What are the forces and pressure involved in causing siphon action?

[R Power]

Simply, what causes siphone action ?

Please don't tell in terms of energy that water tends remain at lower energy so it moves down.

Tell in terms of forces and pressure.
If you say in terms of energy then answer why water would increase it's energy when it rises up in siphone tube before going down in lower container?

I can't find any solid reason on net!

 

[Borek]

In a way this is identical to communicating vessels - they tend to equal level. That's the same process, just now the force acting is sucking the liquid - draw reversed U shaped tube and draw forces acting on liquid, it should become obvious.

 

[R Power]

If the reason were same as that of communicating vessels then siphone effect should work also when both vessels are at same level. But is doesn't. Can u expain?
Secondly,in communicating vessels water flow from one vessel to other along a passage which is horizontal and mainly below the maximum height of water in vessels. But in siphone water flows up in the pipe first doing work against gravity. This is where the problem lies. What makes it do that? The pressure difference between atmospheric pressure and water pressure is not enough to make the water do work against gravity because if it were then water would never stay inside any container, it would just evaporate at atmospheric pressure but it doesn't. So makes it move up in case of siphon?

One expalanation i read on net was that:
Initially gravity pulls the water which is beyond the u shape in the pipe. Now when this water moves down it is creating vacuum where it was earlier in the siphon pipe. So the water behind gets pulled in the vacuum and in this way a continuous flow is seen. But if this were the true reason then then siphon effect should be seen when the vessels are at same level too? But it is not.
It's not that simple. Is it?

 

[R Power]

If the reason were same as that of communicating vessels then siphone effect should work also when both vessels are at same level. But is doesn't. Can u expain?
Secondly,in communicating vessels water flow from one vessel to other along a passage which is horizontal and mainly below the maximum height of water in vessels. But in siphone water flows up in the pipe first doing work against gravity. This is where the problem lies. What makes it do that? The pressure difference between atmospheric pressure and water pressure is not enough to make the water do work against gravity because if it were then water would never stay inside any container, it would just evaporate at atmospheric pressure but it doesn't. So makes it move up in case of siphon?

One expalanation i read on net was that:
Initially gravity pulls the water which is beyond the u shape in the pipe. Now when this water moves down it is creating vacuum where it was earlier in the siphon pipe. So the water behind gets pulled in the vacuum and in this way a continuous flow is seen. But if this were the true reason then then siphon effect should be seen when the vessels are at same level too? But it is not.
It's not that simple. Is it?

 

[Borek]

I think the point that you are missing is that before siphone can act it has to be filled with water - which can mean putting energy into system.

Later pressure difference between both sides is enough to make it work on its own, whatever work has to be done to pull liquid up is recovered when the same liquid goes down on the other side. It is not that the work is FIRST done against gravity, as when liquid in one arm goes up same amount of liquid in other arm goes down, it happens simultaneously.

For the liquid in U shaped tube to be in equilibrium pressures in both arms have to be identical, doesn't matter if they are connected below or above some reference level.

 

[rcgldr]

One way to look at this is to realize that the water is nearly incompressable, and the water in the utube behaves similar to a rope on a frictionless pulley with one side of the rope hanging lower than the other, so that the center of mass is off to one side of the pulley, causing the rope to "fall" towards the lower side. In the same manner, the column of water in the inverted utube has it's center of mass offset towards the longer side, so it also "falls" down the longer side, but the flow is opposed by friction and viscosity and the water in the two tanks, so the flow reaches a terminal velocity.

Another way is to consider the initial condition of an inverted utube filled with water and capped off. The pressure at any point within the inverted utube is lowest at the top of the utube, and increases with vertical distance from the top of the utube, so the pressure is greatest on the lowest end of the inverted utube. The capped ends of the utube rest below the surface of the water in two tanks at different heights. To simplify this, assume that the ends are equally deep in both tanks, so that the pressure just outside the ends of the capped tube is the same.

The pressure inside the caps is greater than the pressure outside the caps, so once both caps are removed, water initially tries to flow downwards from both ends of the inverted utube, reducing the pressure in the inverted utube, with virtually no expansion of the water. The pressure difference is greater on the lower tank, resulting in more of a pressure drop on the lower tank side of the inverted utube. This results in the pressure in the longer side of the inverted utube at the same height at the opening into the higher tank being lower than the pressure at the opening of the higher tank, so the pressure differential causes the water to then accelerate towards the lower tank, until wall friction in the tube, viscosity, and the water in the tanks, causes the flow to reach a terminal velocity.

