Gravity Mill - any comments to this idea?

[prajna]

@ Stefan and ResinRat2,

There is a difference between pumping volume and pumping pressure.  In the ELSA example we have great pumping volume and tiny pumping pressure.  In the Cartesian Pump example the situation is reversed.  The head height is simply a measure of pumping pressure.  You can think of it as an analogy of electricity.  In the first example we have large amperage and low voltage, in the second we have big volts and low amps.

@ wizkycho,
I have added calculations of the total power generated in each cycle and it is disappointingly low compared to the energy required for recompression.  I don't know the energy that would be required for electrolysis but I imagine it would be greater than the energy generated by the cycle.  If you could calculate the energy required to produce 3 litres of H2O2 and the energy contained in that H2O2 then we might be able to evaluate your suggestion.

[FreeEnergy]

[hartiberlin]

@ Stefan and ResinRat2,

There is a difference between pumping volume and pumping pressure.  In the ELSA example we have great pumping volume and tiny pumping pressure.  In the Cartesian Pump example the situation is reversed.  The head height is simply a measure of pumping pressure.  You can think of it as an analogy of electricity.  In the first example we have large amperage and low voltage, in the second we have big volts and low amps.

Hi Pranja,
we could also have a SMALLER exit header tube going sideways under the water already.

So then we will have no problem to pump all the water through this small exit header tube and
thus have there also a lot of pressure inside this smaller diameter pipe.
This way we can avoid the hydrostatic paradoxon.

We can do this easily
by having the exit header tube turn sideways under the sealevel already, so
the buoyancy force of the shuttle can always press the whole water above it
into the small pipe, which drives a water pump or a syringe like piston inside the exit tube.

If we go with a syringe like piston, we can have a look at it,
with how much force it moves and then calculate via multiplication
of the distance s  it moves the output energy W= F x s .

Now if we pull up with this syringe piston a weight,
we could use this later to recompress the shuttle
again.

Will this work ?

[prajna]

Stefan,

At the end of my Cartesian Pump (http://www.DeclarePeace.org.uk/elsa/buoyeng.htm) page I have calculated the energy required to recompress the shuttle and the energy generated by the sinking/buoyancy pumping cycle.  You can ignore all of the mechanics of how the pumping energy is used and just compare it to the recompression energy requirement.  It doesn't come close. So it makes no difference what scheme you come up with, it will never produce enough energy to recompress the shuttle (or compression chamber or whatever else you choose to use.)

I have even added a calculator that lets you change various parameters and then calculates the results (though I have yet to add a recompression section to it.)

[hartiberlin]

Hi Pranja,
for what case is the calculation there ?

I think,
if we stick with the original ELSA,
but just move the water inside a pipe sideways at seawaterlevel and
never go over seawaterlevel with it and
have a syringe piston in the exit tube, so the
piston will pull up a weight, which then
could recompress via leverage the shuttle,
it will indeed work.
When I have more time I will calculate it all through
with an example.

We just have to avoid the hydrostatic paradoxon only
and thus pump the water sideways out into an exit pipe
below seawaterlevel.
Then we will benefit from all the water that is pumped up
via the shuttle.