Doug's continuous distillation

[drmiller100]

It is easy to have ENOUGH plates. 3 is not enough. 72 is enough. 100 is enough. 100 is just as good as 72 and there is no downside to too many plates.

I found 28 inches of half inch marbles to be barely enough(regardless of column diameter), so 36 inches seems to be plenty.

Theory is great and all but there are a lot of great people here who DO and TRY.
I respect those who DO greatly.

[stevea]

I recently found that the Rayleigh batch distillation equation can be used to calculate the energy necessary.
It's necessary to integrate the inverse of the y-x (vapor mol fraction minus liquid mol fraction) from the starting to ending x. I won't bore you with the details, but
Starting with 10%ABV wash:
If you boiloff to 1% ABV lees you need 441204 J/liter.wash or 484 W-hr/gal.wash
If you boiloff to 0.5% ABV lees you need 553206 J/liter.wash or 582 W-hr/gal.wash
That is strictly for heat of vaporization of water and ethanol. Heating the remaining wash as the BP increases will add ~6% or 7% more energy.
To get the vaporization energy down below 200W-hr/gal would require collecting only 4% or 4.3% of the ABV; well under half the total ethanol.

You can get figures close to the oft' quote 404W-hr/gal (I get 416) if you start at ~13%ABV and recoup 90% of the ethanol.

[LWTCS]

Makes sense to me Steve as we arrived at 408 as mentioned.
Edit to add: That Jibes with Manu's number too.

However, practical data has shown that it appears to be less than that. Are you calculating for heat recovery?

You talk to Adam or Brent over at Boundry Oak yet?

By the way,,,what have you built?

[stevea]

In a continuous still why would you count the vaporization of water?
....

B/c you still need to vaporize the ethanol, and that means vaporizing an mount of water related to the remaining wash ABV. Vaporizing either requires energy.

In a pot/batch still the wash drops from say 10%ABV to 1%ABV over time, and this means you boil-off ~20% of the mols of the wash. At 10%ABV you get roughly 3 water molecules vaporized per molecule of EtOH (3:1 ratio). By the time you get to 1%ABV, it's near a 25:1 molecule ratio.

In a continuous stripper system, you still need to vaporize the ethanol, and you necessarily vaporize water in the same way, but the shift from 10%ABV to 1%ABV takes place throughout the height of the stripper. You are doing the same vaporization over space instead of time.

The Successive plates are "powered" by the condensing vapor giving up it's heat of vaporization. So of course the excess water selectively condenses and you end up with more ethanol and less water in the product ... but you don't get the ethanol into vapor phase until you pay the cost by vaporizing a lot of water.

[LWTCS]

Sure.
Takes more energy / heat input to flash 6% than it does to flash 12%

[stevea]

Makes sense to me Steve as we arrived at 408 as mentioned.
Edit to add: That Jibes with Manu's number too.

However, practical data has shown that it appears to be less than that. Are you calculating for heat recovery?

You talk to Adam or Brent over at Boundry Oak yet?

By the way,,,what have you built?

That's only heat of vaporization sufficient to drive off the spec amount of EtOH.
There actually are ways to recover heat from the condenser (~80C) and add this to the 'bottoms' (~100C) but it involves something like a heat pump. Google 'vapor re-compression distillation'. I understand that Hiram Walker installed such a system a couple decades ago - but I doubt it's used on anything small-ish.

[stevea]

You talk to Adam or Brent over at Boundry Oak yet?

By the way,,,what have you built?

Not yet (on either question).

I've worked on design and a test implementation of an automated continuous mashing system, but that's way over-kill for small amounts. It only becomes useful at a several thousand liters of wash per day.

I've recently cobbled together some ideas for a small (<100l), low power batch mashing + temp controlled fermentation system with a fair potential for automation.

I intend to design a distilling system with certain features:

Continuous (semi).
Electrical (simpler control overall)
Relatively low power & throughput (~2kW in as heat), plus liquid-air HX cooling.
Tightly controlled product. I'm only interested in whisky product so <70%ABV, or 0.40 mol-fraction. That makes my life much easier than you azeotrope guys.
Probably reboiler rather than steam injection (yeah, there are practical difficulties).

