Chasing Heat

[Orville]

I’m chasing around distillation heat (mathematically) and it seems the water’s sensible heat stays with the water and the ethanol’s sensible heat either stays with the condensed ethanol or is removed by the output condenser (or split between both).

The latent heat of vaporization for ethanol gets removed by the output condenser, but where does the latent heat of vaporization for the water go, it seems it must be removed through the ethanol by the output condenser, but since there’s about 90% more water, with about 2.5 more energy, the numbers don’t make sense to me.

Can anyone help clear this up???

Thanks.

[kiwistiller]

My understanding (read: I think this, not know this) is as follows.

For a start, maybe the fundamental flaw in your reasoning is that there isn't 90% more water actually boiling off (most of the time). Ethanol is more volatile, so from a 10% wash, you'll be getting something like 50% alcohol in the vapour. if you reflux low wines (which I thougherly reccomend) then you're getting more like 80% ethanol from the boiler.

The fact that the water has more energy is how the reflux can actually work - as (relatively) water rich vapour rises in the column, it mingles with cooler liquid reflux falling. It mingles, and some will condense, giving energy to the reflux, and this is mostly the non-volitile water that will condense. The water in the reflux is less volitile than the ethanol, so tends to stay as reflux, but when the energy is transfered to the reflux, it is more likely to boil off the volitile compounds (ethanol), and the water is likely to stay as a liquid. Of course this isn't happening in neat stages like that, it's pretty stormy in a column. So I guess the answer to where the latent heat is going is some of it is taken away by the reflux condenser, and some of it goes to the product condenser.

does that sort of make sense? It's a bit of a shit explaination, sorry. hopefully it helps.

[Orville]

Kiwistiller,

Your right about me having a fundamental flaw in my reasoning, if I understand your explanation correctly, when we apply heat to the wash it’s sensible heat for both water and ethanol until we reach the boiling temperature of ethanol, then any additional heat added is sensible heat for water and adds to latent heat for the ethanol.

We keep adding heat which continues to be sensible for water and latent for the ethanol until ethanol’s latent heat of vaporization threshold is reached.

Is that closer???

[kiwistiller]

I don't think it really works in neat lines like all the ethanol building latent heat and some building sensible (it's a portion of both), but yes. Have a look at the graph in this thread: http://homedistiller.org/forum/viewtopi ... 1#p6844085 , you can see how the proportions of water and ethanol in the wash and vapour will change as the run progresses.

[airhill]

Orville

If you are very very lucky the ethanol which exits your product condenser will contain 4% water, and it boils at a temperature less than required to boil pure ethanol (which you will not get unless you use other than conventional distilling methods) it does not behave very sensibly

[manu de hanoi]

Kiwistiller,

Your right about me having a fundamental flaw in my reasoning, if I understand your explanation correctly, when we apply heat to the wash it’s sensible heat for both water and ethanol until we reach the boiling temperature of ethanol, then any additional heat added is sensible heat for water and adds to latent heat for the ethanol.

We keep adding heat which continues to be sensible for water and latent for the ethanol until ethanol’s latent heat of vaporization threshold is reached.

Is that closer???

no. In a mix of water + ethanol, the boiling temps is between the the boiling temps of pure water and the boiling temp of pure (95%) ethanol. In the mix both water and ethanol boil at the same time (but in different amounts !)

The latent heat of vaporization for ethanol gets removed by the output condenser, but where does the latent heat of vaporization for the water go, it seems it must be removed through the ethanol by the output condenser, but since there’s about 90% more water, with about 2.5 more energy, the numbers don’t make sense to me.

this is a bit tricky to understand. but basically all the water latent heat is removed by the reflux condenser.
10 % ethanol wash will produce 50% vapor but reflux will return 10% condensate to the boiler. That's how water is removed.


More on the same topic (from a post of mine on another forum ) :

If you consider that
- the vapor leaves your boiler at around 40 to 60 % ABV
- and that if your product output is 95 %

Then it becomes obvious that you must somehow return the excess water to the boiler. This article is about the minimum reflux required to return that water. Calculating that minimum amount of reflux help calculating the minimum opening size for the VM arm (if you agree that the opening is somehow related to the reflux).

Here is a way to calculate the minimum reflux returned to the boiler as a weight fraction of the steam leaving the boiler and entering the column. This is not exactly the same as the reflux returned to the top of the column but maybe someone will finish the calculation or i'll do it another day.
Here we go :
m1= evaporated mass per unit of time (you get that from your power, the abv, ethanol density and the evap heat of ethanol and water, see my article/excel about continuous stripper calculations)
m2= liquid mass returned to the boiler per unit of time (reflux)
m3= mass distilled (output) per unit of time

a1= evaporated ethanol % in mass (see the mc cabe diagram thing)
a2= ethanol % in mass in the liquid returned to the boiler (should be the same or very close to boiler %)
a3=distillate ethanol % in mass

1- total mass conservation :
m1=m2+m3
2- ethanol mass conservation
a1m1=a2m2+a3m3

if you play around the 2 equations above you get :
m2/m1 = (a1-a3)/(a2-a3) if you focus on power
or
m3/(m2+m3) = 1- (a1-a3)/(a2-a3) if you focus on VM opening size (reflux ratio).
todo : calculate mass for the energy equivalent of m2 at the top of the column to get the reflux ratio.

