Most Efficient Rectangular Channel Section – Proportion for Maximum Discharge For a given area and bed slope, a rectangular channel conveys maximum discharge when the breadth is twice the flow depth (b = 2 y), which maximizes hydraulic radius.

Introduction:
The “most economical” (most efficient) channel section minimizes wetted perimeter for a given area, thereby maximizing hydraulic radius and discharge for a given slope and roughness. For rectangles, this yields a simple proportion between breadth and depth.

Given Data / Assumptions:

• Steady, uniform open-channel flow.
• Rectangular prismatic channel with width b and depth y.
• Manning or Chezy friction formulation where Q ∝ A * R^(2/3) (or similar).

Concept / Approach:

For fixed area A = b * y, minimizing wetted perimeter P = b + 2 y maximizes hydraulic radius R = A / P. Using calculus with the constraint b * y = A gives the optimal proportion b = 2 y for a rectangle.

Step-by-Step Solution:

Verification / Alternative check:

Using b = 2 y yields R = (b y)/(b + 2 y) = (2 y^2)/(2 y + 2 y) = y/2, the known efficient-section relation for rectangles.

Why Other Options Are Wrong:

Other ratios increase wetted perimeter for the same area, reducing hydraulic radius and thus discharge capacity under the same slope/roughness.

Common Pitfalls:

Confusing the efficient rectangle with trapezoidal or triangular optimal conditions; assuming the ratio depends on slope or roughness (it depends on geometry for a given A).

Final Answer:

its breadth is twice the depth