Vinegar runs

hornedrhodent · Post by **hornedrhodent** » Sat Jan 06, 2007 2:55 am

What is the purpose of a vinegar run after building a new still? Acetic acid has a higher boiling point than water so presumably the vinegar would just be concentrated in the boiler.

I can understand washing everything in vinegar to dissolve any oxides ,carbonates and fluxes but the reactions happen in the liquid state not the gaseous. I use very dilute H2SO4 for cleaning (rather than vinegar) usually before, during and after construction as it works so well.

Acetates and acetic acid produce ethyl acetate during distillation so why would you pollute your still with it.

Rudi · Post by **Rudi** » Sat Jan 06, 2007 4:32 am

I ran my still with vinegar because I didnt know any better and had read other posts of people doing this after building a new still and thought it must be right

. Would that explain afterward a vineragy/plastic smell that took lots of water flushing to get rid of? Is there a better way to remove the ethyl acetate??

Post by **Grayson_Stewart** » Sat Jan 06, 2007 5:10 am

but the reactions happen in the liquid state not the gaseous.

Why do they happen only in the liquid state?

hornedrhodent · Post by **hornedrhodent** » Sat Jan 06, 2007 5:38 am

="Grayson_Stewart"

Why do they happen only in the liquid state?

Lack of acetic acid in vapour.

Do you boil all the water off first and then distill acetic acid or does a small amount of acetic acid come over like water with ethanol?
How much acetic acid would make it to the top of a reflux still? Do you run it with output full open and no reflux (like a pot still)?

Assuming some acetic acid comes over in vapour would it be as effective as washing it with hot liquid vinegar?

What is the concentration (g/L) of acetic acid in vapour compared with liquid.

Stillhead · Post by **Stillhead** » Sat Jan 06, 2007 1:34 pm

Ethyl acetate formation during distillation is what is known as reactive distillation. Specifically, a reaction with ethanol.

If you were to put vinegar in your boiler without ethanol, this would not happen.

If you were to run vinegar it would evaporate. So too would water. Seperation is easy, but not perfect.

The higher boiling (118C) acetic acid will of course be the first to condense, whereas the water would remain in vapour state longer. This is a method to concentrate acetic acid with careful refluxing/condensing.

solubility and chemical reactivity are two very different things. One is a physical property the other chemical property. Acetic acid is a decent organic solvent and is great at dissolving flux and other impuritites found in a freshly made still.

Rebel_Yell · Post by **Rebel_Yell** » Sat Jan 06, 2007 1:56 pm

Stillhead wrote
The higher boiling (118C) acetic acid will of course be the first to condense, whereas the water would remain in vapour state longer.

Huh?

Stillhead · Post by **Stillhead** » Sat Jan 06, 2007 3:28 pm

Water evaporates with the acetic acid (to a lesser extent). If you cool this mixed vapour stream, the first to condense would be the higher boiling point acetic acid.

The point is that the walls of the still will condense acetic acid which should clean it nice.

Simplest way to see if it cleans good is to try it. Taste the distillate with french fries.

rangaz · Post by **rangaz** » Sat Jan 06, 2007 5:46 pm

dont you mean water would. I did this just yesterday and the first part of output was just water and slowly became more acidic. If it happened like you said, you would get the tails before the heads. The fact that the distilate comes out blue proves the fact that it is cleanig off the oxide layer along with some copper. The acid does not have to be in aqueous state (dissolved) to undergo acid reactions, in fact being in high temp gaseous state as vinegar should speed it up significantly

Stillhead · Post by **Stillhead** » Sun Jan 07, 2007 1:08 pm

The reason you don't get tails before heads is because the higher bp liquids condense first from the gaseous state, thereby not making to the output side.

If you run fast, you can get tails to come over throughout your run. Same with vinegar. Just don't condense when cleaning out your still. Let it run hot and fast and hot vinegar will coat the sides of the still continuously, and so clean it nicely.

Sorry that I did not explain it clearly enough,

B.

TRANSPLANTED HILLBILLY · Sun Jan 07, 2007 3:39 pm

If you run fast, you can get tails to come over throughout your run. Same with vinegar. Just don't condense when cleaning out your still. Let it run hot and fast and hot vinegar will coat the sides of the still continuously, and so clean it nicely.