I'm parsing my way through several textbooks, with a lot of questions unanswered.
I wonder why no one is controlling distillation with pressure (and yes - below atmospheric for safety).. The P-x-y curve look 'tighter' than the T-x-y curve but it should work. How could you make a multi-stage flash distillation system ?
Plate-design (bubble/sieve/valve) looks pretty 'hacky' to me from a theory POV. Is that really the best we can do ?
De-entrainment - getting the 'mist' or 'dew' out of the vapor. IIRC there were some electrostatic precipitator systems to do that in air - what if .... ?

More questions than answers.

[LWTCS]

You talk to Adam or Brent over at Boundry Oak yet?

By the way,,,what have you built?

Not yet (on either question).

I've worked on design and a test implementation of an automated continuous mashing system, but that's way over-kill for small amounts. It only becomes useful at a several thousand liters of wash per day. I would be very interested in that project. Is that for another outfit? Are you able to talk about that at all?

I've recently cobbled together some ideas for a small (<100l), low power batch mashing + temp controlled fermentation system with a fair potential for automation.

I intend to design a distilling system with certain features:

Continuous (semi).
Electrical (simpler control overall)
Relatively low power & throughput (~2kW in as heat), plus liquid-air HX cooling.
Tightly controlled product. I'm only interested in whisky product so <70%ABV, or 0.40 mol-fraction. That makes my life much easier than you azeotrope guys.
Probably reboiler rather than steam injection (yeah, there are practical difficulties).

I'm parsing my way through several textbooks, with a lot of questions unanswered.
I wonder why no one is controlling distillation with pressure (and yes - below atmospheric for safety).. where on the still do you reckon is the best place to control through pressure management? The P-x-y curve look 'tighter' than the T-x-y curve but it should work. How could you make a multi-stage flash distillation system ?
Plate-design (bubble/sieve/valve) looks pretty 'hacky' to me from a theory POV. Is that really the best we can do ? what are the alternatives then?
De-entrainment - getting the 'mist' or 'dew' out of the vapor. IIRC there were some electrostatic precipitator systems to do that in air - what if .... ?

More questions than answers.

[drmiller100]

Sure.
Takes more energy / heat input to flash 6% than it does to flash 12%

Yup. Latent heat of vaporization of water is about 3 times that of etoh.

Change paradigms.

If you had 2 percent etoh for 7 pounds of wash how much energy to boil off all but .1 percent alcohol?

Imagine you have a stripping column and you had 2 percent dripping into a 5 gallon pot of pure water boiling merrily at 212. How much energy to boil the etoh off?

A stripper column is recursive. You don't have to boil off the water of the 10 percent IF YOU HAVE ENOUGH PLATES below the insertion point.

2 gallons per hour, 190 proof, on 4800 watts.

On a pot still, in reflux producing 190 proof at max efficiency, there is liquid dripping back into the pot.

What percentage is that liquid?

On a continuous still, with pure water in the boiler, what is the percentage dripping into the boiler?

[drmiller100]

In a continuous still why would you count the vaporization of water?
....

At 10%ABV you get roughly 3 water molecules vaporized per molecule of EtOH (3:1 ratio). By the time you get to 1%ABV, it's near a 25:1 molecule ratio.

The Successive plates are "powered" by the condensing vapor giving up it's heat of vaporization. So of course the excess water selectively condenses and you end up with more ethanol and less water in the product ... but you don't get the ethanol into vapor phase until you pay the cost by vaporizing a lot of water.

Umm, the first paragraph is incorrect. The vapor of a 10 percent wash is purt near 50 percent etoh. It takes 3 times the energy to do the water as the etoh though.

On the second I'm inserting my wash half way up the column. So the same way successive works up it also works down.

The only WORK done is the 95 percent vapor at the top of the column and heating up the wash being inserted into the column.

[stevea]

In a continuous still why would you count the vaporization of water?
....

