Open channel design — most economical trapezoidal section For the most economical trapezoidal channel, what is the ratio of flow depth y to hydraulic radius R (i.e., y : R)?

Introduction / Context:
“Most economical” sections minimize wetted perimeter for a given area (or maximize discharge for a given cross section) and are widely used in irrigation and stormwater channels. Recognizing the characteristic ratios lets you size channels quickly without lengthy calculus.

Given Data / Assumptions:

• Trapezoidal open-channel flow, steady uniform conditions.
• Hydraulic radius R = A/P, where A is area, P is wetted perimeter.
• We ask for the optimal ratio y:R at the most economical condition.

Concept / Approach:
For the most economical trapezoidal (as for the rectangular optimal case), geometry leads to the result R = y/2. This stems from minimizing P for a fixed A using first-order optimality conditions on the geometric relations for trapezoids with side slope m (horizontal:vertical).

Step-by-Step Solution:
Recall optimal relations: (i) half the top width equals one sloping side; (ii) hydraulic radius at optimum is R = y/2.Therefore, y : R = y : (y/2) = 2 : 1.Hence the required ratio equals 2.

Verification / Alternative check:
Derivations via Manning/Chézy with Lagrange multipliers (minimizing wetted perimeter at fixed area) reproduce R = y/2 for the trapezoidal best section.

Why Other Options Are Wrong:

• 0.5, 1, 1.5: inconsistent with the known optimum R = y/2.

Common Pitfalls:

• Confusing the condition for maximum velocity with that for maximum discharge.
• Applying rectangular-channel results blindly; fortunately, both optimal rectangular and trapezoidal sections share R = y/2.

Final Answer:
2