Note: This is an edited version of the original. I forgot to include the density of ethanol in the original calculations.
Back to the continuous stripper
http://homedistiller.org/forum/viewtopi ... 16&t=33101 .
From experience, but also from calculation
http://homedistiller.org/forum/viewtopi ... 0#p7133969 , the stripped product from a 10%ABV wash is approximately 30%ABV.
For the sake of calculation, suppose that the ratcheting dispenser passes 100ml on each cycle.
So 33.3 ml of product are produced from each 100 ml of wash.
The temperature of the wash is 20*C, say.
In 100 ml of 10%ABV wash there are 90 ml of water and 10 ml of ethanol.
In grams, this is:
10 ml x 0.791 g/ml = 7.91 g ethanol and 90 g water = 97.1 g
To raise 100 ml of wash 1 K would require:
7.91 g x 0.58 cal/g + 90 g x 1 cal/g = 94.5 calories per *C (or K in Kelvins)
So, the specific heat of the wash is 94.5 cal/97.1 g = 0.973 cal/(g K)
The boiling point of the wash, from the usual phase diagram, is 93.1*C.
The product is 30%ABV. So 100 ml of product contains:
30 ml x 0.791 g/ml = 23.7 g ethanol + 70 g water = 93.7 g.
So 33.3 ml would have a mass of 31.2 g.
The LHV of the product is therefore:
23.7 g x 204 cal/g + 70 g x 540 cal/g = 42640 cal per 93.7 g
or 42640/93.7=455 cal/g
The boiling point of the product is 85.7*C, from the phase diagram.
So, we have 97.1 grams of wash going into the liebig and 31.2 g of product being condensed.
The 31.2 grams of vapour will give up 31.2 x 455 cal/g=14196 cal
Now 14196 cal put into 97.1 g of wash would raise its temperature by:
14196 cal/97.1 g/0.973 cal/(g K)=150 K (that's off by 3 K from the erroneous first calculation!)
There is actually more heat available than the wash can take.
So, instead of the usual liebig to condense the product, have two liebigs in series. The first would be cooled by the wash, which would extract as much heat as possible, raising its temperature to 85.7*C, and the second part would finish off the job using water as a coolant.
Now, if instead we had used the waste to warm the incoming wash, we would have: 0.667 x 97.1 = 64.7 grams of water at 100*C. This would result in heating the 97.1 grams of wash to an equilibrium temperature, Teq, of:
97.1 x 0.973 (Teq - 20) = 64.7 (100 - Teq)
Teq = 52.5*C
So, by using the liebig to heat the wash instead of the waste, it would be:
85.7-52.5=33.2*C hotter.
Edit: Interesting; taking into account the density of ethanol makes hardly any difference in the results. To really get accurate, the volume change with %ABV would have to be included, too.
M