Here is the picture from TBird attached.
Many thanks.

Hi Pranja,
if you will work on the javascript,
maybe you can include also the weights in pounds and
the height in inches and the pressure in psi,
so our US friends, who don?t use the metric system
can also easier see their unit results.
Many thanks for this great work !

Regards, Stefan.

@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.

Hi All,
I recalculated this now for to see, if also a mechanical lever action can recompress
this shuttle from 2 bar back to 4 bar ( 400 000 Pascal) pressure and thus shrink the
volume again from 20 cm to 10 cm, so difference 10 cm= 0,1 Meter.

We can use then an external pressure of 4 bar to do this.
As we can also do it via a lever we must see, what force
4 bar means.
This can be calculated this way:

P= F / A ( means: Pressure= Force / Area )
thus
F= P x A ( means: Force= Pressure x Area)

Energy = F x s = P x A x s ( means Energy= Pressure x Area x distance)

So:
400 000 Pa x 0,785398 m^2 x 0,1 meter =8,7 Watthours

This 8.7 Watthours is bigger than the upper only 3 Watthours,
but this is logical, cause we apply already at the start 4 bar
where the shuttle only has 2 bar.
So at the start we could only apply 2,1 bar and so on rising,
until the 4 bar are reached, which would save us some energy
and if you do it with a syringe compression via air pressure,
you would only need the above 3 Watthours, so
doing it with a lever-torquearm  will waste some energy.

But anyway, it is still overunity !
COP = 21,4 / 8,7 =2,46

Enjoy !
Regards. Stefan.

Did somebody else already try, how high
the water can be pressed through a small diameter
exit pipe ?
In my experiment, where I only had a pretty slippery
shuttle, the pressure from the buoyant force was not very big
and thus it did not go very high in a smaller pipe at the top....

I wonder, if the shuttle must not be this high itsself
as high as the exit pipe will be ?

Maybe I just need a tighter fitted shuttle to see how high I can bring
up the water in the exit pipe and I also have to get a much longer main
pipe to play with it....

TBird, what did you see in your experiments ?
Did you try already a smaller exit pipe ontop the main pipe ?

Regards, Stefan.

Hello all
Yesterday I tried the same experiment as you Stefan.
This time with a higher tube (1m High) from the boxing of a Barbiefamily from my daughter and a plastic Cokebottle as a swimmer. I tried both with seal and without, even with valve like function on the upper and lower side of the tube. Tomorrow I will post a photo from my setup. In all cases there was no water lifted up! The water level was always the same. Nor a mm difference.

So 2day I was reviewing your video and saw the difference between yours and my experiment.
I attached a drawing and I hope you all will understand what I?m talking about.

Picture No. 3 is the most important one. It shows that your swimmer body lifted also the tube up, perhaps caused by fiction of the water between the tube and the swimerbody, and that your swimmerbody didn?t reach his whole level as in the startposition.
I assume that the weight of water you "lifted" plus the weight of the tube is equal to bouyancy force of the swimmerbody. So in that moment the setup is in balance. But now we have to understand why the water stands there "lifted". Will it stands there forever?? I think that it won?t!  
Stefan - please try again your experiment and wait a few minutes after the water is lifted. As result I expect that it will flow out between the tube and the body, as long you have no seal there. Or that the tube will slide slowly down and let the water flow out at the upper border of the tube.

If I do my homeworks right noone reported that the tube in the E.L.S.A.setup is a moving part. In the most graphics it is a fixed part, or am I wrong? Otherwise the shuttle have to lift up the hole watercolum and tube against gravity.

This is very important for your experiment Stefan, because when you fix your tube in anykind on the ground, you will not see the same result. Please try again and you will see I?m right.

by
2Tiger

Congratulation Stefan!
Now I am sure that the E.L.S.A. device won?t work as you assume at the beginning of this thread.
But your device will work if you add some things.    

If you let your device lift up with water and tube and try to hold the tube in this upper position (look at my drawing).
Then you have only to open the valve in order to unbalance the system. Some water will flow out and the lowered weight will let the swimmerboddy raise up till he reaches his upper position (floating).

Again CONGRATULATION accidently you have find the right setup.

The best thing in your setup is that it doesn?t have to be a large tube, because the lifting work begins right under the watersurface, the depth will equal to the hight of the lifted watercolum.

Good job!

Now you have to do some maths to calculate if it is still OU.

By
2Tiger

Glad you like it guys.  There is a lot to answer - mostly because the page is not complete yet.

@Stefan - yes I will be adding recompression calculations, COP and metric to imperial conversion.

@tbird - the shuttle weight is calculated as that weight required to sink the shuttle (i.e. balance the volume of the shuttle.)  The volume of the shuttle can be set by changing the pumping cylinder internal diameter or the shuttle height (the most logical thing to do.)  Yes, I have based everything on an ideal shuttle - that is a cylinder that has infinitely thin walls.  There are many improvements that can be made such as to choose material for the shuttle so that we can calculate the internal volume of the shuttle e.g. if the shuttle is made of lead the walls will be Xmm thick and the capacity will be Ycc less than the displacement volume.  We can add a friction calculation to tell us how much energy will be consumed by the seal between the pumping cylinder walls and the shuttle.  We can add viscosity calculations for the water in the header tube. etc.

the first thing you do is "head of water in the feed tube".  this weight is nice to know, but i think just as (if not more) important would be to know what the max diameter of that pipe could be for a given height.  and maybe if given a pipe diamter, what max height could be.

shortly after you get to the shuttle.  it's a bit unclear if you are referring to expanded or compressed state.

The max diameter can be anything you like.  Increasing the diameter (or the height) increases the static head (i.e. the weight of water in the header pipe) and thus the amount of water your shuttle must lift.

When discussing the shuttle volume it might be better to think of it as the capacity of the shuttle.  My shuttle never expands or compresses, rather it is the air that it contains which is compressed at the top of the cycle into the shuttle (which is simply a closed cylinder).  At the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it.  At this point the air under the shuttle will provide lift to the head.  We can ingnore the air inside the shuttle at this point since it is the shuttle's displacement (or physical volume) being equal to its weight (because the calculator sets the weight equal to its displacement) that makes the shuttle bouyancy neutral. In fact the weight of the shuttle needs to be slightly greater than its displacement so that it will sink and the air under the shuttle needs to be slightly more than the static head so that it will 1. lift the head above the top of the header tube and 2. lift the slightly unbouyant shuttle.  I will make a note to include these factors in the calculator. I hope this all makes sense.

from your program "We know that a depth of 10m will compress the air 2 times its volume at the surface.
height / 10 + 1".  not sure if this is worded right.

I can't have worded it right or it would have been crystal clear to you.  If a balloon contains 1 litre of air at sealevel then at 10m depth it will be compressed to a volume of 0.5 litres.  I should probably have said "We know that a depth of 10m will compress the air to half of the volume it occupies at the surface."

Thanks for all the compliments guys.  I will continue working on it.

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.

prajna, if this is the shuttle you plan to use, i think you have a problem.  steps 1 & 2 cancel each other.  if you put air in and then let it out so it will sink, you won't have air for step 3.  maybe you just left out another step that would leave you with enough compress air.

i thank you again for your efforts with the program.  it will be very useful, even if you have read the instructions (i think you have, more so than anyone else)

tbird.