Cartesian diver, overunity concept ?

[hartiberlin]

Hi All,
I am opening up a new thread about this, so it will not get
confused with the gravity mill concept.

Okay, here we have the Cartesian diver concept again:






Now here are some calculations:

Imagine having a 4 Liter shuttle case and attached to it a 1,2 Kg weight.

So the buoyancy force at the top is 2,8 Kg into direction top upwards.

Now we apply via a 1 cm^2 area at the top of the bottle a 3 Kg weight.

This will bring up the pressure at the top from 1 bar outdoor sealevel pressure to 4 bar.

Now due to the gas laws, the volume of the shuttle will be compressed to 1 Liter and
thus have not enough buoyance force to compensate the 1,2 Kg weight, so the shutle is
sinking.

The bottle is maximum 20 Meters deep and at that deepth there is without the 3 Kg weight at the top
a pressure of 3 bar and with the weight applied at the top we have down at 20 Meters a pressure of 6 bars,
due to the water column added above it.


Now when the shuttle has come down to 20 Meters the volume of the shuttle will now be:

P1 x V1 = P2 x V2
thus follows:

4 bar x 1 Liter = 6 bar x 4/6 Liter.
So at 20 Meters deepth the volume is only 4/6 Liter= 0,8 Liter of air inside the shuttle
at 6 bar,
due to the high pressure there.

Now, when we release the 3 Kg weight from the top opening, the pressure at 20 Meters deepth will
be again 3 bar , so that means the shuttle air volume will now be at 4/3 Liter, which is 1,33 Liter and
this gives more buoyance than the 1,2 Kg weight of the shuttle itsself and the shuttle again
rises up and the air in it expands, so that it will rise even faster and have some more upward force,
until it is at the surface, where the airvolume will be back at 4 Liters.

So by placing a 3 Kg weight onto the 1 cm^2 opening we can rise and sink the shuttle so we can
get a work energy out of Integral from 20 to 0 Meters of F(t) x ds

F(t)= Forces depended on time
ds= difference of distance

As the force changes over time with the height of the shuttle, it is probably
much more easy to calculate the output via the gas laws, which I will do next and see,
if there is coming more energy out, that needs to be put in when you just
lift the 3 Kg weight a few millimeters to apply the pressure.

BTW,do you think the 3 Kg weight must be lifted very high,
as the water has to compress the 4 Liters to 1 Liter inside the
shuttle ?
But when you have a big 20 meters deep bottle, I guess you have so much water in there, that it does not need
much pressure on the 3 Kg to move this further into the bottle, so the distance ds is pretty low.
As the energy for this is W= F x s = 3 kg x 9,81 x s, it really now depends on the distance s how big the input enery
must be... Can this calculated somehow ?

I there are e.g. 10000 Liter in the 20 Meter deep bottle, then how far will the 3 Kg weight go into the opening,
when it has to compress the 4 Liters of air down to 1 Liter ?

Many thanks.

Regards, Stefan.

[ooandioo]

Hi Stefan.

You know, first I was negative about your idea. Yesterday I had some time for gooing deeper into it. As water is nearly incompressable, you don't have to care how big the bottle is, as long as its a closed system. The generated pressure on it will always compress the air inside the diver. I also think, your system is not limited to 20m. Sure, the pressure down there is higher than at 5m, but that belongs to the whole diver and so with applying constant pressure to overcome the beginning buoyancy (make the diver havier than the water it displaces) it will sink down till the water absorbs most of the pressure generated (and as water is nearly incompressable that will be a long run). When releasing the pressure, the shuttle will become buoancy, also in 50m deep - as the extra pressure is always the same in the whole system.

Andi.

[hartiberlin]

Hi Andi, there is the problem that the deeper you go with the shuttle the more adds up the compression of the air inside the shuttle via the  deepth pressure of the water. In 50 Meters the deepth pressure is so big, you will not get the shuttle up again, cause the air volume is too much compressed ! Or you would need a compression of 10:1 already at the top, but that would need a very big weight also...

[ooandioo]

Mh, perhaps you are right. I still cannot imagine that you are able to get the diver at 20m depth (perhaps also hold it there) and if its getting at 21m the diver is lost...

[2tiger]

BTW,do you think the 3 Kg weight must be lifted very high,
as the water has to compress the 4 Liters to 1 Liter inside the
shuttle ?
But when you have a big 20 meters deep bottle, I guess you have so much water in there, that it does not need
much pressure on the 3 Kg to move this further into the bottle, so the distance ds is pretty low.
As the energy for this is W= F x s = 3 kg x 9,81 x s, it really now depends on the distance s how big the input enery
must be... Can this calculated somehow ?

I there are e.g. 10000 Liter in the 20 Meter deep bottle, then how far will the 3 Kg weight go into the opening,
when it has to compress the 4 Liters of air down to 1 Liter ?

Hi Stefan

Here is a little drawing.
Take a syringe, cut of the front part so that you can see its own cross-section and glue this on a bottle neck.
With this experimental setup you can simulate how much force you have to apply on the syringe and how far you have to push for make the diver move up or down.

From my point of view the force (pressure) transmision between the syringe and the "bottle neck" of the diver is 1:1 one (in the case of my drawing A1 = A2). Now look at the diver self. It is bottle-shaped and on the widest cross-section it is 3 times (A3) wider than the bottleneck, so there is a ratio from 1:3.

And here is the paradoxon again  . If you push now the syringe 1 cm down it will transmit the force/pressure 1:1 to the bottleneck of the diver, so the water will rise in the bottleneck 1 cm too.
When the water now rises up to the upper cross-section in the diver, remember 1:3 between the cross-sections, the force/ pressure transmission will be 3:1, because of the paradoxon. That means that you have to push the syringe 3 cm down in order to let the water rise 1 cm in the diver.

Nevertheless this setup looks very interesting, because the dimension of the weight and the distance it must been moved depends only of the dimension of the diver and not of the whole watercolum.

CU
2Tiger