The other factor, likely unimportant if you are rounding to only 2 significant digits, is the specific gravity of the finished beer. If the FG of a cask ale is 1.016 and it has 3 g/L of CO2, the mass fraction of CO2 would be 3.0/1016 = 0.00295 = 0.003 (0.3%)

A wheat beer with an FG of 1.008 and 9 g/L of CO2 would have a CO2 fraction of 9/1008 = 0.0089 = 0.009 (0.9%)

]]>Step 1. How much volume is in 1 metre of 4.5” (11.43 cm) pipe?

Dia of 11.43/2 = radius of 5.72 cm

Volume of 1 metre (100 cm) of pipe = pi * r^2 * 100 cm

= 3.14 * 5.72 ^2 * 100 = 10,278 cm^3 = 10,278 mL = 10.278 L

Step 2. If 900 bbl (105,300 L) goes by in an hour, how much volume goes by per second?

Volume / sec = 105,300 L / (60min *60 sec)= 29.25 L / sec

Step 3. Divide volume/sec by volume of 1 metre

Velocity = 29.25 L / sec / 10.278 L/m = 2.85 m/sec

Therefore, the beer flowing through a 4.5″ pipe at a rate of 900 bbl/hr is moving at 2.85 metres per second.

]]>Curious if you have posted those formulae mentioned in the last portion of this post. If so, is there a link to this?

Thank you,

Ely ]]>