At 10%ABV you get roughly 3 water molecules vaporized per molecule of EtOH (3:1 ratio). By the time you get to 1%ABV, it's near a 25:1 molecule ratio.

The Successive plates are "powered" by the condensing vapor giving up it's heat of vaporization. So of course the excess water selectively condenses and you end up with more ethanol and less water in the product ... but you don't get the ethanol into vapor phase until you pay the cost by vaporizing a lot of water.

Umm, the first paragraph is incorrect. The vapor of a 10 percent wash is purt near 50 percent etoh. It takes 3 times the energy to do the water as the etoh though.

On the second I'm inserting my wash half way up the column. So the same way successive works up it also works down.

The only WORK done is the 95 percent vapor at the top of the column and heating up the wash being inserted into the column.

There is no error. I'm talking MOL FRACTION (MF), you're talking ABV. We're both correct w/ that clarification

10%ABV liquid is 3.3%MF. From my VLE calculation, the vapor above 3.3%MF at equilibrium will be 25.3%MF or 54.2%ABV so,
LIQUID: 10%ABV = 3.3%MF
VAPOR: 54.2%ABV = 25.3%MF (~3:1 H2O;EtOH ratio).

1%ABV liquid we have:
LIQUID: 1%ABV = 3.1%MF
VAPOR: 11%ABV = 3.66%MF (~1:26 ratio)

My point is that as you go farther down the stripping section of your column, the ABV & MF drop, and you must vaporize more water to vaporize the ethanol. It takes almost the same energy to vaporize one molecule of ethanol or water but ...
At 10%ABV you must also vaporize 3 molecules of H2O to vaporize 1 molecule of EtOH.
At 1%ABV you must also vaporize 26 molecules of H2O to vaporize 1 molecule of EtOH.
As you can see the energy per molecule of EtOH goes way up, farther down the column.

My *feeling* is that mol fraction is clearer b/c water and ethanol have *about* the same Heat of Vaporization per mol, and 90+%of all the energy for distillation involves that vaporization.

===
Yeah - I know how a stripper works.

>>The only WORK done is the 95 percent vapor at the top of the column and heating up the wash being inserted into the column.

Huh ? You are certainly adding heat (WORK) to the stripping section to get separation. IIRC you add live steam to your base.

[drmiller100]

You don't have to ADD heat at each level in the column. The while point to a packed column is that the water amd etoh trade heat, and the etoh goes up amd water goes down.

Yes, in a traditional pot still the math sucks horribly. To strip the pot you have to boil all the water and alcohol many many times.
With a continuous still the math is very different. Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.
Separating the water from etoh isn't a chemical reaction which is endothermic or exothermic.

The magic of continuous is you only vaporize 95 percent etoh sort of once. Yes water is vaporized at the bottom to carry the heat up the column. Then the column does its magic to separate etoh from water.

[bitter]

wow this is real awesome. Be nice to have for stripping sounds very efficient and fast! 20 GPH of wash nice!

B

[Hügelwilli]

Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.

Perhaps I understand you wrong. But what about reflux? For rectification you need reflux. And reflux costs energy. For a result of 96%abv directly from a wash you clearly have to reflux something like 4/5 of the top vapor regardless how long your column over the feed is or how efficient your packing is. 4/5 means it needs about 5 times the energy you claim it needs, 9/10 would mean you have to invest 10 times the energy. Of course a rectification column saves energy compared to a multi-distillation-potstill, but it is not as extreme you write. Or I understand you wrong.

[jward]

Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.

Perhaps I understand you wrong. But what about reflux? For rectification you need reflux. And reflux costs energy. For a result of 96%abv directly from a wash you clearly have to reflux something like 4/5 of the top vapor regardless how long your column over the feed is or how efficient your packing is. 4/5 means it needs about 5 times the energy you claim it needs, 9/10 would mean you have to invest 10 times the energy. Of course a rectification column saves energy compared to a multi-distillation-potstill, but it is not as extreme you write. Or I understand you wrong.

I may not understand either. If one is running a boiler to push vapor up a column I can't see how that vapor is not adding heat to every level of the column.

