# about the centrifugation - I am newbies (May/03/2007 )

If a rotor with maximum radia distance of 158 mm and at maximum speed of 10,000 rpm (15,000 g) can provide efficient separation of particles from 5 L homogenate in a hour. I am now required to separate 250 L homogenate within four hours. Can i use an enlarged centrifuge rotor to meet the demand? What type/mode of centrifuge would i use to cope with the high capacity? Why (reasoms)?

Thanks

Thanks

g-force is:

g = ((2pi x n)/60)2 x r/9.81

n: rpm

r: radius

time of full sedimentation depends on S (sedimentation coefficient) at a given g-force, and rotor geometry; may be you have the k-value of the rotor you like to use; then calculate time t for full sedimentation:

t = k/S

Thanks

g-force is:

g = ((2pi x n)/60)2 x r/9.81

n: rpm

r: radius

time of full sedimentation depends on S (sedimentation coefficient) at a given g-force, and rotor geometry; may be you have the k-value of the rotor you like to use; then calculate time t for full sedimentation:

t = k/S

i still don't understand , would you mind telling more details to me?\

THanks

what the bearer is referring to is called "clearing time". if you know the sedimentation coefficient of your sample (or component of interest) you can determine how long it will take to pellet it with the rotor you are using.

however, in order for you to centrifuge 250 liters in 4 hours with one centrifuge you will need to use a continuous flow rotor and/or attachment (like the szent-gyorgi attachment for the sorvall ss-34 rotor). with this you run your rotor at or near top speed and adjust the flow rate of your homogenate into the rotor so that the waste comes out clear of your component (we used to use it to collect from large volumes of cyanobacteria culture). you still may not be able to clear the entire 250 liters in the 4 hours you have to complete the process but it will be faster than starting, stopping, emptying and refilling centrifuge bottles.