Fermentation generates nearly 25x the amount of co2 in relation to the wort volume. With some simple fittings you can route your fermentation generated PURE co2 though the keg it is going into. Effectively purging the keg perfectly. Bonus points for using it to purge the lines you use to transfer. Extra bonus points if you use spunding as well.

Clean and sanitize keg as normal, empty, then..

Fermenter out to liquid out on keg

Keg in to blow off bucket (or more kegs)

To get the same level of purge as this, you would have to fully pressurize the keg and fully release it at 25psi at least 20 times.. thats a lot of waster co2 out of your bottle when fermentation can do it for you effortlessly.

If you have blow off you can add another jar inline from the fermenter out to the liquid out. You would then have a tube from the fermenter below the sani level and the tube going to the liquid out as high as possible in the sealed jar.

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“The reaction for fermentation of maltose is:[indent]Maltose + H2O --> 2 Dextrose --> 4 Ethanol + 4 CO2[/indent]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 (ref:

https://en.wikipedia.org/wiki/Brix):[indent]°Plato = -616.868 + 1111.14 * SG - 630.272 * SG^2 + 135.9975 * SG^3 @ 20°C[/indent]Water at 20°C has a density of 0.9982 kg/L, so the weight of 20 L of wort @ 1.050 is:[indent]20 L * 1.050 * 0.9982 kg/L = 20.96 kg[/indent]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 (ref:

https://byo.com/hops/item/408-calculating-alcohol-content-attenuation-extract-and-calories-advanced-homebrewing):[indent]Real_Final_°P = Apparent_Final_°P * 0.8114 + Original_°P * 0.1886[/indent]And, plugging in the numbers for our example:[indent]Real_Final_°P = 2.561 * 0.8114 + 12.39 * 0.1886 = 4.415°P[/indent]Thus the finished beer contains 4.415% by weight of sugar, which works out to:[indent]Final_Sugar_Weight = 20 L * 1.010 * 0.9982 kg/L * 0.04415 = 0.890 kg[/indent]The total sugar fermented works out to:[indent]Fermented_Sugar_Weight = 2.597 kg - 0.890 kg = 1.707 kg[/indent]And the total weight of CO2 created works out to:[indent]CO2_Weight_Created = 1.707 kg_Maltose * 0.5143 kg_CO2/kg_Maltose = 0.878 kg or 878 g of CO2[/indent]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:[indent]

CO2_Velocity = 439000 cm^3 / (4 days * 24 hr/day * 3600 sec/hr * 364 cm^2) = 0.0035 cm/sec[/indent]

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.)

**Second question: 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 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:[indent]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[/indent]If C[0] is the concentration of O2 initially, then after "N" bubbles, the current concentration of O2 is:[indent]C = C[0] * ((V - ΔV) / V)^N[/indent]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:[indent]Final O2 Conc = 210000 ppm * ((25 L - 0.0001 L) / 25 L)^4390000 = 0.005 ppm[/indent]Believe it or not, we reduce the O2 concentration from 21% by volume to

**5 parts per billion** by volume! :smack: :ban: :ban: :ban: Talk about the power of compounding!

We can only conclude that

** using the output of a reasonable size fermentation can very effectively purge a keg of O2.**“

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