# A Math Question

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Or "maths" for those in the UK...

If I have a known volume at a known ABV and then I vaporize/condense an uknown volume can I determine that volume if I know the post-condensing ABV? For example,

I have 1000 liters at 35% ABV, I distill for some period of time and I end up with a volume X at 72% ABV. Can I calculate the volume X with this information?  I want to think it would be 1000(72/35)  but that approach doesn't quite feel right.

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It looks like you're taking the c1v1=c2v2 approach to this, which would be 1000 L*0.32 / 0.72 = 486.1 L. The problem with doing it this way is you'll never get 100% efficiency on your still, so you can't just do simple concentration conversions.

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If I recall high school geometry isn't that called Pearson's Square (or something like that).  I'm unclear on how efficiency plays into this -- yes for each BTU shoved into the still charge Im going to lose some to the environment, but that shouldn't have any bearing on the amount of etoh in the pot and in the receiving vessel. Assuming no leaks I should recover (condense) whatever I vaporize.

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Yes a simple weighted average will give you 486 L at 72% ABV but only if you know what's left in the pot is water at 0% ABV then you will have 486 L at 72% ABV plus about 514 L of 0% ABV of stillage leftover. Otherwise if your stillage is not 0% ABV then you will need to know it's volume and ABV%.

1000 liters at 35% = (volume X at 72%) + (stillage volume at stillage%)

Or if you are only distilling ethanol and water and are using a pot still you could estimate the volume x from the water-ethanol equilibrium curve. If you took the ratio of the integral of the vapor curve from the starting ABV to end ABV of your cut to the total integral of the vapor curve from start ABV to 0 ABV, and multiply by the starting volume, you should get an estimate.

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I agree with WildRoverSpirits and glisades. For c1v1 = c2v2 to work the concentration of alcohol in what is left in the still would have to be zero. Actually there is another problem as well because volumes of alcohol and water are not additive. This means that if you started with 1000 liters and you distilled X liters from it there would not be (1000 - X) liters left behind. So the math(s) just doesn't work.

This non-additive volumes problem can be overcome by using a mass or molar basis rather than volume for the quantity, and then using mass % or mole % respectively for the concentration.  But even when using mass or molar quantities you still have the problem of there being some alcohol left in the still. Theoretically this problem can be solved by using the vapor liquid equilibrium curve and working in small enough increments of the quantity distilled over the top so that we can assume the strength is constant over that increment.

Modifying your example slightly, let's say we start with 1000 kg of spirit at 35 mass %. And let us decide to use 1 kg as the increment for the distilled quantity. Your initial product strength will depending on the number of trays and the reflux ratio you use, but for the sake of example let us assume the product is at 80 mass %. When you have taken off 1 kg (and assuming 1 kg is small enough for the strength to remain at 80 mass %) you will have 1 kg of product containing 0.80 kg of alcohol, and because masses are additive you can say that what is left behind (in the still plus column) is 999 kg containing 349.2 kg of alcohol. Now if the next kg of product comes off at 79 mass % we can say the average strength of the product so far is 79.5% and there is 998 kg of spirit containing 348.41 kg of alcohol left in the still. You could keep on working like this until the avarage strength of the product reached the 72% target. This is the integral procedure suggested by glisade.

But this is very theoretical and it is close to impossible to calculate accurately what the take off strength will be at each step of the way.

All this theory can be neatly side-stepped if you can measure the strength of the feints/stillage. Again, just for the sake of making an easy eaxmple let us assume that in your original example the feints contain 10% ABV (Product still at 72 % ABV). To be accurate we need to work in mass terms so we convert the 1000 liters of 35% ABV to 955.59 kg at 28.91 mass %, the feints to 8.01 mass %, and the product to 64.53 mass % (all at 20 deg C).

If we call the product quantity X kg then the feints will be (955.59 - X) kg.  If M stands for Mass and C for concentration, and the subscript w is for wash, p for product and f for feints then the mass balance is

Mw.Cw = Mf.Cf + Mp.Cp

Mw = 955.59 kg
Cw = 28.91
Mp is unknown
Cp = 64.53
Mf = Mw - Mp = 955.59 - Mp
Cf = 8.01

i.e.
955.59 x 28.91 = (955.59 - Mp) x 8.01 + Mp x 64.53 = 7654.28 - 8.01 x Mp + 64.53 x Mp = 7654.28 + 56.52 x Mp

27626.11 = 7654.28 + 56.52 x Mp
19971.83 = 56.52 x Mp
353.36 = Mp
Mf = 955.59 - 353.36 = 602.23 kg

We can convert the 353.36 kg of product at 64.53 mass % back to volume to get 401.30 liters at 72% ABV. The 608.53 kg of feints at 8.01 mass % converts to 611.58 liters. The 1000 liters of wash that became a total of 1012.88 liters (401.30 + 611.58) of distillate plus feints illustrates that volumes are not additive.

