• Content count

  • Joined

  • Last visited

  • Days Won


meerkat last won the day on February 15

meerkat had the most liked content!

Community Reputation

7 Neutral

About meerkat

  • Rank
    Active Contributor

Contact Methods

  • Website URL

Profile Information

  • Gender
  • Location
    South Africa

Recent Profile Visitors

647 profile views
  1. 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.
  2. 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).
  3. 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 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.
  4. 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.
  5. @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
  6. 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.
  7. @nabtastic- thank you very much for your comments. I am pleased to hear that it is best not to use very high strength spirit for maceration. I have been able to increase my data for the strength of the spirit to around 70% while still maintaining accuracy, but I wouldn't like to push it much further. The way the program has been set up is that after the maceration stage the proof and sugar loading would have to be measured and keyed into the calculator. The blending calculator is now working and if you would like to see a couple of screen shots of it in action they are available at The quantities of the ingredients (and of the product) can be given in mass or volumetric terms - except for the flavoring and any dry sugar which have to be calculated in mass. I believe that your iterative method of blend, measure and calculate is the only way possible. There are two factors that can speed this up and minimize the number of iterations necessary. The most important of course is the measurement stage. If you don't know accurately what is in the liqueur you cannot calculate what to add. Accuracy of measurement only comes with experience and care. The second factor is the calculate (or re-calculate) stage. Hopefully my program will enable distillers to convert their accurate measurements into accurate calculations of what still needs to be added and allow them to avoid too much trial and error. I still need to do a bit more testing of the program and to write up the Help system, but hopefully the release isn't too far off now.
  8. Hi Meerkat,

    I was wondering if you had any progress on the liqueur calulator and a possible release? I'm trying to make an Ouzo with 2.5% by weight sugar as per TTB rules and this would be very helpful to figure out the true proof without a lab still.



