To date, this topic is so hackneyed that many of you, dear distillers, will find it not interesting from the word "at all" and will not attract your precious attention to it at all. Maybe you will be right. However, despite the abundance of various information, it still remains very attractive and no less contradictory. Every self-respecting distiller, in addition to knowing and knowing what and how to smoke, is obliged to know "how much", as well as what this "how much" depends on. For beginners, these questions are always relevant.
We are all well aware that absolute alcohol is anhydrous, one hundred percent ethyl alcohol, the volume of which (AC) is very convenient to use in theoretical calculations of the yield of the final product. The chemical reactions of converting such important carbohydrates as sucrose, starch, glucose and fructose into alcohol are also well known. Finally, if we switch to molar masses in the equations of these chemical reactions, then one kilogram of sucrose, starch and glucose (as well as fructose) gives 0.682, 0.720 and 0.648 liters of absolute alcohol, respectively. These values determine the volume of the theoretical yield of absolute alcohol - an almost unattainable limit of all our efforts and capabilities. We will introduce the notation for these values: AC0.
The inaccessibility of AC0 is due to a number of known losses. Some losses of absolute alcohol are associated with the fermentation of braga, others - with its purification and distillation. Some part of these losses is inevitable, some part comes from the desire to make the product better, and some, of course, is our fault. It is believed that the practical yield of absolute alcohol is less than the theoretical one by 12-28%, subject to compliance with all technologies.
One of the approaches to accounting for losses is that a value close to the average, for example, 18%, is taken as the total value of all losses.
I will give three calculations of the practical AC yield for three different carbohydrates according to the formula:
AC = AC0 * (1 - 0.18) * % Carbohydrates / 100.
1. Morning, sand. AS = 0,682 * 0,82 * 0,998 = 0.558.
2. Barley, grain. AS = 0,720 * 0,82 * 0,564 = 0.333.
3. Grapes, berries. AS = 0,648 * 0,82 * 0,154 = 0.082.
In principle, the idea of taking a single average value for the total amount of losses is logical, simple and understandable. It would be possible to stop at this - for many this is quite enough. But, not for everyone.
And who is not enough, let's figure it out further.
Let us first consider the losses of the first level, that is, the losses occurring at the stage of alcoholic fermentation of the braga. Of course, the biochemical processes of alcoholic fermentation are very complex and unpredictable. In addition to ethyl alcohol and carbon dioxide, the main fermentation products, various by-products are formed - glycerin, acetic aldehyde, various acids, acetone and fusel oil. Part of the carbohydrates is spent on yeast generation, part is consumed by various pathogenic organisms. Some part of the starch does not turn into sugar. Some part of the sugar does not turn into alcohol. Some part of the alcohol evaporates. It is believed that such losses of absolute alcohol are from 7 to 19 % of the volume of the theoretical yield.
Why not at this stage set the value of the practical yield of absolutely alcohol, the same for all types of carbohydrates and equal to, say, 0.6. Let's call this volume the practical yield of absolutely alcohol of the first level:
AC1 = 0.6.
This means that for sucrose, fermentation losses are assumed to be about 12 %: AC0 * 0.88 = 0.682 * 0.88 = 0.60016.
For starch - about 16 %:
AC0 * 0.84 = 0.720 * 0.84 = 0.6048.
For glucose or fructose - about 7.4 %:
AC0 * 0.926 = 0.648 * 0.926 = 0.600048.
Even if this is not the case , it can always be checked in practice.
The value of AC1 = 0.6 turns out to be a very convenient parameter in theory.
Let's assume, for simplicity, that we have launched a sugar braga.
Let the mass of sugar C kg, the hydromodule is 1: M.
Then the volume of sugar dissolved in water will be 0.6 C, and the total volume of braga will be (M + 0.6)C. If the volume of absolute alcohol in the braga is equal to AC1 = 0.6 C, then its strength is determined from the formula:
К =100*0.6С/(М+0.6)С = 60/(М+0.6).
Consider the hydromodule from M = 3 to M = 5, with a step of 0.1. We get the table
The dependence K = 60/(M+0.6) can be depicted on the graph for clarity
Since the strength of the braga at 14° or more is critical for yeast, and the hydromodule at M > 5 is not rational, the graph clearly explains the choice of a hydromodule for sugar braga at the level of M = 4. In this case, the fortress of braga will be about 13°.
We have obtained an excellent way to theoretically determine the fortress of braga by its hydromodule.
For grain braga, the calculations will be a little more complicated, since different cereals increase in volume in different ways. But you can establish a similar dependence and calculate the fortress physically by measuring the total volume of grain braga.
For example, let the mass of grain cereals be 10 kg, the hydromodule is 1 : 3, and the volume of braga after measurement was 42 liters. This means that the volume of cereals occupied in the water was 12 liters (30 + 12 = 42). If the percentage of carbohydrates is known, say 75 %, then the volume of absolute alcohol in the braga will be 10 * 0,75 * 0,6 = 4.5 liters. From here we find the fortress of braga
К = 100 * 4,5 / 42 = 10,7°
and the dependence of the strength of grain braga on the hydromodule
К = 45 / (М + 1,2).
