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:
- 1/8″ ID: 5.4 psi/ft
- 3/16″ ID: 0.71 psi/ft
- 1/4″ ID: 0.19 psi/ft
- 5/16″ ID: 0.069 psi/ft
- 3/8″ ID: 0.026 psi/ft
- 1/2″ ID: 0.0072 psi/ft
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.
Hi, I didn’t make it too far in math (back in the day) so is there a way to simplify all of the above? How about a simple chart with maybe the serving temps along one side and the dispensing psi along the perpendicular side with corresponding tubing lengths in between. I’d probably get all that above if I didn’t let the 70’S get in the way. (if you know what I mean) Thanks!
Chris,
Temperature will actually have almost no effect. So the formula for finding line length is just: L = (P – 6.5)/0.71 where L is in feet and P in psig.
Chart-wise, rounded to the nearest 6 in:
Sean
This is very interesting, how would you solve this for a non-carbonated liquid traveling through a gassing stage. Like Coke syrup for instance in a bag-in-box setup? Assume the liquid travels a distance of 30 feet at 20 ℃ in 3/8″ line (under suction from the Flowjet pump). Then enters a flash filler with 12′ of coiled 1/4″ stainless line to drop temp to 1 ℃. Then enters a carbonating bell and is gassed to a set volume of gas, before traveling only 3′ to the tap.
In this scenario, line length is not the variable, but the gas setting on the carbonator. This might be an interesting setup to think about. Especially since it has the advantage of allowing separation of the gas from the liquid. So theoretically, you could use any gas in the stream of liquid.
How do you think this might apply to beer?
Sean, (wow is that weird to type)
I think some research would have to be done, but I don’t see any reason why inline carbonation/nitrogenation wouldn’t work in beer. My guess would be that additional line length would be required to give the gas time to come to equilibrium solubility. That’s why Coke foams so much coming out of the faucet.
From a practical standpoint, the biggest obstacle might be finding a carb stone small enough to fit in a beer line. If the additional length needed for carbonation had to be combined with a larger line, the length might get excessive. Personally, I would look into carbonating inline between the fermenter and bright tanks.
I take it you’re interested in serving a single bright on both CO2 and N2?
Sean
[…] note. While I was doing some calculations for Two Mile, I decided to expand on a year-old post on draft system balancing, primarily just to include the relevant results for longer draft systems. […]
Hi there,
I’m actually being challenged in serving my beer in a bar – foamy beer on my first customer. I am not sure how to apply all of those calculations. Could you help me if there has a way on which I could troubleshoot the situation. Below some data that may help:
– All grain Premium Lager beer 100% malt.
– Forced Carbonation 2.1 Vol CO2, 14 PSI
– Using soda keg
– The bar informed me they applies 29 PSI to the lines which I found to high based in all I am reading about balancing systems. The room where they keep the kegs connected to the pipes is in the floor immediately below the bar so I guess it is around 19 ft. Which would bring us to 20 psi observing the table above against the 29 PSI they informed to be using.
– The bar also informed the Breweries supply them with Kegs on 21 PSI which I also found to high and would be over-carbonation to most of the beer styles.
– I’ve noticed the room has no cooling and is probably around 68F, so beer is not at the same temp as in the pipes.
Thank you in advance,
Isac
Wow, that is an intensely bad setup. Are they able to pour any beers well?
For a 19 ft rise, 29 psi wouldn’t be too high assuming it’s 3/16″ tubing. You’d need that just to keep the beer carbonated at room temperature.
How long does it take to pour a pint? 19′ of 3/16″ tubing doesn’t even hold half a pint, so to get a proper pour you might have to pour half, wait several minutes for the beer to cool down, then pour the rest.
Sean
Thanks a lot for this thorough explanation. Being a physicist, I find it very interesting.
I have two questions though.
First, how do you find out what is the pressure drop due to the keg and faucet?
Second, how do you know what the flow rate would be? Won’t the flow rate be a function of the pressure drop itself, which is again a function of the flow rate?
Thanks in advance.
Roni
Roni,
I think the pressure drop due to the keg/faucet would probably have to be modeled numerically. They aren’t complex arrangements compared to, say, a nuclear reactor core, but I don’t think they could be accurately modeled as arrangements of sections of smooth pipe either, so lookup tables are probably out. As far as flow rate, you’re right that it’s recursive and would also likely have to be solved for numerically. In reality, I just used a graduated cylinder (or pint glass, for that matter) and a stopwatch.
Cheers,
Sean