir is kept at constant level via seperate water source and float.

Bottom reservor is kept at constant level by being able to overflow. Height difference between reservoirs: 4 metres

Flow rate is thus determined by 4 metre height difference, pipe inside diameter and pipe length.

Total friction including entrance and exit losses will be equal to 4 metres.

Will ignore any friction loss in fittings at top for simplicity.

Expect entrance loss to be 0.78 x velocity head.

Expect exit loss to be 1 x velocity head.

Velocity head = velocity in m/s squared divided by 2 x g (9.8)

Flow rate: 0.97 l/s

Velocity: 2.8 m/s

Velocity head: 2.8 x 2.8 / (2 x 9.8) = 0.4 metres head

Friction loss: 408m per 1000m or 3.26m per 8m or 0.4075m per 1 metre

Entrance loss: 0.78 x 0.4 = 0.312 metres head

Exit loss: 0.4 metres head

Total friction: 3.26 + 0.312 + 0.4 = 3.912 plus sundry for fittings friction = 4 metres.

Operating siphon, actual pressures just outside siphon, just inside siphon, then 1 metre increments.

Then just inside end of siphon and just outside siphon at end.

Atmospheric pressure of 10.3 metres at sea level.

Note: To get water moving, will require velocity head at entrance.

Fit a pressure gauge at each 1 metre.

Readings will be as follows. All readings based on Absolute pressure.

Just outside: 10.3

0: Just inside: 10.3 - 0.312 - 0.4 = 9.588

1: 9.588 - 1 - 0.4075 = 8.18

2: Top of siphon: 8.18 - 1 - 0.4075 = 6.773

3: 6.773 + 1 - 0.4075 = 7.3655

4: 7.3655 + 1 - 0.4075 = 7.958

5: 7.958 + 1 - 0.4075 = 8.5505

6: 8.5505 + 1 - 0.4075 = 9.143

7: 9.143 + 1 - 0.4075 = 9.7355

8: Just Inside: 9.7355 + 1 - 0.4075 = 10.3 (rounded)

Just Outside: 10.3

(Do not need to deduct exit loss, since this is simply velocity head which we have in the moving water)

Now reduce atmospheric pressure to 0.3 metres.

Water in reservoirs will not vapourise, since pressure is still above vapour pressure.

We have thus reduced absolute pressure by 10

All readings based on Absolute pressure.

Just outside: 0.3

0: Just inside: 0.3 - 0.312 - 0.4 = -0.412

1: -0.412 - 1 - 0.4075 = -1.8195

2: Top of siphon: -1.8195 - 1 - 0.4075 = -3.227

3: -3.227 + 1 - 0.4075 = -2.6345

4: -2.6345 + 1 - 0.4075 = -2.042

5: -2.042 + 1 - 0.4075 = -1.4495

6: -1.4495 + 1 - 0.4075 = -0.857

7: -0.857 + 1 - 0.4075 = -0.2645

8: Just Inside: -0.2645 + 1 - 0.4075 = 0.3 (rounded)

Just Outside: 0.3

NOTE: These are Absolute Pressures. You can't have negative absolute pressures in an everyday siphon.

This second siphon will not work.

Mindbuilder is correct, atmospheric pressure is essential for the above siphon to operate it.

The above calculations are the hydraulic calculations used to determine flow, friction and pressures in a pipeline.

In standard pipelines, there are important rules.

1: Avoid negative pressures, i.e below atmosperic pressure, otherwise it can cause the pipe to collapse.

2: Avoid absolute negative pressures, otherwise you will either get no flow,

or a reduced flow to get the absolute pressure back into a positive number.

@Dolphin: Maybe this is the siphon you need to build and test so you can see the results for yourself. Yes density is important, however atmospheric pressure is essential otherwise, as noted above, you are proposing negative absolute pressures.

Also, in reference to your siphon inside a seperate reservoir, you should also note it is not just a matter of the density being the same, but also that the pressure just outside the outlet is higher than the pressure just outside the inlet by the same amount as the height difference between the inlet and outlet. Thus you no longer have a pressure difference to operate the siphon.

Moonshine121.223.82.43 (talk) 23:41, 29 January 2011 (UTC)

Your statement also that the pressure just outside the outlet is higher than the pressure just outside the inlet by the same amount as the height difference between the inlet and outlet needs some explanation. I don't understand what point you are trying to make. Dolphin (t) 07:07, 31 January 2011 (UTC)

Just some extra notes on the above. The above figures detail the actual pressures that would exist in a siphon. They are based on basic hydraulic principles. As water flows in a pipeline, the pressure will increase if the water goes downwards, and decrease if it goes upwards. In addition, the pressure has a consistent decrease due to friction losses, relative to the flow rate/velocity. Thus the figures are accurate, regardless of what people believe operates a siphon. And the 1st siphon will operate.

When you drop the atmospheric pressure to 0.3m, a problem then arises with the figures calculated above.

Those promoting atmospheric pressure plays a role will advise the above siphon will not work.

Those promoting that atmospheric pressure does not play a part, can advise if the siphon will work or not.

If they advise it will work, how do they propose to have negative absolute pressures?

If they advise it will not work, what reason are they suggesting it will not work?

Moonshine124.187.86.46 (talk) 05:24, 31 January 2011 (UTC)

Obviously, if a siphon is to operate, the working fluid must have suitable properties. For example:
  • the freezing point of the working fluid must be less than the ambient temperature otherwise the working fluid in the reservoir will freeze solid
  • the boiling point of the working fluid must be greater than the ambient temperature otherwise the working fluid in the reservoir will boil
  • the density of the working fluid must be sufficiently low that the barometric height is greater than hc, the height of the crest of the apparatus above the fluid surface in the reservoir
  • the viscosity of the working fluid must be sufficiently low that the working fluid will flow through the tube with sufficient speed that we recognise it and agree that it qualifies as a siphon.
These parameters are not required in the fundamental equations of the siphon. Challenging the equations on the grounds that they contain no provision for the density of the working fluid, or its temperature or its viscosity would be unconvincing.
In your second example above, the siphon won't work because you are using a water siphon and yet the atmospheric pressure is so low that water's density is too high. If someone wants to demonstrate a siphon in an evacuated environment of less than one thirtieth of standard atmosphere, it is necessary to find a working fluid with very low density, or use a crest height much lower than 2 metres. Using this situation to support the notion that atmospheric pressure drives a siphon is like suggesting that internal energy also drives a siphon because if the internal energy is inadequate the working fluid freezes. Dolphin (t) 07:07, 31 January 2011 (UTC)


@Dolphin: Re your comment Your statement also that the pressure just outside the outlet is higher than the pressure just outside the inlet by the same amount as the height difference between the inlet and outlet needs some explanation. I don't understand what point you are trying to make.

If we use the example 1 in my recent post, when the siphon is outside the extra reservoir, the pressure just outside the inlet will be 0 metres head gauge pressure, or 10.3 metres head absolute. The pressures just outside the outlet will also be 0 metres head gauge pressure, or 10.3 metres head absolute. And if the siphon inlet/outlets are just below the surface, it will not matter if I measure the pressure just below the surface in the water, or just above the surface in the air.

Lets say we put the complete siphon assembly in another larger reservoir, and put it so the top of the siphon is 0.5 metres below the surface. Thus, the inlet would be 2.5 metres below the surface, the top 0.5 metres as stated, and the outlet 6.5 metres below the surface. I still have a 4 metre height difference between the inlet and outlet. Yet now when I measure the inlet pressure, it will be 2.5 metres gauge pressure or 12.8 (10.3 + 2.5) metres head absolute. The outlet pressure will be 6.5 metres gauge pressure or 16.8 (10.3 + 6.5) metres absolute. So although I have a 4 metre elevation difference, I also have 4 metres more of pressure at the outlet (16.8 absolute) compared to (12.8 absolute) at the inlet. So the elevation difference is countered by the extra pressure at the outlet, and thus we will get no flow.

I was not attempting to argue that it was pressure, rather than density, just that inside the reservoir, we get extra pressure at the outlet that we don't get when we have set up outside the reservoir. It is also of interest to note, that if we now take the siphon set-up back outside the reservoir, but then increase the inlet and outlet pressures to those that we had inside the reservoir, we will agian get no flow. So at the inlet, we add 2.5 metres of extra air pressure. At the outlet, we add 6.5 metres of extra air pressure. We still have 4 metres of elevation difference, we still have water at a similar density, air would be at a similar density (slightly higher due to extra pressure), we still have different densities of the liquid and air, but the siphon does not flow.

The siphon inside the reservoir has the same densities and different pressures, and does not flow. The just mention siphon outside the reservoir has different densities and the same pressures that the one inside the reservoir had, and does not flow.

Yes, of course we need different densities of the water and air. However, the example you provide has you changing not only the densities, but also the pressures at the inlet and outlet. People can then get into an arguement as to what is stopping the flow, the change to density, or the extra pressure cancelling the elevation out.

In relation to my recent post showing 2 exaples of a siphon set-up, please note that my comment

"Mindbuilder is correct, atmospheric pressure is essential for the above siphon to operate it." In this post, I made no reference to "atmospheric pressure drives a siphon" I was just saying that it was essential for a siphon to operate.

In the infamous Hughes article, he opens with the line "... for correcting the common misconception that the operation of a siphon depends on atmospheric pressure" and also he notes "Another seeming ubiquitous misconception is that the maximum height of a siphon is dependent on atmospheric pressure"

In the two examples I provided, we agree that 1 will work, the 2nd will not. In the 2nd one, the only thing that has changed is atmospheric pressure. The density of the water hasn't changed. The heights haven't changed. The density of air, which is less than water would have reduced slightly, which you would think should help our cause. After all, you point out that the lower density of air compared to water helps the siphon work.

So, you have agreed, that when we reduce atmospheric pressure, which is the only change we make, our siphon stops working. Confirming my original statement "Mindbuilder is correct, atmospheric pressure is essential for the above siphon to operate it."

