Kegging With Care: A Guide to Purging.

“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 CO2 purity, we know that some CO2 contains enough oxygen in it alone to stale the beer, so you can see how paramount properly purging actually is.


“PRV Vent” Method


  1. Turn on CO2
  2. Attach CO2 to keg in
  3. Open PRV to vent pressure

How does this work?

When we pressurize the headspace of a keg we produce a burst of CO2 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 CO2. 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 (O2), and the diluent is CO2. For one dilution (purge) cycle, the change in solute concentration is:

New_Conc = Prev_Conc * Starting_Amount / (Starting_Amount + Diluent_Amount)​

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:

New_O2_Conc = Prev_O2_Conc * 14.7 psia / (14.7 psia + Purge_Pressure)​

After we pressurize for the purge, we still have the same amount of O2 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 O2 concentration in the headspace, but since it does reduce the total amount of gas in the headspace, the amount of O2 goes down as well. The pressurize part of the purge cycle reduces the O2 concentration in the gas mix, and the venting then reduces the O2 amount. The effect of additional purge cycles is multiplicative, so the formula for multiple purge cycles is:

Final_O2_Conc = Orig_O2_Conc * (14.7 / (14.7 + Purge_Pressure)) ^ N
Where N = number of purge cycles​

The O2 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 O2 concentration for various numbers of purge cycles at different pressures.


While this method does work in purging the keg of O2, 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*


  1. Completely fill keg with water.
  2. Add a low foaming sanitizer.
  3. 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.
  4. Connect the gas-in post to the CO2 tank and the liquid-out to a bucket or in the drain.
  5. Turn on the CO2 gas to 10 psi and push all the sanitizer out.
  6. Make sure to remove the gas-in and liquid- out posts with pressure still in the keg.
  7. 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 CO2 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 Oin the keg.

 *We assume trimmed gas tubes for this method.

“Fermentation Purge” Method-


  1. 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)
  2. Add another length of tubing to a blow off jar
  3. Allow fermentation generated CO2 to purge your keg for you.

How does this work?


So, what happens if instead of doing pressurize/vent cycles, we flow CO2 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 CO2 inflow is fast enough that CO2 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 O2 per volume of CO2 than if complete mixing of incoming CO2 and existing gas occurred (as it does in the pressurize/vent case.)

The best case for non-mixing of CO2 and headspace would be if there were absolutely no internal “air” currents, such that the only mixing of CO2 with headspace gas would be via diffusion. So the question comes down to: Is the linear CO2 flow rate faster than the diffusion velocity of CO2 in air? If the CO2 flow rate were much faster than diffusion, then mixing would be limited, and continuous flow would be more efficient than purge/vent. If CO2 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 CO2 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 CO2 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 CO2 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 CO2 at STP is about 2 g/L or 0.002 g/cm^3. If we assume 2.5 cm of pure CO2 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 CO2 gradient becomes:

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

The CO2 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 CO2 due to diffusion is:

CO2_Diffusion_Velosity = CO2_Flux / CO2_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 CO2 being fed through a keg from an active fermentation.

The reaction for fermentation of maltose is:

Maltose + H2O –> 2 Dextrose –> 4 Ethanol + 4 CO2

Maltose has a molecular weight of 342.30 g/mol and CO2 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 CO2. So, if we determine how much sugar we ferment over what period of time, we can calculate how much CO2 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 CO2 created works out to:

CO2_Weight_Created = 1.707 kg_Maltose * 0.5143 kg_CO2/kg_Maltose = 0.878 kg or 878 g of CO2

Since CO2 has a density of about 2 g/L, we created about 439 L or 439,000 cm^3 of CO2.

If we push our CO2 through the keg at a constant rate over a four day fermentation, the flow rate of the CO2 over the 364 cm^2 cross section of the keg works out to:

CO2_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 O2 removal assist from the sweeping action of the bulk CO2 flow. As a result of doing this our residual O2 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 CO2 flow probably helps sweep out more O2 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 O2 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 O2 in V “C”. We then have the following:

Total O2 in V before bubble = C * V
O2 lost to bubble = C * ΔV
Total O2 in V after bubble = C (V – ΔV)
Concentration of O2 in V after bubble = C * (V – ΔV) / V​

If C[0] is the concentration of O2 initially, then after “N” bubbles, the current concentration of O2 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 CO2 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 O2 in V, then we have:

Final O2 Conc = 210000 ppm * ((25 L – 0.0001 L) / 25 L)^4390000 = 0.005 ppm​

We reduce the O2 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 CO2 purity post that CO2 alone can contain enough O2 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 O2, with CO2 that has O2 in it, adding fragile beer to it, then either storing it, or even worse carbonating it with CO2 that contains O2! 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 O2 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 CO2 (When we say pure, we mean free of O2.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 CO2 mitigation methods need to be taken. You still have the issue of using the bottle CO2 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.

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