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.
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 CO
2 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:
P
g = 1010 kg/m
3 * 9.81 m/s
2 * 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 CO
2 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:
P
v = [(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.)
Topic: Pulling Keg off of CO2 (Read 4693 times)
