Gravity Mill - any comments to this idea?

[tbird]

good evening stefan,

I try to read your posting, but it sounds to me all pretty confusing.
First you said, you start with 2 Amtmospheres, which is about 2 bar and about 30 psi,
then later you say " since we started with 15psi", so I completely don?t understand your
setup.

you must have missed where i clarified with "this pressure, when the shuttle is compressed so it will desend,".  make sense now?

then later you say " since we started with 15psi",

the line just before sets the stage "even if the shuttle is expanded with 15(+or-)psi at the surface".  the key is "expanded" here as compared to "compressed" before.

Also it would be good, if we would choose and agree on one example of dimensions
we can all work on the same and that not everbody jumps on different
dimensions and sizes so it gets not too confusing.

great suggestion.  of course i like mine best, but seems to be too hard for everyone else to follow, so i think  prajna's example next best.  there are probably other things we should agree on before we get too far along too.  we can chat about it tomorrow.

i understand why you got a little lost, but you did the very best thing.  telling me instantly so i could fix it instead of trying to pick it up as you went along.  well done!!

bed time for me too.  sleep tight everybody.

tbird

[hartiberlin]

@prajna
okay, I will reply later,
but I just found a way it works in all cases !

TBird was totally right, when he said the shuttle must be precompressed !
That is the magic word !

Okay, let us start with 1/10 of the volume of my former example.
We just take a shuttle that is now 10 cm high and 1 meter diameter,
so it has 78,5 Liter volume and start 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 above it through a nozzle,
in the top of the pipe that is just at sealevel , so the 7854 Liter of water will be 1 Meter
above seawaterlevel, which will give us the mentioned  21,4 Watthours of potential
water energy.

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 we only need 3 Watthours of energy to recompress the shuttle to 4 bar pressure and 78,5 Liter
volume and earn 21,4 Watthours by lifting the water up !

This is the real solution !
Many thanks to TBird to getting the idea with the precompression !
That is doing the real trick and we only need to do the compression at the top,
where it is much easier also.
In the deepth at 10 meters the shuttle can run against a rod, which will
switch its internal expansion control mechanism and the shuttle then expands itsself from
4 bar at 78,5 Liter    to    2 bar at 157 Liter just by itsself !

So we now have the final solution and can seek now to optimize the nozzle and
still try to use other and better and more usefull dimensions !

Regards, Stefan.

[hartiberlin]

Now we only have to find out, how high we can sprinkle pump the water through the
nozzle with 755,31 Newton upward force.
If we can get it higher than 1 Meter we will even have a higher COP as about 7 as calculated
above ! Enjoy !
I am happy, I go to be now too ! Have nice dreams !.

[FreeEnergy]

so let me see if i follow here.

after the shuttle reaches the top of the pipe it has saved plenty of water in a tank above that it can run a watermill so we can setup a circuit for the mechanical un/comprssion of the shuttle...the foam/spunge shuttle would get compressed at the top of the pipe this makes the shuttle "heavier" because it has been compressed into a "smaller size/volume" and so it sinks to the buttom of the pipe, once there it activates a siwtch type of thing and the foam/spunge shuttle expands itself and floats up again...and it starts all over

[prajna]

Hope you slept well guys.

@tbird,

Thank you for following my reasoning - what a relief.

Compressing 6 times comes from the fact that we will need 3 litres of air in the shuttle at a depth of 10m in order to achieve the boyancy we require.  Since we will be storing those 3 litres in a tank with a volume of 0.5 litres we will have to compress it by a factor of 6 (0.5 =3/6). That is, in fact, the same as saying the pressure is approximately 2 atmospheres or 2 bar. Remember that there is a difference between absolute pressure and guage pressure.  1 atmosphere = 1 bar = 14.7psi (all approx).  At sealevel we already have 1 atmosphere and at 10m (because we have to add the weight of the column of water to the weight of the column of air) we have 2 atmospheres. In my example we have 1.5 litres to displace at a pressure of 2 atmospheres so that is equivalent to 3 litres at sea level and I have suggested that we will compress all of that at sea level into a tank with 0.5 litres - thus 6 times.

I am sorry to lay metric on you but it makes everything easy to calculate since 1 litre of water is 1000cc and weighs 1kg.  If we work with these quantities then we can easily multiply them by any value once the principle is established.

I am sorry that I didn't make it clear that I am also compressing the air at sea level.  To pump air down would add 2 litres that we have to compress if our tube had an area of 1cm² since the tube has a volume of 1 litre and it would displace 1 litre of water (assuming an ideal tube with walls of zero thickness).  Actually we could get around that by having the compression pump at the bottom rather than at the top but let's keep it simple in engineering terms.

@Steve

Using my units and description we don't have to worry about gravity or newtons or pascals or watthours or W= (P1-1 bar) x V1 x ln (V2 / V1) or adding 80 kilos etc.  It is easy to get confused with all of that.  If you want to later calculate the potential energy in x volume of water at y height above sea level then you can do that but it is not necessary in order to determine whether or not the system will work.  If, after your refreshing sleep, you forget the calculations you have done and work through my model then you should find it simple enough.  If it is simple then it should be easy enough to spot any mistakes I have made.

@FreeEnergy

Yes, more or less