Oxygenation vs. Aeration

Got this guy over the summer:

image

That’s a William’s Brewing Big Oxygen system hooked up to a 20 CF oxygen tank.  So the question becomes how much oxygen to use.  Warning:  Chemistry and algebra follow!

The desired oxygen level is about 8 ppm.  You can almost reach this by using air, either through extended shaking or using a aquarium pump and an airstone, although it’s not sufficient oxygenation for a larger beer.  And it’s inexact.  If you’re shaking or using air and your lag times are variable, look to dissolved oxygen as the cause:  The oxygen running out is the yeast’s signal to switch over to anaerobic fermentation, the one that makes beer.

References call for oxygenating for a minute at a flow rate of 1 lpm.  This makes some awfully big bubbles in the wort.  The bubbles aren’t the problem, their surface area is.  Bigger bubbles mean less gas in contact with the wort and the gas can only dissolve if it’s in contact so smaller bubbles are better.  So I set out to find out how much oxygen for how long I’d need to get first 8 ppm of oxygen in solution, then for later ease of calculation 1 ppm.  I’ll work through the 1 ppm example.

The regulator is adjustable for different flow rates.  Lower flow rates mean smaller bubbles.  Ideally, the wort should dissolve everything and the bubbles should not reach the surface.  Since 1/32 LPM (liters per minute) is the lowest flow rate, I decided to base my calculations on it.  So to start, how many of you remember the ideal gas law?  Thought so, it’s PV=nRT.  I’m interested in a couple of things about oxygen:

A mole of oxygen (if you don’t remember what a mole is, look it up) masses 16 grams and occupies 22.4 liters at standard temperature and pressure (273.15 degrees Kelvin, or zero degrees Celsius, at a pressure of 1 bar).  Units don’t matter so much since we’ll be working with proportions.  I live in Denver at 6,000 feet where the pressure averages 0.81 bar and normally work around 72 degrees Fahrenheit, or 295 degrees Kelvin.  The math is simple proportions from that point on:  22.4 liters*295/273 *1/.81.  This works out to 30 liters per mole, closely enough, or 30 liters per 16 grams of O2.

To achieve 1 ppm, or 1 mg/l, of oxygen in a 20 liter batch (about 5 gallons plus one quart, close enough for homebrew) I need 20 mg of oxygen.  Converting that to liters, multiply by 30 liters per mole, divide by 16,000 mg/mole.  That gives 0.0375 liters.  At a flow rate of 32 minutes/liter, that gives me a requirement of 1.2 minutes of flow to achieve 1 ppm of O2 in solution at a flow rate of 1/32 LPM.  For 8 ppm, I’d need 9.6 minutes, rounded up to 10 because the transfer process isn’t perfect.  I could do it in 5 by upping the flow to 1/16 LPM but at risk of losing more oxygen to air.  May try it if I’m ever in a hurry.

Never thought that high-school chemistry would come in handy, did you?

For reference, at sea level, a minute at that flow rate is enough to get you 1 ppm.

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