Cheers

[kiwistiller]

Well explained manu, thanks. I'm going to need to read that last one a couple of times though

[manu de hanoi]

Well explained manu, thanks. I'm going to need to read that last one a couple of times though

i just figured your kiwi avatar looks like an alambic

[olddog]

With eyes.

OD

[Orville]

Kiwistiller,

I can work with that graph, if I read it correctly, it says that 10% wash boils at 199.5˚F (93˚C), so when heat is applied to the wash both the water and the ethanol absorb sensible heat until the wash reaches 199.5˚F (93˚C) at which time all additional heat is absorbed by both as latent heat until vaporization occurs resulting in vapor that is 54% ethanol and 46% water.

But, the as the ethanol vapor leaves the batch and the condensed water vapor returns to the batch the concentration of ethanol decreases, resulting in increasing the boiling temperature of the wash and decreasing ethanol concentration in the vapor over time.

I’m I sneaking up on this thing yet???

[Orville]

manu de Hanoi,

Kiwistiller,
see my article/excel about continuous stripper calculations

Can you give me a link to the article and or spread-sheet???

Thanks

[manu de hanoi]

http://homedistiller.org/forum/viewtopi ... =1&t=15461

[airhill]

Manu

Pure ethanol is 100%; azeotrope is approx 96% ethanol and 4% water and its boiling point is less than pure ethanol. At least 4% of the water (if your lucky) is exiting the product condenser and does not have the same latent heat as water.
There again I have had a hard day and now a few drinks

[manu de hanoi]

Manu

Pure ethanol is 100%; azeotrope is approx 96% ethanol and 4% water and its boiling point is less than pure ethanol. At least 4% of the water (if your lucky) is exiting the product condenser and does not have the same latent heat as water.
There again I have had a hard day and now a few drinks

My comment about boiling point temperatures was for below azeotropic mixes, the ones you are likely to meet in a distillers boiler.

Thanks for reminding me of the meaning of "pure" and for correcting the numbers, Im pretty sure that 10% wash doesn't boil to 50 % ABV exactly too but I hope you got my point nevertheless, didnt you ?

On the other hand your comment about the latent heat of water changing is quite obscure to me.

[Orville]

manu de Hanoi,

The latent heat of vaporization (LHV) for water (9,198,886 J/hour in your spread-sheet) is required for the water component of the vapor produced by boiling the wash, and there’s no getting around that, so since the LHV of the water component of vapor is heat we’ve already paid for at this point, the best way to utilize that pre-paid heat is to use it to purify the ethanol component of the vapor by re-boiling it (reflux).

The LHV for ethanol (1,453,297 J/hour in your spread-sheet), so the pre-paid LHV of the water component of vapor has (9,198,886 / 1,453,297 =) 6.3 times more heat than what the ethanol component of the boiled vapor needs to be re-boiled. The reflux condenser removes LHV of the water component of vapor, but only after it first uses that heat to purify the ethanol component of the vapor by re-boiling it ~6 more times (reflux).

Is this a proper interpretation of you spread-sheet???

[Orville]

kiwistiller,

Thanks for your time, help, and patients with me, it’s greatly appreciated.

Orville

[manu de hanoi]

manu de Hanoi,

The latent heat of vaporization (LHV) for water (9,198,886 J/hour in your spread-sheet) is required for the water component of the vapor produced by boiling the wash, and there’s no getting around that, so since the LHV of the water component of vapor is heat we’ve already paid for at this point, the best way to utilize that pre-paid heat is to use it to purify the ethanol component of the vapor by re-boiling it (reflux).

correct, except that reflux, even optimized, always costs more energy than plain stripping because the column cannot return 0% abv to the boiler, it can only return the save abv (or higher) as the boiler. So that alcohol returned will have to be reboiled at a cost(see the calculations above).

The reflux condenser removes LHV of the water component of vapor, but only after it first uses that heat to purify the ethanol component of the vapor by re-boiling it ~6 more times (reflux).

In an intricate indirect way yes. I'd rather say that:
all that energy from the water LHV on the bottom of the column will be used to evaporate the ethanol from the reflux. And that the reflux must be adjusted to remove all that LHV from the water.
And that if some of that water LHV remains on top of that column it means that your reflux is too low and that you are getting more water in your distillate than you wanted

Took me 1 hour to make that answer. Reflux is a hard concept to grasp. But I'm glad I could express it clearly.

[Orville]

manu de hanoi,

After buying and reading three books, along with reading everything that search motors turned-up on the internet, reflux has never made much sense to me by following the path of the ethanol, but by following the path of the heat, I feel like I can get a handle on it. I’ve been playing with your spread-sheet which immediately solved the energy balance problems that my spread-sheet had.

Thanks for your help and all your time.

Orville

[manu de hanoi]

Hi Orville, I updated the spreadsheet :
http://homedistiller.org/forum/viewtopi ... 6#p6848456" onclick="window.open(this.href);return false;" rel="nofollow

[Orville]

manu de hanoi,

I got version 2, and I'll pull this one apart too, thanks.

[blanikdog]

Must be something wrong with my eyes cos everytime I see this topic in the new posts I read 'Chasing head'.