Chapter 5, Problem 5.1.1P

To determine

The hydraulic radius and hydraulic depth of the culvert.

Answer to Problem 5.1.1P

The hydraulic radius of the culvertis $11.67\text{in}$ and the hydraulic depth of the culvert is $28.13\text{in}$ .

Explanation of Solution

Given:

The diameter of the culvert is $d_{\text{o}}=36\text{in}$ and the depth of the flow is $y=20\text{in}$.

Concept Used:

The formula to calculate the angle of the circular culvert.

$\theta =2{\cos }^{-1}\left(1-\frac{2y}{d_{\text{o}}}\right)$

The formula to calculate the hydraulic radiusof the circular culvert.

$R=\frac{1}{4}\left(1-\frac{\sin \theta }{\theta }\right)d_{\text{o}}$

The formula to calculate the area of the circular culvert.

$A=\left(\theta +\sin \theta \right)d_{\text{o}}^2$

The formula to calculate thewidth of the circular culvert.

$B=\sin \left(\frac{\theta }{2}\right)d_{\text{o}}$

The formula to calculate the hydraulic depth of the circular culvert.

$D=\frac{A}{B}$

Calculation:

Calculate the angle of the circular culvert.

$\begin{array}{c}\theta =2{\cos }^{-1}\left(1-\frac{2y}{d_{\text{o}}}\right)\\ =2{\cos }^{-1}\left(1-\frac{2\left(20\text{in}\right)}{36\text{in}}\right)\\ =3.364\text{rad}\times \frac{180}{\pi }\\ \text{=}192°74^{\prime }\end{array}$

Calculate the hydraulic radiusof the circular culvert.

$\begin{array}{c}R=\frac{1}{4}\left(1-\frac{\sin \theta }{\theta }\right)d_{\text{o}}\\ =\frac{1}{4}\left(1-\frac{\sin 192°74^{\prime }}{3.364\text{rad}}\right)\left(36\text{in}\right)\\ =11.67\text{in}\end{array}$

Calculate the area of the circular culvert.

$\begin{array}{c}A=\frac{1}{8}\left(\theta -\sin \theta \right)d_{\text{o}}^2\\ =\frac{1}{8}\left(3.364\text{rad}-\sin 192°74^{\prime }\right){\left(36\text{in}\right)}^2\\ =706.92{\text{in}}^2\end{array}$

Calculate thewidth of the circular culvert.

$\begin{array}{c}B=\sin \left(\frac{\theta }{2}\right)d_{\text{o}}\\ =\sin \left(\frac{192°74^{\prime }}{2}\right)\left(36\text{in}\right)\\ =25.13\text{in}\end{array}$

Calculate the hydraulic depth of the circular culvert.

$\begin{array}{c}D=\frac{A}{B}\\ =\frac{706.92{\text{in}}^2}{25.13\text{in}}\\ =28.13\text{in}\end{array}$

Therefore, the hydraulic radius of the culvert is $11.67\text{in}$ and the hydraulic depth of the culvert is $28.13\text{in}$.