*“How do you purge your kegs?”*

*“How do you perform closed transfers on finished beers?”*

*“How do you keep fresh beer on tap for so long?”*

These are questions that are not only asked of us with some regularity, but questions that all brewers have asked their more knowledgeable counterparts at some point in time. The fact of the matter is this: Whether you adhere to the hot-side practices we advocate or not, cold-side oxidation control is of paramount importance to ALL brewers. The issue that arises where keg purging and transferring beer is concerned is that there are many different opinions on how to accomplish them, but only one way to fully purge your vessels of oxygen. If you follow the methods we outline in this post, you will be ready to completely purge your receiving vessels of oxygen and transfer your beer with little to no oxygen pickup. Oxygen is the number one enemy of finished beer. This goes for EVERYONE, including professional brewers.

Anytime beer is transferred it picks up oxygen. How little or how much is directly based off of how efficient you are at purging your receiving vessels. Here again are the industry standards for different stages in the brewery:

As you can see here, even in professional breweries we see a linear pick up of DO at the various stages, this is directly caused by moving the ** finished **beer around in the brewery. What we do know is that we (homebrewers) have the advantage here as we can use spunding to mitigate nearly all those steps and use active yeast as our buffer against oxygen intrusion. Homebrewers don’t usually have to worry about beer traveling and sitting on a shelf somewhere (agitating and staling due to warm temps), the beer is nearly always cold, and we can serve from our spunding vessel. With that said, we still try to get the most oxygen out of the receiving vessels from purging as we can, because the less oxygen in contact with the wort and yeast the better, especially with open transfers (which we do not recommend, but understand if it needs to be done). From our other post on CO

_{2}purity, we know that some CO

_{2}contains enough oxygen in it alone to stale the beer, so you can see how paramount properly purging actually is.

**“PRV Vent” Method**

*Process:*

- Turn on CO
_{2} - Attach CO
_{2}to keg in - Open PRV to vent pressure

*How does this work?*

When we pressurize the headspace of a keg we produce a burst of CO_{2} gas originating at the gas-in tube. This burst creates turbulence in the headspace which very effectively mixes the starting headspace gas and the added CO_{2}. We can safely assume that the gases are well mixed prior to the vent cycle. This means we can use static, or equilibrium, math to determine the amount of dilution. We want to know solute concentration in a solution when additional diluent is added to a solution. In our case the solution is a gas solution, the solute is oxygen (O_{2}), and the diluent is CO_{2}. For one dilution (purge) cycle, the change in solute concentration is:

When working with gases in fixed volume vessels, the “amount” of gas is proportional to the absolute pressure (psia), and absolute pressure equals gauge pressure (psig) plus atmospheric pressure (14.695 psia at sea level.) This follows from the universal gas law: PV = nRT. Thus the original “amount” of gas is 14.7 psia, and the diluent amount of gas is the pressure that we add to the keg, so the dilution per cycle becomes:

_{2}_Conc = Prev_O

_{2}_Conc * 14.7 psia / (14.7 psia + Purge_Pressure)

After we pressurize for the purge, we still have the same amount of O_{2} in the headspace that we started with, but the concentration is lower. Once we vent the headspace, the pressure drops back to 14.7 psia, and we have less total gas than we had before. Venting doesn’t change the O_{2} concentration in the headspace, but since it does reduce the total amount of gas in the headspace, the amount of O_{2} goes down as well. The pressurize part of the purge cycle reduces the O_{2} concentration in the gas mix, and the venting then reduces the O_{2} amount. The effect of additional purge cycles is multiplicative, so the formula for multiple purge cycles is:

_{2}_Conc = Orig_O

_{2}_Conc * (14.7 / (14.7 + Purge_Pressure)) ^ N

The O_{2} concentration in air is 21% or 210,000 ppm. If we assume that the keg headspace starts out as air, then we can calculate and plot the resultant headspace O_{2} concentration for various numbers of purge cycles at different pressures.

While this method does work in purging the keg of O_{2}, it is very inefficient and uses a lot of gas. Also do not forget, this has to be done BEFORE racking any beer into the keg. The other thing to note is no one we know of actually does this to the full amount. 1 cycle is completely fill (allow gas to stop flowing), then complete release. * So a minimum of 16 purges at 30psi is needed*.

