Answer in General Chemistry for simsim #155922

the small size and low density of microbial cells are disadvantageous in centrifugation. Instead of yeast, quartz particles of diameter 0.1mm and specific gravity 2.0 are separated from the culture liquid, by how much is sigma factor reduced?

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


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


 × (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

Now diameter 0.1mm and specific gravity 2.0 µg = 18×1.36×10 −3kg/ms


 × (0.1 × 10-6m)2 × 2 m/s2

=3.084×10-7 m/s

Thus sigma factor is

Σ = 2µg

Q= (1.39 × 10-4 m3s-1)/ (2 × 3.084 × 10-7 m/s)

= 225.35 m2 hence

Σ factor reduces 225.35-110.11

= 115.24 m2

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