Now how you want to word that into the siphon description is another matter, but clearly we need it.

Note also, if we take a siphon, and add air pressure to both the inlet and outlet, say an extra 40 metres head, then we could get this siphon to work over a rise of up to 50 metres. Could we say that atmospheric pressure is driving it. Well, no, since it is only 10.3 metres. Maybe we could mention air pressure. More to the point is pressure differential. We can get a siphon to work, for water to flow in the direction we want, due to pressure differential, that being the pressure at the inlet compared with that at the top of the siphon. Reduce our starting inlet pressure to zero, and we can no longer have a pressure differntial that will get water to move uphill. —Preceding unsigned comment added by 124.177.125.24 (talk) 01:32, 1 February 2011 (UTC)

In the case of a siphon apparatus (with water as the working fluid) that is completely immersed in water, the following expression (which I derived above) is relevant:
In the case of the immersed apparatus, the atmosphere is water rather than air, and the density of the atmosphere is identical to the density of the working fluid. This expression reduces to zero and the math equation shows that ΔP is zero. Consequently we correctly predict that the siphon will not operate.
You wrote People can then get into an arguement as to what is stopping the flow, the change to density, or the extra pressure cancelling the elevation out. I have produced some fundamental equations for the siphon which end up being based on densities. If someone believes that pressure is equally important, or perhaps even more important, someone should produce some equations that demonstrate that point of view. Until I see such an equation I will be assuming it can't be produced.
You have again written that atmospheric pressure is necessary for a siphon to operate. Let's not stop at atmospheric pressure. Let's say the ambient temperature must be higher than the freezing point of the working fluid. And the ambient temperature must be lower than the boiling point of the working fluid. Instead of talking about atmospheric pressure we can say we must choose a working fluid that has a density sufficiently low that the barometric height is higher than the height of the crest of the siphon. The viscosity of the working fluid must also be low enough. And if one of these considerations is not fulfilled we need to find a working fluid that is better suited to the environment in which we are seeking to demonstrate the siphon. But all this is rather elementary, don't you think?
I notice Moonshine is jumping around from one IP address to another. Why not create a User account? Dolphin (t) 02:31, 1 February 2011 (UTC)


Re your comment: If someone believes that pressure is equally important, or perhaps even more important, someone should produce some equations that demonstrate that point of view.

.... you included the formula in your last post: This expression reduces to zero and the math equation shows that ΔP is zero.

And as noted in previous post, keep densities the same, increase the inlet and outlet pressure, and you are able to get a siphon to work over a much greater height. Have we changed the densities, no, not to any great degree. Have we changed pressure. Yes.

Of course we can list as many issues as we want, however for an everyday siphon, some are obvious and don't need to be mentioned. For example, ice is frozen water, which becomes a solid, however we assume no one is trying to siphon ice.

Re IP address. I don't select this, this is automatically listed when I put in the required 4 x ~. Moonshine124.177.125.24 (talk) 03:21, 1 February 2011 (UTC)

My fundamental equation is
It doesn't mention pressure, and certainly not atmospheric pressure. I am saying that if someone disagrees with me and believes pressure is significant, someone should produce a different equation; one that does show the significance of a pressure.
There seems to be general agreement that atmospheric pressure and temperature affect whether a siphon can operate or not, but they don't affect the performance of the siphon such as initial acceleration or steady-state rate of flow. Initial acceleration is determined by ΔP as given by the formula immediately above. Dolphin (t) 04:47, 1 February 2011 (UTC)

Re your comment: There seems to be general agreement that atmospheric pressure and temperature affect whether a siphon can operate or not, but they don't affect the performance of the siphon such as initial acceleration or steady-state rate of flow.

This highlights a common mis-understanding.

When atmospheric pressure is mentioned, many believe this is included because it controls the rate of flow. Hughes has gone to great trouble, in his original article and subsequent postings, that the height difference, not atmospheric pressure, determines the rate of flow. This wasn't really required, since it was already known that all else being equal and understood, water, air, density, pipe size etc, that if you increase the height difference, you increase the flow rate. (Atmospheric pressure doesn't determine rate of flow, except of course if our static lift is too high, where we get no flow which brings us to the 2nd issue)

Seperately, again in your comment, there is general agreement that atmospheric pressure affects whether a siphon can operate.

Somehow these two issues need to be included in any description, without them getting confused with each other. The inclusion of other matters of interest can then be included, if it is deemed necessary, and will assist with the understanding, without causing more confusion. Listing every single parameter obviously will add confusion and have people rush back to the Oxford dictionery for a simple explanation. (I refer here to the opening description that would aim to be a slightly expanded and thus better than an abrievated dictionary).

Re your earlier comment

  • the density of the working fluid must be sufficiently low that the barometric height is greater than hc, the height of the crest of the apparatus above the fluid surface in the reservoir

This relates to and confirms the 2nd issue. If we are dealing with water and air where the densities are know, we can get away without mentioning density, and list a height of 10 metres being the maximum at sea level. Of course, the maximum height is reduced at elevation.

If we are dealing with other liquids, such as mercury, then the density of mercury means that it will not work in the 1st of the two examples I included above. Our siphon limit height in this case is less than 1 metre. Petrol being of a lower density, could be siphoned over a greater height, maybe why people get high sniffing it!

So if height limits in exact metres are mentioned, they probably need to include that it relates to water only, or have a chart that shows the different heights, with a seperate column for density if that is of interest to people.


Moonshine124.177.125.24 (talk) 05:49, 1 February 2011 (UTC)


@Dolphin - Consider the following diagram as a mechanical analogy to a siphon. Imagine that the two bellows in the diagram are of the same dimensions and are filled with pressurized air to simulate atmospheric pressure. Imagine that outside of the bellows is vacuum. The tops of the two weights are not connected to the balance arm, but merely touch, so that the top balance arm cannot exert any upward force on either of the weights. Weight B is heavier than weight A. The balance arm at the bottom left simulates fluid in the upper reservoir of a siphon. If we create an equation like the siphon equation, to calculate the initial acceleration of the weights, we will find that the magnitude of the pressure in the bellows and the force they exert, cancels out of the equation. The acceleration will depend on the difference between the two weights, but not the pressure in the bellows. Actually that's not completely accurate since the pressure in the left bellow must be sufficient to lift weight A or regardless of the pressure in the right bellow, weight A will fall. It is clear in this diagram that it is the left bellows that lifts weight A. The top balance arm cannot lift it because there is no attachment, it is just touching. There is nothing else in the diagram to lift weight A besides the pressure in the bellows. You could of course say that the lower balance arm pushes A up. While true in some strict sense, it is obvious that the lower balance lever can't provide any net lift, but must instead rely on the bellows pressure to provide the necessary force. While it could be said that the fluid at the entrance to a siphon pushes the liquid up, by itself, the liquid of the upper reservoir can only push up the tube to the point where it is level with the rest of the surface of the reservoir.
Mechanical analogy to siphon
So what kind of wording should be used to explain these concepts in an encyclopedia? Here is a quote from Minor(1914)(top of page 153):
"...it is clear that in such a broken siphon the atmospheric pressure is continually active in pushing the liquid up the short arm."
Here is a quote from Potter (1971):
"A typical account of the siphon given in elementary text books might read as follows. ... liquid, pushed in by the atmospheric pressure, enters at A."
Wikipedia is not supposed to be a textbook. But that doesn't mean that there are not a lot of things that Wikipedia should do the same way as a textbook. For example textbooks often use diagrams. Wikipedia also often uses diagrams. This is obviously a good thing. It would be absurd to say that Wikipedia shouldn't use diagrams because textbooks use diagrams and Wikipedia is not a textbook. Likewise textbooks frequently use equations. Wikipedia also uses equations. Textbooks often use language to describe how things work. Wikipedia can, should, and does use similar language to explain how things work where appropriate. It is probably quite appropriate that Wikipedia's language for explaining things should be very similar and often identical to how a textbook explains things. At other times the language used by textbooks is inappropriate to Wikipedia. WP:NOT states that Wikipedia should not use techniques from textbooks like "...leading questions and systematic problem solutions as examples." and then says "Other kinds of examples, specifically those intended to inform rather than to instruct, may be appropriate for inclusion in a Wikipedia article.".
I have just quoted language from a peer reviewed scientific journal by a university physics professor that says the liquid is pushed up the siphon by atmospheric pressure. I have also quoted from a journal article that states that it is a typical textbook account of siphon workings. I have previously quoted similar language by a Phd professor of engineering on his instructional web page, and by other Wikipedia editors in the barometer article. This is clearly a common way to express these ideas.
But is this an encyclopedic way to describe these ideas? WP:NOT states "Texts should be written for everyday readers, not for academics. Article titles should reflect common usage, not academic terminology, whenever possible." and "A Wikipedia article should not be presented on the assumption that the reader is well versed in the topic's field. Introductory language in the lead and initial sections of the article should be written in plain terms and concepts that can be understood by any literate reader of Wikipedia without any knowledge in the given field before advancing to more detailed explanations of the topic."
If saying that atmospheric pressure pushes the liquid up a siphon, is good enough for scientific journal articles, textbooks, and instructional university web pages, I'd say it's probably more than good enough for everyday readers, and Wikipedia. Mindbuilder (talk) 22:39, 1 February 2011 (UTC)
Another experiment to confirm without doubt that it is atmospheric pressure that pushes the liquid up, is to operate a siphon with no reservoirs at all, just a tube. Fill a small diameter tube (I used one of about 3mm) with water and introduce a bubble. Hold a finger over each end and place the tube in a siphon like position and allow the bubble to rise to the top of the siphon. When you release the ends, the siphon will flow and the water will be pushed up the tube by atmospheric pressure. Since there is a bubble of pressurized air at the top of the siphon at the start(at maybe .95 atm for a siphon with a .5m rise) there is nothing pulling the water up. The air bubble at top is pushing down on the water. Since there is no reservoir at the entrance, there is nothing but atmospheric pressure to push the water up. This siphon does rely on the cohesion of water and a small tube to keep the water from flowing around an air bubble at the entrance and draining out the entrance, but the cohesion does not provide any force to push the mass of water up. The cohesive forces will just as happily keep the water in a cohesive falling mass as a rising mass. Mindbuilder (talk) 23:32, 1 February 2011 (UTC)

I’m still thinking about Mindbuilder's mechanical analogue.

