Day 441

You know you’re getting deep into the semester in Brewhouse Calculations when it takes you a solid hour just to review last week’s class. Finished with review, we moved on to some basic fluid mechanics. Liquids and semi-solids are moved all around the brewery in pipes, so an understanding of flow rates and turbulence would seem to be fairly important.

First, we learned that (in the metric world), a force of one newton pressing on one square metre equals a pressure of one pascal. That’s not much pressure, by the way–typical barometric pressure on a nice day is about 101,325 pascals (or about 101.3 kilopascals).

(Isaac Newton, by the way, was an English mathemetician who undoubtedly put many newtons of pressure on the seat of his chair when he sat down despite the fact that he was only one Newton. Blaise Pascal was a French mathemetician, which just goes to show that if you want something named after you, become a mathemetician. But I digress…)

We reviewed the various types of pressure gauges, then moved on to flow meters–the instruments that measure how fast fluid is moving through a pipe.

Due to friction with the pipe walls, fluid near the walls always moves slowest, while fluid in the middle of the pipe always moves fastest. If there is a gradual gradient between the two extremes and little in the way of eddying and break-up, then the flow is said to be smooth or “laminar”. If there is a lot of eddying and mixing of the liquid, the flow is turbulent. This is represented by something called a “Reynolds number”, and the formula for calcualting it is

Re = pud/µ

where

  • Re = the Reynolds number: less than 2100 is laminar, more than 4000 is turbulent, somewhere in between is transitional
  • p =  density of fluid (kg/m3)
  • u = velocity of the liquid (metres per second)
  • d = diameter of the pipe (metres)
  • µ = fluid dynamnic viscosity (kg per metre-second)

If you want to determine the minimum diameter pipe you need to ensure laminar flow, or conversely, the maximum diameter needed for turbulent flow, you can rearrange this formula:

d = (Re µ) / (p u)

For instance, if you want to move some wort that has a density of 1034 kg/m3 and a dynamic viscosity of 5 x 10-3 kg/ms, and your pump is going to move the wort at 5 m/s, then to ensure laminar flow (a Reynolds number of 2100):

d = (2100 x 5 x 103) / (5 x 1034) = 0.002 m = 2 millimetres

So to ensure laminar flow, your pipe can be no smaller than 2 mm in diameter. Since most pipes carrying wort are much larger than 2 mm, it seems you won’t have any trouble maintaining a laminar flow with this particular wort.

In History of Beer, it was more student presentations, including a history of glassware, brewing in Ontario, and the Aztec production and use of pulque, an intoxicating spirit made from the agave plant.

Instructor Bill White also covered brewing in North America in the 20th century (although at times I felt like it was Bill’s history of Labbatt’s in the 20th century.)

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2 Comments on “Day 441”

  1. Fred Says:

    HI
    Interesting postings, but I never got an email, even I’m signed up
    One question
    Beer is pumped with 900 bbl/hr through 4.5 inch pipe, what is the fluid velocity per meter.
    Thanks

    • Alan Brown Says:

      Okay, here’s the answer to your question in three easy-peasy steps, courtesy of Prof. Ray Lansbergen at Niagara College:

      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.


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