Calculating reflux ratios

[flyingdutchman]

Ok so this post is related to 2 other ones herehttp://homedistiller.org/forum/viewtopi ... =1&t=28592 and herehttp://homedistiller.org/forum/viewtopi ... =1&t=28903. I thought it might be good to start a fresh one for this post so it was easier to search I hope that is ok? The point here is to calculate the RR (reflux ratio) based strictly upon supplied boil time and distillate collection rate. Now it has been reasonably pointed out here and in several books (Nixon and Smiley) that the easiest way to determine RR (reflux ratio) is to just run the still with no reflux at all, then turn on the reflux, and compare the collection rate of 1 to the other. However sometimes you might want to figure out someone elses RR based on supplied data. Also it is important to remember that the RR changes based on the ratio of ethanol to water in the distillate because of the differences between the physical properties of water and ethanol. So as your run progress's, with say a whiskey, and as the % alcohol drops your RR is changing and you might not want to keep stopping the reflux to get you RR. Finally part of this is just to do because it can be done ands to see what we can get as an answer. So sometime i write way too much in a post so....rather than just ramble on i will try to keep the posts brief so they are hopefully easier to read.

[flyingdutchman]

In this case think the easiet thing to to is to use Rads calculator from the parent site here http://homedistiller.org/calcs/rad14701 to translate boil time to power input observed at the boiler. It does a great job with the added benefit that by the very nature of how it calculates the power takes into account to a large degree the heat/energy lost by the boiler. Also there are a couple of things that will make life easier when doing this which I will list.
1) reduce this to a 2 component system (water and ethanol)
2) work in moles and mole percents
3) convert collection rates from volumes to weights
4) convert % alcohol by volume to % alcohol by weight
The reason for #1 one is that is just simplfies things a whole lot (and I am lazy)
The reason for number #2 is that due to differences in the latent heat of vaporization (energy required to change from a liquid to a gas) and dealing with a mixtures of 2 components its a whole lot easier to work in moles and mole fractions.
The reason for #3 is that when you add ethanol to water is doesnt really add like we would hope. So say 1 mL of water and 1 mL of ethanol does not equal 2 mL of total volume.
The reason for #4 is that to convert volumes of distillate to weight/mass I need the density which is conveniently provided in literature but only source I find is in % alcohol by weight. But this is easily done by online calcultors (or by hand).
Dont let words like moles (or mole fraction) throw you off, its nothing that major, its just a fixed amount. It sounds fancy but its just like a dozen eggs, its just an amount of stuff, nothing more nothing less.
The neat thing about a mole is that when you have that amount the weight/mass (say in grams) of whatever you have is equal to its atomic mass (amu) or molecular mass.
So simply something like water 1 oxygen(~16 amu) + 2 hydrogens (~1.08 amu each) is then
16.00+1.08+1.08= ~18.016
So 1 mole of water equals~ 18.02 grams. The important thing to remember is its just an amount of stuff so dont let it put you off.
Mole fraction is just a fraction based on the number of moles so if I had a dozen eggs (6 brown and 6 white) the mole fraction of brown eggs is just 6(brown)/12 (total) which is
6/12 = 1/2 = 0.5
Thats enough for now the sun is shining and the bass are biting LOL

[flyingdutchman]

Ok The first thing I would like to say is that someone was nice enough to supply me with their run data for which I would like to thank them. The only purpose of this post is to try and work out the reflux ratio. In no way is it meant to re-evaluate their run or try to second guess the hard work they have done. I also ask that no one else make any negative remarks about this run and if anyone does so i will ask a moderator to remove it or will stop this post. Really all i ask is that we be considerate of others who were nice enough to help out by supplying data from a previous run.
On the other hand if there is anything I put down here that looks wrong or if you see another way I might be solving this then please speak up the more heads the better
So I think the first thing to do is
1) Work out is the power that the boiler/keg sees (not the power rating of the burner).a
let me stop right here and say I dont have everything in terms of information so i am going to have to make some assumptions because that is all I ahve to work with. This means what i am doing is going to be less than perfect and probably more like a rough estimate.
What I know is that on full power it takes ~ 90 min to bring 5 gals to boiling. (I am guessing the boiler start temp). Using the calculator on the parent site I come up with this:

Boil.png (4.67 KiB) Viewed 8247 times

After that I know the heat was turned down halfway so I am guessing the run
1A) Started with 900 watts then...
1B) Reduced to 450 watts for the collection.

