A small tubular bowl centrifuge (thin cake layer with an average radius of r cm

Question

A small tubular bowl centrifuge (thin cake layer with an average radius of r cm and length L cm) was used for a pilot study to separate the biomass from the solubles from the output stream of a chemostat operation . At a volumetric flow rate of Q in L/h , the pilot study found that a rotational speed of ?(rpm) obtained 50% removal of the biomass. % biomass removal (efficiency) is directly proportional to the centrifugal coefficient.Solve symbolically. Allowable variables are: r, L, Q,?,D (dilution rate), V (volume of upstream bioreactor)

a.

Using a larger centrifuge with average radius of 5*r and length 2*L, the

rotational speed

required to process the fermentation broth with 50%

biomass removal at a flow rate of Q is ____________________ in units of

rpm.

b.

The rotational speed of the larger centrifuge needed to keep up with the

upstream

fermentation process

is _____________________ in rpm with

an efficiency of 50%.

c.

To keep up with the

upstream

fermentation process

at a 90% biomass

removal, the

rotational speed of the larger centrifuge needs to be

____________________________ in rpm.

d.

The efficiency of the centrifuge _________________ (pick one:

increases/decreases/does not change) with increasing difference between

densities of particle and fluid.

e.

Increasing the radius of the centrifuge __________________ (pick one:

increases/decreases/does not change) the terminal velocity of the particle

due to gravity.

Explanation / Answer

answer for a

rotational speedrequired to process the fermentation broth with 50% biomass removal at a flow rate of Q is  

s1 percent /s2 percent = q , where s = 50 and s2 is 90 biomass = 1.8 l/hr is the flow rate per rpm

answer for b ;

formula for centripetal force is f =  m v2  / r , f is force , radius =5 , length    =2 , volume = one liter .

so , from the above we can caluculate f = 50 * [1000]2 /   5 = 1581.1/5 = 31.6 rpm is the force , in total

answer for c ;

formula for centripetal force , here the above formula is applied but the mass changes , hence rpm changes

f = mv2 /r = 90* [1000]2 /5 = 2846.0/5 = 569.2 rpm

answer d ;

the efficiency of the centrifuge does not change with increasing difference between densities of particles

answer for e

increasing the radius of the centrifuge increases the terminal velocity of the particle due to gravity

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