Question

In a particular batch of polystyrene microspheres, their diameter is $9.5\mu m$and density $1.06gc{m}^{-3}$. In 20°C water (${\rho }_F=0.998gc{m}^{-3}$), what will be the settling velocity of these particles in a $1G$gravitational field? What time will it take for the particles to sediment $10mm$?

Short answer

The settling velocity of the given batch of polystyrene microspheres is calculated as $3.0435\times {10}^{-4}cm{s}^{-1}$and the time required for sedimentation is$3286s$.

Step-by-step solution

Given information

Diameter of the particle$=d=9.5\mu m=9.5\times {10}^{-3}cm$

Density of the particle$=\rho =1.06gc{m}^{-3}$

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

Sedimentation distance$=h=10mm=1cm$

Calculation of settling velocity

The settling velocity can be calculated from the below Stokes equation:

$u=\frac{\left(\rho -{\rho }_F\right)\times g}{18\times \eta }\times d^2$

Where,

$u=$settling velocity

$\rho =$density of the particle

${\rho }_F=$density of the fluid

$g=$acceleration due to gravitylocalid="1646031316245" $=981cm{s}^{-2}$

$d=$particle diameter

$\eta =$viscositylocalid="1646031335700" $=1.002cP$

Now,

By substituting the given values in the above equation, we get,

localid="1646032711614" $$\begin{gathered} u=\frac{\left(1.06-0.998\right)\times 981}{18\times 1.002}\times {(9.5\times {10}^{-3})}^2 \\ u=304.35\times {10}^{-6}cm{s}^{-1} \\ u=3.0435\times {10}^{-4}cm{s}^{-1} \end{gathered}$$

Calculation of sedimentation time

For calculating the sedimentation time, consider the following relation:

$u=\frac{h}{t}$

Hence, by rearranging, we get,

$t=\frac{h}{u}$

By substituting the values of $h$and $u$in the above equation, we get,

$$\begin{gathered} t=\frac{1}{3.0435\times {10}^{-4}} \\ t=3286s \end{gathered}$$

Final answer

The settling velocity of the polystyrene microspheres is$3.0435\times {10}^{-4}cm{s}^{-1}$, while they take$3286s$to sediment$10mm$.