Gravity Mill - any comments to this idea?

[hartiberlin]

I just saw, the air outside has 1 bar of pressure, so in 10 Meters deep
we need then 2 bar of pressure. this means we need to produce
about 29 psi at 10 Meters deepth
from the 21,4 Watthours we have at the reservoir of water.

Can we do this with the amount of energy ?

[tbird]

hi stefan,

when you get going, you really get going.  i can't even type that fast, much less think and type.  i may miss a thing or two, but here's a little.

you didn't mention the weight of the balloon.  if it's weight is not worth mentioning, then you only have to create enough volume to fill the 1 meter space you described.  once the pressure from the compressor reaches 1 bar, the ballon will start to expand.  it will stay at that pressure until the ballon can't expand any more.  then if more is pumped in, the pressure will rise.  but once the volume is made, no more needs to be pumped.  since the balloon has (basicly) no weight, it will be on its way.  once it reaches the top, with the outside pressure being reduced, the balloon , if allowed, will double in size,  if not allowed to expand, the pressure will be at 2 bar.  does that make sense to you?

also, it will be more productive (more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g  x heightdifference) if you raise the reservoir.  by reducing the exit pipe diameter to 1/2 the volume starting at water level, you can raise the reservoir twice as high.  according to your formula, this doubles "potential-Energy"

are you still with me or do we need more explaning (research)?

tbird

[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

[tbird]

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

[hartiberlin]

hi stefan,

when you get going, you really get going.  i can't even type that fast, much less think and type.  i may miss a thing or two, but here's a little.

Okay, no problem. Yes,I am eager to see, if we can nail it down
also mathematically and thus prove it to every physics teacher, that
this concept just works !

you didn't mention the weight of the balloon.  if it's weight is not worth mentioning, then you only have to create enough volume to fill the 1 meter space you described.  once the pressure from the compressor reaches 1 bar, the ballon will start to expand.  it will stay at that pressure until the ballon can't expand any more.  then if more is pumped in, the pressure will rise.  but once the volume is made, no more needs to be pumped.  since the balloon has (basicly) no weight, it will be on its way.  once it reaches the top, with the outside pressure being reduced, the balloon , if allowed, will double in size,  if not allowed to expand, the pressure will be at 2 bar.  does that make sense to you?

Okay, yes, the ballon will expand ,
but see it more like this:
maybe we should just use a plastic cylinder case, which has a valve at the top and is open at the button.

So it can be filled up at the top via a hose and via the compressed air, so the compressed air is filling up the
water containing plastic cylinder, which weights itsself not very much, so its own weight could
be left out of the calculation.
This plastic cylinder valve is then at the top level made open the valve,
so all aircan escape at the top and then it sinks again due to its own small
weight inside the water and sinks to 10 meters, where the valve at the top is closed
and air is put via a hose into it, so the water goes out of it and it will again have a lifting
buoyancy force.
We need 2 Bars= about 29 PSI to bring the air down there.
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.

also, it will be more productive (more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g  x heightdifference) if you raise the reservoir.  by reducing the exit pipe diameter to 1/2 the volume starting at water level, you can raise the reservoir twice as high.  according to your formula, this doubles "potential-Energy"

?

Well, Okay, maybe if we use a fountain like structure and capture the fountain water inside an upper
reservoir... But remember, the hydrostatic paradoxon requires, that the pressure weight of a water column
is only dependant from the height of the water, and if you make the water column above the seawaterlevel
higher than 1 Meter, also with smaler diameter, it needs also the height of the aircylinder under water at 10 Meters
deepth to be higher, which needs more air to compress and pump down !