Gravity Mill - any comments to this idea?

[hartiberlin]

Maybe we still can use the hydrostatic paradoxon
positively in our case, if we apply
low pressure ( some vaccuum) in the exit pipe.
As there is only very small volume, it also
needs low power to create a vacuum over there
and thus suck the water higher up.

Also the normal 1 bar air pressure at sealevel will help then
to press the water up via the exit pipe.

[tbird]

Sorry tbird, this is like wading through toffee...

So we will need to displace 4.93 litres of water to lift the head and the shuttle. We will call this the displacement volume.

...

We know that the pressure at 10m  depth is 2 times the pressure at water level.
[(height in meters / 10) + 1]

So we will have to compress 4  times our displacement volume into the shuttle
i.e. 19.72 litres
[2 * max pressure * displacement volume]

So the pressure inside our shuttle will be 4.02  bar (remember that most pressure gauges are set to read 0 at sealevel.)
[(total compressed volume / shuttle volume) - 1]

Where do I go wrong in the above and why?

the pressure in your shuttle at 10m that you compressed at surface has a net force of what?  we would figure, 4 inside minus 2 (that includes your absolute) outside equals 2.  one of which is your surface bar.  if you used the net value, you could subtract 1 (at any level).  once the shuttle is allowed to expand you will be left with a net 2 bar (1 of which is still your surface bar).  without the surface bar, your GAUGE pressure will be 1.  so 1 bar expanded, 2 bar compressed. when you change a formula, you have to change it everywhere it is a factor.  if you started with gauge pressure where the absolute bar is not considered, we would see at 10m the pressure is 1 bar.  so to expand a volume 2 times, you will have to have 2 bar.  net pressure then is; 2 bar inside minus outside pressure of 1 bar gauge equals 1 bar net, expanded pressure.

i'm running out of ways to say it.

did that work for you?  i'll keep trying until we get it right.

tbird

[tbird]

stefan,

in your example to prove this wrong, you are only appling this to the area the size of the head pipe (1 cm^2 in diameter).  do you really think this can stop the shuttle with many more time the area pushing up?  did you forget your hydraulics?

i have a hard time with your metric examples, as i have said before.  if you give me an inch example, i'll be more exact.

tbird

[hartiberlin]

TBird,
just do the experiment to attach a smaller diameter exit pipe
at the top of the main tube and see, how high the water will be in it,
if you let the shuttle go up under water inside the main tube.

In my case  the water in the exit pipe was only as high as the shuttle
was itsself high, but okay, I had not so thight tubes and
some leakage, but I already saw the hydrostatic paradoxon
in my experiments to some extent.

Also the buoyance formulas predict you can only bring the water as high in the exit tube
as high as the shuttle is itsself high !
"Too bad to burst our bubble over here..."

[hartiberlin]

All,
if you look at it, that the main water at sealevel and the
maintube with connected exit tube are a "U" shaped tube,
the water column?s heights in it must be equal, also if the volumes
of the 2 legs are much different.

As the shuttle is the partitioning wall between the both
legs, only the shuttle height can be added to one
water column, if the other has much more weight.