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Relationship between Filtration and Pasteurisation

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  • Relationship between Filtration and Pasteurisation

    Hello - looking for guidance on how various levels of filtration influence the amount of pasteurization required for beer. In other words, how does one determine PUs required for an unfiltered beer with ~150 yeast cells/ml vs. PUs required for a heavily filtered beer with ~1-2 yeast cells/ml.

    Thanks!

  • #2
    A bit of a long story. Try and make something of this

    Basically if you work out a spreadsheet with this info, you will find you will need extremely high numbers of PUs with unfiltered beer to ensure good micro quality. I think you are grossly over-estimating how clear your beer is, but since I haven't don't the measurements....

    Have fun


    PU Target Calculation
    The following principles are used in the calculation of pasteurisation units (PUs).

    A Pasteurisation Unit (PU) is a measure of the heat received by a beer during pasteurisation. 1 PU is the lethal effect on microorganisms, of holding beer for 1 minute at 60 oC, or the equivalent time / temperature.

    The Lethal Rate (LT) is considered as the time (in minutes) at 60 oC resulting in the same inactivation as at 1 minute at a different temperature T oC.
    For beer, this is represented by the relationship:
    LT (mins at 60 oC) = 1.393(T-60)

    Pasteurisation unit calculation
    Now LT in mins at 60 oC = PU, for example, at 60oC
    then PU = 1.393(60-60)
    = 1.393o
    = 1/min i.e. 1 minute at 60 oC = 1 PU
    Some useful approximations are as follows:
    An increase of 2 oC increases PU by a factor of approximately 2
    An increase of 7 oC increases PU by a factor of approximately 10

    D Value
    An organism’s resistance to pasteurisation is given by its D value. The D60 value then, is the time taken, in minutes, at 60 oC to reduce a population of the particular organism by 90%, and can be derived experimentally.
    Typical D60 values for beer contaminant microorganisms are as follows:-
    Brewing yeast Typically less than 1 second
    Wild yeast up to 1 minute
    Lactic acid bacteria up to 4 minutes

    Dependant on the specific lactic acid organism present, to effect a decimal reduction may require up to 4 mins at 60 oC, or 4 PUs. If however the lactic organism had a D60 value of 1, then a PU of 1 would suffice to effect a decimal reduction. The D60 value of the specific organism, then can have a major effect on the overall PUs required to be applied.

    Z value
    The temperature dependence of microorganisms is such that increasing the temperature will increase the inactivation, or the D value above will decrease. This means that the D7o value will be lower than the D6o value. This temperature dependence is termed the Z value.
    The Z value is defined as the increase in temperature necessary to reduce the D value by 90% i.e. is a tenfold increase in the rate of thermal inactivation.
    However, as all microorganisms vary in this relationship, the calculation of lethality required would be extremely difficult to compute. As a result a Z value of approximately 7o C is assumed for all microorganisms, from which is derived a Lethal rate chart, which indicates lethal rate values between 50.0o C and 79.9o C.

    Target PU calculation
    In the following example, lactic acid bacteria, with a D60 value of 4 minutes, are occasionally found at levels of up to 100 cells /100 ml. A target reduction in micro-organisms to less than 1 cell / 50 litres is required. The PU calculation is as follows:
    Initial loading (lactic acid bacteria) = 100 cells /100 ml
    = 50,000 cells / 50 litres
    if 4 PUs reduces the population by 90 % (decimal reduction, where D60 = 4), then 5 x 4 PUs are required to reduce the population to 0.5 cells / 50 litre
    Therefore the target PU level required, in this example = 20.
    If however, the organism had a D60 value of 2, then only 2 PUs would be required to cause a decimal reduction, and 5 x 2, or 10 PUs would be enough to reduce the above population to 0.5 cells / 50 litre.
    Therefore the minimum PU level required, in this example = 10
    dick

    Comment


    • #3
      In theory, assuming no other contamination by wild yeasts, bacteria etc., and actual kill rates comply with the above formulae

      150 yeast cells/ml requires a minimum of 6 PUs to reduce the numbers to say 1 viable cell per 50 litre keg

      1-2 yeast cells/ml requires a minimum of 5 PUs to reduce the numbers to say 1 viable cell per 50 litre keg
      dick

      Comment


      • #4
        Another quick update. I found some more Z values

        Brewers yeast 1
        Pediococcus 1
        Lactobacillus 5
        Wild yeast 10

        In other words, wild yeast requires 10 times as many PUs (considerably different from my first source I quoted), and lactobacillus requires 5 times as many PUs to kill off the same number of organisms as brewers yeast. Which is why it is common to apply 15 to 20 PUs minimum, depending on how effective the brewery hygiene is, and filtration is.
        dick

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