I like the concept of oxygen reacting in the beer, making room for more oxygen. Basically showing that the oxidation effect is cumulative if the supply of new oxygen is available. Limiting behavior is the only defense for any of this throughout the entire brewing process as it is impossible to eliminate the threat.
Keg purging with active fermentation
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The post on the keg is pretty close to the same shape as the cup in the disconnect, so most air displaces itself when they physically go together. For all I know when the post pushes the poppet in the disconnect, some CO2 may blast out as well, diluting whatever is left.
To be clear it's not something I worry about
But it's easy enough to purge the line itself a little that I go for it.whattabrau
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Oh I absolutely agree that it doesn't matter. I was using it to illustrate how calculating things to n decimal points gives you ~n decimal points which don't correspond with reality.The post on the keg is pretty close to the same shape as the cup in the disconnect, so most air displaces itself when they physically go together. For all I know when the post pushes the poppet in the disconnect, some CO2 may blast out as well, diluting whatever is left.
To be clear it's not something I worry about
But I was curious when you said you could purge it. I'm not convinced you can, but going full circle: doesn't matter.
ihavenonickname
Well-Known Member
I think this thread is pretty brilliant so I wanted to refresh it and make my contribution.
Here is my ferment in a 10 gallon corny keg. I’ve got almost 9 gallons of 1.060 in it and I’m using the fermentation CO2 to dry purge 7 gallons of serving kegs. I’ve got a spunding valve at the end of the last keg with a few PSI, which helped ensure my corny keg lids stay sealed.
This is my go to set up for most beers, and I’m loving the results. The math makes me confident in purging a volume similar to the volume of beer I’m making. I’m still not confident in how I should set up kegs for my hazy IPA. I prefer to also include a 7gal keg for dry hopping, so that I can get the beer off the yeast and agitate. Timing the The fermentation so that the O2 is one part seems too fussy and too reliant on the math being accurate for my liking so I still wanna figure out a best practice which might include purging liquid sanitizer out with fermentation or separately with commercial CO2. The second photo shows what I’ve done, which was just dry purging, both a dry hop, keg, serving keg, but I’m a little iffy on that.
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To be safe, you can fill the kegs to be purged with sanitizer, and then push the sanitizer out with fermentation CO2. That way the purged volume initially filled with air is quite small. The numbers for trying to purge ~2x keg volume vs. beer volume don't look nearly as good as purging ~1x volume.
Brew on
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That's true for a 10 gallon keg, but if you serialize smaller kegs the math is much more attractive. I've been considering purging 2 kegs and using one for settling/conditioning and the second for serving.To be safe, you can fill the kegs to be purged with sanitizer, and then push the sanitizer out with fermentation CO2. That way the purged volume initially filled with air is quite small. The numbers for trying to purge ~2x keg volume vs. beer volume don't look nearly as good as purging ~1x volume.
Brew on
50L keg purged by 439L of fermentation gas (5 gallon batch output per your superpost)
Keg 1 = 32.3ppm O2
Two 25L kegs purged by 439L of fermentation gas
Keg A = 5 ppb
Keg B = 92 ppb
It appears serializing the kegs ensures a much better sweep.
If I bump the fermentation gas to 439L * 9/5 = 790L*, and attach kegs in this order: 2gal, 5gal, 7gal, assuming actual volumes are 10L, 25L, 35L (matching the original 5L per nominal gallon assumption), and not making any allowances for spunding, dissolved CO2 or fermenter headspace CO2:
Keg X1 = 0.00 ppb
Keg Y1 = 0.00 ppb
Keg Z1 = 0.16 ppb
The last keg will necessarily have the highest O2 concentration. Any pressure will decrease performance.
The results of the above are a bit uglier with the 439L gas:
Keg X2 = 0.00 ppb
Keg Y2 = 8.27 ppb
Keg Z2 = 3,652 ppb
(Obviously infiltration by other means will make the results worse than indicated, but with an ideally impermeable system, it looks like the proposed setup would be fine with the 9 gallons of beer.)
*Did not adjust for gravity. I believe the the 439L is based on a 0.040 gravity drop?
This is a fun thread! I've been purging kegs this way since really trying to nail down my NEIPA recipe and it's helped a ton....I also put half a crushed campden tablet into the sanitized keg when I start to hook up the line from my fermenter to keg to further help reduce oxygen exposure. I've had good enough luck where if a NEIPA hangs out in the keg for a month or more, it's still glowing yellow and the flavors are there. I also put a spunding valve on the gas post so I can spund and keep pressure on the keg seals.
I'm currently using the 15.9G Fermzilla All Rounder and do mostly 10 gallon batches. It's neat that someone's figured out a calculator because I've always wondered how much co2 is produced and what ends up the keg. I'll be following this thread for sure!
I'm currently using the 15.9G Fermzilla All Rounder and do mostly 10 gallon batches. It's neat that someone's figured out a calculator because I've always wondered how much co2 is produced and what ends up the keg. I'll be following this thread for sure!
DuncB
Well-Known Member
The Sankey kegs can be filled to the brim with sanitiser, then insert post with coupler open and top up so that is full. Then purge with ferment gas there should be no minimal gas space that occurs with the corny keg.
The calculations change for these kegs even more so if you used water with less oxygen in, unsure what does of sod met would be needed in 20 litres of water to do that?
