Hydraulically most economical triangular channel section: which triangular geometry maximizes discharge for given area, slope, and roughness?

Introduction / Context:The “most economical” channel section minimizes wetted perimeter P for a given flow area A, thereby maximizing hydraulic radius R = A/P and mean velocity for the same slope and roughness. For a triangular section, this leads to a specific apex (vertex) angle condition.

Given Data / Assumptions:

• Open-channel, steady uniform flow.
• Triangular cross-section with two equal sloping sides (isosceles).
• Optimization under Chezy/Manning framework.

Concept / Approach:

For a triangular section with side slopes symmetric, calculus shows that the wetted perimeter is minimized when each side is inclined at 45° to the channel bed, giving an included vertex angle of 90°. This is commonly expressed as the most economical triangular section. The “right-angled triangle with equal sides” describes precisely this geometry in cross-section.

Step-by-Step Solution:

Verification / Alternative check:

Textbook results: for triangles, most economical section has half-apex angle = 45°, equivalent to vertex angle = 90°.

Why Other Options Are Wrong:

(a) Equilateral gives vertex angle 60°, not optimal. (b) “Right-angled triangle” is too general and could include unequal sides. (c) 45° vertex angle is not correct; the correct vertex angle is 90°. (e) Not applicable.

Common Pitfalls:

Confusing the condition for rectangles (b = 2y) with the triangular case; misreading “45°” as vertex instead of half-apex angle.

Final Answer:

right-angled triangle with equal sides (vertex angle 90°)