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Luke
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Who said anything about tossing stuff out? Goes back into the reflux still to make vodka 
Quote: Originally posted by Zombie  | | Quote: Originally posted by Luke | | Quote: Originally posted by Magpie | I know that making moonshine is a batch process so concentrations must be changing both in the pot and in the thumper as the batch proceeds.
But if the thumper operates as in my last post then it's alcohol content has to be steadily decreasing. This concentration should decrease until the
outgoing vapor in equilibrium with it matches that of the incoming vapor from the pot. At this point a steady state would be reached and no alcohol
concentration increase across the thumper would exist. But the pot is steadily decreasing in alcohol concentration also so maybe a steady state is
never really reached. It seems that this would depend on batch size, thumper size, etc. |
This is exactly right, alcohol content decreases as the batch goes on. This is a good thing in pot distillation though as different compounds come
over at different times.
The still operator will take small samples throughout the run and then decide which ones to put into the final spirit. Usually the first quarter or
third of the run is discarded, and the last third is discarded.
The stuff in the middle of the run is called the hearts and is diluted to 65% abv and then aged on oak. |
We HAVE to hang out Luke. But you have to toss out less hootch...
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Zombie
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| Quote: Originally posted by Luke | Who said anything about tossing stuff out? Goes back into the reflux still to make vodka
| Quote: Originally posted by Zombie | | Quote: Originally posted by Luke | | Quote: Originally posted by Magpie | I know that making moonshine is a batch process so concentrations must be changing both in the pot and in the thumper as the batch proceeds.
But if the thumper operates as in my last post then it's alcohol content has to be steadily decreasing. This concentration should decrease until the
outgoing vapor in equilibrium with it matches that of the incoming vapor from the pot. At this point a steady state would be reached and no alcohol
concentration increase across the thumper would exist. But the pot is steadily decreasing in alcohol concentration also so maybe a steady state is
never really reached. It seems that this would depend on batch size, thumper size, etc. |
This is exactly right, alcohol content decreases as the batch goes on. This is a good thing in pot distillation though as different compounds come
over at different times.
The still operator will take small samples throughout the run and then decide which ones to put into the final spirit. Usually the first quarter or
third of the run is discarded, and the last third is discarded.
The stuff in the middle of the run is called the hearts and is diluted to 65% abv and then aged on oak. |
We HAVE to hang out Luke. But you have to toss out less hootch...
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You're my new BFF!
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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Fulmen
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Blogfast is right, the claims as I understand them violate physics. What comes out of the first stage is appr. 50/50 EtOH/Water (assuming a constant
10% in the boiler), we can all agree on this right? Well, if nothing is returned back into the still, the total product will be 50/50 no
matter what you do. An unheated thumper can split this into two products by partial (or fractionate) condensing, in theory you could get a
final product of ~75% but only by producing a corresponding amount of 25% in the thumper. And if you were to mix these together afterwards you will
end up with the same 50% that came out of the still. There is no way around this as the total product must be the same as the vapor composition
leaving the still.
Now if the distillate from the thumper is returned to the boiler you have a full second stage capable of producing a net product above 50%. But in
order to reboil the condensed liquid you will have to add more energy, so the claim of producing a net purer product without any extra energy is
false.
[Edited on 16-2-15 by Fulmen]
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Zombie
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The first statement is incorrect. re: 50 /50.
The vapor coming from the boiler is going to be more EToH vapor due to it's higher volatility. 60% ABV.
There is math to figure it but from real world experience a 10% ABV mash / wash will always deliver at least 120 proof or 60% ABV. This is true in all
cases due to the Alch evaporating at a lower temp than the water.
The mixture of 10% will boil at approx 189 -190*f. At this temp the water itself can not evaporate so what little does is actually a phase product in
the aclh vapor.
you all have to get the 50 / 50 notion out of your thinking. There is only one time this will ever happen in distilling, and that is when the final
condensate is bottled. It never happens inside a boiler or thumper.
