Gravity Mill - any comments to this idea?

[hartiberlin]

hi

stefan,boy, your mind is really racing!
not sure if i follow your unit, but i get the impression you want to put those electronics under water again,  is that right?

Right, as this is easy and it only needs to be encapsuled, so no water comes
into the generator, that would be not too complicated.
That has the advantage, that you need no seal, to transport any
mechanical push-pull-string-thread action outside the water case.

in this case, do you want to make electricy or move water?  don't forget at depth your "egg" still has to be able to expand enough to be buoyant.  if you are making electricy, won't take as much as pushing water.  do you think you can make a gen that will put out as much at that speed as you would get from the water?  maybe you want to do both at the same time?

You can also build a linear magnet coil-generator to extract energy from the
movement ofthe Cartesian diver directly along its path up and down.
If you use strong neodymium magnets you can also extract quite nice
power, also if the diver does not move too fast...

[tbird]

also if the diver does not move too fast...

did you say what you meant?

tbird

[hartiberlin]

hi stefan,

i could not understand anything on your last link (not my language).

the metric system is pretty cool, but.....iget lost pretty quick with this stuff.

in psi and pounds, we might be able to figure something.  like how much water.  if we know that, you can figure how much work you can do with the leftovers.  would that be good?

Okay, work something out in the PSI and pounds US system and I will
convert it to Metric SI system units.

I really need to now, how much work it takes to compress normal
outside air to 1 Bar pressure = 14,5 psi at 785,4  Liters= 207,5 gallons.

[hartiberlin]

also if the diver does not move too fast...

did you say what you meant?

tbird

Sorry,
It must say: also if it does not move so fast...

so if it moves slowly, you have to at least have big magnet forces
and low coil resistances, so you get big amperage at low speed and convert
this electronically up to higher voltages.

[hartiberlin]

Okay  guys,  lets try again  to calculate it step by step and remove my old bugs.

Pay attention, that I use now the Komma (,) as a seperator to the
decimal point for the numbers below 0 and use NO point between thousands points.

This time I go for a balloon sized swimmer pistion like body,
that gets inflated at 10 Meters deep inside the water via an aircompressor
to pump air into 10 Meters balloon, so it will get a cylindrical balloon
with 1 Meter height and 1 Meter diameter of air and as this is at 10 Meters
deep we need 1 bar of air pressure to do this.


We must try get more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g  x heightdifference,
than needing for the compression of the swimmer unit !

Okay, let?s stay with Andi?s example:

A pipe 12 Meter deep inside the seawater and 1 Meter above the seawater level.
diameter = 1 Meter


Now let us just use the lifting not the sinking to produce the energy.

Okay, now the START condition.

All water in the pipe is in the seawaterlevel position and the
piston like swimmer body must be 10 Meters under seawater level
to be able to go up to the opening of the pipe l Meter above seawater level and thus
pump 10 Meters of volume over the 1 Meter sticking out of the seawaterlevel
high pipe into an upper reservoir 1 Meter above seawaterlevel.
Okay the piston like swimmer body will be also at the START 1 Meter high and have a diameter
of 1 Meter.

Okay, how big will be the potential energy we win, if we succeed to pump this
volume of 10 Meter height of water volume of 1 Meter diameter water up:

height of water above the swimmer unit: 10 Meters= 1000 cm
Volume=Area x height,

Area=r?*Pi=2500cm*Pi= 7853,98 squarecm
Volume=Area x height=7853,98 squarecm x 1000 cm= 7853980cm? this means the water weights 7854 kg
or has a volume of  7854 Liter.


Thus we have a potential energy stored of:

Energy= 7854 Kg x 9,81 x 1 Meter height difference between seawaterlevel and  reservoir = 77048 Joule or Wattseconds.
That means put into Watthours we must divide it through 3600, so
we get about 21,4 Watthours of potential energy stored 1 Meter about sealevel with
about 7854 kg= 7854 Liter of water !


Now let us see, if this would be enough energy to release the
air from the balloon swimmer body, let it sink itsself again to 10 Meters deep
inside the water and then pump air at 1 Bar pressure again down to 10 Meters
deep to inflate the ballon again to  785,4 Liter at 1 Bar pressure so it is 1 Meter high
and 1 Meter in diameter.
Then the whole cycle can start again.


Okay, now I still have to know, how we calculate the compression of the
air to get it down there and build up the balloon for 785,5 Liters at 1 Bar.