I see at least a couple of problems with saying the atmosphere pushes the working fluid up the tube. In the mechanics of rigid bodies we talk about the rigid body being pushed or pulled by another body, or gravity or a force field. For example, we say a tug boat pushes a ship, and a locomotive pulls a train. (Perhaps that is why Professors sometimes use that language to junior students.) However, that is not language used in fluid dynamics. Every parcel of fluid is surrounded by other parcels on all sides, and all sides are applying pressure to it so we invariably talk about a pressure gradient, or the pressures at both ends of the streamline. For example, consider the daily synoptic chart showing at least one high-pressure area (anticyclone) and at least one low pressure area (cyclone). There will be significant winds in between the two. Some people might say these significant winds are caused by the anticyclone pushing the atmosphere. Others might say the atmosphere is pulled by the cyclone. Of course, these significant winds are caused by both systems simultaneously. The winds are of a magnitude that closely matches the pressure gradient. Not surprisingly, the theoretical wind at any location is called the gradient wind because its speed is linked mathematically to the local pressure gradient.

Anyone who says the working fluid is pushed into a siphon by atmospheric pressure is overlooking the fact that the essential tube has two ends, not one. It is true that the entrance to the tube is experiencing pressure equal to atmospheric pressure (plus a tiny bit more due to the depth of the tube entrance, but that is not really significant.) However, it is also true that when the siphon is operating and passing fluid at a steady rate the fluid at the exit is also experiencing pressure equal to atmospheric pressure. The same pressure at both ends of the tube (and reducing pressure approaching the crest of the siphon.) Therefore the average pressure gradient over the full length of the tube is zero.

If someone writes that fluid enters the siphon because it is pushed in by atmospheric pressure, the alert reader should reply “but the same atmospheric pressure also applies at the exit from the tube. If atmospheric pressure at the entrance is sufficient to push fluid into the siphon, why is the same atmospheric pressure applied at the exit not sufficient to push air into the siphon, or at least to stop the fluid from escaping?”

Here is a very simple experiment. Take a short piece of hollow tube, open at both ends. Label one end A and the other end B. Hold the tube horizontal and lower it into water. It quickly fills. The pressure applying at end A is atmospheric pressure (plus a tiny bit more due to the depth of the tube.) That should be more than enough to cause water to flow into the tube at end A and out of the tube at end B. But of course nothing happens. Water doesn’t flow through the tube. Astute readers will say “Aha! Of course water won’t flow through the tube because the pressure applied at end B is the same as the pressure at end A. Viscosity will prevent water moving through a tube that has the same pressure at both ends.”

Now a siphon, when it is operating at a steady rate, has the same pressure at both ends. Do you know why a siphon will carry a steady flow of fluid even though it has the same pressure at both ends? (I do.) Dolphin (t) 07:17, 2 February 2011 (UTC)


Some food for thought:

5 siphon pipes, same pipe size, pipe length, crest height, as per my earlier example, i.e. 2 metre upleg, then 6 metre down leg, 4 metre height difference. Liquid water, reservoir at each end with pipe just below the surfaces

1: Operated per normal in air. Different density, Same pressure, Flow: Yes.

2: Place 1 in another reservoir. Same density, Different pressure, Flow: No

Note for 2: Assume crest 0.5 below surface. Inlet pressure 12.8, Outlet pressure 16.8

3: Take 2 back out into atmosphere, but same pressures as inside the reservoir. Different Density, Different Pressure, Flow: No

Note for 3: Inlet pressure 12.8, Outlet pressure 16.8

4: Take 3, and increase outlet pressure by another 4 metres. Different density, Different Pressure, Flow: Yes (Reverse direction)

Note for 4: Inlet pressure 12.8, Outlet pressure 20.8

5: Take 1 and reduce pressure to 0.3. Different density, Same pressure, Flow: No

So what conclusions can be drawn.

Compare 1 and 2. When we have the same density, we get no flow. So density must drive the siphon. No, not quite. One thing common with the siphons above that have flow, is that they have different densities. When it is the same density, it doesn't flow. So we can draw the conclusion that different densities are essential, but that is not the same as saying it is driving the siphon.

Compare 2 and 3. When we took the siphon out of the reservoir, so we now had the different densities again, but gave it the same pressure as it had in the reservoir, we still did not get it to flow. So comparing 2 and 3, whether we have the same density, or different density, doesn't change the siphon from not flowing to flowing, even though we have not changed the pressure at each end.

Thus we can see now that one of the results of having the same density, is that it causes us to have different pressures at the end, and this extra pressure cancels out the static fall that we were going to rely on to get the water moving.

Still doen't change the fact that we need different density to have a siphon working.

Compare siphon 1 and 3. Siphon 1 has same pressure and flows. Siphon 3 has different pressures and no flow. So we can draw the conclusion that we need the same pressure to get flow, and thus pressures is not driving the siphon. This is not quite the correct conclusion, and we will look at siphon 4 to show why.

Compare siphon 3 and 4. This has differnt pressures and flows, in fact flows in the opposite direction. Again it has different densities, so it still backs up our requirement to have different densities. And it still has water flowing up hill. Both siphon 3 and 4 have different densities, so there is no change there. What has changed is the pressure, which causes it to flow.

Siphon 5: When we remove the pressure (atmospheric), we get no flow, even though we still have different densities.

Overall conclusions. We must have different densities.

We must have atmospheric pressure.

We can have either the same or different pressures, we can still get the siphon to work, except if the different pressures equal exactly the height difference.

When we put the siphon in the reservoir, we changed both the pressure, and made the density the same and flow stopped. but when we took it back out, keeping the pressures the same as in the reservoir, and changing the density, flow did not start. Thus having a different density doesn't drive the siphon.

Having different pressures can cause the water to flow, and the water will flow in whatever direction that allows it to get from where it is to a lower pressure area, even transfering water uphill.

Example 4 might not be considered a siphon, but we haven't changed the nature of the pipe line.

One key issue is pressure difference, and it is involved, for the want of a better word, for driving a siphon.

If I was to tap into the top of a siphon, and start reducing the pressure by taking water out, I can get water to flow in both ends of the siphon at the same time. Water, being the liquid that it is, will simply flow to a lower pressure area when it gets the chance. So it flows into the inlet of a siphon, because the pressure at the crest of the siphon is much lower than the pressure at the siphon inlet, low enough to cover the height difference. Atmospheric pressure allows us to have a pressure difference, however the pressure difference is what causes the water to flow. (And not density that is driving it). The water at the inlet knows nothing about the pressure at the siphon outlet, it just knows or reacts to the fact that the pressure above it is lower, so it moves there. And when it gets there, it moves higher again, all the way to the top of the siphon.

(And the water in the downleg, is doing its thing to get to a lower pressure area, with the aid of gravity, and again not having any involvement or knowledge of the inlet pressure.)

Seperate to the above examples. Set up a siphon with inlet and outlet at same level. Will it flow: No.

How can I make it flow?

Change the density?

Change to a higher density liquid, say mercury. Will not make a difference.

Change to a lower density liquid, say petrol, again will not make a difference.

What if I change to a lower density air. Or a heavier density air. Again will not make a difference.

What if I make the liquid in the reservoirs lighter than the air. Well then the reservoirs will empty as the liquid floats upwards. Again siphon will not work.

So how can I make the siphon work. Simple. Either change the height of the outlet, or change the pressure of inlet (higher) or outlet (lower).

Will the siphon now work: Yes.

Have I changed the density: No. Have I changed the pressure: Yes, since changing the height changes the pressure in the pipeline. Or changing the pressure at either the inlet or outlet, doesn't change the density, but changes the pressure.

Different densities are still required for a siphon to operate, however, only by changing the pressure can I get it to flow. Again, pressure difference, not density that helps to drive a siphon. —Preceding unsigned comment added by 58.165.115.18 (talk) 12:03, 2 February 2011 (UTC) Moonshine124.187.7.65 (talk) 10:35, 2 February 2011 (UTC)

