Question

The diameter of spherical galena particles that have the same settling velocity as spherical quartz particles of diameter 25 μm (both settling in water) is _________ μm (rounded off to one decimal place). Assume Stokes law of settling to be valid. Given: Density of galena = 7400 kg m\(^{-3}\), Density of quartz = 2600 kg m\(^{-3}\), Density of water = 1000 kg m\(^{-3}\).

Hint:

In problems involving settling velocities, Stokes' law allows us to relate the diameters of particles with the same settling velocity by using the ratio of their densities and the density of the fluid. The diameter of a particle is crucial for determining its velocity and behavior in fluids.

Answer 1

Stokes' law for the settling velocity of a spherical particle in a fluid is given by:

\[ v = \frac{2r^2 (\rho_p - \rho_f) g}{9 \eta} \] where:
- \( r \) is the radius of the particle,
- \( \rho_p \) is the density of the particle,
- \( \rho_f \) is the density of the fluid,
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the dynamic viscosity of the fluid (water).

Since the settling velocities of both galena and quartz particles are the same, we can set up the following ratio for the diameters (and thus the radii) of the two particles:

\[ \frac{r_{{galena}}^2 (\rho_{{galena}} - \rho_{{water}})}{r_{{quartz}}^2 (\rho_{{quartz}} - \rho_{{water}})} = 1 \] Using \( r = \frac{d}{2} \), we substitute the given values:

\[ \frac{\left( \frac{d_{{galena}}}{2} \right)^2 (7400 - 1000)}{\left( \frac{25}{2} \right)^2 (2600 - 1000)} = 1 \] Simplifying and solving for \( d_{{galena}} \):

\[ \frac{d_{{galena}}^2 (6400)}{25^2 (1600)} = 1 \] \[ d_{{galena}}^2 = \frac{25^2 (1600)}{6400} \approx 12.5^2 \] Thus, the diameter of the galena particle is approximately:

\[ d_{{galena}} \approx 12.5 \, \mu{m} \]