 

[R Power]

The reason for pressure difference is logical (that when water from one side falls due to gravity then it creates a partial vacuum as it falls down, so water behind the vacuum flows and a continuous flow is obtained) but problem is that why it happens only when the open end of the siphon tube is at certain distance lower than the bottom of water vessel?
If u say it's because the other part is then more longer so gravity has more effect there, then consider: the siphontube's one end which is in water is above in height than the open end(so water in longer part will still be heavier) but now the free end is above the bottom level of the water vessel. Will still siphon effect be seen?

 

[Borek]

why it happens only when the open end of the siphon tube is at certain distance lower than the bottom of water vessel?

Can you post a drawing? Eithe I have troubles grasping what you mean or you are wrong.

 

[R Power]

In the figure , if h2=0, will the siphon action take place?
because i read on net that the open end of siphon should be lower than the upper vessel containing water. h2 is difference between their heights.

According to me only h1 should be required for siphon action to take place! but I read the above thing on net.

 

[Borek]

h2 is completely irrelevant, however, h1 is not that important as well. h1 would be important if there will be no h3, that is, height of the liquid above siphon end. h3 creates pressure at the submerged upper siphon end, that cancels out "negative" pressure of the h3 liquid "hanging" inside the right arm of the siphon.

It is h4 that is a driving force behind liquid motion.

 

[R Power]

but in every video i saw on you tube or any image i see, the second vessel is taken below the the upper vessel(i.e h2 exists). Why?They could also take both vessels at same level(h2=0) and could still maintain h4.Why not?

 

[Borek]

They could and it would still work, perhaps it is kept this way just to make it look more dramatic.

On all three pictures below siphone will work (assuming it was initially started) - although the end situation will be different.

 

[R Power]

I think I've understood now. So, it's gravity plus pressure difference which makes it possible. Am i right?

 

[Borek]

I would say it is pressure difference due to gravity, not one plus the other.

 

[R Power]

consider h4 negative i.e the open end is above the surface of water in first or upper vessel but still first end is under the water in first vessel at a considerable distance where pressure is enough, then what would happen?

 

[russ_watters]

consider h4 negative i.e the open end is above the surface of water in first or upper vessel but still first end is under the water in first vessel at a considerable distance where pressure is enough, then what would happen?

Apply what you learned in the last sentence of post #10 and you tell us!

Follow-up: It seems you think the depth of the bottom of the hose matters. What happens when you put a straw in water? What is the height of the column of water in the straw? What happens to that height when you push the straw further into the water? ...So why would you think that it would have any effect on a siphon?

Remember, (from that last sentence of post #10): water or any fluid flows from an area of high pressure to an area of low pressure. So to calculate which direction (or if) water will flow, calculate the pressure difference between to points in the siphon. Perhaps even use two points right next to each other at the top of the siphon: calculate the pressure on the left side and the pressure on the right side and compare them.

 

[rcgldr]

On all three pictures below siphon will work ... (left image shows exposed end of hose)

The left image case won't work if sufficient air flows upwards through the left side of the tube. This could happen depending on tube diameter and relative height between the left end of the tube and the water level on the right.

 

[R Power]

Apply what you learned in the last sentence of post #10 and you tell us!

Follow-up: It seems you think the depth of the bottom of the hose matters. What happens when you put a straw in water? What is the height of the column of water in the straw? What happens to that height when you push the straw further into the water? ...So why would you think that it would have any effect on a siphon?

Remember, (from that last sentence of post #10): water or any fluid flows from an area of high pressure to an area of low pressure. So to calculate which direction (or if) water will flow, calculate the pressure difference between to points in the siphon. Perhaps even use two points right next to each other at the top of the siphon: calculate the pressure on the left side and the pressure on the right side and compare them.

But just due to pressure difference why would water work against gravity initially?

 

[Borek]

The left image case won't work if sufficient air flows upwards through the left side of the tube. This could happen depending on tube diameter and relative height between the left end of the tube and the water level on the right.

