# Heron fountain antics

1. May 7, 2013

### charizzardd

Let me start off by saying I know it is not perpetual motion. I understand this. I was thinking about the fountain and was wondering if this arrangement would allow you to equalize the compartments and begin the process again, without having to flip the water compartments or pumping it up to the top cylinder again. You would have to open and close some valves...

The system is set up as in the image...
Here is a diagram of a classic design of the fountain. The design I am proposing is similar but the tops of the two cylinders are at the same height, the bottom of one is just lower created the head difference. Adding a valve in between the two to allow for equalization.

Two side by side cylinders with a ball valve in between. As the shorter tube pushes water up the fountain, the water level decreases while the longer tubes water level increases. At some point the flow stops, or you close the outer ball valves in the lines going up to the fountain part. Then opening the middle ball valve would equalize the two cylinders. Upon closing the middle valve again and releasing water from the fountain, the process can start again...

The trick here being that the cylinder on the right is longer than the one on the left. The 'D' is the right cylinder distance below the left cylinder. 'd' is the distance the left cylinder output is higher than the right cylinder input. As long as 'd' is less than 'D' by enough distance to overcome head losses, the right cylinder water column will always be pushing against the one on the left through air pressure (connected at the top of the cylinders. When closing the outermost valves and opening the middle, the cylinders equalize the water height. Then when you close the middle and again open the outside valves, the flow should start again...

What do you think? Will closing the outer valves create some kind of vacuum preventing the water columns from equalizing? Sorry if my paragraph is repetitive...

2. May 7, 2013

### Simon Bridge

To reset the fountain, you have to empty the air chamber of water, and add water to the water chamber. How does your system do this?

In normal operation, a water-pipe delivers water from a reservoir basin to the air chamber. This displaces air through an air pipe into the top of the water chamber... building pressure to displace the water up the fountain tube and back into the reservoir.

In your design, it looks like the air chamber is very big compared with the water chamber.
You have a valve on the fountain tube and the water pipe.
The pipes are poorly drawn - you should draw in the initial or early-operation water levels and draw the pipes their full lengths. eg. It looks like the pipe joining the tops of the chambers is supposed to be the air pipe and there is another one joining the chamber half way down - if the water level is higher than this second pipe, then it will transfer water, otherwise it transfers air ... but you already have an air pipe?

The structure should be quite easy to build using plastic bottles and rubber hose.
Instead of a ball valve, you could use a clamp (clothes peg, locking wrench, pliers) over a short length of hose. Why not build it and see?

Last edited: May 7, 2013
3. May 7, 2013

### charizzardd

The cylinder on the right fills up, the one on the left empties. The cylinder on the right will continue to raise above the level of the water in the cylinder on the left because of the total head created by the outer tube...

Opening the valve in between lowers the water on the right and raises the water in the left cylinder to some equilibrium where the height if both cylinders is the same.

The process can start again because the cylinder on the right is taller and therefore has more head to work with than the cylinder on the left. Because of the arrangement this is always the case

4. May 7, 2013

### charizzardd

Actually. I think moving the outer ball valves to the bottom of the cylinders makes the most sense. This way, when you close those you effectively cut off the rest of the system from the two cylinders and opening the valve between them would force the two cylinders to an equilibrium height. You then close the middle valve and open the side ones releasing the right column of water starting the process all over again

5. May 7, 2013

### Simon Bridge

So the idea of the extra tube is to let water flow from the air chamber to the water chamber ... that wouldn't that have to be uphill, or the fountain won't work? (When you open the connecting pipe, you need the water level in the air chamber to be above that in the water chamber right?)

The idea is that you stop the fountain, twiddle some valves, and conditions are right to restart?
... that would make a perpetual motion machine since automating the valve-fiddling off the energy of the fountain is an engineering task.
i.e. where does the water get it's energy from?

