Cold crashing and oxidation

[Kbar]

I think even if air is sucked back in to the fermenter during crashing, it's still lighter than CO2 so oxidation really isn't a problem.

Agree -O2 will 'float' on top of the CO2 bed you, or I should say, the yeast produced. A protective blanket of sorts.

[brewinhard]

I think even if air is sucked back in to the fermenter during crashing, it's still lighter than CO2 so oxidation really isn't a problem.

Agree -O2 will 'float' on top of the CO2 bed you, or I should say, the yeast produced. A protective blanket of sorts.

That is a common brewer misconception as the two gasses will mix and be evenly distributed otherwise we as humans would be all dead from suffocation of the "heavier" CO2 blanket at head level on our planet earth.

Besides, as temperatures drop the beer itself will absorb more of the CO2 blanket above the beer in the headspace which is why the oxygen can ingress through the airlock to begin with.

Now, I am willing to accept that a 48 hr cold crash (or so) might not be enough time for that small amount of oxygen in the headspace at such cold temps to cause much oxidation especially with a beer that is already saturated with CO2.

I really wish someone would perform some gas measurements on this topic so we can have abetter idea what is really going on in there.
Maybe a Brew Strong topic is in order....I would love to finally get to the bottom of this!

[Kbar]

I think even if air is sucked back in to the fermenter during crashing, it's still lighter than CO2 so oxidation really isn't a problem.

Agree -O2 will 'float' on top of the CO2 bed you, or I should say, the yeast produced. A protective blanket of sorts.

That is a common brewer misconception as the two gasses will mix and be evenly distributed otherwise we as humans would be all dead from suffocation of the "heavier" CO2 blanket at head level on our planet earth.

CO2 is absorbed at ground level and O2 is released. Could this be why the are not separated in our atmosphere? Constant mixing.

[Bad Goat Brewing]

I'm think they mix. Our fire suppression system went off accidently at work once, dumping thousands of pounds of co2. Our building is below ground level, and our ventilation fans are not at lowest point in the building, but running them for a few hours brought the O2 level back 21% in all but the lowest areas. Those areas returned to normal levels overnight.

[BDawg]

What we are talking about here is figuring out the amount of oxygen that is sucked into the headspace as a result of the shrinkage of the CO2 present in the headspace when the temperature drops from room temperature to cold-crashing temperature.

We can get a reasonable estimate using the Ideal Gas law:
Assume:
First, let's assume you have a 6 1/2 gallon carboy and you have 5 1/2 gallons of beer in it.
That means it's 1 gallon of headspace, or we can round that to an even 4 liters which is 4000 cubic centimeters, or 1000000/4000 or 0.004 cubic meters.
Assume 1 standard atmosphere of pressure. No hurricanes or unreasonably high pressure days here.
Assume that the Air is 21% O2, which is close enough
Start at 70F or 294.26 K
Crash at 32F or 273.70 K

The Ideal Gas law states:
pV = nRT

These are linear equations. If we only change one thing, then Changing that one side will result in a linear change in the other side.
We therefore can calculate the ratio of the difference in volume given the difference in temperature,
figure out the amount it shrank by and multiply that amount by .21, to estimate the Volume of 02 that is sucked into the headspace.


pV1 = nRT1
and
pv2 = nRT2

so that means
V1 = nRT1/p
and
V2 = nRT2/p

grouping:
V1 = T1 * (nR/p)
and
V2 = T2 * (nR/p)

Divide them to get the ratio of the two volumes:
V1/V2 = (T1*(nR/p)) / (T2*(nR/p))

Cancel all the like terms, making everything so simple:
V1/V2 = T1/T2

Solve for V2:
V2 = (V1*T2)/T1

V2 = 4L * 273.70 / 294.26

V2 = 3.72 L

Get how much air must flow in to fill the difference:
4 L - 3.72 L = .2794 L or 279.4 cc AIR gets sucked in
279.4 * .21 = 58.674 cc of O2 (the cube root of 58.674 is about 3.886)

58.674 / 4000 = .0146885 or 1.46685 % O2 in the headspace

So the question is, is 1.44885% O2 significant enough to warrant freaking out about this?
Or would the remaining yeast consume a portion of it as they go dormant?

[edit - Before anyone says so, I know that my simplification calculates volumes based on it being pure CO2, but I think the actual difference in volume if we were to go through doing the calculus to take in partial pressures of each of the gasses in air would be low enough that this is close enough for us to approximate how much volume actually changes based on just the temperature change. ]

[Kbar]

Nice Bdawg, but correct me if I am wrong, and I say this on 3 beers

Pressure and Temperature are the changing variables, not Volume. This is not a piston in a car or a compressor or a ballon (etc.), the volume of the Fermentor stays the same. We need to use Gay-Lussac's law, and see that when the temperature drops, so does the pressure, 'drawing' in (based on the atmospheric pressure outside of the Fermentor) the outside air (O2/N2/Misc).

Gay-Lussac's, Charles' and Boyles' laws are all based on the Ideal Gas law you stated above. Keep one constant, while the other two have relationships where they vary.

P1*T2=P2*T1

You are showing a Fermentor changing in volume above. I may be completely wrong. One needs some more steps to correlate the change in pressure to the addition of volume.

[BDawg]

Nice Bdawg, but correct me if I am wrong, and I say this on 3 beers

Pressure and Temperature are the changing variables, not Volume. This is not a piston in a car or a compressor or a ballon (etc.), the volume of the Fermentor stays the same. We need to use Gay-Lussac's law, and see that when the temperature drops, so does the pressure, 'drawing' in (based on the atmospheric pressure outside of the Fermentor) the outside air (O2/N2/Misc).

Gay-Lussac's, Charles' and Boyles' laws are all based on the Ideal Gas law you stated above. Keep one constant, while the other two have relationships where they vary.

P1*T2=P2*T1

You are showing a Fermentor changing in volume above. I may be completely wrong. One needs some more steps to correlate the change in pressure to the addition of volume.

Sort of, but what I'm saying is that at constant pressure, the volume of the original gas will shrink to occupy a smaller portion of the chamber while staying at the same pressure, and is replaced a portion of an infinite supply of air which is coming in at the same pressure. That portion coming in is equal in volume to the amount of shrinkage of the original gas, since its at the same pressure.

P1 (Va1 + VCo21) = nRT1
P2 (Va2 + VCo22) = nRT2

I'm saying that Va1 = 0
and Va2 + VCo22 = VCo21, and P1 = P2 (both at 1 atmosphere), the difference in the volumes will be proportional to the ratio of the temps.

[Stinkfist]

So the question is, is 1.44885% O2 significant enough to warrant freaking out about this?

ahh that is the real question...and the hardest to answer...

Maybe it does matter when combined with all the other places we pick up oxygen...whether it is the most significant I am sure depends on your other processes. But I am certain that any oxygen pickup will eventually have a detrimental effect on your beer.

Personally I think the part about the crash cooling that makes it worse is, that this exchange is all happening at the time the liquid is absorbing as much gases from its environment as it can. When you add oxygen and the temperature is relative constant, it has basically absorbed all it will(unless you are moving it around, which is why it is best to do a closed transfer).

This sounds like something that could be funded from the AHA Money they have for these kind of things..