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DJF90
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) |
Quote: Originally posted by DJF90 ![](images/xpblue/lastpost.gif) |
Yes incompetence. It may pain you to hear it but if you try replicating someone elses work using different solvents/additives, you shouldn't be
suprised when it doesn't work. The experiment should have been performed exactly as Pok had reported, before modifications were made. Several pages of
poor results could have been avoided in such a manner.
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That point shows how LITTLE you know. The experiment has shown itself far less sensitive to materials than you think. All kinds of solvents, an
alternative catalyst, various physical forms of Mg... all work.
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I never disputed that. My point was that if someone had grabbed the right reagents and followed the prep as described, then it ought to have worked
first time, and it would have saved a lot of aggrevation/disappointment. Perhaps you should try listening to what is actually said and not your
narrow-minded mis-interpretation.
Don't you have an experiment to set up?
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blogfast25
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Quote: Originally posted by macckone ![](images/xpblue/lastpost.gif) | 1. Assuming
that the starting and final abv are the same for the pot and thumper that would give 33% enrichment.
2. Then the stream will start to pass through with the
water condensing out and alcohol being absorbed into the vapor stream. |
Point 1: care to explain that?
Point 2: that is part of the thumper fallacy, of course; the idea that the water condenses out. It remains a two-component/two-phase system, as shown
again by delta H.
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blogfast25
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Complaining about timing already? Bit early methinks. The hard work will be carried out by a fellow SMer, as I'm quite badly ill right now. 2 - 3
weeks till results, I think.
Consider yourself ignored now, little troll.
I'll address you if you write something relevant to thumpers.
[Edited on 17-2-2015 by blogfast25]
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blogfast25
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Quote: Originally posted by deltaH ![](images/xpblue/lastpost.gif) |
So the thermodynamic minimum number of ideal stages required to reach a certain concentration can be determined by drawing steps on this diagram
starting from your feed conc. This graph's axis in in mole fraction BTW.
The point is that starting somewhere close to the left-hand-side feed, two or three stages quickly moves you to pretty high alcohol concentrations,
thereafter it starts to get pretty damn hard to enrich further because the curve pinches the steps as you get close to the azeotrope.
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No contest there (but it's basic stuff).
We (Aga/me) found accordingly when comparing EtOH water distillation with and without a Vigreux column.
It explains why relatively simple stills can reach high 'proofs'.
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deltaH
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Quote: Originally posted by macckone ![](images/xpblue/lastpost.gif) | To answer a couple of blogfast's questions. The usual ratio is 1:3 from the original link. Assuming
that the starting and final abv are the same for the pot and thumper that would give 33% enrichment.
note that is simple mass transfer and doesn't account for heat transfer. Starting off with a cold
thumper seems to make more sense thermodynamically. Initially the stream will condense in the
thumper producing a higher abv in the thumper. Then the stream will start to pass through with the
water condensing out and alcohol being absorbed into the vapor stream. Once it reaches steady
state, the amount of water condensed must equal the amount of alcohol evaporated at a heat value
level. Note this is lower than when the thumper is absorbing heat. If the thumper is losing heat
to the environment the enrichment could be greater but never more than the mass value. Hypothetically
the thumper could be alcohol depleted at the end of the run yielding a higher gain than 33%. |
My simulations showed that a cold thumper does no enriching, enrichment only starts from about 73C under my conditions where it peaks rapidly, then
decreases with increasing temperature. In really, the thumper would operate somewhere above 90C I would imagine and reach that pretty quickly. I
would't proof a distillate sample before the thumper gets hot and reaches pseudo steady state.
[Edited on 17-2-2015 by deltaH]
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deltaH
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) | Quote: Originally posted by deltaH ![](images/xpblue/lastpost.gif) |
So the thermodynamic minimum number of ideal stages required to reach a certain concentration can be determined by drawing steps on this diagram
starting from your feed conc. This graph's axis in in mole fraction BTW.
The point is that starting somewhere close to the left-hand-side feed, two or three stages quickly moves you to pretty high alcohol concentrations,
thereafter it starts to get pretty damn hard to enrich further because the curve pinches the steps as you get close to the azeotrope.
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No contest there (but it's basic stuff).
We (Aga/me) found accordingly when comparing EtOH water distillation with and without a Vigreux column.
It explains why relatively simple stills can reach high 'proofs'. |
Exactly, just demonstrating the why to those who want to know.
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blogfast25
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People should consider the following thought experiment.
A thumper is full of almost pure EtOH and at equilibrium temperature. The vapour coming off the still is almost 100 % water, at about 100 C and is led
through the thumper.
What will happen? What will be the mole fraction EtOH coming off the the thumper (vapour) be?
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deltaH
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Nearly pure ethanol comes off if well mixed. As to the how, if you zoomed in on a single bubble of water vapour, water diffuses from the bubble into
the alcohol and alcohol diffuses from the liquid into the bubble. When the water vapour enters the liquid at the edge of the bubble, it releases it's
enthalpy of vaporisation and this is in turn absorbed by the evaporating ethanol. This continuous until equilibrium is reached (but never quite gets
there). But that's not to say that ethanol vapour doesn't also diffuse back into the liquid as well, it is a dynamic system and molecules are
constantly zooting too and fro, all the time, the rate of ethanol evaporating is simply large than it condensing and vice versa for water.
