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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 (ρ) = 1000 kg/m³, dynamic viscosity of water (m) = 0.001 N.s/m², and Stokes law is valid. The approximate minimum size of particles that would be completely removed is
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- 0.01 mm
- 0.02 mm
- 0.03 mm
- 0.04 mm
Correct Option: B
To calculate minimum size of particles equating settling velocity to overflow rate, we get
Vo = Vs
Vo = | ||
18μ |
Now Vo = | = | |||
day.² | 24 × 60 × 60 | s |
⇒ | = | ||
24 × 60 × 60 | 18 × 0.001 |
Solving we get d = 1.965 × 10–5m ≈ 0.02 mm