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meerkat last won the day on July 7

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About meerkat

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  1. Running a seamless glass lab still

    See Part 3 of the TTB video series https://www.ttb.gov/spirits/proofing.shtml
  2. Hello from South Australia

    The Australian Distillers Association group on Facebook seems to be where most of the professional Aussie distillers hang out. There is also the Aussiedistiller forum at http://aussiedistiller.com.au which is very active but seems to be mainly home distillers. And of course there are some very active and knowledgeable Australian distillers right here in the ADI forum.
  3. Still design

    One part of your question which I did not address is whether it is detrimental to have a column that has a larger diameter than necessary. Unfortunately columns that are larger in diameter than necessary do not work as well as correctly designed columns. Depending on the type of column, there will be a narrower or wider range of flows over which it works well. With packed columns if they are too large the liquid will not wet all the packing and there will be zones where the vapor can rise up through the packing without contacting the liquid at all. WIth sieve (perforated plate) columns a minimum vapor velocity is required to prevent the liquid from weeping through the holes. A small amount of weeping does not affect the efficiency, but too much will cause the plates to run dry and in larger columns can lead to mechanical problems like vibration. Correctly designed bubble cap trays will not weep and these columns have the widest range of capacity. But even these trays will fall off in efficiency if they are grossly oversized because the intensity of the bubbling will decrease. This reduces the mixing and the vapor-liquid contact - reducing the separating efficiency of the column. Of course, for all types of columns a detrimental impact of over sizing is the cost of the column. The quantity of liquid held up on each tray also gets to be significant with very large columns and this can affect the recovery of alcohol in batch systems, but is not really a problem for continuous columns.
  4. Still design

    This subject was discussed in Unfortunately that thread got a bit messy with some irrelevant side-issues causing a bit of bickering and hair-splitting. Basically the situation is that the diameter of the column and the size of the pot are determined by different factors, so there is no fixed ratio between them. The column diameter is mainly determined by the vapor velocity up the column. In a small R&D column of around 2" diameter the vapor velocity will be in the region of 6 to 10 inches per second, but on a large vodka column of say 10 ft diameter you can get vapor velocities of 6 to 10 feet per second. So the column diameter is determined by the rate at which you want to run. In a pot still the pot size is determined by the heating method and the size of the batch you are working with. Let us imagine your fermenters produce 100 gallons per batch. You will probably want a pot of around 150 gallons to be able to boil this safely. If you want to process this in 12 hours you will need a column roughly double the diameter (4 x the area) than if you want to process it in 48 hours. And of course you need to put heat into the pot at 4x the rate for the 12 hour scenario. Distillers generally want to be able to process a batch in 8 to 12 hours (one shift) so it turns out that in practice there is an approximately consistent ratio between the pot size and the column diameter (or more correctly the column cross sectional area) but this is a coincidence - as explained above there are different drivers in determining the pot and column sizes.
  5. Volume issues in bottling runs

    Your 1674.8 lbs of 80 proof will give you 804.1 liters at 73.54°F. AlcoDens and the TTB Tables agree on this. But because 750 ml at 60°F grows to 754 ml at 73.54°F you should expect to get 804.1/0.754 = 1066 bottles. This agrees with PeteB’s mass based calculation. The underfill is 4 ml per bottle so you could expect to have 1066 x 4 / 750 = 5.7 extra bottles. The fact that you had 14 too many means that we still need to find where the extra 8 bottles came from. I agree that your scale is unlikely to be the source of error, but keep it in mind to check once you have eliminated all other possible reasons. If your proofing was out and the 1674.8 lbs actually gave you 810.0 liters ( = 1080 x 0.750) instead of the calculated 804.1 then your density at 73.54°F was 7.8262 lb/WG and this would correspond to a proof of 88.2. It is unlikely that you could be this far out. Another possibility for error would be if your bottling temperature was not the 73.54°F in your storage tank. But the temperature would need to be in the region of 90°F to explain the difference. This should be easy to check. As PeteB has said, it would be better to do your quantity checks during the run based on mass rather than volume. If you have an accurate lab scale you could also use it to calibrate your measuring cylinder. Use AlcoDens to calculate the expected weight of 0 proof (i.e. water) when your cylinder is full, and fill it with RO or well filtered water. If you have a bottling machine that uses a fixed head (pressure) and adjustable timer to control the fill quantity then you can set it to give a target weight rather than volume. The weight filled will vary with the temperature of the spirit and if you make a note of the time required and the temperature each time you adjust it you will soon be able to draw up a calibration curve to speed up the job.
  6. Adjusting Proof in Liqueurs