@Dolphin - The pressure at the entrance and the pressure at the exit don't cancel because they are acting on different objects. The liquid in the siphon is made up of many different molecules that are not attached together. In a typical siphon they do not pull on each other, they are all repelling each other. If there was a boat in the middle of a lake and someone was pushing it north and there was a car on the shore and someone was pushing it south, it would be silly to say that the two pushes would cancel.
Lets extend your car analogy to see this principle more clearly. Your car analogy leaves out gravity (which we all agree is important in the siphon) because the weight of the car is supported by the wheels. Lets add gravity into the car analogy in the form of a rocket bolted to the roof of the car. Since the north and south pushes kind of get confused with up and down, and since in the siphon we're working with left and right, lets change the car analogy to a pushing left and right situation. Lets say the rocket is pushing to the right with 200 N of force. Now to spite the guys left and right 1000 N pushes canceling, the car does move. But of course now it is clearly not the guy on the left that is pushing the car to the right, it is the rocket. The rocket can push the entire car because the car is a single rigid body. As stated above the liquid in the siphon is made up of many pieces. We can simplify and say that the siphon is made up of two parts, the liquid in the up side and the liquid in the down side. They abut at the top of the siphon. Now let us change the car model so that there is two cars, bumper to bumper, in between the two guys. Lets say the rocket is mounted on the right hand car and is pushing to the right, again with 200 N force. Now again the cars will move, and it will be the rocket that is pushing the right hand car. But the bumpers are not stuck together with glue, so the right hand car is not pulling the left hand car. In fact, the left hand car is being pushed to the left with 800N force by the right hand car. Yet the left hand car will move to the right because it is being pushed by the guy on the left. It is clear here that even though the pushes of the two guys would seem to cancel, it is not the rocket that is pushing the car on the left, but rather it is the guy on the left. The push of the guy on the left is not canceled by the guy on the right because part of the guy on the right's push is canceled by the rocket, and so isn't available to cancel all of the guy on the left's push.
If using a rocket to represent the force of gravity is confusing, then make the analogy more like the siphon by having the guys push on two cars that straddle a crest in the road, and have one car weigh more than the other, so that if the two guys stopped pushing, the two cars would roll away from each other down opposite sides of the crest. Now if the car on the right is heavier (and the slopes are the same) and the two guys are pushing hard enough (and still equally hard of course), the cars will roll to the right, including the car on the left, which will roll up hill, pushed by the guy on the left. More to come. Mindbuilder (talk) 18:51, 2 February 2011 (UTC)
Though a siphon with a steady flow may have a pressure gradient between the entrance and exit that averages to zero, the atmospheric pressure at the exit is not canceling the atmospheric pressure at the entrance and thereby causing the zero pressure gradient. Consider the poorly analogous but much simpler situation of a rocket flying horizontally through the air at a steady speed. The drag on the rocket will be basically equal to and opposite the thrust of the rocket motor. But we don't say that the motor is not propelling the rocket because the thrust of the motor is canceled by the drag. In the siphon, the atmospheric pressure at the entrance is analogous to the rocket thrust, but we have two more forces in addition to drag, the force of gravity on the liquid and the atmospheric pressure at the entrance. The fluid is pushed through the siphon by atmospheric pressure but is opposed only partially by drag and only partially by atmospheric pressure at the exit. The atmospheric pressure at the exit is pushing up on the fluid just inside the exit hard enough to completely cancel the atmospheric pressure at the entrance, but it doesn't cancel because that upward force of atmospheric pressure on those molecules just inside the entrance is partially canceled by the force of gravity pulling those molecules down. Those molecules don't bump the ones above as hard as they would from atmospheric pressure below them because of the canceling of gravity. A significant part of that force of atmospheric pressure at the exit never bumps its way up the siphon and then down to the entrance to cancel.
You might object that since the pressure gradient averages to zero then the pressure at the exit is equal and must be canceling the pressure at the entrance, but it is not the pressure at the exit that is canceling and bringing the gradient to zero, it is the drag in the tube. If it wasn't for drag in the tube, gravity would cause that pressure gradient to be non-zero, and the atmospheric pressure at the exit would not be sufficient to cancel the pressure at the entrance. If that drag is reduced, the flow rate will increase. In the moments after the the drag is reduced, the pressure gradient will not add to zero and the atmospheric pressure at the exit will not cancel the pressure at the entrance. flow rate will increase even though the atmospheric pressure at the exit hasn't changed and influence of gravity hasn't changed and the atmospheric pressure at the entrance hasn't changed. A reduction of drag does not push the fluid through the siphon. Gravity contributes to accelerating the fluid down the siphon after drag is reduced, but gravity can't pull the liquid up the siphon at an accelerating rate.
Then what does push the fluid up the siphon? If there is a bubble at the top, the fluid is not pulled up. Gravity does not lift the liquid up the siphon, rather the opposite. Gravity prevents atmospheric pressure at the exit from canceling atmospheric pressure at the entrance, but it doesn't lift the liquid up. The movement of molecules on the up side must be in response to forces applied directly to those molecules, not by forces applied to molecules on the other side of the universe or the other side of the siphon. So where does the force come from to lift those molecules up the siphon if not atmospheric pressure? Especially in the case of the siphon with no reservoirs and a bubble at the top, what other upward force is there other than atmospheric pressure to lift that column of water up the siphon? Mindbuilder (talk) 21:06, 2 February 2011 (UTC)
So many words to try to describe such a simple phenomenon!
Here is my way of describing why fluid enters the tube, flows through it and exits at the other end.
If there is a horizontal tube and no fluid is flowing through it, the pressure at either end must be the same. If there is a sloping tube and no fluid is flowing through it, the pressure at the bottom end must be greater than the pressure at the top end according to the following equation:
In the case of a siphon flowing at a steady rate, the pressure at the bottom end of the tube and the pressure at the top are equal. The pressure at the bottom end is not greater than the pressure at the top end by the required ΔP and therefore fluid flows through the tube, entering at the top end and exiting at the bottom end.
With a sloping tube through which no fluid flows, the pressure difference matches the weight of the fluid in the tube. In the case of the siphon, there is no pressure difference but the fluid (in that part of the tube that is below the fluid surface in the reservoir) has significant weight. So it is the weight of the fluid in the tube below the fluid surface in the reservoir that is causing fluid to enter the tube, flow through it, and exit at the bottom end.
If the bottom of the tube is aligned with the fluid surface in the reservoir the siphon won't operate. Obviously, because the weight of the fluid in the tube below the fluid surface ... is zero. Dolphin (t) 21:51, 2 February 2011 (UTC)
What about in a siphon with no reservoirs that starts filled with water except an air bubble at top? Where does the force come from to cause the water on the up side to rise against gravity? Mindbuilder (talk) 22:49, 2 February 2011 (UTC)
I think you are referring to an apparatus that consists of nothing more than an inverted U tube. If the two halves of the U are of equal length the water doesn't move. But if the two legs of the U are of different length water will flow through the U and out the bottom of the longer leg. This is exactly as can be predicted from my previous edit. Dolphin (t) 23:35, 2 February 2011 (UTC)
Yes, but what causes the water on the up side to go up? Those molecules of water on the up side will fall out the entrance by the force of gravity if there is no force to push or pull them up. Where does that force come from? Mindbuilder (talk) 00:03, 3 February 2011 (UTC)
Gravity does indeed exert its influence on every molecule of water. The reason each molecule doesn't accelerate towards the center of the Earth at 9.8 m.s-2 is because hydrostatic pressure above and below cancel the weight of the molecule. Dolphin (t) 00:14, 3 February 2011 (UTC)

@ Mindbuilder. I suggest your recent analogies are getting too complex. There is more effort going into explaining them and getting to understand them, compared to the actual siphon. An analogy is suppose to make something easier to understand.

Why not stick with the siphon and explain that a consequence of gravity is the difference in water levels and the water flowing out the outlet creates a lower pressure at the crest of the siphon.....

A simple vacuum guage shows the reduced pressure.

@ Dolphin. Better too many correct words, than a few incorrect ones.

@ Dolphin. Your comment: So it is the weight of the fluid in the tube below the fluid surface in the reservoir that is causing fluid to enter the tube, flow through it, and exit at the bottom end.????? What the????

The depth of either the inlet tube or outlet tube is placed into each reservoir makes no difference to a siphon's flow. Changing it will not change the whether the siphon will flow or not, and if it was flowing, will not change the rate of flow.

This was a issue that came up re Dr Hughes before he became famous. I was asked to review a proposed siphon in South Australia that was to transfer water in excess of 1 km. There was conflicting information about the height difference, they mentioned 1 metre, but also they mentioned 6 metres of fall. When I asked for clarification, that it had to be one or the other, they replied that the water levels had a 1 metre difference, but they proposed to put the outlet 5 metres under the water level, believing that it would give them extra fall and thus extra flow.

They were also referencing Dr Hughes's siphon paper. Although Dr Hughes's siphon paper was not the cause of this error, it was clear that he had many other errors in his paper in reference to understanding how siphon's work. His supplementary article also had a major flaw, where he had proposed to use siphons to move captured flood water and generate electricty for approx 15,000 homes. He was running quite wild on his idea. Unfortunately, engineering is not his strong point, and it turned out to be only 4 homes. Well, it wasn't even 4, since we then needed to account for the friction loss. So I suggested it we allowed half for elelctricity generation, and 1/2 for friction, we would have enough for only 2 homes, maybe not enough to save the world. He agreed, and dropped the whole power generation thing from his article.

I also pointed out the flaws in his main article, this was before he became famous. Alas, I think the allure of fame in claiming the dictionery was wrong had him rushing to get his story out, even though he referenced Wikipedia and it had pointed out that the chain analogy was flawed, and that he had not conducted any tests to prove his claim that atmospheric pressure wasn't involved.

Moonshine124.187.108.158 (talk) 23:26, 2 February 2011 (UTC)

You wrote So it is the weight of the fluid in the tube below the fluid surface in the reservoir that is causing fluid to enter the tube, flow through it, and exit at the bottom end.????? What the????
Imagine the inverted U, from the surface of the fluid in the reservoir round the crest to the point level again with the surface, contains fluid weighing 20 Newtons. Then the remainder of the tube below the level of the fluid surface in the reservoir contains fluid weighing 10 Newtons. The performance of the siphon is related to the 10 Newtons, not the total weight of 30 Newtons.
If the bottom of this flexible tube is progressively raised so it is level with the surface of the fluid, the figure of 20 Newtons increases to 30 Newtons and the figure of 10 Newtons reduces to zero and the siphon ceases to operate. Dolphin (t) 23:35, 2 February 2011 (UTC)

@ Dolphin: Your comments.....round the crest to the point level again with the surface..

Then .....If the bottom of this flexible tube is progressively raised so it is level with the surface of the fluid.....and the siphon ceases to operate.

Sorry, it will not cease to operate, since in your first example, it would not have been flowing at all. Putting the tube say 5 metres below the surface which contains, say fluid weighting 10 newtons, will not get the siphon to flow. Although you have extra weight of fluid in the tube, you have extra pressure at depth just outside the tube that cancels this out.

Try again!