[bitter]

I could be wrong but I think the bottom tub has the element near the ground. takeoff for waist is above that to ensure enough steam sent up the column to collide with the wash being added after heating up..

[Hügelwilli]

Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.

Perhaps I understand you wrong. But what about reflux? For rectification you need reflux. And reflux costs energy. For a result of 96%abv directly from a wash you clearly have to reflux something like 4/5 of the top vapor regardless how long your column over the feed is or how efficient your packing is. 4/5 means it needs about 5 times the energy you claim it needs, 9/10 would mean you have to invest 10 times the energy. Of course a rectification column saves energy compared to a multi-distillation-potstill, but it is not as extreme you write. Or I understand you wrong.

I may not understand either. If one is running a boiler to push vapor up a column I can't see how that vapor is not adding heat to every level of the column.

Yes, the vapor adds heat upwards. This is very efficient. But then the reflux condenser takes out the heat and sacrifices it for rectification. For making fuel you have to reach very high abv. This demands much reflux what means you will have much heat loss.
You could use a part of the wasted energy to preheat the wash. But this won't help much, because it needs much less energy to heat the wash to its boiling point than to vaporize it. And you cannot use the reflux condenser energy for vaporizing the wash, because the reflux water temp won't reach the boiling point of the wash.

[LWTCS]

You could use a part of the wasted energy to preheat the wash. But this won't help much, because it needs much less energy to heat the wash to its boiling point than to vaporize it.

Help much?
Mate, 10% wash will make vapor at about 190 F (88C).
If you can recover enough heat to get beer injection temps to 172F (for example) from some ambient room temp at any given location, that's not nothing. That is a huge contribution to reduce the temp split (delta t).
We're talking about a savings of (help me out here stevea) 100,000 btu/h on a system that can produce 4 barrels of finished product per shift ( for example ).

[Hügelwilli]

Mate,

for example you have 10%abv wash and 10lt of it gives you 1lt 96%abv. To heat up 10lt of 10%abv wash to its boiling point you need 800Wh.
Then when you condense 1lt 96%abv at the top of the column, you get 209Wh. If you condense it 10 times because of the reflux ratio you get 10 x 209 = 2090Wh. So for the first liter of 96%abv you had to invest 2090 + 800 = 2890Wh, minimum. It is more if your still and your preheater are not perfectly insulated.

The condenser water holds now 2090Wh. This energy you can use to heat up wash. Because the vapor has 78.3°C the cooling water will have maximum 78.3°C. So you can heat wash maximum to around 78°C.
To heat 10lt 10%abv wash from 20°C to 78°C needs 640Wh.

So from the invested 2890Wh you can recover 640. This is 22%. That's what I mean "won't helps much" considering the additional effort to build a still that can recover the heat.
You could recover more, but not to heat up your wash, but perhaps to heat up your distillery or to heat up water for mashing.
And -because you are mentioning barrels- when distilling without as much reflux whiskey or rum the math is different. I would not have written this post in a thread about continuous rum or whiskey stills.

[stevea]

I'm busy studying both a chemE thermo text and Miller's "Whisky Science ...". And I think the correct concepts have been mentioned here.

Almost ALL of the energy goes into vaporization, AND almost same additional energy into condensation of distillate+reflux. We can almost neglect (a) the heating of stuff to `90C in liquid phase, (b) the minor difference in the heat of vaporizing EtOH vs H2O, and (c) the heat of mixing (mixing 0.5 mol EtOH & 0.5mol H20 is slightly exothermic to the tune of ~400J/mol [=3 W-hr/liter of product]). If we used a heat-pump (or thermosiphon) we could recover most of the condensation cooling energy and apply it to the wash, ((then the heat of 'de-mixing' (tiny 3W-hr/liter product) would be the only *theoretical* net input)) but the 3rd law of thermodynamics says that the ~80C condensation energy will NOT flow to the ~95+C wash without a nudge from another source (e.g. heat-pump).

In terms of PRACTICAL still design ...