I still think you need to redraw your diagram showing where the tubes go and the water levels.
But Like I said - this would be inexpensive and easy to build. So just do it ;)

6. May 7, 2013

### Simon Bridge

Lets see if I can be clearer:
Consider the pic:

Here, the lower chamber is the "air chamber", the upper chamber is the "water chamber", the large dish on top is the "basin", the pipe to the left filled with blue is the "water pipe" (includes red and grey sections), the one on the right that looks empty is the "air pipe" and the middle red one with the fountain is, well, the "fountain tube".

The operation is basically that of a siphon. Water is siphoned from the water chamber to the air chamber. You should already have experience of siphons - what are the conditions under which siphoning will work? If you have not had experience with siphoning - get some. I cannot stress too much how important it is to perform these experiments when you get the opportunity.

Resetting the system involves moving water from the air chamber to the water chamber.

Your diagram puts a tall air chamber next to the water chamber (on a lower step so the bottom starts out lower). You have valves to isolate the chambers from the rest of the system. You have an extra pipe communicating between the chambers.

If the system is configured so that, when the water chamber is empty, the air chamber is filled above the level of the communicating tube, then, isolating the chambers and opening the communicating tube should allow water to flow from the air chamber to the water chamber, allowing the process to restart - is that what you were thinking of?

Anyway - it is beginning to sound like this thread falls outside the guidelines for this sort of discussion. In the event it gets locked, you will probably find people to explain things to you at the JREF Forums.

7. May 8, 2013

### charizzardd

First off, sorry my picture is terrible it was just a quick sketch.

Resetting the system involves moving water from the air chamber to the water chamber.

Your diagram puts a tall air chamber next to the water chamber (on a lower step so the bottom starts out lower). You have valves to isolate the chambers from the rest of the system. You have an extra pipe communicating between the chambers. You got it!

If the system is configured so that, when the water chamber is empty, the air chamber is filled above the level of the communicating tube, then, isolating the chambers and opening the communicating tube should allow water to flow from the air chamber to the water chamber, allowing the process to restart - is that what you were thinking of?

I also understand this will mean the water chamber wi never be higher than the air chamber as they will only equalize.
Also you are correct the air communicating pressure tube is the small one at the top of the two chambers

Forget resetting the fountain for a second and consider the arrangement. I guess the main question is whether the fountain would run at all with the short tall chamber scenario. If yes, then I'd like to try and make this thing!

Also, hope it doesn't get loved, I know we are discussing other things, but my intention wasn't to start a perpetual motion discussion, more so about the classical pneumatic and hydraulic operation of this potential modified heron fountain. If it does get removed I would love to continue the conversation in the correct place if this is not it

I do have some experience messing around with siphons, and a pretty decent understanding of pressure and fluid dynamics as well (Mechy), But its been a couple years. Sometimes little things throw me off, but in designing this in my head I was confusing myself about the physics of it. If both chambers are connected to longer tubes then the longer tubes are what dictates the pressure difference in the air and water working chambers.

You would also have to change the shape if the chambers to maximize the volume to contain a lot of working fluid in the equalization area so the pump can run for a while because when you equalize the two, you can only pump until the air chamber is full...

This sort of thing is so much easier to explain not in text. I'll try to come up with a better image but at least after your last post I think you understand the operation I am suggesting!

Thanks for the feedback!

Last edited: May 8, 2013
8. May 8, 2013

### Simon Bridge

Please use quote tags or something so it is easier to tell which is quoted text and which is your reply.
I'll try and fish out the questions but I may miss some.
Well yes it could - just like a siphon will work with a short-tall bucket scenario, and under the same conditions. The size of the chamber does not matter... much. iirc the ratio of volumes is important for the pneumatic coupling between the two parts of the siphon.

Oh don't say that: we love all the posts here :)
Note: figuring if some design concept could be turned into a perpetual motion machine is a valid method for reality-checking it. If it can be, then we've misunderstood the operating principles we are trying to exploit.

In this case, if the fountain could be reset with no more work than the energy of the falling part of the fountain, then it's PMM and impossible.