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Why did I think of that?
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Gin is lovely, I think we can all agree. The standard
methods of production, namely various types of
distillation, are effective but mundane. I propose a new
1. Take a source of low alcohol feed stock, this enters
bottom of the MASSIVE VORTEX CHAMBER (transparent,
Here it encounters some ceramic ultrasonic
see the <link> for examples. You'd need really massive
3. The feed-stock gets ultrasonically forced into making
<5um fog particles. Some of which will be of higher
than the others... just by chance.
4. Once you've got a reasonable fog production going,
time to TURN ON THE MASSIVE VORTEX FANS
5. You will now have a huge man-made tornado of booze.
Because the density of ethanol is 0.789g/cm3, the
of the vortex will have an enrichment of alcohol and the
periphery an enrichment of water, so, simply draw off
center (at the top, let's use gravity as an extra feature)
the good stuff.... and the lower periphery to remove
6. For further concentration, you might want to use
additional vortices in a casdcade, maybe 2 for schnapps,
for gin (no vodka, I'll have no dealings with such
vulgarity). No matter, this process is continuous.
For extra points, the 'distillery' should be located in the
mid-west because of the proximity of grain and their
fondness for tornadoes. I appreciate that this is a
tremendous gimmick... but gimmicks are frequently
deployed in the world of booze...
DO NOT LET JAMES DYSON'S LAWYERS ANYWHERE NEAR
[bs0u0155, Mar 14 2013]
||What you would end up with is a water vapor/alcohol
vapor mixture as the low pressure inside the vortex
accelerated evaporation. So basically this becomes a
big and needlessly complex vacuum still.
||//So basically this becomes a big and needlessly
complex vacuum still//
||The bottle/label designs itself too...
||//5um fog particles. Some of which will be of
higher alcohol than the others... just by
||No, really no, at least to several decimal places of
||Consider a 5µm drop containing, say 10% ethanol.
That's roughly 10e-14 litres of ethanol. That's
about 10^11 molecules of ethanol. Poisson and all
that stuff means that the standard deviation will
be roughly the square root of 10^11, or about
3x10^5 molecules, which is (unsurprisingly) about
1 part in 3x10^5. Very rare droplets will be maybe
three standard deviations away from the mean, or
maybe 1 part in 10^4 away from the mean.
||Thus, the most alcohol-rich droplets will only be
0.01% richer in ethanol than the average.
||Without some equivalent of Maxwell's demon in this monstrous separator - a spirit spirit, or gin gin djinn - it won't work.
||//Thus, the most alcohol-rich droplets will only be
0.01% richer in ethanol than the average//
||So, assuming you're right... what you're saying it
that it will work, it will just take a huge cascade of
machines until the heat-death of the universe to
complete even a modest enrichment. A hilarious
level of inefficiency :-)
||How about improving the front end a little: Yeast
produce ethanol anaerobically from assorted
carbohydrates, however heavy water has a lovely
10% extra density. So, feed the yeast on a regular
source of glucose, but make the general aqueous
environment heavy water. Now, there will be a
little cross talk in the general jiggery-pokery that
is metabolism, but it should yield a 0-79-1.1g/cm3
spread of density.
||I'll admit that the cost may have just skyrocketed
by several hundred fold, but I'm a slave to
||Heavy water will bugger up the yeast's
mitochondrial metabolism, but they don't need
that during the anaerobic fermentation phase, so
you know, disregard that.