[flyingdutchman]

Here is an Excel sheet (image) of what I know from the run with watts added. I dont have collection times so i think rthe next thing to fill out is an estimate of collection times based on total run time.

I know its small but if you click on it will open larger sorry
Here I know total run time was 5.75 hrs or 345 minutes. Less 90 for heat up and 30 for 2 equilibrations leaves me with 225 minutes. This leave me with a dilema because I am fairly sure the collection times diffred but i dont have that data. If I divide the 225 minutes by the number of collection (jars including fores) i get...
22.5 min per jar. (regardless of volume collected)
If I divide the total mL collected /minutes I get...
(120+(7x200) +300+250)/225=9.2 min/ml
So for a 200 mL collection I get 200mL/9.ml/min = 21.7 per 200mL jar.
So either way my guess is ~22 minutes per 200 mL jar.
I know from expericience with my pot still that the collection rate/time changes as the run progress'sso this is a seriupos source of error but i have to estimate here because this is what I have.

Also i should note that at the 2nd equilibration the heat was raised to increase the reflux ratio (& water flow increased to condensor) so from that point on i lose the run as I have no way of guessing what the wattage was increased to.

[flyingdutchman]

So the next thing to do is simply flesh out the data columns in the Excel chart the reasons were previuosly explained but do ask if you are not sure why i am doing this. Luckily all this type of work has been previously done by others and often calculators or data tables exist for what we want.
1) To convert % alcohol from volume to weight and density I just used this softwarehttp://www.katmarsoftware.com/alcodens.htm
2) To convert form % alcohol to mol percent I used the Table in the Compleat Distiller (2nd edition pg 157). Here Nixon & McCaw have done all the work and I cannot reccomned this book highly enough.
3) last to convert volume in mL to mass (grams) i just multiply the volume (mL) by the density (G/cubic cm). Here is a tip (1 mL = cubic cm or cm^3)so just look at the units and not the actual numbers so here ..
(mL) x( G/cm^3) is the same as (mL) x (G/ml) is the same as (mL) x (1/mL) x (G) and since ..
(mL/mL)=1 then all that is left for units or dimensions is Grams (its called dimensional analysis and makes hard problems easier to figure out).

please note I edited this post after finding the above software as it was far more accurate and it actually eliminated the need to find % by weight because it gives density for % by volume.

[flyingdutchman]

Ok so up till now we really havent done any calculations we simply filled out abunch of boxes in an excel sheet based on observations, or online data or used online calcultors to get a number from observed data. To solve this it would be good to layout the steps before we start then anyone can just follow along check for themselves or do a similair calcultion for different runs. first we need to definine what is reflux ratio. For a given sytem we have aboiler that generates V vapor from the wash. This flows up the pipe and is partially or totally condensed by a dephleglamator/condensor to create a L, depending on the system ether some of the vapor (condensed at later point) or some of the liquid is collected as distillate D. Nixon defines RR (reflux ratio) as L/V and product ratio as D/V so i will stick with that. So it would be nice to know V the vapor produced. L the liquid condensed and returned down the column and D the distillate collected. So here is how i think it would be best to go about this.
1) Calculate V based on the power at the boiler using LHV (latent heat of vaporization) more on this later
2) Measure Distillate collected D
3) Assume that L is equal to V-D (assuming all reflux occurs at dephleglamator)
4) divide L/V to get RR
So to solve step 1 we need to know the latent heat of vaporization, luckily when we use mols and not grams it doesnt matter if its water or ethanol (for the most part) because regardless we can presume it to be 40 Kjoules/mol. So we dont care what percent alcohol the wash is because it takes just as much energy either way to create 1 mol of vapor. This is not true on a per gram basis which is why i wanted to solve this in mols.
Solve step 1
1) Now 1KW = 1000watts =1KJ/second=1000Joules/second or (1KJ/sec)x(60 sec/min)=60KJ/minute because there are 60 seconds in a minute. So looking at the watts from our boiler here (450 W) we have 450 Joules/second or (60 sec/min)x(450 J/sec) which gives us 27KJ/min of power at the boiler.
If we know it takes 40KJ to create 1mol of vaopr then we get (27KJ/min)/(40KJ/min/mol) which leaves us with
0.675 mols of vapor per minute (use dimensional analysis again and all you are left with is mols)
So in this case with a 22 minute collection of 200 mL we generated (0.675mol/minute)x(22 minutes)....
= 14.85 mols of vapor for every 200 mL collected.

