Gravity Mill - any comments to this idea?

[hartiberlin]

hi stefan,

i just read my post and thought i might have confused you.  you are right, you do need 2 bar at your compressor for your example.  what you don't need is that much volume.  if you reduced it by half and reduce exit tube diameter above water level, you can still get the same amount of water, but not as fast.

tbird

TBird,
due to the hydrostatic paradoxon we need to keep attention how much water
we have in the tube above sealevel !
If you make this watercolumn too high, also with smaller diameter, it does not matter ,
you need more volume of air down there at 10 Meters deepth !
Reread the Hydrostatic paradoxon...

[tbird]

i need to respond here, but not to everthing.

The volume of the air we have to bring down there depends on the height
of the water cyliner above sealevel !
If we have 1 Meter, we also must have a 1 Meter cylinder of air
under water at 10 Meters deepth.

this is causing you a big problem.  let's think about what is going on.  the reason the water goes up is there is force below.  if you can move 10 pounds of water, it doesn't matter if it's in a 1 by 10 (any measurement) or 10 by 1 area.

see how you can raise the same water higher without changing volume below?

[hartiberlin]

TBird,
I only agree to the water under sealevel.
The water above sealevel puts weight onto the water column
and you have to pay attention to the hydrostatic paradoxon.

[hartiberlin]

Okay, I just found this:

Formula:
P*V=n*R*T => P=n*R*T/V

Isotherm:
T=const => p*v=const

dW=F*dl with  F=P*A
dW=P*A*dl
dW=P*dV
dW=n*R*T*dV/V

Integration yields:

W=n*R*T*ln(V2/V1)
W=P1*V1*ln(V2/V1)

with:
1 bar = 100 kP= 15,4 psi
1 Liter= 1/1000 m^3

We really get a thermodynamic theory guy help us with this.
In the above formula I have missing the pressure P2.

If we take the formula:
W=P1*V1*ln(V2/V1)

W= work-energy, we have to put into compressing the air.This I don?t know yet !

P1= starting air pressure of 1 bar outdoors.
V1= volume of starting air, this I don?t know..
V2= Volume of air compressed at P2= 2 bar = 785,4 Liter


But where can I put P2 ?
If I had a formula including P2 then I could play with it,
as long raise some input energy W until I get to
P2= 2 bar and V2= 785,4 Liter.

Can anybody please help ?
Thanks.

[tbird]

think about it this way.  what stops the piston (ballon, whatever) from rising? either it weighs more than it is displacing or in our case in the tube, a larger weight above water level than the piston is displacing.  right?  so it doesn't matter if it is all placed in the first meter or divided in 2 and placed on top of each other.  both shapes weight the same, but 1 is only half as wide as the other.  so as long as the volume of water in the tube above water level weighs less than the piston is displacing, the piston will rise.  your test was a little awkward for you   but if you could actually put a smaller tube on top at water level, the water will come out higher.  like getting a flat tire on your car, the smaller the hole, the longer it takes to go flat, but it will go flat.