Predicting the ABV of your distillate
Posted: Sun Nov 13, 2022 3:55 am
When you have a binary system of ethanol and water in your still, then it's possible to predict the final alcohol content of your distillate. I'm new to this so I was wondering if i'm doing it correctly. I like calculating stuff and wanna improve my control over the distallation process.
This is the rayleigh equation:
I'm gonna solve this integral numerically by using simpsons rules. But first i need a x(molar fraction ethanol in liquid phase)-y(molar fraction in vapor phase).
For example: 1% ethanol in liquid phase (x) corresponds to about 15% ethanol vapor (y) when the 1% ethanol liquid is boiling.
basically gonna make a table with x values from 1% ethanol in water until about 88% ethanol in water. I make a third column in which 1/(y-x) is calculated.
L1=inital molar content of the still (number of moles ethanol + number of moles water)
L2=total number of moles left after distillation
So of i have to start with a solution which contains 21 mole% of ethanol I integrate the 1/(y-x) column from 21% to 1% ethanol left in the liquid (or 2%, or 3% left).
The number that you get is ln(L1/L2)
From my data going from a 1 mole (L1) 20% ethanol solution to a 1% solution results in 0.51 (L2) moles of a 1% solution left in the still.
This also means that the distillate contains 0,49 moles.
So the amount of ethanol distilled to the distillate is L1*0.2(20%)-(L2*0.01) = 0,2049 moles. The distillate contains a total of 0,49 moles so the distillate. Is 0,2049/0,49*100=42 mole% ethanol in water.
Dont know yet how to convert to ABV but my first try calculating that resulted in 81 V%, which seems a bit high.
So anybody know more about this. And does this calculation look good? Or. Are these numbers not realistic?
This is the rayleigh equation:
I'm gonna solve this integral numerically by using simpsons rules. But first i need a x(molar fraction ethanol in liquid phase)-y(molar fraction in vapor phase).
For example: 1% ethanol in liquid phase (x) corresponds to about 15% ethanol vapor (y) when the 1% ethanol liquid is boiling.
basically gonna make a table with x values from 1% ethanol in water until about 88% ethanol in water. I make a third column in which 1/(y-x) is calculated.
L1=inital molar content of the still (number of moles ethanol + number of moles water)
L2=total number of moles left after distillation
So of i have to start with a solution which contains 21 mole% of ethanol I integrate the 1/(y-x) column from 21% to 1% ethanol left in the liquid (or 2%, or 3% left).
The number that you get is ln(L1/L2)
From my data going from a 1 mole (L1) 20% ethanol solution to a 1% solution results in 0.51 (L2) moles of a 1% solution left in the still.
This also means that the distillate contains 0,49 moles.
So the amount of ethanol distilled to the distillate is L1*0.2(20%)-(L2*0.01) = 0,2049 moles. The distillate contains a total of 0,49 moles so the distillate. Is 0,2049/0,49*100=42 mole% ethanol in water.
Dont know yet how to convert to ABV but my first try calculating that resulted in 81 V%, which seems a bit high.
So anybody know more about this. And does this calculation look good? Or. Are these numbers not realistic?