Gravity Mill - any comments to this idea?

[prajna]

Stefan,

The calculation is for any case - up, down or sideways.  The energy generated by a weight and that generated by buoyancy added together is possible to calculate.  That is what I did. This is the only energy that can be generated by the system regardless of whether it is an ELSA configuration or a Cartesian Pump configuration or even some strange configuration you might invent some time in the future.  This is the only energy in the system regardless of how you store or use it, whether you use it to pump a large volume of water a small distance or a small volume a large distance.  Even using some form of leverage it will not be enough to recompress the shuttle.

[hartiberlin]

Pranja,
maybe you can clarify, what your value
l to r mean ?
How did you calculate them and
why do you give as an energy the pressure
unit Pascal, which is indeed no energy unit.
Pascal is Newton / Meter and Energy is
Newton x Meters !
I don?t understand your calculations there.

Thanks.

[prajna]

Sorry Stefan, typo.  I meant joules.  I'll fix it.  The letters used in the formulae refer to the input values and result values, which have letters next to them: so  'l' represents negative buoyancy and 'r' represents total energy, 'a' would represent shuttle diameter, etc.

[hartiberlin]

Okay All,
now let me recalculate my old example and see if it works
with a horinzontal exit pipe at sealevel :


1. We have a shuttle, that has 1 Meter diameter and
an water column of 10 meters.
The shuttle is 10 cm high and 1 meter diameter,
so it has 78,5 Liter volume and starts at the top at seawaterlevel.
The shuttle is precompressed to 4 Bar !
Now to sink this shuttle we attach a weight of 80 Kg to it.
Now the shuttle together with the weight sinks down.

2. Now at 10 Meter deepth we let the shuttle expand itsself.
From the 4 Bar it had now at 10 Meter deepth there are also
2 bar pressure there due to the water pressure.
So the shuttle expands its volume now to double its size, so
to 157 Liter and being at 2 bar inside, as inside pressure = outside pressure= 2 bar.

Now 157 Liter means 157 Kg buoyant pressure minus the 80 Kg weight is
now 77 Kg or F= m x g = 77 Kg x 9,81 = 755,31 Newton upwards force.

This force now pushes the volume of  7854 Liter of water above the shuttle
through a 90 degrees right turn output pipe,
where a piston is located in it
so as the water is moved up, at the top at seawaterlevel ( 0 meters)
the water goes into a horizontal pipe which has a smaller diameter.
Inside the exit pipe is a piston that is pushed out of the pipe to the side
at seawaterlevel and
the piston is pulling a string(thread) which will
lift a weight up over seawaterlevel.

Now let us calculate, with which force a weight could be lifted this way:
Let us just make the exit pipe area just halfed in diameter, that means diameter= 50 cm,
which would mean a 4 times smaller surface area and a 4 times longer pipe to hold
all the 7854 Liter of water which is pumped into it. So the exit pipe
will be 40 meters long horizontally at seawaterlevel.

As we have now the condition of a hydraulic press, where
F1/F2= A1/A2  and with A2= 4 x A1
we find that
F1= A1 x F2 / A2 = 1/4 x F1= 755,31 Newton / 4 = 188,83 Newton

So we get out of it a force of 188,83 Newton over a distance of 40 Meters
which is an energy of around 7553 Newtonmeters= 7553 Wattseconds / 3600= 2,1 Watthours.

So you see, this is the same energy you would get, if you multiply the
force x distance from our water column the shuttle itsself lifts,
so Energy= 755,31 Newton x 10 Meter = 7553 Wattseconds / 3600= 2,1 Watthours.


3. Now at the top we must recompress the shuttle from now internally at 2 bar and
157 Liter volume ( 20 cm high at 1 Meter diameter)
back to 10 cm and 1 Meter diameter and 78,5 Liter volume at 4 bar.

This can be done again by using a pumping action, this time from the outside.
Now the formula for this is:
W= (P1-1 bar) x V1 x ln (V2 / V1)
So the energy W needed for this pumping is:
(200000 Pa - 100000Pa) x 0,157 m^3 x ln 0,5=-10882 Wattseconds / 3600= 3 Watthours !

So you see, that we need 3 Watthours to compress the shuttle, but only
get about 2.1 Wattshours from our output pipe piston.
So this is only 70 % efficient, so way underunity.

Well, it seems this does not work, as also other
dimensions have the same problems.

So with this I put the ELSA principle to rest.
I will still try to see, if I can get the Cartesian
diver principle to work somehow.

Regards, Stefan.

[FreeEnergy]

you know i think instead of compressing/recompressing the shuttle maybe its better to just lift some kind of small weight that attaches itself (do to some kind of locking mechanism) to the shuttle at the water surface to make the shuttle sink, then once down it dettaches itself (do to some kind of unlocking mechanism) from the weight.

then the process repeats itself over and over again

i think this has been said im not sure.