A/ Columns should be insulated WELL ... esp small columns. You'll lose ~1.5kW per meter^2 of vertical surface from a still to air when there is a 100C vs 20C temp diff and passive convective air flow. A 2inch (5cm) diameter column might push ~10 liter/hr of wash and require ~5kW of heat, but you can lose an easy 10%-15% and more of that, and experience poor stability if a gust of air creates weeping when you have an un-insulated, thermally conductive column . It causes the "supposed" equilibrium vapor to condense on the sides and drop to the lower plate without contributing EtOH or energy to the upper plate. This is a pure loss of energy efficiency.

B/ Columns should be made from very poor thermal conductors, like glass and not copper. I did a calculation a few days ago there a 10%Abv[3.3%MF] wash has a BP around 93C and the vapor and next-plate are ~ 54%ABV[25%MF] with a BP ~83C. So there is a ~10C diff between the lower-vapor and the upper liquid at this plate. Consider a 2"diam copper column - over a vertical length of about ~1cm we have a 10C temperature difference - what does that mean ?

If you have 10C temp differential across 1cm (0.01m) of 2" (5cmD) copper pipe - this requires ~112 W to maintain. (borosilcate glass is ~300x less = negligible).
https://www.petersenproducts.com/Copper ... s/1979.htm
2" USA L copper tube has a thickness of 0.070 inch so a cross-section of (~2*pi*r*thick = 0.29-3m^2). Copper has a thermal conductivity around 386 W/m.K
https://www.sciencedirect.com/topics/en ... oefficient

Power = 386 (W/m.k) * 0.29e-3 m^2 * 10C[K] / 0.01m = 112W

The 112 Watts comes from condensing vapor on the lower vapor section tube (pure loss) and excessively heating the plate-liquor on the upper plate despite not-adding enough EtOH (plate inefficiency by dis-equilibrium). This 112W (~225W over a 2" azeo column) is energy and plate inefficiency.

C/ **MOST IMPORTANT**
Plates should NOT be thermally conductive between lower-vapor and upper liquid. Ideally the vapor from below perfectly transfers heat AND concentrated vapor to the plate liquor above by DIRECT CONACT. If you make a 2" plate from typical 0.05" (1.270mm) copper plate and maintain a 10C difference, you need to flux a shocking 6+kW!

Power = 386 (W/m.k) * (pi*(00254m^2) * 10C[K] / 0.00127m = 6160W !!!

So the reality is that you can't maintain a 10C diff using a 5kW heat source! Instead a LOT of the vapor condenses on the plate-lower adding it's energy but not it's enriching EtOH to the upper plate by direct contact. The plate liquor above therefore has lower EtOH levels from the loss of lower vapor AND also by the increase of heat & boiloff Basically you waste much of the 5kW by vaporizing wash, only to have it return to the pot while transferring energy to a too-dilute plate-liquor above. Note this is plate inefficiency, not energy inefficiency. To create an efficient plate, you want to flux hot, enriched, vapor from below into the plate liquor, and also flux cooler excess plate liquor from above into the lower liquor(reflux). A plate disk made from a few mm thick PTFE or borosilicate glass would be far more efficient (<12W transfer vs 6kW). You want thermal conductivity horizontally, not vertically !

So two concepts apply. Energy efficiency is improved by insulation, or the extraordinary use of a heat-pump from top-vapor condensation to pot-heat. Plate efficiency is improved by only allowing heat transfer from lower-vapor DIRECTLY in contact with plate-liquor. Allowing heat transfer w/o enriched vapor transfer (column or plate conduction) is plate inefficiency.

===

Reflux IS a major energy cost, every bit as energy-expensive as product, so we want to design to minimize reflux for a given amount of product separation. That generally means more plates (and plates are inefficient, as noted above, so *lots* more plates). McCabe-Theile, then Fenske & Fenske-Underwood-Gilliland(FUG) gave us nice ways to estimate the MINIMUM number of theoretical plates necessary. The number of practical plates, given inefficiency, is often 50% higher at industrial scale and perhaps 100% higher at small scale. Reflux is related in complex ways to the number of effective plates, but there is also a minimum reflux required (therefore minimum energy loss required) . The big oil-cracking guys like to keep reflux ratios under 1.0 and as low as 0.50 (so between 33% and 50% of vapor condensate and energy goes "down the tube"). That is their trade off between capital costs (more plates) and energy efficiency. At a small-distillery scale maybe ~10 meter tall columns are practical, and for home operations perhaps ~3 meters. This places a definite limitation on plate-spacing which also impacts efficiency.