This chart might ring the bells for one of you. You'll see as the vapor transfers so do the boiling points. As they transfer so does thr required
power. Thus the magic thump is created. (albeit a short life span)
[Edited on 16-2-2015 by Zombie]
[Edited on 16-2-2015 by Zombie]
[Edited on 16-2-2015 by Zombie]
[Edited on 16-2-2015 by Zombie]
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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Fulmen
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Never mind the specific numbers, they aren't important in this context. Lets just say the mash (at whatever conc) produces 50% EtOH vapor at a certain
point, resulting in 50% alcohol when run straight into a condenser. With me so far? No matter what you do after the vapor leaves the boiler the total
amount of water and alcohol cannot change. So if vapor leaving the boiler is 50% EtOH, you WILL get equal amounts of alcohol and water. Now it is
possible to split this into two products, one being stronger than 50% and one being weaker, but you cannot change the total amounts (unless blowing
vapor to the atmosphere, in which case you loose more alcohol than water).
By returning the distillate from the thumper into the still (turning it into a second stage) you can change this, but this will cost more energy.
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Zombie
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Ok correct...
But that 50% now went into the thumper. By the time you reach a 50/50 ratio the boiler is almost out of alch,
., and the thumper is alch rich. The boiler is now near 200 - 205*f, and the thumper only requires approx 185*f to release it's load.
In the beginning they were equal, and required equal power. This is exactly where the magic "free" power is born.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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Fulmen
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You're missing a quite basic point here. Whatever leaves the boiler leaves. If nothing is returned then your total amount of condensed water and
alcohol will be the same as with a single stage.
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Zombie
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I separated this on purpose.
Every time Alch vapor passes thru heated mash /wash, it collects more alch vapor in the form of micro bubbles about to vaporize.
This is where the rise in ABV occurs in the thump. It is now approx 18% ABV (boiler 2%). This really third round of vaporization (one in the boiler)
(one in the pass under the thumper liquid) (third is the thump output) collects a higher % of alch in the vapor, and leaves more water behind.
Fractional Plate distillation proves this out.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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Zombie
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You have to re read this. I can't explain it any differently.
Maybe this. Run a single stage distillation, and run till all the alch has transfered. The result will be around 50-60 abv.
Do this again, and the result will be around 70 80 abv, and so on.
The thumper IS the again, and the heat is provided from the incoming vapor due to shifting boiling points.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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blogfast25
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| Quote: Originally posted by Oscilllator |
Personally, I think what happens is this:
The hot vapour leaves the still at a concentration of 50% (for example). This vapour travels into the bumper where it condenses, heating the water.
Once enough of the vapour has condensed, the temperature in the bumper will be raised to the extent that it is above the boiling point of ethanol.
When this happens, water will still condense in the bumper, however the the ethanol will not. This means that the ethanol concentration in the vapour
phase has been increased to above 50%, and the energy for this separation has been provided by the condensation of water vapour in the bumper.
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Nope. You're telling yourself tales and falling for them.
An uninsulated box like a thumper BY DEFINITION runs colder than the condensate that forms in it.
Even raising the temperature again by means of incoming vapour means by the Laws of Thermodynamics the consensate's temperature can only reach that of
the incoming vapour, as heat can only flow from hot to cold.
As a result the equilibrium composition leaving the thumper is the same as that entering the thumper, at best.
No amount of mumbo jumbo is going to change that.
I'm surprised people have such trouble seeing such elementary things.
[Edited on 16-2-2015 by blogfast25]
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blogfast25
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| Quote: Originally posted by Fulmen | | You're missing a quite basic point here. Whatever leaves the boiler leaves. If nothing is returned then your total amount of condensed water and
alcohol will be the same as with a single stage. |
Fulmen points to Conservation of Mass, correctly.
Where did all the water go, boys?
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Zombie
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Please Blog. Try the experiment. It will prove you wrong.
Your theories are correct but in practical application it works.
I didn't take a nap so I'm burnt on this one. but do the experiment, and tell us how it goes. It might take you an hour or two at the most.
Brain chemistry will not prove this one out.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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blogfast25
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Fractional distillation works and we know why, pal.
Thumpers are nonsense and we know why too.
