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 ...
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?