[Edited on 17-2-2015 by deltaH]
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Magpie
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The heat is supplied by the condensing of the incoming vapor from the still (latent heat). Efficient condensing would be promoted by having a high
depth of submergence of the still vapor outlet pipe.
Enrichment of the thumper outlet vapor is based on the vapor-liquid equilibrium (VLE) curve for the system ethanol-water. Enrichment is due to the
high relative volatility of alcohol to water in the region of interest, ie, around 20wt% alcohol of the liquid in the thumper. See Zombie's and
deltaH's VLE curves.
This has been stated above several times. I'm just trying to say it more succinctly.
In regard to calling people genitals, Conan O'Brien said "what kind of a dick would call a person a vagina?" ![:D](./images/smilies/biggrin.gif)
![Conan.jpg - 145kB](https://www.sciencemadness.org/whisper/files.php?pid=392777&aid=37561)
[Edited on 17-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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The heat of mixing of ethanol and water is exothermic by - 777 J/mol (Wiki), and all mixing is exoentropic.
So both ΔH and ΔG of mixing are negative (< 0).
To separate, that energy must be expended.
The idea that steam coming off the still provides that energy is fallacious. You need to look at the entire system: unheated thumpers simply introduce
more heat losses. Insulated ones act like steam pipes.
And heated ones are a little more complicated than one might think. Primary still vapour would have to condensed to just below its BP. Once the
thumper is, say, half full, a secondary distillation would be carried out by heating to BP and condensing the thumper vapours. But that wouldn't be a
real thumper...
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Fulmen
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Magpie:
yes, the heat is supplied by the condensing of the incoming vapor. But what condenses? Lets start with .1 mole fraction vapor entering an empty, warm
thumper. What will actually condense? Assuming that only a small fraction of the vapor condenses and full equilibrium between the phases?
We're not banging rocks together here. We know how to put a man back together.
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blogfast25
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My point precisely: EtOH content coming off is lower than it was initially. Gradually EtOH content in thumper and its vapours decrease.
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blogfast25
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Quote: Originally posted by Magpie ![](images/xpblue/lastpost.gif) | The heat is supplied by the condensing of the incoming vapor from the still (latent heat). Efficient condensing would be promoted by having a high
depth of submergence of the still vapor outlet pipe.
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You're creating energy ex nihilo, Magpie. That energy you mention has already been accounted for. Latent heat has to be expended too, to revolatise
the condensate.
[Edited on 17-2-2015 by blogfast25]
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blogfast25
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Assume ALL of it condenses. That condensate's been enriched in EtOH of course, no contest whatsoever.
The fallacy is to consider the primary vapour led into the thumper liquid is just a heat source: it's not, it contains WATER!
The mole fraction EtOH coming off the primary still is of course already lower than that of any condensate formed in the unheated thumper.
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deltaH
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I don't understand what you mean by this exactly. In a multi-trayed distillation column, heat is only inputted at the boiler and taken out in the
condenser (usually), but every tray in the column is enriching more and more. Every tray is not individually heated though. So how does that work
then?
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Etaoin Shrdlu
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Oh god I'm dying of laughter.
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deltaH
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Just saw near the bottom that they offer a two thumpers in series version, was wondering about that. I guess the enrichment effect of a second thumper
is probably much smaller than the first?
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blogfast25
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It doesn't have to be. Heat is provided by the boiler. The column itself, in a ideal scenario, operates adiabatically (ΔH = 0). Things constantly
condense and re-volatise. At the end you get your energy back in the condenser(s), MINUS what was expended for the separation.
My point with regards of Magpie's Latent Heat is precisely that: condensation occurs and latent heat is released but evaporation also occurs and
consumes latent heat.
[Edited on 17-2-2015 by blogfast25]
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Loptr
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They also spell it "Wight Lighting". ![:)](./images/smilies/smile.gif)
[Edited on 17-2-2015 by Loptr]
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blogfast25
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What's also funny is that his refluxer certainly works (despite the pompous name) but does anyone believe his little pickle jar would work? It's
certainly hopelessly under-dimensioned, looking at the size of the pot. That thing would fill up in no time!
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Molecular Manipulations
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Try Gallagher's "Milk of the Gods"
-The manipulator
We are all here on earth to help others; what on earth the others are here for I don't know. -W. H. Auden
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blogfast25
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Quote: Originally posted by deltaH ![](images/xpblue/lastpost.gif) | Just saw near the bottom that they offer a two thumpers in series version, was wondering about that. I guess the enrichment effect of a second thumper
is probably much smaller than the first? |
Two jam jars for the price of one! Triple distillation!
If you put five 'doublers' in do you get 300 % EtOH? ![;)](./images/smilies/wink.gif)
Where's Zomb when we need him? ![:D](./images/smilies/biggrin.gif)
[Edited on 17-2-2015 by blogfast25]
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deltaH
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) |
What's also funny is that his refluxer certainly works (despite the pompous name) but does anyone believe his little pickle jar would work? It's
certainly hopelessly under-dimensioned, looking at the size of the pot. That thing would fill up in no time! |
I was under the impression that the liquid level in the thumper doesn't keep increasing post start-up, no?
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blogfast25
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Link?
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blogfast25
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Quote: Originally posted by deltaH ![](images/xpblue/lastpost.gif) | I was under the impression that the liquid level in the thumper doesn't keep increasing post start-up, no?
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But Delta, HOW CAN THAT POSSIBLY BE?? An uncooled pot like that MUST act in part as a condenser, BY DEFINITION, almost!
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