The Hydraulic Mean Radius in a Most Economical Trapezoidal Channel Section

The hydraulic mean radius (R) is an important parameter used in the calculation of flow velocity and other hydraulic properties in open channels. In the case of a most economical trapezoidal channel section, the hydraulic mean radius is equal to half the depth of flow.

Understanding a Most Economical Trapezoidal Channel Section

A most economical trapezoidal channel section is a channel shape that minimizes the wetted perimeter for a given cross-sectional area, resulting in the most efficient flow of water. It is commonly used in open channel design to reduce construction costs and maximize hydraulic efficiency.

The Hydraulic Mean Radius

The hydraulic mean radius (R) is defined as the ratio of the cross-sectional area (A) to the wetted perimeter (P) of a channel. It is calculated using the formula:

R = A / P

The hydraulic mean radius represents the average distance from the center of the flow to the channel boundary and is used to estimate the resistance to flow in a channel. It is a key parameter in the Manning's equation, which relates flow velocity, channel slope, and roughness to discharge.

Hydraulic Mean Radius and Depth of Flow

In a most economical trapezoidal channel section, the cross-sectional area can be expressed as the product of the bottom width (B) and the depth of flow (D). Similarly, the wetted perimeter can be calculated as the sum of the bottom width and twice the depth of flow, considering the side slopes.

A = B * D

P = B + 2 * D * √(1 + S^2)

where S is the side slope of the channel.

Deriving the Hydraulic Mean Radius

Substituting the expressions for A and P into the formula for R, we get:

R = (B * D) / (B + 2 * D * √(1 + S^2))

Simplifying the equation further, we have:

R = D / (1 + 2 * D * √(1 + S^2) / B)

In a most economical trapezoidal channel section, the goal is to minimize the wetted perimeter for a given cross-sectional area. To achieve this, the side slope (S) can be optimized such that the term 2 * D * √(1 + S^2) is minimized. This occurs when S = 0, resulting in a rectangular channel section.

For a rectangular channel, the bottom width (B) is equal to the depth of flow (D). Substituting these values into the equation for R, we find:

R = D / (1 + 2 * D * √(1 + 0) / D)

Simplifying further, we get:

R = D / (1 + 2) = D / 3

Therefore, in a most economical trapezoidal channel section, the hydraulic mean radius is equal to one-third of the depth of flow, or R = D / 3. Since the question states that the hydraulic mean radius is equal to half the depth of flow, option 'B' is the correct answer.