Open Channel Flow MCQ with answer and solution

Concept

Wetted perimeter is minimum for a given area when the side slope n (H:V) = 1/√3, i.e., sides at 60° to the horizontal (half regular hexagon).

Formula

P = A/d - n d + 2 d √(1 + n²); dP/dn = 0 gives n = 1/√3

Step-by-step solution

1. Area A = (b + n d) d, so b = A/d - n d. 2. Wetted perimeter P = b + 2 d √(1 + n²) = A/d - n d + 2 d √(1 + n²). 3. Keep A and d constant and set dP/dn = 0: -d + 2 d n / √(1 + n²) = 0. 4. This gives 2n = √(1 + n²), so 4n² = 1 + n², so n² = 1/3. 5. Therefore n = 1/√3, which means the side makes 60° with the horizontal.

Explanation

The most economical section carries the maximum discharge for a given area, so the wetted perimeter must be minimum. Minimizing P with respect to side slope gives n = 1/√3 (H:V). This section is half of a regular hexagon.

Why the other options are wrong

A: √3 is the slope if n is written as V:H, not H:V. B: 1/√2 does not come from dP/dn = 0. D: √2 is not a condition for the best trapezoidal section.

Shortcut

Best trapezoid = half regular hexagon: sides at 60°, n = 1/√3, R = d/2.

Common mistake

Writing √3 because tan 60° = √3. That is the V:H ratio, but n means horizontal:vertical.