Yes, but we are talking about idealized case and the pressure difference being the driving force.

We could also say that if the siphon is high enough neither case will work as there will be a vacuum near the (reversed) U top. Also, with correct selection of surface tension, contact angle and tube diameter we can have a hanging drop at the end of the tube. It can all happen, but it doesn't falsify basic principles which the OP has trouble to understand.

 

[Borek]

But just due to pressure difference why would water work against gravity initially?

Pressure difference means - by the very definition of pressure - that there is a force applied. This force works against gravity.

Try to calculate pressures assuming liquid is not moving, just a static initial situation. What forces act of water? What must happen to water in response to these forces?

 

[R Power]

i understand the concept of pressure and forces.
If one vessel is connected to other by a pipe horizontal and below the water level of the top vessel e.g simply a pipe connected between two reservoirs as in figure attached, then it's obvious that water will flow due to pressure difference but here it's inverted u tube and the U is well above water level.
If you were correct that water flows just due to pressure difference then there should be no need to start a siphon. Just immerse siphon pipe into a vessel , atmospheric pressure will act on other free end of it and water should flow as if simply you connect one empty vessel with a filled one with a straight horizontal pipe whhich is connected below the water level in filled vessel as in figure.

If you were correct water left open in a container would go into atmosphere immediately because pressure in container would be more than atmospheric pressure. Then why water stays there?
It's because gravity holds it and prevents it to do work against it.

 

[Borek]

If one vessel is connected to other by a pipe horizontal and below the water level of the top vessel e.g simply a pipe connected between two reservoirs as in figure attached, then it's obvious that water will flow due to pressure difference but here it's inverted u tube and the U is well above water level.

It doesn't matter that U is above the water level, weight of the liquid in both arms cancels out. Starting the siphon means just preparing this part of the tube that contains incompressible liquid weight of which cancels out. This part may require work (although technically, in gravitational field, work is zero). After the tube is filled liquid flows on its own.

But I give up. Siphons don't work. Obviously there is some misconception in your understanding of what is going on, but I have no idea what it is and how to find it out

 

[russ_watters]

But just due to pressure difference why would water work against gravity initially?

The pressure difference is CAUSED BY gravity and the siphon starts full so there is no net work against gravity. You have to treat the siphon as a whole system to see why an apparent contradiction isn't one. And you need to follow a more rigorous process here because you aren't retaining what you have learned: Go back to the diagram and apply what you have learned to it. It will reveal the answer.

 

[rcgldr]

But just due to pressure difference why would water work against gravity initially?

Initially it wouldn't. Assuming the ends of the utube were capped, and then uncapped, then the initial reaction would be some outflow of water in the utbue, and an increase in tension within the water within the utube, since water is nearly incompressable (or non-expandable), the tension of the water in the utube would increase, reducing the pressure of the water everywhere within the utube.

Again, using the unequal lengths of rope hanging over a frictionless pulley analogy, the water in the utube would "fall" towards the end in the lower tank.

At the higher tank end of the utube, the tension would reduce the pressure to below the pressure of the water in the higher tank, and the water would then start accelerating through the utube until wall friction and viscosity caused the flow to reach a terminal velocity.

The initial tension reaction would proably occur in milliseconds. Reaching terminal flow speed would take relatively longer.

 

[ideasrule]

I haven't read the whole thread, but here's my analysis of a siphon. Suppose that the siphon is an inverted U, with two straight tubes and a very small bend at the top.

Gravity acts on the water in the lower leg of the U. It applies a force F1 to the section of water in the lower leg. This force is transmitted through the water because water is incompressible, in the same way that a steel bar transmits force from one end to the other. In order for the water to not move, there has to be a balancing force dragging the water up. Otherwise, the water would accelerate.

Gravity pulls on the water in the higher side with a force F2. F2 is smaller than F1 because the higher leg has less water, so unless there's another force dragging the water down, the water would accelerate. There's no other force, so the water in the higher side does accelerate upwards, and the water in the lower side does accelerate downwards. Water is incompressible, so more water has to enter, or else a gap would be created in the water. F2-F1=dp/dt, so when the force difference starts pulling in water at a rate of dp/dt, the water stops accelerating and moves at a constant speed throughout the siphon.