**“Sanitizer-Purging” Method***

*Process:*

- Completely fill keg with water.
- Add a low foaming sanitizer.
- Attach the gas-in post to the faucet. Open up the PRV. Turn on the water to push any oxygen left in the headspace out through the PRV.
- Connect the gas-in post to the CO
_{2}tank and the liquid-out to a bucket or in the drain. - Turn on the CO
_{2}gas to 10 psi and push all the sanitizer out. - Make sure to remove the gas-in and liquid- out posts with pressure still in the keg.
- When keg has been completely drained of sanitizer there will be a small amount of sanitizer remaining. Invert the keg, making sure that the gas-out is the lowest point in the keg. You may have to trim the gas dip tube as short as possible so it is not protruding into the keg at all. This will force out the remaining sanitizer. (tip:If you keep a slight pressure in the keg what you can do is as you connect your hoses and your lines to the fermenter you can depress the gas-in post and the liquid-out post for a little bit and purge the lines of any oxygen in the lines then connect them to your fermenter).

*How does this work?*

The thought here is you use water instead of gas to fill the void. Pushing it out completely uses considerably less CO_{2} and completely purges the keg. If you do not invert the keg, and leave the residual sanitizer in the keg, you will leave, assuming 1 oz. of residual sanitizer with a 10 ppm DO concentration, 15ppb of O_{2 }in the keg.

**“Fermentation Purge” Method-**

*Process:*

- In place of airlock on fermenter add a length of tubing that connects to your cleaned and sanitized serving keg liquid-out. (Tip: use the same hose you will do the transfer to spund with)
- Add another length of tubing to a blow off jar
- Allow fermentation generated CO
_{2}to purge your keg for you.

*How does this work?*

So, what happens if instead of doing pressurize/vent cycles, we flow CO_{2} into a vessel that originally contains air? Does the flow improve the dilution and removal efficiency of O2 compared to the cyclic process? We can argue that if the CO_{2} inflow is fast enough that CO_{2} comes in faster than it can mix with the air, then it could form a sort of gas piston that would push air ahead of it towards the vent, and that this would push out more O_{2} per volume of CO_{2} than if complete mixing of incoming CO_{2} and existing gas occurred (as it does in the pressurize/vent case.)

The best case for non-mixing of CO_{2} and headspace would be if there were absolutely no internal “air” currents, such that the only mixing of CO_{2} with headspace gas would be via diffusion. So the question comes down to: Is the linear CO_{2} flow rate faster than the diffusion velocity of CO_{2} in air? If the CO_{2} flow rate were much faster than diffusion, then mixing would be limited, and continuous flow would be more efficient than purge/vent. If CO_{2} flow rate were much slower than diffusion, then gases would be mostly mixed, and continuous flow would not be any more efficient than pressurize/vent. If the flow rate and diffusion rates were of the same order of magnitude, then there would be significant, but not complete, mixing, making this the most complex scenario to analyze.

To start we need to get an estimate of the diffusion velocity of CO_{2} in air. If we limit our analysis to one dimensional flow (say from bottom to top of a keg, uniform velocity across the width), things will be much simpler, but still valid. Fick’s first law of diffusion is:

Flux = -D * (ΔConc / ΔDist)

Where Flux is in mass/area-time,

D is the diffusion coefficient, and

ΔConc / ΔDist is the concentration gradient

If we divide Flux [mass/area-time] by density [mass/volume] we get linear velocity [dist/time] which is what we are looking for.

The diffusion coefficient for CO_{2} in air is about 0.15 cm^2/sec. Now if we make some assumptions about gradients we might encounter, we can estimate a linear CO_{2} flow rate due to diffusion. We will use approximate numbers for simplicity, since we are only looking for order of magnitude estimates of velocity.

