Most Efficient Rectangular Channel Section – Proportion for Maximum Discharge For a given area and bed slope, a rectangular channel conveys maximum discharge when the breadth is twice the flow depth (b = 2 y), which maximizes hydraulic radius.
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Aits depth is twice the breadth
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Bits breadth is twice the depth
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Cits depth is thrice the breadth
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Dits breadth is thrice the depth
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Eits breadth equals its depth
Answer
Correct Answer: its breadth is twice the depth
Explanation
Introduction:The “most economical” (most efficient) channel section minimizes wetted perimeter for a given area, thereby maximizing hydraulic radius and discharge for a given slope and roughness. For rectangles, this yields a simple proportion between breadth and depth.
Given Data / Assumptions:
- Steady, uniform open-channel flow.
- Rectangular prismatic channel with width b and depth y.
- Manning or Chezy friction formulation where Q ∝ A * R^(2/3) (or similar).
Concept / Approach:
For fixed area A = b * y, minimizing wetted perimeter P = b + 2 y maximizes hydraulic radius R = A / P. Using calculus with the constraint b * y = A gives the optimal proportion b = 2 y for a rectangle.
Step-by-Step Solution:
1) Let A be fixed; write P = b + 2 y with b = A / y.2) Substitute to get P(y) = A / y + 2 y.3) Minimize P(y) w.r.t. y → dP/dy = −A / y^2 + 2 = 0 → y = sqrt(A / 2).4) Then b = A / y = 2 y, proving the proportion.Verification / Alternative check:
Using b = 2 y yields R = (b y)/(b + 2 y) = (2 y^2)/(2 y + 2 y) = y/2, the known efficient-section relation for rectangles.
Why Other Options Are Wrong:
Other ratios increase wetted perimeter for the same area, reducing hydraulic radius and thus discharge capacity under the same slope/roughness.
Common Pitfalls:
Confusing the efficient rectangle with trapezoidal or triangular optimal conditions; assuming the ratio depends on slope or roughness (it depends on geometry for a given A).
Final Answer:
its breadth is twice the depth