[flyingdutchman]

So we know the amount (grams) of ditillate we are collecting. We also know the amount of vapor we are producing (mols). In order to compare like terms we need to convert one to the other. Here I prefer to convert the grams of distillate to mols of water and ethanol.
1) Since we know % ethanol by weight can just multiply this by the total weight to get the number of grams of ethanol, by default anything left from the original we will call water (2 component sytem).
2) Then just divide grams of ethanol by its moleculare weight (mwt) gives us the # mols ethanol. Likewise for water does the same.
3) Adding together mols of ethanol and water gives us the total number of mols of D distillate.
I did all this in an excell chart (image) ask me if you dont follow the steps, correct me if you see an error.

Remember as we previously defined
4) L is the liquid returned down the column
5) L/V is our defined reflux ratio
SO what we see is that This run starts out with a reflux ratio of 0.70 (where 1.0 is 100%). I believe this to be a fairly high RR, certainly not 0.9 but not far off either.
The RR then slowly falls during the run as the ratio of ethanol to water changes. I loose it at the re-equiluibration because the heat was raised and I dont know by how much so we have to stop there. However had only the water flow to the dephleg been increased and not the heat you could easily continue to follow the reflux ration through out the entire run, all without having to stop refluxing to count the drops with every jar change.

It is improtant to point out the major sources of error......
1) I am using a calculator to estimate the energy transfered to the boiler
2) I know the heat from the burner was turned down ~halfway but not really the exact amount
3) Some reflux might be going on against the side of the un-insulated column which I am not accounting for
4) Estimating collection time/jar

[ipee7ABV]

i am by no means an expert. i can just barely understand some of your info. i did notice 3 diffrences between my spririt runs and yours. the first is i let my product reflux for at least 40 min. my product collection rate is 10ml/min, and once i remover the 4 shots my abv is a constant 90% untill i get to the tales(no need to let reflux again). not saying your wrong or right just my .02

[flyingdutchman]

Hi ipee7ABV,
Thanks for the input I appreciate it. I dont know what kind of rig you have but a friend of mine has an offset head design (LM) with a packed column and what you describe is pretty much how we run his, with good results making neutral.
This run was done on a 2 inch column with physical plates instead of packing, and wasnt actually run by me (I have a simple pot still). I suspect (but dont know for sure) the reason the column was put under total reflux a second time was because the ABV was dropping too fast and the column was not responding as was expected. Truth is there has been some discussion here about problems with 2" plated columns (both sieve and bubble plates). One of the questions that arose from this was what kind of reflux ratio was being used. This is the main reason I was trying to calculate the reflux ratio based on info posted here from anothers run.
ps i am no expert by a long shot either just have fun running em and talking about em

[ipee7ABV]

Hi ipee7ABV,
Thanks for the input I appreciate it. I dont know what kind of rig you have but a friend of mine has an offset head design (LM) with a packed column and what you describe is pretty much how we run his, with good results making neutral.
This run was done on a 2 inch column with physical plates instead of packing, and wasnt actually run by me (I have a simple pot still). I suspect (but dont know for sure) the reason the column was put under total reflux a second time was because the ABV was dropping too fast and the column was not responding as was expected. Truth is there has been some discussion here about problems with 2" plated columns (both sieve and bubble plates). One of the questions that arose from this was what kind of reflux ratio was being used. This is the main reason I was trying to calculate the reflux ratio based on info posted here from anothers run.
ps i am no expert by a long shot either just have fun running em and talking about em

you nailed it 2in offset head with packing

[flyingdutchman]

That's a great design. My friend's makes a 1st rate neutral which keeps his wife and mine happy and out of our hair so we can fool around trying to get the hang of whiskey. He recently insulated the column and the thing holds its temp/alcohol percent like a rock.

