Gravity Mill - any comments to this idea?

[hartiberlin]

@2tiger,
your calculation is a bit hard to understand:
1. What case do you want to calculate,
a)the lifting of the shuttle and pumping up the water
or
b) the downward movement of the shuttle ?

2. You seem to forget, that it is done in big watercase, where the
outer water must be calculated in and which has a big weight.
This is also the reason, why so much water is lifted,
if you look at it, if you disconnect the exit pipe and just look
how much water is lifted to the same height of the seawaterlevel
via the shuttle.

Maybe we should just calculate it all with the hydrostatic pressures,
once with disconnecting the exit pipe and one time with
connceting the exit pipe to the main tube
and only look at the case, that the shuttle
is deep inside the water and will lift the water up
above it.

Regards, Stefan.

[hartiberlin]

@pranja,
please only do the calculations with absolute pressure,
otherwise it is getting too confusing, don?t use the gauge
reading with subtracting 1 bar... it is just only good for
inflating your car tires !

[hartiberlin]

@2tiger,
where is exactly your 1 cm^2 and 5 cm^5 area ?
Seems you have drawn this in the same spot...
This is confusing...

[prajna]

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?

[hartiberlin]

Hi guys,
I had another look at:
http://en.wikipedia.org/wiki/Hydrostatic_pressure

and it seems 2tiger is half way right.
As the buoyancy is only calculated by hydrostatic pressure differences
onto the the upper and lower surface of our shuttle,
only the heights of the water colums above and below it will
give the buoyancy onto the shuttle.

So if the shuttle is with its lower surface at 10 meters deep,
it has a hydrostatic pressure there of
p= rho x g x 10 meter = 100 000 Pascal
and the upper surface has
p= rho x g x  9 meter =  90 000 Pascal,
if our shuttle is 1 meter high and we have no exit pipe,
but the water above the shuttle is only at seawater level.
(assume g= 10 m/s^2 for easier calculation)

Now this is a buoyancy force of 10 000 Newton
if the shuttle has an area of 1 m^2.

Now, if we add water above the shuttle water column via
an exit tube, only the additional height of the water column counts
and not the amount of water, this means, not the weight,
but only the height counts !
Also if the exit pipe is only 1 cm^2 in diameter, if it is also 1 Meter
high, we have no more buoyancy for onto our shuttle,
as now the hydrostatic pressure oonto the shuttle?s upper
surface is the same as the the hydrostatic pressure on the
lower surface of the shuttle= 100 000 Pascal and the shuttle
does not move !
So this is the hydrostatic paradoxon at work making our device
less efficient, as we can only pump the water as high, as the shuttle
itsself is high !

This is too bad.
So the hydrostatic paradoxon limits the height,
the water can be pumped up.
I have to recalculate , if we can find a relationship,
where it still works, also maybe by using both the
up- and downgoing of the shuttle-cycles
for the water transfer energy output.

I guess we have to look for a different case, where we can use
the hydrostatic paradoxon positively and not negatively as in this
case.

Regards, Stefan.