A corny keg has a volume of about 20 L or 20,000 cm^3, and a height of about 55 cm, leaving a cross sectional area of about 20,000 cm^3 / 55 cm = 364 cm^2. The density of CO_{2} at STP is about 2 g/L or 0.002 g/cm^3. If we assume 2.5 cm of pure CO_{2} at the bottom of the keg, and 2.5 cm of air at the top of the keg, and a uniform concentration gradient from the bottom to the top, the CO_{2} gradient becomes:

ΔConc / ΔDist = (0 – 0.002 g/cm^3) / 50 cm = -4.0e-5 g/cm^4

The CO_{2} flux becomes:

Flux = -D * (ΔConc / ΔDist) = -0.15 cm^2/sec * (-4.0e-5 g/cm^4) = 6.0e-6 g/cm^2-sec

And finally the linear velocity of CO_{2} due to diffusion is:

CO_{2}_Diffusion_Velosity = CO_{2}_Flux / CO_{2}_Density = 6.0e-6 g/cm^2-sec / 0.002 g/cm^3 = 0.003 cm/sec

Next we need to determine the linear flow velocity of CO_{2} being fed through a keg from an active fermentation.

The reaction for fermentation of maltose is:

Maltose + H2O –> 2 Dextrose –> 4 Ethanol + 4 CO_{2}

Maltose has a molecular weight of 342.30 g/mol and CO_{2} has a molecular weight of 44.01 g/mol, so each gram of maltose fermented generates 4 * 44.01 / 342.3 = 0.5143 gram of CO_{2}. So, if we determine how much sugar we ferment over what period of time, we can calculate how much CO_{2} we created and calculate an average flow rate over the cross section of a keg.

Let’s work an example assuming 20 L of wort with an OG of 1.050 that achieves 80% apparent attenuation over a four day fermentation. First we have to determine how much sugar we started with. An SG of 1.050 is equivalent to 12.39°Plato, or 12.39% sugar by weight. To convert SG to plato use the following formula:

°Plato = -616.868 + 1111.14 * SG – 630.272 * SG^2 + 135.9975 * SG^3 @ 20°C

Water at 20°C has a density of 0.9982 kg/L, so the weight of 20 L of wort @ 1.050 is:

20 L * 1.050 * 0.9982 kg/L = 20.96 kg

This wort is 12.39% sugar by weight, so the weight of sugar is 2.597 kg. At 80% apparent attenuation, this beer would have an FG of 1.010, or 2.561°Plato. Since the presence of alcohol affects the SG the actual attenuation of the beer is lower (the final °Plato is higher), we must correct the final °Plato using the Balling approximation:

Real_Final_°P = Apparent_Final_°P * 0.8114 + Original_°P * 0.1886

And, plugging in the numbers for our example:

Real_Final_°P = 2.561 * 0.8114 + 12.39 * 0.1886 = 4.415°P

Thus the finished beer contains 4.415% by weight of sugar, which works out to:

Final_Sugar_Weight = 20 L * 1.010 * 0.9982 kg/L * 0.04415 = 0.890 kg

The total sugar fermented works out to:

Fermented_Sugar_Weight = 2.597 kg – 0.890 kg = 1.707 kg

And the total weight of CO_{2} created works out to:

CO_{2}_Weight_Created = 1.707 kg_Maltose * 0.5143 kg_CO_{2}/kg_Maltose = 0.878 kg or 878 g of CO_{2}

Since CO_{2} has a density of about 2 g/L, we created about 439 L or 439,000 cm^3 of CO_{2}.

If we push our CO_{2} through the keg at a constant rate over a four day fermentation, the flow rate of the CO_{2} over the 364 cm^2 cross section of the keg works out to:

CO_{2}_Velocity = 439000 cm^3 / (4 days * 24 hr/day * 3600 sec/hr * 364 cm^2) = 0.0035 cm/sec

Damn, that works out almost the same as our diffusion velocity of 0.003 cm/sec. So, we are in the complex, hard (i.e. infeasible) to analyze regime of relative flow rates. So, what do we do now? Well, we punt, and do the worst case analysis which would assume that we get no O_{2} removal assist from the sweeping action of the bulk CO_{2} flow. As a result of doing this our residual O_{2} levels will be less than we calculate, so we will have a built in safety factor.

So, the answer to our first question is: Yes, the bulk CO_{2} flow probably helps sweep out more O_{2} than do simple pressurize/vent cycles, but the analysis is too difficult, so we’ll just ignore the flow sweep effect, and end up with a pessimistic estimate of our final purged keg O_{2} levels (i.e. things will actually be better than the calculations show.)

