RE: [Distillers] Re: Boiling Points
Increased altitude will lower the atmospheric pressure
Boiling happens when the vapour pressure equals the atmospheric pressure
See http://www.madsci.org/posts/archives/jan99/915378644.Es.r.html as
recommended by Kev
They come up with P = Po x exp (-0.116 x altitude in km)
This URL is really useful, but you then need to know the vapour pressure of
our ethanol/water mixture. This will be different than that of pure water
or pure ethanol.
Here's the detail on calculating it. Its not nice.
At any given temperature, you can work out the individual vapour pressure
for both ethanol and water. There are a couple of ways of doing this ..I
have a calculator at http://www.geocities.com/kiwi_distiller/calc.htm to
work it out, using ...
log (P_sat) = A - B/(T+C)
where P_sat is in units of torr (where torr = Pa / 133.22)
and T is in Centigrade
A, B and C are ...
Ethanol 8.1122, 1592.864, 226.184
Water 8.07131, 1730.63, 233.426
eg at 93C the vapour pressures of ethanol and water are 176.3 kPa and 78.4
You need to convert the % alcohol in the liquid through to its "mole
fraction" (eg the % or fraction, based on the number of molecules present)
Do this by using the molecular weights; ethanol's is 46.0634 g/mole and
waters is 18.0152
and density; ethanol is 0.789 g/mL and water is 1.000 g/mL
eg a 10% alcohol content mash is 10% by volume
ie 100 mL has 10 mL of ethanol and 90 mL of water
turn this into mass % by working out that there's 10 mL * 0.789 g/mL = 7.89
g of ethanol, and 90 mL x 1.00 g/mL = 90 g of water
then ... 7.89 g / 46.0634 g/mole = 0.171 moles of ethanol, and
90 / 18.0152 = 4.996 moles of water.
Thus the "mole fraction" of ethanol = 0.171 / (0.171+4.996) = 0.033
eg 3.3% of the molecules present are ethanol
(thus 96.7% of the molecules are water)
Now the total pressure is equal to the sum of the partial pressure, where
the partial pressures are the pure solvent pressures x their mole fraction
in the liquid,
P tot = mole fraction Eth x Pressure Eth + mole fraction Water x
so ... we know that for a 10% mash, the mole fractions of the ethanol and
water are 0.033 and 1-0.033=0.967 respectively
so, if it were boiling at 90C (where we calculated the individual vapour
pressures ...) the total pressure would be
0.033 x 176.3 + 0.967 x 78.4 = 81.63 kPa
Using the correlation from the URL above, this works out to an altitude of
You need to rearange these equations a little so, or just play with the
numbers, until you get the right temperature that will cause it to boil,
for the given % mash at your known altitude.
An easy(ish) way of doing this would be to draw up a table, where for a
given mole fraction in the liquid (eg mash%), you work out the different
altitudes that correspond to the temperatures from 100 to 90C. Do this for
a couple of different mash % and you'd have the complete story. Then just
look up your own altitude.
If you want, I could do this in a simple spreadsheet for you.