Suggested Response - Q2

Wort with a density of 1,080 kg/m3 enters an empty fermenter, which is vented to atmosphere. The wort enters through a 150 mm internal diameter vertical pipe at a flowrate of 1,000 hL/h. Determine the following:

a)    * The average flow velocity in the 150 mm pipe.

b)    ** An estimate of the height of the initial jet that the wort would form if the wort pressure at the fermenter inlet is 0.05 bar g.

c)    ** Quantify any change in the height of the fountain if the inlet velocity was halved from that in part a.

Response:

a)    The average flow velocity can be determined from consideration of the following equation:

Q space equals space A u

And rearranging for the average flow velocity u:

u space equals space Q over A space equals space fraction numerator begin display style fraction numerator begin display style fraction numerator 1 comma 000 space cross times space 100 over denominator 3 comma 600 space cross times space 1 comma 000 end fraction end style over denominator pi space cross times space 0.15 squared end fraction end style over denominator 4 end fraction

u space equals space 1.57 space m divided by s

b)    To estimate the height of the initial jet in the empty fermenter, we must consider how the energy within the incoming wort stream would be conserved when the wort enters the fermenter. Consideration of Bernoulli’s equation allows us to do this:

rho g h space plus space 1 half space rho u squared space plus space P space equals space c o n s t a n t

Let us apply this to a sketch of the fermenter inlet.

fermenter inlet

Point A is at the entry to the fermenter, where the pressure is 0.05 bar g, and the average wort velocity is 1.57 m/s (from part a) above).

Point B is at the top of the initial fountain (or plume) that the wort would form when it enters an empty fermenter. The pressure here would be 0 bar g, i.e. the same as the atmospheric pressure in the fermenter, and its average vertical velocity would be 0 m/s. The vertical distance h would be the distance between the fermenter inlet and the top of the fountain, relative to a datum, such that h = (hB-hA).

Applying these conditions to Bernoulli’s equation:

rho g h subscript A space plus space 1 half space rho u subscript A squared space plus space P subscript A space equals space rho g h subscript B space plus space 1 half space rho u subscript B squared space plus space P subscript B

And collecting like terms:

rho g space left parenthesis h subscript B space minus space h subscript A space right parenthesis space equals space 1 half space rho space left parenthesis u subscript A squared space minus space u subscript B squared space right parenthesis space minus space left parenthesis P subscript B space end subscript minus space P subscript A right parenthesis

Substituting for known terms:

1 comma 080 space cross times space 9.81 space cross times space h space equals space fraction numerator 1 comma 080 over denominator 2 end fraction space cross times space left parenthesis 1.57 squared space minus space 0 squared right parenthesis space minus space left parenthesis 0 space minus space 5 comma 000 right parenthesis

10 comma 595 space cross times space h space equals space 540 space cross times space 2.47 space plus space 5 comma 000

10 comma 595 space cross times space h space equals space 6 comma 334

h space equals space fraction numerator 6 comma 334 over denominator 10 comma 595 end fraction space equals space 0.6 space m

c)   Recalculating with  = 1.57/2 = 0.79 m/s.

1 comma 080 space cross times space 9.81 space cross times space h space equals space fraction numerator 1 comma 080 over denominator 2 end fraction space cross times space left parenthesis 0.79 squared space minus space 0 squared right parenthesis space minus space left parenthesis 0 space minus space 5 comma 000 right parenthesis

10 comma 595 space cross times space h space equals space 540 space cross times space 0.62 space plus space 5 comma 000

10 comma 595 space cross times space h space equals space 337 space plus space 5 comma 000

h space equals space 0.504 space m

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