oxygenation formula
Where O is your desired oxygen in mg/l(ppm), and B is you batch size in liters. this formula assumes 32g of oxygen per mole, that you get 100% transfer and the ideal gas law that 22.4l per mole at one atmosphere is close enough for brewing.
(((OxB)/1000)/32)22.4=liters of oxygen. example lets say you want 12ppm for 5 gallons (20L) of wort (((12x20)/1000)/32)22.4= ((240/1000)/32)22.4= (.24/32)22.4= (0.0075)22.4=0.168 liters of oxygen using, o2 needed/flow lpm = time, set your flow meter for 0.2lpm and run your O2 for 50.4 seconds. 0.168/0.2 = .84 minutes = 50.4 seconds. Does this help or am I way off my rocker trying to use math to brew beer.:cross: 
Hmmmm. Don't you have to take temperature into account?

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However, I would question that (((12x5)/1000)/32)22.4= ((240/1000)/32)22.4 I never learned the new math, but when I went to school 12x5 = 60, not 240. :D a. 
ajf,
good catch, it should read (12 x 20). 5 gallons is 18.927 liters. there are a few assumptions made with this formula but at room temperature it should be close enough for our needs. I use a welding oxygen tank with regulator and flow meter. I just figured it would be a quick way to get in the ball park for O2 without wasting a lot of O2 or needing to buy a dissolved oxygen meter. 
Temperature comes into it in 2 ways. First, the volume of a mole depends directly on temperature. The volume of a mole of ideal gas is 22.4 L at "STP" which is one atmosphere and 0 °C (273.15 K). The volume of a mole at a different temperature is 22.4*(T + 273.15)/( 273.15) where T is the temperature in °C.
Second, temperature effects the solubility of O2 in wort (but then, so does the gravity). That doesn't matter here because the formula is only calculating the amount of O2 delivered to the wort  not the amount dissolved. There's not much point in applying enough O2 to produce 12 ppm if your wort is 20°P and at 30°C but the number of liters required to get that much, if it would dissolve, isn't effected by the solubility. 
Using the original oxygenation formula and including temp correction for the volume of a mole of O2, I solved for ppm and entered some of my data. The "DO level (ppm)" column is a dissolved oxygen measurement taken right after oxygenation. You can see that only ~518% of the O2 injected is in solution when the reading is taken.
http://www.homebrewtalk.com/gallery/...alculated1.bmp 
true, if i use a value of 20 °C ~ room temperature in my house, then I get 24.040 for the volume of a mole. Using my original example of 12ppm I would need 0.18 liters of oxygen assuming 100% oxygen dissolved.
I guess my question should be, is it useful to try and calculate a ball park figure or just run the oxygen for a set amount of time and call it good? I would think that large brewery's calculate their O2 flow when oxygenating, or do they use DO meters inline and adjust the lpm during transfer. I hope I am not trying to over think the process, but I get bored at work and try to find brewing related things to do. 
dstar26t,
thanks, that is exactly what i have been trying to find. one of these days i will have to see if i can get my hands on a DO meter and try this for myself 
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Clearly if you have lots of bubbles breaking the surface not all the oxygen is dissolving. 
You can see I dropped the flow rate on the last beer from 3 L/min down to 1 L/min and it seems more efficient, although it was only one trial.

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