Will pulling a carbed keg off CO2 for a few days to a week screw things up? Keeping it cold during this time and hooking it back up after said time.
I do that all the time. I’ve never had a problem.
Generally speaking, no, especially not if kept cold.
I dont’ keep my kegs continually hooked to CO2.
I’ll pour a few pints and then repressurize the keg. Worst thing that happens is you might lose some carbonation, but you get that back when you repressurize (not instantly, obviously).
I do it all the time, too, and I don’t necessarily keep it cold. I have five taps, so sometimes I’ll take something off to feature something new or different, and just put it away in the cellar at 60F or thereabouts, and I’ll put it back on at another time, and it is fine, not the worse for wear in any way. Beer keeps well in kegs as long as it is kept away from oxygen and/or extreme temperatures.
The only thing I’d watch out for is leaks in the keg—you’ll be disappointed if the headspace becomes totally depressurized and the beer starts going flat/oxidized. Just do the soapy water trick to detect any leaky seals.
Yep
As others have said, this is fine. Similar to reducing pressure if needed for serving. No worries. Use a lil CIP lube on your gaskets and o-rings for trouble free sealing.
Just ensure you actually remove the black “beer out” connection altogether. Don’t just unscrew the hose from it unless you want one of those things to look back on and laugh about… [emoji16]
Are you talking about removing the post? I removed the gas disconnect but kept the beer out line hooked up to the keg and tap on my kegerator.
Trust me, you’ll know if you’ve done it wrong. You’re fine the way it is. If you unscrew the hose only you’ll have a nice beer fountain until you remove the ball lock disconnect.
OH, who would take the hose off the disconnects? In my case, that would mean ripping off the Oetikir clamps.
I’m guessing he speaks from experience…
Mine screw on. I like it because I can remove the disconnect for cleaning and I use co2 through the beer in line to flush the keg prior to filling. After flushing with co2 and filling with beer I switch the hose over to the co2 post to carb. One must just remember to remove the black disconnect between flushing and carbing.
As a follow up to this, can anyone estimate how long you can actually serve off a carbed keg before it stops pouring? I assume the carbonation of the beer stays, but eventually you lose serving pressure and the beer simply can’t make the way up the tubing.
The way I understand it is that every time you remove CO2 from the keg, some CO2 in the beer will come out of solution until the pressure in the keg is once again equalized. Not sure how much has to be removed and how fast it actually happens, but eventually your beer will have less carbonation.
I can fill a pitcher from head pressure. After one pitcher the pour slows noticeably.
+1
This depends a lot on how much head space is in the keg, too. A very full keg will only pour a short time before there is not enough pressure to push more beer. A half empty keg with the same head pressure will pour lots of pints.
As mentioned, the beer will de-carbonate to equalize the head pressure. Assuming the keg is 60 cm tall and being dispensed at that height, you’d need ~6 kPa (~0.9 psi) head pressure to overcome the static head.
If we assume there’s 19 L of beer and 1 L of headspace at 10 psig (69 kPa), and that we pour quickly enough that no dissolved CO2 leaves the beer, we could pour roughly (10 - 0.9)/10 = 91% of the beer. I say “roughly” because that’s neglecting the remaining beer column’s effect on the static head. Assuming the keg is a cylinder of beer at 1.010 SG, the pressure that needs to be overcome is:
Pg = 1010 kg/m3 * 9.81 m/s2 * 0.6 m * (19 - V) L / 19 L = 313V Pa
Where V is the volume poured in liters. If the beer starts out at 2.5 vol CO2 and 4°C (equilibrium solubility ~1.5 vol) and we pour slowly, such that the beer and headspace are always in equilibrium, the head pressure will be:
Pv = [(1 L * 1.98 g/L * 69/101.325) + (V L * 1.98 g/L * (2.5 - 1.5))] * 101.325 Pa / (20 L * 1.98 g/L) = 244 - 12.7V Pa
Equating and solving gives V = 18.69 L, or a little less than a pint remaining when the flow stops. Starting with a half-and-half keg under the same conditions, V = 9.73 L, so you can indeed pour a little (40 mL) more.
(Edited to flip the definition of V and solve for the half-full keg.)
Very true.