Difficulty: Easy
Correct Answer: trapezoidal
Explanation:
Introduction:
A channel section is termed “most economical” or “most hydraulically efficient” when it carries the maximum discharge for a given area, bed slope, and roughness—equivalently, when the wetted perimeter is minimized, maximizing the hydraulic radius R = A / P.
Given Data / Assumptions:
Concept / Approach:
The theoretically best (maximum R) is a semicircular channel, but it is rarely used at scale. Among common practical forms, the trapezoidal section—optimized with the condition that one-half of the top width equals one of the equal side lengths—yields the least wetted perimeter for a given area, providing superior efficiency over rectangular/triangular forms.
Step-by-Step Solution:
Define hydraulic radius R = A / P; larger R reduces frictional resistance.For a given area A, minimize P. The trapezoid allows both bottom width and side slopes to be tuned to minimize P.Optimal trapezoid (most economical) satisfies: hydraulic mean depth = half the flow depth, and side slopes chosen so one side equals half the top width; this increases R beyond rectangular/triangular shapes.
Verification / Alternative check:
Design handbooks list conditions for most economical trapezoidal and rectangular channels; the trapezoid consistently offers better efficiency and bank stability in practice.
Why Other Options Are Wrong:
Rectangular and square are special cases with higher wetted perimeter for the same area when not at their own “most economical” proportions; triangular channels are typically less efficient and less stable.
Common Pitfalls:
Confusing theoretical semicircular optimum with practical sections; ignoring side-slope stability and erosion criteria.
Final Answer:
trapezoidal
Discussion & Comments