meerkat

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.

@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

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.

@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 http://www.katmarsoftware.com/alcodenslq.htm 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.

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!

Thanks very much for the offer - I will certainly contact you when I need help.

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.

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.

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 http://www.katmarsoftware.com/alcodenslq.htm 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 http://adiforums.com/index.php?/topic/7617-spirit-strength-for-fruit-infusion/

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.

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.

RobertS, I don't think you are grasping at straws at all - I believe you have nailed it. And it would be very easy to verify. If the boiler pressure decreases when the water is fed then that's it.

It confused me as well, but I like the new version. On the top right you will find the "Unread Content" link. This is similar to the old "New Content" except that it only shows new content that you have not read. I like to see all the new content - including stuff I have read. You can fix this by selecting "Read Status" and setting it to "Everything". If you save this as a new stream then you can select it later from the "Activity" option on the main menu where it will be shown as a custom activity. Overall I think the new format has a much cleaner look.

If the size of the charge and the ABV are consistent from run to run, and you are not losing vapor to the atmosphere, then inconsistencies in batch time can only mean that your rate of heat input is varying. The rate of heat input is the product of three factors: Heat rate = heat transfer coefficient x area x temperature difference One of these 3 must be varying. Looking at them in turn: Heat transfer coefficient. This is pretty much fixed by the geometry of your system and the most likely thing to change it would be fouling of the heat transfer surface. You would see this as a gradual decline in heat transfer and I doubt that it would improve by itself to cause inconsistencies. Area. One might think that this cannot change, but if the condensate is not removed quickly enough the condensate builds up and masks some of the area. This method of varying the area is sometimes used to control heat input to the reboiler on large continuous columns. Check your steam trap and strainer for fouling or blockages, and maybe re-check the sizing of the steam trap. A steam trap should typically be sized for 3x the average flow rate. Temperature difference. The temperature of your mash won't vary much between batches, so for the temperature difference to change your steam temperature (and therefore steam pressure) would have to change. Watch the pressure on the steam inlet to the jacket. The problem could also be with the steam trap. If it fails open and passes steam the pressure cannot build up in the jacket and the temperature will be lower than desired. Of course with your spirit runs the rate of reflux will also impact on the batch time.

There have been threads about software in the past that never went anywhere, so I thought I should post a short update. Also, I would like to thank the members that have continued to correspond with me directly with data, advice and encouragement. It gets lonely being shackled to a computer all day so your input is much appreciated. The only way I know to do blending calculations accurately is by having all the quantities in mass terms. So my first task was to be able to convert sugar levels in gram/liter or SG to Mass %, and to convert liqueur alcohol strengths from ABV, Proof or SG to Mass %. These conversions are now working - see the attached screen shot. There is still a lot to do, but at least now I believe we have a fighting chance of getting the math to work. Although the conversion calculator isn't hugely useful, I do see one potential advantage. In the example shown two simple and accurate tests allow us to get a good estimate of the alcohol in the liqueur without having to do a lab distillation. A simple evaporation and weighing exercise can accurately give the sugar level, and an hydrometer or EDM can tell us the SG. The calculator combines these two bits of information and calculates the liqueur alcohol content for us. Here it is shown in Mass %, but we could have chosen ABV or Proof as the display. The final gauging would have to be done with the lab distillation to meet the TTB requirements, but do you think this short cut could save some time during the blending process?

This is correct and is the reason why the failure is often not quickly detected. The pressure in the jacket does not distort the outer part of the jacket, while the inner wall of the jacket (i.e. the shell of the actual tank) can be bulging inwards or even ruptured but it is not visible from outside the tank. I remember one instance where we only discovered the problem when the distorted tank wall fouled and stalled the agitator. My previous post was referring to the pressure in the jacket destroying the shell of the tank from the outside. I agree with you that a cylindrical vessel can withstand a much greater internal pressure than external pressure. The larger the tank diameter the more susceptible it is to this type of damage from the outside, but I have seen relatively small bore (2") jacketed piping where the inner pipe was crushed almost flat by steam in the jacket.

For all practical purposes, 15 psi of vacuum in the tank is the same as 15 psi of pressure in the jacket. The inner shell only "sees" the difference in pressure between the inside and the outside. A clever variant of the external jacket is the dimple jacket. The outer shell is dimpled at about 6" centers and each dimple is welded to the inner shell. This dramatically reduces the unsupported area and allows the use of much thinner inner shells. Google will show you pictures. One of the less predictable failures I have seen with a jacketed vessel was one where only low pressure cooling water was put through the jacket. Unfortunately the inlet and outlet water valves were closed and when hot resin was dumped into the tank the trapped cooling water tried to boil with nowhere to go and totally destroyed the inner shell. For this reason even cooling jackets often have pressure relief valves.

