So I usually have two beer styles on tap in my fridge, currently a lager and pale ale, I even plan for three kegs to be carbonated and dispensed by one regulator and gas line split twice. I don't mind all being carbonated at same levels, however it just occured to me that since the gas lines are open between kegs, it's entirely possible that aroma may mix and mingle between kegs. Is this a problem in reality, and how cheaply can it be solved if so?
2 kegs, one split gas line, aroma mixing?
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Very interesting point!
I‘m not sure how much of an effect it would be, but I could imagine for example a strongly smoked stout having a detectable effect on a very neutral pils (over time).
Normally its more convenient to spilt your gas line between kegs by using a manifold like the one below. As well as having individual gas taps for each keg they also usually have built in non-return (‘check’) valves that prevent gas from one keg getting into another :-)
I‘m not sure how much of an effect it would be, but I could imagine for example a strongly smoked stout having a detectable effect on a very neutral pils (over time).
Normally its more convenient to spilt your gas line between kegs by using a manifold like the one below. As well as having individual gas taps for each keg they also usually have built in non-return (‘check’) valves that prevent gas from one keg getting into another :-)
TETB is absolutely right (How could he be anything else 
)... it is an interesting question though if not using a non return manifold. You'd need to consider vapour pressure and diffusion rates when not using your system, since as soon as you pour a pint the gas in the line will be replaced fresh. If you consider the volume of the tubing, a pressure of perhaps 10psi, and dry bulb temperature of a kegerator at maybe 10°C. Let's assume your vapour from your kegs has reached equilibrium in the pipes.
In a closed keg at equilibrium the relative humidity is always 100%. The liquid evaporates until the partial pressure of the vapour reaches the maximum value for the temperature. So that allows calculation of the absolute humidity for 10psi 10°C at 0.00579 kg/kg. the volume of gas in a line is V = πr^2h. Let's assume you are using 3/8 gas line with an internal diameter of 6.7mm, and the length of tubing from cylinder to each keg is perhaps 2m. So your tubing is volume is going to be 0.00056 m3, or at this pressure 0.01652068 mol of gas. For CO2 with a molar mass of 44.01 so the mass of gas in the tubing is 0.41mg. Your mass of vapour is thus going to be 2.3 µg - which will comprise a mix of both vapours. ie not a lot.
There are some complexities here as there will be some diffusion over time between the two kegs, and while this would be slow, it could mean that a vapour component could eventually, very slowly reach equilibrium in the other keg headspace - this will take a very long time though since both kegs are sealed and at max relative humidity so you're relying on simple diffusion in a still gas. There's also odd things like the pressure drop in one keg while the other is being served that will to a tiny degree suck the vapour from one keg into the other. The masses though as you'll see above are absolutely teeny weeny.
)... it is an interesting question though if not using a non return manifold. You'd need to consider vapour pressure and diffusion rates when not using your system, since as soon as you pour a pint the gas in the line will be replaced fresh. If you consider the volume of the tubing, a pressure of perhaps 10psi, and dry bulb temperature of a kegerator at maybe 10°C. Let's assume your vapour from your kegs has reached equilibrium in the pipes.In a closed keg at equilibrium the relative humidity is always 100%. The liquid evaporates until the partial pressure of the vapour reaches the maximum value for the temperature. So that allows calculation of the absolute humidity for 10psi 10°C at 0.00579 kg/kg. the volume of gas in a line is V = πr^2h. Let's assume you are using 3/8 gas line with an internal diameter of 6.7mm, and the length of tubing from cylinder to each keg is perhaps 2m. So your tubing is volume is going to be 0.00056 m3, or at this pressure 0.01652068 mol of gas. For CO2 with a molar mass of 44.01 so the mass of gas in the tubing is 0.41mg. Your mass of vapour is thus going to be 2.3 µg - which will comprise a mix of both vapours. ie not a lot.
There are some complexities here as there will be some diffusion over time between the two kegs, and while this would be slow, it could mean that a vapour component could eventually, very slowly reach equilibrium in the other keg headspace - this will take a very long time though since both kegs are sealed and at max relative humidity so you're relying on simple diffusion in a still gas. There's also odd things like the pressure drop in one keg while the other is being served that will to a tiny degree suck the vapour from one keg into the other. The masses though as you'll see above are absolutely teeny weeny.
But you only need a teeny weeny amount of a substance to detect it's aroma....TETB is absolutely right (How could he be anything else
In a closed keg at equilibrium the relative humidity is always 100%. The liquid evaporates until the partial pressure of the vapour reaches the maximum value for the temperature. So that allows calculation of the absolute humidity for 10psi 10°C at 0.00579 kg/kg. the volume of gas in a line is V = πr^2h. Let's assume you are using 3/8 gas line with an internal diameter of 6.7mm, and the length of tubing from cylinder to each keg is perhaps 2m. So your tubing is volume is going to be 0.00056 m3, or at this pressure 0.01652068 mol of gas. For CO2 with a molar mass of 44.01 so the mass of gas in the tubing is 0.41mg. Your mass of vapour is thus going to be 2.3 µg - which will comprise a mix of both vapours. ie not a lot.
There are some complexities here as there will be some diffusion over time between the two kegs, and while this would be slow, it could mean that a vapour component could eventually, very slowly reach equilibrium in the other keg headspace - this will take a very long time though since both kegs are sealed and at max relative humidity so you're relying on simple diffusion in a still gas. There's also odd things like the pressure drop in one keg while the other is being served that will to a tiny degree suck the vapour from one keg into the other. The masses though as you'll see above are absolutely teeny weeny.
Yes if you were smelling the gas directly from the line, but you're smelling the vapour from the liquid poured, so that amount of vapour has to enter the keg, be diluted, condense in the liquid, be diluted in the liquid, be released on pouring and then detected..... so several huge dilutional steps extra.
Yes it's quite possibly such tiny amounts that it practically has very little perceptible impact, which probably explains why when I googled it before, seemingly no one else is worried about it or suffers I'll effects. That said, it may be because it's better practice to use a manifold with a valve, and people seem to do as they scale up. I'll think about adding one to my wish list.
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