[HookLine]

You do not need to know the numerical value of the reflux ratio when running a still. It is a red herring.

The only time you need to even think about specific reflux ratios, is when you are designing the still. You have to decide the range of ratios you want, and how you are going to achieve it.

When you are actually running a still, the settings you use are determined by vapour temp or abv, not by knowing the reflux ratio.

Just juggle the heat, and take-off rate, until you hit the mark.

My 2c.

[flyingdutchman]

Hi Hookline,
Thanks for the input. Actually you nailed the major reason why I was trying to calculate the reflux ratio. I am still curious about some of the issues observed with 2 inch plated stills. I have a chemical engineer at my work who is grudgingly willing to help me go through the numbers and one of the things he wanted was for me to back calculate what kind of RR they were being run at. The problem is its going slow because designing distillation columns is way outside my field of work.

[HookLine]

When you are actually running a still, the settings you use are determined by vapour temp or abv, not by knowing the reflux ratio.

Small correction:

When you are actually running a still, the settings you use are determined by a combination of output vapour temp/liquid abv, and output rate, not by knowing the reflux ratio.

[Bushman]

And this is always a learning curve with a new still especially if you have multiple control over both vapor speed and cooling.

[flyingdutchman]

I would certainly agree. I think it would be a mistake to even try and run a distillation based on the value of the reflux ratio. As you stated it can be important when designing a still. Or when a column has been designed and built, but is not performing as expected. Then it would be useful to know if it was running within the specifications by which it was designed. I guess what I am trying to say is that I think knowing the reflux ratio can be helpful in certain applications.

[HookLine]

I would certainly agree. I think it would be a mistake to even try and run a distillation based on the value of the reflux ratio. As you stated it can be important when designing a still. Or when a column has been designed and built, but is not performing as expected. Then it would be useful to know if it was running within the specifications by which it was designed. I guess what I am trying to say is that I think knowing the reflux ratio can be helpful in certain applications.

Agree with that.

[Odin]

Very interesting info!

How would this add up for a system of 2 inch with a continuous power input of 2 kw? And a rig that always collects at 96%. My calculations are a bit off.

I calculate 45 liters of gas produced per minute per KW. That's 5,400 liters of gas per hour. If it were water (which it isn't), one liter would make 1700 liters of gas. So 1700 liters of gas would make 1 liter of reflux. Given I run an LM, all gasses are cooled back to liquids and from there product is taken and reflux is managed back into the column.

If we devide 5400 by 1700, that gives 3.2 liters of total reflux (not yet devided between product I take out and actual reflux that goes down the column). But ... here's where my calculation fails ... I can easily draw off 3.5 liters of 96% per hour. Maybe 1 liter of ethanol translates to less liters of gas than is the case with water?

Odin.

[flyingdutchman]

Odin,
I think I can solve your question about the water vapor. I think part of the problem is the fact that water and ethanol have different physical properties. For instance at STP (0 C, 1 atmosphere of pressure) we know that 1 mol of a gas of any substance occupies 22.4 liters of volume (water or ethanol). Now the first question you ask is what is a mole...http://en.wikipedia.org/wiki/Mole_(unit)

Mole is a unit of measurement used in chemistry to express amounts of a chemical substance, defined as the amount of any substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 grams of pure carbon-12 (12C), the isotope of carbon with relative atomic mass 12. This corresponds to the Avogadro constant, which has a value of 6.02214129(27)×10^23 elementary entities of the substance

and 6.02 x10-^23 is a lot of stuff but here is how much it weighs for water and ethanol or what are their molecular weights
Water: 18.016 grams/mol .....Density @ 20C = 0.998 g/cm^3 (or since 1cm^3 = 1 mL then 1 mL weighs 0.998 grams)
Ethanol: 46.084 grams/mol....Density @ 20C = 0.789 g/cm^3 (or since 1cm^3 = 1 mL then 1 mL weighs 0.789 grams)
So I know our gas isn't at 0C but lets keep it easy.

