Oxidized DIPA -again...

My second kegged DIPA is suffering what appears to be oxidation.
Keg was purged pre-fill, no splashing on transfer.

Here’s my question:
I have been ‘slow’ carbing (11-12 psi over a couple of weeks) my homebrew kegs, as they share a C02 tank with some commercial kegs.

Question:
Does force carbing do anything to eliminate the potential for oxidation?
I know I am clutching at straws, but just trying to see if there is any homebrewer research on this.

I am frustrated, and googling like a banshee.

Quick vs. slow carbing won’t change potential for oxidation; but if it takes a couple of weeks for symptoms to appear, time you’re spending carbing, you’d at least get a week or so of drinking fresher tasting beer in.  That’s all. Oxygen exposure (if you’re sure that’s the problem) is happening somewhere upstream in your process.

^^^^
Just another thought as you trace the problem: you could still have oxygen pickup at kegging.  You say keg was purged, but if you mean you put on gas and “burped” the PRV a few times, there was lots of O2 in that keg.  The only sure way to purge is to fill to the brim with liquid (e.g. sanitizer) and push it all out with CO2.  There was a thread recently on the forum where a member posted calculations on “purging” by burping the PRV and your mind would reel at the number of cycles it would take!  You’re also drawing air into the fermentor as you siphon beer out (if that’s what you do) so you might search here on the forum for threads on “closed transfers.” I’ve just gone that route.

Thanks, Robert!

What psi do you use to do the sanitizer purge with C02?

fwiw the beer tasted great pre-siphon, but I will look into closed transfers, too.

I will investigate…

Psi doesn’t matter, whatever will get liquid out.  You’ll know liquid replaced air, and CO2 replaced liquid.  I say don’t waste gas pressurizimg the keg too high when you’ll just release it to add beer!

If it tasted great out of the fermentor, I bet you’re on the right track here.  This point in the process seemed to be my Achilles heel. Purging the keg is probably the big problem, because air drawn into the fermentor is only making contact at the still surface.  But it’s really easy to do a closed transfer (a carboy cap based rig to push low psi gas in and a racking cane leading to a QD to your keg) and if you like IPA, well, those hops sure do suffer from oxidation don’t they! (Your LHBS has probably already helped others assemble the bits for closed transfer.)

Caution if using fermenters that don’t play well with pressure. Like glass carboys…

ABSOLUTELY! When I say “low psi” I mean like 1 or 2.  I elevate my carboy above the keg, so I just need to apply enough gas to in effect start a siphon and then replace the volume of the draining beer.  It’s slow but O2 free.

These links might help…https://www.experimentalbrew.com/experiments/great-purge-does-full-liquid-purge-keg-protect-hop-aroma-better

https://www.experimentalbrew.com/purging-keg-etc-etc-etc

If anything, it increases it.

https://tapintohach.com/2013/12/02/how-the-purity-of-injected-carbon-dioxide-affects-the-oxygen-concentration-of-beer/

Here are the maths

ppm-o2-after-purge-chart-png.png

ppm-o2-after-purge-chart-2-png.png

You can see it’s much more efficient to use sanitizer and push it out (10-12psi). Also trimming your gas diptube flush with the top wall of the keg and overfilling the keg in a closed transfer will be your best bet.

Remember heat speeds up oxidation or maybe your hops are oxidized.

I’d be curious to see what 5, 10, and 15 seconds of 15psi on the out post would be.

Trimming the gas post… why haven’t I thought of that?

I’ve been looking over some past threads on purging.  It’s been mentioned several times that there may be info on purging with a flow from the bottom, but nobody’s posted a link. Anybody please?
  (Liquid still seems to me the surest thing.  Dalton’ s Law and all.)

Close to full oxygen when just flowing in. Gases don’t stratify, they mix.

You mean like this?

“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: https://en.wikipedia.org/wiki/Diffusion):Flux = -D * (ΔConc / ΔDist)
Where Flux is in mass/area-time,
D is the diffusion coefficient, and
ΔConc / 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: http://compost.css.cornell.edu/oxygen/oxygen.diff.air.html)  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: http://www.engineeringtoolbox.com/gas-density-d_158.html.)  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/sec

Will someone please tell the AHA this ^^^

Thanks, that’s what I’ve always assumed (Dalton again) but never saw it quantified.  Hey, you probably have this covered too:  How much oxygen is introduced by being dissolved in the small residue of sanitizer in the keg?  I’ve never seen numbers on that either.

I have tried to tell everyone but no one listens! [emoji12]

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.

[/quote]

“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: https://en.wikipedia.org/wiki/Brix):°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-calculating-alcohol-content-attenuation-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/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.)

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 * 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) / 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: :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.

We love to discuss this and other advanced topics our forums.  Link in signature.  You will see many active members from here over there.

I don’t know, you lost me.

What I’m talking about would basically resemble a large blichmann beer gun, which many people seem to trust. Only I stead of a bottle it’s a keg. CO2 pushed into bottom of empty keg, with prv open for x amount of time, then fill. I’m just curious what the measured DO numbers are, if anyone has measured. Not trying to say anything about which is better… obviously pushing sanitizer out would seem to be unbeatable. But what if a brewer couldn’t do that for whatever reason. Pure curiosity. It seems like it has to do a better job than CO2 into the short tube IN and out the prv right next to it…