Answer to Question #127906 in General Chemistry for NE

Question

Yeast cells are to be separated from a fermentation broth. Assume that the cells are spherical with diameter 5 µm and density 1.06 g/cm3. The viscosity of the culture broth is 1.36 × 10-3 Ns/m2. At the temperature of separation, the density of the suspending fluid is 0.997 g/cm3. 500 liters of broth must be treated every hour. Specify Σ factor for a suitably-sized disk-stack centrifuge.

Answer 1

The centrifuge sigma factor for separation of yeast cells is 110.11 m2

Explanation:

Given Values are:

Density of cells (ρp) = 1.06 g/cm3 = 1060 kg/m3

Diameter of cell, Dp=5 

µm=5*10-6 m

Density of fluid, ρf =0.997 g/cm3=997 kg/m3

Rate of broth treated, Q=500 L/h=0.5 m3/h=1.39*10-4 m3/s

Viscosity of culture broth, µ = 1.36*10-3 Ns/m2 or 1.36*10-3 kg/m s

 The sigma factor of centrifuge for a disc-stack centrifuge is obtained using the following expression:

Σ = 2µg

Where, where Q is the Rate of broth that must be treated

µg is the Sedimentation velocity:

Sedimentation Velocity

where ρp and ρf is the Density of cell and fluid, respectively

µ is the Viscosity of broth

Dp is the Diameter of cell

g is acceleration due to gravity, g=9.81 m/s2

µg =18×1.36×10 −3kg/ms

(1060−997)kg/m 

 × (5 × 10-6m)2 × 9.81 m/s2

µg = 6.312 × 10-7 m/s

Thus sigma factor is

Σ = 2µg

Q= (1.39 × 10-4 m3s-1)/ (2 × 6.312 × 10-7 m/s) = 110.11 m2

Σ = = 110.11 m2