Air-powered CPS without a regulator

[Drenchenator]

Before I begin, I should say that I do not necessary understand the problems that some are having. I have been reading this thread from its creation and thought that Ben's theory was correct, and I still do.

We should start from what is agreed upon. I believe that we can agree that air is compressed in a chamber of some sort, and that this air is compressed at a pressure above atmospheric pressure. Since Boyle's Law has been proven true, we can assume that the pressure will drop as water is removed from the pressure chamber.

Pressure, by definition, is a force per unit of area. If you break the units of PSI up, you get pounds (force) per square inch (area). A dictionary would define pressure as

Force applied uniformly over a surface

So, now we can assume that the pressure of the air is exerting a force on the surface of the water. The surface of the water would be where the air and water meet in this case, in the chamber. Since pressure is a force per area, we can assume that force is pressure by area. The normal force of pressure is defined as the pressure by the area of the surface it is applied to.

As I have stated earlier, the pressure will definitely drop. But what if we wanted the same force applied to the water through the shot? A CAP soaker (Constant Air Pressure) uses a device to keep the same pressure in the chamber, and since the applied area does not change, the force does not either.

Since the pressure will drop without our control and since F=AP, we would have to vary the surface area that the pressure is applied to to keep a constant force. This is the basis of this "naturally" regulated thing. Though it does not have constant pressure, the force will remain constant, which is just as good.

[Silence]

Originally Posted by joannaardway

Pressure must be constant throughout the soaker.

Any pressure "peaks" will almost instantly "smooth" so that the soaker is isobaric.

If you can find a way of getting a 16 bar jet from a soaker pressurized to 4 bar, you've just solved all the world's energy problems - energy from nothing.

ENERGY IS CONSERVED! (except in energy-mass conversion, but unless your soaker contains a nuclear reactor, we're fine there)

Energy = force * distance.

If we suddenly assume that the mystical sideways chamber idea is correct and force on the pump is less then...

Assuming the same fill, distance must be the same for the same number of pumps?

If we have a 50 newton average over 10 metres, then 500 joules have been stored.

If we have a 100 newton force for the "other chamber", then 1000 joules have been stored.

How come the other chamber has more energy stored? This would make it less efficent for use? How exactly can this even vaguely make sense?

How much physics do I need to shoot across these forums before I can sink this theory for good?

Should I bring out mass-flow computions:

flow = flow co-effiecent * sqrt((pressure in chamber squared - air pressure squared)/temperature)

The flow co-efficent is dependant on NOZZLE area, not chamber area.

And does force even come into that calculation? - NO!

I know this theory is completely flawed - and I will drag out every equation I can to prove it.

EDIT/Sanity revert: This is a post that I would slay other members on sight for, but I can get very, very persistant and aggresive when other people apparantly refuse to at least consider my theory. Sorry, but I've expained my reasons for this before - thus I am leaving the post as it began.

While I definitely support the idea, and while I don't see why it should be impossible, I still don't think my scenario is possible--and hopefully, the original scenario will still be proved possible. joannaardway just summed up my thoughts in a much clearer way than I could, and it indeed doesn't take much to see that you can't get force out of nothing.

However, as Drenchenator says, we should certainly start from what is agreed upon, and the original theory and its basis seems like solid logic to me.

[insanitys_engineer]

Joanna, I have no grasp of fluid dynamics as yet, so feel free to correct mercilessly if the following doesn't follow (so to speak)

When the valve is opened to fire, you have 2 faces of water, connected by a piston of water: the lil nozzle water surface, and the phat PC water surface.

Pressure = Force/Area
The force from the air in the PC will act against the air at the nozzle, because of the non-compressibility of water. lets take an example

20mm diameter nozzle
200mm diameter tank(theyre nice round numbers)
45psi= 300KpA (100 000 pascals =1 bar)

In the tank the pressure is exerting a force of 300k x 0.03m² = 9000N
At the nozzle, the same force must be observed, as the rest of the gun is in equilibrium.
Nozzle: pressure = 9000/.0003m² = 30 000 KpA = 30 000 KpA

Therefore seeing a very powerful jet (100 x the original pressure in this case)

This will be limited by the nozzle and ID, but shows how we can get such power out of a watergun. Multi-PC designs look more promising to me now.

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As for the tank energy creator machine, I can only think that power =force x distnace isnt the only thing in play: to double the cross sectional area, the height halves, therefore the Gravitational potential energy gained by water is also halved. Could this be where the phantom energy of differing tank sizes appears? I'm stumped.

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G'night!

[joannaardway]

Things I can agree on:
- Gravitiatonal energy should be disregarded. - It's 1.5 psi pressure equivalent per metre of water, and we are dealing with 20 centimetres at most - the pressure is negliable.

- Standard pressure and flow calculations - but not necessarily the uses that are being made of them.

Problems I can see:
Water is compressible, but volume wise, the change is miniscule - thus the fact that it compresses can be disregarded for maths, not necessarily theory.

Fluids carry pressure, solids carry force - due to having their own shape, rather than that of the container.

Imagine the following experiment.

If I have a pipe with a metal bar running through it. On each end there is a metal plate - one larger than the other.

If I have a person on each end shoving on the plates, does the size of the plates matter? - no, because force is carried by the metal bar - not pressure.

If I have the same pipe with syringe type things on each end and water in the middle... If one syringe has a larger ID, does this matter?

Yes, - the larger ID will spread the force the user exerts. The force is turned into pressure. Thus the user with the smaller syringe will be able to push the other user's out with theirs.

Force only exists on the division between water and air. After that, the fluid converts it back to pressure.

Force cannot be carried by fluid.

@ Insanitys engineer - Firstly, the force cannot be carried, and secondly, pressure pushes equally on all points.

The equal and opposite force countering it is mostly countered by the bottom of the chamber, rather than being channelled along the piping to the nozzle.