Most economical rectangular channel in open-channel flow For a hydraulically most economical rectangular channel section (maximum discharge for a given area or minimum wetted perimeter), the hydraulic mean depth (R = A / P) is equal to which of the following, in terms of the flow depth?

Introduction / Context:
In open-channel hydraulics, a “most economical” section minimizes wetted perimeter for a given area (or maximizes discharge for a given area) and therefore reduces frictional losses. For a rectangular channel, this geometric condition yields a simple relationship between the width and depth, and thus a specific value of hydraulic mean depth R = A / P relative to the depth y.

Given Data / Assumptions:

• Prismatic rectangular channel with width b and flow depth y.
• Steady, uniform flow so hydraulic radius R = A / P is appropriate.
• “Most economical” means minimum wetted perimeter P for a fixed area A = b * y.

Concept / Approach:

For a rectangle, area A = b * y and wetted perimeter P = b + 2y. Using calculus (optimize P for constant A) or standard results, the condition for most economy is b = 2y. Substituting into R = A / P gives a direct relation between R and y for the optimal section.

Step-by-Step Solution:

Verification / Alternative check:

The same result follows if we maximize R directly at constant A, or minimize wetted perimeter using Lagrange multipliers; both give b = 2y and R = y/2 for a rectangular section.

Why Other Options Are Wrong:

(a) R = y is too large; (c) R = y/3 is too small; (d) is incorrect because a unique relation exists for the optimal rectangle.

Common Pitfalls:

Confusing hydraulic radius R = A / P with hydraulic depth D = A / T (T = top width). For the most economical rectangle, R = y/2 but hydraulic depth D equals y since T = b = 2y and A = b * y.

Final Answer:

half the depth of flow