Question

A polystyrene particle of the same density as in Problem $34-12$settles, moving from a starting radius of $70mm$to an analytical radius of $80mm$in $2.9s$in a centrifugal field of $9000G$. If the particle is in an aqueous solution at $20°C$, what is the Stokes diameter?

Short answer

The Stokes diameter for the given problem can be calculated as $100nm$.

Step-by-step solution

Given

Density of polystyrene particles$=\rho =1.06gc{m}^{-3}=1.06\times {10}^{6}g{m}^3$

Density of water$={\rho }_F=0.998gc{m}^{-3}=0.998\times {10}^{6}g{m}^3$

Initial radius$=r_0=70mm=7cm$

Analytical radius$=r_a=80mm=8cm$

Time$=2.9s$

centrifugal field$=9000G$

viscositylocalid="1646372755741" $=\eta =1.002cP=1.002\times {10}^{-3}Pas$

Conversion of the centrifugal field to the rotational velocity

We know that,

$\mathrm{centrifugal}\mathrm{field}={\omega }^2{r}_a/g$

Where,

$g=$acceleration due to gravity$=981cm{s}^{-2}$

By rearranging the terms, we get,

$\omega =\sqrt{\frac{\mathrm{centrifugal}\mathrm{field}\times g}{r_a}}$

By substituting the values, we get,

$$\begin{gathered} \omega =\sqrt{\frac{9000\times 981}{8}} \\ \omega =1050.54rad/s \end{gathered}$$

Calculation of Stokes diameter

Equation for finding the Stokes diameter is given as:

$d_{Stokes}=\sqrt{\frac{18\eta \ln \left(r_a/r_0\right)}{{\omega }^{2}t\left(\rho -{\rho }_{\mathrm{F}}\right)}}$

By substituting the given values in the above equation, we get,
localid="1646372931628" $$\begin{gathered} d_{Stokes}=\sqrt{\frac{18\times 1.002\times {10}^{-3}\times \ln \left(8/7\right)}{1050.{54}^2\times 2.9\times \left(1.06-0.998\right)\times {10}^6}} \\ {d}_{Stokes}=1.1016\times {10}^{-7}m \\ {d}_{Stokes}=110.16\times {10}^{-9}m \\ {d}_{Stokes}=110.16nm\approx 100nm \end{gathered}$$

Final answer

Hence, for the given values, Stokes diameter is $100nm$.