    No, the original AlcoDens is still available for those who only work with alcohol/water mixtures and do not need the capability of calculating liqueurs. The new product is AlcoDens LQ and it will do everything that the regular AlcoDens does, plus a few calculators that deal with mixtures containing alcohol, water plus sugar.
  7. Adjusting Proof in Liqueurs

    Thanks for the kudos PeteB. It is a complicated calculator to set up the first few times, and if anyone else is having similar problems they are welcome to send me some numbers and I will set up the calculation for them with an explanation of how/why I made the selections. Other users have also reported that after a few runs it becomes much easier, but the learning curve is a bit steep.
  8. How To Proof Rum With Brown sugar Added

    The lab still will allow you to determine the proof and sugar loading of your spirit, but if they are not on target then you still need the math to determine what to add to get to the correct levels.
  9. How To Proof Rum With Brown sugar Added

    If you know the proof and weight (or volume) of the rum and the target sugar loading then you can calculate the quantities of sugar and dilution water required to achieve 80 proof using the calculator downloadable from www.katmarsoftware.com/alcodenslq.htm But the final proof must be verified as described by bluefish_dist
  10. Adjusting Proof in Liqueurs

    The reason for the difference between adding water and sugar is that proof is based on volume and a given mass of sugar has a different volume from the same mass of water. For example. if you have 10 gallons of 100 proof spirit and you add 10 lbs of water the volume will increase to 11.169 gallons. But if you added 10 lbs of sugar to 10 gallons of 100 proof spirit the volume would increase to only 10.749 gallons. In both examples no additional alcohol is added, so both contain the same amount of alcohol after dilution. Adding the water would lower the proof from 100 to 89.54, but adding the sugar would lower it to only 93.03 because the volume has increased by a smaller amount. If you used a standard proof hydrometer to determine the proof of the sample with the sugar added it would read about 30.4 proof because the sugar would have significantly increased the density of the sample and the hydrometer is not calibrated for that.
  11. Evaporation during open fermentation

    The volume will decrease, and part of it is due to the evaporation of the ethanol, but the main reason for the volume decrease is the loss of CO2. If we take the hypothetical example of 1000 kg of water with 200 kg of sugar added we would have a volume of about 1120 litres, since 1 kg of sugar increases the volume by about 0.6 litre. Roughly 100 kg of the sugar is converted to CO2 and is evolved. The other 100 kg will be converted to alcohol with an SG of about 0.8 and ignoring shrinkage we would have a volume of 1080 litres, or a loss of 40 litres. In large commercial distilleries the CO2 that is evolved is taken via a scrubber to recover the alcohol that is carried off. A very rough calculation indicates around 3 or 4% of the ethanol is carried off with the CO2 and in large distilleries this is worth recovering. If you lost 3% of your 80 litres that is another 2.4 litres. So the evaporation losses are real (2.4L) but small compared with the CO2 loss (40L).
  12. Adjusting Proof in Liqueurs

    Towards the end of last year (2016) I thought that I had virtually completed the liqueur blending calculator. But then some of the liqueur makers who have kindly been giving me advice pointed out that sometimes in the proofing process the sugar level is "close enough" to not need any further correction while the alcohol must be adjusted to within the TTB rules. At that stage my calculator required that both the alcohol and sugar be corrected at the same time. I have now added the ability to correct only the alcohol proof while allowing the sugar level to find its own level. Strangely, the math for correcting only alcohol is more complicated than that for correcting both together. A few examples of liqueur blending calculations can be seen at http://www.katmarsoftware.com/alcodenslq.htm I am now drawing a line in the sand and will not make any further changes to the program unless I discover a bug. Hopefully the program will be ready fot public testing by the end of February. Thanks to all of you who have continued giving me advice and encouragement - and for putting up with my broken promises regarding the end date.
  13. A Math Question

    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.
  14. A Math Question

    @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
  15. A Math Question

    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.