Hydraulically equivalent sections in open channel flow: Ratio of the diameter of a circular section to the side of a square section (hydraulically equivalent for equal conveyance).

Introduction / Context:Hydraulically equivalent sections are compared so that their conveyance capacity under the same slope and roughness is equivalent. For uniform flow, discharge is proportional to A * R^(2/3) (Manning–Strickler form), making geometry critical.

Given Data / Assumptions:

• Uniform roughness and bed slope.
• Comparing a full circular section (diameter D) with a full square section (side s).
• Hydraulic conveyance proportional to A * R^(2/3).

Concept / Approach:Set the conveyance of the circle equal to that of the square: A_c * R_c^(2/3) = A_s * R_s^(2/3). For a circle, A_c = π D^2 / 4, P_c = π D, and R_c = A_c / P_c = D/4. For a square, A_s = s^2, P_s = 4 s, and R_s = s/4. Solve for D/s.

Step-by-Step Solution:1) Write equality: (π/4) * D^2 * (D/4)^(2/3) = s^2 * (s/4)^(2/3).2) Cancel the common (1/4)^(2/3) factor to simplify algebra.3) Rearrange to D^(8/3) * (π/4) = s^(8/3).4) Raise both sides to the power 3/8 to isolate D: D * (π/4)^(3/8) = s.5) Hence, D/s = (4/π)^(3/8) ≈ 1.095.

Verification / Alternative check:A quick numerical evaluation of (4/π) equals about 1.273; raising to 0.375 gives approximately 1.095, matching the selected option.

Why Other Options Are Wrong:1.085, 1.075, 1.065, 1.000: These are close but do not match the derived ratio from the conveyance equality. 1.000 would only be correct if hydraulic radius equality were the sole criterion, which it is not here.

Common Pitfalls:

• Equating only hydraulic radius instead of full conveyance A * R^(2/3).
• Using perimeters incorrectly (remember P_c = π D, P_s = 4 s).

Final Answer:1.095