[ExI] Aerographite

[Jeff Davis]

On Thu, Jul 26, 2012 at 9:34 AM, Rafal Smigrodzki
<rafal.smigrodzki at gmail.com> wrote:
> A question for the mathematically proficient people: If you made a
> sphere of aerographite, covered with a gas non-permeable membrane, for
> example a few layers of graphene, and partially evacuated the inside,
> would it float in the air without being crushed by its pressure?

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Start with the ideal gas law:

PV = nRT.

We propose to suck out some of the air from inside the Aerographite
sphere to reduce the mass of (sphere + air) to match the mass of an
equal volume of air at STP..  The reduced air pressure inside the
sphere will be P(2), and the reduced amount of air will be n(2).

V, R, and T are constant in this case, so that

PV = nRT   =>     P/n = RT/V  = a constant.

P/n being a constant means that

P(1)/n(1) = P(2)/n(2), so that

P(2) = n(2)/n(1) x P(1).

density of aerographite = 0.3 g/l
density of air at STP    = 1.3 g/l

Neutral buoyancy requires that:

mass of aerographite sphere + mass of air inside = mass of air displaced

Rearranging gives:

mass of air inside = mass of air displaced - mass aerographite sphere

For a sphere with a volume of one liter, this becomes:

mass of air inside = m(2) = 1.3 g - 0.3 g = 1.0 g

The ratio of m(1) to m(2) is the same as the ratio of n(1) to n(2), so

 n(2)/n(1) = m(2)/m(1) = 1.0/1.3 = 0.769

Then P(2) = n(2)/n(1) x P(1) = 0.769 P(1)

P(1) = Air at STP  =   101   kPa = 14.7 psi
P(2) = 0.769 P(1)  =    77.7 kPa = 11.3 psi

P(2) - P(1) =               23.3 kPa =   3.4 psi

This is the unbalanced pressure, the net pressure on the outside of
the sphere, when using air as the gas inside.

As the sphere is "solid", the above pressure load per unit area -- in
this case, 3.4 psi -- equals the compression load per unit area

So, if the compression strength -- the maximum loading without
deformation, expressed in psi -- is greater than 3.4 psi, the
neutrally-buoyant-sphere will not be crushed.

If the compression strength of the aerographite sphere is greater than
14.7 psi, you could pump all the air out, pump the interior down to a
vacuum, without crush failure.  This would create the condition of
maximum buoyancy, which would allow the sphere to rise to its maximum
altitude, where the air density equals its own 0.3 g/liter.

A little note:Because air density decreases with altitude, the 0.769
sphere will only be neutrally buoyant near sea level.  So only there,
can it displace enough mass to support itself.  And there it will
settle, at the equilibrium altitude, and hang.

I couldn't extract a compressive strength from the information in the article.

 Help, anyone?

That was fun.

Best, Jeff Davis

 "Everything's hard till you know how to do it."
                    Ray Charles