Here the "shrinkage" is negative because we are separating the streams rather than combining them as we do when blending. Because separating streams is just the reverse of a blending operation if you have access to blending software like AlcoDens you can verify this calculation by posing the question as "what quantities of 72% ABV and 10% ABV spirits must be mixed to create 1000 liters of 35% ABV".

For interests sake, we can do the calculation again on a volumetric basis ignoring the contraction to see how large the error would be.  Now we have

Vw = 1000 liters
Cw = 35.00 ABV
Vp is unknown
Cp = 72.00 ABV
Vf = Vw - Vp = 1000 - Vp  (Not true, but ignoring the shrinkage)
Cf = 10.00 ABV

i.e.
1000 x 35.00 = (1000 - Vp) x 10.00 + Vp x 72.00 = 10000 - 10.00 x Vp + 72.00 x Vp = 10000 + 62.00 x Vp
35000.00 = 10000 + 62.00 x Vp
25000 = 62.00 x Vp
403.23 = Vp  (cf 401.30 when calculated via masses = 0.5% error)
Vf = 1000 - 403.23 = 596.77  (cf 611.58 when calculated via masses = -2.4% error)

Depending on the purpose of this calculation these errors may or may not be acceptable.

Edited by meerkat
spelling
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The question I'm actually trying to answer is "is our cold reservoir large enough to condense and cool X lbs of vapor". If I can estimate the volume of condensed distillate based on the starting volume and abv I can answer that question.  meerkat, those figures seem appropriate for this type of calculation, albeit pearsons square is highly unfit for that task.

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@indyspirits - there have been many times that I have done heat balance calculations on numbers less precise than these.  You could probably just take an educated guess for the strength of the stillage and calculate the volume of distillate from the measured distillate strength and then use Pearson's Square to calculate the volume of distillate.

Heat calculateions are almost always done on a mass basis because the latent heat and specific heat data are given that way.

Working with your example with 1000 liters of feed spirit you will have about 790 lbs of distillate (sticking to your units)

Latent heat of condensation for 72% ABV is about 570 btu/lb
Specific heat for 72% ABV is 0.79 btu/lb.°F

Heat load for condensing and cooling (assuming condensing at 175°F and cooling to 100°F)
= Mass x ( Latent Heat + (Specific Heat x Temperature Change))
= 790 x ( 570 + 0.79 x 75 )
= 500 000 btu

Taking the Specific Heat of the cooling water as 1 btu/lb.°F and the allowable  temperature rise of the water as 20°F the mass of water required is
= Heat Load / (Specific Heat x Temperature Change)
= 500 000 / ( 1 x 20 )
= 25 000 lbs of water or 3 000 US gallons

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4 hours ago, meerkat said:

Latent heat of condensation for 72% ABV is about 570 btu/lb

What table are you finding this info in? I run my calcs based on water (970 btu/lb) because (a) my google fu failed me and (b) I'd rather overestimate the amount of cooling water needed.

Many thanks for your input here!

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Wikipedia gives a value for the latent heat of pure ethanol as 38.56 kJ/mol.  This makes it 38.56/46 = 0.838 kJ/gram or 838 kJ/kg.  In my own notes I have a value of 850 kJ/kg.  For water (steam) I always use 2200 kJ/kg, but your 970 btu/lb is more accurate at atmospheric pressure.  I calculated the 570 btu/lb by weighting these two values according to the mass %,

i.e. 0.65 x 850 + 0.35 x 2200 = 1322 kJ/kg or about 570 btu/lb

I agree with you that in this case it is better to have a bit of spare up your sleeve and treating it all as water is not a bad idea.

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• 2 months later...

Good arithmetic, good answer.

How about "heat in = heat out."

If you size the reservoir and delta T to match the BTU you're putting into the wash at the pot over the course of the run, then the resulting heat losses from still radiation give you a nice small margin of error for your res. Personally, I recommend sizing your cooling at 160% of your heating capacity, as there are always errors, and your res can be affected by its environment.

Cuz I'm late for work, so no math from me today!

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