Primary sedimentation design check: A settling tank is 2.4 m deep and must remove particles with settling velocity 0.33 mm/s (for d ≥ 0.06 mm). What is the recommended detention time?

Introduction / Context:Detention time in a settling (clarifier) tank must be sufficient for target particles to settle from the liquid surface to the tank bottom. This problem links settling velocity to tank depth to estimate a suitable detention time for design/operation.

Given Data / Assumptions:

• Tank depth = 2.4 m.
• Target particle settling velocity v_s = 0.33 mm/s = 0.00033 m/s for particles ≥ 0.06 mm.
• Vertical settling distance approximated as the water depth.
• Neglect inlet/outlet effects and turbulence for a first-cut estimate.

Concept / Approach:Time required for settlement is distance/velocity. A conservative design detention time equals or slightly exceeds this value to account for non-idealities. Use t = depth / v_s.

Step-by-Step Solution:1) Convert velocity: 0.33 mm/s = 0.00033 m/s.2) Compute time: t = 2.4 m / 0.00033 m/s = 7272.7 s.3) Convert seconds to hours: 7272.7 s / 3600 s/h ≈ 2.02 h.4) Choose the closest recommended detention period: 2 hours.

Verification / Alternative check:A quick check with minutes: 7272.7 s ≈ 121.2 min ≈ 2.02 h confirms the selection of 2 hours as the nearest practical value.

Why Other Options Are Wrong:30 min, 1 h, 1.5 h: All are less than the calculated 2.02 h and may not ensure removal of the target particles.3 h: Longer than necessary based on the given settling velocity and depth, leading to oversized detention beyond the calculated requirement.

Common Pitfalls:

• Forgetting to convert mm/s to m/s.
• Using surface overflow rate concepts incorrectly when the question focuses on depth-based settling time.
• Ignoring safety factors; here we choose the nearest practical value rounding from 2.02 h.

Final Answer:2 hours