[Edited on 16-2-2015 by blogfast25]
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Zombie
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They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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subsecret
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| Quote: Originally posted by Fulmen | Blogfast is right, the claims as I understand them violate physics. What comes out of the first stage is appr. 50/50 EtOH/Water (assuming a constant
10% in the boiler), we can all agree on this right? Well, if nothing is returned back into the still, the total product will be 50/50 no
matter what you do. An unheated thumper can split this into two products by partial (or fractionate) condensing, in theory you could get a
final product of ~75% but only by producing a corresponding amount of 25% in the thumper. And if you were to mix these together afterwards you will
end up with the same 50% that came out of the still. There is no way around this as the total product must be the same as the vapor composition
leaving the still.
[Edited on 16-2-15 by Fulmen] |
Yes... The efficiency of the thumper depends on how long it's operated before being drained of dilute alcohol. Of course, the water never goes
anywhere, and it's only a matter of time before the concentration of alcohol in the thumper decreases past where it's useful to produce concentrated
alcohol. As long as the temperature of the thumper is kept below the temperature of the pot but just above the BP of alcohol, there would be
some increase in efficiency toward the beginning of the run. This is assuming that the inefficiency comes from the fact that the still is a
good bit hotter than the BP of alcohol, so more water is carried over. The thumper's concentration would begin to decrease immediately, and you'd have
no improvement in separation. If the still is hotter than it needs to be, the thumper is just a fractionating column.
From a practical standpoint, draining the thumper would lead to a lot of loss; it could be recovered.
Fear is what you get when caution wasn't enough.
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blogfast25
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As regards testing a thumper, in principle that is not necessary. It would be like having to prove you can't (by chemical means) make gold from lead:
First Principles show that, so no specific experiment is needed or useful.
Having said that I'm talking to someone on the forum who MIGHT want to run such an experiment (NO guarantees that he'll accept, though).
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Zombie
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Awesomeness,
Your on the right track. BUT MOST of the water is left in the boiler because you are always below the boiling point of the water.
Only a small percentage is carried to the thumper as a phase product in the alcohol..
[Edited on 16-2-2015 by Zombie]
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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blogfast25
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Sigh.
The still CANNOT be hotter 'than it needs to be': the BP of an EtOH/water mixture at atmospheric pressure depends ONLY on mol fraction (ABV if you
prefer) EtOH.
The thumper a 'fractionating column'??? Do you know what a fractionating column actually is?
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Zombie
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| Quote: Originally posted by blogfast25 | As regards testing a thumper, in principle that is not necessary. It would be like having to prove you can't (by chemical means) make gold from lead:
First Principles show that, so no specific experiment is needed or useful.
Having said that I'm talking to someone on the forum who MIGHT want to run such an experiment (NO guarantees that he'll accept, though).
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That will be a blessing.
A constant output rate from the thumper must be maintained throughout the run. Just as in actual conditions.
No matter the scale, nothing more than a broken stream of product.
This rate maintains the equilibrium, just as a fractional column must not be flooded.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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blogfast25
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More obfuscating mumbo jumbo: what is a 'phase product'?
I'll give you this: you'd make a good thumper salesman, blinding people with pseudo-science!
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blogfast25
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| Quote: Originally posted by Zombie | That will be a blessing.
A constant output rate from the thumper must be maintained throughout the run. Just as in actual conditions. |
G-d, your reading difficulties are terrible: WOULD, not will.
I thought it only took a beer can!
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Zombie
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Meaning a product other than pure water molecules. A combined ethanol/water molecule.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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blogfast25
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HAHAHHAHAHAHAHAHAHAHHHAHAHAAH!
Please don't make me piss myself!
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blogfast25
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| Quote: Originally posted by Zombie | A constant output rate from the thumper must be maintained throughout the run. Just as in actual conditions.
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Any special pixie dust I need to get? Do you have a secret formula?
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Luke
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I think you should try this blogfast. Then you can work out why it works. You seem to understand physics and it would be interesting for me to get
that take on it.
Its going to take you you very little to set it up and if it doesnt work, you'll be able to tell everyone how wrong they are (which you seem to enjoy
doing).
Btw, blogfast you kind of come off like a know it all wanker, you should probably work on that.
[Edited on 16-2-2015 by Luke]
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