Moonshine124.187.108.158 (talk) 23:58, 2 February 2011 (UTC)

We must have different images in our minds. If the tube ends 5 metres below the surface of the fluid in the reservoir (and these 5 metres of tube contain fluid weighing 10 Newtons) the siphon should be running like a beauty.
Further to my previous edit, it is the downward movement of fluid weighing 10 Newtons that overcomes fluid friction in the tube and causes fluid in the reservoir to enter the tube and rise towards the crest. Atmospheric pressure is not part of a sound explanation. Dolphin (t) 00:04, 3 February 2011 (UTC)
I didn't see your last reply before my reply further up. Lets continue down here instead of up there. How does the downward movement of the fluid on the down side exert a force on the molecules in the up side when there is a bubble in the top of the siphon? Mindbuilder (talk) 00:13, 3 February 2011 (UTC)
The bubble you see in the tube is actually air. Air is very good at transmitting pressure. The thread of fluid might be discontinuous at the bubble, but the pressure field (or gradient) is continuous. Dolphin (t) 00:17, 3 February 2011 (UTC)
But the pressure of the bubble of air at the top of the siphon will be exerting a downward force on the molecules of water in the up side of the siphon. That would make the siphon go backwards and assist rather than oppose gravity on the up side, would it not? For the molecule at the bottom of the column of water in the up side of the no reservoir siphon, there is no hydrostatic pressure below the molecule. There is nothing but air. Why does it go up? Mindbuilder (talk) 00:29, 3 February 2011 (UTC)


@Dolphin. Please refer to paper: "The pulley analogy does not work for every siphon" by Gorazd Planinsic and Josip Slisko that was printed in the July 2010 Physics Education magazine.

This proved that the chain analogy of the heavier weight water pulling the lighter weight water over the rise was incorrect!

It also proved that the weight of the water is not driving the siphon, since the heavier inlet side lost out to the lighter downleg. The inlet leg was much wider, and thus much heavier, but is was no match for the downleg, even though the downleg had only a 40th of the weight of the upleg.

Moonshine124.187.108.158 (talk) 00:26, 3 February 2011 (UTC)

Atmospheric pressure 2

New thread taking over from the previous one.

@Moonshine: I am aware of the chain analogy and the train analogy. I'm not surprised there is a pulley analogy. I'm no fan of any of them. Scientific phenomenon can be explained by fundamental scientific principles so I always try to focus on the principle rather than getting too distracted by analogies.

When I have a moment I will try to find the document that proves the weight of the fluid below the level of fluid in the reservoir doesn't drive the siphon. It should be fascinating reading. Dolphin (t) 00:41, 3 February 2011 (UTC)

@Moonshine: Now I see the problem. Yes a siphon with a very fat "upleg" will continue to operate. It can be shown, and I have explained, that it is the weight of the fluid below the level of fluid in the reservoir that is pertinent to the performance of the siphon. The very fat "upleg" is above the level of fluid in the reservoir so it is not pertinent. Dolphin (t) 00:48, 3 February 2011 (UTC)


I will leave it with you Dolphin!

Moonshine signing off, and returning to civilisation! 124.187.108.158 (talk) 00:56, 3 February 2011 (UTC)Moonshine

Thanks Moonshine. It has been a pleasure having your participation. We look forward to seeing you back in the future. Dolphin (t) 01:12, 3 February 2011 (UTC)

@Dolphin - Here is a diagram of the no-reservoir siphon with a bubble in the top that I have been referring to.

No-reservoir siphon with bubble at top

Mindbuilder (talk) 01:19, 3 February 2011 (UTC)

Maybe I should ask: How is the force applied to the water on the up side that causes those molecules to go up rather than fall under gravity. Mindbuilder (talk) 01:21, 3 February 2011 (UTC)

Thanks Mindbuilder. I noticed the question when you first asked it an hour or two ago, and I have been thinking about it ever since. Brilliant diagram! I hope to have some intelligent thoughts on the subject very soon. Cheers. Dolphin (t) 01:40, 3 February 2011 (UTC)

Here's a sketch of the trucks on a crest concept.

Note that while it would seem the pushes of the two guys would cancel, they don't. As the big truck starts rolling back down the hill, it does not pull the little truck with it because the bumpers are not stuck together, they're just touching. The little truck is being pushed up the hill by the guy on the left, to spite the fact that you would have expected his push to have been canceled. Mindbuilder (talk) 02:04, 3 February 2011 (UTC)

Excellent question about the siphon, and excellent diagram. First, let's imagine that the two legs of the inverted U are of equal length and have equal amounts of water of height h. The pressure acting on the bottom of both slugs of water is atmospheric pressure Pa. The pressure at the crest of the siphon is Pmin and is equal to:
The pressure difference Pa - Pmin exactly supports each slug of water so neither slug moves and we have a static situation.
Now let's imagine that the legs are of unequal length and have unequal amounts of water. The crest pressure required to support the weight of the longer slug is a lower pressure than is required to support the shorter slug. Imagine the longer slug beginning to move downwards because its weight is not fully supported. Immediately the absolute pressure at the crest reduces even further.
As a result of the reduced pressure at the crest the pressure difference across the shorter slug of water is more than necessary to support the weight of the shorter slug and it begins to move upwards. So the longer slug is accelerating downwards and the shorter slug is accelerating upwards. The bubble is swept along with both of them because its pressure and volume remain almost unchanged. The whole arrangement continues to accelerate until it reaches a speed where viscous forces reduce the acceleration to zero, or the longer slug of water begins to leave the tube.
In summary, the pressure at the bottom of both slugs of water is atmospheric pressure Pa. The pressure at the top of both slugs is Pmin. It is the difference in the weight of the two slugs of water that causes all the water, and the air in the bubble, to accelerate and flow out of the tube.
Perhaps it is a bit like a seesaw with a child on each end, and one child heavier than the other. The end with the heavier child moves downwards and the other end moves upwards as a result. Dolphin (t) 02:18, 3 February 2011 (UTC)
So you do realize that the pressure of the bubble at the top is pushing down on the slug of water in the up side, even after the slug on the down side has started to fall and lowered the pressure at the top of the siphon, right? And you realize that atmospheric pressure is pushing up on the slug of water in the up side of the siphon, right? So do you realize that atmospheric pressure is the only thing pushing up on the slug on the up side? Mindbuilder (talk) 02:25, 3 February 2011 (UTC)
Yes. Atmospheric pressure is pushing upwards on both slugs. It is the only force pushing upwards on the shorter slug. In the case of the longer slug there can be two forces pushing upwards - atmospheric pressure and the viscous forces associated with the speed of the slug relative to the wall of the tube. Dolphin (t) 02:45, 3 February 2011 (UTC)
By the way I recently realized it is easier to do decent sketches in a paint program if you use a canvas with nice large pixel dimensions (about 4000x3000) and then make the sketch fairly small in the middle. A large canvas means you can keep adding to the sketch without bumping into the edges. I zoom in close when drawing freehand with my touchpad to minimize the unsmoothness of the lines. Its quick and easy to crop the image tight after it's done.
Back to siphons. If you prime a siphon like the following sketch and then let it go, is it reasonable to say in layman's terms, that gravity acting on the down side causes a low pressure area at the top which allows atmospheric pressure to push the left hand slug of liquid up the siphon?
Mindbuilder (talk) 07:41, 3 February 2011 (UTC)
Your proposed wording, using layman's terms, looks satisfactory to me. My Fluid Mechanics lecturer would have expected me to explain it something like this:
The weight of the slug of fluid in the vertical part of the tube causes the air at the crest to be at an absolute pressure significantly below atmospheric pressure. The weight of the slug of fluid in the U part of the tube is mostly supported by the tube. It has atmospheric pressure acting on its left end, and a pressure significantly below atmospheric acting on its right end. This imbalance of forces causes the slug in the U part of the tube to accelerate to the right. The slug in the vertical part of the tube accelerates downwards. The accelerations of the two slugs are the same and will continue until either the fluid is gone from the tube or viscous forces reduce the acceleration to zero. Dolphin (t) 11:00, 3 February 2011 (UTC)
Your explanation definitely describes the situation better than mine did. But before we work that out in detail, I want to check how far our understanding is in agreement. Starting with the leftmost siphon in the following sketch, is it fair to say in laymans terms that that when the leftmost siphon is let go, that atmospheric pressure will push the liquid on the left up above its initial level (after the falling column has lowered the pressure at the top of course)? Is it fair to say the same for the rest of the siphons starting at the left and moving to the right?
Mindbuilder (talk) 05:16, 5 February 2011 (UTC)
In the above siphons there is air at the top, so it is clear that there is no pulling at the top. Lets consider next a practical siphon primed to be entirely filled with liquid and no air bubbles. Is it fair to say that whether the fluid at the top of the siphon is air or liquid, it doesn't change the fact that its pressure is still greater than pure vacuum in a practical siphon of moderate height, to spite the fact that its pressure has been lowered by the falling column, so the liquid at the top is still pushing down on the liquid going up the up side? Therefore is it not still atmospheric pressure pushing the liquid up the siphon even in a siphon with no air bubble? And isn't this especially true in a practical siphon where there may be tiny air bubbles that would expand if pulling was going on or where if the pressure got so low that pulling was going on (i.e. zero absolute pressure), then tiny air or vapor bubbles would form and expand? Mindbuilder (talk) 06:15, 5 February 2011 (UTC)
Firstly, my response to your first posting (05:16 am, 5 Feb):
Some laymen would find that explanation satisfactory, especially laymen accustomed to contemplating simple problems in mechanics where the acceleration of a rigid body can often be described as the action of a single force. Other laymen would prefer something like the following. Both slugs of liquid are subjected to two pressures, one greater than the other. The greater of the two is atmospheric pressure at about 100 kilopascals. The other is the pressure of the air trapped between the two slugs. If the liquid is water, and the slug in the vertical tube is one metre long, the pressure of the trapped air will be about 90 kilopascals. The result is a pressure difference of about 10 kilopascals to support the weight of both slugs. This pressure difference is not enough to support the weight of the one metre long slug, but more than enough to support the weight of the slug in the left leg.
Newton's Second Law of Motion states that any body, including any part of a body of liquid, will accelerate in response to the net force acting on it. The net force acting on the one metre long slug of liquid is the resultant of the two pressures (10 kilopascal pressure difference) plus its weight. This slug accelerates downwards.
The net force acting on the slug of liquid in the left leg of the leftmost siphon is also the resultant of the two pressures plus its weight. This slug has a smaller weight and it accelerates upwards towards the crest of the apparatus. Its acceleration up the left leg of the siphon is directly related to the figure of 10 kilopascals. 10 kilopascals is what pushes the liquid up the left leg. 10 kilopascals is the difference between atmospheric pressure and the pressure at the crest of the siphon.
In the rightmost siphon there is a large reservoir of liquid. The weight of the liquid in the reservoir is supported by the floor of the reservoir rather than by the difference in pressure. The net force acting on each of parcel of liquid as it leaves the reservoir and enters the left leg is the same as in the leftmost siphon.
Moving from the leftmost siphon to the rightmost, there is an increasing weight of liquid in the reservoir and an increasing supporting force from the floor of the reservoir. The net force on the liquid in the reservoir is zero so that liquid is stagnant. The dynamics of the thread of liquid actually in the left leg of the siphon appears to be the same in all cases. Dolphin (t) 06:19, 5 February 2011 (UTC)
Secondly, my response to your second posting (06:15 am, 5 Feb):
My position is that there is never a pulling force on a liquid. Whether we are looking at liquid, or air, or a mixture of both, the pressure is always positive and we reasonably think of this as a pushing force. I know there is the technical argument about negative pressures near the top of trees, and in the Z-tube experiments, but I don't buy them. My position is that fluids always push on their neighbours because fluids display only positive pressures (in contrast to solids which regularly display negative pressures called tensile stresses.)
In my previous response I referred to the pressure at the crest of the siphon. I chose this wording deliberately because I am not distinguishing between gas at the crest of the siphon, or liquid. And whatever is at the crest will always be pushing downwards on whatever is below it. Dolphin (t) 06:28, 5 February 2011 (UTC)
If you don't think there can be pulling in liquids, how do you explain the z-tube experiments and experiments where siphons are reported to work at heights exceeding their barometric height in low pressure, such as by Ralph Minor(1914)? Do you suspect they are misreported or fraudulent or maybe true but just not convincing, or what? Mindbuilder (talk) 06:42, 5 February 2011 (UTC)
I don't claim to know anything about Minor's work, and my knowledge of the Z-tube experiment is limited to what I gained during a cursory look at a website. I think the most likely explanation is that there is one or more important aspects of these experiments that I am not aware of, that would explain why these claims are contrary to common practice. For example, we could make a sensational headline out of the fact that Albert Einstein disproved Isaac Newton's model of classical mechanics. That headline would suddenly cease to be sensational when we point out that we are only talking about speeds that are close to the speed of light, entirely out of the realm of our experiences!
What extra do we need to know about Minor's experiments? Were his experiments confined to a special class of fluids, or were they equally applicable to water and mercury? Were his observations available during steady-state conditions, or were they only available momentarily? Were his observations repeatable on a laboratory scale, or were they only observed on a microscopic scale? Were his discoveries achieved at room temperature, or only at very low or very high temperatures? Despite Minor's claims, mercury barometers continue to serve as reliably as they have for centuries. Minor has not persuaded irrigation specialists to implement water siphons that exceed the barometric height for water. Hydraulic equipment such as propellers and pipelines that reduce the local pressure of water to approach the saturation pressure continue to suffer cavitation, sometimes catastrophically. I think I am on safe ground when talking about practical siphons to say that liquids don't exhibit tensile stresses, negative pressures, pull forces etc. Dolphin (t) 11:49, 5 February 2011 (UTC)
The Minor paper can be found for free at reference 32 of the siphon article: "Would a siphon flow in a vacuum! Experimental answers." I thought you already had it. I found it quite interesting. In addition to the main experiment under a bell jar with a vacuum pump, it also shows a cheap simple apparatus consisting of just two bulbs connected by a tube. To make that apparatus you could use small bottles for the bulbs and by just boiling water in the bulbs you could avoid the need for a vacuum pump while still achieving pressures low enough to test a water siphon at heights exceeding the vapor pressure (less than 1m?) It would be more likely to succeed if the tube was glass, which water adheres well to, but it might work with a plastic tube as the AndrewKFletcher video showed his Nylon tube going to 24m. If the water in a plastic tube siphon didn't work, you could repeat the experiment with mineral oil or just motor oil, and use the same trick of boiling a little water in the apparatus to remove the air and get the pressure down far enough so the barometric height of the oil would be low enough to be practical.
I admit that I have a small nagging doubt about the possibility of siphons in vacuum. But between Minor,Nokes,Fletcher,Z-tubes, and Tree research, it appears liquids do have enough cohesion to pull. Of course in the vast majority of practical circumstances such as those you mention, air dissolved in the liquid does appear to defeat the tensile strength. I'm not sure how trees seem to avoid that problem. Maybe they somehow filter the air from the water as it is absorbed by the roots. Mindbuilder (talk) 21:16, 5 February 2011 (UTC)

