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Gentlemen: I want to thank all of you again. The following is stuff I knew many years ago, but the brain cells containing this information went into a deep sleep because of lack of use. Those sleeping brain cells are now for the most part awake because of your help although still somewhat drowsy. I am happy to report that I don't think many of the cells have died over the years.

The following is taken from

http://www.chemguide.co.uk/physical/pha ... d.html#top" onclick="window.open(this.href);return false;" rel="nofollow

and will give a partial explanation of how distillation of water and ethanol works in any type of still for you that are interested.

A large positive deviation from Raoult's Law: ethanol and water mixtures



Using the diagram

Suppose you are going to distil a mixture of ethanol and water with composition C1 as shown on the next diagram. It will boil at a temperature given by the liquid curve and produce a vapour with composition C2.



When that vapour condenses it will, of course, still have the composition C2. If you reboil that, it will produce a new vapour with composition C3.



You can see that if you carried on with this boiling-condensing-reboiling sequence, you would eventually end up with a vapour with a composition of 95.6% ethanol. If you condense that you obviously get a liquid with 95.6% ethanol.

The above is obviously only a partial explanation of the distilling process of water and ethanol alcohol. Go to the above listed site to get a much more detailed explanation if you wish.

Thanks again.

Bert

Umm... That's what we've been trying to explain to you in the other topic... Most all of us who were trying to help you have read all of that, and more, from more than one source already...

I think the issue at hand is more related to whether running a pot still or reflux still than the theory itself... You can control the vapor temperature in a reflux still, via heat input and reflux, but you can't do that with a pot still unless it has a lot of passive reflux, such as when using an upward angled lyne arm or some other means... But even then you can't fully control the vapor temperature in the pot still because of the change in the water:alcohol mixture ratio...

This whole debate goes round and round what seems like a couple times a year and always ends the same... It's physics rearing its ugly head again...

rad wrote

Umm... That's what we've been trying to explain to you in the other topic... Most all of us who were trying to help you have read all of that, and more, from more than one source already...

I fully understand that rad, but as I stated "my brain cells relating to this were asleep". It just did not register. After a good nights sleep and reviewing the chemistry and physics of this on the Internet, the "cells finally woke up". I now remember and understand.

I only posted the graphs and some of the theory in a slightly different manner in hopes that it would help someone, obviously not you fellows that helped me because you already understand[/i], but others. To me the object of forums like this is to help each other. You all helped me, and I, in turn, am trying to help others.

Bert

Fair enough, ammo man... This might actually be better suited for the New distiller reading Lounge where it would be less apt to get lost in the shuffle...

As far as memory goes, I can remember all kinds of information, except for peoples names... That has driven me nuts my entire life...

Heck rad, that may be a good idea. If you wish, do that.

Bert

Hi,

Being new to distilling (pot stilling) I have been pondering this topic and have come to some conclusions in my own 'words'. I've also just purchased a motor controller to control my hotplate for my still. I've read through the relevant threads and have a couple of comments which I don't think were touched upon, or were mentioned in different terms.

• When you are distilling a liquid(wash), you heat it. Apart from the heat lost to the external environment*, the very method by which the wash sheds heat is by the latent heat of evaporation. Ignoring the effect of the reduction in the volume of wash in your still and assuming that the loss of heat to the environment stays the same (at equilibrium once the still has warmed up), if you keep the same input of heat(power) to your still, the heat loss of the wash via evaporation will match the heat input. Given that the boiling points of ethanol and water are not wildly different (only 22°C apart), throughout a run with a constant power input, you will also get a roughly constant liquid output.
  Simply put, the heat(power) you put in to the wash is lost as heat in the form of vapour. So to obtain a constant flow of vapour, and hence liquid condensate, you need to put in a constant amount of heat(power)
  