Pete's suggestion of using the proofing process to confirm the quantity of 192 Proof spirit is excellent, and very easy to do. Let us imagine that you add exactly 3,000 lb of RO water and after mixing and resting the diluted spirit measures 83.24 Proof. You can use the "Blend to Achieve Target Strength" calculator as before, but instead of setting the 192 Proof spirit (i.e. Source 1) as the known quantity you should set Source 2 as the known quantity because you know you added exactly 3,000 lbs of water. Set the strength of the Target Blend to the measured 83.24 Proof and the calculator will tell you that you used 261.5 gallons of 192 Proof spirit. Now the question is "how much additional water must I add to get to 80 Proof?". Click the "Copy Blend" button at the bottom of the window, and then click the "Paste" button in the top left of the Source 1 panel. This tells the calculator that you want to use your results from the previous blending as Source 1 in the subsequent blending. Change the target strength of the final blend to 80 Proof and the calculator shows that you need another 206.56 lbs of water (Source 2).

Yes, provided that your 264 gallons was measured at 60°F your answers are perfect. The 5022 lbs of 80 proof will give you 633.6 gallons of product.

bpsisk, as mentioned by PeteB above I am also trying to get to the bottom of this type of calculation. Please can you give a bit more detail on the 60 ml of sugar and other ingredients you wrote of in post #7. Was this in granular form, or was it a syrup with the sugar and other ingredients dissolved? Do you have the weight of the sugar and other ingredients? This information should help me understand how 60 ml of added ingredients caused only a 50 ml overall volume increase. The "disappearing" 10 ml can be correct under certain conditions - I am trying to find out what those conditions are. Thanks for your help.

Provided that you do not apply too much heat and flood the plates and condenser, the pressure in the pot would not be any different when distilling water than when distilling a wash. The temperatures would be a bit higher and this would lower the heat transfer to the pot, but improve it in the condenser, so unless you abuse the system I would not expect the pressure to rise at all. The question of whether the distilled water would be good enough for proofing can only be answered with relevant experience (which I do not have), but the energy cost and the time lost for product distillation have to be considered.

I agree with James that cavitation can be a real problem with condensate pumps. These pumps always require careful selection and positioning relative to the tank. If you have an existing pump then check the NPSH requirement before you raise the condensate temperature. The amount of heat required to raise the condensate a few degrees is small compared with the heat required to vaporize the water and generate steam, so the economic advantage may not be much in insulating the condensate lines. In my experience, insulation of condensate systems has mostly been for personnel safety and inaccessible lines are often left uninsulated.

The dimensions you are proposing are in the right ballpark for a commercial distillery. There are members here running smaller setups, but I believe your sizes are reasonable. A useful source of information for proprietary items like mixers and pumps is the vendors themselves. They keep information on the applications they have delivered in the past and will be able to advise you on what is practical. There are members here from all around the world. I am sure some of the more hands-on members will be able to address the rest of your questions. As an aside, a "tonne" is an SI unit equal to 1000 kg, making it equivalent to 2204.6 lbs, not 2240.

The closest I have seen to your experiece was a yellow-green heads stream in a continuous neutral spirit plant. The plant wisdom was that this was diacetyl (butanedione). In this plant the fermenters were steel, the stripper column mostly copper and the rectifier fully stainless steel. I would be surprised if a colored copper salt could make it to the top of your column, unless you have significant liquid entrainment up the column. If it is a salt it would have to be formed in the final condenser, but the fact that your pilot condenser is SS tends to rule out that theory. My opinion is that it would have to be formed during fermentation. Did you make any changes in the fermentation temperature or agitation or nutrients that coincided with the appearance of this problem?

The attached graph shows the equilibrium data for ethanol/water mixtures. Unfortunately the scales are in mole fractions, but when expressed as ABV or mass % the conclusions remain the same. The X axis is the concentration of ethanol in the liquid and the Y axis is the concentration of ethanol in the vapor when boiling at atmospheric pressure (760 mmHg). For example, if you boil a 10 ABV% liquid, which is about 0.033 mole fraction the vapor that will be generated will be at a mol fraction of around 0.2, which translates to an ABV of 46%. This is what you would achieve in a simple pot still with no trays and no reflux. As you move across to the right on the X axis the vapor concentration continues to be above the liquid concentration until you get to a mole fraction of 0.8943 (=97.2 ABV%). At this point the vapor generated is at exactly the same concentration as in the liquid. This point where the liquid and vapor have the same concentration is called the azeotrope. Because there is no enriching occuring at this point you can never go beyond this point by normal distillation. The only way to to get beyond this point is to add another chemical as an entrainer. Cyclohexane is often used for this purpose. To get to the azeotropic concentration of 97.2 ABV% would require an infinite number of trays or infinite reflux. The practical limit, with an economic reflux ratio, is about 96% ABV.