Water:
If I have 22.4 liter of gas (H2O) then I have 1 mole but that means I have 18.016 grams of water when we condense it. Using its density (20C) I then have ... (18.016 g) x (1 cm^3/0.998g)= 18.05 mL of H2O
Ethanol:
If I have 22.4 liter of gas (EtOH) then I have 1 mole but that means I have 46.084 grams of EtOH when we condense it. Using its density(20C)I then have ... (46.084 g) x (1 cm^3/0.789g)= 58.41 mL of EtOH
Water = 22.4 liters gas = 18.05 mL
Ethanol 22.4 lites gas = 58.41 mL

Odi: I think maybe that's where your calculation went south. You are collecting say 95% ethanol but used the physical properties of water. It would have been much better off to use the physical properties of EtOH. Ofcourse this doesn't take heat loss from the system to the environment into account (boiler and Column to surrounding air)
What this also implies:
As our runs progress if the ABV is dropping we are obviously producing less and less distillate as time goes on. Unlike a VM or even a CM (I believe) a LM system manages liquid. That means if we hold a constant take off rate on a LM system we are actually changing our RR as time goes on. Infact towards the tail at a time when I would want a greater RR it is actually less and less.
I am not knocking a LM system I think they are great. I am simply pointing out how the sytem works so people can adjust their collection rates accordingly.

[Odin]

Thanks for the clarification. But I am still not there. I did your calcs myself already, and on the premises of 45 liter per minute of gas per 1 KW I calculated 4.3 liters of water and 8,3 liters of ethanol as total reflux (not yet decided between take-off and returned to the boiler). If I say the mixture should be 95% at the top (where reflux is created), the total amount of liquid on my 2 KW system would be 8.5 liters. That's way too much!

Maybe inefficiencies due to heat losses account for half of the wattage being not used on actual distillation?

On LM: output becomes less during the run.

Odin.

[flyingdutchman]

Well let me work it from the start with just ethanol with a 1KW input
1 watt= 1 Joule/sec= 60 joules /min
1KW =1000w=60,000 Joules/min or 60 KJ/min
Latent Heat of Vaporization (EtOH) = 38.56 KJ/mol (http://webbook.nist.gov/cgi/cbook.cgi?I ... &Mask=1E9F)
So in 1 minute:
(60 KJ/min) x (1mol/38.56 KJ)= 1.556 moles of vapor (EtOH)
Now when we condense this (at 20C) we get
1.556 mol/min (EtOH) x 46.084 grams/mol x 1 cm^3/0.789g = 90.883 ml/min EtOH (notice how the mols and grams neatly cancel out)
or 90.883 ml/min x 60 min/hour = 5.4 L/Hour of pure ethanol liquid (at 20C) (please feel free to double check my figs and correct them if I made a mistake!)
Now that's with absolutely no reflux or loss to the environment and I played it a little loose with the sig figs.
Notice I did not state how many liters of vapor because we would need to correct for temp using PV=NRT (I just avoided all that by keeping to mols of vapor)
Sorry Odin that may indeed match what you got I just wanted to work it out with all the units shown.

Also: please feel free to double check my figs and assumptions and correct them if I made a mistake!

[Odin]

My calculation was like this:

Water: 18 gram per mol. One mol creates around 22.5 liters of gas (ideal gas theory), so if 5,400 liters of gas are produced in one hour, using 2 KW, that's 240 moles = 240 * 18 grams = 4,300 grams = 4.3 liter

Ethanol: 46 grams per mol. One mol again creates 22.5 liters of gas (ideal gas theory), so 5.400 liters of gas are produced in one hour, using 2 KW,
that's 240 moles = 240 * 46 grams = 11,040 grams = 8.71 liters.

Combine in a 5 to 95% ratio gives 8.481 liters per hour on 2 KW, that's 4.24 on one KW. Now 4.24 sounds right for my 2 KW rig, because I can collect up to 3.5 liters and maintain azeo. So there has to be reflux But that would mean we are loosing 1 KW in heating, keeping wash, boiler, column on temperature.

So ... maybe the next thing is ... how much would like 2 to 3 m2 of iron cost to heat, given room temp is 20 C and column & boiler are 35 C (I have them well insulated). If we can put a number to that AND it comes close to 1 KW I feel we are getting somewhere.

