Questions: Under local regulation, particles larger than 5 μm (0.0005 cm) in diameter cannot be discharged to the stream. How long after runoff stops entering the retention pond can water be discharged to the stream from the upper 50 cm of the retention pond? Use the same η and ρw as in part a.

Solution Steps

We need to determine the time it takes for particles larger than \(5 \, \mu\text{m}\) in diameter to settle out of the upper 50 cm of a retention pond. This involves calculating the settling velocity of these particles and then using it to find the time required for them to settle 50 cm.

Stokes' Law for the settling velocity \(v_s\) of a spherical particle in a fluid is given by:

\[ v_s = \frac{2}{9} \frac{(\rho_p - \rho_w) g r^2}{\eta} \]

where:

• \(\rho_p\) is the density of the particle,
• \(\rho_w\) is the density of the water,
• \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)),
• \(r\) is the radius of the particle,
• \(\eta\) is the dynamic viscosity of the water.

Assuming \(\rho_p\), \(\rho_w\), and \(\eta\) are given or can be assumed from part (a), we need to calculate \(v_s\) for particles with a diameter of \(5 \, \mu\text{m}\), which means a radius \(r = 2.5 \, \mu\text{m} = 2.5 \times 10^{-4} \, \text{cm}\).

Once we have the settling velocity \(v_s\), the time \(t\) it takes for particles to settle 50 cm can be calculated using:

\[ t = \frac{\text{distance}}{v_s} = \frac{50 \, \text{cm}}{v_s} \]

Final Answer