Thanks for the link to the paper by Minor. I have looked at it closely, particularly the section describing the demonstration of the mercury siphon and mercury barometer inside the evacuated bell jar. It is certainly a fascinating account of a demonstration made before a large audience, especially as it reversed the teaching of a leading science textbook of the day. The only comment I can make about a possible shortcoming in the explanation is the absence of any acknowledgement by Minor of the existence of viscous forces in the siphon. The implication by Minor is that the pressure at the crest of the siphon is less than the saturation pressure of mercury by the equivalent of the distance from the crest of the siphon to the top of the liquid surface in the barometer (distance h1 - B). Viscous forces acting on the moving thread of mercury will cause the absolute pressure at the crest to be higher (or less negative) than would be suggested by the distance from the crest to the top of the mercury in the barometer. (The absolute viscosity of mercury is 70% greater than for water.) However, I see Minor demonstrated some impressive heights (10 - 30cm) so I don't doubt that his results suggest some tensile forces at play in the siphon. Are you aware of any demonstrations of this phenomenon more recent than the ones by Minor? Dolphin (t) 10:45, 6 February 2011 (UTC)

I have re-visited my diagram and convinced myself that viscous forces acting on the moving thread of mercury will cause the absolute pressure at the crest to be lower (or more negative) than would be suggested by the distance h1 - B so I have struck those words from my previous edit. Dolphin (t) 11:21, 6 February 2011 (UTC)
Here is an interesting discussion of vacuum siphons, and it links to the formerly hard to get Nokes(1948) http://www.mindspring.com/~rwramette/siphons.htm Mindbuilder (talk) 18:55, 6 February 2011 (UTC)

New opening paragraph with atmospheric pressure

It seems the old consensus (though not really a consensus, more a truce) has collapsed, and a new, at least partial consensus has formed. So I went ahead and changed the opening paragraph. Improvements? Suggestions? Mindbuilder (talk) 23:04, 6 February 2011 (UTC)

I think we are looking for at least four levels of explanation here. Brief captions for images, The opening paragraph/definition, the laymans theory section, and the technical theory section. The level of detail and the wording in each section should reflect the intended audience and the space constraints of each type of section. Mindbuilder (talk) 23:08, 6 February 2011 (UTC)

@Mike163 - If you still think atmospheric pressure cancels, then what do you think of my sketch above of two trucks pushed towards a crest? Mindbuilder (talk) 23:18, 6 February 2011 (UTC)

Thanks Mindbuilder. Your work on the introduction is very good. It is certainly a good compromise between the various positions we have explored. I think the comment about siphons demonstrated to work in vacuum is given undue prominence. I propose demoting it further down the paragraph as follows:
In practical siphons, atmospheric pressure pushes the liquid up the siphon into the region of reduced pressure at the top of the siphon. The reduced pressure is caused by liquid falling on the exit side. In the laboratory, siphons have been demonstrated to work in a vacuum, indicating the tensile strength of the liquid is contributing to the operation of the siphon.
Dolphin (t) 23:36, 6 February 2011 (UTC)
I don't wish to give vacuum siphons great prominence in the opening paragraph. I only mentioned them in the clause at the beginning of that sentence beacuse I thought it would be slightly better to very briefly address them in the opening paragraph than leave them out. But if you want to move them down the paragraph even further as you suggest, thats fine with me. I would modify your last sentence slightly to make it clear that it is not being stated that tensile strength plays a part in practical siphons: "In the laboratory, some siphons have been demonstrated to work in a vacuum, indicating the tensile strength of the liquid is contributing to the operation of siphons at very low pressures." Mindbuilder (talk) 23:57, 6 February 2011 (UTC)

Moonshine signing back in. Good to see that the recent healthy discussion and debate has had good results. Personally, I like the new description far better than the previous. I am with Mindbuilder, better to include/address vacuum siphons briefly at the start, so the issue does get acknowledged. I did note that's Dolphin's description about tensile strength is contributing to the opeation of the siphon could be wrongly interpreted as referring to tensile strength contributing to practical siphons. Mindbuilder has addressed this

Some other random thoughts. There was a discussion about weight of the water below the reservoir. It appears some of our conflicting views may have resulted from misunderstandings. For example there may be two reservoirs, the upper and lower. If we simply mention reservoir, people may not know which one we refer to.

In the fat upper leg siphon, if you had 6 metre vertical upwards leg, say of 100mm pipe, and 7 metre vertical downward leg of say 4mm tube, there is more weight overall on the upwards side. Dolphin put forward the view that it is the weight of the water below the reservoir level that was important. If we setup the siphon with no water in the 2 metres at the bottom of the downwards leg, and also remove 2 metres of water from the uppwards leg, the siphon will still prime. Yet at the start of priming, there was no water, and thus no weight below the upper reservoir level. From a weight issue, there is clearly far more weight of water in the upwards leg, than there is in the downwards leg, and with no weight of water below the upper reservoir level to have any influence. So why does it prime? Because it is the difference in vertical height levels of the two legs. (It is noted that once primed, there will be extra water below the upper reservoir level)

Refer to the often mentioned formula: Pressure = Density x Gravity x Height.

