A 2.5-m wide rectangular culvert made of ordinary concrete was constructed to carry storm water flow away from a subdivision

Question

A 2.5-m wide rectangular culvert made of ordinary concrete was constructed to carry storm water flow away from a subdivision. The slope of the culvert is 0.002. After a heavy rainstorm, a 1.5 m deep flow of water was measured in the culvert as uniform and steady. Along with the water transport open channel, the culvert also receives particulate matter of approximately 180 µm diameter at the culvert water surface. Assume that the density of particles was 2.5 g/cm3 and that the kinematic viscosity of water is 1.3 x 10-2 cm2/s (at 10°C). What is the minimum distance the particles will travel before settling to the culvert bottom?

Answer 1

To determine the minimum distance the particles will travel before settling to the culvert bottom, we can use Stoke's Law, which relates the settling velocity of a particle in a fluid to its size, density, and the fluid's viscosity. The formula for Stoke's Law is:

\[ v = \frac{{2g(r_p - r_f)}}{9\mu} \]

where:

- \( v \) is the settling velocity of the particle (cm/s),

- \( g \) is the acceleration due to gravity (cm/s²),

- \( r_p \) is the density of the particle (g/cm³),

- \( r_f \) is the density of the fluid (g/cm³), and

- \( \mu \) is the dynamic viscosity of the fluid (cm²/s).

First, let's calculate the settling velocity using the given data:

- \( g = 980 \, \text{cm/s}^2 \) (standard acceleration due to gravity)

- \( r_p = 2.5 \, \text{g/cm}^3 \) (density of particles)

- \( r_f = 1 \, \text{g/cm}^3 \) (density of water)

- \( \mu = 1.3 \times 10^{-2} \, \text{cm}^2/\text{s} \) (kinematic viscosity of water)

Plug these values into Stoke's Law:

\[ v = \frac{{2 \times 980 \times (2.5 - 1)}}{{9 \times 1.3 \times 10^{-2}}} \]

\[ v = \frac{{1960 \times 1.5}}{{11.7 \times 10^{-2}}} \]

\[ v = \frac{{2940}}{{11.7 \times 10^{-2}}} \]

\[ v \approx 25128.2 \, \text{cm/s} \]

Now, we need to find the minimum distance the particles will travel before settling to the culvert bottom. Since the flow is uniform and steady, the settling distance can be calculated using the relationship between velocity, distance, and time:

\[ s = v \times t \]

where:

- \( s \) is the settling distance (cm),

- \( v \) is the settling velocity we just calculated (cm/s), and

- \( t \) is the time for settling.

The time for settling can be estimated using the depth of water in the culvert. Since the water depth is given as 1.5 m (or 150 cm), and assuming the flow is uniform and steady, the time for settling is given by:

\[ t = \frac{d}{v_f} \]

where:

- \( d = 150 \, \text{cm} \) (depth of water in the culvert), and

- \( v_f \) is the velocity of flow in the culvert.

The velocity of flow can be calculated using the Manning's equation for open channel flow:

\[ v_f = \frac{1}{n} \times \frac{1}{R^{2/3}} \times S^{1/2} \]

where:

- \( n \) is the Manning's roughness coefficient (for concrete, it's around 0.012 - 0.015),

- \( R \) is the hydraulic radius (m), and

- \( S \) is the slope of the channel.

First, convert the given values to consistent units:

- \( \text{Culvert width} = 2.5 \, \text{m} = 250 \, \text{cm} \)

- \( \text{Depth of water} = 1.5 \, \text{m} = 150 \, \text{cm} \)

- \( \text{Slope} = 0.002 \)

Calculate the hydraulic radius \( R \):

\[ R = \frac{\text{Area}}{\text{Wet perimeter}} \]

\[ \text{Area} = \text{Culvert width} \times \text{Depth of water} = 250 \times 150 \, \text{cm}^2 \]

\[ \text{Wet perimeter} = \text{Culvert width} + 2 \times \text{Depth of water} = 250 + 2 \times 150 \, \text{cm} \]

\[ R = \frac{250 \times 150}{250 + 2 \times 150} \]

\[ R = \frac{37500}{550} \]

\[ R \approx 68.18 \, \text{cm} \]

Now, calculate the velocity of flow \( v_f \):

\[ v_f = \frac{1}{n} \times \frac{1}{R^{2/3}} \times S^{1/2} \]

Assume \( n = 0.012 \) (using a typical value for concrete channels):

\[ v_f = \frac{1}{0.012} \times \frac{1}{(68.18)^{2/3}} \times (0.002)^{1/2} \]

\[ v_f \approx 277.55 \, \text{cm/s} \]

Finally, calculate the time for settling:

\[ t = \frac{150}{277.55} \]

\[ t \approx 0.54 \, \text{s} \]

Now, substitute this time into the settling distance formula:

\[ s = 25128.2 \times 0.54 \]

\[ s \approx 13568.728 \, \text{cm} \]

Therefore, the minimum distance the particles will travel before settling to the culvert bottom is approximately 13568.73 cm or 135.69 m.