Settling velocity from sedimentation tank geometry:\r For a rectangular sedimentation tank of length L, breadth B, depth D and total discharge Q, the design settling velocity v_s is equal to

Difficulty: Easy

Correct Answer: Q / (L * B)

Explanation:


Introduction / Context:
In ideal sedimentation theory, particle removal depends on the overflow (surface) rate rather than tank depth. Designers therefore use surface loading rate, equated to a critical settling velocity, to size plan area for a given flow and target removal.


Given Data / Assumptions:

  • Rectangular clarifier with length L, breadth B, and water depth D.
  • Steady flow Q and ideal plug-flow behavior (no short-circuiting).
  • Uniform vertical settling behavior without flocculation effects.


Concept / Approach:
In the idealized Hazen model, removal depends on the ratio of settling velocity v_s to overflow rate Q/A, where A is plan area. The critical condition for removal is v_s ≥ Q/A. Thus, v_s equals Q divided by the surface area; tank depth does not affect the theoretical removal efficiency for discrete settling.


Step-by-Step Solution:
Compute plan area: A = L * B.Overflow rate (surface loading) = Q / A.Set design settling velocity v_s = Q / (L * B).Note that depth D influences sludge storage and inlet/outlet hydraulics, not the basic Hazen criterion.


Verification / Alternative check:
Plant performance correlations and tracer studies verify that clarifier depth does not directly change ideal surface-based removal; however, non-idealities (density currents) may necessitate additional depth for hydraulic stability.


Why Other Options Are Wrong:

  • Options including D in the denominator or numerator contradict the surface-overflow concept.
  • Q / (L * B * D) is the average velocity through the tank volume, not settling velocity criterion.


Common Pitfalls:

  • Assuming deeper tanks always improve clarification; without improving hydraulics, depth alone does not increase ideal removal.
  • Ignoring inlet/outlet design that can create short-circuiting and reduce effective surface area.


Final Answer:
Q / (L * B)

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