Let's return to the sugar braga. In one of my articles, I took 4 kg of sugar with a 1:4 hydromodule and after the first distillation I received 5.5 liters of raw alcohol with a strength of 36°. Using the method of determining the fortress of braga "retroactively", it can be established that the fortress of braga in my case was
К = 5,5 * 36 / (16+2,6) = 198 / 18,6 = 10,65°.
But how did it happen that according to theoretical calculations, the strength of sugar braga with a hydromodule of 1:4 should be 13°, and in fact 10.65°. The whole point is the loss of part of the braga during its cleaning. These are second-level losses. By the way, they are easily calculated. First, we calculate the theoretical volume of "pure" braga, that is, the braga that was poured into the cube:
198 / 13 = 15.2 liters.
Therefore, the theoretical volume of the" dirty " braga was
18.6-15.23 = 3.4 liters.
Do you measure the actual volume of "dirty" braga or pour it out? If you pour it out, there will be nothing to compare it with. So, after cleaning, about 3.4 liters of braga with a strength of 13°were lost. And this, in terms of absolute alcohol, is
3,4 * 13 / 100 = 0.44 liters.
Since in our case the volume of absolute alcohol of the first level was 0.6*4 = 2.4 liters, after cleaning the braga, this volume decreased to 2.4-0.44 = 1.96 liters. This is the so-called practical volume of absolute alcohol of the second level. Translated into 1 kg of sugar, this volume (in my case, the purification of braga) is
AC2 = 0.49.
When switching from 0.6 to 0.49, the share of absolute alcohol losses was:
(1 - 0,49/0,6)*100% = 18,3 %.
Of course, this is a very large loss of absolute alcohol. They are directly proportional to the volume of the braga itself. Is it possible to reduce these losses? Probably it is possible. Here everything depends on how we know how to clean the braga and whether we clean it at all. If the quality of the product is important to us, then we can sacrifice its quantity. If quantity is more important to someone, then quality is sacrificed. The main thing is that now we know how these losses are calculated. Finally, the loss of absolute alcohol during fractional distillation. For example, the losses during the selection of goals are, as a rule, another 10% of the AC2. Losses from tails are another 10% . As a result, the practical volume of absolute alcohol of the last, third level:
AC3 = 0.9 * 0.9 *AC2 = 0.8 *AC2.
What does all this give us? Let's say you put sugar braga and peel it, losing 10%. Then after the second fractional distillation, you will get absolute alcohol from each kilogram of sugar
AC3 = 0.8 * AC2 = 0.8 * 0.9 * AC1 = 0,8 *0,9 * 0,6 = 0.43 liters.
If the braga is not cleaned
AC3 = 0.8 * AC2 = 0.8 * AC1 = 0.8 * 0.6 = 0.48 liters.
I will give an example from my own experience. Raw materials: sugar-9 kg. Yeast-ordinary alcohol 200 gr. Water-tap water 36 liters. Braga wandered for 15 days. At the end, its strength on the alcohol meter is 9°. Saccharomer-drowned. The first distillation: the braga is not cleaned and distilled to water (~ 5°). The second distillation: the selection of heads - 7 %, the selection of the body - 5 liters with a strength of 93° at 20°C, the selection of tails -10 % (we do not produce). AC result = 5 * 93 / 100 = 4.65 liters.
And what does our theory say:
AC = 9 (sugar) * 0.93(heads) * 0.9(tails)* 0.6(AC1) = 4.52 liters.
The practical yield of alcohol turned out to be slightly higher than the theoretical one. In this example, the braga fermented really well-instead of the theoretical 13° for the hydromodule 1:4, the strength of the braga actually amounted (after recalculation by raw alcohol) to 13.7°. That is, the actual losses during fermentation were lower than the hypothetical ones.
Next, let's say you decided to put grain braga. The raw material is 10 kg of cereals. The composition of carbohydrates is 75%. After fractional distillation of crude braga, we get absolute alcohol
AC3 = 10 * 0,75 * 0,8 * 0,6 = 3,6 litre.
From 10 kg of corn groats, I get an average of 4 liters of 90° distillate, which is in principle consistent with the theory: AC = 4 * 90 / 100 = 3.6 liters.
Finally, we note that losses also occur during simple distillation. At the end of a simple distillation, part of the alcohol (about 5% ) remains in the bard. "Squeezing" this residue requires additional energy and time and therefore, most often, is impractical. This balance is usually not taken into account. In the end, if the actual yield of alcohol coincides with the theoretical calculations - you did everything right, you are lucky. If more, then at some level there were fewer losses. If less, then, on the contrary, at some stage the losses exceeded the calculated ones.
In any case, now you can calculate the yield of alcohol at the initial stage from almost any type of raw material and, after its output at each level, estimate possible losses. And in addition, you have the dependence of the fortress of braga on its hydromodule, according to which you can "estimate" the final fortress of braga.
Well, and who has read to the end-do not judge strictly. Thank you in advance for your comments. Write and, if you liked my article.