Hey Patrick,
I've never sized a displacement system, but maybe this can serve as a sanity check for your math. Assuming you have one 420-bbl fermentation, your OG is 17 P (starting with no alcohol simplifies the calculations), your gravity drop is linear (which we know isn't true, but it'll be more than offset by the assumption that all fermentations are peaking at the same time), and wort/beer mass is constant:
3 P/day x [1 day/24 hrs] = 0.125 P/hr
OG (SG) = ((0.00000008 x 17 + 0.00001216) x 17 + 0.00387324) x 17 + 1 = 1.06975
Initial density = 1.06975 x 258.7 = 276.74 lb/bbl
Initial mass = 276.74 x 420 = 116,231 lbs
Initial extract = (17/100) x 116,231 = 19,759 lbs
Water mass = 116,231 - 19,759 = 96,472 lbs
Gravity after 1 hr (SG) = ((0.00000008 x 16.875 + 0.00001216) x 16.875 + 0.00387324) x 16.875 + 1 = 1.06921
Density after 1 hr = 1.06921 x 258.7 = 276.6 lb/bbl
ABV after 1 hr = 0.516 x (17 - 16.875) = 0.0645 %
Alcohol volume after 1 hr = (0.0645/100) x 420 x [117,347.8 L/bbl] = 31,790 mL
Alcohol mass after 1 hr = 0.789 x 31,790 / [453.59 g/lb] = 55 lbs
Remaining extract after 1 hr = 116,231 - 96,472 - 55 = 19,704 lbs
Extract to CO2 in 1 hr = 19,759 - 19,704 = 55 lbs
CO2 mass produced in 1 hr = 0.489 x 55 = 26.9 lbs
I'm guessing you assumed that degrees Plato always equals percent extract, which isn't true when alcohol exists. You may have chosen 72.5 lbs/hr of CO2 for a good reason, though, so let's roll with it. Using the ideal gas law for room temperature and atmospheric pressure:
CO2 volume produced in 1 hr = 72.5 x [453.59 g/lb] x 8.314 x 293.15 / (44.01 x 101,325) x [35.315 ft3/m3] = 634.7 ft3
Your number looks good to me there. Assuming that CO2 produced by fermentation displaces the same volume of air (i.e. atmospheric pressure is maintained) results in a CO2 concentration of 1,000,000 x 634 / 14924 = 42,482 ppmv. Again, our numbers are pretty close.
I'm not sure what you mean by "air charges", but here's how I'd proceed:
Max allowable CO2 = 5,000 x 14,924 / 1,000,000 = 74.6 ft3
Min CO2 to remove = 634 - 74.6 = 559.4 ft3/hr
Min flowrate of fans = 14,924 x 559.4 / 634 = 13,168 ft/hr
This approach has a couple of problems:
-If the fans displace cellar air with clean air at a 1:1 ratio, CO2 will accumulate at a rate of 74.6 ft3/hr. Perhaps a better approach would be to simply add 14,924 ft3 of clean air per hour.
-I'd wager the fans won't displace cellar air with clean air at a 1:1 ratio, meaning you'll need a higher flowrate.
Hopefully an HVAC expert can chime in.
Joe Walts
Ale Asylum


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