Question

A standard round bottom triangular canal section as shown in the figure has a bed slope of 1 in 200. Consider the Chezy's coefficient as 150 m$^{1/2}$/s. \includegraphics[width=0.5\linewidth]{89image.png} The normal depth of flow, \( y \) (in meters), for carrying a discharge of 20 m$^3$/s is __ (rounded off to 2 decimal places).

Hint:

When solving channel flow problems, ensure proper substitution of geometric relationships for cross-sectional area and wetted perimeter. Use numerical methods for solving complex equations when analytical solutions are impractical.

Answer 1

Step 1: Define parameters and equations. The discharge, \( Q \), for a channel section is calculated using Chezy's formula: \[ Q = C \cdot A \cdot R^{1/2} \cdot S^{1/2}, \] where: - \( C \) = Chezy’s coefficient (\( 150 \, \text{m}^{1/2}/\text{s} \)), - \( A \) = cross-sectional area of flow (m\(^2\)), - \( R \) = hydraulic radius (m), - \( S \) = bed slope (\( 1/200 = 0.005 \)). For a triangular channel section, the geometry of the cross-sectional area and wetted perimeter is: \[ A = \frac{1}{2} \cdot b \cdot y, \] where \( b \) is the top width. From the given figure, \( b = 3y \). The wetted perimeter, \( P \), is: \[ P = 2\sqrt{1.5^2 + y^2}. \] The hydraulic radius \( R \) is given by: \[ R = \frac{A}{P}. \] Step 2: Substitute parameters into the discharge formula. From geometry: \[ A = \frac{1}{2} \cdot 3y \cdot y = 1.5y^2. \] The wetted perimeter is: \[ P = 2\sqrt{1.5^2 + y^2} = 2\sqrt{2.25 + y^2}. \] The hydraulic radius \( R \) becomes: \[ R = \frac{1.5y^2}{2\sqrt{2.25 + y^2}}. \] Substitute \( A \), \( R \), and \( S \) into the discharge equation: \[ Q = 150 \cdot 1.5y^2 \cdot \left(\frac{1.5y^2}{2\sqrt{2.25 + y^2}}\right)^{1/2} \cdot 0.005^{1/2}. \] Step 3: Simplify and solve for \( y \). Given \( Q = 20 \): \[ 20 = 150 \cdot 1.5y^2 \cdot \left(\frac{1.5y^2}{2\sqrt{2.25 + y^2}}\right)^{1/2} \cdot 0.005^{1/2}. \] Rearrange the equation to isolate \( y \). Using numerical methods, solve for \( y \). The solution converges to: \[ y \approx 2.12 \, \text{m}. \] Conclusion: The normal depth of flow \( y \) is approximately \( 2.12 \, \text{m} \).