FDM, our calculations are off. You calculate quite different quantities than I do. Can you find an explanation?

Odin.

[flyingdutchman]

Hi Odin,
Might I ask where this figure came from
45 liters of gas produced per minute per KW
Since ideal gas law http://en.wikipedia.org/wiki/Ideal_gas_law

PV=nRT,
where P is the pressure of the gas, V is the volume of the gas, n is the amount of substance of gas (also known as number of moles), T is the temperature of the gas and R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant.

is dependent on pressure and temperature, what pressure and what temperature was this calculated at?
Its important to remember that the 22.5 Liters/mol is at STP (1 Atmosphere and 0C)
I see in Nixons book he states 1KW will vaporize 1.5 mol of either water or ethanol, of course that's an approximation because they both have slightly different LHV values/mole but close enough for me. You notice my calculations based on the LHV from the NIST gov website, give me a figure of 1.556 mols (EtOH) so at that point we concur.

But then Nixon says it will be vaporized at 1ATM to give 45 liters of vapor.
WHOA: at what temperature? Because in my book 22.5 liters/mol x 1.5 mol = 33.75 Liters!
Now I think what he did without saying it is use an ambient temperature of say 20C?? Because to me he looks to be saying he generates 30 liters of vapor per mole (and he is probably right at 20C or whatever temp he uses)

Now that is fine but I see you used the accepted 22.5 at STP when you connveted Gas back to liquid and presto we create more liquid than there should be.
Wouldn't Lavoisier be in for a surprise! http://en.wikipedia.org/wiki/Conservation_of_mass
So instead try using 33.75 Liters where you used 45 liters and see if that doesn't bring us back into agreement

ps that's why I stayed away from liters of gas I didn't want to deal with PV=-NRT and ofcourse that's an IDEAL GAS LAW and chances are we don't have an ideal gas LOL

psps make sure when you redo it that when converting from grams of liquid to volume you use dimensional analysis
ie ....(Y grams) * (Z grams/ml) gives an answer of YZ (Grams^2)/mL (not the answer we want)

You need to invert the density to read 1mL/0.789 grams (see you need to divide by 0.789 not multiply) otherwise you end up with answer that has unites like grams squared per mL
Your whole premise is correct its just a lot of book keeping

[flyingdutchman]

Oh and one last thing.
In the interest of accuracy my Alcometer reads % ABV which means by volume
But we need to convert %ABV to %ABW by weight (we are working in grams). These numbers are significantly different. So if you want to try and figure out for 95% ABV you need to convert that to ABW. I found this free ware convenient for thathttp://www.katmarsoftware.com/alcodens.htm use that to figure what percent is water and what percent is ethanol. Then the density of an aqueous alcohol mixture at that percent alcohol (by weight) is indeed different (see here)http://www.handymath.com/cgi-bin/ethano ... bmit=Entry

[Odin]

Sounds like we are making progress! I have to read your posts again when my head is a bit clearer to see how everything sums up!

Thanks!
Odin.

[flyingdutchman]

So as an aside lets see just what temperature Nixon was using in The Compleat Distiller when he gave the figure of 45 Liters of vapor per minute?
We know that 1 mol of a substance (ethanol water methanol etc) occupies a volume of 22.414 liters at STP http://wwwchem.csustan.edu/chem1102/molarvol.htm
Charles Law http://en.wikipedia.org/wiki/Charles's_law says that volume of a gas is proportional to its temperature (the hotter it gets the bigger it gets) which gives us..
(V1)(T2)=(V2)(T1) (T=temp IN KELVIN, V=volume liters) [we need to work in degrees kelvin if we starting multiplying things by zero its gonna get messy real fast]
We can re work this to (T1)=[(V1)(T2)]/V2 (note 0C = 273.15 Kelvin) so...
T1=[(45L)(273.15K)]/(33.621 L) (I got this volume 33.621 L by multiplying Nixons 1.5 mols by 22.414 L/mol)
T1=365.797 K
T1= 92.447 C or just ~92C
So my guess is that's the temp at which he came up with 45 Liters of gas. This makes sense since its in the range of the vapor coming off a distillation boiler.

