I thought to go underwater 100 m below water level and there inflate a leather balloon, volume 1 m3, theoretical work required would be:(adsbygoogle = window.adsbygoogle || []).push({});

Po Vo = ro g h Vo = 980 000 Joule

where

Po pressure 100 m below water level

Vo leather balloon volume

g gravity acceleration

ro water density

h water depth ( 100 m )

Then, I let the balloon go and, pushed by Archimede force Fa, it goes up to the surface, theoretical work done by Archimede force to bring the balloon to the surface will be :

Fa*h= ro*g*V*h = 980 000 joule

We did not violate energy conservation, That’s good !

I thought to go back to 100 m below water surface, but , this time, to use a syringe, a giant one, say 15 m3 max capacity, I thought to remove the needle and install in its place a valve like those for inner tubes, a giant one. Starting with the syringe piston fully down, (syringe volume equal zero), I inflate 1 m3 inside the syringe, work required to inflate the 1 m3 is still 980 000 joule .

I let the syringe go, it goes pushed by Archimede force, but this time, as the syringe move up, outside the syringe the hydrostatic pressure, Px, decreases, compressed air inside the syringe move the piston end expands to volume Vx increasing according to ideal gas law:

Px Vx = costant

When the sysringe gets to the water surface, the air volume inside it is 11 time the volume inside the syringe when it was 100 m below water surface, that is to say at water surface air volume inside the syringe is 11 m3, as well as the archimede force is 11 time the initial force !

Theoretical work done by Archimede force to bring the syringe to the surface looks much bigger tham the one done for the leather ballon, but the work done to inflate the syringe was the same as the one to inflate the leather balloon !

By Inflating and rising the syringe under the water did we gain energy ? how much energy, let see :

x Reference axis , vertical, direction up ward, origin at the syringe in flatting point

h level of water surface in reference to defined x axis

V0 syringe air volume at level zero

Vx syringe air volume at level x

P0 absolute air pressure inside the syringe at level zero

Px absolute air pressure inside the syringe at level x

Fax Archimede force acting on the syringe at level x

ro water density

g gravity acceleration

Infinitesimal work, dLa, done by Archimede force, Fax, to push up the syringe for an infinitesimal path dx is

dLa= Fax dx

considering :

Fax= ro g Vx

From ideal gas law

PxVx = PoVo ; Vx=PoVo/Px ; considerino also : Px= Po-ro g x

We get :

Fax= ro g P0 V0/(Po-ro g x) = - Po Vo /( x-P0/ro g)

Integrating Fax from level zero to level h, ( using notation “ln ass” for natural logarithm of the absolute value)

Work La done by archimede force :

L a= -Po Vo ( ln *** (h-P0/ro g)-lnass(-P0/ro g));

La = P0 V0 ( ln (h+10)-ln 10)

La = Po Vo ln ((h/10)+1)

Say L0 work done to inflate volume Vo at elevation

L0 = Po Vo

Net work obtained is :

L = Po Vo ( ln ((h/10)+1)-1)

Does it violate enegy conservation ?

I think No, because work we called Archimede work was done by earth gravity field, by attracting water heavier than the air, the air goes up. The work to inflate the syringe was done by the compressor.

The only scope of the syringe is to better visualize the air volume, we could have talk about an air bubble.

I can’t find a mistake, any comment would be appreciated.

Thanks, ciao.

Andrea Dall’Olio.

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# Archimede force work

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