Open Channel Flow – Most Economical Rectangular Section For a rectangular channel, the most economical section is the one whose hydraulic radius equals which proportion of the flow depth?

Introduction:
The most economical (most efficient) section minimizes wetted perimeter for a given area, maximizing discharge or minimizing frictional loss. In rectangular channels, there are simple geometric relations at optimum.

Given Data / Assumptions:

• Prismatic, wide rectangular channel is not assumed; exact rectangle is considered.
• Uniform flow conditions with Chezy or Manning resistance concept.
• Hydraulic radius R = A / P, where A is area and P is wetted perimeter.

Concept / Approach:

For a rectangle of breadth b and depth y: A = b * y and P = b + 2y. At the most economical section, dP/dy under constant area or classical optimization yields b = 2y. Substituting b = 2y gives R = A / P = (2y^2) / (2y + 2y) = y/2. Thus R equals half the depth.

Step-by-Step Solution:

Verification / Alternative check:

Using Manning discharge Q proportional to A * R^(2/3) confirms maximizing R increases discharge for given slope and roughness, reinforcing the same geometry b = 2y.

Why Other Options Are Wrong:

Half the breadth: At optimum, R relates to depth, not breadth; breadth equals twice depth.Twice the depth / Twice the breadth / One-third of the depth: None match the derived R = y/2 relation.

Common Pitfalls:

Memorizing b = 2y but forgetting the associated R relationship. Also mixing up hydraulic mean depth with hydraulic radius for non-rectangular sections.

Final Answer:

half the depth