Question

The hydraulic radius for a most economical rectangular open channel with a base width of 5 m is

• A. 1.25 m
• B. 2.5 m
• C. 3.75 m
• D. 5.0 m

Hint:

For a most economical rectangular channel, remember these two simple rules:
1. The depth is half the width: $y = B/2$.
2. The hydraulic radius is half the depth: $R = y/2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the hydraulic radius ($R$) of a "most economical" rectangular channel, given its base width.

Step 2: Key Formula or Approach:
A "most economical" or "most efficient" channel section is one that conveys the maximum discharge for a given cross-sectional area and slope. This corresponds to the section with the minimum wetted perimeter for a given area.
For a

rectangular channel, the condition for the most economical section is that the depth of flow ($y$) must be half of the base width ($B$).
\[ y = \frac{B}{2} \] The hydraulic radius ($R$) is defined as the ratio of the flow area ($A$) to the wetted perimeter ($P$).
For a rectangular channel:
- Area ($A$) = $By$
- Wetted Perimeter ($P$) = $B + 2y$
- Hydraulic Radius ($R$) = $\frac{A}{P} = \frac{By}{B+2y}$

Step 3: Detailed Explanation:
We are given the base width, $B = 5$ m.
First, find the depth of flow ($y$) for the most economical condition:
\[ y = \frac{B}{2} = \frac{5 \text{ m}}{2} = 2.5 \text{ m} \] Now, calculate the hydraulic radius using this depth:
\[ R = \frac{By}{B+2y} = \frac{5 \times 2.5}{5 + 2(2.5)} = \frac{12.5}{5 + 5} = \frac{12.5}{10} \] \[ R = 1.25 \text{ m} \] Alternatively, for a most economical rectangular section, the hydraulic radius is always half the depth of flow: $R = y/2$.
Since $y=2.5$ m, $R = 2.5 / 2 = 1.25$ m.

Step 4: Final Answer:
The hydraulic radius for the most economical rectangular channel is 1.25 m.