  * The only caveat here, besides the shrinking volume of wash, is that as the wash heats up the loss of heat to the environment will increase slightly.
• If the wash did not change in it's proportions of alcohol and water throughout the run - if it remained at, say, 10% alcohol and 90% water, then the temperature of the wash would not change (for a constant power input) through the run, and neither would the vapour temperature. But because alcohol is lost faster than water, owing to it's lower boiling point, the wash becomes lower and lower in ABV.
  Initially, your wash can lose heat more easily, because it is higher in alcohol content and alcohol is easier to evaporate. That is why the wash starts off at a lower temperature - because it is losing heat quite quickly by evaporating alcohol.* As the alcohol disappears it becomes harder and harder for the wash to lose heat and so it heats up. As it heats up more and more water is evaporated, and the method by which the wash loses heat increasingly becomes the evaporation of water.
  
  *One could easily show this if, in the middle of their run they suddenly introduced an amount of pure (liquid) ethanol at the exact same temperature as their wash. The boiling temperature of the mix would suddenly drop, much more ethanol would start to evaporate, the wash would actually cool down, and the liquid output would increase. You would be going backwards down the graph for a bit.
• It is a statistical thing. It is not the case that all of the alcohol will boil off first, and then the still will start to boil off the water. Both fluids are being turned into vapour at all times. The amount of heat in the wash is not evenly distributed. In fact the hotter the temperature the less evenly distributed is the heat; it is a Bolzmann distribution. So throughout the entire running of your still both alcohol and water are being evaporated, even at the very beginning. At 30° C for instance very little of anything will be boiling off, but 30° C is still (most likely) hotter than the ambient temperature, and so left long enough everything will evaporate off, just like a 30° C puddle on the ground would! – though granted through a still with a very small opening this would take a long time.

Hope someone finds this interesting
p

Having grappled with the concepts of controlling the power to the wash rather than working on temps, and understanding the importance of the steady application of heat in order to produce a carefully grauated temp gradient within the column, this has all been most interesting and informative.

My former efforts with my simple pot still produced a very pure spirit, but with a huge expenditure in time and four runs. I am very much hoping to retain my previous quality, very much quicker, and in one run.

I have yet to learn what approximate heat will produce an optimal result in both quality and volume of output, understanding that all these things are trade-offs, and also understand that this will vary according to the size of the boiler, wash, %of alcohol in it and probably other factors as well.

I see on revisiting not much more has been added. Yes Ammo man that was useful as I am still grappling with the issue.

I had managed to get a ss column made but as I had made a move, lost the boiler I was making. Have now after a lot of searching around been able to come up with a decent8 gall Keg, as so often happens we got two on the same day. The smaller one was sadly not much use, but the 8 Gall should be good

We are working on the top section which involves welding the column base to the inverted screw in top valve that was in the Keg, and then finding a suitable element to fit in to the bottom.

Praire Piss has a very nice set up with two elements however my partner and I have been looking into just have one controllable element. We are working with normal 220 V household supply and so looking at 3 kilowatt immersion element powered from a 13Amp fused plug. There is a firm that makes elements that I was talking to and they were proposing a 2 1/4" stainless screw-in base with a coil element of approx 30" length total that would coild down to about 10" long. The width of the keg is about 14" so that should fit OK. they will come back on the cost but I have a horrible feeling it may not be cheap

I do not know much about heating the wash up to a working temp and I would think that to avoid any puking up the column we should possibly be limiting the ammount we begin to run to about six or so galls. I need to try and see where the surface of the wash would come in relation to the neck/exit point, then find if anyone has experience of how full one can safely fill.

Also if this sort of element would bring the wash up to a working temp in a reasonable time, before then being powered back to hold a steady temp gradient within the column. Does heating up the wash too quickly cause problems within the column? Does applying too much power to the element cause any burning/charring or off flavours in the run?

Still so much to learn.

HI Res
Try Ebay. Water Heater Elements. Loads of them here fairly cheap. See what you think.
Regards
Patch

Thanks Patch, I'll do that.