Sigh, enough 4now.

[stevea]

Hmm - I think a write-up I made a few days ago never saw the "submit " button. Miller ""Whisky Science: .." Springer does a complete analysis frm basics of energy for a continuous whisky column, With steam injection, 8ABV%feed, 65%ABV distillate is converts to (from memory) ~630 W-hr applied to the base per gallon of feed. With a re-boiler instead of steam injection the heat drops to ~475 W-hr/gal. Slightly less energy is also required at the condenser for each.

The reflux does NOT just impact the rectification section, but instead you have a nearly constant molar flow of vapor upward, and a different quantitative molar flow of liquid reflux downward. Of course the feed adds to that liquid down below the feed point (or vapor up if super-heated) but the bottom line is that you don't get a free exchange of energy to concentrated EtOH on the stripper-vapor. You are adding energy that must be sacrificed to reflux eventually in order to maintain the separation. The power used in that reflux condenser is "felt" all the way back to the heater at the bottom of the stripper.

AFAIK - no free lunch, 400+ W-hr per gallon feed UNLESS you invest in recycling condenser heat to the bottoms with a heat-pump or something clever.

[Hügelwilli]

stevea, I don't understand everything you write, but there is something I think I can contribute something:

Reflux IS a major energy cost, every bit as energy-expensive as product, so we want to design to minimize reflux for a given amount of product separation. That generally means more plates (and plates are inefficient, as noted above, so *lots* more plates). McCabe-Theile, then Fenske & Fenske-Underwood-Gilliland(FUG) gave us nice ways to estimate the MINIMUM number of theoretical plates necessary. The number of practical plates, given inefficiency, is often 50% higher at industrial scale and perhaps 100% higher at small scale. Reflux is related in complex ways to the number of effective plates, but there is also a minimum reflux required (therefore minimum energy loss required) . The big oil-cracking guys like to keep reflux ratios under 1.0 and as low as 0.50 (so between 33% and 50% of vapor condensate and energy goes "down the tube"). That is their trade off between capital costs (more plates) and energy efficiency. At a small-distillery scale maybe ~10 meter tall columns are practical, and for home operations perhaps ~3 meters. This places a definite limitation on plate-spacing which also impacts efficiency.

Oil cracking seems to be different. In ethanol production to reflux 33 or 50% of the vapor would be not enough by far, regardless how many plates you have. According to McCabe-Thiele and the best VLE-data I can find, for 96%abv with 50 theoretical plates at 100% efficiency from 10%abv in the boiler you need to reflux 73% of the vapor. For 97%abv it is 92% reflux (the aceotrope is 97.2%abv). And if you start with 50%abv it is almost the same.
The conclusion is, there is no way to save energy besides recovering and insulation. High abv needs much energy.

[LWTCS]

High abv needs much energy.

Truth.
And a robust production speed needs even more.

[drmiller100]

You could use a part of the wasted energy to preheat the wash. But this won't help much, because it needs much less energy to heat the wash to its boiling point than to vaporize it.

Help much?
Mate, 10% wash will make vapor at about 190 F (88C).
If you can recover enough heat to get beer injection temps to 172F (for example) from some ambient room temp at any given location, that's not nothing. That is a huge contribution to reduce the temp split (delta t).
We're talking about a savings of (help me out here stevea) 100,000 btu/h on a system that can produce 4 barrels of finished product per shift ( for example ).

Continuous is pretty cool. I run the wash through the upper condenser and it comes out at 150ish degrees.
Then I run the wash through a directional heat exchanger using boiler waste to heat the wash some more. The wash actually starts boiling in this heat exchanger before it enters the column!!!

In this case the higher the abv the LESS energy it takes as the final vaporization at the top has less water.

