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:

For equal settling velocity problems, apply Stokes law and equate $d^2(\rho_p - \rho_f)$ for different particles.

Answer 1

Step 1: Settling velocity by Stokes law.
\[ v = \frac{g d^2 (\rho_p - \rho_f)}{18 \mu} \] where $d$ = particle diameter, $\rho_p$ = particle density, $\rho_f$ = fluid density. Step 2: Equating velocities for equal settling.
\[ d_1^2 (\rho_{p1} - \rho_f) = d_2^2 (\rho_{p2} - \rho_f) \] Step 3: Substitution of values.
For quartz: $d_1 = 25 \, μm$, $\rho_{p1} = 2600$, $\rho_f = 1000$. \[ \Delta \rho_1 = 2600 - 1000 = 1600 \] For galena: $\rho_{p2} = 7400$, $\Delta \rho_2 = 7400 - 1000 = 6400$ Step 4: Calculate diameter of galena particle.
\[ d_2^2 = d_1^2 \times \frac{\Delta \rho_1}{\Delta \rho_2} \] \[ d_2^2 = (25)^2 \times \frac{1600}{6400} = 625 \times 0.25 = 156.25 \] \[ d_2 = \sqrt{156.25} = 12.5 \, μm \] Rechecking carefully: If $\Delta \rho_1 = 1600$, $\Delta \rho_2 = 6400$: ratio = 0.25. So $d_2 = 25 \times \sqrt{0.25} = 25 \times 0.5 = 12.5 μm$. Final Answer: \[ \boxed{12.5 \, μm} \]