The calculations change for these kegs even more so if you used water with less oxygen in, unsure what does of sod met would be needed in 20 litres of water to do that?
Are there any general calculations(rules of thumb) that I can use to determine when to close off my conical fermenter to use the CO2 being generated to carbonate my beer so that I carbonate to a specific volume? Say 2.5 volumes of CO2 @ 20C.Watch this space. Imma write a epic response that will leave you amazed and enthused(or dazed and confused
)
Stay tuned
Brew on
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Ok, you've waited long enough, here is my analysis of the OP's questions. You be the judge, is it epic or not? Worth the wait, or a big let down?
Pull up your waders, it's gonna get deep here. You might want to take this on while drinking coffee rather than beer.
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Let's break this down into manageable pieces, and then look at them one at a time.
First question: Does the continuous flow of CO2 from the fermenter create any different dilution kinetics than the typical multiple cycles of pressurize then vent?
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)) ^ NWhere N = number of purge cyclesThe 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.
View attachment 402029
View attachment 402030
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 (ref: Diffusion - Wikipedia):
Flux = -D * (𝚫Conc / 𝚫Dist)Where Flux is in mass/area-time,D is the diffusion coefficient, and𝚫Conc / 𝚫Dist is the concentration gradientIf 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 (ref: Oxygen Diffusion/Air - Cornell Composting) 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 (ref: Gases - Densities.) 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^4The CO2 flux becomes:
Flux = -D * (𝚫Conc / 𝚫Dist) = -0.15 cm^2/sec * (-4.0e-5 g/cm^4) = 6.0e-6 g/cm^2-secAnd 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/secNext 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 CO2Maltose 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: Brix - Wikipedia):
°Plato = -616.868 + 1111.14 * SG - 630.272 * SG^2 + 135.9975 * SG^3 @ 20°CWater 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 kgThis 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-calcu...ion-extract-and-calories-advanced-homebrewing):
Real_Final_°P = Apparent_Final_°P * 0.8114 + Original_°P * 0.1886And, plugging in the numbers for our example:
Real_Final_°P = 2.561 * 0.8114 + 12.39 * 0.1886 = 4.415°PThus 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 kgThe total sugar fermented works out to:
Fermented_Sugar_Weight = 2.597 kg - 0.890 kg = 1.707 kgAnd 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 CO2Since 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/secDamn, 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:
Total O2 in V before bubble = C * VO2 lost to bubble = C * 𝚫VTotal O2 in V after bubble = C (V - 𝚫V)Concentration of O2 in V after bubble = C * (V - 𝚫V) / VIf C[0] is the concentration of O2 initially, then after "N" bubbles, the current concentration of O2 is:
C = C[0] * ((V - 𝚫V) / V)^NFor 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 ppmBelieve it or not, we reduce the O2 concentration from 21% by volume to 5 parts per billion by volume! :smack:
We can only conclude that using the output of a reasonable size fermentation can very effectively purge a keg of O2.
Coming next, the spreadsheet to allow you to do your own calculations.
Brew on
ihavenonickname
Well-Known Member
It’s 1.003 +\- gravity points. Best to use a spunding valve in case you are a little early.
I look at it a little differently - best to use a spunding valve so you can start a little early on purpose. Each gravity point should get you about 0.5 volumes of CO2, and the beer should be starting with 0.86 volumes. Spunding will protect you and your fermenter from excessive pressure and overcarbonation, but starting a little early will also ensure that you hit your desired carbonation level even if you miss your FG by a point or two on the high side.
2.5 volumes at 20°C corresponds to a pressure of 29 psi (gauge.) Make sure your fermenter is rated for more than this pressure before you try to fully carbonate at room temperature.
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Yes, I would actually start closer to 4C (that is about as low as I can get me system to go), which should put me closer to about 12 psi (gauge), well within the 15 psi max of my fermenter.2.5 volumes at 20°C corresponds to a pressure of 29 psi (gauge.) Make sure your fermenter is rated for more than this pressure before you try to fully carbonate at room temperature.
Brew on
Out of curiosity, where/how did you come with the 0.5 volumes per gravity point? I have seen that before but never a reference on how it was determined.
Well, it's just a rule of thumb so there's some rounding error involved. But you can confirm it by using a priming calculator to determine how much of your favorite fermentable will add a given amount of CO2, then use a recipe builder to find out how much OG and FG that contributes. I get something closer to 0.6 when I average several different fermentables.
That's true, I usually start out at 20C and let it build up to the 15 psi limit (Spike's PRV), and start my cold crash. I have been pretty successful getting close to my carbonation level that way, then I just do a short burst carbonation after it is transferred to the keg.Not gonna get much fermentation (or CO2 generation) at 4°C.
Brew on
The PRV is meant to be an emergency release, not the primary means of pressure control. You should consider getting a spunding valve.
edit: Is this the so-called all-in-one "PRV"? IMO you should still have an emergency relief valve, but I'll leave it at that.
Yes it is the "all in one" they have a spunding valve and a poppet valve mounted horizontally (limits the chance of plugging up).
https://spikebrewing.com/products/pressure-relief-valve
Oh, there is a backup relief valve? That makes me feel a lot better about it. Someone in another thread seemed to be saying there was no emergency relief!
DuncB
Well-Known Member
Yes that thread was quite heated with it's defence of that spunding device.
I'm still in favour of a separate prv somewhere else on the system.
