Surface overflow criterion for sedimentation tank sizing A plant treats Q = 5.7 MLD. For a target to remove all particles with settling velocity v_s = 0.33 mm/s, what is the required surface area A of a rectangular sedimentation tank (ignore end effects)?

Introduction / Context:
For ideal horizontal-flow settling basins, the surface overflow rate (SOR) determines particle removal: all particles with settling velocity v_s greater than or equal to the SOR will be completely removed (in the ideal sense). Thus, sizing is based on matching SOR to the target v_s.

Given Data / Assumptions:

• Total flow Q = 5.7 MLD = 5.7 × 10^3 m^3/day.
• Settling velocity v_s = 0.33 mm/s = 0.00033 m/s.
• Ideal settling theory; short-circuiting and non-idealities neglected.

Concept / Approach:
Use the overflow criterion: SOR = Q/A ≤ v_s ⇒ A ≥ Q/v_s. Convert Q to consistent SI units of m^3/s before calculation to avoid numerical errors.

Step-by-Step Solution:
1) Convert flow: Q = 5,700 m^3/day = 5,700 / 86,400 ≈ 0.06597 m^3/s.2) Compute A = Q / v_s = 0.06597 / 0.00033 ≈ 199.9 m^2.3) Round to a practical value → 200 m^2.4) Select option 200 m^2.

Verification / Alternative check:
Check SOR with the selected area: SOR = Q/A ≈ 0.066 / 200 = 0.00033 m/s = 0.33 mm/s, exactly the target.

Why Other Options Are Wrong:
20 m^2 and 100 m^2 give SORs far above v_s, so particles would not be fully removed; 400 m^2 is conservative but not the nearest correct value.

Common Pitfalls:
Forgetting to convert MLD to m^3/s; mixing mm/s and m/s; incorrectly using detention time instead of SOR for this sizing step.

Final Answer:
200 m^2