Suggested Response - Q1

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a)    A cylindroconical fermenter has a wort depth of 20 m (measured from the base of the cone up to the wort surface) with a headspace pressure of 5 kPa (g). The density of the wort is 1,080 kg/m³, and the atmospheric pressure is 95 kN/m². Calculate the absolute pressure at the base of the fermenter cone.

b)    What would be the external force that a single yeast cell located at the base of the fermenter cone is exposed to? You may assume the yeast cell is a 5 µm diameter (d) sphere and the surface area of a sphere is given by πd².

Response:

a)    The absolute pressure at the base of the fermenter cone will be given by the sum of:

  • Atmospheric pressure of 95 kN/m²

  • The headspace pressure of 5 kPa (g)

  • The hydrostatic pressure due to the 20 m of wort

So:

P subscript C o n e end subscript equals P subscript A t m end subscript plus P subscript H divided by S end subscript plus P subscript W o r t end subscript

P subscript C o n e end subscript equals 95 cross times 10 to the power of ³ plus 5 cross times 10 to the power of ³ plus rho g Z subscript ₁

P subscript C o n e end subscript equals 95 cross times 10 to the power of ³ plus 5 cross times 10 to the power of ³ plus left parenthesis 1 comma 080 cross times 9.81 cross times 20 right parenthesis

P subscript C o n e end subscript equals 100 cross times 10 to the power of ³ plus 212 cross times 10 to the power of ³

P subscript C o n e end subscript equals 312 k P a left parenthesis a right parenthesis

b)    The external force would be:

Pressure defined as:

P r e s s u r e space P equals space space fraction numerator F o r c e space F over denominator A r e a space A end fraction

Rearranging for force F gives:

F o r c e space F equals P r e s s u r e space P cross times A r e a space A

The surface area of a sphere (yeast cell) given by:

A equals pi d ² equals pi cross times left parenthesis 5 cross times 10 minus ⁶ right parenthesis to the power of ²

A equals 7.85 cross times 10 to the power of negative 11 end exponent m to the power of ²

Hence the external force on a yeast cell is given by:

F o r c e space F equals P cross times A equals 312 cross times 10 to the power of ³ cross times 7.85 cross times 10 to the power of negative 11 end exponent

F equals 2.45 cross times 10 to the power of negative ⁵ end exponent N

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