It's full description is: Pressure in pascals = Density in kg per cubic metre x gravitational constant x vertical height in metres.

For water, density = 1000 kg per cubic metre (does vary slightly with temperature) and the gravity constant is 9.8.

So for water, our formula can read: Pressure in Pascals = 1000 x 9.8 x Vertical Height in metres.

Changing to Kilopascals, the formula would read: Pressure in Kilopascals = 9.8 (Constant) x vertical height in metres.

This does not conflict with Dolphins and Mindbuilders recent discussion, merely to confirm it.

The importance of the formula: Pressure in Pascals = 1000 x 9.8 x Vertical Height in metres confims that pressure relates directly to the vertical height of the water. The size of a reservoir or the inside diameter of a tube, be it the upwards leg or the lower leg, does not change the pressure within. (Putting aside less friction that may occur in a larger tube setup). So in Mindbuilders 8 Siphon Image, the pressure in the reservoir doesn't change, which was also stated by Dolphin.

Dr Hughes argued in a seperate blog that a siphon with a bubble in it only showed that it was possible to prime a siphon using atmospheric pressure, and that once the siphon was primed it would revert to tensile strength. A siphon could be operated where small bubbles of air were continually let into the upward leg, say one every second, and only of short length. The siphon would not only prime, but continually operate, confirming that there was no tensile strength at work when this siphon is operating. What is important, for operating any everyday siphon, especially when we add air bubbles into them, is that the vertical height of the water only on the downward leg must always be greater than the vertical height of the water only on the upward leg.

This ensures that we get to create that reduced pressure at the top of the siphon, allowing pressure difference to do it's thing on the upwards leg.

60.231.25.197 (talk) 00:46, 7 February 2011 (UTC)Moonshine

@Mindbuilder: I agree that the demonstrations in vacuum show the presence of tensile strength at low pressure.
@Moonshine: Welcome back. I don't see anything you have written that I disagree with. Good! One very minor point - when I was drafting my fundamental equations of the siphon I wrote that g was the gravitation constant. Then I realised someone would shoot me down on that point. In the hydrostatic equation g is the local acceleration due to gravity. The gravitational constant does not vary with altitude and latitude. Dolphin (t) 01:17, 7 February 2011 (UTC)
Dolphins last edit looks fine. The opening paragraph looks good to me now. I've refined a bunch of references. Mindbuilder (talk) 01:47, 7 February 2011 (UTC)
I managed to mutilate one of the in-line citations but failed to notice it. Thanks for the repair job. Dolphin (t) 02:02, 7 February 2011 (UTC)

Mindbuilder has made some substantial refinements of the article. The article now uses the expression "over the top of an obstacle" to identify a siphon. I see a couple of problems with this expression. Firstly, there may not be an obstacle or at least not one that is readily visible. Secondly, I can imagine an obstacle that is underwater and affects an apparatus that is not a siphon. As we have discussed on this Talk page, the essential feature of a siphon is that liquid flows up to the crest where the pressure is sub-atmospheric, and then down again. There may not be a reservoir, but if there is, to be a siphon the apparatus must cause liquid to flow above the free surface in the reservoir. The present diagrams all show a reservoir so I suggest the expression "over the top of an obstacle" be replaced by the expression above the liquid in the reservoir. Dolphin (t) 07:13, 7 February 2011 (UTC)

I notice Gravity is no longer mentioned in the opening statement. This is what caused all the fuss and recent discussion when it was absent from the dictionary. Surely it can be included in the opening section, since it is such a key issue.

In the theory section, there is the following statement:

This demonstration may fail if the air bubble is so long that as it travels down the lower leg of the siphon it displaces so much liquid that the column of liquid on the longer lower leg of the siphon is no longer heavier than the column of liquid being pushed up the shorter leg of the siphon.

The word heavier needs replacing. As noted in my previous post, and agreed to by Dolphin, it is the vertical height of the water that is important in each leg, not the weight or diameter of the tube. Even in a single size tube, a low angled upleg versus a straight down downleg may have issues using a heavier description. As noted above, the key issue is the vertical height of the water only in the downleg must always be greater than the vertical height of the water only in the upleg.

Moonshine123.211.191.252 (talk) 10:07, 7 February 2011 (UTC)

@Dolphin - I just removed the clause in parenthesis. I put it in there because I wanted to distinguish between the kind of siphons we've been talking about and all the other devices called siphons that work on different principles. But your comments made me realize that my clause really didn't even achieve the distinction I was aiming for since other siphons like siphon bottles raise liquid above their reservoir as well. I also realized that it probably isn't necessary because all the pictures and context should make it clear enough what kind of siphon we're talking about.
@Moonshine - I'll work on putting gravity in the opening paragraph. Mindbuilder (talk) 19:46, 7 February 2011 (UTC)
@Moonshine - Your observation that it is the height of the liquid in the down side compared to the up side that is important, is an insightful observation. I almost changed the wording of the demonstration to mention the height rather than the weight. But after thinking about it I decided to just remove that section. I originally put in that warning about a possible failure because I was afraid people might try the experiment and have it fail. But now I have Micolich and Richert as references to confirm the experiment works. On the off chance somebody tries it and it doesn't work, we can expalin it to them on the talk page. Mindbuilder (talk) 20:21, 7 February 2011 (UTC)


Under the image of the Chain model, there is the comment: The chain model is a useful but flawed analogy to the operation of a siphon.

Suggest the word useful be changed to common or something else more appropriate.

The chain model isn't really useful since it seems to cause more issues and misunderstandings. As a picture, it is not much different to a tube siphon. So it isn't needed to describe how a siphon looks. And it doesn't provide any better understanding, rather the opposite. By all means, keep it in, just get rid of the word: useful.

Moonshine139.168.183.203 (talk) 23:31, 7 February 2011 (UTC)

I think the chain model is very useful. In fact I originally had it near the top of the theory section as basically an introduction to the theory before Dolphin moved it down, and I think it would be a big improvement to move it back up. I think it is useful because I think it is a very simple analogy that makes it immediately obvious why there is no violation of conservation of energy and makes it clear how the siphon is powered. I think the chain model is a more intuitive, practical, and easier to remember mental model than the more accurate pressure model. When setting up a siphon, I think in terms of the chain model unless I run into a situation where it is not sufficient. I think the chain model makes it easier to see where a siphon would be useful and how it would be used. A lot of people can get along without a more technical understanding of the siphon. I'm not the only person who thinks the chain model is a useful analogy. It is very commonly used in various explanations. It's a good starting point. It's only a shame that it's not at the start of the theory section where it would be even more useful. Mindbuilder (talk) 00:00, 8 February 2011 (UTC)

An article that might be of interest, in reference to water, tension and cohesive strength in trees. Although it is not directly related to siphons, it does reference water tension and the maximum height of trees. Do a Google search for the following:

Adam Summers: 2005: How trees get high: and the limit on their height is set by the force that holds water together

As of yesterday, the good doctor is still holding firm on tensile strength, even after having reviewed the following article. ...www.phys.uhh.hawaii.edu/documents/TPT-final.pdf

He still believes all experiments can be explained to tensile strength. I quote:

When the tube is completely filled with a bubble, the water flowing out of the tube stretches the bubble (it can only do this because of molecular cohesion between the water molecules) reducing the pressure in the bubble to below atmospheric pressure enabling water to be pushed into the siphon. Note that the energy to reduce the pressure in the bubble comes from gravity and not the atmosphere.

Moonshine: 58.166.187.102 (talk) 04:30, 8 February 2011 (UTC)

Here is what I call the air launch siphon, a demonstration of a siphon where the liquid on the up side never touches the liquid on the down side.
Air Launch Siphon: Liquid on left never touches liquid on right. No tensile strength needed.
Since the liquid on the down side merely puddles and runs down, no liquid tensile strength is needed. I think Hughes dismissed this siphon as an oddity. But even if it's an oddity, it still proves that at least in some siphons, liquid tensile strength isn't needed. A practical and less extreme variation of the air launch siphon is the siphons that have T-fittings at the top allowing the attachment of a vacuum pump to prime the siphon. Such siphons often run with a bubble at the top in the T-fitting throughout the siphoning process. Liquid tensile strength can't be effective when there is a bubble at the top, even if the bubble doesn't disconnect the two sides, because if there was pulling going on, the bubble would just expand until there was no more pulling. One situation that Hughes still seems to concentrate on is when the siphon has a bubble half way up the up tube. I guess he imagines that the liquid above the bubble must be in tension pulling up on the bubble. An easy demonstration to dispel that theory is simply to observe a siphon with two bubbles, one half way up the up tube and one at the top. Since there is a bubble at the top there can be no pulling on the slug of water that is in the top half of the up tube, yet the siphon will still work, and the bubble will be expanded just as much as if the slug had been able to pull on it. The idea of the bubble being pulled on and expanded is a blatant misconception anyway. Typical bubbles in siphons are filled with air at something like 90kPa pressure. A gas under such pressure will expand on its own to fill a volume far more than the entire volume of the tube, unless the bubble is being squeezed to keep it small. So if there is a column of liquid above a bubble, the column of liquid is not pulling up on the bubble to expand it, The column of liquid is pushing down firmly on the bubble. Mindbuilder (talk) 07:48, 8 February 2011 (UTC)
The second sentence is tediously long - 82 words. It begans as follows But in the English language today, the word siphon is usually understood to be a reference to the device that is commonly in the form of a tube in an inverted U shape, which is used to cause a liquid to flow uphill, above the surface of the reservoir, ... ...
In the interests of reducing its length I suggest we omit the redundant words and leave it as follows:
In the English language today, the word siphon refers to a tube in an inverted U shape which causes a liquid to flow uphill, above the surface of the reservoir, ...
That will remove 19 words and reduce the sentence to 63. Dolphin (t) 22:31, 8 February 2011 (UTC)
You're right that the sentence is undesirably long. Your proposal looks good. I think we need the word "usually" in your version between "siphon" and "refers". Otherwise some will object that siphon doesn't always refer to that kind of device and isn't always a tube or in an inverted U shape. I think we also need the word "But" to transition from the first sentence. Other possibilities might work, such as the word "However". I was previewing your suggestion and several possibilities, so I went ahead and changed it when it looked right. Mindbuilder (talk) 23:51, 8 February 2011 (UTC)
That is a big improvement! Dolphin (t) 00:45, 9 February 2011 (UTC)

Some extra random thoughts.