Confirmed. I ran a vinegar/water solution in my rig way back and the smell of vinegar was very strong coming out. I did this hot and fast for an hour with no condenser water at all. The thing looked like a steam engine on its side.

hornedrhodent · Post by **hornedrhodent** » Tue Jan 09, 2007 2:59 am

Fairy nuff - now it makes sense - no cooling water to the condenser - no insulation - let her rip.

I cant do it though I've got an unpressurised water jacket and cant heat the boiler to more than 100 C so I'll have to be content with washing everything out with dilute sulphuric acid.

masonjar · Post by **masonjar** » Tue Jan 09, 2007 2:59 pm

If you run fast, you can get tails to come over throughout your run.

Something is bugging me about this statement. I understand what you're talking about with the vinegar and not running your condenser - but in a normal pot-still run where you condense everything, how could running fast or slow have any effect on what comes out?

pintoshine · Post by **pintoshine** » Tue Jan 09, 2007 3:37 pm

I agree with you masonjar. The amout of energy put into the system only makes all the stuff boil faster. It does not change the boiling equilibrium of the liquid to gas transition of any component. That wouldn't make sense and is not in accordance with the thermodynamics I studied for my degree. The temperature of the liquid and gas interface has to be the same. The amount of vaporizing is purely a factor of the current heat input and vapor pressure of all the total combined vapor pressures. You can't make one thing boil any earlier than another.

Stillhead · Post by **Stillhead** » Tue Jan 09, 2007 4:02 pm

It is a matter of thermodynamics. The latent heat of vaporization is a specific amount of energy to convert a liquid into steam at a constant temperature.

hearts are ethanol
tails are higher order alcohols (lets assume amyl alcohol)

These two mixed together form what is called a non-ideal mixture, meaning that since they have different intermolecular interactions, that the individual molecules will change into vapour in non-equal proportions.

These proportions can be graphed as phase diagram curves where we can look at the vapour composition curve. We don't need to get that far in depth though.

Just know that the more volatile liquid ethanol will boil off in a mixture with the less volatile amyl alcohol in a quantifiable proportion.

This proportion can be changed, however, if we add more or less heat. If we add just enough heat to boil the higher vapour pressure ehanol, there will not be much amyl alcohol coming off. But if you add a great amount of heat, this proportion changes. This proportion is also determined by the difference in boiling points (and amyl alcohol has a lower vapour pressure than ethanol, which means it takes more energy to raise its pressure to atmospheric and hence boil).

Therefore, when you add a great excess of heat, you will get more 'tails' in your product. This phenomenon is observable by many, and advice is given to dial back the heat and run a little slower for a product that is more pure.

pintoshine · Post by **pintoshine** » Tue Jan 09, 2007 4:25 pm

What you are stating is that the latent heat of vaporation constant for this non-ideal mixture, changes value, proportionally with the amount of heat input. Constants changing. Hmm. I'm quite unconvinced that it is even a possibility.

I am not disputing that the amount of heat that is put into the system will change its internal refluxing(even in a pot still), chemical reaction with the materials(copper salts), surface condensation(counter flow), blow by(vapor out the condenser) and all the other contributing factors. All I'm say is, changing constants, such as latent heat of vaporization constants for any ideal or non-idea mixture, is not possible.
The situation being described here is some other phenomenon but not changing constants.

This proportion can be changed, however, if we add more or less heat.

This is an out right false statment and there is no achedemic or imperical evidence that it is true. The charts you mentioned are based on moving temperature not the energy input. The Dynamics don't work that way.

Stillhead · Post by **Stillhead** » Tue Jan 09, 2007 7:42 pm

Constants certainly do not change.

What I was thinking was that the greater heat input would change the mole fractions in the vapour stream (because of the system not being ideal which causes a change in the intermolecular interactions at the surface of the liquid phase), as the rate of the boiling components increases.

I understand that this change is most likely negligible, but tails are detectable by the palate even at a concentration lower than the sensitivity of a GC.

I guess the biggest factor is in overrunning the separation capacity. The effective vaporizaion and condensation steps required to achieve a condensate of a given composition from a given distillate must not be 'overstepped' by too much heat input. With a pot still, this is very easy to do because the system gets hot right up to the condenser.

masonjar · Post by **masonjar** » Tue Jan 09, 2007 10:23 pm

I'm no chemist, but what you're saying still doesn't seem right to me. Let me set up a thought experiment to illustrate my dilemma.

