This Professor Physics shows that "The Bernoulli Effect" is a misunderstanding:

http://muweb.millersv.edu/~jdooley/macro/macrohyp/eulerap/eulap.htm

Down the page Professor Dooley writes:

"One confusing detail about Bernoulli's equation needs discussion: In
elementary treatments it is often stated that, because of Bernoulli's
equation, high velocity causes low pressure. This is like saying that
the high velocity of bullet leaving a gun caused the low pressure of
the gas outside the barrel of the gun. The field treatment embodied in
Euler's and Bernoulli's equations does not discuss cause and effect.
The field view tells us what parameters "go together" without implying
that one causes the other. If we trace back to the roots of these
equations in Newton's laws, we can extract a cause and effect
statement: Forces are said to cause accelerations. In the same sense,
pressure gradients cause changes in velocity; not the other way
around."

Jan-Olov Newborg

This would be fairly simple to set up and test, which would give a baseline
from which to extrapolate other diameters.
What do you do about the changing ambient pressure, or are you going to feed
this device with a 1st stage regulator?
Constant flow or demand?

There are a number of formulae for calculating gas or liquid flow through
orifice of varying diameters, but these are really
only jumping off points, as I am sure you know.

A mixing tool from Airway, Cyprus, was shown some years ago which should
mix from HP bottles based on the same principle.

It incorporated a sort of Vortex mixing chamber, and a pp-meter for
Oxygen.

Matthias

"mike gray, CID" schrieb:

Hollow gold beads?   How do they make hollow chocolate easter eggs??? Same
way...

rhys

My training in these matters dates back a long time but I will try.

I assume that this is essentially based on Bernouilli s formula as
applied to "perfect fluids" (non-viscous fluids, not sure about the
proper wording in English).

For the application you describe, this should be a correct
approximation. Viscosity will become an issue as pressure increases so
that might explain why no reference is made to Bernouilli s law when
high-pressure gases were used.

I also understand that you can neglect the impact of gases
compressibility for mach numbers below 0.3 which should also be the
case for the application you describe.

I assume that no significant differences in elevation exist in the
circuit you describe.

So the law states that for a steady flow of gas in a tube, at constant
elevation:

static pressure + 1/2 x density x velocity2 = total pressure =
constant

In the tank you may assume that velocity is zero, thus the constant is
know to be equal to the tank s pressure. So we have :

static pressure + 1/2 x density x velocity2  = tank pressure

At the exit point of the tube pressure is whatever it is, but from
your example I assume it is known. So we have :

exit pressure + 1/2 x exit density x velocity2 = tank pressure

exit density x velocity2 = 2 x (tank pressure - exit pressure)

velocity2 = 2 x (tank pressure - exit pressure) / exit density

If I recal correctly density is proportional to pressure, at a given
temperature (we assume a perfect gas) :

exit density = constant1 x exit pressure

velocity2 = 2 x (tank pressure - exit pressure) / (constant1 x exit
pressure)

For a given set of tank pressure and exit pressure, this is a
constant.
If the exit pressure changes but remains low compared to the tank's
pressure, this will be about equal to :

velocity2 = 2 x tank pressure / (constant1 x exit pressure)

velocity = square root of : 2 x constant1 x tank pressure / exit
pressure

So we have a direct relationship (non-linear) between velocity and
pressure ratios

And flow = area x velocity (expressed as a volume per unit of time)

So for a given gas :

flow = area x square root of : 2 x constant1 x tank pressure / exit
pressure

So you can control the flow of each gas, using either tank pressure or
tube diameters of each gas circuit.

Limitations will come from changes in tank pressure as it empties (I
imagine that some people must have tried to link manifolds - area - to
pressure gauges), from the fact that the exit pressure is not actually
constant when you dive, and may not be that small relative to the
tanks pressure.

Sorry if what I wrote is trivial or off-subject. And I may well have
made more than a few mistakes (I am re-discovering it as I write).

I could post some links but they would be in French. I am sure you
have researched that on your side as well.

Hope this helps,

Froggy