Gravity Mill - any comments to this idea?

[hartiberlin]

@2tiger,
you always calculate the case, where the shuttle goes down to its
own weight.
We are discussing always the other case over here,
where the shuttle itsself has no weight ( only a few grams for the plastic case that can be
overlooked) and pushes the whole water column above it up due to its buoyance forces.

Please only calculate for this case.
Also you use only the pressure forumula P= F/A for calculation.
If you have a nozzle, then you need to calculate in water streaming and
water velocities in it and before it , etc.. and also the exit profile of the nozzle counts in,
only then you can calculate how high the water will rise.

So please calculate to have a good nozzle directly at sealevel= 0 Meter
and have a shuttle weight= 0 Kg and have for instance a shuttle volume
of 10 Liter = 10 Kg and have an area of 100 cm^2 surface area of the shuttle
at its top and also the same at the bottom and have the
shuttle started at 10 Meters deepth to go up.

So the shuttle is 1 Meter high and has 100 Liters of water  above it.
the main pipe will then have a diameter of 11,28 cm that is about 4,429 inches.

So how high will the water rise up after the nozzle output ?
Thanks.
Regards, Stefan.

[hartiberlin]

P.S: With the calculation of the hydrostatic pressure paradoxon
to my opinion, if you take any diameter 1 Meter high  exit tube instead
of the nozzle ,then the shuttle would be in equilibrium forces and would not
rise and stay in 10 Meters deepth.

[hartiberlin]

why do you think this will make the water go higher?  will you attach a pump to the nozzle?

Well, maybe if will use a small diameter  exit tube instead of a nozzle,
we might try to evacuate the upper water reservoir to 1/2 bar pressure,
before starting the experiment and see, if this needs less energy, than we
gain by sucking this way the water up maybe also a few meter more ?

I have to calculate this with the gas pressure laws and see, if this will
bring in any gain ?

[hartiberlin]

Well, I had another thought onto how to use the Hydrostatic paradoxon
not negatively as in the Gravity Mill but positively.

Well, if you look again at the Cartesian diver:







Then you can see, that the upper opening of the bottle,
where the blue cap is, is pretty small.
If we need now to pressure the bottle to sink the diver
we could just place  a weight onto the opening of the bottle at the water surface.
The smaller the opening of the bottle is, the smaller must be the weight to
get the same pressure inside the water as:

P= F / A and
F= mass x 9,81

So if we make the opening of the  bottle very small, say only 0.1 cm^2 we can use
a small weight to get a huge water pressure inside the water bottle.

So if we want to sink and rise the diver, which is connected to a generator for output
generation, we only need to lift a small weight up and down a few Millimeters
and thus do for the input only a very small lifting work of the weight that puts
pressure onto the water.


With the Cartesian diver we just convert a STATIC Weight force
into a moving work force=
usefull energy-work output !

What do you think ?

Regards, Stefan,

[tbird]

hi guys,

in for lunch and had to ask.  did you guys lose sight of the fact the elsa is just a pump?