Hi all
I attache some calculations to previews posts, so that you can proof this.

You can change al diameters of all tubes, height of the shuttle and the height of the whole device. The lifted water will always have half volume/height of the shuttle.

You can only lift the water to a higher level when you increase the weight of the shuttle, but the you need more bouyancy force to lift the shuttle back to the top.

Tell me if I?m wrong.

CU
2Tiger
 

2tiger,

be careful with giving definitive statements here you can't really back up.

It doesn?t matter if it is 10 m or 100 m depth, because the watercolums (without shuttle) are canceling its weight eatch other out thrugh the hydrosttic pressure. The only thing that takes effect (unbalance) here is the shuttle with 50% more weight (comparing with same volume of water). As result you?re still lifting half of the shuttle height.
Is the shuttle 1m high then you can lift 50 cm over seawater, but the tube has to have a little bit less (49cm) so the water can flow out.

you are doing ok except for 1 minor detail.  you should use weight instead of height in this sentence, "As result you?re still lifting half of the shuttle height."  they are proportionate, but weight is more appropriate.  i think this is where you start going wrong.

If you choose a lower diameter for the tube over sealevel, let us say the half diameter of the tube under water, then you will have to double the pressure by making the shuttle two times heavier.
ONLY in that case you were able to double the height over sealevel.

That is the hydrostatic paradoxon!

i have covered this extensively in other post, but here goes again.  the doubling the pressure is not needed.  that amount of pressure (created by the weight difference) will pickup a given WEIGHT.  it doesn't matter if it is contained in an area of x by y or y by x.  it exerts the same pressure.  the reason it would stop lifting is because the WEIGHT above water level equals the amount of WEIGHT the shuttle can lift.   the  hydrostatic paradoxon is helpful by providing pressure to the back side of the shuttle to lift the water that is above the shuttle, but only to the surface.  this doesn't apply to the water in the head pipe because hydrostatic paradoxon is based on pressure differences.  this makes them want to be level and equal.  in our case the water outside the pumping pipe has already neutralized  the water above the shuttle, but only to the surface. thus the only weight or pressure the head water can apply to the shuttle is what it WEIGHS.  if the shuttle has more lift, the water will go up.

Tbird
I forgot something.
This will only work by a ration beween "lifting tube" and "shuttlepistontube" from 1:1.
Do the math and see what happen, if you have a ration from 1:5.

By
2Tiger

the ratio you refer to should be comparing the head pipe volume to the pumping pipe volume.  the pressure area (the part that applies the force) of the shuttle will be close to this.  it depends on how well it fits (could have been a big problem with your test).  the other figure is more reliable.  i hope you understand now.

@2tiger,
your calculation is a bit hard to understand:
1. What case do you want to calculate,
a)the lifting of the shuttle and pumping up the water
or
b) the downward movement of the shuttle ?

2. You seem to forget, that it is done in big watercase, where the
outer water must be calculated in and which has a big weight.
This is also the reason, why so much water is lifted,
if you look at it, if you disconnect the exit pipe and just look
how much water is lifted to the same height of the seawaterlevel
via the shuttle.

Maybe we should just calculate it all with the hydrostatic pressures,
once with disconnecting the exit pipe and one time with
connceting the exit pipe to the main tube
and only look at the case, that the shuttle
is deep inside the water and will lift the water up
above it.

Regards, Stefan.

@2tiger,
where is exactly your 1 cm^2 and 5 cm^5 area ?
Seems you have drawn this in the same spot...
This is confusing...

Hi guys,
I had another look at:
http://en.wikipedia.org/wiki/Hydrostatic_pressure

and it seems 2tiger is half way right.
As the buoyancy is only calculated by hydrostatic pressure differences
onto the the upper and lower surface of our shuttle,
only the heights of the water colums above and below it will
give the buoyancy onto the shuttle.

So if the shuttle is with its lower surface at 10 meters deep,
it has a hydrostatic pressure there of
p= rho x g x 10 meter = 100 000 Pascal
and the upper surface has
p= rho x g x  9 meter =  90 000 Pascal,
if our shuttle is 1 meter high and we have no exit pipe,
but the water above the shuttle is only at seawater level.
(assume g= 10 m/s^2 for easier calculation)

Now this is a buoyancy force of 10 000 Newton
if the shuttle has an area of 1 m^2.

Now, if we add water above the shuttle water column via
an exit tube, only the additional height of the water column counts
and not the amount of water, this means, not the weight,
but only the height counts !
Also if the exit pipe is only 1 cm^2 in diameter, if it is also 1 Meter
high, we have no more buoyancy for onto our shuttle,
as now the hydrostatic pressure oonto the shuttle?s upper
surface is the same as the the hydrostatic pressure on the
lower surface of the shuttle= 100 000 Pascal and the shuttle
does not move !
So this is the hydrostatic paradoxon at work making our device
less efficient, as we can only pump the water as high, as the shuttle
itsself is high !

This is too bad.
So the hydrostatic paradoxon limits the height,
the water can be pumped up.
I have to recalculate , if we can find a relationship,
where it still works, also maybe by using both the
up- and downgoing of the shuttle-cycles
for the water transfer energy output.

I guess we have to look for a different case, where we can use
the hydrostatic paradoxon positively and not negatively as in this
case.

Regards, Stefan.

Sorry tbird, this is like wading through toffee...

Quote

So we will need to displace 4.93 litres of water to lift the head and the shuttle. We will call this the displacement volume.

...

We know that the pressure at 10m  depth is 2 times the pressure at water level.
[(height in meters / 10) + 1]

So we will have to compress 4  times our displacement volume into the shuttle
i.e. 19.72 litres
[2 * max pressure * displacement volume]

So the pressure inside our shuttle will be 4.02  bar (remember that most pressure gauges are set to read 0 at sealevel.)
[(total compressed volume / shuttle volume) - 1]

Where do I go wrong in the above and why?

the pressure in your shuttle at 10m that you compressed at surface has a net force of what?  we would figure, 4 inside minus 2 (that includes your absolute) outside equals 2.  one of which is your surface bar.  if you used the net value, you could subtract 1 (at any level).  once the shuttle is allowed to expand you will be left with a net 2 bar (1 of which is still your surface bar).  without the surface bar, your GAUGE pressure will be 1.  so 1 bar expanded, 2 bar compressed. when you change a formula, you have to change it everywhere it is a factor.  if you started with gauge pressure where the absolute bar is not considered, we would see at 10m the pressure is 1 bar.  so to expand a volume 2 times, you will have to have 2 bar.  net pressure then is; 2 bar inside minus outside pressure of 1 bar gauge equals 1 bar net, expanded pressure.

i'm running out of ways to say it.

did that work for you?  i'll keep trying until we get it right.

tbird

TBird,
just do the experiment to attach a smaller diameter exit pipe
at the top of the main tube and see, how high the water will be in it,
if you let the shuttle go up under water inside the main tube.

In my case  the water in the exit pipe was only as high as the shuttle
was itsself high, but okay, I had not so thight tubes and
some leakage, but I already saw the hydrostatic paradoxon
in my experiments to some extent.

Also the buoyance formulas predict you can only bring the water as high in the exit tube
as high as the shuttle is itsself high !
"Too bad to burst our bubble over here..."