Detritus tank sizing: A treatment plant detritus tank carries sewage discharge of 576 L/s at a flow velocity of 0.2 m/s. If the width-to-depth ratio (B:D) is 2, calculate the required depth of the tank.

Introduction / Context:
Detritus tanks remove heavy inorganic particles (grit) by maintaining a controlled, relatively low velocity. Geometry is selected to achieve the target velocity and detention, often expressed via cross-sectional area and width-to-depth proportions.

Given Data / Assumptions:

• Sewage discharge Q = 576 L/s = 0.576 m^3/s.
• Design flow velocity v = 0.2 m/s.
• Width-to-depth ratio B:D = 2 (i.e., B = 2D).
• Steady uniform section with area A = B * D.

Concept / Approach:
The required area A is determined by continuity: A = Q / v. With the ratio B = 2D, substitute into A = B * D to solve for D. This is a straightforward hydraulic sizing step in preliminary design.

Step-by-Step Solution:
Compute area: A = Q / v = 0.576 / 0.2 = 2.88 m^2.Express area via D: A = B * D = (2D) * D = 2D^2.Solve for D: 2D^2 = 2.88 → D^2 = 1.44 → D = 1.20 m.Convert to cm: 1.20 m = 120 cm.

Verification / Alternative check:
Width B = 2D = 2.40 m. Check velocity: v = Q / A = 0.576 / 2.88 = 0.2 m/s (matches design), confirming correctness.

Why Other Options Are Wrong:
100 cm and 110 cm: yield areas too small; velocity would exceed 0.2 m/s.150 cm: produces area larger than required; velocity would drop below 0.2 m/s.90 cm: clearly undersized for the given flow and velocity.

Common Pitfalls:

• Forgetting to convert litres per second to m^3/s.
• Confusing B:D = 2 with D:B = 2; always confirm which dimension is which.

Final Answer:
120 cm