Gravity Mill - any comments to this idea?

[tbird]

hi andi,

sorry to take so long to get to this post.  been thinking about how to get stefan a better internet.  

you wrote:

"hartiberlin, thats what I'm saying. Fixed volumes will not work as the pressure is different at different sea levels. Maybe it should really be a static box with 2 volume states. Lets see, what your calculations are going to look like."

can't imagine what you mean by "fixed volumes".  also, what is a "static box"?  does it look any thing like the shuttle in this drawing?    http://www.icestuff.com/energy/elsa/

as far as having different pressures, that's ok.  did you realize the shuttle in the drawing is pre-pressurised at the surface, before it ever makes it's 1st run?  this pressure, when the shuttle is compressed so it will desend, will be 2 atmospheres (if we are taking it to 1 atmosphere, 10meters +or-).  30psi is a little more than that, but will make the shuttle expand to size, for sure.  to compress a shuttle that is based on a one cubic foot displacement, we'll say the top area is 12 inches square (144sq.in.total), would take 30psi times the 144 square inches which equals 4,320 pounds.  even if the shuttle is expanded with 15(+or-)psi at the surface, 4,320 pounds will compress it to the point where the air being compressed inside will be 30psi.  since we started with 15psi, the volume will be reduced by half.  back to the weight.  if you use a lever to hold the container that the compression water (weight) will be put in, for every foot out, less water will be needed.  if the lever was 5 feet long, the weight would only need to be 1/5th (864lbs) as much.  if you took it out 10 feet, 1/10th (432lbs.).  now at 10ft leverage we've reduced the amount of water needed to recompress from 7.67 cubic feet (4,320lbs.) to 0.767 cubic feet (432lbs.).  since we are goin down to 10 meters (+or-) and delivering water in both directions, we should have an excess of about 65 cubic feet (over 4200 lbs).  not bad for a 1 cubic foot unit, eah?  using the example from this drawing  http://www.icestuff.com/energy/elsa/, the height of the stored energy (water), (if you store it) could be as high as 72 inches using a 1 square inch exit pipe.

do you still want to only use the up stroke?

i have more about the shuttle, if you are not burnt out.

tbird

[ooandioo]

This is really amazing. Some times ago I found elsa on the internet, but nobody would recognize my questions to it. Now its going to be the biggest thread in this forum...

tbird, yes, I'm coming back to the original elsa design. Using a shuttle box that can have 2 different volume states will be the right decision. But, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth. I think first we have to find out if the gained waterpower is able to compress the shuttle. Hartiberlin is on the right way.

[tbird]

andi,

didn't you understand my last post?  do you think my numbers are wrong?

tbird

[tbird]

andi,

btw,

But, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth.

with this shuttle, those pressure differences are not a factor.  when the shuttle expands at the bottom, it will be at it's max volume.  the displacement is the same all the way up.  this brings me to the "i have more about the shuttle" statement.  to take advantage of the pressure left over at the top of the cycle, we could build the shuttle to accomdate more expansion.  it won't deliver any more water in that cycle, but it will do it faster.  so for a given time, we will have more engery (water) available.  the recompression energy will stay the same too.

tbird

[prajna]

Hi guys,

I'm hoping to simplify things a little with the attached drawing of a shuttle.  The displacement area can be full of water or air (depending on whether the shuttle is going up or going down.

Let's work with just one kilo of water and a shuttle weight of 0.5 kilos to keep things simple. We must also consider the volume of our shuttle because it will also displace that same volume of water, let's make things easy and assume it is 0.5 litres. The displacement volume, then, must be 1 litre  - 1 litre to displace 1 kilo of water plus 0.5 to displace the weight of our shuttle minus 0.5 because our shuttle already displaces that much water when the displacement area is full of water. The weight and volume of our shuttle cancel each other out if they are the same.  If the weight is less than the volume then it floats and if the weight is more than the volume it sinks.  Ok so far?

The displacement volume will be the same at any depth but, as Steve pointed out, it requires greater air pressure to displace 1 litre of water as we go deeper because we have to displace that 1 kilo of water plus the weight of a column of water above it (just as at sea level the air pressure is approx 14.7psi - or approx 1 atmosphere - because we are measuring the weight of the column of air above it, from sea level all the way up to the top of the atmosphere.)  At approximately 10m below waterlevel the weight of the column of water is the same as the weight of a column of air the height of our atmosphere!

What our shuttle will do is to contain air in the reservoir sufficient to displace 1.5 litres (i.e. 1.5 kilos, see why we love metric) at a depth of 10m That is 1 litre of water plus the 0.5 litre volume of our shuttle because we will be keeping that volume in our reservoir when we have filled the displacement area. At 10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm².

We can discount 1 litre of air when it comes to recompression of the reservoir because we still have that from the previous cycle. This leaves 2 litres to recompress = 2000cc. If we have a plunger with a cross section of 1 cm² and a length of 20m then a weight of 6.2 kilos will compress the air in the cylinder to that pressure.  If we only have 0.5m to compress in then we will have a cross section size of 40cm² and require a weight of 248 kilos.

The quantity of water we can pump depends on the area of the top of our shuttle.  If it is 20cm² then 1 litre will be 50cm high in the pipe and we can therefore pump 20 litres 0.5m above the water level. Hmmm....

Is my maths wrong or did I miss something or do we have less energy than we thought?

The proposal for the valves, by the way is that V1 is used to fill the reservoir with compressed air, v2 lets the compressed air into the displacement area when the shuttle reaches the bottom and V3 lets the air out when the shuttle reaches the top.

Forgot diagram.  See next message.