Question

You are trying to remove 10.0 µm diameter particles in a water treatment plant. The water is at 20 °C, and the particle density is 1.2 g/mL. The plant treats 0.100 m3 /s of water. It is proposed to use a 3.5 m deep, rectangular sedimentation tank with a length to width ratio of 5:1. What is the minimum required width of the basin?

Answer 1

Answer:

minimum required width of the basin is 42.87 m

Explanation:

given data

diameter particles = 10.0 µm

water temperature =  20 °C

particle density = 1.2 g/mL = 1200 kg/m³

plant treats water =  0.100 m³ /s

deep = 3.5 m

length to width ratio =  5:1

to find out

What is the minimum required width of the basin

solution

we know dynamic viscosity of water = 1.002 ×$10^{-3}$ kg/m-s

and density of water is = 1000 kg/m³

now we apply here stock law for settling velocity that is express as

settling velocity = $\frac{(\rho_p - \rho_w)* gD^2}{18* \mu}$    ..................1

here ρ(p) is particle density and ρ(w) is density of water and µ is dynamic viscosity of water  

so put here value

settling velocity = $\frac{(1200-1000)* 9.81*(10*10^{-6})^2}{18* 1.002*10^{-3}}$

settling velocity = 1.088 × $10^{-5}$  m/s

so now we calculate length of basin

we know length to width ratio is 5:1

so length L = 5b

and

minimum width of basin will be

Q = Area × settling velocity

0.1 = L × b × 1.088 × $10^{-5}$

0.1 = 5b × b × 1.088 × $10^{-5}$

b = 42.87 m

so minimum required width of the basin is 42.87 m