  9. The packing size to column diameter ratio is a compromise between a couple of factors. In general, the larger the packing the lower the pressure drop and the lower the tendency to flood - meaning that you can put more heat into the column and get a higher throughput. On the other hand, the smaller the packing the higher the fractionating power and the lower the HETP (Height Equivalent to a Theoretical Plate) and therefore the greater the separating power of a given height of packing. Smaller packing tends to be more expensive (per cubic foot and per pound) and it generally works out cheaper to get the extra separating power by using a larger packing and a taller column, but of course this is not always possible. The thickness of the wire (or plate) does not play a large part in the mass transfer properties and is mainly set by the physical strength required to support the bed of packing above and for ease of fabrication. The 1/10th rule of thumb is to avoid channelling. In the middle of the column the rings randomly interlock and touch each other and any path up or down through the packing will involve a lot of twists and turns. But against the wall there can be no interlocking and there will be continuous channels that allow the gas to bypass the packings. This lowers the efficiency of the packing but below the 1/10th ratio the impact is small. I haven't seen any hard data indicating the best length for each piece, but most packings (eg Raschig and Pall rings) are "square" with the length close to the diameter. But Cascade Mini-Rings are popular and effective and they have a length roughly 50% of the diameter. The Cascade Mini-Rings have "tongues" that are bent into the center of the ring and this would prevent the rings from interlocking too far and I guess that for SPP having the length and diameter equal would be a good design. PS Fantastic practical machine - totally awsome!
  10. Thanks very much for the offer - I will certainly contact you when I need help.
  11. Thanks to @Silk City Distillers and @captnKB for the replies. The 40 mass% that I chose was an arbitrary number and I definitely have to raise this limit. It makes sense to raise it as far as possible and I will have to spend a day or two squeezing my data to see how far I can take it and still get reliable answers.
  12. Thanks very much Mech and Brian for your responses. @TheMechWarrior - In my preliminary workings I have set the upper limit on the alcohol content at 40 mass%, or about 50 %abv depending on the sugar content. I probably need to increase this a bit for high alcohol liqueur products but my main worry is in the different components that are blended together to make the liqueur. If 90% abv is used to make an infusion and then that is used as part of the final blend then the program must be able to handle whatever abv the infusion finishes at. Thanks for this information as it confirms that whatever the strength the infusions can be, I do need to increase the limit to cope with liqueurs that are actually produced. @brian 73 - Unfortunately I do not have a choice of processes. I want the program to be able to be able to deal with anything (reasonable) that a liqueur maker could throw at it. I'm sure some liqueur makers will use a juice extract to provide the flavor and some of the sugar, but some may use the fruit with spirit to make the substrate for the liqueur. I would like to be able to accommodate both options.
  13. Hi CaptnKB, I'm a bit embarrassed about how long it is taking me to get this done. The bits that have been completed can be seen at but since then I have been working on what Todd (Palmetto Coast) origninally asked above, i.e. how to do the actual blending calculations. I found it very hard to put together a mechanistic formula that I could use in the computer to tackle the wide range of blending options that are possible. With whiskey or vodka you have only alcohol and water to contend with, but although liqueurs only have two more ingredients (flavoring and sugar) the complexity grows by much more than just the doubling in components. But this week brought me my Eureka moment and I believe I have solved all the math and logic problems and now it is simply a case of putting it all together. One aspect that I have not resolved yet is the range of proofs I need to deal with. Although it seems rare for a liqueur to contain more than 90 proof, some of the ingredients may contain higher proofs than that. I have posted a question regarding the proof of spirits used in making fruit infusions and if you can help with that please see my post at
  14. I am developing a computer program for blending of liqueurs containing sugars. I do not run a distillery myself, but I understand one possibility that I should cater for as an ingredient in liqueur production is an infusion made by soaking fruit in rectified spirit. What proof range is used for the rectified spirit? An internet search found only DIY instruction for people who buy a bottle of vodka and make their own infusion at home. I have struggled to find information on how this is done on a commercial scale. Apart from the starting proof of the spirit, I am also interested in knowing how much water and sugar are extracted from the fruit together with the flavors. I don't want to steal anybody's detailed recipes - I only need to know the strength range the program must cover. Any advice will be greatly appreciated.
  15. There are a couple of assumptions behind the TTB method that are not explicitly stated. In practice these assumptions have no effect on the final result, but knowing that they are there does make it easier to understand what is being done. The first is that the interpolation that is being done should strictly be referred to as LINEAR interpolation. If we drew graphs of all the values in Table 1 we would see that the lines are curved. But in the same way that you can build a gently curving wall with straight sided bricks, you can make up a curve with lots of straight elements if the straight sections are all short enough. Because Table 1 is quite fine grained with 1 proof and 1 degree F steps we can safely use linear interpolation. The second assumption is that the corrections for proof and temperature are independent of each other. This is what allows the TTB method to create two separate correction factors (the 0.32 and the -0.144 in their example) and then just add them together. Again, this is not really true, but it only impacts the result in the second or third decimal place and we are not often interested in that. I find it easier to follow these things if I can see pictures. When I studied math our calculus professor told us that at school we had been under the false impression that geometry was a branch of mathematics. Subsequently I found out that the professor was wrong and that great scientists like Paul Dirac and Albert Einstein were able to formulate their mental experiments because of their very strong understanding of geometry. So now I don't feel so bad relying on pictures and graphs. There are two graphs attached. These follow the TTB method of treating the effects of the proof and temperature separately. I apologise to the Americans for my notation - I habitually use commas to mark the decimal but I think you use a full stop. In the first graph it is assumed that the Indicated Proof is 80 and that we are trying to find the true proof if the temperature were 68.36F. From Table 1 we know that (when the indicated proof is 80) for a temperature of 68F the True Proof is 76.4. This is marked as point A. Similarly at a temperature of 69F the True Proof is 76.0. This is marked as point B. We join points A and B with a straight line (remember we are using LINEAR interpolation). Now we can mark the desired temperature of 68.36 on the temperature axis. This would be 36% of the distance between the two marked temperatures. Since we have moved 36% of the distance along the temperature axis we must also move 36% of the distance on the True Proof axis. This is because we are using linear interpolation. The step on the True Proof axis is 0.4 (i.e. 76.4 - 76.0) so the vertical step is 0.36 x 0.4 = 0.144. And because the direction is downwards the 0.4 is negative and the value is taken as -0.144. We can do a similar exercise for when the Indicated Temperature is held at 68F and we are investigating the effect of a change in Indicated Proof. In the second graph the straight line we are using for interpolation goes upwards so we get a positive value of 0.32. Now we can combine these two effects. We know from Table 1 that at an Indicated Proof of 80 and an Indicated Temperature of 68F the True Proof is 76.4. This is our base point. The deviation from the base is -0.144 for the change in temperature from 68F to 68.36F and the deviation from the base is 0.32 for the change in Indicated Proof from 80 to 80.32. Because we have asumed these two effects are independent of each other they can just be added together to give the 76.4 + 0.32 - 0.144 = 76.576 in the TTB example. Plugging all these values into the AlcoDens program gives a True Proof of 76.57. AlcoDens uses the full OIML equation and does not rely on linear interpolation, so the fact that the answers are so similar indicates that the underlying assumptions have not influenced the result at all.