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A settling tank in a water treatment plant is designed for a surface overflow rate of 30 m3/day.m2. Assume specific gravity of sediment particles = 2.65, density of water (?) = 1000 kg/m3, dynamic viscosity of water (µ) = 0.001 N.s/m2, and Stokes' law is valid. The approximate minimum size of particles that would be completely removed is

🗓 Dec 16, 2021

Answer is "0.02mm"

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NNS 🌐 Telangana
To calculate minimum size of particles equating settling velocity to overflow rate, we get
Vo = Vs

Vo =
(Gsl)&gaamma;wd²
18μ

Now Vo =
30 m³
=
30
m
day.² 24 × 60 × 60 s


30
=
(2.65 - 1) × 9.81 × 10³ × d²
24 × 60 × 60 18 × 0.001

Solving we get d = 1.965 × 10–5m ≈ 0.02 mm
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