5.9 caleup of a Disk Centrifuge Based on Labora- tory Data Determine the maximum

Question

5.9 caleup of a Disk Centrifuge Based on Labora- tory Data Determine the maximum flow rate for the clarification of a suspension of lysed Escherichia 2 coli cells by a plant scale disk centrifuge based on laboratory data. The plant centrifuge has a bowl diameter of 25.4 cm and capabilities shown in 5A Table 5.5. For this centrifuge, you also know that 420, R 8 cm, R 20 cm, and number of disks 100 (see Figure 5.5). In a laboratory centrifuge, you determined that it took a minimum of 17 min to clarify the cell lysate at 12,000 rpm. The top of the culture being cen- trifuged was 32 mm from the center of rotation, and

Explanation / Answer

= 2b / 2g ( 3 r22- r12)

G=relative centrifugal force =0.00001118*rotational radius *rotating speed2 = 0.00001118*(25.4/2)*(12000) ^2 = 20445.9

=3.14*12000^2*3.2/ (2*20445.9((3*20^2)-8^2))

=31.15cm2

is the separation capability of the centrifuge. It also represents the area of a gravity settler required to achieve the same amount of separation.

To get separation in a continuous gravity settler, a settler must lodge the particular particle settling velocity (V eq.grav.) with the help of flow rate (Q) and settler cross-sectional area.

V eq.grav. = Q/

Particle settling velocity can be calculated by measuring the distance travelled by the particle in a minute. If 79mm (7.9cm) is the distance moved by the particle in 17 minutes, then in one minute particle velocity is calculated as 7.9/17 = 0.465cm/min

Now, flow rate is calculated as

Q = V *

=> 0.465*31.15

=> 14.48 cubic-centimeter / min. is the flow rate

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