Balancing a draft system is one of those things that should be easy, but inevitably ends up requiring some trial and error. The basic principle seems sound: beverage tubing supplies some characteristic resistance per unit length (2-3 psi/ft for 3/16″ ID tubing), and all one needs to do is divide that into the desired serving pressure to get the length of tubing required. Under typical serving conditions (38°F and 10 psig, yielding 2.4 vol CO2), this seems to provide acceptable results, with 5 ft of 3/16″ beverage tubing giving a balanced pour. For lower- or higher-pressure pouring, however, this approach breaks down. Which is frustrating; after all, the line resistance should be constant.
So let’s take a closer look at what’s going on. Analyzing the draft system, we can see that there are only a few sources of pressure drop:
- The keg tubing, poppet, post, and disconnect;
- The draft line proper;
- Any vertical rise over which gravity will drop pressure;
- The faucet or tap used for dispensing.
With the exception of the beverage tubing, though, all of these should be more or less fixed. So there’s some constant resistance due to the hardware in the system, plus a variable resistance that’s a linear function of the length of the line.
Now we can calculate the actual pressure drop per unit length of tubing. I’m going to assume a flow rate of 40 mL/s. That provides a 12 second pour, which I think is a pretty reasonable average. Given the flow rate, calculating the Reynolds number is trivial:
Re = QDH/νA = (40e-6 m³/s)(4.8e-3 m)/(1e-6 m²/s)(π(2.4e-3 m)²) ≅ 11,000.
We can also work out the velocity, v = Q/A = 2.2 m/s. For a typical home draft system, the beer is covering the distance from the keg to the glass in about a second.
That’s well over the threshold (Re > 2300) for turbulence, so we can assume fully developed turbulent flow throughout the tubing. We’ll further assume a fluid density of 1.015 SG, roughly in the middle of the typical range for beer. Solving for pressure drop using a Moody chart gives a friction factor, f ≅ 0.031, so:
ΔP/l = ½fρv²/d = 0.5*0.031(1015 kg/m³)(2.2 m/s)²/(4.8e-3 m) ≅ 1.6 x 104 Pa/m, or 0.71 psi/ft.
So the pressure drop due to the tubing itself turns out to be much less than the typical assumption. For our prototypical “10 psi, 5 ft” system to be balanced, the other losses in the system – the keg, the faucet, and the vertical distance between them – must be providing approximately 6.5 psi of pressure drop. (In actuality, it will be somewhat less, since a small amount of over-pressure must be present at the tap in order for the beer to pour at a reasonable rate; the important point is that this quantity is irrespective of line length.) Looking at the problem under those conditions, things make a lot more sense.
Let’s take the example of a highly-carbonated beer that we want to serve at 38°F and 3.2 vol CO2. That will require a pressure of about 18.5 psig. With the other hardware dropping 6.5 psi, the length of tubing required is (18.5 – 6.5 psi)/(0.71 psi/ft) ≅ 17 ft. Had we assumed a drop of 2 psi/ft, there would be only 9 ft of tubing in the system, and we would be pouring nothing but foam.
To spare you the math, here are the resistance values for some common tubing sizes:
Bear in mind that these assume a flow rate of one (US) pint every 12 seconds; substantially different flow rates will have different characteristic resistance values. That said, the friction factor varies only slightly with flow rate, so to a first-order approximation a simple ratio should give good results. For example, given an 8-second pour, 3/16″ line should drop about 1.1 psi/ft.
Update: 30 Nov 2012
I’ve expanded on this post a bit, for clarity and to add resistance values for other tubing diameters.