[drmiller100]

I'm busy studying both a chemE thermo text and Miller's "Whisky Science ...". And I think the correct concepts have been mentioned here.

Almost ALL of the energy goes into vaporization, AND almost same additional energy into condensation of distillate+reflux.
Sigh, enough 4now.

The Mol thing is just noise.

I agree ALL of the energy used vaporizes the etoh from the top of the column to the condenser. This is the only real work in the system.

So we agree the plates themselves don't LOSE energy. The plates are a way for etoh to go up and water to go down. Adding more plates doesn't mean it takes more energy.

Instead of a 36 inch reflux column making 95 percent etoh I have that plus ANOTHER 36 inch column stripping the wash.

On my column, at the bottom, what drips into the boiler? 99.6 percent water!!!!

[drmiller100]

Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.

Perhaps I understand you wrong. But what about reflux? For rectification you need reflux. And reflux costs energy. For a result of 96%abv directly from a wash you clearly have to reflux something like 4/5 of the top vapor regardless how long your column over the feed is or how efficient your packing is. 4/5 means it needs about 5 times the energy you claim it needs, 9/10 would mean you have to invest 10 times the energy. Of course a rectification column saves energy compared to a multi-distillation-potstill, but it is not as extreme you write. Or I understand you wrong.

You bring up an excellent point. In a reflux pot still at 10 percent wash, what is the liquid dripping back into the pot?

At BEST case it is 10 percent etoh, same as the wash in the boiler.
What is the vapor coming off the boiler? 50 percent etoh. And 50 percent WATER. and the WATER takes three times as much energy as thr etoh (latent heat of vaporization)

So, 2/3 or (66%) of the energy is wasted lifting the water out of the boiler. Add the 10 percent etoh falling back into the boiler.
That is a total of 76 percent waste.

Do you have less than 50 plate equivalents? Easy to have 5 percent loss.
So one might say a pot reflux still is only about 20 percent efficient at best

There is your 4/5 FOR A POT REFLUX STILL.

On my still, the liquid dripping into the boiler is 99 percent plus water.

[drmiller100]

stevea, I don't understand everything you write, but there is something I think I can contribute something:

Reflux IS a major energy cost, every bit as energy-expensive as product, so we want to design to minimize reflux for a given amount of product separation. That generally means more plates (and plates are inefficient, as noted above, so *lots* more plates). McCabe-Theile, then Fenske & Fenske-Underwood-Gilliland(FUG) gave us nice ways to estimate the MINIMUM number of theoretical plates necessary. The number of practical plates, given inefficiency, is often 50% higher at industrial scale and perhaps 100% higher at small scale. Reflux is related in complex ways to the number of effective plates, but there is also a minimum reflux required (therefore minimum energy loss required) . The big oil-cracking guys like to keep reflux ratios under 1.0 and as low as 0.50 (so between 33% and 50% of vapor condensate and energy goes "down the tube"). That is their trade off between capital costs (more plates) and energy efficiency. At a small-distillery scale maybe ~10 meter tall columns are practical, and for home operations perhaps ~3 meters. This places a definite limitation on plate-spacing which also impacts efficiency.

Oil cracking seems to be different. In ethanol production to reflux 33 or 50% of the vapor would be not enough by far, regardless how many plates you have. According to McCabe-Thiele and the best VLE-data I can find, for 96%abv with 50 theoretical plates at 100% efficiency from 10%abv in the boiler you need to reflux 73% of the vapor. For 97%abv it is 92% reflux (the aceotrope is 97.2%abv). And if you start with 50%abv it is almost the same.
The conclusion is, there is no way to save energy besides recovering and insulation. High abv needs much energy.

At the bottom my boiler has 99 percent water. At the top I have 95 percent etoh. The temp coming off the boiler is 210.

Continuous is a different paradigm. I'm cheating. I'm not following the accepted practice.

But I'm not breaking fizzicks rules. Look up latent heat of vaporization and I'm not breaking that rule. I do not have a perpetual motion machine.

It doesn't intrinsically require energy to separate etoh from h20. It isn't endothermic or expthermic.

