Rosch taking orders on OU Bouyancy device.

[tinman]

290psi is 19.7atm.  So 20x the pressure... which is 20L to 1L ...

a standard cubic foot represents 1.19804 moles  ( http://en.wikipedia.org/wiki/Standard_cubic_foot )
1 cubic foot = 28.3168L

n (moles of gas in 20L) =  (0.706 cubic-feet) * 1.19 = 0.84
R  8.3145 J/mol K    ( http://en.wikipedia.org/wiki/Gas_constant )
T 291.3333K (18.3333C)

ln(V2/v1)  (20L/1L) -2.995732274

= 0.84 * 8.3145 * 291.3 * -3
=
divide by seconds to get watts.

2031.491634J

Well if that is the correct answer,we can take that 2031.491634 joules,and get it to perform 5780.5 joules of work by using a bouyant device. It will be slightly higher than this,as i never took into account temperature rise as the vessel rises closer to sea level-the surface.

[d3x0r]

Well if that is the correct answer,we can take that 2031.491634 joules,and get it to perform 5780.5 joules of work by using a bouyant device. It will be slightly higher than this,as i never took into account temperature rise as the vessel rises closer to sea level-the surface.

temperature falls on decompress.. rises on compression...
(canned air gets cold, decompressing)

[tinman]

temperature falls on decompress.. rises on compression...
(canned air gets cold, decompressing)

Yes,but the ocean water temperature rises as you get closer to the surface. At 200 meters deep,the ocean temperature this time of year in the indian ocean is around 8*C,and near the surfact it is around 18*C-that is a 10*C rise in temperature which will be transfered through the vessel to the gas. In my experiments of late,i have seen how fast enviromental temperatures can raise or lower the temperature of a gas in a sealed vessel/tank.

[markdansie]

In reply regarding jams Qwok h spent jail time. His devices never worked or panned out, I have covered many articles on him over the years.

Mark

[LibreEnergia]

Thank you LE for the civil answer.
I mean to use a standard compressor-a high efficiency one.I know there will be losses in heat,but what is the energy we have in that 1ltr vessel at 290psi gauge pressure. Enviromental temperature will be close to 8*C

You may want to check the numbers but working in SI,

290 psi gauge is 2100.8 kPa absolute.

during isothermal compression PV = constant so 2100.8 * 1 litre = 101.3 * x => starting volume was 20.73 litres.

The amount of air is n = PV/RT where R = 8.313  J  K-1 mol-1,  and T = 280.15 K (8 Celsius) =   (101.3 * 20.73) / (8.313 * 280.15 )  =  0.9 mol of air.

So,  work performed during isothermal compression = -  nRT ln(V/Vo) = -0.9 * 8.313 * 280.15 * ln(1/20.73) = 6354.2 J or 6.35 kJ

You compressor will be less efficient than that so adjust by whatever efficiency factor you think is appropriate,

lets say it is 65% efficient so 6.38 / .65 = 9.8 kJ

The value is moot though. The same maths that allows you to calculate this number can also be used to deduce that you can never achieve more output than input and the overall efficiency is proportional to the temperatures of the hot and cold sources.

There is energy to be extracted by utilising the difference in temperatures between water at the surface and at depth. Overall efficiency is low though because the temperature difference is low. A very large structure that did not pump air using and external compressor, but rather was the compressor itself would be the best for achieving maximum efficiency.