particle size as function of acceleration (centrifugation)

[Hollow]

Hi

can I estimate the size of a particle (dp) which remains in the liquid after centrifugation?
->comparison of two centrifugation conditions e.g. 3000g vs 20000g

dp=f(g, n, rl, rp, t, l)

"Known" paramters would be
acceleration g
viscosity n
density of the liquid rl
density of the particle rp

additional
centrifugation time t
"sedimentation length" l

for simplification: assumption that particle are spheres

[JGK]

Separations based on density

http://en.wikipedia.org/wiki/Sucrose_gr ... rifugation

http://www.science-projects.com/GradCent.htm

[moino]

Analytical Ultracentrifugation: Techniques and Methods, D. Scott, S.E. Harding, and A.J. Rowe, Editors. 2006, Royal Society of Chemistry. ISBN 978-0854045471

Jamison, J.A., et al., Size-Dependent Sedimentation Properties of Nanocrystals. ACS Nano, 2008. 2(2): p. 311-319.
http://pubs3.acs.org/acs/journals/doilo ... /nn700144m

[Hollow]

Thank you for your replies.

I've found some old documents from my study.

There, we get two ways for sedimentation and a dekanter.
we linked the following two equations:

Stokes, laminar
(1) vsed = (dp^2*(rp-rl)*a)/(18*n)

Reynolds
(2) Re = (v*dp*rl)/n

-> v from (2) = vsed (1) leads to (3)

(3) (dp^2*(rp-rl)*a)/(18*n) = (Re*n)/(dp*rl)

assuming a Re = 1 (border of laminar), one can solve for dp

(4) dp = [(18*n^2)/(rl(rp-rl)*a)]^1/3


or the second way leads to an itterative solution:


(5) vsed = [(4*dp*(rp-rl)*a)/(3*rl*cw)]^0.5 (universal sedimentation)

solved for dp:

(6) dp = (3*vsed^2*rl*cw)/(4*(rp-rl)*a)

resistance coefficient
7) cw = 1/3*[1+(72/Re)^0.5]^2

for vsed I take the lenght of the centrifugation tube divided by the centrifugation time

( 8 ) v = l/t

now one can start the itteration (e.g. in excel)

a) select a number for Re (e.g. 1)
b) calculate cw (7)
c) calculate dp (6)
d) calculate Re (2)
e) start again with the new Re in b) until you get the same dp for two consecutive itterations.


Doing so I got the follwing dp:
rp = 1.1 g/ml
rl = 1.0 g/ml (water)
l = 3 cm (microreaction tube)
t = 5 min

@20000g: dp = ca. 300 nm (Re = 30E-6)
@3000g: dp = ca. 780 nm (Re 78E-6)

so I would expect no particles of 1.1g/ml which are greater than the above values.

Would this plausible?