## Calculating power for propane burner

Distillation methods and improvements.

### Re: Calculating power for propane burner

Its raining out here so i thought I fool around a bit more with these numbers. I would like to try and calculate the amount of watts
1) I am losing to the enviroment from my boiler (the aluminum pot from above with 1 gallon of water)
2) See if I can use that to approximate the energy lost when using a halfbarrel keg and compare that to Nixon's figure (courtreosly provided by Usge above)
note: add 500kW for heat losses if using uninsulated boiler.

1)To solve step 1 will use difference of the dt (0-10) temperature change for the 1st ten minutes (when the boiler is close to ambient temp) vs the dt (25-35)temp change in the 25 to 35 minute range which should be close to the range we might be collecting fractions. I will assume the difference in temperature change is simply due to heat lost I wont bother posting all the math again unless someone wants to see it just assume I am finding Q for the Water and Aluminum then convert joules to watts as b-4.
So the dt (0-10) is 21.5C and dt(25-35) is 17.6C. So the difference bet2ween the two is 3.9C which i will call dt(HL) heat lost.
Q for water is then ~3.6810^5 Joules
Q Pot is then~ 2.95*10^4 Joules
Q total is the sum ~ 3.918810^5 Joules
Watts is Q total /(60 x 10 min) =118 Watts
So I am estimating I lose ~ 118 Watts to maintain temperature in my pot which is lost to the air
2) To try and translate this to a keg instaed of my small aluminum pot I will just calculateb the difference in surface area and choose (perhaps erroneously) to ignore the difference in the two materials SS 316 vs Aluminum.
So the Aluminum pot is 5" high with a radius of 5.25" so (skipping the math) for total surface area (top, bottom and sides) I come up with ~ 338 sq inches.
My keg is 20" high witha radius of 7.75" so I come up with a total surface area of ~ 1351 sq inches.
So the keg has 3.99 times more surface area (larger/smaller). So if i multiply the watts lost in the aluminum pot by this factor of 3.99 I should get a rough estimate of heat lost when using an uninsulated keg which I find to be ...
470.82 Watts
This fairly close to Nixon's reported figure of 500 watts[ (500-470.8 )/470.8] x100 = 6.2 % difference
my reason for starting this was to asee the minium energy/power my burner could apply which worked ok i think. From this we were able to calculate how much energy the system loses, I didnt know Nixons number b-4 when i first started. From here I am fairly confident that we should be able to calculate the reflux ratio of a system if we know the actual watts the sytem gets from the burner, the heat lost to the enviorment, and the amount and rate of distillate collection. Does this sound right?
measure twice cut once and if that dont work get a bigger hammer!

flyingdutchman
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### Re: Calculating power for propane burner

I would think so. But, I'm wondering if you are coming at this backwards. Sometimes it's easier to just deal with the actual output (whatever that is) than trying to track all the various parts through multiple systems/places ...any one of which could have losses, variables, etc. I mean how far you want to go? Each curve, etc., has it's own influence on things as well.

The output "already" encompasses the variables in it so why not just use that? For any given heat/power level, measure the max output rate/volume of the distillate coming out. Write that down (so many ml per min). That's your max flowrate/volume. Now...from there, using CM or LM or whatever reflux system you have...adjust the output flow rate down. The more you do that, the higher the reflux rate. If you want to figure the reflux rate...you simply subtract one from the other, and that's your refluxed amount in ml, then make a ratio out of it.. no?

If you want to work that backwards..you can. Start with the reflux ratio you want. And then subtract that from the max flow rate at a given power level, and the rest if your target flow-rate for taking off distillate.

To figure you watt losses...use the formula to determine how much watts you should be using for a given vapor flow/output, then measure how much you actually "use" when to achieve that. The difference is your watt/energy loss.

I think the one thing you'd want to know up front is...what is the optimal vapor flow rate for your column. (min/max range). I think this is where such calcs are handy, as it can give you a very close approximation of what watts/volume you might need based on columns diamater/size. Pick the middle figure and use "that" to perform the above tests using whatever reflux system you have. From there, it seems to me what comes out the other end...will give you the information you need. Trying to follow it through all the bends and turns along the way, etc., might be interesting (makes my head hurt), but I think you'll find in the end that it just confirms what you already know.....ie., there is so much distillate coming out the other end when I apply so much heat. If I leave the heat where it is..and control the take off rate to 1/2. Then 1/2 of that measured max take off rate..is being returned (or available) as reflux in the system. That's how I do it anyway...saves all the brain cells for other things But, perhaps I'm simplifying things too much. Dont' know.

Usge
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### Re: Calculating power for propane burner

Hi Usge,
I definitly do not think you are over simplifying this. There is certainly more than one way to skin a cat. In my case the first goal was to simply see how many watts of power my burner applied at its lowest possible setting. While not an ideal test it seemed that setting the burner on its lowest setting then directly measuring the water temp while tracking time seemed to incorporate the least amount of variables. Aside from that, alot of this applies to our other thread about 2 inch columns which I think you already guessedhttp://homedistiller.org/forum/viewtopic.php?f=1&t=28592.
Now I am not sure if this next part should be posted here or there but by way of explanation I guess I will post it here??
Part of the problem is that I have not built a 2 inch plate column with reflux so some of the ideas you suggest, which would work quite well, I simply cant do. So I am left with two problems.
1)The engineer asked me to provide him with the wattage applied to the boiler for the plated setups.
Here I did not have a wattage but what I did have was the time it took others to bring 5 gals to a boil. I can estimate the wattage using the calculator on the parent website but by directly measuring mine then checking my results against the calculator on the parent website i get a rough idea of how accurately I can work boil time backwards to actual wattage (with the calculator) given things like heat loss.
2)He wants to know the reflux ratios used with these systems. Here all I have to work with is the collection rates provided from previous runs (by others). So i figured if I know the actual wattage and the heat lost at the boiler given a fixed volume of distillate collected with respect to time (as well as % alcohol) I should be able to back calculate the RR based on the LHV of ethanol/water (mol basis).
No doubt this is not the most straight forwards way to do it but its all I have.
Here I think you are dead on
I think the one thing you'd want to know up front is...what is the optimal vapor flow rate for your column. (min/max range).
Now from what was qouted from Nixon
Where optimal range = min 12"/sec — max 20"/sec
Is the optimal vapor flow rate for a bubble or sieve plate the same as a packed column I dont know. There are calculations to figure this out in chemical engineering books like Perry but first I need to know (or assume) things like reflux ratio to use these equations.
From a previous example if it takes 1.5 hours to boil 5 gallons (from say 65F-172F) then the calculator on the parent sitehttp://homedistiller.org/calcs/rad14701 says it should be about `1000Watts. Now given that with respect to a 2 inch column, the other calculator from the parent website says we should generate a vapor speed of 14.57 inches(per second?). So this figure meets the criteria set forth by Nixon pretty well.
measure twice cut once and if that dont work get a bigger hammer!

flyingdutchman
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### Re: Calculating power for propane burner

Here is Nixon's white paper for reference: I think this was originally posted over at the yahoo distillers group

Nixons_Paper-3a.pdf

Usge
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### Re: Calculating power for propane burner

Wow,
Great paper,thanks so much for posting it. I said it before but its worth repeating that the great thing about this site there are so many experienced people here who are willing to share their knowledge. The thing I really like about Nixon's work is that he can approach a complex problem then solve it in a way that makes it seem straight forward and relatively easy.
measure twice cut once and if that dont work get a bigger hammer!

flyingdutchman
Swill Maker

Posts: 241
Joined: Thu Mar 01, 2012 3:02 pm
Location: Ma, USA

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