Suppose we have ten identical pot stills each with an identical wash. Let's neglect energy losses and assume that the stills are perfectly insulated. If we apply exactly the same amount of power to each of the stills then we should get exactly the same vapor coming off of each still, right? Okay, so now lets merge these ten stills so that we have one boiler that is ten times the volume but it has ten equal isolated chambers inside - so effectively there are still ten separate stills but they are heated with one flame. So now we need 10 times the power in our flame in order for the same amount of vapor to come off of each chamber. Okay, nothing has changed. Exactly the same vapor should come off of each of the ten chambers assuming that the flame is equally divided, right? So then, if what you're saying is correct, the instant you remove the walls separating those ten chambers such that you create a single boiler that has ten times the volume, the vapor is suddenly going to change its composition, right? Because if we compare this new, larger still to one of the original stills, the only thing that has changed is the amount of power has increased by a factor of ten.

This seems intuitively incorrect to me, but again, I'm no chemist. The last time I challenged someone on a chemistry statement which I thought was erroneous, it was pintoshine, and he proved me wrong - so I'll admit that I'm on intellectual turf that is not my own. But, I am interested for the sake of education. I thought the reason it was advisable to go slow was because then you could make better cuts. If you had a gigantic swimming-pool-sized commercial pot still, would you only apply 1000 watts to it in order to prevent tails from coming over?

goose eye · Post by **goose eye** » Wed Jan 10, 2007 4:36 am

i dont no nothin about book learnt chemistree an dont no why somethangs
are an aint to sure if im understandin the question.
they say if you crank up 4 hi pressure burners you can get it to puke thru the doublein keg on thru the condencer with cold water in it. how they no im tole is a long story.

Post by **Tater** » Wed Jan 10, 2007 5:01 am

this might help fokes to figure out what hell your talking about

http://www.physchem.co.za/Heat/Latent.htm" onclick="window.open(this.href);return false;" rel="nofollow

pintoshine · Post by **pintoshine** » Wed Jan 10, 2007 6:02 am

they say if you crank up 4 hi pressure burners you can get it to puke thru the doublein keg on thru the condencer with cold water in it.

For his crude way of putting it, this is exactly the senario for the tails.
For 1 quantity of tails and all other things coming off with 1 quantity of heat into the system we get 1 tail condensening on the lid. With 2 units of heat we get 2 units of tails and everything else, with 1 tail condensening in the lyne arm and 1 on the lid. With 3 units of heat we get probably 1 tail on the lid, 1 tail in the lyne arm and one tail in the condenser. But always 1 unit of heat makes one unit of everything and 2 units makes 2 units of everything, all in the same proportions.
Now here is the most confusing thing.
Lets suppose our ratios of things are .5 water .4 alcohol and .1 tails. The rates of each fraction is constant at the mole fraction we are currently at. Adding more heat all the rates increase together.
Here is the confusing part. As soon as the mole fractions of the liquid mixture changes due to more alcohol leaving faster than the amount of water, the rates of evaporation changes for each fraction as well as the temperature. and continues to change over time.
Again adding or removing the heat changes all portions equally.

As far as it is concerned with acetic acid (vinegar) and water mixtures, some vinegar will always be in the mix. The water evaporates earlier in larger fraction at a faster rate than the vinegar. But you could push the vinegar farther toward the output with more heat due to overwelming the capacity and optimal seperation point of the machine.

But all the fractions coming off the boiler stay the same proportional to the heat input.

masonjar · Post by **masonjar** » Wed Jan 10, 2007 6:20 am

But as you crank up the heat, you are less likely to have any condensation on the lid. Or if you insulate the top of the pot, you are less likely to get condensation on the lid. If the condensation on the lid is desirable, then we shouldn't be insulating. Also, condensation on the lyne arm won't make any difference unless it's sloped upwards, which I've never seen. Once something condenses on the downward side, it's going in the product.

I thought puking is referring to the vapors pushing out liquids and solids on foam and bubbles, no?

Stillhead · Post by **Stillhead** » Wed Jan 10, 2007 8:21 am

I must yield under pintoshines heavy hand of truth.

I also believe that the non-ideal system behaves in a way that is more complex than say boiling hexane and cyclohexane (which have the same intermolecular interactions).

But this is an argument that I am not going to win.