Gravity Mill - any comments to this idea?

[prajna]

Here is a new diagram.  I hope it helps.

Picture 1:  the shuttle is at the top and we pressurise the reservoir using valve 1.

Picture 2: the shuttle is at the top and we let the air out of the displacement area by using valve 3.  Now the shuttle descends.

Picture 3: the shuttle is at the bottom and we force the water out (displace it) by opening valve 2.  Now the shuttle ascends in picture 4.

The volume of the shuttle including the reservoir is 1.5 litres: 1 litre in the displacement area and 0.5 litres in the reservoir.  At 10m the pressure is 2 bar so the shuttle will contain 3 litres of air in the space of 1.5 litres.  When the reservoir is fully pressurised it contains 3 litres in a volume of 0.5 litres and is therefore at 6 bar.  When the shuttle is at the top the reservoir still contains 1 litre of air in a volume of 0.5 litres so it is at 2 bar.

Looking at this cycle we need to figure out two things:  how much water will the shuttle lift and how much energy will be needed to recompress the reservoir.  These I have calculated in my earlier message.  I think that model is as simple as we can get.

[ooandioo]

Hi all.
Was out this weekend and am now surprised about the good ideas.

prajna, I like your shuttle design, but the same thing would happen as it happens with my baloon idea. At 10m depth, with your 6bar compressed reservoir you are able to fill the 1liter displacement area. How do you handle the air in the displacement area become double of voume while going upward and the pressure decreases (actually, you filled 3 liters air in your reservoir)?. Thats why I went away from the baloon idea.

stefan, your new calculations are allright, I think thats the way we should keep moving. A closed shuttle, able to have 2 different volume states is the solution and I think Mr. Herring had the same answer.

Andi

[prajna]

ooandioo, yes, as the shuttle rises the volume of air increases and fills the pipe below the shuttle.  Because the shuttle is in a tube it simply increases the boyancy of the shuttle (since the air is now displacing more water than it did at 10m.)  No problem.  Indeed it is somewhat of a benefit since it increases the speed that the shuttle rises.  There is always at least 1 litre of air in the reservoir.  When the shuttle is rising there are always 2 litres of air in the displacement area/below the shuttle.  At 10m those 2 litres take up a volume of 1 litre at 2 bar whilst at the surface those 2 litres take up a volume of 2 litres (1 litre in the displacement area and 1 litre below it) at 1 bar.  No need for a closed shuttle.  If it worries you that air might escape up the sides of the shuttle then just change the volume of the displacement area to 2 litres; nothing else changes except that at the bottom the air only fills half the displacement area.  The pipe that the shuttle goes up is effectively a 10m displacement area with a piston on top.

Let's continue with the open bottomed shuttle because otherwise people will be worried about the mechanics of a two state cylinder rather than the considerably more fundamental problem of whether the sums add up.  As I said in my previous post, I think that this model is conceptually as simple as we can get and that makes analysing the problem easier.

The only problems are 1. How much water can we pump up to what height above water level, and 2. How much energy does it take to compress 2 litres of air to a pressure of 2 bar.  I think that I have answered both of those questions in my initial post but you might like to check my results.

[hartiberlin]

Pranja, I am at the PDA right now and can?t yet see your graphics, will comment later to it, when I am back at the PC.
What we really need to find out now is, how high the water will jump out the nozzle ! This is now extremly important. I have not found yet a formula that states that...
This is the most important factor now for knowing the COP.

[tbird]

hi all,

lots of comments since i went to bed last night.  looks like we might be getting somewhere, but everyone had problems with their arrangements.

first was stefan, then came prajna with basicly the same issue.  the pressure value at 10m.  not sure what caused this unless you forgot the weight still has displacement value that was not considered.  it may not be a positive weight, but it is displacement.

Mr. Herring in most of his examples (we all should read his stuff close) makes the shuttle weight to displace 1 and a half time what it displaces.  that inculdes the area for the compressed air.  now all he has to do is increase the volume of the total shuttle by 2 times.  at 10m the gage pressure (you might use absolute if working with refrigration, but not much else uses it) will read 1 bar (14.7 +or-).  so if we have 1 bar at 10m in the shuttle, our shuttle will have the same shape as it has at the surface without any gage pressure and be netural buoyant.  to make it rise, we only need a little extra expansion to make it positively buoyant.  so you see, we don't need 6 or even 4 bar at that depth.  we only need 1 plus the amount required to give our shuttle the strength (positive pressure) to raise the water above water level the height we want or need to do work.  the 2 things that determine this height is the pressure and diameter of exit tube.

so seeing this mistake in most of the previous posts, anything figured afterwards, based on those numbers, will be wrong.  this brings me to my comment last night  "there are probably other things we should agree on before we get too far along too".  if you make a mistake early on in your calcs, you carry that mistake the rest of the way (lots of wasted time).  if we keep our post to address 1 thing at a time, they will be shorter and easier to correct.

how 'bout it, can we do that?