**What’s the worst case O2 levels left in a keg purged with the output of an active fermentation?**

So, just how do we attack a continuous slow purge flow analytically? Assume a tube runs from the fermenter to the keg liquid post, and an airlock/blow off tube is fitted to the keg gas post. Then every time the airlock bubbles you lose a small volume of the current gas mix (which we are assuming is homogeneous) from the keg and fermenter headspace. Let’s call this volume “ΔV”, and the total volume of the fermenter headspace, keg, tube, etc. “V”. Furthermore, let’s call the current concentration of O_{2} in V “C”. We then have the following:

Total O_{2} in V before bubble = C * V

O_{2} lost to bubble = C * ΔV

Total O_{2} in V after bubble = C (V – ΔV)

Concentration of O_{2} in V after bubble = C * (V – ΔV) / V

If C[0] is the concentration of O_{2} initially, then after “N” bubbles, the current concentration of O_{2} is:

C = C[0] * ((V – ΔV) / V)^N

For V = 25 L and ΔV = 0.0001 L (0.1 mL), (V – ΔV) / V = 0.9999960. We’re not getting much purging action per bubble; this doesn’t look very promising yet.

So, where will we end up at the end of the example fermentation above? Well, we generate 439 L of CO_{2} from fermentation, and if we divide that into 0.0001 L bubbles, we produce a total of 4,390,000 bubbles. If we plug that into our formula above, and start with 210,000 ppm of O_{2} in V, then we have:

Final O_{2} Conc = 210000 ppm * ((25 L – 0.0001 L) / 25 L)^4390000 = 0.005 ppm

**We reduce the O _{2} concentration from 21% by volume to 5ppb.**

So why does this all mater?

Any oxygen left in the keg WILL (it’s a proven fact) end up in your beer. We know from our CO_{2} purity post that CO_{2} alone can contain enough O_{2} in it to stale beer by itself, so anything left in the keg after a purge is only going to add to that.This is also the reason why any transfers of finished beer are not recommended by us.

You are essentially purging a container full of O_{2}, with CO_{2} that has O_{2} in it, adding fragile beer to it, then either storing it, or even worse carbonating it with CO_{2} that contains O_{2}! You have just lost the battle with oxidation at that point.

We have been a huge proponent of spunding for quite some time now, and together with proper purging and closed transfers, you have the best chance possible for the most stable, fresh beer you can manage. With spunding, any minuscule amount of O_{2} left in the keg, or picked up on transfer, is then consumed in the keg, by the yeast. The yeast then naturally carbonate the beer with pure CO_{2} (When we say pure, we mean free of O_{2}.Carbonation happens quite fast here, as you have billions of little carbonation stones in the liquid, and are not just forcing it in as head pressure), and being that you serve directly out of that package, no other CO_{2} mitigation methods need to be taken. You still have the issue of using the bottle CO_{2} to dispense, but honestly so does everyone.

**NOTE: Big shout out to doug293cz from HBT whose very detailed response in the HBT thread listed below forms the basis for this post.**

- http://compost.css.cornell.edu/oxygen/oxygen.diff.air.html
- http://www.engineeringtoolbox.com/gas-density-d_158.html
- https://en.wikipedia.org/wiki/Brix
- https://en.wikipedia.org/wiki/Diffusion
- https://byo.com/hops/item/408-calcu…ion-extract-and-calories-advanced-homebrewing
- source: https://www.homebrewtalk.com/forum/threads/keg-purging-with-active-fermentation.628658/
- https://lifefermented.wordpress.com/2014/01/22/how-much-co2-is-produced-from-brewing/

Thanks for the calculations Bryan, Derek ! I rerun the math and confirmed a fermentation of a 12% (plato) beer produces around 450-500L of CO2. I would assume that the O2 in the fermenter can be easily mixed and consumed. Worst case scenario is that the O2 in the keg does not reach the fermenter and it is just diluted evenly and pushed out. If that is the case 4% of the O2 that was in the keg would remain. Not a minor amount. In other words, I am confident that the fermenter has no O2 but need to be convinced that the O2 in the keg can be pushed out. It seems this method can help finish a previously purged keg though.

Thanks again!!