@Mindbuilder. The pressure in a bubble within a siphon would only be 90kpa if we were dealing with low height siphon experiments. The pressure could go all the way down to close to zero kpa, depending on the crest height and static fall of the siphon.

The original understanding of a siphon was one of atmospheric pressure and gravity being the key issues, even if the dictionary left out gravity. Hughes proposed gravity and tensile strength. Dolpin was proposing density was the key issue via the comment: My thesis is that the driving force of a siphon is the significant difference in density between the working fluid and the surrounding atmosphere.

Sometimes, it is easy to see the flaws in a proposed theory, example Hughes. It is one thing to see the flaws in their theory, it is another matter again to have them see the error of their ways. One idea to attempt to resolve the differences is to assume for a minute they are correct, put aside your own views, then follow their logic and ideas all the way through to see if it all makes sense, or you find flaws or even a better understanding along the way.

I took this approach with Dolphin's density theory. I looked at the siphon inside the reservoir, and agreed that it stopped working when I had the same density. But why did it stop. Then I identified the pressure outside the siphon outlet when inside the reservoir was much higher than it had been, increased by the extra depth of water. This was cancelling out the pressure difference. Maintaining the same pressures that were present inside the reservoir, and taking the siphon back outside to have different densities didn't restart the siphon. But changing the pressures did.

So the conclusion was that yes, different densities was critical for a siphon to operate, but it wasn't simply a matter of gravity and density on its own being the driving force. As noted, I need pressure difference. And this pressure difference actual comes from the different densities. Outside a pipe, if I have air, the air pressure is really no different at the inlet or the outlet. But inside the pipe, the height difference creates extra pressure at the outlet, and allows the water to flow.

The density difference theory also has 2 more issues. I can place the outlet in water, and the water still flows. So there is a matter to conside that the density in the pipe is the same as the one just outside the pipe. We can discount that because we can argue that there is a different density just above the water of the outlet reservoir.

The 2nd issue is more telling. Dolphin made the comment about water accelerating in a siphon until viscous forces reduce the acceleration to zero. In layman's terms, the steady flow rate we will get will be when friction loss equals the available pressure from the static fall. How do we measure viscous forces or friction? Not in density. If density is a driving force, we don't get less of it if we are measuring what is happening in a pipeline. In addition, if density was the sole key issue, how do we explain water going from the inlet to the top of the siphon, when there is no change to the density. In one view, the upleg is the exact opposite of the downleg: the downleg has the siphon travelling from high density (water) to a low density (air) area. The inlet is going from low density (air) to a high density (water) at the crest of the siphon.

When we measure what is happening in a pipeline, how do we measure it? Pressure. We get less pressure as a result of viscous forces/friction.

So this confirms the current deninition. Gravity and density difference create a pressure difference on the downleg. This pressure difference causes the water to flow, and create a lower pressure area at the top of the siphon. The water at the inlet, which is pressurised to atmospheric pressure, will move from it's current pressure area to the lower pressure area at the top of the siphon.

So you may want to include a little information about the importance of the density difference in creating the pressure difference taht allows the siphon to operate. (I am not suggesting this be in the opening)

I am assumming from Dolphin's support of the updated definition that the truth was a combination of part of their thesis, and the acknowledgement that atmospheric pressure then does get to play a part. Personally, I am not so sure about the "Atmospheric pressure pushes" terminology, even if I advocated this in earlier posts. I am not disagreeing with this, just wondering if there are better words to use. Atmospheric pressure creates in effect a pressurised reservoir of water at the siphon inlet. Gravity and the height difference creates the lower pressure at the siphon crest. The atmopshere doesn't do any more or less work when a siphon starts or operates. The water simply moves from it's higher pressure area inside the pipe inlet, to the lower pressure area at the siphon crest.

Moonshine124.186.146.190 (talk) 10:58, 10 February 2011 (UTC)


@Mindbuilder: Re your much earlier comment: The Pipe Friction Handbook cannot be considered a reliable source on this issue. There are a lot of misconceptions among physicists about the workings of siphons. There is no reason to think that the authors of the Pipe Friction Handbook have carefully studied the issue or are aware of all the evidence. Looking at the evidence, it appears the Pipe Friction Handbook is, strictly speaking, wrong. Though perhaps it is right about most practical siphons.

Where exactly did the authors of the Pipe Friction Handbook get it wrong? Why would it not be considered a reliable source, given the faith that people have put into Hughes paper, who simply put forward ideas without any proof or tests, ideas that have been clearly shown to be wrong? The fact that there are lots of misconceptions among physicists about the workings of siphons says more about physicists than it does about siphons. Haven't you heard the one about if you want to argue with a physicists, you need to be one yourself. The authors have clearly studied the issue relating to practical siphons. It seems strange that you would say they have got it wrong, then conclude with the opposite statement, i.e that it is right about most practical siphons. That is what the handbook was written for, not vacuum siphons. And the information proposed was simple and easy to follow, and the latest siphon definition is based around the information and formulas they listed.

Moonshine124.185.231.160 (talk) 23:12, 10 February 2011 (UTC)

The Pipe Friction Handbook appears, strictly speaking, to be wrong in this quote you gave "Theoretically, the limit of possible flow up the inlet leg of a siphon is reached when the pressure Ps at the highest point in the system is absolute zero" Pressures in a siphon can reach negative values as demonstrated by references 1,2,and3 of our siphon article. Handbooks tend to be practical guides and can't be trusted to reliably settle controversial debates of physics theory. I should have used different phrasing rather than "reliable source" since I didn't mean it in the special way it is used here at Wikipedia. I meant the Pipe Friction Handbook can not be trusted to be expert enough to reliably settle the issue or even really be very persuasive on the issue. To many guides to siphons have not even considered liquid tensile strength or considered experiments that tested the theory that liquids will vaporize at negative pressures. In the future I won't use "reliable source" here at Wikipedia except when I intend the special Wikipedia meaning. The latest version of the opening paragraph uses the caveat "In practical siphons", so I wouldn't say it is based around the information from the Pipe Friction Handbook since the quote above you provided from the handbook doesn't seem to include that significant caveat. Mindbuilder (talk) 00:41, 11 February 2011 (UTC)

@ Mindbuilder: There are 2 types of siphons. Practical Siphons and Experimental Siphons. Practical siphons are those used in everyday life, examples: to empty a dam or fish tank or get petrol out of a car. Experimental siphons are those that involve special circumstances, vacuum, low heights and either special liquids or pretreatment of liquids. Unfortunately, many people use the result or draw their own conclusion from one type of siphon, to justify their thesis on the other. For example if an experimental siphon can be made to work in a vacuum, they then claim atmospheric pressure plays no part in a practical siphon.

You are correct, the Pipe Friction Handbook is a practical guide. You claim that the Pipe Friction Handbook doesn't include the significant caveat "In practical siphons". However, if you reread the original post, the next two words after the claimed wrong statement were, "In Practice". I would suggest that "In Practice" and "In practical siphons" have the very same meaning in this context. In addition, when they say "In theory....." they are not writing about experimental siphons, rather the theory of practical siphons.

The Pipe Friction Handbook details and formula was included because Wikipedia had removed all reference to atmospheric pressure from their opening definition of what makes a practical siphon operate. Formulas were listed by others to to show that the siphon rate of flow had nothing to do with atmospheric pressure, and that the atmospheric pressure was the same at the inlet and outlet and thus cancelled each other out. These issues were already known. The Pipe Friction Handbook formula shows the role that atmospheric pressure plays. It wasn't just a matter of maximum height or of keeping the liquid from boiling. To have water moving in any pipe, which includes siphons, we need energy in the form of pressure to get it moving, pressure to overcome any height involved, pressure to overcome friction losses, plus any pressure required at the end.

My use of the words "based around" were inappropriate. They implied that your new description was built on the Pipe Friction Handbook. What I was saying was that the new definition is using the same principles in relation to atmospheric pressure that the Pipe Friction Handbook documented.

Maybe you need to rethink what is a "reliable source". Wikipedia changed the opening definition of a siphon, removing all reference to atmospheric pressure, based on an article from Dr Hughes. The Pipe Friction Handbook made a valuable contribution to the discussion on practical siphons that Wikipedia was having, and is far more persuasive on the issue than the Hughes article.

Moonshine58.164.132.8 (talk) 02:37, 21 February 2011 (UTC)

I don't recall seeing anything in your quotes of the Pipe Friction Handbook that suggest that the authors even knew of the possibility that any siphons could exceed their barometric height. I don't see any reason to believe that its authors analyzed siphon theory thoroughly enough that its pronouncements can be relied on to settle controversial issues of siphon theory. The Pipe Friction Handbook may well be a fine and generally very accurate book, and it may be much more reliable than the Hughes article, but just because it's better than the Hughes article, doesn't mean it can settle our siphon theory issues. And without evidence that it was written with these tricky issues in mind, I don't think it can really even contribute to the debate on the tricky issues. I probably won't have time to discuss the reliability of the PFH any more. However, feel free to invoke the authority of the PFH to support any particular point you want to make. But don't necessarily expect that authority to be persuasive. Mindbuilder (talk) 07:48, 21 February 2011 (UTC)

I would just like to comment on the definition of the siphon (as at 18 june 2011) : I like it! Well done! (Although I can't help myself; would like to see "The reduced pressure is caused by weight and flow of the liquid on the exit side." (not the fall itself) — Preceding unsigned comment added by Paulrho (talk • contribs) 22:51, 17 June 2011 (UTC)

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