Open Channel – Most economical trapezoidal section The discharge through a trapezoidal channel is maximum (most economical section) when which geometric condition holds?

Introduction:
The “most economical” (most efficient) channel section provides maximum discharge for a given cross-sectional area and bed slope, or equivalently minimizes wetted perimeter for a given area. For a trapezoidal section, there is a classical geometric condition for the optimum in which the side length relates to the top width in a specific way.

Given Data / Assumptions:

• Prismatic trapezoidal channel with bed width b, depth y, and side slope z (horizontal:vertical).
• Uniform flow with a given roughness and slope; resistance modeled by Manning or Chezy.
• Comparisons at the same area and slope, seeking maximum hydraulic radius R = A / P.

Concept / Approach:

For a trapezoid: top width T = b + 2z y; wetted perimeter P = b + 2 y sqrt(1 + z^2); area A = y (b + z y). Optimization to maximize R or minimize P for a given A yields two hallmark relations: (i) one of the sloping sides equals half of the top width (s = T / 2), and (ii) the hydraulic radius equals y / 2 at optimum. These relations uniquely characterize the most economical trapezoidal section.

Step-by-Step Solution:

Verification / Alternative check:

Substituting the optimum relations into R confirms R = y / 2, consistent with known textbook results. Geometric sketches also show that equal partition of the top width by the side length occurs at the optimum.

Why Other Options Are Wrong:

Top width twice bottom width: Not a general optimum condition.Depth equal to bottom width: No basis in the optimization.Side equal to bottom width: Dimensional equality does not ensure minimum perimeter.Hydraulic radius equals the depth: At optimum R = y / 2, not y.

Common Pitfalls:

Memorizing one relation (like R = y / 2) without remembering the geometric condition s = T / 2 that is often tested directly in exams.

Final Answer:

the sloping side is equal to half the width at the top