Still searching the forum for any specifics on the application of heat to a wash. I think that a slow steady build up may be desirable for an optimum gradient within the column, but this is simply surmise.

I have been in touch with element manufacturers here in the UK but any custom build is probably going to be more than an off the shelf. However, since the set-up so far has been minimal in cost, I would be prepared to pay a bit if it was felt that this could improve quality and production of our run.

As you said, slow and steady is the way to go. The reason some are using two elements, one larger and one smaller, is that it gives them much finer control of the heat into the pot. A power control on a 1500 watt element is going to have a lot coarser adjustment than the same control on a 250 watt element (think percents).

There are two subjects I haven't seen discussed here that I think might help understand what's going on. The first: Temperature is the measure of average kinetic energy. This means for any mass, lets say of liquid(s), some few molecules will actually be a very low kinetic energy, even freezing, and some will be very high. The average will be whatever temperature your thermometer reads (well, the average near the thermometer, anyway. See the next paragraph). This is why liquids evaporate, even at very cold temperatures. Even water ice will lose some mass to vaporization (look up sublimation). The reason this is important is this: In a standing pot of room temperature water, say, a molecule gets enough energy (we won't go into HOW here..) to become vapor. Almost every time this happens, it will 'collide' with 'colder' molecules and become colder by energy transfer. If this molecule were at the surface of the water, there is a chance it could 'escape' and stay water vapor, but more than likely air molecules would cool it down, and back in the drink it goes. Some do manage to escape, and we get water vapor in the air (a.k.a. humidity!). The same thing happens in a liquid mixture, in all different proportions depending on physics, but the important thing to remember is that even in a perfect (impossible) mixture where everything is at the boiling point (Average Temperature) of alcohol, there is a 50/50 chance that an alcohol vapor molecule will hit something colder and become liquid again before escaping.

Sorry this is so long, but the second topic is just as important. Getting your whole still to the same exact average kinetic energy is physically impossible. It would be really nice if you could - get everything to the exact boiling point of alcohol and bingo! Perfect distillation! Unfortunately, we live in the real world. First: Evaporation removes energy from a liquid by converting it from kinetic (temperature) to potental energy (heat of vaporization). This means that the energy you are adding isn't changing temperature (kinetic energy). Its changing liquid to vapor (potential energy), and the only way to measure that is by the amount of vapor. Second: Everywhere there's a temperature boundary (the walls of your still, the packing, the condenser...) there is a temperature gradient. Your heating element may be 250 deg. The inside wall of your pot will be a degree or more less (the thicker, the cooler) because the liquid is taking heat from it. As that heated liquid circulates, it heats more liquid, but becomes cooler by doing so. This plus vaporizaon makes a temperature gradient in the liquid itself. Those initial bubbles that form on the bottom of a pot before it boils don't look like much, but they are roiling heat transfer factories, with liquid constantly vaporizing and condensing at the bubble wall. All this is to say you can't get the same perfect temperature even in the liquid alone, forget the whole still.

If you're having trouble understanding why you can't control your still by temperature, this is the reason. You cannot know the perfect boiling temperature of your liquid, because you can't know the exact amount of different liquids in it (well.. technically, you could, but it would cost a LOT of money to find out). And even if you did, controlling the temperatures everywhere in your still down to the small part of a degree needed to make your distillation perfect, or even good, is also out of most of our reach. (Both of these are why commercial distilleries are very good at what they do. They can and do know to a very fine degree what's in their mash, and they have the equipment to control heat to a fine degree as well.)

Let me restate something from above: Once your pot is boiling, any added energy is being converted to vapor. This conversion from kinetic to potential energy means that the temperature (average kinetic energy) doesn't change. The only way to measure it is to measure the volume of the vapor being produced. And how do we do that? Yep.. by watching the speed of the condensate drip!

Yep.. simple, really, when it comes down to brass tacks. You may have thought it was magic, or maybe you're doing it by 'feel'. You tell the newbies, "Just watch the drip! It ain't rocket science!" Well, this time, it is!