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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)
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.
Moonshine 121.223.82.43 ( talk) 23:41, 29 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?
Moonshine 124.187.86.46 ( talk) 05:24, 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)
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 ~. Moonshine 124.177.125.24 ( talk) 03:21, 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
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.
Moonshine
124.177.125.24 (
talk)
05:49, 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) Moonshine 124.187.7.65 ( talk) 10:35, 2 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.
Moonshine 124.187.108.158 ( talk) 23:26, 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!
Moonshine 124.187.108.158 ( talk) 23:58, 2 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.
Moonshine 124.187.108.158 ( talk) 00:26, 3 February 2011 (UTC)
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)
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
@Dolphin - Here is a diagram of the no-reservoir siphon with a bubble in the top that I have been referring to.
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)
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)
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)
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)
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 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.
Moonshine 123.211.191.252 ( talk) 10:07, 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.
Moonshine 139.168.183.203 ( talk) 23:31, 7 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)
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.
Moonshine 124.186.146.190 ( 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.
Moonshine 124.185.231.160 ( talk) 23:12, 10 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.
Moonshine 58.164.132.8 ( talk) 02:37, 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)