Thank you all for your interest.
My still does work. I will totally admit I have had to rationalize some of the results and it took a lot of failures to get it to work. And I do still have problems.

Like I need a positive displacement pump that doesn't mind soluble solids and isn't much affected by changes in back pressure.

[drmiller100]

In a continuous still why would you count the vaporization of water?
....

B/c you still need to vaporize the ethanol, and that means vaporizing an mount of water related to the remaining wash ABV. Vaporizing either requires energy.

In a pot/batch still the wash drops from say 10%ABV to 1%ABV over time, and this means you boil-off ~20% of the mols of the wash. At 10%ABV you get roughly 3 water molecules vaporized per molecule of EtOH (3:1 ratio). By the time you get to 1%ABV, it's near a 25:1 molecule ratio.

In a continuous stripper system, you still need to vaporize the ethanol, and you necessarily vaporize water in the same way, but the shift from 10%ABV to 1%ABV takes place throughout the height of the stripper. You are doing the same vaporization over space instead of time.

The Successive plates are "powered" by the condensing vapor giving up it's heat of vaporization. So of course the excess water selectively condenses and you end up with more ethanol and less water in the product ... but you don't get the ethanol into vapor phase until you pay the cost by vaporizing a lot of water.

We've had some disconnects. What does latent heat of vaporization mean to you?
To me it means a pound of water requires 3 times as much energy to phase change to vapor as does a pound of etoh.

To me, a gallon of water weighs about 8 pounds. A gallon. Of Etioh is about 6 pounds.

Does this seem about right to you?

[Hügelwilli]

Ultimately, the only WORK done is lifting the etoh vapors at the top up to the condenser.

Perhaps I understand you wrong. But what about reflux? For rectification you need reflux. And reflux costs energy. For a result of 96%abv directly from a wash you clearly have to reflux something like 4/5 of the top vapor regardless how long your column over the feed is or how efficient your packing is. 4/5 means it needs about 5 times the energy you claim it needs, 9/10 would mean you have to invest 10 times the energy. Of course a rectification column saves energy compared to a multi-distillation-potstill, but it is not as extreme you write. Or I understand you wrong.

You bring up an excellent point. In a reflux pot still at 10 percent wash, what is the liquid dripping back into the pot?

At BEST case it is 10 percent etoh, same as the wash in the boiler.
What is the vapor coming off the boiler? 50 percent etoh. And 50 percent WATER. and the WATER takes three times as much energy as thr etoh (latent heat of vaporization)

So, 2/3 or (66%) of the energy is wasted lifting the water out of the boiler. Add the 10 percent etoh falling back into the boiler.
That is a total of 76 percent waste.

Do you have less than 50 plate equivalents? Easy to have 5 percent loss.
So one might say a pot reflux still is only about 20 percent efficient at best

There is your 4/5 FOR A POT REFLUX STILL.

On my still, the liquid dripping into the boiler is 99 percent plus water.

Yes, 10%abv is dripping back into the boiler at best when you have 10%abv in the boiler of a noncontinuous still.
But there is no wasted energy. The 10%abv looses its energy within the column, where it causes rectification.
The only lost energy is what the condenser eats and the heating up of wash. Both you could recover theoretically. If you don't recover it the energy you need depends mainly on the reflux ratio you use, and the reflux ratio depends mainly on the purity you want.

And at the end of the run when only 0.1%abv is in the boiler also 0.1%abv is dripping back at best. You can get all the alcohol out with an noncontinuous still. Easier than with a continuous still, because you have to add extra plates to it for every bit of additional recovery in a continuous system.

And lifting up water is no lost energy because it's dripping back into the boiler after it has transferred its heat to higher alcohol concentrations in the column.

[LWTCS]

And at the end of the run when only 0.1%abv is in the boiler also 0.1%abv is dripping back at best. You can get all the alcohol out with an noncontinuous still. Easier than with a continuous still, because you have to add extra plates to it for every bit of additional recovery in a continuous system.

Say that again please.
That assertion may apply to some designs. But certainly does not apply to all designs.