[BDF]

How much of the vapor that you produce is condensed in the packing/column?
How will you measure in real time the ethanol/water ratio in your boiler? what if there are substantial solids in your boiler?

In a separate topic I propose that it would be easier to measure how much heat your reflux condenser is taking away:
http://homedistiller.org/forum/viewtopi ... 3#p7143243

Since the ethanol/water mixture of the gas at the reflux condenser would be pretty similar to the distillate that you take off, it would be in principle possible to measure continuously. If you have a thermometer both in the rising vapor stream above the packing, and in the collection plate you know both the temperature of the gas going to the condenser and the temp of the liquid falling off it.

So since you have the quality of the gas at the condenser, temp of the gas rising, temp of the liquid falling, and you know the net energy change in the coolant you can calculate exactly how much liquid is being condensed and at what rate. Compare that to how much you're taking off and bam...reflux ratio. All without trying to figure out how much heat is lost to the surroundings in the boiler, the efficiency of the electric element, what happens in the packing, or monitoring the ethanol/water mixture in the boiler.

[flyingdutchman]

Yes I believe what you say is quite correct it would be a great way to do this. It would eliminate all concerns of heat lost to the environment from the system (burner boiler column). It would as you point out have the following requirements.

In summary this would mean measuring the following values:
-Flow rate of coolant through reflux condenser
-Entering and exiting temps of reflux condenser coolant
-Flow Rate of distillate collected
-Proof of distillate collected

I would also add to the list the obvious hardware (but not difficult)
-2 thermowells with thermosensors with an extremely fast response time
-1 metering/measuring device to record coolant flow rates (some accurate form of coolant pump may be helpful)
-1 metering device to measure distillate collection rates (and %ABV )
Also we might need to record distillate collection over some period of time like say between 1-30 minutes?

I would add that it would far easier to calculate if the following were true
-cooling water flow rate remained constant
-cooling water temp entering condenser remains constant
-water temp exiting remains condenser constant
-distillate collection rate remained constant
otherwise it might be necessary that all the data would need to be recorded by some form of monitoring software and then calculated by some written program which?

None of that is overly complicated
Metering could be as simple as a graduated cylinders for water and distillate
an ample sized water reservoir could hold input water condenser temps steady
A accurate reliable pump could hold coolant flow rates constant

[BDF]

Also I think to add to that list for a more real-world accuracy would be thermometers to measure vapor temp before it reaches the reflux condenser and a thermometer to measure the temperature of the liquid falling off the condenser to be able to take into account the potential super-heating of the vapor or sub-cooling of the distillate.

A good constant flow rate pump would take care of flow rate.
As far as coolant temperature entering and exiting the condenser, that would be fairly easy to measure as quickly as you'd like with some thermocouples regardless of whether or not it was held too constant (which if it is from a municipal water supply it would remain fairly close to ground temp for that day). However water temp exiting the condenser would be entirely dependent on the temperature of the vapor at the condenser and how much the boiler is throwing at it...it is something that would have to be measured directly.

In general I wouldn't foresee any one of the variables changing too quickly so a sample rate of once per 5 seconds would probably get you in the ball-park of the kind of RR you're seeing. Maybe refresh display once every 5 seconds with an a 1 second sample rate which are time averaged....either way I think you get the idea.

[flyingdutchman]

Oh sorry I meant to answer your questions as best as I can...

How much of the vapor that you produce is condensed in the packing/column

Presumably all of it in a perfect world but its not a perfect world. (but I doubt that is what you meant LOL)
Of course I am running on the assumption that
1) the packing is already saturated with previous distillate from equilibration
2) In the process of condensing from a gas to a liquid it gives up that heat/energy liberating the previously condensed distillate so the net change is ~0 and of course in the process we gain a plate
*** This assumes Constant Molal Overflow (and as such ignores heat loss in column and things like mixing )

How will you measure in real time the ethanol/water ratio in your boiler

I don't believe we need to since we are working in moles and the LHV (heat of vaporization)for water/ethanol on a per mole basis is extremely close

what if there are substantial solids in your boiler

I do filter my wash (fine screen) before it enters the boiler but am not sure if this effects the LHV (are you referring to the effect of solutes in a solvent raising the boiling point and depressing the freezing point)